2
$\begingroup$

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 + ... $
$=f(x+ax)$
But this does not seem to the the answer. Can anyone tell me if and where i am wrong?

$\endgroup$
  • 1
    $\begingroup$ This is the dilation operator, not translation. $\endgroup$ – Cameron Williams Feb 12 at 20:55
0
$\begingroup$

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

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


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!

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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}

$\endgroup$
  • $\begingroup$ @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
1
$\begingroup$

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}

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.