Finding a closed formula for calculating $\frac{d^n}{{dx}^n}f\left(x\right)=\frac{d^n}{{dx}^n}e^{x^2}$

Question

For reasons I can't even remember, the other day I wanted to find out if there was a closed formula for calculating the $n$-th derivative $\frac{d^n}{{dx}^n}f\left(x\right)=\frac{d^n}{{dx}^n}e^{x^2}$ for the function $f\left(x\right)=e^{x^2}$. Where I ended up after some trial and error is the formula $$\frac{d^n}{{dx}^n}e^{x^2}=c_n\left(\sum_{0 \leq i \leq \lfloor\frac{n}{2}\rfloor} {p_i x^{n-2i}}\right)e^{x^2}=c_n\left(c_{n-1}x^n + \sum_{1 \leq i \leq \lfloor\frac{n}{2}\rfloor} {p_i x^{n-2i}}\right)e^{x^2},$$ with $c_n=2^{n-\lfloor\frac{n}{2}\rfloor}$.

The $p_{i \geq1}$ turn out to be as follows:

$n$	$p_{i=1}$	$p_{i=2}$	$p_{i=3}$	$p_{i=4}$	$…$
0	–	–	–	–	$…$
1	–	–	–	–	$…$
2	1	–	–	–	$…$
3	3	–	–	–	$…$
4	12	3	–	–	$…$
5	20	15	–	–	$…$
6	60	90	15	–	$…$
7	84	210	105	–	$…$
8	224	840	840	105	$…$
9	288	1512	2520	945	$…$
$…$	$…$	$…$	$…$	$…$	$…$

I have not yet understood the rule behind the $p_i$ sequences $$p_{i=1}:\left(1,3,12,20,60,84,224,288,…\right),$$

$$p_{i=2}:\left(3,15,90,210,840,1512,…\right),$$

$$p_{i=3}:\left(15,105,840,2520,…\right),$$

$$p_{i=4}:\left(105,945,…\right),$$

$$…$$ I suspect it has something to do with binomial coefficients, since the coefficients $p_i$ arise from multiplying binomials during the derivation. One regularity I've noticed so far is that starting at $p_{i=2}$, the first values always correspond to the second ones of the previous $p$ sequence.

Do any of you have an idea how I can formalize the coefficients $p_i$ and the timing of their occurrence and integrate them into the above closed formula? Or do you know if there even already exists a known solution to the problem, namely finding a closed formula to calculate the $n$-th derivative $\frac{d^n}{{dx}^n}f\left(x\right)=\frac{d^n}{{dx}^n}e^{x^2}$?

Thank you and best regards!

Answer 1

Hint: Using Wolfram Alpha we find for small values of $n$ \begin{align*} \frac{d}{dx}e^{x^2}&=2x^2e^{x^{2}}\\ \frac{d^2}{dx^2}e^{x^2}&=2\left(2x^2+1\right)e^{x^{2}}\\ \frac{d^3}{dx^3}e^{x^2}&=4\left(2x^3+3x\right)e^{x^{2}}\tag{1}\\ \frac{d^4}{dx^4}e^{x^2}&=4\left(4x^4+12x^2+3\right)e^{x^{2}}\\ \end{align*}

Another expression besides Faa di Brunos formula is stated as identity (3.56) in H.W. Gould's Tables of Combinatorial Identities, Vol. I and called:

Hoppe Form of Generalized Chain Rule

Let $D_x$ represent differentiation with respect to $x$ and $y=y(x)$. Hence $D^n_x g(y)$ is the $n$-th derivative of $g$ with respect to $x$. The following holds true \begin{align*} D_x^n g(y)=\sum_{k=0}^nD_y^kg(y)\frac{(-1)^k}{k!}\sum_{j=0}^k(-1)^j\binom{k}{j}y^{k-j}D_x^ny^j \end{align*}

In the special case \begin{align*} g(y(x))=e^{y(x)}=e^{x^{2}} \end{align*} we have $$D_y^kg(y)=D_y^k e^y=e^y$$ and obtain \begin{align*} D_x^ne^y=e^y\sum_{k=0}^n\frac{(-1)^k}{k!}\sum_{j=0}^k(-1)^j\binom{k}{j}y^{k-j}D_x^ny^j\tag{2} \end{align*}

With $y=y(x)=x^2$ we obtain from (2) \begin{align*} \color{blue}{D_x^ne^{x^2}}&=e^{x^2}\sum_{k=0}^n\frac{(-1)^k}{k!}\sum_{j=0}^k(-1)^j\binom{k}{j}x^{2(k-j)}D_x^nx^{2j}\tag{3.1}\\ &=e^{x^2}\sum_{k=\left\lfloor\frac{n+1}{2}\right\rfloor}^n\frac{(-1)^k}{k!}\sum_{j=\left\lfloor\frac{n+1}{2}\right\rfloor}^k(-1)^j\binom{k}{j}x^{2(k-j)}D_x^nx^{2j}\tag{3.2}\\ &\,\,\color{blue}{=e^{x^2}\sum_{k=\left\lfloor\frac{n+1}{2}\right\rfloor}^n\frac{(-1)^k}{k!}\sum_{j=\left\lfloor\frac{n+1}{2}\right\rfloor}^k(-1)^j\binom{k}{j}(2j)^{\underline{n}}x^{2k-n}}\tag{3.3}\\ \end{align*}

Comment:

• In (3.1) we apply Hoppe's formula with $y=y(x)=x^2$.

• In (3.2) we observe that we differentiate $x^{2j}$ $n$ times. This implies that terms with indices $k,j< \left\lfloor\frac{n+1}{2}\right\rfloor$ vanish.

• In (3.3) we calculate $D_x^nx^{2j}$ using the falling factorial notation $(2j)^{\underline{n}}=(2j)(2j-1)\cdots(2j-n+1)$.

Let's look at a small example in order to see formula (3.3) in action.

Example: $n=3$ We obtain \begin{align*} e^{x^2}&\sum_{k=2}^{3}\frac{(-1)^k}{k!}\sum_{j=2}^k(-1)^j\binom{k}{j}(2j)(2j-1)(2j-2)x^{2k-3}\\ &=4e^{x^2}\sum_{k=2}^{3}\frac{(-1)^k}{k!}\sum_{j=2}^k(-1)^j\binom{k}{j}j(j-1)(2j-1)x^{2k-3}\\ &=4e^{x^2}\left(\frac{1}{2}\left(\sum_{j=2}^2(-1)^j\binom{2}{j}j(j-1)(2j-1)\right)\right)x\\ &\qquad+4e^{x^2}\left(\frac{(-1)}{6}\left(\sum_{j=2}^3(-1)^j\binom{3}{j}j(j-1)(2j-1)\right)\right)x^3\\ &=4e^{x^2}\left(\frac{1}{2}\binom{2}{2}\cdot 6\right)x +4e^{x^2}\left(-\frac{1}{6}\binom{3}{2}\cdot 6+\frac{1}{6}\binom{3}{3}\cdot 30\right)x^3\\ &\,\,\color{blue}{=4e^{x^2}\left(2x^3+3x\right)} \end{align*} in accordance with (1).