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.

Short answer

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

Step-by-step solution

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 role="math" localid="1649488146257" $\sum _{k=0}^{\infty }\frac{f^{\left(k\right)}{x}_0}{k!}\left(x-x_0\right)$.

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