Find the 4th Taylor polynomial, P4, generated by f(x) = 1/x at center a = 2 ?
just plug and chug
f(x) = f(2) + f'(2)(x-2)/1! + ...
f(2) = 1/2
f'(2) = -1/x^2 = -1/2^2 = -1/4
f"(2) = 2/x^3 = 1/4
f(3)(2) = -6/x^4 = -3/8
f(4)(2) = 24/x^4 = 3/4
f(x) = 1/2 + (-1/4)(x-2)/1! + (1/4)(x-2)^2/2! + (-3/8)(x-2)^3/3! + (3/4)(x-2)^4/4! + ...
= 1/2 - 1/4 (x-2) + 1/8 (x-2)^2 - 1/16 (x-2)^3 + 1/32 (x-2)^4 + ...
To find the 4th Taylor polynomial, P4, generated by f(x) = 1/x at center a = 2, we can use the Taylor series expansion formula:
Pn(x) = f(a) + f'(a)(x - a) + (1/2!)f''(a)(x - a)^2 + (1/3!)f'''(a)(x - a)^3 + ... + (1/n!)f^n(a)(x - a)^n
To apply this formula to f(x) = 1/x at center a = 2, we need to find the derivatives of f(x) and evaluate them at x = 2:
f'(x) = -1/x^2
f''(x) = 2/x^3
f'''(x) = -6/x^4
f^(4)(x) = 24/x^5
Now we can substitute these derivatives into the Taylor series formula:
P4(x) = f(2) + f'(2)(x - 2) + (1/2!)f''(2)(x - 2)^2 + (1/3!)f'''(2)(x - 2)^3 + (1/4!)f^(4)(2)(x - 2)^4
Let's calculate each term:
f(2) = 1/2 = 0.5
f'(2) = -1/(2^2) = -1/4 = -0.25
f''(2) = 2/(2^3) = 2/8 = 0.25
f'''(2) = -6/(2^4) = -6/16 = -0.375
f^(4)(2) = 24/(2^5) = 24/32 = 0.75
Now we can substitute these values into the Taylor polynomial:
P4(x) = 0.5 - 0.25(x - 2) + (1/2!)(0.25)(x - 2)^2 + (1/3!)(-0.375)(x - 2)^3 + (1/4!)(0.75)(x - 2)^4
Simplifying this expression will give you the 4th Taylor polynomial, P4, generated by f(x) = 1/x at center a = 2.