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I want to test linearly dependence of a polynomial f on a list of polynomials F by additional condition on parametric coefficients of linear parametric polynomial (linear for variables not parameters). Please note that:

  1. The polynomialand the members of are always homogenous in the variables.
  2. The coefficients of f, the coefficients of the members of F are all always polynomials in the parameters or contant and the members of N and W are all always polynomials in the parameters.


For example let


(a,b,c,d,e,h are parameters and A1,A2,A3 are variables).

If I use PolyLinearCombo(F,f,{A1,A2,A3}) (see its output is false,[].

Now we let to condition sets for parameters as the following:



The elements of N must be zero means that ebc+ahd=0

and the elements of W are non-zero that is a<>0 and c <>0.

Let a=b=c=d=h=1 and e=-1. This specialization satisfy in the above condition sets N and W. By this specialization we have:


Now if I use PolyLinearCombo(F,f,{A1,A2,A3}) then its output is true,[-1,1].

By this additional two condition sets I have to check that whether f is linearly independent of F or not. How can I do this without specialization? In fact I want an algorithm that its input is (null condition N, not-null condition W, list of polynomials F, a polynomial f, the set of variables) and its output is true and coefficients if f is linearly dependent of F w.r.t. null and not-null conditions N and W, else its output is false.

If the name of new procedure is ExtPolyLinearCombo and 



I want the output of

ExtPolyLinearCombo(N,W,F,f,{A1,A2,A3}) be true,[coefficients]

Thank you very much in advance.



Hi everyone:

I have two functions in terms of theta variable, how can I expressed function Y(X)?

X :=(theta)->cos(theta)+0.8e-1*cos(3.*theta)

Y :=(theta)->-sin(theta)+0.8e-1*sin(3.*theta)

I will earn Y(X) infact.

Thanks alot...





Hi all,

I want to define variables with indices, possibly zero. So for example I define mu[01] and mu[10]... 

My problem is that mu[01] automatically becomes mu[1] in Maple's output, so that I could not distinguish it from mu[001]... 

Is there any way to address it?


Hello. I have an inequality and I need to prove or negate if it is true or false. This inequality has 8 variables. I simplify it and try to see if it is ture or false. I tried "test relation" in maple and it seems I can't say it is always true or false. For some values of the variables it is true and for some others its false. Is there a method I can show if this inequlity is hold under some assumptions? I mean I want to keep some variables as constant and prove it up to a point. My inequlity is below. Thank you for the help in advance.

(P[A]*(p-w)/(1-P[A])-c)*H[A]+(w-P[A]*(p-w)/(1-P[A]))*P[A]*H[A]+w[u]*P[B]*(1-P[A])*H[B] < (P[B]*(p-w)/(1-P[B])-c)*H[B]+(w-P[B]*(p-w)/(1-P[B]))*P[B]*H[B]+w[u]*P[A]*(1-P[B])*H[B]

And this is how it looks on maple:


Basically the problem is that when I try to derive a variable L, which is defined by a very big equation, Maple does not do the maple. But if I copy and paste the big equation it calculates everything perfectly. Could I get any insight on why is that?

The copy paste method is very space consuming.

Thanks in advance.



I am new to Maple and have a problem when solving three equations with three variables. But when  I plug in into solve function then it gives no answer.

eqn1 := 24900 = A*exp(-X*1.293995859*10^22)+A*exp(-Y*1.293995859*10^22)+5852.27;

eqn2 := 6000 = A*exp(-X*1.293995859*10^22)+2422.929937;

eqn3 := 19100 = A*exp(-Y*1.293995859*10^22)+8275.199937;

Variables are [A,X,Y]

I have my question makes any sense. I am from Denmark and not used to write math in english.

I have an characteristic matrix with an variable λ that takes on differen values.

How do I write λ in the matrix so Maple knows that when I call out a row with the variable λ in it and asssign

λ to a specific value, Maple changes λ to the specific value.


Example (I was thinking something like this):




A[1],λ_1               (1-2) 2

A[1],λ_2               (1-4) 2

A[2],λ_1               3 (4-2)

A[2],λ_2               3 (4-4)

Dear friends,

I would like to know the agorithm of isovle for solving 2 order equations for more than 3 variables.

I see that not all the solutions are shown for example for:


 Could you please help me with this?


we always have subscript variable in the math book, but how could this be natral done in maple

I want to get a seq aaa3


but how could I get a  aij



seq(a[ij],i=1..3);  both was not right

i want to plot a arc that starting angle and finishing angle are variable

but it didn't show the arc there is my code

I have denotation like A[0], A[1], A[2], A[3]... But one package doesn't allow to use indexed variables.

I'd like to change denotation. For example, to A0, A1, A2, A3, but I don't know how to do it automatically...

Aslam-u-Alikum, want to write the following in Maple please help









PhD (Scholar)
Department of Mathematics


I'd like to clean up my project a bit. In a chapter of a project I made lots of calculations and declared a many (30) variabeles (Table, Numeric, Formula, ..)

Is there a way to remove all variable's except specified one's? (If it is possible I don't want to use an external file to write it to and read it back after a restart)

Is there a way to do a "restart" and preserve only the one's (2) that I need for my next chapter?
Or if not, without a "restart" and remove all variables except specified one's?

Thanks for your help, 

Is it possible in Maple 15 to solve an equation with a parameter for a given set of parameters? How can this be passed to the solve function, should I use some kind of list?  After obtaining the solution how can I assign the solutions to variables such as x1 for the first value of the parameter, x2 for the second value of the parameters and so on. Furthermore, is this possible with the fsolve command?



Hi guys.

I have variables that look like C[i,j,k,l], where i,j,k,l can range between 1..3 for example. Is there a simple way to create a pattern-based assumption for something like C[i,j,k,l] = C[j,i,k,l] for all i,j,k,l ?

Kind regards,


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