Answer:
(x − π)⁷ / 5040
(x − 1)³ / 16
Explanation:
Taylor series expansion of a function is:
f(x) = ∑ₙ₌₀°° f⁽ⁿ⁾(x₀) / n! (x − x₀)ⁿ
where f⁽ⁿ⁾(x₀) is the nth derivative evaluated at x₀.
For the first problem, f(x) = sin x and x₀ = π. We want the seventh degree term, so n = 7.
The seventh degree term is therefore: f⁽⁷⁾(π) / 7! (x − π)⁷
Find the seventh derivative of sin x:
f(x) = sin x
f⁽¹⁾(x) = cos x
f⁽²⁾(x) = -sin x
f⁽³⁾(x) = -cos x
f⁽⁴⁾(x) = sin x
f⁽⁵⁾(x) = cos x
f⁽⁶⁾(x) = -sin x
f⁽⁷⁾(x) = -cos x
Evaluated at π, f⁽⁷⁾(x) = 1. So the seventh degree term is (x − π)⁷ / 5040.
For the second problem, f(x) = √x and x₀ = 1. We want the third degree term, so n = 3.
The third degree term is therefore: f⁽³⁾(1) / 3! (x − 1)³
Find the third derivative of √x:
f(x) = √x
f⁽¹⁾(x) = ½ x^-½
f⁽²⁾(x) = -¼ x^-³/₂
f⁽³⁾(x) = ⅜ x^-⁵/₂
Evaluated at 1, f⁽³⁾(x) = ⅜. So the third degree term is (x − 1)³ / 16.