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Approximating square roots can be both fun and tedious. Luckily, there are many ways to do it.


The square Root of 2


... If \(x^2\) is \(2\) and \(x\) is to be found, it will be said to be \(\sqrt{2}\). But if you are looking for the approximate root of \(2\), extract the nearest root, which is \(1\), and divide the remainder by twice the root just discovered. Thus the root of the number \(2\) can be said to be \(1\frac{1}{2}\), which is less than the true root. Or add \(1\) to the denominator and the root will said to be \(1\frac{1}{3}\), which is less than the true root. Midway between the two is \(1\frac{5}{8}\), which is very close to the true root. Alternatively, pairs or zeros may be added to the given square up to infinity, and the root may be extracted from [a number] extended in this way as from an exact square number. Thus if you are seeking the root of \(2\), a square, extract, if you you wish, the root of \(\footnotesize 2\, 00\, 00\, 00\, 00\, 00\, 00\, 00\, 00\, 00\, 00\, 00\, 00\, 00\, 00\normalsize\) [which is] \(\footnotesize 141,421,356,237,309,505\normalsize\). So the root of \(2\) is said to be approximately \(\frac{141,421,356,237,309,505}{100,000,000,000,000,000}\).

—François Viéte (1540-1603)


To put this plainly, find the first approximation as \(1\frac{1}{2}\) above, then calculate its square as \((1\frac{1}{2})^2\)\(=2\frac{1}{4}\). Find the difference "remainder", as \(2-2\frac{1}{4}\) \(=-\frac{1}{4}\). Divide this value by twice the current approximation and add it to the approximation: \(1\frac{1}{2}+(-\frac{1}{4}/2(1\frac{1}{2}))=1\frac{1}{2}-\frac{1}{12}=1\frac{5}{12}\). Repeating these steps again gives: \((1\frac{5}{12})^2=(2\frac{1}{144})\);   \(2-2\frac{1}{144}=-\frac{1}{144}\);   \(1\frac{5}{12}+(-\frac{1}{144}\div 2(1\frac{5}{12}))\) \(=1\frac{5}{12}-\frac{1}{408}\)\(=1\frac{169}{408}\) \(\approx 1.414215686\). Since the actual value is approximately \(1.414213562\), this is already accurate to 6 digits.


Viéte also describes similar methods to approximate other various roots in his fascinating book The Analytic Art. I would definitely suggest picking up a copy of this book if you are interested in mathematics. You can get a copy online from Amazon here.


F. Viéte, The Analytic Art, p.316, The Kent State University Press, 1983, Trans. T. Richard Witmer.


Viéte's Method


Viéte's method of approximating roots described above is aptly referred to as Viéte's Method of approximating roots. Verbose and somewhat confusing as it is, it can be simplified to a single, simple algorithmic formula using my own algorithmic notation:

\(x\leftarrow x+\frac{n-x^2}{2x}\bigg]_{i=0}^{\infty}\) \(x\approx\sqrt{n}\)


They say a picture is worth a thousand words, and this visual representation of Viéte's method makes it quite easy to understand, even though the words and worked examples may not appear so simple. We can now reduce this method to its steps:

 1) Let \(n\) be the number of which we want the square root.
 2) Let \(x\) be the first guess (the nearest root).
 3) Subtract \(x^2\) from \(n\), divide this by \(2x\), then add \(x\).
 4) Assign this result to \(x\).
 5) Repeat until infinity (or until some acceptable precision is reached).


Hero's (Heron's) Method


Hero (or Heron) of Alexandria (fl. AD 62) also described a method for approximating square roots using an iterative method where a first guess is divided into the number and the next guess (approximation) is the average of the result and the initial estimate. Algorithmically, this may be written as:

\(x\leftarrow \frac{\frac{n}{x}+x}{2}\bigg]_{i=0}^{\infty}\)

This can be simplified to a form resembling Viéte's method:

\(x\leftarrow \frac{n+x^2}{2x}\bigg]_{i=0}^{\infty}\)


Continued Fractions


\(\sqrt{n} = 1+\frac{n-1}{2+}\,\frac{n-1}{2+}\,\frac{n-1}{2+}\cdots\) \(=1+\Kappa(\frac{n-1}{2})\)

\(\sqrt{n} = a+\frac{n-a^2}{2a+}\,\frac{n-a^2}{2a+}\,\frac{n-a^2}{2a+}\cdots\) \(=a+\Kappa(\frac{n-a^2}{2a})\),   for any \(a\)   (Khovanskii)

\(\sqrt{1+a^2} = a+\frac{1}{2a+}\,\frac{1}{2a+}\,\frac{1}{2a+}\cdots\) \(=a+\Kappa(\frac{1}{2a})\)

\(\sqrt{a^2+b^2} = a+\frac{b^2}{2a+}\,\frac{b^2}{2a+}\,\frac{b^2}{2a+}\cdots\) \(=a+\Kappa(\frac{b^2}{2a})\)

\(\sqrt{a^2+b^2} = a+\frac{b^2}{a+b+}\,\frac{a^2}{a+b+}\,\frac{b^2}{a+b+},\frac{a^2}{a+b+}\cdots\) \(=a+\Kappa(\frac{b^2}{a+b+}\,\frac{a^2}{a+b+})\)




Algebra: Factorials