We have now this equation for working out the number of ones, \(y\)'s, \(y^2\)'s, \(y^3\)'s that you will get no matter how many brackets you start with (or rather, no matter to what power you try to raise those backets).
\[1+my+\frac{m(m-1)}{2\!}y^2+\frac{m(m-1)(m-2)}{3\!}y^3+\ldots\]
With a little thought we can see that this pattern is going to continue for \(y^4\)'s, \(y^5\)'s and every other power of \(y\). Why? Well each higher power of \(y\) is created by multiplying that many \(y\)'s together. To make sure you create every possible combinations of rows you follow the procedure we established last time. First you fix one \(y\), then the rest of the available \(y\)'s, but one, and then you float the remaining \(y\) around all the the other columns available spaces. Once you have used that last \(y\) in every available column, you change position of the \(y\) you fixed last, and then you float the remaining \(y\) again. You keep going until your very first fixed \(y\) has been in every available column.
So the total possible rows you get for any power of \(y\) is the the same number of groups of columns (reducing by one each time to represent all the \(y\)'s slotting into place) multiplied together as the power. So if you want to know how many total rows that generate \(y\)'s to the power of two you have, you do (total columns) multiplied by (total columns less one). If you want to know the same for \(y\) to the power four, you have (total columns) multiplied by (total columns less one) multiplied by (total columns less two) multiplied by (total columns less three). You can see in each multiplication there are the same number of terms as the power of \(y\) you are trying to find.
Of course that just gives you the total amount of rows that will be generated. You still need to get rid of the duplicates. That can be achieved by dividing the total number of rows created in the last step by the total number of ways you can arrange that number of \(y\)'s together. That will always be the total number of \(y\)'s (choices of position for the first \(y\)) multiplied by the choices of position for the second \(y\) placed, and so on down to the last \(y\) for which there is only one position left. Remember we are talking about the different ways of organising the \(y\)'s in the spaces for \(y\)'s already in the lines generated by the procedure in the previous paragraph. We are NOT organising the \(y\)'s in every possible space in those rows. The number that does the dividing is always going to be the same number as we are raising \(y\) to the power of, multiplied by all the whole numbers between it and zero (because we are using up a \(y\) position each time. We show this by using the \(!\) symbol after the number that starts that mutiplication.
We have already written down what the first four terms look like when we multiply \(m\) brackets together. Having thought about the arguments above we can also say what ANY term of that equation will look like. If we want to know the \(k\)th term we would change the symbol \(k\) in the following to the number of the term we wanted to know:
\[\frac{m(m-1)(m-2)\ldots (m-(k-1))}{k\!}y^k\]
That looks a bit tricky. Previously we have seen that the dots in a row mean "and so on" when added onto the end of a series of numbers. Here, albeit it appears between two terms in the top line of a division, it means exactly the same. It just really means "keep putting in brackets with one more being deducted from \(m\) each time, until you get to the bracket where the number being deducted is one less than \(k\)". We do not want to go all the way to \(k\) because then we would have one too many brackets being multiplied.
I am not just being lazy by not writing in all these brackets, I literally cannot do so because I do not know what value \(k\) has, so I do not know how many backets to write in until I pick a \(k\)! \(k\) may even be two, in which case I was wrong in writing in the \((m-2)\) bracket (because I should really have stopped at \((m-1)\) because one is one less than \(k\) when \(k\) is two). The idea behind the way that I have written that line is that I have given all the clues that will be needed for the whole line to be constructed once \(k\) is known, even if I have to add in or subtract brackets.
What happens if we set \(k\) to four? We get this:
\[\begin{align}
&\frac{m(m-1)(m-2)\ldots (m-(4-1))}{4\!}y^4\\
&\frac{m(m-1)(m-2)(m-(3))}{4\!}y^4\\
&\frac{m(m-1)(m-2)(m-3)}{4\!}y^4
\end{align}\]
That is the term that will always tell us the number of \(y\) to the fourth powers once we multiply out our brackets. If we say that we have four brackets, like we had above, \(m\) will be four, and the number of \(y\) to the fourth powers will be:
\[\begin{align}
&\frac{4(4-1)(4-2)(4-3)}{4\!}y^4\\
&\frac{4(3)(2)(1)}{4\!}y^4\\
&\frac{4(3)(2)(1)}{4\cdot 3\cdot 2 \cdot 1}y^4\\
&\frac{24}{24}y^4\\
&1\cdot y^4
\end{align}\]
Or, one \(y^4\). Which if you look above you will find is exactly the number we were expecting. So or logic seems sound. If you are a bit devious of mind, you may well ask, "what happens if I set \(k\) to be more than \(m\)", or in other words, what if I ask this equation how many \(y^4\)'s I will get when I am only multiplying together, say, two brackets. The answer must be zero, because to get to \(y^4\) you must be multiplying four \(y\)'s together. If we only have two brackets, we will only have, at most, two \(y\)'s to multiply together. So what does this look like?
\[\begin{align}
&\frac{2(2-1)(2-2)(2-3)}{4\!}y^4\\
&\frac{2(1)(0)(-1)}{4\!}y^4\\
&\frac{0}{4\!}y^4
\end{align}\]
Anything multiplied by zero is zero. So it doesn't matter what else we have in the top line of that division, as soon as a zero hoves into view the whole thing turns to zero. Zero divided by anything (apart from zero) is also zero. So we end up with zero \(y^4\)'s. Which is what we expected. We can also say that if \(m\) is less than \(k\) we will always end up creating at least one bracket on that line where a number equal to \(m\) is subtracted from \(m\), which reduces the whole thing to zero.
So we are finally able to restate our limiting definition of \(e^x\):
\[\begin{multline}
e^x=\lim_{m \to \infty} 1+my+\frac{m(m-1)}{2!}y^2+\frac{m(m-1)(m-2)}{3!}y^3+\ldots\\
+\frac{m(m-1)(m-2)\ldots (m-(k-1))}{k!}y^k
\end{multline}\]
(Remember that we have defined \(y=\left (\frac{x}{m} \right )\), so we need to replace every \(y\) accordingly)
We have now got rid of the pesky raising to the power, but we still have the problem of this being a limiting function. Let's see if we can get rid of that. How do we do that? What I am going to do is just say, "to hell with approaching infinity let's just make \(m\) infinity and see what happens". In particular I am going to look at each term in turn to see whet effect infinity has.
The first is always one. We agreed that no matter how many times you multiply one by one you still get one.
The second is \(my\) at the moment, but we need to reintroduce \(\left (\frac{x}{m} \right )\). This makes it:
\[m\cdot \frac{x}{m}\]
That is also:
\[\begin{align}
&\frac{m}{1} \cdot \frac{x}{m}\\
&\frac{m\cdot x}{1\cdot m}\\
&\frac{m\cdot x}{m\cdot 1}\\
&\frac{m}{m}\cdot \frac{x}{1}
\end{align}\]
So no matter what \(m\) is, even infinity, once you divide it by itself you get one. So the second term is going to be \(x\) an its own. It has to be because the two infinte \(m\)'s cancel each other out.
Let's look at the third term now:
\[\begin{align}
&\frac{m(m-1)}{2!}y^2\\
&\frac{m(m-1)}{2!}\cdot \left (\frac{x}{m}\right )^2\\
&\frac{m(m-1)}{2!}\cdot \frac{x^2}{m^2}\\
&\frac{m(m-1)}{m^2}\cdot \frac{x^2}{2!}
\end{align}\]
We can just swap around the bottom of those fractions because we are mutiplying them together. Look back up the the discussion about the last term if you want to see the steps. We would just have brought these together into one big division, reoganised the order of the terms and then separated out into these two fractions. Ok. Let's set \(m\) equal to infinity.
look at the first part of that multiplication. We are setting \(m\) to infinity. If we do that then \((m-1)\) is still infinity because infinity minus one is still infinity. Agreed? Given that is the case, we get:
\[\frac{\infty(\infty-1)}{\infty^2}\cdot \frac{x^2}{2!}\]
All together now, what is infinity minus one? Yes! Infinity. So we actually have:
\[\frac{\infty(\infty)}{\infty^2}\cdot \frac{x^2}{2!}\]
Which is:
\[\frac{\infty^2}{\infty^2}\cdot \frac{x^2}{2!}\]
What is anything (even infinity squared) divided by itself? Yes - one.
\[1\cdot \frac{x^2}{2!}\]
So the horrible looking third term actually just dissolved down to a nice simple looking fraction. This is also going to be our pattern. Think about it. We have exacly the same number of brackets on the top line of each division as the power we raise the \(y\) to which is multiplied by that division. So here we had two brackets and we squared \(y\). Next time we add a power and a bracket, because we have three brackets and we cube \(y\). This pattern continues. It means we can always be sure, no matter which term in this infinitely long series that we are looking at, that the power of \(m\) on the bottom of the fraction exactly balances the number of brackets containing \(m\) on the top of the fraction. When we set \(m\) to infinity that makes all the brackets on the top contain infinity, no matter what number they try to decuct from infinity. And that means that we have the same number of infinities multiplying together on the top and bottom of the division. If we have the same on the top and bottom we know that just equals one.
So we can say that when we set \(m\) to infinity, the \(k\)th term (assuming that the term at the very beginning is the zeroth term) of the series is \(\frac{x^k}{k!}\), which looks very neat indeed. So we can now write out \(e^x\) as an infinite series:
\[e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\]
If only we could write the first two terms of the series in the form \(\frac{x^k}{k!}\) we could even write out a nice summation symbol for all this. Well, let's just see. The first term (as opposed to the zeroth term) is \(x\). What would happen if we tried to work out the first term using our rules? Well, it would be \(\frac{x^1}{1!}\). One factorial (\(1!\)) is just one. And \(x\) to the power of one is just \(x\). So we end up with \(\frac{x}{1}\) which is just \(x\) which is exactly what we want - so that works fine.
How about the zeroth term? Well we agreed ages ago that anything to the power of zero is one. That applies just as much to \(x^0\), so our rule \(\frac{x^0}{0!}\) becomes \(\frac{1}{0!}\). But surely \(0!\) is zero, right?
Nah. We assumed that anything to the power zero was going to be zero, but it turned out to be one. Same thing happens for much the same reason here. Here's what happens.
Any normal factorial, \(3!\) or \(2!\) for instance, is just the factorial of the number one smaller mutiplied by the number of the factorial. So \(3!=3\cdot 2!\), and \(2!=2\cdot 1!\). If \(0!\) was zero, then this would not work with \(1!\), because that would turn into \(1\cdot 0!=1\cdot 0=0\) when we know that it is actually one. So to preserve the way the whole factorial system works, \(0!\) has to be one.
Going back to the zeroth term (\(\frac{1}{0!}\)) this gives us \(\frac{1}{1}\), or one. Which is the first term in our series. Woopee!
This means we CAN write the sum which generates the series like this:
\[\sum_{k=0}^\infty \frac{x^k}{k!}\]
What we are trying to do here is state what \(e^x\) is. So let's test all this with \(e^1\). Looking at our sum we can see that the \(x\)'s all appear on the top of the fractions. So all we need to do is replace them all with ones. So:
\[e^1=1+1+\frac{1^2}{2!}+\frac{1^3}{3!}+\frac{1^4}{4!}+\ldots\]
And of course one to any power is just one, so this simplifies to:
\[e^1=1+1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\ldots\]
Let's work out those factorials:
\[e^1=1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\ldots\]
So we can see roughly what is going on, we'll stick to the first five terms here. We'll also convert them all into twenty fourths:
\[\begin{align}
e^1 &\approx \frac{24}{24}+\frac{24}{24}+\frac{12}{24}+\frac{4}{24}+\frac{1}{24}\\
e^1 &\approx \frac{65}{24}\\
e^1 &\approx 2.708333333
\end{align}\]
Well, that's not bad at all after only five terms. It is definitely heading in the right direction. Let's sum up then. We have gone from:
\[e=\lim_{n \to \infty} (1+\tfrac{1}{n})^n\]
to
\[\sum_{k=0}^\infty \frac{x^k}{k!}\]
or
\[e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\]
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