Author Topic: Test 1 Question 2  (Read 280 times)

Jiayu Chen

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Test 1 Question 2
« on: November 16, 2020, 11:20:31 AM »
Can somebody help me with question2? I lost mark in calculating R.

RunboZhang

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Re: Test 1 Question 2
« Reply #1 on: November 16, 2020, 01:47:37 PM »
$
\textbf{For part (a):} \\\\
\text{By ratio test, we have : }\\\\
$

$
\begin{gather}
\begin{aligned}

\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| &= \lim_{n \to \infty}|\frac{(1+ i\sqrt{3})^{n+1}}{4^{n+1}(n+1)log^{3}(n+1)} \cdot \frac{n 4^{n} log^{3}(n)}{(1+i\sqrt{3})^{n}}| \\\\
&= \lim_{n \to \infty} |\frac{1+i\sqrt{3}}{4} \cdot \frac{nlog^{3}(n)}{(n+1)log^3(n+1)}| \\\\
&= \frac{1}{2} \text{(by further expanding this absolute value)}\\\\
&= \frac{1}{R}

\end{aligned}
\end{gather}
$

$\text{Therefore the radius of convergence is 2.}$


$
\textbf{For part (b):} \\\\
\text{We can follow the similar method. By ratio test, we have : }\\\\
$

$
\begin{gather}
\begin{aligned}

\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| &= \lim_{n \to \infty} |\frac{(i+\sqrt{3})^{n+1} \cdot \frac{e^{2(n)+1}-1}{e^{2(n+1)}+1}}{4^{n+1}} \cdot \frac{4^n}{(i+\sqrt{3})^{n}\cdot \frac{e^{2n}-1}{e^{2n}+1}}| \\\\
&= \lim_{n \to \infty} |\frac{i+\sqrt{3}}{4} \cdot \frac{e^{2n+2}-1}{e^{2n+2}+1} \cdot \frac{e^{2n}+1}{e^{2n}-1}| \\\\
&= \frac{1}{2} \text{(by further expanding this absolute value, and most terms can be canceled out.)}\\\\
&= \frac{1}{R}

\end{aligned}
\end{gather}
$

$\text{Therefore the radius of convergence is 2.} \\\\$

$\text{Above is how I got my answers. Correct me if I made any mistakes. Feel free to comment below :) }$
« Last Edit: November 16, 2020, 01:53:28 PM by RunboZhang »