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Laurent series (COMPLICATED, URGENT)
Find the Laurent series for [tex]f(z) = \frac{{2{z^2}}}{{2{z^2} - 5z + 2}} + \sin (\frac{3}{{{z^2}}})[/tex] around z = 0, in which the series absolutely converges for z = -i.
I have several problem with this. I tell you what I tried.
First: I separate the polinomial fraction in partial fractions. I don't know if the procedure is OK: [tex]\frac{{2{z^2}}}{{2{z^2} - 5z + 2}} = 2{z^2}\left( {\frac{2}{{3(z - 2)}} - \frac{2}{{3(z - 1/2)}}} \right)[/tex]. I leave the 2z2 there because it's its own Laurent Series around z=0, for all z.
Second: I compare the function with the geometrical function (considering that I have to find the Laurent series in a vecinity of infinite for the fraction with the (z - 1/2)) and I use a variable change for the sine:
a) [tex]\frac{2}{{3(z - \frac{1}{2})}} \to \frac{2}{{3z(1 - \frac{1}{{2z}})}}[/tex]
[tex]\frac{2}{{3z(1 - \frac{1}{{2z}})}} = \frac{2}{{3z}}\sum\limits_{n = 0}^\infty {\frac{1}{{{2^n}{z^n}}}} {\rm{ }}\forall z/\left| z \right| > \frac{1}{2}[/tex]
b)
[tex]\frac{{ -1}}{{3(z - \frac{1}{2})}} = \frac{2}{{3(\frac{1}{2} - z)}} = \frac{4}{{3(1 - 2z)}}[/tex]
c) Since [tex]\sin (u) = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{u^{2n + 1}}}}{{(2n + 1)!}}}[/tex], then, using the change of variable u = 3/z2, I get that:
[tex]\sin (\frac{3}{{{z^2}}}) = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{3^{2n + 1}}}}{{(2n + 1)!{z^{4n + 2}}}}} {\rm{ }}\forall z/\left| z \right| > 0[/tex]
Then, I have that:
[tex]f(z) = 2{z^2}\left( {\frac{2}{{3z}}\sum\limits_{n = 0}^\infty {\frac{1}{{{2^n}{z^n}}}} - \frac{1}{3}\sum\limits_{n = 0}^\infty {\frac{{{z^n}}}{{{2^n}}}} } \right) + \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{3^{2n + 1}}}}{{(2n + 1)!{z^{4n + 2}}}}} {\rm{, }}\forall z/\frac{1}{2}< \left| z \right|< 2[/tex]
My question is, how can I come up with a sole expresion for the coefficients? I couldn't even find one for the series inside the parenthesis...
Thanks
Homework Statement
Find the Laurent series for [tex]f(z) = \frac{{2{z^2}}}{{2{z^2} - 5z + 2}} + \sin (\frac{3}{{{z^2}}})[/tex] around z = 0, in which the series absolutely converges for z = -i.
The Attempt at a Solution
I have several problem with this. I tell you what I tried.
First: I separate the polinomial fraction in partial fractions. I don't know if the procedure is OK: [tex]\frac{{2{z^2}}}{{2{z^2} - 5z + 2}} = 2{z^2}\left( {\frac{2}{{3(z - 2)}} - \frac{2}{{3(z - 1/2)}}} \right)[/tex]. I leave the 2z2 there because it's its own Laurent Series around z=0, for all z.
Second: I compare the function with the geometrical function (considering that I have to find the Laurent series in a vecinity of infinite for the fraction with the (z - 1/2)) and I use a variable change for the sine:
a) [tex]\frac{2}{{3(z - \frac{1}{2})}} \to \frac{2}{{3z(1 - \frac{1}{{2z}})}}[/tex]
[tex]\frac{2}{{3z(1 - \frac{1}{{2z}})}} = \frac{2}{{3z}}\sum\limits_{n = 0}^\infty {\frac{1}{{{2^n}{z^n}}}} {\rm{ }}\forall z/\left| z \right| > \frac{1}{2}[/tex]
b)
[tex]\frac{{ -1}}{{3(z - \frac{1}{2})}} = \frac{2}{{3(\frac{1}{2} - z)}} = \frac{4}{{3(1 - 2z)}}[/tex]
c) Since [tex]\sin (u) = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{u^{2n + 1}}}}{{(2n + 1)!}}}[/tex], then, using the change of variable u = 3/z2, I get that:
[tex]\sin (\frac{3}{{{z^2}}}) = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{3^{2n + 1}}}}{{(2n + 1)!{z^{4n + 2}}}}} {\rm{ }}\forall z/\left| z \right| > 0[/tex]
Then, I have that:
[tex]f(z) = 2{z^2}\left( {\frac{2}{{3z}}\sum\limits_{n = 0}^\infty {\frac{1}{{{2^n}{z^n}}}} - \frac{1}{3}\sum\limits_{n = 0}^\infty {\frac{{{z^n}}}{{{2^n}}}} } \right) + \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{3^{2n + 1}}}}{{(2n + 1)!{z^{4n + 2}}}}} {\rm{, }}\forall z/\frac{1}{2}< \left| z \right|< 2[/tex]
My question is, how can I come up with a sole expresion for the coefficients? I couldn't even find one for the series inside the parenthesis...
Thanks