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Question: Questions on applying identities from the FunctionAdvisor to expressions

Question:Questions on applying identities from the FunctionAdvisor to expressions

C_R 3317 Maple 2023
substitution identity functionadvisor
September 02 2023
2

I make more and more use of the FunctionAdvisor. I have started to apply rules from the advisor to expressions. Here are two examples with questions:

NULL

Expression to apply an identiy to

JacobiSN(sin((1/2)*`ϕ__0`)*t, csc((1/2)*`ϕ__0`)) = JacobiSN(z, k)

JacobiSN(sin((1/2)*varphi__0)*t, csc((1/2)*varphi__0)) = JacobiSN(z, k)

 (1)

map(op, JacobiSN(sin((1/2)*varphi__0)*t, csc((1/2)*varphi__0)) = JacobiSN(z, k))

(sin((1/2)*varphi__0)*t, csc((1/2)*varphi__0)) = (z, k)

 (2)

solve([(rhs-lhs)((sin((1/2)*varphi__0)*t, csc((1/2)*varphi__0)) = (z, k))], {k, z})[]

k = csc((1/2)*varphi__0), z = sin((1/2)*varphi__0)*t

 (3)

Using the following identity from Maples FunctionAdvisor and the correspondence in (3)

FunctionAdvisor(identities, JacobiSN(z, 1/k))[5]

JacobiSN(z, k) = JacobiSN(z*k, 1/k)/k

 (4)

convert(subs(k = csc((1/2)*varphi__0), z = sin((1/2)*varphi__0)*t, JacobiSN(z, k) = JacobiSN(z*k, 1/k)/k), sincos)

JacobiSN(sin((1/2)*varphi__0)*t, 1/sin((1/2)*varphi__0)) = sin((1/2)*varphi__0)*JacobiSN(t, sin((1/2)*varphi__0))

 (5)

That worked. Q1: But is it a good way to do so?

Now  a new example: Converting InverseJacobinAM to InverseJacobiSN

NULL

NULL

FunctionAdvisor(identities, InverseJacobiSN(z, k))[3]

InverseJacobiSN(z, k) = InverseJacobiAM(arcsin(z), k)

 (6)

InverseJacobiAM((1/2)*`ϕ__0`, sqrt(2)/sqrt(1-cos(`ϕ__0`))) = rhs(InverseJacobiSN(z, k) = InverseJacobiAM(arcsin(z), k))

InverseJacobiAM((1/2)*varphi__0, 2^(1/2)/(1-cos(varphi__0))^(1/2)) = EllipticF(z, k)

 (7)

map(op, InverseJacobiAM((1/2)*varphi__0, 2^(1/2)/(1-cos(varphi__0))^(1/2)) = EllipticF(z, k))

((1/2)*varphi__0, 2^(1/2)/(1-cos(varphi__0))^(1/2)) = (z, k)

 (8)

solve({(rhs-lhs)(((1/2)*varphi__0, 2^(1/2)/(1-cos(varphi__0))^(1/2)) = (z, k))}, {k, z})

{k = 2^(1/2)/(1-cos(varphi__0))^(1/2), z = (1/2)*varphi__0}

 (9)

This is of course wrong since comparing the InverseJacobiAM expression in (6) and (7) z should be

(1/2)*`ϕ__0` = arcsin(z)

(1/2)*varphi__0 = arcsin(z)

 (10)

solve((1/2)*varphi__0 = arcsin(z), {z})

{z = sin((1/2)*varphi__0)}

 (11)

Q2: How to avoid simplification of InverseJacobiAM(arcsin(z), k)to EllipticF(z, k)


Any advice?

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