If I have a vector like u:=<-2+m,3+m> and then I put m:=5 (for exemple) when I repeat u it doesnot replace the value of m by 5,

instead repeats the initial expression. I would like to find a way to achieve this in order to automate calculations.

This website suggests that the viewer write equations containing differentials and use them to plot topographical lines.

https://mathcurve.com/courbes3d.gb/topographic/topographic.shtml

I am basically familiar with equations containing derivatives, but not with equations containing differentials.

How can one employ these to draw the lines shown in the website's various plots of surfaces?

S := proc (n) local i; for i to n do if i <= n and type(i, divisors(n)) and type(n/i, odd) then sum(i) end if end do end proc

Hello!
I solved a simple PDE numerically and found that I couldn't implement a simple thing!
For example:
restart:
PDE := diff(u(x,t),t)=1/10*diff(u(x,t),x,x);
IBC := {u(x,0)=1, u(0,t)=0, D[1](u)(1,t)=0};
pds := pdsolve(PDE,IBC,numeric);
p1 := pds:-plot(t=0): p2 := pds:-plot(t=1/10):p3 := pds:-plot(t=1/2):p4 := pds:-plot(t=1):
plots[display]({p1,p2,p3,p4});
I got some spatial dependencies for different fixed t. But I want to get opposite! How to get time dependence for fixed x?
Replacing t with x results in an error.
Update.
This equation is example, really I solve this:
PDE := diff(u(x,t),t)+v*(1+k*sin(2*Pi*f*t))*diff(u(x,t),x)=a*diff(u(x,t),x,x);
IBC := {u(x,0)=T__0,u(0,t)=sin(2*Pi*f*t),u(L,t)=T__0};
So sin makes this equation nonlinear, so I need numerical procedure.
 

Hello, 

when trying to solve equation using fsolve, it returns just what i wrote and didn't solve. Can anyone help? I tried all i know. File is attached below.

Thanks a lot.

II.01.mw

restart;
f:=(x,y)-> x*y^3-x^2/y^3;
eq:= subs([x=alpha*x,y=alpha^p*y],f(x,y));
lis:=[]:
for tmp in op(eq) do
    if has(tmp,alpha) then
       lis:=[op(lis),op(2,simplify(select(has,tmp,alpha)))];
    fi;
od;
lis;
PDEtools:-Solve(lis[1]=lis[2],p)

This method happend to work for this example, but it does not work in general. This is because the function can be anything. Here is a second example

restart;
f:=(x,y)-> -2/3*x+1/3*(x^2+3*y)^(1/2);
eq:= subs([x=alpha*x,y=alpha^p*y],f(x,y));

We see by inspection if p=2 then we can write the above as, by setting p=2 as 

But the Maple code I had above does not work on this

restart;
f:=(x,y)-> -2/3*x+1/3*(x^2+3*y)^(1/2);
eq:= subs([x=alpha*x,y=alpha^p*y],f(x,y));
lis:=[]:
for tmp in op(eq) do
    if has(tmp,alpha) then
       lis:=[op(lis),simplify(select(has,tmp,alpha))];
    fi;
od;
lis;

The problem is finding all powers of alpha in each term and setting up an equation to find p such that all terms have same numerical value.

So I abandoned this method as too messy to program (it works well for hand solution, this is an example where solving something by hand is easier than on the computer) and then tried solve directly, like this (on the first example)

restart;
f:=(x,y)-> x*y^3-x^2/y^3;
eq:= subs([x=alpha*x,y=alpha^p*y],f(x,y))- alpha^r*f(x,y)=0;

And here is where I am stuck. I need to ask Maple if it can find alpha, r that are rational numbers, such that the above equation is solved.

PDEtools:-Solve(eq,[p,r])

But we know from above that p=1/6 and r=3/2 is solution

simplify(subs([p=1/6,r=3/2],eq))

             0=0

The question is: Can Maple solve such an equation for  p,r? I remember now reading something about parameteric solver in Maple. I need to look that up to see if it helps. I tried SolveTools:-SemiAlgebraic on this, but it did not help.

Are there better methods to determine in Maple if function is isobaric and to find the index p and r?

Reference: book

Notice, some places define isobaric function as one in which we can find p such that

But this is not correct. I've seen it on 2 pages on the net. The correct definition is 

Also note: When p=1 isobaric function becomes a homogeneous function which is special case of isobaric and r is now called the degree of homogeneity 

But for isobaric, p do not have to be 1. This is the main difference.

f(x,y) will always be a function of x,y.  I will add more examples of isobaric functions to test against.

edit. Simplified question is restated below

After looking more into this. I found I actually wanted to solve the simpler problem, which is

So please ignore the general case of isobaric where r<>p-1 which is harder. 

I only need to look at this case where r=p-1 which is actually simpler. It turned out this is what I need for my solver and not the general case.

I will now give my solution to this and show where the problem I am having.

I am using Maple solve to find p. So no r any more. Only solving for one variable. But Maple solve fails sometimes. And this is the problem.

The problem is that solve can find p for many cases, but fails on some, where there is clearly a solution. I will show 5 examples below.

#example 1. WORKS
restart;
f:=(x,y)-> 3*sqrt(x*y);
eq:= subs([x=alpha*x,y=alpha^p*y],f(x,y))- alpha^(p-1)*f(x,y)=0;
PDEtools:-Solve(eq,p) assuming alpha>0;

#example 2. WORKS
restart;
f:=(x,y)-> 4*(x*y)^(1/3);
eq:= subs([x=alpha*x,y=alpha^p*y],f(x,y))- alpha^(p-1)*f(x,y)=0;
PDEtools:-Solve(eq,p) assuming alpha>0;

#example 3. WORKS
restart;
f:=(x,y)-> (-3*x^2*y-y^2)/(2*x^3+3*x*y);
eq:= subs([x=alpha*x,y=alpha^p*y],f(x,y))- alpha^(p-1)*f(x,y)=0;
PDEtools:-Solve(eq,p) assuming alpha>0;

#example 4 WORKS
restart;
f:=(x,y)-> (-(x*y)/2+sqrt(x^2*y^2-4*y)/2)*y;
eq:= subs([x=alpha*x,y=alpha^p*y],f(x,y))- alpha^(p-1)*f(x,y)=0;
PDEtools:-Solve(eq,p) assuming alpha>0;

And here is the one that does not work

#Example 5. Does not work. How to make it work?
restart;
f:=(x,y)-> -2*x/3+sqrt(x^2+3*y)/3;
eq:= subs([x=alpha*x,y=alpha^p*y],f(x,y))- alpha^(p-1)*f(x,y)=0;
PDEtools:-Solve(eq,p) assuming alpha>0;

There is a solution for this. p=2 works. Why solve did not find it? I mean, it gives RootOf, which does not help me decide if this is isobaric or not. Why it did not find the solution p=2?

simplify(subs(p=2,eq)) assuming alpha>0

                 0=0

Using Mathemtica, it found p=2 solution. How to make Maple find this solution is my question.

Sorry for the long question. I first thought I needed the general isobaric case, but found later i needed the simpler one.

Plot the wedge cut from the cylinder x²+y²=1 by the planes z=-y and z=0.

Plot the solid that lies between the cylinder z= y^2 and the xy -plane that is bounded by the planes x = 0, x =1, y = −1, y =1.

Hi, 
I want to solve a simple newsvendor problem for general Demand Probability Distribution with maple code.
let, 
Underage cost 
                              c[u]

Overage cost 
                              c[o]

Demand (
D;
) follows pdf 
f(.);
 and CDF 
F;

(.);

Decision varible is quantity 
q;

So cost function is 

TC := piecewise(q-D >= 0, c[o]*(q-D), q-D < 0, c[u]*(-q+D));
 piecewise(0 <= q - D, c[o] (q - D), q - D < 0, c[u] (-q + D))

After taking expectation of TC and derivating it w.r.t q we will get 
q;
* as critical fractile i.e.

q*= F^(-1) (
                             c[u]    
                          -----------
                          c[u] + c[o]
)      i.e. F inverse (
c[u]/(c[u]+c[o]);
)

Can someone please help me how to find q* in MAPLE using solve() or any other function?

Thanks in Advance!

This might be something basic. But looking at help I do not see it now.

why has(expr,sqrt) do not give true?  is something special about sqrt function?

restart;
mysol :=sin(x);
has(mysol,sin); #works

mysol :=ln(x);
has(mysol,ln);  #works

mysol :=sqrt(x);
has(mysol,sqrt);  #does not work
hasfun(mysol,sqrt);  #does not work

Just wondering why, that is all. Only this worked

hastype(mysol,sqrt)

                true

I am sure there is a good reason for this. sqrt seems to be special function.

Maple 2020.2

Hello,

I'm trying to solve the following ode system:

odes := diff(a0(t), t) - diff(a1(t), t) + diff(a2(t), t) = 0, diff(a0(t), t) - diff(a2(t), t) - 16*a2(t) = 0, diff(b0(t), t) + diff(b1(t), t) + diff(b2(t), t) = 0, diff(b0(t), t) - diff(b2(t), t) - 48*b2(t) = 0, diff(a1(t), t) + 4*diff(a2(t), t) - 3*diff(b1(t), t) + 12*diff(b2(t), t) = 0, diff(a0(t), t) + diff(a1(t), t) + diff(a2(t), t) - diff(b0(t), t) + diff(b1(t), t) - diff(b2(t), t) = 0;
   
ics := a0(0) - a2(0) = 0, b0(0) - b2(0) = 1, a0(0) - a1(0) + a2(0) = 0, b0(0) + b1(0) + b2(0) = 1, a1(0) + 4*a2(0) - 3*b1(0) + 12*b2(0) = 0, a0(0) + a1(0) + a2(0) - b0(0) + b1(0) - b2(0) = 0;

dsdef := dsolve({ics, odes}, numeric);

There are 6 ordinary differential equations with 6 initial conditions. The functions are named a0(t), a1(t), a2(t), b0(t), b1(t), b2(t).

The question is that Maple solve the system analliticaly (exact dsolve) but when I run the "numerical dsolve" it returns: "Error, (in DEtools/convertsys) invalid specification of initial conditions", and I can't recognize the error/s.

Thanks.

How to show that the function z0(t)=t²+exp(-t)  is solution of z"(t)+2z'(t)+z(t)=2exp(-t); Thank you.

I tried to draw a normal boxplot for the a simple data using Maple. My attempt is in below.

X := [13, 15, 16, 16, 19, 20, 21, 21, 22, 22, 25, 25, 25, 25, 30, 33, 33, 35, 35, 35, 35, 36, 40, 45, 46, 52, 70]:
P1 := Statistics[BoxPlot](X, offset = 1, distance = 1, width = 1, orientation = horizontal, outliers = true);
plots[display](P1, view = [1 .. 80, 0.5 .. 2.5]);

But I am not happy from the result.

1. It seems the box-plot range doesn't include the max, 70, instead it puts a circle there!

      2. What are the small rectangles and circles? I just need a simple box-plot with the line-range starting from min and ending at max, then a box starting from the first quartile and ending at the third quartile and a vertical line at the median. No extra feature and not ignoring any data.

      3. I don't want the vertical axis which is for the index of the data, I just have one data and that vertical axis is just distracting the audience.

How to resolve these three issues?

all in real domain, sqrt(A)*sqrt(B) can be combined to sqrt(A*B) when both A,B are non-negative.

Is there a way in Maple to  find the conditions when sqrt(A)*sqrt(B)=sqrt(A*B) ? i.e. the conditions on A,B where this is true?

A,B will only be functions of x,y.   

An example will make things clear.

restart;
expr1:=sqrt(x^2*y - 4)*sqrt(x^2*y);
expr2:=sqrt(  (x^2*y - 4)*(x^2*y));

By looking at the above, we see that expr1 = expr2 when  x^2*y-4>=0 and x>0 or x^2*y-4>=0 and x<0. Actually I think only x^2*y-4>=0 is needed, since x is being squared anyway.

How to make Maple show this? I can't get Maple to show this

solve(expr1=expr2,[x,y]) assuming real;

But this is wrong. it says it is true for all x and all y?.   Mathematica can do it using Reduce command

I know I can force the combination by using the command

combine(expr1,sqrt,symbolic);

ps. Maple took the x outside the sqrt. So x>0 is assumed here.

pps. I do not understand why simplify(expr1,symbolic) did not work here, and neither  simplify(expr1,symbolic,size=false) worked. Only combine worked.

But I wanted to see if Maple could tell the condition when this is allowed, so I can write these down.

It would be nice if the command above would also tell the conditions under which it combined the sqrts. But this information is not given.

This is all done non-interactive in a program without being able to look at the screen and decide what to do. Only thing I know is that if an expression has sqrts and functions of x,y.

Is there a way in Maple to have tell conditions when expr1=expr2?

Maple 2020.2

Hi, 
The result below surprises me.
Why does  rgf_findrecur return a result instead of saying that there is no homogeneous linear recurrence of order 1?

interface(version)

Standard Worksheet Interface, Maple 2015.2, Mac OS X, December 21 2015 Build ID 1097895


genfunc:-rgf_findrecur(1, [1, 2, 3, 4], t, n);
                       t(n) = 2 t(n - 1)

PS: maybe this was a problem in Maple 2015, if it has been corrected since just let me know