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        riddles >> putnam exam (pure math) >> f(x) = 1/(1+e^x)
        
(Message started by: william wu on Oct 24^th, 2004, 12:06am)

Title: f(x) = 1/(1+e^x)
Post by william wu on Oct 24^th, 2004, 12:06am

Given f(x) = 1/(1+e^x), find f⁽¹⁰⁰⁾(0).

[ The f⁽ⁿ⁾ notation denotes repeated application of the function. So f⁽²⁾(x) = f(f(x)). ]

Edit: corrected a typo

Title: Re: f(x) = 1/(1+e^x)
Post by Barukh on Oct 24^th, 2004, 12:49am

Isn't it just [hide]the solution of the equation x = 1/(1+e^x)[/hide]?

:-/

Title: Re: f(x) = 1/(1+e^x)
Post by william wu on Oct 24^th, 2004, 10:26am

Sorry, I had a typo: it should read f⁽¹⁰⁰⁾(0).

I should warn beforehand that I haven't been able to solve this myself; hopefully the problem is legit.

Title: Re: f(x) = 1/(1+e^x)
Post by Grimbal on Oct 24^th, 2004, 3:26pm

It still tends quite fast to the solution of x=1/(1+e^x). It is not there yet, but close enough.

Title: Re: f(x) = 1/(1+e^x)
Post by Icarus on Oct 24^th, 2004, 3:26pm

According to Excel, it is 0.401058138.... Repeated application converges to this value fairly fast. As Barukh indicates, this is the solution to x = 1/(1+e^x). I would imagine it is highly unlikely that f⁽¹⁰⁰⁾(0) is exactly the value of x solving the equation. It is very close, however.

Whether or not there is "nice" expression for the number is another matter.