🎯⭐ INTERACTIVE LESSON

Lagrange Error Bound

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Lagrange Error Bound - Complete Interactive Lesson

Part 1: Error Bound Formula

Lagrange Error Bound

Part 1 of 7 — The Formula

Taylor's Theorem with Remainder

f(x)=Pn(x)+Rn(x)f(x) = P_n(x) + R_n(x)

Lagrange Error Bound

|R_n(x)| leq rac{M}{(n+1)!}|x - a|^{n+1}

where M=maxf(n+1)(c)M = max|f^{(n+1)}(c)| for cc between aa and xx.

This tells you: how good is your Taylor polynomial approximation?

Lagrange Error 🎯

Key Takeaways — Part 1

RnMxan+1(n+1)!|R_n| \leq \frac{M|x-a|^{n+1}}{(n+1)!} where MM bounds f(n+1)|f^{(n+1)}|.

Part 2: Finding Maximum Error

Lagrange Error Bound

Part 2 of 7 — Finding MM

Strategy for Bounding MM

M=maxcextbetweenaextandxf(n+1)(c)M = max_{c ext{ between } a ext{ and } x} |f^{(n+1)}(c)|

For common functions:

FunctionAll derivatives bounded by
sinxsin x, cosxcos xM=1M = 1 (always!)
exe^x on [0,b][0, b]M=ebM = e^b
lnxln x near aaNeed careful analysis

Finding MM 🎯

Key Takeaways — Part 2

sinx\sin x and cosx\cos x always have M=1M = 1. This makes error bounds very clean.

Part 3: Alternating Series Error

Lagrange Error Bound

Part 3 of 7 — How Many Terms Do We Need?

Determining Degree for Desired Accuracy

"How many terms of exe^x centered at 00 are needed for Rn(1)<0.001|R_n(1)| < 0.001?"

Need: rac{M}{(n+1)!} < 0.001 where Mleq3M leq 3 (bound for ece^c on [0,1][0,1]).

rac{3}{(n+1)!} < 0.001

(n+1)!>3000(n+1)! > 3000

7!=5040>30007! = 5040 > 3000

So n=6n = 6 (use P6P_6, i.e., 7 terms).

How Many Terms? 🎯

Key Takeaways — Part 3

Solve (n+1)!>M/ϵ(n+1)! > M/\epsilon to find the required degree.

Part 4: Choosing Degree

Lagrange Error

Part 4 of 7 — Alternating Series Error vs Lagrange

When to Use Which?

ScenarioUse
Alternating seriesAS Error Bound ($
General Taylor polynomialLagrange Error Bound
Either appliesAS bound is usually tighter!

Key Insight

For alternating Taylor series (like sinx\sin x, cosx\cos x, exe^{-x}), the alternating series error bound is often easier and tighter.

AS vs Lagrange 🎯

Key Takeaways — Part 4

AS bound is simpler for alternating Taylor series. Lagrange works for all Taylor polys.

Part 5: Applications

Lagrange Error

Part 5 of 7 — AP FRQ Practice

Typical AP Question Format

"Let ff be a function with f(4)(x)6|f^{(4)}(x)| \leq 6 for all xx in [1,1][-1, 1]. Show that the error of P3(x)P_3(x) at x=0.5x = 0.5 is less than 0.020.02."

Solution: R3(0.5)64!(0.5)4=624116=1640.0156<0.02|R_3(0.5)| \leq \frac{6}{4!}(0.5)^4 = \frac{6}{24} \cdot \frac{1}{16} = \frac{1}{64} \approx 0.0156 < 0.02

AP Practice 🎯

Key Takeaways — Part 5

Show your formula, substitute values, simplify. State conclusion clearly.

Part 6: Problem-Solving Workshop

Lagrange Error

Part 6 of 7 — Practice Workshop

Error Bound Practice 🎯

Workshop Complete!

Part 7: Review & Applications

Lagrange Error Bound — Review

Part 7 of 7 — Final Assessment

Final Assessment 🎯

Lagrange Error Bound — Complete! ✅