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The Root of the Matter: Approximating Roots with the Greeks - Method, Motivation, and Verification

Author(s): 
Matthew Haines and Jody Sorensen (Augsburg University)

Method and Motivation

In his work On Mathematics Useful for the Understanding of Plato [4], Theon proposed a method which can be used to provide ever closer rational approximations of \(\sqrt{2}.\) This method is sometimes referred to as Theon's ladder. The method starts with \(x_0=1,\) \(y_0=1.\) Our estimate of \(\sqrt{2},\) which will be the ratio \(\displaystyle{\frac {y} {x}},\) thus starts as 1. The numbers \(x\) and \(y\) are sometimes referred to as side and diagonal numbers – more on that to come. [5] To get a better estimate of \(\sqrt{2},\) we let \(x_1=x_0+y_0\) and \(y_1 = 2x_0+y_0.\) Thus we get that \(x_1=2\) and \(y_1=3,\) so our new estimate is \(\displaystyle{\frac {3} {2}}  = 1.5.\) The recursive formula is \[ x_{n+1} = x_n+y_n \ \ \ \ \ \ \ \ y_{n+1} = 2x_n+y_n .\] Continuing in this way for a few more steps gives us the results in Table 1.

\[\begin{array}{r|r|r|r} n & x_n & y_n & y_n/x_n \\\hline 0 & 1 & 1 & 1 \\\hline 1 & 2 & 3 & 1.5 \\\hline 2 & 5 & 7 & 1.4 \\\hline 3 & 12 & 17 & 1.4167 \\\hline 4 & 29 & 41 & 1.4138\\\hline 5 & 70 & 99 & 1.4143\\\hline 6 & 169 & 239 & 1.4142\end{array}\]

Table 1. Side and diagonal numbers

So it appears that Theon's method approximates \(\sqrt{2} = 1.4142....\) Note that the denominators (or sides), \(x_n,\) are the Pell numbers, and their properties are well-studied by number theorists.

We cannot be sure how (or why) Theon came up with this procedure. One possible origin is numerical. [1] If \(\sqrt{2}\) were rational, then \(\displaystyle\sqrt{2} = {\frac{y}{x}}\) for some positive integers \(x\) and \(y.\) Squaring and simplifying gives us \(2x^2 =y^2.\) So, does that equation have integer solutions? Let's look at the list of possibilities for \(y^2\) and \(2x^2\) in Table 2.

\[\begin{array}{r|cccccccc} n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\\hline n^2 & 1 & 4 & \fbox{9} & 16 & 25 & 36 & \fbox{49} & 64 \\\hline 2n^2 & 2 & \fbox{8} & 18 & 32 & \fbox{50} & 72 & 98 & 128\end{array}\]

Table 2. Values of \(y^2\) and \(2x^2\)

From this tiny table we don't see any places where a square is equal to two times a square, but we do see some places where they are close. For example if \(x=5\) and \(y=7,\) then \(2x^2 = 50 = y^2+1.\) This indicates that \(\displaystyle{\frac {7}{5}}\) is a decent approximation of \(\sqrt{2}.\) If we continue this table and list the possibilities as fractions, we get this list: \[{\frac{1}{1}}, \,{\frac{3}{2}},\, {\frac{7}{5}},\, {\frac{17}{12}},\, \dots.\] These are all values of \(x\) and \(y\) where \(2x^2=y^2 \pm 1.\) If you look for a pattern here, you see that the new denominator is the sum of the previous numerator and denominator, or \(x_{n+1} = x_n+y_n.\) The new numerator is this new denominator plus the old denominator, or \(y_{n+1} = x_{n+1}+x_n = x_n+y_n+x_n = 2x_n+y_n.\) Testing this out gives the next term as \(\displaystyle{\frac{41}{29}}\) which satisfies \(2\cdot29^2 = 1682=41^2+1.\) So Theon could have found this pattern by inspection.

There are also possible geometric motivations for this process, which explain why \(x\) and \(y\) are referred to as "side" and "diagonal" numbers. Suppose we start with an isosceles right triangle \(ABC\) whose leg (side) is of length \(x\) and hypotenuse (diagonal) of length \(y,\) as in Figure 1. Note that this figure has the added point \(H\) to create the square \(ABHC\) to emphasize the names "side and diagonal numbers." Extend the two sides by \(y\) to create an isosceles right triangle with sides of length \(x+y.\)

Figure 1. An isosceles right triangle with sides of length \(x+y\) and diagonal \(\overline{DG}\) of length \(2x+y\)


We can see that the larger triangle has a hypotenuse of length \(2x+y\) by noticing that \[\triangle BCA \cong \triangle DBE \cong \triangle GCF\] and that \(BCFE\) is a rectangle. Thus \(DE=FG=x\) and \(EF=y.\) 

This triangle motivates the idea that if the side is extended from \(x\) to \(x+y,\) then the diagonal should be extended from \(y\) to \(2x+y,\) which fits Theon's method. We should note that no integers \(x\) and \(y\) exist that are sides and diagonals of the isosceles right triangle. This approach is simply a motivation and not a proof.

Verification

Let's prove that Theon's method gives a method for finding better and better rational approximations of \(\sqrt{2}.\)  Suppose we start with \(x\) and \(y\) satisfying \(2x^2-y^2 = \pm 1\) (note that \(x_0=y_0=1\) satisfies this) and let \(x^*=x+y, y^*=2x+y.\) Then

\begin{eqnarray*} 2(x^*)^2-(y^*)^2 &=& 2(x+y)^2-(2x+y)^2 \\ &=& -2x^2+y^2 = \mp 1.\end{eqnarray*}

This shows that as we iterate, \(2x^2-y^2\) remains \(\pm 1.\) Now if \(2x_n^2-y_n^2 = \pm 1,\) then  \(\displaystyle 2 = {\frac {y_n^2} {x_n^2}} \pm {\frac {1} {x_n^2}}.\) If we start with positive values for \(x_0\) and \(y_0,\) then as we iterate, \(x_n\) gets bigger by at least one at each step, so \(x_n\) goes to \(\infty.\) Therefore we can say that \(\displaystyle {\frac {1} {x_n^2}}\) will go to \(0^{+}.\) This means that \(\displaystyle\left({\frac{y_n} {x_n}} \right)^2\) will tend to 2. Assuming that \(\displaystyle \lim_{n\rightarrow \infty} {\frac {y_n} {x_n}}\) exists, then \(\displaystyle {\frac {y_n} {x_n}}\) goes to \(\sqrt 2,\) as desired.

So, in about 100 CE, Theon of Smyrna developed (for an unknown reason) an iterative method that can be used to approximate \(\sqrt 2\) by rational numbers. With our modern mathematical tools, we will look at Theon's method through the lenses of Geometry and Linear Algebra.

Matthew Haines and Jody Sorensen (Augsburg University), "The Root of the Matter: Approximating Roots with the Greeks - Method, Motivation, and Verification," Convergence (June 2018)

Dummy View - NOT TO BE DELETED