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

FullSolution Problems. In questions 511, 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 ,

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 , that is the value , matches the middle term 1x^{2}/2. Then use the Taylor remainder formula given by
where c is a value in the interval , and ƒ^{(n+1)} is the n+1st 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 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. Therefore, T_{4}(x) is given by 
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.

Solution 

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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: We see that T_{2}(x) = T_{3}(x) = 1x^{2}/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 , we have that for x values in , where c is an unknown value in . As , we have that (drawing a picture helps confirm these computations  the cosine function is always positive on this interval). Also, we have that and so we conclude that
as required. 