Maclaurin Series in Depth: Error Bounds, Limits & Series-Based Approximations

First, rewrite the function to match the geometric series form 1/(1-r). Factor out a 1/2 to get (½) * [1/(1 - (-x/2))]. Using the geometric series sum with r = -x/2, we get (½) * Sum of (-x/2)ⁿ. This simplifies to (½) * (-1)ⁿ * xⁿ / 2ⁿ = (-1)ⁿ xⁿ / 2ⁿ⁺¹.

cos(x) ≈ 1 - x²/2! + x⁴/4!. For x=0.1: 1 - 0.01/2 + 0.0001/24 = 1 - 0.005 + 0.000004167 ≈ 0.995004167.

We divide (x - x³/6 + ...) by (1 - x²/2 + ...). The first term is x/1 = x. Multiplying x by the denominator gives x - x³/2. Subtracting this from the numerator: (-x³/6) - (-x³/2) = -x³/6 + 3x³/6 = 2x³/6 = x³/3. The next term in the quotient is (x³/3)/1 = x³/3. Thus, tan x = x + x³/3 + …

Using cos(x) = 1 - x²/2! + x⁴/4! - x⁶/6! + ..., the numerator 1 - cos(x) equals x²/2! - x⁴/4! + x⁶/6! - ... . The coefficient of x² is 1/2! = 1/2. This coefficient becomes the constant term after dividing by x² and thus determines the limit value. This question isolates the specific step of extracting the critical coefficient from the series.

This series represents the even terms of the Maclaurin series for eˣ and e⁻ˣ. We know that cosh(x) = (eˣ + e⁻ˣ)/2 = n=0 x²n/(2n)! . Therefore, at x = 1, cosh(1) = (e + e^(-1))/2 = Σ 1/(2n)! from n=0 to ∞.

Start by substituting the series: (eˣ - 1)/x = [(1 + x + x²/2! + x³/3! + x⁴/4! + ...) - 1]/x. The 1 and -1 cancel, leaving (x + x²/2! + x³/3! + x⁴/4! + ...)/x. Now divide each term in the numerator by x: x/x = 1, (x²/2!)/x = x/2!, (x³/3!)/x = x²/3!, and so on. This gives the simplified series 1 + x/2! + x²/3! + x³/4! + ... Option B is the numerator before dividing by x. Option C is the original series for eˣ. Option D incorrectly separates 1/x.

The Maclaurin series for arctan(x) is Sum of (-1)ⁿ x^(2n+1) / (2n+1). To find the series for x * arctan(x), we multiply the arctan series by x. This adds 1 to the exponent of x, changing x^(2n+1) to x^(2n+2). The coefficients and denominators remain unchanged.

The series converges to the function inside the radius of convergence (|x| < 1), and because it is alternating, the error when stopping at any term is less than the absolute value of the next term.

After substituting the series for eˣ and canceling the 1 and x terms, the numerator becomes x²/2! + x³/3! + x⁴/4! + ... . We can write this sum using a general index n starting from 0. For the first term, when n=0, we need x^(0+2) / (0+2)! which is x²/2!. For the second term, when n=1, we need x^(1+2) / (1+2)! which is x³/3!. This pattern continues for all subsequent terms. Therefore, the general term is x^(n+2) / (n+2)! for n ≥ 0. Option A would incorrectly include the canceled 1 and x terms. Option C shifts the exponent and factorial by 1 instead of 2. Option D would only include even powers of x, but the series has all powers starting from x².

To find the Maclaurin series for e²ˣ, one must substitute (2x) for x in the standard series 1 + x + x²/2! + .... This yields 1 + (2x) + (2x)²/2! = 1 + 2x + 2x². The student incorrectly expanded the quadratic term (likely writing x² or x²/2 instead of 2x²), which is a common error when composing series.

The numerator cancels up to x², so we need the next term in the series to get a non-zero result after division by x³. The eˣ series must include at least the x³ term: 1 + x + x²/2! + x³/3! + ... .

The differential equation y'' + y = 0 with y(0) = 0 and y'(0) = 1 describes the function y(x) = sin(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 a₁ = 1, a_0 = 0, and the pattern a_(2n+1) = (-1)ⁿ/(2n+1)!, a_(2n) = 0, giving sin(x) = Σ (-1)ⁿ x^(2n+1)/(2n+1)!.

sin(t²) = t² - (t²)³/3! + ... = t² - t⁶/6 + ... . Multiplying by t gives t³ - t⁷/6 + ... . The first non-zero term is t³.

eˣ = 1 + x + x²/2 + x³/6 + ... and e⁻ˣ = 1 - x + x²/2 - x³/6 + ... . Adding gives 2 + x² + x⁴/12 + ... . Subtracting 2 leaves x² + x⁴/12 + ... . The lowest power is x².

For sin x / x, substituting just the first term (x) of the sin x series is sufficient: (x + ...)/x → 1. The others require more terms before non-zero constants emerge after division.