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Use the Newton’s method to approximate the solution of the equation: 1-x-tan x = 0 starting at x0 = 0.5 (the initial value). i.e. Find x1,x2,x3.. .

Guest Oct 28, 2014

Best Answer 

 #1
avatar+92775 
+5

Newton's method

You are probably supposed to know the formula but I can't remember formulas so I work it out ever time.

$$$Remember, the gradient of the tangent at $x=x_1 $ is given by$ f'(x_0)\\
so\\\\
f'(x_0)=\frac{f(x_0)-0}{x_0-x_1}\\\\
f'(x_0)=\frac{f(x_0)}{x_0-x_1}\\\\
x_0-x_1=\frac{f(x_0)}{f'(x_0)}\\\\
-x_1=-x_0+\frac{f(x_0)}{f'(x_0)}\\\\
x_1=x_0-\frac{f(x_0)}{f'(x_0)}\\\\
$Now you have the formula$\\\\
\boxed{x_1=x_0-\frac{f(x_0)}{f'(x_0)}}\\\\$$

 

1-x-tan x = 0 starting at x0 = 0.5 

Let

$$\\f(x)=1-x-tanx\\
f'(x)=-1-sec^2x\\\\
f(0.5)=1-0.5-tan0.5\\
f(0.5)=0.5-0.546\\
f(0.5)=-0.046\\\\
f'(0.5)=-1-sec^2(0.5)\\
f'(0.5)=-1-(1/cos^2(0.5))\\
f'(0.5)=-2.298\\
so\\
x_1=0.5-\frac{f(0.5)}{f'(0.5)}\\\\
x_1=0.5-\frac{-0.046}{-2.298}\\\\
x_1=0.480$$

 

I have applied newtons method just once.

If you want more accuracy you can apply it again and again until you are happy. :)

Melody  Oct 28, 2014
 #1
avatar+92775 
+5
Best Answer

Newton's method

You are probably supposed to know the formula but I can't remember formulas so I work it out ever time.

$$$Remember, the gradient of the tangent at $x=x_1 $ is given by$ f'(x_0)\\
so\\\\
f'(x_0)=\frac{f(x_0)-0}{x_0-x_1}\\\\
f'(x_0)=\frac{f(x_0)}{x_0-x_1}\\\\
x_0-x_1=\frac{f(x_0)}{f'(x_0)}\\\\
-x_1=-x_0+\frac{f(x_0)}{f'(x_0)}\\\\
x_1=x_0-\frac{f(x_0)}{f'(x_0)}\\\\
$Now you have the formula$\\\\
\boxed{x_1=x_0-\frac{f(x_0)}{f'(x_0)}}\\\\$$

 

1-x-tan x = 0 starting at x0 = 0.5 

Let

$$\\f(x)=1-x-tanx\\
f'(x)=-1-sec^2x\\\\
f(0.5)=1-0.5-tan0.5\\
f(0.5)=0.5-0.546\\
f(0.5)=-0.046\\\\
f'(0.5)=-1-sec^2(0.5)\\
f'(0.5)=-1-(1/cos^2(0.5))\\
f'(0.5)=-2.298\\
so\\
x_1=0.5-\frac{f(0.5)}{f'(0.5)}\\\\
x_1=0.5-\frac{-0.046}{-2.298}\\\\
x_1=0.480$$

 

I have applied newtons method just once.

If you want more accuracy you can apply it again and again until you are happy. :)

Melody  Oct 28, 2014

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