Mixing
Is the textbook solution wrong by a sign? Laurent Series
$begingroup$
Find the Laurent series of $frac{e^z}{z^2 -1}$ about $z = 1$.
Here is my solution:
Factor denominator
$frac{e^z}{(z-1)(z+1)}$
let $w = z - 1$, and so $z = w + 1$, substitute in
$frac{e^{w+1}}{w(w+2)}$
Do partial fraction decomposition to get rid of the exponential in numerator
$frac{e^{w+1}}{w(w+2)}$ = $frac{A}{w} + frac{B}{w+2}$
$e^{w+1} = A(w+2) + Bw$
If $w=0$, $e=2A$, $A = e/2$. If $w = -2$, $frac{1}{e} = -2B$, $B = frac{-1}
{2e}$
Thus our new equation is
$frac{e}{2w} + frac{-1}{2e(w+2)}$
Because the first term $frac{e}{2w}$ is already in terms of $w = (z-1), we leave it be. The second term we can expand using geometric series expansion
$frac{-1}{4e} cdot frac{1}{1 - (frac{-w}{2})}$
and so the Laurent series we get is
$$frac{e}{2w} - frac{1}{4e} cdot {1 - frac{w}{2} + frac{w^2}{4} - frac{w^3}{8} ... } $$
However, the textbook solution has same terms, but no negative sign. Where did I go wrong?
complex-analysis laurent-series
edited Jan 23 at 6:37
Robert Z
100k1069140
asked Jan 23 at 5:52
MinYoung KimMinYoung Kim
907
$endgroup$
add a comment |
$begingroup$
Find the Laurent series of $frac{e^z}{z^2 -1}$ about $z = 1$.
Here is my solution:
Factor denominator
$frac{e^z}{(z-1)(z+1)}$
let $w = z - 1$, and so $z = w + 1$, substitute in
$frac{e^{w+1}}{w(w+2)}$
Do partial fraction decomposition to get rid of the exponential in numerator
$frac{e^{w+1}}{w(w+2)}$ = $frac{A}{w} + frac{B}{w+2}$
$e^{w+1} = A(w+2) + Bw$
If $w=0$, $e=2A$, $A = e/2$. If $w = -2$, $frac{1}{e} = -2B$, $B = frac{-1}
{2e}$
Thus our new equation is
$frac{e}{2w} + frac{-1}{2e(w+2)}$
Because the first term $frac{e}{2w}$ is already in terms of $w = (z-1), we leave it be. The second term we can expand using geometric series expansion
$frac{-1}{4e} cdot frac{1}{1 - (frac{-w}{2})}$
and so the Laurent series we get is
$$frac{e}{2w} - frac{1}{4e} cdot {1 - frac{w}{2} + frac{w^2}{4} - frac{w^3}{8} ... } $$
However, the textbook solution has same terms, but no negative sign. Where did I go wrong?
complex-analysis laurent-series
edited Jan 23 at 6:37
Robert Z
100k1069140
asked Jan 23 at 5:52
MinYoung KimMinYoung Kim
907
$endgroup$
$begingroup$
Ummm… that's not a rational function… How could you do partial fraction decomposition?
$endgroup$
– xbh
Jan 23 at 6:16
add a comment |
$begingroup$
Find the Laurent series of $frac{e^z}{z^2 -1}$ about $z = 1$.
Here is my solution:
Factor denominator
$frac{e^z}{(z-1)(z+1)}$
let $w = z - 1$, and so $z = w + 1$, substitute in
$frac{e^{w+1}}{w(w+2)}$
Do partial fraction decomposition to get rid of the exponential in numerator
$frac{e^{w+1}}{w(w+2)}$ = $frac{A}{w} + frac{B}{w+2}$
$e^{w+1} = A(w+2) + Bw$
If $w=0$, $e=2A$, $A = e/2$. If $w = -2$, $frac{1}{e} = -2B$, $B = frac{-1}
{2e}$
Thus our new equation is
$frac{e}{2w} + frac{-1}{2e(w+2)}$
Because the first term $frac{e}{2w}$ is already in terms of $w = (z-1), we leave it be. The second term we can expand using geometric series expansion
$frac{-1}{4e} cdot frac{1}{1 - (frac{-w}{2})}$
and so the Laurent series we get is
$$frac{e}{2w} - frac{1}{4e} cdot {1 - frac{w}{2} + frac{w^2}{4} - frac{w^3}{8} ... } $$
However, the textbook solution has same terms, but no negative sign. Where did I go wrong?
complex-analysis laurent-series
edited Jan 23 at 6:37
Robert Z
100k1069140
asked Jan 23 at 5:52
MinYoung KimMinYoung Kim
907
$endgroup$
Find the Laurent series of $frac{e^z}{z^2 -1}$ about $z = 1$.
Here is my solution:
Factor denominator
$frac{e^z}{(z-1)(z+1)}$
let $w = z - 1$, and so $z = w + 1$, substitute in
$frac{e^{w+1}}{w(w+2)}$
Do partial fraction decomposition to get rid of the exponential in numerator
$frac{e^{w+1}}{w(w+2)}$ = $frac{A}{w} + frac{B}{w+2}$
$e^{w+1} = A(w+2) + Bw$
If $w=0$, $e=2A$, $A = e/2$. If $w = -2$, $frac{1}{e} = -2B$, $B = frac{-1}
{2e}$
Thus our new equation is
$frac{e}{2w} + frac{-1}{2e(w+2)}$
Because the first term $frac{e}{2w}$ is already in terms of $w = (z-1), we leave it be. The second term we can expand using geometric series expansion
$frac{-1}{4e} cdot frac{1}{1 - (frac{-w}{2})}$
and so the Laurent series we get is
$$frac{e}{2w} - frac{1}{4e} cdot {1 - frac{w}{2} + frac{w^2}{4} - frac{w^3}{8} ... } $$
However, the textbook solution has same terms, but no negative sign. Where did I go wrong?
complex-analysis laurent-series
complex-analysis laurent-series
edited Jan 23 at 6:37
Robert Z
100k1069140
asked Jan 23 at 5:52
MinYoung KimMinYoung Kim
907
edited Jan 23 at 6:37
Robert Z
100k1069140
asked Jan 23 at 5:52
MinYoung KimMinYoung Kim
907
edited Jan 23 at 6:37
Robert Z
100k1069140
edited Jan 23 at 6:37
Robert Z
100k1069140
edited Jan 23 at 6:37
Robert Z
100k1069140
100k1069140
asked Jan 23 at 5:52
MinYoung KimMinYoung Kim
907
asked Jan 23 at 5:52
MinYoung KimMinYoung Kim
907
asked Jan 23 at 5:52
MinYoung KimMinYoung Kim
907
907
$begingroup$
Ummm… that's not a rational function… How could you do partial fraction decomposition?
$endgroup$
– xbh
Jan 23 at 6:16
add a comment |
$begingroup$
Ummm… that's not a rational function… How could you do partial fraction decomposition?
$endgroup$
– xbh
Jan 23 at 6:16
$begingroup$
Ummm… that's not a rational function… How could you do partial fraction decomposition?
$endgroup$
– xbh
Jan 23 at 6:16
$begingroup$
Ummm… that's not a rational function… How could you do partial fraction decomposition?
$endgroup$
– xbh
Jan 23 at 6:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You considered $e^{w+1}$ as a rational function (which is not!). Instead you should expand it at $w=0$ as
$$e^{w+1}=ecdot e^w=e left(1+w+frac{w^2}{2}+frac{w^3}{6}+dotsright).$$
Hence, after decomposing the rational function $frac{1}{w(w+2)}$,
$$frac{e^{w+1}}{w(w+2)}=e^{w+1}left(frac{1}{2w} - frac{1/4}{1+w/2}right)\=
eleft(1+w+frac{w^2}{2}+frac{w^3}{6}+dotsright)left(frac{1}{2w} - frac{1}{4}left(1-frac{w}{2}+frac{w^2}{4}+dotsright)right).$$
Can you take it from here?
Finally the Laurent expansion should be
$$frac{e}{2w} + frac{e}{4} left(1 + frac{w}{2} + frac{w^2}{12} +dotsright).$$
answered Jan 23 at 6:22
Robert ZRobert Z
100k1069140
$endgroup$
$begingroup$
So there's no way to avoid the tedious multiplication of series terms? Also why is $e^z$ not a rational function?
$endgroup$
– MinYoung Kim
Jan 23 at 15:54
1
$begingroup$
No. But how many terms of the expansion do you need? A rational function is by definition a ratio of polynomials.
$endgroup$
– Robert Z
Jan 23 at 16:00
$begingroup$
not alot , but would prefer to avoid it if possible. Thank you.
$endgroup$
– MinYoung Kim
Jan 23 at 16:37
1
$begingroup$
In order to have the Laurent expansion up to $w^2$ you need the expansion of $e^w$ up to $w^3$.
$endgroup$
– Robert Z
Jan 23 at 16:40
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You considered $e^{w+1}$ as a rational function (which is not!). Instead you should expand it at $w=0$ as
$$e^{w+1}=ecdot e^w=e left(1+w+frac{w^2}{2}+frac{w^3}{6}+dotsright).$$
Hence, after decomposing the rational function $frac{1}{w(w+2)}$,
$$frac{e^{w+1}}{w(w+2)}=e^{w+1}left(frac{1}{2w} - frac{1/4}{1+w/2}right)\=
eleft(1+w+frac{w^2}{2}+frac{w^3}{6}+dotsright)left(frac{1}{2w} - frac{1}{4}left(1-frac{w}{2}+frac{w^2}{4}+dotsright)right).$$
Can you take it from here?
Finally the Laurent expansion should be
$$frac{e}{2w} + frac{e}{4} left(1 + frac{w}{2} + frac{w^2}{12} +dotsright).$$
answered Jan 23 at 6:22
Robert ZRobert Z
100k1069140
$endgroup$
$begingroup$
So there's no way to avoid the tedious multiplication of series terms? Also why is $e^z$ not a rational function?
$endgroup$
– MinYoung Kim
Jan 23 at 15:54
1
$begingroup$
No. But how many terms of the expansion do you need? A rational function is by definition a ratio of polynomials.
$endgroup$
– Robert Z
Jan 23 at 16:00
$begingroup$
not alot , but would prefer to avoid it if possible. Thank you.
$endgroup$
– MinYoung Kim
Jan 23 at 16:37
1
$begingroup$
In order to have the Laurent expansion up to $w^2$ you need the expansion of $e^w$ up to $w^3$.
$endgroup$
– Robert Z
Jan 23 at 16:40
add a comment |
$begingroup$
You considered $e^{w+1}$ as a rational function (which is not!). Instead you should expand it at $w=0$ as
$$e^{w+1}=ecdot e^w=e left(1+w+frac{w^2}{2}+frac{w^3}{6}+dotsright).$$
Hence, after decomposing the rational function $frac{1}{w(w+2)}$,
$$frac{e^{w+1}}{w(w+2)}=e^{w+1}left(frac{1}{2w} - frac{1/4}{1+w/2}right)\=
eleft(1+w+frac{w^2}{2}+frac{w^3}{6}+dotsright)left(frac{1}{2w} - frac{1}{4}left(1-frac{w}{2}+frac{w^2}{4}+dotsright)right).$$
Can you take it from here?
Finally the Laurent expansion should be
$$frac{e}{2w} + frac{e}{4} left(1 + frac{w}{2} + frac{w^2}{12} +dotsright).$$
answered Jan 23 at 6:22
Robert ZRobert Z
100k1069140
$endgroup$
$begingroup$
So there's no way to avoid the tedious multiplication of series terms? Also why is $e^z$ not a rational function?
$endgroup$
– MinYoung Kim
Jan 23 at 15:54
1
$begingroup$
No. But how many terms of the expansion do you need? A rational function is by definition a ratio of polynomials.
$endgroup$
– Robert Z
Jan 23 at 16:00
$begingroup$
not alot , but would prefer to avoid it if possible. Thank you.
$endgroup$
– MinYoung Kim
Jan 23 at 16:37
1
$begingroup$
In order to have the Laurent expansion up to $w^2$ you need the expansion of $e^w$ up to $w^3$.
$endgroup$
– Robert Z
Jan 23 at 16:40
add a comment |
$begingroup$
You considered $e^{w+1}$ as a rational function (which is not!). Instead you should expand it at $w=0$ as
$$e^{w+1}=ecdot e^w=e left(1+w+frac{w^2}{2}+frac{w^3}{6}+dotsright).$$
Hence, after decomposing the rational function $frac{1}{w(w+2)}$,
$$frac{e^{w+1}}{w(w+2)}=e^{w+1}left(frac{1}{2w} - frac{1/4}{1+w/2}right)\=
eleft(1+w+frac{w^2}{2}+frac{w^3}{6}+dotsright)left(frac{1}{2w} - frac{1}{4}left(1-frac{w}{2}+frac{w^2}{4}+dotsright)right).$$
Can you take it from here?
Finally the Laurent expansion should be
$$frac{e}{2w} + frac{e}{4} left(1 + frac{w}{2} + frac{w^2}{12} +dotsright).$$
answered Jan 23 at 6:22
Robert ZRobert Z
100k1069140
$endgroup$
You considered $e^{w+1}$ as a rational function (which is not!). Instead you should expand it at $w=0$ as
$$e^{w+1}=ecdot e^w=e left(1+w+frac{w^2}{2}+frac{w^3}{6}+dotsright).$$
Hence, after decomposing the rational function $frac{1}{w(w+2)}$,
$$frac{e^{w+1}}{w(w+2)}=e^{w+1}left(frac{1}{2w} - frac{1/4}{1+w/2}right)\=
eleft(1+w+frac{w^2}{2}+frac{w^3}{6}+dotsright)left(frac{1}{2w} - frac{1}{4}left(1-frac{w}{2}+frac{w^2}{4}+dotsright)right).$$
Can you take it from here?
Finally the Laurent expansion should be
$$frac{e}{2w} + frac{e}{4} left(1 + frac{w}{2} + frac{w^2}{12} +dotsright).$$
answered Jan 23 at 6:22
Robert ZRobert Z
100k1069140
edited Jan 23 at 6:51
answered Jan 23 at 6:22
Robert ZRobert Z
100k1069140
answered Jan 23 at 6:22
Robert ZRobert Z
100k1069140
answered Jan 23 at 6:22
Robert ZRobert Z
100k1069140
100k1069140
$begingroup$
So there's no way to avoid the tedious multiplication of series terms? Also why is $e^z$ not a rational function?
$endgroup$
– MinYoung Kim
Jan 23 at 15:54
1
$begingroup$
No. But how many terms of the expansion do you need? A rational function is by definition a ratio of polynomials.
$endgroup$
– Robert Z
Jan 23 at 16:00
$begingroup$
not alot , but would prefer to avoid it if possible. Thank you.
$endgroup$
– MinYoung Kim
Jan 23 at 16:37
1
$begingroup$
In order to have the Laurent expansion up to $w^2$ you need the expansion of $e^w$ up to $w^3$.
$endgroup$
– Robert Z
Jan 23 at 16:40
add a comment |
$begingroup$
So there's no way to avoid the tedious multiplication of series terms? Also why is $e^z$ not a rational function?
$endgroup$
– MinYoung Kim
Jan 23 at 15:54
1
$begingroup$
No. But how many terms of the expansion do you need? A rational function is by definition a ratio of polynomials.
$endgroup$
– Robert Z
Jan 23 at 16:00
$begingroup$
not alot , but would prefer to avoid it if possible. Thank you.
$endgroup$
– MinYoung Kim
Jan 23 at 16:37
1
$begingroup$
In order to have the Laurent expansion up to $w^2$ you need the expansion of $e^w$ up to $w^3$.
$endgroup$
– Robert Z
Jan 23 at 16:40
$begingroup$
So there's no way to avoid the tedious multiplication of series terms? Also why is $e^z$ not a rational function?
$endgroup$
– MinYoung Kim
Jan 23 at 15:54
$begingroup$
So there's no way to avoid the tedious multiplication of series terms? Also why is $e^z$ not a rational function?
$endgroup$
– MinYoung Kim
Jan 23 at 15:54
1
1
$begingroup$
No. But how many terms of the expansion do you need? A rational function is by definition a ratio of polynomials.
$endgroup$
– Robert Z
Jan 23 at 16:00
$begingroup$
No. But how many terms of the expansion do you need? A rational function is by definition a ratio of polynomials.
$endgroup$
– Robert Z
Jan 23 at 16:00
$begingroup$
not alot , but would prefer to avoid it if possible. Thank you.
$endgroup$
– MinYoung Kim
Jan 23 at 16:37
$begingroup$
not alot , but would prefer to avoid it if possible. Thank you.
$endgroup$
– MinYoung Kim
Jan 23 at 16:37
1
1
$begingroup$
In order to have the Laurent expansion up to $w^2$ you need the expansion of $e^w$ up to $w^3$.
$endgroup$
– Robert Z
Jan 23 at 16:40
$begingroup$
In order to have the Laurent expansion up to $w^2$ you need the expansion of $e^w$ up to $w^3$.
$endgroup$
– Robert Z
Jan 23 at 16:40
add a comment |
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$begingroup$
Ummm… that's not a rational function… How could you do partial fraction decomposition?
$endgroup$
– xbh
Jan 23 at 6:16