What are the applications of continued fractions?

Question

What is the most motivating way to introduce continued fractions? Are there any real life applications of continued fractions?

Answer 1

One very nice elementary application is Gosper's batting average problem: if a baseball player's (3-digit rounded) batting average is $.334$, what's the smallest number of at-bats that player could have? (Batting average is computed as (number of hits)/(at-bats).)

The solution proceeds by noting that a rounded average of $.334$ corresponds to an actual number in the range $[.3335, .3345)$; finding the continued fractions for these values yields $.3335 = 667/2000 = [0; 2, 1, 666]$ and $.3345 = 669/2000 = [0; 2, 1, 94, 1, 1, 3]$. This implies that the 'simplest' number within the range is $[0; 2, 1, 95] = \frac{96}{287}\approx 0.334495$.

Answer 2

Since no one mention it, I think it's really pleasant knowing that every rational tangle correspond to some continued fraction(Conway showed that).

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For further information see:

• An Enumeration of Knots and Links, and Some of Their Algebraic Properties by Conway

• Another proof On the classification of rational tangles

Here is good expository article:

• The Power of Mathematics by Conway

• Conway’s Rational Tangles, a good educational article

Note:Image source: http://rationaltangle.wordpress.com/what-are-tanglesrational-tangles/

Answer 3

There is this interesting article about the application of continued fractions in Phyllotaxis, the research on leaves, which I found pretty interesting. Maybe you can find more over google? ${}{}{}{}{}{}$

Answer 4

Here are some properties/uses that might be used to motivate an investigation of Continued Fractions:

1. Surprisingly good rational approximations to real numbers
2. Finite for rational numbers and repeating for quadratic algebraic numbers
3. Used to solve linear Diophantine equations
4. Used to solve Pell-type equations

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Continued fractions are extremely good rational approximations to real numbers. If $\frac pq$ is a continued fraction approximation for the real number $r$, we have $$ \left|\,\frac pq-r\,\right|\le\frac1{q^2}\tag{1} $$ Furthermore, if we have $$ \left|\,\frac pq-r\,\right|\le\frac1{2q^2}\tag{2} $$ then $\frac pq$ is a continued fraction approximation for $r$.

In this manner, continued fractions offer the best rational approximations to real numbers.

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The continued fraction for a real number $r$ terminates if and only if $r$ is rational. Furthermore, the continued fraction for a real number $r$ is periodic if and only if $r$ is not rational and $ar^2+br+c=0$ where $a,b,c\in\mathbb{Z}$ are not all $0$.

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Successive convergents, $\frac{p_j}{q_j}$ and $\frac{p_{j+1}}{q_{j+1}}$ of a continued fraction have the property $$ p_jq_{j+1}-q_jp_{j+1}=(-1)^j\tag{3} $$ Note that $(3)$ implies $$ \frac{p_j}{q_j}-\frac{p_{j+1}}{q_{j+1}}=\frac{(-1)^j}{q_jq_{j+1}}\tag{4} $$ Thus, successive convergents alternate above and below their limit. $(4)$ shows why $(1)$ is true.

Given $a,b\in\mathbb{Z}$ relatively prime, $(3)$ is very useful in solving the Diophantine equation $$ ax+by=1\tag{5} $$ For example, we can solve $23x+17y=1$ by computing the continued fraction for $\frac{23}{17}$: $$ \frac{23}{17}=1+\cfrac1{2+\cfrac1{1+\cfrac1{5}}}\tag{6} $$ The next to last convergent is $\frac43$: $$ \frac43=1+\cfrac1{2+\cfrac1{1}}\tag{7} $$ thus, $(3)$ says $$ 4\cdot17-3\cdot23=-1\tag{8} $$ Therefore, we have the solution $$ 3\cdot23-4\cdot17=1\tag{9} $$

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The accuracy of Continued Fraction approximations allow the solution of Pell's equation. For example, consider the continued fraction for $\sqrt3$: $$ \sqrt3=(1,1,2,1,2,\overbrace{1,2,}^{\text{repeats}}\dots)\tag{10} $$ Since $\left|\frac pq-\sqrt3\right|\le\frac1{q^2}$, we have $$ \begin{align} \left|p^2-3q^2\right| &=\left|p-q\sqrt3\right|(p+q\sqrt3)\\ &\lesssim\frac1q\cdot2q\sqrt3\\ &=2\sqrt3\tag{11} \end{align} $$ Thus, no matter how big $p$ and $q$ get, $p^2-3q^2$ is bounded; in fact, it is periodic. The sequence of convergents for $(10)$ is $$ \frac11,\color{#C00000}{\frac21},\frac53,\color{#C00000}{\frac74},\frac{19}{11},\color{#C00000}{\frac{26}{15}},\frac{71}{41},\dots\tag{12} $$ and the sequence of $p^2-3q^2$ is $$ -2,\color{#C00000}{1},-2,\color{#C00000}{1},-2,\color{#C00000}{1},-2,\dots\tag{13} $$ Thus, the convergents in red give solutions for the Pell equation $p^2-3q^2=1$.

Answer 5

There are at least three integer factorization algorithms based (one way or other) on properties of continued fractions. The one by Lehman is a fairly elementary (but delightfully clever) deterministic method.

1. Shanks
2. Brillhart and Morrison
3. Lehman

Answer 6

These may not be the best ways to motivate continued fractions, but here are some applications.

There is a well-known method for "encoding" a geodesic on the modular surface via continued fractions by looking at the continued fraction expansions of the "endpoints" of the geodesic on the real line. Studying the continued fraction associated to a geodesic is one way of understanding how the geodesic "winds around" the surface. For example, if the continued fractions are periodic, then the geodesic will be periodic (a closed loop) as well. If the sequences of integers appearing in the continued fraction expansion contain all finite sequences of integers, then the geodesic is dense. To my knowledge the first time this is used is in an old paper of Emil Artin, Ein mechanisches System mit quadiergodischen Bahnen. This idea is expanded upon and clarified in a series of papers by Caroline Series. See, for instance, Non-Euclidean geometry, continued fractions, and ergodic theory.

More recently, continued fractions have come up in studying the dynamics of flows on translation surfaces -- a special case of which is studying billiards in a rational polygon. For example, continued fractions and Diophantine approximations make an appearance in studying billiards in the wind-tree. That is, imagine taking the plane and placing identical rectangular obstacles centered at each integer lattice point. Now consider an ideal point-mass which bounces between the obstacles, reflecting off the obstacles according to the rule from geometric optics that angle of incidence equals angle of reflection. Studying the motion of that billiard (e.g., are there periodic orbits, dense orbits, escaping orbits, recurrent orbits, etc.) can, at least in some special cases, boil down to studying continued fractions. See The Ehrenfest wind-tree model: periodic directions, recurrence, diffusion by Hubert, Leliévre, Troubetzkoy and Divergent directions in some periodic wind-tree models by Delecroix.

Again, those probably aren't the best ways to motivate continued fractions if you're seeing them for the first time, but they do have plenty of applications.

Answer 7

Application #1: The first proof that $e$ is irrational, due to Euler, was based on showing its continued fraction expansion is infinite: $e = [2,1,2,1,1,4,1,1,6,1,1,8,1,1,10,...]$ with the obvious repeating pattern of even numbers separated by two $1$'s.

Application #2: A year consists of $365.24219$ days. If we want to create a calendar with a block of $Y$ years containing $L$ leap years, then $Y$ years will contain $365Y+L$ days. The average number of days in a year of each block is $(365Y+L)/Y = 365 + L/Y$, so we want $L/Y$ to be a good rational approximation to $.24219$, whose continued fraction expansion is $[0,4,7,1,3,24,6,2,2]$. The first six continued fraction convergents are $0$, $1/4$, $7/29$, $8/33$, $31/128$, and $752/3105$. The convergent $1/4$ leads to the Julian calendar having $1$ leap year every $4$ years. The choice $L/Y = 8/33$ is used in Iran: $8$ leap years in a block of $33$ years is part of the Jalil calendar proposed by Omar Khayyam in the year $1079$. If we look at intermediate convergents between $31/128$ and $752/3105$, which by definition are fractions $(31a + 8)/(128a+33)$ for $a = 1,2,\dots,23$, then at $a = 6$ we get $194/801$, and $801$ is nearly $800$, a nice round $8$ centuries. So let's modify this a little to consider $194/800 = 97/400$: this is the Gregorian calendar, which has $97$ leap years every $400$ years, that is a leap year every $4$th year except for years divisible by $100$ but not by $400$ (a total of $100 - 3 = 97$ leap years).

Application #3: As spin wrote in a comment, being able to make a good guess for the value of a rational number when you know only some approximation to it (even if the approximation is itself rational!) is a key part of Shor's algorithm for factoring integers on a quantum computer. Guessing rational numbers from numerical approximations is also important in doing computations related to elliptic curves, as William Stein explained to me: if an elliptic curve $E$ is defined over $\mathbf Q$ and you use complex analysis to compute a coefficient of a Weierstrass equation for $E$, or to compute $L(E,1)/\Omega$, all you get is a decimal estimate for a number that is supposed to be rational and you want to decide what that rational number ought to be (which might then be rigorously proved by other methods). Typically the decimal approximation of the rational number may have far fewer digits than the full period length, so you can't guess the fraction as simply as you can if someone asks you what fraction $.333333333$ probably is (namely $1/3$).

Here is an example of how continued fractions let you guess rational numbers from decimal expansions, which I did as a trick at the end of a talk I gave today on continued fractions. I asked a student in the audience to compute a fraction as a decimal on a calculator and tell me what digits showed up on the calculator, but make sure there was no obviously periodic part shown by the calculator and not to tell me the original fraction and make sure to write down the original fraction for comparison with my guess later. The student gave me $.1391200951$. Using a computer algebra package, I converted $.1391200951$ into a continued fraction: it is $[0,7,5,3,6,1,56891,1,103,2]$. The $56891$ is so huge compared to what comes before that my educated guess for the student's original fraction is $[0,7,5,3,6,1]$, which is $117/841$. That indeed was the number the student wrote down earlier. The audience was impressed. The actual decimal period of $117/841$ is $812$, so I was able to figure out a fraction having a repeating decimal period length over $800$ knowing only the first $10$ digits of its decimal expansion.

Answer 8

There are many beautiful theorems about continued fractions. For example, a real number is rational if and only if its continued fraction expansion is finite (however, this is not the case for decimal system, since $\frac{1}{3}=0.3\cdots$ is infinite and rational). Also, continued fraction is the best possible rational approximation to any given irrational number. Hence it can be used to approximate irrationals (for example, $\pi$).

For other applications, one can use continued fraction to solve the Pell's equation or to generate a chaotic process (by the Gaussian map: $x_{n+1}=\frac{1}{x_n}-\lfloor{\frac{1}{x_n}}\rfloor$).

You may want to read the chapter in continued fractions in Hardy & Wright's book An Introduction to the Theory of Numbers. Or you can also read a short article titled Chaos in Numberland: The secret life of continued fractions by John Barrow.

Answer 9

Two ideas which could be useful for you:

$1$. Proposal: Combine continued fractions with the concepts of golden ratio and fibonacci numbers

I think starting with the simplest continued fraction

$$\Phi=\frac{1}{1+\frac{1}{1+\frac{1}{1+\cdots}}}=[1,1,1,\dots]$$

offers charming possibilities. With this prototype you can bring together continued fractions with the golden ratio and with the fibonacci sequence as fundamental and beautiful concepts.

Fibonacci Sequence and continued fractions: Let $(F_n)_{n\geq0}$ the Fibonacci sequence, then

$$\Phi=\lim_{n\rightarrow\infty}\frac{F_n}{F_{n-1}}=[1,1,1,\dots]$$

Golden ratio and continued fractions:

$$\Phi=\frac{\sqrt{5}-1}{2}=[1,1,1,\dots]$$

You could talk about the mystic pentagram, detect the golden ratio between successive length and diagonals of inscribed pentagrams and reveal the representation of $\Phi$ as continued fraction. You could also talk about golden rectangles and their applications by artists and architects like Le Corbusier and Dali or the classic Parthenon with its many geometric relations resulting seemingly in golden ratios.

I do appreciate the hint from BIS HD. Phyllotaxis as form of evolutionary development which has optimized the positions of the seeds of sunflowers, daisies or pine cones using spirals is a great example how simple and fundamental concepts $[1,1,1,\dots]$ iteratively applied are behind seemingly complex structures of nature. And, counting the number of seeds of successive spirals will give you successive Fibonacci numbers!

$2.$ Proposal: Use continued fractions and lattice paths as picturesque concept

I think, that lattice paths are a fine, picturesque example of an application of continued fractions. Lattice paths are structures we can see and we can look what happens, when they are manipulated.

Here I would like to point to a wonderful presentation of continued fractions summarizing the contribution to this subject of the great Flajolet and also providing some historical aspects. For me this is a superb (and funny!) example to familiarize the audience with continued fractions. At least the first part of this presentation could also be useful for your needs. On page $149$ you will find a reference to a (rather complex) real life application in theoretical physics Planar maps and continued fractions from J. Bouttier and E. Guitter.

Here is an article Combinatorial aspects of continued fractions from Flajolet containing lattice pathes and maybe you could also find some motivating material in the survey Continued fractions from Euclid to the present day.

Answer 10

Musical notes follow a geometric sequence, for their frequency $f_n$ relative to a base frequency $f_0$ :

$$f_n = f_0 \times \left(\sqrt[12]{2}\right)^n$$

For example, an A note has defined frequency 440 Hz, so a note 5 chromatic steps above it (D note) is $$f_5 = 440 \sqrt[12]{2}^5 = 440 \times 1.3348...$$

But a musician will often want to play a notes that are small ratios of each other because it sounds better. So what are the best rational approximations for $2^{5/12}$ ? Look at the continued fraction:

$$2^{5/12} = [1; 2, 1, 73, ...] \approx \left\{1, \frac 43, \frac 75 \frac{299}{244}, ... \right\}$$

So we can see that the musician is probably playing a $\frac 43$ ratio. It was obvious for this interval but you can try other intervals.