Q. 15 - Calculus

Question

What is $x_0$ if the power series $\sum _{k=0}^{\mathrm{\infty }} a_k{\left(x-x_0\right)}^k$ converges conditionally at both $x=-4$ and $x=8$.

Step-by-Step Solution

Answer

Ans:  $x_0=2$

Step 1. Given information.

given,

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

Step 2. To find the interval of convergence for the power series, let us first assume b k = a k x − x 0 k , so b k + 1 = a k + 1 x − x 0 k + 1

  Therefore,

 $\begin{array}{r}\underset{k\to \mathrm{\infty }}{lim} \left|\frac{b_{k+1}}{b_k}\right|=\underset{k\to \mathrm{\infty }}{lim} \left|\frac{a_{k+1}{\left(x-x_0\right)}^{k+1}}{a_k{\left(x-x_0\right)}^k}\right|\\ =\underset{k\to \mathrm{\infty }}{lim} \frac{a_{k+1}}{a_k}\left|x-x_0\right|\\ =\frac{a_{k+1}}{a_k}\underset{k\to \mathrm{\infty }}{lim} \left|x-x_0\right|\end{array}$

Step 3. Now,

Here the limit is $\left|x-x_0\right|$. So, by the ratio test of absolute convergence, we know that the series will converge absolutely, when $\left|x-x_0\right|<1$, that is $-1<x-x_0<1$

Implies that

  $-1+x_0<x<1+x_0$

Step 4. Thus,

Now, to find the value of $x_0$, such that the series $\sum _{k=0}^{\mathrm{\infty }} a_k{\left(x-x_0\right)}^k$ convergence conditionally at both $x=-4$ and $x=8$, the interval of convergence must satisfy

$-1+x_0=-4\text{ and }1+x_0=8$

Hence, solving for $x_0$

  $x_0=2$