Linear Approximation Formulae


Sometimes we can approximate complicated functions with simpler ones that give the accuracy we want for specific applications and are easier to work with. The approximating functions are called linearization. They are based on tangents. We introduce new variables dx and dy and define them in a way that gives new meaning to the Leibniz notation dy/dx. We will use dy to estimate error in measurement and sensitivity to change.

Linear approximation: If we see the graph of y = x^2 and y = 2x – 1, the tangent to a curve y = f(x) lies close to the curve near the point of tangency. For a brief interval to either side, the y-values along the tangent line give a good approximation to the y-values on the curve. The more we magnify the graph of a function near a point where the function is differentiable, the flatter the graph becomes and the more it resembles its tangent. In the graph of the function y = f(x), the tangent line passes through the point (a, f(a)), so its point slope equation is y = f(a) + f ‘(a) (x – a). Thus, the tangent is the graph of the function L(x) = f(a) + f ‘ (a) (x – a). For as long as the line remains close to the graph of f, L(x) gives a good approximation to f(x). According to the definition, Linear approximation formula can be given as below: If f is differentiable at x = a, then the approximating function L (x) = f(a) + f ‘ (a) (x – a)is the linearization of f at a. The approximation f (x) ˜ L (x) of f by L is the standard linear approximate of ‘f’ at a” and is also known as Local Linear Approximation.The point x = a is the centre of the approximation. Let us take some Linear approximation examples

Example: Find the linearization of f(x) = sqrt(1 + x) at x = 0.

Solution: With f ‘ (x) = ½ (1 + x)^-1/2,

We have f(0) = 1, f ‘ (0) = ½, and

L (x) = f (a) + f ‘ (a) (x – a) = 1 + ½ (x – 0) = 1 + x/2.

The approximation sqrt(1 + x) ˜ 1 + x/2 gives

Sqrt(1.2) ˜ 1 + 0.2/2 = 1.10

Sqrt(1.05) ˜ 1 + 0.05/2 = 1.025

Sqrt(1.005) ˜ 1 + 0.005/2 = 1.00250

A linear approximate normally loses accuracy away from its centre. The approximation sqrt(1 + x) ˜ 1 + x/2 will probably be too crude to be useful near x = 3. There, we need linearization at x = 3.