Code: Select all
a = x/(1-x)
b = 1/a
f(x) = 0 | x ≤ 0
f(x) = a^b * b^a | 0 < x < 1
f(x) = 0 | x ≥ 1
I believe that this function is infinitely differentiable at x=0 and x=1. How would I go about proving this?
Moderators: gmalivuk, Moderators General, Prelates
Code: Select all
a = x/(1-x)
b = 1/a
f(x) = 0 | x ≤ 0
f(x) = a^b * b^a | 0 < x < 1
f(x) = 0 | x ≥ 1
Qaanol wrote:At x=0, can you prove that your function…
• is continuous?
• is differentiable?
• is twice-differentiable?
What do you imagine a proof that it is infinitely-differentiable would look like?
MostlyHarmless wrote:It’s also worth noting that your title doesn’t match your question, which is good because the title is impossible. A bump function can be infinitely differentiate, but it cannot be analytic.
measure wrote:MostlyHarmless wrote:It’s also worth noting that your title doesn’t match your question, which is good because the title is impossible. A bump function can be infinitely differentiate, but it cannot be analytic.
When I wrote the title, I thought "analytic" just meant that it consisted of compositions of "ordinary" functions. I didn't realize that the piecewise definition excluded it. Thanks! (Fixed)
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