 Welcome to the screencast where we're going to look at some variations on the idea of mathematical induction that will allow us to use that technique on a larger set of statements. So first let's recall what the principle of mathematical induction says. Induction is an argumentation technique that we've used for proving propositions that claim that something at predicate, which we'll call P of n, is true for all natural numbers n. That part of the setup for all natural numbers is really important and that's the main point of distinction between what we've learned so far and what we're about to learn. Anyway, we're trying to prove that P of n is true for all natural numbers n. So following the example of the weary traveler trying to climb the staircase, we proceed in two stages. First, we prove the base case, that P of 1 is true. 1, of course, is the smallest natural number there is. And so if we can prove that P of 1 is true, we've proven that the proposition is true in the quote-unquote smallest possible case. Then we assume that for some natural number k, which we don't specify, that P of k is true. That is, we've made it up the stairs to some unspecified point. We called that assumption the inductive hypothesis. And from the inductive hypothesis, we then proceed to prove that P of k plus 1 is true. That is, we can make it from the step we're on to the step in front of us. If we do all this, then P of n is true for all natural numbers because there's no limit to the step that we can reach. And induction is very powerful, but it does have limits as stated. So let's, to see this, let's explore this predicate. 3 to the nth is bigger than 1 plus 2 to the nth. And the 2 is the only thing being raised to the nth power here. That's a predicate. And it may be true for some natural numbers and false for others. We don't know unless we play with it. So let's do that for a minute. Let's try some natural numbers n, into the left-hand side and into the right-hand side. Actually, I'm not going to do this. I'm going to give this to you as a concept check. So assume that the nth is playing around with some natural number values of n. Put them in for n and fill in the blank. For blank, we have 3 to the nth bigger than 1 plus 2 to the nth. And here are your choices. And once you think you have the answer, come on back from a paused video and let's see what we've got. So it doesn't take much playing around to get the idea that B is right. If n is equal to 1, we would have 3 on the left side of this inequality and 1 plus 2 or 3 on the right. And that actually makes P of 1 false because this isn't an equation. It's a strict inequality. Since 3 is not bigger than 3, the inequality is not true for all positive integers. If n is equal to 2 on the other hand, then we get 3 squared or 9 on the left and 1 plus 2 squared, which is 5 on the right. And 9 is bigger than 5, of course. So P of 2 is true. If n is equal to 3, we get 27 on the left. And 1 plus 2 to the 3rd, that's 9 on the right. So P of 3 is true. If n equals 4, we get 81 on the left and 1 plus 2 to the 4th, which is 17 on the right. So P of 4 is true. So these are just examples, but it's enough to make us believe that B is the right answer. But now we would need to prove this proposition that we believe it's true. Induction seems like a natural approach here, except for if we use the straight principle of mathematical induction that we've learned earlier, it wouldn't work because P of 1, the traditional base case, is not true. So what do we do? Well, this kind of seems like a technicality. I mean, the proposition appears to be true in all cases except 1. So shouldn't we just be able to ignore that one case where it doesn't work? Well, the answer actually turns out to be yes. And we formalize that in what we call the extended principle of mathematical induction. The extended principle of mathematical induction is a modification of the straight principle of mathematical induction that says if we're trying to prove a predicate of n is true, not necessarily for all natural numbers, but for all natural numbers bigger than or equal to some integer m, then we're going to proceed as follows. First, we can make the base case to prove not P of 1, but P of m is true. m is like a threshold. The predicate is true not necessarily for every natural number, but for every natural number at or beyond this threshold. And we take the base case, not to be 1, but the threshold value m. So we're going to prove P of m is true in the base case this time. Then we assume the inductive hypothesis that the predicate is true for some natural number k bigger than or equal to our threshold. And then we prove that P of k plus 1 is true. So in other words, the extended principle of mathematical induction just says if we think the proposition is true, not for all natural numbers, but for all natural numbers past a certain point, then we can just start at that point rather than at 1. Otherwise it's exactly the same as the principle of induction. So let's go back and see how this works with our proposition from earlier, where we're going to try to prove that 3 to the nth is bigger than 1 plus 2 to the nth. And we think that this is true for n all natural numbers n bigger than or equal to 2. We know it's not true for n equals 1, but we think that if we're at 2 or higher, then this should be okay. Let's use our new extended principle to get there. So for the base case, we're going to start at n equals 2, because again, we know for a fact that n equals 1 doesn't work, but we do think the proposition works for all n bigger numbers. So we just need to prove p of 2 is true. So let's demonstrate the inequality holds for n equals 2. Actually, we already did this work, but let's do it again here just for completeness. So put in 2 to the left-hand side and get 3 squared equals 9. And put 2 in on the right to get 1 plus 2 squared equals 5. And just note that 9 is bigger than 5. And that's that. That demonstrates another reason why you should always play with the problem first. You might end up doing some essential work here. So that's what we're going to take care of now. Now we move on to the inductive step. For the inductive hypothesis, we're going to assume that 3 to the k is bigger than 1 plus 2 to the k for some integer k that is bigger than or equal to 2. Again, there's the threshold value. And we want to prove, as usual, that 3 to the k plus 1 is bigger than 1 plus 2 to the k plus 1. Because somehow it makes it easier for me to think about this, I'm going to switch this around and write 1 plus 2 to the k plus 1 less than 3 to the k plus 1. It's the same inequality. So let's start with the left hand side of this inequality, 1 plus 2 to the k plus 1. Now what do you do first? Keep in mind that we want to get this less than 3 to the k plus 1. So somewhere some threes need to show up. And somewhere in this process, I should try to use the inductive hypothesis, which is the assumption that 1 plus 2 to the k is less than 3 to the k. This inductive hypothesis, remember, is a fact that we can use. The k plus 1 stage is something we're going to give. And so we can't touch that. So one thing I want to do here is somehow get 1 plus 2 to the k in the picture. Now, a partial step in that direction is to factor off a 2 from 2 to the k plus 1. So that gives us 1 plus 2 times 2 to the k. I can kind of see the inductive hypothesis forming here, but I can't just pull out a factor of 2. But keeping in mind that what I'm trying to prove is an inequality, and this expression you see here is going to be at the very bottom of that inequality. It's okay to start introducing items here that create inequalities. So I'm going to make two totally trivial observations. First, 1 is less than 3. And second, 2 is less than 3. I think we all get that. But this allows me to make two replacements that get me going in the right direction. First of all, I'm going to replace the 1 here with a 3 to get 1 plus 2 times 2 to the k less than 3 plus 2 times 2 to the k. Definitely the thing on the right is larger. Now let's replace the factor of 2 you see here with a factor of 3. And that's going to give me all this stuff is less than 3 plus 3 times 2 to the k. And now I can factor off a 3 and get all this final right side is equal to 3 times 1 plus 2 to the k. And like magic, there's my inductive hypothesis. The inductive hypothesis tells me that this item in parentheses is less than 3 to the k. So I can write this is less than 3 times 3 to the k. And this last term, of course, is equal to 3 to the k plus 1. So that proves the inductive step because the 1 plus 2 to the k plus 1 in the end is less than 3 to the k plus 1 and the proof is done. If you wonder where all that stuff came from, just realize a couple of things. First of all, there's a certain amount of creativity involved and a good amount of messing around with calculations that you have to do. Calculations that don't work until you hit up on the right idea. I'll tell you that in making the screen cast out, I had to mess around for about 10 minutes or so before I found an approach that actually works. And on any given day, it might have taken me 10 times that amount. So this is not a straight line solution that you should see immediately. It does take experimentation. And second of all, though, this isn't just blind experimentation. There are two principles that guide my thinking here. One is that I need to get the inductive hypothesis in the picture somehow. So I'm working towards getting a 1 plus 2 to the k to appear. Another thing that guides me is that I'm working towards proving that 1 plus 2 to the k plus 1 is less than 3 to the k plus 1. So while I can't assume this conclusion, I can definitely keep it in mind and let it guide my selection and my decision making throughout the proof. I know what the goal is, so while it might take a while to find a series of calculations that works, it's not just blindly throwing things at the problem and hoping something sticks. So that's an example of the extended principle of mathematical induction. Thanks for watching.