Newton Raphson

**juxi** · November 11th 2010, 12:50 PM

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

**Educated** · November 11th 2010, 11:32 PM

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.

**Archie Meade** · November 12th 2010, 02:42 AM

Originally Posted by juxi

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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