I wanted to change  eq:= 1/2 * sqrt(-2*lambda)  to 1/2 %* sqrt(-2*lambda)  using a rule.

It works outside of rule ofcourse. But when I put %* in the RHS of the rule, maple hangs. It seems it is going into infinite loop.

I tried the trick of using '%*' but this gives syntax error.

Same problem happens when using %. and not just %*

Is there a workaround?

Attached worksheet. Make sure to save all work before trying it.
 

Download maples_hangs_applyrule_sept_10_2024.mw

This is an excersise in one of Mathematica's tutorials. The problem is to find from all persumtations of [1,2,3,4] those lists which has 2 in them before 3.

But the idea is that 2 can be anywhere before the 3, with possibly zero or more numbers between.

Will show how to do this in that other software, and ask if there is a way using Maple pattern matching or better way to do this in Maple better than what I did.

L=Permutations[{1,2,3,4}]
Cases[L,{___,2,___,3,___}]

gives

{{1, 2, 3, 4}, {1, 2, 4, 3}, {1, 4, 2, 3}, {2, 1, 3, 4}, 
{2, 1, 4,  3}, {2, 3, 1, 4}, {2, 3, 4, 1}, {2, 4, 1, 3}, 
{2, 4, 3, 1}, {4, 1,  2, 3}, {4, 2, 1, 3}, {4, 2, 3, 1}}

We see in all of the above, 2 is before 3.

In that other software, the ___ means there is zero or more things.  I could not find how to do this in Maple's pattern matching. So had to use has and then find the index of 2 and 3 in each list and check if the index of 2 is smaller than the index of 3. Which is kinda awakard and not as elegent as using a pattern, but it gives same result.

L:=combinat:-permute([1,2,3,4]);
map(X->`if`(has(X,2) and has(X,3) and ListTools:-Search(2,X)<ListTools:-Search(3,X),X,NULL) ,L);

Gives

[[1, 2, 3, 4], [1, 2, 4, 3], [1, 4, 2, 3], [2, 1, 3, 4], 
[2, 1, 4, 3], [2, 3, 1, 4], [2, 3, 4, 1], [2, 4, 1, 3], 
[2, 4, 3, 1], [4, 1, 2, 3], [4, 2, 1, 3], [4, 2, 3, 1]]

Can you suggest a way using either patmatch or applyrules in Maple to do the same?  

Could this be done easier than what I did using evalindents without the need for pattern?

I find patterns easier to work with myself.

ps. converting each list to string, and then using regular expression or string matching, is not what I am looking for. 

ps. the check in Maple code of has(X,2) and has(X,3) is not really needed here, since we know each list will have 2 and 3 in them. But I kept these to make it more general for other type of problems where these extra checks might be needed.

applyrule is useful but seems limited. I can't do operation such as expand in the RHS of the rule.

For example, I wanted to applyrule that says to take sin(x::anything) and change it to cos(expand(x)), but it does not expand x. 

It seems because in the RHS of the rule, at the instance applyrule sees expand, x remains a symbol and not evaluated. So there is nothing to expand. It gets evaluated at later time, but by then too late for expand to do anything, it is gone. Just a guess.

But is there a trick to allow one to do more things in RHS of applyrule, such as expand or simplify? This would make applyrule much more useful. Otherwise, as it is, applyrule is of limited use. 

I know I can use evalindets ofcourse for this. 

Download apply_rule.mw

Using other software, there is no such problem using other operations on RHS of a rule

You see, expand worked on RHS of rule. 

I'd like to do same thing using applyrule in Maple. Is there a trick to allow this?

Maple 2024.1

I was trying to odetest one of my solutions to this ode when I got this internal Maple error I have not seen before.

Is this a legitimate error? my solution could be wrong ofcourse but strange to get internal error in this case.

Does this happen on earlier versions of Maple?

Download internal_error_PDEtools_NumerDenom_sept_4_2024.mw

Here is another ode solution by Maple, I am not able to get zero from odetest. 

If someone could come up with a trick or method to verify this, that will be great. I tried all sort of assumptions and not able to get zero. Even coulditbe() could not give zero. 

It is a Chini ode and also homogeneous, `class G`.

Solving using either method, odetest do not returns zero.  There are no initial conditions. I am sure the solutions are correct, but odetest need some help or may be this requires different method to verify not using odetest?

Worksheet below.

Download challenge_odetest_sept_1_2024.mw