3 ms·
> He works with some family of polynomial functions who agree on the sets K[a] that have open interior (2.1). Of course, two polynomials that agree on infinitel
by shoo 8y ago
> He works with some family of polynomial functions who agree on the sets K[a] that have open interior (2.1). Of course, two polynomials that agree on infinitely many points are identical.
Consider f(x, y) := xy and g(x, y) := xy^2
Fixing x=0 note that f and g agree along {(x, y) | x = 0, y in R}. But f is not identical to g. There is no open subset of R^2 such that f and g agree throughout the subset.
Would rewording "two polynomials that agree on infinitely many points are identical" as "two polynomials that agree on any open set" fix this? Or restrict the statement to polynomials in one variable only?
- wbl 8y agoC is not R^2. Neither f nor g is a polynomial over C, which is why they aren't a counterexample.
- marcinmozejko 8y agoAnd x = 0 is not an open set.
- shoo 8y agoagreed, x=0 is not an open set for n>1 dimensions, but it is "infinitely many points"
- m00n 8y agoYou're right, I was thinking about univariate polynomials. The "preprint" is only concerned with functions of one complex argument, so this should suffice.