Find the nth Taylor polynomial for the function, centered at c. f(x) = ln(x), n = 4, c = 3 P4(x) =

P₄(x) = ln3 + 1/3 (x-3) - 1/9*2! (x-3)² + 2/27*3! (x-3)³ - 2/27*4! (x - 3)⁴

Step-by-step explanation:

Given:

f(x) = ln(x)

n = 4

c = 3

To find:

nth Taylor polynomial for the function, centered at c

Solution:

The Taylor series for f(x) = ln x centered at 3 is:

Since c = 3 So,

Now

f(3) = ln 3

f'(x) = 1/x ⇒ f'(3) = 1/3

f''(x) = -1/x² ⇒ f''(3) = -1/3² = -1/9

f'''(x) = 2/x³  ⇒ f'''(3) = 2/3³ = 2/27

f (x) = -6/x⁴ ⇒ f (3) = -6/3⁴ = -6/81 = - 2/27

So Taylor polynomial for n=4 is:

P₄(x) = ln3 + 1/3 (x-3) - 1/9*2! (x-3)² + 2/27*3! (x-3)³ - 2/27*4! (x - 3)⁴

this was already solved by me with a different person. if you look it up in the search bar at the top of the page, you will get the answer.

step-by-step explanation: