MATH103 April 2014
• Q1 (a) • Q1 (b) i • Q1 (b) ii • Q1 (c) i • Q1 (c) ii • Q1 (c) iii • Q1 (d) i • Q1 (d) ii • Q1 (d) iii • Q1 (e) i • Q1 (e) ii • Q2 (a) • Q2 (b) • Q3 (a) • Q3 (b) • Q3 (c) • Q4 (a) • Q4 (b) • Q4 (c) • Q5 (a) • Q5 (b) • Q5 (c) • Q6 (a) • Q6 (c) • Q7 (a) • Q7 (b) • Q7 (c) • Q8 (a) • Q8 (b) • Q9 (a) • Q9 (b) • Q9 (c) • Q10 (a) • Q10 (b) • Q10 (c) • Q11 •
Question 07 (a)
. Determine and .
(Note: the sum refers to a series expansion centred at .)
Work must be shown for full marks. Simplify fully.
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.
This is a Taylor series problem.
Use the series expansion of .
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.
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Note: If you have forgotten the series expansion of (at ), recall that, in general, such an expansion has the form
For , we have and it follows that for all . Thus, and we get
Now replacing by in the above expansion yields
Since only even powers of appear in this series, whenever is odd. In particular, .
On the other hand, for . But the coefficient of is . For , this yields .
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