Author Topic: TT2A Problem 2  (Read 557 times)

Victor Ivrii

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TT2A Problem 2
« on: November 24, 2018, 04:55:21 AM »
(a) Find the decomposition into power series at ${z=0}$ of $$f(z)=(1-z)^{-1/3}.$$ What is the radius of convergence?

(b) Plugging in $z^2$ instead of $z$, integrating and multi[lying by $z^{-1}$, obtain a decomposition at $z=0$ of 
$$F(z)=\frac{1}{z}\int_0^z (1-z^2)^{-1/3}\,dz$$ (which is the special case of the generalized hypergeometric function).

Jihang Yu

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Re: TT2A Problem 2
« Reply #1 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}

Jihang Yu

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Re: TT2A Problem 2
« Reply #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*}