Let $\operatorname{cn}(u\mid m)$ be Gudermann's notation for the Jacobi elliptic function $\operatorname{cn}$. It is well known to be doubly periodic. For $0<m<1$ the two periods, $4 K(m)$ and $4 i K(1-m)$, are purely real and purely imaginary respectively. For any $n_1, n_2 \in \mathbb{Z}$: $$ \operatorname{cn}\left(u + 4 n_1 K(m) + 4 n_2 i K(1-m) \mid m \right) = \operatorname{cn}\left(u \mid m \right) $$ and where $K(m)$ denotes the complete elliptic integral of the first kind (in that link $m=k^2$).

For $m=\frac{1}{2}$, $f(x,y) = \operatorname{cn}(x+i y \mid m)$ is doubly periodic in a square with sidelength $4K\left(\frac{1}{2}\right) \approx 7.4163$.

By looking at Peirce quincuncial projection I noticed empirically that, for $x\in \mathbb{R}$, $f(x,x)$ is of unit magnitude: $$ \left| f(x,x) \right| = \left| \operatorname{cn}\left(x+i x \mid \frac{1}{2}\right) \right| = 1 $$ enter image description here

I am looking for either a proof, or a reference to one.

Thank you.


Jacobi's imaginary transformation $$ \mathrm{cn}(iu|m)=\frac{1}{\mathrm{cn}(u|m')},$$ and parity of $\mathrm{cn}(u)$ will do the job. $\blacksquare$

  • $\begingroup$ Thank you! This is exactly it. I will post my solution shortly $\endgroup$ – Sasha Apr 29 '13 at 2:19
  • 1
    $\begingroup$ Yours is quite a lot neater! $\endgroup$ – Sasha Apr 29 '13 at 2:47

Using the addition theorem: $$ \operatorname{cn}\left(x+ i x \mid m\right) = \frac{\operatorname{cn}(x \mid m) \operatorname{cn}(i x \mid m) - \operatorname{sn}(x \mid m) \operatorname{sn}(i x \mid m) \operatorname{dn}(x \mid m) \operatorname{dn}(i x \mid m) }{1-m \operatorname{sn}^2(x \mid m) \operatorname{sn}^2(i x \mid m)} $$ Using the Jacobi's imaginary transformations: $$ \operatorname{cn}(i x \mid m) = \frac{1}{\operatorname{cn}(x \mid 1-m)} \quad \operatorname{sn}(i x \mid m) = i \frac{\operatorname{sn}(x \mid 1-m)}{\operatorname{cn}(x \mid 1-m)} \quad \operatorname{dn}(i x \mid m) = \frac{\operatorname{dn}(x \mid 1-m)}{\operatorname{cn}(x \mid 1-m)} $$ we get $$ \operatorname{cn}\left(x+ i x \mid m\right) = \frac{\operatorname{cn}(x \mid 1-m) \operatorname{cn}(x\mid m)-i \operatorname{dn}(x \mid 1-m) \operatorname{dn}(x\mid m) \operatorname{sn}(x\mid 1-m) \operatorname{sn}(x \mid m)}{\operatorname{cn}^2(x\mid 1-m) + m \operatorname{sn}^2(x \mid 1-m) \operatorname{sn}^2(x \mid m)} $$ Substituting $m={1 \over 2}$ we get $$ \operatorname{cn}\left(x+ i x \mid {1 \over 2}\right) = \frac{\operatorname{cn}^2(x\mid {1\over 2}) - i \operatorname{sn}^2(x\mid {1\over 2}) \operatorname{dn}^2(x\mid {1\over 2}) }{\operatorname{cn}^2(x\mid {1\over 2}) + {1\over 2} \operatorname{sn}^4(x\mid {1\over 2})} $$ The absolute value now follows to be 1: $$ \left| \operatorname{cn}\left(x+ i x \mid {1 \over 2}\right) \right|^2 = \frac{\operatorname{cn}^4(x\mid {1\over 2}) + \operatorname{sn}^4(x\mid {1\over 2}) \operatorname{dn}^4(x\mid {1\over 2}) }{\left( \operatorname{cn}^2(x\mid {1\over 2}) + {1\over 2} \operatorname{sn}^4(x\mid {1\over 2}) \right)^2} $$ Using defining relations $\operatorname{cn}^2(x \mid 1/2) + \operatorname{sn}^2(x \mid 1/2) = 1$ and $\operatorname{dn}^2(x \mid 1/2) + {1 \over 2} \operatorname{sn}^2(x \mid 1/2) = 1$ the right hand side equals one.


Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.