Taylor Expansion for $f(x) = \frac{x}{(1+x^{2})^{2}}$

Question

I found the expansions for $$f(x) = \frac{1}{1+x^{2}} = \sum_{k=0}^{\infty}(-1)^{k}x^{2k}$$ $$g(x) = \frac{1}{(1+x^{2})^{2}} = \sum_{k=0}^{\infty}(-1)^{k}x^{4k}$$ $$h(x) = \frac{x}{1+x^{2}} = \sum_{k = 0}^{\infty}(-1)^{k}x^{2k+1}$$

But what would be the proper approach of finding an expansion for $$w(x) = \frac{x}{(1+x^{2})^{2}}$$

I think I can either go with: (1) x* $\frac{1}{(1+x^{2})^{2}}$ which would give me $\sum_{k = 0}^{\infty} (-1)^{k} x ^ {4k + 1}$ or (2) squaring the denominator of $\frac{x}{1 + x^{2}}$ which would give me $\sum_{k = 0}^{\infty} (-1)^{k} x ^ {4k + 2}$.

I think that approach (1) is more correct because it deals with just multiplying by x rather than squaring (1+$x^{2}$) but am I right? Is (1) correct and is my reasoning for its correctness correct?

Answer 1

Start from

$$\frac1{1+x}=\sum_{k=0}^\infty(-x)^k.$$

Differentiating both members,

$$-\frac1{(1+x)^2}=\sum_{k=1}^\infty k(-x)^{k-1}=-\sum_{k=0}^\infty (k+1)(-x)^k.$$

Now you can substitute $x^2$ for $x$,

$$\frac1{(1+x^2)^2}=\sum_{k=0}^\infty (k+1)(-x^2)^k$$

and multiply by $x$ (your approach (1)),

$$\frac x{(1+x^2)^2}=\sum_{k=0}^\infty (k+1)(-x^2)^kx=\sum_{k=0}^\infty (k+1)(-1)^kx^{2k+1}.$$

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To "square the denominator" as in your approach (2), you should perform the product of two series, which is not straightforward.

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Alternatively to the "derivative trick" (which needs to be justified by convergence analysis), you can obtain the second series as follows: let

$$h(x)=(1+x)^{-1}\to h(0)=1,\\ h'(x)=(-1)(1+x)^{-2}\to h'(0)=-1,\\ h''(x)=(-1)(1+x)^{-3}\to h''(0)=(-1)(-2)=2!,\\ h'''(x)=(-1)(1+x)^{-4}\to h'''(0)=(-1)(-2)(-3)=-3!,\\\cdots $$

This obviously gives you the development

$$h(x)=\frac1{1+x}=\sum_{k=0}^\infty\frac{(-1)^kk!}{k!}x^k=\sum_{k=0}^\infty(-x)^k.$$

But for the same "price", you obtain the development of $i(x):=h'(x)$ by shifting the coefficients by one position ($i(0)=h'(0), i'(0)=h''(0), i''(0)=h'''(0),\cdots$):

$$i(x)=h'(x)=-\frac1{(1+x)^2}=\sum_{k=0}^\infty\frac{(-1)^{k+1}(k+1)!}{k!}x^k=-\sum_{k=0}^\infty (k+1)(-x)^k.$$

Answer 2

I think the proper approach is as follows. Note that: $$\frac{d}{dx}\left( \frac{-1}{2(1+x^2)} \right)= \frac{x}{(1+x^2)^2}.$$

The power series for $$\frac{1}{1+x^2}=\sum_{n=0}^{\infty} (-1)^n x^{2n},$$ as you found. Hence differentiating the power series and then multiplying by the $-\frac{1}{2}$ gives the desired result for your function in consideration.

If I were you, I wouldn't try to square a series. Trying to square series usually complicates more things. For example, I refer to a problem in analysis, namely the Basel Problem, and a particular solution, which does something like you are trying to do. Can the Basel problem be solved by Leibniz today?

Answer 3

Your expansion of $g (x) $ doesn't seem correct.

try this

$$\frac {1}{(1+x^2)^2}=\frac {1}{1-(-x^4-2x^2)}$$

$$=1+t+t^2+t^3+.... $$ with $$t^n=(-1)^n (x^4+2x^2)^n $$ $$=(-1)^nx^{2n}\sum_{k=0}^n \binom {n}{k}x^{2k}2^{n-k} $$