• \(f(n) = \sum_{k=1}^n 1/(\sqrt{k}+\sqrt{k+1})\).
  • This sum telescopes: rationalize the denominator by multiplying by 1/1, where \(1  = \sqrt{k} - \sqrt{k+1}\).
• \(\int_0^\inftty \ln(x) dx / (x^2 + 9)\).
  • We tried several approaches. We have a product so we tried integrating by parts, but couldn't make that work. Turns out though there is a lot of symmetry, which frequently happens with integral. First we change variables to replace the 9 with 1. If we let \(x = 3t\) we get \(\int_0^\infty \ln(3t) 3dt / (9t^2 + 9)\). This then gives us \(\int_0^\infty \ln(t) dt / (9t^2 + 9) + \int_0^\infty \ln(3)dt / (9t^2+9)\). The second integral can be done and is arctan, and can be evaulated easily. The first however is hard and we don't have a nice anti-derivative. Fortunately for many problems like this we don't need an anti-derivative; this is the difference between definite and indefinite integrals. Frequently the definite integrals have a nice answer, and that's the case here. We now exploit symmetry and break it into integrating from 0 to 1 and then 1 to \(\infty\). When we do this we then can make the integrals over the same region by letting \(t = 1/u\) in the second integral, and then the two integrals from 0 to 1 cancel! This is a common result -- these problems have a great deal of symmetry that can be exploited.
  • Mathematica can solve!  Integrate[Log[x] / (x^2 + 1), {x,0,Infinity}]
• \(\sum_{n=1}^\infty (2^{\langle \sqrt{n} \rangle} + 2^{-\langle \sqrt{n} \rangle}) / 2^n\), where \(\langle \sqrt{n} \rangle\) is the closet integer to \(\sqrt{n}\).
  • We did some work on this, figuring out where the function \(\langle \sqrt{n} \rangle\) is constant. One of the big issues in problems like this is whether or not it's better to keep the two terms in the numerator together or to split. You can make arguments for both viewpoints, but often there is a reason they give you the problem -- either to hide something or to allow for some cancellation. For this problem it turns out that it is better to have them together.
  • See page 5 of http://www.cs.berkeley.edu/~wkahan/MathH90/SPutn01.pdf for a solution.
• We also did a problem from the 2013 Problem, B1. http://www.eecis.udel.edu/~mckennar/academics/competitions/putnam/Solution.pdf
  • We saw the value of looking at small cases and building up intuition. We looked at the sum from 1 to n for various n, and saw that it was always -1 when n was odd. This suggested what the answer should be. We then proceeded by induction (solution above).
• Useful inequalities:
  • http://www.math.wustl.edu/~victor/putnam/01ineq.pdf
  • http://www.math.union.edu/activities/local/putnam/Inequalities.pdf

• http://en.wikipedia.org/wiki/Look-and-say_sequence
• Proof of cosmological theorem: http://www.ams.org/journals/era/1997-03-11/S1079-6762-97-00026-7/home.html
• More on cosmological: http://www.nathanieljohnston.com/2010/10/a-derivation-of-conways-degree-71-look-and-say-polynomial/

• Josephus problem: http://en.wikipedia.org/wiki/Josephus_problem
• Mathematica program: Josephusk2014.nb    Josephusk2014.pdf   

• Geometric series: http://en.wikipedia.org/wiki/Geometric_series
• Memoryless property: http://en.wikipedia.org/wiki/Memorylessness

• Checkers is solved:

• Catalan's conjecture: http://en.wikipedia.org/wiki/Catalan's_conjecture
• Fermat's last theorem: http://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem
• Beal's conjecture: http://en.wikipedia.org/wiki/Beal's_conjecture
• Notice how many nice, elementary items this problem gives us, from looking at algebra in good ways to divisibility properties of numbers to passing to solutions of certain Diophantine equations.

• Comment: as we've seen many times, it often helps to introduce structure. Make someone special! See if you can solve the problem if you have additional information, then see what is needed to get that information!
• Comment: interestingly, there's a danger of putting on blinders. The problem never asks us to find the FIRST time that everyone has been in, only SOME time. It's very easy to assume something that isn't stated. Here's the nine dots in a 3x3 grid with three straight lines, never lifting pen from paper: https://www.youtube.com/watch?v=a-6-2kZZe2w . Here's a nice description: http://www.archimedes-lab.org/How_to_Solve/9_dots.html
• Comment: Part of the goal of the class is to connect to real math problems. This riddle connects with many issues in number theory: often the way we prove something exists is we prove there are infinitely many, and thus there must be at least one! A great example is Dirichlet's Theorem of Primes in Arithmetic Progression.

• Comment: never give up on induction! Break into small cases. Build intuition.

Here is a differentiating identies handout (this became the nucleus for a section in a probability book I'm writing). If the sums are finite there is no difficulty justifying the method, but for infinite sums it is very important to check to make sure we can do this interchange (interchanging the sum and the derivative); it is frequently referred to as differentiating under the integral sign. Differentiating identities is a powerful technique; it creates infinitely many more identities from a given one.

We have to be very careful in interchanging orders of operations. We can talk about interchanging two integrals, or a derivative and an integral (click here for conditions on when this is permissible; this is called differentiating under the integral sign). In general we cannot interchange orders of operations (\(\sqrt{a+b}\) is typically not \(\sqrt{a} + \sqrt{b}\)), but sometimes we're fortunate (click here for a nice article on Wikipedia on when this is permissible).

• It is not always possible to interchange orders of integration (see Fubini's Theorem for when this may be done). The main take-away is that we must be careful interchanging.

• Fubini's theorem is one of the most important results in integration theory in several variables. There really isn't an analogue in one dimension, as there we have no choice in how to integrate!
  • Whenever you are given a theorem, it is worthwhile to remove a condition and see if it is still true. Typically the answer is no (or if it is still true, the proof is frequently much harder). There are many functions and regions where the order of integration matters. The simplest example is looking at double sums rather than double integrals, though with a little work we can convert this example to a double integral. We give a sequence \(a_{mn}\) such that \(\sum_{m = 0}^{\infty} \sum_{n = 0}^{\infty} a_{m,n}\) is not equal to \(\sum_{n = 0}^{\infty} \sum_{m = 0}^{\infty} a_{m,n}\). For \(m, n \ge 0\) let \(a_{m,n} = 1\) if \(m = n\), \(-1\) if \(n=m+1\) and \(0\) otherwise. We showed that the two different orders of summation yield different answers. The reason for this is that the sum of the absolute value of the terms diverges.
  • Click here for another example where we cannot interchange the order of integration; a more involved example is available here.
  • Click here for a video by Cameron on how he applies Fubini's theorem to change the order of operations (he does a double sum instead of a double integral, but the principle is the same).

• What's so nice about the geometric series is that the tail is a geometric series itself.
  • One approach is to say if \(f(x) = \sum_{n=0}^\infty x^n\), then \(f(x) = \sum_{n=0}^{N-1} x^n + \sum_{n=N}^\infty x^n = \sum_{n=0}^{N-1} x^n + x^N / (1-x)\). We now have a finite sum, and the derivative of the sum is the sum of the derivatives.
  • Arguing as above, we can use the fact that the tail of the geometric series is a geometric series, and thus \(f(x) = \sum_{n=0}^{N-1} x^n + x^N f(x)\). Solving yields \(f(x) = \sum_{n=0}^{N-1} x^n / (1 - x^N)\). We can now find \(f'\) by using the quotient rule, as the derivative of the numerator is fine as it is a finite sum. I strongly urge you to do this. What's nice is that we'll then send \(N\) to infinity, and the denominator \((1-x^N)^2\) will tend rapidly to 1.

We studied one of my favorite problems, given \(S = a_1 + \cdots + a_n\) with each \(a_i\) a positive integer, maximize the product of the \(a_i\). We quickly see the optimal is when each \(a_i\) is 2 or 3, and since \(2\ast 2 \ast 2 < 3 \ast 3\) we want \(3\)'s over \(2\)'s. We converted to a real problem and assumed there were \(n\) summands, each a real number. We got a function defined on the integers to maximize, replaced it with a function defined on the reals so calculus would be applicable. We then curve sketched and saw the function was increasing to its maximum and decreasing past it, so the optimal integer soln was either to the left or right of the optimal real soln (here optimal soln is referring to the number of summands). It's unusual to be this fortunate.

We had to maximize \(a_1 \ast \cdots \ast a_n\) given \(a_1 + \cdots + a_n = S\) and each \(a_i > 0\). We can do this with Lagrange multipliers, or since each \(a_i\) is in \([1, S]\) we can appeal to the \(n=2\) case because a real continuous function on a compact set attains its max and min. What is nice is that this existence result from real analysis improves to being constructive; if we were at the optimal point and all coordinates were not equal, we could simply replace two of them with the average and improve the product.

A nice application of this problem is that for disk storage (see radix economy), base 3 has advantages over base 2, though base 2 has the very fast binary search. Another nice example of base 3 occurs with the Cantor set.

We talked a lot about how to use logarithms to make an analysis easier, or to exponentiate. For example, \((S/x)^x = \exp(x \log(S/x))\).

Here is a warning that you may have been fooled into believing you learned certain derivatives when you hadn't. For example:

• \(f(x) = x^n\) has derivative \(n x^{n-1}\). This follows from the definition of the derivative and the binomial theorem to expand \((x+h)^n\) when \(n\) is a positive integer.

• \(f(x) = x^{p/q}\) has derivative \(\frac{p}{q} x^{p/q-1}\). This follows by setting \(g(x) = f(x)^q = x^p\) and then differentiating, which gives \(g'(x) = q f(x)^{q-1} f'(x) = p x^{p-1}\), and then substituting and solving for \(f'(x)\). We cannot get it the same was as the derivative of \(x^n\), as that would require knowing the binomial theorem for non-integral exponents.

• \(f(x) = x^{\sqrt{2}}\) has derivative \(\sqrt{2} x^{\sqrt{2}-1}\). This follows from using the exponential function and the chain rule: \(x^{\sqrt{2}} = \exp(\sqrt{2} \log x)\).

• Thus, \(x^r\) does have derivative \(r x^{r-1}\), but the proof for general \(r\) goes through the exponential function.

• \(e^\pi\) vs \(\pi^e\): take logarithms of both sides.
• \(\pi \log e = \pi\) vs \(e \log \pi = e \log(e + \pi - e)\), where the last follows form adding zero. We do this so we can exploit log laws (log of a product is sum of the logs).
• \(\pi\) vs \(e \log e + e \log\left(1 + \frac{\pi - e}{e}\right)\), and now we use \(\log(1+x) = x - x^2/2 + x^3/3 - x^4/4 = x\) plus a negative number (decreasing in absolute value, alternating).
• \(\pi\) vs \(e + e \frac{\pi -e}{e}\) + negative number.
• \(\pi\) vs \(\pi\) + negative number; thus it must be \(\pi \ge \pi\) + negative number, and thus working upward we find \(e^\pi \ge \pi^e\).

• I'm writing a module on streaming information; if you want to see that email me and I'll send.
• Below are links to some of the key points in streaming vieo. My slides are available here.Here is the link to the lecture: https://www.youtube.com/watch?v=WjnHNHPhJtU
  • Error detection
  • Error correction
  • Check digit schemes
  • UPC Symbols (25th anniversary article)
  • Target story
  • Hamming codes (click here for the Hamming (7,4) code and click here for the general case)
  • Hamming codes are related to sphere packing.
  • Fourier Series (and a wonderful generalization, wavelets)
  • Digital signatures

• Consider the following problem: take the numbers 1, 2, ..., \(n\) and randomly sort them. If the number \(k\) is on top, reverse the order of the first \(k\) numbers. Continue as long as the top number isn't a 1. Prove that eventually this process terminates with a 1 at the top. (So, if we start with the ordering 4 2 1 3 5, the next ordering is 3 1 2 4 5, then 2 1 3 4 5 and finally 1 2 3 4 5).
• In class we searched for several different monoinvariants. I think the following solution works. It clearly suffices to show that eventually either \(1\) or \(n\) is at the top, or \(n\) is at the bottom (if it's 1 we win, while if \(n\) is at the top then on the next turn it's on the bottom, and an \(n\) on the bottom reduces us to the case of \(n-1\) and we are done by induction).
• I claim that either 1 is eventually on top, or \(n\) is eventually on the bottom (note \(n\) on top implies \(n\) is on the bottom at the next step). Let us assume this does not happen for some \(n\), and without loss of generality let \(N\) be the smallest integer such that \(1\) is never on top and \(N\) is never on the bottom. As there are only finitely many ways to order \(N-1\) numbers, the ordering must be cycle (and in all the elements of the cycle, \(N\) is never on the top or bottom).
• We have \(N\) is never on the top or the bottom; thus whatever card is on the bottom must always stay on the bottom; let's call that card \(k\). Since it is only the top card that causes change to happen, note that we could safely switch \(N\) and \(k\). If we do this, we would still have an infinite cycle but now it would only involve the numbers 1, 2, ..., \(N-1\). This contradicts the minimality of \(N\) and thus our assumption is wrong.

• http://www.cut-the-knot.org/do_you_know/pigeon.shtml
• http://www.math.ust.hk/~mabfchen/Math391I/Pigeonhole.pdf (more extensive notes)
• http://zimmer.csufresno.edu/~larryc/proofs/proofs.pigeonhole.html (a friendly version)

• We talked about circular orderings. This leads to a difficult but fun problem, the napkin problem (I'm not the greatest at manners, and as a group mathematicians are often taken to be a bit clueless).
• We then turned to the cookie problem, counting the number of ways to divide 10 identical cookies among 5 distinct people (here's the classic clip where Cookie Monster meets the Count, whose full name is Count von Count -- they changed how he appears!). Counting is very important; we talked a bit about counting 'good' games in tic-tac-toe, and ignoring counting obviously bad ones. (Consider the error in the classic Princess Bride `Chess' Match or in the Princess Bridge Battle of Wits). It's usually called the stars and bars problem. What I love here is the power of changing your perspective -- we go from a very painful brute force approach to being able to solve it in one line.
• I have posted the cookie problem on my math riddles page (email me if you want to contribute); someone with far more patience than I solved it by brute force. Here's their solution. The final number, tabbed from the others, are how many distinct rearrangements we have of this basic configuration. The total  is 1001, or (10+5-1 choose 5-1). We saw this problem lead to a discussion of multinomial coefficients. When there are two people getting cookies, say 8 and 2, there aren't (5 choose 2) ways to assign people, but (5 choose 2) * 2! (we choose the 2 people, then there are 2! ways to choose which gets the 8 and which gets the 2). If we have 8 1 1 it's more involved. In that case it's (5 choose 3) to choose the three people, 3! ways to order which of the people gets which number, but then we must divide by 2! (as the two people getting 1 are indistinguishable). A better way to view 3!/2! is 3! / (2! 1!) (note the numbers on the bottom sum to the top). This is an example of a multinomial coefficient, a generalization of binomial coefficients. For example, if we have MISSISSIPPI, there would be 11! ways to order the letters (order matters) if the letters are distinguishable, but they're not. So let's put subscripts on the letters: MI₁S₁S₂I₂S₃S₄I₃P₁P₂I₄. We then have 4! ways of placing the four marked S's in the four S positions, and so on, giving 11! / (4! 4! 2! 1!) (I like including the final 1 so that the bottom sums to the top).
• Below are the person's solutions, arranged differently than we did in class.
• 0 0 0 0 10         5
  
  0 0 0 1 9         20
  0 0 0 2 8         20
  0 0 0 3 7         20
  0 0 0 4 6         20
  0 0 0 5 5         10
  
  0 0 1 1 8         30
  0 0 1 2 7         60
  0 0 1 3 6         60
  0 0 1 4 5         60
  0 0 2 2 6         30
  0 0 2 3 5         60
  0 0 2 4 4         30
  0 0 3 3 4         30
  
  0 1 1 1 7         20
  0 1 1 2 6         60
  0 1 1 3 5         60
  0 1 1 4 4         30
  0 1 2 2 5         60
  0 1 2 3 4         120
  0 1 3 3 3         20
  0 2 2 2 4         20
  0 2 2 3 3         30
  
  1 1 1 1 6         5    
  1 1 1 2 5         20
  1 1 1 3 4         20
  1 1 2 2 4         30
  1 1 2 3 3         30
  1 2 2 2 3         20
  
  2 2 2 2 2          1
• What we're really doing is solving the equation \(x_1 + \cdots + x_5 = 10\) in non-negative integers. This is a very special type of Diophantine equation. It's actually a special case of Waring's problem, which looks at solving \(x1^k + \cdots + x_s^k = n\) for fixed \(s\) and \(k\). These problems are in general not accessible through combinatorics; the case \(k=1\) is special. The general approach proceeds via generating functions, which we will cover in great detail later in the semester (it's one of the key concepts of the class).
• Our solution to the cookie problem is quite elegant, and in some respects reminiscent of geometry class (remember all those proofs where the teacher cleverly addsauxiliary lines; the difference here is we just add more cookies). While it is possible to solve many combinatorial problems by brute force in principle, in practice this is not a good way to go -- it is time consuming, and quite likely that one makes a mistake. Typically one finds a way to interpret a given quantity two ways; we can compute one of them and thus we obtain a formula for the other. For example, we showed the number of ways of dividing C cookies among P people is (C + P - 1 choose P-1); here all the identical cookies are divided. What if we don't assume all the cookies are divided -- what is the answer now? It is just \(\sum_{c = 0}^{C} (c + P - 1\ {\rm choose}\ P - 1)\); this is because we are just going through all the cases (we give out no cookies, 1 cookie, ...). What does this sum equal? Imagine now we have another person, say the Cookie Monster (this is one of Cameron's favorite clips), who gets all the remaining cookies. Then dividing at most C cookies among P people is the same as dividing exactly C cookies among P+1 people, and hence our sum equals (C + P+1 - 1 choose P+1 - 1).
• Related to the cookie problem is the partition problem: for the cookie problem we consider 2+3+3+1+1 different from 1+2+3+3+1 as the five people are distinct; if we don't consider these distinct then we have a partition problem. It's a lot more complicated to count these, but some great mathematics (such as Young tableau).
• Finally, we ended with a brief discussion of the lottery problem. Say there are 50 numbers. If we cannot use any of the 50 numbers more than once, there are (50 choose 6) = 15,890,700 ways. What if we can use the same number multiple times -- how many combinations are there now? Writing the answer cleanly would give it away, so I'll just say that if we have to choose 6 numbers from {1,...,50} and we can use each number up to 6 times, and if order doesn't matter, then the number of combinations is 28,989,675, which is less than a factor of two more! For comparison, note that (300 choose 6) is the significantly larger 962,822,846,700, which is over 60,000 times larger than (50 choose 6)! If you want to see the solution, let me know.