C_R

3412 Reputation

21 Badges

5 years, 316 days

MaplePrimes Activity


These are questions asked by C_R

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?

Download Applying_identities_from_FunctionAdivisor.mw

Q1: Why does Maple use the bar as a delimiter for certain elliptic expressions and the comma for others?

Is that in line with: Gradshteyn and Ryzhik (G&R) and in the popular "Handbook of Mathematical Functions" edited by Abramowitz and Stegun (A&S), as stated in help(JacobiAM)?      

Q2: Can I have the comma instead of the vertical bar for the Jacobian functions?

Q3: If not, how to get a bit more space between the symbols and the bar for better readability?

 

interface(version)

`Standard Worksheet Interface, Maple 2023.0, Windows 10, March 6 2023 Build ID 1689885`

(1)

FunctionAdvisor(relate, EllipticF, InverseJacobiSN)

EllipticF(z, k) = InverseJacobiSN(z, k)

(2)

Typesetting:-EnableTypesetRule(Typesetting:-SpecialFunctionRules)

{}

(3)

EllipticF(z, k) = InverseJacobiSN(z, k)

EllipticF(z, k) = InverseJacobiSN(z, k)

(4)

NULL


 

Download Argument_delimiter.mw

restart; interface(version)

`Standard Worksheet Interface, Maple 2023.0, Windows 10, March 6 2023 Build ID 1689885`

(1)

y(L) = (EllipticK((1/2)*sqrt(2*sin(`ϕ__0`)+2))-EllipticF(1/sqrt(sin(`ϕ__0`)+1), (1/2)*sqrt(2)*sqrt(sin(`ϕ__0`)+1))-2*EllipticE((1/2)*sqrt(2)*sqrt(sin(`ϕ__0`)+1))+2*EllipticE(1/sqrt(sin(`ϕ__0`)+1), (1/2)*sqrt(2)*sqrt(sin(`ϕ__0`)+1)))/sqrt(C)

y(L) = (EllipticK((1/2)*(2*sin(varphi__0)+2)^(1/2))-EllipticF(1/(sin(varphi__0)+1)^(1/2), (1/2)*2^(1/2)*(sin(varphi__0)+1)^(1/2))-2*EllipticE((1/2)*2^(1/2)*(sin(varphi__0)+1)^(1/2))+2*EllipticE(1/(sin(varphi__0)+1)^(1/2), (1/2)*2^(1/2)*(sin(varphi__0)+1)^(1/2)))/C^(1/2)

(2)

Typesetting:-EnableTypesetRule({"EllipticE", "EllipticF", "EllipticK"})

{}

(3)

y(L) = (EllipticK((1/2)*(2*sin(varphi__0)+2)^(1/2))-EllipticF(1/(sin(varphi__0)+1)^(1/2), (1/2)*2^(1/2)*(sin(varphi__0)+1)^(1/2))-2*EllipticE((1/2)*2^(1/2)*(sin(varphi__0)+1)^(1/2))+2*EllipticE(1/(sin(varphi__0)+1)^(1/2), (1/2)*2^(1/2)*(sin(varphi__0)+1)^(1/2)))/C^(1/2)

y(L) = (EllipticK((1/2)*(2*sin(varphi__0)+2)^(1/2))-EllipticF(1/(sin(varphi__0)+1)^(1/2), (1/2)*2^(1/2)*(sin(varphi__0)+1)^(1/2))-2*EllipticE((1/2)*2^(1/2)*(sin(varphi__0)+1)^(1/2))+2*EllipticE(1/(sin(varphi__0)+1)^(1/2), (1/2)*2^(1/2)*(sin(varphi__0)+1)^(1/2)))/C^(1/2)

(4)

NULL

Download Typesetrule.mw

I want to partially execute a large worksheet using to a point where I want to proceed with step-by-step execution . I am looking for a command that brings execution to a halt.

Threads:-Sleep(big number) and works but is there anything else that I could use.

In the same context: Execute worksheet removes all output. I would like to keep the output in the worksheet. Is that possible?

 

So far I have only noticed them for special functions.

For the case of Airy in 2D: Is possible the get for the input Ai(x) the output Ai(x) and not AiryAi(x)?

First 25 26 27 28 29 30 31 Last Page 27 of 44