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Series Applications

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Series Applications - Complete Interactive Lesson

Part 1: Core Concepts

Series Applications — Using Taylor Series

Part 1 of 7 — Approximating Functions

Why Use Series?

Taylor and Maclaurin series convert functions into polynomials, making them useful for:

Approximating difficult function values
Evaluating limits
Computing integrals that have no closed-form antiderivative
Solving differential equations

Key Series to Know

Function	Maclaurin Series	Interval
$e^x$	$\sum_{n=0}^\infty \frac{x^n}{n!}$	$(-\infty, \infty)$
$\sin x$	$\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}$
$\cos x$	$\sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}$
$\frac{1}{1-x}$	$\sum_{n=0}^\infty x^n$
$\ln(1+x)$	$\sum_{n=1}^\infty \frac{(-1)^{n+1} x^n}{n}$
$\arctan x$	$\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2n+1}$

$\boxed{\text{Memorize these six — they appear on every BC exam}}$

Approximating Function Values

To approximate $e^{0.1}$ using a 3rd-degree Maclaurin polynomial:

$e^x \approx 1 + x + \frac{x^2}{2} + \frac{x^3}{6}$

Check Your Understanding

Series Construction

Practice

Key Techniques

$\boxed{\text{Known series} + \text{substitution} = \text{new series}}$

Memorize the six standard Maclaurin series
Create new series by substituting into known ones
Polynomial approximations are most accurate near the center

Next: Part 2 — Series for Computing Integrals

Part 2: Worked Examples

Series for Computing Integrals

Part 2 of 7 — Integrating the "Unintegrable"

The Power of Term-by-Term Integration

Some functions have no elementary antiderivative, but their Taylor series can be integrated term by term:

$\boxed{\int \sum_{n=0}^\infty a_n x^n\,dx = \sum_{n=0}^\infty \frac{a_n x^{n+1}}{n+1} + C}$

Part 3: Problem-Solving Patterns

Series for Evaluating Limits

Part 3 of 7 — An Alternative to L'Hôpital's Rule

The Taylor Series Approach to Limits

For $0/0$ indeterminate forms, substitute the Taylor series and simplify:

$\lim_{x \to 0} \frac{\sin x - x}{x^3}$

Part 4: Graphs and Interpretation

Differentiation of Power Series

Part 4 of 7 — Generating New Series from Old

Term-by-Term Differentiation

A power series can be differentiated term by term within its interval of convergence:

$\boxed{\text{If } f(x) = \sum_{n=0}^\infty a_n x^n, \text{ then } f'(x) = \sum_{n=1}^\infty n a_n x^{n-1}}$

Part 5: Applications

AP Exam Strategies — Series Applications

Part 5 of 7 — How Series Questions Appear on the BC Exam

Series FRQ Structure

The AP BC exam typically has one full FRQ dedicated to Taylor/Maclaurin series. Common parts:

Part	Typical question
(a)	Write the first 4 nonzero terms and general term
(b)	Find the interval of convergence
(c)	Use the series to approximate an integral
(d)	Bound the error of the approximation

AP Tip: This FRQ is one of the most predictable on the BC exam. Practice the pattern and you can earn nearly full credit.

FRQ Answer Templates

"Write the first four nonzero terms of the Taylor series for $f$ about $x = 0$ ."

Part 6: Exam Strategy

Problem-Solving Workshop — Series Applications

Part 6 of 7 — Guided Practice Problems

Work through these AP-style problems. Each targets a key series application skill.

Warm-Up: Quick Checks

Problem 1: Full FRQ Walkthrough

Let $f(x) = \ln(1 + x)$ .

Problem 2

Problem 3

Key Takeaways

Substitution into known series is faster than computing derivatives
For integrals, integrate the series and evaluate — simpler than FTC with complex antiderivatives
For limits, cancel the leading terms to find the dominant behavior
Always simplify fractions on the exam — AP readers check exact answers

Part 7: Mixed Review

Comprehensive Review — Series Applications

Part 7 of 7 — Putting It All Together

Core Skills Summary

Skill	Technique
Approximate $f(x)$	Build Taylor polynomial from known series
Compute $\int f(x)\,dx$	Integrate series term by term
Evaluate

$e^{0.1} \approx 1 + 0.1 + 0.005 + 0.000167 = 1.105167$

Actual value: $e^{0.1} = 1.105171...$ Error $< 0.000004$ .

Creating New Series by Substitution

To find the series for $e^{-x^2}$ , substitute $-x^2$ for $x$ in $e^x$ :

$e^{-x^2} = \sum_{n=0}^\infty \frac{(-x^2)^n}{n!} = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{n!}$

$= 1 - x^2 + \frac{x^4}{2} - \frac{x^6}{6} + \cdots$

AP Tip: Substitution into a known series is the fastest way to build new series on the AP exam.

This is valid within the interval of convergence.

Key Fact: Term-by-term integration is the ONLY way to handle $\int e^{-x^2}dx$ , $\int \frac{\sin x}{x}dx$ , etc. on the AP exam.

Classic Example: $\int_0^1 e^{-x^2}\,dx$

We cannot find an antiderivative, but we can use the series:

$e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \frac{x^8}{4!} - \cdots$

Integrate term by term:

$\int_0^1 e^{-x^2}\,dx = \left[x - \frac{x^3}{3} + \frac{x^5}{10} - \frac{x^7}{42} + \frac{x^9}{216} - \cdots\right]_0^1$

$= 1 - \frac{1}{3} + \frac{1}{10} - \frac{1}{42} + \frac{1}{216} - \cdots \approx 0.7468$

How Many Terms?

By the Alternating Series Estimation Theorem, the error is bounded by the first omitted term:

Using 4 terms: error $< 1/216 \approx 0.0046$
Using 5 terms: error $< 1/1320 \approx 0.00076$

Another Classic: $\int \frac{\sin x}{x}\,dx$

$\frac{\sin x}{x} = \frac{1}{x}\left(x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots\right) = 1 - \frac{x^2}{6} + \frac{x^4}{120} - \cdots$

$\int_0^1 \frac{\sin x}{x}\,dx = \left[x - \frac{x^3}{18} + \frac{x^5}{600} - \cdots\right]_0^1 = 1 - \frac{1}{18} + \frac{1}{600} - \cdots$

AP Tip: When asked to "write the first four nonzero terms and use them to approximate the integral," this is exactly the technique to use.

General Pattern

To integrate	Use the series for	Then integrate
$e^{-x^2}$	$e^u$ with

Check Your Understanding

Series Integration Practice

Key Technique

$\boxed{\int f(x)\,dx = \int \sum a_n x^n\,dx = \sum \frac{a_n x^{n+1}}{n+1} + C}$

When to use: The integrand has no elementary antiderivative OR the problem specifically asks for a series approach.

Error bound: For alternating series, error $<$ first omitted term.

Next: Part 3 — Series for Evaluating Limits

Substitute: $\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots$

$= \lim_{x \to 0} \frac{(x - x^3/6 + \cdots) - x}{x^3} = \lim_{x \to 0} \frac{-x^3/6 + \cdots}{x^3} = -\frac{1}{6}$

Key Fact: One substitution replaces multiple L'Hôpital applications. This limit would require three rounds of L'Hôpital's Rule.

More Examples

Example 1: $\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}$

$e^x = 1 + x + x^2/2 + x^3/6 + \cdots$

$\frac{(1 + x + x^2/2 + \cdots) - 1 - x}{x^2} = \frac{x^2/2 + x^3/6 + \cdots}{x^2} = \frac{1}{2} + \frac{x}{6} + \cdots \to \frac{1}{2}$

Example 2: $\lim_{x \to 0} \frac{1 - \cos x}{x^2}$

$\cos x = 1 - x^2/2 + x^4/24 - \cdots$

$\frac{1 - (1 - x^2/2 + \cdots)}{x^2} = \frac{x^2/2 - x^4/24 + \cdots}{x^2} = \frac{1}{2}$

When Series Beat L'Hôpital

Scenario	L'Hôpital	Series
$\frac{\sin x - x}{x^3}$	3 applications	One substitution

Check Your Understanding

Limit Computation

Key Technique

$\boxed{\text{Substitute Taylor series} \to \text{Cancel leading terms} \to \text{Read off the limit}}$

This is especially powerful when the limit would require 3+ applications of L'Hôpital's Rule.

Next: Part 4 — Differentiation of Power Series

The radius of convergence stays the same (though endpoint behavior may change).

$\frac{1}{1-x} = \sum_{n=0}^\infty x^n = 1 + x + x^2 + x^3 + \cdots$

$\frac{1}{(1-x)^2} = \sum_{n=1}^\infty n x^{n-1} = 1 + 2x + 3x^2 + 4x^3 + \cdots$

AP Tip: This technique generates series that would be hard to derive from scratch.

Combining Operations

You can chain substitution, differentiation, and integration:

Find the series for $\ln(1+x)$ :

Start with $\frac{1}{1+x} = 1 - x + x^2 - x^3 + \cdots$

Integrate: $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots$

Find the series for $\frac{x}{(1-x)^2}$ :

$\frac{x}{(1-x)^2} = x \cdot \sum_{n=1}^\infty n x^{n-1} = \sum_{n=1}^\infty n x^n = x + 2x^2 + 3x^3 + \cdots$

Operations Summary

Operation	Effect on $\sum a_n x^n$
Differentiate	$\sum n a_n x^{n-1}$

Check Your Understanding

Key Rules

$\boxed{f(x) = \sum a_n x^n \implies f'(x) = \sum n a_n x^{n-1}}$

$\boxed{\int f(x)\,dx = \sum \frac{a_n x^{n+1}}{n+1} + C}$

Both operations preserve the radius of convergence. Chain these with substitution and multiplication to build nearly any series.

Next: Part 5 — AP Exam Strategies for Series Applications

$f(0) = \ldots$ , $f'(0) = \ldots$ , $f''(0) = \ldots$ , $f'''(0) = \ldots$

$P_3(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3$

OR (if built from known series):

Since $\sin x = x - x^3/6 + x^5/120 - \cdots$ , $x\sin(x^2) = x^3 - x^7/6 + x^{11}/120 - \cdots$

"Use the series to approximate $\int_0^{1/2} f(x)\,dx$ ."

$\int_0^{1/2} (\text{first few terms})\,dx = [\text{antiderivatives}]_0^{1/2} = \text{value}$

"Show the error is less than $1/100$ ."

By the Alternating Series Estimation Theorem, the error is less than the absolute value of the first omitted term: $|a_{n+1}| = \ldots < 1/100$ . ✓

AP-Style Questions

FRQ Practice

Let $f(x) = e^{-x}$ .

AP Series Checklist

✓ Know the six standard Maclaurin series
✓ Build new series via substitution, differentiation, integration
✓ Write correct general term with proper index
✓ Determine radius/interval of convergence
✓ Integrate series to approximate definite integrals
✓ Use AST error bound for alternating series
✓ Use Lagrange error bound for non-alternating series

Next: Part 6 — Problem-Solving Workshop

Next: Part 7 — Comprehensive Review

Comprehensive Check

Mixed Application Review

Series Applications — Complete ✓

You've mastered:

Approximating functions — build from 6 known Maclaurin series via substitution
Computing integrals — integrate term by term when no antiderivative exists
Evaluating limits — expand, cancel, read off the coefficient
Differentiating series — power rule term by term, generates new functions
Error analysis — AST and Lagrange bounds

Key Formula Reference:

$e^x = \sum \frac{x^n}{n!}, \quad \sin x = \sum \frac{(-1)^n x^{2n+1}}{(2n+1)!}, \quad \cos x = \sum \frac{(-1)^n x^{2n}}{(2n)!}$

$\frac{1}{1-x} = \sum x^n, \quad \ln(1+x) = \sum \frac{(-1)^{n+1} x^n}{n}, \quad \arctan x = \sum \frac{(-1)^n x^{2n+1}}{2n+1}$

Series Applications topic complete!

[x−3x3+10x5−42x7+216x9−⋯]01

[x−18x3+600x5−⋯]01=

\frac{e^x - 1 - x - x^2/2}{x^3}

∑n(−1)n+1xn,arctanx=

Series Applications

Series Applications - Complete Interactive Lesson

Part 1: Core Concepts

Series Applications — Using Taylor Series

Why Use Series?

Key Series to Know

Approximating Function Values

Key Techniques

Part 2: Worked Examples

Series for Computing Integrals

The Power of Term-by-Term Integration

Part 3: Problem-Solving Patterns

Series for Evaluating Limits

The Taylor Series Approach to Limits

Part 4: Graphs and Interpretation

Differentiation of Power Series

Term-by-Term Differentiation

Part 5: Applications

AP Exam Strategies — Series Applications

Series FRQ Structure

FRQ Answer Templates

Part 6: Exam Strategy

Problem-Solving Workshop — Series Applications

Key Takeaways

Part 7: Mixed Review

Comprehensive Review — Series Applications

Core Skills Summary

Creating New Series by Substitution

Classic Example: ∫01e−x2 dx\int_0^1 e^{-x^2}\,dx∫01​e−x2dx

How Many Terms?

Another Classic: ∫sin⁡xx dx\int \frac{\sin x}{x}\,dx∫xsinx​dx

General Pattern

Key Technique

More Examples

When Series Beat L'Hôpital

Key Technique

Key Example

Combining Operations

Operations Summary

Key Rules

AP Series Checklist

Series Applications — Complete ✓

Classic Example: $\int_0^1 e^{-x^2}\,dx$

Another Classic: $\int \frac{\sin x}{x}\,dx$