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:

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

  • $\begingroup$ I think $\,(1 - \frac{1}{4}m_{1n}\text{sinh}(2u) -1)\,$ must be a typo because the $1-1=0$ and vanish. $\endgroup$
    – Somos
    Apr 10, 2021 at 23:18
  • $\begingroup$ 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! $\endgroup$
    – Brad
    Apr 12, 2021 at 8:52

2 Answers 2


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.

Perhaps there are some more details which are not available in the question.

  • $\begingroup$ Thanks for your answer. I think a problem with this approach is that it assumes a fixed value of k=1, rather than taking the limit/expansion as k approaches 1. I have managed to obtain the answer but with some leftover terms, which I can physically try to argue are negligible. Thank you for your time! $\endgroup$
    – Brad
    Mar 1, 2021 at 10:04

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 $$ \operatorname{sd}(z,k)^2 =\sinh^2 z-\frac{{k^{\prime}}^{2}}{4}(z\sinh 2z% +(-3+\cosh 2z)\sinh^2z)+O\left({k^{\prime}}^{4}\right). \tag{4}$$


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