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- Thread starter matqkks
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- Mar 5, 2012

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The thing that made me interested in continued fractions, is how rapidly $\pi$ is approximated by it with simple fractions - made it seem magical!What is the most motivating way to introduce continued fractions? Are there any real life applications of continued fractions?

\begin{array}{}

[3] &=& 3 &=& 3.00000000000000 \\

[3;7] &=& \frac{22}{7} &=& 3.14285714285714 \\

[3;7;15] &=& \frac{333}{106} &=& 3.14150943396226 \\

[3;7;15;1] &=& \frac{355}{113} &=& 3.14159292035398 \\

[3;7;15;1;292] &=& \frac{103993}{33102} &=& 3.14159265301190 \\

\pi &=& && 3.14159265358979

\end{array}

- Mar 22, 2013

- 573

I'd think it is not at all very rapid. Each to his own, perhaps?I like Serena said:The thing that made me interested in continued fractions, is how rapidly $\pi$ is approximated by it with simple fractions - made it seem magical!

- Feb 13, 2012

- 1,704

The main drawback of the continued fraction [3; 7, 15, 1, 292, 1, 1, ...] is that the sequence doesn't follow a precise scheme, so that computation cannot proceed autonomously. The following continued fraction follow precise schemes...I'd think it is not at all very rapid. Each to his own, perhaps?

$\displaystyle \pi = 3 + \frac{1}{6 + \frac{3^{2}}{6 + \frac{5^{2}}{6 + ...}}}\ (1)$

$\displaystyle \pi = \frac{4}{1 + \frac{1}{3 + \frac{2^{2}}{5 + \frac{3^{2}}{7+...}}}}\ (2)$

Kind regards

$\chi$ $\sigma$

- Mar 22, 2013

- 573

Right.chisigma said:The main drawback of the continued fraction [3; 7, 15, 1, 292, 1, 1, ...] is that the sequence doesn't follow a precise scheme.

PS : I'd like to add another CF to your list :

$$1 + \frac{1^{2}}{2 + \frac{3^{2}}{2 + \frac{5^{2}}{2 + \frac{7^{2}}{2 + \cdots}}}}$$

- Aug 18, 2013

- 76

$$\varphi =1+\frac{1}{1+\frac{1}{1+\frac{1}{1+...}}}$$

$$\sqrt{2}=1+\frac{1}{2+\frac{1}{2+\frac{1}{2+...}}}$$

$$e=2+\frac{1}{1+\frac{1}{2+\frac{1}{1+...}}}=[2;1,2,1,1,4,1,1,6,1,1,8,...]$$

- Mar 22, 2013

- 573

Uh, but they are not for calculating $\pi$ are they?

- Aug 18, 2013

- 76

The OP never asked for specific numbers.

- Feb 13, 2012

- 1,704

... and till now no satisfactory answer has been given. Regarding the possible use of CF for the computation of a constant like $\pi$ we can compare a series solution and a Cf solution. The series solution can be based on the following McLaurin expansion...

$\displaystyle \sin^{-1} x = x + \sum_{n=1}^{\infty} \frac{(2 n -1)!!}{(2 n + 1)\ (2 n)!!}\ x^{2 n + 1} (1)$

... and from (1) we derive...

$\displaystyle \pi = 3 + 6\ \sum_{n=1}^{\infty} \frac{(2 n -1)!!}{2^{2 n + 1}\ (2 n + 1)\ (2 n)!!}\ (2)$

One possible CF solution is...

$\displaystyle \pi = 3 + \frac{1}{6 + \frac{3^{2}}{6 + \frac{5^{2}}{6 + ...}}}\ (3)$

Which allows a more comfortable computation of $\pi$?... the (2) allows at each step to verify in some way the accuracy of computation and if necessary a further step can be done without problems. In the (3) the computation proceeds 'backward' in the sense that the last term is the first to be computed and to perform a further step the entire procedure must be repeated from the beginning... this way isn't comfortable!...

An example of application to CF to the 'real life' derives from my past professional experience. When I was involved in telecom equipment design it was not unusually to generate a 'funny frequency' like 84080 Hz frequency locked to a 'more conventional' frequency like 16000 Hz. In order to realize that it was necessary to implement a non integer frequency divider by something like $\frac{1051}{100}$ and the hardware scheme was directly derived by the CF expansion...

$\displaystyle \frac{1051}{100} = 11 - \frac{1}{2 + \frac{1}{25 - \frac{1}{2}}}\ (4)$

Kind regards

$\chi$ $\sigma$

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- Feb 7, 2012

- 2,725

It's not exactly a "real life" application, but continued fractions are a key tool in attacking Diophantine equations such as Pell's equation. See MarkFL's notes on that topic.What is the most motivating way to introduce continued fractions? Are there any real life applications of continued fractions?

- Feb 15, 2012

- 1,967

$x^2 - bx - c = 0$, where $b,c$ are integers.

Elementary algebra shows that:

$x = b + \dfrac{c}{x} = b + \dfrac{c}{b + \frac{c}{x}} = b + \dfrac{c}{b + \frac{c}{b+\frac{c}{x}}} = \dots$

allowing any such equation to be "solved" by use of continued fractions.

One can use an adaptation of this to calculate $\sqrt{n}$:

First one determines $a_0 = \lfloor{\sqrt{n}}\rfloor$, and writes:

$\sqrt{n} = a_0 + \dfrac{1}{x_1}$.

Thus leads to:

$x_1 = \dfrac{1}{\sqrt{n} - m} = \dfrac{\sqrt{n} + a_0}{n - a_0^2}$

One then repeats this process using:

$a_1 = \lfloor{\dfrac{\sqrt{n} + a_0}{n - a_0^2}}\rfloor$ so that:

$x_1 = a_1 + \dfrac{1}{x_2}$, leading to the continued fraction:

$\sqrt{n} = a_0 + \dfrac{1}{a_1 + \frac{1}{a_2 + \frac{1}{a_3 + \frac{1}{a_4 + \dots}}}}$

It can be shown that the sequence $a_0,a_1,a_2,\dots$ is "eventually periodic" which means that square roots stand in the same relationship to continued fractions as rational numbers do to decimal expansions.