idle math musing

[learnedax]

Suppose there is a function f(x), which can be differentiated an infinite number of times. Suppose that a certain f(x₀) evaluates to 0, and so do f'(x₀), f''(x₀), etc. for every finite number of differentials. Now suppose that we can determine what the Nth differential of f(x) is in the limit as N goes to infinity, and that this limit differential is non-zero at x₀.

Essentially this would mean all finite rates of change would be nil, but the function would still not actually be flatlined (though unless the limit differential is infinite it'd take infinitely long for f(x) to show any change).

Is there any reason such a function could not exist? Do you know of any functions that have similar characteristics?

Comments

[londo.livejournal.com]

I'm unaware of anything which could remain exactly zero through every finite number of iterations, but becomes non-zero in the infinite case.

[mesatchornug.livejournal.com]

yeah. f'(0) is, by definition, 0.

and df(0) is likewise a constant.

[learnedax.livejournal.com]

Er, as you have stated that it is not really true. If f(x) is sin(x), then f(0) = 0, f'(0) = 1.

[metahacker.livejournal.com]

Hmm.

I see where you're going with this. You're asking if inf*0=finite, in a weird way.

(dang I need htmlTeX)

You've posited a function g(x, n), where g(x, 0) = f(x), g(x, 1) = f'(x), etc.
You are asking whether there is any such function g() such that:
  g(x, n) = 0 for finite n, and
  lim[n->inf][g(x,n)] > 0 (for some x)

I think we can ignore how you got there, and ask if there is any series s which goes
   [0, 0, 0, 0 .. >0], i.e.,
     for all i s_i = 0, but
     lim[i->inf](s_i) > 0

I think the answer is no, but I'm not sure.

A typical way to do the first part of the definition is
   s_0 = 0
   s_(i+1) = s_(i)
ahe thing is, aleph_0 is right next to the integers, and I can't see making any distinction
between i+1 and aleph_0 in any sort of mathematical definition.

[learnedax.livejournal.com]

(Six Apart should totally adopt MediaWiki T_ex formatting.)

I think you're correct in your recasting of the question. The only way that springs to mind to construct such a series would be to have s_i = f(i)/inf, such that lim[i->inf] f(i)->inf, except that I don't believe you can just stick in an arbitrary constant inf there. You would need to include a divisor with an infinite evaluation regardless of the value of i. If you could, of course, it'd be trivial to make f(i) such that applying l'Hôpital's rule would give us a finite result.

So the question might reduce to "can there be a constant infinite term?"