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Supposing a square root $\sqrt{X}$, let $x$ be the approximation of $\sqrt{X}$, then we get these 2 formulas to estimate $\sqrt{X}$:

$x_{n+1}=\frac{x_n+\frac{X}{x_n}}{2}$ and

$x_{n+2}=\frac{x_{n}x_{n+1}+X}{x_{n}+x_{n+1}}$.

Question 1. How do we prove these formula works and why does they work?

For example, when we try to use these two formulas to show his approximation of $\sqrt{3}$, supposing $x_1=\frac{5}{3}$ (which is a lower bound),

from the first formula we get

$x_{2}=\frac{26}{15}$ and $x_{4}=\frac{1351}{780}$ (Archimedes' upper bound);

from the second formula we get

$x_{3}=\frac{265}{153}$ (Archimedes' lower bound) and $x_{5}=\frac{13775}{7953}$.

Question 2. Why does the first formula give upper bounds and the second one give lower bounds (in this circumstance)?

Thanks.

Aiden
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  • The sequence ${x_n}$ given by $x_{n+1}=\frac12(x_n+\frac a{x_n})$ converges to $\sqrt a$ for any $x_1>0$ (see here for the proof). ${x_n}$ is a non-increasing sequence, so it gives the upper bound. – Cm7F7Bb Aug 12 '14 at 08:16
  • the second sequence does not give only lower bounds. Starting with $x_1=x_2=\frac{5}{3}$ one gets: "lower bound","lower bound","upper bound","lower bound","lower bound","upper bound",... – miracle173 Aug 12 '14 at 09:31

1 Answers1

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Assume X is a rational number for which you would like to approximate its square root with a rational number. Say you have an estimate $x_1$ and would like a better estimate. You know that if $x_1$ is an estimate for $\sqrt{X}$ then $\frac{X}{x_1}$ is an estimate as well, and their geometric mean is your answer. You cannot calculate the geometric mean of $x_1$ and $\frac{X}{x_1}$ so you calculate the arithmetic mean instead, obtaining a new estimate $x_2 = \frac{x_1 + \frac{X}{x_1}}{2}$. Since the arithmetic mean is greater than the geometric mean, this means that $x_2 > \sqrt{X}$.
This method is known as the Babylonian method, and produces only upper estimates.
Now, take the parabola defined as $y = x^2$. Finding $\sqrt{X}$ is equivalent to finding the point at which the parabola and the horizontal line $y = X$ intersect. We cannot calculate when they intersect, but we can use an approximation. Assume we have estimates $x_1$ and $x_2$. Draw a straight line that intersects the parabola at points $x = x_1$ and $x = x_2$. This straight line intersects the horizontal line $y = X$ at a point $x = x_3$. It is not difficult to show that $x_3 = \frac{X + x_1 x_2}{x_1 + x_2}$. This new estimation can be a lower or upper estimate depending on $x_1$ and $x_2$. It is an upper estimate when both inputs are lower or upper, and a lower estimate when one input is lower and the other one is upper.
I think it is reasonable to assume that Archimedes used these two methods to approximate $\sqrt{3}$ as $\frac{265}{153} < \sqrt{3} < \frac{1351}{780}$. However, both Internet and the mathematical journals seem to overlook this simple method. We can call the second method "secant method".
By the way, if $x_1 = x_2$, then the Babylonian method is a particular case of the secant method. Also, it is possible to relate our estimates as $$2 x_1 (x_2 - \sqrt{X}) = (x_1 - \sqrt{X})^2$$ $$(x_1 + x_2) (x_3 - \sqrt{X}) = (x_1 - \sqrt{X}) (x_2 - \sqrt{X})$$ This makes clear why bounds are upper or lower. Also, if we define $a_n$ and $b_n$ as $(x_1 - \sqrt{X})^n = a_n \sqrt{X} + b_n$ we can see that $x_n = - \frac{b_n}{a_n}$. As a result, your equations should have been written as $$x_{2n} = \frac{x_n + \frac{X}{x_n}}{2}\\ x_{n+m} = \frac{X + x_n x_m}{x_n + x_m}$$

user3209698
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