SOLVED! Does a simple function exixt with these properties?

| > || Is there a simply-expressible function F(x) defined on [0,infinity) with
| > these properties?
| > ||
| > || F(0) = 0
| > || F(infinity) = 1 (in the usual sense of a limit)
| > || F(1) = alpha, 0 < alpha < 1
| > || (can restrict alpha > 1-exp(-1) if that helps any)
| > || dF/dx(0) = 1
| > || dF/dx(infinity) = 0
| > || dF/dx(x) is monotone and nicely smooth
| >
| >
| > | F(x) = 1-exp(-x-a*x^2)
| >
| > Another typo or? A quick look shows this meets only 2 of the 5 conditions...
|
| It meets all 5 conditions for 1 > alpha > 1-exp(-1), and meets
| all 6 conditions for 1-exp(-3/2) >= alpha > 1-exp(-1). I'm not
| sure what you're thinking here.

Nor am I! (Possibly writing to a different suggested formula?) <sigh>

Nonetheless, it doesn't meet all six conditions, even for the given restriction
on alpha. The sole fault being F' is non-monotonic as shown below:

F(x,a) = 1 - exp(-x-a*x^2) so
F'(x,a) = exp(-x-a*x^2)(x-2*a*x)

F(1,0.7) = 0.817 So "a" is not alpha (I lost your original post, sorry.
No doubt you provided a formula for "a" as a function of alpha) We
require alpha = F(1,a) which implies a = -ln(1-alpha)

Consider alpha = 0.7 1-exp(-3/2) > 0.7 > 1-exp(1) A = -ln(1-0.7)
F'(0.2,A) ~ 1.156 > 1 Thus F' is not monotone.

Nonetheless, we have two winners:

| > | Try F(x) = (1 - exp(-x^k))^(1/k) where k = ln(1-exp(-1))/ln(alpha).
| >
| > Beautiful, simple, elegant. I love it. Alas, it only works (meets all 6
| > conditions) for alpha = 1-exp(-1).
|
| Why do you say this? Rob Johnson claims monotonicity of F' for
| all alpha in (0,1); the other 5 conditions are easy to verify for
| such alpha.

Apologies! So sorry. Rob is correct. Dunno how I decided F' was
non-monotone. <sigh> And it clearly reproduces 1-exp(-x), for
alpha = 1-exp(-1), an unstated but necessary condition for my use.

| Also, you might want to look at x/(1+x^k)^(1/k), k =
| ln(1/2)/ln(alpha). It's easy to check that for any alpha in
| (0,1), all 6 conditions hold.

Indeed they do, so we have two solutions. FWIW It has the property
that for alpha = 1-exp(-1) it approaches the limit of 1 for large x much
more slowly than does 1-exp(-x), a drawback for my application - but
that wasn't in my "rules" was it?

Thanks to all, greatly appreciate you patience with my stupidity.
Chas



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