Math is easy

Here is a problem for you guys: is there a function that takes two natural numbers (a,b) and yields the number of connected components of Top[S^a, S^B]?

explain your notation

>
m a t h

i heard about that but how does this even work? isnt it just first principle?

The connected components of the space of maps from the a-sphere to the b-sphere (which is automatically a CW-complex).

They tried to formalize philosophical abstract concepts to build math.

Prove or give a counter-example of the following statement:

In three space dimensions and time, given an initial velocity field, there exists a vector velocity and a scalar pressure field, which are both smooth and globally defined, that solve the Navier–Stokes equations.

NOW THAT'S WHAT I CALL MATH, volume 1.

Would anyone buy such a book?

Is it possible to do this by inscribing a triangle in the circle?

Only if it had a Pug smoking a cigar while reading a paper about the Banach-Tarski theorem on the cover

Couldn't you also do this with vertical lines?

1/1(1*1)1/1(1*1)1/1

NP=P :^)

At the very least, I would pirate it.

>doing it in the easy way

pshhh

There was a user on /g/ that said his professor was reviewing a paper that might prove that P != NP

His professor shouldn't be sharing that information.

That information is hardly enough to get "scooped" -- I suspect the majority of mathematicians are convinced that P != NP and it's not helping them prove it any faster.

No, you aren't supposed to talk about papers you're reviewing.

intergral of (e^x^e^x^e^x^e^x^e^x^e^x)/(666*(-69)^sqrt(-i))

No, you can't get equal area pieces if you inscribe a triangle.

No, if you only used vertical chords that don't intersect then the pieces on the side must be congruent in order to have the same area.

The area a chord divides the circle into determines its position relative to the center of the circle. For example, if a chord divides a circle into parts that are 1/3 and 2/3 of the circle's area, then the angle between the radii which touch the endpoints of the chord is the solution to the equation

x - sin(x) = 2pi(1/3)

Such an equation will always have only one solution between 0 and 2pi, but can only be solved numerically. Let's call it the "position angle".

In order to describe any combination of chords, we must determine a similar kind of angle that describes the separation between the radii touching the endpoints of intersecting chords based on the area they together delineate. Let us call this the "separation angle" This is a bit more complicated. The separation angle is the solution to

cos(z/2) cos(x-y/2) csc(x-y/2-z/2) (1-cos(y/2) sec(x-y/2))^2-sin(x) cos(y/2) sec(x-y/2)+x = 2pi F

Where y and z are position angles of the two intersecting chords and F is the fraction of the total area delineated by the intersecting chords. This of course can only be solved numerically.

So it seems rather difficult to prove that it is impossible for such a set of chords to exist. I have proven by exhaustion that it is impossible for this to be done in 5 chords or less.

Divide the circle with a horizontal line in two. For the left side use the horizontal line trick, for the right side draw chords from a single point at the top of the circle, and end of each chord in appropriate points on circumference of the right half circle.

Could be done in 3 chords I suppose?

Yeah did you not read what I said? Look at the two chords at the ends. They must be congruent with each other since the only thing that determines their shape is the area they delineate, which is equal.

I'm trying to find the name of this problem (or just formulate it better).

A circuit is a graph of components and traces. For a given circuit, find the configuration that has the least jumpers over traces (in 2D space).

Dunno about a name, but I believe you're asking for the circuit with the least self-intersections in 2d space?

So if you're allowed to cut the circle using only chords that don't intersect, doesn't that mean you'll always end up with at least 2 figures being circular segments, thus proving such division impossible?

>So if you're allowed to cut the circle using only chords that don't intersect
Where did I say they aren't allowed to intersect? Of course they are, or the problem would be trivially impossible.

So is the seperation angle, the angle of the intersection of the two position angles of the intersecting chords?

So you basically created these equations for all groups of three from the five chords and then ran all of the equations through some program to show there is no solution for 5 chords?

Groups of two*

>So is the seperation angle, the angle of the intersection of the two position angles of the intersecting chords?
Yes, exactly.

>So you basically created these equations for all groups of three from the five chords and then ran all of the equations through some program to show there is no solution for 5 chords?
Groups of two you mean. Yes, essentially that's what I did. There are some things that help reduce the problem space. Anything where a cycle of intersections is made, like three chords intersecting to form a triangle in the middle, doesn't work because it's an overdetermined system.

Getting there
A circuit is a planar graph of components and traces. Find the _____ that minimizes self-intersections.
What do you call a transformation where edges still connect to the same vertices?

>intergral of (e^x^e^x^e^x^e^x^e^x^e^x)/(666*(-69)^sqrt(-i))

easy peasy m8

Why does an overdetermined system mean there are no solutions?

Here is the simplest division that breaks symmetry. It would be a solution if that small chord at the top left didn't cross that other chord.

Because it has negative degrees of freedom. In other words, it has properties that can't all be true at the same time.

A stupid example of an overdetermined system:
x = 5
x = 2

There is no such x, that solves this system.
The only way such a system (x=a, x=b) is if two of the equations where linearly dependent (if a=b)

I'm probably explaining this horribly

Okay, makes sense.

Yes, you are both explaining this horribly. Because you are both wrong.

en.wikipedia.org/wiki/Overdetermined_system

Nothing says an overdetermined system can't have a solution.

>Nothing says an overdetermined system can't have a solution.
It's generally intended that way by anyone with a brain, as an overdetermined system with a solution has unnecessary constraints that can be simply discarded to yield an equivalent system that is not overdetermined.

If you want to be a pedant, be less of a brainlet asshole.

>interpreted
The definition seems pretty clear to me.

>greentexting with a different word than i used
I didn't disagree with the definition, see:
>If you want to be a pedant, be less of a brainlet asshole.

You can show an inverse square law with triangular numbers sequence.

Show us how it's done.