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I was trying to find the solution for two theta variables in a couple of simultaneous equations (infact this is an iverse kinematics problem for a two link system pendulum).
The following are the initial inputs/equations to be manipulated:

Then I use the folowing command to rearrange for the theta values which I am after:

which gives me the result:

This is all fine until I give in values for l1, l2, x and y:


I have a RootOf in there with a _Z term poping up here and there. I know that this configuration of the two link mechanism in fact dows have a solution and that these numbers are reasonable. Thus I have three questions:

Why does this happen?
What does the "signum" mean here?
how do I go about getting the nummerical values?

Many thanks,
- pjf


I have 20 by 20 matrix (with only two symbolic components, say a and b), other entries of the matrix are either populated with zeros or real numbers. I am trying to obtain the inverse.


All the command I have tried returned an error message. In particular M^(-1) yields:

Error, (in radnormal/ifactors) too many levels of recursion.


I would appreciate any suggestion that could solve this issue. Thank you


ps: I am essentially trying to solve Ax=b. The commands within LinearAlgebra all yielded similar error messages.

where A^T is the transpose of A and it's given that (A^T).A is not invertible.

I am stuck as to how to arrive at the solution for x in this case. I initially thought I could multiply both sides by the inverse of A^T reducing it to Ax=b but that was obviously wrong since A^T is itself not invertible(it is singular).

Dear All


I have a question:

How can I computet the laplace inverse of a function that is below!

My version of Maple is 13:

It should be noted that r>0, \rho>0.5, y\in R.


> restart;
> with(MTM);
print(??); # input placeholder
> ilaplace(r^(rho-.5)*abs(y)^(rho-1)*exp(-r*y)/GAMMA(rho-.5), s);

Dear all,

I have a question, why is the output of the inverse Laplace transformation if the signal is multiplied by itself not just convoluted in time domain:









assume system of polynomials are above, how to test whether inverse exist

and find inverse of them




I am trying to find the inverse of 8x8 generic symbolic matrix. Everytime I evaluate the program I get the following error:


Error, (in expand/bigprod) Maple was unable to allocate enough memory to complete this computation. Please see ?alloc


Can anyone guide me how to increase the memory allocation? And would that solve the problem for me?


Any help is appreciated.



Dear all,

It's very convenient to define a DE or PDE through Differential Operator D, for example,

((D[1, 1]+D[1, 2]+D[2, 2])(z))(x, y) = exp(x)*sin(y)

Is it possible to realize Inverse Operator Method of Operator D? How to solve the following equation if we rewrite the pde through inverse operator method?

(z)(x, y)=((D[1, 1]+D[1, 2]+D[2, 2])^(-1))exp(x)*sin(y)


Thanks a lot.

How to find inverse function of a multivariable function

for example

f := x^2 + y + z^3

f := x^2 + y^3

I am trying to invert a function and, during the course, wind up with something like,


a_1*V^P*V + a_2*V^P*V^2 + a_3*V^P*V^3 + a_4*V^P*V^3 + a_5*V^P*V^3 + a_6*V^P*V^4 + \cdots


is there a way to get the poly into a form 


(a_1+a_2+...)*V^P + (b_1+b_2+...)*V^(P+1) + (c_1+c_2+...)*V^(P+2) + .....



Thanks. Attached is my Maple 16 sheet.




main := V^P*(V-1)^Q = c^(P-Q)*X^Q*(V-c^2)^Q

V^P*(V-1)^Q = c^(P-Q)*X^Q*(V-c^2)^Q



assume(Q, real); additionally(Q > 0); additionally(Q, integer)

assume(P, real); additionally(P > 0); additionally(P, integer)``



you := proc (s, n) V[1] := s; for i to n do V[i+1] := (`@`(simplify, expand))(c^((P-Q)/P)*X^(Q/P)*(-c^2+V[i])^(Q/P)/(V[i]-1)^(Q/P), symbolic) end do end proc

proc (s, n) local V, i; V[1] := s; for i to n do V[i+1] := (`@`(simplify, expand))(c^((P-Q)/P)*X^(Q/P)*(V[i]-c^2)^(Q/P)/(V[i]-1)^(Q/P), symbolic) end do end proc








Rearranging "main" leads to the following.


main2 := X^Q = V^P*(V-1)^Q/(c^(P-Q)*(V-c^2)^Q)

X^Q = V^P*(V-1)^Q/(c^(P-Q)*(V-c^2)^Q)



It can't take a series due to the series about zero, so I split it into


split := X^Q = V^P*(a[1]+a[2]*V+a[3]*V^2+O(V^3))

X^Q = V^P*(a[1]+a[2]*V+a[3]*V^2+O(V^3))



Then I'll get a polynomial expression, switch places of the two variables' leading terms, and substitute into the "original" polynomial (split).



  one := (V-1)^Q/(c^(P-Q)*(V-c^2)^Q)




two := `assuming`([simplify(`assuming`([series(one, V, 3)], [V > c^2, c > 1]))], [real])




three := convert(two, polynom)




four := simplify(expand(V^P*three))




five := collect(four, V, factor)



collect(five, V)



factor(five, V)

Error, (in factor) 2nd argument, V, is not a valid algebraic extension











La aplicacion de las matrices en su más claro ejemplo, dirigido explicitamente a la criptografia; solo hay que tener conocimiento de algebra de matrices y calculo de matriz inversa. (versión español).

The application of the matrices in its most obvious example, explicitly directed to cryptography; just have to have knowledge of matrix algebra and inverse matrix calculation. (version english).


Lenin Araujo Castillo

Physics Pure

Computer Science


I am trying to obtain an inverse of a symbolical 4 by 4 matrix in order to diagonalize it. I have noticed that the command 'Inverse' seems to find a matrix inverse almost instantaneously, whereas the Matrix Inverse command from the Linear Algebra package takes far longer - on the order of hours. Can anyone explain the difference between them? It seems like 'Inverse' is roughly speaking, an approximation to 'Matrix Inverse', however I have not been able to find...

Hi, I am trying to do the calculation with a matrix, and I have hard time to find a way to calculate the Moore-Penrose generalized inverse of a matrix in Maple. Could anybody help me?

Hello Everyone,

I got a question about the inverse laplace transform of the classic second order system:


zeta < 1

if I take the inverse laplace transform, it will give me root of (zeta^2-1) term, how can I force maple to consider the range of zeta before doing inverse laplace transform?


I have an inverse kinematics model for a double inverted pendulum. There are multiple solutions for each joint angles due to symmetry and I have to choose the appropriate ones. I thought I can use the previous choices and compare them with the current multiple solutions, so I could choose the ones which are closer to the previous ones. However, since the double inverted pendulum moves in in both region at the equilibrium, this aproach does not work. Could you advise...

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