AP Calculus BC 10.14 Finding Taylor or Maclaurin Series for a Function Study Notes - New Syllabus
AP Calculus BC 10.14 Finding Taylor or Maclaurin Series for a Function Study Notes- New syllabus
AP Calculus BC 10.14 Finding Taylor or Maclaurin Series for a Function 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:
- Finding Taylor or Maclaurin Series for a Function
Finding Taylor or Maclaurin Series for a Function
Finding Taylor or Maclaurin Series for a Function
A Taylor series is a way of representing a function as an infinite sum of polynomial terms determined from the function’s derivatives at a specific point \( a \). Each term of the series becomes a better approximation of the function near \( x = a \). A Maclaurin series is a special case of the Taylor series centered at \( a = 0 \).
These series are very useful in approximating complicated functions, solving differential equations, and in mathematical modeling when exact functions are difficult to work with. They also play an important role in physics, engineering, and numerical computation.
General Formula for Taylor Series about \( x = a \):
$ f(x) = \sum_{n=0}^{\infty} \dfrac{f^{(n)}(a)}{n!} (x-a)^n $
where:
- \( f^{(n)}(a) \) is the \( n \)th derivative of \( f(x) \) evaluated at \( x = a \).
- \( n! \) is the factorial of \( n \).
- The variable term is \( (x-a)^n \).
Maclaurin Series (special case \( a = 0 \)):
$ f(x) = \sum_{n=0}^{\infty} \dfrac{f^{(n)}(0)}{n!} x^n $
Step-by-Step Process for Finding a Taylor or Maclaurin Series:
- Identify the function \( f(x) \) and the center \( a \) for the expansion.
- Find successive derivatives \( f'(x), f”(x), f^{(3)}(x), \dots \) until a pattern emerges.
- Evaluate each derivative at \( x = a \).
- Substitute values into the formula: $ \sum_{n=0}^{\infty} \dfrac{f^{(n)}(a)}{n!} (x-a)^n $
- Write out several terms to see the pattern clearly.
- Use known series expansions when possible to avoid heavy derivative computation.
Common Maclaurin Series Expansions:
| Function \( f(x) \) | Maclaurin Series | Interval of Convergence |
|---|---|---|
| \( e^x \) | \( \displaystyle \sum_{n=0}^{\infty} \dfrac{x^n}{n!} \) | All real \( x \) |
| \( \sin x \) | \( \displaystyle \sum_{n=0}^{\infty} (-1)^n \dfrac{x^{2n+1}}{(2n+1)!} \) | All real \( x \) |
| \( \cos x \) | \( \displaystyle \sum_{n=0}^{\infty} (-1)^n \dfrac{x^{2n}}{(2n)!} \) | All real \( x \) |
| \( \dfrac{1}{1-x} \) | \( \displaystyle \sum_{n=0}^{\infty} x^n \) | \( |x| < 1 \) |
| \( \ln(1+x) \) | \( \displaystyle \sum_{n=1}^{\infty} (-1)^{n+1} \dfrac{x^n}{n} \) | \( -1 < x \le 1 \) |
| \( \arctan x \) | \( \displaystyle \sum_{n=0}^{\infty} (-1)^n \dfrac{x^{2n+1}}{2n+1} \) | \( |x| \le 1 \) |
Example
Find the Maclaurin series for \( f(x) = \cos x \) and its interval of convergence.
▶️ Answer/Explanation
Derivatives:
\( f(x) = \cos x \), \( f'(x) = -\sin x \), \( f”(x) = -\cos x \), \( f^{(3)}(x) = \sin x \), repeating every 4 terms.
At \( x = 0 \): \( f(0) = 1 \), \( f'(0) = 0 \), \( f”(0) = -1 \), \( f^{(3)}(0) = 0 \).
Only even terms survive:
$ \cos x = \sum_{n=0}^{\infty} (-1)^n \dfrac{x^{2n}}{(2n)!} $
Interval of Convergence: All real \( x \).
Example
Find the Taylor series for \( f(x) = e^x \) about \( a = 2 \).
▶️ Answer/Explanation
All derivatives of \( e^x \) are \( e^x \).
At \( x = 2 \): \( f^{(n)}(2) = e^2 \) for all \( n \).
$ e^x = \sum_{n=0}^{\infty} \dfrac{e^2}{n!} (x – 2)^n $
First terms: $ e^x \approx e^2 \left[ 1 + (x-2) + \dfrac{(x-2)^2}{2!} + \dfrac{(x-2)^3}{3!} + \cdots \right] $
Interval of Convergence: All real \( x \).
Example
Find the Maclaurin series for \( f(x) = \ln(1+x) \) and determine its interval of convergence.
▶️ Answer/Explanation
Using standard expansion: $ \ln(1+x) = \sum_{n=1}^{\infty} (-1)^{n+1} \dfrac{x^n}{n} $
First few terms: $ x – \dfrac{x^2}{2} + \dfrac{x^3}{3} – \dfrac{x^4}{4} + \cdots $
Radius \( R = 1 \).
- \( x = 1 \): converges (\( \ln 2 \), alternating harmonic).
- \( x = -1 \): diverges (harmonic series).
Interval: \( -1 < x \le 1 \).