Calculate $f^{(n)}(1) $ for $ e^{2z^2} $

Question

Taylor theorem : $$ f(z) = \sum_{n=0}^{\infty} a_n (z-a)^n $$ $$ a_n = \frac{1}{2\pi i} \int_{\gamma} \frac{f(w)}{(w-a)^{n+1}}dw = \frac{f^{(n)}(a)}{n!} $$

So we need to write this thing around $1$ :

$$ 2z^2 = 2(z-1)^2 +4(z-1) + 2 $$

$$ f(z) = e^{2(z-1)^2 +4(z-1) + 2} = e^2 \sum_{n=0}^{\infty} \frac{w^{2k}}{k!} \sum_{m=0}^{\infty} \frac{4^mw^{m}}{m!} $$ where $ w = z-1 $. I wrote this thing as

$$ f(z) = e^2 \sum_{n=0}^{\infty} c_n w^n $$

where $c_n$ is

$$ \sum_{2k+m=n}^{} \frac{1}{k!} \frac{4^m}{m!} = \sum_{k=0}^{[n/2]} \frac{1}{k!} \frac{4^{n-2k}}{(n-2k)!} $$

Now I simply multiply $ c_n $ with $n!$ and I get the result? Am I doing this right or is there any other way? Thanks for the time.

Answer 1

You can apply the formula directly: $$ f^{(n)}(1)=\frac{n!}{2\pi i}\int_{\partial R\Bbb D }\frac{e^{2z^2}}{(z-1)^{n+1}}\,\mathrm d z,\quad R>1\\ =\frac{n!}{2\pi i}\int_{\partial R\Bbb D }\frac{\sum_{k\geqslant 0}(2z^2)^k/k!}{(z-1)^{n+1}}\,\mathrm d z\\ =\sum_{k\geqslant 0}\frac{2^k}{k!}\frac{n!}{2\pi i}\int_{\partial R\Bbb D }\frac{z^{2k}}{(z-1)^{n+1}}\,\mathrm d z $$ And now note that $$ \frac{n!}{2\pi i}\int_{\partial R\Bbb D }\frac{z^{2k}}{(z-1)^{n+1}}\,\mathrm d z=[D^n z^{2k}]_{z=1}=(2k)^{\underline{n}} $$ where $D$ is the derivative operator (respect to $z$) and $a^{\underline{b}}$ is a notation for falling factorials. Therefore $$ f^{(n)}(1)=\sum_{k\geqslant 0}\frac{2^k(2k)^{\underline{n}}}{k!} $$ Im not sure by now if the last series have a closed form. Note that we can get the last series just differentiating directly the exponential series $\sum_{k\geqslant 0}\frac{(2z^2)^k}{k!}$ and after taking the value at $z=1$.

Answer 2

Let's see if a more straight-forward method could work, here. \begin{align*} f^{(0)}(z)&=e^{2z^2}\\ f^{(0)}(1)&=e^{2}\\ f^{(1)}(z)&=4ze^{2z^2}\\ f^{(1)}(1)&=4e^{2}\\ f^{(2)}(z)&=\left(4+16z^2\right)e^{2z^2}\\ f^{(2)}(1)&=20e^2\\ f^{(3)}(z)&=\left(48z+64z^3\right)e^{2z^2}\\ f^{(3)}(1)&=112e^{2}\\ f^{(4)}(z)&=\left(48+384z^2+256z^4\right)e^{2z^2}\\ f^{(4)}(1)&=688e^2\\ f^{(5)}(z)&=\left(960z+2560z^3+1024z^5\right)e^{2z^2}\\ f^{(5)}(1)&=4544e^2 \end{align*} Interestingly, the coefficients $1, 4, 20, 112, 688, 4544,\dots$ form a recognized integer sequence. Here is a tabulation for a very large number of these coefficients. And here is an approximate formula for the $n$th coefficient of $e^2:$ $$a(n) \approx 2^{n - 1/2} n^{n/2} \exp(-1 + 2\sqrt n - n/2) $$