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November 9[edit]

Differential geometry textbook[edit]

Hello. I'm looking for a good introduction to differential geometry; can anyone suggest a particular author or text? Thanks in advance. —Anonymous Dissident^Talk 12:27, 9 November 2010 (UTC)[reply]

I hear Lee's introduction to smooth manifold is pretty good, but I havent looked at it yetMoney is tight (talk) 14:22, 9 November 2010 (UTC)[reply]

Spivak's Comprehensive Introduction to Differential Geometry is well-used and praised. [1] Depending on your background, you may find it more comprehensive than introductory :) I also recommend his book on introductory real analysis, which is unfortunately named `Calculus'. SemanticMantis (talk) 16:43, 9 November 2010 (UTC)[reply]

It depends on what are your interests and where you want to go. Very beautiful is Foundations of Differentiable Manifolds and Lie Groups, by F.Warner. --pm a 20:14, 9 November 2010 (UTC)[reply]

are these the same thing[edit]

Toroidal coordinates Bispherical coordinates —Preceding unsigned comment added by 129.67.37.227 (talk) 18:33, 9 November 2010 (UTC)[reply]

No — they rotate their common bipolar coordinates base system around two different axes. --Tardis (talk) 20:57, 9 November 2010 (UTC)[reply]

Probability[edit]

If a number between 0 and 1 is chosen at random, then what is the probability that exactly 5 of the first 10 digits are less than 5?

My textbook says (1/2)^10, but I would think that you would have to take into account permutations (ie 10/5!5! * (1/2)^10). Why am I wrong? And suppose the question asked to find the probability that at least 5 of the first 10 digits are less than 5? My first guess would be 10/5!5! * (1/2)^5, but the actual number is absurd (ie >1). —Preceding unsigned comment added by 76.68.247.201 (talk) 20:52, 9 November 2010 (UTC)[reply]

Your textbook is wrong on the first point: it is

{10 \choose 5}\left({\frac {1}{2}}\right)^{10}

(see combinations (not permutations) for the notation). It's easier to see the pattern if you say "7 of them less than 4" instead:

{10 \choose 7}\left({\frac {4}{10}}\right)^{7}\left({\frac {6}{10}}\right)^{3}

. Just 2^-10 is the probability that, say, the first five are less than 5 and the next five are not (e.g., 0.4124165869...). But for the "at least" question, it's not enough to drop the power to 5; you have to consider each possibility that's sufficient (5, 6, …, 10). So it's

\sum _{n=5}^{10}{10 \choose n}\left({\frac {1}{2}}\right)^{10}

. (Again, it's clearer if the probabilities aren't even: "at least 5 less than 4" is

\sum _{n=5}^{10}{10 \choose n}\left({\frac {4}{10}}\right)^{n}\left({\frac {6}{10}}\right)^{10-n}

. --Tardis (talk) 21:07, 9 November 2010 (UTC)[reply]

Okay, thanks. 76.68.247.201 (talk) 23:59, 9 November 2010 (UTC)[reply]

That exactly 5 of the first 10 digits are less than 5 is not a property of a real number, because 0.00000999999... = 0.0000100000.. . Bo Jacoby (talk) 19:31, 10 November 2010 (UTC).[reply]

But the set of troublemakers is countable, hence of probability zero, so it does not matter.—Emil J. 19:47, 10 November 2010 (UTC)[reply]

Hah, made me laugh! It's finite, even, since most troublemakers have representations which differ only after the first 10 digits. 67.158.43.41 (talk) 02:38, 11 November 2010 (UTC)[reply]

Determining all odd primes for which 15 is a quadratic residue modulo p[edit]

Hello everyone,

Pretty much as the question says, what is the best way to find all odd primes p such that 15 is a quadratic residue modulo p? I have been playing around with Jacobi symbols, but I tend to get into about 10 different special cases: $({\frac {15}{p}})=({\frac {p}{15}})\cdot (-1)^{\frac {p-1}{2}}=({\frac {p}{3}})({\frac {p}{5}})\cdot (-1)^{\frac {p-1}{2}}$ and then I seem to get into lots of different cases based on whether each of these terms is ±1, and even then for each of those cases there are some subcases and it seems to be a bit of a mess. Could anyone suggest a better way, or must it be done on a case-by-case basis?

I am happy using standard Legendre/Jacobi symbols, but outside of that I'd prefer not to use too much machinery (though a few modular arithmetic results like Gauss' lemma, FLT etc. are fine) - if that's possible, of course. Thankyou for the help in advance! Otherlobby17 (talk) 20:54, 9 November 2010 (UTC)[reply]

The expression you wrote is exactly what you need to think about. If you can evaluate all three factors, you know whether or not 15 is a quadratic residue mod p. You need to think about cases, but there are more than 10-- do you see from your formula what the cases are?

If we didn't know about quadratic reciprocity, we might think there were infinitely many cases, since there are infinitely many odd primes. But, in fact, there is a pattern to the answer, and the cases you must consider reveal the shape of that pattern. If you can see the simplest way to state the answer, it's not that messy. (At least, it's a vast improvement on infinitely many cases!) 140.114.81.55 (talk) 02:25, 10 November 2010 (UTC)[reply]

I would guess the cases are just dependent on the values of p modulo 3, 4 and 5? The 3/5 come from the Legendre symbols, and the 4 from the (-1) factor? Otherlobby17 (talk) 03:10, 10 November 2010 (UTC)[reply]

Small typesetting thing: \left( and \right) can make your Legendre symbols prettier, on the off chance you didn't know it already. 67.158.43.41 (talk) 19:48, 10 November 2010 (UTC)[reply]

Yes, those are the cases. (And you can combine those three 'modulos' into one big 'modulo'.) Once you write out the whole answer, you can check a few p's to test it. (For example, 15 is clearly a square mod 11.)140.114.81.55 (talk) 02:27, 11 November 2010 (UTC)[reply]