If $f(x) = e^{x^{2}}$, show that $f^{(2n)}(0)=(2n)!/n!$
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3Perhaps examine $f$'s Taylor Series about $x=0$ (which can be obtained by substitution into the series for $e^x$). – David Mitra Jun 09 '14 at 10:18
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1a simple induction using the Leibnitz formula should be okay. – mookid Jun 09 '14 at 10:19
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I'm not familiar with the Leibnitz formula. – Sally G Jun 09 '14 at 10:21
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1it is: $(fg)^{(n)} = \sum \binom n k f^{(k)} g^{(n-k)}$ – mookid Jun 09 '14 at 10:24
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$\displaystyle\bigg(e^{x^k}\bigg)^{(kn)}(0)=\frac{(kn)!}{n!}$ – Lucian Jun 09 '14 at 11:21
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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.
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1How did you go through the step: $$e^{x^2}=\sum_{n=0}^\infty\frac {x^{2n}}{n!}=\sum_{n=0}^\infty\frac {f^{(2n)}(0)}{(2n)!}{x^{2n}}$$ ? – Sally G Jun 09 '14 at 10:26
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For the first equality I replaced $u$ by $x^2$ as I explained in my answer and for the second I used the general form of the power series: If $f$ is $\mathscr{C}^\infty$ then $$f(x)=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n$$ – Jun 09 '14 at 10:28
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You need a bit more than $C^\infty$ for that step; you need analyticity. – not all wrong Jun 09 '14 at 10:55
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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 $\dfrac{F''(0)}{2!}=1$.
mathse
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$f(x)=e^{x^2}\Rightarrow f'(x)=2xf(x)$ Now use Leibnitz's repeated differentiation $2n-1$ times to get $$f^{(2n)}(x)=2xf^{(2n-1)}(x)+2\binom{2n-1}{1}f^{(2n-2)}\\ \Rightarrow f^{(2n)}(0)=2(2n-1)f^{(2n-2)}(0)=\cdots=2^n(2n-1)(2n-3)\cdots 3\cdot 1\\ =2^n\frac{(2n)!}{2^n n!}=\frac{(2n)!}{n!}$$
Samrat Mukhopadhyay
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