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♾️AP Calculus AB/BC

Key Taylor Series Expansions

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Why This Matters

Taylor series are the bridge between the calculus you've learned—derivatives, integrals, limits—and the powerful idea that any smooth function can be rebuilt from its derivatives at a single point. On the BC exam, you're being tested on whether you understand how polynomials can approximate transcendental functions, why those approximations work, and how to quantify their accuracy. This isn't just abstract theory: Taylor series let you evaluate impossible integrals, solve differential equations, and understand function behavior in ways that direct computation simply can't.

The key concepts you'll encounter include polynomial approximation, convergence and divergence, error bounds, and series manipulation. Don't just memorize the expansions for exe^x, sinx\sin x, and cosx\cos x—know why they take the forms they do (alternating signs, factorial denominators, odd vs. even powers). When you understand the underlying structure, you can derive any series on the spot and tackle FRQ questions that ask you to build, manipulate, or analyze Taylor polynomials.


The Foundation: Taylor and Maclaurin Polynomials

Taylor polynomials capture a function's behavior by matching its value and derivatives at a center point. The more derivatives you match, the better your approximation becomes near that point.

Taylor Series Definition

  • General form: f(x)=n=0f(n)(a)n!(xa)nf(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n—each term encodes one derivative's contribution
  • Coefficient formula f(n)(a)n!\frac{f^{(n)}(a)}{n!} ensures the polynomial's nnth derivative at x=ax = a matches f(n)(a)f^{(n)}(a)
  • Polynomial approximation means replacing complicated functions with sums of powers—the foundation of numerical analysis

Maclaurin Series

  • Special case where a=0a = 0, giving the simpler form f(x)=n=0f(n)(0)n!xnf(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n
  • Computational advantage—evaluating derivatives at zero often yields clean integer or zero values
  • Most common expansions (exe^x, sinx\sin x, cosx\cos x, ln(1+x)\ln(1+x)) are Maclaurin series because their centers at zero produce elegant patterns

Compare: Taylor series centered at aa vs. Maclaurin series—both use the same coefficient formula f(n)(a)n!\frac{f^{(n)}(a)}{n!}, but Maclaurin fixes a=0a = 0. If an FRQ gives you derivatives at a non-zero point, you're building a Taylor series; if derivatives are at zero, it's Maclaurin.


The Essential Expansions You Must Know

These four series appear constantly on the BC exam. Memorize their forms, but understand their patterns—odd/even powers, alternating signs, and factorial growth.

Exponential Function exe^x

  • Expansion: ex=1+x+x22!+x33!+=n=0xnn!e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots = \sum_{n=0}^{\infty} \frac{x^n}{n!}
  • All positive terms—since every derivative of exe^x equals exe^x, and e0=1e^0 = 1
  • Converges for all real xx (infinite radius of convergence), making it the most well-behaved series

Sine Function sinx\sin x

  • Expansion: sinx=xx33!+x55!=n=0(1)nx2n+1(2n+1)!\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}
  • Odd powers only—reflects that sinx\sin x is an odd function (sin(x)=sinx\sin(-x) = -\sin x)
  • Alternating signs come from the derivative cycle: sincossincos\sin \to \cos \to -\sin \to -\cos

Cosine Function cosx\cos x

  • Expansion: cosx=1x22!+x44!=n=0(1)nx2n(2n)!\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}
  • Even powers only—reflects that cosx\cos x is an even function (cos(x)=cosx\cos(-x) = \cos x)
  • Same derivative cycle as sine produces the alternating pattern; converges for all real xx

Natural Logarithm ln(1+x)\ln(1+x)

  • Expansion: ln(1+x)=xx22+x33=n=1(1)n+1xnn\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n}
  • No factorials—coefficients are simply 1n\frac{1}{n}, making this series grow more slowly
  • Limited convergence: only valid for 1<x1-1 < x \leq 1—a common exam trap

Compare: sinx\sin x vs. cosx\cos x—both alternate signs and have factorial denominators, but sine uses odd powers (starts with xx) while cosine uses even powers (starts with 11). Remember: sine is odd, cosine is even—the function's symmetry determines its series structure.


Error Analysis and the Lagrange Remainder

Understanding error bounds separates students who can use Taylor series from those who truly understand them. The Lagrange remainder tells you exactly how wrong your approximation might be.

Taylor's Theorem

  • Core guarantee—any sufficiently smooth function equals its Taylor polynomial plus a remainder term
  • Truncation creates error: stopping at degree nn means ignoring infinitely many higher-order terms
  • Foundation for approximation—the theorem justifies why Taylor polynomials work at all

Lagrange Remainder Form

  • Formula: Rn(x)=f(n+1)(c)(n+1)!(xa)n+1R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1} for some cc between aa and xx
  • The mystery cc—you don't know its exact value, so you bound f(n+1)(c)|f^{(n+1)}(c)| by its maximum on the interval
  • Error bound strategy: find MM such that f(n+1)(c)M|f^{(n+1)}(c)| \leq M, then Rn(x)Mxan+1(n+1)!|R_n(x)| \leq \frac{M \cdot |x-a|^{n+1}}{(n+1)!}

Compare: Taylor's Theorem vs. Mean Value Theorem—both guarantee the existence of some point cc in an interval. MVT finds where the derivative equals the average rate of change; Taylor's Theorem uses cc to express approximation error. FRQs often ask you to bound error using the Lagrange form—always identify the maximum of f(n+1)|f^{(n+1)}| on your interval.


Convergence: Where Does the Series Work?

A Taylor series isn't useful if it doesn't converge. The radius of convergence defines the "safe zone" where your series actually represents the function.

Radius of Convergence

  • Definition: the value RR such that the series converges for xa<R|x - a| < R and diverges for xa>R|x - a| > R
  • Ratio test is your primary tool: if limnan+1an=L\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = L, then R=1LR = \frac{1}{L}
  • Three possibilities: R=0R = 0 (converges only at center), R=R = \infty (converges everywhere), or finite RR

Interval of Convergence

  • More than just radius—you must check endpoints separately using other convergence tests
  • Endpoint behavior varies: ln(1+x)\ln(1+x) converges at x=1x = 1 but diverges at x=1x = -1
  • Common exam question: find the interval, then determine whether each endpoint is included

Compare: exe^x (converges for all xx) vs. ln(1+x)\ln(1+x) (converges only for 1<x1-1 < x \leq 1)—the exponential's derivatives stay bounded relative to factorial growth, while the logarithm's derivatives explode as you move away from the center. Always state the interval of convergence when writing a series.


Building and Manipulating Series

You won't always be handed a series—sometimes you need to construct one or modify a known series. These techniques turn memorized formulas into flexible problem-solving tools.

Finding Coefficients

  • Coefficient formula: an=f(n)(a)n!a_n = \frac{f^{(n)}(a)}{n!} requires computing derivatives at the center
  • Pattern recognition—after finding the first few coefficients, look for a general formula
  • Derivative tables on FRQs often provide f(a),f(a),f(a),f(a), f'(a), f''(a), \ldots so you can build the polynomial directly

Series Operations

  • Addition/subtraction: combine term by term when series have the same center
  • Multiplication: use the Cauchy product or multiply out first few terms for polynomial approximations
  • Substitution: replace xx with another expression (e.g., substitute x2-x^2 into exe^x to get ex2e^{-x^2})

Composition and Substitution

  • Most efficient technique—rather than computing derivatives of ex2e^{-x^2}, substitute x2-x^2 into the known series for eue^u
  • Convergence changes: substituting x2x^2 for xx in ln(1+x)\ln(1+x) gives ln(1+x2)\ln(1+x^2), valid for x1|x| \leq 1
  • Integration/differentiation of series: integrate or differentiate term by term within the interval of convergence

Compare: Computing derivatives directly vs. substitution—for cos(x2)\cos(x^2), finding f(n)(0)f^{(n)}(0) is tedious, but substituting x2x^2 into the cosine series instantly gives 1x42!+x84!1 - \frac{x^4}{2!} + \frac{x^8}{4!} - \cdots. Use substitution whenever a function is a composition involving a known series.


Power Series Connection

Taylor series are a specific type of power series. Understanding this relationship helps you see Taylor series as part of a larger framework.

Relationship to Power Series

  • Every Taylor series is a power series of the form n=0cn(xa)n\sum_{n=0}^{\infty} c_n(x-a)^n
  • Not every power series is Taylor—Taylor series have coefficients determined by derivatives; general power series don't require this
  • Uniqueness theorem: if a function has a power series representation centered at aa, it must be the Taylor series

Quick Reference Table

ConceptBest Examples
Maclaurin series (center at 0)exe^x, sinx\sin x, cosx\cos x, ln(1+x)\ln(1+x)
Infinite radius of convergenceexe^x, sinx\sin x, cosx\cos x
Finite radius of convergenceln(1+x)\ln(1+x) (R=1R = 1), 11x\frac{1}{1-x} (R=1R = 1)
Odd function series (odd powers)sinx\sin x, arctanx\arctan x
Even function series (even powers)cosx\cos x
Alternating seriessinx\sin x, cosx\cos x, ln(1+x)\ln(1+x)
Error bound (Lagrange remainder)Any truncated Taylor polynomial
Series by substitutionex2e^{-x^2}, cos(x2)\cos(x^2), ln(1+x2)\ln(1+x^2)

Self-Check Questions

  1. Which two series among exe^x, sinx\sin x, cosx\cos x, and ln(1+x)\ln(1+x) share the property of converging for all real numbers? What makes them different from ln(1+x)\ln(1+x)?

  2. If you're given a table of f(2)f(2), f(2)f'(2), f(2)f''(2), and f(2)f'''(2), write the formula for the third-degree Taylor polynomial centered at x=2x = 2.

  3. Compare and contrast the series for sinx\sin x and cosx\cos x: How do their terms reflect the odd/even nature of each function?

  4. You need to approximate e0.1e^{0.1} with error less than 0.00010.0001. How would you use the Lagrange remainder to determine how many terms are sufficient?

  5. Explain why substituting x2-x^2 into the series for exe^x is more efficient than computing derivatives of ex2e^{-x^2} directly. What is the resulting series for ex2e^{-x^2}?