So last time we worked out HOW to raise the number \(e\) to an imaginary power - specifically the imaginary power that we are focussed on namely \(e^{x}\). We were still using our definition of e as the limit of a function when one of the variables increased in size towards infinity. This function:
\[e^{x}=\lim_{m \to \infty} (1+\tfrac{x}{m})^{m}\]
That function was a little different to our first definition of \(e\) because it is actually the function that gives us \(e\) to the power of \(x\). Which is handy. When we ran through everything I decided that while the function seemed to work perfectly it did not actually tell us anything about WHY these numbers combine in this fashion.
So let's take a different tack. Let's stop defining \(e\) as the limit of a function as you change a variable. Let's try to define it purely in terms of a one variable function. In otherwords, what I want to do is get rid of the \(m\) in the above function.
Right-o. How do we do that? Well, what we have is this:
\[(1+\tfrac{x}{m})^{m}\]
We want to get rid of the \(m\). The \(m\) just tells us how many times we need to multiply together the stuff in the brackets. Let's say that \(m\) was two. The function would look like this:
\[(1+\tfrac{x}{2})^{2}\]
Which would expand into this:
\[(1+\tfrac{x}{2})\cdot (1+\tfrac{x}{2})\]
Now we have the same kind of problem as the red and blue baskets with apples and oranges in them. Only this time we have exactly the same things in each bracket. it doesn't change the way we deal with it though, we just need to unpack the first bracket like this:
\[1\cdot (1+\tfrac{x}{2})+\tfrac{x}{2}\cdot (1+\tfrac{x}{2})\]
And then we can unpack the second bracket like this:
\[1\cdot 1+1\cdot \tfrac{x}{2}+\tfrac{x}{2}\cdot 1+\tfrac{x}{2}\cdot \tfrac{x}{2}\]
Which turns into:
\[1+ \tfrac{x}{2}+\tfrac{x}{2}+\tfrac{(x)^2}{2^2}\]
And finally:
\[1+ 2\cdot \tfrac{x}{2}+\tfrac{x^2}{4}\]
That's fair enough, and perfectly logical. However, what we want to do is not just multiply two brackets together, but to multiply \(m\) brackets together. How on earth do we do that? What we are going to do is to look closely at exactly what we get out of the brackets when we multiply them together, and how that changes when we add more and more brackets to the mix. We want to see if we can work out a general rule to tell us what we will get without having to go through all that tedious multiplication and addition that I just did above.
To help us with this process I am going to simplify our brackets a bit. Instead of \((1+\tfrac{x}{m})\), which is a bit of a mouthful, I am going to replace the \(\tfrac{x}{m}\) bit with the letter \(y\). So our bracket now looks like this: \((1+y)\). That's much nicer to look at. All we have to do is remember that when we are finished, we need to replace \(y\) where ever we find it with \(\tfrac{x}{m}\).
Right, to get a feel for what we are talking about lets look at three examples, \((1+y)^2\), \((1+y)^3\), and \((1+y)^4\). I won't do much talking in between, let's just have a look at the logical steps.
First of all \((1+y)^4=(1+y)\cdot(1+y)^3\), and \((1+y)^3=(1+y)\cdot(1+y)^2\). So we just need to work out the power two answer and then multiply through by another bracket, and so on. So, power two:
\[\begin{align}
&(1+y)^2\\
&(1+y)(1+y)\\
&1(1+y)+y(1+y)\\
&1\cdot1+1\cdot y+y\cdot1+y\cdot y\\
&1+y+y+y^2\\
&1+2y+y^2
\end{align}\]
Now power three:
\[\begin{align}
&(1+y)(1+2y+y^2)\\
&1(1+2y+y^2)+y(1+2y+y^2)\\
&1\cdot 1+1\cdot 2y+1\cdot y^2+y\cdot 1+y\cdot 2y+y\cdot y^2\\
&1+2y+y^2+y+y(y+y)+y^3\\
&1+2y+y^2+y+y^2+y^2+y^3\\
&1+2y+y^2+y+2y^2+y^3\\
&1+2y+y+3y^2+y^3\\
&1+3y+3y^2+y^3
\end{align}\]
Now power four:
\[\begin{align}
&(1+y)(1+3y+3y^2+y^3)\\
&1(1+3y+3y^2+y^3)+y(1+3y+3y^2+y^3)\\
&(1+3y+3y^2+y^3)+y(1+3y+3y^2+y^3)\\
&1+3y+3y^2+y^3+y\cdot 1+y\cdot 3y+y\cdot 3y^2+y\cdot y^3\\
&1+3y+3y^2+y^3+y+3y^2+3y^3+y^4\\
&1+4y+3y^2+y^3+3y^2+3y^3+y^4\\
&1+4y+6y^2+y^3+3y^3+y^4\\
&1+4y+6y^2+4y^3+y^4
\end{align}\]
Let's summarise what we have found so far:
\[\begin{align}
(1+y)^2&=1+2y+y^2\\
(1+y)^3&=1+3y+3y^2+y^3\\
(1+y)^4&=1+4y+6y^2+4y^3+y^4
\end{align}\]
Can we start to describe our results? Well, first of all each result starts with a one. Secondly each result ends with a single \(y\) to the same power as we were raising the bracket to. Thirdly, inbetween, the one and the single \(y\) we have some amount of every power of \(y\). In other words, reading from left to right, you have no \(y\)'s, then some number of just \(y\) and then some number of \(y^2\) then \(y^3\) and so on. We never miss a power of \(y\).
These are all fairly obvious. The last observation may have passed you by. Look at the numbers we multiply the \(y\)'s by. They are symmetrical! In the third power row you get three three, and in the fourth power row you get four - six - four. This is promising, because it suggests that there is a pattern to be discovered! Let's have a try at working out how we get to those numbers.
Why does each line start with a one? Remember the process that we used to generate these lines. We are multiplying together numerous brackets. The way we multiply brackets together is to multiply the individual terms in each bracket by every other term in all the other brackets. That's easy for only two brackets, because in each multiplication you only have two things being multiplied. With three or four brackets, we are going to have three or four items being multiplied.
To generate the first row above we went throught the whole term by term routine, but then when generating the other lines, we just mutlpied the last line by another bracket as a short cut. Let's actually look at what would have happened with three brackets, from scratch as it were:
\[(1+y)(1+y)(1+y)\]
First we multlply the first term in each of the brackets:
\[1\cdot 1\cdot 1\]
That gives us our number one. The more brackets you have the more ones go into that multiplication, but even with \(m\) number of ones in that multiplication the result is still going to be one. That's where the one comes from. What next? Lets have the first terms of the first two brackets and then the second from the last:
\[1\cdot 1\cdot y\]
That gives us a y on it's own (the ones just collapse down to tell us we end up with one multiplied by \(y\)). And the same thing will happen when the \(y\) in the middle bracket is multiplied by the ones in the other two. That looks like this:
\[1\cdot y\cdot 1\]
And finally we will also get the same result from choosing the \(y\) in the first bracket and ones in the last two:
\[y\cdot 1\cdot 1\]
If we then add up all those \(y\)'s we end up with, drumroll, \(3y\), which is what we were expecting. If you think about it there is no way for us to get any other individual \(y\)'s. There are only two options from each bracket a one or a \(y\). And if you multiply two or more \(y\)'s together you cannot end up with a \(y\) as opposed to a \(y^2\) or a \(y^3\). So the only options for getting \(y\)'s are to pick only one \(y\). We can only do that for as many brackets as we have. So that means if we have \(m\) brackets we will get \(m\) \(y\)'s. So far, with \(m\) brackets, we can say that we will have:
\[1+my+\ldots\]
If we move onto \(y^2\)'s things get a bit trickier. We have to keep track of all the combinations we have tried so far, and that is getting complicated. So I am going to use a table to keep track of things.
That shows us the result for the first line. Now let's add on the other options that we have worked through so far:
In order to do things as systematically as possible when using multiple \(y\)'s, I am going to fix one, and then move the other one around. This means that I won't accidentally miss any of the possible arrangements. I'll fix one in the third bracket first, putting the other bracket in the second and then first brackets:
Good. Now lets put the fixed \(y\) in the second bracket, and we'll use the last and then first for the other \(y\):
And last of all we will fix the \(y\) in the first bracket and use the last and then second brackets for the other \(y\):
Hang on. We have ended up with six \(y^2\)'s. We were only expecting three. What has gone wrong? If you look at the first and third \(y^2\) rows you will see that they are identical. They both represent a one in the first bracket mutiplied by a \(y\) in the second and third brackets. But once we have done that combination once, we can't do it again. So we have over estimated. If you look closely, you will see that each \(y^2\) line is duplicated once. Because we have twice as many \(y^2\) as we need we need to divide the result by two. Six divided by two is three, which IS what we were expecting. Why did we end up with twice as many as we were expecting? Look at this line:
How did we create that twice? Well once we had the fixed \(y\) in the first column, and then we had it in the second column. But both times we generated the same row. There are two \(y\)'s in the row, so there have to be two copies of the row in our whole table. Why? You have one copy for when the first \(y\) is fixed and one copy for when the second \(y\) is fixed. Good.
Next question. How did we end up with six rows? Well we were creating our rows by a careful systematic process. FIrst we picked a row for the fixed \(y\). We had three rows to choose from. Then we placed our floating \(y\). We no longer had three rows to choose from, because the fixed \(y\) was taking up a row. So we had one less that three, or two, rows to choose from. For each of the rows the fixed \(y\) was in, we had two options for the floating. That gave us three multiplied by two, or six total rows. Can we then come to any conclusions about the number of \(y^2\)'s we will have with \(m\) number of brackets?
We will have \(m\) choices of rows for our first \(y\), and then \(m-1\) options for our second \(y\). We will, however, end up with more rows then we need, because we will get one for each possible rearrangement of the fixed and floating \(y\)'s. With two possible \(y\)'s (fixed and floating) there are two possible ways to position them in each different row. So if the total variety of rows we get are \(m\) multiplied times \(m-1\), we then need to divide that by the number repeated rows, which for two types of \(y\) will be two. So we can say that the total number of \(y^2\)'s we will get is \(\frac{m(m-1)}{2}y^2\). We can add that on to our running total like this:
\[1+my+\frac{m(m-1)}{2}y^2+\ldots\]
Let's pause for a moment and check that this works with two brackets. With two brackets \(m=2\) so we get:
\[\begin{align}
1+2\cdot y+\frac{2\cdot (2-1)y^2}{2}+\ldots\\
1+2y+\frac{2\cdot (1)y^2}{2}+\ldots\\
1+2y+\frac{2\cdot y^2}{2}+\ldots\\
1+2y+\tfrac{2}{2}\cdot y^2+\ldots\\
1+2y+1\cdot y^2+\ldots\\
1+2y+y^2+\ldots
\end{align}\]
So far so good - that's what we got when we did it manually - but what about the \(\ldots\) bit? Given that we do not need it (because the first three terms are all we need) it must be zero, so we will see why next time.
Just before we go onto the next one, with our current table, there will be one final line to add:
That just confirms for three brackets multiplied together that we will end with one \(y^3\), becuase with only three brackets to choose from there will only ever be one way of multiplying one item from each to get a \(y^3\) and that is by choosing \(y\) in each bracket.
Right. On to four brackets. Let's think about our table. The first row is going to be all ones, representing the only way you can multiply all the ones together. Next will be the rows with one \(y\) in each row. There will be four rows of them, because there are four sets of brackets to choose an individual \(y\) from. Lets show that now:
Same old, same old. Now lets do the \(y^2\)'s. Just like last time, we will fix one \(y\) and then float one \(y\) around. This looks like this:
So you can see that for every \(y_{fixed}\) we have the other \(y\) in each of the other columns. So the total number of rows should be four multiplied by four minus one (three). That comes to twelve which is exactly the number of rows we find above. Excellent. As with the last time, if you ignore the disctinction of \(y_{fixed}\) as opposed to \(y\), you have twice as many rows than you need, because you can rearrange \(y\) and \(y_{fixed}\) in two ways for each possible row. So again you need to divide by two. Twelve divided by two is six, which is just what we found when we did this manually.
Now. What about \(y^3\)? Before we start, lets think through logically that should happen. We will have three \(y\)'s to choose from the brackets available. If we do the same as last time we will fix one \(y\), from four different options, but then we will have two other \(y\)'s to pop in. For the second \(y\) you have one less brackets to choose from. And for the third \(y\) we have one less bracket again to choose from. So the total number of rows we will generate is four multiplied by three multiplied by two. That's going to be twenty four in total. How many duplicates are we going to get? Well, for each possible unique row we have three places for \(y\)'s. Does that mean we get triple the number of rows we need instead of double like last time? No.
Obviously, we can't be right, because we already know the answer is four (from when we did it manually above). But twenty four divided by three is eight, not four. Four is twenty four divided by six not three. So where does the six come from? If you have a row with three \(y\)'s in it, how many different ways can you order the \(y\)'s? With one \(y\) there was only one way to put it in a row with one space for it, so the answer is one. With two spaces and two \(y\)'s, you can place the first \(y\) in two places, with the other one falling into the only remaining place. So the answer is two multiplied by one, or two. For three \(y\)'s you have three choices for the first \(y\), then two places for the second \(y\) and only one for the last \(y\). The answer is then three mutiplied by two mutiplied by one. That comes to six, which is the number we predicted. So we can say that the total possible number of rows is going to be \(m\) (for the first \(y\)) multiplied by \(m-1\) (for the second \(y\)) multiplied by \(m-2\) (the places left for the last \(y\)). Each row generated by that process has three \(y\)'s in it. There are three multiplied by two mutiplied by one ways to organise three \(y\)'s in three places, so we end up with six times as many rows as we need.
The number of \(y^3\)'s generated from \(m\) brackets is going to be \(\frac{m(m-1)(m-2)}{6}y^3\). We get the six from three times two times one. So we can expand our formula to:
\[1+my+\frac{m(m-1)}{2}y^2+\frac{m(m-1)(m-2)}{6}y^3+\ldots\]
If it wasn't obviously already, when we are looking at higher and higher powers of \(y\) from more and more brackets, this pattern is going to continue. For whatever power of \(y\) we are interested it, we are always going to have the total number of possible columns available for each \(y\) you need to place. That will always be one fewer as you place the \(y\)'s. For each of the unique lines that process generates, there will the same number of copies as there are ways to organise the number of \(y\)'s that you are placing.
We are stuck with the complicated \(m(m-1)(m-2)\ldots (m-(m-1)\) mess, but can we do anything with the three mutiplied by two mutiplied by one stuff? The answer is yes. This kind of stuff crops up from time to time, and there is a simple notation for it. You just write \(3!\) to represent three mutiplied by two mutiplied by one and \(4!\) is four multiplied by three mutiplied by two mutiplied by one. And so on. That turns our formula into:
\[1+my+\frac{m(m-1)}{2!}y^2+\frac{m(m-1)(m-2)}{3!}y^3+\ldots\]
Which is a bit neater. We'll see where this takes us next time.
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