We continue to work to the power rule for derivatives, and we will be using the counting techniques we introduced in Counting, Permutations and Combinations. In particular, we will need the binomial theorem, so we provide the theorem and an explanation of why it is true. Once we have this, we will be able to move on to working with derivatives.
Binomial theorem
The binomial theorem states that
That is, if we expand the expression (x+y)^{n}, we get that the coefficients of the terms x^{i}y^{ni} can be determined by finding the number of combinations of n objects i at time. I have found that as written, especially in summation notation, this theorem often confuses and scares students. They often don’t even attempt to get a grasp the idea because of this initial shock in the notation. However, taking the time to go through and explain it, makes it much more accessible.
To do so, if we are to expand (x+y)^{n} we will take
where we have the product of n monomials. When trying to explain counting techniques, I went through an example with you and used that to explain the general situation. While you can take the product for n=2 then n=3 and try to see a pattern, this will actually make the process more difficult because the way you would normally multiply these out doesn’t easily give our result. Instead, in this case, I work on the general case to begin with.
Therefore, if we do take the product of these n terms, note that we will multiply either x or y in the first case by x or y in the second by x or y in the third and so on up to the nth case. If I do this I will end up with something like xxyxyyxyyx…yxyx. That is, I am paying attention to the order they came out in. In so doing, I note that in multiplying out all of the terms, I will get all possible outcomes where an entry is either x or y and they are the same entry if and only if they are the same in order.
Note that the coefficient in front of each of these terms is 1. Since, multiplication is commutative, we can rewrite any such term as x^{i}y^{ni} for some i. Therefore, if we want to find the coefficient in front of this term, we need to count the number of distinct outcomes above that correspond to rearrangement. In order to get something equivalent to this, note that from the n different monomials, we would have had to choose x in i of them and y in the rest. Therefore, if we can count the total ways of doing this by saying, pick i of the monomials to be x out of the n total options. Since choosing the first and second to be x is the same as choosing the second and first to be x, we note that in this manner, order doesn’t matter. Therefore, there are exactly n choose i ways to do this, which is precisely what the binomial theorem states.
While we may not have multiplied out n terms, we have given a very detailed description of why the theorem is true. In so doing, we have also taken a scary looking equation and found that all it’s really telling me is that if I multiply these terms out, the number of ways to end up with a certain number of xs is precisely n choose i. I find in class that this seems significantly less scary to students and is worth the effort to go through.
Pascal’s Triangle
Pascal’s Triangle is an organization of the coefficients in front of the expansion of terms in (x+y)^{n}. That is, we take the coefficients and organize them into a triangle pictorially and get something like
Here I have numbered the rows on the left using green and the diagonals on the left using purple. How we use this table is, if we want to find the coefficient for x^{i}y^{ni} in the expansion of (x+y)^{n}, we look at the nth row and ith column. The corresponding entry is then the coefficient.
Another way to look at Pascal’s Triangle, is that it gives the number of combinations of n objects i at a time as the number in the nth row and ith diagonal (note that we start counting at 0). Since the binomial theorem states these are indeed the same thing, both interpretations are equivalent.
The interesting thing about this triangle, however, is the ease with which we can write it up. That is, we don’t have to individually calculate the values, instead we can always place 1s on the beginning and end of a row and we can find the other number by adding the entry in the previous row and the same diagonal to the entry in the previous row and the previous diagonal.
For example, if I want to find the 6th row (which is not pictured above) we would start on the left with a 1, then the next entry would be 1+5=6, followed by 5+10=15, then 10+10=20 and so on until reaching the end of the row where we put another 1. This really is quite handy. I regularly use this since it makes finding expansions of monomial products much quicker. This particularly comes in handy in class because I don’t have the time to actually multiply out such things.
Even though this pattern is correct, why does it hold? Just noting that its there doesn’t give us any real insight into what’s going on, so we have this handy tool, but no idea how it works. As a curious mathematician, this is what I want an answer to. Therefore, what is the pattern actually saying. Well, if we want to find an entry in the triangle in the nth row and ith column, we find this by taking the entry in the (n1)th row and ith diagonal and adding the entry in the (n1)th row and (i1)th diagonal. Since these entry are defined as the number of combinations, what we are really saying is that
This seems interesting. We in fact know how to find the number of combinations using factorials, and we can indeed prove that this is correct algebraically. If you’d like to see that, I’ve provided a write up of the proof. However, this really doesn’t provide any useful insight into why it’s true. Instead, let’s think about this another way.
Suppose that you have n elements, and you want to pick i of them when order doesn’t matter.
 Then the number of ways to do this is n choose i, as we saw earlier.
 On the other hand, choose any element in the set (there must be at least one since n is greater than or equal to 1) and call it a.

 Then, if I pick i elements from the larger set, a is either in the set of i elements, or it isn’t, but it can’t be both in and not in at the same time.

 The number of ways to have a in the set, is


 We have one choice to put a in the set.



 We can then choose the rest of the elements.



 There are n1 choose i1 ways to do this.


 On the other hand, the number of ways we can have a set without a is,


 We choose the elements, without allowing a to be chosen.



 There are n1 choose i ways to do this.

 The total number of sets with i elements is the total number of sets with i elements and a in it plus the total number of set with i elements but a not in it,
 Hence both sides of the above equation provide two ways to count the same thing.
I do enjoy going through this with students, since it provides insight into what is going on instead of just showing a lot of arithmetic. As such, I feel it helps understand what is going on much better, and they seem to enjoy the argument. While we won’t use Pascal’s Triangle in our explanation for the power rule, it was difficult to mention the binomial theorem without mentioning Pascal’s Triangle.
Conclusion
I should note that in Calculus I give the students the binomial theorem without this justification. While I feel they could follow the argument presented, I suggest that if they are interested in such a proof, they can come by during office hours so that I can explain it to them. However, I do go through this argument in my introduction to proofs class. It is a great example of a combinatorial proof. Furthermore, since I mention the result in earlier classes, I am referring back to earlier material for the students, which helps reinforce the idea and promote retention. Since I do primarily have math majors in this course, this is one of the more popular topics in this section. While students with other majors enjoy talking about cards, dice, or other games, the math majors get excited by the mathematical result.
Next time, we will continue to work toward the power rule by talking about limits. If you enjoyed the post or the series, make sure to like it. Also, I would greatly appreciate it if you’d take the time to share the post or webpage on Social Media.
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