Term Test 2 sample P2

[Victor Ivrii]

(a) Find the decomposition into power series at ${z=0}$ of $f(z)=2(1-z)^{1/2}$. What is the radius of convergence?

(b) Plugging in $z^2$ instead of $z$ and integrating, obtain a decomposition at $z=0$ of  $F(z)=z\sqrt (1-z^2)+\arcsin (z)$.

[ZhenDi Pan]

First we have
\begin{equation}
f(z)=2(1-z)^{1/2}
\end{equation}
Then the $nth$ derivative of $f(z)$ can be derived as
\begin{equation}
f^\prime(z) = -(1-z)^{-1/2} \\
f''(z) = -\frac{1}{2}(1-z)^{-3/2} \\
f'''(z) = -\frac{1}{2} \times  \frac{3}{2}(1-z)^{-5/2} \\
f^{(n)}(z) =-(\frac{1}{2} \times \frac{3}{2} \times \dots \times \frac{2n-3}{2})(1-z)^{-(2n-1)/2}
\end{equation}
At $z=0$
\begin{equation}
f(0) = 2
f'(0) = -1 \\
f''(0) = -\frac{1}{2} \\
f'''(0) = -\frac{1}{2} \times \frac{3}{2} \\
f''''(0) =  -\frac{1}{2} \times \frac{3}{2} \times \frac{5}{2} \\
f^{(n)}(0) =   -\frac{1}{2} \times \frac{3}{2} \times \dots \times \frac{2n-3}{2}
\end{equation}
Thus we have the power series
\begin{equation}
f(z)= \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}z^n = 2 - z -\frac{1}{4}z^2 - \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}

[Ende Jin]

we know
\begin{align*}
    \int \sqrt{1 - z^2} dz &= z \sqrt{1 - z^2} - \int z d \sqrt{1-z^2} \\
    & = z \sqrt{1-z^2}  + \int \frac{1}{\sqrt{1 - z^2}}dz - \int \sqrt{1 - z^2} dz
\end{align*}
Thus
\begin{align*}
    2 \int \sqrt{1 - z^2}dz = z\sqrt{1-z^2} + \arcsin z
\end{align*}

Thus
\begin{align*}
    F(z) &= \int f(z^2) \\
    & = \int (\sum_{n=1}^\infty \frac{f^{(n)}(0)}{n!}z^{2n} +2 )dz \\
    & = 2z + \sum_{n=1}^\infty \frac{f^{(n)}(0)}{n!} \frac{1}{2n+1} z^{2n+1} \\
    & = 2z + \sum_{n=1}^\infty \frac{1}{n!} \frac{1}{2n+1} (-1) \frac{1}{2^{n-1}} \prod_{j=1}^{n-1}(2j-1) z^{2n+1} \\
    &= 2z -  \sum_{n=1}^\infty \frac{1}{n!} \frac{1}{2n+1} \frac{\prod_{j=1}^{n-1}(2j-1) }{2^{n-1}} z^{2n+1}
\end{align*}

[Mengyang Li]

For question b, you forgot to plus constant C when you do integral

[Victor Ivrii]

Use $k!!=k(k-2)\cdots l$ where $s=1$ for $k$ odd and $s=2$ for $k$ even; then product becomes $(2n-3)!!$ . However, neither makes sense as $n=1$ and you need to write the term with $n=1$ separately and start summation from $n=2$;
since $\sqrt{1-z^2}= 1-\frac{1}{2}z^2+\ldots$, $F(z)= z-\frac{1}{6}z^3+\ldots$ where $...$ denote terms with $n\ge 2$ you wrote

There will be no constant since integral is definite (from $0$)