 In 1735, Euler tackled the problem of finding the sum of the reciprocal squares using the factorization of the Taylor series. In any factorization of a polynomial, the coefficient of x in the expansion is the sum of the coefficients of x in the factors, and so that says that 1 over y is the sum of the reciprocals, 1 over a, 1 over b, and so on. We can, and Euler did, go further since the coefficients in the expansion are the sums of products of coefficients in the factors. So how can we use this? Consider a sum of terms, which we'll call p. If we want to find the sum of the squares of the terms, let alpha be the sum of the terms, and beta be the sum of the pairwise products of the terms, then with a little bit of effort, we can show that q is alpha squared minus 2 beta. So now let's return to our expansion, where a, b, c, d, and so on are the solutions to y equals sine x, and we'll let p equal the sum of the reciprocals, q be the sum of the reciprocal squares, and let alpha be this sum. The coefficient of x in the expansion is the sum of the coefficients of x in the factors, and so comparing our coefficients gives us, this was Euler's first result, and it gives us that alpha is 1 over y. And since we know alpha is the sum of the reciprocals, then the sum of the reciprocals is 1 over y. But wait, there's more. The coefficient of x squared in the expansion is the sum of the pairwise products of the coefficients of x in the factors. We don't have an x squared term in the expansion, so our coefficient is 0, and so 0 is the sum of the reciprocals of the pairwise products, but remember this sum of the pairwise products is what we were calling beta, and so we know that beta is 0. And since alpha is 1 over y, that tells us that q, the sum of the reciprocal squares, is... So, let's talk specifics. If a is the least positive solution to y equals sine x, our positive solutions will be, and our negative solutions will be, and if we take p to be the reciprocals of these values, Euler's formulas give us p is equal to 1 over y. Suppose y equals 1, then we have, and if we do a little bit of arithmetic we can simplify this to, which is the series that Leibniz found. Now in the next step, it's important to keep in mind that the solutions to y equals sine x are the denominators of the series for p, so for sine x equal to 1, the series p is, and not any simplified form, because the relationship between p and q that we're using is dependent on the coefficients of the factorization. So from our sum for p, remember q is the sum of the squares, Euler's formulas give us q equals 1 over y squared. Since y equals 1, this means q is equal to 1, and so we have, and multiplying by pi squared over 4 and simplifying gives us, and as you can see we're very close to the basal problem, the only difference here is that we're missing the reciprocals of the squares of the even numbers. Which is not a problem, suppose s is this sum of reciprocal squares, if we do a little bit of algebra we find, and here we have the sum of the reciprocals of the squares of the odd numbers, which we know to be pi squared over 8, and so that says that s is pi squared over 6.