Minimization (especially in several dimensions) is in general a tough task, and it is usually better to facilitate the task for Mathematica. So normally in calculations in Mathematica it is easier to replace the minimization procedure by an equivalent one, i.e. to take the derivative. The following example
function[x_, m_] := x^m;
derivative = D[function[x, m], x];
Solve[derivative == 0, x]
gives the desired analytical result
{{x -> 0^(1/(-1 + m))}}
for any parameter m
. It tells you that the only extremum point point of the function is 0
(for m<>1
). This means that if you take the interval x>=1 for minimization, your function does not have extremum points inside it and hence attains its minimum at the boundary.
You can also directly write
Solve[{derivative == 0,x>1,m>0}, x]
to see that no extremum points exist inside the defined interval.
To conclude, you can analytically minimize complicated expressions in Mathematica, but you should wisely choose the way you formulate the problem for the computer.
Thanks to gwr
for pointing out that my previous answer didn't work.
$Assumtpions
will only be used by functions which themselves have anAssumptions
option. I know this doesn't solve your problem, but at least it clarifies why this doesn't work. It seems sometimes it's possible to include assumptions on the parameters in the constraint list, but in this case it doesn't work. $\endgroup$mminimize x^m when x >= 1 and m>=0
andminimize x^m when x >= 1 and m<0
in Wolfram Alpha! The first question is bestowed with a better fitting answer than the second one. As for $0>m>-\infty$ the function $x^m$ has no global minimum the second answer suffers from numerical defects. $\endgroup$Minimize[{x^m,x>=1&&m>0},x]
(additional condition explicitly given) andMinimize[{x^(Abs[m]+1/10),x>=1},x]
(the exponent being obviously positive).Minimize[{x^2,x>=1},x]
works fine, however. $\endgroup$