AP Calculus BC 10.15 Representing Functions as Power Series Study Notes - New Syllabus
AP Calculus BC 10.15 Representing Functions as Power Series Study Notes- New syllabus
AP Calculus BC 10.15 Representing Functions as Power Series 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:
- Representing Functions as Power Series
Representing Functions as Power Series
Representing Functions as Power Series
A power series is an infinite sum of the form:
$ \sum_{n=0}^{\infty} c_n (x-a)^n $
where:
- \( c_n \) are constants (coefficients),
- \( a \) is the center of the series,
- The variable term is \( (x-a)^n \).
The graph shows a function and three approximations of it by partial sums of a power series.
Power series allow us to represent functions as infinite polynomials within a certain interval of convergence.
Once we know a few standard series (like the common Maclaurin expansions from Topic 10.14), we can use algebraic manipulations to find the series of more complicated functions without calculating many derivatives.
Key Concept: If we know the series for a basic function, we can find series for related functions by:
- Substituting a different expression for \( x \) (shifting or scaling),
- Multiplying or dividing the series by constants or powers of \( x \),
- Differentiating or integrating the series term by term.
Step-by-Step Process:
- Start with a known standard series (from the common Maclaurin table).
- Apply algebraic transformations (substitution, multiplication, division, differentiation, or integration) to match the given function.
- Write the resulting series explicitly, showing how the transformation changes the coefficients and powers.
- Determine the new interval of convergence, adjusting for any transformations.
Common Transformations and Their Effect on Power Series:
| Transformation | Effect on Series | Example Starting Series |
|---|---|---|
| Replace \( x \) with \( kx \) | Multiply every power term by \( k^n \) | From \( \displaystyle \sum x^n \) to \( \displaystyle \sum (kx)^n \) |
| Replace \( x \) with \( x-a \) | Series is centered at \( a \) instead of 0 | From \( \displaystyle \sum x^n \) to \( \displaystyle \sum (x-a)^n \) |
| Multiply by \( x^m \) | Add \( m \) to each power of \( x \) | From \( \displaystyle \sum x^n \) to \( \displaystyle \sum x^{n+m} \) |
| Differentiate term-by-term | Each term \( c_n (x-a)^n \) becomes \( n c_n (x-a)^{n-1} \) | From \( \displaystyle \sum x^n \) to \( \displaystyle \sum n x^{n-1} \) |
| Integrate term-by-term | Each term \( c_n (x-a)^n \) becomes \( \dfrac{c_n}{n+1} (x-a)^{n+1} \) | From \( \displaystyle \sum x^n \) to \( \displaystyle \sum \dfrac{x^{n+1}}{n+1} \) |
Example
Represent \( \dfrac{1}{1-2x} \) as a power series.
▶️ Answer/Explanation
We know the standard series: $ \dfrac{1}{1-x} = \sum_{n=0}^{\infty} x^n, \quad |x| < 1 $
Replace \( x \) with \( 2x \): $ \dfrac{1}{1-2x} = \sum_{n=0}^{\infty} (2x)^n = \sum_{n=0}^{\infty} 2^n x^n $
Convergence condition: \( |2x| < 1 \) ⇒ \( |x| < \dfrac{1}{2} \).
Final Answer: $ \dfrac{1}{1-2x} = \sum_{n=0}^{\infty} 2^n x^n, \quad |x| < \dfrac{1}{2} $
Example
Find the power series representation of \( \ln(1-x^2) \).
▶️ Answer/Explanation
We know: $ \ln(1+u) = \sum_{n=1}^{\infty} (-1)^{n+1} \dfrac{u^n}{n}, \quad |u| < 1 $
Here \( u = -x^2 \):
$ \ln(1 – x^2) = \sum_{n=1}^{\infty} (-1)^{n+1} \dfrac{(-x^2)^n}{n} $
Simplify signs: $ \ln(1 – x^2) = -\sum_{n=1}^{\infty} \dfrac{x^{2n}}{n} $
Convergence condition: \( |x^2| < 1 \) ⇒ \( |x| < 1 \), and diverges at \( x = \pm 1 \).
Example
Find the power series for \( \arctan(3x) \).
▶️ Answer/Explanation
We know: $ \arctan u = \sum_{n=0}^{\infty} (-1)^n \dfrac{u^{2n+1}}{2n+1}, \quad |u| \le 1 $
Here \( u = 3x \): $ \arctan(3x) = \sum_{n=0}^{\infty} (-1)^n \dfrac{(3x)^{2n+1}}{2n+1} $
That is: $ \arctan(3x) = \sum_{n=0}^{\infty} (-1)^n \dfrac{3^{2n+1} x^{2n+1}}{2n+1} $
Convergence condition: \( |3x| \le 1 \) ⇒ \( |x| \le \dfrac{1}{3} \) (endpoint behavior depends on alternating test).