Newton's Method Solution 4

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SOLUTION 4: We know from trigonometry that

$\ \sin \pi = 0$ , so I will choose to solve the equation

$\ \sin x = 0 \$ for

$x$ . Let's define function

$f(x) = \sin x$ , whose graph is given below.

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The derivative of

$f$ is

$f'(x) = \cos x$ . Now use Newton's Method:

$x_{n+1} = x_{n} - { f(x_{n}) \over f'(x_{n}) } \ \ \ \ \longrightarrow$

$x_{n+1} = x_{n} - { \sin x_{n} \over \cos x_{n} } \ \ \ \ \longrightarrow$

$x_{n+1} = x_{n} - { \tan x_{n} }$ I will choose to let

$x_{0}=2$ . Using Newton's Method formula for 7 iterations in a spreadsheet results in :

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Thus

$\ \pi \$ to ten decimal places is

$\ \pi \approx 3.1415926536$ .

IMPORTANT NOTE: Choosing a different initial guess can lead to other solutions to the equation

$\ \sin x = 0 \$ . For example, if I choose to let

$x_{0}= 1$ , then Newton's Method leads to the solution

$\ x=0 \$ . Using Newton's Method formula for iterations in a spreadsheet results in :

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