Taylor & Maclaurin Series

Taylor Series (centered at $x = a$)

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

Maclaurin Series ($a = 0$)

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

Essential Maclaurin Series (Memorize These!)

$e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \quad (\text{all } x)$

$\sin x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \quad (\text{all } x)$

$\cos x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \quad (\text{all } x)$

$\frac{1}{1-x} = \sum_{n=0}^\infty x^n = 1 + x + x^2 + x^3 + \cdots \quad (|x| < 1)$

$\ln(1+x) = \sum_{n=1}^\infty \frac{(-1)^{n+1} x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \quad (-1 < x \leq 1)$

Radius and Interval of Convergence

Use the Ratio Test to find the radius of convergence $R$.

• If $R = \infty$: converges for all $x$.
• If $R = 0$: converges only at the center.
• Check endpoints separately.

Lagrange Error Bound

$|R_n(x)| \leq \frac{M|x-a|^{n+1}}{(n+1)!}$ where $M$ is the maximum of $|f^{(n+1)}|$ on the interval.

Operations on Power Series

• Substitution: Replace $x$ with $g(x)$ in a known series.
• Differentiation: Differentiate term by term (same radius of convergence).
• Integration: Integrate term by term.

Problem: Find the Maclaurin series for $e^{-x^2}$ through the $x^6$ term.

Solution: Start with $e^u = 1 + u + u^2/2! + u^3/3! + \cdots$. Substitute $u = -x^2$:

$e^{-x^2} = 1 + (-x^2) + \frac{(-x^2)^2}{2!} + \frac{(-x^2)^3}{3!} + \cdots$ $= 1 - x^2 + \frac{x^4}{2} - \frac{x^6}{6} + \cdots$

• Memorize the Maclaurin series for $e^x$, $\sin x$, $\cos x$, $1/(1-x)$, $\ln(1+x)$.
• Build new series by substitution, differentiation, or integration.
• Radius of convergence: use the Ratio Test.
• Lagrange error bound estimates the truncation error.