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# 2011 Calculus BC free response #6d

## Video transcript

Part D. Let p sub 4 of x be the fourth degree Taylor polynomial for f about x equals 0. Using information from the graph of y is equal to the absolute value of the fifth derivative of f of x, shown above-- and that's the graph right over here, I put it on the side to save space-- show that the absolute value of the difference between the polynomial-- the fourth degree polynomial evaluated at 1/4 and the function evaluated at 1/4 is less than one over 3,000. So this is, it seems like quite an interesting problem here. So first, let's just think about what they're asking. They're asking for the difference between the polynomial and the function evaluated 1/4. In other words, they're asking us to-- they're asking us to bound the error of the function approximation at x is equal to 1/4. So let me draw a little graph here just so we can visualize it a little better. So if that's our y-axis and-- let me draw it like this. So this is our y-axis and this is our x-axis. And our function, let's say our function looks something like this. Our polynomial is centered at 0. Our fourth degree polynomial is centered at 0. And so it'll be equal to the function at 0. And then it'll become probably a worse approximation as we get further and further away from 0. And what they're saying is, let's bound how bad of an approximation that is when x is equal to 1/4. So if we say that this is x is equal to 1/4 right over here, they're saying, look, take the value of the polynomial at 1/4, subtract from that the value of the function at 1/4, and then take the absolute value and then that'll essentially give us the absolute value of the error. Or that will give us the absolute value of the remainder or give us the distance between the two functions there. And they want us to bound that distance. Now, to do this problem there is something that you just kind of have to know. And I've proven it in another series of videos-- two videos I think it took me-- in the Khan Academy Calculus playlist. If you do a search for bounded or error or approximation or Taylor, you should probably find it. And that there's a general idea here which we've proved that if you have some function-- so I'm going to show the general-- so if you have some function f of x and that you have some polynomial approximation of it of degree n--- some polynomial approximation of degree n. And let's say that this polynomial approximation is centered at a. In our particular case, a is 0. So that you're centered at a. So we could write-- we could say this is about a. It's an approximation about x equals a. We can bound the error. So let me first define this error function here. Sometimes it's called a remainder. So we could say, the error function for the nth degree polynomial about a-- the n-th degree Taylor polynomial approximation about a-- it will be a function of x. So for any other x that you pick, what will this error be? Is equal to the difference between these two things. You could say it's f of x minus that or that minus f of x. But let me just write it this way. This is equal to f of x minus the polynomial approximation of x. And if you want to take the absolute value of this, that's equal to absolute value of that. And if you're taking absolute values, then you can switch the order. So this is the same thing as the absolute value of the polynomial approximation at x minus f of x. So the thing that you just need to know is, we can bound this. We can bound this if we know some properties about the n plus 1. So this is the nth degree approximation. If we know some properties about the n plus 1th derivative of f. So if we know that the n plus 1th is the derivative of f is less than or equal to some maximum value-- and in particular, the absolute value of the n plus 1th derivative of f-- is less than or equal to some-- we could call this some maximum value-- over the interval, over x is in the interval between a and some b, where b is greater than a. Then we can make the statement-- and this is, once again, something that I prove in that video. I think I call it the boundedness of the error remainder for Taylor approximation. Something like that. Then we know that the error function at any given x where x is greater than or-- let me say this in particular. Is any given x that's part of this interval-- where x is part of this interval between a and b-- is going to be less than or equal to M times x minus a to the n plus 1th power over n plus 1. So this is the thing that you really have to know ahead of time to be able to do this problem. It would be nearly impossible in the time constraints of the AP exam to prove this from first principles. I encourage you to watch that video so you can understand how this proof happens from first principles. But going into the AP exam, based on the facts that this did show up in the 2011 exam, it's probably good to know this property, to know how to bound it. And this is clearly what they're asking for because they're giving us-- so we know what the fourth-- we know what p sub four is. p sub 4 of x-- it's up to the fourth degree term of the polynomial approximation of f about 0. In part B, we got all the way the sixth degree term. So if you want to go just to the fourth degree term, this is what p sub 4 is. So this is 1 plus x squared over 2 plus x to the fourth over 4. So we know that. And we want to bound this thing right over here. So if we can find an M-- so this is the fourth degree polynomial. This is the fourth degree error that we're talking about in this case. Let me write it this way. This is the nth error in the general case. But we want to bound the fourth degree error. So if we want bound the fourth degree error-- so we can say the fourth degree error-- at b. In this case, we're going to say-- or at 1/4 I should say, not b. I never mentioned a b before. I mentioned the b when I prove it. The fourth degree error at 1/4 is going to be less than or equal to-- I did mention a b here so if you take this to be 1/4, and 1/4 is part of that interval so you can apply it here. So the fourth degree error, I guess you could say, at 1/4 is going to be less than or equal to some M times b-- I should say, 1/4 minus a. a in our particular case for this problem is a 0. So 1/4 minus is just 0 to the n plus 1th power. So this is 4-- this is the fourth degree polynomial. n is 4 so the n plus 1 power is the fifth power. All of that over-- and sorry, this was supposed to be n plus 1-- all of that over n plus 1 factorial. So this is going to be over 5 factorial. Don't want to confuse you. This was-- all of that is over n plus 1 factorial. So all of this is over 5 factorial. So we can make this statement if we know that the fifth derivative-- the n plus 1th derivative-- if we know that the fifth derivative of f is less than or equal to some value M-- the absolute value of the fifth derivative of f, I should say--- is less than or equal to some value at M over the interval x is between 0 and 1/4. Those are our a's and b's in this particular case. So when you look at the graph-- and that's exactly what they gave us. They gave us the graph of the absolute value of the fifth derivative of x. And 1/4 is sitting right over here. So this is 1/4. We don't know exactly what we hit over here. This looks like it's roughly, I don't know, 31 or 32. But we know it's definitely less than 40 over the interval between 0 and 1/4. So we can pick M to be 40. We know that this is less than or equal to 40 over that interval. So we can make this statement. And so we know that this thing right over here is going to be less because this over here is the same as this over here. We know this is less than-- let me write it all out. This is-- and I know of people who got fives on this example without being able to do this. So don't stress out if this seems really bizarre too. This is really so they can separate out who happened to know this, really. So this we can say is based on this property, we can say is less than or equal to 40 times 1/4 to the fifth power. So this is 40 times 1/4 to the fifth power. So I could just say divided by 4 to the fifth power times-- and then the denominator over here-- we have 5 factorial. I'll just write it out. 5 times 4 times 3 times 2. And we don't have to write the 1. That doesn't change the value. And now we can simplify this. We can divide the numerator and the denominator by 4. So this becomes 10, that just becomes a 1. We can divide it by 5. This becomes a 2, that becomes a 1. We can divide by 1. That becomes a-- oh, sorry, we can divide by 2. That becomes a 1, that becomes a 1. And you say that this quantity over here is going to be equal to 1. So let me write it. So all of this stuff is less than or equal to 1 over 4 to the fifth times 3. That's all we have left in the denominator. And so what is 4 to the fifth? You may or may not know. You can work it out by hand if you like, but 4 to the fifth is the same thing as 2 to the 10th power because 4 is 2 squared. 2 to the 10th power, if you're familiar with computer science, is 1,024. But you can work it out yourself. So this is going to be-- this is less than or equal to 1,024 times 3. So what is that? That is less than or equal to 1 over 3. Oh, what is it? 72? 3,000. 1 over 3,072. So this quantity right over here is less than 1 over 3,072. We know. We know that for a fact. Well, if something is less than 1 over 3,072, it's definitely going to be less than 1 over 3,000. 1 over 3,000 is a larger number. It has a smaller denominator. That's definitely less than or equal to 1 over 3,000. And then we are done. We've proven that this is less than 1 over 3,000. We've proven that the error at 1/4 is less than 1 over 3,000.
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