So this is how pi is generated from the lengths of hypotenuses

Don't you just love how neatly pi is generated by nested square roots of 2? Just as lovely as n/((n!)^1/n) approaching e.
My autistic mind is so satisfied by this.

So sqrt(2+sqrt(2+sqrt(2+...)))=2. So the nested radical expression actually tends to zero on its own but the 2^n makes it just bigger enough for it to approach pi.

Please answer with some quality posts about the subject matter, this is the most interesting thread on the front page after all.

The raadical is clearly irrational and yet it feeds the exponent all the decimal values of pi? I csll bullshit on that.

>>8876669
Yes it's called a limit.

I wonder what other sequences of form [math]\alpha^n \sqrt {\alpha - \sqrt{\beta + \sqrt{\beta + \sqrt{ \beta+ ...}}}} [/math] where [math]\alpha[/math] is the limit of [math] \sqrt{\beta + \sqrt{\beta + \sqrt{ \beta+ ...}}}[/math], n is the amount of recursions, and beta is any value that meets the convergence criteria, converge to. I tried with the golden ratio and slowly got closer to 1, but I haven't really shown that the limit is 1.

>>8877100
prove that the radical is irrational, faggot

https://math.stackexchange.com/questions/85217/why-is-this-series-of-square-root-of-twos-equal-pi

Care to share your sympathies for this endearing expression?

File: untitled.png (8KB, 522x367px) Image search: [Google]

8KB, 522x367px

>>8877245
why should i have to prove it? i'm an engineer. my job is to tell you to prove it so i can implement it in a new type of screw that costs more for consumers while doing the same task.