2011 Calculus BC free response #6a Taylor Series approximation of sin(x)
2011 Calculus BC free response #6a
- Problem number six
- Let f of x is equal to the sine of x-squared plus cosine of x
- The graph of y is equal to the absolute value of the fifth derivative of f at x is shown above
- and I haven't shown it here just so we have some space. I'll show it when we need to show it
- I think we have to show it in part D.
- So first let's do part A right over here.
- Write the first four nonzero terms of the Taylor series for sin of x
- about x=0, and write the first four nonzero terms of the Taylor series for the sin of x-squared
- about x equals zero. So let's do this first part.
- And just as a reminder, a Taylor series is a polynomial approximation
- of a function. So just to give us a quick reminder--we go into much more detail
- in this on the videos about Taylor series.
- But if you have a function that looks like this, and you wanted to approximate it with a Taylor series
- around zero, you could, if you only have one term in your Taylor series, you would just literally
- it would just be a constant, just like that. If you have two terms in your Taylor series
- it would be a line that looks like that.
- If you have three terms in your Taylor series, you'd get to a second-degree term.
- You'd start approximating it something like that.
- If you get to the third degree term, it might start looking like that
- And as you add more and more terms, you get a better and better approximation of your function
- and if you add an infinite number, it might actually converge to your function.
- So let's just remind ourselves. If I had a function, if I have f of x
- its Taylor series: I can approximate it with a Taylor series.
- And if we want to center that approximation around zero,
- it'll be equal to f of zero plus f-prime of zero (the derivative at zero) times x
- plus the second derivative at zero times x-squared over 2 factorial.
- (You could have divided this term right here by 1 factorial, which is just 1)
- 2 factorial is just 2.
- And then plus the third derivative of f at zero times x to the third over 3 factorial
- plus the fourth derivative . . . I think you get the idea here.
- the fourth derivative at zero times x to the fourth divided by four factorial
- and so on and so forth.
- So what they want us to do is to find the first four non-zero terms of the
- Taylor series for sine of x
- And some of you might already know this. We actually covered this
- in the video where we show Euler's identity, but we'll do it again right over here.
- So if we just take, I'll call it g(x), because they already
- defined f(x) up here. So let's say that g(x)
- And let's put this in a new color, just to ease the monotony
- Let's say that g of x is equal to sin of x. Then we know that g of zero
- is going to be equal to zero.
- And then if we take the derivative, g-prime of x is going
- to be equal to negative, no it's going to be positive cosine of x
- Positive cosine of x. g-prime of zero is now going to be equal to 1
- cosine of zero is one. Then if you take the second derivative.
- The derivative of cosine of x is negative sine of x
- And the second derivative at zero, once again, is going
- to be equal to zero. Sine of zero is zero.
- And now let's take the third derivative
- The third derivative of our function, g, the derivative of
- negative sine of x is negative cosine of x.
- And the third derivative at zero is now going to be
- equal to negative one. And we could keep going.
- You can already guess where this might lead to.
- But I'll just keep going, just in case.
- The fourth derivative is once again going to be equal to sine of x
- is now going to be equal to sine of x. And then the
- fourth derivative at zero, because it's the same thing as the function
- is now going to be zero as well.
- So this is going to be equal to g the fourth at zero.
- And then we could keep going. This is going to be the same thing
- as g to the fifth because we start cycling
- as we take more and more derivatives. So this is going to be
- the same thing as the fifth derivative at zero
- This is going to be the same thing as the sixth derivative at zero
- And this is going to be the same--this is an equals sign right here
- equals one. And this is going to be the same thing as the seventh
- derivative at zero. So if you want the first four
- non-zero terms. Let's just work through it.
- So first of all, f of zero--I'll do this in a new color
- if we want to approximate g of x
- if we want to approximate sine of x
- so we'll say sine of x is going to be approximately equal to
- so this first term right over here
- that is sine of 0, that's g of zero
- that's going to be zero, so we don't even have to write it
- then we go to this term right over here. Well the first derivative
- f prime at zero, or in this case g prime of zero
- is going to be equal to one
- So it's going to be one times x. So you're going to have an x there
- And the next term is going to be zero, we see that over there
- because the second derivative of our function evaluated at zero is zero
- our third derivative of our function evaluated at zero is 1
- so that term's going to show up again
- Actually, that term is negative one--I don't want to make a mistake here
- that is negative one. It was negative cosine of x
- if you evaluate negative cosine of x at zero you get negative one
- right over here.
- So the third derivative, this thing right over here, is negative one
- So then we have minus x to the third over 3 factorial
- The fourth derivative, once again, is zero
- The fifth derivative evaluated at zero is one. So then
- you're going to have plus 1 times x to the fifth over five factorial
- Then the sixth derivative is zero, so that term is going to disappear
- I didn't even write it up here.
- And then the seventh term, the coefficient is negative one
- Or, the seventh derivative evaluated at zero is negative one
- So you have negative one times x to the seventh over seven factorial
- And we had to go all the way to the seventh degree term to find
- the first four non-zero terms for our Taylor series
- And so, we're done with at least the first part.
- We found the first four non-zero terms of sine of x.
- So now what about sine squared of x
- Or sine of x-squared. We have to be careful here. Because
- you might just say, okay, let me just apply this formula
- And you're going to find very quickly
- when you start taking the second and third derivatives
- of this thing right over here, it's going to get really messy
- but what you can say is look, sine of x is approximately this thing over here
- What happens if I just replace the x with x-squared?
- Then I get sine of x-squared is approximately equal to
- instead of an x here, I'm going to put the x-squared
- instead of an x to the third, I'm going to put x-squared to the third over here, over three factorial
- Instead of an x to the fifth, I can put an x-squared to the fifth power
- over five factorial
- and instead of an x to the seventh, I can put an x-squared to the seventh
- power over seven factorial
- So this is a very important thing to realize.
- Because, if you had started to directly take the Taylor series around zero
- of this thing right here, you'd have taken up all your time
- to take the derivatives, and you probably wouldn't have been able
- to do it anyways, because it would have gotten really messy.
- And the key here is just to realize that if you just substitute
- x-squared for x, you're then going to get your approximation for
- sine of x-squared.
- And we can simplify this a little bit
- This is going to be approximately equal to
- x-squared minus x to the sixth over three factorial
- plus x to the tenth over five factorial
- minus x to the fourteenth over seven factorial
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