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Applications of Differentiation

3.8 Newton’s Method

Focus on newton iteration, repeated approximation, invalid starting points.

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Ideas in this lesson

01Newton iterationWorked steps · five questions02Repeated approximationWorked steps · five questions03Invalid starting pointsWorked steps · five questions
Idea 1 of 3

Newton iteration

The next approximation is the xx-intercept of the tangent at the current approximation.

Worked example 1

Use Newton’ss Method for x2−2=0x^{2} - 2 = 0 with x0=2x_{0} = 2. Find x1x1.
  1. xn+1=xn−xn2−22xn=xn+2xn2x_{n+1} = x_{n} - \dfrac{x_{n}^{2} - 2}{2x_{n}} = \dfrac{x_{n} + \dfrac{2}{x_{n}}}{2}.
  2. x1=32≈1.5x1 = \dfrac{3}{2} \approx 1.5

Answer: 32≈1.5\dfrac{3}{2} \approx 1.5

Worked example 2

Use Newton’ss Method for x2−3=0x^{2} - 3 = 0 with x0=3x_{0} = 3. Find x2x2.
  1. xn+1=xn−xn2−32xn=xn+3xn2x_{n+1} = x_{n} - \dfrac{x_{n}^{2} - 3}{2x_{n}} = \dfrac{x_{n} + \dfrac{3}{x_{n}}}{2}.
  2. x1=2x1 = 2
  3. x2=74≈1.75x2 = \dfrac{7}{4} \approx 1.75

Answer: 74≈1.75\dfrac{7}{4} \approx 1.75

Idea 2 of 3

Repeated approximation

Use the previous iterate without premature rounding. Check the residual or successive-iterate change for convergence.

Worked example 1

Use Newton’ss Method for x2−2=0x^{2} - 2 = 0 with x0=2x_{0} = 2. Find x1x1.
  1. xn+1=xn−xn2−22xn=xn+2xn2x_{n+1} = x_{n} - \dfrac{x_{n}^{2} - 2}{2x_{n}} = \dfrac{x_{n} + \dfrac{2}{x_{n}}}{2}.
  2. x1=32≈1.5x1 = \dfrac{3}{2} \approx 1.5

Answer: 32≈1.5\dfrac{3}{2} \approx 1.5

Worked example 2

Use Newton’ss Method for x2−3=0x^{2} - 3 = 0 with x0=3x_{0} = 3. Find x2x2.
  1. xn+1=xn−xn2−32xn=xn+3xn2x_{n+1} = x_{n} - \dfrac{x_{n}^{2} - 3}{2x_{n}} = \dfrac{x_{n} + \dfrac{3}{x_{n}}}{2}.
  2. x1=2x1 = 2
  3. x2=74≈1.75x2 = \dfrac{7}{4} \approx 1.75

Answer: 74≈1.75\dfrac{7}{4} \approx 1.75

Idea 3 of 3

Invalid starting points

Newton’ss formula divides by f′f'. A zero derivative can make an update impossible, and a nonzero derivative does not guarantee convergence.

Worked example 1

For f(x)=x3−2f(x) = x^{3} - 2, why does Newton’ss Method fail with x0=0x_{0} = 0?
  1. f′(x)=3x2  ⟹  f′(0)=0f'(x) = 3x^{2} \implies f'(0) = 0
  2. x1=0−−20x_{1} = 0 - \dfrac{-2}{0} is undefined

Answer: The derivative is zero at the starting point

Worked example 2

For f(x)=x3−3f(x) = x^{3} - 3, why does Newton’ss Method fail with x0=0x_{0} = 0?
  1. f′(x)=3x2  ⟹  f′(0)=0f'(x) = 3x^{2} \implies f'(0) = 0
  2. x1=0−−30x_{1} = 0 - \dfrac{-3}{0} is undefined

Answer: The derivative is zero at the starting point

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §3.8,p3.8, p. 225225. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.