Given the function, f(x)=1/x, find the P1(x), P2(x), P(3), and P(4) for interval [1,4]. Find the error of each approximation when x=4.

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You need to find Taylor polynomials, `P_1(x), P_2(x), P_3(x), P_4(x)` at x = 4, such that:

`P_1(4) = f(4) + f'(4)(x - 4)`

`P_2(4) = f(4) + f'(4)(x - 4) + (f''(4))/(2!)(x - 4)^2`

`P_3(4) = f(4) + f'(4)(x - 4) + (f''(4))/(2!)(x - 4)^2 + (f'''(4))/(3!)(x - 4)^3`

`P_4(4) = f(4) + f'(4)(x - 4) + (f''(4))/(2!)(x - 4)^2 + (f'''(4))/(3!)(x - 4)^3 + (f^(4)(4))/(4!)(x - 4)^4`

Evaluating `f'(x) ` yields:

`f'(x) = (1/x)' = -1/(x^2)`

Evaluating `f'(4)` yields:

`f'(4) = -1/16`

Evaluating `f''(x)` yields:

`f''(x) = (-1/(x^2))'`

`f''(x) = 2x/(x^4) => f''(x) = 2/x^3 => f''(4) = 2/64 = 1/32`

Evaluating `f'''(x)` yields:

`f'''(x) = (2/x^3)'`

`f'''(x) = -6x^2/(x^6) => f'''(x) = -6/(x^4) => f'''(4) = -6/(16*16) = -3/(8*16)`

Evaluating ` f^(4)(x)` yields:

`f^(4)(x) = (-6/(x^4))' => f^(4)(x) = 24x^3/(x^8)`

`f^(4)(x) = 24/(x^5) => f^(4)(4) = 24/(4*16^2) = 6/(16^2) = 3/(8*16)`

Replacing the results in Taylor polynomials, yields:

`P_1(4) = 1/4 - 1/16(x - 4)`

`P_2(4) = 1/4 - 1/16(x - 4)+ 1/(64)(x - 4)^2`

`P_3(4) =1/4 - 1/16(x - 4)+ 1/(64)(x - 4)^2 - 3/(3!*8*16)(x - 4)^3`

`P_3(4) =1/4 - 1/16(x - 4)+ 1/(64)(x - 4)^2 - 1/(2*8*16)(x - 4)^3`

`P_4(4) = 1/4 - 1/16(x - 4)+ 1/(64)(x - 4)^2 - 1/(2*8*16)(x - 4)^3 + 3/(4!*8*16)(x - 4)^4`

`P_4(4) = 1/4 - 1/16(x - 4)+ 1/(64)(x - 4)^2 - 1/(2*8*16)(x - 4)^3 + 1/(64*16)(x - 4)^4`