Maple 15 Questions and Posts

These are Posts and Questions associated with the product, Maple 15

yVal := 0.01

xVal := 0.01

p8 := plot([fdiff(('rhs')((pds:-value(f(x, y)))(x, yVal)[3]), [y, y], x = z)], z = 0 .. 20, color = [red])

I cant seem to plot the graph for yVal, but if I were to switch it around to (xVal,y), the graph works perfectly. How should I input the code such that I can obtain the f''(x,0) graph. 

Any help will be greatly appreciated :) thanks

fyp3.mw

hello,

Given f(x)=2x^2-4x, how do I define and plot the functions f(x), 2f(x), 4f(x), -f(x) and -4f(x) in maple 15?

Thank You.

 

The distance from the point to the surface easily calculated using the NLPSolve of Optimization package. If the point is not special, we will find for it a point on the surface, the distance between these two points is the shortest between the selected point and the surface.
Two examples:  the implicit surface and the parametric surface.
To test, we restore the normals from the  calculated  points (red) by using analytical equations.
DISTANCE_TO_SURFACE.mw

I can't seem to plot the second derivative graph for f'' versus y. Is it possible to use the fdiff command twice in a line?

Open to all opinions. Any help would be greatly appreciated :)

fyp3.mw

The Mobius strip  Mobius_strip_rolling.mw

Variant:


The line and the curve on the surface.

 

Hi everyone.

I'm using Maple 15 to try to give random values to a matrix with some columns fixed, the problem is that it has to be unitary, so I managed to create a system of equations (with 36 linear equations and 24 variables) which should give me the information i require.

I have been trying with fsolve, etc. but i haven't been able to solve it yet, as I usually get the error: "Error, (in fsolve) number of equations, 36, does not match number of variables, 24"

Is there any way to find out which equations are neccesary and get rid of the rest of them? Or can Maple give ramdon values to some of the variables and calculate the others?

Thanks in advantage!

Hi!

I have seen th following procedure to compute the image of the points of [0,1] under the so called Peano space-filling curve (sorry, I have to pasted the code in "text plane" mode):

P[0] := (x, y) -> ((1/3)*y, (1/3)*x);

P[1] := (x, y) -> (-(1/3)*x+1/3, (1/3)*y+1/3); 

P[2] := (x, y) -> ((1/3)*x, (1/3)*y+2/3); 

P[3] := (x, y) -> ((1/3)*x+1/3, -(1/3)*y+1);

P[4] := (x, y) -> (2/3-(1/3)*y, 2/3-(1/3)*x); 

P[5] := (x, y) -> ( (1/3)*x+1/3, 1/3-(1/3)*y));

 P[6] := (x, y) -> (1/3)*x+2/3, (1/3)*y);

P[7] := (x, y) -> (-(1/3)*x+1, (1/3)*y+1/3);

P[8] := (x, y) -> ((1/3)*x+2/3, (1/3)*y+2/3);

peano := proc (t::numeric, depth::integer)

local q, r; global P;

if depth = 0 then return 0, 0 end if;

q := floor(9*t); r := 9*t-q;

return P[q](peano(r, depth-1))

end proc;

 

Now, I need to use the procedure "peanofun" as a function. For instance, if we define f:=(x,y)->x+y, I need to use (plot, compute, etc) for instance, the function f(peanofun(t,5))

Can you help me with this issue, please?

Many thanks for your time!

 

 

can anyone run my code?
please do it and send it for me.
thanks.

restart;
with(plottools): with(LinearAlgebra): with(plots):
ode := `assuming`([diff(Y(X), `$`(X, 2))+2*alpha*(diff(Y(X), X))+beta^2*Y(X) = 0], [alpha >= 0, beta >= 0, alpha+beta > 0]):
F := unapply(solve(subs({X = x, Y(X) = y, diff(Y(X), X) = yp, diff(Y(X), `$`(X, 2)) = yz}, ode), yz), x, y, yp):
Fp := unapply(solve(subs({X = x, Y(X) = y, diff(Y(X), X) = yp, diff(Y(X), `$`(X, 2)) = yz, diff(Y(X), `$`(X, 3)) = yt}, diff(ode, X)), yt), x, y, yp, yz):
Ni := seq(i, i = 0 .. 9), 15, 20:
for ni in Ni do
print(ni);
st := time[real]();
f[0, ni] := F(x[0], y[0, ni], yp[0, ni]);
fp[0, ni] := Fp(x[0], y[0, ni], yp[0, ni], f[0, ni]);
f[1, ni] := F(x[1], y[1, ni], yp[1, ni]);
fp[1, ni] := Fp(x[1], y[1, ni], yp[1, ni], f[1, ni]);
y[2, 0] := y[0, ni]+2*h*yp[0, ni]+(6/5)*f[0, ni]*h^2+(4/15)*fp[0, ni]*h^3+(4/5)*f[1, ni]*h^2+(4/15)*fp[1, ni]*h^3;
yp[2, 0] := yp[0, ni]+2*f[0, ni]*h+(2/3)*fp[0, ni]*h^2+(4/3)*fp[1, ni]*h^2;
for j to ni do
f[2, j-1] := F(x[2], y[2, j-1], yp[2, j-1]);
fp[2, j-1] := Fp(x[2], y[2, j-1], yp[2, j-1], f[2, j-1]);
y[2, j] := y[1, ni]+h*yp[1, ni]+(7/20)*f[1, ni]*h^2+(1/20)*fp[1, ni]*h^3+(3/20)*f[2, j-1]*h^2-(1/30)*fp[2, j-1]*h^3;
yp[2, j] := yp[1, ni]+(1/2)*f[1, ni]*h+(1/12)*fp[1, ni]*h^2+(1/2)*f[2, j-1]*h-(1/12)*fp[2, j-1]*h^2;
end do:
Ms := Matrix(4, 4); Ms[1, 3] := 1; Ms[2, 4] := 1;
y[2, ni] := collect(algsubs(h*alpha = H1, expand(algsubs(h*beta = H2, expand(y[2, ni])))), {y[0, ni], y[1, ni], yp[0, ni], yp[1, ni]});
Ms[3, 1] := coeff(y[2, ni], y[0, ni]);
Ms[3, 2] := coeff(y[2, ni], yp[0, ni])/h;
Ms[3, 3] := coeff(y[2, ni], y[1, ni]);
Ms[3, 4] := coeff(y[2, ni], yp[1, ni])/h;
hyp[2, ni] := collect(algsubs(h*alpha = H1, expand(algsubs(h*beta = H2, expand(h*yp[2, ni])))), {y[0, ni], y[1, ni], yp[0, ni], yp[1, ni]});
Ms[4, 1] := coeff(hyp[2, ni], y[0, ni]);
Ms[4, 2] := coeff(hyp[2, ni], yp[0, ni])/h;
Ms[4, 3] := coeff(hyp[2, ni], y[1, ni]);
Ms[4, 4] := coeff(hyp[2, ni], yp[1, ni])/h;
sol := Eigenvalues(Ms);
print(time[real]()-st);
st := time[real]();
SR[ni, 1] := implicitplot(max(seq(abs(sol[ii]), ii = 1 .. numelems(sol))) <= 1, H1 = 0 .. 3, H2 = 0 .. 3, filledregions, gridrefine = 3, axes = Boxed, view = [-2 .. 3, -3 .. 3], labels = [H[1], H[2]], labeldirections = ["horizontal", "vertical"]);
SR[ni, 2] := implicitplot(max(seq(abs(sol[ii]), ii = 1 .. numelems(sol))) <= 1, H1 = -2 .. 3, H2 = -2 .. 3, gridrefine = 3, axes = Boxed, view = [-2 .. 3, -3 .. 3], labels = [H[1], H[2]], labeldirections = ["horizontal", "vertical"]);
print(time[real]()-st);
end do;
for i in Ni do
i;
display({SR[i, 1], SR[i, 2], line([-1, 0], [3, 0], color = red, linestyle = dash), line([0, -3], [0, 3], color = red, linestyle = dash)});
end do;
display({seq(SR[i, 2], i = 0 .. Ni)});

    My profile picture was formerly animation and looked like this: 


  It would be interesting to paint a triangle on a sphere. How can I do that?

Parametric equation of second-order curve in 3d. Draghilev method.
PLAN_CURVE_3d_1.mw
Examples:
x1^2+x1*x3+13*x2^2+x3-1=0;
x1+x2+x3=0;


 x1^2+0.1*x2^2+x3^2-9=0;
 x1+3*x3+1=0;


 x1^2-0.1*x2^2+x3^2-9=0;
 x1+3*x3+1=0;

Parametric equation of a circle in 3d by three points. Draghilev method.

CIRCLE_3_POINTS_geom3d_2.mw

One way is coloring a surface on both sides. We build equidistant surface with very small radius and stain the equidistant surface in color different from the color of the original surface.
Examples coloring of surfaces on both sides.  Radius equal to abs (0.0001).
x3-0.5*exp(sin(x1+2.5*x2+x3))=0;
(x1^2+x2^2-0.4)^2+(x3+sin(x1*x2+x3))^4-0.1=0;

2_COLORS.mw


Dears

Hope everything fine with you. I want to solve the attached problem by numarically and want to plot it but failed. Please see the attachement and correct it. I am waiting your positive respone. 

System_of_ODEs.mw

With my best regards and sincerely.

Muhammad Usman

School of Mathematical Sciences 
Peking University, Beijing, China

Hello dear! Hope you will be fine. I want to simplify the equation (4) like equation (3) in the attached file please correct it. I shall be thankful to you.

Help.mw

With my best regards and sincerely.

Muhammad Usman

School of Mathematical Sciences 
Peking University, Beijing, China

Hello! Hope everything fine with you. I am try to solve the three-point differential by numerical method in attached file but failed. Please see the attachement and solve my problem. I am very thankful your kind effort. Please take care.

three-point.mw

With my best regards and sincerely.

Muhammad Usman

School of Mathematical Sciences 
Peking University, Beijing, China

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