1 · The error term, properly
After Pₙ(x), the error is f(x) − Pₙ(x) = f⁽ⁿ⁺¹⁾(ξ)·(x−a)ⁿ⁺¹/(n+1)!. If we can bound |f⁽ⁿ⁺¹⁾| ≤ M on our interval, then |error| ≤ M·|x−a|ⁿ⁺¹/(n+1)! — a hard guarantee.
2 · First example — estimating e^{0.1}
Use P₂(0.1) = 1.10517. Remainder ≤ e^{0.1}(0.1)³/6 ≈ 1.84×10⁻⁴. So e^{0.1} = 1.10517 ± 0.00018. Three correct decimals, proven.
3 · Second example — a trig estimate
Estimate sin(0.3) using P₃(x) = x − x³/6: P₃(0.3) = 0.3 − 0.0045 = 0.2955. The next term is +x⁵/120 = 0.00002, so error ≤ 2.1×10⁻⁵. Answer: 0.2955 ± 0.00002.
4 · Third example — logarithms
Estimate ln(1.1) ≈ 0.1 − 0.005 + 0.00033 ≈ 0.09533. Remainder over the geometric-like alternating tail is below the first omitted term 0.1⁴/4 = 2.5×10⁻⁵. So 0.09533 ± 0.00003.
5 · Fourth example — an integral by series
∫₀^{0.5} e^{−x²}dx. Expand e^{−x²} = 1 − x² + x⁴/2 − ⋯, integrate term-by-term: 0.5 − 0.125/3 + 0.03125/10 − ⋯ ≈ 0.4615. Bounding the tail by the first omitted term keeps error < 0.0001.
Estimate ∫₀^{0.5} e^{−x²}dx by integrating the series. Expand e^{−x²} = 1 − x² + x⁴/2 − ⋯, integrate term-by-term: 0.5 − 0.125/3 + 0.03125/10 − ⋯ ≈ 0.4615. Bounding the tail by the first omitted term keeps the error under 0.0001 — a concrete instance of an estimation problem.
6 · The alternating-series shortcut
If a series alternates with terms decreasing to 0, the error is less than the first term you omitted. No M-hunting needed — often the fastest rigorous bound in the course.
Declaring “the series converges” is not an error bound. You must exhibit M and the factorial denominator, or cite the alternating test with decreasing terms. A value without a bound is a rumor, not a result.
Alternating: |f(x) − Pₙ(x)| ≤ |first omitted term|
Pick M generously (e^{|x|} ≤ e^{0.5} on |x|≤0.5, |sin| ≤ 1, |cos| ≤ 1)
Rule of thumb: 3–4 terms usually buy 4+ decimals near 0
Common mistakes
- Waving “the series converges” without bounding M — convergence ≠ enough accuracy.
- Forgetting the factorial in the remainder — off by a factor of (n+1) is a failed proof.
- Using an alternating bound when the series does not alternate — the shortcut is licensed only there.
- Confusing “error” sign: remainder can be positive or negative; the bound is on its size.