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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 ( x ) = 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 ( x ) = 0 . If f is the first-degree polynomial f ( x ) = a x + b , then the solution of f ( x ) = 0 is given by the formula x = b a . If f is the second-degree polynomial f ( x ) = a x 2 + b x + c , the solutions of f ( x ) = 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 ( x ) = x 5 + 8 x 4 + 4 x 3 2 x 7 .

No formula exists that allows us to find the solutions of f ( x ) = 0 . Similar difficulties exist for nonpolynomial functions. For example, consider the task of finding solutions of tan ( x ) 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 ( x ) = 0 . By sketching a graph of f , we can estimate a root of f ( x ) = 0 . Let’s call this estimate x 0 . We then draw the tangent line to f at x 0 . If f ( x 0 ) 0 , this tangent line intersects the x -axis at some point ( x 1 , 0 ) . 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 ( x 1 ) 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 ,… . Typically, the numbers x 0 , x 1 , x 2 ,… quickly approach an actual root x * , as shown in the following figure.

This function f(x) is drawn with points (x0, f(x0)), (x1, f(x1)), and (x2, f(x2)) marked on the function. From (x0, f(x0)), a tangent line is drawn, and it strikes the x axis at x1. From (x0, f(x0)), a tangent line is drawn, and it strikes the x axis at x2. If a tangent line were drawn from (x2, f(x2)), it appears that it would come very close to x*, which is the actual root. Each tangent line drawn in this order appears to get closer and closer to x*.
The approximations x 0 , x 1 , x 2 ,… approach the actual root x * . The approximations are derived by looking at tangent lines to the graph of f .

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

y = f ( x 0 ) + f ( x 0 ) ( x x 0 ) .

Therefore, x 1 must satisfy

f ( x 0 ) + f ( x 0 ) ( x 1 x 0 ) = 0 .

Solving this equation for x 1 , we conclude that

x 1 = x 0 f ( x 0 ) f ( x 0 ) .

Similarly, the point ( x 2 , 0 ) is the x -intercept of the tangent line to f at x 1 . Therefore, x 2 satisfies the equation

x 2 = x 1 f ( x 1 ) f ( x 1 ) .

In general, for n > 0 , x n satisfies

x n = x n 1 f ( x n 1 ) f ( x n 1 ) .

Next we see how to make use of this technique to approximate the root of the polynomial f ( x ) = x 3 3 x + 1 .

Questions & Answers

find the equation of the tangent to the curve y=2x³-x²+3x+1 at the points x=1 and x=3
Esther Reply
derivative of logarithms function
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sidra
ex 2.1 question no 11
khansa
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Rasul
ex 2.1 question no. 11
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khansa
Find the derivative of g(x)=−3.
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How can I help you?
Tlou
evaluate the following computation (x³-8/x-2)
Murtala Reply
teach me how to solve the first law of calculus.
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teach me also how to solve the first law of calculus
Bilson
what is differentiation
Ibrahim Reply
only god knows😂
abdulkadir
f(x) = x-2 g(x) = 3x + 5 fog(x)? f(x)/g(x)
Naufal Reply
fog(x)= f(g(x)) = x-2 = 3x+5-2 = 3x+3 f(x)/g(x)= x-2/3x+5
diron
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jimmy
how to use fundamental theorem to solve exponential
JULIA Reply
find the bounded area of the parabola y^2=4x and y=16x
Omar Reply
what is absolute value means?
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Chicken nuggets
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MM
🐔🦃 nuggets
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(mathematics) For a complex number a+bi, the principal square root of the sum of the squares of its real and imaginary parts, √a2+b2 . Denoted by | |. The absolute value |x| of a real number x is √x2 , which is equal to x if x is non-negative, and −x if x is negative.
Ismael
find integration of loge x
Game Reply
find the volume of a solid about the y-axis, x=0, x=1, y=0, y=7+x^3
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Brad Reply
Can calculus give the answers as same as other methods give in basic classes while solving the numericals?
Cosmos Reply
log tan (x/4+x/2)
Rohan
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Rohan
y=(x^2 + 3x).(eipix)
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Ismael
Practice Key Terms 2

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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