1

The iteration formula $$x_{n+1}=x_n-(\cos x_n)(\sin x_n)+R\cos^2x_n$$

where $R$ is a positive constant is obtained by applying Newton's method to some function $f(x)$. What is $f(x)$? What can this formula be used for?

I got the solution for $f(x)$ to be $\dfrac{\sin x}{x\cos x} = \dfrac{1}{x}\tan x$ (I can show the work if needed), but I'm struggling to see what the iteration formula is useful for? I would guess finding roots of $\dfrac{1}{x}\tan x$ at certain locations. Thanks for any help!

K. Rmth
  • 1,749
Drake
  • 851
  • The problem tells me to find the original function from that iteration formula...I found it to be (1/x)tanx...I can show my work getting to there if that helps. I will plot the function and use it to find the nearby root. I was just wondering if there is anything special about this iteration formula that the problem seems to suggest. – Drake Sep 29 '13 at 00:39
  • The iteration formula appears to locate all the zeros of f(x). I'm wondering if the formula is useful because of the R? – Drake Sep 29 '13 at 02:01
  • Newton's method doesn't find all the zeros, only one zero based on your initial guess. Your solution for $f(x)$ should have $R$ appearing explicitly. – Emily Dec 13 '13 at 19:41

1 Answers1

2

If $f(x) = \tan x - R$ then $f'(x) = \sec^2 x$, and plugging these into Newton's method gives $$ x_{n+1} = x_n - \frac{\tan x_n - R}{\sec^2 x_n} = x_n - \sin x_n \cos x_n + R\cos^2 x_n. $$ This would be used to solve $\tan x - R = 0$, or $x = \tan^{-1}(R)$ for a given value of $R$.

Peter Phipps
  • 3,065