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According to this page: https://proofwiki.org/wiki/Definition:Bernoulli_Numbers

The series is : $\dfrac{x}{e^x-1}=1-\dfrac{x}{2}+\dfrac{B_1x^2}{2!}+\dfrac{B_2x^4}{4!}+\dfrac{B_3x^6}{6!}...$ where $B_1=-\dfrac{1}{2}, B_2=\dfrac{1}{6}, B_3=0, B_4=-\dfrac{1}{30}$

How can I expand obtain the Bernoulli coefficient as indicated in the website?

Using the series above, I have:

$\dfrac{x}{e^x-1}=1-\dfrac{x}{2}+\dfrac{(-\frac{1}{2})x^2}{2!}+\dfrac{\frac{1}{6}x^4}{4!}+\dfrac{(0)x^6}{6!}...$

$=\dfrac{x}{e^x-1}=1-\dfrac{x}{2}+\dfrac{x^2}{4}+\dfrac{x^4}{144}+0...$

This looks wrong since the series expansion according to Wolfram is:

$\dfrac{x}{e^x-1}=1-\dfrac{x}{2}+\dfrac{x^2}{12}-\dfrac{x^4}{720}...$

But then the series of Wolfram doesn't show the Bernoulli number. I am confused, can you help me here?

James Warthington
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  • You may want to check the sign of $x^2 $ in the expansion of the 2nd equation. Also, on the same equation, where did the zero come from in the coefficient of $x^6$? – NoChance Nov 04 '19 at 00:15
  • @NoChance: the third Bernoulli number is $0$ – James Warthington Nov 04 '19 at 00:21
  • OK, May be you want to check the post below, look for the image with the formula #1541. The formula uses Even powers only - This may justify the inconsistent representation you found: https://math.stackexchange.com/questions/1885685/important-identities-that-can-be-obtained-by-manipulating-the-function-fracx?rq=1 – NoChance Nov 04 '19 at 00:26
  • @NoChance: Do you know how can I expand this function into the series like Wolfram? Is there a way to expand this function and obtain the Bernoulli number before simplifying to get the equation in Wolfram? – James Warthington Nov 04 '19 at 00:31
  • I could only do Taylor's expansion, but since the calculations are tedious, I am not sure what we will get. This link may be relevant: https://math.stackexchange.com/questions/3420676/series-expansion-for-fx-fracxex-1?rq=1 – NoChance Nov 04 '19 at 00:38
  • @NoChance How about just expanding the series in the denominator? I have tried to directly use the Maclaurin formula and it is very tedious, not to mention that the derivatives are undefined. – James Warthington Nov 04 '19 at 00:42
  • you can use Taylor's at points other than zero. Expanding the denominator will not help since you can't manipulate $\frac{x}{a+b+c+d+....} $ – NoChance Nov 04 '19 at 00:45
  • @NoChance can you do this in the answer section? – James Warthington Nov 04 '19 at 00:46
  • @NoChance, the link above is asked by me. Unable to get a new answer, I post this question to get more people involved. – James Warthington Nov 04 '19 at 00:47
  • @NoChance Do you know how metamorphy get his answer? – James Warthington Nov 04 '19 at 00:48
  • I am not sure what the source of the answer is really. – NoChance Nov 04 '19 at 00:56
  • @I think I can figure out the Bernoulli number part, I just need to know how to expand the series into Wolfram's form. – James Warthington Nov 04 '19 at 00:58
  • How about the Taylor series for $\dfrac{1}{e^x-1}$, I think we are close to get to the answer. – James Warthington Nov 04 '19 at 01:08
  • Yeh...its getting late, I missed the sign. The right expansion is very ugly - I tried on www.Symbolab.com (type Taylor followed by the function you wan to expand). – NoChance Nov 04 '19 at 01:13
  • According to the page you linked, $\dfrac{x}{e^x-1}=1-\dfrac{x}{2}+\dfrac{B_1x^2}{2!}\color{red}-\dfrac{B_2x^4}{4!}+\dfrac{B_3x^6}{6!}...$, and that is the archaic definition of Bernoulli numbers, according to which $B_3$ is not zero – J. W. Tanner Nov 04 '19 at 01:54
  • @J.W.Tanner: can you have a look at this thread and help me out? https://math.stackexchange.com/questions/3420982/how-to-expand-the-series-dfrac1ex-1 – James Warthington Nov 04 '19 at 01:58

1 Answers1

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Note, the generating function for the Bernoulli numbers is defined as \begin{align*} \frac{x}{e^x-1}=\sum_{n=0}^\infty\frac{B_nx^n}{n!} \end{align*} with $\frac{B_\color{blue}{n}}{n!}$ the coefficient of $x^\color{blue}{n}$.

Since we have $B_0=1, B_1=-\frac{1}{2}, B_2=\frac{1}{6}, B_3=0, B_4=-\frac{1}{30},\ldots$ we obtain \begin{align*} \color{blue}{\frac{x}{e^x-1}}&=\sum_{n=0}^\infty\frac{B_nx^n}{n!}\\ &=B_0+\frac{B_1x}{1!}+\frac{B_2x^2}{2!}+\frac{B_3x^3}{3!}+\frac{B_4x^4}{4!}+\cdots\\ &=1+\frac{\left(-\frac{1}{2}\right)x}{1}+\frac{\left(\frac{1}{6}\right)x^2}{2}+\frac{0x^3}{6}+\frac{\left(-\frac{1}{30}\right)x^4}{24}+\cdots\\ &\,\,\color{blue}{=1-\frac{x}{2}+\frac{x^2}{12}-\frac{x^4}{720}+\cdots} \end{align*}

Markus Scheuer
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