What is a Taylor Series? — BUders Math Concepts

Simple Definition

A Taylor series is an infinite sum of polynomial terms that exactly represents a smooth function. Each term is built from the function's value and its derivatives at a chosen center point $a$:

The idea: if you know everything about a function at one point (its value, slope, curvature, jerk, …), you can reconstruct the function everywhere nearby.

When the center is $a = 0$, the Taylor series is also called a Maclaurin series — a common special case.

Taylor Series centered at $a$

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

Expanded form

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

Maclaurin Series — special case $a = 0$

$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}\,x^n = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \cdots$$

Intuition

🎯

Matching derivatives one by one

Imagine trying to impersonate a function with a polynomial. First you match the value at point $a$. Then you match the slope. Then the curvature. Each new term you add forces the polynomial to agree with the original function in one more way at that point.

With infinitely many terms, the polynomial becomes a perfect copy — at least within some radius around $a$.

📡

Why polynomials?

Polynomials are the simplest functions computers and humans can evaluate — just multiply and add. If you can convert $\sin x$, $e^x$, or $\ln(1+x)$ into a polynomial, you can compute them to any desired precision. That's exactly what calculators do internally using Taylor series.

0th order

$P_0(x) = 1$

matches value

1st order

$P_1(x) = 1 + x$

+ slope

2nd order

$P_2(x) = 1 + x + \dfrac{x^2}{2}$

+ curvature

4th order

$P_4(x) = 1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6} + \dfrac{x^4}{24}$

≈ $e^x$ near 0

Key Components

Each term in the Taylor series has three parts working together:

Part	Expression	Role
$f^{(n)}(a)$	The $n$-th derivative of $f$ evaluated at $a$	Encodes the function's behavior at $a$
$n!$	Factorial: $1 \cdot 2 \cdot 3 \cdots n$	Cancels the derivative of $(x-a)^n$, keeping coefficients correct
$(x-a)^n$	Power of the distance from center	Higher powers are small near $a$, large far away

Radius of Convergence — a Taylor series doesn't always converge for all $x$. It converges within some interval $|x - a| < R$, where $R$ is the radius of convergence. Outside this interval, the series diverges.

Radius of convergence — Ratio Test

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

Common Maclaurin Series

These are the most frequently used Taylor series, all centered at $a = 0$. Memorizing them saves enormous time.

Function	Series	Converges for
$e^x$	$\displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$	all $x$
$\sin x$	$\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n\,x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots$	all $x$
$\cos x$	$\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n\,x^{2n}}{(2n)!} = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots$	all $x$
$\ln(1+x)$	$\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}\,x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$	$-1 < x \le 1$
$\dfrac{1}{1-x}$	$\displaystyle\sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots$	$\|x\| < 1$
$(1+x)^k$	$\displaystyle 1 + kx + \frac{k(k-1)}{2!}x^2 + \frac{k(k-1)(k-2)}{3!}x^3 + \cdots$	$\|x\| < 1$

Taylor vs Maclaurin

Taylor Series

General — centered at $a$

Use when $f$ is simpler or better-behaved near some $a \ne 0$. E.g., expanding $\ln x$ near $x = 1$ instead of $x = 0$ (where $\ln$ is undefined).

Maclaurin Series

Special case — centered at $0$

The most common choice when $f(0)$ is defined. All the standard series ($e^x$, $\sin x$, $\cos x$) are Maclaurin series.

Partial Sum

$n$-th degree polynomial $P_n(x)$

Truncating at $n$ terms gives an approximation. The error is bounded by Taylor's Remainder Theorem (Lagrange form).

Power Series

General $\sum c_n (x-a)^n$

A Taylor series is a power series whose coefficients $c_n = \frac{f^{(n)}(a)}{n!}$ are determined by the function's derivatives.

Worked Examples

Example 1 — Maclaurin of $e^x$

Find the Maclaurin series for $f(x) = e^x$.

$f^{(n)}(x) = e^x$ for all $n$
$f^{(n)}(0) = 1$ for all $n$

$\displaystyle e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$

$= 1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6} + \cdots$

Example 2 — Maclaurin of $\sin x$

Find the Maclaurin series for $f(x) = \sin x$.

$f(0)=0,\; f'(0)=1,\; f''(0)=0,\; f'''(0)=-1, \ldots$

Only odd powers survive:

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

$= x - \dfrac{x^3}{6} + \dfrac{x^5}{120} - \cdots$

Example 3 — Taylor at $a=1$

Find the Taylor series for $f(x) = \ln x$ centered at $a = 1$.

$f(1)=0,\; f'(1)=1,\; f''(1)=-1,\; f'''(1)=2!$

$f^{(n)}(1) = \dfrac{(-1)^{n+1}(n-1)!}{1}$ for $n \ge 1$

$\displaystyle \ln x = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(x-1)^n$

Example 4 — Approximation

Approximate $e^{0.1}$ using the first 4 terms of the Maclaurin series.

$e^{0.1} \approx 1 + 0.1 + \dfrac{(0.1)^2}{2} + \dfrac{(0.1)^3}{6}$

$= 1 + 0.1 + 0.005 + 0.000167$

$\approx 1.105167$

Exact: $e^{0.1} \approx 1.10517$ ✅

Remainder & Error

When you stop at the $n$-th term, the error is given by Taylor's Remainder Theorem (Lagrange form). This lets you bound how accurate your approximation is.

Lagrange Remainder

$$R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$$

for some $c$ between $a$ and $x$

In practice you bound $|f^{(n+1)}(c)|$ by its maximum on the interval, giving a worst-case error estimate. This is how you know how many terms you need for a desired precision.

Common Mistakes

✗ Forgetting the factorial in the denominator

The $n!$ is not optional — it's what makes the derivative of the $n$-th term match $f^{(n)}(a)$. Writing $\frac{f^{(n)}(a)}{n}(x-a)^n$ gives completely wrong coefficients.

✗ Assuming convergence everywhere

A Taylor series converges only within its radius of convergence $R$. For example, $\ln(1+x)$ converges only for $-1 < x \le 1$. Plugging in $x = 2$ gives a divergent series, not $\ln 3$.

✗ Confusing Taylor and Maclaurin

Maclaurin is just Taylor with $a = 0$. When the center is not zero, every $x^n$ must become $(x-a)^n$ and derivatives are evaluated at $a$, not at 0.

✗ Differentiating the wrong number of times

To find the coefficient of $(x-a)^n$, you need the $n$-th derivative at $a$. A common error is stopping one derivative too early or late. Always index carefully: $n=0$ is the function itself, $n=1$ is the first derivative, and so on.

Related Concepts

Part	Expression	Role
\(f^{(n)}(a)\)	The \(n\)-th derivative of \(f\) evaluated at \(a\)	Encodes the function's behavior at \(a\)
\(n!\)	Factorial: \(1 \cdot 2 \cdot 3 \cdots n\)	Cancels the derivative of \((x-a)^n\), keeping coefficients correct
\((x-a)^n\)	Power of the distance from center	Higher powers are small near \(a\), large far away

Function	Series	Converges for
\(e^x\)	\(\displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\)	all \(x\)
\(\sin x\)	\(\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n\,x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots\)	all \(x\)
\(\cos x\)	\(\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n\,x^{2n}}{(2n)!} = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots\)	all \(x\)
\(\ln(1+x)\)	\(\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}\,x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots\)	\(-1 < x \le 1\)
\(\dfrac{1}{1-x}\)	\(\displaystyle\sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots\)	\(\|x\| < 1\)
\((1+x)^k\)	\(\displaystyle 1 + kx + \frac{k(k-1)}{2!}x^2 + \frac{k(k-1)(k-2)}{3!}x^3 + \cdots\)	\(\|x\| < 1\)

a Taylor Series?

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