4 ms·

Not a math student, but I tried it anyway: 0 for x<=0 1 for x>=1 e^(1 - 1/(1 - (x - 1)^2)) else Does this work? Does the last condition mean smoothness

• S4M 10y ago
  That one will not work because the function is not C∞ [I didn't know how to make that symbol] in 1. it's first derivative is 0 in 1, but that's only because the derivative for 1-1/(1-(x-1)^2) is 0 in 1, that will not be the case for the second derivative. If you look at what's going on in 0, the derivative of any order is something like phi(x) e^(1 - 1/(1 - (x - 1)^2)) where phi is a rational function, and thus it will always be 0 in 0 (the exponential always beats a rational function). so the answer is, similarly to what kmill posted: 0 for x <=0 1 for x >= 1 1/(1+exp(K(x))) else With K(x) = (x-1/2)/(x(x-1)) (x-1/2 is added to create a change of sign between 0 and 1) in 0, K(x) -> ∞, so 1/(1+exp(K(x)) -> 0, and in 1, K(x) -> -∞, so 1/(1+exp(K(x)) -> 1. To see that the function is C∞, you can check that the n-th derivative 1/(1+exp(K(x))) is of the form phi(x) exp(K(x)) 1/(1+exp(K(x)))^(n+1) where phi is a rational function, and then verify that the n-th derivative is 0 in 1 and 0.