# Thread: A function such that...

1. ## A function such that...

I need to find/make a function, f(x), such that:

f(0)=h
f(x) > 0 along x(0,1)
f ' (x) < 0 along x:(0,1)
f(1) is negligible (f(1)<<h/10000) or f(1)=0
f ' (1) is negligible or 0
Integral from 0 to 1 of f(x) dx = 1

f ' ' (c)=0 c:(0,1) (that's second deriv if you can't read it)
f ' ' (x)<0 x:(0,c)
f ' ' (x)>0 x:(c,1)

So basically a nice smooth curve along x:[0,1] starting at f(0)=h and headed slowly downward, quickly downward at x=c, then slowly downward to essentially 0 when x=1.

It looks so nice when I draw it that I have to guess something simple will satisfy the above, but I'm not finding it so far.

2. Originally Posted by Rescription
I need to find/make a function, f(x), such that:

f(0)=h
f(x) > 0 along x(0,1)
f ' (x) < 0 along x:(0,1)
f(1) is negligible (f(1)<<h/10000) or f(1)=0
f ' (1) is negligible or 0
Integral from 0 to 1 of f(x) dx = 1

f ' ' (c)=0 c:(0,1) (that's second deriv if you can't read it)
f ' ' (x)<0 x:(0,c)
f ' ' (x)>0 x:(c,1)

So basically a nice smooth curve along x:[0,1] starting at f(0)=h and headed slowly downward, quickly downward at x=c, then slowly downward to essentially 0 when x=1.
Second attempt (the first one was wrong):

Choose $\displaystyle \displaystyle\alpha$ with $\displaystyle \alpha>1$. Then the function $\displaystyle f(x) = (1-x^\alpha)^2h$ has all these properties, where $\displaystyle c = \Bigl(\dfrac{\alpha-1}{2\alpha-1}\Bigr)^{1/\alpha}$. As $\displaystyle \alpha$ goes from 1 to ∞, c goes from 0 to 1. The only snag is that, given c, it is not possible to get an explicit expression for $\displaystyle \alpha$ in terms of c.