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Let $P(x), Q(x)$ be two polynomials with real coefficients and set $F(x) = \frac{P(x)}{Q(x)}$. Consider a table which has the function $e^{\int_0^x F \, dx}$ on it. The table has the set of rules that the sum of any two functions on the table is on the table, multiplying any function on the table by a polynomial results in a function on the table and differentiating a function on the table gives another function on the table. Describe the set of functions on the table.

Attempts at a solution: (heuristic)

The integral of $P(x)/Q(x)$ will lead to the logarithm of some rational function with fractional exponents. Thus, the initial function is rational, and so the set must include all rational functions.

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What are your thoughts on this problem? Did you try writing some arbitrary functions that have this form, and then seeing how the "rules" restrict that arbitrariness? – rajb245 Feb 9 '14 at 15:41
I did. However, they do seem to be fairly arbitrary at my stage in this problem, so I didn't include those musings in the original post. (I'll do so now though). – Ayesha Feb 9 '14 at 15:51

We begin with $e^{\int_0^x F \, dx}$. Adding this function to itself just gives it an integer coefficient, which could also be done by multiplying it by a polynomial. Differentiating it will multiply it by a rational function (specifically, by $F$). So at the start, the most general thing we can hope to do is multiply it by a rational function.

Suppose that after $n$ steps, all the functions we can generate are equal to $e^{\int_0^x F \, dx}$ times a rational function. Adding two such functions creates another such function. Multiplying such a function by a polynomial produces such a function. Differentiating such a function, by applying the product rule and gathering terms, produces yet another such function. So all of the functions we can ever generate are of the form $e^{\int_0^x F \, dx}R(x)$, with $R$ a rational function.

Note that $e^{\int_0^x F \, dx}$ isn't really contributing anything to the difficulty here. Each function is uniquely identified by its rational function factor. When we sum two elements, we simply sum their rational functions, when we multiply by a polynomial we simply multiply the rational function by that polynomial, and when we differentiate we... well, let me write the rules down:

Where $A$ and $B$ are rational functions generated previously, we can generate:

  1. $A + B$
  2. $AP$, with $P$ any polynomial
  3. $A'+AF$ (by differentiating)

From these rules we can further determine that all of the functions we can generate are polynomial combinations of derivatives of $F$, that is, functions of the form $\sum_0^nP_iF^{(i)}$.

I'll try to continue this later if I have the time, but the key insight here is that you're really just generating rational functions, the $e^{\int_0^x F \, dx}$ doesn't matter.

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Ah, thank you! It's nice that you took time to do this well after it was posted. – Ayesha Feb 10 '14 at 20:12
@Ayesha No further ideas yet, although I did do some thinking about how iteration of rule (3) would affect a polynomial combination. Be sure to post your own answer if you figure it out! – Jack M Feb 10 '14 at 22:14

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