find the radius of convergence for 1st derivative of f.

**ewkimchi** · June 2nd 2010, 12:10 AM

If f = n=0, infinite ((-1)^(n+1))(x-1)^(n))/n, find the radius of convergence for 1st derivative of f.

**Prove It** · June 2nd 2010, 12:29 AM

Originally Posted by ewkimchi

If f = n=0, infinite ((-1)^(n+1))(x-1)^(n))/n, find the radius of convergence for 1st derivative of f.

$f(x) = \sum_{n = 0}^{\infty}\frac{(-1)^{n + 1}(x - 1)^n}{n}$

$f'(x) = \sum_{n = 0}^{\infty}\frac{(-1)^{n + 1}n(x - 1)^{n - 1}}{n}$

$= \sum_{n = 0}^{\infty}(-1)^{n + 1}(x - 1)^{n - 1}$ .

The radius of convergence of $f'(x)$ is given where

$\lim_{n \to \infty}\left|\frac{t_{n + 1}}{t_n}\right| < 1$

$\lim_{n \to \infty}\left|\frac{(-1)^{n + 2}(x - 1)^{n}}{(-1)^{n + 1}(x - 1)^{n - 1}}\right| < 1$

$\lim_{n \to \infty}\left|x - 1\right|< 1$

$|x - 1| < 1$

$-1 < x - 1 < 1$

$0 < x < 2$ .

The radius of convergence is $1$ , and the interval of convergence is $0 < x < 2$ .

Thread: find the radius of convergence for 1st derivative of f.

LinkBack

Thread Tools

Display

find the radius of convergence for 1st derivative of f.

Similar Math Help Forum Discussions

derivative and radius of convergence

Find the radius of convegence and the radius of convergence

find radius of convergence

find its radius of convergence

find radius of convergence

Search Tags