I tried Taylor expanding the following function for small $x \ll 1$:
\begin{align*}
f(x) = \frac{1}{\sqrt{2\pi}} \int\limits^{+\infty}_{-\infty} e^{\frac{-xy^4}{24}-\frac{y^2}{2}}\,\mathrm{d}y.
\end{align*}
I already know that $f$ takes on finite values for $x \geq0$. My goal is to get a series expression for $f$ of the following form:
\begin{align*}
f_N(x) = \sum^N_{n = 0}a_n x^n.
\end{align*}
So after using the formula (for $a = 0$)
\begin{align*}
f(x) = \sum^\infty_{n = 0} \frac{f^{(n)}(a)}{n!}(x-a)^n,
\end{align*}
I get up to the $N$-th term:
\begin{align*}
f_N(x) = \sum^N_{n = 0} \frac{1}{n!}\frac{(-x)^n}{24^n}(4n-1)!!
\end{align*}
However, when plugging in a small value for $x$ (for example $x = 0.1$) in Wolfram Alpha, I get that till $N = 25$ the approximation is very good, but after around $N = 30$ the series diverges away from the true value ($\approx 0.988306$ ). So for $N \rightarrow + \infty$, the series seems to diverge.
My questions are:
How can this be ? I thought that for larger and larger $N$ the Taylor approximation would be more and more better ? (well, for small $x$ atleast)
Does this mean that the radius of convergence is equal to $0$ and that the interval of convergence is just the point $x = 0$ ? If not, then what is the radius of convergence ? If yes, doesn't this contradict the fact that $f$ is finite for $x \geq1$ ?
Best Answer
Indeed the radius of convergence of the function $f$ is zero, which means that the series
$$ \sum_{k=0}^{\infty} \frac{(-1)^k (4k-1)!!}{k! 24^k} z^k $$
diverges for all $z \neq 0$. Although this can be directly verified by using Stirling's approximation, a more fundamental reason for this is that $0$ is a branch point of $f(z)$. Consequently, there is no way $f(z)$ extends to an analytic function about $0$. For instance, the closed-form
$$ f(z) = \sqrt{\frac{3}{2\pi z}} \, e^{3/4z} K_{1/4}(3/4z) \tag{*} $$
in Claude Leibovici's answer has the branch cut $(-\infty, 0]$, as seen from the domain coloring plot of $\text{(*)}$ on the rectangle with the corners $\pm 5 \pm 5i$:
That said, your best hope for an asymptotic expansion of polynomial form is
$$ f(z) = \sum_{k=0}^{N-1} \frac{(-1)^k (4k-1)!!}{k! 24^k} z^k + \mathcal{O}(z^N) \qquad \text{as } z \to 0 \tag{1} $$
within Stolz angle for each fixed $N \geq 0$.
Addendum. Here is a proof that $\text{(1)}$ holds at least within the region $\operatorname{Re}(z) \geq 0$, although I suspect that this holds within larger sectors.
Note that for each fixed $N$, Taylor's theorem tells that for $\operatorname{Re}(z) \geq 0$,
\begin{align*} &f(z) - \sum_{k=0}^{N-1} \frac{(-1)^k (4k-1)!!}{k! 24^k} z^k \\ &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \Biggl[ \exp\left(-\frac{zy^4}{24}\right) - \sum_{k=0}^{N-1} \frac{1}{k!} \left( -\frac{zy^4}{24} \right)^k \Biggr] e^{-\frac{y^2}{2}} \, \mathrm{d}y \\ &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \Biggl( \int_{0}^{1} \frac{z^N (1-t)^{N-1}}{(N-1)!} \left(-\frac{y^4}{24}\right)^N \exp\left(-\frac{zty^4}{24}\right) \, \mathrm{d}t \Biggr) e^{-\frac{y^2}{2}} \, \mathrm{d}y, \end{align*}
where the last line follows from the Lagrange form of the remainder. Taking absolute value,
\begin{align*} &\left| f(z) - \sum_{k=0}^{N-1} \frac{(-1)^k (4k-1)!!}{k! 24^k} z^k \right| \\ &\leq \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \Biggl( \int_{0}^{1} \frac{\left| z \right|^N (1-t)^{N-1}}{(N-1)!} \frac{y^{4N}}{24^N} \, \mathrm{d}t \Biggr) e^{-\frac{y^2}{2}} \, \mathrm{d}y \\ &= \frac{(4N-1)!!}{N!24^N} \left|z\right|^N. \end{align*}
This proves that the error term in $\text{(*)}$ is indeed $\mathcal{O}(z^N)$, at least within the region $\operatorname{Re}(z) \geq 0$.