Wikipedia:Reference desk/Archives/Mathematics/2012 April 5

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

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 $f(x)+ce^{x}+e^{x}\int _{0}^{x}f(y)e^{-y}dy$, 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. ${\frac {|1|}{1}}(e^{1}-1)=|1|+ce^{1}+e^{1}\int _{0}^{1}|y|e^{-y}dy\implies e-1=(c+1)e-1$. 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:

$\ker \left({\frac {\operatorname {d} ^{n+1}}{\operatorname {d} \!x^{n+1}}}\right)=\{c_{0}+c_{1}x+\cdots +c_{n}x^{n}\}\,.$

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

$\int \left(\int |x|\,\operatorname {d} \!x\right)\operatorname {d} \!x={\frac {|x|x^{2}}{3!}}+c_{1}x+c_{0}\,.$

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

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

It's the leading term in [σ] that I'm interested in, i.e. the class member corresponding to the zero power series (0 ∈ R[[x]]). — 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

$(Jf)(x)=\int _{0}^{x}f(t)\,dt.$

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:

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

(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)