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Iterated trigonometric differentiation

  1. Jul 1, 2006 #1

    These are some equations that I recently developed and submitting for review.

    Evaluations?, comments?

    Iterated trigonometric differentiation:
    [tex]\frac{d^n}{dx^n} \sin x = \sin \left(x + \frac{n \pi}{2} \right)[/tex]

    [tex]\frac{d^n}{dx^n} \cos x = \cos \left(x + \frac{n \pi}{2} \right)[/tex]

    Iterated trigonometric integration:
    [tex](I^n \sin)(x) = \sin \left( x - \frac{n \pi}{2} \right)[/tex]
    [tex](I^n \cos)(x) = \cos \left( x - \frac{n \pi}{2} \right)[/tex]
     
    Last edited: Jul 1, 2006
  2. jcsd
  3. Jul 1, 2006 #2
    How do you understand math like that what does that all mean.
     
  4. Jul 1, 2006 #3
    Looks fine to me. Are you trying to find the half-derivative of trig functions?
     
  5. Jul 1, 2006 #4

    Iteration is a 'repeated' process used in higher order mathematics, for example:
    [tex]\frac{d^2}{dx^2} x^3 = \frac{d}{dx} \left( \frac{d}{dx} x^3 \right) = \frac{d}{dx} \left( 3x^2 \right) = 6x[/tex]

    Therefore, the second iterated derivative of this function is:
    [tex]\frac{d^2}{dx^2} x^3 = 6x[/tex]

    The derivative is repeatedly derived twice, or 'iterated' twice.

    Negative, are you referring to 'fractional' calculus?
    Is this a demonstration of a half-derivative trig function?
    [tex]\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}} \sin x = \sin \left(x + \frac{\pi}{4} \right)[/tex]
     
    Last edited: Jul 1, 2006
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