$\textbf{Problem2:} \\ \\
\text{Find the power series expansion at z=0 of}$
$$f(z) = \frac{1}{z} \int_{0}^{z}{\frac{1}{z}\int_{0}^{z}{cos(w^2)\,dw}\,dz}$$
$\text{What is the radius of convergence?}$
$\textbf{Solution:} \\ \\
\text{Firstly, we have power series,}$
$$cos(w) = \sum_{n=0}^{\infty}{(-1)^{n}\frac{w^{2n}}{(2n)!}} \\ $$
$\text{with a radius of convergence of } \infty \\$
$\text{Then substitute } w \text{ by } w^2$
$$cos(w^2) = \sum_{n=0}^{\infty}{(-1)^{n}\frac{w^{4n}}{(2n)!}}$$
$\text{Now substitute the series in the integral and calculate the integral}$
$
\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}
$
$\text{Hence the power series expansion is }$
$$f(z) = \sum_{n=0}^{\infty}{(-1)^{n} \frac{z^{4n}}{(2n)! \cdot (4n+1)^{2}}}$$
$\text{Since the radius of convergence of } cos(z) \text{ is } \infty \text{. Thus } f(z) \text{ has the same radius of convergence of } R=\infty.$
