Maple 2017 Questions and Posts

These are Posts and Questions associated with the product, Maple 2017

I was using the following idea to get the computation time of my computations in Maple;

my computation

But it seems if I use

Grid:-Map(m->My computation(m),[some sequence]);

Where the sequence has more than one element, this idea doesn't work! It seems Maple time() remains fixed during this Grid:-Map when the sequence has more than one element. So then how should I check the computation time?

In the following pictures you can see when the sequence has one element, it works normal, but when it has two elements, it doesn't work the same.

Hello everybody,

I am quiet new to Maple and just have to program a small tool.

I need to show a conculison in a pop-up Window which should contain a matrix and a plot.

I tried different ways but they didn't work.

Thanks in advance


it seems that Maple2017 handles the noncommutative product wrongly (while it used to do it properly up to maple 17 afaik). 



Q^2*P*Q + Q*P*Q^2;


gives a correct result: BUT 

Q(t)^2*P(t)*Q(t) + Q(t)*P(t)*Q(t)^2;


gives the result:

2Q(t)^3*P(t)  (which is wrong).

It used to work fine in Maple 17. I need to differentiate noncommutative polynomials in P(t), Q(t), which was done without problem in Maple 17 but now seems to be broken. 


Any explanation/workaround/fix? Is it fixed in Maple 2018?



Digits := 4



a1 := int((diff(phi(x), x, x, x, x))*phi(x), x = 0 .. L)




a3 := int(phi(x)*phi(x), x = 0 .. L)




lambda := 0.170e-1



B := 0.223e11



A := 0.346e11



k[n] := W*(4*R*G*L^2+pi^2*(2*n-1)^2)/(C*(pi^2*(2*n-1)^2+4*R*L^2*(G+W)))



b[n] := 4*pi*(2*n-1)/(4*R*G*L^2+pi^2*(2*n-1)^2)



U := Heaviside(t)



w := 1



L := 3.5



Ra := 9



Rb := 5



W := 1



G := 0.5e-3



R := Ra+Rb



C := 0.2e-1



h := 0.250e-1



nu := .22



I1 := (1/3)*w*h^3



E[0] := A+B



nu1 := (1-nu)/((1+nu)*(1-2*nu))




proc (x, t) options operator, arrow; (W*(sum(-(1/8)*Pi^3*b[n]*(2*n-1)^3*cos((1/2)*(2*n-1)*Pi*x/L)*exp(-k[n]*t)/(k[n]*L^3), n = 1 .. 8))+R*G*(R*G)^.5*sinh((R*G)^.5*(x-L))*t/cosh((R*G)^.5*L)+C*U*R*G*(R*G)^.5*sinh((R*G)^.5*(x-L))/cosh((R*G)^.5*L))*exp(-W*t/C)/(w*h) end proc


eq := -nu1*I1*a1*((1/2)*E[0]*q(t)+B*lambda*(int(exp(-lambda*(t-s))*q(s), s = 0 .. t)))+2*w*h^2*beta(x, t)

-0.4603e11*q(t)-0.6134e9*(int(exp(-0.170e-1*t+0.170e-1*s)*q(s), s = 0 .. t))+0.5000e-1*(-0.7235e-2*pi*(pi^2+686.)*cos(.4488*x)*exp(-50.*(.3430+pi^2)*t/(pi^2+686.))/(.3430+pi^2)^2-.5860*pi*(9.*pi^2+686.)*cos(1.346*x)*exp(-50.*(.3430+9.*pi^2)*t/(9.*pi^2+686.))/(.3430+9.*pi^2)^2-4.522*pi*(25.*pi^2+686.)*cos(2.244*x)*exp(-50.*(.3430+25.*pi^2)*t/(25.*pi^2+686.))/(.3430+25.*pi^2)^2-17.37*pi*(49.*pi^2+686.)*cos(3.142*x)*exp(-50.*(.3430+49.*pi^2)*t/(49.*pi^2+686.))/(.3430+49.*pi^2)^2-47.47*pi*(81.*pi^2+686.)*cos(4.039*x)*exp(-50.*(.3430+81.*pi^2)*t/(81.*pi^2+686.))/(.3430+81.*pi^2)^2-105.9*pi*(121.*pi^2+686.)*cos(4.937*x)*exp(-50.*(.3430+121.*pi^2)*t/(121.*pi^2+686.))/(.3430+121.*pi^2)^2-206.6*pi*(169.*pi^2+686.)*cos(5.834*x)*exp(-50.*(.3430+169.*pi^2)*t/(169.*pi^2+686.))/(.3430+169.*pi^2)^2-366.3*pi*(225.*pi^2+686.)*cos(6.732*x)*exp(-50.*(.3430+225.*pi^2)*t/(225.*pi^2+686.))/(.3430+225.*pi^2)^2+0.5616e-3*sinh(0.8367e-1*x-.2928)*t+0.1123e-4*Heaviside(t)*sinh(0.8367e-1*x-.2928))*exp(-50.00*t)





Hello everyone,


I am currently trying to plot lines from different lists.

I got 3 lists with points and another 3 lists with points (Connect each point from one list with the other), and another list with my x-axis.


I tried something like that

(nply in this case is 4)

for i from 1 to nply do

sigma1P1[i] := display(line([grenzeu[i], sigma1unter[i]], [grenzeo[i], sigma1ober[i]])):

sigma2P1[i] := display(line([grenzeu[i], sigma2unter[i]], [grenzeo[i], sigma2ober[i]])): 

tau12P1[i] := display(line([grenzeu[i], tau12unter[i]], [grenzeo[i], tau12ober[i]])):

end do:

The for loop creates 3 tables with 4 line plots, but the plot:-display(sigma1P1,sigma2P1,tau12P1);

gives me this Error message:

Error, `plot` does not evaluate to a module

Have anyone an idea how to get these 3 table with plots in one plots?

And if yes is it possible to implement this in a EMbedded Plot Window?


Many thanks in advance!




Dear all

I want to solve a set of parametric inequalities with constraints but it takes too much time to processing. do you know any faster solution? 
thanks in advance

Dear friends, i want to solve inequalities as follow:

solve({0 < Q, 0 < delta, 0 < t, 0 < E[0], 0 < lambda[1], 0 < (lambda[1]^2*(Q+2*t+4*delta-4*E[0])+lambda[1]*(Q+2*t+3*delta-7*E[0])-3*E[0])/((lambda[1]^2-6*lambda[1]-6)*t), Q < -(3*delta*lambda[1]+2*t*lambda[1]-4*E[0]*lambda[1]+6*delta-3*E[0])/lambda[1], delta < E[0]*(4*lambda[1]+3)/(3*(lambda[1]+2)), t < -(3*delta*lambda[1]-4*E[0]*lambda[1]+6*delta-3*E[0])/(2*lambda[1]), lambda[1] < 1, (lambda[1]^2*(Q+2*t+4*delta-4*E[0])+lambda[1]*(Q+2*t+3*delta-7*E[0])-3*E[0])/((lambda[1]^2-6*lambda[1]-6)*t) < (lambda[1]*(-2*Q-4*t-2*delta+3*E[0])-lambda[1]^2*delta+lambda[1]^2*E[0]+2*E[0]-2*Q-4*t)/((lambda[1]^2-6*lambda[1]-6)*t) and (lambda[1]*(-2*Q-4*t-2*delta+3*E[0])-lambda[1]^2*delta+lambda[1]^2*E[0]+2*E[0]-2*Q-4*t)/((lambda[1]^2-6*lambda[1]-6)*t) < 1})

But, processing is very heavy and takes too much time! is there any alternative solution?

Thank you in advance.

rho := -2*K*(s+K)^2*k[e]*sinh((s+K)*x/d)/(s*d^2*(4*(s+K)^3*cosh((s+K)*h/d)*h/d^2-4*K*(s+K)*cosh((s+K)*h/d)*h/d+4*sinh((s+K)*h/d)*K))

All variables are constant except s and x

I just noticed something funny. If I use the left side tool of Maple to define a matrix with zero entries and then change some entries and take determinant, then Maple treats zeros as symbold!!!! If I have to define my matrix completely myself, then what is the use of this tool?

I want to compute the following supremum of the function.

I'll write in Latex code.

I have the following term which I want to estimate:

\delta_1(\epsilon):= \sup_{|x|<100}\sup_{ 0<= t<1/\epsilon} \epsilon \cdot |\int_0^t [ f(x,\cos s , \sin s , p_0(t)+q_0(s))-g(x,t)]ds|

where p_0(t) is an unknown function that depends on t alone.


g(x,t):= \lim_{T\to \infty} 1/T \int_0^T f(x,\cos s , \sin s , p_0(t)+q_0(s))ds

q_0(s):= \exp(-s)(q_0(0)+\int_0^s (h(x_0(0),\cos(r), \sin(r))-p_0(0))dr)

where h(x,y_1,y_2):=y_1^2.

Can someone lend me a hand?



How to do a full reinstall so that no settings are kept?

I tried deleting the maple 2017 folder in program files after I uninstalled but that did not help.

Hi everyone 

I need help, I have downloaded Maple, and I have read about the student package, which can show the step by step solution, I would like to see the step by step solution than just a final solution to the equation problem. 

I have the maple 2017, but does anyone know, how to download this package, and use it in maple? 

There is several links to this, but I cant find it where to download the package, I hope that someone is able to help me 


Greetings, I am trying to make augmented matrix in the form of A, B and F where A is the tridiagonal matrix, B is the vector and F is the non-linear part of the system. I already tried (A,B,F):=augment(X,output='A','B','F'); but it is an error. Can someone help me? Thank you in advance!



The matrix X is as follows:


X:=[[[-lambda/(h^2),-u/(2 h)+lambda/(h^2),0,0],

       [u/(2 h)+lambda/(h^2),-lambda/(h^2),-u/(2 h)+lambda/(h^2),0],

       [0,u/(2 h)+lambda/(h^2),-lambda/(h^2),-u/(2 h)+lambda/(h^2)],

       [0,0,u/(2 h)+lambda/(h^2),-lambda/(h^2)]]];

I have made a plots[multiple] with 3 different functions, but I wish to connect the ends with vertical lines. Does anybody have a solution to this? I have uploaded screenshot Thank you in advance. BR Jens

Hello people in mapleprimes,

I am writing this sentences, hoping to be given any answer from you, to the 
question I have, about conversion of Array to List.


brings a response of Array as, which I call BB,

Array(1 .. 3, 1 .. 2, 1 .. 3, {(1, 2, 3) = HFloat(1.), (2, 2, 1) = HFloat(.540302305868139765), (2, 2, 2) = HFloat(.841470984807896505), (2, 2, 3) = HFloat(1.), (3, 2, 1) = HFloat(-.832293673094284814), (3, 2, 2) = HFloat(1.81859485365136342), (3, 2, 3) = HFloat(1.)}, datatype = float[8])

And, I converted it to List, with 


The response of this is, which I call CC,

[[[0., 0., 0.], [0., 0., 1.]], [[0., 0., 0.], [.540302305868140, .841470984807897, 1.]], [[-0., 0., 0.], [-.832293673094285, 1.81859485365136, 1.]]]

, which is composed of three lists:
[[0., 0., 0.], [0., 0., 1.]]

[[0., 0., 0.], [.540302305868140, .841470984807897, 1.]]

[[-0., 0., 0.], [-.832293673094285, 1.81859485365136, 1.]]

Then, my question is following:

Seeing CC, I can see three segments are shown.
The first one is  the segment between [0., 0., 0.] and [0., 0., 1.].
But, I can't find any elements of Array BB corresponding to [0,0,0] above in CC.
And as for 0, 0 in [0, 0, 1] in CC above, the first 0 should be (1,2,1) and the second one should be (1,2,2).
But, there are no such (1,2,1) and (1,2,2) in BB. 
Despite of these, BB is converted to CC, from any reason I want to know and ask this question for.

As for [[0., 0., 0.], [.540302305868140, .841470984807897, 1.]] in BB, the circumstances are the same.
I don't know from where [0,0,0] appeared. Surely, as MESH requires three points, you might say that
such things might occur. But, it is only after I converted Array BB to List CC, with plot3d or MESH then having no relation 
to the present circumstances. So, I think that it is not possible to say that Maple added [0,0,0] or such things to Array BB

 as it is about plot 3D or about MESH.

So, my question is how the conversion of BB to CC could have occurred.

Thanks in advance.

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