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Expert-verified Found in: Page 703 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # The Calculus of Power Series: Let $$\sum_{k=0}^{\infty }a_{k}\left ( x-x_{0} \right )^{k}$$ be a power series in $$x-x_{0}$$ that converges to a function $$f(x)$$ on an interval $$I$$.The derivative of $$f$$ is given by $$f^{\prime}\left( x \right )=$$____.The interval of convergence for this new series is ______, with the possible exception that _____.

The derivative of $$f$$ is given by $$f^{\prime}\left( x \right )=\sum_{k=1}^{\infty }ka_{k}\left ( x-x_{0} \right )^{k-1}$$.

The interval of convergence for this new series is $$I$$, with the possible exception that if $$I$$ is a finite interval, the convergence at the endpoints of $$I$$ may be different than it was for the initial series.

See the step by step solution

## Step 1. Derivative of $$f$$

The given function is $$\sum_{k=0}^{\infty }a_{k}\left ( x-x_{0} \right )^{k}$$.

So, the derivative of the function is

$$f^{\prime}\left ( x \right )=\frac{d}{dx}\left ( \sum_{k=0}^{\infty }a_{k}\left ( x-x_{0} \right )^{k} \right )$$

$$f^{\prime}\left ( x \right )=\sum_{k=0}^{\infty }a_{k}\frac{d}{dx}\left ( \left ( x-x_{0} \right )^{k} \right)$$

$$f^{\prime}\left ( x \right )=\sum_{k=1}^{\infty }ka_{k}\left ( x-x_{0} \right )^{k-1}$$

## Step 2. Derivative of $$f$$

So, the derivative of $$f$$ is given by $$f^{\prime}\left( x \right )=\sum_{k=1}^{\infty }ka_{k}\left ( x-x_{0} \right )^{k-1}$$.

The interval of convergence for this new series is $$I$$, with the possible exception that if $$I$$ is a finite interval, the convergence at the endpoints of $$I$$ may be different than it was for the initial series. ### Want to see more solutions like these? 