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Hello people in mapleprimes,


I have a question. I hope someone give an answer to me.




brings an error

Error, invalid input: limit expects its 2nd argument, p, to be of type Or(name = algebraic, set(name = algebraic)), but received ((x-a)^2+(y-b)^2)^(1/2) = 0

Isn't other way than the following?



Best wishes.


I am working in Document mode, 2D input.  I have doing some complex number evalutions

Lets say I have a complex number Z= 2+3i

when I ask maple for the "abs(Z)" I get the display of "abs(2+3i)" rather than the numberic answer. This occurs with all the complex operation arg, abs, polar conversion, 

polar(abs(2+3*i), argument(2+3*i))   this is from Maple. 

How do I display the numeric value for these functions in my document


Thanks Bill


I've got a load of data that contains a lot of noise in the form of sinusoidal interference patterns. This is quite thick and it disturbes what I am trying to look at. I've uploaded a picture to represent what I am trying to show. I know I need a cut-off frequency I just don't know how to implement it. Thank you in advance for your help!

Gambia Man

In maple18 has:


but how to see step-by-step factor ??

(Sorry! I'm not very good at English.)

Thank you!

Dear Community,

I've made a nonlinear curve fit with the Minimize routine (see attachment). What would be an easy and elegant way to rerun the model (Model) with the fitted values of a, b, c and plot the result together with the measured points in the same chart? I'm stuck here.

Tx in advance,

best regards




Hey there,

I am using the fsolve command in order to solve numerically a system of equations with N equations and N unknowns. According to my discretization the number of equations changes. If I have a small number of equations it all works out perfectly. But if I increase the number of equations I just get something like that:

Sorry, for the long post, but for a small number of unknowns Ai it works. It seems that maple doesnt try to compute? Has anyone encountered the same problems?


Any help is appreciated.


Greetings to all. The title describes it well, I am writing about testing the limits of the Maple integration engine. A recent discussion at features a family of integrals that involve the product of a power of the natural logarithm and a rational function, more precisely,

int((log(x))^n/(x^3+1), x=0..infinity);

These integrals can be evaluated recursively as described at the MSE link using a technique that generalizes to other types of rational factors. Unfortunately Maple apparently only finds a simple closed form for a few small initial values of n. The following transcript of a Maple session illustrates the problem. Mathematica was successful here. Also observe the memory allocation in the Maple session.

    |\^/|     Maple 18 (X86 64 LINUX)
._|\|   |/|_. Copyright (c) Maplesoft, a division of Waterloo Maple Inc. 2014
 \  MAPLE  /  All rights reserved. Maple is a trademark of
 <____ ____>  Waterloo Maple Inc.
      |       Type ? for help.
> restart; read `cl.maple`;
alpha := (n, k) ->

    -1/3 exp(1/3 I Pi + 2/3 I Pi k) (1/3 I Pi + 2/3 I Pi k)

Q := proc(n)
local res;
option remember;
    if n = 0 then return 2/9*sqrt(3)*Pi end if;
    res := -add(alpha(n + 1, k), k = 0 .. 2)/(n + 1) - add(
        binomial(n + 1, p)*(2*I*Pi)^(n - p)*Q(p),
        p = 0 .. n - 1)/(n + 1);
end proc

                              /               n
                             |          log(x)
              VERIF := n ->  |          ------- dx
                             |           3
                            /           x  + 1

> Q(6);
                                7  1/2
                          910 Pi  3

> VERIF(6);
memory used=3.8MB, alloc=40.3MB, time=0.18
       7  1/2
9890 Pi  3       490    5  1/2
------------- + ----- Pi  3    Psi(1, 1/3)
   177147       19683

        490    5  1/2                10    3  1/2            2
     + ----- Pi  3    Psi(1, 2/3) + ---- Pi  3    Psi(1, 1/3)
       19683                        2187

        20   1/2   3
     + ---- 3    Pi  Psi(1, 2/3) Psi(1, 1/3)

        10    3  1/2            2    40                 4
     + ---- Pi  3    Psi(1, 2/3)  + ----- Psi(2, 2/3) Pi
       2187                         19683

        10   1/2               3
     + ---- 3    Pi Psi(1, 1/3)

       10               1/2               2
     + --- Psi(1, 2/3) 3    Pi Psi(1, 1/3)

       10   1/2                           2
     + --- 3    Pi Psi(1, 1/3) Psi(1, 2/3)

        10   1/2               3    40     4
     + ---- 3    Pi Psi(1, 2/3)  - ----- Pi  Psi(2, 1/3)
       2187                        19683

        20             2  1/2
     + ---- Psi(2, 2/3)  3    Pi

        40               1/2
     - ---- Psi(2, 2/3) 3    Psi(2, 1/3) Pi

        40    2
     + ---- Pi  Psi(2, 2/3) Psi(1, 1/3)

        40    2
     + ---- Pi  Psi(2, 2/3) Psi(1, 2/3)

        20   1/2            2
     + ---- 3    Psi(2, 1/3)  Pi

        40    2
     - ---- Pi  Psi(1, 1/3) Psi(2, 1/3)

        40    2
     - ---- Pi  Psi(1, 2/3) Psi(2, 1/3)

> evalf(Q(6));

> evalf(VERIF(6));

> quit
memory used=22.4MB, alloc=44.3MB, time=0.47
user@host:~/complex-logint$ math
Mathematica 10.0 for Linux x86 (64-bit)
Copyright 1988-2014 Wolfram Research, Inc.

In[1]:= Integrate[Log[z]^6/(1+z^3), {z, 0, Infinity}]

          910 Pi
Out[1]= ------------
        2187 Sqrt[3]

In[2]:= N[Out[1]]

Out[2]= 725.573


My question for you all is what the appropriate techniques would be to get Maple to at least simplify the rather involved output from the integration engine to obtain a match of the closed form from the recursive equation.

Best regards, Marko Riedel.


Below z is made using different complex values on polar form, and I then need to express the resulting z on polar form with numeric values for length and angle.  However, I had no luck using evalc, evalf, or other I could find.

How can I convert z to a polar form with numeric arguments like shown below ?

Hi everybody,

I would like to define a function with random values to be used in pdsolve (numeric) as a initial condition.

Any help?



Trying to solve this IVP of the SHO  (second order linear costant-coefficient).

Everything works fine until I come to the solving even after using dsolve with initial conditions (even using the differential operator D in the initial conditions)  , the answer still contains _C1, an unknown constant.

The full worksheet is below.  The code for dsolve is:

sol3 := dsolve(subs(par1, {de1, D(x)*0 = 0, x(0) = 1}), x(t));


Hoping you can help with a solution.





Hi guys.

           if an expression complicated as the following,  

sqrt(sqrt(9)*sqrt((1+(b+1)^2*c^2+((10/3)*b-2)*c)*(1+(b+1)^2*c^2+(2*b-2)*c))+3+(3*b^2+6*b+3)*c^2+(8*b-6)*c) , where b>0 c>0

is it possible to tell whether it could be positive? 

I used coulditbe command, however, it returned 'FAIL'.

Having a function where the value is for example only defined when abs(x) <= 1, then how can I specify that the value is otherwise undefined, the replacing "How_to_specify_undefined_value" below?

I am using Maple 2015.2 on a Windows 10 machine.  I use the plot command to generate a simple graph.  I then use the Manipulator Pan tool to change the axes limits.  The system does not redraw the function with the new range limits.  The parameters in the Axis properties have been changed appropriately but the graph does not display for the new limits.  Even if I change the parameters without using Manipulator Pan tool, the system does not redraw the function with the new range limits.  Any guidance about what I am doing wrong?  I am attaching an example file in case the behavior continues on other systems.



Dear All

It is well known that the package "PDEtools" is helpful in finding infinitesimal transformations for PDEs which I illustrate as follow:


DepVars := [u(x, y, t)]

[u(x, y, t)]


declare(u(x, y, t)):

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


U := diff_table(u(x, y, t)):

PDE1 := U[t, x]+(3/2)*u(x, y, t)*U[x, x]+(3/2)*U[x]^2+(1/4)*U[x, x, x, x]+(3/4)*sigma*U[y, y] = 0:

G := [seq(xi[j](x, y, t, u), j = [x, y, t]), seq(eta[j](x, y, t, u), j = [u])]:


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


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


DetSys := DeterminingPDE(PDE1, G, integrabilityconditions = false):


{eta[u](x, y, t, u) = (1/9)*(-2*(diff(diff(diff(_F1(t), t), t), t))*y^2-4*(diff(diff(_F2(t), t), t))*y+6*sigma*(-(3/2)*(diff(_F1(t), t))*u+(1/2)*(diff(diff(_F1(t), t), t))*x+diff(_F3(t), t)))/sigma, xi[t](x, y, t, u) = (3/2)*_F1(t)+_C1, xi[x](x, y, t, u) = (1/6)*(-2*(diff(diff(_F1(t), t), t))*y^2-4*(diff(_F2(t), t))*y+3*sigma*((diff(_F1(t), t))*x+2*_F3(t)))/sigma, xi[y](x, y, t, u) = (diff(_F1(t), t))*y+_F2(t)}


The set (4) gives infinitesimal transformations. How we can write  vector fields corresponding to arbitrary constant C1and arbitrary functions "F1(t), F2(t), F3(t) "?"" 




I am having 26th degree polynomial univariate equation , I used Isolate to get the roots. but I am getting some extra roots which are not true they I even tried to substitute those roots in original equation then I got non zero answer instead of getting nearly zero answer.How is it possible??


equation looks like:


Solutions i got:

[t = -4.162501845, t = -2.295186769, t = -1.300314688, t = -.8048430445, t = -0.6596008501e-1, t = -0.4212510777e-1, t = 0.4212510777e-1, t = 0.6596008501e-1, t = .8048430445, t = 1.300314688, t = 2.295186769, t = 4.162501845]

t=4.162501845 give me non zero answer when I substitute it in the equation given above:

I got this answer: 4.750212083*10^39


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