Chapter 8 — Lagrange Error Bound

Why this chapter matters: this is the most challenging topic in BC series — but every problem follows the same template. Memorize the formula and the 3-step method, and you'll handle any AP error problem.

By the end of this chapter you will:

1. State the Lagrange error bound formula
2. Compute an upper bound on the truncation error of a Taylor polynomial
3. Use the error bound to prove a Taylor series equals the original $f$
4. Solve "how many terms?" inverse problems on the AP

8.1 Why Lagrange?

Recall Chapter 6: the Taylor polynomial $P_n(x)$ approximates $f(x)$. The natural question is: how big is the error?

Lagrange gives both an exact formula (with one unknown $c$) and a practical upper bound.

🎵 Audio 1 — Why we need this — audio/ch8_01_motivation.mp3

8.2 The Lagrange Remainder Formula

Taylor's theorem (with remainder): if $f$ is $(n+1)$-times differentiable on an interval $I$ containing $a$ and $x$, then

$$f(x) = P_n(x) + R_n(x)$$

where the remainder is

$$R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x - a)^{n+1}$$

for some $c$ between $a$ and $x$ (existence guaranteed; exact location unknown).

Lagrange error bound (the practical form):

$$\boxed{\lvert R_n(x)\rvert \le \frac{M}{(n+1)!}\lvert x - a\rvert^{n+1}}$$

where $M$ is an upper bound for $\lvert f^{(n+1)}(t)\rvert$ on the interval between $a$ and $x$.

🎵 Audio 2 — Formula breakdown — audio/ch8_02_formula.mp3

🔑 Three things to remember:

• $M$ bounds the $(n+1)$-th derivative — not the $n$-th
• All three slots use $n+1$ (factorial, derivative order, distance power) — easy to off-by-one
• The closer $x$ is to $a$, the smaller the error

8.3 The 3-Step Method

Step 1: identify $n$, $x_0$, $a$

The problem will say something like "use the degree-$n$ Taylor polynomial to estimate $f(x_0)$"; that gives you $n$, $x_0$, $a$.

Step 2: find $M$ — the key step

Find an upper bound for $\lvert f^{(n+1)}(t)\rvert$ on the interval $[a, x_0]$ (or $[x_0, a]$).

Practical shortcuts (memorize):

Step 3: plug into the formula

$$\lvert R_n(x_0)\rvert \le \frac{M}{(n+1)!}\lvert x_0 - a\rvert^{n+1}$$

🎵 Audio 3 — 3-step method — audio/ch8_03_three_steps.mp3

8.4 Example 1 — Estimate $\sin(0.5)$

Problem: use the degree-5 Maclaurin polynomial of $\sin x$ to estimate $\sin(0.5)$. Bound the error.

Step 1: $n = 5$, $x_0 = 0.5$, $a = 0$.

Step 2: $f^{(6)}(t) = -\sin t$, so $\lvert f^{(6)}(t)\rvert \le 1 = M$.

Step 3:
$$\lvert R_5(0.5)\rvert \le \frac{1}{6!}(0.5)^6 = \frac{1/64}{720} \approx 2.17 \times 10^{-5}$$

Estimate: $P_5(0.5) = 0.5 - \frac{0.5^3}{6} + \frac{0.5^5}{120} \approx 0.4794271$.

True $\sin(0.5) \approx 0.4794255$, actual error $\approx 1.6 \times 10^{-6}$ — well within the bound. ✓

🎵 Audio 4 — sin(0.5) walkthrough — audio/ch8_04_sin_example.mp3

8.5 Example 2 — Inverse Problem (AP-favorite)

Problem: estimate $e^{0.5}$ using the Maclaurin polynomial of $e^x$. What's the smallest degree $n$ that guarantees error below $0.001$?

Approach: find smallest $n$ with $\frac{M \cdot 0.5^{n+1}}{(n+1)!} < 0.001$.

Step 1: $f^{(n+1)}(t) = e^t \le e^{0.5} \le \sqrt{e} < 2$. Take $M = 2$.

Step 2: solve $\frac{2 \cdot 0.5^{n+1}}{(n+1)!} < 0.001$ by trial:

• $n = 4$: $\frac{2 \cdot 0.5^5}{5!} = \frac{1}{1920} \approx 5.2 \times 10^{-4}$ ✓
• $n = 3$: $\frac{2 \cdot 0.5^4}{4!} = \frac{1}{192} \approx 5.2 \times 10^{-3}$ ✗

Smallest degree: $n = 4$.

🎵 Audio 5 — Inverse / smallest-degree problem — audio/ch8_05_precision_search.mp3

🔑 Strategy: increment $n$ from small values until the inequality holds. Don't try to solve $(n+1)!$ inequalities algebraically.

8.6 Proving a Taylor Series Equals $f$

Theorem: if $\lim_{n\to\infty} R_n(x) = 0$ for every $x$ in an interval $I$, then the Taylor series equals $f$ on $I$:

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

Example: prove $\sum \frac{x^n}{n!} = e^x$ for all $x$.

Fix $x$. $\lvert f^{(n+1)}(t)\rvert = e^t \le e^{\lvert x\rvert} = M$ on $[0, x]$ or $[x, 0]$.

$$\lvert R_n(x)\rvert \le \frac{e^{\lvert x\rvert}}{(n+1)!}\lvert x\rvert^{n+1} \to 0$$

since $\frac{\lvert x\rvert^{n+1}}{(n+1)!} \to 0$ for any fixed $x$.

So the Taylor series converges to $e^x$ for every real $x$. ✅

🎵 Audio 6 — Proving Taylor convergence to $f$ — audio/ch8_06_convergence_proof.mp3

8.7 Comparing AST and Lagrange

🔑 Choosing: if the series alternates, AST is faster. Otherwise use Lagrange.

8.8 Mixed Example (AP-style)

Problem: $f(x) = \cos x$, $a = 0$, $P_4(x) = 1 - \frac{x^2}{2} + \frac{x^4}{24}$. Bound $\lvert R_4(0.3)\rvert$.

Step 1: $n = 4$, $x_0 = 0.3$.

Step 2: $f^{(5)}(t) = -\sin t$, so $\lvert f^{(5)}(t)\rvert \le 1 = M$.

Step 3:
$$\lvert R_4(0.3)\rvert \le \frac{1}{5!}(0.3)^5 = \frac{0.00243}{120} \approx 2.025 \times 10^{-5}$$

Estimate $P_4(0.3) = 1 - 0.045 + 0.0003375 = 0.9553375$ vs true $\cos(0.3) \approx 0.9553365$, agreement to $10^{-6}$. ✓

🎵 Audio 7 — Mixed example + recap — audio/ch8_07_examples_recap.mp3

8.9 Trap Alerts ⚠️

1. Off-by-one on $n+1$. The formula uses $f^{(n+1)}$, $(n+1)!$, and $\lvert x-a\rvert^{n+1}$ — all three.
2. $M$ bounds the $(n+1)$-th derivative, not the $n$-th.
3. $M$ is over $[a, x_0]$ only, not the entire real line.
4. $\sin, \cos$ derivatives all bounded by 1 — easiest case. $e^x$ requires $e^{\max}$.
5. AP free response wants you to show how $M$ was found — don't skip the justification.

8.10 Mnemonic

"$(n+1)!$ in the denominator, $M$ caps the next derivative, distance to the $(n+1)$-th power."

All three slots use $n+1$.

Media Inventory

End of chapter. End of the BC series deep-dive notes.