"e" and the Binomial Theorem
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Written by Bryce Ringwood   


In this article, we will look at the binomial theorem and its relationship to "e". Euler's number "e", like pi (and phi) is one of the 3 most important mathematical constants. You will see we used it in the article on capacitors and we will also see why it is so important when we look at calculus. Indeed, "e" is normally introduced as part of calculus, but this is awkward because we use it as a base for natural logarithms and I want to explain these as the last topic before we move on.

Explaining "e" without calculus means that we have to know something about the binomial theorem and that, in turn, becomes easy to master once we know about permutations and combinations.S Ramanujan courtesy Oberwolfach Photo Collection

Now, I have to admit that I have had two "mental blocks" in mathematics. The first has been permutations and combinations, and the second has been logarithms. If you look for explanations in textbooks or the Internet on permutations, you will find very little to help you understand how the formulae are derived or what they really mean. Its as if the subject is so simple, you should just know.

Well, we are not all like the great mathematician Srinivasa Ramanujan (1897-1920) who was almost able to pluck mathematical theorems from the air. For colleagues trying to understand his work, there was something of a stumbling block. He would prove his theorems with very few intermediate steps, making verification of his work difficult, to say the least.

"Mr. Ramanujan's methods were so terse and novel and his presentation so lacking in clearness and precision, that the ordinary [mathematical reader], unaccustomed to such intellectual gymnastics, could hardly follow him."

This is how I feel about the explanations of permutations and combinations. I can't guarantee to do any better, but I will try.

Arrangements of ObjectsBlue? Pepper

The other day, my wife brought home a set of peppers - one red, the other green and another yellow. I found a blue one, (Not really - it was almost black) making four in all. The question is - how many ways can I arrange them ? The diagram below should give some insight:

Possible Arrangements ofFour Capsicums

These possible "arrangements" are actually the possible permutaions of the four peppers. In "natural english" we don't discriminate between the words "choice", "combination" , "arrangement" and so on. In each case, the peppers are in a different order and we consider the order to be important. You can see that with four peppers, I can put any of the four first. With three - I can put any of the three giving me 4X3, and with two peppers, I can put either one first giving me 4X3X2 and so on.

The number $4\times 3\times 2\times1 = 4!$ or "factorial 4". As we will see, it can also be written ${^4}P_4$, the $"P"$ standing for "Permutation".

By the way, a "combination" lock is really a "permutation" lock. If it weren't anyone could open it - think about it.


The following picture describes permutations - choices of k things from a set of n things. Let's take a look at how our peppers are doing:


Well, I hope that illustrates permutations. I think you can see that 4 peppers chosen two at a time will yield 12 possible permutations or choices where the order of the peppers is important. We write this as ${^2}P_4 = 4\times 3$ You can read it out loud as "two permutations of 4 objects". Now note that  ${^2}P{_4} ={ {4\times 3\times 2 \times 1}\over {2 \times 1} }={{4!}\over {2!}}$.

This leads us to express $ {^n}P{_k} ={ n(n-1)(n-2) ... (n - k + 1)} ={ n!\over{(n - k)!}}$.

Now - take a look at your calculator. If it is a recent scientific calculator, it probably has a "nPr" button.  If I have 52 playing cards - how many choices of three cards can I make taking three cards at a time, assuming the order is important (KQJ doesn't equal JQK for example)?

How many ways can you arrange 52 playing cards ? - 8,066 X 10 with 67 zeroes after it.


In order to complete this brief journey, we need to understand combinations. This time the order of our choice doesn't matter. In other words, we could work out our permutations first and then simply divide by the number of arrangements of the objects we are choosing.:

Combinations of peppers

This means that in our pepper problem, we now can make 6 choices. We can verbalise this as "6 choose 2", and it is written as ${^6}C{_2} = {(1/{2!)}\times {^6}P{_2}}$ or, when we look at the next section $\displaystyle{ {6 \choose 2}}$.

Note the relationships $\displaystyle{{{^n}C{_k}} = {  {1\over{k!}  }{{^n}P{_k} } } } ={ {^n}C{_{(n-k)}}} $ .

You will see on your calculator, a button "nCr". Now how many choices of three cards do we have from a 52 card pack, assuming order is unimportant, i.e. JQK = QJK = KQJ etc)?

The Binomial Theorem

If you have managed to follow the preceding three sections - you will be an instant genius, and understand how to prove the binomial theorem at a single blow. Almost. If you have not managed to understand combinations or choices - then I have not explained it all in a clear way and you need a clearer explanation from someone else. I won't be offended - I also found it very difficult to follow at first. Well - OK - it took me 54 years before I understood it. I think.

I'm going to skip all that stuff about Pascal's Triangle and get right to it. Here is the conjecture:

${\displaystyle{  {(a+b)^n}={  {{n\choose 0} {a^n}{b^0} } +  {{n\choose 1} {a^{n-1}}{b^1} }...  + ... {{n\choose k} {a^{n-k}}{b^{k}} }  ...+...        {{n\choose n} {a^0}{b^n} }               }          }}$

For example:

${\displaystyle{  {(a+b)^3}={  {{3\choose 0} {a^3}{b^0} } +  {{3\choose 1} {a^{(3-1)}}{b^1} }+{{3\choose 2} {a^{(3-2)}}{b^{2}} } + {{3\choose 3} {a^0}{b^3} }               }          }}$

Which can be rewritten:

${\displaystyle{  {(a+b)^3}={  a^3 + 3 {a^2}b + 3 ab^2 + b^3 }  }} $ because mathematicians have all agreed the $0!=1$.

Let's write the above a slightly different way:

$\displaystyle{ {{\color{red}{(a+b)}}{\color{green}{(a+b)}}{\color{orange}{(a+b)}} }={    {\color{red}a}{\color{green}a}{\color{orange}a} +(\color{red}{a} {\color{green}a}{\color{orange}b} + {\color{red}b}{\color{green}a}{\color{orange}a} +{\color{red}a} {\color{green}b}{\color{orange}a}) + (\color{red}{b} {\color{green}b}{\color{orange}a} + {\color{red}a}{\color{green}b}{\color{orange}b} +{\color{red}b} {\color{green}a}{\color{orange}b})  + {\color{red}b}{\color{green}b}{\color{orange}b}} }$

From the colours, you can see the combinations. Of course, $aab=aba=baa$, which is how we arrive at the combination formula. You can see that if you multiply by another $(a+b)$ the conjecture still holds, and intuitively, I am sure that you are convinced the formula at the start of this section must be true. Agreed, this is not a formal "proof". You can easily prove it for yourself by showing that if the result is true for n, then it is true for n+1 and since it is true for 3 (we just showed that), then it must be true for 4 ... and for all values of n.


Now for that magic number "e". This was first "discovered" by Jacob Bernoulli who attempted to find the value of the expression $\displaystyle{{\left(1 + {1\over n}\right)}^n}$ as $n\rightarrow \infty$. As we shall see, this defines "e". It was Leonhard Euler and John Napier who first proposed its used for tables of logarithms.

I will, of course, be mentioning "e" after the tea break, when we look at calculus, but for now, we will be using it in our discoveries into the world of logarithms.

"e" itself is an irrational number like pi. The first few digits are 2.718 281 828 ... and so on (according to a calculator). Its value has been calculated to over a billion decimal places. (How do they do that?)

Now that we know all about the binomial theorem, we can work out the value of the k-th term of the series as:

$\displaystyle{{n \choose k}{\bigg( {1\over {k!}}\bigg) }{{{ \bigg( {1\over n}} \bigg) }^k}} $.Recall that this can be expressed as:

$\displaystyle{  {\left( {1\over{k!}}\right) }   {\left( {{n(n-1)(n-2)...(n-k+1)}\over {n^k}}\right) }            }$. When n becomes very large, this value becomes simply $\displaystyle{   {1\over{k!}}  }$(Prove it with your calculator). This means we can work out e from the expression:

$\displaystyle{  e= {1\over {0!}} + {1\over {1!}} + {1\over{2!}} + ... {1\over{k!}} ....        }$ (try this also on your calculator ;)


Here we reminded ourselves about combinations, the binomial theorem and where the important number "e" comes from. Next we will find out why Napier and Euler got so excited about it when we look at logarithms.

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