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Assuming the Riemann hypothesis $\rho=\frac12+i\gamma$, then $$1-\frac{1}{\rho}=-\frac{\frac12-i\gamma}{\frac12+i\gamma}=e^{2i\theta},\qquad \theta=\arctan\frac{1}{2\gamma}.$$ $$\sum_\rho\Bigl[1-\Bigl(1-\frac{1}{\rho}\Bigr)^n\Bigr]=\sum_\gamma\bigl(1-e^{2in\theta}\bigr).$$ Here $n$ is fixed and for $\gamma\to+\infty$ $$1-e^{2in\theta}\sim 2n\theta\sim \frac{... 0 One way to get a handle on the fact that X\mapsto [X,X] is of 'mixed variance' was invented by Dana Scott around 1970. He showed how to solve the 'domain equation' X\cong [X,X] to obtain a model of untyped lambda calculus. The basic idea is to replace the category \cal C by a category that has the same objects, but has as arrows 'embedding-projection ... 12 It seems worth giving the cup-length argument, as it's relatively short and sweet. Suppose \mathbb{R}P^n=U_1\cup\cdots\cup U_n, with each U_i\approx\mathbb{R}^n, and let c\in H^1(\mathbb{R}P^n;\mathbb{Z}/2) be the generator. For each i the inclusion-induced map H^1(\mathbb{R}P^n;\mathbb{Z}/2)\to H^1(U_i;\mathbb{Z}/2) is trivial, so by the long ... 1 The concept first appeared in Shelah's paper Independence results The theorem you have stated should be folklore, but you may see Tapani Hyttinen and Mika Rautila, The canary tree revisited for a proof. Of course in the above cited papers, the definitions are not given for some stationary set, but that is clear how to modify them for this case. Finally, a ... 3 Suppose that H^0(L^{\otimes n})|_ Y is non-zero for all Y and n sufficiently big, and has a section which vanishes somewhere on Y. Then it follows that the base set of L is trivial: indeed, L has a non-zero section on any complex subvariety, which includes the base set. This implies that the natural map P_n:\; X\rightarrow {\mathbb P}(H^0(X, L^{\... 32 Expanding on the comment by @user127776, the key reference is Palais, "Lusternik-Schnirelman Theory on Banach Manifolds", Topology 5 (1966), where it is proved that if X can be covered by n contractible closed sets, then the cup-length of X is strictly less than n. (Here the cup-length is the largest n such that for some field F and ... 0 I found the answer in theorem 8.7 in the survey article on periodic algebras by Erdmann and Skowronski. They are certain (twisted) mesh algebras of Dynkin type. 0 These lecture notes by Piotr Hajłasz might have the introductory level you are looking for: The lectures will be divided into two almost independent streams. One of them is the theory of Sobolev spaces with numerous aspects which go far beyond the calculus of variations. The second stream is just calculus of variations. The theory of Sobolev spaces is a ... 0 Here's another, perhaps more probabilistic, approach. It's known that \pi can be represented in terms of the mean occupation times as follows. Fix a state, for convenience r. Then, for any state j,$$ \pi_j=m_{rr}^{-1} E^r\left[ \sum_{n=0}^{T_r-1}1_{\{X_n=j\}}\right], $$where E^r denotes expectation for the chain started in r, and T_r is the ... 7 Just a partial answer for the definite version, name it \beta_0. Using Laplace Transform from this Maple 2021 output$$ I(p) = \int_{0}^{\infty} e^{-px} (I_{0}(2 \sqrt{x}) - 1) dx = \frac{e^{1/p}-1}{p}$$Since$$\frac{1}{e^{x}-1}-\frac{1}{e^x} = \frac{e^{-2x}}{1-e^{-x}}= \sum_{p=2}^{\infty} e^{-px}$$the integral reduces to the following series$$\beta_0 =...

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There is a nice criterion that is usually applied in this situation. A is flat over $R$ if the Krull dimension of the fiber ring $A_{\mathfrak{P}}/\mathfrak{P}$ for all prime ideals $\mathfrak{P} \subset R$ is $n-r$ (or the fiber is empty). In scheme theoretic language, the fibers either have the expected dimension (are complete intersections) or are empty. ...

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Aiming to add to Pólya's content, I wrote a book about a technique to solve any problem in STEM. It's titled "The Top-Down Approach to Problem Solving", ISBN 979-8464073296. You can find it on amazon. I hope this is a good fit for you.

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To find $f(1)$ to high precision, we will expand $f$ as a Laurent series in $(z+1)^{-1}$, and solve for the coefficients. Setting $f(z)=g(z+1)$, we want to find $$g(z)=z+0+a_1z^{-1}+a_2z^{-2}+\cdots$$ satisfying $$g(z)=z-1+\frac{g(2z-1)}{g(2z)},$$ or $(g(z)-z+1)g(2z)=g(2z-1)$. Hence $$\begin{array}{rcrcrcrcrl} \bigg(1&+&a_1z^{-1}&+&a_2z^{-2}&... 4 It actually isn't true that the complexity of \theta could be meaningfully bounded in the term of complexities of \varphi and \psi. Let us fix an arbitrary recursive ordinal \alpha. Below I sketch a construction of infinitary \Pi_n formulas \varphi,\psi, for some finite n such that there are no \Pi_\alpha interpolant for them. Consider the ... 0 There are several results in topology about metrics taking values in (the positive cone of) a partially ordered Abelian group \mathbb{G}=(G, <, +, 0). The first occurrence of this spaces I'm aware of is dated 1950 by Sikorski (MR0040643 - R. Sikorski, Remarks on some topological spaces of high power). He literally started a new branch of topology that ... 30 Taking the fact that Catalan numbers C_n measure the number of binary trees on n nodes, we can find an involution on the set of these trees: choose the lexicographically first node in the tree that's unbalanced (i.e., where the number of nodes in the left subtree is different from the number in the right subtree). Then partner this tree with the one ... 34 Apparently not mentioned yet, though surely not new: use the quadratic equation satisfied by the generating function. Since we look for n+1 to be a power of 2, we shift the index by 1 and consider$$ F = \sum_{n=0}^\infty C_n x^{n+1} = x + x^2 + 2 x^3 + 5 x^4 + 14 x^5 + 42 x^6 + 132 x^7 + 429 x^8 + \cdots. Then F = x + F^2. Instead of solving ... 5 Here is one I've thought of a while ago for my combinatorics classes, but never ended up using. Let C_{0},C_{1},C_{2},\ldots be the Catalan numbers. Let \mathbb{N} = \left\{0,1,2,\ldots\right\}. Theorem 1. For each n\in\mathbb{N}, we have \begin{align*} C_{n+1}=\sum\limits_{k=0}^{\left\lfloor n/2\right\rfloor }2^{n-2k}\dbinom{n}{2k} C_{k}. \end{align*}... 9 A proof by Koshy and Salmassi starting from Segner's recurrence formula for the Catalan numbers. A proof by Ji and Wilf using the method of recursive palindromes. 2 The projective space PE of a topological vector space E is Hausdorff but in general is not Tychonoff, not functionally Hausdorff and even not Urysohn (let us recall that a topological space is Urysohn if any distinct points have disjoint closed neighborhoods). As a suitable counterexample, consider the countale product of lines E=\mathbb R^\omega. The ... 4 Maybe of interest, in the paper "Orbispaces as differentiable stratified spaces." Lett. Math. Phys. 108 (2018), no. 3, 805–859, Crainic and Mestre prove in proposition 26 that for proper Lie groupoid G\rightrightarrows M $$M and the orbit space X=M/G together with the canonical stratifications, are differentiable stratified spaces. ... 3 Say that L=\partial_x^2 (the heat equation). Your problem is ill-posed in all reasonnable context : it does not admit a solution for generic data taken in spaces L^2, C^\infty or even in distributional spaces, although it does when the data are analytic (Cauchy-Kowalevska). Let us consider the simplified situation where F\equiv0 and (0,T) is replaced ... 1 Apart from the obscure Czech journal, Pospíšil published his result in the Annals of Mathematics: Bedřich Pospíšil: Remark on bicompact spaces, Annals of Mathematics 38 (1937), no. 4, pp. 845–846, doi 10.2307/1968840. 3 This inequality is known. Indeed, inequality (1.1) by Kwapień and Szulga states that$$\Big(\int_S\Big(\int_T|f(s,t)|^q\mu(dt)\Big)^{p/q}\nu(ds)\Big)^{1/p} \\ \le \Big(\int_T\Big(\int_S|f(s,t)|^p\nu(ds)\Big)^{q/p})\mu(dt)\Big)^{1/q}if 0<q<p, where \mu and \nu are measures. The case p<q<0 is obtained from this by replacing f (which ... 3 The answer is yes, at least over \mathbb{C}, since 2-dimensional (cyclic) quotient singularities are taut (starr, in German), namely, they are uniquely characterized, up to biholomorphisms, by their resolution graph. In other words, every 2-dimensional normal singularity, having the same resolution graph of a (cyclic) quotient singularity, is itself a ... 4 In general, holomorphic maps f: \mathbb{C} \rightarrow \mathbb{C} have no fixed points; but using Picard theorem we can show that f\circ f always have fixed point. Theorem (Fixed-point theorem) Let f: \mathbb{C} \rightarrow \mathbb{C} be holomorphic. Then f \circ f: \mathbb{C} \rightarrow \mathbb{C} always has a fixed point unless f is a ... 0 Here is a proof that is pretty standard for existence and unicity in such geometric setting by convergence, in the figure the initiale cevians intersect at P. Arrange the segments to the sides in increasing order, here PF\le PE\le PG. It is easy to see that the orange parts are smaller than their corresponding segments (PE,PF) and the green ones are ... 1 In addition to @Abdelmalek Abdesselam's and @Johannes Hahn's good points, I can add a few things from my own context: One very non-negotiable thing is that Eisenstein series, with a significant role in the spectral theory of automorphic forms, are not in the relevant L^2 space, but are only (in an appropriate sense) "of moderate growth". The ... 7 If this were true, then any short exact sequence of vector bundles would split. Indeed, if 0 \to \mathscr E_1 \to \mathscr E_2 \to \mathscr E_3 \to 0 is a short exact sequence of vector bundles, then both \begin{align*} K^\bullet = \cdots \to 0 \to \mathscr E_1 \to \mathscr E_2 \to 0 \to \cdots \end{align*} and L^\bullet = \mathscr E_3 are resolutions ... 0 This is a comment, but too long for that. One can try to apply the p,q method, i.e., assume that the three vertices are (0,0), (1,0) and (p,q) respectively. Then if P has barycentric coordinates (\lambda_1,\lambda_2,\lambda_3) with respect to the vertices, the equality of the squares of the lengths from P to the vertices of the cevian ... 7 No. The simplest example is given by the following two resolutions of the structure sheaf of a point P \in \mathbb{P}^1: 0 \to \mathcal{O}_{\mathbb{P}^1}(-1) \to \mathcal{O}_{\mathbb{P}^1} \to \mathcal{O}_{P} \to 0 $$and$$ 0 \to \mathcal{O}_{\mathbb{P}^1}(-2) \to \mathcal{O}_{\mathbb{P}^1}(-1) \to \mathcal{O}_{P} \to 0 $$(the second is obtained ... 0 Strictly speaking, my question is about the 1d case but I'm really interested in the 2d case so I'm leaving it open. However, I'm giving here a (partial) answer for the 1d case. Consider the 1d problem of minimizing$$ \int_{-1}^1 g(x) u(x) + \phi(u'). All of the problems in the question can be reformulated in this form. Denote by p = u', and note that ... 6 I've only just seen this rather old thread. I've recently been computing with cochains on BG for G a finite group in characteristic p, and have some rather surprising conclusions. If G has either semidihedral or generalised quaternion Sylow 2-subgroups and no normal subgroup of index two, then the cochains on BG with mod two coefficients is ... 8 I am not that strong on the representation-theory side, but know more about the combinatorics side. If you want to get an overview of the symmetric functions and the associated combinatorics (crystal bases, RSK etc), then one starting point (with references!) is www.symmetricfunctions.com. I am the admin for this site, so all errors and issues are completely ... 4 It's right in the beginning of De Bruijin's paper. More generally, \log p(rn) \sim \frac 1 {2 \log r} \log^2 \frac n {\log n} where p(rn) is the number of partitions of rn into powers of r. The result is attributed to Kurt Mahler's paper "On a special functional equation", published in Journ. London Math. Soc. in 1940. 11 M. Haiman "Notes on Macdonald polynomials and the geometry of the Hilbert scheme of points on \mathbb{P}^2". By one of the greatest specialists of interactions between combinatorics and algebraic geometry. 1 The quadratic form whose matrix is \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & I_n \\ 0 & I_n & 0 \end{pmatrix} gives an embedding of \operatorname{SO}(2n + 1, \mathbb C) in \operatorname{GL}(2n + 1, \mathbb C) whose derivative is your specified embedding \mathfrak{so}(2n + 1, \mathbb C) \to \mathfrak{gl}(2n + 1, \mathbb C). Under ... 1 I have recently discovered this YouTube channel https://youtube.com/channel/UC-ssgSLLaFYn1LARgEAqWBg which contains lots of mathematical physics and mathematics course. 1)Introduction to topological field theory, link: https://youtube.com/playlist?list=PLqX5gFCSJtMC7tju8pAoZZ_RqYeXoEYxl 2)Quantum Mechanics for Mathematicians, link: https://youtube.com/... 4 Following @Abdelmalek's great advice in the comments above it is enough to compute the following triple integral: \begin{align*} I=\frac{\pi^s}{\Gamma(s)}\int_{0}^{\infty}\left(\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} (x_1+x_2 i)^{\ell}\cdot e^{-\pi t(x_1^2+x_2^2+a^2)} e^{2\pi i (x_1\zeta_1+x_2\zeta_2)} dx_1 dx_2 \right) ... 6 This is precisely Proposition 1.120 on p.49 in Besse's "Einstein Manifolds" (I am using the reprint of the 1987 edition, so the numbering may be different in the older edition): A 3-dimensional pseudo-Riemannian manifold is Einstein iff it has constant (sectional) curvature. Note that the proof is before the proposition. -2 Reflection principles are a good starting point, but applying them, well... Although worldly cardinals are the "smallest type" of ZFC-unprovable cardinals, let's jump up to the uncountable inaccessibles. The classical argument is that V is (strongly) inaccessible, i.e. it is not the successor of any number, it is cofinal with itself, and is not the ... 17 Here is a bijective proof that equates both a(n) and b(n+1) to the quantityp(n)+2p(n-1)+\cdots+np(1)+(n+1)p(0) \tag1$$where p(n) is the number of partitions of n. For a(n): each partition with three types of 1's corresponds to a partition of k together with (n-k) parts from \{1',1''\}. This second gadget can be chosen in (n-k+1) ways, ... 6 For what it's worth, here is a non-combinatorial proof. That \frac{1}{(1-x)^2}\prod_{k=1}^{\infty} \frac{1}{1-x^k} is the generating function for partitions of n with three flavors of 1's is clear. So I will focus on the sum of all the irredundant parts of partitions of n. Let me use ||\lambda|| to denote this sum of irredundant parts (in contrast ... 5 Yes, this is true. In fact something even stronger is known: For arbitrary positive integers r,j, and prime \ell\geq 5, there are infinitely many values of n for which$$p(n)=r\pmod{\ell^j}.$$This is a special case of a conjecture of Newman and is proved in the paper "Coefficients of half-integral weight modular forms modulo \ell^j" by ... 6 Lowry-Duda, Tanagichi, and Thorne proved that if K is a number field with absolute discriminant D_K, d=[K:\mathbb{Q}] = r_1+2r_2, where r_1 (resp. r_2) is the number of real (resp. complex) embeddings into \mathbb{C}, h the order of the ideal class group of K, regulator R, w roots of unity, then there exists an effectively computable ... 13 Using the fact that 1-q^{3n}=(1-q^{2n})+q^{2n}(1-q^n), we can write$$W_n(q)=\binom{2n}{n}_q+q^{2n}\binom{2n-1}{n-1}_q$$and then the result follows from the fact that the degree of W_n(q) is n^2+n, together with the fact that q-binomial coefficients are polynomials in q with positive coefficients (unimodal even). 2 Consider the metric g on the domain$$ Q = \{ (x^0, \dots, x^n)\ :\ x^0, x^1, \dots, x^n > 0\} $$given by$$ g_{ij} = \frac{\delta_{ij}}{x^i},\ 0 \le i, j \le n. $$Let$$ \Delta = \{ x^0 + \cdots x^n = 1 \} \subset Q. $$On \Delta, x^0 = 1-x^1-\cdots - x^n. Therefore$$ dx^0 = -dx^1 - \cdots - dx^n, $$and the metric g restricted to \... 2 When T\le\sqrt x and x=\frac12+\mathbb Z^+, it is possible to show using Perron's formula that$$ \pi(x)={1\over2\pi i}\int_{k-iT}^{k+iT}{x^s\over s}\log\zeta(s)\mathrm ds+\mathcal O\left(x\log x\over T\right)  After applying certain analytic properties of $\zeta(s)$ in the critical strip, the task turns into evaluating the following residue integral: $... 1 I was eventually able to get the answer (only) for$n = 1$, building on a calculation of the Christoffel symbols for arbitrary$n$. But this still seems far from a full answer. Let$(\mu_1,\dots,\mu_n)$parametrize the standard simplex$\Delta_{n} := \{(\mu_1,\dots,\mu_{n+1}) \in \mathbb{R}^{n+1}_{\ge 0}: \sum_i \mu_i = 1\}$, so that$\mu_{n+1} = 1 - \sum_{i ...

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Update: I had an email exchange with Victor Guba. He has kindly confirmed that there is indeed a typo in (the Russian and English versions of) his paper: a "кусок" (as per his paper) and an "$s$-piece" (as per Kashintsev's paper) are meant to be one and the same thing. The part below was written before hearing from Guba. (A recent ...

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