Consider the operator $D= e^{ax*d/dx} $ operating on an infinitely differentiable function f(x).

My approach:
$Df(x)= f(x) + ax*df(x)/dx + (ax)^2*d^2f(x)/dx^2 + ... $
But this does not seem to the the answer. Can anyone tell me if and where i am wrong?

  • 1
    $\begingroup$ This is the dilation operator, not translation. $\endgroup$ – Cameron Williams Feb 12 at 20:55

It appears that the OP interpreted the operator $e^{\left((ax)\frac{d}{dx}\right)}$ on $f(x)$ incorrectly to mean $\sum_{n=0}^\infty \frac{(ax)^n}{n!}\frac{d^nf(x)}{dx^n}$.

This misinterpretation in the OP is due to applying incorrectly successive applications of the operator $ax\frac{d}{dx}$. One application reveals $ \left((ax)\frac{d}{dx}\right)f(x)=axf'(x)$. A subsequent application yields

$$\begin{align} \left((ax)\frac{d}{dx}\right)^2 f(x)&=\left((ax)\frac{d}{dx}\right)\left(axf'(x)\right)\\\\ &=(ax)^2f''(x)+a^2xf'(x)\\\\ &\ne (ax)^2f''(x) \end{align}$$

We will show in the following that $\sum_{n=0}^\infty \frac{(ax)^n}{n!}\frac{d^nf(x)}{dx^n}=f(ax+a)$ while $e^{\left((ax)\frac{d}{dx}\right)}=f(e^a x)$.

We begin by analyzing the operator $e^{a\frac{d}{dx}}$ on $C^\infty$ as defined by

$$\left(e^{a\frac{d}{dx}}\right)\{f(x)\}=\sum_{n=0}^\infty \frac{f^{(n)}(x)}{n!}a^n\tag1$$

Note that the Taylor series of $f(x+a)$ around $x$ can be written

$$f(x+a)=\sum_{n=0}^\infty \frac{f^{(n)}(x)}{n!}a^n\tag2$$

Comparing $(1)$ and $(2)$ reveals


Next, we transform the operator $x\frac{d}{dx}$ by enforcing the substitution $x=e^y$. Then, denoting $f(x)=f(e^y)=g(y)$, we see that

$$\begin{align} \left(ax\frac{d}{dx}\right) f(x)&= \left(ae^y\frac{dy}{dx}\frac{d}{dy}\right) f(e^y)\\\\ &=\left(ae^ye^{-y}\frac{d}{dy}\right) f(e^y)\\\\ &=\left(a\frac{d}{dy}\right) g(y)\tag3 \end{align}$$

Finally, using $(2)$ and $(3)$ reveals

$$\begin{align} \left(e^{(ax)\frac{d}{dx}}\right) f(x)&=\sum_{n=0}^\infty \frac1{n!}\left(\left(ax\frac{d}{dx}\right)^n \right)f(x)\\\\ &=\sum_{n=0}^\infty \frac{g^{(n)}(y)}{n!}a^n\\\\ &=g(y+a)\\\\ &=f(e^{y+a})\\\\ &=f(e^ae^y)\\\\ &=f(e^ax) \end{align}$$

And we are done!

  • $\begingroup$ Oh, yes. Now i see my mistake. Thanks and there is a mistake in the 7th line where there should be a '+' in place of a '=". $\endgroup$ – D.K. Feb 13 at 17:18
  • $\begingroup$ You're welcome. My pleasure. And thank you for catching the typographical error. I've edited it accordingly. $\endgroup$ – Mark Viola Feb 13 at 18:14

Sketch: Define \begin{align} u(t, x) = \exp\left(tx\frac{\partial}{\partial x}\right)f(x) \end{align} then we see that $u$ solves \begin{align} \partial_t u - x \partial_x u =0, \ \ \text{ with }\ \ u(0, x) = f(x). \end{align} By the method of characteristics, we see that \begin{align} u(t, x) = f(xe^{t}) \end{align} which means \begin{align} \exp\left( ax\frac{\partial}{\partial x}\right) f = u(a, x) = f(xe^a). \end{align}

  • $\begingroup$ @MarkViola $\exp(tL)$ is the semigroup for $u_t=Lu$ unless I made some error somewhere. $\endgroup$ – Jacky Chong Feb 12 at 22:33
  • $\begingroup$ Consider f(x)=exp(x) . Then expanding the operator in the exponential series, I am getting Df(x)= exp(x+ax)= f(x+ax) which is not equal to f(xe^a). Is there something wrong in my approach of expanding the operator in the exponential series? $\endgroup$ – D.K. Feb 13 at 8:37

Remember that $$ e^x = 1 +x +\frac{1}{2!} x^2 + \cdots +\frac{1}{n!} x^n + \cdots $$ Hence, $$ \exp(ax \partial_x) f = f +ax\partial_xf +\frac{1}{2}ax\partial_x(ax\partial_x f)+\cdots $$ Let $x=e^y$ so that $x\partial_x f(x) = \partial_y f(e^y)$. Therefore, \begin{align} \exp(ax \partial_x)f &= \sum_{n=0}^\infty \frac{1}{n!} (\partial_y^n f(e^y)) a^n \\ &= f(e^{y+a}) \\ &= f(xe^a) \end{align}

  • $\begingroup$ Yes. I expanding the operator the same way in my approach. $\endgroup$ – D.K. Feb 13 at 8:38

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