Find the taylor polynomial t3(x) for the function f centered at the number f(x) = 1/ x , a = 2.

Solution:

Let f be a function with derivatives of all orders throughout some interval containing a as an interior point.

Then the taylor series generated by f at x = a is

\( \sum_{k=0}^{\infty}\frac{f^{(k)}(a)}{k!}(x - a)^{k} = f(a) + f^{'}(a)(x - a) + \frac{f^{"}(a)}{2!}(x - a)^{2} + .....+ \frac{f^{(n)}(a)}{n!}(x - a)^{n} + .... \)

t₃(x) = f’’(a)/2!)(x - a)²

f(x) = 1/x

= x^-1

f’(x) = - x^-2

f’’(x) = 2x^-3

f’’(x = a =2)

= 2(2)^-3

= 2/8 = 1/4

Hence,

t₃(x) = f’’(2) / 2!) (x - a)²

= (1/4) (1/2!) (x - 2)²

= (1/8) (x - 2)² or (x - 2)² / 8

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Find the taylor polynomial t3(x) for the function f centered at the number f(x) = 1/ x , a = 2.

Summary:

The third term of the taylor series is (x-2)² . 8