Worked example: estimating e_ using Lagrange error bound | AP Calculus BC | Khan Academy
TLDRThis instructional video explores estimating the value of e^x using a Taylor polynomial centered at x=2 with an error margin less than 0.001. The instructor reviews Taylor's remainder theorem and applies it to e^x, noting that all derivatives of e^x are e^x itself. By bounding the (n+1)th derivative by e^2 within the interval of interest, the video guides through algebraic manipulation to find the least degree of the polynomial required. Through trial and error with successive values of n, the conclusion is that a fifth-degree polynomial is sufficient to ensure the error is below the specified threshold.
Takeaways
- 📚 The task is to estimate \( e^{1.45} \) using a Taylor polynomial centered at \( x = 2 \) and determine the least degree of the polynomial to ensure an error less than 0.001.
- 🔍 The LaGrange error bound or Taylor's remainder theorem is the method to be used for bounding the error of the Taylor polynomial approximation.
- 📉 The remainder of the nth degree Taylor polynomial is less than a specific expression involving \( n \), \( x \), \( c \), and an upper bound \( M \) on the \( (n+1) \)th derivative of the function.
- 📈 The function to be approximated is \( e^x \), and its derivatives of all orders are \( e^x \), which simplifies the process of bounding the \( (n+1) \)th derivative.
- 📊 The value of \( M \) is bounded by \( e^2 \) since \( x \) is within the interval [0, 2], which is the range relevant for the approximation.
- 🔢 The error bound formula involves \( e^2 \), the difference between \( x \) and the center \( c \) raised to the \( n+1 \) power, and divided by \( (n+1)! \).
- 🧩 The goal is to find the smallest \( n \) such that the error bound is less than 0.001, which involves algebraic manipulation and calculation.
- 📝 The process involves dividing 0.001 by \( e^2 \) to find a threshold value for \( 0.55^{n+1} / (n+1)! \) to be less than.
- 🔧 Trial and error with increasing values of \( n \) is used to find the smallest \( n \) that satisfies the error condition.
- 🎯 The least degree of the polynomial that ensures an error smaller than 0.001 is found to be 5, as \( n = 5 \) makes the error bound sufficiently low.
Q & A
What is the primary goal of the example provided in the script?
-The primary goal is to estimate e to the 1.45 using a Taylor polynomial about x equals two, ensuring the error is smaller than 0.001.
What method is suggested for determining the necessary degree of the Taylor polynomial?
-The LaGrange error bound or Taylor's remainder theorem is suggested for determining the necessary degree of the Taylor polynomial.
What function is being approximated in this example?
-The function being approximated is f(x) = e^x.
What are the key variables involved in applying the LaGrange error bound?
-The key variables are n (the degree of the polynomial), x (the value at which the error is calculated, which is 1.45), c (the center of the Taylor polynomial, which is 2), and M (an upper bound on the absolute value of the n+1 derivative of the function).
Why is the n+1 derivative of the function f(x) = e^x particularly convenient in this example?
-The n+1 derivative of f(x) = e^x is convenient because it remains e^x for all derivatives, making it easier to establish a bound.
What interval is considered for establishing the bound on e^x?
-The interval considered is 0 < x ≤ 2, which contains both the value 1.45 (where we are calculating the error) and 2 (where the polynomial is centered).
What is the upper bound M chosen for e^x in the interval [0, 2]?
-The upper bound M chosen for e^x in the interval [0, 2] is e^2.
What inequality must be satisfied to ensure the error is less than 0.001?
-The inequality to be satisfied is (e^2 / (n+1)!) * |0.55^(n+1)| < 0.001.
How is the inequality simplified to make it easier to solve for n?
-The inequality is simplified by dividing both sides by e^2, resulting in |0.55^(n+1) / (n+1)!| < 0.001 / e^2.
What is the smallest degree of the Taylor polynomial that satisfies the error condition?
-The smallest degree of the Taylor polynomial that satisfies the error condition is n = 5.
Outlines
📚 Introduction to Taylor Polynomials and Error Bounds
The instructor begins by introducing the concept of estimating the value of e to the power of 1.45 using a Taylor polynomial centered at x equals two. The goal is to determine the minimum degree of the polynomial required to ensure the error is less than 0.001. The LaGrange error bound and Taylor's remainder theorem are suggested as tools for this task. A review of Taylor's remainder theorem is provided, explaining the formula for the absolute value of the remainder of an nth degree Taylor polynomial. The instructor emphasizes the importance of identifying an upper bound M for the absolute value of the (n+1)th derivative of the function, which in this case is e to the power of x, and is conveniently bounded by e squared within the interval of interest.
🔍 Calculating the Degree for a Specified Error Bound
The instructor proceeds to calculate the degree of the Taylor polynomial that will ensure the error is less than 0.001 when estimating e to the power of 1.45. The process involves using the LaGrange error bound formula with the determined M value, e squared, and the specific x value of 1.45. The instructor simplifies the formula and sets up an inequality to solve for the degree n, which involves raising 0.55 to the power of (n+1) and dividing by (n+1) factorial. The goal is to find the smallest n such that this expression is less than 0.001 divided by e squared. The instructor demonstrates the use of a calculator to test different values of n, starting with n=3 and increasing until the condition is met. After evaluating n=3 and n=4, it is determined that n=5 is the smallest degree that satisfies the error bound condition, thus ensuring an error smaller than the specified threshold.
Mindmap
Keywords
💡Taylor Polynomial
💡Error Bound
💡Lagrange Error Bound
💡Taylor's Remainder Theorem
💡Derivative
💡Exponential Function
💡Interval
💡M (Upper Bound)
💡Factorial
💡Algebraic Manipulation
💡Calculator
Highlights
Estimating e to the 1.45 using a Taylor polynomial about x equals two.
The importance of determining the least degree of the polynomial to assure an error smaller than 0.001.
Introduction to the LaGrange error bound and Taylor's remainder theorem for approximating functions.
Explanation of the absolute value of the remainder for the nth degree Taylor polynomial.
Identification of 'M' as an upper bound on the absolute value of the (n+1)th derivative of the function.
The convenience of e^x as the derivative of any order for the function f(x) = e^x.
Bounding e^x by e^2 for x in the interval [0, 2] to simplify the problem.
Using e^2 as the value for 'M' in the LaGrange error bound formula.
Setting up the inequality to find the least degree 'n' of the Taylor polynomial.
Algebraic manipulation to solve for 'n' where the error is less than 0.001.
Dividing both sides of the inequality by e^2 to isolate the term involving 'n'.
Using a calculator to find the smallest 'n' that satisfies the inequality.
Starting the trial with n equals three to find the appropriate degree of the polynomial.
Incrementing 'n' to four and then five to find when the error bound is less than 0.001.
Conclusion that the least degree of the polynomial that assures an error smaller than 1/1000 is five.
Transcripts
Browse More Related Video
Avon High School - AP Calculus BC - Topic 10.12 - Example 2
Worked example: estimating sin(0.4) using Lagrange error bound | AP Calculus BC | Khan Academy
2023 AP Calculus BC FRQ #6
2011 Calculus BC free response #6d | AP Calculus BC | Khan Academy
Avon High School - AP Calculus BC - Topic 10.12 - Example 3
Approximating functions with polynomials (part 3)
5.0 / 5 (0 votes)
Thanks for rating: