A proof that π < 22/7

Note: this proof has a fatal error.

There’s a reasonably well-known proof that 22/7 > \pi, which can be written in one line:

0 < \int_0^1 {x^4 (1-x)^4 \over 1+x^2} dx = {22 \over 7} - \pi But I’ve always found this one unsatisfying because what does that integral have to do with \pi anyway? As it turns out, \pi enters through the integral \int_0^1 {1 \over (1+x^2)} = \pi/4. But let’s say I’m a purist and think that \pi is about circles. Can I do better? (Of course I can. If I couldn’t I wouldn’t be writing this post.) Start by observing that (21/11)^3 > 7, which can be shown by explicit computation: 11^3 = 1331 and 21^3 / 7 = 21^2 \times 3 = 1323.

Edited to add, November 30: of course I just showed here that (21/11)^3 < 7. This is what happens when you do arithmetic in your head…

Now, the sine function is given by the alternating series sin(x) = x - {x^3 \over 3!} + {x^5 \over 5!} - {x^7 \over 7!} + \cdots and in particular \sin(x) > x - x^3/6 by the alternating series test. Applying this with x = 11/21 gives \sin(11/21) > 11/21 - (11/21)^3/6 > 11/21 - (1/7)/6 = 1/2.

Taking the inverse sine of both sides, 11/21 > \sin^{-1} 1/2.

Finally, we have \pi = 6 \sin^{-1} (1/2). This is a geometric fact that goes back to Euclid’s construction of the hexagon. So \pi < 22/7.

On related notes:

  •  Noam Elkies, Why is π2 so close to 10?
  • Alejandro Morales, Igor Pak, and Greta Panova, Why is π < 2φ?  (This is not a particularly good approximation, but it actually admits a combinatorial proof in terms of Fibonacci and Euler numbers.)
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