I am reading a paper and they mention a hump function maximization: I am trying to prove the point of maximization: m = ( 1 − x ) 1 − σ x σ where x , σ ∈ [ 0 , 1 ]It is said that m is a hump-shaped function of x maximized at x = σ, where σ is a parameter that is assumed to be fixed here.My attempt:First derivative with respect to x: ( 1 − x ) − σ x σ + ( 1 − x ) 1 − σ x σ − 1 = 0 1 + ( 1 − x ) 1 x − 1 = 0Second derivative with respect to x: − ( 1 − x ) − σ − 1 x σ + ( 1 − x ) − σ x σ − 1 − ( 1 − x ) − σ x σ − 1 + ( 1 − x ) 1 − σ x σ − 2 I couldn't get to the given result based on the above. Any clarification would be appreciated!

Question

I am reading a paper and they mention a hump function maximization: I am trying to prove the point of maximization:  m  =  (  1  −  x      )          1      −      σ            x    σ   where   x  ,  σ  ∈  [  0  ,  1  ]It is said that m is a hump-shaped function of x maximized at   x  =  σ, where   σ is a parameter that is assumed to be fixed here.My attempt:First derivative with respect to x:   (  1  −  x      )          −      σ            x    σ    +  (  1  −  x      )          1      −      σ            x          σ      −      1        =  0  1  +  (  1  −  x      )          1            x          −      1        =  0Second derivative with respect to x:   −  (  1  −  x      )          −      σ      −      1            x    σ    +  (  1  −  x      )          −      σ            x          σ      −      1        −  (  1  −  x      )          −      σ            x          σ      −      1        +  (  1  −  x      )          1      −      σ            x          σ      −      2      I couldn&#039;t get to the given result based on the above. Any clarification would be appreciated!

Gornil2 · Accepted Answer

If x maximizes   m then it will also maximize  log  ⁡  (  m  )  =  (  1  −  σ  )  log  ⁡  (  1  −  x  )  +  σ  log  ⁡  (  x  )because   log is strictly monotonic increasing.So let us try to find the maximum by setting   0  =      m    ′   and trying to solve for   x:                    0                                      =          !                                      ∂            log            ⁡            (            m            )                                ∂            x                                                            =        (        1        −        σ        )                  1                      x            −            1                          +        σ                  1          x                                            ⟺                0                            =        (        1        −        σ        )        x        +        σ        (        x        −        1        )                                          =        x        −        σ                                    ⟺                x                            =        σ            Now it should be easy to reason that   m is maximal at   x  =  σ.

I am reading a paper and they mention a hump function maximization: I am trying to prove the point o

Answered question

Answer & Explanation

New Questions in High school geometry