[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.