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April 5

Series

Let's say you have a function, and you can find anti-derivatives of any order. (The example that I have in mind if $f(x)=|x|$.) Next, you sum all of these anti-derivatives to give, hopefully, a new function. In the case of $f(x)=|x|$ you get

$\sigma (x)=\frac{|x|}{x}(e^x-1).$

Is there a name for this kind of construction? Can anyone point me towards any interesting references? — Fly by Night (talk) 01:55, 5 April 2012 (UTC)

If f is differentiable then σ satisfies the first order differential equation σ' - σ = f'. Rckrone (talk) 04:36, 5 April 2012 (UTC)

Anti-derivatives are not unique and therefore neither will be your resulting $\sigma$. I guess you are implicitly assuming initial conditions such as ${\sigma }^{(n)}(0)=k_n$ for any n. I've never encountered this before.Widener (talk) 07:17, 5 April 2012 (UTC)

Good point, and the solution to that differential equation is $f(x)+c{e}^x+e^x\underset{0}{\overset{x}{\int }}f(y)e^{-y}dy$, which seems to give the desired answer for $c=1$.--Itinerant1 (talk) 09:12, 5 April 2012 (UTC)

I think you must mean, when $c=0$. I just tested it with $x=1$ using the example Fly By Night gave. $\frac{|1|}{1}(e^1-1)=|1|+c{e}^1+e^1\underset{0}{\overset{1}{\int }}|y|e^{-y}dy⟹e-1=(c+1)e-1$. This is what you get if you assume $\sigma (0)=f(0)$ for a general sigma. Widener (talk) 10:32, 5 April 2012 (UTC)

That's right, I'd be setting all of the constants of integration equal to zero. After all:

$\ker \left(\frac{{\mathrm{d}}^{n+1}}{\mathrm{d}{x}^{n+1}}\right)=\{c_0+c_{1}x+\cdots +c_n{x}^n\}.$

When we find the anti-derivatives of a function, we get a function plus an arbitrary polynomial, e.g.

$\int (\int |x|\mathrm{d}x)\mathrm{d}x=\frac{|x|x^2}{3!}+c_{1}x+c_0.$

If we work out all of the anti-derivatives and then sum, we get a class of functions:

$[\sigma ]=\frac{|x|}{x}(e^x-1)+ℝ[[x]]$

It's the leading term in [σ] that I'm interested in, i.e. the class member corresponding to the zero power series Template:Nowrap — Fly by Night (talk) 11:25, 6 April 2012 (UTC)

Define a mapping $J$ on a suitable function space (say $L^1$, although you can do this with spaces of measures too) by

$(Jf)(x)=\underset{0}{\overset{x}{\int }}f(t)dt.$

You want to compute the resolvent operator $R=(I-J{)}^{-1}$ (by the sum of the geometric series). A concrete formula for this is possible using the Fourier transform:

$Rf(x)={ℱ}_x^{-1}\left(\frac{2\pi i\xi }{1+2\pi i\xi }{ℱ}_{\xi }f\right).$

(This may be up to a constant like $f(0)$. I didn't keep careful track of delta functions when computing this.) Sławomir Biały (talk) 12:35, 6 April 2012 (UTC)