Day 41, MAT115R

• Today's quiz will be over knot/link identification (can you draw any of our knots or links?), as well as Hilbert's Hotel.
• In spite of my begging, several of you haven't submitted your one-paragraph description of your project and a one paragraph description of your logo. Your grade has begun to suffer....

  Don't let it suffer anymore! Get me your descriptions!

• You should email me your logo before Monday at 2pm.

• We continued our adventure at Hilbert's Hotel: where they'll always leave a light on for you, and they can accomodate you even if the hotel is full! Unless you're irrational....

  We began by reviewing what we'd learned last time about its ability to accomodate big groups. It seemed to handle everything that we could throw at it!

• I tried to convince you that the natural numbers and the real numbers cannot be put into one-to-one correspondence -- which means that they are not the same size.

  We tried to assign every real number a room at the Hilbert Hotel, but it turned out that there was at least one real number that didn't have a room. The reals are simply too big.

  Some of you remain unconvinced.... If you've kept up with your readings you've seen that others have also been trying to convince you -- are you getting there?

• Finally, we discussed sets, and their associated "power sets" (which are sets of subsets of a set!).

  We saw that the power set is always bigger than the set itself -- even the empty set, which has zero elements, has a power set with one element.

  The general rule is that if the cardinality of a set is \(n\), then the power set of that set will have cardinality \(2^n\).

• And I asserted that this is true even for infinite sets. Thus \[ Card(\mathbb{N}) < Card(\wp({\mathbb{N}})) < Card(\wp({\wp({\mathbb{N}})})) < \ldots \] There is an infinite family of bigger and bigger infinities!

Question of the day:

• Can we add up an infinite number of things and get a sensible answer? The answer is sometimes yes, as we saw when discussing the Dichotomy of Zeno:

  \[ \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\ldots=1 \]

• The reason for us to consider this problem seriously is contained in Pascal's triangle.

  Let's make ourselves a copy of our old friend, but let's leave a little room at the top. Skip four rows before starting in with the "1", and start a couple of spaces right of center.

• Now what does this have to do with Pascal's triangle? Let's first add all the invisible zeros. (If you did your homework, you know that Vi Hart helped you to see the invisible!)

  But a great mathematician named Jacob Bernoulli long ago made them visible:

  

  but only from one side!:) Let's put them on both sides.

• Having added all the infinite zeros, we're going to figure out how to continue Pascal's triangle in the backwards sense! We know how to go forward, but how do you go backward? It turns out that we have infinitely many options, which can be good or bad.

  What are some options?

• What might guide us as we try to figure out what two numbers add to 1?
  • Symmetry is a fundamental property of the triangle, and it seems like a useful thing to preserve. That leads us to just one choice for those two entries above 1 -- and once we make that choice, the entire row above the first one is complete.

    However, we're not going to preserve symmetry! You're shocked, I know, but there are other things to preserve....

  • Patterns: there are several patterns as we scan the "diagonals" of Pascal's triangle, and, while we can't preserve them all, we can preserve those on one side. So we'll choose to work the left side.

  • Ironically, this will create a certain type of "reflective symmetry" that we might not have noticed before.

• The first pattern we'll preserve is the line of ones running down the right hand side of the triangle.

  Once this choice is made, all the rest is determined! In particular, we have to put zeros on all the entries to the right above.

• Let's fill in the first entirely "backwards row", using the property of Pascal's triangle that the value below is given by the sum of the two elements above.

• Once we've done that, let's recall what the entries of Pascal's triangle add up to: powers of 2. Perhaps that pattern will continue. But we've been adding finite sums down below, on the original Pascal's triangle. We're now in the presence of an infinite sum: \[ S_1 = 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - \ldots \]

  What do you think that should add up to?

  If Pascal's "powers of two" pattern continues, then you get \[ S_1 = \frac{1}{2} \] I can show you why that one makes sense. What we're doing is adding that row to itself, and if you do that in a careful way you see that you will get that the row is equal to \(\frac{1}{2}\).

  Which is exactly what Euler got. Wait, what?

• What does the next row look like?

  Let's now focus on the sum \[ S_2 = 1-2+3-4+5-6+7-8+\ldots \] The power of the pattern of the "powers of two" allows us to calculate what that sum "should be".

  (And Euler agrees again! \(\frac{1}{4}\)....)

• One more row, at which point we focus on \[ S_3 = 1-3+6-10+15-21+28+\ldots \] Let's try the same trick again, and we'll see that it should equal \(\frac{1}{8}\).

• But now to the big story: \[ S = 1+2+3+4+5+6+7+8+\ldots \] What should that add up to? I think that you're going to be surprised at what Euler found².

• A final note. This "Backwards Pascal" triangle preserves another pattern: Fibonacci numbers!

  The Fibonacci numbers can also be extended backwards, and if we check out our extended Pascal's triangle, we see that the usual diagonals preserve the pattern of Fibonacci numbers!

• John Boaz's presentation My favorite number: 24, featuring "Euler's crazy calculation": \[ 1+2+3+4+5+6+ \ldots = \frac{-1}{12} \]

• The "Greatest Mathematician Alive", Terence Tao, also describes the same surprising result, in the article The Euler-Maclaurin formula, Bernoulli numbers, the zeta function, and real-variable analytic continuation.

• My collection of knots and links

• My Borromean knots (and links)

• Tricolor These Handout: just two of the knots/links are not tricolorable; all the rest are!
• A much more thorough introduction to both Reidemeister moves and tricolorability is in this article: Tricolorability of Knots, by Kayla Jacobs.