Hi

A padel group organises doubles tournaments with the following structure: 16 players participate in four rounds, with each round consisting of four doubles matches. Players switch partners after each round.

So what I require is some code to generate the first round matches, the second round, the third, the nth .....

note the worksheet I bastardised from Tom Leslie's work. 

2_man_teams_doubles.mw

it took me hrs to find this as my solution was failing verification and I did not know why.

What logic do you think Maple used to simplify this:

expr:=sqrt(1 + sin(x))/x;
simplify(expr)

To this

How could the above be simpler than

?

Compare to Mathematica

And this is what I expected. I am now scared to use simplify in Maple as I do not know what I will get back.

Is there a way to tell Maple not to do such strange "simplification"? I am doing this in code, and the code does not know what the expression is.

To see an example of the side effect of this, here is one, where if solution to an ode is simplified first, it no longer verifies by odetest without adding extra assumptions:

One does not expect that simplified solution no longer verfiies the ode.

Sure, I can do

          odetest(simplify(sol),ode) assuming real;

and now it gives 0. But the point is that the first one did not need assumptions.

Download simplify_with_odetest.mw

Maple 2024 on windows 10.

I am trying to tidy up cases where a proc returns multiple values. Have being trying Tabulate. I can get it to work when called after the results are returned. I would the procedure to do this but  keep acces to the name(s) assigned to the returned values.

A,B,C:= proc(...)  .....  return a, b, c    end proc. 

So basically display a tabulated of a, b, c.
 

Download 2024-03-18_Q_Format_Returned_Results_into_a_Table..mw

Hi,

I calculate the standard deviation using Maple, which differs from the standard deviation obtained by the calculator (TI). Can you provide an explanation for this difference?

Thanks

StDevQ.mw

I am trying to get Maple to simplify the following trigonometric expressions (for "generic" parameters) as much as possible: 

sineExpr(3) := (
   sin(a[2] - b[1])*sin(a[3] - b[1]))/(
   sin(b[2] - b[1])*sin(b[3] - b[1]))*sin(a[1] - b[1]) + (
   sin(a[3] - b[2])*sin(a[1] - b[2]))/(
   sin(b[3] - b[2])*sin(b[1] - b[2]))*sin(a[2] - b[2]) + (
   sin(a[1] - b[3])*sin(a[2] - b[3]))/(
   sin(b[1] - b[3])*sin(b[2] - b[3]))*sin(a[3] - b[3]);
 = 
   


sineExpr(4) := (
   sin(a[2] - b[1])*sin(a[3] - b[1])*sin(a[4] - b[1]))/(
   sin(b[2] - b[1])*sin(b[3] - b[1])*sin(b[4] - b[1]))*
   sin(a[1] - b[1]) + (
   sin(a[3] - b[2])*sin(a[4] - b[2])*sin(a[1] - b[2]))/(
   sin(b[3] - b[2])*sin(b[4] - b[2])*sin(b[1] - b[2]))*
   sin(a[2] - b[2]) + (
   sin(a[4] - b[3])*sin(a[1] - b[3])*sin(a[2] - b[3]))/(
   sin(b[4] - b[3])*sin(b[1] - b[3])*sin(b[2] - b[3]))*
   sin(a[3] - b[3]) + (
   sin(a[1] - b[4])*sin(a[2] - b[4])*sin(a[3] - b[4]))/(
   sin(b[1] - b[4])*sin(b[2] - b[4])*sin(b[3] - b[4]))*
   sin(a[4] - b[4]);
 = 

So far, all of my attempts have failed: 
 

Download sinIdentity.mw

Note that because zero testing is frequently considerably easier, combine always succeeds in showing that the difference between the simplest possible and the original version is zero. 

combine(sin((a[1]+a[2]+a[3])-(b[1]+b[2]+b[3]))-sineExpr(3));
 = 
                               0

combine(sin((a[1]+a[2]+a[3]+a[4])-(b[1]+b[2]+b[3]+b[4]))-sineExpr(4));
 = 
                               0

However, I wonder if Maple can thoroughly simplify them without knowing those known “simplest possible” form beforehand. 
I also tried some other functions like rationalize, radnormal, and `convert/trig`, yet Maple appears to have not been capable of completely simplifying even the sub-simplest case 𝑚＝2. Is there any workaround? 

Of note, it can be demonstrated inductively that ∀m∈ℕ_₊ 

where none of the denominators is 0. Nevertheless, as mentioned above, is it possible to transform  and  (as well as , if possible) into potentially more elegant forms (Ideally,  is rewritten into ,  is rewritten into , and  is rewritten into .) without any such a priori knowledge? 
In Mma, these may be done using TrigReduce directly (cf. ); unfortunately, I cannot found a Maple equivalent to such functionality. 

Good everyone,

I am solving a pde problem and I wanted to get the table values for u(0,0.1) but it's just returning the pds. Attach below is the maple worksheet for the code. 

Anyone with suggestions, please. 

Test.mw

 I am writing help pages for a package. The inital Overview should be at the top of the listing like in other Maple help directories. 

How to I do this? Mine is listing purely alphabetically.
 

Download Help_Edit_to_Database_2023.mw

Dear all 

I have an ideal, and code its definition  and I compute the height  but no result return. There is an error. 

height_ideal.mw

Thank you for you help

Hello

I am looking for an efficient code to divide a given integer n by another integer d as many times as possible.

For example:

For n=294912 and d=8 the answer shoud be 9, because 294912/8^5=9.

Thank you for your help.

THis ode looks complicated

ode := (2*x^(5/2) - 3*y(x)^(5/3))/(2*x^(5/2)*y(x)^(2/3)) + ((-2*x^(5/2) + 3*y(x)^(5/3))*diff(y(x), x))/(3*x^(3/2)*y(x)^(5/3)) = 0;

But is actually a simple first order linear ode:

RHS:=solve(ode,diff(y(x),x));
new_ode:=diff(y(x),x)=RHS;

Whose solution is 

But Maple gives this very complicated answer as shown below. When asking it to solve as linear ode, it now gives the much simpler solution.  

Maple complicated solutions are all verified OK. But the question is, why did it not give this simple solution?

Attached worksheet.  All on Maple 2024

Download why_missed_simple_solution_march_17_2024.mw

I am wondering why Maple simplifies (x^(1/3))^3 to x ,  but not (x^3)^(1/3) .
I even tried the surd function. I believe the surd function is for real number arguments, so it should simplify to x.

Download inverse1.mw

Please, Help me to fix this error..

Download Grafik_R0kurang1.mw

I cannot figure out which operand(?) is substituded here

subs(1 = 2, a*b);
                              2  2
                             a  b 

Same for

subs(1 = 3, a + b);
                           3 a + 3 b

but

subs(1 = 2, a/b);
                                2
                               a 
                               --
                               b 

subs(1 = 3, a - b);
                            3 a - b

Is this by design?

In Maple 2022

restart;

res := t^3 - 3*t^2*sqrt(t^2*(12*sqrt(2)*ln(t) + 9*ln(t)^2 + 8)^(1/3))*ln(t)/(12*sqrt(2)*ln(t) + 9*ln(t)^2 + 8)^(2/3) - 2*t^2*sqrt(t^2*(12*sqrt(2)*ln(t) + 9*ln(t)^2 + 8)^(1/3))*sqrt(2)/(12*sqrt(2)*ln(t) + 9*ln(t)^2 + 8)^(2/3);
plot(res,t=-5..1)

gives

The same exact code in Maple 2024 gives

Worksheet is below. 

Both on same PC. Windows 10.

Will report to Maplesoft, but thought to check also here is others have seen such problem before.

Btw, Maple 2022 plot is the correct one.

Download wrong_plot_V_2024_march_15_2024.mw

Maple's coulditbe  is useful. But unfortunately it does not return back to the user the conditions under which the proposition was found true. This could make it much more useful. It seems in way similar to Mathematica' Reduce but Reduce returns the conditions.

Is there a way to find the conditions which makes it true? 

I use coulditbe alot. I use it to verify that the result of odetest (I call it the residue) is zero or not. Maytimes, odetest does not return zero. And using simplify, or evalb or is to check if the residue is zero, all fail. But many times, coulditbe returns true, meaning the residue is zero. But I do not know under what conditions. In Mathematica's Reduce, it tells me the conditions. 

Here is one of hundreds of examples I have

restart;
ode:=(t^3+y(t)^2*sqrt(t^2+y(t)^2))-(t*y(t)*sqrt(t^2+y(t)^2))*diff(y(t),t)=0;
ic:=y(1)=1;
sol:=dsolve([ode,ic]);
the_residue:=odetest(sol,[ode,ic]);

You see, odetest says it could not verify the solution (the first entry above) but it did verify the solution against the initial conditions. 

Using simplify, evalb and is all also could not verify it

simplify(the_residue[1]);
evalb(the_residue[1]=0);
is(the_residue[1]=0);

Now coulditbe does:

_EnvTry:='hard':
coulditbe(the_residue[1]=0);

So the solution is correct, but I do not know under what conditions. Using Mathematica's Reduce I can find this:

So now back in Maple, I can do this

simplify(the_residue[1]) assuming t>exp(-2*sqrt(2)/3);

                      0

Actually in this example, just using assume t>0 also gives zero. But I am using Mathematica's result for illustration.

You might ask, why do I need to know for what values of the independent variable is the residue zero?

Because in some cases, the residue is zero only at single point! So it does not make sense to say the solution is verified to be correct only at one single point of the domain, right?

it needs to be some finite range at least. Here is an example of an ode whose solution is correct only at x=0

ode:=diff(y(x),x)=3*x*(y(x)-1)^(1/3);
ic:=y(3)=-7;
sol:=dsolve([ode,ic]);
the_residue:=odetest(sol,[ode,ic]);

And simplify, evalb, is all fail to verifiy this, but coulditbe says true

simplify(the_residue[1]);
evalb(the_residue[1]=0);
is(the_residue[1]=0);
_EnvTry:='hard':
coulditbe(the_residue[1]=0);

So now, we ask, is this solution then correct or not? It turns out to be zero but only at origin x=0

plot(abs(the_residue[1]),x=-1..1)

If I knew that residue is zero only at single point, then I would say this solution is not correct, right?

And that is why I need to know under what conditions coulditbe retruned true.

I tried infolevel[coulditbe]:=5 but nothing more was displayed on the screen.

Mathematica's Reduce confirms that when x=0 the residue is zero.

So my question is simply this: Can one obtain the conditions used by coulditbe to determine when result is true?

It will be useful if Maple could in future version return the value/range which makes it true.