SandorSzabo

I want to obtain the exact (symbolic) solution of

240*t^3 + 144*t^2 - 135*t -52 =0

in the form  a+ b*I, where a, b are (symbolic) real numbers.

It is possible if I understand Wikipedia well.

"solve" gives "RootOf" and the "convert(......, radical)"  gives quantities such as

(9522 + 45*I* squarerootsymbol(226511))^(1/3)

Thanks.

I'm at the very beginning of Goebner bases, so sorry if my question is too elementary or I don't use the correct terminology.

I have polynomials, p1(x1,...,x9), ..., p8(x1,...,x9), and q(x1,...,x9).

I need an algebraic representation of q by p1,....  . Since the answer can be long, it would be useful to obtain the "coefficients" separately also.  One (a bit stupid) example is:

p1(x1,x2):= x1 + x2,  p2:= x1^2 + x2^2,  q(x1,x2):= x1^3 + x2^3.

Maple gave the following answer

x=RootOf( _Z^2*a+1, label=_L1)*b

Does it mean that

(x/b)^2*a+1=0 ?

Thanks.

The question is simple.

How could I find an earlier (not Active) forum topic?

(I use Firefox 3.5 on xp.)

Thanks,    Sandor

Maple produces a very long answer which I can see on the monitor, but on the printer it is truncated.

Is there a way somehow wrapping the too long lines?

Thanks

How could I prove by Maple (11) that the following matrices are unitarily equivalent?

first row           cos(theta)    -sin(theta)

second row      sin(theta)      cos(theta)

and

first row       exp(i*theta)       0

second row         0           exp(-i*theta)

First of all: I'm so sorry, I posted my question to a wrong place ( into the poll).

So I copy here the question and the answer of jakubi and my reply to jakubi.

Question:

I would like to solve the following system

x*(2*sin(x)*y^2+x^3*cos(x)+x*cos(x)*y^2) = y*(2*sin(y)*x^2+y*cos(y)*x^2+y^3*cos(y)),

-x*sin(x) = y*sin(y);

Until I have founded only the solutions

x = k*Pi,  y = +/- k*Pi;    k is any integer.

jakubi's answer:

more solutions

I have 5 matrices, Hmatrix[1], and so on.

I can test

map(IsDefinite,Hmatrix)

Maple gives the 5 answers.

How can I test using map the negative_semidefinite property?

Thanks.

Assume n is positive integer.

dsolve((1-x^2)*(diff(y(x), x, x))-2*x*(diff(y(x), x))+(n*(1+n)-n^2/(1-x^2))*y(x), y(x))

Maple says

y(x) = _C1*LegendreP(n, n, x)+_C2*LegendreQ(n, n, x)

However

 dsolve((1-x^2)*(diff(y(x), x, x))-2*x*(diff(y(x), x))+(n*(1+n)-n^2/(1-x^2))*y(x), y(x), 'formal_solution', 'coeffs' = 'mhypergeom')

Maple says

y(x) = _C1*(Sum((-1)^(2*_n1)*GAMMA(_n1-(1/2)*n)*x^(2*_n1)/GAMMA(_n1+1), _n1 = 0 .. infinity))/GAMMA(-(1/2)*n)

I want to prove by Maple if  -1<x<1 and n is positive integer then

sum(k^n*x^k, k = 1 .. infinity) = (sum([sum((-1)^(m+1)*(binomial(n+1, m-1))(r-m+1)^n, m = 1 .. r)]*x^r, r = 1 .. n))/(1-x)^(n+1)

I would appreciate for any help.

By Maple

convert(LegendreQ(1,x),hypergeom)

hypergeom([1, 3/2], [5/2], 1/x^2)/(3*x^2)

LegendreQ(1,0.)

-1.000000000-3.141592654*10^(-15)*I

The 2F1 form suggests for me that in x=0 LegendreQ is infinite or at least undetermined. So what is the truth?

I work with LegendreQ(1,x) where x is real.

Thanks,

                      Sandor

convert( (z)_n, GAMMA)  does not give the desired  GAMMA(z+n)/GAMMA(z).

Instead of Maple gives the answer   (z)_n.

How could I obtain the GAMMA form?

Thanks,

                   Sandor

In Maple11 odeadvisor does not know the Euler type differential equation.

For example, R(r)+ r R'(r)+r^2 R''(r)=0.

It is the same situation in Maple12?

Thanks, Sandor

I have the expression

p(x,a) w''(x) + q(x,a) w'(x) + r(x,a) w(x)

How could I obtain p(x,a), etc.?

Thanks,

              Sandor

I have a matrix,  T:=[cos(x)  -sin(x); sin(x)  cos(x)].

Maple11 can calculate the eigenvalues symbolically. Nice property. However

(v,e):=Eigenvectors(T)

gives

Error, (in LinearAlgebra:-LA_Main:-Eigenvectors) expecting either Matrices of rationals, rational functions, radical functions, algebraic numbers, or algebraic functions, or Matrices of complex(numeric) values.

Is there any possibility to calculate the eigenvectors symbolically?

Thanks,