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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
So that's our second part of the problem.