3
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$\pi\approx 3.141592654$

Why is it so close to $3$?

I find this intriguing, this cannot be a coincidence.

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closed as off-topic by Brian Fitzpatrick, Claude Leibovici, mookid, copper.hat, Gyu Eun Lee Mar 28 '14 at 6:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is not about mathematics, within the scope defined in the help center." – Brian Fitzpatrick, Claude Leibovici, mookid, copper.hat
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 15
    $\begingroup$ Would you prefer it farther away? $\endgroup$ – Will Jagy Mar 28 '14 at 5:36
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    $\begingroup$ Because a regular hexagon is "so close to" a circle? $\endgroup$ – Blue Mar 28 '14 at 5:36
  • 1
    $\begingroup$ And why is the golden ratio $\frac{1 + \sqrt{5}}{2} \approx 1.62$? The answer? Because that's what it is. $\endgroup$ – Jared Mar 28 '14 at 5:58
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    $\begingroup$ I should also comment that some consider $\pi$ to be the wrong value. Why not define $\pi$ as the ratio of the circumference to the radius? Then, instead of $C = 2\pi r$, we would have $C = \pi r$. People tend to use the radius more often than the diameter. Furthermore, then there would be $\pi$ "radians" in a full circle (why should it be $2\pi$, why multiply by $2$)? Then you'd be asking why is $\pi$ so close to $6$? And the answer would be the same, because it is. $\endgroup$ – Jared Mar 28 '14 at 6:02
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    $\begingroup$ Why is $e$ so close to ${ 2718281828 \over 1000000000} $? An accident perhaps? $\endgroup$ – copper.hat Mar 28 '14 at 6:48
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As you can clearly see from this figure, $2\pi\simeq6\iff\pi\simeq3$, since the side of the inscribed regular polygon is equal to the radius of the circle:

$\qquad\qquad\qquad\qquad$1

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22
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It's even closer to 3.1 ${}{}$!

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    $\begingroup$ The series $$\pi=3\sum_{k=0}^{\infty}\frac{1}{(1+k)(1+2k)(1+4k)}$$ yields both $3$ and $3.1$ when taking one and two terms: $$\pi=3+3\sum_{k=1}^{\infty}\frac{1}{(1+k)(1+2k)(1+4k)}$$ $$\pi=3+\frac{1}{10}+3\sum_{k=2}^{\infty}\frac{1}{(1+k)(1+2k)(1+4k)}$$ $\endgroup$ – Jaume Oliver Lafont Feb 2 '16 at 12:29
7
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Let $a<b$ be integers. Pick a number, which we'll call $\pi'$ in $[a,b]$ uniformly at random. The chance that $\pi'$ is within $\pi-3\approx .1415$ of some integer is $2(\pi-3)\approx .283$.

If something has a near $30\%$ chance of occuring at random, I would say that it could definitely just be a coincidence.

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