Toronto Math Forum
MAT334-2018F => MAT334--Tests => Term Test 2 => Topic started by: Victor Ivrii on November 24, 2018, 04:28:29 AM
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(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 $\arcsin (z)$.
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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}
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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.