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March 29

Kuratowski's theorem and Petersen's graph

hello. I want to use Kuratowski's theorem to prove the non-planarity of the Petersen graph. Now as every vertex in the Petersen graph has degree 3, I assume it can't contain a subdivision of K₅. So I have been trying to find a subdivision of K_3,3 in it. But whenever I try to find a subdivision the following problem occurs: The subdivisions I find always involve replacing edges by paths which have vertices in common. That is to say if X={a,b,c} and Y={x,y,z} are the vertices making up my K_3,3 then a subdivision should replace edges ax,ay,az,bx,by,bz,cx,cy,cz by paths which are distinct. But the nature of the Petersen graph is such that I always find paths where at least two of them have a vertex in common. For example avx and bvy. My question is: Is such a subdivision valid? If not, then how can I use Kuratowski's theorem to prove the non-planarity of the Petersen graph. Thanks--Shahab (talk) 06:44, 29 March 2009 (UTC)[reply]

If you remove one vertex from the Peterson graph, and perform all three possible un-subdivisions (I hope that makes sense?) then you get K3,3. Eric. 131.215.158.238 (talk) 08:56, 29 March 2009 (UTC)[reply]

Interperating xkcd webcomic: "My normal approach is useless here"

Someone showed me this webcomic from the (relatively) popular webcomic xkcd.com: here which contains several math functions with hearts representing "love".

Can anybody explain to me these functions and what they mean? Ultimately I'm not actually trying to learn the math, just trying to understand the jokes.

I see the first one is "Square root of Love". That's fine. "cos Love" obviously implies trigonometry. The 4th one I take it means "Set of love = ?". The 3rd one seems arbitrary. And the rest I don't get. Rfwoolf (talk) 08:23, 29 March 2009 (UTC)[reply]

Not sure how you're numbering them but have a look at derivative, identity matrix, and normal distribution. J kasd 08:37, 29 March 2009 (UTC)[reply]

Do you mean Fourier transform? Eric. 131.215.158.238 (talk) 08:59, 29 March 2009 (UTC)[reply]

I too have been seeing it as a Fourier transform, but it actually isn't, is it? Could it be a mistake? —130.237.45.207 (talk) 09:00, 30 March 2009 (UTC)[reply]

There are several different standard choices of how to normalize the Fourier transform, so dividing by sqrt(2 pi) isn't a problem. Same with failing to multiply by 2 pi in the exponent. As for notation, both "F(heart)" (here "F" is the Fourier transform of "f") and "F{f}(heart)" (here "F" is the Fourier transform operator, and F{f} is the Fourier transform of "f") would make more sense than "F{heart}", but eh. Eric. 131.215.158.238 (talk) 19:36, 30 March 2009 (UTC)[reply]

Yeah, Eric's right. Sorry about that. J kasd 01:42, 31 March 2009 (UTC)[reply]

(I am the 130.237.45.207 above.) To explore the properties of ♡, the Fourier transform of ♡ could be useful, and the notation "F{♡}" I would interpret precisely as that, as the Fourier transform of ♡. But then, as you said, he goes on and takes the Fourier transform of f instead, treating ♡ as if it was a frequency variable, which is what I found strange. This is how I would have done it (save for varying definitions, as you mention):

F\{\heartsuit \}(\omega )={\frac {1}{\sqrt {2\pi }}}\int \limits _{-\infty }^{\infty }\heartsuit (t)\,\mathrm {e} ^{-\mathrm {i} \omega t}\,\mathrm {d} t

—Bromskloss (talk) 09:44, 2 April 2009 (UTC)[reply]

xkcd is obviously not trying hard enough. Barnes Wallis showed how to do it in the book 'Mathematics with Love'. Dmcq (talk) 10:24, 29 March 2009 (UTC)[reply]

The joke is just that his "normal approach" to understanding some aspect of the world is to use mathematics. The particular symbols aren't part of the joke, they're just meant to suggest "using math". -- BenRG (talk) 11:42, 29 March 2009 (UTC)[reply]

To make it more of an inside joke for mathematicians, they could substitute

2r\left(\sin t-{1 \over 2}\sin 2t\right)\,

(one of the ~~equations~~ expressions for a cardoid) for the heart symbol. StuRat (talk) 16:38, 29 March 2009 (UTC)[reply]

Um.... what you called "one of the equations for the cardioid" is not an equation. Michael Hardy (talk) 19:50, 29 March 2009 (UTC)[reply]

Sorry. It was when I wrote that, but then I went back and took out the equals sign and left side, without changing "equation" to "expression". It's now fixed. StuRat (talk) 04:14, 30 March 2009 (UTC)[reply]

And it's an expression for one coordinate of a cardioid. —Tamfang (talk) 16:27, 30 March 2009 (UTC)[reply]

Comparative statistics subset of Inferential statistics

Would I be wrong if I were to describe comparative statistics as a subset of inferential statistics in a statistics coursework? --212.120.248.41 (talk) 14:36, 29 March 2009 (UTC)[reply]

It depends on what you mean by comparative statistics - it could be argued that the phrase is wider than inferential statistics, which puts some mathematics into the comparison, in which case the latter could be a subset of the former.…86.164.72.165 (talk) 19:43, 29 March 2009 (UTC)[reply]

Triangle sides

If the length of one side a of a triangle is equal to one third the sum of the lengths of the other two sides b and c, how would you show that the angle opposite side a is the smallest angle? Thanks, 76.248.244.232 (talk) 22:26, 29 March 2009 (UTC)[reply]

Without being too obvious (since this sounds like a homework problem) lake a look at the "See also" links in the triangle article. One of the laws referenced (hint hint) may be the answer you are looking for. If you still can't find the answer, add a follow-up post describing what you have looked at so far and someone will help you further. -- Tcncv (talk) 23:27, 29 March 2009 (UTC)[reply]

I see Law of sines, law of cosines, and law of tangents. I'm not very good at trigonometry, so I need a little more help, please. —Preceding unsigned comment added by 76.248.244.232 (talk) 23:42, 29 March 2009 (UTC)[reply]

Actually on rereading the problem I missed part of the problem. First, you need to determine if it is possible for either of the other two sides to be shorter than a. Draw a few pictures and then ask if b is smaller than a, how long must c be. Rremember that the lengths are related by a = (b + c) / 3. Then ask if this is geometrically possible. After you've identified the relationship between a and wither of the other two sizes, one of the above laws gives you a proportional relationship between the lengths and one of the trigonometric functions. If you apply this relationship, you are almost there. -- Tcncv (talk) 00:01, 30 March 2009 (UTC)[reply]

The side of length a is the shortest one if and only if the angle opposite that side is the smallest one. So the problem is: is that side the shortest one? If not, call the shortest one b. Then

3a = b + c,

c = 3a − b > 3a − a = 2a.

The triangle inequality tells us that

c ≤ a + b.

But

a + b < a + a = 2a.

So we've got

c ≤ 2a

but also

c > 2a.

Can't happen. So our assumption that a is not the smallest one must have been wrong. Michael Hardy (talk) 00:34, 30 March 2009 (UTC)[reply]