Toronto Math Forum
MAT3342018F => MAT334Lectures & Home Assignments => Topic started by: Kathy Ngo on November 20, 2018, 12:18:52 AM

Can anyone show me how they did part (a)?
I got a series but it's not the same as the solutions in the back of the textbook.

Here is what I did for part (a):
For $1<z<2$, if we follow equation (12) and (13), we have
$Res(\frac{z+2}{\frac{z2}{z+1}};1) = \frac{z+2}{z2}_{z=1}= \frac{1}{3}$
$Res(\frac{z+2}{\frac{z+1}{z2}};2) = \frac{z+2}{z+1}_{z=2}= \frac{4}{3}$
By (12) and (13):
$\sum_{\infty}^{\infty}a_nz^n$,where
$a_n = \sum\frac{1}{3}(1)^n,n=1,2,...$
$and$
$a_n = \sum\frac{4}{3}(2)^{n1}=\sum\frac{4}{3}(\frac{1}{2^{n+1}}),n=0,1,2,...$
Similar for when $2<z<\infty$
You can read example 12 from 2.5

Can you guys explain why you can use (12) for z = 1, and (13) for z=2?
Following the definition in the textbook you can't use either (12) or (13) because z_j < r and z_j > R aren't satisfied.

RE: Min Gyu Woo
If you only want a legitimate procedure for the answer to the question, you can just "instantiate" the proof.
Since
\begin{align*}
\frac{z + 2}{ (z2) (z + 1)} & = \frac{4}{3}(\frac{1}{2} \frac{1}{1  \frac{z}{2}})  \frac{1}{3} (\frac{1}{z}\frac{1}{1+\frac{1}{z}}) \\
&=  \frac{4}{3} \frac{1}{2} (\sum (\frac{z}{2})^n)  \frac{1}{3} \frac{1}{z}(\sum (\frac{1}{z})^n)
\end{align*}
The second line is correct because the convergence relies on the geometric series instead of (12) and (13).
However, it is reasonable to ask if (12) and (13) can be extended to less than or equal to.
The second part of the question can solve similarly.

Is it just me or is the entirety of Q23 answer key wrong...

I was also wondering if the answer key is wrong.
I don't understand how to get the part $a_n=\frac{4}{3^{n+2}}$.

Ende
Fix parenthesis