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April 23[edit]

Rejection Method for simulating random values from a density function and Conditioning Approach for reducing the variance in a parameter estimate[edit]

I checked the wiki page but I'm still confused. Please give an intuitive explanation and also give a theoretical proof of why it works. Any illustrations that help me better understand it will be appreciated.

Also, how do you use a conditioning approach to reduce the variance in an estimate of a population parameter? —Preceding unsigned comment added by 70.68.120.162 (talk) 00:11, 23 April 2010 (UTC)[reply]

Please name the exact titles of the articles in question. Bo Jacoby (talk) 09:42, 23 April 2010 (UTC).[reply]

Cofree object[edit]

A free group is a free group. A freespace on a set X is the discrete space on X. A cofree space on X is the trivial space on X. What's a cofree group? Money is tight (talk) 13:15, 23 April 2010 (UTC)[reply]

There aren't any. Well, the trivial group is cofree on a one-element set, but that's it. Algebraist 15:55, 23 April 2010 (UTC)[reply]

Well that's funny... Is it reflected in the fact that the forgetful functor for Grp doesn't even preserve coproducts? It's obvious that the coproduct of groups is not the disjoint union of their underlying sets. Whereas this is true for Top. Still find this stuff quite confusing Money is tight (talk) 02:39, 24 April 2010 (UTC)[reply]

Yes, that's related. A functor with a right adjoint is cocontinuous, so if cofree groups always existed, then the forgetful functor from Grp to Set would have to preserve colimits, which it does not. Algebraist 02:43, 24 April 2010 (UTC)[reply]

Thanks for your help :D Money is tight (talk) 04:11, 24 April 2010 (UTC)[reply]

prime numbers[edit]

Is there an algorithm where I can determine the number of prime numbers found between 1 and X? I am trying to find out if the number of primes between 1 and 1,000,000 is itself a prime number. Googlemeister (talk) 19:13, 23 April 2010 (UTC)[reply]

Prime-counting function. 76.230.7.121 (talk) 19:16, 23 April 2010 (UTC)[reply]

You can just use http://primes.utm.edu/nthprime/ which says: "There are 78,498 primes less than or equal to 1,000,000." PrimeHunter (talk) 19:20, 23 April 2010 (UTC)[reply]

Cool, so 1 million does not, but 1-10million has a prime number of prime numbers. Googlemeister (talk) 19:46, 23 April 2010 (UTC)[reply]

Yes. Primality tests of the π(x) column in Prime-counting function#Table of π(x), x / ln x, and li(x) (copied from oeis:A006880) shows that the only n values from 1 to 23 for which π(10ⁿ) is prime are 4 and 7. PrimeHunter (talk) 20:47, 23 April 2010 (UTC)[reply]

The largest π(x) which can be found at http://primes.utm.edu/nthprime/ is π(3×10¹³) = 1000121668853 which is prime. The π(x) tables at http://www.ieeta.pt/~tos/primes.html include all x with a single non-zero digit up to 10²³. Testing shows the largest of these where π(x) is prime is π(3×10¹³)! PrimeHunter (talk) 21:16, 23 April 2010 (UTC)[reply]

Constructing prime ideals[edit]

If I is a proper ideal in a commutative, unital ring R, define J to be $J=\{r\in R\mid \exists s\in R:rs\in I,{\mbox{ }}r,s{\mbox{ non-units}}\}$ . I've shown that when J is proper, it's a prime ideal (I think) - what extra conditions do I need to impose to ensure J is always proper? (Is it ever?) For non-prime ideals I in Z, J is always equal to Z, but, as far as I can see, for $I=\mathbf {Z} [x]\cdot x^{5}$ in $\mathbf {Z} [x]$ , $J\neq \mathbf {Z} [x]$ , because $x+1\not \in J$ . Thanks, Icthyos (talk) 22:06, 23 April 2010 (UTC)[reply]

Woops, disregard that second example, 1+x does lie in J. Icthyos (talk) 22:48, 23 April 2010 (UTC)[reply]

You appear to have some typos. If r in J implies r is a non-unit, then J is always proper since it does not contain 1, though J is not an ideal in most cases. Also note that s=0 is not a unit, so that every r in R (that is a non-unit) is in J, since rs=0 is in I. JackSchmidt (talk) 00:07, 24 April 2010 (UTC)[reply]

Yes...I don't really know what I was trying to say. I came back hoping to correct before anyone had responded, but you beat me to it! Thanks for trying to talk some sense into me, Icthyos (talk) 00:13, 24 April 2010 (UTC)[reply]

No problem. Colon ideals have a definition similar to yours and are often useful. JackSchmidt (talk) 00:38, 24 April 2010 (UTC)[reply]

My problem was, I neglected the fact that the sum of two non-units isn't necessarily a non-unit. I shoved in the 'non-unit' part after I'd checked J (without it) was an ideal, in an attempt to force J to be proper, and forgot that would be an issue. I'm tempted to keep tinkering, in an attempt to produce a J that is either prime, or R itself - or is this an exercise in futility? Thanks, Icthyos (talk) 17:32, 25 April 2010 (UTC)[reply]