From Wikipedia, the free encyclopedia
Welcome to the Wikipedia Mathematics Reference Desk Archives |
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
[edit] September 16
[edit] What does this mean?
I came across this in an (non-wikipedia) article. I have no idea what it means.
Can someone explains what it actually means. 220.237.181.98 01:57, 16 September 2007 (UTC)
- Which article? I don't know a meaning. Perhaps it is a failed attempt to write the summation
- PrimeHunter 03:37, 16 September 2007 (UTC)
-
- No, it's definately minus t cube on two which is . There is no typo. 220.237.181.98 04:45, 16 September 2007 (UTC)
-
-
- If you shift the index by 1, then you get what you want. So maybe it is a failed attempt to write this
- --Spoon! 05:52, 16 September 2007 (UTC)
- Can you give us some more context? What was the article about? Maelin (Talk | Contribs) 05:02, 16 September 2007 (UTC)
-
- The article is entitled "The Difference Calculus". It's about a collection of mathematical tools for solving difference equations. At least that is what it says. 211.28.126.201 09:19, 16 September 2007 (UTC)
-
-
- Could you provide a link or other info on where to find it? —Bromskloss 11:47, 16 September 2007 (UTC)
-
-
- Perhaps they are showing how to solve the recurrence relation at + 1 = at + t3 --Spoon! 15:20, 16 September 2007 (UTC)
- I don't know the article, but in this context Σ and Δ are possibly operators turning functions on integers into other functions on integers, defined as follows. If F is defined on the integers, then the forward difference ΔF is the function f such that f(t) = F(t+1) – F(t). For example, if F(t) = t4 – 2t3 + t2, then (ΔF)(t) = 4t3. The operator Σ is such that if Σf = F, then ΔF = f. This allows an indefinite summation constant for Σ, which can be fixed by agreeing that (Σf)(0) = 0. Other definitions are possible; in particular the "backward difference" (∇F)(t) = F(t) – F(t–1). See also Difference operator. --Lambiam 18:47, 16 September 2007 (UTC)
-
- I agree with Lambiam, it must be the summation operator as defined in Knuth's Concrete Mathematics. – b_jonas 21:10, 17 September 2007 (UTC)