MaplePrimes Questions

Hi there, I'm a really noob when it comes to maple.
My teacher wants me to solve a differential equation using this software and the only problem I have is that I dont know which command to use to solve the equation with boundary conditions.
I've already tried the dsolve command but then I realized that it only works with initial conditions, and the conditions I have are boundary (initial-end). So how do I actually solve the differential equation ?

When I download the help document here

https://www.maplesoft.com/support/help/maple/view.aspx?path=int%2Fdetails

I get a bug when I execute the page.

I copied pasted the last line using 1d math notation to replicate the error.

There is no addition symbol between x^2 + O(x^5)

On the help page I don't see an error. 

 I uploaded the help file   int-details_(2).mw 

Hello people in mapleprimes,

I hope you will help me about modifying an expression.

How can I factor the expression with w and Delta__1 in the attached file?

e2_3 := w^(1-sigma)*((f__11*sigma*Delta__1/L__1)^(-1/(sigma-1))/w)^(k-sigma+1)*k/(Delta__1*(k-sigma+1)*a__0^k);

Thank you for advance.

Modifying_expression_factor.mw

how can i simplify with definition two new parameter 

i show in this picture

ii := evalr(csch(INTERVAL(-infinity .. -1, 1 .. infinity)));
             INTERVAL(-csch(1) .. 0, csch(1) .. 0)

evalr(ii-ii);
                            -csch(1)

(Make sure you don't have the Typesetting level set to Extended, or ii won't be displayed correctly.)

There are no functions to test for interval membership (or to compute intersections/unions of intervals), so it's unclear how Maple treats the interval csch(1)..0, but apparently evalr is getting tripped up by it.

Also:

evalr(sech(INTERVAL(-infinity .. -1, 1 .. infinity)));
                               0

evalr(arccsch(INTERVAL(-infinity .. -1, 1 .. infinity)));
Error, (in evalr/arccsch) too many levels of recursion

 

Hello,

For example, let's say we have the following function

 

Now if we want to calculate the value of a function we have to

subs([x1=10,x2=20],eval(hf));

My question is the following: How to substitute values for x1 and x2 using values from vector? something like that

x:=[10,20];

subs([x],eval(hf));

 

Best,

Rariusz

 

 Hello,

I solve a partial differential equation using the THETA METHOD on Maple17, but what is also of essence is the skin friction and Nusselt Number which are {diff(u(x, t), x)} at x = 0 and {diff(T(x, t), x)} at x = 0 respectively. How can I achieve the two using the THETA METHOD? This is the PDE in it full flesh. 

 PDE := {diff(phi(x, t), t) = (diff(phi(x, t), x, x))/S__c, diff(u(x, t), t) = diff(u(x, t), x, x)-M^2*(u(x, t)+m*w(x, t))/(m^2+1)-u(x, t)/`ϰ`-2*Omega^2*w(x, t), diff(w(x, t), t) = diff(w(x, t), x, x)+M^2*(m*u(x, t)-w(x, t))/(m^2+1)-w(x, t)/`ϰ`+2*Omega^2*u(x, t), diff(theta(x, t), t) = lambda*(diff(theta(x, t), x, x))/P__r+D__r*(diff(phi(x, t), x, x))}

With conditions: BC := {phi(0, t) = 1, phi(12, t) = 0, phi(x, 0) = 0, u(0, t) = t, u(12, t) = 0, u(x, 0) = 0, w(0, t) = 0, w(12, t) = 0, w(x, 0) = 0, theta(0, t) = 1, theta(12, t) = 0, theta(x, 0) = 0}

Where the parameter like P_r, S_c  e.t.c are constants to assign value to.

Thanks.

I want to plot the compex solutions of a single equation. I do

> plot([fsolve(x^3 = 1, complex)])

The output consists of a warning:

"Warning, unable to evaluate 2 of the 3 functions to numeric values in the region; complex values were detected"

and a horizontal line.

What am I doing wrong?

Thank you!

The linked worksheet displays the geodesic between two points on a surface z(x,y).

Surfacepath.mw

Gradient descent will find a stepped approximate path of fastest descent between the points, but is there a way to find a function defining an exact i.e. smooth path of fastest descent? If so, how can a precise time of fastest descent be determined?

Namely, I mean

solve({y >= 4*x^4+4*x^2*y+1/2, sqrt((1/2)*(x-y)^2-(x-y)^4) = -2*x^2+y^2}, [x, y]);
                               []

The answer (no solution) is not correct in view of 

eval({y >= 4*x^4+4*x^2*y+1/2, sqrt((1/2)*(x-y)^2-(x-y)^4) = -2*x^2+y^2}, [x = 0, y = 1/2]);
      {(1/8)*sqrt(4) = 1/4, 1/2 <= 1/2}
eval({y >= 4*x^4+4*x^2*y+1/2, sqrt((1/2)*(x-y)^2-(x-y)^4) = -2*x^2+y^2}, [x = -1, y = -3/2]);
      {(1/8)*sqrt(4) = 1/4, -3/2 <= -3/2}            

 

Hi all, I want to ask how to use for loop reverse in Maple.

Example, i have array . In Java then

for(int i = arr.size(); i > 0; i--) {

    //do something

}

What is simillar for loop in Maple?

Thank you very much.

Hi, when I wan to generate results, it shows this; What should I do?

Error, (in solve) a constant is invalid as a variable, I. 

I have a system of equations that can't be solved analytically, but can be solved using fsolve. I would like to plot the solution for one of the unknowns as a function of a parameter "DV" that is in a few of the equations. There are two issues at hand. First, even if I give DV a numerical value, the solutions are of the form { x = 0.23, y = 4.56 }, a form that plot( ) obviously can't handle directly. Generally I have dealt with this by using the op( ) command to pick out the solution I want; is there a better way?

But even after I have the solution I want, fsolve( ) won't work until I give it a value for DV. Thus a plot( ) command like

   plot(op(2, op(2, fsolve({e1, e2}, {x, y})), DV = -1 .. 1)

fails with an "Error, (in fsolve) DV is in the equation, and is not solved for" error.

There's got to be an esy way to do this!

Here Maple treats the intervals as independent, as I think it should:

INTERVAL(1 .. 2)+INTERVAL(1 .. 2);
              INTERVAL(1 .. 2) + INTERVAL(1 .. 2)

INTERVAL(1 .. 2)-INTERVAL(1 .. 2);
              INTERVAL(1 .. 2) - INTERVAL(1 .. 2)

(even though for the first one an autosimplification to 2*INTERVAL(1..2) wouldn't matter). But as soon as there are non-linear terms, Maple ignores that rule:

INTERVAL(1 .. 2)^2-INTERVAL(1 .. 2)^2;
                               0

INTERVAL(1 .. 2)/INTERVAL(1 .. 2);
                               1

INTERVAL(-1 .. 1)*INTERVAL(-1 .. 1);
                                        2
                       INTERVAL(-1 .. 1) 

That means I cannot substitute intervals for variables in a formula. In fact, I don't know how INTERVAL(-1 .. 1)*INTERVAL(-1 .. 1) can be computed as a product of two independent intervals at all, there doesn't seem to be any way to pass it to evalr.

 

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