Q5 TUT 5201

[Victor Ivrii]

Find the power-series expansion about the given point for the given function; find the largest disc in which the series is valid:
 $$\frac{z+2}{z+3}\qquad\text{about}\; z_0 = -1.$$

[Tianfangtong Zhang]

\begin{align*}
 \frac{z+2}{z+3} &= \frac{z+3-1}{z+3} \\ &= 1-\frac{1}{z+3}\\
 &= 1- \frac{1}{z+1+2}\\ &= 1 - \frac{1}{2} \frac{1}{1+\frac{z+1}{2}}\\
 &= 1 - \frac{1}{2} \frac{1}{1-\frac{-(z+1)}{2}}\\
 &= 1 - \sum_{n=0}^{\infty}(\frac{-(z+1)}{2})^n \\
 &= 1 - \sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}}(z-(-1))^n
\end{align*}

[ZhenDi Pan]

We can first rewrite $\frac{z+2}{z+3}$ as
\begin{equation}
\frac{z+2}{z+3} = \frac{z+3-1}{z+3}
\end{equation}
\begin{align*}
\frac{z+3-1}{z+3} & = 1 - \frac{1}{z+3} \\
& = 1 - \frac{1}{z+1+2} \\
& = 1 - \frac{1}{2} \times \frac{1}{1+\frac{(z+1)}{2}} \\
& = 1 - \frac{1}{2} \times \frac{1}{1-\frac{-(z+1)}{2}}
\end{align*}
The power series expansion is
\begin{equation}
1 - \frac{1}{2} \times \frac{1}{1-\frac{-(z+1)}{2}} = 1 - \frac{1}{2} \times \sum_{n=0}^{\infty} (\frac{-(z+1)}{2})^n \\
= 1 - \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{n+1}}(z-(-1))^n = 1 - \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{n+1}}(z+1)^n
\end{equation}
The largest disc in which the series is valid:
\begin{equation}
\mid -\frac{z+1}{2}\mid < 1 \\
\mid z+1\mid < 2
\end{equation}

[Heng Kan]

See the attatched scanned picture. A different way to get the expansion.