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

Solution 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:

Solution 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 $f^{(n)}$.

Solution 3:

Let $y=e^{x^2}$. We want to find $\frac{d^n y}{dx^n}$.

Note that $\ln y = x^2$ and so $\frac{1}{y} \frac{dy}{dx} = 2x\iff \frac{dy}{dx}=2xy$.

Differentiating again yields:

$\frac{d^2y}{dx^2}=2x\frac{dy}{dx}+2y$,

$\frac{d^3 y}{dx^3} = 2x \frac{d^2 y}{dx^2} + 4\frac{dy}{dx}$,

$\frac{d^4 y}{dx^4} = 2x \frac{d^3 y}{dx^3} + 6\frac{d^2 y}{dx^2}$.

We might guess that $\frac{d^n y}{dx^n} = 2x \frac{d^{n-1} y}{dx^{n-1}}+2(n-1) \frac{d^{n-2} y}{dx^{n-2}}$, which is easily proved by straightforward induction (we leave this as an exercise to the reader. ;) )

Let $u_k=\frac{d^k y}{dx^k}$ for easier notation. We thus have established

$u_n = 2x u_{n-1}+2(n-1)u_{n-2}$.

It is easy to find that the coefficient of $x^n y$ in $u_n$ is $2^n$, but I'm not seeing any nice way of solving the recursion at the moment.

Solution 4:

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 $g_n(x)=f^{(n)}(x)/f(x)$ 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 $$g_n(x)=\sum_{m+2k=n} \frac{(2x)^m n!}{m!k!}=\sum_{k=0}^{\lfloor n/2\rfloor}\frac{2^{n-2k}n!}{(n-2k)!k!}x^{n-2k}$$

And $f^{(n)}(x)=f(x)g_n(x)$.