AP Calculus BC 10.12 Lagrange Error Bound Study Notes - New Syllabus
AP Calculus BC 10.12 Lagrange Error Bound Study Notes- New syllabus
AP Calculus BC 10.12 Lagrange Error Bound Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- Power series allow us to represent associated functions on an appropriate interval.
Key Concepts:
- Lagrange Error Bound
Lagrange Error Bound
Lagrange Error Bound
The Lagrange Error Bound (or remainder estimate) gives an upper bound on the error when a Taylor polynomial \( P_n(x) \) is used to approximate a function \( f(x) \) near \( x = a \). It quantifies how close the polynomial approximation is to the actual function.
Formula:
If \( f \) is \( (n+1) \) times differentiable on an interval containing \( a \) and \( x \), then the remainder term is
$ R_n(x) = \dfrac{f^{(n+1)}(c)}{(n+1)!} (x-a)^{n+1} $ for some \( c \) between \( a \) and \( x \).
The Lagrange Error Bound is found by taking a bound \( M \) on \( |f^{(n+1)}(t)| \) for all \( t \) between \( a \) and \( x \):
$ |R_n(x)| \le \dfrac{M}{(n+1)!} |x-a|^{n+1} $
Interpretation:
- \( M \) is the maximum absolute value of the \((n+1)\)th derivative over the interval.
- The bound tells us the worst-case error for approximating \( f(x) \) by \( P_n(x) \).
- Smaller \( |x-a| \) and larger \( n \) usually make the error smaller.
Purpose:
This bound is used to ensure that the approximation meets a desired accuracy. In AP Calculus, it often appears in questions asking how many terms are needed to guarantee the error is less than a given tolerance.
Alternating Series Error Bound and Taylor Polynomials
If the Taylor series for a function is an alternating series (terms alternate in sign) and the absolute value of the terms decreases to zero, then the Alternating Series Error Bound applies to the Taylor polynomial approximation.
Theorem:
Let $ S = \sum_{n=0}^{\infty} (-1)^n b_n $ be an alternating series with \( b_{n+1} \le b_n \) and \( \lim_{n\to\infty} b_n = 0 \).
If \( S_N \) is the sum of the first \( N \) terms (the \( N \)th-degree Taylor polynomial), then the remainder (error) satisfies: $ |R_N| \le b_{N+1} $
In Taylor Polynomial Context:
- The remainder \( R_N(x) \) is the difference between the actual function value \( f(x) \) and the \( N \)th-degree Taylor polynomial \( P_N(x) \).
- If the Taylor series is alternating and decreasing in absolute value for \( x \) in the interval, then: $ | f(x) – P_N(x) | \le \text{magnitude of next term} $
- This gives a quick and easy way to estimate how accurate the approximation is, without computing higher derivatives or using the full Lagrange error bound.
Example
Use the Lagrange Error Bound to estimate the error when approximating \( e^x \) at \( x = 0.1 \) using the second-degree Taylor polynomial centered at \( a = 0 \).
▶️ Answer/Explanation
The Maclaurin polynomial of degree 2 for \( e^x \) is \( P_2(x) = 1 + x + \dfrac{x^2}{2} \).
Here, \( f^{(n+1)}(x) = f^{(3)}(x) = e^x \). Over \( 0 \le t \le 0.1 \), \( e^t \) is maximized at \( t = 0.1 \), so \( M = e^{0.1} \approx 1.10517 \).
By the error bound formula:
$ |R_2(0.1)| \le \dfrac{1.10517}{3!} (0.1)^3 = \dfrac{1.10517}{6} (0.001) \approx 0.0001842 $
Final Answer: The error is at most \( \mathbf{0.0001842} \).
Example
Approximate \( \sin(0.3) \) using the third-degree Maclaurin polynomial and use Lagrange Error Bound to estimate the error.
▶️ Answer/Explanation
The Maclaurin polynomial of degree 3 for \( \sin x \) is \( P_3(x) = x – \dfrac{x^3}{6} \).
We need the 4th derivative: \( f^{(4)}(x) = \sin x \). Over \( 0 \le t \le 0.3 \), \( |\sin t| \le 0.3 < 1 \), so the maximum is \( M = 1 \) (since sine’s absolute maximum is 1).
By the formula:
$ |R_3(0.3)| \le \dfrac{1}{4!} (0.3)^4 = \dfrac{1}{24} (0.0081) \approx 0.0003375 $
Final Answer: The error is at most \( \mathbf{0.0003375} \).
Example
Approximate \( \sin(0.5) \) using the 3rd-degree Maclaurin polynomial and bound the error using the alternating series error bound.
▶️ Answer/Explanation
Maclaurin series for \( \sin x \): $ \sin x = x – \dfrac{x^3}{3!} + \dfrac{x^5}{5!} – \cdots $
3rd-degree polynomial: $ P_3(x) = x – \dfrac{x^3}{6} $
For \( x = 0.5 \): $ P_3(0.5) = 0.5 – \dfrac{(0.5)^3}{6} = 0.5 – \dfrac{0.125}{6} \approx 0.47917 $
Next term (magnitude): $ b_{4} = \dfrac{(0.5)^5}{5!} = \dfrac{0.03125}{120} \approx 0.0002604 $
Error bound: $ | \sin(0.5) – P_3(0.5) | \le 0.0002604 $
So the approximation is accurate to within about \( 2.6 \times 10^{-4} \).