Q. 4 - Calculus

Question

If the series $\sum _{k=0}^{\infty }{b}_k{\left(x-3\right)}^k$ converges to the function $g\left(x\right)$ on the interval (−2, 8), provide a formula for $b_k$ in terms of the function g. 

Step-by-Step Solution

Answer

The formula for $b_k$ is $\frac{f^{\left(k\right)}\left(3\right)}{k!}$.

Step 1. Given Information.

The series $\sum _{k=0}^{\infty }{b}_k{\left(x-3\right)}^k$ is converges to $g\left(x\right)$on $\left(-2,8\right)$. 

Step 2. Explanation.

The series is converges to the function so, Taylor polynomial for $g\left(x\right)$ is in the form $\sum _{k=0}^{\infty }\frac{f^{\left(k\right)}{x}_0}{k!}\left(x-x_0\right)$.

For series $\sum _{k=0}^{\infty }{b}_k{\left(x-3\right)}^k$, So, $b_k=\frac{f^k\left(3\right)}{k!}$