Week 10 Discussion Notes

Table of Contents

• Power Series

• Taylor Series

  • Taylor Polynomials
  • Taylor Series

• Examples

Power Series

Here is probably the most important thing you learn in this class:

Definition

Let $\displaystyle F\p{x} = \sum_{n=0}^\infty a_n\p{x - c}^n$. $F$ is called a power series and we call $c$ the center of $F$.

Power series are great because they have a bunch of nice properties:

Proposition (properties of power series)

Let $\displaystyle F\p{x} = \sum_{n=0}^\infty a_n\p{x - c}^n$ be a power series.

1. There exists a number $R \in \br{0, \infty}$ called the radius of convergence of $F$ such that if $\abs{x - c} < R$, then $F$ converges absolutely and if $\abs{x - c} > R$, then $F$ diverges. If $R = 0$, then the series only converges at its center and if $R = \infty$, then $F$ converges absolutely everywhere.

2. The coefficients can be expressed in terms of the derivatives of $F$:

   $$a_n = \frac{F^{\p{n}}\p{c}}{n!}$$

3. You can differentiate term-by-term, i.e., you can pretend the sum isn't there and take the derivative like normal:

   $$F'\p{x} = \sum_{n=1}^\infty na_n\p{x - c}^{n-1}$$

   $F'$ is also a power series and it has the same radius as $F$.

4. You can also integrate term-by-term:

   $$\int F\p{x} \,\diff{x} = C + \sum_{n=0}^\infty \frac{a_n}{n+1} \p{x - c}^{n+1}$$

   Like $F'$, the antiderivative of $F$ is a power series with the same radius of convergence as $F$.

The reason why power series are so important are because they're used for Taylor series:

Taylor Series

Taylor Polynomials

If you look in the previous section, given a power series $F$, we know its derivatives right away from the coefficients. So, it's natural to wonder what happens if you go the other way, which is what Taylor series are all about.

Definition

Let $f$ be a function with $n$ derivatives. Then the $n$-th Taylor polynomial centered at $x = c$ for $f$ is

$$\begin{aligned} T_n\p{x} &= f\p{c} + f'\p{c}\p{x - c} + \frac{f''\p{c}}{2}\p{x - c}^2 + \cdots + \frac{f^{\p{n}}\p{c}}{n!}\p{x - c}^n \\ &= \sum_{k=0}^n \frac{f^{\p{k}}\p{c}}{k!}\p{x - c}^k. \end{aligned}$$

If $c = 0$, then the Taylor polynomial is also called the Maclaurin polynomial.

If $n = 1$, then you get $T_1\p{x} = f\p{c} + f'\p{c}\p{x - c}$, which is the tangent line to $f$ at $x = c$, which is the linear function that best approximates $f$ near $x = c$. Building off of that, we can interpret $T_n\p{x}$ as the $n$-th degree polynomial that best approximates $f$ near $x = c$.

With this interpretation, a natural question is how good is this approximation? This is answered with the following error bound:

Theorem

Suppose $f^{\p{n+1}}$ exists and is continuous. Then if $\abs{f^{\p{n+1}}\p{u}} \leq K$ for all $u$ between $c$ and $x$, then

$$\abs{T_n\p{x} - f\p{x}} \leq \frac{K}{\p{n + 1}!}\abs{x - c}^{n+1}.$$

Notice that $K$ depends on $x$, i.e., if you change $x$, you have to recalculate $K$. Also notice that if $x$ is really far away from $c$, then $\abs{x - c}^{n+1}$ becomes very large, and if $x$ is really close to $c$, then $\abs{x - c}^{n+1}$ becomes very small, which aligns with the interpretation that $T_n\p{x}$ best approximates $f$ near $x = c$.

Here's a helpful mnemonic to help you remember the error bound formula: if you have $T_n\p{x}$, then the "next term" in the polynomial would look like

Put everything in absolute values and replace $\abs{f^{\p{n+1}}\p{c}}$ with the upper bound $K$, and that gives you

Taylor Series

Taylor polynomials already look like power series, except Taylor polynomials are finite sums. Once you turn it into an infinite sum, you get a Taylor series:

Definition

Let $f$ be an infinitely differentiable function. Then the Taylor series of $f$ at $x = c$ is

$$T\p{x} = \sum_{n=0}^\infty \frac{f^{\p{n}}\p{c}}{n!}\p{x - c}^n.$$

If $c = 0$, then $T\p{x}$ is also called the Maclaurin series of $f$.

You would hope that because Taylor polynomials are good approximations of $f$ that Taylor series should be an "infinitely good" approximation of $f$, i.e., the Taylor series of $f$ converges to $f$. Unfortunately, that isn't always the case.

Example 1.

Let

$$f\p{x} = \begin{cases} e^{-1/x^2} & \text{if } x \neq 0, \\ 0 & \text{if } x = 0. \end{cases}$$

Then $f$ has the (unfortunate) property that $f^{\p{n}}\p{0} = 0$ for all $n \geq 1$. This means that the Taylor series of $f$ is given by $T\p{x} = 0$, which means that $T\p{x}$ only converges to $f$ at the center $x = 0$.

On the other hand, there are still a lot of functions whose Taylor series do converge to their associated function.

Proposition (common Taylor series expansions)

The following functions have Taylor series which converge to them on the specified interval:

$$\begin{array}{rcllll} f\p{x} & = & T\p{x} \\[1ex] \hline \\[-1ex] e^x & = & \displaystyle\sum_{n=0}^\infty \frac{x^n}{n!} & = & \displaystyle 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots & \text{for } x \in \R \\[3ex] \sin{x} & = & \displaystyle\sum_{n=0}^\infty \p{-1}^n\frac{x^{2n+1}}{\p{2n+1}!} & = & \displaystyle x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots & \text{for } x \in \R \\[3ex] \cos{x} & = & \displaystyle\sum_{n=0}^\infty \p{-1}^n\frac{x^{2n}}{\p{2n}!} & = & \displaystyle 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots & \text{for } x \in \R \\[3ex] \displaystyle\frac{1}{1 - x} & = & \displaystyle\sum_{n=0}^\infty x^n & = & \displaystyle 1 + x + x^2 + x^3 + \cdots & \text{for } x \in \p{-1, 1} \end{array}$$

Here, $n! = 1 \cdot 2 \cdot 3 \cdots \p{n-1} \cdot n$ with $0! = 1$.

From these, you can derive a lot of other Taylor series as well.

Example 2.

Find the Maclaurin series for $\arctan{x}$.

Solution.

We can start with the geometric series:

$$\frac{1}{1 - x} = \sum_{n=0}^\infty x^n.$$

If we replace $x$ with $-x^2$, then we get

$$\frac{1}{1 + x^2} = \sum_{n=0}^\infty \p{-1}^n x^{2n}.$$

Integrating on both sides,

$$\arctan{x} = C + \sum_{n=0}^\infty \p{-1}^n\frac{x^{2n+1}}{2n + 1}.$$

To solve for the constant, we can plug in $0$ on both sides which gives $C = 0$, so

$$\arctan{x} = \sum_{n=0}^\infty \p{-1}^n \frac{x^{2n+1}}{2n + 1} \quad\text{for } \abs{x} < 1.$$

Examples

Example 3.

Calculate $\displaystyle \sum_{n=1}^\infty nx^n$.

Solution.

We can start with the geometric series and differentiate term-by-term:

$$\frac{1}{1 - x} = \sum_{n=0}^\infty x^n \implies \frac{1}{\p{1 - x}^2} = \sum_{n=1}^\infty nx^{n-1}.$$

We're really close--we're just off by a factor $x$, which we can just multiply in:

$$\sum_{n=1}^\infty nx^n = \boxed{\frac{x}{\p{1 - x}^2}}.$$

Example 4.

Find the interval of convergence of $\displaystyle \sum_{n=1}^\infty \frac{3^nx^n}{n^3}$.

Solution.

When calculating intervals of convergence, you should always start with the root test or ratio test. In this problem, I'm going to use the ratio test:

$$\lim_{n\to\infty} \abs{\frac{3^{n+1}x^{n+1}}{\p{n+1}^3} \cdot \frac{n^3}{3^nx^n}} = \lim_{n\to\infty} \abs{3x} \cdot \p{\frac{n}{n + 1}}^3 = \abs{3x}.$$

Thus, the series converges if $\abs{3x} < 1 \iff \abs{x} < \frac{1}{3}$ and diverges if $\abs{3x} > 1 \iff \abs{x} > \frac{1}{3}$.

After calculating the radius of convergence, you always want to check the endpoints next:

$x = \frac{1}{3}$: Substituting in, the series becomes

$$\sum_{n=1}^\infty \frac{3^n\frac{1}{3^n}}{n^3} = \sum_{n=1}^\infty \frac{1}{n^3},$$

which converges by $p$-series.

$x = -\frac{1}{3}$: Similarly,

$$\sum_{n=1}^\infty \frac{3^n\frac{\p{-1}^n}{3^n}}{n^3} = \sum_{n=1}^\infty \frac{\p{-1}^n}{n^3},$$

which converges absolutely by $p$-series (or if you like, it converges by the alternating series test).

So, overall, the interval of convergence is

$$\boxed{\br{-\frac{1}{3}, \frac{1}{3}}}.$$

Example 5.

Find $n$ such that $\abs{e - T_n\p{1}} < 10^{-8}$, where $T_n$ is the $n$-th Maclaurin polynomial for $f\p{x} = e^x$.

Solution.

We want to use the error bound. The first step would be to figure out what $K$ is in this case: $f^{\p{n+1}}\p{u} = e^x$. Since $e^x$ is an increasing function, it is bounded at the largest value of $x$ possible, which is, in this case, is at $x = 1$, so we can use $K = e^1 = e$. The theorem then tells us that

$$\abs{e - T_n\p{1}} \leq \frac{e\abs{1}^{n+1}}{\p{n + 1}!} = \frac{e}{\p{n + 1}!} < 10^{-8}.$$

To finish off the problem, we need to solve for $n$ that makes the last inequality true. You can do this by plugging in numbers into your calculator (e.g., try $n = 1, 2, 3, \ldots$ since $n$ is an integer). You should end up finding that the first $n$ that works is

$$\boxed{n = 11}.$$

Example 6.

Calculate $\displaystyle 1 - \frac{\pi^2}{4^2 \cdot 2!} + \frac{\pi^4}{4^4 \cdot 4!} - \frac{\pi^6}{4^6 \cdot 6!} + \cdots$.

Solution.

Whenever you see a problem like, you always want to figure out the pattern. After some trial and error, you should end up with

$$\sum_{n=0}^\infty \p{-1}^n \frac{\pi^{2n}}{4^{2n}\p{2n}!}.$$

The $2n$'s should hopefully be familiar to you--compare the sum to

$$\cos{x} = \sum_{n=0}^\infty \p{-1}^n \frac{x^{2n}}{\p{2n}!}.$$

If you close enough, you'll see that the sum is just the power series of $\cos{x}$ evaluated at $x = \frac{\pi}{4}$, i.e.,

$$\sum_{n=0}^\infty \p{-1}^n \frac{\pi^{2n}}{4^{2n}\p{2n}!} = \cos\frac{\pi}{4} = \boxed{\frac{\sqrt{2}}{2}}.$$

Example 7.

Calculate $\displaystyle \sum_{n=2}^\infty \frac{2^{n+3}}{n!}$.

Solution.

Like the previous problem, we're going to want to use a Taylor series to calculate this. The $n!$ should alert you to $e^x$, which has the expansion

$$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}.$$

It looks close, but we need to fiddle with our series first to get it into the right form. For example, the index is off.

$$\begin{aligned} \sum_{n=2}^\infty \frac{2^{n+3}}{n!} = 2^3 \sum_{n=2}^\infty \frac{2^n}{n!} &= 8\p{\sum_{n=0}^\infty \frac{2^n}{n!} - \frac{2^0}{0!} - \frac{2^1}{1!}} \\ &= 8\p{e^2 - 1 - 2} \\ &= \boxed{8\p{e^2 - 3}}. \end{aligned}$$