How to expand the series $\dfrac{1}{e^x-1}$

Question

I am looking for a way to expand this series $\dfrac{1}{e^x-1}$. The goal is to be able to finally derive the expansion for $\dfrac{x}{e^x-1}$ by multiplying $x$ with the numerator of the first series.

So far, I am stuck. I try to do this as followed:

$1+\dfrac{(-1)(-e^x)}{1!}+\dfrac{(-1)(-2)(-e^x)^2}{}+\dfrac{(-1)(-2)(-3)(-e^x)^3}{3!}...$

This form doesn't look right, Wolfram indicates that the power series for

$\dfrac{1}{e^x-1}=\dfrac{1}{x}-\dfrac{1}{2}+\dfrac{x}{12}-\dfrac{x^3}{720}...$

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

I think my approach to get the series is right, but I don't know how to make $\dfrac{1}{e^x-1}$ equals the series given by Wolfram.

Can you show me a way to expand these series?

https://math.stackexchange.com/questions/681783/maclaurin-series-of-frac1ex-1 — MafPrivate, Nov 04 '19 at 02:04
It would be helpful to know why you want this series. Do you want to compute $\int \frac{x}{e^x - 1} , dx$? — Hans Engler, Nov 04 '19 at 02:11
@IsaacYIUMathStudio So how do you compute the Laurent series? I haven't studied complex analysis so this may be above my head. The thread you gave me gives quite a lot of sigma notation formula that I don't underestand. — James Warthington, Nov 04 '19 at 02:11
@Hans Engler: I'm interested in the generating function of Bernoulli numbers — James Warthington, Nov 04 '19 at 02:12
All of those posts use the Bernoulli number in advance. I wish to derive them afterwards. — James Warthington, Nov 04 '19 at 02:13

Matt · Answer 1 · 2019-11-04T02:43:55.847

1

Use

$$ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$$

So that

$$ \dfrac{1}{e^x-1} = \dfrac{1}{\sum_{n=0}^{\infty} \frac{x^n}{n!}-1} = \dfrac{1}{\sum_{n=1}^{\infty} \frac{x^n}{n!}}.$$

Then, what you're looking for is the (multiplicative) inverse of

$${\sum_{n=1}^{\infty} \frac{x^n}{n!}} = x + \frac{x^2}{2} + \frac{x^3}{3!} + ...$$

That is, some expression

$$\sum_{m= - \infty}^{\infty} \alpha_m x^m$$

such that

$$(\sum_{m=-\infty}^{\infty} \alpha_m x^m)({\sum_{n=1}^{\infty} \frac{x^n}{n!}}) = 1$$

In particular, for integers $k$ with $k \neq 0$ we have

$$\sum_{m+n = k} \frac{\alpha_m}{n!} = 0,$$

and also

$$\sum_{m+n = 0} \frac{\alpha_m}{n!} = 1.$$

This firstly forces that $\alpha_m = 0$ for $m<{-1}$. Is this enough of a hint for you to do the rest?

edited Nov 04 '19 at 02:43

answered Nov 04 '19 at 02:29

Matt

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From this part below I don't understand: $$\sum_{n= - \infty}^{\infty} \alpha_n x^n$$ – James Warthington Nov 04 '19 at 02:36
Perhaps you could elaborate on which bit you aren't understanding? – Matt Nov 04 '19 at 02:43
I have altered the indices. Please let me know if that makes anything clearer – Matt Nov 04 '19 at 02:44
Why does it go from negative infinity to positive infinity? – James Warthington Nov 04 '19 at 02:44
Because there may be negative powers of $x$, like $x^{-1}$, $x^{-2}$, and so on. – Matt Nov 04 '19 at 02:48

J. W. Tanner · Answer 2 · 2019-11-04T03:13:54.067

0

If $\dfrac 1{e^x-1}=\dfrac{a_{-1}}x+a_0+a_1x+a_2x^2+a_3x^3+...$

then $\left(\dfrac{a_{-1}}x+a_0+a_1x+a_2x^2+a_3x^3+...\right)\left(x+\dfrac{x^2} 2+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{x^5}{5!}...\right)=1$

so $a_{-1}+\left(a_0+\dfrac {a_{-1}}2\right)x+\left(a_1+\dfrac{a_0}2+\dfrac{a_{-1}}{3!}\right)x^2+\left(a_2+\dfrac{a_1}2+\dfrac{a_0}{3!}+\dfrac{a_{-1}}{4!}\right)x^3$

$+\left(a_3+\dfrac{a_2}2+\dfrac{a_1}{3!}+\dfrac{a_0}{4!}+\dfrac{a_{-1}}{5!}\right)x^4+...=1$

so $a_{-1}=1$ and $ a_0+\dfrac {a_{-1}}2=a_1+\dfrac{a_0}2+\dfrac{a_{-1}}{3!}=a_2+\dfrac{a_1}2+\dfrac{a_0}{3!}+\dfrac{a_{-1}}{4!}$

$=a_3+\dfrac{a_2}2+\dfrac{a_1}{3!}+\dfrac{a_0}{4!}+\dfrac{a_{-1}}{5!}=...=0.$

Thus $a_{-1}=1$, so from $a_0+\dfrac {a_{-1}}2=0$ we get $ a_0=-\dfrac12$,

so from $a_1+\dfrac{a_0}2+\dfrac{a_{-1}}{3!}=0$ we get $a_1=\dfrac1{12}$,

so from $a_2+\dfrac{a_1}2+\dfrac{a_0}{3!}+\dfrac{a_{-1}}{4!}=0$ we get $a_2=0$,

so from $a_3+\dfrac{a_2}2+\dfrac{a_1}{3!}+\dfrac{a_0}{4!}+\dfrac{a_{-1}}{5!}=0 $ we get $a_3=-\dfrac1{720},...$

edited Nov 04 '19 at 03:13

answered Nov 04 '19 at 02:24

J. W. Tanner

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Remark for future readers: I've gone ahead below and proven that we have no terms below $x^{-1}$ in the series expansion of the inverse, but I suppose it's obvious enough when written out. – Matt Nov 04 '19 at 02:35
@J.W.Tanner I am lost in this part: $a_{-1}+\left(a_0+\dfrac {a_{-1}}2\right)x+\left(a_1+\dfrac{a_0}2+\dfrac{a_{-1}}{3!}\right)x^2+\left(a_2+\dfrac{a_1}2+\dfrac{a_0}{3!}+\dfrac{a_{-1}}{4!}\right)x^3$ – James Warthington Nov 04 '19 at 02:39
@J.W.Tanner What are you doing with the coefficients there? – James Warthington Nov 04 '19 at 02:39
are you lost getting to that line or the line after it? – J. W. Tanner Nov 04 '19 at 02:41
I multiplied the power series – J. W. Tanner Nov 04 '19 at 02:43
@I thought you multiply by using the distributive property? I don't know what is Cauchy product. Sorry! – James Warthington Nov 04 '19 at 02:46
@J.W.Tanner How do you get $\dfrac{a_-1}{x}$? I understand the rest are variables of the unknown coefficients. – James Warthington Nov 04 '19 at 02:47
@J.W.Tanner So you group them differently, how do you know $\dfrac{a_-1}{x}=1?$ – James Warthington Nov 04 '19 at 02:51
$a_{-1}=1$ because the constant terms (those without postive powers of $x$) on both sides of the equation must be equal – J. W. Tanner Nov 04 '19 at 02:54
@J.W.Tanner I am watching the Cauchy product right now on youtube, so bear with me a second ok? – James Warthington Nov 04 '19 at 02:57
Can you do this part in more details in your answer, I don't know how do you get those coefficients correctly:Thus $a_{-1}=1, a_0=-\dfrac12, a_1=\dfrac1{12}, a_2=0, a_3=-\dfrac1{720},...$ – James Warthington Nov 04 '19 at 03:06
I edited accordingly – J. W. Tanner Nov 04 '19 at 03:14
How do you solve for $a_0$, I don't understand this part $a_{-1}=1$, so from $a_0+\dfrac {a_{-1}}2=0$ we get $ a_0=-\dfrac12$ – James Warthington Nov 04 '19 at 03:22
that is correct – J. W. Tanner Nov 04 '19 at 03:24
@J.W.Tanner How? – James Warthington Nov 04 '19 at 03:27
2

@JamesWarthington It is mildly apparent to at least myself from your rather prompt replies that you aren't making much effort to work on this yourself. I suggest working for a while, and coming back with a more coherent question if you're still stuck, as opposed to "I don't understand how you go from line $x$ to line $y$. – Matt Nov 04 '19 at 03:30
@Matt I will post an answer that contains all the points I don't understand. – James Warthington Nov 04 '19 at 03:32
@J.W.Tanner I just have one last question, why do you choose the first term of the series of unknown coefficient to be $\frac{a_{-1}}{x}$ – James Warthington Nov 04 '19 at 15:31

How to expand the series $\dfrac{1}{e^x-1}$

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