Q. 2 - Calculus

Question

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 indefinite integral of \(f\) is given by \(\int f\left ( x \right )dx =\) _____.

The interval of convergence for this new series is ______, with the possible exception that _____.

Step-by-Step Solution

Answer

The indefinite integral of \(f\) is given by \(\int f\left ( x \right )dx =\sum_{k=0}^{\infty }\frac{a_{k}}{k+1}\left ( x-x_{0} \right )^{k+1}+C\).

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.

Step 1. The indefinite integral of \(f\)

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

So, the indefinite integral of the function is

\(\int f\left ( x \right )dx =\int \left ( \sum_{k=0}^{\infty }a_{k}\left ( x-x_{0} \right )^{k} \right )dx\)

\(\int f\left ( x \right )dx =\sum_{k=0}^{\infty }a_{k}\int \left ( x-x_{0} \right )^{k}dx\)

\(\int f\left ( x \right )dx =\sum_{k=0}^{\infty }\frac{a_{k}}{k+1}\left ( x-x_{0} \right )^{k+1}+C\)

Step 2. The indefinite integral of \(f\)

Therefore, the indefinite integral of \(f\) is given by \(\int f\left ( x \right )dx =\sum_{k=0}^{\infty }\frac{a_{k}}{k+1}\left ( x-x_{0} \right )^{k+1}+C\).

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.