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reviewed Reviewed Is there a way to do that integration?
Apr
21
comment Rational function with absolute value $1$ on unit circle
Great, thank you very much. I guess this stronger identity theorem is implicit in his "counting of poles/zeros" of a rational function, which for him are always finite determined by the degrees of the polynomials in the numerator/denominator. $M (z) - 1$ is rational and has infinitely many zeros (at least on the unit circle), so it has to be the zero function. (I totally forgot about $\frac{a−b}{1 - \overline a b}$, thank you! Now there are no more loose ends.)
Apr
21
comment Rational function with absolute value $1$ on unit circle
The book is Ahlfors Complex analysis (I have the second edition). The Lagrange interpolation says that for any $n$ values, you can find a polynomial P of degree $< n$ such that P takes the specified values. If one could show that the points in the unit circle are "generic enough", one might argue that there couldn't be any other rational function, satisfying the uncountably many "constraints" $M (z) = 1$ for $\lvert z \rvert = 1$... (But I don't know if I'm on the right track...)
Apr
21
comment Rational function with absolute value $1$ on unit circle
The previous section has as exercise "verify the Cauchy-Riemann equations for $z^2$ and $z^3$". In other words, almost no "tools", only the definition of a holomorphic/rational function. There is a theorem "If all zeros of a polynomial lie in a half plane, then all zeros of the derivative lie in the same half plane", but I don't think it's relevant. The reader also learned how to develop partial fractions (and construct the Lagrange interpolation polynomial by proving $\frac{P (z)}{Q (z)} = \sum_{k=1}^n \frac{P (a_k)}{Q' (a_k)(z-a_k)}$, where $\deg P < n$ and $a_i$'s are the roots of $Q$.)
Apr
21
comment Rational function with absolute value $1$ on unit circle
@DanielFischer I don't quite understand the "identity theorem" you cite. We have that $M (z) = R (z) . \overline{R (1/\overline z)}$ satisfies $M (z) = 1$ for $\lvert z \rvert = 1$. But the set of such $z$ is not open, so the usual identity theorem does not apply. Also, at the point of the book, the identity theorem hasn't been mentioned, so I was wondering if there is an easier argument. (The previous exercise asks to construct the Lagrange interpolation polynomial.)
Apr
15
reviewed Reviewed Finding a distribution whose KL Divergence from a given distribution is a constant $\alpha$
Apr
15
reviewed Reviewed Does sum of all natural numbers contradict another rule?
Feb
12
answered How does advancing through the math major work?
Feb
11
comment What is the flaw of this proof (largest integer)?
@TripeHound 13 fingers on each hand or on both hands? Wait... does God have two hands?
Feb
9
awarded  Popular Question
Sep
11
answered $\mathbb{R}^\ast$, $\mathbb{R}_+$, $\mathbb{R}^\ast_+$ “deprecated”?
Sep
11
asked How to determine the type of singularities in affine varieties
Jun
17
awarded  Yearling
Jun
1
comment Is this a known formula? $\det (id, \partial_x, \partial_y)^t (f, g, h)$
@BCLC I found it while calculating automorphisms of rank-two vector bundles on a non-commutative deformation of a surface. This expression comes from rewriting three expressions containing Poisson structures as this determinant. The expressions with Poisson structures don't simplify to anything nice, so I am wondering if maybe this determinant expression is something nice I just don't know.
May
30
asked Is this a known formula? $\det (id, \partial_x, \partial_y)^t (f, g, h)$
May
29
awarded  Inquisitive
May
28
asked Computing the cotangent complex: what's the ring?
May
17
revised A proof for Atiyah-Macdonald Exercise I.21.iii
added 217 characters in body
May
17
comment A proof for Atiyah-Macdonald Exercise I.21.iii
@TakumiMurayama That is where my confusion lies. Maybe it is a shorthand for the set of all the elements in the contractions of the prime ideals containing $\mathfrak b$ (or a shorthand for the ideal generated by all these elements). The first equality in the last chain of equalities says $r (E^c) = r (E)^c$, which I only know to hold when $E$ is an ideal. I'm asking if there is a way to make sense of this "proof", because if the answer is yes, it is shorter than the other proof I found. But if the answer is no, then I will simply ignore it.
May
17
asked A proof for Atiyah-Macdonald Exercise I.21.iii