Sin/Taylor Series - Githubissues

TheCrazyT commented 3 years ago

Need to investigate if this leads to something:

for p=prime number:

$\prod_{n=2}^{p-1}{\sin{\frac{\pi p}{n}}} \neq 0$

Since sin(pi) is zero, and every multiplication with "0" equals "0", the formula checks if "p" is divideable through any "n" (for 2 <= n < p).

SIN can be written as Taylor Series:

$\sin{x}=\sum_{n=0}^{\infinity}{\frac{(-1)^n}{(2n+1)!}x^{2n+1}}$

Note:

the Taylor Series is only an approximation that can contain errors
simplier formula without sin: $\product_{x=2}^{x=p-1}{(\prod_{i=2}^{p-1}{(\frac{p}{x}- i)})}\neq0$
of course one could use modulus in a formula, but that shurely cannot be reduced

Not so shure about this, but maybe every prime number could be somehow related to PI.

Edit: (actually ... NO since PI can be used as constant and the sin-formula outputs 0 for every p/n = natural number)

TheCrazyT commented 3 years ago

Example wolfram alpha links:

TheCrazyT commented 3 years ago

Sum-formula's can be shortened by formula's from gauss. Maybe there exist something similar for products? Maybe the pochhammer function

Maybe this:

$x! = (\frac{x}{2}!)^2\frac{x!}{(\frac{x}{2}!)^2}$

will be similar (if done by hand).

Example:

$1*2*3*4*5*6 = (1*2*3)^2\frac{1*2*3*4*5*6}{(1*2*3)(1*2*3)} = 36*2*5*2= 720$

TheCrazyT commented 3 years ago

$\prod_{x=1}^{(p-1)!}(\frac{(p-1)!}{p}-x)\gt0$

TheCrazyT commented 3 years ago

basically modified Wilson's theorem

TheCrazyT commented 3 years ago

TheCrazyT / factorization-maps