Series Applications

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

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

Approximating Function Values

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

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

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

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.

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

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:

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

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

Check Your Understanding

Series Integration Practice

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

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:

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

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

Check Your Understanding

Limit Computation

Practice

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

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:

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

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

Key Example

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

Differentiate:

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

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

Check Your Understanding

Building Series

Practice

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

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:

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 $x = 0$."

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

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

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

Practice

AP Series Checklist

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

Next: Part 6 — Problem-Solving Workshop

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

Next: Part 7 — Comprehensive Review

Comprehensive Review — Series Applications

Part 7 of 7 — Putting It All Together

Core Skills Summary

Comprehensive Check

Mixed Application Review

Final Challenge

Series Applications — Complete ✓

You've mastered:

1. Approximating functions — build from 6 known Maclaurin series via substitution
2. Computing integrals — integrate term by term when no antiderivative exists
3. Evaluating limits — expand, cancel, read off the coefficient
4. Differentiating series — power rule term by term, generates new functions
5. 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!