Viète à grande Vitesse or Accelerated Viète (kind of)

[Albert Chan]

(06-18-2024 10:58 PM)Albert Chan Wrote:  

(06-18-2024 08:05 PM)Gerson W. Barbosa Wrote:  Starting with Viète's formula

\(\frac2\pi = \frac{\sqrt 2}2 \cdot \frac{\sqrt{2+\sqrt 2}}2 \cdot \frac{\sqrt{2+\sqrt{2+\sqrt 2}}}2 \cdots\)

pi/2 * cos(pi/4) * cos(pi/2^3) * cos(pi/2^4) ... = c = 1 + ε      // n cosines

We can run a few numbers for ε, and guess the pattern ... ε = c-1 = x^2/6 = (pi/2/2^n)^2 / 6

With n cosines, cos denominator goes from 2^2 to 2^(n+1)
If we have infinite cosinse, LHS is just 1

1 = c * cos(pi/2^(n+2)) * cos(pi/2^(n+3)) * cos(pi/2^(n+4)) ...      // infinite cosines

(06-04-2024 08:44 PM)Albert Chan Wrote:  cos(pi*z/2) = (1-(z/1)²) * (1-(z/3)²) * (1-(z/5)²) * (1-(z/7)²) ...

cos(pi*z/2^2) = (1-(z/1)²/4¹) * (1-(z/3)²/4¹) * (1-(z/5)²/4¹) * (1-(z/7)²/4¹) ...
cos(pi*z/2^3) = (1-(z/1)²/4²) * (1-(z/3)²/4²) * (1-(z/5)²/4²) * (1-(z/7)²/4²) ...
cos(pi*z/2^4) = (1-(z/1)²/4³) * (1-(z/3)²/4³) * (1-(z/5)²/4³) * (1-(z/7)²/4³) ...
...

1/4 + 1/4² + 1/4³ + ... = 1/4 / (1-1/4) = 1/3

Multiply all factors columnwise, ignore tiny z^4 or higher terms

product(cos(pi*z/2^k, k=2 .. inf) ≈ (1-(z/√3/1)²) * (1-(z/√3/3)²) * (1-(z/√3/5)²) ... = cos(pi*z/2/√3)

Let z = 1/2^n, we have:

cos(pi/2^(n+2)) * cos(pi/2^(n+3)) * cos(pi/2^(n+4)) ... ≈ cos(pi/2^(n+1)/√3)

cos(x) = 1 - x^2/2! + x^4/4! - x^6/6! + ...      → 1/cos(x) ≈ 1 + x^2/2, if x is small

c = 1 + ε ≈ 1/cos(pi/2^(n+1)/√3) ≈ 1 + (pi/2^(n+1)/√3)^2 / 2 = 1 + (pi/2^(n+1))^2/6      QED