Elementary Approach to Modular Equations: Jacobi's Transformation Theory 1

To recapitulate the basics of elliptic integral theory (details here) we have $$K = K(k) = \int_{0}^{\pi/2}\frac{d\theta}{\sqrt{1 - k^{2}\sin^{2}\theta}} = \int_{0}^{1}\frac{dx}{\sqrt{(1 - x^{2})(1 - k^{2}x^{2})}}$$ $$E = E(k) = \int_{0}^{\pi/2}\sqrt{1 - k^{2}\sin^{2}\theta}\,d\theta = \int_{0}^{1}\frac{\sqrt{1 - k^{2}x^{2}}}{\sqrt{1 - x^{2}}}\,dx$$
We have here $0 \leq k \leq 1$ and the complementary modulus $k' = \sqrt{1 - k^{2}}$ or in other words $k^{2} + k'^{2} = 1$. Also we denote $K' = K(k'), E' = E(k')$ and we have the Legendre's Identity $$KE' + K'E - KK' = \frac{\pi}{2}$$

The Function $K'/K$

From the definition it is clear that $K$ is a strictly increasing function of $k$ and moreover $K \to \infty$ as $k \to 1$. Thus function $K(k)$ maps interval $[0, 1)$ to $[\pi/2, \infty)$. Similarly $E$ is strictly decreasing and maps $[0, 1]$ to $[1, \pi/2]$. Therefore it follows that $K'$ is a strictly decreasing function of $k$ and $E'$ is a strictly increasing function of $k$.

From the above arguments it follows that the function $K'/K$ is a strictly decreasing function of $k$. When $k \to 0$, $K'/K \to \infty$ and as $k \to 1, K'/K \to 0$. Thus the function $K'/K$ maps the interval $(0, 1)$ to the interval $(0, \infty)$. For any given positive number $\alpha$ we have a unique positive $k \in (0, 1)$ such that $K'/K = \alpha$. Let $p$ be any positive number and then $p\alpha$ is also positive and therefore there is a unique number $l \in (0, 1)$ such that $K(l')/K(l) = p\alpha$. We traditionally denote $L = K(l), L' = K(l')$. Therefore we can write $$\frac{L'}{L} = p\alpha = p\,\frac{K'}{K}$$
Again let's think in the following way. Suppose a number $k \in (0, 1)$ and a positive number $p$ is given. From $k$ we can calculate $K'/K$ and hence $pK'/K$ corresponding to which there is a unique number $l \in (0, 1)$ such that $L'/L = pK'/K$. In this fashion $l$ turns out to be a function of $k$. Therefore the equation $L'/L = pK'/K$ determines $l$ as a function of $k$ which is one-one and invertible. From the continuity of $K$ the function turns out to be continuous. To derive these results formally we need the fundamental formulas: $$\frac{dk'}{dk} = -\frac{k}{k'},\,\, \frac{dK}{dk} = \frac{E - k'^{2}K}{kk'^{2}},\,\, \frac{dE}{dk} = \frac{E - K}{k}$$ (for the proofs of these results read here)

Using these we have $$\frac{dK'}{dk} = \frac{dk'}{dk}\frac{dK'}{dk'} = -\frac{k}{k'}\frac{E' - k^{2}K'}{k^{2}k'} = \frac{k^{2}K' - E'}{kk'^{2}}$$ and therefore $$\begin{align}\frac{d}{dk}\left(\frac{K'}{K}\right) &= \dfrac{K\dfrac{dK'}{dk} - K'\dfrac{dK}{dk}}{K^{2}} = \frac{1}{K^{2}}\left(\frac{K(k^{2}K' - E')}{kk'^{2}} - \frac{K'(E - k'^{2}K)}{kk'^{2}}\right)\notag\\ &= \frac{KK' - KE' - K'E}{kk'^{2}K^{2}} = -\frac{\pi}{2kk'^{2}K^{2}}\notag\end{align}$$ Thus we have finally another fundamental relation $$\frac{d}{dk}\left(\frac{K'}{K}\right) = -\frac{\pi}{2kk'^{2}K^{2}}$$ which shows that $K'/K$ is strictly decreasing. On differentiating the equation $L'/L = pK'/K$ we get $$\frac{dl}{ll'^{2}L^{2}} = p\frac{dk}{kk'^{2}K^{2}}$$ or $$\frac{dl}{dk} = p\,\frac{ll'^{2}}{kk'^{2}}\left(\frac{L}{K}\right)^{2}$$ and therefore $l$ is a strictly increasing function of $k$ and maps $(0, 1)$ to $(0, 1)$. Also $p <(=,>) 1 \Rightarrow l > (=,<) k$. We also see that the ratio $L/K$ is a function of $k, l$ and we traditionally call it the multiplier and denote by $M_{p}(l, k)$. Thus we have $$pM_{p}^{2}(l, k) = p\left(\frac{L}{K}\right)^{2} = \frac{kk'^{2}}{ll'^{2}}\frac{dl}{dk}$$ Also it should be noted that if $p < (=, >) 1 \Rightarrow M_{p}(l, k) > (=, <) 1$.

Jacobi's Transformation Theory

Jacobi showed that the relation between $l$ and $k$ is algebraic when $p$ is rational. First of all as an example we can take $p = 2$. Then by Landen's transformation $l = (1 - k')/(1 + k') < k$ we see that $L'/L = 2K'/K$. Clearly here the relation between $l$ and $k$ is algebraic and is also given by $k = 2\sqrt{l}/(1 + l)$. Also $$\frac{dl}{dk} = \frac{(1 + k')(k/k') + (1 - k')(k/k')}{(1 + k')^{2}} = \frac{2k}{k'(1 + k')^{2}}$$ and therefore $$2M_{2}^{2}(l, k) = \frac{kk'^{2}}{ll'^{2}}\frac{dl}{dk} = kk'^{2}\frac{1 + k'}{1 - k'}\frac{(1 + k')^{2}}{4k'}\frac{2k}{k'(1 + k')^{2}} = \frac{k^{2}}{2}\frac{(1 + k')^{2}}{1 - k'^{2}}$$ so that $L/K = M_{2}(l, k) = (1 + k')/2 = 1/(1 + l)$ as should have been the case.

Next let's understand that it is necessary only to study the case when $p$ is a positive prime. For, if for positive prime $p$ the relation $L'/L = pK'/K$ is leading to an algebraic relation between $l$ and $k$ then we can show that the same kind of relation holds when $p$ is any positive rational number. First suppose $p$ is composite positive integer and $p = p_{1}p_{2}$ where both $p_{1}, p_{2}$ are primes. Then $L'/L = pK'/K$ can be written $L'/L = p_{1}\Gamma'/\Gamma = p_{1}(p_{2}K'/K)$. Since the relation between $l, \gamma$ and $\gamma, k$ is algebraic by assumption, the relation between $l, k$ is algebraic. The same reasoning can be extended to the case when $p$ is a product of any number of primes. Thus the case of positive integers is handled. In case of $p = a/b$ where $a, b$ are positive integers we can write $L'/L = pK'/K$ as $bL'/L = aK'/K = \Gamma'/\Gamma$. Since the relation between $\gamma, k$ and $\gamma, l$ is algebraic therefore the relation between $k, l$ is algebraic.

So from now on let's assume that $p$ is a positive prime. The algebraic relation between $l, k$ implied by the equation $L'/L = pK'/K$ is called a modular equation of degree $p$. The way Jacobi obtained these modular equations is related to the transformation of elliptic integrals. Consider the differential $$\frac{dy}{\sqrt{(1 - y^{2})(1 - l^{2}y^{2})}}$$ By suitable rational transformation $y = f(x)/g(x)$ where $f(x), g(x)$ are polynomials it is expected to reduce the differential to the form $$\frac{dx}{M\sqrt{(1 - x^{2})(1 - k^{2}x^{2})}}$$ Thus in effect we wish to establish a relation of the form $$\frac{Mdy}{\sqrt{(1 - y^{2})(1 - l^{2}y^{2})}} = \frac{dx}{\sqrt{(1 - x^{2})(1 - k^{2}x^{2})}}$$ where $y$ is a rational function of $x$ and $M$ is some constant. That such a transformation is possible is not so obvious. Jacobi started his Fundamenta Nova by describing this elegant theory of transformation and showed that such a transformation exists for every value of the positive prime $p$ which is called the order of the transformation. In what follows we shall assume that $p$ is an odd prime ($p = 2$ is covered by the famous Landen Transformations).

The constant $M$ is called multiplier and depends upon the values of $l, k$ and is related to the multiplier $M_{p}(l, k)$. To simplify the process of transformation the rational function $y = U(x)/V(x)$ is chosen to be of very specific form. With the requirement that $y$ vanishes with $x$ we take $y = xf(x)/g(x)$ where $f(x), g(x)$ are polynomials of degrees $(p - 1)$. Also $f(x), g(x)$ are supposed to be even functions so that they are functions of $x^{2}$ and then we can write $$y = \frac{xN(1, x^{2})}{D(1, x^{2})}$$ where $N, D$ are homogeneous polynomials of degree $(p - 1)/2$. Another condition which we impose is that the relation between $y$ and $x$ does not change when $(x, y)$ is replaced by $(1/kx, 1/ly)$. That this is possible needs to be demonstrated. First of all we can see that $$N\left(1, \frac{1}{k^{2}x^{2}}\right) = \left(\frac{1}{kx}\right)^{p - 1}N(k^{2}x^{2}, 1)$$ We can next choose $D(1, x^{2})$ such that $N(k^{2}x^{2}, 1) = \Delta D(1, x^{2})$ where $\Delta$ is a constant. Thus the coefficients of $D(1, x^{2})$ are same as those of $N(1, x^{2})$, but in reverse order and multiplied by suitable powers of $k$. It thus follows that $$N\left(1, \frac{1}{k^{2}x^{2}}\right) = \Delta\left(\frac{1}{kx}\right)^{p - 1}D(1, x^{2})$$ Replacing $x$ by $1/kx$ we get $$N(1, x^{2}) = \Delta x^{p - 1}D\left(1, \frac{1}{k^{2}x^{2}}\right)$$ Multiplying the above two equations we get $$N(1, x^{2})N\left(1, \frac{1}{k^{2}x^{2}}\right) = \frac{\Delta^{2}}{k^{p - 1}}D(1, x^{2})D\left(1, \frac{1}{k^{2}x^{2}}\right)$$ Now let's suppose that in the defining equation if we put $1/kx$ in place of $x$ then the value of $y$ changes to $y_{1}$. Thus $$y_{1} = \dfrac{\dfrac{1}{kx}N\left(1, \dfrac{1}{k^{2}x^{2}}\right)}{D\left(1, \dfrac{1}{k^{2}x^{2}}\right)} = \frac{1}{kx}\frac{\Delta^{2}}{k^{p - 1}}\frac{D(1, x^{2})}{N(1, x^{2})} = \frac{\Delta^{2}}{k^{p}}\frac{D(1, x^{2})}{xN(1, x^{2})} = \frac{\Delta^{2}}{k^{p}}\frac{1}{y}$$ Clearly this reduces to $1/ly$ provided $l = k^{p}/\Delta^{2}$.

Hence a transformation of the form $y = U(x)/V(x)$ with $U$ of degree $p$ and $V$ of degree $(p - 1)$ with $U$ being odd function and $V$ being even function is possible which remains invariant under the change of variables $(x, y) \to (1/kx, 1/ly)$.

Next we would like to have further restrictions on the form of $U$ and $V$. Putting $y = U/V$ leads to $$dy = \frac{VU' - UV'}{V^{2}}\,dx,\,\, \sqrt{(1 - y^{2})(1 - l^{2}y^{2})} = \frac{1}{V^{2}}\sqrt{(V^{2} - U^{2})(V^{2} - l^{2}U^{2})}$$ and hence $$\frac{dy}{\sqrt{(1 - y^{2})(1 - l^{2}y^{2})}} = \frac{VU' - UV'}{\sqrt{(V^{2} - U^{2})(V^{2} - l^{2}U^{2})}}\,dx$$ If it is possible that $V + U = (1 + x)A^{2}, V - U = (1 - x)B^{2}$ for some polynomials $A, B$ then we can see that $$1 + y = (1 + x)A^{2}/V,\, 1 - y = (1 - x)B^{2}/V$$ and by invariance of relation between $x$ and $y$ under the transformation $$(x, y) \to (1/kx, 1/ky)$$ we can see that we must also have the relations $$1 + ly = (1 + kx)C^{2}/V,\, 1 - ly = (1 - kx)D^{2}/V$$ or $$V +lU = (1 + kx)C^{2},\, V - lU = (1 - kx)D^{2}$$ Now we can see that if $(x - \alpha)$ is a factor of $A$ then $(x - \alpha)^{2}$ divides $V + U$ and hence $(x - \alpha)$ divides $(V + U)U' - U(V + U)' = VU' - UV'$. From the same logic it follows that $A, B, C, D$ each divide $VU' - UV'$ and since the degrees of $VU' - UV'$ and $ABCD$ are same (equal to $(2p - 2)$) it follows that the ratio $ABCD/(VU' - UV')$ is a constant which we denote by $M$. Thus we arrive at $$\frac{Mdy}{\sqrt{(1 - y^{2})(1 - l^{2}y^{2})}} = \frac{dx}{\sqrt{(1 - x^{2})(1 - k^{2}x^{2})}}$$ The idea is now to choose polynomials $A, B$ in a suitable form. We keep $A = P + Qx, B = P - Qx$ such that $P, Q$ are even functions of $x$ and the degree of $P \pm Qx$ is $(p - 1)/2$. In general if $p = (4n - 1)$ then the degrees of $P, Q$ are $(2n - 2)$ and if $p = 4n + 1$, degree of $P$ is $2n$ and that of $Q$ is $(2n - 2)$.

Thus we arrive at the following transformation $$\frac{1 - y}{1 + y} = \frac{1 - x}{1 + x}\frac{(P - Qx)^{2}}{(P + Qx)^{2}}$$ so that $$y = \frac{x(P^{2} + 2PQ + Q^{2}x^{2})}{P^{2} + 2PQx^{2} + Q^{2}x^{2}}$$ and then we obtain $$\begin{align}1 - y &= \frac{(1 - x)(P - Qx)^{2}}{P^{2} + 2PQx^{2} + Q^{2}x^{2}}\notag\\ 1 + y &= \frac{(1 + x)(P + Qx)^{2}}{P^{2} + 2PQx^{2} + Q^{2}x^{2}}\notag\\ 1 - ly &= \frac{(1 - kx)(P' - Q'x)^{2}}{P^{2} + 2PQx^{2} + Q^{2}x^{2}}\notag\\ 1 + ly &= \frac{(1 + kx)(P' + Q'x)^{2}}{P^{2} + 2PQx^{2} + Q^{2}x^{2}}\notag\end{align}$$ The actual methodology of finding the polynomials $U, V$ (or $P, Q$) would be dealt with in next post by showing examples with $p = 3$ and $p = 5$. From the theoretical investigation above it is clear that the process of finding these polynomials (i.e. their coefficients) is totally algebraical and hence the relation between $k$ and $l$ is algebraic (note that $l = k^{p}/\Delta^{2}$, and that $\Delta$ itself would be an algebraic function of $l, k$). It should also be observed that the multiplier $M$ is also an algebraical function of $l$ and $k$ (calculation of $M$ is easy if we observe that $1/M = dy/dx \text{ at } x = 0$).

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