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Derivative theorem

  1. Aug 16, 2005 #1

    Any calculus researchers interested in disproving this theorem with some simple integers?

    Orion1 derivative integer factorial theorem:
    [tex]\frac{d^n}{dx^n} (\ln x) = (-1)^{n + 1} \frac{(n - 1)!}{x^n}[/tex]

    Is this theorem correct?
     
    Last edited: Aug 16, 2005
  2. jcsd
  3. Aug 16, 2005 #2
    Is this homework? You can prove it by combining the two basic identities,

    [tex]\frac{d}{dx} \ln x = \frac{1}{x} \equiv x^{-1}[/tex]

    [tex]\frac{d}{dx} x^{m} = m x^{m-1} \, (m \neq 0) [/tex]
     
    Last edited by a moderator: Aug 16, 2005
  4. Aug 18, 2005 #3
    Expired Post...

    Expired Post...
     
    Last edited: Aug 19, 2005
  5. Aug 18, 2005 #4

    TD

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    You could prove it by induction on n.
    Check for n = 1, assume valid for n = k, prove then for n = k+1.
     
  6. Aug 18, 2005 #5

    matt grime

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    They prove it because they do, as any one who knows about talyor series for instance could tell you. Again, do not name things after yourself, that is not allowed in mathematics, not least when it is this elementary and well known.
     
  7. Aug 23, 2005 #6
    Generic Iteration...


    Is this the correct iterated integral of this function?

    iterated integration theorem: (generic)
    [tex](I^n f)(x) = \int_0^x (-1)^{n + 1} \frac{(n - 1)!}{t^n} dt = \ln x[/tex]
     
  8. Aug 23, 2005 #7

    Tide

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    Orion's Conjecture Belted


    No, it is not correct. It fails when n = 0.
     
  9. Aug 24, 2005 #8

    Galileo

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    That integral doesn't converge.
     
  10. Aug 25, 2005 #9
    theorem...


    theorem:
    [tex]\frac{d^n}{dx^n} (\ln x) = \left\{ \begin{array}{rcl} \ln x & \text{if} & n = 0 \\ (-1)^{n + 1} \frac{(n - 1)!}{x^n} & \text{if} & n > 0 \end{array} \right.[/tex]
     
    Last edited: Aug 26, 2005
  11. Aug 26, 2005 #10

    Tide

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    No, the zeroth derivative of the natural log is -- the natural log!
     
  12. Aug 26, 2005 #11

    Then this theorem abstains at [tex]n > 0[/tex].

    theorem:
    [tex]\frac{d^n}{dx^n} (\ln x) = (-1)^{n + 1} \frac{(n - 1)!}{x^n} \; \; n > 0[/tex]

    [tex]\frac{d^n}{dx^n} (\ln x) = \left\{ \begin{array}{rcl} \ln x & \text{if} & n = 0 \\ (-1)^{n + 1} \frac{(n - 1)!}{x^n} & \text{if} & n > 0 \end{array} \right.[/tex]

    Are there any more challenges to this theorem?
     
    Last edited: Aug 26, 2005
  13. Aug 26, 2005 #12

    Hurkyl

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    I'm worried about nonintegral values of n.
     
  14. Aug 26, 2005 #13


    iterated integration theorem:
    [tex](I^n f)(x) = \int_1^x (-1)^{n + 1} \frac{(n - 1)!}{t^n} dt = \ln x[/tex]

    Is this theorem correct? Does this theorem converge?
     
    Last edited: Aug 26, 2005
  15. Aug 26, 2005 #14

    matt grime

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    no, that doesn't work since the integral you wrote down evaluates to

    (-1)^n(n-2)!x^{-n+1} +(-1)^n(n-2)!

    it isn't an "iterated" integral, it is just an integral.

    and theorems do not converge. integrals can converge, sequences can converge.
     
  16. Aug 26, 2005 #15

    lurflurf

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    Orion1 seems to be attempting to make use a a result due to Cauchy that if I is the operator defined by
    [tex](If)(x):=\int_0^x f(t)dt[/tex]
    then if the operator is applied n time the result is
    [tex](I^nf)(x)=\frac{1}{\Gamma(n)}\int_0^x (x-t)^{n-1}f(t)dt[/tex]
    This result is often used as the foundation for the so called fractional calcus when operators of nonintegral order are considered. Since the right hand side is well defined for n not an integer this is taken as a definition of the left hant side when n is not an integer.
     
  17. Aug 26, 2005 #16

    matt grime

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    i had guessed what he was *trying* to do, but we can't be sure, and i was pointing out what he *had* done
     
  18. Sep 2, 2005 #17
    foundation theorem...


    That is correct, I was uncertain how the symbolic method described such a theorem, it certainly is not listed in my calculus book.

    As I understand this theorem, the integer, or non-integer must be defined 'prior' to performing the actual integration for each successive iteration.

    I suppose the next question, does this theorem obey the fractional calculus foundation theorem?

    [tex](I^nf)(x)=\frac{1}{\Gamma(n)}\int_0^x (x-t)^{n-1}f(t)dt[/tex]

    [tex]\frac{d^n}{dx^n} (\ln x) = \left\{ \begin{array}{rcl} \ln x & \text{if} & n = 0 \\ (-1)^{n + 1} \frac{(n - 1)!}{x^n} & \text{if} & n > 0 \end{array} \right.[/tex]

    fractional calculus theorem: (correct symbolic method?)
    [tex](I^nf)(x) = \frac{1}{\Gamma(n)} \int_0^x (x-t)^{n-1}f(t)dt = \frac{1}{\Gamma(n)} \int_0^x (-1)^{n + 1} \frac{(n - 1)!}{t^n} dt = \ln x \; \; n > 0[/tex]

    In response to Tide's post, [tex]n = 0[/tex] appears to be special case:
    [tex](I^0f)(x) = \frac{1}{\Gamma(0)} \int_0^x \frac{f(t)}{(x-t)}dt \; \; n = 0[/tex]
     
    Last edited: Sep 2, 2005
  19. Oct 9, 2005 #18

    benorin

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    Fractional Integrals are (by the given operator) only defined for n>0, atleast per the following link:

    http://mathworld.wolfram.com/FractionalIntegral.html
     
  20. Oct 12, 2005 #19
    Derivative Dogma...


    derivative integer factorial theorem:
    [tex]\frac{d^n}{dx^n} (\ln x) = (-1)^{n + 1} \frac{(n - 1)!}{x^n} \; \; n > 0[/tex]
    Then the original theorem is correct, because like Fractional Integrals, Fractional Derivatives are only defined for n > 0?

    I disagree that n = 0 derivatives are actually derivatives at all!:
    [tex]\frac{d^n}{dx^n} f(x) = g(x) \; \; n > 0[/tex]

    However when n = 0:
    [tex]\frac{d^n}{dx^n} f(x) = f(x) \; \; n = 0[/tex]

    It just seems that any derivative theorem would always fail at n = 0, because n = 0 is not a derivative!

    Can anyone actually state a 'Fractional Calculus' derivative theorem that does not fail at n = 0?
     
    Last edited: Oct 12, 2005
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