Maple 15 Questions and Posts

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


Let C a square in the n-diemnsional Euclidean space. Somebody know how to divide C into 2^{n} congruent subsquares? 

For instance, for n=2 and  say C:=[0,1]x[0,1], the unit closed square, we will obtain the 2^{2}=4 subsquares [0,1/4]x[0,1/4], [0,1/4]x[1/2], [1/2,1]x[0,1/4] and [1/2,1]x[1/2,1].  

Many thanks in advance for your comments!!

Dear all

Hope everything is fine with everything. I want to draw the graph of the u(x,0.5) and T(x,0.5) for different values of alpha like alpha =0.4,0.6,0.8 while keeping Gr, R and Pr are fixed. Please solve the following problem I shall be vary thankful to you. Thanks in advance

with the following BCs

Hello dearz.

Hope you will be fine with everything. I am facing in plotting the set of points like seq(u[i,20] $ i=1..25) in the attached file. Please see the problem and fix it. I shall be vary thankful. Waiting quick and positive response. 

I want to find an approximation for a 3-dim vector y(t)=(y1,y2,y3) at multiple times t, so as to get:

y(t1)=[b0,0,0](y1(t1))^0(y2(t1))^0(y3(t1))^0 + [b0,0,1](y1(t1))^0(y2(t1))^0(y3(t1))^1 + ... + [b3,0,0](y1(t1))^3(y2(t1))^0(y3(t1))^0

y(t2)=[b0,0,0](y1(t2))^0(y2(t2))^0(y3(t2))^0 + [b0,0,1](y1(t2))^0(y2(t2))^0(y3(t2))^1 + ... + [b3,0,0](y1(t2))^3(y2(t2))^0(y3(t2))^0


So I want 20 b coefficients with quaternary-base subscripts (I belive it is called) for multiple values of t.

I want to have enough approximations to solve for the the coefficients b and then perform a Least Squares method Calculation thereafter. 

Can anyone help me please?


I hope my question is not to general. I have a polynomial of 8th order

expression:=a8(z) * x^8 + .... + a1(z) * x + a0(z) = 0

on which I am using solve/RootOf


Now when I plot it against z the solution has a jump, why?

When deriving the polynomial I could as well have used another variable instead of x above, say y. These two are related by a function...Then when I write down the 8th order polynomial in y and use RootOf/solve, then no jump occurs.

Is there a way to handle this because left of the jump the solution is not correct while right of it, it is...


Assume that we hace a set points in the plane, put X:=[a1,a2,...,aN] where each ai is given by its coordinates [x,y]. The commnad "convexhull(X)" give us the points of the convex hull of X, but how I can find to "lower-right" of these points? Please, see the attached image. I need to findo the points A,C,E and F, marked with a solid circle.

Many thanks in advances for your comments.



with(PDEtools); declare(u(x, y, z, t), U(X, Y, Z, T)); interface(showassumed = 0); assume(a > 0, p > 0); W := diff_table(u(x, y, z, t)); E := 6*W[]*W[x]+W[t]+W[x, y, z] = 0; InvE := proc (PDE) local Eq1, Eq2, Eq3, Eq4, tr1, tr2, tr3, tr4, term1, term2, term3, term4, sys1; tr1 := {t = T/a^beta, x = X/a^alpha, y = Y/a^mu, z = Z/a^nu, u(x, y, z, t) = U(X, Y, Z, T)/a^zeta}; tr2 := eval(tr1, zeta = 1); Eq1 := combine(dchange(tr2, PDE, [X, Y, Z, T, U])); Eq2 := map(lhs, PDE = Eq1); term1 := select(has, select(has, select(has, rhs(Eq2), a), beta), a); term2 := expand(rhs(Eq2)/term1); term3 := select(has, select(has, term2, a), a); sys1 := {select(has, op(1, term3), a) = 1, select(has, op(2, term3), a) = 1}; tr3 := solve(sys1, {alpha, beta, mu, nu}); tr4 := subs(tr3, tr2); print(tr3, tr4); Eq3 := dchange(tr4, PDE, [X, Y, Z, T, U]); term4 := select(has, op(1, lhs(Eq3)), a); Eq4 := expand(Eq3/term4); PDE = simplify(Eq4) end proc; InvE(E)

how to find the contour of time series data? and how to find curvature function of this contour?

The representation of the tangent plane in the form of a square with a given length of the side at any point on the surface.

The equation of the tangent plane to the surface at a given point is obtained from the condition that the tangent plane is perpendicular to the normal vector. With the aid of any auxiliary point not lying on this normal to the surface, we define the direction on the tangent plane. From the given point in this direction, we lay off segments equal to half the length of the side of our square and with the help of these segments we construct the square itself, lying on the tangent plane with the center at a given point.

An examples of constructing tangent planes at points of the same intersection line for two surfaces.

i am trying to write the differential equation 

u_{t}=u_{xx}+2u^{2}(1-u) in my maple 15. 

but it shows error,

Error, empty number and  1 additional error.


I need to solve system of 6 non linear equations. 

Down here you can see the code I wrote and at the end used to fsolve function, and it is not running. I get an error about the const 'V': Error, (in fsolve) V is in the equation, and is not solved for.

What is the right way to solve this system?

Thank you very much!



omega1 := 1.562;
omega2 := 2.449;
omega3 := 3.325;
y1 := c1*sin(omega1*t+phi1)+c2*sin(omega2*t+phi2)+c3*sin(omega3*t+phi3);



y2 := .1019*c1*sin(omega1*t+phi1)+.75*c2*sin(omega2*t+phi2)+.4608*c3*sin(omega3*t+phi3);



y3 := .407*c1*sin(omega1*t+phi1)+(0*c2)*sin(omega2*t+phi2)+1.844*c3*sin(omega3*t+phi3);
eq1 := subs(t = 0, y1) = 0;
eq2 := subs(t = 0, y2) = 0;
eq3 := subs(t = 0, y3) = 0;
eq4 := subs(t = 0, diff(y1, t)) = V;
eq5 := subs(t = 0, diff(y2, t)) = 0;
eq6 := subs(t = 0, diff(y3, t)) = 0;



eqs := [eq1, eq2, eq3, eq4, eq5, eq6];
vars := [c1, c2, c3, phi1, phi2, phi3];
fsolve({eq1, eq2, eq3, eq4, eq5, eq6}, {c1, c2, c3, phi1, phi2, phi3});


Dear, I am facing the problem for solvin the attached file. Solution obtained for L=1 successfully but i need the solution for large value of L e.g., L =10. Please find the attachement and fix the problem. I am waiting positive response. Thanks in advance.


Dear I want to solve the system of ODEs in attached file for different values of m1 i.e., 3,4,5 how I will use the value of m1 in dsolve. I am waiting your positive response thanks in advance.



Hope you would be fine. I want to solve the following PDEs by numerically for v[nf]=alpha[nf]=Ec=mu[nf]=C=1 and Pr=6.2

Eq1 := diff(u(x, t), t) = v[nf]*(diff(u(x, t), x, x));

Eq2 := diff(u(x, t), t) = alpha[nf]*(diff(theta(x, t), x, x))/Pr+Ec*mu[nf]*C*(diff(u(x, t), x))^2;

ICs := u(x, 0) = 0, theta(x, 0);

BCs := u(0, t) = 1, theta(0, t) = 1, u(10, t) = 0, theta(10, t) = 0;

and find the values of (diff(u(0, t), x))/(1-phi)^2.5 for different values of phi. Thanks in advace 

With my best regards and sincerely.

Muhammad Usman

School of Mathematical Sciences 
Peking University, Beijing, China

any idea for my problem?


> k1 := sum(X[h, t], t = 1 .. 23) >= 9;
9 <= X[h, 1] + X[h, 2] + X[h, 3] + X[h, 4] + X[h, 5] + X[h, 6]

   + X[h, 7] + X[h, 8] + X[h, 9] + X[h, 10] + X[h, 11] + X[h, 12]

   + X[h, 13] + X[h, 14] + X[h, 15] + X[h, 16] + X[h, 17]

   + X[h, 18] + X[h, 19] + X[h, 20] + X[h, 21] + X[h, 22]

   + X[h, 23], h = 1 .. 6

why 'h' still 'h'. from my textbooks the formula must be like this :

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