Science:Math Exam Resources/Courses/MATH100/December 2010/Question 07

MATH100 December 2010

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[hide] Question 07
Full-Solution Problems. In questions 2-8, justify your answers and show all your work. If a box is provided, write your final answer there. Simplification of answers is not required unless explicitly requested. Determine and carefully justify an upper bound for the absolute value of the error that would result if the second degree Maclaurin polynomial $T_{2}(x)$ was used to estimate $f(1)$ , where $f(x)=e^{x}(x^{2}-7x+15)$ Your upper bound may be left unsimplified ("Calculator-ready") but it must be a specific numerical value. NOTE: You are not required to find $T_{2}(x)$ and no credit will be given for finding $T_{2}(x)$ .

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[show] Hint
We want to know the value of (or at least an upper bound of) $\displaystyle \|T_{2}(1)-f(1)\|$ Taylor's remainder formula says that there is a $c\in [0,1]$ such that $\displaystyle \|T_{2}(1)-f(1)\|={\frac {\|f'''(c)\|\|x-a\|^{3}}{3!}}\leq {\frac {M\|1-0\|^{3}}{6}}={\frac {M}{6}}$ where M is an upper bound for the absolute value of the third derivative on $\displaystyle [0,1]$ .

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[show] Solution 1
Taylor's remainder formula says that $\displaystyle \|T_{2}(1)-f(1)\|={\frac {\|f'''(c)\|\|x-a\|^{3}}{3!}}\leq {\frac {M\|1-0\|^{3}}{6}}={\frac {M}{6}}$ where M is an upper bound for the absolute value of the third derivative on the interval $\displaystyle [0,1]$ . Computing the third derivative via the product rule gives ${\begin{aligned}f(x)&=e^{x}(x^{2}-7x+15)\\f'(x)&=e^{x}(x^{2}-7x+15)+e^{x}(2x-7)\\&=e^{x}(x^{2}-7x+15+2x-7)\\&=e^{x}(x^{2}-5x+8)\\f''(x)&=e^{x}(x^{2}-5x+8)+e^{x}(2x-5)\\&=e^{x}(x^{2}-5x+8+2x-5)\\&=e^{x}(x^{2}-3x+3)\\f'''(x)&=e^{x}(x^{2}-3x+3)+e^{x}(2x-3)\\&=e^{x}(x^{2}-3x+3+2x-3)\\&=e^{x}(x^{2}-x)\\\end{aligned}}$ Notice that $\displaystyle e^{x}$ is increasing on $\displaystyle [0,1]$ and thus obtains its maximum at x = 1, i.e. $\|e^{x}\|=e^{x}\leq e^{1}=e$ for $x\in [0,1]$ . The function $\displaystyle x^{2}-x$ is a parabola with roots at 0 and 1. Hence its minimum occurs at the midpoint of the roots, namely at $\displaystyle x=1/2$ . Since the value at the endpoints x = 0 and x = 1 is zero, the minimum value of the parabola maximizes its absolute value. Thus, $\displaystyle \|x^{2}-x\|\leq \|(1/2)^{2}-(1/2)\|=1/4$ Since we want to bound the absolute value of the third derivative, we know that $\displaystyle \|f'''(x)\|=\|e^{x}\|\|x^{2}-x\|\leq e\cdot 1/4=e/4=M$ . Hence, we have $\displaystyle \|T_{2}(1)-f(1)\|\leq {\frac {M}{6}}={\frac {e}{24}}$ as required.

[show] Solution 2
Proceed as in solution 1 to see that ${\begin{aligned}f'''(x)&=e^{x}(x^{2}-x)\end{aligned}}$ We now need a bound on this function on $\displaystyle [0,1]$ . Notice that the third derivative is neither monotonically increasing nor is it monotonically decreasing on the interval. Hence we are not guaranteed to find the maximum value by just looking at the endpoints. In fact, at the endpoints we have $\displaystyle f'''(0)=0=f'''(1)$ , which is certainly not the maximum of the absolute value $\displaystyle \|f'''(x)\|$ . Instead we do it the proper way: Take the derivative $\displaystyle f''''(x)=e^{x}(x^{2}-x)+e^{x}(2x-1)=e^{x}(x^{2}+x-1)$ and set it to zero $\displaystyle 0=e^{x}(x^{2}+x-1)$ Recall that $\displaystyle e^{x}\neq 0$ . Using the quadratic formula, we see that $\displaystyle x={\frac {-1\pm {\sqrt {1^{2}-4(1)(-1)}}}{2(1)}}={\frac {-1\pm {\sqrt {5}}}{2}}$ only the positive such root is in $\displaystyle [0,1]$ . Let $\displaystyle \phi :={\frac {-1+{\sqrt {5}}}{2}}$ . Then, our function $\displaystyle f'''(x)$ obtains an extreme value at $\displaystyle \phi$ . Since the function is nonzero at this point, we know that $\displaystyle \|f'''(x)\|$ is maximal at $\displaystyle \phi$ . Thus, we have that $\displaystyle \|T_{2}(1)-f(1)\|\leq {\frac {M}{6}}={\frac {\|f'''(\phi )\|}{6}}={\frac {e^{\phi }\|\phi ^{2}-\phi \|}{6}}$ completing the solution.