# Thread: Laurent's series for a function

1. ## Laurent's series for a function

Hi. Can you help me with deriving the Laurent series for $\displaystyle f(z) = e^{1/z}$ . I can derive it by substituting $\displaystyle 1/z$ for $\displaystyle z$ in the taylor series expression for $\displaystyle f(z) = e^z$ but I was trying to derive the Laurent Series using the integral formulae for the coefficients. When I do that, the integrals of the coefficients of the positive powers of z fail to vanish.

I would appreciate it a lot if someone took the trouble of pointing out how they vanish. Thank you!

2. Originally Posted by sashikanth
Hi. Can you help me with deriving the Laurent series for $\displaystyle f(z) = e^{1/z}$ . I can derive it by substituting $\displaystyle 1/z$ for $\displaystyle z$ in the taylor series expression for $\displaystyle f(z) = e^z$ but I was trying to derive the Laurent Series using the integral formulae for the coefficients. When I do that, the integrals of the coefficients of the positive powers of z fail to vanish.

I would appreciate it a lot if someone took the trouble of pointing out how they vanish. Thank you!

I don't understand: this section is abstract algebra and this questions doesn't belong here...is that very hard to understand?!

Tonio

3. Originally Posted by sashikanth
Hi. Can you help me with deriving the Laurent series for $\displaystyle f(z) = e^{1/z}$ . I can derive it by substituting $\displaystyle 1/z$ for $\displaystyle z$ in the taylor series expression for $\displaystyle f(z) = e^z$ but I was trying to derive the Laurent Series using the integral formulae for the coefficients. When I do that, the integrals of the coefficients of the positive powers of z fail to vanish.

I would appreciate it a lot if someone took the trouble of pointing out how they vanish. Thank you!
Are you aware of using the 'residue at infinity' for computation of integrals?

4. There is a kind of vicious circle about the question. Consider:

$\displaystyle f(z)=e^{1/z}$

then, $\displaystyle z=0$ is the only singularity of $\displaystyle f$. Besides, $\displaystyle \nexists\;{\lim_{z \to 0 }f(z)$

this means that $\displaystyle z=0$ is an essential singulariy and its residue is usually found by means of the Laurent expansion. For example if:

$\displaystyle f(z)=\ldots+\dfrac{A_{-2}}{z^2}+\dfrac{A_{-1}}{z}+A_0+A_1z+A_2z^2+\ldots$

then,

$\displaystyle \dfrac{f(z)}{z}=\ldots+\dfrac{A_{-2}}{z^3}+\dfrac{A_{-1}}{z^2}+\dfrac{A_0}{z}+A_1+A_2z+\ldots$

and

$\displaystyle A_0=\dfrac{1}{2\pi i}\displaystyle\int_{|z|=R}\dfrac{f(z)dz}{z}=\dfra c{1}{2\pi i}\cdot 2\pi i \;\textrm{Res}(f,z=0)=\textrm{coef}\left(\dfrac{1} {z}\right)=A_0$

Fernando Revilla