Mixing
How to find Laurent series Expansion
$begingroup$
$f(z)$ is defined like this:
$$
f(z) = frac{z}{(z-1)(z-3)}
$$
I need to find a series for $f(z)$ that involves positive and negative powers of $(z-1)$, which converges to $f(z)$ when $0 leq |z - 1| leq 2$.
What I understand from question is I must expand $f(z)$ Laurent series.
$$ f(z) = sum_{m=0}^{infty}a_{m}(z-1)^{m} + sum_{m=1}^{infty}b_{m}(z-1)^{-m}$$
where,
$$ a_{m} = frac{1}{j2pi}oint_{C}frac{f(z)}{(z-1)^{m+1}}dz $$
$$ b_{m} = frac{1}{j2pi}oint_{C}frac{f(z)}{(z-1)^{1-m}}dz $$
This is what theory tells me.
But I apply partial fraction method to this function like this:
$$ f(z) = frac{z}{(z-1)(z-3)} = frac{z^{-1}}{(1-z^{-1})(1-3z^{-1})} = frac{-1/2}{(1-z^{-1})} + frac{1/2}{(1-3z^{-1})} $$
And I know this series expansion from z-transform like this:
$$ f(z) = -frac{1}{2} sum_{k=0}^{infty}z^{-k} + frac{1}{2} sum_{k=0}^{infty}3^{k}z^{-k}$$
I obtain a series expansion but it looks like Mclaurin series not a Laurent series.
Here, my first question an expression may have different type of series expansion?
And second, how to find a Laurent series for $ f(z) $
complex-analysis power-series laurent-series
asked Mar 8 '15 at 10:24
mehmetmehmet
1741110
$endgroup$
add a comment |
$begingroup$
$f(z)$ is defined like this:
$$
f(z) = frac{z}{(z-1)(z-3)}
$$
I need to find a series for $f(z)$ that involves positive and negative powers of $(z-1)$, which converges to $f(z)$ when $0 leq |z - 1| leq 2$.
What I understand from question is I must expand $f(z)$ Laurent series.
$$ f(z) = sum_{m=0}^{infty}a_{m}(z-1)^{m} + sum_{m=1}^{infty}b_{m}(z-1)^{-m}$$
where,
$$ a_{m} = frac{1}{j2pi}oint_{C}frac{f(z)}{(z-1)^{m+1}}dz $$
$$ b_{m} = frac{1}{j2pi}oint_{C}frac{f(z)}{(z-1)^{1-m}}dz $$
This is what theory tells me.
But I apply partial fraction method to this function like this:
$$ f(z) = frac{z}{(z-1)(z-3)} = frac{z^{-1}}{(1-z^{-1})(1-3z^{-1})} = frac{-1/2}{(1-z^{-1})} + frac{1/2}{(1-3z^{-1})} $$
And I know this series expansion from z-transform like this:
$$ f(z) = -frac{1}{2} sum_{k=0}^{infty}z^{-k} + frac{1}{2} sum_{k=0}^{infty}3^{k}z^{-k}$$
I obtain a series expansion but it looks like Mclaurin series not a Laurent series.
Here, my first question an expression may have different type of series expansion?
And second, how to find a Laurent series for $ f(z) $
complex-analysis power-series laurent-series
asked Mar 8 '15 at 10:24
mehmetmehmet
1741110
$endgroup$
add a comment |
$begingroup$
$f(z)$ is defined like this:
$$
f(z) = frac{z}{(z-1)(z-3)}
$$
I need to find a series for $f(z)$ that involves positive and negative powers of $(z-1)$, which converges to $f(z)$ when $0 leq |z - 1| leq 2$.
What I understand from question is I must expand $f(z)$ Laurent series.
$$ f(z) = sum_{m=0}^{infty}a_{m}(z-1)^{m} + sum_{m=1}^{infty}b_{m}(z-1)^{-m}$$
where,
$$ a_{m} = frac{1}{j2pi}oint_{C}frac{f(z)}{(z-1)^{m+1}}dz $$
$$ b_{m} = frac{1}{j2pi}oint_{C}frac{f(z)}{(z-1)^{1-m}}dz $$
This is what theory tells me.
But I apply partial fraction method to this function like this:
$$ f(z) = frac{z}{(z-1)(z-3)} = frac{z^{-1}}{(1-z^{-1})(1-3z^{-1})} = frac{-1/2}{(1-z^{-1})} + frac{1/2}{(1-3z^{-1})} $$
And I know this series expansion from z-transform like this:
$$ f(z) = -frac{1}{2} sum_{k=0}^{infty}z^{-k} + frac{1}{2} sum_{k=0}^{infty}3^{k}z^{-k}$$
I obtain a series expansion but it looks like Mclaurin series not a Laurent series.
Here, my first question an expression may have different type of series expansion?
And second, how to find a Laurent series for $ f(z) $
complex-analysis power-series laurent-series
asked Mar 8 '15 at 10:24
mehmetmehmet
1741110
$endgroup$
$f(z)$ is defined like this:
$$
f(z) = frac{z}{(z-1)(z-3)}
$$
I need to find a series for $f(z)$ that involves positive and negative powers of $(z-1)$, which converges to $f(z)$ when $0 leq |z - 1| leq 2$.
What I understand from question is I must expand $f(z)$ Laurent series.
$$ f(z) = sum_{m=0}^{infty}a_{m}(z-1)^{m} + sum_{m=1}^{infty}b_{m}(z-1)^{-m}$$
where,
$$ a_{m} = frac{1}{j2pi}oint_{C}frac{f(z)}{(z-1)^{m+1}}dz $$
$$ b_{m} = frac{1}{j2pi}oint_{C}frac{f(z)}{(z-1)^{1-m}}dz $$
This is what theory tells me.
But I apply partial fraction method to this function like this:
$$ f(z) = frac{z}{(z-1)(z-3)} = frac{z^{-1}}{(1-z^{-1})(1-3z^{-1})} = frac{-1/2}{(1-z^{-1})} + frac{1/2}{(1-3z^{-1})} $$
And I know this series expansion from z-transform like this:
$$ f(z) = -frac{1}{2} sum_{k=0}^{infty}z^{-k} + frac{1}{2} sum_{k=0}^{infty}3^{k}z^{-k}$$
I obtain a series expansion but it looks like Mclaurin series not a Laurent series.
Here, my first question an expression may have different type of series expansion?
And second, how to find a Laurent series for $ f(z) $
complex-analysis power-series laurent-series
complex-analysis power-series laurent-series
asked Mar 8 '15 at 10:24
mehmetmehmet
1741110
asked Mar 8 '15 at 10:24
mehmetmehmet
1741110
edited Mar 12 '15 at 23:15
mehmet
asked Mar 8 '15 at 10:24
mehmetmehmet
1741110
asked Mar 8 '15 at 10:24
mehmetmehmet
1741110
asked Mar 8 '15 at 10:24
mehmetmehmet
1741110
1741110
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
No need for contour integrals, just give a name to the quantity you want a Laurent series in, and expand. So with $x=z-1$:
$$
frac z{(z-1)(z-3)}=frac{x+1}{x(x-2)}
=x^{-1}left(1-frac3{2-x}right)
\=x^{-1}left(1-frac32sum_{igeq0}bigl(frac x2bigr)^iright)
=-frac12x^{-1}+sum_{igeq0}frac{-3}{4times2^i}x^i.
$$
You can now substitute $x:=z-1$ if you like.
answered Mar 8 '15 at 18:15
Marc van LeeuwenMarc van Leeuwen
86.5k5106220
$endgroup$
$begingroup$
This is a good one except "-2" part (I think I must be directly "2"). Thanks.
$endgroup$
– mehmet
Mar 9 '15 at 17:45
1
$begingroup$
Soryy, I forgot to say I mean in sum operation. I believe it must be positive sign of 2.
$endgroup$
– mehmet
Mar 9 '15 at 18:16
$begingroup$
Also the sign of the sum operation must be negative, I think?
$endgroup$
– mehmet
Mar 9 '15 at 18:26
1
$begingroup$
@mehmet: OK, I see now. Corrected that, thanks.
$endgroup$
– Marc van Leeuwen
Mar 10 '15 at 5:28
$begingroup$
Out of interest how would you do Laurent Expansions with Integrals if I may ask
$endgroup$
– Zophikel
Sep 11 '17 at 20:39
add a comment |
$begingroup$
The problem is that if you use $frac 1{1-z} = sum z^n$ you are essentially writing the Laurent expansion in a neighborhood of $0$. Since you want powers of $z-1$, it means that you want an expansion in a neighborhood of $1$!
Hence the $frac {1}{z-1} = (z-1)^{-1}$ term is already "good" (just like $frac 1z$ would be in a Laurent expansion in a neighborhood of $0$).
You have to expand in a neighborhood of $1$ the expression $frac z{z-3}$.
You can set $t = z - 1 implies z = t + 1$ so your expression becomes $frac{t+1}{t-2}$.
Now use partial fractions, and find the Laurent series in a neighborhood of $0$ with respect to $t$ of $frac{t+1}{t-2}$ (This means you can use the geometric series formula), substitute back $t = z-1$ and divide everything by $z-1$ to get the final result.
edited Jan 1 at 19:43
amWhy
192k28225439
answered Mar 8 '15 at 10:32
AntAnt
17.4k22873
$endgroup$
$begingroup$
Thanks for your explanation! (:
$endgroup$
– mehmet
Mar 9 '15 at 18:19
1
$begingroup$
@mehmet You're welcome! :)
$endgroup$
– Ant
Mar 9 '15 at 18:20
add a comment |
$begingroup$
First, yes. The same function can have different Laurent series, depending on the center of annulus in question.
Consider now the given function. Clearly, the question will be solved with ease once we find the series of $$g(z)=frac{z}{z-3}.$$Note that$$g(z)=1+frac{3}{z-3}=1+frac{3}{(z-1)-2}=1+frac{3/2}{frac{z-1}{2}-1},$$and the last expression can be represented as the sum of a geometric series.
answered Mar 8 '15 at 10:38
Amitai YuvalAmitai Yuval
15.1k11126
$endgroup$
$begingroup$
Thank you for your answer (:
$endgroup$
– mehmet
Mar 9 '15 at 17:42
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No need for contour integrals, just give a name to the quantity you want a Laurent series in, and expand. So with $x=z-1$:
$$
frac z{(z-1)(z-3)}=frac{x+1}{x(x-2)}
=x^{-1}left(1-frac3{2-x}right)
\=x^{-1}left(1-frac32sum_{igeq0}bigl(frac x2bigr)^iright)
=-frac12x^{-1}+sum_{igeq0}frac{-3}{4times2^i}x^i.
$$
You can now substitute $x:=z-1$ if you like.
answered Mar 8 '15 at 18:15
Marc van LeeuwenMarc van Leeuwen
86.5k5106220
$endgroup$
$begingroup$
This is a good one except "-2" part (I think I must be directly "2"). Thanks.
$endgroup$
– mehmet
Mar 9 '15 at 17:45
1
$begingroup$
Soryy, I forgot to say I mean in sum operation. I believe it must be positive sign of 2.
$endgroup$
– mehmet
Mar 9 '15 at 18:16
$begingroup$
Also the sign of the sum operation must be negative, I think?
$endgroup$
– mehmet
Mar 9 '15 at 18:26
1
$begingroup$
@mehmet: OK, I see now. Corrected that, thanks.
$endgroup$
– Marc van Leeuwen
Mar 10 '15 at 5:28
$begingroup$
Out of interest how would you do Laurent Expansions with Integrals if I may ask
$endgroup$
– Zophikel
Sep 11 '17 at 20:39
add a comment |
$begingroup$
No need for contour integrals, just give a name to the quantity you want a Laurent series in, and expand. So with $x=z-1$:
$$
frac z{(z-1)(z-3)}=frac{x+1}{x(x-2)}
=x^{-1}left(1-frac3{2-x}right)
\=x^{-1}left(1-frac32sum_{igeq0}bigl(frac x2bigr)^iright)
=-frac12x^{-1}+sum_{igeq0}frac{-3}{4times2^i}x^i.
$$
You can now substitute $x:=z-1$ if you like.
answered Mar 8 '15 at 18:15
Marc van LeeuwenMarc van Leeuwen
86.5k5106220
$endgroup$
$begingroup$
This is a good one except "-2" part (I think I must be directly "2"). Thanks.
$endgroup$
– mehmet
Mar 9 '15 at 17:45
1
$begingroup$
Soryy, I forgot to say I mean in sum operation. I believe it must be positive sign of 2.
$endgroup$
– mehmet
Mar 9 '15 at 18:16
$begingroup$
Also the sign of the sum operation must be negative, I think?
$endgroup$
– mehmet
Mar 9 '15 at 18:26
1
$begingroup$
@mehmet: OK, I see now. Corrected that, thanks.
$endgroup$
– Marc van Leeuwen
Mar 10 '15 at 5:28
$begingroup$
Out of interest how would you do Laurent Expansions with Integrals if I may ask
$endgroup$
– Zophikel
Sep 11 '17 at 20:39
add a comment |
$begingroup$
No need for contour integrals, just give a name to the quantity you want a Laurent series in, and expand. So with $x=z-1$:
$$
frac z{(z-1)(z-3)}=frac{x+1}{x(x-2)}
=x^{-1}left(1-frac3{2-x}right)
\=x^{-1}left(1-frac32sum_{igeq0}bigl(frac x2bigr)^iright)
=-frac12x^{-1}+sum_{igeq0}frac{-3}{4times2^i}x^i.
$$
You can now substitute $x:=z-1$ if you like.
answered Mar 8 '15 at 18:15
Marc van LeeuwenMarc van Leeuwen
86.5k5106220
$endgroup$
No need for contour integrals, just give a name to the quantity you want a Laurent series in, and expand. So with $x=z-1$:
$$
frac z{(z-1)(z-3)}=frac{x+1}{x(x-2)}
=x^{-1}left(1-frac3{2-x}right)
\=x^{-1}left(1-frac32sum_{igeq0}bigl(frac x2bigr)^iright)
=-frac12x^{-1}+sum_{igeq0}frac{-3}{4times2^i}x^i.
$$
You can now substitute $x:=z-1$ if you like.
answered Mar 8 '15 at 18:15
Marc van LeeuwenMarc van Leeuwen
86.5k5106220
edited Mar 2 '16 at 11:06
answered Mar 8 '15 at 18:15
Marc van LeeuwenMarc van Leeuwen
86.5k5106220
answered Mar 8 '15 at 18:15
Marc van LeeuwenMarc van Leeuwen
86.5k5106220
answered Mar 8 '15 at 18:15
Marc van LeeuwenMarc van Leeuwen
86.5k5106220
86.5k5106220
$begingroup$
This is a good one except "-2" part (I think I must be directly "2"). Thanks.
$endgroup$
– mehmet
Mar 9 '15 at 17:45
1
$begingroup$
Soryy, I forgot to say I mean in sum operation. I believe it must be positive sign of 2.
$endgroup$
– mehmet
Mar 9 '15 at 18:16
$begingroup$
Also the sign of the sum operation must be negative, I think?
$endgroup$
– mehmet
Mar 9 '15 at 18:26
1
$begingroup$
@mehmet: OK, I see now. Corrected that, thanks.
$endgroup$
– Marc van Leeuwen
Mar 10 '15 at 5:28
$begingroup$
Out of interest how would you do Laurent Expansions with Integrals if I may ask
$endgroup$
– Zophikel
Sep 11 '17 at 20:39
add a comment |
$begingroup$
This is a good one except "-2" part (I think I must be directly "2"). Thanks.
$endgroup$
– mehmet
Mar 9 '15 at 17:45
1
$begingroup$
Soryy, I forgot to say I mean in sum operation. I believe it must be positive sign of 2.
$endgroup$
– mehmet
Mar 9 '15 at 18:16
$begingroup$
Also the sign of the sum operation must be negative, I think?
$endgroup$
– mehmet
Mar 9 '15 at 18:26
1
$begingroup$
@mehmet: OK, I see now. Corrected that, thanks.
$endgroup$
– Marc van Leeuwen
Mar 10 '15 at 5:28
$begingroup$
Out of interest how would you do Laurent Expansions with Integrals if I may ask
$endgroup$
– Zophikel
Sep 11 '17 at 20:39
$begingroup$
This is a good one except "-2" part (I think I must be directly "2"). Thanks.
$endgroup$
– mehmet
Mar 9 '15 at 17:45
$begingroup$
This is a good one except "-2" part (I think I must be directly "2"). Thanks.
$endgroup$
– mehmet
Mar 9 '15 at 17:45
1
1
$begingroup$
Soryy, I forgot to say I mean in sum operation. I believe it must be positive sign of 2.
$endgroup$
– mehmet
Mar 9 '15 at 18:16
$begingroup$
Soryy, I forgot to say I mean in sum operation. I believe it must be positive sign of 2.
$endgroup$
– mehmet
Mar 9 '15 at 18:16
$begingroup$
Also the sign of the sum operation must be negative, I think?
$endgroup$
– mehmet
Mar 9 '15 at 18:26
$begingroup$
Also the sign of the sum operation must be negative, I think?
$endgroup$
– mehmet
Mar 9 '15 at 18:26
1
1
$begingroup$
@mehmet: OK, I see now. Corrected that, thanks.
$endgroup$
– Marc van Leeuwen
Mar 10 '15 at 5:28
$begingroup$
@mehmet: OK, I see now. Corrected that, thanks.
$endgroup$
– Marc van Leeuwen
Mar 10 '15 at 5:28
$begingroup$
Out of interest how would you do Laurent Expansions with Integrals if I may ask
$endgroup$
– Zophikel
Sep 11 '17 at 20:39
$begingroup$
Out of interest how would you do Laurent Expansions with Integrals if I may ask
$endgroup$
– Zophikel
Sep 11 '17 at 20:39
add a comment |
$begingroup$
The problem is that if you use $frac 1{1-z} = sum z^n$ you are essentially writing the Laurent expansion in a neighborhood of $0$. Since you want powers of $z-1$, it means that you want an expansion in a neighborhood of $1$!
Hence the $frac {1}{z-1} = (z-1)^{-1}$ term is already "good" (just like $frac 1z$ would be in a Laurent expansion in a neighborhood of $0$).
You have to expand in a neighborhood of $1$ the expression $frac z{z-3}$.
You can set $t = z - 1 implies z = t + 1$ so your expression becomes $frac{t+1}{t-2}$.
Now use partial fractions, and find the Laurent series in a neighborhood of $0$ with respect to $t$ of $frac{t+1}{t-2}$ (This means you can use the geometric series formula), substitute back $t = z-1$ and divide everything by $z-1$ to get the final result.
edited Jan 1 at 19:43
amWhy
192k28225439
answered Mar 8 '15 at 10:32
AntAnt
17.4k22873
$endgroup$
$begingroup$
Thanks for your explanation! (:
$endgroup$
– mehmet
Mar 9 '15 at 18:19
1
$begingroup$
@mehmet You're welcome! :)
$endgroup$
– Ant
Mar 9 '15 at 18:20
add a comment |
$begingroup$
The problem is that if you use $frac 1{1-z} = sum z^n$ you are essentially writing the Laurent expansion in a neighborhood of $0$. Since you want powers of $z-1$, it means that you want an expansion in a neighborhood of $1$!
Hence the $frac {1}{z-1} = (z-1)^{-1}$ term is already "good" (just like $frac 1z$ would be in a Laurent expansion in a neighborhood of $0$).
You have to expand in a neighborhood of $1$ the expression $frac z{z-3}$.
You can set $t = z - 1 implies z = t + 1$ so your expression becomes $frac{t+1}{t-2}$.
Now use partial fractions, and find the Laurent series in a neighborhood of $0$ with respect to $t$ of $frac{t+1}{t-2}$ (This means you can use the geometric series formula), substitute back $t = z-1$ and divide everything by $z-1$ to get the final result.
edited Jan 1 at 19:43
amWhy
192k28225439
answered Mar 8 '15 at 10:32
AntAnt
17.4k22873
$endgroup$
$begingroup$
Thanks for your explanation! (:
$endgroup$
– mehmet
Mar 9 '15 at 18:19
1
$begingroup$
@mehmet You're welcome! :)
$endgroup$
– Ant
Mar 9 '15 at 18:20
add a comment |
$begingroup$
The problem is that if you use $frac 1{1-z} = sum z^n$ you are essentially writing the Laurent expansion in a neighborhood of $0$. Since you want powers of $z-1$, it means that you want an expansion in a neighborhood of $1$!
Hence the $frac {1}{z-1} = (z-1)^{-1}$ term is already "good" (just like $frac 1z$ would be in a Laurent expansion in a neighborhood of $0$).
You have to expand in a neighborhood of $1$ the expression $frac z{z-3}$.
You can set $t = z - 1 implies z = t + 1$ so your expression becomes $frac{t+1}{t-2}$.
Now use partial fractions, and find the Laurent series in a neighborhood of $0$ with respect to $t$ of $frac{t+1}{t-2}$ (This means you can use the geometric series formula), substitute back $t = z-1$ and divide everything by $z-1$ to get the final result.
edited Jan 1 at 19:43
amWhy
192k28225439
answered Mar 8 '15 at 10:32
AntAnt
17.4k22873
$endgroup$
The problem is that if you use $frac 1{1-z} = sum z^n$ you are essentially writing the Laurent expansion in a neighborhood of $0$. Since you want powers of $z-1$, it means that you want an expansion in a neighborhood of $1$!
Hence the $frac {1}{z-1} = (z-1)^{-1}$ term is already "good" (just like $frac 1z$ would be in a Laurent expansion in a neighborhood of $0$).
You have to expand in a neighborhood of $1$ the expression $frac z{z-3}$.
You can set $t = z - 1 implies z = t + 1$ so your expression becomes $frac{t+1}{t-2}$.
Now use partial fractions, and find the Laurent series in a neighborhood of $0$ with respect to $t$ of $frac{t+1}{t-2}$ (This means you can use the geometric series formula), substitute back $t = z-1$ and divide everything by $z-1$ to get the final result.
edited Jan 1 at 19:43
amWhy
192k28225439
answered Mar 8 '15 at 10:32
AntAnt
17.4k22873
edited Jan 1 at 19:43
amWhy
192k28225439
edited Jan 1 at 19:43
amWhy
192k28225439
edited Jan 1 at 19:43
amWhy
192k28225439
192k28225439
answered Mar 8 '15 at 10:32
AntAnt
17.4k22873
answered Mar 8 '15 at 10:32
AntAnt
17.4k22873
answered Mar 8 '15 at 10:32
AntAnt
17.4k22873
17.4k22873
$begingroup$
Thanks for your explanation! (:
$endgroup$
– mehmet
Mar 9 '15 at 18:19
1
$begingroup$
@mehmet You're welcome! :)
$endgroup$
– Ant
Mar 9 '15 at 18:20
add a comment |
$begingroup$
Thanks for your explanation! (:
$endgroup$
– mehmet
Mar 9 '15 at 18:19
1
$begingroup$
@mehmet You're welcome! :)
$endgroup$
– Ant
Mar 9 '15 at 18:20
$begingroup$
Thanks for your explanation! (:
$endgroup$
– mehmet
Mar 9 '15 at 18:19
$begingroup$
Thanks for your explanation! (:
$endgroup$
– mehmet
Mar 9 '15 at 18:19
1
1
$begingroup$
@mehmet You're welcome! :)
$endgroup$
– Ant
Mar 9 '15 at 18:20
$begingroup$
@mehmet You're welcome! :)
$endgroup$
– Ant
Mar 9 '15 at 18:20
add a comment |
$begingroup$
First, yes. The same function can have different Laurent series, depending on the center of annulus in question.
Consider now the given function. Clearly, the question will be solved with ease once we find the series of $$g(z)=frac{z}{z-3}.$$Note that$$g(z)=1+frac{3}{z-3}=1+frac{3}{(z-1)-2}=1+frac{3/2}{frac{z-1}{2}-1},$$and the last expression can be represented as the sum of a geometric series.
answered Mar 8 '15 at 10:38
Amitai YuvalAmitai Yuval
15.1k11126
$endgroup$
$begingroup$
Thank you for your answer (:
$endgroup$
– mehmet
Mar 9 '15 at 17:42
add a comment |
$begingroup$
First, yes. The same function can have different Laurent series, depending on the center of annulus in question.
Consider now the given function. Clearly, the question will be solved with ease once we find the series of $$g(z)=frac{z}{z-3}.$$Note that$$g(z)=1+frac{3}{z-3}=1+frac{3}{(z-1)-2}=1+frac{3/2}{frac{z-1}{2}-1},$$and the last expression can be represented as the sum of a geometric series.
answered Mar 8 '15 at 10:38
Amitai YuvalAmitai Yuval
15.1k11126
$endgroup$
$begingroup$
Thank you for your answer (:
$endgroup$
– mehmet
Mar 9 '15 at 17:42
add a comment |
$begingroup$
First, yes. The same function can have different Laurent series, depending on the center of annulus in question.
Consider now the given function. Clearly, the question will be solved with ease once we find the series of $$g(z)=frac{z}{z-3}.$$Note that$$g(z)=1+frac{3}{z-3}=1+frac{3}{(z-1)-2}=1+frac{3/2}{frac{z-1}{2}-1},$$and the last expression can be represented as the sum of a geometric series.
answered Mar 8 '15 at 10:38
Amitai YuvalAmitai Yuval
15.1k11126
$endgroup$
First, yes. The same function can have different Laurent series, depending on the center of annulus in question.
Consider now the given function. Clearly, the question will be solved with ease once we find the series of $$g(z)=frac{z}{z-3}.$$Note that$$g(z)=1+frac{3}{z-3}=1+frac{3}{(z-1)-2}=1+frac{3/2}{frac{z-1}{2}-1},$$and the last expression can be represented as the sum of a geometric series.
answered Mar 8 '15 at 10:38
Amitai YuvalAmitai Yuval
15.1k11126
answered Mar 8 '15 at 10:38
Amitai YuvalAmitai Yuval
15.1k11126
answered Mar 8 '15 at 10:38
Amitai YuvalAmitai Yuval
15.1k11126
answered Mar 8 '15 at 10:38
Amitai YuvalAmitai Yuval
15.1k11126
15.1k11126
$begingroup$
Thank you for your answer (:
$endgroup$
– mehmet
Mar 9 '15 at 17:42
add a comment |
$begingroup$
Thank you for your answer (:
$endgroup$
– mehmet
Mar 9 '15 at 17:42
$begingroup$
Thank you for your answer (:
$endgroup$
– mehmet
Mar 9 '15 at 17:42
$begingroup$
Thank you for your answer (:
$endgroup$
– mehmet
Mar 9 '15 at 17:42
add a comment |
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