Science:Math Exam Resources/Courses/MATH100/December 2012/Question 11

MATH100 December 2012

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Question 11

Full-Solution Problems. In questions 5-11, justify your answers and show all your work. If a box is provided, write your final answer there. Unless otherwise indicated, simplification of numerical answers is required in these questions. Give a complete proof that for all x satisfying $-1\leq x\leq 1$, $0\leq \cos(x)-\left(1-{\frac {x^{2}}{2}}\right)\leq {\frac {1}{24}}$ Hint: A question on Taylor polynomials has not yet appeared on this exam.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
The hint in the given question is a good start. Find the largest n so that the function minus the Taylor polynomial centred at $x=0$, that is the value $f(x)-T_{n}(x)$, matches the middle term 1-x2/2. Then use the Taylor remainder formula given by $R_{n}={\frac {f^{(n+1)}(c)}{(n+1)!}}(x-0)^{n+1}$ where c is a value in the interval $[-1,1]$, and ƒ(n+1) is the n+1-st derivative of ƒ.

Hint 2
Recall the definition of the Taylor polynomial. As an example, the 4th degree Taylor polynomial of ƒ(x), centered around a = 0, is given by $f(x)\approx T_{4}(x)=f(0)+f'(0)x+f''(0){\frac {x^{2}}{2}}+f'''(0){\frac {x^{3}}{3!}}+f''''(0){\frac {x^{4}}{4!}}$ So we need to calculate a couple derivatives of cos(x).

Hint 3
We calculate the 4th order Taylor polynomial of ƒ(x), centered around a = 0. ${\begin{aligned}f(x)&=\cos(x)\quad &f(0)&=1\\f'(x)&=-\sin(x)\quad &f'(0)&=0\\f''(x)&=-\cos(x)\quad &f''(0)&=-1\\f'''(x)&=\sin(x)\quad &f'''(0)&=0\\f''''(x)&=\cos(x)\quad &f''''(0)&=1\end{aligned}}$ Therefore, T4(x) is given by $T_{4}(x)=1-{\frac {x^{2}}{2}}+{\frac {x^{4}}{24}}$

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Following the hints, we calculate the Taylor polynomials up to degree four: ${\begin{aligned}T_{0}(x)&=1\\T_{1}(x)&=1\\T_{2}(x)&=1-{\frac {x^{2}}{2}}\\T_{3}(x)&=1-{\frac {x^{2}}{2}}\\T_{4}(x)&=1-{\frac {x^{2}}{2}}+{\frac {x^{4}}{24}}\end{aligned}}$ We see that T2(x) = T3(x) = 1-x2/2 indeed matches the middle term in the question statement. Hence we can use the Taylor remainder formula either for n=2 or n=3. We choose n=3 because this gives us a smaller error term here; we divide the error by 4! instead of 3!. Our interval is $\displaystyle [-1,1]$, we have that for x values in $\displaystyle [-1,1]$, ${\begin{aligned}\cos(x)-\left(1-{\frac {x^{2}}{2}}\right)&=f(x)-T_{3}(x)=R_{3}(x)\\&={\frac {f''''(c)}{4!}}(x-0)^{4}={\frac {\cos(c)}{24}}x^{4}\end{aligned}}$ where c is an unknown value in $\displaystyle [-1,1]$. As $\displaystyle -{\frac {\pi }{2}}\leq -1\leq c\leq 1\leq {\frac {\pi }{2}}$, we have that $\displaystyle 0\leq \cos(c)\leq 1$ (drawing a picture helps confirm these computations - the cosine function is always positive on this interval). Also, we have that $\displaystyle 0\leq x^{4}\leq 1$ and so we conclude that $\displaystyle 0\leq \cos(x)-\left(1-{\frac {x^{2}}{2}}\right)=R_{3}(x)={\frac {\cos(c)}{24}}x^{4}\leq {\frac {1}{24}}$ as required.

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