Video: Estimating a Function Using the Taylor Polynomial of 𝑛th Degree

Estimate the function 𝑓(π‘₯) = 2 sin (3π‘₯) with a third-degree Taylor polynomial at π‘₯ = πœ‹.

05:20

Video Transcript

Estimate the function 𝑓 of π‘₯ is equal to two times the sin of three π‘₯ with a third-degree Taylor polynomial at π‘₯ is equal to πœ‹.

The question is asking us to find a Taylor polynomial approximation of third degree at π‘₯ is equal to πœ‹ of the function 𝑓 of π‘₯ is equal to two times the sin of three π‘₯. We recall an 𝑁th-degree Taylor polynomial approximation of the function 𝑔 of π‘₯ at π‘₯ is equal to π‘Ž is given by 𝑇 𝑁 of π‘₯ is equal to 𝑔 evaluated at π‘Ž. Plus the first derivative of 𝑔 evaluated at π‘Ž divided by one factorial multiplied by π‘₯ minus π‘Ž. Plus the second derivative of 𝑔 evaluate at π‘Ž divided by two factorial multiplied by π‘₯ minus π‘Ž squared.

And we keep adding terms of this form all the way up to the 𝑁th derivative of 𝑔 evaluated at π‘Ž divided by 𝑁 factorial multiplied by π‘₯ minus π‘Ž to the power of 𝑁. We call 𝑁 the degree of our Taylor polynomial because it is also the highest power of π‘₯ which appears in the Taylor polynomial. We call π‘Ž the center of our Taylor polynomial.

We’re asked to find the third-degree Taylor polynomial of our function. This means we’ll need to find the third derivative of our function 𝑓 of π‘₯. So, we’ll set 𝑁 equal to three. And since we want to find our polynomial at π‘₯ is equal to πœ‹, we’ll set our center π‘Ž equal to πœ‹. To find the third-degree Taylor polynomial approximation of our function, we need to find the first, second, and third derivative of our function 𝑓 of π‘₯.

Since we need to evaluate this at π‘₯ is equal to our center πœ‹, we’ll start by finding the first derivative of 𝑓 of π‘₯. To differentiate 𝑓 of π‘₯, which is two times the sin of three π‘₯, we recall, for constants π‘Ž and 𝑏, the derivative of π‘Ž sin of 𝑏π‘₯ with respect to π‘₯ is equal to π‘Žπ‘ times the cos of 𝑏π‘₯. Applying this, we get 𝑓 prime of π‘₯ is equal to two times three times the cos of three π‘₯. And we can simplify two times three to just be six.

Now, to find the second derivative of our function 𝑓 of π‘₯, we need to differentiate the first derivative of our function 𝑓 of π‘₯. That’s the derivative of six times the cos of three π‘₯ with respect to π‘₯. And to differentiate this, we recall, for constants π‘Ž and 𝑏, the derivative of π‘Ž cos of 𝑏π‘₯ with respect to π‘₯ is equal to negative π‘Žπ‘ times the sin of 𝑏π‘₯. Applying this to our function six times the cos of three π‘₯ gives us our second derivative of 𝑓 of π‘₯ is equal to negative six times three times the sin of three π‘₯. And we can simplify negative six times three to be negative 18.

To find our third derivative of 𝑓 of π‘₯, we need to differentiate our second derivative of 𝑓 of π‘₯. That’s the derivative of negative 18 times the sin of three π‘₯ with respect to π‘₯. Using our first rule for differentiating trigonometric functions, we have that this is equal to negative 18 times three times the cos of three π‘₯. And we have negative 18 times three can be simplified to negative 54.

We see, in our Taylor polynomial expansion, we want to evaluate each of our derivative functions at the value of the center π‘Ž. In our case, the center of our Taylor approximation is πœ‹. So, let’s evaluate 𝑓, 𝑓 prime, 𝑓 double prime, and 𝑓 triple prime at π‘₯ is equal to πœ‹. We have that 𝑓 evaluated at πœ‹ is equal to two times the sin of three πœ‹. However, the sin of three πœ‹ is just equal to zero, so 𝑓 evaluated at πœ‹ is equal to zero.

Next, we have 𝑓 prime evaluated at π‘₯ is equal to πœ‹ is equal to six times the cos of three πœ‹. The cos of three πœ‹ is equal to negative one. So, this evaluates to give us negative six. Our second derivative function evaluated at π‘₯ is equal to πœ‹ gives us negative 18 multiplied by the sin of three πœ‹. We know the sin of three πœ‹ is zero, so this evaluates to give us zero.

Finally, our third derivative of 𝑓 of π‘₯ evaluated at π‘₯ is equal to πœ‹ gives us negative 54 times the cos of three πœ‹. And the cos of three πœ‹ is negative one. So, this evaluates to give us 54. We’re now ready to find our third-degree Taylor approximation of two times the sin of three π‘₯ at π‘₯ is equal to πœ‹. It’s equal to 𝑓 of πœ‹. Plus 𝑓 prime of πœ‹ over one factorial multiplied by π‘₯ minus πœ‹. Plus 𝑓 double prime of πœ‹ over two factorial times π‘₯ minus πœ‹ squared. Plus 𝑓 triple prime of πœ‹ over three factorial times π‘₯ minus πœ‹ cubed.

We already found the values of 𝑓 evaluated at πœ‹, 𝑓 prime evaluated at πœ‹, 𝑓 double prime evaluated at πœ‹, and 𝑓 triple prime evaluated at πœ‹. Substituting these values into our Taylor approximation, we see that the first term and the third term have a factor of zero. So, they’re equal to zero. So, this simplifies to give negative six over one factorial multiplied by π‘₯ minus πœ‹ plus 54 over three factorial multiplied by π‘₯ minus πœ‹ cubed.

One factorial is just equal to one. So, negative six divided by one factorial is negative six. And three factorial is equal to six. So, 54 divided by three factorial is equal to nine. Therefore, we’ve shown that the first-degree [third-degree] Taylor polynomial approximation at π‘₯ is equal to πœ‹ of the function 𝑓 of π‘₯ is equal to two times the sin of three π‘₯ is equal to negative six times π‘₯ minus πœ‹ plus nine times π‘₯ minus πœ‹ cubed.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.