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Could use some help with a proof

  1. Dec 19, 2008 #1

    I've been wondering about something for a while now. As a computer scientist, I never had much in the way of analysis or advanced calculus, but there is something I have always assumed was true but have never been able to prove.

    Say that you have a function f : R -> R which is infinitely differentiable (and continuous). The derivatives may eventually equal zero, but they must also be infinitely differentiable (and continuous) according to this same definition.

    Further, assume that on some finite closed interval [a, b] the function is constant; that is, for any x in [a, b], f(x) = c for some finite value of c (with c constant).

    Is it true that f(x) = c for all x in R?

    Also, as a computer scientist, if the bottom line does not follow, I would definitely like an example of a function satisfying my requirements but which is not constant. No ad absurdum proofs, please. Thanks in advance for any help I may get in this matter.
  2. jcsd
  3. Dec 19, 2008 #2

    "infinitely differentiable (and continuous)"

    The heaviside function is not differentiable at x = 0, last time I checked. Nor is it continuous there.
  4. Dec 19, 2008 #3


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    I know, I missed that part of your post (which is why I deleted it just before you replied) :rolleyes:
  5. Dec 19, 2008 #4
    lol. I reread my post to make sure I said that. I agree the problem would have been a little flippant if I hadn't specified that. ;D

    Thanks for your interest, though. It seems obvious to me, but is it so obvious? If it were so obvious, you'd think I'd have a proof by now...
  6. Dec 19, 2008 #5


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    Perhaps I misinterpretted - or misunderstand - your post but wouldn't the piecewise funcion:

    c, a < x < b
    f = (x - a)^2 + c, x < a
    (x - b)^2 + c, x > b

    satisfy the conditions? f(x) = c in some finite interval [a,b], is continuous for all values x, and I think is differentiable; moreover f(x) =/ c for all values x.
  7. Dec 19, 2008 #6
    Well, the problem is that the second derivative,

    f ' ' (x) = {2, x < a} or {0, a < x < b} or {2, b < x}

    Therefore, the second derivative is neither continuous nor differentiable for x = a or x = b, which I require.
  8. Dec 19, 2008 #7
  9. Dec 19, 2008 #8
    Oh, wow. Well, I suppose that does it. Clever...

    Thanks, Moo of Doom, for showing me that... it's a perfect counterexample. Thanks!
  10. Dec 19, 2008 #9
    Arby, I'd take a look at Moo of Doom's link. Taylor's theorem only applies to analytic functions, I believe. I already knew it was true for analytic functions, but... I thought it may have been true more generally.

    Thanks for your help, everybody.
  11. Dec 19, 2008 #10
    Sorry didn't see it before I blinked, then I deleted it, thanks.
  12. Dec 19, 2008 #11
    What about a piecewise defined function:

    f(x) =

    (x+1)^2 for -inf < x < -1
    0 for -1 <= x <= 1
    (x-1)^2 for 1 < x < +inf

    this is continuous and analytic and all those nice things isnt it?

    Whoops i'm not doing very well today. 2nd deriv problems...
    Last edited: Dec 19, 2008
  13. Dec 20, 2008 #12


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    The function defined by
    [tex]f(x)= e^{-1/x^2}[/tex] for x< 0
    [tex] f(x)= 0[/tex] for [itex]0\le x\le 1[/itex]
    [tex] f(x)= e^{-1/(x-1)^2}[/itex] for x> 1
    is, I believe, infinitely differentiable for all x.
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