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Math Help - Newtons method?

  1. #1
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    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?
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Newtons method?

    Quote Originally Posted by homeylova223 View Post
    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?




    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...
    Last edited by Also sprach Zarathustra; August 3rd 2011 at 10:29 AM. Reason: LaTeX...
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  3. #3
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    Re: Newtons method?

    Quote Originally Posted by homeylova223 View Post
    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.

    So Did I do this correctly?
    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.
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