Taylor and Power Series

When the Mathematics, Surveying and Transportation Engineering (MSTE) paper asks you to estimate e^0.2, cos(0.3), or a small-angle offset without leaning on a calculator button, it is really testing whether you can build a function out of powers of x and stop after a few terms with a known error. That is exactly what a Taylor series does: it rebuilds a smooth function from its value and its derivatives at one point. This lesson covers the Taylor and Maclaurin formulas, the standard series worth memorizing, how far from the center they are allowed to be used (radius and interval of convergence), and two clean ways to bound the truncation error so your hand estimate comes with a guarantee.

From tangent line to Taylor polynomial

A tangent line matches a curve in value and slope at one point. A Taylor polynomial does the same with more derivatives, so it hugs the curve over a wider stretch. The Taylor series of a function f centered at x = a is:

f(x) = f(a) + f'(a)(x - a) + f''(a)/2! (x - a)^2 + f'''(a)/3! (x - a)^3 + ...

so its general term is f^(n)(a)/n! times (x - a)^n, summed from n = 0 to infinity. When the center is a = 0 the series is called a Maclaurin series, and every (x - a) becomes just x:

f(x) = f(0) + f'(0) x + f''(0)/2! x^2 + f'''(0)/3! x^3 + ...

The two ideas that make this useful on the board are that the first few terms already approximate f well near the center, and that a truncated series comes with an error you can bound.

Worked example (derive a Maclaurin series): Build the Maclaurin series of f(x) = cos(x). List the derivatives and read them off at x = 0: f = cos so f(0) = 1; f' = -sin so f'(0) = 0; f'' = -cos so f''(0) = -1; f''' = sin so f'''(0) = 0; f'''' = cos so f''''(0) = 1, and the pattern repeats every four steps. Dropping the zero terms leaves cos(x) = 1 - x^2/2! + x^4/4! - x^6/6! + ... = 1 - x^2/2 + x^4/24 - x^6/720 + ..., an alternating series in even powers only.

The standard series worth memorizing

Five Maclaurin series cover almost every board item. Memorize the first three (which converge for every x) and know how the last two are built, because the board loves to derive one series from another.

Function	Maclaurin series	Converges for
e^x	1 + x + x^2/2! + x^3/3! + x^4/4! + ...	all real x
sin(x)	x - x^3/3! + x^5/5! - x^7/7! + ...	all real x
cos(x)	1 - x^2/2! + x^4/4! - x^6/6! + ...	all real x
ln(1 + x)	x - x^2/2 + x^3/3 - x^4/4 + ...	-1 < x <= 1
(1 + x)^p	1 + p x + p(p - 1)/2! x^2 + ...	|x| < 1

Two derivations are worth knowing on sight. The sine and cosine series are the odd-power and even-power halves of the exponential pattern, which is why sin(x) carries only x, x^3, x^5 and cos(x) carries only 1, x^2, x^4. The logarithm series comes from integrating the geometric series term by term.

Worked example (derive ln(1 + x)): Start from the geometric series 1/(1 + x) = 1 - x + x^2 - x^3 + ..., valid for |x| < 1. Integrate both sides from 0 to x. The left side integrates to ln(1 + x), and the right side integrates term by term to x - x^2/2 + x^3/3 - x^4/4 + .... So ln(1 + x) = x - x^2/2 + x^3/3 - x^4/4 + ..., which also converges at the right endpoint x = 1 to give the famous ln(2) = 1 - 1/2 + 1/3 - 1/4 + ....

Radius and interval of convergence

A power series only equals its function inside a symmetric window around the center. The half-width of that window is the radius of convergence R, and the fastest way to find it is the ratio test: form L = limit as n goes to infinity of |a_(n+1) / a_n|, and the series converges wherever L < 1. Solving L < 1 for x gives |x - a| < R.

The catch is the two endpoints |x - a| = R. There the ratio test gives L = 1 and says nothing, so each endpoint must be tested separately with another convergence test. That is what turns a radius into an interval of convergence, which may be open, closed, or half-open.

Worked example (radius and interval): Find where the power series sum from n = 1 of x^n / n converges. Here a_n = x^n / n, so |a_(n+1) / a_n| = |x^(n+1) / (n + 1)| divided by |x^n / n| = |x| times n / (n + 1). The limit as n goes to infinity is |x| times 1 = |x|, so the series converges when |x| < 1, giving R = 1. Now test the endpoints. At x = 1 the series is the harmonic series 1 + 1/2 + 1/3 + ..., which diverges. At x = -1 it is the alternating harmonic series -1 + 1/2 - 1/3 + ..., which converges. So the interval of convergence is -1 <= x < 1, written [-1, 1).

Truncation error and the Lagrange remainder

Stopping a Taylor series after the term in (x - a)^n leaves the Taylor polynomial P_n, and the leftover is the remainder R_n = f(x) - P_n(x). Taylor's theorem gives that remainder an exact form, the Lagrange remainder:

R_n(x) = f^(n+1)(c) / (n + 1)! times (x - a)^(n+1), for some c between a and x.

You do not know c, so you bound the size of f^(n+1) over the interval and get a guaranteed ceiling on the error. When the series alternates and its terms shrink in magnitude, there is an even simpler bound: the truncation error is at most the magnitude of the first omitted term. Both bounds turn a hand estimate into a number you can trust.

Worked example (Lagrange error bound): Approximate e^0.2 with the quadratic Taylor polynomial P2(x) = 1 + x + x^2/2. Then P2(0.2) = 1 + 0.2 + (0.2)^2 / 2 = 1 + 0.2 + 0.04/2 = 1 + 0.2 + 0.02 = 1.22. The remainder is R2 = f'''(c) / 3! times x^3 = e^c / 6 times (0.2)^3, for some c in (0, 0.2). Since e^c grows with c, e^c < e^0.2, and a safe over-estimate is e^0.2 < 1.25. So |R2| < 1.25 / 6 times 0.008 = 0.010 / 6 = 0.00167. The estimate is e^0.2 approximately 1.22 with error under 0.00167, and indeed the true value 1.2214 differs from 1.22 by 0.0014.

Approximating a value to a stated number of terms

Board items usually fix the number of terms and ask for the value plus an error bound. The recipe is the same every time: write the series, plug in x, add the stated terms, then quote the first omitted term (for an alternating series) or a Lagrange bound as the guaranteed error.

Worked example (stated terms with error bound): Estimate cos(0.3) using the terms up to x^4, then bound the error. cos(0.3) approximately 1 - (0.3)^2 / 2 + (0.3)^4 / 24 = 1 - 0.09/2 + 0.0081/24 = 1 - 0.045 + 0.0003375 = 0.9553375, so cos(0.3) approximately 0.955338 (a pure number, since the cosine of an angle in radians is dimensionless). The series alternates, so the error is at most the first omitted term x^6 / 6! = (0.3)^6 / 720 = 0.000729 / 720 = 0.00000101. The true value 0.9553365 sits well inside that bound.

Worked example (small-angle offset in meters): A surveyor turns a 50 m sight line through a small angle theta = 0.02 rad and needs the perpendicular offset L times sin(theta). Using two terms of the sine series, sin(0.02) approximately 0.02 - (0.02)^3 / 6 = 0.02 - 0.000008 / 6 = 0.02 - 0.00000133 = 0.01999867. Then the offset is 50 m times 0.01999867 = 0.99993 m. The crude linear estimate L times theta = 50 m times 0.02 = 1.00000 m overshoots by about 0.07 mm, which the cubic term corrects.

Worked example (series in an applied correction, seconds): The period of a pendulum at a moderate swing amplitude theta0 uses the series T = T0 times [1 + theta0^2 / 16 + ...], where T0 = 2 pi times sqrt(L / g) is the small-swing period. For L = 1 m and g = 9.81 m/s^2, T0 = 2 pi times sqrt(1 / 9.81) = 2 pi times 0.31928 = 2.00607 s. At an amplitude theta0 = 0.3 rad the correction factor is 1 + (0.3)^2 / 16 = 1 + 0.09/16 = 1.005625, so T = 2.00607 s times 1.005625 = 2.017 s, about 11 ms longer than the small-swing value.

Exam-day strategy

• Name the center first: an expansion about x = 0 is a Maclaurin series, and every (x - a) collapses to x, which is where most board items live.
• Memorize e^x, sin(x), and cos(x) cold and remember they converge for every real x; sin(x) carries only odd powers and cos(x) only even powers.
• Build ln(1 + x) by integrating the geometric series 1/(1 + x) term by term, and read the binomial series (1 + x)^p straight off the pattern 1 + p x + p(p - 1)/2! x^2.
• Get the radius R from the ratio test (converges when the limit of |a_(n+1) / a_n| is < 1), then test the two endpoints separately, because the ratio test gives no verdict when that limit equals 1.
• For an alternating series with shrinking terms, bound the truncation error by the magnitude of the first omitted term; it is the fastest guarantee on the paper.
• For a non-alternating series, fall back on the Lagrange remainder f^(n+1)(c)/(n + 1)! times (x - a)^(n+1), bounding the derivative over the interval to get a ceiling.
• Keep x small: the closer x is to the center, the fewer terms you need, so a two- or three-term estimate of e^x, sin(x), or cos(x) is usually accurate enough for the choices offered.