Try reviewing these fundamentals firstHigher order derivativesIntroduction to infinite seriesFunctions expressed as power series

Still don't get it?Review these basic concepts…Higher order derivativesIntroduction to infinite seriesFunctions expressed as power seriesNope, I got it.

You are watching: Find a function whose maclaurin expansion is

## Taylor Series and Maclaurin Series

In order to understand Taylor and Maclaurin Series, we need to first look at power series.

## What are Power Series?

A power series is basically a series with the variable x in it. Formally speaking, the power series formula is:

Formula 1: Power Series

where cnc_{n}cn are the coefficients of each term in the series and aaa is a constant. Power series are important because we can use them to represent a function. For example, the power series representation of the function f(x)=1(1−x)(for∣x∣f(x) = frac{1}{(1-x)} (for |x|f(x)=(1−x)1(for∣x∣ 1)1)1) is:

Formula 2: Geometric Series Representation

where a=1a = 1a=1 and cn=1c_{n} = 1cn=1.However, what if I want to find a power series representation for the integral of 1(1−x)frac{1}{(1-x)}(1−x)1? All you have to do is integrate the power series.

## Find a Power Series Representation for the function

Question 1: Find a power series representation for the integral of the function

Equation 1: Power Series Representation integral pt.1

Equation 1: Power Series Representation integral pt.2

## Power Series to a Taylor Series

Now this is where Taylor and Maclaurin Series come in. Taylor Series and Maclaurin Series are very important when we want to express a function as a power series. For example, exe^{x}ex and cosxcos xcosx can be expressed as a power series! First, we will examine what Taylor Series are, and then use the Taylor Series Expansion to find the first few terms of the series. Then we will learn how to represent some function as a Taylor series, and even differentiate or integrate them. Lastly, we will look at how to derive Taylor Polynomials from Taylor Series, and then use them to approximate functions. Note that we will also look at Maclaurin Series.

## What is a Taylor Series

So what exactly are Taylor Series? If possible (not always), we can represent a function f(x)f(x)f(x) about x=ax=ax=a as a Power Series in the form:

where fn(a)f^{n} (a)fn(a) is the nthn^{th}nth derivative about x=ax = ax=a. This is the Taylor Series formula. If it is centred around x=0x = 0x=0, then we call it the **Maclaurin Series**. Maclaurin Series are in the form:

Formula 4: Maclaurin Series

Here are some commonly used functions that can be represented as a Maclaurin Series:

Formula 5: Common Taylor Series

We will learn how to use the Taylor Series formula later to get the common series, but first let's talk about Taylor Series Expansion.

## Taylor Series Expansion

Of course if we expand the Taylor series out, we will get:

This is known as the **Taylor Expansion Formula**. We can use this to compute an infinite number of terms for the Taylor Series.

## Finding the First Few Terms

For example, let's say I want to compute the first three terms of the Taylor Series exe^{x}ex about x=1x = 1x=1.

**Question 2**: Find the first three terms of the Taylor Series for f(x)=exf(x) = e^{x}f(x)=ex.

We will use the Taylor Series Expansion up to the third term. In other words, the first three terms are:

## Finding the Taylor Series

Instead of finding the first three terms of the Taylor series, what if I want to find all the terms? In other words, can I find the Taylor Series which can give me all the terms? This is possible; however it can be difficult because you need to notice the pattern. Let's try it out!

**Question 3**: Find the Taylor Series of f(x)=exf(x) = e^{x}f(x)=ex at x=1x = 1x=1.

Recall that the Taylor Expansion is:

We know the first three terms, but we don't know any terms after. In fact, there are an infinite amount of terms after the third term. So how is it possible to figure what the term is when nn n→∞ infty∞? Well, we look for the pattern of the derivatives. If we are able to spot the patterns, then we will be able to figure out the nthn^{th}nth derivative is. Let's take a few derivatives first. Notice that:

The more derivatives you take, the most you realize that you will just get exe^{x}ex back. Hence we can conclude that the nthn^{th}nth derivative is:

Notice that this Taylor Series for exe^{x}ex is different from the Maclaurin Series for exe^{x}ex. This is because this one is centred at x=1x=1x=1, while the other is centred around x=0x=0x=0.You may have noticed that finding the nthn^{th}nth derivative was really easy here. What if the nthn^{th}nth derivative was not so easy to spot?

**Question 4**: Find the Taylor Series of f(x)=sinxf(x) = sin xf(x)=sinx centred around a=0a = 0a=0.

Notice that if we take a few derivatives, we get:

Now the nthn^{th}nth derivative is not easy to spot here because the derivatives keep switching from cosine to sine. However, we do notice that the 4th4^{th}4th derivative goes back sinxsin xsinx again. This means that if we derive more after the 4th4^{th}4th derivative, then we are going to get the same things again. We may see the pattern, but it doesn't tell us much about the nthn^{th}nth derivative. Why don't we plug a=0a = 0a=0 into the derivatives?

Now we are getting something here. The values of the nthn^{th}nth derivative are always going to be 0, -1, or 1. Let's go ahead and find the first six terms of the Taylor Series using these derivative.

Equation 4: Taylor Series of sinx pt.3

If we are to add all the terms together (including term after the sixth term), we will get:

Equation 4: Taylor Series of sinx pt.4

This is the Taylor Expansion of sinxsin xsinx. Notice that every odd term is 0. In addition, every second term has interchanging signs. So we are going to rewrite this equation to:

Even though we have these three terms, we can pretty much see the patterns of where this series is going. The powers of xxx are always going to be odd. So we can generalize the powers to be 2n+12n+12n+1. The factorials are also always odd. So we can generalize the factorials to be 2n+12n+12n+1. The powers of -1 always go up by 1, so we can generalize that to be nnn. Hence, we can write the Taylor Series sinxsin xsinx as

which is a very common Taylor series. Note that you can use the same strategy when trying to find the Taylor Series for y=cosxy = cos xy=cosx.

**Question 5**: Find the Taylor Series of f(x) = cosx centred around.

Notice that if we take a few derivatives, we get:

Again, the nthn^{th}nth derivative is not easy to spot here because the derivatives keep switching from cosine to sine. However, we do notice that the 4th4^{th}4th derivative goes back cosxcos xcosx again. This means if we derive more after the 4th4^{th}4th derivative, then we are going to get a loop. Now plugging in a=0a=0a=0 we have

Again, the values of the nthn^{th}nth derivative are always going to be 0, -1, or 1. Let's find the first six terms of the Taylor Series using the derivatives from above.

Equation 5: Taylor Series of cosx pt.3

If we are to add all the terms together (including term after the sixth term), we will get:

Equation 5: Taylor Series of cosx pt.4

Notice that this time all even terms are 0 and every odd term have interchanging signs. So we are going to rewrite this equation to:

We pretty much know the pattern here. The powers of x are always even. So we can generalize the powers to be 2n2n2n. The factorials are always even, so we can generalize them to be 2n2n2n. Lastly, the powers of -1 goes up by 1. So we can generalize that to be n. Hence, we can write the Taylor Series cosxcos xcosx as:

## Taylor Expansion Relationship of cosx and sinx

Notice that the Maclaurin Series of cosxcos xcosx and sinxsin xsinx are very similar. In fact, they only defer by the powers. If we were to expand the Taylor series of cosxcos xcosx and sinxsin xsinx, we see that:

We can actually find a relationship between these two Taylor expansions by integrating. Notice that we were to find the integral of cosxcos xcosx, then

which is the Taylor Expansion of cosxcos xcosx.

See more: How Much Is 1.5 Million

## Taylor Series of Harder Functions

Now that we know how to use the Taylor Series Formula, let's learn how to manipulate the formula to find Taylor Series of harder functions.

**Question 6**:Find the Taylor Series of f(x)=sinxxf(x) = frac{sin x}{x}f(x)=xsinx.

So we see that the function has sinxsin xsinx in it. We know that sinxsin xsinx has the common Taylor series:

and so we just found the Taylor series for sinxxfrac{sin x}{x}xsinx. Let's do a harder question.

**Question 7**: Find the Taylor Series of

Equation 8: Taylor Series of 2x^3cos(3x^4) pt.1

Notice that cosine is in the function. So we probably want to use the Taylor Series:

Equation 8: Taylor Series of 2x^3cos(3x^4) pt.2

See that inside the cosine is 3x43x^{4}3×4. So what were going to do is replace all the xxx's, and make them into 3x43x^{4}3×4. In other words,

Thus we are done and this is the Taylor Series of 2x3cos(3×4)2x^{3} cos (3x^{4})2x3cos(3×4). If you want to do more practice problems, then I suggest you look at this link.

http://tutorial.math.lamar.edu/Problems/CalcII/TaylorSeries.aspx

Each question has a step-by-step solution, so you can check your work!

## Taylor Series Approximation

Note that the Taylor Series Expansion goes on as nn n→ intfy, but in practicality we cannot go to infinity. As humans (or even computers) we cannot go on forever, so we have to stop somewhere. This means we need to alter the formula for us so that it is computable.

We alter the formula will be:

Notice that since we stopped looking for terms after n, we have to make it an approximation instead. This formula is known as the **Taylor approximation**. It is a well known formula that is used to approximate certain values.

Notice on the right hand side of the equation that it is a polynomial of degree n. We actually call this the **Taylor polynomial** Tn(x)T_{n} (x)Tn(x). In other words, the Taylor polynomial formula is:

Let's do an example of finding the Taylor polynomial, and approximating a value.

**Question 8**: Find the 3rd3^{rd}3rd degree Taylor Polynomial of f(x)=ln(x)f(x) = ln (x)f(x)=ln(x) centred at a=1a = 1a=1. Then approximate ln(2)ln (2)ln(2).

If we are doing a Taylor Polynomial of degree 3 centred at a=1a = 1a=1, then use the formula up to the 4th4^{th}4th term:

Just in case you forgot, ln1ln 1ln1 gives us 0. That's why f(a)=0f(a) = 0f(a)=0. Now plugging everything into the formula of the 3rd3^{rd}3rd degree Taylor polynomial gives:

Now we have to approximate ln(2)ln (2)ln(2). In order to do this, we need to use the Taylor polynomial that we just found. Notice that according to the Taylor approximation:

So ln(2)ln (2)ln(2) is approximately around 56frac{5}{6}65. See that 56frac{5}{6}65 in decimal form is 0.833333…

Now if you pull out your calculator, we are actually pretty close. The actual value of ln(2)ln (2)ln(2) is 0.69314718056….

## The Error Term

We know that Taylor Approximation is just an approximation. However, what if we want to know the difference between the actual value and the approximated value? We call the difference the **error term**, and it can be calculated using the following formula:

Keep in mind that the zzz variable is a value that is between aaa and xxx, which gives the largest possible error.

Let's use the **error term formula** to find the error of our previous question.

**Question 9**: Find the error of ln(2)ln (2)ln(2).

See more: Theme Of Where Are You Going Where Have You Been Themes, Access Denied

In order to find the error, we need to find

Notice from our previous question that we found the Taylor polynomial of degree 3. So we set n=3n = 3n=3. This means we need to find:

Equation 10: Taylor Series Error term ln(2) pt.2

See that the fourth derivative of the function is:

Equation 10: Taylor Series Error term ln(2) pt.3

Now our function is in terms of xxx, but we need it in term of zzz. So we just set z=xz = xz=x. This means that:

Since we are talking the **error** of our approximation, the negative sign doesn't matter here. So realistically we are looking at:

Now recall that zzz is a number between aaa and xxx which makes the error term the largest value. In other words, zzz must be:

because a=1a=1a=1, and x=2x=2x=2. Now what zzz value must we pick so that our error term is the largest?

Notice that the variable zzz is in the denominator. So if we pick smaller values of zzz, then the error term will become bigger. Since the smallest value of zzz we can pick is 1, then we set z=1z = 1z=1. Thus,

is our error.

## Taylor's Theorem

Now think of it like this. If we were to add the error term and the approximated value together, wouldn't I get the actual value? This is correct! In fact, we can say this formally. If the Taylor polynomial is the approximated function and Rn(x)R_{n} (x)Rn(x) is the error term, then adding them gives the actual function. In other words,