# Jacobi $sd(u,m)$ hyperbolic approximation for $m \to 1$

I am following a paper (Fink 1976); the details are a mathematical problem.

We are given a solution (paraphrased from the paper for clarity)

$$y^2 = \alpha m_{1n} sd^2(u,\sqrt{m_n}),$$

where $$sd$$ is a Jacobi Elliptic Function, and $$m_n + m_{1n} = 1$$.

In a particular limit, where $$m_n \to 1$$, we are told that we have an approximation for this equation:

$$y^2 = \alpha m_{1n} \left[\text{cosh}^2(u) \left(1 - \frac{1}{4}m_{1n}\text{sinh}(2u)\right) -1\right].$$

To my eye, it appears that this is a combination of the hyperbolic expansions as given for example in this Wikipedia page or this page on JE functions from NIST:

$$\operatorname{sn}\left(z,k\right)=\tanh z-\frac{{k^{\prime}}^{2}}{4}(z-\sinh z% \cosh z){\operatorname{sech}}^{2}z+O\left({k^{\prime}}^{4}\right),$$ $$\operatorname{dn}\left(z,k\right)=\operatorname{sech}z+\frac{{k^{\prime}}^{2}} {4}(z+\sinh z\cosh z)\tanh z\operatorname{sech}z+O\left({k^{\prime}}^{4}\right).$$

No such expression is given as an approximation for $$sd$$.

What I cannot understand is how the factors of $$\pm z$$ in the terms can possibly be removed from these expressions. Any advice or guidance would be appreciated - I have tried using other approximations since $$m_{1n}$$ is small such as $$1/(1-x)$$ expansions to deal with the denominators etc.

EDIT: An additional check is given by solving this resulting expression for $$m_{1n}$$ and the limit that $$u$$ is large:

$$m_{1n} \approx 4 e^{-2u}[ 1 - \sqrt{1-y^2}].$$

• I think $\,(1 - \frac{1}{4}m_{1n}\text{sinh}(2u) -1)\,$ must be a typo because the $1-1=0$ and vanish. Apr 10, 2021 at 23:18
• Dear Somos - thank you for your keen eye! Yes, there is a small typo here; I have missed a bracket after the sinh, such that the cosh does not multiply the final 1. I have corrected above; thank you!
The definition of Jacobi elliptic functions is given in terms of amplitude $$\phi$$ by the relation \begin{align}u&=\int_0^{\phi}\frac{dx}{\sqrt{1-k^2\sin^2x}}\tag{1}\\ \operatorname {sn} (u, k) &=\sin\phi\tag{2}\\ \operatorname {cn} (u, k) & =\cos\phi\tag{3}\\ \operatorname {dn} (u, k) & =\sqrt{1-k^2\sin^2\phi}\tag{4} \end{align} The function $$\operatorname {sd}$$ in your question is defined by $$\operatorname {sd} (u, k) =\frac{\operatorname {sn} (u, k)} {\operatorname {dn} (u, k)} \tag{5}$$ When $$k=1$$ then we have $$u=\log(\sec\phi+\tan\phi), \sec\phi=\cosh u, \tan\phi=\sinh u\tag{6}$$ And then using $$(5)$$ we have $$\operatorname {sd}^2 (u, k) =\tan^2\phi=\sinh^2 u$$ However this does not resemble the expression in question.
We are given that $$\operatorname{sn}\left(z,k\right)=\tanh z-\frac{{k^{\prime}}^{2}}{4}(z-\sinh z% \cosh z){\operatorname{sech}}^{2}z+O\left({k^{\prime}}^{4}\right), \tag{1}$$ and $$\operatorname{dn}\left(z,k\right)=\operatorname{sech}z+\frac{{k^{\prime}}^{2}} {4}(z+\sinh z\cosh z)\tanh z\operatorname{sech}z+O\left({k^{\prime}}^{4}\right). \tag{2}$$ Divide the power series to get $$\operatorname{sd}(z,k)=\sinh z-\frac{{k^{\prime}}^{2}}{4}\left(z\cosh z% +\frac12(-3+\cosh 2z)\sinh z\right)+O\left({k^{\prime}}^{4}\right). \tag{3}$$ Square this power series to get $$\operatorname{sd}(z,k)^2 =\sinh^2 z-\frac{{k^{\prime}}^{2}}{4}(z\sinh 2z% +(-3+\cosh 2z)\sinh^2z)+O\left({k^{\prime}}^{4}\right). \tag{4}$$