Science:Math Exam Resources/Courses/MATH103/April 2010/Question 01 (c)

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MATH103 April 2010

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q1 (f)  •  Q1 (g)  •  Q1 (h)  •  Q2  •  Q3 (a)  •  Q3 (b)  •  Q4 (a)  •  Q4 (b)  •  Q4 (c)  •  Q5 (a)  •  Q5 (b)  •  Q6 (a)  •  Q6 (b)  •  Q7 (a)  •  Q7 (b)  •  Q7 (c)  •  Q8  •  

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Question 01 (c)

Multiple Choice Question: Exactly ONE of the answers provided is correct. There is no partial credit in this question. The expression $\displaystyle S_{3}=1+x^{3}+{\frac {x^{6}}{2}}$ is the first three terms of a Taylor series for: $\displaystyle {\begin{aligned}{\textit {i.}}\quad &\sin(x^{3})\\{\textit {ii.}}\quad &\cos(x^{\frac {3}{2}})\\{\textit {iii.}}\quad &\cos(x)\\{\textit {iv.}}\quad &\cos(x^{3})\\{\textit {v.}}\quad &e^{x^{3}}\end{aligned}}$

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
Recall the Taylor series expansion for $\sin(x)$, $\cos(x)$, and $e^{x}$.

Hint 2

${\begin{aligned}\sin x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)!}}x^{2n+1}=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-\cdots \quad {\text{ for all }}x\!\\\cos x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n)!}}x^{2n}=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-\cdots \quad {\text{ for all }}x\!\\e^{x}&=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}=1+x+{\frac {x^{2}}{2!}}+\cdots \quad {\text{ for all }}x\!\end{aligned}}$

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 1

Following the hints we can just plug in the argument of the functions and calculate ${\begin{aligned}\sin(x^{3})&=x^{3}-{\frac {x^{9}}{3!}}+{\frac {x^{15}}{5!}}-\cdots \\\cos(x^{3/2})&=1-{\frac {x^{3}}{2!}}+{\frac {x^{6}}{4!}}-\cdots \\\cos(x)&=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-\cdots \\\cos(x^{3})&=1-{\frac {x^{6}}{2!}}+{\frac {x^{12}}{4!}}-\cdots \\e^{x^{3}}&=1+x^{3}+{\frac {x^{6}}{2!}}+\cdots \end{aligned}}$ Hence, the correct answer is v.

Solution 2

A clever way of doing this problem is to consider the following: $S_{3}(0)=1$, hence the candidate function needs to also satisfy this condition. Since $\sin(0)=0\not =1$, choice i. can be eliminated. $S_{3}$ is increasing for small positive values of x, since, for small x, the term x3 dominates. In particular, $S_{3}(x)>1$ for small positive values of x. But since the cosine is decreasing for small positive values of x, and can also never be more than 1, the choices ii, iii, and iv can all be eliminated. This only leaves v. Notice that this choice makes sense: $e^{0}=0$, and $e^{x^{3}}$ is increasing.

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