Here’s a very simple function:
You give it a positive real number, it gives you the square root. It maps 81 to 9, 100 to 10, 2 to 1.414…, and so on and so forth. It’s pretty much the only one-argument function that’s built into most pocket calculators, which says something about its utility and ubiquity. But let’s say we did want to come up with an easy way to get approximate square roots. Maybe our calculator is broken, or (much more likely) maybe we’re looking for an analytic approximation that’s easier to deal with than the square root in and of itself. To generate our approximation, let’s take a look at the function:
It starts off rather dramatically curved, but as x gets larger and larger, the increasing value of the square root smooths out and begins to look more and more like a straight line of constant slope. We can imagine that if we drew a straight line tangent to this curve at a point of exactly known square root, we could use that line to estimate the value of the square root in the immediate neighborhood of this point. For instance, we know that the square root of 64 is 8. Therefore if we draw a line just touching that point on the graph, we might just have developed a pretty good approximation for finding the square root of numbers near 64. Actually carrying this out, we have a graph that looks like this:
To actually do anything useful with this, we need to find the equation of that line. Using the point-slope form of the equation of a line, we know that our line will have the following form. I’ll explain the constants after I print it:
The line itself is a function, and since f is already taken we’ll call it g. The constant m is just the slope of the line, and the constants x0 and y0 are the coordinates of the point on the original curve to which the line is tangent. (x0 = 64 and y0 = 8, because 8 is the square root of 64)
The slope m is the part that we don’t yet know a priori. To find it we need a little bit of calculus. Calculus tells us that the slope of the tangent line is equal to the derivative of the original function, evaluated at that point. So calculating the derivative, we find that:
Putting x = 64 in that expression, we find that our particular m is 1/16. Therefore for numbers near 64, their square root is roughly g(x):
Let’s check it. Plugging in (say) x = 65, we get that the square root of 65 is about 8.0625. The actual answer is 8.062257748… Which is pretty close.
The procedure works anywhere, you are by no means limited to the vicinity of x = 64. You can redo the calculation and put the tangent line anywhere convenient. For instance, if you want to calculate square roots in the vicinity of x = 100, just recognise that the new x0 = 100, the new y0 = 10, and the new m = 1/(2*10) = 1/20. In fact it will turn out that the higher x0, the broader the range of accuracy of this approximation about that point.
I admit this application for a linear approximation is a little contrived. But it does work, and if it didn’t physics would have been very hard-pressed indeed to have developed past the pre-Newton dark ages.