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 $\alpha\in\mathbb C$?

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    $\begingroup$ $f(a\cdot x)=f(e^{\ln a+\ln x})$ but this is for $a>0$ and $x>0$ $\endgroup$
    – Norbert
    Mar 5, 2012 at 10:37
  • $\begingroup$ @Norbert thanks, good point! $\alpha\neq 0$ and $x\neq 0$ should be sufficient when complex numbers are ok (the branch cut of $\ln$ shouldn't matter due to the re-exponentiation). So the answer might involve $\frac{d}{d\ln x}$ $\endgroup$ Mar 5, 2012 at 10:40
  • $\begingroup$ @Norbert thanks, I extended this to an answer. I wonder about it's general validity though $\endgroup$ Mar 5, 2012 at 10:51
  • $\begingroup$ Tobias Kienzler, not at all :) $\endgroup$
    – Norbert
    Mar 5, 2012 at 10:58
  • $\begingroup$ Please @MrNobody, don't repost your low quality answer, which was already deleted twice. $\endgroup$ Jun 16, 2023 at 17:14

3 Answers 3


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.

  • $\begingroup$ interesting, so the $x^n$ form the Eigenbase of scaling $\endgroup$ Mar 5, 2012 at 12:27
  • $\begingroup$ @joriki This is so enlightening! $\endgroup$
    – Pedro
    Apr 1, 2012 at 2:48
  • $\begingroup$ @Peter: I'm glad to hear that :-) $\endgroup$
    – joriki
    Apr 1, 2012 at 2:49
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    $\begingroup$ @Peter: I learned about this in physics, where it's part of the treatment of continuous symmetry operations like translations and rotations; in mathematics I think it's part of the theory of Lie groups and Lie algebras, and specifically the exponential map. $\endgroup$
    – joriki
    Apr 1, 2012 at 3:05
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    $\begingroup$ @TobiasKienzler I have posted the question here. $\endgroup$ Apr 1, 2012 at 19:16

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 $e^{\ln\alpha}=\alpha$).


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 Möbius, 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 $exp[(e^t-1)x]=exp[t\phi_{.}(x)]$.

Edit 2/2014: Also more simply, $exp[(a-1):zd/dz:]f(z)=f(a·z)$. And, the next logical extension is to treat $a$ as an umbral variable, i.e., $a^n=a_n$, as Blissard did.

Edit 6/2014: An equivalent op, when acting on fcts. analytic at the origin, is $exp(a:xD_{x=0}:)$.

Applying the last two ops. with $a$ an umbral variable to $exp(x)$ gives the Euler or binomial transformation for exp. generating fcts., which can then be related to the Euler transform for ordinary generating functions through the Borel-Laplace transform. Then evaluating at $x=1$ gives the Euler summation for a series. To me, the differential ops. make these relations transparent.

Edit July 20, 2021:

In 1852 Charles Graves (brother of John Graves, the inventor of octonions in 1843) published "On a generalization of the symbolic statement of Taylor's theorem" in the Proc. of the Royal Irish Academy, presenting $e^{g(x)D} \; x = f^{(-1)}(1+f(x))$ where $g(x) = 1/f'(x)$, giving the examples for $g(x) = x^n$ for $n \geq 0$.

  • $\begingroup$ thanks, interesting point. The last identity can be seen from $z^2\frac d{dz}=-\frac d{d\frac1z}$, so there's an even more general expression for $m\ge1$ using $\frac d{dz^{-m}} = -\frac1m z^{-(m+1)}\frac d{dz}$ $\endgroup$ Apr 1, 2012 at 9:18
  • $\begingroup$ actually, this way around, all $m\in\mathbb C\backslash\{0\}$ are acceptable $\endgroup$ Apr 1, 2012 at 9:35
  • $\begingroup$ @Tobias,thanks for adding the Wiki links to my note on the Witt and Virasoro algebras. See my MF notes for general m, and the Addendum as well as oeis.org/A094638. Wolfdieter Lang has also explored the Witt algebra (cs.uwaterloo.ca/journals/JIS/VOL12/Lang/lang.pdf). $\endgroup$ Apr 2, 2012 at 6:42
  • $\begingroup$ normal ordering? this is getting related to QFT more than I expected... $\endgroup$ Apr 3, 2012 at 11:55
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    $\begingroup$ See also mathoverflow.net/questions/117287/… and mathoverflow.net/questions/145555/… $\endgroup$ Jan 15, 2021 at 17:05

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