# $n$th derivative of $e^{1/x}$

I am trying to find the $n$'th derivative of $f(x)=e^{1/x}$. When looking at the first few derivatives I noticed a pattern and eventually found the following formula

$$\frac{\mathrm d^n}{\mathrm dx^n}f(x)=(-1)^n e^{1/x} \cdot \sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} x^{-2 n+k}$$

I tested it for the first $20$ derivatives and it got them all. Mathematica says that it is some hypergeometric distribution but I don't want to use that. Now I am trying to verify it by induction but my algebra is not good enough to do the induction step.

Here is what I tried for the induction (incomplete, maybe incorrect)

\begin{align*} \frac{\mathrm d^{n+1}}{\mathrm dx^{n+1}}f(x)&=\frac{\mathrm d}{\mathrm dx}(-1)^n e^{1/x} \cdot \sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} x^{-2 n+k}\\ &=(-1)^n e^{1/x} \cdot \left(\sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} (-2n+k) x^{-2 n+k-1}\right)-e^{1/x} \cdot \sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} x^{-2 (n+1)+k}\\ &=(-1)^n e^{1/x} \cdot \sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k}((-2n+k) x^{-2 n+k-1}-x^{-2 (n+1)+k)})\\ &=(-1)^{n+1} e^{1/x} \cdot \sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k}(2n x-k x+1) x^{-2 (n+1)+k} \end{align*}

I don't know how to get on from here.

• If your calculus was good enough to get the first few derivates, surely you can differentiate the RHS once wrt x? After that you don't need any calculus. Commented Jan 20, 2011 at 11:38
• @Peter The problem is merging the sums by shifting the index so it fits and handling the binomial coefficients, not differentiating. Commented Jan 20, 2011 at 11:40
• Sorry I should have said that my algebra was not good enough instead of calculus. I updated the question to show how far I get. Commented Jan 20, 2011 at 12:08
• Don't merge your two sums so early. Instead do a variable substitution in one to get a sum from $k^\prime = 1$ to $n$, then expand both sums to the range $0$ to $n$ using $\binom{n}{-1} = \binom{n}{n+1} = 0$. Expand in terms of factorials if necessary. Commented Jan 20, 2011 at 12:34
• mathcs.pugetsound.edu/~mspivey/Exp.pdf Commented Jul 10, 2013 at 6:19

How's this?

$$\left(\sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} \left(-2n+k\right) x^{-2 n+k-1}\right) - \left(\sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} x^{-2 \left(n+1\right)+k}\right) =$$

$$= \left(\sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} \left(-2n+k\right) x^{-2\left(n+1\right)+k+1}\right) - \left(\sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} x^{-2 \left(n+1\right)+k}\right) =$$

$$= \left(\sum _{k'=1}^{n} \left(k'-1\right)! \binom{n}{k'-1} \binom{n-1}{k'-1} \left(-2n+k'-1\right) x^{-2\left(n+1\right)+k'}\right) - \left(\sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} x^{-2 \left(n+1\right)+k}\right) =$$

$$= \left(\sum _{k'=0}^{n} \left(k'-1\right)! \binom{n}{k'-1} \binom{n-1}{k'-1} \left(-2n+k'-1\right) x^{-2\left(n+1\right)+k'}\right) - \left(\sum _{k=0}^{n} k! \binom{n}{k} \binom{n-1}{k} x^{-2 \left(n+1\right)+k}\right) =$$

$$= \sum _{k=0}^{n} \left(\left(k-1\right)! \binom{n}{k-1} \binom{n-1}{k-1} \left(-2n+k-1\right) - k! \binom{n}{k} \binom{n-1}{k}\right) x^{-2 \left(n+1\right)+k}$$

Then $$\left(k-1\right)! \binom{n}{k-1} \binom{n-1}{k-1} \left(-2n+k-1\right) - k! \binom{n}{k} \binom{n-1}{k} =$$

$$= \frac{\left(k-1\right)!n!\left(n-1\right)!\left(-2n+k-1\right)}{\left(n-k+1\right)!\left(k-1\right)!\left(n-k\right)!\left(k-1\right)!} - \frac{k!n!\left(n-1\right)!}{\left(n-k\right)!k!k!\left(n-k-1\right)!} =$$

$$= \frac{n!\left(n-1\right)!\left(-2n+k-1\right)k}{\left(n-k+1\right)!\left(n-k\right)!k!} - \frac{n!\left(n-1\right)!\left(n-k\right)\left(n-k+1\right)}{\left(n-k\right)!k!\left(n-k+1\right)!} =$$

$$= \frac{n!\left(n-1\right)!}{\left(n-k+1\right)!\left(n-k\right)!k!} \left(\left(-2n+k-1\right)k - \left(n-k\right)\left(n-k+1\right)\right) =$$

$$= \frac{-n\left(n+1\right)n!\left(n-1\right)!}{\left(n-k+1\right)!\left(n-k\right)!k!} =$$

$$= -\frac{\left(n+1\right)!}{\left(n-k+1\right)!} \binom{n}{k} =$$

$$= -k! \binom{n+1}{k} \binom{n}{k}$$

• I just saw something nasty in the calculation. In line 5 the following expression occurs: $\sum _{k=0}^{n} ((k-1)!\ldots$ but $(-1)!$ is not defined. I think the result is still correct but this is odd Commented Jan 21, 2011 at 20:35
• Yes, I should have been more explicit there. It's actually $(k-1)!\binom{n}{k-1}\binom{n-1}{k-1}$ so there's one $(k-1)!$ in the numerator and two in the denominator. If you take limits you can prove that the term as a whole converges to zero. Commented Jan 21, 2011 at 20:44
• @Listing One approach is to separate out the first term and handle it separately. In the exercise I was working through, it balances out in the end. He doesn't show it either, but my paper and pencil checks. math.stackexchange.com/questions/2111260/… Commented Mar 26 at 9:46

Peter's and Mike's answers have clearly settled this question; I'll just explain the OP's mention of "Mathematica says that it is some hypergeometric distribution". More specifically, one wonders how Mathematica might have arrived at the Kummer confluent hypergeometric function ${}_1 F_1\left({{a}\atop{b}}\mid x\right)$.

We start with the transformed Maclaurin series:

$$\exp\left(\frac1{x}\right)=1+\sum_{k=1}^\infty \frac1{x^k k!}$$

and then differentiate the series termwise, using the formula

$$\frac{\mathrm d^n}{\mathrm dx^n}x^{-k}=\frac{(-1)^n (n+k-1)!}{(k-1)! x^{k+n}}$$

to yield

$$\frac{\mathrm d^n}{\mathrm dx^n}\exp\left(\frac1{x}\right)=\frac{(-1)^n}{x^n}\sum_{k=1}^\infty \frac{(n+k-1)!}{k!(k-1)! x^k}$$

which can be reindexed like so:

$$\frac{\mathrm d^n}{\mathrm dx^n}\exp\left(\frac1{x}\right)=\frac{(-1)^n}{x^n}\sum_{k=0}^\infty \frac{(n+k)!}{(k+1)!k! x^{k+1}}$$

and rewritten in terms of Pochhammer symbols to yield

$$\frac{\mathrm d^n}{\mathrm dx^n}\exp\left(\frac1{x}\right)=\frac{(-1)^n}{x^n}\sum_{k=0}^\infty \frac{n!(n+1)_k}{(2)_k k! x^{k+1}}$$

or, expressed hypergeometrically,

$$\frac{\mathrm d^n}{\mathrm dx^n}\exp\left(\frac1{x}\right)=\frac{(-1)^n n!}{x^{n+1}}{}_1 F_1\left({{n+1}\atop{2}}\mid \frac1{x}\right)$$

It's not immediately obvious that this one can be expressed in terms of $\exp(1/x)$ multiplied by a finite sum, but we can then apply the Kummer transformation to the hypergeometric function like so:

$$\frac{\mathrm d^n}{\mathrm dx^n}\exp\left(\frac1{x}\right)=\frac{(-1)^n n!}{x^{n+1}}\exp\left(\frac1{x}\right){}_1 F_1\left({{1-n}\atop{2}}\mid -\frac1{x}\right)$$

and find that the hypergeometric function has a nonpositive numerator parameter $1-n$ for $n > 0$, which means it can be expressed as a terminating series (since the Pochhammer symbol $(1-n)_k$ is $0$ for $k \geq n$):

$$\frac{\mathrm d^n}{\mathrm dx^n}\exp\left(\frac1{x}\right)=\frac{(-1)^n n!}{x^{n+1}}\exp\left(\frac1{x}\right)\sum_{k=0}^{n-1}\frac{(-1)^k (1-n)_k}{k!(k+1)!x^k}$$

Rewriting the Pochhammer symbol in terms of factorials, we get

$$\frac{\mathrm d^n}{\mathrm dx^n}\exp\left(\frac1{x}\right)=\frac{(-1)^n n!}{x^{n+1}}\exp\left(\frac1{x}\right)\sum_{k=0}^{n-1}\frac{(n-1)!}{(n-k-1)!k!(k+1)!x^k}$$

which is equivalent to the formula you obtained.

My point here, I guess, is that hypergeometric functions are nothing to be uncomfortable with; if it helps, just consider them as shorthand for certain (in)finite sums.

• Of course, the good thing about having the hypergeometric function expression is that you can then consider derivatives of $\exp(1/x)$ for positive noninteger $n$... $n=1/2$ for instance (the semiderivative) yields an expression in terms of Bessel functions. Commented May 3, 2011 at 12:17

You can get the same result using the Faà di Bruno formula and Bell polynomials.

The Faà di Bruno formula says that $$\frac{d^n}{dx^n} f(g(x)) = \sum_{k=0}^n f^{(k)}(g(x)) B_{n,k}\left(g'(x),g''(x),\dots,g^{(n-k+1)}(x)\right),$$

where $$B_{n,k}(x_1,x_2,\dots,x_{n-k+1})$$ is a partial Bell polynomial.

Since $$f(x) = e^x$$, and $$g(x) = 1/x$$, we have $$\frac{d^n}{dx^n} e^{1/x} = \sum_{k=0}^n e^{1/x} B_{n,k}\left(-x^{-2},2x^{-3},\dots,(-1)^{n-k+1}(n-k+1)!x^{-n+k-2}\right).$$

It is known that $$B_{n,k}(1!,2!,\dots,(n-k+1)!) = \binom{n}{k}\binom{n-1}{k-1} (n-k)!$$, a Lah number (see Wikipedia on Bell polynomials or Comtet's Advanced Combinatorics, p. 135). Thus (keeping track of signs and powers of $$x$$), $$\frac{d^n}{dx^n} e^{1/x} = (-1)^n e^{1/x} \sum_{k=0}^n \binom{n}{k}\binom{n-1}{k-1} (n-k)! x^{-n-k},$$ which is just your formula with an index change (and where $$\binom{n-1}{-1} = 0$$).

Here's a completely elementary argument that I like better than my previous answer that uses the Faà di Bruno formula. All that is required are the Maclaurin series for $e^x$ and the definition of the Lah numbers $L(n,k)$ as the coefficients arising when converting rising factorial powers to falling factorial powers via $$x^{\overline{n}} = \sum_{k=1}^n L(n,k) x^{\underline{k}}.$$

First, substitute $1/x$ for $x$ in the Maclaurin series for $e^x$ and differentiate $n$ times. This yields $$\frac{d^n}{dx^n} e^{1/x} = (-1)^n x^{-n} \sum_{j=0}^{\infty} \frac{j^{\overline{n}} x^{-j}}{j!}.$$

Then convert rising powers to falling powers to obtain
$$\frac{d^n}{dx^n} e^{1/x} = (-1)^n x^{-n} \sum_{j=0}^{\infty} \sum_{k=1}^n L(n,k) \frac{j^{\underline{k}} x^{-j}}{j!}.$$

Working with the summation in $j$, we have $$\sum_{j=0}^{\infty} \frac{j^{\underline{k}} x^{-j}}{j!} = \sum_{j=k}^{\infty} \frac{j^{\underline{k}} x^{-j}}{j!} = \sum_{j=k}^{\infty} \frac{j! x^{-j}}{(j-k)! j!} = \sum_{i=0}^{\infty} \frac{x^{-i-k}}{i!} = x^{-k} e^{1/x},$$ where the last step uses, once again, the Maclaurin series for $e^x$.

Thus we have $$\frac{d^n}{dx^n} e^{1/x} = (-1)^n e^{1/x} \sum_{k=1}^n L(n,k) x^{-n-k},$$ which is just your formula with an index change, as $$L(n,k) = \binom{n}{k}\binom{n-1}{k-1} (n-k)!.$$

(From another perspective, this is computing the rising factorial moments of a Poisson $(1/x)$ distribution. With this in mind, converting to falling factorial powers seems reasonable, as the falling factorial moments of a Poisson distribution have a known simple expression. See also this related answer to a previous math.SE question.)

We may obtain a recursive formula, as follows: \begin{align} f\left( t \right) &= e^{1/t} \\ f'\left( t \right) &= - \frac{1}{{t^2 }}f\left( t \right) \\ f''\left( t \right) &= - \frac{1}{{t^2 }}f'\left( t \right) + f\left( t \right)\frac{2}{{t^3 }} \\ &= - \frac{1}{{t^2 }}\left\{ {\frac{1}{{t^2 }}f\left( t \right)} \right\} + f\left( t \right)\frac{2}{{t^3 }} \\ &= \left( {\frac{1}{{t^4 }} + \frac{2}{{t^3 }}} \right)f\left( t \right) \\ \ldots \end{align} Inductively, let us assume $f^{(n-1)}(t)=P_{n-1}(\frac{1}{t})f(t)$ is true, for some polynomial $P_{n-1}$. Now, for $n$, we have \begin{align} f^{\left( n \right)} \left( t \right) &= - \frac{1}{{t^2 }}P'_{n - 1} \left( {\frac{1}{t}} \right)f\left( t \right) + P_{n - 1} \left( {\frac{1}{t}} \right)f'\left( t \right) \\ &= - \frac{1}{{t^2 }}P'_{n - 1} \left( {\frac{1}{t}} \right)f\left( t \right) + P_{n - 1} \left( {\frac{1}{t}} \right)\left\{ { - \frac{1}{{t^2 }}f\left( t \right)} \right\} \\ &= - \frac{1}{{t^2 }}\left\{ {P'_{n - 1} \left( {\frac{1}{t}} \right) + P_{n - 1} \left( {\frac{1}{t}} \right)} \right\}f\left( t \right) \\ &= P_n \left( {\frac{1}{t}} \right)f\left( t \right) \end{align} Thus, \begin{align} f^{\left( n \right)} \left( t \right) = P_n \left( {\frac{1}{t}} \right)f\left( t \right) \end{align} where, $P_n \left( x \right): = x^2 \left[ {P'_{n - 1} \left( x \right) - P_{n - 1} \left( x \right)} \right]$, $P_0=1$.

For $$i\in\mathbb{N}$$ and $$t\ne0$$, we have $$$$\label{exp-frac1x-expans}\tag{1} \bigl(\textrm{e}^{1/t}\bigr)^{(i)}=\textrm{e}^{1/t}\frac{(-1)^i}{t^{2i}}\sum_{k=0}^{i-1}\binom{i}{k}\binom{i-1}{k}{k!}t^{k}.$$$$ To the best of my knowledge, an inductive proof of the derivative formula \eqref{exp-frac1x-expans} was published on pages 123--124, Section 2.1, Theorem 2.1 of the paper [1] below. Hereafter, there have been a number of literature, for example, the papers [2, 3] below and those collected at a site on wordpress, dedicated to the investigation of the function $$\textrm{e}^{1/t}$$, the derivative formula \eqref{exp-frac1x-expans}, and their applications to several areas in mathematics.

References

1. Xiao-Jing Zhang, Feng Qi, and Wen-Hui Li, Properties of three functions relating to the exponential function and the existence of partitions of unity, International Journal of Open Problems in Computer Science and Mathematics 5 (2012), no. 3, 122--127; available online at https://doi.org/10.12816/0006128.
2. Siad Daboul, Jan Mangaldan, Michael Z. Spivey, and Peter J. Taylor, The Lah numbers and the $$n$$th derivative of $$e^{1/x}$$, Mathematics Magazine 86 (2013), no. 1, 39–47; available online at https://doi.org/10.4169/math.mag.86.1.039.
3. Khristo N. Boyadzhiev, Lah numbers, Laguerre polynomials of order negative one, and the $$n$$th derivative of $$\exp(1/x)$$, Acta Universitatis Sapientiae Mathematica 8 (2016), no. 1, 22–31; available online at https://doi.org/10.1515/ausm-2016-0002.
• mathoverflow.net/q/405015/147732 Commented Sep 28, 2021 at 13:26
• Your second reference was actually inspired by this Math.SE question and its first few answers. Compare the authors in the second reference with the authors of the first few answers. Commented Mar 9, 2023 at 0:02