Question

If the series $\underset{k-0}{\overset{\infty }{\sum a_k{x}^k}}$converges to the function $f\left(x\right)$ on the interval (−2, 2), provide a formula for $a_k$in terms of the function f .

Short answer

The formula for$a_k$is$\frac{f^{\left(k\right)}\left(0\right)}{k!}$

Step-by-step solution

Step 1. Given information

The series is $\underset{k-0}{\overset{\infty }{\sum a_k{x}^k}}$is convergent to f(x) on interval (-2,2).

Step 2. Explanation.

The series is converges to the function so, Maclaurin polynomial for f(x) is in the form $P_n\left(x\right)=\sum _{k=0}^n\frac{f^{\left(k\right)}\left(0\right)}{k!}{x}^k$.

For series$\underset{k-0}{\overset{\infty }{\sum a_k{x}^k}}$,$a_k=\frac{f^k\left(0\right)}{k!}$