AP Calculus BC 10.13 Convergence Intervals Overview

Review power-series intervals by finding the radius, testing both endpoints and reporting the correct interval notation.

[Maximum number: 6]

The Taylor series for a function f about x=4 is given by

$\sum_{n=1}^{\infty} \frac{(x-4)^{n+1}}{(n+1) 3^{n}}=\frac{(x-4)^{2}}{2 \cdot 3}+\frac{(x-4)^{3}}{3 \cdot 3^{2}}+\frac{(x-4)^{4}}{4 \cdot 3^{3}}+\cdots+\frac{(x-4)^{n+1}}{(n+1) 3^{n}}+\cdots$ and converges to f(x) on

its interval of convergence.

(a)

Using the ratio test, find the interval of convergence of the Taylor series for f about x=4.

Justify your answer.

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A response that presents any of these reciprocal ratios earns P1 and is eligible for P2, P3, and P4,

but not P5.

- A response that does not present a ratio is not eligible for P2.

- A response that does not use the absolute value of a ratio can earn both P1 and P2.

- A response that does not use absolute value in computing the limit is eligible for P3 if the expression

$-1<\frac{x-4}{3}<1$ or an equivalent inequality is presented.

- A response that presents an incorrect limit of form $\frac{|x-4|}{b}$ , where b>0, is eligible for P3.

$\mathbf{P 3}$ is then earned for correctly finding an interval with interior (1,7) or (4-b, 4+b).

Note: |x-4|<b is not sufficient to earn P3. x-4<b can earn P3 if this is resolved to

-b+4<x<b+4.

- P4 is earned for considering both endpoints of the correct interval or both endpoints of an incorrect

interval that has earned P3.

- To earn P5, a response must correctly analyze the series at x=1 and x=7, and present the correct

interval of convergence. Naming of an appropriate test is sufficient for the analysis at each endpoint.

In addition to the tests listed in the model solution, the direct comparison test to an appropriate series

or the integral test may also be used.

(b)

It is known that the radius of convergence of the Taylor series for f about x=4 is the same as

the radius of convergence of the Taylor series for $f^{\prime}$ about x=4. Does the Taylor series for

$f^{\prime}$ described in part B converge to $f^{\prime}(x)=\frac{x-4}{7-x}$ at x=8 ? Give a reason for your answer.

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D It is known that the radius of convergence of the Taylor series for f about x=4 is the same as the radius

of convergence of the Taylor series for $f^{\prime}$ about x=4. Does the Taylor series for $f^{\prime}$ described in

part B converge to $f^{\prime}(x)=\frac{x-4}{7-x}$ at x=8 ? Give a reason for your answer.

It follows from the work in part A that the interior of the interval Answer with reason Point 9 (P9)

of convergence of the Taylor series for $f^{\prime}$ is 1<x<7.

Therefore, x=8 would be outside the interval of convergence

of $f^{\prime}$ , and the Taylor series for $f^{\prime}$ would not converge to

$f^{\prime}(x)=\frac{x-4}{7-x}$ at x=8.

Scoring Notes for Part D

- A response of "no, x=8 is outside the interval of convergence" is sufficient to earn P9.

- P9 can be earned with a response consistent with an incorrect interval of convergence imported from

part A.

- Alternate solutions:

○ Because the series for $f^{\prime}(x)$ is geometric, this converges to f(x) for all values of x such that

the common ratio $\frac{x-4}{3}$ is between -1 and 1.

$-1<\frac{x-4}{3}<1 \Rightarrow-3<x-4<3 \Rightarrow 1<x<7$

○ Because x=8 is outside the interval 1<x<7, the series for $f^{\prime}$ does not converge to

$f^{\prime}(x)=\frac{x-4}{7-x}$ at x=8.

○ For x=8, the series is given by $\sum_{n=1}^{\infty}\left(\frac{4}{3}\right)^{n}$ , which is a geometric series with $r=\frac{4}{3}>1$ .

Therefore, the series diverges for x=8.

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Covered syllabus points

10.13 Radius and Interval of Convergence of Power Series question 1

Question (a)

Question (b)