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Wize University Calculus 1 Textbook > Applications of Differentiation

Newton's Method

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Newton's Method

Often we can not find solutions to equations algebraically. A powerful iterative process for finding roots is Newton's Method.

Newton's Method

If xnx_n is an approximate solution to f(x)=0f(x)=0, and f′(x)≠0f'(x) \ne0 then the next approximation, xn+1x_{n+1} , is given by

xn+1=xn−f(xn)f′(xn)\boxed{x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}}

Watch Out!
When f′(xn)f'(x_n) is close to 00 you may obtain bad approximations!

Example: Newton's Method

Apply Newton’s Method three times to the function f(x)=x3−4x+5f\left(x\right)=x^3-4x+5 to approximate a solution to the equation x3−4x=−5.x^3-4x=-5.
Start with x0=−2.x_0=−2.

xn+1=xn−f(xn)f′(xn)\boxed{x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}}

f(−2)=(−2)3−4(−2)+5=5f\left(-2\right)=\left(-2\right)^3-4\left(-2\right)+5=5

f′(x)=3x2−4            f′(−2)=3(−2)2−4=8f'\left(x\right)=3x^2-4\ \ \ \ \ \ \ \ \ \ \ \ f'\left(-2\right)=3\left(-2\right)^2-4=8

xn+1=xn−xn3−4xn+53xn2−4x_{n+1}=x_n-\frac{x_n^3-4x_n+5}{3x_n^2-4}

x1=−2−(−2)3−4(−2)+53(−2)2−4=−218=−2.625x_1=-2-\frac{\left(-2\right)^3-4\left(-2\right)+5}{3\left(-2\right)^2-4}=\frac{-21}{8}=-2.625

x2=−2.625−(−2.625)3−4(−2.625)+53(−2.625)2−4=−2.470x_2=-2.625-\frac{\left(-2.625\right)^3-4\left(-2.625\right)+5}{3\left(-2.625\right)^2-4}=-2.470

x3=−2.470−(−2.470)3−4(−2.470)+53(−2.470)2−4=−2.457x_3=-2.470-\frac{\left(-2.470\right)^3-4\left(-2.470\right)+5}{3\left(-2.470\right)^2-4}=-2.457
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Example: Newton's Method

Approximate 3\sqrt{3} to 3 decimal places using Newton's method.

Notice that 3\sqrt{3} is a solution to the equation x2−3=0x^2-3=0. Take f(x)=x2−3f(x)=x^2-3; then f′(x)=2xf'(x)=2x, and for n≥0n \geq 0,
xn+1=xn−xn2−32xn.x_{n+1}=x_n-\dfrac{x_n^2-3}{2x_n}.
Since we know 12=1≤3≤4=22→1≤3≤2,1^2=1 \leq 3 \leq 4=2^2 \rightarrow1 \leq \sqrt{3} \leq 2,let's take x0=1x_0=1. Since we're looking for three decimal places of accuracy, I'll round all results to four decimal places:

x1=1−12−32(1)=2x2=2−22−32(2)=1.75x3=1.75−1.752−32(1.75)=1.7321x4=1.7321−1.73212−32(1.7321)=1.7321\begin{array}{l} x_1=1-\dfrac{1^2-3}{2(1)} = 2 \\[+1em] x_2 = 2-\dfrac{2^2-3}{2(2)}=1.75\\[+1em] x_3=1.75-\dfrac{1.75^2-3}{2(1.75)}=1.7321\\[+1em] x_4=1.7321-\dfrac{1.7321^2-3}{2(1.7321)}=1.7321 \end{array}
Since x3,x4x_3,x_4 agree to (at least) three decimal places, we take 3≈1.732.\sqrt{3} \approx 1.732.

Practice: Newton's Method

Use Newton's Method to approximate a solution to f(x)=x4−x−10f(x)=x^4-x-10

Starting with x1=1x_1=1 , x2=?x_2=?
Given y=15x5+4x3+36x−1000y=\frac{1}{5}x^5+4x^3+36x-1000, what starting points would make Newton's method fail? Find a formula for the iterate xn+1x_{n+1} in term of xnx_n.
Extra Practice
Newton's Method: Estimate Intersection Point
Use Newton's method to estimate the point of intersection of y=e−x2\displaystyle y=e^{-x^{2}} and y=xy=x.
Newton's Method: Inverse Trigonometric Function
Starting from x0=1x_0=1, Newton's method is used to approximate the solutions of the equation arctan⁡x−12x2=0\displaystyle \arctan x-\frac{1}{2x^2}=0. Find the next approximation x1x_1.
Newton's Method: Estimate Intersection Point
Use Newton's method to estimate the point of intersection of y=e−x2\displaystyle y=e^{-x^{2}} and y=xy=x.
Practice: Approximating a Ln
Q.\textbf{Q.} Approximate ln⁡2\ln{2} to 4 decimal places using Newton's method.
Newton's Method
Use Newton's Method to approximate a solution to f(x)=x4−x=10f(x)=x^4-x=10
Starting with x1=1x_1=1 , x2=?x_2=?
Newton's Method
Using x0=1x_0=1 use Newton's Method to approximate a solution to cos⁡x=x\cos x=x.
Newton's Method
Using Newton's Method with x0=0x_0=0, find x1,…,x4x_1, \dots, x_4 for f(x)=2x3−x+1f(x)=2x^3-x+1.
Newton's Method
Find 23\sqrt[3]{2} to three decimal places.
Newton's Method
Take x0=0x_{0}=0 and find x1,x2,x3x_1,x_2,x_3 for f(x)=x3−3x2+x+1f(x)=x^{3}-3x^{2}+x+1. If necessary, round your answer to the fourth decimal place.
Given y=15x5+4x3+36x−1000y=\dfrac{1}{5}x^5+4x^3+36x-1000, what starting points would make Newton's Method fail? Find a formula for xn+1x_{n+1} in terms of xnx_n.
Practice: Approximating a Ln
Q.\textbf{Q.} Approximate ln⁡2\ln{2} to 4 decimal places using Newton's method.
Newton's Method: Inverse Trigonometric Function
Starting from x0=1x_0=1, Newton's method is used to approximate the solutions of the equation arctan⁡x−12x2=0\displaystyle \arctan x-\frac{1}{2x^2}=0. Find the next approximation x1x_1.
Newton's Method: Estimate Intersection Point
Use Newton's method to estimate the point of intersection of y=e−x2\displaystyle y=e^{-x^{2}} and y=xy=x.
Use one iteration of Newton's method to approximate 93\sqrt[3]{9} where the initial position is x=2.
Newton's Method
Which formula should be used in order to approximate arctan⁡(0.1)\arctan(0.1) using Newton's Method?
Newton's Method
Which of the following points on the following function should you NOT use in order to use Newton's Method?
f(x)=x3−2x+2f(x) = x^3 - 2x + 2
Choose all that apply.
Newton's Method
Consider the curve generated by the equation
x=−5x2x = - \frac{5}{x^2}
Which polynomial would you use Newton's Method on to estimate a solution to the above equation, and what integer should you choose as a starting point to best estimate the solution?
Given y=15x5+4x3+36x−1000y=\frac{1}{5}x^5+4x^3+36x-1000, what starting points would make Newton's method fail? Find a formula for the iterate xn+1x_{n+1} in term of xnx_n.