I wonder what a function f(x) with the property that f(f(x)) = e^x might look like. (My usual caveat: DON’T TELL ME & RUIN THE WONDERING STAGE.)
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Replying to @CmonMattTHINK @divbyzero
I'm not sure, but whatever it is has to solve this nonlinear differential equationpic.twitter.com/E0CxIFwjLR
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I’ll solve this numerically tomorrow and see if it looks like anything I recognize, and/or if curve fitting to the solution gives something correct. I don’t see any reason f can’t be holomorphic though, and I’m pretty sure Picard-Lindelöf gives us existence and uniqueness
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Replying to @JakobsZane @InertialObservr and
In particular, the derivative is never zero and everywhere continuous, so we also have a guarantee that f is invertible (by whatever that theorem is called that says a function is locally invertible if its derivative is continuous and nonzero in that neighborhood)
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But doesn’t that just say that f(f(x)) is is invertible? Pardon my analysis is a bit rusty
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Replying to @InertialObservr @JakobsZane and
From the expression you have above in the box, if you assume f is smooth enough and defined for all R then f' can never vanish. And then it's inverse is found by just drawing the graph and reflecting on the x=y line (I mean there are theorems, but that's even easier :) )
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Replying to @Quasilocal @InertialObservr and
Of course, the assumption that it is well-defined on all of R needs to be justified
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Replying to @Quasilocal @InertialObservr and
Yeah, I think Picard-Lindelöf gives you that, no? I may be trying to extend it too far with the function composition but I think it’s still valid
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Replying to @JakobsZane @InertialObservr and
I'm not actually sure. For example consider: f(f(x))=x f'(f(x))f'(x)=1 But f(x)=1/x solves this
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I love this problem
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Replying to @InertialObservr @Quasilocal and
So, it is perhaps that this problem got a family of solutions spanning on its "solution surface"?
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