Video: Maclaurin and Taylor Series of Common Functions

In this video, we will learn how to find the Taylor/Maclaurin series representation of common functions such as exponential and trigonometric functions and binomial expansion.

16:29

Video Transcript

In this video, we will learn how to find the Maclaurin series representation of common functions such as exponential and trigonometric functions and binomial expansion. We’ll then see some important uses of Maclaurin expansions, such as using the Maclaurin series expansion to estimate certain values of functions, find Maclaurin expansions of different forms of common functions, and to estimate integrals.

We know that this is the Taylor series of a function 𝑓 at π‘Ž. But when we center our approximation at π‘Ž equals zero, we get this series which has a special name. We call this the Maclaurin series. We use this to approximate functions by firstly finding the first, second, third, and so on derivatives of 𝑓, and then evaluating these at π‘₯ equals zero. We then substitute these values back into the Maclaurin series and simplify. One application of the Maclaurin series is that we can use our approximation to estimate our function for different values of π‘₯. Let’s see an example.

Consider 𝑔 of π‘₯ equals 𝑒 to the π‘₯ power. There’s two parts to this question. The first part says, find the Maclaurin series of 𝑔 of π‘₯. And the second part says, use the first three terms of this series to find an approximate value of 𝑒 to the power of 0.4 to two decimal places.

Let’s start by writing out the Maclaurin series expansion for a general function 𝑓 of π‘₯. So, let’s begin by finding some of the derivatives of our function 𝑔 of π‘₯. 𝑔 of π‘₯ equals 𝑒 to the π‘₯ power, so 𝑔 of zero equals 𝑒 to the zero power, which is one. We differentiate 𝑒 to the π‘₯ to get the first derivative. So, we recall the fact that the derivative of 𝑒 to the power of π‘₯ with respect to π‘₯ is just 𝑒 to the power of π‘₯. So, 𝑔 prime of π‘₯ equals 𝑒 to the π‘₯ power. And so, it follows that 𝑔 prime of zero is one. And in fact, we see that all the derivatives of 𝑔 of π‘₯ are 𝑒 to the power of π‘₯.

So, now let’s make the substitutions. As our function is defined as 𝑔 of π‘₯, we can replace 𝑓 with 𝑔 and begin to make some substitutions. We found that 𝑔 of zero is one. And we found that 𝑔 prime of zero is one. And we found that 𝑔 double prime of zero is one. And 𝑔 triple prime of zero is one. And this will continue in this way for all the future terms. We can bring the π‘₯, the π‘₯ squared, and the π‘₯ cubed, and so on on to the top of the fraction. And then, we can spot that we can actually write this as the sum from 𝑛 equals zero to ∞ of π‘₯ to the 𝑛th power over 𝑛 factorial. So, this gives us the Maclaurin series expansion of 𝑒 to the π‘₯ power.

The second part of this question asks us to use the first three terms of this series to find an approximate value of 𝑒 to the 0.4 power to two decimal places. Here are the first three terms of our series. And we want to approximate 𝑒 to the π‘₯ when π‘₯ equals 0.4. So, we replace π‘₯ with 0.4, and we find that this is one plus 0.4 over one factorial. But we know that one factorial is just one. So, this is 0.4. And 0.4 squared is 0.16. And that’s over two factorial, which is two multiplied by one, which is two. But 0.16 over two is just 0.08. So, adding these up, we find that our approximation is 1.48. We could then use a calculator to check how good our approximation is. Using a calculator gives us the 𝑒 to the power of 0.4 is 1.4918 to four decimal places. So, we can see that even though we only use the first three terms of our series, our approximation is actually quite good.

Being able to find the Maclaurin series expansion of common functions is really useful because this opens the door to finding Maclaurin series of more difficult forms of common functions. For instance, in the last example, we saw that the Maclaurin series of 𝑒 to the π‘₯ power is the sum from 𝑛 equals zero to ∞ of π‘₯ to the 𝑛th power over 𝑛 factorial. So, if we want to find the Maclaurin expansion of 𝑒 to the two π‘₯ power, rather than start from scratch to find this expansion, we can replace π‘₯ in the series with two π‘₯. We must be careful here because, remember, we’ve got to raise both of the terms in the parentheses to the power. And we then see that we can write the series as the sum from 𝑛 equals zero to ∞ of two to the 𝑛th power multiplied by π‘₯ to the 𝑛th power over 𝑛 factorial. Let’s now see another example.

Find the Maclaurin series of the hyperbolic sin of three π‘₯ equals 𝑒 to the three π‘₯ power minus 𝑒 to the negative three π‘₯ power over two.

Let’s start by writing out the general form for the Maclaurin series expansion for a function 𝑓. To make things easier for ourselves, let’s start by finding the Maclaurin series expansion for the hyperbolic sin of π‘₯. We’ll then be able to substitute π‘₯ with three π‘₯ to get the Maclaurin series of the hyperbolic sin of three π‘₯. So, we’re going to need to take our function, 𝑓 of π‘₯, to be the hyperbolic sin of π‘₯, and we’re going to need to evaluate this at zero and its derivatives at zero. So, we’ll also have to remember that the derivative with respect to π‘₯ of the hyperbolic sin of π‘₯ is the hyperbolic cos of π‘₯. And the derivative of the hyperbolic cos of π‘₯ with respect to π‘₯ is the hyperbolic sin of π‘₯.

Let’s create a table for the derivatives of 𝑓 of π‘₯ equals the hyperbolic sin of π‘₯. So, when 𝑛 is zero, we just have 𝑓 of π‘₯, which is the hyperbolic sin of π‘₯. So, to evaluate this when π‘₯ is zero, we use the fact that the hyperbolic sin of π‘₯ is equal to 𝑒 to the π‘₯ power minus 𝑒 to the negative π‘₯ power over two. So, the hyperbolic sin of zero is 𝑒 to the zero power minus 𝑒 to the negative zero power over two. But as 𝑒 to the zero power is just one, this is one minus one over two, which is zero. When 𝑛 is equal to one, we have the first derivative of 𝑓, which we’ve seen already is the hyperbolic cos of π‘₯.

In order to evaluate this at zero, we recall that the hyperbolic cos of π‘₯ is equal to 𝑒 to the π‘₯ power add 𝑒 to the negative π‘₯ power over two. So, the hyperbolic cos of zero is equal to 𝑒 to the zero power add 𝑒 to the negative zero power over two. But we know that 𝑒 to the zero power is just one. So, this is one add one over two, which is just one. When 𝑛 is two, we’re looking for the second derivative of 𝑓 of π‘₯. We can get this by differentiating the first derivative, which was the hyperbolic cos of π‘₯. And because the hyperbolic cos of π‘₯ differentiates to give us the hyperbolic sin of π‘₯, this is the hyperbolic sin of π‘₯. We know what this is; when we evaluate it at zero, it just gives us zero.

And when 𝑛 is three, we want to find the third derivative of 𝑓, which is the derivative of the second derivative, which we found to be the hyperbolic sin. So, this is the hyperbolic cos of π‘₯, which we know when π‘₯ is zero, this is one. And We can start to see a bit of a pattern here. Each derivative of 𝑓 evaluated at zero alternates between zero and one. So, we’re going to use this table to write out the Maclaurin series for the hyperbolic sin of π‘₯. When we substitute in our values, we see that every other term is zero. We’re only left with the odd powers of π‘₯. We want to write this as a series, but we need to make sure we only end up with those odd powers of π‘₯.

So, our series is the sum from 𝑛 equals zero to ∞ of π‘₯ to the power of two 𝑛 add one over two 𝑛 add one factorial. We can use this now to find the Maclaurin series for the hyperbolic sin of three π‘₯. We do this by replacing π‘₯ with three π‘₯. And we can see that this gives us three π‘₯ over one factorial add three π‘₯ cubed over three factorial add three π‘₯ to the fifth power over five factorial, and so on. And we can see that we can write this as the sum from 𝑛 equals zero to ∞ of three π‘₯ raised to the power of two 𝑛 add one over two 𝑛 add one factorial.

One use of Maclaurin series is that we can integrate functions which would normally be hard to integrate much more easily if we can find the Maclaurin series expansion for the function. We can compute both definite and indefinite integrals. If we do have limits of integration, we can use the Maclaurin series expansion to estimate the value of the integral. Let’s see an example.

Approximate the integral between zero and one of sin of π‘₯ squared with respect to π‘₯ using the first two terms of an appropriate series.

Let’s begin by writing out the Maclaurin series expansion for a function 𝑓. Now, the question asks us to use an appropriate series. But what would an appropriate series be? Well, if we can find the Maclaurin series expansion for sin of π‘₯ squared, we could then integrate from there. To find the Maclaurin series expansion of a function, we need 𝑓 and a few derivatives of 𝑓. Sin of π‘₯ squared could get a little bit complicated trying to find the derivatives. So, let’s start by finding the Maclaurin series expansion of sin of π‘₯. We’ll let 𝑓 of π‘₯ equal sin of π‘₯. We’ll find the first five derivatives of 𝑓 of π‘₯. To do this, we recall that the derivative with respect to π‘₯ of sin of π‘₯ is cos of π‘₯. And the derivative with respect to π‘₯ of cos of π‘₯ is negative sin of π‘₯.

So, we differentiate sin of π‘₯ to get the first derivative, which is cos of π‘₯. We differentiate the first derivative to get the second derivative, which is negative sin of π‘₯. And we differentiate again to get the third derivative, and so on. But what we actually need to do is evaluate each of these at zero. Well, sin of zero is zero, and cos of zero is one. So, negative cos of zero is negative one. So, now we can write out the first few terms of the Maclaurin series expansion of sin of π‘₯. We use the general form of Maclaurin series expansion and replace 𝑓 and its derivatives with the ones that we found. And we find that every other term is zero. In fact, all the terms with positive powers of π‘₯ disappear here. So, we’re left with only the odd powers of π‘₯. And also notice how the terms alternate between positive and negative.

So, this is the Maclaurin expansion for sin of π‘₯. We’ll leave it in this form for now and use it to find the Maclaurin series for sin of π‘₯ squared. So, let’s clear some space to do this. We use the series expansion that we found for sin of π‘₯ in order to find the Maclaurin series expansion for sin of π‘₯ squared. All we need to do is replace π‘₯ with π‘₯ squared. We then remember the index law that tells us that π‘₯ raised to the power of 𝑛 raised to the power of π‘š is equal to π‘₯ raised to the power of 𝑛 multiplied by π‘š. So, π‘₯ squared and then raised to the power of three is π‘₯ raised to the sixth power. π‘₯ squared and then raised to the fifth power is just π‘₯ raised to the 10th power.

So, we can rewrite the integral between zero and one of sin of π‘₯ squared with respect to π‘₯ as the integral between zero and one of π‘₯ squared minus π‘₯ to the sixth power over three factorial add π‘₯ to the 10th power over five factorial and so on with respect to π‘₯. A few things to notice at this point is that I rewrote π‘₯ squared over one factorial as π‘₯ squared because one factorial is just one and dividing by one gives us that same number. Also note that it’s okay that we’ve only written out three terms here. Remember that in the question, we’re asked to approximate this integral using only the first two terms. So, we’re actually only going to need the first two terms here.

From here, It’s just a case of integrating using integration rules that we know. We use the power rule for integration, which tells us that we add one to the power and then divide by the new power. So, π‘₯ squared integrates to π‘₯ cubed over three. π‘₯ to the sixth power over three factorial just integrates to π‘₯ to the seventh power over seven multiplied by three factorial. And π‘₯ to the 10th power over five factorial integrates to π‘₯ to the 11th power over 11 multiplied by five factorial, and so on. And we need to evaluate this for our limits. We firstly substitute in our upper limit, which is one. And then, we subtract our function with zero substituted in.

But notice that if we substitute zero, all of these terms will just be zero. So, we’d just be subtracting zero. So, from here, let’s just simplify what we’ve got. One raised to anything will just give us one. So, all these numerators will be one. We can also calculate our denominators by remembering that the factorial of a number is the product of that number and all the integers below it to one. And remember, we’ve been asked to approximate this integral using only the first two terms. So, our approximation for this integral is one over three minus one over 42, which we calculate to be 13 over 42.

There’s another really important result that we get from the Maclaurin series expansion, and it comes from the binomial theorem. We know that the binomial theorem gives us a way of expanding π‘Ž plus 𝑏 to the π‘˜th power, where π‘Ž and 𝑏 are real numbers and π‘˜ is a positive integer, where these are called the binomial coefficients. For example, π‘Ž plus 𝑏 raised to the power of two is π‘Ž squared plus two π‘Žπ‘ plus 𝑏 squared. Or, π‘Ž plus 𝑏 raised to the third power is π‘Ž cubed plus three π‘Ž squared 𝑏 plus three π‘Žπ‘ squared plus 𝑏 cubed. So, this expands a polynomial when π‘˜ is a positive integer with a finite sum.

But there’s a special case of the binomial theorem which comes from the case when π‘Ž equals one and 𝑏 equals π‘₯. So, this is one plus π‘₯ to the π‘˜th power. In 1665, Isaac Newton generalize this theorem for powers which are fractional or negative or both. However, when π‘˜ is not positive, one plus π‘₯ to the π‘˜th power is no longer a polynomial, so we can’t find a finite sum of terms that equals 𝑓 of π‘₯. So, when π‘˜ is fractional or negative, this is not a finite sum anymore. It’s an infinite series. To find the series, we compute the Maclaurin series of one plus π‘₯ to the π‘˜th power. To do this, we need to find 𝑓 and derivatives of 𝑓 evaluated at zero. 𝑓 of π‘₯ is one plus π‘₯ to the π‘˜th power. We differentiate this to get the first derivative.

Using the chain rule, this is π‘˜ multiplied by one plus π‘₯ raised to the power of π‘˜ minus one. And we continue in this way to find the second, third, and so on derivatives. But we need to evaluate each of these at zero, which we do by substituting in π‘₯ equals zero. And then, we substitute these values into our general form for Maclaurin series expansion, which gives us the Maclaurin series expansion of one plus π‘₯ to the π‘˜th power. This gives us the important series known as the binomial series. This can be shown to converge if the absolute value of π‘₯ is less than one and diverge if the absolute value of π‘₯ is greater than one. Therefore, the binomial expansion is only valid when the absolute value of π‘₯ is less than one.

Let’s summarize the main points from this lesson. Here is the Maclaurin series expansion for a function 𝑓 of π‘₯. We can use the Maclaurin series expansion of common functions such as the exponential function or trig functions to approximate these functions for different values of π‘₯. We can use the Maclaurin series expansion of common functions to find the Maclaurin series expansion of different forms of the function with a simple substitution and manipulation. Maclaurin series expansions allow us to approximate integrals which would normally be tricky to compute. And finally, Maclaurin series expansion gives us the binomial series derived from the binomial theorem with π‘Ž equal to one and 𝑏 equal to π‘₯ and works for π‘˜ being negative or fractional or both.

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