C_R

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These are questions asked by C_R

Is there an easier or shorter way to do the following? Any handy package command?

RootOf(-(Int(1/sqrt(2*cos(_a)+RootOf(-(Int(1/sqrt(2*cos(_a)+_Z), _a = 0 .. Pi))+_a+c__2)), _a = 0 .. _Z))+x+c__2)

RootOf(-(Int(1/(2*cos(_a)+RootOf(-(Int(1/(2*cos(_a)+_Z)^(1/2), _a = 0 .. Pi))+_a+c__2))^(1/2), _a = 0 .. _Z))+x+c__2)

(1)

indets(RootOf(-(Int(1/(2*cos(_a)+RootOf(-(Int(1/(2*cos(_a)+_Z)^(1/2), _a = 0 .. Pi))+_a+c__2))^(1/2), _a = 0 .. _Z))+x+c__2), 'specfunc(anything, RootOf)')

{RootOf(-(Int(1/(2*cos(_a)+_Z)^(1/2), _a = 0 .. Pi))+_a+c__2), RootOf(-(Int(1/(2*cos(_a)+RootOf(-(Int(1/(2*cos(_a)+_Z)^(1/2), _a = 0 .. Pi))+_a+c__2))^(1/2), _a = 0 .. _Z))+x+c__2)}

(2)

subs(_a = _b, {RootOf(-(Int(1/(2*cos(_a)+_Z)^(1/2), _a = 0 .. Pi))+_a+c__2), RootOf(-(Int(1/(2*cos(_a)+RootOf(-(Int(1/(2*cos(_a)+_Z)^(1/2), _a = 0 .. Pi))+_a+c__2))^(1/2), _a = 0 .. _Z))+x+c__2)}[1])

RootOf(-(Int(1/(2*cos(_b)+_Z)^(1/2), _b = 0 .. Pi))+_b+c__2)

(3)

subs({RootOf(-(Int(1/(2*cos(_a)+_Z)^(1/2), _a = 0 .. Pi))+_a+c__2), RootOf(-(Int(1/(2*cos(_a)+RootOf(-(Int(1/(2*cos(_a)+_Z)^(1/2), _a = 0 .. Pi))+_a+c__2))^(1/2), _a = 0 .. _Z))+x+c__2)}[1] = RootOf(-(Int(1/(2*cos(_b)+_Z)^(1/2), _b = 0 .. Pi))+_b+c__2), RootOf(-(Int(1/(2*cos(_a)+RootOf(-(Int(1/(2*cos(_a)+_Z)^(1/2), _a = 0 .. Pi))+_a+c__2))^(1/2), _a = 0 .. _Z))+x+c__2))

RootOf(-(Int(1/(2*cos(_a)+RootOf(-(Int(1/(2*cos(_b)+_Z)^(1/2), _b = 0 .. Pi))+_b+c__2))^(1/2), _a = 0 .. _Z))+x+c__2)

(4)

NULL

Related question: Is that substitution mathematically correct or does the _a belong to the outermost RootOf?

Download parameters_in_nested_RootOf.mw

To find the recently introduced new command that compares two expressions, I prompted the large model: 

Answer:

A useful Maple command specifically for comparing two expressions is verify.
Use it like this:
verify(expr1, expr2, equal);
This asks Maple whether the two expressions are mathematically equal.
If you want a more robust algebraic check, a very common alternative is:
is(simplify(expr1 - expr2) = 0);
For polynomial expressions, this is also common:
evalb(expand(expr1 - expr2) = 0);

This is not what I was looking for. I was looking for the command that visually highlights differences.
Can the prompt be improved?

I am asking because I could neither find a type nor a property for nonzero integers. However, with the assume facility it is possible to combine properties 

assume(x<>0,x::integer)
about(x)
Originally x, renamed x~:
  is assumed to be: AndProp(integer,OrProp(RealRange(-infinity,Open(0)),RealRange(Open(0),infinity)))

Or:

restart;
assume(x::Non(0),x::integer)
about(x)
Originally x, renamed x~:
  is assumed to be: AndProp(integer,OrProp(RealRange(-infinity,Open(0)),RealRange(Open(0),infinity)))

Or:

assume(x::NonZero,x::integer)
about(x)
Originally x, renamed x~:
  is assumed to be: AndProp(integer,OrProp(RealRange(-infinity,Open(0)),RealRange(Open(0),infinity)))

Non(0) is also accepted by the type command which raises the question whether Non(0) is a property, a type or both.

On the other hand type does not recognise NonZero but "::NonZero" is accepted in the assume command (note the :: which is the operator for type definitions). 

The function Non, which is also listed under types, is also accepted in the context of properties. This is somewhat confusing since types and properties are different in nature. Can somone clarify?

By the way, the above examples also work when Non is replaced with Not because they are equivalent. However showstat(Non) works but showstat(Not) throws an error.

MAple 2026.0

Maple 2025.2 craches on SupportTools.-Update()

Can anyone verify? 

Solutions?

FYI: The user interface issues I was having with Maple 2026  have now been resolved.

Documents imported with the AI-assistant can containt private and confidential information.

I was wondering how privacy is handeled by third party AI services that are called by the AI assistant.

From https://openai.com/enterprise-privacy/ it is not clear which product is running and how Maplesoft has set it up.

Anyone knows more?

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