## how to find the roots of the equations ...

det_roots.docx

How to find the roots of the equation ?

like if the roots of the equations are m ,n,o,p then the value of these roots in terms of λ0  λ1  λ2  λ3  µ1  µ2 and  µ3

## Solve for x in determinants...

Calculate the values of the following determinants

## 2024 possible bug...

Windows 10 64bit

If I double click on a  file to open it from explorer  i.e. launch Maple, it is 50/50 whether Maple hangs on opening the file and I have to kill it in the Task Manager. This only happens the 1st time I try to open the file. Subsequent clicks on it will open it. If Maple is already open the problem does not happen.  The problem willl also be there after the PC is restarted. Has anyone else noticed this?

## Customize surface plot...

Dear Users!

I hope everyone here is fine. In the attached file I have a list of points in three dimensions. I want to plot surfaceplot (also in Dimension=2) of only those points which are less than 1.

But I want to plot the density plot for all points with the range on x-axes 5 to 15 and the range on y-axes 50-1000

Help.mw

## suggestion for simpler form of solution to this od...

May be someone can come up with a way to simplify this ode solution? I used the option useInt but the solution can be written in much simpler way than Maple gives.  Below is worksheet showing Maple's 2024 solution and my hand solution.

 > ode:=diff(y(x),x)^3=y(x)+x

 > maple_sol:=dsolve(ode,useInt): maple_sol:=Vector([maple_sol]);

 > mysol1:= Intat(1/(_a^(1/3) + 1), _a = (y(x) + x))=x+_C1: mysol2:= Intat(1/( -(-1)^(1/3)*_a^(1/3) + 1), _a = (y(x) + x))=x+_C1: mysol3:= Intat(1/( (-1)^(2/3)*_a^(1/3) + 1), _a = (y(x) + x))=x+_C1: mysol:=Vector([mysol1,mysol2,mysol3]);
 >

 > map(X->odetest(X,ode),mysol)
 >

 >

I keep losing the edits I do. I post screen shot. Click submit, then find all my changes are lost. Will try one more time and give up:

This is Maple solution

This is implified version

Both versions are verified correct by odetest. The question is there is a way to obtain the simpler form from Maple.

## Incomplete math in maple calculator...

When I calculate the edge values of a matrix the result is lengthy expression that could be simplified if evaluated numerically, Why is that not done?

## why dsolve does not give this simple solution to f...

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

 > restart;

 > interface(version);

 > Physics:-Version();

 > 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;

 > #why such complicated solutions? sol:=[dsolve(ode)];

 > #all solution are correct map(X->odetest(X,ode),sol);

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

 > dsolve(new_ode);

 > #force it to solve it as first order linear ode dsolve(ode,y(x),[`linear`])

## What is substituted here...

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?

## how to obtain conditions coulditbe used to obtain ...

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.

## What is the policy on re-asking a question? ...

I asked a question a while ago

https://www.mapleprimes.com/questions/235734-How-Can-The-Functional-Derivativevariation

Which did not get any immediate responses but one late one which i had replied to, but nothing else since then.

Is it appropriate to delete and repost and try to get more responses/exposure and see if anyone else has any ideas?

Since @ecterrab is the one who responded I consider this an interesting problem that I would like to continue discussing.

Hi,

I am trying to convert from MapleSim to Simulink Matlab by using the S-function code Generation connector, but I got the above message when I uploaded the selected subsystem and I have no idea how can I fix it. Please help me

## A numerical algorithm for SFC and its pseudo-inver...

Hi!

I am trying to implement the algorithms given in this paper (free for download) in Maple 2015

https://www.researchgate.net/publication/374636058_A_simple_algorithm_for_computing_a_multi-dimensional_the_Sierpinski_space-filling_curve_generalization

Such algorithms, apparently very easy, provides an approximation of  a sapce-filling curve and its pseudo-inverse. I am interesting in this space--filling curve for its properties. Please, find attached the Maple file, I am not sure if the code of the paper is not fine or I am doing something wrong.

Sierp_v1.mw

## How do I solve ODE system in RK method?...

How to solve and plot a ODE system in RK method.
eq1 := diff(f(x), x, x, x)-(1/2)*Sc*sin(alpha)*g(x)*(diff(g(x), x, x))+(1/2)*x*cos(alpha)*(diff(f(x), x, x))+(1/2)*sin(alpha)*f(x)*(diff(f(x), x, x)) = 0; eq2 := (diff(g(x), x, x, x))/Pm+(1/2)*x*cos(alpha)*(diff(g(x), x, x))+sin(alpha)*f(x)*(diff(g(x), x, x))-sin(alpha)*(diff(f(x), x, x))*g(x) = 0; eq3 := (diff(theta(x), x, x))/Pr+(1/2)*x*cos(alpha)*(diff(theta(x), x))+(1/2)*x*(diff(f(x), x))*(diff(theta(x), x))+sin(alpha)*(x*(diff(f(x), x))-f(x))*(diff(theta(x), x))-Nb*(diff(s(x), x))*(diff(theta(x), x))-Nt*(diff(theta(x), x))^2+(1/4)*Sc*Br*sin(alpha)^2*(diff(f(x), x))^2*(x*(diff(g(x), x))-g(x))+(diff(g(x), x))^2*(x*(diff(f(x), x))-f(x)) = 0; eq4 := diff(s(x), x, x)+S*((1/2)*cos(alpha)*x*(diff(s(x), x))+(1/2)*sin(alpha)*f(x)*(diff(s(x), x)))+Nt*(diff(theta(x), x, x))/Nb = 0

ics := f(0) = 0, (D(f))(0) = 1, g(0) = 0, (D(g))(0) = 1, theta(0) = 1, s(0) = 1; bcs := (D(f))(100) = 0, (D(g))(100) = 0, theta(100) = 0, s(100) = 0

alpha = - 30 degree, Sc = 1.0, Pm = .1, Pr = 6.2, Nb = .1, Nt = .1, Br = .5, S = 1