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.

Short 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-by-step solution

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

role="math" localid="1649392578291" $\underset{k\to \mathrm{\infty }}{lim} \left|\frac{b_{k+1}}{b_k}\right|=\underset{k\to \mathrm{\infty }}{lim} \left|\frac{a_{k+1}}{a_k}x\right|$

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

role="math" localid="1649392734689" $\left|x\right|=\frac{a_k}{a_{k+1}}$

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 ∑k=0∞ akx−x0k , 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 role="math" localid="1649393278712" $\frac{a_k}{a_{k+1}}=p$, 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 $p$.