Is there a formula similar to $f(x+a) = e^{a\frac{d}{dx}}f(x)$ to express $f(\alpha\cdot x)$?

Question

Using the Taylor expansion

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

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

$$f(x+a) = e^{a\frac{d}{dx}}f(x).$$

But, does a linear operator $\hat A$ exist such that

$$ f(\alpha\cdot x) = \hat A(\alpha) f(x)$$

for some $\alpha\in\mathbb C$?

Answer 1

The exponential representation of the Taylor series that you mention is related to the fact that the operator $\mathrm d/\mathrm dx$ is the generator of translations. That is, a translation can be viewed as the exponentiation of an infinitesimal translation, and the corresponding transformation of a function can likewise be viewed as the exponentiation of the transformation corresponding to an infinitesimal translation. To first order, the translation $x\to x+\epsilon$ has the effect $f(x)\to f(x+\epsilon)=f(x)+\epsilon\mathrm d/\mathrm dx f(x)$ and thus corresponds to multiplication with the operator $1+\epsilon\mathrm d/\mathrm dx$; applying this operation $a/\epsilon$ times leads to $ x\to x+a$ and $f\to(1+\epsilon\mathrm d/\mathrm dx)^{a/\epsilon}f$, which goes to $\exp(a\mathrm d/\mathrm dx)f$ for $\epsilon\to0$.

Applying the same approach, we can write $\alpha=\exp(\log\alpha)$ and consider the scaling $x\to\alpha x$ as $\log\alpha/\epsilon$ applications of the infinitesimal scaling $x\to(1+\epsilon)x$. The corresponding infinitesimal transformation of $f$ is $f(x+\epsilon x)=f(x)+\epsilon x\mathrm d/\mathrm dxf(x)=(1+\epsilon x\mathrm d/\mathrm dx)f$, and applying this $\log\alpha/\epsilon$ times and taking $\epsilon\to0$ yields $f\to\exp(\log\alpha x\mathrm d/\mathrm dx)f$. For instance, you can check this for $f(x)=x^n$; in this case $x\mathrm d/\mathrm dx f=nf$, so this yields $f\to\exp(n\log\alpha)f=\alpha^n f$ as expected.

Answer 2

Based on Norbert's comment the answer should be

$$ f(\alpha\cdot x) = \alpha^{x\frac{d}{d x}}f(x)$$

To see this, just define $g(y):=f(e^y)$ such that $f(x) = g(\ln x)$, then one obtains

$$ f(\alpha x) = g(\ln\alpha+\ln x) = e^{\ln\alpha\frac{d}{d\ln x}}g(\ln x)$$

Then apply $\frac{d}{d\ln x}=\frac{dx}{d\ln x}\frac{d}{dx}=x\frac{d}{dx}$.

I guess there are some restrictions though, especially $x\neq 0$ since the Logarithm is not defined then ($\alpha\neq0$ is caught by $e^{\ln\alpha}=\alpha$).

Answer 3

More generally, there are the special linear conformal transformations SL(2,R) associated with the differential operators

$S_{-1}f(z)=\exp\left(a\frac d{dz}\right)f(z)=f(z+a)$

$S_{0}f(z)=\exp\left(bz\frac d{dz}\right)f(z)=f(e^b z)$

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

The $z^{m+1}\frac d{dz}$ (m=-1,0,1) are a representation of a subgroup of the infinite Witt Lie algebra associated with the Virasoro algebra, and their exponential maps can be used to construct Mobius, or linear fractional, transformations.

For more info (combinatorics, generalizations), see my notes "Mathemagical Forests" (pages 13-15) at my little "arxiv".

Also refer to this question at Physics Forum.

(Update) Another way to look at the the scaling operator is

$S_{0}f(z)= exp[(e^t-1):zd/dz:]f(z)=exp[t\phi_{.}(:zd/dz:)]f(z)=exp(tzd/dz)f(z)$

where $(:zd/dz:)^n=z^n(d/dz)^n$ and $(\phi_{.}(x))^n=\phi_{n}(x)$ is the n’th Bell/Touchard/exponential polynomial with the exponential generating function $exp[(e^t-1)x]=exp[t\phi_{.}(x)]$.