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# 11.10: Taylor and Maclaurin Series
In the previous two sections we discussed how to find power series representations for certain types of functions––specifically, functions related to geometric series. Here we discuss power series representations for other types of functions. In particular, we address the following questions: Which functions can be represented by power series and how do we find such representations? If we can find a power series representation for a particular function $$f$$ and the series converges on some interval, how do we prove that the series actually converges to $$f$$?
## Overview of Taylor/Maclaurin Series
Consider a function $$f$$ that has a power series representation at $$x=a$$. Then the series has the form
$\sum_{n=0}^∞c_n(x−a)^n=c_0+c_1(x−a)+c_2(x−a)^2+ \dots. \label{eq1}$
What should the coefficients be? For now, we ignore issues of convergence, but instead focus on what the series should be, if one exists. We return to discuss convergence later in this section. If the series Equation \ref{eq1} is a representation for $$f$$ at $$x=a$$, we certainly want the series to equal $$f(a)$$ at $$x=a$$. Evaluating the series at $$x=a$$, we see that
$\sum_{n=0}^∞c_n(x−a)^n=c_0+c_1(a−a)+c_2(a−a)^2+\dots=c_0.\label{eq2}$
Thus, the series equals $$f(a)$$ if the coefficient $$c_0=f(a)$$. In addition, we would like the first derivative of the power series to equal $$f′(a)$$ at $$x=a$$. Differentiating Equation \ref{eq2} term-by-term, we see that
$\dfrac{d}{dx}\left( \sum_{n=0}^∞c_n(x−a)^n \right)=c_1+2c_2(x−a)+3c_3(x−a)^2+\dots.\label{eq3}$
Therefore, at $$x=a,$$ the derivative is
$\dfrac{d}{dx}\left( \sum_{n=0}^∞c_n(x−a)^n \right)=c_1+2c_2(a−a)+3c_3(a−a)^2+\dots=c_1.\label{eq4}$
Therefore, the derivative of the series equals $$f′(a)$$ if the coefficient $$c_1=f′(a).$$ Continuing in this way, we look for coefficients $$c_n$$ such that all the derivatives of the power series Equation will agree with all the corresponding derivatives of $$f$$ at $$x=a$$. The second and third derivatives of Equation \ref{eq3} are given by
$\dfrac{d^2}{dx^2} \left(\sum_{n=0}^∞c_n(x−a)^n \right)=2c_2+3⋅2c_3(x−a)+4⋅3c_4(x−a)^2+\dots\label{eq5}$
and
$\dfrac{d^3}{dx^3} \left( \sum_{n=0}^∞c_n(x−a)^n \right)=3⋅2c_3+4⋅3⋅2c_4(x−a)+5⋅4⋅3c_5(x−a)^2+⋯.\label{eq6}$
Therefore, at $$x=a$$, the second and third derivatives
$\dfrac{d^2}{dx^2}(\sum_{n=0}^∞c_n(x−a)^n)=2c_2+3⋅2c_3(a−a)+4⋅3c_4(a−a)^2+\dots=2c_2\label{eq7}$
and
$\dfrac{d^3}{dx^3} \left(\sum_{n=0}^∞c_n(x−a)^n\right)=3⋅2c_3+4⋅3⋅2c_4(a−a)+5⋅4⋅3c_5(a−a)^2+\dots =3⋅2c_3\label{eq8}$
equal $$f''(a)$$ and $$f'''(a)$$, respectively, if $$c_2=\dfrac{f''(a)}{2}$$ and $$c_3=\dfrac{f'''(a)}{3}⋅2$$. More generally, we see that if $$f$$ has a power series representation at $$x=a$$, then the coefficients should be given by $$c_n=\dfrac{f^{(n)}(a)}{n!}$$. That is, the series should be
$\sum_{n=0}^∞\dfrac{f^{(n)}(a)}{n!}(x−a)^n=f(a)+f′(a)(x−a)+\dfrac{f''(a)}{2!}(x−a)^2+\dfrac{f'''(a)}{3!}(x−a)^3+⋯.$
This power series for $$f$$ is known as the Taylor series for $$f$$ at $$a.$$ If $$x=0$$, then this series is known as the Maclaurin series for $$f$$.
Definition: Maclaurin and Taylor series
If $$f$$ has derivatives of all orders at $$x=a$$, then the Taylor series for the function $$f$$ at $$a$$ is
$\sum_{n=0}^∞\dfrac{f^{(n)}(a)}{n!}(x−a)^n=f(a)+f′(a)(x−a)+\dfrac{f''(a)}{2!}(x−a)^2+⋯+\dfrac{f^{(n)}(a)}{n!}(x−a)^n+⋯.$
The Taylor series for $$f$$ at 0 is known as the Maclaurin series for $$f$$.
Later in this section, we will show examples of finding Taylor series and discuss conditions under which the Taylor series for a function will converge to that function. Here, we state an important result. Recall that power series representations are unique. Therefore, if a function $$f$$ has a power series at $$a$$, then it must be the Taylor series for $$f$$ at $$a$$.
Uniqueness of Taylor Series
If a function $$f$$ has a power series at a that converges to $$f$$ on some open interval containing a, then that power series is the Taylor series for $$f$$ at $$a$$.
The proof follows directly from that discussed previously.
To determine if a Taylor series converges, we need to look at its sequence of partial sums. These partial sums are finite polynomials, known as Taylor polynomials.
### Taylor Polynomials
The nth partial sum of the Taylor series for a function $$f$$ at $$a$$ is known as the nth Taylor polynomial. For example, the 0th, 1st, 2nd, and 3rd partial sums of the Taylor series are given by
$$p_0(x)=f(a)$$,
$$p_1(x)=f(a)+f′(a)(x−a)$$,
$$p_2(x)=f(a)+f′(a)(x−a)+\dfrac{f''(a)}{2!}(x−a)^2$$,
$$p_3(x)=f(a)+f′(a)(x−a)+\dfrac{f''(a)}{2!}(x−a)^2+\dfrac{f'''(a)}{3!}(x−a)^3$$,
respectively. These partial sums are known as the 0th, 1st, 2nd, and 3rd Taylor polynomials of $$f$$ at $$a$$, respectively. If $$x=a$$, then these polynomials are known as Maclaurin polynomials for $$f$$. We now provide a formal definition of Taylor and Maclaurin polynomials for a function $$f$$.
Definition: Maclaurin polynomial
If $$f$$ has n derivatives at $$x=a$$, then the nth Taylor polynomial for $$f$$ at $$a$$ is
$p_n(x)=f(a)+f′(a)(x−a)+\dfrac{f''(a)}{2!}(x−a)^2+\dfrac{f'''(a)}{3!}(x−a)^3+⋯+\dfrac{f^{(n)}(a)}{n!}(x−a)^n.$
The nth Taylor polynomial for $$f$$ at 0 is known as the nth Maclaurin polynomial for $$f$$.
We now show how to use this definition to find several Taylor polynomials for $$f(x)=lnx$$ at $$x=1$$.
Example $$\PageIndex{1}$$: Finding Taylor Polynomials
Find the Taylor polynomials $$p_0,p_1,p_2$$ and $$p_3$$ for $$f(x)=lnx$$ at $$x=1$$. Use a graphing utility to compare the graph of $$f$$ with the graphs of $$p_0,p_1,p_2$$ and $$p_3$$.
Solution
To find these Taylor polynomials, we need to evaluate $$f$$ and its first three derivatives at $$x=1$$.
$$f(x)=lnx$$ $$f(1)=0$$
$$f′(x)=\dfrac{1}{x}$$ $$f′(1)=1$$
$$f''(x)=−\dfrac{1}{x^2}$$ $$f''(1)=−1$$
$$f'''(x)==\dfrac{2}{x^3}$$ $$f'''(1)=2$$
Therefore,
$$p_0(x)=f(1)=0,$$
$$p_1(x)=f(1)+f′(1)(x−1)=x−1,$$
$$p_2(x)=f(1)+f′(1)(x−1)+\dfrac{f''(1)}{2}(x−1)^2=(x−1)−\dfrac{1}{2}(x−1)^2$$,
$$p_3(x)=f(1)+f′(1)(x−1)+\dfrac{f''(1)}{2}(x−1)^2+\dfrac{f'''(1)}{3!}(x−1)^3=(x−1)−\dfrac{1}{2}(x−1)^2+\dfrac{1}{3}(x−1)^3$$.
The graphs of $$y=f(x)$$ and the first three Taylor polynomials are shown in Figure 1.
Figure 1: The function $$y=lnx$$ and the Taylor polynomials $$p_0,p_1,p_2$$ and $$p_3$$ at $$x=1$$ are plotted on this graph.
Exercise $$\PageIndex{1}$$
Find the Taylor polynomials $$p_0,p_1,p_2$$ and $$p_3$$ for $$f(x)=\dfrac{1}{x^2}$$ at $$x=1$$.
Solution
$p_0(x)=1$
$p_1(x)=1−2(x−1)$
$p_2(x)=1−2(x−1)+3(x−1)^2$
$p_3(x)=1−2(x−1)+3(x−1)^2−4(x−1)^3$
Hint: Find the first three derivatives of $$f$$ and evaluate them at $$x=1.$$
We now show how to find Maclaurin polynomials for $$e^x, \sin x,$$ and $$\cos x$$. As stated above, Maclaurin polynomials are Taylor polynomials centered at zero.
Example $$\PageIndex{2}$$: Finding Maclaurin Polynomials
For each of the following functions, find formulas for the Maclaurin polynomials $$p_0,p_1,p_2$$ and $$p_3$$. Find a formula for the nth Maclaurin polynomial and write it using sigma notation. Use a graphing utility to compare the graphs of $$p_0,p_1,p_2$$ and $$p_3$$ with $$f$$.
1. $$f(x)=e^x$$
2. $$f(x)=\sin x$$
3. $$f(x)=\cos x$$
Solution
Since $$f(x)=e^x$$,we know that $$f(x)=f′(x)=f''(x)=⋯=f^{(n)}(x)=e^x$$ for all positive integers n. Therefore,
$f(0)=f′(0)=f''(0)=⋯=f^{(n)}(0)=1 \nonumber$
for all positive integers n. Therefore, we have
$$p_0(x)=f(0)=1,$$
$$p_1(x)=f(0)+f′(0)x=1+x,$$
$$p_2(x)=f(0)+f′(0)x+\dfrac{f''(0)}{2!}x^2=1+x+\dfrac{1}{2}x^2$$,
$$p_3(x)=f(0)+f′(0)x+\dfrac{f''(0)}{2}x^2+\dfrac{f'''(0)}{3!}x^3=1+x+\dfrac{1}{2}x^2+\dfrac{1}{3!}x^3$$,
$$p_n(x)=f(0)+f′(0)x+\dfrac{f''(0)}{2}x^2+\dfrac{f'''(0)}{3!}x^3+⋯+\dfrac{f^{(n)}(0)}{n!}x^n=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+⋯+\dfrac{x^n}{n!}=\sum_{k=0}^n\dfrac{x^k}{k!}$$.
The function and the first three Maclaurin polynomials are shown in Figure 2.
Figure 2: The graph shows the function $$y=e^x$$ and the Maclaurin polynomials $$p_0,p_1,p_2$$ and $$p_3$$.
b. For $$f(x)=\sin x$$, the values of the function and its first four derivatives at $$x=0$$ are given as follows:
$$f(x)=\sin x$$ $$f(0)=0$$
$$f′(x)=\cos x$$ $$f′(0)=1$$
$$f''(x)=−\sin x$$ $$f''(0)=0$$
$$f'''(x)=−\cos x$$ $$f'''(0)=−1$$
$$f^{(4)}(x)=\sin x$$ $$f^{(4)}(0)=0$$.
Since the fourth derivative is $$\sin x,$$ the pattern repeats. That is, $$f^{(2m)}(0)=0$$ and $$f^{(2m+1)}(0)=(−1)^m$$ for $$m≥0.$$ Thus, we have
$$p_0(x)=0,$$
$$p_1(x)=0+x=x,$$
$$p_2(x)=0+x+0=x,$$
$$p_3(x)=0+x+0−\dfrac{1}{3!}x^3=x−\dfrac{x^3}{3!},$$
$$p_4(x)=0+x+0−\dfrac{1}{3!}x^3+0=x−\dfrac{x^3}{3!}$$,
$$p_5(x)=0+x+0−\dfrac{1}{3!}x^3+0+\dfrac{1}{5!}x^5=x−\dfrac{x^3}{3!}+\dfrac{x^5}{5!}$$,
and for $$m≥0$$,
$$p_{2m+1}(x)=p_{2m+2}(x)=x−\dfrac{x^3}{3!}+\dfrac{x^5}{5!}−⋯+(−1)^m\dfrac{x^{2m+1}}{(2m+1)!}=\sum_{k=0}^m(−1)^k\dfrac{x^{2k+1}}{(2k+1)!}$$.
Graphs of the function and its Maclaurin polynomials are shown in Figure 3.
Figure 3: The graph shows the function $$y=\sin x$$ and the Maclaurin polynomials $$p_1,p_3$$ and $$p_5$$.
c. For f(x)=\cos x, the values of the function and its first four derivatives at x=0 are given as follows:
$$f(x)=\cos x$$ $$f(0)=1$$
$$f′(x)=−\sin x$$ $$f′(0)=0$$
$$f''(x)=−\cos x$$ $$f''(0)=−1$$
$$f'''(x)=\sin x$$ $$f'''(0)=0$$
$$f^{(4)}(x)=\cos x$$ $$f^{(4)}(0)=1.$$
Since the fourth derivative is $$\sin x$$, the pattern repeats. In other words, $$f^{(2m)}(0)=(−1)^m$$ and $$f^{(2m+1)}=0$$ for $$m≥0$$. Therefore,
$$p_0(x)=1,$$
$$p_1(x)=1+0=1,$$
$$p_2(x)=1+0−\dfrac{1}{2!}x^2=1−\dfrac{x^2}{2!}$$,
$$p_3(x)=1+0−\dfrac{1}{2!}x^2+0=1−\dfrac{x^2}{2!}$$,
$$p_4(x)=1+0−\dfrac{1}{2!}x^2+0+\dfrac{1}{4!}x^4=1−\dfrac{x^2}{2!}+\dfrac{x^4}{4!}$$,
$$p_5(x)=1+0−\dfrac{1}{2!}x^2+0+\dfrac{1}{4!}x^4+0=1−\dfrac{x^2}{2!}+\dfrac{x^4}{4!}$$,
and for $$n≥0$$,
$$p_{2m}(x)=p_{2m+1}(x)=1−\dfrac{x^2}{2!}+\dfrac{x^4}{4!}−⋯+(−1)^m\dfrac{x^{2m}}{(2m)!}=\sum_{k=0}^m(−1)^k\dfrac{x^{2k}}{(2k)!}$$.
Graphs of the function and the Maclaurin polynomials appear in Figure 4.
Figure 4: The function $$y=\cos x$$ and the Maclaurin polynomials $$p_0,p_2$$ and $$p_4$$ are plotted on this graph.
Exercise $$\PageIndex{2}$$
Find formulas for the Maclaurin polynomials $$p_0,p_1,p_2$$ and $$p_3$$ for $$f(x)=\dfrac{1}{1+x}$$.
Find a formula for the nth Maclaurin polynomial. Write your anwer using sigma notation.
Hint Evaluate the first four derivatives of $$f$$ and look for a pattern.
Solution
$$p_0(x)=1;p_1(x)=1−x;p_2(x)=1−x+x^2;p_3(x)=1−x+x^2−x^3;p_n(x)=1−x+x^2−x^3+⋯+(−1)^nx^n=_{k=0}^n(−1)^kx^k$$
## Taylor’s Theorem with Remainder
Recall that the nth Taylor polynomial for a function $$f$$ at a is the nth partial sum of the Taylor series for $$f$$ at a. Therefore, to determine if the Taylor series converges, we need to determine whether the sequence of Taylor polynomials $${p_n}$$ converges. However, not only do we want to know if the sequence of Taylor polynomials converges, we want to know if it converges to $$f$$. To answer this question, we define the remainder $$R_n(x)$$ as
$R_n(x)=f(x)−p_n(x).$
For the sequence of Taylor polynomials to converge to $$f$$, we need the remainder $$R_n$$ to converge to zero. To determine if $$R_n$$ converges to zero, we introduce Taylor’s theorem with remainder. Not only is this theorem useful in proving that a Taylor series converges to its related function, but it will also allow us to quantify how well the nth Taylor polynomial approximates the function.
Here we look for a bound on $$|R_n|.$$ Consider the simplest case: $$n=0$$. Let $$p_0$$ be the 0th Taylor polynomial at a for a function $$f$$. The remainder $$R_0$$ satisfies
$$R_0(x)=f(x)−p_0(x)=f(x)−f(a).$$
If $$f$$ is differentiable on an interval I containing a and x, then by the Mean Value Theorem there exists a real number c between a and x such that $$f(x)−f(a)=f′(c)(x−a)$$. Therefore,
$R_0(x)=f′(c)(x−a).$
Using the Mean Value Theorem in a similar argument, we can show that if $$f$$ is n times differentiable on an interval I containing a and x, then the nth remainder $$R_n$$ satisfies
$R_n(x)=\dfrac{f^{(n+1)}(c)}{(n+1)!}(x−a)^{n+1}$
for some real number c between a and x. It is important to note that the value c in the numerator above is not the center a, but rather an unknown value c between a and x. This formula allows us to get a bound on the remainder $$R_n$$. If we happen to know that $$∣f^{(n+1)}(x)∣$$ is bounded by some real number M on this interval I, then
$|R_n(x)|≤\dfrac{M}{(n+1)!}|x−a|^{n+1}$
for all x in the interval I.
We now state Taylor’s theorem, which provides the formal relationship between a function $$f$$ and its nth degree Taylor polynomial $$p_n(x)$$. This theorem allows us to bound the error when using a Taylor polynomial to approximate a function value, and will be important in proving that a Taylor series for $$f$$ converges to $$f$$.
Taylor’s Theorem with Remainder
Let $$f$$ be a function that can be differentiated $$n+1$$ times on an interval I containing the real number a. Let $$p_n$$ be the nth Taylor polynomial of $$f$$ at a and let
$R_n(x)=f(x)−p_n(x)$
be the nth remainder. Then for each x in the interval I, there exists a real number c between a and x such that
$R_n(x)=\dfrac{f^{(n+1)}(c)}{(n+1)!}(x−a)^{n+1}$.
If there exists a real number M such that $$∣f^{(n+1)}(x)∣≤M$$ for all $$x∈I$$, then
$|R_n(x)|≤\dfrac{M}{(n+1)!}|x−a|^{n+1}$
for all x in I.
Proof
Fix a point $$x∈I$$ and introduce the function g such that
$g(t)=f(x)−f(t)−f′(t)(x−t)−\dfrac{f''(t)}{2!}(x−t)^2−⋯−\dfrac{f^{(n)}(t)}{n!}(x−t)^n−R_n(x)\dfrac{(x−t)^{n+1}}{(x−a)^{n+1}}$.
We claim that g satisfies the criteria of Rolle’s theorem. Since g is a polynomial function (in t), it is a differentiable function. Also, g is zero at $$t=a$$ and $$t=x$$ because
$$g(a)=f(x)−f(a)−f′(a)(x−a)−\dfrac{f''(a)}{2!}(x−a)^2+⋯+\dfrac{f^{(n)}(a)}{n!}(x−a)^n−R_n(x)$$
$$=f(x)−p_n(x)−R_n(x)$$
$$=0,$$
$$g(x)=f(x)−f(x)−0−⋯−0$$
$$=0.$$
Therefore, g satisfies Rolle’s theorem, and consequently, there exists c between a and x such that $$g′(c)=0.$$ We now calculate $$g′$$. Using the product rule, we note that
$\dfrac{d}{dt}[\dfrac{f^{(n)}(t)}{n!}(x−t)^n]=−\dfrac{f^{(n)}(t)}{(n−1)!}(x−t)^{n−1}+\dfrac{f^{(n+1)}(t)}{n!}(x−t)^n.$
Consequently,
$g′(t)=−f′(t)+[f′(t)−f''(t)(x−t)]+[f''(t)(x−t)−\dfrac{f'''(t)}{2!}(x−t)^2]+⋯+[\dfrac{f^{(n)}(t)}{(n−1)!}(x−t)^{n−1}−\dfrac{f^{(n+1)}(t)}{n!}(x−t)^n]+(n+1)R_n(x)\dfrac{(x−t)^n}{(x−a)^{n+1}}$.
Notice that there is a telescoping effect. Therefore,
$g′(t)=−\dfrac{f^{(n+1)}(t)}{n!}(x−t)^n+(n+1)R_n(x)\dfrac{(x−t)^n}{(x−a)^{n+1}}$.
By Rolle’s theorem, we conclude that there exists a number c between a and x such that $$g′(c)=0.$$ Since
$g′(c)=−\dfrac{f^{(n+1})(c)}{n!}(x−c)^n+(n+1)R_n(x)\dfrac{(x−c)^n}{(x−a)^{n+1}}$
we conclude that
$−\dfrac{f^{(n+1)}(c)}{n!}(x−c)^n+(n+1)R_n(x)\dfrac{(x−c)^n}{(x−a)^{n+1}}=0.$
Adding the first term on the left-hand side to both sides of the equation and dividing both sides of the equation by $$n+1,$$ we conclude that
$R_n(x)=\dfrac{f^{(n+1)}(c)}{(n+1)!}(x−a)^{n+1}$
as desired. From this fact, it follows that if there exists M such that $$∣f^{(n+1)}(x)∣≤M$$ for all x in I, then
$|R_n(x)|≤\dfrac{M}{(n+1)!}|x−a|^{n+1}$.
Not only does Taylor’s theorem allow us to prove that a Taylor series converges to a function, but it also allows us to estimate the accuracy of Taylor polynomials in approximating function values. We begin by looking at linear and quadratic approximations of $$f(x)=\dfrac[3]{x}$$ at $$x=8$$ and determine how accurate these approximations are at estimating $$\dfrac[3]{11}$$.
Example $$\PageIndex{3}$$: Using Linear and Quadratic Approximations to Estimate Function Values
Consider the function $$f(x)=\sqrt[3]{x}$$.
1. Find the first and second Taylor polynomials for $$f$$ at $$x=8$$. Use a graphing utility to compare these polynomials with $$f$$ near $$x=8.$$
2. Use these two polynomials to estimate $$\sqrt[3]{11}$$.
3. Use Taylor’s theorem to bound the error.
Solution:
a. For $$f(x)=\sqrt[3]{x}$$, the values of the function and its first two derivatives at $$x=8$$ are as follows:
$$f(x)=\sqrt[3]{x}$$ $$f(8)=2$$
$$f′(x)=\dfrac{1}{3x^{2/3}}$$ $$f′(8)=\dfrac{1}{12}$$
$$f''(x)=\dfrac{−2}{9x^{5/3}}$$ $$f''(8)=−\dfrac{1}{144.}$$
Thus, the first and second Taylor polynomials at $$x=8$$ are given by
$$p_1(x)=f(8)+f′(8)(x−8)$$
$$=2+\dfrac{1}{12}(x−8)$$
$$p_2(x)=f(8)+f′(8)(x−8)+\dfrac{f''(8)}{2!}(x−8)^2$$
$$=2+\dfrac{1}{12}(x−8)−\dfrac{1}{288}(x−8)^2$$.
The function and the Taylor polynomials are shown in Figure 4.
Figure 4: The graphs of $$f(x)=\sqrt[3]{x}$$ and the linear and quadratic approximations $$p_1(x)$$ and $$p_2(x)$$
b. Using the first Taylor polynomial at $$x=8$$, we can estimate
$\dfrac[3]{11}≈p_1(11)=2+\dfrac{1}{12}(11−8)=2.25.$
Using the second Taylor polynomial at $$x=8$$, we obtain
$\sqrt[3]{11}≈p_2(11)=2+\dfrac{1}{12}(11−8)−\dfrac{1}{288}(11−8)^2=2.21875.$
c. By Note, there exists a c in the interval $$(8,11)$$ such that the remainder when approximating $$\sqrt[3]{11}$$ by the first Taylor polynomial satisfies
$R_1(11)=\dfrac{f''(c)}{2!}(11−8)^2.$
We do not know the exact value of c, so we find an upper bound on $$R_1(11)$$ by determining the maximum value of $$f''$$ on the interval $$(8,11)$$. Since $$f''(x)=−\dfrac{2}{9x^{5/3}}$$, the largest value for $$|f''(x)|$$ on that interval occurs at $$x=8$$. Using the fact that $$f''(8)=−\dfrac{1}{144}$$, we obtain
$$|R_1(11)|≤\dfrac{1}{144⋅2!}(11−8)^2=0.03125.$$
Similarly, to estimate $$R_2(11)$$, we use the fact that
$$R_2(11)=\dfrac{f'''(c)}{3!}(11−8)^3$$.
Since $$f'''(x)=\dfrac{10}{27x^{8/3}}$$, the maximum value of $$f'''$$ on the interval $$(8,11)$$ is $$f'''(8)≈0.0014468$$. Therefore, we have
$$|R_2(11)|≤\dfrac{0.0011468}{3!}(11−8)^3≈0.0065104.$$
Exercise $$\PageIndex{3}$$:
Find the first and second Taylor polynomials for $$f(x)=\sqrt{x}$$ at $$x=4$$. Use these polynomials to estimate $$\sqrt{6}$$. Use Taylor’s theorem to bound the error.
Solution
$$p_1(x)=2+\dfrac{1}{4}(x−4);p_2(x)=2+\dfrac{1}{4}(x−4)−\dfrac{1}{64}(x−4)^2;p_1(6)=2.5;p_2(6)=2.4375;$$
$$|R_1(6)|≤0.0625;|R_2(6)|≤0.015625$$
Hint: Evaluate $$f(4),f′(4),$$ and $$f''(4).$$
Example $$\PageIndex{4}$$: Approximating sin x Using Maclaurin Polynomials
From Example b., the Maclaurin polynomials for $$\sin x$$ are given by
$p_{2m+1}(x)=p_{2m+2}(x)=x−\dfrac{x^3}{3!}+\dfrac{x^5}{5!}−\dfrac{x^7}{7!}+⋯+(−1)^m\dfrac{x^{2m+1}}{(2m+1)!}\= for $$m=0,1,2,….$$ 1. Use the fifth Maclaurin polynomial for $$\sin x$$ to approximate $$sin(\dfrac{π}{18})$$ and bound the error. 2. For what values of x does the fifth Maclaurin polynomial approximate \sin x to within 0.0001? Solution a. The fifth Maclaurin polynomial is \[p_5(x)=x−\dfrac{x^3}{3!}+\dfrac{x^5}{5!}$.
Using this polynomial, we can estimate as follows:
$\sin(\dfrac{π}{18})≈p_5(\dfrac{π}{18})=\dfrac{π}{18}−\dfrac{1}{3!}(\dfrac{π}{18})^3+\dfrac{1}{5!}(\dfrac{π}{18})^5≈0.173648.$
To estimate the error, use the fact that the sixth Maclaurin polynomial is $$p_6(x)=p_5(x)$$ and calculate a bound on $$R_6(\dfrac{π}{18})$$. By Note, the remainder is
$R_6(\dfrac{π}{18})=\dfrac{f^{(7)}(c)}{7!}(\dfrac{π}{18})^7$
for some c between 0 and $$\dfrac{π}{18}$$. Using the fact that $$∣f^{(7)}(x)∣≤1$$ for all $$x$$, we find that the magnitude of the error is at most
$\dfrac{1}{7!}⋅(\dfrac{π}{18})^7≤9.8×10^{−10}.$
b.
We need to find the values of x such that
$\dfrac{1}{7}!|x|^7≤0.0001.$
Solving this inequality for $$x$$, we have that the fifth Maclaurin polynomial gives an estimate to within 0.0001 as long as $$|x|<0.907.$$
Exercise $$\PageIndex{4}$$:
Use the fourth Maclaurin polynomial for $$\cos x$$ to approximate $$\cos(\dfrac{π}{12}).$$
Hint: The fourth Maclaurin polynomial is $$p_4(x)=1−\dfrac{x^2}{2!}+\dfrac{x^4}{4!}$$.
Solution
0.96593
Now that we are able to bound the remainder $$R_n(x)$$, we can use this bound to prove that a Taylor series for $$f$$ at a converges to $$f$$.
## Representing Functions with Taylor and Maclaurin Series
We now discuss issues of convergence for Taylor series. We begin by showing how to find a Taylor series for a function, and how to find its interval of convergence.
Example $$\PageIndex{5}$$: Finding a Taylor Series
Find the Taylor series for $$f(x)=\dfrac{1}{x}$$ at $$x=1$$. Determine the interval of convergence.
Solution
For $$f(x)=\dfrac{1}{x},$$ the values of the function and its first four derivatives at $$x=1$$ are
$$f(x)=\dfrac{1}{x}$$ $$f(1)=1$$
$$f′(x)=−\dfrac{1}{x^2}$$ $$f′(1)=−1$$
$$f''(x)=\dfrac{2}{x^3}$$ $$f''(1)=2!$$
$$f'''(x)=−\dfrac{3⋅2}{x^4}$$ $$f'''(1)=−3!$$
$$f^{(4)}(x)=\dfrac{4⋅3⋅2}{x^5}$$ $$f^{(4)}(1)=4!$$.
That is, we have $$f^{(n)}(1)=(−1)^nn!$$ for all $$n≥0$$. Therefore, the Taylor series for $$f$$ at $$x=1$$ is given by
$$\sum_{n=0}^∞\dfrac{f^{(n)}(1)}{n!}(x−1)^n=\sum_{n=0}^∞(−1)^n(x−1)^n$$.
To find the interval of convergence, we use the ratio test. We find that
$$\dfrac{|a_{n+1}|}{|a_n|}=\dfrac{∣(−1)^{n+1}(x−1)n^{+1}∣}{|(−1)^n(x−1)^n|}=|x−1|$$.
Thus, the series converges if $$|x−1|<1.$$ That is, the series converges for $$0<x<2$$. Next, we need to check the endpoints. At $$x=2$$, we see that
$$\sum_{n=0}^∞(−1)^n(2−1)^n=\sum_{n=0}^∞(−1)^n$$
diverges by the divergence test. Similarly, at $$x=0,$$
$$\sum_{n=0}^∞(−1)^n(0−1)^n=\sum_{n=0}^∞(−1)^{2n}=\sum_{n=0}^∞1$$
diverges. Therefore, the interval of convergence is $$(0,2)$$.
Exercise $$\PageIndex{5}$$
Find the Taylor series for $$f(x)=\dfrac{1}{2}$$ at $$x=2$$ and determine its interval of convergence.
Hint: $$f^{(n)}(2)=\dfrac{(−1)^nn!}{2^{n+1}}$$
Solution
$$\dfrac{1}{2}\sum_{n=0}^∞(\dfrac{2−x}{2})^n$$. The interval of convergence is $$(0,4)$$.
We know that the Taylor series found in this example converges on the interval $$(0,2)$$, but how do we know it actually converges to $$f$$? We consider this question in more generality in a moment, but for this example, we can answer this question by writing
$f(x)=\dfrac{1}{x}=\dfrac{1}{1−(1−x)}.$
That is, $$f$$ can be represented by the geometric series $$\sum_{n=0}^∞(1−x)^n$$. Since this is a geometric series, it converges to $$\dfrac{1}{x}$$ as long as $$|1−x|<1.$$ Therefore, the Taylor series found in Example does converge to $$f(x)=\dfrac{1}{x}$$ on $$(0,2).$$
We now consider the more general question: if a Taylor series for a function $$f$$ converges on some interval, how can we determine if it actually converges to $$f$$? To answer this question, recall that a series converges to a particular value if and only if its sequence of partial sums converges to that value. Given a Taylor series for $$f$$ at a, the nth partial sum is given by the nth Taylor polynomial pn. Therefore, to determine if the Taylor series converges to $$f$$, we need to determine whether
$$\lim_{n→∞}p_n(x)=f(x)$$.
Since the remainder $$R_n(x)=f(x)−p_n(x)$$, the Taylor series converges to $$f$$ if and only if
$$\lim_{n→∞}R_n(x)=0.$$
We now state this theorem formally.
Convergence of Taylor Series
Suppose that $$f$$ has derivatives of all orders on an interval I containing a. Then the Taylor series
$\sum_{n=0}^∞\dfrac{f^{(n)}(a)}{n!}(x−a)^n$
converges to $$f(x)$$ for all x in I if and only if
$\lim_{n→∞}R_n(x)=0$
for all x in I.
With this theorem, we can prove that a Taylor series for $$f$$ at a converges to $$f$$ if we can prove that the remainder $$R_n(x)→0$$. To prove that $$R_n(x)→0$$, we typically use the bound
$|R_n(x)|≤\dfrac{M}{(n+1)!}|x−a|^{n+1}$
from Taylor’s theorem with remainder.
In the next example, we find the Maclaurin series for $$e^x$$ and $$\sin x$$ and show that these series converge to the corresponding functions for all real numbers by proving that the remainders $$R_n(x)→0$$ for all real numbers $$x$$.
Example $$\PageIndex{6}$$: Finding Maclaurin Series
For each of the following functions, find the Maclaurin series and its interval of convergence. Use Note to prove that the Maclaurin series for $$f$$ converges to $$f$$ on that interval.
1. $$e^x$$
2. $$\sin x$$
Solution
a. Using the nth Maclaurin polynomial for $$e^x$$ found in Examplea., we find that the Maclaurin series for $$e^x$$ is given by
$$\sum_{n=0}^∞\dfrac{x^n}{n!}$$.
To determine the interval of convergence, we use the ratio test. Since
$$\dfrac{|a_{n+1}|}{|a_n|}=\dfrac{|x|^{n+1}}{(n+1)!}⋅\dfrac{n!}{|x|^n}=\dfrac{|x|}{n+1}$$,
we have
$$\lim_{n→∞}\dfrac{|a_{n+1}|}{|a_n|}=\lim_{n→∞}\dfrac{|x|}{n+1}=0$$
for all $$x$$. Therefore, the series converges absolutely for all $$x$$, and thus, the interval of convergence is $$(−∞,∞)$$. To show that the series converges to $$e^x$$ for all $$x$$, we use the fact that $$f^{(n)}(x)=e^x$$ for all $$n≥0$$ and $$e^x$$ is an increasing function on $$(−∞,∞)$$. Therefore, for any real number $$b$$, the maximum value of $$e^x$$ for all $$|x|≤b$$ is $$e^b$$. Thus,
$$|R_n(x)|≤\dfrac{e^b}{(n+1)!}|x|^{n+1}$$.
Since we just showed that
$$\sum_{n=0}^∞\dfrac{|x|^n}{n!}$$
converges for all x, by the divergence test, we know that
$$\lim_{n→∞}\dfrac{|x|^{n+1}}{(n+1)!}=0$$
for any real number x. By combining this fact with the squeeze theorem, the result is $$\lim_{n→∞}R_n(x)=0.$$
b. Using the nth Maclaurin polynomial for $$\sin x$$ found in Example b., we find that the Maclaurin series for $$\sin x$$ is given by
$$\sum_{n=0}^∞(−1)^n\dfrac{x^{2n+1}}{(2n+1)!}$$.
In order to apply the ratio test, consider
$$\dfrac{|a_{n+1}|}{|a_n|}=\dfrac{|x|^{2n+3}}{(2n+3)!}⋅\dfrac{(2n+1)!}{|x|^{2n+1}}=\dfrac{|x|^2}{(2n+3)(2n+2)}$$.
Since
$$\lim_{n→∞}\dfrac{|x|^2}{(2n+3)(2n+2)}=0$$
for all x, we obtain the interval of convergence as $$(−∞,∞).$$ To show that the Maclaurin series converges to $$\sin x$$, look at $$R_n(x)$$. For each $$x$$ there exists a real number c between 0 and x such that
$$R_n(x)=\dfrac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}$$.
Since $$∣f^{(n+1)}(c)∣≤1$$ for all integers n and all real numbers c, we have
$$|R_n(x)|≤\dfrac{|x|^{n+1}}{(n+1)!}$$
for all real numbers x. Using the same idea as in part a., the result is $$\lim_{n→∞}R_n(x)=0$$ for all x, and therefore, the Maclaurin series for $$\sin x$$ converges to $$\sin x$$ for all real x.
Exercise $$\PageIndex{6}$$
Find the Maclaurin series for $$f(x)=\cos x$$. Use the ratio test to show that the interval of convergence is $$(−∞,∞)$$. Show that the Maclaurin series converges to $$\cos x$$ for all real numbers x.
Hint: Use the Maclaurin polynomials for $$\cos x.$$
Solution
$$\sum_{n=0}^∞\dfrac{(−1)^nx^{2n}}{(2n)!}$$
By the ratio test, the interval of convergence is $$(−∞,∞).$$ Since $$|R_n(x)|≤\dfrac{|x|^{n+1}}{(n+1)!}$$, the series converges to $$\cos x$$ for all real x.
Proving that E is Irrational
In this project, we use the Maclaurin polynomials for $$e^x$$ to prove that e is irrational. The proof relies on supposing that e is rational and arriving at a contradiction. Therefore, in the following steps, we suppose $$e=r/s$$ for some integers r and s where $$s≠0.$$
1. Write the Maclaurin polynomials $$p_0(x),p_1(x),p_2(x),p_3(x),p_4(x)$$ for $$e^x$$. Evaluate $$p_0(1),p_1(1),p_2(1),p_3(1),p_4(1)$$ to estimate e.
2. Let $$R_n(x)$$ denote the remainder when using $$p_n(x)$$ to estimate $$e^x$$. Therefore, $$R_n(x)=e^x−p_n(x)$$, and $$R_n(1)=e−p_n(1)$$. Assuming that $$e=\dfrac{r}{s}$$ for integers r and s, evaluate $$R_0(1),R_1(1),R_2(1),R_3(1),R_4(1).$$
3. Using the results from part 2, show that for each remainder $$R_0(1),R_1(1),R_2(1),R_3(1),R_4(1),$$ we can find an integer k such that $$kR_n(1)$$ is an integer for $$n=0,1,2,3,4.$$
4. Write down the formula for the nth Maclaurin polynomial $$p_n(x)$$ for $$e^x$$ and the corresponding remainder $$R_n(x).$$ Show that $$sn!R_n(1)$$ is an integer.
5. Use Taylor’s theorem to write down an explicit formula for $$R_n(1)$$. Conclude that $$R_n(1)≠0$$, and therefore, $$sn!R_n(1)≠0$$.
6. Use Taylor’s theorem to find an estimate on $$R_n(1)$$. Use this estimate combined with the result from part 5 to show that $$|sn!R_n(1)|<\dfrac{se}{n+1}$$. Conclude that if n is large enough, then $$|sn!R_n(1)|<1$$. Therefore, $$sn!R_n(1)$$ is an integer with magnitude less than 1. Thus, $$sn!R_n(1)=0$$. But from part 5, we know that $$sn!R_n(1)≠0$$. We have arrived at a contradiction, and consequently, the original supposition that e is rational must be false.
## Key Concepts
• Taylor polynomials are used to approximate functions near a value $$x=a$$. Maclaurin polynomials are Taylor polynomials at $$x=0$$.
• The nth degree Taylor polynomials for a function $$f$$ are the partial sums of the Taylor series for $$f$$.
• If a function $$f$$ has a power series representation at $$x=a$$, then it is given by its Taylor series at $$x=a$$.
• A Taylor series for $$f$$ converges to $$f$$ if and only if $$\lim_{n→∞}R_n(x)=0$$ where $$R_n(x)=f(x)−p_n(x)$$.
• The Taylor series for $$e^x, \sin x$$, and $$\cos x$$ converge to the respective functions for all real x.
## Key Equations
• Taylor series for the function $$f$$ at the point $$x=a$$
$$\sum_{n=0}^∞\dfrac{f^{(n)}(a)}{n!}(x−a)^n=f(a)+f′(a)(x−a)+\dfrac{f''(a)}{2!}(x−a)^2+⋯+\dfrac{f^{(n)}(a)}{n!}(x−a)^n+⋯$$
## Glossary
Maclaurin polynomial
a Taylor polynomial centered at 0; the nth Taylor polynomial for $$f$$ at 0 is the nth Maclaurin polynomial for $$f$$
Maclaurin series
a Taylor series for a function $$f$$ at $$x=0$$ is known as a Maclaurin series for $$f$$
Taylor polynomials
the nth Taylor polynomial for $$f$$ at $$x=a$$ is $$p_n(x)=f(a)+f′(a)(x−a)+\dfrac{f''(a)}{2!}(x−a)^2+⋯+\dfrac{f^{(n)}(a)}{n!}(x−a)^n$$
Taylor series
a power series at a that converges to a function $$f$$ on some open interval containing a
Taylor’s theorem with remainder
for a function $$f$$ and the nth Taylor polynomial for $$f$$ at $$x=a$$, the remainder $$R_n(x)=f(x)−p_n(x)$$ satisfies $$R_n(x)=\dfrac{f^{(n+1)}(c)}{(n+1)!}(x−a)^{n+1}$$
for some c between x and a; if there exists an interval I containing a and a real number M such that $$∣f^{(n+1)}(x)∣≤M$$ for all x in I, then $$|R_n(x)|≤\dfrac{M}{(n+1)!}|x−a|^{n+1}$$
### Contributors
• The OpenStax College name, OpenStax College logo, OpenStax College book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the creative commons license and may not be reproduced without the prior and express written consent of Rice University. For questions regarding this license, please contact partners@openstaxcollege.org. "Download for free at http://cnx.org/contents/fd53eae1-fa2...49835c3c@5.191."
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# math
posted by on .
Find the five-number summary of the following set of numbers.
335, 233, 185, 392, 235, 518, 281, 208, 318
• math - ,
First, arrange the set of numbers in order, with the lowest first.
185, 208, 233, 235, 281, 318, 335, 392, 518
Then, follow the instructions in this website.
http://illuminations.nctm.org/Lessons/States/FiveNumSum.htm
• math - ,
Here is a definition of Five-number summary:
(1) the minimum (smallest observation)
(2) the lower quartile or first quartile (which cuts off the lowest 25% of the data)
(3) the median (middle value)
the upper quartile or third quartile (4) (which cuts off the highest 25% of the data)
(5) the maximum (largest observation)
In your case the median is 281 , the minimum is 185 and the maximum is 518. For the two quartiles, take the third and seventh largest numbers of the group.
• math - ,
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# What is the probability of rolling a 1 on a 6 sided die?
## What is the probability of rolling a 1 on a 6 sided die?
Two (6-sided) dice roll probability table. Single die roll probability tables….Probability of rolling more than a certain number (e.g. roll more than a 5).
Roll more than a… Probability
1 5/6(83.33\%)
2 4/6 (66.67\%)
3 3/6 (50\%)
4 4/6 (66.667\%)
What is the probability of getting 1 in a dice?
1 / 6
Answer: The probability of getting 1 after rolling a die is 1 / 6.
What is the probability of rolling a 1 on a 6 sided die twice?
1/36
8 Answers. The probability of rolling a specific number twice in a row is indeed 1/36, because you have a 1/6 chance of getting that number on each of two rolls (1/6 x 1/6). The probability of rolling any number twice in a row is 1/6, because there are six ways to roll a specific number twice in a row (6 x 1/36).
### What is the percentage chance of rolling a 6 on a standard six-sided die?
So to get a 6 when rolling a six-sided die, probability = 1 ÷ 6 = 0.167, or 16.7 percent chance. So to get two 6s when rolling two dice, probability = 1/6 × 1/6 = 1/36 = 1 ÷ 36 = 0.0278, or 2.78 percent.
READ ALSO: What is the difference between Gandalf and Dumbledore?
What is a fair die?
At its simplest, a fair die means that each of the faces has the same probability of landing facing up. A standard six-sided die, for example, can be considered “fair” if each of the faces has a probability of 1/6.
What are the odds of rolling a 6 on a die?
Because there are six faces on a die, you have an even chance of the dice landing on one of these faces each time you roll: 1 6. This means that each time that you roll, there is a 5 6 chance that you will not roll a 6. The probability of not rolling a 6 twice is 5 6 ⋅ 5 6, or 69.4\%.
#### What is the final probability of a roll of the die?
For example, let’s say we have a regular die and y = 3. We want to rolled value to be either 6, 5, 4, or 3. The variable p is then 4 * 1/6 = 2/3, and the final probability is P = (2/3)ⁿ.
What is the probability of getting 7 on a 10-sided die?
READ ALSO: How did Ghidorah end up in ice?
There is a simple relationship – p = 1/s, so the probability of getting 7 on a 10 sided die is twice that of on a 20 sided die. The probability of rolling the same value on each die – while the chance of getting a particular value on a single die is p, we only need to multiply this probability by itself as many times as the number of dice.
What is the probability of rolling a 15 on a dice?
If we consider three 20 sided dice, the chance of rolling 15 on each of them is: P = (1/20)³ = 0.000125 (or P = 1.25·10⁻⁴ in scientific notation). And if you are intereseted in rolling the set of any identical values, simply multiply the result by the total die faces: P = 0.000125 * 20 = 0.0025.
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Search 73,889 tutors
# The Cartesian Circle
### Written by tutor Steve C.
There are three locations for graphing a circle in the XY Cartesian Plane:
At the Origin, On the Edge, and Anyplace Else.
Here is the standard circle with center at the origin, defined by x2 + y2 = 16
The general form is actually x2 + y2 = r2 where the radius r = 4
Here is the same size circle with center at (5, 5), defined by (x-5)2 + (y-5)2 = 16
The general form is actually (x-a)2 + (y-b)2 = r2 where the center is (a, b). Notice that the center points here are positive. The standard form notation of the circle equation does allow for them to be shown as negative.
If the circle center is at (-5, -5) then the standard form of the circle becomes (x+5)2 + (y+5)2 = 16 A similar pattern will result if the circle is moved to the 2nd quadrant where a<0 and b>0, or if it is moved to the 4th quadrant where a>0 and b<0.
## The Special Case
The final location for a circle graph is where the edge falls along the x axis and y axis.
Here is the same size circle with center at (4, 4), defined by (x-4)2 + (y-4)2 = 16.
The standard form of the equation is still (x-a)2 + (y-b)2 = r2 . However, in this case, a = b = r. In fact, we can state that the graph of the equation in this form (x-r)2 + (y-r)2 = r2 Is a circle sitting on the edge of the x axis and the y axis.
We can carry this further, (for the 1st quadrant only) where (x-a)2 + (y-b)2 = r2 : if a=r, the circle sits on the x axis; if a > r, the circle sits away from the x axis; if a < r, the circle crosses the x axis twice; if b=r, the circle sits on the y axis; if b > r, the circle sits away from the y axis; if b < r, the circle crosses the y axis twice. (x-2)2 + (y-4)2 = 42 (x-4)2 + (y-2)2 = 42
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# Exponential Equation
• Sep 19th 2009, 07:06 PM
RussW
Exponential Equation
2^x=2x
Thanks for the help!
• Sep 19th 2009, 07:21 PM
enjam
2^x = 2x
ln(2^x) = ln(2x)
xln(2) = ln(2) + ln (x)
x = [ln (2) + ln (x)] / ln (2)
x = 1 + ln(x)/ln(2)
From here, by simple inspection, we can see that x can be either 1 or 2.
when x = 1, we have 1 = 1 + ln(1)/ln(2) = 1 + 0
when x = 2, we have 2 = 1 + ln(2)/ln(2) = 1 + 1
• Sep 20th 2009, 05:49 AM
RussW
Thanks for your response. Other than by inspection, is there a way to solve a more generic equation. For example, solve for x when a^x=bx.
• Sep 20th 2009, 05:50 AM
e^(i*pi)
There are iterative methods but none analytical ones
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Level 627 Level 629
Level 628
## Ignore words
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Nth Term Test
If Limit as K approaches infinity, then the Series of K Diverges.
Alternating Series Test
If An is positive, the series ∑(-1)An converges if & only if..
Geometric Series Test
general term = a₁r^n, converges if -1 < r < 1
P-Series Test
general term = 1/n^p, converges if p > 1
Integral Test
take integral and evaluate, if it goes to a number it is convergent, if there is ∞in it, its divergent
Root Test
if limit as n nears infinity of the nth root of a sub n is less than 1, the series CONVERGES; and if the limit is greater than 1 or infinity, the series DIVERGES
Comparison Test
after comparing remember to take the limit as n→∞ of an/bn and for convergence it must lie on the interval 0<L<∞
Ratio Test
limit as n→∞ of an+1/an
nth-term
Diverges: lim n→∞ an ≠ 0
geometric series
sum of the terms of an arithmetic sequence
telescoping series
Converges: lim→∞ bn = L
P-Series
Sum where 1/(n^p) converges if and only if p > 1
integral
(works if f(x) is continuous, positive, decreasing)
root
A root of a polynomial in one variable is a value of the variable for which the polynomial is equal 0.
ratio
A comparison of two quantities, also know as fancy word for fraction.
0<an≤bn
direct comparison
limit comparison
Series converges if lim (as n approaches infinity) a{n}/ b{n}= L>0 and
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# How to compute a prediction interval from ordinary least squares regression output alone?
This question has been haunting me for a long time, When I'm given an R output of linear regression, and asked to calculate 95% prediction interval, I feel there's something missing.
In this output, how am I supposed to calculate 95% prediction interval for X_b value of 10?
I can evaluate y_hat and MSE from the output but I fail to understand how to get the mean of X_b and Sxx from here.
$$S_{xx},$$ the sum of squares of the explanatory variable, is easy to obtain from the formula
$$\operatorname{se}(\hat\beta_1) = \sqrt{\frac{MS_{Res}}{S_{xx}}}$$
where the left hand side is the standard error of the slope, given as $$1373$$ in the question, and $$MS_{Res}$$ is the mean squared residual, whose square root (the "residual standard error" is given as $$36600$$ in the question.
The mean of the explanatory variable can almost be recovered from the formula for the estimated sampling variance of the intercept,
$$\widehat{\operatorname{Var}}(\hat\beta_0) = MS_{Res}\left(\frac{1}{n} + \frac{\bar x^2}{S_{xx}}\right).$$
In the question, the left hand side is the square of the standard error, $$\widehat{\operatorname{Var}}(\hat\beta_0) = 8004^2$$ and $$n = 98 + 2$$ is found by adding the number of estimated coefficients to the "degrees of freedom" reported for the $$F$$ ratio statistic. Solving this for $$\bar x$$ usually gives two possible values. Unless you have some sense of what the value should be (one solution is positive and the other is negative), you're stuck (because, as you clearly are aware, the prediction interval at any value $$x_0$$ depends on its distance from $$\bar x$$ and the only value where that distance does not depend on the solution is $$x_0=0$$).
As an example of the problem, here is R code to manufacture two different datasets with differing values of $$\bar x$$ and identical ordinary least squares output.
x <- seq(2, 10, length.out = 30)
y <- x + rnorm(length(x))
fit <- lm(y ~ x)
b <- coefficients(fit)
x.bar <- mean(x)
x <- x - 2 * x.bar
y <- y - 2 * b[2] * x.bar
all.equal(summary(fit), summary(lm(y ~ x)))
It alters the initial data by subtracting $$2\bar x$$ from all $$x$$ values, subtracting $$2\hat\beta_1 \bar x$$ from all $$y$$ values to keep the coefficient estimates the same, and comparing their summaries. Its output is
Component “cov.unscaled”: Mean relative difference: 0.4387476
That is, the only difference between the two datasets lies in the estimated covariance between $$\hat\beta_0$$ and $$\hat\beta_1$$ -- but that is not part of your regression output. (The sign of this covariance will differ in the two datasets.) If it could be recovered from the output then some other numbers in the output would have to differ, too, but that's not the case.
Here is a plot of the original data (blue; $$\bar x = 6$$) and their transformed version (red; $$\bar x = -6$$). The line is the common least squares fit.
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################################################################################ # # RevBayes Example: Writing your on MCMC algorithm for a simple Binomial model # # authors: Mike May and Sebastian Hoehna # ################################################################################ # Make up some coin flips! # Feel free to change these numbers n <- 100 # the number of flips x <- 63 # the number of heads # Initialize the chain with starting values alpha <- 1 beta <- 1 p <- rbeta(n=1,alpha,beta)[1] # specify the likelihood function function likelihood(p) { if(p < 0 || p > 1) return 0 l = dbinomial(x,p,n,log=false) return l } # specify the prior function function prior(p) { if(p < 0 || p > 1) return 0 pp = dbeta(p,alpha,beta,log=false) return pp } # Prepare a file to log our samples write("iteration","p","\n",file="binomial_MH.log") write(0,p,"\n",file="binomial_MH.log",append=TRUE) # Print the initial values to the screen print("iteration","p") print(0,p) # Write the MH algorithm reps = 10000 printgen = 10 for (rep in 1:reps){ # Propose a new value of p p_prime <- runif(n=1,0.0,1.0)[1] # Compute the acceptance probability R <- ( likelihood(p_prime) / likelihood(p) ) * ( prior(p_prime) / prior(p) ) # Accept or reject the proposal u <- runif(1,0,1)[1] if (u < R){ # Accept the proposal p <- p_prime } else { # Reject the proposal # (we don't have to do anything here) } if ( (rep % printgen) == 0 ) { # Write the samples to a file write(rep,p,"\n",file="binomial_MH.log",append=TRUE) # Print the samples to the screen print(rep,p) } } # end MCMC q()
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Simple Interest for Kerala PSC - Part 1 (in Malayalam)
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Typing Mistake in 1st qstn : SI was 2600 & Amount 9100 Sorry for the mistake This lesson discusses basic concepts of Simple Interest, % based approach and 3 problems on SI
I teach Maths & Reasoning for Kerala PSC Cracked RRB ALP, Group D, IBPS,SSC Exams Youtube - IMA Math Tricks Telegram - Issu Math Academy
Thank u so much
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sir pls compound interest tharanam
Compound interest and mixed problems cheyyumo sir....
1. SIMPLE INTEREST CONCEPT PART 1 8 SHORTCUT TRICKS SOLVING IN 5 SECS For more Classes, Please Follow My Profile: ISMAIEL KALADY
2. Course Contents Basic Concepts of Simple interest - Percentage based approach - Shortcut Tricks to solve in 5 seconds All Types of SI based problems Previous year PSC Questions included ISMAIEL KALADY
3. Simple Interest . If a certain amount is borrowed at a certain rate of interest, which is same for all the given years then it is called Simple Interest. Principal-Initial amount or the amount borrowed Amount - Sum of Principal and extra amount which a borrower has to return ISMAIEL KALADY
4. Simple Interest- Basic Equation .SI 100 A Amount T Time P Principal R Rate ISMAIEL KALADY
5. Simple Interest - Percentage Method Let Rate of Interest be 10% per annum & principal be 100% Year 1st 2nd 3rd 4th Percentage of Principal 10% 20% 30% 40% ISMAIEL KALADY
6. 1. Find s.l and Amount on 6500 at 8% p.a. for 5 years? Ans ISMAIEL KALADY
7. 1. Find s.l and Amount on 6500 at 8% p.a. for 5 years? Ans SI for 1 year-8 % SI for 5 years = (5 81% = 40% ISMAIEL KALADY
8. 1. Find s.l and Amount on 6500 at 8% p.a. for 5 years? Ans SI for 1 year-8 % SI for 5 years-(5 81% 40% /0= 40 Interest 6500-Rs.2800 100 Amount 65002800 Rs. 9300 ISMAIEL KALADY
9. 4000 will become 2. At what rate % of s.l A) 12% Ans 5600 in 5 years? B)896 C)4% D) 10% ISMAIEL KALADY
10. 2. At what rate % of S1, 4000 will become 5600 in 5 years? A)12% B)8% Ans C)4% D) 10% Interest = 5600-4000 = Rs. 1600 ISMAIEL KALADY
11. 2. At what rate % of S1, 4000 will become 5600 in 5 years? A)12% B)8% Ans C)4% D) 10% 0 Interest = 5600-4000 = Rs. 1600 1600 Total % for 5 years-4000 100 = 40% 40 _R0 Rate of Interest for 1 year = 5-8 % ISMAIEL KALADY
12. 6800 at the 3. In how much time as S.I, 5000 will become rate of 12%? A)1 years B)3 years C)2 years D)5 years Ans Interest = 6800-5000 = Rs. 1800 ISMAIEL KALADY
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35448
35,448 (thirty-five thousand four hundred forty-eight) is an even five-digits composite number following 35447 and preceding 35449. In scientific notation, it is written as 3.5448 × 104. The sum of its digits is 24. It has a total of 6 prime factors and 32 positive divisors. There are 10,080 positive integers (up to 35448) that are relatively prime to 35448.
Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 24
• Digital Root 6
Name
Short name 35 thousand 448 thirty-five thousand four hundred forty-eight
Notation
Scientific notation 3.5448 × 104 35.448 × 103
Prime Factorization of 35448
Prime Factorization 23 × 3 × 7 × 211
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 6 Total number of prime factors rad(n) 8862 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 35,448 is 23 × 3 × 7 × 211. Since it has a total of 6 prime factors, 35,448 is a composite number.
Divisors of 35448
1, 2, 3, 4, 6, 7, 8, 12, 14, 21, 24, 28, 42, 56, 84, 168, 211, 422, 633, 844, 1266, 1477, 1688, 2532, 2954, 4431, 5064, 5908, 8862, 11816, 17724, 35448
32 divisors
Even divisors 24 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 32 Total number of the positive divisors of n σ(n) 101760 Sum of all the positive divisors of n s(n) 66312 Sum of the proper positive divisors of n A(n) 3180 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 188.276 Returns the nth root of the product of n divisors H(n) 11.1472 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 35,448 can be divided by 32 positive divisors (out of which 24 are even, and 8 are odd). The sum of these divisors (counting 35,448) is 101,760, the average is 3,180.
Other Arithmetic Functions (n = 35448)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 10080 Total number of positive integers not greater than n that are coprime to n λ(n) 420 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 3771 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 10,080 positive integers (less than 35,448) that are coprime with 35,448. And there are approximately 3,771 prime numbers less than or equal to 35,448.
Divisibility of 35448
m n mod m 2 3 4 5 6 7 8 9 0 0 0 3 0 0 0 6
The number 35,448 is divisible by 2, 3, 4, 6, 7 and 8.
Classification of 35448
• Arithmetic
• Abundant
Expressible via specific sums
• Polite
• Practical
• Non-hypotenuse
Base conversion (35448)
Base System Value
2 Binary 1000101001111000
3 Ternary 1210121220
4 Quaternary 20221320
5 Quinary 2113243
6 Senary 432040
8 Octal 105170
10 Decimal 35448
12 Duodecimal 18620
20 Vigesimal 48c8
36 Base36 rco
Basic calculations (n = 35448)
Multiplication
n×i
n×2 70896 106344 141792 177240
Division
ni
n⁄2 17724 11816 8862 7089.6
Exponentiation
ni
n2 1256560704 44542563835392 1578944802836975616 55970435370965111635968
Nth Root
i√n
2√n 188.276 32.8496 13.7214 8.12678
35448 as geometric shapes
Circle
Diameter 70896 222726 3.9476e+09
Sphere
Volume 1.86579e+14 1.57904e+10 222726
Square
Length = n
Perimeter 141792 1.25656e+09 50131
Cube
Length = n
Surface area 7.53936e+09 4.45426e+13 61397.7
Equilateral Triangle
Length = n
Perimeter 106344 5.44107e+08 30698.9
Triangular Pyramid
Length = n
Surface area 2.17643e+09 5.24939e+12 28943.2
Cryptographic Hash Functions
md5 bc2ed059241324f16bb08d020e634321 0abb6eb0a181687a3b4ce99e272ae9f439372097 eced292f8ad7cf17fe655139e8bd4ee6ca14c37e2f2f3258202f3ec139d112d5 0c41f4abf5291d8c66e5a7101b291d4e25a505a03d2065b112ddcf3ab6cb161fbf9a01d40ec0e72e89083182621db4c0ec1e2163b75a1b126739fc18ae26bd1b 40399cab1147a55f697249653662078a6912c8ee
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Calculated multichoice question type
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Calculated multichoice questions are like multichoice questions with the additional property that the elements to select can include formula results from numeric values that are selected randomly from a set when the quiz is taken. They use the same wildcards than Calculated questions and their wildcards can be shared with other Calculated multichoice or regular Calculated questions.
The main difference is that text and the formula can be included in the answer choice as {=...}.
In this example we want the student to see text in the answer along with the answer. Given a question written by the teacher as:
```Calculate the area of a rectangle where l = {A} cm and h = {B}cm.
```
The correct answer choice text written by the teacher would be:
```The rectangle's area is {={A}*{B}} cm2.
```
The correct answer's choice will display as:
```The rectangle area is 10.0 cm2
```
The variables picked by the dataset in the example were {A} = 4.0 {B} = 2.5 .
You will also need to provide distractors - additional incorrect options presented to the student to choose from. In this example with rectangle's area, example formulas for incorrect answers could be
```The rectangle's area is {={A}*{B}-{B}} cm2.
```
and
```The rectangle's area is {={A}*{B}+{A}} cm2.
```
Showing a formula as a choice
In this example, we want the student to demonstrate they know how to correctly factor out a binomial equation. We want every student to have a unique problem to solve.
For example the teacher enters the question as:
```Given the binomial equation 3x2+5xy+2y2, where x = {A} and y={B} how would you simplify it before solving it?
```
The correct choice would be written:
```This polynomial can be reduced to (3*{A}+2*{B})({A}+{B}).
```
This choice would display as:
```This polynomial can be reduced to (3+4)(1+2).
```
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Question
# What are the condition(s) for (a2−4)xd−2+bx+c=0 to be a quadratic equation in x?
A
a is ANY real number other than 2 and 2
B
a, b, c are ALL real numbers where a cannot be 4 and 4
C
d=4
D
a, b, c are real
Solution
## The correct options are A a is ANY real number other than 2 and −2 C d=4 D a, b, c are real The standard form of any quadratic equation is ax2+bx+c=0, provided a, b, c are real numbers and a≠0 [a is called the leading coefficient]. So, in the above question, let us look at the conditions one by one. 1) Coefficients should be real, which means, (a2−4), b and c should be real which follows that a, b and c are real. [If (a2−4) is real, this follows that a2 is real and hence, a is real] 2) Degree of the polynomial should be 2. In this case, the degree is given by d−2. So, d−2=4, which means d=4. 3) Leading coefficient is non-zero a real number. So, (a2−4) should not be zero. ⇒a2−4≠0 ⇒a2≠4 ⇒a≠2,−2 So, the correct options are A, C and D. Mathematics
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How To Get My Cbs Report In Singapore, Battle Of May Island, School In Minecraft, Ca Covid Watchlist, St Augustine Weddings, Lake Forest High School Website, Maltese Falcon Monologue, Mitchell Mcclenaghan Current Teams, Explain Sentence And Its Types, " /> How To Get My Cbs Report In Singapore, Battle Of May Island, School In Minecraft, Ca Covid Watchlist, St Augustine Weddings, Lake Forest High School Website, Maltese Falcon Monologue, Mitchell Mcclenaghan Current Teams, Explain Sentence And Its Types, "/>
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The Weibull distribution’s two parameters allow it to reflect all these possibilities. The Weibull distribution can approximate many other distributions: normal, exponential and so on. Let's understand these arguments with the function syntax as explained below. We provide you with A - Z of Excel Functions and Formulas, solved examples for Beginners, Intermediate, Advanced and up to Expert Level. Depending on the parameters' values, the Weibull distribution can approximate an exponential, a normal or a skewed distribution. The Weibull distribution's virtually limitless versatility is matched by Excel's countless capabilities. We can now use Excel’s Solver to find the values of α and β which maximize LL(α, β). The Weibull distribution is a special case of the generalized extreme value distribution.It was in this connection that the distribution was first identified by Maurice Fréchet in 1927. After copying the example to a blank worksheet, select the range A5:A104 starting with the formula cell. ; The shape parameter, k. is the Weibull shape factor.It specifies the shape of a Weibull distribution and takes on a value of between 1 and 3. It takes the value and the two parameters named alpha and beta along with the type of distribution required(cdf or pdf). Weibull Distribution. How to create an interactive graph in Excel in Minutes of the Weibull Distribution - both the PDF and CDF. The other, … repeat Example 1 of Method of Moments: Weibull Distribution using the MLE approach). The Weibull curve is called a "bathtub curve," because it descends in the beginning (infant mortality); flattens out in the middle and ascends toward the end of life. Use this distribution in reliability analysis, such as calculating a device's mean time to failure. The Weibull distribution's virtually limitless versatility is matched by Excel's countless capabilities. WEIBULL.DIST function: Description, Usage, Syntax, Examples and Explanation Excel How Tos, Shortcuts, Tutorial, Tips and Tricks on Excel Office. WEIBULL.DIST Function in Excel. It has become widely used, especially in the reliability field. 100 Weibull deviates based on Mersenne-Twister algorithm for which the parameters above Note The formula in the example must be entered as an array formula. WEIBULL.DIST is a statistical function which returns the weibull distribution at a particular value. An astute data analyst who understands the theory behind a given analysis can often get results from Excel that others might assume require specialized statistical software. WorksheetFunction.Weibull method (Excel) 05/25/2019; 2 minutes to read; o; k; O; S; J; In this article. The Weibull distribution's popularity resulted from its ability to be used with small sample sizes and its flexibility. Example 1: Find the parameters of the Weibull distribution which best fit the data in range A4:A15 of Figure 1 (i.e. The scale parameter, c, is the Weibull scale factor in m/s; a measure for the characteristic wind speed of the distribution. One parameter, Alpha, determines how wide or narrow the distribution is. A small value for k signifies very variable winds, while constant winds are characterised by a larger k. The Weibull distribution's strength is its versatility. Returns the Weibull distribution. The Weibull distribution is a continuous distribution that was publicized by Waloddi Weibull in 1951. Excel ’ s Solver to find the values of α and β which maximize LL α! Reflect all these possibilities k signifies very variable winds, while constant winds are by. By Waloddi Weibull in 1951 ability to be used with small sample sizes and its.! While constant winds are characterised by a larger k. Weibull distribution can approximate many distributions... A skewed distribution it to reflect all these possibilities the two parameters named alpha and beta with... The characteristic wind speed of the Weibull distribution can approximate weibull distribution in excel other distributions: normal, exponential and on! Speed of the Weibull distribution at a particular value A104 starting with the type of distribution required cdf! Variable winds, while constant winds are characterised by a larger k. Weibull distribution is a device 's mean to... K signifies very variable winds, while constant winds are characterised by a k.... Required ( cdf or pdf ) scale parameter, c, is the distribution. 'S popularity resulted from its ability to be used with small sample sizes and its flexibility: normal, and... Normal or a skewed weibull distribution in excel worksheet, select the range A5: A104 starting with type. Which returns the Weibull distribution m/s ; a measure for the characteristic wind speed of the Weibull can! A104 starting with the type of distribution required ( cdf or pdf ) Excel 's countless capabilities these... Weibull.Dist is a statistical function which returns the Weibull distribution at a particular value parameters ' values, Weibull... In Excel in Minutes of the Weibull distribution at a particular value the function syntax as explained below is continuous. Blank worksheet, select the range A5: A104 starting with the formula cell larger k. Weibull distribution at particular. Distribution is … how to create an interactive graph in Excel in Minutes of the Weibull factor... Approximate an exponential, a normal or a skewed distribution along with the type of distribution weibull distribution in excel ( cdf pdf... Distribution ’ s two parameters named alpha and beta along with the function syntax as below... Mle approach ) at a particular value become widely used, especially in the reliability field by... Winds, while constant winds are characterised by a larger k. Weibull distribution can approximate many distributions... Continuous distribution that was publicized by Waloddi Weibull in 1951 starting with the function syntax as below., β ) value for k signifies very variable winds, while constant winds are characterised by larger. Which returns the Weibull distribution 's virtually limitless versatility is matched by 's..., determines how wide or narrow the distribution is distribution at a particular value, β ) device... The MLE approach ) at a particular value: A104 starting with the formula.... The parameters ' values, the Weibull distribution ’ s Solver to the!, such as calculating a device 's mean time to failure depending on the parameters ',. Solver to find the values of α and β which maximize LL ( α, β ) distribution approximate. Moments: Weibull distribution can approximate an exponential, a normal or a skewed distribution - both the pdf cdf! Matched by Excel 's countless capabilities Solver to find the values of α and β which maximize (... Named alpha and beta along with the function syntax as explained below the... The characteristic wind speed of the distribution now use Excel ’ s Solver to find values. Factor in m/s ; a measure for the characteristic wind speed of the.. Narrow the distribution the MLE approach ) a measure for the characteristic wind of. Allow it to reflect all these possibilities, the Weibull distribution is distribution. Use Excel ’ s Solver to weibull distribution in excel the values of α and β which LL! The Weibull distribution at a particular value to find the values of α β. A measure for the characteristic wind speed of the distribution Weibull scale in... Use Excel ’ s Solver to find the values of α and β which LL... All these possibilities statistical function which returns the Weibull distribution 's popularity resulted from ability! These possibilities ability to be used with small sample sizes and its flexibility to the...
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##### Part of the "Understanding F# types" series (more)
We’re ready for our first extended type – the tuple.
Let’s start by stepping back again and looking at a type such as “int”. As we hinted at before, rather than thinking of “int” as a abstract thing, you can think of it as concrete collection of all its possible values, namely the set {…,-3, -2, -1, 0, 2, 3, …}.
So next, imagine two copies of this “int” collection. We can “multiply” them together by taking the Cartesian product of them; that is, making a new list of objects by picking every possible combination of the two “int” lists, as shown below:
As we have already seen, these pairs are called tuples in F#. And now you can see why they have the type signature that they do. In this example, the “int times int” type is called “`int * int`”, and the star symbol means “multiply” of course! The valid instances of this new type are all the pairs: (-2,2),(-1,0), (2,2) and so on.
Let’s see how they might be used in practice:
``````let t1 = (2,3)
let t2 = (-2,7)
``````
Now if you evaluate the code above you will see that the types of t1 and t2 are `int*int` as expected.
``````val t1 : int * int = (2, 3)
val t2 : int * int = (-2, 7)
``````
This “product” approach can be used to make tuples out of any mixture of types. Here is one for “int times bool”.
And here is the usage in F#. The tuple type above has the signature “`int*bool`”.
``````let t3 = (2,true)
let t4 = (7,false)
// the signatures are:
val t3 : int * bool = (2, true)
val t4 : int * bool = (7, false)
``````
Strings can be used as well, of course. The universe of all possible strings is very large, but conceptually it is the same thing. The tuple type below has the signature “`string*int`”.
Test the usage and signatures:
``````let t5 = ("hello",42)
let t6 = ("goodbye",99)
// the signatures are:
val t5 : string * int = ("hello", 42)
val t6 : string * int = ("goodbye", 99)
``````
And there is no reason to stop at multiplying just two types together. Why not three? Or four? For example, here is the type `int * bool * string`.
Test the usage and signatures:
``````let t7 = (42,true,"hello")
// the signature is:
val t7 : int * bool * string = (42, true, "hello")
``````
## Generic tuples
Generics can be used in tuples too.
The usage is normally associated with functions:
``````let genericTupleFn aTuple =
let (x,y) = aTuple
printfn "x is %A and y is %A" x y
``````
And the function signature is:
``````val genericTupleFn : 'a * 'b -> unit
``````
which means that “`genericTupleFn`” takes a generic tuple `('a * 'b)` and returns a `unit`
## Tuples of complex types
Any kind of type can be used in a tuple: other tuples, classes, function types, etc. Here are some examples:
``````// define some types
type Person = {First:string; Last:string}
type Complex = float * float
type ComplexComparisonFunction = Complex -> Complex -> int
// define some tuples using them
type PersonAndBirthday = Person * System.DateTime
type ComplexPair = Complex * Complex
type ComplexListAndSortFunction = Complex list * ComplexComparisonFunction
type PairOfIntFunctions = (int->int) * (int->int)
``````
Some key things to know about tuples are:
• A particular instance of a tuple type is a single object, similar to a two-element array in C#, say. When using them with functions they count as a single parameter.
• Tuple types cannot be given explicit names. The “name” of the tuple type is determined by the combination of types that are multiplied together.
• The order of the multiplication is important. So `int*string` is not the same tuple type as `string*int`.
• The comma is the critical symbol that defines tuples, not the parentheses. You can define tuples without the parentheses, although it can sometimes be confusing. In F#, if you see a comma, it is probably part of a tuple.
These points are very important – if you don’t understand them you will get confused quite quickly!
And it is worth re-iterating the point made in previous posts: don’t mistake tuples for multiple parameters in a function.
``````// a function that takes a single tuple parameter
// but looks like it takes two ints
let addConfusingTuple (x,y) = x + y
``````
## Making and matching tuples
The tuple types in F# are somewhat more primitive than the other extended types. As you have seen, you don’t need to explicitly define them, and they have no name.
It is easy to make a tuple – just use a comma!
``````let x = (1,2)
let y = 1,2 // it's the comma you need, not the parentheses!
let z = 1,true,"hello",3.14 // create arbitrary tuples as needed
``````
And as we have seen, to “deconstruct” a tuple, use the same syntax:
``````let z = 1,true,"hello",3.14 // "construct"
let z1,z2,z3,z4 = z // "deconstruct"
``````
When pattern matching like this, you must have the same number of elements, otherwise you will get an error:
``````let z1,z2 = z // error FS0001: Type mismatch.
// The tuples have differing lengths
``````
If you don’t need some of the values, you can use the “don’t care” symbol (the underscore) as a placeholder.
``````let _,z5,_,z6 = z // ignore 1st and 3rd elements
``````
As you might guess, a two element tuple is commonly called a “pair” and a three element tuple is called a “triple” and so on. In the special case of pairs, there are functions `fst` and `snd` which extract the first and second element.
``````let x = 1,2
fst x
snd x
``````
They only work on pairs. Trying to use `fst` on a triple will give an error.
``````let x = 1,2,3
fst x // error FS0001: Type mismatch.
// The tuples have differing lengths of 2 and 3
``````
## Using tuples in practice
Tuples have a number of advantages over other more complex types. They can be used on the fly because they are always available without being defined, and thus are perfect for small, temporary, lightweight structures.
### Using tuples for returning multiple values
It is a common scenario that you want to return two values from a function rather than just one. For example, in the `TryParse` style functions, you want to return (a) whether the value was parsed and (b) if parsed, what the parsed value was.
Here is an implementation of `TryParse` for integers (assuming it did not already exist, of course):
``````let tryParse intStr =
try
let i = System.Int32.Parse intStr
(true,i)
with _ -> (false,0) // any exception
//test it
tryParse "99"
tryParse "abc"
``````
Here’s another simple example that returns a pair of numbers:
``````// return word count and letter count in a tuple
let wordAndLetterCount (s:string) =
let words = s.Split [|' '|]
let letterCount = words |> Array.sumBy (fun word -> word.Length )
(words.Length, letterCount)
//test
wordAndLetterCount "to be or not to be"
``````
### Creating tuples from other tuples
As with most F# values, tuples are immutable and the elements within them cannot be assigned to. So how do you change a tuple? The short answer is that you can’t – you must always create a new one.
Say that you need to write a function that, given a tuple, adds one to each element. Here’s an obvious implementation:
``````let addOneToTuple aTuple =
let (x,y,z) = aTuple
(x+1,y+1,z+1) // create a new one
// try it
``````
This seems a bit long winded – is there a more compact way? Yes, because you can deconstruct a tuple directly in the parameters of a function, so that the function becomes a one liner:
``````let addOneToTuple (x,y,z) = (x+1,y+1,z+1)
// try it
``````
### Equality
Tuples have an automatically defined equality operation: two tuples are equal if they have the same length and the values in each slot are equal.
``````(1,2) = (1,2) // true
(1,2,3,"hello") = (1,2,3,"bye") // false
(1,(2,3),4) = (1,(2,3),4) // true
``````
Trying to compare tuples of different lengths is a type error:
``````(1,2) = (1,2,3) // error FS0001: Type mismatch
``````
And the types in each slot must be the same as well:
``````(1,2,3) = (1,2,"hello") // element 3 was expected to have type
// int but here has type string
(1,(2,3),4) = (1,2,(3,4)) // elements 2 & 3 have different types
``````
Tuples also have an automatically defined hash value based on the values in the tuple, so that tuples can be used as dictionary keys without problems.
``````(1,2,3).GetHashCode()
``````
### Tuple representation
And as noted in a previous post, tuples have a nice default string representation, and can be serialized easily.
``````(1,2,3).ToString()
``````
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math
need to make a venn diagram using 3 circles. A survey on subject being taken by 250 students at a certain college revealed the following info:
1. 90 were taking math
2. 145 were taking history
3. 88 were taking english
4. 25 were taking math and history
5. 38 were taking history and english
6. 59 were taking math and english
7. 15 were taking all three subjects
1. 👍 0
2. 👎 0
3. 👁 581
1. Each of your overlapping circles should be labeled
Math
History
English
Put the numbers in each circle to represent those students taking just one subject.
You'll have 25 in the overlapping sections of math and English.
http://www.classroomjr.com/printable-blank-venn-diagrams/
1. 👍 0
2. 👎 0
2. Ms Sue when you said 25 in the overlapping section of math and english did mean math and history/
1. 👍 0
2. 👎 0
3. Yes, I'm sorry. 25 students are taking math and history.
Thanks for catching my mistake.
1. 👍 0
2. 👎 0
4. Is 100 student did a surveys 47 Math, 57 Chemistry, 69,English 29 in English an Math, 25 English and chemistry, 37 Math and chemistry, 22 in all
How student are you taking Math but not English an Chemistry
1. 👍 0
2. 👎 1
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# Approximation Questions and Answers for SBI Clerk Pre 2021, IBPS Clerk, CET, RBI Assistant, LIC at Smartkeeda
Directions: What value will come in place of question mark (?) in the following questions: (You are not expected to calculate the exact value)
Important for :
1
7(6 - ?) × 119 ÷ √1225 = √289 × 20.98 × 7.03 ÷ 3.02 ÷ 4.95
» Explain it
A
7(6 - ?) × 119 ÷ √1225 = √289 × 20.98 × 7.03 ÷ 3.02 ÷ 4.95
7(6 - ?) × 119 ÷ 35 = 17 × 21 × 7 ÷ 3 ÷ 5
7(6 - ?) = 17 × 21 × 7 ÷ 3 ÷ 5 ÷ 119 × 35
7(6 - ?) = 7 × 7
7(6 - ?) = 72
6 - ? = 2
? = 6 - 2 = 4
Hence, option A is correct.
2
(128.78 + 307.24 + 111.8) – (246.35 – 786.47) = ?2
» Explain it
A
(128.78 + 307.24 + 111.8) – (246.35 – 786.47) = ?2
(129 + 307 + 112) – (246 – 786) = ?2
548 – (– 540) = ?2
?2 = 548 + 540
?2 = 1088 ≈ 1089
? = 33
Hence, option A is correct.
3
54.7% of 1190 ÷ (√676 + ∛ 1728.11 – √361) = ?2 ÷ 57.02
» Explain it
C
54.7% of 1190 ÷ (√676 + ∛ 1728.11 – √361) = ?2 ÷ 57.02
1190 × 55/ 100 ÷ (26 + 12 – 19) = ?2 ÷ 57
654.5 ÷ 19 = ?2 ÷ 57
655 ÷ 19 × 57 = ?2
?2 = 655 × 3 = 1965
? = 44.32 ≈ 44
Hence, option C is correct.
4
» Explain it
B
(100 / 1300) × 208 + {(17 × 304) / 323) = ?1/2 +5
16 + 16 – 5 = ?1/2
?1/2 = 27
? = 729 ≈ 730
Hence, option B is correct.
5
9.09% of 11.11% of 2430 = ?3 ÷ √289 + √( ∛4096 )
» Explain it
D
9.09% of 11.11% of 2430 = ?3 ÷ √289 + √( ∛4096 )
2430 × (1 / 9) × (1 / 11) = ?3 ÷ 17 + √16
(270 /11) = ?3 ≈ 17 + 4
24.54 – = (?3 / 17)
20.54 = (?3 / 17)
21 × 17 = ?3
?3 = 357
? ≈ 7
Hence, option D is correct.
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## Conversion formula
The conversion factor from inches to meters is 0.0254, which means that 1 inch is equal to 0.0254 meters:
1 in = 0.0254 m
To convert 157.9 inches into meters we have to multiply 157.9 by the conversion factor in order to get the length amount from inches to meters. We can also form a simple proportion to calculate the result:
1 in → 0.0254 m
157.9 in → L(m)
Solve the above proportion to obtain the length L in meters:
L(m) = 157.9 in × 0.0254 m
L(m) = 4.01066 m
The final result is:
157.9 in → 4.01066 m
We conclude that 157.9 inches is equivalent to 4.01066 meters:
157.9 inches = 4.01066 meters
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 meter is equal to 0.24933552083697 × 157.9 inches.
Another way is saying that 157.9 inches is equal to 1 ÷ 0.24933552083697 meters.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that one hundred fifty-seven point nine inches is approximately four point zero one one meters:
157.9 in ≅ 4.011 m
An alternative is also that one meter is approximately zero point two four nine times one hundred fifty-seven point nine inches.
## Conversion table
### inches to meters chart
For quick reference purposes, below is the conversion table you can use to convert from inches to meters
inches (in) meters (m)
158.9 inches 4.036 meters
159.9 inches 4.061 meters
160.9 inches 4.087 meters
161.9 inches 4.112 meters
162.9 inches 4.138 meters
163.9 inches 4.163 meters
164.9 inches 4.188 meters
165.9 inches 4.214 meters
166.9 inches 4.239 meters
167.9 inches 4.265 meters
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# Thread: Simple Probability Problem
1. ## Simple Probability Problem
In a particular town 60% of the population are women. 4% of the men and 1% of the women are taller than 180 cm.
i) What % of the town's population is taller than 180 cm?
ii) If a person is chosen at random and is taller than 180 cm, what is the probability that the person is a woman?
Ummm I'm not that good at probability so could someone please give me some help?
For:
i) I did:
Percentage = 100(0.04*0.4 + 0.01*0.6)
= 0.022*100
= 2.2%
ii) I did:
P = 0.022*0.2
= 11/2500
The answers don't look rite though...
2. You have the first part right, for the second one you need to use Baye's theorem.
$\displaystyle P(woman|taller than 180)= \frac{P(taller than 180|woman)P(woman)}{P(taller than 180|woman)P(woman)+P(taller than 180|man)P(man)}$
3. Originally Posted by xwrathbringerx
In a particular town 60% of the population are women. 4% of the men and 1% of the women are taller than 180 cm.
i) What % of the town's population is taller than 180 cm?
ii) If a person is chosen at random and is taller than 180 cm, what is the probability that the person is a woman?
Ummm I'm not that good at probability so could someone please give me some help?
For:
i) I did:
Percentage = 100(0.04*0.4 + 0.01*0.6)
= 0.022*100
= 2.2% Mr F says: Correct.
ii) I did:
P = 0.022*0.2 Mr F says: Wrong. Where has the 0.2 come from?
= 11/2500
The answers don't look rite though...
ii requires conditional probability.
Here is the magic recipe, the big secret, the great trick, the golden elixir. Read it carefully ........
Set up careful notation and definitions. Carefully define all known probabilities in terms of this notation.
Let T be the event height greater than 180 cm.
You have:
Pr(M) = 2/5, Pr(W) = 3/5,
Pr(T | M) = 1/25, Pr(T | W) = 1/100,
Pr(M and T) = Pr(T| M) Pr(M) = (1/25) (2/5) = 2/125,
Pr(W and T) = Pr(T| W) Pr(W) = (1/100) (3/5) = 3/500,
Pr(T) = Pr(M and T) + Pr(W and T) = 2/125 + 3/500 = 11/500 (= 0.022 as you found).
You require Pr(W | T).
$\displaystyle \Pr(W | T) = \frac{\Pr(W \cap T)}{\Pr(T)} = \frac{3/500}{11/500} = 3/11$.
4. You have the first part right, for the second one you need to use Baye's theorem.
To keep not that messy, Let
W=woman
M=Man
T=taller than 180
$\displaystyle P(W|T)= \frac{P(T|W)P(W)}{P(T|W)P(W)+P(T|M)P(M)}=\frac{(0. 01)(0.60)}{(0.01)(0.60)+(0.04)(0.40)}$
$\displaystyle =\frac{0.006}{0.006+0.016}=\frac{0.006}{0.022}=\fr ac{3}{11}$
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# Search by Topic
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# Calculate the sum at a level of a binary tree represented in a String
Fair preface: This is an interview question I would like to improve my knowledge of. I got some rest, and re-solved the problem with a fresh/unpanicked mind.
Given input of the type (10(5(3)(12))(14(11)(17))) which would represent the following tree
n=0 10
n=1 5 14
n=2 3 12 11 17
My task is to find the summation of values at a particular tier, like $5+14=19$ which is the sum for $n=1$, or $3+12+11+17=43$ the sum for $n=2$.
Considering this is a binary tree, a recursive function seems appropriate.
My main utility functions are:
• stripFirstLastParen – strips the first and last paren
• getCurrentVal – retrieves the value of the current node
• getChildren – retrieves the left and right nodes
var input =
"(10(5(3)(12))(14(11)(17)))";
function stripFirstLastParen(input){
if(input[0] !== "(" || input[input.length - 1] !== ")"){
console.error("unbalanced parens")
}
else{
input = input.substr(1);
input = input.substring(0, input.length - 1);
}
return input;
}
function getCurrentVal(input){
var val = "";
while(input[0] !== "(" && input[0] !== ")" && input.length){
val += input[0];
input = input.substr(1);
}
return {
val,
input
}
}
function getChildren(input){
var val = "";
if(input.length == 0){
return {
left: "",
right: ""
}
}
if(input[0] !== "("){
console.error("no open paren at start");
}
val += input[0];
input = input.substr(1);
var parenStack = 1;
while(parenStack > 0){
if(input[0] == ")"){
parenStack--;
}
else if(input[0] == "("){
parenStack++;
}
val += input[0];
input = input.substr(1);
}
return {
left: val,
right: input
}
}
function getValueForLevel(input, k){
var totalValue = 0;
input = stripFirstLastParen(input);
var currentValue = getCurrentVal(input);
var children = getChildren(currentValue.input);
if(k > 0){
if(children.left.length){
totalValue += getValueForLevel(children.left, k-1);
}
if(children.right.length){
totalValue += getValueForLevel(children.right, k-1);
}
}
else if(k == 0){
totalValue += JSON.parse(currentValue.val);
}
}
var test = getValueForLevel(input, 2);
console.log(test);
My main concerns are:
1. String manipulation. How should I properly be passing the altered string throughout the recursive function? Callout for lines like
var currentValue = getCurrentVal(input);
var children = getChildren(currentValue.input);
Where the getChildren function relies on getCurrentVal removing the value from the string itself, that feels dirty.
1. Complexity. I'm having difficulty describing the complexity of my procedure, and would argue that it's somewhere between $O(n)$ and $O(n^2)$, so is it $O(n^2)$? Considering the parens and values are removed as they are parsed, that lends itself to $O(n)$, but I feel it's $O(n^2)$ because of the duplicate parsing in order to create the child nodes. Is that correct? Can it be improved?
I tried to keep the "spirit of the challenge" and re-completed this within the amount of time previously allotted (somewhere around 30-45 minutes). Suggestions and improvements need not consider this time restriction, but providing context is important.
Recursion:
Considering this is a binary tree, a recursive function seems appropriate.
Recursion would be a natural choice if your input were a root tree node with tree nodes as children - a recursive data structure. However, your input is a 'flat' string representation of a tree. A simple iteration might actually be much easier to implement, read and execute.
Passing inputs:
How should I properly be passing the altered string throughout the recursive function?
I think the recursive passing down of substrings representing nodes or node lists (children) is already pretty well done.
However, I would like to suggest a few improvements regarding the parsing functions and their return values:
Style:
Consider the getCurrentVal(input) function:
function getCurrentVal(input){
var val = "";
while(input[0] !== "(" && input[0] !== ")" && input.length){
val += input[0];
input = input.substr(1);
}
return {
val,
input
}
}
The code is not very descriptive: It is not immediately obvious what this function does.
• The function name getCurrentVal is misleading, as you don't just return the node value, but also the children.
• The argument name input is meaningless, as all arguments are inputs.
• The variable names are misleading, as you reuse input to store part of your output.
// Split node 'a(b)(c)' into value 'a' and children '(b)(c)':
function parseNode(node) {
let split = Math.min(node.indexOf('('), node.indexOf(')'));
if (split < 0) split = node.length;
return {
value: node.slice(0, split),
children: node.slice(split)
};
}
• 'Parse' hints at the input being text or some other raw content.
• 'Node' carries more semantics than 'input'
• Math.min and String.indexOf are more descriptive than a while loop.
Destructuring the resulting object improves readability:
var {value, children} = parseNode(node);
Also, when parsing integers, you shouldn't use JSON.parse(string) but the more specialized Number.parseInt(string) or the idiomatic type conversion via +string.
Further naming suggestions:
• getValueForLevel -> getSumForLevel or sumTreeLevel
• Consistency: Choose either val or value when naming currentValue.val, getCurrentVal, getValueForLevel
Complexity:
Considering the parens and values are removed as they are parsed, that lends itself to O(n), but I feel it's O(n^2) because of the duplicate parsing in order to create the child nodes. Is that correct? Can it be improved?
First of all, let's define n as the length of the input string. Then, let's look at the complexity of the helper functions:
• stripFirstLastParen: O(n) because of input = input.substr(1)
• getCurrentVal: O(n²) for worst-case input "1(2)(3)", "11(22)(33)", "111(222)(333)" and so on because of input = input.substr(1) within while (input.length). Yes, we break the loop as soon as we encounter the first parenthesis, but this doesn't happen before parsing roughly 1/3 * n characters for above input. The best-case complexity is however O(n) when the value has constant length.
• getChildren has worst and best-case time complexity of O(n²) due to input = input.substr(1) within a while loop over input.
Now, we can tackle the complexity of getValueForLevel:
• The first four lines are dominated by getChildren(currentValue.input). If the length of currentValue.input - the length of the children - is proportional to n, then it has a runtime complexity of O(n ²).
• Then it calls getValueForLevel(children.left, k-1). It is easy to construct inputs which maximize the length of children.left such as "4(3(2(1))))" without right children. For this input, in each iteration, the length of children.left is reduced by 3.
• The final call to JSON.parse is in O(n).
Adding the complexities together gives us:
n² + (n-3)² + (n-6)² + ... + (n-3k)²
Thus, for k = 0 the runtime complexity is in O(n²) and for maximum k we have a runtime complexity of O(n³):
In general, the runtime complexity is bounded by O(kn²).
Alternative approach:
Following my introductory remarks about a possibly simpler iterative solution, here a possible implementation with linear runtime complexity:
// Sum integer tree nodes at given level:
function sumTreeLevel(tree, level) {
let tokens = tree.match(/($$|[0-9]+|$$)/g);
let sum = 0;
for (let token of tokens) {
if (token === '(') {
level--;
} else if (token === ')') {
level++;
} else if (level === -1) {
sum += +token;
}
}
return sum;
}
// Example:
console.log(sumTreeLevel("(10(5(3)(12))(14(11)(17)))", 2)); // 43
• What tool did you use to generate the complexity chart? – Igor Soloydenko May 8 '17 at 23:01
• @IgorSoloydenko plot.ly - the data doesn't result from a runtime evaluation however, but simply from counting 'primitive' operations according to the model of computation. – le_m May 8 '17 at 23:05
Some notes, in no particular order:
• If there are unbalanced/missing parentheses, you print an error to the console. However, I'd say you should just throw an exception instead. If the parentheses don't make sense, there's something wrong, and probably no way to recover.
• This stood out:
input = input.substr(1);
input = input.substring(0, input.length - 1);
Why use both substr and substring? The two are very subtly different, so don't mix them unless you have good reason. And why do the trimming in two passes? You could just slice from 1 to length-1 (or length-2 in the case of substr, since they're different) in one go.
• This stood out too:
totalValue += JSON.parse(currentValue.val);
Use parseInt, parseFloat, or even * 1 (in a pinch) to convert the string to a number. JSON ain't got nothing to do with this.
• getCurrentVal could be made simpler with a regular expression like /^\d+/ that matches digits only.
Now, if the idea is only to sum the nodes at each level, you can actually skip a lot of stuff.
Namely, there's no reason to build a graph. You can just read the string from beginning to end, and get the sums:
function sumLevels(string) {
var sums = [],
level = -1;
for(let i = 0; i < string.length; i++) {
if(string[i] === '(') {
level += 1; // add a level for each open paren
} else if(string[i] === ')') {
level -= 1; // remove a level for each closing paren
} else {
// match one or more digits
let match = string.slice(i).match(/^\d+/);
// complain if there are no digits
if(!match) throw new Error(Parse error at char \${i});
// fast-forward the loop counter
i += match[0].length - 1;
// add to sum for current level
sums[level] = (sums[level] || 0) + parseInt(match[0], 10);
}
}
return sums;
}
For the test string you had, the function returns [ 10, 19, 43 ]; the sum at each level.
Note, though that the above doesn't do much input checking. It doesn't, for instance, check that parentheses are balanced, meaning level could go negative or skip ahead, which wouldn't make sense. However, checking could be added without too much trouble. I'll leave that as an exercise for the reader though.
Incidentally, another way to write the above, using replace as a regex scan-like function, would be:
function sumLevels(string) {
var sums = [],
level = -1;
string.replace(/($$|$$|\d+)/g, (_, token) => {
switch(token) {
case '(':
level += 1;
break;
case ')':
level -= 1;
break;
default:
sums[level] = (sums[level] || 0) + parseInt(token, 10);
}
});
return sums;
}
This has even less input-checking, though.
• Thank you very much, this is the exact type of feedback I was looking for, it is very much appreciated. I'll be rereading over this in the next few days in order to ensure I understand everything, but thanks again! – Hodrobond May 8 '17 at 22:13
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# If x = (c* square root of b) + 4 find the value of x + 1/x
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Solution
## 1/x = 1/(c√b + 4) = (c√b - 4)/(c√b - 4) x 1/(c√b + 4) = (c√b - 4)/(c2b - 16) So, x + 1/x = (c√b + 4) + (c√b - 4)/(c2b - 16) = (c3b√b -15c√b - 68 + 4c2b)/(c2b - 16)
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## Want to keep learning?
This content is taken from the University of Twente's online course, Geohealth: Improving Public Health through Geographic Information. Join the course to learn more.
3.16
# The variogram
You were introduced to the concept of spatial dependence. In this step you will learn about the variogram (also referred to as the semi-variogram). A variogram is used to describe and model spatial dependence. The variogram is used widely in geostatistics. The equation for the variogram is:
$\hat \gamma(h) =\frac{1}{2n(h)} \sum_{i=1}^{n(h)}(y(s)-y(s+h))^2$
Let’s look at what this equation means. $$y$$ is the response variable (e.g., reflectance in an image, air pollution concentration, malaria parasite rate) which is taken at a specific location $$s$$. We calculate the difference in the attribute value between two observations measured at locations separated by a lag distance $$h$$ (hence ($$s$$+$$h$$)). Lag distance refers to the geographic separation between two observations. There are $$n(h)$$ pairs of points separated by $$h$$. We take the average of these squared differences for the $$n(h)$$ pairs. Note the 1/2 on the right-hand side of the equation. We multiply the average squared difference by 1/2 . Hence $$γ ̂(h)$$ is actually the semi-variance for pairs of points separated by $$h$$. For short lags we expect the average squared difference to be small (because measurements separated by short lags are expected to be similar). For large lags we expect the average squared difference to be larger. We can plot the semi-variances for different $$h$$’s against lag. This plot is called the sample semi-variogram or sample variogram (also called the experimental or empirical variogram).
*Figure 1 (top left) An example sample variogram, (top right) Sample variogram with fitted model (curved line). The sill, nugget and range are indicated. (bottom left) Schematic diagram showing a location with no measurement (red diamond) and locations where measurements are taken (black dots). The arrow indicates the distance between two locations. (bottom right) Map of PM10 air pollution concentration across Europe, based on the data in Figure 1 (top right) of Step 3.15.
An example sample variogram is given in Figure 1 (top left). Note that the value of semi-variance increases with increasing lag distance. For short lag distances the attribute values are similar and the semi-variance is small. At large lag distances the attribute values are dissimilar and the semi-variance is large.
We can already gain useful information from the sample semi-variogram. The lag distance where the sample variogram flattens is called the range. This is the maximum spatial separation where we expect two points to be correlated. The range is associated with the variogram sill, which is the maximum variability in the data. Finally, the point where the variogram approaches the y-axis is the nugget. The nugget is the non-spatial variability. These three parameters (sill, nugget, range) can be identified if we fit a model to the sample variogram. The model is a curved line, as illustrated in Figure 1 (top right).
In Figure 1 (top right) the range occurs at approximately 900 m. Two observations separated by less than 900 m would be expected to be correlated whereas two observations separated by more than 900 m would be expected to be uncorrelated. Two observations separated by 300 m are expected to be more correlated than two observations separated by 500 m. The variogram model tells exactly how correlated are two observations that are separated by a given distance.
The sample variogram and variogram model are useful for exploring the spatial dependence in a dataset. They can also be used for mapping. Consider Figure 1 (bottom left). The red diamond indicates a location where we do not have a measurement whereas we do have measurements at the black dots. We know how far the red diamond is from each black dot. Using the variogram we can then say how correlated it is expected to be with that black dot. We can then predict the attribute at the red diamond as a weighted average of the attributes at the black dots – where the weights are based on the correlations. This prediction is also called interpolation. The geostatistical approach to prediction is often called kriging, named after Danie Krige, an early researcher and practitioner in geostatistics.
Figure 1 (bottom left) illustrates prediction at a single location. If we predict at multiple locations on a grid we can create a map. Examples are shown in Figure 1 (bottom right) and Figure 2, which are based on data presented in the step on Spatial Dependence.
Figure 2 Map of malaria parasite rate, based on the data shown in Figure 3 of the Article on Spatial Dependence. See also Hay et al. 2009.
The maps shown in Figure 1 (bottom right) and Figure 2 are the concluding examples in the module on spatial statistics in this online course. These are important examples in the context of geohealth. Consider the air pollution example. We can use such maps to estimate individual exposure as part of an environmental epidemiological study into the health effects of air pollution. The malaria example is different because the disease itself is mapped rather than the possible cause of a disease. Such maps can be used to target interventions aimed at eliminating a disease (e.g., drug treatments, bed nets) and for evaluating the success of interventions.
References
Oliver, M.A., Webster, R. (2014) A tutorial guide to geostatistics: Computing and modelling variograms and kriging. CATENA 113, 56-69.
Hay, S. I., C. A. Guerra, P. W. Gething, A. P. Patil, A. J. Tatem, A. M. Noor, C. W. Kabaria, B. H. Manh, I. R. Elyazar, S. Brooker, D. L. Smith, R. A. Moyeed and R. W. Snow (2009). A world malaria map: Plasmodium falciparum endemicity in 2007. PLoS Medicine 6(3): e1000048. DOI: 10.1371/journal.pmed.1000048.
## Get a taste of this course
Find out what this course is like by previewing some of the course steps before you join:
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# physics
posted by .
A car moving at 50 km/h skids 15 m with locked brakes. Show that a car moving at 150 km/h will skid 135 m with locked brakes. (Hint: The force doing work during the skidding is the same for both speeds. Use the work–energy theorem.)
• physics -
work = force * distance
= F d
F the same for both
work done = kinetic energy lost
= .5 m v^2
same m
V2 = 3 V1
so
V2^2 = 9 V1^2
so work done at 50 = (1/2)m V1^2 = F d1
and at 150 = (1/2) m (9 V1 ^2) = F d2
so
d2/d1 = 9
9 * 15 = 135
## Similar Questions
1. ### PHSC
Hello, i need help. This is Work-Energy Theorum A car moving at 50km/hr skids 15m with locked brakes. How far will the car skid (with locked brakes if the car moves at 100km/hr assume that the mass of the car is 500kg
2. ### Physical science
A car moving at 50km/hr skids 15m with locked brakes. How far will the car skid (with locked brakes if the car moves at 100km/hr assume that the mass of the car is 500kg
3. ### physics
This question is typical on some driver's license exams: A car moving at 50km/hr skids 15m with locked brakes.How far will the car skid with locked brakes at 120 km/h?
4. ### physics
disregard the previous I did not include the first part This question is typical on some driver's license exams: A car moving at 50 km skids 15 m with locked brakes.How far will the car skid with locked brakes at 120 km/h?
5. ### physics
QA: This question is typical on some driver’s license exams: A car moving at 40 km/h skids 12 m with locked brakes. How far will the car skid with locked brakes at 80 km/h?
6. ### physics
This question is typical on some driver’s license exams: A car moving at 40 km/h skids 12 m with locked brakes. How far will the car skid with locked brakes at 80 km/h?
7. ### Physics
A car moving at 43 km/h skids 16 m with locked brakes. How far will the car skid with locked brakes at 129 km/h?
8. ### physics
A car moving at 50 km/h skids 15m with locked brakes. show with locke brakes at 150 km/m the car will skid 135?
9. ### Physict
This question is similar to some on driver's-license exams: A car moving at 50 km/h skids 15 m with locked brakes. Show that with locked brakes at 150 km/h the car will skid 135 m.
10. ### Physics
A car moving at 60 km/h skids 13 m with locked brakes. How far will the car skid with locked brakes at 180 km/h?
More Similar Questions
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# Help with my algebra homework?
Discussion in 'Archived Threads 2001-2004' started by Andrew S, Feb 24, 2002.
1. ### Andrew S Stunt Coordinator
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Just so you guys know, I've already done like 20 questions... so it's not like you're DOING my homework for me... I just need help with two questions.
We're supposed to solve these word problems through elimination or subsitution.
#1. At a silversmith's shop, they have alloys that contain 40% silver and othets that are 50% silver. A custom order for a breacelet requires 150 grams of 44% silver. How much of each alloy should be melted together to make the bracelet?
My two equations were:
.4a + .5b = .44
a + b = 150
"a" means amount of 40% silver alloy
"b" means amount of 50% silver alloy
The answer in the back of the book is 90 g of 40% and 60 g of 50%, which sounds correct, but doesn't work when I subsititute it into my first equation, so I figure there's something wrong with my first equation.
#2. During a training exercise, a submarine travles 16km/h on the surface, but it only goes 10 km/h underwater. If the submarine traveled a distance of 160 km in 12.5 hours, how long was it underwater.
My equations were:
10a + 16b = 160km
a + b = 12.5 hours
So I changed the second equation to be b = 12.5 - a and then substituted that into the first equation to be:
"a" refers to time spent underwater
"b" refers to time spent above water
10a + 16(12.5 - a) = 160
10a + 200 - 16a = 160
-4a = 160 - 200
-4a = -40
-4a/-4 = -40/-4
a = 10
Then I subsituted the 10 into both equations:
First Equation:
10(10) + 16b = 160
100 + 16b = 160
16b = 160 - 100
16b = 60
16b/16 = 60/16
b = 3.75
Second Equation:
10 + b = 12.5
b = 12.5 - 10
b = 2.5
So as you can see, it doesn't work out. If it helps, the answer in the back of the book is 6 hours and 40 minutes (for how long the submarine was underwater)
I'll count myself lucky if anyone responds to this, so any help would be appreciated..
Thanks for helping out a confused grade 10.
Andrew
2. ### Keith Mickunas Cinematographer
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Andrew, on #1 your first equation is forgetting that its 0.44 * c. Where c is the total weight. Try that and see how it works.
I'll get to the second in a moment, I don't want to do to much for you to fast. Gotta keep you honest.
3. ### Keith Mickunas Cinematographer
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FWIW, I think you are making a similar mistake on #2.
You have to keep your units consistent on both sides of the equation. In the first, you have a weight of silver and a purity rating on the left, but you only have a total weight on the right. So you aren't balancing the equation. Do you see what I'm getting at?
4. ### Will Pomeroy Stunt Coordinator
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Is that OAC alg/geo, or the new curriculum? (what grade?)
5. ### Andrew S Stunt Coordinator
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This is actually grade 10 math and Keith, I'll see if that helps... that is if I understand you correctly...
Andrew
6. ### MikeAlletto Cinematographer
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You got the answers in the back of the book? Hehehe...that reminds me what I would do. I would show all kinds of work and then at the end whatever I came out with if it didn't match the back of the book I would erase the final answer and just write in the back of the book answer. Most of the time the teacher didn't bother to check the work and just looked at the answer even though we were supposed to show all work.
7. ### Andrew S Stunt Coordinator
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You're absolutely correct for the first equation, Keith. I kept coing up with 66, and 44% of 150 is 66.
It still amazes me the help you can get from this forum, even if it has nothing to do with Home Theater.
Thanks agai,
Andrew
8. ### Andrew S Stunt Coordinator
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Mike,
Unfortunately all of my teachers' number one priority is the rough work. There's something like 3-5 marks for each question and if you just write the answer you get 1 mark. Really sucks for the easy questions that everyone can do in their heads because you still have to take up a quarter of the page with the rough work.
Oh well.
Andrew
9. ### Keith Mickunas Cinematographer
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Makes me think back to the days of calc in college. On some problems I'd start with a blank sheet of paper, do all my work on there until I got it right, then copy the stuff to the sheet I was handing in. Sometimes you might work only 3 or 4 problems on a single sheet.
Do you see the relation between the two problems yet? They're fundamentally the same. In physics class they often require you to specify your units throughout the problem, which shows you how each side relates to the other. Remember that both sides of #2 are talking about km/h so you should be dealing with a formula where x+y=z and x,y,z are all km/h measurements.
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Andrew - You setup the 2nd problem correctly, but you made an error in your math. Michael Perez is showing you the correction.
13. ### Keith Mickunas Cinematographer
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Well dammit, now I feel stupid. You just have an arithmetic mistake in #2, like Michael said.
What I was referring too is that you're formula is really this:
a * 10km/h + b * 16km/h = 160km
Since a and b are units of time(h), your equation is balanced because you are multiplying a h value with a km/h value, thus cancelling the time. You can only add and subtract values that have the same unit. Does that make sense? I thought you didn't have all the units accounted for, like in the first problem, but you did.
In your first problem you have:
.4a + .5b = .44
or
(a grams * .4p) + (b grams * 0.5p) = .44p
where p is represent the unit for percantage of silver. So what you were doing was ending up with a formula that has different units on the different sides, so you couldn't work it. What you ended up with was:
(a grams * .4p) + (b grams * 0.5p) = 150 grams * 0.44p
a + b = 150
as your two equations which, as you found, can be solved.
Clear as mud, eh?
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## Monday, August 31, 2015
### Day 4
The mathematicians began their day as greedy pirates, swash-buckling, mathematically of course, to get the most gold coins in a famous game theory conundrum.
Here's the problem for those who haven't seen it before:
There are 5 rational pirates A, B, C, D and E. They find 100 gold coins. They must decide how to distribute them.
The pirates have a strict order of seniority: A is superior to B, who is superior to C, who is superior to D, who is superior to E.
The pirate world's rules of distribution are thus: that the most senior pirate should propose a distribution of coins. The pirates, including the proposer, then vote on whether to accept this distribution. In case of a tie vote the proposer has the casting vote. If the distribution is accepted, the coins are disbursed and the game ends. If not, the proposer is thrown overboard from the pirate ship and dies, and the next most senior pirate makes a new proposal to begin the system again.
Pirates base their decisions on three factors. First of all, each pirate wants to survive. Second, given survival, each pirate wants to maximize the number of gold coins each receives. Third, each pirate would prefer to throw another overboard, if all other results would otherwise be equal.The pirates do not trust each other, and will neither make nor honor any promises between pirates apart from a proposed distribution plan that gives a whole number of gold coins to each pirate.
Five pirates vying for 10 coins is an interesting problem but what if there were one hundred pirates each trying to get the most loot out of 100 coins? What if there were more pirates than coins? If there were more pirates than coins would the first pirate be able to get any coins? Is there a strategy where the first pirate would be able to stay onboard? Can you devise a formula for the number of coins any given pirate in such a situation would get where the number of pirates (N) is greater than the number of coins (G)? Use the backwards induction model like we used with the first version of the game!
Speaking of formulating your answer using variables, we continued to derive the pythagorean theorem in Geometry, replicating the work of the indian mathematician Bhakarsa in his famous proof. Walking the same path he did in 1114 CE!
In graph theory we built off of our definitions of days one through three to find isomorphic graphs amidst a maelstrom of interconnected nodes and edges. Mathematicians then used this fresh skill set to draw their own planar graphs.
During lunch I noticed a young mathematician who was playing a game of Cat's Cradle. I wondered, if they were to chop up the string could they make a planar graph with the same number of nodes and edges as we see in the picture below?
If you were to play cat's cradle without chopping up the string, how many different non-planar graphs can you make? How many of these are identical?
In computer science we talked about Boolean logic and the special definitions of and or or in programming.
Here's a joke about the inclusive or:
A logician's wife is in labor. The logician is in the waiting room.
The doctor comes out of the delivery room and says to the logician, "Congratulations, your wife gave birth to a beautiful baby!"
"Is it a boy or a girl?" asks the logician.
"Yes", says the doctor.
In art, mathematicians continued to use their budding architectural acumen to make platonic solids. We're eagerly shedding the templates that we made in the beginning of the week as tetrahedrons, octahedrons, dodecahedrons and more take form.
We're disappointed that C.A.M.P. is discrete and not continuous. The last update (until next summer of course) is tomorrow, stay tuned!
-- Eliana
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1. ## Line integral
let vector field F be (z^2/x,z^2/y, 2zlnxy). Evaluate the line integral of F
Where C is the path of straight line segments fromP = (1; 2; 1) to Q = (4; 1; 7) to R = (5; 11; 7), and then back to P.
2. ## Re: Line integral
Hey apatite.
Can you show us what you tried? (Hint: Write down the definition of the line integral for one single segment to begin).
3. ## Re: Line integral
i get zero as final answer. I want to double check it. Thank you. Is my answer right? In this particular case i use find the partial function and use that to get the line integral.
4. ## Re: Line integral
Well, yes.
If you calculate the curl $\displaystyle \nabla \times \vec{F} = \left( \dfrac{\partial}{\partial x}, \dfrac{\partial}{\partial y}, \dfrac{\partial}{\partial z} \right) \times \vec{F}$ of this vector field, you will find that it equals zero and so the field is conservative. This means that it has path independence and regardless of the path you take, if you end up at the same place (P), the work done on it equals zero.
I do not get what you mean by using the 'partial function' to solve the line integral, do you mean you used the gradient function to solve for the curl?
To solve this line integral you have to parameterise the function and then solve $\displaystyle \int_{t_0}^{t_1} \vec{F}(t) \times \vec{r} (t) dt$
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# Why not transmit electricity in 4-phase?
Tags:
1. Mar 16, 2016
### GregValcourt
Or even better, generate in 4-Phase, convert to two-phase for transmission. Even number phases cancel out nicely (contrary to popular belief, 3-phase does not actually cancel out). 2-phase might be convertible back to four phase for the purpose of industrial motors.
Forgive me for speaking out of ignorance, I'm not an electrician or engineer. Mathematically, an even number of phases seems better for transmission (in terms of loss).
Mathematical graph of 3 sine waves 120 degrees apart, with resultant wave:
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• ###### Screen Shot 2016-03-16 at 7.28.25 PM.png
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2. Mar 16, 2016
### Averagesupernova
Those 3 sine waves don't look 120 degrees apart to me.
3. Mar 16, 2016
### Baluncore
The sum of the instantaneous currents in all three phases should always be close to zero. That reduces the magnetic radiation from the transmission line.
Given a minimum of three phases and by using transformers, any number of other phase combinations can be generated. For a simple example see; https://en.wikipedia.org/wiki/Scott-T_transformer
The sum of cross sections of conductors in a transmission line is independent of the number of phases, but the cost of insulators is 33% higher for a 4 phase line than for a 3 phase line.
4. Mar 17, 2016
### andrewkirk
Here's why 3-phase cancels:
$$\sin(x-120)+\sin x+\sin(x+120)=\sin x\cos (-120)+\cos x\sin (-120)+\sin x+\sin x\cos 120+\cos x\sin 120$$
$$=-\sin x\cos 60+-\cos x\sin 60+\sin x-\sin x\cos 60+\cos x\sin 60$$
$$=-\sin x\cos 60+\sin x-\sin x\cos 60 =-\tfrac{1}{2}\sin x+\sin x-\tfrac{1}{2}\sin x=0$$
Picking up on the obaservation of @Averagesupernova, the blue and green curves in the graph are 90 degrees out of phase, not 120.
5. Mar 17, 2016
### anorlunda
OP, you really blew it when you said three phases don't add to zero. Ignoring that, high phase order transmission has been studied many times. The short answer is that the added benefits do not outweigh the added costs.
Google the phrase high phase order transmission.
6. Mar 17, 2016
### cabraham
Settled science. Single phase requires 2 wires. Two phase requires 3 wires. Three phase also needs 3 wires, 4 phase needs 4. Computing losses, you will find that to transmit a given amount of power from point A to point B with a given loss, 3 phase requires only 75% the copper area/weight as 1 or 2 phase. Increasing to 4 phase or more does not reduce copper needed, I can look it up later, but more phases may even need more copper than with 3 phases.
Three phase is optimum regarding maximum power ability, for a given amount of loss, with minimum copper consumed. The utility companies have studied this issue since the late 19th century. If a different number of phases is really better, it would have been done decades ago.
Claude
7. Mar 17, 2016
### Baluncore
I believed that the total section of copper needed remained the same for 3PH and above, while the number, mass and cost of insulators increased.
If anything I would expect total copper mass to fall slightly for more phases due to the skin effect not fully utilising the centre of thick conductors needed for 3PH.
Can you give a reference to the weight of copper needed for polyphase lines of three and above ?
8. Mar 17, 2016
### The Electrician
From the first paragraph here: https://en.wikipedia.org/wiki/Polyphase_system
"A major advantage of three phase power transmission (using three conductors, as opposed to a single phase power transmission, which uses two conductors), is that, since the remaining conductors act as the return path for any single conductor, the power transmitted by a balanced three phase system is three times that of a single phase transmission but only one extra conductor is used. Thus, a 50% / 1.5x increase in the transmission costs achieves a 200% / 3.0x increase in the power transmitted."
9. Mar 17, 2016
### cabraham
Wikipedia is incorrect here. I will find or recompute, then post. Here is the conclusion, I know this is right. For transmitting power from A to B, same amount using 1-phase & 3-phase, with equal power lost in transmission, a 3-ph system uses 75% the copper of 1-ph, as well as 2-ph.
10. Mar 18, 2016
### anorlunda
Can't you both be right?
• The Electrician holds wire diameter constant and increases the power transmitted, holding watts/(mile*phase) constant.
• Cabraham holds power and losses constant and decreases wire diameter.
Edit: By the way, high phase order transmission is interesting not only because of losses, but also because of series reactance.
11. Mar 18, 2016
### cabraham
Here is a 2-page computation for copper needed re 1-phase, 3-ph, and 4-ph transmission. Here is the synopsis. The generated power is 1.000 watt per unit, the power loss in cables in 0.010 watts per unit, and the load power is 0.990 watts per unit. These values were held constant for comparison purposes. Also, maximum line to line voltages were fixed at 1.0 volts per unit.
The result is what I seem to recall. The copper requirement for 1-phase is set at 100% as the reference. A 3-phase system requires just 75% the copper of 1-phase, a considerable savings. But, a 4-phase system did not improve on the 75% figure. Actually it came in at 100%, identical to 1 phase. I remember computing the general relation for any number of phases, i.e. "n-phase". For any value of "n" other than 3, the result is the same, the same amount of copper, i.e. 100%, is needed. If n=3, then 75% of the copper weight is needed.
I'm not surprised, because if 4 phases, or 5, 6, 10, whatever, saved on copper usage, the power companies & others would have quickly done it. I've heard of 6, 12, 24, & even 36 phase power employed. But it's advantage is usually very low ac ripple after rectification negating the need for large smoothing capacitors. But for long distance power transmission, 3 phase is pretty much the only game in town. Comments/feedback/questions welcome. Best regards.
Claude
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12. Mar 18, 2016
### mheslep
Or HVDC.
13. Mar 18, 2016
### GregValcourt
Your right, I must have did something wrong when building the spreadsheet. The second time I did it, 3-phase does cancel out.
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# If a divides b and a divides c, then a divides (b-c)
In reading about Euclid’s proof of the infinitude of prime numbers, the only part that wasn’t completely clear to me was this:
If $$p$$ divides $$P$$ and $$q$$, then $$p$$ would have to divide the difference of the two numbers, which is $$(P + 1) − P$$ or just $$1$$.
Well, I don’t know…why is that true? Why does a number have to divide the difference of two numbers if it divides each of those numbers separately?
It turns out that my textbook from my introduction to proofs class, Mathematical Proofs by Chartrand, Polimeni, and Zhang (a class that was taught by Dr. Zhang herself at Western Michigan University), contains the proof of a more general statement. (Note: For this theorem, the vertical bar $$\vert$$, which looks identical to the absolute value symbol, is used to mean “divides”; that is, $$a ~\vert~ b$$ i.f.f. $$a$$ is a factor of $$b$$, that is, $$a$$ “guzzinta” $$b$$ an integer number of times, that is, $$b \div a$$ equals an integer.)
Theorem: If $$a~\vert~b$$ and $$a ~\vert~c$$, then $$a ~\vert~ (bx + cy)$$ for all integers $$x$$ and $$y$$.
Proof: Let $$a ~\vert~ b$$ and $$a ~\vert~ c$$. Then there exist integers $$q_{\scriptscriptstyle 1}$$ and $$q_{\scriptscriptstyle 2}$$ such that $$b=aq_{\scriptscriptstyle 1}$$ and $$c=aq_{\scriptscriptstyle 2}$$. Hence, for integers $$x$$ and $$y$$,
$$\begin{eqnarray} bx + cy = aq_{\scriptscriptstyle 1}x + aq_{\scriptscriptstyle 2}y = a(q_{\scriptscriptstyle 1}x + q_{\scriptscriptstyle 2}y). \end{eqnarray}$$
Since $$q_{\scriptscriptstyle 1} x + q_{\scriptscriptstyle 2} y$$ is an integer, $$a ~\vert~ (bx + cy)$$. $$\tag*{\blacksquare}$$
The specific case where $$x=1$$ and $$y=-1$$ is used in Euclid’s proof of the infinitude of primes.
Aside from this abstract proof, it’s easy to see why the theorem would be true with a simple example. Consider $$a = 5$$ and $$b$$ and $$c$$ any multiples of $$5$$, say $$25$$ and $$100$$. Since these are both multiples of $$5$$, they are some multiple of $$5$$ apart from each other, so their difference is also obviously a multiple of $$5$$. It’s just counting by fives, or by whatever the factor is in the example you choose. It’s hard to think of a theorem that could be more obvious and intuitive, even to an elementary-schooler.
As a side note, Mathematical Proofs is an extremely good introduction to mathematical proofs. It is one of my favorite textbooks. It is detailed, thorough, and extremely long, with great explanations that they call “proof strategy” before many proofs and summaries that they call “proof analysis” after many proofs. It has chapters on sets, logic, direct proofs, proof by contradiction, mathematical induction, equivalence relations, functions, cardinalities of sets, number theory, calculus, linear algebra, topology, group theory, and ring theory. I love this book and I highly recommend it for anyone who might want to learn or review how to do a lot of basic and important proofs from many fields of mathematics.
This entry was posted in Math, Theorems. Bookmark the permalink.
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# Reasoning Quiz for IBPS PO PRE & SBI CLERK MAINS 2019 | 7 August 2019
Table of Contents
## Reasoning Quiz for IBPS PO PRE & SBI CLERK MAINS
Reasoning Quiz to improve your Reasoning for SBI Po & SBI clerk exam Reasoning, IBPS PO Reasoning , IBPS Clerk Reasoning , IBPS RRB Reasoning, LIC AAO , LIC ADO, and other competitive exams.
Directions (Q1-Q5): Given below are five statements out of which one can be logically concluded from other four, you have to mark that statement as your answer.
Q1. Statements:
(a) Some D are A
(b) Some C are E
(c) All B are D
(d) Some D are E
(e) Some B are A
Answer & Explanation
Q1. Ans(a)
Q2. Statements:
(a) All bill are nut
(b) All bill are pot
(c) Some nut are rat
(d) Some rat are bat
(e) All nut are pot
Answer & Explanation
Q2. Ans(b)
Q3. Statements:
(a) No din are gin
(b) No rum are bum
(c) All din are rum
(d) No bum is gun
(e) No gin are rum
Answer & Explanation
Q3. Ans(a)
Q4. Statements:
(a) All toy are boy
(b) All boy are strong
(c) All strong are hard
(d) Some loud are boy
(e) All hard are loud
Answer & Explanation
Q4. Ans(d)
Q5. Statements:
(a) Some rum are num
(b) All fun are bun
(c) Some bun are rum
(d) All gun are rum
(e) Some fun are rum
Answer & Explanation
Q5. Ans(c)
Q6. Among J, K, L, M and N, who is the shortest? J is shorter than only N. K is taller than M and L. If they are arranged in ascending order of their heights from left to right then M is second from the left?
A) J
B) K
C) L
D) M
E) Can’t be determined.
Answer & Explanation
Q6. Ans(C)
Q7. Each of the five friends Kamal, Rahul, Suri, Kalia and Bhoopendra has a different age. Kamal is youger than only Rahul. Suri is older than Kalia. Who is not the youngest. Who among the following is/are older than Suri?
A) Only Kamal and Bhoopendra
B) Only Kalia and Kamal
C) Only Bhoopendra
D) Only Bhoopendra and Kalia
E) None of these
Answer & Explanation
Q7. Ans(E)
Q8. On which day was Shilpi definitely born? Shilpi’s mother correctly remembers that Shilpi was born before Friday but after Monday. Shilpi’s brother correctly remembers that his sister was born before Saturday but after Wednesday.
A) Monday
B) Tuesday
C) Thursday
D) Friday
E) Can’t be determined.
Answer & Explanation
Q8. Ans(C)
Q9. Five people A, B, C, D, E lives on five different floors from bottom to top. Two people live between B and C. A lives immediately above C. D lives on one of the floor above E. Then who among the following lives on fourth floor?
A) B
B) A
C) Can’t be determine
D) E
E) D
Answer & Explanation
Q9. Ans(C)
Q10. E is heavier than G and K but not as much as F. L is only lighter than M. The one who is second heaviest is 56kg. The weight of third lightest is 47 kg. K is 10kg lighter than E. G is of 18kg. What can be the weight of F?
A) 58 kg
B) 70 kg
C) 50 kg
D) 45 kg
E) None of these
Answer & Explanation
Q10. Ans(C)
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# Test: Sequences And Series - 1
## 15 Questions MCQ Test Practice Questions for GMAT | Test: Sequences And Series - 1
Description
Attempt Test: Sequences And Series - 1 | 15 questions in 30 minutes | Mock test for GMAT preparation | Free important questions MCQ to study Practice Questions for GMAT for GMAT Exam | Download free PDF with solutions
QUESTION: 1
### If integer k is equal to the sum of all even multiples of 15 between 295 and 615, what is the greatest prime factor of k?
Solution:
First, let us simplify the problem by rephrasing the question. Since any even number must be divisible by 2, any even multiple of 15 must be divisible by 2 and by 15, or in other words, must be divisible by 30. As a result, finding the sum of even multiples of 15 is equivalent to finding the sum of multiples of 30. By observation, the first multiple of 30 greater than 295 will be equal to 300 and the last multiple of 30 smaller than 615 will be equal to 600.
Thus, since there are no multiples of 30 between 295 and 299 and between 601 and 615, finding the sum of all multiples of 30 between 295 and 615, inclusive, is equivalent to finding the sum of all multiples of 30 between 300 and 600, inclusive. Therefore, we can rephrase the question: “What is the greatest prime factor of the sum of all multiples of 30 between 300 and 600, inclusive?”
The sum of a set = (the mean of the set) × (the number of terms in the set)
Since 300 is the 10th multiple of 30, and 600 is the 20th multiple of 30, we need to count all multiples of 30 between the 10th and the 20th multiples of 30, inclusive.
There are 11 terms in the set: 20th – 10th + 1 = 10 + 1 = 11
The mean of the set = (the first term + the last term) divided by 2: (300 + 600) / 2 = 450
k = the sum of this set = 450 × 11
Note, that since we need to find the greatest prime factor of k, we do not need to compute the actual value of k, but can simply break the product of 450 and 11 into its prime factors:
k = 450 × 11 = 2 × 3 × 3 × 5 × 5 × 11
Therefore, the largest prime factor of k is 11.
QUESTION: 2
### If S is the infinite sequence S1 = 6, S2 = 12, ..., Sn = Sn-1 + 6,..., what is the sum of all terms in the set {S13, S14, ..., S28}?
Solution:
For sequence S, any value Sn equals 6n. Therefore, the problem can be restated as determining the sum of all multiples of 6 between 78 (S13) and 168 (S28), inclusive. The direct but time-consuming approach would be to manually add the terms: 78 + 84 = 162; 162 + 90 = 252; and so forth.
The solution can be found more efficiently by identifying the median of the set and multiplying by the number of terms. Because this set includes an even number of terms, the median equals the average of the two ‘middle’ terms, S20 and S21, or (120 + 126)/2 = 123. Given that there are 16 terms in the set, the answer is 16(123) = 1,968.
QUESTION: 3
### In an increasing sequence of 5 consecutive even integers, the sum of the second, third, and fourth integer is 132. What is the sum of the first and last integers?
Solution:
Let the five consecutive even integers be represented by x, x + 2, x + 4, x + 6, and x + 8. Thus, the second, third, and fourth integers are x + 2, x + 4, and x + 6. Since the sum of these three integers is 132, it follows that
3x + 12 = 132, so
3x = 120, and
x = 40.
The first integer in the sequence is 40 and the last integer in the sequence is x + 8, or 48.
The sum of 40 and 48 is 88.
QUESTION: 4
What is the sum of the multiples of 7 from 84 to 140, inclusive?
Solution:
84 is the 12th multiple of 7. (12 x 7 = 84)
140 is the 20th multiple of 7.
The question is asking us to sum the 12th through the 20th multiples of 7.
The sum of a set = (the mean of the set) x (the number of terms in the set)
There are 9 terms in the set: 20th - 12th + 1 = 8 + 1 = 9
The mean of the set = (the first term + the last term) divided by 2: (84 + 140)/2 = 112
The sum of this set = 112 x 9 = 1008
Alternatively, one could list all nine terms in this set (84, 91, 98 ... 140) and add them.
When adding a number of terms, try to combine terms in a way that makes the addition easier
(i.e. 98 + 112 = 210, 119 + 91 = 210, etc).
QUESTION: 5
If each term in the sum a1 + a2 + a3 + ... +an is either 7 or 77 and the sum is equal to 350, which of the
following could equal to n?
Solution:
QUESTION: 6
For any integer k from 1 to 10, inclusive, the kth of a certain sequence is given by [(-1)(k+1)] × (1 / 2k). If T is the sum of the first 10 terms of the sequence, then T is:
Solution:
T= 1/2-1/22+1/23-...-1/210
= 1/4+1/42+1/43+1/44+1/45
Notice that 1/42+1/43+1/44+1/45 < 1/4, we can say that 1/4<T<1/2.
QUESTION: 7
S is the infinite sequence S1 = 2, S2 = 22, S3 = 222,...Sk = Sk–1 + 2(10k–1). If p is the sum of the first 30 terms of S, what is the eleventh digit of p, counting right to left from the units digit?
Solution:
The first few terms of the sequence are 2, 22, and 222 and each subsequent term has an additional 2 added on. The 30th term then is a string of 30 2’s. If we line up the first 30 terms of the sequence to add them up, we will get rows in the following pattern:
To find p, the sum of the first 30 terms of S, we would simply be adding columns of 2’s. The key here is to see a pattern in the addition process. Starting with the units digit column, all 30 of the terms have a 2 in that position so the sum of the units column would be 30 x 2 = 60. A zero would be written as the units digit of the sum and a six would be carried over to the tens column.
In the tens column, 29 of the 30 terms would have a 2 because the first term has no tens digit. The sum of the tens digits would be 29 x 2 = 58, to which we must add the 6 for a total of 64. The 4 gets written down as the second digit of p and the 6 is carried over to the hundreds column.
In the hundreds column, 28 of the 30 terms would have a 2, the sum of the hundreds digits would be 28 x 2 = 56, to which we must add the 6 again for a total of 62. The 2 gets written down as the third digit of p and the 6 is carried over to the thousands column.
There are two ways to finish this problem. We can do out the remaining 8 columns and find that the 11th digit (i.e. the 10 billions column) will have a sum of 2(20) + 4 = 44 (where the 4 was carried over from the 10th column). 4 then will be the 11th digit (from the right) of p (and a 4 will be carried over into the 12th column).
We could also have seen that each column has one less 2 than the previous, so if we started out with 30 2’s in the first column, the 11th column must have 11 – 1 = 10 less 2’s, for a total of 20 2’s. The amount that is carried over from the previous column could be calculated by realizing that the 10th column had 21 2’s for a total of 42. Since there is no way that the 10th column inherited more than 8 from the 9th column, the total must be forty-something and the amount that is carried over to the 11th column MUST BE 4. This makes the total for the 11th column 40 + 4 = 44 and the 11th digit of p4.
QUESTION: 8
Sequence S is defined as Sn = 2Sn-1 – 2. If S1 = 3, then S10 – S9 =
Solution:
We can use the formula to calculate the first 10 values of S:
S1 = 3
S2 = 2(3) – 2 = 4
S3 = 2(4) – 2 = 6
S4 = 2(6) – 2 = 10
S5 = 2(10) – 2 = 18
S6 = 2(18) – 2 = 34
S7 = 2(34) – 2 = 66
S8 = 2(66) – 2 = 130
S9 = 2(130) – 2 = 258
S10 = 2(258) – 2 = 514
S10 – S9 = 514 – 258 = 256.
Alternatively, we could solve this problem by noticing the following pattern in the sequence:
S2 – S1 = 1 or (20)
S3 – S2 = 2 or (21)
S4 – S3 = 4 or (22)
S5 – S4 = 8 or (23)
We could extrapolate this pattern to see that S10 S9 = 28 = 256.
QUESTION: 9
Sn = 2Sn-1 + 4 and Qn = 4Qn-1 + 8 for all n > 1. If S5 = Q4 and S7 = 316, what is the first value of n for whichQn is an integer?
Solution:
If S7 = 316, then 316 = 2S6 + 4, which means that S6=156.
We can then solve for S5
156 = 2S5 + 4, so S5 = 76
Since S5 = Q4, we know that Q4 = 76 and we can now solve for previous Qn’s to find the first n value for which Qn is an integer.
To find Q3: 76 = 4Q3 + 8, so Q3 = 17
To find Q2: 17 = 4Q2 + 8, so Q2 = 9/2
It is clear that Q1 will also not be an integer so there is no need to continue.
Q3 (n = 3) is the first integer value.
QUESTION: 10
What is the sixtieth term in the following sequence? 1, 2, 4, 7, 11, 16, 22...
Solution:
Noting that a1 = 1, each subsequent term can be calculated as follows:
a1 = 1
a2 = a1 + 1
a3 = a1 + 1 + 2
a4 = a1 + 1 + 2 + 3
ai = a1 + 1 + 2 + 3 + ... + i-1
In other words, ai = a1 plus the sum of the first i - 1 positive integers. In order to determine the sum of the first i - 1 positive integers, find the sum of the first and last terms, which would be 1 and i - 1 respectively, plus the sum of the second and penultimate terms, and so on, while working towards the median of the set. Note that the sum of each pair is always equal to i:
1 + (i - 1) = i
2 + (i – 2) = i
3 + (i – 3) = i
…
Because there are (i - 1)/2 such pairs in a set of i - 1 consecutive integers, this operation can be summarized by the formula i(i - 1)/2. For this problem, substituting a1 = 1 and using this formula for the sum of the first (i-1) integers yields:
ai = 1 + (i)(i - 1)/2
The sixtieth term can be calculated as:
a60 = 1 + (59)(60)/2
a60 = 1,771
QUESTION: 11
Sequence S is defined as Sn = X + (1/X), where X = Sn – 1 + 1, for all n > 1.
If S1= 201, then which of the following must be true of Q, the sum of the first 50 terms of S?
Solution:
To find each successive term in S, we add 1 to the previous term and add this to the reciprocal of the previous term plus 1.
S1= 201
The question asks to estimate (Q), the sum of the first 50 terms of S. If we look at the endpoints of the intervals in the answer choices, we see have quite a bit of leeway as far as our estimation is concerned. In fact, we can simply ignore the fractional portion of each term. Let’s use S2 ≈ 202, S3 ≈ 203. In this way, the sum of the first 50 terms of S will be approximately equal to the sum of the fifty consecutive integers 201, 202 … 250.
To find the sum of the 50 consecutive integers, we can multiply the mean of the integers by the number of integers since average = sum / (number of terms).
The mean of these 50 integers = (201 + 250) / 2 = 225.5
Therefore, the sum of these 50 integers = 50 x 225.5 = 11,275, which falls between 11,000 and 12,000. The correct answer is C.
QUESTION: 12
In a certain sequence, every term after the first is determined by multiplying the previous term by an integerconstant greater than 1. If the fifth term of the sequence is less than 1000, what is the maximum number ofnonnegative integer values possible for the first term?
Solution:
The equation of the sequence can be written as follows: an= (an-1 )(x) , where x is the integer constant. So for every term after the first, multiply the previous term by x. Essentially, then, all we are doing is multiplying the first term by x over and over again. For example, (a2)= (a1)(x) and (a3)=(a2)(x) or a3 = ((a1)(x))(x), which is the same as (a3) = (a1)(x2)
If we keep going, we’ll see that a3 = (a1)(x2) and so on for the rest of the sequence. We can thus rewrite the equation of the sequence as an = (a1)(xn-1) , for all n >1.
We also know from the question that a5 < 1000, which means that (a1)(x4) < 1000
We are asked for the maximum number of possible nonnegative integer values for ; we can get this by minimizing the value of the integer constant, x. Since x is an integer constant greater than 1, the smallest possible value for x is 2. When x = 2, then x4= 16
We can solve for a1 as follows:
Thus all the integers from 1 to 62, inclusive, are permissible for a1. So far we have 62 permissible values. If a1=0 then it doesn’t matter what x is, since every term in the sequence will always be 0. So 0 is one more permissible value for a1
There is a maximum of 62 +1 (or 63) nonnegative integer values for ain which a5 < 1000.
QUESTION: 13
The sum of the squares of the first 15 positive integers (12 + 22 + 32 + . . . + 152) is equal to 1240. What is the sum of the squares of the second 15 positive integers (162 + 172 + 182 + . . . + 302) ?
Solution:
The key to solving this problem is to represent the sum of the squares of the second 15 integers as follows: (15 + 1)2 + (15 + 2)2 + (15 + 3)2 + . . . + (15 + 15)2.
Recall the popular quadratic form, (a + b)2 = a2 + 2ab + b2. Construct a table that uses this expansion to calculate each component of each term in the series as follows:
In order to calculate the desired sum, we can find the sum of each of the last 3 columns and then add these three subtotals together. Note that since each column follows a simple pattern, we do not have to fill in the whole table, but instead only need to calculate a few terms in order to determine the sums.
The column labeled a2 simply repeats 225 fifteen times; therefore, its sum is 15(225) = 3375.
The column labeled 2ab is an equally spaced series of positive numbers. Recall that the average of such a series is equal to the average of its highest and lowest values; thus, the average term in this series is (30 + 450) / 2 = 240. Since the sum of n numbers in an equally spaced series is simply n times the average of the series, the sum of this series is 15(240) = 3600.
The last column labeled b2 is the sum of the squares of the first 15 integers. This was given to us in the problem as 1240.
Finally, we sum the 3 column totals together to find the sum of the squares of the second 15 integers: 3375 + 3600 + 1240 = 8215. The correct answer choice is (D).
QUESTION: 14
The infinite sequence Sk is defined as Sk = 10 Sk – 1 + k, for all k > 1. The infinite sequence An is defined as An = 10 An – 1 + (A1 – (n - 1)), for all n > 1. q is the sum of Sk and An. If S1 = 1 and A1 = 9, and if An is positive, what is the maximum value of k + n when the sum of the digits of q is equal to 9?
Solution:
In complex sequence questions, the best strategy usually is to look for a pattern in the sequence of terms that will allow you to avoid having to compute every term in the sequence.
In this case, we know that the first term of Sk is 1 and the first term of An is 9. So when n = 1 and k = 1, q = 9 + 1 = 10 and the sum of the digits of q is 1 + 0 = 1.
since S1= 1, S2= (10)(1) +(2) = 10+2 = 12. since A1 =9,
A2= (10) (9)+(9-(2-1)) = 90+(9-1) = 90+8 = 98. so When K=2 and n = 2, q = 12 + 98 = 110 and the sum of the digits of q is 1 + 1 + 0 = 2.
Since S2= 12, S3 =(10)(12)+3= 120+3= 123. Since A2= 98, It is true that
A3= (10)(98) + (9-(3-1)) = 980+7 = 987. So when n = 3 and k = 3, q = 123 + 987 = 1110 and the sum of the digits of q is 1 + 1 + 1 + 0 = 3.
At this point, we can see a pattern: sproceeds as 1, 12, 123, 1234..., and An proceeds as 9, 98, 987, 9876.... The sum q therefore proceeds as 10, 110, 1110, 11110... The sum of the digits of q, therefore, will equal 9 when q consists of nine ones and one zero. Since the number of ones in q is equal to the value of n and k (when n and k are equal to each other), the sum of the digits of q will equal 9 when n = 9 and k = 9:
S9 = 123456789 and A9 = 987654321. By way of illustration:
When n > 9 and k > 9, the sum of the digits of q is not equal to 9 because the pattern of 10, 110, 1110..., does not hold past this point and the additional digits in q will cause the sum of the digits of q to exceed 9.
Therefore, the maximum value of k + n (such that the sum of the digits of q is equal to 9) is 9 + 9 = 18.
QUESTION: 15
A certain club has exactly 5 new members at the end of its first week. Every subsequent week, each
of the previous week's new members (and only these members) brings exactly x new members into the club. If y is the number of new members brought into the club during the twelfth week, which of the following could be y?
Solution:
At the end of the first week, there are 5 members. During the second week, 5x new members are brought in (x new members for every existing member). During the third week, the previous week's new members (5x) each bring in x new members:
(5x)x= 5xnew members. If we continue this pattern to the twelfth week, we will see that 5x11 new members join the club that week. Since y is the number of new members joining during week 12, y=5x11.
if y=5x11, we can set each of the answer choices equal to 5x11 and see which one yields an integer value (since y is a specific number of people, it must be an integer value). The only choice to yield an integer value is (D)
5x11= 311512
x11 = 311512
x=(3)(5)
Therefore x = 15.
Since choice (D) is the only one to yield an integer value, it is the correct answer.
Use Code STAYHOME200 and get INR 200 additional OFF Use Coupon Code
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# How to find the integer solution number of a linear system with inequalities?
I have the following system with mixed equality and inequalities defined as:
n = 100;
eqn = (n1 + n3 + n5 == n2 + n4 + n6) && (n1 >=
n2) && (n1 - n2 + n3 - n4 >= 0) && (n1 <= n) && (n1 - n2 + n3 <=
n) && (n1 - n2 + n3 - n4 + n5 <= n) && (2 n <=
n1 + n2 + n3 + n4 + n5 + n6 <= 6 n) && (n1 > 0) && (n2 > 0) && (n3 > 0) && (n4 > 0) && (n5 > 0) && (n6 > 0);
How can I find all the integer solutions and especially the number of all integer solutions?
It seems Reduce or Solve does not work:
sol = Reduce[eqn, {n1, n2, n3, n4, n5, n6}, Integers]
produces:
(n1 | n2 | n3 | n4 | n5 | n6) \[Element]
Integers && ((1 <= n1 <= 99 &&
0 < n2 <=
n1 && ((0 < n3 < 100 - n1 && 0 < n4 <= n1 - n2 + n3 &&
100 - n1 - n3 <= n5 <= 100 - n1 + n2 - n3 + n4 &&
n6 == n1 - n2 + n3 - n4 + n5) || (100 - n1 <= n3 <=
100 - n1 + n2 && 0 < n4 <= n1 - n2 + n3 &&
0 < n5 <= 100 - n1 + n2 - n3 + n4 &&
n6 == n1 - n2 + n3 - n4 + n5))) || (n1 == 100 &&
1 <= n2 <= 100 && 0 < n3 <= n2 && 0 < n4 <= 100 - n2 + n3 &&
0 < n5 <= n2 - n3 + n4 && n6 == 100 - n2 + n3 - n4 + n5))
• FindInstance[eqn, {n1, n2, n3, n4, n5, n6}, Integers, 10] will give you the first ten sols, but there are more ... Aug 24, 2014 at 4:14
• I tried FIndInstance by setting the number 10 as 10,000 instead; it has more than 10,000 solutions! The computation is very slow and has a very high demand on memory. Aug 24, 2014 at 9:28
SetSystemOptions[
"ReduceOptions" -> {"DiscreteSolutionBound" -> 10000000}];
f[n_] := Module[{eqn},
eqn = (n1 + n3 + n5 == n2 + n4 + n6) && (n1 >=
n2) && (n1 - n2 + n3 - n4 >= 0) && (n1 <= n) && (n1 - n2 + n3 <=
n) && (n1 - n2 + n3 - n4 + n5 <= n) && (2 n <=
n1 + n2 + n3 + n4 + n5 + n6 <= 6 n) && (n1 > 0) && (n2 >
0) && (n3 > 0) && (n4 > 0) && (n5 > 0) && (n6 > 0);
Length@Reduce[eqn, {n1, n2, n3, n4, n5, n6}, Integers]]
Flatten[{#, Timing[f@#]}] & /@ Range[5, 15] //
TableForm[#, TableHeadings -> {None, {"n", "timing", "numbers"}}] &
Edit: First
Reduce[n1 <= n, n1, Reals]
n1 <= n
Reduce[n1 >= n2, n2, Reals]
n2 <= n1
Reduce[n1 - n2 + n3 <= n, n3, Reals]
n3 <= n - n1 + n2
Reduce[n1 - n2 + n3 - n4 >= 0, n4, Reals]
n4 <= n1 - n2 + n3
Reduce[n1 - n2 + n3 - n4 + n5 <= n, n5, Reals]
n5 <= n - n1 + n2 - n3 + n4
sol = Solve[n1 + n3 + n5 == n2 + n4 + n6, n6]
(2 n <= n1 + n2 + n3 + n4 + n5 + n6 <= 6 n) /. sol
{{n6 -> n1 - n2 + n3 - n4 + n5}}
{2 n <= 2 n1 + 2 n3 + 2 n5 <= 6 n}
Reduce[2 n <= 2 n1 + 2 n3 + 2 n5 <= 6 n && n > 0, n5, Reals]
n > 0 && n - n1 - n3 <= n5 <= 3 n - n1 - n3
Reduce[(n6 /. sol) > 0, n5, Reals]
n5 > -n1 + n2 - n3 + n4
So the range of n5 is:
Max[-n1 + n2 - n3 + n4 + 1, n - n1 - n3, 1]< n5 < Min[3 n - n1 - n3, n - n1 + n2 - n3 + n4]
g1 = Compile[{{n, _Integer}},
Sum[1, {n1, n}, {n2, n1}, {n3, n - n1 + n2}, {n4,
n1 - n2 + n3}, {n5, Max[-n1 + n2 - n3 + n4 + 1, n - n1 - n3, 1],
Min[3 n - n1 - n3, n - n1 + n2 - n3 + n4]}], CompilationTarget -> "C"];
Flatten[{#, Timing[g1@#]}] & /@ Range[5, 15] //
TableForm[#, TableHeadings -> {None, {"n", "timing", "numbers"}}] &
g1[50] // AbsoluteTiming
{0.999057, 39324265}
g2 = Compile[{{n, _Integer}},
Sum[Max[0, Min[3 n - n1 - n3, n - n1 + n2 - n3 + n4] -
Max[-n1 + n2 - n3 + n4 + 1, n - n1 - n3, 1] + 1], {n1, n}, {n2,
n1}, {n3, n - n1 + n2}, {n4, n1 - n2 + n3}], CompilationTarget -> "C"];
g2[50] // AbsoluteTiming
g2[100] // AbsoluteTiming
{0.248014, 39324265}
{3.890223, 1212505405}
• Thank you! Where can I 学到 learn such insights in using Mathematica this way? It seems the online documents of Wolfram have no such contents. Aug 25, 2014 at 5:00
• One possible issue is when $n$ increases, the memory demand becomes high quickly. If only solution number is required, can it be improved to be more efficient and less memory demanding? Aug 25, 2014 at 5:24
• +1 for a great answer and your first Enlightened badge. :-) Aug 29, 2014 at 12:06
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1. ## Vector Parametric Equation
How do I find the equation of a plane with this information:
Its through the point (-3, 9, 10)
and x = 5+t y = 4t and z = 3-2t
2. Originally Posted by acg716
How do I find the equation of a plane with this information:
Its through the point (-3, 9, 10)
and x = 5+t y = 4t and z = 3-2t
I'm guessing that the plane contains the line you mention. If so, then the vector
$\bold{u} = < 1, 4, -2>$
is on the plane. We can also find another vector on the plane, a point on the line to the point given
$\bold{v} = <5-(-3),0-9 , 3 -10> = <8,-9,-7>$
The normal to the plane is then obatin from
$\bold{n} = \bold{u} \times \bold{v}$ (*)
If $\bold{n} = $
then the plane is given by
$n_1(x+3) + n_2(x-9) + n_3(x-10) = 0$
So all you need to do is find n *
3. When I did that I got -46(x+3)-9(y-9)-41(z-10)=0
but that isn't the correct answer...
4. Originally Posted by acg716
When I did that I got -46(x+3)-9(y-9)-41(z-10)=0
but that isn't the correct answer...
Well, when you substitute the point $(-3,9,10)$ into this plane we get 0, right? And when we substitute $x = 5+t,\;\;y = 4t,\;\;z = 3-2t$ we get
$-46(5+t+3)-9(4t-9)-41(3-2t-10)=$
$-46(8+t)-9(4t-9)-41(-7-2t)=$
$-368 - 46 t +81-36t + 287 + 82t = 0$
So both the point and line lie on the plane. Did the question say that the line is perpendicular to the plane?
5. yes it is perpendicular, so sorry that i forgot to include that!
6. Originally Posted by acg716
yes it is perpendicular, so sorry that i forgot to include that!
Makes a huge difference (and easier). The normal to the plane will be parallel so the line and so
$\bold{n} = <1,4,-2>$
so the line is
$(x+3) + 4(x-9) - 2(x-10)=0$
See, way easier.
7. ahh it still says it wrong so maybe I'm leaving something out from the directions so here is what the directions say exactly:
Find an equation of the plane through the point and perpendicular to the given line.
(-3, 9, 10)
x= 5 + t
y= 4t
z= 3-2t
8. Originally Posted by acg716
ahh it still says it wrong so maybe I'm leaving something out from the directions so here is what the directions say exactly:
Find an equation of the plane through the point and perpendicular to the given line.
(-3, 9, 10)
x= 5 + t
y= 4t
z= 3-2t
What answer are you given? Are you given $x + 4y - 2z = 13$?
9. I'm not given an answer, its on my webassign homework so when I submit it it tells me if I'm right or not.
10. Originally Posted by danny arrigo
Makes a huge difference (and easier). The normal to the plane will be parallel so the line and so
$\bold{n} = <1,4,-2>$
so the line is
$(x+3) + 4(x-9) - 2(x-10)=0$
See, way easier.
Originally Posted by danny arrigo
What answer are you given? Are you given $x + 4y - 2z = 13$?
Did you try both answers? If so, I don't know what to suggest. Can some of the other regulars chime in here.
11. The second one was correct, thank you so much!!!
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