Sure, to compute the Taylor polynomials centered at a given point, say a, we first need to compute the derivatives of the function at that point and then use these to create the polynomial.
Let the function f(x) = 14*sin(x). Then, the first derivative of the function f with respect to x is f'(x) = 14*cos(x). The second derivative of the function f with respect to x is f''(x) = -14*sin(x), and the third derivative is f'''(x) = -14*cos(x).
We are centered at a = 2π. Therefore, we compute these derivatives at this point:
f(2π) = 14*sin(2π) = 0,
f'(2π) = 14*cos(2π) = 14,
f''(2π) = -14*sin(2π) = 0,
f'''(2π) = -14*cos(2π) = -14.
The Taylor polynomial of degree 2 is given by:
T2(x) = f(a) + f'(a)*(x - a) + f''(a)*(x - a)^2/2
= 0 + 14*(x - 2π) + 0
= 14*x - 28π.
and the Taylor polynomial of degree 3 is:
T3(x) = T2(x) + f'''(a)*(x - a)^3/6
= 14*x - 28π + (-14)*(x - 2π)^3/6
= 14*x - 28π
(the third degree term vanishes in this case because (x - 2π)^3 is cubic, and will still be 0 when x = 2π).
Therefore, both the second and third degree Taylor polynomials centered at a = 2π for the function f(x) =14sin(x) are T2(x) = T3(x) = 14*x - 28π.