As in title I'm trying to write the Laurent series of $\displaystyle 1/z$ for $\displaystyle 0<|z|<1$, but I'm stuck...
$\displaystyle \sum_{n=0}^{+\infty} z^{n-1}0^n$ ???
If $\displaystyle n=0$ I have $\displaystyle \frac{1}{z}$, while if $\displaystyle n$ is not $\displaystyle 0$ I have $\displaystyle 0$. Right?
Edit: Sorry $\displaystyle 0^0$ is indeterminate... I really have no idea...
Don't complicate things ( ) the solution is exactly:
$\displaystyle \dfrac{1}{z}=\ldots +\dfrac{0}{z^3}+\dfrac{0}{z^2}+\dfrac{1}{z}+0+0z+0 z^2+0z^3+\ldots\quad (0<|z|<+\infty)$
And of course, the expansion is also valid for $\displaystyle 0<|z|<1$ .
Ok, another last thing, the exercise was at the beginning: Laurent series of $\displaystyle \frac{z^3+1}{z^2(z-1)}$ for $\displaystyle 0<|z|<1$,
then I wrote partial fractions: $\displaystyle \frac{z^3+1}{z^2(z-1)} = \frac{z}{z-1}-\frac{1}{z^2}-\frac{1}{z}+\frac{1}{z-1}$ and then I wrote the Laurent Series for every fraction:
$\displaystyle \frac{1}{z-1} = -\sum_{n=0}^{+\infty}z^n$
$\displaystyle \frac{z}{z-1}= -\sum_{n=0}^{+\infty}z^{n+1}$
At this point my question is: I can leave $\displaystyle -\frac{1}{z}$ and $\displaystyle -\frac{1}{z^2}$ as they are?
So the final result will be:
$\displaystyle \frac{z^3+1}{z^2(z-1)} = -\sum_{n=0}^{+\infty}z^n-\sum_{n=0}^{+\infty}z^{n+1} -\frac{1}{z^2}-\frac{1}{z}$
If it is allowed to consider a Taylor expansion a particular case of Laurent expansion where all the negative coefficients are 0, then the expansion of $\displaystyle f(z)=\frac{1}{z}$ around $\displaystyle z=\frac{1}{2}$ ...
$\displaystyle \frac{1}{z} = \sum_{n=0}^{\infty} (-1)^{n} \frac{z^{n}}{2^{n+1}}$ (1)
... is a Laurent expansion. The (1) converges for $\displaystyle |1-2 z|<1$ but with a procedure called 'analytic extension' the analyticity of $\displaystyle \frac{1}{z}$ can be 'extended' to the region $\displaystyle 0<|z|<1$. Of course a spontaneous question is: what's the practical utility of all that?...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
The final answer is correct, but it looks a bit clumsy because of the two separate summations, which could easily be combined into one. In fact, it would have been better to write the partial fraction expression using only one term with denominator $\displaystyle z-1:$
$\displaystyle \frac{z}{z-1}+\frac{1}{z-1} = \frac{z+1}{z-1} = \frac{(z-1)+2}{z-1} = 1 + \frac2{z-1}.$
That way, you get a constant term 1 for the Laurent series, plus a single series arising from the z-1 denominator.
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Yes, you are. A "Laurent series" typically cannot be written in that form- that's what distinguishes it from a Taylor's series. A Laurent series has negative powers. You can write a Laurent series in the form $\displaystyle \sum_{n=-\infty}^\infty a_nz^n$.
In this case just define $\displaystyle a_{-1}= 1$, $\displaystyle a_n= 0$ for all other n.