Continued Fraction Algorithm/Examples/Root 2
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Example of Use of Continued Fraction Algorithm
The Continued Fraction Algorithm is a method for finding the continued fraction expansion for any irrational number to as many partial denominators as desired.
Let $x_0$ be the irrational number in question.
The steps are:
| \(\text {(1)}: \quad\) | \(\ds k\) | \(=\) | \(\ds 0\) | initialise | ||||||||||
| \(\text {(2)}: \quad\) | \(\ds a_k\) | \(=\) | \(\ds \floor {x_k}\) | the $k$th partial denominator (that is, $a_k$) is the integer part of $x_k$ | ||||||||||
| \(\text {(3)}: \quad\) | \(\ds x_{k + 1}\) | \(=\) | \(\ds \frac 1 {x_k - a_k}\) | the subsequent term is the reciprocal of the fractional part of $x_k$ | ||||||||||
| \(\text {(4)}: \quad\) | \(\ds k\) | \(=\) | \(\ds k + 1\) | increase $k$ by $1$ then go to step $(2)$ |
Then $x_0 = \sqbrk {a_0, a_1, a_2, \ldots}$ is the required continued fraction expansion.
Example
Applying the Continued Fraction Algorithm to $\sqrt 2$:
| \(\text {(1)}: \quad\) | \(\ds x_0\) | \(=\) | \(\ds \sqrt 2\) | |||||||||||
| \(\text {(2)}: \quad\) | \(\ds a_0\) | \(=\) | \(\ds \floor {x_0}\) | \(\ds = \floor {\sqrt 2}\) | step $(2)$ | |||||||||
| \(\text {(3)}: \quad\) | \(\ds \) | \(=\) | \(\ds 1\) | integer part of $\sqrt 2$ is $1$ | ||||||||||
| \(\text {(4)}: \quad\) | \(\ds x_{0 + 1}\) | \(=\) | \(\ds \frac 1 {x_0 - a_0}\) | \(\ds = \frac 1 {\sqrt 2 - 1}\) | step $(3)$ | |||||||||
| \(\text {(5)}: \quad\) | \(\ds x_1\) | \(=\) | \(\ds \frac 1 {\sqrt 2 - 1} \times \paren {\frac {\sqrt 2 + 1} {\sqrt 2 + 1} }\) | multiply by $1$ | ||||||||||
| \(\text {(6)}: \quad\) | \(\ds \) | \(=\) | \(\ds \sqrt 2 + 1\) | |||||||||||
| \(\text {(7)}: \quad\) | \(\ds a_1\) | \(=\) | \(\ds \floor {x_1}\) | \(\ds = \floor {\sqrt 2 + 1}\) | step $(2)$ | |||||||||
| \(\text {(8)}: \quad\) | \(\ds \) | \(=\) | \(\ds 2\) | integer part of $\paren {\sqrt 2 + 1 }$ is $2$ | ||||||||||
| \(\text {(9)}: \quad\) | \(\ds x_{1 + 1}\) | \(=\) | \(\ds \frac 1 {x_1 - a_1}\) | \(\ds = \frac 1 {\paren {\sqrt 2 + 1 } - 2}\) | step $(3)$ | |||||||||
| \(\text {(10)}: \quad\) | \(\ds x_2\) | \(=\) | \(\ds \frac 1 {\sqrt 2 - 1} \times \paren {\frac {\sqrt 2 + 1} {\sqrt 2 + 1} }\) | multiply by $1$ | ||||||||||
| \(\text {(11)}: \quad\) | \(\ds \) | \(=\) | \(\ds \sqrt 2 + 1\) | lines $7$ through $11$ repeat ad infinitum: $\sqrt 2 = \sqbrk {1, \sequence 2}$ |
$\blacksquare$