Hey guyz, I am in trouble with calculation attached integral. it is a simple function but a bit long. I can't solve it with maple, Do U have any idea?

intg.mw

hello, i just try to plot the relation between my outputs (u, and phat) with i from 0 to 10 , but i have aproblem any suggestions? 
 

Download hw_4_structural.mw

Hello everyone,

     I am having trouble trying to solve a system of differential equations. The modeling was made from the equilibrium equations of a pressure vessel. The set of equations is shown below:

     As you see it is a set of two second-order partial differential equations. So, we need four boundary conditions. This one is the first. It means that the left end of the pressure vessel is fixed.

This one is the second boundary condition. It means that the right end of the pressure vessel is free.

This one is the third boundary condition. It means that the inner surface of the pressure vessel is subject to an external load:

At last, we have the fourth boundary condition. It means that the outer surface of the pressure vessel is free.

     The first test I have been trying to do is the static case. In this case, the time terms (the right side of the two equations shown) is zero.

    The maple commands that I am using are the following:

restart; E := 200*10^9; nu := .33; G := E/(2*(1+nu)); RI := 0.254e-1; RO := 2*RI; p := proc (x) options operator, arrow; 50000000 end proc; sys := [E*(nu*(diff(v(x, r), x))/r+nu*(diff(diff(v(x, r), x), r))+(1-nu)*(diff(diff(u(x, r), x), x)))/(-2*nu^2-nu+1)+G*(diff(diff(u(x, r), r), r)+diff(diff(v(x, r), x), r)+(diff(u(x, r), r))/r+(diff(v(x, r), x))/r) = 0, E*((1-nu)*(diff(diff(v(x, r), r), r))+nu*(diff(diff(u(x, r), x), r))+(1-nu)*(diff(v(x, r), r))/r-(1-nu)*v(x, r)/r^2)/(-2*nu^2-nu+1)+G*(diff(diff(u(x, r), r), x)+diff(diff(v(x, r), x), x)) = 0]; BCs := {E*(nu*v(L, r)/r+nu*(D[2](v))(L, r)+(1-nu)*(D[1](u))(L, r))/(-2*nu^2-nu+1) = 0, E*(nu*v(x, RI)/RI+(1-nu)*(D[2](v))(x, RI)+nu*(D[1](u))(x, RI))/(-2*nu^2-nu+1) = -p(x), E*(nu*v(x, RO)/RO+(1-nu)*(D[2](v))(x, RO)+nu*(D[1](u))(x, RO))/(-2*nu^2-nu+1) = 0, u(0, r) = 0}

sol := pdsolve(sys, BCs, numeric)

I am getting the following error:

Error, (in pdsolve/numeric/process_IBCs) initial/boundary conditions must depend upon exactly one of the independent variables: 0.1459531181e12*v(L, r)/r+0.1459531181e12*(D[2](v))(L, r)+0.2963290579e12*(D[1](u))(L, r) = 0

In this case, my boundary conditions do depend on more than one independent variable. How do I proceed?

Thank you in advance,

Pedro Guaraldi

Is there anyone who has seen maple 2017 provide some details about what new features are being introduced. Is there a platform where we can suggest what features we would like to be added or enhanced?

I'm trying to define some multilinear forms to study differential geometry. What I need is only symbolic. My intention is symplify computations involving multilinear forms.

For example, to create an symbolic inner product "g" I used the command "define" like in this post:

http://www.mapleprimes.com/questions/203480-Define-And-Use-Abstract-Linear-Operator

So I tipped:

define(g, orderless, multilinear);

My doubt is: how can I declare that g(x,y) is always scalar?

With it I would simplify things like g(g(z,w)*x,y) = g(z,w)*g(x,y)

In my case, specifically, I type:

v:=(X,Y,Z)->g(Y,Z)*X-g(X,Z)*Y;

r:=(X,Y,Z,W)->g(v(X,Y,Z),W)-g(Y,T)*g(v(X,T,Z),W)+g(X,T)*g(v(Y,T,Z),W);

expand(r(X,Y,Z,W));

and the result is:

g(W,g(Y,Z)*X)-g(W,g(X,Z)*Y)-g(T,Y)*g(W,g(T,Z)*X)+g(T,Y)*g(W,g(X,Z)*T)+g(T,X)*g(W,g(T,Z)*Y)-g(T,X)*g(W,g(Y,Z)*T)

But I would enjoy that it were:

g(Y,Z)*g(W,X)-g(X,Z)*g(W,Y)-g(T,Z)*g(T,Y)*g(W,X)+g(X,Z)*g(T,Y)*g(W,T)+g(T,Z)*g(T,X)*g(W,Y)-g(Y,Z)*g(T,X)*g(W,T)

Is there a way to declare that g(x,y) is always scalar?

Thanks.

I have a problem using dchange when my variable depend on two (or more variables) and I would like to apply the chain rule.

For example, when I use the command

I would expect something like 

But I get an error saying that the number of new variables and transformation equations must be the same.

Any idea how I could solve it? 

Thanls a lot for your help.

Hi I have this code

 psi:=proc(n,x);
 (1/sqrt(sqrt(pi)*2^n*factorial(n))*exp(-x^2/2)*HermiteH(n,x))
 end proc;
 psi := proc(n, x) exp(-1/2*x^2)*HermiteH(n, x)/sqrt(sqrt(pi)*2^n*n!) end proc

 psi2=proc(a,x);
 psi(a,x):=(1/sqrt(sqrt(pi)*2^a*factorial(a))*exp(-x^2/2)*HermiteH(a,x))
 end proc;
psi2 = (proc(a, x)
    psi(a, x) := exp(-1/2*x^2)*HermiteH(a, x)/sqrt(sqrt(pi)*2^a*a!)

end proc)
 for n from 0 to 2 do;
 for a from 0 to 2 do;
 result=proc(n,a);
 result(n,a)=psi*psi2
 end proc;
print(evalf(int(result(n,a),x=0..infinity)));
od;
od;

it returns

I know the results for (0,0), (1,1) and (2,2) should be 1 and the rest should be 0.

Can anybody help fix this please

When function is given as two variables, how to find extremas? help thanks

example: z=x³+y³-3xy

and alse the function as given condition z=x+2y, x²+y²=5

i need to solve for u[i+1] as i attached i wrote the equations but i cant get any answers for it, the delta t is 0.1 and i need to go for ten steps, thank you
 

Download hw_4_structural.mw

I need some help. I'm trying to solve this system of equations, but maple says the solutions may have been lost. Here are the equations:

Hi all,

I tried to fit my data (x,y) with a model by using Minimize the Chisquare. By example the model is y=a*x+b, Chisquare is (y-yexp)^2. And I performed a function Minimize(Chisquare) to have a and b.

I need to extract the error of parameter like a±aerror, and b±berror.

Thank you for your helps,

Best regards,

As you know the matrix calculations of the Matlab is more powerfull than Maple, My question is that is it possible to call Matlab solver from Maple worksheet? I mean, in part of my Maple worksheet (for loop) I need to compute the eigenvalue and square root of a big matrix (order of 1000) that Maple couldnt while Matlab can do it. So it will be so nice if I can call Matlab directly from my ws rather than I export my data and import it to Matlab and calculate and return back to Maple (manualy It is tedious).

Is it possible???

Usually maple displays legend as either a colored line or symbol. If pointline is used in the structure of the command the plot is displayed as both point and line while the legends appear as a colored line, Is there a way in which the curve will combine both point and line and the legend(s) will be strictly point?

Thank you and kind regards

Hi everybody, 

I have a continuous function f of a single variable (all the details can be found in the attached file) and I want to build a more regular approximation of it (let's say F). The construction process ensures that F is C-infinite.

When I plot f and F (command "plot"), for visual comparisons, the F curve presents "holes", that is intervals where there is no plot.
However, the value of F(x) for any x in those void ranges is a real just as F(y) is for any y in a plotted interval.

Note that this phenomenon does not appear  when I use PLOT(CURVES(...)) for F (attached file)

I guess I probably use "plot" in a wrong way, or maybe some option I don't know could prevent it to happen ?

Could you please have a look to the attached file ?
I look forward for your response.


LacunaryPlot.mw 

Assume you have a matrix A and somewhere you want to make a copy of it like B and working with them independently. Let's say you have a loop and after doing a proc on B again you have to make it equal to A. So changes on B shouldn't effect on A. What is the common way of taking such copies of A in Maple?

The following methods don't work.
 

A:=Matrix(3);
B:=A;
B(1,1):=1;
A;

A:=Matrix(3);
B:=subs(B=A,B);
B(1,1):=1;
A;

What I came up with is the following but it will look weird if one really needs to write something meaningless like *2/2.

A:=Matrix(3);
B := (1/2)*subs(B = 2*A, B);
B(1,1):=1;
A;