Continued Fraction

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Call the extra x.

Now we can take $\frac{1}{x}.$ That too is a whole number, plus a bit left over. Again, that extra bit might be zero, in which case we're done, but it might not be. Subtract off the whole number, and repeat.

• Note: This will terminate if and only if the starting number is rational.
• Not hard to prove, and an interesting exercise.

$\pi$	=	3	+	0.1415926...
1/0.1415926...	=	7	+	0.0625133...
1/0.0625133...	=	15	+	0.9965944...
1/0.9965944...	=	1	+	0.0034172...
1/0.0034172...	=	292	+	0.6345908...

$\Large{\pi=3+\frac{1}{7+\frac{1}{15+\frac{1}{1+\frac{1}{292+...}}}}}$

We can write this as $\pi=[3;7,15,1,292,...]$

Cutting this off at different stages gives us rational approximations.

• $[3]=3=3.0000...$
• $[3;7]=22/7\approx3.142857$
• $[3;7,15]=333/106\approx3.14151$
• $[3;7,15,1]=355/113\approx3.141593$
• $[3;7,15,1,292]=103993/33102\approx3.1415926530$

As you can see, the large number in the expansion causes a sudden jump in the terms used in the rational approximation.

However, a large number in the continued fraction implies a small error in the previous step. That means that cutting off the continued fraction just before a large number will give an unreasonably good approximation.

Hence $\pi\approx\frac{355}{113}$

Using this technique gives an approximation with an error that is "best" given a limit on the size of the denominator.

1 2 2 2 2 2 ... 0 1 1 3 7 17 41 99 ... 1 0 1 2 5 12 29 70 ...

So we have $\frac{41}{29}$ and $\frac{99}{70}$ as approximations to $\sqrt{2}.$

Each term is the product of the quotient above it and the term on its left, plus the term two to the left. Here it is for $\pi$

3 7 15 1 292 ... 0 1 3 22 333 355 103933 ... 1 0 1 7 106 113 33102 ...

So 333=15*22+3 and 113=1*106+7, and we get $\frac{355}{113}$ as an approximation of $\pi.$