WEBVTT mathematics/pre-calculus/selhorst-jones
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Hi--welcome back to Educator.com.
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Today, we are going to talk about the binomial theorem.
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Long ago, when we studied polynomials, we learned that a **binomial** is an expression that has two terms.
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For example, each of the below is a binomial, because it has two separate terms:
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x + 7; 2l³ - 5; 3y² + √2z.
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A binomial is anything that we can put in the form a + b, some a and some b.
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All we are looking for to be a binomial is just having two terms.
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What if we wanted to see what happens if we raise a binomial to some power?
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If it was small, like (a + b)², we could FOIL it; we could put one factor next to another factor,
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and then we would just multiply them out by distribution and just work it out.
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We are used to doing that; we have been doing that for years--it wouldn't be that difficult.
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But what if it was really large, like (a + b)⁷, or even larger?
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We could do it by just putting all of the factors out, but it is going to get really big and really messy really fast.
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It is not going to be easy for us to multiply out (a + b)⁷, and it would be even worse if it was a larger value than 7.
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So, in this lesson, we are going to learn how to expand a binomial for any arbitrary power n.
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If we have any integer n, we will be able to use this to figure out how to expand (a + b) raised to the n.
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To help us understand what we are working with, let's look at (a + b)^n expanded for various values of n.
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(a + b)⁰ will just come out to be 1, because if you raise anything to the 0, you get 1.
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(a + b)¹ comes out to be a + b; (a + b)² would come out as a² + 2ab + b².
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(a + b)³ would be a³ + 3a²b + 3ab² + b³.
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(a + b)⁴ equals a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴.
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And (a + b)⁵ would equal a⁵ + 5a⁴b + 10a³b² + 10a²b³ + 5ab⁴ + b⁵.
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We could keep going in this way; but we are starting to see a pattern at this point.
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We can realize that there is a pattern going on, and we can use this pattern to let us figure out general things about a + b, to just any n whatsoever.
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Let's look specifically at (a + b)⁴ and (a + b)⁵ above
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to help us see some properties about how binomials work.
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We will be able to talk about (a + b)^n, the general expansion of a binomial, by looking at these in specific.
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So, looking at these two as stand-ins for the general way that (a + b)^n expands, we notice various things.
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The first thing is that the expansion of (a + b)^n has n + 1 terms.
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What I mean here is that (a + b)⁴--well, that would be n = 4;
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we count how many terms come out in the expansion: 1, 2, 3, 4, 5: there are 5 terms total.
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We started with n = 4; 4 + 1, n + 1, gets us 5; so we have 5 terms coming out of it.
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The same thing happens with (a + b)⁵; we have 1, 2, 3, 4, 5, 6 terms total: 5 + 1 is 6 terms.
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So, we end up getting n + 1 as the number of terms that come out.
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The expression also has symmetry; powers of a decrease by 1 each step, every time we move forward a term, while powers of b will increase by 1.
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If we have (a + b)⁴, for example, we start at a⁴, because (a + b)⁴...
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well, the largest kind of a that it will be able to produce (largest in the sense that it will have the largest exponent) will be a⁴.
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We manage to get a⁴ out of it; and then the next one--we will end up having a³
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(it has just stepped down 1), and then the next term would be a² (step down again), a (step down again),
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and finally it turns blank; and a blank is another way of saying a⁰, because a⁰ is just 1, so it disappears.
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The same thing happens with the b's, except that it does the reverse process.
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It increases; since it is blank, it doesn't appear here; that is b⁰, and then we have b¹;
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and then b², and then b³, and then finally b⁴.
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Our a's are decreasing by 1 each step, at the same time that our b's are increasing.
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a starts at n; b starts at 0; and then click, click, click, click, click, click...until finally a is at 0 and b is at n,
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and we have finished their expansion in terms of the a and b exponents.
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The sum of the powers for each term comes out to be n.
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Here, our n is 4; if we grab any term inside of this...we will end up seeing a³b¹:
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if we add up the powers on these terms, 3 + 1 gets us 4.
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If we grab another one, like a²b², 2 + 2 gets us 4.
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Even at the extremes...a⁰ and b⁴: well, 4 + 0 gets us 4.
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This ends up happening with (a + b)⁵, as well; if we grab any one of the terms here...
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this is a 4; this is a 1; 4 + 1 gets us 5; so if we add the powers, we always end up getting n.
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Whatever our (a + b)^n is, adding the powers of the a and the b will always come out to be equal to the value of n.
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And finally, the coefficients vary symmetrically; there is symmetry in how the coefficients are.
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If we start in the middle, there is a 6 here by itself, because it is in the very middle, so it can't be symmetric with anything else.
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And then, we have 4's matching on either side, and then we have 1's (effectively, the coefficient here is a 1,
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because it doesn't appear at all, so its coefficient is 1); and the 1's match on either side, as well.
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The same thing here: the 10's match, and the 5's match, and the invisible 1's match.
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All right, so we end up seeing that the coefficients vary symmetrically; great.
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With all of these ideas in mind, these properties, we see that any expansion of a binomial, (a + b)^n,
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is going to have this form of a^n, a^n - 1, a^n - 2,
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until we finally work down to a², a, and then just a⁰ in here.
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And we reverse that with the b's; we have b⁰ at the beginning, and then it works up:
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b¹, b², working up...b^n - 2, b^n - 1, and b^n, finally.
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So, we are always going to have this form; the only question is what goes in the blanks in front of it.
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We call these blanks binomial coefficients, because they are the coefficients of the binomial expansion.
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They are the number multiplying against that binomial having been expanded--
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each one of the coefficients for the expansion of some a, some b, and powers on those a's and b's.
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The only question we have at this point is what those coefficients will be.
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We get this with the binomial theorem; if we want to know the binomial coefficients of some binomial expansion, we can use the binomial theorem.
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The binomial theorem says that expanding some (a + b)^n, the coefficient of the term a^n - kb^k
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will be n choose k (and n choose k is just the same thing as n! over (n - k)! times k!)...remember, we talked about this n choose k,
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this parentheses with the values n and k above each other; this is n choose k--we have talked about this before...
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You can also write it as nCk, or c of n,k; but we still pronounce all of these spoken aloud the same way: "n choose k."
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We first learned about this idea in the lesson Permutations and Combinations.
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So, if you really want a lot more experience with how that works, you can go and look that up.
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But really, we have enough, just for what we are looking for here, from what we will be able to get in just this lesson.
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Also, remember: the exclamation mark means factorial, and factorial is the value that we get
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when we multiply a number by all of the positive integers below it.
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For example, 5! is equal to 5 times 4 times 3 times 2 times 1; we multiply those all together, and we get the value of 5!.
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Finally, we define 0! as equal to 1; we just say 0! = 1; that is the end of the story, because it will help us out.
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It makes things easier, and there are also some other reasons that we don't really need to go into.
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You have to memorize that 0! = 1; that is just that.
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All right, let's see the binomial theorem get used in an example, so we can see how it works.
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For any (a + b)^n, the coefficient of the a^n - kb^k term is n choose k
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(which is the same thing as saying n!/(n - k)!(k!)).
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For example, if we had (a + b)⁷, and we wanted to know the term containing a²b⁵,
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when expanded; that is going to show up, because 2 + 5 is 7;
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so we know that a²b⁵ will show up in the expansion.
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We could figure out what the term would be with the binomial theorem.
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We know that it is going to be n choose k; so the only thing that we have to do is figure out what our n is;
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well, it is (a + b)⁷, so it means that n = 7.
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What is our k? Well, it is b⁵, so our k equals 5.
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a^n - k, b^k...we can also check, because it is n - k here, and this is 2;
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well, if we take 7 - 5, sure enough, we get 2, which is this value right here.
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So, it checks out; we are following the method of the binomial theorem.
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We see how the binomial theorem works: we have a^7 - 5 times b⁵; that is a²b⁵.
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We know that the coefficient is going to be n choose k; our n is 7; our k is 5; so we have 7 choose 5.
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That expands to 7!/2!(5!) times a²b⁵.
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And that will simplify to 21 times a²b⁵; so we have been able to figure out what the coefficient is,
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what the term would look like when it is fully expanded: we get 21a²b⁵.
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If you are not quite sure how we get the value of 21 out of that, we can figure it out,
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because 7 choose 5...well, we wrote here already that that would be 7! divided by 7 - 5 (that is 2), factorial, times 5!.
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7! is the same thing as 7 times 6 times 5 times 4 times 3 times 2 times 1.
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We can also write 5 times 4 times 3 times 2 times 1 as just 5!.
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And now we have things we can cancel out; 2! is just 2 times 1, so we can just leave that as 2; so it is 2 times 5!.
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We have 5! and 5!; they cancel out; 7 times 6 over 2...6 cancels into 3 times 2, so our 2's cancel out; we have 7 times 3, or 21;
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and that is how we end up getting the 21 for our coefficient.
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And we know it will be a²b⁵, because we figured out that it was the coefficient of the a²b⁵ term.
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Great; at this point, we can now know the coefficient of any term, which means that we can expand (a + b)^n with relative ease.
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(a + b)^n will end up being n choose 0 a^n, because it is a^k of 0 there,
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because b hasn't even shown up, so it is b⁰.
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And then, next we would have n choose 1: a^n - 1b¹; our k there is 1.
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Next, we would have n choose 2 times a^n - 2b².
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And we would continue on in this fashion, until we have n choose n - 2; n minus (n - 2) comes out to be +2;
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the minus on the -2 becomes positive when we get just +2 on our a; and k is n - 2 b^n - 2.
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n choose n - 1: ab^n - 1; and finally, n choose n: it is now a^n - n, or a⁰, so it just disappears; and b^n.
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We can now expand out the whole thing with relative ease.
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It will definitely still take some effort; we will have to figure out what each of these "choose" values comes out to be.
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But we can write the whole thing out, expanded, and we will be able to get it out.
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And it will still be easier than doing each term, and then multiplying (a + b) and (a + b) and (a + b) and (a + b), and trying to FOIL the whole thing.
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Expanding factors like that would just be a real pain, if it was a large thing, like (a + b)^10.
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So, this is really useful if we are trying to expand a very large exponent, n.
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Also, we can condense this in a really nice way.
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This is a monster of an expansion; it would be a pain if we had to write this repeatedly.
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We can use sigma (or summation) notation to condense this into a much smaller thing.
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(a + b)^n is equal to the sum of k = 0 to n of n choose k times a^n - kb^k.
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If you want to check and make sure that that works--well, if we plug in k = 0, sure enough, we end up getting this term.
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n choose 0: a^n - 0 is a^n; b⁰ just disappears, so we end up getting that.
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And we work all the way through; here would be k = n; n choose n; a^n - nb^n; we end up getting n choose n b^n.
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And we end up seeing all of the terms here show up; there would be k = 1, k = 2, k = 3 next, k = 4...until we finally get to k = n - 2, k = n - 1, k = n.
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We can write the whole thing with sigma/summation notation.
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If this is really confusing, if you really haven't done much work with sigma/summation notation,
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remember to check out the lesson Introduction to Series; we will talk about this.
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We will have a much better understanding of sigma/summation notation (that is a mouthful) and how it works through.
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So, make sure that you check out that lesson if this is really confusing and you have to use it.
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Proof of the binomial theorem: Proving the binomial theorem is within our reach.
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We could prove this; it is quite challenging, but it is an interesting proof.
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And it will give us the chance to try out our shiny new proof-by-induction.
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We will use proof by induction to prove this thing.
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Working through it will really deepen our understanding of how mathematics works.
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We will get a much better understanding of how some really complicated stuff in math works, by just working through it.
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However, proving it won't directly help us see how to use the theorem.
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If we work through it, it won't directly help us in any way for the kind of problems that you are likely to have to work on right now.
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It will help us understand why the theorem works, but it is not likely to actually make anything easier that we actually have to do in class.
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As such, the proof for the binomial theorem is in this lesson; it is here; but it is at the end, after the examples.
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If you are interested, I encourage you to go ahead and check it out; it is some cool stuff.
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You will get the chance to really flex your math muscles.
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We are basically going to be looking at a college-level proof.
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But I believe that, if you are interested in this sort of thing, and you put a little bit of effort
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into watching it and working through it, you will totally be able to understand it.
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So, if you are interested in this kind of math, go ahead and check it out; it is a really cool thing.
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We will spend a while working through it and explaining it carefully.
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But at the same time, if you are busy, or you don't really care--you are not that interested in math--
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you are just working through this so that you can get a grade in your class, well, I would like for you to be interested in math;
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but I am not going to be able to always make you interested in math.
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So, I just want you to do well; but to be honest, you don't actually have to understand how this proof works
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to do well in a math class at this level, or even in the next level.
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This is more something for if you think you might be really interested in serious math or serious-level science courses.
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You probably want to check this out; but that is still even going to be a couple of years away, if you are currently in a precalculus class like this.
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All right, so check it out if you are interested; but if you are not, don't worry about it.
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Pascal's triangle: we can arrange the binomial coefficients in a triangular pattern to make a thing called Pascal's triangle.
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It is named after one of the people to have found it and discovered it.
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We can have n = 0 as our first row (well, the 0th row, we will call it), n = 1 as the next row (the first row),
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n = 2 as the next row (what we will call the second row), n = 3, n = 4, and so on, and so on.
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Notice: we have our n value at the top each time; so it is n = 0 and then n = 1, n = 2, n = 3, and so on and so on...
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And then, we always have 0, and then 0, 1, 0, 1, 2, 0, 1, 2, 3...until it finally ends up having the number on top match the number on bottom.
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And that is how we end up making the triangle.
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Why is this interesting? We are going to see a really interesting pattern emerge when we replace each n choose k,
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each of these like this, with what the value is that ends up coming out of it.
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What does 2 choose 0 come out to be? We replace it, and we start to see a pattern emerge.
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We replace it, and we end up getting this: n = 0 gets 1 for its row; n = 1 gets 1 and 1; n = 2 gets 1 2 1.
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n = 3 is 1 3 3 1; n = 4 is 1 4 6 4 1; you might have seen this before.
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Notice how each row can be created from the row above through diagonal addition of the terms.
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For example, if we go out from n = 0, and we go out diagonally left and diagonally right,
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1 added down would end up getting us 1 and 1.
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Then, if we add diagonally again, down left, down right, and then the same thing for the other one--
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down left and down right, well, 1 +...only by itself would come out as 1.
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1 + 1 would get us 2; it is this one and this one, combined together, because it has both of the diagonals coming in through here.
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And then, 1 only on itself would get us 1 here.
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And we can keep going along with this pattern: 1 diagonally out, 2 diagonally out, 1 diagonally out...1 just by itself would get 1;
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1 + 2 would get us 3; 2 + 1 would get us 3; 1 by itself would get us 1.
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And we can keep going in this pattern: 1 on itself would get 1; 1 on 3 is 4; 3 on 3 is 6; 3 on 1 is 4; 1 by itself is 1.
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So, by doing this, we can extend the triangle down to whatever value of n we are interested in considering.
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Whatever value of n we are interested in considering, we can just get to by diagonal addition of terms.
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For example, if we are interested in knowing what n = 5 is, then we just expand out diagonally like this, through diagonal addition.
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And we can work it out: 1 added to itself just gets 1; 1 added to 4 gets us 5; 4 added to 6 gets us 10;
00:17:53.600 --> 00:17:59.300
6 added to 4 gets us 10; 4 added to 1 gets us 5; 1 added to itself gets us 1.
00:17:59.300 --> 00:18:04.000
And we have been able to create the row n = 5, the fifth row.
00:18:04.000 --> 00:18:09.000
Since the first row is based off of n = 0, it is called the 0th row.
00:18:09.000 --> 00:18:14.400
So, here is the 0th row; this will help keep us from having confusion later on.
00:18:14.400 --> 00:18:21.100
The next row is called the first row, because it is based off of n = 1; so here is the first row.
00:18:21.100 --> 00:18:24.600
We might think that the first row would be the one on top, but not with Pascal's triangle,
00:18:24.600 --> 00:18:28.800
because it is really based off of n = 0, and then n = 1.
00:18:28.800 --> 00:18:34.500
So, the second row would be the one connected to n = 2.
00:18:34.500 --> 00:18:40.900
It is based on the number of the n, not the actual location of the row, so much as if you were just counting location of rows.
00:18:40.900 --> 00:18:48.300
It is based on the n that it is connected to: n = 4 would produce the fourth row, and so on and so forth.
00:18:48.300 --> 00:18:51.600
Why do we care about Pascal's triangle--why is this thing useful?
00:18:51.600 --> 00:18:57.900
Because the nth row gives the coefficients to expanding (a + b)^n.
00:18:57.900 --> 00:19:01.700
The way that we set this up in the first place was by using those binomial coefficients.
00:19:01.700 --> 00:19:07.700
Remember: all of those binomial coefficients, the n choose k--we had that set up based off of which n we were on.
00:19:07.700 --> 00:19:11.900
So, what we did was put down all of the binomial coefficients for a given n value,
00:19:11.900 --> 00:19:18.700
which means what (a + b)^n is...we have just talked about all of the coefficients for expanding some (a + b)^n.
00:19:18.700 --> 00:19:22.200
It is really handy, if we have to expand the entire thing.
00:19:22.200 --> 00:19:24.400
For example, since the nth row gives the coefficients
00:19:24.400 --> 00:19:29.500
to expanding (a + b)^n, we can expand (x + y)⁴ using the triangle.
00:19:29.500 --> 00:19:36.200
If it is (x + y)⁴, then that means we are dealing with n = 4; so we care about this row of the triangle.
00:19:36.200 --> 00:19:38.800
We work this out, and that means that it is going to be (x + y)⁴.
00:19:38.800 --> 00:19:44.600
Well, we know that it is going to be some blank, and it will start with x⁴y⁰...
00:19:44.600 --> 00:19:54.500
plus _x³y¹ (y goes up by 1; x goes down by 1) + _x²y²
00:19:54.500 --> 00:20:02.200
+ _x¹y³ + _x⁰y⁴.
00:20:02.200 --> 00:20:13.300
At this point, we can now plug in each of our binomial coefficients: 1 will go here; 4 here; 6 here; 4 here; 1 here.
00:20:13.300 --> 00:20:15.400
And we can also write the thing a little bit more simply.
00:20:15.400 --> 00:20:21.100
1x⁴y⁰: well, 1...we normally don't write a coefficient of 1; so we can just write this as x⁴,
00:20:21.100 --> 00:20:24.900
because y⁰ also just turns into a 1, and we don't normally write out the 1's.
00:20:24.900 --> 00:20:32.000
x⁴ + 4x³y¹...we normally just write that as y, so we can write that like that;
00:20:32.000 --> 00:20:38.800
plus 6x²y² + 4x¹ (we normally just write it as x) y³,
00:20:38.800 --> 00:20:47.100
plus 1x⁰...well, we normally don't write 1's; x⁰ just becomes 1, so it will disappear as well;
00:20:47.100 --> 00:20:52.200
and we are left with just y⁴; so we end up being able to expand all of (x + y)⁴,
00:20:52.200 --> 00:20:58.800
using Pascal's triangle, which tells us the binomial coefficients for the associated n row.
00:20:58.800 --> 00:21:02.300
We can use that triangle, and we can expand things pretty easily.
00:21:02.300 --> 00:21:06.400
If we had tried to expand (x + y)⁴ by hand, we would probably still be doing it.
00:21:06.400 --> 00:21:10.200
It takes a while; and this was even through explaining it.
00:21:10.200 --> 00:21:12.800
If we were just punching it out, we would be able to do this really quickly.
00:21:12.800 --> 00:21:21.800
So, Pascal's triangle is really useful for when we have to expand some binomial quickly and easily, without having to take a whole bunch of time--pretty useful.
00:21:21.800 --> 00:21:23.600
All right, we are ready for some examples.
00:21:23.600 --> 00:21:31.500
Use the binomial theorem to give the coefficients for the term containing s⁴t⁷ from (s + t)^11.
00:21:31.500 --> 00:21:34.600
The first thing that we want to identify is what the n we are working with is.
00:21:34.600 --> 00:21:37.600
The n that we are working with is 11.
00:21:37.600 --> 00:21:47.700
All we care about is s⁴t⁷; we are asked to find the binomial coefficient for that term when we expand (s + t)^11.
00:21:47.700 --> 00:21:52.000
We don't really care about all of the other terms; all we really care about is s⁴t⁷.
00:21:52.000 --> 00:21:58.100
What is that going to be? That is going to end up being some n choose k.
00:21:58.100 --> 00:22:01.600
And if we wanted to have that, it would also be s⁴t⁷.
00:22:01.600 --> 00:22:05.200
So really, at this point, all that we need to find is what k is.
00:22:05.200 --> 00:22:18.100
k is what? Well, it is normally...from the binomial theorem, we would have n choose k; a^n - kb^k.
00:22:18.100 --> 00:22:26.200
So, in this case, we have n choose k, but we are not using a and b; we are using s and t.
00:22:26.200 --> 00:22:31.500
So, it will be s^n - kt^k; what is our n?
00:22:31.500 --> 00:22:38.200
We have 11 choose k, times s^11 - kt^k.
00:22:38.200 --> 00:22:47.800
Well, if we have t⁷ and s⁴, then it must be that k has to be equal to 7.
00:22:47.800 --> 00:22:52.700
11 - 7 gets us 4, and k = 7 gets us t⁷.
00:22:52.700 --> 00:22:56.200
So, we now know what k is; we now know what n is; so we can plug this in.
00:22:56.200 --> 00:23:02.800
We have 11 choose 7, times s⁴t⁷; that will be the s⁴t⁷ term.
00:23:02.800 --> 00:23:06.500
What comes out to be 11 choose 7? Let's go do that in a sidebar.
00:23:06.500 --> 00:23:18.400
11 choose 7: well, that is going to be 11! over (11 - 7), 4, factorial, times 7!.
00:23:18.400 --> 00:23:24.700
11!...well, we can write that as 11 times 10 times 9 times 8 times 7....
00:23:24.700 --> 00:23:29.900
But we can also just write it as times 7! now; that is convenient, because we have 7! on the bottom.
00:23:29.900 --> 00:23:38.900
So, 4! times 7!...we cancel out the 7!'s, and we have 11 times 10 times 9 times 8.
00:23:38.900 --> 00:23:43.900
On the bottom, we can expand 4! as 4 times 3 times 2 times 1.
00:23:43.900 --> 00:23:51.300
We will just leave the 1 off; 2 and 4 become 8, which cancels the 8 on top; 3 cancels with part of the 3 of the 9,
00:23:51.300 --> 00:23:54.400
so we have times 3, because we cancel out one of the 3's from the 9.
00:23:54.400 --> 00:24:06.600
11 times 10 times 3 comes out to be 330; 330 is our coefficient, so we get 330 times s⁴t⁷.
00:24:06.600 --> 00:24:10.600
That is what we get as the coefficient of the term containing s⁴t⁷.
00:24:10.600 --> 00:24:19.900
The coefficient is just the 330 part, and the whole term would be 330 times s⁴t⁷, when we expand (s + t)^11.
00:24:19.900 --> 00:24:22.600
Now, could you imagine how difficult that would be to do by hand?
00:24:22.600 --> 00:24:27.200
That would be practically impossible for us to do by hand, because it would take so much time and effort.
00:24:27.200 --> 00:24:33.300
But by using this n choose k business, we are able to figure out what the coefficient is relatively easily.
00:24:33.300 --> 00:24:40.600
Use the binomial theorem to give the coefficient for the term containing x^10y^12 from (x² + y³)⁹.
00:24:40.600 --> 00:24:46.600
The first thing that I want to point out: x^10y^12: well, what we talked about before was:
00:24:46.600 --> 00:24:51.600
if you take the exponents, and you add them together, well, what is our n?
00:24:51.600 --> 00:24:59.800
Our n here is n = 9; but if we add 10 and 12 together, 10 + 12 is 22; what?
00:24:59.800 --> 00:25:02.400
Wait, that doesn't make sense; what is going on here?
00:25:02.400 --> 00:25:08.200
The issue isn't that it is x^10y^12, because what we really have is a and b.
00:25:08.200 --> 00:25:17.900
But what is our a and b? Our a and b are x² and y³; a = x²; b = y³.
00:25:17.900 --> 00:25:34.900
So, it is (a + b)⁹; so when we expand this out, what we are looking at is n choose k on a^n - kb^k.
00:25:34.900 --> 00:25:41.000
But now we have called out a and b as x² and y³, because that is what the actual two terms of it are.
00:25:41.000 --> 00:25:45.700
It is not just x, and it is not just y; it is x² and y³.
00:25:45.700 --> 00:25:48.800
So, we have some n choose k; we will plug those in later.
00:25:48.800 --> 00:26:00.700
And so, it is x² to the n - k, and b--what is b?--b is y³, to the k.
00:26:00.700 --> 00:26:09.200
So now, we need to figure out what value k has to be for us to be able to get x^10y^12.
00:26:09.200 --> 00:26:17.100
Well, if we have y³ to the k, and we know that that ends up being the same thing as y^12,
00:26:17.100 --> 00:26:23.700
well, then k has to be equal to 4, because (y³)⁴ would be 3 times 4, so y^12.
00:26:23.700 --> 00:26:27.500
So, we end up getting y^12 from setting k equal to 4.
00:26:27.500 --> 00:26:40.700
And similarly, 9 minus 4, since it is a^n - k here...9, our n value here, minus 4, our k value here, would get us 5 for n - k.
00:26:40.700 --> 00:26:47.200
And x² to the fifth does come out to be x^10; 2 times 5 is 10.
00:26:47.200 --> 00:26:55.400
At this point, we now see what our k is; our k is 4; our n is 9; so we can plug those in.
00:26:55.400 --> 00:27:10.200
And what we are looking at is 9 choose 4, times x² to the n - k, so in this case 5, y³, to the fourth.
00:27:10.200 --> 00:27:15.100
So, that is x² to the 5; that does become n; y³ to the fourth does become 12.
00:27:15.100 --> 00:27:19.300
And 5 and 4 put together do become 9; so everything checks out--this makes sense.
00:27:19.300 --> 00:27:23.200
At this point, we only need to figure out what 9 choose 4 is.
00:27:23.200 --> 00:27:33.400
9! over (9 - 4), the top minus the bottom, that is 5, factorial, times just the bottom, factorial:
00:27:33.400 --> 00:27:43.000
9!--we can write that as 9 times 8 times 7 times 6 times 5!, over 5!; look, that is convenient; they cancel;
00:27:43.000 --> 00:27:49.700
times 4; and now let's expand 4!, as well: 4 times 3 times 2...we will omit the 1, because it doesn't do anything.
00:27:49.700 --> 00:27:57.000
5! and 5! cancel; 3 and 2 cancel out the 6; 8 cancels with a 4 to make a 2 on top.
00:27:57.000 --> 00:28:03.800
So, we have 9 times 2 times 7; 9 times 2 times 7 comes out to be 126.
00:28:03.800 --> 00:28:09.700
There is our coefficient right there; or if we wanted to write out the whole thing, we would end up getting 126,
00:28:09.700 --> 00:28:16.100
times x...not to the fifth; not squared; but x^10, what it was, because it said the term containing
00:28:16.100 --> 00:28:24.500
x^10y^12, not the one where we have raised that part of the binomial to the fifth, but x^10y^12.
00:28:24.500 --> 00:28:30.800
And what it asked for was the coefficient; the coefficient is 126; times x^10y^12.
00:28:30.800 --> 00:28:40.000
Great; the third example: Find the value of the coefficient c on the term cx⁸ in the expansion of (2x² - 3)⁷.
00:28:40.000 --> 00:28:47.700
Once again, we need to realize that this is 8, but what we have is an n = 7.
00:28:47.700 --> 00:28:52.500
So, we realize that it is 2x²; that is the thing we are dealing with.
00:28:52.500 --> 00:29:02.700
So, we have a = 2x², and b equals...not just 3, but it has to be including the negative as well,
00:29:02.700 --> 00:29:11.100
because normally it is a + b; so if it is a + b normally, but now it is a - b, then it must be that the b also has a negative inside of it.
00:29:11.100 --> 00:29:14.600
So, our b is equal to -3; OK.
00:29:14.600 --> 00:29:23.700
At this point, we want to figure out what number we have to raise 2x² to, to get x⁸ to pop out of that.
00:29:23.700 --> 00:29:31.200
What number do we raise that to? If we have x² (we will consider just the x² for a moment,
00:29:31.200 --> 00:29:41.100
because the 2 won't affect what exponent the x has) to the ?, is equal to x⁸;
00:29:41.100 --> 00:29:45.800
well, then our question mark has to be 4, because 2 times 4 comes out to be 8.
00:29:45.800 --> 00:29:50.700
So, x², raised to the 4, equals x⁸.
00:29:50.700 --> 00:30:01.200
We know that whenever we do this expansion, n choose k, times 2x², our whole a, to the n - k,
00:30:01.200 --> 00:30:06.900
times our whole b, -3, to the k; now we need to figure out what our k is.
00:30:06.900 --> 00:30:14.900
Well, our n was 7, so our k must be 3, if we are going to produce a 4 there.
00:30:14.900 --> 00:30:23.500
n - k has to be 4, because we know that this value here has to match up with n - k here.
00:30:23.500 --> 00:30:34.600
n - k there has to match up, as well; so our 7 is what we have for n; 7 - k = 4; 7 - k = 4 means that k has to be 3.
00:30:34.600 --> 00:30:39.500
Now, we have figured out that k has to be 3; we are ready to plug everything in.
00:30:39.500 --> 00:30:55.200
n choose k: 7 choose 3, times 2x² to the n - k (comes out to be 4); -3 to the 3.
00:30:55.200 --> 00:31:02.000
n - k at 4...2x² to the fourth; sure enough, that would make some constant times x⁸.
00:31:02.000 --> 00:31:06.300
-3 to the 3: that is just going to also be part of that constant term, c.
00:31:06.300 --> 00:31:12.600
At this point, let's figure out what it is; let's solve 7 choose 3; what does that value come out to be?
00:31:12.600 --> 00:31:24.700
7! over (7 - 3), 4, factorial, times 3!...we have 7 times 5 times 6...sorry, 7 times 6 times 5; I'm not sure why I confused that order;
00:31:24.700 --> 00:31:28.300
it doesn't matter, but I don't want to make you think that something weird happened;
00:31:28.300 --> 00:31:35.200
7 times 6 times 5, over 4!, cancelled out with all of the rest of those...we have 3! on the bottom; that is just 3 times 2 times 1.
00:31:35.200 --> 00:31:40.100
We will omit that; 3 times 2 cancels out with the 6; 7 times 5 gets us 35.
00:31:40.100 --> 00:31:45.000
We can now swap that in for our 7 choose 3, and we have 35 times...
00:31:45.000 --> 00:31:52.700
2x² becomes 2 to the fourth, x² to the fourth; 2⁴ is 16; x² is x⁸.
00:31:52.700 --> 00:32:09.900
-3 cubed is -27 (-3 times -3 times -3); we use a calculator to multiply 35, 16, and -27, and we get -15,120x⁸.
00:32:09.900 --> 00:32:17.100
That means that our c must be equal to -15,120, because what we are looking for
00:32:17.100 --> 00:32:23.500
was the coefficient c on the expansion of cx⁸, once the thing is fully expanded.
00:32:23.500 --> 00:32:27.600
All right, great; we are ready to move on to Pascal's triangle, the fourth example.
00:32:27.600 --> 00:32:31.200
Use Pascal's triangle to expand (p - q)⁶.
00:32:31.200 --> 00:32:36.100
If we are going to the sixth, then that means n = 6; we are going to have to get down to the sixth row,
00:32:36.100 --> 00:32:45.100
which is really the row that is 7 down (if we say that the row at the very top is one down); but we will just write n = 0,
00:32:45.100 --> 00:32:55.200
so we don't get confused by this; so our 0th is at a 1, and then n = 1 is the next one: 1 1;
00:32:55.200 --> 00:33:18.700
n = 2: 1 2 1; n = 3: 1 3 3 1; n = 4: 1 4 6 4 1; n = 5: 1...1 + 4 is 5...4 + 6 is 10; 6 + 4 is 10; 4 + 1 is 5; 1 on itself is 1.
00:33:18.700 --> 00:33:31.600
And finally, the one we care about, n = 6: 1 on itself is 1; 1 + 5 is 6; 5 + 10 is 15; 10 + 10 is 20; 10 + 5 is 15; 5 + 1 is 6; 1 on itself is 1.
00:33:31.600 --> 00:33:36.300
Those are all of the coefficients that we care about at this expansion.
00:33:36.300 --> 00:33:42.800
All right, if we are going to expand (p - q)⁶, we know that it is going to end up having some expansion
00:33:42.800 --> 00:33:56.900
that looks like blank, this one first...well, this one is p; and this one, though, is actually -q; so it is going to be _p⁶
00:33:56.900 --> 00:34:04.900
times (-q)⁰; and then we work the q part up and the p part down, one step at a time:
00:34:04.900 --> 00:34:10.600
p⁶ becomes p⁵ on our next step; (-q)⁰ becomes (-q)¹;
00:34:10.600 --> 00:34:23.800
plus _p⁴(-q)²...and just continue on the next line...+ p³ times (-q)³...
00:34:23.800 --> 00:34:36.900
oh, there should be a blank after that plus, right here; + _p²(-q)⁴ +
00:34:36.900 --> 00:34:46.200
(I will just bring this down to the next line, once again) _p¹(-q)⁵ +
00:34:46.200 --> 00:34:54.100
_p⁰(-q)⁶; great--at this point, we can now swap in our values for our binomial coefficients.
00:34:54.100 --> 00:35:08.300
So, we know 1 will go here; a 6 will go here; a 15 will go here; a 20 will go here; a 15 will go here; a 6 will go here; a 1 will go here.
00:35:08.300 --> 00:35:10.600
See how they come in symmetrically?
00:35:10.600 --> 00:35:14.700
At this point, we will now write it all on this line around here.
00:35:14.700 --> 00:35:20.100
And we will also simplify it: 1p⁶(-q)⁰: well, 1 just disappears;
00:35:20.100 --> 00:35:24.100
we have p⁶; and (-q)⁰--well, if you raise anything to the 0,
00:35:24.100 --> 00:35:26.600
it just becomes 1; so we will have that disappear, as well.
00:35:26.600 --> 00:35:31.800
6p⁵(-q)¹: well, that -q is going to show up now; it is going to come in.
00:35:31.800 --> 00:35:41.000
So, it is not plus; it is now minus, because it is -q to the 1; so - 6p⁵q.
00:35:41.000 --> 00:35:44.900
That -q here doesn't get cancelled out, because there is just one negative.
00:35:44.900 --> 00:35:56.400
And next, we will have + 15p⁴(-q)²; negative on negative cancels out, so we do have positive q².
00:35:56.400 --> 00:36:07.500
Plus 20p³(-q)³; negative, negative, negative means we still have a negative.
00:36:07.500 --> 00:36:11.500
So, it is not a plus; it is a minus, because of the (-q)³.
00:36:11.500 --> 00:36:19.200
So, we have q³; next, plus 15...that was the last one...this is what we are working on now:
00:36:19.200 --> 00:36:26.400
15p²(-q)⁴: well, the negatives will end up canceling there, because it is an even number, q⁴.
00:36:26.400 --> 00:36:36.200
Plus 6p¹(-q)⁵; well, (-q)⁵ is an odd, so it will be minus 6pq⁵.
00:36:36.200 --> 00:36:41.000
And then finally, (-q)⁶, with a coefficient of 1, is our very last term.
00:36:41.000 --> 00:36:47.400
We just did this one here as our last term: 1p⁰...those things both disappear, and we have +q⁶.
00:36:47.400 --> 00:36:52.000
And the negative disappears, because it has an even exponent; great.
00:36:52.000 --> 00:37:00.900
And there is our full expansion; it is still not super fast, but it is much, much better than if we had tried to expand that whole thing out by hand.
00:37:00.900 --> 00:37:07.100
If we did that whole thing by hand, it would be taking forever; but this goes relatively quickly, by using Pascal's triangle.
00:37:07.100 --> 00:37:15.600
The final example: Use Pascal's triangle to expand (√3u² + 1/2√p)⁴.
00:37:15.600 --> 00:37:21.000
At this point, what is our n? Our n equals 4, so we need to get down to the n = 4 row.
00:37:21.000 --> 00:37:44.100
We start at n = 0: 1; n = 1: 1 1; n = 2: 1 2 1; n = 3: 1 3 3 1; n = 4 (finally, the row that we actually care about): 1 4 6 (3 + 3 is 6); 3 + 1 is 4; and 1.
00:37:44.100 --> 00:37:48.600
Great; so that is the row that we will actually end up using, because our n equals 4.
00:37:48.600 --> 00:37:51.600
So that we don't end up getting cramped, I am not going to end up writing here.
00:37:51.600 --> 00:37:54.700
I am going to end up doing the expansion down here.
00:37:54.700 --> 00:37:59.000
Our first thing will be some blank, times...well, what is our a?
00:37:59.000 --> 00:38:06.800
That is the good thing to identify: √3u² is what our a is equal to; a + b to the some n...
00:38:06.800 --> 00:38:16.600
so in this case, the a is all of √3u², and the b is all of 1/2√p; OK.
00:38:16.600 --> 00:38:23.800
We are now ready to do this; the first one will be √3u² to the n = 4.
00:38:23.800 --> 00:38:27.800
And then, 1/2√p to the 0...I won't even write that one there;
00:38:27.800 --> 00:38:45.700
plus _(√3u²)³(1/2√p)¹ + _(√3u²)²(1/2√p)²
00:38:45.700 --> 00:38:51.500
+ _...let me break that down to the next line, just so that we can have plenty of room...
00:38:51.500 --> 00:39:07.900
+ _(√3u²)¹(1/2√p)³ + _(√3u²)⁰,
00:39:07.900 --> 00:39:17.300
so we can just make that entire thing disappear; times (1/2√p), all to the 4.
00:39:17.300 --> 00:39:22.200
Great; I want you to notice here: Don't get confused by the fact that it is u² or √p.
00:39:22.200 --> 00:39:25.700
The fact that there are exponents on stuff inside of the binomial doesn't matter.
00:39:25.700 --> 00:39:34.100
We still do this part where the one exponent steps down with each step, and the other exponent,
00:39:34.100 --> 00:39:39.000
the one that starts at 0, steps up with each step of the terms that we work through.
00:39:39.000 --> 00:39:41.900
Great; at this point, we can now plug in our binomial coefficients:
00:39:41.900 --> 00:39:50.300
a 1 for our first blank, a 4 for the next blank, a 6 for the next blank, a 4 for the next blank, and a 1 for our final blank.
00:39:50.300 --> 00:39:55.100
At this point, we can now finally actually simplify this one; well, let's do it in two steps.
00:39:55.100 --> 00:40:03.100
1...we will just have that disappear; (√3u²)⁴: well, √3 squared is 3; so √3 so the fourth is 3².
00:40:03.100 --> 00:40:05.800
So, √3 to the fourth is the same thing as 9.
00:40:05.800 --> 00:40:14.200
u² to the fourth is u⁸; plus 4 times...√3 cubed is going to be 3√3.
00:40:14.200 --> 00:40:21.600
(u²)³ is u⁶; (1/2√p)¹ is just 1/2√p.
00:40:21.600 --> 00:40:30.300
Plus...6(√3u²)²: √3 squared is 3; u² squared is u⁴.
00:40:30.300 --> 00:40:41.400
Times...1/2√p squared will be 1/4 (1/2 squared is 1/2 times 1/2, so 1/4); √p squared is p.
00:40:41.400 --> 00:40:52.600
Break the line once again; plus...next is 4(√3u²)¹; well, that just stays as √3u².
00:40:52.600 --> 00:41:02.100
Times...(1/2√p)³: 1/2 to the 3 is 1/8; times √p cubed is p√p.
00:41:02.100 --> 00:41:12.800
And then finally, plus 1 times 1/2√p to the fourth; that is going to be 1/16; (√p)⁴ is going to come out as p²; great.
00:41:12.800 --> 00:41:16.700
Now, we can simplify this whole thing and do the last of the simplification.
00:41:16.700 --> 00:41:25.700
9u⁸ +...4 times 3√3 times 1/2...1/2 cancels the 4 down to a 2;
00:41:25.700 --> 00:41:39.400
we are left with 6√3u⁶√p, plus...we have 1/4 here, so this will become 1/2;
00:41:39.400 --> 00:41:54.000
this will become 3; 6 times 1/4 becomes 3 times 1/2; so 3 times 3 is 9, divided by 2...we have 9/2 as the coefficient there, u⁴p.
00:41:54.000 --> 00:42:02.700
Plus...4√3u²1/8p√p...that is a lot of things; so 1/8...that will cancel to 2,
00:42:02.700 --> 00:42:11.600
when it knocks out this 4; and so we have (√3/2)u²p√p.
00:42:11.600 --> 00:42:16.900
And finally, our last term is 1/16p².
00:42:16.900 --> 00:42:24.300
And there is the full expansion; it is about as difficult as any expansion with Pascal's triangle will end up being.
00:42:24.300 --> 00:42:29.400
But still, notice how much faster that ended up making it than if we had tried to do this whole thing by hand--
00:42:29.400 --> 00:42:33.400
doing √3u² + 1/2√p, times √3u² + 1/2√p,
00:42:33.400 --> 00:42:37.100
times √3u² + 1/2√p, times √3u² + 1/2√p...
00:42:37.100 --> 00:42:39.300
We would still be working through it, and still have a lot more to go.
00:42:39.300 --> 00:42:40.500
So, it makes things faster.
00:42:40.500 --> 00:42:44.000
We have to be careful, and make sure that...there are a couple of things that we have to pay attention to.
00:42:44.000 --> 00:42:50.500
We really have to make sure that our n, the row that we are using of Pascal's triangle, matches to the exponent we are raising to.
00:42:50.500 --> 00:42:54.700
We have to pay attention to what our whole a is and what our whole b is.
00:42:54.700 --> 00:43:01.800
It has to be the entire term, not just the u, not just the p, but the entire term of something + something.
00:43:01.800 --> 00:43:05.700
And if it is a minus, it has to be something plus a negative something.
00:43:05.700 --> 00:43:10.700
And then, we set it up with this step down/step up pattern, the blanks; and then, we can just slot everything in.
00:43:10.700 --> 00:43:14.600
If it is a simpler problem, you might be able to do it without having to take such careful steps.
00:43:14.600 --> 00:43:18.500
But if it is a big, complicated problem, I really recommend doing the careful steps.
00:43:18.500 --> 00:43:21.900
It will make it easy to not make a mistake and make the problem just slow and easy.
00:43:21.900 --> 00:43:27.900
All right, that finishes the examples; if you are leaving now, we will see you at Educator.com later; goodbye!
00:43:27.900 --> 00:43:40.000
But if you are staying around for the proof to the theorem, it is time for the "bonus round."
00:43:40.000 --> 00:43:43.500
All right, we are ready to tackle a proof of the binomial theorem.
00:43:43.500 --> 00:43:52.400
First, let's phrase the theorem as, "For any n = 0, 1, 2...and so on and so on" so that is all natural numbers,
00:43:52.400 --> 00:44:01.100
"(a + b) to the n is equal to the sum, going from k = 0 to n, of n choose k times a^n - kb^k."
00:44:01.100 --> 00:44:06.200
So, notice that that is just putting all of the binomial coefficients added together, for a fully expanded term.
00:44:06.200 --> 00:44:10.800
We are going to prove this by induction; so, just in case you haven't watched the previous lessons,
00:44:10.800 --> 00:44:15.300
but you are really interested in this proof, we will be using mathematical induction.
00:44:15.300 --> 00:44:19.500
And we will also be using a fair bit if sigma (that is, summation) notation.
00:44:19.500 --> 00:44:24.400
So, if you are not familiar with mathematical induction or sigma/summation notation, watch the lesson on mathematical induction,
00:44:24.400 --> 00:44:28.900
and watch the lesson on Introduction to Series, where we will talk about sigma/summation notation,
00:44:28.900 --> 00:44:31.500
because those things will definitely be necessary prerequisites.
00:44:31.500 --> 00:44:36.400
And before we get into this, I really want to stress: what we are about to go through is a pretty challenging proof.
00:44:36.400 --> 00:44:42.400
This is definitely a college-level proof for a reasonably good math class.
00:44:42.400 --> 00:44:47.700
So, don't be shocked if you find this a little bit challenging to work through.
00:44:47.700 --> 00:44:52.000
This isn't going to be super easy to understand the first time; and that is perfectly OK.
00:44:52.000 --> 00:44:55.700
Math is a puzzle to be worked through and understood over time.
00:44:55.700 --> 00:44:59.600
So, watch through this; if you have difficulty understanding what is going on, moment-to-moment,
00:44:59.600 --> 00:45:04.900
I highly recommend that you take a piece of paper and actually write what is happening on-screen
00:45:04.900 --> 00:45:07.500
as we are working through it in the lesson.
00:45:07.500 --> 00:45:11.000
By working through it yourself, you will be able to understand the steps better,
00:45:11.000 --> 00:45:15.300
because you will have to internalize them, because you yourself will have to be understanding and working through it.
00:45:15.300 --> 00:45:18.400
If at some point a step happens where you have no idea how that step happens,
00:45:18.400 --> 00:45:21.700
pause the video and try to understand what just happened on your paper.
00:45:21.700 --> 00:45:24.500
Figure out how you can get from one step to the next step.
00:45:24.500 --> 00:45:27.300
If you still can't figure it out, back it up; hopefully I will explain through it.
00:45:27.300 --> 00:45:30.800
I will explain a lot of this; we will be working through it slowly and carefully.
00:45:30.800 --> 00:45:33.300
But a really great trick is to work through it on paper.
00:45:33.300 --> 00:45:39.300
It really makes things make a lot more sense, just getting it down on paper, so that is not just "rattling around" in your head.
00:45:39.300 --> 00:45:42.100
You can actually see it, think about it, and have it in front of you.
00:45:42.100 --> 00:45:43.800
It is a really great way to work.
00:45:43.800 --> 00:45:48.200
I really recommend this any time that you are trying to work through a math textbook and things get confusing in the math textbook.
00:45:48.200 --> 00:45:51.500
Just take a piece of scratch paper and write what is going on in the math textbook.
00:45:51.500 --> 00:45:56.100
Writing it out for yourself, understanding it step-by-step, when you write it out in your own hand,
00:45:56.100 --> 00:46:00.400
will end up making most issues that you end up having clear up so much faster,
00:46:00.400 --> 00:46:04.600
because you are having to work through it yourself; you are not just trying to read someone else's language.
00:46:04.600 --> 00:46:07.900
You are putting it in your own language before you are moving on.
00:46:07.900 --> 00:46:14.100
All right, let's get started; this is really good--this is what math is really about, in my opinion.
00:46:14.100 --> 00:46:19.100
Math lets us do all sorts of things; but for me, the cool thing about math is the fact that is puzzles,
00:46:19.100 --> 00:46:26.000
that it is logic; it is something that we can believe in, and it is just there--real logic, real proof.
00:46:26.000 --> 00:46:29.500
This stuff actually comes together and can be shown very clearly.
00:46:29.500 --> 00:46:36.500
All right, let's get to it: proof by induction: begin by noticing how nicely the theorem lends itself to being written
00:46:36.500 --> 00:46:40.800
as some p<font size="-6">n</font> statement, some statement for some value of n.
00:46:40.800 --> 00:46:48.000
We have n = 0, 1, 2...so it is just stepping up, one at a time, at (a + b)^n equals this sum here.
00:46:48.000 --> 00:46:55.200
That is what the binomial theorem says; so we can just say that the nth statement is (a + b)^n equals that sum; great.
00:46:55.200 --> 00:46:58.800
It is really easy to write out in this statement idea.
00:46:58.800 --> 00:47:03.500
Our base case: now, before we prove the base case, notice that there is this slight issue.
00:47:03.500 --> 00:47:08.800
The theorem doesn't start at n = 1; it starts at n = 0.
00:47:08.800 --> 00:47:13.500
When we first talked about mathematical induction, we always set our base case at n = 1.
00:47:13.500 --> 00:47:18.600
But with this one, we can't set our base case at n = 1; we actually need to start at p₀.
00:47:18.600 --> 00:47:22.900
So, it is not p₁, but p₀, because the theorem starts at n = 0.
00:47:22.900 --> 00:47:26.000
If we think about this, though, we realize that this isn't really an issue.
00:47:26.000 --> 00:47:29.400
It is just that we need a starting point to work out of.
00:47:29.400 --> 00:47:34.500
It needs to be the first stepping-stone that we then walk forward from, using that inductive step.
00:47:34.500 --> 00:47:37.500
That is what the inductive step is: that is taking one step after another,
00:47:37.500 --> 00:47:41.700
and guaranteeing that future stepping-stones will be there, as long as we have this first stepping-stone.
00:47:41.700 --> 00:47:47.100
So, it doesn't really matter if this stepping stone starts at n = 0 or starts at n = 15 or starts at n = 1.
00:47:47.100 --> 00:47:51.400
It just needs to be some integer value that we can then step forward from.
00:47:51.400 --> 00:47:55.800
n = 0 is a perfectly fine thing to set our base case at.
00:47:55.800 --> 00:48:01.200
We work through this: a base case of p₀ would end up giving us (a + b)⁰.
00:48:01.200 --> 00:48:05.600
Is that equal to (we are saying this as a question; we don't know if this is true yet;
00:48:05.600 --> 00:48:09.100
we are checking to make sure that the two sides are true)...it would be equal to the sum
00:48:09.100 --> 00:48:16.500
(if this is true) of k = 0 summed to 0 of 0 choose k, a^0 - kb^k.
00:48:16.500 --> 00:48:21.800
So, our left side...well, (a + b)⁰...anything raised to the 0 just comes out to be 1.
00:48:21.800 --> 00:48:27.800
Our sum, k = 0, going up to 0...well, that means that where we start is where we stop.
00:48:27.800 --> 00:48:32.600
So, there is only one term in this series when we expand this summation, this sigma notation.
00:48:32.600 --> 00:48:40.400
It just turns into 0 choose 0; we plug in our k; we plug in the k here, 0 - 0, and plug in the k here, b to the 0.
00:48:40.400 --> 00:48:41.800
So, what does that come out to be?
00:48:41.800 --> 00:48:46.800
Well, a^0 - 0 just comes out to be 1; b⁰ just comes out to be 1.
00:48:46.800 --> 00:48:52.300
0 choose 0 is going to be 0!/(0 - 0)!, so 0! times 0!.
00:48:52.300 --> 00:49:00.900
So, we have this here, 0!/0!(0!): well, remember: when we set up 0!, we just stated 0! = 1.
00:49:00.900 --> 00:49:10.100
We simply define it that way; that means 0!/0!(0!) turned out to be 1, so our entire right side is 1.
00:49:10.100 --> 00:49:13.000
Our entire left side is 1, as we already figured out early on.
00:49:13.000 --> 00:49:18.900
So that means, indeed, that this does check out; the base case is true; we are good with the base case.
00:49:18.900 --> 00:49:26.600
On to the inductive step: now, here is where things get a little bit tricky, just for understanding notation.
00:49:26.600 --> 00:49:37.700
There is a minor issue with notation here: when we first saw induction, we always had some general statement, p<font size="-6">n</font>, this (a + b)^n = stuff.
00:49:37.700 --> 00:49:44.500
Now, what we did for the inductive step was assumed...we said, "If p<font size="-6">k</font> is true,
00:49:44.500 --> 00:49:47.600
then we have to show that p<font size="-6">k + 1</font> is true."
00:49:47.600 --> 00:49:51.300
If this thing is true, then the next one must be true.
00:49:51.300 --> 00:49:58.200
But we have this problem; the way that we usually write the binomial theorem, we use k.
00:49:58.200 --> 00:50:02.800
k shows up as the index in the sum; we are using this all over.
00:50:02.800 --> 00:50:09.800
So, we cannot use p<font size="-6">k</font>; this means we cannot use p<font size="-6">k</font>...
00:50:09.800 --> 00:50:16.100
if we use p<font size="-6">k</font>, we will end up having to re-index and change around variables.
00:50:16.100 --> 00:50:19.200
So, we cannot use p<font size="-6">k</font> without re-indexing and changing around variables.
00:50:19.200 --> 00:50:22.700
If we had to use p<font size="-6">k</font> and then p<font size="-6">k + 1</font>, we would have to come out with some new variable
00:50:22.700 --> 00:50:26.400
to swap out for all of these k's up here, and then it would get confusing,
00:50:26.400 --> 00:50:29.300
because that is not how we are already used to ending up working with the theorem.
00:50:29.300 --> 00:50:32.800
And so, it is going to change its ways, and it just makes things confusing.
00:50:32.800 --> 00:50:37.500
What we are going to do, to avoid confusing ourselves with new variables:
00:50:37.500 --> 00:50:41.600
to avoid confusion, we are going to prove the following for this step:
00:50:41.600 --> 00:50:48.100
if p<font size="-6">n</font> is true for some n, then we will show that p<font size="-6">n + 1</font> is true, as well.
00:50:48.100 --> 00:50:54.600
So, up here, this is the general statement of p<font size="-6">n</font>, and we are trying to show that it is true for all n.
00:50:54.600 --> 00:50:58.100
Any natural number n is what we are trying to show for our general statement.
00:50:58.100 --> 00:51:03.200
But what we are going to do is say, "Sure, let's assume that it is true for some specific n."
00:51:03.200 --> 00:51:06.600
We don't need to name what that specific n is; but it is just one value of n.
00:51:06.600 --> 00:51:11.900
And then, we are going to show that, if it is true for that one value, it has to be true for the next value, n + 1, as well.
00:51:11.900 --> 00:51:16.300
This is OK; before, we talked about p<font size="-6">k</font> and then p<font size="-6">k + 1</font>.
00:51:16.300 --> 00:51:25.200
But it doesn't really matter what symbol we use; all that matters is that, if it is true at one step, it must be true at the next step.
00:51:25.200 --> 00:51:27.900
If it is true here, it has to be true at the next step.
00:51:27.900 --> 00:51:31.200
If it is true at p<font size="-6">k</font>, then it has to be true at p<font size="-6">k + 1</font>.
00:51:31.200 --> 00:51:34.900
If it is true at p<font size="-6">z</font>, it has to be true at p<font size="-6">z + 1</font>.
00:51:34.900 --> 00:51:37.900
If it is true at p<font size="-6">n</font>, it has to be true at p<font size="-6">n + 1</font>.
00:51:37.900 --> 00:51:41.900
That is different than our general statement of "it is true for all n, forever."
00:51:41.900 --> 00:51:48.800
We are saying, "Let's say that it is true for some specific n; and then let's show that it has to be true for the next one in line."
00:51:48.800 --> 00:51:53.300
This might be a little bit confusing; but really, it is OK to end up using this p<font size="-6">n</font>, then p<font size="-6">n + 1</font>,
00:51:53.300 --> 00:51:56.800
because what we are not doing is assuming that it is always true for all n.
00:51:56.800 --> 00:52:02.200
We are just saying, "Let's say it is true for some n," and then we will show what would happen for the next n, n + 1.
00:52:02.200 --> 00:52:05.300
And that is what we will do to avoid having to get all of this notation confusion
00:52:05.300 --> 00:52:09.300
of having to use letters other than n, of having to use letters other than k,
00:52:09.300 --> 00:52:11.100
because then it would be not what we are used to seeing.
00:52:11.100 --> 00:52:16.300
And while we could work through it, it makes it a lot easier if it is something that we are used to seeing, used to working with.
00:52:16.300 --> 00:52:20.200
We will end up avoiding notation confusion by keeping n and k in this,
00:52:20.200 --> 00:52:24.100
and just sort of knowing what is going on, on a personal level.
00:52:24.100 --> 00:52:25.900
All right, we are ready to actually work through this.
00:52:25.900 --> 00:52:31.500
Let's set this up: for our inductive step, we start by assuming our inductive hypothesis.
00:52:31.500 --> 00:52:43.300
We assume that, for some n, p<font size="-6">n</font> is true; that is, (a + b)^n is equal to the sum of k = 0 to n of n choose k times a^n - kb^k.
00:52:43.300 --> 00:52:49.700
Now, what we want to show is that p<font size="-6">n + 1</font> must also be true of p<font size="-6">n</font> is true.
00:52:49.700 --> 00:52:54.100
Assuming p<font size="-6">n</font>, p<font size="-6">n + 1</font> must be true; that is what we are working towards.
00:52:54.100 --> 00:52:56.100
We don't know it yet; we are working towards it.
00:52:56.100 --> 00:53:02.200
So, how would p<font size="-6">n + 1</font> be stated? Well, that would be stated as (a + b)^n + 1
00:53:02.200 --> 00:53:10.300
equals...and now we swap out the n here for n + 1, and the n here for n + 1, and the n here for n + 1;
00:53:10.300 --> 00:53:14.600
and that would be the p<font size="-6">n + 1</font> version; up here is the p<font size="-6">n</font> version.
00:53:14.600 --> 00:53:19.200
So, what we want to show is that this here is true.
00:53:19.200 --> 00:53:26.800
Thus, if we can show that the left side of p<font size="-6">n + 1</font> is just the same thing as the right side--
00:53:26.800 --> 00:53:31.500
that they are just two different ways...writing the left side and the right side, that they are clearly equivalent,
00:53:31.500 --> 00:53:35.500
by manipulating it through "Well, we can do this, and we can do this, and we can do this..."
00:53:35.500 --> 00:53:41.000
and they are all clearly equal, and we can make it from one side to the other side; we will have shown that this statement is true.
00:53:41.000 --> 00:53:46.800
If we can show that the left side is the same thing as the right, we will have shown that, indeed, the statement is true.
00:53:46.800 --> 00:53:51.000
So, what we want to do is see how we can manipulate (a + b)^n + 1.
00:53:51.000 --> 00:53:54.400
Well, in a proof by induction, we always want to toss our inductive hypothesis into this.
00:53:54.400 --> 00:54:03.700
So, our first question is, "Is there a way to turn (a + b)^n + 1 into something where we can use our hypothesis?"
00:54:03.700 --> 00:54:09.400
We want to see if there is a way to turn (a + b)^n + 1 into the right side of p<font size="-6">n + 1</font>.
00:54:09.400 --> 00:54:17.100
Our first question is going to be, "How can we apply our hypothesis--how can we apply p<font size="-6">n</font>, (a + b)^n = that sum?"
00:54:17.100 --> 00:54:23.900
We look at it: if we look at (a + b)^n + 1, it is pretty easy to get (a + b) to the n to show up; we just split it:
00:54:23.900 --> 00:54:28.700
(a + b)^n + 1 is just the same thing as (a + b)(a + b)^n.
00:54:28.700 --> 00:54:36.200
At this point, we can now swap out (a + b)^n for the other side of our inductive hypothesis.
00:54:36.200 --> 00:54:42.500
We swap it out; we now have (a + b) times the sum k = 0 to n of n choose k (a^n - kb^k).
00:54:42.500 --> 00:54:44.900
We have just swapped out for our inductive hypothesis.
00:54:44.900 --> 00:54:53.300
Now, at this point, we have a + b over here; we can distribute that: a + b distributes onto that sum.
00:54:53.300 --> 00:54:59.700
And we will have a times the sum, plus b times the sum.
00:54:59.700 --> 00:55:04.400
Well, a and b don't vary once we have expanded the thing; they just end up being the same thing.
00:55:04.400 --> 00:55:07.600
So, what we can do is actually bring them inside of the sum.
00:55:07.600 --> 00:55:12.000
They have no direct connection to the index, so we can bring them inside of the sum.
00:55:12.000 --> 00:55:27.500
So, a will apply onto the a^n - k; a^n - k times a will be a^n - k + 1, which we can write as a^n + 1 - k.
00:55:27.500 --> 00:55:35.600
The same thing happens over here: b applied to b^k becomes just b^k + 1; great.
00:55:35.600 --> 00:55:38.100
At this point, let's expand these sigma notations.
00:55:38.100 --> 00:55:43.700
The sigma notations are a great way of keeping things compact, but it is a little hard to tell what is going on inside of the notation.
00:55:43.700 --> 00:55:46.300
So, let's expand them, so we can see what is going on better.
00:55:46.300 --> 00:55:51.900
If we expand this, we will have k = 0 here; we will plug that in here first, so we will have n choose 0
00:55:51.900 --> 00:55:57.700
times a^n + 1 - 0, so a^n + 1, and b⁰ (that just disappears).
00:55:57.700 --> 00:56:02.400
And then, next, we will go up one step, and choose 1; a^nb...great.
00:56:02.400 --> 00:56:07.000
And we will work our way up, until finally we get up to n; and so, we will have n choose n.
00:56:07.000 --> 00:56:11.700
And it will be a^n + 1 - n when we are plugging in n.
00:56:11.700 --> 00:56:20.800
a^n + 1 - n gets us just 1, so we have a to the 1 here; and b^n now, because we are swapping in n for k; and we have b^n here.
00:56:20.800 --> 00:56:26.400
Great; so that is what we end up getting from this sum on this side.
00:56:26.400 --> 00:56:28.600
And we can do basically the same thing and expand this sum over here.
00:56:28.600 --> 00:56:37.300
We start at k = 0; we will get n choose 0, a^nb¹ (if we plug in k at 0, we will have 0 + 1, so b¹).
00:56:37.300 --> 00:56:40.900
We work our way up; we get to n - 1, one before our last one.
00:56:40.900 --> 00:56:48.600
n choose n - 1; we will have a^n - n - 1; if you are not sure why that comes out as n - n - 1,
00:56:48.600 --> 00:56:52.400
well, this becomes minus n plus 1, so we get a¹ here.
00:56:52.400 --> 00:57:02.000
k + 1...well, if we are plugging in n - 1 for k, n - 1 + 1 gets us ab^n.
00:57:02.000 --> 00:57:10.500
And then finally, our last one, when we plug in n: swap that in for our k; n - k (n - n for our a here) will end up just disappearing entirely.
00:57:10.500 --> 00:57:12.300
We will have a to the 0, so we just omit that.
00:57:12.300 --> 00:57:19.600
We will have n choose n, and b, plugging in k for n...n + 1 here, so we get n + 1 here.
00:57:19.600 --> 00:57:23.500
So now, we have expanded these two sums; now we can start working with them.
00:57:23.500 --> 00:57:27.400
What we want to do first is notice that there are some similarities between these things.
00:57:27.400 --> 00:57:34.400
In fact, we have this really great similarity going on here: a^nb and a^nb match up.
00:57:34.400 --> 00:57:38.900
ab^n and ab^n match up; in fact, we would end up having this match-up
00:57:38.900 --> 00:57:44.000
between all of the things inside of those ellipses; the pattern of matching will continue throughout.
00:57:44.000 --> 00:57:54.400
The only things who don't fit this matching pattern are the ones at the far ends, the a^n + 1 and the b^n + 1.
00:57:54.400 --> 00:58:01.200
Notice: the only terms that don't match up of these a to the something, b to the something terms are the first and last terms, respectively.
00:58:01.200 --> 00:58:04.600
If we separate them out, we can relate the sums; we can have this connection.
00:58:04.600 --> 00:58:12.100
We pull them out: n choose 0, a^n + 1; n choose n, b^n + 1 on the two ends.
00:58:12.100 --> 00:58:16.600
And then, we end up having the matching going on here: n choose 1, a^nb;
00:58:16.600 --> 00:58:19.100
that will match up to n choose 0, a^nb.
00:58:19.100 --> 00:58:27.500
And then, n choose n, ab^n, will match up to ab^n with an n choose n - 1 times ab^n.
00:58:27.500 --> 00:58:30.200
We have this nice matching going on here.
00:58:30.200 --> 00:58:34.200
The matching will also continue in here with the same thing of the top
00:58:34.200 --> 00:58:39.900
being above the bottom by 1 on its bottom and above the bottom by 1 on its bottom, every time; great.
00:58:39.900 --> 00:58:42.800
So, if that is the case, we can combine the parts in brackets.
00:58:42.800 --> 00:58:52.100
If n choose 1, n choose 0, are both multiplied by a^nb, well, then we can combine them: a^nb...
00:58:52.100 --> 00:58:59.200
and we end up having n choose 1 plus n choose 0, all of that times a^nb.
00:58:59.200 --> 00:59:01.500
We pull out the a^nb when we put them together.
00:59:01.500 --> 00:59:06.700
The same thing happens over here: n choose n, ab^n, n choose n - 1, ab^n:
00:59:06.700 --> 00:59:13.600
we pull out the ab^n, and we end up getting n choose n plus n choose n - 1, that quantity, times ab^n here.
00:59:13.600 --> 00:59:20.900
And inside of it, we will end up having the same pattern go on with the n choose something plus n choose (1 less than that something),
00:59:20.900 --> 00:59:25.100
times the ab stuff; so that pattern will end up continuing throughout.
00:59:25.100 --> 00:59:31.500
And we still have our things on the end; if you are not sure how they transformed from n choose 0 and n choose n,
00:59:31.500 --> 00:59:37.800
to just the a and b, well, n choose 0 is just 1 (if you are not sure about that n choose 0, well,
00:59:37.800 --> 00:59:47.000
that is equal to n! over (n - 0)!, so n!, times 0!; 0! is just 1; n! and n! cancel out;
00:59:47.000 --> 00:59:51.500
so we end up just getting a 1 here, so it disappears and we are just left with a^n + 1).
00:59:51.500 --> 01:00:00.600
The exact same reasoning works for n choose n becoming 1, as well, on the b^n + 1; and we get b^n + 1 here, as well.
01:00:00.600 --> 01:00:03.700
At this point, we can put this giant sum in sigma notation.
01:00:03.700 --> 01:00:07.100
a^n + 1 just stays around; b^n + 1 just stays around.
01:00:07.100 --> 01:00:13.900
The thing we care about is this giant sum here, which we broke on lines, because it is too much to write on a single line.
01:00:13.900 --> 01:00:26.900
We can do that as the summation from k = 1 to n of n choose k plus n choose k - 1, as a quantity, times a^n + 1 - k times b^k.
01:00:26.900 --> 01:00:28.300
Let's check and make sure that this works.
01:00:28.300 --> 01:00:37.200
If we did, indeed, plug in k = 1, yes, that checks out; 1 for k here, k - 1, 1 - 1; that does get us 0.
01:00:37.200 --> 01:00:44.000
a^n + 1 - 1 would get us a^n; b¹ would get us b; so that checks out there.
01:00:44.000 --> 01:00:46.600
If we were to do n, that would end up checking out here, as well.
01:00:46.600 --> 01:00:48.900
This is probably one of the hardest steps to see.
01:00:48.900 --> 01:00:54.200
If you work through it slowly, you can see that, yes, what we are doing is that this whole sum
01:00:54.200 --> 01:01:02.900
is just the same thing as what we have this in these giant brackets, this giant bracketed-off portion.
01:01:02.900 --> 01:01:06.800
That is the same thing here; so we can put the whole thing in sigma notation.
01:01:06.800 --> 01:01:08.900
At this point, let's check in with our original goal.
01:01:08.900 --> 01:01:17.200
We have gotten pretty far with this thing; what we have is a^n + 1 plus the big sum plus b^n + 1.
01:01:17.200 --> 01:01:20.700
Now, compare that to what we are trying to get the whole thing to look like.
01:01:20.700 --> 01:01:24.300
We started with the left side, and we are trying to get it to turn into what its right side was.
01:01:24.300 --> 01:01:33.500
And its right side was a summation: k = 0 to n + 1 of n + 1 choose k, times a^n + 1 - kb^k.
01:01:33.500 --> 01:01:36.800
This stuff here matches really closely to this.
01:01:36.800 --> 01:01:40.600
And n + 1, k is not too far off from what is here.
01:01:40.600 --> 01:01:43.800
And n and k = 1...that is not too far off, either.
01:01:43.800 --> 01:01:46.400
There are some similarities; we are starting to get there.
01:01:46.400 --> 01:01:54.100
What would be great is if we could somehow show that what we have here is the same thing as what we have here,
01:01:54.100 --> 01:02:01.300
that n choose k + n choose k - 1 is just the same thing as n + 1 choose k.
01:02:01.300 --> 01:02:05.800
If we could somehow show that that is the same thing, we would be really close
01:02:05.800 --> 01:02:10.300
to showing what we are trying to show-- what we want to show--getting to our goal.
01:02:10.300 --> 01:02:19.300
So, since that would be so useful if it were true--this n choose k plus n choose k - 1 equals n + 1 choose k-- let's go ahead and try to prove it.
01:02:19.300 --> 01:02:23.800
What we will do is create a lemma; a lemma is basically just a mini-theorem.
01:02:23.800 --> 01:02:29.100
A mini-theorem is something that we use to prove another larger theorem that we are interested in.
01:02:29.100 --> 01:02:34.100
It is basically like a thing that we can use to make a short hop to some more useful conclusion.
01:02:34.100 --> 01:02:39.300
It deserves being proven on its own, because it is not just something we can do in a single line.
01:02:39.300 --> 01:02:42.600
But it makes sense, and it is something that we will use towards another theorem.
01:02:42.600 --> 01:02:45.100
It is something that we make on the way towards another theorem.
01:02:45.100 --> 01:02:49.600
That is a lemma; so what we want to do is show that that is true, as a lemma.
01:02:49.600 --> 01:02:57.600
Our lemma is that, for any natural numbers r and s, r choose s plus r choose s - 1 is equal to r + 1 choose s.
01:02:57.600 --> 01:03:01.300
Now, if we want to use our lemma, we have to prove anything that we want to use.
01:03:01.300 --> 01:03:04.300
This is a proof, after all; so let's get to proving it.
01:03:04.300 --> 01:03:08.800
Let's investigate the left side, r choose s plus r choose s - 1.
01:03:08.800 --> 01:03:21.700
We break that down: our r choose s becomes r! over (r - s)!(s!), plus...r choose s - 1 becomes r! over (r + 1 - s)!(s - 1)!.
01:03:21.700 --> 01:03:24.000
And the only thing that you might be confused about is that r + 1 - s.
01:03:24.000 --> 01:03:32.000
Well, remember: r - (s - 1), because it is this quantity here, would end up coming out to be...
01:03:32.000 --> 01:03:38.400
and that part will be factorial; that is what matches up here...r - (s - 1)...that becomes plus; that becomes minus; that becomes positive.
01:03:38.400 --> 01:03:44.200
So, that is why we end up getting the r + 1 and the - s there; that is how we get that.
01:03:44.200 --> 01:03:48.500
What we want to do now is get these things over a common denominator.
01:03:48.500 --> 01:03:51.200
Any time we are trying to simplify and work with expressions, if we have two fractions,
01:03:51.200 --> 01:03:54.800
and we are trying to show that they are something else, well, we put them together, and we see what happens.
01:03:54.800 --> 01:03:57.700
If we are going to get these over a common denominator, let's look at them for a second.
01:03:57.700 --> 01:04:09.000
Well, we have (r - s)!(s!) and (r + 1 - s)!(s - 1)!; well, notice: (r + 1 - s)! is just one more than (r - s)!.
01:04:09.000 --> 01:04:11.900
r - s + 1 is the same thing as r + 1 - s.
01:04:11.900 --> 01:04:18.100
Similarly, (s - 1)! is the same thing as one more going up to s.
01:04:18.100 --> 01:04:26.800
So, we are trying to get s - 1 up by 1, and r - s up by 1; we want to get this to go up, and we want to get this to go up.
01:04:26.800 --> 01:04:31.600
If we can get them to each increase by 1 inside of the factorial, we will be able to have a common denominator.
01:04:31.600 --> 01:04:42.800
And notice: based on how a factorial works, we have the following way to increase a factorial by multiplication: x! times (x + 1) equals (x + 1)!.
01:04:42.800 --> 01:04:45.100
In case you don't see that, consider if we had 6!.
01:04:45.100 --> 01:04:51.000
Well, that would be the same thing as 6 times 5 times 4 times 3 times 2 times 1.
01:04:51.000 --> 01:04:58.400
Well, if we came along and multiplied 6 by 6 + 1, that is, 7, times 6!; well, that would be the same thing as 7 times...
01:04:58.400 --> 01:05:03.200
the expansion of 6!, 6 times 5 times 4 times 3 times 2 times 1.
01:05:03.200 --> 01:05:07.700
Well, now we have 7 all the way down to 1; so we can write that as 7!.
01:05:07.700 --> 01:05:12.300
So, if we have x!, and we multiply it by one more than that x, we will end up increasing.
01:05:12.300 --> 01:05:16.400
we will hop up one more step for our factorial, and we will go (x + 1)!.
01:05:16.400 --> 01:05:29.400
That idea is what we will use to increase (s - 1)! up to s!, and to increase (r - s)! up to (r - s + 1)!, or (r + 1 - s)!.
01:05:29.400 --> 01:05:34.500
All right, to do this, we will multiply the left side by (r + 1 - s)/(r + 1 - s).
01:05:34.500 --> 01:05:38.100
We can't just multiply the bottom; we have to multiply with a fraction on top and bottom.
01:05:38.100 --> 01:05:46.300
And we will multiply over here by s and s, because they match up, too; increasing s - 1 + 1 gets us s; r - s + 1 gets us r + 1 - s.
01:05:46.300 --> 01:05:53.700
OK, we work that out; on the top, r! times r + 1 - s will just be r! times (r + 1 - s).
01:05:53.700 --> 01:06:02.500
On the bottom, the (r - s) times one more will end up increasing to (r + 1 - s)!; the s! remains the same.
01:06:02.500 --> 01:06:09.300
On the other side, the r! times s is just still r! times s; they can't really combine.
01:06:09.300 --> 01:06:15.700
On the bottom, though, we have s - 1 times s; that is s - 1 + 1, so that becomes s!.
01:06:15.700 --> 01:06:18.700
And the left side doesn't change: r + 1 - s!.
01:06:18.700 --> 01:06:24.200
At this point, we now have a common denominator; the denominator here is the exact same as the denominator here.
01:06:24.200 --> 01:06:32.200
So, we can combine them; we combine them, and we have r! times (r + 1 - s), r! times s, up here;
01:06:32.200 --> 01:06:36.800
our denominators are just the same thing, because they just combined by addition.
01:06:36.800 --> 01:06:46.700
But on the top, what we can do is pull out the r!'s; they pull out to r! times what the factors they were being multiplied against were.
01:06:46.700 --> 01:06:56.700
That is (r + 1 - s), and then + s; the - s cancels with the + s, and we are left with r!(r + 1).
01:06:56.700 --> 01:07:03.100
And r + 1 times r!...well, that just means it bumps up by 1; so now we have (r + 1)!.
01:07:03.100 --> 01:07:06.500
We can also toss some fancy parentheses around this to help us see it.
01:07:06.500 --> 01:07:18.900
We have (r + 1)! on the top, and (r + 1 - s)!(s!), which is just the exact same thing as r + 1 choose s.
01:07:18.900 --> 01:07:28.600
(r + 1)! on the top, times (r + 1 - s)!(s!)--that is how r + 1 choose s works.
01:07:28.600 --> 01:07:34.100
So, what we have done is: we have now shown that the left side and the right side are exactly the same thing.
01:07:34.100 --> 01:07:39.100
We have set out to show that the left side and right side of our original thing are equivalent.
01:07:39.100 --> 01:07:46.200
Our original lemma is that they are the same thing: so r choose s plus r choose s - 1 is equal to r + 1 choose s.
01:07:46.200 --> 01:07:51.100
We have now completed our lemma; great; let's go and apply that lemma.
01:07:51.100 --> 01:07:54.500
So, here is our lemma, right here; and where were we before?
01:07:54.500 --> 01:08:01.500
Where we left off with our proof was: we had a^n + 1 and b^n + 1 on the two extremes,
01:08:01.500 --> 01:08:05.100
and then the big one that we really care about the most, the sum in the middle:
01:08:05.100 --> 01:08:16.400
the sum of k = 1 to n of (n choose k plus n choose k - 1) times a^n + 1 - kb^k; great.
01:08:16.400 --> 01:08:25.100
But at this point, we now have that nice, handy lemma; we have r choose s plus r choose s - 1.
01:08:25.100 --> 01:08:31.400
Well, that matches up to n choose k and n choose k - 1; so we will get n + 1 choose k out of it.
01:08:31.400 --> 01:08:36.700
We just used our lemma; we are on our way to ending this thing--just a little bit more effort, and we will be done.
01:08:36.700 --> 01:08:43.400
At this point, we are really, really close; this one here looks practically just like what we want to show in the end.
01:08:43.400 --> 01:08:49.000
The only difference is that we need to have our upper and lower summation limits be 0 and n + 1.
01:08:49.000 --> 01:08:54.400
We want to get this to become a 0, and we want to get this to become n + 1.
01:08:54.400 --> 01:08:56.200
How are we going to get that to happen?
01:08:56.200 --> 01:09:02.600
Well, on the way to doing that, we also have to deal somehow with this a^n + 1 and this b^n + 1.
01:09:02.600 --> 01:09:05.700
They are also in the way; they weren't in our finished thing.
01:09:05.700 --> 01:09:10.700
Maybe these things can be done together; maybe we can somehow shove these into that sigma notation--
01:09:10.700 --> 01:09:16.900
shove them into the sum, and in doing so, get the 0 and get the n + 1 that we want to appear in that.
01:09:16.900 --> 01:09:19.000
Indeed, we can figure out a way to do that.
01:09:19.000 --> 01:09:25.500
Notice: 1 is equal to n + 1 choose 0; also, 1 is equal to n + 1 choose n + 1.
01:09:25.500 --> 01:09:29.700
Now, you can go along and multiply anything you want by n.
01:09:29.700 --> 01:09:34.100
So, since you can multiply anything by 1, then we can multiply a 1 here and a 1 here.
01:09:34.100 --> 01:09:38.800
Well, we know that 1 is just the same thing as n + 1 choose 0; so we swap that out for n + 1 choose 0.
01:09:38.800 --> 01:09:44.900
And 1 is also just the same thing as n + 1 choose n + 1, so we swap that out here, as well.
01:09:44.900 --> 01:09:48.300
So now, we have something that we can mash into our summation.
01:09:48.300 --> 01:09:58.000
They fit the format of the summation; so we can fit them into the sigma notation, to obtain k = 0 to n + 1.
01:09:58.000 --> 01:10:05.400
If we plug in k = 0 on this n + 1 choose k, well, that would be n + 1 choose 0 for k = 0.
01:10:05.400 --> 01:10:11.300
And a^n + 1 - k, a^n + 1 - 0, would be a^n + 1b^k.
01:10:11.300 --> 01:10:15.600
b⁰ would be something that doesn't exist.
01:10:15.600 --> 01:10:23.300
And then, the same exact thing happens over here: if we had k equal to n + 1, then we would have n + 1 choose n + 1.
01:10:23.300 --> 01:10:29.000
And we would have a to the n + 1 minus...k is n + 1, so n + 1 - (n + 1).
01:10:29.000 --> 01:10:32.100
Well, that would cancel; a⁰...and so we have the thing that doesn't show up here.
01:10:32.100 --> 01:10:34.900
It doesn't show up, either; it just disappears.
01:10:34.900 --> 01:10:37.300
And b^n + 1 for our k...
01:10:37.300 --> 01:10:40.700
And so, we see that they mash in; we can pull them in.
01:10:40.700 --> 01:10:45.000
Everything else ends up staying the same; n and k = 1 aren't affected, because they do that.
01:10:45.000 --> 01:10:50.700
It is just that these fit the format of the summation, so we can pull them in on the bottom and pull them in on the top.
01:10:50.700 --> 01:10:53.400
We can fit them into the sigma notation, and we obtain this.
01:10:53.400 --> 01:10:56.600
We have managed to show what we were trying to show from the beginning.
01:10:56.600 --> 01:11:05.000
After much effort, we have finally shown that, assuming our inductive hypothesis, p<font size="-6">n</font>, we now know that p<font size="-6">n + 1</font> is true.
01:11:05.000 --> 01:11:14.700
That is, (a + b)^n + 1 is equal to the sum of k = 0 to n + 1 of n + 1 choose k times a^n + 1 - kb^k.
01:11:14.700 --> 01:11:18.100
That checks out; we have finished showing our inductive step.
01:11:18.100 --> 01:11:21.800
At this point, we know that the base case is true--we showed that from the beginning;
01:11:21.800 --> 01:11:24.500
and we have now finished showing that the inductive step is true.
01:11:24.500 --> 01:11:27.900
So, with some p<font size="-6">n</font>, we know that p<font size="-6">n + 1</font> must be true.
01:11:27.900 --> 01:11:35.100
Since that works as our inductive step, we now know, combining that with our base case, that p<font size="-6">n</font> is true for all n.
01:11:35.100 --> 01:11:42.500
In other words, for any n equal to 0, 1, 2...we have the binomial theorem completely finished proving.
01:11:42.500 --> 01:11:47.200
We have finished proving the binomial theorem; we are done--pretty cool.
01:11:47.200 --> 01:11:51.900
So, once again, I want to say: what you just saw is not easy.
01:11:51.900 --> 01:11:56.700
From the precalculus-level work that you are currently doing (or whatever the class that you are currently working in is),
01:11:56.700 --> 01:12:01.000
this is 2 or 3 years ahead--doing these kinds of proofs.
01:12:01.000 --> 01:12:05.800
But I honestly believe that, if you are really interested in this sort of stuff, you should be exposed to it now.
01:12:05.800 --> 01:12:08.900
You totally have the chance to be able to understand this stuff and work through it.
01:12:08.900 --> 01:12:12.500
It won't be easy like the sort of work that you are used to doing.
01:12:12.500 --> 01:12:15.900
If you are in the kind of position where you want to understand this sort of thing,
01:12:15.900 --> 01:12:19.900
the work that you are normally doing in your precalculus class probably isn't terribly difficult for you.
01:12:19.900 --> 01:12:25.600
But this right here is a really interesting thing that lets us see how far math gets out--
01:12:25.600 --> 01:12:31.100
what kind of stuff math is really like at the college level, when we are really studying serious-level math.
01:12:31.100 --> 01:12:37.800
I think that this stuff is really cool, absolutely beautiful, and a great chance to just see logic in its purest form--
01:12:37.800 --> 01:12:41.900
being able to make an argument that is really ironclad; I think that is really cool.
01:12:41.900 --> 01:12:45.100
It is not for everybody; if you think this is just absolutely awful, and you watched the whole thing,
01:12:45.100 --> 01:12:48.400
just because...I don't know why you watched the whole thing...but if you watched through it,
01:12:48.400 --> 01:12:51.600
and you really didn't like it, that is OK; there are lots of things to do out there in the world.
01:12:51.600 --> 01:12:55.300
But if you thought this was really cool, that is awesome; I think it is really cool, too.
01:12:55.300 --> 01:12:57.600
And you can go ahead and study math, if you are interested.
01:12:57.600 --> 01:13:03.200
You can study something like computer science, where you talk about algorithms like that, as well.
01:13:03.200 --> 01:13:10.600
There are lots of things that will let you have the same idea of well-argued proof, of things building on top of each other-- these really interesting conceptual ideas.
01:13:10.600 --> 01:13:13.000
All right, we will see you at Educator.com later--goodbye!