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I really like how differentiation is introduced for polynomials:

Let $P(t) \in A[t]$ :

$$D_P(t,s) = \frac{P(t) - P(s)}{t-s} \;\; \in A[t,s]$$

and the derivative of $P$ is

$$P'(t) = D_P(t,t).$$

It resembles a definition from calculus and it doesn't involve coefficients of a polynomial. Is there a way to introduce an antiderivative for polynomials in such a manner? That is it should be constructive, should not mess with coefficients and should resemble integral (which?) somehow.

Obviously not every polynomial have an antiderivative in the same ring, e.g. $x^2$ doesn't seem to have an antiderivative in $\mathbb Z[x]$.

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Perhaps we want (a) a multivariable rational function $\ell=f/g\in k(x_1,\cdots)$, (b) $k$-algebra embeddings $\alpha_1,\cdots,\alpha_m:k[t]\hookrightarrow k(x_1,\cdots)$ and (c) an evaluation map $e:k(x_1,\cdots)\to k(t)$ such that $\varphi=e\circ\ell(\alpha_1,\cdots,\alpha_m,x_1,\cdots)$ has image $k[t]$ and satisfies $\partial(\varphi(p))=p$. I feel it's impossible but I'm not sure how it could be proven. Maybe a contradiction based on valuations? – anon May 9 '12 at 7:00
See also this answer. – Bill Dubuque Aug 4 '14 at 2:22
up vote 0 down vote accepted

UPD Nah, that won't work out, see Interpretation of $\sum_{i=1}^n i$ I assume the answer is it is impossible.

OLD I'm quite amazed, but the essential part is done and it turned out to be really easy in some sense. Now I just don't understand why I didn't get it earlier.

First, a bit of reminder:

$$\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$$

Straightforward analogy with calculus to find the antiderivative of $x^s$ for $s \in \{1,2,3\}$:

$$\sum_{i=1}^{\frac{x}{\Delta x}} (i \, \Delta x) \, \Delta x = \frac{x(x + \Delta x)}{2} \to \frac{x^2}{2}$$

$$\sum_{i=1}^{\frac{x}{\Delta x}} (i \, \Delta x)^2 \, \Delta x = \frac{x(x + \Delta x)(2x + \Delta x)}{6} \to \frac{x^3}{3}$$

$$\sum_{i=1}^{\frac{x}{\Delta x}} (i \, \Delta x)^3 \, \Delta x = \frac{x^2 (x + \Delta x)^2}{4} \to \frac{x^4}{4}$$

Thus the candidate rule to find algebraic antiderivative of polynomial $P(t)$:

$$I_P(t, \Delta t) = \sum_{i=1}^{\frac{t}{\Delta t}} P(i \, \Delta t) \, \Delta t $$

$$\int P(t) \, dt = I_P(t,0)$$

However I don't know yet how to interpret this formula. Actually I don't even know what is for example

$$1,2,3, \ldots, n,$$

and what is

$$\sum_{i=1}^n i^2$$

The only thing I know it gives something from $\mathbb Q[n]$, or maybe $\mathbb Z(n)$:


Basically I need to learn what algebraic structures are the things mentioned belong to and what operations from this structures were engaged. By the way these questions are valid also for the differentiation.

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