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Assume we have the exponential-series $ \small \exp(x) = 1+ {x \over 1!} + {x^2 \over 2! } + \cdots = \sum\limits_{k=0 }^\infty {x^k \over k!} $ modified with a polynomial in the coefficients
say $ \qquad \displaystyle f_1(x)= \sum_{k=0}^\infty { k^2 + k \over 2} {x^k \over k!} $
or $ \qquad \displaystyle f_2(x)= \sum_{k=0}^\infty ( 15 k^3 + 15 k^2 - 10k - 8) {x^k \over k!} $
or in general
$ \qquad \displaystyle f_3(x)= \sum_{k=0}^\infty ( d k^3 + c k^2 + b k + a) {x^k \over k!} $

-: is there a good formula/algorithm/scheme how this has to be expressed as composition of the $\exp(x)$-function? (I know this can be solved using the derivatives and cancelling of k's in the polynomial with the factorials in the denominator - I'm asking for a handy/memorizable translation-formula )

Because I can factor my examples under study: is there possibly a special handy scheme, if the polynomials are given in a form like this
$ \qquad \displaystyle f_2(x)= \sum_{k=0}^\infty ( k-1)(k-2) {x^k \over k!} $ ?

J. M. ain't a mathematician
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Gottfried Helms
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4 Answers4

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\begin{eqnarray} \sum_{k=0}^\infty k\frac{x^k}{k!}&=&\sum_{k=1}^\infty \frac{x^k}{(k-1)!}=x\sum_{k=0}^\infty\frac{x^k}{k!}=xe^x;\cr \sum_{k=0}^\infty k^2\frac{x^k}{k!}&=&x\sum_{k=1}^\infty k\frac{x^{k-1}}{(k-1)!}=x\sum_{k=0}^\infty(k+1)\frac{x^k}{k!}=x(x+1)e^x\cr \sum_{k=0}^\infty k^3\frac{x^k}{k!}&=&\sum_{k=1}^\infty k^2\frac{x^k}{(k-1)!}=x\sum_{k=0}^\infty(k+1)^2\frac{x^k}{k!}\cr &=&x\sum_{k=0}^\infty(k^2+2k+1)\frac{x^k}{k!}=x[x(x+1)+2x+1]e^x\cr &=&x(x^2+3x+1)e^x. \end{eqnarray} Hence \begin{eqnarray} \sum_{k=0}^\infty (dk^3+ck^2+bk+a)\frac{x^k}{k!}&=&d\sum_{k=0}^\infty k^3\frac{x^k}{k!}+c\sum_{k=0}^\infty k^2\frac{x^k}{k!}+b\sum_{k=0}^\infty k\frac{x^k}{k!}+a\sum_{k=0}^\infty \frac{x^k}{k!}\cr &=&dx(x^2+3x+1)e^x+c(x^2+x)e^x+bxe^x+ae^x\cr &=&[a+bx+c(x^2+x)+d(x^3+3x^2+x)]e^x\cr &=&[a+(b+c+d)x+(c+3d)x^2+dx^3]e^x \end{eqnarray}

HorizonsMaths
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  • Very nice, thanks to all respondents. I'll accept this answer, because it starts a conversion-formula (...to be generalized...) for the coefficients for the polynomials $a+bk+ck^2+dk^3+...$ at the terms of the series into their equivalents at the exponential-function composition $A=a$,$B=(b+c+d)$ ... to become $(A+Bx+Cx^2+Dx^3+...)e^x)$ That's just what I'm after. – Gottfried Helms Aug 03 '12 at 08:19
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Dobiński's formula for the Bell polynomials is easily rearranged for your purposes:

$$\mathscr{B}_n(x)\exp\,x=\sum_{k=0}^\infty \frac{k^n x^k}{k!}$$

A convenient representation for the Bell polynomials is

$$\mathscr{B}_n(x)=\sum_{j=0}^n\left\{n\atop k\right\}x^k$$

where $\left\{n\atop k\right\}$ is a Stirling subset number (Stirling number of the second kind).

Thus, in principle, one can always express a series of the form

$$\sum_{k=0}^\infty p(k)\frac{x^k}{k!}$$

for a polynomial $p(k)$ in terms of a polynomial multiplied by $\exp\,x$.

J. M. ain't a mathematician
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$$P_3=k(k-1)(k-2)=k^3-3k^2+2k$$ $$P_2=k(k-1)=k^2-k$$ $$P_1=k$$

And $\sum P_i \frac{x^k}{k!}=x^ie^x$

$$\begin{array}{ccl} 15k^3+15k^2-10k-8&=&15P_3+60k^2-40k-8\\ &=&15P_3+60P_2+20k-8\\ &=&15P_3+60P_2+20P_1-8P_0\\ \end{array}$$

Hence $$\sum(15k^3+15k^2-10k-8)\frac{x^k}{k!}=(15x^3+60x^2+20x-8)e^x$$

Xoff
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  • Note that this can be derived by differentiating the power series expansion for $e^x$. – Alex Becker Aug 03 '12 at 07:34
  • @AlexBecker:True, that was part of my first attempt, but it was too complicated for paper&pen. I just missed a handy general translation recipe for the polynomial at the terms into that at the $e^x$. – Gottfried Helms Aug 03 '12 at 08:28
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Well, with the given answers of @mercy and the other respondents I found now a very simple transformation scheme, which fits my needs perfectly.

If we denote the coefficients of the polynomial at the series-terms as row-vector $S = [a,b,c,d,...z] $, the coefficients at the x for the composition of the exponential-series with the vector $T=[A,B,C,D,...,Z] $ and the matrix of the Striling numbers 2'nd kind as S2 where

$S2 = \small \begin{bmatrix} 1 & . & . & . & . & . & . & \cdots \\ 0 & 1 & . & . & . & . & . & \cdots \\ 0 & 1 & 1 & . & . & . & . & \cdots \\ 0 & 1 & 3 & 1 & . & . & . & \cdots \\ 0 & 1 & 7 & 6 & 1 & . & . & \cdots \\ 0 & 1 & 15 & 25 & 10 & 1 & . & \cdots \\ 0 & 1 & 31 & 90 & 65 & 15 & 1 & \cdots \\ \vdots &\vdots &\vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{bmatrix}$ of appropriate size,

then by $ S \cdot S2 = T $ we get the required coefficients for the composition in T.

This helps also to solve the first step in my earlier question, because the polynomials in x, which describe the compositions of the exponential-series have all the factor $(x-1)$ in them. Then, if I set x=1 it means, that all coefficients in the first column of the assumed "Null-matrix" are indeed zero.
My current main question asks this in a different way; but if I change order of summation accordingly then I have by this, that the formal power series for the function $f_1(x)$ has just that zeros in all coefficients at x, so that indeed the function $f_1(x)$ must be zero for all $x$. Cute...

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Gottfried Helms
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