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Let $E_P$ be a polynomial operator, defined by: $$E_P(x) := \frac{p_0x^0}{0!}+\frac{p_1x^1}{1!}+\frac{p_2x^2}{2!}+\cdots$$ (where $P(x) = p_0x^0+p_1x^1+p_2x^2+\cdots$)

Is there any conventional way to express this in terms of $P(x)$?

What if we know $P(x)=\ln'(Q(x))$?

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    Hint: $\;\;p_0 = \frac{1}{0\,!}P(0)\,, \;p_1 = \frac{1}{1\,!}P'(0)\,, \;p_2 = \frac{1}{2\,!} P''(0), \;\cdots$ – 2017-01-26
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    not sure how that hint helps.... – 2017-01-26
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    @GregMartin Maybe $\sum \frac{P^{(k)}(0)}{(k\,!)^2}x^k$ helps. Maybe it doesn't, since it's not clear to me what is being asked. – 2017-01-26
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    It can be expressed in terms of $P(x)$ using the inverse Laplace transform. – 2017-01-27
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    What I was looking for is some formula that doesn't use any iteration or sums. E.g. if $P(x) = \frac{1}{1-x} = 1+x+x^2+...$, then $E_P(x) = e^x = e^{1-\frac{1}{P}}$ Perhaps there is some general pattern? – 2017-01-27

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