text:
GAVI appears to be out of room for more funding (good news)

We’ve always been interested in GAVI, a large funding vehicle for immunizations (which we consider to be one of the best interventions out there for accomplishing good).

Until recently, GAVI projected a need for $3.7 billion between 2011-2015 (archived). However, yesterday there was an announcement that GAVI had raised$4.3 billion, more than enough to cover this need (archived).

In the past, we’ve refrained from recommending GAVI because we have trouble fully understanding its activities, but we’ve been continuing to revisit it and think about how we might gain better understanding. Now we are fairly confident in not recommending GAVI because it appears to have all the funding it needs (which, given its area of focus, we consider very good news).

This situation illustrates one of the trickier room for more funding-related challenges of individual giving. Even if a nonprofit has significant funding needs today, are there big donors right around the corner about to swoop in and render your donation irrelevant?

Before we recommend a charity, we seek as good an understanding as possible of its room for more funding, and this includes asking it about what revenue it expects in the future. This is the best way we know to avoid recommending a charity just before its funding gap is closed by megadonors, but we don’t think this approach is foolproof. We continue to find the issue of room for more funding – and in particular, the possibility of a GAVI-like situation – to be very difficult to deal with.

• College Student on June 15, 2011 at 1:08 am said:

“the possibility of a GAVI-like situation – to be very difficult to deal with.”

Only here, Holden, can I find the juxtaposition of the GAVI situation and “very difficult.” But, that’s exactly why I love Givewell. It’s original and important work. I really appreciate the team at Givewell and I don’t know if you get thanked enough. Thank you so much!

• Jonah on June 15, 2011 at 9:55 am said:

Maybe a good question for charities is “if you receive donations that exceed your current funding gap, what will these donations be used for?” Ideally they would be regranted to an effective alternate charity with room for more funding (isn’t this what GiveWell does with extra donations?) and if such an answer were standard the issue under discussion would be rendered moot, but I suppose the portion of organizations that do this is negligible. Still, this might be worthwhile for GiveWell to push for.

• Ian Turner on July 17, 2011 at 7:31 pm said:

Jonah: In general, you can expect the response to that question to be whatever the charity thinks that you want to hear. It can vary depending on the expectations set up by different donors and can change over time. In other words, while a direct answer would be very useful, it’s hard to tell if the answer you have is indeed candid.

• Jonah S on July 20, 2011 at 8:38 pm said:

Ian:

What you say is currently true for almost all charities. But in principle charities could lay out budget plans ahead of time and report back a year later on how much funding they received, whether they used the money as originally intended or whether their plans changed and how they used funds beyond those needed. Potentially the accuracy of such a report could be checked relatively inexpensively.

This way donors could check back in a year after donating to a given organization to see whether the additional funds were used as intended and charities could build up records of credibility/discredibility on this axis.

• Ian Turner on July 21, 2011 at 12:03 pm said:

Jonah: Yes, if charities made public statements about their future intentions, then it would be much more difficult to tell different donors different things, or to quietly backtrack. Publicly stating one’s future project plans would increase accountability, which is one reason why most charities would never do it unless they were forced to.

The same situation exists in the private sector in the management-shareholder relationship, by the way.

Another issue is that charities sometimes do need to radically change their plans in response to outside events. If civil war broke out in Mozambique, I would not criticize VillageReach for making major program changes, even though my donation to them was on the presumption that they would be deploying the Cabo Delgado model nationwide. This one is less of an issue as long as charities provide rationale for changes and donors are reasonable, but unfortunately there are a lot of unreasonable donors out there.


text:
# $\int_{0}^{\infty}\frac{1}{kx^{2}+1} dx=\int_{0}^{\infty} dx -\int_{0}^{\infty}\frac{kx^{2}}{kx^{2}+1} dx$ diverges?

Please, compare and contrast different spaces, it depends on the space! I cannot formulate this mathematically but I think that $\mathbb R$ extended with infinities behave totally differently.

$\int_{0}^{\infty}\frac{1}{kx^{2}+1} dx=\int_{0}^{\infty} dx -\int_{0}^{\infty}\frac{kx^{2}}{kx^{2}+1}$ where the term $\int_{0}^{\infty} dx$ diverges.

But

Let's consider $k=1$ then $\int_{0}^{\infty}\frac{1}{x^{2}+1} dx= Arctan(x)_{0}^{\infty} =\frac{\pi}{2}$ so I must have some mistake in

$$\int_{0}^{\infty}\frac{1}{kx^{2}+1} dx=\int_{0}^{\infty}\frac{(kx^{2}+1)-kx^{2}}{kx^{2}+1} dx=\int_{0}^{\infty} dx -\int_{0}^{\infty}\frac{kx^{2}}{kx^{2}+1}$$

but I cannot see it, where am I doing a mistake?

[Update]

Let' s take simpler example:

$$\int_{0}^{\infty} \frac{1+x+x^{2}+x^{3}}{x^{k}} dx= \int_{0}^{\infty} x^{-k} dx+\int_{0}^{\infty} x^{-k+1} dx+ \int_{0}^{\infty} x^{-k+2} dx+ \int_{0}^{\infty} x^{-k+3}dx$$

So this is also wrong? I cannot break it up like this? What does it mean to be $indefinite$? If I have extended real space $\mathbb R\cup\{\infty\}\cup\{-\infty\}$, what about now? Now look $\infty$ is just a normal number now, it is not indefinite so does it converge?

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$\int_0^\infty 0dx=\int_0^\infty (x-x)dx=\int_0^\infty xdx-\int_0^\infty xdx$ –  Savinov Evgeny Feb 11 '12 at 23:54

Suppose $A=B+C$. If the integrals of $A,B,C$ all exist, then $\int A=\int B+\int C$. The trouble is, it is possible that one of the integrals exists and the other two don't. For example, $${1\over3(x-3)}={1\over3x}+{1\over x(x-3)}$$ is an algebraic identity, but if you write $$\int_{-1}^1{dx\over3(x-3)}=\int_{-1}^1{dx\over3x}+\int_{-1}^1{dx\over x(x-3)}$$ you're in trouble because the two integrals on the right don't exist. (Actually, in this case the difficulty already arises before you even integrate; the left side of the identity makes sense for $x=0$, while the right side doesn't.)

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I like this answer because it is a generic and simple answer -- if I have understood now this (after consultation), I need to be particularly careful with any singular point (where zero is a divisor or where the border is mock-infinite while it actually nears some big M) -- so besides the zero -divisors (not sure whether I can scale this to algebraic zero-divisors $ab=0$ where $a\not=0 \vee b\not=0$ -- probably not, do not be misguided by this term now then here). Thanks. –  hhh Feb 15 '12 at 19:08

$\infty-\infty$ is undefined (the right hand side of your equation).

If $\int_0^\infty f(x)\, dx$ converges and if $\int_0^\infty g(x)\, dx$ converges, then you can write $\int_0^\infty[ f(x)+g(x)]\, dx=\int_0^\infty f(x)\, dx +\int_0^\infty g(x)\, dx$.

But you can't split a convergent improper integral into two pieces as you did unless both (or just one) pieces converge, for essentially the reason I gave at the outset. (Note that both the integrals on the right hand side of your equation are infinite. But you can't do anything with them, the right hand side is not defined, and the step you made going from the left hand side to the right hand side is not justified.)

That $\infty-\infty$ is undefined should become clear if you consider the limits of the expressions: $(n)-(n+1)$, or $(n+17)-(n-42)$, or $(n^2)-(n)$.

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...can you somehow break up the point or elaborate on it? (I know that I can solve this with basic u -substitution but I cannot understand this undefined thing yet, thinking...) –  hhh Feb 11 '12 at 23:39
Look at my updated question -- I cannot understand this at all. Please, compare to extended real space. I think it depends on your space. I am making now a case that I can do what I am doing in an extented space but I cannot still see why I cannot do it in non-extended space...there must be some theory about this...not just $indefinite$ -word. –  hhh Feb 12 '12 at 0:03
To expand on David's point: you know $\lim_{n\to\infty}4=4$ and you know $4=(n+4)-n$ but if you try to evaluate $\lim_{n\to\infty}4$ by writing $\lim_{n\to\infty}4=\lim_{n\to\infty}(n+4)-\lim_{n\to\infty}n$ then you've traded in something that converges for two things that don't. That's essentially what you've done with your integrals. –  Gerry Myerson Feb 12 '12 at 0:05
Now look $\infty$ is just a normal number now... To expand on @David's answer on this specific point: $\infty$ is not a normal number, whatever normal means. It is true that some arithmetic properties of the real line can be extended, for example, one can safely use $\infty+3=\infty-11=\infty+\infty=\infty$ and $4-\infty=-7-\infty=-\infty$. But other properties cannot be extended, in any way whatsoever. For example, $\infty-\infty$ has no meaning, as the basic counterexamples provided by others show. And the example in your title is based on this case. –  Did Feb 12 '12 at 11:21
By the way, your second example is kosher because all the functions involved are nonnegative, and addition (but not subtraction) is well defined on $[0,+\infty)\cup\{+\infty\}$. This means that all the parts on the RHS are finite iff the LHS is, and then the (finite) sum of the parts on the RHS equals the LHS. The other case is when the LHS is infinite, then at least one part on the RHS is infinite as well. Note finally that, at a more advanced level, one can manipulate finite negative values as well, as long as $\infty-\infty$ does not enter the picture. –  Did Feb 12 '12 at 11:30

You're missing the fact that for $k >0$

$$\int\limits_0^\infty {\frac{{dx}}{{k{x^2} + 1}} < } \int\limits_0^\infty {\frac{{dx}}{{{x^2} + 1}} = \frac{\pi }{2}}$$

so the integral converges.

And that putting

$$\sqrt k x = u$$

$$\int\limits_0^\infty {\frac{{dx}}{{k{x^2} + 1}}} = \frac{1}{{\sqrt k }}\int\limits_0^\infty {\frac{{du}}{{{u^2} + 1}}} = \frac{1}{{\sqrt k }}\frac{\pi }{2}$$

There is not "mistake" so to call it, but rather that you're not thinking about

$$\int\limits_0^\infty {f\left( x \right)dx} = \mathop {\lim }\limits_{u \to \infty } \int\limits_0^u {f\left( x \right)dx}$$

and that when the function is not continuous at $x=c\in(a,b)$ then the integral should be interpreted as:

$$\int\limits_a^b {f\left( x \right)dx} = \mathop {\lim }\limits_{m \to c} \int\limits_a^m {f\left( x \right)dx} + \int\limits_m^d {f\left( x \right)dx}$$

In the case of the polynomials you have a discontinuity at $x=0$ so when you take limits you have

$$\int\limits_0^\infty {{x^{1 - k}}dx} + \int\limits_0^\infty {{x^{2 - k}}dx} + \int\limits_0^\infty {{x^{3 - k}}dx} = \left( {\mathop {\lim }\limits_{u \to \infty } \frac{{{u^{2 - k}}}}{{2 - k}} + \frac{{{u^{3 - k}}}}{{3 - k}} + \frac{{{u^{4 - k}}}}{{4 - k}}} \right) - \left( {\mathop {\lim }\limits_{u \to 0} \frac{{{u^{2 - k}}}}{{2 - k}} + \frac{{{u^{3 - k}}}}{{3 - k}} + \frac{{{u^{4 - k}}}}{{4 - k}}} \right)$$

And you can radily check that the limit at infinity is zero for all three expressions but for zero things go to infinity. If $k<0$ then the problem would be at infinity.

-

format: plain


text:
# Two different bibliographies with different styles and sortings, Vol.2

The answer of moewe appearing in this question is almost what I want to achieve:

• one bibliography containing online references which are numbered and sorted in order of appearence (nothing to change here, moewes code is perfect!)
• one bibliography containing all book and article references, sorted alphabetically (almost what I want!)

For the second bib, instead of only showing three letters of the authors followed by the last two numbers of the year, e.g [Sut68], I would instead like to have the full last name of the first author, followed by a comma, space and then the full year, e.g [Sutherland, 1968].

I have literally no idea how to achieve this, I tried messing around with moewes provided code, but could not find any solution apart from breaking it. I would highly appreciate if someone could just post the changed part to achieve this!

Edit: Here is the code moewe provided in the answer from the question above, if that helps:

\documentclass{article}

\usepackage{filecontents}

\begin{filecontents}{biblatextest1.bib}
@BOOK{BookA03,
author    = {Author Aaa},
title     = {Some Title},
publisher = {Some Publisher},
year      = 2003,
}
@BOOK{BookB02,
author    = {Author Bbb},
title     = {Some Title},
publisher = {Some Publisher},
year      = 2002,
}
\end{filecontents}

\begin{filecontents}{biblatextest2.bib}
author  = {Author Ccc},
title   = {Some Title},
year    = 2004,
url     = {www.test1.com/bild.jpg},
}
author  = {Author Ddd},
title   = {Some Title},
year    = 2001,
url     = {www.test2.com/bild.jpg},
}
\end{filecontents}

\usepackage[style = alphabetic, labelnumber, defernumbers = true,  backend = biber]{biblatex}

\usepackage{hyperref}

%Append keywords to identify different bibliography entries.
\DeclareSourcemap{
\maps[datatype=bibtex, overwrite]{
\map{
\perdatasource{biblatextest1.bib}
\step[fieldset=KEYWORDS, fieldvalue=primary, append]
}
\map{
\perdatasource{biblatextest2.bib}
\step[fieldset=KEYWORDS, fieldvalue=secondary, append]
}
}
}

\DeclareFieldFormat{labelnumberwidth}{\mkbibbrackets{#1}}
\renewbibmacro*{cite}{%
\printtext[bibhyperref]{%
\printfield{labelprefix}%
\ifkeyword{secondary}
{\printfield{labelnumber}}
{\printfield{labelalpha}%
\printfield{extraalpha}}}}

\defbibenvironment{bibliographyNUM}
{\list
{\printtext[labelnumberwidth]{%
\printfield{prefixnumber}%
\printfield{labelnumber}}}
{\setlength{\labelwidth}{\labelnumberwidth}%
\setlength{\leftmargin}{\labelwidth}%
\setlength{\labelsep}{\biblabelsep}%
\setlength{\itemsep}{\bibitemsep}%
\setlength{\parsep}{\bibparsep}}%
\renewcommand*{\makelabel}[1]{\hss##1}}
{\endlist}
{\item}

\assignrefcontextkeyws[sorting=none]{secondary}
\begin{document}

The first two citations \cite{LinkD01} and \cite{BookB02}.
The others are \cite{LinkC04} and \cite{BookA03}.

\printbibliography[title=Bibliography, keyword=primary]

\newrefcontext[sorting=none]
\printbibliography[env=bibliographyNUM,title=References, keyword=secondary, resetnumbers]
\end{document}

This yields following result (also provided by moewe):

As I already said, I only really want [EXA97] to be changed to [Example, 1997]. Though, it would be much much nicer to not have the whole brackets in the bibliography at all! Here a optimal solution I would love to have (I edited this in photoshop, as I have no Idea how to achieve this):

• Questions seeking debugging help ("why isn't this code working?") must include the desired behavior, a specific problem or error and the shortest code necessary to reproduce it in the question itself. Questions without a clear problem statement are not useful to other readers. See minimal working example (MWE). – Henri Menke Sep 24 '18 at 1:51
• I edited the post and hope all information is now provided! – LaMahuni Sep 27 '18 at 14:39

The approach is similar to Biblatex: Two bibliographies with different styles and sortings, but this time we take authoryear as a base and copy its definition of cite.

\documentclass{article}
\usepackage[style = authoryear, labelnumber, defernumbers = true, backend = biber, autocite=inline]{biblatex}
\usepackage{hyperref}

\DeclareFieldFormat{labelnumberwidth}{\mkbibbrackets{#1}}

\renewbibmacro*{cite}{%
\iffieldundef{shorthand}
{\ifkeyword{secondary}
{\printtext[bibhyperref]{%
\printfield{labelprefix}%
\printfield{labelnumber}}}
{\ifthenelse{\ifnameundef{labelname}\OR\iffieldundef{labelyear}}
{\usebibmacro{cite:label}%
\setunit{\printdelim{nonameyeardelim}}}
{\printnames{labelname}%
\setunit{\printdelim{nameyeardelim}}}%
{\usebibmacro{cite:shorthand}}}

\DeclareCiteCommand{\parencite}[\mkbibbrackets]
{\usebibmacro{prenote}}%
{\usebibmacro{citeindex}%
\usebibmacro{cite}}
{\multicitedelim}
{\usebibmacro{postnote}}

\DeclareCiteCommand*{\parencite}[\mkbibbrackets]
{\usebibmacro{prenote}}
{\usebibmacro{citeindex}%
\usebibmacro{citeyear}}
{\multicitedelim}
{\usebibmacro{postnote}}

\defbibenvironment{bibliographyNUM}
{\list
{\printtext[labelnumberwidth]{%
\printfield{labelprefix}%
\printfield{labelnumber}}}
{\setlength{\labelwidth}{\labelnumberwidth}%
\setlength{\leftmargin}{\labelwidth}%
\setlength{\labelsep}{\biblabelsep}%
\setlength{\itemsep}{\bibitemsep}%
\setlength{\parsep}{\bibparsep}}%
\renewcommand*{\makelabel}[1]{\hss##1}}
{\endlist}
{\item}

\assignrefcontextkeyws[sorting=none]{secondary}

\usepackage{filecontents}
\begin{filecontents}{\jobname.bib}
@BOOK{BookA03,
author    = {Author Aaa},
title     = {Some Title},
publisher = {Some Publisher},
year      = 2003,
keywords  = {primary},
}
@BOOK{BookB02,
author    = {Author Bbb},
title     = {Some Title},
publisher = {Some Publisher},
year      = 2002,
keywords  = {primary},
}
author    = {Author Ccc},
title     = {Some Title},
year      = 2004,
url       = {www.test1.com/bild.jpg},
keywords  = {secondary},
}
author    = {Author Ddd},
title     = {Some Title},
year      = 2001,
url       = {www.test2.com/bild.jpg},
keywords  = {secondary},
}
\end{filecontents}

\begin{document}

The first two citations \autocite{LinkD01} and \autocite{BookB02}.
The others are \autocite{LinkC04} and \autocite{BookA03}.

\printbibliography[title=Bibliography, keyword=primary]

\newrefcontext[sorting=none]
\printbibliography[env=bibliographyNUM,title=References, keyword=secondary, resetnumbers]
\end{document}

A version that works for the outdated biblatex version on Overleaf can be found at https://v1.overleaf.com/read/cppjsymdshwd. The necessary changes were minute: It is enough to replace cite:labeldate+extradate with cite:labelyear+extrayear.

• You, Sir, are the best!!! – LaMahuni Sep 27 '18 at 16:04
• @LaMahuni If the answer solved your question you may want to consider accepting it tex.stackexchange.com/help/someone-answers – moewe Sep 27 '18 at 16:21
• I was probably happy to early.. If I copy your code into a new project, it gives me the error "Package biblatex Error: Bibliography macro 'cite:labeldate+extradate' undefined." Also one more change if possible: I would like to still have the brackets around the numbers and name/year in the text itself! Just not in the first bibliography. – LaMahuni Sep 27 '18 at 16:28
• @LaMahuni Then your biblatex version is too old. Can you make an update? – moewe Sep 27 '18 at 16:34
• I am using the online version of overleaf, this should be the newest version, shouldn't it? – LaMahuni Sep 27 '18 at 17:05


text:
Chinese Journal of Chemical Physics  2020, Vol. 33 Issue (1): 125-128

#### The article information

Hai-hua Zhou, Zeng-kui Liu, Zi-qiu Chen, Ming Sun, Qian Chen, Sheng-wen Duan, Chao Jiao

Pure Rotational Spectrum of Dibenzofuran in Range of 2-6 GHz
6 GHz范围内二苯并呋喃的纯旋转光谱
Chinese Journal of Chemical Physics, 2020, 33(1): 125-128

http://dx.doi.org/10.1063/1674-0068/cjcp1912219

### Article history

Accepted on: December 24, 2019
Pure Rotational Spectrum of Dibenzofuran in Range of 2-6 GHz
Hai-hua Zhoua , Zeng-kui Liua , Zi-qiu Chena , Ming Sunb , Qian Chenb , Sheng-wen Duanb , Chao Jiaob
Dated: Received on December 3, 2019; Accepted on December 24, 2019
a. College of Chemistry and Chemical Engineering, Lanzhou University, Lanzhou 730000, China;
b. School of Electronic and Optical Engineering, Nanjing University of Science and Technology, Nanjing 210094, China
Abstract: We report the observation and assignment of the rotational spectra of dibenzofuran measured in the range of 2$-$6 GHz with a newly constructed broadband chirped-pulse Fourier transform microwave (cp-FTMW) spectrometer. An analysis of the microwave spectra led to the assignment of 40 $b$-type transitions, resulting in the accurate determination of the rotational constants $A$ = 2278.19770(38) MHz, $B$ = 601.12248(10) MHz, and $C$ = 475.753120(98) MHz.
Key words: Dibenzofuran    Polycyclic aromatic hydrocarbons    Broadband rotational spectroscopy
Ⅰ. INTRODUCTION

The origin of ubiquitous unidentified infrared emission (UIR) features seen in the interstellar medium (ISM) remains a mystery. Although polycyclic aromatic hydrocarbons (PAHs) are widely believed to be the carriers for the UIR bands [1, 2], the molecular specific identification has been proven difficult using the commonly employed rotational spectroscopy in radio astronomy due to the often small or zero dipole moment that many PAHs possess. On the other hand, the search for simple aromatic molecules that are likely linked to the formation of PAHs but easier to observe, seems to be a valid approach to constrain models in interstellar chemistry. Dibenzofuran has been proposed to lead to PAH formation at temperatures above 800 ℃ through oxidation pathways [3]. In this context, as a simple oxygenated PAH with considerable dipole moment, it would serve as a candidate for the detection of PAHs in the ISM.

Dibenzofurans have been observed in ambient air [4] and its formation can be attributed to the same source as unsubstituted PAHs such as incomplete combustion in waste incineration [5, 6]. As an oxygenated PAH, dibenzofurans are produced through secondary oxidation of PAHs by chemical and microbiological processes [7, 8]. Higher toxicity such as mutagenicity and carcinogenicity has been shown over the parent PAHs in oxygenated PAHs [9]. Dibenzofuran in particular, undergoes oxidation reactions with OH radicals, NO$_3$ radicals and O$_3$ in the atmosphere [10], making the information pertaining to its gas-phase structure and spectroscopy of fundamental interest. Dibenzofuran can serve as a prototype of this class of substituted PAHs to be studied by spectroscopy as this tricyclic moiety is present in this family of compounds.

Up to date, dibenzofuran has been investigated by vibrational spectroscopy [11, 12] and rotationally-resolved electronic spectroscopy [13-15]. This work aims to extend the range of laboratory spectroscopy to support the astrophysical observation of potential precursor molecules for multi-ring PAHs through a microwave investigation of its pure rotational spectrum. We report the measurement and analysis of the observed spectra of dibenzofuran in the vibrational ground state in the range of 2$-$6 GHz.

Ⅱ. METHODS A. Experiments

The rotational spectra of dibenzofuran were recorded using a chirp-pulse Fourier Transform microwave (cp-FTMW) spectrometer working between 1 and 18 GHz recently developed at Nanjing University of Science and Technology (NJUST) described in detail [16, 17]. In short, a chirped pulse is produced by mixing linear frequency sweeps generated by an arbitrary wave form generator (Tabor WX1281, 1.25 GS/s, 1 GHz) and the cw output of a microwave synthesizer (Anapico Apsyn420, 1$-$18 GHz). This broadband pulse is then amplified using a solid state amplifier (BONN, 5W) before being coupled to the custom-built vacuum chamber which contains a pair of high gain horn antenna for broadcasting and receiving the microwave signal. The vacuum chamber is pumped through a molecular pump with a background pressure around 1$\times$10$^{-5}$ Pa. A gas mixture of noble gas seeded with sample molecules was allowed to enter the vacuum chamber through a pulsed solenoid valve with a diameter of 1.0 mm (Parker Series 9). The pulsed valve was designed to heat up to 250 ℃ to obtain sufficient vapor pressure for certain solid chemicals.

The molecular emission is first amplified by a low-noise amplifier (Miteq AFS44 LNA, 1$-$18 GHz) before being down converted using the microwave synthesizer, which is subsequently digitized using a 2.5 GHz bandwidth oscilloscope (Lecroy WaveRunner 6 Z). Fast Fourier transformation of the resulting free induction decay leads to the broadband spectrum over the full bandwidth of the chirp. The whole sequence can be repeated to produce multiple free induction decays (FIDs) so that the resulting spectrum gives improved signal-to-noise ratio. The microwave synthesizer, the arbitrary wave form generator and the oscilloscope are referenced to a 10 MHz rubidium standard (Stanford Research Systems, FS725) for external stability.

As dibenzofuran is a non-volatile molecule with a vapour pressure of ca. 27 Pa at room temperature, a simple heating unit was included within the pulsed nozzle assembly to increase the temperature to 200 ℃ at which a vapour pressure higher than 120 Pa is expected. A pressure of 0.6 MPa argon was then allowed to pass over the sample as carrier gas for the supersonic jet expansion.

The excitation frequency from the microwave synthesizer was mixed with a 5 $\mathtt{μ}$s linear frequency sweep, or "chirp", from 0 to 500 MHz, leading to a 1 GHz bandwidth of excitation. The resultant free induction decay (FID) was collected at a sampling rate of 40 GSa/s for 40 $\mathtt{μ}$s, giving a spectral resolution (full-width-at-half-maximum) of 80 kHz. The data acquisition sequence was repeated and a total of 1, 000, 000 FIDs were averaged to produce the spectrum via fast Fourier transformation. A total of four 1 GHz scans were used to cover the 2$-$6 GHz region. An overview spectrum of one such 1-GHz window is presented in FIG. 1 with a section showing assigned transitions in FIG. 2.

 FIG. 1 An overview of the microwave spectrum of dibenzofuran between 4.5 and 5.5 GHz measured with 1, 000, 000 FIDs
 FIG. 2 A zoomed-in spectrum of the microwave spectrum of dibenzofuran with three assigned transitions
B. Computation

Both ab initio and density functional theory (DFT) methods were employed to optimise the ground state structure of dibenzofuran. The ab initio calculation was carried out using the second order Møller-Plesset perturbation theory (MP2) [18] while the DFT calculation was performed using the Becke, 3-parameter, Lee-Yang-Parr (B3LYP) [19, 20], exchange-correlation functional with the correlation consistent polarised valence n-tuple $\zeta$ (cc-pVnZ, $n$ = D and T here) basis sets. The augmentation of such basis sets was not used as Treitel and co-workers concluded that the inclusion of diffuse functions does not improve the calculated geometrical parameters for PAH anions significantly [21]. All geometric optimisations were carried out using the Gaussian 09 package [22], and the calculated rotational constants and dipole moment are shown in Table Ⅰ. The Cartesian coordinates of the optimised structures calculated using each level of theory and basis set can be found in supplementary materials.

Table Ⅰ Computational results for dibenzofuran (∆a = 0)
Ⅲ. ASSIGNMENT AND RESULTS A. Effective Hamiltonian

The analyses of the pure rotational spectra in the GHz and THz regions were carried out using Watson's $A$-reduced effective Hamiltonian [23] in the $I^{\rm{r}}$ representation, which is shown here including up to sextic centrifugal distortion constants:

 $\begin{array}{l} \hat H_{{\rm{rot}}}^{v,v} = {A_v}\hat J_z^2 + {B_v}\hat J_x^2 + {C_v}\hat J_y^2 - \Delta _J^v{{\hat J}^4} - \Delta _{JK}^v{{\hat J}^2}\hat J_z^2 + \\ \;\;\;\;\;\;\;\;\;\Delta _K^v\hat J_z^4 - \frac{1}{2}{\left[ {\left( {\delta _J^v{{\hat J}^2} + \delta _K^v\hat J_z^2} \right),\left( {\hat J_ + ^2 + \hat J_ - ^2} \right)} \right]_ + } + \\ \;\;\;\;\;\;\;\;\;\phi _J^v{{\hat J}^6} + \phi _{JK}^v{{\hat J}^4}\hat J_z^2 + \phi _{KJ}^v{{\hat J}^2}\hat J_z^4 + \phi _K^v\hat J_z^6 + \\ \;\;\;\;\;\;\;\;\;\frac{1}{2}{\left[ {\left( {\eta _J^v{{\hat J}^4} + \eta _{JK}^v{{\hat J}^2}\hat J_z^2 + \eta _K^v\hat J_z^4} \right),\left( {\hat J_ + ^2 + \hat J_ - ^2} \right)} \right]_ + }\quad \end{array}$ (1)

where with i = $\sqrt{(-1)}$, we have the angular momentum operators:

 $\begin{eqnarray} {\hat J^2} = \hat J_x^2 + \hat J_y^2 + \hat J_z^2, \quad {\hat J_ \pm } = {\hat J_x} \pm \rm{i}{\hat J_y} \end{eqnarray}$

All observed transitions were fitted using the $I^{\rm{r}}$ representation in the Pickette's SPFIT/SPCAT spectral fitting programme [24].

B. Assignment of the pure rotational spectra in the ground state

Dibenzofuran is an asymmetric rotor of $C_{\rm{2v}}$ symmetry. Its permanent electric dipole moment is expected to lie along the $b$-axis allowing transitions to obey $b$-type selection rules (eo$\leftrightarrow$oe and oe$\leftrightarrow$eo, e for even and o for odd values of the quantum numbers $K_{\rm{a}}$ and $K_{\rm{c}}$) [25]. The principal axis system for dibenzofuran is given in FIG. 3.

 FIG. 3 Structure of dibenzofuran (carbon atoms are labeled in grey and oxygen in red). The $C_2$ symmetry axis coincides with the $b$-axis. Axes definition in the principal inertial axis system is shown at the top. The $c$-axis is perpendicular to the $ab$ plane

The assignment of the observed spectra in the range of 2$-$6 GHz was assisted by a simulation of the spectrum using the calculated rotational constants at MP2/cc-pVTZ level. A total of 40 $b$-type transitions were assigned using the PGOPHER programme [26, 27] in the range of 2$-$6 GHz with $J_\max$ = 19 and $K_{\rm{a}}\ _\max$ = 5. The assigned transitions were fitted and the resulting spectroscopic parameters are given in Table Ⅱ. The root-mean-square deviation $d_{\rm{rms}}$ of the fit is 4 kHz, with the typical linewidths (full-width at half-maximum) of about 80 kHz. A full list of the assigned transitions is provided in supplementary materials.

Table Ⅱ Spectroscopic constants for the ground state of dibenzofuran
Ⅳ. DISCUSSION

Pure rotational transitions were measured and assigned from 2 GHz to 6 GHz, providing the first study in the microwave range. This allowed for an accurate determination of the rotational constants. Comparing the rotational constants with values obtained from reported high resolution electronic spectroscopy studies [14, 15], the accuracy of our values is improved by at least two orders of magnitude as shown in Table Ⅱ, reflecting the high resolution of Fourier transform microwave (FTMW) over techniques in the optical range. The root-mean-square deviations ($\sigma_{\rm{rms}}$) of the fit of 4.6 kHz is considerably less than one-tenth of the $\sim$80 kHz observed linewidths, suggesting the effective Hamiltonian employed here provides an accurate description of observed spectral features in this range. As a result, the spectroscopic parameters reproduce the experimental spectrum very well. In comparison, the $\sigma_{\rm{rms}}$ values in Ref.[14] and Ref.[15] are a few MHz. It is worth mentioning that such an agreement was achieved by including only three rotational constants in the fit while centrifugal distortion constants were found not sensitive to the observed transitions in this study.

All calculations in this study show a planar structure for dibenzofuran with $C_{\rm{2v}}$ symmetry, indicating the only non-zero dipole moment being along the $C_2$ axis which coincides with the $b$-axis through the oxygen atom as shown in FIG. 3. This is consistent with the experimental findings that only $b$-type transitions were observed. Although the experimentally obtained rotational constants alone were not enough for accurate structural determination, one can determine the inertial defect to verify the expected planar cyclic structure. Dibenzofuran with a $C_{\rm{2v}}$ is expected to have a zero inertial defect as it is planar (see Table Ⅱ). However, our value of the inertial defect calculated using the experimental rotational constants is $-$0.285652. As it is a slightly negative number, we can attribute it to the averaging of vibrational levels of the out-of-plane, low-lying, large amplitude motions that are often associated with PAHs. In particular, the "butterfly" motions are ubiquitous in tricyclic molecules and the fundamental of such motion in dibenzofuran is at only $\sim$100 cm$^{-1}$ [28].

Ⅴ. CONCLUSION

In the current study, we presented the first spectroscopic study of dibenzofuran in the microwave region. It has been demonstrated that cp-FTMW is capable of measuring such a non-volatile molecule in a supersonic jet. The results from this work have the potential to guide the astrophysical and atmospheric searches for this molecule via remote sensing. It has been observed that the equilibrium structure of this tricyclic molecule has coupled with the low-lying, out-of-plane modes and our accurately determined rotational constants can provide the basis for the rovibrational studies of such modes in the far-infrared region.

Supplementary materials: The line list of assigned transitions as well as the theoretical geometric parameters of dibenzofuran is included in the electronic supplementary data.

Ⅵ. ACKNOWLEDGMENTS

This work was supported by the National Natural Science Foundation of China (No.61627802 and No.U1531107), the Fundamental Research Funds for Chinese Central Universities (No.lzujbky-2018-k08, No.lzujbky-2019-65, and No.lzujbky-2019-ct05).

Supplementary Materials for

The following content is included as the Supplementary Materials of this publication:

Table S1 Observed pure rotational transitions (in MHz) of the ground state of dibenzofuran
Table S2 Dibenzofuran's B3LYP/cc-PVDZ Optimized Geometry
Table S3 Dibenzofuran's B3LYP/cc-PVTZ Optimized Geometry
Table S4 Dibenzofuran's MP2/cc-PVDZ Optimized Geometry
Table S5 Dibenzofuran's MP2/cc-PVTZ Optimized Geometry
Reference
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6 GHz范围内二苯并呋喃的纯旋转光谱

a. 兰州大学化学化学工程学院，兰州 730000;
b. 南京理工大学电子工程与光电技术学院，南京 210094


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# Finding minimum fly-by radius between Mars and spacecraft from interpolating function

I've written an interplanetary trajectory solver/plotter that plots the path taken by a spacecraft on an Earth-Mars mission, but have run into a little trouble when the spacecraft actually reaches Mars. I'm hoping to be able to get the spacecraft into an orbit around Mars upon arrival, but before I can do that I need to find its closest approach in order to apply the correct delta-v at the right time (using the wonders of NDSolve's WhenEvent function). The trajectory plot relies on the output of an interpolating function from NDSolve, and it is this interpolating function that I am trying to work with. So far I know when the closest approach occurs to within a day, and it is between days 254 and 255 of the trip. I have put the Heliocentric x-y positions of Mars and the spacecraft (between days 254 and 255) into separate tables using the following:

MarsPosition = Table[{x[2][t], y[2][t]} /. Soln, {t, 254*86400, 255*86400, 86400/100}]
SpaceCraftPosition = Table[{x[3][t], y[3][t]} /. Soln, {t, 254*86400, 255*86400, 86400/100}]

And have then calculated their relative x-y positions using the following:

dxy = Sqrt[(MarsPosition - SpaceCraftPosition)^2]

But I am now trying to calculate the minimum approach radius (i.e. combining the x-y positions into a magnitude) from the above bits of code. My futile attempt of trying

dr = Min[Sqrt[dxy[1]^2 + dxy[2]^2]]

and other similar attempts have unfortunately ended in failure. Some sample output looks like this:

  dxy = {{{2.00946*10^8, 9.69241*10^7}}, {{2.03432*10^8,
9.83663*10^7}}, {{2.05917*10^8, 9.98081*10^7}},...}

Where, for example, the first entry shows the x and y relative position as {2.00946*10^8, 9.69241*10^7} and I'm trying to combine those two components into a radius. Combining the x-y position pairs into a radius using Sqrt[x^2 + y^2] for the first 3 position pairs given above I would get

 dr = {{2.231*10^8},{2.25965*10^8},{2.2883*10^8},...}

But this was done manually for each entry and is not practical for hundreds of entries. From this list of radius positions, I can then use Min[] to find the closes approach point for the spacecraft relative to Mars.

As can be seen below, the spacecraft gets very close to Mars (where the inner semi-circle is Earth's orbit, the outer semi-circle is Mars' orbit and the line joining the two is the spacecraft's trajectory):

And if propagated further we can see that it gets close enough for Mars' gravity to affect its trajectory:

So it looks pretty close, but exactly how close it what I'm hoping to find out. Any help would be greatly appreciated.

-
You need to give us Soln first or all your flybys are belong to us :D PS: Look up Norm. – Yves Klett Mar 14 '14 at 12:33
I was hoping someone would know a general way to work with tables where each entry gives an x-y position pair. I can give the code for Soln if needed, but it is a ton of code what will take some time to edit into a "stackexchange friendly" format. – user7388 Mar 14 '14 at 12:36
yeah, but some data to work with would be great, otherwise (how) are we supposed to make that up? Did you already look up Norm or Nearest? – Yves Klett Mar 14 '14 at 12:38
I would think you would be able to use Soln and FindMinimum. You could probably find the closest point with WhenEvent, too, if Soln is produced by NDSolve. Hard to tell. – Michael E2 Mar 14 '14 at 13:11
Very true, my apologies, I've edited the original post and given some example output for the first 3 x-y position pairs from the dxy table as well as what I hope to get as a table for dr. Doing that now I realise that I should have just give a general example in the first place instead of writing out what I'm using it for. – user7388 Mar 14 '14 at 13:13

I guess I'll answer my own question in case someone else needs to use it for future reference. The following code was used to find the minimum approach radius:

    MarsPosition = Table[{x[2][t], y[2][t]} /. Soln, {t, 253*86400, 256*86400, 60}] ;
SpaceCraftPosition = Table[{x[3][t], y[3][t]} /. Soln, {t, 253*86400, 256*86400, 60}] ;
(*xy positions for spacecraft relative to Mars*)
dxy = Sqrt[(MarsPosition - SpaceCraftPosition)^2];
(*Table used to hold approach radii*)
dr = Table[Norm[dxy[[i]]], {i, 1, Length[dxy]}]

Yup, I agree it was very vague. I have edited it to include test data for the dxy table. – user7388 Mar 14 '14 at 14:25


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# What is the “rate factor” or “second-step rate constant” in the reaction rate equation?

The reaction rate equation can be written as:

$$r=r_0 K \prod_n a_n$$

where the $a_n$s are the activities of each reactant in the reaction equation (in the case where we find the reaction rate from reactants to products, of course. Else the other way around). These activities are multiplied together. (This is often written as concentrations instead, I have seen, which seems to be an idealization.)

$K$ is the equillibrium constant, $K=\exp\left(\frac{-\Delta G}{RT}\right)$.

I am curious as to what exactly the $r_0$ is? In the wikipedia article it is denoted $k_2$ and called the "rate constant for the second step." The second step here refers to the intermediate step at the actived-complex.

So far so good, but what is this quantity at this intermediate step? Is it a reaction rate from the activated-complex to the final product (in that case, how can this be different from the reaction rate from the reactants to this state)?

• $k_2$ is used in a particular example (reaction of hydrogen and nitric oxide) in the Wikipedia article, not "in general" for a formula derivation. – MaxW Nov 20 '15 at 19:37
• You seem to be talking about Michaelis-Menten kinetics? If so, it's the rate constant for the production of the product and the release of the enzyme from the enzyme-substrate complex. It may have many steps embedded into it though. – timaeus222 Nov 21 '15 at 1:05
• If not, then you can take that comment about MM kinetics as just an example; we have: $$E + S ⇌ ES → E + P$$ And $k_2$ would be for the non-elementary, potentially multi-step process for going from $ES$ complex to free enzyme $E$ and product $P$. – timaeus222 Nov 21 '15 at 1:10

The implication is that the reaction mechanism for any particular reaction, if written like this:

$$A + B → C$$

might not proceed as one step. Having a $k_2$ indicates that there is a second step to the mechanism, and potentially even several more. $k_{obs}$ is probably a better name for it, because sometimes you see it written as a simplified "one-step" process, such as in Michaelis-Menten Kinetics, when really, it might be $1$ more step, or it might even be $20$, depending on the reaction (I'm exaggerating a bit, but you get the point).

If we take Michaelis-Menten enzyme kinetics as an example:

$$E+S⇌ES→E+P$$

The first step is an equilibrium in which the concentration of the intermediate is assumed to not change because the intermediate is being produced about as fast as it is being consumed. This would be the Steady State Approximation. In other words:

$\frac{d[ES]}{dt} = 0 = k_1[E][S] - k_{-1}[ES] - k_2[ES]$

Eventually...

$[ES] = \frac{k_1}{k_{-1} + k_2}[E][S]$

The second step is where the $ES$ complex does something in such a way that the enzyme unbinds somehow, and the product is made... somehow. Depending on the reactants and products, the mechanism is different.

Anyways, the point is that $k_2$ describes this second step in this type of reaction (which is what you were wondering---why the "second step"?), hence the $2$.

In any case, $k_2$ corresponds to the reaction step(s) that proceed after an intermediate product is formed on the first step for a reaction mechanism.


text:
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Try Another Version of These Questions

Question 1
Suppose that the price of Jackets is $1. In addition, suppose that the firm's total costs are$5 and that the firm currently sells 20 Jackets.

Given this information, what is this firm's total revenue?

Hint

Question 2
Suppose that the price of Pineapples is $4. In addition, suppose that the firm's total costs are$49 and that the firm currently sells 120 Pineapples.

Given this information, what is this firm's profits?

Hint

Question 3

The table below shows data for the production of Avocados for an individual firm operating in a perfectly competitive market.

0030
103050
206062
309071
4012080
5015089
60180100
70210130
80240200

Given this data, what is the per unit price of Avocados?

Hint

Question 4

The table below shows data for the production of pineapples for an individual firm operating in a perfectly competitive market.

Quantity of pineapplesTotal RevenueTotal Costs
0010
104030
208040
3012060
4016090
50200130
60240180
70280240
80320310

Given this data, what are this firm's profits if it chooses to produce 80 pineapples?

Hint

Question 5

The table below shows data for the production of avocados for an individual firm operating in a perfectly competitive market.

0010
104030
208040
3012060
4016090
50200130
60240180
70280240
80320310

Given this data, complete the table:

 Quantity of avocados Profit 0 -10 10 10 20 40 30 60 40 70 50 70 60 60 70 40 80 10

At what quantity does this firm maximize its profit?

NOTE: If there are two quantities with the same level of profits, pick the larger of the two quantities!

Hint


text:
# Probability density functions and probability of X in an interval

## Probability density functions must meet two specific criteria

Probability density refers to the probability that a continuous random variable ???X??? will exist within a set of conditions.

It follows that using the probability density equations will tell us the likelihood of an ???X??? existing in the interval ???[a,b]???.

Hi! I'm krista.

A probability density function ???f(x)??? must meet these conditions:

1. ???f(x)\ge0??? for all values of ???x???

2. ???\int^\infty_{-\infty}f(x)\ dx=1???

The equation for probability density is

???P(a\le{X}\le{b})=\int^b_af(x)\ dx???

where ???P(a\le{X}\le{b})??? is the probability that ???X??? exists in ???[a,b]???.

## Showing that f(x) is a probability density function, then finding the probability that X lies in an interval

Example

Let ???f(x)=\left(\frac{x^3}{5,000}\right)(10-x)??? for ???0\le{x}\le{10}??? and ???f(x)=0??? for all other values of ???x???. Show that ???f(x)??? is a probability density function and find ???P(1\le{X}\le{4})???.

The first thing we need to do is show that ???f(x)??? is a probability density function. We can see that the interval ???0\le{x}\le{10}??? is positive. For all other possibilities we know that ???f(x)=0???. This means we’ve satisfied the first criteria for a probability density equation. Now we need to verify that

???\int^\infty_{-\infty}f(x)\ dx=1???

We can set the interval to ???[0,10]??? since it’s only in this interval that the equation doesn’t equal ???0???.

???\int^\infty_{-\infty}f(x)\ dx=\int^{10}_0\left(\frac{x^3}{5,000}\right)(10-x)\ dx???

???\int^\infty_{-\infty}f(x)\ dx=\int^{10}_0\frac{x^3}{500}-\frac{x^4}{5,000}\ dx???

???\int^\infty_{-\infty}f(x)\ dx=\int^{10}_0\frac{x^3}{500}\ dx+\int^{10}_0-\frac{x^4}{5,000}\ dx???

???\int^\infty_{-\infty}f(x)\ dx=\frac{x^4}{2,000}-\frac{x^5}{25,000}\bigg|^{10}_0???

???\int^\infty_{-\infty}f(x)\ dx=\left[\frac{(10)^4}{2,000}-\frac{(10)^5}{25,000}\right]-\left[\frac{(0)^4}{2,000}-\frac{(0)^5}{25,000}\right]???

???\int^\infty_{-\infty}f(x)\ dx=1???

The equation has met both of the criteria, so we’ve verified that it’s a probability density function.

using the probability density equations will tell us the likelihood of an x existing in the interval [a,b].

In order to solve for ???P(1\le{X}\le{4})???, we’ll identify the interval ???[1,4]??? and plug it into the probability density equation.

???P(a\le{X}\le{b})=\int^b_af(x)\ dx???

???P(1\le{X}\le{4})=\int^4_1\left(\frac{x^3}{5,000}\right)(10-x)\ dx???

???P(1\le{X}\le{4})=\int^4_1\frac{x^3}{500}-\frac{x^4}{5,000}\ dx???

???P(1\le{X}\le{4})=\int^4_1\frac{x^3}{500}\ dx+\int^{10}_0-\frac{x^4}{5,000}\ dx???

???P(1\le{X}\le{4})=\frac{x^4}{2,000}-\frac{x^5}{25,000}\bigg|^4_1???

???P(1\le{X}\le{4})=\left[\frac{(4)^4}{2,000}-\frac{(4)^5}{25,000}\right]-\left[\frac{(1)^4}{2,000}-\frac{(1)^5}{25,000}\right]???

???P(1\le{X}\le{4})=0.0866???

The answer tell us that the probability of ???X??? existing between ???1??? and ???4??? is about ???8.66\%???.


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### The Infamous Monty Hall

A (Brief) History:

The Monty Hall problem is arguably the most infamous probability puzzle in recent history. It was originally proposed in its current form in 1975, but only really surged into the public spotlight in 1990 when in appeared in a Parade column written by Marilyn vos Savant. For more details on the history of this problem see Wikipedia: Monty Hall problem.

The Original:

As given in Marilyn's column the problem read:
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1 [but the door is not opened], and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
The phrasing here is ambiguous. Does Monty always show you a goat? Does he only show you a goat when switching would lead to a win? Does he only show you a goat when switching would lead to a loss? In fact, Monty could assign you any probability $$p$$ of winning when switching that he wanted. All Monty needs to do is with probability $$p$$ only show you a goat if switching wins; otherwise, he'd only show you a goat if switching loses. Some of the unstated assumptions undoubtedly led to additional confusion in what's already a conceptually difficult problem for many people.

Unambiguous Phrasing (Krauss and Wang 2003):
Suppose you're on a game show and you're given the choice of three doors [and will win what is behind the chosen door]. Behind one door is a car; behind the others, goats [unwanted booby prizes]. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one [uniformly] at random. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you "Do you want to switch to Door Number 2?" Is it to your advantage to change your choice?
Solution:

Standing, of course, gives us a $$1/3$$ probability of winning. But what if we switch? Well, the car is behind one of the doors, and if it's behind Door 1 with probability $$1/3$$ it must be behind one of the remaining two doors with probability $$2/3$$. But it's not behind Door 3 because Monty just showed us a goat. Aha, so the probability that the prize is behind Door 2 must be $$2/3$$ now, so we should switch! Another way of seeing that this is correct is to pretend that you close your eyes and cover your ears right before Monty Hall reveals the goat. Now by switching you'll win the prize if it's behind either Door 2 or Door 3, so your probability of winning increases from $$1/3$$ to $$2/3$$ by switching.

My next entry will be a solution using Bayes' Theorem.

### A Bayes' Solution to Monty Hall

For any problem involving conditional probabilities one of your greatest allies is Bayes' Theorem. Bayes' Theorem says that for two events A and B, the probability of A given B is related to the probability of B given A in a specific way.

Standard notation:

probability of A given B is written $$\Pr(A \mid B)$$
probability of B is written $$\Pr(B)$$

Bayes' Theorem:

Using the notation above, Bayes' Theorem can be written: $\Pr(A \mid B) = \frac{\Pr(B \mid A)\times \Pr(A)}{\Pr(B)}$Let's apply Bayes' Theorem to the Monty Hall problem. If you recall, we're told that behind three doors there are two goats and one car, all randomly placed. We initially choose a door, and then Monty, who knows what's behind the doors, always shows us a goat behind one of the remaining doors. He can always do this as there are two goats; if we chose the car initially, Monty picks one of the two doors with a goat behind it at random.

Assume we pick Door 1 and then Monty sho…

### Notes on Setting up a Titan V under Ubuntu 17.04

I recently purchased a Titan V GPU to use for machine and deep learning, and in the process of installing the latest Nvidia driver's hosed my Ubuntu 16.04 install. I was overdue for a fresh install of Linux, anyway, so I decided to upgrade some of my drives at the same time. Here are some of my notes for the process I went through to get the Titan V working perfectly with TensorFlow 1.5 under Ubuntu 17.04.

Old install:
Ubuntu 16.04
EVGA GeForce GTX Titan SuperClocked 6GB
2TB Seagate NAS HDD
+ additional drives

New install:
Ubuntu 17.04
Titan V 12GB
/ partition on a 250GB Samsung 840 Pro SSD (had an extra around)
/home partition on a new 1TB Crucial MX500 SSD
New WD Blue 4TB HDD
+ additional drives

You'll need to install Linux in legacy mode, not UEFI, in order to use Nvidia's proprietary drivers for the Titan V. Note that Linux will cheerfully boot in UEFI mode, but will not load any proprietary drivers (including Nvidia's). You'll need proprietary drivers for Tens…

### Mixed Models in R - Bigger, Faster, Stronger

When you start doing more advanced sports analytics you'll eventually starting working with what are known as hierarchical, nested or mixed effects models. These are models that contain both fixed and random effects. There are multiple ways of defining fixed vs random random effects, but one way I find particularly useful is that random effects are being "predicted" rather than "estimated", and this in turn involves some "shrinkage" towards the mean.

Here's some R code for NCAA ice hockey power rankings using a nested Poisson model (which can be found in my hockey GitHub repository):
model <- gs ~ year+field+d_div+o_div+game_length+(1|offense)+(1|defense)+(1|game_id) fit <- glmer(model, data=g, verbose=TRUE, family=poisson(link=log) ) The fixed effects are year, field (home/away/neutral), d_div (NCAA division of the defense), o_div (NCAA division of the offense) and game_length (number of overtime periods); off…


text:
The First Principles

1. Apr 18, 2008

roam

[SOLVED] The First Principles

Hello!
This is my question:

1) Write down the definition of ƒ'(3), where $$f(x) = \sqrt{x + 1}$$

2) Use the definition to evaluate ƒ'(3).

This isn't an exercise from a book etc.. so I don't know what's the correct solution to this question but here's my attempt:

By the first principles( $$f'(x) = lim h\rightarrow0 \frac{f(x+h) - f(x)}{h}$$ ):
∴ $$\frac{\sqrt{(x+h)} - \sqrt{x+1}}{h}$$

So would the definition of ƒ'(3) be like this:

$$f'(x) = \frac{\sqrt{3+h} - \sqrt{3+1}}{h}$$

Simplifying gives:
$$= 1$$

But another time I didn't use the first principles:

$$f(x) = \sqrt{x + 1}$$

$$1/2(x+1)^-1/2$$

$$\frac{1}{1/2.\sqrt{x+1}}$$

$$\frac{1}{1/2.\sqrt{3+1}} = 1$$

(?) - "?"

I got the same answer using both methods (=1). Am I on the right track? What else do I need to show?

Thank you.

Last edited: Apr 18, 2008
2. Apr 18, 2008

Tedjn

Please recheck what you get for this:

$$f'(x) = \frac{\sqrt{(x+h)} - \sqrt{x+1}}{h}$$

Remember you need to substitute x+h for x.

3. Apr 18, 2008

roam

You mean like:
$$\frac{(x+h) - \sqrt{x+1}}{h}$$

$$\frac{(3+h) - \sqrt{3+1}}{h}$$

$$3 - 2$$ , → $$= 1$$

But is the rest of my above working correct? Did I show everything that was needed?
What would be the right solution to my question?

4. Apr 18, 2008

Gib Z

No thats not what he meant :( Recheck your working, if f(x) = sqrt( x+1), what is f(x+h) ? Its not sqrt(x+h).

5. Apr 18, 2008

HallsofIvy

Staff Emeritus

No, for two reasons (1) you forgot the "+1" in the first term and (2) you forgot the limit !!
$$f'(x)= \lim_{h\rightarrow 0}\frac{\sqrt{3+h+1}-\sqrt{3+1}}{h}$$

No, it does not! Even forgetting the "+1" $\sqrt{3+h}- \sqrt{3+1}$ is NOT equal to h and so the fraction is not equal to 1!

No,
$$\frac{1}{2\sqrt{x+1}}$$

$$\frac{1}{2\sqrt{3+1}}= \frac{1}{4}$$


text:
# Are convolution algebras ever "topologically noetherian"?

For finite groups $G$, we have the group ring $k[G]$, and we can think of $G$-representations as $k[G]$-modules. It is known that for $G$ virtually polycyclic, $k[G]$ is a Noetherian ring, which means that if $V$ is a finitely generated $G$-rep, then all its submodules are also finitely generated.

If $G$ is a topological group with an interesting topology, then $k[G]$ is almost never what you actually want to look at, since it ignores the topology on $G$. Indeed, what we care about are continuous representations of $G$, and there are various continuous group algebras in this case - $C(G)$, $L^{1,2,∞}(G)$, the algebra of Radon measure $M(G)$, and many variants - with multiplication again given by convolution.

Here the analogous property should be "topologically Noetherian". We want a topologically Noetherian module to be one all of whose closed submodules are topologically finitely generated, and a topologically Noetherian ring to be one all of whose t.f.g. modules are top. Noeth. I'm pretty sure this reduces to the usual property of ACC on closed ideals, but if there's any subtlety, this seems like the right definition.

So, are any of the various group algebras (absolutely including any not mentioned above) topologically Noetherian in this sense for nice $G$, say for $G$ a compact Lie group? My sense is that this is probably too much to hope for.

• When you say "closed", are you assuming your convolution algebra is complete in an appropriate topology? Oct 29 '14 at 23:05
• Good point - hm, I guess so? I mean, really what I care about is that the category of continuous G-reps is Noetherian in the sense described. But since things are much more commonly phrased in terms of the ring, I figured people might know more about it from this angle. Oct 31 '14 at 4:59
• In a separable C*-algebra, each closed left/right ideal is topologically singly generated. Nov 25 '14 at 13:53

The completed group ring of a compact $$p$$-adic Lie group with coefficients in a field of characteristic $$p$$ is topologically Noetherian. In fact, it is even abstractly Noetherian. This is a theorem of M. Lazard from the 1960s, and you can read more about these rings here.


text:
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text:
# How do you read the Electron Configuration Periodic Table? For example, if yu wanted to find Mg what would its configuration be?

Jun 20, 2018

You can start by writing the subshell notation of the element you want to find

#### Explanation:

Subshell notation of magnesium is 2.8.2.

We know that magnesium is in Group 2, Period 3. Hence it has an energy level of 3.

By doing this, we can start listing out the electron configuration of magnesium.

$1 {s}^{2}$$2 {s}^{2}$$2 {p}^{6}$$3 {s}^{2}$

For any other elements, simply write down the total number of electrons. You will realize that once you finish writing the electron configuration, the numbers will add up the the number of electrons of the element you want to find.


text:
Algebra and Trigonometry 10th Edition

$\ln z(z-1)^2=\ln z+2\ln(z-1)$
Use the Product and Power Properties: $\ln z(z-1)^2=\ln z+\ln(z-1)^2=\ln z+2\ln(z-1)$


text:
# Why is that if every row of a matrix sums to 1, then the rows of the inverse matrix sums to 1 too?

Why is that if every row of a matrix sums to $1$ then the rows of its inverse matrix sum to $1$ too?

For example, consider

$$A=\begin{pmatrix} 1/3 & 2/3 \\ 3/4 & 1/4 \end{pmatrix}$$

then its inverse is

$$A^{-1}=\begin{pmatrix} -3/5 & 8/5 \\ 9/5 & -4/5 \end{pmatrix},$$

which satisfies the condition. Is it true for every such matrix?

Let $v = (1, 1, \ldots , 1)'$ be a column vector of all $1$s. Then the rows of $A$ adding to $1$ is equivalent to saying $Av = v$. So when $A$ is invertible, we will have $$A^{-1}v = A^{-1}Av = v$$ Thus $A^{-1}$ has rows summing to $1$ as well. (Note that $A$ will not always be invertible.)

• To extend the assertion we can say that property of row sum equal 1 holds also for any matrix $A^k$ ($A$ invertible) when $k$ is an integer $(...,-2,-1,0,1,2, ..)$ – Widawensen Jan 25 '18 at 10:33
• The name of such a matrix is a right stochastic matrix (en.wikipedia.org/wiki/Stochastic_matrix), and it has a few more interesting properties. – Victor May Jan 25 '18 at 16:40
• @VictorMay The condition here is more general than that of stochastic. These entries don't need to be non-negative. – John Coleman Jan 25 '18 at 17:25
• @VictorMay This is probably one of those cases where different definitions can be found in different sources, e.g.: en.wiktionary.org/wiki/right_stochastic_matrix . Some authors require it, some don't. I suspect that most do (since it is frequently defined in the context of Markov chains where the elements are probabilities) but haven't done any systematic survey. I was mistaken in assuming that the definition that I was familiar with was universal. – John Coleman Jan 25 '18 at 23:27
• @David That meshes with my impression. Perhaps I was trying too hard to be diplomatic. I did find a couple of definitions online which didn't explicitly contain the phrase "nonnegative" (which caused me to think that Victor's use is not unprecedented) but I suspect now that those cases were just somewhat sloppy and assumed nonnegative as being clear from context. Nonnegative is both natural and at least nearly universal in the definition. – John Coleman Jan 30 '18 at 20:45


text:
# KISS - Getaway Tab

Key:
/    = slide up
\    = slide down
b    = bend (whole step)
b^   = bend (1/2 step)
b^^  = bend (1 1/2 steps)
pb   = pre-bend
r    = release-bend
t    = tap with righthand finger
h    = hammer-on
p    = pull-off
~    = Vibrato
*    = Natural Harmonic
#(#) = Trill
**   = Artificial Harmonic
x    = Dead notes (no pitch)
P.M. = Palm mute (- -> underneath indicates which notes)
(\)  = Dive w\bar
(/)  = Release w\bar
Tp   = Tap w\plectrum

Rhythm Fig. 1a

Eb
Bb
Gb  7      9    7      7      9    7
Db  5      5    5      5      5    5
Ab
Eb

Eb
Bb
Gb
Db  7      9    7      5    5
Ab  5      5    5      3    3    5
Eb                               3

Rhythm Fig. 1b

Eb
Bb
Gb  7                 (7)
Db  5                 (5)
Ab
Eb

Eb
Bb
Gb  7              5    5
Db  7              5    5    5
Ab  5              3    3    5
Eb                           3

Rhythm Fig. 1c

Eb
Bb
Gb  12                 12
Db  12                 12
Ab  10                 10
Eb

Eb
Bb
Gb  14              12    12    12
Db  12              10    10    12
Ab                              10
Eb

Rhythm Fig. 2a

Eb
Bb
Gb  7       9     10     9  7       9     10     9
Db  5    5  5  5  5   5  5  5    5  5  5  5   5  5
Ab
Eb

Eb
Bb
Gb
Db  7     9     10     9  5    5
Ab  5  5  5  5  5   5  5  3    3    5
Eb                                  3

Rhythm Fig. 2b

Eb
Bb
Gb  7       9     7  7  9  7       9     7  7  9
Db  5    5  5  5  5  5  5  5    5  5  5  5  5  5
Ab
Eb

Eb
Bb
Gb                       5    5
Db  7     9     7  7  9  5    5    5
Ab  5  5  5  5  5  5  5  3    3    5
Eb                                 3

Rhythm Fig. 2c

Eb
Bb  12        13      12  12  13  12
Gb  12    12  12  12  12  12  12  12
Db
Ab
Eb
< - - - - - - - 2x - - - - - - - - - >

Eb
Bb  7     8     7  7  8
Gb  7  7  7  7  7  7  7  5    5    4
Db                       5    5    5
Ab
Eb

Rhythm Fig. 3a

Eb
Bb
Gb  2                           2    2    2
Db  2                           2    2    2
Ab  0                           0    0    0
Eb

Eb
Bb
Gb  2
Db  2              3            5
Ab  0              3            5
Eb                 1            3

Eb
Bb
Gb  2                         2    2    2
Db  2                         2    2    2
Ab  0                         0    0    0
Eb

Eb
Bb
Gb  2              5          7
Db  2              5          7
Ab  0              3          5
Eb

Rhythm Fig. 3b

Eb
Bb
Gb  9                         9    9    9
Db  7                         7    7    7
Ab  0                         0    0    0
Eb

Eb
Bb
Gb  9              10          12
Db  7              10          12
Ab  0              8           10
Eb

Eb
Bb
Gb  9                         9    9    9
Db  7                         7    7    7
Ab  0                         0    0    0
Eb

Eb
Bb
Gb  9              5          7
Db  7              5          7
Ab  0              3          5
Eb

Rhythm Fig. 3c

Eb
Bb
Gb
Db  14      16      14      14
Ab  12  12  12  12  12      12
Eb

Eb
Bb
Gb
Db  14      16      14      10            12
Ab  12  12  12  12  12      8             10
Eb

Eb
Bb
Gb
Db  14      16      14      14
Ab  12  12  12  12  12      12
Eb

Eb
Bb
Gb                          12            14
Db  14      16      14      10            12
Ab  12  12  12  12  12
Eb

Rhythm Fig. 4

Eb
Bb
Gb
Db  5  6  7              8  9  10
Ab  5  6  7              8  9  10
Eb  3  4  5              6  7  8

Eb
Bb
Gb           10
Db  8        10
Ab  8        8
Eb  6

Rhythm Fig. 5a

Eb
Bb
Gb  3
Db  3        3         5
Ab  1        3         5
Eb           1         3

Rhythm Fig. 5b

Eb
Bb
Gb  10        10
Db  8         10       12      14      12    14
Ab            8        10  10  10  10  10    10  10
Eb

Eb
Bb
Gb
Db  12  12  14  12
Ab  10  10  10  10
Eb

Rhythm Fig. 6

Eb
Bb                        5b 5  3
Gb  3                     5b 5  3  5  3
Db  3        3                           5  3  5  3
Ab  1        3
Eb           1

Eb
Bb
Gb
Db                              5\
Ab  5                           5\
Eb     3                        3\

Solo

Eb
Bb                                          11      11
Gb                                          13b     13b
Db     /12b           12b~      12b~
Ab
Eb
Hold bend - - - - - - - - - >

Eb
Bb  11     11     11      13   13 13  13   13 16 13h16p13
Gb  13b    13b    13b     15b         15b
Db
Ab
Eb
< - 3x - >

Eb
Bb      13h16p13    13   16b^^   16b^^   16b^^   16b^^
Gb  15           15
Db
Ab
Eb

Eb                                           11p8
Bb  16b^^   16b^^  r16        8h11p8     8        11p8
Gb                                   10
Db
Ab
Eb
< - 3x - >

Eb
Bb  11  11 8h11p8     8
Gb                10     10b r10p8
Db                                 10  10b
Ab
Eb

GETAWAY
Ace Frehley

[Rhy. Fig 1a]
[Rhy. Fig 2a w/ 1b and 1c]
[Rhy. Fig 2a w/ 2b and 2c] (2x)

[Rhy. Fig 3a w/ 3b and 3c] (2x)
I've been tryin' to get away, to get away
To leave this old town
I've been tryin' to see my way, to see my way
So I won't be found
Many times I've felt alone, I've felt alone
But that's all behind
One thing I know, I know
One thing is that

[Rhy. Fig 1a w/ 1b and 1c]
I know I have to go
Got no dough
Should I stay or should I go
[Rhy. Fig 2a w/ 1b and 1c]
Gotta get away
Don't want to stay
Leavin' tomorrow by subway

[Rhy. Fig 3a w/ 3b and 3c] (2x)
Some people they tell me, they tell me
I've been here to long
It don't compel me, they tell me
I just don't belong
Sometimes I wonder, I wonder
What's right or what's wrong
One thing I know, I know
One thing is that

[Rhy. Fig 1a w/ 1b and 1c]
I know I have to go
Got no dough
Should I stay or should I go
[Rhy. Fig 2a w/ 1b and 1c]
Gotta get away
Don't want to stay
Leavin' tomorrow by subway

[Solo over Rhy. Fig 3a and Rhy. Fig 3b][Rhy. Fig 1a]
[Rhy. Fig 1a w/ 1b and 1c]
[Rhy. Fig 2a w/ 2b and 2c] (2x)

[Rhy. Fig 5a w/ Rhy. Fig 5b] (3x)
I gotta get away
Get away, get away, get away
I gotta get away
Yeah, baby get away, gotta, gotta, gotta get away
[Rhy. Fig 6]
Whoo, you've gotta get away


text:
## Old Zortzi

### (Bilbao)

El Viejo Zortzi restaurant, founded in 1981 , where you can enjoy a pleasant seasonal cuisine, avant-garde and flavor. You can also enjoy in our bar a unique place in Bilbao to have tapas made at the moment and the possibility of tasting a large selection of wines, cavas and champagnes by the glass (Coravin system) that exceeds 50 references.

Facilities: Terrace
Price: 40€ - 60€
Route to which it belongs (Eat): Wine destination
Type of cuisine: Pintxos / Tapas, Traditional cuisine

#### Opening Hours

Monday: 1:00 p.m. to 5:00 p.m. – 6:30 p.m. to 10:30 p.m. h
Tuesday: 1:00 p.m. to 5:00 p.m. – 6:30 p.m. to 10:30 p.m. h
Wednesday: 1:00 p.m. to 5:00 p.m. – 6:30 p.m. to 10:30 p.m. h
Thursday: 1:00 p.m. to 5:00 p.m. – 6:30 p.m. to 10:30 p.m. h
Friday: 1:00 p.m. to 5:00 p.m. – 6:30 p.m. to 10:30 p.m. h
Saturday: 1:00 p.m. to 5:00 p.m. – 6:30 p.m. to 10:30 p.m. h
Closing periods: Sunday closed


text:
5

# SpontaneousTAS onspontaneous AH= TASthe follornng diagram on theSVL5TemperatureHV...

## Question

###### SpontaneousTAS onspontaneous AH= TASthe follornng diagram on theSVL5TemperatureHV

Spontaneous TAS onspontaneous AH= TAS the follornng diagram on the SVL 5 Temperature HV

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text:
Amazon SageMaker is a fully-managed service that enables data scientists and developers to build, train, and deploy machine learning (ML) models at 50% lower TCO than self-managed deployments on Elastic Compute Cloud (Amazon EC2). Elastic Inference is a capability of SageMaker that delivers 20% better performance for model inference than AWS Deep Learning Containers on EC2 by accelerating inference through model compilation, model server tuning, and underlying hardware and software acceleration technologies.

Inference is the process of making predictions using a trained ML model. For production ML applications, inference accounts for up to 90% of total compute costs. Hence, when deploying an ML model for inference, accelerating inference performance on low-cost instance types is an effective way to reduce overall compute costs while meeting performance requirements such as latency and throughput. For example, running ML models on GPU-based instances provides good inference performance; however, selecting the right instance size and optimizing GPU utilization is challenging because different ML models require different amounts of compute and memory resources.

Elastic Inference Accelerators (EIA) solve this problem by enabling you to attach the right amount of GPU-powered inference acceleration to any Amazon SageMaker ML instance. You can choose any CPU instance type that best suits your application’s overall compute and memory needs, and separately attach the right amount of GPU-powered inference acceleration needed to satisfy your performance requirements. This allows you to reduce inference costs by using compute resources more efficiently. Along with hardware acceleration, Elastic Inference offers software acceleration through SageMaker Neo, a capability of SageMaker that automatically compiles ML models for any ML framework and to any target hardware. With SageMaker Neo, you don’t need to set up third-party or framework-specific compiler software or tune the model manually for optimizing inference performance. With Elastic Inference, you can combine software and hardware acceleration to get the best inference performance on SageMaker.

This post demonstrates how you can use hardware and software-based inference acceleration to reduce costs and improve latency for pre-trained TensorFlow models on Amazon SageMaker. We show you how to compile a pre-trained TensorFlow ResNet-50 model using SageMaker Neo and how to deploy this model to a SageMaker Endpoint with Elastic Inference.

## Setup

First, we need to ensure we have SageMaker Python SDK  >=2.32.1 and import necessary Python packages. If you are using SageMaker Notebook Instances, select conda_tensorflow_p36 as your kernel. Note that you may have to restart your kernel after upgrading packages.

import numpy as np
import time
import json
import requests
import boto3
import os
import sagemaker

Next, we’ll get the IAM execution role and a few other SageMaker specific variables from our notebook environment so that SageMaker can access resources in your AWS account. See the documentation for more information on how to set this up.

from sagemaker import get_execution_role
from sagemaker.session import Session role = get_execution_role()
sess = Session()
region = sess.boto_region_name
bucket = sess.default_bucket()

## Get pre-trained model for compilation

SageMaker Neo supports compiling TensorFlow/Keras, PyTorch, ONNX, and XGBoost models. However, only Neo-compiled TensorFlow models are supported on EIA as of this writing. TensorFlow models should be in SavedModel format or frozen graph format. Learn more here.

### Import ResNet50 model from Keras

We will import ResNet50 model from Keras applications and create a model artifact model.tar.gz.

import tensorflow as tf
import tarfile tf.keras.backend.set_image_data_format('channels_last')
pretrained_model = tf.keras.applications.resnet.ResNet50()
saved_model_dir = '1'
tf.saved_model.save(pretrained_model, saved_model_dir) with tarfile.open('model.tar.gz', 'w:gz') as tar: tar.add(saved_model_dir)

### Upload model artifact to S3

SageMaker Neo expects a path to the model artifact in Amazon S3, so we will upload the model artifact to an S3 bucket.

from sagemaker.utils import name_from_base prefix = name_from_base('ResNet50')
print('S3 path for input model: {}'.format(input_model_path))

## Compile model for EI Accelerator using SageMaker Neo

Now the model is ready to be compiled by SageMaker Neo. Note that ml_eia2 needs to be set for target_instance_family field in order for the model to be optimized for EI accelerator deployment. If you want to compile your own model for EI accelerator, refer to Neo compilation API. In order to compile the model, you also need to provide the model input_shape and any optional compiler_options to your model. Note that 32-bit floating-point types (FP32) are the default precision mode for ML models. We include this here to be explicit versus compiling with lower precision models. Learn more about advantages of different precision types here.

from sagemaker.tensorflow import TensorFlowModel # Create a TensorFlow SageMaker model
tensorflow_model = TensorFlowModel(model_data=input_model_path, role=role, framework_version='2.3') # Compile the model for EI accelerator in SageMaker Neo
output_path = '/'.join(input_model_path.split('/')[:-1])
compilation_job_name = prefix + "-fp32"
compiled_model_fp32 = tensorflow_model.compile(target_instance_family='ml_eia2', input_shape={"input_1": [1, 224, 224, 3]}, output_path=output_path, role=role, job_name=compilation_job_name, framework='tensorflow', compiler_options={"precision_mode": "fp32"})

## Deploy compiled model to an Endpoint with EI Accelerator attached

Deploying a model to a SageMaker Endpoint uses the same deploy function whether or not a model is compiled using SageMaker Neo. The only change required for utilizing EI Accelerator is to provide an accelerator_type parameter, which determines the type of EI accelerator to be attached to your endpoint. All supported types of accelerators can be found here.

predictor_compiled_fp32 = compiled_model_fp32.deploy(initial_instance_count=1,
instance_type='ml.m5.xlarge', accelerator_type='ml.eia2.large')

## Benchmarking endpoints

Once the endpoint is created, we will benchmark to measure latency. The model expects input shape of 1 x 224 x 224 x 3, so we expand the dog image (224x224x3) with a batch size of 1 to be compatible with the model input. The benchmark first runs a series of 100 warmup inferences, and then runs 1000 inferences to make sure that we get an accurate estimate of latency ignoring startup times. Latency percentiles are reported from these 1000 inferences.

import numpy as np
import matplotlib.image as mpimg data = mpimg.imread('dog.jpg')
data = np.expand_dims(data, axis=0)
print("Input data shape: {}".format(data.shape)) import time
import numpy as np def benchmark_sm_endpoint(predictor, input_data): print('Doing warmup round of 100 inferences (not counted)') for i in range(100): output = predictor.predict(input_data) time.sleep(3) client_times = [] print('Running 1000 inferences') for i in range(1000): client_start = time.time() output = predictor.predict(data) client_end = time.time() client_times.append((client_end - client_start)*1000) print('Client end-to-end latency percentiles:') client_avg = np.mean(client_times) client_p50 = np.percentile(client_times, 50) client_p90 = np.percentile(client_times, 90) client_p99 = np.percentile(client_times, 99) print('Avg | P50 | P90 | P99') print('{:.4f} | {:.4f} | {:.4f} | {:.4f}\n'.format(client_avg, client_p50, client_p90, client_p99)) benchmark_sm_endpoint(predictor_compiled_fp32, data)

From the benchmark above, the output will be similar to the following:

Doing warmup round of 100 inferences (not counted)
Running 1000 inferences
Client end-to-end latency percentiles:
Avg | P50 | P90 | P99
103.2129 | 124.4727 | 129.1123 | 133.2371

## Compile and benchmark model with quantization

Quantization based model optimizations represent model weights in lower precision (e.g. FP16) which increases throughput and offers lower latency. Using FP16 precision in particular provides faster performance than FP32 with effectively no drop (<0.1%) in model accuracy. When you enable FP16 precision, SageMaker Neo chooses kernels from both FP16 and FP32 precision. For the ResNet50 model in this post, we are able to compile the model along with FP16 quantization by setting the precision_mode under compiler_options.

# Create a TensorFlow SageMaker model
tensorflow_model = TensorFlowModel(model_data=input_model_path, role=role, framework_version='2.3') # Compile the model for EI accelerator in SageMaker Neo
output_path = '/'.join(input_model_path.split('/')[:-1])
compilation_job_name = prefix + "-fp16"
compiled_model_fp16 = tensorflow_model.compile(target_instance_family='ml_eia2', input_shape={"input_1": [1, 224, 224, 3]}, output_path=output_path, role=role, job_name=compilation_job_name, framework='tensorflow', compiler_options={"precision_mode": "fp16"}) # Deploy the compiled model to SM endpoint with EI attached
predictor_compiled_fp16 = compiled_model_fp16.deploy(initial_instance_count=1, instance_type='ml.m5.xlarge', accelerator_type='ml.eia2.large') # Benchmark the SageMaker endpoint
benchmark_sm_endpoint(predictor_compiled_fp16, data)

Benchmark data for model compiled with FP16 will appear as follows:

Doing warmup round of 100 inferences (not counted)
Running 1000 inferences
Client end-to-end latency percentiles:
Avg | P50 | P90 | P99
91.8721 | 112.8929 | 117.7130 | 122.6844

## Compare latency with unoptimized model on EIA

We could see that model compiled with FP16 precision mode is faster than the model compiled with FP32, now let’s get the latency for an uncompiled model as well.

# Create a TensorFlow SageMaker model
tensorflow_model = TensorFlowModel(model_data=input_model_path, role=role, framework_version='2.3') # Deploy the uncompiled model to SM endpoint with EI attached
predictor_uncompiled = tensorflow_model.deploy(initial_instance_count=1, instance_type='ml.m5.xlarge', accelerator_type='ml.eia2.large') # Benchmark the SageMaker endpoint
benchmark_sm_endpoint(predictor_uncompiled, data)

Benchmark data for uncompiled model will appear as follows:

Doing warmup round of 100 inferences (not counted)
Running 1000 inferences
Client end-to-end latency percentiles:
Avg | P50 | P90 | P99
117.1654 | 137.9665 | 143.5326 | 150.2070

### Clean up endpoints

Having an endpoint running will incur some costs. Therefore, we would delete the endpoint to release the resources after finishing this example.

session.delete_endpoint(predictor_compiled_fp32.endpoint_name)
session.delete_endpoint(predictor_compiled_fp16.endpoint_name)
session.delete_endpoint(predictor_uncompiled.endpoint_name)

## Performance comparison

To understand the performance improvement from model compilation and quantization, you can visualize differences in percentile latency for models with different optimizations in following plot. For our model, we find that adding model compilation improves latency by 13.5% compared to the unoptimized model. Adding quantization (FP16) to the compiled model results in 27.5% improvement in latency compared to the unoptimized model.

## Summary

SageMaker Elastic Inference is an easy-to-use solution for adding model optimizations to improve inference performance on Amazon SageMaker. With Elastic Inference accelerators, you can get GPU inference acceleration and remain more cost-effective than standalone SageMaker GPU instances. With SageMaker Neo, software-based acceleration provided by model optimizations further improves performance (27.5%) over unoptimized models.

If you have any questions or comments, use the Amazon SageMaker Discussion Forums or send an email to [email protected].


text:
It estimates the R2 for a continuous predicted-observed dataset.

## Usage

R2(data = NULL, obs, pred, tidy = FALSE, na.rm = TRUE)

## Arguments

data

(Optional) argument to call an existing data frame containing the data.

obs

Vector with observed values (numeric).

pred

Vector with predicted values (numeric).

tidy

Logical operator (TRUE/FALSE) to decide the type of return. TRUE returns a data.frame, FALSE returns a list; Default : FALSE.

na.rm

Logic argument to remove rows with missing values (NA). Default is na.rm = TRUE.

## Value

an object of class numeric within a list (if tidy = FALSE) or within a data frame (if tidy = TRUE).

## Details

The R2 is one of the most widely used metrics evaluating models performance. R2 is appropriate to estimate the strength of linear association between two variables. It is positively bounded to 1, but it may produce negative values. The closer to 1 the better linear association between predictions and observations. However, R2 presents a major flaw for prediction performance evaluation: it is not sensitive to lack of accuracy (additive or proportional bias). Thus, R2 only measures precision, but it does not account for accuracy of the predictions. For the formula and more details, see online-documentation

## References

Yang et al. (2014). An evaluation of the statistical methods for testing the performance of crop models with observed data. Agric. Syst. 127, 81-89. doi:10.1016/j.agsy.2014.01.008

## Examples

# \donttest{
set.seed(1)
X <- rnorm(n = 100, mean = 0, sd = 10)
Y <- X + rnorm(n=100, mean = 0, sd = 3)
R2(obs = X, pred = Y)
#> \$R2
#> [1] 0.9070934
#>
# }