Welcome back, everyone. Today is the day we’ve been waiting for since Lesson 11! Now that we’ve learned about power series in Lesson 19 (link here) we can finally get back to what motivated us to study sequences and series in the first place: “infinite-degree” Taylor polynomials, also known as Taylor series!

Lesson 20: Taylor and Maclaurin Series & Working with Taylor Series

Learning goals

- Describe the procedure for finding a Taylor series for a function.
- Find a Taylor series for a function
- Recognize the Taylor series expansions of common functions.
- Recognize and apply techniques to find the Taylor series for a function.

**Topic**:

- Volume 2, Section 6.3
*Taylor and Maclaurin Series*(link to textbook section) - Volume 2, Section 6.4
*Working with Taylor Series*(link to textbook section)

**WeBWorK**:

- Series – Taylor and Maclaurin Series

**Motivating question**

Before asking our motivating question, let’s recall two facts that we know from previous lessons:

- Recall from Lesson 11 (link here) that a degree $N$ Taylor polynomial $T_N(x)$ approximates a $f(x)$ function near the center $x=a$. Recall also that as the degree $N$ gets higher, the approximation gets better. At the time we asked what the “best approximation” of $f(x)$ should be and we hinted at a hypothetical “infinite-degree polynomial.”
- Recall in our motivating example from Lesson 19 (link here) we saw that the function $f(x) = \frac{1}{1-x}$ is represented by the power series $P(x) = \sum_{n=0}^\infty x^n$ on the interval of convergence $(-1,1)$. We used this example to build power series representations of other functions, but it’s not clear yet how to build series representations of other functions.

Today we ask: how can we put thees two facts together?

Determine the degree 4 Taylor polynomial for $f(x) = \ln(1+x)$ centered at $a=0$.

We saw this exercise back in Lesson 11. You can review the solution in the video linked here.

## Motivating example

In Warmup exercise 1, we saw that the degree 4 Taylor polynomial for $f(x) = \ln(1+x)$ centered at $a=0$ is $p_4(x) = x – \frac{1}{2}x^2 + \frac{1}{3}x^3-\frac{1}{4}x^4$.

In Lesson 19 we saw that the function $f(x) = \ln(1+x)$ is represented by the power series $P(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1}$. Notice that the 4th partial sum of $P(x)$ is $\sum_{n=0}^3(-1)^n \frac{x^{n+1}}{n+1} = x – \frac{1}{2}x^2 + \frac{1}{3}x^3-\frac{1}{4}x^4 = p_4(x)$.

In fact, if we had been asked to find the “infinite-degree” Taylor polynomial for $f(x) = \ln(1+x)$ centered at $a=0$ we would have just found…$P(x)$! So the “best approximation” of $f(x)$ by Taylor polynomials is in fact a power series…we’ll call it the Taylor series $T(x)$! Recall that in Lesson 19, we saw that the interval of convergence for this power series is $(-1,1)$. Click here to review the interactive Desmos graph that showed the graphs of the first several Taylor polynomials converging toward the graph of $f(x) = \ln(1+x)$ on the interval $(-1,1)$.

**Conclusion:** the Taylor series for $f(x) = \ln(1+x)$ centered at $a=0$ is \[T(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1}.\] At times, we may shift the index of summation to rewrite $T(x)$ as $\sum_{n=1}^\infty (-1)^{n+1} \frac{x^{n}}{n}.$

## Taylor series

**Definition:** Let $f(x)$ be a function and let $x=a$ be a point in its domain. Assume that the derivatives $f^{(n)}(a)$ exist for all $n = 0, 1, 2, \dots$. Then the *Taylor series for $f(x)$ centered at $x=a$* is the power series:

\[T(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n\] \[= f(a) + f'(a)(x-a)+\frac{f”(a)}{2}(x-a)^2 + \frac{f”'(a)}{3!}(x-a)^3 + \frac{f^{(4)}}{4!}(x-a)^4 + \cdots\]

If the center $a=0$ then the Taylor series is sometimes called the Maclaurin series (this was the case for Taylor and Maclaurin polynomials too).

## Examples

Finding Taylor series is not terribly different from finding Taylor polynomials, since partial sums of Taylor series *are* Taylor polynomials.

One difference, though, is that the domain of a polynomial is always $(-\infty, \infty)$, while the interval of convergence for a power series might be smaller. So after we’ve found a Taylor series, we may have to then find its interval and radius of convergence.

Another difference is that for most Taylor polynomials that we’re asked to find, the degrees are usually low, so we can just find all the terms and we don’t have to figure out the pattern to find a formula for the general term $\frac{f^{(n)}}{n!}(x-a)^n$ in terms of $n$. However, if we need to find the radius or interval of convergence for a Taylor series using the ratio test, we will have to find a formula for the general term.

**Video 1 **below shows how to find the first few terms of the Taylor series for centered at .

**Video 2** below shows how to find the Taylor/Maclaurin series for at .

**Video 3** is sort of backwards compared to the previous two examples: we’re given a power series and asked to determine which function it’s the Taylor series for.

**Video 4** below shows how to find a Taylor series for $f(x) = \ln(x)$ at $a=2$ in summation notation.

One you already know the Taylor series for at , you can use this knowledge to build Taylor series of more complicated functions at . **Video 5** below shows how to find the Taylor/Maclaurin series for without having to take a bunch of derivatives.

**Video 6** below shows how to find the Taylor series for $f(x) = 5 \cos(3x^2)$ centered at $x=0$. This is similar to the example in Video in that finding the Taylor coefficients directly would be more work than modifying the Taylor series for a simpler function.

Don’t forget that we may need to find the radius or interval of convergence of a Taylor series, just like we did for a regular ol’ power series in Lesson 19 (link here).

## Summary & applications

On the interval of convergence, the Taylor series $T(x)$ for a function $f(x)$ centered at $x=a$ converges to the function $f(x)$ itself. This means that on the interval of convergence, we can use $T(x)$ in place of $f(x)$.

Why might we want to use $T(x)$ instead of $f(x)$? Well, for example, imagine that we were trying to integrate $f(x)$. As we saw in Lessons 1 to 9, integrating functions is not always easy and it’s not always even possible with the tools we have in our integration techniques toolbox. But if we knew the Taylor series $T(x)$ that represents $f(x)$, we could easily integrate it term by term. If you take a course in differential equations (MAT 2680 at CityTech) you’ll see how to use these Taylor series representations to solve differential equations.

### Additional Video Resources

Find the Taylor series for the function $f(x) = \ln(x)$ centered at $a=1$. Leave your answer in summation notation.