All 4 entries tagged Maths

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July 13, 2011

Circles and Triangles

Writing about web page http://blog.computationalcomplexity.org/2007/08/is-pi-defined-in-best-way.html

And then suddenly it was two weeks later.

I found an article a few weeks ago that I meant to record somewhere because it had an interesting idea that I’d not considered before. To wit:

“A circle, if you think about it, is effectively a triangle that has been wrapped around the origin”

and this is then extrapolated to higher dimensions. I think it’s a nice connection between areas of circles and triangles.

Christopher Midgley : 13 Jul 2011 10:44 | Tags: Maths | Comments (0) | Close comments | Report a problem

June 20, 2011

Geometric proof of the Leibniz series for π/4

This proof comes from A Garden of Integrals by Frank E. Burk; some additions have been made to make it more comprehensible, one diagram has been poorly copied.

We seek to prove Leibniz’s series for $\frac{\pi}{4}$ , that is:
$\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}-\frac{1}{3}+\cdots$

Consider the quarter unit circle centred at (1,0). It has area $\frac{\pi}{4}$ . Determine the area of the sector OD by splitting it into infinitesimal triangles OAB and summing.

We observe that, from the diagram, the area of $\triangle OAB$ is roughly equal to the area of $\triangle OAB'$ , with B’ the intersection of the perpendicular to the x-axis through B and the tangent to the circle at A. Taking AB’ as the base and OC as the height, we observe $\triangle OAB\approx\frac{1}{2}AB\times OC$ .

By similar triangles, $\frac{AB'}{dx}=\frac{z}{OC}$ , so $\triangle OAB\approx\frac{1}{2}z dx$ . Observe also:
$x=1-\cos\theta = 2\sin^2(\frac{\theta}{2})$ and $z=\tan(\frac{\theta}{2})$ . Now, observe:
$1+\tan^2(\frac{\theta}{2})=\sec^2(\frac{\theta}{2})\\ 1+z^2=\frac{1}{\cos^2(\frac{\theta}{2})}\\ \frac{1}{1+z^2}=\cos^2(\frac{\theta}{2})\\ \therefore x=\frac{2z^2}{1+z^2}$
Alternatively (longer but easier to see), note that:
$z^2 =\tan^2(\frac{\theta}{2})=\frac{\sin^2(\frac{\theta}{2})}{\cos^2(\frac{\theta}{2})} = \frac{\frac{1}{2}x^2}{1-\frac{1}{2}x^2}= \frac{x}{2-x}$
and rearrange to get the previous result. Also note $xz=\int z\mathrm{d}x+\int x\mathrm{d}z$ (by integration by parts for Riemann-Stieltjes integrals, say)

Hence the area of the quarter circle is:
$\int_0^1\frac{1}{2}z\mathrm{d}x =\frac{1}{2}\left[xz|_0^1-\int_0^1x\mathrm{d}z\right] = \frac{1}{2}\left[1-\int_0^1\frac{2z^2}{1+z^2}\mathrm{d}z\right] = \frac{1}{2} - \int_0^1z^2(1-z^2+z^4-\cdots)\mathrm{d}z =\frac{1}{2} - \frac{1}{3} + \frac{1}{5} + \cdots$
by long division followed by termwise integration. The left integral can be evaluated by substitution or by considering the geometric meaning – it’s the area of the sector which is the quarter circle subtract the right isosceles triangle with two sides equal to 1. Either way, the answer is $\frac{\pi}{4}-\frac{1}{2}$ .
Therefore, by adding $\frac{1}{2}$ to both sides, we obtain Leibniz’s series:
$\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\ldots$

Christopher Midgley : 20 Jun 2011 13:07 | Tags: Maths | Comments (0) | Close comments | Report a problem

June 19, 2011

Trigonometric sum–to–product identities, and the reverse

The first time I encountered these identities was when my tutor told me they’d probably be on the Foundations Basics exam (they were), as they weren’t on the core MEI syllabus. As such, I haven’t internalised them at all – I haven’t even memorised them – for the exam, I memorised the general structure and how to derive them. Fortunately, they haven’t come up again.

I’ll leave a quick, single example derived. Begin with the angle addition formulae:
$\sin(A+B)=\sin A\cos B+\cos A\sin B\\\sin(A-B)=\sin A\cos B-\cos A\sin B$
Sum them.
$\sin(A+B)+\sin(A-B)=2\sin A\cos B\\ \sin A\cos B=\frac{1}{2}\left[\sin(A+B)+\sin(A-B)\right]\\ \text{Let } A=\frac{C+D}{2}\text{, }B=\frac{C-D}{2}\\ \sin(C)+\sin(D)=2\sin\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right)$

Christopher Midgley : 19 Jun 2011 16:38 | Tags: Maths | Comments (0) | Close comments | Report a problem

June 16, 2011

The Triangle Inequality

Before I came to university, I had no idea what the triangle inequality was – I’d never heard of anything specifically referred to as “the triangle inequality” and, as it turned out, had never run across the actual equation, either. Wikipedia only mentions it halfway down the page, in a page containing many other inequalities it could be. I’d heard it referred to before, and found myself somewhat confused. But here, it’s almost always this one:

$|x+y|\leq|x|+|y|$
Rarely it’s a natural extension of this, such as $|x-y|\geq\left||x|-|y|\right|$ .

This was first introduced to me in the first Analysis lecture. It’s then been used only in Analysis that I can remember, but it seems like it should be useful elsewhere, as well. It finds use in showing a strictly contracting sequence is Cauchy, as well as in other induction proofs.

The proof for the first part is quite simple:
$|x+y|^2=|x|^2\pm2|x||y|+|y|^2\leq|x|^2+2|x||y|+|y|^2=(|x|+|y|)^2$
There was also a different, neat proof I found online:
We know $-|a|\leq a\leq|a|$ and similarly for b. Thus, adding the two gives:
$-(|a|+|b|)\leq a+b\leq|a|+|b|$ from which the result follows by the definition of modulus.

For the second half, apply the inequality to $|a|=|a-b+b|\leq|a-b|+|b|$ . So $|a|-|b|\leq|a-b|$ . As this is symmetric in a and b, the general inequality follows.