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I want to find out the MacLaurin series of this function and find out for which $x$ it equals the original function: $f(x)=\frac{x}{1+3x^2}$

AFAIK I can use this equation: $(1+x)^\alpha=\sum_{n=0}^{\infty}\binom{\alpha}{n}*x^n$. So the MacLaurin series should be $f(x)=x*\sum_{n=0}^{\infty}\binom{-1}{n}*(3x^2)^n$

I think the series equals the function in it's interval of convergence.

So, to find out the convergence radius, I used this approach: $\sum_{n=0}^{\infty}a_n*(x-x_0)^n \Rightarrow r=(\lim\limits_{n \rightarrow \infty}\sqrt[n]{|a_n|})^{-1}$

Then I transformed the series into the form I need:

$ \sum_{n=0}^{\infty}\binom{-1}{n}*(3^n*x^{2^n})\\ \sum_{n=0}^{\infty}(\binom{-1}{n}*3^n)*(x^{2n})\\ \sum_{n=0}^{\infty}(\binom{-1}{n}*3^n*x^{n})*(x-0)^n\\ a_n=-3^n*x^n$

And used the formula:

$ r=(\lim\limits_{n \rightarrow \infty}\sqrt[n]{|a_n|})^{-1}\\ r=(\lim\limits_{n \rightarrow \infty}\sqrt[n]{|-3^n*x^n|})^{-1}\\ r=(\lim\limits_{n \rightarrow \infty}\sqrt[n]{|-3^n|}*\lim\limits_{n \rightarrow \infty}\sqrt[n]{|x^n|})^{-1}\\ r=(|3|*|x|)^-1\\ r=\frac{1}{3|x|} $

But that can't be true, can it? What is the $x$ in $r$? I can't determine the interval of convergence because I don't know what $x$ is.

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  • $\begingroup$ The series has a radius of convergence equals $\frac{1}{\sqrt{3}}$. $\endgroup$ – science Mar 2 '15 at 21:22
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This is very easy to answer by knowing just a little about complex analysis and meromorphic functions. We have: $$f(z)=\frac{z}{1+3z^2}=\frac{1}{6}\left(\frac{1}{z-\frac{i}{\sqrt{3}}}+\frac{1}{z+\frac{i}{\sqrt{3}}}\right)$$ hence $f(z)$ has two simple poles with residue $\frac{1}{6}$ at $z=\pm\frac{i}{\sqrt{3}}$ and the radius of convergence of the Taylor series of $f(z)$ around $z=0$ cannot be bigger than $\frac{1}{\sqrt{3}}$, the distance from the origin of the closest singularity. Since the radius of convergence of: $$\sum_{n\geq 0} \alpha^n\,z^n $$ for any $\alpha\in\mathbb{C}\setminus\{0\}$ is exactly $\frac{1}{|\alpha|}$, and for every $z\in\mathbb{C}$ such that $|z|<\frac{1}{|\alpha|}$ we have: $$\sum_{n\geq 0} \alpha^n\,z^n = \frac{1}{1-\alpha z},$$ it follows that the radius of convergence of the Taylor series of $f(z)$ around $z=0$ is exactly $\color{red}{\frac{1}{\sqrt{3}}}$.

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The problem with your argument is that in $$ \sum_{n=0}^{\infty}\binom{-1}{n}3^n x^n x^n $$ doens't have Taylor coefficients $\binom{-1}{n}3^nx^n$ because the coefficients cannot be functions of $x$. What you want to do is define $a_m$ by $$ a_m = \cases{0 & $m$ odd \\ \binom{-1}{m/2} 3^{m/2}& $m$ even} $$ Then we have $$ \sum_{n=0}^{\infty}\binom{-1}{n}3^n x^{2n} = \sum_{m=0}^\infty a_m x^m $$ and $$ \limsup|a_m|^{1/m} = \limsup|a_{2n}|^{1/2n} = \limsup|\binom{-1}{n}3^n|^{1/2n} = \sqrt{3} $$

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