## Thursday, 15 December 2011

### Infinite Circles Problem

I recently encountered a mathematical problem from NRich Maths about inscribed circles in an equilateral triangle and it really intrigued and after a little bit of intense thought I managed to figure it out.

For those of who wish to know what the problem is without clicking the link:
A circle of radius 1 cm is inscribed in an equilateral triangle. A smaller circle is inscribed at each vertex touching the first circle and tangent to the two 'containing' sides of the triangle. This process is continued ad infinitum...

What is the sum of the circumferences of all the circles?
What is the sum of their areas?
Adding all the circumferences or adding all the areas, which sum grows faster?
Now this might not be immediately obvious as the best way to approach this problem, so we need to think about what we know, what we need to know and the best way to approach this.

What We Know:

1. The radius of the largest circle is 1cm.
2. All of the triangles angles are 60° as the triangle is an equilateral.
3. The area of the first circle is π, the circumference is 2π.
What We Need To Know:
1. The ratio of the radii from each circle to the next.
2. The height of the triangle.
3. The area of all the circles.
4. The circumference of all the circles.
The height of the triangle may seem like a bit of a strange necessity, but if you know the diameter of the main circle (2cm) then it helps to know what the sum of the diameters of all of the circles will be (height-2).

Now if the the radius were arranged so it was at a right angle to the triangle and a line was drawn from the centre of the largest circle the corner of the side the radius touches the angle would be half of the original angle which is 60° so the new angle is 30°. This is hard to picture but an angle will help that.
Now we have two angles and one side, so we can use the Sin rule to find the size of the line from the centre of the circle to the corner of the triangle.
This means that the radius of the largest circle plus the diameters of all of the other circles is 2. So the height of the triangle 2 plus the radius of the larger circle, which equals 3. It also means that the sum of the diameters of all of the other circles equals 1 too.

Now we can begin to actually tackle the problem of the sum of the circumferences of all the circles. We already know that the circumference of the first circle is 2π, if the sum of all the diameters of the other circles in one line is 1 we can see that the area is then π, but we still have two other sets of circles. So we have the total circumference of 2π+π+π+π which means the total circumference is 5π. Problem one solved.

The second problem is slightly more awkward as the radius is not as easy to find and although the way I am about to explain does work in may not be the most efficient, but it does work and it utilises some very nice Core 2 techniques.

As you may have noticed, there will be an infinite number of circles going into any of the corners (this is caused by the curved shape of a circle against the straight side of the triangle). If we exclude the large circle then the sum of the diameters of into one corner is 1.

The fact that the triangles always get smaller, means that the rate at which they 'increase' is less than 1, we will call this ratio 1/n, the radius of the second triangle will also be 1/n because of the fact that the first radius is 1.

We know that the sum of the diameters equals 1, which means that the sum of 2*radius is also equal to 1. We also know the first term of this series (1/n), the ratio of the series (1/n) and the sum of the series (1). As our ratio is less than 0 we can use the formula covered in C2 for that:
Using that we can rearrange to find what n equals and thus find the ratio. I have included the original equilateral triangle image along with some labelling to help to explain my notation.
So we have that the ratio from radius to radius is 1/3, so to find the sum of the area of all the triangles we must use the sum of an infinite series again. Given that the first term is π (from πr^2 and r = 1), the ratio is r^2, which gives 1/9. We have three of the series so we will times the sum of these by 3, but then we have included the largest triangle three times, so we must subtract this two times (-2π).
This means that the area of all the circles is 11π/8, problem two solved.

The last problem is considerably easy to handle, it simply asks which sum grows faster, this is the one that has a larger ratio. Well the ratio of the area is 1/9, whereas the circumference is 2/3 (r is 1/3, but we want twice this). So this means the circumference increases faster.

This problem really is a lovely one, it combines some relatively simple maths in an advanced form, pieces them all together and leaves you to solve the puzzle. Maths is fun. Maths is really, really fun!

I realise I may have explained fair chunks of this poorly, it is very difficult to convey what is happening and without being in front of you. So if you are left with any questions as to what I have done, or why, simply leave a comment and I will explain or email me at lewis.mead@eloquentmath.com for more information.

Also to let you know, I will be completing a Core 2 revision guide pretty soon (give me a week or so), so keep checking back here for updates on that.

## Wednesday, 14 December 2011

### Core 2 Numerical Integration: Trapezium Rule

Taken from my upcoming Core 2 Revision Guide. This is a brief preview and an explanation on one of the methods of estimating a definite integral. Others include the mid-ordinate rule, Newton-Cotes formula and the Simpsons rule. However none of these need to be dealt with until later units.

Sometimes you may be given an equation that seems impossible to integrate given our current known techniques, and that is because they are. So we need a method of determining the area beneath a curve at given limits. Which is actually not too difficult, if we take shapes that are sort of like the curve, find the area of all of them and then add them up we will get roughly the area beneath the curve at those points.

As you may have guessed by now the area will be estimated using trapeziums. The reason for this is that trapeziums naturally slope and can emulate the shape of a curve (given enough of them). The best way to explain how to do this is explain this is to use an example that we can integrate, which I will do as y = x2.

So we have to start by knowing what our limits are (in our case they are at 1 and 4), we also need to know what the height of each of our trapeziums are (the horizontal bit) and what the size of each of the two parallel lines, in our case these are at the points when x = 1, x = 2, x = 3 and x = 4

Once we know each of these we can then find the area of each of the trapeziums, add them together and voila we have an estimate for the area beneath the curve. If you don't remember the area of a trapezium is 0.5(a+b)h where a and b are the parallel sides and h is the height. So the area of each of the trapezium respectively is 0.5*(1+4)*1, 0.5*(4+9)*1 and 0.5*(9+16)*1 which is 2.5, 6.5 and 12.5 so if we add these together (which is 21.5) we get the rough area beneath the graph of y = x2.

If we look at how we have found the area of our three trapeziums [0.5*(1+4)*1 + 0.5*(4+9)*1 + 0.5*(9+16)*1] we can see that there are two things that multiply by each other in each trapezium (0.5 and 1) so if we take them out we get: 0.5*1*(1+4+4+9+9+16), this can be further simplified as there are two 4's and two 9's so we get 0.5*1*[(1+16)+2(4+9)] which is were we derive the formula for the rough area beneath the curve at given limits.

Words you need to know is that an ordinate is the y-coordinate at the values of x we have, so for example when x = 1 the ordinate is also 1, when x = 2 the ordinate is 4, etc. And a strip is a trapezium, so if we have 4 strips we have 4 trapezium, also you should note that if there are n strips there will be n+1 ordinates.

So the word equation for an estimate of the definite integration between limits a and b is:

The formula for this (that you will be given in the exam) is:
Just so you know y0 is the first ordinate, y1 is the second and yn is the last.

This is particularly useful when you need to find the integral of something that our current rules for calculus can not give an answer for it, with √(2x), is an example of where the trapezium rule is useful. In fact using our formula I will solve this between the limits of 3 and 1 with 4 strips.

You will need to know how to make our estimate more accurate, to do this you simply use more strips as the more there are the closer to trapeziums will be to making the actual shape.

If anything I have wrote does not appear to make sense please comment and I will do my best to explain.

## Sunday, 11 December 2011

### Do numbers exist?

This seems like a bit of an odd topic for a maths fanatic to discuss, and really it isn't very mathematical to even think about it, as you will you soon find out it is more philosophical.

You'd think it is obvious that the answer is "yes, they do exist", but it really isn't that simple. The fundamental idea of numbers started with counting a number of items, animals, children, etc. and from there it spiralled off. In fact the definition of a number is: "an arithmetical value, expressed as a word or symbol representing a particular quantity".

So numbers 'exist' to serve a purpose for counting, arithmetic and calculations. So if we have 'two' dogs, do we count the number of hairs they have as being them? Do we count the number of bugs on them? Do we need to count the number of atoms in the dog towards the 'two'? So is it really 'two' or is it some uncountable number of atoms? Or quarks? But just because we can not clearly state what two is in real life, it does not mean they do not exist.

Numbers do make interpreting what is happening an awful lot easier, but that doesn't mean they 'exist'. They easily could have been created by man to help with problems, the problem arises when you need to define what 'existing' is. To 'exist' you need to have reality or being, do numbers really have this? They are not a physical, touchable thing and could anyone really argue the case of a number having a reality to it?

You could then argue that nothing really, truly exists, as what is reality? But that is a completely different tangent for a future post. I mean, do thoughts really exist? Do words that are spoke? But anyway, I digress...

Numbers can be thought of as tools developed by us a civilization to help understand the world we live in. But if there was a race of intelligent aliens there is almost no doubt that they would have a form of maths, they may work in a different base (in fact the Aztec's worked in base 6, opposed to our base 10), or some of their basic principles may be different (a negative times a negative could still be a negative, for example). But there is almost no doubt that basic principles of maths will be there. And once they are in place more advanced concepts start to develop like how are these numbers distributed? What about a number between 1 and 2? How do we add numbers? How do we multiply numbers? And so on.

So whether they are invented or they exist, they are a necessity. It takes away the "what if?" factor from so many elements and provides concrete and quantifies otherwise incomprehensible things and data. Numbers are so useful for everything, so really it doesn't matter if they exist or not because what they accomplish is real, the data, the facts and the information they provide about the world we live in is real, they have a real impact.

So rather than getting bogged down in the semantics of whether numbers 'exist' or not should not detract from the joy and beauty of mathematics. Maths is fun, do not let your philosophical stand point alter any of that.

## Wednesday, 7 December 2011

### Riemann Hypothesis

Now I am definitely not an expert in this field, and in fact even the experts aren't really experts in the conventional sense. No one is an expert on it in the conventional sense, it is still unsolved. Over 150 years old and it still remains unsolved not for the lack of trying! In fact it is so important to mathematicians that the Clay Mathematics Institute has put a $1,000,000 bounty on its head (that is, you get$1,000,000 if you manage to solve it).

But what actually is the Riemann Hypothesis? It is a conjecture about the location of the non-trivial zeros of the Riemann Zeta function, it states that all the zeros should lie on the critical strip 0.5+it. "Oh yeah!", I hear you cry, now you get it, obviously. I will explain what this means properly later on in this post. But first I will state what it means. If true it implies a lot of things about the distribution of prime numbers, and as you may or may not know they are very irregular and very difficult to find as the numbers get very, very big.

To track back to my earlier point, what is the Riemann Zeta Function ( it is denoted as ζ(s), ζ being the Greek lower case from which z was derived)?
 Where s is an imaginary number, a+ib.
This requires that you understand sequences and seriesimaginary numbers and imaginary exponents. The real intrigue of this comes from the fact that it can be represented by Euler's product.
As you may or may not notice this is comprised of the prime numbers, this means that there is a sort of subliminal link between the natural numbers and the prime numbers. This showed that the prime numbers were not just positioned randomly and are not merely the building blocks to numbers but there is an actual link between them and the natural numbers.

The Riemann Zeta Function on its face doesn't look too difficult, I mean it is just an infinite sequence, even with a complex power you'd expect this to be possible and even pretty easy. But that is not the case at all, part of the reason is how sporadic complex exponents can be, and although it is not too difficult to find solutions (using a high powered computer thousands can be found each hour) it is incredibly, incredibly hard to find a proof for all the solutions.
 The plot of the Riemann Zeta Function, the red line is the real part, the blue part is the imaginary part.
You can see, this function seems to have little to no consistency to it, but a fair amount is known about the function. A lot of the zeros do actually satisfy the hypothesis, over 10 trillion of them in fact. And you'd think that is a proof alone, but as it often involves an iterated log (a log of a log, log(log(x)) and this increases very, very, very slowly in fact log(log(10,000,000,000)) = 1, so 10 trillion really isn't anything. If it is still holding true for log(log(x))>40 there may be a greater unanimous opinion on the truth of the hypothesis.

Every mathematician worth his salt has had an encounter with the Riemann Hypothesis and it has withheld every single attempt thus far. The maths used to try and tackle the problem is so complex that entirely new branches of mathematics have been created to deal with it, this maths to laymen has literally nothing, at all, to do with the prime numbers. It is so complex and far away from the problem that it almost boggles mathematicians minds, but it consumes them, it is their passion and life.

Prime numbers are the passion for many and the Riemann Hypothesis is merely an extension of that, and hopefully it will be solved in my life time.

If you have caught the prime number bug I suggest you read the excellent book by Karl Sabbagh called Dr Riemann's Zeros.

## Sunday, 4 December 2011

### A-Level Maths Revision: AQA Core 1

Now this took an awful lot of work, like so much, it is 23 pages long and I really hope that it will help you a lot. Tell your friends about it and just spread the word. There are over 50 questions in the revision guide and I meticulously detail over every aspect of Core 1. I use both graphics and words to explain all terminologies and chapters.
If you need any help on anything in this revision guide or anything else comment here or email me at lewis.mead@eloquentmath.com.

Current Corrections:
1.) d.) (-16-11 sqrt(2))/14
"knew" incorrectly in place of "new"
General formatting errors

## Wednesday, 30 November 2011

### AS Maths Core 1: Polynomials

Poly, as you may or may not know, means many, and nomial as you likely will not know means terms. Put this together and we get many terms, and that really is the definition of a polynomial. This topic is reasonably large, but none of it is too difficult and it is easy marks to gain in the exam as long as you are careful and know your stuff!

Discriminant

The discriminant of a quadratic equation is something that can tell you how many real roots a quadratic equation will have. If we look at the quadratic equation we know that that you can only find the square root of a positive integer (and zero). The discriminant is what is being square rooted in the quadratic equation (b2-4ac).

If the discriminant is greater than 0 then the quadratic would have two distinct real roots (it crosses the x-axis twice), if the discriminant is equal to 0 then there is one repeated real root (it just touches the x-axis once, at its vertex) and if the discriminant is less than 0 then there are no real roots (it will not touch the x-axis).

If b2-4ac > 0 then the quadratic has two distinct real roots.
If b2-4ac = 0 then the quadratic has one repeated real root.
If b2-4ac < 0 then the quadratic has no real roots.

Now, quadratic inequalities can be pretty tricky if you never spent enough time learning them, but if you do spend enough time doing you will be able to get full marks on them every time. First of all the general form of a quadratic inequality is: ax2+bx+c > 0 or ax2+bx+c < 0, note that the signs could also be or , but this makes no difference in the method but you just need to ensure you use the correct sign in your final answer.

The first thing you need to do is solve the equation equal to zero to get the points that the quadratic crosses the x-axis. Once you have this you need to sketch the quadratic and the shade the side of the quadratic where it is either <0 or 0<. If the quadratic has the sign like this:  ax2+bx+c > 0, you will need to shade only about the x-axis (as that is where y>0), if the inequality is the other way,  ax2+bx+c < 0, then you need to shade below the x-axis (as that is where y<0). It is then easy to see what values the inequality will satisfy.

Example:
Find the range of values that satisfy the inequality,  x2+6x+8 < 0
We’ll begin by solving the equation equal to 0.
x2+6x+8, this is then factorised to (x+2)(x+4) = 0, so the quadratic crosses the x-axis at (-2,0) and (-4,0).
Plotting this graph and then shading where the graph is less than 0.

This then means that the inequalities range of results is when x>-4 and x<-2, this can be wrote as a continuous set of values, -4<x<-2.

Note that if the inequality is of the form ax2+bx+c < 0, it will always be a continuous set of values. If it is of the form ax2+bx+c > 0, it will not be a continuous set of values, ever.

Complete the Square
This method of solving quadratics is itself not exactly too efficient and if the quadratic is an awkward one it is often much simpler to use the quadratic formula (which was derived from completing the square) to solve it. But it does have other very useful properties, most of which you will be required to learn.

I will begin by talking you through the steps of how you complete the square of anything.
ax2+bx+c = 0
If we divide the expression through by ‘a’ we get:

x2+(b/a)x+(c/a) = 0
Now if we want to get a perfect square (hence complete the square) we must half the second term, square it, add it on and take it away:

x2+(b/a)x+(b/2a)2-(b/2a)2+(c/a) = 0
Which we can then simplify into the following form:
(x+(b/2a))2-(b/2a)2+(c/a) = 0

Now if you are observant you will see that what is in the bracket with the x is the original number we halved. This will always be the case. Also you should note that -(b/2a)2+(c/a) will be combined as they will both be real numbers.

Now we have our general form of a quadratic in a completed square form, (x+(b/2a))2-(b/2a)2+(c/a) = 0, what can we actually do with it? We can use it to find the vertex of a quadratic equation, going from our example the vertex would be (-(b/2a), -(b/2a)2+(c/a)) . Now this looks pretty confusing but when you apply it to an actual equation it is incredibly easy.

Example:
Find the vertex of 2x²+8x+4=0.
We begin by dividing through by 2 to get, x
²+4x+2=0
Halving 4 gives us 2, so our new equation must be:
(x+2)² - 2² + 2 = 0
And on simplifying we get: (x+2)² - 2 = 0
So the vertex must be at (-2, -2)

And it is as easy as that! The only other thing you must know for completing the square is how you change x² into x²+bx+c, and this really is just an extension of completing the square. The vertex of x² is (0,0), so if the vertex of a quadratic is say, (a, b) then the graph of x² is TRANSLATED by

Using our worked example again:
Find the transformation that maps 2x² onto 2x²+8x+4=0
Using our vertex from the previous example, (-2,-2)
This means that x
² is translated by

Remainder Theorem
This again, is very, very easy to do. You just have to watch out for very small errors when doing your calculations and that you do not make an error straight away. To check the remainder when f(x) is divided by (x+a) we get the notation of f(-a), so wherever there is an x in the expression we put a ‘-a’. Whatever this then equals to is the remainder when f(x) is divided by (x+a). This really is all there is to it.

Example:
Find the remainder when f(x) = x³+3x²-6x+4 is divided by (x+2).
f(-2) = (-2)³+3(-2)²-6(-2)+4
f(-2) = -8+12+12+4
f(-2) = 20
Therefore when
x³+3x²-6x+4 is divided by (x+2) the remainder is 20.

Factor Theorem
The Factor Theorem is just a special case of the remainder theorem; this is when the remainder of that is equal to 0. For example 6/2 has a remainder of 0 therefore 2 is a factor of 6 (obviously). So (x+a) is a factor of f(x) when f(-a) = 0.

Example:
Verify that (x+2) is a factor of f(x) = x³+4x²+x-6.
f(-2) = (-2)³+4(-2)²+(-2)-6
f(-2) = -8+16-2-6 = 0
f(-2) = 0, therefore (x+2) is a factor of f(x)

Algebraic Long Division
Often you will have one factor of a cubic equation and need to find the other factors. Now there are other ways to do this (factor theorem, solving by inspection, etc.) but personally I find that algebraic long division is the most efficient and accurate way of finding the other factors.

You will only be dealing with factors of the form (x+a), it will never be (2x+a), etc. So it never gets too complicated. I will explain what you need to do with an example.
Now, this is something that might be quite hard to get your head around at first, and you really need to just practice as much as you can. You also need to know that often the quadratic formed can be further factorised. This is true in our example, so representing x³-2x²-5x+6 as a product of three linear factors is: (x+2)(x-1)(x-3). If you have an equation where there is no say 'x' term or 'x²' term you must when dividing it make it 0x² or 0x.

Often you will be required to sketch the graph of a quadratic or a cubic once you have found the roots of the polynomial. There are a few things you have to make sure you do, if you do them you will always get the marks for the quadratic.
• Always mark the points where the polynomial crosses the x-axis and the y-axis.
•  Draw the axes with a ruler.
• Draw the polynomial smoothly; do not draw it bit by bit.
• If you need to draw a line on the graph as well, mark all points of intersection (with graph and axes).
• Make sure you use a pencil in case of mistakes.
• Draw the graph a reasonable size, (I’d go for at least a third of a page if not more).
• Make sure that the shape is roughly correct and resembles the correct polynomial.
• Get the minimum point in roughly the right area.

My advice would be roughly work out where each point is going to be, draw it with a fluid action and then mark the points of intersection on.

And that is it for my guide on polynomials in Core 1, I have taken this from my upcoming eBook revision guide for Core 1, the eBook will also include questions, answers and far more detail. Keep checking back for more.

## Saturday, 26 November 2011

### Euler's Formula

As I mentioned in my previous post on imaginary exponents (read that here), I would explain why in fact Euler's beautiful and immense formula actually works. First of all, I'll actually say what the formula is:
eix = Cos(x)+iSin(x)

But this is not exactly intuitive why this is the case. The answer lies in a brilliant piece of maths devised by Brook Taylor, it is called the Taylor series. You can represent any function as the sum of an infinite series of polynomials. This is incredibly useful when it comes to Sin(x), Cos(x) and ex, and when you delve into the Taylor series of these you can begin to see where eix = Cos(x)+iSin(x) comes from:

Now, it might not be immediately obvious how those are related, but all the right terms are there we just need to piece it together. If we begin to manipulate the Taylor series for ex and we replace 'x' with 'ix' we will begin to see our proof. For the purposes of this I must mention how ix works i1 = i, i2 = -1, i3 = -i, i4 = 1, i5 = i,... and it continues in this fashion for all integer powers of i.

Now we have that e^ix = (1 - x^2/2!+ x^4/4!- ((x^6)/6!)+ ...) + i (x - x^3/3!+ (x^5/5!)- x^7/7!+ ...), and these look awfully familiar. In fact if I refer you back to the Taylor series' of Cos(x) and Sin(x):

You can see that these are apparent in what we have now discovered  eix to equal. This then means that:

And hence we have our proof of Euler's formula and that eix = Cos(x)+iSin(x).

You may have also seen Euler's formula in action as Euler's identity which is often described as the most beautifully profound equation in maths. Euler's identity is eiπ + 1 = 0, and it is so beautiful because it incorporates the five most important numbers in maths: π, e, 1, 0 and i.

Why does eiπ + 1 = 0? Well if we look at our proof of Euler's formula, eix = Cos(x)+iSin(x) and we input π we get: eiπ = Cos(π)+iSin(π). Sin(π) = 0, Cos(π) = -1. Therefore  eiπ = -1, so  eiπ + 1 = 0.

### Imaginary Exponents: x^i

To learn how we do this I first need to explain a very, very useful mathematical formula. It is called "Euler's formula", and this formula gives us a way to find the value of the imaginary exponential function (e^ix) using methods that we already have well defined and are easy to deal with. Euler's formula is: e^ix = Cos(x)+iSin(x), where x is an angle in radians. I may do a post on the actual maths behind why this is the case, but for the purpose of this post it has no relevance.

Now, we can find e^ix, but what use is this if we want to find 2^i, i^i or just anything raised to the power of i, let's call this a^i. So, we want to find a^x where x = i, so we need to try and remember a^x as something involving e raised to the power of something. So this means we have, a^x = e^y.
This then means that a^x = e^[xln(a)]. So we now have a^x in a form involving e raised to a power. So now we can input when x = i. Now by simply placing this into the equation we get, a^i = e^[i*ln(a)]. We can then turn this into something we can solve using Euler's formula, e^[i*ln(a)] = Cos[ln(a)]+iSin[ln(a)].
So now to actually input some numbers to this. Let's say I want to find 2^i, so from our previously defined formula we now have that: e^[i*ln(2)] = Cos[ln(2)]+iSin[ln(2)]. Using our calculators we will find that this is roughly Cos(0.693147)+iSin(0.693147), which then equates to roughly 0.76924+0.63896i. So 2^i ≈ 0.76924+0.63896i

As you can see, this is a complex number and it will be a lot of the time when we deal with imaginary exponents, but (as you may have thought) there are times when the solution to a^i will be a real answer. This is when iSin(x) = 0, and this will happen at Sin(x) = 0 and if you know your Sine curves you will know that this is at Sin(kπ), where k is any integer. Using our formula derived from Euler's, e^[i*ln(a)] = Cos[ln(a)]+iSin[ln(a)], we can see that if a^i is a real number, ln(a) = kπ. If we make both sides to the power of e, we can clear our logarithm to get: a = e^kπ. This should also then mean that the solution that is real (where a = e^kπ) should be equivalent to Cos(kπ).

Therefore if this is correct, then (e^3π)^i should produce a real value. (e^3π)^i = Cos[ln(e^3π)]+iSin[ln(e^3π)]. And when you do work this out, low and behold you get the answer of -1 (which incidentally is the same as Cos(3π)).

## Thursday, 24 November 2011

### Maths and the Real World: Bayes' Theorem

Bayes' Theorem is one of the most practical theorems to apply to everyday life and if used correctly it can be an indispensable decision making tool. In a nutshell what the Bayes' Theorem does is measure the confidence that something is true. It takes the uncertainty before and after observing the modelled system and links the two.

We shall use an example to help explain what the Bayes' Theorem is and how it works. Let's consider the example that you have had a persistent headache for a week now and you're not certain what the cause it. But you do believe that it is caused either by stress (hypothesis A) or by having caffeine (hypothesis B).

So to test if stress is the key to the chronic headaches you have a day of relaxation whilst you've got a headache and have had coffee on the same day. By the end your headache has gone, so this can be considered as evidence. This should have some relation to how much more likely is A than B. But how strong is this evidence exactly? And how do we show which hypothesis it supports? Bayes' Theorem tells us that these answers lie in what is called the Bayes' Factor.

The Bayes' Factor is the question: "How much more likely would it be for this evidence to occur if A were true than if B were true?". This question must lead to one of three conclusions:

1. The evidence would be more likely to occur if A were true than if B were true. This means that the evidence supports A rather than B.
2. The evidence would be just as likely to occur if A or B were true. This means that the evidence has no real weight to whether A or B is more likely to be correct. That means that the "evidence" is not actually evidence at all.
3. The evidence would be more likely to occur if B were true than if A were true. This means that the evidence supports B rather than A.
In our example of chronic headaches the Bayes' factor becomes: "How much more likely would it be for the headache to disappear after having a day of relaxation if stress were the cause compared to if caffeine was the cause?".

Now we do do not know the precise answer to this, but we can give a rough approximation to it. A day of relaxation could have some effect at stopping a headache if caffeine was the cause, but it shouldn't have too much of an effect, no more than a 1 in 5 chance for a persistent headache. On the other hand if the factor of stress is dealt with and the headache disappears, that is a pretty good indication that stress is the key cause, so the chances that stress is the main cause is about 1 in 2.

How likely the headache would have stopped given A is 1/2. How likely the headache would have stopped given B is 1/5. Hence the Bayes' factor, how likely would it be for the headache to stop given A compared to how likely it was to stop given B, is at least (1/2) / (1/5) = 2.5.

This means that given our evidence we should now believe that A is at least 2.5 times more likely compared to B, this is compared to what we used to think. The Bayes' factor tells us how much more our new evidence should cause us to believe the likelihood of one of our hypotheses.

Now let's suppose that you already suspected that stress was twice as likely to be the main cause (as you had recently taken on more responsibility causing more stress). Now we know that the Bayes' Factor is at that A is at least 2.5 times more likely than B, but as we already believe A to be twice as likely as B we know that A is now at least 5 times more likely than B.

Bayes' Theorem is useful because it tells us the correct sort of question to ask ourselves and then it uses maths and statistics to provide us with a suitable answer and easy to understand conclusion. Bayes' Theorem can also provide an answer when looking at just one variable, you simply change B to to A' (not A).

However as humans we tend to have a very poor ability at distinguishing what is or isn't evidence. If we're expecting a particular result we're far more likely to apply whatever evidence we've got and assess it with bias.

So the important part of interpreting the evidence we now have is to always use the question "How much more likely would it be for this evidence to occur if A were true than if B were true?". In the next post I do I will be writing about the maths behind Bayes' theorem.

## Saturday, 19 November 2011

### Estimating Detectable Alien Life: Drake Equation

To begin thinking about what equation we would need to construct to show the number of potentially detectable alien civilizations we must first begin by considering the factors that will affect this number.

Before I begin explaining how we go about constructing the equation I want to first say that this equation is a slightly altered version of the Drake equation, it will yield the exact same results (comment for an explanation as to why) but I personally feel this version is more intuitive and far easier to grasp as a concept. I also need to say that although this is an equation of sorts, it does not have an implicit use that will give a correct answer, the reason for this is just we do not know enough about a lot of the variables to make them constants, so they will change depending on each interpretation.

Well let's start by thinking about what we need as the variables in the equation. The obvious first thing to consider is the number of star systems in our galaxy, we will denote this S. Now our best estimate for this is anything from 200 billion to 600 billion stars, with the increase in the power of our telescopes we get more and more accurate estimates constantly.

A lot of these star systems are simply devoid of any planets at all, they are just a star with no planets at all. So the next variable that we need to consider is the fraction of stars with planets, we will denote this as P. Now this really is a pretty massive estimate, we could never know the exact fraction of stars that have planets, but again we do have methods of checking if a star has a planet and currently it is thought that around 50% of stars have planets too.

Now we currently will have an approximation of the number of stars with planets orbiting them, the next thing we will want to do to narrow down this number to the number of detectable alien life is for what fraction of these star systems lies a planet that is capable of supporting life, moreover we want a planet that could be Earth-like, this reason is for what we know now is that life can only develop on planets that are Earth-like. However for all we now their may be incredibly intelligent gaseous beings on a distant planet that has developed a highly technical civilization, we just do not know. However, we will denote this variable as E, for Earth-like planets. To put a number onto this I will just pluck a number completely out of the air and say only 10% of planets could be capable of sustaining life.

Just because the tools are there it does not mean that it will result in their being life. This variable is the fraction of Earth-like planets that do develop life, we will denote this L. However, I think a lot of the planets that are Earth-like will evolve life in some form, there is life at the very deepest depths of the ocean, there is life where a human would be completely obliterated within seconds. For this reason I think that if the planet has the means to sustain life, it often will, I estimate it will at least half of the time, so to be conservative I will put this variable at 0.5.

Again, just because there is life it doesn't mean that it will ever become 'intelligent' enough. Life may not have evolved as well on Earth had the dinosaurs not been wiped out, they may have never developed to an intelligent enough state to communicate using radio waves, they may have just stayed as a less intelligent being. This variable will then be the fraction of life that will develop into intelligent life, we will denote this as I. Now this will be far more rare than their just being life, and in fact there is not even a rough number we can apply to this, but just for the sakes of this let's say that 1% of life will at one stage become intelligent.

The next variable we need to consider is the fraction of these civilizations that communicate via a means that we will be able to detect, we will denote this variable as C. For example humans have been around and intelligent for around thousands of years, but only for the last 80 or so would we be able to be detected, this is because of the discovery of radio waves. So before that we were, as a civilization, undetectable. Also we may discover that in a few hundred years there are far more efficient and productive methods to communicate, other alien civilizations may already be using this. So let's say again, pulling numbers completely at random, that 10% of intelligent civilizations develop a means of communication that we can detect.

The last variable that we need to consider is the fraction the average time the civilization is able to communicate takes up of the average age of star system, we will call this T. How long a civilization is able to communicate is something that although we do not know (as we still exist, just!) we can estimate. We have only been able to communicate via radio waves for 80 years, and every single one of those has been riddled with war. However, I remain optimistic that we, and all intelligent life, should be able to last about 10,000 years in the state of communication. The average age of a start system is around 10 billion years. So the calculation to find T is 10,000/10,000,000,000, which as a decimal is 0.000001.

The equation that we now have, after considering all the things we need to look for is:
Number of Alien Civilizations = Number of Stars * Fraction of stars with Planets * Fraction of Earthlike Plants * * Fraction of Planets with Life * Fraction of Intelligent Planets * Ability to Communicate * (Lifetime of Planet)/(Lifetime of Star)

This is a lot of text for maths, so to put it algebraically:
N = S*P*E*L*I*C*T

Now if we input the estimates that I designated earlier we get:
N = 200,000,000,000 * 0.5 * 0.01 * 0.5 * 0.1 * 0.1 * 0.000001

Now if we do this calculation we get that in this case N = 5. So there should be about 5 alien civilizations that are detectable from my very estimated estimates.

But the point of this equation is not to come up with a concrete number of civilizations that we must be able to communicate with right now. But this equation gives us the types of data that we should be looking for if we want to know the potential amount of aliens in the galaxy. If we know what to be looking for to know how many potentially intelligent aliens could be in our galaxy we can hone our efforts in on what data we need to look for. Also, it is pretty cool to be able to estimate how many aliens are out there and able to be detected in our own galaxy, and what happens if we change certain variables, etc.

If you fancy estimating how many detectable lives our out there using the original Drake equation, try it out for yourself at WolframAlpha.

## Wednesday, 16 November 2011

### Maths and the Real World: Linear Programming

Linear Programming may be bread and butter to you or it may be an entirely new concept. But it is one of the most applicable pieces of maths that is used in every day life by business and companies alike, this of course is minimising costs and maximising profits.

I will propose a problem to you, you are a company that sells two types of fruit drinks that consists of fruit juice and sugar syrup. Juice A consists of 0.3 litres of fruit juice and 0.5 litres of syrup and Juice B consists of 0.6 litres of juice and 0.4 litres of syrup. You have 30,000 litres of juice and 40,000 litres of syrup already in your stock. The profit for Juice A is 20p and the profit for Juice B is 30p. Given this scenario, you wish to maximise your profit.

How would you go about doing this? Well let's begin by putting the information we have in a table and go from there.

 Fruit Juice (in litres) Syrup (in litres) Profit (in pence) A 0.3 0.5 20 B 0.6 0.4 30 Total 20,000 30,000

Now, from this information we need to construct the constraints of the problem into mathematical terms. What inequality will represent the amount of fruit juice that is allowed to be used? Well it must be less than or equal to 30,000 that is clear, it also depends on how much of it is used by Juice A and Juice B, so if 0.3 of A is used each time Juice A is created, and 0.6 of B is used when Juice B is created. This then means that 0.3A+0.6B ≤ 30,000. Using the same rules we must concur that 0.5A+0.4B ≤ 40,000. Also we want to maximise the amount of profit that we make, this means that P = 20A+30B. But there are other less obvious constraints that we must consider. We can not use a negative amount of juice or syrup so A ≥ 0 and B ≥ 0.

To get an idea of what sort of values we can have we plot these inequalities onto a graph which will give us an idea of what values of A and B are actually possible to obtain.

 The blue shaded region is the answers that are within the constraints of our inequalities. This is called the region of feasibility.
So we have the region that the answers must be within, now we want to go about maximising the profit which has an equation of, P = 20A+30B. This will be the last point that the line P = 20A+30B touches on the region of feasibility, this means that what P actually equals is arbitrary as we only need to gradient of that line and it will then be moved until it touches the last point it possible can on the region of feasibility. So we will choose a number that is convenient to plot for us, I'll be using P = 600.

 The line begins to gain opacity as it moves closer to the further point on the region of feasibility. Point 'A' is the maximum point within the reason of feasibility, thereforethis is the maximum value.
We could try to read this point of the graph but it would far more accurate to solve this using where the point is met by the two equations and solve simultaneously. So we are solving the simultaneous equations of 0.3A+0.6B = 30,000 and 0.5A+0.4B = 40,000.

This then in the context means that to optimise the profit within our constraints we should make 200,000/3 litres of Juice A and 50,000/3 of Juice B. This then equates to about £6333, which is the maximum profit we can achieve from the circumstances we have been given.

This is just an example of how to use linear programming to optimise finances, but it is easily transferable to almost any situation. The only thing you may need to watch out for is if A and B are number of items that need to be sold, they must be whole numbers (obviously) so you may need to round and then check that this will still lie in the region of feasibility.

Again, I hope you find this interesting and in fact very applicable to real life. If you have any questions on anything I have done, how it works or even how I create my images, please comment and I will reply.