TITLE: Show that $\lim\limits_{z \to a}\frac{\log|f(z) - f(a)|}{\log |z - a|}$ is an integer.
QUESTION [0 upvotes]: Let $f$ be analytic in a neighborhood of $a$.  Show that $\lim\limits_{z \to a}\frac{\log|f(z) - f(a)|}{\log |z - a|}$ exists and is an integer.
We have $$\frac{\log|f(z) - f(a)|}{\log |z - a|} = \frac{\log|\frac{f(z) - f(a)}{z-a}| + \log |z-a| }{\log |z-a|} = \frac{\log | \frac{f(z)-f(a)}{z-a}|}{\log |z-a|} + 1$$ which goes to $\frac{\log |f'(a)|}{-\infty} + 1 = 1$ as $z$ goes to $a$, provided that $f'(a) \neq 0$. But what if $f'(a) = 0$?

REPLY [4 votes]: You can choose a neighbourhood $\mathcal{V}$ of $a$ small enough that $f(z)-f(a)\neq0$ $\forall x\in\mathcal{V}-\{a\}$, then you can write
$$f(z)-f(a)=(z-a)^ng(z)$$
with $g(z)\neq0$ in $\mathcal{V}$.
From here we just have to put that on the limit,
$$\lim_{z\rightarrow a}\frac{\log|f(z)-f(a)|}{\log|z-a|}=\lim_{z\rightarrow a}\frac{\log|z-a|^n|g(z)|}{\log|z-a|}=\lim_{z\rightarrow a}\frac{n\log|z-a|+\log|g(z)|}{\log|z-a|}=$$
$$n+\lim_{z\rightarrow a}\frac{|g(z)|}{-\infty}=n$$

REPLY [1 votes]: Since $f$ is holomorphic in a neighborhood of $a$, then we can expand $f$ in a Taylor series about $a$:
$$f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(z-a)^n.$$
Then
$$f(z) - f(a) = \sum\limits_{n=1}^{\infty} \frac{f^{(n)}(a)}{n!}(z-a)^n = f'(a)(z-a) + \frac{1}{2} f''(a)(z-a)^2 + \cdots$$
If $f'(a) = \cdots = f^{(n)}(a) = 0$, then consider
$$\log|f(z) - f(a)| = \log\left|\frac{f(z)-f(a)}{(z-a)^{n+1}}\right| + (n+1)\log|z-a|.$$
I hope you can take it from here.


TITLE: Quine's NF and paradoxes
QUESTION [2 upvotes]: It's well known it's impossible to solve the Russel paradox using only the axioms of the Cantor set theory. The Quine's New foundation $(NF)$ is able to prevent this paradox and others like for example, the Burali - Forti. Are there paradoxes impossible to solve inside the axioms of $NF$ that are solved inside $NFU$?
Thanks.

REPLY [2 votes]: All the "classical" paradoxes with naive set theory fail to go through in NF; however, the consistency of NF even relative to ZFC is open (well, Holmes has a claimed consistency proof but I don't think it's been vetted yet).

Interestingly, it is known that NFU and its various familiar extensions are consistent, relative to very weak theories:

NFU itself is relatively consistent with PA.
NFU + Infinity + Choice is relatively consistent with a particular weak set theory called "MacLane set theory."

Moreover these consistency proofs are not very hard. So there's an interesting jump in strength going from NFU to NF.
Another manifestation of this jump in strength is the remarkable fact that NF disproves the axiom of choice while choice is consistent with ZFU! There doesn't seem to be a snappy reason for this; Specker's proof is quite complicated (although see here for an outline of the proof). The disproof of choice obviously isn't a paradox in the sense of a proof of inconsistency, but it is a very odd result, especially given that a rejection of choice isn't implicit in the motivation for NF.
This is the closest thing to a paradox which is impossible to solve inside the axioms of NF that is solved inside NFU; it's not really a "paradox" in the modern sense, but it is a proof of an arguably highly counterintuitive result (the negation of choice). And this argument dies a horrible death if we try to run it in NFU.
So I would say the following:

While we are confident that NFU is paradox-free (since we are confident, I hope!, in the consistency of Peano arithmetic), the evidence in favor of the consistency of NF is much weaker; there is a possible proof of consistency relative to a fragment of ZFC, which is still (I think) unvetted, and there are fundamental weirdnesses about NF which do not apply to NFU which suggest (at least to me) that we shouldn't expect NF to be consistent just because NFU is.


TITLE: On the meaning of being algebraically closed
QUESTION [28 upvotes]: The definition of algebraic number is that $\alpha$ is an algebraic number if there is a nonzero polynomial $p(x)$ in $\mathbb{Q}[x]$ such that $p(\alpha)=0$.
By algebraic closure, every nonconstant polynomial with algebraic coefficients has algebraic roots; then, there will be also a nonconstant polynomial with rational coefficients that has those roots.
I feel uncomfortable with the idea that the root of a polynomial with algebraic coefficients is again algebraic; why are we sure that for every polynomial in $\mathbb{\bar{Q}}[x]$ we could find a polynomial in $\mathbb{Q}[x]$ that has the same roots?
I apologize if I'm asking something really trivial or my question comes from a big misunderstanding of basic concepts.

REPLY [4 votes]: HINT $\ $ It boils down to the transitivity of algebraic extensions, which boils down to
$\rm\qquad\quad\ K\ =\   \mathbb Q(\alpha)\: =\ \mathbb Q\langle1,\alpha,\:\ldots\:,\alpha^{m-1}\rangle,\quad\ \ F\ =\ K(\beta)\: =\ K\langle1,\beta,\:\ldots\:,\beta^{\:n-1}\rangle $
$\rm\quad \Rightarrow\ \ F\ =\ \mathbb Q(\alpha)(\beta)\: =\ \mathbb{Q}  \big\langle\{1,\alpha,\:\ldots\:,\alpha^{m-1} \}\cdot \{1,\beta,\:\ldots\:,\beta^{\:n-1}\}\big\rangle\: =\ \mathbb Q\big\langle\alpha^i\:\beta^j\big\rangle_{i\:<\:m,\ j\:<\:n}$
which follows simply be employing the minimal polynomials of $\rm\:\alpha,\:\beta\:$ as rewrite rules $\rm\:\alpha^m \to\ f(\alpha),\ \ \beta^{\:n}\to\ g(\beta)\:$ to reduce all powers of $\rm\:\alpha,\:\beta\:$ to powers $\rm\:< m,\:n\:,\:$ respectively.
Therefore, since $\rm\:[F:\mathbb Q]\:$ is finite, we deduce that $\rm\:\beta\:$ is algebraic over $\rm\:\mathbb Q\:.$ More explicitly, since, by above, $\rm\:dim\ F/\mathbb Q\ =\ m\:n\:,\: $ the $\rm\ m\:n+1\ $ elements $\rm\:1,\:\beta,\:\ldots\:,\beta^{\:m\:n}$ are linearly dependent over $\mathbb Q\:,\:$ yielding a polynomial $\rm\:0\ne h(x)\in \mathbb Q[x]\:$ such that $\rm\:h(\beta) = 0\:.\:$ Therefore $\rm\:\beta\:$ is algebraic over $\rm\:\mathbb Q\:.$


TITLE: invariant measure under irrational rotation on $S^1$
QUESTION [2 upvotes]: Prove that if $T:S^1 \to S^1$ is an irrational rotation, then the only  probability measure on $S^1$ that is $T-$invariant is the lebesgue measure or a multiple or it. We are considering the lebesgue measure times $\frac{1}{\pi}$ (in this way it's a probability measure) and we are considering the $\sigma-$algebra of the borel sets restricted to $S^1$.

$S^1=\{u\in \mathbb R^2:||u||=1\}$. If $T$ is said to be an irrational rotation, if there exist an irrational number $\alpha\in \mathbb R-\mathbb Q$ such that if $u=(cos t,sin t)$ then $Tu=(cos (t+\alpha),sin ((t+\alpha)))$.
I have no idea how to attack this problem :S. Clearly Lebesgue measure is invariant under rotations, but I don't know how to proceed.
EDITED: Now all the assumptions are right. Thanks!

REPLY [0 votes]: Let $T_{\alpha}(z) = \exp(2\pi i\alpha)$ be where $\alpha \notin\mathbb{Q}$ and $[v,w)\subset S^1$, with $v = exp(2\pi ip)$ y $w = \exp(2\pi iq)$, a semi-opened interval in $S^1$ that has taken counterclockwise such that $l([v,w)) = q-p = \frac{1}{m}$ for some $m \in\mathbb{N}$.
We take $\mu\in M_{T_{\alpha}}(X)$, by regularity, given $\epsilon > 0$ there exists $F\subset(v,w)$ closed subset and $U\supset[v,w]$ opened subset such that $\mu((v,w)\backslash F) < \epsilon$ and $\mu(U\backslash [v,w]) < \epsilon$. Since $v, w\in U\backslash F$ there exists $\epsilon > \delta > 0$ such that $[v-\delta,v+\delta]\cup[w-\delta,w+\delta]\subset U\backslash F$ and we take $u\in [v-\delta,v)$. We know that the orbit of $u$ is dense in $S^1$, then there exist $n_1,n_2,n_3\in\mathbb{N}$ with $n_1<n_2<n_3$ such that $T_{\alpha}^{n_1}u\in(v,v+\delta], T_{\alpha}^{n_2}u\in[w-\delta,w)$ and $T_{\alpha}^{n_3}u\in(w,w+\delta]$. Then we have
        $$\displaystyle\bigcup_{k=0}^{m-1}[T_{\alpha}^{n_1+k(n_3-n_1)}u,T_{\alpha}^{n_2+k(n_3-n_1)}u]\subset \bigcup_{k=0}^{m-1}[v.e^{2\pi ik(q-p)},v.e^{2\pi i(k+1)(q-p)})\subset\bigcup_{k=0}^{m-1}[T_{\alpha}^{kn_2}u,T_{\alpha}^{n_3+kn_2}u]$$
        Notice that the family $\{[v.e^{2\pi ik(q-p)},v.e^{2\pi i(k+1)(q-p)})\}_{0\leq k\leq m-1}$ form a partition of $S^1$. Thus we have
        $$\displaystyle\sum_{k=0}^{m-1}\mu([T_{\alpha}^{n_1+k(n_3-n_1)}u,T_{\alpha}^{n_2+k(n_3-n_1)}u])\leq 1\leq\sum_{k=0}^{m-1}\mu([T_{\alpha}^{kn_2}u,T_{\alpha}^{n_3+kn_2}u])$$
        Since $\mu$ is $T_{\alpha}-$invariant it follows that
        $$m\mu([T_{\alpha}^{n_1}u,T_{\alpha}^{n_2}u]) \leq 1\leq m\mu([u,T_{\alpha}^{n_3}u])$$
        Notice that $F\subset[T_{\alpha}^{n_1}u,T_{\alpha}^{n_2}u]$ and $[u,T_{\alpha}^{n_3}u]\subset U$. Then from the regularity condition it follows that
        $$m\mu([v,w))-m\epsilon < m\mu(U)-m\epsilon < m\mu(F)\leq1\leq m\mu(U) < m\mu(F) + m\epsilon < m\mu([v,w))+m\epsilon$$
        So
        $$\dfrac{1}{m} - \epsilon < \mu([v,w)) < \dfrac{1}{m} + \epsilon$$
        From this $\mu([v,w)) = \dfrac{1}{m}$.
        When $l([v,w)) = \dfrac{s}{m}$ for some $s,m \in\mathbb{N}$ we divide the interval in $s$ semi-opened intervals of equal length and we conclude by the above that $\mu([v,w)) = \frac{s}{m}$.
        To the last, if $[v,w)$ is any semi-opened interval, we can write it as embedded union of semi-opened intervals with rational length. By continuity we have $\mu([v,w)) = l([v,w))$.
Finaly by Caratheodory theorem it follows the $\mu$ must be the Haar measure.
If we have a finite measure, we use the probability measure obtained from divide this measure by the measure of the total space.


TITLE: How to solve $2 \tan x / (1 - (\tan x)^2) = (\sin 2x)^2$?
QUESTION [2 upvotes]: $$\frac {2\tan {x}}{1-(\tan {x})^2} = (\sin {2x})^2$$
I tried a lot but I get nowhere

REPLY [1 votes]: $\frac{2\tan x}{1-\tan^2x}=(\sin 2x)^2$
$\tan 2x=\sin^2 2x$
$\frac{\sin 2x}{\cos 2x}=\sin^2 2x$
$\frac{\sin 2x}{\cos 2x}-\sin^2 2x=0$
$\frac{\sin 2x}{\cos 2x}(1-\sin 2x \cos 2x)=0$
either $\sin 2x=0$ or $1-\sin 2x \cos 2x=0$
$\sin 2x=0$ gives $x=\frac{n\pi}{2}$
$1-\sin 2x \cos 2x=0\Rightarrow\sin 4x=2$ which is not possible as $\sin$ cannot be greater than 1.

REPLY [0 votes]: $\sin{2x}=\frac{2\tan x}{1+\tan^2{x}}$
One solution is instantly, $x=n\pi$..
For other solutions, $\tan{x} \ne 0$, you can cross multiply and obtain a quartic equation in $\tan{x}$. I am typing on my mobile, so not working the whole solution out right away. But hope this enough for you to proceed.

REPLY [0 votes]: $$\frac {2\tan {x}}{1-(\tan {x})^2} = \frac{2\times \dfrac{\sin(x)}{\cos(x)}}{1-\dfrac{\sin^2(x)}{\cos^2(x)}}=\frac{2\times \dfrac{\sin(x)}{\cos(x)}\times \cos^2(x)}{\cos^2(x)-\sin^2(x)}$$
$$=\frac{\sin(x)\cos(x) + \sin(x)\cos(x)}{\cos(x)\cos(x)-\sin(x)\sin(x)}=\frac{\sin(2x)}{\cos(2x)}=\tan(2x)=\sin^2(2x)$$
$$\sin(2x)(1-\sin(2x)\cos(2x))=0$$
So : $ 1-\sin(2x)\cos(2x)=1-\frac 12 \sin(4x)=0 \implies \sin(4x)=2$ which is impossible.
$\color{red}{\text{or}}$ $\sin(2x)=0\implies x=\dfrac 12 n\pi, n\in \mathbb{Z}$


TITLE: What are conjectures that are true for primes but then turned out to be false for some composite number?
QUESTION [10 upvotes]: Note: This is an update formulation since many people misunderstood the question before.
Of course it is easy to make a statement like "Every n is a prime or at most 1000", which is true for every prime $n$ and every small $n$ but fails for $n=1002$. What are "real" conjectures that were known to hold for primes and small values, then turned out to be false?
An excellent example about cyclotomic polynomials was given by Aaron in the comments. Here the conjecture was that the coefficients are $0, \pm 1$ for every $n$. This holds for primes and small $n$'s, but fails for $105$.
Also the existence of Carmichael numbers comes close, but here the problem itself involves primes, I would like something less "primey". I know conjuctures that are or were known only for primes. Recently solved is Colorful Tverberg, still unknown is Evasiveness or this little MO problem.

REPLY [6 votes]: An aliquot sequence is a recursive sequence in which each term is the sum of the proper divisors of the previous term. Catalan conjectured that no aliquot sequence is unbounded. The conjecture is trivially true for primes, since the aliquot sequence for $p$ is $p,1,0$ (terminating, since the sum of the proper divisors of zero is undefined). The conjecture is also true, by computation, for all $n\le275$. Strictly speaking, this doesn't qualify as an answer to the question, since there is no $n$ for which the conjecture has been proved to be false, but there are so many integers for which the aliquot sequence shows no sign of ever reaching a bound that it seems most plausible to conclude that the conjecture is false.
Here is a website dedicated to the problem.

REPLY [4 votes]: An important question in matroid structure theory is given a matroid $M$, how many inequivalent representations does it have over a fixed finite field $\mathbb{F}$?  Two $\mathbb{F}$-matrices $A$ and $B$ are equivalent if they are related by a projective transformation, in which case they clearly represent the same matroid.
In 1988, Kahn proved that $3$-connected $GF(4)$-representable matroids have a unique representation over $GF(4)$.  In the same paper, he conjectured that for every prime power $q$, there is an $n(q)$, such that every $3$-connected $GF(q)$-representable matroid has at most $n(q)$ inequivalent representations. Oxley, Vertigan and Whittle proved Kahn's conjecture for $GF(5)$, but also showed that it fails for all larger finite fields.  This naturally led to the question of what happens for $4$-connectivity, and the following weakening of Kahn's Conjecture.
Conjecture 1. For every prime power $q$, there is an $n(q)$ such that every $4$-connected $GF(q)$-representable matroid has at most $n(q)$ inequivalent representations over $GF(q)$.
One would expect that Conjecture $1$ holds for all finite fields, or fails for sufficiently large finite fields. However, remarkably there is a dichotomy depending on whether $q$ is prime!
Theorem 2. (Geelen and Whittle) For every prime $p$, there is an $n(p)$ such that every $4$-connected $GF(p)$-representable matroid has at most $n(p)$ inequivalent representations over $GF(p)$.
The proof of Theorem $2$ is not easy; the paper is $175$ pages.
Theorem 3. (Geelen, Gerards and Whittle)
For every non-prime power $q \geq 9$, there is a $4$-connected $GF(q)$-representable matroid with arbitrarily many inequivalent representations over $GF(q)$.


TITLE: Is there any sort of higher-order SVD (quadratic and above) for dimensionality reduction?
QUESTION [7 upvotes]: (Posted this on math.stackexchange and cross.correlated over more than a week ago, but didn't get an answer, and this is a question in my research so this seems like it might have been the better place to ask...)
I'm wondering if there exists any higher-order SVD for dimensionality reduction.  Note that I do not mean multilinear PCA, which operates on data tensors, but some form of SVD which can produce, say, a quadratic approximation of a dataset.
Intuitively, SVD takes in a set of vectors as a datset, and computes the ranked eigenvalues and eigenvectors which correspond to a reduced subspace and a "weight" representing their importance.  Given a decomposition:
$$A = U\Sigma V^T$$
Where $A$ is a matrix where the columns are data vectors, the truncated matrix $U$ (keeping only the leftmost $n$ columns; call this truncated version $\mathcal{U}$) provides a reduced approximate basis for the original dataset $A$.  However, this approximation is purely linear.  It can thus be thought of as a linear approximation of the data on some $n$ reduced variable set.  For a particular column of $a_i$ of $A$, $a_i \approx \mathcal{U} x_i$ for some dimensionality-reduced vector $x_i$.
I am wondering if it is possible to do something akin to a Taylor expansion here, and recover a higher-order approximation of the data.  For example, for a quadratic system of order $n$, I'd like to generate the $\mathcal{U}$ matrix, but also a third-order tensor $\mathcal{W}$ such that a "quadratic" approximation of the data could be written, as, say:
$a_i \approx \mathcal{U} x_i + \frac{1}{2} x_i^T (\mathcal{W} ~ \vdots ~ x_i)$
where $\vdots$ is a tensor-vector contraction.  (This is a form I made up now for discussion's sake; it is possible that the true form of some quadratic approximation is slightly different).  Here, the best $x_i$ the quadratic approximation will be better than the linear one.
Ideally, I'd be searching for some general theory which would allow for arbitrary order approximation.
Is there any general established method for this (or something similar)?  If so, is it tractable?  If not, is there a mathematically grounded reason why not?

REPLY [2 votes]: A Non-linear Generalization of Singular Value Decomposition proposes a non-linear extension of the singular value decomposition by appending additional columns to the data matrix which are non-linearly derived from the existing columns.
Section V gives an application to the logistic map, $X_{n+1}=\lambda X_n(1-X_n)$, where $\lambda$ is an unknown parameter that we wish to retrieve from the data. When SVD is applied to a matrix that contains columns with the time series $X_n$ as well as the square $X_n^2$, the appearance of zero singular values indicates a linear relationship between columns, from which an estimate for $\lambda$ is obtained.