Ho to find function the Superimum of function (sup f(x) ), for example, I want to find the supremum of the exponential distribution i.e. f(x)=lambda*exp(-lambda*x)
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Dear Muhammad Ahsan-ul-Haq
To find a supremum of one variable function is an easy problem.
Assume that you have y = f(x): (a,b) into R, then compute the derivative dy/dx.
If dy/dx>0 for all x, then y = f(x) is increasing and the sup at b and the inf at a.
If dy/dx 0, the y = f(x) has minimum (inf.)at ( x0, y0 ).
d2y/dx2 (x0, y0 ) < 0, the y = f(x) has maximum (sup.) at ( x0, y0 ).
d2y/dx2 (x0, y0 ) = 0, the y = f(x) has inflection point at ( x0, y0 ).
For y = λe^(- λx) , then dy/dx= - (λ)2 e^(- λx) < 0
If y is defined over a bounded interval say (a,b), then sup is at a.
If the domain is R, then y goes to zero as x goes to + infinity ( if λ >0)
and y goes to infinity as x goes to + infinity ( if λ < 0).
We do a similar argument when x goes to - infinity.
Therefore, the answer is based on the value of λ and the domain of definition.
Best regards
I do not agree. Just imagine an oscillating function in one variable over a closed interval. The oscillations can be tighter and tighter around some points, and the highs and the lows can also vary in an uncontrolled fashion - like frantic radio waves. There is no chance that you can guarantee to find a global maximum or minimum of a function like that. If you say that we can bisect, I say let's tighten the oscillations. It's a game that you will lose. :-) At least you will have to wait until the Sun swallows the Earth.
Oscillations show boundedness. By applying the completeness property of the real numbers, infimum and supremum always exist.
I guess you mean: searching analytically the smallest real value M such that f(x)
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