Square roots and logarithms without a calculator (Part 12)

I recently came across the following computational trick: to estimate \sqrt{b}, use

\sqrt{b} \approx \displaystyle \frac{b+a}{2\sqrt{a}},

where a is the closest perfect square to b. For example,

\sqrt{26} \approx \displaystyle \frac{26+25}{2\sqrt{25}} = 5.1.

I had not seen this trick before — at least stated in these terms — and I’m definitely not a fan of computational tricks without an explanation. In this case, the approximation is a straightforward consequence of a technique we teach in calculus. If f(x) = (1+x)^n, then f'(x) = n (1+x)^{n-1}, so that f'(0) = n. Since f(0) = 1, the equation of the tangent line to f(x) at x = 0 is

L(x) = f(0) + f'(0) \cdot (x-0) = 1 + nx.

The key observation is that, for x \approx 0, the graph of L(x) will be very close indeed to the graph of f(x). In Calculus I, this is sometimes called the linearization of f at x =a. In Calculus II, we observe that these are the first two terms in the Taylor series expansion of f about x = a.

For the problem at hand, if n = 1/2, then

\sqrt{1+x} \approx 1 + \displaystyle \frac{x}{2}

if x is close to zero. Therefore, if a is a perfect square close to b so that the relative difference (b-a)/a is small, then

\sqrt{b} = \sqrt{a + b - a}

= \sqrt{a} \sqrt{1 + \displaystyle \frac{b-a}{a}}

\approx \sqrt{a} \displaystyle \left(1 + \frac{b-a}{2a} \right)

= \sqrt{a} \displaystyle \left( \frac{2a + b-a}{2a} \right)

= \sqrt{a} \displaystyle \left( \frac{b+a}{2a} \right)

= \displaystyle \frac{b+a}{2\sqrt{a}}.

One more thought: All of the above might be a bit much to swallow for a talented but young student who has not yet learned calculus. So here’s another heuristic explanation that does not require calculus: if a \approx b, then the geometric mean \sqrt{ab} will be approximately equal to the arithmetic mean (a+b)/2. That is,

\sqrt{ab} \approx \displaystyle \frac{a+b}{2},

so that

\sqrt{b} \approx \displaystyle \frac{a+b}{2\sqrt{a}}.

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