Taylor Series Question: f(x) = x^4 at x = 3, approx at x =3.1

[mroberts]

Question
Determine the function value and the first and second derivative of f (x) = x⁴ at x = 3. Then, estimate the function value at x = 3.1, using a first-order and second-order Taylor series approximation. How big is the error? Illustrate with a graph.

My attempt......

	f (x) = x4 f (3) = 81	f ’ (x) = 4x3 f ’ (3) = 108	f ” (x) = 12x2 f ” (3) = 108
First order Taylor series Approximation. ∆B ≈ dB/dy x ∆y ∆B ≈ 4(3)^3 x 0.1 ∆B ≈ 10.8 f(3.1) ≈ 91.8	Second Order Taylor Series ∆B≈ dB/dy x ∆y + 1d^2B/2dy^2 x ∆y^2 ∆B≈ 4(3)^3 x (0.1) + (0.5) x 12(3)^2 x(0.1)^2 ∆B≈ 10.8 + 0.54 ∆B≈ 11.34 f(3.1) = 92.34

ERROR = 11.34-10.8 =0.54

>>Not sure if i'm on the right track. Any help much appreciated.
Regards

 

[Deleted member 4993]

Question
Determine the function value and the first and second derivative of f (x) = x⁴ at x = 3. Then, estimate the function value at x = 3.1, using a first-order and second-order Taylor series approximation. How big is the error? Illustrate with a graph.

My attempt......

	f (x) = x4 f (3) = 81	f ’ (x) = 4x3 f ’ (3) = 108	f ” (x) = 12x2 f ” (3) = 108
First order Taylor series Approximation. ∆B ≈ dB/dy x ∆y ∆B ≈ 4(3)^3 x 0.1 ∆B ≈ 10.8 f(3.1) ≈ 91.8	Second Order Taylor Series ∆B≈ dB/dy x ∆y + 1d^2B/2dy^2 x ∆y^2 ∆B≈ 4(3)^3 x (0.1) + (0.5) x 12(3)^2 x(0.1)^2 ∆B≈ 10.8 + 0.54 ∆B≈ 11.34 f(3.1) = 92.34

ERROR = 11.34-10.8 =0.54

>>Not sure if i'm on the right track. Any help much appreciated.
Regards

As I interpret it, calculation of error should be made from the actual value of the function.

Thus error in first order approximation = |(3.1)^4 -91.8|/(3.1)^4 * 100%

 

[mroberts]

I interpreted it as the error between using first-order and second-order Taylor series approximations because
the second-order approximation would be more accurate...