Short on time? This page is the recap. For the reasoning, the number-line diagram and worked examples, use the full study guide. BC-only material. You need Taylor polynomials (10.11) and the alternating series error bound (10.10); see the Calculus BC roadmap.
Recap
- The error of a Taylor estimate is f(x) − Pₙ(x).
- Lagrange error bound: if |f⁽ⁿ⁺¹⁾(z)| ≤ M for all z between a and x, then |f(x) − Pₙ(x)| ≤ M · |x − a|ⁿ⁺¹ / (n + 1)!
- It is “the next Taylor term with the unknown derivative replaced by its worst case”.
- The true value lies in [Pₙ(x) − bound, Pₙ(x) + bound].
- The bound is a guarantee, not the actual error.
Key relationships
| Step | What to do | Watch for |
|---|---|---|
| Degree | Identify n, the degree of the polynomial used | If the next coefficient is 0, Pₙ = Pₙ₊₁ and you may use the higher n |
| Derivative | Find f⁽ⁿ⁺¹⁾ | The next derivative, not f⁽ⁿ⁾ |
| M | Largest size of f⁽ⁿ⁺¹⁾ between a and x, or anything bigger | Check both ends; sin and cos give M = 1; for eᶻ round up (e.g. 3) |
| Bound | M times the distance to the power n + 1, divided by (n + 1)! | (n + 1)!, not n! |
| Conclusion | Write “bound < target” in words or symbols | The comparison earns the point |
| Degree needed | Increase n until the bound is below the target | Bound guarantees; actual error may already be smaller |
| Alternating option | Error at most the first omitted term | Only if terms alternate, shrink and tend to 0 |
Assumptions
- f has an (n + 1)th derivative on the interval between a and x.
- Trig functions use radians.
Mistakes to avoid
- Using f⁽ⁿ⁾ or n! instead of f⁽ⁿ⁺¹⁾ and (n + 1)!.
- Taking M at the centre when the derivative is bigger at the other end.
- Treating the bound as the exact error.
- Writing the bound but not the comparison sentence.
- Using the alternating series bound when the terms do not alternate (for example eˣ at a positive x).
Quick self-check
- P₂ is used about x = a, the size of f‴ is at most 4 between a and x, and x − a = 0.3. What is the Lagrange bound? (4(0.3)³/3! = 0.018)
- P₂(x) = 1 + x + x²/2 estimates e^0.2. Using M = 3, what is the bound? (3(0.2)³/6 = 0.004)
- Why is M = 3 allowed in question 2? (For 0 ≤ z ≤ 0.2, eᶻ ≤ e^0.2 < e < 3.)
Next: practice questions.