Science:Math Exam Resources/Courses/MATH101/April 2013/Question 02 (c)

MATH101 April 2013

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Question 02 (c)
Short-Answer Questions. Question 1-3 are short-answer questions. Put your answer in the box provided. Simplify your answer as much as possible. Full Marks will be awarded for a correct answer placed in the box. Show your work, for part marks. The first two terms in the Maclaurin series for $x^{2}\sin(x^{3})$ are $ax^{5}+bx^{11}$, where a and b are constants. Find the values of a and b.

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Work with the MacLaurin series for $\sin x$ first, then manipulate it.

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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. Recall the MacLaurin series for sin(x) is given by $\sin(x)=x-{\frac {x^{3}}{3!}}+\ldots$ so ${\begin{aligned}x^{2}\sin(x^{3})&=x^{2}\left(x^{3}-{\frac {x^{9}}{3!}}+\ldots \right)\\&=x^{5}-{\frac {x^{11}}{3!}}+\ldots \end{aligned}}$ so $a=1,b=-{\frac {1}{6}}$

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