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Math Help - A function such that...

  1. #1
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    Lightbulb 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.

    I appreciate any help you might have.
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  2. #2
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    Quote Originally Posted by Rescription View Post
    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\alpha with \alpha>1. Then the function f(x) = (1-x^\alpha)^2h has all these properties, where c = \Bigl(\dfrac{\alpha-1}{2\alpha-1}\Bigr)^{1/\alpha}. As \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 \alpha in terms of c.
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