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Ernesto's question at Yahoo! Answers regarding finding the simple continued fraction for sqrt(3)

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MarkFL

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Feb 24, 2012
13,775
Here is the question:

Simple continued fraction of √3?


Find the simple ( all numerators have to be 1) continued fraction of √3 (square root of three)

Explain.

Thanks.
I have posted a link there to this thread so the OP can view my work.
 
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MarkFL

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Feb 24, 2012
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Hello Ernesto,

The minimal polynomial having \(\displaystyle 1+\sqrt{3}\) as a root is:

\(\displaystyle x^2-2x-2=0\)

Using this as our characteristic equation, we obtain the recursion:

(1) \(\displaystyle A_{n+1}=2A_{n}+2A_{n-1}\)

Where we will have:

\(\displaystyle \lim_{n\to\infty}\left(\frac{A_{n+1}}{A_{n}} \right)=\sqrt{3}+1\)

Therefore, we may approximate $\sqrt{3}$ with:

\(\displaystyle \sqrt{3}\approx\frac{A_{n+1}}{A_{n}}-1=\frac{A_{n+1}-A_{n}}{A_{n}}\)

Using our recursion in (1), we may write:

\(\displaystyle \sqrt{3}\approx\frac{A_{n}+2A_{n-1}}{A_{n}}=1+\frac{1}{\dfrac{A_{n}}{2A_{n-1}}}\)

Now, since the recursion in (1) may be written in $A_{n}$ as:

\(\displaystyle A_{n}=2A_{n-1}+2A_{n-2}\)

We may write:

\(\displaystyle \sqrt{3}\approx1+\frac{1}{\dfrac{2A_{n-1}+2A_{n-2}}{2A_{n-1}}}=1+\frac{1}{1+\dfrac{1}{\dfrac{A_{n-1}}{A_{n-2}}}}\)

Using the recursion for \(\displaystyle A_{n-1}\) we may state:

\(\displaystyle \sqrt{3}\approx1+\frac{1}{1+\dfrac{1}{\dfrac{2A_{n-2}+2A_{n-3}}{A_{n-2}}}}=1+\frac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{A_{n-2}}{2A_{n-3}}}}}\)

Hence, we will find that by allowing $n\to\infty$ and repeating the above process ad infinitum, we will obtain:

\(\displaystyle \sqrt{3}=1+\dfrac{1}{1+\dfrac{1}{2+ \dfrac{1}{1+\dfrac{1}{2+\cdots}}}}\)

Written in linear form, we have:

\(\displaystyle \sqrt{3}=\left[1;\overline{1,2} \right]\)
 

soroban

Well-known member
Feb 2, 2012
409
Hello, Erne4sto!

Here is a primitive solution.
It requires a calculator and some stamina.


Find the simple continued fraction for [tex]\sqrt{3}.[/tex]

[tex]\sqrt{3} \;=\; 1 + 0.732050808[/tex]

. . . .[tex]=\;1 + \frac{1}{1 + 0.366025404}[/tex]

. . . .[tex]=\;1 + \frac{1}{1 + \dfrac{1}{2 +0.732050808}} [/tex]

. . . .[tex]=\;1 + \frac{1}{1 + \dfrac{1}{2 + \dfrac{1}{1+0.366025404}}} [/tex]

. . . .[tex]=\;1 + \frac{1}{1 + \dfrac{1}{2 + \dfrac{1}{1+\dfrac{1}{2 + 0.732050808}}}} [/tex]


We note the repeating pattern: .[tex]1,1,2,1,2,1,2,\;.\;.\;.[/tex]

Therefore: .[tex]\sqrt{3} \;=\;[1,\overline{1,2}] [/tex]