Laurent series of a fractional cosine and sine

Question

i was wondering how i can write the laurent series of

$$\cos(\dfrac{1}{z-2})$$ in z=2 ?

and if i want the laurent series of $$\sin(\dfrac{1}{z})$$ ( 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 )

score 2 · Accepted Answer · answered Jul 26 '18 at 13:11

2

Since$$\cos(z)=1-\frac{z^2}{2!}+\frac{z^4}{4!}-\cdots,$$you have$$\cos\left(\frac1{z-2}\right)=1-\frac1{2!(z-2)^2}+\frac1{4!(z-2)^4}-\cdots$$

answered Jul 26 '18 at 13:11

José Carlos Santos

440,053

BUT is this not centered in 0 ? i need the one centered in 2 and i'm not understanding :( – xmaionx Jul 26 '18 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 '18 at 13:31
so if for instance i have cos (1/(z-3)) so the laurent series centered in 3 is $$\cos\left(\frac1{z-3}\right)=1-\frac1{2!(z-3)^2}+\frac1{4!(z-3)^4}-\cdots$$ – xmaionx Jul 26 '18 at 13:32
1

@xmaionx In fact. – José Carlos Santos Jul 26 '18 at 13:33
so (z-3) is like the translation of the serie ? the same if i have cos ( 1//z+4)) the sere would be $$\cos\left(\frac1{z+4}\right)=1-\frac1{2!(z+4)^2}+\frac1{4!(z+4)^4}-\cdots$$ ?? – xmaionx Jul 26 '18 at 13:34
@xmaionx In fact (once again). – José Carlos Santos Jul 26 '18 at 13:39
Ok i have just another question... if i want to write the laurent (taylor) series of sin(1/z) but in 2 not in 0 , how i have to do ? – xmaionx Jul 26 '18 at 13:52
1

@xmaionx There's no simple answer for that question. – José Carlos Santos Jul 26 '18 at 13:54
why i just cant sostitute sin(1/z) with 1/(z-2) just as we have done before ? – xmaionx Jul 26 '18 at 14:00
@xmaionx That's not what I did before, since I started with the Taylor series of $\cos(z)$, not of $\cos\left(\frac1z\right)$. – José Carlos Santos Jul 26 '18 at 14:03
mmm sorry i 'm not understanding this... :( you started from cos(z) then substitute z with 1/(z-2) , , right ? and why here i can't start from sin(z) and substitute 1/(z-2) ? it isn't the same ? – xmaionx Jul 26 '18 at 14:09
@xmaionx You can start with $\sin(z)$ and replace with $\frac1{z-2}$. But if you start with $\sin\left(\frac1z\right)$ and you replace $z$ with $\frac1{z-2}$, then what you'll get is $\sin(z-2)$, which is not what you are interested in. – José Carlos Santos Jul 26 '18 at 14:11
ahhh ok thk you so much you realy helped me ! so if i want the series of a function like sin(1/(z)) but for instance ** centered in 3 ** not in 0 i just substitute the z in the series of sin(z) with 1/(z-3) right ? – xmaionx Jul 26 '18 at 14:28
@xmaionx I think that I provided an answer to that question more than enough times. – José Carlos Santos Jul 26 '18 at 14:31
Thk so much you have been so much for me ! – xmaionx Jul 26 '18 at 14:38

Laurent series of a fractional cosine and sine

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