The graph of this equation $\frac{dA}{dt}=A(2-\frac{A}{5000})$, at first seems exponential but then it becomes a constant equal to 10000 and the reason it's because of this term $-A^2/5000$ right?
Somehow you have to smash togather the saddle points if you want to do it for a torus... im sure the 3rd critical point (apart from the max/min) is going to be a Monkey saddle
what are necessary conditions for $f:\Bbb R \to \Bbb R$ such that for every metric space $(X,d)$, $(X,f \circ d)$ is a metric space with the same open sets?
Does polarization give you a functor from normed vector spaces (with morphisms being continuous maps) to inner product spaces (ditto)? @MatheinBoulomenos
Let $n$ be an integer.
Then any prime factor of
$$ 5 n^4 - 70 n^3 + 380 n^2 - 945 n + 911 $$
Must be congruent to 1 mod 10.
Also
Let $n$ be an integer.
Then any prime factor of
$$ 5 n^4 - 10 n^3 + 20 n^2 - 15 n + 11 $$
Must be congruent to 1 mod 10.
How to prove these ?
How to find such i...
There's a singularity which looks like three lines intersecting each other transversely in every pair (like -A-), which closes up and becomes three lines intersecting each other at a single triple point (like -X-) and then drifts apart to become -A- again
That triple point is where the Monkey saddle is at
So realize that if $f : T^2 \to \Bbb R$ is a map on the torus with one maxima, one minima and one fucked saddle, then $\nabla f$ (the gradient) gives a vector field on $T^2$ with one source, one sink and one monkey saddle (the index -2 picture).
There is a way to get such a vector field on the torus. Think of the hexagonal tiling of the plane and quotient by the lattice isometries. This gives a torus as a quotient of a hexagon
The 1-skeleton of this picture has a "Y" bit in it, as you can see in the picture
There is a vector field along that Y with a Monkey saddle at the center of the Y
I am 90% sure you can produce a vector field on the hexagon which quotients to a vector field with one source, one sink and one Monkey saddle like that
It'd be conservative, and it's antiderivative is going to be the 3cp map
Consider the following iterations :
$x_0 = z$
Where $z$ is complex.
$x_n = \frac{ x_{n-1}^2 - 1}{n}$
It is well known that for real $z > 3$ the sequence grows double exponentially.
It is known that for $z = 3$ the sequence grows linear ; in fact like $3,4,5,6,7,...$.
In fact When considering...
There's actually a trouble. I don't think there's a vector field with a singularity which has a local picture like a "Y". Also notice that the picture I posted has two Y's in it.
I was too hasty. There was a problem Ted came up with once which is, come up with a vector field on the torus which has a single source and a single saddle
My way of seeing it was take the usual square picture on the torus and consider the vector field on the square which is like $(-y, x)/\|x\|$ and stays tangential to the boundary of the square
When it quotients down the four corners of the square forms a "X" picture" where the saddle lies and the interior of the square has a source
its so hard to tell when to use hyperbolic trig sub or just ordinary trig sub in the integrals i don't see when i should use one over the over. does any one know a good rule of thumb for this?
Your formula clearly doesn't work for $y = x^2$, then eg. It gives $\pi \int_0^{\sqrt{r}} x^4 dx$ for a paraboloid of radius $r$, which is wrong, isn't it?
The general idea: when x gets large, the largest power will dominate the others. Find the convergance/divergance of that dominating function and tweak it into a proof of convergance/divergance for the original function.
Well, crap. OK, yes, x^(-3/2) does converge-- it doesn't matter that it blows up near 0 because the function is well-defined at 0, we're only concerned as x -> infinity.
The second one is like x^(-1/2) which diverges. My mistake if I said it wrong.
We only need the upper bounding condition eventually. So, for x>1, we know x^(-3/2) converges, and thus the original function converges. For 0 < x < 1, we know the original function converges for a different reason.
I may have had it some time back, I got sick in Texas right after my aunt did and it was nasty, but the main feature was really flu-style rather than... stomach-type things, aside from mild diarrhea, so I dunno... Anyway came here, recovered quickly, now two of my friends got sick back to back but it's very stomach virus-y
Let $n$ be an integer.
Then any prime factor of
$$ 5 n^4 - 70 n^3 + 380 n^2 - 945 n + 911 $$
Must be congruent to 1 mod 10.
Also
Let $n$ be an integer.
Then any prime factor of
$$ 5 n^4 - 10 n^3 + 20 n^2 - 15 n + 11 $$
Must be congruent to 1 mod 10.
How to prove these ?
How to find such i...
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