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🎯⭐ INTERACTIVE LESSON

Taylor & Maclaurin Series

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Taylor & Maclaurin Series - Complete Interactive Lesson

Part 1: Core Concepts

Taylor & Maclaurin Series — The General Formula

Part 1 of 7 — Taylor Polynomial Construction

Taylor Series Centered at x=cx = c

f(x)=∑n=0∞f(n)(c)n!(x−c)n\boxed{f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(c)}{n!}(x - c)^n}

=f(c)+f′(c)(x−c)+f′′(c)2!(x−c)2+f′′′(c)3!(x−c)3+⋯= f(c) + f'(c)(x-c) + \frac{f''(c)}{2!}(x-c)^2 + \frac{f'''(c)}{3!}(x-c)^3 + \cdots

Maclaurin Series (Special Case: c=0c = 0)

f(x)=∑n=0∞f(n)(0)n!xn=f(0)+f′(0)x+f′′(0)2!x2+⋯f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \cdots

Taylor Polynomials

The nnth-degree Taylor polynomial is the partial sum:

Tn(x)=∑k=0nf(k)(c)k!(x−c)kT_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(c)}{k!}(x-c)^k

DegreePolynomialApproximation Quality
T0T_0f(c)f(c)Constant (matches value)
T1T_1f(c)+f′(c)(x−c)f(c) + f'(c)(x-c)Linear (matches slope)
T2T_2+f′′(c)(x−c)2/2+ f''(c)(x-c)^2/2Quadratic (matches concavity)

AP Tip: "Write the nnth-degree Taylor polynomial" means Tn(x)T_n(x). "Write the first four nonzero terms of the Taylor series" may give a higher-degree polynomial.

Example: Taylor Series for exe^x at c=0c = 0

f(x)=ex  ⟹  f(n)(x)=ex  ⟹  f(n)(0)=1f(x) = e^x \implies f^{(n)}(x) = e^x \implies f^{(n)}(0) = 1 for all nn

ex=∑n=0∞xnn!=1+x+x22+x36+x424+⋯e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \cdots

Example: Taylor Series for sin⁡x\sin x at c=0c = 0

nnf(n)(x)f^{(n)}(x)f(n)(0)f^{(n)}(0)
0sin⁡x\sin x00
1cos⁡x\cos x11
2−sin⁡x-\sin x00
3−cos⁡x-\cos x−1-1
4sin⁡x\sin x00

Pattern repeats with period 4: 0,1,0,−1,0,1,0,−1,…0, 1, 0, -1, 0, 1, 0, -1, \ldots

sin⁡x=x−x33!+x55!−x77!+⋯=∑n=0∞(−1)nx2n+1(2n+1)!\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}

Taylor Polynomial Basics

Building Taylor Polynomials

Derivative Extraction

Summary

  • Taylor series: ∑f(n)(c)(x−c)n/n!\sum f^{(n)}(c)(x-c)^n/n!
  • Maclaurin series: Taylor at c=0c = 0
  • Tn(x)T_n(x) = partial sum through degree nn
  • Know the difference between "degree nn" and "first kk nonzero terms"

Next: Part 2 — Computing Taylor Series from Scratch.

Part 2: Worked Examples

Taylor & Maclaurin Series — Computing from Scratch

Part 2 of 7 — Derivative Tables & Non-Zero Centers

Method: Derivative Table

To build the Taylor series at cc:

  1. Compute f(c),f′(c),f′′(c),f′′′(c),…f(c), f'(c), f''(c), f'''(c), \ldots
  2. Form coefficients an=f(n)(c)/n!a_n = f^{(n)}(c)/n!
  3. Write ∑an(x−c)n\sum a_n(x-c)^n

Example: f(x)=xf(x) = \sqrt{x} at c=4c = 4

nnf(n)(x)f^{(n)}(x)f(n)(4)f^{(n)}(4)an=f(n)(4)/n!a_n = f^{(n)}(4)/n!
0x1/2x^{1/2}2222
112x−1/2\frac{1}{2}x^{-1/2}14\frac{1}{4}14\frac{1}{4}
2−14x−3/2-\frac{1}{4}x^{-3/2}−132-\frac{1}{32}−164-\frac{1}{64}
338x−5/2\frac{3}{8}x^{-5/2}3256\frac{3}{256}3256⋅6=1512\frac{3}{256 \cdot 6} = \frac{1}{512}

x≈2+14(x−4)−164(x−4)2+1512(x−4)3−⋯\sqrt{x} \approx 2 + \frac{1}{4}(x-4) - \frac{1}{64}(x-4)^2 + \frac{1}{512}(x-4)^3 - \cdots

AP Tip: This is the method for functions NOT in the "big six" list. You compute derivatives until a pattern emerges or until you have enough terms.

Taylor Series at Non-Zero Centers

Example: exe^x centered at c=1c = 1

f(n)(1)=ef^{(n)}(1) = e for all nn.

ex=∑n=0∞en!(x−1)n=e[1+(x−1)+(x−1)22!+⋯ ]e^x = \sum_{n=0}^\infty \frac{e}{n!}(x-1)^n = e\left[1 + (x-1) + \frac{(x-1)^2}{2!} + \cdots\right]

Example: sin⁡x\sin x centered at c=π/2c = \pi/2

nnf(n)(π/2)f^{(n)}(\pi/2)
011
100
2−1-1
300

sin⁡x=1−(x−π/2)22!+(x−π/2)44!−⋯\sin x = 1 - \frac{(x - \pi/2)^2}{2!} + \frac{(x-\pi/2)^4}{4!} - \cdots

Notice this looks like cos⁡(x−π/2)\cos(x - \pi/2), which makes sense since sin⁡x=cos⁡(x−π/2)\sin x = \cos(x - \pi/2)!

Computing Taylor Series

Derivative Computation

Taylor at Non-Zero Center

Summary

  • Build Taylor series via derivative tables
  • Non-zero centers: evaluate at cc, use (x−c)n(x-c)^n
  • Look for patterns in derivatives (cyclic, factorial, powers)
  • On the AP exam, you typically need 3-4 terms, not the general formula

Next: Part 3 — Known Series and Manipulation Techniques.

Part 3: Problem-Solving Patterns

Taylor & Maclaurin — Known Series & Manipulation

Part 3 of 7 — Using Known Series to Build New Ones

The Big Six (Must Memorize)

11−x=∑n=0∞xn,∣x∣<1ex=∑n=0∞xnn!,all xsin⁡x=∑n=0∞(−1)nx2n+1(2n+1)!,all xcos⁡x=∑n=0∞(−1)nx2n(2n)!,all xln⁡(1+x)=∑n=1∞(−1)n+1xnn,−1<x≤1arctan⁡x=∑n=0∞(−1)nx2n+12n+1,−1≤x≤1\boxed{\begin{aligned} \frac{1}{1-x} &= \sum_{n=0}^\infty x^n, \quad |x| < 1 \\ e^x &= \sum_{n=0}^\infty \frac{x^n}{n!}, \quad \text{all } x \\ \sin x &= \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}, \quad \text{all } x \\ \cos x &= \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}, \quad \text{all } x \\ \ln(1+x) &= \sum_{n=1}^\infty \frac{(-1)^{n+1} x^n}{n}, \quad -1 < x \le 1 \\ \arctan x &= \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2n+1}, \quad -1 \le x \le 1 \end{aligned}}

Manipulation Toolkit

TechniqueExample
Substitutione−x2e^{-x^2}: replace xx with −x2-x^2 in exe^x
Multiplicationxcos⁡xx\cos x: multiply cos⁡x\cos x series by xx
Differentiation1/(1−x)21/(1-x)^2: differentiate 1/(1−x)1/(1-x)
Integrationarctan⁡x\arctan x: integrate 1/(1+x2)1/(1+x^2)
Additioncosh⁡x=(ex+e−x)/2\cosh x = (e^x + e^{-x})/2: add two series

AP Tip: Building from known series is MUCH faster than computing derivatives from scratch. Always try this first.

Worked Examples

Example 1: xe−xxe^{-x}

e−x=∑(−x)nn!=∑(−1)nxnn!e^{-x} = \sum \frac{(-x)^n}{n!} = \sum \frac{(-1)^n x^n}{n!}

xe−x=∑n=0∞(−1)nxn+1n!=x−x2+x32−x46+⋯xe^{-x} = \sum_{n=0}^\infty \frac{(-1)^n x^{n+1}}{n!} = x - x^2 + \frac{x^3}{2} - \frac{x^4}{6} + \cdots

Example 2: cos⁡2x\cos^2 x (using identity)

cos⁡2x=1+cos⁡2x2=12+12∑n=0∞(−1)n(2x)2n(2n)!\cos^2 x = \frac{1 + \cos 2x}{2} = \frac{1}{2} + \frac{1}{2}\sum_{n=0}^\infty \frac{(-1)^n (2x)^{2n}}{(2n)!}

=12+12[1−2x2+2x43−⋯ ]= \frac{1}{2} + \frac{1}{2}\left[1 - 2x^2 + \frac{2x^4}{3} - \cdots\right]

Wait: cos⁡(2x)=1−(2x)2/2!+(2x)4/4!−⋯=1−2x2+2x4/3−⋯\cos(2x) = 1 - (2x)^2/2! + (2x)^4/4! - \cdots = 1 - 2x^2 + 2x^4/3 - \cdots

cos⁡2x=1−x2+x4/3−⋯\cos^2 x = 1 - x^2 + x^4/3 - \cdots

Key Insight: Using trig identities + known series is often the fastest approach.

Manipulation Practice

Series Construction

Quick Coefficient

Summary

  • Always try manipulating known series before computing derivatives
  • Substitution, multiplication, and differentiation/integration are the main tools
  • Trig identities can simplify series construction
  • Only even/odd powers? Pay attention to symmetry

Next: Part 4 — Taylor's Theorem and the Remainder.

Part 4: Graphs and Interpretation

Taylor & Maclaurin — Taylor's Theorem & Remainder

Part 4 of 7 — The Lagrange Error Bound

Taylor's Theorem

If ff has (n+1)(n+1) continuous derivatives, then:

f(x)=Tn(x)+Rn(x)f(x) = T_n(x) + R_n(x)

where Rn(x)=R_n(x) = the remainder (error of approximation).

The Lagrange Error Bound

∣Rn(x)∣≤M⋅∣x−c∣n+1(n+1)!\boxed{|R_n(x)| \le \frac{M \cdot |x - c|^{n+1}}{(n+1)!}}

where M=max⁡t∣f(n+1)(t)∣M = \max_{t}|f^{(n+1)}(t)| on the interval between cc and xx.

Comparison of Error Bounds

BoundWhen to UseFormula
LagrangeAny Taylor polynomial$M
ASTAlternating Taylor series$

AP Tip: Use the AST error bound when the series alternates — it's simpler. Use Lagrange when it doesn't alternate or when specifically asked.

Example: Bound the Error of ex≈T3(x)e^x \approx T_3(x) at x=0.5x = 0.5

T3(0.5)=1+0.5+0.125+0.02083‾=1.64583‾T_3(0.5) = 1 + 0.5 + 0.125 + 0.0208\overline{3} = 1.6458\overline{3}

For the Lagrange bound: f(4)(x)=exf^{(4)}(x) = e^x

M=max⁡0≤t≤0.5et=e0.5≈1.649M = \max_{0 \le t \le 0.5} e^t = e^{0.5} \approx 1.649

∣R3(0.5)∣≤1.649⋅(0.5)44!=1.649⋅0.062524≈0.00429|R_3(0.5)| \le \frac{1.649 \cdot (0.5)^4}{4!} = \frac{1.649 \cdot 0.0625}{24} \approx 0.00429

Actual: e0.5≈1.6487e^{0.5} \approx 1.6487, error ≈0.0029\approx 0.0029. The bound (0.00430.0043) is correct and conservative.

Common MM Values

Functionf(n+1)(x)f^{(n+1)}(x)MM on [0,x0][0, x_0]
exe^xexe^xex0e^{x_0} (or use e1=3e^1 = 3 as crude bound)
sin⁡x\sin x±sin⁡x\pm\sin x or ±cos⁡x\pm\cos xM=1M = 1 always!
cos⁡x\cos x±sin⁡x\pm\sin x or ±cos⁡x\pm\cos xM=1M = 1 always!

Lagrange Error Practice

Error Bound Decisions

Lagrange Computation

Summary

  • Lagrange Error: ∣Rn(x)∣≤M∣x−c∣n+1/(n+1)!|R_n(x)| \le M|x-c|^{n+1}/(n+1)!
  • Find MM by bounding ∣f(n+1)∣|f^{(n+1)}| on the interval
  • For sin⁡/cos⁡\sin/\cos: M=1M = 1 always
  • For exe^x: M=e∣x∣M = e^{|x|} (or use crude bound like 33)
  • Use AST when series alternates (it's tighter and easier)

Next: Part 5 — AP FRQ Strategies for Taylor Series.

Part 5: Applications

Taylor & Maclaurin — AP FRQ Strategies

Part 5 of 7 — Exam Techniques

The FRQ Taylor Series Question

This appears on virtually EVERY BC exam. The typical structure:

Part (a): Write the first 4 nonzero terms and the general term of the Taylor/Maclaurin series for ff.

Part (b): Find the interval/radius of convergence.

Part (c): Use the series to approximate a value or integral.

Part (d): Show the approximation has error less than some bound.

Part (a) Strategy

If ff is...Strategy
A known function (exe^x, sin⁡x\sin x, etc.)Write the known series directly
A composition/productManipulate known series
An unfamiliar functionCompute derivatives at center
Given as a DE solutionMatch coefficients

AP Tip: "General term" means write ∑\sum notation with nn. This is where students lose the most points — verify your general term by checking it produces the first few terms correctly.

Part (b): Interval of Convergence

Always use Ratio Test → test endpoints.

Write your answer as an interval with proper notation: [−1,1)[-1,1), not "−1-1 to 11."

Part (c): Approximation

Substitute the given value into your series: sin⁡(0.5)≈0.5−(0.5)3/6+(0.5)5/120=0.5−0.02083+0.00026=0.47943\sin(0.5) \approx 0.5 - (0.5)^3/6 + (0.5)^5/120 = 0.5 - 0.02083 + 0.00026 = 0.47943

Part (d): Error Bound

Choose between:

  • AST Error Bound if the series alternates (simpler!)
  • Lagrange Error Bound if not alternating or specifically asked

Template for AST: "Since the series is alternating with decreasing terms converging to 00, the error is bounded by the first omitted term: ∣R∣≤∣aN+1∣=…<ϵ|R| \le |a_{N+1}| = \ldots < \epsilon."

Template for Lagrange: "By Taylor's theorem, ∣Rn(x)∣≤M∣x−c∣n+1/(n+1)!|R_n(x)| \le M|x-c|^{n+1}/(n+1)! where M=max⁡∣f(n+1)∣=…M = \max|f^{(n+1)}| = \ldots"

Scoring Insight

Each part is typically worth 2-3 points. Justification language matters — use precise mathematical statements.

AP Question Types

FRQ Decisions

FRQ Practice

Summary

  • The Taylor FRQ has a predictable structure: series → IOC → approx → error
  • Use known series when possible; compute derivatives as last resort
  • For error bounds: AST when alternating, Lagrange otherwise
  • Verify your general term reproduces the terms you wrote
  • Show ALL work — the AP graders need to see your reasoning

Next: Part 6 — Problem-Solving Workshop.

Part 6: Exam Strategy

Taylor & Maclaurin — Problem-Solving Workshop

Part 6 of 7 — Mixed Practice

Workshop Focus

This workshop covers the full range of Taylor series tasks:

  • Computing series from scratch
  • Manipulating known series
  • Finding intervals of convergence
  • Error bound calculations
  • Integrating/differentiating series

Workshop Problems

Series Identification

Derivative from Series

Workshop Takeaways

  • Derivative table method for unfamiliar functions (like tan⁡x\tan x)
  • Binomial series for (1+x)p(1+x)^p when pp is not a positive integer
  • Series evaluation by identifying the function
  • Integration of series for functions with no elementary antiderivative

Next: Part 7 — Comprehensive Review.

Part 7: Mixed Review

Taylor & Maclaurin — Comprehensive Review

Part 7 of 7 — Complete Topic Review

Master Reference

ConceptKey Formula
Taylor series∑f(n)(c)(x−c)n/n!\sum f^{(n)}(c)(x-c)^n/n!
MaclaurinTaylor at c=0c = 0
Tn(x)T_n(x)Partial sum through degree nn
Lagrange error$
Coefficient ↔ derivativean=f(n)(c)/n!a_n = f^{(n)}(c)/n! ↔ f(n)(c)=n!⋅anf^{(n)}(c) = n! \cdot a_n

The Six Essential Series

11−x,ex,sin⁡x,cos⁡x,ln⁡(1+x),arctan⁡x\boxed{\frac{1}{1-x},\quad e^x,\quad \sin x,\quad \cos x,\quad \ln(1+x),\quad \arctan x}

Key Fact: Taylor/Maclaurin series is the single most heavily tested BC topic. Expect 4+ MC questions and a full FRQ.

Comprehensive Review MC

More Review

Final Drill

Final Challenge

Taylor & Maclaurin — Complete Summary

You've mastered:

  • Taylor formula — ∑f(n)(c)(x−c)n/n!\sum f^{(n)}(c)(x-c)^n/n!
  • Computing from scratch — derivative tables
  • Known series manipulation — substitution, products, differentiation, integration
  • Lagrange error bound — M∣x−c∣n+1/(n+1)!M|x-c|^{n+1}/(n+1)!
  • AP FRQ strategies — the 4-part structure
  • Coefficient-derivative connection — f(n)(c)=n!⋅anf^{(n)}(c) = n! \cdot a_n

Up Next: Lagrange Error Bound — deeper exploration of remainder estimation.