Author Topic: TT2 Problem 2  (Read 736 times)

Victor Ivrii

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TT2 Problem 2
« on: November 24, 2018, 04:28:29 AM »
(a) Find the decomposition into power series at ${z=0}$ of $f(z)=(1-z)^{-\frac{1}{2}}$. What is the radius of convergence?

(b) Plugging in $z^2$ instead of $z$ and integrating, obtain a decomposition at $z=0$ of  $\arcsin (z)$.

ZhenDi Pan

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Re: TT2 Problem 2
« Reply #1 on: November 24, 2018, 04:47:23 AM »
For question a, we have
\begin{equation}
f(z)=(1-z)^{-1/2} \\
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}{2}(1-z)^{-3/2} \\
f''(z) = \frac{3}{4}(1-z)^{-5/2} \\
f'''(z) = \frac{15}{8} \times  (1-z)^{-7/2} \\
f''''(z) =\frac{105}{16} \times (1-z)^{-9/2}
\end{equation}
At $z=0$
\begin{equation}
f(0) = 1
f'(0) = \frac{1}{2} \\
f''(0) = \frac{3}{4} \\
f'''(0) = \frac{15}{8} \\
f''''(0) =  \frac{105}{16} \\
f^{(n)}(0) =  \frac{1 \times 3 \times \dots \times (2n-1)}{2^n} \\
a_n = \frac{1 \times 3 \times 5 \times \dots \times (2n-1)}{2^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}{2} + \frac{3z^2}{8}+ \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{2n+1}{2(n+1)} \times \frac{1}{1} \mid \\
\frac{1}{R} = \lim_{n \to \infty} \mid \frac{2n+1}{2n+2} \mid = 1 \\
R = 1
\end{equation}

For question b, let
\begin{equation}
F(z) = \arcsin(z) \\
F'(z)=\frac{1}{\sqrt{1-z^2}} = (1-z^2)^{-1/2}
\end{equation}
Note that
\begin{equation}
f(z^2)=(1-z^2)^{-1/2} \Rightarrow F'(z)=f(z^2) \\
F(z) = \int f(z^2)
\end{equation}
Then
\begin{equation}
f(z^2) = \sum_{n=0}^{\infty}\frac{1\times 3 \times 5 \dots \times (2n-1)}{2^n n!}z^2n \\
F(z) = \int f(z^2) = \int \sum_{n=0}^{\infty}\frac{1\times 3 \times 5 \dots \times (2n+1)}{2^n n!}z^{2n} dz
F(z) = (\sum_{n=0}^{\infty}\frac{1\times 3 \times 5 \dots \times (2n-1)}{2^n n! \cdot (2n+1)}z^{2n+1}) +C
\end{equation}
Since $F(0) = 0 \Rightarrow C=0$
\begin{equation}
F(z) = \sum_{n=0}^{\infty}\frac{1\times 3 \times 5 \dots \times (2n+1)}{2^n n! \cdot (2n+!)}z^{2n+1}
\end{equation}

hanyu Qi

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Re: TT2 Problem 2
« Reply #2 on: November 24, 2018, 05:33:36 PM »
Can we use geometric series on f(z) and assume |z|<1, then we can write 1/√(1-z) directly into Laurent series.

How?

 However the series decomposition may be different but the radius of convergence is the same.

We are not looking for just radius. Basically this post was a flood. V.I.
« Last Edit: November 29, 2018, 07:34:53 AM by Victor Ivrii »