Dror Bar-Natan: Classes: 2003-04: Math 157 - Analysis I: (115) Next: Class Home Previous: Final Exam

# Solution of the Final Exam

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Problem 1. We say that a set of real numbers is dense if for any open interval , the intersection is non-empty.

1. Give an example of a dense set whose complement is also dense.
2. Give an example of a non-dense set whose complement is also not dense.
3. Prove that if is an increasing function ( for ) and if the range of is dense, then is continuous.

Solution.

1. Take for example , the set of rational numbers. Then is the set of irrational numbers. We've seen in class that between any two (different) numbers (i.e., within any open interval) there is a rational number and there is an irrational number. Hence both and are dense.
2. Take for example , the set of non-negative numbers. Then is the set of negative numbers. The set is not dense because, for example, it's intersection with the interval is empty. The set is not dense because, for example, it's intersection with the interval is empty.
3. We have to show that for every and for every there is a so that implies . So let be given. By the density of we know that we can find an element of in the interval and another element of in the interval . That is, we can find and so that and . It follows from the monotonicity of that and that . Now set (this is a positive number because and ). Finally if then is in the interval . By the monotonicity of it follows that is in the interval , and so , 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. Our function is defined for all . As goes to exponentials dominate polynomials, and so certainly gets much bigger than . So . Solving the equation we see that the only intersection of the graph of with the axes is at . We can compute and . Solving we see that the only critical points are when . That is, at . As , the point is a local max. As , the point is a local min. As there are no other critical points and the behaviour of near the ends of its domain of deifnition is mute (as determined before), is actually a global max and is actually a global min. Thus overall the graph is:

Problem and Solution 3. Compute the following derivative and the following integrals:

1. Using the fundamental theorem of calculus in the form and the chain rule with we get

2. We make the substitution (and thus and ) to compute

3. Integrating by parts twice we get

4. We make the substitution (and thus and ) to compute

5. We use the factorization to get

Problem 4. In solving this problem you are not allowed to use any properties of the exponential function .

1. Two differentiable functions, and , defined over the entire real line , are known to satisfy , , and for all and also . Prove that and are the same. That is, prove that for all .
2. A differentiable function defined over the entire real line is known to satisfy and for all and also . Prove that for all .

Solution.

1. Set (this is well defined because is never 0) and compute

So is a constant. But , so that constant is 1 and for all . This means that .
2. Fix and set and . Then and and . All the other conditions of the first part of this question are even easier to verify, and so the conclusion of that part holds. Namely, , which means .

Problem 5. In solving this problem you are not allowed to use any properties of the trigonometric functions.

1. A twice-differentiable function defined over the entire real line is known to satisfy for all and also . Write out the degree Taylor polynomial of at .
2. Write a formula for the remainder term . (To keep the notation simple, you are allowed to assume that is even or even that is divisible by 4).
3. Prove that is the zero function: for all .

Solution.

1. From it is clear that and that . So and and hence all the coefficients of are 0. In other words, .
2. If is divisible by then and so the remainder formula says that for any there is a between 0 and for which

3. Factorials grow faster then exponentials, so in the remainder formula the denominator grows faster then the term , while the numerator is bounded (by the theorem that a continuous function on a closed interval is bounded). So the remainder goes to 0 when goes to , and hence . But for all , so necessarily .

Remark 1. Two alternative forms of the remainder formula are

and

Either one of those could equaly well be used to solve part 3 of the problem.

Remark 2. There is an alternative approach to the whole problem; start with part 3 and go backwards. To do part 3, consider the function . We have , so is a constant function. But , so must be the 0 function. But is a sum of squares, and the only way a sum of squares can be 0 is if each summand is 0. So and hence as required in part 3. But if is the 0 function then its Taylor polynomials are all 0 and the remainder terms are also all 0, solving parts 1 and 2 as well. This is not the solution I had in mind when I wrote the problem, but people who solved the problem this way got full credit.

Problem 6. In solving this problem you are not allowed to use the irrationality of , but you are allowed, indeed advised, to borrow a few lines from the proof of the irrationality of .

Is there a non-zero polynomial defined on the interval and with values in the interval so that it and all of its derivatives are integers at both the point 0 and the point ? In either case, prove your answer in detail.

Solution. There is no such polynomial. Had there been one, we would have

but also, by repeated integration by parts (an even number of times, for simplicity),

For any the first term in this formula involves only integers (as , , , , and are all integers), and if is larger than the degree of , the second term is 0. So is an integer. But by the first formula it is in . That can't be.

The results. 80 students took the exam; the average grade was 69.33/120, the median was 71.5/120 and the standard deviation was 26.51. The overall grade average for the course (of ) was 68.5, the median was 71.57 and the standard deviation was 18.64. Finally, the transformation was applied to the grades, with . This made the average grade 70.41, the median 73.5 and the standard deviation 17.77. There were 30 A's (grades higher or equal to 80) and 12 failures (grades below 50).

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Dror Bar-Natan 2004-05-10