Question

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

Answer 1

Answer:

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:

$P_{n}(x) = f(c) + \frac{f'(c)}{1!}(x-c)+\frac{f''(c)}{2!}(x-c)^{2} +\frac{f'''(c)}{3!}(x-c)^{3}+...+\frac{f^{n} (c)}{n!}(x-c)^{n}$

Since c = 3 So,

$P_{4}(x) = f(3) + \frac{f'(3)}{1!}(x-3)+\frac{f''(3)}{2!}(x-3)^{2} +\frac{f'''(3)}{3!}(x-3)^{3}+...+\frac{f^{n} (3)}{n!}(x-3)^{n}$

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$^{(4)}$ (x) = -6/x⁴ ⇒ f$^{(4)}$ (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