Q. 20 - Calculus

Question

Let $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^k$ be a power series in x with an interval of convergence$[-2,2)$. What is the radius of convergence of the power series $\sum _{k=0}^{\mathrm{\infty }} a_k(x-3{)}^k$? Justify your answer. 

Step-by-Step Solution

Answer

Ans:  The radius of convergence of the power series  $\sum _{k=0}^{\mathrm{\infty }} a_k(x-3{)}^k$ is $2$.

Step 1. Given information.

given,

    $\sum _{k=0}^{\mathrm{\infty }} a_k(x-3{)}^k$

Step 2. Solution:

  So, the radius of convergence of the series is $2$

Therefore, we try to evaluate the constant term in the power series using the radius of convergence. 

Let us consider $b_k=a_k{x}^k so b_{k+1}=a_{k+1}{x}^{k+1}$

Now apply the ratio test for absolute convergence, that is  

  $\underset{k\to \mathrm{\infty }}{lim} \left|\frac{b_{k+1}}{b_k}\right|=\underset{k\to \mathrm{\infty }}{lim} \left|\frac{a_{k+1}}{a_k}x\right|$

So according to the ratio test for absolute convergence, the series will converge only when  $\left|\frac{a_{k+1}}{a_k}x\right|<1$

Implies that $\left|x\right|=\frac{a_k}{a_{k+1}}$

 where, $\left|x\right|=\frac{a_k}{a_{k+1}}$ is the radius of convergence of the power series $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^k$ 

  Since we have already considered the radius of convergence of the power series   $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^k$ is $2$.

Therefore,

   $\frac{a_k}{a_k+1}=2$

Step 3. Now, to find the radius of convergence of the power series &#8721; k = 0 &#8734; &#8202; a k ( x &#8722; 3 ) k

  Again apply the ratio test. So here,

  $\begin{array}{r}\underset{k\to \mathrm{\infty }}{lim} \left|\frac{b_{k+1}}{b_k}\right|=\underset{k\to \mathrm{\infty }}{lim} \left|\frac{a_{k+1}(x-3{)}^{k+1}}{a_k(x-3{)}^k}\right|\\ =\underset{k\to \mathrm{\infty }}{lim} \left|\frac{a_{k+1}}{a_k}(x-3)\right| \end{array}$

So according to the ratio test for absolute convergence, the series will converge only when $\left|\frac{a_{k+1}}{a_k}\left(x-x_0\right)\right|<1$

Implies that 

    $|x-3|<\frac{a_k}{a_{k+1}}$

Plug $\frac{a_k}{a_{k+1}}=2$, from the previous power series

Thus,

  $|x-3|<2$

Step 4. Therefore,

The radius of convergence of the power series $\sum _{k=0}^{\mathrm{\infty }} a_k(x-3{)}^k$ is $2$.