Use the definition of Taylor series to find the taylor series (centered at c) for the function. f(x) = 1/x, c = 1.

Solution:

Let f be a function with derivatives of all orders throughout some interval containing c as an interior point. Then the taylor series generated by f at x = c is

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

To find the Taylor series we need to find f(1), f’(1), f’’(1)...... Taking the derivatives we get

f(x) = x^-1, f(1) = 1^-1 = 1/1

f’(x) = -x^-2, f’(1) = - 1/1²

f”(x) = 2!x^-3, f”(1)2! = 1^-3 = 1/1³

f”’(x) = -3!x^-4, f’’’(1)/3! = 1^-4 = 1/1⁴

fⁿ(x) = (-1)ⁿn!x^{-(n + 1)}, fⁿ(1)/n! = (-1)ⁿ/1^{(n + 1)}

The Taylor series is

f(1) + f’(1)(x - 1) + f”(1)(x - 1)² 2!  +...+  fⁿ(1)(x - 1)ⁿ / n! +…

= 1 - (x - 1) + (x -1)² 1³ + … + (-1)ⁿ(x - 1)ⁿ (1)^{(n + 1)} + …

= 1 - (x - 1) + (x - 1)² + …+ (x - 1)ⁿ

Use the definition of Taylor series to find the taylor series (centered at c) for the function. f(x) = 1/x, c = 1.

Summary:

Using the definition of Taylor series (centered at c) for the function. f(x) = 1/x, c = 1 is given by 1 - (x - 1) + (x - 1)² + …+ (x - 1)ⁿ.