Expand e^{\frac{z}{z-2}} Laurent Series: z=2

[Stumped1]

expand \(\displaystyle e^{\frac{z}{z-2}}\) in a Laurent series about \(\displaystyle z=2\)

I cannot start this.

my attempt so far has been

\(\displaystyle e^\frac{z}{z-2}=1 + \frac{z}{z-2} + \frac{z^2}{(z-2)^2 2!} + \frac{z^3}{(z-2)^3 3!}\)

This is unlike the other problems I have worked. Seems I need to manipulate this equation some way before plugging into \(\displaystyle e^z\)'s maclaurin series

Thanks for any help with this!

 

[Prove It]

What you have done seems fine...

 

[chisigma]

expand \(\displaystyle e^{\frac{z}{z-2}}\) in a Laurent series about \(\displaystyle z=2\)

I cannot start this.

my attempt so far has been

\(\displaystyle e^\frac{z}{z-2}=1 + \frac{z}{z-2} + \frac{z^2}{(z-2)^2 2!} + \frac{z^3}{(z-2)^3 3!}\)

This is unlike the other problems I have worked. Seems I need to manipulate this equation some way before plugging into \(\displaystyle e^z\)'s maclaurin series

Thanks for any help with this!

Because is $\displaystyle \frac{z}{z-2} = 1 + \frac{2}{z-2}$ You obtain...

$\displaystyle e^{\frac{z}{z-2}} =e\ \{ 1 + \frac{2}{z- 2} + \frac{4}{2\ (z-2)^{2}} + ... + \frac{2^{n}}{n!\ (z-2)^{n}} + ...\}\ (1)$

Kind regards

$\chi$ $\sigma$

 

[alyafey22]

Note that $f$ must be written in the form

\(\displaystyle \sum_{n=-\infty}^\infty a_n (z-2)^n\)

where $a_n$'s are independent of $z$ as chisigma stated.

 

[blue_raver22]

I would approach this problem by first understanding the concept of a Laurent series. A Laurent series is a representation of a complex function as a power series, where the power can be both positive and negative. It is typically used to expand functions that have singularities, such as poles or branch points.

In this case, the function e^{\frac{z}{z-2}} has a singularity at z=2, so we can use a Laurent series to expand it about that point. The general form of a Laurent series is:

f(z) = \sum_{n=-\infty}^{\infty} a_n (z-z_0)^n

Where z_0 is the center of the series and a_n are the coefficients.

To expand e^{\frac{z}{z-2}} about z=2, we can first rewrite the function as e^{z/(z-2)} = e^{1 + \frac{z-2}{z-2}} = e(e^1)^{\frac{z-2}{z-2}} = e(e^1)^{1 + \frac{z-3}{z-2}}. Now, we can use the Maclaurin series for e^x to expand e^{1 + \frac{z-3}{z-2}}:

e^{1 + \frac{z-3}{z-2}} = e\left(1 + \frac{1}{1!}(1 + \frac{z-3}{z-2}) + \frac{1}{2!}(1 + \frac{z-3}{z-2})^2 + \frac{1}{3!}(1 + \frac{z-3}{z-2})^3 + ...\right)

= e\left(1 + (1 + \frac{1}{1!})\frac{z-3}{z-2} + (1 + \frac{1}{1!} + \frac{1}{2!})\left(\frac{z-3}{z-2}\right)^2 + (1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!})\left(\frac{z-3}{z-2}\right)^3 + ...\right)

= e\left(1 + \frac{z