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Thread: 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 $\displaystyle 0=\cos(x)-2x$

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


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

    Start off with $\displaystyle x_0$ as where you might think the root might be. If you have no idea, start off with $\displaystyle x_0 = 0$ and just substitute in the values into the formula to get $\displaystyle x_1, x_2, x_3 ...$ 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.

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

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

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

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

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

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

    You could use a value of x around $\displaystyle 0\rightarrow1$ for $\displaystyle x_0$
    Attached Thumbnails Attached Thumbnails Newton Raphson-cosx-2x.jpg  
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