## Monday Math 26

Recall the formula we learned for the sine function in terms of an infinite product:

Thus:
.
Now, let us compare this to the Taylor series for sine:

Thus

and so
.

Now, if you consider the expansion of the infinite product , we see that all x2 terms arise from multiplying one term from the right of one parenthesis, and the ones from the rest of the parentheses: thus, we have terms  etc.
Thus we have the coefficient of x2 as , but via the Taylor series, the coefficient is , thus telling us .

Note that this sum is , where  is the Riemann zeta function introduced in this previous post. The problem of finding the exact value of this series and proving it was known as the Basel problem, and was first solved by Euler at age 28.

Now, let us consider the x4 terms: here, the result from the product formula are terms of the form , with k and l distinct, and only one such for each pair of distinct numbers. Thus, the coefficient is
, and by examining the Taylor series, we have
.

Consider: . First, we see that this sum is:
.
However, we can also break up the sum into three parts: , , and :

As the sum for  is the same as for . We already found the value for these, and so:
.
And thus, we have the exact value for .

In fact, we can find the exact value for the zeta function at any positive even integer. More specifically, , where the  are the Bernoulli numbers, which I have mentioned earlier in the context of Faulhaber’s Formula.

### 4 Responses to “Monday Math 26”

1. Physics Friday 33: Blackbody Radiation (Part 2) « Twisted One 151’s Weblog Says:

[…] integral, I showed in Monday Math 29 that , we use x=4 to see that . Now, , and we found in Monday Math 26 that , so , and thus , where is known as the Stefan-Boltzmann constant, and the relation is […]

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

[…] Math 58 By twistedone151 Recall the product formula for the sine function (here and here): Thus: or with z=πx, Taking the logarithm of both sides: . Now, , so taking the derivative […]

3. Monday Math 60 « Twisted One 151’s Weblog Says:

[…] , we see that , and so . For s=2, this then answers our original question: , and using and (see here), we find . Similarly, (see here), and so . Possibly related posts: (automatically generated)Euler […]

4. Monday Math 78 « Twisted One 151’s Weblog Says:

[…] us: . Now, we proved here that , so the above is just this for the case n=2, so since and (see here), then our integral is […]