How would you do this problem, I am stuck.
Find the Taylor series of f(x)=1/x centered at x=2.
1/2 - (x-2)/4 + (x-2)^2/8 - (x-2)^3/16+ ...
To find the Taylor series of a function centered at a specific point, we can apply the Taylor series formula:
f(x) = f(a) + f'(a)(x-a)/1! + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ...
In this case, f(x) = 1/x and we want the Taylor series centered at x = 2.
Step 1: Find the value of f(a).
Evaluate f(x) at x = a:
f(a) = 1/a
Step 2: Find the value of f'(a).
Differentiate f(x) with respect to x:
f'(x) = -1/x^2
Evaluate f'(x) at x = a:
f'(a) = -1/a^2
Step 3: Substitute the values of f(a) and f'(a) into the Taylor series formula.
f(x) ≈ 1/a - (1/a^2)(x-a)/1! + ...
Therefore, the Taylor series of f(x) = 1/x centered at x = 2 is approximately:
f(x) ≈ 1/a - (1/a^2)(x-a)/1! + (higher-order terms)
Note that the higher-order terms involve additional derivatives evaluated at the center point x = a, but since we have only determined f(x) and f'(x), we cannot calculate those terms explicitly without additional information or calculations.