# Mean Value theorem and Newton's Method

I am trying to prove that: given $x_0, x_1, x_2 \ldots$ the sequence of approximations to $\pi$, use the mean value theorem to show that $|\pi-x_{j+1}| = |\tan c_j||\pi - x_j|$, where $c_j$ is some number between $x_j$ and $\pi$.

So I do the following, but I am not sure if they are right. Let $g(x) = \sin x$. Then let $f(x) = x - \frac{g'(x)}{g(x)} = x - \tan x$. As $f(x)$ is continuous on $(\pi/2,3\pi/2)$ and differentiable on this interval excluding the endpoints, then the mean value theorem says that

$\exists c \in [\frac{\pi}{2}+\epsilon, \frac{3\pi}{2}-\epsilon]$ such that $f'(c) = \frac{f(b) - f(a)}{b-a}$ where $b = \frac{3\pi}{2} - \epsilon$, $a= \frac{\pi}{2} +\epsilon$ for some $\epsilon>0$. Now (is this right?) that if $x_0$ is in this interval, then $\exists c_0$ such that $f'(c_0) = \frac{f(\pi)-f(x_0)}{\pi - x_0}$, as $(x_0,\pi)\subset(\frac{\pi}{2} +\epsilon,\frac{2\pi}{2}+\epsilon)$.

(b) Now as $x_0,x_1,\ldots$ are the sequence of approximations of $\pi$ then either $x_0 < x_1 \ldots$ or $x_0 > x_1 \ldots$, depending on whether the $x_i's$ approach $\pi$ from the left or right, as $x - \tan(x)$ is strictly increasing on $(\pi/2,3\pi/2)$. In other words, this means that $(x_0,\pi) \supset (x_1,\pi) \supset \ldots$. (Can I deduce this from the above?)

So here are several bits that are bothering me. If I look at (b) above, it is sort of assuming that my sequence is strictly increasing and bounded above, and hence is convergent (which is what i am supposed to prove later on on). Also, if $f(x)$ is continuous and differentiable on $(\pi/2,3\pi/2)$, then does it mean that $f(x)$ is continuous and differentiable on a subinterval of $(\pi/2,3\pi/2)$?

Thanks, Ben

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Just to add, how do I know as well that all my $x_i's$ will stay in the interval $(\pi/2,3\pi/2)$? – user38268 Apr 17 '11 at 5:51
How are the $x_i$ defined? – Sam Lisi Apr 17 '11 at 13:56

Below is my interpretation of what the question might be about, and calculations that should lead you to a complete solution of all the parts of the homework question you were asked, and, I hope, to a better understanding of the situation. In dealing with a question, it is often crucially important to find out "what's really happening." Your posts show perhaps too great a concentration on "what standard tool shall I use." The editorial is over, on to the mathematics!

We are using the Newton Method to approximate the solution $x=\pi$ of the equation $\sin x=0$. Where we start matters a great deal. Suppose that we start with the estimate $x_0$, where for now we only assume that $x_0$ is in the interval $(\pi/2,3\pi/2)$. (Later we will see that the initial estimate $x_0$ needs to be closer to $\pi$.)

The usual Newton Method iteration produces a sequence $x_0$, $x_1$, $x_2$, and so on, with $$x_{k+1}=f(x_k)$$ where $$f(x)=x-\frac{\sin x}{\cos x}=x-\tan x$$ Note that $f'(x)=1-\sec^2 x=-\tan^2 x$. This is defined in our interval and negative, so $f(x)$ is decreasing (you had it increasing) on our interval. By the Mean Value Theorem, $$\frac{f(\pi)-f(x_k)}{\pi -x_k}=f'(c_k)$$ for some $c_k$ between $x_k$ and $\pi$. But $f(\pi)-f(x_k)=\pi-x_{k+1}$. It follows that $$\pi-x_{k+1}=-\tan^2(c_k)(\pi-x_k)$$ There is obvious bad behaviour if $|\tan(c_k)|$ is large. For example, if $x_0$ is only a tiny bit bigger than $\pi/2$, $x_1$ will be far away from our interval.

So now assume that $3\pi/4<x_0<5\pi/4$. The square of the tangent function is less than $1$ on this interval. Since $$\pi-x_{k+1}=-\tan^2(c_k)(\pi-x_k)$$ it follows that the distance from $x_{k+1}$ to $\pi$ is less than the distance from $x_k$ to $\pi$.

Now we can see that if $3\pi/4<x_0<5\pi/4$, and if $x_0$ is in the interval $(3\pi/4,5\pi/4)$, then so is $x_1$, but then so is $x_2$, and so on. From this we can see that the sequence $(x_k)$ has limit $\pi$, by a general "bounded monotone" argument that you seem to have good control of. (Note that the $x_k$ bounce from too high to too low, so be careful!)

But we can do better than that! Since $$|\pi-x_{k+1}|\le \tan^2(c_0)|\pi-x_k|$$ for all $k$, we can see that $$|\pi-x_n|\le \tan^{2n}(c_0)|\pi-x_0|$$ so the convergence to $\pi$ is at least as fast as the convergence of a geometric series. (The convergence is much faster than that, for the attraction term $\tan^2(c_k)$ approaches $0$ as $x_k$ approaches $\pi$.) When the Newton Method is good, it is very very good.

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@user6312 Ok thanks; here's something i've had to deal with as well: Showing that for any $x_0$ in the interval $(3\pi/4,5\pi/4)$, any $x_j$ will not be of the form $(2k+1)\pi/2$, otherwise i'm stuffed. I've shown that already, but now what about part (b) above that I have written?? I sort of know that it is incorrect as it is tacitly assuming that my sequence is convergent. – user38268 Apr 17 '11 at 22:51
@David Benjamin Lin: I showed above that $\pi-x_{k+1}=-\tan^2(c_k)(\pi-x_k)$. If $x_0$ is in the interval in your comment above, I used this to show that $x_{1}$ is closer to $\pi$ than $x_0$ is. Of course this continues, everything is closer to $\pi$ than $x_0$ was, and in particular we can't hit any of the "bad" numbers you are worried about, since they are not in this interval. Try to read my solution carefully. All the (b) stuff is irrelevant. – André Nicolas Apr 18 '11 at 1:00
@user6312 Ok Thanks. I've figured out some other way to deal with the above, without invoking convergence or anything. – user38268 Apr 18 '11 at 1:41