Foxmaths! 2.0

January 30, 2009

Pi, Geometric and Recursive

Filed under: Maths — Fox @ 3:41 am

Consider the following picture.

piagram

We have a circle, we’ll say radius of 1. A is the center of the circle. So, given an angle DAB, we make the line DB of length x. Bisect the angle, intersecting the circle at point C, and then ‘pop’ x out to form two isosceles triangles, of base length x’.

So, quick geometric question – what is x’ in terms of x?

A few applications of Pythagoras yields,

AE = \sqrt{1 - (x/2)^2}

AE + CE = 1

(x/2)^2 + CE^2 = (x')^2

All this combines to yield

x' = \sqrt{ 2 - 2 \sqrt{1 - \frac{x^2}{4}  } }

Or, better

x' = \sqrt{ 2 - \sqrt{4 - x^2} }

This is a simple enough result. But we can do something neat with it. Imagine you start with a simple polygon embedded in the circle, a square of side length x, for example. When you ‘pop’ the sides out, you get an octagon, 8 sides, each of length x’. But the perimeter of the octagon is closer to laying along the circle than the perimeter of the square. The circumference of the circle is 2*\pi , so the perimeters of these successive polygons will converge on 2*\pi . This is the basic idea behind things like Archimedes method for calculating \pi . And the formula we just derived allows us to calculate the successive side lengths.

Starting with the inscribed square, it has side length \sqrt{2} . So define the sequence,

x_0 = \sqrt{2}

x_{n+1} = \sqrt{ 2 - \sqrt{4 - {x_n}^2}  }

Then, how many sides does each polygon have? Starting at 4, the number of sides doubles each time. So on the n-th step, there are 4*2^n many sides. Letting A_n be the total perimeter, we have,

A_n = 4*2^n*x_n

And A_n converges to the circle, or 2\pi ! Of course, we might as well divide A by 2, so it converges to just \pi . Thus,

B_n = 2^{n+1}*x_n

Now, I think this is really pretty neat, because if you look at the structure of successive values of x, it looks really very nice.

x_0 = \sqrt{2}

x_1 = \sqrt{2 - \sqrt{2}}

x_2 = \sqrt{2 - \sqrt{2 + \sqrt{2}}}

x_3 = \sqrt{2 - \sqrt{2 + \sqrt{2 + \sqrt{2}}}}

Etc. Which is kind of neat. Multiply that by the appropriate power of 2, and you get a good approximation of \pi .

Of course, it has its drawbacks. It converges very slowly – the 10th term of the sequence only gives you 6 correct digits, the 20th term giving only 12. But the structure of this sequence I just find fascinating and beautiful.

Looking at it from a more generalized point of view, suppose the angle traced out by DAB is \theta , for some angle \theta . Then, using basic trig definitions, x = 2*Sin(\theta/2) . Keeping just that triangle, DAB, and popping out sides again and again, the result is that you successively approximate that arc of the circle. But the length of that arc, given that we defined the angle to be \theta , and the radius is 1, is also given by \theta

So, defining the related sequence, for some initial length x,

x_0 = x

x_{n+1} = \sqrt{ 2 - \sqrt{4 - {x_n}^2}  }

A_n = 2^n*x_n

A_n approximates the length of that arc, and therefore converges to \theta .

This seems entirely uninteresting, except for the fact that we can invert everything and get \theta in terms of x, as

\theta = 2*ArcSin(\frac{x}{2})

This shows that A_n is actually converging to 2*ArcSin(\frac{x}{2}) . So, on the off chance that you need to calculate an ArcSin sometime, and are good with roots, there you go!

January 26, 2009

A Small Piece of Pi, and Something Deeper

Filed under: Maths — Fox @ 5:14 am

With the appropriate application of a unit circle you can derive the following recursive sequence.

A_1 = 2 \sqrt{2}

A_{n+1} = 2^{n+1}\sqrt{2 - \sqrt{4 - \frac{{A_n}^2}{4^n}}}

The resulting terms have a very nice form. For example

A_4 = 2^4 \sqrt{2 - \sqrt{2+\sqrt{2+\sqrt{2}}}}

Importantly, A_n converges to \pi as n goes to infinity.

However! I don’t really care about that. Instead, consider if you define A_1 to just be a general value x. Then A_n converges very quickly to a function that looks something like this.

pigraph

And it just so happens that evaluating the above function at 2 \sqrt{2} gives \pi .

My question is, is there a better way to express that function?

Don’t know … will ponder on this.

UPDATE: Didn’t take too long… A_n for A_1 = x converges to f(x) = 4 ArcSin(x/4) . Interesting.

January 23, 2009

Euler-Mascheroni: A Curious Identity

Filed under: Maths — Fox @ 1:26 am

There’s this curious mathematical constant, known as the Euler-Mascheroni Constant. In brief, consider the sum of the first n reciprocals, 1 + 1/2 + 1/3 + … + 1/n. As n goes to infinity, this approaches ln(n). They get very close, with just a slight offset. And that offset converges as n increases. Thus we have,

\gamma = lim_{n \rightarrow \infty} ( \sum_{k = 1}^n \frac{1}{k} - ln(n) )

Numerically, it has the value

\gamma = 0.577215664901532860606512090082...

So, I was playing around earlier on Mathematica, as is my wont, and discovered the following curious identity, which I present without proof.

\int_0^1 Sin(\frac{1}{x}) dx + \sum_{n = 1}^\infty \frac{ (-1)^n }{ (2n+1)!(2n) } = 1 - \gamma

January 21, 2009

Back in Business!

Filed under: Maths — Fox @ 3:29 am

So, back to classes once more. Once more into the breach? And many other overused metaphors as well?

But things are very good. I feel good about the classes I’m in, for the most part. Physics should be interesting, I hope (Nuclear and Particle, Electronics, and Thermal II). Math should certainly give me something to think about (Graph Theory, PDE). So! Here is to a bright future. And an excellent present.

And yet again, I find myself in that murky region of mathematics, where I have an overabundance of questions, and absolutely no answers.

Currently of interest is the following. Let’s define a sequence in the following way. For n = 1, 2, ... ,

A_n = \int_0^1 Sin(2 \pi \frac{n (n + 1)}{y + n}) dy

The problem is this. Given the above, express A_i nicely in terms of the previous terms of the sequence.

How do you feel about that? How are your thoughts?

January 12, 2009

New Semester!

Filed under: Maths — Fox @ 3:47 am

Back to classes tomorrow : ) I’m excited. Such promise in the air. I love buying school supplies. Looking at all those fresh, clean notebooks – just imagine everything that could be written them. And the pencils and pens to do that writing. Such potential! Very exciting times. There should be some interesting math on the horizon, but for the time being I will leave you with this, I find, fascinating identity. For positive x, we have

({\int_0^x e^{-s^2} ds})^2 + \int_0^1 \frac{e^{-x^2(1+t^2)}}{1+t^2} dt = \frac{\pi}{4}

How do you like that? What are your thoughts?

January 7, 2009

The Omega Constant

Filed under: Maths — Fox @ 3:51 pm

The Omega Constant has the approximate value

w = 0.56714329040978387299996866221035554975381578718651...

More to the point, it is defined, using the W-Function to be the principle solution to the equation

1 = w*e^w

It is not easily solvable or invertible – no satisfying way to isolate w at least. My question is, given the above definition of w, how best to calculate it? And, in the process, we find a nice solution to a problem that’s bothered me for years.
(more…)

January 3, 2009

False Dichotomy and Calculating the Invisible

Filed under: Maths — Tags: , , — Fox @ 8:13 am

People generally like to divide math in to pure math and applied math. Pure math is generally taken to exist in this world of abstract objects and ideas, with little basis in whatever is taken to be ‘reality’ in math. My roommate loves that kind of thing. Diving into and swimming through this world of perfect, pure thought. Among people like him, applied math is generally thought of as something dirty, and generally might better be referred to as ‘calculation’. In general, a problem applied math could simply be a specific implementation or problem. Or perhaps a calculation. Being specific, it is therefore uninteresting.

Or so the general concensus seems to be. To which I say, bogus.

To me, math is about being able to figure things out. How things work. What’s really going on at the heart of the matter. To that end, pure math is as applied as applied math – utilizing tools, genius, and creativity to reach a desired goal. Similarly, in a specific problem relegated to applied math, it could easily be that the techniques needed to solve it are as elegant, abstract, or beautiful as anything you might find in the ‘pure’ maths. The idea that math could be divided into either, or that either is more worthy than the other, is simply ridiculous. Each feeds into the other. Each relies on the other. To imagine they are separate, or to limit yourself to one, is to deprive yourself of a deep and fascinating world.

Just something that was on my mind.

But, consider the function,

f(x) = \sum_{n = 0}^\infty x^{2^n}

For x between -1 and 1, the summation converges rapidly to f(x). What fascinates me about this function is my relative inability to calculate anything about it. For example, it looks like f(1/2) should have an interesting value, but it doesn’t seem to come out to anything recognizable. And the function itself doesn’t seem to be anything specific.

Now, if you plot the function, you find it has one real zero, at approximately x = -0.6. The question I have is, given that we can’t really calculate the function itself, how can we calculate the value of this zero? The summation converges very quickly, so one possibility would be to approximate the zero by calculating the real zero of partial sums. But, for high order partial sums, the best that can be done, in calculating the real zero, is just another approximation, using something like Newton’s Method. How to be sure that this approximation of an approximation is really converging on what we’re after? Is there anything better?

January 1, 2009

Happy New Year!

Filed under: Personal — Fox @ 9:05 am

Happy New Year, dear reader, to you and yours.

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