Video Transcript
Find the error bound when using the fourth Taylor Polynomial for the function 𝑓 of 𝑥 is equal to the sin of negative 𝜋𝑥 at 𝑥 is equal to one to approximate the value of 𝑓 evaluated at four over five. Give you answer in scientific form to three significant figures.
In this question, we’re trying to approximate 𝑓 evaluated at four over five, where 𝑓 of 𝑥 is the sin of negative 𝜋𝑥. We want to do this by using the fourth Taylor polynomial centered at 𝑥 is equal to one. The question wants us to find the bound on the error in this approximation. We need to give our answer in scientific form to three significant figures.
To do this. Let’s start by recalling how we would find a bound on the error for the 𝑛th Taylor polynomial approximation of a function. To start, recall the function 𝑓 of 𝑥 can be approximated by its 𝑛th Taylor polynomial 𝑇 𝑛 of 𝑥 plus the remainder term 𝑅 𝑛 of 𝑥. And we know if we can approximate a function 𝑓 of 𝑥 by its 𝑛th Taylor polynomial, we can actually find a bound for the remainder term.
We recall if we’re approximating a function 𝑓 at a value of 𝑥 by using its 𝑛th Taylor polynomial centered at 𝑎, then the absolute value of the remainder term, written 𝑅 𝑛, is less than or equal to the absolute value of 𝑀 times 𝑥 minus 𝑎 all raised to the power of 𝑛 plus one divided by 𝑛 plus one factorial, where our value of 𝑀 is an upper bound on the 𝑛 plus oneth derivative of our function 𝑓 of 𝑥 on an interval which contains both 𝑥 and 𝑎.
And it is worth pointing out, we can’t always do this. There are some assumptions being made. For example, we need to be able to differentiate our function 𝑓 𝑛 plus one times. In fact, there are more conditions to do with the continuity of the 𝑛 plus one derivatives of 𝑓 of 𝑥. However, we won’t go through these in this video since the question assumes that we can do this anyway.
The question wants us to find our error bound. And we can see this is a very complicated-looking expression. However, we can simplify this expression by finding the values of 𝑥, 𝑎, and 𝑛 given to us in the question. First, we’re approximating 𝑓 at four over five, so we’ll set our value of 𝑥 equal to four over five. And in fact, we can update our bounds to have 𝑥 set to be four over five.
Next, we can see the question wants us to use the fourth Taylor polynomial. If we’re using the fourth Taylor polynomial to approximate our function 𝑓 of 𝑥, then we should set our value of 𝑛 equal to four. Similarly, we’ll update the value of 𝑛 to be four. Finally, the question tells us we’re approximating our function by using a Taylor polynomial centered at 𝑥 is equal to one. And if our Taylor polynomial is centered at one, we’ll set our value of 𝑎 equal to one. And of course, we’ll update our value of 𝑎 set to be one.
Now, when we look at our error bound, we can see there’s only one unknown. We don’t know the value of 𝑀. We can see that 𝑀 is an upper bound on the fifth derivative of 𝑓 of 𝑥 on an interval containing four over five and one. So, to find our error bound, we need to find the value of 𝑛. First, let’s choose our interval containing both four over five and one. We’ll choose the closed interval from four over five to one. The reason we choose this is it’s the smallest interval containing both of these values. And the smaller the interval we choose, the better our error bound will be.
Next, we see to find the value of 𝑀, we’re going to need to find the fifth derivative of 𝑓 of 𝑥 with respect to 𝑥. To do this, we’re going to need to differentiate 𝑓 of 𝑥 five times. Let’s start by finding the first derivative of 𝑓 of 𝑥 with respect to 𝑥. First, we need to multiply it by the coefficient of 𝑥, which is negative 𝜋. Next, recall when we differentiate the sine function, we get the cosine function. This gives us 𝑓 prime of 𝑥 is equal to negative 𝜋 times the cos of negative 𝜋𝑥.
We’ll do the same to find the second derivative of 𝑓 of 𝑥 with respect to 𝑥. First, we multiply it by the coefficient of 𝑥, which is negative 𝜋. This gives us negative 𝜋 multiplied by negative 𝜋. Next, when we differentiate the cosine function, we need to remember we get negative the sine function. This gives us 𝑓 double prime of 𝑥 is equal to negative 𝜋 times negative 𝜋 multiplied by negative sin of negative 𝜋𝑥. And of course, we can simplify this to negative 𝜋 squared times the sin of negative 𝜋𝑥.
We can do exactly the same to find 𝑓 triple prime of 𝑥. Doing this and simplifying, we get 𝑓 triple prime of 𝑥 is equal to 𝜋 cubed times the cos of negative 𝜋𝑥. We’ll do exactly the same thing again to find the fourth derivative of 𝑓 of 𝑥 with respect to 𝑥. It’s equal to 𝜋 to the fourth power times the sin of negative 𝜋𝑥. And we need to do this one more time to find the fifth derivative. We get the fifth derivative of 𝑓 of 𝑥 with respect to 𝑥 is equal to negative one times 𝜋 to the fifth power multiplied by the cos of negative 𝜋𝑥.
To find our value of 𝑀, we need to maximize the absolute value of this expression. So, we’ll take the absolute value of both sides of our expression. We get the absolute value of the fifth derivative of 𝑓 of 𝑥 with respect to 𝑥 is equal to the absolute value of negative 𝜋 to the fifth power times the cos of negative 𝜋𝑥. And we can simplify this expression. The absolute value of negative 𝜋 to the fifth power is just 𝜋 to the fifth power. So, we’ll take this outside of our absolute value symbol as just 𝜋 to the fifth power.
Remember, to find the value of 𝑀, we need to maximize this expression on the closed interval from four over five to one. And of course, 𝜋 to the fifth power is a constant; it doesn’t vary as our value of 𝑥 varies. So, we need to maximize the absolute value of the cos of negative 𝜋𝑥. Of course, we know the cos of negative 𝜋𝑥 will vary from one to negative one. So, the absolute value of this will have a maximum value of one. So, the biggest this can be for any real value of 𝑥 is one.
Let’s see if this is also true on our interval. And we can in fact see this straight away. When 𝑥 is equal to one, we have the cos of negative 𝜋. And the cos of negative 𝜋 is negative one. Therefore, the maximum value of the absolute value of the fifth derivative of 𝑓 of 𝑥 on the closed interval from four over five to one is 𝜋 to the fifth power. And this will be the value of our constant 𝑀. And now that we found the value of the constant 𝑀, we can just find our error bound.
Substituting 𝑀 is equal to 𝜋 to the fifth power into our error bound, we get the absolute value of 𝑅 four will be less than or equal to the absolute value of 𝜋 to the fifth power times four over five minus one all raised to the power of four plus one all divided by four plus one factorial. Now, we need to evaluate our expression for the error bound. And if we calculate this expression and write out its decimal expansion, we get 0.00081605 and this continues.
But the question doesn’t want us to find this error bound exactly. It wants us to find it to three significant figures in scientific form. So, we’ll start by finding this to three significant figures. Remember, the initial bound of zeroes don’t count. We need to find the first three digits after this, which is eight, one, and six. Next, we need to look at the first omitted digit to see if we need to round up. Our first omitted digit is zero, so we don’t need to round up in this case.
Then, we need to approximate this in scientific form to three significant figures. Our three significant figures are eight, one, and six. And in scientific form, we need to write this as 8.16 times 10 to the 𝑚th power for some integer 𝑚. And to find the value of 𝑚, we see we need to multiply our error bound by 10 to the fourth power to get approximately 8.16. This means our value of 𝑚 is negative four since 8.16 times 10 to the negative four is approximately equal to our error bound. And this is our final answer. The error bound for our approximation is approximately equal to 8.16 times 10 to the power of negative four.
And it’s worth pointing out one thing. This is an incredibly small number. And remember this represents an error bound for our approximation. And when our error bound is small, that means our approximation is very accurate. And because our approximation is accurate, this means we don’t need to take more terms in our Taylor polynomial. The fourth Taylor polynomial was good enough for this approximation. Therefore, we were able to show the error bound when using the fourth Taylor polynomial for the function 𝑓 of 𝑥 is equal to the sin of negative 𝜋𝑥 at 𝑥 is equal to one to approximate the value of 𝑓 of four over five in scientific form to three significant figures is 8.16 times 10 to the power of negative four.
