# Newtons method?

• Aug 3rd 2011, 09:27 AM
homeylova223
Newtons method?
Find the zeroes of the function using Newtons method.

f(x)=(1/4)x^3-3x^2+(3/4)x-2

I find that when x=11.803 y=0

Then I do Newtons method ((xn))-((f(x)n))/((f'(x)n))

((x))-((1/4x^3-3x^2+3/4x-2))/((3/4x^2-6x+3/4))

I find that when x=11.803 y=11.803

So Did I do this correctly?(Crying)
• Aug 3rd 2011, 10:12 AM
Also sprach Zarathustra
Re: Newtons method?
Quote:

Originally Posted by homeylova223
Find the zeroes of the function using Newtons method.

f(x)=(1/4)x^3-3x^2+(3/4)x-2

I find that when x=11.803 y=0

Then I do Newtons method ((xn))-((f(x)n))/((f'(x)n))

((x))-((1/4x^3-3x^2+3/4x-2))/((3/4x^2-6x+3/4))

I find that when x=11.803 y=11.803

So Did I do this correctly?(Crying)

$f(x)=\frac{1}{4}x^3-3x^2+\frac{3}{4}x-2$

$f(10)<0, f(12)>0$, hence by Rolle's theorem we know that there is a root $x_0 \in [10,12]$.

$f'(x)=\frac{3}{4}x^2-6x+\frac{3}{4}=\frac{3}{4}(x-(4-\sqrt{15}))(x-(4+\sqrt{15}))$

$f''(x)=\frac{3}{2}x-6$

$f'(x)$ and $f''(x)$ are greater than $0$ for all $x \in [10,12]$ (Why this fact is important?)

Now we start:

$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=x_n-\frac{\frac{1}{4}x^3_n-3x^2_n+\frac{3}{4}x_n-2}{\frac{3}{4}x^2_n-6x_n+\frac{3}{4}}$

After the algebraic calculations:

$x_{n+1}=\frac{2x^3_n-12x^2_n+8}{3x^2_n-24x_n+3}$

We take $x_1=12$ (Why we took 12 ?),

$x_2=\frac{2(12)^3-12(12)^2+8}{3(12)^2-24(12)+3}=\frac{248}{21}\approx 11.8095238$

$x_3=\frac{2(11.8095238)^3-12(11.8095238)^2+8}{3(11.8095238)^2-24(11.8095238)+3}\approx 11.803263...$
• Aug 3rd 2011, 10:48 AM
HallsofIvy
Re: Newtons method?
Quote:

Originally Posted by homeylova223
Find the zeroes of the function using Newtons method.

f(x)=(1/4)x^3-3x^2+(3/4)x-2

I find that when x=11.803 y=0

Then I do Newtons method ((xn))-((f(x)n))/((f'(x)n))

((x))-((1/4x^3-3x^2+3/4x-2))/((3/4x^2-6x+3/4))

I find that when x=11.803 y=11.803

What do you mean by that? Newton's method does not give you "y" it gives you $x_{n+1}$, the next number in a sequence hopefully converging to a zero of the function. If you mean you put 11.803 into the Newton's method formula, for $x_n$, and got back $x_{n+1}= 11.803$, yes, of course! If [tex]f(x_n)= 0, then the formula is $x_{n+1}= x_n- 0/f'(x_n)$ which is just $x_{n+1}= x_n- 0= x_n$.

That's essentially how you tell when you have arrived at a solution- when the iteration starts giving you the same number again and again.

Quote:

So Did I do this correctly?(Crying)
I am not sure what you did! You say found that "when x= 11.803, f(x)= 0". How did you find that value of x? And once you have found it why then put that value into the Newton's method formula?

I ran a quick Newton's method starting with $x_0= 0$
and got $x_1= 2.6666$, $x_2= 0.9935$, $x_3= .1053$, $x_4= 15.5345$, $x_5= 13.0675$, $x_6= 11.8091$, $x_7= 11.8032$, $x_8= 11.8032$

Since we are now getting the same result, we can say that one solution, to three decimal places, is 11.803.