Applications of Taylor Series
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 6.4: “Working with Taylor Series” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/6-4-working-with-taylor-series |
| Textbook used in class | Stewart, Calculus, Section 11.10--11.11: Taylor Series and Applications |
Opening Scenario
The series $1 - 1/3 + 1/5 - 1/7 + \cdots$ is just $\arctan(1) = \pi/4$. The series $1 - 1/2 + 1/3 - 1/4 + \cdots$ is $\ln 2$. Taylor series connect abstract infinite sums to the familiar functions of calculus, and they also provide the binomial series, which generalizes the binomial theorem to fractional and negative exponents.
Quick Reference
Binomial Series. For any $k$ (not necessarily a positive integer) and $|x| < 1$: \[ (1 + x)^k = \sum_{n=0}^{\infty} \binom{k}{n} x^n, \] where the generalized binomial coefficients are \[ \binom{k}{0} = 1, \qquad \binom{k}{n} = \frac{k(k-1)(k-2)\cdots(k-n+1)}{n!}. \] The series terminates (becomes a polynomial) when $k$ is a non-negative integer.
Using known series to evaluate sums:
- $\arctan(1) = \pi/4 = \sum_{n=0}^\infty (-1)^n/(2n+1)$ (Leibniz formula)
- $\ln 2 = \sum_{n=1}^\infty (-1)^{n-1}/n$ (alternating harmonic)
Key Concepts
1. The Binomial Series
Example 1. Find the Maclaurin series for $\dfrac{1}{\sqrt{1-x^2}}$ and its interval of convergence. (Stewart 11.10, binomial example.)
Write $\dfrac{1}{\sqrt{1-x^2}} = (1 - x^2)^{-1/2}$. Apply the binomial series with $k = -1/2$ and replace $x$ by $-x^2$:
\[ (1+u)^{-1/2} = 1 - \frac{1}{2}u + \frac{(-1/2)(-3/2)}{2!}u^2 + \frac{(-1/2)(-3/2)(-5/2)}{3!}u^3 + \cdots \] \[ = 1 - \frac{u}{2} + \frac{3u^2}{8} - \frac{5u^3}{16} + \cdots \]
Substitute $u = -x^2$: \[ \frac{1}{\sqrt{1-x^2}} = 1 + \frac{x^2}{2} + \frac{3x^4}{8} + \frac{5x^6}{16} + \cdots \]
This series converges for $|-x^2| < 1$, i.e., $|x| < 1$.
Boxed answer: $\dfrac{1}{\sqrt{1-x^2}} = \displaystyle\sum_{n=0}^\infty \dbinom{-1/2}{n}(-1)^n x^{2n}$, for $|x| < 1$.
2. Evaluating Infinite Sums Using Known Series
Example 2. Evaluate $\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}$.
This is the series $\arctan x = \sum_{n=0}^\infty (-1)^n x^{2n+1}/(2n+1)$ evaluated at $x = 1$: \[ \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} = \arctan(1) = \frac{\pi}{4}. \]
Example 3. Find the sum $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \cdot 2^n}$.
Use $\ln(1+x) = \sum_{n=1}^\infty (-1)^{n-1} x^n/n$ and substitute $x = 1/2$: \[ \ln\left(\frac{3}{2}\right) = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n \cdot 2^n}. \]
Boxed answer: $\ln(3/2)$.
3. Binomial Series for Square Roots and Reciprocals
The binomial series is the natural tool for series involving $(1+x)^k$ when $k$ is not a positive integer. Common cases:
| $k$ | $(1+x)^k$ |
|---|---|
| $1/2$ | $\sqrt{1+x}$ |
| $-1/2$ | $1/\sqrt{1+x}$ |
| $-1$ | $1/(1+x)$ (reduces to the geometric series) |
| $1/3$ | $(1+x)^{1/3}$ |
For each, compute $\binom{k}{n} = k(k-1)\cdots(k-n+1)/n!$ for the first few $n$ needed.
applying the binomial series outside $|x| < 1$. The series $\sum \binom{k}{n} x^n$ converges for $|x| < 1$ regardless of $k$. At the endpoints $x = \pm 1$ the behavior depends on $k$: the series converges at $x = 1$ when $k > -1$, and at $x = -1$ when $k > 0$. Do not assume convergence at the endpoints without checking.
Common Errors Summary
| Error | Correction |
|---|---|
| Computing $\binom{k}{n}$ as $k!/(n!(k-n)!)$ for non-integer $k$ | Use the falling-factorial form: $k(k-1)\cdots(k-n+1)/n!$; factorials are not defined for non-integers |
| Applying the binomial series when $|x| > 1$ | The series requires $|x| < 1$ |
| Confusing the binomial series with the finite binomial theorem | For positive integer $k$, the series terminates; for other $k$, it is infinite |
Leveled Practice
Problem 1. Find the first three terms of the binomial series for $\sqrt[3]{1+x}$.
Show answer
$k = 1/3$. $\binom{1/3}{0} = 1$, $\binom{1/3}{1} = 1/3$, $\binom{1/3}{2} = (1/3)(1/3-1)/2! = (1/3)(-2/3)/2 = -1/9$.
$(1+x)^{1/3} \approx 1 + \dfrac{x}{3} - \dfrac{x^2}{9} + \cdots$, for $|x| < 1$.
Problem 2. Use the series for $\ln(1+x)$ to evaluate $\displaystyle\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n \cdot 3^n}$.
Show answer
Substitute $x = 1/3$ into $\ln(1+x) = \sum_{n=1}^\infty (-1)^{n-1} x^n/n$:
$\sum_{n=1}^\infty \dfrac{(-1)^{n-1}}{n \cdot 3^n} = \ln(1 + 1/3) = \ln(4/3)$.