Wikipedia:Reference desk/Archives/Mathematics/2018 June 17

= June 17 =

Convergence of Taylor expansion
I asked Wolfram Alpha to show me the Taylor expansion of x^.5 at 1 so I could check my math. But it included a region of convergence. Can someone tell me how to determine the region of convergence? RJFJR (talk) 23:45, 17 June 2018 (UTC)
 * For one-variable Taylor series, the Cauchy–Hadamard theorem provides a straightforward formulaic way to determine the radius of convergence. In this case, we have a Taylor series given by $$\sqrt{x} = -\sum_{n = 0}^\infty\frac{(2n - 3)!!}{n!(-2)^n}(x-1)^n = -\sum_{n = 0}^\infty\frac{(2n - 3)!!}{(2n)!!(-1)^n}(x-1)^n$$ from which we clearly see that the limit superior of the nth root of the absolute value of the nth term is 1 as $$n \to \infty$$, yielding a radius of convergence of 1. This can perhaps more easily be demonstrated using the ratio test. Since we took the expansion about $$x = 1$$, the result follows.--Jasper Deng (talk) 00:14, 18 June 2018 (UTC)
 * Thank you. RJFJR (talk) 03:09, 18 June 2018 (UTC)