# How to solve an equation using Newton's method with and without backtracking?

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?

Thanks

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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

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 $f(c)=0$.

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