Why and how are pi and nested square roots of 2 related? How to derive Vieta's formula from this for example, or vice versa?
Please give insight.
Why and how are pi and nested square roots of 2 related? How to derive Vieta's formula from this for example...
Owen Perez
Ryan Turner
I think both stem from iterating sin- and cos- relations to each other.
Levi Morales
the last squareroot is unclear to me. it says sqrt(2 + ... + sqrt(2)). what happens at the "..."?
Nathan Reyes
Pi is the arc length of the curve y = sqrt(x^2 + 1), so you take an arc length integral to get this interesting thing as an intermediate step if you work the integral in a certain verbose way.
Jonathan Phillips
I understant it as :
[math] \pi = \lim 2^k \cdot U_k [/math]
Where [math] U_1 = \sqrt 2 \\ U_{n+1} = \sqrt {U_n + 2} [/math]
Leo Bennett
Actually, there's yet another relation between them.
The cosine of pi/(2^k) can be expressed with sqrt 2s and you can sum them up to build the area of a quarter circle from the inside with triangles.
Connor Ward
edit: you can sum their reciprogs up
Luis Lewis
Can someone provide a step by step (at least 5 steps) guide on how to calculate the hypotenuses gotten from dividing the circle into 2^n parts and connecting the dots?
I can't calculate past the third? step which yields sqrt(2-sqrt(2))
I'm so retarded
Tyler Walker
And this only because I know sin(pi/4) yields sqrt(2)/2.
How should I "know" that sin(pi/8) is sqrt(2-sqrt(2))/2 and so forth?
Isaac Price
>and you can sum them up to build the area of a quarter circle from the inside with triangles
Is it beneficial to calculate the areas of the triangles when their hypotenuses will yield pi already, with the 2^n multiplier.
Ian Phillips
Just what is it with these circles and square roots of two? It's so pretty.
Brody Smith
Oh, didn't notice this post. Would you have the time to work out the integral in a certain verbose way? Or someone else? This is fundamental and interesting after all.
Connor Allen
Doesn't that actually lead to 2^(k+1)
Colton Green
Note, sqrt(2+sqrt(2+sqrt(2+...)))=2.
Oliver Gomez
Veeky Forums why aren't you more interested, these are really interesting limits for pi.
Please be more interested and do that wonky arc length integral in a verbose way for the fun of it. And explain how it leads to the result in OP image.
Brandon Evans
mathhelpboards.com
Very thorough post. Interesting stuff.
Eli Brooks
Wtf are you mathfags even saying
Brayden Hill
You start with the double-angle formula.
[eqn] \sin(2x) = 2 \sin(x) \cos(x) \\
\sin^2(2x) = 4 \sin^2(x) (1 - \sin^2(x)) \\
4 \sin^4(x) - 4 \sin^2(x) + \sin^2(2x) = 0 \\
\sin^2(x) = \frac{1}{2} \left(1 - \sqrt{1 - \sin^2(2x)} \right)
[/eqn]
If you let x=pi/8 you get
[eqn] \sin^2(\frac{\pi}{8}) = \frac{1}{2} \left(1 - \sqrt{1 - \sin^2(\frac{\pi}{4})}\right)
= \frac{1}{2} \left(1 - \sqrt{1 - \frac{1}{2}}\right) \\
= \frac{1}{4} \left( 2 - \sqrt{2} \right) [/eqn]
Now you can plug in x = pi/16 to find sin(pi/16) and so on.
Kayden Taylor
There is a minus there somewhere.
Nathan Lewis
As long as you can establish that:
[eqn]\sin{\frac{\pi}{2^{k+1}}}=\frac{1}{2}\underbrace{\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots\sqrt{2}}}}}}}_{k\text{ square roots}}[/eqn]
(As demonstrated in )
Then the result follows from the fact that:
[eqn]\lim_{x\rightarrow0}{\frac{\sin{x}}{x}}=1[/eqn]
Lincoln Jackson
Check out an analysis book newfriend.
Nolan Bailey
I came up with a way to do that with just complex numbers
Alexander Butler
ITT:brainlets
en.wikipedia.org