Take the 2-minute tour ×
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.

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 $1.567769039908$

share|improve this question
@J.M. Thank you for improving my Latex attempt :) –  Paulpro Oct 12 '11 at 0:18
add comment

2 Answers

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 $r$.

share|improve this answer
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
add comment

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 $\xi>0$.

share|improve this answer
add comment

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.