Consider Newton's method for finding the root of f(x) = x-sin(x). Run it on Matlab to find what is the rate of convergence. What is the value $\lambda$ such that $|x_{n+1} - \alpha| = \lambda|x_n - \alpha|$ where $\alpha$ is the root. Find the multiplicity m of this root $\alpha$.

Using Newton's method for finding roots (octave) I have found that this f(x) converges at 2.0236e-08, and it does so in close to 50 iterations. My code is as follows $x = x - ((x-\sin(x))/(1-\cos(x)))$

How do I go about solving for $\lambda$ and the multiplicity of m?

Thanks for your time!

  • $\begingroup$ If you have the numbers still, then you can examine $$\lambda=\frac{|x_{n+1}-\alpha|}{|x_n-\alpha|}$$ and see what they look like. (The equality you gave suggests they should all be the same, although I would expect a limiting value.) If you don't still have the numbers, then I guess you get to regenerate the numbers. $\endgroup$
    – tabstop
    Commented Feb 27, 2014 at 0:57

1 Answer 1


We have $f(x) = x- \sin x~$ and it is clear that $x = 0$ is a root.

  • $f(x) = x- \sin x~$ and $~f(0) = 0$
  • $f'(x) = 1 - \cos x$ and $~f'(0) = 0$
  • $f''(x) = \sin x~$ and $~f''(0) = 0$
  • $f'''(x) = \cos x~$ and $~f'''(0) = 1$

Hence, we have a triple root, so $m = 3$.

We can now write (do you know where this result comes from):

$$\lambda = \dfrac{m-1}{m} = \dfrac{2}{3}$$

Now, if we use $x_0 = 1$ and want to figure out how many steps it will take to get eight digits of accuracy, we have:

$$\left(\dfrac{2}{3}\right)^n < \dfrac{1}{2} 10^{-8} \implies n \ge 48$$

Compare this to your numerical results using $e_{n+1} \approx \dfrac{2}{3} e_n$. In other words, use your numerical results and add a column which shows $~|x_{n+1} - \alpha| = \lambda|x_n - \alpha|~$ where $\alpha$ is the root

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