Q.4 - Calculus

Question

Let$f\left(x\right)=\sum _{k=0}^{\infty }{a}_k{\left(x-x_0\right)}^k$and let G be an antiderivative for f. Explain why we do not have enough information to determine $G\left(x_0\right).What is G^{\text{'}}\left(x_0\right)?$ What is $G^{\text{'}\text{'}}\left(x_0\right)?$ What is $G^{\left(k\right)}\left(x_0\right)?$

Step-by-Step Solution

Answer

Let us consider the function such that that

$f\left(0\right)=1\text{ and }{f}^{\text{'}}\left(x\right)=-3$

Consider the definite integral

$I={\int }_0^{05}\frac{dx}{1+x^3}$

step 1: given information

Let us consider the power series for the function

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

Since G is the antiderivative for the function $f\left(x\right)$, so take the integration of the function with respect to x to calculate G

Therefore,

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

Implies that

$G=\int \left[\sum _{k=0}^{\infty }{a}_k{\left(x-x_0\right)}^k\right]$

$=\sum _{k=0}^{\infty }{a}_k\int {\left(x-x_0\right)}^{k}dx$

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

That is,

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

Where C is the constant of integration

So, to calculate the value of the constant of integration C, we must have additional information about the initial value of the function G.

Since the power series can also be written as

$f\left(x\right)=a_0+a_1{\left(x-x_0\right)}^1+a_2{\left(x-x_0\right)}^2+\dots$

Therefore, $G^{\text{'}}\left(x_0\right)=f\left(x_0\right)$ is

$G^{\text{'}}\left(x_0\right)=a_0$

step 2: given information

Similarly, $G^m\left(x_0\right)$ can be calculated by takin the derivative of the function $G^{\text{'}\text{'}}\left(x\right)$ and then substituting $x=x_0$

That is,

$G^{\text{'}\text{'}}\left(x\right)=\frac{d}{dx}\left[\sum _{k=0}^{\infty }{a}_{k+1}(k+1){\left(x-x_0\right)}^k\right]$

$=\sum _{k=0}^{\infty }{a}_{k+1}(k+1)\frac{d}{dx}{\left(x-x_0\right)}^k$

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

Change the index

Thus,

$G^{\text{'}\text{'}}\left(x\right)=\sum _{k=0}^{\infty }{a}_{k+2}(k+2)(k+1){\left(x-x_0\right)}^2$

Therefore,

$G^m\left(x_0\right)=2a_2$

Similarly, $G^{\left(k\right)}\left(x_0\right)$can be calculated by differentiating $G\left(x\right)$k number of times and then substituting $x=x_0$

Therefore,

$G^{\left(k\right)}\left(x_0\right)=(k-1)!a_k$