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Taylor and Maclaurin Series, Intuitively

A Taylor series is the natural end of an idea you met on the first day of derivatives. A tangent line approximates a function near a point using its value and its slope. If matching two pieces of information gives a decent approximation, matching three should give a better one, and matching all of them should give something remarkable. That is exactly what a Taylor series is: the polynomial that agrees with a function in every derivative at a single point.

This guide explains where the formula comes from, why the factorials are unavoidable, which standard series you should know cold, and how to say something honest about the size of the error.

From Tangent Lines to Infinite Polynomials

The tangent line at a point is the unique first-degree polynomial whose value and first derivative match the function there. Add an x squared term and you can also match the second derivative, which fixes the concavity and makes the approximation hug the curve noticeably longer. Add a cube term and you match the third derivative, and so on. Each new term buys you agreement in one more derivative, and each new term dies off faster near the center, so it refines the fit without ruining what the earlier terms already got right.

Push that process to infinity and you get the Taylor series. The intuition to carry is that a Taylor series reconstructs a function entirely from local information at one point: everything it knows, it learned from the derivatives at the center. That is a strong claim, and it comes with fine print.

The Formula, and Why the Factorial Is There

The Taylor series of f centered at a is the sum over n of the nth derivative of f at a, divided by n factorial, times the quantity x minus a raised to the n. The factorial is not decoration; it is a correction that undoes differentiation. Differentiating x minus a to the n exactly n times produces n factorial, so dividing the coefficient by n factorial is precisely what makes the nth derivative of the nth term come out equal to the nth derivative of the function.

That also explains the shape of the answer. If you differentiate the whole series n times and then set x equal to a, every term with a lower power has been differentiated away to zero, every term with a higher power still contains a factor of x minus a and vanishes at the center, and exactly one term survives. Match that surviving term to the nth derivative and the coefficient formula falls out. Nothing about it is arbitrary.

A Maclaurin series is nothing new: it is the Taylor series centered at a equals zero. Center at zero whenever the problem allows it, since the powers of x are simpler than powers of x minus a, and the standard series are all tabulated in that form.

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

The Taylor series of f centered at a; setting a equal to zero gives the Maclaurin series.

The Standard Series Worth Memorizing

A handful of Maclaurin series appear so often that deriving them each time is wasted effort. The exponential function e to the x is the sum of x to the n over n factorial, valid for every real x, and it is the easiest to derive because every derivative of e to the x equals one at zero. Sine is the sum of negative one to the n times x to the two n plus one, over two n plus one factorial, again for all x; cosine is the sum of negative one to the n times x to the two n, over two n factorial, for all x. Sine keeps only odd powers and cosine only even powers, which matches the fact that sine is odd and cosine is even.

Three more are essential. The geometric series says one over the quantity one minus x is the sum of x to the n, valid only when the absolute value of x is less than one. The logarithm series says the natural logarithm of one plus x is the sum, starting at n equals one, of negative one to the n plus one times x to the n over n, valid for x greater than negative one and less than or equal to one. Arctangent is the sum of negative one to the n times x to the two n plus one, over two n plus one, valid for the absolute value of x less than or equal to one.

Notice how the intervals differ. The exponential and trigonometric series converge everywhere, while the last three are limited by trouble sitting at distance one from the origin. Memorize the interval alongside the series: a series quoted outside its interval of validity is simply false.

e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}, \qquad \sin x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}, \qquad \cos x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}

Three Maclaurin series valid for every real x.

\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \;(|x|<1), \qquad \ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n} \;(-1 < x \le 1)

Two series whose validity stops at distance one from the center.

Build New Series Instead of Differentiating

Computing a Taylor series straight from the definition means finding a general formula for the nth derivative, which is painful for anything but the simplest functions. In practice almost nobody does that. Instead you start from a known series and transform it, because substitution, multiplication by a power, differentiation, and integration all pass through an infinite sum legally inside its interval of convergence.

Substituting is the fastest trick. Replacing x with x squared in the exponential series gives e to the x squared as the sum of x to the two n over n factorial, with no derivatives computed. Multiplying by a power shifts every exponent: x times the sine series gives x sine x in one line.

Differentiation and integration are just as useful. Differentiating the sine series term by term produces one minus x squared over two factorial plus x to the fourth over four factorial and so on, which is exactly the cosine series, as it must be. Substituting negative x squared into the geometric series gives one over one plus x squared as the alternating sum of x to the two n, and integrating that term by term gives the arctangent series, complete with its odd powers and its two n plus one denominators. Recognizing which known series to start from is most of the work on these problems.

How Wrong Is a Truncated Series?

Cutting a Taylor series off after the term of degree n leaves the Taylor polynomial of degree n, and the difference between the function and that polynomial is the remainder. Lagrange's form of the remainder says it equals the derivative of order n plus one, evaluated at some unknown point between the center and x, divided by n plus one factorial, times x minus a raised to the n plus one. You never learn that unknown point, but you rarely need it: bounding the derivative of order n plus one by a constant M on the interval turns the equality into a usable error bound.

Try it on sine near zero. The degree-three Taylor polynomial is x minus x cubed over six, which at x equals 0.1 gives 0.09983333. Every derivative of sine is bounded by one, so with M equal to one the bound on the remainder is 0.1 to the fourth over four factorial, about 4.2 times ten to the negative sixth. The true value of sine of 0.1 is 0.09983342, so the actual error is roughly 8.3 times ten to the negative eighth, comfortably inside the bound. Since the degree-four term of the sine series is zero, the same polynomial is also the degree-four approximation, and the sharper bound of 0.1 to the fifth over five factorial matches the observed error almost exactly.

One caveat completes the picture. Having a Taylor series is not the same as equalling it. Some functions are infinitely differentiable with every derivative zero at the origin, so their Maclaurin series is identically zero while the function is not. What bridges that gap is the remainder going to zero.

|R_n(x)| \le \frac{M}{(n+1)!}\,|x-a|^{n+1}, \quad M \ge \max \left|f^{(n+1)}\right|

The Lagrange error bound for a degree-n Taylor approximation.

What These Series Are Actually For

Taylor series are not a curiosity at the end of the syllabus; they are how transcendental functions get computed and how hard limits and integrals get tamed. A few uses justify the memorization.

Evaluate limits that resist algebra by replacing each function with its first few terms.
Integrate functions with no elementary antiderivative, such as e to the negative x squared, term by term.
Justify the small-angle approximation: sine of x is approximately x, with an error on the order of x cubed over six.
Approximate values numerically to a guaranteed accuracy using the Lagrange bound.
Recognize a mystery series as a known function evaluated at a specific point.

Frequently asked questions

What is the difference between a Taylor series and a Maclaurin series?

None, beyond the center. A Maclaurin series is a Taylor series centered at zero. Every Maclaurin series is a Taylor series, but a Taylor series centered anywhere else is not a Maclaurin series.

Why do the coefficients have factorials in them?

Differentiating the term with x minus a to the n exactly n times produces a factor of n factorial. Dividing by n factorial cancels it, which is what forces the nth derivative of the series to match the nth derivative of the function at the center.

Do I have to compute derivatives to find a series?

Usually not. Start from a known series and substitute, multiply by a power, differentiate, or integrate term by term. All of these are valid inside the interval of convergence and are far faster than finding a general nth derivative.

How do I bound the error of a Taylor approximation?

Use the Lagrange remainder: bound the derivative of order n plus one by a constant M on the interval between the center and your x, then the error is at most M times the distance to the n plus first power, divided by n plus one factorial.