Find Laurent series of $\frac{x^2}{(1+x^2)^2}$

Question

I would like to find the Laurent series of the function $$ f(x)=\frac{x^2}{(1+x^2)^2}. $$

I already know that there are two poles of order $2$, namely $x=\pm i$. And I also know that, by partial fraction decomposition, $$ \frac{x^2}{(1+x^2)^2}=\frac{1}{x^2+1}-\frac{1}{(1+x^2)^2}. $$

Hint: $(1+y)^{-2}=\sum_{n\ge0}(-1)^n(n+1)y^n$ by Newton's generalized binomial theorem. Take $y=x^2$, then multiply by $x^2$. — J.G., Jan 04 '23 at 15:18
There are many Laurent series associated to that function. They depend the annulus where the expansion is expected to converge, in particular on the center of that annulus. After the partial fraction decomposition, the rest of the work will depend on that information. — plop, Jan 04 '23 at 15:29
That is a more specific problem and doesn't necessarily need to pass through a Laurent series. If a Laurent series is to be used, it must converge in an annulus centered at $x=i$, with inner radius $0$ and positive outer radius. The latter we know it can go all the way to the nearest singularity (x=-i), which is at distance $2$. You can write the $1+x^2$ as $(x-i)(x+i)=(x-i)((x-i)+2i)=2i(x-i)(1+(x-i)/(2i))$. Now, the part $(1+(x-i)/2i)^k$, which we have for $k=-1$ and $k=-2$ in the partial fraction decomposition, you can expand using the binomial series linked above. — plop, Jan 04 '23 at 15:39
But take into account that all the work of the Laurent series is not needed. We know the pole at $x=i$ has order $2$. So, we can just multiply by $(x-i)^2$, take derivative of order $1$, and compute the limit as $x\to i$, to get the residue. This is because we expect the Laurent series to be $\frac{a_{-2}}{(x-i)^2}+\frac{a_{-1}}{(x-i)}+[\text{the regular part}]$. So, those operations (the multiplication, the derivative and the limit) will give us the $a_{-1}$. — plop, Jan 04 '23 at 15:41
Okay, then I need to compute $$ \lim_{x\to i}\frac{2x(x-i)(ix^2+2x-i)}{(x^2+1)^3} $$ — Rhjg, Jan 04 '23 at 15:49
Something like that. I haven't computed it. Expect all the $x-i$ in the numerator and denominator to cancel each other. In your expression it looks like they do, giving $\frac{2ix}{(x+i)^3}\to \frac{-2}{(2i)^3}$ — plop, Jan 04 '23 at 15:53

score 1 · Answer 1 · answered Jan 09 '23 at 22:53

Here are two variations to calculate the residue of $f$ at $x=i$. We already know the pole of $f$ at $x=i$ is of order $2$. A common way is to use the

Limit formula for higher order poles:

We obtain \begin{align*} \color{blue}{\mathrm{res}_{x=i}f(x)}&=\mathrm{res}_{x=i}\frac{x^2}{\left(1+x^2\right)}\\ &=\lim_{x\to i}\frac{d}{dx}\left((x-i)^2f(x)\right)\\ &=\lim_{x\to i}\frac{d}{dx}\left(\frac{x^2}{(x+i)^2}\right)\\ &=\lim_{x\to i}\frac{2ix}{(x+i)^3}\\ &=\frac{2i^2}{(2i)^3}\\ &\,\,\color{blue}{=-\frac{1}{4}i}\tag{1} \end{align*}

It is somewhat more cumbersome, but also not so difficult, to calculate the remainder of $f$ at $x=i$ by consequent expansion at $x=i$.

Manual expansion in terms of $(x-i)$:

We start with \begin{align*} f(x)&=\frac{x^2}{\left(1+x^2\right)^2}=\frac{x^2}{(x+i)^2(x-i)^2}\\ &=\frac{(x-i+i)^2}{(x-i+2i)^2(x-i)^2}\\ &=\frac{(x-i)^2+2i(x-i)-1}{(2i)^2\left(1+\frac{x-i}{2i}\right)^2(x-i)^2}\\ &=-\frac{1}{4}\frac{1}{\left(1+\frac{x-i}{2i}\right)^2} +\frac{1}{2i(x-i)}\,\frac{1}{\left(1+\frac{x-i}{2i}\right)^2}\\ &\qquad+\frac{1}{4(x-i)^2}\,\frac{1}{\left(1+\frac{x-i}{2i}\right)^2}\tag{2} \end{align*} The expansion of the square of the geometric series in (2) gives \begin{align*} \frac{1}{\left(1+\frac{x-i}{2i}\right)^2} &=\left(1-\frac{x-i}{2i}+\cdots\right)\left(1-\frac{x-i}{2i}+\cdots\right)\\ &=1-\frac{x-i}{i}+\mathrm{higher\ order\ terms}\tag{3} \end{align*}

We derive from (2) and (3) \begin{align*} \color{blue}{\mathrm{res}_{x=i}f(x)}=0+\frac{1}{2i}+\frac{1}{4}\left(-\frac{1}{i}\right)\,\,\color{blue}{=-\frac{1}{4}i} \end{align*} in accordance with (1).

Find Laurent series of $\frac{x^2}{(1+x^2)^2}$

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