 One of Euler's great early achievements was finding the sum of the series of reciprocal squares. This is known as the Basel problem because Leipnitz and the brothers Bernoulli of Basel attempted to solve it. The approach of Leipnitz and the Bernoulli's was to let y equal the power series, then the sum of the reciprocal squares would be the value of y when x is equal to 1. Leipnitz then leads to a differential equation, and fortunately neither Leipnitz nor Bernoulli's were able to solve the resulting differential equation. Euler, who was Johann's student, used a different strategy in his On the Summs of series of reciprocals presented to the St. Petersburg Academy on December 5, 1735. He found the sum, then he found the sum again in a different way, and then gave a third solution. Then in 1741 he gave a fourth solution. Did we mention Euler was the most prolific mathematician in history? If you really want to learn how to do mathematics, you could do far worse than to study the works of Euler intensively. Euler's solutions rely on the McLaren series for sine of x. Let y equal sine of x, and so we have. Now Euler didn't use the factorial notation, but we will just to simplify our expressions. Dividing by y and rearranging gives us, if we assume we can factor the series as an infinite degree polynomial, which we can, although that's not known until much later, then we note the following. Since the constant term is 1, each factor can be expressed as 1 minus x over r, where r is a root of the polynomial. So we'll write that this way. Now if we expand this polynomial, we get, but since this is supposed to be equal to our power series, we can compare the coefficients of x to get Euler's result that 1 over y is the sum of the reciprocals of a, b, c, and d, and so on, and a, b, c, and d satisfy the equation sine x equals y. At this point, Euler wrote, and for those of us who can't read Latin, what this says is something like the following, let the least positive solution to sine x equal y be a, let half the circumference of the unit circle be p, then the solutions to sine x equal to y are and. And note that Euler's p is the same as our pi. In other words, here are the positive and negative solutions to sine x equal to y. Euler doesn't give an example until later, but one will be helpful at this point, considering the equation sine x equal to 1 half. So again, from our McLaren series expansion, we could write the equation, and we can factor the right-hand side. And again, if we expand the right-hand side, it's supposed to be, and comparing our coefficients gives our relationship. And consequently, two is the sum of the reciprocals, where a, b, c, d, and so on, are the solutions to sine x equal to 1 half. Now those solutions are, and so our series gives us, which we can simplify to.