I think that the Marsden example really was "easy to see", since I did think of Darboux when I saw it (and I'm not exactly well-versed in symplectic stuff).
Hey everyone, I put up a proof problem that I need help with (induction question). If anyone can answer it or help me, that would be awesome! math.stackexchange.com/questions/2355579/…
Prove that no matter how James splits the piles (starting with a single pile of n stones), the sum of the numbers on the blackboard at the end of the procedure is always the same.
So the question is actually pretty clear about the splitting of the piles. He chooses one pile of stones and splits it into two smaller piles, writes the product of the sizes on the board, and then he repeats this process until there is only one stone in each pile
@JohnLocke My suggestion: If you start with a pile of $n$ stones, and select $m$ of those stones, you'll end up with a pile of $n-m$ stones and a pile of $m$ stones.
Sorry about the "prime-numbers" tag. I thought it should have the same realm of contributors.
This question surely prompts subjective answers without further explanation of 'beautiful' in terms of numbers. By a beautiful number, I mean one which has notable properties and patterns between the di...
@Typhon looks like emma put up a pretty good answer so far, just didnt finish the inductive step part. I understand it from there, but I am wondering whether we need to now think of further cases?
The answer is 4 8 15 16 23 42
Joke aside what you actually need is iterated induction (sorta).
Some setup:
The ordering of two piles is irrelevant. A pile of 2 and a pile of 3 is the same as a pile of 3 and a pile of 2. This is important, and we need not replicate these cases when resolving...
Typhon, how mad were you about deciphering the stones and piles problem? It was really getting you riled up right? Imagine me here as a student looking at it...
Well, if you studied the topology of dynamical systems on manifolds, there's a lot of geometry and interesting analysis. But that's more sophisticated.
i actually hope that amie wilkinson teaches something in the vein of dynamics on manifolds in a quarter where i dont have a lot going on, id like to learn something about it
Let $y=\sqrt{4x+x^2}-\sqrt{2x+x^2}$ (multiplying both sides of the above approximation by $\sqrt{x})$.
We want to show that $y\to 1$ as $x\to\infty$
One way to do that would be to factor $x$ from the square roots and get $y=x\left(\sqrt{1+4/x}-\sqrt{1+2/x}\right)$. But $4/x,2/x$ are small for large $x$, so we can do a binomial expansion and proceed from there.
alternatively, we can try the idea I was spouting above once more
rearranging, we have $$y+\sqrt{2x+x^2}=\sqrt{4x+x^2}\implies y^2+2y\sqrt{2x+x^2}+2x+x^2=4x+x^2\implies 2y \sqrt{2x+x^2}=2x-y^2$$
Squaring both sides again, that gives $4y^2(2x+x^2)=4x^2-4xy^2+y^4$
If I now divide both sides by $x^2$ and take $x\to\infty$ (presuming that $y$ goes to a constant) then this becomes $4y^2=4\implies y=\pm 1$
And it'll have to be $y\to 1$ because $y>0$ for all $x>0$.
[To be expanded later when I get back home] Seeking for maths device on a possible art like effect: the impression of passing by horizontally a row of lamps, how to make the impression more dramatic
The answer is 4 8 15 16 23 42
Joke aside what you actually need is iterated induction (sorta).
Some setup:
The ordering of two piles is irrelevant. A pile of 2 and a pile of 3 is the same as a pile of 3 and a pile of 2. This is important, and we need not replicate these cases when resolving...
