### Author Topic: LEC5101-TT3-ALT-F-Q2  (Read 725 times)

#### RunboZhang

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##### LEC5101-TT3-ALT-F-Q2
« on: November 27, 2020, 08:03:20 PM »
$\textbf{Problem2:} \\ \\ \text{Find the power series expansion at z=0 of}$
$$f(z) = \frac{1}{z} \int_{0}^{z}{\frac{1}{z}\int_{0}^{z}{cos(w^2)\,dw}\,dz}$$
$\text{What is the radius of convergence?}$

$\textbf{Solution:} \\ \\ \text{Firstly, we have power series,}$
$$cos(w) = \sum_{n=0}^{\infty}{(-1)^{n}\frac{w^{2n}}{(2n)!}} \\$$
$\text{with a radius of convergence of } \infty \\$

$\text{Then substitute } w \text{ by } w^2$
$$cos(w^2) = \sum_{n=0}^{\infty}{(-1)^{n}\frac{w^{4n}}{(2n)!}}$$

$\text{Now substitute the series in the integral and calculate the integral}$

\begin{gather} \begin{aligned} f(z) &= \frac{1}{z} \int_{0}^{z}{\frac{1}{z}} \int_{0}^{z}{cos(w^2)\,dw}\,dz \\\\ &= \frac{1}{z} \int_{0}^{z}{\frac{1}{z}}(\int_{0}^{z} {\sum_{n=0}^{\infty}{\frac{(-1)^{n}w^{4n}}{(2n)!}\,}}\,dw) \,dz \\\\ &= \frac{1}{z} \int_{0}^{z}{\frac{1}{z}}\sum_{n=0}^{\infty}{ \int_{0}^{z}{(-1)^{n} \cdot \frac{w^{4n}}{(2n)!}}\,dw}\,dz \\\\ &=\frac{1}{z} \int_{0}^{z}{\frac{1}{z}} \sum_{n=0}^{\infty}{(-1)^{n}\cdot \frac{z^{4n+1}}{(2n)! \cdot (4n+1)}}\,dz \\\\ &= \frac{1}{z} \int_{0}^{z}{\sum_{n=0}^{\infty}{(-1)^{n} \frac{z^{4n}}{(2n)! \cdot (4n+1)}}} \,dz\\\\ &= \frac{1}{z} \sum_{n=0}^{\infty}{\int_{0}^{z}{(-1)^{n} \frac{z^{4n}}{(2n)! \cdot (4n+1)}}}\,dz \\\\ &= \frac{1}{z} \sum_{n=0}^{\infty}{(-1)^{n} \frac{z^{4n+1}}{(2n)! \cdot (4n+1)^{2}}} \\\\ &= \sum_{n=0}^{\infty}{(-1)^{n} \frac{z^{4n}}{(2n)! \cdot (4n+1)^{2}}} \end{aligned} \end{gather}

$\text{Hence the power series expansion is }$
$$f(z) = \sum_{n=0}^{\infty}{(-1)^{n} \frac{z^{4n}}{(2n)! \cdot (4n+1)^{2}}}$$

$\text{Since the radius of convergence of } cos(z) \text{ is } \infty \text{. Thus } f(z) \text{ has the same radius of convergence of } R=\infty.$