# 4.9 Newton’s method

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• Describe the steps of Newton’s method.
• Explain what an iterative process means.
• Recognize when Newton’s method does not work.
• Apply iterative processes to various situations.

In many areas of pure and applied mathematics, we are interested in finding solutions to an equation of the form $f\left(x\right)=0.$ For most functions, however, it is difficult—if not impossible—to calculate their zeroes explicitly. In this section, we take a look at a technique that provides a very efficient way of approximating the zeroes of functions . This technique makes use of tangent line approximations and is behind the method used often by calculators and computers to find zeroes.

## Describing newton’s method

Consider the task of finding the solutions of $f\left(x\right)=0.$ If $f$ is the first-degree polynomial $f\left(x\right)=ax+b,$ then the solution of $f\left(x\right)=0$ is given by the formula $x=-\frac{b}{a}.$ If $f$ is the second-degree polynomial $f\left(x\right)=a{x}^{2}+bx+c,$ the solutions of $f\left(x\right)=0$ can be found by using the quadratic formula. However, for polynomials of degree $3$ or more, finding roots of $f$ becomes more complicated. Although formulas exist for third- and fourth-degree polynomials, they are quite complicated. Also, if $f$ is a polynomial of degree $5$ or greater, it is known that no such formulas exist. For example, consider the function

$f\left(x\right)={x}^{5}+8{x}^{4}+4{x}^{3}-2x-7.$

No formula exists that allows us to find the solutions of $f\left(x\right)=0.$ Similar difficulties exist for nonpolynomial functions. For example, consider the task of finding solutions of $\text{tan}\left(x\right)-x=0.$ No simple formula exists for the solutions of this equation. In cases such as these, we can use Newton’s method to approximate the roots.

Newton’s method    makes use of the following idea to approximate the solutions of $f\left(x\right)=0.$ By sketching a graph of $f,$ we can estimate a root of $f\left(x\right)=0.$ Let’s call this estimate ${x}_{0}.$ We then draw the tangent line to $f$ at ${x}_{0}.$ If ${f}^{\prime }\left({x}_{0}\right)\ne 0,$ this tangent line intersects the $x$ -axis at some point $\left({x}_{1},0\right).$ Now let ${x}_{1}$ be the next approximation to the actual root. Typically, ${x}_{1}$ is closer than ${x}_{0}$ to an actual root. Next we draw the tangent line to $f$ at ${x}_{1}.$ If ${f}^{\prime }\left({x}_{1}\right)\ne 0,$ this tangent line also intersects the $x$ -axis, producing another approximation, ${x}_{2}.$ We continue in this way, deriving a list of approximations: ${x}_{0},{x}_{1},{x}_{2}\text{,…}.$ Typically, the numbers ${x}_{0},{x}_{1},{x}_{2}\text{,…}$ quickly approach an actual root $x*,$ as shown in the following figure.

Now let’s look at how to calculate the approximations ${x}_{0},{x}_{1},{x}_{2}\text{,…}.$ If ${x}_{0}$ is our first approximation, the approximation ${x}_{1}$ is defined by letting $\left({x}_{1},0\right)$ be the $x$ -intercept of the tangent line to $f$ at ${x}_{0}.$ The equation of this tangent line is given by

$y=f\left({x}_{0}\right)+{f}^{\prime }\left({x}_{0}\right)\left(x-{x}_{0}\right).$

Therefore, ${x}_{1}$ must satisfy

$f\left({x}_{0}\right)+{f}^{\prime }\left({x}_{0}\right)\left({x}_{1}-{x}_{0}\right)=0.$

Solving this equation for ${x}_{1},$ we conclude that

${x}_{1}={x}_{0}-\frac{f\left({x}_{0}\right)}{f\prime \left({x}_{0}\right)}.$

Similarly, the point $\left({x}_{2},0\right)$ is the $x$ -intercept of the tangent line to $f$ at ${x}_{1}.$ Therefore, ${x}_{2}$ satisfies the equation

${x}_{2}={x}_{1}-\frac{f\left({x}_{1}\right)}{f\prime \left({x}_{1}\right)}.$

In general, for $n>0,{x}_{n}$ satisfies

${x}_{n}={x}_{n-1}-\frac{f\left({x}_{n-1}\right)}{f\prime \left({x}_{n-1}\right)}.$

Next we see how to make use of this technique to approximate the root of the polynomial $f\left(x\right)={x}^{3}-3x+1.$

#### Questions & Answers

determine the area of the region enclosed by x²+y=1,2x-y+4=0
Gerald Reply
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MP
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CALCULUS
For example any questions...
CALCULUS
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for the function f(x)={x^2-7x+104 x<=7 7x+55 x>7' does limx7 f(x) exist?
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Integral of e^x/(1+e^2x)tan^-1 (e^x)
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fg[[(45)]]²+45⅓x²=100
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find the values of c such that the graph of f(x)=x^4+2x^3+cx^2+2x+2
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Debdoot
A conical container of radius 10 ft and height 30 ft is filled with water to a depth of 15 ft. How much work is required to pump all the water out through a hole in the top of the container if the unit weight of the water is 62.4 lb/ft^3?
Milca Reply
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IBRAHIM
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a^6÷8
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Muzamil

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