Maclaurin series practice questions.

Determine the Maclaurin series of sin x, up to and including the term in x⁵.

• 1 − x²/2 + x⁴/24
• x + x³/6 + x⁵/120
• x − x³/6 + x⁵/120
• x − x³/3 + x⁵/5

Answer: x − x³/6 + x⁵/120

sin x = x − x³/3! + x⁵/5! − … with alternating signs and odd powers; 3! = 6 and 5! = 120. The last option is the series for arctan x, a classic mix-up.

Calculate the coefficient of x³ in the Maclaurin series of e^(2x).

• 4/3
• 1/6
• 8/3
• 2/3

Answer: 4/3

Substituting u = 2x into eᵘ = Σ uⁿ/n! gives the x³ term (2x)³/3! = 8x³/6, so the coefficient is 4/3. Ignoring the factor 2 altogether gives 1/6, and squaring instead of cubing it gives 2/3.

State the first three non-zero terms of the Maclaurin series of ln(1 + x).

• 1 + x + x²/2
• x + x²/2 + x³/3
• x − x³/6 + x⁵/120
• x − x²/2 + x³/3

Answer: x − x²/2 + x³/3

ln(1 + x) = x − x²/2 + x³/3 − … with alternating signs and no factorials. The all-positive version x + x²/2 + x³/3 is the series of −ln(1 − x).

Using Maclaurin series, determine lim (x→0) of (1 − cos x)/x².

• 0
• 1/2
• 1
• 2

Answer: 1/2

cos x = 1 − x²/2 + x⁴/24 − …, so 1 − cos x = x²/2 − x⁴/24 + …. Dividing by x² gives 1/2 − x²/24 + … → 1/2 as x → 0.

Which of the following is the Maclaurin series of arctan x?

• x − x³/3 + x⁵/5 − …
• x + x³/3 + x⁵/5 + …
• x − x³/6 + x⁵/120 − …
• 1 − x²/2 + x⁴/24 − …

Answer: x − x³/3 + x⁵/5 − …

Integrating 1/(1 + x²) = 1 − x² + x⁴ − … term by term gives arctan x = x − x³/3 + x⁵/5 − …; the denominators are the odd integers, not factorials.

Determine the first two non-zero terms of the Maclaurin series of tan x.

• x − x³/3
• x + x³/6
• x + x³/3
• x − x³/6

Answer: x + x³/3

With f(x) = tan x: f(0) = 0, f′(0) = 1, f″(0) = 0, f‴(0) = 2, so tan x = x + 2x³/3! + … = x + x³/3 + …. The option x − x³/3 is arctan x.