Hello, If you have an aperiodic function that attains a maximum at some later time say tau, what command in maple would allow you to find it? Mathematica, as far as I know, is only limited to specifying where you think the greatest maximum might be and the only alternative to automate this would only find the first maximum not the greatest maximum.

Reduce::nsmet: This system cannot be solved with the methods \
available to Reduce. >>

 

If I have some element tau in my matrix, how do I declare it to be real?

 

> with(LinearAlgebra);
> A := `<|>`(`<,>`(-sqrt(5/3)*tau, -tau, 0, 0), `<,>`(-tau, -sqrt(3/5)*tau, -2*tau/sqrt(15), 0), `<,>`(0, -2*tau/sqrt(15), sqrt(3/5)*tau, 0), `<,>`(0, 0, -tau, sqrt(5/3)*tau));

 

How do I find for which t is the following equation true?

 

_EnvAllSolutions := true; solve([((1/24)*sqrt(1/11*(819+210*sqrt(5)-126*sqrt(11)-90*sqrt(55)))*cos(sqrt(2/5*(7-2*sqrt(11)))*t*tau/h)+(1/8)*sqrt(1/33*(273+70*sqrt(5)+42*sqrt(11)+30*sqrt(55)))*cos(sqrt(2/5*(7+2*sqrt(11)))*t*tau/h))^2+(-(1/8)*sqrt(1/11*(91+39*sqrt(5)-2*sqrt(22*(47+21*sqrt(5)))))*sin(sqrt(2/5*(7-2*sqrt(11)))*t*tau/h)+(1/8)*sqrt(1/11*(91+39*sqrt(5)+2*sqrt(22*(47+21*sqrt(5)))))*sin(sqrt(2/5*(7+2*sqrt(11)))*t*tau/h))^2 = 1, t > 0, h > 0, tau > 0], t)