Taylor Series Expansion to Compute Derivatives

In summary, the homework statement is to find the Taylor series expansion of about x=1. The first step is to simplify by letting u= x- 1. Now the function is u/(1- u^2). The sum of the geometric series \sum_{n=0}^\infty ar^n is a/(1- r) so let a= u and u= r^2. 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)^
  • #1
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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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  • #2
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]
 
  • #3
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?
 
  • #4
Please find my attempt at the solution attached
 

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  • #5
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.
 
  • #6
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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  • #7
Like I said, compare the series you got with the generic series for the Taylor expansion.
 
  • #8
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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  • #9
Can someone please help me with this? It is very important.
 

What is a Taylor series expansion?

A Taylor series expansion is a mathematical tool used to approximate a function by representing it as an infinite sum of simpler functions. It is based on the idea that any differentiable function can be approximated by a polynomial function.

Why is Taylor series expansion used to compute derivatives?

Taylor series expansion is used to compute derivatives because it provides a way to approximate the value of a function and its derivatives at a specific point. This allows for the calculation of derivatives of complicated functions that cannot be easily evaluated using traditional methods.

What is the formula for a Taylor series?

The formula for a Taylor series is given by:
f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ...

How many terms are needed in a Taylor series to accurately compute a derivative?

The number of terms needed in a Taylor series to accurately compute a derivative depends on the complexity of the function and the desired level of accuracy. Generally, the more terms included in the series, the more accurate the approximation will be.

Are there any limitations to using Taylor series expansion to compute derivatives?

Yes, there are limitations to using Taylor series expansion. It can only be used for differentiable functions, and the function must have a finite number of derivatives at the point of expansion. Additionally, the series may not converge for certain values of x, resulting in an inaccurate approximation.

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