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 *x*^{2} 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 *x*^{2} 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 *x*^{4} 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.

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Tags: Math, Monday Math, Riemann Zeta Function, Sine Product Formula, Taylor Series

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