Calculus 2, Fall 2016 (Edited Spring 2021)

Motivation

  • Over the next few sections, we will be learning about power series, Taylor series, and Taylor polynomials.
  • You might be asking "why are we learning about these things? And how does this relate to all of the things we have learned about in the past 5 sections?"
  • Let's start with an example that will hopefully motivate our discussion over the next 3 sections.
  • Take a few minutes to compute the following:

    \(\displaystyle \int e^{-x^2} \, dx\).

Motivation (continued)

  • You couldn't compute \(\int e^{-x^2} \, dx\).
  • It turns you can't express the antiderivative of \(e^{-x^2}\) using elementary functions.
  • However, it can be shown that

    \(\displaystyle e^{-x^2} = \sum_{n=0}^\infty \frac{(-x^2)^n}{n!} = \sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{n!}\)
    \(\displaystyle \, \, \, \, \, \, = 1 - x^2 + \frac{x^4}{2} - \frac{x^6}{6}+\frac{x^8}{24} - \frac{x^{10}}{120}+\cdots\)
  • The expression on the right is called a power series.
  • It is also called the Taylor series for \(e^{-x^2}\) centered at 0.
  • We will see later why these are equal.

Motivation (continued 2)

  • We have that

    \(\displaystyle e^{-x^2} = \sum_{n=0}^\infty \frac{(-x^2)^n}{n!} = \sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{n!}\)
  • Let's look at some graphs to see what is going on: Desmos e^(-x^2) Plot

Motivation (continued 3)

  • We have that

    \(\displaystyle e^{-x^2} = \sum_{n=0}^\infty \frac{(-x^2)^n}{n!} = \sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{n!}\)
  • Notice that

    \(\displaystyle \lim_{n\rightarrow\infty} \left|\frac{(-x^2)^{n+1}}{(n+1)!}\cdot\frac{n!}{(-x^2)^n}\right| =\lim_{n\rightarrow\infty} \left|\frac{-x^2}{n+1}\right| =0 \)

    no matter what \(x\) is.
  • So the radius of convergence is \(\infty\).
  • That means that the power series "works" everywhere.

Motivation (continued 4)

  • Now, if \(\displaystyle e^{-x^2} = 1 - x^2 + \frac{x^4}{2} - \frac{x^6}{6}+\frac{x^8}{24} - \frac{x^{10}}{120}+\cdots\)
  • It stands to reason that

    \(\displaystyle \int e^{-x^2} \, dx = x - \frac{x^3}{3} + \frac{x^5}{10} - \frac{x^7}{42}+\frac{x^9}{216} - \frac{x^{11}}{1320}+\cdots\)
  • Thus, we kind of have a way to express the antiderivative of \(e^{-x^2}\) using elementary functions. So what's the problem?
  • Given a value of \(x\), how do you compute \(\displaystyle \sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{n!}\)?

Motivation (summarized)

Over the next few sections, we will answer questions like:

  • How do we express a given function as a power series?
  • If we express a function as a power series, for what values of \(x\) is it valid?
  • How do we evaluate a power series at a given value? (It has infinitely many terms.)
  • Can we estimate the value of a power series at a given value? If so, how do we know how accurate our estimate is?

Idea of a power series

Recall from calc 1: the derivative of a function at a single point \(f'(a)\) is a special case of the derivative function \(f'(x)\) obtained by evaluating at \(x=a\).

Now in calc 2: the sum of a series \(\displaystyle \sum_{n=0}^{\infty} a_n\) is a special case of the sum of a power series \(f(x) = \displaystyle \sum_{n=0}^{\infty} a_n x^n\) obtained by evaluating at \(x=1\).

Given a power series, does it converge to function we already know and love? For what \(x\) does the series converge?

Power series

Definition: A power series in \(x\) centered at \(0\) is a function of the form \[f(x) = \sum_{n=0}^{\infty} a_n x^n = a_0 + a_1 x + a_2 x^2 + \cdots.\] A power series in \(x\) centered at \(c\) is a function of the form \[f(x) = \sum_{n=0}^{\infty} a_n (x-c)^n = a_0 + a_1(x-c) + a_2(x-c)^2 + \cdots.\]

Convergence

Let \(\displaystyle f(x) = \sum_{n=0}^{\infty} x^n\).

  1. Is this a power series? If so, what is \(a_n\)?

  2. For what values of \(x\) does \(f(x)\) converge?

  3. For what values of \(x\) does \(f(x-4)\) converge?

  4. For what values of \(x\) does \(\frac{1}{5}f(x/5)\) converge?

Geometric series convergence

\(\displaystyle f(x) = \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}\) for \(-1 < x < 1\)

Geometric series convergence

\(\displaystyle f(x-4) = \sum_{n=0}^{\infty} (x-4)^n = \frac{1}{1-(x-4)} = \frac{1}{5-x}\) for \(3 < x < 5\)

Geometric series convergence

\(\displaystyle \frac{1}{5} f\left(\frac{x}{5}\right) = \frac{1}{5} \sum_{n=0}^{\infty} \left(\frac{x}{5}\right)^n = \frac{1}{5} \cdot \frac{1}{1-\frac{x}{5}} = \frac{1}{5-x}\) for \(-5 < x < 5\)

Convergence

Convergence

Find the interval of all \(x\) values for which the series converges. That is, find the center, determine the radius of convergence using the ratio test, and check for convergence at the points where the ratio test is inconclusive.

  1. \(\displaystyle \sum_{n=1}^{\infty} \frac{(x+1)^n}{n \cdot 3^n}\). Remember to check endpoints!

  2. \(\displaystyle \sum_{n=0}^\infty \frac{(-2)^n x^n}{(2n)!}\). What do you think this converges to? Plot a partial sum with many terms!

Example

\(\displaystyle \cos(\sqrt{2x}) = \sum_{n=0}^\infty \frac{(-2x)^n}{(2n)!}\) for \(x \geq 0\).