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

#1**+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**+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