Science:Math Exam Resources/Courses/MATH220/April 2005/Question 04

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MATH220 April 2005

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Question 04
Define S to be the set of all real numbers x such that exactly one of the two series $\sum _{k=1}^{\infty }\left(x-{\frac {1}{2}}\right)^{k}\quad {\text{ and }}\quad \sum _{k=1}^{\infty }\left(x+{\frac {1}{2}}\right)^{k}$ converges. Write down (with justification) a simple expression for S, using interval notation and set operations such as $\cup$, $\cap$ and / or -.

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
The series $\sum _{k=1}^{\infty }x^{k}$ converges for all values of x between -1 and 1.

Checking a solution serves two purposes: helping you if, after having used the hint, 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

The first series $\sum _{k=1}^{\infty }\left(x-{\frac {1}{2}}\right)^{k}$ converges for all values of x such that $\displaystyle -1<x-{\frac {1}{2}}<1$ which we can rewrite as $\displaystyle -1+{\frac {1}{2}}<x<1+{\frac {1}{2}}$ that is $\displaystyle -{\frac {1}{2}}<x<{\frac {3}{2}}$ The second series $\sum _{k=1}^{\infty }\left(x+{\frac {1}{2}}\right)^{k}$ converges for all values of x such that $\displaystyle -1<x+{\frac {1}{2}}<1$ which we can rewrite as $\displaystyle -1-{\frac {1}{2}}<x<1-{\frac {1}{2}}$ that is $\displaystyle -{\frac {3}{2}}<x<{\frac {1}{2}}$ So values of x in the interval (-1/2,3/2) will make the first series converge and values in the interval (-3/2,1/2) will make the second series converge. Hence the values of x which make exactly one of the series convergent cannot be in the intersection of these two intervals. We can now conclude that that the set S is $\displaystyle S=\left(-{\frac {3}{2}},-{\frac {1}{2}}\right]\cup \left[{\frac {1}{2}},{\frac {3}{2}}\right)$

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