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Ernesto's question at Yahoo! Answers regarding finding the simple continued fraction for sqrt(3)
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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:
\(\displaystyle \sqrt{3}=\left[1;\overline{1,2} \right]\)
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:
\(\displaystyle \sqrt{3}=\left[1;\overline{1,2} \right]\)
Hello, Erne4sto!
Here is a primitive solution.
It requires a calculator and some stamina.
[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]
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]