Hi all,

While solving a différential equation, I encounter an integral:

 

Anyone know a slick way to change the arguments of a function to possibly different things depending on what they are?

 

Let's say I have objects a, b and c.  a and b are of type function and c is of type set.  F is another function object.

 

F(a,b,c) is defined.

 

What I want to do is construct F(g(a),g(b), h(c)).....

 

Ideas?

solve used to be one of Maple's strongest commands -- it even subsumed simplify in power.  But, over the years, dsolve slowly took over as the most powerful comand.  At the same time, people started realizing that within the framework of differential equations, the toolbox was actually larger than the one for algebraic equations (and most algebraic tools are still available).  So many tasks that one thinks of doing purely algebraically can also be done using differential equations, with perhaps the most surprising one is to factor multivariate polynomials via partial differential equations.

Reduce::nsmet: This system cannot be solved with the methods \
available to Reduce. >>

 

Hi, I'd like to know where's maple directory by default on my computer. I just found a program that would really help which is run on maple. I will join the read me file of the program i want to install.
The unclear part for me is "  (b) Copy the file dot.mapleinit in the Bethe root directory into your
    home directory where it is read in from MAPLE      cp dot.mapleinit ---> [user-home]/.mapleinit"

For t := x^(1-I), f := arcsin(t) + arcsin(1/t) i have problems with

  dd:= PDEtools[dpolyform](y(z)=f,no_Fn);

                           d
                    dd := [-- y(x) = 0] &where []
                           dx

Hi all

Hi everybody.

 

Hey guys,

I am struggling again trying to solve very similar pde in Maple V.

pdesolve(-a*u(t,x)+diff(u(t,x),t)+(c^2)*b*(1/t)*[log(2*T/t)]^(-b-1)diff(u(t,x),x,x)=0,u(t,x))

Could anybody try running it in the older version? Are there any solutions?

Thank you,

Vilen

Hallo Everybody,

I hope someone can help me.

I want to solve numerically the diffusion equation,

          diff(csh(rh, t), t) = Dif*(diff(rh^2*(diff(csh(rh, t), rh)), rh))/(r0*rh^2)

          where "Dif" is the diffusivity and "r0" is the initial value of rh (the equations were normalized).

subject to the following boundary and initial conditions:

In Classic, this input:

753406.29+42098.70;

yields this output:

795504.99

But in Standard, the same input yields

7.9550499 10

How do I set my options in Standard so that the output format is like it is in Classic?

I poked around "display options" etc. but did not find anything.

 

If I have some element tau in my matrix, how do I declare it to be real?

 

> with(LinearAlgebra);
> A := `<|>`(`<,>`(-sqrt(5/3)*tau, -tau, 0, 0), `<,>`(-tau, -sqrt(3/5)*tau, -2*tau/sqrt(15), 0), `<,>`(0, -2*tau/sqrt(15), sqrt(3/5)*tau, 0), `<,>`(0, 0, -tau, sqrt(5/3)*tau));

 

The time function doesn't seem to be working consistantly.

Following the example in the help page for the time() function I can't get as low as .041.  The best I get is .046 and sometimes .047 .062 and .063.  Are there services running in the background that can affect the run performance like a firewall for example?

Can anyone else get .041?  For comparison I'm using a P4 - 3 GHz. and 2.5 Gb RAM @ 533 MHz. with 48 processes running in the task manager.

So I'd like to take two ranges of numbers and simlify their combination.  Example:  RealRange(2,5) union RealRange(3,6) should yield RealRange(2,6), but the union function doesn't accept RealRange data types as its argument.   Is there an available command that will produce the desired result?

Dear All, I'm trying to substitute a series ansatz into a coupled couple of equations to find coefficients. The equations are quite cumbersome, but the principle shouldn't be too difficult. The equations are f3A(alpha,mu,Omega,delta,m) =0 f4A(alpha,mu,Omega,m) = 0 where f3A and f4A are complicated functions of the several variables. The region I'm working in is delta=0, m=1 Now, I have reason to believe that the following ansatz will work: alpha = A0 + A2/Omega^2 + A4/Omega^4 +.... -mu=M0 + M2/Omega^2 + M4/Omega^4 +.....