Hello

I would like to get Maple to solve a system of 3 equations, nonlinear. 

I use the solve command to get the correct solutions.

However I would like to see the steps, is there a way to do this?

Thanks

I am trying to setup a general metric with the Physics package.  The metric is composed of the Minkowski metric plus the product of two null vectors. Here is the code:

retstart;

with(Physics);

Define(l[mu],eta[mu,nu]);

eta[mu,nu] := rhs(g_[]);

Setup(g[mu,nu]=eta[mu,nu]+l[mu]*l[nu]);

I get the following error:

Error, (in Physics:-Setup) wrong argument: g[mu, nu] = l[mu]*l[nu]+(Matrix(4, 4, {(1, 1) = -1, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (2, 2) = -1, (2, 3) = 0, (2, 4) = 0, (3, 3) = -1, (3, 4) = 0, (4, 4) = 1}, storage = triangular[upper], shape = [symmetric]))

My plan is to apply the rule g[~mu,~nu]*l[mu]*l[nu] = 0 and calculate the Christoffel symbols using the metric.

I am trying to use the Physics package because the DifferentialGeometry package seemed focused on Newman-Penrose.  I will not be using NP for the calculations, only a strict calculation of the Einstein field equations from the given metric.

Thank you.

Hello,

I have to simplify a piecewise function and Maple gets a more complicated solution than needed.




I don't know how to handle this kind of problems with Maple?
I don't understand why Maple doesn't see this?
Am I doing something wrong?

Thanks in advance for your help / advice.


# the code of my example
restart:
Mf(x):=piecewise(x<=L/2,1/2*x*F,x>1/2*L,1/2*x*F-F*(x-1/2*L));
# Make a dimensionless function:
# -    Mf(x):= Mf(xi)*F*L
# -    variable ξ  ( xi:=x/L )
eq[1]:=Mf(xi)*F*L=Mf(x);
Mf(xi):=solve(eq[1],Mf(xi));
Mf(xi):=subs(x=xi*L,Mf(xi));
# F is the Force and L is the Length of the beam:
Mf(xi):=simplify(Mf(xi)) assuming F>0,L>0;
print("When I simplify this function by hand it will be");
Mf(xi):=piecewise(xi<=1/2,1/2*xi,xi>1/2,-1/2*xi+1/2);

Hi everybody,

I am using maple to solve differential equations.  I have been using maple for some time and came across the infolevel[dsolve] command or option.  I tried finding the answer to my question online but cant seem to come across a straight answer.

My question is what does the integer do that you have to equal the infolevel[dsolve] command?  

For example I am using a book that states

"An even more important diagnostic tool is the infolevel[dsolve] command, which will give information on what methods are used in attempting to solve the ODE when dsolve is applied, even if unsuccessful.  An integer between 1 and 5 must be specified, with generally more detailed information being provided as the number is increased. On applying the dsolve command to ode2, the method of attack is summarized in the following output and, in this case, the general solution y(x) given with two arbitrary coefficients C1 and C2.

What does the integer between 1 and 5 mean or do? In the example above the book uses 5, why 5?.

Thanks,

Matt 

I’m trying to figure out how to find a basis for a subspace, V, of R³ defined by V = {(x, y, z)l(2x-3y+6z=0)}

I’m using the student linear algebra module for maple 17

I’ve tried defining the subspace and asking for the basis of V but I always get an error code.

I’ve tried consulting the maple website and looking through their help menu, but can’t find anything that answers how to find a basis... At least a basis from the subspace defined in my problem.

I know how to find a basis for the subspace by hand but not with maple.

Any help will be greatly appreciated. 

restart:assume(M>0);

Eq1 := diff(psi(y), y, y, y, y)-M*(diff(psi(y), y, y))-Gr*b*y = 0;

bcs1:=psi(0)=0,(D@@2)(psi)(0)=0,psi(h)=-F/2,D(psi)(h)=A;

res1:=(dsolve(Eq1));

res2:=(dsolve({Eq1,bcs1},psi(y)));

match(rhs(res2)=rhs(res1),y,s);

s:

C3:=eval(_C3, s);

I am unable to find the constants. Anyway around this?

This has been buggin me all day, but if i define a variable with an expresion say "a:=3 x+5" and i then want to assign this expression to a function like so "f:=x->a". if i then call f(7), maple return with "3 x+5" so just returning the value of a without substituting x with 7.

Is there any way in which i can define a function this way?

I am using the following commands to create an animation of a simple cardioid being traced out. I would like to add to this an animation of the filling of the region bounded by the curve and the coordinate axes as the curve is being traced out. How might one achieve this?

with(plots):

animatecurve([2-2*sin(theta),theta,theta=0.. 2*Pi],coords=polar,axiscoordinates=polar,frames=250, numpoints=300, scaling=constrained)

Thanks.

I have a plot of a sinusoidal function(cos_phi) which changes with incrementing values of variable k. When this function goes above 1 or below -1 I would like to have the range of k for which it occurs outputed somehow. Would anybody know how to do this?

Given the lengths of the three sides of a triangle, write a procedure, triType, to produce an output which shows whether the triangle is acute, right or obtuse with the three sides given??

Hello,

I am using Maple 17 to solve some equations. I am solving these equations with the solve command. There are some bounds on the parameters so I am including these inequalities. I have 4 variables, so I solve for 3 to get them in terms of the last one.

It returns piecewise solutions as per the inequalties as I hoped.

However for two ranges of the inequalities I get [] which I assume means no solution.

For another range I get an expression for each of my paramters.

For the final range I get 0 . What does this 0 mean and what do the [] mean?

I wondered if 0 meant that all paremeters equalling zero was a solution, but this is false.

Thanks

Good afternoon sir.

Please let me know how to upgrade mu current version Maple17 to Maple 18.

With thanks & Regards

M.Anand

Assistant Professor in Mathematics

SR International Institute of Technology,

Hyderabad, Andhra Pradesh, INDIA.

Good afternoon sir.

I request your kind support to the above cited query.

With thanks & Regards

M.Anand

Assistant Professor in Mathematics

SR International Institute of Technology,

Hyderabad, Andhra Pradesh, INDIA.

Good afternoon sir.

I request your kind suggestion to the above cited query.

With thanks & Regards

M.Anand

Assistant Professor in Mathematics

SR International Institute of Technology,

Hyderabad, Andhra Pradesh, INDIA.

firstly apologies in advance for stuff in this question such as "triangle symbol",  my computer is pretty old. 

ok so i was confused a bit here, what i'm trying to do is write a maple procedure that computes Af for a given f contained in V . except we only need to correct the bug in the script below. This script demonstrates such a procedure in the case that omega is a square. The domain is given here as the negative set of a function F contained in V .  I have left in notes where/what i think we need to do but i dunno how to...

N:=10 ; # Global Var
F:=(x,y)->sgn(abs(x-N/2)+abs(y-N/2)-N/4);
Average := proc(F, f0) local f, i, j;
f := f0; # !!!!!!!!!!!!!! something is bad here...
for i to N do for j to N do
if F(i, j) < 0 then
f[i, j] := (f0[i - 1, j] + f0[i + 1, j] + f0[i, j + 1] + f0[i, j - 1])/4 ;
end if;
end do;end do;
return f;
end proc;
f0:=Matrix(N,F); # just to have something to test the procedure
Average(F,f0); # does not return the expected average, modifies f0

the necessary information we were given to produce this so far was..

Let N be a positive integer and [N] = {i contained in N | 1<= i <=N }  Let "Omega" C {(i,j) contained in [N] x [N] | 2<=i,j<=N-1} be a subset. Let V = R^([N]x[N]) be the vector space of real valued functions [N]x[N] -> R
and A,
"triangle symbol":V->V (average) and "triangle symbole" (Laplacian) be the linear maps such that
[Af](i; j) = f(i; j)      if (i; j) not contained in "Omega"   OR

                             [f(i, j + 1) + f(i, j - 1) + f(i + 1, j) + f(i - 1, j)]/4 if (i,j) is contained in "Omega"

["traingle symbol"f](i,j) =  0 if (i,j) isnt contained in "Omega"   OR

                            ( f(i,j) - [f(i, j + 1) + f(i, j - 1) + f(i + 1, j) + f(i - 1, j)]/4 )    if (i,j) is contained in "Omega"

 Please and thank you for any help in advance <3