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No, because a set must have unique elements; sin(t+pi/2) is basically cos(t). | No, because a set must have unique elements; sin(t+pi/2) is basically cos(t). | ||
The union of both sets is a set with elements from both S1 and S2. | The union of both sets is a set with elements from both S1 and S2. | ||
| − | S1 U S2 = {sin(t),cos(t),sin(t | + | S1 U S2 = {sin(t),cos(t),sin(t/2)} |
===Answer 2=== | ===Answer 2=== | ||
Revision as of 15:51, 10 January 2013
Contents
Practice Problemon set operations
Consider the following sets:
$ \begin{align} S_1 &= \left\{ \sin (t), \cos (t)\right\}, \\ S_2 & = \left\{ \sin (\frac{t}{2}), \sin (t+\frac{\pi}{2})\right\}. \\ \end{align} $
Write $ S_1 \cup S_2 $ explicitely. Is $ S_1 \cup S_2 $ a set?
You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!
Answer 1
No, because a set must have unique elements; sin(t+pi/2) is basically cos(t). The union of both sets is a set with elements from both S1 and S2. S1 U S2 = {sin(t),cos(t),sin(t/2)}
Answer 2
$ S_1 \cup S_2 = \left\{ \sin (t),\sin (\frac{t}{2}), \cos (t)\right\} $
$ S_1 \cup S_2 $ is a set because the union of two sets is the set of all distinct elements from those two sets. In this case because $ \sin (t+\frac{\pi}{2}) $ and $ \cos (t) $ are part of the same equivalence class, we only need to include one of these elements in our union set.
Instructor's suggestion: Can anyone illustrate the answer using a Venn diagram? -pm
Answer 3
Write it here.
