I have a sum like this
$$\sum_{k=0}^\infty (-1)^k \frac{(2k)!}{k!k!} x^k$$
I wonder if this can be a Taylor expansion of a function. What could be the function here
This form is quite similar to Taylor expansion of $arcsinh$ $$arcsinh(x) = \sum_{k=0}^\infty \frac{(-1)^k(2k)!}{4^k k!k! (2k+1)}x^{2k+1}$$
Thanks.
On
You made a good catch thinking about $$\sinh ^{-1}(x) = \sum_{k=0}^\infty (-1)^k\frac{(2k)!}{4^k k!k! (2k+1)}x^{2k+1}$$
Consider $$y=\sum_{k=0}^\infty (-1)^k\frac{(2k)!}{4^k k!k! (2k+1)}x^{2k+1}\implies y'=\sum_{k=0}^\infty (-1)^k\frac{(2k)!}{4^k k!k! }x^{2k}=\frac{1}{\sqrt{1+x^2}}$$ Now, rewrite $$y'=\sum_{k=0}^\infty (-1)^k\frac{(2k)!}{ k!k! }\left(\frac{x^2}{4}\right)^k$$ and you are almost at the solution.
Note that $(2k)!/(2^kk!)$ is the product of the odd integers from $1$ to $2k-1$ inclusive, which go in steps of $2$, whereas $(2k)!/(4^kk!)$ is the product of the half-odd integers from $\frac12$ to $k-\frac12$ inclusive, which go in steps of $1$. So the falling Pochhammer symbol $(-\frac12)_k$ is $(2k)!/((-4)^kk!)$, whence $(2k)!/((-4)^kk!^2)$ is the $(-x)^k$ coefficient in $(1+x)^{-1/2}$ by the binomial theorem. So the original series is $(1+4x)^{-1/2}$ instead.