Periods of Jacobi Elliptic functions

  • Thread starter Thread starter binbagsss
  • Start date Start date
  • Tags Tags
    Functions Jacobi
binbagsss
Messages
1,291
Reaction score
12

Homework Statement



I have that ##(\psi(z)-e_j)^{1/2}=e^{\frac{-n_jz}{2}}\frac{\sigma(z+\frac{w_j}{2})}{\sigma(\frac{w_j}{2})\sigma(z)}##

has period ##w_i## if ##i=j##
and period ##2w_i## if ##i\neq j##

where ##i,j=1,2,3## and ##w_3=w_1+w_2## (*)
where ##e_j=\psi(\frac{w_j}{2})##

I have the following (two) identities:

##\sigma(z+w_j)=-\sigma(z)e^{n_j(z+\frac{w_j}{2})}## (1)

From which we can get:

##\sigma(z+\frac{w_j}{2})=-e^{n_j z}\sigma(z-\frac{w_j}{2})## (2)

With the above information I need to deduce that:

##S(z)## has period lattice with ##(2w_1,w_2)##
##C(z)## has period lattice with ##(2w_1,w_1+w_2)##
##D(z)## has period lattice with ##(w_1,2w_2)##

where ##S(z), C(z), D(z)## are given as attached:

attachhere.png


Homework Equations



see above, see below, look up, look down, look all around.

The Attempt at a Solution



[/B]

My book says this is straightforward with the result (*) stated above. But I can't see it for ##C(z) ## and ##D(z)##...

For example,

##C(z)## numerator, using (*), has periods ##(w_1,2w_2)##, whilst the denominator has periods ##(2w_1,w_2)##..

So I can't see how a obvious conclusion can be made? (Similarly for ##D(z)##)...
All I can see to do is work through a tonne of algebra, each period in turn, using (1) and (2), however then I'm not making use of (*), from which my book says the result simply follows, so I don't think I should be needing to do this anyway...

Many thanks in advance.
 
Physics news on Phys.org
bump.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top