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Consider the following sum:

$$ \sum_{i =0}^{j} \left( \frac{(j-i)^ix^i \ln(x)^{(j-i)}\ln(x)^i}{(j-i)!i!} \right) $$

I can simplify the sum to:

$$ \ln(x)^j\sum_{i =0}^{j} \left( \frac{(j-i)^ix^i}{(j-i)!i!} \right) $$ Furthermore I can observe that

$$ \frac{1}{(j-i)!(i!)} = \frac{1}{j!} \begin{pmatrix} j \\ i\end{pmatrix} $$

Thus:

$$ \frac{\ln(x)^j}{j!}\sum_{i =0}^{j} \left( \begin{pmatrix}j \\ i \end{pmatrix}(j-i)^ix^i \right) $$

But I don't know how to go in for the kill.

Sidharth Ghoshal
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3 Answers3

5

Note: This answer does not provide a closed expression but a generating function which might also be helpful for further calculations.

Let's consider OPs sum by exchanging for (my) convenience $i,j$ with $n,k$ and ignoring the factor $\ln(x)^n$.

\begin{align*} \frac{1}{n!}&\sum_{k=0}^n\binom{n}{k}(n-k)^kx^k\tag{1}\\ &=\frac{1}{n!}\sum_{k=0}^n\binom{n}{k}k^{n-k}x^{n-k} \end{align*}

Let $A(z)$ denote the generating function of OPs expression (1). We show

The following is valid:

\begin{align*} A(z) &:= \sum_{n=0}^{\infty}\left(\frac{1}{n!}\sum_{k=0}^n\binom{n}{k}k^{n-k}x^{n-k}\right)z^n\tag{2}\\ &=\exp\left(ze^{xz}\right) \end{align*}

Intermezzo: Bell polynomials

According to Louis Comtet's Advanced Combinatorics section 3.3 ([3a']) the partial Bell polynomials $B_{n,k}=B_{n,k}(x_1,x_2,\ldots,x_{n-k+1})$ are defined via

\begin{align*} B_{n,k}=\frac{n!}{k!}[t^n]\left(\sum_{m\geq 1}x_m\frac{t^m}{m!}\right)^k \end{align*}

We use the coefficient of operator $[t^n]$ to denote the coefficient of $t^n$ in the formal power series.

The specific case $B_{n,k}=B_{n,k}(1,2,\ldots,n-k+1)$ yields:

\begin{align*} B_{n,k}&=\frac{n!}{k!}[t^n]\left(\sum_{m\geq 1}m\frac{t^m}{m!}\right)^k\\ &=\frac{n!}{k!}[t^n]\left(t\sum_{m\geq 0}\frac{t^{m}}{m!}\right)^k\\ &=\frac{n!}{k!}[t^{n}]t^k\exp(kt)\tag{3}\\ &=\frac{n!}{k!}[t^{n-k}]\sum_{m\geq 0}\frac{(kt)^{m}}{m!}\\ &=\frac{n!}{k!}\frac{k^{n-k}}{(n-k)!}\\ &=\binom{n}{k}k^{n-k}\tag{4} \end{align*}

We observe the partial Bell polynomials $B_{n,k}(1,2,\ldots,n-k+1)=\binom{n}{k}k^{n-k}$ are the link between OPs expression and the path to finding the generating function $A(z)$.

Using the expression (3) in (1) we obtain

\begin{align*} \frac{1}{n!}\sum_{k=0}^n&\binom{n}{k}k^{n-k}x^{n-k}\\ &=\sum_{k=0}^n\frac{1}{k!}[t^{n-k}]e^{kt}x^{n-k}\\ &=x^n[t^n]\sum_{k=0}^{\infty}\left(\frac{te^t}{x}\right)^k\frac{1}{k!}\tag{4}\\ &=x^n[t^n]\exp\left(\frac{te^t}{x}\right)\tag{5} \end{align*}

In (4) we changed the upper limit of the index $k$ from $n$ to $\infty$ which does not contribute anything (just adding $0$'s).

From the last expression (5) we finally get

\begin{align*} A(z) &=\sum_{n=0}^{\infty}\left(\frac{1}{n!}\sum_{k=0}^n\binom{n}{k}k^{n-k}x^{n-k}\right)z^n\\ &=\sum_{n=0}^{\infty}\left(x^n[t^n]\exp\left(\frac{te^t}{x}\right)\right)z^n\tag{6}\\ &=\sum_{n=0}^{\infty}\left([t^n]\exp\left(\frac{te^t}{x}\right)\right)(xz)^n\\ &=\exp\left(\frac{xze^{xz}}{x}\right)\tag{7}\\ &=\exp\left(ze^{xz}\right)\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\Box \end{align*}

and the claim follows.

Comment:

  • In (6) we use expression (5)

  • In (7) we use a substitution rule for formal power series:

$$A(z)=\sum_{n=0}^{\infty}a_nz^n=\sum_{n=0}^{\infty}\left([t^n]A(t)\right)z^n$$

epi163sqrt
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  • So i'm guessing there isn't a closed form for this? If there is, what is the natural way to convert a generating function into a closed form? – Sidharth Ghoshal Mar 17 '15 at 20:27
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    @frogeyedpeas: Maybe not a satisfactory answer, but the best I can give at the moment: I recommend reading the first section in R.P.Stanleys *[Enumerative Combinatorics](http://www.amazon.com/Enumerative-Combinatorics-Cambridge-Advanced-Mathematics/dp/1107602629)* "How to count" - He provides us with about 10 small examples and only the simplest has a closed form. The others are represented via generating functions or recurrence relations or by other means and we are invited to accept these more complex structures as *basic building* blocks for further analysis. – epi163sqrt Mar 17 '15 at 21:48
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    @frogeyedpeas: I also think that there is no *unique natural* way to find a closed form. It mostly depends on the existing knowledge, routine and techniques. Let's have a look at the three answers so far. They all are different, interesting and could be part of one's repertoire. So each derivation could be regarded as *natural approach*. The same holds for different ways to derive a closed form (in case a closed form really exists). – epi163sqrt Mar 17 '15 at 22:01
  • @frogeyedpeas: Thanks a lot for accepting my answer and granting the bounty. A hint which might be of interest for you: Today I've answered *[this question](http://math.stackexchange.com/questions/1140281/how-to-deal-with-this-double-summation/1200287#1200287)* which provides some different techniques around how to find a closed formula. Best regards, – epi163sqrt Mar 21 '15 at 23:37
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Here is an alternate derivation of the generating function by @MarkusScheuer.

Suppose we are trying to evaluate $$A(z) = \sum_{n\ge 0} \frac{z^n}{n!} \sum_{k=0}^n {n\choose k} (kx)^{n-k}.$$

Introduce the integral representation $$(kx)^{n-k} = \frac{(n-k)!}{2\pi i} \int_{|w|=R} \frac{1}{w^{n-k+1}} \exp(kxw) \; dw.$$ which certainly holds for $0\lt R\lt \infty.$

This yields $$A(z) = \sum_{n\ge 0} \frac{z^n}{n!} \sum_{k=0}^n {n\choose k} \frac{(n-k)!}{2\pi i} \int_{|w|=|z|+\epsilon} \frac{1}{w^{n-k+1}} \exp(kxw) \; dw$$

or $$\frac{1}{2\pi i} \int_{|w|=|z|+\epsilon} \sum_{n\ge 0} \frac{z^n}{w^{n+1}} \sum_{k=0}^n \frac{1}{k!} \exp(kxw) w^k \; dw.$$

This is $$\frac{1}{2\pi i} \int_{|w|=|z|+\epsilon} \sum_{k\ge 0} \frac{1}{k!} \exp(kxw) w^k \sum_{n\ge k} \frac{z^n}{w^{n+1}} \; dw.$$ or $$\frac{1}{2\pi i} \int_{|w|=|z|+\epsilon} \sum_{k\ge 0} \frac{1}{k!} \exp(kxw) w^k \frac{z^k}{w^k} \sum_{n\ge 0} \frac{z^n}{w^{n+1}} \; dw.$$ Note that the second sum converges in the chosen annulus $|w|\gt |z|$ which means we may continue to simplify to obtain

$$\frac{1}{2\pi i} \int_{|w|=|z|+\epsilon} \sum_{k\ge 0} \frac{1}{k!} \exp(kxw) z^k \frac{1}{w} \frac{1}{1-z/w} \; dw \\ = \frac{1}{2\pi i} \int_{|w|=|z|+\epsilon} \frac{1}{w-z} \exp(z\exp(xw))\; dw.$$

By the Cauchy Residue Theorem the pole at $w=z$ contributes $$\exp(z e^{xz})$$ which was to be shown.

Marko Riedel
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Let us write $y=\ln x$. The question is equivalent to compute $$a_j(x)=\sum_{i=0}^j\frac{(j-i)^i\;x^i}{(j-i)!\;i!}.$$ An elegant method is to use, as already said, the generating function $$f(x,y)=\sum_{j=0}^\infty a_j(x) y^j=\sum_{j=0}^\infty y^j\sum_{i=0}^j \frac{(j-i)^i}{(j-i)!}\frac{x^i}{i!}.$$

Using the little diagram $$j\uparrow \begin{array}{cccc}\bullet&\bullet&\bullet&\bullet\\ \bullet&\bullet&\bullet&\\ \bullet&\bullet&&\\ \bullet&&&\\&\overrightarrow{i}&&\end{array}$$ we can swap the sums as follows $$\sum_{j=0}^\infty\sum_{i=0}^j=\sum_{i=0}^\infty\sum_{j=i}^\infty,$$ and obtain, after setting $k=j-i$ $$f(x,y)=\sum_{i=0}^\infty\sum_{k=0}^\infty \frac{k^ix^iy^i}{i!}\frac{y^k}{k!}.$$ Performing the sum over $i$ first we obtain $f(x,y)=\sum_k\exp(xyk)\frac{y^k}{k!}$ and easily get the final result $$f(x,y)=\exp\left(y\mathrm{e}^{xy}\right).$$ We note that if $y=\ln x$, $f(x,\ln x)=x^{x^x}$. The required result can be expressed as $$y^j \left[y^j\right]\exp(y\mathrm{e}^{xy}).$$

Tom-Tom
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  • Thats very interesting! I had arrived at this sum via the exponent tower $$x^{x^x}$$ I was hoping I could find a closed form to re-express $$x^{x^x}$$ in terms of 3 and x solely, based on the previous two responses it appears that won't be trivial to do – Sidharth Ghoshal Mar 17 '15 at 20:28