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Question: Can Maple help to confirm equality of these elliptic expressions (or the opposite)

Question:Can Maple help to confirm equality of these elliptic expressions (or the opposite)

C_R 2530 Maple
integration simplification elliptic-integral
February 05 2023
2

 

 

queryequal(expr1, expr2)

queryequal(Int(1/((x__0-x)^(1/2)*(-x^2+1)^(1/2)), x = 0 .. x__0) = 2^(1/2)*EllipticK((1/2)*(2+2*x__0)^(1/2))-2^(1/2)*EllipticF(1/(1+x__0)^(1/2), (1/2)*(2+2*x__0)^(1/2)), 2^(1/2)*EllipticF((1-1/(1+x__0))^(1/2)*2^(1/2), (1/2)*(1+x__0)^(1/2)*2^(1/2)))

 (1)

(This question has its origin here).
Plotting the difference of rhs - lhs seems to indicate equality over the range [0..1)

difference := rhs(op(queryequal(Int(1/((x__0-x)^(1/2)*(-x^2+1)^(1/2)), x = 0 .. x__0) = 2^(1/2)*EllipticK((1/2)*(2+2*x__0)^(1/2))-2^(1/2)*EllipticF(1/(1+x__0)^(1/2), (1/2)*(2+2*x__0)^(1/2)), 2^(1/2)*EllipticF((1-1/(1+x__0))^(1/2)*2^(1/2), (1/2)*(1+x__0)^(1/2)*2^(1/2))))[1])-op(queryequal(Int(1/((x__0-x)^(1/2)*(-x^2+1)^(1/2)), x = 0 .. x__0) = 2^(1/2)*EllipticK((1/2)*(2+2*x__0)^(1/2))-2^(1/2)*EllipticF(1/(1+x__0)^(1/2), (1/2)*(2+2*x__0)^(1/2)), 2^(1/2)*EllipticF((1-1/(1+x__0))^(1/2)*2^(1/2), (1/2)*(1+x__0)^(1/2)*2^(1/2))))[2]; plot(difference, x__0 = 0 .. 1, title = 'difference', color = RED)

  

Try simplify

`assuming`([simplify(difference)], [0 < x__0 and x__0 < 1])

(EllipticK((1/2)*(2+2*x__0)^(1/2))-EllipticF(x__0^(1/2)*2^(1/2)/(1+x__0)^(1/2), (1/2)*(1+x__0)^(1/2)*2^(1/2))-EllipticF(1/(1+x__0)^(1/2), (1/2)*(1+x__0)^(1/2)*2^(1/2)))*2^(1/2)

 (2)

Try combine

`assuming`([combine(difference)], [0 < x__0 and x__0 < 1])

-2^(1/2)*EllipticF((x__0/(1+x__0))^(1/2)*2^(1/2), (1/2)*(2+2*x__0)^(1/2))-2^(1/2)*EllipticF(1/(1+x__0)^(1/2), (1/2)*(2+2*x__0)^(1/2))+2^(1/2)*EllipticK((1/2)*(2+2*x__0)^(1/2))

 (3)

Try conversion to integral form

convert(difference, Int)

2^(1/2)*(Int(2/((-_alpha1^2+1)^(1/2)*(-2*_alpha1^2*x__0-2*_alpha1^2+4)^(1/2)), _alpha1 = 0 .. 1))-2^(1/2)*(Int(2/((-_alpha1^2+1)^(1/2)*(-2*_alpha1^2*x__0-2*_alpha1^2+4)^(1/2)), _alpha1 = 0 .. 1/(1+x__0)^(1/2)))-2^(1/2)*(Int(2/((-_alpha1^2+1)^(1/2)*(-2*_alpha1^2*x__0-2*_alpha1^2+4)^(1/2)), _alpha1 = 0 .. (1-1/(1+x__0))^(1/2)*2^(1/2)))

 (4)

`assuming`([simplify(%)], [0 < x__0 and x__0 < 1])

-2*2^(1/2)*(Int(1/((-_alpha1^2+1)^(1/2)*(4+(-2*x__0-2)*_alpha1^2)^(1/2)), _alpha1 = 0 .. 1/(1+x__0)^(1/2))-(Int(1/((-_alpha1^2+1)^(1/2)*(4+(-2*x__0-2)*_alpha1^2)^(1/2)), _alpha1 = x__0^(1/2)*2^(1/2)/(1+x__0)^(1/2) .. 1)))

 (5)

plot(-2*2^(1/2)*(Int(1/((-_alpha1^2+1)^(1/2)*(4+(-2*x__0-2)*_alpha1^2)^(1/2)), _alpha1 = 0 .. 1/(1+x__0)^(1/2))-(Int(1/((-_alpha1^2+1)^(1/2)*(4+(-2*x__0-2)*_alpha1^2)^(1/2)), _alpha1 = x__0^(1/2)*2^(1/2)/(1+x__0)^(1/2) .. 1))), x__0 = .999 .. 1, color = RED)

  

evalf(eval(-2*2^(1/2)*(Int(1/((-_alpha1^2+1)^(1/2)*(4+(-2*x__0-2)*_alpha1^2)^(1/2)), _alpha1 = 0 .. 1/(1+x__0)^(1/2))-(Int(1/((-_alpha1^2+1)^(1/2)*(4+(-2*x__0-2)*_alpha1^2)^(1/2)), _alpha1 = x__0^(1/2)*2^(1/2)/(1+x__0)^(1/2) .. 1))), x__0 = 1), 40); evalf(eval(-2*2^(1/2)*(Int(1/((-_alpha1^2+1)^(1/2)*(4+(-2*x__0-2)*_alpha1^2)^(1/2)), _alpha1 = 0 .. 1/(1+x__0)^(1/2))-(Int(1/((-_alpha1^2+1)^(1/2)*(4+(-2*x__0-2)*_alpha1^2)^(1/2)), _alpha1 = x__0^(1/2)*2^(1/2)/(1+x__0)^(1/2) .. 1))), x__0 = .9999999999), 40)

-0.8716953127866940896552122384831392143938e-30

 (6)

There seems to be a finite difference at x__0 = 1

``

Try an addition theorem from DLMF

Since all elliptic expression have the same modulus k, the following addition theorem could be applied under the condition that the corresponding case is fulfilled https://dlmf.nist.gov/19.11#E7

Check if the following case applies (i.e. is ψ=π/2?)

 

NULL

NULL

queryequal(Int(1/((x__0-x)^(1/2)*(-x^2+1)^(1/2)), x = 0 .. x__0) = 2^(1/2)*EllipticK((1/2)*(2+2*x__0)^(1/2))-2^(1/2)*EllipticF(1/(1+x__0)^(1/2), (1/2)*(2+2*x__0)^(1/2)), 2^(1/2)*EllipticF((1-1/(1+x__0))^(1/2)*2^(1/2), (1/2)*(1+x__0)^(1/2)*2^(1/2)))

queryequal(Int(1/((x__0-x)^(1/2)*(-x^2+1)^(1/2)), x = 0 .. x__0) = 2^(1/2)*EllipticK((1/2)*(2+2*x__0)^(1/2))-2^(1/2)*EllipticF(1/(1+x__0)^(1/2), (1/2)*(2+2*x__0)^(1/2)), 2^(1/2)*EllipticF((1-1/(1+x__0))^(1/2)*2^(1/2), (1/2)*(1+x__0)^(1/2)*2^(1/2)))

 (7)

theta = 1/sqrt(1+x__0), `&varphi;` = sqrt(1-1/(1+x__0))*sqrt(2)

theta = 1/(x__0+1)^(1/2), varphi = (1-1/(x__0+1))^(1/2)*2^(1/2)

 (8)

k = (1/2)*sqrt(1+x__0)*sqrt(2)

k = (1/2)*(x__0+1)^(1/2)*2^(1/2)

 (9)

tan((1/2)*psi) = (sin(theta)*Delta(`&varphi;`)+sin(`&varphi;`)*Delta(theta))/(cos(theta)+cos(`&varphi;`))

tan((1/2)*psi) = (sin(theta)*Delta(varphi)+sin(varphi)*Delta(theta))/(cos(theta)+cos(varphi))

 (10)

Delta(theta) = sqrt(1-k^2*sin(theta)^2)

Delta(theta) = (1-k^2*sin(theta)^2)^(1/2)

 (11)

subs(theta = `&varphi;`, Delta(theta) = (1-k^2*sin(theta)^2)^(1/2))

Delta(varphi) = (1-k^2*sin(varphi)^2)^(1/2)

 (12)

subs(Delta(theta) = (1-k^2*sin(theta)^2)^(1/2), Delta(varphi) = (1-k^2*sin(varphi)^2)^(1/2), k = (1/2)*(1+x__0)^(1/2)*2^(1/2), theta = 1/(1+x__0)^(1/2), varphi = (1-1/(1+x__0))^(1/2)*2^(1/2), tan((1/2)*psi) = (sin(theta)*Delta(varphi)+sin(varphi)*Delta(theta))/(cos(theta)+cos(varphi)))

tan((1/2)*psi) = (sin(1/(x__0+1)^(1/2))*(1-(1/2)*(x__0+1)*sin((1-1/(x__0+1))^(1/2)*2^(1/2))^2)^(1/2)+sin((1-1/(x__0+1))^(1/2)*2^(1/2))*(1-(1/2)*(x__0+1)*sin(1/(x__0+1)^(1/2))^2)^(1/2))/(cos(1/(x__0+1)^(1/2))+cos((1-1/(x__0+1))^(1/2)*2^(1/2)))

 (13)

`assuming`([simplify(subs(psi = (1/2)*Pi, tan((1/2)*psi) = (sin(1/(1+x__0)^(1/2))*(1-(1/2)*(1+x__0)*sin((1-1/(1+x__0))^(1/2)*2^(1/2))^2)^(1/2)+sin((1-1/(1+x__0))^(1/2)*2^(1/2))*(1-(1/2)*(1+x__0)*sin(1/(1+x__0)^(1/2))^2)^(1/2))/(cos(1/(1+x__0)^(1/2))+cos((1-1/(1+x__0))^(1/2)*2^(1/2)))))], [0 < x__0 and x__0 < 1])

1 = (sin(1/(x__0+1)^(1/2))*((2+2*x__0)*cos(x__0^(1/2)*2^(1/2)/(x__0+1)^(1/2))^2+2-2*x__0)^(1/2)+sin(x__0^(1/2)*2^(1/2)/(x__0+1)^(1/2))*((2+2*x__0)*cos(1/(x__0+1)^(1/2))^2+2-2*x__0)^(1/2))/(2*cos(1/(x__0+1)^(1/2))+2*cos(x__0^(1/2)*2^(1/2)/(x__0+1)^(1/2)))

 (14)

plot([lhs(1 = (sin(1/(1+x__0)^(1/2))*((2+2*x__0)*cos(x__0^(1/2)*2^(1/2)/(1+x__0)^(1/2))^2+2-2*x__0)^(1/2)+sin(x__0^(1/2)*2^(1/2)/(1+x__0)^(1/2))*((2+2*x__0)*cos(1/(1+x__0)^(1/2))^2+2-2*x__0)^(1/2))/(2*cos(1/(1+x__0)^(1/2))+2*cos(x__0^(1/2)*2^(1/2)/(1+x__0)^(1/2)))), rhs(1 = (sin(1/(1+x__0)^(1/2))*((2+2*x__0)*cos(x__0^(1/2)*2^(1/2)/(1+x__0)^(1/2))^2+2-2*x__0)^(1/2)+sin(x__0^(1/2)*2^(1/2)/(1+x__0)^(1/2))*((2+2*x__0)*cos(1/(1+x__0)^(1/2))^2+2-2*x__0)^(1/2))/(2*cos(1/(1+x__0)^(1/2))+2*cos(x__0^(1/2)*2^(1/2)/(1+x__0)^(1/2))))], x__0 = 0 .. 1, legend = [lhs, rhs])

  

lhs and rhs are not equal -> Case psi = (1/2)*Pi does not apply (provided that the formulas have been applied correctly).
Perhaps this is why Maple does not simply the difference to zero.  NULL


Anything else that could be tried in Maple (maybe with other formulas)?

Download Equality_of_Elliptic_expressions.mw
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