Maple 18 Questions and Posts

These are Posts and Questions associated with the product, Maple 18

I am trying to calculate the integral


Maple cannot calculate the integral. I tried to expand theta in the series form and substitute in the integral, still cannot calculate it.

any suggestion to tackle this problem whould be helpful.

Thank you



Which {b} is an eigenvector


z0:= a0^2/(4*lambda);  










I have plotted some figures and saved them yesterfay!

now once i opened them some nonsence digits appear on the figure! see the picture please. anyone has similar experience? how to solve it!

Dont make me disappointed maple! two days work is invain now !


I want to do matrix operation,vis-a-vis,addition,inverse of a 15 x 15 matrix. I already typed the matrix,but remains the operation.

Thank you




I solve a set of equations in this way and I have three set of answers ,but I don`t know wich one is true.

and I have another question ,how can I assume v[0] like a constant?


alpha[2]:= 2.727272728*10^5: alpha[4]:= 3738.685337: alpha[6]:= -30.18675539: alpha[7] := -4.116375735*10^6: alpha[8] := 1.859504132*10^10: alpha[9]:= 2.489142857*10^(-12):

l10:=(alpha[7]*v[0]^2+1)*gamma[i*n]^4+(-alpha[4]*beta[n]^2+alpha[8]*v[0]^2-alpha[9])*gamma[i*n]^2+(2*I)*gamma[i*n]*alpha[2]*beta[n]*v[0]+(2*I)*gamma[i*n]^3*alpha[6]*beta[n]*v[0]-beta[n]^2 = 0:

l11 := subs(i = 1, l10);

l12 := subs(i = 2, l10);

l13 := subs(i = 3, l10);

l14 := subs(i = 4, l10);

l15 := (exp(I*(gamma[n]+gamma[2*n]))+exp(I*(gamma[3*n]+gamma[4*n])))*(gamma[3*n]^2-gamma[4*n]^2)*(gamma[n]^2-gamma[2*n]^2)+(exp(I*(gamma[n]+gamma[4*n]))+exp(I*(gamma[2*n]+gamma[3*n])))*(gamma[2*n]^2-gamma[3*n]^2)*(gamma[n]^2-gamma[4*n]^2)+(exp(I*(gamma[2*n]+gamma[4*n]))+exp(I*(gamma[n]+gamma[3*n])))*(gamma[2*n]^2-gamma[4*n]^2)*(-gamma[n]^2+gamma[3*n]^2) = 0;

l1 := combine(expand(evalc(l15)), trig):

l2 := combine(expand(evalc(Re(l15))), trig):

l3 := combine(expand(evalc(Im(l15))), trig): v[0] := 1; 1

fsolve({l1, l11, l12, l13, l14}, {beta[n], gamma[n], gamma[2*n], gamma[3*n], gamma[4*n]}):

fsolve({l11, l12, l13, l14, l2}):

solve({l11, l12, l13, l14, l3}):



I want to calculate the ratio of the length of day and night for every latitude on earth ?
but i confused on using Maple in a wise way for finding the formula !
this is my demonstration :



the grat circle that divides the earth's surface into two dark and bright sides

[sin(t)*cos(tilt), cos(t), sin(t)*sin(tilt)]

[sin(t)*cos(tilt), cos(t), sin(t)*sin(tilt)]


circle of revolving of a point on earth in 24 hours

[sin(t)*cos(Latitude), cos(t)*cos(Latitude), sin(Latitude)]

[sin(t)*cos(Latitude), cos(t)*cos(Latitude), sin(Latitude)]


Visualization of dark and bright side the of earth


Explore(plots[display](plots[spacecurve]({[sin(t)*cos(tilt), cos(t), sin(t)*sin(tilt), color = red], [sin(t)*cos(Latitude), cos(t)*cos(Latitude), sin(Latitude), color = blue]}, t = 0 .. 2*Pi, scaling = constrained, thickness = 4, labels = [x, y, Latitudez], labeldirections = [horizontal, horizontal, vertical], axes = frame), plottools[rotate](plottools[hemisphere]([0, 0, 0], 1, capped = false, color = green, grid = [10, 10], style = surface), 0, tilt, 0), plottools[rotate](plottools[hemisphere]([0, 0, 0], 1, capped = false, color = black, grid = [10, 10], style = surface), 0, Pi+tilt, 0)), parameters = [tilt = 0 .. Pi, Latitude = -(1/2)*Pi .. (1/2)*Pi], initialvalues = [tilt = (1/2)*Pi+.409, Latitude = 1.16])





Hi dears,

How can I draw the solution of the following 3d linear inequalities in Maple?

A:={-x-y+3*z >= 0, -x+2*y >= 0, 3*x-2*y-z >= 0, x > 0, y > 0, z > 0}

I am looking forward to hearing from you

Sincerely yours.


Hi all, I have a problem someone can help me

F := {a^2, b^2, c^2, ab, bc, ca}

G := [a^2, b^2, c^2, ab, bc, ca]

How to convert F to G and G to F ?

Thanks you very much.

Q1: Pascal’s Matrix of order n is given by:
Sij =(i + j)!/ i!*j!
Use Mable to produce Pascal’s Matrix of order 8.

Q2: Study the Matrix decomposition (i.e. QR, LU, and LLT), then use Maple to produce these decompositions for a random Matrix of order 6.

Q3: Write one paragraph of your own to explain Moore-Penrose Inverse of a Matrix. Use Maple to find Moore-Penrose Inverse for a random Matrix of order 8.

Q4: Use Maple to find Jordan Canonical form for a random Matrix of order 10.

Q5: Use the seq command to generate the triple [i,j,k] for all possible values for 1 ≤ i,j,k ≤ 10, then plot this triple. i.e. Use nested seq .

Q6: Let F[n] be the set:
F[n] = {p / q: 1 ≤ q ≤ n,p ≤ p ≤ q}
Use Maple to find F[6].

To keep upgrading Maple in our university we need to justify it's usage. We have an unlimited floating license. How can we track the usage of Maple by our students, faculty, etc?

Dear All

I am trying to use differential operator two times in sucession but I am not getting desired differentiation. Please see content below:



DepVars := [u(x, t), v(x, t), phi(x, t), psi(x, t)]; 1; declare(u(x, t), v(x, t), phi(x, t), psi(x, t))

[u(x, t), v(x, t), phi(x, t), psi(x, t)]


u(x, t)*`will now be displayed as`*u


v(x, t)*`will now be displayed as`*v


phi(x, t)*`will now be displayed as`*phi


psi(x, t)*`will now be displayed as`*psi



I := phi(x, t)*(diff(v(x, t), x)+b*(diff(u(x, t), x, x)))+psi(x, t)*(diff(u(x, t), x, x, x)+d*(diff(v(x, t), x, x)))

phi(x, t)*(diff(v(x, t), x)+b*(diff(diff(u(x, t), x), x)))+psi(x, t)*(diff(diff(diff(u(x, t), x), x), x)+d*(diff(diff(v(x, t), x), x)))


ToJet(I, DepVars)

phi*(b*u[x, x]+v[x])+psi*(d*v[x, x]+u[x, x, x])


T[1] := proc (f) options operator, arrow; diff(f, u[x]) end proc; 1; T[2] := proc (f) options operator, arrow; diff(f, u[x, x]) end proc; 1; T[3] := proc (f) options operator, arrow; diff(f, u[x, x, x]) end proc; 1; U[1] := proc (f) options operator, arrow; diff(f, v[x]) end proc; 1; U[2] := proc (f) options operator, arrow; diff(f, v[x, x]) end proc

proc (f) options operator, arrow; diff(f, u[x]) end proc


proc (f) options operator, arrow; diff(f, u[x, x]) end proc


proc (f) options operator, arrow; diff(f, u[x, x, x]) end proc


proc (f) options operator, arrow; diff(f, v[x]) end proc


proc (f) options operator, arrow; diff(f, v[x, x]) end proc


d := proc (f) options operator, arrow; diff(f, x)+u[x]*(diff(f, u))+u[x, x]*(diff(f, u[x]))+u[x, x, x]*(diff(f, u[x, x])) end proc

proc (f) options operator, arrow; diff(f, x)+u[x]*(diff(f, u))+u[x, x]*(diff(f, u[x]))+u[x, x, x]*(diff(f, u[x, x])) end proc





Why this is giving zero result ?

If I change Ito jet notation then I get result non zero result like:

T[2](phi*(b*u[x, x]+v[x])+psi*(d*v[x, x]+u[x, x, x]))



d(T[2](phi*(b*u[x, x]+v[x])+psi*(d*v[x, x]+u[x, x, x])))



I was expecting this to be "phi[x]*b, "but instead I am getting zero result.``



G := 6.6743*10^(-8);

a := 1.9501*10^24;

b := .3306;

c := 2.99792458*10^10;

d := 2.035;

pi := 3.143;

eps := 3.8220*10^35;

g(r) = 1-s(r)/0.06123;

j(r) = e^(-(1/2)*w(r))*(1-2*G*v(r)/(r*c^2))^.5

sys := diff(v(r), r) = 4*pi*r^2*eps/c^2, ics=v(0)=0

diff(u(r), r) = -G*(eps+u(r))*(v(r)+4*Pi*r^3*u(r)/c^2)/(c^2*(r^2-2*G*r*v(r)/c^2)),u(0)=1.3668*10^34

diff(w(r), r) = 1.485232054*10^(-28)*(v(r)+4.450600224*10^(-21)*pi*r^3*u(r))/(r^2-2*G*r*v(r)/c^2), conditions: w(0)=0,iterate it to find w(688240)=-2.05684, it solve must satistfy the both conditions.

diff(r^4*j(r)*(diff(g(r), r)), r)+4*r^3*g(r)*(diff(j(r), r)) = 0, conditions dg(r)/dr =0  at r=0, g(688240) =0.87214

diff(J(r), r) = (8*pi*(1/3))*(eps/c^2+u(r)/c^2)*(g(r)*j(r).(r^4))/(1-2*G*v(r)/(r*c^2)) condition J(0)=0.

I am trying to apply window functions from the SignalProcessing package to arrays. However, some window functions (e.g. Hann) appear to operate on a [0..N-1] index basis, others (e.g. Welch) work on [1..N].

I don't know how to make the latter ones work correctly, since after applying the windowing function the first entry in the array is not zero, as it should be.

Best regards

I have made an algorithm for producing random walks (only possible to walk one step to either direction except downwards(EDIT: It is possible to go downwards, but it has to be when making a turn to the right or left). The walks are determined by the rand function:

R3:=rand(1..3): (1: go straight on; 2: turn right; 3: turn left)
M:=15; N:=1500;


  local i,j,r,X,Y,L;
  for j from 1 to M do
    X[0,j]:=0;                                # Initialization
    X[1,j]:=1;                                # The first step should still be taken to the point (1,0)
    for i from 2 to N do
      if r=1 then X[i,j]:=2*X[i-1,j]-X[i-2,j]; Y[i,j]:=2*Y[i-1,j]-Y[i-2,j];                   # go straight on
      elif r=2 then X[i,j]:=X[i-1,j]+Y[i-1,j]-Y[i-2,j]; Y[i,j]:=Y[i-1,j]-X[i-1,j]+X[i-2,j];    # turn right
      else X[i,j]:=X[i-1,j]-Y[i-1,j]+Y[i-2,j]; Y[i,j]:=Y[i-1,j]+X[i-1,j]-X[i-2,j];             # turn left
      end if;
      if (X[i,j]=X[j,j] and Y[i,j]=Y[j,j]) then L[j]:=i; break; end if; (This is wrong)
    end do;
  end do:
  return [X,Y,L];
end proc:

The question from is like this:
Modify the algorithm such that it stops at r[i] if r[i] = r[j] for any 0 <= j <= i-2. r=(xi,yi). M is the number of random walks and N is the number of steps. The length of the paths should be stored in L[m] (m=1..M). How do I implement the if-test correctly?

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