Question

$Letf\left(x\right)=\infty$$k=0ak(x-x0)$k and let G be an antiderivative

for f . Explain why we do not have enough information to

determine $G(x0).$What is G $(x0)WhatisG$$(x0)WhatisG\left(k\right)(x0)$?

Short answer

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

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

$G^{\text{'}\text{'}}\left(x_0\right)=a_1$

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

Step-by-step solution

given infomation

Let us consider the power series for the function

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

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

Implies that

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

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

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

That is,

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

Where $\mathrm{C}$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 $G$

Now, take the derivative of the function$f\left(x\right)$to find$f^{\text{'}}\left(x\right)$

Thus,

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

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

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

Change the index

So,

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

That is,

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

$=a_1·1·{\left(x-x_0\right)}^0+a_2·2·{\left(x-x_0\right)}^1+a_3·3·{\left(x-x_0\right)}^1+\dots$

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

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

solution

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,

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

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

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

Change the index

Thus,

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

Therefore,

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

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