@nm I can see that working for a signal that has a starting point other than 0 and no other shifts involved, but I am wondering about signals built from shifted steps / ramps / etc.  If the forcing function is something like r(t) - u(t-1) - r(t-1) with u(t)=Heaviside(t) and r(t) = t Heaviside(t).  I won't have time to see if I can break Maple with that until this weekend, but I plan to try!

Hi,

I would like to plot this function from x= 2pi to 4pi.  I entered this into the plotting command, and nothing happened.  How do I plot this from 2pi to 4pi?

plot_from_two_pi_to_4_pi.mw

Is it possible to have the results of a MapleFlow container wrap to the next line as opposed to just extending off the page?

Thanks.

restart;
kp := .3;

Pr := .3; N := .5; g := .5; A := 1; B := 0; M := .5; lambda := .5; Ec := .5;

rf := 997.1; kf := .613; cpf := 4179; `σf` := 0.5e-1;
p1 := 0.1e-1; sigma1 := 2380000; rs1 := 4250; ks1 := 8.9538; cps1 := 686.2;
p2 := 0.5e-1; sigma2 := 3500000; rs2 := 10500; ks2 := 429; cps2 := 235;

NULL;
a1 := (1-p1)^2.5*(1-p2)^2.5;
a2 := (1-p2)*(1-p1+p1*rs1/rf)+p2*rs2/rf;
a3 := 1+3*((p1*sigma1+p2*sigma2)/`σf`-p1-p2)/(2+(p1*sigma1+p2*sigma2)/((p1+p2)*`σf`)-((p1*sigma1+p2*sigma2)/`σf`-p1-p2));

a4 := (1-p2)*(1-p1+p1*rs1*cps1/(rf*cpf))+p2*rs2*cps2/(rf*cpf);
a5 := (ks1+2*kf-2*p1*(kf-ks1))*(ks2+2*kf*(ks1+2*kf-2*p1*(kf-ks1))/(ks1+2*kf+p1*(kf-ks1))-2*p2*(kf*(ks1+2*kf-2*p1*(kf-ks1))/(ks1+2*kf+p1*(kf-ks1))-ks2))/((ks1+2*kf+p1*(kf-ks1))*(ks2+2*kf*(ks1+2*kf-2*p1*(kf-ks1))/(ks1+2*kf+p1*(kf-ks1))+2*p2*(kf*(ks1+2*kf-2*p1*(kf-ks1))/(ks1+2*kf+p1*(kf-ks1))-ks2)));

OdeSys := (diff(U(Y), Y, Y))/(a1*a2)+Theta(Y)+N*(Theta(Y)*Theta(Y))-a3*(M*M)*U(Y)/a2-(kp*kp)*U(Y)/(a1*a2), a5*(diff(Theta(Y), Y, Y))/a4+Pr*Ec*((diff(U(Y), Y))^2+U(Y)^2*(kp*kp))/(a1*a2); Cond := U(0) = lambda*(D(U))(0), Theta(0) = A+g*(D(Theta))(0), U(1) = 0, Theta(1) = B; Ans := dsolve([OdeSys, Cond], numeric, output = listprocedure);
U := proc (Y) options operator, arrow, function_assign; (eval(U(Y), Ans))(0) end proc;
                 U := Y -> (eval(U(Y), Ans))(0)
Theta := proc (Y) options operator, arrow, function_assign; (eval(Theta(Y), Ans))(0) end proc;
             Theta := Y -> (eval(Theta(Y), Ans))(0)
Theta_b := (int(U(Y)*Theta(Y), Y = 0 .. 1))/(int(U(Y), Y = 0 .. 1));
Error, (in Theta) too many levels of recursion
Q := int(U(Y), Y = 0 .. 1, numeric);
Error, (in Theta) too many levels of recursion
NUMERIC := [(eval((diff(U(Y), Y))/a1, Ans))(0), (eval(-(diff(Theta(Y), Y))/(Theta_b*a5), Ans))(0)];
Error, (in Theta) too many levels of recursion

i need the solution  for Y=0 and Y=1

Hello,

How to factor the following polynomial : n*xⁿ - 2*n*x^{(n - 1)} + xⁿ

I can't find a command to write : x^n-1*((n+1)*x-2n)

Thank you for your help.

While I was elaborating on a math problem, I came across the following expression which actually should be equal to one. Maple unfortunately was unable to fully provide a simplified expression. Is there a way to do that? 

Thank you

Streamlines, isotherms and microrotations for Re = 1, Pr = 7.2, Gr = 10⁵ and (a) Ha = 0 (b) Ha = 30 (c) Ha = 60 (d) Ha = 100.

for Ra = 10⁵, Ha = 50, Pr = 0.025 and θ = 1 − Y

eqat := {M . (D(theta))(0)+2.*Pr . f(0) = 0, diff(phi(eta), eta, eta)+2.*Sc . f(eta) . (diff(phi(eta), eta))-(1/2)*S . Sc . eta . (diff(phi(eta), eta))+N[t]/N[b] . (diff(theta(eta), eta, eta)) = 0, diff(g(eta), eta, eta)-2.*(diff(f(eta), eta)) . g(eta)+2.*f(eta) . (diff(g(eta), eta))-S . (g(eta)+(1/2)*eta . (diff(g(eta), eta)))-1/(sigma . Re[r]) . ((1+d^%H . exp(-eta))/(1+d . exp(-eta))) . g(eta)-beta^%H . ((1+d^%H . exp(-eta))^2/sqrt(1+d . exp(-eta))) . g(eta) . sqrt((diff(f(eta), eta))^2+g(eta)^2) = 0, diff(theta(eta), eta, eta)+2.*Pr . f(eta) . (diff(theta(eta), eta))-(1/2)*S . Pr . eta . (diff(theta(eta), eta))+N[b] . Pr . ((diff(theta(eta), eta)) . (diff(phi(eta), eta)))+N[t] . Pr . ((diff(theta(eta), eta))^2)+4/3 . N . (diff((C[T]+theta(eta))^3 . (diff(theta(eta), eta)), eta)) = 0, diff(f(eta), eta, eta, eta)-(diff(f(eta), eta))^2+2.*f(eta) . (diff(f(eta), eta))+g(eta)^2-S . (diff(f(eta), eta)+(1/2)*eta . (diff(f(eta), eta, eta)))-1/(sigma . Re[r]) . ((1+d^%H . exp(-eta))/(1+d . exp(-eta))) . (diff(f(eta), eta))-beta^%H . ((1+d^%H . exp(-eta))^2/sqrt(1+d . exp(-eta))) . (diff(f(eta), eta)) . sqrt((diff(f(eta), eta))^2+g(eta)^2) = 0, g(0) = 1, g(6) = 0, phi(0) = 1, phi(6) = 0, theta(0) = 1, theta(6) = 0, (D(f))(0) = 1, (D(f))(6) = 0};
sys1 := eval(eqat, {M = 0, N = 2, Pr = .8, S = -2.5, Sc = .5, d = 2, sigma = .2, C[T] = .5, N[b] = .4, N[t] = .4, Re[r] = 1.1, beta^%H = .2, d^%H = 1.5});
sys2 := eval(eqat, {M = 0, N = 2, Pr = .8, S = -2.5, Sc = .5, d = 2, sigma = .2, C[T] = .5, N[b] = .4, N[t] = .4, Re[r] = 1.1, beta^%H = .4, d^%H = 1.5});
sys3 := eval(eqat, {M = 0, N = 2, Pr = .8, S = -2.5, Sc = .5, d = 2, sigma = .2, C[T] = .5, N[b] = .4, N[t] = .4, Re[r] = 1.1, beta^%H = .6, d^%H = 1.5});
sys4 := eval(eqat, {M = 0, N = 2, Pr = .8, S = -2.5, Sc = .5, d = 2, sigma = .2, C[T] = .5, N[b] = .4, N[t] = .4, Re[r] = 1.1, beta^%H = .8, d^%H = 1.5});
sys5 := eval(eqat, {M = .5, N = 2, Pr = .8, S = -2.5, Sc = .5, d = 2, sigma = .2, C[T] = .5, N[b] = .4, N[t] = .4, Re[r] = 1.1, beta^%H = .2, d^%H = 1.5});
sys6 := eval(eqat, {M = .5, N = 2, Pr = .8, S = -2.5, Sc = .5, d = 2, sigma = .2, C[T] = .5, N[b] = .4, N[t] = .4, Re[r] = 1.1, beta^%H = .4, d^%H = 1.5});
sys7 := eval(eqat, {M = .5, N = 2, Pr = .8, S = -2.5, Sc = .5, d = 2, sigma = .2, C[T] = .5, N[b] = .4, N[t] = .4, Re[r] = 1.1, beta^%H = .6, d^%H = 1.5});
sys8 := eval(eqat, {M = .5, N = 2, Pr = .8, S = -2.5, Sc = .5, d = 2, sigma = .2, C[T] = .5, N[b] = .4, N[t] = .4, Re[r] = 1.1, beta^%H = .8, d^%H = 1.5});
sys9 := eval(eqat, {M = 1, N = 2, Pr = .8, S = -2.5, Sc = .5, d = 2, sigma = .2, C[T] = .5, N[b] = .4, N[t] = .4, Re[r] = 1.1, beta^%H = .2, d^%H = 1.5});
sys10 := eval(eqat, {M = 1, N = 2, Pr = .8, S = -2.5, Sc = .5, d = 2, sigma = .2, C[T] = .5, N[b] = .4, N[t] = .4, Re[r] = 1.1, beta^%H = .4, d^%H = 1.5});
sys11 := eval(eqat, {M = 1, N = 2, Pr = .8, S = -2.5, Sc = .5, d = 2, sigma = .2, C[T] = .5, N[b] = .4, N[t] = .4, Re[r] = 1.1, beta^%H = .6, d^%H = 1.5});
sys12 := eval(eqat, {M = 1, N = 2, Pr = .8, S = -2.5, Sc = .5, d = 2, sigma = .2, C[T] = .5, N[b] = .4, N[t] = .4, Re[r] = 1.1, beta^%H = .8, d^%H = 1.5});
sys13 := eval(eqat, {M = 1.5, N = 2, Pr = .8, S = -2.5, Sc = .5, d = 2, sigma = .2, C[T] = .5, N[b] = .4, N[t] = .4, Re[r] = 1.1, beta^%H = .2, d^%H = 1.5});
sys14 := eval(eqat, {M = 1.5, N = 2, Pr = .8, S = -2.5, Sc = .5, d = 2, sigma = .2, C[T] = .5, N[b] = .4, N[t] = .4, Re[r] = 1.1, beta^%H = .4, d^%H = 1.5});
sys15 := eval(eqat, {M = 1.5, N = 2, Pr = .8, S = -2.5, Sc = .5, d = 2, sigma = .2, C[T] = .5, N[b] = .4, N[t] = .4, Re[r] = 1.1, beta^%H = .6, d^%H = 1.5});
sys16 := eval(eqat, {M = 1.5, N = 2, Pr = .8, S = -2.5, Sc = .5, d = 2, sigma = .2, C[T] = .5, N[b] = .4, N[t] = .4, Re[r] = 1.1, beta^%H = .8, d^%H = 1.5});
 

Hi,

I ploted the step response of a MIMO system in Maple using DynamicSystems object.

The plot is incorrect.

What am I doing wrong?

Thanks for your help

I have a system described by 

I want to plot Y(s)/Z(s) = ((C . (1/((s . I) - A))) . B) + D with stepped inputs on both inputs

The system above evaluates to 

My commands are 

ss_a := A__m;
ss_b := B__m;
ss_c := C__m;
ss_d := D__m;
sys4 := StateSpace(ss_a, ss_b, ss_c, ss_d);
plots:-display([ResponsePlot(sys4, [Step(), Step()], 'duration' = 5, color = red)]);

Maple is returning the incorrect plot

The correct plot is 

SYSTEM

Correct plot

Hi Everyone 

I want to to an iteration of an expression with 100 steps. To be honest I have no idea how to handle this in maple. I also didnt find much infomation on maplesoft.com 

The expression i want to iterate looks like this: 

Has someone an idea how to do this?

Thanks in advance!

I would like to solve this system of PDEs along the x-interval [0,1] in three different subintervals: from 0 to 0.35, from 0.35 to 0.6, and from 0.6 to 1. I tried to solve the system by setting these same subintervals as you might see in my script, however it is now what I need. Any help would be very appreciated.

restart;
d1 := 0.05;
d2 := 0.3;
AA := 0.2;
BB := 0.1;
PDE1 := diff(u(x, t), t) = d1*diff(u(x, t), x, x) + w(x, t)*exp(AA*u(x, t) - BB*v(x, t));
PDE2 := diff(v(x, t), t) = d2*diff(v(x, t), x, x) - w(x, t)*exp(AA*u(x, t) - BB*v(x, t));
PDE3 := 0.0001*diff(w(x, t), t) = diff(w(x, t), x) - 0.8*x + 3.3;
IBC1 := u(0, t) = 1, u(1, t) = 0, u(x, 0) = piecewise(x < 0.35, -(4*x)*x + 1, 0.35 < x and x < 0.65, 1.32958 - 1.29167*x, 0.65 < x, 4*(x - 1)^2);
IBC2 := v(0, t) = 0, v(1, t) = 1, v(x, 0) = piecewise(x < 0.35, (4*x)*x + 1, 0.35 < x and x < 0.65, 1.32958 - 1.29167*x, 0.65 < x, -4*(x - 1)^2);
IBC3 := w(0, t) = 0.5, w(x, 0) = 1 - (0.3*x)*x;
pds := pdsolve([PDE1, PDE2, PDE3], [IBC1, IBC2, IBC3], numeric, time = t, range = 0 .. 1);
p1 := pds:-plot(t = 0, numpoints = 50);
p2 := pds:-plot(t = 1/8, numpoints = 50, color = blue);
p3 := pds:-plot(t = 1/4, numpoints = 50, color = green);

[I split this off from here into a separate question. dharr]

@dharr Thanks. For the second one, an output is . Is it possible to compel Maple to attempt to simplify the second algebraic number to a less complicated expression automatically? 
Its minimal polynomial can be computed. However, this is not so convenient for the specific purpose. Actually, I want something like this: 

evalA(-4*RootOf(_Z^3 - 3*_Z^2 - 10*_Z - 1)^2 + 19*RootOf(_Z^3 - 3*_Z^2 - 10*_Z - 1) + 3);
 = 
                       /  3        2           \
               5 RootOf\_Z  + 10 _Z  + 3 _Z - 1/

Mathematica has an additional function RootReduce to do so directly, but I cannot find such functionality in Maple.

Remark. A fairly complicated one: 

evalA(-45658*RootOf(37*_Z^6 - 382*_Z^5 + 1388*_Z^4 - 2188*_Z^3 + 1475*_Z^2 - 406*_Z + 37, index = 6)^5 + 417257*RootOf(37*_Z^6 - 382*_Z^5 + 1388*_Z^4 - 2188*_Z^3 + 1475*_Z^2 - 406*_Z + 37, index = 6)^4 - 1252087*RootOf(37*_Z^6 - 382*_Z^5 + 1388*_Z^4 - 2188*_Z^3 + 1475*_Z^2 - 406*_Z + 37, index = 6)^3 + 1463384*RootOf(37*_Z^6 - 382*_Z^5 + 1388*_Z^4 - 2188*_Z^3 + 1475*_Z^2 - 406*_Z + 37, index = 6)^2 - 558475*RootOf(37*_Z^6 - 382*_Z^5 + 1388*_Z^4 - 2188*_Z^3 + 1475*_Z^2 - 406*_Z + 37, index = 6) + 69230);
 = 
            /
17991 RootOf\

       6         5          4          3          2                
  37 _Z  - 406 _Z  + 1475 _Z  - 2188 _Z  + 1388 _Z  - 382 _Z + 37, 

           \
  index = 4/

Convert a table in a form 

output≔table([(2,4)=["O-H",0.97234632],(1,2)=["O-O",1.44940000],(1,3)=["O-H",0.97232285]])

Table can have more elements in a similar form at (1,2) and (2,1) position 

In the new matrix converted

We should get a square matrix

With at (2,4) and (4,2) position 

0.97234632

Similarly at (1,2) and (2,1) position 1.44940000

And so on 

Remember a square matrix and it is symmetric 

(Deleted because not reproducible on a different PC)

With 1D

With 2D

The root cause might be the same as for this open question.

int_warning_2D.mw

int_warning_1D.mw

Fix a plot output image size to a certain big size instead of going to each plot and then trying to maginify it where output runs through pages.

How to insert pagebreak using code itself where I want as i run code 

As if I get more lines of output difficut to run through output give ctrl+Enter each place later 

Kind help