By Fatskills Exam Guides Team — the exam nerds behind 28,500+ quizzes and 2.1M practice questions across 500+ global exams.
Mastering Taylor and Maclaurin series lets you approximate any function—like predicting projectile motion in physics or acing the 10-15% of your AP Calc BC exam that covers infinite series. Miss this, and you’re leaving easy points on the table.
Formula: [ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n ] Variables: - f(x) = function you’re approximating - a = center point (where you’re expanding) - f⁽ⁿ⁾(a) = n-th derivative of f at x=a - n! = factorial of n
Memorise This. (but the AP exam often provides it).
Formula: [ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n ] Variables: - Same as Taylor, but a=0.
Memorise This. (it’s just Taylor at a=0).
MEMORIZE THESE (AP exams test these constantly).
Formula: [ R = \lim_{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right| ] Variables: - aₙ = general term of the series - R = radius of convergence
GIVEN ON EXAM SHEET (but you must know how to use it).
Step 1: Choose the center (a) - If the problem says "Maclaurin," a=0. - If it says "Taylor at x=a," use that a.
Step 2: Compute derivatives at x=a - Find f(a), f'(a), f''(a), f'''(a), etc., until you see a pattern.
Step 3: Plug into the Taylor formula - Write the series: (\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n).
Step 4: Simplify the general term - Look for patterns (e.g., alternating signs, factorials, powers).
Step 5: Write the first 3-4 terms explicitly - Exams often ask for the "first four non-zero terms."
Step 1: Identify the general term (aₙ) - For a series (\sum a_n (x - a)^n), extract aₙ.
Step 2: Apply the Ratio Test - Compute ( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ). - If L=0, R=∞ (converges everywhere). - If L=∞, R=0 (converges only at x=a). - Otherwise, R = 1/L.
Step 3: Check endpoints (if asked for interval) - Plug x = a + R and x = a - R into the series. - Test convergence (e.g., p-series, alternating series test).
Problem: Find the Maclaurin series for f(x) = eˣ and its radius of convergence.
Step 1: a=0 (Maclaurin). Step 2: Compute derivatives at x=0: - f(x) = eˣ → f(0) = 1 - f'(x) = eˣ → f'(0) = 1 - f''(x) = eˣ → f''(0) = 1 - All derivatives at x=0 are 1.
Step 3: Plug into Taylor formula: [ eˣ = \sum_{n=0}^{\infty} \frac{1}{n!} x^n ]
Step 4: Simplify: [ eˣ = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots ]
Step 5: Radius of convergence (Ratio Test): - General term: aₙ = 1/n! - ( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{1/(n+1)!}{1/n!} = \lim_{n \to \infty} \frac{1}{n+1} = 0 ) - Since L=0, R=∞.
Answer: [ eˣ = \sum_{n=0}^{\infty} \frac{x^n}{n!}, \quad R = \infty ]
What we did and why: We used the definition of Maclaurin series (Taylor at a=0) and the Ratio Test to confirm it converges everywhere. This is a must-know series for the AP exam.
Problem: Find the Taylor series for f(x) = ln(x) centered at a=1 and its radius of convergence.
Step 1: a=1. Step 2: Compute derivatives at x=1: - f(x) = ln(x) → f(1) = 0 - f'(x) = 1/x → f'(1) = 1 - f''(x) = -1/x² → f''(1) = -1 - f'''(x) = 2/x³ → f'''(1) = 2 - f⁽⁴⁾(x) = -6/x⁴ → f⁽⁴⁾(1) = -6 - Pattern: f⁽ⁿ⁾(1) = (-1)ⁿ⁺¹ (n-1)!
Step 3: Plug into Taylor formula: [ \ln(x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} (n-1)!}{n!} (x - 1)^n ] Simplify: [ \ln(x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} (x - 1)^n ]
Step 4: First 4 terms: [ \ln(x) = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4} + \dots ]
Step 5: Radius of convergence (Ratio Test): - General term: aₙ = (-1)ⁿ⁺¹ / n - ( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{n}{n+1} = 1 ) - So, R = 1/L = 1.
Step 6: Check endpoints: - At x=2: (\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}) (converges by Alternating Series Test). - At x=0: (\sum_{n=1}^{\infty} \frac{-1}{n}) (diverges, harmonic series).
Answer: [ \ln(x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} (x - 1)^n, \quad R = 1 ] Interval of convergence: (0, 2].
What we did and why: We derived the series by computing derivatives, simplified the general term, and used the Ratio Test to find R. Checking endpoints is critical—exams often ask for the interval, not just R.
Problem: Use the Maclaurin series for sin(x) to approximate sin(0.1) with 3 non-zero terms. Estimate the error.
Step 1: Recall Maclaurin series for sin(x): [ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots ]
Step 2: First 3 non-zero terms: [ \sin(x) \approx x - \frac{x^3}{6} + \frac{x^5}{120} ]
Step 3: Plug in x=0.1: [ \sin(0.1) \approx 0.1 - \frac{(0.1)^3}{6} + \frac{(0.1)^5}{120} ] [ = 0.1 - 0.0001667 + 0.000000083 ] [ \approx 0.0998334 ]
Step 4: Estimate error (Lagrange remainder): - The next term is (-\frac{x^7}{7!}). - Error ≤ (\left| \frac{(0.1)^7}{5040} \right| \approx 2 \times 10^{-10}) (negligible).
Answer: [ \sin(0.1) \approx 0.0998334 ] Error ≤ (2 \times 10^{-10}).
What we did and why: We used the known Maclaurin series for sin(x), truncated it, and plugged in x=0.1. The error bound ensures the approximation is accurate enough for full credit.
"Listen up—this is your 60-second cheat sheet for Taylor and Maclaurin series. First, memorize the big four: eˣ, sin(x), cos(x), and 1/(1-x). For any other function, compute derivatives at the center a, plug into the Taylor formula, and simplify. To find the radius of convergence, use the Ratio Test on the general term—if the limit is L, R=1/L. Always check endpoints unless R=∞. On the exam, if they ask for an approximation, truncate the series and estimate the error. You’ve got this—now go ace that test!
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