Can a repeated integral be simplified into a single integral?

[Jhenrique]

If a repeated integral can be expressed how an unique integral:

https://en.wikipedia.org/wiki/Cauchy_formula_for_repeated_integration

So is possible express the nth derivative with an unique differentiation?

 

[Mark44]

If a repeated integral can be expressed how an unique integral:

https://en.wikipedia.org/wiki/Cauchy_formula_for_repeated_integration

So is possible express the nth derivative with an unique differentiation?

What are your thoughts on this?

Given a function f, do you think it would be possible to express f'' as a single differentiation?

 

[Jhenrique]

What are your thoughts on this?

Given a function f, do you think it would be possible to express f'' as a single differentiation?

Yeah! Maybe, if is possible to find the nth antiderivative with an unique integral, so should be possible to find the nth derivative with an unique differentiation through of algebraic manipulation.

 

[lurflurf]

There is Cauchy's differentiation formula
$$\mathrm{f}^{(n)}(x)=
\frac{n!}{2\pi \imath}\oint \frac{\mathrm{f}(z) \mathrm{d}z } {(z-x)^{n+1}}$$
and some other related formula, but I do not recall any of the form
$$\left( \dfrac{d}{dx}\right) ^n \mathrm{f}(x)=\dfrac{d}{dx} \mathrm{g}_n (x)\mathrm{f}(x)$$
which given the rules of differentiation seems impossible for general f
https://en.wikipedia.org/wiki/Cauchy's_integral_formula

 

[Jhenrique]

$$\left( \dfrac{d}{dx}\right) ^n \mathrm{f}(x)=\dfrac{d}{dx} \mathrm{g}_n (x)\mathrm{f}(x)$$

Yeah! I'm looking for something like this!

 

[bigfooted]

We only have Leibniz' rule for the nth derivative of a product of functions:
http://en.wikipedia.org/wiki/General_Leibniz_rule

 

[Mark44]

We only have Leibniz' rule for the nth derivative of a product of functions:
http://en.wikipedia.org/wiki/General_Leibniz_rule

So who was this General Leibniz they are referring to?

 

[Jhenrique]

We only have Leibniz' rule for the nth derivative of a product of functions:
http://en.wikipedia.org/wiki/General_Leibniz_rule

It's more comprehensible express the derivative of a produt like way:
$$(f\times g)^{(2)} = f^{(2)}g^{(0)} + 2f^{(1)}g^{(1)} + f^{(0)}g^{(2)}$$