Q.14 - Calculus

Question

Perform the following steps for the power series in x − x 0 in

Exercises 11–16:

(a) Find the interval of covergence, I, for the series.

(b) Let f be the function to which the series converges on I.

Find the power series in x − x 0 for f 

(c) Find the power series in uncaught exception: Http Error #500

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x 0 f(t) dt

$\sum _{k=0}^{\infty }\frac{1}{k+2}{x}^{3k+1}$

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

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

part (a) step 1 : solution

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

To find the interval of convergence of the power series, use the ratio test for absolute convergence

Let us first assume $b_i=\frac{1}{k+2}{x}^{3+1}$, so $b_{k+1}=\frac{1}{k+3}{x}^{3k+4}$

Therefore,

$\underset{k\to \infty }{\lim }\left|\frac{b_{k+1}}{b_k}\right|=\underset{k\to \infty }{\lim }\left|\frac{\frac{1}{k+3}{x}^{3k+4}}{\frac{1}{k+2}{x}^{3+1}}\right|$

$=\underset{k\to \infty }{\lim }\frac{k+2}{k+3}\left|x^3\right|$

Now, by the ratio test for absolute convergence, the series will converge only when $|x{|}^3<1$ Therefore,

$x\in (-1,1)$

Now, we check the series at the end points

So, when $x=-1$

${\left.\sum _{k=0}^{\infty }\frac{1}{k+2}{x}^{3+1}\right|}_{x=-1}=\sum _{k=0}^{\infty }\frac{1}{k+2}(-1{)}^{3k+1}$

$=\sum _{k=0}^{\infty }\frac{1}{k+2}$

Therefore, the interval of convergence of the power series

$f\left(x\right)=\sum _{k=0}^{\infty }\frac{1}{k+2}{x}^{3+1}is[-1,1)$

part (b) step 1 : solution

 Since$f\left(x\right)=\sum _{k=0}^{\infty }\frac{1}{k+2}{x}^{3+1},$ so to find the power series in $x-x_0$ for $f^{\text{'}},$ let us take the derivative of the function $f\left(x\right)$

Therefore,

$f^{\text{'}}\left(x\right)=\frac{d}{dx}\left[\sum _{k=0}^x\frac{1}{k+2}{x}^{3k+1}\right]$

$=\sum _{k=0}^{\infty }\frac{1}{k+2}\frac{d}{dx}{x}^{3k+1}$

$=\sum _{k=0}^{\infty }\frac{1}{k+2}(3k+1)x^{3k}$

$=\sum _{k=0}^{\infty }\frac{3k+1}{k+2}{x}^{3k}$

Now, we change the index in the final step

So, the power series in $x-x_{0}f$or $f^{\text{'}}$ is

$f^{\text{'}}\left(x\right)=\sum _{k=1}^{\infty }\frac{3k+4}{k+3}{x}^{3k+1}$

part (c) step 1 :solution

Also, to find the power series in $x-x_0$ for F, let us integrate the function $f\left(x\right)$ from $x_0$ to x

Therefore,

$F\left(x\right)={\int }_{x_0}^x\left[\sum _{k=0}^{\infty }\frac{1}{k+2}{x}^{u+1}\right]dt$

$=\sum _{k=0}^{\infty }\frac{1}{k+2}{\int }_{x_0}^x\left[t^{3k+1}\right]dt$

$=\sum _{k=0}^{\infty }\frac{1}{k+2}{\left[\frac{t^{3k+2}}{3k+2}\right]}^x$

Thus,

$F\left(x\right)=\sum _{k=0}^{\infty }\frac{1}{(k+2)(3k+2)}{x}^{3+2}$

Now, we change the index in the final step

So, the power series in $x-x_0$ for $F\left(x\right)={\int }_{x_0}^{x}f\left(t\right)dt$ is

$F\left(x\right)=\sum _{k=1}^{\infty }\frac{1}{(k+2)(3k+2)}{x}^{3+2}$