Wikipedia:Reference desk/Archives/Mathematics/2007 September 16

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[edit] September 16

[edit] What does this mean?

I came across this in an (non-wikipedia) article. I have no idea what it means.

\Sigma \, t^3 = \frac{t^4}{4}-\frac{t^3}{2}+\frac{t^2}{4}

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
\sum_{x=1}^t x^3 = \left(\frac{t(t+1)}{2}\right)^2 = \frac{t^4}{4}+\frac{t^3}{2}+\frac{t^2}{4}
PrimeHunter 03:37, 16 September 2007 (UTC)
No, it's definately minus t cube on two which is -\frac{t^3}{2} . 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
\sum_{x=1}^{t-1} x^3 = \left(\frac{t(t-1)}{2}\right)^2 = \frac{t^4}{4}-\frac{t^3}{2}+\frac{t^2}{4}
--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)