Taylor Series: How a Function Becomes an Infinite Polynomial

A polynomial is the simplest kind of function. Add some powers of $x$ together with chosen coefficients and you have a curve you can evaluate, differentiate, integrate, or graph with no fuss. A function like $\sin x$, $e^x$, or $\ln(1+x)$, by contrast, is defined by a rule that no finite calculation can carry out exactly. The remarkable discovery, made in the early days of calculus, is that under reasonable conditions the second kind of function is secretly the first. Any sufficiently smooth function can be written as a polynomial — just one with infinitely many terms — and its coefficients are determined by a single piece of information: all the function’s derivatives at one chosen point.

This infinite polynomial is the Taylor series, and it is one of the most far-reaching ideas in all of analysis. This article explains where it comes from, what it looks like, and why it works so well — and where it fails.

The recipe

Pick a point, call it $a$, where you know everything about the function $f$: its value $f(a)$, its slope $f'(a)$, its curvature $f''(a)$, and every higher derivative $f^{(n)}(a)$. The Taylor series of $f$ centred at $a$ is

$f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n.$

The pattern is fixed and mechanical. The $n$-th term is the $n$-th derivative at $a$, divided by $n!$, multiplied by $(x-a)^n$. The factorial appears as the natural counter-weight to repeated differentiation: differentiating $(x-a)^n/n!$ exactly $n$ times leaves you with $1$, so the construction is rigged so that the $n$-th derivative of the polynomial at $a$ comes out equal to $f^{(n)}(a)$ on the nose. The series is the unique polynomial — infinite in length but finite in pattern — that matches $f$ and all its derivatives at the point $a$.

When the centre is $a = 0$ the formula simplifies and is traditionally called a Maclaurin series. Two beautiful examples:

$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots,$

$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots.$

The series for $e^x$ has every power; the series for $\sin x$ has only odd powers, with alternating signs. Both converge for every real number $x$ — and indeed for every complex number — which is why these particular functions are so well behaved.

Watching the polynomial catch up

The most vivid way to see why the formula works is to draw the function and the partial polynomials together. Take $\sin x$ and start adding terms. The first-degree polynomial is just $x$, a straight line tangent to the sine curve at the origin. It tracks the curve well for small $x$ and quickly drifts away. Add the cubic correction $-x^3/6$ and the bend appears: the new curve hugs $\sin x$ much further out before peeling off. Add the fifth-degree term and the agreement extends further still. Each new term steals a little more of the sine wave’s behaviour.

What the picture shows is a single fact dressed up in many guises. Each successive polynomial agrees with $\sin x$ in one more derivative at $x = 0$. The straight line matches the value and slope. The cubic also matches the curvature and the rate of change of curvature. The quintic matches still more. Match enough derivatives at one point and the curve has nowhere left to wiggle: it must, in the limit, equal $\sin x$ everywhere.

The fine print: convergence

The miracle does not work for every function or every distance from the centre. Two issues can spoil it. The first is that the infinite sum might not converge: each Taylor series has a radius of convergence $R$, and outside the disk $|x - a| < R$ the partial sums fly off to infinity. For $\sin x$ and $e^x$ that radius is infinite — the series works for every $x$. But for $\ln(1+x)$ centred at $0$, the radius is exactly $1$. Try to use the series at $x = 2$ and it diverges, even though $\ln 3$ is a perfectly ordinary number. The issue is that $\ln(1+x)$ has a problem at $x = -1$, where it shoots to $-\infty$, and the radius of convergence reaches out to the nearest such bad point — even when, as here, that point sits behind you on the number line.

The second issue is subtler. The series can converge, yet not to the original function. The classic example is

$f(x) = \begin{cases} e^{-1/x^2}, & x \neq 0, \\ 0, & x = 0. \end{cases}$

This function is infinitely differentiable, and every one of its derivatives at $0$ equals zero. So its Maclaurin series is the polynomial $0 + 0x + 0x^2 + \cdots$, which converges to $0$ everywhere — yet $f(x)$ is clearly not zero for $x \neq 0$. The series converges, beautifully and unconditionally, to the wrong answer. Such functions are called smooth but not analytic; they sit just outside the class for which Taylor’s theorem returns the function intact.

Functions that do equal their Taylor series in a neighbourhood of every point in their domain are called analytic. Polynomials are analytic; so are $e^x$, $\sin x$, $\cos x$, $\ln x$, and most of the named functions of mathematics. The pathology of $e^{-1/x^2}$ is unusual enough that most working scientists never meet it — but its existence is why analysts state the convergence theorem so carefully.

What it is good for

The first answer is computation. Calculators and computers compute $\sin x$, $e^x$, $\ln x$, and friends by truncating their Taylor series after a few terms and bounding the leftover. The convergence is fast enough — especially after a smart change of variable — that double precision is achieved in well under a dozen multiplications.

The second answer is approximation. In physics, the equation governing almost any system is too complicated to solve exactly, but near an equilibrium the system can be replaced by the first one or two terms of its Taylor expansion — and the resulting linear problem can be solved by hand. This is why so many physical laws (Hooke’s law for a spring, the small-angle pendulum, the SIR model near disease-free equilibrium) are linear: not because nature is linear, but because we are looking at it close enough to a quiet point that the nonlinear corrections do not matter yet.

The third answer is theoretical. Taylor series turn the calculus of smooth real functions into something close to algebra: differentiating becomes shifting coefficients, integrating becomes the same shift in the other direction, multiplying functions becomes multiplying power series. In complex analysis, the corresponding objects — power series in a complex variable — are so well behaved that an entire branch of mathematics is built on them. One of the great results of that subject is that a complex function differentiable in some region is automatically analytic there, which is to say: it equals its Taylor series. There is no analogue for real functions. The complex world is, in this technical sense, simpler than the real one — and Taylor series are the bridge.

Taylor’s theorem captures, in one formula, the idea that a smooth curve can be reconstructed from purely local information: the position, slope, and all higher rates of change at a single point. It is a striking statement about how much a function tells about itself in any one place, and it remains, three centuries after Brook Taylor wrote it down, one of the workhorses of applied and pure mathematics alike.

Frequently asked

Who came up with Taylor series?

The English mathematician Brook Taylor published the general formula in 1715, though special cases were known earlier — Madhava of Sangamagrama discovered the series for sine, cosine, and arctangent in 14th-century India, and Newton, Gregory, and Leibniz had all used similar expansions. The special case centred at zero is called a Maclaurin series, after Colin Maclaurin who popularized it in the following decades.

Why does a polynomial built from derivatives equal the function?

Because at the chosen centre point, the polynomial is engineered to match the function's value, slope, curvature, and every higher rate of change exactly. For a smooth function, knowing all those numbers at a single point pins down the function completely — there is only one analytic curve that fits an infinite list of matching derivatives, so the infinite polynomial must reproduce the original.

Does every function have a convergent Taylor series?

No. The Taylor series only converges within a certain radius around the centre, called the radius of convergence, and even where it converges it doesn't always converge to the original function. The classic counterexample is e^(−1/x²), which is perfectly smooth but whose Taylor series around zero is identically zero. Functions whose Taylor series do converge back to them are called analytic, and they form the well-behaved core of analysis.

Where are Taylor series actually used?

Everywhere a function needs to be replaced by something simpler. Calculators compute sin, cos, log, and exp using truncated Taylor (or related) polynomials. Physics linearises a system near equilibrium by keeping only the first two terms. Numerical methods build integration rules and differential equation solvers from Taylor expansions. Complex analysis is built on the fact that every analytic function is locally a power series. The technique is one of the workhorses of applied mathematics.

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