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Given the recursion relationship $n_{i+1}=(e^{-x})\frac{dn_{i}}{dx}$ for $i=0,1,2,\ldots,$ where $n_{0}$ is a known function of $x$, what is the solution to $n_{k}$ in terms of $x$ and the derivatives of $n_{0}$, where $k$ is any positive integer?

Re what I've tried, things simplify a bit if I express $n_0$ as $n_0=e^{y}$ and look for the solution in terms of the derivatives of $y$ (which may be expressed in terms of $x$, $n_0$, and the derivatives of $n_0$ as required by the question). We have then,

$n_1=e^{y-x}\frac{dy}{dx}$

$n_{2}=e^{y-2x}\left( \frac{d^{2}y}{dx^{2}}+\left( \frac{dy}{dx}\right)^{2}-\frac{dy}{dx}\right)$

$n_{3}=e^{y-3x}\left( \frac{d^{3}y}{dx^{3}}+3\frac{dy}{dx}\frac{d^{2}% y}{dx^{2}}+\left( \frac{dy}{dx}\right) ^{3}-3\left( \frac{dy}{dx}\right) ^{2}-3\frac{d^{2}y}{dx^{2}}+2\frac{dy}{dx}\right) $

This gives a feel for where things are headed. The solution comes down to figuring out the integer coefficients of all the resultant derivatives and products of derivatives.

Edward L. Ruden
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  • Hey! Could you elaborate what you tried? Also don't post the question in the title, it gets a bit cramped – Lourenco Entrudo Nov 19 '22 at 22:47
  • Hint: find $f(x)$ so $e^{-x}\frac{d}{dx}=\frac{d}{df}$, then write $n_0$ as a function of $f$. – J.G. Nov 19 '22 at 22:47
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    "Also don't post the question in the title, it gets a bit cramped". I disagree: the title should be as informative and specific to the problem as possible. – Adam Rubinson Nov 19 '22 at 23:01
  • @Lourenco Entrudo - Re what I've tried. Things simplify a bit if I express $n_0$ as $n_0=e^{y}$ and look for the solution in terms of the derivatives of $y$ (which may be expressed in terms of $x$, $n_0$, and the derivatives of $n_0$ as required by the question). We have then, $n_1=e^{y-x}\frac{dy}{dx}$ and $n_{2}=e^{y-2x}\left( \frac{d^{2}y}{dx^{2}}+\left( \frac{dy}{dx}\right)^{2}-\frac{dy}{dx}\right)$. This gives a feel for where things are headed. The solution comes down to figuring out the integer coefficients of all the resultant combinations of derivatives and powers of derivatives. – Edward L. Ruden Nov 20 '22 at 00:14
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    Edward: Please edit your post to include the information in your comment above. – amWhy Nov 20 '22 at 00:41
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    @amWhy, done. I am new to this forum and was not familiar with its expectation of showing such attempts. I did not do so initially since sometimes providing ones own ideas for pursuing a solution that didn't work out steer people in the wrong direction. – Edward L. Ruden Nov 20 '22 at 17:28
  • Have you tried using generating functions, i.e, looking at $U(x,t)=\sum u_k(x)t^k$? I think this delivers the perhaps not helpful $u_k(x)=(e^{-x}D)^k u_0(x)$ (where $D$ is the differentiation wrt $x$ operator). [Oh sorry, I've written $u$ for $n$, I can't do series where $n$ is a function!] – ancient mathematician Nov 21 '22 at 07:52
  • Using the Laplace transform, the recurrence has a clean representation. – Cesareo Nov 21 '22 at 10:15

1 Answers1

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I will solve the slightly more general recurrence problem

$$n_k=e^{\lambda x}\frac{d}{dx}n_{k-1}$$

Denote $f(x):=e^{\lambda x}, D:=d/dx$ for brevity. Clearly, $Df=\lambda f$ for exponentials are eigenfunctions of the derivative operator. By induction, it can be easily shown that the solution to the recurrence takes the form

$$n_k(x)=f^k(x)\lambda^k~ \left[\Omega_k \left(\frac{D}{\lambda}\right)\right]n_0(x)$$

where $\Omega_k$ are polynomials of degree $k$. The quantity in brackets is a finite sum of powers of the derivative operator, and hence well defined. To generate a recurrence for these polynomials, act with $fD$ on both sides of the equation and reuse it for the $k+1$-th term of the recurrence to obtain the very simple formula

$$\Omega_{k+1}(x)=(x+k)\Omega_k(x)$$

which is solved by the family of polynomials given by the rising factorial:

$$\Omega_k(x)=x^{(k)}=\prod_{m=1}^{k-1}(x+m)=\sum_{m=0}^k(-1)^{k-m}s(k,m)x^m$$

where $s(k,m)$ denote the Stirling numbers of the first kind. The coefficients in your expansion are proportional to them. More precisely, for $\lambda=-1$, it is seen that

$$n_k(x)=e^{-kx}\sum_{m=0}^k s(k,m)\frac{d^m}{dx^m}n_0(x)$$

and this concludes the analysis.

DinosaurEgg
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  • Thanks, @DinosaurEgg. There is an error, though. Checking the first three solutions using the recursion:

    $n_{2}=e^{-2x}\left( -\frac{d}{dx}n_{0}+\frac{d^{2}}{dx^{2}}n_{0}\right)$

    $n_{3}=e^{-3x}\left( 2\frac{d}{dx}n_{0}-3\frac{d^{2}}{dx^{2}}n_{0}+\frac{d^{3}}{dx^{3}}n_{0}\right)$

    Your solution is missing the negative signs. Should there be a $(-1)^{k-m}$ in there?

     – Edward L. Ruden Nov 22 '22 at 00:18
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    I'm following the convention where the stirling numbers of the first kind are signed (they are generated by the falling factorial). – DinosaurEgg Nov 22 '22 at 00:39
  • Thank you for the clarification. I've posted a follow-up question regarding what turns out to be the second half of my problem: https://math.stackexchange.com/questions/4582268 . It turns out that my assumption that expressing $y$ derivatives in terms of $n_0$ derivatives where $n_0=e^y$ is not as trivial as I assumed. I actually need the converse now. – Edward L. Ruden Nov 22 '22 at 02:25