Huge thanks are extended to Sam who is now hosting my cgi Latex script!! Which means that all the goodness of Latex is once again working on my Blog!
Hopefully at some stage my host PlusNet will sort out this little niggle! (Apart from this problem I can’t really fault them!)

[tex]x^2 + y^2 = r^2[/tex] - Just because I can!!There are a few other odds and ends I want to implement on the site and I will continue to leave updates as these come along!

I am having issues with getting the Latex script and some other scripts to play nicely together…

So until I get that fixed the latex will just appear as the code…

Hope to get this fixed ASAP! If anyone knows anything about Apache mod_rewrites and why this may be preventing me to running cgi scripts please let me know!!

Or as a temporary fix if you happen to have some cgi enables webspace going spare that I could host this one file on, let me know!

I came across an interesting problem today.

Consider a Circle, inside the circle inscribe a triangle, inside that triangle inscribe a circle, then inside that circle inscribe a square, then another circle, then a pentagon and so on…

The question is does this sequence of shapes converge to a point, or a circle? And if it converges to a circle what is it’s radius?

Now we can start to build up our solution as outlined below

So we have now got a recursive definition for the radius of the circle, in the form:

[tex]R_{n+1} = sin(\frac{n \pi- 2 \pi}{2n})R_{n}[/tex]

If we now apply the subtraction formula for Sine:

[tex]sin(a-b) = sin(a)cos(b)-sin(b)cos(a)[/tex]

Then we obtain:

[tex]sin (\frac{n \pi- 2\pi}{2n}) = sin(\frac{\pi}{2}- \frac{\pi}{n}) = sin(\frac{\pi}{2})cos(\frac{\pi}{n})- sin(\frac{\pi}{n})cos(\frac{\pi}{2})[/tex]

But [tex]sin(\frac{\pi}{2}) = 1 and cos(\frac{\pi}{2}) = 0[/tex]

so the above simplifies to

[tex]sin(\frac{\pi}{2}- \frac{\pi}{n}) = cos(\frac{\pi}{n})[/tex]

So our recursive formula is now:

[tex]R_{n+1} = cos(\frac{\pi}{n})R_{n}[/tex]

So we now have a formula for the radius of each circle based up the radius of the last one.

So the radius of the nth Circle (i.e. the circle inscibed in the polygon of side n) can be expressed as:

[tex]R_{n} = \LARGE \displaystyle\prod_{t=3}^{n} \Large cos(\frac{\pi}{t}) R_0[/tex] where [tex]R_0[/tex] represents the initial radius.

We start at [tex]n=3[/tex] becuase this represent the first polygon - the triangle.

The next question though is does this tend to a limit?

Using Excel I checked the first 5000 terms of [tex]\LARGE \displaystyle\prod_{t=3}^{\infty} cos(\frac{\pi}{t})[/tex] and it seems to tend to 0.115038839…

Next I need to prove that the function actually tends to a non-zero limit! But that will have to wait till I have worked out how to prove it!!

Look out for the next installation in this interesting problem…

So having played with it a bit longer (as well as some helpful prods in the right direction - Thanks Jenni!) we can now make a little more progress…

The trick is to convert the infinite product into an infinite sum… To do this we need to find a why to turn multiplication into addition. What do we know that does this? The LOG function!!

So: [tex]log \Bigg[ \LARGE\prod_{n=3}^{\infty} \Large cos(\frac{\pi}{n}) \Bigg] = \LARGE \sum_{n=3}^{\infty} \Large log \Big[ cos(\frac{\pi}{n}) \Big][/tex]

So now that we have an infinite sum, what do we need to do to show that it converges to a limit?

After much struggling and racking of my brain, I suddenly remembered the Integral Test.

This tells us that if our function is:
* Continuous
* Positive
* Decreasing
Then if the integral of the function:

[tex]\Bigint_1^\infty f(x) dx[/tex]

is convergent then,

[tex]\LARGE \sum_{1}^{\infty}\Large a_i[/tex]

where [tex]a_i=f(i)[/tex] is also convergent.

So what do we need to do?

Well our function [tex]f(x) = log(cos(\frac{\pi}{n}))[/tex] is continuous but it is negative and increasing…

So let our new function [tex]g(x) =-f(x) =-log(cos(\frac{\pi}{n}))[/tex]

So [tex]g(x)[/tex] is now continuous, positive and decreasing. Which means we can now apply the Integral Test!

Now we could fudge our function sothat we evaluated it between [tex]1[/tex] and [tex]\infty[/tex], but this wouldn’t effect the result. So instead we will evaluate it between [tex]3[/tex] and [tex]\infty[/tex].

Using my new best friend Mathematica (downloaded a 15 day trial from their website) I was able to evaluate the definite integral

[tex]\Bigint_3^\infty -log(cos(\frac{\pi}{n})) dx \approx 1.76859[/tex] which is convergent!

Which by the *Integral Test* tells us our sum

[tex]\LARGE \sum_{n=3}^{\infty} \Large log \Big[ cos(\frac{\pi}{n}) \Big][/tex] is convergent to a non-zero value (all values are non-zero, so the summost be non-zero)

Hence our initial product:

[tex]\LARGE\prod_{n=3}^{\infty} \Large cos(\frac{\pi}{n})[/tex]

Must also be convergent to a non-zero value.

So we have now proved that if we start with a Circle and inscribe a triangle, then a circle, then a sqaure,…

Then this does not tend to a point. Instead it tends to a circle with a radius that is non-zero.

To evaluate this radius numerically we need to use a computer as it not possible (as far as I am aware) to express the constant in terms of any other nice constants such as [tex]\pi, e[/tex]…

Using *Mathematica* again it was possible to evaluate it more accuratley as:

0.1149420448532962

So the limit of the product which is approximated by the above number represents the ratio of the radius of the outermost circle to the radius of the circle which marks out the limit of this geometric sequence…

If you spot any mistakes or have anything to add stick it in the comments!

I was reading through September 2004 issue of MT (Mathematics Teacher - published by the ATM), and came across an article on teaching “Subtituting into Formula”

In the article Colin Foster (from our local Henry VII’s) talks about trying to find ways to make the task of subtituting in to formulas seem more interesting and relevant.

Whilst looking at the some sequnces with his Y9 group he stumbled across the idea that any finite sequnce can be expressed by a sufficiently complicated equation…

So he decided to set the students a more complicated task. He devised a formula which when you substitute in the integers 1,..,8 produces the schools phone number!!

The formula is quite complicated:

[tex]f(x) = \frac{107n^2}{5040}- \frac{241n^6}{360} + \frac{1547n^5}{180}+\frac{4157n^4}{72}+\frac{156179n^3}{720}-\frac{81017n^2}{180}[/tex][tex]+\frac{197387n}{420}- 180[/tex]
So subsituting in you get:[tex]f(1) = 7, f(2) = 6, [/tex] … , [tex] f(8) = 1[/tex]

He talk about the way in which he uses the task in the lesson by allocating pairs of students to look at substitutin in each number and then trying to build up a coherent sequnce and seeing if anyone can guess what it is… He also talks about the fact that their are inaccuracies from the claculator and how he asks students to think about why these might be the case…

I found the whole idea quite interesting and so decided to try to find the corresponding formula for Ruth’s school.

The article gave no clues as to how you might find your own formula… So with a but of hunting around and a couple of failed experiments I releasied that it is _just_ a matter of solving a system of 8 simultaneous equations. Now being the lasy person I am I decided to do this by forming a Matrix and getting excel to claculate its invervse and hence solve the system. (excel file attached at the end of the post)

So now we have a Formula, Ruth said she would have a go at using it with her Year 9 group. When she has had a go I will report back as to how well it worked.

Download Excel File Here

*Note:* This will produce the decimal co-efficents of your equation, however you will need to turn these into fractions yourself!