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I am trying to expand the function $f(x)=\dfrac{x}{e^x-1}$ into Maclaurin series and I encounter this problem. The first derivative is $f'(x)=-\dfrac{e^x(x-1)+1}{(e^x-1)^2}$. When $f(0)=\dfrac{1(0-1)+1}{(1-1)}^2$, which is undefined. Can I continue to expand the series using the second derivative.

Is there a faster way to expand this series without directly employing the Maclaurin formula?

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 $B_1=-\dfrac{1}{2}, B_2=\dfrac{1}{6}, B_3=0$

How can I expand the series as is shown in that page?

James Warthington
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    The first derivative is undefined but you could use L'hopital's rule to find the limit – J. W. Tanner Nov 03 '19 at 20:55
  • @J.W.Tanner: Is there a way to expand this series without going through so many derivatives? Each derivative just gets more complicated, not simpler. – James Warthington Nov 03 '19 at 20:58
  • @metamorphy: can you do in a few steps so I can learn? So there is no faster way and I have to go through the derivatives? This is going to be very tedious. – James Warthington Nov 03 '19 at 21:18
  • The third derivative is also undefined. This makes expanding this function so inconvenient. Do I have to use L'Hopital for each derivative? – James Warthington Nov 03 '19 at 21:24
  • @metamorphy: this convinces me that expanding this series at $x=0$ is not feasible, since every derivative is going to be undefined. Is there a better way to do this? – James Warthington Nov 03 '19 at 21:26
  • Converted it to an answer (I'm slow with my phone...), which now looks redundant. – metamorphy Nov 03 '19 at 21:45
  • The series you gave matches neither the regular definition of Bernoulli numbers $\left[\dfrac x{e^x-1}=\sum\limits_{n=0}^\infty \dfrac {B_nx^n}{n!}\right]$ nor the archaic definition $\left[=1-\dfrac x2+\sum\limits_{n=1}^\infty (-1)^{n-1}\dfrac {B^*_n x^{2n}}{(2n)!}\right]$ given on the page you linked – J. W. Tanner Nov 04 '19 at 01:23
  • @ J. W. Tanner, you mean the series from proofwiki? – James Warthington Nov 04 '19 at 01:30

2 Answers2

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$\dfrac{e^x-1}x=\dfrac{x+\dfrac{x^2}2+\dfrac{x^3}6+...}x=1+\dfrac12x+\dfrac16x^2...$

$\therefore\left(B_0+B_1x+B_2\dfrac{x^2}2+...\right) \left(1+\dfrac12x+\dfrac16x^2...\right)=1$

$\therefore B_0+\left(B_1+\dfrac12B_0\right)x+\left(\dfrac{1}2B_2+\dfrac12B_1+\dfrac16B_0\right)x^2+...=1$

$\therefore B_0=1, B_1+\dfrac12B_0=0, \dfrac12B_2+\dfrac12B_1+\dfrac16B_0=0, ...$

$\therefore B_0=1, B_1=-\dfrac12, B_2=\dfrac16, ...$

J. W. Tanner
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  • Yeah, expanding the numerator and denominator separately is what I am reading right now, it can be found here: https://math.stackexchange.com/questions/3214465/maclaurin-expansion-of-expx-1-expx2?rq=1 – James Warthington Nov 03 '19 at 21:35
  • Comparing this with expansion above, it seems that something is off, the series expanding on the website I cite at the beginning of this thread has an alternating sign. – James Warthington Nov 03 '19 at 21:48
  • The way the other answer is written, the $n^{th}$ Bernoulli number $B_n=b_n n!$ – J. W. Tanner Nov 03 '19 at 21:53
  • I don't understand what you mean? – James Warthington Nov 03 '19 at 22:22
  • I mean the other answer to this question says, for example, $b_2=1/12,$ and that corresponds to $B_2=2!/12=1/6$ – J. W. Tanner Nov 03 '19 at 22:24
  • there is something wrong with this expansion. The coefficient for the third term is $\frac{1}{24}x^{3}$, not a Bernoulli number. Besides, the sign should be alternating instead of being all positive. For example, the second term should be $-\frac{1}{2}$ instead of $\frac{1}{2}$ – James Warthington Nov 03 '19 at 22:26
  • The next equation would be $B_3\dfrac16+B_2\dfrac14+B_1\dfrac16+B_0\dfrac1{24}=0,$ which implies $B_3=0$ – J. W. Tanner Nov 03 '19 at 22:35
  • Did you realize that the question you linked above was about a different function? – J. W. Tanner Nov 03 '19 at 22:36
  • Also, I think I used the older definition of Bernoulli numbers – J. W. Tanner Nov 03 '19 at 22:40
  • The series I expand by manipulating the way you do is $1+\frac{2!}{x}+\frac{3!}{x^2}+\frac{4!}{x^3}...$ I don't see where is the Bernoulli number of this series. Also, it should have alternating sign to match that Bernoulli coefficients. – James Warthington Nov 03 '19 at 22:46
  • I honestly don't understand your answer above. I thought you just need to expand the series by manipulating the original series $e^x$. – James Warthington Nov 03 '19 at 22:58
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One (not aware of the Bernoulli numbers) can compute the reciprocal series for $$\frac{e^x-1}{x}=\sum_{n=0}^{\infty}a_n x^n,\qquad a_n=\frac{1}{(n+1)!}.$$ That is, assuming $x/(e^x-1)=\sum_{n=0}^{\infty}b_n x^n$, we get the system of equations for unknowns $b_n$ from $\left(\sum_{n=0}^{\infty}a_n x^n\right)\left(\sum_{n=0}^{\infty}b_n x^n\right)=1$, i.e. \begin{align}\color{gray}{[x^0]}&\quad a_0 b_0=1,\\\color{gray}{[x^1]}&\quad a_0 b_1+a_1 b_0=0,\\\color{gray}{[x^2]}&\quad a_0 b_2+a_1 b_1+a_2 b_0=0,\\\color{gray}{[x^3]}&\quad a_0 b_3+a_1 b_2+a_2 b_1+a_3 b_0=0,\\&\ldots\end{align} which gives \begin{align}b_0&=1/a_0=1,\\b_1&=-a_1 b_0/a_0=-1/2,\\b_2&=-(a_2 b_0+a_1 b_1)/a_0=1/12,\\b_3&=-(a_3 b_0+a_2 b_1+a_1 b_2)/a_0=0,\\b_4&=-(a_4 b_0+a_3 b_1+a_2 b_2+a_1 b_3)/a_0=-1/720,\end{align} and so on.

metamorphy
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