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This seems counterintuitive, but $22/7$ is closer to $\pi$ than $3.14=314/100$ which has a significantly greater denominator.

Why is $22/7$ a better approximation for $\pi$ than $3.14$?

This has important implications: e.g. Should "$\pi$-day" be the $14^{th}$ of March or $22^{nd}$ of July?

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    Yes; that was just an attempt at a joke. @Louis: (a) My question clearly implied that I already knew it to be true, and (b) Compare this to Q: "why is the sky blue?" A: "just look at it; it's blue." [The main question was more between-the-lines; i.e. property X seems counter-intuitive; how can I fix my intuition so as to not walk into property X traps in the future (be they related to \pi or not).]2011-08-22

4 Answers 4

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It only seems odd to you because you are used to representing numbers in base 10. What if you used base 7?

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    Yes, but 3.1 would look nicer than 313/101.2010-08-21
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Well, just measure $|\pi - 22/7|$ and $|\pi-3.14|$ ...

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    measure or calculate?2014-11-06
21

Just for fun...

Here is a proof that $\displaystyle \frac{22}{7}$ is a better approximation than $\displaystyle 3.14$.

First we consider the amazing and well known integral formula for $\displaystyle \frac{22}{7} -\pi$ (for instance see this page: Proof that 22/7 exceeds pi).

$\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^2}dx = \frac{22}{7} -\pi$

We will to show that

$0 < \frac{22}{7} -\pi < \pi - 3.14$

That $\displaystyle 0 < \frac{22}{7} - \pi$ follows trivially from the above integral.

We will now show that

$\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^2}dx < \frac{1}{700}$

We split this up as

$\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^2}dx = \int_{0}^{\frac{1}{2}}\frac{x^{4}(1-x)^{4}}{1+x^2} + \int_{\frac{1}{2}}^{1}\frac{x^{4}(1-x)^{4}}{1+x^2}dx$

The first integral can be upper-bounded by replacing $\displaystyle x$ in the denominator with $\displaystyle 0$ and the second integral can be upper-bounded by replacing $\displaystyle x$ in the denominator with $\displaystyle \frac{1}{2}$.

Thus we have that

$\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^2}dx < \int_{0}^{\frac{1}{2}}x^{4}(1-x)^{4}dx + \int_{\frac{1}{2}}^{1} \frac{4x^{4}(1-x)^{4}}{5}dx $

Now $\int_{0}^{\frac{1}{2}}x^{4}(1-x)^{4}dx = \int_{\frac{1}{2}}^{1}x^{4}(1-x)^{4}dx$ as $\displaystyle x^4(1-x)^4$ is symmetric about $\displaystyle x = \frac{1}{2}$

It is also known that $\int_{0}^{1}x^{4}(1-x)^{4}dx = \frac{1}{630}$ (see the above page again)

Thus we have that

$\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^2}dx < \frac{1}{2*630} + \frac{4}{5*2*630} = \frac{1}{700}$

Thus we have that

$\frac{22}{7} - \pi < \frac{1}{700}$

i.e

2\pi > 2(\frac{22}{7} - \frac{1}{700})

2\pi > \frac{22}{7} + \frac{22}{7} - \frac{2}{700}

2\pi > \frac{22}{7} + \frac{2200}{700} - \frac{2}{700}

2\pi > \frac{22}{7} + \frac{2198}{700}

2\pi > \frac{22}{7} + \frac{314}{100}

Thus we have that

$0 < \frac{22}{7} - \pi < \pi - \frac{314}{100}$

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It has to do with the continued fraction expansion of $\pi$. Suppose $[a_1, a_2, \ldots]$ is the continued fraction of an irrational number $\alpha$ -- that is, if $a_n$ is the (essentially unique) sequence of natural numbers such that if we define partial convergents by $x_1 = a_1$, $x_2 = a_1 + 1/a_2$, $x_3 = a_1 + 1/(a_2 + 1/a_3)$, $x_4 = a_1 + 1/(a_2 + 1/(a_3 + 1/a_4))$, and so on, then $\alpha = \lim_{n\to\infty} x_n$. Then the partial convergents $x_n$ are rational numbers that approximate $\alpha$ better than anything that is not a partial convergent, in the following sense: a rational number $\frac pq$ satisfies the inequality $|\alpha - \frac pq| < \frac 1{2q^2}$ if and only if $\frac pq$ is one of the convergents $x_n$. (One could, of course, come up with different notions of what constitutes a "good" approximation.)

The continued fraction expansion of $\pi$ is $[3,7,15,1,292,1,1,\dots]$, so the first few convergents are $3$, $\frac{22}{7}$, $\frac{333}{106}$, $\frac{355}{113}$, etc. Thus $\frac{22}{7}$ is a better approximation than $\frac{314}{100}$ (in the above sense) because it appears in the list of partial convergents, while $\frac{314}{100}$ does not.

Incidentally, the approximation $x_n$ is best when the coefficient $a_{n+1}$ is quite large, so the size of $a_5 = 292$ means that $x_4 = \frac{355}{113}$ is a particularly good approximation.

At the risk of self-promotion, I wrote a brief exposition of all this in a bit more detail -- you can find it on my website if you're interested, at http://www.math.psu.edu/climenha/contfrac.html.

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    I've just stumbled back on this thread from elsewhere and noticed that it looks somewhat disjointed, as Aryabhata's username has changed from what it was when we had this e$x$change of comments (at the time it was "Moron")... I figured I should add this by way of historical e$x$planation so that the exchange would make a little more sense, since I can't go back and edit my comments to use Aryabhata's current username.2012-03-14