Taylor Polynomials, Taylor Series, and Power Series

Advanced ≈ 50 min Chapter 24 8 exercises
Why this mattersNear any point, a smooth curve looks like its tangent line — and better, like a polynomial. That single fact explains error estimates, numerical methods, and why your calculator can answer harder questions than you can integrate.

1 · Linearization, revisited

Near a, f(x) ≈ f(a) + f′(a)(x−a) = L(x). The tangent line is the degree-1 Taylor polynomial. Adding the next term gives a quadratic that hugs the curve longer; adding every term gives a series — not a polynomial, but something even more powerful.

2 · Higher-degree approximations

The Taylor polynomial of degree n at a is

Pₙ(x) = f(a) + f′(a)(x−a) + f″(a)(x−a)²/2! + ⋯ + f⁽ⁿ⁾(a)(x−a)ⁿ/n!

Each term irons out one more piece of local behavior: value, slope, curvature, jerk. More terms, wider trust region (usually).

3 · Taylor’s theorem and the remainder

If f is smooth, the error after n terms is

f(x) − Pₙ(x) = f⁽ⁿ⁺¹⁾(ξ)·(x−a)ⁿ⁺¹/(n+1)! for some ξ between x and a.

This Lagrange remainder is what lets us prove an approximation good before we trust it — not assert it.

Worked example 1

Approximate e^{0.1} to 4 decimals. Use P₂(x) = 1 + x + x²/2 at x = 0.1: gives 1.10517. Remainder ≤ e^{ξ}(0.1)³/6 ≤ e^{0.1}(0.001)/6 ≈ 1.84×10⁻⁴. So the true value is 1.10517 ± 0.00018 — a guarantee, not a hope.

4 · Maclaurin series — Taylors at 0

Set a = 0 and you get the famous five:

eˣ = Σ xⁿ/n! · sin x = Σ (−1)ⁿx^{2n+1}/(2n+1)! · cos x = Σ (−1)ⁿx^{2n}/(2n)! · 1/(1−x) = Σ xⁿ (|x|<1) · ln(1+x) = Σ (−1)^{n−1}xⁿ/n.

Each is a polynomial machine that sums to a transcendental function inside its radius.

5 · Power series in general

A power series Σ cₙ(x−a)ⁿ converges inside an interval centered at a, diverges outside, and needs checking at the endpoints. Differentiating or integrating term-by-term inside the interval stays valid — a calculus superpower we'll exploit in Chapter 26.

Watch outNot every smooth function equals its Taylor series. The classic villain f(x) = e^{−1/x²} (with f(0)=0) has every derivative 0 at 0, so its Taylor series is all zeros — while the function isn't. Convergence does not imply equality.
Chapter formula sheet L(x) = f(a) + f′(a)(x−a)
Pₙ(x) = Σ_{k=0}^{n} f⁽ᵏ⁾(a)(x−a)ᵏ/k!
Rem. |f(x) − Pₙ(x)| ≤ M·|x−a|ⁿ⁺¹/(n+1)!  (M = max |f⁽ⁿ⁺¹⁾|)
Maclaurin: sin x ≈ x − x³/6 + x⁵/120  ·  cos x ≈ 1 − x²/2 + x⁴/24
eˣ ≈ 1 + x + x²/2 + x³/6  ·  ln(1+x) ≈ x − x²/2 + x³/3  ·  1/(1−x) ≈ 1 + x + x² + ⋯

Common mistakes

  • Forgetting the /k! — the factorial is not optional decoration, it’s the slope correction per order.
  • Mismatching center: an a=0 series cannot approximate well around a=2 unless you shift.
  • Claiming series = function everywhere, even though endpoints/radius may forbid it.
  • Confusing linearization (degree-1 Taylor) with differentials (dy) — related but not identical.
  • Using eˣ’s series where |x|≫1; the polynomials may oscillate wildly outside a modest interval.

✎ Practice — marked instantly

The degree-1 Taylor polynomial of f at a is:
  • A. f(a) + f′(a)(x−a)
  • B. f′(a) + f(a)(x−a)
  • C. f(a) + f′(a)
  • D. f(x) + f′(a)(x−a)

Multiple choice · interactive marking is enabled when JavaScript loads.

Degree-2 Maclaurin approximation of cos x at x = 0.5 is 1 − x²/2. Compute it at x = 0.5 (give 4 decimals, accept within 0.01).

Numeric answer · interactive marking is enabled when JavaScript loads.

The Maclaurin series of eˣ starts 1 + x + __. Give the next term in the form c·x².

Short answer · interactive marking is enabled when JavaScript loads.

Taylor’s theorem says the error after n terms is:
  • A. f⁽ⁿ⁺¹⁾(ξ)(x−a)ⁿ⁺¹/(n+1)!
  • B. fⁿ(x)/(n−1)!
  • C. (x−a)ⁿ only
  • D. zero for every smooth f

Multiple choice · interactive marking is enabled when JavaScript loads.

Using sin x ≈ x − x³/6, estimate sin(0.2). Accept 0.005.

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The geometric power series Σ xⁿ equals 1/(1−x) exactly when:
  • A. |x| < 1
  • B. all real x
  • C. x < −1
  • D. only x = 0

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Maclaurin polynomial of degree 1 for ln(1+x) is:

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A smooth function with all derivatives 0 at a point can still differ from its Taylor series there — example:
  • A. f(x) = e^{−1/x²}, f(0)=0
  • B. f(x) = x⁵
  • C. f(x) = eˣ
  • D. f(x) = sin x

Multiple choice · interactive marking is enabled when JavaScript loads.