MHB Sqrt(3) Simple Continued Fraction: Find & Explain

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The simple continued fraction for √3 is expressed as [1; overline(1, 2)], indicating a repeating pattern of 1 and 2 after the initial integer part. The derivation involves using the minimal polynomial x² - 2x - 2, leading to a recursive relation that approximates √3. By iterating this recursion, the continued fraction representation is confirmed. The process showcases the relationship between the approximations and the continued fraction structure. This method effectively illustrates how to derive continued fractions for square roots.
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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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Hello Ernesto,

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

$$x^2-2x-2=0$$

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

(1) $$A_{n+1}=2A_{n}+2A_{n-1}$$

Where we will have:

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

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

$$\sqrt{3}\approx\frac{A_{n+1}}{A_{n}}-1=\frac{A_{n+1}-A_{n}}{A_{n}}$$

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

$$\sqrt{3}\approx\frac{A_{n}+2A_{n-1}}{A_{n}}=1+\frac{1}{\dfrac{A_{n}}{2A_{n-1}}}$$

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

$$A_{n}=2A_{n-1}+2A_{n-2}$$

We may write:

$$\sqrt{3}\approx1+\frac{1}{\dfrac{2A_{n-1}+2A_{n-2}}{2A_{n-1}}}=1+\frac{1}{1+\dfrac{1}{\dfrac{A_{n-1}}{A_{n-2}}}}$$

Using the recursion for $$A_{n-1}$$ we may state:

$$\sqrt{3}\approx1+\frac{1}{1+\dfrac{1}{\dfrac{2A_{n-2}+2A_{n-3}}{A_{n-2}}}}=1+\frac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{A_{n-2}}{2A_{n-3}}}}}$$

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

$$\sqrt{3}=1+\dfrac{1}{1+\dfrac{1}{2+ \dfrac{1}{1+\dfrac{1}{2+\cdots}}}}$$

Written in linear form, we have:

$$\sqrt{3}=\left[1;\overline{1,2} \right]$$
 
Hello, Erne4sto!

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

Find the simple continued fraction for \sqrt{3}.
\sqrt{3} \;=\; 1 + 0.732050808

. . . .=\;1 + \frac{1}{1 + 0.366025404}

. . . .=\;1 + \frac{1}{1 + \dfrac{1}{2 +0.732050808}}

. . . .=\;1 + \frac{1}{1 + \dfrac{1}{2 + \dfrac{1}{1+0.366025404}}}

. . . .=\;1 + \frac{1}{1 + \dfrac{1}{2 + \dfrac{1}{1+\dfrac{1}{2 + 0.732050808}}}} We note the repeating pattern: .1,1,2,1,2,1,2,\;.\;.\;.

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