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)