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Use Newton’s Method to find the absolute minimum value of the function...

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user3498849 | eNoter

Posted May 13, 2013 at 5:51 AM via web

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Use Newton’s Method to find the absolute minimum value of the function
f(x)=x^2+sinx to four decimal places.

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted May 13, 2013 at 6:47 AM (Answer #1)

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`f(x) =` `x^2+sinx`

`f'(x) = 2x+cosx`

At absolute minimum/maximum f'(x) = 0

Let `f'(x) = p(x)`

Then;

`p'(x) = 2-sinx`

Using Newton's method for the first approximation `x_0` we will get a better approximation `x_1` by;

`x_1 = x_0-(p(x_0))/(p'(x_0))`

 

Let us say `x_0 = 30`

Then we can get `x_1` as;

`x_1 = 30-(2xx30+cos30)/(2-sin30) = -10.5774`

 

Now the second approximation `x_2` can be given by;

`x_2 = x_1-(p(x_1))/(p'(x_1))`

 

`x_2 = -10.5774-(2xx-10.5774+cos(-10.5774))/(2-sin(-10.5774)) `

`x_2 = -1.3394`


Similarly we can get;

`x_3 = -0.0509`

`x_4 = -0.5000`

`x_5 = -0.4999`

`x_6 = -0.4999`

 

So at `x = -0.4999` we have a stationary point.

 

`f'(x) = 2x+cosx`

`f''(x) = 2-sinx`

`f''(x)_((x = -0.4999)) = 2.0087 > 0`

So f(x) has a minimum at x = -0.4999

 

So the absolute minimum of f(x) occurs at x = -0.4999

 

Sources:

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pramodpandey | College Teacher | Valedictorian

Posted May 13, 2013 at 6:49 AM (Answer #2)

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We have given

`f(x)=x^2+sin(x)`  ,minimum of this function f will occure if

f'(x)=0

`f'(x)=2x+cos(x)`

i.e we have to find zero of function f'(x).

Let

`G(x)=2x+cos(x)`

Newton method is

`x_(n+1)=x_n-(G(x_n))/(G'(x_n)), G'(x_n)!=0`

`G'(x)=2-sin(x)`

`` `x_(n+1)=x_n-{2x_n+cos(x_n)}/{2-sin(x_n)}`

Let 

`x_0=0`

`x_1=0-{1/2}=-.5`

`x_2=(-.5)-{-1+cos(-.5)}/(2+sin(.5)}=-.4999833`

`x_3=(-.4999833)-{2xx(-.4999833)+cos(.4999833)}/{2+sin(.4999833)}`

`=-.499978717`

`` Thus absolute minimum value of f(x) is

f(x)=.241252553

f(x)=.2413  ( corrected to four decimal place)

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