Find an expression for the $n$-th derivative of $f(x)=e^{x^2}$

Question

I need to find an expression for $n$th derivative of $f(x) = e^{x^2}$. Really need help.

Answer 1

Here is a less elementary solution.

We consider the exponential generating function

$$G(t,x) = \sum_{n=0}^{\infty} \frac{t^n}{n!} \frac{d^n}{dx^n}e^{x^2}.$$

Then we can identify this series as the McLaurin series of $e^{(x+t)^2}$ near $t = 0$. Thus we must have

$$G(t,x) = e^{(x+t)^2} = e^{x^2} e^{2xt} e^{t^2} = e^{x^2}\left( \sum_{k=0}^{\infty} \frac{(2x)^k}{k!} t^k \right)\left( \sum_{l=0}^{\infty} \frac{1}{l!} t^{2l} \right). $$

Expanding this series and comparing, we have

$$ \frac{d^n}{dx^n} e^{x^2} = \left( \sum_{j=0}^{\lfloor n/2 \rfloor} \frac{n!}{j!(n-2j)!}(2x)^{n-2j} \right) e^{x^2}. $$

Here is an example:

Answer 2

I'm not sure if this is what you are after, but there is a recursion that you can establish. Each derivative will be the product of a polynomial in $x$ with $e^{x^2}$. If $f^{(n)}(x)=p_n(x)e^{x^2}$, then we have $$p_0(x)=1$$ and $$p_n(x)=2x\,p_{n-1}(x)+p^\prime_{n-1}(x)$$

This recursion has the following associated infinite matrix:

$$M=\begin{bmatrix} 0 & 1 & 0 & 0 & \cdots\\ 2 & 0 & 2 & 0 & \cdots\\ 0 & 2 & 0 & 3 & \cdots\\ 0 & 0 & 2 & 0 & \cdots\\ \vdots & \vdots & \vdots & \vdots & \ddots\\ \end{bmatrix}$$

viewed as a transformation on the vector space of polynomials in $x$ with basis $\left\{1,x,x^2,x^3,\ldots\right\}$. The first column of $M^n$ gives you the $n$th derivative of $f$. If there is any hope to solve the recursion explicitly (which I'm not optimisitic for) then I'd recommend further study of this matrix. Maybe it can be diagonalized, and that would give way to a formula for $f^{(n)}$.

Answer 3

Another approach is to write:$$f(x+y)=\sum_{n=0}^\infty \frac{f^{(n)}(x)}{n!} y^n$$

Now $f(x+y)=f(x)f(y)e^{2xy}$. So if we let $g_n(x)=f^{(n)}(x)/f(x)$m then we see that:

$$f(y)e^{2xy} = \sum_{n=0}^\infty \frac{g_n(x)}{n!}y^n$$

But $$f(y)=\sum_{k=0}^\infty \frac{1}{k!} y^{2k}$$ and $$e^{2xy} = \sum_{m=0}^\infty \frac{(2x)^m}{m!}y^m$$

So:

$$f(y)e^{2xy} = \sum_{n=0}^\infty \frac{y^n}{n!} \sum_{m+2k=n} \frac{(2x)^m n!}{m!k!}$$

which gives us $$g_n(x)=\sum_{m+2k=n} \frac{(2x)^m n!}{m!k!}$$

Answer 4

A related problem 1, related 2. I already posted a solution on this website related to this one.

Here is the first formula, which gives only the $n$th derivative of integer order of ${\rm e}^{x^2}$

$$ {{\rm e}^{{x}^{2}}}\sum _{s=0}^{n} \left( {x}^{2\,s-n}\sum _{k=0 }^{n}{2}^{k+s} \left[\matrix{n\\k+s}\right] \left\{\matrix{k+s\\s}\right\} \right)\,, $$

where $\left[\matrix{n\\k+s}\right]$ and $\left\{\matrix{k+s\\s}\right\}$ are the Stirling numbers of the first kind and the second kind respectively.

The second formula is more general. It is a unified formula for the $n$th derivative and the $n$th anti derivative of real orders (including integer orders) of ${\rm e}^{x^2}$ in terms of the Meijer $G$-function,

$$ \left( -1 \right)^{\frac{n}{2}}{2}^{n} G^{1, 2}_{2, 3}\left(-{x}^{2}\, \Big\vert\,^{-\frac{n}{2}, -\frac{n}{2}+\frac{1}{2}}_{-\frac{n}{2}, \frac{1}{2}, 0}\right)\,.$$

Note that,

(i) if $n > 0$, then the formula gives derivatives of order $n$ ($n$ can be integer or real ).

(ii) if $ n<0 $, then the formula gives anti-derivatives of order $n$ ($n$ can be integer or real).

(iii) if $n=0$, then it gives the original function.

See here for the details.

Answer 5

The derivatives at $0$ may be found simply using the Taylor expansion: $$e^x = \sum_{k=0}^\infty {\frac{x^k}{k!}}$$ So that: $$e^{x^2} = \sum_{k=0}^\infty {\frac{x^{2k}}{k!}} = \sum_{k=0}^\infty {\frac{x^{k}}{k!}f^{(k)}(0)}$$ Leading to: $$f^{(2n)}(0) = {(2n)! \over n!} , \ f^{(2n+1)}(0) = 0$$ This means that for $f^{(n)}(x) = p_n(x) e^{x^2} = (\sum a^{(n)}_k x^k) e^{x^2}$:

$$a^{(2n)}_0 = {(2n)! \over n!}, \ \ a^{(2n + 1)}_0 = 0$$ And using alex's recursion: $$a^{(2n - 1)}_1 = {(2n)! \over n!}$$