Please respond me by email, thanks.

wingwatson7@gmail.com

Good evening, dear experts. 

I want to solve the system of PDEs, but i get a mistake:

"Error, (in pdsolve/numeric/xprofile) unable to compute solution for t>HFloat(0.0):

solution becomes undefined, problem may be ill posed or method may be ill suited to solution"

How  can I determine the method for pdsolve?
Thanks for answers.

It's my file with the functions:
restart;
alias(X = x(t, tau), Y = y(t, tau));
xt, yt := map(diff, [X, Y], t)[];
xtau, ytau := map(diff, [X, Y], tau)[];
M := (xtau*(-Y+.1*X)+ytau*X)/(xtau^2+ytau^2);
pde1 := xt = -Y+.1*X-xtau*M;
pde2 := yt = -M*ytau+X;

cond := {x(0, tau) = 1, x(t, 0) = 1, y(0, tau) = 1, y(t, 0) = 1};
Sol := pdsolve({pde1, pde2}, cond, numeric, time = t, range = 0 .. 1);
Sol:-value(t = 1);
%;
Error, (in pdsolve/numeric/xprofile) unable to compute solution for t>HFloat(0.0):
solution becomes undefined, problem may be ill posed or method may be ill suited to solution

I have  AX=B , I want to find X , A is 3*3 matrix  <.9,-.3,-.6>|<-.2,.6,-.4>|<-.4,-.4,.8>   B=matrix(3, 1, [600, 600, 600]) ? I using  linearslove, but it is not work.

Hello, I am trying to do a fourier transfrom using the package < DiscreteTransfroms >.

The function is an gaussian function for now,

Here is the code I tried

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

restart

with(DiscreteTransform):

> X := Vector(1000, proc (k) options operator, arrow; (1/200)*k-5/2 end proc);
> Y := Vector(1000, proc (k) options operator, arrow; evalf(exp(-10*((1/100)*k-5)^2)) end proc);

> X2, Y2 := FourierTransform(X, Y);
Vector[column](%id = 18446744080244879358),

Vector[column](%id = 18446744080244879478)
> plot(X2, Re(Y2));

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

The program returns two vector, X2 and Y2 who are supposed to be the fourier transforme of a gaussian so.. a gausian but when I plot the result X2 on the horizontal and Y2 on vertical, the graph doesn't resemble a gaussian function or any function at all.

Please help!!

Alex

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.

My attempt to export array data using the Browse>Export option when looking at the data fails to work as needed.

The exported data always start in cell A1 of the Excel worksheet even a different starting cell is entered into the Matrix Browser export window.

In summary, my experience is that the Matrix Browser matrix export in Maple 16 and Maple 17 ignores the information about the intended starting destination cell in the Excel file.

Does anyone find that the Matrix Browser exports to a starting Microsoft Excel cell other than A1? 

This exporting with Matrix Browser worked find in Maple 15.  It has been broken in Maple 16 and 17, including Maple 17.02.

Thanks.

so i am trying to plot some limit cycles and it won't plot. I don't know what's wrong.

> restart;
with (DEtools);
> L := -4.80; MU := 0.1e-1;
DE13 := {(D(x))(t) =
y(t)*(1-y(t)*y(t))+MU*x(t)*(x(t)*x(t)-3*y(t)*y(t)-L), (D(y))(t) = -x(t)*(1-2*x(t)*x(t))+MU*y(t)*(x(t)*x(t)-3*y(t)*y(t)-L)};
DEplot(DE13, [x(t), y(t)], t = 0 .. 20, [[x(0) = 0.1e-1, y(0) = .99], [x(0) = -.1, y(0) = -.9], [x(0) = 1.1, y(0) = 0], [x(0) = 0, y(0) = .2], [x(0) = 0, y(0) = .6], [x(0) = .6, y(0) = 0], [x(0) = .75, y(0) = 1], [x(0) = .1, y(0) = .1], [x(0) = .5, y(0) = 1.0], [x(0) = -.5, y(0) = 1], [x(0) = .5, y(0) = -1], [x(0) = -.5, y(0) = -1], [x(0) = -0.1e-1, y(0) = .99], [x(0) = 0.1e-1, y(0) = -.99], [x(0) = -0.1e-1, y(0) = -.99], [x(0) = .5, y(0) = -1], [x(0) = -.5, y(0) = -1], [x(0) = 0.1e-1, y(0) = .9]], stepsize = 0.1e-1, scene = [x(t), y(t)], title = "Phaseplane 3 Prime Plot", linecolor = black, thickness = 1);

I am to program a computation on maple.
But I get the message :

Error, unable to match delimiters

followed by the full sequence of command.

How can I do to fix this problem?

I want my graphic to be like this (file: Hansevi's Graphic.mw ):

instead of this (file: My graphic.mw):

Does anyone know why both graphics are different?

*The graphics were supposed to be similar -_-

Good afternoon sir.

I request your kind support to the above cited question.

With thanks & Regards

M.Anand

Assistant Professor in Mathematics

SR International Institute of Technology,

Hyderabad, Andhra Pradesh, INDIA.

restart;

Lfh := proc(numoflevel, h, fx, var)

if numoflevel = 1 then

        hello := 0;

        for i from 1 to nops(var) do

                hello := hello + diff(h[i], var[i])*fx[i];

        od;

        return hello;

else

        hello := 0;

        for i from 1 to nops(var) do

                hello := hello + diff(Lfh(numoflevel-1, h, fx, var), var[i])*fx[i];

        od;

        return hello;

end if;

end proc:

f:=[x3-x2^3,-x2,x1^2-x3];

g:=[0,-1,1];

h:=[x1,0,0];

variables := [x1,x2,x3];

Lf1h := Lfh(1,h,f,variables);

Lgf1h := Lfh(1,[seq(Lf1h,n=1..nops(variables))],g,variables);

Lf2h := Lfh(2,h,f,variables);

Lgf2h := Lfh(1,[seq(Lf2h,n=1..nops(variables))],g,variables);

Lf3h := Lfh(3,h,f,variables);

Lgf3h := Lfh(1,[seq(Lf3h,n=1..nops(variables))],g,variables);

i doubt that my book example is wrong

it said relative rank is 2, but i can not find any zero when Lgf3h

if i continue to Lgf4h, Lgf5h, Lgf6h...etc, still no zero what does it mean?

Example

Diff(x1,t) = x2^2,
Diff(x2,t) = u,
y = x1

result in

Diff(y,t$2)-4*u^2*Diff(y,t) = 0
2*u*x2 - Diff(y,t^2) = 0
x1 - y = 0

is it possible a general and generic method to apply some steps on system below

Diff(u1, t) = u1*u2-u1*u3,
Diff(u2, t) = -u1*u2+u2*u3-(1/2)*(conjugate(u2-u3)*(u1*u2-u1*u3)+conjugate(u1)*(-u1*u2+u2*u3)-conjugate(u1)*(u1*u3-u2*u3))/conjugate(u1),
Diff(u3, t) = u1*u3-u2*u3+(1/2)*(conjugate(u2-u3)*(u1*u2-u1*u3)+conjugate(u1)*(-u1*u2+u2*u3)-conjugate(u1)*(u1*u3-u2*u3))/conjugate(u1),
y = u1

Here is the question:Prior to this question I was given f(z)=z^2+1, N(z)=(z^2-1)/(2z), T(z)=z-I/z+I such that T(N^k (z))=(T(z))^2^k. And L is a set of number on the real axis. Now the question is that given we have two regions of the complext plane as follow:

R₊ = {z : N^k{z) -> i as k -> ∞}; R_- = {z : N^k(z) -> -i as k -> ∞}.

Draw a diagram to illustrate these regions, the line L and the roots i and -i. We call R₊ the basin of attractionfor the root +i, and similarly R_-is the basin of attraction for the root -i.

 Show that if z is on the set L (the common boundary of the two regions R₊ and R_{_}, then N^k(z) stays on L for all values of k. (This is easy once you identify what L is.) So in this case iteration does not produce a root at all.

So basically my problem is that the fact I'm not very familar with the commands to draw such diagram, and I don't know much about Newton's method to compute complex roots. It would be appreciated if anyone can help me how to get start with the question. Thanks.

when a user liked a particular photo for the first launch of a mobile app, and the time different where by the event started and stoped was k seconds which is at constant (does not change). At this constant thus triggered the like application operation or function that was connected to the application server. Then later the user unlike the photo and the same event and operation was carried out at of (k + 2) seconds. The time variation that was carried out during the sequential or running loop event carried out by 2 seconds.

check if this derived equation is correct first f(X_n) = e_n(X_nⁿ  + X_n⁰ + 1) . if its correct do the second question.

what is the rate at which the operation will be finish given an equation f(X_n) = e_n(X_nⁿ  + X_n⁰ + 1) 

The dot product of vector v and u= ||v|| ||u||fosters also vector u*v=XvXu+Yv=A, v=<2,5>, u=<4,9>.