Monday Math 57

Previously, we introduced the Bernoulli polynomials, and the Bernoulli numbers. In particular, we found that the Bernoulli numbers obey (and along with B0=1 can be defined by) the relation .

Now, let us examine the Maclaurin series for the function (the singularity at z=0 is removable, and is also removable in the derivatives of all order).
Let , so that our series is then

Now, the Maclaurin series for ex is , so the series for ez-1 is

Thus, we have:
Now, we use the Cauchy product for infinite series:
, where .
This tells us

But the product of the two series is unity, so we have c0=1, cn=0 for n≥1. Thus

for n≥1
This, however, is the relation for the Bernoulli numbers: , and so the Maclaurin series is
and .


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One Response to “Monday Math 57”

  1. Monday Math 58 « Twisted One 151’s Weblog Says:

    […] , where is the Riemann zeta function. Now, we use Euler’s formula: as , , , and . Previously, we showed that the function can be given by the series . Thus , and so . Now, as and , we can […]

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