[Sasha Volokh, August 31, 2006 at 12:21pm] Trackbacks Math bleg: Can anyone recommend to me any simple functions with the following properties: f(0)=0, f(infinity)=1, and f'(0)=infinity? I.e., a function that could be used for the probability of getting a policy change as a function of your lobbying expenditures; but I'd like to have that first-derivative condition so I'm guaranteed to have an interior solution when I maximize af(x)-x for any a. UPDATE: Oops! I meant f(0)=0, not f(x)=0. That's corrected now. UPDATE 2: Thanks, folks. Aaron Bergman suggested 1 + e^(-x^2)(sqrt(x)-1). Unfortunately, I forgot to specify that I also wanted f'(x)>0 for all x; and that function goes above 1 and then dips back down so f(infinity)=1. While I'm at it, I also wanted f"(x)<0 for all x. Syd suggested sqrt(x)/(sqrt(x)+1)), which works fine. Chrismn suggested 1-e^(-sqrt(x)), which also works fine. Maniakes suggested the logistic function, but I don't think I can make that match all three of my conditions. Chrismn suggested the constant absolute risk aversion -e^(-ax)+1, but that violates f(0)=0. Aaron Bergman, finally, suggests (2/pi) atan(sqrt(x)), which looks cool. UPDATE 3: Aaron Bergman also suggests an ingenious method, which I was unaware of, to generate all the functions like that you want! Just take a function g such that g(0)=0, g(infinity)=1, and g'(0) is finite (he doesn't say it, but perhaps you also need it to be nonzero). Then for any r in the interval (0,1), define f(x)=g(x^r). That way, first, f(0) = g(0^r) = g(0) = 0. Second, f(infinity) = g(infinity^r) = g(infinity) = 1. And, third, f'(x) = r x^(r-1) g'(x^r), so f'(0) = r 0^(r-1) g'(0^r) = r infinity g'(0) = infinity. All three functions I liked above are special cases of this. Thanks, Aaron! UPDATE 4: Reader Paul Edelman suggests another function: sqrt(x/(1+x)). This illustrates a similar way of generating such functions: Take a function g defined as in Update 3, then define f(x)=[g(x)]^r. That way, f(0)=0 and f(infinity)=1 obviously. Also, f'(x) = r g'(x) [g(x)]^(r-1), so f'(0)=infinity. Now that's not the way Paul found the function. He used yet another way, which is also nice: Look for a function g such that g(0)=0, g'(0)=0, and g has a vertical asymptote at 1. Then take the inverse of that function, and that's your f. So x^2 definitely has value and first derivative 0 at 0, and if you stick a (1-x^2) denominator on it, you get yourself an asymptote. So g(x)=x^2/(1-x^2) works, and f(x)=sqrt(x/(1+x)) is its inverse function. UPDATE 5: Sadly, these functions don't have closed-form solutions, if, say, I want to take the inverse of the derivative!