In our introductory post we had talked about the similarities of the elliptic functions with the circular functions. At the same time we discussed the properties which were quite unlike those of circular functions (like double periodicity). In this post we are going to discuss some further properties of elliptic functions which have no analogue in the theory of circular functions.

Landen's Transformation is a set of formulas relating the elliptic functions of one modulus (the parameter $ k$) with the elliptic functions of another but related modulus. There is a general theory of transformation for the elliptic functions which we will not discuss in detail here, but it is worth mentioning that the Landen's transformation which we describe is actually a special kind of transformation and it is known as the transformation of

Then we have \begin{align} k\sin\theta &= \sin 2\psi\cos\theta - \cos 2\psi\sin\theta\notag\\ \Rightarrow \cos\theta\sin 2\psi &= (k + \cos 2\psi)\sin\theta\notag\\ \Rightarrow \tan\theta &= \frac{\sin 2\psi}{k + \cos 2\psi}\tag{2} \end{align} We can now see that \begin{align} \sin\theta &= \frac{\tan\theta}{\sqrt{1 + \tan^{2}\theta}}\notag\\ &=\dfrac{\dfrac{\sin 2\psi}{k + \cos 2\psi}}{\sqrt{1 + \left(\dfrac{\sin 2\psi}{k + \cos 2\psi}\right)^{2}}}\notag\\ &= \frac{\sin 2\psi}{\sqrt{(k + \cos 2\psi)^{2} + \sin^{2}2\psi}}\notag\\ &= \frac{\sin 2\psi}{\sqrt{1 + k^{2} + 2k\cos 2\psi}}\notag\\ &= \frac{2\sin\psi\cos\psi}{\sqrt{1 + k^{2} + 2k - 4k\sin^{2}\psi}}\notag\\ &= \dfrac{2\sin\psi\cos\psi}{(1 + k)\sqrt{1 - \left(\dfrac{2\sqrt{k}}{1 + k}\right)^{2}\sin^{2}\psi}}\notag\\ \Rightarrow \sin\theta &= \frac{2}{1 + k}\frac{\sin\psi\cos\psi}{\sqrt{1 - k_{1}^{2}\sin^{2}\psi}}\text{ where } k_{1} = \frac{2\sqrt{k}}{1 + k}\tag{3} \end{align} Similarly $$\cos\theta = \frac{k + \cos 2\psi}{\sqrt{(k + \cos 2\psi)^{2} + \sin^{2}2\psi}}\tag{4}$$ and \begin{align} \sqrt{1 - k^{2}\sin^{2}\theta} &= \sqrt{1 - \sin^{2}(2\psi - \theta)} = \cos(2\psi - \theta)\notag\\ &= \cos 2\psi\cos\theta + \sin 2\psi\sin\theta\notag\\ &= \frac{\cos 2\psi(k + \cos 2\psi) + \sin^{2}2\psi}{\sqrt{(k + \cos 2\psi)^{2} + \sin^{2}2\psi}}\notag\\ &= \frac{1 + k\cos 2\psi}{\sqrt{(k + \cos 2\psi)^{2} + \sin^{2}2\psi}}\tag{5} \end{align} Differentiating equation $(2)$ and using equation $(4)$ we get \begin{align} \sec^{2}\theta\,d\theta &= \frac{(k + \cos 2\psi)2\cos 2\psi + 2\sin^{2}2\psi}{(k + \cos 2\psi)^{2}}\,d\psi\notag\\ \Rightarrow d\theta &= \frac{2(1 + k\cos 2\psi)}{(k + \cos 2\psi)^{2}}\frac{(k + \cos 2\psi)^{2}}{(k + \cos 2\psi)^{2} + \sin^{2}2\psi}\,d\psi\notag\\ &= \frac{2(1 + k\cos 2\psi)}{(k + \cos 2\psi)^{2} + \sin^{2}2\psi}\,d\psi\tag{6} \end{align} Hence from equation $(5)$ and $(6)$ we get \begin{align} \frac{d\theta}{\sqrt{1 - k^{2}\sin^{2}\theta}} &= \frac{2d\psi}{\sqrt{(k + \cos 2\psi)^{2} + \sin^{2}2\psi}}\notag\\ &= \frac{2}{1 + k}\frac{d\psi}{\sqrt{1 - k_{1}^{2}\sin^{2}\psi}}\notag \end{align} so that \begin{align} u &= F(\phi, k) = \int_{0}^{\phi}\frac{d\theta}{\sqrt{1 - k^{2}\sin^{2}\theta}}\notag\\ &= \frac{2}{1 + k}\int_{0}^{\phi_{1}}\frac{d\psi}{\sqrt{1 - k_{1}^{2}\sin^{2}\psi}} = \frac{2}{1 + k}F(\phi_{1}, k_{1})\notag \end{align} and we thus have $$\frac{(1 + k)u}{2} = F(\phi_{1}, k_{1})$$ which is equivalent to $$\text{sn}\left(\frac{(1 + k)u}{2}, \frac{2\sqrt{k}}{1 + k}\right) = \sin\phi_{1}$$ From equation $(3)$ we have $$\sin\phi = \frac{2}{1 + k}\frac{\sin\phi_{1}\cos\phi_{1}}{\sqrt{1 - k_{1}^{2}\sin^{2}\phi_{1}}}$$ hence it follows that $$\text{sn}(u, k) = \dfrac{2}{1 + k}\dfrac{\text{sn}\left(\dfrac{(1 + k)u}{2}, \dfrac{2\sqrt{k}}{1 + k}\right)\text{cn}\left(\dfrac{(1 + k)u}{2}, \dfrac{2\sqrt{k}}{1 + k}\right)}{\text{dn}\left(\dfrac{(1 + k)u}{2}, \dfrac{2\sqrt{k}}{1 + k}\right)}$$ This is the celebrated Landen's transformation for elliptic functions. It express the elliptic function of modulus $ k$ in terms of elliptic functions of another modulus $ k_{1} = 2\sqrt{k} / ( 1 + k)$ which is greater than $ k$. The above equation can be written in a slightly more convenient form by letting $ \lambda = k_{1} = 2\sqrt{k} / ( 1 + k)$ so that $ k = (1 - \lambda') / (1 + \lambda')$ and $ 2/(1 + k) = 1 + \lambda'$. We then have $$\text{sn}\left((1 + \lambda')u, \frac{1 - \lambda'}{1 + \lambda'}\right) = \frac{(1 + \lambda')\,\text{sn}(u, \lambda)\,\text{cn}(u, \lambda)}{\text{dn}(u, \lambda)}$$ Replacing $ \lambda$ by $ k$ we get $$\text{sn}\left((1 + k')u, \frac{1 - k'}{1 + k'}\right) = \frac{(1 + k')\,\text{sn}(u, k)\,\text{cn}(u, k)}{\text{dn}(u, k)}\tag{7}$$ We can now be a bit more systematic and starting with an initial modulus $ k = k_{0}$ with $ 0 < k_{0} < 1$ we can form a sequence of increase moduli $ \{k_{n}\}$ given by the ascending Landen transformation $$ \displaystyle k_{n + 1} = \frac{2\sqrt{k_{n}}}{1 + k_{n}}$$ We have a corresponding descending Landen transformation of the moduli by the equation $$k_{n} = \frac{1 - k_{n + 1}'}{1 + k_{n + 1}'}$$ so that we can get the sequence of moduli extended to negative indexes $$ \ldots < k_{-n} < k_{-n + 1} < \ldots < k_{-2} < k_{-1} < k_{0} = k < k_{1} < k_{2} < \ldots < k_{n} < \ldots$$ and for all integers $ n$ we have $$k_{n + 1} = \frac{2\sqrt{k_{n}}}{1 + k_{n}}\,\,\, k_{n} = \frac{1 - k_{n + 1}'}{1 + k_{n + 1}'}$$ It is clear that $ k_{n} \to 1$ as $ n \to \infty$ and $ k_{-n} \to 0$ as $ n \to \infty$ (and the convergence is damn fast, to be precise it is quadratic).

The transformation is quite useful as it allows us to replace the modulus by a smaller or a greater one and ultimately to the special cases $ k = 0$ (when the elliptic function reduce the circular functions) and $ k = 1$ (when they reduce to the hyperbolic functions).

Another Landen transformation is obtained by the substitution

$$\sin\psi = \frac{(1 + k)\sin\theta}{1 + k\sin^{2}\theta}\tag{8}$$ and this is applied on the integral $$u_{1} = F(\phi_{1}, k_{1}) = \int_{0}^{\phi_{1}}\frac{d\psi}{\sqrt{1 - k_{1}^{2}\sin^{2}\psi}}$$ and the interval of integration changes from $ [0, \phi_{1}]$ to $ [0, \phi]$. Clearly we have $$\sqrt{1 - k_{1}^{2}\sin^{2}\psi} = \sqrt{1 - \frac{4k}{(1 + k)^{2}}\frac{(1 + k)^{2}\sin^{2}\theta}{(1 + k\sin^{2}\theta)^{2}}} = \frac{1 - k\sin^{2}\theta}{1 + k\sin^{2}\theta}$$ and by differentiation of equation $(8)$ we get \begin{align} \cos\psi\,d\psi &= \frac{(1 + k\sin^{2}\theta)(1 + k)\cos\theta - (1 + k)\sin\theta \cdot 2k\sin\theta\cos\theta}{(1 + k\sin^{2}\theta)^{2}}\,d\theta\notag\\ \Rightarrow d\psi &= \frac{(1 + k)\cos\theta (1 - k\sin^{2}\theta)}{(1 + k\sin^{2}\theta)^{2}}\frac{1 + k\sin^{2}\theta}{\sqrt{(1 + k\sin^{2}\theta)^{2} - (1 + k)^{2}\sin^{2}\theta}}\,d\theta\notag \end{align} which leads to \begin{align} \frac{d\psi}{\sqrt{1 - k_{1}^{2}\sin^{2}\psi}} &= \frac{(1 + k)\cos\theta}{\sqrt{(1 + k\sin^{2}\theta)^{2} - (1 + k)^{2}\sin^{2}\theta}}\,d\theta\notag\\ &= \frac{(1 + k)\cos\theta}{\sqrt{(1 + k - k\cos^{2}\theta)^{2} - (1 + k)^{2} + (1 + k)^{2}\cos^{2}\theta}}\,d\theta\notag\\ &= \frac{(1 + k)\cos\theta}{\sqrt{\cos^{2}\theta(1 + k^{2} + 2k - 2k - 2k^{2} + k^{2}\cos^{2}\theta)}}\,d\theta\notag\\ \Rightarrow \frac{d\psi}{\sqrt{1 - k_{1}^{2}\sin^{2}\psi}} &= (1 + k) \cdot \frac{d\theta}{\sqrt{1 - k^{2}\sin^{2}\theta}}\tag{9} \end{align} Thus we get $$u_{1} = F(\phi_{1}, k_{1}) = (1 + k)F(\phi, k) = (1 + k)u$$ where $$\sin\phi_{1} = \frac{(1 + k)\sin\phi}{1 + k\sin^{2}\phi}$$ or $$\text{sn}\left((1 + k)u, \frac{2\sqrt{k}}{1 + k}\right) = \frac{(1 + k)\,\text{sn}(u, k)}{1 + k\,\text{sn}^{2}(u, k)}\tag{10}$$ Also note that when $ \phi_{1} = \pi / 2$ then $ \phi = \pi / 2$ so that we have $ K(k_{1}) = (1 + k)K(k)$ or in a simpler notation $ K_{1} = (1 + k)K$ which can also be written as $ K = ((1 + k_{1}') / 2) K_{1}$.

An interesting point to note here is that the moduli $ k$ and $ k_{1}$ have exactly the same relation as the moduli $ k_{1}'$ and $ k'$ have. Hence we can easily see that $$K' = (1 + k_{1}')K_{1}^{\prime} = \frac{2K}{K_{1}}K_{1}'$$ or $$\frac{K'}{K} = 2\cdot \frac{K_{1}'}{K_{1}}$$ We can now summarize as follows:

Thus we can first note that $ 2 = (1 + k)(1 + \gamma')$ and set $ v = (1 + k)u$ so that $ 2u = (1 + \gamma')v$ and then \begin{align} \text{sn}(2u, k) &= \text{sn}\left((1 + \gamma')v, \frac{1 - \gamma'}{1 + \gamma'}\right) = \frac{(1 + \gamma')\,\text{sn}(v, \gamma)\,\text{cn}(v, \gamma)}{\text{dn}(v, \gamma)}\notag\\ &= \frac{(1 + \gamma')\,\text{sn}((1 + k)u, \gamma)\,\text{cn}((1 + k)u, \gamma)}{\text{dn}((1 + k)u, \gamma)}\notag \end{align} and we can replace the functions of modulus $ \gamma$ by functions of modulus $ k$ using the descending Landen transformation.

The corresponding descending transformation looks like $$\text{sn}\left(\frac{u}{M_{1}}, \lambda_{1}\right) = h(\text{sn}(u, k))$$ where $ \lambda_{1} > k$ which gives us $ \Lambda_{1} = K / M_{1}$. The Jacobi's imaginary transformation then gives a relation of the form $$\text{sn}\left(\frac{u}{M_{1}}, \lambda_{1}'\right) = p(\text{sn}(u, k'))$$ and we then get $ \Lambda_{1}' = K' / (nM_{1})$. Finally we have the relation $$\frac{1}{n}\frac{\Lambda'}{\Lambda} = \frac{K'}{K} = n\frac{\Lambda_{1}'}{\Lambda_{1}}$$ We shall not be discussing these general transformations any further and will use the Landen's transformation to get inifnite product expansions for the elliptic functions. This is slated for the next post.

Landen's Transformation is a set of formulas relating the elliptic functions of one modulus (the parameter $ k$) with the elliptic functions of another but related modulus. There is a general theory of transformation for the elliptic functions which we will not discuss in detail here, but it is worth mentioning that the Landen's transformation which we describe is actually a special kind of transformation and it is known as the transformation of

*second order*. The general theory deals with the transformation of order $ n$ where $ n$ is a positive integer greater than unity.### Landen's Transformation

We will begin with the equation $$u = \int_{0}^{\phi} \frac{d\theta}{\sqrt{1 - k^{2}\sin^{2}\theta}}$$ and $ \text{sn}(u, k) = \sin\phi$. Let us substitute $$ \sin(2\psi - \theta) = k\sin\theta\tag{1}$$ so that the limits of the interval of integration changes from $ [0, \phi]$ to $ [0, \phi_{1}]$Then we have \begin{align} k\sin\theta &= \sin 2\psi\cos\theta - \cos 2\psi\sin\theta\notag\\ \Rightarrow \cos\theta\sin 2\psi &= (k + \cos 2\psi)\sin\theta\notag\\ \Rightarrow \tan\theta &= \frac{\sin 2\psi}{k + \cos 2\psi}\tag{2} \end{align} We can now see that \begin{align} \sin\theta &= \frac{\tan\theta}{\sqrt{1 + \tan^{2}\theta}}\notag\\ &=\dfrac{\dfrac{\sin 2\psi}{k + \cos 2\psi}}{\sqrt{1 + \left(\dfrac{\sin 2\psi}{k + \cos 2\psi}\right)^{2}}}\notag\\ &= \frac{\sin 2\psi}{\sqrt{(k + \cos 2\psi)^{2} + \sin^{2}2\psi}}\notag\\ &= \frac{\sin 2\psi}{\sqrt{1 + k^{2} + 2k\cos 2\psi}}\notag\\ &= \frac{2\sin\psi\cos\psi}{\sqrt{1 + k^{2} + 2k - 4k\sin^{2}\psi}}\notag\\ &= \dfrac{2\sin\psi\cos\psi}{(1 + k)\sqrt{1 - \left(\dfrac{2\sqrt{k}}{1 + k}\right)^{2}\sin^{2}\psi}}\notag\\ \Rightarrow \sin\theta &= \frac{2}{1 + k}\frac{\sin\psi\cos\psi}{\sqrt{1 - k_{1}^{2}\sin^{2}\psi}}\text{ where } k_{1} = \frac{2\sqrt{k}}{1 + k}\tag{3} \end{align} Similarly $$\cos\theta = \frac{k + \cos 2\psi}{\sqrt{(k + \cos 2\psi)^{2} + \sin^{2}2\psi}}\tag{4}$$ and \begin{align} \sqrt{1 - k^{2}\sin^{2}\theta} &= \sqrt{1 - \sin^{2}(2\psi - \theta)} = \cos(2\psi - \theta)\notag\\ &= \cos 2\psi\cos\theta + \sin 2\psi\sin\theta\notag\\ &= \frac{\cos 2\psi(k + \cos 2\psi) + \sin^{2}2\psi}{\sqrt{(k + \cos 2\psi)^{2} + \sin^{2}2\psi}}\notag\\ &= \frac{1 + k\cos 2\psi}{\sqrt{(k + \cos 2\psi)^{2} + \sin^{2}2\psi}}\tag{5} \end{align} Differentiating equation $(2)$ and using equation $(4)$ we get \begin{align} \sec^{2}\theta\,d\theta &= \frac{(k + \cos 2\psi)2\cos 2\psi + 2\sin^{2}2\psi}{(k + \cos 2\psi)^{2}}\,d\psi\notag\\ \Rightarrow d\theta &= \frac{2(1 + k\cos 2\psi)}{(k + \cos 2\psi)^{2}}\frac{(k + \cos 2\psi)^{2}}{(k + \cos 2\psi)^{2} + \sin^{2}2\psi}\,d\psi\notag\\ &= \frac{2(1 + k\cos 2\psi)}{(k + \cos 2\psi)^{2} + \sin^{2}2\psi}\,d\psi\tag{6} \end{align} Hence from equation $(5)$ and $(6)$ we get \begin{align} \frac{d\theta}{\sqrt{1 - k^{2}\sin^{2}\theta}} &= \frac{2d\psi}{\sqrt{(k + \cos 2\psi)^{2} + \sin^{2}2\psi}}\notag\\ &= \frac{2}{1 + k}\frac{d\psi}{\sqrt{1 - k_{1}^{2}\sin^{2}\psi}}\notag \end{align} so that \begin{align} u &= F(\phi, k) = \int_{0}^{\phi}\frac{d\theta}{\sqrt{1 - k^{2}\sin^{2}\theta}}\notag\\ &= \frac{2}{1 + k}\int_{0}^{\phi_{1}}\frac{d\psi}{\sqrt{1 - k_{1}^{2}\sin^{2}\psi}} = \frac{2}{1 + k}F(\phi_{1}, k_{1})\notag \end{align} and we thus have $$\frac{(1 + k)u}{2} = F(\phi_{1}, k_{1})$$ which is equivalent to $$\text{sn}\left(\frac{(1 + k)u}{2}, \frac{2\sqrt{k}}{1 + k}\right) = \sin\phi_{1}$$ From equation $(3)$ we have $$\sin\phi = \frac{2}{1 + k}\frac{\sin\phi_{1}\cos\phi_{1}}{\sqrt{1 - k_{1}^{2}\sin^{2}\phi_{1}}}$$ hence it follows that $$\text{sn}(u, k) = \dfrac{2}{1 + k}\dfrac{\text{sn}\left(\dfrac{(1 + k)u}{2}, \dfrac{2\sqrt{k}}{1 + k}\right)\text{cn}\left(\dfrac{(1 + k)u}{2}, \dfrac{2\sqrt{k}}{1 + k}\right)}{\text{dn}\left(\dfrac{(1 + k)u}{2}, \dfrac{2\sqrt{k}}{1 + k}\right)}$$ This is the celebrated Landen's transformation for elliptic functions. It express the elliptic function of modulus $ k$ in terms of elliptic functions of another modulus $ k_{1} = 2\sqrt{k} / ( 1 + k)$ which is greater than $ k$. The above equation can be written in a slightly more convenient form by letting $ \lambda = k_{1} = 2\sqrt{k} / ( 1 + k)$ so that $ k = (1 - \lambda') / (1 + \lambda')$ and $ 2/(1 + k) = 1 + \lambda'$. We then have $$\text{sn}\left((1 + \lambda')u, \frac{1 - \lambda'}{1 + \lambda'}\right) = \frac{(1 + \lambda')\,\text{sn}(u, \lambda)\,\text{cn}(u, \lambda)}{\text{dn}(u, \lambda)}$$ Replacing $ \lambda$ by $ k$ we get $$\text{sn}\left((1 + k')u, \frac{1 - k'}{1 + k'}\right) = \frac{(1 + k')\,\text{sn}(u, k)\,\text{cn}(u, k)}{\text{dn}(u, k)}\tag{7}$$ We can now be a bit more systematic and starting with an initial modulus $ k = k_{0}$ with $ 0 < k_{0} < 1$ we can form a sequence of increase moduli $ \{k_{n}\}$ given by the ascending Landen transformation $$ \displaystyle k_{n + 1} = \frac{2\sqrt{k_{n}}}{1 + k_{n}}$$ We have a corresponding descending Landen transformation of the moduli by the equation $$k_{n} = \frac{1 - k_{n + 1}'}{1 + k_{n + 1}'}$$ so that we can get the sequence of moduli extended to negative indexes $$ \ldots < k_{-n} < k_{-n + 1} < \ldots < k_{-2} < k_{-1} < k_{0} = k < k_{1} < k_{2} < \ldots < k_{n} < \ldots$$ and for all integers $ n$ we have $$k_{n + 1} = \frac{2\sqrt{k_{n}}}{1 + k_{n}}\,\,\, k_{n} = \frac{1 - k_{n + 1}'}{1 + k_{n + 1}'}$$ It is clear that $ k_{n} \to 1$ as $ n \to \infty$ and $ k_{-n} \to 0$ as $ n \to \infty$ (and the convergence is damn fast, to be precise it is quadratic).

The transformation is quite useful as it allows us to replace the modulus by a smaller or a greater one and ultimately to the special cases $ k = 0$ (when the elliptic function reduce the circular functions) and $ k = 1$ (when they reduce to the hyperbolic functions).

Another Landen transformation is obtained by the substitution

$$\sin\psi = \frac{(1 + k)\sin\theta}{1 + k\sin^{2}\theta}\tag{8}$$ and this is applied on the integral $$u_{1} = F(\phi_{1}, k_{1}) = \int_{0}^{\phi_{1}}\frac{d\psi}{\sqrt{1 - k_{1}^{2}\sin^{2}\psi}}$$ and the interval of integration changes from $ [0, \phi_{1}]$ to $ [0, \phi]$. Clearly we have $$\sqrt{1 - k_{1}^{2}\sin^{2}\psi} = \sqrt{1 - \frac{4k}{(1 + k)^{2}}\frac{(1 + k)^{2}\sin^{2}\theta}{(1 + k\sin^{2}\theta)^{2}}} = \frac{1 - k\sin^{2}\theta}{1 + k\sin^{2}\theta}$$ and by differentiation of equation $(8)$ we get \begin{align} \cos\psi\,d\psi &= \frac{(1 + k\sin^{2}\theta)(1 + k)\cos\theta - (1 + k)\sin\theta \cdot 2k\sin\theta\cos\theta}{(1 + k\sin^{2}\theta)^{2}}\,d\theta\notag\\ \Rightarrow d\psi &= \frac{(1 + k)\cos\theta (1 - k\sin^{2}\theta)}{(1 + k\sin^{2}\theta)^{2}}\frac{1 + k\sin^{2}\theta}{\sqrt{(1 + k\sin^{2}\theta)^{2} - (1 + k)^{2}\sin^{2}\theta}}\,d\theta\notag \end{align} which leads to \begin{align} \frac{d\psi}{\sqrt{1 - k_{1}^{2}\sin^{2}\psi}} &= \frac{(1 + k)\cos\theta}{\sqrt{(1 + k\sin^{2}\theta)^{2} - (1 + k)^{2}\sin^{2}\theta}}\,d\theta\notag\\ &= \frac{(1 + k)\cos\theta}{\sqrt{(1 + k - k\cos^{2}\theta)^{2} - (1 + k)^{2} + (1 + k)^{2}\cos^{2}\theta}}\,d\theta\notag\\ &= \frac{(1 + k)\cos\theta}{\sqrt{\cos^{2}\theta(1 + k^{2} + 2k - 2k - 2k^{2} + k^{2}\cos^{2}\theta)}}\,d\theta\notag\\ \Rightarrow \frac{d\psi}{\sqrt{1 - k_{1}^{2}\sin^{2}\psi}} &= (1 + k) \cdot \frac{d\theta}{\sqrt{1 - k^{2}\sin^{2}\theta}}\tag{9} \end{align} Thus we get $$u_{1} = F(\phi_{1}, k_{1}) = (1 + k)F(\phi, k) = (1 + k)u$$ where $$\sin\phi_{1} = \frac{(1 + k)\sin\phi}{1 + k\sin^{2}\phi}$$ or $$\text{sn}\left((1 + k)u, \frac{2\sqrt{k}}{1 + k}\right) = \frac{(1 + k)\,\text{sn}(u, k)}{1 + k\,\text{sn}^{2}(u, k)}\tag{10}$$ Also note that when $ \phi_{1} = \pi / 2$ then $ \phi = \pi / 2$ so that we have $ K(k_{1}) = (1 + k)K(k)$ or in a simpler notation $ K_{1} = (1 + k)K$ which can also be written as $ K = ((1 + k_{1}') / 2) K_{1}$.

An interesting point to note here is that the moduli $ k$ and $ k_{1}$ have exactly the same relation as the moduli $ k_{1}'$ and $ k'$ have. Hence we can easily see that $$K' = (1 + k_{1}')K_{1}^{\prime} = \frac{2K}{K_{1}}K_{1}'$$ or $$\frac{K'}{K} = 2\cdot \frac{K_{1}'}{K_{1}}$$ We can now summarize as follows:

*Let modulus $ k$ with $ 0 < k < 1$ be given. Then we have two related moduli $ \lambda, \gamma$ such that $ \lambda < k < \gamma$ and*\begin{align} \lambda &= \frac{1 - k'}{1 + k'},\,\, k = \frac{2\sqrt{\lambda}}{1 + \lambda}\notag\\ k &= \frac{1 - \gamma'}{1 + \gamma'},\,\, \gamma = \frac{2\sqrt{k}}{1 + k}\notag\\ \gamma' &= \frac{1 - k}{1 + k},\,\, k' = \frac{1 - \lambda}{1 + \lambda}\notag \end{align}*then the moduli $ \lambda, k, \gamma$ form an ascending Landen sequence. It is then also seen (surprise!) that the complementary moduli in reverse order $ \gamma', k', \lambda'$ form an ascending Landen sequence.**The ascending Landen's Transformation relates the elliptic functions of a given modulus with those of a greater modulus:*$$\text{sn}\left((1 + k')u, \frac{1 - k'}{1 + k'}\right) = \frac{(1 + k')\,\text{sn}(u, k)\,\text{cn}(u, k)}{\text{dn}(u, k)}$$*or*$$\text{sn}((1 + k')u, \lambda) = \frac{(1 + k')\,\text{sn}(u, k)\,\text{cn}(u, k)}{\text{dn}(u, k)}$$*The descending Landen's Transformation relates the elliptic functions of a given modulus with those of a smaller modulus:*$$\text{sn}\left((1 + k)u, \frac{2\sqrt{k}}{1 + k}\right) = \frac{(1 + k)\,\text{sn}(u, k)}{1 + k\,\text{sn}^{2}(u, k)}$$*or*$$\text{sn}((1 + k)u, \gamma) = \frac{(1 + k)\,\text{sn}(u, k)}{1 + k\,\text{sn}^{2}(u, k)}$$*Moreover we have the following relation between the complete elliptic integrals $ \Lambda, K, \Gamma$ corresponding to moduli $ \lambda, k, \gamma$:*$$(1 + \lambda)\Lambda = K = \frac{1}{1 + k}\Gamma$$ $$\frac{1}{2}(1 + \lambda)\Lambda' = K' = \frac{2}{1 + k}\Gamma'$$ $$\frac{1}{2}\frac{\Lambda'}{\Lambda} = \frac{K'}{K} = 2\frac{\Gamma'}{\Gamma}$$ The above two transformation can be used to express the elliptic functions of argument $ 2u$ in terms of those of argmument $ u$ (which is more easily done by the addition formulas). Such a relation between the functions of argument $ 2u$ and $ u$ is called a duplication formula. The Landen transformations help us break the duplication formula into two parts by switching to a larger modulus first and then back to the original smaller modulus (or alternatively first switching to a smaller modulus and then going back to the original larger modulus).Thus we can first note that $ 2 = (1 + k)(1 + \gamma')$ and set $ v = (1 + k)u$ so that $ 2u = (1 + \gamma')v$ and then \begin{align} \text{sn}(2u, k) &= \text{sn}\left((1 + \gamma')v, \frac{1 - \gamma'}{1 + \gamma'}\right) = \frac{(1 + \gamma')\,\text{sn}(v, \gamma)\,\text{cn}(v, \gamma)}{\text{dn}(v, \gamma)}\notag\\ &= \frac{(1 + \gamma')\,\text{sn}((1 + k)u, \gamma)\,\text{cn}((1 + k)u, \gamma)}{\text{dn}((1 + k)u, \gamma)}\notag \end{align} and we can replace the functions of modulus $ \gamma$ by functions of modulus $ k$ using the descending Landen transformation.

### General Transformations of order $ n$

We will end this post by mentioning some remarks about the general theory of transformation of elliptic functions. Like in the case of the second order there exist transformations of the $ n^{\text{th}}$ order where we have a relation between the given modulus and a new modulus (which could be greater or smaller depending upon whether the transformation is ascending or descending) and then we can relate the elliptic functions of the given modulus with those of the newer modulus. For odd values of $ n$ the relation between the elliptic functions can be expressed rationally, but for $ n$ even the relation includes some irrationalities (mainly square roots). These relations are of the form (let's just take the case when $ n$ is odd) $$\text{sn}\left(\frac{u}{M}, \lambda\right) = f(\text{sn}(u, k))$$ where $ f$ is a rational function, $ M$ (called*multiplier)*is a constant depending only on $ k, \lambda$ and $ \lambda < k$ is the new modulus. This also leads to the relation between the complete elliptic integrals $ \Lambda = K / (nM)$. Applying Jacobi's imaginary transformation we obtain a relation $$\text{sn}\left(\frac{u}{M}, \lambda'\right) = g(\text{sn}(u, k'))$$ where $ g$ is a rational function and we get the relation $ \Lambda' = K' / M$.The corresponding descending transformation looks like $$\text{sn}\left(\frac{u}{M_{1}}, \lambda_{1}\right) = h(\text{sn}(u, k))$$ where $ \lambda_{1} > k$ which gives us $ \Lambda_{1} = K / M_{1}$. The Jacobi's imaginary transformation then gives a relation of the form $$\text{sn}\left(\frac{u}{M_{1}}, \lambda_{1}'\right) = p(\text{sn}(u, k'))$$ and we then get $ \Lambda_{1}' = K' / (nM_{1})$. Finally we have the relation $$\frac{1}{n}\frac{\Lambda'}{\Lambda} = \frac{K'}{K} = n\frac{\Lambda_{1}'}{\Lambda_{1}}$$ We shall not be discussing these general transformations any further and will use the Landen's transformation to get inifnite product expansions for the elliptic functions. This is slated for the next post.

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