Taylor Series Expansion to Compute Derivatives

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Homework Help Overview

The problem involves finding the Taylor series expansion of the function f(x) = (x-1)/(1+(x-1)^2) about x=1, with the goal of computing the 9th and 10th derivatives at that point.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss simplifying the function by substituting u = x - 1 and explore the use of geometric series for expansion. There are attempts to derive the Taylor series and confusion regarding the behavior of derivatives at x=1.

Discussion Status

Some participants have provided guidance on using series expansion techniques and comparing derived series to the Taylor expansion form. There is ongoing exploration of how to accurately compute the derivatives, with some expressing uncertainty about the necessity of directly calculating the derivatives of the original function.

Contextual Notes

Participants are grappling with the implications of their series expansions and the behavior of terms when evaluated at x=1, leading to questions about the correctness of their approaches and assumptions regarding the derivatives.

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Homework Statement



Find the Taylor series expansion of

f(x) = (x-1)/(1+(x-1)^2)

about x=1 and use this to compute f(9)(1) and f(10)(1)

Homework Equations



The sum from n=0 to infinity of f(k)(c)/(k!) (x-c)k

The Attempt at a Solution



I'm not sure how to approach this problem. Using the expansion formula is clearly incorrect as the derivative keeps on expanding. Any help would be greatly appreciated.
 
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First, simplify by letting u= x- 1. Now the function is [itex]u/(1- u^2)[/itex].

The sum of the geometric series [itex]\sum_{n=0}^\infty ar^n[/itex] is a/(1- r) so let a= u and [itex]u= r^2[/itex]
 
I'm a little confused. I believe the bottom would become 1+u2.

I tried using the power series for 1/(1-x) (sum from n = 0 to infinity of x^n).

This left me with the sum from n = 0 to infinity of (x-1)^(2n+1).

I then wrote out the Taylor Polynomial until the 9th power because I wanted to evaluate f(9)(1).

When I plug in 1, all terms go to zero. I don't think this is correct.

Any thoughts?
 
Please find my attempt at the solution attached
 

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You're fine up to
[tex]\frac{u}{1+u^2} = u \sum_{n=0}^\infty (-u^2)^n[/tex]You want to pull the sign out first before pulling the factor of u from out front into the series.
[tex]\frac{u}{1+u^2} = u \sum_{n=0}^\infty (-1)^n(u^2)^n = \sum_{n=0}^\infty (-1)^n u^{2n+1} = \sum_{n=0}^\infty (-1)^n (x-1)^{2n+1}[/tex]
Now compare this series to the Taylor expansion about x=1:
[tex]f(x) = \sum_{m=0}^\infty \frac{f^{(m)}(1)}{m!} (x-1)^m[/tex]You should be able to read off what the 9th and 10th derivatives are simply by looking at the series you obtained above.
 
I'm sorry, could you please elaborate further on how to find the derivatives?

I have attached my attempt but I don't think it's correct because I'm getting 0 for both the 9th and 10th derivatives.
 

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Like I said, compare the series you got with the generic series for the Taylor expansion.
 
I think I am missing something.

Do I actually have to compute the 9th derivative of the original function?

Another attempt is attached.
 

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Can someone please help me with this? It is very important.
 

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