Examples 3.1.10(c): Computing Square Roots
xn+1 = 1/2 (xn + a / xn)Show that this sequence converges to the square root of a regardless of the starting point x0 > 0.
|Term||Exact Value||Approximate Value||x0||2||x1 = 1/2 (x0 + a / x0)||1/2 (2 + 2/2) = 3/2||1.5||x2 = 1/2 (x1 + a / x1)||1/2 (3/2 + 2/3/2) = 17/12||1.416666667||x3 = 1/2 (x2 + a / x2)||1/2 (17/12 + 2/17/12) = 577/408||1.414215686|
|x4 = 1/2 (x3 + a / x3)||1/2 (577/408 + 2/577/408) = 665857/470832||1.414213562375|
After only 4 steps our sequence has approximated the 'true' value of with 11 digits accuracy.
Now that we have seen the usefulness of this particular recursive sequence we need to prove that it really does converge to the square root of a.
First, note that xn > 0 for all n so that the sequence is bounded below.
Next, let's see if the sequence is monotone decreasing, in which case it would have to converge to some limit. Compute
xn - xn+1 = xn - 1/2 (xn + a / xn) = 1/2 (xn2 - a) / xnNow let's take a look at xn2 - a:
xn2 - a = 1/4 (xn-1 + a / xn-1)2 - aBut that means that xn - xn+1 0, or equivalently xn xn+1. Hence, the sequence is monotone decreasing and bounded below by 0 so it must converge.
= 1/4 xn-12 + 1/2 a + 1/4 a2 / xn-12 - a
= 1/4 xn-12 - 1/2 a + 1/4 a2 / xn-12
= 1/4 (xn-1 - a / xn-1)2
We now know that xn = L. To find that limit, we could try the following:
(*) L = xn = xn+1 = 1/2 (xn + a/xn) = 1/2 (L + a / L)Solving the equation L = 1/2 (L + a / L) gives
2 L2 = L2 + aor equivalently
L2 = awhich means that the limit L is indeed the square root of a, as required.
However, our proof contains one small caveat. In order to take the limit inside the fraction in equation (*) we need to know that L is not zero before we can write down equation (*). We already know that xn is bounded below by zero, but that is not good enough to exclude the possibility of L = 0. But we have already shown that
xn2 - a = 1/4 (xn-1 - a / xn-1)2 0so that xn2 a. That implies that the limit of the sequence (which we already know exists) is strictly positive since a > 0. Therefore equation (*) is justified and we have completed the proof.