Questions about Cosine

[Kharakov]

I've noticed in various 2d and 3d fractal forms, that one can apply reflections to the cosine portion of the fractal formulas and get continuous fractals (fractals that evolve smoothly with iteration).

The reflection is the following:

\(-a^n \times \cos ({n\Theta}) + a \cos ({\Theta}) + i [a^n \times \sin ({n\Theta}) + a \sin ({\Theta}) ]\)
to
\(a^n \times \cos ({n\Theta}) + a \cos ({\Theta}) + i [a^n \times \sin ({n\Theta}) + a \sin ({\Theta}) ]\)

The reflection is of the first term \(a^n\times \cos (n\Theta)\) to \(-a^n \times \cos (n\Theta) \).

You can apply the reflection uniformly, without any conditional, and get a continuous fractal. This works for every kind of rotation based fractal formulas that I've tested it on.

However, you can only get continuous fractals when you apply the following conditionals to specific multipliers for angles (once again, this works for every simple rotation based fractal formula):

For n=2,6,10,14.... with n being the angular multiplier (and for some fractals, magnitude) the formula can have the following reflection applied to it with the following specific condition and remain continuous:

if \(\cos(\Theta) > |\sin(\Theta)|\) then

\(-a^n \times \cos ({n\Theta}) + a \cos ({\Theta}) + i [a^n \times \sin ({n\Theta}) + a \sin ({\Theta}) ]\)

else

\(a^n \times \cos ({n\Theta}) + a \cos ({\Theta}) + i [a^n \times \sin ({n\Theta}) + a \sin ({\Theta}) ]\)

For n= anything else besides 2 + 4*integer, you get a discontinuous fractal if you use the conditional \(\cos(\Theta) > |\sin(\Theta)|\) to apply the reflection.

For n=4,8,12... I have not found a conditional that creates a discontinuous fractal.

For n=3,5,7,9... n= 1+2*integer, the following conditional produces continuous results:

if \(\cos(\Theta) > 0 \) then reflect the sign of the first term, the same as above.


What I'm trying to figure out is a generic formula, based on n (the angular rotation multiplier), that generates the conditional test smoothly, for all n. I assuming it's something very simple, that I've missed, that will tie it all together.

 

[Kharakov]

Sheesh, I've wondered about that for years, and now that I posted it and wrote it out in a way that makes the question clear, I see what I missed. Funny that...

If x > |y| flip the sign of x^2-y^2, if x < |y|, x^2-y^2 is already negative.... doy, it's just constraining the value of that part to a negative value. It's the same thing as doing \(-|\cos(n\Theta)|\)

So now what I need is the name of a general formula for the binomial series expansion, something like the gamma function, so that I can do:

\({(a+b)}^n\) for non integer n. Anyone?

 

[beero1000]

Sheesh, I've wondered about that for years, and now that I posted it and wrote it out in a way that makes the question clear, I see what I missed. Funny that...

If x > |y| flip the sign of x^2-y^2, if x < |y|, x^2-y^2 is already negative.... doy, it's just constraining the value of that part to a negative value. It's the same thing as doing \(-|\cos(n\Theta)|\)

So now what I need is the name of a general formula for the binomial series expansion, something like the gamma function, so that I can do:

\({(a+b)}^n\) for non integer n. Anyone?

You want the generalized binomial theorem.

\((1 + x)^\alpha = \sum_{k = 0}^\infty {{\alpha}\choose{k}} x^k\), where \(|x| < 1\) and \({{\alpha}\choose{k}}= \frac{\alpha(\alpha-1)\dots(\alpha-k+1)}{k!}\).

 

[Kharakov]

You want the generalized binomial theorem.

\((1 + x)^\alpha = \sum_{k = 0}^\infty {{\alpha}\choose{k}} x^k\), where \(|x| < 1\) and \({{\alpha}\choose{k}}= \frac{\alpha(\alpha-1)\dots(\alpha-k+1)}{k!}\).

Thanks. The term helped me find a way to use non-integer binomial coefficients over at Wolfram's Mathworld.

\((x+y)^r =\sum_{k=0}^\infty {r \choose k} x^{r-k} y^k \)

\( {r\choose k}= \frac{\Gamma(r+1)}{\Gamma(k+1) \Gamma(r-k+1)} \)