The entropy function has its maximum at x=1/2
Yet when one uses solve:
x=var('x')
assume(x>0)
solve(derivative(-x*log(x) -(1-x)*log(1-x),x)==0,x)
sage returns something that is correct, but curiously non-specific
[log(x) == log(-x + 1)]
It almost seems like sage can't figure out that x=1/2 solves the above even after the assumption on x.
Instead,
solve(e^derivative(-x*log(x) -(1-x)*log(1-x),x)==1,x)
gives
[x == (1/2)]
Alternatives :
sage: Ex=-(x*log(x)+(1-x)*log(1-x))
sage: solve(Ex.diff(x),x)
[log(x) == log(-x + 1)]
Fails indeed. but :
sage: solve(Ex.diff(x),x, algorithm="sympy")
[x == (1/2)]
sage: solve(Ex.diff(x),x, algorithm="giac")
Warning, argument is not an equation, solving -ln(sageVARx)+ln(-sageVARx+1)=0
[1/2]
sage: solve(Ex.diff(x),x, algorithm="fricas")
[log(x) == log(-x + 1)]
Not directly translatable as of 9.7.beta6 :
sage: mathematica.Solve(Ex.diff(x)==0,x)
{{x -> 1/2}}
Somewhat trichodynamic :
sage: foo=solve(Ex.diff(x),x)[0]
sage: foo.operator()(*map(exp, foo.operands())).solve(x)
[x == (1/2)]
HTH,
Asked: 2022-08-04 12:25:34 +0100
Seen: 190 times
Last updated: Aug 04 '22
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