Series Representation for Function

[KDawgAtsu]

Homework Statement

Fin the Taylor series about x = 0 for:
f(x) = 1 / (1-x²)²

Homework Equations

g(x) = 1 / (1-x²)

The Attempt at a Solution

Differentiating g(x), the series representation of g'(x) is Ʃn(x²)^n-1
Since f(x) = g'(x)/2x
f(x) = Ʃ(n/2)x^2n-1
I'm pretty sure that's right, but what I don't get is that the solution for this problem is also written as Ʃn(x²)^n-1, which is just g'(x).

What am I misunderstanding here?

 

[SammyS]

Homework Statement

Fin the Taylor series about x = 0 for:
f(x) = 1 / (1-x²)²

Homework Equations

g(x) = 1 / (1-x²)

The Attempt at a Solution

Differentiating g(x), the series representation of g'(x) is Ʃn(x²)^n-1
Since f(x) = g'(x)/2x
f(x) = Ʃ(n/2)x^2n-1
I'm pretty sure that's right, but what I don't get is that the solution for this problem is also written as Ʃn(x²)^n-1, which is just g'(x).

What am I misunderstanding here?

The Taylor series about x = 0 for g(x) = 1 / (1-x²) is: $\displaystyle g(x)=\sum_{n=0}^{\infty}(x^2)^n=\sum_{n=0}^{\infty}(x^{2n})\,,$ where x⁰=1 .

Differentiating term by term gives: $\displaystyle g'(x)=\sum_{n=1}^{\infty}2n(x^{2n-1})=\sum_{n=0}^{\infty}2(n+1)x(x^{2n})\,,\text{ or equivalently }\sum_{n=1}^{\infty}\frac{2n}{x}(x^2)^n$

Dividing by 2x gives: $\displaystyle f(x)=\sum_{n=1}^{\infty}n(x^2)^{n-1}=\sum_{n=0}^{\infty}(n+1)(x^{2})^n\,.$

They're all somewhat different.

 

[KDawgAtsu]

$\displaystyle g'(x)=\sum_{n=1}^{\infty}2n(x^{2n-1})=\sum_{n=0}^{\infty}2(n+1)x(x^{2n})\,,\text{ or equivalently }\sum_{n=1}^{\infty}\frac{2n}{x}(x^2)^n$



Where did the x in front of the x^2n come from? I know you just shifted n=1 to n = 0, but I can't see where the x comes from?

 

[SammyS]

$\displaystyle g'(x)=\sum_{n=1}^{\infty}2n(x^{2n-1})=\sum_{n=0}^{\infty}2(n+1)x(x^{2n})\,,\text{ or equivalently }\sum_{n=1}^{\infty}\frac{2n}{x}(x^2)^n$

Where did the x in front of the x^2n come from? I know you just shifted n=1 to n = 0, but I can't see where the x comes from?

$\displaystyle g'(x)=\sum_{n=1}^{\infty}2n(x^{2n-1}) =\sum_{n=0}^{\infty}2(n+1)(x^{2n+1})=\sum_{n=0}^{ \infty}2(n+1)(x\cdot x^{2n})$

Change n to n+1 so 2n - 1 becomes 2(n+1)-1 = 2n + 2 - 1 = 2n+1 .

 

[vela]

Differentiating g(x), the series representation of g'(x) is Ʃn(x²)^n-1

If you were wondering why your attempt didn't work out, it's because you forgot to use the chain rule when you differentiated (x²)ⁿ. That's where the factor of 2x went missing in your calculation.