Wikipedia:Reference desk/Archives/Mathematics/2011 July 4

= July 4 =

Nilpotent groups
Let G a finite group, D and N normal subgroups of G, and D ≤ N ∩ Φ(G). How to prove that if N/D nilpotent, then N is also nilpotent? --84.62.199.166 (talk) 05:59, 4 July 2011 (UTC)


 * I'm sure that you read the large, highlighted, advice section at the top of the Reference Desk page, and in particular the section entitled "How do I word my question for the best results?". The fifth point in that section is
 * "If your question is homework, show that you have attempted an answer first, and we will try to help you past the stuck point. If you don't show an effort, you probably won't get help. The reference desk will not do your homework for you."
 * Sorry to be a pain, but could you please play the game and show that you have attempted an answer first? We will try to help you past the stuck point. Remember that the Reference Resk will not do your homework for you. — Fly by Night  ( talk )  19:59, 4 July 2011 (UTC)
 * As a suggestion, try proving a special case first, i.e. if G/Φ(G) is nilpotent then G is nilpotent.--RDBury (talk) 23:06, 4 July 2011 (UTC)

Convergence of a generated Sequence
Please help me to understand this relating issue of Convergence of a Sequence. I read whole day and night but still cannot entirely comprehend the notion. Thanks in advance.Torment273 (talk) 16:47, 4 July 2011 (UTC)

http://farm6.static.flickr.com/5080/5901705444_898eb4336a_b.jpg
 * The statement does not seem quite correct. What it should probably say is that if v is a fixed point of f (i.e. f(v)=v), |f'(u)|<1 in a neighborhood of v, and u0 is in this neighborhood, then the sequence u1=f(u0), u2=f(u1), ... converges to v. if |f'(u)|>1 in a neighborhood of v then the sequence does not converge to v, unless of course you happen to pick u0=v. |f'(u)|<1 includes the case where f'(u)=0 for some values of u, so I'm a bit confused why f'(u)>0 is listed as a requirement. In any case, it's much easier to prove using the mean value theorem then trying to make sense of the diagrams for each case.--RDBury (talk) 23:00, 4 July 2011 (UTC)


 * See our articles on fixed point (mathematics) and fixed point iteration for more details. Cobweb plot has some animations. Gandalf61 (talk) 14:04, 5 July 2011 (UTC)

Definition in topology
What are pairwise non-homeomorphic connected infinite genus smooth surfaces in R3?

Thanks. — Preceding unsigned comment added by 131.111.222.12 (talk) 17:10, 4 July 2011 (UTC)


 * Note: the question comes from here (question 11) -- it is an "example sheet" for a past course, so this does not appear to be a homework problem. Looie496 (talk) 18:34, 4 July 2011 (UTC)


 * Asking to prove there are countably many pairwise non-homoemorphic surfaces with these properties just means you need to show that there is a countable number of such surfaces, no two of which are homeomorphic to each other (or else the question would be trivial). Often in topology we only care about spaces up to homeomorphism.  Show there are countably many different ones up to homeomorphism.  For "connected" or "smooth" see the articles if those are points of confusion (although smoothness is sort of moot since every topological 2 manifold has a smooth structure).  "Infinite genus" is defined (more or less) in the problem sheet Looie496 linked. Rckrone (talk) 19:55, 4 July 2011 (UTC)

This is a nice example of why the more traditional way of using hyphens is good: It means
 * infinite-genus smooth surfaces

and not anything like
 * infinite genus-smooth surfaces

(whatever you might try to guess that means). I tripped over a bit of a mental speed bump for all of a tenth of a second, that I wouldn't have hit if the hyphen had there, as it should be. Michael Hardy (talk) 21:13, 4 July 2011 (UTC)

pie
Taylor is making a mini apple pie with a radius of 3.14 in for her math class. What is the area of her mini pie? — Preceding unsigned comment added by 71.84.112.169 (talk) 21:27, 4 July 2011 (UTC)
 * Just apply the standard formula for area of a circle. Michael Hardy (talk) 01:48, 5 July 2011 (UTC)
 * Umph, it's a stupid joke whose punchline ends up being "No, pie round" or something of that sort. Looie496 (talk) 02:47, 5 July 2011 (UTC)
 * $$\pi \times 3.14^2 = 30.97$$ or $$\frac{24649 \pi}{2500}$$ to be exact. Widener (talk) 05:43, 5 July 2011 (UTC)