Lagrange Error Bound

Lagrange Error Bound — The Formula

Part 1 of 7 — Understanding the Remainder

Taylor's Theorem with Remainder

If $f$ has $(n+1)$ continuous derivatives on an interval containing $c$ and $x$, then:

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

where the Lagrange remainder (error) satisfies:

$\boxed{|R_n(x)| = |f(x) - T_n(x)| \le \frac{M \cdot |x - c|^{n+1}}{(n+1)!}}$

and $M = \max_{t \in I} |f^{(n+1)}(t)|$ on the interval $I$ between $c$ and $x$.

Breaking Down Each Piece

AP Tip: Finding $M$ is where most students struggle. You need to bound $|f^{(n+1)}(t)|$ for all $t$ between $c$ and $x$ — not just at the endpoints.

Finding $M$: The Critical Step

Strategy 1: Direct bound (trig functions)

For $\sin x$ and $\cos x$: ALL derivatives are $\pm \sin$ or $\pm \cos$, so $|f^{(n+1)}(t)| \le 1$ for all $t$.

$M = 1 \text{ for sine and cosine — always!}$

Strategy 2: Monotone bound ($e^x$)

$f^{(n+1)}(x) = e^x$ is increasing, so $M = e^{x_0}$ where $x_0$ is the endpoint farther from center.

For crude bounds: $e < 3$, so use $M = 3$ when $x \le 1$.

Strategy 3: Given information

AP problems often state: "Let $|f^{(n+1)}(t)| \le K$ on $[a, b]$..."

Example

Bound the error of $T_4(x)$ for $\cos x$ at $x = 0.5$:

$|R_4(0.5)| \le \frac{1 \cdot (0.5)^5}{5!} = \frac{1/32}{120} = \frac{1}{3840} \approx 0.000260$

Lagrange Basics

Finding M Practice

Error Bound Computation

Summary

• Lagrange Error: $|R_n(x)| \le M|x-c|^{n+1}/(n+1)!$
• $M$ = max of $|f^{(n+1)}|$ on the interval between $c$ and $x$
• For $\sin/\cos$: $M = 1$ always
• For $e^x$: $M = e^{\max|x|}$ (or crude bound like $3$)

Next: Part 2 — Finding $n$ for a Given Accuracy.

Finding n for Desired Accuracy

Part 2 of 7 — How Many Terms Do You Need?

The Key Question

AP problems often ask: "How many terms of the Taylor series are needed to approximate $f(x)$ within $\varepsilon$?"

Method: Solve $\frac{M \cdot |x - c|^{n+1}}{(n+1)!} < \varepsilon$ for $n$.

Worked Example: $\cos(0.5)$ to Within $10^{-6}$

Using $T_n(0.5)$ centered at $c = 0$. Since $M = 1$ for cosine:

$\frac{(0.5)^{n+1}}{(n+1)!} < 10^{-6}$

So we need at least $T_8$ (though for cosine, only even-powered terms are nonzero, so effectively $5$ nonzero terms).

Key Fact: For trig/exponential functions, the factorial in the denominator eventually dominates any fixed $|x - c|^{n+1}$, guaranteeing convergence.

Systematic Approach for $e^x$

Approximate $e^1$ using $T_n(1)$ centered at $0$ to within $10^{-4}$.

$M = e^1 < 3$, so: $\frac{3 \cdot 1^{n+1}}{(n+1)!} < 10^{-4}$, i.e., $(n+1)! > 30000$.

So $T_7(1)$ approximates $e$ to within $10^{-4}$.

Special Case: Alternating Series

If the series alternates and satisfies the conditions of the AST, the alternating series error bound is tighter:

$|R_n| \le |a_{n+1}| \quad \text{(first omitted term)}$

AP Tip: On the AP exam, if the series alternates, the AST error bound is usually simpler. Use Lagrange when the series does NOT alternate or when explicitly asked.

Determining Sufficient n

Term Count Practice

Finding n

Summary

• To find $n$: solve $M|x-c|^{n+1}/(n+1)! < \varepsilon$
• Build a table — try different $n$ until the bound is small enough
• For alternating series, the AST error bound may require fewer terms
• The factorial always eventually dominates, so the series converges

Next: Part 3 — Bounding Derivatives Strategically.

Bounding Derivatives Strategically

Part 3 of 7 — Finding $M$ for Different Functions

The Core Challenge

The Lagrange bound $|R_n(x)| \le M|x-c|^{n+1}/(n+1)!$ requires:

$M = \max_{t \in I} |f^{(n+1)}(t)|$

This is straightforward for $\sin, \cos, e^x$. For other functions, you need strategy.

Function-by-Function Guide

Key Fact: On the AP exam, complicated derivatives are often given to you. You just plug into the formula.

Example 1: $\ln(1+x)$ at $c = 0$, $x = 0.5$, $n = 3$

$f(x) = \ln(1+x)$. Derivatives:

• $f'(x) = (1+x)^{-1}$
• $f''(x) = -(1+x)^{-2}$
• $f'''(x) = 2(1+x)^{-3}$
• $f^{(4)}(x) = -6(1+x)^{-4}$

$|f^{(4)}(t)| = \frac{6}{(1+t)^4}$

On $[0, 0.5]$: decreasing, so max at $t = 0$: $M = 6$.

$|R_3(0.5)| \le \frac{6(0.5)^4}{4!} = \frac{6 \cdot 0.0625}{24} = \frac{0.375}{24} = 0.015625$

Example 2: When $M$ is Given

"$f$ has derivatives of all orders. It is known that $|f^{(5)}(t)| \le 12$ for all $t$ in $[2, 3]$."

$|R_4(3)| \le \frac{12 \cdot |3 - 2|^5}{5!} = \frac{12}{120} = 0.1$

AP Tip: When the bound on a derivative is stated in an FRQ, that IS the value of $M$. Don't second-guess it.

Bounding Strategies

M-Value Practice

Computing the Bound

Summary

• For $\sin/\cos$: $M = 1$
• For $e^x$: use monotonicity to find max on interval
• For quotient-type derivatives: max where denominator is smallest
• When AP provides the bound, just plug in

Next: Part 4 — Lagrange vs. AST Error Bounds.

Lagrange vs. AST Error Bounds

Part 4 of 7 — Choosing the Right Error Bound

Two Error Bound Tools

When to Use Each

$\boxed{\text{Alternating series} \Rightarrow \text{AST bound (simpler)}}$ $\boxed{\text{Non-alternating or "Use Lagrange"} \Rightarrow \text{Lagrange bound}}$

Key Fact: Even when a series alternates, the AP may say "Use the Lagrange error bound" — then you MUST use Lagrange, not AST.

Side-by-Side Comparison

Approximate $\cos(0.5)$ using $T_4(0.5)$ centered at $0$.

$\cos x = 1 - x^2/2 + x^4/24 - x^6/720 + \cdots$

AST Bound: First omitted term: $|a_5| = (0.5)^6/720 \approx 2.17 \times 10^{-5}$

Actually, $T_4$ includes terms through $x^4$. The next nonzero term is $-x^6/720$: $|\text{error}| \le (0.5)^6/720 = 1/46080 \approx 2.17 \times 10^{-5}$

Lagrange Bound: $|R_4(0.5)| \le M(0.5)^5/5! = 1 \cdot (0.5)^5/120 = 1/3840 \approx 2.60 \times 10^{-4}$

Comparison: AST gives $2.17 \times 10^{-5}$; Lagrange gives $2.60 \times 10^{-4}$.

The AST bound is about 12× tighter because it accounts for the fact that the $x^5$ coefficient is $0$ in the cosine series, while Lagrange does not.

AP Tip: When both apply, the AST bound is usually better — but read the problem carefully. "Use Lagrange" means Lagrange, even if AST is tighter.

Choosing the Right Bound

Bound Selection Practice

Bound Comparison

Summary

• AST bound: simpler, tighter, only for alternating series
• Lagrange bound: universal, requires $M$
• AP exam: use whichever is specified; default to AST when series alternates
• When both apply, AST ≤ Lagrange (AST never overestimates worse)

Next: Part 5 — AP Exam FRQ Strategies.

AP Exam FRQ Strategies

Part 5 of 7 — Earning Full Credit

How Lagrange Appears on the AP Exam

Common FRQ Patterns:

1. "Use the Lagrange error bound to show that..."

   • Given: $f$, some derivatives, $c$, $x$
   • Write the formula, identify $M$, compute, compare to target

2. "Show the approximation is within $\varepsilon$ of the actual value"

   • Same setup but you must conclude with an inequality

3. "Find the minimum degree $n$ such that..."

   • Try successive $n$ values in the bound

Template for Full Credit

Step 1: State the formula: $|R_n(x)| \le M|x - c|^{n+1}/(n+1)!$

Step 2: Identify each component:

• $n =$ __, $c =$ __, $x =$ __

Step 3: Find or state $M$:

• "Since $|f^{(n+1)}(t)| \le M$ on $[c, x]$..."

Step 4: Compute and conclude:

• "$|R_n(x)| \le \text{value} < \varepsilon$. Therefore..."

AP Tip: The graders look for the formula stated, $M$ identified with justification, and the final inequality. Missing any one of these costs a point.

Model FRQ Response

"Let $f(x) = \sin x$. Use the Lagrange error bound to show that $T_5(1)$ approximates $\sin(1)$ within $0.002$."

Response:

By the Lagrange error bound:

$|R_5(1)| \le \frac{M \cdot |1 - 0|^6}{6!}$

Since all derivatives of $\sin x$ satisfy $|f^{(k)}(t)| \le 1$ for all $t$, we have $M = 1$.

$|R_5(1)| \le \frac{1 \cdot 1^6}{720} = \frac{1}{720} \approx 0.00139$

Since $0.00139 < 0.002$, the approximation $T_5(1)$ is within $0.002$ of $\sin(1)$. $\square$

Common Mistakes That Lose Points

FRQ Strategy

FRQ Setup Practice

FRQ Computation

Summary

• State formula → identify $M$ → compute → conclude with inequality
• Use the $M$ given in the problem when provided
• $(n+1)!$ not $n!$ — get the formula exactly right
• Always include the final comparison: "bound $< \varepsilon$, therefore..."

Next: Part 6 — Problem-Solving Workshop.

Problem-Solving Workshop

Part 6 of 7 — Mixed Practice

Work through these problems combining all Lagrange error bound skills.

Workshop Problems — Multiple Choice

Workshop — Strategy Selection

Workshop — Computation

Workshop Summary

• Always identify: alternating → AST; non-alternating → Lagrange
• For Lagrange: find $M$ using the function's derivative behavior
• $M = 1$ for trig, $M = e^{\text{endpoint}}$ (or crude bound) for $e^x$
• Final step: state the inequality explicitly

Next: Part 7 — Comprehensive Review.

Comprehensive Review

Part 7 of 7 — Lagrange Error Bound Mastery

Complete Formula Reference

$\boxed{|R_n(x)| \le \frac{M \cdot |x - c|^{n+1}}{(n+1)!}, \quad M = \max_{t \in [c,x]} |f^{(n+1)}(t)|}$

Decision Flowchart

1. Is the series alternating?

   • Yes → Use AST bound (unless told otherwise)
   • No → Use Lagrange

2. Is $M$ given in the problem?

   • Yes → Use it directly
   • No → Find max of $|f^{(n+1)}|$ on the interval

3. Plug into the formula and conclude.

Quick $M$ Reference

Review — Conceptual

Review — Computation

Review — Identify the Error

Review — Final Challenge

Topic Complete!

You've mastered the Lagrange Error Bound:

• The formula and what each part means
• Finding $M$ for common functions
• Determining how many terms you need
• Lagrange vs. AST: when to use each
• AP FRQ response format for full credit

Key Takeaway: The Lagrange Error Bound is one of the most tested BC topics. Master the formula, practice finding $M$, and always conclude with an explicit inequality.