Can I find a power series expansion of this function without a Taylor series?

Question

I was asked to find the power series expansion of $f(x) = \frac{x}{\sqrt{x^2+4}}$ about $x = 0$. Is there a way to do a power expansion without finding the Taylor series? Deriving this function multiple times seems extremely tedious.

Answer 1

Since$$\frac x{\sqrt{4+x^2}}=\frac{x/2}{\sqrt{1+(x/2)^2}}=\frac x2\left(1+\left(\frac x2\right)^2\right)^{-1/2}$$and since$$(1+x)^{-1/2}=\sum_{n=0}^\infty\binom{-1/2}nx^n,$$you have\begin{align*}\frac x{\sqrt{4+x^2}}&=\frac x2\sum_{n=0}^\infty\binom{-1/2}n\frac{x^{2n}}{2^n}\\&=\sum_{n=0}^\infty\frac1{2^{n+1}}\binom{-1/2}nx^{2n+1}.\end{align*}

Answer 2

For $$\tag1 \frac{f(2x)}x=\sum_{n=0}^\infty a_nx^n,$$ we immediately see that $a_0=1$ because $f(x)\approx \frac x{\sqrt 4}$ for $x\approx 0$. Also, $2_n=0$ for odd $n$ because $f$ is odd. We have $$\frac{f(2x)^2}{x^2}=\frac{1}{1+x^2}=\sum_{n=0}^\infty (-1)^{n}x^{2n}.$$ On the other hand, squaring $(1)$ gives $$1+2a_2x^2+(a_2^2+2a_4)x^4+(2a_2a_4+2a_6)x^6+\cdots $$ We read off $a_2=-\frac12$, $a_4=\frac12(1-\frac14)=\frac38$, $a_6=-\frac5{16}$, etc. You can find a few more coefficients without much work, make a guess for the general term, and prove it by induction. From there, it is straightforward to obtain the series for $f(x)$ itself.

Answer 3

Notice that $\frac{x}{\sqrt{x^2+4}}=\frac{\mathrm d}{\mathrm dx}\sqrt{x^2+4}$. So if you find the Taylor series for $\sqrt{x^2+4}$, you just have to take its derivative and are done. That one can be derived from the Taylor series of $\sqrt{x+1}$, which is significantly easier to find.

Answer 4

If you want a truncated series, you can $$f(x) = \frac{x}{\sqrt{x^2+4}}$$ Make $x=2y$ to get $$f(y)=\frac{y}{\sqrt{1+y^2}}=\frac{y}{\left(1+y^2\right)^{\frac 12}}$$ Now, use the binomial expansion of the denominator $$f(y)=\frac{y}{1+\frac{y^2}{2}-\frac{y^4}{8}+\frac{y^6}{16}+\cdots }$$ and now long division to get $$f(y)=y-\frac{y^3}{2}+\frac{3 y^5}{8}-\frac{5 y^7}{16}+O\left(y^8\right)$$ Now, replace $y$ by $\frac x 2$.