Q. 18 - Calculus

Question

Let $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^k$ be a power series in x with a radius of convergence $p$. What is the radius of convergence of the power series $\sum _{k=0}^{\mathrm{\infty }} a_k{\left(x-x_0\right)}^k$? Make sure you justify your answer.

Step-by-Step Solution

Answer

Ans: The radius of convergence of the power series $\sum _{k=0}^{\mathrm{\infty }} a_k{\left(x-x_0\right)}^k$ is $p$.

Step 1. Given information.

given,

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

Step 2. Solution.

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

Therefore, 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, $\frac{a_k}{a_{k+1}}$ 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 $p$.

Therefore,

    $\frac{a_k}{a_{k+1}}=p$

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

  So, here

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

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

   $\left|x-x_0\right|<\frac{a_k}{a_{k+1}}$

 

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

Thus,

  $|x-x_0|<p$

Step 4. Thus,

Therefore, the radius of convergence of the power series $\sum _{k=0}^{\mathrm{\infty }} a_k{\left(x-x_0\right)}^k$ is $p$.