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I got another one (a bit more complicated): find a function f that verifies: - f(x) = 0 for x <= 0 - f(x) = 1 for x >= 1 - f admits a derivative of any order

by S4M 10y ago

I got another one (a bit more complicated): find a function f that verifies: - f(x) = 0 for x <= 0 - f(x) = 1 for x >= 1 - f admits a derivative of any order The last condition rules out functions like f(x) = 0 if x<=0,x if x in [0,1] and 1 if x >=0. (Note: piecewise defined functions are still OK)

johnp_ 10y ago
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 (C∞)? (Not a native english or math speaker, so please forgive my ignorance) edit: HN ate the link :/ wolframalpha code: Piecewise[{{0, x <= 0}, {1, x >=1}}, {e^(1 - 1/(1 - (x - 1)^2))}]
- 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.
kmill 10y ago
https://en.wikipedia.org/wiki/Bump_function https://en.wikipedia.org/wiki/Bump_function
GFK_of_xmaspast 10y ago
My real analysis is extremely rusty, but could you use Urysohn's lemma to get an arbitrary continuous "step" function and then integrate that.

by S4M 10y ago

johnp_ 10y ago

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 (C∞)? (Not a native english or math speaker, so please forgive my ignorance) edit: HN ate the link :/ wolframalpha code: Piecewise[{{0, x <= 0}, {1, x >=1}}, {e^(1 - 1/(1 - (x - 1)^2))}]

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.

kmill 10y ago

https://en.wikipedia.org/wiki/Bump_function https://en.wikipedia.org/wiki/Bump_function

GFK_of_xmaspast 10y ago

My real analysis is extremely rusty, but could you use Urysohn's lemma to get an arbitrary continuous "step" function and then integrate that.