Laurent Series $e^frac11-z$, $|z|>1$.

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I tried to find Laurent expansion for:



$e^frac11-z$, $|z|>1$.



I tried next: $frac11-z=-sum_n=1^inftyfrac1z^n$,



then using $e^frac11-z=1+frac11-z+frac12!(1-z)^2+frac13!(1-z)^3+...$



Then I have $e^frac11-z$ = $1$ + $(-frac1z-frac1z^2-frac1z^3-...) + (frac12!)(frac1z^2+frac1z^4+…)+...$ = $1 - frac1z-frac12z^2-...$.



The result is wrong, since in the book the answer is: $1 - frac1z+ frac12z^2-frac16z^3+frac124z^4-frac19120z^5...$



What I did wrong?




complex-analysis laurent-series




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  • @DavidC.Ullrich ... Yes, using invalid steps can make any problems easier.
    – GEdgar
    Jul 30 at 16:56










  • @GEdgar Omg I can't believe I said that...
    – David C. Ullrich
    Jul 30 at 17:00














up vote
0
down vote

favorite
1












I tried to find Laurent expansion for:



$e^frac11-z$, $|z|>1$.



I tried next: $frac11-z=-sum_n=1^inftyfrac1z^n$,



then using $e^frac11-z=1+frac11-z+frac12!(1-z)^2+frac13!(1-z)^3+...$



Then I have $e^frac11-z$ = $1$ + $(-frac1z-frac1z^2-frac1z^3-...) + (frac12!)(frac1z^2+frac1z^4+…)+...$ = $1 - frac1z-frac12z^2-...$.



The result is wrong, since in the book the answer is: $1 - frac1z+ frac12z^2-frac16z^3+frac124z^4-frac19120z^5...$



What I did wrong?




complex-analysis laurent-series




share|cite|improve this question





















  • @DavidC.Ullrich ... Yes, using invalid steps can make any problems easier.
    – GEdgar
    Jul 30 at 16:56










  • @GEdgar Omg I can't believe I said that...
    – David C. Ullrich
    Jul 30 at 17:00












up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





I tried to find Laurent expansion for:



$e^frac11-z$, $|z|>1$.



I tried next: $frac11-z=-sum_n=1^inftyfrac1z^n$,



then using $e^frac11-z=1+frac11-z+frac12!(1-z)^2+frac13!(1-z)^3+...$



Then I have $e^frac11-z$ = $1$ + $(-frac1z-frac1z^2-frac1z^3-...) + (frac12!)(frac1z^2+frac1z^4+…)+...$ = $1 - frac1z-frac12z^2-...$.



The result is wrong, since in the book the answer is: $1 - frac1z+ frac12z^2-frac16z^3+frac124z^4-frac19120z^5...$



What I did wrong?




complex-analysis laurent-series




share|cite|improve this question













I tried to find Laurent expansion for:



$e^frac11-z$, $|z|>1$.



I tried next: $frac11-z=-sum_n=1^inftyfrac1z^n$,



then using $e^frac11-z=1+frac11-z+frac12!(1-z)^2+frac13!(1-z)^3+...$



Then I have $e^frac11-z$ = $1$ + $(-frac1z-frac1z^2-frac1z^3-...) + (frac12!)(frac1z^2+frac1z^4+…)+...$ = $1 - frac1z-frac12z^2-...$.



The result is wrong, since in the book the answer is: $1 - frac1z+ frac12z^2-frac16z^3+frac124z^4-frac19120z^5...$



What I did wrong?





complex-analysis laurent-series





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 30 at 16:53









David C. Ullrich

53.8k33481




53.8k33481









asked Jul 30 at 16:31









Metso

906




906











  • @DavidC.Ullrich ... Yes, using invalid steps can make any problems easier.
    – GEdgar
    Jul 30 at 16:56










  • @GEdgar Omg I can't believe I said that...
    – David C. Ullrich
    Jul 30 at 17:00
















  • @DavidC.Ullrich ... Yes, using invalid steps can make any problems easier.
    – GEdgar
    Jul 30 at 16:56










  • @GEdgar Omg I can't believe I said that...
    – David C. Ullrich
    Jul 30 at 17:00















@DavidC.Ullrich ... Yes, using invalid steps can make any problems easier.
– GEdgar
Jul 30 at 16:56




@DavidC.Ullrich ... Yes, using invalid steps can make any problems easier.
– GEdgar
Jul 30 at 16:56












@GEdgar Omg I can't believe I said that...
– David C. Ullrich
Jul 30 at 17:00




@GEdgar Omg I can't believe I said that...
– David C. Ullrich
Jul 30 at 17:00










2 Answers
2






active

oldest

votes

















up vote
1
down vote



accepted










With $z=dfrac1w$ we have $|w|<1$ then by $e^z$ expansion
beginalign
e^frac11-z
&= e^frac-w1-w \
&= sum_ngeq0dfrac1n!left(frac-w1-wright)^n \
&= sum_ngeq0dfrac(-w)^nn!left(1-wright)^-n \
&= sum_ngeq0dfrac(-w)^nn!left(1+nw+dfracn(n+1)2w^2+cdotsright) \
&= sum_ngeq0dfrac(-1)^nn!left(w^n+nw^n+1+dfracn(n+1)2w^n+2+cdotsright)
endalign
after calculation some terms we let $w=dfrac1z$.



Edit: One may find
beginalign
&1 \
&-(w+w^2+w^3+w^4+w^5+cdots) \
&+frac12(w^2+2w^3+3w^4+4w^5+cdots) \
&-frac16(w^3+3w^4+6w^5+cdots) \
&+cdots\
&=1-w-fracw^22-fracw^36+fracw^424+frac19 w^5120+cdots\
&=colorblue1-dfrac1z-frac12z^2-frac16z^3+frac124z^4+frac19120z^5+cdots
endalign






share|cite|improve this answer























  • the answer in the book: $1 - frac1z+ frac12z^2-frac16z^3+frac124z^4-frac19120z^5…$. I can't get it.
    – Metso
    Jul 30 at 18:27











  • See the series and let $n=0$
    – user 108128
    Jul 30 at 18:32










  • so the first is $1$ right?
    – user 108128
    Jul 30 at 18:33










  • yes, the first is 1. I'll try other numbers
    – Metso
    Jul 30 at 18:37











  • now for $n=1$ it is $-(w+w^2)$
    – user 108128
    Jul 30 at 18:39

















up vote
2
down vote













Note that$$left(-frac1z-frac1z^2-frac1z^3-frac1z^4-cdotsright)^2=frac1z^2+frac2z^3+frac3z^4+cdots,$$that$$left(-frac1z-frac1z^2-frac1z^3-frac1z^4-cdotsright)^3=-frac1z^3-frac3z^4-frac6z^5-cdots,$$and so on. This explains why you got the wrong result.






share|cite|improve this answer





















  • sorry, but I can't get the answer from the book. I have again $1 - frac1z-frac12z^2-...$, but not $1 - frac1z+frac12z^2-...$
    – Metso
    Jul 30 at 17:48











  • What's the smallest $n$ for which your coefficient of $frac1z^n$ is different from the one from the book?
    – José Carlos Santos
    Jul 30 at 17:51










  • the answer: $1 - frac1z+ frac12z^2-frac16z^3+frac124z^4-frac19120z^5…$ As you can see it starts from $frac12z^2$, $n=2$
    – Metso
    Jul 30 at 17:52











  • @Metso You forgot the $frac12!$ that you wrote yourself in the original question.
    – José Carlos Santos
    Jul 30 at 18:02











  • result is $1-frac1z -frac12!z^2$
    – Metso
    Jul 30 at 18:07











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










With $z=dfrac1w$ we have $|w|<1$ then by $e^z$ expansion
beginalign
e^frac11-z
&= e^frac-w1-w \
&= sum_ngeq0dfrac1n!left(frac-w1-wright)^n \
&= sum_ngeq0dfrac(-w)^nn!left(1-wright)^-n \
&= sum_ngeq0dfrac(-w)^nn!left(1+nw+dfracn(n+1)2w^2+cdotsright) \
&= sum_ngeq0dfrac(-1)^nn!left(w^n+nw^n+1+dfracn(n+1)2w^n+2+cdotsright)
endalign
after calculation some terms we let $w=dfrac1z$.



Edit: One may find
beginalign
&1 \
&-(w+w^2+w^3+w^4+w^5+cdots) \
&+frac12(w^2+2w^3+3w^4+4w^5+cdots) \
&-frac16(w^3+3w^4+6w^5+cdots) \
&+cdots\
&=1-w-fracw^22-fracw^36+fracw^424+frac19 w^5120+cdots\
&=colorblue1-dfrac1z-frac12z^2-frac16z^3+frac124z^4+frac19120z^5+cdots
endalign






share|cite|improve this answer























  • the answer in the book: $1 - frac1z+ frac12z^2-frac16z^3+frac124z^4-frac19120z^5…$. I can't get it.
    – Metso
    Jul 30 at 18:27











  • See the series and let $n=0$
    – user 108128
    Jul 30 at 18:32










  • so the first is $1$ right?
    – user 108128
    Jul 30 at 18:33










  • yes, the first is 1. I'll try other numbers
    – Metso
    Jul 30 at 18:37











  • now for $n=1$ it is $-(w+w^2)$
    – user 108128
    Jul 30 at 18:39














up vote
1
down vote



accepted










With $z=dfrac1w$ we have $|w|<1$ then by $e^z$ expansion
beginalign
e^frac11-z
&= e^frac-w1-w \
&= sum_ngeq0dfrac1n!left(frac-w1-wright)^n \
&= sum_ngeq0dfrac(-w)^nn!left(1-wright)^-n \
&= sum_ngeq0dfrac(-w)^nn!left(1+nw+dfracn(n+1)2w^2+cdotsright) \
&= sum_ngeq0dfrac(-1)^nn!left(w^n+nw^n+1+dfracn(n+1)2w^n+2+cdotsright)
endalign
after calculation some terms we let $w=dfrac1z$.



Edit: One may find
beginalign
&1 \
&-(w+w^2+w^3+w^4+w^5+cdots) \
&+frac12(w^2+2w^3+3w^4+4w^5+cdots) \
&-frac16(w^3+3w^4+6w^5+cdots) \
&+cdots\
&=1-w-fracw^22-fracw^36+fracw^424+frac19 w^5120+cdots\
&=colorblue1-dfrac1z-frac12z^2-frac16z^3+frac124z^4+frac19120z^5+cdots
endalign






share|cite|improve this answer























  • the answer in the book: $1 - frac1z+ frac12z^2-frac16z^3+frac124z^4-frac19120z^5…$. I can't get it.
    – Metso
    Jul 30 at 18:27











  • See the series and let $n=0$
    – user 108128
    Jul 30 at 18:32










  • so the first is $1$ right?
    – user 108128
    Jul 30 at 18:33










  • yes, the first is 1. I'll try other numbers
    – Metso
    Jul 30 at 18:37











  • now for $n=1$ it is $-(w+w^2)$
    – user 108128
    Jul 30 at 18:39












up vote
1
down vote



accepted







up vote
1
down vote



accepted






With $z=dfrac1w$ we have $|w|<1$ then by $e^z$ expansion
beginalign
e^frac11-z
&= e^frac-w1-w \
&= sum_ngeq0dfrac1n!left(frac-w1-wright)^n \
&= sum_ngeq0dfrac(-w)^nn!left(1-wright)^-n \
&= sum_ngeq0dfrac(-w)^nn!left(1+nw+dfracn(n+1)2w^2+cdotsright) \
&= sum_ngeq0dfrac(-1)^nn!left(w^n+nw^n+1+dfracn(n+1)2w^n+2+cdotsright)
endalign
after calculation some terms we let $w=dfrac1z$.



Edit: One may find
beginalign
&1 \
&-(w+w^2+w^3+w^4+w^5+cdots) \
&+frac12(w^2+2w^3+3w^4+4w^5+cdots) \
&-frac16(w^3+3w^4+6w^5+cdots) \
&+cdots\
&=1-w-fracw^22-fracw^36+fracw^424+frac19 w^5120+cdots\
&=colorblue1-dfrac1z-frac12z^2-frac16z^3+frac124z^4+frac19120z^5+cdots
endalign






share|cite|improve this answer















With $z=dfrac1w$ we have $|w|<1$ then by $e^z$ expansion
beginalign
e^frac11-z
&= e^frac-w1-w \
&= sum_ngeq0dfrac1n!left(frac-w1-wright)^n \
&= sum_ngeq0dfrac(-w)^nn!left(1-wright)^-n \
&= sum_ngeq0dfrac(-w)^nn!left(1+nw+dfracn(n+1)2w^2+cdotsright) \
&= sum_ngeq0dfrac(-1)^nn!left(w^n+nw^n+1+dfracn(n+1)2w^n+2+cdotsright)
endalign
after calculation some terms we let $w=dfrac1z$.



Edit: One may find
beginalign
&1 \
&-(w+w^2+w^3+w^4+w^5+cdots) \
&+frac12(w^2+2w^3+3w^4+4w^5+cdots) \
&-frac16(w^3+3w^4+6w^5+cdots) \
&+cdots\
&=1-w-fracw^22-fracw^36+fracw^424+frac19 w^5120+cdots\
&=colorblue1-dfrac1z-frac12z^2-frac16z^3+frac124z^4+frac19120z^5+cdots
endalign







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 30 at 19:27


























answered Jul 30 at 18:02









user 108128

19k41544




19k41544











  • the answer in the book: $1 - frac1z+ frac12z^2-frac16z^3+frac124z^4-frac19120z^5…$. I can't get it.
    – Metso
    Jul 30 at 18:27











  • See the series and let $n=0$
    – user 108128
    Jul 30 at 18:32










  • so the first is $1$ right?
    – user 108128
    Jul 30 at 18:33










  • yes, the first is 1. I'll try other numbers
    – Metso
    Jul 30 at 18:37











  • now for $n=1$ it is $-(w+w^2)$
    – user 108128
    Jul 30 at 18:39
















  • the answer in the book: $1 - frac1z+ frac12z^2-frac16z^3+frac124z^4-frac19120z^5…$. I can't get it.
    – Metso
    Jul 30 at 18:27











  • See the series and let $n=0$
    – user 108128
    Jul 30 at 18:32










  • so the first is $1$ right?
    – user 108128
    Jul 30 at 18:33










  • yes, the first is 1. I'll try other numbers
    – Metso
    Jul 30 at 18:37











  • now for $n=1$ it is $-(w+w^2)$
    – user 108128
    Jul 30 at 18:39















the answer in the book: $1 - frac1z+ frac12z^2-frac16z^3+frac124z^4-frac19120z^5…$. I can't get it.
– Metso
Jul 30 at 18:27





the answer in the book: $1 - frac1z+ frac12z^2-frac16z^3+frac124z^4-frac19120z^5…$. I can't get it.
– Metso
Jul 30 at 18:27













See the series and let $n=0$
– user 108128
Jul 30 at 18:32




See the series and let $n=0$
– user 108128
Jul 30 at 18:32












so the first is $1$ right?
– user 108128
Jul 30 at 18:33




so the first is $1$ right?
– user 108128
Jul 30 at 18:33












yes, the first is 1. I'll try other numbers
– Metso
Jul 30 at 18:37





yes, the first is 1. I'll try other numbers
– Metso
Jul 30 at 18:37













now for $n=1$ it is $-(w+w^2)$
– user 108128
Jul 30 at 18:39




now for $n=1$ it is $-(w+w^2)$
– user 108128
Jul 30 at 18:39










up vote
2
down vote













Note that$$left(-frac1z-frac1z^2-frac1z^3-frac1z^4-cdotsright)^2=frac1z^2+frac2z^3+frac3z^4+cdots,$$that$$left(-frac1z-frac1z^2-frac1z^3-frac1z^4-cdotsright)^3=-frac1z^3-frac3z^4-frac6z^5-cdots,$$and so on. This explains why you got the wrong result.






share|cite|improve this answer





















  • sorry, but I can't get the answer from the book. I have again $1 - frac1z-frac12z^2-...$, but not $1 - frac1z+frac12z^2-...$
    – Metso
    Jul 30 at 17:48











  • What's the smallest $n$ for which your coefficient of $frac1z^n$ is different from the one from the book?
    – José Carlos Santos
    Jul 30 at 17:51










  • the answer: $1 - frac1z+ frac12z^2-frac16z^3+frac124z^4-frac19120z^5…$ As you can see it starts from $frac12z^2$, $n=2$
    – Metso
    Jul 30 at 17:52











  • @Metso You forgot the $frac12!$ that you wrote yourself in the original question.
    – José Carlos Santos
    Jul 30 at 18:02











  • result is $1-frac1z -frac12!z^2$
    – Metso
    Jul 30 at 18:07















up vote
2
down vote













Note that$$left(-frac1z-frac1z^2-frac1z^3-frac1z^4-cdotsright)^2=frac1z^2+frac2z^3+frac3z^4+cdots,$$that$$left(-frac1z-frac1z^2-frac1z^3-frac1z^4-cdotsright)^3=-frac1z^3-frac3z^4-frac6z^5-cdots,$$and so on. This explains why you got the wrong result.






share|cite|improve this answer





















  • sorry, but I can't get the answer from the book. I have again $1 - frac1z-frac12z^2-...$, but not $1 - frac1z+frac12z^2-...$
    – Metso
    Jul 30 at 17:48











  • What's the smallest $n$ for which your coefficient of $frac1z^n$ is different from the one from the book?
    – José Carlos Santos
    Jul 30 at 17:51










  • the answer: $1 - frac1z+ frac12z^2-frac16z^3+frac124z^4-frac19120z^5…$ As you can see it starts from $frac12z^2$, $n=2$
    – Metso
    Jul 30 at 17:52











  • @Metso You forgot the $frac12!$ that you wrote yourself in the original question.
    – José Carlos Santos
    Jul 30 at 18:02











  • result is $1-frac1z -frac12!z^2$
    – Metso
    Jul 30 at 18:07













up vote
2
down vote










up vote
2
down vote









Note that$$left(-frac1z-frac1z^2-frac1z^3-frac1z^4-cdotsright)^2=frac1z^2+frac2z^3+frac3z^4+cdots,$$that$$left(-frac1z-frac1z^2-frac1z^3-frac1z^4-cdotsright)^3=-frac1z^3-frac3z^4-frac6z^5-cdots,$$and so on. This explains why you got the wrong result.






share|cite|improve this answer













Note that$$left(-frac1z-frac1z^2-frac1z^3-frac1z^4-cdotsright)^2=frac1z^2+frac2z^3+frac3z^4+cdots,$$that$$left(-frac1z-frac1z^2-frac1z^3-frac1z^4-cdotsright)^3=-frac1z^3-frac3z^4-frac6z^5-cdots,$$and so on. This explains why you got the wrong result.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 30 at 16:42









José Carlos Santos

112k1696172




112k1696172











  • sorry, but I can't get the answer from the book. I have again $1 - frac1z-frac12z^2-...$, but not $1 - frac1z+frac12z^2-...$
    – Metso
    Jul 30 at 17:48











  • What's the smallest $n$ for which your coefficient of $frac1z^n$ is different from the one from the book?
    – José Carlos Santos
    Jul 30 at 17:51










  • the answer: $1 - frac1z+ frac12z^2-frac16z^3+frac124z^4-frac19120z^5…$ As you can see it starts from $frac12z^2$, $n=2$
    – Metso
    Jul 30 at 17:52











  • @Metso You forgot the $frac12!$ that you wrote yourself in the original question.
    – José Carlos Santos
    Jul 30 at 18:02











  • result is $1-frac1z -frac12!z^2$
    – Metso
    Jul 30 at 18:07

















  • sorry, but I can't get the answer from the book. I have again $1 - frac1z-frac12z^2-...$, but not $1 - frac1z+frac12z^2-...$
    – Metso
    Jul 30 at 17:48











  • What's the smallest $n$ for which your coefficient of $frac1z^n$ is different from the one from the book?
    – José Carlos Santos
    Jul 30 at 17:51










  • the answer: $1 - frac1z+ frac12z^2-frac16z^3+frac124z^4-frac19120z^5…$ As you can see it starts from $frac12z^2$, $n=2$
    – Metso
    Jul 30 at 17:52











  • @Metso You forgot the $frac12!$ that you wrote yourself in the original question.
    – José Carlos Santos
    Jul 30 at 18:02











  • result is $1-frac1z -frac12!z^2$
    – Metso
    Jul 30 at 18:07
















sorry, but I can't get the answer from the book. I have again $1 - frac1z-frac12z^2-...$, but not $1 - frac1z+frac12z^2-...$
– Metso
Jul 30 at 17:48





sorry, but I can't get the answer from the book. I have again $1 - frac1z-frac12z^2-...$, but not $1 - frac1z+frac12z^2-...$
– Metso
Jul 30 at 17:48













What's the smallest $n$ for which your coefficient of $frac1z^n$ is different from the one from the book?
– José Carlos Santos
Jul 30 at 17:51




What's the smallest $n$ for which your coefficient of $frac1z^n$ is different from the one from the book?
– José Carlos Santos
Jul 30 at 17:51












the answer: $1 - frac1z+ frac12z^2-frac16z^3+frac124z^4-frac19120z^5…$ As you can see it starts from $frac12z^2$, $n=2$
– Metso
Jul 30 at 17:52





the answer: $1 - frac1z+ frac12z^2-frac16z^3+frac124z^4-frac19120z^5…$ As you can see it starts from $frac12z^2$, $n=2$
– Metso
Jul 30 at 17:52













@Metso You forgot the $frac12!$ that you wrote yourself in the original question.
– José Carlos Santos
Jul 30 at 18:02





@Metso You forgot the $frac12!$ that you wrote yourself in the original question.
– José Carlos Santos
Jul 30 at 18:02













result is $1-frac1z -frac12!z^2$
– Metso
Jul 30 at 18:07





result is $1-frac1z -frac12!z^2$
– Metso
Jul 30 at 18:07













 

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