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.