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### How do I approximate an irrational square root without using a calculator?

Although this might look like a smart trick at the beginning, the truth is that as a physicist this kind of approximations turn out to be useful very often. This is quite likely to come up in the PAT as well!

It is important to get used to understanding symbolic expressions (no numbers!), so I will first derive the maths and then apply the resulting expression to a couple of illustrative examples.

Our aim is to obtain an approximate value for the inexact square root x of an arbitrary number n, i.e. to solve x = n1/2. In other words, we want to solve the equation f(x) = x2 = n. Suppose we know the exact root x0 of another number n0 which is very close to n, i.e. f(x0) = x02 = n0 (this is key!). We can do a Taylor series expansion of f(x) about x0 up to first order (first derivative f'(x)), which is simply:

f(x) = f(x0) + f'(x0)(x - x0).

The first derivative f'(x) = 2x, which evaluated at x0 is f'(x0) = 2x0. Substituting f(x) = n and f(x0) = n0 leads to:

n = n0 + 2x0(x - x0).

We are almost done! We know all the values of n, n0 and x0 and just need to solve for x. Rearranging:

x = x0 + (n - n0)/(2x0)

That's it! Remember that this expression is just an approximation for the actual square root of n, but a quite good one the closer n0 is to n.

Let's plug in some numbers. Suppose you were asked to approximate the square root of n = 66. Despite the solution being irrational, there is a close number n0 = 64 with a nice exact square root x0 = 8. Using the expression above we find:

x = 8 + (66 - 64)/(2*8) = 8 + 2/16 = 8 + 1/8 = 8 + 0.125 = 8.125

How good is this? It's just 0.01% off the actual value 8.12404...!

What if we chose n0 = 81 instead, the square root of which is x0 = 9? This would give x = 8.167 - worse, but still a reasonable result considering how far 81 is from 66. Never forget to quote both the positive and negative square roots in your final solution!

Would you be able to derive a similar expression for approximating a cubic root?

2 years ago

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