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This is similar to this question, except I am finding the radius now using the Bisection method and then Newton's method for finding a zero.

This is a computer science for a Numerical Methods course. In the question the arc length is 4, and the chord length is 3.

I derived a function such that $f(r) = 0$ when $r$ is the radius like this:

$$L = r\theta$$ $$4 = r\theta$$ Using the Law of Cosines:

$$3^2 = 2r^2(1 - \cos\theta)$$ $$ \Rightarrow \theta = \cos^{-1}(1 - 9/2r^2)$$ $$ \Rightarrow 4 = r\cos^{-1}(1 - 9/2r^2)$$

So $f(r) = 4 - r\,\cos^{-1}(1 - 9/2r^2)$ will be $0$ when $r$ is the radius.

This all works well so far, but for the two methods I need an initial approximation. For bisection I need a left and a right $x$, and for newtons method I should use the average of my two $x$ values from the bisection. It's easy to show that a lower bound for the radius is $1.5$ since the chord length is 3. If the radius was less than $1.5$ the chord length would be greater than the diameter, which is a contradiction. I am having a lot of difficulty finding an upper bound though. If I just pick a value like $1.7$ or $\pi/2$ then I get the correct answer, but although I'm not sure that it's required for my assignment, I'd like to know how I can find an upper bound that I can show is actually provably greater than the radius.

The actually radius assuming my program and equation is correct is $1.567769039908$

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@J.M. Thank you for improving my Latex attempt :) –  Paulpro Oct 12 '11 at 0:18

2 Answers 2

up vote 2 down vote accepted

The arc length can be larger than the diameter. Even if you don't allow angles greater than $\pi$ radians the arc length can be $\pi r$, so you should take the minimum $r$ as arclength$/\pi$ (or maybe even divide by $2\pi$-safer and it only costs one bisection). Having found the minimum, you can just keep doubling it until $f(r)$ changes sign.

Added: If $c$ is the chord, $\frac{c}{2r}=\sin \frac{\theta }{2}$. As we want $r$ large, we want $\sin \frac{\theta }{2}$ to be small, so use $\frac{\theta }{2}-\frac{\theta ^3}{48}$ to get an upper bound (alternating series theorem). This gives $c=L-\frac{L\theta ^2}{24}$ and with $L=r\theta$ you get $r$.

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I based the minimum off the chord length, not the arc length. I'm assuming the chord length ($3$) can not be larger than the diameter. The angle is < $\pi$ radians. We can assume that. –  Paulpro Oct 12 '11 at 0:36
    
I made a couple typos in my question, I see where you got that the arc length was 3 from now. I edited it my question to be what I originally intended. –  Paulpro Oct 12 '11 at 0:40
    
Thank you! I haven't followed through the math to see how it works quite yet (I will later when I'm home), but I used the formulae you gave me an got $\sqrt(16/6)$ which is a great upper bound! –  Paulpro Oct 12 '11 at 23:38

One has to look at this problem from a global perspective. If $(0,\pm1)$ are the endpoints of the chord and $(x,0)$, $\ -\infty<x<\infty$, is the center of the circle $\gamma$ through these two points then the right side arc of $\gamma$ has length $$f(x):=\sqrt{1+x^2}\bigl(\pi+2\arctan(x)\bigr)\ .$$ We have to solve the equation $f(x)=\ell$ for a given $\ell>2$. Graphing $f$ one sees that $f$ is monotonically increasing and, what is important, convex on all of ${\mathbb R}$; furthermore $f(0)=\pi$. This suggests the following procedure:

If $\ell<\pi$ then start Newton-Raphson with $x_0:=0$, and you will get a monotonically decreasing sequence $(x_n)_{n\geq0}$ converging to the unique solution $\xi<0$. The radius $\rho$ of the circle $\gamma$ is then given by $\rho=\sqrt{1+\xi^2}$. But note that if $\ell$ is only slightly larger than $2$ then neither $\xi$ nor $\rho$ are "well defined" numerically, and maybe subtler methods are necessary in this case.

If $\ell>\pi$ then start Newton-Raphson with $x_0:=0$ as well. The next point $x_1$ will then be too far to the right, but from then on the $x_n$ will again decrease to the unique solution $\xi>0$.

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