# 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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