Messages

1

Term Test 2 / Re: TT2A Problem 2

« on: November 24, 2018, 06:03:47 AM »

For question b
\begin{equation}
F(z) = \frac{1}{z}\int^z_0 f(z^2)dz \\
\end{equation}
Note that
\begin{equation}
f(z) = \sum_{n=0}^{\infty}\frac{1\times 4 \times 7 \dots \times (3n-2)}{3^n n!}z^{n} \\
f(z^2)= \sum_{n=0}^{\infty}\frac{1\times 4 \times 7 \dots \times (3n-2)}{3^n n!}z^{2n} \\
\end{equation}
Then
\begin{align*}
F(z) &= \frac{1}{z}\int  \sum_{n=0}^{\infty}\frac{1\times 4 \times 7 \dots \times (3n-2)}{3^n n!}z^{2n} \\
 &= \frac{1}{z} \sum_{n=0}^{\infty} \int  \frac{1\times 4 \times 7 \dots \times (3n-2)}{3^n n!}z^{2n} \\
 &= \frac{1}{z} \sum_{n=0}^{\infty}  \frac{1\times 4 \times 7 \dots \times (3n-2)}{3^n n!(2n+1)}z^{2n+1} \\
 &= \sum_{n=0}^{\infty}  \frac{1\times 4 \times 7 \dots \times (3n-2)}{3^n n!(2n+1)}z^{2n}
\end{align*}

2

Term Test 2 / Re: TT2A Problem 2

« on: November 24, 2018, 05:53:17 AM »

For question a, we have
\begin{equation}
f(z)=(1-z)^{-1/3} \\
a_n = \frac{f^{(n)}(z_0)}{n!} =  \frac{f^{(n)}(0)}{n!}
\end{equation}
Then the $nth$ derivative of $f(z)$ can be derived as
\begin{equation}
f^\prime(z) = \frac{1}{3}(1-z)^{-4/3} \\
f''(z) = \frac{4}{9}(1-z)^{-7/3} \\
f'''(z) = \frac{28}{7} \times  (1-z)^{-10/3} \\
f''''(z) =\frac{280}{81} \times (1-z)^{-13/3}
\end{equation}
At $z=0$
\begin{equation}
f(0) = 1
f'(0) = \frac{1}{3} \\
f''(0) = \frac{4}{9} \\
f'''(0) = \frac{28}{27} \\
f''''(0) =  \frac{280}{81} \\
f^{(n)}(0) =  \frac{1 \times 4 \times \dots \times (3n-2)}{3^n} \\
a_n = \frac{1 \times 4 \times 7 \times \dots \times (2n-1)}{3^n n!}
\end{equation}
Thus we have the power series
\begin{equation}
f(z)= \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}z^n = 1 + \frac{z}{3} + \frac{2z^2}{9}+ \dots
\end{equation}
The radius of convergence is
\begin{equation}
\frac{1}{R} = \lim_{n \to \infty} |\frac{a_{n+1}}{a_{n}}| \\
\frac{1}{R} = \lim_{n \to \infty} \mid \frac{f^{(n+1)}(0)}{(n+1)!} \times \frac{n!}{f^{(n)}(0)} \mid \\
\frac{1}{R} = \lim_{n \to \infty} \mid \frac{3n+1}{3(n+1)} \times \frac{1}{1} \mid \\
\frac{1}{R} = \lim_{n \to \infty} \mid \frac{3n+1}{3n+3} \mid = 1 \\
R = 1
\end{equation}

3

Term Test 2 / Re: TT2 Problem 3

« on: November 24, 2018, 05:21:40 AM »

Now let $z=\frac{1}{w}$, we have
\begin{equation}
f(z)=f(\frac{1}{w})=\frac{1}{w^2}(\frac{1}{w^2}-\pi^2)\frac{\cos^2(\frac{1}{w})}{\sin^2(\frac{1}{w})} \\
\lim_{w \to 0} f(\frac{1}{w}) = \infty \cdot \infty \cdot a \qquad a = \infty, a = constant, a = 0 \\
\lim_{w \to 0} f(\frac{1}{w}) = \infty,0 \Rightarrow \lim_{z \to \infty} f(z) = \infty,0
\end{equation}
So it is an not isolated singularity

4

Term Test 2 / Re: TT2 Problem 5

« on: November 24, 2018, 05:12:43 AM »

There is a typo in (b) where it should be k>=0 rather than k not equal to zero. Sorry about it.

You could correct in the text

5

Term Test 2 / Re: TT2 Problem 5

« on: November 24, 2018, 04:51:46 AM »

We have
\begin{equation}
f(z) = \frac{5}{(z-2)(z+3)} \\
Res(f;2)=\frac{5}{z+3}\bigg\rvert_{z=2} = 1 \\
Res(f;-3)=\frac{5}{z-2}\bigg\rvert_{z=-3} = -1
\end{equation}

Question a
As $\mid z \mid < 2$, $r=0, R=2$, so we have
\begin{equation}
a_k = -(2^{-k-1} \times 1+(-3)^{-k-1} \times -1) = -2^{-k-1} + (-3)^{-k-1}, k \geq 0 \\
f(z) =  \sum_{k=0}^{\infty} a_k z^k = \sum_{k=0}^{\infty} (-2^{-k-1}+(-3)^{-k-1}) z^k
\end{equation}

Question b
As $2 < \mid z \mid < 3$, $r=2, R=3$, so we have $$a_k =
\begin{cases}
2^{k-1} \times 1 & \text{for } k \leq -1 \\
-(-3)^{-k-1} \times (-1) = -3^{-k-1} & \text{for } k \neq 0
\end{cases}
$$
$$a_k =
\begin{cases}
2^{k-1} & \text{for } k \leq -1 \\
-3^{-k-1} & \text{for } k \neq 0
\end{cases}
$$
Therefore
\begin{equation}
f(z) = \sum_{k=-\infty}^{\infty} a_k z^k = \sum_{k=-\infty}^{-1} 2^{k-1} z^k + \sum_{0}^{\infty} -3^{-k-1}z^k
\end{equation}

Question c
As $\mid z \mid >3$, $r=3, R=\infty$. So we have
\begin{align*}
a_k &=2^{k-1} \times 1 + (-3)^{k-1} \times (-1) \qquad k \leq -1 \\
& = 2^{k-1} - (-3)^{k-1} \qquad k \leq -1
\end{align*}
Therefore Line above is correct, line below is wrong (substitution error). V.I.
\begin{equation}
f(z) = \sum_{k=-\infty}^{-1} a_k z^k = \sum_{k=-\infty}^{-1} (2^{k-1} -(-1)^{k-1}) z^k
\end{equation}

6

Term Test 2 / Re: TT2 Problem 3

« on: November 24, 2018, 04:48:57 AM »

We have
\begin{equation}
f(z) = z^2(z^2-\pi^2)\cot^2(z) \\
f(z) = z^2(z^2-\pi^2)\frac{\cos^2(z)}{sin^2(z)} \Rightarrow \sin^2(z) \neq 0 \Rightarrow \sin(z) \neq 0 \Rightarrow z \neq n\pi, n \in \mathbb{Z}
\end{equation}
So the singular points are $z_0 = n\pi, n \in \mathbb{Z}$.

Now consider the part
\begin{equation}
\frac{\cos^2(z)}{\sin^2(z)} = \frac{1-\sin^2(z)}{\sin^2(z)} =\frac{1}{\sin^2(z)} - 1\\
\lim_{z \to z_0} \frac{\cos^2(z)}{\sin^2(z)} = \infty - 1 = \infty
\end{equation}
Notice when $z_0 = n\pi, n=0$, we have $z^2=0$.
When $z_0 = n\pi, n=\pm1$, we have $(z^2 -\pi^2) = 0.$ Therefore for $z_0=n\pi, n=0,\pm1$
\begin{equation}
\lim_{z \to z_0} f(z) = 0
\end{equation}
So there are three removable singularities.

For $z_0=n\pi, n \neq 0,\pm1$
\begin{equation}
\lim_{z \to z_0} f(z) = \infty
\end{equation}
So there are all poles of order 2.