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Searching for a smooth function that goes from flat zero to exponential growth (lemmy.world)

submitted 1 month ago* (last edited 1 month ago) by surrealpartisan@lemmy.world to c/math@lemmy.world

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Last night an old idea came back to me, an idea about a function where all the derivatives start from zero and then grow smoothly. I thought it would be impossible, but then I found some interesting stuff on Wikipedia. So, I learned to use SymPy and wasted a lot of time with it. Here's a report of my (non-)findings.

(UPDATE: I did some numerical differentiation, which showed that h(x) does have negative derivatives. See details in this comment. A disappointment, although perhaps not a surprising one. It doesn't however, necessarily mean the goal is impossible.)

So, if anyone knows whether such a function exists and what it looks like, please tell me.

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[–] e0qdk@reddthat.com 4 points 1 month ago (2 children)

all the derivatives start from zero and then grow smoothly.

I think you need to have a discontinuity in a derivative at some level to have a function like this where the lower derivatives grow smoothly. If you have zero at all levels and no discontinuities... nothing should ever change, right?

[–] surrealpartisan@lemmy.world 7 points 1 month ago* (last edited 1 month ago) (1 children)

That's what I thought too, but it turns out that is not true! Here's a couple of interesting links: Non-analytic smooth function, Bump function, Flat function. EDIT: Fixed the links and added another one.

[–] e0qdk@reddthat.com 2 points 1 month ago

Thanks for the links. This is outside the area of math I usually deal with, but I agree it's interesting. I think I understand what you're asking for now, but I've hit my mental limit for today trying to work it out. Good luck in your search!

[–] kogasa@programming.dev 2 points 1 month ago* (last edited 1 month ago)

f(x) = e^(-1/x^2) for x != 0, f(0) = 0. It's relatively easy to show this is infinitely differentiable at x=0 and every derivative is 0.

The intuition that an infinitely differentiable function can be described globally by its derivatives locally is actually true for complex differentiable functions, and this property is sometimes referred to as "rigidity" of complex-differentiable (or analytic/holomorphic) functions. It doesn't hold for functions that are only differentiable along the real axis.

[–] WolfLink@sh.itjust.works 2 points 1 month ago (1 children)

Could you approximate derivatives by finite differences?

Could you write your own code implementing the the derivatives?

[–] surrealpartisan@lemmy.world 1 points 1 month ago* (last edited 1 month ago) (1 children)

Could you approximate derivatives by finite differences?

Yes. I will try that.

Could you write your own code implementing the the derivatives?

No, I don't think I could.

[–] surrealpartisan@lemmy.world 1 points 1 month ago* (last edited 1 month ago)

I did some numerical differentiation, with ten thousand points between 0 and 10. Negative values appeared in the third derivative. The attached figure zooms into them. While I think those sudden spikes may very well be numerical artifacts caused by float rounding errors or something like that, there is a clear negative slope around them, further confirmed in the fourth derivative. So, this function is not what I hoped it to be.

[–] wabasso@lemmy.ca 2 points 1 month ago (1 children)

I won’t be able to help you, but was wondering if you could help me understand what tau is in the equations. I got lost when that showed up.

[–] surrealpartisan@lemmy.world 2 points 1 month ago* (last edited 1 month ago)

You can think of the convolution as a process to smooth the function g by making its values at points around each t affect that at t. So, tau is the distance between t and another point, and Psi(tau) tells how much the other point contributes to the smoothing at point t. In a more decent situation, the integral in (7) would have been properly solved and tau would have disappeared, never to bother us again.