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Square Root

\begin{figure}\begin{center}\BoxedEPSF{SquareRoot.epsf scaled 700}\end{center}\end{figure}

\begin{figure}\begin{center}\BoxedEPSF{SqrtReIm.epsf scaled 720}\end{center}\end{figure}

A square root of $x$ is a number $r$ such that $r^2=x$. This is written $r=x^{1/2}$ ($x$ to the 1/2 Power) or $r=\sqrt{x}$. The square root function $f(x)=\sqrt{x}$ is the Inverse Function of $f(x)=x^2$. Square roots are also called Radicals or Surds. A general Complex Number $z$ has two square roots. For example, for the real Positive number $x=9$, the two square roots are $\sqrt{9}=\pm 3$, since $3^2=(-3)^2=9$. Similarly, for the real Negative number $x=-9$, the two square roots are $\sqrt{-9}=\pm 3i$, where i is the Imaginary Number defined by $i^2=-1$. In common usage, unless otherwise specified, ``the'' square root is generally taken to mean the Positive square root.


The square root of 2 is the Irrational Number $\sqrt{2}\approx 1.41421356$ (Sloane's A002193), which has the simple periodic Continued Fraction 1, 2, 2, 2, 2, 2, .... The square root of 3 is the Irrational Number $\sqrt{3}\approx
1.73205081$ (Sloane's A002194), which has the simple periodic Continued Fraction 1, 1, 2, 1, 2, 1, 2, .... In general, the Continued Fractions of the square roots of all Positive integers are periodic.


The square roots of a Complex Number are given by


\begin{displaymath}
\sqrt{x+iy}=\pm\sqrt{x^2+y^2}\left\{{\cos\left[{{1\over 2}\t...
...[{{1\over 2}\tan^{-1}\left({y\over x}\right)}\right]}\right\}.
\end{displaymath} (1)

As can be seen in the above figure, the Imaginary Part of the complex square root function has a Branch Cut along the Negative real axis.


A Nested Radical of the form $\sqrt{a\pm b\sqrt{c}}$ can sometimes be simplified into a simple square root by equating

\begin{displaymath}
\sqrt{a\pm b\sqrt{c}} = \sqrt{d} \pm \sqrt{e}\,.
\end{displaymath} (2)

Squaring gives
\begin{displaymath}
a\pm b\sqrt{c}=d+e\pm 2\sqrt{de}\,,
\end{displaymath} (3)

so
$\displaystyle a$ $\textstyle =$ $\displaystyle d+e$ (4)
$\displaystyle b^2c$ $\textstyle =$ $\displaystyle 4de.$ (5)

Solving for $d$ and $e$ gives
\begin{displaymath}
d,e = {a \pm \sqrt{a^2-b^2 c}\over 2}.
\end{displaymath} (6)


A sequence of approximations $a/b$ to $\sqrt{n}$ can be derived by factoring

\begin{displaymath}
a^2-nb^2=\pm 1
\end{displaymath} (7)

(where $-1$ is possible only if $-1$ is a Quadratic Residue of $n$). Then
\begin{displaymath}
(a+b\sqrt{n}\,)(a-b\sqrt{n}\,)=\pm 1
\end{displaymath} (8)


\begin{displaymath}
(a+b\sqrt{n}\,)^k(a-b\sqrt{n}\,)^k=(\pm 1)^k=\pm 1,
\end{displaymath} (9)

and
$\displaystyle (1+\sqrt{n}\,)^1$ $\textstyle =$ $\displaystyle 1+\sqrt{n}$ (10)
$\displaystyle (1+\sqrt{n}\,)^2$ $\textstyle =$ $\displaystyle (1+n)+2\sqrt{n}$ (11)
$\displaystyle (1+\sqrt{n}\,)(a+b\sqrt{n}\,)$ $\textstyle =$ $\displaystyle (a+bn)+\sqrt{n}\,(a+b).$ (12)

Therefore, $a$ and $b$ are given by the Recurrence Relations
$\displaystyle a_i$ $\textstyle =$ $\displaystyle a_{i-1}+b_{i-1}n$ (13)
$\displaystyle b_i$ $\textstyle =$ $\displaystyle a_{i-1}+b_{i-1}$ (14)

with $a_1=b_1=1$. The error obtained using this method is
\begin{displaymath}
\left\vert{{a\over b}-\sqrt{n}\,}\right\vert = {1\over b(a+b\sqrt{n}\,)} < {1\over 2b^2}.
\end{displaymath} (15)

The first few approximants to $\sqrt{n}$ are therefore given by
\begin{displaymath}
1, {\textstyle{1\over 2}}(1+n), {1+3n\over 3+n}, {1+6n+n^2\over 4(n+1)}, {1+10n+5n^2\over 5+10n+n^2}, \ldots.
\end{displaymath} (16)

This Algorithm is sometimes known as the Bhaskara-Brouckner Algorithm. For the case $n=2$, this gives the convergents to $\sqrt{2}$ as 1, 3/2, 7/5, 17/12, 41/29, 99/70, ....


Another general technique for deriving this sequence, known as Newton's Iteration, is obtained by letting $x=\sqrt{n}$. Then $x=n/x$, so the Sequence

\begin{displaymath}
x_k={1\over 2}\left({x_{k-1}+{n\over x_{k-1}}}\right)
\end{displaymath} (17)

converges quadratically to the root. The first few approximants to $\sqrt{n}$ are therefore given by


\begin{displaymath}
1, {\textstyle{1\over 2}}(1+n), {1+6n+n^2\over 4(n+1)}, {1+28n+70n^2+28n^3+n^4\over 8(1+n)(1+6n+n^2)}, \ldots.
\end{displaymath} (18)

For $\sqrt{2}$, this gives the convergents 1, 3/2, 17/12, 577/408, 665857/470832, ....

See also Continued Square Root, Cube Root, Nested Radical, Newton's Iteration, Quadratic Surd, Root of Unity, Square Number, Square Triangular Number, Surd


References

Sloane, N. J. A. Sequences A002193/M3195 and A002194/M4326 in ``An On-Line Version of the Encyclopedia of Integer Sequences.'' http://www.research.att.com/~njas/sequences/eisonline.html and Sloane, N. J. A. and Plouffe, S. The Encyclopedia of Integer Sequences. San Diego: Academic Press, 1995.

Spanier, J. and Oldham, K. B. ``The Square-Root Function $\sqrt{bx+c}$ and Its Reciprocal,'' ``The $b\sqrt{a^2-x^2}$ Function and Its Reciprocal,'' and ``The $b\sqrt{x^2+a}$ Function.'' Chs. 12, 14, and 15 in An Atlas of Functions. Washington, DC: Hemisphere, pp. 91-99, 107-115, and 115-122, 1987.

Williams, H. C. ``A Numerical Investigation into the Length of the Period of the Continued Fraction Expansion of $\sqrt{D}$.'' Math. Comp. 36, 593-601, 1981.



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© 1996-9 Eric W. Weisstein
1999-05-26