WEBVTT
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Today is going to be one of the more difficult lectures of the term.
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So put on your thinking caps as they say in elementary school.
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The topic is going to be what's called the convolution.
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The convolution is something very peculiar that you do to two functions to get a third function.
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It has its own special symbol, F of T.
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Asterisk is the universal symbol that's used for that.
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So this is a new function of T, which bears very little resemblance to the one's F of T that you started with.
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I'm going to give you the formula for it, but I first,
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there are two ways of motivating it, and both are important.
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There's a formal motivation, which is why it's tucked into this section on the plus transform.
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And the formal motivation is the following.
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Suppose we start with a Laplace transform of two different, those two functions.
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Now, the most natural question to ask is since Laplace transforms are really a pain to calculate,
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is from old Laplace transforms, is it easy to get new ones?
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And the first thing, of course, summing functions is easy.
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That gives you the sum of the transforms.
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But a more natural question would be, suppose I want to multiply F of T and G of T.
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Is there some, hopefully, some meat formula, if I multiply the product of the,
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take the product of these two, is there some meat formula for the Laplace transform of that product?
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That would simplify life greatly.
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And the answer is, there is no such formula, and there never will be.
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Well, we will not give up entirely.
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Suppose we ask the other question.
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Suppose, instead, I multiply the Laplace transforms.
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Could that be related to something I cook up out of F of T and G of T?
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Could it be the transform of something I cooked up out of F of T and G of T?
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And that's what the convolution is for.
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The answer is that F of S times G of S turns out to be the Laplace transform of the convolution.
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The convolution, and that's one way of defining it, is the thing that you,
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the function of T that you should put in there in order that it's Laplace transform,
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turn out to be the product of F of S times G of S.
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Now, I'll give you in a moment the formula for it, but I think I'll,
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I'll give you one and a quarter minutes of, well, two minutes of motivation as to why
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this should be such a formula.
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Now, I won't calculate this out to the end because I don't have time.
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But here's the reason why this should be such a formula, and you might suspect,
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and therefore it will be worth looking for.
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It's because remember I told you where the Laplace transform came from,
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that the Laplace transform was the continuous analog of a power series.
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So when you ask a general question like that, the place to look for is if you know
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an analogous idea, say, does it work something like that work there?
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So here I have a power series, summation A n x to the n.
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Remember, you can write this in computer notation as A of n to make it look like
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F of n, F of t.
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And the analog is between n is turned into t when you turn a power series into
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the Laplace transform and x gets turned into e to the negative S and one formula
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just turns into the other.
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OK, so there's a formula for F of x.
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This is the analog of the Laplace transform.
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And similarly, G of x, here is summation B n x to the n.
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Now, again, the naive question would be, well, suppose I multiply the two
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coefficients together and add up that power series, summation A n B n times x to the
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n.
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Is that somehow related to F and G?
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And of course, everybody knows the answer to that is no.
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That's no relation, whatever.
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But suppose instead I multiply these two guys, well, in that case, I'll get a new
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power series.
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I don't know what it's coefficient so, but let's write them down.
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Let's just call them c n's.
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So what I'm asking is, if this corresponds to the product of the two Laplace
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transforms, and what I want to know is, is there a formula, which says that c n
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is equal to something that can be calculated out of the A i and the B j.
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Now the answer to that is yes, there is.
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And the formula for c n is called the convolution.
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Now, you could figure out this formula yourself.
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You figure that.
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Anyone who's smart enough to be interested in the question of the first place is
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smart enough to figure out what that formula is.
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And it will give you great pleasure to see that it's just like the formula for
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the convolution that I'm going to give you now.
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So what is that formula for the convolution?
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OK, hang on.
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Now you're not going to like it.
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But you didn't like the formulas with a little plus transform either.
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You felt wiser, grown up, getting it.
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But it's a mouthful to swallow.
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It's something you get used to slowly.
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Now you'll get used to the convolution equally slowly.
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So what is the convolution of f of t and g of t?
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It's a function calculated according to the corresponding formula.
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It's a function of t.
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It's the integral from 0 to t of f of u.
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U is a dummy variable because it's going to be integrated out when I do the
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integration, g of t minus u dt.
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That's it.
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I didn't make it up.
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I'm just bearing the bad news.
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Well, what do you do when you see a formula that seems to, you know, it's, well,
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you know, well, the first thing you can do, of course, is try calculating.
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Just to get some feeling for what kind of a thing.
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You know, what's this going to do?
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Let's try some examples.
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Let's see.
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Let's calculate what would be a modest beginning.
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Let's calculate the convolution of t with itself.
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Or better yet, let's calculate the convolution just so that you can tell the
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difference.
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t with t squared.
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t squared with t make it a little easier.
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Finally, the convolution is symmetric.
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f star g is the same thing as g star f.
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Let's put that down explicitly.
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I forgot to last period.
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So tell all the guys who came to the 1 o'clock lecture that you know something
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that they don't.
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Now, that's a theory.
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It's a commutative.
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This operation is commutative in other words.
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Now, that has to be a theorem because the formula is not symmetric.
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The formula does not treat f and g equally.
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And therefore, this is not obvious.
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It's at least not obvious if you look at it that way.
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But it is obvious if you look at it that way.
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Why?
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In other words, f star g is the guy whose Laplace transform is f of s times g of s.
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Well, what would g star f be?
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That would be the guy whose Laplace transform is g times f.
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But capital f times capital g is the same as capital g times capital f.
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So it's because the Laplace transforms are commutative.
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Ordinary multiplication is commutative.
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It follows that this has to be commutative, commutative 2.
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So I'll write that down.
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Since f times g is equal to g f.
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And you have to understand that here, I mean, these are the Laplace transforms.
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So it's not obvious if you look at the formula of those guys.
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But it's not obvious from the formula.
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OK, let's calculate the Laplace transform of, sorry, the convolution of t star.
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Let's do it by the formula.
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By the formula, I calculate integral 0 to t.
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Now, I take the first function, but I change its variable to the dummy variable u.
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So that's u squared.
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The second function, I replace its variable by u minus t.
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So this is times u t minus u.
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Sorry.
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OK?
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Do you see that this, to calculate that this is what I have to write down, that's what
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the formula becomes.
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Any question?
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Anything wrong?
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Oh, sorry, the integration is respect to u, of course.
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This should be it.
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Thanks very much.
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OK, let's do it.
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So it is.
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U squared t is, remember, it's integrated with respect to u.
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So it's u cubed over 3 times t.
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The rest of it is the integral of u cubed, which is u to the fourth over 4.
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All this is to be evaluated between 0 and t.
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At the upper limit, so I put u equal t, I get t to the fourth over 3 minus t to the
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fourth over 4.
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Of course, at the lower limit, u is 0.
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So both of these terms of 0, there's nothing there.
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And the answer is, therefore, t to the fourth divided by a third minus a quarter is a
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12.
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So let's do a get from the formula.
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But, of course, there's an easier way to do it.
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We can cheat and use the Laplace transform instead.
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If I Laplace transform it, the Laplace transform of t squared is what?
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Is 2 factorial divided by s cubed?
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The Laplace transform of t is 1 divided by s squared.
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And so because this is the convolution of these, it should correspond to the product
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of the Laplace transforms, which is 2 over s to the fifth power.
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Well, is that the same as this?
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What's the Laplace transform of?
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In other words, what's the inverse Laplace transform of 2 over s to the fifth?
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Well, the inverse Laplace transform of 4 factorial over s to the fifth is how much?
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That's t to the fourth, right?
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Now, how does this differ?
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Well, to turn that into that, I should divide by 4 times 3.
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So this should be 1 12t to the fourth, 1 over 4 times 3, because this is 24 and that's 2.
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So divide by 12 to turn what constant is again.
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So it works, at least in that case.
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But now, notice, this is not an ordinary product.
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The convolution of t squared and t is not something like t cubed, it's something like t to the fourth power.
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And there's a funny constant in there, too.
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Very unpredictable.
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Let's look at the convolution.
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Let's take another example of a convolution.
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Let's do something really humble, just to show you that this, even the simplest example, this is not trivial.
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Let's take the convolution of f of t with 1.
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Can you take, yeah, what is a function just like any function?
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But you get something out of the convolution?
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Yes, yes.
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Let's just write down the formula.
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Now, I can't use a little plus transform here because you won't know what to do with it.
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You don't have that formula yet.
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It's a secret one that only I know.
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So let's calculate it out the way we're supposed to.
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So it's the integral from 0 to t of f of u.
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And now, what do I do with that 1?
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I'm supposed to take, that 1 is the function g of t.
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Wherever I see a t, I'm supposed to log in t minus u.
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Well, I don't see any t there, but that's something for rejoicing.
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There's nothing to do to make the substitute.
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It's just 1.
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So the answer is it's the, this curious thing, the convolution of a function with 1,
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u integrated from 0 to t.
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Well, as they said, now, it's the one to land, the positive things are getting curious
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or an curious, sir.
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I mean, what is going on with this crazy function?
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And where are we supposed to start with it?
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Well, I've got to prove this for you, mostly because the proof is easy.
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In other words, I'm going to prove that that's true.
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And as I give the proof, you'll see where the convolution is coming from.
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That's number 1.
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And number 2, the real reason I'm giving you the proof because it's a marvelous exercise
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in changing the variables in a double integral.
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Now, that's something you all know how to do, even the ones who are taking 1802
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concurrently and I didn't advise you to do that.
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But I've arranged the course so it's possible to do.
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But I knew that by the time we got to this, you would already know how to change variables
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in a double integral.
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So, and in fact, you'll have the advantage of remembering how to do it because you just
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had it a week or two ago, whereas all the other guys did something dim in their distance.
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So I'm reviewing how to change variables in a double integral.
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I'm showing you it's good for something.
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So what we're out to try to prove is this formula.
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Let's put that down and say you understand.
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Okay, let's do it.
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Now, we'll use the desert island method.
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So you have as much time as you want your desert island.
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In fact, I'm going to even go at the opposite way.
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I'm going to start with, you got a lot of time on your hand and say,
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gee, I wonder if there's a, if I take the product of the plus transforms,
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I wonder if there's some crazy function I could put in there which would make things work.
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You've never heard of the convolution.
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You're going to discover it all by yourself.
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Okay, so how do you begin?
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So we'll start with the left hand side.
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We're looking for some nice way of calculating that as a plus transform of a single function.
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So the way to begin is by writing out the definitions.
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We couldn't use anything else since we don't have anything else to use.
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Now, looking ahead, I'm going to not use T.
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I'm going to use two neutral variables when I calculate.
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After all, the T is just a dummy variable anyway.
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You'll see in a minute the wisdom of doing this.
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So it's this times the integral which gives the plus transform of g.
239
00:16:47.600 --> 00:16:54.640
So that's e to the negative s, v, let's say, times g of v, v.
240
00:16:54.640 --> 00:16:57.720
Okay, everybody can get that far.
241
00:16:57.720 --> 00:17:03.400
And now we have to start with the, well, this is a single integral,
242
00:17:03.400 --> 00:17:08.600
an 1801 integral involving u, and this is an 1801 integral involving v.
243
00:17:08.600 --> 00:17:11.920
Well, would you take the product of two integrals like that?
244
00:17:11.920 --> 00:17:17.560
Remember, when you evaluate a double integral, there's an easy case where you don't,
245
00:17:17.560 --> 00:17:19.800
it's much easier than any other case.
246
00:17:19.800 --> 00:17:25.560
If you could write the inside, if you're integrating over a rectangle, for example.
247
00:17:25.560 --> 00:17:31.840
And you can write the integral as a product of a function just of u,
248
00:17:31.840 --> 00:17:36.840
and the product of a function just as v, then the integral is very easy to evaluate.
249
00:17:36.840 --> 00:17:38.080
You can forget all the rules.
250
00:17:38.080 --> 00:17:42.600
Just take all the u part out, all the v part out, and integrate them separately,
251
00:17:42.600 --> 00:17:44.560
a to b, c to d.
252
00:17:44.560 --> 00:17:47.400
That's the easy case of evaluating a double integral.
253
00:17:47.400 --> 00:17:50.160
It's what everybody tries to do, even when it's not appropriate.
254
00:17:52.120 --> 00:17:56.440
Now here it is appropriate, except I'm going to use it backwards.
255
00:17:56.440 --> 00:17:59.400
This is the result of having done that.
256
00:17:59.400 --> 00:18:04.640
If this is the result of having done it, what was the step just before it?
257
00:18:04.640 --> 00:18:11.640
Well, I must have been trying to evaluate a double integral as u runs from zero to infinity
258
00:18:11.640 --> 00:18:14.200
and v runs from zero to infinity of what?
259
00:18:14.200 --> 00:18:16.920
Well, of the product of these two functions.
260
00:18:16.920 --> 00:18:17.920
Now what is that?
261
00:18:17.920 --> 00:18:21.560
E to the minus s u times e to the minus s v.
262
00:18:21.560 --> 00:18:26.240
Well, I must surely want to combine those.
263
00:18:26.240 --> 00:18:29.680
E to the minus s u times e to the minus s v.
264
00:18:29.680 --> 00:18:30.400
And what's left?
265
00:18:30.400 --> 00:18:34.400
Well, what gets dragged along?
266
00:18:34.400 --> 00:18:40.360
d u d v. This is the same as that because of that law I just gave you.
267
00:18:40.360 --> 00:18:44.200
This is the product of a function just of u and a function just of v.
268
00:18:44.200 --> 00:18:47.960
And therefore, it's okay to separate the two integrals out that way.
269
00:18:47.960 --> 00:18:51.720
Because I'm integrating over sort of a rectangle that goes to infinity that way,
270
00:18:51.720 --> 00:18:52.960
and infinity that way.
271
00:18:52.960 --> 00:19:00.120
But what I'm integrating is over the plane, in other words, this region of the plane.
272
00:19:00.120 --> 00:19:07.040
As u v goes from zero to infinity, zero to infinity.
273
00:19:07.040 --> 00:19:10.440
Now, let's take a look.
274
00:19:10.440 --> 00:19:12.680
What are we looking for?
275
00:19:12.680 --> 00:19:21.480
Well, we're looking for, we would be very happy if u plus v were t.
276
00:19:21.480 --> 00:19:26.520
Let's make it t.
277
00:19:26.520 --> 00:19:31.080
In other words, I'm introducing a new variable, t, u plus v.
278
00:19:31.080 --> 00:19:35.160
And it's suggested by the form in which I'm looking for the answer.
279
00:19:35.160 --> 00:19:38.840
Now, of course, we need another variable.
280
00:19:38.840 --> 00:19:40.640
We could keep either u or v.
281
00:19:40.640 --> 00:19:42.800
Oh, let's keep u.
282
00:19:42.800 --> 00:19:47.600
That means v, we've just gave a musical chairs, v got dropped out.
283
00:19:47.600 --> 00:19:49.360
Well, we can't have three variables.
284
00:19:49.360 --> 00:19:50.600
We only have room for two.
285
00:19:50.600 --> 00:19:59.800
But we will remember it, rest in piece v was equal to t minus u in case we ever need him again.
286
00:20:04.520 --> 00:20:07.240
OK, let's now put in the limits.
287
00:20:07.240 --> 00:20:09.960
Let's put in the integral, the rest of the change of variable.
288
00:20:09.960 --> 00:20:15.840
So I'm now changing it to these new variables, t and u.
289
00:20:15.840 --> 00:20:17.920
So it's e to the negative s t.
290
00:20:17.920 --> 00:20:21.880
Well, f of u, I don't have to do anything, too.
291
00:20:21.880 --> 00:20:25.120
But g of v, I'm not allowed to keep v.
292
00:20:25.120 --> 00:20:30.720
So v has to be changed to t minus u.
293
00:20:30.720 --> 00:20:34.600
Amazing things are happening.
294
00:20:34.600 --> 00:20:37.080
du.
295
00:20:37.080 --> 00:20:40.120
Now, I want to change this to an integral du dt.
296
00:20:40.120 --> 00:20:42.400
Now, for that, you have to be a little careful.
297
00:20:42.400 --> 00:20:47.720
We have two things to do to figure out this, what goes with that.
298
00:20:47.720 --> 00:20:51.600
And we have to put in the limits also.
299
00:20:51.600 --> 00:20:57.320
Now, those are the two non-trivial operations when you change integrals in a change variables
300
00:20:57.320 --> 00:20:58.200
in a double integral.
301
00:20:58.200 --> 00:21:00.320
So let's be really careful.
302
00:21:00.320 --> 00:21:01.600
Let's do the easier the two.
303
00:21:01.600 --> 00:21:10.040
First, I want to change from du dv to du dt.
304
00:21:10.040 --> 00:21:14.120
And now, to do that, you have to put in the Jacobian matrix.
305
00:21:14.120 --> 00:21:16.120
The Jacobian's a term.
306
00:21:16.120 --> 00:21:19.920
Ah, you have any of you forgot that?
307
00:21:19.920 --> 00:21:21.320
I won't even bother asking.
308
00:21:21.320 --> 00:21:25.600
Ah, come on.
309
00:21:25.600 --> 00:21:27.200
You only lose two points.
310
00:21:27.200 --> 00:21:30.360
What's the, it doesn't matter if you put in the Jacobian.
311
00:21:30.360 --> 00:21:33.520
Well, as you see, you know, the gridded forget something.
312
00:21:33.520 --> 00:21:36.560
You lose less credit for forgetting that than anything else.
313
00:21:36.560 --> 00:21:41.720
So it's the Jacobian of u in v with respect to u and t.
314
00:21:41.720 --> 00:21:49.640
So to calculate that, you write u equals u v equals t minus u.
315
00:21:49.640 --> 00:21:59.600
And then the Jacobian is, the Jacobian is blah, blah, blah, blah, blah.
316
00:21:59.600 --> 00:22:05.920
The partial, the matrix, the determinant of partial derivatives.
317
00:22:05.920 --> 00:22:10.760
So it's the determinant whose entries are the partial of u with respect to u,
318
00:22:10.760 --> 00:22:14.720
the partial of u with respect to t, but these are independent variables.
319
00:22:14.720 --> 00:22:15.800
So that's zero.
320
00:22:15.800 --> 00:22:19.240
The partial of v with respect to u is negative 1.
321
00:22:19.240 --> 00:22:22.720
The partial of v with respect to t is 1.
322
00:22:22.720 --> 00:22:25.520
So the Jacobian is 1.
323
00:22:25.520 --> 00:22:27.120
So if you forgot it, no harm.
324
00:22:30.920 --> 00:22:34.960
So the Jacobian is 1.
325
00:22:34.960 --> 00:22:38.000
Now, more serious, and in some ways, I think for most of you,
326
00:22:38.000 --> 00:22:42.320
the most difficult part of the operation is putting in the new limits.
327
00:22:43.600 --> 00:22:47.160
Now, for that, you look at the region over which you're integrating.
328
00:22:47.160 --> 00:22:49.520
I think I'd better do that carefully.
329
00:22:49.520 --> 00:22:51.080
Let's, I need a bigger picture.
330
00:22:51.080 --> 00:22:53.040
That's really what I'm trying to say.
331
00:22:53.040 --> 00:22:55.920
So here is the u v coordinates.
332
00:22:55.920 --> 00:23:00.040
And I want to change these to u v to u t coordinates.
333
00:23:00.040 --> 00:23:03.240
The integration is over the first quadrant.
334
00:23:03.240 --> 00:23:07.240
So what you do is, when you do the integral,
335
00:23:07.240 --> 00:23:13.240
the first step is u is varying and t is held fixed.
336
00:23:13.240 --> 00:23:21.240
So in the first integration, u varies t is held fixed.
337
00:23:21.240 --> 00:23:24.240
Now, what is holding t fixed in this picture mean?
338
00:23:24.240 --> 00:23:29.240
Well, t is equal to u plus v, so u plus v is fixed.
339
00:23:29.240 --> 00:23:33.240
It's a constant, in other words.
340
00:23:33.240 --> 00:23:36.240
Now, where are the curves along which u plus v is a constant?
341
00:23:36.240 --> 00:23:40.240
Well, they are these lines.
342
00:23:40.240 --> 00:23:45.240
These are the lines along which u plus v equals a constant.
343
00:23:45.240 --> 00:23:47.240
Or t is a constant.
344
00:23:47.240 --> 00:23:51.240
The reason I'm holding t at constant is because the first integration
345
00:23:51.240 --> 00:23:55.240
only allows you to change t is held fixed.
346
00:23:55.240 --> 00:23:59.240
Okay, you let u increase.
347
00:23:59.240 --> 00:24:03.240
As u increase and t is held fixed,
348
00:24:03.240 --> 00:24:08.240
I'm traversing these lines in this direction.
349
00:24:08.240 --> 00:24:11.240
That's the direction on which u is increasing.
350
00:24:11.240 --> 00:24:15.240
I integrate from the point, from the u value where they leave the region,
351
00:24:15.240 --> 00:24:19.240
into the end of the region, what's the u value where they enter the region?
352
00:24:19.240 --> 00:24:22.240
u is equal to zero.
353
00:24:22.240 --> 00:24:24.240
Everybody would know that.
354
00:24:24.240 --> 00:24:31.240
Not so many people would be able to figure out what to put for where it leaves the region.
355
00:24:31.240 --> 00:24:35.240
What's u, the value of u, when it leaves the region?
356
00:24:35.240 --> 00:24:39.240
Well, this is the curve v equals zero.
357
00:24:39.240 --> 00:24:45.240
But v equals zero is, in another language, u equals t.
358
00:24:45.240 --> 00:24:48.240
t minus u equals zero, or u equals t.
359
00:24:48.240 --> 00:24:52.240
In other words, they enter the region where u equals zero,
360
00:24:52.240 --> 00:24:57.240
and they leave where u is t, has the value t.
361
00:25:03.240 --> 00:25:05.240
And how about the other guys?
362
00:25:05.240 --> 00:25:07.240
For which t's do I want to do this?
363
00:25:07.240 --> 00:25:10.240
Well, I want to do it for all these t values.
364
00:25:10.240 --> 00:25:14.240
Well, now the t value here, that's the starting one.
365
00:25:14.240 --> 00:25:18.240
Here t is zero, and here t is not zero.
366
00:25:18.240 --> 00:25:24.240
And if I go out and cover the whole first quadrant, I'll be going letting t increase to infinity.
367
00:25:24.240 --> 00:25:27.240
The sum of u and v, I will be letting increase to infinity.
368
00:25:27.240 --> 00:25:29.240
So it's zero to infinity.
369
00:25:29.240 --> 00:25:36.240
So all this is an exercise in changing, taking this double integral in uv co-ordinates,
370
00:25:36.240 --> 00:25:41.240
and changing it to this double integral, an equivalent double integral over the same region,
371
00:25:41.240 --> 00:25:44.240
but now in u t co-ordinates.
372
00:25:44.240 --> 00:25:49.240
And now that's the answer.
373
00:25:49.240 --> 00:25:57.240
So here is the answer, because look, if I, since the first integration is with respect to u,
374
00:25:57.240 --> 00:26:06.240
this guy can migrate outside because it doesn't involve u.
375
00:26:06.240 --> 00:26:11.240
That only involves t, and t is only caught by the second integration.
376
00:26:11.240 --> 00:26:14.240
So I can put this outside, and what do I end up with?
377
00:26:14.240 --> 00:26:21.240
The integral from zero to infinity of e to the negative s t times what's left?
378
00:26:21.240 --> 00:26:25.240
A funny expression, but you're on your desert island, and found it.
379
00:26:25.240 --> 00:26:32.240
It's funny expression, integral from zero to t, f of u, g of t minus u, du,
380
00:26:32.240 --> 00:26:35.240
ensure the convolution.
381
00:26:35.240 --> 00:26:40.240
Exactly the convolution.
382
00:26:40.240 --> 00:26:44.240
So all you have to do is get the idea that there might be a formula, sit down,
383
00:26:44.240 --> 00:26:48.240
change variables, the double integral, and echo, you got your formula.
384
00:26:48.240 --> 00:26:58.240
Well, I would like to spend much of the rest of the period.
385
00:26:58.240 --> 00:27:02.240
In other words, so that's how it relates to the Laplace transform.
386
00:27:02.240 --> 00:27:05.240
That's how it comes out of the Laplace transform.
387
00:27:05.240 --> 00:27:08.240
Here's how to use it, you know, calculated either with the Laplace transform
388
00:27:08.240 --> 00:27:11.240
or directly from the integral.
389
00:27:11.240 --> 00:27:16.240
And of course you'll solve problems Laplace transform problems differential equations using the convolution.
390
00:27:16.240 --> 00:27:20.240
But I have to tell you that most people, convolution is very important,
391
00:27:20.240 --> 00:27:24.240
and most people who use it don't use it in connection with the Laplace transform.
392
00:27:24.240 --> 00:27:27.240
They use it for its own sake.
393
00:27:27.240 --> 00:27:35.240
The first place I learned that, you know, outside of MIT people use the convolution was actually
394
00:27:35.240 --> 00:27:41.240
a daughter who, you know, was, she's an environmental engineer, a consultant,
395
00:27:41.240 --> 00:27:44.240
environmental consultant, she does risk assessment and stuff like that.
396
00:27:44.240 --> 00:27:48.240
But anyway, she had this paper on acid range, she was trying to read for a client.
397
00:27:48.240 --> 00:27:56.240
And she said, you know, it was something about calculated acid rain falls on soil men
398
00:27:56.240 --> 00:28:01.240
from there, the stuff leaches into a river, but things happen to it on the way.
399
00:28:01.240 --> 00:28:07.240
And then she said, well, that they calculated the end, how much the river gets polluted.
400
00:28:07.240 --> 00:28:10.240
But she said, but it's a convolution.
401
00:28:10.240 --> 00:28:13.240
She said, well, what's the convolution?
402
00:28:13.240 --> 00:28:17.240
So I told her she was too young to learn about the convolution.
403
00:28:17.240 --> 00:28:21.240
And, you know, so I thought I'd better look it up first.
404
00:28:21.240 --> 00:28:25.240
I mean, I, of course, knew what the convolution was, but I was a little puzzled with that application.
405
00:28:25.240 --> 00:28:31.240
So I read the paper, it was interesting. And I, thinking about it, I then, other people have come to me, you know,
406
00:28:31.240 --> 00:28:40.240
some guy with a problem about the drill dice cores of the North Pole and from the radioactive carbon and so on,
407
00:28:40.240 --> 00:28:44.240
with deducing various things about the climate, you know, 60 billion years ago.
408
00:28:44.240 --> 00:28:48.240
And it was all convolution. He asked me if I could explain that to him.
409
00:28:48.240 --> 00:28:52.240
So let me give you a second question.
410
00:28:52.240 --> 00:29:00.240
So let me give you a sort of an all-purpose thing, a simple all-purpose model, which can be adapted,
411
00:29:00.240 --> 00:29:07.240
which is a very good way, I think, of the convolution in my opinion.
412
00:29:07.240 --> 00:29:16.240
It's a problem of radioactive dumping.
413
00:29:16.240 --> 00:29:24.240
It's in the notes, by the way. So I'm just, if you want to take a chance and just listen to what I'm saying,
414
00:29:24.240 --> 00:29:30.240
rather than scribbling everything down, maybe you'll be able to figure it out in the notes also.
415
00:29:30.240 --> 00:29:35.240
So the problem is we have some factory or some nuclear plant or something like that.
416
00:29:35.240 --> 00:29:42.240
It's producing nuclear material, radioactive waste, a certain not always at the same rate.
417
00:29:42.240 --> 00:29:47.240
And then it quarts, it dumps it on a pile somewhere.
418
00:29:47.240 --> 00:29:55.240
So radioactive waste is dumped and there's a dumping function.
419
00:29:55.240 --> 00:30:01.240
I'll call that F of T. The dump rate.
420
00:30:01.240 --> 00:30:04.240
That's the dumping rate. Let's say T is in years.
421
00:30:04.240 --> 00:30:11.240
You like to have units and quantity of kilograms, I don't know, whatever you want.
422
00:30:11.240 --> 00:30:22.240
Now, what does the dumping rate mean? The dumping rate means that if I are the time axis,
423
00:30:22.240 --> 00:30:27.240
if I have two times which are close together, for example, two successive days,
424
00:30:27.240 --> 00:30:33.240
the new midnight and two successive days, then there's a time interval between them.
425
00:30:33.240 --> 00:30:40.240
I'll call that delta T. To say the dumping rate is F of T means that the amount,
426
00:30:40.240 --> 00:30:52.240
the amount dumped in this time interval, in the time interval from Ti to Ti plus 1,
427
00:30:52.240 --> 00:31:00.240
is approximately not exactly because the dumping rate isn't even constant within this time interval.
428
00:31:00.240 --> 00:31:11.240
But it's approximately the dumping rate times the time over which the dumping was taking place.
429
00:31:11.240 --> 00:31:20.240
That's what I mean by the dump rate. And it gets more and more accurate the small of the time interval you take.
430
00:31:20.240 --> 00:31:32.240
Now, here's my problem. The problem is you start dumping at time T equals 0,
431
00:31:32.240 --> 00:31:54.240
at time T equals T, how much radioactive waste is in the pile?
432
00:31:54.240 --> 00:32:00.240
Now, what makes that problem slightly complicated is radioactive waste decays.
433
00:32:00.240 --> 00:32:06.240
If I put some at certain day and then go back several months later, I don't,
434
00:32:06.240 --> 00:32:13.240
and nothing's happened in between, I don't have the same amount that I dumped because some of a fraction of a decayed.
435
00:32:13.240 --> 00:32:23.240
I have less. And our answer to the problem must take account of for each piece of waste how long it has been in the pile
436
00:32:23.240 --> 00:32:30.240
because that takes account of how long it had to decay and what it ends up as.
437
00:32:30.240 --> 00:32:42.240
So the calculation, the essential part of the calculation will be that if you have an initial amount of this substance
438
00:32:42.240 --> 00:32:59.240
and it decays for a time T, this is the amount left at time T.
439
00:32:59.240 --> 00:33:04.240
You know, this is a law of radioactive decay. You knew that coming into 1803.
440
00:33:04.240 --> 00:33:09.240
Although, it's of course a simple differential equation which produces it, but I'll assume you simply know the answer.
441
00:33:09.240 --> 00:33:17.240
K depends on the material, so I'm going to assume that the nuclear plant dumps the same radioactive substance each time.
442
00:33:17.240 --> 00:33:23.240
It's only one substance I'm calculating and K is itself. Assume the K is fixed.
443
00:33:23.240 --> 00:33:30.240
I don't have to change from one K from one material to a K for another because it's mixing up the stuff.
444
00:33:30.240 --> 00:33:35.240
Just one material. Okay, and now let's calculate it.
445
00:33:35.240 --> 00:33:43.240
Here's the idea. I'll take the T axis, but now I'm going to change its name to the U axis. You'll see why it does the second.
446
00:33:43.240 --> 00:33:51.240
Starts at zero. I'm interested in what's happening at the time T. How much is left at time T?
447
00:33:51.240 --> 00:34:03.240
So I'm going to divide up the interval from zero to T on this time axis into, well here's U zero, the starting point U1, U2.
448
00:34:03.240 --> 00:34:19.240
Let's make this U1. U2, U3, and so on. Let's call this UN.
449
00:34:19.240 --> 00:34:30.240
Okay, now the amount.
450
00:34:30.240 --> 00:34:37.240
So what I'm going to do is look at the amount, take a time interval from UI to UI plus one.
451
00:34:37.240 --> 00:34:45.240
This is a time interval delta U divided up into equal time intervals.
452
00:34:45.240 --> 00:35:00.240
So the amount dumped in the time interval from UI to UI plus one is equal to approximately F of UI, the dumping function there, times delta U.
453
00:35:00.240 --> 00:35:06.240
We calculated that before. That's what the meaning of the dumping rate.
454
00:35:06.240 --> 00:35:21.240
By time T, how much is it decayed to? It has decayed. How much is left in other words?
455
00:35:21.240 --> 00:35:35.240
Well, this is the starting amount. So the answer is going to be, it's F of UI times delta U times this factor which tells how much it decayed.
456
00:35:35.240 --> 00:35:44.240
Now, so time. So this is the starting amount at time UI. That's what it was first dumped.
457
00:35:44.240 --> 00:35:52.240
Now this is the amount that was dumped. Time multiplied that by E to the minus K times. Now what should I put up in there?
458
00:35:52.240 --> 00:36:00.240
I have to put the length of time that it had to decay. What is the length of time that it had to decay?
459
00:36:00.240 --> 00:36:18.240
It was dumped at UI. I'm looking at time T. It's decayed for the time length T minus UI.
460
00:36:18.240 --> 00:36:41.240
The length of time it had on the pile. So the stuff that was dumped in this time interval, when I, at time T, when I come to look at it, this is how much of it is left.
461
00:36:41.240 --> 00:36:53.240
And now all I have to do is add up that quantity for this time, the stuff that was dumped in this time interval plus the stuff dumped in, plus and so on, all the way up to the stuff that was dumped yesterday.
462
00:36:53.240 --> 00:37:11.240
And the answer will be there for the answer will be the total amount left at time T. That is not yet decayed.
463
00:37:11.240 --> 00:37:23.240
So it will be approximately, you add up the amount coming from the first time interval plus the amount coming in, so on. So it will be F of UI, I'll save the delta U for the end.
464
00:37:23.240 --> 00:37:42.240
And then E to the minus K times T minus UI times delta U. So these two parts represent the amount dumped, and this is the decay factor. And I add those up as all I runs from well where I start from 1 to N, let's say.
465
00:37:42.240 --> 00:38:00.240
And then the delta T goes to zero. In other words, make this more accurate by taking finer and finer sub-divide. In other words, instead of looking every month to see how much was dumped, let's look every week, every day, and so on, to make this calculation more accurate.
466
00:38:00.240 --> 00:38:22.240
So this approach is the exact amount, which will be the integral. This sum is a reman sum. It approaches the integral from zero to, well, I'm adding it up from U1 equals zero to UN equals T, the final value.
467
00:38:22.240 --> 00:38:42.240
So this will be the integral from the starting point to the ending point of F of UI to the minus K times T minus U, du. That's the answer to the problem. It's given by this rather funny looking integral, but from this point of view, it's entirely natural.
468
00:38:42.240 --> 00:39:01.240
So this combination of the dumping function, this doesn't care what the material was. It only wants to know how much was put on every day. And this part, which doesn't care how much was put on each day, it just is an intrinsic constant of the material involving its decay rate.
469
00:39:01.240 --> 00:39:18.240
So this total thing represents the total amount, and that is, what is it? It's the convolution of F of T with what function? E to the minus KT.
470
00:39:18.240 --> 00:39:28.240
It's the convolution of the dumping function, and the decay function. And the convolution is exactly the operation that you have to have to do that.
471
00:39:28.240 --> 00:39:45.240
So I think this is the most intuitive approach, physical approach to the meaning of this of the convolution. In this particular, you said, wow, that's very special. Okay, so it tells you what the meaning of a convolution with an exponential is, but what about the convolution with all the other functions?
472
00:39:45.240 --> 00:39:57.240
We're going to have to use in this course. They can all be interpreted just by being a little flexible in your approach. I'll give you two examples of this.
473
00:39:57.240 --> 00:40:17.240
Well, you'll three. First of all, I use it for the problem that I ask you about a bank account. That's not something any of you are interested in. Okay, so how about, suppose that's how I dump garbage?
474
00:40:17.240 --> 00:40:35.240
So, something that doesn't decay at all. What's the answer going to be? Well, the calculation will be exactly the same. It'll be the convolution of the dumping function. The only difference is that now the garbage isn't going to decay.
475
00:40:35.240 --> 00:40:51.240
So, no matter how long it's left, the same amount is going to be left at the end. In other words, I don't want the exponential decay function. I want the function one, the constant function one, because when I stick it on the pile, nothing happens to it. It just stays there.
476
00:40:51.240 --> 00:41:04.240
So it's going to be the convolution of this one, because this is constant. It's undecaying.
477
00:41:04.240 --> 00:41:29.240
By the identical reasoning. So, what's the answer going to be? It's going to be the integral from zero to T of f of u du. Now, that's an 801 problem. If I dump with a dumping rate f of u, and I dump from time zero to time T, how much do is on the pile? They don't give it. They always give velocity problems and problems of how to slice up bread, lows, and stuff like that.
478
00:41:29.240 --> 00:41:44.240
But this is a real life problem. If that's the dumping rate, you dump for T days from zero to time zero to time T, how much do you have left at the end? Answer the integral of f of u du from zero to T.
479
00:41:44.240 --> 00:41:59.240
I'll give you another example. Suppose I wanted a function, wanted to interpret something which grows like T, for instance. All I want is a physical interpretation.
480
00:41:59.240 --> 00:42:14.240
Well, I have to think of making a pile of something, metaphorical pile. We don't know what you have to make a physical pile. And the thing should be growing like T. Well, what grows like T? Not bacteria, they grow exponentially.
481
00:42:14.240 --> 00:42:42.240
I was before the lecture I was trying to think so. I came up with chickens. I'm sure on a chicken farm. Little baby chickens grow linearly. All little animals. Anyway, I have to observe that babies grow linearly, at least for a while. Thank God. After a while they taper off. But at the beginning, they eat every four hours or whatever. And they eat the same amount pretty much. And that adds up.
482
00:42:42.240 --> 00:42:56.240
So let's suppose this represents the growth, linear growth of chickens. Baby chicks. Baby chick, that makes me sound cuter.
483
00:42:56.240 --> 00:43:11.240
Let's say a farmer, a chicken farmer, is starting a new brood. Anyway, the hens lay at a certain rate. And each of those are incubated. And after a couple of months, they turn into a little baby chicks come out.
484
00:43:11.240 --> 00:43:34.240
So this will be the production rate for new chickens. Okay. And it will be the convolution, which will be tell you at time t, the number of kilograms. We better do this in kilograms, I'm afraid.
485
00:43:34.240 --> 00:44:01.240
Now, that's not as heartless as it seems. The number of kilograms of chickens. It's time t. It really isn't heartless. Because after all, why would the farmer want to know that? Well, because a certain number of pounds of chicken eat a certain number of pounds of chicken feed. And that's how much he has to dump us have to give them every day. That's how we calculate his expenses.
486
00:44:01.240 --> 00:44:16.240
He will have to know what the convolution is or better yet he will hire you who knows what the convolution is and you'll be able to tell him. Okay, why don't we stop there and go to recitation tomorrow because I'll be doing important things.