Talk:Cauchy formula for repeated integration

This entire page needs a lot of work, particularly in the scalar "proof" section. KlappCK (talk) 17:49, 27 July 2011 (UTC)

Well, that's overstating things a bit. Is there a lot to say about the Cauchy formula for repeated integration? :P I can fill in the rest of the proof sometime today, but it's a straightforward application of the multidimensional chain rule. Hell, I'll put that much in right now, that takes no work... Sniffnoy (talk) 06:31, 28 July 2011 (UTC)

Huh, now that I look closer, it's worse than I realized -- it pretty much totally ignores basepoint issues, concerning itself solely with the antiderivative part of the result. OK, I'll fix that later. Sniffnoy (talk) —Preceding undated comment added 06:37, 28 July 2011 (UTC).

Agreed Sniffnoy, I will give you another 48 hours to improve upon this (arguably) more elegant proof by induction before I go in and simply demonstrate that the result (Cauchy's formula) is correct by working "backwards" via repeated differentiation. I think that this section would also benefit from a rigorous explanation of when this rule applies. KlappCK (talk) 12:49, 28 July 2011 (UTC)

Uh... how does that differ from the approach currently there? I'm not following. This *is* a proof by repeated differentiation. Sniffnoy (talk) 17:23, 28 July 2011 (UTC)

Saying "Applying the chain rule, we can determine that ${\frac {d}{;dx}}f^{(-n)}(x)={\frac {d}{dx}}{\frac {1}{(n-1)!}}\int _{a}^{x}\left(x-y\right)^{n-1}f(y)\,dy=f^{-(n-1)}(x).$" is NOT repeated differentiation. If any thing it is a pointer to repeated differentiation, which is not satisfactory for a proof. Furthermore, this proof starts with a single integral and purportedly uses proof by induction to show the result comes from repeated used of the formula, ${\frac {d}{dx}}f^{(-1)}(x)={\frac {d}{dx}}\int _{a}^{x}f(y)\,dy=f(x).$, except it fails to use the base case to demonstrate the result for every subsequent case. This is fundamentally different than starting with the assertion: $\int _{x_{0}}^{x}\int _{x_{0}}^{x_{1}}\dots \int _{x_{0}}^{x_{n-1}}f(x_{n})\,dx_{n}\dots \,dx_{2}\,dx_{1}=\int _{x_{0}}^{x}f(t){\frac {(x-t)^{n-1}}{(n-1)!}}\,dt.$ and repeatedly differentiating until you arrive at an explicitly equivalent result.KlappCK (talk) 13:10, 29 July 2011 (UTC)

Repetition is just a special case of induction. Every proof by "repeated" something, if written out formally and explicitly, is actually a proof by induction, because the whole definition of "doing something n times" is recursive (inductive). There isn't a meaningful distinction. I personally find the explicit induction easier to understand in this case, but regardless, they're the same. Sniffnoy (talk) 18:45, 29 July 2011 (UTC)

Disagreements over what constitutes a proof by induction aside, I believe your changes are a step in the right direction.KlappCK (talk) 14:44, 2 August 2011 (UTC)