Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Here is the definition:

Bernoulli numbers arise in Taylor series in the expansion $$\frac{x}{e^x-1}=\sum_{k=0}^\infty B_k \frac{x^k}{k!}$$

I've tried to naively expand $\frac{x}{e^x-1}$ around $x_0=0$ and didn't quite understand what should be done with all the $(e^x-1)^{k+1}$ in the denominators of $f^{(k)}(0)$?

How does one get to $$1 - \frac{x}{2} + \frac{x^2}{12} - \frac{x^4}{720} + \ ...$$ directly from the definition (without rearranging the whole thing into the recursive formula)?

Thanks a lot!

share|improve this question
    
This is the exponential generating function of Bernoulli numbers –  Mhenni Benghorbal May 4 '13 at 12:39
add comment

2 Answers

up vote 4 down vote accepted

Expand $e^{x}$ around $x=0$ and then reexpand \begin{align}\frac{1}{1+\frac{x}{2!}+\frac{x^2}{3!}+\ldots}=1-\left(\frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}\ldots\right)+\left(\frac{x}{2!}+\frac{x^2}{3!}+\ldots\right)^2-\left(\frac{x}{2!}+\ldots\right)^3+\ldots=\\ =1-\left(\frac{x}{2}+\frac{x^2}{6}+\frac{x^3}{24}+\ldots\right)+\frac{x^2}{4}\left(1+\frac{x}{3}+\ldots\right)^2-\frac{x^3}{8}\left(1+\ldots\right)^3+\ldots=\\ =1-\left(\frac{x}{2}+\frac{x^2}{6}+\frac{x^3}{24}\right)+\frac{x^2}{4}\left(1+\frac{2x}{3}\right)-\frac{x^3}{8}+O\left(x^4\right)=\\ =1-\frac{x}{2}+\left(-\frac{1}{6}+\frac14\right)x^2+\left(-\frac{1}{24}+\frac{1}{6}-\frac{1}{8}\right)x^3+O\left(x^4\right)=\\ =1-\frac{x}{2}+\frac{x^2}{12}+O\left(x^4\right).\end{align}

share|improve this answer
    
Could you please elaborate how to get from your RHS to $1 - \frac{x}{2} + \frac{x^2}{12} - \frac{x^4}{720} + \ ...$? –  Leo Schmidt May 4 '13 at 12:42
add comment

I know that the question asks for a method not using recursion, but this is just so much easier and cleaner than computing the series for $\dfrac{x}{e^x-1}$.

Given $$ \frac{x}{e^x-1}=\sum_{k=0}^\infty B_k\frac{x^k}{k!}\tag{1} $$ we can rewrite $(1)$ as $$ \begin{align} 1 &=\sum_{j=0}^\infty\frac{x^j}{(j+1)!}\sum_{k=0}^\infty B_k\frac{x^k}{k!}\\ &=\sum_{k=0}^\infty x^k\sum_{j=0}^k\frac{B_{k-j}}{(k-j)!\,(j+1)!}\\ &=\sum_{k=0}^\infty\frac{x^k}{(k+1)!}\sum_{j=0}^k\binom{k+1}{j+1}B_{k-j}\tag{2} \end{align} $$ which yields $B_0=1$ and the recurrence $$ B_k=-\frac1{k+1}\sum_{j=1}^k\binom{k+1}{j+1}B_{k-j}\tag{3} $$ for $k\ge1$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.