1. ## [SOLVED] about taylor and laurent series relation

Hello, I know a Taylor series for any function must be the same that laurent series of this function with a real variable, but do they have the same form always? I mean:

We know:

$sin x = \sum \frac{(-1)^n}{(2n+1)!} x^{2n+1}$

and:

$sin z =\sum \frac{(-1)^{n}}{(2n+1)!}z^{2n+1}$

If we know taylor series of f(x) and we want calculate laurent of f(z), can we just replace x for z?

Thank you.

2. Originally Posted by Bop
Hello, I know a Taylor series for any function must be the same that laurent series of this function with a real variable, but do they have the same form always? I mean:

We know:

$sin x = \sum \frac{(-1)^n}{(2n+1)!} x^{2n+1}$

and:

$sin z =\sum \frac{(-1)^{n}}{(2n+1)!}z^{2n+1}$

If we know taylor series of f(x) and we want calculate laurent of f(z), can we just replace x for z?

Thank you.
If a Laurent series has no negative powers of $z$ in it, it is the same as the Taylor series.
If a Laurent series has negative powers of $z$ in it, then a Taylor series does not exist (with radius of convergence > 0).

3. There is any way to find if all negative powers are zero without resolving all integrals?

Thank you!

4. Originally Posted by Bop
There is any way to find if all negative powers are zero without resolving all integrals?

Thank you!
Well, if the function has a finite value at $z_0$ and is analytic there, there are certainly no negative powers appearing in the expansion of the function in a neighbourhood of $z_0$.