Taylor polynomial of a complex function

• Oct 17th 2009, 08:15 PM
arbolis
Taylor polynomial of a complex function
Find Taylor's polynomial of the following function in $z_0=-1$ : $f(z)=\frac{1}{z^2}$.
Determine the radius of convergence of the series.

My attempt :
$f(-1)=\sum _{n=0}^{\infty} \frac{(-1)^n(n+1)}{z^{n+2}}(z+1)^n$. Which answers the first question.

Now trying to figure out the second question :
So $a_n=\frac{(-1)^n(n+1)}{z^{n+2}}$.
I apply the ratio test : $\left| \frac{a_{n+1}}{a_n} \right| =\frac{(-1)(n+2)}{z}$ which tends to $-\frac{1}{z}$ when $n$ tends to $\infty$.
Hence the radius of convergence is ...? It makes no sense at all.
I'd like to know where is(are) my error(s). Thanks a lot!
• Oct 17th 2009, 08:27 PM
mr fantastic
Quote:

Originally Posted by arbolis
Find Taylor's polynomial of the following function in $z_0=-1$ : $f(z)=\frac{1}{z^2}$.
Determine the radius of convergence of the series.

My attempt :
$f(-1)=\sum _{n=0}^{\infty} \frac{(-1)^n(n+1)!}{z^{n+2}}(z+1)^n$. Which answers the first question.

[snip]

No. This is definitely not a Taylor Series. Do you see why?

Using the formula given in the theorem, the required Taylor Series is $1 + 2(z + 1) + 3(z + 1)^2 + 4(z + 1)^3 + .... = \sum_{n = 0}^{+ \infty}(n + 1) (z + 1)^n$. Do you see why?
• Oct 17th 2009, 08:35 PM
arbolis
Quote:

Originally Posted by mr fantastic
No. This is definitely not a Taylor Series. Do you see why?

Using the formula given in the theorem, the required Taylor Series is $1 + 2(z + 1) + 3(z + 1)^2 + 4(z + 1)^3 + .... = \sum_{n = 0}^{+ \infty}(n + 1) (z + 1)^n$. Do you see why?

I made a mistake, I've edited my post. However I still don't reach the Taylor series you provided. I don't see why the series I found is not a Taylor series.

I've used the fact that $f(z)=\sum _{n=0}^{\infty} \frac{f^{(n)}(z-z_0)}{n!}$.
I've found out that $f^{(n)}=\frac{(-1)^n (n+1)!}{z^{(n+2)}}$.

Now you make me doubt the formula holds for the function $z \mapsto \frac{1}{z^2}$ due to the singularity at $z=0$.
Edit : I understand the equality you gave, but I don't know how to reach it. I mean I don't know how to reach that the Taylor Series is $1 + 2(z + 1) + 3(z + 1)^2 + 4(z + 1)^3 + ...$

Edit 2 : Sorry, I just found my error. I forgot to evaluate $f^{(n)}$ in $z_0$...
I think I'll do it. Thanks a lot!
• Oct 17th 2009, 08:56 PM
redsoxfan325
Quote:

Originally Posted by arbolis
I made a mistake, I've edited my post. However I still don't reach the Taylor series you provided. I don't see why the series I found is not a Taylor series.

I've used the fact that $f(z)=\sum _{n=0}^{\infty} \frac{f^{(n)}(z-z_0)}{n!}$.
I've found out that $f^{(n)}=\frac{(-1)^n (n+1)!}{z^{(n+2)}}$.

Now you make me doubt the formula holds for the function $z \mapsto \frac{1}{z^2}$ due to the singularity at $z=0$.
Edit : I understand the equality you gave, but I don't know how to reach it. I mean I don't know how to reach that the Taylor Series is $1 + 2(z + 1) + 3(z + 1)^2 + 4(z + 1)^3 + ...$

Edit 2 : Sorry, I just found my error. I forgot to evaluate $f^{(n)}$ in $z_0$...
I think I'll do it. Thanks a lot!

Yes, $\frac{(-1)^n}{z^{n+2}}$ evaluated at $z=-1$ always equals $1$.
• Oct 18th 2009, 04:36 AM
HallsofIvy
The radius of convergence of a function will extend to the nearest point at which the function is not analytic. Here, the function is $\frac{1}{z^2}$ which is not analytic at z= 0. Since the center of the region of convergence is at $z_0= -1$, the radius of convergence is |0-(-1)|= 1.

This can be used even for real functions. For example, the function $\frac{1}{1+ x^2}$ is analytic for all x except -i or -i. If it is expanded in a power series about x= 0, then its radius of convergence is |0- i|= 1.

If it is expanded in a power series about x= 1, its radius of convergence is $|1- i|= \sqrt{2}$.
• Oct 18th 2009, 07:50 AM
arbolis
Quote:

Originally Posted by HallsofIvy
The radius of convergence of a function will extend to the nearest point at which the function is not analytic.

So an entire function always has an infinite radius of convergence?