
The Work of John Milnor, a giant in modern mathematics [pdf] - Outdoorsman
http://www.abelprize.no/c53720/binfil/download.php?tid=53562
======
droffel
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.

~~~
dang
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.

~~~
MakingBreakfast
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.

~~~
waterhouse
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:

    
    
      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
    

And then let's check the other rule.

    
    
      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
    

(I'm using approximately the proof format described in EWD 1300:
[https://www.cs.utexas.edu/users/EWD/transcriptions/EWD13xx/E...](https://www.cs.utexas.edu/users/EWD/transcriptions/EWD13xx/EWD1300.html))

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](https://en.wikipedia.org/wiki/Fibonacci_number#Matrix_form)

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kkylin
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.

------
michael_nielsen
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...](http://mathoverflow.net/questions/58061/how-can-there-be-
topological-4-manifolds-with-no-differentiable-structure/59406)

------
kdavis
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.

------
appleflaxen
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?

~~~
jeffcoat
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](https://en.wikipedia.org/wiki/Spherical_geometry)

------
jmalicki
[2011]

