January 23 Information

Is there a website that shows the numbers in digit form like 13.7 billion? Thanks. -- Donmust90 ( talk) 01:10, 23 January 2013 (UTC)Donmust90 — Preceding unsigned comment added by Donmust90 ( talk • contribs) 00:56, 23 January 2013 (UTC) reply

Do you mean:

1) 13,700,000,000

Or:

2) 1.37 × 10¹⁰ ? StuRat ( talk) 01:20, 23 January 2013 (UTC) reply

Either way, Wolfram Alpha at http://www.wolframalpha.com/ expands the term into both normal and scientific notation. -- Stephan Schulz ( talk) 01:31, 23 January 2013 (UTC) reply

Hmm, the only thing that should be added is to note that 'billion' can be an ambiguous term - in some contexts 10^9, others 10^12 ---- _nonsense ferret 02:20, 23 January 2013 (UTC) reply

Yeah, but the 10^12 meaning is almost obsolete in English. When The Economist started using the "short billion", the game was over. -- Trovatore ( talk) 02:23, 23 January 2013 (UTC) reply

Yes, Harold Wilson was responsible for changing the meaning of "billion" in the UK, but the 10^12 (million million) meaning cannot be obsolete whilst I still use it (in common with most of the European continent who have a similar word). Dbfirs 07:57, 24 January 2013 (UTC) reply

I did say "almost". I also said in English, which makes the point about Continental Europe irrelevant. -- Trovatore ( talk) 09:25, 24 January 2013 (UTC) reply

OK, point taken, though millions (not billions) in Continental Europe do speak English to some extent. Dbfirs 08:22, 25 January 2013 (UTC) reply

${\displaystyle \sum _{\vec {\alpha }}{\frac {1}{\prod _{k=1}^{K}\alpha _{k}}}}$ where the sum is taken over all K-dimensional vectors whose entries are nonnegative integers which sum to ${\displaystyle A}$? -- AnalysisAlgebra ( talk) 16:20, 23 January 2013 (UTC) reply

If K > 1 and A is a positive integer, the answer is ${\displaystyle \infty }$. There always will be an entry equal to 0.

So you may have meant to say that the components of the vectors are positive integers (but then I don't know the answer to the question). Icek ( talk) 16:41, 23 January 2013 (UTC) reply

(Assuming positive integers, as pointed out by Icek.) Already for K = 2, the sum equals

${\displaystyle \sum _{\alpha =1}^{A-1}{\frac {1}{\alpha (A-\alpha )}}={\frac {1}{A}}\sum _{\alpha =1}^{A-1}{\frac {1}{\alpha }}+{\frac {1}{A-\alpha }}={\frac {2}{A}}H_{A-1},}$

where H_n is the nth harmonic number. There is no closed-form expression for that under the usually understood meanings of “closed-form”.— Emil  J. 17:05, 23 January 2013 (UTC) reply