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I saw this almost answered here:

Exponential of the differential operator

(it is the unaccepted answer)

What I am looking to "solve" is $$ \sum_{j=0}^d\; \left( e^{\epsilon\,\partial_x} \right)^j \left[ e^{\sum_{a=0}^{\infty} \epsilon^a\,f_a(x)} \right], $$ where $\epsilon >0$ and $\epsilon<<1$ and $d < \infty$. Also, $f_a(x)$ are smooth simple, but nonsimilar functions. Putting aside convergence issues, and looking at just the operator at first, $\sum_{j=0}^d\; \left( e^{\epsilon\,\partial_x} \right)^j$, from the above paper it is confirmed to me that, $$ e^{\epsilon\,\partial_x} \equiv \sum_{q=0}^{\infty} \frac{\epsilon^q\,\partial_x^{(q)}}{q!}. $$

But my question asks about $$ \sum_{j=0}^d\; \left( e^{\epsilon\,\partial_x} \right)^j, $$

could it be just, $$ \sum_{j=0}^d\,\sum_{q=0}^{\infty}\, \frac{\epsilon^{qj}\,\partial_x^{(qj)}}{{(q!)}^j} \hspace{4mm} ???? $$

Also, as I don't really see a nice way (no pattern, etc.) of calculating $$ \frac{d^{(j)}}{dx^{(j)}}\left[ e^{f(x)} \right] $$ would it be a good/bad idea to expand that also into a Maclaurin series? (Sorry, this is in regards to what the operator is acting on):

$$ \sum_{j=0}^d\,\sum_{q=0}^{\infty}\, \frac{\epsilon^{qj}\,\partial_x^{(qj)}}{{(q!)}^j} \left[ e^{\sum_{a=0}^{\infty} \epsilon^a\,f_a(x)} \right] $$

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nate
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    I believe $\left(e^{\epsilon \partial_x}\right)^j = \sum_{q\geq 0}\frac{(\epsilon j)^q \partial_x^q}{q!}$. – Jonathan Y. Aug 18 '13 at 22:26
  • Hmm... Is this with the summation over $j$, right? Well I know that $$ \sum_{j=1}^d \left(e^{\epsilonf(x)}\right)^j = \frac{(e^{\epsilonf(x)})^{d+1}}{e^{\epsilonf(x)}-1} - \frac{e^{\epsilonf(x)}}{e^{\epsilon*f(x)}-1}$$ but I didn't want to expand them all out and collect... If I can just substitute in the operator $\partial_x$ instead of the function $f(x)$ then I have an expression which I guess I could Taylor series and collect on... Am I missing something obvious about yours? – nate Aug 18 '13 at 22:47
  • Hmm, I've the problem, that the formula in your question is an unviable jungle to me (which does not say anything against your formula, only about my abilities...) , but as I've seen in your linked question that is easily discussed in terms of the Pascal-matrix - separated into its subdiagonals, when seen as exponential series using the powers of the differential operator - and unseparated when seen as operator of the "+1". Since the pascal-matrix is a very smooth object, the translation of your problem into the matrix-language as well might be helpful and leading to a simple solution... – Gottfried Helms Aug 18 '13 at 22:58
  • No, summation (well, this is a series, not a sum) over $q$ (I only expanded every element of the outer sum). In any event, when to operators $A,B$ commute one has $e^A e^B=e^{A+B}$, implying that in particular $(e^A)^n = e^{nA}$. – Jonathan Y. Aug 18 '13 at 23:00
  • I 'understand' the Baker-Campbell-Hausdorff formulation, and these do commute (being the same), but doesn't what you say mean, for 2, $e^{\epsilon,\partial_x + \epsilon,\partial_x} = e^{\epsilon(\partial_x + \partial_x)}$ which you say means $e^{2,\epsilon,\partial_x}$. Likewise, for 3: $e^{\epsilon(\partial_x + \partial_x + \partial_x)} = e^{3,\epsilon,\partial_x}$. Maybe I am being too literal? – nate Aug 18 '13 at 23:23
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    Certainly, addition of operators is defined via addition in the image space; put differently, in the (linear-)space of operators, $A+A=2A$. – Jonathan Y. Aug 18 '13 at 23:33
  • Thanks much! If you make it an answer I'll accept it. Need to someday maybe take an operator theory class :) – nate Aug 18 '13 at 23:46
  • Sure thing, but it's unfit to be an answer. I didn't differentiate that monster of a function ;) – Jonathan Y. Aug 19 '13 at 00:48

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