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December 19

sin(infinity)

Look at Indeterminate form. Is $\sin(\infty )$ indeterminate?? If not, please explain why. Georgia guy (talk) 01:47, 19 December 2013 (UTC)[reply]

It is. The limit does not exist, because for every epsilon, not every delta satisfies the condition of an epsilon-delta definition.--Jasper Deng (talk) 01:50, 19 December 2013 (UTC)[reply]

Nevertheless, this method of evaluating the Fresnel integrals at infinity seems to work if we assume this limit is 0 (which of course it is not).--Jasper Deng (talk) 01:54, 19 December 2013 (UTC)--Jasper Deng (talk) 01:54, 19 December 2013 (UTC)[reply]

No, it's not an "indeterminate form". The term "indeterminate form" is a historical one and is restricted to an enumerated list of, I think, seven, and

\sin(\infty )

did not happen to get included.

However, like the enumerated indeterminate forms, it is a function applied to a point where there is no continuous extension of that function to that point. So it could have been an indeterminate form. It just didn't make the list, because no one thought about it. There is no point in adding it now, because we have better ways of talking about such functions. --Trovatore (talk) 07:53, 19 December 2013 (UTC)[reply]

Indeterminate form is applied to the same sort of thing but using operations like + and × not written in functional form like add(1,2). The problem with those straightforward operators is people tend to do transformations on them without thinking whereas with functions they don't normally, they apply general rules for limits with functions. Dmcq (talk) 13:04, 19 December 2013 (UTC)[reply]

sin(z) has an essential singularity at infinity. Count Iblis (talk) 11:44, 27 December 2013 (UTC)[reply]

Product of normal distrubution and another distrubution

Hi all,

I know this is probably a really simple question, but it's been while since my statistics A level... So performance of employees at a company is assumed to be standard-ly normally distributed, with each employee given a rating according to following table:

Bottom 3%	20%	55%	20%	Top 3%
Significantly Below	Below	Average	Above	Significantly Above

They are then given a bonus amount depending on rating:

Significantly Below	Below	Average	Above	Significantly Above
0%	3%	6%	8.4%	11.25%

Clearly this second distributed is not symmetrical - how would you find the product of these 2 distributions to get a 3rd so that you could compare the bonus of one emploee with the rest of the company/compared to the average bonus? What would be a measure of it's asymmetry?

Thanks for your help! 80.254.147.164 (talk) 14:32, 19 December 2013 (UTC)[reply]

If you pick an employee at random, there's a 3% chance his bonus is 0%, a 20% chance his bonus is 3%, a 55% chance his bonus is 6%, a 20% chance his bonus is 8.4%, and a 3% chance his bonus is 11.25%. That's the distribution. 150.203.188.53 (talk) 16:57, 19 December 2013 (UTC)[reply]

Your second distribution is certainly skewed (see Skewness and Skew normal distribution), but the fixed steps of the bonuses mean that it is no longer a normal distribution. Dbfirs 08:23, 20 December 2013 (UTC)[reply]

Let ${\mathcal {N}}$ be Baire space. How to prove that ${\mathcal {N}}^{\omega }$ is homeomorphic to ${\mathcal {N}}$ ?

Let ${\mathcal {N}}$ be Baire space, i.e. the space $\omega ^{\omega }$ of all infinite sequences of natural numbers, $\langle a_{n};n\in \mathbb {N} \rangle$ .For every finite sequence $s=\langle a_{k};k<n\rangle$ , let:

$O(s)=\{f\in {\mathcal {N}};s\subset f\}=\{c_{k};k\in \mathbb {N} ;(\forall k<n),c_{k}=a_{k}\}$

be the basis sets for the topology.

My question: How can I prove that, ${\mathcal {N}}^{\omega }$ is homeomorphic to ${\mathcal {N}}$ ?

Thank you! Topologia clalit (talk) 19:31, 19 December 2013 (UTC)[reply]

Not homework, right? It's actually really easy. An element of

{\mathcal {N}}^{\omega }

— is an ω-sequence of ω-sequences of natural numbers. Just interleave them coordinate-wise so that they become a single ω-sequence, in a uniquely reversible way. Now check that the mapping is bicontinuous, which is easy. --Trovatore (talk) 20:42, 19 December 2013 (UTC)[reply]

I am not sure I quite understand what interleave means.. for example, in the mapping you sugested, what is the image of the following sequence of sequences?
1,2,3,4,5,6,...
2,4,6,8,10,12,...
3,6,9,12,15,18,...
.
.
n,2n,3n,4n,5n,6n,...
.
.
Thank you! (and.. not homework.. I am trying to prepare myself for starting a Ph.D and right now reading chapter 4 of Jech's book: "Set Theory") Thanks again for your help. Topologia clalit (talk) 20:54, 19 December 2013 (UTC)[reply]

Just pick any old bijection between ω×ω and ω, and use it to make sure all the values you have listed show up. For example, you could use the back-and-forth-diagonal-zigzag schedule that's often used to show people the rationals are countable. If you used that one, you'd get 1,2,2,3,4,3,4,6,6,4,5,8,9,8,5,.... --Trovatore (talk) 20:59, 19 December 2013 (UTC)[reply]

Got it thanks. I am not sure yet about how to show the homeomorphism, but i'll try thinking about it first. Thanks! Topologia clalit (talk) 08:03, 20 December 2013 (UTC)[reply]

I've added some formatting, also replaced some less-than and greater-than with angle bracekts. --CiaPan (talk) 22:50, 20 December 2013 (UTC)[reply]

December 19

sin(infinity)

Product of normal distrubution and another distrubution

Let N {\displaystyle {\mathcal {N}}} be Baire space. How to prove that N ω {\displaystyle {\mathcal {N}}^{\omega }} is homeomorphic to N {\displaystyle {\mathcal {N}}} ?

Let ${\mathcal {N}}$ be Baire space. How to prove that ${\mathcal {N}}^{\omega }$ is homeomorphic to ${\mathcal {N}}$ ?