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Math Help - Newton's method

  1. #1
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    Newton's method

    Use Newton's method to approximate to within 10^-4, the value of x that produces the point on the graph of y = 1/x that is closest to (2, 1).

    I tried setting p0 = 2 and using the algorithm p_{n} = p_{n-1} - f(p_{n-1})/f'(p_{n-1}) where f(x) = 1/x and f'(x) = -1/x^2 but the number is actually increasing? It goes 4, 8, 16, 32, ...

    I'm not sure what I'm doing wrong here.
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  2. #2
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    Quote Originally Posted by Makall View Post
    Use Newton's method to approximate to within 10^-4, the value of x that produces the point on the graph of y = 1/x that is closest to (2, 1).

    I tried setting p0 = 2 and using the algorithm p_{n} = p_{n-1} - f(p_{n-1})/f'(p_{n-1}) where f(x) = 1/x and f'(x) = -1/x^2 but the number is actually increasing? It goes 4, 8, 16, 32, ...

    I'm not sure what I'm doing wrong here.
    Hi Makall,

    You could develop a function for the distance from (2,1) to points on the graph.
    Then differentiate that function, setting it equal to zero to find the value of x
    producing that minimum distance.

    A distance equation is

    d^2=(x-2)^2+(y-1)^2=x^2-4x+4+y^2-2y+1

    Now substitute y=\frac{1}{x}

    to get

    d^2=x^2-4x+5+-2x^{-1}+x^{-2}

    Differentiate this, set it equal to zero and use Newton's method to find the root near x=2.
    Attached Thumbnails Attached Thumbnails Newton's method-1-x.jpg  
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  3. #3
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    So I got this f = {\left(x - 2\right)}^2 + {\left(\frac{1}{x} - 1\right)}^2 and differentiated it to get this f' = 2\, x + \frac{2}{x^2} - \frac{2}{x^3} - 4

    Then I used this equation x = x - \frac{{\left(x - 2\right)}^2 + {\left(\frac{1}{x} -  1\right)}^2}{2\, x + \frac{2}{x^2} - \frac{2}{x^3} - 4}

    Used initial value as 2 and got these numbers.

    x = 2
    >> for i=1:10, x = x - ((x - 2)^2 + (1/x - 1)^2)/(2*x + 2/x^2 - 2/x^3 - 4), end
    x = 1
    x = 1.500000000000000
    x = 2.013157894736842
    x = 1.090411751549197
    x = 1.587061684245663
    x = 2.164611788880735
    x = 1.598178106645094
    x = 2.188802634131807
    x = 1.641701289644409
    x = 2.300866462141473

    The solution says p0 = 2, p2 = 1.866760 and the point is (1.866760, 0.535687). The number seems to be alternating so I'm not sure what's wrong with it.
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  4. #4
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    Hi Makall,

    to get the distance, you need to take the square root of the square of the distance,
    differentiate that, equate to zero,
    find the value of x using Newton's method that causes the derivative to be zero.
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  5. #5
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    d^2=x^2-4x+5-\frac{2}{x}+\frac{1}{x^2}

    d=\sqrt{x^2-4x+5-\frac{2}{x}+\frac{1}{x^2}}

    To find the co-ordinate x at the minimum distance from (2,1)
    differentiate d with respect to x and equate to zero
    (finds x at the minimum value of the curve of d).

    d'=\frac{1}{2}\left(x^2-4x+5-\frac{2}{x}+\frac{1}{x^2}\right)^{-\frac{1}{2}}\left(2x-4+\frac{2}{x^2}-\frac{2}{x^3}\right)

    =\frac{x-2+\frac{1}{x^2}-\frac{1}{x^3}}{\sqrt{x^2-4x+5-\frac{2}{x}+\frac{1}{x^2}}}

    This =0 when the numerator =0

    We need to find x for which

    x-2+\frac{1}{x^2}-\frac{1}{x^3}=0

    g(x)=x^4-2x^3+x-1=0

    We use Newton's method to find the root of this near x=2.

    g'(x)=4x^3-6x^2+1

    X_{n+1}=X_n-\frac{g(X_n)}{g'(X_n)}

    Taking x=2 as the initial estimate, X_n

    X_{n+1}=2-\frac{16-16+2-1}{32-24+1}=2-\frac{1}{9}
    Attached Thumbnails Attached Thumbnails Newton's method-d.jpg   Newton's method-g-x-.jpg  
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