# Newton Raphson

• Nov 11th 2010, 12:50 PM
juxi
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
• Nov 11th 2010, 11:32 PM
Educated
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.
• Nov 12th 2010, 02:42 AM
Quote:

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.

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

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 $\displaystyle x_0$