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I have found the following formula: $$\frac{\operatorname d^n}{\operatorname dx^n}\left(\frac{x}{e^x-1}\right)=(-1)^n\,\frac{n\sum\limits_{k=0}^{n}e^{kx}\sum\limits_{i=0}^{k}(-1)^i\binom{n+1}{i}(k-i)^{n-1}+x\sum\limits_{k=0}^{n}e^{kx}\sum\limits_{i=0}^{k}(-1)^i\binom{n+1}{i}(k-i)^n}{\left(e^x-1\right)^{n+1}}. $$ My proof of this formula is complicated. Can somebody find some simple proof?

Did
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Marek
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    What is your complicated proof? Does it take more than 2 pages? If you don't feel like writing it in $\LaTeX$ you can put it as an image. – JMCF125 Dec 20 '13 at 12:44
  • My proof is natural, but has too many steps. It's via many recurrences, and prooving some complicated combinatorial identities. – Marek Dec 20 '13 at 12:51
  • Then maybe you find some of those identities online, link them, and put the rest of it in here. Or (assuming you have it on paper and a multifunction printer), put the images regarding all steps on a cloud service and the main steps here, along with the links. I didn't downvote (I ran out of votes), but I agree you have to show your steps and your effort. – JMCF125 Dec 20 '13 at 13:17
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    Link to my proof: https://drive.google.com/file/d/0B4Bs_2_T-WoAb2R0MnoyQ0pWYU0/edit?usp=sharing But, excuse me, my english is not so good. – Marek Dec 20 '13 at 13:37
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    Not so good? I've seen worse from native speakers. I suggested an edit to add the link to your post. That must have been a lot of work! I'll upvote as soon as I can. – JMCF125 Dec 20 '13 at 13:46
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    I completely don't understand the downvotes. The question certainly shows effort, and is quite a natural one. – Jakub Konieczny Dec 20 '13 at 14:03
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    Check this. – Mhenni Benghorbal Dec 20 '13 at 14:11
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    I upvoted the question. – Mhenni Benghorbal Dec 20 '13 at 14:17
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    @Feanor, initially, there was no link to that document. – JMCF125 Dec 20 '13 at 14:26
  • Doesn't this come directly from the series in terms of Bernoulli numbers by definition? See http://en.wikipedia.org/wiki/Bernoulli_number#Generating_function. – lhf Dec 20 '13 at 15:28
  • Your “proof” needs permission to open. – Тyma Gaidash Oct 22 '21 at 17:40

5 Answers5

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Here's a starting point.

Using the general Leibniz rule,

$$\frac{d^n}{dx^n}\frac{x}{e^x-1}=\sum_{k=0}^n\binom{n}{k}\left(\frac{d^k}{dx^k}x\right)\left(\frac{d^{n-k}}{dx^{n-k}}\frac{1}{e^x-1}\right).$$

Now, the "0-th" derivative of $x$ is $x$, and the 1st derivative of $x$ is $1$, and all 2nd and higher order derivatives vanish. Then, the only non-zero terms in the sum above are the k=0,1 terms, so we get:

$$\frac{d^n}{dx^n}\frac{x}{e^x-1} = \binom{n}{0}\left(\frac{d^0}{dx^0}x\right)\left(\frac{d^{n-0}}{dx^{n-0}}\frac{1}{e^x-1}\right)+\binom{n}{1}\left(\frac{d^1}{dx^1}x\right)\left(\frac{d^{n-1}}{dx^{n-1}}\frac{1}{e^x-1}\right)\\ = x\left(\frac{d^{n}}{dx^{n}}\frac{1}{e^x-1}\right) + n\left(\frac{d^{n-1}}{dx^{n-1}}\frac{1}{e^x-1}\right).$$

At the very least this should obviate the need for complicated products of summations as you currently have.

David H
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  • Did you not forget $\binom{n}{k}$ in the first equation? – Sasha Dec 20 '13 at 14:14
  • @Sasha Thank you for catching that little gaffe, and shame on the people who upvoted me. $\binom{n}{0}=1$ doesn't change much, so it comes to missing factor of $\binom{n}{1}=n$ in the second term. – David H Dec 20 '13 at 14:23
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i hope this would help you in some way $$\frac{x}{e^x-1}=\frac{x}{\cos(ix)-i\sin (ix)-1}$$ $$=\frac{x}{-(1-\cos (ix))-\sin (ix)}$$ $$=\frac{x}{-2\sin^2(\frac{ix}{2})-\sin (ix)}$$ $$=\frac{x}{-2\sin (\frac{ix}{2})(\sin (\frac{ix}{2})+\cos (\frac{ix}{2}))}$$ $$=\frac{x}{-2\sin (\frac{ix}{2})e^{\frac{-x}{2}}}$$ $$=\frac{xe^{\frac{x}{2}}}{i\sinh (\frac{x}{2})}$$

Suraj M S
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Related problems: (I), (II). Here is a formula

$$ \sum _{k=0}^{n} \sum _{m=0}^{n-k} ( -1 )^{m} m! {n\choose k} {n-k \brace m}(2-k)_k\,\frac{{x}^{1-k}\,{\rm e}^{mx}}{ \left( {{\rm e}^{x}}-1 \right)^{m+1}}, $$

where $(a)_b$ is the Pocchammer symbol and $ {n \brace k} $ is the Stirling numbers of the second kind.

Community
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Mhenni Benghorbal
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The Taylor series of $\frac{x}{e^x-1}$ is well-known:

$$\frac{x}{e^x-1}=\sum_{k=0}^\infty \frac{B_k}{k!} x^k$$

$$\Longrightarrow \frac{\operatorname d^n}{\operatorname dx^n}\left(\frac{x}{e^x-1}\right)=\sum_{k=0}^\infty \frac{B_k}{k!} \frac{\operatorname d^n}{\operatorname dx^n}x^k$$

$$=\sum_{k=0}^\infty \frac{B_k}{k!} \frac{k!}{(k-n)!}x^{k-n}=\sum_{k=0}^\infty \frac{B_k}{(k-n)!}x^{k-n}.$$

Ali Shadhar
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The formula \begin{equation*} \biggl(\frac{1}{\operatorname{e}^x-1}\biggr)^{(n)} =(-1)^n\sum_{j=1}^{n+1}(j-1)!S(n+1,j)\biggl(\frac1{\operatorname{e}^x-1}\biggr)^j \end{equation*} is perhaps useful. This formula is a consequence of the following

References

  1. Bai-Ni Guo and Feng Qi, Explicit formulae for computing Euler polynomials in terms of Stirling numbers of the second kind, Journal of Computational and Applied Mathematics 272 (2014), 251--257; available online at https://doi.org/10.1016/j.cam.2014.05.018.
  2. Bai-Ni Guo and Feng Qi, Some identities and an explicit formula for Bernoulli and Stirling numbers, Journal of Computational and Applied Mathematics 255 (2014), 568--579; available online at https://doi.org/10.1016/j.cam.2013.06.020.
  3. C.-F. Wei and B.-N. Guo, Complete monotonicity of functions connected with the exponential function and derivatives, Abstr. Appl. Anal. 2014 (2014), Article ID 851213, 5 pages; available online at https://doi.org/10.1155/2014/851213.
  4. A.-M. Xu and Z.-D. Cen, Some identities involving exponential functions and Stirling numbers and applications, J. Comput. Appl. Math. 260 (2014), 201--207; available online at https://doi.org/10.1016/j.cam.2013.09.077.
qifeng618
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