Infinite Nested Radical Product

Calculus Level 5

Find p p : (Report your answer to 5 places after decimal)

2 5 1 3 ϕ lim s 1 2 s k = 1 s ( Δ k ) = 1 p \large \displaystyle \frac 2{5} \cdot \frac 1{\sqrt{3-\phi}} \cdot \lim_{s \to \infty}\frac 1{2^s}\cdot \prod_{k=1}^s (\Delta _{k})=\frac 1{p}

where, Δ m = 2 + 2 + 2 + + 2 + 2 + ϕ m r o o t s \Delta_m = \underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt{2+\sqrt{2+\phi}}}}}}_{m \ roots}

and, ϕ = 1 + 5 2 \phi = \displaystyle \frac {1+\sqrt{5}}{2} is the Golden Ratio.

For example: Δ 1 = 2 + ϕ , Δ 3 = 2 + 2 + 2 + ϕ , \Delta_1=\sqrt{2+\phi}, \Delta_3=\sqrt{2+\sqrt{2+\sqrt{2+\phi}}}, etc.


The answer is 3.14159.

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1 solution

We know,

sin θ θ = cos θ 2 cos θ 4 cos θ 8 cos θ 16 cos θ 32 \displaystyle \frac {\sin \theta}\theta=\cos {\frac {\theta}2} \cdot \cos {\frac {\theta}4} \cdot \cos {\frac {\theta}8} \cdot \cos {\frac {\theta}{16}} \cdot \cos {\frac {\theta}{32}} \cdots

Putting θ = π 5 \displaystyle \theta= \frac{\pi}5 we get:

5 3 ϕ 2 π = cos π 10 cos π 20 cos π 40 cos π 80 cos π 160 \displaystyle \frac{5\cdot \sqrt{3-\phi}}{2 \cdot \pi}=\cos {\frac {\pi}{10}} \cdot \cos {\frac {\pi}{20}} \cdot \cos {\frac {\pi}{40}} \cdot \cos {\frac {\pi}{80}} \cdot \cos {\frac {\pi}{160}} \cdots

Also we know, cos π 10 n = 1 2 Δ m = 1 2 2 + 2 + 2 + + 2 + 2 + ϕ m r o o t s \displaystyle \cos{\frac{\pi}{10n}}=\frac 12 \Delta_m = \frac 12\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt{2+\sqrt{2+\phi}}}}}}_{m \ roots}

hence,

2 5 1 3 ϕ lim s 1 2 s k = 0 s ( Δ 2 k ) = 1 π \displaystyle \frac 2{5} \cdot \frac 1{\sqrt{3-\phi}} \cdot \lim_{s \to \infty}\frac 1{2^s} \cdot \prod_{k=0}^s (\Delta _{2^k})=\frac 1{\pi} p = π 3.14159 \therefore p=\pi \approx 3.14159

The 2 equations you have written after "we know" is amazing. May you send proof for those Equation??

Adarsh Singh - 3 years, 7 months ago

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Here is the proof of Euler's Formula:

Mrigank Shekhar Pathak - 3 years, 7 months ago

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