Maclaurin Series Basics: Key Expansions & Geometric Links

The Maclaurin series for eˣ is derived from evaluating all derivatives of eˣ at x=0. Since every derivative of eˣ is eˣ, and e⁰= 1, we have f⁽ⁿ⁾(0) = 1 for all n. The general Maclaurin series formula is Σ  from n=0 to infinity of f⁽ⁿ⁾(0) * xⁿ / n!. Substituting f⁽ⁿ⁾(0) = 1 gives us Σ  from n=0 to infinity of xⁿ / n!. This expands to 1 + x + x²/2! + x³/3! + x⁴/4! + ... Option B is a geometric series. Option C represents cos x. Option D does not match any standard series.

The series for 1/(1-x) is Σ  from n=0 to infinity of xⁿ. Writing out the terms: when n=0, x⁰= 1. When n=1, x^1 = x. When n=2, x². When n=3, x³. So the first four terms are 1 + x + x² + x³. Option B alternates signs, which is incorrect for this series. Option C includes factorial denominators, which belong to eˣ, not this geometric series. Option D only includes even powers.

ln(1+x) = x - x²/2 + x³/3 - ... For x=0.1: 0.1 - (0.01)/2 + (0.001)/3 = 0.1 - 0.005 + 0.000333... ≈ 0.095333.

Using the general formula Σn=0 ∞ (-1)ⁿ x^(2n+1) / (2n+1)!, we compute the first three terms. For n=0: (-1)⁰x^1 / 1! = x. For n=1: (-1)^1 x³ / 3! = -x³/6. For n=2: (-1)² x⁵ / 5! = x⁵/120. The series is x - x³/6 + x⁵/120.

The first three terms of eˣ are 1 + x + x²/2. For x = 0.2: 1 + 0.2 + (0.04)/2 = 1 + 0.2 + 0.02 = 1.22.

Using the general formula Σ  from n=0 to infinity of (-1)ⁿ x^(2n) / (2n)!, we compute the first four terms. For n=0: (-1)⁰x⁰/ 0! = 1. For n=1: (-1)^1 x² / 2! = -x²/2. For n=2: (-1)² x⁴ / 4! = x⁴/24. For n=3: (-1)³ x⁶ / 6! = -x⁶/720. The series is 1 - x²/2 + x⁴/24 - x⁶/720. Option A shows these numerical values. Option C shows the same values using factorial notation (2! = 2, 4! = 24, 6! = 720). Since both represent the same series, D is correct.

A Maclaurin series is simply a Taylor series centered at a = 0. All the formulas are identical except that (x - 0)ⁿ = xⁿ and all derivatives are evaluated at x = 0.

To find the series for e^(x²), we substitute x² in place of x in the standard eˣ series: Σ  from n=0 to infinity of (x²)ⁿ / n!. Since (x²)ⁿ = x^(2n), this equals Σ  from n=0 to infinity of x^(2n) / n!. Therefore both representations A and C are mathematically equivalent. Option B is the original eˣ series without substitution. The expansion gives 1 + x² + x⁴/2! + x⁶/3! + x^8/4! + …

Starting with the sin x series: n=0 (-1)ⁿ x^(2n+1) / (2n+1)!. Replacing x with 2x gives n=0 (-1)ⁿ (2x)^(2n+1) / (2n+1)!. Expanding (2x)^(2n+1) gives 2^(2n+1) x^(2n+1), so option B is equivalent. Option C shows the first few terms explicitly. Since all three represent the same series, D is the correct answer. The series expands to 2x - 8x³/6 + 32x⁵/120 - …

The series for cos x is n=0 (-1)ⁿ x^(2n) / (2n)!. Substituting x³ for x gives n=0 (-1)ⁿ (x³)^(2n) / (2n)!. Since (x³)^(2n) = x^(6n), this equals n=0 (-1)ⁿ x^(6n) / (2n)!. Therefore both A and B represent the same series. Option C shows the expanded first four terms, which matches this pattern: x⁰= 1, x⁶ = x^(61), x^12 = x^(62), x^18 = x^(6*3).

The Maclaurin series for eˣ is Σ  from n=0 to infinity of xⁿ / n!. Using the ratio test, we examine the lim_{n 🠒∞}  |aₙ₊₁/aₙ| = |xⁿ⁺¹/(n+1)! * n!/xⁿ| = |x/(n+1)|. As n approaches infinity, this limit equals 0 for any finite value of x, which is less than 1. Since the ratio test shows the limit is 0 regardless of x, the series converges for all real numbers. Option B is the interval of convergence for 1/(1-x). Option C is too restrictive. Option D incorrectly limits the radius.

If we differentiate the sin x series term by term, we get d/dx[x - x³/3! + x⁵/5! - x⁷/7! + ...] = 1 - 3x²/3! + 5x⁴/5! - 7x⁶/7! + ... = 1 - x²/2! + x⁴/4! - x⁶/6! + ..., which is exactly the Maclaurin series for cos x. This matches the calculus fact that d/dx(sin x) = cos x. Option B is false because the derivative of cos x is -sin x, not sin x. Option C is false as they are clearly related through differentiation. Option D is incorrect substitution.

The Maclaurin series for eˣ is 1 + x + x²/2! + x³/3! + x⁴/4! + ... When substituting -x for x, we get 1 - x + x²/2! - x³/3! + x⁴/4! - ... The student's series is missing the factorial denominators, which are essential for the correct coefficients. Each term's denominator must be n! to match the derivatives of eˣ evaluated at 0. Option B is wrong because the signs correctly alternate for e⁻ˣ. Option C is wrong because e⁻ˣ has all powers. Option D is wrong because Maclaurin series are infinite.

Looking at the pattern: only even powers appear, with alternating signs, and denominators are 0!, 2!, 4!, 6! (remember 0! = 1, 2! = 2, 4! = 24, 6! = 720). This matches the structure of the cos x series: Σ  from n=0 to infinity of (-1)ⁿ x^(2n) / (2n)!, which expands to exactly these terms. Option B (sin x) would have odd powers. Option C (e^(-x²)) would have all terms positive. Option D (ln(1+x²)) would have denominators 1, 2, 3, 4,... not factorials.

The Maclaurin series for sin x has only odd powers. The term x⁷ appears when 2n+1 = 7, which gives n = 3. The general term is (-1)ⁿ x^(2n+1) / (2n+1)!. For n=3, we get (-1)³ x⁷ / 7! = -x⁷ / 5040. Therefore the coefficient is -1/5040. Option A has the wrong sign. Option C is missing the factorial. Option D would be correct for an even power like x⁶, but 7 is odd so the coefficient is non-zero.