i have a probability function f(x,y,z)=x^3*y*z and constraint on its ranges x<y , y<z. how can i plot it fot f(x,y,z)=0.5

I am using Maple 2017 and the following equations gives me in correct result when I run `maple m.mpl` in terminal, however, when I run in using the GUI, the result is correct. (one result is postive while one is negative)

res := solve({
T000000=1/(1*4.57*10^(-06)+1*2.07*10^(-06)+1*2.83*10^(-06)) + T100000*1*4.57*10^(-06)/(1*4.57*10^(-06)+1*2.07*10^(-06)+1*2.83*10^(-06)) + T010000*1*2.07*10^(-06)/(1*4.57*10^(-06)+1*2.07*10^(-06)+1*2.83*10^(-06)) + T000010*1*2.83*10^(-06)/(1*4.57*10^(-06)+1*2.07*10^(-06)+1*2.83*10^(-06)) ,
T000010=1/(1*4.57*10^(-06)+1*2.07*10^(-06)+4) + T100010*1*4.57*10^(-06)/(1*4.57*10^(-06)+1*2.07*10^(-06)+4) + T010010*1*2.07*10^(-06)/(1*4.57*10^(-06)+1*2.07*10^(-06)+4) + T000000*4/(1*4.57*10^(-06)+1*2.07*10^(-06)+4) ,
T010000=1/(1*4.57*10^(-06)+1*2.83*10^(-06)+4) + T110000*1*4.57*10^(-06)/(1*4.57*10^(-06)+1*2.83*10^(-06)+4) + T010010*1*2.83*10^(-06)/(1*4.57*10^(-06)+1*2.83*10^(-06)+4) + T000000*4/(1*4.57*10^(-06)+1*2.83*10^(-06)+4) ,
T010010=1/(1*4.57*10^(-06)+4) + T000010*4/2/(1*4.57*10^(-06)+4) + T010000*4/2/(1*4.57*10^(-06)+4) ,
T100000=1/(1*2.07*10^(-06)+1*2.83*10^(-06)+4) + T110000*1*2.07*10^(-06)/(1*2.07*10^(-06)+1*2.83*10^(-06)+4) + T100010*1*2.83*10^(-06)/(1*2.07*10^(-06)+1*2.83*10^(-06)+4) + T000000*4/(1*2.07*10^(-06)+1*2.83*10^(-06)+4) ,
T100010=1/(1*2.07*10^(-06)+4) + T000010*4/2/(1*2.07*10^(-06)+4) + T100000*4/2/(1*2.07*10^(-06)+4) ,
T110000=1/(1*2.83*10^(-06)+4) + T010000*4/2/(1*2.83*10^(-06)+4) + T100000*4/2/(1*2.83*10^(-06)+4) }, { T000000,T000010,T010000,T010010,T100000,T100010,T110000 }):
T0 := subs(res, T000000):
printf("%g\n", T0);

Hi, in attached code pauldaas_05.mw, the procedure 'aFshear' and 'aFmoment' must be called within procedure ´aF0011´.

The procedures ´aF0011´, 'aFshear', 'aFmoment' build the matrixes R, C1 and C2 respectively.

Unfortunately ('aFshear' and 'aFmoment') gives 'Error, invalid left hand side in assignment'.

It is strange that ´aF0011´ can call them separately ('aFshear' or 'aFmoment') but not both ('aFshear' and 'aFmoment').

Maybe there should be an 'unassign' statement for 'L'  somewhere. I tried that but without result.

Is there anyone who can help me.

restart:
with(linalg):
L:=Vector[row]([''RA'',''MA'',''thetaA'',''yA'',''RB'',''MB'',''thetaB'',''yB'']):
R:=Matrix(0,8,0):
C1:=Matrix(0,10,0):
C2:=Matrix(0,10,0):

aF0011:=proc(nr,a,l,F,EI)
local B,RA,MA,thetaA,yA,RB,MB,thetaB,yB,aF;
global R;
R:=extend(R,1,0,0);
B[RA]:=0;
B[MA]:=0;
B[thetaA]:=F*(l-a)^2/(2*EI);
B[yA]:=-F*(a^3-3*a*l^2+2*l^3)/(6*EI);
B[RB]:=F;
B[MB]:=-F*(l-a);
B[thetaB]:=0;
B[yB]:=0;
B:=Matrix( [seq(B[i],i in [RA,MA,thetaA,yA,RB,MB,thetaB,yB])]);
copyinto(B,R,nr,1);
aFshear(nr,R);   #nested procedure
aFmoment(nr,R);  #nested procedure
end:

aFshear:=proc(nr,R)
local i,c;
global C1;
C1:=extend(C1,1,0,0);
R:=convert(R,Matrix);
zip(assign,L,R[nr,..]);
c[0]:=RA;
c[1]:=0;
c[2]:=0;
c[3]:=0;
c[4]:=-F;
c[5]:=0;
c[6]:=0;
c[7]:=0;
c[8]:=0;
c[9]:=0;
c:=Matrix([seq(c[i],i in [0,1,2,3,4,5,6,7,8,9])]);
copyinto(c,C1,nr,1);
end proc:

aFmoment:=proc(nr,R)
local i,c;
global C2;
C2:=extend(C2,1,0,0);
R:=convert(R,Matrix);
zip(assign,L,R[nr,..]);
c[0]:=MA;
c[1]:=RA;
c[2]:=0;
c[3]:=0;
c[4]:=0;
c[5]:=-F;
c[6]:=0;
c[7]:=0;
c[8]:=0;
c[9]:=0;
c:=Matrix([seq(c[i],i in [0,1,2,3,4,5,6,7,8,9])]);
copyinto(c,C2,nr,1);
end proc:

aF0011(1,a__1,l,F__1,EI):
aF0011(2,a__2,l,F__2,EI):
aF0011(3,a__3,l,F__3,EI):
R:=convert(R,Matrix);
C1:=convert(C1,Matrix);
C2:=convert(C2,Matrix);

l:=8000:
F__1:=15000:
F__2:=20000:
F__3:=35000:
a__1:=2000:
a__2:=2000:
a__3:=6000:
EI:=2E+5*2.49E+7:
R;
C1;
C2;

I have a system of ODEs with parameters, p[i], and variables, x[i].

f := [
-p[1]*x[1]^2+x[2],
-2*p[1]^2*x[1]^3+2*p[1]*x[1]*x[2]+x[1]+1
];

associated with the innitial conditions:
[x[1](0) = p[2], x[2](0) = p[3]].

I have sets of parameters that i am interested in

[p[1] = 1, p[2] = 2, p[3] = 3]
[p[1] = 2, p[2] = 2, p[3] = 7]
[p[1] = 3, p[2] = 2, p[3] = 11]
[p[1] = 4, p[2] = 2, p[3] = 15]
[p[1] = 5, p[2] = 2, p[3] = 19]
[p[1] = 6, p[2] = 2, p[3] = 23]
and so on.


The solutions for these equations appear to be anisotropic scalings of each other.

i.e.

/xh[1]\ =/1 0\ /x[1]\      
\xh[2]/   \0 q/ \x[2]/

where x is the variable for one parameter vector and xh is the variable for another; and q is a nuimber I am interesting in working out (and think may be p[1]).

how do you work out q?

[this is in fact a complicated way of asking when I numerically integrate an ODE on maple how do i get access to the sequence of numbers in the result]

Hi,

It might be a very basic question, but I am having trouble with the substitution of multiplication of symbols.  I have a equation in the format of     , and I need to substitute xy=1 in this equation without interferring in f(x,y).

When I am using subs commnad in maple, it is not recognizing xy nad giving me the same equation.

I really appreciate help in this matter.

Thanks in advance

Regards

Sunit

I have the following PDE

                         ...(1)

My Question: I have done manually following calculations. I want to verify the following operations are right, or not by MAPLE. Could you help me, please?

The METHOD:

Suppose that

                         ...(2)

In here,  is an Nx1 matrix, P, C are NxN matrices. (N is an integer and superscript ^T   denotes the transpose of the matrix.)  and P are given matrices. But the matrix C is ungiven I will find it in the final step. But my question doesn' t include all steps. I just wonder how to calculate the first two steps by Maple.

( If Maple doesn' t do matrix algebra, we can treat them as if , P, C were not matrix. I think the result won' t be changed. We will get again equation 9 by Maple.)

First Step

We will find the followings 

 in terms of the matrices  ,  P and C.

So, if we integrate Equation (2) with respect to x (from 0 to x), and  by using the following two assumptions

we have 

   ...(3)
 

substituting x=1  in Equation (3)

 ...(4)

if rewrite Eqn. (4), we have

...(5)

substituting Eqn. (5) to Eqn. (3), we have 

...(6)

integrating Equation (3) with respect to t,

we have 

...(7)

If we integrate Equation (2) from 0 to x with respect to t, we have

...(8)

Second Step We will substitute the terms to the pde ( Equation 1)

Substituting Eqn. (6), (7), (8) to Eqn. (1), we have finally

...(9)

I want to do the above calculations by Maple.  

Because I have more complex questions than above, I want to write a Maple code in order to avoid calculation errors.  

Final Step for curious: it's hard to explain the whole method here. Briefly, we will discretize equation 9 for some collocation points t and x. And after doing it, we will have a system of an algebraic equation. (N equation and N unknown ( C is Nx1 unknown vector to be find) )

And then we will substitute vector C to Eqn. 7

Code for Matrices , P^1, P^2,etc.

code.mw 

Best Regards...

Dear Users!

Hope you are doing well. Below is the code to solve system of equations

restart;M:=3;
u[0, 0] := tau[1]-2*tau[2]+3*tau[3] = 1;
u[0, 1] := tau[1]-tau[3] = 1/2;
u[0, 2] := tau[1]+2*tau[2]+3*tau[3] = 2;
SOL[0] := fsolve({seq(`$`(u[0, l1], l1 = 0 .. M-1))});

the solutions SOL[0] := {tau[1] = .7500000000, tau[2] = .2500000000, tau[3] = .2500000000}
I want to define a vector of order M*1 having the solution of of tau's like that

[.7500000000  .2500000000  .2500000000]

I'm waiting for you positive response. Thanks in advance.
Special request to:
@acer @Carl Love @Kitonum @Preben Alsholm

Hello;
Anyone can guide to make the code of this expression

Based on the equation at here https://www.mapleprimes.com/questions/209660-Problem-With-RealDomainsolve

I tried solve the equation (x-1)*sqrt(x^2 - 4)=0 in Real domain. My code
restart;
RealDomain:-solve((x-1)*sqrt(x^2-4) = 0, x);

I got there solutions are 1, 2, -2.  I think, If we solve the given in RealDomain, we only get two solutions -2 and 2.

My question is: How many solutions are there in the equation (x-1)*sqrt(x^2 - 4)=0 by RealDomain:-solve?

Two pictures by using Mathematica.

Hi,
I would like to trace periodic functions. I saw that this was possible with the old 'FourierSeries' package with the "Rept" command. How to reproduce the same thing in the Maple 19 environment? Thank you

fs_examples.mw

I am trying to get the exponential Fourier series versiion  with C_n seperate and it appears this package https://www.maplesoft.com/applications/view.aspx?SID=4857&view=html seems to do this

While I'm not particular to this package https://www.maplesoft.com/applications/view.aspx?SID=4857&view=html I cannot get it to work. It could be that FourierSeries is not built into maple in which case only the worksheet appears to be availble on the link above with no application in which case it should be removed.

If not an alternative would do. For example I got this https://www.maplesoft.com/applications/view.aspx?SID=33406 to work but I cannot get it in  format like Khanshan's package see example output (1.7).

1. y''(x)+10y(x)=99sin(x), y(0)=1, y'(0)=11 in the interval [0,100]

the exact solution is y(x)=cos(10x)+sin(10x)+sin(x)

2. y'=z, y(0)=1

    z'=-y(x)+x, z(0)=2

    the exact solutions are y(x)=cos(x)+sin(x)+x, z(x)=cos(x)-sin(10x)+1

Hello all, 

I know this question is a little bit of stretch but is there any way to shuffle terms in an equation set, making those equations written in a standard form?

For example, here are three equations:

eq1:=(L1/n12 + n12*L2)*diff(i2(t), t) + L1*diff(i3(t), t)/n13 = -v1(t) - R1*i3(t)/n13 - R1*i2(t)/n12 + n12*v2(t) - n12*R2*i2(t);

eq2:=L1*diff(i2(t), t)/n12 + (L1/n13 + n13*L3)*diff(i3(t), t) = -v1(t) - R1*i3(t)/n13 - R1*i2(t)/n12 + n13*v3(t) - n13*R3*i3(t);

eq3:=-L2*diff(i2(t), t) + L1*diff(i1(t), t)/n12 = -v2(t) + R2*i2(t) + v1(t)/n12 - R1*i1(t)/n12;

In this equation set, it can be inferred that 'i1(t)', 'i2(t)' and 'i3(t)' would be the state variables and 'v1(t)', 'v2(t)' and 'v3(t)' are external input variables. 

Then, probably the set can be rewritten in the state space equation form, i.e., diff(X) = A*X + B*U, where X is the state variables vector and U is a vector of external input variables. 

Is there any chance to make Maple to rewrite the equations in this way?

Thank you in advance.