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# Solution of the Final Exam

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Problem 1. Let and denote functions defined on some set .

1. Prove that

2. Find an example for a pair , for which

3. Find an example for a pair , for which

Solution.

1. For any , and and hence . Thus is an upper bound for on , and hence it is no smaller than the least upper bound for on , which is .
2. Take say and to be the constant functions 0, and then and are both 0.
3. Take say and on . Then and hence while and and hence . Thus as required.

Problem 2. Sketch the graph of the function . Make sure that your graph clearly indicates the following:

• The domain of definition of .
• The behaviour of near the points where it is not defined (if any) and as .
• The exact coordinates of the - and -intercepts and all minimas and maximas of .

Solution. is defined for , and the following limits are easily computed: , and . The only solution for is , hence the only intersection of the graph of with the axes is at . Other than at , the numerator of is always positive, hence the sign of the function is determined by the sign of the denominator . Thus for and for . Finally and thus is positive and is increasing (locally) for and is negative and is decreasing (locally) for . Thus overall the graph is:

Problem 3. Compute the following integrals:

1. Solution. By long division of polynomials, . Thus we can rewrite our integral as a sum of two terms as follows

2. Solution. Again we rewrite the integral as a sum of two terms. On the first we perform the substitution ; the second is elementary:

3. Solution. We integrate by parts twice, as follows:

4. Solution. Set and then and and so and

5. Solution.

Or using a shorter and less precise notation, but good enough --

Problem 4. Agents of the CSIS have secretly developed a function that has the following properties:

• for all .
• is differentiable at 0 and .
Prove the following:
1. is everywhere differentiable and .
2. for all . The only lemma you may assume is that if a function satisfies for all then is a constant function.

Solution.

1. The given fact that means that . Hence, using we get

This proves both that is differentiable at and that .
2. Consider . Differentiating we get

Hence is a constant function. But , hence this constant must be . So and thus .

Problem 5.

1. Prove that if a sequence of continuous functions converges uniformly to a function on some interval , then is continuous on .
2. Prove that the series converges on and that its sum is a continuous function of .

Solution.

1. See Spivak's Theorem 2 of Chapter 24.
2. and converges. Hence by the Weierstrass M-Test the series converges uniformly. As each of the terms is continuous, the first part of this question implies that so is the sum.

Problem 6. Prove that the complex function is everywhere continuous but nowhere differentiable.

Solution. The key point is that for every complex number . Let and set . Now if then . This proves the continuity of . Let us check if this function is differentiable:

If we restrict our attention to real then the latter quotient is always , so the limit would be . If we restrict our attention to imaginary , with real , then that quotient is so the limit would be . Hence the limit cannot exist and is not differentiable at (an arbitrary) .

The results. 76 students took the exam; the average grade was 72.66/120, the median was 71.5/120 and the standard deviation was 25.5. The overall grade average for the course (of ) was 66.92, the median was 64.9 and the standard deviation was 17.16. Finally, the transformation was applied to the grades, with . This made the average grade 71.55, the median 70 and the standard deviation 15.31. There were 25 A's (grades above 80) and 5 failures (grades below 50).

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Dror Bar-Natan 2003-05-01