Question: Suppose we have a function f defined by the power series f(x)=ak(x-a)k k=0 on its interval of convergence I. It turns out that f



Suppose we have a function f defined by the power series f(x)=ak(x-a)k k=0 on its interval of convergence I. It turns out that f is differentiable on I, and f'(x) = kak(x a)-1, for x I. k=1 That is, differentiating a function defined by a power series is the same as differentiating the power series term-by-term inside the interval of convergence. We can use this to obtain closed-form expressions for some new power series from old ones we already know. Likewise, if we again have f defined by the power series f(x)=ak(x-a)k k=0 on the interval of convergence I, then is integrable on I, and [f(x) dx = ak -(x a) * +1 + C, for x I, k+1 and C is a constant of integration. That is, as with differentiation, integrating a function defined by a power series is the same as integrating the power series term-by-term inside the interval of convergence. In the following, use the fact that we know 1 and some clever substitutions to obtain closed-form expressions for the following related infinite series: (i) 1-2+22 +...= (-1)** k=0 (iii) 1-2+z +...= (-1)2k k=0 Now, if we integrate the last formula above (noticing that there is no constant of integration on either side), we get: 25 x7 + 3 5 7 k=0 (-1)*2k+1 2k+1 This series was originally called Gregory's series, named after the Scottish mathematician James Gregory (1638- 1675). Since it was first discovered by the Indian mathematician Madhava of Sangamagrma (c.1340-c.1425), it is also referred to as the Madhava-Gregory series. When we substitute x = 1 into the Madhava-Gregory series, we get the famous and surprising formula 11 1- + 3 5 1 7 (-1)* 2k+1 which has many names, one of which is the Madhava-Leibniz formula for *, named for German mathematician Gottfried Leibniz (1646-1716).
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