Dror Bar-Natan: Classes: 2003-04: Math 157 - Analysis I: (27) Next: Homework Assignment 7 Previous: Term Exam 1

# Solution of Term Exam 1

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Problem 1. All that is known about the angle is that . Can you find and ? Explain your reasoning in full detail.

Solution. (Graded by C.-N. (J.) Hung) In class we wrote the formula . Also using and taking we get

Dividing the numerator and denominator by this becomes

Likewise using we get

Problem 2.

1. State the definition of the natural numbers.
2. Prove that every natural number has the property that whenever is natural, so is .

1. The set of natural numbers is the smallest set of numbers for which
• ,
• if then .
Alternatively, the set of natural numbers is the intersection of all sets of numbers satisfying
• ,
• if then .

2. Let be the assertion whenever is natural, so is ''. We prove by induction on :
1. asserts that whenever is natural, so is ''. This is true by the second bullet in the definition of .
2. Assume , that is, assume that whenever is natural, so is . Let be a natural number. Then is a natural number because by the number is natural and because adding one to a natural number gives a natural number by the second bullet in the definition of . So we have shown that whenever is natural so is , and this is the assertion .

Problem 3. Recall that a function is called even'' if for all and odd'' if for all , and let be some arbitrary function.

1. Find an even function and an odd function so that .
2. Show that if where and are even and and are odd, then and .

1. Set and . Then while (so is even) and (so is odd).

2. Assume where is even and is odd. Then

So necessarily . Now if as above, then both and can play the role of in this argument, so they are both equal to and in particular they equal each other. Likewise,

and arguing like before, .

Problem 4. Sketch, to the best of your understanding, the graph of the function

(What happens for near 0? Near ? For large ? Is the graph symmetric? Does it appear to have a peak somewhere?)

If then and so ; furthermore, the larger is (while ), the larger is and hence the smaller is. When approaches from above, approaches 0 from above and hence becomes larger and larger. If the and so . When , and when approaches from below, approaches from below and approaches 0 from below, and so becomes more and more negative. In summary, the graph looks something like:

Problem 5.

1. Suppose that for all , and that the limits and both exist. Prove that .
2. Suppose that for all , and that the limits and both exist. Is it always true that ? (If you think it's always true, write a proof. If you think it isn't always true, provide a counterexample).

Solution. (Graded by C.-N. (J.) Hung)

1. Let and and assume by contradiction that ; that is, that . Use the existence of the two limits to find and so that

and

Now choose some specific for which both and . But then and so while and so . Therefore remembering that for all we get

or

or

which is a contradiction. Thus the assumption that must be incorrect and thus .

2. Take for all and for all and . Then for all but . So it isn't always true that if for all and the limits exist, then .

The results. 105 students took the exam; the average grade was 67.19, the median was 70 and the standard deviation was 21.12.

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Dror Bar-Natan 2004-10-18