Taylor series centered at c = 1

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Homework Help Overview

The discussion revolves around finding the Taylor series of the function 1/x centered at c = 1. Participants are examining the derivatives of the function and attempting to establish a pattern for the coefficients in the series expansion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are listing derivatives of the function and evaluating them at the center point. There is an attempt to identify a pattern in the derivatives to formulate the Taylor series. One participant questions the validity of their derived series and seeks confirmation from others.

Discussion Status

Some participants express uncertainty about the correctness of their answers, noting discrepancies with the textbook solution. There is a recognition of potential issues with the textbook's answer, leading to a decision to set it aside for further discussion in class.

Contextual Notes

Participants mention previous problems with the textbook's answers, indicating a lack of trust in the provided solutions. There is also a note that the textbook's answer does not converge at the center point, raising questions about its relevance to the current problem.

DrummingAtom
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Homework Statement



Find the Taylor Series of 1/x centered at c = 1.

Homework Equations



[tex]\sum_{n=0}^{\infty} f^n (c) \frac{(x-c)^n}{n!}[/tex]

The Attempt at a Solution



I made a list of the derivatives:

f(x) = 1/x
f'(x) = -1/x2
f''(x) = 2/x3
f'''(x) = -6/x4

f(1) = 1
f'(1) = -1
f''(1) = 2
f'''(1) = -6

From this I see the pattern fn(c) = (-1)n(n!)

[tex]\sum_{n=0}^{\infty} (-1)^n(n!) \frac{(x-1)^n}{n!}[/tex]

Then I canceled the factorials and I'm left with

[tex]\sum_{n=0}^{\infty} (-1)^n (x-1)^n[/tex]

Checked my answer and it's way off.. Thanks for any help.
 
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DrummingAtom said:

Homework Statement



Find the Taylor Series of 1/x centered at c = 1.

Homework Equations



[tex]\sum_{n=0}^{\infty} f^n (c) \frac{(x-c)^n}{n!}[/tex]

The Attempt at a Solution



I made a list of the derivatives:

f(x) = 1/x
f'(x) = -1/x2
f''(x) = 2/x3
f'''(x) = -6/x4

f(1) = 1
f'(1) = -1
f''(1) = 2
f'''(1) = -6

From this I see the pattern fn(c) = (-1)n(n!)

[tex]\sum_{n=0}^{\infty} (-1)^n(n!) \frac{(x-1)^n}{n!}[/tex]

Then I canceled the factorials and I'm left with

[tex]\sum_{n=0}^{\infty} (-1)^n (x-1)^n[/tex]

Checked my answer and it's way off.. Thanks for any help.

How did you check your answer to conclude "it's way off"? I agree with your answer. I'll probably disagree with your "check".
 
Hmm well that's reassuring. We've had a couple problems with the answers in this book being off.

The back of the book said the answer is:

[tex] \sum_{n=0}^{\infty} \frac {(-1)^{n+1}(x-5)^n} {4^{n+1}}[/tex]

I'll just go ahead and ignore that answer for now until next class. Thanks for your help.
 
DrummingAtom said:
Hmm well that's reassuring. We've had a couple problems with the answers in this book being off.

The back of the book said the answer is:

[tex] \sum_{n=0}^{\infty} \frac {(-1)^{n+1}(x-5)^n} {4^{n+1}}[/tex]

I'll just go ahead and ignore that answer for now until next class. Thanks for your help.

The book's answer doesn't even converge at x=1 and it's not expressed in powers of (x-1). It looks like the answer to some completely different exercise. Yes, ignore it.
 

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