To recapitulate the basics of elliptic integral theory (details here) we have
K=K(k)=∫π/20dθ√1−k2sin2θ=∫10dx√(1−x2)(1−k2x2)
E=E(k)=∫π/20√1−k2sin2θdθ=∫10√1−k2x2√1−x2dx
We have here 0≤k≤1 and the complementary modulus k′=√1−k2 or in other words k2+k′2=1. Also we denote K′=K(k′),E′=E(k′) and we have the Legendre's Identity KE′+K′E−KK′=π2
From the above arguments it follows that the function K′/K is a strictly decreasing function of k. When k→0, K′/K→∞ and as k→1,K′/K→0. Thus the function K′/K maps the interval (0,1) to the interval (0,∞). For any given positive number α we have a unique positive k∈(0,1) such that K′/K=α. Let p be any positive number and then pα is also positive and therefore there is a unique number l∈(0,1) such that K(l′)/K(l)=pα. We traditionally denote L=K(l),L′=K(l′). Therefore we can write L′L=pα=pK′K
Again let's think in the following way. Suppose a number k∈(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∈(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: dk′dk=−kk′,dKdk=E−k′2Kkk′2,dEdk=E−Kk (for the proofs of these results read here)
Using these we have dK′dk=dk′dkdK′dk′=−kk′E′−k2K′k2k′=k2K′−E′kk′2 and therefore ddk(K′K)=KdK′dk−K′dKdkK2=1K2(K(k2K′−E′)kk′2−K′(E−k′2K)kk′2)=KK′−KE′−K′Ekk′2K2=−π2kk′2K2 Thus we have finally another fundamental relation ddk(K′K)=−π2kk′2K2 which shows that K′/K is strictly decreasing. On differentiating the equation L′/L=pK′/K we get dlll′2L2=pdkkk′2K2 or dldk=pll′2kk′2(LK)2 and therefore l is a strictly increasing function of k and maps (0,1) to (0,1). Also p<(=,>)1⇒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 Mp(l,k). Thus we have pM2p(l,k)=p(LK)2=kk′2ll′2dldk Also it should be noted that if p<(=,>)1⇒Mp(l,k)>(=,<)1.
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=p1p2 where both p1,p2 are primes. Then L′/L=pK′/K can be written L′/L=p1Γ′/Γ=p1(p2K′/K). Since the relation between l,γ and γ,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=Γ′/Γ. Since the relation between γ,k and γ,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 dy√(1−y2)(1−l2y2) 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 dxM√(1−x2)(1−k2x2) Thus in effect we wish to establish a relation of the form Mdy√(1−y2)(1−l2y2)=dx√(1−x2)(1−k2x2) 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 Mp(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 x2 and then we can write y=xN(1,x2)D(1,x2) 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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We have here 0≤k≤1 and the complementary modulus k′=√1−k2 or in other words k2+k′2=1. Also we denote K′=K(k′),E′=E(k′) and we have the Legendre's Identity KE′+K′E−KK′=π2
The Function K′/K
From the definition it is clear that K is a strictly increasing function of k and moreover K→∞ as k→1. Thus function K(k) maps interval [0,1) to [π/2,∞). Similarly E is strictly decreasing and maps [0,1] to [1,π/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→0, K′/K→∞ and as k→1,K′/K→0. Thus the function K′/K maps the interval (0,1) to the interval (0,∞). For any given positive number α we have a unique positive k∈(0,1) such that K′/K=α. Let p be any positive number and then pα is also positive and therefore there is a unique number l∈(0,1) such that K(l′)/K(l)=pα. We traditionally denote L=K(l),L′=K(l′). Therefore we can write L′L=pα=pK′K
Again let's think in the following way. Suppose a number k∈(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∈(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: dk′dk=−kk′,dKdk=E−k′2Kkk′2,dEdk=E−Kk (for the proofs of these results read here)
Using these we have dK′dk=dk′dkdK′dk′=−kk′E′−k2K′k2k′=k2K′−E′kk′2 and therefore ddk(K′K)=KdK′dk−K′dKdkK2=1K2(K(k2K′−E′)kk′2−K′(E−k′2K)kk′2)=KK′−KE′−K′Ekk′2K2=−π2kk′2K2 Thus we have finally another fundamental relation ddk(K′K)=−π2kk′2K2 which shows that K′/K is strictly decreasing. On differentiating the equation L′/L=pK′/K we get dlll′2L2=pdkkk′2K2 or dldk=pll′2kk′2(LK)2 and therefore l is a strictly increasing function of k and maps (0,1) to (0,1). Also p<(=,>)1⇒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 Mp(l,k). Thus we have pM2p(l,k)=p(LK)2=kk′2ll′2dldk Also it should be noted that if p<(=,>)1⇒Mp(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√l/(1+l). Also dldk=(1+k′)(k/k′)+(1−k′)(k/k′)(1+k′)2=2kk′(1+k′)2 and therefore 2M22(l,k)=kk′2ll′2dldk=kk′21+k′1−k′(1+k′)24k′2kk′(1+k′)2=k22(1+k′)21−k′2 so that L/K=M2(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=p1p2 where both p1,p2 are primes. Then L′/L=pK′/K can be written L′/L=p1Γ′/Γ=p1(p2K′/K). Since the relation between l,γ and γ,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=Γ′/Γ. Since the relation between γ,k and γ,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 dy√(1−y2)(1−l2y2) 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 dxM√(1−x2)(1−k2x2) Thus in effect we wish to establish a relation of the form Mdy√(1−y2)(1−l2y2)=dx√(1−x2)(1−k2x2) 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 Mp(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 x2 and then we can write y=xN(1,x2)D(1,x2) 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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y=x*f(x)/g(x) Why f(x),g(x) should be polynomials of degrees p-1?
pokhkitro
December 20, 2021 at 6:09 PM