anonymous_user
Compute the 3^{rd} Taylor Polynomial for f(x) = x^5-2x^4+x^3-9x^2+x-1, expanded about x_0=1.
An archive of Mark's Spring 2018 Numerical Analysis course.
Section 1.2 Question 5
anonymous_user
We may compute the 3^{rd} Taylor Polynomial of a function f(x) with the formula
T_3(x)=f(x_0) + \sum_{j=1}^3 \left(\frac{f^{(j)}(x_0)}{j!}(x-x_0)^j\right).
We note that
f(x)=x^5-2x^4+x^3-9x^2+x-1, \\
f'(x)=5x^4-8x^3+3x^2-18x+1, \\
f''(x)=20x^3-24x^2+6x-18,\\
f'''(x)=60x^2-48x+6.
Evaluating these at x_0=1, we have f(x_0)=-9, f'(x_0)=-17,f''(x_0)=-16,f'''(x_0)=18. Plugging these values into the formula for T_3(x), we have
T_3(x)=-9+\left(\frac{-17}{1!}(x-1)\right)+\left(\frac{-16}{2!}(x-1)^2\right)+\left(\frac{18}{3!}(x-1)^3\right)\\
=-9 -17(x-1)-8(x-1)^2+3(x-1)^3.
mark
Close! I think you might have made a silly little mistake when you computed f'''(x_0).