with(combinat):
list1 := permute([a, b, a, b, a, b], 3);
list1a := subs(b=1,subs(a=0, list1));
list1a := permute([seq(seq(k,k=0..1),k2=1..3)], 3);
list2 := permute([a, b, c, d, e, f, g, h, a, b, c, d, e, f, g, h, a, b, c, d, e, f, g, h], 3);
list3 := subs(h=18,subs(g=17,subs(f=16,subs(e=15,subs(d=14,subs(c=13,subs(b=12,subs(a=11,list2))))))));
list3 := permute([seq(seq(k,k=11..18),k2=1..3)], 3);
list5 := Matrix(nops(list1a)*nops(list3), 1);
count := 1;
for n from 1 to nops(list3) do
temp1 := subs(1=list1a[1],list3[n]);
for k from 11 to nops(list1a)+10 do
temp1 := subs(k=list1a[k-10],temp1);
od;
list5[count] := temp1;
count := count + 1;
od;
list6 := permute(list5, 2);

Error, (in combinat:-permute) 1st argument must be a list, set or a non-negative integer

I have a bit of an issue.

I'm trying to solve this:

eq1:= exp(-lambda)*exp(k)/factorial(k) > .95

lambda:=2

solve(eq1,k)

Warning, solutions may have been lost

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.

Consider the differential equation zZ'' + Z' + a²Z = 0,  where Z = Z(z).  Using the change of variables x = \sqrt{z/b}with b a constant,  obtain the differential equation Z'' + (1/x)Z' + c²Z = 0, where Z = Z(x) and c = 2a \sqrt{b}.

I tried Maple help and it offers the dchange command, and what I have tried is shown below;

with(PDEtools):

DE:= ...

tr:= {z = x²b}

dchange(tr, DE)

This did not return anything however.  I am thinking I need to specify that b is a constant, however, I am a little unsure on how to do this. Is the above the correct way to proceed?  I don't see how I have specified anywhere that in the final PDE, I require Z=Z(x).

Thanks for any help.  This is my first post here, so apologies for the typesetting. If there is inbuilt latex, I will use it next time.

Hi;

Trying to find the maximum values of the implicit function.. I tree the following commands.

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

(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)));

with(plots):

implicitplot((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))),a=0..100,u=0..20,numpoints=9000000,color=black,axes=boxed,font=[1,1,18],thickness=2,tickmarks=[3,3],view=[0..12,0..12],labels=[a,u]);

maximize((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))), N=0.5..1, location);

Any help I'll appriciated 

hi guys i want to solve this equation with maple please help me

eq[1]:=0.223569c_1+2.35589c_2*c_1^2+0.002356c_1*c_2^2;

eq[2]:=1.277899c_1*c_3-2.350023c_2*c_3^2+7.5856c_3*c_2^2;

eq[3]:=3.225989c_1^2+-2.35589c_3*c_1^2-7.28356c_3*c_2^3;

i want solve those equations with newton method

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.

Hello every one,

I am facing problem with for loop. Please have a look.

let say I have T[0]:= 99 ; X[0]:=0; f(x):=108 + 2 x

  for i from 1 to 2 do f(x[i-1] ):

                              y[i]:=x[i-1] +1 ;

if f(y[i])<f(x[i-1]) then x[i]:=y[i] elif 0.98< e^0.99/T[0]  then x[i]:=y[i] else x[i]:=x [i-1]

end if;

end do;

after I got the answer from this loop, now I want to continue the loop for T[1]:= 0.1 T[0], 

 how  to do another loop by using the same  information but in different value of T? in other words, the only change in the looping is the value T so that the next loop I have the set of answer.

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 have been struggling (reading Ore/Weyl Algebra documentation) to understand how to input a PDE system with polynomial coeff. in Weyl algebra notation so I can compute a Groebner basis for it. I would be very grateful if someone could  show, using the simple example below, which differential operators in Ore_algebra[diff_algebra] should one declare to express the system in Weyl algebra notation. The systems I'm working are more complicated but all have many dependent variables, f and g functions in this example:

pdesys:= [ x*diff( f(x,y,z),x)- z*diff( g(x,y,z),y) = 0, (x^2-y)*diff( f(x,y,z),z)- y*diff( g(x,y,z),z) = 0 ]

I am struggling to get a single output value for my piece wise function. I am trying to plot this piecewise function. All the B-values and L value are previously defined.

My end goal is to plot the function.

Unfortunately, I cannot get the piecewise function to output a single value. I have tried s(2) which gives me the entire piecewise function with the correct numbers plugged in.

Here is the file in its full form.

damshw9.mw

I am trying to expand  a rational function that is in the form:

P(z) = (1 + z⁻¹)² (1 + z)²· (r1z + r0 + r1z⁻¹)

to the form:

P(z) = r1z³ + (4r1 + r0)z² + (7r1 + 4r0)z + (8r1 + 6r0)

+(7r1 + 4r0)z⁻¹ + (4r1 + r0)z⁻² + r1z⁻³

Can someone show me how to do this please?

I am getting an error dialog when I try to Browse an array I created programmatically.  The title of the dialog is RTable Browse Error with and error that says Empty RTable Structure. 

Now the data is 0..20 x 0..1 Array Datatype :anything Storage: rectangular

Order: Fortran_Order. 

If I do a showall, I can see the data, I just cannot browse it.  Any help would be appreciated.

Code is:

for i from 1 to N do
 t:=h*i:
 X[i,0]:=x1+h*(-x1)*x2+x1:
 X[i,1]:=x2+h*x1*x2-x2:
 x1:=X[i,0]:
 x2:=X[i,1]:
 print('Iteration',i,'AtTime',t,x1,x2)
end do;
print(X);

Thanks

Melvin

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?

Hi everyone, I'm trying to find the value of f but I always have this error message : Error, (in BesselJ) too many levels of recursion.

Here is the function:

f(t)=BesselJ(0, t)+int(BesselJ(0, x)*f(t-x), x = 0 .. t)

Is there any way that I can isolate my f without having to rewrite the equation using laplace transform properties?

Thanks,

Frank