Maclaurin Series from Differential Equations: Modeling with sin, cos, sinh, cosh & Beyond

The Maclaurin series for cos(x) contains only even powers of x with alternating signs. The general term is (-1)ⁿ x²n/(2n)!, where n starts from 0: n=0 gives 1, n=1 gives -x²/2!, n=2 gives x⁴/4!, etc.

We know that 1/(1-x) = Sum of xⁿ. Differentiating both sides with respect to x, the derivative of 1/(1-x) is 1/(1-x)². The derivative of the series Sum of xⁿ is Sum of n*xⁿ⁻¹. Note that the index starts at n=1 because the constant term (n=0) becomes 0 upon differentiation. This can also be written as Sum from n=0 to infinity of (n+1)xⁿ.

Since d/dx[ln(1-x)] = -1/(1-x), we integrate both sides:

ln(1-x) = ∫[-1/(1-x)]dx = -∫0∞[Σn=0∞ xⁿ]dx. Integrating term by term:

ln(1-x) = -Σn=0∞[xⁿ⁺¹/(n+1)] + C. To determine C, we substitute x = 0. So

ln(1-0) = ln(1) = 0 = -Σ[0] + C, so C = 0. Therefore, ln(1-x) = -Σn=0∞[xⁿ⁺¹/(n+1)]. Reindexing by letting k = n+1 gives ln(1-x) = -Σk=1∞[xᵏ/k]. This can be written as:

ln(1-x) = -x - x²/2 - x³/3 - x⁴/4 - … This series converges for |x| < 1, and also at x = -1 (by the alternating series test), but diverges at x = 1.

The Maclaurin series for sin(x) is x - x³/3! + x^5/5! - x^7/7! + ... When we substitute, sin(x) - x + x³/6 = (x - x³/6 + x^5/120 - ...) - x + x³/6 = x^5/120 - ... Dividing by x^5 gives 1/120 - ..., so we need terms up to x^5 in the sin(x) series to get a non-zero constant term after division.

Starting with the Maclaurin series for sin(x) = x - x³/3! + x⁵/5! - x⁷/7! + ..., we substitute x → x² to get sin(x²) = x² - (x²)³/3! + (x²)⁵/5! - (x²)⁷/7! + ... = x² - x⁶/6 + x¹⁰/120 - x¹⁴/5040 + x¹⁸/362880.

The Maclaurin series for e^x = 1 + x + x²/2 + x³/6 + ... So e^(2x) = 1 + 2x + (2x)²/2 + ... = 1 + 2x + 2x² + ... and 2e^x = 2(1 + x + x²/2 + ...) = 2 + 2x + x² + ... Therefore e^(2x) - 2e^x + 1 = (1 + 2x + 2x²) - (2 + 2x + x²) + 1 + higher terms = (1 - 2 + 1) + (2x - 2x) + (2x² - x²) + ... = x² + ... If we only used two terms (up to x), we would get e^(2x) - 2e^x + 1 = (1 + 2x) - 2(1 + x) + 1 = 1 + 2x - 2 - 2x + 1 = 0, which is incorrect. We need at least three terms to get the x² term.

Using the binomial series (1 + x)^p = Σ [p(p-1)...(p-n+1)/n!]xⁿ from n=0 to ∞, with p = -1/2, we get [(-½)(-3/2)(-5/2)...(-(2n-1)/2)/n!]xⁿ = [(-1)ⁿ (1·3·5·...·(2n-1))/(2ⁿ n!)]xⁿ. Since 1·3·5·...·(2n-1) = (2n)!/(2·4·6·...·2n) = (2n)!/(2ⁿ n!), we get [(-1)ⁿ (2n)!/(2ⁿ n! · 2ⁿ n!)]xⁿ = [(-1)ⁿ (2n)!/(4ⁿ (n!)²)]xⁿ.

Starting with the Maclaurin series for eˣ = Σ xⁿ/n! from n=0 to ∞, we substitute x → -x² to get e^(-x²) = Σ (-x²)ⁿ/n! = Σ (-1)ⁿ x^(2n)/n! from n=0 to ∞.

To find the Maclaurin series for arcsin(x), we integrate the series for 1/√(1-x²) = Σ [(2n)!/(4ⁿ (n!)²)]x²n from n=0 to ∞. Integrating term by term: ∫ Σ [(2n)!/(4ⁿ (n!)²)]x²n dx = Σ [(2n)!/(4ⁿ (n!)²)] x^(2n+1)/(2n+1) + C. At x = 0, arcsin(0) = 0, so C = 0.

The hyperbolic sine function is defined as sinh(x) = (eˣ - e⁻ˣ)/2. Using the Maclaurin series for eˣ = Σ xⁿ/n! and e⁻ˣ = Σ (-1)ⁿ xⁿ/n!, we get sinh(x) = (½)[Σ xⁿ/n! - Σ (-1)ⁿ xⁿ/n!] = (½)Σ [1 - (-1)ⁿ]xⁿ/n!. This gives non-zero terms only when n is odd, so sinh(x) = Σ x^(2n+1)/(2n+1)! from n=0 to ∞.

The differential equation y'' - y = 0 with y(0) = 1 and y'(0) = 0 describes the function y(x) = cosh(x). Assuming a Maclaurin series y = Σ aₙ xⁿ, we find y' = Σ n aₙ xⁿ⁻¹ and y'' = Σ n(n-1)aₙ x^(n-2). Substituting into y'' - y = 0 gives Σ n(n-1)aₙ x^(n-2) - Σ aₙ xⁿ = 0. Reindexing and equating coefficients shows that a_0 = 1, a₁ = 0, and the pattern a_(2n) = 1/(2n)!, a_(2n+1) = 0, giving cosh(x) = Σ x²n/(2n)!.

We divide each term in the numerator by x². The first term is (x²/2) / x² = 1/2, which is a constant. The second term is (x³/6) / x² = x/6. The third term is (x⁴/24) / x² = x²/24. The resulting series is 1/2 + x/6 + x²/24 + ... . The question asks for the coefficient of the x term, which is the number multiplying x. In the term x/6, the coefficient is 1/6. Option A would be correct if there were no x term at all. Option B is the constant term, not the x term. Option D is the coefficient of the x² term in the final series.

1 − cos(x) = x²/2 − x⁴/24 + ... lowest power is x².

This is the alternating series representation of arctan(1). Using the Maclaurin series for arctan(x) = Σ (-1)ⁿ x^(2n+1)/(2n+1) from n=0 to ∞, we substitute x = 1 to get arctan(1) = Σ (-1)ⁿ (1)^(2n+1)/(2n+1) = Σ (-1)ⁿ/(2n+1) from n=0 to ∞. Since arctan(1) = π/4, this series converges to π/4.

For cos(3x), substitute 3x into the series: cos(3x) = 1 - (3x)²/2! + (3x)⁴/4! - ... = 1 - 9x²/2 + 81x⁴/24 - ... . The numerator becomes 1 - (1 - 9x²/2 + ...) = 9x²/2 - ... . The coefficient of x² is 9/2.