Talk:Indefinite sum

Usually do not use constant C in expressions, but assume $\sum _{x}f(x)=\sum _{0}^{x-1}f(x)$ .--MathFacts (talk) 20:56, 2 April 2009 (UTC)[reply]

Indefinite article[edit]

Reading this article for the first time, I see at least one major flaw: it never states what kind of function are acted upon. The phrase "the linear operator, inverse of the forward difference operator" (never mind that operator is not invertible) implies a reference to Difference operator to find this out, and I see nothing obvious in that article that implies that is operator operates on sequences, or functions defined on Z. Indeed the difference operators involve a shift parameter h that is real, and the central difference operator even implies existence of values at distance h/2. So I any presume that in fact the operation is on a space of functions defined on R. But then the phrase "If F(x) is a solution of this functional equation for a given f(x), then so is F(x)+C for any constant C. Therefore each indefinite sum actually represents a family of functions, differing by an additive constant" is false, since many more functions then those differing by a constant have the same finite difference (for a given h), namely one can introduce independent constants for every class in R/hZ. Marc van Leeuwen (talk) 12:42, 30 March 2010 (UTC)[reply]

That indefinite sum is inverse to difference is not a definition but property. There are different definitions which give the same results (up to a constant). I agree this is not clarified in the article.--MathFacts (talk) 22:12, 30 March 2010 (UTC)[reply]

Yes, there are

\beth _{2}

solutions -- as many solutions as there are functions on the real line, since

G(x)=F(x)+\theta (x)

also satisfies

G(x+1)-G(x)=f(x)

for any 1-periodic function

\theta (x)

. mike4ty4 (talk) 01:54, 23 June 2010 (UTC)[reply]

Much ado about nothing. Look at www.oddmaths.info . —Preceding unsigned comment added by Ascold1 (talk • contribs) 21:59, 23 May 2011 (UTC)[reply]

new indefinite sum[edit]

This article is outdated. Now indefinite sum is completely reduced to indefinite integral. To be sure of it look at http://www.oddmathes.info/indefinitesum . I recommend to begin with examples. There is at the bottom of a page a button “Examples”. In a drop-down menu click “Instruction”, read it and then open “Example1”. The example is a live one. You can insert an analytic function at your choice and calculate it. In the example there are detailed instructions in commentaries before each step. When you become sure that the formula works you can look at the article to see why it works.

Good luck.Ascold1 (talk) 23:37, 24 June 2011 (UTC)[reply]

The link seems to be broken, could you either correct it or show what it leads to? 173.73.170.4 (talk) 01:58, 19 July 2022 (UTC)[reply]

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Antidifference of exponential functions[edit]

The formula ∑ a^x = a^x / (a - 1) + C does not seem right

How about ∑ a^x = (a^(x+1) - 1) / (a - 1) ? That avoids the 'C' constant.

Title[edit]

I feel like the title Antidifference makes more sense then Indefinite sum TheTrueSauce (talk) 21:09, 17 April 2023 (UTC)[reply]

Agreed. But the term "indefinite sum" is used in Wolfram products.--Reciprocist (talk) 04:33, 13 December 2023 (UTC)[reply]