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## Fraction calculator

Дробью в математике называется число, представляющее часть единицы или несколько её частей. Обыкновенная дробь записывается в виде двух чисел, разделенных обычно горизонтальной чертой, обозначающей знак деления. Нажимая на вычисленных дробях открывается меню (Рис.2), что позволяет записать данную дробь в исходные дроби a и b, а также преобразовать на месте дроби в обыкновенную дробь, смешанную дробь или в десятичное число. Android market Казахстан - скачать бесплатно программы, игры, виджеты, приложения для Андроид устройств. Free online fraction calculator to add, subtract, multiply, divide or simplify/reduce fractions. The fraction calculator automatically reduces the result Use our Fraction Calculator to solve fraction problems, including improper fractions, mixed numbers and whole numbers. Online fraction calculator for adding, subtracting, multiplying and dividing mixed and improper fractions. To convert fraction to percent divide numerator by denominator and multiply by 100. Calculator to change fraction into percent or find percentage form of a fraction. To convert fraction to decimal number divide numerator by denominator. Calculator to find decimal form of a fraction or to change fractions into decimals Convert a fraction to a decimal value using a simple calculator. Plus, learn two methods for converting a fraction to a decimal, with examples. Convert inch fractions to decimal and back, convert inches to metric measurements and back, convert to feet and back, and find how to calculate inches. Shows you step-by-step how to add, subtract, multiply, divide your fractions! This calculator will solve your problems. Fraction as a decimal percent. Fraction to decimal number conversion calculator, how to convert and conversion table. Онлайн калькулятор дробей предназначен для расчета простых и смешанных дробей с целыми. Decimal to fraction number conversion: calculator, how to convert and conversion table. It is designed to return the approximate fraction equivalent of a decimal number. An improper fraction is a fraction whose nominator is greater than its denominator. For example, $${5 \over 4}$$. A mixed fraction is a fraction. This online calculator will find the partial fraction decomposition of the rational function, with steps shown. Fractions and decimals represent the same things: numbers that are not whole numbers. This page will show you how to convert from a fraction, into a decimal. Convert From a Decimal To a Fraction - powered by WebMath. Directions Type the Numerator in the Numerator box. Type the Denominator in the Denominator box and then click Click to Calculate If the number Decimal Fraction and Percent Conversion Calculator. Decimal to percent and percent to decimal conversion. Calculate Cost with the Discount, Multiple Discounts Please use this form if you would like to have this math solver on your website, free of charge. Additional Information There are four different types of fractions: common fractions, decimal fractions, percentages and ratios. When the word 'fraction' Subwoofer wiring and Subwoofer enclosure calculators. Fraction to Decimal, Parallel wiring, Series wiring, Port Length and Volume Calculators. Play Fraction Circles Now at hoodamath.com where Free Online Cool Math Games are Unblocked at School. Easily generate monthly and yearly amortiztion schedules for a proposed loan with our loan amortization calculator. Partial Fractions Calculator - find the partial fractions of a fractions step-by-step. Use these colorful fraction strips to model fraction relationships and solve mixed fraction problems. Free to use, priceless for teaching. Time Master II - Time Calculator for Pilots, Accounting, Broadcasting; Ultra Measure Master - Most Complete, Easy-to-Use Metric Conversion Calculator. Find the percent of a slope using rise over run with this Slope Percent Calculator. In mathematics, a continued fraction is an expression obtained through an iterative process of representing a number as the sum of its integer part and the reciprocal. R duire Pour ne pas bloquer votre navigateur, il ne faut pas introduire des nombres de plus de trois chiffres surtout. Maths is always daunting, there’s no way around it. But thanks to our collection of maths calculators, everyone can perform and understand useful mathematical.


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Encyclopedia > Boundary layer

In physics and fluid mechanics, a boundary layer is that layer of fluid in the immediate vicinity of a bounding surface. In the Earth's atmosphere the planetary boundary layer is the air layer near the ground affected by diurnal heat, moisture or momentum transfer to or from the surface. On an aircraft wing the boundary layer is the part of the flow close to the wing. The boundary layer effect occurs at the field region in which all changes occur in the flow pattern. The boundary layer distorts surrounding nonviscous flow. It is a phenomenon of viscous forces. This effect is related to the Reynolds number. Physics (Greek: (phÃºsis), nature and (phusikÃ©), knowledge of nature) is the science concerned with the discovery and characterization of universal laws which govern matter, energy, space, and time. ... Fluid mechanics is the subdiscipline of continuum mechanics that studies fluids, that is, liquids and gases. ... A fluid is defined as a substance that continually deforms (flows) under an applied shear stress regardless of the magnitude of the applied stress. ... Layers of Atmosphere - not to scale (NOAA)[3] Earths atmosphere is a layer of gases surrounding the planet Earth and retained by the Earths gravity. ... The planetary boundary layer (PBL) is also known as the atmospheric boundary layer (ABL). ... Look up aircraft in Wiktionary, the free dictionary. ... A Laughing Gull with its wings extended in a gull wing profile Aircraft wing planform shapes: a swept wing KC-10 Extender (top) refuels a trapezoid-wing F/A-22 Raptor A wing is a surface used to produce lift and therefore flight, for travel in the air or another... flOw is a Flash game created by Jenova Chen and Nicholas Clark. ... A pattern is a form, template, or model (or, more abstractly, a set of rules) which can be used to make or to generate things or parts of a thing, especially if the things that are generated have enough in common for the underlying pattern to be inferred or discerned... Viscosity is a measure of the resistance of a fluid to deform under shear stress. ... In physics, force is an influence that may cause an object to accelerate. ... In fluid mechanics, the Reynolds number is the ratio of inertial forces (vsÏ) to viscous forces (Î¼/L) and consequently it quantifies the relative importance of these two types of forces for given flow conditions. ...

Laminar boundary layers come in various forms and can be loosely classified according to their structure and the circumstances under which they are created. The thin shear layer which develops on an oscillating body is an example of a Stokes layer, whilst the Blasius boundary layer refers to the well-known similarity solution for the steady boundary layer attached to a flat plate held in an oncoming unidirectional flow. When a fluid rotates, viscous forces may be balanced by Coriolis effects, rather than convective inertia, leading to the formation of an Eckman layer. Thermal boundary layers also exist in heat transfer. Multiple types of boundary layers can coexist near to a surface simultaneously. A Blasius boundary layer describes the steady two-dimensional boundary layer that forms on a semi-infinite plate which is held parallel to a constant unidirectional flow . ... Several equivalence relations in mathematics are called similarity. ... In standard boundary-layer theory, the effects of viscous diffusion are usually balanced by convective inertia. ...

## Contents

The aerodynamic boundary layer was first defined by Ludwig Prandtl in a paper presented at the third Congress of Mathematicians in Heidelberg, Germany. It allows aerodynamicists to simplify the equations of fluid flow by dividing the flow field into two areas: one inside the boundary layer, where viscosity is dominant and the majority of the drag experienced by a body immersed in a fluid is created, and one outside the boundary layer where viscosity can be neglected without significant effects on the solution. This allows a closed-form solution for the flow in both areas, which is a significant simplification over the solution of the full Navier-Stokes equations. The majority of the heat transfer to and from a body also takes place within the boundary layer, again allowing the equations to be simplified in the flow field outside the boundary layer. This article does not cite any references or sources. ... Ludwig Prandtl (4 February 1875 - 15 August 1953) was a German physicist. ... Map of Germany showing Heidelberg Heidelberg (halfway between Stuttgart and Frankfurt) is a city in Baden-Württemberg, Germany. ... Viscosity is a measure of the resistance of a fluid to deform under shear stress. ... An object falling through a gas or liquid experiences a force in direction opposite to its motion. ... The Navier-Stokes equations, named after Claude-Louis Navier and George Gabriel Stokes, are a set of equations which describe the motion of fluid substances such as liquids and gases. ... In physics, heat, symbolized by Q, is defined as transfer of thermal energy [1] Generally, heat is a form of energy transfer associated with the different motions of atoms, molecules and other particles that comprise matter when it is hot and when it is cold. ...

The thickness of the velocity boundary layer is normally defined as the distance from the solid body at which the flow velocity is 99% of the freestream velocity, that is, the velocity that is calculated at the surface of the body in an inviscid flow solution. The no-slip condition requires that the flow velocity at the surface of a solid object is zero and that the fluid temperature is equal to the temperature of the surface. The flow velocity will then increase rapidly within the boundary layer, governed by the boundary layer equations, below. The thermal boundary layer thickness is similarly the distance from the body at which the temperature is 99% of the temperature found from an inviscid solution. The ratio of the two thicknesses is governed by the Prandtl number. If the Prandtl number is 1, the two boundary layers are the same thickness. If the Prandtl number is greater than 1, the thermal boundary layer is thinner than the velocity boundary layer. If the Prandtl number is less than 1, which is the case for air at standard conditions, the thermal boundary layer is thicker than the velocity boundary layer. In fluid dynamics, the no-slip condition states that fluids stick to surfaces past which they flow. ... The Prandtl Number is a dimensionless number approximating the ratio of momentum diffusivity and thermal diffusivity. ...

In high-performance designs, such as sailplanes and commercial transport aircraft, much attention is paid to controlling the behavior of the boundary layer to minimize drag. Two effects must to be considered. First, the boundary layer adds to the effective thickness of the body, through the displacement thickness, hence increasing the pressure drag. Secondly, the shear forces at the surface of the wing create skin friction drag. Gliders are un-powered heavier-than-air aircraft. ... Displacement thickness is distance by which a surface would have to be moved parallel to itself towards the reference plane in an ideal fluid stream of velocity to give the same volumetric flow as occurs between the surface and the reference plane in a real fluid. ... In physics and mechanics, shear refers to a deformation that causes parallel surfaces to slide past one another (as opposed to compression and tension, which cause parallel surfaces to move towards or away from one another). ... In Aerodynamics, skin friction is the component of parasitic drag arising from the friction of the fluid against the skin of the object that is moving through it. ...

At high Reynolds numbers, typical of full-sized aircraft, it is desirable to have a laminar boundary layer. This results in a lower skin friction due to the characteristic velocity profile of laminar flow. However, the boundary layer inevitably thickens and becomes less stable as the flow develops along the body, and eventually becomes turbulent, the process known as boundary layer transition. One way of dealing with this problem is to suck the boundary layer away through a porous surface (see Boundary layer suction). This can result in a reduction in drag, but is usually impractical due to the mechanical complexity involved. In fluid mechanics, the Reynolds number is the ratio of inertial forces (vsÏ) to viscous forces (Î¼/L) and consequently it quantifies the relative importance of these two types of forces for given flow conditions. ... laminar and turbulent water flow over the hull of a submarine In fluid dynamics, laminar flow is a flow regime characterized by high momentum diffusion, low momentum convection, and pressure and velocity independence from time. ... Turbulent flow around an obstacle; the flow further away is laminar Laminar and turbulent water flow over the hull of a submarine Turbulence creating a vortex on an airplane wing In fluid dynamics, turbulence or turbulent flow is a flow regime characterized by low-momentum diffusion, high momentum convection, and... The process of a laminar boundary layer becoming turbulent is known as boundary layer transition. ... A pore, in general, is some form of opening, usually very small. ... Boundary layer suction is technique in which an air pump is used to extract the boundary layer at the wing of an aircraft. ...

At lower Reynolds numbers, such as those seen with model aircraft, it is relatively easy to maintain laminar flow. This gives low skin-friction, which is desirable. However, the same velocity profile which gives the laminar boundary layer its low skin friction also causes it to be badly affected by adverse pressure gradients. As the pressure begins to recover over the rear part of the wing chord, a laminar boundary layer will tend to separate from the surface. Such separation causes a large increase in the pressure drag, since it greatly increases the effective size of the wing section. In these cases, it can be advantageous to deliberately trip the boundary layer into turbulence at a point prior to the location of laminar separation, using a turbulator. The fuller velocity profile of the turbulent boundary layer allows it to sustain the adverse pressure gradient without separating. Thus, although the skin friction is increased, overall the drag is decreased. This is the principle behind the dimpling on golf balls, as well as vortex generators on light aircraft. Special wing sections have also been designed which tailor the pressure recovery so that laminar separation is reduced or even eliminated. This represents an optimum compromise between the pressure drag from flow separation and skin friction the induced turbulence. In fluid mechanics, the Reynolds number is the ratio of inertial forces (vsÏ) to viscous forces (Î¼/L) and consequently it quantifies the relative importance of these two types of forces for given flow conditions. ... An adverse pressure gradient occurs when the static pressure increases in the direction of the flow. ... Airflow separating from a wing which is at a high angle of attack All solid objects travelling through a fluid (or alternatively a stationary object exposed to a moving fluid) acquire a boundary layer of fluid around them where friction between the fluid molecules and the objects rough surface... Parasitic drag (also called parasite drag) is drag caused by moving a solid object through a fluid. ... A turbulator is a device for improving the flow of air over a wing. ... 1967 Model Cessna 182K in flight showing after-market vortex generators on the wing leading edge A vortex generator is an aerodynamic surface, basically a small vane, that creates a vortex. ...

## Boundary layer equations

The deduction of the boundary layer equations was perhaps one of the most important advances in aerodynamics. Using an order of magnitude analysis, the well-known governing Navier-Stokes equations of viscous fluid flow can be greatly simplified within the boundary layer. Notably, the characteristic of the partial differential equations (PDE) becomes parabolic, rather than the elliptical form of the full Navier-Stokes equations. This greatly simplifies the solution of the equations. By making the boundary layer approximation, the flow is divided into an inviscid portion (which is easy to solve by a number of methods) and the boundary layer, which is governed by an easier to solve PDE. The Navier-Stokes equations for a two-dimensional steady incompressible flow in cartesian coordinates are given by The Navier-Stokes equations, named after Claude-Louis Navier and George Gabriel Stokes, are a set of equations which describe the motion of fluid substances such as liquids and gases. ... Viscosity is a measure of the resistance of a fluid to deformation under shear stress. ... A fluid is defined as a substance that continually deforms (flows) under an applied shear stress regardless of the magnitude of the applied stress. ... flOw is a Flash game created by Jenova Chen and Nicholas Clark. ... In linear algebra, the characteristic equation of a square matrix A is the equation in one variable λ where I is the identity matrix. ... In mathematics, a partial differential equation (PDE) is a relation involving an unknown function of several independent variables and its partial derivatives with respect to those variables. ... In mathematics, a partial differential equation (PDE) is a relation involving an unknown function of several independent variables and its partial derivatives with respect to those variables. ... In mathematics, an incompressible surface is a kind of two-dimensional surface inside of a 3-manifold. ...

${partial uoverpartial x}+{partial voverpartial y}=0$
$u{partial u over partial x}+v{partial u over partial y}=-{1over rho} {partial p over partial x}+{nu}({partial^2 uover partial x^2}+{partial^2 uover partial y^2})$
$u{partial v over partial x}+v{partial v over partial y}=-{1over rho} {partial p over partial y}+{nu}({partial^2 vover partial x^2}+{partial^2 vover partial y^2})$

where u and v are the velocity components, ρ is the density, p is the pressure, and ν is the kinematic viscosity of the fluid at a point. Viscosity is a measure of the resistance of a fluid to deformation under shear stress. ...

The approximation states that, for a sufficiently high Reynolds number the flow over a surface can be divided into an outer region of inviscid flow unaffected by viscosity (the majority of the flow), and a region close to the surface where viscosity is important (the boundary layer). Let u and v be streamwise and transverse (wall normal) velocities respectively inside the boundary layer. Using asymptotic analysis, it can be shown that the above equations of motion reduce within the boundary layer to become In fluid mechanics, the Reynolds number is the ratio of inertial forces (vsÏ) to viscous forces (Î¼/L) and consequently it quantifies the relative importance of these two types of forces for given flow conditions. ... In mathematics and applications, particularly the analysis of algorithms, asymptotic analysis is a method of classifying limiting behaviour, by concentrating on some trend. ...

${partial uoverpartial x}+{partial voverpartial y}=0$
$u{partial u over partial x}+v{partial u over partial y}=-{1over rho} {partial p over partial x}+{nu}{partial^2 uover partial y^2}$

and the remarkable result that

${1over rho} {partial p over partial y}=0$

The asymptotic analysis also shows that v, the wall normal velocity, is small compared with u the streamwise velocity, and that variations in properties in the streamwise direction are generally much lower than those in the wall normal direction.

Since the static pressure p is independent of y, then pressure at the edge of the boundary layer is the pressure throughout the boundary layer at a given streamwise position. The external pressure may be obtained through an application of Bernoulli's Equation. Let u0 be the fluid velocity outside the boundary layer, where u and u0 are both parallel. This gives upon substituting for p the following result In fluid dynamics, Bernoullis equation, derived by Daniel Bernoulli, describes the behavior of a fluid moving along a streamline. ...

$u{partial u over partial x}+v{partial u over partial y}=u_0{partial u_0 over partial x}+{nu}{partial^2 uover partial y^2}$

with the boundary condition

${partial uoverpartial x}+{partial voverpartial y}=0$

For a flow in which the static pressure p also does not change in the direction of the flow then

${partial poverpartial x}=0$

so u0 remains constant.

Therefore, the equation of motion simplifies to become

$u{partial u over partial x}+v{partial u over partial y}={nu}{partial^2 uover partial y^2}$

These approximations are used in a variety of practical flow problems of scientific and engineering interest. The above analysis is for any instantaneous laminar or turbulent boundary layer, but are used mainly in laminar flow studies since the mean flow is also the instantaneous flow because there are no velocity fluctuations present. laminar and turbulent water flow over the hull of a submarine In fluid dynamics, laminar flow is a flow regime characterized by high momentum diffusion, low momentum convection, and pressure and velocity independence from time. ... Turbulent flow around an obstacle; the flow further away is laminar Laminar and turbulent water flow over the hull of a submarine Turbulence creating a vortex on an airplane wing In fluid dynamics, turbulence or turbulent flow is a flow regime characterized by low-momentum diffusion, high momentum convection, and... In statistics, mean has two related meanings: Look up mean in Wiktionary, the free dictionary. ...

## Turbulent boundary layers

The treatment of turbulent boundary layers is far more difficult due to the time-dependent variation of the flow properties. One of the most widely used techniques in which turbulent flows are tackled is to apply Reynolds decomposition. Here the instantaneous flow properties are decomposed into a mean and fluctuating component. Applying this technique to the boundary layer equations give the full turbulent boundary layer equations not often given in literature, viz. In fluid dynamics and turbulence, Reynolds decomposition is a mathematical technique to separate the average and fluctuating parts of a quantity. ...

${partial overline{u}overpartial x}+{partial overline{v}overpartial y}=0$
$overline{u}{partial overline{u} over partial x}+overline{v}{partial overline{u} over partial y}=-{1over rho} {partial overline{p} over partial x}+{nu}({partial^2 overline{u}over partial x^2}+{partial^2 overline{u}over partial y^2})-frac{partial}{partial y}(overline{u'v'})-frac{partial}{partial x}(overline{u'^2})$
$overline{u}{partial overline{v} over partial x}+overline{v}{partial overline{v} over partial y}=-{1over rho} {partial overline{p} over partial y}+{nu}({partial^2 overline{v}over partial x^2}+{partial^2 overline{v}over partial y^2})-frac{partial}{partial x}(overline{u'v'})-frac{partial}{partial y}(overline{v'^2})$

Using the same order-of-magnitude analysis as for the instantaneous equations, these turbulent boundary layer equations generally reduce to become in their classical form:

${partial overline{u}overpartial x}+{partial overline{v}overpartial y}=0$
$overline{u}{partial overline{u} over partial x}+overline{v}{partial overline{u} over partial y}=-{1over rho} {partial overline{p} over partial x}+{nu}{partial^2 overline{u}over partial y^2}-frac{partial}{partial y}(overline{u'v'})$
${partial overline{p} over partial y}=0$

The additional term $overline{u'v'}$ in the turbulent boundary layer equations is known as the Reynolds shear stress and is unknown a priori. The solution of the turbulent boundary layer equations therefore necessitate the use of a turbulence model, which aims to express the Reynolds shear stress in terms of known flow variables or derivatives. The lack of accuracy and generality of such models is the single major obstacle which inhibits the successful prediction of turbulent flow properties in modern aerodynamics. The terms a priori and a posteriori are used in philosophy to distinguish between two different types of propositional knowledge. ... Turbulence modeling is the area of physical modeling where a simpler mathematical model than the full time dependent Navier-Stokes Equations is used to predict of the effects of turbulence. ...

## Boundary layer turbine

This effect was exploited in the Tesla turbine, patented by Nikola Tesla in 1913. It is referred to as a bladeless turbine because it uses the boundary layer effect and not a fluid impinging upon the blades as in a conventional turbine. Boundary layer turbines are also known as cohesion-type turbines, bladeless turbine, and Prandtl layer turbine (after Ludwig Prandtl). The Tesla turbine is a bladeless turbine design patented by Nikola Tesla in 1913. ... Nikola Tesla (1856-1943)[1] was a world-renowned Serbian inventor, physicist, mechanical engineer and electrical engineer. ... A Siemens steam turbine with the case opened. ... Ludwig Prandtl (4 February 1875 - 15 August 1953) was a German physicist. ...

Boundary layer suction is technique in which an air pump is used to extract the boundary layer at the wing of an aircraft. ...

Results from FactBites:

 Boundary Layer (785 words) The details of the flow within the boundary layer are very important for many problems in aerodynamics, including wing stall, the skin friction drag on an object, and the heat transfer that occurs in high speed flight. For lower Reynolds numbers, the boundary layer is laminar and the streamwise velocity changes uniformly as one moves away from the wall, as shown on the left side of the figure. The effects of the boundary layer on lift are contained in the lift coefficient and the effects on drag are contained in the drag coefficient.
More results at FactBites »

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data_mining:neural_network:loss_functions

# Loss functions

Difference between two probability distributions

$-(y log(\hat{y}) + (1-y) log(1-\hat{y})$

or

$-1/N \sum_{i=1}^N (y_i log(\hat{y_i}) + (1-y_i) log(1-\hat{y_i})$

Recap entropy:

In bin classification entropy for distribution q(y) with 50:50 classes is: $H(q) = log(2)$

For other distrubition (and in general) with C classes, entropy of distribution is $H(q) = - \sum_{c=1}^C q(y_c) * log(q(y_c))$

Small value of $q(y_c)$ leads to large negative log (multiple times of q): math.log(0.01) = -4.6

Large value of $q(y_c)$ leads to small negative log: math.log(0.99) = -0.01

More classes, higher entropy

Recap cross-entropy:

$H_p(q) = - \sum_{c=1}^C q(y_c) * log(p(y_c))$ where p is other distribution

When $p == q$, then $H_p(q) == H(q)$, so cross-entropy >= entropy.

Recap KL-Divergence:

Is difference between cross-entropy and entropy.

$D_{KL}(q||p) = H_p(q) - H(q) = \sum_{c=1}^C q(y_c) (log(q(y_c) - log(p(y_c)))$

For an algo we need to find closes $p$ for $q$ by minimizing cross-entropy.

In training we have $N$ samples. For one particular example the distribution is known to which class it belongs. The loss function should minimize the average cross-entropy.

Outcome: Scalar [0,1] using sigmoid

$-\sum_{i}^C(y_i log(\hat{y_i})$ $C$ is number of classes

Outcome: Vector [0,1] using softmax

$-\sum_{i}^C(y_i log(\hat{y_i}) + (1-y_i) log(1-\hat{y_i})$

Ouctome: Vector [0,1] using sigmoid

• data_mining/neural_network/loss_functions.txt


text:
# How to make biblatex treat the prenote like an author list?

I am trying to customize a biblatex citation style (adapted from biblatex-juradiss). I have the problem that in my text I use many \cite commands which have two or more authors given in the prenote, and I want them formatted similarly to the editors of the publication that is referenced. So if I write

\cite[Smith, John and Johnson, Henry][23]{biblabel}

I would like the output to look like:

Smith / Johnson in: Editora / Editorb, other bibliographical data, p. 23.

where Editora and Editorb are the editor's surnames from the entry in my .bib file, and "other bibliographical data" stands for whatever biblatex would normally output after the editors. Biblatex does the editor list nicely, and I'd like the authors to look the same.

Please note that the slashes are not in italics, so I can't just give the additional author names inline with the slash in the prenote. Also, biblatex uses whitespace in the editor list which is somewhat narrower than usual.

• The issue with this is that the prenote argument would have to be passed to Biber from within the document, as Biber has to parse name list like these for biblatex to make sense of them - and I'm not sure how that would work. You could create a dummy entry for "Smith, John and Johnson, Henry" though like @misc{smithjohnson, author{Smith, John and Johnson, Henry}} in the .bib file and then do \cite[\citeauthor{smithjohnson}][23]{biblabel}. (If you go for this option, biblatex. comfortable author citing might be of interest to you.) – moewe May 23 '14 at 5:10
• Instead of using @misc for the dummy entry smithjohnson, it might be better to create a new entry type @person/@author that does not appear in the bibliography even if cited and just holds the names. See also Create a register of persons with biblatex. – moewe May 23 '14 at 5:14
• Since last coming here, I have understood more of the issue. The core problem is biblatex's data model: I was going to use the prenote for displaying a list of authors. This would require to treat the data entry for the prenote as a name list, while it is a field, which has no index. The "obvious" solution should be to go to the data model definition and change the type of the prenote entry from "field" to "names", and then in the .cbx file copy the author format into that of the prenote field. Sadly, this looks like more than I can chew at this time, as submission date is drawing closer. – Michael Aug 9 '14 at 20:21
• The problem with this approach though is that postnote and prenote are treated differently from the other fields by biblatex. While the standard fields are consumed (and digested) by Biber, Biber does not get to see either postnote or prenote, neither are the two fields mentioned in any of the data model files (so one would not be able to change the type from field to name; and it would be necessary for Biber to be able to see the post/prenote fields). – moewe Aug 10 '14 at 8:49

The main problem is that name lists have to be parsed by Biber/BibTeX so they can be used properly. But the prenote and postnote field are local to the .tex file and are not passed to Biber for consumption.

So we either have to pass the prenote to Biber by means of writing to a temporary file, or we have to fake Biber's name parsing from within the document with biblatex/LaTeX itself.

The first method is can be implemented with limited extra work. The command \formatasname can be used anywhere to return a list of names as they would be written in a .bib file as biblatex would print them.

\formatasname writes the names it is given to a temporary .bib file and then cites them. In the process the names are parsed by Biber and are formatted just like other names.

Just be careful that you have no .bib file ending with -formatnames.bib in your folder. That file is automatically created by the code below and will be overwritten without warning.

\documentclass[british]{report}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{babel}
\usepackage{csquotes}

\usepackage[style=authoryear, backend=biber]{biblatex}

\makeatletter
\DeclareCiteCommand{\nfblx@formatasname}
{}
{\printnames[labelname]{author}}
{}
{}

\def\nfblx@bibfile@name{\jobname -formatnames.bib}
\newwrite\nfblx@bibfile
\immediate\openout\nfblx@bibfile=\nfblx@bibfile@name
\newcounter{nfblx@name}
\setcounter{nfblx@name}{0}
\newcommand*{\formatasname}[1]{%
\stepcounter{nfblx@name}%
\edef\nfblx@tmp@cite{%
\noexpand\nfblx@formatasname{nfblx@name@\the\value{nfblx@name}}%
}%
\nfblx@tmp@cite
\immediate\write\nfblx@bibfile{%
@misc{nfblx@name@\the\value{nfblx@name}, author = {\unexpanded{#1}}, %
options = {dataonly=true},}%
}%
}

\AtEndDocument{%
\closeout\nfblx@bibfile}


text:
RE: UAS: Seems idiotic regulation isn't without its compensation! :-)

Kermit2

Joined Feb 5, 2010
4,162
G I Joe would be better off with civilian health care.
I wouldn't let the docs at the VA hospital wipe his shiney plastic a$$. jpanhalt Joined Jan 18, 2008 10,982 Just wait until the same idiocy infects the DOT! Four-year olds will need a license to drive a drone "vehicle." And of course, laws against DWI will be strictly enforced. John wayneh Joined Sep 9, 2010 16,546 I think in many cases the "castle doctrine" will trump some silly FAA ruling. If a device is in your airspace and you have reasonable claim of being threatened, no jury is going to send you to jail for eliminating the threat. It'll get really messy when the property line is unclear, and the elevation is "high". A device hovering 10 feet above your patio and unquestionably over your property is very different than one passing over the edge of your property at 100 feet elevation. The law will evolve, slowly. In all matters, bad events make for bad laws. This will be no different. Thread Starter Hypatia's Protege Joined Mar 1, 2015 3,226 G I Joe would be better off with civilian health care. I wouldn't let the docs at the VA hospital wipe his shiney plastic a$$.
No worries of that! -- Where the VA is concerned he'd succumb (if only to old age) long before they laid hands on him!

Best regards
HP

Hypatia's Protege

Joined Mar 1, 2015
3,226
Just wait until the same idiocy infects the DOT! Four-year olds will need a license to drive a drone "vehicle." And of course, laws against DWI will be strictly enforced.

John
Re: the DMV --- In a (somewhat) parallel vein 'we' recently witnessed a conviction for intoxicated operation of an 'easy chair'
http://www.mprnews.org/story/2009/10/22/la-z-boy-dwi

Curiously, the authorities raised no objection to his not-infrequent (and, quite-likely, 'less than sober') operation of the (unregistered and unregistrable) contrivance on public roadways (including I-35!) in the months preceding the incident... --- Guess there's something sacred about pub car-parks?

Best regards
HP

Last edited:

Aleph(0)

Joined Mar 14, 2015
597
Curiously, the authorities raised no objection to his not-infrequent (and, quite-likely, 'less than sober') operation of the (unregistered and unregistrable) contrivance on public roadways (including I-35!) in the months preceding the incident... --- Guess there's something sacred about pub car-parks?
Anyhow the crash was behind the pharmacy where lounge patrons aren't supposed to park! I say that was the real reason Denny's luck ran out! But what do I know?

Hypatia's Protege

Joined Mar 1, 2015
3,226
I think in many cases the "castle doctrine" will trump some silly FAA ruling. If a device is in your airspace and you have reasonable claim of being threatened, no jury is going to send you to jail for eliminating the threat.

It'll get really messy when the property line is unclear, and the elevation is "high". A device hovering 10 feet above your patio and unquestionably over your property is very different than one passing over the edge of your property at 100 feet elevation.

The law will evolve, slowly. In all matters, bad events make for bad laws. This will be no different.
Indeed! A 'straight answer' as to the lower limits of the NAS seems rather elusive --- Though consensus puts it at 200 feet 'above average terrain' -- I've yet to locate an official citation...

Best regards
HP

Hypatia's Protege

Joined Mar 1, 2015
3,226
Anyhow the crash was behind the pharmacy where lounge patrons aren't supposed to park! I say that was the real reason Denny's luck ran out! But what do I know?
Ewwwwwe!!! At a break-neck 2 MPH it must have been grisly! Wrenched metal, blood, gibs --- Best not to dwell on it

Best regards
HP

jpanhalt

Joined Jan 18, 2008
10,982
Unfortunately, you will never get a straight answer as to what airspace ownership means to the fee simple property owner. Two-hundred feet may sound reasonable, but there are obvious exceptions. One is the overhanging branch from a neighbor's tree. Better check the "tree laws"in your state before you do anything about it. Another is a low flying crop duster that may need to extend its turn over your property to be safe.

And then there is the EPA, which by some interpretations extends it authority from the limits of outer space to well beneath the surface. Here is a statement from the Ohio bar on fee simple (https://www.ohiobar.org/General Resources/LawandYou/TLAY_Chapter08.pdf )
Today, the airspace above the parcel is not exclusive; it is considered to be part of the public domain and subject to rights of navigation, which means that members of the general public can travel through or use that airspace.

Does an antenna on your property constitute something that is reasonably necessary? One had better check with local zoning before doing that.

Personally, I would consider anything I could swat or knock down with a high-volume, high-pressure water hose (about 30 feet vertical with my sprayer) or below the roof line of my house is in my protected space. Do I have the right to use a rifle to shoot it down? Probably not.

As for being the subject of the average voyeur with a drone, there are certain protections, but you need to identify the operator of the drone. You probably have a case for trespass, but not for use of deadly force. The Castle Doctrine does not apply outside one's home, and in some jurisdictions does not even apply to an attached garage. The trespass offense could be a powerful disincentive to the drone operator depending on what images are captured, particularly if children are involved.

John

Hypatia's Protege

Joined Mar 1, 2015
3,226
Unfortunately, you will never get a straight answer as to what airspace ownership means to the fee simple property owner. Two-hundred feet may sound reasonable, but there are obvious exceptions. One is the overhanging branch from a neighbor's tree. Better check the "tree laws"in your state before you do anything about it. Another is a low flying crop duster that may need to extend its turn over your property to be safe.

And then there is the EPA, which by some interpretations extends it authority from the limits of outer space to well beneath the surface. Here is a statement from the Ohio bar on fee simple (https://www.ohiobar.org/General Resources/LawandYou/TLAY_Chapter08.pdf )

View attachment 104878

Does an antenna on your property constitute something that is reasonably necessary? One had better check with local zoning before doing that.

Personally, I would consider anything I could swat or knock down with a high-volume, high-pressure water hose (about 30 feet vertical with my sprayer) or below the roof line of my house is in my protected space. Do I have the right to use a rifle to shoot it down? Probably not.

As for being the subject of the average voyeur with a drone, there are certain protections, but you need to identify the operator of the drone. You probably have a case for trespass, but not for use of deadly force. The Castle Doctrine does not apply outside one's home, and in some jurisdictions does not even apply to an attached garage. The trespass offense could be a powerful disincentive to the drone operator depending on what images are captured, particularly if children are involved.

John
The willingness with with which many cast aside their liberty in the name of "The greater public interest" and (alleged) "safety" is nothing short of chilling!

Does an antenna on your property constitute something that is reasonably necessary? One had better check with local zoning before doing that.
At my Um... 'request' The FCC 'schooled' my zoning commission on that score -- to the tune of \$50K --- Not often, but sometimes even Feds can be beautiful!

Best regards
HP

jpanhalt

Joined Jan 18, 2008
10,982
I agree. Fundamental rights are constantly being eroded by the increasing need to petition the government before the exercise of those rights.

Unfortunately, the government has cost on its side. Most of us just cave in and do it to avoid the ruinous and potentially non-recoverable costs.

John

OBW0549

Joined Mar 2, 2015
3,566
I think there's another consideration here, aside from any issues of property rights, airspace rights or privacy rights: as any responsible gun owner will tell you, firing a gun up into the air, for any reason whatsoever, is a really, REALLY stupid thing to do. Basic gun safety demands that you always be mindful of where your bullet is going to land if it's not stopped by your target.

Hypatia's Protege

Joined Mar 1, 2015
3,226
I think there's another consideration here, aside from any issues of property rights, airspace rights or privacy rights: as any responsible gun owner will tell you, firing a gun up into the air, for any reason whatsoever, is a really, REALLY stupid thing to do. Basic gun safety demands that you always be mindful of where your bullet is going to land if it's not stopped by your target.
I hear that! - I suspect the return velocity of even a 1 Oz slug would likely be very dangerous -- to say nothing of the hazard to legitimate air traffic...

Best regards
HP

wayneh

Joined Sep 9, 2010
16,546
As for being the subject of the average voyeur with a drone, there are certain protections, but you need to identify the operator of the drone. You probably have a case for trespass, but not for use of deadly force.
The drone is not alive, so I don't think "deadly" force has any relevance. If I can legally and safely fire my 12-gauge into the air above my house, if there happens to be a drone in the way, too bad for the drone.

jpanhalt

Joined Jan 18, 2008
10,982
The drone is not alive, so I don't think "deadly" force has any relevance. If I can legally and safely fire my 12-gauge into the air above my house, if there happens to be a drone in the way, too bad for the drone.
Not if the FAA gets involved. My use of "deadly force" was ambiguous. I meant shoot it down.

BTW, should I be so inclined, I would use a shotgun too. Much less problem stray projectiles. If it were within 25 feet, my 18-1/8" barreled, 12 gauge would the job, and the sound alone might discourage the operator from investigating further.

John

wayneh

Joined Sep 9, 2010
16,546
If I had a capture net, I'd use that. Free drone!

boatsman

Joined Jan 17, 2008
186
A follow up from #12's post. Click on the site he quotes. Go to the bottom of the page to 'youtube video' click on it and go to 'prototype quadrotor with machine gun. This is the ultimate in repeling unwanted aerial intruders; every household should have one!

#12

Joined Nov 30, 2010
18,216
Go to the bottom of the page to 'youtube video'
That wasn't easy to find, so here's the link:


text:
#### black in color structures (24.10.01)

##### Dr. U. Kasper

I have some PLOT (not plot) structures with diﬀerent COLORs and I am unable to give one of them the color black. What are the values (or the value) for RGB, HUE or HSV? Or is there an other option for this?

##### Theodore Kolokolnikov (25.10.01)

The trick I use is the following. Suppose that ’pic’ is your plot structure, for example:

Then you can do this:

This will turn everything in your picture black.

##### Robert Israel (25.10.01)

Black isCOLOUR(RGB,0,0,0). It doesn’t have a HUE. In HSV it would be COLOUR(HSV,h,s,0) where h and s can be anything from 0 to 1.

You can get the RGB values for the named colours from the table plot/colortable. So, for example:


text:
User ppyang - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T07:36:22Z http://mathoverflow.net/feeds/user/19399 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/122615/derivative-of-the-regularized-upper-incomplete-gamma-function Derivative of the regularized upper incomplete gamma function ppyang 2013-02-22T08:22:40Z 2013-02-27T00:27:58Z
Hello everyone!

I have a question about the derivative of the regularized upper incomplete gamma function. Considering the gamma function and the incomplete gamma function \begin{eqnarray} \Gamma(x)&=&\int_0^\infty{t^{x-1}e^{-t}\mathrm{d}t},\\ \Gamma(x,\lambda)&=&\int_\lambda^\infty{t^{x-1}e^{-t}\mathrm{d}t},\end{eqnarray} where the quotient of $\Gamma(x,\lambda)$ divided by $\Gamma(x)$ is usually denoted as regularized upper incomplete gamma function $$Q(x,\lambda)=\frac{\Gamma(x,\lambda)}{\Gamma(x)},$$ there seems to be a precise estimation for the derivative $\frac{\partial Q(x,\lambda)}{\partial x}$, but I cannot prove it.

With numerical experiment, I found that given $\lambda$ is not small, $\frac{\partial Q(x,\lambda)}{\partial s}$ is almost proportional to $$\frac{\lambda^xe^{-\lambda}}{x^xe^{-x}}=\exp\left(-\left(x\log x-x\log\lambda-x+\lambda\right)\right).$$ Denoting $$I(\lambda)=\int_0^\infty{\frac{\lambda^xe^{-\lambda}}{x^xe^{-x}}\mathrm{d}x},$$ the partial derivative $\frac{\partial Q(x,\lambda)}{\partial x}$ seems to have the following approximation $$\frac{\partial Q(x,\lambda)}{\partial x} \approx\frac{\lambda^xe^{-\lambda}}{x^xe^{-x}}/I(\lambda).$$

I give the comparison of these two formulas from the numerical experiments with Matlab in following figures, where the blue solid line and green dotted line represent $\frac{\partial Q(x,\lambda)}{\partial x}$ and $\frac{\lambda^xe^{-\lambda}}{x^xe^{-x}}/I(\lambda)$ respectively.

From these figures, it's obvious that as the increase of $\lambda$, the two lines approach to be equal. However, I cannot prove such a relationship due to the complexity of the derivative $\frac{\partial Q(x,\lambda)}{\partial x}$.

Could you help me with some comments about this question? Any suggestions will be helpful and thank you very much!
 http://mathoverflow.net/questions/122200/are-there-a-group-of-mappings-from-n-1-dim-space-to-an-n-1-sphere-guaranteein Are there a group of mappings from (n-1)-dim space to an (n-1)-sphere guaranteeing the orthogonality of images? ppyang 2013-02-18T17:26:56Z 2013-02-18T17:26:56Z
Hello, everyone.

As we know that in an $n$-dimensional Euclidean space $\mathbb{R}^n$, there exists a continuous bijective mapping from a subset $V^{n-1}\subseteq\mathbb{R}^{n-1}$ to a unit $(n-1)$-sphere $S^{n-1}=\{\mathbf{x}\in\mathbb{R}^n:\|\mathbf{x}\|=1\}$ since the $(n-1)$-sphere is indeed an $(n-1)$-dimensional manifold.

The most straightforward example is obtained from the spherical coordinate transformation by setting $r=1$, which bijectively maps a $(n-1)$-dimenional cuboid $V^{n-1}=[0,\pi]\times\ldots\times[0,\pi]\times[0,2\pi)$ to $S^{n-1}$. (Refer to http://en.wikipedia.org/wiki/N-sphere#Spherical_coordinates)

Then I wonder if there exist $n$ such mappings $\{\mathbf{p}_i:V^{n-1}\mapsto S^{n-1}\subset\mathbb{R}^n\}_{i=1}^n$ so that the $n$ images $\mathbf{p}_1(\theta),\ldots,\mathbf{p}_n(\theta)$ corresponding to any same inverseimage $\theta\in V^{n-1}$ are orthogonal to each other.

Consider the case where $n=2$. In $\mathbb{R}^2$, we can map a line segment $[0,2\pi)\subseteq\mathbb{R}^1$ to a unit circle $S^1=\{\mathbf{x}\in\mathbb{R}^2:\|\mathbf{x}\|=1\}$ by either of the following two maps:

\begin{eqnarray} \mathbf{p}_1(\theta)&=&\left[\cos(\theta),\sin(\theta)\right]^\top;\\ \mathbf{p}_2(\theta)&=&\left[-\sin(\theta),\cos(\theta)\right]^\top; \end{eqnarray} where $\mathbf{p}_1,\mathbf{p}_2:[0,2\pi)\mapsto S^1$ are bijections. What is more important, it is easy to verify that $\mathbf{p}_1^\top(\theta)\mathbf{p}_2(\theta)=0,\forall\theta\in[0,2\pi)$, and thus we obtain such a group of orthogonal mappings in $\mathbb{R}^2$.

Then, what about when $n>2$? Although we can map $V^{n-1}$ to $S^{n-1}$ using the spherical coordinate transformation as mentioned above, I cannot construct $n$ such mappings with orthogonality yet. Does there exist such a group of mappings?

Any suggestion will be welcome and thank you very much!

Comment

I asked another question yesterday about if it is possible to obtain a group of orthogonal vectors also orthogonal to a given one by orthogonal transformation. If there was such a transformation, this question would be solved as follows:

We can define $\mathbf{p}_1$ as the spherical coordinate transformation and then generate $\mathbf{p}_i(i\geq 2)$ by these orthogonal transformations. The orthogonality of $\mathbf{p}_i$'s is guaranteed by the definition of these transformations. Then, since $\mathbf{p}_i(i\geq 2)$ is obtained by an orthogonal transformation on $\mathbf{p}_1$, if $\mathbf{p}_1(\theta)$ is a bijection to $S^{n-1}$, so does $\mathbf{p}_i(i\geq 2)$.

Unfortunately, there does not exist such a orthogonal transformation in general. Then, can the question I asked today be solved in another way? Thank you very much!
 http://mathoverflow.net/questions/122063/is-it-possible-to-obtain-the-vectors-orthogonal-to-a-given-one-by-orthogonal-tran Is it possible to obtain the vectors orthogonal to a given one by orthogonal transformations? ppyang 2013-02-17T14:34:47Z 2013-02-17T15:38:20Z
Hello, everyone!

Supposing that there is a unit vector in $n$-dimensional real space $\mathbf{x}_1\in\mathbb{R}^n$, I want to get a group of $n-1$ vectors to form an orthogonal basis with $\mathbf{x}_1$. One way to achieve this goal is to firstly randomly generate $n-1$ linear independent vectors and then orthogonalize them use the Gram-Schmidt process to obtain the orthogonality.

Then, I wonder if it is possible to achieve this goal by a group of orthogonal transformations, i.e., if there are $n-1$ orthogonal matrices $\{A_i\}_{i=2}^n$ so that $\mathbf{x}_i^\top\mathbf{x}_j=0,\forall i\neq j$ where $\mathbf{x}_i=A_i\mathbf{x}_1,i=2,\ldots,n$.

Is there any result about this question, please? Any suggestion will be welcome. Thank you very much!

Comment

In the special case where $n=2$, it is possible by $A_2=\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$. It is straightforward to check that the orghotonality of $A_2$ and $(A_2\mathbf{x}_1)^\top\mathbf{x}_1=0,\forall\mathbf{x}_1$. It is easy to comprehend because $A_2$ is indeed a ration of 90 degree in a 2-dimensional plane.

However, it is not so simple in the case of $n=3$. First, to obatin the orthogonality by a transformation $A_2$, i.e., $\mathbf{x}^\top A_2\mathbf{x}=0,\forall\mathbf{x}\in\mathbb{R}^n$, the following equation need to be satisfied. \begin{align} &a_{11}=a_{22}=a_{33}=0\\ &a_{12}+a_{21}=a_{13}+a_{31}=a_{23}+a_{32}=0 \end{align} where $a_{ij}$ is the $i,j$-th element of $A_2$.

With the conditions above, it is straightforward to check that $|A_2|=0$ and thus $A_2$ cannot be an orthogonal matrix, which means there does not exist an orthogonal transformation to obtain an orthogonal vector to a given one.

In the case of $n=4$, it is possible to find an orthogonal transformation to obatin an orthogonal vector to a given one. But I did not find how to find 3 such transformations to form an orthogonal basis including the given vector.
 http://mathoverflow.net/questions/116163/how-to-solve-this-optimization-with-the-orthogonal-constraint How to solve this optimization with the orthogonal constraint? ppyang 2012-12-12T09:37:28Z 2012-12-19T07:24:59Z
Problem

Supposing that $A$ is a symmetric real matrix and $\{\mathbf{w}_i\}_{i=1}^n$ is any orthogonal basis on $\mathbb{R}^n$ such that $W^\top W=WW^\top=\mathbf{I}_n$ where $W=\left[\mathbf{w}_1\;\mathbf{w}_2\;\ldots\;\mathbf{w}_n\right]$, what is the maximum of $\sum_{i=1}^n{\left(\mathbf{w}_i^\top A\mathbf{w}_i\right)^2}$?

To be concise, we can reformulate it as the following optimization problem: $$\begin{split} \max_W&S(W)=\sum_{i=1}^n{\left(\mathbf{w}_i^\top A\mathbf{w}_i\right)^2}\\ \mathrm{s.t.}\,&W^\top W=WW^\top=\mathbf{I}_n, \end{split}$$ where $W=\left[\mathbf{w}_1\;\mathbf{w}_2\;\ldots\;\mathbf{w}_n\right]$.

With help of MATLAB, I found that the maximum of $S(W)$ is always obtained when $W=\bar{W}$ such that $\{\mathbf{w}_i\}_{i=1}^n$ are the eigenvectors of $A$ and the optimal value is $S(\bar{W})=\sum_{i=1}^n{\lambda_i^2}=\|A\|_\mathrm{F}^2$. I had ever given a proof but recently I found it was wrong. However, I think the conclusion should be correct. The constraint that $W$ is orthogonal makes the problem difficult.

Since it is obvious that the assumed optimal value $S(W)=\|A\|_\mathrm{F}^2$ is always obtained by choosing $W$ to be the eigenvectors of $A$, if we can prove that $S(W)\leq\|A\|_\mathrm{F}^2$ for $\forall W$, the problem is solved. However, this seems to be also difficult.

Comments

Actually, the problem may be furthermore simplified as follows.

Assume the spectral decomposition of $A$ is $A=V\Lambda V^\top$ and let $\tilde{W}=V^\top W$ (or $\tilde{\mathbf{w}}_i=V^\top\mathbf{w}_i$), then we have $S(W)=\sum_{i=1}^n{\left(\tilde{\mathbf{w}}_i^\top\Lambda\tilde{\mathbf{w}}_i\right)^2}$ where $\Lambda$ is a diagonal matrix contaning the eigenvalues of $A$.

Since $V$ is orthogonal, we can solve for $\tilde{W}$ instead of $W$. Then the problem turns to be $$\begin{split} \max_\tilde{W}&S(\tilde{W}) =\sum_{i=1}^n{\left(\tilde{\mathbf{w}}_i^\top \Lambda\tilde{\mathbf{w}}_i\right)^2} =\sum_{i=1}^n{\left(\sum_j^n{\lambda_j\tilde{w}_{ij}^2}\right)^2}\\ \mathrm{s.t.}\,&\tilde{W}^\top\tilde{W}=\tilde{W}\tilde{W}^\top=\mathbf{I}_n, \end{split}$$ where $\Lambda$ is diagonal and $\tilde{\mathbf{w}}_i=\left[\tilde{w}_{i1}\;\tilde{w}_{i2}\;\ldots\;\tilde{w}_{in}\right]^\top$.

However, this problem seems to be also difficult to solve. Please help me to give some suggestions about this problem.

Thank you very much!
 http://mathoverflow.net/questions/116163/how-to-solve-this-optimization-with-the-orthogonal-constraint/116277#116277 Answer by ppyang for How to solve this optimization with the orthogonal constraint? ppyang 2012-12-13T13:41:08Z 2012-12-13T13:41:08Z
I am sorry to trouble you all and I found the answer to this question finally which is very simple.

Due to the orthogonality of $\tilde{W}$, it is straightforward that $\sum_{i=1}^n{\tilde{w}_{ij}^2}=1,\forall j$ and $\sum_{j=1}^n{\tilde{w}_{ij}^2}=1,\forall i$.

Then for $\forall i$, from the convexity of $f(x)=x^2$, we have \begin{equation*} \left(\sum_{j=1}^n{\tilde{w}_{ij}^2\lambda_j}\right)^2\leq\sum_{j=1}^n{\tilde{w}_{ij}^2\lambda_j^2}, \end{equation*} and thus \begin{equation*} S(\tilde{W}) =\sum_{i=1}^n{\left(\sum_{j=1}^n{\lambda_j\tilde{w}_{ij}^2}\right)^2} \leq\sum_{i=1}^n{\sum_{j=1}^n{\tilde{w}_{ij}^2\lambda_j^2}} =\sum_{j=1}^n{\sum_{i=1}^n{\tilde{w}_{ij}^2\lambda_j^2}} =\sum_{j=1}^n{\lambda_j^2} =\|A\|_\mathrm{F}^2 \end{equation*}

Since $\|A\|_\mathrm{F}^2$ provides an upper bound for $S(\tilde{W})$ and this value is always obtainable by choosing $\left\{\mathrm{w}_i\right\}_{i=1}^n$ as the eigenvectors of $A$, this is the maximum value.
 http://mathoverflow.net/questions/89159/what-is-the-geometry-of-the-intersection-of-some-cones-defined-by-generalized-ine What is the geometry of the intersection of some cones defined by generalized inequalities? ppyang 2012-02-22T04:25:55Z 2012-11-27T16:46:59Z
Hello, considering that for real numbers, the intersection of intervals defined by simple inequalities has a quite simple form as $$\bigcap_i\{x|x\leq a_i\}=\{x|x\leq\min_i{a_i}\}$$

However, what is the case if the variables are chosen as Hermitian matrices, and the interval defined by inequality is replaced with the convex cone defined by the generalized inequality?

All variables in following are assumed to be Hermitian matrices.

To be specific, define the generalize inequality $X\preceq A_i$ to denote that $X-A_i$ is negative semi-definite, then $\{X|X\preceq A_i\}$ defines a convex cone in the Hermitian matrix space.

Is there any result about the intersection of these cones? To say, can the following set be simplified? $$\bigcap_i\{X|X\preceq A_i\}$$

When does there exist such an $A$ to satisfy $\{X|X\preceq A\}=\bigcap_i\{X|X\preceq A_i\}$?

Or how to describe the geometry of the intersection of such cones?

Any suggestion or comment on this question will be appreciated and thanks very much for your help!

==================================================================================

Acknowledgement and more questions about @Suvrit's comment:

Thanks to @Suvrit for your suggestion! Your comment provides a good way to think about this problem. However, I thought about your method but the problem seems to be more complicated than expected.

Take an example for illustration. Denote $\mathcal{C}(A)=\{X|X\preceq A\}$, then if I want to solve \begin{eqnarray} \min_X&&f(X)\\ \mathrm{s.t.}&&X\in\mathcal{C}(A_1)\cap\mathcal{C}(A_2)\cap\mathcal{C}(A_3) \end{eqnarray} by first solve $\min_{X_1\in\mathcal{C}(A_1)\cap\mathcal{C}(A_2)} f(X_1)$ and then $\min_{X_2\in\mathcal{C}(X_1)\cap\mathcal{C}(A_3)}f(X_2)$, the solution in deed satisfies the constraints due to $$\mathcal{C}(X_1)\cap\mathcal{C}(A_3)\subseteq\mathcal{C}(A_1)\cap\mathcal{C}(A_2)\cap\mathcal{C}(A_3).$$ However, these two sets are not identical, and thus the optimal solution in $\mathcal{C}(X_1)\cap\mathcal{C}(A_3)$ is not guaranteed to be also optimal in $\mathcal{C}(A_1)\cap\mathcal{C}(A_2)\cap\mathcal{C}(A_3)$.

I think the difficulty of this problem results from the complex structure of the intersections of cones $\bigcap_i\mathcal{C}(A_i)$. Do you have some more suggestions about this problem?

Thank you very much for your help!
 http://mathoverflow.net/questions/99299/how-to-calculate-the-inverse-of-the-sum-of-an-identity-and-a-kronecker-product-ef How to calculate the inverse of the sum of an identity and a Kronecker product efficiently? ppyang 2012-06-11T13:50:21Z 2012-06-15T23:50:43Z
I have a matrix $K$ which is the sum of a identity and a Kronecker product of two symmetric matrices as following and I want to calculate the inverse of it $K^{-1}$. \begin{eqnarray} K=\mathbf{I}_{mn}+XX^\top\otimes YY^\top \end{eqnarray} where $X$ and $Y$ are $m\times p$ and $n\times q$ matrices respectively and $m>p,n>q$.

Since the size of Kronecker product of two matrices is the product of their sizes, the matrix to be inverted is very large. Can this calculation be simplified?

Using Woodbury matrix identity, the matrix to be inverted can be reduced from $mn\times mn$ to $pq\times pq$ as \begin{eqnarray} \left(\mathbf{I}+XX^\top\otimes YY^\top\right)^{-1} &=&\left(\mathbf{I}+(X\otimes Y)(X\otimes Y)^\top\right)^{-1}\\ &=&\mathbf{I}-(X\otimes Y)\left(\mathbf{I}+(X\otimes Y)^\top(X\otimes Y)\right)^{-1}(X\otimes Y)^\top \end{eqnarray} However, in my application, $p$ and $q$ are not much less than $m$ and $n$. Thus inverting a $pq\times pq$ matrix is still a consuming work and I want to find a more simple calculation of $K^{-1}$.

Consider the following problem. If there is not the identity matrix $\mathbf{I}_{m\times n}$, we can invert $XX^\top$ and $YY^\top$ respectively and then calculate the Kronecker product as $$(XX^\top\otimes YY^\top)^{-1}=(XX^\top)^{-1}\otimes(YY^\top)^{-1}$$ using the property of Kronecker product. However, if the identity item exists, this property cannot be utilized.

Then my question is that if there is any way to decompose the calculation of $K^{-1}$ into inverses of some smaller matrices whose size is linear with $m$ and $n$ but not their products? Or is it possible to extract the Kronecker product $\otimes$ out of the inverse?

If you have any suggestion or idea, please let me know. Thank you very much for your help!
 http://mathoverflow.net/questions/93149/is-it-possible-to-extend-this-inequality-about-euclidean-distance-frobenius-norm Is it possible to extend this inequality about Euclidean distance &Frobenius norm to more general Bregman divergence such as relative entropy & von Neumann divergence? ppyang 2012-04-04T17:40:44Z 2012-04-18T05:44:09Z
Motivation- A Special Case

Supposing $A,B\in\mathbb{S}^{m\times m}$ are symmetric positive semi-definite (SPD) matrices and $\mathbf{x}\in\mathbb{R}^m$ is a unit vector where $\|\mathbf{x}\|=1$, we found that the squared Euclidean distance of two quadratic forms $\left(\mathbf{x}^\top A\mathbf{x}-\mathbf{x}^\top B\mathbf{x}\right)^2$ is bounded by the squared Frobenius norm of difference of the two matrices $\|A-B\|_F^2$.

Denoting the spectral decomposition of $A-B$ as $A-B=W\Phi W^\top$ where $\Phi=\mathrm{diag}\left(\phi_1,\phi_2,\ldots,\phi_m\right)$ is a diagonal matrix of eigenvalues, we have \begin{eqnarray} &&\left(\mathbf{x}^\top A\mathbf{x}-\mathbf{x}^\top B\mathbf{x}\right)^2 =\left(\mathbf{x}^\top(A-B)\mathbf{x}\right)^2 =\left(\mathbf{x}^\top W\Phi W^\top\mathbf{x}\right)^2\\ =&&\left(\sum_i{x_{W,i}^2\phi_i}\right)^2 \leq\max_i{\phi_i^2}\leq\sum_i{\phi_i^2} =\|A-B\|_F^2 \end{eqnarray} where $W^\top\mathbf{x}=\mathbf{x}_W=\left[x_{W,1}\;x_{W,2}\;\ldots\;x_{W,m}\right]^\top$ and $\mathbf{x}_W^\top\mathbf{x}_W=\sum_i{x_{W,i}^2}=1$.

Therefore, for $\forall \mathbf{x}\in\mathbb{R}^m\;\mathrm{s.t.}\;\|\mathbf{x}\|=1$, we have $\left(\mathbf{x}^\top A\mathbf{x}-\mathbf{x}^\top B\mathbf{x}\right)^2\leq\|A-B\|_F^2$.

Question- Could this be generalized?

However, the squared Euclidean distance is a special case of Bregman divergence $$D_\varphi(\mathbf{x},\mathbf{y})=\varphi(\mathbf{x})-\varphi(\mathbf{y})-\nabla\varphi(\mathbf{y})^\top(\mathbf{x}-\mathbf{y})$$ where $\varphi$ is the convex seed function.

On the other hand, the squared Frobenius norm of difference of two matrices is a special case of Bregman matrix divergence $$D_\phi(A,B)=\phi(A)-\phi(B)-\mathrm{tr}\left((\nabla\phi(B))^\top(A-B)\right)$$ where $\phi(A)=(\varphi\circ\lambda)(A)$ is a compound matrix function in which $\lambda$ is the function that lists the eigenvalues of $A$ and $\varphi$ is the convex seed function.

In the example above, the seed function is $\varphi(\mathbf{x})=\mathbf{x}^\top\mathbf{x}$ and we can rewrite the inequality as $$D_\varphi\left(\mathbf{x}^\top A\mathbf{x},\mathbf{x}^\top B\mathbf{x}\right) \leq D_\phi(A,B)$$ where $\|\mathbf{x}\|=1$ and $\phi=\varphi\circ\lambda$. The function $\lambda$ lists the eigenvalues of the matrix argument.

With the property of Bregman matrix divergence, the inequality can also be written as \begin{eqnarray} D_\varphi\left(\mathbf{x}^\top\mathbf{V}\Lambda\mathbf{V}^\top\mathbf{x}, \mathbf{x}^\top\mathbf{U}\Theta\mathbf{U}^\top\mathbf{x}\right) &=&D_\varphi\left(\sum_i(\mathbf{v}_i^\top\mathbf{x})^2\lambda_i,\sum_j(\mathbf{u}_j^\top\mathbf{x})^2\theta_j\right)\\ &\leq&\sum_i\sum_j{(\mathbf{v}_i^\top\mathbf{u}_j)^2D_\varphi(\lambda_i,\theta_j)} \end{eqnarray} where $A=V\Lambda V^\top,B=U\Theta B^\top$ are spectral decompositions and $\mathbf{v}_i,\mathbf{u}_j$ are columns of $V,U$ respectively.

My Question is: can this inequality be extended to general Bregman divergence and Bregman matrix divergence with different seed functions chosen?

Or under what condition such an inequality exists?

For example, if $\varphi(\mathbf{x})=\sum_i{x_i\log x_i-x_i}$,

then $D_\varphi$ is the relative entropy (KL-divergence) $$\mathrm{KL}(\mathbf{x},\mathbf{y})=\sum_i\left(x_i(\log x_i-\log y_i)-x_i+y_i\right),$$

and $D_\phi$ is the von Neumann divergence $$D_{vn}(A,B)=\mathrm{tr}\left(A\log A-A\log B-A+B\right).$$

In this case, does the following inequality holds for $\forall\mathbf{x}\in\mathbb{R}^m$ satisfying $\|\mathbf{x}\|=1$? $$\mathrm{KL}\left(\mathbf{x}^\top A\mathbf{x},\mathbf{x}^\top B\mathbf{x}\right) \leq D_{vN}(A,B)$$

I did many experiments about this inequality about relative entropy and von Neumann divergence with random generalized SPD matrices using Matlab and it always holds. However, does it really hold?

Could anyone please give me some help for this question or recommend some relevant papers? Any suggestion will be appreciated. Thank you very much!
 http://mathoverflow.net/questions/93149/is-it-possible-to-extend-this-inequality-about-euclidean-distance-frobenius-norm/94306#94306 Answer by ppyang for Is it possible to extend this inequality about Euclidean distance &Frobenius norm to more general Bregman divergence such as relative entropy & von Neumann divergence? ppyang 2012-04-17T17:11:05Z 2012-04-18T05:44:09Z
I found a proof of this problem for the case of $\varphi(x)=\sum_i{x_i\log x_i-x_i}$. If you find there is anything mistake in the proof, please let me know. Thank you!

The case of $\varphi(x)=\sum_i{x_i\log x_i-x_i}$ can be proved with the method similar to Lindblad, Completely positive maps and entropy inequalities, 1975 and Lindblad, Expectations and entropy inequalities for finite quantum systems, 1974. The inequality can be strengthened as $$\sum_i{\mathrm{KL}\left(\mathbf{x}_i^\top A\mathbf{x}_i,\mathbf{x}_i^\top B\mathbf{x}_i\right)}\leq D_{vN}\left(A,B\right)$$

Actually, a very similar result has been proposed in some papers about quantum information theory, such as the two papers referred above. The referred result is that for any trace preserving map $\Phi$, given by $\Phi(A)=\sum_{i=1}^n{V_iAV_i^\top}$ and $\sum_{i=1}^n{V_i^\top V_i}=1$, we have that $\mathrm{tr}\left(\Phi(A),\Phi(B)\right)\leq D_\phi(A,B)$, where $A,B$ are both density operators which are Hermitian positive definite matrices satisfying $\mathrm{tr}A=\mathrm{tr}B=1$ and $\varphi(x)=x\log x$.

We found that if the trace constraints $\mathrm{tr}A=\mathrm{tr}B=1$ are dropped and $\varphi(x)=x\log x$ is replaced with $\varphi(x)=x\log x-x$, the inequality still holds.

The proof is outlined as following:

The von Neumann divergence has the following additivity property with Kronecker product: $$D_{vN}(A\otimes P,B\otimes P)=D_{vN}(A,B)\cdot\mathrm{tr}P$$

Using the joint convexity and the additivity, we can prove that the von Neumann divergence has the monotonicity with partial trace as \begin{equation*} \begin{split} D_{vN}(\mathrm{tr}_2(A),\mathrm{tr}_2(B)) =&D_{vN}\left(\mathrm{tr}_2(A)\otimes\frac{\mathbf{I}_2}{m}, \mathrm{tr}_2(B)\otimes\frac{\mathbf{I}_2}{m}\right) /\mathrm{tr}\left(\frac{\mathbf{I}_2}{m}\right)\\ =&D_{vN}\left(\sum_{j=1}^N{p_jW_jAW_j^+},\sum_{j=1}^N{p_jW_jBW_j^+}\right)\\ \leq&\sum_{j=1}^{N}{p_jD_{vN}\left(W_jAW_j^+,W_jBW_j^+\right)}\\ =&D_{vN}(A,B)\end{split} \end{equation*}

For any trace preserving map $\Phi$, given by $\Phi(A)=\sum_{i=1}^n{V_iAV_i^\top}$ and $\sum_{i=1}^n{V_i^\top V_i}=1$, it can be represented as a unitary operation+partial tracing. Therefore, we have that \begin{equation*} \begin{split} D_{vN}\left(\Phi(A),\Phi(B)\right) =&D_{vN}\left(\mathrm{tr}_2(W(A\otimes\mathbf{s}\mathbf{s}^\top)W^\top), \mathrm{tr}_2(W(B\otimes\mathbf{s}\mathbf{s}^\top)W^\top)\right)\\ \leq&D_{vN}\left(W(A\otimes\mathbf{s}\mathbf{s}^\top)W^\top, W(B\otimes\mathbf{s}\mathbf{s}^\top)W^\top\right)\\ =&D_{vN}\left(A,B\right) \end{split} \end{equation*}

Then the sum of relative entropy of the quadratic forms could be represented as matrix divergence and bounded. \begin{equation*} \begin{split} \sum_i{\mathrm{KL}\left(\mathbf{x}_i^\top A\mathbf{x}_i,\mathbf{x}_i^\top B\mathbf{x}_i\right)} =&\sum_{i,j}{(\mathbf{x}_i^\top\mathbf{x}_j)^2 \mathrm{KL}(\mathbf{x}_i^\top A\mathbf{x}_i,\mathbf{x}_j^\top B\mathbf{x}_j)}\\ =&D_{vN}(\sum_i{X_iAX_i^\top},\sum_i{X_iBX_i^\top})\\ \leq&D_{vN}\left(A,B\right) \end{split} \end{equation*} where $X_i=\mathbf{x}_i\mathbf{x}_i^\top$.

If there is any mistake in the proof, please let me know. Any other suggestions are also welcomed. Thank you very much!
 http://mathoverflow.net/questions/93448/how-to-calculate-the-maximum-of-the-relative-entropy-of-two-quadratics How to calculate the maximum of the relative entropy of two quadratics? ppyang 2012-04-07T19:33:37Z 2012-04-07T19:33:37Z
Hello, everyone!

Suppose that there are two Hermitian matrices $A,B$ and a unit vector $\mathbf{x}$, then how to calculate the relative entropy of the quadratic forms determined by them $\mathbf{x}^\top A\mathbf{x}$ and $\mathbf{x}^\top B\mathbf{x}$?

It can be equivalent to the following problem: \begin{eqnarray} \max_{\mathbf{x}}&&\mathrm{KL}\left(\mathbf{x}^\top A\mathbf{x},\mathbf{x}^\top B\mathbf{x}\right)\\ \mathrm{s.t.}&&\mathbf{x}^\top\mathbf{x}=1 \end{eqnarray} where $\mathrm{KL}\left(p,q\right)=p\log p-p\log q-p+q$.

Does there exist an analytic solution for this optimization problem? Please give me some help. Any suggestion will be appreciated!

Thank you very much!
 http://mathoverflow.net/questions/93232/does-this-inequality-of-negative-relative-entropy-and-quantum-relative-entropy-ho Does this inequality of negative relative entropy and quantum relative entropy hold? ppyang 2012-04-05T17:18:01Z 2012-04-06T05:28:59Z
Hello, everyone!

Question

I have a question about the relationship between general relative entropy and general quantum relative entropy: Given a unit vector $|i\rangle$ and two Hermitian matrices $A,B$, does the following inequality holds? $$H\left(\langle i|A|i\rangle\parallel\langle i|B|i\rangle\right)\leq S(A\parallel B)$$ where $H(\cdot\parallel\cdot)$ and $S(\cdot\parallel\cdot)$ are general relative entropy and general quantum relative entropy respectively defined as following.

Denote the general negative relative entropy as $$H(p\parallel q)=\sum_i\left(p_i\log\frac{p_i}{q_i}-p_i+q_i\right),$$ and the general quantum relative entropy (von Neumann divergence) as $$S(A\parallel B)=\mathrm{tr}\left(A(\log A-\log B)-A+B\right),$$ where $A,B$ are both positive semi-definite matrices which are not necessarily to be density matrices.

I have repeated the experiments about this inequality with more than 100,000 random generalized SPD matrices using Matlab, and it always holds. However, does it really hold in theory?

Motivation

What motivates this conjecture is the similar inequality of squared Euclidean distance and squared Frobenius norm. Specifically, given a unit vector $|i\rangle$ and two Hermitian matrices $A,B$, the following inequality holds with simple matrix calculations. $$\left\|\langle i|A|i\rangle-\langle i|B|i\rangle\right\|^2\leq\|A-B\|_\mathrm{F}^2.$$ In context of Bregman divergence, squared Euclidean distance and squared Frobenius norm are Bregman divergence and Bregman matrix divergence with the same seed function $\varphi(\mathbf{x})=\sum_ix_i^2$, while general relative entropy and general quantum relative entropy with the same seed function $\varphi(\mathbf{x})=\sum_i\left(x_i\log x_i-x_i\right)$.

Could anyone please give me some help for this question or recommend some relevant papers? Any suggestion will be appreciated.

Thank you very much!
 http://mathoverflow.net/questions/92908/is-there-relationship-between-f-left-sum-i-mathbfv-i-top-mathbfx2-lamb Is there relationship between $f\left({\sum_i(\mathbf{v}_i^\top\mathbf{x})^2\lambda_i},\sum_j{(\mathbf{u}_j^\top\mathbf{x})^2\theta_j}\right)$ and $\sum_i\sum_j{(\mathbf{v}_i^\top\mathbf{u}_j)^2f(\lambda_i,\theta_j)}$ if $f$ is jointly convex? ppyang 2012-04-02T16:25:56Z 2012-04-03T10:02:13Z
Hello, everyone!

As we know that by Jensen's inequality, for jointly convex function $f$ and $\sum_ix_i^2=1$, we have $$f(\sum_i{x_i^2\lambda_i},\sum_i{x_i^2\theta_i)}\leq\sum_i{x_i^2f(\lambda_i,\theta_i)}\leq\max_if(\lambda_i,\theta_i)\leq\sum_if(\lambda_i,\theta_i),$$ and we get a bound of $f(\sum_i{x_i^2\lambda_i},\sum_i{x_i^2\theta_i)}$ independent of $\{x_i\}$.

However, I wonder if this inequality can be extended to the case where the probability distribution $\{x_i\}$ on the two variables of $f$ are not identical but just constrained.

To be more specifically, suppose that $f:\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}$ is jointly convex with both its arguments, $V=[\mathbf{v}_1\;\mathbf{v}_2\;\ldots\;\mathbf{v}_m]$ and $U=[\mathbf{u}_1\;\mathbf{u}_2\;\ldots\;\mathbf{u}_m]$ are orthogonal matrices and thus $\{\mathbf{v}_i\},\{\mathbf{u}_j\}$ consist an orthonormal basis in $\mathbb{R}^m$ respectively.

Then for any $\mathbf{x}\in\mathbb{R}^m$ satisfying $\|\mathbf{x}\|=1$, I wonder if there is a relationship between $L_1$ and $L_2$ shown in the following two formulas. \begin{eqnarray} L_1&=&f\left({\sum_i(\mathbf{v}_i^\top\mathbf{x})^2\lambda_i},\sum_j{(\mathbf{u}_j^\top\mathbf{x})^2\theta_j}\right)\\ L_2&=&\sum_i\sum_j{(\mathbf{v}_i^\top\mathbf{u}_j)^2f(\lambda_i,\theta_j)} \end{eqnarray}

In language of matrix, $L_1$ can also be formulated as $f\left(\mathbf{x}^\top V\Lambda V^\top\mathbf{x},\mathbf{x}^\top U\Theta U^\top\mathbf{x}\right)$.

Considering that $\sum_i(\mathbf{v}_i^\top\mathbf{x})^2=\sum_j(\mathbf{v}^\top\mathbf{x})^2=1$, my question is that does there exist an inequality about $L_1$ and $\gamma L_2$ where $\gamma$ is any constant independent of $\mathbf{x}$?

How should I consider about this problem? Or are there any papers about this topic for reference?

Could anyone be so kind to help me about this question? Any suggestion will be appreciated! Thank you very much!

Remark:

I tried to simply apply the Jensen's inequality to $L_1$ and get the result $$L_1\leq\sum_i\sum_j{(\mathbf{v}_i^\top\mathbf{x})^2(\mathbf{u}_j^\top\mathbf{x})^2f(\lambda_i,\theta_j)}.$$ Does there exists any relationship between this formula and $L_2$?

Any suggestion will be appreciated! Thank you very much!
 http://mathoverflow.net/questions/92079/given-a-jointly-convex-function-f-what-is-the-bound-of-f-left-sum-ip-i2x-i Given a jointly convex function $f$, what is the bound of $f\left(\sum_ip_i^2x_i,\sum_jq_j^2y_j\right)$if $\mathbf{p},\mathbf{q}$ are constrained in a manifold? ppyang 2012-03-24T15:11:05Z 2012-03-24T15:11:05Z
Suppose there is a jointly convex function $f:\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}$, $\mathbf{x},\mathbf{y}\in\mathbb{R}^m$ and $\mathbf{p}=[p_1\;\ldots\;p_m]^\top,\mathbf{q}=[q_1\;\ldots\;q_m]^\top$ satisfying $\|p\|^2=\sum_ip_i^2=\|q\|^2=\sum_jq_j^2=1$, I want to find the bound of $f\left({\mathbf{p}^2}^\top\mathbf{x},{\mathbf{q}^2}^\top\mathbf{y}\right)$ by decomposing it into weighted sum of $f(x_i,y_j)$ if $\mathbf{p},\mathbf{q}$ are constrained in a manifold.

The case without constraints of $\mathbf{p}$ and $\mathbf{q}$

Denote the element-wise squared vector of $\mathbf{p},\mathbf{q}$ as $$\mathbf{p}^2=[p_1^2\;\ldots\;p_m^2]^\top,\mathbf{q}^2=[q_1^2\;\ldots\;q_m^2]^\top,$$ then by Jensen's inequality for jointly convex function, we have \begin{eqnarray} f\left({\mathbf{p}^2}^\top\mathbf{x},{\mathbf{q}^2}^\top\mathbf{y}\right) &=&f\left(\sum_ip_i^2x_i,\sum_jq_j^2y_j\right)\\ &=&f\left(\sum_i\left(p_i^2\sum_jq_j^2x_i\right),\sum_j\left(q_j^2\sum_ip_i^2y_j\right)\right)\\ &=&f\left(\sum_i\sum_jp_i^2q_j^2x_i,\sum_i\sum_jp_i^2q_j^2y_j\right)\\ &\leq&\sum_i\sum_j{p_i^2q_j^2f(x_i,y_j)}\\ &=&{\mathbf{p}^2}^\top F\mathbf{q}^2 \end{eqnarray} where $F_{ij}=f(x_i,y_j)$. This provides a general bound for $f\left({\mathbf{p}^2}^\top\mathbf{x},{\mathbf{q}^2}^\top\mathbf{y}\right)$.

Suppose $C=[c_{ij}]_{m\times m}$ is an orthogonal matrix and the element squared matrix of $C$ is $$C^2=[c_{ij}^2]_{m\times m},$$ then $C^2$ is a doubly stochastic matrix where $C^2\mathbf{1}=\mathbf{1}$and ${C^2}^\top\mathbf{1}=\mathbf{1}$.

The case if with constraint ${\mathbf{q}^2}={C^2}^\top\mathbf{p}^2$

If an additional constraint $\mathbf{q}^2={C^2}^\top\mathbf{p}^2$ is applied so that $\mathbf{p}$ and $\mathbf{q}$ are constrained on a manifold, we can substitute the constraint into the inequality above, we get the bound \begin{eqnarray} f\left({\mathbf{p}^2}^\top\mathbf{x},{\mathbf{q}^2}^\top\mathbf{y}\right) \leq{\mathbf{p}^2}^\top F{C^2}^\top\mathbf{p}^2 \end{eqnarray}

On the other hand, applying Jensen's inequality directly on the original formular, we can obtain another bound as \begin{eqnarray} f\left({\mathbf{p}^2}^\top\mathbf{x},{\mathbf{q}^2}^\top\mathbf{y}\right) &=&f\left(\sum_ip_i^2x_i,\sum_j\sum_ic_{ij}^2p_i^2y_j\right)\\ &=&f\left(\sum_ip_i^2x_i,\sum_i\left(p_i^2\sum_jc_{ij}^2y_j\right)\right)\\ &\leq&\sum_ip_i^2f\left(x_i,\sum_jc_{ij}^2y_j\right)\\ &\leq&\sum_i\sum_j{p_i^2c_{ij}^2f(x_i,y_j)}\\ &=&{\mathbf{p}^2}^\top(F\circ C^2)\mathbf{1} \end{eqnarray} where $\circ$ is the Hadamard (element-wise) product.

It is notable that this bound is simpler than the one obtained by substitution and has $\mathbf{p}$ is homogeneous in both sides of the inequality.

What is the case if with constraint $\mathbf{q}=C^\top\mathbf{p}$?

My question is that if the constraint is replaced with $\mathbf{q}=C^\top\mathbf{p}$, could we obtain the bound of $f\left({\mathbf{p}^2}^\top\mathbf{x},{\mathbf{q}^2}^\top\mathbf{y}\right)$ with Jensen's equality? If such a bound exists, could it have the simple form where $\mathbf{p}$ similar to ${\mathbf{p}^2}^\top(F\circ C^2)\mathbf{1}$?

Could anyone be so kind to give me some help on this problem? Any suggestion will be appreciated! Thank you very much!
 http://mathoverflow.net/questions/89939/given-mathbfx-i-top-a-mathbfx-i-for-a-spd-matrix-a-and-orthonormal-base Given $\mathbf{x}_i^\top A\mathbf{x}_i$ for a SPD matrix $A$ and orthonormal bases $\mathbf{x}_i$, what is the bound of its eigenvalues? ppyang 2012-03-01T08:38:26Z 2012-03-01T09:59:19Z
Assume that $A_{d\times d}$ is a symmetric positive semi-definite matrix, and $\{\mathbf{x}_1,\ldots,\mathbf{x}_d\}$ composes a group of orthogonal bases of $\mathbb{R}^d$ where $\mathbf{x}_i\bot\mathbf{x}_j,\forall i\neq j$ and $\|\mathbf{x}_i\|=1$.

Then, my question is that given $$\mathbf{x}_i^\top A\mathbf{x}_i=a_i,\forall i$$ how to estimate the bound of the eigenvalues of $A$?

Using eigenvalue decomposition $A=U\Lambda U^\top$ and let $\tilde{\mathbf{x}}_i=U^\top\mathbf{x}_i$, this question can be simplified as to estimate the bound of the elements of the diagonal matrix $\Lambda$ given $$\tilde{\mathbf{x}}_i^\top\Lambda\tilde{\mathbf{x}}_i=a_i,\forall i.$$ Since $U$ is orthogonal, we also have $\tilde{\mathbf{x}}_i\bot\tilde{\mathbf{x}}_j,\forall i\neq j$ and $\|\tilde{\mathbf{x}}_i\|=1$.

Denote $\Lambda=\mathrm{diag}[\lambda_1\;\ldots\;\lambda_d]^\top$ and $\tilde{\mathbf{x}}_i=[\tilde{x}_{i1}\;\ldots\;\tilde{x}_{id}]$, the equations above can be represented as $$\sum_j\lambda_j\tilde{x}_{ij}^2=a_i,\forall i$$

However, this question seems to be not so simple yet.

In the case of $d=2$, I found the following method to solve it.

Denote $\tilde{\mathbf{x}}_1=[\cos\theta\;\sin\theta]^\top$ and $\tilde{\mathbf{x}}_2=[-\sin\theta\;\cos\theta]^\top$, we have \begin{eqnarray} \lambda_1\cos^2\theta+\lambda_2\sin^2\theta&=&a_1\\ \lambda_1\sin^2\theta+\lambda_2\cos^2\theta&=&a_2 \end{eqnarray} Solve the equation system and we get \begin{eqnarray} \lambda_1&=&\frac{1}{\Delta}(a_1\cos^2\theta-a_2\sin^2\theta) =\frac{a_1-a_2\tan^2\theta}{1-\tan^2\theta}=a_2+\frac{a_1-a_2}{1-\tan^2\theta}\\ \lambda_2&=&\frac{1}{\Delta}(-a_1\sin^2\theta+a_2\cos^2\theta) =\frac{-a_1\tan^2\theta+a_2}{1-\tan^2\theta}=a_1-\frac{a_1-a_2}{1-\tan^2\theta} \end{eqnarray} where $\Delta=\cos^4\theta-\sin^4\theta=\cos^2\theta-\sin^2\theta$.

Without loss of generality, assume that $a_1\geq a_2$ and $\lambda_1>\lambda_2$, we can deduce that $1-\tan^2\theta>0$, then from $\lambda_i\geq 0$, we have $$\tan^2\theta\leq\min\{\frac{a_2}{a_1},\frac{a_1}{a_2}\}=\frac{a_2}{a_1}$$ Therefore it is obvious that when $\tan^2\theta=\frac{a_2}{a_1}$, $\lambda_1$ and $\lambda_2$ take its maximum and minimum respectively as \begin{eqnarray} \lambda_1&=&a_1+a_2\\ \lambda_2&=&0 \end{eqnarray} and we get the bound of the eigenvalues of $A$.

However, how to solve the problem if $d\geq3$?

Any suggestion will be appreciated! Thank you very much!
 http://mathoverflow.net/questions/89125/what-is-the-minimum-of-the-frobenius-norm-in-the-intersection-of-positive-semidef What is the minimum of the Frobenius norm in the intersection of positive semidefinite cones? ppyang 2012-02-21T17:56:31Z 2012-02-23T02:27:48Z
For scalar variables $x$, we have a simple solution for the following problem. \begin{eqnarray} \min_x&&\alpha(x-a)^2+\beta(x-b)^2 \\ \mathrm{s.t. }&&x\leq a\\ &&x\leq b \end{eqnarray} where $\alpha,\beta>0$.

The optimal solution $x=\min(a,b)$ is straightforward and independent of $\alpha,\beta$.

However, in the case of real symmetrical matrix variables, the problem seems to be much more complex because the relationship $\leq$ in constraints has to be replaced with $\preceq$ which is a more complex relationship defined as $$A\preceq B\Leftrightarrow A-B\preceq0$$ where $A\preceq0$ means $A$ is negative semi-definite.

Then the problem above is reformulated as following with matrix variables. Assume that all matrices in this problem are real symmetrical. \begin{eqnarray} \min_X&&\alpha\|X-A\|_F^2+\beta\|X-B\|_F^2\\ \mathrm{s.t.}&&X\preceq A\\ &&X\preceq B \end{eqnarray} where $\alpha,\beta\geq0$.

If $A$ and $B$ can be diagonalized by the same orthogonal matrix $U$, the problem reduces to a the problems of eigenvalues since $$\alpha\|X-A\|_F^2+\beta\|X-B\|_F^2=\alpha\|\tilde{X}-\Lambda\|_F^2+\beta\|\tilde{X}-\Theta\|_F^2$$ and \begin{eqnarray} X\preceq A\Leftrightarrow\tilde{X}\preceq\Lambda\\ Y\preceq B\Leftrightarrow\tilde{X}\preceq\Theta \end{eqnarray} where $\tilde{X}=U^\top XU$, $\Lambda=U^\top AU$, $\Theta=U^\top BU$. Since $\Lambda$ and $\Theta$ are diagonal matrices, the problem can decompose to sum of some scalar variables and solved independently.

However, how to solve it if $A$ and $B$ have different eigenvectors? Is the solution independent of $\alpha$ and $\beta$ yet?

Could any one be so kind to help me with the question or give some suggestions? Thank you very much!
 http://mathoverflow.net/questions/87534/how-to-calculate-the-integral-of-the-exponential-of-negative-unnormalized-relativ How to calculate the integral of the exponential of negative unnormalized relative entropy? ppyang 2012-02-04T16:32:55Z 2012-02-04T16:32:55Z
I want to calculate the integral of the exponential of negative unnormalized relative entropy shown as follows. This integral seems to be finite but I don't know how to calculate it. $$\int_0^{+\infty}{e^{-(x\log x-x\log y-x+y)}dx}$$ where $y$ is a known as a positive constant.

Could anyone be so kind to help me with some suggestions? Thank you very much!
 http://mathoverflow.net/questions/83323/how-to-deal-with-the-vector-norm-item-as-a-denominator-in-this-expectation How to deal with the vector norm item as a denominator in this expectation? ppyang 2011-12-13T07:24:50Z 2011-12-13T07:24:50Z
Hello, everyone. I want to calculate the expectation shown in the following formula, where $X$ follows a standard $d$-dimensional multi-variable normal distribution as $X\sim\mathbb{N}(\mathbf{0},\mathbb{I}_{d\times d})$ $$E_X\left[\frac{X^\top\mathbf{A}XX^\top\mathbf{B}X}{\|X\|^2}\right]$$ where $\|X\|^2=X^\top X$, and $\mathbf{A},\mathbf{B}$ are both real symmetric matrices.

The expectation of the numerator is straightforward and there was result for it. However, the denominator seems to make the problem more difficult. Are there any simple approaches to deal with this problem? Thank you very much!
 http://mathoverflow.net/questions/81531/how-to-calculate-this-expectation-with-logarithm How to calculate this expectation with logarithm? ppyang 2011-11-21T17:10:10Z 2011-11-22T00:42:23Z
If $\mathbf{x}\sim\mathbf{N}(\mathbf{0},\mathbf{I})$, and assume that $A$ is a symmetric positive definite matrix, how can I calculate the following two expectations where there is a logarithm in it? $$\mathrm{E}_\mathbf{x}\left[\log(\mathbf{x}^\top A\mathbf{x})\right]$$ and

$$\mathrm{E}_\mathbf{x}\left[(\mathbf{x}^\top A\mathbf{x})\log(\mathbf{x}^\top A\mathbf{x})\right]$$

Using the decomposition $A=U\Lambda U^\top$ and $\mathbf{y}=U^\top\mathbf{x}$, the first expectation can be reduced to $$\mathrm{E}_\mathbf{y}\left[\log(\mathbf{y}^\top \Lambda\mathbf{y})\right]$$

$$=\mathrm{E}_\mathbf{y}\left[\log\left(\sum_i{\lambda_i\mathbf{y}_i^2}\right)\right]$$ where $y$ has the same distribution as $x$ because $U$ is a orthogonal matrix.

However, since the random variables $y_i$ are in the logarithm function, I cannot decompose the expectation further and such a expectation seems to be difficult to calculate.

Could you please help me with this problem or give some suggestions about it? Thank you very much!
 http://mathoverflow.net/questions/81419/how-to-calculate-this-expectation-where-the-random-variable-is-restricted-on-a-sp How to calculate this expectation where the random variable is restricted on a sphere? ppyang 2011-11-20T13:35:33Z 2011-11-20T22:42:50Z
Hello! I have a question about how to calculate the expectation of a quadratic form as follows, where $X$ is a random variable that uniformly distributed on the unit sphere: $$E_X[(\mathbf{x}^\top A\mathbf{x})^2] =\int_{\mathbf{x}\in S}{p(\mathbf{x})(\mathbf{x}^\top A\mathbf{x})^2dS(\mathbf{x})} =\int_{\mathbf{x}\in S}{\frac{1}{4\pi}(\mathbf{x}^\top A\mathbf{x})^2dS(\mathbf{x})}$$ where $S={\mathbf{x}\in\mathcal{R}^N|\mathbf{x}^\top\mathbf{x}=1}$.

If $\mathbf{x}$ is Gaussian, there are some conclusions about the expectation of the quadratic forms. However, I find it difficult to deal with when the variable is distributed on a sphere. When the dimension is 2 or 3, this problem can be solved by representing it with polar coordinate. However, when the dimension is high, such a representation will be rather redundant, how can I calculate this integral then? Please give some help for this problem if you have any idea. Thank you very much!
 http://mathoverflow.net/questions/122615/derivative-of-the-regularized-upper-incomplete-gamma-function Comment by ppyang ppyang 2013-02-25T00:35:27Z 2013-02-25T00:35:27Z @Suvrit Sorry to have made confusion by using different variables in the definition of $Q$. Thank you for your attention, but I cannot understand your last sentense clearly. Can you explain how to use this formula to solve my question in more detail? Thank you very much! http://mathoverflow.net/questions/122615/derivative-of-the-regularized-upper-incomplete-gamma-function Comment by ppyang ppyang 2013-02-23T12:30:59Z 2013-02-23T12:30:59Z @Suvrit Thank you for your attention! Do you mean there is a formula for $\frac{\partial Q(x,λ)}{\partial x}$ in wikipedia? I am sorry that I did not find such a formula but only the formula $\frac{\partial\Gamma(x,λ)}{\partial x}=\ln\lambda\Gamma(x,\lambda)+\lambda T(3,x,\lambda)$ instead. If you found the formula for the derivative of $Q(x,\lambda)$ w.r.t. $x$, could you please show me the url of the wikipedia? Thank you very much! http://mathoverflow.net/questions/122063/is-it-possible-to-obtain-the-vectors-orthogonal-to-a-given-one-by-orthogonal-tran/122069#122069 Comment by ppyang ppyang 2013-02-18T09:32:33Z 2013-02-18T09:32:33Z It's helpful of your answer. However there are some points in your answer that I cannot understand clearly since I do not major in mathematics. Could you please recommend some references or papers to explain your answer in more detail, please? Thank you very much! http://mathoverflow.net/questions/122063/is-it-possible-to-obtain-the-vectors-orthogonal-to-a-given-one-by-orthogonal-tran/122068#122068 Comment by ppyang ppyang 2013-02-18T09:17:01Z 2013-02-18T09:17:01Z I think both of your answers are very helpful. However, since I cannot accept both of your answers in the system and the answer given by Charles Rezk is in more detail, I choose to set his answer as the accepted answer. Anyway, it's very kind of you and thank you very much! http://mathoverflow.net/questions/122063/is-it-possible-to-obtain-the-vectors-orthogonal-to-a-given-one-by-orthogonal-tran/122068#122068 Comment by ppyang ppyang 2013-02-18T08:12:28Z 2013-02-18T08:12:28Z It's very kind of you for your help! Could you tell me the title of Adams's paper mentioned in your answer, please? I would like to understand how the result is obtained. Thank you! http://mathoverflow.net/questions/116163/how-to-solve-this-optimization-with-the-orthogonal-constraint Comment by ppyang ppyang 2012-12-13T13:43:28Z 2012-12-13T13:43:28Z @Robert Bryant, I am sorry to trouble you all with this simple question and I have solved it. However, I have to say that this is indeed not my homework or exam problem. I will think more by myself before posting the question next time. Thank you very much! http://mathoverflow.net/questions/89159/what-is-the-geometry-of-the-intersection-of-some-cones-defined-by-generalized-ine/114670#114670 Comment by ppyang ppyang 2012-11-29T02:22:45Z 2012-11-29T02:22:45Z I am reading the papers you suggested and they indeed provide an interesting standpoint for this question. Thank you very much! http://mathoverflow.net/questions/99299/how-to-calculate-the-inverse-of-the-sum-of-an-identity-and-a-kronecker-product-ef/99760#99760 Comment by ppyang ppyang 2012-06-17T11:38:42Z 2012-06-17T11:38:42Z You provide an excellent idea that reduce the most complex calculation of an $mn\times mn$ matrix into a $m\times m$ matrix and a $n\times n$ matrix, where the inverse is taken on only a diagonal. I think it solve my problem. Thank you very much! http://mathoverflow.net/questions/99299/how-to-calculate-the-inverse-of-the-sum-of-an-identity-and-a-kronecker-product-ef Comment by ppyang ppyang 2012-06-12T17:34:29Z 2012-06-12T17:34:29Z @Felix Goldberg: I am still interested in whether there exists some way to simplify this calculation but I have no idea yet. http://mathoverflow.net/questions/99299/how-to-calculate-the-inverse-of-the-sum-of-an-identity-and-a-kronecker-product-ef Comment by ppyang ppyang 2012-06-12T02:12:02Z 2012-06-12T02:12:02Z @Federico Poloni, thank you for your suggestion. What is more, the purpose that I took that example is to show that I want to simplify the calculation utilizing the property of Kronecker product, however, there is some mistake in it. Thank you! http://mathoverflow.net/questions/99299/how-to-calculate-the-inverse-of-the-sum-of-an-identity-and-a-kronecker-product-ef Comment by ppyang ppyang 2012-06-12T02:07:20Z 2012-06-12T02:07:20Z @Federico Poloni, yes, you are right! $(XX^\top)^{-1}$ really does not exist. Thank you! http://mathoverflow.net/questions/93232/does-this-inequality-of-negative-relative-entropy-and-quantum-relative-entropy-ho Comment by ppyang ppyang 2012-04-17T17:39:45Z 2012-04-17T17:39:45Z I found a proof of this problem and the inequality does holds. A related problem has been studied several years ago. I have presented the outline of the proof in another question as the following url: mathoverflow.net/questions/93149/… If there is any mistake in the proof, please let me know. Thank you! http://mathoverflow.net/questions/93149/is-it-possible-to-extend-this-inequality-about-euclidean-distance-frobenius-norm Comment by ppyang ppyang 2012-04-17T14:45:48Z 2012-04-17T14:45:48Z I have found the proof of the case of $\varphi(x)=\sum_i{x_i\log x_i-x_i}$ and the inequality $\mathrm{KL}\left(\mathbf{x}^\top A\mathbf{x},\mathbf{x}^\top B\mathbf{x}\right)\leq D_{vN}(A,B)$ indeed holds. I will present the proof soon. http://mathoverflow.net/questions/92908/is-there-relationship-between-f-left-sum-i-mathbfv-i-top-mathbfx2-lamb/92937#92937 Comment by ppyang ppyang 2012-04-03T10:01:34Z 2012-04-03T10:01:34Z Thank you for your counter example! http://mathoverflow.net/questions/92908/is-there-relationship-between-f-left-sum-i-mathbfv-i-top-mathbfx2-lamb Comment by ppyang ppyang 2012-04-03T02:29:37Z 2012-04-03T02:29:37Z @Marc, OK! I will remember this next time. Thanks!


text:
The Unapologetic Mathematician

Enriched Categories II

So we have the basic data of a category $\mathcal{C}$ enriched over a monoidal category $\mathcal{V}$. Of course, what I left out were the relations that have to hold. And they’re just the same as those from categories, but now written in terms of $\mathcal{V}$ instead of $\mathbf{Set}$: associativity and identity relations, as encoded in the following commutative diagrams:

Notice how these are very similar to the axioms for a monoidal category or a monoid object. And this shouldn’t be unexpected by now, since we know that a monoid is just a (small) category with only one object. In fact, if we only have one object in a $\mathcal{V}$-enriched category we get back exactly a monoid object in $\mathcal{V}$!

Now, often we’re thinking of our hom-objects as “hom-sets with additional structure”. There should be a nice way to forget that extra structure and recover just a regular category again. To an extent this is true, but for some monoidal categories $\mathcal{V}$ the “underlying set” functor isn’t really an underlying set at all. For now, though, let’s look at a familiar category of “sets with extra structure” and see how we get the underlying set out of the category itself.

Again, the good example to always refer back to for enriched categories is $\mathbf{Ab}$, the category of abelian groups with tensor product as the monoidal structure. We recall that the functor giving the free abelian group on a set is left adjoint to the forgetful functor from abelian groups to sets. That is, $\hom_\mathbf{Ab}(F(S),A)\cong\hom_\mathbf{Set}(S,U(A))$. We also know that we can consider an element of the underlying set $U(A)$ of an abelian group as a function from a one-point set into $U(A)$. That is, $\hom_\mathbf{Set}(\{*\},U(A))\cong U(A)$. Putting these together, we see that $U(A)\cong\hom_\mathbf{Ab}(\mathbb{Z},A)$, since $\mathbb{Z}$ is the free abelian group on one generator.

But $\mathbb{Z}$ is also the identity object for the tensor product! The same sort of argument goes through for all our usual sets-with-structure, telling us that in all these cases the “underlying set” functor is represented by the monoidal identity $\mathbf{1}$, which is the free object on one generator. We take this as our general rule, giving the representable functor $V(\underline{\hphantom{X}})=\hom_{\mathcal{V}_0}(\mathbf{1},\underline{\hphantom{X}}):\mathcal{V}_0\rightarrow\mathbf{Set}$. In many cases (but not all!) this is the usual “underlying set” functor, but now we’ve written it entirely in terms of the monoidal category $\mathcal{V}$!

As time goes by, we’ll use this construction to recover the “underlying category” of an enriched category. The basic idea should be apparent, but before we can really write it down properly we need to enrich the notions of functors and natural transformations.


text:
# To Infinity...

Image from Wikipedia

The time has come to post some real material to this site—a simple exploration of physics that will (hopefully) build audience interest and help me learn techniques of generating math-laden web content. After spending the last couple of days mulling over which one of my little projects would be the first one I share, the answer came to me late last night in the form of a text message from a friend:

Excellent question! And, the perfect metaphor for getting this website off the ground. Thanks, JB!

The answer is yes, and we shall examine the formula for planetary escape velocity ($$v_e$$) to see why:

$v_e=\sqrt{\frac{2 G M}{R}} \label{080717:ve} .$

$$M$$ and $$R$$ represent our planet's mass and radius, respectively. $$G$$ is the gravitational constant.

Using Earth's values, one finds that

$v_e=11.186\text{km}/\text{s}.$

(For residents of Liberia, Myanmar, and the United States of America, that's $$6.951\text{mi}/\text{s}$$.)

Now, here comes the fun part: we can learn just as much about $$v_e$$ by noting what \eqref{080717:ve} does not contain, i.e. references to anything that might be escaping. The object's mass, it's size and shape, it's direction of travel—none of that enters the equation. It is the environment (planet) that determines $$v_e$$, and as long as an object's speed is greater than $$v_e$$, it will leave that environment and never return!

Finally, there is one more subtle (and unfortunately necessary) point to be made: the escape "velocity," $$v_e$$, is actually an escape speed.

In physics, speed and velocity have two different meanings. Speed describes the distance an object travels within some amount of time (or in other words, how fast it's going), while velocity is the combination of an object's speed and the direction in which it is moving. Speed and velocity are related, but they are different! Nevertheless, the terms are often used interchangeably, even amongst physicists, which is how we wind up with things such as $$v_e$$ being called "escape velocity" (even though it's a speed!)

That point, the difference between velocity and speed, cannot be made enough. Why? Because knowing this difference allows one to easily recognize that \eqref{080717:ve}, which represents speed, says absolutely nothing about the direction of the object's motion. Therefore, it does not matter whether the object is moving straight up and away from the planet's surface, off towards the horizon, or anywhere else in between; if its speed is greater than $$v_e$$, then it is not coming back.

To summarize, once again, the answer to the original question is yes—no matter what the object is and in which direction it is traveling, having a speed greater than $$7\text{mi}/\text{s}$$ means it leaves the Earth forever.*