|Dror Bar-Natan: Classes: 2002-03: Math 157 - Analysis I:||(286)||
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Problem 1. Let and denote functions defined on some set .
Problem 2. Sketch the graph of the function . Make sure that your graph clearly indicates the following:
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:
Solution. By long division of polynomials, . Thus we can rewrite our integral as a sum of two terms as follows
Solution. Again we rewrite the integral as a sum of two terms. On the first we perform the substitution ; the second is elementary:
Solution. We integrate by parts twice, as follows:
Solution. Set and then and and so and
Problem 4. Agents of the CSIS have secretly developed a function that has the following properties:
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:
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).