Intermission

OK, we have been distracted by shiny things in the form of the mandelbrot set. Now we can get back to our task in hand, working out why \(e^{i\pi}+1=0\).

We have now met our cast of characters. Let's have a look at them all lined up on the complex plane:

So you take the number at \(\pi\) and multiply it by \(i\). That's easy enough to visualise now. We look at both numbers in polar form. \(\pi\) is just \(\pi\angle 0\) and \(i\) is \(1\angle \tfrac{\pi}{2}\). We just need to multiply the absolute size (\(\pi\cdot 1\)) and we add the angles (\(0+\tfrac{\pi}{2}\)). That gets us to \(\pi\angle \tfrac{\pi}{2}\). That looks like this:

So we are now left with the job of trying to work out how to raise the number \(e\) to this new number \(i\cdot \pi\). To do this, we are going to have to work out how to multiply \(e\) to a complex number. To start with, lets go back and look at our definition of \(e\).

\[e=\lim_{n \to \infty} (1+\tfrac{1}{n})^n\]

Lets look only at the function bit:

\[(1+\tfrac{1}{n})^n\]

Let's now raise that bit to a power:

\[((1+\tfrac{1}{n})^n)^x\]

We now know that when raising something to a power and then to a power again, we can just multiply the powers. In other words the expression above really means take the bit in brackets and multiply \(n\) of them together. Now take that group of things being multiplied together and multiply \(x\) of those groups together. If you did that you would just end up with \(x\) groups of \(n\) things all multiplied together. So essentially you end up with \(n\cdot x\) numbers of the stuff in brackets being multiplied together. So we can just write this instead:

\[(1+\tfrac{1}{n})^{n\cdot x}\]

How does this help me work out \(e\) to a complex power? Well, what I know how to do is to multiply, divide and add complex numbers. I do not know how to raise a number to a complex exponent. So what I really want to do is to get that \(x\) away from the exponent. How do I do that?

Let's create a new variable \(m\). Let's define this to be \(n\cdot x\):

\[m=n\cdot x\]

Now I will multiply both sides of that equation by \(\tfrac{1}{n}\):

\[\begin{align*}
\tfrac{1}{n}\cdot m&=\tfrac{1}{n}\cdot n\cdot x\\
\tfrac{m}{n}&=\tfrac{n}{n}\cdot x\\
\tfrac{m}{n}&=1\cdot x
\end{align*}\]

We also need to multiply both sides by \(\tfrac{1}{m}\):

\[\begin{align*}
\tfrac{m}{n}\cdot \tfrac{1}{m}&=x\cdot \tfrac{1}{m}\\
\tfrac{m\cdot 1}{n\cdot m}&=\tfrac{x}{m}\\
\tfrac{m\cdot 1}{m\cdot n}&=\tfrac{x}{m}\\
\tfrac{m}{m}\cdot \tfrac{1}{n}&=\tfrac{x}{m}\\
1\cdot \tfrac{1}{n}&=\tfrac{x}{m}
\end{align*}\]

We now have two defintions to use:

\[\begin{align*}
n\cdot x&=m\\
\tfrac{1}{n}&=\tfrac{x}{m}\\
\end{align*}\]

And we just need to fit them into this equation:

\[(1+\tfrac{1}{n})^{n\cdot x}\]

We have both a \(\tfrac{1}{n}\) and a \(n\cdot x\), so let's have at it:

\[(1+\tfrac{x}{m})^{m}\]

There we go - mission accomplished. We have manged to get the \(x\) AWAY from the exponent. We can now say that:

\[e^x=\lim_{m \to \infty} (1+\tfrac{x}{m})^{m}\]

We can now see what \(e\) raised to a complex power is. First, though, let's just run through this with real powers to check it is working. If we square \(e\), then according to my calculator we should get \(7.389056099\). So let's try:

\[e^2=\lim_{m \to \infty} (1+\tfrac{2}{m})^{m}\]

If you set \(m\) to \(1000000\) you get an equation that looks like this:

\[\begin{align*}
\left (\frac{1000000}{1000000}+\frac{2}{1000000}\right )^{1000000}\\
\left (\frac{1000002}{1000000}\right )^{1000000}\\
\frac{1000002^{1000000}}{1000000^{1000000}}
\end{align*}\]

If you work out that horror you get \(7.389041321\), which agrees to four decimal places with the "real" answer. So we are on the right track! So what we can do now is to replace the \(2\) with \(i\pi\). That looks like this:

\[e^{i\pi}=\lim_{m \to \infty} (1+\tfrac{i\pi}{m})^{m}\]

Let's set \(m\) to a hundred to see how we get on. That makes our equation:

\[e^{i\pi}\approx (1+\tfrac{i\pi}{100})^{100}\]

Let's deal with the bit in brackets first. How do we do the division? It is unsurprisingly the opposite of multiplication. So instead of multiplying the absolute values together you divide the first by the second. Then you deduct the second angle from the first. Remember the number we are dividing looks like this:

In polar form (for division) that is \(\pi\angle \tfrac{\pi}{2}\). The polar form for the number doing the dividing (the one on the bottom) is \(100\angle 0\). So the maths looks like this:

\[\frac{\pi\angle \tfrac{\pi}{2}}{100\angle 0}\]
\[(\tfrac{\pi}{100})\angle (\tfrac{\pi}{2}-0)\]
\[(\tfrac{\pi}{100})\angle \tfrac{\pi}{2}\]

So it is going to be at the same angle but only a hundredth the distance away from the origin. We then want to add one onto that. That just moves the number one unit in the positive real direction. That looks like this:

We can now see the number on the complex plane. This is the thing that we are going to raise to the power one hundred. Raising to a power is all about multiplication, so we are going to want to put the number into polar form. So what is it in polar form? First the absolute value is (again using Pythagoras) the square root of one squared plus \(\left (\frac{\pi}{100}\right )^2\). So the absolute value is:

\[\sqrt{1^2+\left(\frac{\pi}{100}\right )^2}\]
\[\sqrt{1^2+\left(\frac{\pi^2}{100^2}\right )}\]
\[\sqrt{1+\frac{\pi^2}{10000}}\]
\[\sqrt{\frac{10000}{10000}+\frac{\pi^2}{10000}}\]
\[\sqrt{\frac{10000+\pi^2}{10000}}\]

Just looking at that, you can see that it is a number just a little big bigger than one. Why? Well ten thousand and a little bit divided by ten thousand is pretty close to one. And the square root of something pretty close to one, is even closer to one. You can compare that with the diagram:

That makes sense. What about the size of the angle? The sine function of the angle gives us the height of the point above the real axis, divided by the absolute value (to scale everything to the unit circle). If we work out that scaled down height, we can use the inverse of sine function to give us the distance around the unit circle to that point, and hence the size of the angle. The scaled height is \(\frac{\frac{\pi}{100}}{\sqrt{\frac{10000+\pi^2}{10000}}}\). Hmm. That looks like a nightmare, but it really isn't. We have already seen that the bit on the bottom (the absolute value) is pretty close to one. Any number divided by one is just itself. So really, we are interested in the bit on top. That tells us that the sine of the angle we are looking for is roughly \(\tfrac{\pi}{100}\). In fact, if you work it out it is \(0.0314004349\).

Now to actually DO the inverse sine function we could start drawing our unit circle, and then make very very precise measurements, or we could just use a calculator. Calculator it is then. The distance round the unit circle to a point \(0.0314004349\) above the real axis is \(0.0314055972\). So that's our angle in radians.

(If you think long enough about this you may well ask 'How the fuck does my calculator KNOW that this is the angle size that corresponds with that sine?' The calculator doesn't draw a circle and get out a ruler. Does it come with all the possible sine values for all possible angles to whatever number of decimal places? No. And before we have finished with this, you will find out what your calculator did to get this precise result).

If you look at the size of the angle, you may notice that it is pretty bloody close to the sine of the angle. In other words the distance around the unit circle to the point is almost exactly the same size as the height of the point above the real axis. Apparently this happens when your angle is very small. Why? Well, let's look at the diagram:

That's our number. Let's scale it to the unit circle (this does not change much because we know that the dashed line there is pretty close to one anyway), and use our old friends Imogen, Polly and Abby to see what is going on:

We can see the very small angle we are dealing with. What I am going to do next is to get rid of Raul because that is not relevant to this particular discussion. I am then going to move Imogen so that it is directly under Polly:

Now let's zoom in on the interesting bit:

Can you see that the orange line of Imogen is practically the same length as the section of circle round to Polly that it is pretty much obstructing? Let's move closer:

Can you see how the orange line and the blue line up to Polly are very close in length? That's exactly what I mean when I say that the sine of an angle is very close to the size of the angle in radians, when the angle is very small.

Anyway, enough distractions. We have our polar form number: \(\sqrt{\frac{10000+\pi^2}{10000}}\angle 0.0314055972\).

What are we going to do with it? The absolute value of the number is going to be multiplied by itself one hundred times. The angle is going to be added to itself one hundred times. Where does that take us? Well, the size is actually a number very close to one. If it was the square root of ten thousand divided by ten thousand, it would be one. The only thing that stops it being one is the \(pi^2\) on the top line. So it is roughly the square root of ten thousand and nine over ten thousand. That is very very close to one. So when we multiply it by itself one hundred times it should still be pretty close to one unit long. In fact it turns out to be a bit over \(1.05\), but not much. Now the angle. If you look closely you will see that the angle is, to four decimal places, one hundredth of \(\pi\). So what do you get if you multiply one hundredth of \(\pi\) by one hundred? \(\pi\)! In fact for our numbers you get to \(99.99\%\) of \(\pi\). So, our end result, is:

\[\left(\sqrt{\frac{10000+\pi^2}{10000}}\angle 0.0314055972\right)^{100}\approx 1.05\angle99.99\%\pi\]

Well, the angle \(\pi\) is of course half a circle, which makes the result, to within \(5\%\), negative one. So we can say, to within \(5\%\) that:

\[e^{i\pi}\approx (1+\tfrac{i\pi}{100})^{100}\approx -1\]

To generate the identity that started this whole thing, all we do is add one, an equals sign, and zero. So it looks like we are definitely on the right track. What about increasing the value for \(m\)? Well, I can't be bothered running though all the arithmetic again for a start. But lets try to imagine what would happen.

First the multiplication of \(\pi\) and \(i\) would proceed unchanged. But then we would divide that number by a much larger number, say a million. That would bring the point down to a millionth of \(\pi\) away from the real axis. We would still add one, which would take us out to a point very, very close to one. The angle would be much smaller and the absolute size would by much much closer to one.

Secondly, when we then raised the absolute size to the power of a million, it would stay much closer to one. And although the angle would be much smaller, we would then add a million of them together, getting us even closer to \(\pi\) as the total. So as the \(m\) number gets larger, the result gets closer and closer to negative one.

In fact, if you put set \(m\) to infinity, you will divide \(i\pi\) by infinity, getting an infinitely small number. One plus an infinitely small number is infinitely close to one. That number's imaginary part would be infinitely small. When you raised the absolute value of the number to infinity, it would stay at one. And the angle, which would be an infinith fraction of \(\pi\) multiplied by infinity would be exactly \(\pi\)! Which is exactly what we wanted to prove isn't it? Let off the fireworks and start the band we're done!

No. No we're not. This:

\[e^{i\pi}=\lim_{m \to \infty} (1+\tfrac{i\pi}{m})^{m}=-1\]

:gives us no real sense of what is going on here. In other words, sure, this formula tells us that raising \(e\) to \(i\pi\) gives you negative one, but it does not tell us WHY.

That's why this section is just called the intermission. This has been a test of all the concepts and tools that we have built up along so far. The good news is that they work. So we are doing the right thing. The problem at the moment is this whole 'limit' feature of our definition of \(e\). What we are going to do next, is to try to get rid of the whole limit approach to the puzzle altogether.

Such an Imagination

OK. Lets have a go at imaginary numbers. These are NOT straightforwards. The name is also very annoying because, they are very real indeed, and I will try to satisfy you that that is the case in due course.

First of all what the hell are they? We have already seen four different types of numbers. We have seen whole numbers such as one, two, three, ten and so on. These are obvious in our every day world. How many pens are on the desk? Three. How many cows are in that field? Twenty. They are used to count separate and distinct objects. Looked at formally, the \(x\) in the following equation is a whole number:

\[x+5=7\]

We can see without too much effort that \(x\) is two. Two is obviously a whole number so we are quite happy with that as a value of \(x\) that makes the equation work.

When we looked at the opposite of addition, we found a different type of number, a negative number. We agreed that saying you had negative five cows in a field did not make sense. The concept was so odd that for a long time mathematicians refused to accept that these numbers really existed. Nowadays we are perfectly happy with the concept (unless your bank account is very, very, negative, in which case you will be perfectly unhappy with the concept).

So what about a slightly different equation which works with a negative value of x?

\[x+5=2\]

To work, we need to set \(x\) equal to negative three. \(x\) cannot be a positive number, because there are no positive numbers that are five smaller than two. Again, we are quite happy with this, but weirdly even just a few hundred years ago the best mathematicians in the world would have said that there was NO ANSWER to that problem.

OK, next we considered fractions. We agreed that while you could have half a cow, it would not be very pleasant to look at. More fundamentally we realised that fractions are ratios between two whole numbers. So if I have five apples, and my friend has ten apples then I have half as many apples as my friend. That is a statement about the ratio of our apple collections to each other. What would this look like stated as an algebra question?

\[3\cdot x=2\]

This is just a little bit trickier because it is saying three what's are two? Or, what is two divided by three? We are quite happy with the answer two thirds. And in general people have always been comfortable with this idea. After all, you are just comparing two whole numbers.

OK, moving on we then learned about irrational numbers. These are numbers that cannot be written as one whole number divided by another whole number. We satisfied ourselves that the number which you multiply by itself to get two is one of these numbers. The equation which has one of these as an answer looks like this:

\[x^2-2=0\]

For the hard of thinking, you add two to both sides and then take the square root of both sides getting \(x\) equal to the square root of two. The ancient greeks really did not like this. They felt that all numbers should be rational, and they were really disturbed to find out that was not the case. it's a bit more abstract today, but we are generally not bothered by the idea that there are some numbers which would just go on for ever if you tried to write them out.

Now then. What is the answer to the following puzzle. What number, if you multiply it by itself, and then add one, gives zero? Or algebraically:

\[x^2+1=0\]

It is very similar to the square root of two equation just above isn't it? So what do we do? We subtract one from each side, getting:

\[x^2=-1\]

And we then take the square root of each side:

\[x=\sqrt{-1}\]

So the answer is the number, that if you multiply it by itself, makes negative one. OK, so what would that be then? One multiplied by itself is one, so is negative one multiplied by itself negative one? We need to think about multiplying negative things to get an answer to that.

We said that multiplying is just a special type of addition. So that three multiplied by four is:

\[3+3+3+3=12\]

Notice that there are four threes there. So what would negative three multiplied by four look like? Well, we would just add together four negative threes. That would look like this:

\[(-3)+(-3)+(-3)+(-3)=-12\]

Remember that adding a negative is the same as subtracting a positive. And also remember that you do stuff in brackets first. So while you have plus signs in between each set of brackets, the fact that there the numbers INSIDE the brackets are negative means that you end up subtracting. Didn't we say though that it doesn't matter which way round you multiply things? So what does four multiplied by negative three look like as addition? Well, sticking with our definitions, it is four added together negative three times. How do you add something a negative amount of times? Remember that a negative something is the same as the something subtracted from zero. So, for positive multiplication you add up a group of things, but for negative multiplication you subtract your number from zero the same amount of times you are supposed to negatively multiply it by. So it looks like this:

\[(0)-(4)-(4)-(4)=-12\]

(The zero looks a bit odd there, and I suppose it could be implied in the same way that positive one is implied to be zero plus one. So we could have written the four threes above as zero plus the four threes.)

As we would expect that is also negative twelve. So we have now looked at a positive multiplied by a positive (where both the number and the sign between the numbers are positive). That gives a positive result. We have looked at a negative multiplied by a positive (the numbers are negative but the sign between them is positive). And we just looked at a positive multiplied by a negative (the numbers are positive but the sign in between them changes to a subtraction instead of addition sign). What about the last option, multiplying a negative number, a negative amount of times? What does that look like as an addition?

Well, you will be multiplying a negative number, so the numbers IN the brackets are going to be negative. And we are going to be doing it a negative amount of times so the numbers BETWEEN the brackets will also be negative. The sum looks like this:

\[(0)-(-3)-(-3)-(-3)-(-3)=12\]

Because we are duplicating a negative number a negative amount of times we end up with two negative signs. Adding a negative is the same as subtracting the number, so subtracting a negative is the same as adding the number. Sound weird? Not really, if I lend you £10, then you owe me £10 (lets call that negative £10). If I then subtract, or cancel, the debt I have effectively gifted you £10. So I turn a negative obligation (you have to give me £10) into a positive benefit (I have given you £10).

So what you actually get when you subtract all those negative numbers is a positive number. In my minds eye, I see the two minus signs combine into a plus sign, with one of them rotating through ninety degrees. So for every two minus signs you create a plus.

So what about our hypothesis that if you multiply negative one by itself negative one times you get negative one? That would mean that you subtract (-) negative (-) one (1) from zero once (1). That would look like this:

\[(0)-(-1)\]

The two minus signs combine to form a plus, and you get positive one. So the square root of negative one cannot be negative one. It cannot be positive one either, because two positive numbers multiplied together always give a positive answer. So if the answer cannot be negative and cannot be positive, then what the hell is it? All the numbers that we know about - all of the numbers above, are either less than zero or greater than zero (or zero itself, I grant you). So how can a number be neither positive or negative?

Oh dear. It appears that we are stumped. And indeed for a long time people treated equations which produced answers that were the square roots of negative numbers in the same way as they treated equations which gave negative numbers themselves as answers. In other words they ignored them.

We do not do that any more though. Instead we say that there IS a number which is the square root of negative one, or a number which if you multiply it by itself gives negative one. We have a symbol for that number, and the symbol is \(i\). What we say, and just work with me on this, is that each number has a 'real' part which is a multiple of one, and an 'imaginary' part which is a multiple of \(i\). The number that is actually the square root of negative one is a number with no real part (or technically a real part multiplied by zero), and an imaginary part which is \(i\) multiplied by one. We say that \(i\) is not positive or negative in the sense of being more or less than zero. Instead we say that it has a whole positive and negative spectrum all to itself. So the following equation makes perfect sense:

\[i-2i=-i\]

This all sounds a bit abstract. What would such a number look like, and how does it relate to the other numbers that we are familiar with? I'll try and make it more visual next time.

e (by gum)

Now that we have got that fairly abstract stuff about functions and limits out of the way we can finally talk about \(e\). This is another number like \(\pi\) in that it is a constant. \(e\) is always the same number like 2 is two or 5 is five. It does not change like the \(x\)'s we have been looking at.

So what number is it? Well, lets talk about the interest on your bank account. We want to arrive at an exact figure for \(e\), not a multiple of something else, so we will start with a unit value in our bank account. If you remember a unit is always one. So lets say you have £1 in your bank account. It could be Euros, Dollars, Yen, or Sestertii. What is important is that we have one of them.

Your bank pays you interest on the money that it holds for you. it does this by giving your money to other people to set up businesses or buy houses, and charging them for the service. Because the money was yours it passes on some of the payment for this service to you. In these financially desperate times, of course, what the bank does is charge a large amount to lend out money, gives you a tiny amount back and keeps the vast amount of the charge for itself. Anyway, what we want to know is how much you will have in your account at the end of a period of investment. Traditionally the period of investment used to compare different methods of saving is one year, so let's use that.

Now, we also need to know what interest rate is going to be applied to your account. Lets say that you have pictures of your bank manager in flagrante delicto with a local farm animal, and you have been given an interest rate of 100%. (In reality, by contrast, an interest rate of one half of one percent would be more likely.)

So how much do you have in your account at the end of the year? The answer seems to be obvious - if you get 100% and you start with £1 you will have your original £1 and 100% of that again (£1) in interest, giving a total of £2 at the end of the year.

Things, however, are not that simple. We have made an assumption that your interest will be calculated and paid at the end of the year. That may not necessarily be the case. Instead, let's consider what happens if you negotiate with the bank to be paid your interest in two lump sums, one halfway through the year and then at the end of the year.

After six months of the year have passed, you will get one half of 100% of the interest on £1. That would be 50% of the total of £1 or 50p. Just what we would have expected. BUT WAIT a moment, because when you come to get your interest for the second six months, you are not earning interest on £1 any more, you now have £1.50 in your account. So you get half of 100% of the interest on £1.50. 100% of that interest would be £1.50 so you get 75p. So at the end of the year you have £1.50 plus 75p, or £2.25.

Hey, there is an extra 25p in interest over and above what you get if the interest is worked out once at the end of the year. Where did that come from? It comes from getting paid interest on the interest that you have already earned. 25p is half of 100% of the interest that you could have got on the 50p, had you started the year with that in your account. That sounds excellent, more money from seemingly nothing. Can you try that trick again?

Let's go back to the bank manager and show him the CCTV footage you have of him attending a very private party with the coach of the local college's ladies volleyball team. And the whole team. He agrees to calculate your interest payments every month. So at the end of the first month you get one twelfth of 100% of the interest on £1 or about 8p. At the end of the second month you get one twelfth of 100% of the interest on £1.08, or about 9p. At the end of the third month you get one twelfth of 100% of the interest on £1.17 (£1.08 plus 9p), which is nearly 10p. Note that each month the interest you get is a little bit more than what you got the previous month, because each month you get a little bit of interest paid on the interest you earned in the previous month.

So what do you get at the end of the year? If you do all of the adding up, you get a little bit more than £2.61. So that's pretty good. An extra 36p over the 25p we earned by calculating it twice per year. But notice - we are into the territory of diminishing returns. We haven't earned twelve lots of 25p extra. What is happening is that we are squeezing smaller and smaller amounts of interest out each time we calculate. So the increase over calculating it one per year is not that much more.

Can we work out what is happening here mathematically? Let's go back to payment once a year. How to we mathematically represent payment of 100% of the interest at the end of the year? We need to show that we still have our original £1, and we have been given an extra £1. That would be:

\[1+1=2\]

We could also write that down as:

\[1\cdot 2=2\]

because we end up with twice as many ones as we started with. Adding 100% to something means multiplying that thing by two, because you have your original thing, and you have a 100% copy of it, so you have two of your original thing.

Right. What happens with payment twice a year, after six months and then after twelve? Well, after the first six months you get half of 100% of the annual interest. How do we show that part? Well, we are not doubling our original thing, but we are adding half of it to the original. If you want to add half of something you would multiply it by one and a half. The one shows you are keeping your original, and the half gives you the half. So after six months we get:

\[1\cdot 1\tfrac {1}{2} = 1\tfrac{1}{2}\]

That's right, because we had £1.50 after the first six months. So what happens at the end of the year? Well, we want to take what we had at the halfway point and add half of 100% of the annual interest rate on that amount. Again that means we need to multiply the amount we had after six months by one (to show that that amount stays in our account) and a half (to show we are getting half of 100% of the annual rate). This looks like this:

\[1\tfrac{1}{2}\cdot 1\tfrac{1}{2}=2\tfrac{1}{4}\]

And that works out again, because you had £2.25 at the end of the year where you got interest after each six months. Let's try to work out what is happening. We wrote out the maths for the second interest payment above as the amount you had after six months multiplied by one and a half. But we already know that the amount you had after six months was one and a half times one. Remember that it does not make a difference in what order you multiply things. So we could have written the whole year out as:

\[1\cdot 1\tfrac{1}{2}\cdot 1\tfrac{1}{2}=2\tfrac{1}{4}\]

That works out mathematically because:

\[1\tfrac{1}{2} = \tfrac{3}{2}\]

So (remembering that to multiply two fractions together you just multiply the bottoms together and the tops together):

\[1\cdot 1\tfrac{1}{2}\cdot 1\tfrac{1}{2}=2\tfrac{1}{4}\]
\[1\cdot \tfrac{3}{2}\cdot \tfrac{3}{2}=2\tfrac{1}{4}\]
\[1\cdot \tfrac{9}{4}=2\tfrac{1}{4}\]
\[1\cdot (\tfrac{8}{4}+\tfrac{1}{4})=2\tfrac{1}{4}\]
\[1\cdot (2+\tfrac{1}{4})=2\tfrac{1}{4}\]
\[(2+\tfrac{1}{4})=2\tfrac{1}{4}\]

Anyway, back to the result. We also know how to write one number multiplied by itself don't we? That's just the square of the number, or the number raised to power two. That would make it:

\[1\cdot (1\tfrac{1}{2})^2=2\tfrac{1}{4}\]

Notice that we now have two twos on the left hand side of the equals sign. One is the denominator of the fraction, and the other is the power to which we raise the number in the brackets. We also get our interest paid twice a year. Coincidence? No! The fraction shows how much of the total annual interest we get applied each time, and the power number shows how many times we get paid the interest. So they are always going to be the same. If we get paid interest twice we get half each time. If we get paid four times, we get a quarter each time. Any if we get paid monthly we get a twelfth each time. If we write out the sum for monthly interest payments, it looks like this:

\[\begin{multline}
1\cdot 1\tfrac{1}{12}\cdot 1\tfrac{1}{12}\cdot 1\tfrac{1}{12}\cdot 1\tfrac{1}{12}\cdot 1\tfrac{1}{12}\cdot 1\tfrac{1}{12}\cdot 1\tfrac{1}{12}\cdot\\
1\tfrac{1}{12}\cdot 1\tfrac{1}{12}\cdot 1\tfrac{1}{12}\cdot 1\tfrac{1}{12}\cdot 1\tfrac{1}{12}=\tfrac{23298085122481}{8916100448256}
\end{multline}\]

The fraction at the end looks insane because:

\[1\tfrac{1}{12} = \tfrac{13}{12}\]
and when you multiply \(\tfrac{13}{12}\) by itself twelve times you get \(\frac{13^{12}}{12^{12}}\) which is the fraction at the end of that line. The maths works though, because if you divide the number on the top of the fraction by the number on the bottom of the fraction you get about 2.61, and you got £2.61 at the end of the year with monthly payments. That all looks like a hell of a mess though, so lets tidy it up a bit.

\[1\cdot (1\tfrac{1}{12})^{12}=\tfrac{23298085122481}{8916100448256}\]

Ladies and gentleman we have ourselves a pattern. If we calculate and pay interest \(n\) times a year, then at the end of the year we will have \(1\cdot(1\tfrac{1}{n})^n\) pounds, dollars, euros or whatever. We can describe those instructions as function, and give it the name \(f\), and the variable in question is \(n\), so:

\[f(n)=1\cdot(1\tfrac{1}{n})^n\]

Which we can tidy up a bit. The 1 at the beginning is just redundant. Look at the examples above - it does nothing. It just hangs about telling you you have one of whatever comes. Remember when we decided to use a unit value - this is why. We can just get rid of the one. If we started with £2 in our account we would need to bring the two in at this point in place of the one, but we are just working in units of one now. So removing the one we get:

\[f(n)=(1\tfrac{1}{n})^n\]

Also as we saw above, one and a half is the same as one plus a half. So to complete the tidying:

\[f(n)=(1+\tfrac{1}{n})^n\]

So we can calculate what we would get if we got paid every day, by sticking in 365 for \(n\):

\[f(365)=(1+\tfrac{1}{365})^{365}\]
\[f(365)\approx 2.71456748\]

The \(\approx\) just means approximately. As we saw, just going to \(n=12\) generates stupidly big fractions which have very long decimal expansions, so I am just rounding off the result. We could also calculate the amount we get if interest was paid every second of every minute of every hour of every day of the year:

\[60\cdot60\cdot24\cdot365=31536000\]
\[f(31536000)=(1+\tfrac{1}{31536000})^{31536000}\]
\[f(31536000)\approx 2.7182817\]

I should point out at this stage that there is nothing special about the choice of a year as the period we are using. It could just as well have been a week or an hour or a decade. The important thing is how many times during that period we pay interest.

As you should hopefully be able to see, the bigger the number that we plug into our function the closer we get to a fixed number. We have gone from \(2 \to 2.25 \to 2.61 \to 2.71456748 \to 2.7182817\). We are getting closer and closer to a number just a little bit bigger than 2.7. If we were paid interest infinitely often during the year, in other words if interest constantly accrued into our account every moment, then the amount we would have at the end of the year would be the number that we are approaching. This number, then, is the limit of the function that we have worked out, which is approached as the size of the number that we plug in gets larger and larger. Remember that we write this as:

\[\lim_{n \to \infty} (1+\tfrac{1}{n})^n\approx2.7182817\]

I still have to use the \(\approx\) symbol because the number just keeps going and going. Just like \(\pi\). Just like \(\pi\) we use a letter instead to represent the number exactly. That letter is \(e\). So:

\[e=\lim_{n \to \infty} (1+\tfrac{1}{n})^n\]

What does that actually mean? It means that if you have an interest rate of 100% and you get paid interest infinitely often during any period, then you will end up with \(e\) times your original balance. It is not a ratio like \(\pi\) is, so we have to reach it by a more circuitous route, but there it is an actual number represented by an actual symbol.

It's All Greek to Me

Yes \(\pi\) is a letter. In actual fact it is a letter of the greek alphabet. Quite a lot of mathematical place markers are. \(\theta\) (pronounced theta) is commonly used to represent and unknown or varying angle in problems. \(\pi\) is a bit different because it does not represent an unknown or fluctuating quantity, but instead it represents a single, solitary number. The number it represents is a ratio, an unchanging ratio - at least in our universe.

Let us consider a circle. A useful definition of a circle is the set of all points which lie the same distance from a single point. You can easily make a circle by using a pointy sharp thing with a fixed length of string attached to it, and with some sort of device that leaves a mark attached to the end of the string. You also need a flat surface. You can draw circles on non-flat surfaces, like footballs or saddles, but that way lies a special kind of madness called non euclidean geometry, which I am not touching with a bargepole. You stick the pointy sharp bit into the surface you wish to draw a circle on, pull the string tight, and place the marking device on the surface. What you have marked is a point which is the string's length away from the pointy bit. Now if you lift the marking device up and place it down (with the string still held taught) ANYWHERE else, you will mark another point the string's length away from the pointy bit. If you keep doing this randomly, you will eventually see the outline of a circle start to form - defined by all the individual points.

Of course, that is not the sensible way to draw a circle. Instead of lifting the marking device up every time, you just leave it touching the surface and, again keeping the string tight, you move it in either direction. I say either direction because you will find that you are constrained to only move two ways, clockwise or counter clockwise. Once you have moved back to the point you started from you have drawn a circle. Well done. Now, we can say five things about the circle you have marked. We can of course say what colour it is, but that is irrelevant for geometrical purposes. Secondly we can say how thick the line is that marks out the circle. This will depend on your choice of marking device. A felt tip pen or highlighter will leave a thicker line than a biro, and a crayon or a piece of chalk will leave a thicker line that all of the foregoing.

In pure geometry though, lines do not have thickness, and points do not have an area. A line, including the outline of the circle we drew, only extends in one direction. This makes sense, because of our definition., The circle is the set of all points EXACTLY the same distance from another point, not sort of the same distance depending on how big your marking device is. So the thickness of the line marking the circle, as with its colour, is irrelevant to us.

Thirdly, we can say how big the area is inside the circle. This is going to be measured on a two dimensional surface, so the answer will be some number of whatever units you like to the power of two. By that I mean the units are squared, not the number of them. So if your chosen measure of length is the flangit, then your chosen measurement of area will be square flangits. A square flangit is just a square whose length is exactly a flangit. Squaring a number is the same as raising it to the power of two. The area in the circle is therefor some number of square flangits.

Fourthly, we could talk about the length of the line we have drawn on the surface. If we took it in our minds eye and straightened it out, and measured it, how long would it be? Fifthly and finally we can talk about the length of the bit of string between the pointy sharp bit and the marking device.

As it turns out, the area in the circle and the length of the line you draw are related ONLY to the length of that bit of string. If that seems remarkable, remember that it doesn't matter where on your surface you poke your pointy sharp thing your circle will still look the same. The only determining factor of the size of the circle you create is the length of string you allow between the pointy sharp bit and the marking device.

So what does all this have to do with \(\pi\)? Well, if you take your length of string, and multiply it by \(2\pi\) you get the length of the line you draw. So \(\pi\) is the RATIO between the length of the string and the length of the line around the outside of the circle you draw with that string. It matters not a jot how long your bit of string is, the circle that results will ALWAYS have a line that is \(2\pi\) times the length of that string long around its outside.

So why do we say \(2\pi\) and not just \(\pi\) for this ratio? Good question. It turns out that it is a lot more useful to work with \(2\pi\) because \(\pi\) itself turns up in far more places on its own. If we went with \(\pi\) then we would keep having to half it. For example, the area of the circle that we can describe? It is the length of the string times itself (this makes the units squares remember) multiplied by \(\pi\). If we defined \(\pi\) as the direct ratio between the length of the string and the line, that number used to find the area would have to be half of \(\pi\). Which looks messy.

The line which forms the circle is called the circumference of the circle. Another word for something that runs around the outside of something else is the perimeter. The word for perimeter in ancient greek started with a letter in the ancient greek alphabet. Can you guess which letter? The length of the bit of string is called the radius of the circle. The point we made with the pointy sharp thing is the centre of the circle.

There is a line which is double the length of the radius, and is described as the straight line from one point on the circle, through the centre, to another point on the circle (which will inevitably be exactly opposite the starting point). This line is the diameter of the circle. Because this line is the radius multiplied by two, and because (as we have seen) it does not matter which order you multiply things in, you can get rid of the two from the \(2\pi\) when you talk about the ratio between the diameter and the circumference. The circumference is therefor just \(\pi\) times the diameter. This is because the radius times two times \(\pi\) is the same as two times the radius times \(\pi\), and two times the radius is the diameter.

What does this all look like? Glad you asked. Have a gander at this snazzy diagram:

I don't like the diameter though, because it has nothing to do with the construction of the circle. The diameter comes about once a circle has been made, while the radius is used in making the circle in the first place. So I prefer to think of \(\pi\) as it relates to the radius not the diameter.

The number itself is a bit more than three. It is not a whole number you can count on your fingers, and neither is it a fraction (although \(\tfrac{22}{7}\) comes pretty close. If you were to write out \(\pi\) as a decimal number as three point something something something, you would be writing "somethings" for ever, because there are an infinite amount of them. So instead of worrying about all those "somethings" we just write \(\pi\). This is a fantastic animation of the circumference "unrolling" to show it is \(pi\) times the diameter:

That animation was created (and GPL licensed) by wikipedia user John Reid.

What is Zero?

If you look at Roman Numerals, i.e. I for 1, II for 2, V for 5, VIII for 8, X for 10 and so on, you should spot two things. One thing is easy to spot, and the other is quite hard.

It is easy to spot that the complexity of Roman Numerals is not obviously connected to their size. For instance 37 is XXXVII, while 52 is LII. It is quite hard to spot that this is because their system does not place varying values on the symbols (I, X, L, V etc) depending on where they fall in the sequence. Instead the value is always the same for each symbol, X always is 10 and L is always 50.

The whole number being written is calculated by applying a sequence of rules, which basically says, read from left to right adding each symbol's value unless the symbol to the right of it is greater than it, in which case you deduct its value from that symbol. So, XI is 10+1=11 and IX is 10-1 which is 9. Taken a little further, 1998 is MCMXCVIII which is M(1000)+CM(1000-100)+XC(100-10)+V(5)+I(1)+I(1)+I(1). What is important to note is that the two [C] symbols BOTH mean 100 despite them falling in different positions.

Our system of numerals is different. We also use different symbols to represent the constituent parts of big numbers. So far, so Roman. What is different is that the symbols have no relation to each other but their value depends on where in the sequence they fall. In one sense we could be said to read our sequence of symbols from right to left. The right most symbol (if we are dealing with whole numbers only) tells us how many one's are in the total number. The next rightmost symbol tells us how many ten's, the next how many hundred's and so on. So if we see 123, we quickly add 3 one's, two ten's and one hundred.

If it seems odd that we read the number from right to left, remember it is done subconsciously. If read from left to right, one can not know what the first symbol in the number represents (be it millions, hundreds of thousands or so on) until you have counted how MANY symbols are in the number. So logically you start at the right, where you KNOW the symbol means one's, and then add up moving from right to left. In reality though, it doesn't really matter which direction you read the symbols in, as long as you know what value each symbol position has in advance. This is unlike Roman Numerals, where the end result would be very different if you started at one end rather than the other - see XI and IX for example. Also, note that if you have the number 144, the two [4] symbols mean different things. The middle one means 4x10 or 40, and the right most one means 4x1 or 4. So the symbols increase in relative value the further left you go because they are multiplied by larger and larger sums. So the complexity of our numbers is directly proportionate to their numerical size. 237 is a bigger sequence of symbols than 84 and so on. The Romans were no thickies, built loads of stuff, and conquered the majority of their world (just don't tell the Chinese), so why were they using this cumbersome system where thirty seven takes longer to write than fifty two, and both are longer than one thousand and one?

The answer is rather profound. If we want to write the number one hundred and nine, then we start with the symbol [9] at the right, followed by the symbol [0] to show there are no tens, followed by the symbol [1] to show there is one hundred: 109. The Romans could not do this because they had no symbol which meant nothing. They had I, V, X, L, C, D, and M, each of which represents a specific number, but they had nothing which represented zero, because they did not consider that to BE a number. For us it would be like asking "What is the wavelength of black light?" (I do not mean the purpley ultraviolet stuff that shows up washing powder residue and dandruff with equal aplomb). Black light does not have a wavelength because there are no waves, it is the absence of light. Or it would be like asking "How does the sound of a violin not playing differ from the sound of a trumpet not playing?". It is a meaningless question. The absence of something does not need to be measured, and we do not need a symbol to represent it.

You may say that they could represent nothing by just not writing anything, which makes a sort of sense, but the problem is then distinguishing between 10 and 100, or 11, 101 and 1001. No, for a system of numbers based on symbols having different values depending on their position you need a clear symbol to represent "none of this value needed". A Roman would be puzzled by this; it would be like making up a shopping list with one amphora of wine, three buckets of milk, two handfuls of berries, no apples, and five bags of nuts. Why do you even bother to mention the apples if you want none of them?

So we are forced to have the symbol [0] because of the simpler way we write out numbers. But just the fact that we got [0] makes us think differently about the concept. Once it is a symbol, is it a number? I do not propose to try to answer that, but I will say that we can sometimes use it like a number. We can add it to other numbers (we get the number we started with), we can multiply other numbers by it (we get zero every time), we can even multiply other numbers by themselves [0] times - meaning raise them to the power 0 - (oddly we get one and not zero every time). We cannot divide numbers by zero. That breaks things.

For our present purposes then, [0] is as fundamental as [1].

What is One?

Let's start at the very beginning. What is '1'? The obvious answer is that it is the first number. But is it? (See '0' next). Or is it even a number? Y'see every other number you can count on your fingers, toes, abacus etc etc, is made up of lots of '1's. So there is a school of thought that '1' isn't actually a number, but is instead the building blocks of all numbers.

One can now wander off down a long and winding path that is a career in Number Theory. Note that that is a sensible academic discipline which attracts seriously minded professional people dedicated to deep study and contemplation. Numerology is NOT to be confused with Number Theory. THAT is a nutty mystical system (in the loosest possible sense) of superstitions which attracts people with serious mental health issues dedicated to deep confusion and picking lucky lottery numbers.

Let us not wander down that path though. For our purposes here, let us instead contemplate 1 as a unit of something. The concept of a unit of something is a tricky one, and one which was never satisfactorily explained to me in school. For length, for instance, you have to decide what "units" you are working in - which is not quite the same as saying [imperial] or [metric]. No, the units will be miles or kilometres, yards or metres, inches or centimetre and so on. You need to get this bit right so that if you measure length 2 and add a length 3, what you get is a total length of 5. If you add 2 metres to 3 miles you do not get 5 of anything, you get a mess, and your probe fails to enter orbit. As I said though, this is not just imperial versus metric. If you add 2 metres to 3 kilometres, again you do not end up with 5 of anything.

So when you are deciding on which units you are using to measure you are really deciding an absolute reference length. ALL other lengths you are going to be measuring are then multiples (or maybe fractions) of that reference length. That means that you can do normal maths on these lengths - addition or multiplication for instance - and you will get sensible answers WHICH ARE themselves multiples or fractions of the SAME reference length.

I remember being very confused in school trying to answer the question "How many square centimetres are in a cubic centimetre?". I kept imagining pieces of paper cut into squares a centimetre on each side and then stacked one on top of the other, and being confused because I could not work out how many would fit to a cubic centimetre. I now know that I was confused about this because I was dealing with two completely different units. A cubic centimetre is a unit in its own right, and is not some multiple of units of square centimetres. There is no exchange rate. A cubic centimetre is a unit of volume and represents a cube (funny that) with sides which are one centimetre in length. A square centimetre is a unit of area and is a square with sides which are one centimetre in length. No one had properly explained the concept of units to me.

So 1 can be seen as the ultimate unit. Even centimetres can be broken down into smaller divisions. For instance a centimetre is actually one hundredth of a metre, which in turn is actually the distance travelled by light in a vacuum in roughly one three hundred thousandth of a second. But 1 does not need to be broken down any more, nor can it be. It is the absolute basic unit, from which everything else is referenced. So when I say I am adding 83 to 32, I actually MEAN I am adding 83 ones to 32 ones.

At a very basic level, and I LIKE very basic levels, 1 is the basic unit which every other thing is measured relative to. It therefore represents the difference between something and nothing. Which takes us neatly on to...