From Taylor Polynomial to Taylor Series: Infinite Expansions Around a Point

The Taylor series for 1/x centered at 1 is 1 - (x-1) + (x-1)² - (x-1)³... If we plug in x=1.1, then (x-1) = 0.1, resulting in the series 1 - 0.1 + (0.1)²... Thus, the sum is 1/1.1.

We start with the general term for the Taylor series, where the coefficient of (x-a)ⁿ is f⁽ⁿ⁾(a)/n!. For f(x) = eˣ, all derivatives f⁽ⁿ⁾(x) = eˣ, so f⁽ⁿ⁾(1) = e. For n=4, the coefficient is e/4! = e/24.

To find the Taylor polynomial of degree 2 centered at a = π/3, we use the general formula: T(x) = f(a) + f'(a)(x-a) + f''(a)/2!². We need to calculate the function value and its first two derivatives at x = π/3.

First, we evaluate the function f(x) = sin(x) at a = π/3. Since sin(π/3) is √3/2, the first term (the constant term) is √3/2.

Second, we find the first derivative f'(x). The derivative of sin(x) is cos(x). Evaluating this at a = π/3 gives cos(π/3), which equals 1/2. This gives us the coefficient for the linear term: (1/2)(x-π/3).

Third, we find the second derivative f''(x). The derivative of cos(x) is -sin(x). Evaluating this at a = π/3 gives -sin(π/3), which equals -√3/2. We must divide this value by 2! (which is 2) according to the Taylor formula. This gives us the quadratic term: (-√3/2)/2 * (x-π/3)², which is written in Option A as -(√3/2)(x-π/3)²/2.

Combining these three parts, we get: (√3/2) + (½)(x-π/3) - (√3/2)(x-π/3)²/2.

The 3rd-degree Taylor polynomial for ln(x) at 1 is (x-1) - (x-1)²/2 + (x-1)³/3. At x=1.2, let h=0.2, so 0.2 - (0.04)/2 + (0.008)/3 = 0.2 - 0.02 + 0.0026667 = 0.1826667, which approximates 0.1827.

We start with f(x) = eˣ, so f(1) = e. Then f'(x) = eˣ, f'(1) = e. f''(x) = eˣ, f''(1) = e. f'''(x) = eˣ, f'''(1) = e. f^(4)(x) = eˣ, f^(4)(1) = e. The polynomial is f(1) + f'(1)(x-1) + f''(1)(x-1)²/2! + f'''(1)(x-1)³/3! + f^(4)(1)(x-1)⁴/4!. Substituting the values gives e + e (x-1) + e (x-1)²/2 + e (x-1)³/6 + e (x-1)⁴/24. Factoring out e gives e [1 + (x-1) + (x-1)²/2 + (x-1)³/6 + (x-1)⁴/24].

The 4th-degree Taylor polynomial for eˣ centered at 1 is e [1 + (x-1) + (x-1)²/2 + (x-1)³/6 + (x-1)⁴/24]. At x=1.1, x-1=0.1, so e [1 + 0.1 + (0.01)/2 + (0.001)/6 + (0.0001)/24]. Compute inside: 1 + 0.1 + 0.005 + 0.0001667 + 0.000004167 ≈ 1.1051708. Then multiply by e ≈ 2.7182818 * 1.1051708 ≈ 3.004166, which rounds to 3.0042.

We start with f(x) = cos(x), so f(π/3) = cos(π/3) = 1/2. Then we find f'(x) = -sin(x), so f'(π/3) = -√3/2. Next we find f''(x) = -cos(x), so f''(π/3) = -1/2. Then we find f'''(x) = sin(x), so f'''(π/3) = √3/2. The Taylor polynomial is f(π/3) + f'(π/3)(x-π/3) + f''(π/3)(x-π/3)²/2! + f'''(π/3)(x-π/3)³/3!. Substituting the values gives 1/2 + (-√3/2)(x-π/3) + (-½)(x-π/3)²/2 + (√3/2)(x-π/3)³/6.

The 3rd-degree Taylor polynomial for sin(x) centered at π/3 is √3/2 + (½)(x-π/3) - (√3/2)(x-π/3)²/2 - (½)(x-π/3)³/6. At x = π/3 + 0.1, (x-π/3) = 0.1, so √3/2 + (½)(0.1) - (√3/2)(0.01)/2 - (½)(0.001)/6. First √3/2 ≈ 0.8660, (½)(0.1) = 0.05, (√3/2)(0.01)/2 ≈ 0.8660 * 0.005 = 0.00433, (½)(0.001)/6 ≈ 0.5 * 0.001 /6 = 0.0000833. So 0.8660 + 0.05 - 0.00433 - 0.0000833 ≈ 0.9115867, which rounds to 0.9116.

The Lagrange remainder for degree 3 is |R3(x)| ≤ max |f^(4)(ξ)| (|x-1|⁴ / 4! ) for ξ between 1 and 1.1. For f(x) = ln(x), f^(4)(x) = -6/x⁴, so |f^(4)(ξ)| = 6/ξ⁴. Since ξ in (1,1.1), the max is at ξ=1 (6/1⁴ =6) since 6/ξ⁴ is a decreasing function. So the error bound is (6/24)(|1.1-1|)⁴ =  0.000025.

The Taylor series for ln(x) centered at 1 is Σ (-1)^{k+1} (x-1)^k / k for k=1 to ∞. For x=1.5, h=0.5, the series is alternating and the terms decrease in magnitude for |h|<1. The first 4 terms are up to k=4, the next term is the 5th term |(-1)^{5+1} (0.5)⁵ /5| = (0.5)⁵ /5. By the alternating series error bound, the error is less than the absolute value of the first omitted term, which is (0.5)⁵ /5.

The Taylor series for eˣ centered at 1 is e Σ (x-1)ⁿ / n! from n=0 to ∞. The series for e^(x-1) is Σ (x-1)ⁿ / n!, which converges for all real x by the ratio test, since lim |a_{n+1}/aₙ| = |x-1|/(n+1) → 0 <1 for all x. Therefore, as the degree increases, the polynomial approximates eˣ over the entire real line.

The Taylor theorem states that the remainder Rₙ(x) = f(x) - Pₙ(x) = fⁿ⁺¹(ξ) (x-a)ⁿ⁺¹ / (n+1)! for some ξ between a and x. For f(x) = cos(x), a = π/3, the remainder is fⁿ⁺¹(ξ) (x-π/3)ⁿ⁺¹ / (n+1)! where ξ is between x and π/3.

The 3rd-degree Taylor polynomial for cos(x) centered at π/3 is 1/2 - (√3/2)(x-π/3) - (½)(x-π/3)²/2 + (√3/2)(x-π/3)³/6. At x = π/3 + 0.1, (x-π/3) = 0.1, so 0.5 - (0.8660)(0.1) - (0.5)(0.01)/2 + (0.8660)(0.001)/6. Compute -0.86600.1 = -0.0866, -0.50.005 = -0.0025, +0.8660*0.001 /6 ≈ 0.000866 /6 ≈ 0.000144. So 0.5 -0.0866 -0.0025 +0.000144 ≈ 0.411044, which rounds to 0.4110.

The Lagrange remainder for degree 4 is R4(x) = f^(5)(c) (x-1)⁵ /5! for some c between 1 and x. For f(x) = eˣ, f^(5)(c) = e^c, x=1.1, (x-1)=0.1, 5! =120, so e^c (0.1)⁵ /120 for some c between 1 and 1.1.

We are using a third-degree Taylor polynomial (n = 3), so the Lagrange remainder uses the fourth derivative. The distance from the center a = 2 to x = 3 is |3−2| = 1. The error bound is |R₃(3)| ≤ (max |f^{(4)}(z)|) · |3−2|^{4}/4! ≤ 12 · 1⁴/24 = 12/24 = 1/2, which is exactly (1⁴/4!) · 12.