Explanation:
1. f(x) = cos(x) → f(π/3) = ½
f'(x) = -sin(x) → f'(π/3) = -½√3
P₁(x) = f(a) + f'(a)/1! (x − a)
P₁(x) = ½ − ½√3 (x − π/3)
2. f(x) is a third order polynomial, so any third order Taylor series will equal f(x).
The remainder is the difference between f(x) and the partial sum. So R₅ = 0 and R₂ = -5x² + 4x³.
I and II only
3. eˣ = 1 + x/1! + x²/2! + x³/3! + ...
e^(0.1) = 1 + 0.1/1! + 0.1²/2! + 0.1³/3! + ...
0.1ⁿ⁺¹/(n+1)! < 0.00001
By trial and error:
0.1¹⁺¹/(1+1)! = 0.005
0.1²⁺¹/(2+1)! = 0.00017
0.1³⁺¹/(3+1)! = 0.000004
n = 3
4. f(x) = ∑ aₙ xⁿ
g(x) = ∫₀ˣ f(t) dt
g(x) = ∫₀ˣ a₀ dt + ∫₀ˣ a₁t dt + ∫₀ˣ a₂t² dt + ... + C
g(x) = a₀x + a₁x²/2 + a₂x³/3 + ... + g(0)
g(x) = 3 + ∑ aₙ xⁿ⁺¹/(n+1)
5. f(x) = √x = x^½ → f(4) = 2
f'(x) = ½ x^(-½) → f'(4) = ¼
f''(x) = -¼ x^(-³/₂) → f''(4) = -¹/₃₂
f⁽³⁾(x) = ⅜ x^(-⁵/₂) → f⁽³⁾(4) = ³/₂₅₆
f⁽⁴⁾(x) = -¹⁵/₁₆ x^(-⁷/₂) → f⁽⁴⁾(4) = -¹⁵/₂₀₄₈
P₄(x) = f(4) + f'(4) (x−4) / 1! + f''(4) (x−4)² / 2! + f⁽³⁾(4) (x−4)³ / 3! + f⁽⁴⁾(4) (x−4)⁴ / 4!
P₄(x) = 2 + ¼ (x−4) − ¹/₆₄ (x−4)² + ¹/₅₁₂ (x−4)³ − ⁵/₁₆₃₈₄ (x−4)⁴
P₄(4.2) = 2 + ¼ (0.2) − ¹/₆₄ (0.2)² + ¹/₅₁₂ (0.2)³ − ⁵/₁₆₃₈₄ (0.2)⁴
P₄(4.2) = 2.049390137