5
$\begingroup$

I've read that Bernoulli Numbers are defined by the series $$ \frac{z}{e^z-1}\equiv \sum\limits_{n=0}^{\infty}B_n\frac{z^n}{n!},$$
So if I start with $0$ I get $$ B_0\frac{1}{1}=B_0{1}. $$ My question is, why is there a $B_0$ in the term...is it of any significance? Or just a "marker" or something to indicate that this is the $B_0$ term?

If I find the second term I get $$ B_1\frac{z}{1}=B_1z $$ What's the $z$? I've read it must be $\left|z\right|<2\pi$, but how does one get $\frac{-1}{2}$?

$\endgroup$
2
  • $\begingroup$ In other words, couldn't this second term be virtually anything in $-2\pi<z<2\pi$? $\endgroup$ – bjd2385 Oct 2 '14 at 12:29
  • $\begingroup$ I think that $B_0=1$ and $B_1=-\frac{1}{2}$ are defined by convention for the first Bernoulli numbers. $B_0=1$ and $B_1=+\frac{1}{2}$ define the second Bernoulli numbers. $\endgroup$ – Claude Leibovici Oct 2 '14 at 12:40
6
$\begingroup$

First of all, using the Taylor series for $e^z$ we have $$ \frac{e^z-1}{z} = 1 + \frac{z}{2} + \frac{z^2}{6} + \frac{z^3}{24} + \cdots. $$ Multiplying this by the power series for $z/(e^z-1)$ and comparing coefficients (the product should be $1$) we get $$ \begin{align*} 1 &= B_0 \\ 0 &= B_1 + 1/2 \\ 0 &= (2B_2) + (1/2) B_1 + (1/6) B_0 \\ 0 &= (6B_3) + (1/2) (2B_2) + (1/6) B_1 + (1/24) B_0 \end{align*} $$ and so on. Therefore $B_0 = 1$, $B_1 = -1/2$, $B_2 = 1/6$, $B_3 = -1/30$, and so on.

If you look at the function $$ f(z) = \frac{z}{e^z-1} + \frac{z}{2} $$ then you find out that $$ \begin{align*} f(-z) = \frac{-z}{e^{-z}-1} - \frac{z}{2} = \frac{ze^z}{e^z-1} - \frac{z}{2} = \frac{z}{e^z-1} + z - \frac{z}{2} = f(z). \end{align*} $$ Therefore $f(z)$ is even and all the odd coefficients in its power series vanish. This shows that apart from $B_1 = -1/2$, all other odd-indexed Bernoulli numbers vanish. Why do we need them, then? They're just the sequence whose exponential generating series is $z/(e^z-1)$.

$\endgroup$
1
  • $\begingroup$ I see, so they're simply examining the power series, i.e. the sum is a power series. I can't believe I missed that :p Oh well lol. Thanks for the answers guys, you've been a great help! $\endgroup$ – bjd2385 Oct 2 '14 at 13:01
5
$\begingroup$

The statement defining the Bernoulli numbers says that the power series expansion of $\frac{z}{e^z-1}$ is equal to the power series on the right; that is, the coefficient of $x^i$ in the power series expansion is $B_i$. The first few terms of the expansion are $$\frac{z}{e^z-1} = 1-\frac{z}{2}+\frac{z^2}{12}-\frac{z^4}{720}+\frac{z^6}{30240}+\cdots.$$ Thus \begin{align*} B_0&=1 \\ B_1&=-\frac{1}{2}\\ B_2&=2!\cdot\frac{1}{12}=\frac{1}{6}\\ B_3&=0\\ B_4 &= -4!\cdot\frac{1}{720} = -\frac{1}{30} \end{align*} and so forth.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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