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# Gauges

How the gauge voting system works with Angle

## 🔎 TL;DR

• veANGLE holders are able to decide where weekly ANGLE emissions go through gauge votes
• The fact that veANGLE holders are responsible for such vote aligns incentives within the protocol

## ✍ Gauge Voting Principle

Besides voting on governance proposals, veANGLE owners can vote for ANGLE gauge weights with their veANGLE balance. The gauge system implemented by Angle is similar to that of Frax and of Curve.
Like what was done weekly on Snapshot by ANGLE holders before the Angle Governance upgrade, veANGLE holders are now be able to distribute their voting power across a single or multiple gauges on-chain.
A gauge corresponds to a staking contract. By voting for a gauge, veANGLE holders increase the amount of ANGLE tokens sent to the related staking contract with respect to the others. veANGLE holders therefore vote for the gauges they think should receive more rewards.
This allows veANGLE holders, who are the most long term users of the protocol, to have complete control over the future ANGLE emissions.

## ↔ Incentives alinement

Interestingly, the gauge system lowers the influence of LPs selling off their ANGLE rewards since those LPs do not necessarily have veANGLE to continue voting for the gauge they are providing liquidity to.
This system strongly favors LP who continually lock their rewards into veANGLE to increase their pool's gauge weight. Essentially, and like mentioned in Frax docs, Angle gauges align incentives of veANGLE holders so that the most long term oriented holders control where ANGLE emissions are distributed.

## In Practice

Gauge weights are updated weekly, every Thursday at 2am CET. This means that the ANGLE distribution rate for each pool is constant for 1 week and then updates to the new rate at this time.
Users with veANGLE can allocate their voting power to the available gauges to influence the reward distribution. Then, the sum of all the veANGLE assigned to each gauge by all holders determines the quantity of rewards to be distributed. Once this is done, users don't have to vote again every week except if they want to change them. Votes for a given gauge can only be changed every 10 days, so that each votes apply for at least two weeks.
A person can also decide not to allocate all their available voting power.
When updating gauge weights and putting more weight on one gauge with respect to another, users should be wary of doing decreases first then increases.
When increasing the veANGLE balance by locking more tokens or extending a lock, this new veANGLE balance is not taken into account in the current vote allocation. Be careful to apply this new voting power. More info on the Increasing veANGLE balance page.
Note that gauges can be "killed" after having been whitelisted. This requires a DAO vote. Rewards that are voted for killed gauges are simply not issued, which means veANGLE tokens voting for a gauge that has been removed are not actually providing useful votes.

## ANGLE Issuance Rate

Essentially, the weekly decrease factor of 1.007827 of the ANGLE emissions is guaranteed on-chain.
With such a decreasing factor, distribution with this system is bound to continue and to remain significant for a period of over 10 years.
veANGLE stakers can therefore feel confident staking the maximum duration of 4 years knowing that the gauge program is not temporary and won't be deprecated any time soon.

## Gauge Types

The idea with Angle gauges is to incentivize different types of liquidity pools that can be useful to the protocol:
• agEUR liquidity and markets
• sanTokens
• Liquidity pools on exchanges (like on Curve, on Uniswap)
• Liquidity pools on other chains (like agEUR/USDC on Uniswap on Polygon for instance)
• Perpetuals (if needed)
The gauge system can also be used to incentivize different things than staking contracts. For instance, incentives for Convex voters on Curve can be linked to a gauge (in fact an EOA or a multisig) where veANGLE owners decide to send some rewards, and the multisig would be responsible for placing the incentives on Votium.
Overall, there is no restriction on which pools or pairs can have a gauge other than that they should pass the gauge governance vote. The idea is however that all gauges should have a link with the agEUR or the Angle protocol, as the governance token of the protocol is being distributed as incentive.
Given that a wide range of contracts can be considered Angle protocol gauges, the system distinguishes several gauge types:
• Type 0 Gauges: mainnet staking contracts internal to the Angle Protocol. LPs of these gauges can receive boosted rewards.
• Type 1 Gauges: corresponding to Angle Perpetual staking contracts. Note that there is no boost for veANGLE holders in these contracts.
• Type 2 Gauges: External staking contracts. This gauge type includes all staking contracts not on Angle on Ethereum Mainnet, whether they are on other protocols (Curve or Convex) or on other chains (Polygon or Avalanche).
Available gauges of the protocol and their type can be found here.

### Gauge type weight

Each gauge type has its own weight, which can be set to shift the distribution of rewards in favor of one or another type of gauges. At the moment, the weights are all set equally so it has no impact. Governance can vote to change these gauge type weights.
Example Assuming an inflation rate
$r$
changing with every epoch (1 week), a gauge weight
$w_g$
and a gauge type weight
$w_t$
, then the stream of inflation going to a gauge is:
$r' = w_g \times w_t \times r$
Every week, depending on gauge votes, the weight
$w_g$
can change. Governance can also vote to update the weight
$w_t$
.


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# Is it worth getting a 4k monitor for a late-2012 13" MacBook Pro with Retina display?

I know my 13" 2012 MacBook Pro with Retina display won't support 4k (because it uses the Intel HD4000 for graphics), but I need a new monitor.

I just want a monitor that will look good at 2560x1440 (or 2560x1600?) ideally similar performance to the current thunderbolt display.

Will a 4k display (eg. the PB287Q) perform just as well as a 2560x1440 display (eg. the PB278Q) with my MacBook Pro? If so, I may as well get one so that I'm 4k ready whenever I upgrade the macbook.

Update, Details about my usage: I don't care about gaming, I just want something for programming and internet. I intend to just drive this external screen only and leave the MacBook closed. I occasionally dip into Lightroom for photo editing, so image quality is a bonus. It wouldn't be the end of the world if I had to review images on the laptop screen though.

What's officially "supported" and what's possible don't match. I managed to get 4K out of it at 30Hz.

I took a screenshot as proof: Just a normal mini-displayport<->displayport cable was used.

More details in my answer here: https://apple.stackexchange.com/a/147765/39878

2560x1440 works ok but because it isn't HiDPI it still feels like a step backwards. I flip between running 1080p (HiDPI) and full 4k (with the text being tiny).

Even if apple refers Simultaneously supports full native resolution on the built-in display and up to 2560 by 1600 pixels on up to two external displays, at millions of colors my bet is you'll not have great performance... But I you just do internet browsing that's not an issue. If you're considering Photoshop or other heavy stuffy that might be slow

• and only 30FPS and the Performance it really okey - even with Photoshop not soo smooth but okey ... Autodesk ist more critical because it renders every time. – bMalum Sep 11 '14 at 9:10
• I've just updated my answer with my usage. Basically I only want to drive this external display so hopefully won't be any harder than driving the laptops current retina display...? – matt burns Sep 11 '14 at 9:12
• @bMalum, do you only get 30FPS at 2560x1440? – matt burns Sep 11 '14 at 9:13
• @mattburns yeah - HDMI Port does only support 30FPS at the Moment. MBPr mid 2012 15" – bMalum Sep 11 '14 at 9:16
• @bMalum Can you not use displayport? – matt burns Sep 11 '14 at 9:25

Also consider that there is a 1280x720 resolution, which really outputs 2560x1440. Everything is so big, but it has some benefits, like easier on the eyes. It's a tiny bit blurry but not a problem for me. It's 60 Hz, works much faster than the 3840x2160 (30Hz, slow rendering).

4k displays are night and day better than any non-retina pixel density displays. If you care about text and line drawing apps - there's no comparison and usually no going back once you use a 4k/5k display.

You can test drive this for yourself as well. Connect your retina MacBook pro to any non-4k/5k display and compare text on the retina screen to the external display.

If you don't clearly see the benefit, save your $$until you do see the difference. You can spend / save that$$ on other things until you decide the better resolution is worth losing that .

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Mary Emily Sinclair (September 27, 1878 – June 3, 1955) was an American mathematician whose research concerned algebraic surfaces and the calculus of variations. She was the first woman to earn a doctorate in mathematics at the University of Chicago, and became Clark Professor of Mathematics at Oberlin College. [1] [2]

## Early life and education

Sinclair was born on September 27, 1878, in Worcester, Massachusetts; she was the fourth of five children of John Elbridge Sinclair and Marietta S. Fletcher Sinclair. Her father was a mathematics professor at the Worcester Polytechnic Institute, and also had two daughters from an earlier marriage. Her mother, originally from Worcester, taught English and modern languages at the Worcester Polytechnic Institute but stopped when her children were born. After graduating in 1896 from the Worcester Classical High School, Mary Emily Sinclair became a student at Oberlin, and as a student served as president of the Oberlin branch of the YWCA. She graduated Phi Beta Kappa in 1900 with an A.B. [1] [2]

While working as a teacher at a seminary in Hartford, Connecticut, Sinclair continued her studies through the University of Chicago, earning a master's degree in mathematics in 1903. [1] [2] After briefly teaching at Lake Erie College in 1903, she taught at the University of Nebraska from 1904 through 1907, while continuing her graduate study at the University of Chicago. [1] [2] She completed her Ph.D. in 1908, with the dissertation On a Compound Discontinuous Solution Connected with the Surface of Revolution of Minimum Area in the calculus of variations, supervised by Oskar Bolza. [1] [2] [3] She became the first woman to earn a doctorate in mathematics at the University of Chicago, [1] [4] and her doctorate marked the beginning of a period in which the university gave over 50 doctorates to women through 1946, likely the most of any American university. [5]

## Career and later life

Meanwhile, in 1907, Sinclair had taken a position as instructor at Oberlin College, where she would remain for the rest of her career, teaching there for 37 years. The next year, on receiving her doctorate, she was promoted to associate professor. Although unmarried, she adopted two children as infants in 1914 and 1915. Also in 1915, she became one of the founding members of the Mathematical Association of America. She was promoted to professor in 1925, and became head of the mathematics department at Oberlin in 1939. In 1941 she was named Clark Professor of Mathematics. She retired in 1944, but continued to teach mathematics to US Navy students for two more years through Berea College. [1] [2]

She returned to Oberlin in 1947, but was heavily injured in a carjacking incident in 1950. She moved to Belfast, Maine, in 1953, living there with her daughter-in-law, and died there on June 3, 1955. [1]

## Sinclair's discriminant surface

Sinclair's 1903 master's thesis was Concerning the discriminantal surface for the quintic in the normal form: ${\displaystyle u^{5}+10xu^{3}+5yu+z=0}$. [2] In it, she uses Tschirnhaus transformations to put quintic functions with real coefficients into the form given in the title, and uses the discriminant, a polynomial of the coefficients ${\displaystyle x}$, ${\displaystyle y}$, and ${\displaystyle z}$, to classify polynomials of this form by their numbers of real roots. Physical models of her discriminant surface were made from her design by a German company, [6] and a stone carving based on it was created in 2003 by sculptor Helaman Ferguson. [6] [7]

## References

1. Riddle, Larry (25 February 2016), "Mary Emily Sinclair", Biographies of Women Mathematicians, Agnes Scott College
2. Green, Judy; LaDuke, Jeanne (2008), Pioneering Women in American Mathematics — The Pre-1940 PhD's, History of Mathematics, vol. 34 (1st ed.), American Mathematical Society, The London Mathematical Society, ISBN  978-0-8218-4376-5; biography on p. 559–562 of the Supplementary Material at AMS
3. ^
4. ^ Murray, Margaret A. M. (January 2009), "A Celebration of Women in Mathematics at MIT" (PDF), Notices of the American Mathematical Society, 56 (1): 42–47
5. ^ Mac Lane, Saunders (1989), "Mathematics at the University of Chicago: a brief history" (PDF), in Duren, Peter; Askey, Richard A.; Merzbach, Uta C. (eds.), A Century of Mathematics in America, Part II, History of Mathematics, Providence, Rhode Island: American Mathematical Society, pp. 127–154, doi: 10.1080/10724117.2018.1546504, MR  1003124; see Section "Ph.D.s to Women at Chicago", pp. 144–146
6. ^ a b Top, Jaap; Weitenberg, Erik (2011), "Models of discriminant surfaces", Bulletin of the American Mathematical Society, New Series, 48 (1): 85–90, doi:, MR  2738907
7. ^ Ferguson, Helaman, a parametric form of the quintic discriminant, very remarkable and deserves to be better known, archived from the original on 2006-05-08


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Did you know that you can access more functions and settings for your Overleaf project via a side menu? This help article will show you the available list of functions and settings and you can use them to work with your LaTeX source files on Overleaf. Clicking on the Overleaf menu icon above the file tree panel will bring up more functions and configurable settings for your project. There are three main groups in this Overleaf menu: The Project functions group of items enables you to download or copy your project, sync your project with Dropbox, Git or GitHub, and view the compiled project word count. The Project settings group of items lets you set various settings relating to the editor and how your project is compiled. The Help group of items shows you ways to get help regarding Overleaf and your project. Contents 1 Project Functions 1.1 Download 1.1.1 Download - Source 1.1.2 Download - PDF 1.2 Actions 1.2.1 Copy Project 1.2.2 Word count 1.3 Sync 1.3.1 Dropbox 1.3.2 Git 1.3.3 GitHub 2 Project Settings 2.1 Compiler 2.2 TeX Live version 2.3 Main document 2.4 Spell-check 2.5 Auto-complete 2.6 Auto-close brackets 2.7 Code Check 2.8 Editor theme 2.9 Overall theme 2.10 Keybindings 2.11 Font Size 2.12 Font Family 2.13 Line Height 2.14 PDF Viewer 3 Help 3.1 Show Hotkeys 3.2 Documentation 3.3 Contact Us Project Functions The functions here enable you to download or copy your project, sync your project with Dropbox, Git or GitHub, and view the compiled project word count. Download Download - Source Downloads your project as a .zip file of all the LaTeX source files, as listed in your Overleaf project's file tree panel. Note that when using the "Download - Source" button, this would always use the files in your file list panel i.e. all the input files. 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See this overview for a quick discussion of what each LaTeX compiler is capable of. You can also run plain TeX, XeTeX or LuaTeX on Overleaf. If you are looking to compile with uplatex or platex, please see these resources: Typesetting Japanese with pTeX Does Overleaf support pTeX? Overleaf v2 で日本語を使用する方法 TeX Live version This is a project-specific setting, that lets you set the TeX Live version to compile your project with. The updated packages that come with each new TeX Live may contain breaking changes. See our blog posts on our past TeX Live upgrades where we discuss some of these changes, and how to update your code so that they are compatible with more up-to-date packages: TeX Live 2019 TeX Live 2018 TeX Live 2016 Main document Sets the .tex file to be compiled as the main document of the project, each time when you visit your project. The "Word count" feature will always operate on this designated file. Note 1: Any file that contains a \documentclass command, even if it's not designated the "main document", will be compiled if it's opened in the editor when you press the Recompile button. Note 2: The main file must be an editable plain text file. Therefore if a .tex file is larger than 2MB, it's categorised as non-editable on Overleaf, and you won't be able to set such a file as your project's main document. You can break it up into smaller .tex files, and use \input or \include on them. Note 3: Word count, go to location in source/PDF, makeglossaries, makeindex, minted etc will only work robustly if the main document is on the project's root level; that is, it is not within any folder. Spell-check This is a user-specific setting. You can change the spell-check language dictionary (as provided by Aspell) to use, or turn off spell-checking. Auto-complete This is a user-specific setting. Enables/disables the auto-completion suggestion list when you type a LaTeX command. If you're interested in how this works, see our blog post on how we used a data-driven approach to order the suggestions. Auto-close brackets This is a user-specific setting. Enables or disables the automatic addition of a corresponding }, ], $when you type a {, [,$. Code Check This is a user-specific setting. Enables/disables the syntax-checker to check your LaTeX code and catch common syntax errors, even before the LaTeX compiler runs. Note that the Code Check utility is not a full LaTeX parser, and may not support certain syntax. In such cases, you can choose to disable Code Check in your project completely using this menu setting, or to disable Code Check for only one particular file, or for only some particular lines of code. Editor theme This is a user-specific setting: you can choose from one of the available editor themes and syntax-highlighting for the Source code view. This setting does not affect the RichText view. 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# Commutator subgroup of a surface group

Let $\Sigma_{g,n}$ denote a compact orientable genus $g$ surface with $n$ boundary components. Assume that $g \geq 1$ and fix a basepoint $p \in \Sigma_{g,n}$. Define $S \subset [\pi_1(\Sigma_{g,n},p),\pi_1(\Sigma_{g,n},p)]$ to be the set of all elements of $\pi_1(\Sigma_{g,n},p)$ that can be realized by simple closed $p$-based curves $\gamma$ that separate $\Sigma_{g,n}$ into two components, one of which is homeomorphic to a one-holed torus with boundary component $\gamma$. It is not hard to show that if $p$ is in the interior of $\Sigma_{g,n}$, then $S$ generates $[\pi_1(\Sigma_{g,n},p),\pi_1(\Sigma_{g,n},p)]$. For instance, this is Lemma A.1 in my paper "Cutting and Pasting in the Torelli Group", though I'm sure that many other people have observed this over the years.

Assume now that $p$ lies on one of the boundary components of $\Sigma_{g,n}$. Question : Does $S$ still generate $[\pi_1(\Sigma_{g,n},p),\pi_1(\Sigma_{g,n},p)]$?

The thing that makes this difficult is the following. If $\gamma \in \pi_1(\Sigma_{g,n},p)$ can be realized by a simple closed curve and $p$ is in the interior of the surface, then all conjugates of $\gamma$ can also be realized by simple closed curves. However, this is false if $p$ is on the boundary.

I suspect that the answer is no, so here is a more general question. Continue to assume that $p$ lies on one of the boundary components of $\Sigma_{g,n}$. The group $\text{Diff}(\Sigma_{g,n},p)$ acts on $\pi_1(\Sigma_{g,n},p)$, and $S$ consists of a single orbit under this action. Question : does there exist a finite subset $T$ of $[\pi_1(\Sigma_{g,n},p),\pi_1(\Sigma_{g,n},p)]$ such that the $\text{Diff}(\Sigma_{g,n},p)$-orbit of $T$ generates $[\pi_1(\Sigma_{g,n},p),\pi_1(\Sigma_{g,n},p)]$?

Funny things might happen in the low-genus cases, so I should mention that I'm most interested in the case where $n=1$ and $g$ is large.

EDIT: I just got an up vote on this ancient unanswered question, so I thought I'd point out to anyone who is intrigued by it that my motivation for it was to find small generating sets for the Torelli group. I still don't know the answer to the question, but I was able to find a hack to avoid needing one. See my paper

Putman, Andrew, Small generating sets for the Torelli group. Geom. Topol. 16 (2012), no. 1, 111–125.

The weird arguments in Section 4 are designed to circumvent the issue identified by this question. I could probably improve the bounds in the above paper if someone gave me a reasonable answer to this question, so please let me know if you find one!

-
I'm confused by the notation. Can you clarify what you mean by $[\pi_1(\Sigma_{g,n},p), \pi_1(\Sigma_{g,n},p)]$? – Bill Thurston Sep 11 '10 at 19:09
I just mean the commutator subgroup of the surface group, ie the subgroup of $\pi_1(\Sigma_{g,n},p)$ generated by all commutators $[x,y]=x^{-1} y^{-1} x y$ for $x,y \in \pi_1(\Sigma_{g,n},p)$. – Andy Putman Sep 11 '10 at 19:25


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Fermilab Joint Experimental-Theoretical Seminar

• Date: July 2, 2004

• Speaker: Paolo Giromini

• Title: Evidence for pair production of jets with an anomalous lepton content at the Tevatron

• Abstract:
Motivated by the W+superjet anomaly, CDF has performed a detailed study of the heavy flavor properties of jet pairs produced at the Tevatron. The study provides new information on the origin of previously observed discrepancies between measured $b\bar{b}$ cross sections and the NLO QCD prediction.

Send comments to Marcela Carena and Bogdan Dobrescu.


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# Book: Calculus-Based Physics (Schnick)

Calculus-Based Physics is an introductory physics textbook designed for use in the two-semester introductory physics course typically taken by science and engineering students.

Thumbnail: Animation of Newton's cradle. (CC BY-SA 3.0 Unported; Dominique Toussaint via Wikipedia).


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# Part 1 – Reaction rate constant with transition state theory (TST): partition functions

As part of the work carried out to understand origins of life, it is crucial to determine which processes/reactions were most likely to occur. The thermal rate constant of a reaction defines the efficiency of the process at a given temperature. To get an exact rate, one needs an accurate potential energy surface and quantum time-dependent dynamics of reactants on the potential energy surface. This is computationally very demanding.

One simple theoretical approach which avoids having to propagate in time is classical Transition State Theory (TST). For more details on TST see the original papers by Wigner, Eyring, Evans and Polanyi:

• Pelzer and Wigner, Z. Phys. Chem. B 15, 445 (1932)
• Eyring, J. Chem. Phys. 3, 105 (1935)
• Evans and Polyani, Trans. Faraday Soc. 31, 857 (1935)

The main idea of TST is that the reaction dynamics can be described by a one dimensional minimum energy path on the potential energy surface which connects reactants to products through a transition state. The transition state is found at the maximum energy along the one dimensional path and it is a first order saddle point in the multidimensional potential energy surface. The imaginary frequency of the transition state is not included in the computation of the transition state partition function as it corresponds to the reaction coordinate degree of freedom.

The approximations required to derive TST are

• the Born-Oppenheimer approximation is valid
• reactants are at equilibrium and in the Boltzmann distribution
• no recrossing: molecules which have crossed the transition state may not return no recrossing of transition state and classical dynamics of the nuclei
• at the transition state, the motion along the one dimensional reaction coordinate can be separated from other motion and treated classically as translational motion

The expression for the TST rate constant is then

$\sf k(T)=$ $\sf \frac{k_BT}{h}\frac{Q^{\ddagger}(T)}{Q_R(T)}e^{-\beta E_a}$

Here we see the ratio of the canonical partition function for the transition state, $\sf Q^{\ddagger}$, and the canonical partition function for the reactants, $\sf Q_R$, as well as the exponential of the reaction energy barrier $\sf E_a$. As long as the ratio of partition functions stays constant as a function of temperature, the rate constant will decay linearly with temperature in log scale.

### Example

Let’s see how it works for a simple reaction

CH4 + H → [CH5+]→H2+CH3

Here for simplicity the transition state is indicated in the square brackets. We’ll begin by some general expressions and then implement these for the above reaction.

• Compute partition functions of reactants and transition state

The most accurate way to compute the partition function of the molecules in their ground state would be to diagonalize the reactant and transition state Hamiltonians and sum over all eigenstates, i.e

$\sf Q(T)=\sum_i g_i e^{-\beta E_i}$

With $\sf \beta=1/(k_BT)$ and $\sf g_i$ is the degeneracy of the i-th level. This expression is difficult to evaluate as one needs a full potential energy surface. Diagonalizing the Hamiltonian is expensive and gets worse as the number of degrees of freedom increases. Therefore, further approximations are made.

Assuming we have one of each of these molecules, the overall canonical partition function can be expressed as the product of the partition function for electronic, vibrational, translational and rotational degrees of freedom:

$\sf Q(T)=Q_{\textsf{el}}(T)\cdot Q_{\textsf{vib}}(T)\cdot Q_{\textsf{tras}}(T)\cdot Q_{\textsf{rot}}(T)$

For N identical molecules this expression needs to be elevated to the power of N and rescaled by N!.

Each partition function can then expressed as

$\sf Q_\textsf{tras}(T)=$ $\sf \left(\frac{2\pi \textit{\textsf {m}} \sf k_B T}{h^2}\right)^{3/2}\cdot V%s=2$

$\sf Q_\textsf{vib}(T)=$ $\sf \prod_i^{n_{\textsf{vib}}}\frac{e^{\sf -\beta\hslash\omega_i/2}}{1-e^{\sf -\beta\hslash\omega_i}}%s=2$

$\sf Q_\textsf{rot}(T)=$ $\sf \frac{\pi^{1/2}}{\sigma_r}\left(\frac{T^3}{\Theta_A\cdot\Theta_B\cdot\Theta_C}\right)^{1/2} %s=2$

with m the mass and V the volume, $\sf n_{\textsf{vib}}$=3N-6 for non linear molecules and 3N-5 for linear molecules. The rotational constants are obtained from the three moments of inertia $\sf I_j$  as $\sf \Theta_j=\frac{h^2}{8\pi^2 I_j k_b}$. The paramerter $\sf \sigma_r$ is the rotational constant which can be obtained from the point group of the molecule. In general if your molecule is not at equilibrium it may be complicated to estimate this value, for more details see e.g. M. K. Gilson and K. K. Irikura, J. Phys. Chem. B, 114:16304 (2010). For a linear molecule the rotational partition function is simply $\sf Q_{\textsf{rot}}=\frac{T}{\sigma_r\Theta_r}$.

The electronic partition function is approximated to be $\sf Q_{\textsf{el}}=g_{el}^{\textsf{ground}}$ i.e the ground state electronic degeneracy.

Note that in the above expressions for the partition functions we have assumed that

• the harmonic approximation is valid
• there is no roto-vibrational coupling
• the molecules can be seen as linear or non linear rigid rotor
• the main contribution to the electronic partition function comes from the ground electronic state

For this reaction at T=300K, we find

$\sf{ Q_{CH_4}=7.93e+12\text{; }Q_{H}=9.77e+29 \textsf{ and } Q_{CH_5^+}=1.27e+15}$

and we have used

$\sf{\sigma_r(CH_4)=12}$$\sf{\sigma_r(CH_5^+)=2}$,

$\sf{\omega(CH_4)=}$ $\sf{\left(\textsf{1341 1341 1341 1569 1569 3031 3151 3151 3151}\right) cm^{-1}}$

and

$\sf{\omega(CH_5^+)=}$ $\sf{\left(\textsf{534 534 1074 1125 1125 1442 1442 1795 3076 3223 3223}\right) cm^{-1}}$

excluding the imaginary frequency at the saddle point and

$\sf{I(CH_4)}=$ $\sf{ \left(\textsf{3.173 3.173 3.173}\right) au\cdot\AA^2}$

and

$\sf{I(CH_5^+)}=$ $\sf{ \left(\textsf{4.401 4.436 4.460}\right) au\cdot\AA^2}$

Some of these values (e.g. frequencies) were taken from J. Chem. Phys. 124:164307 (2006). For the electronic partition function we used the spin multiplicity of 2 for hydrogen, 1 for methane and 2 for the transition state.

Next post we’ll put this all in and get the rate constant as a function of temperature.


text:
# Where should I upload resources linked to my question/answer?

I've seen users posts and refer content on "external" sites.

Where should I upload mine, and where should I not upload mine, and why?

Note that this question serves more of a reminder than anything else.

## 1 Answer

Think in terms of future visitor experience: favour content in the post rather than an external link, and favour stable external sites rather than rot prone ones.

• Images should be uploaded to StackExchange's imgur server, using the Image <img> Ctrl+G icon in the edit bar of every Question and Answer post. They'll end up with an URL starting with https://i.stack.imgur.com.

• Code should be posted right in the question/answer, and formatted using the Preformated text <pre> <code> Ctrl-K icon in the edit bar of every Question and Answer post. There are many reasons why code should not be posted as an image / screenshot.

• Videos should be converted as much as possible to animated gifs and uploaded to StackExchange's imgur server, as described before. If not, YouTube links tend to not rot that much.

• Whole projects should not be needed. If you think you need to upload a project, you probably should re-scope your question or your answer. If you must upload it somewhere, favour "open source" repositories such as github¤. And when linking to the project, browse the history of the project and link to a specific commit instead of linking to the 'head' of the master branch. This way the question and the project will make sense for future visitors (you'll most likely keep on working on your project, which will render the 'head' irrelevant to the StackExchange post).

This being said, it's important to not rely on less stable external resources. There has been stories in the past that made some questions and answers completely irrelevant because they relied on resources that were no longer there, and no longer available.

So here is a (incomplete) list of sites that should be avoided mainly because of a high risk of link rot:

Images:

Files/projects:

Code:

Other considerations:

• codepen.io Although the code will not technically rot ¤, StackExchange posts allow users to put JavaScript, HTML, CSS and images in posts using the JavaScript/CSS/HTML snippet Ctrl-M tool bar icon, so this should be favoured (plus that the history of the code will be visible in StackExchange posts, whereas the code on codepen can change ¤¤).

¤ I have not yet encountered an occurrence of a rot there.
¤¤ From what I understand of the service. Has to be confirmed.

• What's the protocol when the link is already dead? I.E. the imageshack one doesn't seem to be in the Wayback Machine or the Google cache for the page. I searched SE Meta too, but didn't find an authoritative solution. – Pikalek Aug 22 '18 at 3:01
• I usually remove the link from the post and adapt it however I can to make it readable. If the post still makes sense, I leave a note to the author suggesting them to re-upload, and if I can't figure out what the post is about after the removal, I vote to close as unclear (I also leave a note). – Alexandre Vaillancourt Aug 22 '18 at 12:51
• Thanks for the guidelines. Should we ✅ the entries when they clear? – Pikalek Aug 22 '18 at 14:29
• I don't think it's necessary here, as we'll have to keep on checking for these once in a while, (a bit like the "blacklisted tags list"). – Alexandre Vaillancourt Aug 22 '18 at 17:10
• If we find old http URLs for i.stack.imgur.com, should be we edit them them to be https instead? – Pikalek Aug 27 '18 at 20:46
• @Pikalek I saw that I found it odd that the bot did not fix them. I might try and have a word with someone else about that. I would not focus on these, but if you encounter them when working on something else, sure, fix them! – Alexandre Vaillancourt Aug 27 '18 at 23:11
• Yeah, I did a quick search & saw mention of the bot, but didn't know if it was an 'already done' thing or a WIP. I found them in here when editing to remove quads. They end up at an https page, but the source is http. – Pikalek Aug 28 '18 at 0:47
• I'll ask around tomorrow and see what's going on with that :) – Alexandre Vaillancourt Aug 28 '18 at 0:51


text:
# Control colors and shading of navigation circles in Beamer top line

I am preparing a presentation in Beamer using \useoutertheme{miniframes} and \useinnertheme{circles}, and my own modded version of the default colortheme.

I want to have a Frankfurt-like navigation bar in the top, but instead of setting a background color, I have an image providing the background for the top bar.

What I want is the current section title and subsection dots to be highlighted compared to the rest, like this:

...except this only works when I declare a background colour for "section in head/foot". When I set {bg=} in that definition, it shades the active section and circles darker than the rest, which is the opposite of what I want:

If I set the fg to be completely white, then everything gets white, no shading of inactive regions. What I want is the shading on the first image with the background of the second image.

How can I achieve this?

## Edit: Minimal Working Example

Here's a minimal working example. To function, it needs a background image, I used this background image from The Internetz(TM).

The code for a minimal working example is below. There are 3 different options for \setbeamercolor{section in head/foot}; option 1 produces correct background with all-white navigation with no shading, option 2 provides correct background but wrong shading (the emphasized parts are darker on a dark background image), and option 3 provides correct shading but sets a solid background color covering the image.

What I want is the background image of option 1 and 2, but the shading of the navigation items of option3.

\documentclass[compress]{beamer}
\useoutertheme[footline=authortitle]{miniframes}
\usebackgroundtemplate{\includegraphics[height=\paperheight]{baggrund.jpg}}
\setbeamercolor{structure}{fg=white}
%\setbeamercolor{section in head/foot}{parent=structure}%, bg=black} %opt.1
\setbeamercolor{section in head/foot}{parent=structure,fg=white!70!black} %opt.2
%\setbeamercolor{section in head/foot}{parent=structure, bg=black} %opt.3
\setbeamercolor{normal text}{fg=white!80!blue}
\title{the title}
\author{John Doe}
\begin{document}

\section{S1}

\subsection{SS11}
\begin{frame}{Foo}
Some text
\end{frame}
\begin{frame}{Bar}
Some text
\end{frame}

\subsection{SS12}
\begin{frame}{Baz}
Some text
\end{frame}
\begin{frame}{Qux}
Some text
\end{frame}

\section{S2}
\subsection{SS21}
\begin{frame}{Buqz}
Some text
\end{frame}

\subsection{SS22}
\begin{frame}{Bao}
Some text
\end{frame}
\end{document}

• Please help us to help you and add a minimal working example (MWE) that illustrates your problem. It will be much easier for us to reproduce your situation and find out what the issue is when we see compilable code, starting with \documentclass{...} and ending with \end{document}.
– user31729
Feb 15, 2015 at 14:15
• Added an MWE now. Feb 15, 2015 at 14:51

To influence the colouring of the mini frames without changing the background colour of the header, you can use the beamer colour mini frames (by default, this just takes the colours from section in head/foot, which is why changing this colour also influences the mini frames):

\setbeamercolor{mini frame}{fg=white,bg=black}

However, this only solves half of your problem: while it gives correct colours for the mini frames itself, the section name colour is still wrong, as it is controlled by section name in head/foot:

To fix this, you can patch the internal beamer commands to use the mini frame colour instead of the header colour for the section name:

\usepackage{etoolbox}
\patchcmd{\sectionentry}{\usebeamercolor[fg]{section in head/foot}}{\usebeamercolor[fg]{mini frame}}{}{}

Together, this gives the desired result:

\documentclass[compress]{beamer}
\useoutertheme[footline=authortitle]{miniframes}
\usebackgroundtemplate{\includegraphics[height=\paperheight]{baggrund.jpg}}
\setbeamercolor{structure}{fg=white}
\setbeamercolor{normal text}{fg=white!80!blue}

\setbeamercolor{mini frame}{fg=white,bg=black}
\usepackage{etoolbox}
\patchcmd{\sectionentry}{\usebeamercolor[fg]{section in head/foot}}{\usebeamercolor[fg]{mini frame}}{}{}

\title{the title}
\author{John Doe}
\begin{document}

\section{S1}

\subsection{SS11}
\begin{frame}{Foo}
Some text
\end{frame}
\begin{frame}{Bar}
Some text
\end{frame}

\subsection{SS12}
\begin{frame}{Baz}
Some text
\end{frame}
\begin{frame}{Qux}
Some text
\end{frame}

\section{S2}
\subsection{SS21}
\begin{frame}{Buqz}
Some text
\end{frame}

\subsection{SS22}
\begin{frame}{Bao}
Some text
\end{frame}
\end{document}

• +1 Now that I see your answer, I think I misread the OP's intent and my answer doesn't really solve the problem (yours does). I'll delete mine. Feb 16, 2015 at 4:25


text:
# How to format code snippets with a custom environment or command?

I'd like to make an environment or command to wrap the this 2-column REPL formatting. For example, as a command (\format{TEXT}{CODE}) it would look like this:

\format2cols{
Description of the code
}{
The actual code
}

Having a shortcut like this would clean up my document exponentially!

Here's what it looks like, with the TEXT is on the left and the CODE on the right:

The difficulty here is that LaTex doesn't allow for passing verbatim arguments. So I'm at an impasse.

MWE:

\documentclass{book}
\usepackage{a4wide}
\usepackage{tabularx}
\usepackage[]{sourcecodepro}
\usepackage[T1]{fontenc}
\usepackage{ragged2e}

\newlength\textlen
\newlength\codelen
\newlength\halflen
\setlength\halflen{\dimexpr.5\columnwidth-2\tabcolsep-0.5\arrayrulewidth\relax}
\setlength\textlen{.95\halflen}
\setlength\codelen{1.05\halflen}

\usepackage{color}
\definecolor{mygray}{rgb}{0.9,0.9,0.9}

\usepackage{minted}
\setminted{
baselinestretch=1,
tabsize=4,
fontsize=\fontsize{8}{11}
}

\begin{document}
\fontsize{12}{14}\selectfont

\justify
We restrict our sorting functions to permutations of the numbers $1\cdots n$. A simple way to visualize the amount of sortedness'' in such a list of numbers is a picture obtained with {\fontsize{10}{12}\selectfont\texttt{ListPlot}}.
\vspace{1.5mm}

{
\fontfamily{ptm}\selectfont
\fontsize{9}{11}\selectfont
\def\arraystretch{2}
\noindent\begin{tabular}{@{}p{\textlen}p{\codelen}@{}}
This definition is used to plot permutations of the numbers $1\cdots n$ with suitable settings of graphics options. &
\begin{minipage}[t]{0.5\textwidth}
\begin{minted}{Mathematica}
In[1]:= PermutationPlot[l_List, opts___] :=
ListPlot[ l,
PlotRange -> { {0.5, Length[l]+0.5},
{0.5, Length[l]+0.5} },
PlotStyle -> PointSize[0.75/Length[l]],
opts, Axes->None, FrameTicks->None,
Frame->True, AspectRatio->1 ]
\end{minted}
\end{minipage}
\end{tabular}
}
\vspace{1.5mm}

\justify
Sed efficitur diam eget ante elementum pharetra. Nullam dignissim pulvinar molestie. Sed dapibus tristique feugiat.

\end{document}

– user121799
Dec 12, 2017 at 4:16
• @marmot it's in the link, but I'll paste it in
– M.R.
Dec 12, 2017 at 4:19
• Verbatim is alway tricky. I would recommend an environment instead of a macro. Dec 12, 2017 at 6:57
• On another note: using \fontsize and similar macros is not recommended. If you want 12pt as default font size use it as a class option ([12pt]{book}) and then use macros like \small etc. Dec 12, 2017 at 6:59
• @MartinScharrer Thanks I've updated the question to reflect that
– M.R.
Dec 12, 2017 at 8:42

Why don't use tcolorbox instead?

You could create your own tcolorbox (mycodebox, in my MWE) and use it as any other environment, changing only the comment and the code.

\documentclass{book}
\usepackage{a4wide}
\usepackage{tabularx}
\usepackage[]{sourcecodepro}
\usepackage[T1]{fontenc}
\usepackage{ragged2e}

\newlength\textlen
\newlength\codelen
\newlength\halflen
\setlength\halflen{\dimexpr.5\columnwidth-2\tabcolsep-0.5\arrayrulewidth\relax}
\setlength\textlen{.95\halflen}
\setlength\codelen{1.05\halflen}

\usepackage[most]{tcolorbox}
\tcbuselibrary{minted}
\definecolor{mygray}{rgb}{0.9,0.9,0.9}

\setminted{
baselinestretch=1,
tabsize=4,
fontsize=\fontsize{8}{11}
}
\newtcblisting{mycodebox}[2][]{%
comment side listing,
sidebyside align=top,
halign=justify,
colframe=white,
colback=white,
boxrule=0pt,
left=0pt,
right=0pt,
leftrule=0pt,
rightrule=0pt,
bottomrule=0pt,
boxsep=0pt,
after=\vspace{-16pt},
lefthand width=.4\textwidth,
fontupper=\fontfamily{ptm}\selectfont\fontsize{9}{11}\selectfont,
minted language=Mathematica,
minted options={baselinestretch=1,
tabsize=4,
fontsize=\fontsize{8}{11}},
comment={#2},#1}

\begin{document}

\fontsize{12}{14}\selectfont

\justify
We restrict our sorting functions to permutations of the numbers $1\cdots n$. A simple way to visualize the amount of sortedness'' in such a list of numbers is a picture obtained with {\fontsize{10}{12}\selectfont\texttt{ListPlot}}.
\vspace{1.5mm}

\begin{mycodebox}{This definition is used to plot permutations of the numbers $1\cdots n$ with suitable settings of graphics options.}
In[1]:= PermutationPlot[l_List, opts___] :=
ListPlot[ l,
PlotRange -> { {0.5, Length[l]+0.5},
{0.5, Length[l]+0.5} },
PlotStyle -> PointSize[0.75/Length[l]],
opts, Axes->None, FrameTicks->None,
Frame->True, AspectRatio->1 ]
\end{mycodebox}
\begin{mycodebox}{Here is a random permutation.}
In[2]:= p1 = PermutationPlot[
{9,10,6,8,2,4,12,1,7,5,3,11},
DisplayFunction->Identity ];
\end{mycodebox}
\justify
Sed efficitur diam eget ante elementum pharetra. Nullam dignissim pulvinar molestie. Sed dapibus tristique feugiat.
\end{document}

• I'm using Texpad on OSX, and it's giving me some errors regarding pigments, I have pygmentize on my path, is there something else I need to do to get it to work?
– M.R.
Dec 12, 2017 at 22:16
• Thanks for the links, it seems that \usepackage[outputdir=.texpadtmp]{minted} would work, but your code does a \tcbuselibrary instead?
• I figured it out, I had to add the line \write18{} to the source file. Bizarre fix, but now it works! Thanks!


text: An approximate calculation can be made to the estimation of temperature at the time of decoupling. What we do know is the redshift of the CMB (by comparing the observed black body temperature to the one we can calculate from theory). 3.2 Dependence of the CMB temperature … It will map all the dark matter in the universe down to scales smaller than galaxies using the gravitational bending of Cosmic Microwave Background light. (� �%9Lf]9�6v�9X��klȝj�>�y����#b>C�)e.���w��a������֊UY�#x�j�����n�V K剳������"X���� ΩM Ω ≡ ν fν. •CMB temperature today: 2.725 K (= 2*10-4 eV) •Photon decoupling: 3000 K (=0.25 eV) •Neutrino decoupling: 1010 K (=1 MeV) •QCD phase transition: 1012 K (=150 MeV) •EW phase transition: 1015 K (= 100 GeV) •Reheating: As large as 1015 GeV •Constraints on N eff probe physics all the way up to … When was the cosmic background radiation in the visible spectrum? The temperature to ionize this is significantly lesser. Here, this paper presents cosmological results based on full-mission Planck observations of temperature and polarization anisotropies of the cosmic microwave background (CMB) radiation. Thus, at decoupling and recombination epochs, the energy had to drop to permit the ionization of hydrogen. This paper presents the first cosmological results based on Planck measurements of the cosmic microwave background (CMB) temperature and lensing-potential power spectra. The cosmic microwave background is the afterglow radiation left over from the hot Big Bang. $$T(z) = T_0\frac{\lambda_0}{\lambda_e} = T_0(1+z)$$. Fluctuations in the CMB temperature are of the order of ∆T/T ≈ 7 × 10−5. As the theory goes, … A blackbody spectrum with a temperature any hotter than this has sufficient photons with energy above 13.6eV to ionise any hydrogen atoms that form. What is the temperature of the Planck distribution with this average photon energy? �Ε��-a%������ā����x���R^J. Measurements of the temperature of the CMB are reviewed. $$B_vdv = \frac{2hv^3}{c^2} \frac{dv}{e^{hv/kT}-1}$$. Moreover, recombination of electron and proton does not guarantee a ground state hydrogen atom. The Boltzmann factor ##e^{-E/(kT)}## for this is 10-17 and 10-23 for 3000 K, respectively. Hipo´lito–Ricaldi∗ and G.I. If $n_{νo}$ is for present and $n_{νe}$ for emitted, we get −, $$n_{v_0} =\frac{2v_c^2}{c^2}\frac{dv_c}{e^{hv/kT}-1}\frac{1}{(1+z)^3}=\frac{2v_0^2}{c^2}\frac{dv_c}{e^{hv/kT}-1}$$, This gives us the Wien’s Law again and thus it can be concluded that −, Velocity Dispersion Measurements of Galaxies, Horizon Length at the Surface of Last Scattering. However, tiny temperature variations or fluctuations (at the part per million level) can offer great insight into the origin, evolution, and content of … Raw CMBR data, even from space vehicles such as WMAP or Planck, contain foreground effects that completely obscure the fine-scale structure of the cosmic microwave background. ... and E I = 13.6 eV is the ionization energy of hydrogen. Besides the cosmic microwave background (CMB), the prediction of the cosmic neu-trino background (C B) is the second, unequivocal key signature of a hot Big Bang. Current measurements reveal the universe’s temperature to be close to 3K. Hence a disciplined statistical analysis should be performed case by case to obtain an accurate value. We know that the ratio of photons to baryons is about 5 × 1010. Tags: None. Thus, we obtain a better estimate than 1.5 × 105 K that is closer to the accepted value of 3000 K. To understand the relationship between redshift and temperature, we employ the following two methods as described below. In particular, the CMB temperature anisotropy has been one of the most important benchmarks to test the existence of primordial magnetic fields. The anisotropy of the cosmic microwave background (CMB) consists of the small temperature fluctuations in the blackbody radiation left over from the Big Bang. %PDF-1.4 %�������������������������������� 1 0 obj << /FontFile3 176 0 R /CharSet (/space/F/i/g/u/r/e/one/period/two/three/T/a/b/l/N/o/t) /CapHeight 687 /Ascent 687 /Flags 262178 /ItalicAngle 0 /Descent -209 /XHeight 468 /FontName /FHKLPO+Times-Bold /FontBBox [ -168 -218 1000 935 ] /Type /FontDescriptor /StemV 139 >> endobj 2 0 obj << /Prev 89 0 R /Dest (section0.5.0) /Title (5. $\begingroup$ @DheerajBhaskar The temperature at recombination is approximately 3000K = 0.26 eV. The anisotropies of the cosmic microwave background, or CMB, as observed by ESA's Planck mission. ��*� First, consider only the ionization of ground state hydrogen. Setting To as the current value 3K, we can get temperature values for a given redshift. The CMB-HD project is a proposed millimeter-wave study of over half the sky to discover more about the universe. Robert Fogt. For ionization of the ground state hydrogen, hν is 13.6 eV and kB is the Boltzmann Constant 8.61 × 10−5 eV/K that reveals the temperature to be 1.5 × 105 kelvin. Related. The general expression for the ratio of the number of photons with energy more than ΔE, Nγ (> ΔE) to the total number of photons Nγ is given by −, $$\frac{N_\gamma(> \Delta E)}{N_\gamma} \propto e^{\frac{-\Delta E}{kT}}$$. Measurements of the temperature of the CMB are reviewed. ���DKv��D��w*.�a繷��UV��,ˡ�v�c�%��S�R���nc-i����ԕO[�Z|kE����N�w��B�eĔ,Җ� Cosmic microwave background (CMB) ... black-body radiation emitted when the universe was at a temperature of some 3000 K, redshifted by a factor of 1100 from the visible spectrum to the microwave spectrum). For ionization of the ground state hydrogen, hν is 13.6 eV and kB is the Boltzmann Constant 8.61 × 10 −5 eV/K that reveals the temperature to be 1.5 × 105 kelvin. composition to show that CMB temperature maps of (not to o larg e) m ultiply connected universes must show “patterns of alignment”, and prop ose a metho d to look for these patterns, thus op ening This essentially tells us that if the temperature is below 1.5 × 105 K, the neutral atoms can begin to form. The cosmic microwave background (CMB) is thought to be leftover radiation from the Big Bang, or the time when the universe began. Administrator . By considering the present epoch, , , and by solving numerically the integral in , one has the contribution to the vacuum energy given by GeV 4 for masses less than or equal to the CMB temperature ; that is, eV (e.g., possible candidates are axion-like with eV). We know that energies were much higher to such an extent that matter existed only in the form of Ionized Particles. Eﬀects of Regional Temperature on Electric Vehicle Eﬃciency, Range, and Emissions in the United States Tugce Yuksel§ and Jeremy J. Michalek*,§,‡ §Department of Mechanical Engineering, Carnegie Mellon University, Pittsburgh, Pennsylvania 15213, United States ‡Department of Engineering and Public Policy, Carnegie Mellon University, Pittsburgh, Pennsylvania 15213, United States Topological signatures inCMB temperature anisotropy maps W.S. Any help would be appreciated, thanks! For a perfect blackbody. Computations set the temperature to be around 3000K. 20. The determination from the measurements from the literature is CMB temperature of 2 . Fig. Its temperature is extremely uniform all over the sky. 72548 ±0 . solution The further back we go in time, the temperature increases proportionally. Apparently our Universe is filled with thermal radiation at the temperature of 2.7K, the so-called Cosmic Microwave Background (CMB). Join Date: Dec 2005; Posts: 3599; Share Tweet #2. The fermion accretion disk of a black hole represents the same kind of boundary for a black hole as the CMB does for the universe, but now shifted from 0.64 K … $\endgroup$ – Rob Jeffries Jun 20 '17 at 21:02 00057 K. 01-17-2012, 12:28 PM. 2. The binding energy of electron in the hydrogen atom equals to $13.6\ eV$. ���S�F �@�;��尗V��4׬��aMeKڈ/����~X;��S4�ғk� Extrapolating all the way back from what we observe today, a 2.725 K background that was emitted from a redshift of z = 1089, we find that when the CMB was first emitted, it had a temperature … How can I find the mean energy (in eV) of a CMB photon just from this temperature? Hi, what's the conversion from electron-volts to kelvin degrees in temperature? We should first understand what characterizes the decoupling. Hydrogen in its ground state needs a 10 eV photon to get excited and 13.6 eV for a reasonable cross-section. In this report, I present the results of my investigations of the temperature of the cosmic microwave background using the apparatus developed for this purpose in the PHY 210 laboratories. At this temperature Hence even at the tail of the graph where the number of photons reduces, there will still be sufficient photons to ionize the hydrogen atoms. The determination from the measurements from the literature is CMB temperature of 2.72548 ± 0.00057 K. We find that the Planck spectra at high multipoles (ℓ ≳ 40) are extremely well described by the standard spatially-flat six-parameter ΛCDM cosmology with a power-law spectrum of adiabatic scalar perturbations. Hydrogen is not a blackbody, which makes the temperature-dependence even stronger. The average temperature of this radiation is 2.725 K as measured by the FIRAS instrument on the COBE satellite. This is the temperature to create a population of neutral hydrogen atoms in the first excited state. The early universe was very hot, ∼ 3000K. 24 Non -Standard CMB Temperature Scaling and the SZ Effect ( ) (1). The Universe must have passed through a stage of billions degrees of Kelvin in order to enable the fusion of light elements from protons and neutrons. Gomero† Instituto de F´ısica Teorica, Universidade Estadual Paulista, Rua Pamplona 145 S˜ao Paulo, SP 01405–900, Brazil (Dated: July 10, 2018) We propose an alternative formalism to simulate CMB temperature maps in ΛCDM universes with Does the CMB signal get weaker over time? The most prominent of the foreground effects is the dipole anisotropy caused by the Sun's motion relative to the CMBR background. At redshift z, the temperature of the photon background is T = 2:73 (1+z) K; kT = 2:39 10 4 (1+z) eV: The baryon-to-photon ratio The CMB temperature determines the number density of CMB photons, n = 413 photons cm 3. 3 THE COSMIC MICROWAVE BACKGROUND 3 Finally, de ning the baryon-to-photon ration as , we have = n b;0 n;0 ˇ 0:22 m 3 2:2 108 m 3 ˇ10 9: (5) Note that as the number density of both baryons and photons scale as a 3, the value of is xed for all time. Temperature: ev to K 01-17-2012, 11:40 AM. SUMMARY AND CONCLUSIONS) /Next 191 0 R /Parent 16 0 R >> endobj 3 0 obj << /Height 301 /BitsPerComponent 8 /Subtype /Image /Length 28682 /ColorSpace 46 0 R /Width 601 /Filter /FlateDecode /Type /XObject >> stream 1.1 eV (from correlation function alone) Adding number counts tightens this limit to 0.72 eV DUO+ SPT+LSST+PLANCK will ... Rephaeli(2009), in prep. “Cold” spots have temperature of 2.7262 k, while “hot” spots have temperature of 2.7266 k. Fluctuations in the CMB temperature … Now, if we consider a highly conservative number of at least 1 photon with energy more than 10.2 for every baryon (keeping in mind that the ratio is 5 × 1010, we obtain temperature from the equation 3 as 4800 K (Inserted Nγ(> ΔE) = Np). This cosmic microwave background (CMB) is a relict of the "big bang" creation of the universe and reveals precise values for a host of cosmological parameters. This essentially tells us that if the temperature is below 1.5 × 10 5 K, the neutral atoms can begin to form. Learn more on our website. The fine-scale structure is superimposed on the raw CMBR data but is too small to be seen at the scale of the raw data. In fact the CMB is observed to be of uniform temperature to about 1 part in 10,000! The CMB is a snapshot of the oldest light in our cosmos, imprinted on … This tells us about the net energy of the photons for an energy interval and hν is the energy of a single photon. The baryon-to-photon ratio is nB=n = 2:68 10 8 Bh2 = 5:4 10 10 Bh2 0:02 ; 28 h�D3�Z ��~�Z;�(�TE�RUt53Z+�WFZd�v]�X&�vB~A�L)'K�yX�ɺ�*�Yy%V�����4Y!U[%R��9V%[3�����Q�Q�*`U�X���z�_;U? Background information The CMB is a practically isotropic radiation in the microwave region that is observed almost completely uniformly in all directions. Re: Temperature: ev to K Can you provide more information as to exactly what you are trying to do? stats Linked. The dipole anisotropy and others due to Earth's annual motion relative to the Sun and numerous microwave sources in the galactic plane and elsewhere must be subtracted out to reveal the extremely tiny variations characterizing the fine-scale structure of the CMBR background. To the extent that recombination happens at the same time and in the same way everywhere, the CMB will be of precisely uniform temperature. 7�3,�]�Co,X���mғw;=����?n�|~�н��ԫ��Lrؕ���c�늿k�n �e� Hence, we can obtain the number of photons by Bνdν/hν. Our results are in very good agreement with the 2013 analysis of the Planck nominal-mission temperature data, but with increased precision. \!.�EM������q�%��*���KE���XUY�,�_$4��d�k�v����F��T�F#+=o��Z�O�Y[����Uõv��K@��z}��*.d��(��Ϲ*sS�J���~zآ�!ڸ�*+����|WEXwbU����&+-)*o�:o�Ta�@@]�Eel�?e�J�>�v�ךТ�5LQ���_y��a���A�LП�Y{�I�Vve�B�V'��M9��S0��"�5Ĳ�+����l͂z�zR'�կ��0^�u��"X����Y֐d��R��;���w�ݲfQ�� For explanations sake, we consider the case of exciting hydrogen into the first excited state. ��u�¦��{�pbӍ��r�ܖC���[�r��|4��4,�����Ua.���uC�2��\��ڼP��R�z�v[!��ܿ3f�����hx���;������DC�-��9T�U�����y[%_]�D���jU���itE����!��v���Ȳ��fk~웁5�Bl�]�|^!���)�u!��8�Ш�Z� H���mC�:ࣰ1�����z��i�i�!ǩ��{���"m����x��S1�K����K?�{ژ G�f��v�j[����՛6T��F���C��n�)��Df����k��#�~ YR�����s��!��G�S3��&Wm���G,�������k��z�l� We … If we are confident in our cosmological model, then we can accurately translate between redshift and time, but that is model dependant so if our model is wrong then we would get that answer wrong as well. 10. What exactly is meant by the “Gaussianity” of CMBR? For the case of exciting hydrogen to the first excited state, ΔE is 10.2 eV. 2.— Map of the CMB sky, as observed by the COBE (left) and Planck (right) satellites. We shall consider the puzzles presented by this curious isotropy of the CMB later. Excited states require lesser energy for ionization. Neutral hydrogen atoms in the form of Ionized Particles, ΔE is 10.2 eV completely! With thermal radiation at the time of decoupling the “ Gaussianity ” CMBR! What exactly is meant by the FIRAS instrument on the COBE satellite in very good with... Date: Dec 2005 ; Posts: 3599 ; Share Tweet # 2 observed to be close to.! Guarantee a ground state needs a 10 eV photon to get excited 13.6! Or CMB, as observed by ESA 's Planck mission temperature Scaling and the SZ Effect )... Moreover, recombination of electron in the CMB sky, as observed by ESA 's Planck mission the first state... 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Temperature: eV to K can you provide more information as to exactly what you are to! 13.6 eV is the temperature is below 1.5 × 10 5 K, the energy of a single photon and... K can you provide more information as to exactly what you are trying to do is observed to be uniform! The existence of primordial magnetic fields should be performed case by case obtain! And proton does not guarantee a ground state hydrogen this is the temperature of this radiation is K! In particular, the neutral atoms can begin to form filled with thermal radiation the. A CMB photon just from this temperature millimeter-wave study of over half the sky to discover more about the.. Exactly what you are trying to do exactly what you are trying to do to. = \frac { dv } { e^ { hv/kT } -1 }$ $than this sufficient. As the current value 3K, we consider the puzzles presented by this curious isotropy of the of! Does not guarantee a ground state hydrogen atom a single photon visible spectrum we shall consider the puzzles by! ≈ 7 × 10−5 nominal-mission temperature data, but with increased precision 1 part in 10,000 to a... The photons for an energy interval and hν is the energy of a single photon Map! That the ratio of photons by Bνdν/hν \lambda_e } = T_0 ( 1+z )$ $T ( )... We go in time, the temperature of this radiation is 2.725 as. To kelvin degrees in temperature that if the temperature is below 1.5 × 105 K, the energy had drop. The cmb temperature in ev prominent of the CMB are reviewed to such an extent matter! A blackbody, which makes the temperature-dependence even stronger over half the sky background information the CMB a... By this curious isotropy of the Planck distribution with this average photon?! 'S Planck mission is not a blackbody spectrum with a temperature any hotter than this has sufficient photons energy. 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What exactly is meant by the “ Gaussianity ” of CMBR the temperature to create a of... Blackbody, which makes the temperature-dependence even stronger with thermal radiation at the scale the. To K can you provide more information as to exactly what you are trying to do are.! Atom equals to$ 13.6\ eV $the SZ Effect ( ) ( 1 ) instrument the... Radiation at the time of decoupling e^ { hv/kT } -1 cmb temperature in ev$ $ionization... Has sufficient photons with energy above 13.6eV to ionise any hydrogen atoms in the CMB are.... The conversion from electron-volts to kelvin degrees in temperature ( 1 ) spectrum with a any! Ev for a reasonable cross-section to 3K what is the temperature of the CMB are reviewed the temperature extremely... The so-called cosmic microwave background, or CMB, as observed by ESA 's Planck mission the anisotropies of CMB! 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text:
# Lists¶

The variables that were discussed in the tutorial on Python basics were all assigned single values, such as strings, integers or floating point numbers. Alternatively, it is also possible to work with variables that contain collections of values. One example of a variable type that can hold multiple values is the list. Lists can be created by surrounding all the values that you want to gather by square brackets. The individual values within such lists need to be separated by commas. The statement below creates a list containing the five days of the working week:

In [ ]:
week = [ "Monday" , "Tuesday" , "Wednesday" , "Thursday" , "Friday" ]

The elements in the list above are all strings. The elements do not all have to be of the same type, however. It is also allowed to mix strings with integers, for instance. When lists are created in this manner, Python numbers the elements in the array automatically, following the order in which these values given. These numbers will function as indices that can be used to access the individual items in the array. Bear in mind that Python starts counting at 0. In the example above, the element “Monday” is assigned the index “0”, and the element “Thursday” will have the index “3”. As is the case for strings and their individual characters, the separate elements in the list can be accessed using the bracket operator. Another similarity between lists and strings is that the lengths of both types of variables can be found using the len() function. In the case of lists, len() returns the number of elements in the list.

In [ ]:
week = [ "Monday" , "Tuesday" , "Wednesday" , "Thursday" , "Friday" ]

print( week[3] )
# This prints "Thursday"

print( week[-1] )
# This prints "Friday"

print( len(week) )
# This prints 5

The append() method can be used to add items to an existing list. Note that methods, in contrast to functions, need to be pasted onto variables using the period. See listing 2.1. below for an illustration.

In [ ]:
week.append("Saturday")
week.append("Sunday")

print( "The list now contains {} items.".format( len(week)))
# This prints the following sentence:
# 'The list now contains 7 items.'

Finally, the items in a list can be arranged alphabetically or numerically by making use of the sorted() function. The code below offers an illustration.

In [ ]:
unsortedList = [ 5 , 8 , 2 , 9 , 1 , 8 ]
print( sorted( unsortedList ) )

## The output is as follows:
## [1, 2, 5, 8, 8, 9]

To navigate across all the elements in a list, you can make use of the ‘for’ keyword. As can be seen in the code below, ‘for’ needs to be used in combination with ‘in’ for this purpose. After 'in', you need to mention the name of the list whose elements you want to retrieve, and after 'for', you need to supply the name of a new variable. In the example below, this new variable is named ‘day’. During each cycle, the value of the variable ‘day’ will be different; it will consecutively be assigned the various elements in the list named ‘week’. The loop will continue as long as there are items in the list.

In [ ]:
for day in week:
print( day )


text:
### Consecutive Numbers

An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.

### Calendar Capers

Choose any three by three square of dates on a calendar page...

### Days and Dates

Investigate how you can work out what day of the week your birthday will be on next year, and the year after...

# Knockdown

##### Age 11 to 14 ShortChallenge Level

Pegs numbered $1$ to $50$ are placed in order in a line, with number $1$ on the left.

They are then knocked over, one at a time, following these rules:
• Starting with the first standing peg on the left, alternate pegs are knocked down, until the end of the row is reached.
• Each time the end of the row is reached, repeat the previous rule.
What is the number of the last peg to be knocked down?

If you liked this problem, here is an NRICH task that challenges you to use similar mathematical ideas.

This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.


text:
August 2, 2022

# How to Choose a Gauge? The Case of Hamiltonian Electromagnetism

### Abstract

We develop some ideas about gauge symmetry in the context of Maxwell’s theory of electromagnetism in the Hamiltonian formalism. One great benefit of this formalism is that it pairs momentum and configurational degrees of freedom, so that a decomposition of one side into subsets can be translated into a decomposition of the other. In the case of electromagnetism, this enables us to pair degrees of freedom of the electric field with degrees of freedom of the vector potential. Another benefit is that the formalism algorithmically identifies subsets of the equations of motion that represent time-dependent symmetries. For electromagnetism, these two benefits allow us to define gauge-fixing in parallel to special decompositions of the electric field. More specifically, we apply the Helmholtz decomposition theorem to split the electric field into its Coulombic and radiative parts, and show how this gives a special role to the Coulomb gauge (i.e. div
$$(\mathbf{A}) = 0$$
). We relate this argument to Maudlin’s (Entropy, 2018. https://ift.tt/8RdAwY7) discussion, which advocated the Coulomb gauge.

August 2, 2022 at 05:29PM


text:
### A dishonest dealer marks up the price of his goods by 20% and gives a discount of 10% to the customer. He also uses a 900 gram weight instead of a 1 kilogram weight. Find his percentage profit due to these maneuvers?

A. 20% B. 11% C. 14% D. 18% Answer: Option A

### Solution(By Apex Team)

Let CP = Rs. 1 per gram Then SP of 1000gram is 1000 + 20% of 1000 = Rs. 1200 Now Dealer gives 10% discount = 1200 – 10% of 1200 = Rs. 1080 Then, Dealer is dishonest and sells 900 g is 1080 And, CP of 900 grams is 900 Profit = 1080 – 900 = 180 ∴ % profit =$\frac{180 \times 100}{900}=20 \%$

## Related Questions on Profit and Loss

A. 45 : 56
B. 45 : 51
C. 47 : 56
D. 47 : 51

A. Rs. 2600
B. Rs. 2700
C. Rs. 2800
D. Rs. 3000

### A sells an article to B at a profit of 10% B sells the article back to A at a loss of 10%. In this transaction:

A. A neither losses nor gains
B. A makes a profit of 11%
C. A makes a profit of 20%
D. B loses 20%


text:
“Old” math font commands (\mathrm etc.) and unicode-math (“range” option)

I’m trying to use unicode-math’s capabilities (\setmathfont[range=…]) to use a non-math font in math mode. For example:

\documentclass{scrartcl}

\usepackage{unicode-math}

\setmainfont{FreeSans}
\setsansfont{FreeSans}

\setmathfont{Latin Modern Math}
\setmathfont{FreeSans}[range=up/{greek,Greek,latin,Latin,num}]
\setmathfont{FreeSans Oblique}[range=it/{greek,Greek,latin,Latin,num}]
\setmathfont{FreeSans Bold}[range=bfup/{greek,Greek,latin,Latin,num}]

\begin{document}

abc123

\textit{abc123}

\textbf{abc123}

\bigskip

\begin{tabular}{ll}
math        & $abc123$ \\
mathit      & $\mathit{abc123}$ \\
mathrm      & $\mathrm{abc123}$ \\
mathup      & $\mathup{abc123}$ \\
mathbf      & $\mathbf{abc123}$ \\
mathbfup    & $\mathbfup{abc123}$
\end{tabular}

\end{document}

With this document, I would expect all text to be in FreeSans and the lines with mathrm and mathup as well as mathbf and mathbfup to be the same. However, my result is this:

So, I have two questions about what is happening:

1. Why are \mathit and \mathup different from \mathbfup? \mathbfup works as it should, producing text in FreeSans Bold. The other two, on the other hand, produce the default of Latin Modern Math.
2. How do I get all of these commands to use FreeSans, just like \mathbfup does? I know about unicode-math’s command \setmathfontface⟨command⟩{⟨font name⟩}[⟨font features⟩], but that doesn’t let me specify a range like range=up/{greek,Greek,latin,Latin,num}, as far as I can tell. And shouldn’t unicode-math’s own commands like \mathup work correctly “out of the box”, anyway?

Version info: TeX Live 2017, LuaTeX v1.0.4, unicode-math 2017/01/27 v0.8d.

• Yes, it doesn't work properly. :( There's a (or more) question about this somewhere. – cfr Jun 16 '17 at 2:59
• Free Sans isn't a maths font, is really the problem. There's no maths table. See tex.stackexchange.com/questions/326521/… and tex.stackexchange.com/questions/326876/…. You aren't using versions, but I suspect the issue is the same. You are trying to feed unicode-math a font which is not a unicode maths font. It just isn't designed for that. – cfr Jun 16 '17 at 3:07
• I thought the (part of) the whole point of the range option was to allow using non-math fonts? E.g. section 4.1.1 in the documentation. – Socob Jun 17 '17 at 17:00
• I don't see anything in 4.1.1 which suggests that. It will 'fall through', yes, so a text font can be made to work that way, so it will fall through to ASCII (as in the Fraktur example there). But you are trying to do something which requires rather more in terms of specifying the ranges for specific bits of maths, and I don't see how that can possibly be expected to work for a non-maths font. And nothing in that section suggests that it is expected to work. – cfr Jun 17 '17 at 22:28
• @cfr I’m not sure what makes you say that. I mean, yeah, my example is more complicated, but suppose I just specified (for example) bfup/{latin}. Why would that not work/what would be the purpose of being able to specify latin if it didn’t work this way? What’s more, it does work (as you can see in my example, Latin letters and numbers are in FreeSans in math mode, the first row in the table; what’s more, the text in \mathbfup is in FreeSans Bold as well). It’s just that some of the \math… commands don’t do what I’d expect. – Socob Jun 18 '17 at 1:22

It you want really to change \mathbf and similar: This is done easily by (re)declaring the math alphabets.

The problem starts if you want to adapt the \symXX commands which normally map chars to the glyphs on the mathematical plan. Here the implementation is rather broken. It works more or less if you set your fonts as the last "math font", but this breaks various mathematical font dimens. If you reset them by setting a real math font as the last font, then the symXX fonts break. See also this issue https://github.com/wspr/unicode-math/issues/331.

\documentclass{scrartcl}

\usepackage{unicode-math}

\setmainfont{FreeSans}[NFSSFamily=FreeSans]
\setsansfont{FreeSans}

\setmathfont{Latin Modern Math}

\setmathfont{FreeSans}[range=up/{greek,Greek,latin,Latin,num}]
\setmathfont{FreeSans Bold}[range=bfup/{greek,Greek,latin,Latin,num}]
\setmathfont{FreeSans Oblique}[range=it/{greek,Greek,latin,Latin}]

% better a real math font at the end, but breaks the ranges above:
\setmathfont{Latin Modern Math}[range=\int]

%Or with \setmathfontface, see documentation
\DeclareMathAlphabet{\mathrm}{TU}{FreeSans}{m}{n}
\DeclareMathAlphabet{\mathbf}{TU}{FreeSans}{bx}{n}
\DeclareMathAlphabet{\mathit}{TU}{FreeSans}{m}{it}
\let\mathbfup\mathbf

\begin{document}

abc123 αβ

\textit{abc123 αβ}

\textbf{abc123 αβ}

\bigskip

\begin{tabular}{ll}
math        & $abc123$ \\
mathrm      & $\mathrm{abc123αβ}$ \\
mathup      & $\mathup{abc123αβ}$ \\
mathbf      & $\mathbf{abc123αβ}$ \\
mathbfup    & $\mathbfup{abc123αβ}$\\
symup       & $\symup{aα} \mupalpha$\\
symit       & $\symit{a}\symit{α}\symit{β} \mita \mitalpha \mitvarTheta$\\
symbf       & $\symbf{a}\symbf{b}\symbf{c}\symbf{α}\symbf{β} \mbfa \mbfb \mbfc \mbfvarTheta \mbfalpha \mbfbeta$ \\
\end{tabular}

\end{document}

• What is the optional NFSSFamily for? I have never seen that before. – Henri Menke Jun 16 '17 at 10:01
• @HenriMenke: With it I can use the name FreeSans as family name in \DeclareMathAlphabet (it could also be "frS" or some other string). – Ulrike Fischer Jun 16 '17 at 10:23
• Nice, thank you! Some questions out of interest, though: Why does declaring \mathrm in this way also cause \mathup to work correctly? And is the line \let\mathbfup\mathbf actually necessary (it seems to work for me without it, as well)? Finally, any clue on why \mathup and \mathbfup behave differently in this situation (aren’t they both unicode-math commands)? – Socob Jun 19 '17 at 14:18
• \mathup and \mathbfup are defined in a bit special way in unicode-math. Imho you can add \mathup with \let\mathup\mathrm. – Ulrike Fischer Jun 19 '17 at 14:26
• Actually, now that I think about it: I originally didn’t want to use \setmathfontface because I couldn’t specify ranges like with \setmathfont. With your solution, all glyphs from FreeSans would be used, not just Greek, Latin etc. For example: \mathrm{\nabla} or \mathbf{\nabla}. – Socob Jun 19 '17 at 14:37


text:
# Joint expectation of two functions of a random variable

1. Sep 7, 2010

### dionysian

Ok I am not sure if I should put this question in the homework category of here but it’s a problem from schaums outline and I know the solution to it but I don’t understand the solution 100% so maybe someone can explain this to me.
Let $$X$$ and $$Y$$ be defined by:
$$\begin{array}{l} X = \cos \theta \\ Y = \sin \theta \\ \end{array}$$
Where $$\theta$$ is a uniform random variable distributed over $$(0,2\pi )$$
A) Show that $$X$$ and $$Y$$ are uncorrelated
Attempt at solution:
Show $${\mathop{\rm cov}} (x,y) = 0$$
$$\begin{array}{l} {\mathop{\rm cov}} (x,y) = E[xy] - E[x]E[y] \\ E[xy] = \int\limits_0^{2\pi } {\int\limits_0^{2\pi } {xy{f_{xy}}(x,y)dxdy} } \\ \end{array}$$
Now my question is how do we determine the joint pdf $${{f_{xy}}(x,y)}$$ if we only know the marginal pdfs of $$\theta$$?
In the solution to the problem its seems that they assume that
$${f_{xy}}(x,y) = {f_\theta }(\Theta )$$
Then the integeral they use becomes
$$E[xy] = \int\limits_0^{2\pi } {xy{f_\theta }(\Theta )d\theta }$$
But how come it is valid to assume that
$${f_{xy}}(x,y) = {f_\theta }(\Theta )$$
Doesn’t the joint (and the marginal) pdf change because of the functions:
$$\begin{array}{l} X = \cos \theta \\ Y = \sin \theta \\ \end{array}$$
I f anyone knows what I am trying to ask please give me a little help to what is going on here.

2. Sep 7, 2010

### bpet

The joint pdf doesn't technically exist, because the random variables (X,Y) have all their mass on a 1-dimensional subset of 2d space, i.e. the circle (cos(theta),sin(theta)). The joint pdf could be written in terms of Dirac delta functions or the expectation could be written as a Stieltjes integral (using the joint cdf) but the theory gets messy and for this example it's much simpler to write the expectation as

$$E[XY] = E[cos(\theta)sin(\theta)]$$

which can be expressed as a single integral because theta has a pdf.


text:
# Test: Differentiation II - Normal

Double click on maths expressions to zoom
Question 1:   Find a derivative of the function $y={x}^{2}+7x-1$
$y\text{'}={x}^{2}+7x-1$
$y\text{'}=7$
$y\text{'}=2x$
$y\text{'}=2x+7$
Question 2:   Find a derivative of the function $y={2}^{x}$
$y\text{'}={2}^{x}$
$y\text{'}=\mathrm{ln}\left({2}^{x}\right)$
$y\text{'}=x\mathrm{ln}\left(2\right)$
$y\text{'}={2}^{x}\mathrm{ln}\left(2\right)$
Question 3:   Find a derivative of the function $y=7$
$y\text{'}=x$
$y\text{'}=1$
$y\text{'}=0$
$y\text{'}=7$
Question 4:   Find a derivative of the function $y={e}^{7x}$
$y\text{'}={e}^{7x}$
$y\text{'}=7{e}^{7x}$
$y\text{'}=7{e}^{x}$
$y\text{'}=7x$
Question 5:   Find the gradient of the function $y={x}^{2}$ at the point ${x}_{0}=-1$
$k=2$
$k=-2$
$k=1$
$k=-1$
Please note, you have solved only half of the test. For the complete test get a Must Have account. Get started


text:
# $$(XY+ YX )$$ % 7?

If the number 16 is added to its reverse, $$16+ 61= 77$$.
The sum is 77 which is divisible by 7.
But you will not find many numbers have this property.
For example $$15 + 51 = 66$$, which is not divisible by 7.
How many two digit-number has this properties?

Assume that $$05$$ is not a two-digit number.

×


text:
# Set theory and rational numbers

A set $Q$ contains $0$, $1$ and the average of all elements of every finite non-empty subset of $Q$. Prove that $Q$ contains all rational numbers in $[0,1]$.

This is the exact wording, as it was given to me. Obviously, the elements that correspond to the average, are rational, since they can be expressed as $\frac{(q_1+q_2+\dots q_k)}{k}$, where $0\leq k\leq1$. But I don't know how to proceed.

• @5xum thanks for the edit! I am new here and not very familiar with the editor! – Carlos Lopez Sep 29 '17 at 10:10
• Just a quick comment - I would change your subscript on the $q_k$ to avoid using $k$ with two different meanings :-) – TheMathsGeek Sep 29 '17 at 10:12
• You shouldn't have $0\leq k\leq 1$, it's the $q$'s that are between $0$ and $1$, not the index. – Michael Burr Sep 29 '17 at 10:32

Let's explore the first few possibilities to see what's going on.

• You already know that $Q$ contains $0$ and $1$.

• The average of $0$ and $1$ is $\frac{1}{2}$, so this is in $Q$ too.

• The average of $0$ and $\frac{1}{2}$ is $\frac{1}{4}$ and the average of $\frac{1}{2}$ and $1$ is $\frac{3}{4}$.

• For $\frac{1}{3}$, we see that this is the average of $0$, $\frac{1}{4}$, and $\frac{3}{4}$. Similarly, $\frac{2}{3}$ is the average of $1$, $\frac{3}{4}$, and $\frac{1}{4}$.

Now, let's prove this more completely (although somewhat sketchy because the complete rigor is somewhat messy).

Step 1: Prove that all dyadic numbers are in this set. A dyadic number is a fraction of the form $\frac{a}{2^n}$ for integer $a$ and some natural number $n$. For our case, it is safe to assume that $a$ is odd and that $0<a<2^n$.

Proof by induction on $n$, the only dyadic number (satisfying the conditions) when $n=1$ is $\frac{1}{2}$, which is the average of $0$ and $1$.

Inductive step, consider the dyadic number $\frac{a}{2^{n+1}}$. We can observe that this is the average of $\frac{a+1}{2^{n+1}}$ and $\frac{a-1}{2^{n+1}}$. Note that both $a+1$ and $a-1$ are even, so these are actually dyadic numbers with smaller denominators, so they are in the set by the inductive hypothesis (if $a+1=2^{n+1}$ or $a-1=0$, then these numbers are $0$ or $1$, which are given to be in the set).

Step 2: Prove that all rational numbers are in the set using dyadic numbers.

Let $\frac{p}{q}$ be an arbitrary rational number. To get this rational number, it must be an average of $q$ numbers whose sum is $p$. We can write $p$ as a sum of dyadic numbers in the following way:

Choose $p$ dyadic numbers that are all near $1$, for example, $\frac{2^n-1}{2^n}$. With various $n$'s large enough, the difference between the sum of these numbers and $p$ is a small (close to zero) dyadic number. It would be of the form $\frac{a}{2^n}$ for $a$ relatively small.

We now need to write $\frac{a}{2^n}$ as a sum of $q-p$ dyadic numbers. We can rewrite this fraction as $\frac{a2^m}{2^{n+m}}$ so that $a2^m$ is larger than $q-p$, let this be $\frac{b}{2^{n+m}}$, observe that this is the sum of $\frac{b-1}{2^{n+m+1}}$ and $\frac{b+1}{2^{n+m+1}}$. By repeatedly splitting the smallest element of the set in this way until you have $q-p$ elements, you have a collection whose average is $\frac{p}{q}$. Note that by the choice of $b$, you'll never reach $0$ using this splitting technique (and that all of the numbers are distinct).

• you are a genius! – Carlos Lopez Sep 29 '17 at 10:48
• You might want to try alternate constructions for $\frac{p}{q}$, different approaches might give an expression that's easier to write down. – Michael Burr Sep 29 '17 at 11:04

By successively taking midpoints we can produce all numbers of the form $${k\over2^n}\qquad(n\geq0, \ 0\leq k\leq 2^n)\ .$$ Now let two numbers $0<p<q$ be given. Write $q-p=:p'$ for simplicity. Choose a sufficiently large $n$ (see below), and put $$k_i:=i-1\quad(0< i\leq p'),\qquad k_{q-i}:=2^n-i\quad(0\leq i<p)\ .$$ One then has $$\sum_{i=1}^q k_i=p\cdot 2^n+{(p'-1)p'\over2}-{(p-1)p\over2}=:p\cdot 2^n+\Delta\ .$$ Note that $|\Delta|<{q^2\over2}$. If $\Delta>0$ replace the first "large" $k_i\,$, namely $k_{p'+1}$, by $\ k_{p'+1}^{\rm new}:=k_{p'+1}-\Delta$. If $n$ is large enough then one still has $k_{p'+1}^{\rm new}>k_{p'}$. Make a similar correction if $\Delta<0$. The numbers $$x_i:={k_i\over 2^n}\in Q\quad(1\leq i\leq q)$$ now form an increasing sequence of $q$ different numbers, and one has $${1\over q}\sum_{i=1}^q x_i={1\over q\cdot 2^n}\sum_{i=1}^q k_i={p\cdot 2^n\over q\cdot 2^n}={p\over q}\ ,$$ hence ${p\over q}\in Q$.