Finding Taylor or Maclaurin Series for a Function: Practice Questions (Calculus BC 10.14)

Answer: (B). The derivatives of cos x at 0 cycle 1, 0, −1, 0, so only even powers appear, with alternating signs and the matching factorial: 1 − x²/2! + x⁴/4! − … .

• (A) is the series for sin x: odd powers, starting with x. But cos 0 = 1, so the series must start with 1.
• (C) has no alternating sign. Its x² coefficient is +1/2, which would make the second derivative at 0 equal to +1; for cos x it is −cos 0 = −1.
• (D) uses n! instead of (2n)!. Its x⁴ coefficient is 1/2!, but cos x needs 1/4! = 1/24.

Answer: (B). π^(2n)/4ⁿ = (π/2)^(2n), so the series is Σ (−1)ⁿ(π/2)^(2n)/(2n)!. The even powers, the (2n)! and the alternating signs match the cos x series with x = π/2. So the sum is cos(π/2) = 0.

• (A) is cos π: it misses the 4ⁿ in the denominator and uses x = π.
• (C) is cos(π/4): it divides π by 4 instead of combining π^(2n)/4ⁿ = (π/2)^(2n).
• (D) is sin(π/2): it matches the wrong series. The sin series has odd powers and (2n + 1)!.

Answer: (A). Every derivative of eˣ is eˣ, so f⁽ⁿ⁾(3) = e³ for every n. The coefficients are e³/n!, and the powers are (x − 3)ⁿ.

• (B) uses the derivative values at 0 (all equal to 1) instead of at 3. Its value at x = 3 is 1, not e³.
• (C) has the right coefficients but powers of x, which belong to a series about 0.
• (D) forgets to divide by n!. Its second derivative at 3 would be 2e³, not e³.

Answer: (A). There is no x term, so a₁ = 0 and f′(0) = 1! · 0 = 0. The x² coefficient is −4, so f″(0) = 2! · (−4) = −8 < 0. By the second derivative test, f has a local maximum at x = 0.

• (B) reverses the second derivative test: f″(0) < 0 means concave down, which gives a maximum.
• (C) reads the coefficient as the derivative. f″(0) = 2! × (−4) = −8.
• (D) does the same for the cubic term. f‴(0) = 3! × 7 = 42.

(a) Derivatives: cos x, −sin x, −cos x, sin x, cos x, … At x = π, cos π = −1 and sin π = 0, so the values are:

f(π) = −1, f′(π) = 0, f″(π) = 1, f‴(π) = 0, f⁽⁴⁾(π) = −1, f⁽⁵⁾(π) = 0, f⁽⁶⁾(π) = 1, …

Only even orders survive, with values −1, 1, −1, 1, … . Dividing by the factorials:

cos x = −1 + (x − π)²/2! − (x − π)⁴/4! + (x − π)⁶/6! − … = Σ from n = 0 to ∞ of (−1)ⁿ⁺¹(x − π)^(2n)/(2n)!

Check: cos x = −cos(x − π), so the answer is minus the cos series with x − π in place of x. ✓

(b) f⁽⁵⁾(π) = −sin π = 0, so the (x − π)⁵ term is 0. Adding a zero term does not change the polynomial, so P₅ = P₄ = −1 + (x − π)²/2 − (x − π)⁴/24.

Suggested mark points (3): 1 for correct derivative values at π; 1 for the four nonzero terms with factorials and signs; 1 for a correct general term and the explanation in (b).

Common error: writing the general term with (−1)ⁿ, which gives +1 as the first term. Check it at n = 0: the series must start with f(π) = −1.

(a) aₙ = f⁽ⁿ⁾(2)/n! = 1/((n² + 1)5ⁿ). So a₀ = 1, a₁ = 1/10, a₂ = 1/125, a₃ = 1/1250.

Taylor series: 1 + (x − 2)/10 + (x − 2)²/125 + (x − 2)³/1250 + … = Σ from n = 0 to ∞ of (x − 2)ⁿ/((n² + 1)5ⁿ)

(b) Ratio: |uₙ₊₁/uₙ| = (|x − 2|/5) · (n² + 1)/((n + 1)² + 1) → |x − 2|/5. So |x − 2| < 5, R = 5, open interval (−3, 7).

• x = 7: x − 2 = 5, so the series is Σ 1/(n² + 1). For n ≥ 1, 0 < 1/(n² + 1) < 1/n², and Σ 1/n² is a convergent p-series (p = 2). By the direct comparison test it converges.
• x = −3: x − 2 = −5, so the series is Σ (−1)ⁿ/(n² + 1). Its absolute values form the convergent series from x = 7, so it converges absolutely.

Interval of convergence: [−3, 7].

(c) P₂(x) = 1 + (x − 2)/10 + (x − 2)²/125. With x − 2 = 0.5: P₂(2.5) = 1 + 0.05 + 0.25/125 = 1.052 (= 263/250).

(d) f′(2) = a₁ · 1! = 1/10 > 0, so f is increasing at x = 2.

Total: 6 points. In (a), forgetting to divide by n! (writing f⁽ⁿ⁾(2) as the coefficient) loses both (a) points, but later parts can earn credit if they follow correctly from the student’s series.

(a) sin(π/4) = cos(π/4) = √2/2. Derivatives: sin x, cos x, −sin x, −cos x. So

f(π/4) = √2/2, f′(π/4) = √2/2, f″(π/4) = −√2/2, f‴(π/4) = −√2/2.

sin x = √2/2 + (√2/2)(x − π/4) − (√2/4)(x − π/4)² − (√2/12)(x − π/4)³ + …

(The last two coefficients are (−√2/2)/2! and (−√2/2)/3!.)

(b) The derivatives repeat every 4 steps: sin, cos, −sin, −cos. At π/4 the values follow the sign pattern +, +, −, −, repeating, always with size √2/2. Since 10 = 4 · 2 + 2, the tenth derivative is −sin x, so f⁽¹⁰⁾(π/4) = −√2/2. The coefficient is (−√2/2)/10! = −√2/(2 · 10!) (about −1.95 × 10⁻⁷).

(c) Every coefficient has size (√2/2)/n!, and none is 0. The ratio of the sizes of consecutive terms is |x − π/4| · n!/(n + 1)! = |x − π/4|/(n + 1), which tends to 0 for every x. By the ratio test, the series converges for every real x (R = ∞).

(d) With h = 0.1: P₃ = (√2/2)(1 + 0.1 − 0.01/2 − 0.001/6) ≈ 0.774164. Calculator: sin(π/4 + 0.1) ≈ 0.774167. They differ by about 0.000003, so the third-degree partial sum is already very close.

Total: 6 points. Acceptable alternative for (c): sin x = sin(π/4)cos(x − π/4) + cos(π/4)sin(x − π/4), and the cos and sin series converge for all real numbers. For (d), a difference of 1 in the sixth decimal place caused by rounding during the working is acceptable with a correct method.