Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Lets assume I have this equation: $$\log(e^x+e^{-x})=2x+5,\quad x \in (-50,50).$$

As always we have to pick a starting point to solve this by Newton's method, but how can i know for what initial values this equation will or will not converge?

I think i need to use Matlab here, but I am not good at it...

Also, if I have $\log(e^x+e^{-x})=0.1x+5$, how can i specify the region of convergence for each possible solution?

At the end, I know that Newton's method has a quadratic convergence, can we show that it is true for our problem cases?


share|cite|improve this question
Oh so this is how you find the answer for my homework. Remember, it's due monday outside my office by 5pm. – user51956 Dec 6 '12 at 1:48
@inter milan. Oh cool do you really like Italian football? About hw, I just double check with my solutions and want to get more insight from community. The book I am using has very few examples to understand all the problems and questions I have :) – ASROMA Dec 7 '12 at 3:03
@inter milan: At least I am trying to find things I do not understand clearly on my own. Not like others that copy everything from each other :) – ASROMA Dec 7 '12 at 3:11
up vote 1 down vote accepted

We have the function $f(x)=\log(e^x+e^{-x})-(2x+5),\quad x \in (-50,50).$ and it's first derivative $f'(x)$ and to solve using the Newton's method, we use the sequence (for an $x_0$) $$x_{n+1}=x_n-\cfrac{f(x_n)}{f'(x_n)}$$

  • It's sufficient to look for a point $x_0$ close to the root and the tangent at that point exists, non zero i.e. $f'(x_0\ne0)$ and finite.

Look at the function $$g(x)=x-\cfrac{f(x)}{f'(x)}$$ where $$g'(x)=\cfrac{f(x)f''(x)}{(f'(x))^2}$$ and $$g''(x)=\cfrac{[f'(x)f''(x)+f(x)f'''(x)]\cdot(f'(x))^2-2f(x)f'(x)(f''(x))^2}{(f'(x))^4}$$

To have convergence for a point $x_0$ such that $f(x_0) \approx f(r)=0$, $r$ is the root, observe that $g'(x_0) \approx 0$ so we can't set a condition there but with at the second derivative of $g(x)$ we have $$g''(x_0)=\cfrac{f''(x_0)}{f'(x_0)}$$ Observe that $g''(x_0)$ has no prioritized reason to be zero, so the condition stated above is necessary but not always sufficient to have a quadratic convergence.

EDIT The most optimal way to guess a good $x_0$ is to use the Intermediate Value Theorem.

If $f$ is a real valued continuous function in the interval $[a,b]$ such that $f(a)<0$, $f(b)>0\quad$ OR $\quad f(a)>0$, $f(b)<0$ then $\exists c\in [a,b]$ such that $f(c)=0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.