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$(\pi+2)$-Day: Buffon's Noodle
Maxim Arnold and Nathan Williams
Department of Mathematical Sciences
University of Texas at Dallas
$\pi$: the ratio of the circumference of a circle to its diameter.
Mathematics: an Experimental Science
In 1773, Georges Louis Leclerc, le Comte de Buffon devised a
mathematical experiment...
Buffon's Needle
To perform this experiment at home, you'll need:
- Needles. (Preferably substitute something less sharp.)
- A grid of equally spaced parallel lines (unit distance apart).
Buffon's Needle
How often does a randomly tossed needle cross a line?
Call this probability $E$ (depends on length $\ell$ of a needle).
Now compute
\[\frac{2}{\text{average number of crossings}} \approx \frac{2}{E}\]
Now compute
\[\frac{2}{\text{average number of crossings}} \approx \frac{2}{E} = \pi?\]
Buffon's Needle
By breaking a needle in two, one checks that \[E=C \ell,\] where $\ell$ is the length of the needle and $C$ is
some constant.
But how can we determine this constant?
Buffon's Noodle
$E=C\ell$ only depends on $\ell$, and doesn't change for a "noodle".
Buffon's Noodle
$E=C\ell$ only depends on $\ell$, and doesn't change for a "noodle".
Since we can approximate a noodle by straight needles.
Buffon's Noodle
$E=C\ell$ only depends on $\ell$, and doesn't change for a "noodle".
Use a circular noodle of unit diameter.
The length $\ell$ of such a noodle is our
definition of $\pi$.
Such a noodle will always have exactly two intersections:
Use a circular noodle of unit diameter.
The length $\ell$ of such a noodle is our
definition of $\pi$.
Such a noodle will always have exactly two intersections:
Since $E = C \ell$, we can now determine our constant $C$:
\[2 = C \pi, \text{ or } C = \frac{2}{\pi}.\]
We conclude that if $\ell=1$
\[\frac{2}{\text{average number of crossings}} \approx \frac{2}{E} = \frac{2}{C\ell} = \frac{2}{\pi/2} = \pi.\]
How to compute (digits of) $\pi$?
Approximating $\pi$ using Buffon's needles is slow. There are more efficient ways....
- Inscribe/circumscribe regular polygons in a circle.
- Infinite series.
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