Lesson Video: Maclaurin Series | Nagwa Lesson Video: Maclaurin Series | Nagwa

# Lesson Video: Maclaurin Series Mathematics • Higher Education

In this video, we will learn how to find Maclaurin series of a function and find the radius of convergence of the series.

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### Video Transcript

In this video, weβll learn the definition of the Maclaurin series and its relationship to the Taylor series. Weβll look at how to apply this definition to represent an arbitrary function as the Maclaurin series and how we can find the radius of convergence of these types of series.

We might recall that if a function π has a power series expansion at π, the Taylor series of a function π about π is given by π of π₯ equals the sum of the πth derivative of π evaluated π over π factorial times π₯ minus π to the πth power for values of π between zero and β. This can be written as π of π plus π prime of π over one factorial times π₯ minus π plus π double prime of π over two factorial times π₯ minus π squared and so on. For the special case where π is equal to zero, the Taylor series becomes the sum from π equals zero to β of the πth derivative evaluated at zero over π factorial times π₯ to the πth power. Now, this case arises frequently enough that itβs given the special name the Maclaurin series. And weβre going to have a look at how we might apply this formula.

Find the Maclaurin series of the function π of π₯ equals π to the power of π₯.

We recall that the Maclaurin series of a function π of π₯ is given by the sum of the πth derivative of π evaluated at zero over π factorial times π₯ to the πth power for values of π between zero and β. So weβll begin by differentiating π of π₯ to evaluate π prime of zero, π double prime of zero, and so on. Ultimately, weβre hoping to find what the πth derivative of our function is at π₯ equals zero. We know that π of π₯ is equal to π to the power of π₯. The derivative of π to the power of π₯ is simply π to the power of π₯. So we find that π prime of π₯ is π to the power of π₯. π double prime of π₯ is the derivative of this, which is also π to the power of π₯. And we can continue this. And weβll see that the πth derivative of π will also be equal to π to the power of π₯.

We need to work out what π of zero, π prime of zero, and so on is. So π of zero is π to the power of zero, which is one. π prime of zero is also π to the power of zero, which is one. And since we know that the πth derivative of π at π₯ is π to the power of π₯, we know that the πth derivative of π at zero is π to the power of zero, which is one. We therefore replace the πth derivative of π evaluated at zero with one. And we find that our summand is one over π factorial times π₯ to the πth power. We can, of course, simplify this to π₯ to the πth power over π factorial. And so we see that the Maclaurin series of the function π of π₯ equals π to the power of π₯ is the sum from π equals zero to β of π₯ to the πth power over π factorial. And this is one plus π₯ plus π₯ squared over two factorial plus π₯ cubed over three factorial and so on.

So what about the radius of convergence of this series? We recall that if a series converges on the open interval from negative π to π, its radius of convergence is the number denoted by π. And of course, a power series always converges absolutely within its radius of convergence. Similarly, the interval of convergence is the open, closed, or semiclosed range of values of π₯, for which the Maclaurin series converges to the value of the function.

So letβs have a look at how we can find the radius of convergence for a Maclaurin series.

A Maclaurin series is given by the sum from π equals one to β of negative one to the power of π plus one times π times π₯ to the power of π minus one. Find the radius of convergence for the series.

We recall that we can use the ratio test to help us find the radius of convergence of a series. The part of the ratio test weβre interested in says, suppose we have the series given by the sum of π π. If the limit as π approaches β of the absolute value of π π plus one over π π is less than one, then the series is absolutely convergent and hence convergent. In our case, we define π π to be equal to negative one to the power of π plus one times π times π₯ to the power of π minus one. This means that π π plus one is negative one to the power of π plus two times π plus one times π₯ to the πth power. And so to find the radius of convergence, we need to work out where the limit as π approaches β of the absolute value of the quotient of these is less than one.

Now, we can actually simplify everything inside our limit. We recall that when we divide two numbers with an equal base, we subtract the exponents. This means negative one to the power of π plus two divided by negative one to the power of π plus one is simply negative one. Similarly, π₯ to the power of π divided by π₯ to the power of π minus one is simply π₯. Letβs rewrite this as the limit as π approaches β of the absolute value of negative one π₯ times π plus one over π. π₯ is independent event. So we can take the absolute value of negative π₯ outside of our limit.

And we can now look to evaluate our limit. We divide each term on the numerator by π. And weβre looking to find the limit as π approaches β of one plus one over π. Well, as π gets larger, one over π gets smaller. And it approaches zero. This means that the limit as π approaches β of the absolute value of one plus one over π is simply one. So we have that the absolute value of negative π₯ times one is less than one. Or indeed, the absolute value of negative π₯ is less than one.

We can rewrite this as the absolute value of negative one times the absolute value of π₯. And then, of course, the absolute value of negative one is simply one. So by the ratio test, the absolute value of π₯ must be less than one for our series to be convergent. And therefore, the radius of convergence is equal to one.

Weβll now look at how we can derive a Maclaurin series for a trigonometric function.

Consider the function π of π₯ equals cos of π₯. What are the first four derivatives of π with respect to π₯? Write the general form for the πth derivative of π with respect to π₯. And hence, derive the Maclaurin series for cos of π₯.

There is a fourth part to this question, which asks us to find the radius π of convergence of the Maclaurin series for cos of π₯. And weβll consider that at the very end. So we begin by finding the first four derivatives of π with respect to π₯. Weβre told that π of π₯ is equal to cos of π₯. And so we quote the general result for the first derivative of cos of π₯ is negative sin π₯. And therefore, π prime of π₯, the first derivative of π with respect to π₯, is negative sin π₯.

Next, we recall the general result for the derivative of sin π₯. Itβs cos π₯. And this means the derivative of negative sin π₯ must be negative cos of π₯. By returning to the first general result we quoted, weβll be able to differentiate negative cos of π₯. Itβs negative, negative sin of π₯, which is, of course, simply sin of π₯. So the third derivative is sin of π₯. And finally, we obtain the fourth derivative to be equal to cos of π₯. And so we have the first four derivatives. They are negative sin π₯, negative cos π₯, sin π₯, and cos π₯.

The second part of this question asks us to find a general form for the πth derivative of π with respect to π₯. And so we need to spot a pattern with our derivatives. Firstly, we should notice that if we differentiate cos of π₯, we go back to negative sin π₯. And this cycle will continue. Next, we recall that sine and cosine are horizontal translations of one another. And so if we sketch the curve of π¦ equals negative sin of π₯, that, of course, is a reflection on the π₯-axis for the graph of π¦ equals sin π₯. We see that we can write π¦ equals negative sin π₯ as π¦ equals cos of π₯ plus π by two. Itβs a horizontal translation of the graph of π¦ equals cos of π₯, left by π by two radians. So π prime of π₯ is equal to cos of π₯ plus π by two.

Weβll repeat this for the graph of π¦ equals negative cos of π₯. Itβs a reflection again in the π₯-axis of the graph π¦ equals cos of π₯. And it, therefore, also can be represented by a horizontal translation of the graph of π¦ equals cos of π₯ left by π radians. So we can write π double prime of π₯, which was negative cos of π₯, as cos of π₯ plus π. By applying a similar thought process, we find that the third derivative, sin of π₯, can be written as cos of π₯ plus three π by two. And the fourth derivative can be written as cos of π₯ plus two π.

Remember, cos itself is periodic with a period of two π radians. So cos of π₯ plus two π is exactly the same as cos of π₯. And we might be starting to spot a pattern. In fact, if we write the second derivative, cos of π₯ plus π, as cos of π₯ plus two π over two and the fourth derivative as cos of π₯ plus four π over two, we see we can write the πth derivative as cos of π₯ plus ππ over two. And so the general form for the πth derivative of π with respect to π₯ is cos of π₯ plus ππ over two.

The third part of this question asks us to derive the Maclaurin series for cos of π₯. And so we recall this is the sum from π equals zero to β of the πth derivative of π evaluated at zero over π factorial times π₯ to the πth power. Well, in this case, the πth derivative of π evaluated at zero is cos of zero plus ππ over two. When π is zero, we have π of zero, which is cos of zero which is one. When π is one, we have π prime of zero, which is cos of π by two, which is zero. When π is two, we have cos of π, which is negative one. And when π is equal to three, we have cos of three by two, which is once again zero. And of course, as we saw in the second part of this question, this cycle continues.

And this means the first few terms will be one minus π₯ squared over two factorial plus π₯ to fourth power over four factorial minus π₯ to the six over six factorial, which we can write as the sum from π equals zero to β of negative one to the πth power. Remember, this part will just give us the alternating signs. We go from positive to negative and back to positive again. This is all over two π factorial. This bit gives us the even factorials and the denominator times π₯ to the power of two π. And so weβve derived the Maclaurin series for cos of π₯. Itβs the sum from π equals zero to β of negative one to the πth power over two π factorial times π₯ to the power of two π. And, of course, if we chose to alternatively write this as shown, that would be absolutely fine too.

Weβre now going to clear some space and consider the fourth and final part of this question. Weβll keep the Maclaurin series for cos of π₯ on screen as weβre going to use that in a moment. And this question says, what is the radius π of convergence of the Maclaurin series for cos of π₯? We recall that, in general, thereβs an open interval from negative π to π, in which a power series converges. And that number π is called the radius of convergence. And we can use the ratio test to find this value. The part of the test weβre interested in says that, given a series of sum of π π, if the limit as π approaches β of the absolute value of π π plus one over π π is less than one, then the series is absolutely convergent and hence convergent.

In our case, we let π π be equal to negative one to the πth power times π₯ to the power of two π over two π factorial. Then, π π plus one is negative one to the power of π plus one times π₯ to the power of two times π plus one over two times π plus one factorial. Weβll distribute the parentheses to make the next step easier. Two times π plus one is two π plus two. According to the ratio test, we want to find where the limit as π approaches β of the absolute value of the quotient of these is less than one.

We know that when dividing by a fraction, we simply multiply by the reciprocal of that fraction. So we can say that we want the limit as π approaches β of the absolute value of negative one to the power of π plus one times π₯ to the power of two π plus two over two π plus two factorial times two π factorial over negative one to the πth power times π₯ to the power of two π. And then, we recall that when dividing two numbers with equal basis, we simply subtract their exponents so that π₯ to the power of π divided by π₯ to the power of π is π₯ to the power of π minus π. Weβll use this to simplify negative one to the power of π plus one divided by negative one to the power of π. Itβs simply negative one. Similarly, weβll be able to simplify π₯ to the power of two π plus two divided by π₯ to the power of two π. Itβs π₯ squared.

And what about these factorials? Well, we know that two π plus two factorial is the same as two π plus two times two π plus one times two π times two π minus one and so on. Alternatively, thatβs the same as two π plus two times two π plus one times two π factorial. And that allows us to divide through by two π factorial. When we do, weβre left with two π plus two times two π plus one on the denominator of our fraction. So we want the limit as π approaches β of the absolute value of this fraction to be less than one. We spot that negative one times π₯ squared is independent event. And so we can take the absolute value of negative π₯ squared outside of the limit. And we have the absolute value of negative π₯ squared times the limit as π approaches β of one over two π plus two times two π plus one.

Weβre now ready to evaluate this limit. As π grows larger, one over two π plus two times two π plus one approaches zero. And so, in fact, weβre looking to find the values of π₯ such that the absolute value of negative π₯ squared times zero is less than one. Well, when we multiply the absolute value of negative π₯ squared by zero, weβll always get zero. And this means our Maclaurin series converges for all values of π₯. And we can therefore say that π equals β or positive β.

In this video, weβve learned that the Maclaurin series is a special case of the Taylor series, where π is equal to zero. Itβs given by the sum from π equals zero to β of the πth derivative of π evaluated at zero over π factorial times π₯ to the πth power. And finally, we saw that if a series converges on the open interval negative π to π, its radius of convergence is the number denoted by π. And we can use the ratio test to find this value.

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