What is a power series?

A series whose terms carry successive powers of a variable: the sum of aₙ(x − x₀)ⁿ. For a fixed x it is an ordinary number series, so its convergence can be examined — and because the answer differs from one x to another, the whole thing behaves like a function defined wherever the series converges.

What is the radius of convergence?

A number R such that the series converges for every x closer to the centre than R and diverges for every x further away. It is usually computed with the ratio test, as the limit of |aₙ/aₙ₊₁|. The extreme cases occur too: R = 0 means convergence only at the centre, and R = ∞ means convergence on the whole line.

How does the radius differ from the interval of convergence?

The radius gives the reach; the interval gives the exact set of arguments, endpoints included. Inside the radius convergence is certain and outside it divergence is, but at both endpoints the test says nothing and each has to be examined on its own. For Σ xⁿ/n the interval runs from −1 included to 1 excluded.

Where do the coefficients of a Taylor series come from?

From the derivatives of the function at the point it is expanded about. If a function is the sum of a power series, then differentiating that series n times and setting x = x₀ leaves exactly n!·aₙ, so aₙ = f⁽ⁿ⁾(x₀)/n!. The coefficients are therefore not chosen — they are forced by the function itself.

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

The point they are expanded about. A Maclaurin series is simply a Taylor series at x₀ = 0, so it runs in powers of x itself rather than of x − x₀. All four canonical expansions — eˣ, sine, cosine and the logarithm — are quoted in that version.

Why does the sine expansion contain only odd powers?

Because the sine is an odd function and the series has to inherit that. The computation shows it too: the successive derivatives of the sine at zero run 0, 1, 0, −1, 0, 1, …, so every coefficient of an even power comes out zero. The cosine, an even function, has only even powers for the same reason.

How large is the error of a Taylor polynomial?

Bounded by M·|x − x₀|ⁿ⁺¹/(n + 1)!, where M bounds the derivative of order n + 1 on the stretch examined. The practical upshot is that the approximation is better the closer you are to the centre of the expansion — and that the factorial in the denominator lets one extra term improve the result by an order of magnitude.

Can every function be expanded into a Taylor series?

No. The function must have derivatives of all orders at the point of expansion, and even then the series need not equal it anywhere but at the centre — there are smooth functions whose Maclaurin series is identically zero while they themselves are not. That the expansions in this lesson do equal their functions is a separate theorem, not an obvious fact.

Advanced level

Power series and the Taylor series

Put a letter x into a series and it stops being a number and becomes a function. It turns out that eˣ, sine, cosine and the logarithm can all be written this way — as infinite polynomials. Their finite pieces are what a calculator actually evaluates those functions with.

Before you start

This topic builds on earlier ideas. Before you start, it's worth working through the lessons below — they'll make everything click:

Number seriesA series is the limit of its sequence of partial sums — and only in a few lucky cases can that limit be computed outright. So instead of the sum, a humbler question is asked first: does the series have a sum at all. The tests answer it: comparison, ratio, root and Leibniz.
The derivativeA derivative measures how fast a function changes at a point — it is the slope of the tangent to the graph. Meet the definition through the difference quotient and the six rules that compute it without taking a limit every time.

Where this is used

Real situations where you count exactly the way this lesson teaches:

How a calculator evaluates a sine
A processor holds no table of sines — it sums a few terms of a series. For x = 0.5 the polynomial x − x³/6 + x⁵/120 alone gives 0.4794255, and the true sine is 0.4794255 — seven digits of agreement from three terms. The next term would change the result by less than 10⁻⁹, so it is simply never computed.
A pendulum at small angles
The formula for the period of a pendulum, T = 2π√(L/g), comes from replacing sin θ by θ, i.e. from truncating the sine series after its first term. At a swing of 5 degrees, or 0.0873 radians, the sine is 0.08716 — a difference of 0.15 percent, so a 1-metre pendulum has a period of 2.006 seconds instead of 2.007. At 30 degrees the error is already 4.5 percent and the formula must not be used.
Continuously compounded interest
A deposit with a nominal rate of 3 percent compounded continuously grows over a year by a factor of e^0.03. The first two terms of the series give 1 + 0.03 = 1.03, and the third adds 0.00045, i.e. 1.03045 — really 3.045 percent rather than 3. On 20,000 that is 9 of difference, and it is exactly that quadratic term which separates continuous compounding from ordinary.
Logarithmic returns
Analysts measure price changes as ln(1 + r) rather than r, because such returns may be added. The logarithm series says when the difference is negligible: a 2 percent rise gives 0.0198, an error of 1 percent, while a 50 percent rise gives 0.405 instead of 0.5 — off by 19 percent. The first term of the expansion is a good approximation only near zero.

All formulas

Power series
$\sum_{n=0}^{\infty} a_n (x - x_0)^n$
a function of its argument x
Radius of convergence
$R = \lim_{n \to \infty} \left|\frac{a_n}{a_{n+1}}\right|$
convergent for |x − x_0| < R
Taylor series
$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(x_0)}{n!}\,(x - x_0)^n$
the coefficients are derivatives at x_0
Maclaurin series
$f(x) = f(0) + f'(0)\,x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots$
the Taylor series about zero
Expansion of eˣ
$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$
convergent for every x
Expansion of sine
$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$
odd powers only
Expansion of cosine
$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$
even powers only
Expansion of the logarithm
$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$
only for −1 < x ≤ 1
Approximation error
$\big|R_n(x)\big| \leq \frac{M\,|x - x_0|^{n+1}}{(n+1)!}$
M bounds the derivative of order n+1

Every number series in the previous lesson summed to a number. Now put a variable into the terms:

\sum_{n=0}^{\infty} a_n (x - x_0)^n = a_0 + a_1(x - x_0) + a_2(x - x_0)^2 + \dots

For each fixed $x$ this is an ordinary number series — convergent for some arguments, not for others. A power series is therefore a function of its argument, defined wherever it converges. You have known the first example for a long time: the geometric series

1 + x + x^2 + x^3 + \dots = \frac{1}{1-x} \qquad \text{for } |x| < 1

Infinitely many powers on the left, a short formula on the right. This lesson is about when such a trade is legal and how to find the coefficients it needs.

The radius of convergence

For the geometric series the condition read $|x| < 1$ , and that was no accident. Every power series converges on an interval symmetric about its centre, and its reach is called the radius of convergence:

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

$\lvert x - x_0 \rvert$	what happens
less than $R$	series convergent (absolutely, at that)
greater than $R$	series divergent
equal to $R$	must be examined separately, endpoint by endpoint

The formula for $R$ is nothing but the ratio test applied to $|a_n x^n|$ and solved for $|x|$ .

Find the radius and the interval of convergence of Σ xⁿ/n.

The interval of convergence of Σ xⁿ/n. The filled circle at −1 and the hollow one at 1 are the results of two separate computations, which the radius by itself does not settle.

Two extreme cases are worth knowing:

$\sum \tfrac{x^n}{n!}$ has $R = \lim \tfrac{(n+1)!}{n!} = \infty$ — it converges for every $x$ ;
$\sum n!\,x^n$ has $R = \lim \tfrac{n!}{(n+1)!} = 0$ — it converges only at $x = 0$ , so as a function it is useless.

Where the coefficients come from

Suppose a function $f$ is the sum of some power series about zero:

f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \dots

Setting $x = 0$ kills everything but the first term, so $a_0 = f(0)$ . Now differentiate both sides:

f'(x) = a_1 + 2a_2 x + 3a_3 x^2 + \dots \quad \Longrightarrow \quad a_1 = f'(0)

f''(x) = 2a_2 + 6a_3 x + \dots \quad \Longrightarrow \quad a_2 = \frac{f''(0)}{2}

Each time the constant $n!\,a_n$ is what remains, so in general $a_n = \tfrac{f^{(n)}(0)}{n!}$ . Hence the Taylor formula, in the version about an arbitrary point $x_0$ :

f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(x_0)}{n!}\,(x - x_0)^n

The version with $x_0 = 0$ is called the Maclaurin series and is the one used most often. Note that the coefficients are not chosen — they are forced by the derivatives of the function.

Four expansions

Find the Maclaurin series of eˣ.

Sine and cosine go the same way, except that their derivatives at zero cycle through $0, 1, 0, -1$ (for the sine) and $1, 0, -1, 0$ (for the cosine) — which is why one expansion keeps only odd powers and the other only even ones.

function	Maclaurin expansion	$R$
$e^x$	$1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dots$	$\infty$
$\sin x$	$x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dots$	$\infty$
$\cos x$	$1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \dots$	$\infty$
$\ln(1+x)$	$x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \dots$	$1$

The last row differs from the rest, and not by chance: $\ln(1+x)$ does not exist for $x \leq -1$ , so no series has any business representing it there. The radius $1$ is exactly the distance from the centre of the expansion to that obstacle.

Approximating with a polynomial

Truncating the series after the term of degree $n$ gives the Taylor polynomial $T_n$ — and that is the practical payoff of the whole theory.

T_1 = x, \qquad T_3 = x - \frac{x^3}{6}, \qquad T_5 = x - \frac{x^3}{6} + \frac{x^5}{120}

The sine and three of its Maclaurin polynomials. They all set off from the same point and hold on to the curve the longer the higher their degree — and then each of them breaks away, because a polynomial cannot oscillate forever.

The drawing shows two things at once. Near zero every one of the polynomials hugs the sine, and that is the content of the Taylor formula. Far from zero each of them breaks away — because a polynomial runs off to infinity while the sine oscillates forever between $-1$ and $1$ . An expansion is local by nature, even when the radius of convergence is infinite.

The approximation error

How much exactly is lost by truncating? The remainder bound answers that:

\big|R_n(x)\big| \leq \frac{M\,|x - x_0|^{n+1}}{(n+1)!}

where $M$ bounds the derivative of order $n+1$ on the stretch examined. The crucial part is the factorial in the denominator: one extra term can improve the result by an order of magnitude.

Bound the error of approximating sin 0.5 by T₃ and compare it with reality.

The same computation lies behind the formula for the period of a pendulum: replacing $\sin\theta$ by $\theta$ is a truncation of the sine series after its first term, allowed only at small angles — and at angles expressed in radians, because the whole theory rests on the derivative of the sine being the cosine, which stops being true in degrees. The angle converter will turn a value in degrees into radians for you.

Exercises

The set asks four things: the radius of convergence (a number or a fraction), the coefficient of xⁿ in a Maclaurin expansion (a fraction, zero included — a missing power is a full answer), the Maclaurin polynomial of a given degree (typed as a formula, e.g. 1+x+x^2/2), and the approximate value that polynomial produces, rounded to four places.

Practice

Work through a set of exercises — they get harder as you go. At the end you'll see your score and the mistakes worth reviewing.

Exercise 1 of 8Score: 0

Radius of convergence: Σ(n = 1 … ∞) xⁿ/3ⁿ

The set draws only series with a finite radius, and it never asks for the interval of convergence: that answer is a pair of endpoints together with which of them is open, and that is not a number. Both cases are worked out in the examples above.

Common mistakes

Confusing the radius with the interval — the radius is a single number, the interval is a set with endpoints; those endpoints have to be examined separately and can give different answers.
Skipping the endpoints — the ratio test says nothing at $|x - x_0| = R$ , so leaving those two points out leaves the exercise unfinished.
Assuming that $R = \infty$ makes a Taylor polynomial good everywhere — it is the whole series that converges; its finite truncation still breaks away far from the centre.
Expanding $\ln x$ instead of $\ln(1+x)$ — the logarithm has no value at zero, so it is the shifted version that is expanded; this is not a cosmetic choice of notation.
Losing the factorial in the denominator — the coefficient is $f^{(n)}(x_0)/n!$ , not the derivative alone; at $n = 5$ that is a factor of 120.
Approximating far from the centre of the expansion — the error grows like $|x - x_0|^{n+1}$ , so doubling the distance can multiply it several dozen times.

Formula card

Topic: Power and Taylor series

Power series
$\sum_{n=0}^{\infty} a_n (x - x_0)^n$
a function of its argument x
Radius of convergence
$R = \lim_{n \to \infty} \left|\frac{a_n}{a_{n+1}}\right|$
convergent for |x − x_0| < R
Taylor series
$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(x_0)}{n!}\,(x - x_0)^n$
the coefficients are derivatives at x_0
Maclaurin series
$f(x) = f(0) + f'(0)\,x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots$
the Taylor series about zero
Expansion of eˣ
$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$
convergent for every x
Expansion of sine
$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$
odd powers only
Expansion of cosine
$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$
even powers only
Expansion of the logarithm
$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$
only for −1 < x ≤ 1
Approximation error
$\big|R_n(x)\big| \leq \frac{M\,|x - x_0|^{n+1}}{(n+1)!}$
M bounds the derivative of order n+1

The sine and three of its Maclaurin polynomials. Each one stays with the curve longer than the last: the line y = x only near zero, the degree-5 polynomial across the whole visible hump.

The interval of convergence of Σ xⁿ/n. The radius is 1, so inside (−1; 1) the series converges for certain; each endpoint is settled separately and they give different answers.

Frequently asked questions

A luminous parabola and a straight line crossing above a faint coordinate grid, the vertex flaring like a star, against a cosmic nebula in violet and magenta

Functions and analysis

Why a function's graph can never double back

The whole definition of a function comes down to one word: uniqueness. One argument, one value — and not a single one more. That seemingly minor condition decides why a circle is not the graph of a function, why a line and a parabola describe so much of the world, and why four datasets with identical summary statistics can look nothing alike.

12 min