Calculus II: Series Cheat Sheet

Taylor Series Expansion

A Taylor series represents a function as an infinite sum of terms, calculated from the values of the function's derivatives at a single point a.

$$ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n $$

Expanded, this looks like:

$$ f(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots $$

Here, f⁽ⁿ⁾(a) is the n^th derivative of f evaluated at the point a.

A Maclaurin series is a special case of the Taylor series, centered at a = 0. This is the most common type of power series expansion.

$$ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n $$

Expanded, this looks like:

$$ f(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots $$

Memorizing or having these handy is crucial. Most problems involve manipulating these known series rather than deriving them from scratch.

Function	Series Expansion	Interval of Convergence
$$ \frac{1}{1-x} $$	$$ \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots $$	$$ (-1, 1) $$
$$ e^x $$	$$ \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots $$	$$ (-\infty, \infty) $$
$$ \sin(x) $$	$$ \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots $$	$$ (-\infty, \infty) $$
$$ \cos(x) $$	$$ \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots $$	$$ (-\infty, \infty) $$
$$ \ln(1+x) $$	$$ \sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots $$	$$ (-1, 1] $$
$$ \arctan(x) $$	$$ \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots $$	$$ [-1, 1] $$
$$ (1+x)^k $$ (Binomial Series)	$$ 1 + kx + \frac{k(k-1)}{2!}x^2 + \frac{k(k-1)(k-2)}{3!}x^3 + \cdots $$	$$ (-1, 1) $$

You can use the known series above to find new ones through simple algebraic manipulation.

Substitution: Replace x with another function.
Example: To find the series for e^{-x^3}, substitute u = -x^3 into the series for e^u.

$$ e^{(-x^3)} = 1 + \frac{(-x^3)}{1!} + \frac{(-x^3)^2}{2!} + \cdots = 1 - x^3 + \frac{x^6}{2} - \cdots $$
Multiplication: Multiply the entire series by a constant or a power of x.
Example: To find the series for x^2 \cos(x), multiply the series for cos(x) by x^2.

$$ x^2 \cos(x) = x^2 \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \right) = x^2 - \frac{x^4}{2!} + \frac{x^6}{4!} - \cdots $$
Differentiation & Integration: You can differentiate or integrate a known series term-by-term.
Example: The series for \frac{1}{(1-x)^2} is the derivative of the series for \frac{1}{1-x}.

$$ \frac{d}{dx} \left( \sum_{n=0}^{\infty} x^n \right) = \sum_{n=1}^{\infty} nx^{n-1} = 1 + 2x + 3x^2 + \cdots $$