Hello! In this video, I am going to explain binomial expansion which incidentally means
I will be talking about the binomial theorem and Pascal’s Triangle.
This video is first in a series of three videos and therefore the easiest.
The target audience is basically anyone with the desire to learn but from an academic perspective
high school students and undergraduate students.
In the next video, we’ll talk about the multinomial theorem
which is geared toward graduate students and in the final video we will delve into 3D figures
and some super cool formulae.
First of all, what is a binomial?
‘Bi’ as is 2 and ‘nomial’ from polynomial. Therefore, a binomial is a polynomial with two terms.
So, a binomial is basically the addition or subtraction of two terms.
(a + b) is a binomial and so is (a - b).
We won’t be worrying about (a – b) because (a + b) includes it.
I know that doesn’t sound right but let me explain by using this example:
Here you might compare it with (a – b) and say a is 2 and b is 5. Which
is perfectly correct, but you can also compare it with (a + b) by rewriting it like so. (2+(-5))^2
is the same as (2-5)^2. Now here, you compare it with a + b. So a is 2 and b is negative 5.
This is also the reason why we don’t discuss (-a + b) and (-a -b). Ok, with that
out of the way, let’s look at some binomials raised to some power.
(a + b)^0 is 1. That is a property – any number raised to 0 is equal to 1.
Now I am required by the law of mathematics to inform you that this property is true only if the
number itself is not 0. Because 0 raised to 0 is actually undefined.
This topic gets into calculus and is completely unnecessary for this video.
Next, (a+b) raised to the first power is a + b
but I am going to write it as one a plus one b.
Typically we don’t write 1 as the co-efficient in mathematics but I am going to write it here
and the reason will be apparent very soon.
Next (a+b) squared is a squared + 2ab + b squared.
In India, where I grew up, students are expected to memorize this as a formula because math instructors
are like – pop quiz.
[Instructor enters. First thing he does is:-]
Instructor: Roll no. 8, what is a +b the whole squared?
… Viraj: It is a squared plus 2ab plus b squared.
I am not one of those who memorizes every formula. I should be ashamed.
In fact, I am just good at deriving formulas quickly. Let’s see what would have happened if the
instructor had asked me.
Instructor: Roll no. 12, what is (a + b) the whole squared?
[Why me?!]
A square is basically number times itself. So, (a+b) squared is a+b times a+b. That is
a times a+b + b times a+b. Which simplifies to:
Viraj: It is a squared + b squared + 2ab
After doing this derivation tens of times, it just gets committed to memory.
And, if the operation is computationally intense,
I grab a piece of paper and derive the formula by scribbling.
Anyway, that was about India but here in the United States, they use something called the FOIL technique.
I can go off on a rant here but I am planning to make a separate video expressing why I believe FOIL is one
of the worst thing students are taught over here.
Anyway, (a+b) cubed is a cubed + 3a squared b + 3ab squared + b cubed.
If you don’t have this stuff memorized, then you can distribute
this expression like so and arrive at the same final step. But if you try this on your
own, you’ll realize that it is a lot of work. I promise I don’t enjoy dragging this
but I want you try to simplify (a + b) raised to four by distributing this expression properly.
[5 seconds]
I know most of you did not try it – shame on you for being so lazy; but
if you did then there are two of the potential things that you learnt. Firstly, steps increase
exponentially and secondly, if you make one mistake, it cascades all the way to the end.
So it is quite clear that this method is very inefficient. Let’s discover something better together.
[Transition] If you look closely, there is actually a very
nice pattern going on here. There’s a pattern with the numbers, i.e. the co-efficients, and
also a pattern with the variables. Let’s start with the pattern for co-efficients.
And to see it clearly, I am going to create a similar table but without the variables.
So, on the left column, we have the powers also known as ‘index’ or ‘indices’ for plural.
And on the right we have just the co-efficients but I have arranged the
numbers in this triangular pattern. It is a simple yet interesting arrangement of numbers
in the form of a triangle where every number is the sum of the two right above it.
For instance, take this 4 here, it is the sum of 1 and 3. This 6 here is the sum of 3 and
3. This triangle has a name in mathematics, it is called Pascal’s Triangle.
You probably guessed that this is the case because some ancient mathematician named Pascal discovered
it. And your guess is actually pretty good but still lacking so let me fill you in.
Blaise Pascal is a French mathematician who lived in the 1600s. But Chinese, Persian and Indian
mathematicians had been using this triangle centuries before Pascal was even born.
[Stop ranting Viraj! You can make a separate video for that!]
Ok, [clears throat]
Returning to mathematics. This property is, incidentally, called the Pascal’s Identity.
At this point I would like to tell you that I haven’t closed this triangle because Pascal’s
Triangle does not have an end, per se. You can follow this pattern and keep creating
the consecutive rows. I want you to create the next row by following the pattern.
[5 seconds]
1 5 10 10 5 1, simple right? And if you wanted, you could find the 6th row,
the 7th row and keep doing this forever. And just a side note for those that are super
curious – in mathematics such operations are said to have a countably infinite number
of steps. Yes! There are different sizes of infinity. Check the link if you want to learn
more. The Pascal’s Triangle is full of interesting
patterns like triangular numbers, powers of 2, powers of 11, the Fibonacci series and
lots more. Again, for more information, check the link in the video description.
Ok before jumping to the next phase of the video, I am going to very quickly talk about
two topics. 1) factorials and 2) combinatorics. Now don’t worry, even if you don’t know
anything about these topics. You just have to remember a couple of things that I am about
to tell you. First of all, factorial is represented by the exclamation symbol and the formula
is n! = n*(n-1)*(n-2)*…*1. That might look scary, but the application is quite simple.
5! = 5*4*3*2*1 = 120 and 0! = 1 by definition. I am not going to explain why 0! Is 1 because
that is a topic that deserves its own independent video. Go check out this video by Eddie Woo
if you are curious. Ok, now moving to combinatorics. (n/r) this stands for n choose r or as I like
to call it ‘n C r’. In mathematics you can find chapters or even books on this topic
because combinatorics is very extensively used in probability and statistics. But we
don’t need any of that for this video. We just care about the formula. It is: n choose
r is equal to n! over (n-r)! times r! So I just want you to know these three things:
n! expansion. 0! Is 1 by definition and the combinatorics formula. Here’s another sidenote
for the curious ones: all four of these are the same thing represented differently. Textbooks
here in the united states use this representation and scientific calculators use one of these
two and finally this representation is what I used back when I was in India. Anyway, for
the purpose of this video I will only have this representation.
Now let’s go back to our Pascal’s Triangle. This exact information can be presented using
the combinatorics symbols like so. Don’t believe me? Ok, let’s see it in action by
picking any number from this triangle and checking if it is equal to the corresponding
element from this other triangle. How about 6 from this one? The corresponding element
is 4 choose 2. We will have to use the formula to simplify this. 4 choose 2 equals 4!/((4-2)!*2!)
= 4!/(2!*2!)
4! in the numerator simplifies to 4*3*2*1
2! in the denominator is 2*1
So in the numerator we have 24 and in the denominator we have 4.
That simplifies to 6. And you can use this process to check a couple more on
your own but since I am recording this as a video I’ll stop at this and say that these
triangles are equal. So, we have figured out the pattern of the
numbers which are the co-efficients. We still have to find the pattern for variables. Don’t
worry, the pattern for the variables is very distinct. In fact, I am going to ask you to
pause and find it on your own.
[5 seconds]
I will just copy this (a+b) cubed equation to this section.
We know that the co-efficients come from this row of Pascal’s Triangle.
Now if you look closely the pattern that a follows is that it starts at 3 and keeps decrementing
by 1. Some of you may be wondering what happened to a in the last term. Well a raised to 0
is 1. So this expression is the same as this: 1a^3b^0+3a^2b+3ab^2+1a^0b^3. And b starts
at 0 and increments by 1 until it reaches 3. So (a + b)^3 gets its co-efficients from
this row of pascal’s triangle and a starts with power of 3 and then a squared then a
raised to 1 and finally a raised to 0. B starts with a power of 0 then b raised to 1 then
b raised to 2 and finally b raised to 3. Wow! That was quite a lot but with this information
you can expand (a + b) raised to any natural number power. A couple of facts for you to
remember: For each term, the power of a + the power of b is equal to what you raised
the binomial to. As you can see, 3+0=3, 2+1=3, 1+2=3 and finally 0+3=3. Second fact is that
the total number of terms is one more than the power of the binomial. This was (a + b)
raised to 3, therefore its expansion has 3+1 terms. Well 4 terms. How about you try (a
+b) raised to 7 manually.
[5 seconds]
There is an obvious problem with this method
and if you actually tried to do what I asked you then you to by raising (a+b) to the seventh power
then you most likely experienced it firsthand.
In order to find the required row, you had to find all the previous ones too. And if you made any mistake,
it cascaded down all the way. So, clearly this method is better than the previous one where we used
distribution but still it is not the most efficient one. How do we solve this issue?
Well, by using the other version of Pascal’s triangle – the one that had the combinatorics
symbols instead of numbers. Let me explain, using this combinatorics method, we don’t
care about the previous rows. We know that the required row is 7 c 0, 7 c 1 , 7 c 2 all the
way up to 7 c 7. And to find the numeric values we can use the formula or we could use the
nCr function which you can find on any scientific calculator. This way we are completely independent
of the previous rows. Now, you just have to align these co-efficients
with their respective variables. A starts at maximum which is 7 in this case and decrements
by 1 until it hits 0 whereas b starts at 0 and increments by 1 all the way to 7.
Now you have to put all this stuff into one equation
and there you have it - binomial expansion of (a+b) to the 7th power. This is quite self-explanatory but let me
put this out there – this combinatorics method is also resistant to that cascading
error effect – where one error messes up everything. All things considered, for expanding
a binomial – this combinatorics method is the way to go.
[Viraj copying Michael from Vsauce]
Mathematicians love to put things in a general form. You have probably seen general forms
for quadratic equations, linear equations, matrices, etc. So why not put this idea into
a general form: (a + b)^n where n is any natural number like 55, 100 or even 500. Let’s do this in pieces.
First just the co-efficients. Using the combinatorics method we know that the
co-efficients will be nC0, nC1, nC2, … nCn. Now lets look at the variables: we know that
a will start at the maximum which is n and keep decrementing by 1 until it reaches 0.
So a^n, then a^(n-1), then a^(n-2) all the way up to a^(n-n) which is essentially a^0
which in turn is 1. B Will start at 0 and increment by 1 until it reaches the maximum
which is n. So b^0, then b^1, then b^2 all the way to b^n.
Now we just have to put these together in one equation. (a + b)^n = nC0*a^n*b^0 + nC1*a^(n-1)*b^1
+ nC2*a^(n-2)*b^2 + … + nCn*a^0*b^n.
This incidentally has a name in mathematics
– it is called the binomial theorem. Most books use something called proof by induction
to prove it and I am planning to make a separate video just proving the binomial theorem using
induction. I will put a card here, in case you are interested. Anyway, most textbooks
also take it one step further in terms of simplifying. They do it by using the summation
operator. To incorporate summation into this equation
we need to find another pattern which I am sure some of you have already spotted. But
those that haven’t figured it out, here’s a hint. Try to relate the power of a to its
combinatorics symbol. Likewise try to relate the power of b to its combinatorics symbol.
[5 seconds]
Ok the power of b is the easier one. It basically gets this number. As for
the power of a, it is always this number minus this number. And using this newly acquired
information, we can rewrite this equation of binomial theorem like so: Summation r goes
from 0 to n of n choose r * a raised to (n-r) times b raised to r. If you understand this,
Awesome! But if you don’t have the foggiest idea of what this is, that is totally fine
because I did not even try to explain it. The idea is quite simple, my aim is to explain
binomial expansion using the binomial theorem and you don’t need to know this summation
stuff to understand and use the binomial theorem. But still, I am willing to do a sloppy job here.
This means r takes integral values from 0 all the way up to n and after each value
you put a plus sign. Since r starts at 0, you put 0 in place of r and get nC0*a^(n-0)*b^0
then you addition sign and then the next value of r which is 1 and plug in it and keep doing
this process until you reach n. You get this expression which is the binomial theorem.
That is all that I have for you for this video. I am going to create a video where I solve
a couple problems using the binomial theorem just to give you an opportunity to apply this
new knowledge and originally the idea was to include it in this video but this one just got too long
and therefore, I decided to separate it.
I will be posting a link in the video description with resources and additional interesting stuff.
If you enjoyed the video consider becoming a subscriber and giving this video a thumbs up and maybe
even turn toggle the bell icon. And as always thank you for watching!
[Thank you for surviving this cringefest]