Q. 5 - Calculus

Question

Let $f\left(x\right)=\sum _{k=0}^{\infty }{a}_k{\left(x-x_0\right)}^k$ and let $G$ be the antiderivative for $f$ with the property that $G\left(x_0\right)=7$. Find the Taylor series in$x_0$ for $G$.

Step-by-Step Solution

Answer

The Taylor series in $x_0$ for G is  $G\left(x\right)=\sum _{k=0}^{\infty }\frac{a_k}{k+1}{\left(x-x_0\right)}^k+7$.

Step 1. Given information

The function is $f\left(x\right)=\sum _{k=0}^{\infty }{a}_k{\left(x-x_0\right)}^k$.

Find the Taylor series in $x_0$ for $G$?

Step 2. Simplification

Let's take a look at the function's power series$f\left(x\right)=\sum _{k=0}^{\infty }{a}_k{\left(x-x_0\right)}^k$

Given that $G$ is the antiderivative of f.

$G\left(x\right)=\int f\left(x\right)dx$

Where $G\left(x_0\right)=7$

Thus, the Taylor series in $x_0$ for$G$ is

$\begin{array}{rcl}G& =& \int \left[\sum _{k=0}^{\infty }{a}_k{\left(x-x_0\right)}^k\right]dx\\ & =& \sum _{k=0}^{\infty }{a}_k\int {\left(x-x_0\right)}^{k}dx\\ & =& \sum _{k=0}^{\infty }{a}_k\frac{{\left(x-x_0\right)}^{k+1}}{k+1}+C\end{array}$

Where C is the constant of integration,

Step 3. Find the Taylor series

Change the index of the power series you've created now.

So,

$G=\sum _{k=0}^{\infty }\frac{a_k}{k+1}{\left(x-x_0\right)}^k+C$

The value C can be found in this case by using$G\left(x_0\right)=7$

Thus,

$G\left(x_0\right)=\sum _{k=0}^{\infty }\frac{a_k}{k+1}{\left(x_0-x_0\right)}^k+C$

It signifies that

$7=C$

Therefore,$G\left(x\right)=\sum _{k=0}^{\infty }\frac{a_k}{k+1}{\left(x-x_0\right)}^k+7$