Prove that a function is concave

  • Thread starter Thread starter blaah
  • Start date Start date
  • Tags Tags
    Concave Function
blaah
Messages
2
Reaction score
0

Homework Statement


f:R->R, c'
prove that f is concave iff f(x*)+(x-x*)f'(x*)>=f(x)


Homework Equations


assume the function is only once differentiable


The Attempt at a Solution


i have no idea how to approach this question...:confused:
 
Physics news on Phys.org
blaah said:

Homework Statement


f:R->R, c'
prove that f is concave iff f(x*)+(x-x*)f'(x*)>=f(x)


Homework Equations


assume the function is only once differentiable


The Attempt at a Solution


i have no idea how to approach this question...:confused:
Are x* and x any two values of x? Are there any restrictions on the values of x?

To prove your statement you need to prove two things:
  1. f is concave ==> f(x*) + (x - x*) f'(x*) >= f(x)
  2. f(x*) + (x - x*) f'(x*) >= f(x) ==> f is concave
For the first, what does it mean for a function to be concave?
For the second, one approach would be a proof by contradiction. Suppose that f(x*) + (x - x*) f'(x*) >= f(x) is true and that f is not concave. If you arrive at a contradiction, it means that your original assumption was incorrect, and therefore f must be concave.

Mark
 
for all x, x*

i know that for the function to be concave all the points on the tangent need to be on or below the function...but i doesn't help...i've been staring at the problems for days now, with no result...
 
Looks to me like the mean value theorem would be useful here.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top