text:
# Homotopy of algebroid paths

In Lectures on Integrability of Lie Brackets, Proposition 3.15, Fernandes and Crainic motivate what it means to say that two paths on an algebroid are homotopic. I spent quite some time trying to grasp the proof, but it still eludes me. I will write down some definitions as to make this question self-contained, but everything I write down can be looked up in their document.

• For an algebroid $$\pi : A \to M$$ with bracket $$[\cdot,\cdot]$$ and anchor $$\rho$$, they define the notion of an $$A$$-path as a $$C^1$$-path $$a : [0,1] \to A$$ such that $$\rho \circ a = \frac{d \gamma}{dt}$$, where $$\gamma = \pi \circ a : [0,1] \to M$$ is the corresponding base-path.

• A variation of $$A$$-paths is a family of $$A$$-paths $$(a_\epsilon)_{\epsilon \in [0,1]}$$ which is $$C^2$$ in $$\epsilon$$ and so that the end points of the base paths are fixed.

• Lastly, if $$\alpha(s,\cdot) \in \Gamma(A)$$, $$s \in [0,1]$$ are time-dependent sections, their flow is the the unique maximal family of $$A$$-automorphisms $$\{\phi^{t,s}_\alpha\}_{t,s}$$ with $$\phi^{t,s}_\alpha \phi^{s,u}_\alpha = \phi^{t,u}_\alpha$$, $$\phi^{t,t}_\alpha = \text{id}$$ and \begin{align*} \frac{d}{dt} \Big \vert_{t = s} \left(\phi^{t,s}_\alpha\right)^\star (\beta) = [\alpha^s,\beta] \quad \forall \beta \in \Gamma(A) \end{align*} with $$(\phi^{t,s}_\alpha)^\star (\beta)(x) = \phi^{s,t}_\alpha \beta(\phi^{t,s}_{\rho(\alpha)}(x))$$, where $$\phi^{t,s}_{\rho(\alpha)}$$ is the usual time-dependent flow on $$M$$.

With these definitions in place, in Proposition 3.15 they are given a variation of $$A$$-paths $$(a_\epsilon)$$ and a time-dependent section $$\xi_\epsilon$$ so that $$\xi_\epsilon(t,\gamma(t)) = a_\epsilon(t)$$. In the proof, they define $$\eta (\epsilon,t,x) := \int_0^t \phi^{t,s}_{\xi_\epsilon} \frac{d \xi_{\epsilon}}{d \epsilon} (x,\phi^{s,t}_{\rho(\xi_\epsilon)} (x)) ds \in A_x.$$

They prove that the so-defined section $$\eta(\epsilon,t,\cdot) \in \Gamma(A)$$ has the property $$\frac{d \eta}{d t } - \frac{d \xi_\epsilon}{d \epsilon} = [\eta,\xi].$$

Up to here, I think I have been able to follow the proof, but now I am lost: They define $$X := \rho(\xi), Y := \rho(\eta)$$ with the anchor of the algebroid, and say these fulfil a similar equation, presumably $$\frac{dY}{dt} - \frac{dX}{d\epsilon} = [Y,X],$$ simply by applying the anchor map.
They then claim that because $$X(\epsilon,t,\gamma_\epsilon(t)) = \rho( a_\epsilon(t)) = \frac{d \gamma_\epsilon}{d t}(t)$$ (true as property of an $$A$$-path), it follows that $$Y(\epsilon,t,\gamma_\epsilon(t)) = \frac{d \gamma_\epsilon}{d \epsilon}$$. I have no idea how that follows and have tried a couple different unsuccessful approaches. It seems like this has something to do with $$[Y,X](t,\epsilon,\gamma_\epsilon(t)) = 0$$, which may well be true, but I cannot see a good reason. Apologies for this long, convoluted and very specific question, but I am kinda hoping that there is an absolute algebroid-pro running around who can instantly see the solution to my troubles. Thank you!


text:
# NAG Library Routine Document

## 1Purpose

s21daf returns the value of the general elliptic integral of the second kind $F\left(z,{k}^{\prime },a,b\right)$ for a complex argument $z$, via the function name.

## 2Specification

Fortran Interface
 Function s21daf ( z, akp, a, b,
 Complex (Kind=nag_wp) :: s21daf Integer, Intent (Inout) :: ifail Real (Kind=nag_wp), Intent (In) :: akp, a, b Complex (Kind=nag_wp), Intent (In) :: z
#include nagmk26.h
 Complex s21daf_ (const Complex *z, const double *akp, const double *a, const double *b, Integer *ifail)

## 3Description

s21daf evaluates an approximation to the general elliptic integral of the second kind $F\left(z,{k}^{\prime },a,b\right)$ given by
 $Fz,k′,a,b=∫0za+bζ2 1+ζ21+ζ21+k′2ζ2 dζ,$
where $a$ and $b$ are real arguments, $z$ is a complex argument whose real part is non-negative and ${k}^{\prime }$ is a real argument (the complementary modulus). The evaluation of $F$ is based on the Gauss transformation. Further details, in particular for the conformal mapping provided by $F$, can be found in Bulirsch (1960).
Special values include
 $F z, k ′ ,1,1 = ∫ 0 z d ζ 1 + ζ 2 1 + k′ 2 ζ 2 ,$
or ${F}_{1}\left(z,{k}^{\prime }\right)$ (the elliptic integral of the first kind) and
 $Fz,k′,1,k′2=∫0z1+k′2ζ2 1+ζ21+ζ2 dζ,$
or ${F}_{2}\left(z,{k}^{\prime }\right)$ (the elliptic integral of the second kind). Note that the values of ${F}_{1}\left(z,{k}^{\prime }\right)$ and ${F}_{2}\left(z,{k}^{\prime }\right)$ are equal to ${\mathrm{tan}}^{-1}\left(z\right)$ in the trivial case ${k}^{\prime }=1$.
s21daf is derived from an Algol 60 procedure given by Bulirsch (1960). Constraints are placed on the values of $z$ and ${k}^{\prime }$ in order to avoid the possibility of machine overflow.

## 4References

Bulirsch R (1960) Numerical calculation of elliptic integrals and elliptic functions Numer. Math. 7 76–90

## 5Arguments

1:     $\mathbf{z}$ – Complex (Kind=nag_wp)Input
On entry: the argument $z$ of the function.
Constraints:
• $0.0\le \mathrm{Re}\left({\mathbf{z}}\right)\le \lambda$;
• $\mathrm{abs}\left(\mathrm{Im}\left({\mathbf{z}}\right)\right)\le \lambda$, where ${\lambda }^{6}=1/{\mathbf{x02amf}}$.
2:     $\mathbf{akp}$ – Real (Kind=nag_wp)Input
On entry: the argument ${k}^{\prime }$ of the function.
Constraint: $\mathrm{abs}\left({\mathbf{akp}}\right)\le \lambda$.
3:     $\mathbf{a}$ – Real (Kind=nag_wp)Input
On entry: the argument $a$ of the function.
4:     $\mathbf{b}$ – Real (Kind=nag_wp)Input
On entry: the argument $b$ of the function.
5:     $\mathbf{ifail}$ – IntegerInput/Output
On entry: ifail must be set to $0$, $-1\text{ or }1$. If you are unfamiliar with this argument you should refer to Section 3.4 in How to Use the NAG Library and its Documentation for details.
For environments where it might be inappropriate to halt program execution when an error is detected, the value $-1\text{ or }1$ is recommended. If the output of error messages is undesirable, then the value $1$ is recommended. Otherwise, if you are not familiar with this argument, the recommended value is $0$. When the value $-\mathbf{1}\text{ or }\mathbf{1}$ is used it is essential to test the value of ifail on exit.
On exit: ${\mathbf{ifail}}={\mathbf{0}}$ unless the routine detects an error or a warning has been flagged (see Section 6).

## 6Error Indicators and Warnings

If on entry ${\mathbf{ifail}}=0$ or $-1$, explanatory error messages are output on the current error message unit (as defined by x04aaf).
Errors or warnings detected by the routine:
${\mathbf{ifail}}=1$
 On entry, $\mathrm{Re}\left({\mathbf{z}}\right)<0.0$, or $\mathrm{Re}\left({\mathbf{z}}\right)>\lambda$, or $\left|\mathrm{Im}\left({\mathbf{z}}\right)\right|>\lambda$, or $\left|{\mathbf{akp}}\right|>\lambda$, where ${\lambda }^{6}=1/{\mathbf{x02amf}}$.
${\mathbf{ifail}}=2$
The iterative procedure used to evaluate the integral has failed to converge. The result is returned as zero.
${\mathbf{ifail}}=-99$
See Section 3.9 in How to Use the NAG Library and its Documentation for further information.
${\mathbf{ifail}}=-399$
Your licence key may have expired or may not have been installed correctly.
See Section 3.8 in How to Use the NAG Library and its Documentation for further information.
${\mathbf{ifail}}=-999$
Dynamic memory allocation failed.
See Section 3.7 in How to Use the NAG Library and its Documentation for further information.

## 7Accuracy

In principle the routine is capable of achieving full relative precision in the computed values. However, the accuracy obtainable in practice depends on the accuracy of the standard elementary functions such as atan2 and log.

## 8Parallelism and Performance

s21daf is not threaded in any implementation.

None.

## 10Example

This example evaluates the elliptic integral of the first kind ${F}_{1}\left(z,{k}^{\prime }\right)$ given by
 $F1z,k′=∫0zdζ 1+ζ21+k′2ζ2 ,$
where $z=1.2+3.7i$ and ${k}^{\prime }=0.5$, and prints the results.

### 10.1Program Text

Program Text (s21dafe.f90)

### 10.2Program Data

Program Data (s21dafe.d)

### 10.3Program Results

Program Results (s21dafe.r)

© The Numerical Algorithms Group Ltd, Oxford, UK. 2017

format: plain


text:
Role of solvents in the electronic transport properties of single-molecule junctions

2016
SFB 767 - TP C02
Journal article
Published
Published in
Beilstein Journal of Nanotechnology ; 7 (2016). - pp. 1055-1067. - eISSN 2190-4286
Abstract
We report on an experimental study of the charge transport through tunnel gaps formed by adjustable gold electrodes immersed into different solvents that are commonly used in the field of molecular electronics (ethanol, toluene, mesitylene, 1,2,4-trichlorobenzene, isopropanol, toluene/tetrahydrofuran mixtures) for the study of single-molecule contacts of functional molecules. We present measurements of the conductance as a function of gap width, conductance histograms as well as current–voltage characteristics of narrow gaps and discuss them in terms of the Simmons model, which is the standard model for describing transport via tunnel barriers, and the resonant single-level model, often applied to single-molecule junctions. One of our conclusions is that stable junctions may form from solvents as well and that both conductance–distance traces and current–voltage characteristics have to be studied to distinguish between contacts of solvent molecules and of molecules under study.
530 Physics
Keywords
electrochemical environment; mechanically controllable break junction; molecular electronics; polar solvent; single-molecule junctions
Cite This
ISO 690LUKA-GUTH, Katharina, Sebastian HAMBSCH, Andreas BLOCH, Philipp EHRENREICH, Bernd M. BRIECHLE, Filip KILIBARDA, Torsten SENDLER, Dmytro SYSOIEV, Thomas HUHN, Artur ERBE, Elke SCHEER, 2016. Role of solvents in the electronic transport properties of single-molecule junctions. In: Beilstein Journal of Nanotechnology. 7, pp. 1055-1067. eISSN 2190-4286. Available under: doi: 10.3762/bjnano.7.99
BibTex
@article{LukaGuth2016solve-34913,
year={2016},
doi={10.3762/bjnano.7.99},
title={Role of solvents in the electronic transport properties of single-molecule junctions},
volume={7},
journal={Beilstein Journal of Nanotechnology},
pages={1055--1067},
author={Luka-Guth, Katharina and Hambsch, Sebastian and Bloch, Andreas and Ehrenreich, Philipp and Briechle, Bernd M. and Kilibarda, Filip and Sendler, Torsten and Sysoiev, Dmytro and Huhn, Thomas and Erbe, Artur and Scheer, Elke},
note={This work was supported by the Deutsche Forschungsgemeinschaft through SFB767 and by the Research Network “Functional Nanostructures” of the Baden-Württemberg Foundation.}
}

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<dcterms:abstract xml:lang="eng">We report on an experimental study of the charge transport through tunnel gaps formed by adjustable gold electrodes immersed into different solvents that are commonly used in the field of molecular electronics (ethanol, toluene, mesitylene, 1,2,4-trichlorobenzene, isopropanol, toluene/tetrahydrofuran mixtures) for the study of single-molecule contacts of functional molecules. We present measurements of the conductance as a function of gap width, conductance histograms as well as current–voltage characteristics of narrow gaps and discuss them in terms of the Simmons model, which is the standard model for describing transport via tunnel barriers, and the resonant single-level model, often applied to single-molecule junctions. One of our conclusions is that stable junctions may form from solvents as well and that both conductance–distance traces and current–voltage characteristics have to be studied to distinguish between contacts of solvent molecules and of molecules under study.</dcterms:abstract>
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Comment on publication
This work was supported by the Deutsche Forschungsgemeinschaft through SFB767 and by the Research Network “Functional Nanostructures” of the Baden-Württemberg Foundation.
Yes


text:
# Properties

 Label 1170.2.bs.e Level $1170$ Weight $2$ Character orbit 1170.bs Analytic conductor $9.342$ Analytic rank $0$ Dimension $4$ CM no Inner twists $2$

# Related objects

## Newspace parameters

 Level: $$N$$ $$=$$ $$1170 = 2 \cdot 3^{2} \cdot 5 \cdot 13$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 1170.bs (of order $$6$$, degree $$2$$, minimal)

## Newform invariants

 Self dual: no Analytic conductor: $$9.34249703649$$ Analytic rank: $$0$$ Dimension: $$4$$ Relative dimension: $$2$$ over $$\Q(\zeta_{6})$$ Coefficient field: $$\Q(\zeta_{12})$$ Defining polynomial: $$x^{4} - x^{2} + 1$$ Coefficient ring: $$\Z[a_1, a_2]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 390) Sato-Tate group: $\mathrm{SU}(2)[C_{6}]$

## $q$-expansion

Coefficients of the $$q$$-expansion are expressed in terms of a primitive root of unity $$\zeta_{12}$$. We also show the integral $$q$$-expansion of the trace form.

 $$f(q)$$ $$=$$ $$q + ( -\zeta_{12} + \zeta_{12}^{3} ) q^{2} + ( 1 - \zeta_{12}^{2} ) q^{4} -\zeta_{12}^{3} q^{5} + 2 \zeta_{12} q^{7} + \zeta_{12}^{3} q^{8} +O(q^{10})$$ $$q + ( -\zeta_{12} + \zeta_{12}^{3} ) q^{2} + ( 1 - \zeta_{12}^{2} ) q^{4} -\zeta_{12}^{3} q^{5} + 2 \zeta_{12} q^{7} + \zeta_{12}^{3} q^{8} + \zeta_{12}^{2} q^{10} + ( -4 + 3 \zeta_{12} + 2 \zeta_{12}^{2} - 3 \zeta_{12}^{3} ) q^{11} + ( -1 + 4 \zeta_{12}^{2} ) q^{13} -2 q^{14} -\zeta_{12}^{2} q^{16} + ( -4 + 4 \zeta_{12}^{2} ) q^{17} + ( 2 - 4 \zeta_{12} + 2 \zeta_{12}^{2} ) q^{19} -\zeta_{12} q^{20} + ( -3 + 2 \zeta_{12} + 3 \zeta_{12}^{2} - 4 \zeta_{12}^{3} ) q^{22} + ( -\zeta_{12} + 2 \zeta_{12}^{2} - \zeta_{12}^{3} ) q^{23} - q^{25} + ( -3 \zeta_{12} - \zeta_{12}^{3} ) q^{26} + ( 2 \zeta_{12} - 2 \zeta_{12}^{3} ) q^{28} + ( \zeta_{12} + 2 \zeta_{12}^{2} + \zeta_{12}^{3} ) q^{29} + ( -1 + 2 \zeta_{12}^{2} ) q^{31} + \zeta_{12} q^{32} -4 \zeta_{12}^{3} q^{34} + ( 2 - 2 \zeta_{12}^{2} ) q^{35} + ( 6 - 4 \zeta_{12} - 3 \zeta_{12}^{2} + 4 \zeta_{12}^{3} ) q^{37} + ( 4 - 4 \zeta_{12} + 2 \zeta_{12}^{3} ) q^{38} + q^{40} + ( 2 \zeta_{12} - 2 \zeta_{12}^{3} ) q^{41} + ( -5 + 4 \zeta_{12} + 5 \zeta_{12}^{2} - 8 \zeta_{12}^{3} ) q^{43} + ( -2 + 4 \zeta_{12}^{2} - 3 \zeta_{12}^{3} ) q^{44} + ( 1 - 2 \zeta_{12} + \zeta_{12}^{2} ) q^{46} + ( -2 + 4 \zeta_{12}^{2} + 7 \zeta_{12}^{3} ) q^{47} -3 \zeta_{12}^{2} q^{49} + ( \zeta_{12} - \zeta_{12}^{3} ) q^{50} + ( 3 + \zeta_{12}^{2} ) q^{52} + ( 6 + 8 \zeta_{12} - 4 \zeta_{12}^{3} ) q^{53} + ( 2 \zeta_{12} - 3 \zeta_{12}^{2} + 2 \zeta_{12}^{3} ) q^{55} + ( -2 + 2 \zeta_{12}^{2} ) q^{56} + ( -1 - 2 \zeta_{12} - \zeta_{12}^{2} ) q^{58} + ( -2 + 5 \zeta_{12} - 2 \zeta_{12}^{2} ) q^{59} + ( -6 \zeta_{12} + 12 \zeta_{12}^{3} ) q^{61} + ( -\zeta_{12} - \zeta_{12}^{3} ) q^{62} - q^{64} + ( 4 \zeta_{12} - 3 \zeta_{12}^{3} ) q^{65} + ( -4 + 8 \zeta_{12} + 2 \zeta_{12}^{2} - 8 \zeta_{12}^{3} ) q^{67} + 4 \zeta_{12}^{2} q^{68} + 2 \zeta_{12}^{3} q^{70} + ( 6 - 2 \zeta_{12} + 6 \zeta_{12}^{2} ) q^{71} + 2 \zeta_{12}^{3} q^{73} + ( 4 - 3 \zeta_{12} - 4 \zeta_{12}^{2} + 6 \zeta_{12}^{3} ) q^{74} + ( 4 - 4 \zeta_{12} - 2 \zeta_{12}^{2} + 4 \zeta_{12}^{3} ) q^{76} + ( 6 - 8 \zeta_{12} + 4 \zeta_{12}^{3} ) q^{77} + ( -7 + 8 \zeta_{12} - 4 \zeta_{12}^{3} ) q^{79} + ( -\zeta_{12} + \zeta_{12}^{3} ) q^{80} + ( -2 + 2 \zeta_{12}^{2} ) q^{82} + ( -4 + 8 \zeta_{12}^{2} - 2 \zeta_{12}^{3} ) q^{83} + 4 \zeta_{12} q^{85} + ( -4 + 8 \zeta_{12}^{2} - 5 \zeta_{12}^{3} ) q^{86} + ( -2 \zeta_{12} + 3 \zeta_{12}^{2} - 2 \zeta_{12}^{3} ) q^{88} + ( -4 - 4 \zeta_{12} + 2 \zeta_{12}^{2} + 4 \zeta_{12}^{3} ) q^{89} + ( -2 \zeta_{12} + 8 \zeta_{12}^{3} ) q^{91} + ( 2 - 2 \zeta_{12} + \zeta_{12}^{3} ) q^{92} + ( -2 \zeta_{12} - 7 \zeta_{12}^{2} - 2 \zeta_{12}^{3} ) q^{94} + ( -4 + 2 \zeta_{12} + 4 \zeta_{12}^{2} - 4 \zeta_{12}^{3} ) q^{95} + ( -2 - 4 \zeta_{12} - 2 \zeta_{12}^{2} ) q^{97} + 3 \zeta_{12} q^{98} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$4q + 2q^{4} + O(q^{10})$$ $$4q + 2q^{4} + 2q^{10} - 12q^{11} + 4q^{13} - 8q^{14} - 2q^{16} - 8q^{17} + 12q^{19} - 6q^{22} + 4q^{23} - 4q^{25} + 4q^{29} + 4q^{35} + 18q^{37} + 16q^{38} + 4q^{40} - 10q^{43} + 6q^{46} - 6q^{49} + 14q^{52} + 24q^{53} - 6q^{55} - 4q^{56} - 6q^{58} - 12q^{59} - 4q^{64} - 12q^{67} + 8q^{68} + 36q^{71} + 8q^{74} + 12q^{76} + 24q^{77} - 28q^{79} - 4q^{82} + 6q^{88} - 12q^{89} + 8q^{92} - 14q^{94} - 8q^{95} - 12q^{97} + O(q^{100})$$

## Character values

We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/1170\mathbb{Z}\right)^\times$$.

 $$n$$ $$911$$ $$937$$ $$1081$$ $$\chi(n)$$ $$1$$ $$1$$ $$1 - \zeta_{12}^{2}$$

## Embeddings

For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.

For more information on an embedded modular form you can click on its label.

Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
361.1
 0.866025 + 0.500000i −0.866025 − 0.500000i 0.866025 − 0.500000i −0.866025 + 0.500000i
−0.866025 + 0.500000i 0 0.500000 0.866025i 1.00000i 0 1.73205 + 1.00000i 1.00000i 0 0.500000 + 0.866025i
361.2 0.866025 0.500000i 0 0.500000 0.866025i 1.00000i 0 −1.73205 1.00000i 1.00000i 0 0.500000 + 0.866025i
901.1 −0.866025 0.500000i 0 0.500000 + 0.866025i 1.00000i 0 1.73205 1.00000i 1.00000i 0 0.500000 0.866025i
901.2 0.866025 + 0.500000i 0 0.500000 + 0.866025i 1.00000i 0 −1.73205 + 1.00000i 1.00000i 0 0.500000 0.866025i
 $$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles

## Inner twists

Char Parity Ord Mult Type
1.a even 1 1 trivial
13.e even 6 1 inner

## Twists

By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 1170.2.bs.e 4
3.b odd 2 1 390.2.bb.b 4
13.e even 6 1 inner 1170.2.bs.e 4
15.d odd 2 1 1950.2.bc.b 4
15.e even 4 1 1950.2.y.c 4
15.e even 4 1 1950.2.y.f 4
39.h odd 6 1 390.2.bb.b 4
39.h odd 6 1 5070.2.b.o 4
39.i odd 6 1 5070.2.b.o 4
39.k even 12 1 5070.2.a.y 2
39.k even 12 1 5070.2.a.bg 2
195.y odd 6 1 1950.2.bc.b 4
195.bf even 12 1 1950.2.y.c 4
195.bf even 12 1 1950.2.y.f 4

By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
390.2.bb.b 4 3.b odd 2 1
390.2.bb.b 4 39.h odd 6 1
1170.2.bs.e 4 1.a even 1 1 trivial
1170.2.bs.e 4 13.e even 6 1 inner
1950.2.y.c 4 15.e even 4 1
1950.2.y.c 4 195.bf even 12 1
1950.2.y.f 4 15.e even 4 1
1950.2.y.f 4 195.bf even 12 1
1950.2.bc.b 4 15.d odd 2 1
1950.2.bc.b 4 195.y odd 6 1
5070.2.a.y 2 39.k even 12 1
5070.2.a.bg 2 39.k even 12 1
5070.2.b.o 4 39.h odd 6 1
5070.2.b.o 4 39.i odd 6 1

## Hecke kernels

This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(1170, [\chi])$$:

 $$T_{7}^{4} - 4 T_{7}^{2} + 16$$ $$T_{11}^{4} + 12 T_{11}^{3} + 51 T_{11}^{2} + 36 T_{11} + 9$$

## Hecke characteristic polynomials

$p$ $F_p(T)$
$2$ $$1 - T^{2} + T^{4}$$
$3$ $$T^{4}$$
$5$ $$( 1 + T^{2} )^{2}$$
$7$ $$16 - 4 T^{2} + T^{4}$$
$11$ $$9 + 36 T + 51 T^{2} + 12 T^{3} + T^{4}$$
$13$ $$( 13 - 2 T + T^{2} )^{2}$$
$17$ $$( 16 + 4 T + T^{2} )^{2}$$
$19$ $$16 + 48 T + 44 T^{2} - 12 T^{3} + T^{4}$$
$23$ $$1 - 4 T + 15 T^{2} - 4 T^{3} + T^{4}$$
$29$ $$1 - 4 T + 15 T^{2} - 4 T^{3} + T^{4}$$
$31$ $$( 3 + T^{2} )^{2}$$
$37$ $$121 - 198 T + 119 T^{2} - 18 T^{3} + T^{4}$$
$41$ $$16 - 4 T^{2} + T^{4}$$
$43$ $$529 - 230 T + 123 T^{2} + 10 T^{3} + T^{4}$$
$47$ $$1369 + 122 T^{2} + T^{4}$$
$53$ $$( -12 - 12 T + T^{2} )^{2}$$
$59$ $$169 - 156 T + 35 T^{2} + 12 T^{3} + T^{4}$$
$61$ $$11664 + 108 T^{2} + T^{4}$$
$67$ $$2704 - 624 T - 4 T^{2} + 12 T^{3} + T^{4}$$
$71$ $$10816 - 3744 T + 536 T^{2} - 36 T^{3} + T^{4}$$
$73$ $$( 4 + T^{2} )^{2}$$
$79$ $$( 1 + 14 T + T^{2} )^{2}$$
$83$ $$1936 + 104 T^{2} + T^{4}$$
$89$ $$16 - 48 T + 44 T^{2} + 12 T^{3} + T^{4}$$
$97$ $$16 - 48 T + 44 T^{2} + 12 T^{3} + T^{4}$$


text:
# What Polygon Has 24 Sides?

What is the exterior angle of a 24 sided polygon?

A 24 -gon has 24 exterior angles. Each exterior angle measures 36024=15 . The interior angle measures 180−15=165 .

What is a 27 sided polygon called?

What is the name of a polygon with?

# Name of the Polygon + Geometric Drawing
27 sides heptaicosagon
28 sides octaicosagon
29 sides enneaicosagon
30 sides triacontagon

Can 24 be an interior angle of a regular polygon Why?

We know that the measure of each interior angle of a regular polygon the formula is given by $\dfrac180^\circ \left( n - 2 \right)n$. Thus, the regular polygon with interior angle each measuring $165^\circ$ is a 24-sided polygon. Hence, our option A is correct.

## Related Question What polygon has 24 sides?

### What do you call a polygon with 20 sides?

In geometry, an icosagon or 20-gon is a twenty-sided polygon. The sum of any icosagon's interior angles is 3240 degrees.

### What is a polygon with 30 sides called?

In geometry, a triacontagon or 30-gon is a thirty-sided polygon. The sum of any triacontagon's interior angles is 5040 degrees.

### What is a 1 billion sided shape called?

Regular megagon
Regular megagon
Schläfli symbol 1000000, t500000, tt250000, ttt125000, tttt62500, ttttt31250, tttttt15625
Coxeter–Dynkin diagrams
Symmetry group Dihedral (D1000000), order 2×1000000
Internal angle (degrees) 179.99964°

### What type of polygon has 12 sides?

A dodecagon is a 12-sided polygon. Several special types of dodecagons are illustrated above. In particular, a dodecagon with vertices equally spaced around a circle and with all sides the same length is a regular polygon known as a regular dodecagon.

### How many sides does a regular polygon have if each angle is 24?

=> Number of sides of polygon with each angle of 24 is 15.

Pentagon

### How many sides does a regular polygon have if the exterior angles each measure 24 a 12 sides B 15 sides C 24 sides D 30 sides?

The polygon has 15 sides.

### How many sides does a septagon?

A heptagon is a seven-sided polygon. It is also known as a septagon. The word heptagon comes from two words: 'hepta', meaning seven and 'gon' meaning sides. So, if you put this together, the word 'heptagon' directly translates to '7 sided shape'.

### What is after Hexagon?

Shapes – Polygons – Pentagon, Hexagon, Heptagon, Octagon, Nonagon, Decagon- 11 Worksheets / FREE Printable Worksheets – Worksheetfun.

### Is there a 2 sided shape?

In geometry, a digon is a polygon with two sides (edges) and two vertices. A regular digon has both angles equal and both sides equal and is represented by Schläfli symbol 2.

### How many triangles are in a Enneagon?

Properties of all nonagons

Number of diagonals 27
Number of triangles 7
Sum of interior angles 1260°

### What does a Chiliagon look like?

A chiliagon or 1000-gon is a shape with 1000 sides and 1000 corners.

### What is the most complex shape?

• In geometry, the rhombicosidodecahedron, is an Archimedean solid, one of thirteen convex isogonal nonprismatic solids constructed of two or more types of regular polygon faces.
• It has 20 regular triangular faces, 30 square faces, 12 regular pentagonal faces, 60 vertices, and 120 edges.
• ### What do you call a polygon with 7 sides?

A heptagon is a seven-sided polygon. It is also sometimes called a septagon, though this usage mixes a Latin prefix sept- (derived from septua-, meaning "seven") with the Greek suffix -gon (from gonia, meaning "angle"), and is therefore not recommended.

### What is after Dodecagon?

Number of sides Name
10 decagon
11 hendecagon
12 dodecagon
13 triskaidecagon or tridecagon

### What is a 99 sided shape called?

Regular enneacontagon
Regular enneacontagon
Properties Convex, cyclic, equilateral, isogonal, isotoxal

### What do you call a 50 sided polygon?

In geometry, a pentacontagon or pentecontagon or 50-gon is a fifty-sided polygon. The sum of any pentacontagon's interior angles is 8640 degrees.

### What is a 80 sided shape called?

In geometry, an octacontagon (or ogdoëcontagon or 80-gon from Ancient Greek ὁγδοήκοντα, eighty) is an eighty-sided polygon. The sum of any octacontagon's interior angles is 14040 degrees.

### Is a myriagon a circle?

A myriagon, is a polygon with ten thousand sides, and cannot be visually distinguished from a circle.

### Is a Megagon a circle?

A Megagon is a polygon with 1,000,000 sides and angles. Even if drawn at the size of the earth, it would still be very hard to visually distinguish from a circle.

### Is there an infinite sided shape?

In geometry, an apeirogon (from the Greek words "ἄπειρος" apeiros: "infinite, boundless", and "γωνία" gonia: "angle") or infinite polygon is a generalized polygon with a countably infinite number of sides. Apeirogons are the two-dimensional case of infinite polytopes.

### What do you call a 9999 sided polygon?

What do you call a 9999-sided polygon? A nonanonacontanonactanonaliagon.

### What is a polygon with 23 sides called?

In geometry, an icositrigon (or icosikaitrigon) or 23-gon is a 23-sided polygon.

### What is a 72 sided shape called?

Regular heptacontagon
Regular heptacontagon
Coxeter–Dynkin diagrams
Symmetry group Dihedral (D70), order 2×70
Internal angle (degrees) ≈174.857°
Properties Convex, cyclic, equilateral, isogonal, isotoxal

### What polygon has 13 sides?

In geometry, a tridecagon or triskaidecagon or 13-gon is a thirteen-sided polygon.

### What is a polygon with 11 sides called?

In geometry, a hendecagon (also undecagon or endecagon) or 11-gon is an eleven-sided polygon. (The name hendecagon, from Greek hendeka "eleven" and –gon "corner", is often preferred to the hybrid undecagon, whose first part is formed from Latin undecim "eleven".)

### Which of the following polygon has 8 sides?

Regular octagon
Regular octagon
A regular octagon
Type Regular polygon
Edges and vertices 8
Schläfli symbol 8, t4

### How many sides does a polygon with an exterior angle of 162?

Solution: Given: Each interior angle of a polygon is 162°. We know that the sum of an interior angle and an exterior angle of a polygon is equal to 180°. Hence, the number of sides of the given polygon is 20.

### How many sides a regular polygon have?

Thus, the regular polygon has 24 sides.

### How many sides are there?

2D Shapes

Triangle - 3 Sides Square - 4 Sides
Pentagon - 5 Sides Hexagon - 6 sides
Heptagon - 7 Sides Octagon - 8 Sides
Nonagon - 9 Sides Decagon - 10 Sides
More

Decagon

Triangle

### How much sides does a hexagon have?

A hexagon is a 2D geometric polygon that has six sides and six angles. It has no curved sides and all the lines are closed. The internal angles of a regular hexagon add up to 720 degrees.

### How many sides does a 540 degree polygon have?

If the sum of the interior angles of a polygon is 540 degrees, the polygon has 5 sides.

### How much is one outer angle for a 24 sided regular polygon B how much is its one inner angle c What is the sum of the inner angles?

So, for a 24-sided regular polygon, each exterior angle = 360°/24 = 15°. Consequently, each interior angle of a 24-sided regular polygon = 180° - 15° = 165°.

### What is a irregular polygon?

Irregular polygons are polygons that have unequal angles and unequal sides, as opposed to regular polygons which are polygons that have equal sides and equal angles. For example, this rug can be described as a single irregular polygon.

### What is a 200 sided shape called?

What is the name of a polygon with?

# Name of the Polygon + Geometric Drawing
200 sides dihectogon
300 sides trihectogon
400 sides tetrahectogon
500 sides pentahectogon
Posted in FAQ


text:
# Why doesn't my derivation of the arcsec derivative formula not work?

We would like to find the derivative of $$\operatorname{arcsec}(x) = y$$. Rearranging this, we get $$x = \sec(y)$$. Taking the derivative of both sides, we get $$1 = \sec(y)\tan(y)y^\prime$$. Thus, $$y^\prime = \frac{1}{\sec(y)\tan(y)}.$$ Now, draw a right triangle with an acute angle $$y$$, hypotenuse $$x$$, and a side adjacent to angle $$y$$ with length $$1$$. By the Pythagorean theorem, the side opposite angle $$y$$ is equal to $$\sqrt{x^2-1}$$. Thus, we have $$x = \sec(y)$$, which we already knew and got from the triangle, and $$\tan(y) = \frac{1}{\sqrt{x^2-1}}$$ from the triangle we drew. We substitute these values in to our derivative expression to get $$y' = \frac{1}{x\sqrt{x^2-1}}.$$

However, $$y^\prime$$ should equal $$\frac1{|x|\sqrt{x^2-1}}$$ (note the absolute value). Since the difference occurred in the $$|x|$$ part, that means I must've done something wrong when I said $$\sec(y) = x$$, but I do not know why this assumption is wrong. Please correct my proof.

• If your hypotenuse $x$ has a negative length, you have to give the other side a negative length to compensate, making it $-\sqrt{x^2-1}$. (I can't make this rigorous, but I think it's the source of your problem.) Jan 25 at 20:09
• Why do I have to make the other side a negative length? By pythagoras we have a^2 + b^2 = c^2, so a,b,c can be negative, or positive, while satisfying this equation. Jan 25 at 20:12
• In that case, why do you think it should be positive? I'm just pointing out one way to resolve your contradiction. Jan 25 at 20:13

The sloppy reasoning occurs with the right triangle you wrote. You must be careful about the range (values) of the arcsec function. $$y=\text{arcsec}(x)$$ can lie either in $$[0,\pi/2)$$ or in $$(\pi/2,\pi]$$. (These correspond, respectively, to $$x>0$$ and $$x<0$$.)

$$\tan(y)<0$$ when $$y\in (\pi/2,\pi]$$, and so $$\tan(y)=-\sqrt{x^2-1}$$ in that event. Working with your formula for $$y'$$, we note that when $$x<0$$, $$\sec(y)\tan(y)= x(-\sqrt{x^2-1})= (-x)\sqrt{x^2-1} =|x|\sqrt{x^2-1},$$ and this explains the formula. To be honest, it's a bit sneaky to move the negative to the other term, but it allows us to write a single formula, rather than writing down cases. (There's no problem, of course, when $$x>0$$.)

• @ZaWarudo I was not meaning anything too serious, certainly nothing pejorative. Obviously, there's nothing too creative about reassociating the $-1$ factor. But a beginning calculus student might not think to do this. Jan 25 at 20:58
Ted Shifrin has already explained the problem with your derivation, but if you're looking for a 'safer' way of finding the derivative then consider this: if $$\DeclareMathOperator{\arcsec}{arcsec} y = \arcsec x$$, then using the identity $$\arcsec x = \arccos 1/x$$, we see that \begin{align} \frac{dy}{dx}=\frac{d}{dx}\left(\arccos 1/x\right) &= -\frac{1}{\sqrt{1-\left(\frac{1}{x}\right)^2}} \cdot -\frac{1}{x^2} \\[4pt] &= \frac{1}{x^2\sqrt{1-\frac{1}{x^2}}} \, . \end{align}


text:
# How to remove parenthesis in an equation number created by \tag?

This is probably simple, but I can't figure it out. How can I remove the parentheses around a \tag in an align equation? I would like the second line (and only that) to contain [2] instead of ([2]). A workaround would be to give it no label, to see the label [2] as part of the equation, and move it to the right text margin by hand. But how? \hfill does not insert space here.

\documentclass{scrartcl}

\usepackage[T1]{fontenc}
\usepackage[american,ngerman]{babel}
\usepackage{amsmath}

\begin{document}
\begin{align}
a &= b \\
c &= d \tag{[2]}
\end{align}
\end{document}

• Use \tag*, see p. 3 of amsmath documentation. Jun 20 '12 at 11:27
• Ahh... I even used that before some time ago... Thanks, egreg! Jun 20 '12 at 11:31
• @egreg: I made a CW anser so Marius can accept it an this question won’t stay unanswered. Hope you don’t mind :-)
– Tobi
Jun 20 '12 at 11:38

As egreg said, use the starred version \tag* which is the same as \tag except that it does not automatically enclose the tag in parentheses.
\documentclass{article}
$$a + b = c \tag*{[2]}$$

• Does not work. TeX capacity exceeded, sorry [grouping levels=255]. \end{equation*} Works fine with normal \tag. But not \tag*{\eqref{eqlabel}} Feb 1 at 16:10


text:
# An infinite cardinal agrees with all its well-orders on sets of full size.

Suppose $\kappa$ is an infinite von Neumann cardinal (well ordered by $\in$), and take ${<}$ a well-order on $\kappa$. Does there necessarily exists a subset $X\subset\kappa$ of full size (in bijective correspondence with $\kappa$ as a set) on which $<\upharpoonright X^2=\in\upharpoonright X^2$ (i.e. ${<}$ and ${\in}$ agree on $X$)?

Attempt:

Taking $(\kappa, <)\approx_f(\alpha, \in)$, where $f$ is an order-isomorphism and $\alpha$ is an ordinal yields $\alpha\in\kappa$ or $\alpha=\kappa$ (since $\kappa$ is the least ordinal order-isomorphic to a well order on $\kappa$). I feel something like $X=f^{-1}(\alpha)$ should work, but I don't see how I'm using that $\kappa$ is infinite in such a proof, and the statement is clearly false for finite cardinals. I guess this problem could be put "the collection of fixed points of $f$ has the same cardinality as $\kappa$"?

-

## 1 Answer

Marcel, the standard argument for this problem uses choice. One argues first for regular cardinals, by an easy recursion, and then for singular cardinals, where some care is needed. A second proof uses partition calculus, the Erdős-Dushnik-Miller theorem. Both these proofs use somewhat more sophisticated machinery than one would expect.

One can prove the result without using choice; there is a metamathematical trick for this, but I asked in MathOverflow for a combinatorial proof. Clinton Conley found a very pretty argument, and you can read the details here.

-
Thank you! It is also nice to know this appears as a problem in Kunen. I thought it might but didn't spot it skimming through. – Marcel T. Nov 6 '12 at 2:49


text:
## Essential University Physics: Volume 1 (3rd Edition)

We use the equation that relates frequency to velocity and wavelength to find: $f = \frac{v}{\lambda} = \frac{343 \ m/s}{11 \ m} \approx 31 \ Hz$


text:
TL;DR

I’ve become curious about how to organize tournaments (in the sense of allocating games in it) when games have more than two players.

In these days of Coronavirus, there’s been a surge in online playing, and board games make no exception. In particular, BoardGameArena is a very nice place to play online, and they saw about a 6x increase in their traffic (which gave them a few issues…).

One thing that always tickled me is that there is a tournament system. Alas, as it is today their system only caters for two-players games, i.e. even when games would allow for additional players, instances in tournaments only allow two. This is sub-optimal in a lot of games that I like (e.g. Tokaido) and that are better played with three or more players.

Two-players tournaments are easy…

… and there are a lot of ways to set them up. One of the easiest ways is to get a number of players that is a power of 2, then half them at each round with direct eliminations.

One consequence of direct-elimination matches is that half of the people will play only a single game in the tournament. Which might be good for the competition, less for people who wants to play 🤨

In the two-players space the answer to this issue is round robin tournaments, in which each participant plays against each other. This is the same as, for example, sport leagues (although often in this case they play two matches against each other participant).

Direct elimination for more players?

One easy extension is to organize $N$-players games in direct-elimination matches, assuming that there are $N^p$ players. This means having $p$ rounds. Alas, this has the same drawback of the direct elimination tournaments for two-players games, with the additional negative aspects that:

• at each round, more people end playing at the same time ($\frac{2}{3}$ for three-players games, $\frac{3}{4}$ for four-players games, and so on…)
• more players are needed for the same number of rounds (e.g. a three-rounds tournament for two-players games would require $2^3=8$ participants, for three-players games would require $3^3=27$ participants, i.e. more than three times).

One possibility is to let more players through each round. For example, letting two players pass to the next round in four-player games basically means keeping the same structure as in two-player direct-elimination tournaments:

 1-1 1-2 1-3 1-4   1-5 1-6 1-7 1-8

2-1 2-2           2-3 2-4

winner

At any rate, this does not solve the need for people to play more!

k-players leagues?

Another solution would be to find a way to extend the “league” approach to $k$-players games (out of a total population of $n$ players participating in the tournament).

One straightforward way to do this is to form all $n \choose k$ sub-sets of $k$ players out of all $n$ participants, where each of them will play in ${n-1} \choose {k-1}$ matches. This might mean a few too many matches though: a tournament with 8 players overall and 4-player matches would mean 70 matches overall, where each player competes in 35 of them. Ooops.

One observation is that many of those matches are… redundant. From the example with 4-players matches out of 8 total participants, we have the following matches (among others):

1 2 3 4
1 2 3 5
1 2 3 6
1 2 3 7
1 2 3 8

They are somehow… pretty similar, in that players 1, 2, and 3 are playing five matches with each other inside.

Hence, while it’s interesting that each participant plays against every other one at some time, we can probably remove a lot of the redundant games and enjoy the tournament. In general, we should aim for an arrangement where we have a limit on how many times the same people play at the same table.

Block Designs

One answer to the challenge in the previous section is to leverage Block Designs. This is the definition for t-designs (slightly adapted):

Given a finite set $X$ (of $v$ elements called points) and integers $t$, $k$, $r$, $\lambda \geq 1$, we define a $t$-design $B$ to be a family of $k$-element subsets of $X$, called blocks, such that any $x$ in $X$ is contained in $r$ blocks, and any subset of $t$ of distinct points is contained in $\lambda$ blocks. The number of elements in family $B$ is $b$.

Uh? Translated in tournamentese:

• we have a set $X$ of $v$ participants to the tournament;
• we want to organize matches with $k$ players inside;
• each player competes in $r$ matches;
• we want that $t$ players compete in exactly $\lambda$ matches in which all are present at the same time.

In case we want to limit the number of times pairs of players compete at the same table, we set $t = 2$, deal with 2-designs, and call them BIBD (Balanced Incomplete Block Designs). Actually, block design usually refers to 2-designs, and we will stick with them in the following.

Easy right? Next problem pleaaaaase!

Well, not so fast.

Block Designs for arranging matches?

There are several different ways to create BIBDs, not all totally amenable for tabletop games. For example, consider the BIBD induced by the Fano plane (numbers are player identifiers, from 0 up to 6):

(1, 3, 5)
(0, 3, 4)
(2, 3, 6)
(0, 1, 2)
(1, 4, 6)
(0, 5, 6)
(2, 4, 5)

It might be applicable to a tournament of 7 players for games that accept 3 players at a time. Everyone plays against anybody else, but only once. Yay!

There is a big defect though. As any other block design based on finite projective planes, it has the characteristic that any two blocks always share exactly one point. Which means: if you want to limit the number of games that are played at the same time, e.g. so that each player is only on a single table at the same time (like a real-time table would more or less require), then you can only play a very limited number of games at the same time. In particular, if you limit to one game per player at most, at any time only a single game can run, where three participants play and four wait.

Finite affine planes to the rescue

On the other hand, BIBDs induced by finite affine plaines do not have this limitation. It’s easy to get one from a finite projective plane: just get rid of a block and all items in it, and what you’re left with is a finite affine plane. In the Fano plane case, let’s get rid of the group (0, 5, 6), as well as players 0, 5, and 6:

(1, 3, 5) -> (1, 3)
(0, 3, 4) -> (3, 4)
(2, 3, 6) -> (2, 3)
(0, 1, 2) -> (1, 2)
(1, 4, 6) -> (1, 4)
(0, 5, 6) ->
(2, 4, 5) -> (2, 4)

This should look familiar: it’s basically a round-robin arrangement for four players, which allows two real-time games to go on at the same time, for a total of 3 rounds:

1st round: (1, 2) (3, 4)
2nd round: (1, 3) (2, 4)
3rd round: (1, 4) (2, 3)

This is quite amenable now:

• everybody plays against each other, but no more than once
• everybody plays at each round
• everybody plays a reasonable amount of games

Alternative paths of investigation

Another promising path for investigating is to explore the so-called Social Golfer Problem (see also here), which is formulated as follows:

The task is to schedule $g \cdot p$ golfers in $g$ groups of $p$ players for $w$ weeks such that no two golfers play in the same group more than once.

The goal is to find the minimum number of weeks $w$ where this can happen, and I’m also not sure at the moment if this also requires that every golfer still plays with every other one… but it seems promising to find out something to code in the future, and also to overcome some rigidity in the schemas that we will investigate in the short time.


text:
# Understanding the manipulation of Laplacian

1. Jun 25, 2009

### h_cet

Hi;

I am trying to understand the rytov approximation... and when I was studying that, I could not understand a manipilation...

ΔeØ + k2eØ = 0

▼[▼ØeØ] + k2eØ = 0

2ØeØ + (▼Ø)2eØ+k2eØ = 0

I can not understand these manipilations... for a long time, I have searched the properties of laplecian but I could not find any propertiy in connection with laplecian...

So, please explain how this property is working...

be well...

2. Jun 25, 2009

### Mute

It's an application of the rule

$$\nabla \cdot (\phi \mathbf{F}) = \nabla\phi \cdot \mathbf{F} + \phi \nabla \cdot \mathbf{F}$$,

where phi is a scalar function and F is a vector. For your case, the scalar function is $\exp[\Theta]$ and the vector is $\nabla \Theta$.

This comes from treating the laplacian as div grad: grad acts on exp(Theta) to give (grad Theta) exp[Theta] by the chain rule, and then the div acts on (grad Theta) exp[Theta].


text:
# Math-Linux.com

Knowledge base dedicated to Linux and applied mathematics.

Home > Mathematics > Interpolation > Newton’s interpolation polynomial

## Newton’s interpolation polynomial

In this section, we shall study the polynomial interpolation in the form of Newton. Given a sequence of (n+1) data points and a function f, the aim is to determine an n-th degreee polynomial which interpolates f at these points. We shall resort to the notion of divided differences.

### Interpolation

Given points , the points defined by are called points of interpolation. The points defined by are the values of interpolation. To interpolate a function , the values of interpolation are defined as follows:

### Reminders about Lagrange interpolation polynomials

We have seen that Lagrange interpolation polynomial of degree was:

where the ’s are polynomials of degree forming a basis of

Such polynomials are not convenient, since numerically, it is difficult to deduce from . For this reason, we introduce Newton’s interpolation polynomial.

### Newton’s interpolation polynomial and Newton’s basis properties

The polynomials of Newton’s basis, , are defined by:

with the following convention:

Moreover

The set of polynomials (Newton’s basis) are a basis of , the space of polynomials of degree at most equal to . Indeed, they constitute an echelon-degree set of polynomials.
Newton’s interpolation polynomial of degree related to the subdivision is:

where

We shall see how to determine the coefficients in the following section entitled the divided differences.

### Divided differences

Newton’s interpolation polynomial of degree , , evaluated at , gives:

Generally speaking, we write:

is called a zero-order divided difference.

Newton’s interpolation polynomial of degree , , evaluated at , gives:

Hence

is called -order divided difference.

Newton’s interpolation polynomial of degree , , evaluated at , gives:

Therefore:

The following form is generally preferred:

Hence

is called-order divided difference.
By recurrence, we obtain:

is thus called a -order divided difference.
In practice, when we want to determine the -order divided difference
for instance, we need the following quantities

Hence

Properties. Let and be a permutation of . Then

### Newton’s interpolation polynomial of degree

Newton’s interpolation polynomial of degree is obtained via the successive divided differences:

### Error of interpolation

Assume that and . Let be the closed set defined by (the smallest closed set containing and the ’s).

Theorem.

Proof. Let We have:

By successively applying Rolle’s theorem ( times), equals zero at a given point :

Therefore, we have:

Since the coefficient of in is ,

hence

We now assume that and . Let be the closed set defined by (the smallest closed set containing and the ’s).

Theorem.

Proof. There are two possible ways: (1) the one encountered in Lagrange polynomial interpolation and (2) the following.

If , the problem is over!

Let and assume that .

Consider the unique polynomial of degree which interpolates at the points . verifies

The polynomial can be written as:

According to the previous theorem

By setting , ,

Since this result is valid regardless of , the case is made!

### An example of computing Newton’s interpolation polynomial

Given a set of 3 data points , we shall determine Newton’s interpolation polynomial of degree 2 which passes through these points.

Consequently:

### Scilab: computing Newton’s interpolation polynomial

Scilab function newton.sci determines Newton’s interpolation polynomial. contains the points of interpolation and the values of interpolation. is Newton’s interpolation polynomial computed by means of divided differences.sées.

newton.sci

function[P]=newton(X,Y)//X nodes,Y values;P is the numerical Newton polynomial
n=length(X);// n is the number of nodes. (n-1) is the degree
for j=2:n,
for i=1:n-j+1,Y(i,j)=(Y(i+1,j-1)-Y(i,j-1))/(X(i+j-1)-X(i));end,
end,
x=poly(0,"x");
P=Y(1,n);
for i=2:n, P=P*(x-X(i))+Y(i,n-i+1); end
endfunction;

Therefore, we obtain:

-->X=[0;2;4]; Y=[1;5;17]; P=newton(X,Y)
P = 1 + x^2


text:
[Theorem 79, MatCA]

Lemma 15.50.14. Let $A$ be a G-ring. Let $I \subset A$ be an ideal and let $A^\wedge$ be the completion of $A$ with respect to $I$. Then $A \to A^\wedge$ is regular.

Proof. The ring map $A \to A^\wedge$ is flat by Algebra, Lemma 10.97.2. The ring $A^\wedge$ is Noetherian by Algebra, Lemma 10.97.6. Thus it suffices to check the third condition of Lemma 15.41.2. Let $\mathfrak m' \subset A^\wedge$ be a maximal ideal lying over $\mathfrak m \subset A$. By Algebra, Lemma 10.96.6 we have $IA^\wedge \subset \mathfrak m'$. Since $A^\wedge /IA^\wedge = A/I$ we see that $I \subset \mathfrak m$, $\mathfrak m/I = \mathfrak m'/IA^\wedge$, and $A/\mathfrak m = A^\wedge /\mathfrak m'$. Since $A^\wedge /\mathfrak m'$ is a field, we conclude that $\mathfrak m$ is a maximal ideal as well. Then $A_\mathfrak m \to A^\wedge _{\mathfrak m'}$ is a flat local ring homomorphism of Noetherian local rings which identifies residue fields and such that $\mathfrak m A^\wedge _{\mathfrak m'} = \mathfrak m'A^\wedge _{\mathfrak m'}$. Thus it induces an isomorphism on complete local rings, see Lemma 15.43.9. Let $(A_\mathfrak m)^\wedge$ be the completion of $A_\mathfrak m$ with respect to its maximal ideal. The ring map

$(A^\wedge )_{\mathfrak m'} \to ((A^\wedge )_{\mathfrak m'})^\wedge = (A_\mathfrak m)^\wedge$

is faithfully flat (Algebra, Lemma 10.97.3). Thus we can apply Lemma 15.41.7 to the ring maps

$A_\mathfrak m \to (A^\wedge )_{\mathfrak m'} \to (A_\mathfrak m)^\wedge$

to conclude because $A_\mathfrak m \to (A_\mathfrak m)^\wedge$ is regular as $A$ is a G-ring. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


text:
## AC Circuit

Ex. 21.5, chpt. 21.

Repeat this problem, using the system described in Exercise 21.4


text:
# Philip Hofmann

## Observation of Electrically Tunable van Hove Singularities in Twisted Bilayer Graphene from NanoARPES

Publikation: Bidrag til tidsskrift/Konferencebidrag i tidsskrift /Bidrag til avisTidsskriftartikelForskningpeer review

### DOI

The possibility of triggering correlated phenomena by placing a singularity of the density of states near the Fermi energy remains an intriguing avenue toward engineering the properties of quantum materials. Twisted bilayer graphene is a key material in this regard because the superlattice produced by the rotated graphene layers introduces a van Hove singularity and flat bands near the Fermi energy that cause the emergence of numerous correlated phases, including superconductivity. Direct demonstration of electrostatic control of the superlattice bands over a wide energy range has, so far, been critically missing. This work examines the effect of electrical doping on the electronic band structure of twisted bilayer graphene using a back-gated device architecture for angle-resolved photoemission measurements with a nano-focused light spot. A twist angle of 12.2° is selected such that the superlattice Brillouin zone is sufficiently large to enable identification of van Hove singularities and flat band segments in momentum space. The doping dependence of these features is extracted over an energy range of 0.4 eV, expanding the combinations of twist angle and doping where they can be placed at the Fermi energy and thereby induce new correlated electronic phases in twisted bilayer graphene.


text:
Courtney L.

# Math Equations Math Problem Math Help

define variables, write an equation, and then use the equation in the solving process.

Let C(x)=500+40x be the cost to manufacture x items. Find the average cost per item to produce 70 items.

By:


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# Explaining the stability of a numerical scheme through functional analysis

In this paper, the stability of a discretization $$L_hx=y~~~~~~~~~~~(*)$$on a continuous problem $$Lx=y~~~~~~~~~~~(**)$$ where $$L\in BL(X,Y)$$ and $$h$$ represents the mesh parameters, is given by the uniform boundedness of the inverse operator $$L_h^{-1}$$. That is, the numerical scheme $$(*)$$ is said to be stable if $$\lVert L_h^{-1} \lVert \leq C,$$ for some $$C>0$$ irrespective of the family of mesh parameters represented by $$h$$.

My doubts:

How can we ensure the boundedness of the most often partial differential operators appear in $$(*)$$ as we have seen that derivative operators are not continuous?

How can we practically explain the connection between stability and the 'uniform boundedness'?

• Most derivative operators are continuous in the appropriate norms (e.g. $\frac{\partial}{\partial x}\in W^{1,1}\to L^1$); see Sobolev spaces. As for why this is a useful definition, see Thm. 1.1 in the paper you cite. Jun 23 at 9:54

One notion of stability is that "for small differences in input, we get small differences in output". Mathematically, it could be, e.g., for two pieces of data $$y_1, y_2$$ and their respective solutions $$x_1,x_2$$, you would want $$\|x_1-x_2\| \leq C \|y_1-y_2\|.$$

If $$y_1$$ and $$y_2$$ are 'close' in some sense, then the solutions to the inverse problem should also be 'close', perhaps up to an extra multiplicative factor $$C$$. The uniform boundedness of $$L_h^{-1}$$ gives precisely this notion of stability, since (for this system) a solution is produced via $$x=L_h^{-1}y$$. So

$$\|x_1-x_2\|=\|L_h^{-1}y_1-L_h^{-1}y_2\|\leq\|L_h^{-1}\|\|y_1-y_2\|\leq C\|y_1-y_2\|.$$

Why is $$L_h$$ bounded?: Since (as far as I can tell) $$L_h$$ is acting on a finite dimensional discretization. Even though it represents differentiation, it is still bounded since it is acting on a finite dimensional space (and all linear operators in finite dimensional spaces are bounded).

Concerning $$(**)$$: this is a very good observation, and in general, a linear operator involving differentiation will not be bounded in an infinite-dimensional space (and its inverse may not even exist)! So very careful analysis must be done in order to ensure that, as the grid $$h$$ refines, you actually converge to a solution of the continuous problem. Check out the literature/proofs on Galerkin methods for more info on this :-)

• Sir, Which is that finite dimensional vector space domain of $L_h$? 2 days ago
• @MessiLio The discrete spaces mentioned at the start of section 2 of the paper in your post
– Zim
2 days ago
• Sir, I couldn't understand how the discrete space $X_h$ is a finite dimensional subspace of $X$ ('in some cases'). Can you make it clear ? 2 days ago
• @MessiLio ok a very simple example is as follows: Consider all continuous functions on $[0,1]$. This is an infinite-dimensional space. However, if we fixed a particular grid, say $h=\{0,0.5,1\}$, then a "function" defined on $X_h$ is just a mapping from each point in $h$ to $\mathbb{R}$, i.e., we represent a function $f$ with a vector in (the finite dimensional space) $\mathbb{R}^3$ which represents $(f(0), f(0.5), f(1))$. So, in general in 1D, the cardinality of your discretization is the dimension of your space.
– Zim
2 days ago
• @MessiLio a lot of folks use a slight abuse of notation here. For any function in $X$, you can construct is representation in $X_h$ via just taking its values on the grid. For any real numbers representing the values of $f$ at the gridpoints, you can construct a representative function in $X$ in the following procedure: For every grid point, define a "hat function" on $X$ which takes the value of $1$ at that gridpoint, and $0$ at other gridpoints.The linear combination of the function values and the hat functions is a 'representation' of the finite dimensional "function" in $X$.
– Zim
2 days ago


text:
# zbMATH — the first resource for mathematics

Lower bounds for the energy of unit vector fields and applications. (English) Zbl 0908.58004
J. Funct. Anal. 152, No. 2, 379-403 (1998); erratum ibid. 171, No. 1, 233 (2000).
The author gives a lower bound for the Dirichlet energy of a unit vector field defined in a perforated domain of $$\mathbb{R}^2$$ with nonzero degree on the outer boundary in terms of the total diameter of the holes. Using this, the author also gives lower bounds and compactness results for Ginzburg-Laudau functionals.

##### MSC:
 5.8e+16 Variational problems concerning extremal problems in several variables; Yang-Mills functionals
Full Text:
##### References:
 [1] Almeida, L.; Bethuel, F., Seminaire X EDP, (1995-1996) [2] N. André, I. Shafrir, Asymptotic behaviour of minimizers for the Ginzburg- Landau functional with weight, Arch. Rational Mech. Anal. [3] Beaulieu, A.; Hadiji, R., On a class of ginzburg – landau equations with weight, Panamer. math. J., 5, 1-33, (1995) · Zbl 0843.49003 [4] Bethuel, F.; Brezis, H.; Hélein, F., Asymptotics for the minimization of a ginzburg – landau functional, Calc. var. partial differential equations, 1, 123-148, (1993) · Zbl 0834.35014 [5] Bethuel, F.; Brezis, H.; Hélein, F., Ginzburg – landau vortices, Progress in nonlinear differential equations and their applications, (1994), Birkhauser Basel [6] Bethuel, F.; Demengel, F., Extensions for Sobolev maps between manifolds, Calc. var. partial differential equations, 3, 475-491, (1995) · Zbl 0846.46021 [7] Bethuel, F.; Rivière, T., Vortices for a variational problem related to supra- conductivity, Ann. inst. H. Poincaré anal. nonlinéaire, 12, 243-303, (1995) · Zbl 0842.35119 [8] Boutet de Monvel-Berthier, A.; Georgescu, V.; Purice, R., A boundary value problem related to the ginzburg – landau model, Comm. math. phys., 141, 1-23, (1991) · Zbl 0742.35045 [9] Brezis, H.; Merle, F.; Rivière, T., Quantisation effects for −δuuu22, Arch. rational mech. anal., 126, 35-58, (1994) [10] Han, Z.G.; Li, Y.Y., Degenerate elliptic systems and application to ginzburg – landau type equations, part I, Calc. var. partial differential equations, 4, 171-202, (1996) · Zbl 0847.35055 [11] Han, Z.G.; Shafrir, I., Lower bounds for the energy ofS1, J. anal. math., 68, 295-305, (1995) [12] R. Jerrard [13] Lin, F.H., Solutions of the ginzburg – landau equations and critical points of the normalized energy, Ann. inst. H. Poincaré anal. nonlinéaire, 12, 599-622, (1995) · Zbl 0845.35052 [14] Rivière, T., Lignes de tourbillons dans le modèle de Higgs abélien, C.R. acad. sci. Paris. Série I, 395, 73-76, (1995) · Zbl 0840.35109 [15] Schoen, R.; Uhlenbeck, K., Boundary regularity and the Dirichlet problem for harmonic maps, J. differential geom., 18, 253-268, (1983) · Zbl 0547.58020 [16] Struwe, M., On the asymptotic behavior of minimizers of the ginzburg – landau model in 2 dimensions, J. differential integral equations, 7, (1994) · Zbl 0809.35031
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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# factor.scope

0th

Percentile

##### Compute Allowed Changes in Adding to or Dropping from a Formula

add.scope and drop.scope compute those terms that can be individually added to or dropped from a model while respecting the hierarchy of terms.

Keywords
models
##### Usage
add.scope(terms1, terms2)drop.scope(terms1, terms2)factor.scope(factor, scope)
##### Arguments
terms1

the terms or formula for the base model.

terms2

the terms or formula for the upper (add.scope) or lower (drop.scope) scope. If missing for drop.scope it is taken to be the null formula, so all terms (except any intercept) are candidates to be dropped.

factor

the "factor" attribute of the terms of the base object.

scope

a list with one or both components drop and add giving the "factor" attribute of the lower and upper scopes respectively.

##### Details

factor.scope is not intended to be called directly by users.

##### Value

For add.scope and drop.scope a character vector of terms labels. For factor.scope, a list with components drop and add, character vectors of terms labels.

add1, drop1, aov, lm

##### Aliases
library(stats) # NOT RUN { add.scope( ~ a + b + c + a:b, ~ (a + b + c)^3) # [1] "a:c" "b:c" drop.scope( ~ a + b + c + a:b) # [1] "c" "a:b" # }


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# Controlling the robotiq_85_gripper (2-fingered gripper) in a Gazebo simulation

Hello all,

I am trying for some time now to control (open/close) the two fingered robotiq_85_gripper, as it is described in ( https://github.com/DualUR5Husky/robot... ) but augmented with the transmissions and inertial described in ( https://github.com/waypointrobotics/r... ). The problem is that I am not sure how to transmit open - close commands in a Gazebo 7 environment. Do I need an extra plugin to transmit commands to the controllers? I found the following package: ( https://github.com/DualUR5Husky/robot... ), but it doesn't seam to do anything.


text:
# Showing a Norm is not an Inner Product

1. Mar 27, 2012

### Punkyc7

Show the taxicab norm is not an IP.

taxicab norm is v=(x$_{1}$.....x$_{n}$)
then ||V||= |x$_{1}$|+.....+|x$_{n}$|)

I was thinking about using the parallelogram law

but I would get this nasty thing

(|x$_{1}$+w$_{1}$|+...........+|x$_{n}$+w$_{n}$|)$^{2}$+(|x$_{1}$-w$_{1}$}+...........+|x$_{n}$-w$_{n}$|$)^{2}$=2(|x$_{1}$|+.....+|x$_{n}$|)$^{2}$+(|w$_{1}$|+.....+|w$_{n}$|)$^{2}$

Also there might be some typos with the absolute value signs the latex get messy

2. Mar 27, 2012

### dirk_mec1

Hint: the norm doesn't satisfy linearity (it does satisfy the triangle inequality).

3. Mar 27, 2012

### clamtrox

Do you mean that ||v+w||is not an inner product? This is easy to show! Just go through the definition of inner product -- one of the axioms is clearly not satisfied in this case.


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# Approximating Square Root of Numbers by Averaging

There are several algorithms developed to manually calculate the square root of a number. My favorite is the method of averaging, which is probably the least complicated.  In this method, the number is divided by the nearest square root estimate, and then the quotient and the square root estimate are averaged. The result is the new estimate.

Sounds complicated? Not really.

Example: Approximate $\sqrt{24}$.

Note: Using a calculator, $\sqrt{24} \approx 4.8989794855$.

Solution

We know that $\sqrt{24}$ is between $4^2$ and $5^2$. Between the two, it is nearer to $5^2$. So, the first estimate is $5$. Next, we divide $24$ by the first estimate and “average” it with the quotient .

Second Estimate

Divide: $\frac{24}{5} = 4.8$.

Average: $\frac{4.8 + 5}{2} = 4.9$

Second estimate: $4.9$ (2 significant digits)

Third Estimate

Divide: $\frac{24}{4.9} = 4.898$

Average: $\frac{4.9 + 4.897}{2} = 4.899$

Third estimate: $4.899$ (4 significant digits)

Fourth Estimate

Divide: $\frac{24}{4.899} = 4.8989589$

Average: $\frac{4.899 + 4.8989589}{2}$

Fourth estimate: $4.8989794$ (8 significant digits)

Notice that the third estimate is already correct up to 8 significant digits which means that this method is quite accurate.  In the fifth, sixth,   and seventh estimates, 16, 32 and 64 digits should be kept respectively.  Of course, repeating the process over and over will obtain more accurate approximations.

Update: Don Cohen has a short Mathematica code for this method in his site.

Reference: Modern Elementary Mathematics

## 5 comments on “Approximating Square Root of Numbers by Averaging”

1. Guillermo,

The interesting thing about this method is that it is an example of iterating a function and finding infinite sequences. You can see this method using Mathematica, on my site at http://www.mathman.biz/html/sqrtbyav&iter.html with various links to other ways to use iteration to find infinite sequences with young people, and solving equations (Peter Farrell on Math 2.0 mentioned Newton’s method using iteration as another way to solve quadratic equations). This file is also on my A Map To Calculus at Pi and sqrts.

Don

• Guillermo Bautista on said:

Thank you for the information Don. I updated the blog and mentioned it above.

You may use these HTML tags and attributes:   <acronym title="">
      <q cite="">


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# What is the purpose of row normalization

I understand the reasoning behind column normalization, as it causes features to be weighted equally, even if they are not measured on the same scale – however, often in the nearest neighbour literature, both columns and rows are normalized. What is the row normalization for/why normalize rows? Specifically, how does the result of the row normalization affect the similarity/distance between row vectors?

This is a relatively old thread but I recently encountered this issue in my work and stumbled upon this discussion. The question has been answered but I feel that the danger of normalizing the rows when it is not the unit of analysis (see @DJohnson’s answer above) has not been addressed.

The main point is that normalizing rows can be detrimental to any subsequent analysis, such as nearest-neighbor or k-means. For simplicity, I will keep the answer specific to mean-centering the rows.

To illustrate it, I will use simulated Gaussian data at the corners of a hypercube. Luckily in R there is a convenient function for that (the code is at end of the answer). In the 2D case it is straightforward that the row-mean-centered data will fall on a line passing through the origin at 135 degrees. The simulated data is then clustered using k-means with correct number of clusters. The data and the clustering results (visualized in 2D using PCA on the original data) look like this (the axes for the leftmost plot are different). The different shapes of the points in the clustering plots refer to the ground-truth cluster assignment and the colors are result of the k-means clustering.

The top-left and bottom-right clusters get cut in half when the data is row-mean-centered. So the distances after row-mean-centering get distorted and are not very meaningful (at-least based on the knowledge of the data).

Not so surprising in 2D, what if we use more dimensions? Here is what happens with 3D data. The clustering solution after row-mean-centering is “bad”.

And similar with 4D data (now shown for brevity).

Why is this happening? The row-mean-centering pushes the data into some space where some features come closer than otherwise they are. This should be reflected in the correlation between the features. Let’s look at that (first on the original data and then on the row-mean-centered data for 2D and 3D cases).

[,1] [,2] [1,] 1.000 -0.001 [2,] -0.001 1.000 [,1] [,2] [1,] 1 -1 [2,] -1 1 [,1] [,2] [,3] [1,] 1.000 -0.001 0.002 [2,] -0.001 1.000 0.003 [3,] 0.002 0.003 1.000 [,1] [,2] [,3] [1,] 1.000 -0.504 -0.501 [2,] -0.504 1.000 -0.495 [3,] -0.501 -0.495 1.000
So it looks like row-mean-centering is introducing correlations between the features. How is this affected by number of features? We can do a simple simulation to figure that out. The result of the simulation are shown below (again the code at the end).

So as the number of features increase the effect of row-mean-centering seems to diminish, at-least in terms of the introduced correlations. But we just used uniformly distributed random data for this simulation (as is common when studying the curse-of-dimensionality).

So what happens when we use real data? As many times the intrinsic dimensionality of the data is lower the curse might not apply. In such a case I would guess that row-mean-centering might be a “bad” choice as shown above. Of course, more rigorous analysis is needed to make any definitive claims.

Code for clustering simulation

palette(rainbow(10))
set.seed(1024)
require(mlbench)
N <- 5000
for(D in 2:4) {
X <- mlbench.hypercube(N, d=D)
sh <- as.numeric(X$classes) K <- length(unique(sh)) X <- X$x

Xc <- sweep(X,2,apply(X,2,mean),"-")
Xr <- sweep(X,1,apply(X,1,mean),"-")

show(round(cor(X),3))
show(round(cor(Xr),3))

par(mfrow=c(1,1))

k <- kmeans(X,K,iter.max = 1000, nstart = 10)
kc <- kmeans(Xc,K,iter.max = 1000, nstart = 10)
kr <- kmeans(Xr,K,iter.max = 1000, nstart = 10)
pc <- prcomp(X)
par(mfrow=c(1,4))

lim <- c(min(min(X),min(Xr),min(Xc)), max(max(X),max(Xr),max(Xc)))
plot(X[,1], X[,2], xlim=lim, ylim=lim, xlab="Feature 1", ylab="Feature 2",main="Data",col=1,pch=1)
points(Xc[,1], Xc[,2], col=2,pch=2)
points(Xr[,1], Xr[,2], col=3,pch=3)
legend("topleft",legend=c("Original","Center-cols","Center-rows"),col=c(1,2,3),pch=c(1,2,3))
abline(h=0,v=0,lty=3)

plot(pc$x[,1], pc$x[,2], col=rainbow(K)[k$cluster], xlab="PC 1", ylab="PC 2", main="Cluster original", pch=sh) plot(pc$x[,1], pc$x[,2], col=rainbow(K)[kc$cluster], xlab="PC 1", ylab="PC 2", main="Cluster center-col", pch=sh)
plot(pc$x[,1], pc$x[,2], col=rainbow(K)[kr\$cluster], xlab="PC 1", ylab="PC 2", main="Cluster center-row", pch=sh)
}

Code for increasing features simulation

set.seed(2048)
N <- 1000
Cmax <- c()
Crmax <- c()
for(D in 2:100) {
X <- matrix(runif(N*D), nrow=N)
C <- abs(cor(X))
diag(C) <- NA
Cmax <- c(Cmax, max(C, na.rm=TRUE))

Xr <- sweep(X,1,apply(X,1,mean),"-")
Cr <- abs(cor(Xr))
diag(Cr) <- NA
Crmax <- c(Crmax, max(Cr, na.rm=TRUE))
}
par(mfrow=c(1,1))
plot(Cmax, ylim=c(0,1), ylab="Max. cor.", xlab="#Features",col=1,pch=1)
points(Crmax, ylim=c(0,1), col=2, pch=2)
legend("topright", legend=c("Original","Center-row"),pch=1:2,col=1:2)

EDIT

After some googling ended up on this page where the simulations show similar behavior and proposes that the correlation introduced by row-mean-centering to be $-1/(p-1)$.


text:
# What must be a distance (in meters) between a point charge of -5 times 10^{-7} C and another...

## Question:

What must be a distance (in meters) between a point charge of {eq}-5 \times 10^{-7}\ C {/eq} and another charge of {eq}+3 \times 10^{-7}\ C {/eq} so that the magnitude of the force between them will be equal to {eq}0.058\ N {/eq}? {eq}(k = 9 \times 10^9\ N \cdot m^2 / C^2) {/eq}

## Electrostatic Force between the two stationary charges:

The electrostatic force is defined as the force acting between the stationary charged particles. According to Coulomb's law, the electrostatic force is expressed as such;

{eq}F_{C} \propto q_{1} q_{2} \,\,\,\,\,\, {/eq}, {eq}F_{C} \propto \dfrac{1}{r^{2}} {/eq}

Therefore,

{eq}F_{C} = \dfrac{kq_{1}q_{2}}{r^{2}} {/eq}

Where;

• k is the electrostatic force constant.
• r is the separation between two stationary charges.

Note:

• The electrostatic force between two charged particles or objects is either attractive or repulsive in nature. It depends on the polarity of the stationary charges.

Become a Study.com member to unlock this answer! Create your account

We are given the following information:

{eq}q_{1} = - 5.0 \times 10^{-7} \, C. {/eq}

• {eq}q_{2} =+3.0 \times 10^{-7} \, C. {/eq}
• The electrostatic...

Coulomb's Law: Variables Affecting the Force Between Two Charged Particles

from

Chapter 6 / Lesson 3
39K

In the 18th century, Charles Coulomb uncovered the secrets of electrostatic force between charged particles. The results of his experiments led to what is now known as Coulomb's Law, which tells us how force, charge, and distance are all related.


text:
Measurement of the associated production of a Higgs boson decaying into $b$-quarks with a vector boson at high transverse momentum in $pp$ collisions at $\sqrt{s} = 13$ TeV with the ATLAS detector
2020.
The collaboration

Abstract
The associated production of a Higgs boson with a $W$ or $Z$ boson decaying into leptons and where the Higgs boson decays to a $b\bar{b}$ pair is measured in the high vector-boson transverse momentum regime, above 250 GeV, with the ATLAS detector. The analysed data, corresponding to an integrated luminosity of 139 fb$^{-1}$, were collected in proton-proton collisions at the Large Hadron Collider between 2015 and 2018 at a centre-of-mass energy of $\sqrt{s} = 13$ TeV. The measured signal strength, defined as the ratio of the measured signal yield to that predicted by the Standard Model, is $0.72 ^{+0.39}_{-0.36}$ corresponding to an observed (expected) significance of 2.1 (2.7) standard deviations. Cross-sections of associated production of a Higgs boson decaying into $b$ quark pairs with a $W$ or $Z$ gauge boson, decaying into leptons, are measured in two exclusive transverse momentum regions, 250-400 GeV and above 400 GeV, and interpreted as constraints on anomalous couplings in the framework of a Standard Model effective field theory.


text:
Is it possible to use deep learning to generate a 2D image from a few numerical values?

Is it possible to train a DL model that will generate a full resolution 2D image based on few numbers describing this image and what type of model or architecture would that be?

What I want to achieve is that I deliver to the model some numbers for example describing positions of objects on the screen and number describing how lit the scene is and I get back a 2D image with objects in their correct positions and proper lighting, but for one set of input data values I will get always one same image (see image above). These input data also could be anything else than positions and lighting, these are only examples helping to visualize what I mean.

This all, of course, assuming that I have a lot of annotated training data that consists of images and labels of the objects' positions and scene lighting values.

EDIT: The final model would be trained on real images taken from Full HD camera, not some simple shapes like presented here, that I did only to explain better my question.

• I do not see the reason to refer to any kind of deep learning for this problem. You have objects and properties of the scene and just render the scene and output view from a certain angle. Jul 23, 2021 at 9:08
• @spiridon_the_sun_rotator Well, the reason is that this is only simplification of my problem that I done to help visualise what I am asking for. Later, I want to train suggested model on data of real images taken with Full HD camera and make sort of a realistic scene renderer, generated by DL. Jul 23, 2021 at 11:01
• Also, I don't see why, for experimental/research reasons, I should restrain myself from using other methods of doing tasks that can be done with regular/simple methods. My question was about whether such DL models exist, not how to render scene. Jul 23, 2021 at 11:05
• Did you take a look at GANs? Jul 23, 2021 at 18:00

Probably closely related to the problem of interest would be something akin Neural Radiance Fields (NeRF for short) https://www.matthewtancik.com/nerf.

The model takes several images from different angles and view of the scene and learns a 3d representation of the scene, that can be used to sample novel views of this scene (not present in the training data).

The knowledge of the scene is encoded in the weights of MLP. Namely, the renderer is a function that takes coordinates $$(x, y, z)$$ and viewing angle $$(\theta, \phi)$$ as an input, and outputs the color (RGB) and the density $$\sigma$$.

For your problem, one can modify, the network, that is takes in addition to the coordinates and angles, other properties of the scene, such as lightning (some scalar), e.t.c and outputs again (RGB) vector and density $$\sigma$$.

In order for procedure to be successful, I would expect, that model needs data with different lighting, and provided there are enough data points, one could synthesize the view with given lightning.

The presence of multiple objects can be probably taken into account, by passing a set of coordinates $$(x_i, y_i, z_i)$$ - but this seems to complicate things a lot, and the training procedure.

• This is very close to what I need indeed, with the different output, I think their model outputs some data like points in cloud, and only later they make a scene from this by use of traditional rendering methods. What I want to do is to make such a renderer to produce ready 2D images directly from input data. But I think I will try something like reversed FC network or reversed CNN network, that instead of making predictions from images it makes images from predictions, sort of thing. Jul 24, 2021 at 6:25


text: A charged cylinder with homogeneous charge density $$\rho_0$$ and radius $$R$$ generates an electric field $$E$$ inside and outside the cylinder.


text:
# Odd bounded function limit at $x=0$

$$f:\mathbb{R} \to \mathbb{R}$$ is a bounded odd function. What can be said about $$\lim_{x \to 0}f(x)$$?

1. Must be $$0$$
2. If exists then $$0$$
3. Must exist but may not be $$0$$

I am stuck at this question. I know, for continuous odd function we must have value $$0$$ at $$x=0$$. But we don't know if this function is continuous or not. How to proceed? Thanks in advance.

$$\lim_{x\to0^-}f(x)=-\lim_{x\to0^+}f(x).$$
• @user587126 Take the function which is $0$ at $x = 0$, $1$ on $\mathbb R^+$, and $-1$ on $\mathbb R^-$. – Alex Provost Jul 18 '19 at 12:29
• @user587126 Consider $$f(x) = \begin{cases} 1, &\text{ if x\notin \mathbb{Q}, x > 0}\\ -1, &\text{ if x\notin \mathbb{Q}, x < 0}\\ 0, &\text{ if x\in \mathbb{Q}}\\ \end{cases}$$ Then $\lim_{x\to 0^+} f(x)$ and $\lim_{x\to 0^-} f(x)$ both don't exist. – mechanodroid Jul 18 '19 at 12:59


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# Browse Dissertations and Theses - Kinesiology and Community Health by Title

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The major purpose of this study was to determine the effects of a generalized self-control skill training program on the health knowledge, habits, attitudes, and pertinent clinical variable measures of ninth grade students.

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Several researchers have investigated the effects of jogging, and aerobic exercise, on state and trait anxiety. The role of jogging in selectively reducing somatic anxiety as has been postulated in the psychobiological ...

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Effects of acute exercise and chronic exercise training on activities of five hepatic lipogenic enzymes were studied in rats fed different high-carbohydrate diets. In Study 1, ATP citrate lyase (CL), pyruvate kinase (PK), ...

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# Energy of SHM

1. Dec 20, 2006

### mbrmbrg

1. The problem statement, all variables and given/known data

(a) When the displacement in SHM is one-half the amplitude xm, what fraction of the total energy is kinetic energy?

(b) What fraction is potential energy?

(c) At what displacement, in terms of the amplitude, is the energy of the system half kinetic energy and half potential energy?

2. Relevant equations

$$U=\frac{1}{2}kx^2$$

$$K+U=E$$

$$E=\frac{1}{2}kx_m^2$$

3. The attempt at a solution

I got parts (a) and (b) quite neatly, and I'm convinced that I got (c) also, only WebAssign isn't so convinced.

WORK FOR PART C:
$$U=\frac{1}{2}E$$

$$\frac{1}{2}kx^2=\frac{1}{2}(\frac{1}{2}kx_m^2)$$

some halfs and k's cancel, leaving

$$x^2=\frac{1}{2}x_m^2$$

$$x=\frac{x_m}{\sqrt{2}}$$

Last edited: Dec 20, 2006
2. Dec 20, 2006

### neutrino

That's correct, and so is -xm/sqrt(2), since the question asked for displacement.

How did you enter the answer? Maybe it was expecting the answer in some other format.

3. Dec 20, 2006

### mbrmbrg

Thanks; that was it. I entered xm/sqrt2, but they wanted xm/sqrt(2). Looks the same when you click the eyeball to ask what it'll look like!


text:
# Supersingular curves with small non-integer endomorphisms

### Jonathan Love (Stanford) and Dan Boneh (Stanford)

29-Jun-2020, 21:10-21:40 (23 months ago)

Abstract: We introduce a special class of supersingular curves over $\mathbb{F}_{p^2}$, characterized by the existence of non-integer endomorphisms of small degree. A number of properties of this set is proved. Most notably, we show that when this set partitions into subsets in such a way that curves within each subset have small-degree isogenies between them, but curves in distinct subsets have no small-degree isogenies between them. Despite this, we show that isogenies between these curves can be computed efficiently, giving a technique for computing isogenies between certain prescribed curves that cannot be reasonably connected by searching on $\ell$-isogeny graphs.

cryptography and securityalgebraic geometrynumber theory

Audience: researchers in the topic

( chat | paper | slides | video )

Comments: The slides used in the pre-recorded video can be found here.

Chairs: Steven Galbraith and Christophe Petit

Series comments: Registration is now open. Registration is free but required to access the chat and livestream.

This is a hybrid synchronous/asynchronous conference with several ways to participate.

• Click the "paper" link to view contributed papers and posters (open to all).
• Click the "video" link to view pre-recorded talks of accepted papers (open to all).
These are 15-20 minutes aimed at a general algorithmic number theory audience.
• Click the "slides" link to view slides used in the pre-recorded video when available (open to all).
• Click the "chat" link to access the chat stream related to the talk or poster before, during, and after the live event (registration required).
• Click the "livestream" button to join the live event when it is taking place (registration required and you must be logged in).
For accepted papers the audience will be expected to have watched the pre-recorded video and have the paper in front of them.

The invited talks will be recorded and made available via the "video" link after the talk is over. None of the other sessions will be recorded.

 Organizer: Steven Galbraith* Curator: Andrew Sutherland* *contact for this listing

Export talk to


text:
# Proth number

A Proth number $p$ is a number of the form $k2^{m}+1$, where $k$ is odd (e.g., $k=2n+1$ with an integer $n>(-1)$), $m$ is also an integer and $2^{m}>k$. The first few Proth numbers are 3, 5, 9, 13, 17, 25, 33, 41, 49, 57, 65, 81, 97, 113, 129, 145, 161, 177, 193, etc. (listed in A080075 of Sloane’s OEIS). The Fermat numbers are a subset of the Proth numbers, namely with $k=1$.

Title Proth number ProthNumber 2013-03-22 17:21:09 2013-03-22 17:21:09 PrimeFan (13766) PrimeFan (13766) 4 PrimeFan (13766) Definition msc 11A51


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# How reimport a package in UDK with command line?

sometimes ago, I saw a command to re-import a UDK package content without opening the UDK Editor. (Especially, it was used for reimporting a Flash package in a scaleform tutorial, but it doesn't matter now )

It's very useful, because every time I change my .swf file, I should open the heavy UDK and just press re-import.

So, can anyone give me that solution again ?

I found the solution For packages containing .swf files here :

gfximport Commandlet

To improve iteration time, the gfximport commandlet, run from your Unreal Engine 3 executable (e.g UDKLift.exe) allows for content importing and reimporting without loading Unreal Editor. The commandlet assumes that content has been arranged in the manner described above.

In order to import a file, you must run gfximport commandlet while specifying relative paths to all of the .SWF files within the GameDir\Flash directory that you wish to import separated by spaces.

UDKLift.exe gfximport [path_1] [path_2] [path_3]

In order to re-import files that are already imported into UE3, simply run the commandlet with no arguments.

UDKLift.exe gfximport

The commandlet will reimport all the SwfMovies for which the .SWF file on disk is newer than the one imported.

Example:

Suppose you want to import these files.

C:\UDK\UDK-2010-07\UDKGame\Flash\UI\MainMenu.swf
C:\UDK\UDK-2010-07\UDKGame\Flash\UI\Resources.swf

You would execute the following command:

UDKLift.exe gfximport UI/MainMenu.swf UI/Resources.swf

They will be imported into the UI package as SwfMovie'UI.MainMenu' and SwfMovie'UI.Resources'. The package will be saved as GameDir\Content\GFx\UI.upk.


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# 50105. Seesaw Chandelier Tree

## I'm a slow walker, but I never walk backwards.

Write a function to construct a seesaw chandelier tree.

A seesaw chandelier is a hierarchy of seesaws in which every seesaw is balanced. A seesaw chandelier can be recursively defined as follow. A balancing point is defined as in the previous tasks.

• If there is a balancing point, then the balancing point links to another seesaw chandelier with a "left" pointer, and links to the other seesaw chandelier with a "right" pointer.
• If there is no balancing point, then a seesaw chandelier has a sequence of nodes linked by the "left" pointer as a linked list.

We want to build a seesaw chandelier tree from a sequence of $n$ non-negative integers, indexed from $0$ to $n-1$. For the first level, we consider the whole sequence as a seesaw and find the balancing point at index $i$ between $0$ to $n-1$, as we did in the previous tasks. Now the root of the tree is the balancing point at index $i$ and two partitions -- one on the left tree and one on the right tree. The one on the left has index from $0$ to $i-1$, and the one on the right has index from $i+1$ to $n-1$. Now we consider these two partitions as two independent seesaw hierarchies and find the balancing points for them. We can recursively do this until the length is less than $3$, or we cannot find a balancing point for a partition. In both cases we stop the recursion and use left pointer to link all the nodes in the partition from right to left. Please refer to the second condition in the previous recursive definition.

For example, we now want to build a seesaw chandelier with a sequence of $8$ non-negative integers, $(2, 5, 2, 0, 7, 1, 7, 4)$. At the first level, we can find the balancing point at $7$ (with index $4$), because $2 \times 4 + 5 \times 3 + 2 \times 2 + 0 \times 1 = 1 \times 1 + 7 \times 2 + 4 \times 3$. Then we partition the sequence into two sequences, $(2, 5, 2, 0)$ and $(1, 7, 4)$. Then we recursively find the balancing points for them. We can find $5$ (index $1$) for the partition on the left, because $2 \times 1 = 2 \times 1 + 0 \times 2$, but no balancing point on the right. After that the recursion stops and we link the nodes in each partition from right to left since we cannot find a balancing point for the new partitions or the length is less than $3$.

p10192

Note that if the seesaw has balancing point, then there is only one balancing point since there is only one center of mass for every seesaw. In addition, the torque needs to be stored in a “long long int” to prevent integer overflow. All pointers must be set to NULL if they are not used to link the seesaws together.

Please construct the seesaw chandelier tree in the function, Node *ConstructTree( … ), then return the pointer of the root. The main program and the prototype of function ConstructTree are as followed:

construct.h

123456789#ifndef CONSTRUCT#define CONSTRUCTtypedef struct Node{    int value;    struct Node *left, *right;} Node; Node *ConstructTree(int T[], int N);#endif

main.c

12345678910111213141516171819202122232425262728#include <stdio.h>#include "construct.h" #define MAXN 16384 int T[MAXN]; void PrintTree( Node *root ){    if (root == NULL)        return;     printf("%d\n", root->value);    PrintTree(root->left);    PrintTree(root->right);    return;} int main(){    int N;    scanf("%d", &N);    for (int i=0; i<N; i++)        scanf("%d", &T[i]);     Node *tree = ConstructTree( T, N );     PrintTree( tree );    return 0;}

## Compile

gcc -std=c99 -O2 -c construct.c -lmgcc -std=c99 -O2 -c main.c -lmgcc -std=c99 -O2 main.o construct.o -lm

• 10 points: It is guaranteed you can only find the balancing point at the top level, and none at the second level.
• 20 points: It is guaranteed you can only find the balancing point at the top level and the second level, but none at the third level.
• 70 points: It is guaranteed you can find the balancing point at the top, but nothing is guaranteed at other levels.

## Input Format

The input contains only one test case. The first line contains one integers $n$, representing the number of sequence. The second line contains $n$ integers, $w_{1}$, $w_{2}$, … , $w_{n}$, representing the weights of index.

• $0\lt n\lt 16384$

## Output Format

Print all the values of nodes in the tree by pre-ordering, and each node in a line.

## Sample Input 0

31 0 0

## Sample Output 0

100

## Sample Input 1

42 1 5 5

## Sample Output 1

5125

## Sample Input 2

92 5 2 0 7 1 1 4 3

## Sample Output 2

752024113

## Sample Input 3

82 5 2 0 7 1 7 4

## Sample Output 3

75202471