Iterated trigonometric differentiation

1. Jul 1, 2006

Orion1

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

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

$$\frac{d^n}{dx^n} \cos x = \cos \left(x + \frac{n \pi}{2} \right)$$

Iterated trigonometric integration:
$$(I^n \sin)(x) = \sin \left( x - \frac{n \pi}{2} \right)$$
$$(I^n \cos)(x) = \cos \left( x - \frac{n \pi}{2} \right)$$

Last edited: Jul 1, 2006
2. Jul 1, 2006

vkgoku2012

How do you understand math like that what does that all mean.

3. Jul 1, 2006

gnomedt

Looks fine to me. Are you trying to find the half-derivative of trig functions?

4. Jul 1, 2006

Orion1

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

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

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?
$$\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}} \sin x = \sin \left(x + \frac{\pi}{4} \right)$$

Last edited: Jul 1, 2006