We can finish this paper with the famous integral for the ellipse arc length, which give the name to all elliptic group. Arc of ellipse: ds=∫dz√(A-x2)/√(1-x2) A=m/(m-1)>1 The transformation of the integrand with application of projective variable change of Moebius x=(az+b)/(cz+d) (1) will give I=√[(z2+2mz+n)/√[(zq)] denominator of first degree (*). It will be no more elliptic if: A) the green trinomial has a double root, id est n=m2, and (z-m) comes out of radical. B) both trinomials have a common root, id est q has to be root for the green trinomial, then q2+2mq+n=0 Only one condition: and we have 2 freedom degrees. In my blog there is the B case (paper 5) Here we expose the A case ********************** There is not any periodicity double or alternate, we go from x= 0 to 1. It is a normal curve, the only problem is an improper point for x=1 (vertical tangent). Then when we want to calculate one quart of ellipse, the definite integration, we have to make from x=0 to x=1/2 (by instance), and a second integral at derivative to y from x=1/2, to x=1 See figure, the second tintegration will be from x=1/2 to x=1 (y=0) at the derivative respect to Y;ds/dy (it is in page end). We’ll do a double try … ***************** 2 2 ds=∫dz√(A-x )/√(1-x ) The projective transformation of Moebius x=(z+b)/(cz+d) dx=dz/(z+d/c)2
A>1 is a data calling d/x=h
gives
x2=[z2+2bz+b2]/[z2+2hz+h2] A-x2=[z2(A-1)+2z(Ah+b)+Ah2-b2]/(z+h)2 1-x2=(2z(h-b)+h2-b2]//z+h)2=(z+q)/(z+h)2 I=∫dz√(A-x2)/√(1-x2)/(z+h)-2
the two (z+h) compensate.
The blue is (z+q) q=(h+b)/2
b=2q-h
we do d=0 h=0 q=b/2
b is 2q
the numerator is z2(A-x2)=(Α−1)z2+2bz-b2
the roots are z=-b+/-b√[1+(A-1)]/(A-1)=b(-1+/-√A)/(A-1) one have to be q remind that A-1=√A-1)(√A+1) then z=1/(√(A+1)=q b(+)=2q
b=2/(1+√A)
b’(-)=2/(√A-1) is the other root I=∫dz√[(z+2(1-√A)]/z2 we do z=t2-b dz=tdt