text:
# zbMATH — the first resource for mathematics

Autour des conjectures de Bloch et Kato. III: Le cas général. (On the conjectures of Bloch and Kato. III: The general situation). (French. Abridged English version) Zbl 0758.14001
[For part I and II see ibid. No. 5, 189-196 and No. 6, 349-356 (1991; Zbl 0749.11052 and 752.14001).]
This paper is the third of a series on the Bloch-Kato conjectures about the values at $$s=0$$ of the $$L$$-function of a ‘motive’ (or a ‘motivic structure’) over a number field. We use concepts and notation defined in the two foregoing articles I and II cited above. Let $$E$$ be a number field, and define a premotivic category $${\mathcal M}$$ to be a full subcategory of the $$\mathbb{Q}$$-linear neutral Tannakian category $$SPM_ \mathbb{Q}(E)$$ of premotivic structures over $$E$$ such that $${\mathcal M}$$ is stable under subobject, quotient, direct sum, tensor product and dual. Such $${\mathcal M}$$ is an abelian category. For an object $$M$$ of $${\mathcal M}$$ one defines the $$\mathbb{Q}$$-vector spaces $$H^ 0_ M(E,M):=\text{Hom}(1,M)$$ and $$H^ 1_ M(E,M):=\text{Ext}^ 1(1,M)$$. One has maps $$u_ M:H^ 0(E,M)_ \mathbb{R}\to\text{Ker}(\alpha_ M)$$ and $$u_ M':\text{Coker}(\alpha_ M)\to H^ 0(E,M^*(1))^*_ \mathbb{R}$$.
The
‘$$\ell$$-adic realization functor’ induces a $$\mathbb{Q}$$-linear map $$H^ 1_ M(E,M)\to H^ 1(E,M_ \ell)$$, $$x\mapsto x_ \ell$$. Write $$H^ 1_{M,f}(E,M)$$ for the $$\mathbb{Q}$$-subspace of $$H^ 1_ M(E,M)$$ consisting of all of $$x$$ such that $$x_ \ell\in H^ 1_ f(E,M_ \ell)$$ for all $$\ell$$. Then $${\mathcal M}$$ is called $$f$$-admissible if
(i) for any object $$M$$ of $${\mathcal M}$$, $$\dim_ \mathbb{Q} H^ 1_{M,f}(E,M)<\infty$$;
(ii) $$H^ 1_{M,f}(E,1)=0$$;
(iii) for any short exact sequence $$0\to M\to N\to 1\to 0$$, whose class in $$H^ 1_ M(E,M)$$ is contained in $$H^ 1_{M,f}(E,M)$$, the map $$H^ 1_{M,f}(E,N^*(1))\to H^ 1_{M,f}(E,M^*(1))$$ is surjective;
(iv) the map $$u_{M^*(1)}:H^ 0(E,M^*(1))_ \mathbb{R}\to\text{Ker}(\alpha_{M^*})$$ is surjective for any object $$M$$ of $${\mathcal M}$$ such that $$H^ 1_{M,f}(E,M)=0$$ and the sequence $0\to H^ 0(E,M)_ \mathbb{R}\to\text{Ker}(\alpha_ M)\to H^ 1_{M,f}(E,M^*(1))^*_ \mathbb{R}\to 0$ is exact.
One also defines $$f$$-equivalence on $${\mathcal M}$$ as the strongest equivalence relation on $${\mathcal O}b({\mathcal M})$$ such that (i) if $$0\to M\to N\to 1\to 0$$ is a short exact sequence with class in $$H^ 1_{M,f}(E,M)$$, then $$M$$ and $$N$$ are $$f$$-equivalent; (ii) if $$M$$ and $$N$$ are $$f$$-equivalent on $${\mathcal M}$$, then $$M^*(1)$$ and $$N^*(1)$$ are also $$f$$-equivalent.
In the sequel it is assumed that one has at one’s disposal the category $$SM_ \mathbb{Q}(E)$$ of motivic structures, and we put $$H^ 1(E,M)=H^ 1_{SM_ \mathbb{Q}(E)}(E,M)$$ and $$H^ 1_ f(E,M)=H^ 1_{SM_ \mathbb{Q}(E),f}(E,M)$$. A motivic category is defined as a premotivic subcategory of $$SM_ \mathbb{Q}(E)$$. The importance of the notion of motivic category lies in the fact that, once one has a motivic structure $$M$$, one may hope to construct a motivic category containing $$M$$ without knowing the whole $$SM_ \mathbb{Q}(E)$$. Let $$M$$ be an object of the motivic category $${\mathcal M}$$. One defines $L_ f(M):=\text{det}_ \mathbb{Q} H^ 0(E,M)\otimes(\text{det}_ \mathbb{Q} H^ 1_ f(E,M))^{-1}$ and the fundamental line of $$M$$: $\Delta_ f(M):=L_ f(M)\otimes L_ f(M^*(1))\otimes(\text{det}_ \mathbb{Q} M^ +_ B)^{- 1}\otimes\text{det}_ \mathbb{Q}(t_ M(E)).$ Several conjectures can be stated, but one must be careful because they all involve the badly known category $${\mathcal M}$$, e.g. “$$C_{L,\text{weak}}(M,{\mathcal M})$$”:
(i) $$r_ M=\dim_ \mathbb{Q} H^ 1_ f(E,M^*(1))-\dim_ \mathbb{Q} H^ 0(E,M^*(1))$$;
(ii) if $$b$$ is a $$\mathbb{Q}$$-basis of $$\Delta_ f(M)$$, then one has $$L^*(M,0)\cdot| b|_{EP,\infty}\in\mathbb{Q}^ \times$$.
Also, “$$C_{\text{Gal}}(M,{\mathcal M})$$”:
(iii) One has isomorphisms $$\mathbb{Q}_ \ell\otimes_ \mathbb{Q} H^ 0(E,M)@>\sim>> H^ 0(E,M_ \ell)$$ and $$\mathbb{Q}_ \mathbb{Q}\otimes_ \mathbb{Q} H^ 1_ f(E,M)@>\sim>> H^ 1_ f(E,M_ \ell)$$ for all primes $$\ell$$ and similarly for $$M^*(1)$$.
Finally, under the assumption of “$$C_{\text{Gal}}(M,{\mathcal M})$$”, “$$C_ L(M,{\mathcal M})$$”:
If $$b$$ is a basis of $$\Delta_ f(M)$$, one has $$| b|_{EP,\ell}=1$$ for almost all $$\ell$$, and with $$\{\text{prime numbers}\}\cup\{\infty\}=\overline P$$, $$L^*(M,0)\cdot\prod_{\ell\in\overline P}| b|_{EP,\ell}\in\{1,-1\}$$.
This last “conjecture” can be translated into an equivalent conjecture in terms of Tamagawa measures. In case $$M$$ is $$f$$-closed, these “conjectures” amount to true conjectures, i.e. without knowledge of the whole category of motivic structures. One gets the following true conjecture $$C_{\text{ord}}(M,\ell)$$: For a prime $$\ell$$, one has $$r_ M=\dim_{\mathbb{Q}_ \ell}H^ 1_ f(E,M^*(1)_ \ell)-\dim_{\mathbb{Q}_ \ell}H^ 0(E,M^*(1)_ \ell)$$ . The “conjectures” are invariant under $$f$$-equivalence. The main result of the paper now says that one has a well defined Weil restriction functor $$\text{Res}_{E/\mathbb{Q}}$$ to reduce problems and conjectures over $$E$$ to corresponding ones over $$\mathbb{Q}$$, and, fortunately, over $$\mathbb{Q}$$ every object $$M$$ of a motivic category $${\mathcal M}$$ is $$f$$-equivalent to an $$f$$-closed object. This leads to an algorithm to reduce “conjectures” to true conjectures for $$f$$- closed objects.

##### MSC:
 14A20 Generalizations (algebraic spaces, stacks) 11G09 Drinfel’d modules; higher-dimensional motives, etc. 14G10 Zeta functions and related questions in algebraic geometry (e.g., Birch-Swinnerton-Dyer conjecture) 19E99 $$K$$-theory in geometry 14C35 Applications of methods of algebraic $$K$$-theory in algebraic geometry


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# zbMATH — the first resource for mathematics

H$$^p$$ spaces of several variables. (English) Zbl 0257.46078
This paper is a major contribution to the study of $$H^p$$ spaces, singular integrals, and harmonic analysis an $$\mathbb R^n$$. Classically the theory of $$H^p$$ spaces arose from analytic function theory. $$H^p$$ was defined as the space of analytic functions in the upper half plane with boundary values in $$L^p(\mathbb R)$$. More recently this definition has been generalised to $$\mathbb R^n$$ by introducing generalised conjugate harmonic functions in $$\mathbb R_+^{n+1}=\{(x,t): x\in\mathbb R^n$$, $$t>0\}$$. The authors present several intrinsic descriptions of $$H^p$$, of a real variable nature, not involving conjugate functions. These results greatly clarify the meaning of $$H^p$$, as well as throwing new light on the behaviour of convolution operators an $$L^p$$. The following is a summary of some of the main results. The first main result is the description of the dual of $$H^1$$. $$H^1$$ is the Banach space of all functions $$f$$ in $$L^1(\mathbb R^n)$$ such that $$R_jf\in L^1(\mathbb R^n)$$, $$j=1,\dots,n$$, where $$R_j$$ is the $$j$$-th Riesz transform. (In terms of Fourier transforms, $$(R_jf)^\wedge(y)= y_j\hat f(y)/|y|$$. When $$n=1$$ the definition says that the Hilbert transform of $$f$$ is in $$L^1$$, or equivalently $$\int f(x)\,dx=0$$ and $$f=g+\bar h$$ where $$g$$ and $$h$$ are in the classical “analytic” $$H^1$$). The authors prove that the dual of $$H'$$ is the space of all functionals of the form $$\varphi\to\int f\varphi$$ (suitably interpreted if $$f\varphi\not\in L^1$$), where $$\varphi$$ is a function of bounded mean oscillation (BMO), which means that there is a constant $$C>0$$ (depending an $$\varphi$$) such that $$\int_Q |f-f_Q|\leq C|Q|$$ for any cube $$Q$$ in $$\mathbb R^n$$, where $$f_Q= |Q|^{-1}\int_Q f$$. The proof of this remarkable and deep result depends an an inequality of Carleson and some clever manipulations of Littlewood Paley functions. The hard part is to show that a BMO function defines a bounded functional on $$H^1$$. The essential difficulties are already present in the case $$n=1$$. This duality leads to a new approach to convolution operators which brings out the usefulness of BMO as a substitute for $$L^\infty$$. If $$T$$ is a convolution operator (i.e. $$Tf= K*f$$ for some distribution $$K$$ an $$\mathbb R^n$$) mapping $$L^\infty$$ into BMO, then the authors show, using the duality and a description of $$L^p$$ related to BMO, that $$T$$ maps $$H^1$$ into $$H^1$$, BMO into BMO, and $$L^p$$ into $$L^p$$ for $$1<p<\infty$$. The condition that $$T$$ map $$L^\infty$$ into BMO is relatively easy to verify for singular integral operators of Calderón-Zygmund type. A more refined version of this result, involving Calderón’s complex method of interpolation, enables the authors to prove new results on $$L^p$$ multipliers. The authors then turn to $$H^p$$ spaces for general $$p$$ $$(0<p<\infty)$$. They define these first as spaces of harmonic functions an $$\mathbb R_+^{n+1}$$, without reference to boundary values. Specifically, a harmonic function $$u_0$$ is in $$H^p$$ if there exist harmonic functions $$u_1,\dots,u_n$$ satisfying
$\partial u_j/\partial x_i=0, \quad \sum_{i=0}^n \partial u_i/\partial x_i \quad\text{and}\quad \sup_{t>0} \int_{\mathbb R^n}|u(x,t)|^p\,dx<\infty,$
where $$|u|^2= \sum_{i=0}^n |u_i|^2$$. This definition is appropriate if $$p>(n-1)/n$$. In general a more elaborate version (here omitted) is needed (the point is that $$|u|^p$$ is subharmonic only if $$p\geq (n-1)/n$$). The main result is as follows: let $$u$$ be harmonic in $$\mathbb R_+^{n+1}$$ and define $$u^*(x)= \sup_t |u(x,t)|$$. Then $$u\in H^p$$ if and only if $$u^*\in L^p$$. (For $$n=1$$ this was proved by Burkholder, Gundy and Silverstein in 1971). Finally the authors consider boundary values. If $$u\in H^p$$ then $$u(x,t)\to f(x)$$ as $$t\to0$$, in the distribution sense, where $$f$$ is a tempered distribution on $$\mathbb R^n$$. Denote the set of such $$f$$ also by $$H^p$$. Then $$H^p= L^p$$ for $$p>1$$ and for $$p=1$$ this definition is consistent with the earlier one. The last result above characterises $$H^p$$ in terms of Poisson integrals (as $$u$$ is the Poisson integral of $$f$$). The authors show that the Poisson kernel can be replaced by any smooth approximate identity – more precisely, fix a smooth function $$\varphi$$ on $$\mathbb R^n$$, decreasing rapidly at $$\infty$$, with $$\int\varphi=1$$. Put $$\varphi_t(x)= t^{-n} \varphi(x/t)$$, and for any tempered disribution $$f$$ write $$f^*(x)= \sup_{t>0} |\varphi_t*f(x)|$$. Then (for $$0<p<\infty$$) $$f\in H^p$$ if and only if $$f^*\in L^p$$. (“Non-tangential” versions of this and the preceding result are also given). This result implies for example that one could define $$H^p$$ in terms of the wave equation rather than Laplace’s and get the same space of functions on $$\mathbb R^n$$. The paper concludes with a proof that certain singular integral operators map $$H^p$$ to itself.
Reviewer: Alexander M. Davie

##### MSC:
 46J15 Banach algebras of differentiable or analytic functions, $$H^p$$-spaces 46E30 Spaces of measurable functions ($$L^p$$-spaces, Orlicz spaces, Köthe function spaces, Lorentz spaces, rearrangement invariant spaces, ideal spaces, etc.) 42B25 Maximal functions, Littlewood-Paley theory 30D55 $$H^p$$-classes (MSC2000) 42A50 Conjugate functions, conjugate series, singular integrals 44A35 Convolution as an integral transform 46F10 Operations with distributions and generalized functions
Full Text:
##### References:
 [1] Burkholder, D. L. & Gundy, R. F., Distribution function inequalities for the area integral. To appear in Studia Math. · Zbl 0219.31009 [2] Burkholder, D. L., Gundy, R. F. &Silverstein, M. L., A maximal function characterization of the classH p .Trans. Amer. Math. Soc., 157 (1971), 137–153. · Zbl 0223.30048 [3] Calderón, A. P., Commutators of singular integral operators.Proc. Nat. Acad. Sci. U.S.A., 53 (1965), 1092–1099. · Zbl 0151.16901 · doi:10.1073/pnas.53.5.1092 [4] –, Intermediate spaces and interpolation, the complex method.Studia Math., 24 (1964), 113–190. · Zbl 0204.13703 [5] Calderón, A. P. &Zygmund, A., On the existence of certain singular integrals.Acta Math., 88 (1952), 85–139. · Zbl 0047.10201 · doi:10.1007/BF02392130 [6] –, On higher gradients of harmonic functions.Studia Math., 24 (1964), 211–226. · Zbl 0168.37002 [7] Duren, P. L., Romberg, B. W. &Shields, A. L., Linear functionals onH p spaces with 0
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.

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# \rowcolor in a command with optional arguments does not work ('Misplaced \noalign…')

I'm trying to create a very modular table for a project, where each row is constructed by a command. If I try to set a row color for a table row generated by a command with optional argument, gives the error message:

! Misplaced \noalign.
\rowcolor ->\noalign
{\ifnum 0=}\fi \global \let \CT@do@color \CT@@do@color...
l.9     \myRowA{bbbb}

Using \cellcolor instead works, just as well as converting the optional argument to a mandatory. Only row colors in combination with optional arguments give that error.

Here is a minimal example showing the problem:

\documentclass{article}
\usepackage[table]{xcolor}

\begin{document}
\newcommand{\myRowA}[2][]{
\rowcolor{blue} #1 & #2 \\ \hline
}
\begin{tabular}{ll}\hline
\myRowA{bbbb}
\end{tabular}

\end{document}
`

Any idea what I'm doing wrong or how I can make this work properly?

• commands with optional arguments are not expandable, and this means that \rowcolor is not the first command in the row. – Ulrike Fischer Jan 29 at 22:27


text:
# Zeros of the Riemann zeta function

Symbol: RiemannZeta $\zeta\!\left(s\right)$ Riemann zeta function
Symbol: RiemannZetaZero $\rho_{n}$ Nontrivial zero of the Riemann zeta function
Symbol: RiemannHypothesis $\operatorname{RH}$ Riemann hypothesis

## Main properties

$\left(\operatorname{RH}\right) \iff \left(\operatorname{Re}(s) = \frac{1}{2} \;\text{ for all } s \in \mathbb{C} \text{ with } 0 \le \operatorname{Re}(s) \le 1 \;\mathbin{\operatorname{and}}\; \zeta\!\left(s\right) = 0\right)$
$\left(\operatorname{RH}\right) \iff \left(\operatorname{Re}\!\left(\rho_{n}\right) = \frac{1}{2} \;\text{ for all } n \in \mathbb{Z}_{\ge 1}\right)$
$\mathop{\operatorname{zeros}\,}\limits_{s \in \mathbb{R}} \zeta\!\left(s\right) = \left\{ -2 n : n \in \mathbb{Z}_{\ge 1} \right\}$
$\mathop{\operatorname{zeros}\,}\limits_{s \in \mathbb{C},\,0 \le \operatorname{Re}(s) \le 1} \zeta\!\left(s\right) = \left\{ \rho_{n} : n \in \mathbb{Z} \;\mathbin{\operatorname{and}}\; n \ne 0 \right\}$
$\mathop{\operatorname{zeros}\,}\limits_{s \in \mathbb{C}} \zeta\!\left(s\right) = \left\{ -2 n : n \in \mathbb{Z}_{\ge 1} \right\} \cup \left\{ \rho_{n} : n \in \mathbb{Z} \;\mathbin{\operatorname{and}}\; n \ne 0 \right\}$
$0 < \operatorname{Re}\!\left(\rho_{n}\right) < 1$
$\operatorname{Re}\!\left(\rho_{n}\right) = \frac{1}{2}$
$\rho_{-n} = \overline{\rho_{n}}$

## Numerical values

$\rho_{1} \in \frac{1}{2} + \left[14.134725141734693790457251983562470270784257115699 \pm 2.44 \cdot 10^{-49}\right] i$
$\rho_{2} \in \frac{1}{2} + \left[21.022039638771554992628479593896902777334340524903 \pm 2.19 \cdot 10^{-49}\right] i$
Table of $\operatorname{Im}\!\left(\rho_{n}\right)$ to 50 digits for $1 \le n \le 50$
Table of $\operatorname{Im}\!\left(\rho_{n}\right)$ to 10 digits for $1 \le n \le 500$
Table of $\operatorname{Im}\!\left(\rho_{{10}^{n}}\right)$ to 50 digits for $0 \le n \le 16$

Related topics: Riemann zeta function

Copyright (C) Fredrik Johansson and contributors. Fungrim is provided under the MIT license. The source code is on GitHub.

2020-08-27 09:56:25.682319 UTC


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# Is there an isomorphism between universal domains $\mathcal{P}\omega$ and the interval domain $\mathbf{I}\mathbb{R}$?

Is there a constructive way to one-to-one associate elements $$x\in\mathcal{P}\omega$$ with elements $${\scriptstyle\mathbf I}\in\mathbf{I}\mathbb{R}$$ ?
I assumed there should be since they're both universal (so kind of an embedding-embedding pair, so to speak), but haven't been able to figure out how it could be done, nor google a solution.

Edit: Re comment How do you constructively embed $$\mathcal{P}(\{0,1,2\})$$ into the interval domain?...
Well, here's what I'd been thinking to do for $$\mathcal{P}\omega$$ in general, which doesn't work as a bijection, but which maybe vaguely works as an embedding. It's all kind of fast-and-loose in the spirit of Stoy's remark about "fiddling with sets of integers" (last paragraph on pg.116 in his book).
First let's consider your definition $$\mathbf{I}\mathbb{R} = \{[a,b] \mid a, b \in \mathbb{R}_{[0,1]} \land a \leq b \}$$ restricted to $$\mathbb{R}_{[0,1]}=\{r\in\mathbb{R}\mid 0\leq r\leq1\}$$. Then for $$x\in\mathcal{P}\omega$$, first, for each $$i\in x$$, use the inverse pairing function $$\mathbb{N}\times\mathbb{N}\to\mathbb{N}$$, e.g., https://en.wikipedia.org/wiki/Pairing_function, to get $$x\to x_1,x_2$$. Now use the usual construction $$x\subseteq\mathbb{N}\to r_x=\sum_{i\in x}\frac1{2^i}\in\mathbb{R}_{[0,1]}$$. So now we have $$x\in\mathcal{P}\omega\to r_{x_1},r_{x_2}$$, with which we associate the interval $$[r_{_\mbox{lo}},r_{_\mbox{hi}}]$$, choosing the lesser of $$r_{x_1},r_{x_2}$$ for $$r_{_\mbox{lo}}$$ and ditto for hi. And those $$r$$'s will be rational if $$x$$ is finite, and computable/constructive if it's r.e.
Second, if it's really necessary to consider $$\mathbb{R}$$ rather than just $$\mathbb{R}_{[0,1]}$$, then for any $$x\in\mathcal{P}\omega$$, first get $$x_1,x_2$$ as above, but now separate each into its even and odd numbers, i.e., $$x_1^e=\{\frac i2\mid i\in x_1\land i\mbox{ even}\}$$ and $$x_1^o=\{\frac{i+1}2\in x_1\mid i\mbox{ odd}\}$$. And now use $$x_1^e$$ to determine the integer part of $$r_{x_1}$$ as $$\sum_{i\in x_1^e}2^i$$, and use $$x_1^o$$ to determine the decimal part as above.

• Is this supposed to be an isomorphism of underlying sets (any bijection), or an isomorphism of posets (monotone), or an isomorphism of domains (continuous)? Nov 5 '21 at 13:02
• Sorry, I misread, you're looking for embeddings, not isomorphisms. Nov 5 '21 at 14:25
• Can you embed $P(\{0,1,2\})$ into $\mathrm{I}\mathbb{R}$? Also, please write down the exact definition of $\mathrm{I}\mathbb{R}$. Nov 5 '21 at 14:55
• Really? How do you constructively embed $P(\{0,1,2\})$ into the interval domain? And I think it's still important that you write down what you think the constructive interval domain is. Do you use Cauchy reals, Dedekind reals, or lower and upper Dedekind cuts? It may matter. Nov 6 '21 at 8:30
• Your edit is not using $\mathcal{P}(\omega)$ but $2^{\mathbb{N}}$. Nov 7 '21 at 8:49

This is only half an answer, but allow me to clear up a constructive point about the interval domain.

The usual definition of the interval domain is $$\mathrm{I}\mathbb{R} = \{[a,b] \mid a, b \in \mathbb{R} \land a \leq b \} \cup \{\mathbb{R}\},$$ ordered by reverse inclusion. If we try to use this definition constructively, we get stuck when showing that it is a dcpo. Even showng that an increasing sequence has a supremum is impossible (because constructively it may not be the case that a bounded increasing sequence of reals has a supremum).

The correct constructive definition of the interval domain goes as follows. An element of $$\mathrm{I}\mathbb{R}$$ is a pair $$(L, U) \in P(\mathbb{Q}) \times P(\mathbb{Q})$$ such that

1. $$L$$ is lower and rounded: $$p \in L \iff \exists q \in L \,.\, p < q$$.
2. $$U$$ is upper and rounded: $$q \in U \iff \exists p \in U \,.\, p < q$$.
3. $$L$$ and $$U$$ are disjoint.

The ordering is defined by $$(L,U) \sqsubseteq (L', U') \iff L \subseteq L' \land U \subseteq U'.$$ Think of $$L$$ and $$U$$ as the endpoints of the interval, given respectively as a lower and an upper Dedekind cut. Note that $$\emptyset$$ is allowed as a cut. It represents either $$-\infty$$ or $$\infty$$, for instance $$(\emptyset, \emptyset)$$ is the bottom element, $$(\emptyset, U)$$ represents an interval $$(-\infty, u]$$, and $$(L, \emptyset)$$ represents an interval $$[\ell, \infty)$$.

Clearly, we have an inclusion $$\mathrm{I}\mathbb{R} \subseteq P(\mathbb{Q}) \times P(\mathbb{Q})$$, and since $$P(\mathbb{Q}) \times P(\mathbb{Q}) \cong P(\mathbb{Q} + \mathbb{Q}) \cong P(\omega),$$ we have an inclusion $$\mathrm{I}\mathbb{R} \to P(\omega)$$. The other direction I am still thinking about.

It is also quite clear that the two domains are not isomorphic posets, as $$P(\omega)$$ is an algebraic lattice but $$\mathrm{I}\mathbb{R}$$ is not.

Supplemental: I cannot tell from your question what precisely you are trying to do, but perhaps you will find Cartesian closed categories of separable Scott domains relevant. There we find invariants, namely coherence numbers, that determine when one universal domain embeds into another.

• Thanks. Okay, I see your point that the computable subset of $\mathbf{I}\mathbb{R}$ won't be a subdomain (i.e., the "whether or not" part of my comment above should just be "not"). And thanks also for the more formal analysis than anything that had come to mind for me. It all makes me realize that I should have omitted "constructive" in the original question (I should maybe just be interested in the computable elements of each domain, and maybe an isomorphism between those subsets). Nov 7 '21 at 7:57
• That won't work either. The computable elements form effective domains (you get the c.e. sets instead of $P(\omega)$ and the computble interval domain). These are not computably isomorphic as domains because, again, one is algebraic and the other is not. Nov 7 '21 at 8:47
• Do you have an idea that any two universal domains ought to be isomorphic? Because that is not true. Nov 7 '21 at 8:48
• I linked to a paper that classifies universal separable Scott domains up to retraction, perhaps that will be relevant to your work. Nov 7 '21 at 9:23
• Re "idea that any two universal domains ought to be isomorphic": isomorphic is apparently too strong, but what had crossed my mind is that given any two universal domains $\mathcal{U}_1$ and $\mathcal{U}_2$, and any domain $D$, then there always exist retractions $\hat{r}_1:\mathcal{U}_1\to D$ and $\hat{r}_2:\mathcal{U}_2\to D$. So we can always say $\hat{r}_1\sim\hat{r}_2$ in some way, whereby the two collections of retractions are related in some way, hopefully a strong way, which hopefully extends to $\mathcal{U}_1$ and $\mathcal{U}_2$ themselves. But I hadn't carried this idea any further. Nov 8 '21 at 6:04


text:
# Does not show categories or products in homepage

I need to list all the categories on the home page and is not working, I used:

{{block type="catalog/navigation" name="catalog.category" template="catalog/category/list_home.phtml"}}

I also just in case tried using other calls on the homepage and does not work as well:

{{block type="catalog/product_list" name="home.catalog.product.list" alias="products_homepage" category_id="2" template="catalog/product/list.phtml"}}

{{block type="catalog/product_list" name="home.catalog.product.list" alias="products_homepage" category_id="3" template="catalog/product/list.phtml"}}

{{block type="catalog/product_list" column_count="4" category_id="3" template="catalog/product/list.phtml"}}

Someone can explain me wrong? I am using the theme RWD.

I have never seen that in my life in magento, but are experiencing this error.

I need to show the categories on the home page as the error first, i test try a workaround with two examples.

If your Magento version is Magento 1.9.2 or greater, then you need to whitelist catalog/product_list block type. You can do that through

System > Permissions > Blocks

After do that, the last block directive is enough to show category in the Home page.

{{block type="catalog/product_list" column_count="4" category_id="3" template="catalog/product/list.phtml"}}

Here what I did to get mine to work, I just add these below codes to Homepage->Design->Layout Update XML

<reference name="content">
<block type="catalog/product_list" name="featured1" template="catalog/product/list.phtml">
<action method="setCategoryId"><category_id>5</category_id></action>
<action method="setColumnCount"><count>5</count></action>
</block>
</reference>
<category_id>5</category_id>
<reference name="content">
<block type="catalog/product_list" name="featured2" template="catalog/product/list.phtml">
<action method="setCategoryId"><category_id>6</category_id></action>
<action method="setColumnCount"><count>5</count></action>
</block>
</reference>
<category_id>6</category_id>

I only have two categories in my case, but if you have more you can keep adding the block and change the category id.

</div><div class="list-title">


text:
# Creationnism with series.

1. ### alingy1

324
Hi, I saw this "proof" on some entertainment website:

0=0
0=0+0+0+0+0+0...
0=1-1+1-1+1-1+1-1+1-1...
0=1+0+0+0+0+0+0+0....
0=1

We create something from nothing!

Can someone explain what is wrong here? There has to be an illegal step!

2. ### SammyS

9,273
Staff Emeritus
The infinite series, 1-1+1-1+1-1+1-1+1-1... does not converge.

Last edited: May 17, 2014
3. ### micromass

20,069
Staff Emeritus
The illegal step is that you're working with infinite sums. These are not at all well behaved. What you do is essentially the following:

$$(1-1) + (1-1) + (1-1) + ... = 1 + (-1+1) + (-1 + 1) + (-1 + 1) + ...$$

which is just rearranging the brackets. This is the illegal step. You can easily rearrange the brackets in finite sums ##a+(b+c) = (a+b)+c##, but not in infinite sums. You need to rigorously prove in infinite sums that you can do this. You can't in this case. You can in others.

4. ### alingy1

324
The next question is: why can't I rearrange them?

5. ### micromass

20,069
Staff Emeritus
I was expecting this. This shows a wrong attitude towards mathematics. Don't worry, you're in good company as most ancient mathematicians had this attitude.

Your attitude is: I can do anything in mathematics as long as it looks right. For example, rearranging brackets looks right in the finite case, so it must work in the infinite case. This attitude has brought many paradoxes and contradictions to mathematics which took hundreds of years to resolve.

The modern attitude is quite the opposite. It is: you can't do anything in mathematics without first showing rigorously that you can. So if we look at this question with the modern attitude, then we notice that we can rearrange brackets for finite sums. But we also immediately remark that doesn't imply anything for infinite sums. There is no reason to expect that a property like this holds if we have no rigorous proof of it. In fact, this example alone suffices to show that the property does not hold, it is a counterexample!

This ancient versus modern attitude is very pervasive. For example, in ancient times it was thought that since stones fall towards the earth, that would imply that everything falls to the earth. Thus the earth is the center of the universe and everything falls towards it.
The modern attitude is that there is no reason to expect things that hold for stones that they also hold for planets. It must be demonstrated somehow first. In fact, it has been demonstrated that it does not hold.

Some things to consider:

1 person likes this.
6. ### DrewD

515
Just to point out how much of a counter example this is, let's see what would happen if one assumes that such a rearrangement is acceptable. Then surely we can also rearrange

##0=2+2-2+2-2\ldots## and ##0=3+3-3+3-3\ldots##

and thus for any other number. Since ##0=1## and ##0## is the additive identity,

##1=1+0=1+1=2##

which will also work with any choice of numbers! So in a mathematical world where arbitrary rearrangements of infinite series are acceptable, numbers really lose all meaning. It isn't just a weird conclusion that we choose to sweep under the rug. One either concludes that every number is equal and math is pointless, or this particular property does not hold for infinite sums.

7. ### disregardthat

1,817
You have to get analytical, and ask what do we mean by $1-1+1-1+1-1+...$?

As it stands, it's ambiguous. When dealing with infinite series, we define them as the limit of the partial sums of an infinite sequence of terms $a_1,a_2,a_3...$ The partial sums are $S_n = \Sigma^n_{k=1}a_k=a_1+a_2+...+a_n$.

The infinite sum is the limit $\lim_{n \to \infty} S_n$ (if it exists) which we write as $\Sigma^{\infty}_{k=1}a_k$. This is what the infinite sum means. It only makes sense if the limit exists. In that case we say that the infinite sum converges. If it does not exist, then we say it diverges.

So, what do we mean by $1-1+1-1+...$ ?

Is it the infinite sum formed by the sequence of terms $a_1 = 1, a_2 = -1, a_3 = 1, a_4 = -1, ...$ ?

In that case, the partial sums will be $S_1 = 1, S_2 = 0, S_3 = 1,...$ The sequence $S_n$ is alternating between 1 and 0 and will not converge. So the limit does not exist.

However, by adding parentheses as such: $1+(-1+1) + (-1+1) + (-1+1) + ...$, and saying that $a_1 = 1, a_2 = -1+1 = 0, a_3 = -1+1 = 0, ...$ Then we are talking about an entirely different sequence of terms, and we get an entirely different sequence of partial sums $S_n$. In this case, $S_n = 1$ for all n, and the limit is 1. The infinite sums are not the same, because the sequence we used to define what the sum meant is not the same.

As to your question on why we can't rearrange the terms as such, what needs to be understood is that by adding the parantheses we are not talking about the same sum. There are circumstances where we may rearrange the sum as such and end up with the same result, but those are not satisfied here. In fact, rather view this example as a proof of that rearranging in such a way does not produce the same result in general.

Last edited: May 17, 2014
8. ### Rakstarflame

1

Pls can u xplain wt it means in simpler language????? I m just a grade eight student

### Staff: Mentor

In layman terms: infinite series is a sum of all terms in an infinite sequence.

Such a sum can converge - that is, the more terms you add, the closer you get to some number.

For example if you add all terms

$$\frac 1 {2^1} + \frac 1 {2^2} + \frac 1 {2^3} + \frac 1 {2^4} ...$$

you will get 1. You can easily check that the sum gets closer and closer to 1 using just a calculator.

But not every series converges. 1+1+1+1... doesn't - it just gets larger and larger.

1-1+1-1... is one of those series that don't converge, although it does it in a different way - it doesn't just grow infinitely, it oscillates between 1 and 0.


text:
VS2013 Update 2 RC – Error APPX3210

If you stumble across the following error message with Visual Studio 2013 Update 2 RC:

1>[…]\Package.appxmanifest(19,64): error APPX3210: App manifest references the image ‘Assets/Icon150.png’ which does not have a candidate in main app package.

have a look a at the manifest file. In my case I had been using forward slashes in some of the asset paths. While this used to work before update 2 it now leads to an error. So to fix this I had to replace this:

<m2:DefaultTile Square310x310Logo="Assets\Icon310.png"   Wide310x150Logo="Assets/Icon150.png">

with this:

<m2:DefaultTile Square310x310Logo="Assets\Icon310.png"   Wide310x150Logo="Assets\Icon150.png">

and the error went away.


text:
## BloombergNEF annual battery price survey finds prices fell 6% from 2020 to 2021

##### 01 December 2021

Lithium-ion battery pack prices, which were above $1,200 per kilowatt-hour in 2010, have fallen 89% in real terms to$132/kWh in 2021, according to a new report from BloombergNEF (BNEF). (All prices are in real 2021 $, unless stated as nominal.) This marks a 6% drop from$140/kWh in 2020. Continuing cost reductions bode well for the future of electric vehicles; however, the impact of rising commodity prices and increased costs for key materials such as electrolytes has put pressure on the industry in the second half of the year.

These prices are an average across multiple battery end-uses, including different types of electric vehicles, buses and stationary storage projects. For battery electric vehicle (BEV) packs in particular, prices were $118/kWh on a volume-weighted average basis in 2021, according to the report. At the cell level, average BEV prices were just$97/kWh. This indicates that on average, cells account for 82% of the total pack price. Over the past two years, the cell-to-pack cost ratio has diverged from the traditional 70:30 split, a result of changes to pack design, such as the introduction of cell-to-pack designs.

On a regional basis, battery pack prices were cheapest in China, at $111/kWh. Packs in the US and Europe cost 40% and 60% higher, respectively. This reflects the relative immaturity of these markets, the diverse range of applications and, for the higher end of the range, low volume and bespoke orders. Prices continued to fall in 2021 as the adoption of the lower-cost lithium iron phosphate (LFP) cathode chemistry increased, and as the use of expensive cobalt in nickel-base cathodes continued to slide. On average, LFP cells were almost 30% cheaper than NMC cells in 2021. However, even lower-cost chemistries such as LFP—which is particularly exposed to lithium carbonate prices—have felt the bite of rising costs throughout the supply chain. Since September, Chinese producers have raised LFP prices by between 10-20%. Based on historical trends, BNEF’s 2021 Battery Price Survey, which was launched in time for the virtual BNEF Summit Shanghai, predicts that by 2024 average pack prices should be below$100/kWh. It is at around this price point that automakers should be able to produce and sell mass-market EVs at the same price (and with the same margin) as comparable internal combustion vehicles in some markets. This assumes no subsidies are available, but actual pricing strategies will vary by automaker and geography.

However, higher raw material prices mean that in the near-term, average pack prices could rise to $135/kWh in 2022 in nominal terms. In the absence of other improvements that can mitigate this impact, this could mean that the point at which prices fall below$100/kWh could be pushed back by two years. This would impact EV affordability or manufacturers’ margins and could hurt the economics of energy storage projects.

Although battery prices fell overall across 2021, in the second half of the year prices have been rising. We estimate that on average the price of an NMC (811) cell is $10/kWh higher in the fourth quarter than it was in the first three months of the year, with prices now closing in on$110/kWh. This creates a tough environment for automakers, particularly those in Europe, which have to increase EV sales in order to meet average fleet emissions standards. These automakers may now have to make a choice between reducing their margins or passing costs on, at the risk of putting consumers off purchasing an EV.

—James Frith, BNEF’s head of energy storage research and lead author of the report

The path to achieving $100/kWh is clear, although the timing now looks more uncertain. In 2021, a wave of automakers released battery technology roadmaps outlining how prices can be reduced below$100/kWh. Companies like Renault and Ford have publicly announced targets of $80/kWh by 2030. Prices for lithium have risen substantially this year as a result of constraints within global supply chains, rising demand in China and Europe and the recent production curbs in China. Although we expect demand to keep growing in 2022, other factors such as global supply-chain constraints and China’s production curbs should have been resolved by 1Q 2022. This will help to ease lithium prices. —Kwasi Ampofo, head of metals and mining at BloombergNEF Continued investment in R&D alongside capacity expansion across the supply chain will help to improve battery technology and reduce costs over the next decade. BloombergNEF expects next-generation technologies, such as silicon and lithium metal anodes, solid-state electrolytes and new cathode material and cell manufacturing processes, to play an important role in enabling these price reductions. ### Comments These seem somewhat off. When Nissan introduced the Leaf in 2010 I remember Nissan saying they were paying<$400 /Kwh at the cell level.
The figure from the comentariat was \$375/Kwh at the cell level.
Now these were LMO cells ,not a good choice but LFPs would be the same price.

Try to find cells or packs at these prices. Battery vendors are cashing in big time. The markups are huge.

The comments to this entry are closed.


text:
Principle of superpostion

Principle of superpostion

This principle tells us that if charge Q is acted upon by several charges q1 , q2 …. qn , then the net force on Q can be found out by calculating separately the forces F1 , F2 … Fn , exerted by q1 , q2 , …. qn respectively on Q and then adding these forces vectorially.
Their resultant F is the total force on q due to the collection of charges.

Thus the net force on a charge Q is the vector sum of forces on Q due to all the remaining charges.

$\displaystyle \vec{F} = \vec{F_{1}} + \vec{F_{2}} +\vec{F_{3}} + ….. + \vec{F_{n}}$

Example : It is required to hold equal charges, q in equilibrium at the corners of a square. What charge when placed at the centre of the square will do this ?

Solution: Let ABCD be a square of side a , and Q be the charge placed at the centre.

$\displaystyle \vec{F_{BA}} = k \frac{q^2}{a^2}\hat{i}$

$\displaystyle \vec{F_{BC}} = -k \frac{q^2}{a^2}\hat{j}$

$\displaystyle \vec{F_{BD}} = k \frac{q^2}{(a\sqrt2)^2}(cos45\hat{i}-sin45\hat{j})$

$\displaystyle \vec{F_{BQ}} = k \frac{q Q}{(a/\sqrt2)^2}(cos45\hat{i}-sin45\hat{j})$

$\displaystyle F_x = k[\frac{q^2}{a^2}+ \frac{q^2}{(a\sqrt2)^2 }.\frac{1}{\sqrt2} +\frac{q Q}{(a/\sqrt2)^2}.\frac{1}{\sqrt2} ]$

$\displaystyle F_y = -k[\frac{q^2}{a^2}+ \frac{q^2}{(a\sqrt2)^2 }.\frac{1}{\sqrt2} +\frac{q Q}{(a/\sqrt2)^2}.\frac{1}{\sqrt2} ]$

Net force on the charge at B is

$\displaystyle \vec{F}= F_x\hat{i} + F_y\hat{j}$

For charge, q to be in equilibrium at B, the net force on it must be zero.

$\displaystyle F_x = 0 , F_y = 0$

$\displaystyle F_x = k[\frac{q^2}{a^2}+ \frac{q^2}{(a\sqrt2)^2 }.\frac{1}{\sqrt2} +\frac{q Q}{(a/\sqrt2)^2}.\frac{1}{\sqrt2} ] = 0$

$\displaystyle Q = -\frac{q}{4}(1 + 2\sqrt2)$

Similarly , If Fy=0

$\displaystyle Q = -\frac{q}{4}(1 + 2\sqrt2)$


text:
# How to show that $⟨a,b | aba=bab⟩$ is not the trivial group?

I want to show that G = $$⟨a,b | aba=bab⟩$$ is not the trivial group

I tried to find homomorphism $$\phi$$ from $$G$$ to $$\mathbb Z$$ which maps $$a$$ to $$0$$ and $$b$$ to $$1$$ (or $$b$$ to $$0$$ and $$a$$ to $$1$$) and if such a homomorphism exists , $$\phi(b)$$ is non-trivial and thus b is non-trivial. but I didn't found such a homomorphism.

I'm also tried to conclude it directly from the relation and I failed again

Thanks

• If you add the relation $ab=ba$ then you quickly deduce $a=b$, so there is a surjective map to $\mathbb Z$.
– lulu
Dec 7 '18 at 13:13
• You can define your homomorphism to ${\mathbb Z}$ by mapping $a$ and $b$ to $1$. Dec 7 '18 at 15:12

Note that $$\mathbb Z$$ satisfies this, with $$a=b$$.

Phrased differently, adding the relation $$ab=ba$$ we quickly deduce that $$a=b$$, so the new relation gives a surjective map to $$\mathbb Z$$.

• Also $S_3$ with $a=(1\ 2)$ and $b=(2\ 3)$.
– bof
Dec 7 '18 at 14:41
• @lulu OK, so we can "extend" $G$ with another relation and get a presentaion of $Z$ , can you please explain how this helps in my case ? Tnx
– R.P
Dec 7 '18 at 15:38
• If a set surjects onto an infinite set, it is obviously infinite.
– lulu
Dec 7 '18 at 19:50

The deficiency of a finite presentation $$$$ is defined to be $$|X|-|R|$$. If $$deg(G)>0$$ then group $$G$$ is of order infinite.


text:
# Question 78c76

Jun 17, 2016

1.430 × 10^6color(white)(l) "g Fe"_2"O"_3#

#### Explanation:

1. Find the number of moles of Fe you need to make by dividing mass by atomic mass.

Since you need grams at the end, convert kilograms to grams first.

1000 kg = 1 000 000 g.

Divide by 55.85 g/mol Fe.

1 000 000 g ÷ 55.85 g/mol Fe = 17 905 mol Fe.

2. The ratio of $F {e}_{2} {O}_{3}$ to $F e$ is 2:4, so you'll need half as much $F {e}_{2} {O}_{3}$ as $F e$.

${\text{17 905 mol Fe" ÷ "2 mol Fe/mol Fe"_2"O"_3 = "8952.6 mol Fe"_2"O}}_{3}$
3. Multiply this number by the formula mass of $F {e}_{2} {O}_{3}$:
${\text{8952.6 mol Fe"_2"O"_3 × "((55.85×2)+(16.00×3)) g/mol Fe"_2"O"_3 = 1.430 × 10^6 "g Fe"_2"O}}_{3}$.


text:
# How do I find K and DeltaH for the overall reaction? The overall reaction is: "Ni"("CO")_4(g) + 4"H"_2"O"(g) rightleftharpoons "Ni"(s) + 4"CO"_2(g) + 4"H"_2(g)

## How do I find $K$ and $\Delta H$ for the overall reaction? $\text{Ni"("CO")_4(g) rightleftharpoons "Ni"(s) + 4"CO} \left(g\right)$, ${K}_{1} = {1.1}^{A}$, $\Delta {H}_{1} = 4 D$ ${\text{MnO"(s) + "CO"(g) rightleftharpoons "Mn"(s) + "CO}}_{2} \left(g\right)$, ${K}_{2} = {B}^{2.7}$, $\Delta {H}_{2} = \frac{340}{E}$ ${\text{Mn"(s) + "H"_2"O"(g) rightleftharpoons "MnO"(s) + "H}}_{2} \left(g\right)$, ${K}_{3} = {1.84}^{C}$, $\Delta {H}_{3} = - 5.77 F$ The overall reaction is: ${\text{Ni"("CO")_4(g) + 4"H"_2"O"(g) rightleftharpoons "Ni"(s) + 4"CO"_2(g) + 4"H}}_{2} \left(g\right)$ $A = 2$ $B = 3$ $C = 4$ $D = 5$ $E = 6$ $F = 7$

Jul 8, 2017

It really makes the most sense to identify which species are intermediates and catalysts first... That's how anyone can blindly achieve the overall reaction, simply by trying to cancel out intermediates and catalysts, or even just trying to get the products to match up.

INTERMEDIATES + CATALYSTS

Intermediates are as they seem to be; they appear in the middle of the mechanism, and go away after the reaction is over.

Catalysts come in to speed up the reaction, but are not consumed by the time the reaction is over.

By inspection, you should find:

• $\text{Mn} \left(s\right)$ is an intermediate; appears in step 2, consumed in step 3.
• $\text{CO} \left(g\right)$ is an intermediate. Appears in step 1, must be consumed in step 2 in order to not be in the overall reaction.
• $\text{MnO} \left(s\right)$ is a catalyst. Consumed in step 2, reformed in step 3.

So, all you have to do is multiply each reaction step's coefficient by a constant in order to cancel out the intermediates and catalyst. This isn't nearly as complicated as it seems, because no steps need to be reversed.

FORMING OVERALL REACTION

Since you want $4 {\text{H}}_{2} \left(g\right)$ as a product, multiply step 3 by $4$. Since you want $\text{CO} \left(g\right)$ to go away, multiply step 2 by $4$ so that $4 \text{CO} \left(g\right)$ is a reactant in step 2 that gets used up from step 1.

Then by magic, the rest works itself out...

"Ni"("CO")_4(g) rightleftharpoons "Ni"(s) + cancel(4"CO"(g))

4(cancel("MnO"(s)) + cancel("CO"(g)) rightleftharpoons cancel("Mn"(s)) + "CO"_2(g))

4(cancel("Mn"(s)) + "H"_2"O"(g) rightleftharpoons cancel("MnO"(s)) + "H"_2(g))

$\text{----------------------------------------------------------------}$

${\text{Ni"("CO")_4(g) + 4"H"_2"O"(g) rightleftharpoons "Ni"(s) + 4"CO"_2(g) + 4"H}}_{2} \left(g\right)$

EQUILIBRIUM CONSTANTS AND ENTHALPIES

Now, it's just a matter of recalling the general rules of ${K}_{e q}$ and $\Delta H$ with regards to scaling.

• For all thermodynamic functions, scaling a reaction multiplies the function by the constant.
• For all equilibrium constants, scaling the reaction by a constant multiplies all the exponents by that constant (since equilibria are elementary reactions), i.e. raise $\boldsymbol{{K}_{e q}}$ to that constant.

Thus...

${K}_{1} ' = {K}_{1}$ $\text{ "" }$ $\Delta {H}_{1} ' = \Delta {H}_{1}$
${K}_{2} ' = {\left({K}_{2}\right)}^{4}$ $\text{ }$ $\Delta {H}_{2} ' = 4 \Delta {H}_{2}$
${K}_{3} ' = {\left({K}_{3}\right)}^{4}$ $\text{ }$ $\Delta {H}_{3} ' = 4 \Delta {H}_{3}$

And so... noting that adding sequential steps multiplies equilibrium constants and adds thermodynamic functions for each step by Hess's Law...

$\textcolor{b l u e}{{K}_{t o t} '} = {K}_{1} ' {K}_{2} ' {K}_{3} ' = {K}_{1} {\left({K}_{2} {K}_{3}\right)}^{4}$

$= {1.1}^{A} \cdot {\left({B}^{2.7} {1.84}^{C}\right)}^{4}$

$= \textcolor{b l u e}{{1.1}^{A} {B}^{10.8} {1.84}^{4 C}}$

And...

$\textcolor{b l u e}{\Delta {H}_{t o t} '} = \Delta {H}_{1} ' + \Delta {H}_{2} ' + \Delta {H}_{3} '$

$= \Delta {H}_{1} + 4 \left(\Delta {H}_{2} + \Delta {H}_{3}\right)$

$= 4 D + 4 \left(\frac{340}{E} - 5.77 F\right)$

$= \textcolor{b l u e}{4 \left(D + \frac{340}{E} - 5.77 F\right) \text{ kJ}}$

You're given the constants at the top of the page. So, plug them in and you can get your answer.


text:
# Experimental evaluation of ultrasound-guided 3D needle steering in biological tissue

• Published 2014 in
International Journal of Computer Assisted…

#### Abstract

In this paper, we present a system capable of automatically steering bevel tip flexible needles under ultrasound guidance toward stationary and moving targets in gelatin phantoms and biological tissue while avoiding stationary and moving obstacles. We use three-dimensional (3D) ultrasound to track the needle tip during the procedure. Our system uses a fast sampling-based path planner to compute and periodically update a feasible path to the target that avoids obstacles. We then use a novel control algorithm to steer the needle along the path in a manner that reduces the number of needle rotations, thus reducing tissue damage. We present experimental results for needle insertion procedures for both stationary and moving targets and obstacles for up to 90 mm of needle insertion. We obtained a mean targeting error of $$0.32\pm 0.10$$ 0.32 ± 0.10 and $$0.38\,\pm \,0.19$$ 0.38 ± 0.19  mm in gelatin-based phantom and biological tissue, respectively. The achieved submillimeter accuracy suggests that our approach is sufficient to target the smallest lesions ( $$\phi$$ ϕ  2 mm) that can be detected using state-of-the-art ultrasound imaging systems.

DOI: 10.1007/s11548-014-0987-y

#### Statistics

Citations per Year

#### Citation Velocity: 10

Averaging 10 citations per year over the last 3 years.


text:
Integration By Parts

1. Sep 19, 2007

danago

Hi. When evaluating an integral such as:

$$\int {x^4 e^x dx}$$

Is integrating by parts 4 times the best method, or is there a more efficient way?

Dan.

2. Sep 19, 2007

JonF

I think you are gonna have to do parts 4 times.

3. Sep 19, 2007

rock.freak667

Well if the problem was a definite integral for example say:
$\int_{0}^{1} x^4e^x dx$

you could easily make a reduction formula for easy calculations. But that is a different question altogether. As previously stated you would have to integrate by parts at 4 times

4. Sep 19, 2007

danago

Alright, thanks for the replies guys


text:
Hobbled rook tour - Hamiltonian cycle on square grid

Consider square grid of even sides ($2n \times 2n$). It is easy to see that there must exist Hamiltonian cycle on the corresponding grid graph. Such a cycle is called balance if the number of vertical edges equal the number of horizontal edges. It is easy to construct balance Hamiltonian cycles for odd $n$. But for even $n$ I could not contruct such balance cycles nor can I prove that those cycles don't exist. Question : Does there exist balance Hamiltonian cycle on the grid for even $n$?

Curiously I could not find a lot of research material on this, despite it being quite a natural object. This is sometimes called meander http://oeis.org/wiki/Meanders_filling_out_an_n-by-k_grid_%28not_reduced_for_symmetry%29

There is also an article that characterize and enumerate (by an algorithm) these cycles. http://www.mat.univie.ac.at/~slc/s/s34erlangen.pdf

The cycle divides the squares grid into regions (and closes one of them), and all the regions are tree (no cycle of squares that share edge). I feel that the answer to the original question will come from a clever insight into those trees.

Edit : I've made changes to make it more appropriate for MO.

-
the reason for the question being...? –  Anthony Quas Mar 24 '13 at 20:42
It is a well-known olympiad problem. –  Ilya Bogdanov Mar 24 '13 at 20:49
A "hobbled Rook" is otherwise known as a Wazir, "a very old piece, appearing in some very early chess variants, such as Tamerlane chess." en.wikipedia.org/wiki/Wazir_(chess) –  Noam D. Elkies Mar 24 '13 at 21:10
Do you mean, "visit each square exactly once in an 8 by 8 chessboard?" If not, it seems very easy to do. –  Noah S Mar 24 '13 at 21:20
@Noah. yes, it's essentially a Hamiltonian cycle on the square grid –  John Mar 24 '13 at 21:26


text:
# Math Help - basic

1. ## basic

let a,b,c real no's such that a+2b+c=4 then find max value of (ab + bc + ac)

(ans=4)

2. $a+2b+c=4\,\implies\,b=\frac{4-(a+c)}2$

If you substitute this into $ab+bc+ca$ and simplify, I think you should get

$ab+bc+ca\ =\ 4-\frac{(a-2)^2}2-\frac{(c-2)^2}2$

Hence $ab+bc+ca\,\le\,4$ for all $a,b,c\in\mathbb R.$

3. ## favour

sir can you please add the simplification part actually i could not manage to get to the simplified form

4. $ab+bc+ca\,=\,b(a+c)+ac$

$=\,\frac{(4-(a+c))(a+c)}2+ac$

$=\,\cdots$

Carry on from there.


text:
networks(deprecated)/icosahedron - Maple Help

networks

 icosahedron
 creates an icosahedron graph

 Calling Sequence G:=icosahedron()

Parameters

 G - is returned as the icosahedron

Description

 • Important:The networks package has been deprecated. Use the superseding command GraphTheory[SpecialGraphs][IcosahedronGraph]instead.
 • The simple graph known as a icosahedron is generated and returned as the value of the procedure call.
 • This routine is normally loaded via the command with(networks) but may also be referenced using the full name networks[icosahedron](...).

Examples

Important:The networks package has been deprecated. Use the superseding command GraphTheory[SpecialGraphs][IcosahedronGraph]instead.

 > $\mathrm{with}\left(\mathrm{networks}\right):$
 > $G≔\mathrm{icosahedron}\left(\right):$
 > $\mathrm{edges}\left(G\right)$
 $\left\{{\mathrm{e1}}{,}{\mathrm{e10}}{,}{\mathrm{e11}}{,}{\mathrm{e12}}{,}{\mathrm{e13}}{,}{\mathrm{e14}}{,}{\mathrm{e15}}{,}{\mathrm{e16}}{,}{\mathrm{e17}}{,}{\mathrm{e18}}{,}{\mathrm{e19}}{,}{\mathrm{e2}}{,}{\mathrm{e20}}{,}{\mathrm{e21}}{,}{\mathrm{e22}}{,}{\mathrm{e23}}{,}{\mathrm{e24}}{,}{\mathrm{e25}}{,}{\mathrm{e26}}{,}{\mathrm{e27}}{,}{\mathrm{e28}}{,}{\mathrm{e29}}{,}{\mathrm{e3}}{,}{\mathrm{e30}}{,}{\mathrm{e4}}{,}{\mathrm{e5}}{,}{\mathrm{e6}}{,}{\mathrm{e7}}{,}{\mathrm{e8}}{,}{\mathrm{e9}}\right\}$ (1)
 > $\mathrm{vertices}\left(G\right)$
 $\left\{{0}{,}{1}{,}{2}{,}{3}{,}{4}{,}{5}{,}{6}{,}{7}{,}{8}{,}{9}{,}{10}{,}{11}\right\}$ (2)


text:
# Removing extra space with \pmod command

When using the \pmod command as a subscript to a sum, I often encounter some extra unwanted space. For example, writing

$\sum_{n \equiv 1 \pmod{k}} ...$

produces the following image:

Does anyone have any ideas to delete the extra white space between the 1 and the open parenthesis?

-

I'd define a variant of \pmod:

\makeatletter
\let\@@pmod\pmod
\DeclareRobustCommand{\pmod}{\@ifstar\@pmods\@@pmod}
\def\@pmods#1{\mkern4mu({\operator@font mod}\mkern 6mu#1)}
\makeatother

You can use \pmod as before in all other situations and write

$\sum_{n \equiv 1 \pmod*{k}} ...$

when you need it as a subscript to a sum.

Another way might be to load amsmath and redefine \pod (\pmod is defined in terms of it):

\usepackage{amsmath}
\makeatletter
\renewcommand{\pod}[1]{\allowbreak\mathchoice
{\if@display \mkern 18mu\else \mkern 8mu\fi (#1)}
{\if@display \mkern 18mu\else \mkern 8mu\fi (#1)}
{\mkern4mu(#1)}
{\mkern4mu(#1)}
}

This will avoid having to say \pmod* and the result will be the same as shown.

-
This works great. Thanks for your help! – JavaMan Dec 24 '11 at 3:10
egreg, shouldn't the \allowbreak go before the \mathchoice? I was just looking at your comment on Andrew Swann's answer here: tex.stackexchange.com/questions/255584/… – justin Aug 3 '15 at 20:23
@justin You're perfectly right! – egreg Aug 3 '15 at 20:28


text:
Skip to main content

# Square Root

Let’s take a look at the concept of taking the square root of a number.

## Understanding the Square Root

To square a number, just multiply that number by itself. For example, 32 = 9. A square root works in the opposite way. For instance, if you square 3, you get 9, and if you "take the square root of 9", you get 3 (i.e. 32 = 9 so √9 = 3). In general, a square root of a number is a value that can be multiplied by itself to give the original number. The “√” symbol is called the "radical" symbol. The expression "√9" is either read as "root nine", "radical nine", or "the square root of nine".

Example:

Find the square root of the number 25.

Solution:   √25 = 5 since (5)2 = 25

The largest misconception about the use of a square root is this: √9 = ± 3. This is not correct. Square roots are always positive, so the correct value is √9 = 3. Note that the value of the simplified radical is positive and this is the only value of the square root and this positive result is called the “principle” root. While either of +3 and –3 might have been squared to get 9, "the square root of nine" is defined to be only the positive option, +3. If we want the -3 then we do the following:

-√9 = - (√9 ) = - (3) = -3.

Notice that the parentheses are used to make the point on just how the minus sign is appearing. In general the middle two steps are omitted. So, if we want the negative value we have to actually put in the minus sign.

This misconception arises because we are sometimes asked to solve things like x2 = 9. When you solve the equation     x2 = 9,  you are trying to find all possible values that might have been squared to get 9. Here, the answer is x2= ± 3 and often this will be solved by “taking the square root” of both sides. Here is the proper solution for this problem:

x2 = 9

x = ±√9

x = ± 3

Note that the ± shows up in the second step before the square root is evaluated. It does not show up as part of taking the square root.

For all real numbers x, we have the following piecewise relationship:

Example:

Solve the equation 2x2 + 8 = 40

Solution:

2x2 + 8 = 40

2x2 = 32

x2 = 16

x =  ±√16

x = ± 4

Avoiding Common Math Mistakes with Square Roots:


text:
Roland Waters

2022-06-20

Does there exist a positive irrational number $\alpha$, such that for any positive integer $n$ the number $⌊n\alpha ⌋$ is not a prime?

Abigail Palmer

This is called a Beatty sequence. There will indeed always be a prime in the sequence. The bound for the OP's sequence (provided $\alpha >1$) is
$p\le {L}^{35-16ϵ}{\alpha }^{2\left(1-ϵ\right)}{p}_{m+l}^{1+ϵ}$
where $L=\mathrm{log}\left(2\alpha \right)$, ${p}_{n}$ denotes the numerator of the ${n}^{\text{th}}$ convergent to the regular continued fraction expansion of $\alpha$, and $m$ is the unique integer such that ${p}_{m}\le {L}^{16}{\alpha }^{2}<{p}_{m+1}$. $ϵ$ can be chosen arbitrarily small, but l depends on $ϵ$.

Do you have a similar question?


text:
# Ellipse: geometric equivalence of two definitions

1. Jul 29, 2008

### lemma28

I've been stuck with this problem:

An ellipse can be defined as
1) locus of points for which is constant the sum of the distances from two fixed points (foci)
2) locus of points for which is constant the ratio between the distances from a fixed point (focus) and a fixed line (directrix). The ratio value is the eccentricity of the ellipse.

I would like to find a pure geometric demonstration that property 1) implies 2) and/or vice-versa.

It's easy to demonstrate the equivalence between the two definitions if one translates the properties into analytical relationships between the coordinates of the locus points (cartesian or polar). I'm not interested in that.

I would like to find a (possibly elegant) proof of the equivalence between definition 1) and 2) via pure geometric arguments (a la Euclide): with no coordinates but just geometric relationships, based on ruler and compass.

I've searched the net and various textbook but haven't find nothing similar (even if I have found out several interesting geometrical constructions based on properties 1) or 2) that could be the first half of the bridge to proving the equivalence, i.e. the http://gallica.bnf.fr/ark:/12148/bpt6k99630k/f240" seems interesting).

Any help from a true geometry lover will be appreciated.

Last edited by a moderator: Apr 23, 2017
2. Jul 29, 2008

### tiny-tim

Last edited by a moderator: Apr 23, 2017
3. Jul 29, 2008

### HallsofIvy

Staff Emeritus
You can't prove the equivalence of those- they are not equivalent. (2) is true for any conic section. To get an ellipse you have to add that that ratio is larger than 0 and less than 1.

Last edited by a moderator: Apr 23, 2017
4. Jul 29, 2008

### lemma28

You're right. Naturally I intended
...
2) locus of points for which the ratio between the distances from a fixed point (focus) and a fixed line (directrix) has a constant value that is larger than 0 and less than 1.
...

5. Aug 5, 2008

### lemma28

Thanks tiny-tim for the suggestion about the Dandelin spheres. They do the work in the simplest way.
But I was not completely satisfied since I looked for demonstrations using "plane" geometry.
So I’ve worked a little bit more on it and I think that these are the “geometric” proofs in 2 dimensions).
I post them in case someone else is interested and in case someone can show me a better and simpler way to do it, or some mistakes.
Especially the point 1 of part 2 (just sketched here) seems quite awkward to me, it should be simpler than that, but I haven’t found a better way to do it.
The attached file contains the figure.

First part
Given the locus of points for which is constant the sum of the distances from two fixed points F1 and F2 then for any point belonging to the locus it is
PF1=e*PH1,
where PH1 is the distance form an appropriate fixed line (directrix) perpendicular to the line joining F1 and F2.

Demonstration
Let it be
F1F2=2c;
PF1+PF2=2a;
PF1=f1;
PF2=f2;
PH1=d1;
PH2=d2;

First consider the triangle F1F2P, and construct the circumscribed circle around it.
The perpendicular bisector of the segment F1F2 meets the circle in V (and the quadrilateral VF1F2P is a cyclic quadrilateral). Let’s call N the intersection of this line with the line joining F1 and F2, and we’ll call
NF1=m1;
NF2=m2;
Since F1V=F2V, it follows, for the chord properties, that the segment PV bisects the angle F1PF2, so that F1PN=NPF2=F1F2V=,F2F1V=$$\alpha$$.
Applying the angle bisector theorem to the triangle F1PF2 we have
f1:m1=f2:m2
and also
(f1+f2):f1=(m1+m2):m1 –> 2a:f1=2c:m1 –> m1/f1=c/a and
(f1+f2):f2=(m1+m2):m2 –> 2a:f2=2c:m2 –> m2/f2=c/a

The triangles H1PF1 and F1PN are similar (they have the same angle $$\alpha$$ and two alternate interior angles). So are the triangles H2PF2 and F2PN.
So it is
m1:f1=f1:d1=c/a and m2:f2=f2:d2=c/a
whence f1+f2=c/a*(d1+d2) (that proves the statement, since d1+d2 is the length of the fixed segment H1H2 and c/a=e is constant).
Combining the proportions it is
m1:d1=m2:d2=(c/a)2 and the same for the sums.
So the sum of the distances PF1+PF2 is mean proportional between the lengths H1H2 and F1F2

Second part
Given the locus of points for which the ratio between the distances from a fixed point F1 (focus) and a fixed line (directrix) has a constant value that is larger than 0 and less than 1, then
1) There exists a second focus F2 and a second directrix for which the same relation applies
2) For any point P belonging to the locus the sum of the distances of P to the foci F1 and F2 has a constant value.

Demonstration of 1)
Basically, starting from the focus F1, the directrix, a point P belonging to the locus, and calling H1 the perpendicular projection of P on the directrix it’s possible to build the triangle PF1H1, and the similar triangle PF1N. The lines PN and H1F1 meets at a point V. The second focus F2 is such that the line PV bisects the angle F2PF1.
Having built the same figure as in the demonstration of the first part it follows that the focus F2 is the second focus and the line VF2 inrtersects the line H1P in the point H2 that defines the position of the second directrix.

Demonstration of 2)
Since f1:d1=f2:d2=e, it follows that f1=e*d1 and f2=e*d2 and summing f1+f2=e*(d1+d2)=e*H1H2. So the sum of the distances of P to the foci F1 and F2 has a constant value.

File size:
26.3 KB
Views:
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6. Aug 7, 2008

### tiny-tim

splendid!

Hi lemma28!
Quite right!

I particularly like the way you proved the angles are the same.

(I've spent some time trying to find a shortcut to Part I that doesn't involve the circle, but I can't find one … so your idea of using the circle is most helpful … how did you come up with it?

My only objection to Part I is that I think you laboured the ratios a bit. Otherwise it's excellent.

Part II seems fine to me … the diagram works in reverse … I'd just leave it at that. Nice statement of the Part II theorem, also.)

Cool diagram, too!

(Have you done diagrams for hyperbola and parabola? The hyperbola proof is probably the same. The parabola diagram is presumably going to be a bit trickier … though the proof is trivial, if you're allowed to subtract infinities. )

Final suggestion: Can you prove … same way … that line PV is perpendicular to the tangent to the ellipse?

With our without that, since I've not seen anything like this before, I think the proof and diagram are good enough to send to one of the more amateur-friendly maths journals!

7. Aug 7, 2008

### lemma28

Thanks again tiny-tim.

I must admit that it was hard work to get to the circle. I've tried several other directions before.... It's very time consuming to get to the right path in geometry (unless you have some sort of genial mind of course, which is not my case... :grumpy:)

I've just discovered GeoGebra. Very good program to make some geometry analysis and diagrams

Don't provoke me! . I've absolute no intention of being absorbed again in the conics, where everything seems easy and turns out not to be. I'm content with the ellipse, for the time being.

That would be equivalent to proving the reflection property of the ellipse. I've already seen that demonstration somewhere before. If I remember well it goes like assuming that the perpendicular to PV has a second different point of intersection with the ellipse and then proving that this second point cannot lie on the elllipse since it wouldn't respect the definition (Reductio ad absurdum).

8. Aug 8, 2008

### Doodle Bob

There are two problems with Part One:

1. The line from which H1 and H2 are derived, is not itself defined. This makes any claims about H1 and H2 -- particularly that the triangles H1PF1 and F1PN are similar -- suspect.

2. The point V is defined as the intersection of a circle and a nontangential line -- giving us a choice of two possible points. Which one do you choose? Or does it not matter?

9. Aug 8, 2008

### tiny-tim

Hi Doodle Bob!

lemma28 defines H1 and H2 as the points where the "horizontal" line through P meets VF1 and VF2.

And the other "possible point" on the circle is P itself.

So its all ok!

10. Aug 8, 2008

### lemma28

No tiny-tim, I think Doodle Bob was right in pointing out that V was not properly defined.
Originally it was defined as the intersection between the perpendicular bisector of the segment F1F2 and the circle.
After Doodle Bob remark it should be corrected as "intersection between the perpendicular bisector of the segment F1F2 and the circle on the arc F1F2 not including P. Otherwise the point N goes outside the segment F1F2 and everything goes astray.
After that we can define H1 and H2 as "the points where the "horizontal" line through P meets VF1 and VF2".
Note that the distance of the line horizontal H1H2 from the horizontal axis F1F2 changes with P.

I think with these adjustments the demonstration still holds.
Thanks Doodle Bob for your remark.

11. Aug 8, 2008

### tiny-tim

Hi lemma28!
Or more simply as the intersection of the bisector of F1PF2 with the circle … since equal angles are subtended by equal arcs or chords, that proves that VF1 = VF2.

I must admit that I took one look at the diagram today, and automatically assumed that that was how it had been defined, without checking.

(btw, I had earlier tried to use that bisector to define N, and to then proceed without using the circle at all, and eventually had to admit that your idea of using the circle was best, since it's a clever way of proving that the other relevant angles are equal. )

Though I suspect your original proof works for either choice of V anyway, with the other choice just needing a larger piece of paper!


text:
In this note, we are interested in the following problem: given a chessboard and a set of dominoes, is it possible to tile our chessboard using the dominoes we have? For example, let $C(m,n)$ be the following chessboard

and let $D_2$ denote the set of the “true” dominoes of size 2, that is

Is it possible to tile $C(m,n)$ using only $D_2$? Surprisingly, this problem can be solved thanks to the van Kampen diagrams used in group theory.

Let $\Pi = \langle g_1,\ldots, g_n \mid r_1= \cdots = r_m=1 \rangle$ be a group presentation. A van Kampen diagram $\Gamma \subset \mathbb{R}^2$, relative to $\Pi$, is a finite connected planar graph whose edges are oriented and labelled by the generators $g_1, \ldots, g_n$, such that the word labelling the boundary of any cell of $\Gamma$ (ie., any bounded component of $\mathbb{R}^2 \backslash \Gamma$) is a relator $r_1, \ldots, r_m$. For convenience, we call the word labelling the boundary of the unbounded component of $\mathbb{R}^2 \backslash \Gamma$ the boundary word of $\Gamma$.

For example, if we consider the presentation $\Pi = \langle x,y \mid x^2=y^2, \ y=xyx \rangle$, the following graph defines a van Kampen diagram (relative to $\Pi$) whose boundary word is $x^4$.

Theorem 1: Let $G$ be a group with a presentation $\Pi = \langle \Sigma \mid \mathcal{R} \rangle$. A given word $w$ written over $\Sigma$ is equal to $1$ in $G$ if and only there exists a van Kampen diagram $\Gamma$, relative to $\Pi$, whose boundary word is $w$.

Proof. Suppose that there exists a van Kampen diagram $\Gamma$ whose boundary word is $w$. We easily prove that $w=1$ in $G$ by induction on the number of cells of $\Gamma$. If $\Gamma$ has just one cell, then its boundary word is a relator of $\Pi$ by definition, so that that $w=1$ trivially. If $\Gamma$ has at least two cells, then we can write $w=uv$ where $u$ corresponds to the intersection between the boundary path of $\Gamma$ and the boundary of one of its cells $c$. Let $uz=1$ be the relator associated to $c$. Removing $c$ to $\Gamma$, we obtain a new van Kampen diagram whose boundary word is $z^{-1}v$. By our induction hypothesis, we conclude that

$w=uv=z^{-1}v=1$

in our group $G$.

Conversely, suppose that $w=1$ in $G$. Thus, as an element of the free group $F(\Sigma)$, the word $w$ belongs to the normal subgroup generated by $\mathcal{R}$, so we can write

$w = g_1r_1^{\pm 1} g_1^{-1} \cdot g_2r_2^{\pm 1} g_2^{-1} \cdots g_k r_k^{\pm 1} g_k^{-1}$

for some $r_j \in \mathcal{R}$ and $g_j \in F(\Sigma)$. Now consider the following van Kampen diagram:

Clearly, it defines a van Kampen diagram whose boundary word is $w$. $\square$

Now, we associate a group presentation to any set of dominoes $D$. Let $x,y$ be two fixed letters. To any dominoe $d \in D$, we define a word $w(d)$ written over $\{ x^{\pm 1}, y^{\pm 1} \}$ following the rule illustrated by

Let $\Pi(D)$ denote the group presentation $\langle x,y \mid w(d)=1, d \in D \rangle$ and $G(D)$ the underlying group. For example,

$\Pi(D_2)= \langle x,y \mid [x,y^2]=[y,x^2]=1 \rangle$

with the notation $[a,b]=aba^{-1}b^{-1}$. Now, we can notice that a tiling of some chessboard using $D$ produces a van Kampen diagram relative to $\Pi(D)$. For example, the tiling

of $C(5,2)$ using $D_2$ produces the van Kampen diagram

Thus, if we defined the boundary word of a chessboard as the word obtained by following the same rule as the one used for the dominoes, we deduce from Theorem 1:

Theorem 2: If a chessboard can be tiled by a set of dominoes $D$, then its boundary word is equal to $1$ in the group $G(D)$.

As a concrete application, we are now able to solve our initial problem:

Theorem 3: $C(m,n)$ can be tiled by $D_2$ if and only if $m$ and $n$ don’t have the same parity.

$\Pi(D_2)= \langle x,y \mid [x,y^2]=[x^2,y]=1 \rangle$.

The boundary path of $C(m,n)$ is

$w(m,n) = y^{n-1}xyx^{m-1}y^{-n+1}x^{-1}y^{-1}x^{-m+1}$.

Notice that $x^2$ and $y^2$ belongs to the center of $G(D_2)$. Thus, if $m$ and $n$ are both odd,

$w(m,n)= xyx^{-1}y^{-1}=[x,y]$,

and if they are both even,

$w(m,n)=y^{-1}xyx^{-1}yx^{-1}y^{-1}x=[y^{-1},x][y,x^{-1}]$.

Now we want to prove that $[x,y] \neq 1$ and $[y^{-1},x][y,x^{-1}] \neq 1$ in $G(D_2)$. For this purpose, let us consider the morphism

$\varphi : G(D_2) \twoheadrightarrow \langle x,y \mid x^2=y^2=1 \rangle$

onto the infinite dihedral group $D_{\infty}$. It is clear that $[x,y] \neq 1$ and

$[y^{-1},x][y,x^{-1}]=(yx)^4 \neq 1$

in $D_{\infty}$. Therefore, these words are not equal to $1$ in $G(D_2)$ as well. According to Theorem 2, we conclude that $C(m,n)$ cannot be tiled by $D_2$ if $m$ and $n$ have the same parity.

Conversely, suppose without loss of generality that $m$ is even and $n$ odd. Now, the particular tiling

clearly yields a tiling of $C(m,n)$ by $D_2$. $\square$

In fact, Theorem 3 can be deduced from the following tricky argument: First, it is obvious that if $C(m,n)$ can be tiled by $D_2$ then $C(m,n)$ must have an odd number of squares. Thus, if $m$ and $n$ are both odd, $C(m,n)$ cannot be tiled by $D_2$. Secondly, if we color the squares of $C(m,n)$ in black and white exactly as a true chessboard and if $m,n$ are both even, then $C(m,n)$ is obtained from a $m \times n$ chessboard by removing two squares of the same color, say two black squares. But, if we color the dominoes of $D_2$ so that they contain a square of each color, we deduce that, if $C(m,n)$ could be tiled by $D_2$, it would contain the same number of black and white squares. Therefore, we conclude that $C(m,n)$ cannot be tiled by $D_2$ if $m$ and $n$ are both even.

In the same way, if we denote by $K(m,n)$ the chessboard

then we can prove:

Theorem 4: $K(m,n)$ can be tiled by $D_2$ if and only if $m$ or $n$ is odd.

However, the argument using van Kampen diagrams can be adapted successfully to other tiling problems, where our previous trick does not work any more. Let $D_3$ denote the following set of dominoes:

Then, the associated group presentation is

$\Pi (D_3) = \langle x,y \mid [x,y^3]=[x^3,y]=1 \rangle$.

Following the proof of Theorem 3, we get:

Theorem 5: $K(m,n)$ can be tiled by $D_3$ if and only if $m=n=2 \ \text{mod} \ 3$.

The proof is left as an exercice.

This point of view was introduced by Conway and Lagarias, in their article Tiling with polyominoes and combinatorial group theory, and was also exploited by Thurston in Conway’s tiling groups. See also Hitchman’s article, The topology of tile invariants, and references therein for more information.


text:
# Shortcut for inserting matching \end{something} in VIM

I remember some time ago I always used a shortcut in VIM that allowed me to type this (<X> is indication of the location of the cursor):

... rest of document ...

\begin{<ENVIRONMENT>}
some text
more text
<X>

... document continues ....

and then I would key in some shortcut at the location of <X> and the matching \end{<ENVIRONMENT>} would be inserted:

... rest of document ...

\begin{<ENVIRONMENT>}
some text
more text
\end{<ENVIRONMENT>}

... document continues ....

Here, of course, <ENVIRONMENT> can be anything, like align or itemize.

I know about VIM's <C-x> + <C-p>, <C-x + <C-n> and <C-x> + <C-o>. The last one works very well for closing HTML tags, but it doesn't work in LaTeX.

I forgot what it was and I hope someone here knows it. It might have been a plugin but I don't think so. If anyone can tell me what it is, I will be forever grateful.

EDIT: Of course I am using LaTeX-Suite (or vim-latex, wish they'd stick to a name).

• I'm not referring to that, but nice to know, it's basically as useful as what I'm looking for, thank you! Jul 4 '12 at 19:04
• Perhaps you are thinking of the closeb script. Jul 4 '12 at 22:17
• Tested the closeb script, works very well. Feb 4 '14 at 0:22

## 6 Answers

When using VimTeX, you can type ]] in insert mode to close the current environment.

Something that is pretty much as useful is to type nameofenvironment then F5 which gives

\begin{nameofenvironment}

\end{nameofenvironment}<++>

and puts the cursor in the body of the environment. You can leave the environment easily by pressing CTRL J

• Thank you, this works. Is there any shortkey that also autocomplete the nameofenviron to nameofenvironment? Oct 27 '17 at 4:19
• Does not work for Neovim without configuration. At least it should be pointed out what should be configured for the shortcut F5 to work. Mar 23 '19 at 8:46
• This does not work for me.
– Soap
May 9 '20 at 14:17

All of the solutions above close an environment for you after you have typed \begin{<environment name>}... but I prefer to just type the name of the environment and then have vim fill in the \begin{...} and \end{...} tags for me. To do this I have the following lines in ~/.vim/ftplugin/tex.vim:

" put \begin{} \end{} tags tags around the current word
map  <C-B>      YpkI\begin{<ESC>A}<ESC>jI\end{<ESC>A}<esc>kA
map! <C-B> <ESC>YpkI\begin{<ESC>A}<ESC>jI\end{<ESC>A}<esc>kA

This allows me to type a word at the start of a line, such as align, itemize, theorem, ... then control-B changes it into the mark-up for the environment. For example, if I type align* followed by control-B then vim translates this into:

\begin{align*}
\end{align*}

A useful feature of my macro is that it preserves indentation of the environment name.

EDIT As pointed out is the comments, this is a standard part of vim-latex.

• cmhughes' answer describes the same function as this, which is built into vim-latex Aug 16 '16 at 11:30
• @DaiBowen Thanks. I missed this. I won't delete the post as it provides the details of one way to do this without using vim-latex, which I at least don't use.
– user30471
Aug 16 '16 at 11:54

You search for the closeb plugin. It does exactly what you describe.

If you enable it and press Ctrl+_ in insert mode, it closes the current environment.

put in your vimrc:

inoremap \end <><Esc>?begin<Enter>Vy/<><Enter>pfbcwend<Esc>kddo

How it works: innoremap create a mapping keys in insert mode when you insert \end. put a "<>" string, find(/) the word "begin" and copy the entire line (Vy), then back to the "<>" (\<>) and paste the line (p), find the first b in the line and change the word (cw) for "end" after that delete the "<>" string and put in insert mode.

So: The text must be like that:

\begin{something}
some texot o comands

In diferents lines. It must work.

If you use the following incantation

map! <C-A> <ESC>"ayy"aPi\begin{<ESC>Ea}<ESC><CR>i\end{<ESC>Ea}<ESC>O

then, using A while in insert mode, the line

abc

turns into

\begin{abc}

\end{abc}

with the cursor in between (and still in insert mode).

• The characters got interpreted and not printed (:-): Jun 29 '19 at 20:17
• :map! <C-A> <ESC>"ayy"aPi\begin{<ESC>Ea}<ESC><CR>i\end{<ESC>Ea}<ESC>O Jun 29 '19 at 20:22


text:
# Free vibration analysis of laminated plate/shell structures based on FSDT with a stabilized nodal-integrated quadrilateral element

Nguyen-Van, H. and Mai-Duy, N. and Tran-Cong, T. (2008) Free vibration analysis of laminated plate/shell structures based on FSDT with a stabilized nodal-integrated quadrilateral element. Journal of Sound and Vibration, 313 (1-2). pp. 205-223. ISSN 0022-460X

 Preview
PDF (Accepted Version)
Nguyen-Van_Mai-Duy_Tran-Cong_JSV_v313n1-2_AM.pdf

This paper reports numerical analyses of free vibration of laminated composite plate/ shell structures of various shapes, span-to-thickness ratios, boundary conditions and lay-up sequences. The method is based on a novel four-node quadrilateral element, namely MISQ20, within the framework of the first-order shear deformation theory (FSDT). The element is built by incorporating a strain smoothing method into the bilinear four-node quadrilateral finite element where the strain smoothing operation is based on mesh-free conforming nodal integration. The bending and membrane stiffness matrices are based on the boundaries of smoothing cells while the shear term is evaluated by $2\times2$ Gauss quadrature. Through several numerical examples, the capability, efficiency and simplicity of the element are demonstrated. Convergence studies and comparison with other existing solutions in the literature suggest that the present element is robust, computationally inexpensive and free of locking.


text:
# CARlasso

Published in CRAN, 2021

R implementation of

Shen, Yunyi, and Claudia Solis-Lemus. “Bayesian Conditional Auto-Regressive LASSO Models to Learn Sparse Networks with Predictors.” arXiv preprint arXiv:2012.08397 (2020).

Documentations can be found: https://yunyishen.ml/CAR-LASSO/

The package is on CRAN, to install it, use:

install.packages("CARlasso")

To install from Github:

devtools::install_github("YunyiShen/CAR-LASSO")

Example run, we generated data from a 5-node AR1 model with each node has a specific treatment then use CAR-ALASSO to reconstruct such network and plot the result.

set.seed(42)
dt <- simu_AR1(n=100,k=5, rho=0.7)
car_res <- CARlasso(y1+y2+y3+y4+y5~x1+x2+x3+x4+x5, data = dt, adaptive = TRUE)
plot(car_res,tol = 0.05)
# with horseshoe inference
car_res <- horseshoe(car_res)
plot(car_res)

To run a reduced version of the analysis on human gut microbiome (with less predictors and responses), try:

gut_res <- CARlasso(Alistipes+Bacteroides+
Eubacterium+Parabacteroides+all_others~
BMI+Age+Gender+Stratum,
adaptive = TRUE, n_iter = 5000,
n_burn_in = 1000, thin_by = 10)
# horseshoe will take a while, as it's currently implemented in R rather than C++
gut_res <- horseshoe(gut_res)
plot(gut_res)

It might take a little while due to the sampling process of the latent normal variable

Though we don’t recommend treating compositional data as counts, as a illustration, we can run the counting model:

gut_res <- CARlasso(Alistipes+Bacteroides+
Eubacterium+Parabacteroides+all_others~
BMI+Age+Gender+Stratum,
r_beta = 0.1, # default sometimes cause singularity in Poisson model due to exponential transformation, slightly change can fix it.
n_iter = 5000,
n_burn_in = 1000, thin_by = 10)
# horseshoe will take a while, as it's currently implemented in R rather than C++
gut_res <- horseshoe(gut_res)
plot(gut_res)


text:
Calc6.23

Discuss the convergence or divergence of the series $\sum _{{n=1}}^{{\infty }}{\frac {n+4}{(n+2)(n+1)}}$.
Another comparison test we can use is the limit comparison test. To do this test, we compare two series by looking at the ratio of their terms as $n\rightarrow \infty$. If this term is a finite positive number, then both series either converge or diverge.
$\lim _{{n\rightarrow \infty }}{\frac {{\frac {1}{n}}}{{\frac {n+4}{(n+2)(n+1)}}}}=\lim _{{n\rightarrow \infty }}{\frac {(n+2)(n+1)}{n(n+4)}}=1\,$
Since this is a finite, positive number, then both $\sum _{{n=1}}^{{\infty }}{\frac {n+4}{(n+2)(n+1)}}$ and $\sum _{{n=1}}^{{\infty }}{\frac {1}{n}}$ converge or both diverge. We know that $\sum _{{n=1}}^{{\infty }}{\frac {1}{n}}$ diverges so our series also diverges.


text:
2 fixed definition of $P_N$; added 18 characters in body

So, when people say, "the moduli problem of classifying elliptic curves over $\mathbb{C}$ with level $N$ structure", there are usually two associated functors I've seen:

1. $P_N : \textbf{Ell}\rightarrow\textbf{Sets}$, where $\textbf{Ell}$ is the category of elliptic curves $E\rightarrow S$ over $S$ and morphisms are cartesian squares, and

$P_N(E/S) = \text{set of isomorphisms } \alpha : E[N]\rightarrow \mathbb{Z}/N\mathbb{Z} (\mathbb{Z}/N\mathbb{Z})^2 \text{ of determinant 1}$

2. $F_N : \textbf{Sch}\rightarrow\textbf{Sets}$, where

$F_N(S) = \text{set of isomorphism classes of pairs } (E/S,\alpha) \text{ with } \alpha\in P_N(E/S)$

I apologize for the length of this post, but this has been terribly confusing for me.

Alright, so I know that for $N\ge 3$, both functors are representable by the modular scheme $Y(N) := \Gamma(N)\backslash\mathcal{H}$ which are fine moduli schemes, and that's got something to do with the fact that there are no automorphisms of elliptic curves $E/S$ fixing any $\alpha\in P_N(E/S)$.

However, in the case $N = 1,2$, the modular curve $Y(2) := \Gamma(2)\backslash\mathcal{H}$ only gives you a coarse moduli scheme.

How should I think of the relation between the two above functors? In a way, a representing object $E/S$ for $P_N$ gives you both the universal elliptic curve $E$ and the base moduli scheme $S$ in one fell swoop. However, the functor $P_N$ doesn't seem like a naturally phrased moduli problem, since being able to represent $P_N$ just says:

"there is an elliptic curve $E/S$ such that for any other elliptic curve $E'/S'$, a level structure on $E'/S'$ is equivalent to a morphism $S'\rightarrow S$ such that $E'\cong E\times_S S'$."

(On the other hand, it seems that after thinking about it for a bit, it seems you can show that $P_N$ representable $\Longrightarrow$ $F_N$ representable)

On the other hand, the functor $F_N$ is much more natural, in that a representing object for $F_N$ much more clearly parametrizes elliptic curves with level structure. However, Peter Bruin's article (http://user.math.uzh.ch/bruin/moduli.pdf) and Katz/Mazur's book (specifically thm's 3.6 and 4.7) both seem to imply that if $P_N$ is not rigid (eg, $N = 1,2$), then even if $F_N$ is representable by an object $M$, the object $M$ might not carry a universal family. I am further confused by the fact that wikipedia (in the section on Fine Moduli Spaces) says that the universal family exists and must correspond to $\text{id}_M\in\text{Hom}(M,M)$.

I'm assuming wikipedia is wrong.

If wikipedia is wrong, then in the case $N = 2$, is the functor $F_2$ representable? If it is, is $\Gamma(2)\backslash\mathcal{H}$ the representing object?

When people talk of the moduli problem of classifying elliptic curves with full level $N$ structure, which functor are they referring to?

...onto stacks...

For $N = 1,2$, there is no fine moduli scheme, and hence at least $P_N$ is not representable. However, is there a fine moduli stack? (does that mean anything?)

In general, would I be correct in saying that a stack for a moduli problem is basically just the moduli functor itself? Can you make this more precise? (though I guess you'd have to replace "isomorphism classes of..." with the objects themselves)

Are there meaningful moduli problems that aren't stacks?

Thanks for bearing with me

• will
1

# what exactly is the moduli functor for classifying elliptic curves with (full) level N structure?

So, when people say, "the moduli problem of classifying elliptic curves with level $N$ structure", there are usually two associated functors I've seen:

1. $P_N : \textbf{Ell}\rightarrow\textbf{Sets}$, where $\textbf{Ell}$ is the category of elliptic curves $E\rightarrow S$ over $S$ and morphisms are cartesian squares, and

$P_N(E/S) = \text{set of isomorphisms } \alpha : E[N]\rightarrow \mathbb{Z}/N\mathbb{Z} \text{ of determinant 1}$

2. $F_N : \textbf{Sch}\rightarrow\textbf{Sets}$, where

$F_N(S) = \text{set of isomorphism classes of pairs } (E/S,\alpha) \text{ with } \alpha\in P_N(E/S)$

I apologize for the length of this post, but this has been terribly confusing for me.

Alright, so I know that for $N\ge 3$, both functors are representable by the modular scheme $Y(N) := \Gamma(N)\backslash\mathcal{H}$ which are fine moduli schemes, and that's got something to do with the fact that there are no automorphisms of elliptic curves $E/S$ fixing any $\alpha\in P_N(E/S)$.

However, in the case $N = 1,2$, the modular curve $Y(2) := \Gamma(2)\backslash\mathcal{H}$ only gives you a coarse moduli scheme.

How should I think of the relation between the two above functors? In a way, a representing object $E/S$ for $P_N$ gives you both the universal elliptic curve $E$ and the base moduli scheme $S$ in one fell swoop. However, the functor $P_N$ doesn't seem like a naturally phrased moduli problem, since being able to represent $P_N$ just says:

"there is an elliptic curve $E/S$ such that for any other elliptic curve $E'/S'$, a level structure on $E'/S'$ is equivalent to a morphism $S'\rightarrow S$ such that $E'\cong E\times_S S'$."

(On the other hand, it seems that after thinking about it for a bit, you can show that $P_N$ representable $\Longrightarrow$ $F_N$ representable)

On the other hand, the functor $F_N$ is much more natural, in that a representing object for $F_N$ much more clearly parametrizes elliptic curves with level structure. However, Peter Bruin's article (http://user.math.uzh.ch/bruin/moduli.pdf) and Katz/Mazur's book (specifically thm's 3.6 and 4.7) both seem to imply that if $P_N$ is not rigid (eg, $N = 1,2$), then even if $F_N$ is representable by an object $M$, the object $M$ might not carry a universal family. I am further confused by the fact that wikipedia (in the section on Fine Moduli Spaces) says that the universal family exists and must correspond to $\text{id}_M\in\text{Hom}(M,M)$.

I'm assuming wikipedia is wrong.

If wikipedia is wrong, then in the case $N = 2$, is the functor $F_2$ representable? If it is, is $\Gamma(2)\backslash\mathcal{H}$ the representing object?

When people talk of the moduli problem of classifying elliptic curves with full level $N$ structure, which functor are they referring to?

...onto stacks...

For $N = 1,2$, there is no fine moduli scheme, and hence at least $P_N$ is not representable. However, is there a fine moduli stack? (does that mean anything?)

In general, would I be correct in saying that a stack for a moduli problem is basically just the moduli functor itself? Can you make this more precise? (though I guess you'd have to replace "isomorphism classes of..." with the objects themselves)

Are there meaningful moduli problems that aren't stacks?

Thanks for bearing with me

• will


text:
# Zero Sums in a $q$-Deformation Remain Zero for $q=1$

Looking at previous question, I begun to think and came upon the following more solid question: Let $SU_q(N)$ be the usual quantized coordinated algebra of $SU(N)$. Now consider, for some fixed value of $q$, the element $$v := \sum_{i=1}^k \lambda^i(q) v_i \in SU_q(N),$$ where each $\lambda^i(q)$ is some polynomial function in $q$, and each $v_i$ is a product of some subset of the generators of $SU_q(N)$. It has the obvious corresponding element in ${\cal O}(SU(N))$: $$v' := \sum_{i=1}^k \lambda^i(1) v_i.$$ It seems to be for me obvious that if $v =0$, then we must necessarily to have $v' = 0$. However, I can't see how to prove this with care?

Edit: I should say transcendental $q$ to make this question non-trivial

-

I think so (unless i missed a subtility), at least for N=2 (and the argument is probably similar in general). Namely, recall that there's a basis of the vector space $SL_q(2)$ consisting of $\{a^nb^mc^r,b^mc^rd^s,m,r,s\in N_o, n\in N\}$. So you can, wlog, assume that the $v_i's$ are of this form in your expression in $SL_q(2)$, with coefficient that will be polynomials in $q$. If you assume that the expression $v(q)$ vanishes for generic $q$'s, then these polynomials must be identically 0 (because you assumed they are polynomials, not any function).