Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 5

Applications of the Derivative

Newton’s Method

As we mentioned earlier, it is not always possible to solve algebraically the equation $f(x)=0$ for some function $f$. But we have methods to locate or approximate the value of the exact solution. Newton’s method is one of them. The idea is based on linear approximation, discussed in Unit ${\displaystyle 4}$. We use the lines that are tangent to the curve to construct a sequence that converges to a solution to the equation $f(x)=0$.

A sequence is a set of numbers listed in a specific order. For instance, the sequence ${\displaystyle \left\{1\text{,} 2\text{,} 3\text{,} 4\text{,} 5 \text{.} \text{.} \text{.}\right\}}$ is not equal to the sequence ${\displaystyle \left\{2\text{,} 1\text{,} 3\text{,} 4\text{,} 5\text{,} 6 \text{.} \text{.} \text{.}\right\}}$—the set of numbers may be the same, but the order is different. The n-th term of a sequence is denoted with a subindex $n$; for example, ${\displaystyle x_n}$ or ${\displaystyle s_n\text{.}}$

For the sequence

${\displaystyle \left\{1\text{,} {\displaystyle \frac{1}{2}}\text{,} {\displaystyle \frac{1}{3}}\text{,} {\displaystyle \frac{1}{4}}\text{,} \text{.} \text{.} \text{.}\right\}\text{,} }$

the first term is ${\displaystyle x_1=1}$, the second is ${\displaystyle x_2=1∕2}$, the third is ${\displaystyle x_3=1∕3}$, and the n-th term is ${\displaystyle x_n=1∕n\text{.}}$

Intuitively, we say that a sequence ${\displaystyle \left\{x_n\right\}}$ converges to a number ${\displaystyle s}$ if, for a large number $n$, the distance between ${\displaystyle s}$ and ${\displaystyle x_n}$ is very small. That is ${\displaystyle s\approx x_n}$ for a large $n$. The larger the $n$, the better the approximation, and we write

${\displaystyle \underset{n\to \infty }{lim}{x}_n=s\text{.}}$

The difficult part in Newton’s method is to choose the first term of the sequence (i.e., ${\displaystyle x_1}$). In making our choice, we may want to use the graph of the function, the Intermediate Value Theorem, or trial and error. Once the first term is chosen, the others are deduced recursively by applying the formula

${\displaystyle x_{n+1}=x_n-{\displaystyle \frac{f\left(x_n\right)}{f^{\prime }\left(x_n\right)}}\text{.}}$

Example 5.56. Finding the value of ${\displaystyle \sqrt[3]{30}}$ is the same as finding an $x$ such that ${\displaystyle x^3=30}$; that is, $x$ is the solution of ${\displaystyle x^3-30=0\text{.}}$

Let us consider the function ${\displaystyle f\left(x\right)=x^3-30}$. To apply Newton’s method, we must choose ${\displaystyle x_1}$.

Since ${\displaystyle f\left(3\right)=27-30=-3}$ we choose ${\displaystyle x_1=3}$.

We have ${\displaystyle f^{\prime }\left(x\right)=3x^2}$; hence,

${\displaystyle x_{n+1}=x_n-{\displaystyle \frac{{\left(x_n\right)}^3-30}{3{\left(x_n\right)}^2}}}$

and

${\displaystyle x_2=x_1-{\displaystyle \frac{f\left(x_1\right)}{f^{\prime }\left(x_1\right)}}=3-{\displaystyle \frac{\left(3^3\right)-30}{3{\left(3\right)}^2}}={\displaystyle \frac{28}{9}}\approx 3\text{.}111111\text{.}}$

${\displaystyle x_3={\displaystyle \frac{28}{9}}-{\displaystyle \frac{{\left({\displaystyle {\displaystyle \frac{28}{9}}}\right)}^3-30}{3{\left({\displaystyle {\displaystyle \frac{28}{9}}}\right)}^2}}={\displaystyle \frac{32887}{10584}}\approx 3\text{.}107237\text{.}}$

You can check that ${\displaystyle x_4\approx 3\text{.}10723251\text{.}}$

Example 5.57. Let us find a solution of ${\displaystyle x^3-3x+6=0\text{.}}$

At first, we consider ${\displaystyle f\left(x\right)=x^3-3x+6}$, and we choose ${\displaystyle x_1=1}$. We have ${\displaystyle f^{\prime }\left(x\right)=3x^2-3}$. But ${\displaystyle f^{\prime }\left(1\right)=0}$, and we cannot divide by 0; thus, Newton’s method is not applicable for this value of ${\displaystyle x_1\text{.}}$

However, we can see that ${\displaystyle f\left(-3\right)=-12<0<10=f\left(-2\right)}$, and by the Intermediate Value Theorem, we know that one solution is in the interval ${\displaystyle \left(-3\text{,} -2\right)}$. Hence, we choose ${\displaystyle x_1=-3}$, and

${\displaystyle x_{n+1}=x_n-{\displaystyle \frac{x_n^3-3x_n+6}{3x_n^2-3}}\text{.}}$

${\displaystyle x_2=-3-{\displaystyle \frac{{\left(-3\right)}^3-3\left(-3\right)+6}{3{\left(-3\right)}^2-3}}=-{\displaystyle \frac{5}{2}}\text{.}}$

So,

${\displaystyle x_3=-{\displaystyle \frac{5}{2}}-{\displaystyle \frac{{\left(-2\text{.}5\right)}^3-3\left(-2\text{.}5\right)+6}{3{\left(-2\text{.}5\right)}^2-3}}\approx -2\text{.}365079\text{.}}$

So one solution is approximately equal to ${\displaystyle -2\text{.}3650}$.

You may want to find ${\displaystyle x_4}$.

In the examples above, we illustrated the application of Newton’s method. Now you should read the section titled “Newton’s Method,” on pages 252-256 of the textbook to see how this method was developed.

Finishing This Unit

1. Review the objectives of this unit and make sure you are able to meet all of them.
2. If there is a concept, definition, example or exercise that is not yet clear to you, go back and reread it, then contact your tutor for help.
3. Do the exercises in “Learning from Mistakes” section for this unit.
4. You may want to do Exercises 1, 5, 13, 15, 17, 21, 23, 26, 38, 41, 49 and 51 from the “Review” (pages 260-262 of the textbook).