Items tagged with symbolic

I can not find a solution to the integral of the function below the maple, can anyone help me?



assume(-1 < rho and rho < 1, alpha1 > 0, beta1 > 0, alpha2 > 0, beta2 > 0, t1 > 0, t2 > 0)

f := proc (t1, t2, alpha1, beta1, alpha2, beta2, rho) options operator, arrow; (1/4)*(sqrt(beta1/t1)+(beta1/t1)^(3/2))*(sqrt(beta2/t2)+(beta2/t2)^(3/2))*exp(-((sqrt(t1/beta1)-sqrt(beta1/t1))^2/alpha1^2+(sqrt(t2/beta2)-sqrt(beta2/t2))^2/alpha2^2-2*rho*(sqrt(t1/beta1)-sqrt(beta1/t1))*(sqrt(t2/beta2)-sqrt(beta2/t2))/(alpha1*alpha2))/(2-2*rho^2))/(alpha1*beta1*alpha2*beta2*Pi*sqrt(1-rho^2)) end proc

int(int(f(t1, t2, alpha1, beta1, alpha2, beta2, rho), t2 = 1 .. infinity), t1 = 0.1e-2 .. y)



Is there a way to solve matrices symbolically?

an example would be A*X=B

where the answer would be 



I have tried to look for a thing in maple that will do this but so far i had no luck. Does anyone know ?

Thanks in advance


I have encountered some strange issue with symbolic summation. Would be grateful for any help.

Here is the code (inserted as image):

Symbolic summation of powers of sines & cosines


Code in text:



The issue is that symbolic summation produces the formula (for general n) which contradicts the particular case (n=4).

Could somebody explain why this is happening? Is it a bug or am I missing something here?

I have tried all versions of Maple downto 14 - same situation.
Also Mathematica givers the same answer.

Thank you in advance,
Pavel Holoborodko.

Multiprecision Computing Toolbox

I'm running into a very simple problem with the way that Maple integrates Heaviside functions. Naively, it should act like a step function, but it is not integrating properly. See the attached document.

int(int(Heaviside(-x^2-y^2+1), x = -1 .. 1), y = -1 .. 1)



evalf(Int(Heaviside(-x^2-y^2+1), [x = -1 .. 1, y = -1 .. 1]))



int(piecewise(-x^2-y^2+1 > 0, 1, 0), [x = -1 .. 1, y = -1 .. 1])




Note that the symbolic integration of the Heaviside function (defined to be 1 inside the unit circle and 0 outside) gives zero, whereas it should clearly give the area of the unit circle, which the numerical integration does. I even checked that the (suposedly equivalent) piecewise definition symbolically evaluates to the area, and it, too, gets the right answer.

Anyone have any clue as to why the symbolic integration of this Heaviside function is so wrong? My understanding is that if we do the integral as two nested 1D integrals, the returned function (as a function of y) is zero everywhere except at y=0, but that result cannot be right either.





I would like to determine a closed form solution (=analytical solution) of the following trigonometric equations system.

The unknowns are :

ListAllUnknowns := [Psi(t), Theta[1](t), Theta[2](t), x[1](t), x[2](t), z[1](t), z[2](t)]

Do you have ideas so as to conduct the symbolic resolution of this trigonometric equations system ?

I have been told that the use of Grobner basis could be useful but I have never try this.

Thanks a lot for yours feedbacks.

Hello everyone,
I would like to get a symbolic result of each variable x,y and z for the following 3 nonlinear equations. Maple does not respond to the following code at all. (Not even an error report.)


eq1 := x^2+y^2+z^2-134*x+800*y-360*z+31489, 2;
eq2 := x^2+y^2+z^2-934*x+900*y-370*z+321789, 2;
eq3 := x^2+y^2+z^2-614*x+1350*y-1110*z+70048, 97;
solve({eq1, eq2, eq3}, {x, y, z});

Thanks in advance.

P.S: Afterwards my intention is to solve these equaitons numerically for different variable values, and transfer to MatLab in order to plot animations and graphs. 

Hello guys,

I was just playing around with differential equations, when I noticed that symbolic solution is  different from the numerical.What is the reason for this strange behavior?

ODE := (diff(y(x), x))*(ln(y(x))+x) = 1

sol := dsolve({ODE, y(1) = 1}, y(x))

a := plot(op(2, sol), x = .75 .. 2, color = "Red");
sol2 := dsolve([ODE, y(1) = 1], numeric, range = .75 .. 2);

b := odeplot(sol2, .75 .. 2, thickness = 4);
display({a, b});

Mariusz Iwaniuk


I have been trying to solve the following equation with respect to y, but I have not been successful. In fact, I always get answer RootOf(...). I should mention that all variables and parameters are real non-negative. I have also tested with "assume", but it did not help. Any suggestion would be appreciated. 


eq := -((y-b)*mu-y)*x^beta*alpha+y^beta*varepsilon*(x-a) = 0

-((y-b)*mu-y)*x^beta*alpha+y^beta*varepsilon*(x-a) = 0


solve(eq, y)




-x^beta*alpha*b*mu = 0








I would like to symbolically determine the rank of a jacobian matrix. In the help, I have seen that the Rank function of the LinearAlgebra can be used for this purpose. However, when I use this function, the function doesn't allow to find the different singularities that can occur on my jacobian matrix.

Here a exemple of a jacobian matrix that I obtain on a slidercrank mechanism:

Phi := Matrix(2, 3, {(1, 1) = -l1*sin(theta(t)), (1, 2) = -1, (1, 3) = l2*cos(beta(t)), (2, 1) = l1*cos(theta(t)), (2, 2) = 0, (2, 3) = l2*sin(beta(t))})

The rank of this jaobian (Phi) gives 2 whatever the values of theta(t) and beta(t). However, if the values of  theta(t) and beta(t) are :theta(t)=Pi/2,beta(t)=0. The rank shouldn't be 2 but 1.

Is a way to obtain the symbolic calculation of the rank of a jacobian matrix which can distinguish different cases following the values of the parameters ? In others words, my dream will be to have a Rank function (or another algorithm) which can gives :
the rank is 2 if theta(t) different of Pi/2 [Pi] and beta(t)=0 [Pi] 
and otherwise 1 if ...
and perhaps 0 if ...

Thanks a lot for your help.

I let a piece of code with an example of calculation of the rank

Hi guys,
I want to import symbolic matrix from matlab to Maple, How I can do that ? 

I solve a linear system of equations which is rank deficient. Naturally, when Maple solves it symbolically, it chooses some of its variables to use them as a basis to express the solution. 

In a specific problem I'm solving, the basis chosen by Maple is -very- smart, showing a good exploitation of the problem structure. 

I'm curious as to what kind of factorization is used by default, or if there's a lot of by hand "black magic" involved, what are its general characteristics. 


Best regards


Sorry for the uninformative title. I've never used Maple, but I'm willing to buy a student license and learn it. But before spending too much effort and money I need to know if it suits my needs.

Basically what I need to do is:

1) I have a positive definite symmetric matrix of size nxn, where n can range from 2 to inf. I don't know the elements, except the fact that the diagonal has ones everywhere. All I know is that the elements out of the diagonal are in the range [0,1)

2) I have to compute the lower triangular cholesky decomposition of this matrix, lets call it L.

3) I need to subtract from each element of L the mean of the elements in the respective column. Lets call this matrix L*

4) Then I need to evaluate another nxn matrix computed from the elements of L* following a simple pattern.

5) Finally I need to find the eigenvalues of this last matrix.

What I would ideally want is to get a symbolic representation of the n eigenvalues as symbolic functions of the (unknown) elements of the matrix at point 1.

I can drop the assumption of n being unknown, i.e. fix n=3 and get the 3 functions that, after replacing the right values, give me the eigenvalues, then fix n=4 and get 4 functions, etc.

Is this possible to do in maple?

Thank you

I am able to get unlimeted numbers of equations describing my system. These equations are generally relate quotients of multivariate polynomials. Each additional equation I get is generally less than twice the length of the last, and it is not always the case that an equation is independant of the previous equations. Although I can get unlimited numbers of equations describing the system, it is not overdetermined.

I am interested in solving these equations for their variables. There are about 30 cases I am working on, the smallest number of evariables is six, the largest would be twenty.

I want to be able to solve these equations in the minimal time possible. But I don't understand the function solve well enough to do so.

How do I choose the equations to minimise the time taken for the command solve to proccess them?
How does the command solve work?


  1. if I process the command solve([Eq1,Eq2,Eq3...Eqn],variables) would the command solve([Eq[1],Eq[2],Eq[3]...Eq[n],Eq[n+1]],variables) take longer if Eq[n+1] is not indipendant of the previous equations? 
  2. Is there a way of checking whether Eq[n+1] is independant of the previous vequations, fast enough for it to be useful to check the equations before they are processed?
  3. Does the ordering of the equations affect the speed of solve?
  4. Is there a way of pre processing the equations before they are put into solve that will save it time? (for example factorising them, simplifying them etc...)



I try to find the exact (symbolic) value of


I tried 'simplify' with different options and 'convert'. It would be pi=3.141... as numerical approximation suggests.

Many thanks.

when the variable is in limits of integration?

there exist an error : 

Error, (in int) unable to compute a numeric answer for symbolic limits, q = -3.141592654+arccos(11./(1.+.1000000000*t)) .. 3.141592654-1.*arccos(11./(1.+.1000000000*t))

How to make this code work?

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