Use Newton’s Method to find the absolute minimum value of the function f(x)=x^2+sinx to four decimal places.
2 Answers
jeew-m | College Teacher | (Level 1) Educator Emeritus
`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
pramodpandey | College Teacher | (Level 3) Valedictorian
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)