text:
# [texhax] biblatex and tex4ht

Thu Aug 20 13:12:03 CEST 2009

On 20-Aug-09, at 6:21 AM, <NAME> wrote:

> 2009/8/17 Radhakrishnan CV <cvr at river-valley.org>:

[...]

>> take a look
>> at:
>>
>>  http://tug.org/~cvr/christoph.zip
>>
>> which has test.tex, test.html and test.odt. Seems like oolatex
>> behaves as
>> expected.
>
> thanks for testing this. so here is what I get when using your
> test.tex together with the mybib.bib:
>
> 1. trying to run latex on it gives me an error on \ConfigureEnv
> (undefined control sequence)

Sorry, that was my mistake, please forget it.

> 2. oolatex accepts the command but it only renders the citations in
> the first section, not the one in the subsection.

Wors fine here. Please take a look at the contents of the archive at:

http://tug.org/~cvr/test.zip

> 3. I have no Idea where the second and third reference in your
> odt-file are from. I don't have these.

Just copied the first item twice in mybib.bib and slightly changed the
keys. That was to test footnote numbering.

> 4. on the "positive" side, you are right that the stub-environment is
> rendered properly. I have no idea, how it can be working for me on one
> document but not in the other (I doublechecked!). Maybe something else
> in the preamble of my other document? If you want, I can send it to
> you.

OK, please do off the list.

> 5. also your starred subsection works. the problem is that I was
> referring to a starred sub*sub*section, so if you substitute
> \subesection* with \subsubsection* you should get the error, At least
> I am still getting it.

That is true. Eitan provided 'ooffice' option as a proof of concept
that even word processor formats can be generated from TeX sources
with the help of TeX4ht. It is not perfect, though. If you want to
make high fidelity output, you need to extend ooffice.4ht which is the
package loaded in the background when you call 'oolatex'. You need to
add extra hooks in a private configuration, which is the recommended
way to do, where you can define new environments or modify existing
ones to suit your requirements. I have done an example for you to fix
the problem of \subsubsection* in a file namely, chris.cfg which is
also included in the test.zip and listed below:

------------ begin private configuration ------------
\Preamble{ooffice}

\begin{document}
\catcode:=11
\catcode@=11

\Configure{likesubsubsection}
{\IgnorePar\EndP \HCode{<!--start of subsubsection-->}}
{\IgnorePar\EndP \HCode{<!--end of subsubsection-->}}
{\HCode{<text:h
\text:outlinelevel="4"   ><text:span
text:style-name="subsubsection"  >}%
\HCode{</text:span>}\IgnoreIndent}
{\HCode{</text:h>\Hnewline}\par\ShowPar}
\catcode:=12
\catcode@=12

\EndPreamble
--------------- end private configuration ------------

You might invoke oolatex in the following fashion:

oolatex test "chris"

and see the output. It should be OK.

> What next?
>
> BTW, I took a chance today in using oolatex on my thesis, which is
> quie a bit more complicated than the other documents. The result was
> sobering. I gut so many errors (mainly "undefined control sequence"

This is quite imaginable. You have to create a custom package
configuring all the environments, commands which were disliked by
TeX4ht.

> that I couldn't even count them. It ended where I had included a
> pdf-graphic, which of course doesn't work with latex and hence with
> oolatex. I haven't retried it without that figure, but if you are
> you at least a selection. But I suppose, you can take any bigger book
> project and oolatex will not work...

Those problems can be fixed without much difficulties. However,
pressures at work don't allow me to spend more time on your project.
Someone with more deeper knowledge of Open Office XML format will be
the best candidate to perfect ooffice.4ht.

Best regards
--


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# Down but not out

ISSN: 0043-8022

Article publication date: 1 June 2000

32

## Citation

(2000), "Down but not out", Work Study, Vol. 49 No. 3. https://doi.org/10.1108/ws.2000.07949cab.008

## Publisher

:

Emerald Group Publishing Limited

Late last year, Ernst & Young suggested that Internet retailing for the period mid-October through the end of December 1999, would amount to between US$12 billion and$15 billion. In fact, results were somewhat less than expected and revised figures are between $10 billion and$13 billion. Accurate figures for the whole of 1999 should emerge about now - Ernst & Young is maintaining its full year online retail sales estimates of $25 billion to$30 billion (after its survey of 1,283 Internet users). This is big business! (Amazon.com was the most-visited shopping site for the Xmas season, the survey found. In a separate survey by Media Metrix, Amazon.com averaged 5,693 daily unique visitors from 22 November to 26 December, followed by auction site eBay, with 4,073 average daily visitors, and eToys with 1,662 average daily visitors.)

format: plain


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# Math Help - More peano arithmetic.

1. ## Just One Last ?

Thank you soooooooo muchhhhh Captainblack. Without you, I won't be able to prove these laws of algebra. I have ONE LAST prove I can't prove. Will you pleaseeee helpp me on this one too? The axioms are still needed to prove this problem as well.

For every natural number i, for every natural number j, for every natural number k, i(j + k) = ij + ik.

2. Originally Posted by jenjen
Thank you soooooooo muchhhhh Captainblack.
---
Do you let me prove instead,
$(j+k)i=(ji)+(ki)$?--->Because CaptainBlank shown it was commutative.
If yes, then I have a prove (involving induction axiom) which I will post tomorrow because it is late here.

3. Originally Posted by jenjen
Thank you soooooooo muchhhhh Captainblack. Without you, I won't be able to prove these laws of algebra. I have ONE LAST prove I can't prove. Will you pleaseeee helpp me on this one too? The axioms are still needed to prove this problem as well.

For every natural number i, for every natural number j, for every natural number k, i(j + k) = ij + ik.
The first seven pages of this pdf document provideds the proofs of the basic
properties of arithmetic for Peano Arithmetic. Though with a slightly different
notation you can find the basic ideas needed to prove all the properties you


text:
Please use this identifier to cite or link to this item: http://arks.princeton.edu/ark:/88435/dsp011r66j3892
 Title: Increase of Issue: Poetry and Succession in Elizabethan England Authors: <NAME>visors: <NAME> Contributors: English Department Keywords: Early Modern PoetryElizabeth IShakespeareSidneySpenserSuccession Subjects: English literature Issue Date: 2018 Publisher: Princeton, NJ : Princeton University Abstract: The controversy over who would take the throne after Elizabeth I’s death arose from her refusal to do what was expected of her as both a monarch and a woman: namely, marry and produce an heir. The great uncertainty over what would come next for the Elizabethans after the death of their monarch, a subject of debate from the time of Elizabeth’s accession in 1558 to her death in 1603, provided an opportunity for poets to explore uncertainty as it related to ideas of continuity in verse. Coupled with an enduring national anxiety about the possibility of civil war after Elizabeth’s death was the Parliamentary gag order, in the form of the 1571 Treasons Act, which banned discussion of the succession. Increase of Issue: Poetry and Succession in Elizabethan England explores the relationship between monarchical succession and poetic innovation during the reign of Elizabeth I, arguing that a scepticism of the ability of the past to make secure the future emerged in English poetry during this period. The first chapter explores the relationship between gender, rhetoric, and debates about the succession in the works of George Puttenham. Analyzing his lesser-known writings, this chapter argues that the succession debates shaped Puttenham’s poetics manual, The Arte of English Poesy. Chapter Two examines how Elizabeth’s demand for silence on the matter of the succession led Sir <NAME> to imagine alternate forms of counsel in the pastoral eclogues of his Arcadia. The third chapter focuses on the genealogical cantos of The Faerie Queene, arguing that <NAME> used the stanza and canto structures of this poem to imagine a version of order in reaction to the disrupted succession. Chapter Four concludes the project by considering the implications of the succession debate on Shakespeare’s sonnets. In tracing the multiple chronologies of these poems, this chapter focuses in particular on the fact that reproduction could no longer provide an answer to the uncertain succession in the late period of Elizabeth’s reign. Shakespeare’s sonnet speaker debates how forms of reproduction and increase operate in relation to the desire for political, personal, and poetic continuity. URI: http://arks.princeton.edu/ark:/88435/dsp011r66j3892 Alternate format: The Mudd Manuscript Library retains one bound copy of each dissertation. Search for these copies in the library's main catalog: catalog.princeton.edu Type of Material: Academic dissertations (Ph.D.) Language: en Appears in Collections: English

Files in This Item:
This content is embargoed until 2020-09-28. For more information contact the Mudd Manuscript Library.


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# Wave-propagation algorithms¶

## One space dimension¶

Clawpack can be used to solve a system of equations of the form

$\kappa(x)q_t+f(q)_x = \psi(q,x,t),$

where $$q=q(x,t)\in{\cal R}^m$$. The standard case of a homogeneous conservation law has $$\kappa\equiv 1$$ and $$\psi\equiv 0$$,

(1)$q_t+f(q)_x=0.$

The flux function $$f(q)$$ can also depend explicitly on $$x$$ and $$t$$ as well as on $$q$$. Hyperbolic systems that are not in conservation form, e.g.,

$q_t+A(x,t)q_x=0,$

can also be solved. See [LeVeque-FVMHP] for more details about the wave-propagation algorithms that are briefly described here.

The basic requirement on the homogeneous system is that it be hyperbolic in the sense that a Riemann solver can be specified that, for any two states $$q_{i-1}$$ and $$Q_i$$, returns a set of $$M_w$$ waves $${\cal W}^p_{i-1/2}$$ and speeds $$s^p_{i-1/2}$$ satisfying

$\sum_{p=1}^{M_w} {\cal W}^p_{i-1/2} = Q_i-Q_{i-1} \equiv \Delta Q_{i-1/2}.$

The Riemann solver must also return a left-going fluctuation $${\cal A}^-\Delta Q_{i-1/2}$$ and a right-going fluctuation $${\cal A}^+\Delta Q_{i-1/2}$$. In the standard conservative case (1) these should satisfy

(2)${\cal A}^-\Delta Q_{i-1/2}+{\cal A}^+\Delta Q_{i-1/2} = f(Q_i)-f(Q_{i-1})$

and the fluctuations then define a “flux-difference splitting”.

${\cal A}^-\Delta Q_{i-1/2} = \sum_p (s^p_{i-1/2})^- {\cal W}^p_{i-1/2}, \qquad {\cal A}^+\Delta Q_{i-1/2} = \sum_p (s^p_{i-1/2})^+ {\cal W}^p_{i-1/2},$

where $$s^- = \min(s,0)$$ and $$s^+ = \max(s,0)$$. In the nonconservative case eqn{claw_1dnoncon}, there is no “flux function” $$f(q)$$, and the constraint (2) need not be satisfied.

## Godunov’s method¶

Only the fluctuations are used for the first-order Godunov method, which is implemented in the form

$Q_i^{n+1} = Q_i^n - \frac{\Delta t}{\Delta x}\left[{\cal A}^+\Delta Q_{i-1/2} + {\cal A}^-\Delta Q_{i+1/2}\right],$

assuming $$\kappa \equiv 1$$.

The Riemann solver must be supplied by the user in the form of a subroutine rp1, as described in Specifying the Riemann solver.

Typically the Riemann solver first computes waves and speeds and then uses these to compute $${\cal A}^+Q_{i-1/2}$$ and $${\cal A}^-Q_{i-1/2}$$ internally in the Riemann solver.

## High-resolution methods¶

The waves and speeds must also be returned by the Riemann solver in order to use the high-resolution methods described in [LeVeque-FVMHP], which reduce to a second-order accurate Lax-Wendroff type method when limiters are not used. By introducing wave limiters, non-physical oscillations near discontinuities or steep gradients in the solution can be avoided. The limiters are based on the theory of TVD methods for nonlinear scalar equations and extended in a natural way to systems of equations.

These methods take the form

$Q_i^{n+1} = Q_i^n - \frac{\Delta t}{\Delta x}\left[{\cal A}^+Q_{i-1/2} + {\cal A}^-Q_{i+1/2}\right] - \frac{\Delta t}{\Delta x}(\tilde F_{i+1/2} - \tilde F_{i-1/2})$

where

$\tilde F_{i-1/2} = \frac 1 2 \sum_{p=1}^{M_w} |s^p_{i-1/2}| \left( 1-\frac{\Delta t}{\Delta x} |s^p_{i-1/2}|\right) \tilde{\cal W}_{i-1/2}^p.$

Here $$\tilde{\cal W}_{i-1/2}^p$$ represents a limited version of the wave $${\cal W}_{i-1/2}^p$$, obtained by comparing $${\cal W}_{i-1/2}^p$$ to $${\cal W}_{i-3/2}^p$$ if $$s^p>0$$ or to $${\cal W}_{i+1/2}^p$$ if $$s^p<0$$.

## f-wave formulation¶

For equations with spatially-varying flux functions, such as

(3)$q_t+f(q,x)_x=0.$

it is often convenient to use the f-wave formulation of the algorithm, as proposed in [BaleLevMitRoss02]. Rather than decomposing the jump $$Q_i-Q_{i-1}$$ into waves, the idea of the f-wave approach is to decompose the flux difference $$f(Q_i,x_i) - f(Q_{i-1},x_{i-1})$$ into f-waves using appropriate eigenvectors of the Jacobian matrices to either side of the interface. See f-wave Riemann solvers for more details.

## Capacity functions¶

When a capacity function $$\kappa(x)$$ is present, the Godunov method becomes

$Q_i^{n+1} = Q_i^n - \frac{\Delta t}{\kappa_i \Delta x} \left[{\cal A}^+Q_{i-1/2} + {\cal A}^-Q_{i+1/2}\right],$

See [LeVeque-FVMHP] for discussion of this algorithm and its extension to the high-resolution method. Capacity functions are useful in particular for solving equations on a mapped grid.

## Source terms¶

If the equation has a source term, a routine src1 must also be supplied that solves the source term equation $$q_t=\psi$$ over a time step. A fractional step method is used to couple this with the homogeneous solution, as described in Using src for source terms.

## Boundary conditions¶

Boundary conditions are imposed by setting values in ghost cells each time step, as described in Boundary conditions.

TODO: Continue description – 2d and 3d, transverse solvers.


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# monkey's uncle

## Trying out the LaTeX Plugin

#### September 11th, 2008 · No Comments · Uncategorized

This is an experiment to see how the WordPress $\LaTeX$ plugin works.  What better equation to try out than the Euler-Lotka equation:

where $\alpha$ is age at first reproduction, $\beta$ is age at last reproduction, $r$ is the intrinsic rate of population increase, $l(a)$ is the proportion surviving to exact age $a$, and $m(a)$ is the fertility rate of women age $a.$

Tags:

### No Comments so far ↓

There are no comments yet...Kick things off by filling out the form below.


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## Stream: maths

### Topic: algebra structure on lc

#### <NAME> (Mar 21 2019 at 07:29):

If A is an R-algebra, then there is a natural algebra structure on lc R A, which is just "expanding the product". It turns lc.total into an algebra hom. Should this go in mathlib. (I think it is valuable.) If so, where?
https://github.com/leanprover-community/lean-perfectoid-spaces/blob/master/src/for_mathlib/lc_algebra.lean

Last updated: May 11 2021 at 16:22 UTC


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# Excercise on distinguishable particles interacting with Hamiltonian

1. Feb 11, 2013

### venetiano77

Please, help me with this problem!

Two distinguishable particles of spin 1/2 interact with Hamiltonian
H=A*S1,z*S2,x
with A a positive constant. S1,z and S2,x are the operators related to the z-component of the spin of the first particle and to the x-component of the spin of the second particle, respectively.
If at t=0 the system is in its simultaneous eigenket of S1,z for the eigenvalue h/4pi (ħ/2) and of S2,x for the eigenvalue h/4pi (ħ/2), FIND THE STATE OF THE SYSTEM AT THE TIME t

Attempt of solution:
Firstly, one should use the formulas for temporal evolution. Please, help me!

#### Attached Files:

• ###### buttare.GIF
File size:
36.6 KB
Views:
83
2. Feb 11, 2013

### TSny

Re: Excercise on distinguishable particles interacting with Hamiltonia

Note that the problem in the attachment states that the initial state is an eigenket for S2,z rather than S2,x.

Last edited: Feb 11, 2013


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# What is the chemical name for "Na"_2"O" ?

Mar 24, 2018

Sodium oxide.

#### Explanation:

The first thing to notice here is that this compound contains a metal, sodium, and a nonmetal, oxygen, so right from the start, you know that you're dealing with an ionic compound.

Now, sodium is located in group 1 of the Periodic Table, which means that an atom of sodium has $1$ valence electron. In order to complete its octet, sodium will lose this valence electron and form the sodium cation, ${\text{Na}}^{+}$.

Oxygen, on the other hand, is located in group 16 of the Periodic Table, which means that an atom of oxygen has $6$ valence electrons. In order to complete its octet, oxygen takes in $2$ electrons and forms the oxide anion, ${\text{O}}^{2 -}$.

In order to name this compound, use the name of the cation first and the name of the anion last.

You will have

"Na"_ 2"O" -> 2color(blue)(["Na"]^(+)) + color(red)(["O"]^(2-)) -> color(blue)("sodium") quad color(red)("oxide")

Notice that the sodium cation carries a $1 +$ charge and the oxide anion carries a $2 -$ charge, which is why you need $2$ sodium cations to balance the overall negative charge coming from $1$ oxide anion.


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# TaxGNI Fair Income Tax

Income tax represents the most fair and honest tax systems. While VAT, sales, and excise taxes are regressive, hide tax rates, and aren’t paid by people who put their money in savings or financial investments, income taxes are taken up-front and apply to all income.

Traditional income tax systems require adjustments to tax brackets, rates, and standard deductions, along with multiple filing statuses and other complexities. This not only makes tax filing complex, but also hides effective tax rates and changes to the distribution of tax burden from taxpayers.

The TaxGNI system replaces this with adult equivalence factors and a simple effective tax rate computation. Tax rates are based on the per-capita income, and so adjust for an equitable society rather than for one with a growing poverty base: TaxGNI is designed for societies using a structural minimum wage.

## Calculating Taxes

In practice, an online or smartphone application would compute the taxes owed with a simple form like below.

 Persons age 14 or above: 3 Persons younger than age 14: 1 Highest individual’s income: $45,000 Total household income:$85,000 Your Taxes Computed Below Equivalence Factor: 3.1 Tax Rate: 11.34% Federal Taxes: 9,639 A taxpayer can work out their taxes with a pocket calculator, their income, and a few published numbers. 1. Divide your household income by the income baseline to get your Income Share. 2. Count your household members and add up the Equivalence Value. 3. Compute the Equivalence Factor. 4. Divide your Income Share by the Equivalence Factor to get your Modified Income Share. 5. Compute the Effective Tax Rate. 6. Multiply the Effective Tax Rate by your household income to find your income tax liability. ## Income Share TaxGNI computes tax rates based on a taxpayer’s income as a portion of the per-capita income or GNI/C. \begin{aligned} I &= \frac{Income}{per-capita\ Income} \end{aligned} ## The Equivalence Factor Traditional systems give joint-filing households higher initial tax brackets, and remove this advantage at high incomes. They also allow exemptions for the taxpayer and each dependent meeting a complex set of rules and phasing out at higher incomes. TaxGNI replaces filing statuses, brackets, and exemptions with an adult equivalence factor. Economists use equivalence values to adjust a household’s income to the income of a single-adult household at the same standard of living. Our equivalence factor phases out at higher incomes, removing the tax advantage. TaxGNI modifies the household income with this factor, and uses that to determine the effective tax rate. Any combination of taxpayers living in the same household may file as a unit, which helps multi-generational households and other complex households. ### Calculating the Equivalence Factor Under TaxGNI, the adult earning the highest income is the Head Earner. If the Head Earner represents ⅞ or less of the total household income, one additional income-earning adult also becomes a Head Earner. The values below apply to the household. Head EarnersAdults (age 14+)Children 10.70.5 TaxGNI modifies this value using your household income (I) and the per-capita income or GNI/C (G) with the below formula, which a pocket calculator can compute. \begin{aligned} E(q) &= 1 + \frac{q-1}{1+e^{2(I-3)}} \end{aligned} If the GNI/C is60,000, an \$85,000 household with two working adults, a 14-year-old child, and a 13-year-old child would have an equivalence factor as below.

\begin{aligned} I &= 1.42 \\ q &= 2 + 0.7 + 0.5 \\ &= 3.2 \\ \\ E(q) &= 1 + \frac{q-1}{1+e^{2(I-3)}} \\ &= 3.11 \end{aligned}

The household’s modified income becomes their income divided by the equivalence factor.

\begin{aligned} I_{[m]} &= \frac{I}{E(q)} \\ &=0.457 \end{aligned}

## The Effective Tax Rate

TaxGNI computes the effective tax rate using a continuous curve multiplied by the top tax rate (R). The formula must be published for transparency reasons.

Our proposed formula is the one below, with fixed values for (a) and (b). Altering these values can make the curve flatter or more-progressive.

\begin{aligned} T(I_{[m]}) &= (1 – \frac{1}{a \times I_{[m]}+b}) \times R \end{aligned}

The initial implementation should match up as best as possible to the effective tax rates of a traditional bracketed system.

## Credits and Deferred Income

We prefer credits to deductions, or the use of a program like a universal citizen’s dividend to offset taxes. The mortgage deduction, for example, would either cease to exist or become a credit of some portion of mortgage interest modified by the equivalence factor.

Deferred income raises further questions. Research suggests 401(k) and IRA deductions favor higher-income earners, and stronger Social Security benefits would better support lower- and middle-income households.

Other deductions should be eliminated where appropriate. A universal healthcare system should simply cover more out-of-pocket expenses rather than providing a deduction. Universal college and vocational education obsoletes the incredibly-weak lifetime learning credit.

Deductions sound great to policymakers and constituents, but most working-class Americans don’t benefit. Our society should replace such clutter with better programs, credits, and fair taxes to begin with.


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# GMAT Quantitative University of Management Development Institute of Singapore (MDIS) 2017

Subject: ; Class: ; with 37 questions; test in 75 minutes; update 26/10/2017
 Time 75 minutes Time to take the test Start exam Click button start to test. Guide to the test Subjects Gmat test Update 26/10/2017 Class Level 3 Number of questions 37 View 934 Tested 0

Question 1.

A gambler began playing blackjack with $110 in chips. After exactly 12 hands, he left the table with$320 in chips, having won some hands and lost others. Each win earned $100 and each loss cost$10. How many possible outcomes were there for the first 5 hands he played? (For example, won the first hand, lost the second, etc.)

 (A) 10 (B) 18 (C) 26 (D) 32 (E) 64

Question 2.

In a 4 person race, medals are awarded to the fastest 3 runners. The first-place runner receives a gold medal, the second-place runner receives a silver medal, and the third-place runner receives a bronze medal. In the event of a tie, the tied runners receive the same color medal. (For example, if there is a two-way tie for first- place, the top two runners receive gold medals, the next-fastest runner receives a silver medal, and no bronze medal is awarded). Assuming that exactly three medals are awarded, and that the three medal winners stand together with their medals to form a victory circle, how many different victory circles are possible?

 (A) 24 (B) 52 (C) 96 (D) 144 (E) 648

Question 3.

The organizers of a week-long fair have hired exactly five security guards to patrol the fairgrounds at night for the duration of the event. Exactly two guards are assigned to patrol the grounds every night, with no guard assigned consecutive nights. If the fair begins on a Monday, how many different pairs of guards will be available to patrol the fairgrounds on the following Saturday night?

 (A) 9 (B) 7 (C) 5 (D) 3 (E) 2

Question 4.

Larry, Michael, and Doug have five donuts to share. If any one of the men can be given any whole number of donuts from 0 to 5, in how many different ways can the donuts be distributed?

 (A) 21 (B) 42 (C) 120 (D) 504 (E) 5040

Question 5.

How many different combinations of outcomes can you make by rolling three standard (6-sided) dice if the order of the dice does not matter?

 (A) 24 (B) 30 (C) 56 (D) 120 (E) 216

Question 6.

A certain league has four divisions. The respective divisions had 9, 10, 11, and 12 teams qualify for the playoffs. Each division held its own double-elimination tournament -- where a team is eliminated from the tournament upon losing two games -- in order to determine its champion. The four division champions then played in a single-elimination tournament -- where a team is eliminated upon losing one game -- in order  todetermine the overall league champion. Assuming that there were no ties and no forfeits, what is the maximum number of games that could have been played in order to determine the overall league champion?

 (A) 79 (B) 83 (C) 85 (D) 87 (E) 88

Question 7.

You have a bag of 9 letters: 3 Xs, 3 Ys, and 3 Zs. You are given a box divided into 3 rows and 3 columns for a total of 9 areas. How many different ways can you place one letter into each area such that there are no rows or columns with 2 or more of the same letter? (Note: One such way is shown below.)

 X Y Z Y Z X Z X Y
 (A) 5 (B) 6 (C) 7 (D) 12 (E) 18

Question 8.

Eight women of eight different heights are to pose for a photo in two rows of four. Each woman in the second row must stand directly behind a shorter woman in the first row. In addition, all of the women in each row must be arranged in order of increasing height from left to right. Assuming that these restrictions are fully adhered to, in how many different ways can the women pose?

 (A) 2 (B) 14 (C) 15 (D) 16 (E) 18

Question 9.

You have a six-sided cube and six cans of paint, each a different color. You may not mix colors of paint. How many distinct ways can you paint the cube using a different color for each side? (If you can reorient a cube to look like another cube, then the two cubes are not distinct.)

 (A) 24 (B) 30 (C) 48 (D) 60 (E) 120

Question 10.

A group of four women and three men have tickets for seven adjacent seats in one row of a theatre. If the three men will not sit in three adjacent seats, how many possible different seating arrangements are there for these 7 theatre-goers?

 (A) 7! – 2!3!2! (B) 7! – 4!3! (C) 7! – 5!3! (D) 7 × 2!3!2! (E) 2!3!2!

Question 11.

Anthony and Michael sit on the six-member board of directors for company X. If the board is to be split up into 2 three-person subcommittees, what percent of all the possible subcommittees that include Michael also include Anthony?

 (A) 20% (B) 30% (C) 40% (D) 50% (E) 60%

Question 12.

A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there?

 (A) 28 (B) 32 (C) 48 (D) 60 (E) 120

Question 13.

Six mobsters have arrived at the theater for the premiere of the film “Goodbuddies.” One of the mobsters, Frankie, is an informer, and he's afraid that another member of his crew, Joey, is on to him. Frankie, wanting to keep Joey in his sights, insists upon standing behind Joey in line at the concession stand. How many ways can the six arrange themselves in line such that Frankie’s requirement is satisfied?

 (A) 6 (B) 24 (C) 120 (D) 360 (E) 720

Question 14.

A student committee that must consist of 5 members is to be formed from a pool of 8 candidates. How many different committees are possible?

 (A) 5 (B) 8 (C) 40 (D) 56 (E) 336

Question 15.

How many ways are there to award a gold, silver and bronze medal to 10 contending teams?

 (A) 10 × 9 × 8 (B) 10! / 3! 7! (C) 10! / 3! (D) 360 (E) 300

Question 16.

On Tuesday, Kramer purchases exactly 3 new shirts, 2 new sweaters, and 4 new hats, On the following day and each subsequent day thereafter, Kramer wears one of his new shirts together with one of his new sweaters and one of his new hats. Kramer avoids wearing the exact same combination of shirt, sweater, and hat for as long as possible. On which day is this no longer possible?

 (A) Tuesday (B) Wednesday (C) Thursday (D) Friday (E) Saturday

Question 17.

A certain stock exchange designates each stock with a one- , two-, or three-letter code, where each letter is selected from the 26 letters of the alphabet. If the letter may be repeated and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes?

 (A) 2951 (B) 8125 (C) 15600 (D) 15302 (E) 18278

Question 18.

A certain law firm consists of 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formed in which at least one member of the group is a senior partner? (Two groups are considered different if at least one group member is different.)

 (A) 48 (B) 100 (C) 120 (D) 288 (E) 600

Question 19.

Pat will walk from intersection X to intersection y along route that is confined to the square grid of four streets and three avenues shown in the map above. How many routes from X to Y can Pat take that have the minimum possible length?

 (A) 6 (B) 8 (C) 10 (D) 14 (E) 16

Question 20.

A fair coin is flipped three times. What is the probability that the coin lands on heads exactly twice?

 (A) 1/8 (B) 3/8 (C) 1/2 (D) 5/8 (E) 7/8

Question 21.

There are 10 women and 3 men in room A. One person is picked at random from room A and moved to room B, where there are already 3 women and 5 men. If a single person is then to be picked from room B, what is the probability that a woman will be picked?

 (A) 13/21 (B) 49/117 (C) 15/52 (D) 5/18 (E) 40/117

Question 22.

If the probability of rain on any given day in Chicago during the summer is 50%, independent of what happens on any other day, what is the probability of having exactly 3 rainy days from July 4 through July 8, inclusive?

 (A) 1/32 (B) 2/25 (C) 5/16 (D) 8/25 (E) 3/4

Question 23.

In a shipment of 20 cars, 3 are found to be defective. If four cars are selected at random, what is the probability that exactly one of the four will be defective?

 (A) 170/1615 (B) 3/20 (C) 8/19 (D) 3/5 (E) 4/5

Question 24.

A certain bag of gemstones is composed of two-thirds diamonds and one-third rubies. If the probability of randomly selecting two diamonds from the bag, without replacement, is 5/12, what is the probability of selecting two rubies from the bag, without replacement?

 (A) 5/36 (B) 5/24 (C) 1/12 (D) 1/6 (E) 1/4

Question 25.

Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 competitors in the triathlon and medals are awarded for first, second, and third place, what is the probability that at least two of the triplets will win a medal?

 (A) 3/14 (B) 19/84 (C) 11/42 (D) 15/28 (F) 3/4

Question 26.

Set S is the set of all prime integers between 0 and 20. If three numbers are chosen randomly from set S and each number can be chosen only once, what is the positive difference between the probability that the product of these three numbers is a number less than 31 and the probability that the sum of these three numbers is odd?

 (A) 1/336 (B) 1/2 (C) 17/28 (D) 3/4 (E) 301/336

Question 27.

A random 10-letter code is to be formed using the letters A, B, C, D, E, F, G, H, I and I (only the “I” will be used twice). What is the probability that a code that has the two I’s adjacent to one another will be formed?

 (A) 1/10 (B) 1/8 (C) 1/5 (D) 1/4 (E) 1/2

Question 28.

If p2 – 13p + 40 = q, and p is a positive integer between 1 and 10, inclusive, what is the probability that q < 0?

 (A) 1/10 (B) 1/5 (C) 2/5 (D) 3/5 (E) 3/10

Question 29.

A box contains three pairs of blue gloves and two pairs of green gloves. Each pair consists of a left-hand glove and a right-hand glove. Each of the gloves is separate from its mate and thoroughly mixed together with the others in the box. If three gloves are randomly selected from the box, what is the probability that a matched set (i.e., a left- and right-hand glove of the same color) will be among the three gloves selected?

 (A) 3/10 (B) 23/60 (C) 7/12 (D) 41/60 (E) 5/6

Question 30.

A football team has 99 players. Each player has a uniform number from 1 to 99 and no two players share the same number. When football practice ends, all the players run off the field one-by-one in a completely random manner. What is the probability that the first four players off the field will leave in order of increasing uniform numbers (e.g., #2, then #6, then #67, then #72, etc)?

 (A) 1/64 (B) 1/48 (C) 1/36 (D) 1/24 (E) 1/16

Question 31.

Bill and Jane play a simple game involving two fair dice, each of which has six sides numbered from 1 to 6 (with an equal chance of landing on any side). Bill rolls the dice and his score is the total of the two dice. Jane then rolls the dice and her score is the total of her two dice. If Jane’s score is higher than Bill’s, she wins the game. What is the probability the Jane will win the game?

 (A) 15/36 (B) 175/432 (C) 575/1296 (D) 583/1296 (E) 1/2

Question 32.

Kate and Danny each have $10. Together, they flip a fair coin 5 times. Every time the coin lands on heads, Kate gives Danny$1. Every time the coin lands on tails, Danny gives Kate $1. After the five coin flips, what is the probability that Kate has more than$10 but less than \$15?

 (A) 5/16 (B) 1/2 (C) 12/30 (D) 15/32 (E) 3/8

Question 33.

There is a 10% chance that it won’t snow all winter long. There is a 20% chance that schools will not be closed all winter long. What is the greatest possible probability that it will snow and schools will be closed during the winter?

 (A) 55% (B) 60% (C) 70% (D) 72% (E) 80%

Question 34.

There are y different travelers who each have a choice of vacationing at one of n different destinations. What is the probability that all y travelers will end up vacationing at the same destination?

 (A) 1/n! (B) n/n! (C) 1/ny (D) 1/ny -1 (E) n/yn

Question 35.

A small, experimental plane has three engines, one of which is redundant. That is, as long as two of the engines are working, the plane will stay in the air. Over the course of a typical flight, there is a 1/3 chance that engine one will fail. There is a 75% probability that engine two will work. The third engine works only half the time. What is the probability that the plane will crash in any given flight?

 (A) 7/12 (B) 1/4 (C) 1/2 (D) 7/24 (E) 17/24

Question 36.

Ms. Barton has four children. You are told correctly that she has at least two girls but you are not told which two of her four children are those girls. What is the probability that she also has two boys? (Assume that the probability of having a boy is the same as the probability of having a girl.)

 (A) 1/4 (B) 3/8 (C) 5/11 (D) 1/2 (E) 6/11

Question 37.

Laura has a deck of standard playing cards with 13 of the 52 cards designated as a "heart." If Laura shuffles the deck thoroughly and then deals 10 cards off the top of the deck, what is the probability that the 10th card dealt is a heart?

 (A) 1/4 (B) 1/5 (C) 5/26 (D) 12/42 (E) 13/42


text:
# Compare Gram-Schmidt and Householder Orthogonalization Algorithms

Posted by <NAME>,

Classical Gram-Schmidt and Modified Gram-Schmidt are two algorithms for orthogonalizing a set of vectors. Householder elementary reflectors can be used for the same task. The three algorithms have very different roundoff error properties.... read more >>

# The Graeffe Root-Squaring Method for Computing the Zeros of a Polynomial1

Posted by <NAME>,

At a minisymposium honoring Charlie Van Loan this week during the SIAM Annual Meeting, I will describe several dubious methods for computing the zeros of polynomials. One of the methods is the Graeffe Root-squaring method, which I will demonstrate using my favorite cubic, $x^3-2x-5$.... read more >>

# 19 Dubious Ways to Compute the Zeros of a Polynomial2

Posted by <NAME>,

During the SIAM Annual Meeting this summer in Boston there will be a special minisymposium Wednesday afternoon, July 13, honoring <NAME>, who is retiring at Cornell. (I use "at" because he's not leaving Ithaca.) I will give a talk titled "19 Dubious Way to Compute the Zeros of a Polynomial", following in the footsteps of the paper about the matrix exponential that Charlie and I wrote in 1978 and updated 25 years later. I really don't have 19 ways to compute polynomial zeros, but then I only have a half hour for my talk. Most of the methods have been described previously in this blog. Today's post is mostly about "roots".... read more >>

# Math and Music2

Posted by <NAME>,

What does $\sqrt[12]{2}$ have to do with music? What are equal temperament and just intonation? How can the MATLAB function rats help tune a piano? (This post is based in part on the Music chapter in my online book, Experiments in MATLAB.)... read more >>

# Modernization of Numerical Integration, From Quad to Integral

Posted by <NAME>,

The MATLAB functions for the numerical evaluation of integrals has evolved from quad, through quadl and quadgk, to today's integral.... read more >>

# Strang and Moler Video Course on Differential Equations

Posted by <NAME>,

G<NAME> has produced a MOOC-style video course on Differential Equations and Linear Algebra. I have added some videos about the MATLAB ODE suite. The series is available from the MathWorks Web site, MIT OpenCourseWare and several other popular sources.... read more >>

# Further Twists of the Moebius Strip2

Posted by <NAME>,

The equations generating a surf plot of the Moebius strip can be parameterized and the parameters allowed to take on expanded values. The results are a family of surfaces that I have been displaying for as long as I have had computer graphics available.... read more >>

# The Eigenwalker Model of the Human Gait

Posted by <NAME>,

A model of the human gait, developed by <NAME>, is a five-term Fourier series with vector-valued coefficients that are the principal components for data obtained in motion capture experiments involving subjects walking on a treadmill.... read more >>

# Dark Energy Gravitational Waves

Posted by <NAME>,

Recent theoretical, observational and computational results establish the possibility that gravitational waves produced by the dark energy created at the dawn of the universe affect the clock rate of silicon digital processors operating at very low temperatures.... read more >>

# Piet Hein, Super Ellipses and Soma Cubes3

Posted by <NAME>,

An extraordinarily creative Danish mathematician, inventor, and poet who often wrote under the Old Norse pseudonym "Kumbel" meaning "tombstone." A direct descendant of the Dutch naval hero of the 16th century who had the same name, <NAME> was born in Copenhagen and studied at the Institute for Theoretical Physics of the University of Copenhagen (later the Niels Bohr Institute) and the Technical University of Denmark. ... read more >>

These postings are the author's and don't necessarily represent the opinions of MathWorks.


text:
mersenneforum.org (https://www.mersenneforum.org/index.php)
-   Homework Help (https://www.mersenneforum.org/forumdisplay.php?f=78)
-   -   Help with exercise questions from Elementary Number Theory (https://www.mersenneforum.org/showthread.php?t=27067)

 charybdis 2021-11-27 16:32

[QUOTE=R. Gerbicz;594026]...furthermore: k*binomial(p,k)=p*binomial(p-1,k-1) is also true...[/QUOTE]

And we can do this combinatorially too: both are the way of choosing k elements from a set of p elements, where we label one of those k elements as "special".
We get k*binomial(p,k) by choosing the set of k elements and then picking the special element, and p*binomial(p-1,k-1) by first picking the special element and then choosing the other k-1 elements from the remaining p-1 elements of our set.

More famously, binomial(p-1,k-1)+binomial(p-1,k)=binomial(p,k) is easiest to see combinatorially: we partition the binomial(p,k) choices of k elements from the numbers {1,...,p} into two, depending on whether they contain 1 or not. binomial(p-1,k-1) is the number of choices containing 1, and binomial(p-1,k) is the number of choices not containing 1.

 Dr Sardonicus 2021-11-29 16:05

[QUOTE=R. Gerbicz;594026]Modified your idea, close to a pure combinatorial proof:

The k=0 case is trivial, so assume that 0<k<p,
it is known: binomial(p-1,k-1)+binomial(p-1,k)=binomial(p,k) [for here you don't need that p is prime]
<snip>[/QUOTE]Of course! And for 0 < k < p, $$\frac{p!}{k!(p-k)!}$$ clearly is divisible by p because the denominator is composed of factors less than p.

The result can be extended slightly. Since p divides binomial(p,k) for 0 < k < p when p is prime, we have the "freshman's dream" polynomial identity in [b]F[/b][sub]p[/sub][x,y]

$$(x\;+\;y)^{p}\;=\;x^{p}\;+\;y^{p}$$

Repeatedly raising to the p[sup]th[/sup] power, we see that in [b]F[/b][sub]p[/sub][x,y] for any positive integer n,

$$(x\;+\;y)^{p^{n}}\;=\;x^{p^{n}}\;+\;y^{p^{n}}$$

which shows that binomial(p[sup]n[/sup],k) is divisible by p for 0 < k < p[sup]n[/sup].

Then the above argument shows that for p prime and any positive integer n,

[TEX]{p^{n}-1\choose k}\equiv -1^k \;\pmod p[/TEX]

 MattcAnderson 2021-12-03 13:29

more of this

*smile*

 bur 2021-12-10 17:33

Thanks a lot for all the replies, it will take me while to get through all that.

By "reduced" I meant, there would be no smaller integer congruent to them mod p. I forgot about that smaller could also mean "negative".

All times are UTC. The time now is 08:12.


text:
PREPRINT
F785BBE2-874C-4104-B2E3-57022B63F5DA

# Searching for compact objects with Gaia DR3

<NAME>, <NAME>, <NAME>, <NAME>, <NAME>, <NAME>, <NAME>
arXiv:2207.05434

Submitted on 12 July 2022

## Abstract

We search for compact objects in binaries based on Gaia DR3. A sample of ten targets is derived under the conditions: low temperature (${T}_{\mathrm{e}\mathrm{f}\mathrm{f}}<6000$ K), high radial velocity variation ($\mathrm{\Delta }{V}_{\mathrm{r}}>200$ km s${}^{-1}$), high mass function ($f\left({M}_{2}\right)>1{M}_{\odot }$), and ellipsoidal-like light curves. Two targets have LAMOST spectroscopic observations, one of which is a double-lined spectroscopic binary. The observational data of seven targets are not self-consistent, since their photometric periods are even shorter than the theoretical minimum orbital periods calculated by the stellar parameters from Gaia DR3. According to the gathered data, two targets may contain compact objects and are worth follow-up observations. This work may serve as an example to demonstrate the feasibility of searching for compact objects in the massive Gaia data.

## Preprint

Comment: To be submitted

Subjects: Astrophysics - Solar and Stellar Astrophysics; Astrophysics - Astrophysics of Galaxies


text:
# How to figure out binominal distribution

• June 4th 2009, 05:41 PM
krzyrice
How to figure out binominal distribution
Here is the question:

I need help figuring out what is the success rate of this. Thanks for your help.
• June 4th 2009, 05:57 PM
pickslides
$
P(A\cup B) = P(A)+P(B) = \frac{1}{4} + \frac{1}{8} = \frac{2}{8} + \frac{1}{8} = \frac{3}{8}
$

(Rofl)(Rofl)(Rofl)
• June 4th 2009, 06:12 PM
krzyrice
Thank you for help pickslides.

However, I am now stuck on figuring out part a)
It tells you it is spun once.
So is the answer 1C1 x (.375)^1
• June 4th 2009, 06:22 PM
pickslides
My answer is correct for 1 trial.

Using the binomial theorem

$
P(A \cup B) = \binom{1}{1}\left(\frac{3}{8}\right)^1\left(\frac{ 5}{8}\right)^0 = \frac{3}{8}
$
• June 4th 2009, 06:44 PM
krzyrice
Excellent explanation. Thanks!


text:
# Data Capture Server

While technically not an addon, the Data Capture server (or dcserver in short) is a separate product that can be deployed on a Windows or Linux server as a service or daemon.

The dcserver is used to schedule and perform bulk imports, freeing the Phaedra clients of this resource-intensive task. To facilitate this, the dcserver has a web interface that allows users to:

• Schedule new scan jobs that scan specific network locations for new data and may trigger capture jobs to import this data.
• Track the queue of capture jobs. For each capture job, users can see the progress, and the submitter may choose to cancel the job.
• Register themselves to receive e-mail notifications on various events, such as capture job completion or errors.

In addition, the dcserver can also be used to host a scriptable RESTful API, allowing other systems to consume data from Phaedra.

## Setup

Technically, the dcserver is a product very similar to the Phaedra client, except that it runs in headless mode (optionally with a console attached to it) instead of in a user interface, i.e. a workbench.

-vmargs
-Xmx8G
-Ddatacapture.server=true
-Djetty.http.port=8080
-Dphaedra.config=smb://hostname/sharename/config.xml
-Dphaedra.env=EnvironmentName
-Dphaedra.fs.path=/path/to/fileshare

• -Xmx: the maximum amount of memory the dcserver is allowed to use. Make sure to set this value sufficiently high as capture jobs may run concurrently and consume a high amount of memory.
• -Ddatacapture.server: set to true to enable all the server features such as the datacapture task queue and the web interface.
• -Djetty.http.port: set the port where the web interface will be exposed on.
• -Dphaedra.config: set the smb or file path pointing to the environment configuration file.
• -Dphaedra.env: set the name of the environment to connect to

If the dcserver runs on a Linux host, one additional setting is required: -Dphaedra.fs.path. In the config.xml file, the environment's file share component will specify an smb:// path. Due to a technical limitation, Phaedra instances running on Linux cannot access such smb paths with write access (which is needed to be able to import data). The setting Dphaedra.fs.path is used to provide an alternative path to the file share, either via a writable mount point or a local file path.

In addition to the above server settings, you can also configure various other system settings via the ini file:

• -Djava.io.tmpdir: set the directory that will be used to store temporary files.
• -Djetty.https.port: set the SSL port where the web interface will be exposed on.
• -Djavax.net.ssl.keyStore: the location of a Java keystore file (.jks) containing certificates, if you enable https.
• -Djavax.net.ssl.keyStorePassword: the password that is needed to access the keystore file.

Note that all of these settings should come after the "-vmargs" line.

## Running the server

To launch the server in a console, execute the dcserver binary in the main folder.

Note: This executable can be wrapped into a service just like any regular executable file. The configuration of a service or daemon is beyond the scope of this document.

## Using the web interface

When the dcserver is up and running, it will log a message mentioning the HTTP port being listened on, for example:

[jetty.server.AbstractConnector] Started ServerConnector@4fa5cc73{HTTP/1.1,[http/1.1]}{0.0.0.0:8080}

You can then open a browser and point it to the following URL:

http://hostname:8080/ui/dc/index.html

## Capture Jobs

The Capture Jobs page shows the activity of the dcserver: it lists data capture jobs that are currently running, or have recently finished. By default, the jobs from the past 2 weeks are listed. You can use the From and Until filters to show jobs from a different time window.

The Status column shows the outcome of the job:

• Queued: The job is in a queue, and will run as soon as the server has free resources to do so
• Running: The job is currently being executed
• Cancelled: The job has been cancelled by a user or administrator
• Error: The job has failed with an error message
• Finished: The job has completed and its results have been imported in Phaedra

To view more details for a job (such as the error message of a failed job), click on the Details button to the right of the job.

### Launching Capture Jobs

New capture jobs can be launched manually, using the Submit new Capture Job button in the top right of the screen.

To launch a capture job, enter the following information:

• Source Path: The path to the folder that should be captured. Note that this should be a network path (e.g. a Windows UNC path) that the dcserver can access.
• Protocol: The protocol in Phaedra where the data should be imported into. Selecting a protocol will automatically select a Capture Config as well (see below).
• Capture Config: The capture configuration that should be used by this capture job. Usually, this is selected automatically by selecting a Protocol, and there is no need to select this field manually.
• Experiment: A new or existing experiment in the selected Protocol where the plates will be imported into.

## Scan Jobs

While launching capture jobs manually may be desirable in some cases, in other situations it is more convenient to automate this procedure. For example, when your team is performing the primary screen phase for a large assay, you may have instruments generating hundreds of plates every day, and you don't want to submit dozens of capture jobs to get them all into Phaedra!

That is the purpose of Scan Jobs: they are defined once, and then run on a regular interval, e.g. every hour or every day. They will look for new data to import, and will automatically submit Capture Jobs for any new data that is found.

To create a new Scan Job, select the Create New Scan Job button in the top right of the screen.

Enter the required information for the new Scan Job:

• Name: A unique name for this scan job
• Description: A description for this scan job
• Type: A scanner type. By default, Phaedra comes with 2 scanner types: folderScanner, which can be used to scan network locations such as shared drives, and columbusScanner which can be used to scan the contents of a Columbus server.
• Schedule: A scheduling interval, telling the dcserver how often to run the scan job. For more information about the scheduling syntax, see http://www.quartz-scheduler.org/documentation/quartz-2.x/tutorials/crontrigger.html
• Scanner configuration: Detailed XML configuration for the scan job, which depends on the selected scanner Type. Example:
    <config>
<path>\\example.server.com\share1\dataToImport</path>
<experimentPattern>AssayX-experiment[0-9]{4}.*</experimentPattern>
<plateInProgressFlag>processing</plateInProgressFlag>
<protocolId>123</protocolId>
</config>

Below is a full list of available settings:

• forceDuplicateCapture: Set to 'true' to force the dcserver to capture experiments, even if they have already been captured previously. Default: false.
• createMissingWellFeatures: Set to 'true' to automatically add new well features to the target protocol if they are found in the data but do not exist in the protocol yet. Default: false.
• createMissingSubWellFeatures: Set to 'true' to automatically add new subwell features to the target protocol if they are found in the data but do not exist in the protocol yet. Default: false.
• protocolId: Specify the ID of the protocol to import new data into. This will also determine which capture configuration to use, as each protocol has a default capture configuration.
• captureConfig: Specify a capture configuration to use to capture any data found by this scan job. Default: none (use the protocol's default, see above).

Settings that are specific to folderScanners:

• path: The network location to scan.
• experimentPattern: The pattern that folder names must match to be considered for import. One capture job will be submitted per experiment folder. Default: .* (matches any name)
• platePattern: The pattern that subfolders of experiment folders must match to be considered for import. If a plate folder name does not match, it will be skipped during the capture job. Default: .* (matches any name)
• plateInProgressFlag: In an experiment folder, if a file or folder is found matching this name, the experiment folder will be considered incomplete, and skipped.

Settings that are specific to columbusScanners:

• host: The hostname of the Columbus server to scan.
• port: The port number of the Columbus server to scan.


text:
## anonymous one year ago Given f(x)=5cos(x) and c=7π/6, determine whether the limit as h approaches 0 of (f(c+h)−f(c))/h exists, and give the value if the limit does exist. a) 152 b) −52 c) 52 d) The limit does not exist. e) 54

what do you mean by $$\sf c=\dfrac{7\pi}{6}$$ ?


text:
# Why is the prospective new kilogram standard a sphere?

I can understand the choice of material, silicon 28, but why is it a sphere rather than (say) a cube? Article here

I would have thought that a sphere would have been the hardest shape to machine accurately.

• shape doesn't matter after all? i mean whether or not the sphere has a perfect shape Mar 23, 2015 at 13:55
• To the person voting to close as "opinion based" - clearly the people who chose this shape had some idea of what they were doing; the OP is asking what their reasoning was. That is not a matter of opinion - the reasoning used is what it is. Mar 23, 2015 at 14:07
• Spheres are by far the most common shape machined with extremely close tolerances: for ball bearings. I can go onto McMaster's website and buy some steel balls with a tolerance of better than 1 part in 10000. Commodity-grade ball bearings generally have tolerances from around 10 to 1 micrometers (hair is about 10x larger). Mar 23, 2015 at 15:59
• I'd expect the corners and edges of a cube to be more susceptible to erosion (evaporation?) than the surface of a sphere. Mar 23, 2015 at 16:58
• @Michael on the subject of rolling it, 1 kg is about the right weight for candlepins. Mar 25, 2015 at 14:22

If you know the diameter of the sphere, you know everything you need to know about the dimensions. It all comes down to one single value.

Any other shape requires multiple dimensions and thus multiple values. Further, measuring a cube or another shape for accuracy is harder than measuring a sphere.

Making very accurate spheres is not as difficult as you might think - it's no different than making optical glass or mirrors using grinding techniques, and, in fact, they are measured much the same way with lasers for very high accuracy.

This video goes into a little more detail as to why they are doing this, how they achieved it, and how the sphere is made.

• Upvoting for the awesome video! Mar 23, 2015 at 16:15
• Not sure about the 'only need 1 measurement' thing. Assuming it is a perfect sphere, you only need to know diameter. Assuming a perfect cube, you only need to know 1 side length. The problem is if the sphere is oblate or the cube is actually rectangular, not square then you need to know more than 1 Mar 23, 2015 at 20:54
• @Scott if you watch the process video, I think you'll find that making it from one value is (relatively) easy with a sphere. The process, by design, produces a perfect sphere (within a degree of error). However, there's no process that produces a perfect cube. You have to grind six faces separately. The error will necessarily be much larger simply due to the fact that each face is cut and found independently of the other faces. Mar 23, 2015 at 22:24
• Then you have to measure it. That's just as complicated as making it, and requires more surface area to measure. Rotating machinery is easier to build with a high tolerance than translating machinery. This all doesn't mean that it's impossible, but spheres are easier on a number of levels compared to other solids. Mar 23, 2015 at 22:29
• Also - they measured from 60,000 directions to confirm it is "flat everywhere". So while you end up specifying the object with a single parameter that doesn't mean a single measurement suffices. Mar 24, 2015 at 1:06

There is a nice article in the New Scientist that describes how these spheres were made.

Sphere can be made very precisely (and their shape measured accurately) simply because of their symmetry - and therefore their volume can be determined most accurately. The video clip in the above article shows this in detail.

Of course a sphere also has the lowest surface area (less contamination) but I don't think that comes into it.

UPDATE

I wonder how much the sphere distorts under its own weight... if you use the Hertz equation (see for example here)

$$F(x) = \frac{2E\sqrt{R}}{3(1-\sigma^2)}\cdot x^{3/2}$$

and we set $F = mg = 10 N$, then for the sphere in question, with $E= 150 GPa$ (sources vary, this is for estimating only), $\sigma=0.3$ and $R = 50 mm$ (again, rounding 96.75/2) we get x = 870 nm. That would be the distance by which the surface is indented when the sphere rests on a flat surface; the contact area would be extremely small at $\pi R d = 0.14 mm ^2$.

There will also be a much smaller distortion of the diameter of the sphere. Using simple dimensional arguments, the stress of the weight will be proportional to the cross sectional area of the sphere, which would make the distortion along the diameter on the order of

$$\frac{\Delta x}{x} \propto \frac{mg}{\pi R^2 E} \approx 10^{-8}$$

If that is correct, it implies a distortion of less than a nm for the sphere. Which is a reassuringly small number. I suspect that, on the scale of this experiment, it does not need to be taken into account (once somebody has done the calculation more accurately...)

• Given that it's a single crystal, the distortion should be very low, but even further, it's supported by a ring and a much larger surface area than a single point support on a flat surface. Mar 23, 2015 at 16:14
• @AdamDavis - the fact that it is a single crystal is captured in the use of a Young's modulus of 150 GPa (I think); and yes, they don't support it on a flat surface so the Hertzian estimate is much larger than the actual distortion. However, the "sagging" under its own weight should not depend on how it is supported and I think that the last part of the calculation (< 1 nm change in diameter) is instructive. I find that it's helpful to put numbers on these things - if only to understand their approximate magnitude. Mar 23, 2015 at 16:17
• I think that measurement to under a nm will allow calculation of the number of atoms in the sphere that will correspond to a mass variation well below the variations seen between the prototype kilogram and its comparison control kilograms. So even a number quite a bit bigger than this would allow for a more workable kilogram definition than we already have. Also note that one could get quite sophisticated and measure the sag versus orientation accurately with the sphere in an interferometric test setup. Mar 25, 2015 at 3:28

I'd like to add a bit more to Adam Davis's concise answer.

Newton's Sphere Production Method

Firstly, note that the lathing process shown in The video that Adam Davis's Answer links to with the sphere rotated and an oscillating abrasive cup sliding over its surface about pseudo-random rotation axes is the standard technique invented by <NAME> that is used today to make precise spherical surfaces. The point about the technique is that even if the abrading cup is not quite spherical, the system settles down to a state where both the cup and machined sphere are spherical, as they grind one another until the pressure between them is uniformly distributed across their contact region for all orientations of the pseudo-randomly oriented cup. This surface is of course spherical. One begins, naturally, with an oversize rough sphere and, through practice, tweaks the process so that the steady state is reached precisely when the desired radius is achieved.

Spherical Surfaces are Really Easy to Measure Extremely Well

Now I'd like to talk further about how amazingly accurately one can measure a sphere's radius and its deviation from ideal sphericalhood. The video briefly shows a simple laser ranging verification that would be done as a rough measurement during the sphere's making. To get really serious, one can do waaaaaay better than this, and so the following is my understanding (gleaned from questions at conferences) of how the CSIRO in Australia envisage verification would be done reasonably cheaply and certainly repeatably for the production of national and regional test standard kilograms.

This is done this through absolute interferometry - a completely self-calibrating measurement technique that not only measures the sphere's actual surface, but calibrates the interferometer at the same time. In principle, you can easily understand how interferometry can be self calibrating (this is NOT how it's done): you are counting numbers of wavelengths so "all you need to do" is to count the number of fringes by adjusting the pathlength from zero in an interferometer trained on one point on the sphere, and then shift to another point and do the same. So you can then describe the sphere's surface in terms of co-ordinates grounded on numbers of wavelengths. Not only would this be horribly time consuming, it would be impracticable mechanically, but the idea of characterizing the surface in terms of wavelength grounded co-ordinates is clear.

Here is where the sphere makes it easy - through the amazing Three-Sphere-Test. The paper by its inventor (<NAME> WARNING- Do not read this man's CV - Severe brain envy hazard)):

describes the technique fully, but the principle is as follows.

I've sketched here a Fizeau interferometer, which is the commonest test setup used to verify spheres. Here we an incident lightfield $I$ which is focussed to a spherical wave by an extremely well designed antireflexion coated focussing surface $AR$, so that it becomes a spherical wave converging on the tested sphere $TS$ through a spherical reference surface $SR$. One adjusts the interferometer until the tested sphere and the reference surface are nearly concentric, and the reflexions $R_R$ from the reference surface and $R_T$ pass back out through the converging surface $AR$ to become interfering, nominally plane waves that make the interferogram on the interferometer's CCD array. However, the "reference" reflexion $R_T$ is only as spherical as the reference surface $SR$ and converging surface $AR$ will make them, and we don't at the outset know exactly what the forms of these surfaces are. Moreover the probe reflexion $R_T$ is also "contaminated" by imperfections in $AR$ and $SR$.

So the three sphere test takes (1) one measurement with spheres as shown, then (2) one rotates either the tested sphere, or the reference lens (wontedly the latter) through 180 degrees to take a second interferogram and (3) lastly, one then focusses the interferometer onto the surface of the tested sphere and takes a third interferogram. This last interferogram is essentially making a near-diffraction limited spot on the surface of the tested sphere, so even rough quality surfaces are excellent over the submicron lateral scales of the spot. Therefore, the third interferogram is influenced by the surfaces $SR$ and $AR$ alone.

It can be fairly straightforwardly shown that one can deduce from these three measurements the shape of the tested sphere's surface and the wavefront distortion imparted by the interferometer, thus not only getting an absolute measurement but also a calibration of the interferometer.

There is an analogous "three flat test". However, it is more time consuming and complicated. The three sphere test captures and wholly characterizes as much of the sphere as the interferometer's numerical aperture will allow (typically about 0.4NA, so we can image a solid angle $\pi\times NA^2$ or about a twentieth of the sphere's surface at a time and then make and link together thirty or so charts of the surface), and the movements between images are easy to automate as this kind of test is best done by a robotic, hands-off procedure.

With instantaneous phase shifting interferometry (where one uses tricks with the light's polarization) one can simultaneously take interferograms with relative phase delays of $\exp\left(\frac{2\,k\,\pi\,i}{N}\right);\,k=0,1,2,\cdots,N-1$ between the reference and probe beams (where $N$ is 3 or 4 in the implementations I have seen), so one can reliably reconstruct the test sphere induced wavefront map to within a tiny fraction of a wavelength. If you use high optical power (to get good SNR notwithstanding quantum noise) and average measurements, well under nanometre accuracy can nowadays be gotten by a test setup costing less than $USD100K. • This is a really really good answer! Mar 28, 2015 at 14:43 • @Floris Thanks, Floris. If you're interested in this stuff, you might enjoy reading a bit QED Technologies qedmrf.com . In particular, check out their "stitching" technology qedmrf.com/metrology/ssi-technology Mar 29, 2015 at 0:12 • @Floris BTW, I shouldn't be surprised, if the kilogram grounded on a number of$Si\$ atoms is adopted, if QED Technologies ends up playing a major role in defining "implementation" standards and technology for different labs to set up their own mass standards. Mar 29, 2015 at 2:39

Another reason that nobody <NAME> mentioned in a comment, is durability. The corner of a cube would be easy to accidentally break because a small force on it would equate to a large pressure (its area gets smaller as the shape of the cube gets more accurate). In fact I think it would be likely to break during machining, resulting in truncated corners (even if only microscopically).

• @Floris: thanks, didn't see that. Edited answer. Mar 29, 2015 at 0:57

A sphere might be harder to machine, but easiest to verify the accuracy of, especially when accounting for slight changes due to temperature.

It should be noted that any standard like this must not only have a mass of 1 kg (or whatever) but also have some secondary method of verifying mass, in this case, being able to count (to some accuracy) the number of silicon 28 atoms within the sphere.

• Spheres are the easiest to machine. Mar 23, 2015 at 16:01
• @NickT, really? I'd need a bandsaw and a right angle to machine a cuboid. Spheres might be the easiest to machine to ultra-high accuracy, but I think I said that. Mar 23, 2015 at 16:09
• @jiminion All I need to make a sphere to ultra-high accuracy is any chunk of homogeneous and isotropic material and an abrasive (not sure if sand will suffice). And (lots of) time. Machine tools are strictly optional. Mar 24, 2015 at 4:04
• @BerndJendrissek More like infinite time. Without precision tooling, you will make mistakes and ruin your work. You'd need tooling just to measure it. (This differs from grinding lenses, which CAN be done accurately without special tooling.) Mar 24, 2015 at 13:22


text:
OpenStudy (anonymous):

Car with mass 940kg is moving at speed 48km/h. How much force do you need to stop it in 15m distance?

5 years ago
OpenStudy (ghazi):

first of all find the retardation $(48)^2=2*a*D$ and $a=\frac{ (48)^2 }{ 2*15 }=76.8m/s^2$ now F=m*a F=940*76.8 N @duck123 you are done

5 years ago
OpenStudy (ash2326):

@duck123 Did you understand?

5 years ago
OpenStudy (anonymous):

I'm trying to figure out where 2*a*d part is coming from

5 years ago
OpenStudy (ghazi):

$V^2=U^2-2a*D$ V is final velocity which is zero and U is = 48 i have taken -a because it is retardation hope that helps

5 years ago
OpenStudy (anonymous):

oh, now i get it. thanks!

5 years ago
OpenStudy (ghazi):

:)

5 years ago