I was given that solitary initial state to see how it will deform as time goes on. am struggling to get my code so that I can get video frames. please help on how I can generate my code

u_o(x)= a₀x²(1-x)² for x (less than or equal to) x (less than or equal) 1

u₀(x)    = 0 for x > 1

the video clips will be representing the function u(*,t) :x to u(x,t)

for a sequence of choices of t such as t=0; t=0,5...t=3

 “Error, too many levels of recursion” ,but there’s just a single expression.

How can this happen?

The help documents read，

“

 The function unames returns an expression sequence consisting of all the active names in the current Maple session which are ``unassigned names''.

“

But what unames() returns is obviously not the contents one expects:

Hello,

It is not a very important question... but i like when i have a nice worksheet.

I would like to change the color of the maple input. Unfortunately, i manage to change the color for the inputs already created but not for the new maple input.

How can i change the color for the new maple input i want to create ?

Thank you for your help.

How can I solve raychaudhuri equations numerically using GRtensor?

g := (-12*y^2+1)/(4*y^2+1)^3 * ln(abs(-1+2^(1-x-I*y)));

Compute Int(g, [x=1/2 .. infinity, y=0 .. infinity]).

PS: this stems from How to calculate hard integral?, 
but it is not related to the RH.

Hi. I want to generate a Julia set, and teh first instruction in the demo, applications fractal is

GenerateComplex(a,b,c,d,e)

I have Maple13 and I can not find such instruction. Could you tell me where can I get it, or how to define it?

Thanks

Help Please! :)
As it is seen in the picture, I can not integrate the power series. In contrast, the differentiation works!
what is wrong?

 support type error when plot, and moreover how to animate this plot

restart;
with(ExcelTools):
with(ListTools):
with(DynamicSystems):
filename := "0257.HK";
open3 := Import(cat(cat("C://Temp//HK//Bank//",filename),".xls"), filename, "B2:B100");
high3 := Import(cat(cat("C://Temp//HK//Bank//",filename),".xls"), filename, "C2:C100");
low3 := Import(cat(cat("C://Temp//HK//Bank//",filename),".xls"), filename, "D2:D100");
close3 := Import(cat(cat("C://Temp//HK//Bank//",filename),".xls"), filename, "E2:E100");
with(CurveFitting):
n := 31;
f := Vector(n);
f2 := Vector(n);
open2 := Vector(n);high2 := Vector(n);gain2 := Vector(n);algebra2 := Vector(n);creative2 := Vector(n);creative3 := Vector(n);
upper2 := Vector(n);lower2 := Vector(n);upperloweratio := Vector(n);
deltaopen2 := Vector(n); deltahigh2 := Vector(n); deltalow2 := Vector(n); deltaclose2 := Vector(n);
logn := Vector(n);
for i from 0 to n-4 do
open2[i+1] := PolynomialInterpolation([[0,open3[n-i][1]],[1,open3[n-(i+1)][1]],[2,open3[n-(i+2)][1]],[4,open3[n-(i+3)][1]]],t):
high2[i+1] := PolynomialInterpolation([[0,high3[n-i][1]],[1,high3[n-(i+1)][1]],[2,high3[n-(i+2)][1]],[4,high3[n-(i+3)][1]]],t):
low2[i+1] := PolynomialInterpolation([[0,low3[n-i][1]],[1,low3[n-(i+1)][1]],[2,low3[n-(i+2)][1]],[4,low3[n-(i+3)][1]]],t):
if (close3[i+1][1]/close3[i+2][1]-1) < 0 then
gain2[i+1] := -1*round(100*abs(close3[i+1][1]/close3[i+2][1]-1)):
else
gain2[i+1] := round(abs(100*(close3[i+1][1]/close3[i+2][1]-1))):
end if;
od;
n := 31;
newclose := Vector(n);
for j from 0 to n-4 do
for i from 0 to n-4 do
x1 := close3[i+1];
y1 := close3[i+1];
newclose[i+1] := subs(y=y1, subs(x=x1, (1/2)*(-y+sqrt(-3*y^2-4*y*x))/y))
od;
close3 := newclose;
plot(close3(x), x=1..31);
od;

Hi, I have a bivariate generating function that looks like this

where x is enumerating by binary strings by length and k is counting a number patterns in the string. I would like to convert the series into partial fractions. Convert[parfrac] only seems to work when k is given a value, which I did for several small small choices, from which I guessed the general partial fraction decomposition. Would someone have an idea on how to extract the partial fractions directly in terms of x and k?

Thanks,

best, Luke

Hello all!

I have to solve 1D Heat equation with Neumann B.C. using implicit scheme.

I have: 

I have my code in Maple for the solution of this problem using explicit sceme for Neumann B.C.. And I also have the solution of the problem using implicit scheme(but for Dirichle B.C.).

implicit_method_Dirichle_B.C..mws
explicit_method_Neumann_B.C..mws

I know that my Neumann B.C. for implicit scheme will be written like this.
I determined the ghost points and then got the final view of the B.Cs.:

But I can not imagine how I should put my Neumann  B.C. for implicit scheme in the code. 

Please, help me! I will be very grateful!

Hi,

Can anyone help me to solve this system of equations in Maple?

solve({
(-(1*Rfd*(1-(Ladssec/Lfd)+((((1.007033 +1*(Laqssec+Ll))*Ladssec)/(((Ra+0.04527646)^2+(1.007033 +1*(Ladssec+Ll))*(1.007033 +1*(Laqssec+Ll)))*Lfd)) *Ladssec)))/(Lfd))=-0.0069
,(-(1*Rfd*(-(Ladssec/Lfd)+((((1.007033 +1*(Laqssec+Ll))*Ladssec)/(((Ra+0.04527646)^2+(1.007033 +1*(Ladssec+Ll))*(1.007033 +1*(Laqssec+Ll)))*L1d)) *Ladssec)))/(Lfd))=0.002689
,(-(1*Rfd*(-((Ra+0.04527646)/((Ra+0.04527646)^2+(1.007033 +1*(Ladssec+Ll))*(1.007033 +1*(Laqssec+Ll))))*(Laqssec/L1q))*Ladssec)/(Lfd))=0.00002647
,(-(1*Rfd*(-((Ra+0.04527646)/((Ra+0.04527646)^2+(1.007033 +1*(Ladssec+Ll))*(1.007033 +1*(Laqssec+Ll))))*(Laqssec/L2q)) *Ladssec)/(Lfd))=-0.00001362

,(-(1*R1d*(-(Ladssec/Lfd)+((((1.007033 +1*(Laqssec+Ll))*Ladssec)/(((Ra+0.04527646)^2+(1.007033 +1*(Ladssec+Ll))*(1.007033 +1*(Laqssec+Ll)))*Lfd))*Ladssec)))/(L1d))=0.1052
,(-(1*R1d*(1-(Ladssec/Lfd)+((((1.007033 +1*(Laqssec+Ll))*Ladssec)/(((Ra+0.04527646)^2+(1.007033 +1*(Ladssec+Ll))*(1.007033 +1*(Laqssec+Ll)))*L1d)) *Ladssec)))/(L1d))=-0.2585
,(-(1*R1d*(-((Ra+0.04527646)/((Ra+0.04527646)^2+(1.007033 +1*(Ladssec+Ll))*(1.007033 +1*(Laqssec+Ll))))*(Laqssec/L1q))*Ladssec)/(L1d))=0.0009122
,(-(1*R1d*(-((Ra+0.04527646)/((Ra+0.04527646)^2+(1.007033 +1*(Ladssec+Ll))*(1.007033 +1*(Laqssec+Ll))))*(Laqssec/L2q)) *Ladssec)/(L1d))=-0.0005093

,(-(1*R1q*(((Ra+0.04527646)/((Ra+0.04527646)^2+(1.007033 +1*(Ladssec+Ll))*(1.007033 +1*(Laqssec+Ll))))*(Ladssec/Lfd))*Laqssec)/(L1q))=-0.000292
,(-(1*R1q*(-((Ra+0.04527646)/((Ra+0.04527646)^2+(1.007033 +1*(Ladssec+Ll))*(1.007033 +1*(Laqssec+Ll))))*(Ladssec/L1d))*Laqssec)/(L1q))=0.0008507
,(-(1*R1q*(1-(Laqssec/L1q)+((((1.007033 +1*(Ladssec+Ll))*Laqssec)/(((Ra+0.04527646)^2+(1.007033 +1*(Ladssec+Ll))*(1.007033 +1*(Laqssec+Ll)))*L1q))*Laqssec)))/(L1q))=-0.04423
,(-(1*R1q*(-(Laqssec/L2q)+((((1.007033 +1*(Ladssec+Ll))*Laqssec)/(((Ra+0.04527646)^2+(1.007033 +1*(Ladssec+Ll))*(1.007033 +1*(Laqssec+Ll)))*L2q))*Laqssec)))/(L1q))=0.0003831

,(-(1*R2q*(((Ra+0.04527646)/((Ra+0.04527646)^2+(1.007033 +1*(Ladssec+Ll))*(1.007033 +1*(Laqssec+Ll))))*(Ladssec/Lfd))*Laqssec)/(L2q))=-0.001785
,(-(1*R2q*(-((Ra+0.04527646)/((Ra+0.04527646)^2+(1.007033 +1*(Ladssec+Ll))*(1.007033 +1*(Laqssec+Ll))))*(Ladssec/L1d))*Laqssec)/(L2q))=-0.03134
,(-(1*R2q*(-(Laqssec/L1q)+((((1.007033 +1*(Ladssec+Ll))*Laqssec)/(((Ra+0.04527646)^2+(1.007033 +1*(Ladssec+Ll))*(1.007033 +1*(Laqssec+Ll)))*L1q))*Laqssec)))/(L2q))=4.239
,(-(1*R2q*(1-(Laqssec/L2q)+((((1.007033 +1*(Ladssec+Ll))*Laqssec)/(((Ra+0.04527646)^2+(1.007033 +1*(Ladssec+Ll))*(1.007033 +1*(Laqssec+Ll)))*L2q))*Laqssec)))/(L2q))=-4.868

}, {L1d, L1q, Ladssec, Laqssec, Lfd, R1d, R1q, Rfd, Ra, Ll},useassumptions)

assuming Lfd ~= 0.1632, R1d~=0.0269, L1d~=0.0750, L1q~=-0.1350, Ladssec~=0.0500, Laqssec~= -0.1500, Rfd~=0.0016, R1q~= 0.0215, Ra~=0.005, Ll~=0.15

which theory can explain the interaction of composition of group for combination of composition of group

if succeed to search a list of groups, what is the next step research them?

Hi,

I have a linear system to solve.

mm:=proc(a,x,h,i)
local A,Z1,Z2,Z,F,result;  # to declare the local variable
A:=array(1..2,1..2,[[1,1],[a,a+h]]);
Z1:=evalf(int(1/(abs(y-x)+.000000001),y=a..a+h));
Z2:=evalf(int(y/(abs(y-x)+.000000001),y=a..a+h));
Z:=array([Z1,Z2]);
F:=evalm(inverse(A)&*Z);
result:=F[i]
end:

My questions: 

1) My exact Z1 is Z1:=evalf(int(1/(abs(y-x)),y=a..a+h)); but I ask if can I put

Z1:=evalf(int(1/(abs(y-x)+.000000001),y=a..a+h));

the same for Z2.

2) Can I writte in a simple form the vector Z.  Because, later, il have a second system contains Z1,Z2, Z3, Z4,Z5.  The difference between Z1 and Z2 is the variable "y" added in the integral of Z2.

Many thinks.

I wish to  plot 2D animate for the soltion of this equation here is the code

restart;
with(PDEtools):
with(ArrayTools):
with(plots):

f:=u->sech(u):
g:=v->sech(v):
h:=1/10:
N:=20:
M:=20:
V:=x->x^2:
psi:=Array(0..N/h+1,0..M/h+1):
for i from 0 to N/h do
psi[i,0]:=evalf(f(i*h)):
od:
for j from 0 to M/h do
psi[0,j]:=evalf(g(j*h)):
od:
for i from 1 to N/h do
for j from 1 to M/h do
psi[i,j]:=-psi[i-1, j-1]+(1-(1/8)*h^2*V((1/2)*h*(j-i-1)))*psi[i, j-1]+(1-(1/8)*h^2*V((1/2)*h*(j- i+1)))*psi[i-1,j]:
od:
od:
ls:=[seq([seq([i*h,j*h,psi[i,j]],i=0..N/h)],j=0..M/h)]:

surfdata((ls),axes=boxed,labels=[`u`,`v`,`psi(u,v)`],shading=zhue,style=patchcontour);