A Taylor series rebuilds a function from what it is doing at a single point. You record the function's value, slope, curvature and every higher derivative at $x = a$, and you write down the one power series that matches all of them: $$\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots$$ When $a = 0$ it is called a Maclaurin series. That is the definition in OpenStax Calculus Volume 2, Section 6.3.
Why the coefficients are $f^{(n)}(a)/n!$
Suppose $f(x) = c_0 + c_1(x-a) + c_2(x-a)^2 + \cdots$. Put $x = a$: every term after the first vanishes, so $c_0 = f(a)$. Differentiate once and put $x = a$: you get $c_1 = f'(a)$. Differentiate twice: the $(x-a)^2$ term gives $2c_2$, so $c_2 = f''(a)/2$. Differentiate $n$ times and the $(x-a)^n$ term leaves $n!\,c_n$, so $c_n = f^{(n)}(a)/n!$. The factorial is there to undo the factors the power rule produces. OpenStax adds that power series are unique, so if $f$ has any power series at $a$ converging to it, it is this one.
Taylor polynomials: stopping early
Cutting the series after the $(x-a)^n$ term gives the $n$th Taylor polynomial $p_n(x)$. The first one, $p_1(x) = f(a) + f'(a)(x-a)$, is the tangent line. Each extra term bends the polynomial to match one more derivative, so it hugs the curve over a wider stretch.
The three Maclaurin series to know
From OpenStax's worked examples:
- $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots = \sum_{n=0}^{\infty} \frac{x^n}{n!}$, because every derivative of $e^x$ is $e^x$, which equals 1 at 0.
- $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$, odd powers only, because the derivatives at 0 cycle $0, 1, 0, -1$.
- $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$, even powers only.
OpenStax uses the ratio test to show all three converge for every real $x$, and the remainder bound below to show they converge to the function.
Worked example: estimating $\sqrt[3]{11}$
OpenStax centres $f(x) = \sqrt[3]{x}$ at $a = 8$, where $f(8) = 2$, $f'(8) = \frac{1}{12}$ and $f''(8) = -\frac{1}{144}$. Then $$p_1(11) = 2 + \tfrac{1}{12}(3) = 2.25, \qquad p_2(11) = 2 + \tfrac{1}{12}(3) - \tfrac{1}{288}(3)^2 = 2.21875.$$ The quadratic is the better estimate, and the remainder bound below says how much better.
How big is the error?
Taylor's theorem with remainder: if $f$ can be differentiated $n+1$ times on an interval containing $a$ and $x$, then $$R_n(x) = f(x) - p_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$$ for some $c$ between $a$ and $x$. If $|f^{(n+1)}| \le M$ there, then $|R_n(x)| \le \frac{M}{(n+1)!}|x-a|^{n+1}$. OpenStax applies this to $\sin\left(\frac{\pi}{18}\right) \approx 0.173648$ using the fifth Maclaurin polynomial and finds the error is at most about $9.8 \times 10^{-10}$. The $(n+1)!$ in the denominator grows so fast that a few terms often give many correct digits near the centre.
A Taylor series converges to $f$ exactly when $R_n(x) \to 0$. Matching every derivative at one point is not by itself enough; the remainder has to shrink.
Checkpoints
- Write $p_0$ to $p_3$ for $\ln x$ at $a = 1$.
- Use the fourth Maclaurin polynomial for $\cos x$ to approximate $\cos\left(\frac{\pi}{12}\right)$.
- Find the Taylor series of $\frac{1}{x}$ at $a = 1$ and its interval of convergence.
- Use $\sin x \approx x - \frac{x^3}{6}$ to evaluate $\lim_{x\to 0} \frac{\sin x - x}{x^3}$.
Common mistakes
- Dropping the factorials. The coefficient is $f^{(n)}(a)$ divided by $n!$.
- Centring at the wrong point. Use powers of $(x-a)$, not $x$, unless $a = 0$.
- Assuming convergence everywhere. The series for $\frac{1}{x}$ at 1 only converges on $(0,2)$.
- Using $a$ instead of $c$ in the remainder. The derivative in the error term is evaluated at an unknown point between $a$ and $x$; bound it over that interval.
Related reading
- Sequences, series and infinite sums and convergence tests for series
- What a derivative measures and the mean value theorem
- L'Hôpital's rule intuition, a second way to handle the limits above
- Calculus foundations topic map, all explainers and the sources page
Related tools
The Taylor polynomial calculator builds $p_n(x)$ for you. Work one by hand first, then compare.
Sources
Last reviewed: 12 November 2026.
Common questions
What is a Taylor series?
It is a power series built from a function's derivatives at one point a. Its terms are f^(n)(a)/n! times (x - a)^n, chosen so that the series matches the function's value and every derivative at a.
What is the difference between a Taylor series and a Maclaurin series?
A Maclaurin series is a Taylor series centred at a = 0.
What is a Taylor polynomial?
The nth partial sum of the Taylor series. It is a polynomial of degree n that approximates the function near a.
How accurate is a Taylor polynomial?
Taylor's theorem with remainder says the error is f^(n+1)(c)/(n+1)! times (x - a)^(n+1) for some c between a and x. If the (n+1)th derivative is bounded by M, the error is at most M/(n+1)! times |x - a|^(n+1).