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I recently found out that starting with $\frac{1}{1}$ and letting $p$ be the numerator and $q$ the denominator, then you can generate a sequence that approximates $\sqrt 2$ using:

$$\frac{p+2q}{p+q}$$

The first few terms of the sequence are:

$\frac{1}{1}$, $\frac{3}{2}$, $\frac{7}{5}$, $\frac{17}{12}$, $\frac{41}{29}$, $\frac{99}{70}$, $...$

But why is this case?

scoopfaze
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Jamminermit
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    Hint: $p^2-2q^2=\pm1$ – J. W. Tanner Jul 14 '19 at 18:45
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    Also related: https://math.stackexchange.com/q/730349/42969, https://math.stackexchange.com/q/818845/42969 – Martin R Jul 14 '19 at 18:50
  • If you are familiar with eigenvectors and eigenvalues, you can also start with $\begin{pmatrix}p_{n+1} \ q_{n+1}\end{pmatrix} = \begin{pmatrix} 1 & 2 \ 1 & 1 \end{pmatrix} \begin{pmatrix}p_{n} \ q_{n}\end{pmatrix}$ from which we derive $\begin{pmatrix}p_{n} \ q_{n}\end{pmatrix} = \begin{pmatrix} 1 & 2 \ 1 & 1 \end{pmatrix}^n \begin{pmatrix} 1 \ 0 \end{pmatrix}$. – Reinhard Meier Jul 14 '19 at 19:29
  • Also related: the secant method and group laws on conics - see here on Rudin's example. – Bill Dubuque Jul 14 '19 at 20:00

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