# Series expansion for $f(x)=\frac{x}{e^x-1}$

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?

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?

• The first derivative is undefined but you could use L'hopital's rule to find the limit – J. W. Tanner Nov 3 '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 3 '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 3 '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 3 '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 3 '19 at 21:26

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.

• you mean $x/(e^x-1)=\sum_{n=0}^{\infty}B_n x^n/n!$ ? – J. W. Tanner Nov 3 '19 at 21:51
• @J.W.Tanner Can you further explain to me how metamorphy get his answer? I don't understand his answer. – James Warthington Nov 4 '19 at 2:20
• @JamesWarthington: I think I have to explain it myself (see the edit; basically, after "That is, ...", I'm trying to rephrase in simple words the referenced part of the wiki article, which you don't seem to grasp). Can we go step by step on what remains unclear? – metamorphy Nov 4 '19 at 22:06
• @J.W.Tanner: I'm trying to avoid the $B_n$ ;) – metamorphy Nov 4 '19 at 22:08
• @metamorphy: How do you obtain $a_n=\dfrac{1}{(n+1!)}$ I have to say I am awe with your ability to calculate the coefficient so exactly. I am envious. Thank you! – James Warthington Nov 4 '19 at 22:10

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

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

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

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

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

• Yeah, expanding the numerator and denominator separately is what I am reading right now, it can be found here: math.stackexchange.com/questions/3214465/… – James Warthington Nov 3 '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 3 '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 3 '19 at 21:53
• I don't understand what you mean? – James Warthington Nov 3 '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 3 '19 at 22:24