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December 13

Did the Mayans or Babylonians invent zero first?

It's obvious that the Mayans and Babylonians developed zero independently, but which of them developed it first? --75.28.52.54 (talk) 14:47, 13 December 2009 (UTC)[reply]

Looking at the article on the Mayan civilisation, there is a subsection on their mathematics, see here. Looking at the article on the Babylonian civilisation, there is a subsection on their mathematics, see here. My guess would be that the Babylonians used zero first; they are a much older people than the Maya. The Maya were using zero by 36 BC, "the earliest inscriptions in an identifiably-Maya script date back to 200–300 BC". But there is a well edited mathematical text from the Old Babylonian period (1830-1531 BC). ~~ Dr Dec (Talk) ~~ 15:24, 13 December 2009 (UTC)[reply]

Did the Babylonians use zero, I could not see a reference in our Babylonian mathematics articles. --Salix (talk): 13:51, 16 December 2009 (UTC)[reply]

Good point. The OP said that they did, so I was taking it as given that they did. ~~ Dr Dec (Talk) ~~ 01:48, 18 December 2009 (UTC)[reply]

More on finite differences

Thanks to everyone who answered my previous question, I now know the finite difference for n! (based on a difference of 1), and from that I can calculate the finite difference for (n-1)! or (n-j)! etc. But to turn things upside-down, what please is the finite difference (based on a difference of 1) for "1/n!"? For "1/((n-j)!(n!))"? —Preceding unsigned comment added by ImJustAsking (talk • contribs) 19:54, 13 December 2009 (UTC)[reply]

It's pretty basic algebra to work it out - have you tried to do it yourself? You have the difference of two fractions, so your first step should be to put them over a common denominator (this is really easy) and then simplify it (which is also easy). --Tango (talk) 20:36, 13 December 2009 (UTC)[reply]

To quantify "easy": I just did it in 5 lines and 20 seconds. --Tango (talk) 20:38, 13 December 2009 (UTC)[reply]

AS in your previous question, it's not clear if you are just happy with the first difference, or you want in general the k-th iterated difference (in which case the answer is to be expressed in terms of the k-th Charlier polynomial). --pma (talk) 23:12, 13 December 2009 (UTC)[reply]

Scratch the question about 1/(n!(n-j)!) - it was downright careless of me not to realize that the two terms can separated using partial fractions. Acccording to an example I worked out, the first difference of 1/n! is "1/(n+1)! - 1/n!", which seems almost too simple to be true. Is it? Finally, I am happy with just the first difference. —Preceding unsigned comment added by ImJustAsking (talk • contribs) 13:14, 14 December 2009 (UTC)[reply]

That is the first difference, by definition. You then need to simplify it. I suspect you don't really understand what finite differences are - I suggest you learn about them a little more before trying to solve your own problems. --Tango (talk) 14:46, 14 December 2009 (UTC)[reply]

Anyway, if you just want the first difference, your answer is correct. You may also like to write it -n/(n+1)! as Tango suggests. --pma (talk) 15:49, 14 December 2009 (UTC)[reply]