Linear Approximation

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Linear Approximation

Derivatives allow us to approximate changes to a function over a small interval.

Some notation:

Linear Approximation of \(\Delta f\): If \(f\) is differentiable at \(a\), and \(\Delta x\) is small, then: \[\Delta f \approx f'(a) \Delta x\] In other words, we can approximate the actual change \(\Delta f\) with the quantity \(f'(a)\Delta x\).

In graphical terms, the linear approximation says that the tangent line is a good approximation to the function as long as we keep close to the point of tangency.

Example: Suppose \(f(x) = \frac{1}{x}\), \(a=5\) and \(\Delta x = 0.1\). Then: \[\Delta f = f(5.1) - f(5) = \frac{1}{5.1} - \frac{1}{5} = -0.003921569\] Instead we could estimate this quantity via: \[f'(5)\cdot 0.1 = \frac{-1}{5^2} \cdot 0.1 = -0.004\] The error we have made is only \(-0.004-(-0.003921569) = -0.000078431\), which is about a \(2\%\) error relative to the actual value.

Differential Notation

In this notation we write \(dx=\Delta x\) and \(dy = f'(a)dx\). Then: \[\Delta y \approx dy\]

We can write the same approximation in terms of \(f(x)\). For that we write \(x = a + \Delta x\), so \(\Delta f = f(x) - f(a)\), and \(dy = f'(a)(x-a)\). We then have the following approximation:

Approximation of \(f(x)\) by its linearization

If f(x) is differentiable at \(x=a\), and \(x\) is a number is close to \(a\), then: \[f(x) \approx L(x) = f(a) + f'(a)(x-a)\]

Example: If \(f(x) = \sqrt{x}\) and \(a=4\). Then \(f'(x) = \frac{1}{2\sqrt{x}}\), and we can compute the linearization: \[L(x) = f(4) + f'(4)(x-4) = 2 + \frac{1}{4}(x-4)\] As an example, let us estimate the value of \(\sqrt{4.1}\), which is in reality very close to \(2.024846\). Our linearization would instead give us: \[L(x) = 2 + \frac{1}{4}(4.1-4) = 2.025\]

Linearizations have numerous applications in other sciences.