After working out the poles to:
I have the answer z1=2i and z2=3
Given this how would we find the taylor series if:
First you do a partial fraction decomposition by requiring thatGiven this how would we find the taylor series if:
which gives, if I am not misaken, that . (I might be mistaken, however, so please to the math yourself.)
Now you can develop each of these two fractions with the help of a geometric series like this:
You can get it, starting from the same partial fraction decomposition, like this
Well, yes, the condition for the geometric series expansion to converge is that holds.its just aint ? For the first part when we divide by we have
which is equal to
Therefore, if , we have that , and thus the series converges. But if , we have , thus diverges.
By a purely algebraic transformation of that first term of the partial fraction decomposition we can use the (for ) convergent geometric series , instead.
Simililarly, if we were told to develop the given function for , we would have to transform the second term of the partial fraction decomposition in such a way that we can develop it in a, for , convergent series of the form .
The first term of your partial fraction decomposition is already (almost) in that form, namely
For the second term of your partial fraction decomposition you can substitute for and try to transform it into a power series in terms of .
Note that this series converges, as required, if , because in that case we have that