Find the Maclaurin Series for $\frac{x}{x^2 + 1}$

Question

How to find the Maclaurin Series for $$\frac{x}{x^2 + 1}?$$

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I know that the Maclaurin Series is given by:

$$f(x) = f(0) + f'(0)x + \dfrac{f''(0)}{2!}x^2 + \dots + \dfrac{f^{(k)}(0)}{k!}x^k.$$

So from repeatedly deriving $f(x)$ until there was a pattern. I have found that the Maclaurin Series for $\frac{1}{1+x}$ is given as $1-x + x^2 - x^3 + x^4 - x^5 + \dots$

In sigma notation: $$\sum_{n=0}^{\infty} (-1)^n \cdot x^n.$$

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Now I need to get the result for the new function $\frac{x}{x^2 + 1}$. Because $x^2+1$ is irreducible is there any way to do this without getting into imaginary numbers? Formally, the question says "Using your series from $\frac{1}{1+x}$, find the Maclaurin series for $\frac{x}{x^2+1}$" Perhaps I have to multiply through by $x^2$?

Answer 1

$${\frac {1}{1+x}}=\sum _{n=0}^{\infty }(-1)^nx^{n}$$ let $x\rightarrow x^2$ $${\frac {1}{1+x^2}}=\sum _{n=0}^{\infty }(-1)^nx^{2n}$$ then multiply by $x$ $${\frac {x}{1+x^2}}=\sum _{n=0}^{\infty }(-1)^nx^{2n+1}$$

Answer 2

Just another way to do it.

Consider that $${\frac {x}{1+x^2}}=\frac 12 \frac d {dx} \log(1+x^2)$$ Now, using $$\log(1+t)=\sum_{n=1}^\infty(-1)^{n+1}\frac{t^n}n$$ $$\log(1+x^2)=\sum_{n=1}^\infty(-1)^{n+1}\frac{x^{2n}}n$$ $$\frac d {dx} \log(1+x^2)=\sum_{n=1}^\infty(-1)^{n+1}\frac{2nx^{2n-1}}n=2\sum_{n=1}^\infty(-1)^{n+1}x^{2n-1}=2\sum_{m=0}^\infty(-1)^{m}x^{2m+1}$$ and then the result.