November 11 Information

See: Wikipedia:Reference desk/Computing - Subset symbol doesn't render on Wikipedia. Hopefully someone here can follow up on the problem. -- Tcncv ( talk) 01:18, 11 November 2008 (UTC) reply

Apparently it's been fixed now - it was just a broken image. -- Tango ( talk) 11:43, 11 November 2008 (UTC) reply

a+b+c=0 —Preceding unsigned comment added by 119.154.23.100 ( talk) 06:55, 11 November 2008 (UTC) reply

The cosine law can be proven by considering right triangles imbedded inside the triangle in question. Once you have got these right triangles you can apply the Pythogorean theorem, perform some trigonometric calculations to obtain the identity given by the cosine rule (try it for an equilateral triangle by dividing the triangle into two right triangles of equal area. Use trigonometry and the Pythogorean theorem to obtain the result. Generalize to an arbitrary triangle).

Topology Expert ( talk) 07:22, 11 November 2008 (UTC) reply

Which cosine law do you want? a+b+c=0 isn't a format I know. Law of cosines has proofs, or Law of cosines (spherical).-- Maltelauridsbrigge ( talk) 12:46, 12 November 2008 (UTC) reply

Let X be a topological space. Give necessary and sufficient conditions for X to be prametrizable.

I was just interested to know whether there was a purely topological characterization of pra-metric spaces (formulated only in terms of separation, countability, compactness-related, connectedness-related etc... axioms (and not other metrization conditions)). Could someone please provide this condition (no reference necessary; just the condition)? I think that I have found conditions that imply prametrizability but I haven't yet checked the converse (these conditions are quite weak so it is likely that the converse holds but I have not yet verified it).

Thankyou very much for your help.

Topology Expert ( talk) 08:23, 11 November 2008 (UTC) reply

If I understand, the property of your topological space X of being prametrizable, stated in another way is: for any ${\displaystyle x\in X}$ there exists a function ${\displaystyle d_{x}:X\to [0,1]}$ which is continuous at ${\displaystyle y=x}$ and vanishes there. Is it so or am I missing somehting? If so,(no good) It is a property of the ndb system of each point, right? For instance, being "locally Tychonov" should imply prametrizability, but maybe you have something weaker in mind? -- PMajer ( talk) 11:01, 11 November 2008 (UTC) reply

Thankyou for the reply. I did have something weaker in mind; basically I was just wondering whether there are necessary and sufficient conditions that imply pra-metrizability. Locally Tychonov could work but I think that it is a bit too strong (I was thinking along the lines of first countability and the T_1 axiom?).

Topology Expert ( talk) 12:33, 11 November 2008 (UTC) reply

Prametric spaces need not be T₁. I don't think they have to be first countable, but I'm not sure yet. Algebraist 13:07, 11 November 2008 (UTC) reply

Thanks for the response. I know that prametric spaces don't need to be T_1 (haven't verified first countability), but I was thinking whether you could weaken first countability to perhaps 'almost get necessary and sufficient conditions' (except for T_1). Maybe this problem is not as easy as it sounds. How does finding necessary and sufficient conditions for a separating prametric to induce the topology of a topological space X sound (in this case the T_1 axiom is necessary)?

Topology Expert ( talk) 04:22, 12 November 2008 (UTC) reply

Hello. I'm having some difficulty with this integral. The book itself says it's "nasty" and doesn't include answers, nor working. I've managed to split it up into fractions, but this has only helped a little.

${\displaystyle \int {\frac {1}{x(x^{2}+x+1)}}dx}$

${\displaystyle =\int {\frac {-x-1}{(x^{2}+x+1)}}dx+\int {\frac {1}{x}}dx}$

${\displaystyle =\int {\frac {-x}{(x^{2}+x+1)}}dx+\int {\frac {-1}{(x^{2}+x+1)}}dx+\int {\frac {1}{x}}dx}$

Then I get stuck with the first two, but I can obviously get the third. According to Wolfram, the first one is this and the second part is this. I'm quessing they've used some sort of substitution, ${\displaystyle u=x^{2}+x+1...du=2x+1dx}$

Any futher help would be greatly appreciated. Thanks! 86.159.225.61 ( talk) 17:51, 11 November 2008 (UTC) reply

One way to a solution is to factorize x²+x+1. Dmcq ( talk) 18:32, 11 November 2008 (UTC) reply

Go back to your second step (before you split the sum of fractions), and complete the square in the denominator. You'll have to adjust the numerator a bit. Now you want to split the fractions like you did in your third step. One should be a normal substitution, and the other a trig substitution (or fiddle around and make it look like the derivative of arctan). 76.126.116.54 ( talk) 19:41, 11 November 2008 (UTC) reply

Don't separate them like that. You have this:

${\displaystyle \int {\frac {-x-1}{x^{2}+x+1}}\,dx}$

So let

${\displaystyle u=x^{2}+x+1\,}$

so that

${\displaystyle du=(2x+1)\,dx}$

and then ${\displaystyle {du \over 2}=\left(x+{1 \over 2}\right)\,dx.}$ So then you write

${\displaystyle \int {\frac {-x-1}{x^{2}+x+1}}\,dx=\int {\frac {-\left(x+{1 \over 2}\right)}{x^{2}+x+1}}\,dx+\int {\frac {-1/2}{x^{2}+x+1}}dx}$

Do the first of the above by using the substitution above.

Next:

${\displaystyle \int {dx \over x^{2}+x+1}=\int }$${\displaystyle {dx \over \left(x+{1 \over 2}\right)^{2}+{3 \over 4}}.}$

That is a routine case of completing the square. Remember: the point of completing the square is always to reduce a quadratic polynomial to another quadratic polynomial with only a "squared" term and a constant term, but NO linear term. Then:

${\displaystyle {4 \over 3}\int {dx \over {4 \over 3}\left(x+{1 \over 2}\right)^{2}+1}={4 \over 3}\int {dx \over \left({2x+1 \over {\sqrt {3}}}\right)^{2}+1}}$

Then use the substitution

${\displaystyle u={2x+1 \over {\sqrt {3}}}}$

and you get an arctangent. Michael Hardy ( talk) 00:16, 14 November 2008 (UTC) reply