10
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Let $$\alpha = \sqrt{2\sin^2 1+\sqrt{2\sin^2 2 + \sqrt{2\sin^2 3 + \cdots}}} =\sqrt{3.1415...}$$ Prove that $\alpha^2 \neq \pi$. It is a remarkable approximation though.

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  • $\begingroup$ Is it till infinity? $\endgroup$ – Zenix Oct 24 at 14:47
  • $\begingroup$ $\sin^2 1$, that's $1$ radian? $\endgroup$ – GEdgar Oct 24 at 14:47
  • $\begingroup$ @GEdgar: yes, radians. $\endgroup$ – Vincent Granville Oct 24 at 14:58
  • $\begingroup$ @Zenix: the nested radicals go indefinitely, but only the first 4 digits match those of $\pi$. $\endgroup$ – Vincent Granville Oct 24 at 15:00
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    $\begingroup$ A cool find, maybe it was just random just as with $\sqrt{2}+\sqrt{3}\approx \pi$ $\endgroup$ – user712576 Oct 24 at 15:22
7
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This is easy numerically. It is well known that $$\sqrt{2+\sqrt{2+\cdots}}=2$$ so the tail in the nested radical is always less than $2$. We find that $$\alpha < \sqrt{2\sin^2 1+\sqrt{2\sin^2 2 + \sqrt{2\sin^2 3 + \cdots +\sqrt{2\sin^2 13+2}}}}\approx 1.7724371077589929$$ so that $$\alpha^2 < 3.1415333009610635 < \pi$$

Here's my python script:

from math import sqrt, sin

def f(n):
    answer = 2+sqrt(2*sin(n)**2)
    for k in range(n-1,0,-1):
        answer = sqrt(2*(sin(k)**2)+answer)
    return answer

alpha = f(13)
print(alpha, alpha**2)

This prints

1.7724371077589929 3.1415333009610635
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  • 3
    $\begingroup$ That doesn't look $< \pi$ to me...? $\endgroup$ – Mees de Vries Oct 24 at 15:18
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    $\begingroup$ However, changing 7 into 13 in your script gives $\alpha^2 < \pi$ as desired. $\endgroup$ – Mees de Vries Oct 24 at 15:19
  • 1
    $\begingroup$ @MeesdeVries Wow. I must be getting cross-eyed. Thanks $\endgroup$ – saulspatz Oct 24 at 15:23

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