The more terms we have in a Taylor polynomial approximation of a function, the closer we get to the function. But HOW close? In this video, we prove the Lagrange error bound for Taylor polynomials. Created by Sal Khan.
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- I dont understand how Sal goes from the integral of E(n+1)(x) dx. to just E(n)(x), around the 10 minute mark.(20 votes)
- The function is not E^(n+1). The (n+1) is actually how many derivatives we've taken. So when we take the antiderivative, we "lose" one derivative and go to (n+1). It's not a power, it's a derivative :). Hope this helps!(41 votes)
- Can anyone explain to me why at10:00, where -ma<=c,why we have to take the lower bound of c, making c=-ma? I am really confused at that part.(15 votes)
- This part actually has an error, which can be fixed by using a definite integral between x and a (or a and x, depending on whether x is less than a), rather than an indefinite integral. I have sent a message to Mr. Khan about this error, so hopefully this gets corrected soon. Anyway, a definite integral with these bounds gets you from step to step quite smoothly :) Hope that helps!(18 votes)
- Sal just assumed we know M in this video, but how would one actually find it?(16 votes)
- Since the function is continuous in the interval of [a, b], you know there will be a maximum value. It can be f(b), f(a), or f(c) where f`(c) = 0 (extreme value).(12 votes)
- In the video when you integrate both sides isn't the value of n supposed to increase by one therefore making it n+1 instead of n-1?(2 votes)
- This confused me at first as well. But like Sebastian pointed out, it's the derivative degree that is decreasing, not an exponent. There's no power rule or anything going on here.(10 votes)
- Are there any videos that instruct on situations where M is given?(8 votes)
- There are subsequent examples that demonstrate how to find M for specific functions:
- Why are we free to choose the integration constant valur in the limit bounds of the error function?? can we always choose any value of c while finding a function(7 votes)
- we aren't "free" to choose c, it must be within certain constraints that arise when trying to calculate it. we choose the smallest value we can so that the potential difference between the rhs and lhs is as small as possible.(3 votes)
- Is it possible to make a better bound?(5 votes)
- In a general sense, for any given n, there is no better bound. You can prove this to yourself by constructing examples where E(x) is exactly equal to the bound shown in the video. Here is one such example. Let's say that f(x) = x + x^2 / 2 and that one takes a Taylor polynomial approximation with degree 1 ( n = 1 ) at zero ( a = 0). Then, the polynomial approximation is P(x) = x; the error function is E(x) = x^2 / 2; E''(x) = 1 and, thus, M = 1; and M (x-a)^(n+1) / (n+1)! = 1 * x^2 / 2! = E(x).
Of course, it is true that you can get a better approximation by increasing n. In the above example, you get a perfect approximation ( E(x) = 0 ) by increasing n from 1 to 2.(5 votes)
- When taking the antiderivative of the remainder at8:10why does that not have a constant added to it?
Shouldn't anti[E^(n+1)(x)] be equal to E^(n)(x) + C ?(2 votes)
- Thanks I get it now. Since its an unknown constant. Its aggregated to a single C on the right side of the equal sign.(4 votes)
- so how do you determine the M in the final formula?(4 votes)
- You need to figure it out yourself. Let's say you have the function
e^xand you have a Maclaurin polynomial with degree 3. The polynomial is
1 + 1/2 * x^2 + 1/6 * x^3 + 1/24 * x^4. Let's also say that you want to approximate it at
0.1. The biggest the function can get in the interval
[0, 0.1]is obviously
e^xis an increasing function. This means that the
Min this case will be
- How do you find M for the error function?(4 votes)
- For sin(x) , M is 1 because no matter how many times you take derivatives it always lies between -1 and 1. However, for unbounded functions like e^x, you have to find the greater value between a and b, and M is e^a or e^b. Otherwise it's the same.(2 votes)
In the last video, we started to explore the notion of an error function. Not to be confused with the expected value because it really does reflect the same notation. But here E is for error. And we could also thought it will some times here referred to as Reminder function. And we saw it's really just the difference as we, the difference between the function and our approximation of the function. So for example, this, this distance right over here, that is our error. That is our error at the x is equal to b. And what we really care about is the absolute value of it. Because at some points f of x might be larger than the polynomial. Sometimes the polynomial might be larger than f of x. What we care is the absolute distance between them. And so what I want to do in this video is try to bound, try to bound our error at some b. Try to bound our error. So say it's less than or equal to some constant value. Try to bound it at b for some b is greater than a. We're just going to assume that b is greater than a. And we saw some tantalizing, we, we got to a bit of a tantalizing result that seems like we might be able to bound it in the last video. We saw that the n plus 1th derivative of our error function is equal to the n plus 1th derivative of our function. Or their absolute values would also be equal to. So if we could somehow bound the n plus 1th derivative of our function over some interval, an interval that matters to us. An interval that maybe has b in it. Then, we can, at least bound the n plus 1th derivative our error function. And then, maybe we can do a little bit of integration to bound the error itself at some value b. So, let's see if we can do that. Well, let's just assume, let's just assume that we're in a reality where we do know something about the n plus1 derivative of f of x. Let's say we do know that this. We do it in a color I that haven't used yet. Well, I'll do it in white. So let's say that that thing over there looks something like that. So that is f the n plus 1th derivative. The n plus 1th derivative. And I only care about it over this interval right over here. Who cares what it does later, I just gotta bound it over the interval cuz at the end of the day I just wanna balance b right over there. So let's say that the absolute value of this. Let's say that we know. Let me write it over here, let's say that we know. We know that the absolute value of the n plus 1th derivative, the n plus 1th. And, I apologize I actually switch between the capital N and the lower-case n and I did that in the last video. I shouldn't have, but now that you know that I did that hopefully it doesn't confuse it. N plus 1th, so let's say we know that the n plus 1th derivative of f of x, the absolute value of it, let's say it's bounded. Let's say it's less than or equal to some m over the interval, cuz we only care about the interval. It might not be bounded in general, but all we care is it takes some maximum value over this interval. So over, over, over the interval x, I could write it this way, over the interval x is a member between a and b, so this includes both of them. It's a closed interval, x could be a, x could be b, or x could be anything in between. And we can say this generally that, that this derivative will have some maximum value. So this is its, the absolute value, maximum value, max value, m for max. We know that it will have a maximum value, if this thing is continuous. So once again we're going to assume that it is continuous, that it has some maximum value over this interval right over here. Well this thing, this thing right over here, we know is the same thing as the n plus 1 derivative of the error function. So then we know, so then that, that implies, that implies that, that implies that the, that's a new color, let me do that in blue, or that green. That implies that the, the, the end plus one derivative of the error function. The absolute value of it because these are the same thing is also, is also bounded by m. So that's a little bit of an interesting result but it gets us no where near there. It might look similar but this is the n plus 1 derivative of the error function. And, and we'll have to think about how we can get an m in the future. We're assuming that we some how know it and maybe we'll do some example problems where we figure that out. But this is the m plus 1th derivative. We bounded it's absolute value but we really want to bound the actual error function. The 0 is the derivative, you could say, the actual function itself. What we could try to integrate both sides of this and see if we can eventually get to e, to get to e of x. To get our, to our error function or our remainder function so let's do that. Let's take the integral, let's take the integral of both sides of this. Now the integral on this left hand side, it's a little interesting. We take the integral of the absolute value. It would be easier if we were taking the absolute value of the integral. And lucky for us, the way it's set up. So let me just write a little aside here. We know generally that if I take, and it's something for you to think about. If I take, so if I have two options, if I have two options, this option versus and I don't know, they look the same right now. I know they look the same right now. So over here, I'm gonna have the integral of the absolute value and over here I'm going to have the absolute value of the interval. Which of these is going to be, which of these can be larger? Well, you just have to think about the scenarios. So, if f of x is always positive over the interval that you're taking the integration, then they're going to be the same thing. They're, you're gonna get positive values. Take the absolute of a value of a positive value. It doesn't make a difference. What matters is if f of x is negative. If f of x, if f of x is negative the entire time, so if this our x-axis, that is our y-axis. If f of x is, well we saw if it's positive the entire time, you're taking the absolute value of a positive, absolute value of positive. It's not going to matter. These two things are going to be equal. If f of x is negative the whole time, then you're going to get, then this integral going to evaluate to a negative value. But then, you would take the absolute value of it. And then over here, you're just going to, this is, the integral going to value to a positive value and it's still going to be the same thing. The interesting case is when f of x is both positive and negative, so you can imagine a situation like this. If f of x looks something like that, then this right over here, the integral, you'd have positive. This would be positive and then this would be negative right over here. And so they would cancel each other out. So this would be a smaller value than if you took the integral of the absolute value. So the integral, the absolute value of f would look something like this. So all of the areas are going to be, if you view the integral, if you view this it is definitely going to be a definite integral. All of the areas, all of the areas would be positive. So when you it, you are going to get a bigger value when you take the integral of an absolute value. Then you will, especially when f of x goes both positive and negative over the interval. Then you would if you took the integral first and then the absolute value. Cuz once again, if you took the integral first, for something like this, you'd get a low value cause this stuff would cancel out. Would cancel out with this stuff right over here then you'd take the absolute value of just a lower, a lower magnitude number. And so in general, the integral, the integral, sorry the absolute value of the integral is going to be less than or equal to the integral of the absolute value. So we can say, so this right here is the integral of the absolute value which is going to be greater than or equal. What we have written over here is just this. That's going to be greater than or equal to, and I think you'll see why I'm why I'm doing this in a second. Greater than or equal to the absolute value, the absolute value of the integral of, of the n plus 1th derivative. The n plus 1th derivative of, x, dx. And the reason why this is useful, is that we can still keep the inequality that, this is less than, or equal to this. But now, this is a pretty straight forward integral to evaluate. The indo, the anti-derivative of the n plus 1th derivative, is going to be the nth derivative. So this business, right over here. Is just going to the absolute value of the nth derivative. The absolute value of the nth derivative of our error function. Did I say expected value? I shouldn't. See, it even confuses me. This is the error function. I should've used r, r for remainder. But this all error. The, noth, nothing about probability or expected value in this video. This is. E for error. So anyway, this is going to be the nth derivative of our error function, which is going to be less than or equal to this. Which is less than or equal to the anti-derivative of M. Well, that's a constant. So that's going to be mx, mx. And since we're just taking indefinite integrals. We can't forget the idea that we have a constant over here. And in general, when you're trying to create an upper bound you want as low of an upper bound as possible. So we wanna minimize, we wanna minimize what this constant is. And lucky for us, we do have, we do know what this, what this function, what value this function takes on at a point. We know that the nth derivative of our error function at a is equal to 0. I think we wrote it over here. The nth derivative at a is equal to 0. And that's because the nth derivative of the function and the approximation at a are going to be the same exact thing. And so, if we evaluate both sides of this at a, I'll do that over here on the side, we know that the absolute value. We know the absolute value of the nth derivative at a, we know that this thing is going to be equal to the absolute value of 0. Which is 0. Which needs to be less than or equal to when you evaluate this thing at a, which is less than or equal to m a plus c. And so you can, if you look at this part of the inequality, you subtract m a from both sides. You get negative m a is less than or equal to c. So our constant here, based on that little condition that we were able to get in the last video. Our constant is going to be greater than or equal to negative ma. So if we want to minimize the constant, if we wanna get this as low of a bound as possible, we would wanna pick c is equal to negative Ma. That is the lowest possible c that will meet these constraints that we know are true. So, we will actually pick c to be negative Ma. And then we can rewrite this whole thing as the absolute value of the nth derivative of the error function. The nth derivative of the error function. Not the expected value. I have a strange suspicion I might have said expected value. But, this is the error function. The nth der. The absolute value of the nth derivative of the error function is less than or equal to M times x minus a. And once again all of the constraints hold. This is for, this is for x as part of the interval. The closed interval between, the closed interval between a and b. But looks like we're making progress. We at least went from the m plus 1 derivative to the n derivative. Lets see if we can keep going. So same general idea. This if we know this then we know that we can take the integral of both sides of this. So we can take the integral of both sides of this the anti derivative of both sides. And we know from what we figured out up here that something's that's even smaller than this right over here. Is, is the absolute value of the integral of the expected value. Now [LAUGH] see, I said it. Of the error function, not the expected value. Of the error function. The nth derivative of the error function of x. The nth derivative of the error function of x dx. So we know that this is less than or equal to based on the exact same logic there. And this is useful because this is just going to be, this is just going to be the nth minus 1 derivative of our error function of x. And of course we have the absolute value outside of it. And now this is going to be less than or equal to. It's less than or equal to this, which is less than or equal to this, which is less than or equal to this right over here. The anti-derivative of this right over here is going to be M times x minus a squared over 2. You could do U substitution if you want or you could just say hey look. I have a little expression here, it's derivative is 1. So it's implicitly there so I can just treat it as kind of a U. So raise it to an exponent and then divide that exponent. But once again I'm taking indefinite integrals. So I'm going to say a plus C over here. But let's use that same exact logic. If we evaluate this at A, you're going to have it. If you evaluate this while, let's evaluate both sides of this at A. the left side, evaluated at A, we know, is going to be zero. We figured that out, all, up here in the last video. So you get, I'm gonna do it on the right over here. You get zero, when you valued the left side of a. The right side of a, if you, the right side of the value of a you get m times a menus a square over 2. So you are gonna get 0 plus c, so you are gonna get, 0 is less or equal to c. Once again we want to minimize our constant, we wanna minimize our upper boundary up here. So we wanna pick the lowest possible c that we talk constrains. So the lowest possible c that meets our constraint is zero. And so the general idea here is that we can keep doing this, we can keep doing exactly what we're doing all the way, all the way, all the way until. And so we keep integrating it at the exact same, same way that I've done it all the way that we get and using this exact same property here. All the way until we get, the bound on the error function of x. So you could view this as the 0th derivative. You know, we're going all the way to the 0th derivative, which is really just the error function. The bound on the error function of x is going to be less than or equal to, and what's it going to be? And you can already see the pattern here. Is that it's going to be m times x, minus a. And the exponent, the one way to think about it, this exponent plus this derivative is going to be equal to n plus 1. Now this derivative is zero so this exponent is going to be n plus 1. And whatever the exponent is, you're going to have,a nd maybe I should have done it, you're going to have n plus one factorial over here. And if say wait why, where does this n plus 1 factorial come from? I just had a two here. Well think about what happens when we integrate this again. You're going to raise this to the third power and then divide by three. So your denominator is going to have two times three. Then when you integrate it again, you're going to raise it to the fourth power and then divide by four. So then your denominator is going to be two times three times four. Four factorial. So whatever power you're raising to, the denominator is going to be that power factorial. But what's really interesting now is if we are able to figure out that maximum value of our function. If we're able to figure out that maximum value of our function right there. We now have a way of bounding our error function over that interval, over that interval between a and b. So for example, the error function at b. We can now bound it if we know what an m is. We can say the error function at b is going to be less than or equal to m times b minus a to the n plus 1th power over n plus 1 factorial. So that gets us a really powerful, I guess you could call it, result, kinda the, the math behind it. And now we can show some examples where this could actually be applied.