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January 23[edit]

website regarding numbers like 13.7 billion[edit]

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]

Is there a closed form expression for this quantity?[edit]

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

If K > 1 and A is a positive integer, the answer is

\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

\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]