Question

Find the coefficients of the first 4 terms of Taylor series of the function (f(x) = ln(4+x)) centered at 2, and write the corresponding Taylor polynomial.

a) (b₀ = 2, b₁ =1/3, b₂ = -1/9, b₃ =1/27)

b) (b₀ = 1, b₁ =1/2, b₂ =1/8, b₃ =1/16)

c) (b₀ = 0, b₁ = 1, b₂ =1/2, b₃ =1/3)

d) (b₀ = 2, b₁ = 1, b₂ =1/2, b₃ =1/4)

Answer 1

The coefficients of the first 4 terms of the Taylor series of the function ln(4+x) centered at 2 are b₀ = ln(6), b₁ = 1/6, b₂ = -1/36, b₃ = 1/54. The corresponding Taylor polynomial is T(x) = ln(6) + (1/6)(x-2) - (1/36)(x-2)^2 + (1/54)(x-2)^3.

Step-by-step explanation:

The Taylor series expansion of a function f(x) centered at a value a is given by:

f(x) = f(a) + f'(a)(x-a)/1! + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ...

In this case, the function is f(x) = ln(4+x) and we want to find the coefficients of the first 4 terms of the Taylor series centered at 2. To find the coefficients, we can calculate the derivatives of f(x) and evaluate them at x=a=2. The first few derivatives are:

f'(x) = 1/(4+x), f''(x) = -1/(4+x)^2, f'''(x) = 2/(4+x)^3

Evaluating these at x=a=2, we get:

f'(2) = 1/6, f''(2) = -1/36, f'''(2) = 1/54

So the coefficients of the first 4 terms of the Taylor series centered at 2 are b₀ = ln(4+2) = ln(6), b₁ = f'(2) = 1/6, b₂ = f''(2) = -1/36, b₃ = f'''(2) = 1/54. The corresponding Taylor polynomial is:

T(x) = ln(6) + (1/6)(x-2) - (1/36)(x-2)^2 + (1/54)(x-2)^3