text: (CC BY-NC-SA 3.0; anonymous) When those atoms are given energy, the electrons absorb the energy and move to a higher energy level. Likewise, when the atoms relaxed back to a lower energy state, any amount of energy could be released. | VLBI, Parkes In Figure 12.8 you can see the line emission spectrum of hydrogen. If you look closely at the curve you will notice that the object emits some radiation at every wavelength including in the ultraviolet and infrared wavebands. As it absorbs energy it heats up and re-radiates the energy as electromagnetic radiation. The luminous regions have a sharp “cut-off” at one end but are diffuse at the other. | ATCA These energy levels of the electrons in atoms are quantized, meaning again that the electron must move from one energy level to another in discrete steps rather than continuously. The gas absorbs certain wavelengths of energy and allows others to pass through. When it returns back to the ground state, it releases the energy that it had previously gained in the form of electromagnetic radiation. The Lyman Series involves transitions down to the n = 1 orbit and involve higher frequency photons in the UV region whilst the Paschen Series (to n = 3) produces IR spectral lines. (a) When a hydrogen atom absorbs a photon of light, an electron is excited to an orbit that has a higher energy and larger value of n. (b) Images of the emission and absorption spectra of hydrogen are shown here. It happen when light makes an electron jump to a higher orbital and light energy is absorbed. Have questions or comments? Figure 12.9 shows the absorption spectrum. An electron orbit a nucleus in a stable energy level. | InterSat This plot is called the blackbody curve or the Planck curve, after the German physicist Max Planck who first postulated that electromagnetic radiation was quantised. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. R.A. EDWARDS M.A. Cosmic engine for senior students Virtual Radio Interferometer | ATELIB These two types are in fact related and arise due to quantum mechanical interactions between electrons orbiting atoms and photons of light. Radial Velocities When the string and arrow are pulled back, we now have a situation where kinetic energy (pulling of the string) has been converted to potential energy (the tension on the string). If a photon of a specific frequency interacts with the electron, it can gain sufficient energy to "jump up" one or more levels. These downward transitions of the excited electrons back to the ground state (the lowest energy) produced the line spectrum. In the sealed lamp are two plates across which a high voltage is applied and a stream of e⁻s is produced across the plates. "Neon" signs are familiar examples of gas discharge tubes. Classical theory was unable to explain the existence of atomic emission spectra, also known as line-emission spectra. Thus, emission spectra are produced by thin gases in which the atoms do not experience many collisions (because of the low density). | VLBI, Astrophysics overview It is used in astronomy to study the elements present in stars and in chemical analysis. If we just wanted to compare the peak wavelengths we can plot them using a normalised energy output in which the peak wavelength for each corresponds to an intensity = 1.0. Australia Telescope Compact Array | Planets An absorption spectrum is formed when white light passes through a cool gas. Australia Telescope Online Archive (ATOA) This is shown schematically in the diagram below for the Bohr model of the atom. Thus, as all the photons of different energies (or wavelengths or colors) stream by the hydrogen atoms, photons with thisparticular wavelength can be absorbed by those atoms whose … Engineering research The continuous spectrum produced by a black body is distinctive and can be shown as an intensity plot of intensity against emitted wavelength. | Visitor list Public observatories Simple examples in the visible wavebands are shown below. Parkes Observatory online store Astrophysics for senior students These bombard the gaseous Na … Atomic Spectra: {eq}\\ {/eq} Atomic spectra is the spectrum of the electromagnetic radiation emitted or absorbed by an atom. Astrophysics graduate student programs How are atomic emission spectra produced in this exercise? The emission spectrum of a chemical element or chemical compound is the spectrum of frequencies of electromagnetic radiation emitted due to an atom or molecule making a transition from a high energy state to a lower energy state. Adjust the interface to make it easier to use for different conditions. As electrons jump down to the n = 2 orbit, they emit photons of specific frequency (hence colour) that can be seen as emission lines in the visible part of the em spectrum. The simplest method is to heat the sample to a high temperature, after which the excitations are produced by collisions between the sample … The 6,000 K object clearly peaks in the visible part of the spectrum whereas the peak of the 4,000 K object borders the visible and the infrared regions. | IDL [external link] The archer releases the arrow and the potential energy is translated into kinetic energy as the arrow moves. Staff list | Student list SuperMongo (SM) It is found to consist of a series of sharp lines in UV, visible and Infrared region. There are multiple energy states available to an electron, so there are many possible transitions, leading to the numerous wavelengths that … Imagine heating a solid steel sphere with a blowtorch. Engineering education program ATCA Live, CSIRO Radio Astronomy Image Archive The emission lines correspond to photons of discrete energies that are emitted when excited atomic states in the gas make transitions back to lower-lying levels. | Parkes This procedure is called emission spectroscopy and is a very useful scientific tool. ATNF Technical Memos, Astronomical tools & software overview it is produced with an electron moves from one energy level to a lower energy level. AIPS | RPFITS Emission spectra of chemical compounds, or of substances containing atoms grouped together to form molecules, may produce emission spectra in which regions of the spectrum appear, separated by dark spaces. This visible set of lines is called the Balmer series. Since the electron energy levels are unique for each element, every gas discharge tube will glow with a distinctive color depending on the identity of the gas (see below). Emission spectrum … Archery as a sport or a means of defense has existed for centuries. Events, Technology overview 1 Ångstrom = 0.1 nanometers. Describe how an emission spectrum is formed Electrons that have been excited are unstable and fall back to their ground state. This loads a font easier to read for people with dyslexia. At rest, there is no tension on the bowstring and no force on the arrow. So how do atoms gain energy in the first place? One way is to pass an electric current through an enclosed sample of a gas at low pressure. ), in Physics for O.N.C. Suppose a beam of white light (which consists of photons of all visible wavelengths) shines through a gas of atomic hydrogen. | ASAP Courses, 1970. CASApy The emitted light corresponds to energies of the specific electrons. The emitted light corresponds to energies of the specific electrons. This renders the document as white on black. Visiting Australia Telescope Compact Array, Parkes radio telescope webcam Let’s look at the hydrogen atom from the perspective of the Bohr model. Arrangement of these wavelength are known as a emission spectra of atom. Email discussion lists, On-Line Proposal Applications and Links (OPAL) When you remove the torch you can feel the heat being re-radiated by the sphere. Figure $$\PageIndex{6}$$: Absorption and Emission Spectra. The plot below shows Planck curves for an object at four different temperatures from 6,000 K to 4,000 K. Note the wavelength here is expressed in units of Ångstroms. Every element gives off a unique fingerprint of light, so analyzing the frequencies of this light helps identify the chemical that generated it. This can help those with trouble processing rapid screen movements. It turns out that electrons behave the same way when energy is put into the system or released from the system. Planck curves for black bodies of differing temperatures. What gives rise to a continuous spectrum? The emitted light corresponds to energies of the specific electrons. The different colors of light produced by emission spectra of different elements allows them to be identified. According to classical physics, a ground state atom would be able to absorb any amount of energy rather than only discrete amounts. Watch the recordings here on Youtube! An excited state of an atom is a state where its potential energy is higher than the ground state. However, only signs that glow with the red-orange color seen in the figure are actually filled with neon. 4. Work experience for senior students | livedata/gridzilla | ASKAPSoft How an Emission Spectrum Is Produced . When matter is very hot it emits light. That is: an emission spectrum is a series of bright lines on a dark background. Missed the LibreFest? Computing: Getting started guide [internal access] MRO Support Facility | Mopra Spectra can be simplified to one of three basic types. Visualisation software Become an astronomer | TCS Atomic emission spectra are produced when excited electrons return to ground state. Eventually it starts to glow dimly. White light viewed through a prism and a rainbow are examples of continuous spectra. Astronomical images Line spectra appear in two forms, absorption spectra, showing dark lines on a bright background, and emission spectra with bright lines on a dark or black background. Work experience for school students, Marsfield headquarters (CANTAB. | MONICA How do the curves compare? Karma An atom in the excited state is not stable. Management | Director Teacher workshops During transition, different wavelength of photon are emitted by the atom. Contact us, Marsfield From n = 5, the possible emissions are 5->4, 5->3, 5->2, and 5->1.that makes 4 lines. CSIRO Radio Astronomy Image Archive, Visiting Parkes radio telescope Newsletters About ATNF overview Accommodation & computing reservations A photon of wavelength 656 nanometers has just the right energy to raise an electron in a hydrogen atom from the second to the third orbit. When an atom or molecule absorbs energy, electrons are bumped into a higher energy state. You can see clearly from the plot that a 10,000 K star would have its peak wavelength in the ultraviolet part of the em spectrum whilst a 3,000 K star would emit most of its radiation in the infrared part. Annual reports Describes atomic emission spectra and the source of the emission lines. | Tempo2 In the real world some objects approximate the behaviour of blackbodies. Let's see how Bohr's model of the atom explains the three types of spectra. Now what happens if the temperature of the black body source is different? CSIRO ATNF Data Archives | Duchamp When the electron drops to a lower energy state, a photon is released equal to the energy between the two states. | ATCA Secondly, the hotter the object, the shorter the wavelength of the peak of the curve. The emission lines correspond to photons of discrete energies that are emitted when excited atomic states in the … As already mentioned, stars approximate black body objects and can vary in their effective temperatures from around 2,000 K to about 30,000 K. If you tried to plot the intensity of two stars with these extremes on a plot like the one above it would be extremely difficult to show them on the same linear scale. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. The electrons in an atom tend to be arranged in such a way that the energy of the atom is as low as possible. | MRO Tips on holding a viewing night, Universe@CSIRO blog This plot is called the blackbody curve or the Planck curve, after the German physicist Max Planck who first postulated that electromagnetic radiation was quantised. The emission spectrum of a chemical element or chemical compound is the spectrum of frequencies of electromagnetic radiation emitted due to an atom or molecule making a transition from a high energy state to a lower energy state. Receivers & dishes Graduate student programs CASS Diversity Committee, Publications overview As you continue heating it the sphere glows first red, then orange, yellow then white hot. As the energy levels have different values, each of the possible electron transitions within an atom will produce a photon with a different energy. Australian Square Kilometre Array Pathfinder There are many possible electron transitions for each atom, and each … This would result in what is known as a continuous spectrum, where all wavelengths and frequencies are represented. Atomic emission spectra are produced when excited electrons return to the ground state. Contact us, Governance overview Wallpapers, What is radio astronomy? Both emission and absorption techniques can be used to get the same information about the energy levels of an atom. Specific examples are discussed on another page. Absorption spectrum is a gap in the overall spectrum. The Balmer series of visible lines for atomic hydrogen are caused by transitions from the n = 2 orbit to and from higher orbits. | Coord CK-12 Foundation by Sharon Bewick, Richard Parsons, Therese Forsythe, Shonna Robinson, and Jean Dupon. Miriad Visitor programs Absorption of light by a hydrogen atom. The number of spectral lines that can be produced is vast given the permutations of atoms, molecules and orbital transitions possible. Postdoctoral fellowships Practice This is shown below for six different temperatures. Emission spectra are bright-line spectra, absorption spectra are dark-line spectra. Electromagnetic radiation can be described in terms of its wavelength the distance between th… What is a pulsar? An atomic emission spectrum is the pattern of lines formed when light passes through a prism to separate it into the different frequencies of light it contains. Because each element has an exactly defined line emission spectrum, scientists are able to identify them by the color of flame they produce. Parkes radio telescope Atomic emission spectra were more proof of the quantized nature of light and led to a new model of the atom based on quantum theory. | Mathematica | PSRCat When a narrow beam of this light was viewed through a prism, the light was separated into four lines of very specific wavelengths (and frequencies since $$\lambda$$ and $$\nu$$ are inversely related). Engineering education Photons of light each have a specific frequency. An emission spectrum is the opposite of an absorption spectrum. Emission Spectra: An emission spectrum is produced when atoms release energy. Observers archives tools & information | Arch Australia Telescope Steering Committee Otherwise, ultraviolet and infrared photons could also be produced. | CDSCC An emission spectrum is the electromagnetic radiation (EMR), such as visible light, a substance emits. Normalised plots for 6 different black bodies. The energy of a photon is a function of its frequency and is determined by: E = hf where f is the frequency of the photon, E is the energy and h is Planck's constant (= 6.626 x 10-34J.s). Email discussion lists, Careers overview News | Events Line emission and absorption spectra Studying the line spectra produced by hot gases and absorbed by cooler gases allows us to identify the elements in stars. One means by which a continuous spectrum can be produced is by thermal emission from a black body. When is an absorption spectrum produced? The electron then "de-excites" and jumps back down to a lower energy orbit, emitting a photon of specific frequency. emission spectra are produced by thin gases in which the atoms do not experience many collisions (because of the low density). It is the exact opposite of the emission spectrum! Instead of getting light with a few colors missing, in an emission spectrum, those are the only colors we get. The continuous spectrum produced by a black body is distinctive and can be shown as an intensity plot of intensity against emitted wavelength. | WCSLIB, ATNF Technical Memos These must be sources of thermal energy and must be sufficiently opaque that light interacts with the material inside the source. | ATCA, Parkes Canberra Deep Space Communication Complex Very Long Baseline Interferometry, Applying for observing time A spectral line is a dark or bright line in an otherwise uniform and continuous spectrum, resulting from emission or absorption of light in a narrow frequency range, compared with the nearby frequencies. Not only does the shape of the curve determine the relative intensity of the different components of the continuous spectrum produced by the star, it also determines the colour of the star. This renders the document in high contrast mode. This photon, however, could be emitted in any direction, not just in the same direction as the original incident photon. If the transition involved an electron dropping from a higher level into the n = 2 state, the photon was visible. Remember rule #3! Two key points should be apparent. Mopra radio telescope Band spectra. Click, Australian Square Kilometre Array Pathfinder, Canberra Deep Space Communication Complex, On-Line Proposal Applications and Links (OPAL), Australia Telescope Online Archive (ATOA), Computing: Getting started guide [internal access], Visiting Australia Telescope Compact Array, Historical Introduction to Spectroscopy. Astrophysics staff Firstly, a hotter object emits more energy at every wavelength than a cooler one. Answer link. On-Line Proposal Applications and Links (OPAL) | Get more help from Chegg. You said you know the theory, but what you might not understand is how much a good experimenter must really know theory in depth. Scientists studied the distinctive pink color of the gas discharge created by hydrogen gas. A black body radiator is a theoretical object that is totally absorbent to all thermal energy that falls on it, thus it does not reflect any light so appears black. Electron transitions in the Bohr model of the Hydrogen atom. Scientific support of facilities The ground state of an atom is the lowest energy state of the atom. Thus, emission spectra are produced by thin gases in which the atoms do not experience many collisions (because of the low density). One use of this technique is to identify the elements present in distant stars. For example, copper produces a blue flame, lithium and strontium a red flame, calcium an orange flame, sodium a yellow flame, and barium a green flame. When is an emission spectrum produced? An emission line is produced by an atom in a excited'' energy state---the electron is not in as low an energy orbit as possible. The photon energy of the emitted photon is equal to the energy difference between the two states. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. A 10,000 K star appears blue-white whilst a 3,000 K star appears red. it is produced when an electron moves from one energy level to a higher energy level. Observing schedules Teacher resources Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. The figure below shows the atomic emission spectrum of hydrogen. This way emission spectra produce. | MSF Current telescope status | CASAcore This is particularly relevant in astronomy and is discussed in the next section. Murchison Radio-astronomy Observatory An emission spectrum is the line pattern formed when an element is excited and gives off energy. This sphere approximates what physicists call a black body. Sources of continuous, emission, and absorption spectra Kirchhoff's Laws Thus,emission spectraare produced by thin gases in which the atoms do not experience many … You should also notice that the amount of energy emitted is not the same for all wavelengths and that in this case, the peak wavelength falls within the region of visible light. Time Assignment Committee These frequencies of photon of light are then missing in the spectrum produced, instead there are dark lines showing the frequencies absorbed. Examples of such objects include the tungsten filaments of incandescent lamps and the cores of stars. If you were able to keep heating it sufficiently it may even glow blue hot. PULSE@Parkes Australia Telescope 20GHz Survey Digital systems Publications & acknowledgements The photon is absorbed by the electron so cannot continue on to be detected by an observer. [ "article:topic", "showtoc:no", "license:ccbync", "program:ck12" ], information contact us at info@libretexts.org, status page at https://status.libretexts.org. The emission spectra of different elements are unique, so the spectra given off by hydrogen will be different than that of helium. GIPSY CSIROpedia. Why can atomic emission spectra be used to identify what elements are present in a sample? Astronomical spectra can be combination of absorption and emission lines on a continuous background spectrum. Australia Telescope User Committee, Our people overview Atomic emission spectra are produced when excited electrons return to the ground state. Applying the torch again puts more energy into the sphere - it gets hotter. Summer vacation program Signs of other colors contain different gases or mixtures of gases. The emission lines correspond to photons of discrete energies that are emitted when excited atomic states in the gas make transitions back to lower-lying levels. 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# Real Numbers : Exercise 1.2 (Mathematics NCERT Class 10th)

Q.1     Express each number as product of its prime factors:
(i) 140      (ii) 156      (iii) 3825      (iv) 5005      (vi) 7429

Sol.       (i) We use the division method as shown below :

Therefore, 140 = 2 × 2 × 5 × 7 $= {2^2} \times 5 \times 7$

(ii) We use the division method as shown below:

Therefore, 156 = 2 × 2 × 3 × 13  $= {2^2} \times 3 \times 13$

(iii) We use the division method as shown below :

Therefore, 3825 = 3 × 3 × 5 × 5 × 17 $= {3^2} \times {5^2} \times 17$

(iv) We use the division method as shown below :

Therefore, 5005 = 5 × 7 × 11 × 13

(v) We use the division method as shown below :

Therefore, 7429 = 17 × 19 × 23

Q.2     Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.
(i) 26 and 91     (ii) 510 and 92     (iii) 336 and 54
Sol.       (i) 26 and 91

26 = 2 × 13 and 91 = 7 × 13
Therefore, LCM of 26 and 91 = 2 × 7 × 13 = 182
and HCF of 26 and 91 = 13

Now, 182 × 13 = 2366 and 26 × 91 = 2366
Since, 182 × 13 = 26 × 91
Hence verified.

(ii) 510 and 92

510 = 2 × 3 × 5 × 17 and 92 = 2 × 2 × 23
Therefore, LCM of 510 and 92 = 2 × 2 × 3 × 5 × 17 × 23 = 23460
and HCF of 510 and 92 = 2
Now, 23460 × 2 = 46920 and 510 × 92 = 46920

Since 23460 × 2 = 510 × 92
Hence verified.

(iii) 336 and 54

336 = 2 × 2 × 2 × 2 × 3 × 7

and 54 = 2 × 3 × 3 × 3
Therefore, LCM of 336 and 54 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 = 3024
and HCF of 336 and 54 = 2 × 3 = 6
Now, 3024 × 6 = 18144
and 336 × 54 = 18144
Since, 3024 × 6 = 336 × 54
Hence verified.

Q.3       Find the LCM and HCF of the following integers by applying the prime factorisation method
(i) 12,15 and 21      (ii) 17, 23 and 29      (iii) 8, 9 and 25
Sol.         (i) First we write the prime factorisation of each of the given numbers.
12 = 2 × 2 × 3 = ${2^2}$ × 3, 15 = 3 × 5 and 21 = 3 × 7
Therefore, LCM = ${2^2}$ × 3 × 5 × 7 = 420
and, HCF = 3

(ii) First we write the prime factorisation of each of the given numbers.
17 = 17, 23 = 23 and 29 = 29
Therefore, LCM = 17 × 23 × 29 = 11339
and HCF = 1

(iii)  First we write the prime factorisation of each of the given numbers.
8 = 2 × 2 × 2
$= {2^3},9 = 3\, \times 3 = {3^2}$, 25 = 5 × 5 = ${5^2}$
Therefore, LCM $= {2^3} \times {3^2} \times {5^2} = 8 \times 9 \times 25 = 1800$
and HCF = 1

Q.4     Given that HCF (306, 657) = 9, find LCM (306, 657).
Sol.       We know that the product of the HCF and the LCM of two numbers is equal to the product of the given numbers.
Therefor, HCF (306,657) × LCM (306,657) = 306 × 657
$\Rightarrow$ 9 × LCM (306 × 657) = 306 × 657
$\Rightarrow$ LCM (306,657) $= {{306 \times 657} \over 9}$  = 22338

Q.5.     Check whether ${6^n}$ can end with the digit 0 for any natureal number n.
Sol.        If the number ${6^n}$, for any n ends with the digit zero, then it is divisible by 5. That is, the prime factorisation of ${6^n}$ contains the prime 5. That is, not possible as the only prime in the factorisation of ${6^n}$ is 2 and 3 and the uniqueness of the Fundamental Theorem of Arihmetic guarantees that there are no other primes in the factorisation of ${6^n}$. So, there is no $n \in N$ for which ${6^n}$ ends with the digit zero.

Q.6    Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Sol.     Since, 7 × 11 × 13 + 13 = 13 × (7 × 11 × 1 + 1)
= 13 × (77 + 1)
= 13 × 78
$\Rightarrow$ It is a composite number.
Again, 7 × 6 × 5 × 4 × 3 × 1 × 1 × 1 + 5 = 5 × (7 × 6 × 4 × 3 × 1 × 1 + 1)
$\Rightarrow$ It is a composite number.

Q.7     There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the path, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they again at the starting point ?
Sol.      To find the LCM of 18 and 12, we have

18 = 2 × 3 × 3 and 12 = 2 × 2 × 3
LCM of 18 and 12 = 2 × 2 × 3 × 3 = 36
So, Sonia and Ravi will meet again at the starting point after 36 minutes.

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# How to prove that $\langle Au_n, u_n-u\rangle_{X^*\times X}\to 0$

Let $$X$$ be a reflexive Banach space and consider a linear and continuous operator $$A\colon X\to X^*$$. Suppose that $$(u_n)$$ is a sequance of elements from $$X$$ and assume that $$u_n\to u$$ weakly in $$X$$. How to deduce that $$\langle Au_n, u_n-u\rangle_{X^*\times X}\to 0\,?$$

Let $$x\in X$$. Consider $$\lim_{n\to \infty}\langle Au_n-Au,x\rangle_{X^*\times X}=\langle\lim_{\ n\to \infty}A(u_n-u),x\rangle_{X^*\times X}=0.$$ This means $$Au_n\to Au$$ weakly$$^*$$ in $$X^*$$. As $$X$$ is reflexive, so is its dual. Since in reflexive Banach spaces weak and weak$$^*$$ topologies coincide, we deduce that $$Au_n\to Au$$ weakly in $$X^*$$. How to use that and linearity of $$A$$ to deduce $$\langle Au_n, u_n-u\rangle_{X^*\times X}\to 0?$$

This is false. Take $$X=L^2(\mathbb R)$$, and $$A=I$$, the identity operator. Fix $$f\in L^2$$, $$f\ne 0$$ and define $$u_n(x):=f(x-n).$$ Then $$u_n \rightharpoonup 0$$, so $$u=0$$ in the notation of the original question, but $$\langle u_n | u_n\rangle = \int_{-\infty}^\infty f(x-n)^2\, dx = \lVert f\rVert_{L^2}^2\ne 0.$$
(See here for a proof that $$u_n\rightharpoonup 0$$).
• Thank you! So this ($\langle Au_n, u_n-u\rangle_{X^*\times X}\to 0$) would be true if $u_n\to u$ strongly in $X$ only (I mean without any additional assumptions). – zorro47 Mar 4 '19 at 11:31
• @zorro47 the proof works with any sequence weakly converging to $0$ but not doing so in norm. The specific example does converge weakly to zero, you need to check that $$\int_{\Bbb R}f(x-n)g(x)\,dx$$ converges to zero for any $f,g\in L^2$. This is a measure theory exercise. – s.harp Mar 4 '19 at 12:34
I have another counter example. Let $$X$$ be a Hilbert space and $$A\colon X\to X^*$$ be defined by: $$\langle Au,v\rangle=\langle-u,v\rangle_{X}.$$ Let $$(u_n)$$ be an orthonormal set of vecotrs from $$X$$. Thus, for every $$v\in X$$, by the Bessel inequality $$\sum_{n=1}^{\infty}|\langle u_n,v\rangle|^2\le \|v\|^2_X$$ we deduce that $$\lim\langle u_n,v\rangle=0$$. Since $$X$$ is Hilbert, we have that $$u_n\to0$$ weakly in $$X$$. Then $$\langle Au_n,u_n-u\rangle=\langle Au_n,u_n\rangle=-\langle u_n, u_n\rangle_{X}=-\|u_n\|^2=-1\ne0$$ for all $$n\in \mathbb{N}$$.
• Yes; this is another case of a sequence $u_n$ that converges weakly and does not converge strongly. The operator $A=I$ furnishes a counterexample in this case, too. – Giuseppe Negro Mar 4 '19 at 22:51


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# Overview

## What is NetMonitor?

NetMonitor is a humble Netspeed replacement for gnome-shell.

## What it looks like?

Everybody loves screenshots, right?

## Disclaimer

As I couldn't find any real documentation for writing gnome-shell extensions, I based my code on better or worse snippets and tutorials found on internet. Some of the sources are mentioned below:

## How it works?

The extension connects to NetworkManager via dbus and gets list of available network devices. Then it parses /proc/net/dev file for interfaces' statistics.

I should have used imports.gi.NetworkManager instead of writing all the code myself, but I've found the module too late (did I mention lack of documentation?). Moreover I should have used imports.gi.GTop instead of parsing /proc/net/dev/, but I don't believe it's supported in gnome-shell 3.0.

## Instalation

In order to install gsettings schema, copy org.gnome.shell.extensions.net-monitor.gschema.xml file to /usr/share/glib-2.0/schemas and invoke glib-compile-schemas on that direcory. Note that you need to be root or use sudo to do that:

# cp org.gnome.shell.extensions.net-monitor.gschema.xml /usr/share/glib-2.0/schemas
# glib-compile-schemas /usr/share/glib-2.0/schemas

The NetMonitor@zdyb.tk directory should be copied to /usr/share/gnome-shell/extensions or ~/.local/share/gnome-shell/extensions/:

# cp NetMonitor\@zdyb.tk /usr/share/gnome-shell/extensions

or:

Have fun!

## License

Copyright 2011 Aleksander Zdyb

This program is free software: you can redistribute it and/or modify it under the terms of the GNU General Public License as published by the Free Software Foundation, either version 3 of the License, or (at your option) any later version.

This program is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License for more details.

You should have received a copy of the GNU General Public License along with this program. If not, see http://www.gnu.org/licenses/.


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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# 22.10: Balancing Redox Reactions: Half-Reaction Method

Difficulty Level: At Grade Created by: CK-12
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Practice Balancing Redox Reactions: Half-Reaction Method
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What makes the paint color orange?

The picture above shows one of the two Thunder Dolphin amusement ride trains. This train has an orange stripe while its companion train has a yellow stripe. Pigments of these colors are often made with a dichromate salt (usually sodium or potassium dichromate). These brightly colored compounds serve as strong oxidizing agents in chemical reactions.

### Balancing Redox Equations: Half-Reaction Method

Another method for balancing redox reactions uses half-reactions. Recall that a half-reaction is either the oxidation or reduction that occurs, treated separately. The half-reaction method works better than the oxidation-number method when the substances in the reaction are in aqueous solution. The aqueous solution is typically either acidic or basic, so hydrogen ions or hydroxide ions are present.

In general, the half-reactions are first balanced by atoms separately. Electrons are included in the half-reactions. These are then balanced so that the number of electrons lost is equal to the number of electrons gained. Finally, the two half-reactions are added back together. The example is the oxidation of Fe2+ ions to Fe3+ ions by dichromate (Cr2O72−) in acidic solution. The dichromate ions are reduced to Cr3+ ions.

Potassium dichromate (K2Cr2O7)

Step 1: Write the unbalanced ionic equation.

Fe2+(aq)+Cr2O27(aq)Fe3+(aq)+Cr3+(aq)\begin{align*}\text{Fe}^{2+}(aq)+\text{Cr}_2\text{O}^{2-}_7(aq) \rightarrow \text{Fe}^{3+}(aq)+\text{Cr}^{3+}(aq)\end{align*}

Notice that the equation is far from balanced, as there are no oxygen atoms on the right side. This will be resolved by the balancing method.

Step 2: Write separate half-reactions for the oxidation and the reduction processes. Determine the oxidation numbers first, if necessary.

Oxidation: Fe2+(aq)Fe3+(aq)Reduction: Cr2+6 O27(aq)Cr3+(aq)\begin{align*}& \text{Oxidation:} \ \text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) \\ & \text{Reduction:} \ \overset{\underset{\mathrm{+6}}{}}{\text{Cr}_2} \ \text{O}^{2-}_7(aq) \rightarrow \text{Cr}^{3+}(aq)\end{align*}

Step 3: Balance the atoms in the half-reactions other than the hydrogen and oxygen. In the oxidation half-reaction above, the iron atoms are already balanced. The reduction half-reaction needs to be balanced with the chromium atoms.

Cr2O27(aq)2Cr3+(aq)\begin{align*}\text{Cr}_2\text{O}^{2-}_7(aq) \rightarrow 2\text{Cr}^{3+}(aq)\end{align*}

Step 4: Balance oxygen atoms by adding water molecules to the appropriate side of the equation. For the reduction half-reaction above, seven H2O molecules will be added to the product side.

Cr2O27(aq)2Cr3+(aq)+7H2O(l)\begin{align*}\text{Cr}_2\text{O}^{2-}_7(aq) \rightarrow 2\text{Cr}^{3+}(aq)+7\text{H}_2\text{O}(l)\end{align*}

Now the hydrogen atoms need to be balanced. In an acidic medium, add hydrogen ions to balance. In this example, fourteen H+ ions will be added to the reactant side.

14H+(aq)+Cr2O27(aq)2Cr3+(aq)+7H2O(l)\begin{align*}14\text{H}^+(aq)+\text{Cr}_2\text{O}^{2-}_7(aq) \rightarrow 2\text{Cr}^{3+}(aq)+7\text{H}_2\text{O}(l)\end{align*}

Step 5: Balance the charges by adding electrons to each half-reaction. For the oxidation half-reaction, the electrons will need to be added to the product side. For the reduction half-reaction, the electrons will be added to the reactant side. By adding one electron to the product side of the oxidation half-reaction, there is a 2+ total charge on both sides.

Fe2+(aq)Fe3+(aq)+e\begin{align*}\text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq)+e^-\end{align*}

There is a total charge of 12+ on the reactant side of the reduction half-reaction (14 – 2). The product side has a total charge of 6+ due to the two chromium ions (2 × 3). To balance the charge, six electrons need to be added to the reactant side.

6e+14H+(aq)+Cr2O27(aq)2Cr3+(aq)+7H2O(l)\begin{align*}6e^-+14\text{H}^+(aq)+\text{Cr}_2\text{O}^{2-}_7(aq) \rightarrow 2\text{Cr}^{3+}(aq)+7\text{H}_2\text{O}(l)\end{align*}

Now equalize the electrons by multiplying everything in one or both equations by a coefficient. In this example, the oxidation half-reaction will be multiplied by six.

6Fe2+(aq)6Fe3+(aq)+6e\begin{align*}6\text{Fe}^{2+}(aq) \rightarrow 6\text{Fe}^{3+}(aq)+6e^-\end{align*}

Step 6: Add the two half-reactions together. The electrons must cancel. Balance any remaining substances by inspection. If necessary, cancel out H2O or H+ that appear on both sides.

6Fe2+(aq)6e+14H+(aq)+Cr2O27(aq)14H+(aq)+6Fe2+(aq)+Cr2O27(aq)6Fe3+(aq)+6e2Cr3+(aq)+7H2O(l)6Fe3++2Cr3+(aq)+7H2O(l)\begin{align*}6\text{Fe}^{2+}(aq) & \rightarrow 6\text{Fe}^{3+}(aq)+ \cancel{6e^-} \\ \cancel{6e^-}+14\text{H}^+(aq)+\text{Cr}_2\text{O}^{2-}_7(aq) & \rightarrow 2\text{Cr}^{3+}(aq)+7\text{H}_2\text{O}(l) \\ \hline 14\text{H}^+(aq)+6\text{Fe}^{2+}(aq)+\text{Cr}_2\text{O}^{2-}_7(aq) & \rightarrow 6\text{Fe}^{3+}+2\text{Cr}^{3+}(aq)+7\text{H}_2\text{O}(l)\end{align*}

Step 7: Check the balancing. In the above equation, there are 14 H, 6 Fe, 2 Cr, and 7 O on both sides. The net charge is 24+ on both sides. The equation is balanced.

#### Summary

• The half-reaction method of balancing redox equations is described.

#### Practice

Do all half-reaction equation balancing problems at the link below:

#### Review

Questions

1. Why do you need the unbalanced ionic equation?
2. What does balancing each half-reaction separately tell you?
3. How do you balance hydrogens in an acidic medium?

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## Friday, September 18, 2009

### EZ-Link Giro top-up - Part 8 (EZ-Link top-up far from easy)

New scheme, called EZ-Reload By Giro, charges 25 cents a top-up
The Sunday Times SEPTEMBER 20, 2009, HOT, 3
By Akshita Nanda, akshitan@sph.com.sg

Giro scheme for new cards is not only more costly, but it is also less convenient, say frustrated commuters

It may be called ez-link, but the updated travel fare card that replaces the old one from next month has got commuters feeling decidedly "un-ez".

The "old" ez-link cards cease to be valid from Oct 1 and commuters have been able to exchange them for new ones since January.

However, several of the new card's features introduced in stages over that time have prompted complaints - particularly its Giro top-up system, which users say is more costly and less convenient.

Under the old system, Giro top-up was free. Users could apply ,for this service Instantly at general ticketing machines in any MRT station.

But the new scheme, called EZ-Reload By Giro, charges 25 cents a top-up. Applying for the scheme also involves filling in paperwork and a wait of up to three weeks for the request to be processed.

Frustrated commuters have written in to The Straits Times with their gripes.

Insurance adviser Albert Lim, 60, says: "I find it very unfair. No merchant has charged for Giro before.'' He has opted out of EZ-Reload and will now do his top-ups via machines or ticket offices at transport hubs.

Retired teacher Lim Pok Beng, 70, fears that charging for Giro top-ups will open the floodgates. He asks: ''Will others follow to charge for Giro also?"

He is eligible for a Senior Citizen Concession Card, and happily for him, the Giro top-up fees do not apply to his card. All concession cards, including those for students and children, are still being handled by TransitLink, the transport firm that managed the old Giro top-up system.

The EZ-Reload scheme is being managed by a different transport firm, EZLink.

Another irate commuter is 24-year-old Fiona Seah, who is used to automatic top-ups from her bank account costing her nothing. The marketing executive was a faithful user of TransitLink's Bank Giro Top-up Scheme, which topped up her card every time it was empty adding $20 each time. This would happen about four times a month for her. When EZ-Reload replaced the old system late last month, she queued at a ticketing office to switch over - and then quit the queue on discovering the new charges, which do not stop at the Giro gyration. Apart from the 25-cent fee, a refundable deposit equal to the top-up value a commuter chooses will be deducted from his bank account under the Giro scheme. They also pay a$1.50 application charge, but EZLink is waiving this until February. No such fee was charged under the old system.

If Ms Seah were to opt for EZ-Reload now, she would have to pay an extra $12 a year for automatic top-ups, plus a one-time$20 deposit. She makes the point: ''Is it necessarv for tne refundable deposit to be equal to the reload value?"

EZLink told LifeStyle that it pays first when the top-up value is added to a fare card. Tha. amount is collected only a day later from the customer's account. The one-time deposlt of the reload value is required to manage the risk that it may not be able to get back the cash it paid out.

Automatic top-ups have become more important to users as the new card requires a minimum cash balance of $3 before it can be used to travel on the MRT. The old one did not, as it stored$3 deposit could be used for travel.

Ms Seah is now considering switching to Giro top-up via credit card. There is still a 25-cent fee for each transaction, but she says: "At least I'll get rebates."

Some credit cards give users points for ez-link top-ups that they can use for rebates, discounts or gifts. Commuters who use their credit card to top-up their fare card need not put down a deposit.

On the other hand, the new ez-link card does have some attractive features in that it is Cepas-compliant, which means it can be used as cashless payment for a range of services from bus and train rides to retail purchases and paying Electronic Road Pricing fees. Cepas stands for Contactless e-Purse Application Standard.

After reader's complaints, EZLink defended its charges, saying that the 25-cent fee is to recover the processing and operating costs of the Giro service. It told LifeStyle that 98.5 per cent of the commuters use "free top-up channels" such as ticketing machines to add value to their cards.

"The Giro top-up facility is thus a value-added service offered to commuters to give them more choices", its spokesman said.

However, Mr. Seah Seng Choon, executive director of consumer watchdog Case said: "The 25-cent charge is unnecessary as we believe EZLink should be able to cover this cost through returns generated from float which consumers top up, like TransitLink was able to do before."

He added that the 25-cent "convenience fee" is unfair because commuters need not pay the fee under the old system.
akshitan@sph.com.sg

Picture is obtained from http://cheryltay.files.wordpress.com/2009/08/bmw_z4_design_exterior.jpg

Reference


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## Envision Math 4th Grade Textbook Answer Key Topic 2.9 Subtracting Across Zeros

Draw a Picture and Write an Equation

The mass of a human brain is how much greater than the mass of a chimpanzee brain?

Guided Practice*

Do you know HOW?

Question 1.
In one week, Sandy earned $36 from her babysitting job. She got$15 more for doing her chores. How much money did Sandy earn?

Question 2.
How can you show that 930 grams is a reasonable answer for the question asked above?

Question 3.
Write a Problem Write a problem using the table at the top. Solve. Draw a picture to help you.

Independent Practice

Question 4.
Four cities are on the same road that runs east and west. Fleming is west of Bridgewater but east of Clinton. Union is between Fleming and
Bridgewater. It is 21 miles from Fleming to Union. It is 55 miles from Clinton to Union. How far is it from Clinton to Fleming?

Question 5.
Scott and his friends walk to school together. Scott leaves his home at 7:00 A.M. He meets Johnny and Zach at the end of the block. Next, they meet Paul, Tim, and Pete. Dan and Torey join them one block before the school. How many friends walk to school altogether?

What do I know?
The average mass of a chimpanzee brain is 420 grams. The average mass of a human brain is 1,350 grams.

What am I asked to find?

The difference between the masses

Plan and Solve

Draw a picture.

Write an equation. Use subtraction to solve.
1,350 – 420 =
The human brain has a mass that is 930 grams more than the chimpanzee brain.

Question 6.
The American Kennel Club recognizes 17 breeds of herding dogs and 26 breeds of terriers. Draw a picture that could help find the total number of herding dogs and terriers.

Question 7.
Using the information in Exercise 6, write an equation to find how many more breeds of terriers than herding dogs there are.

For 8 through 10, use the table to the right.

Question 8.
There are about 200 more animals in the Minnesota Zoo than in the Phoenix Zoo. About how many species of animals are in the Minnesota Zoo?

Question 9.
About how many more species are in the Indianapolis Zoo than the Phoenix Zoo?

Question 10.
How can you find the number of species of animals at the San Francisco Zoo?

Question 11.
A parking lot had a total of 243 cars in one day. By 6:00 A.M., there were 67 cars in the lot. In the next hour, 13 more cars joined these. How many more cars would come to the lot by the end of the day?

Question 12.
A shoe store sold 162 pairs of shoes. The goal was to sell 345 pairs. How many pairs of shoes did they NOT sell?

Independent Practice

For 13 and 14, use the table at the right.

Question 13.
What equation can you write to help find the cost of the shoes and socks together?

Question 14.
What equation can you write to help find the difference between the cost of the shirt and the shorts?

Question 15.
Byron spent $7.75 on popcorn and a drink at the movie theater. The popcorn was$4.25. How much was the cost of the drink?

Question 16.
Each school day, Mikaela sold the same number of tickets to the school play. On Monday she sold 4 tickets. How many tickets did she sell all together in 5 days?

Question 17.
Writing to Explain Ken makes 2 nametags in the time it takes Mary to make 5 nametags. When Mary has made 15 nametags, how many has Ken made?

Question 18.
Mr. Lee had 62 pencils at the beginning of the school year. At the end of the school year he had 8 pencils left. How many pencils were given out during the year?

Question 19.
Carlene bought a book for $13.58. She paid with a$10 bill and a $5 bill. Which expression would find the amount of change Carlene would receive? A.$15 – $13.58 B.$15 – $1.42 C.$10 + $5 D.$13.58 + \$1.42

Question 20.
Terrence rode 15 rides before lunch at the county fair. He rode 13 rides after lunch. Each ride requires 3 tickets. Which expression
represents the number of rides he rode during the day?
A. 15 – 13
B. 15 + 13
C. 15 – 3
D. 13 – 3

Stop and Practice

Find each sum. Estimate to check if the answer is reasonable.

Question 1.

Question 2.

Question 3.

Question 4.

Question 5.
5,474 + 723

Question 6.
47,090 + 2,910

Question 7.
6,685 + 37

Find each difference. Estimate to check if the answer is reasonable.

Question 8.

Question 9.

Question 10.

Question 11.

Question 12.

Question 13.
5,302 – 88

Question 14.
2,249 – 51

Question 15.
8,001 – 4,832

Error Search Find each sum or difference that is not correct. Write it correctly and explain the error.

Question 16.

Question 17.

Question 18.

Question 19.

Question 20.

Number Sense

Estimating and Reasoning Write whether each statement is true or false. Explain your answer.

Question 21.
The number 213,753 is ten thousand more than 223,753.

Question 22.
The sum of 6,823 and 1,339 is greater than 7,000 but less than 9,000.

Question 23.
The sum of 42,239 and 11,013 is less than 50,000.


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In many books and derivations of the Black-Scholes PDE one sees that

$$\Pi=V-\Delta F \Rightarrow d\Pi=dV-\Delta dF$$

which implicitly assumes that $d\Delta=0$. Somewhere down the road one then deduces that

$$\Delta=\frac{\partial V}{\partial F}$$

to simplify the equation. Doesn't this contradict the initial assumption that $d\Delta=0$? If one performs a full differentiation

$$\Pi=V-\Delta F \Rightarrow d\Pi=dV-\Delta dF - F d\Delta$$

the rest of the story goes wrong. Isn't it true that $\Delta = \Delta(t, S)$, i.e. is depending on time and the underlying stochastic process and hence has to be differentiated?

-
The delta hedge ensures that the portfolio is self-financing, i.e. there is not change in portfolio value. There is an explanation on the Wikipedia page: en.wikipedia.org/wiki/Black-Scholes_equation –  user13247 May 21 at 11:13


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# Math Help - Easy Problem made frustrating

1. ## Easy Problem made frustrating

d^2y/dx^2 = 0.5dy/dx where y=0 and dy/dx=3 when x=0
We have to find what y is when x=5
My solution keeps turning out to be y=6e^5/2 - 6
However the back of my textbook says y=6e^5/2 - 6e^2
Here is how i did it
let v = dy/dx
therefore d^2y/dx^2 = vdv/dy
vdv/dy = 0.5v
dv=0.5dy ... intergrate both sides
v=0.5y + C
3=0.5(0) + C
C=3
v=0.5y + 3
dy/dx=0.5y + 3
2dy/y+6 = dx
2loge(y+6) = x + B
B = 2loge(6)
2loge(y+6/6)=x
y+6/6=e^x/2
y=6e^5/2 - 6
Where have I gone wrong and what is the correct way to do it if i am wrong???

2. Ok, so 2nd derivative is .5 times first derivative, means if that was your differential equation, then y=C1*e^(.5*x) is one fundamental solution, and y=C2 is the other. At x=0, there are the boundary conditions y=0 and y'=3. You stated it really badly by the way. The first few times I read it, I thought you meant d^2y/dx^2 = 0.5dy/dx was true only at y=0, and dy/dx=3 is true when x=0, which would be 2 boundary conditions and no differential equation. So obviously C1=-C2 so that y(0)=0. Then y'=.5*C1=3 at x=0, so C1=6. y=6*(e^(.5*x)-1). If x=5, then y=6*(e^(2.5)-1). Apparently you didn't go wrong, since you got the same answer I did, although I never involved the natural log.

3. sorry about mix up, thks for clearing this up, i don't understand your method, (i'm pretty useless), but i'm relieved you got the same answer. now i can sleep in peace lol.


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# Smooth Molecular Decompositions of Functions and Singular Integral Operators

by
Paperback
Under minimal assumptions on a function $\psi$ we obtain wavelet-type frames of the form $\psi_{j,k}(x) = r^{(1/2)n j} \psi(r^j x - sk), j \in \integer, k \in \integer^n,$ for some $r > 1$ and $s > 0$. This collection is shown to be a frame for a scale of Triebel-Lizorkin spaces (which includes Lebesgue, Sobolev and Hardy spaces) and the reproducing formula converges in norm as well as pointwise a.e. The construction follows from a characterization of those operators which are bounded on a space of smooth molecules. This characterization also allows us to decompose a broad range of singular integral operators in terms of smooth molecules.


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## Advice on data and code

I have been reading a few papers on how to structure data and code. In the post, I provide a list of the papers I have found together with the main advice/rules offered in the respective papers (do consult the individual papers for examples and explanations).

Noteworthy, there is an overlap in the advice the papers give. I do not agree with everything (and as you can see, they are written for people working with tabular data and definitely not people working with more modern workflows with several gigabytes of data stored in NoSQL databases). However, overall, I can highly recommend that you take most of these recommendations on board.

Last, another good resource is the supplementary material to A Practical Guide for Transparency in Psychological Science. This paper deals with the details of folder structure, data documentation, analytical reproducibility, etc.

Here we go:

#### Nagler (1995): Coding Style and Good Computing Practice

1. Maintain a lab book from the beginning of a project to the end.
2. Code each variable so that it corresponds as closely as possible to a verbal description of the substantive hypothesis the variable will be used to test.
3. Correct errors in code where they occur, and rerun the code.
4. Separate tasks related to data manipulation vs. data analysis into separate files.
5. Design each program to perform only one task.
6. Do not try to be as clever as possible when coding. Try to writecode that is as simple as possible.
7. Set up each section of a program to perform only one task.
8. Use a consistent style regarding lower- and upper-case letters.
9. Use variable names that have substantive meaning.
10. Use variable names that indicate direction where possible.
11. Use appropriate white space in your programs, and do so in a consistent fashion to make the programs easy to read.
12. Include comments before each block of code describing the purpose of the code.
13. Include comments for any line of code if the meaning of the line will not be unambiguous to someone other than yourself.
14. Rewrite any code that is not clear.
15. Verify that missing data are handled correctly on any recode or creation of a new variable.
16. After creating each new variable or recoding any variable, produce frequencies or descriptive statistics of the new variable and examine them to be sure that you achieved what you intended.
17. When possible, automate things and avoid placing hard-wired values (those computed “by hand”) in code.

#### Broman and Woo (2018): Data Organization in Spreadsheets

1. Be consistent
– Use consistent codes for categorical variables.
– Use a consistent fixed code for any missing values.
– Use consistent variable names.
– Use consistent subject identifiers.
– Use a consistent data layout in multiple files.
– Use consistent file names.
– Use a consistent format for all dates.
– Use consistent phrases in your notes.
– Be careful about extra spaces within cells.
2. Choose good names for things
– Don’t use spaces, either in variable names or file names.
– Avoid special characters, except for underscores and hyphens.
– Keep names short, but meaningful.
– Never include “final” in a file name
3. Write dates as YYYY-MM-DD
4. Fill in all cells and use some common code for missing data.
5. Put just one thing in a cell
6. Don’t use more than one row for the variable names
7. Create a data dictionary
8. No calculations in the raw data files
9. Don’t use font color or highlighting as data
10. Make backups
11. Use data validation to avoid errors
12. Save the data in plain text files

#### Balaban et al. (2021): Ten simple rules for quick and dirty scientific programming

1. Think before you code
3. Look for opportunities for code reuse
5. Avoid premature optimization
6. Use automated unit testing for critical components
7. Refactor frequently
8. Write self-documenting code for programmers and a readme file for users
10. Go explore and be rigorous when you publish

#### Wilson et al. (2017): Good enough practices in scientific computing

1. Data management
– Save the raw data.
– Ensure that raw data are backed up in more than one location.
– Create the data you wish to see in the world.
– Create analysis-friendly data.
– Record all the steps used to process data.
– Anticipate the need to use multiple tables, and use a uniquei dentifier for every record.
– Submit data to a reputable DOI-issuing repository so that others can access and cite it.
2. Software
– Place a brief explanatory comment at the start of every program.
– Decompose programs into functions.
– Be ruthless about eliminating duplication.
– Always search for well-maintained software libraries that do what you need.
– Test libraries before relying on them.
– Give functions and variables meaningful names.
– Make dependencies and requirementsexplicit.
– Do not comment and uncomment sections of code to control a program’s behavior.
– Provide a simple example or test dataset.
– Submit code to a reputable DOI-issuing repository.
3. Collaboration
– Create an overview of your project.
– Create a shared “to-do” list for theproject.
– Decide on communication strategies.
– Make the project citable.
4. Project organization
– Put each project in its own directory, which is named after the project.
– Put text documents associated with the project in the doc directory.
– Put raw data and metadata in a data directory and files generated during cleanup and analysis in a results directory.
– Put project source code in the src directory.
– Put external scripts or compiled programs in the bin directory.
– Name all files to reflect their contentor function.
5. Keeping track of changes
– Backup (almost) everything created by a human being as soon as it is created.
– Keep changes small.
– Share changes frequently.
– Create, maintain, and use a checklist for saving and sharing changes to the project.
– Store each project in a folder that is mirrored off the researcher’s working machine.
– Add a file called CHANGELOG.txt to the project’s docs subfolder.
– Copy the entire project whenever a significant change has been made.
– Use a version control system.
6. Manuscripts
– Write manuscripts using online tools with rich formatting, change tracking, and reference management.
– Write the manuscript in a plain text format that permits version control.

#### Gentzkow and Shapiro (2014): Code and Data for the Social Sciences

1. Automation
– Automate everything that can be automated.
– Write a single script that executes all code from beginning to end.
2. Version Control
– Store code and data under version control.
– Run the whole directory before checking it back in.
3. Directories
– Separate directories by function.
– Separate files into inputs and outputs.
– Make directories portable.
4. Keys
– Store cleaned data in tables with unique, non-missing keys.
– Keep data normalized as far into your code pipeline as you can.
5. Abstraction
– Abstract to eliminate redundancy.
– Abstract to improve clarity.
– Otherwise, don’t abstract.
6. Documentation
– Don’t write documentation you will not maintain.
– Code should be self-documenting.
7. Management
– E-mail is not a task management system.
8. Code style
– Keep it short and purposeful.
– Use descriptive names.
– Pay special attention to coding algebra.
– Make logical switches intuitive.
– Be consistent.
– Check for errors.
– Write tests.
– Profile slow code relentlessly.
– Separate slow code from fast code.

## Replace equations with code

Here is a suggestion: In empirical research, academics should move equations from the methods section to the appendix and, if anything, show the few lines of code used to estimate the model(s) in the software being used (ideally with citations to the software and statistical packages). Preferably, it should be possible to understand the estimation strategy without having to read any equations.

Of course, I am talking about the type of work that is not primarily interested in developing a new estimator or a formal theory that can be applied to a few case studies (or shed light on the limitations of empirical models). I am not against the use of equations or abstractions of any kind to communicate clearly and without ambiguity. I am, however, skeptical towards how empirical research often include equations for the sake of … including equations.

I have a theory that academics, and in particular political scientists, put more equations in their research to show off their skills rather than to help the reader understand what is going on. In most cases, equations are not needed and are often there only to impress reviewers and peers, which of course are the same people (hence, peer-review). The use of equations are excluding readers rather than including readers.

I am confident that most researchers spend more time in their favourite statistical IDE than they do writing and reading equations. For that reason, I also believe that most researchers will find it easier to read actual code instead of equations. Take this example on the equation and code for a binomial regression model (estimated with glmer()) from Twitter:

Personally, I find it much easier to understand what is going on when I look at the R code instead of the extracted equation. Not only that, I also find it easier to think of potential alternatives to the regression model, e.g., that I can easily change the functional form and see how such changes will affect the results. This is something I rarely consider when I only look at equations.

The example above is from R, and not all researchers use or understand R. However, I am quite certain that everybody that understands the equation above will also be able to understand the few lines of code. And when people use Stata, it is often even easier to read the code (even if you are not an avid Stata user). SPSS syntax is much more difficult to read but that says more about why you should not use SPSS in the first place.

I am not against the use of equations in research papers. However, I do believe empirical research would be much better off by showing and citing code instead of equations. Accordingly, please replace equations with code.


text:
# Generic Repository Configuration

 This page describes Configuration for Generic repository. Since midPoint 4.4 we recommend using the new Native repository. Configuration for Native repository is described here.

By default, the repository configuration is stored in the $HOME/midpoint directory of the user the application server runs as. This directory is further referenced to as midpoint.home. This directory will be created upon the very first start of midPoint. The configuration file config.xml will be generated if it does not exist and it will be pre-configured to use the embedded H2 repository. You can first start midPoint with embedded H2 repository and then reconfigure the created config.xml to use another database, or you can prepare config.xml before the midPoint starts for the very first time using the sample configurations (see the child pages below). ## Configuration options Default values are used only if repository is in embedded mode, otherwise configuration validation fails. Option Description Default database Simplified option for repository setup. Possible values are h2, mysql, oracle, sqlserver, postgresql. This option will set defaults for other options, for example embedded, hibernateHbm2ddl, hibernateDialect and driverClassName based on the selected database. These defaults can be overridden by specifying custom values in configuration. h2 dropIfExists Drops database files if they exist during start. Useful for tests. false baseDir Directory where H2 files will be saved if we’re running in embedded mode. <baseDir>${midpoint.home}</baseDir> can be used if we want to store H2 db files in midpoint.home directory

current folder "."

fileName

Database filename. Name for H2 files if we’re running in embedded mode.

midpoint

embedded

Embedded H2 mode.

true

asServer

This option can be used if we’re running in H2 embedded mode. If the server mode is turned on, H2 runs with TCP server. Other applications/services can connect to H2 server. If false, H2 runs in file mode.

false

tcpSSL

Embedded H2 server mode SSL.

false

port

Embedded H2 server mode port.

5437

hibernateHbm2ddl

Automatically validates or exports schema DDL to the database when the SessionFactory is created. E.g. validate | update | create | create-drop | none. With create-drop the database schema will be dropped when the SessionFactory is closed explicitly. For production environments, validate should be used (before midPoint 3.9) or none (3.9 and later). Please see Schema creation and updating section later.

for H2 update, otherwise none (was validate before 3.9)

hibernateDialect

SQL dialect based on chosen DB Supported hibernate dialects.

org.hibernate.dialect.H2Dialect

dataSource

Example: <dataSource>java:comp/env/jdbc/midpoint</dataSource>

sa (if embedded=true), null (otherwise)

empty string (if embedded=true), null (otherwise)

jdbcUrl

URL for JDBC connection.

if embedded=true url is computed from previous parameters

driverClassName

Driver class name for JDBC connection.

org.h2.Driver (if embedded=true)

fullObjectFormat

Property specifies format (or language) used to store serialized object representation (m_object.fullObject). Supported values are xml and json. This is safe to change any time, objects are read properly regardless of the format they are stored in.

xml

useZip

Property provides optional compression for column storing serialized object representation (m_object.fullObject).

Please, consult your RDBMS documentation as some databases may do this transparently behind the scenes or even in addition to this setting which is likely undesired.

false

useZipAudit

(Since 4.0.3, 4.1.1, 4.2.1, 4.3.) Property controlling whether serialized delta and its operation result in audit is compressed (m_audit_delta.delta/fullResult).
Important: Because this was on by default since 3.8 and the option was introduced much later, its default (true) reflects the situation in 3.8 and later.
If separate audit repository is used, this option must be used there. If single repository is used (by default), it must be used in main repository configuration.

Please, consult also your RDBMS documentation to see if this is necessary and not useless or harmful.

true

minPoolSize

Minimal # of connections in connection pool, if connection pool is not provided through dataSource.

8

maxPoolSize

Maximum # of connections in connection pool, if connection pool is not provided through dataSource.

20

initializationFailTimeout

Hikari pool initialization failure timeout, in milliseconds. It is there to allow midPoint to wait until the repository is up and running and therefore to avoid failing prematurely. Introduced in midPoint 3.9.

1 ms (effectively keeping the behavior as it was before midPoint 3.9)

 Since 4.2 This functionality is available since version 4.2.

It is possible to connect to the database without specifying password or username or both. Examples are PostgreSQL trust authentication (though definitely not recommended for serious deployments) or Microsoft SQL Server with integrated security.

Simply skip configuration elements jdbcUsername and jdbcPassword. If everything is configured as expected, connection will be successful, otherwise JDBC driver will throw an exception and midPoint will not start.

See MID-5342.

## Schema creation and updating

 Since 3.9 This functionality is available since version 3.9.

In midPoint 3.9 we have implemented a more flexible and powerful approach to schema validation and maintenance. It replaces the standard Hibernate ORM approach. It is enabled by setting hibernateHbm2ddl parameter to none, which is now the default for non-H2 databases.

What it does:

1. First, it determines the state of the database schema by:

1. running standard Hibernate schema validation procedure (just like validate option for hibernateHbm2ddl would do),

2. examining explicit schema version by looking at parameter databaseSchemaVersion in m_global_metadata table. This is a new table introduced in midPoint 3.9.

2. Then it acts upon these data, either by

1. continuing with the midPoint startup process,

2. stopping the midPoint startup process with an appropriate error message,

3. or trying to remediate the situation e.g. by running a schema creation or schema upgrade SQL script.

Schema validation and maintenance is the driven by these configuration options:

Option Description Default

skipExplicitSchemaValidation

Whether to skip this process of explicit schema validation.

• true (i.e. "skip") if hibernateHbm2ddl is validate, update, create, or create-drop;

• false (i.e. "do not skip") otherwise (e.g. if it is none which is the default for non-H2 databases)

missingSchemaAction

What to do if the database schema is not present:

• stop: midPoint startup process is stopped with an appropriate explanation message.

• warn: midPoint startup process continues (with a warning message), very probably to be crashed soon because of a repository access failure. This option is therefore not recommended;

• create: midPoint tries to create the schema using appropriate SQL script. Then it checks the schema for validity again and stops if it’s (still) invalid.

stop

What to do if the database schema is present but it is outdated and it seems to be upgradeable:

• stop: midPoint startup process is stopped with an appropriate explanation message;

• warn: midPoint startup process continues (with a warning message), very probably to be crashed sooner or later because of a repository access failure. This option is therefore not recommended.

• If possible, midPoint tries to upgrade the schema by running appropriate SQL script. Then it checks the schema for validity again and stops if it’s (still) invalid.

• If not possible, midPoint acts as in stop case: outputs a message and stops.

Note that currently (as of 3.9) the only supported automated upgrade is from 3.8 to 3.9. Please consider carefully whether you want to run this automatic upgrade also for the production environment. It is perhaps better to still run the upgrade manually in such a situation.

stop

incompatibleSchemaAction

What to do if the database schema is present, is not compatible and not upgradeable. A typical example is when the schema is newer than the current version of midPoint.

• stop: midPoint startup process is stopped with an appropriate explanation message.

• warn: midPoint startup process continues (with a warning message), very probably to be crashed sooner or later because of a repository access failure. This option is therefore not recommended.

stop

schemaVersionIfMissing

If the schema version cannot be determined from m_global_metadata table e.g. because the table does not exist, it is possible to specify it using this parameter. It applies only if the version is missing in the database.

(none)

schemaVersionOverride

Overrides any schema version information in the m_global_metadata table.

(none)

schemaVariant

Used to specify what schema variant is to be used for automated creation or upgrade of the database schema. Currently, the only known variant is utf8mb4 for MySQL/MariaDB. Beware: it is the administrator’s responsibility to choose the correct variant! MidPoint does not try to determine the variant present in the database. So be sure to avoid applying e.g. mysql-upgrade-3.8-3.9-utf8mb4.sql if the database is not in utf8mb4 character set, or vice versa.

(none)

createMissingCustomColumns

(Since 4.2) If true, midPoint tries to alter tables with custom columns (currently only in audit) if the column is missing. Intended for test, not for production usage.

false (no columns are created)

### Schema creation and updating (before 3.9)

In earlier versions of midPoint the schema creation and update is driven solely by the hibernateHbm2ddl parameter. For production environments it is strongly recommended setting it to validate that is the default value for non-H2 databases. Then you have to maintain it manually using SQL scripts which are located in the distribution package.

For current unreleased MidPoint SQL scripts are located in our git.

## Data source configuration

Instead of putting JDBC configuration to config.xml, you can use data source of the application server.

 Obsolete This is obsolete functionality. It is no longer supported or maintained. It is relevant only for WAR deployments which are not recommended deployment model since midPoint 4.0.

There are two steps for configuring data sources. Data source configuration is common for all supported databases.

1. First step is DB resource configuration in application server. Here is example for Tomcat 7. This XML part is located in <tomcat-location>/conf/server.xml, resource will be available for all applications in the container.

<GlobalNamingResources>
<Resource name="jdbc/mysql" auth="Container" type="javax.sql.DataSource"
url="jdbc:mysql://localhost:3306/midpoint-big"
driverClassName="com.mysql.jdbc.Driver"
accessToUnderlyingConnectionAllowed="true"
initialSize="5" maxWait="5000"
maxActive="30" maxIdle="5"
validationQuery="select 1"
poolPreparedStatements="true"/>
</GlobalNamingResources>

Also configure <tomcat-location>/conf/context.xml file:

<?xml version='1.0' encoding='utf-8'?>
<Context>
<WatchedResource>WEB-INF/web.xml</WatchedResource>
global="jdbc/mysql"
type="javax.sql.DataSource"/>
</Context>
2. Next step is configuration in file config.xml located in midpoint.home folder. hibernateDialect depends on your DB choice, dataSource is based on resource name.

<configuration>
<midpoint>
<repository>
<repositoryServiceFactoryClass>com.evolveum.midpoint.repo.sql.SqlRepositoryFactory</repositoryServiceFactoryClass>
<embedded>false</embedded>
<hibernateDialect>com.evolveum.midpoint.repo.sql.util.MidPointMySQLDialect</hibernateDialect>
<hibernateHbm2ddl>validate</hibernateHbm2ddl>
<dataSource>java:comp/env/jdbc/mysql</dataSource>
</repository>
</midpoint>
</configuration>

## Index tuning

Anything that is externalized into columns and related tables (like extensions, references, etc.) is effectively searchable using Query API. It is not possible to search for information stored only in the serialized form of the object. But searchable and efficiently searchable are two different things.

For some tables (object types) and some columns no indexes are needed, but for others they typically are. MidPoint is provided with all essential indexes out of the box. Despite that it is possible to come up with real-life queries that will perform badly. While it is possible to cover nearly all needs by more and more indexes, it is not necessarily a good idea to have them all created by default. Indexes also take space and if not necessary only add cost to insert/update operation without really helping.

For any non-trivial installation it is recommended to check the database performance logs/statistics regularly to identify sluggish queries. When identified check existing (predefined) indexes whether they should have covered the case and investigate why they didn’t. If the existing index does not cover the case, don’t be afraid to add the index according to your specific needs.

Following notes and tips can be helpful:

• Don’t index each column separately if the critical query uses multiple where conditions, use multi-column index instead. Order of columns is important.

• Searching using like (especially with % at the start of the value) or case-insensitive search often require specific indexes. Consult your database resources; some databases don’t offer function-based index and indexing the column using lower/upper (depending on the used query) may not be possible. Some databases offer specialized indexes, e.g. PostgreSQL trigram indexes that can significantly boost performance.

• In general, don’t index columns with low-cardinality (e.g. boolean or integer representing enum with just a few distinct values) alone. Leave the column unindexed and let other indexes do the job first. Searching only by such a column is not recommended. It is however possible to use low-cardinality column in multi-column index, and it may be beneficial when it is the first column (of course, only selects using the column in WHERE use such an index).

• It is possible to use where clause in an index when only specific values are selected often , e.g. value indicating active user. This is called partial index. This is typical for low variability columns (booleans, enums), using them in where part of the index is good, and it also makes the index size smaller.

 This technique is not possible with midPoint and Oracle, because Oracle emulates partial indexes with function based indexes and midPoint cannot generate WHERE clause using the needed function.