The Work of John Milnor, a giant in modern mathematics [pdf](abelprize.no)
abelprize.no
The Work of John Milnor, a giant in modern mathematics [pdf]
http://www.abelprize.no/c53720/binfil/download.php?tid=53562
18 comments
I think this is said every time Milnor is mentioned somewhere, but I'll repeat it anyway: the first few pages of _Morse Theory_ is some of the nicest mathematical writing I've seen anywhere. (The rest of the book is very good too.) _Topology from the Differentiable Viewpoint_ is likewise very clear and accessible.
In the PDF, the author says something to the effect of parallel lines on the sphere always cross, and points to the longitude lines in the graphic as an example. But the same example shows lattitude lines that appear parallel on that surface, and don't cross.
Does anyone know why there is an apparent contradiction?
Does anyone know why there is an apparent contradiction?
It's because "latitude lines" aren't really lines in this context (except the one at the equator).
If you took two points from one of those latitude "lines", you'd always be able to find a shorter path between them than the path along the latitude. If you take the shortest path between the two points and extend it all the way around the sphere, you'll end up with a great circle. That great circle is an actual "line" in this geometry.
https://en.wikipedia.org/wiki/Spherical_geometry
If you took two points from one of those latitude "lines", you'd always be able to find a shorter path between them than the path along the latitude. If you take the shortest path between the two points and extend it all the way around the sphere, you'll end up with a great circle. That great circle is an actual "line" in this geometry.
https://en.wikipedia.org/wiki/Spherical_geometry
[2011]
What is the next term in the sequence
[1, 1, 28, 2, 8, 6, 992, 1, 3, 2, 16256, 2, 16, 16, 523264]?
The answer is 24, if anyone was wondering.
Coming to this answer as a "logical conclusion" requires knowing the context of the question, that is, the reader is assumed to have (or somehow discover) the knowledge of how many differential structures there are in N dimensions, for an N starting at 5, and going up to 19.
While the paper is fascinating, I doubt that any sane person would think that this is any kind of a realistic "IQ" test question. The title feels like clickbait, and even more so since this "question" wasn't even the topic of the paper at all.
Edit: The title has been altered to no longer be clickbait. You can disregard that part of my comment.
The answer is 24, if anyone was wondering.
Coming to this answer as a "logical conclusion" requires knowing the context of the question, that is, the reader is assumed to have (or somehow discover) the knowledge of how many differential structures there are in N dimensions, for an N starting at 5, and going up to 19.
While the paper is fascinating, I doubt that any sane person would think that this is any kind of a realistic "IQ" test question. The title feels like clickbait, and even more so since this "question" wasn't even the topic of the paper at all.
Edit: The title has been altered to no longer be clickbait. You can disregard that part of my comment.
Ok, we took out the bit about an IQ question and replaced it with other subtitle. If Gowers says someone is a giant in modern mathematics, that's no clickbait.
For the sequence 1,2,5,14,41,122, I didn't notice that you gained each new term by multiplying by three and then subtracting one. However, the pattern I did notice is each additional term is obtained by adding 3^(n-1) where n is the term number. 1 + 3^0 = 2. 2 + 3^1 = 5. 5 + 3^2 = 14. Since these two methods provide the same result, they are probably related somehow, but I can't seem to figure it out.
Let f(n) = the nth term of the sequence (where f(1) = 1, f(2) = 2, f(3) = 5, etc). You're saying that f(n) = 3^(n-1) + f(n-1). It also happens that f(n) = 3 * f(n-1) - 1. If we started this sequence with a different value, like 1000, these rules would yield totally different sequences; but it happens that our choice of starting value means these rules are equivalent.
Why is this so? It will help to observe that f(n) = (3^(n-1) + 1)/2. Let's check if your rule is then correct:
The totally formal thing to do, rather than "observing" the formula, would be to prove by induction that the formula was correct (i.e. that the given rule, "n -> 3n - 1" with n starting at 1, yields it). How would one come up with the formula in the first place? Generally, if your rule involves multiplying by k every time, then the nth term will probably have k^n in it--though if you started with, say, n = 1/2, you wouldn't get anywhere. Is there a totally generic, plug-and-chug procedure that would yield a formula like this? Approximately, yes. For now, I'll just link to the Fibonacci series: https://en.wikipedia.org/wiki/Fibonacci_number#Matrix_form
Why is this so? It will help to observe that f(n) = (3^(n-1) + 1)/2. Let's check if your rule is then correct:
f(n) ?= 3^(n-1) + f(n-1)
<=> (plug in observed formula for f)
(3^(n-1) + 1)/2 ?= 3^(n-2) + (3^(n-2) + 1)/2
<=> (multiply by 2)
3^(n-1) + 1 ?= 2*3^(n-2) + 3^(n-2) + 1
<=> (subtract 1)
3^(n-1) ?= 2*3^(n-2) + 3^(n-2)
<=> (add up multiples of 3^(n-2) on right)
3^(n-1) ?= 3*3^(n-2)
<=> (identity)
T
f(n) ?= 3*f(n-1) - 1
<=> (plug in formula for f)
(3^(n-1) + 1)/2 ?= 3*(3^(n-2) + 1)/2 - 2
<=> (multiply by 2, distribute 3 on right)
3^(n-1) + 1 ?= 3*3^(n-2) + 3*1 - 2
<=> (simplify right)
3^(n-1) + 1 ?= 3^(n-1) + 1
<=> (identity)
T
The totally formal thing to do, rather than "observing" the formula, would be to prove by induction that the formula was correct (i.e. that the given rule, "n -> 3n - 1" with n starting at 1, yields it). How would one come up with the formula in the first place? Generally, if your rule involves multiplying by k every time, then the nth term will probably have k^n in it--though if you started with, say, n = 1/2, you wouldn't get anywhere. Is there a totally generic, plug-and-chug procedure that would yield a formula like this? Approximately, yes. For now, I'll just link to the Fibonacci series: https://en.wikipedia.org/wiki/Fibonacci_number#Matrix_form
Note: this appreciation is from 2011, when Milnor won the Abel Prize.
It’s safe to say that anyone who wins an Abel Prize is a giant in mathematics.
It’s safe to say that anyone who wins an Abel Prize is a giant in mathematics.
I like this variant: Complete this sequence: 1, 1, 1, ?, 1, 1, ...
If you said infinity, you're right! Number of exotic differentiable structures on R^n.
https://en.wikipedia.org/wiki/Exotic_R4
If you said infinity, you're right! Number of exotic differentiable structures on R^n.
https://en.wikipedia.org/wiki/Exotic_R4
Here is OEIS, a general service for solving these types of puzzles and much more: http://oeis.org/search?q=1%2C+1%2C+28%2C+2%2C+8%2C+6%2C+992%... You can always just look it up.
Also, correction: this is a number of classes of oriented differentiable structures on N-spheres, not just "in N dimensions".
Also, correction: this is a number of classes of oriented differentiable structures on N-spheres, not just "in N dimensions".
According to this paper (http://arxiv.org/pdf/1601.02184v1.pdf), the entry for n=56 should also be one. Apparently, these values come from a computer-assisted calculation from 1990, and the error wasn't discovered until 2014.
When I look at this sequence I see that for from 7 on, there seem nearly always exist more than one class of oriented differentiable structures on N-spheres, except for (up to 63) for 12 and 61. Does anybody know what makes the dimensions 12 and 61 so special here?
That looks to be addressed here: http://www.ams.org/notices/201106/rtx110600804p.pdf
Right. IQ is generally supposed to measure reasoning and problem solving abilities. While some knowledge is a prerequisite, this "test" would simply be testing your familiarity with a particular set of knowledge.
Of course, if you taught students this knowledge first, then having them figure it out is a good test of problem solving ability.
Of course, if you taught students this knowledge first, then having them figure it out is a good test of problem solving ability.
IQ tests attempt to measure how many "contexts", human "constructs" across various disciplines, a person is familiar with and fluidly capable in...
The fewer the contexts...well, I'll leave that to individual interpretation....
The fewer the contexts...well, I'll leave that to individual interpretation....
The hard question is not to exend this series up but to extend it down by only one entry. That's the real hardest IQ question ever.
There's a fascinating short companion (of sorts) to this, which is a MathOverflow question written by Gowers: http://mathoverflow.net/questions/58061/how-can-there-be-top...