Video: Finding the Taylor Series Representation of a Polynomial

Consider the function 𝑓(π‘₯) = (2 + π‘₯)ΒΉΒ². Find the Taylor series representation of the function 𝑓(π‘₯) = (2 + π‘₯)ΒΉΒ² at π‘₯ = βˆ’1. Are the terms of the Taylor series representation for the function 𝑓 finite or infinite?

10:13

Video Transcript

Consider the function 𝑓 of π‘₯ is equal to two plus π‘₯ all raised to the 12th power. Find the Taylor series representation of the function 𝑓 of π‘₯ is equal to two plus π‘₯ all raised to the 12th power at π‘₯ is equal to negative one. Are the terms of the Taylor series representation for the function 𝑓 finite or infinite?

We’re given a function 𝑓 of π‘₯. The first thing we need to do is determine the Taylor series representation of this function at π‘₯ is equal to negative one. So let’s start by recalling what the Taylor series representation of a function 𝑓 of π‘₯ at π‘₯ is equal to π‘Ž is. It’s the power series the sum from 𝑛 equals zero to ∞ of the 𝑛th derivative of 𝑓 of π‘₯ with respect to π‘₯ at π‘₯ is equal to π‘Ž divided by 𝑛 factorial multiplied by π‘₯ minus π‘Ž all raised to the 𝑛th power. We call π‘Ž the center of our Taylor series. In this case, we’re told to use the value π‘Ž is equal to negative one.

So to find our Taylor series representation of our function, we need to find the 𝑛th derivatives of our function evaluated at π‘₯ is equal to π‘Ž. So in our case, we need to find the 𝑛th derivative of our function 𝑓 of π‘₯ is equal to two plus π‘₯ all raised to the 12th power with respect to π‘₯ at π‘₯ is equal to negative one. Our series starts from 𝑛 is equal to zero. So let’s start with the zeroth derivative of 𝑓 of π‘₯ at π‘₯ is equal to negative one. Remember, when we say the zeroth derivative of a function, we just mean that function. So this is just 𝑓 evaluated at negative one.

So we substitute π‘₯ is equal to negative one into our function 𝑓 of π‘₯. This gives us two plus negative one all raised to the 12th power. If we evaluate this, we just get one. To find the next term in our Taylor series, we’re going to need to know the first derivative of 𝑓 of π‘₯ with respect to π‘₯ at π‘₯ is equal to negative one. But we can’t do this directly. We first need to differentiate our function 𝑓 of π‘₯. The first derivative of 𝑓 of π‘₯ with respect to π‘₯ is equal to the derivative of two plus π‘₯ all raised to the 12th power with respect to π‘₯. And this is the composition of two functions, so we could differentiate this by using the chain rule. We could also do this but distributing our exponent over the parentheses by using binomial expansion.

However, the easiest way to do this is by using the general power rule. We recall this tells us for differentiable function 𝑔 of π‘₯ and real constant π‘˜, the derivative of 𝑔 of π‘₯ raised to the power of π‘˜ with respect to π‘₯ is equal to π‘˜ times 𝑔 prime of π‘₯ multiplied by 𝑔 of π‘₯ raised to the power of π‘˜ minus one. In this case, our exponent π‘˜ is equal to 12 and our function 𝑔 of π‘₯ is equal to two plus π‘₯. And 𝑔 of π‘₯ is a linear function. So its derivative with respect to π‘₯ will just be the coefficient of π‘₯, which in this case is equal to one. So 𝑔 prime of π‘₯ is just one.

So by using the general power rule, we get the first derivative of 𝑓 of π‘₯ with respect to π‘₯ is equal to 12 times one multiplied by two plus π‘₯ raised to the power of 11. And of course we can simplify this to give us 12 times two plus π‘₯ all raised to the 11th power. We can now use this to find the first derivative of 𝑓 of π‘₯ with respect to π‘₯ evaluated at negative one. We substitute π‘₯ is equal to negative one into this expression. This gives us 12 times two plus negative one all raised to the 11th power. But of course, two plus negative one all raised to the 11th power is just equal to one. So this simplifies to just give us 12. So we found this value to be 12.

To find the next term in our Taylor series representation, we’re going to need to find the second derivative of 𝑓 of π‘₯ with respect to π‘₯ evaluated at negative one. And of course we can’t find this directly. We’re going to need to differentiate 12 times two plus π‘₯ all raised to the 11th power. So we’re going to need to differentiate 12 times two plus π‘₯ all raised to the 11th power with respect to π‘₯. And in fact, to do this, we have the exact same options we did before. We could do this by using the chain rule or distributing the power over our parentheses and using the power rule. Or we could do this by using the general power rule.

We’ll do this by using the general power rule. This time, our exponent π‘˜ is equal to 11 and 𝑔 of π‘₯ is once again equal to two plus π‘₯. And once again, we have 𝑔 prime of π‘₯ is equal to one. So by using the general power rule, we get the second derivative of 𝑓 of π‘₯ with respect to π‘₯ is equal to 12 times 11 multiplied by one times two plus π‘₯ all raised to the 10th power. And we’ll simplify this and write it as 12 times 11 multiplied by two plus π‘₯ all raised to the 10th power.

We can now use this to find the second derivative of 𝑓 of π‘₯ with respect to π‘₯ evaluated at negative one. We substitute π‘₯ is equal to negative one into our expression for the second derivative of 𝑓 of π‘₯ with respect to π‘₯. We get 12 times 11 multiplied by two plus negative one all raised to the 10th power. But two plus negative one is just equal to one. So this simplifies to just give us 12 times 11. And we will leave this as it is because we’re starting to see a pattern.

To find the next term in our Taylor series representation, we need to find the third derivative of 𝑓 of π‘₯ with respect to π‘₯ evaluated at negative one. To do this, we’re going to need to differentiate the second derivative of 𝑓 of π‘₯ with respect to π‘₯. But we can just do this by using the general power rule. This time, our exponent π‘˜ is 10 and 𝑔 of π‘₯ is still two plus π‘₯. In fact, we can see each time we apply the general power rule, we reduce the value of π‘˜ by one, but we don’t change our function 𝑔 of π‘₯. We still have 𝑔 prime of π‘₯ is equal to one.

Using the general power rule, we now get the third derivative of 𝑓 of π‘₯ with respect to π‘₯ is equal to 12 times 11 times 10 multiplied by two plus π‘₯ all raised to the ninth power. And then, if we were to use this to find our fourth derivative of 𝑓 of π‘₯ with respect to π‘₯, we would now have 𝑔 of π‘₯ being two plus π‘₯. But then our exponent π‘˜ will just be equal to nine. So by applying the general power rule again, we would get the fourth derivative of 𝑓 of π‘₯ with respect to π‘₯ is equal to 12 times 11 times 10 times nine multiplied by two plus π‘₯ all raised to the eighth power. And in fact we can see we can keep doing this.

We can now use these two expressions to find our third and fourth derivatives of 𝑓 of π‘₯ with respect to π‘₯ evaluated at negative one. In our third derivative, we substitute π‘₯ is equal to negative one. We get 12 times 11 times 10 multiplied by two plus negative one all raised to the ninth power, which simplifies to give us 12 times 11 times 10. And we can do the same for the fourth derivative of 𝑓 of π‘₯ with respect to π‘₯ evaluated at negative one. This simplifies to give us 12 times 11 times 10 times nine.

Now that we found these values, we can substitute these into the summand of our Taylor series representation for our function 𝑓 of π‘₯. So let’s start with the first term in our Taylor series representation for 𝑓 of π‘₯. That’s when 𝑛 is equal to zero and π‘Ž is equal to negative one. Remember, we already showed that the zeroth derivative of 𝑓 of π‘₯ with respect to π‘₯ at negative one is one. So our first term is one divided by zero factorial times π‘₯ minus negative one all raised to the zeroth power. And of course, we can simplify this. First, zero factorial is just equal to one. Next, in this term and all of our subsequent terms, we can write π‘₯ minus negative one as π‘₯ plus one.

However, this doesn’t help us simplify this term since this is raised to the zeroth power. And we know any number raised to the zeroth power is just equal to one. So this whole term simplifies to give us one divided by one, which we know is equal to one. Let’s now move on to the second term in our Taylor series representation. That’s when 𝑛 is equal to one. Remember, we already showed the first derivative of 𝑓 of π‘₯ with respect to π‘₯ at negative one is 12. So we get 12 divided by one factorial multiplied by π‘₯ plus one all raised to the first power. And we can simplify this in exactly the same way we did before.

First, one factorial is just equal to one. Next, π‘₯ plus one all raised to the first power is just equal to π‘₯ plus one. Finally, 12 divided by one is just equal to 12. So our second term simplifies to give us 12 times π‘₯ plus one. Let’s now move on to the third term in our Taylor series representation. That’s when 𝑛 is equal to two. This time, we get 12 times 11 divided by two factorial multiplied by π‘₯ plus one all squared. And in this instance, the only bit of simplification we’ll do is notice that two factorial is just equal to two. And we can do exactly the same to find our fourth and fifth terms where 𝑛 is equal to three and four, respectively.

And we’ll do the same simplification we did before. We’ll write three factorial as six and four factorial as 24. And remember, there will be more terms to this Taylor series representation. We only found a finite number of these. This means we were able to show the Taylor series representation of our function 𝑓 of π‘₯ is given by the following expression. But what about the second part of our question? We need to determine whether there are a finite or infinite number of terms in the Taylor series representation of our function 𝑓. There’s a few different ways of noticing this. For example, remember when we were using the general power rule.

If we kept applying this, we would always reduce our value of π‘˜ by one. Eventually, our value of π‘˜ would be equal to zero. Then, we would be multiplying by zero. So our derivatives would eventually just be equal to zero. But then, if the derivative is equal to zero, from this point on, the rest of the terms in our Taylor series representation will just be equal to zero. Another way of seeing this is to just look at our function 𝑓 of π‘₯. It’s two plus π‘₯ all raised to the 12th power. Think what would happen if we distributed 12 over our parentheses. We could’ve done this by using binomial distribution. However, we only need to know that we get a degree 12 polynomial. And what would happen if we differentiated this polynomial 13 times with respect to π‘₯?

We would differentiate this term by term by using the power rule for differentiation. We multiply by the exponent of π‘₯ and then reduce this exponent by one. But if we’re differentiating this 13 times, we’ll then be multiplying every term by zero. So the 13th derivative of this function must be equal to zero. And then by the same logic, at this point, the rest of the terms in our Taylor series representation will be zero. Therefore, we saw two different ways to see that the Taylor series representation of the function 𝑓 of π‘₯ is equal to two plus π‘₯ all raised to the 12th power at π‘₯ is equal to negative one must have a finite number of terms.

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