Maple 2018 Questions and Posts

These are Posts and Questions associated with the product, Maple 2018

I am trying to figure out which computational method is used in function Issimilar, (which determines similarity of matrices) in Maple 2018 and whether or not its a rational one.

Dear experts

I am using Maple to solve a complex equation. My idea is to separate real and imaginary parts and then solve a set of the equation when both real and imaginary parts are zero. the following are the equation and the way I made real and imaginary parts;

(K*( Q*sinh(K)*cosh(Q)-K*cosh(K)*sinh(Q))*(1+s*K^2)    +p*(-4*K^2*Q*(K^2+Q^2)        +Q*(Q^4+2*K^2*Q^2+5*K^4)*cosh(K)*cosh(Q)        -K*(Q^4+6*K^2*Q^2+K^4)*sinh(K)*sinh(Q)))/(K^2*Q*cosh(Q))

eq:= (K*( Q*sinh(K)*cosh(Q)-K*cosh(K)*sinh(Q))*(1+s*K^2)+p*(-4*K^2*Q*(K^2+Q^2)+Q*(Q^4+2*K^2*Q^2+5*K^4)*cosh(K)*cosh(Q)-K*(Q^4+6*K^2*Q^2+K^4)*sinh(K)*sinh(Q)))/(K^2*Q*cosh(Q)):

so the K and the Q are both complex variables and p and s are constant.

p := 0.1019367992e-3, s := 7.135575943      K:=Kr+I*Kim    Q:= sqrt(K^2-I*h^2*2*Pi/1.0e-6)


Q:= sqrt(K^2-I*h^2*2*Pi/1e-6);

therefore the real and imaginary parts of the equation are 

A:=evalc(Re(eq)):   B:=evalc(Im(eq)):

finally, I tried to solve it as following






the problem is that Maple can not solve it and returns the command. I  know that there is solutions. How can I solve this equation?


the maple file is


I guess Maple use Newton method to solve equation or system of equations. Is there an alternative? I mean what are the possible methods?

Hello users. I have a question on my work.

I'm trying to construct the equation and plot it. And I got 2 errors(warning).

Please help me how to solve this problem check the image below and attached file.


Maple can easily solve the B4 problem of the Putnam Mathematical Competition 2019  link


B4.  Let F be the set of functions f(x,y) that are twice continuously differentiable for x≥1, y≥1 and that satisfy the following two equations:
    x*(diff(f(x, y), x))+y*(diff(f(x, y), y)) = x*y*ln(x*y)

x^2*(diff(f(x, y), x, x))+y^2*(diff(f(x, y), y, y)) = x*y


For each f2F, let


"m(f) = min[s>=1]  (f(s+1,s+1)-f(s+1,s)-f(s,s+1)+f(s,s))"


Determine m(f), and show that it is independent of the choice of f.


# Solution

x*diff(f(x,y),x)+y*diff(f(x,y),y) = x*y*ln(x*y),
x^2*diff(f(x,y),x,x)+y^2*diff(f(x,y),y,y) = x*y

{f(x, y) = (1/2)*(x*y+2*_C1)*ln(x*y)-(1/2)*x*y-2*_C1*ln(x)+_C2}


f:=unapply(rhs(%[]), x,y);

proc (x, y) options operator, arrow; (1/2)*(y*x+2*_C1)*ln(y*x)-(1/2)*y*x-2*_C1*ln(x)+_C2 end proc


h := f(s+1, s+1) - f(s+1, s) - f(s, s+1) + f(s, s);



minimize(h, s=1..infinity);



answer = simplify(%);

answer = 2*ln(2)-1/2





I have a question regarding the dsolve function in Maple. I am trying to solve a system of 5 first order ODE's. First I solve the differential equations to arrive at the general solution using the dsolve command. Here Maple already produces very large and slow output while I would actually expect a much more compact output. 

Then I want to evaluate the final solution by using the initial conditons. At this point Maple keeps 'evaluating..' and does not come up with a solution. Only when I fill in all the parameters Maple finds a solution ( still very slow ). However as I want to be able to play with the input parameters after evaluation I want a solution without having to fill in the parameters prior to solving the system.

My question is whether I am presenting the system of ODE's in the right way to Maple. Should I rewrite the system for Maple to be able to produce a more dense solution? Or should I for instance use Laplace Transform to simplify the equations prior to solving?

Please help!




eq1 := sig_total-sig2(t)-sig3(t)-sig4(t)-sig5(t)+T1*(-(diff(sig2(t), t))-(diff(sig3(t), t))-(diff(sig4(t), t))-(diff(sig5(t), t))) = u1*(diff(eps(t), t));
eq2 := sig2(t)+T2*(diff(sig2(t), t)) = u2*(diff(eps(t), t));
eq3 := sig3(t)+T3*(diff(sig3(t), t)) = u3*(diff(eps(t), t));
eq4 := sig4(t)+T4*(diff(sig4(t), t)) = u4*(diff(eps(t), t));
eq5 := sig5(t)+T5*(diff(sig5(t), t)) = u5*(diff(eps(t), t));

dsolve({eq1, eq2, eq3, eq4, eq5}, {eps(t), sig2(t), sig3(t), sig4(t), sig5(t)});
sig_total := sig_0;
desys := {eq1, eq2, eq3, eq4, eq5}; ic := {eps(0) = sig_0/(E1+E2+E3+E4+E5), sig2(0) = sig_0/(1+(E1+E3+E4+E5)/E2), sig3(0) = sig_0/(1+(E1+E2+E4+E5)/E3), sig4(0) = sig_0/(1+(E1+E3+E2+E5)/E4), sig5(0) = sig_0/(1+(E1+E3+E4+E2)/E5)};

solution := combine(dsolve(desys union ic, {eps(t), sig2(t), sig3(t), sig4(t), sig5(t)})); assign(solution);

E_total := 33112; a := .1; b := .2; c := .15; d := .2; e := .35;

E1 := a*E_total; E2 := b*E_total; E3 := c*E_total; E4 := d*E_total; E5 := e*E_total; T1 := 1; T2 := 10; T3 := 100; T4 := 1000; T5 := 10000; u1 := T1*E1; u2 := T2*E2; u3 := T3*E3; u4 := T4*E4; u5 := T5*E5; sig_0 := 2;

plot(eps(t), t = 0 .. 672, y = 0 .. 0.12e-3);
y := eval(eps(t), t = 672);



Here are two sequences of commands that should give the same kind of plots. But, while the first one returns the expected display, the second doesn't (look to the labels on the histogram plot).

There is no hidden character that could explain this second display. Just that I proceeded this way:

  1. I executed the second sequence once.
  2. Then I told myself that displaying the histogram was superfluous, so I replaced its final semicolon with a colon.
  3. And I finally thought that, no, the histogram had a real interest and that I should display it; and I restored the semcolon (this is what you can see on the second sequence).
    And this add to the histogram the labels inherited from the second plot ...

Nothing dramatic here, but was the development team aware of this curiosity?



N := 10:
S := Statistics:-Sample(Normal(0, 1), 10)^+:
plots:-logplot(<  < [$1..N]> | S >, labels=["A", "B"]);




N := 10:
S := Statistics:-Sample(Normal(0, 1), 10)^+:
plots:-logplot(<  < [$1..N]> | S >, labels=["A", "B"]);






I am studying the motion of a beam coupled to piezoelectric strips. This continuous system is modelled by two DE:

YI*diff(w(x,t), x$4)-N[0]*cos(2*omega*t)*diff(w(x,t), x$2)+c*diff(w(x,t), t)+`&rho;A`*diff(w(x,t), t$2)+C[em,m]*v(t) = 0;


C[p]*diff(v(t), t)+1/R[l]*v(t) = C[em,e]*(D[1,2](w)(0,t)-D[1,2](w)(ell,t));

where "w(x,t)" stands for the beam's vibration and "v(t)" means the electric voltage, which is constant throught the beam. I would like to numerically solve both DE simultaneosly, but maple will not let me do it. I would like to know why. I am getting the following error:

Error, (in pdsolve/numeric/process_PDEs) number of dependent variables and number of PDE must be the same

I suppose it is because "w(x,t)" depends on "x" and "t", while "v(t)" depends solely on time, but I am not sure. Could someone help me out? Here is my current code:

declare(w(x,t), v(t)):

YI*diff(w(x,t), x$4)-N[0]*cos(2*omega*t)*diff(w(x,t), x$2)+c*diff(w(x,t), t)+`&rho;A`*diff(w(x,t), t$2)+C[em,m]*v(t) = 0;
pde1:= subs([YI = 1e4, N[0] = 5e3, c = 300, omega = 3.2233993, C[em,m] = 1], %):
ibc1:= w(0,t) = 0, D[1,1](w)(0,t) = 0, w(ell,t) = 0, D[1,1](w)(ell,t) = 0, D[2](w)(x,0) = 0, w(x,0) = sin(Pi*x/ell):

C[p]*diff(v(t), t)+1/R[l]*v(t) = C[em,e]*(D[1,2](w)(0,t)-D[1,2](w)(ell,t));
pde2:= subs([C[p] = 10, R[l] = 1000, C[em,e] = 1, ell = 5], %):
ibc2:= v(0) = 0:

pdsolve({pde1, pde2}, {ibc1, ibc2}, numeric);


with(LinearAlgebra); A := `

You can graph Maple in 4 dimensions ... help...





I have a problem using the NonlinearFit function from the Statistics package in Maple 2018.

I want to fit an exponential function which is non-linear in the parameters. The function in itself is working fine but i want to implement an extra condition on the parameters that are fitted. I already implemented the range of each parameter which is from 0 to 1, but I also want to implement the following condition:

a + b + c = 1.0

This is the code that i am using:

X := Vector([0, 100, 200, 300, 400, 500], datatype = float);
Y := Vector([0.2e-2, 0.5e-2, 0.7e-2, 0.75e-2, 0.77e-2, 0.8e-2], datatype = float);
nlfit := NonlinearFit(epsfunc, X, Y, t, parameterranges = [a = 0 .. 1, b = 0 .. 1, c = 0 .. 1], initialvalues = [a = .2, b = .2, c = .2], output = [parametervalues, leastsquaresfunction]);


It there a way to implement the additional condition that a+b+c=1.0?







Epsfunc is the result of solving an ODE using dsolve:

the following code is used:


eq1 := x(t)+(t1+t2)*(diff(x(t), t))+t1*t2*(diff(x(t), t, t)) = (n1+n2)*(diff(eps(t), t))+(n1*t2+n2*t1)*(diff(eps(t), t, t));
tr := n1*n2*(E1+E2)/((n1+n2)*E1*E2);
x := proc (t) options operator, arrow, function_assign; x0 end proc;
solution := dsolve({eq1, eps(0) = x0/(E1+E2), (D(eps))(0) = x0*((n1/E1+n2/E2)/(n1+n2)-1/(E1+E2))/tr}, eps(t)); assign(solution);

E := 500;
E1 := a*E; E2 := b*E; t1 := 100; t2 := c*t1; n1 := E1*t1; n2 := E2*t2; x0 := 2;
epsfunc := eval(eps(t));


Here is an example.  How would I solve this non-homogeneous system of ODEs


I have just upgraded my laptop from Windows 7 to Windows 10.  On starting up Maple 2018,  I receive the attached message on screen.  This is after previously loading the worksheet successfully.   Today,  I am not able to do so.  I need to permanently register my firewall to allow Maple to run; can anyone help?



Dear all 

I need to animate the following function in two dimensions

,u:=  (x, y) -->x^2+y^2+3

((gradplot(u(x, y), x = -10 .. 10, y = -2 .. 2, grid = [8, 8], thickness = 3, arrows = SLIM, color = u(x, y



expand((x-c)^2+(y-d)^2-R^2) = 0; algsubs(-R^2+c^2+d^2 = f, %); P := proc (x, y) options operator, arrow; -2*x*c-2*y*d+x^2+y^2+f = 0 end proc; 2 2 P := (x, y) -> -2 x c - 2 y d + x + y + f = 0 P(a*cos(theta), b*sin(theta)); G := unapply(%, theta); #usage des formules d'Euler simplify(expand(4*(exp(I*theta))^2*subs(cos(theta) = (exp(I*theta)+exp(-I*theta))*(1/2), sin(theta) = (exp(I*theta)-exp(-I*theta))/(2*I), G(theta)))); poly := sort(subs(exp(I*theta) = X, exp((2*I)*theta) = X^2, exp((3*I)*theta) = X^3, exp((4*I)*theta) = X^4, %)); coeff(lhs(poly), X^4)/tcoeff(lhs(poly)); # exp(I*theta1),exp(I*theta2),exp(I*theta3),exp(I*theta4) sont les racines de ce polynôme unitaire : exp(I*theta1)*exp(I*theta2)*exp(I*theta3)*exp(I*theta4) =1 exp(I*(θ1+θ2+θ3+θ4)=1 d'où θ1+θ2+θ3+θ4 ≡ 2*Pi


I'm currently solving an equations where the boundary conditions is at infinity. I'm trying to solve it by using dsolve but i can't seem to find a solutions. Here is my equations:

ode1 := diff(f(eta), eta$3)+(diff(f(eta), eta$2))*f(eta)-(diff(f(eta), eta))^2-M . (diff(f(eta), eta))-A . (diff(f(eta), eta)+(1/2)*(eta . (diff(f(eta), eta$2)))) = 0;

ode2 := diff(theta(eta), eta$2)+Pr*(f(eta) . (diff(theta(eta), eta))-(diff(f(eta), eta)) . theta(eta)-A . (theta(eta)+1/2 . eta . (diff(theta(eta), eta)))) = 0;

and my boundary conditions are:

bcs := f(0) = 0, (D(f))(0) = 1, ((D@@2)(f))(inf) = 0, theta(0) = 1, theta(inf) = 0;

The value of Pr=7, M=1 and A=[0,1,2,4]

I really need your help, please. 

Thank you :)

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