How do you do these two questions?

Answer:

(x − π)⁷ / 5040

(x − 1)³ / 16

Step-by-step 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.

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