[Calculus] V. Linear Approximations and Differentials

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Linear Approximations

Definition 5.1.1.  The approximation

\( \displaystyle f(x) \approx f(a) + f'(a)(x - a) \qquad \cdots \qquad \text{(i)} \)

is called the linear approximation or tangent line approximation of \( f \) at \( a \). The linear function whose graph is this tangent line, that is,

\( \displaystyle L(x) = f(a) + f'(a)(x - a) \qquad \cdots \qquad \text{(ii)} \)

is called the linearization of \( f \) at \( a \).

Example 5.1.2.  Find the linearization of the function \( f(x) = \sqrt{x + 3} \) at \( a = 1 \) and use it to approximate the number \( \sqrt{3.98} \).

Solution.  The derivative of \( f(x) = (x + 3)^{1/2} \) is

\( \displaystyle f'(x) = \frac{1}{2}(x + 3)^{-1/2} = \frac{1}{2\sqrt{x + 3}} \)

and so we have \( f(1) = 2 \) and \( \displaystyle f'(1) = \frac{1}{4} \). Putting these values into Equation \( \text{(ii)} \), we see that the linearization is

\( \displaystyle L(x) = f(1) + f'(1)(x - 1) = 2 + \frac{1}{4}(x - 1) = \frac{7}{4} + \frac{x}{4} \)

The corresponding linear approximation \( \text{(i)} \) is

\( \displaystyle \sqrt{x + 3} \approx \frac{7}{4} + \frac{x}{4} \quad \text{when} \ x \ \text{is near} \ 1 \)

In particular, we have

\( \displaystyle \sqrt{3.98} \approx \frac{7}{4} + \frac{0.98}{4} = 1.995 \)

\( \Box \)

Theorem 5.1.3. (Newton-Raphson Method)  Suppose \( y = f(x) \) is differentiable and \( f(r) = 0 \). If we define \( x_{n} \) (\( n \in \mathbb{N} \)) as

\( \displaystyle x_{n+1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})} \)

Then, under certain circumstances, \( x_{n} \to r \) as \( n \to \infty \).

Remark 5.1.4.  Although the function \( f(x) \) is differentiable on \( \mathbb{R} \), Theorem 5.1.3 might not work properly if the properties of \( f(x) \) are not appropriate. Now we will explain how this method works, partly as an application of the idea of linear approximation.

Figure 1

Let's say that we want to find the approximation of a root of \( y =f(x) \). Since we wish to solve an equation of the form \( f(x) = 0 \), the roots of the equation correspond to the \( x \)-intercepts of the graph of \( f \). The root that we are trying to find is labeled \( r \) in the figure. We start with a first approximation \( x_{1} \). Consider the tangent line \( L \) to the curve \( y = f(x) \) at the point \( \left(x_{1}, f(x_{1}) \right) \) and look at the \( x \)-intercept of \( L \), labeled \( x_{2} \). The idea behind Theorem 5.1.3 is that the tangent line is close to the curve and so its \( x \)-intercept, \( x_{2} \), is close to the \( x \)-intercept of the curve (namely, the root \( r \) that we are seeking). Because the tangent is a line, we can easily find its \( x \)-intercept.

To find a formula for \( x_{2} \) in terms of \( x_{1} \) we use the fact that the slope of \( L \) is \( f'(x_{1}) \), so its equation is

\( \displaystyle y - f(x_{1}) = f'(x_{1})(x - x_{1}) \)

Since the \( x \)-intercept of \( L \) is \( x_{2} \), we know that the point \( \left( x_{2}, 0 \right) \) is on the line, and so

\( \displaystyle 0 - f(x_{1}) = f'(x_{1})(x_{2} - x_{1}) \)

If \( f'(x_{1}) \neq 0 \), we can solve this equation for \( x_{2} \):

\( \displaystyle x_{2} = x_{1} - \frac{f(x_{1})}{f'(x_{1})} \)

We use \( x_{2} \) as a second approximation to \( r \).

Next we repeat this procedure with \( x_{1} \) replaced by the second approximation \( x_{2} \), using the tangent line at \( \left( x_{2}, f(x_{2}) \right) \). This gives a third approximation:

\( \displaystyle x_{3} = x_{2} - \frac{f(x_{2})}{f'(x_{2})} \)

If we keep repeating this process, we obtain a sequence of approximations \( x_{1},\ x_{2},\ x_{3},\ x_{4}, \cdots \ \). In general, if the \( n \)th approximation is \( x_{n} \) and \( f'(x_{n}) \neq 0 \), then the next approximation is given by

\( \displaystyle x_{n+1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})} \)

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Differentials

Definition 5.2.1.  If \( y = f(x) \), where \( f \) is a differentiable function, then the differential \( \operatorname{d}\!x \) is an independent variable; that is, \( \operatorname{d}\!x \) can be given the value of any real number. The differential \( \operatorname{d}\!y \) is then defined in terms of \( \operatorname{d}\!x \) by the equation

\( \displaystyle \operatorname{d}\!y = f'(x)\operatorname{d}\!x \)

So \( \operatorname{d}\!y \) is a dependent variable; it depends on the values of \( x \) and \( \operatorname{d}\!x \). If \( \operatorname{d}\!x \) is given a specific value and \( x \) is taken to be some specific number in the domain of \( f \), then the numerical value of \( \operatorname{d}\!y \) is determined.

In the notation of differentials, the linear approximation \( \text{(i)} \) can be written as

\( \displaystyle f(a + \operatorname{d}\!x) \approx f(a) + \operatorname{d}\!y \)

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The body text and the figures were excerpted from James Stewart, Calculus 8E.