1
$\begingroup$

Given the elliptic modulus $k$,such that the complementary modulus is defined by $k'\equiv \sqrt{1-k^2}$,where $\phi\equiv am(u|k)$ is the jacobi amplitude and $K(k)$ is the complete elliptic integral of the first kind.

The elliptic generalisation of euler's formula in complex analysis(which expresses jacobi's elliptic functions in terms of the exponential function) is

$e^{i \phi}=\text{cn(u|k)}+i\text{sn(u|k)}$

The formula is valid multiples of $4K(k)$ and also multivalued.

How would we prove the formula using the theory of elliptic functions instead of elliptic functions defined in terms of trigonometric functions?

$\endgroup$
1
  • $\begingroup$ What exactly do you mean by "using the theory of elliptic functions"? Without defining what that means, it is impossible to give a good answer to your question. Please include this definition or a link to where this is defined in your question itself. $\endgroup$
    – Somos
    Mar 7 at 0:20

1 Answer 1

Reset to default
0
$\begingroup$

By Jacboi's original definition, $\text{cn(u|k)}:=\cos\phi$ and $\text{sn(u|k)}:=\sin\phi$. But $e^{i\phi}=\cos\phi+i\sin\phi$ by Euler and the result follows.

$\endgroup$
4
  • $\begingroup$ Thanks for the answer,but the question asked for a proof using the the theory of elliptic functions instead of elliptic functions defined in terms of trigonometric functions $\endgroup$
    – Nicco
    Aug 15, 2017 at 2:55
  • $\begingroup$ Jacobi invented his theory of elliptic functions and that was his original definition. That is, sine amplitude $\sin\textrm{am}.(u|k) = \sin(am(u|k))$. I can look up the original equations in his "Fundamenta Nova Theoriae Functionum Ellipticarum" for you if you wish. Or just look in Jacobi Elliptic Function in Wikipedia. $\endgroup$
    – Somos
    Aug 15, 2017 at 3:00
  • $\begingroup$ I agree and that's exactly what motivated me that the formula is true,but I was looking for a proof using ,for example taylor series of the jacobi elliptic functions or some other tools fron the theory and not directly using their definition $\endgroup$
    – Nicco
    Aug 15, 2017 at 3:28
  • 1
    $\begingroup$ @Nicco Then I suggest being explicit about that in your question, and being explicit what you consider the definition of $sn()$ and $cn()$ because there are multiple ways to define them. $\endgroup$
    – Somos
    Aug 15, 2017 at 3:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .