10
$\begingroup$

On Wolfram Alpha, I see continued fractions being listed in the results. Although I understand continued fractions, and how they can be used for approximations, what is a better approximation than a decimal representation of the number?

For example,

31.999999999664

tells me so much more than

enter image description here

So, I do not get why continued fractions are important. Are they supposed to tell me something which I am missing all these days?

  • 0
    Large entries indicate particular good approximations. The reason why $\pi\approx\frac{355}{113}$ is good to $6$ decimal digits is that the continued fraction starts with $[3,7,15,1,292,1,\cdots]$. A matter of taste how useful or interesting such stuff is ...2017-04-05

5 Answers 5

6

I once encountered a situation where I had a floating-point number, probably 10 or 12 decimal places long, but I suspected that it was actually a rounded-off version of some rational number with not-too-big denominator (less than 1000). How can I find (or at least guess) the rational number? What I did was to start computing (in Excel, which tells you something about my programming skills) the continued fraction of my floating-point number. For a while, the integers I got were reasonably small ones, but then I got one that was way bigger than the previous ones. So I said: That must be the rounded-off version of $\infty$. In other words, I replaced that big integer with $\infty$, which amounts to cutting off the continued fraction at that step. The resulting finitely long continued fraction is, of course, rational, and it gave the answer I wanted. (Later, I checked with a number theorist, and he confirmed that this is how one should attack such questions.)

  • 0
    @KCd I don't recall details, but I'm pretty sure the issue arose in connection with secret-sharing schemes, probably dealing with estimates on their rates. Some of these estimates come from solving a big system of linear inequalities with integer coefficients; that would be why I expected decent rational numbers to be involved.2017-04-17
5

How are continued fractions cool? Let me count the ways:

  1. If you know how to get the partial terms of the continued fraction expansion of an irrational number, you essentially have a way to find approximate values of the irrational number. This gives you one possible way to get a floating-point estimate for $\sqrt{2}$, $\pi$, etc.

  2. It is known that a number is irrational if and only if it has an infinite continued fraction expansion. This makes it a sort of irrationality test.

  3. Continued fraction expansions of irrational numbers exhibit sometimes surprising regularity. $\pi$ and $e$ have continued fraction representations that are simple, which is strange seeing as they are transcendental numbers. This can sometimes tell you things about these numbers, and other times the existence of such a regular expansion is just plain cool.

  4. As mentioned in the comments, continued fractions can be used to solve certain equations like the Pell equation.

And others. Continued fractions solve some other problems - a famous story relates that Mahalanobis and Ramanujan shared a room, and one day Mahalanobis posed Ramanujan a problem which he instantly solved for all cases via continued fractions.

2

Some commonly-encountered irrational numbers have "nice" representations as continued fractions:

  • $e = [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, 1, 1, 12, ...]$
  • $\sqrt{2} = [1; 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, ...]$
  • $\phi = [1; 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...]$

By continuing the pattern to the desired level of precision, you can obtain an accurate rational approximation of the number.

1

Continued fractions are useful for converting decimals into fractions, or reducing fractions to their common divisor. Often finding fractions between X and Y, like 8/11 for the logrithms of the chords of the heptagon, means that (in this case), you only need to consider 11 cases.

0

In your example, the huge number $2997044470$ tells you that you will make very little error stopping before it. At the third step you know you are very close to $32$.