TT's Maths Thread

[pooshwaltzer]

http://www.cs.sunysb.edu/~skiena/combinatorica/animations/euler.html

[TrueTears]

then how about my example?

[pooshwaltzer]

then how about my example?

All I can say is, you need to (1) use ALL vertices at least ONCE and (2) include all the edges during tracement. Your example is not an EULER cycle by Konisberg's definition.

[TrueTears]

But if you follow the wikipedia definition: "An Eulerian cycle, in an undirected graph is a cycle that uses each edge exactly once." my graph satisfies that criteria.

a) graph is undirectied. tick

b) it is a cycle. tick

c) Uses each edge exactly once. tick


But anyway I think I get it now.

If a graph has a Euler cycle (following my books definition that the graph is connected and every vertex has an even degree) then a cycle which includes all the edges exactly once implies that each vertex will be touched at least once.

By being connected ensures there is no isolated vertex and to prove that each vertex has to be touched at least once, since the graph is connected, by traversing each edge once you will arrive at a vertex and by traversing ALL edges ensures that you will have touched every single vertex.

Can anyone confirm this?

[Ahmad]

I wouldn't worry about it, the connected case is the interesting one.

[TrueTears]

"If a connected, planar graph is drawn on a plane then the plane can be divided into continuous regions called faces. A face is characterised by the cycle that forms its boundary."

This is what my book says, and then it goes on to illustrate an example by saying this graph:

[IMG]http://img245.imageshack.us/img245/3914/planargraph.jpg[/img]

... has 4 faces, namely A, B, C, D (D is the face bounded by the cycle (1,2,3,4,6,1)

however shouldn't there be at least 6 faces? ie a face bounded by (1,2,5,4,6,1) and (1,2,3,4,5,1)...?

[TrueTears]

Find if it exists.

After testing different paths, the limit appears to be 0, so let's try an proof. I got a bit rusty at this, I just don't know how to finish it off lol.

We need to prove if for every there exists a such that if then

We know that

But

Thus we can take but then how do we relate this back to again? I forgot lols

[kamil9876]

That's really simple, remember that the distance from to is . So you need to make sure that . So if i believe the rest of that garbage u posted (ill assume u are right), it is enough to set .

Because look:

I'll assume this is true, i cbf checking if u derived it properly. Now the interesting thing that this is saying is that:



Where is the distance between and .

So indeed if then as desired.









Btw: it is more economical to do it like this:

and apply the sandwhich theorem (or if u have real fetish for e-d, you can easily translate it to that if u UNDERSTAND it properly, i'm sick of how ppl treat e-d as some alien symbolic manipulation)

[TrueTears]

lololol ah yeah, i get it now, thanks heaps kamil

[TrueTears]

The total bill for a house is $260.65 over a 93 day period.

6 people live in this house.

F has lived 93 days (from the first day till the 93rd day which is 20/08)

L, Z, J all 3 of them has lived 30 days they all moved in from 21/07 to 20/08

W has lived 32 days moved in from 19/07 to 20/08

WG has lived 10 days 19/07 to 29/07

How much should they pay respectively? Obviously the person living longer should pay a bigger proportion.

[Mao]

Total people-day: F: 93 person-day, L+Z+J: 90 person-day, W+WG: 42 person-day, total: 225 person-day

Total bill is 260.65, thus each person-day is 260.65/224 = s

F pays 93*s, L pays 30*s and so on. [which is how I work my bills.. if that helps?]

[pooshwaltzer]

Ever had a landlord charge on the basis of a different cost driver? That is, one not based on "person-days"?

I've encountered,

* Time in apartment (down to the minute) based on facility entry cards
* floor area of let room (sq ft) excluding common areas
* Use historical cost info to work out average bill expenditure per month and charge as overhead plus inflation premium
* Divide bill by number of tenants as LL failed grade 3 math.
* Just charge a weekly flat rate irrespective of actual use; LL never attended primary.

[TrueTears]

For the change of variables in a double integral involving the Jacobian ie:



Do we strictly have to evaluate BEFORE THE ? Or can we also do it like this:



Ofcourse after taken into consideration changing the terminals.

I ask this question because sometimes evaluating with respect to first yields an non elementary integral so reversing the order of the and and changing the respective terminals works much better, but I'm not sure if the general Fubini's Theorem here applies to transformed double integrals.

Thanks.

[Mao]

That really depends on what S is.

I'm fairly sure you can prove that



But of course, if the boundary conditions of a pair of u or v are functions of the other variable, then they'd have to be the inner integral.

[TrueTears]

Ah okay thanks Mao, yeah I'm actually not that interested on a proof here, but I was wondering if Fubini's Theorem still applies after a Jacobian transformation.