Taylor Series Calculator

A Taylor series represents a function as an infinite polynomial built from the function's derivatives at a single point $a$:

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

When $a = 0$, the series is called a Maclaurin series:

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

Why this matters: Taylor series convert calculations on possibly hard functions ($\sin x$, $e^x$, $\ln x$, $\sqrt{1 + x}$) into calculations on polynomials, which computers and humans can handle. They're the foundation of numerical methods, asymptotic expansions, and approximation theory.

The Taylor polynomial of degree $n$ is the partial sum keeping terms up to $(x-a)^n$. It's the best polynomial approximation of $f$ near $a$ in a precise sense (matching value and first $n$ derivatives).

Step 1: Compute Derivatives at the Expansion Point

For $f(x)$ and expansion point $a$, compute $f(a), f'(a), f''(a), \ldots, f^{(n)}(a)$.

Step 2: Plug into the Formula

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

Common Maclaurin Series to Memorize

$e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$

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

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

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

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

Radius of Convergence

A Taylor series converges only within a radius of convergence $R$ around $a$. Find it using the ratio test:

$R = \lim_{n \to \infty} \left|\frac{a_n}{a_{n+1}}\right|$

Outside this radius, the series diverges and does not represent the function. Inside, convergence is usually uniform on compact subsets.

Manipulating Known Series

For speed, substitute, differentiate, or integrate known series instead of computing derivatives from scratch:

• $e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots$ (substitute $-x^2$ into $e^x$)
• $\frac{1}{(1-x)^2} = \frac{d}{dx}\frac{1}{1-x} = \sum_{n=1}^\infty n x^{n-1}$

• Forgetting the factorial: The $n$th term has a $\frac{1}{n!}$, not just the derivative. Skipping this gives a wildly wrong answer.
• Using the series outside its radius of convergence: $\frac{1}{1-x}$ does not equal $\sum x^n$ when $|x| > 1$ — the series diverges there.
• Forgetting to center at $a$: A Taylor series around $a$ uses powers of $(x-a)$, not $x$.
• Confusing degree and number of terms: A degree-$n$ Taylor polynomial has $n+1$ terms (degrees $0$ through $n$).
• Substitution sign errors: $\sin(-x) = -\sin(x)$, so the series of $\sin(-x)$ has alternating signs flipped compared to $\sin(x)$.