Local Linear Approximation

Local Linear Approximation

  May 08, 2021    Roy John

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You might have seen the famous approximation $\sin(x) \approx x$ for $x$ near 0. We can use derivatives to approximate non-linear functions by simpler linear functions. We can better understand this by looking at the graph a function $f$ under magnification.

The blue curve is the graph of $f(x) = \sin(x)$ and the yellow line is its tangent at $x = 0$. We can see that, under magnification of the curve at the point $x = 0$, the tangent line to the curve at $x = 0$ is a close approximation of the function for $x$ near 0. In general, for a function $f$: if $f$ is differentiable at a point $x_0$, then stronger and stronger magnifications at $(x_0,f(x_0))$ eventually make the curve containing $(x_0,f(x_0))$ look more and more like a non-vertical line segment, that line being the tangent line to the graph of $f$ at $(x_0,f(x)_0))$. Thus, a function $f$ that is differentiable at $x_0$ is said to be locally linear at the point $(x_0,f(x_0))$ . 

By contrast, the graph of a function that is not differentiable at $x_0$ due to a corner at the point $(x_0,f(x_0))$ cannot be magnified to resemble a straight line segment at that point.

The function $f(x) = \lvert x \rvert - 1$ is not differentiable at $x = 0$. We can see the corner at the point $(0,-1)$. The magnification of the graph at the point clearly shows that $f$ is not locally linear at $(0.-1)$.

Now we can discuss this idea analytically. Assume that a function $f$ is differentiable at $x_0$. The equation of the tangent line to the graph of the function through $(x_0,f(x_0))$ is $y = f(x_0) + f'(x_0)(x - x_0)$, where $f'(x_0)$ is the derivative of $f(x)$ at $x_0$. Since this line closely approximates the graph of $f$ for values of $x$ near $x_0$, it follows that

                                    $$f(x) \approx f(x_0) + f'(x_0)(x - x_0)$$

provided that $x$ is close to $x_0$. We call this the local linear approximation of $f$ at $x_0$.

For example,

let $f(x) = \sin(x)$, $f'(x) = \cos(x)$, $f(x) = \sin(0) = 0$ and $f'(x) = \cos(0) = 1$. Therefore,

$$\sin(x) \approx sin(0) + \cos(0)(x - 0)$$

$$\sin(x) \approx 0 + 1(x - 0)$$

$$\sin(x) \approx x.$$