If $f(x) = e^{x^{2}}$, show that $f^{(2n)}(0)=(2n)!/n!$

Question

If $f(x) = e^{x^{2}}$, show that $f^{(2n)}(0)=(2n)!/n!$

Answer 1

We have

$$e^u=\sum_{n=0}^\infty\frac {u^n}{n!}$$ so with $u=x^2$ we have $$e^{x^2}=\sum_{n=0}^\infty\frac {x^{2n}}{n!}=\sum_{n=0}^\infty\frac {f^{(2n)}(0)}{(2n)!}{x^{2n}}$$ and we deduce the result by the unicity of the power series.

Answer 2

You can also, as a funny exercise, apply Faa di Bruno's formula for the $n$th derivative of a composite function, which says that

$$\dfrac{d^{m}}{dx^m}(g\circ F)(x)=\sum \dfrac{m!}{b_1!b_2!\cdots b_m!}g^{(k)}(F(x))\prod_{i=1}^m\left(\dfrac{F^{(i)}(x)}{i!}\right)^{b_i},$$

where the sum is over all nonnegative integers $b_1,\ldots,b_m$ such that $\sum_{i=1}^m ib_i=m$ and where $k=\sum_{i=1}^m b_i$.

Now, let $m=2n$ and choose $g(x)=e^x$, $F(x)=x^2$, then $f(x)=e^{x^2}=(g\circ F)(x)$. Note that $g^{(k)}(x)=e^x$ for all $k$ and that $F'(x)=2x$, $F''(x)=2$ and $F'''=F^{(iv)}(x)=\cdots=0$. Thus, evaluated at $x=0$, the only non-vanishing summands arise when $b_1=b_3=b_4=\cdots=b_{2n}=0$ and $b_2=n$ such that $f^{(2n)}(0)$ is immediately given as $\dfrac{m!}{b_1!b_2!\cdots b_m!}=\dfrac{(2n)!}{n!}$, as required, since $g^{(k)}(F(0))=1$ and $\dfrac{F''(0)}{2!}=1$.

Answer 3

$f(x)=e^{x^2}\Rightarrow f'(x)=2xf(x)$ Now use Leibnitz's repeated differentiation $2n-1$ times to get $$f^{(2n)}(x)=2xf^{(2n-1)}(x)+2\binom{2n-1}{1}f^{(2n-2)}\\ \Rightarrow f^{(2n)}(0)=2(2n-1)f^{(2n-2)}(0)=\cdots=2^n(2n-1)(2n-3)\cdots 3\cdot 1\\ =2^n\frac{(2n)!}{2^n n!}=\frac{(2n)!}{n!}$$