Elliptic Functions: Addition Formulas

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After having dealt with the basic properties of elliptic functions in the previous post we shall now focus on the addition formulas for them. These are used to express the functions of sum of two arguments in terms of functions of each argument separately. The additions formulas are algebraic in nature and in fact, in general any function with an algebraic addition formula is necessarily an elliptic function or a limiting case of it. We will not prove this general result here as it requires the use of theory of functions, but we shall be content to derive the formulas for the specific elliptic function which we are considering here.

Addition Formula

Let us assume that the variables u1 and u2 vary in such a way that their sum u1+u2 is a constant. And we set: s1=sn(u1,k),c1=cn(u1,k),d1=dn(u1,k)s2=sn(u2,k),c2=cn(u2,k),d2=dn(u2,k)u1+u2=a so that du2du1=1 Then we have ddu1(s1+s2)=c1d1c2d2ddu1(s1c2+s2c1)=s1s2d2+c1d1c2s2s1d1c2d2c1=(d1d2)(c1c2s1s2)ddu1(d1+d2)=k2(s1c1s2c2)=k2s1c1(c22+s22)+k2s2c2(c21+s21)=k2(c1c2s1s2)(s1c2s2c1) Since we have k2(s21c22s22c21)=k2(s21(1s22)s22(1s21))=k2(s21s22)=d21d22 It therefore follows that (d1+d2)ddu1(s1c2+s2c1)=(s1c2+s2c1)ddu1(d1+d2) From the above it is clear that the ratio (s1c2+s2c1)/(d1+d2) is a constant independent of u1 as long as the sum u1+u2=a is a constant. Therefore value of this ratio is obtained by putting u1=0 and u2=a and we have s1c2+s2c1d1+d2=sn(a,k)1+dn(a,k)=sn(u1+u2,k)1+dn(u1+u2,k) Again we can see that s1c2s2c1d1d2=1k2d1+d2s1c2+s2c1= a constant independent of u1=sn(a,k)dn(a,k)1 by putting u1=a,u2=0=sn(u1+u2,k)dn(u1+u2,k)1 We therefore have the following results: dn(u1+u2,k)+1sn(u1+u2,k)=d1+d2s1c2+s2c1dn(u1+u2,k)1sn(u1+u2,k)=d1d2s1c2s2c1 Subtracting the above equations we get 2sn(u1+u2,k)=d1+d2s1c2+s2c1d1d2s1c2s2c1=2(s1c2d2s2c1d1)s21c22s22c21 and therefore sn(u1+u2,k)=s21s22s1c2d2s2c1d1 Adding the equations (A) and (B) we get 2ds(u1+u2,k)=d1+d2s1c2+s2c1+d1d2s1c2s2c1=2(s1c2d1s2c1d2)s21s22 and therefore dn(u1+u2,k)=s1c2d1s2c1d2s1c2d2s2c1d1 In similar fashion one can obtain the following formulas: cn(u1+u2,k)+1sn(u1+u2,k)=c1+c2s1d2+s2d1cn(u1+u2,k)1sn(u1+u2,k)=c1c2s1d2s2d1 Adding the above equations we get 2cs(u1+u2,k)=2(s1c1d2s2c2d1)s21s22 and finally we have cn(u1+u2,k)=s1c1d2s2c2d1s1c2d2s2c1d1 To summarize we have the following addition formulas: sn(u1+u2,k)=s21s22s1c2d2s2c1d1cn(u1+u2,k)=s1c1d2s2c2d1s1c2d2s2c1d1dn(u1+u2,k)=s1c2d1s2c1d2s1c2d2s2c1d1 The above forms are hard to remember and don't look similar to addition formulas for circular functions. However they can be transformed by algebraic manipulations into a form which is similar to those of circular functions. This we will do in the next post. For the time being we can use these results to obtain expressions for sn(u+K,k),cn(u+K,k),dn(u+K,k).
Putting u1=u and u2=K we get sn(u+K,k)=cn(u,k)dn(u,k)=cd(u,k)cn(u+K,k)=ksn(u,k)dn(u,k)=ksd(u,k)dn(u+K,k)=kdn(u,k)=knd(u,k) We can now put u1+K in place of u1 in the addition formulas to get sn(u1+K+u2,k)=c21d21s22c1c2d2d1+k2s1s2d21cn(u1+u2,k)dn(u1+u2,k)=c21d21s22c1c2d1d2+k2s1s2 and dn(u1+K+u2,k)=kc1c2d21+ks1s2d2d1c1c2d2d1+k2s1s2d21kdn(u1+u2,k)=kc1c2+ks1s2d1d2c1c2d1d2+k2s1s2dn(u1+u2,k)=c1c2d1d2+k2s1s2c1c2+s1s2d1d2cn(u1+u2,k)=c21d21s22c1c2+s1s2d1d2=1s21s22+k2s21s22c1c2+s1s2d1d2 Again cn(u1+K+u2,k)=ks1c1d2d21ks2c2d1c1c2d2d1+k2s1s2d21sn(u1+u2,k)dn(u1+u2,k)=s1c1d2+s2c2d1c1c2d1d2+k2s1s2 Thus we can now write the addition formulas in a different form: sn(u1+u2,k)=s1c1d2+s2c2d1c1c2+s1s2d1d2cn(u1+u2,k)=1s21s22+k2s21s22c1c2+s1s2d1d2dn(u1+u2,k)=c1c2d1d2+k2s1s2c1c2+s1s2d1d2 We have so far developed the addition formulas in two different forms and we note that in each of the forms the formulas have a common denominator. Using them we have also obtained the expressions for sn(u+K,k) etc.

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