Finding the Interval of Convergence for f'(x) in a Power Series - Homework Help

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Homework Help Overview

The discussion revolves around finding the interval of convergence for the derivative of a power series function, specifically f'(x), where f(x) is defined as a series involving terms of the form \((x-5)^n\) and \((-1)^n\) divided by \(n5^n\).

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore how to differentiate the power series, questioning whether to treat \(n\) as a constant. There is also discussion about the form of the derivative and the correctness of the signs in the series terms.

Discussion Status

The discussion is ongoing, with participants providing guidance on treating \(n\) as a constant during differentiation and cautioning about potential issues when differentiating series. Some participants suggest that finding the radius of convergence for the original function may suffice, indicating a possible direction for further exploration.

Contextual Notes

There is a mention of potential issues when differentiating series, particularly regarding the starting index of summation and the implications for convergence. Participants are also clarifying the correct form of the series terms.

bcjochim07
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Homework Statement


Find the interval of convergence of f'(x)

f(x) Sum from n=1 to infinity [(x-5)^n*(-1)^n]/[n5^n]



Homework Equations





The Attempt at a Solution



My problem is I am unsure how to take the derivative with the n's and x's should I treat n as a constant?

After that, I think I can get the interval of convergence.
 
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Is the derivative the sum from n=1 to infinity of [(x-5)^(n-1) * (-1)^(n+1)]/[5^n]?
 
Last edited:
bcjochim07 said:
Is the derivative the sum from n=1 to infinity of [(x-5)^(n-1) * (-1)^(n+1)]/[5^n]?

Almost. How did (-1)^n become (-1)^(n+1)?
 
Oh... it should be (-1)^(n+1) in the original function, I just typed it out wrong.
 
It looks fine then.
 
yes you just treat n as a constant.

Be careful when taking the derivative of a series though. Here the issue didn't come up, but if you have x^n where n starts at 0 and end up with x^(n-1) where n starts at 0 then you would get x^(-1) for n=0 which is a no-no so you'd have to move you n up to starting at 1 do solve that problem.
 
I think finding R for the original function should be enough; differentiation does not change R.

Using this way, you get R = 5?
 

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