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This question was motivated by a question by Tobias Kienzler and its wonderful answers.

I begin as in the linked question...

Using the Taylor expansion

$$f(z+a) = \sum_{k=0}^\infty \frac{a^k}{k!}\frac{d^k }{dz^k}f(z)$$

one can formally express the sum as the linear operator $e^{a\frac{d}{dz}}$ to obtain

$$f(z+a) = e^{a\frac{d}{dz}}f(z).$$

Other relationships were given in an answer by Tom Copeland:

$$ f(e^b z) = \exp\left(bz\frac d{dz}\right)f(z), $$

$$ f\left(\frac z{1-cz}\right) = \exp\left(c z^{2}\frac d{dz}\right)f(z). $$

My question is about the reverse. What if we start on the right-hand side with a function different from $\exp$ in the operator?

I asked about the specific case for $\sin$ in the comments of joriki's answer and Tobias found that

$$ \sin\!\left(a\frac{d}{dz}\right)f(z) = \frac{1}{2i}(f(z+ia) - f(z-ia)), $$

and similarly that

$$ \cosh\!\left(a\frac{d}{dz}\right)f(z) = \frac{1}{2}(f(z+a) - f(z-a)). $$

He also conjectured that the symmetrization of a function might be obtained from an operator like

$$\exp\left(i\frac\pi2\frac d{d\ln z}\right)\cosh\left(i\frac\pi2\frac d{d\ln z}\right)$$

I don't know much about Lie algebras so I apologize if this is too broad:

For which operators $\text{D}$ like these is $\text{D}f$ something 'nice' as in these examples?

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  • $\begingroup$ nice question! Suggestion for a title: "How to determine the action of an operator $D\left(z, \frac d{dz}\right)$?" $\endgroup$ Apr 1, 2012 at 19:38
  • $\begingroup$ Thanks @TobiasKienzler, you knew what I was looking for better than I did! $\endgroup$ Apr 1, 2012 at 19:44
  • $\begingroup$ @PedroTamaroff Thanks, I hope you also checked the answers :) $\endgroup$ Nov 10, 2016 at 15:23
  • $\begingroup$ Revisiting this question, I think Fractional Calculus and formal power series are strongly related to this as well... $\endgroup$ Nov 10, 2016 at 15:29

4 Answers 4

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I obtained those identities by using $\sin(x) = \frac{e^{ix}-e^{-ix}}{2i}$ etc. (without worrying about convergence of operators, I admit).

In general, you can use the Fourier-Transform of your operator as follows:

$$\begin{array}{rl} D\left(z,\frac d{dz}\right)f(z) &= D\left(z,\frac d{dz}\right)\frac1{2\pi}\int_{-\infty}^\infty dk \int_{-\infty}^\infty dw\ f(w)e^{ik(z-w)} \\ &= \int_{-\infty}^\infty dw \underbrace{\frac1{2\pi}\int_{-\infty}^\infty dk\ D\left(z,\frac d{dz}\right)e^{ik(z-w)}}_{=:W(z,w)}\ f(w) \end{array}$$

edit[ Note that you can replace $\frac d{dz}$ by $ik$ if it does not act on the $z$ in $D$ anymore, however for $\exp\left(bz\frac d{dz}\right)$ you can't! ]

So (by, once again, ignoring detailed discussions on convergence, whether swapping the integration is valid etc. (sorry, I'm a Physicist...)) you obtain a $W$ that for each $z$ gives you a weight distribution.

For demonstration, take $D=e^{a\frac d{dz}}$ to obtain $W(z,w)=\delta(w-(z+a))$.

On the other hand, if you want to know $\int_{-\infty}^\infty f(z)\,dz$ you can require $W=1$ and therefore $D = 2\pi\delta\left(\frac d{dz}\right)$. More generally, for a given weight $W(z,w)$ one obtains a differential operator $D\left(z,\frac d{dz}\right)$ via the inverse Fourier transform, so in summary:

$$ W(z,w) = \frac1{2\pi}\int_{-\infty}^\infty D\left(z,\frac d{dz}\right)e^{ik(z-w)}\,dk, \\D(z,ik) = \frac1{2\pi}\int_{-\infty}^\infty e^{-ikw}W(z,w+z)\,dw$$

The latter formula gives you $D$ such that $\frac d{dz}$ only acts to the right of it, so if you apply this to $\exp\left(bz\frac d{dz}\right)$ you will obtain a different expression (that can be reformulated).

Using $W(z,w) = \chi_{[-\infty,z]}(w)$ one can therefore also express the antiderivative via an operator $\frac1{2\pi}\int_{-\infty}^ze^{-w\frac d{dz}}\,dw$.

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  • $\begingroup$ This is really cool. $\endgroup$ Apr 1, 2012 at 19:39
  • $\begingroup$ Thanks! I hope you're happy about making me use my brains on a Sunday :-) $\endgroup$ Apr 1, 2012 at 19:43
  • $\begingroup$ I added the inversion, hope it's correct despite it being midnight... $\endgroup$ Apr 1, 2012 at 21:30
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Another angle on characterizing the action of D(z,d/dz):

With f(z) expressible as a Taylor series, you could also look at the umbral operator

$U(c.y;d/dz) f(z) = exp(c.yd/dz) f(z) = f(z+c.y)$, formally, with the special cases,

A) $exp(c.d/dz)|_{z=0} f(z) = f(c.)$ ,

B) $exp[-(1-c.) d/dz]|_{z=1} f(z) = f(c.)$,

C) $exp(c.:zd/dz:) f(z) = f[(1+c.)z]$, and

D) $exp[-(1-c.) :zd/dz:] f(z) = f(c. z)$

where $c.^n=c_n$, $(:zd/dz:)^n=z^n(d/dz)^n$, and, e.g., $(z+c.y)^n=\sum_{j=0}^n \binom{n}{j} c_j y^j z^{n-j}$.

For the sine operator above, in U let $y=a$ and $c_n=sin(\pi n/2)$.

For the cosh operator above, in U let $y=a$ and $c_n=|cos(\pi n/2)|$.

For the scaling operator, in D let $c.= e^b$ and see my notes in MSQ 116633 on $S_0$.

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Along a different vein, complementing Tobias' formulation:

For $z>0$ and appropriate $\sigma$, formally

$$K \left(z\frac{d}{dz} \right)f(z)=\displaystyle\frac{1}{2\pi i} \int_{\sigma-i\infty}^{\sigma+ i\infty} \frac{\pi}{\sin(\pi s)} K(-s)g(-s) \frac{z^{-s}}{(-s)!} ds$$

with

$$\displaystyle\int^{\infty}_{0}{f(z) \frac{z^{s-1}}{(s-1)!} dz} = g(-s)$$

using a modified Mellin transform and its inverse.

For action on $f(z)=e^{-z}$ with $K(\omega)=\binom{\omega+\alpha+\beta}{\beta}$, see my notes "The Inverse Mellin Transform, Bell Polynomials, a Generalized Dobinski Relation, and the Confluent Hypergeometric Functions".

Also formally,

$$K \left(z\frac{d}{dz} \right)f_{LPT}(z)=\displaystyle\frac{1}{2\pi i} \int_{\sigma-i\infty}^{\sigma+ i\infty} \frac{\pi}{\sin(\pi s)} K(-s)f_{MT}(1-s) \frac{z^{-s}}{(-s)!} ds$$

for $f_{LPT}(z)$ the Laplace transform of $f(x)$ and $f_{MT}(s)$ the unmodified Mellin transform of $f(x)$; i.e.,

$$\displaystyle\int^{\infty}_{0}{f(x) e^{-xz} dx} = f_{LPT}(z)$$ and $$\displaystyle\int^{\infty}_{0}{f(x) x^{s-1} dx} = f_{MT}(s)$$

This is derivable formally from

$$e^{-xz}=\displaystyle\frac{1}{2\pi i} \int_{\sigma-i\infty}^{\sigma+ i\infty} \frac{\pi}{\sin(\pi s)} \frac{(xz)^{-s}}{(-s)!} ds \text{ for }\sigma>0$$

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  • $\begingroup$ @Peter, thanks for improving the LaTex code. I'll try to make good use of it in the future. $\endgroup$ Apr 10, 2012 at 22:27
  • $\begingroup$ You're welcome. Great triad of answers by the way. I hope someday I'll understand them fully! $\endgroup$
    – Pedro
    Apr 10, 2012 at 22:32
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Consider a compositional inverse pair of functions, $h$ and $h^{-1}$, analytic at the origin with $h(0)=0=h^{-1}(0)$.

Then with $\omega=h(z)$ and $g(z)=1/[dh(z)/dz]$,

$$\exp \left[ {t \cdot g(z)\frac{d}{{dz}}} \right]f(z) = \exp \left[ {t\frac{d}{{d\omega }}} \right]f[{h^{ - 1}}(\omega )] = f[{h^{ - 1}}[t + h(z)]] = f[L(t,z)]$$

(see OEIS A145271 and A139605),

so with $D_{FT}(\alpha)$ the Fourier transform of $D(z)$, formally

$$D\left( {t \cdot g(z)\frac{d}{{dz}}} \right)f(z) = \int\limits_{ - \infty }^\infty {{D_{FT}}} (\alpha )\exp \left[ {2\pi i\alpha t\cdot g(z)\frac{d}{{dz}}} \right]d\alpha f(z)$$

$$ = \int\limits_{ - \infty }^\infty {{D_{FT}}} (\alpha )f\left\{ {{h^{ - 1}}\left[ {2\pi i\alpha t + h(z)} \right]} \right\}d\alpha = \int\limits_{ - \infty }^\infty {{D_{FT}}} (\alpha )f\left[ {L\left( {2\pi i\alpha t,z} \right)} \right]d\alpha $$

For the special case $D(z)=\sin(2\pi a z)$, $D_{FT}=\dfrac{\delta(\alpha-a)- \delta(\alpha+a)}{2i}$,

and so

$$sin\left( {2\pi a \cdot g(z)\frac{d}{{dz}}} \right)f(z) = \frac{{f\left\{ {{h^{ - 1}}\left[ {h(z) + 2\pi ia} \right]} \right\} - f\left\{ {{h^{ - 1}}\left[ {h(z) - 2\pi ia} \right]} \right\}}}{{2i}}$$

(For a consistency check, try $h(z)=z$.)

Similarly, switch to the inverse Laplace transform to obtain formally

$$D\left( {t \cdot g(z)\frac{d}{{dz}}} \right)f(z) = \frac{1}{{2\pi i}}\int\limits_{\sigma - i\infty }^{\sigma + i\infty } {{D_{LPT}}} (p)\exp \left[ {pt \cdot g(z)\frac{d}{{dz}}} \right]dpf(z)$$

$$ = \frac{1}{{2\pi i}}\int\limits_{\sigma - i\infty }^{\sigma + i\infty } {{D_{LPT}}} \left( p \right)f\left\{ {{h^{ - 1}}\left[ {pt + h(z)} \right]} \right\}dp = \frac{1}{{2\pi i}}\int\limits_{\sigma - i\infty }^{\sigma + i\infty } {{D_{LPT}}} \left( p \right)f\left[ {L\left( {pt,z} \right)} \right]dp$$

For the special case $D(z)=\cosh(az)$, ${{\text{D}}_{LPT}}{\text{ = }}\frac{1}{2}\left[ {\frac{1}{{p - a}}{\text{ + }}\frac{1}{{p + a}}} \right]$,

and purely formally

$${\text{cosh}}\left[ {ag(z)\frac{d}{{dz}}} \right]f(z) = \frac{1}{{2\pi i}}\int\limits_{\sigma - i\infty }^{\sigma + i\infty } {\frac{1}{2}} \left[ {\frac{1}{{p - a}} + \frac{1}{{p + a}}} \right]f\left\{ {{h^{ - 1}}\left[ {p + h(z)} \right]} \right\}dp$$

$=\frac{1}{2}[f[h^{-1}[a+h(z)]]+ f[h^{-1}[-a+h(z)]]$.

Examples can be constructed from

$g(z)=(1+z)^{m+1}$, $h^{-1}(z)=(1-mz)^{-1/m}-1$, $h(z) = - \dfrac{{{{(1 + z)}^{ - m}} - 1}}{m}$, and

$L(t,z)=h^{-1}[t+h(z)]=[(1+z)^{-m}-mt]^{-1/m}-1$

with the limiting case for $m=0$ being

$g(z)=(1+z)$, $h^{-1}(z)= \exp(z)-1$, $h(z)= \log(1+z) $, and

$L(t,z)=h^{-1}[t+h(z)]=(1+z)e^{t}-1$.

Note for the Witt algebra that the actions are given by

$exp[tz^{m+1}d/dz]f(z)=f[z(1-mtz^{m})^{-1/m}]$.

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  • $\begingroup$ I think it would be best to condense your three answers into one. $\endgroup$ Apr 6, 2012 at 23:54
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    $\begingroup$ @Antonio, why? Three different approaches, three different paragraphs (four including Tobias') makes it easier to comment on the different approaches and express interest or disinterest in each, and also suggests how broad the question and answers are. $\endgroup$ Apr 8, 2012 at 1:04
  • $\begingroup$ @Peter, thanks for the better formatting. (However, you also introduced a substantial math error that I had to undo.) $\endgroup$ Apr 10, 2012 at 22:20

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