Laurent series of a fractional cosine and sine

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0
down vote

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i was wondering how i can write the laurent series of

$$cos(dfrac1z-2)$$ in z=2 ?

and if i want the laurent series of $$sin(dfrac1z)$$ ( for Z=2 too , how it is because here 2 is not a singualrity and i don't know how to write it if it's not a singularity )

complex-analysis taylor-expansion laurent-series

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asked Jul 26 at 13:01

xmaionx

52

add a comment | 

up vote
0
down vote

favorite

i was wondering how i can write the laurent series of

$$cos(dfrac1z-2)$$ in z=2 ?

and if i want the laurent series of $$sin(dfrac1z)$$ ( for Z=2 too , how it is because here 2 is not a singualrity and i don't know how to write it if it's not a singularity )

complex-analysis taylor-expansion laurent-series

share|cite|improve this question

asked Jul 26 at 13:01

xmaionx

52

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

i was wondering how i can write the laurent series of

$$cos(dfrac1z-2)$$ in z=2 ?

and if i want the laurent series of $$sin(dfrac1z)$$ ( for Z=2 too , how it is because here 2 is not a singualrity and i don't know how to write it if it's not a singularity )

complex-analysis taylor-expansion laurent-series

share|cite|improve this question

asked Jul 26 at 13:01

xmaionx

52

i was wondering how i can write the laurent series of

$$cos(dfrac1z-2)$$ in z=2 ?

and if i want the laurent series of $$sin(dfrac1z)$$ ( for Z=2 too , how it is because here 2 is not a singualrity and i don't know how to write it if it's not a singularity )

complex-analysis taylor-expansion laurent-series

share|cite|improve this question

asked Jul 26 at 13:01

xmaionx

52

share|cite|improve this question

share|cite|improve this question

asked Jul 26 at 13:01

xmaionx

52

asked Jul 26 at 13:01

xmaionx

52

asked Jul 26 at 13:01

xmaionx

52

52

add a comment | 

add a comment | 

1 Answer
1

active

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votes

up vote
-1
down vote



accepted

Since$$cos(z)=1-fracz^22!+fracz^44!-cdots,$$you have$$cosleft(frac1z-2right)=1-frac12!(z-2)^2+frac14!(z-2)^4-cdots$$

share|cite|improve this answer

answered Jul 26 at 13:11

José Carlos Santos

113k1696173

• 
  
  
  

  
  

  
  
  
  
  
  

  
  BUT is this not centered in 0 ? i need the one centered in 2 and i'm not understanding :(
  – xmaionx
  Jul 26 at 13:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  How could this not be centered at $2$!? Don't you see all those $(z−2)^k$? Do you see any power of $z$ in my answer?
  – José Carlos Santos
  Jul 26 at 13:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  so if for instance i have cos (1/(z-3)) so the laurent series centered in 3 is $$cosleft(frac1z-3right)=1-frac12!(z-3)^2+frac14!(z-3)^4-cdots$$
  – xmaionx
  Jul 26 at 13:32
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @xmaionx In fact.
  – José Carlos Santos
  Jul 26 at 13:33
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @xmaionx There's no simple answer for that question.
  – José Carlos Santos
  Jul 26 at 13:54
  

  
  
  
  

 | 
show 10 more comments

Your Answer

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
-1
down vote



accepted

Since$$cos(z)=1-fracz^22!+fracz^44!-cdots,$$you have$$cosleft(frac1z-2right)=1-frac12!(z-2)^2+frac14!(z-2)^4-cdots$$

share|cite|improve this answer

answered Jul 26 at 13:11

José Carlos Santos

113k1696173

• 
  
  
  

  
  

  
  
  
  
  
  

  
  BUT is this not centered in 0 ? i need the one centered in 2 and i'm not understanding :(
  – xmaionx
  Jul 26 at 13:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  How could this not be centered at $2$!? Don't you see all those $(z−2)^k$? Do you see any power of $z$ in my answer?
  – José Carlos Santos
  Jul 26 at 13:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  so if for instance i have cos (1/(z-3)) so the laurent series centered in 3 is $$cosleft(frac1z-3right)=1-frac12!(z-3)^2+frac14!(z-3)^4-cdots$$
  – xmaionx
  Jul 26 at 13:32
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @xmaionx In fact.
  – José Carlos Santos
  Jul 26 at 13:33
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @xmaionx There's no simple answer for that question.
  – José Carlos Santos
  Jul 26 at 13:54
  

  
  
  
  

 | 
show 10 more comments

up vote
-1
down vote



accepted

Since$$cos(z)=1-fracz^22!+fracz^44!-cdots,$$you have$$cosleft(frac1z-2right)=1-frac12!(z-2)^2+frac14!(z-2)^4-cdots$$

share|cite|improve this answer

answered Jul 26 at 13:11

José Carlos Santos

113k1696173

• 
  
  
  

  
  

  
  
  
  
  
  

  
  BUT is this not centered in 0 ? i need the one centered in 2 and i'm not understanding :(
  – xmaionx
  Jul 26 at 13:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  How could this not be centered at $2$!? Don't you see all those $(z−2)^k$? Do you see any power of $z$ in my answer?
  – José Carlos Santos
  Jul 26 at 13:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  so if for instance i have cos (1/(z-3)) so the laurent series centered in 3 is $$cosleft(frac1z-3right)=1-frac12!(z-3)^2+frac14!(z-3)^4-cdots$$
  – xmaionx
  Jul 26 at 13:32
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @xmaionx In fact.
  – José Carlos Santos
  Jul 26 at 13:33
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @xmaionx There's no simple answer for that question.
  – José Carlos Santos
  Jul 26 at 13:54
  

  
  
  
  

 | 
show 10 more comments

up vote
-1
down vote



accepted

up vote
-1
down vote



accepted

Since$$cos(z)=1-fracz^22!+fracz^44!-cdots,$$you have$$cosleft(frac1z-2right)=1-frac12!(z-2)^2+frac14!(z-2)^4-cdots$$

share|cite|improve this answer

answered Jul 26 at 13:11

José Carlos Santos

113k1696173

Since$$cos(z)=1-fracz^22!+fracz^44!-cdots,$$you have$$cosleft(frac1z-2right)=1-frac12!(z-2)^2+frac14!(z-2)^4-cdots$$

share|cite|improve this answer

answered Jul 26 at 13:11

José Carlos Santos

113k1696173

share|cite|improve this answer

share|cite|improve this answer

answered Jul 26 at 13:11

José Carlos Santos

113k1696173

answered Jul 26 at 13:11

José Carlos Santos

113k1696173

answered Jul 26 at 13:11

José Carlos Santos

113k1696173

113k1696173

• 
  
  
  

  
  

  
  
  
  
  
  

  
  BUT is this not centered in 0 ? i need the one centered in 2 and i'm not understanding :(
  – xmaionx
  Jul 26 at 13:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  How could this not be centered at $2$!? Don't you see all those $(z−2)^k$? Do you see any power of $z$ in my answer?
  – José Carlos Santos
  Jul 26 at 13:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  so if for instance i have cos (1/(z-3)) so the laurent series centered in 3 is $$cosleft(frac1z-3right)=1-frac12!(z-3)^2+frac14!(z-3)^4-cdots$$
  – xmaionx
  Jul 26 at 13:32
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @xmaionx In fact.
  – José Carlos Santos
  Jul 26 at 13:33
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @xmaionx There's no simple answer for that question.
  – José Carlos Santos
  Jul 26 at 13:54
  

  
  
  
  

 | 
show 10 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  BUT is this not centered in 0 ? i need the one centered in 2 and i'm not understanding :(
  – xmaionx
  Jul 26 at 13:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  How could this not be centered at $2$!? Don't you see all those $(z−2)^k$? Do you see any power of $z$ in my answer?
  – José Carlos Santos
  Jul 26 at 13:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  so if for instance i have cos (1/(z-3)) so the laurent series centered in 3 is $$cosleft(frac1z-3right)=1-frac12!(z-3)^2+frac14!(z-3)^4-cdots$$
  – xmaionx
  Jul 26 at 13:32
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @xmaionx In fact.
  – José Carlos Santos
  Jul 26 at 13:33
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @xmaionx There's no simple answer for that question.
  – José Carlos Santos
  Jul 26 at 13:54
  

  
  
  
  

BUT is this not centered in 0 ? i need the one centered in 2 and i'm not understanding :(
– xmaionx
Jul 26 at 13:19

BUT is this not centered in 0 ? i need the one centered in 2 and i'm not understanding :(
– xmaionx
Jul 26 at 13:19

How could this not be centered at $2$!? Don't you see all those $(z−2)^k$? Do you see any power of $z$ in my answer?
– José Carlos Santos
Jul 26 at 13:31

How could this not be centered at $2$!? Don't you see all those $(z−2)^k$? Do you see any power of $z$ in my answer?
– José Carlos Santos
Jul 26 at 13:31

so if for instance i have cos (1/(z-3)) so the laurent series centered in 3 is $$cosleft(frac1z-3right)=1-frac12!(z-3)^2+frac14!(z-3)^4-cdots$$
– xmaionx
Jul 26 at 13:32

so if for instance i have cos (1/(z-3)) so the laurent series centered in 3 is $$cosleft(frac1z-3right)=1-frac12!(z-3)^2+frac14!(z-3)^4-cdots$$
– xmaionx
Jul 26 at 13:32

1

1

@xmaionx In fact.
– José Carlos Santos
Jul 26 at 13:33

@xmaionx In fact.
– José Carlos Santos
Jul 26 at 13:33

1

1

@xmaionx There's no simple answer for that question.
– José Carlos Santos
Jul 26 at 13:54

@xmaionx There's no simple answer for that question.
– José Carlos Santos
Jul 26 at 13:54

 | 
show 10 more comments

 

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