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I read that $e^\mathrm{D}f(x)=f(x+1)$, were $\mathrm{D}$ is the differential operator and $f(x)$ is an analytical function.

I tried writing out the definitions as I'd expected them to be (I haven't read much about operator theory):

$$e^\mathrm{D}=\sum_{n=0}^\infty \frac{\mathrm{D}^n}{n!}$$ and $$f(x)=\sum_{m=0}^\infty \frac{\mathrm{D}^m f(x)|_{x=0}}{m!},$$ so $$e^\mathrm{D}f(x)=\sum_{n=0}^\infty \frac{\mathrm{D}^n}{k!}\left(\sum_{m=0}^\infty \frac{\mathrm{D}^m f(x)|_{x=0}}{m!}\right).$$

How do I proceed from here?

Thanks.

Bobson Dugnutt
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    No, $$\bigl(e^D f\bigr)(x) = \sum_{k = 0}^\infty \frac{(D^k f)(x)}{k!} = \sum_{k = 0}^\infty \frac{f^{(k)}(x)}{k!}\cdot 1^k.$$ Now assume that $f$ is analytic on a disk with radius greater than $1$ around $x$. – Daniel Fischer Nov 27 '16 at 19:37
  • You are correct about the first definition (of $e^{D}$ as a power-series), however the second one is not correct: there should be no evaluation at $x=0$. It's simply $D^mf(x) = f^{(m)}(x)$ - the $m$'th derivative of $f$ evaluated at $x$. To see the direct link of this expression to Taylor-series (as hinted to in the comment above) it might help to try to prove the slightly different statement $f(x+h) = e^{hD}f(x)$ where $h$ is a constant. – Winther Nov 27 '16 at 20:15
  • @Winther I'm not sure I follow. Would you start from $f(x+h)$, write out it's Taylor (or McLaurin) series (and why shouldn't it be evaluated?) and then somehow rewrite that as being $f(x)+hf^1(x)+h^2f^2(x)/2+\dots$? – Bobson Dugnutt Nov 27 '16 at 20:42
  • I might have misread your question on that point. I though you meant $e^{D}f = \sum \frac{\left.f^{(n)}(x)\right|_{x=0}}{n!}$ however I now think you meant for the second line to be the Taylor expansion right? What you need to do is to Taylor expand about $x$ instead as expanding about $0$, i.e. $f(x+h) = f(x) + hf'(x) + \ldots$ – Winther Nov 27 '16 at 20:55
  • Related : Exponential of a function times derivative . – Han de Bruijn Oct 22 '17 at 18:52

1 Answers1

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Call $\epsilon = 1 / n$, for some integer $n$, then

$$ e^D = (e^{\epsilon D})^n \tag{1} $$

Now, note that for large $n$

$$ f(x + \epsilon) \approx f(x) + \epsilon f'(x) \tag{2} $$

therefore

$$ e^{\epsilon D}f(x) = \left(1 + \epsilon D + \cdots\right)f(x) \approx f(x) + \epsilon Df(x) = f(x) + \epsilon f'(x) \stackrel{(2)}{=} f(x + \epsilon) \tag{3} $$

Going back to Eq. (1)

$$ e^Df(x) = \lim_{n\to \infty}(e^{\epsilon D})^nf(x) = \lim_{n\to\infty}f(x + n\epsilon ) = f(x + 1) $$

caverac
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