Hi every one,

Q1:

I tried to get the max $ min of a following function:

l:=1:alpha:=1:b:=100:k:=20:

eq1 := (alpha+(l+alpha)*u+alpha*k*u^2)*a =
u*(alpha+(l+alpha)*u+alpha*k*u^2)*(1+l*alpha*b/((alpha+(l+alpha)*u+alpha*k*u^2))):

I did this code but it seems it didnt work for this equation

maximize(eq, u=1..12, location);

minimize(eq, u=1..12, location);

Also, I think about solving the cubic i feel i'm so close to the solve but couldn't

factor((rhs-lhs)(eq1));

eq:=collect(%,u);

Q:=(a,u)->eq;sol:=evalf(solve(Q(a,u),u)): S:=array([],1..3): S[1]:=sol[1]:S[2]:=sol[2]:S[3]:=sol[3]:

Q2:

the same thing wanted to get the maximum and the minimum of the function v

here the code

restart;
eq1:=(alpha+(l+alpha)*u+alpha*k*u^2)*a=
u*(alpha+(l+alpha)*u+alpha*k*u^2)*(1+l*alpha*b/((alpha+(l+alpha)*u+alpha*k*u^2)));
eq2:=v=alpha*b*(1+u+k*u^2)/(alpha+(l+alpha)*u+alpha*k*u^2);
factor((rhs-lhs)(eq1));
eq1:=collect(%,u);
params:={l=10,alpha=0.5,b=100,k=20};
U:=[solve(eval(eq1,params),u)]; #3 solutions for u
#plots:-complexplot(U,a=0..20,style=point); #plot in the complex u-plane
vua:=eval(solve(eq2,v),params): #v expressed in terms of u and a
V:=eval~(vua,u=~U): #the 3 solutions for v in terms of a

## PLOT the function V
plot(V,a=0..75,v=0..100,color=black,labels=[a,v],axes=boxed,numpoints=90,linestyle=1,font=[1,1,18],thickness=2,tickmarks=[4,4],view=[0..65,25..100]);

I do appricaited any advises

function [y g] = ques5
% using Simpson Formula to approximate the integration
% input:
% f(x): [a b]


end
% use Euler formula to compute function y
for i = 1:N
    if i ==1

legend('numerical y','exact y','numerical g','exact g')

function g = f2g(a,b)
% f(x) = x
g = (b-a)/6*(a + 4*(a+b)/2 + b);


ican use matlab to solve this problem but not maple
please help

http://homepages.lboro.ac.uk/~makk/MathRev_Lie.pdf

ode1 := Diff(f(x),x$2)+2*Diff(f(x),x)+f(x);
with(DEtools):
with(PDETools):
gen1 := symgen(ode1);
with(PDEtools):
DepVars := ([f])(t);
NewVars := ([g])(r);
SymmetryTransformation(gen1, DepVars, NewVars);

Error, invalid input: too many and/or wrong type of arguments passed to PDEtools:-SymmetryTransformation; first unused argument is [_xi = -x, _eta = f*x]

generator1 := rhs(sym1[3][1])*Diff(g, x)+ rhs(sym1[3][2])*Diff(g, b)

what is X1 and X2 so that [X1, X2] = X1*X2 - X2*X1 = (X1(e2)-X2(e1))*Diff(g, z) ?

is it possible to use lie group to represent a differential equation, and convert this group back to differential equation ? how do it do?

how to find symmetry z + 2*t*a, when you do not know before taylor calcaulation?

fza := z + 2*t*a;
fza := x;
fza := z + subs(a=0, diff(fza,a))*a;

Two questions:

The algortihms that Groebner[Basis] uses at each step computes some "tentative" or "pseudo-basis". The "tentative" basis is not a Groebner basis but it is in the ideal generated by the original system of polynomial eq.

1) Is this correct ? Provided this is correct, then

2) How can one retrive the last "tentative" basis?
 If I just use timelimit I can abort the computations but how can one retrive the last computation?

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?