Taylor Series¶

Proposition¶

Let $a$ be an interior point of $E\subseteq \mathbb{R}\text{,}$ and let $f\colon E\to\mathbb{R}$ be real analytic at $x=a\text{.}$ Then, for some $r>0$ with $(a-r,a+r)\subseteq E$ $E$ , $f$ has the power series expansion around $x=a$ $a$ :

f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (x-a)^n\quad \text{for all } x \in (a-r,a+r),

known as the Taylor series of $f$ around $x=a\text{.}$

Interpretation¶

If a function is so smooth that it behaves exactly like an infinite polynomial near a point, then that polynomial is built entirely from its derivatives at that point. The Taylor series makes this precise: for a real analytic $f$ , knowing $f(a),f'(a),f''(a),\dots$ determines $f$ on a whole interval around $a$ .

Sin x on the interval from minus pi to pi with its Taylor polynomials of degree 1, 3, 5 and 7 around 0, each following the curve further out.

The figure shows the partial sums $P_1,P_3,P_5,P_7$ of the Taylor series of $\sin x$ at $0$ ; only odd powers occur, because the even derivatives of $\sin$ vanish at $0$ . Each partial sum is a Taylor polynomial and follows $\sin x$ over a wider interval than the previous one: $P_7$ stays within $0.08$ of $\sin x$ on all of $[-\pi,\pi]$ . Further out, every partial sum leaves the band $-1\le y\le1$ $1$ , but for each fixed $x$ the partial sums converge to $\sin x$ as the degree grows.

For $\ln(1+x)$ and $1/(1-x)$ the Taylor polynomials converge to $f$ inside the shaded interval and not beyond it; for $\mathrm{e}^x$ and $\sin x$ everywhere.

Real analyticity is the hypothesis that makes the series equal $f$ . Being smooth is enough to write the series down, but not for it to converge to $f$ , as the smooth function in the Key Results shows. Even for an analytic function the equality is limited by the interval of convergence: $\ln(1+x)$ is defined for all $x>-1$ $- 1$ , but its Taylor series at $0$ converges only for $-1<x\le1$ $1$ .

Taylor series are how exponentials and trigonometric functions are handled as sums. The Poisson distribution has total probability $1$ because $\displaystyle \sum_{k=0}^{\infty}\lambda^k/k!=\mathrm{e}^{\lambda}$ $e^{λ}$ . The exponential series returns in Math V, where the same sum with a square matrix in place of the number defines the matrix exponential.

Key Results¶

The Taylor series centred at the origin is called the Maclaurin series.
For any smooth function you can write down its Taylor series around $a$ , but it need not converge to the function: it equals $f$ on a neighbourhood of $a$ if and only if $f$ is real analytic at $a\text{.}$ For example, $f(x)=\mathrm{e}^{-1/x^2}$ for $x\neq0$ $0$ , $f(0)=0$ $0$ , is smooth, but its Taylor series at $0$ is identically $0\text{.}$

Examples¶

The Taylor series of the natural exponential function at the origin:

\displaystyle \mathrm{e}^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots.

The generalized binomial expansion (binomial series) at the origin, which extends the binomial theorem to real exponents, valid for $|x|<1$ and any real exponent $\alpha$ (and for all $x$ when $\alpha$ is a nonnegative integer, since the series then terminates):

(1+x)^\alpha = \sum_{i=0}^\infty \binom{\alpha}{i} x^i = 1 + \alpha x + \frac{\alpha(\alpha-1)}{2}x^2 + \frac{\alpha(\alpha-1)(\alpha-2)}{6}x^3 + \cdots,

where $\displaystyle\binom{\alpha}{i}=\frac{\alpha(\alpha-1)\cdots(\alpha-i+1)}{i!}\text{.}$

Prerequisites¶

Real Analytic Function

Sources¶

Math III: Analysis · Lecture 3

Aliases: Taylor expansion Keywords: Maclaurin series, exponential series Symbols: alpha, element of, infinity, lambda, pi, real numbers, subset, sum

Last updated: 22 October 2026 19:04