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Math Help - Newton Raphson

  1. #1
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    Newton Raphson

    I am trying to determine the real number x with cos(x)=2x, Newton Method with f(x)=cos(x)-2x.
    How do I try and solve this?
    Thanks,
    Joan
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  2. #2
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    You are trying to solve 0=\cos(x)-2x

    The function is: f(x)=\cos(x)-2x therefore the derivative is f'(x)=...


    Newton Raphson method: x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}

    Start off with x_0 as where you might think the root might be. If you have no idea, start off with x_0 = 0 and just substitute in the values into the formula to get x_1, x_2, x_3 ... and so on.
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  3. #3
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    Quote Originally Posted by juxi View Post
    I am trying to determine the real number x with cos(x)=2x, Newton Method with f(x)=cos(x)-2x.
    How do I try and solve this?
    Thanks,
    Joan
    The Newton-Raphson method finds close approximations to points at which a graph crosses the x-axis.

    cosx=2x\Rightarrow\ cosx-2x=0,\;\;2x-cosx=0

    The graph of the function cosx-2x crosses the x-axis when cosx=2x

    f(x)=2x is a straight line, passing through the origin (0,0) and having a slope of 2

    g(x)=cosx is periodic, with a local maximum at x=0 and is 0 at x=0.5{\pi}

    Since 2x=1 when x=\frac{1}{2}, then since cosx is decreasing from x=0\rightarrow\ x=0.5{\pi}

    then there is a solution to cosx=2x slightly to the left of x=\frac{1}{2}

    You could use a value of x around 0\rightarrow1 for x_0
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