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# 4.5 Gravitational field due to rigid bodies

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A rigid body is an aggregation of small elements, which can be treated as point mass.

## Gravitational field of rigid bodies

We shall develop few relations here for the gravitational field strength of bodies of particular geometric shape without any reference to Earth’s gravitation.

Newton’s law of gravitation is stated strictly in terms of point mass. The expression of gravitational field due to a particle, as derived from this law, serves as starting point for developing expressions of field strength due to rigid bodies. The derivation for field strength for geometric shapes in this module, therefore, is based on developing technique to treat a real body mass as aggregation of small elements and combine individual effects. There is a bit of visualization required as we need to combine vectors, having directional property.

Along these derivations for gravitational field strength, we shall also establish Newton’s shell theory, which has been the important basic consideration for treating spherical mass as point mass.

The celestial bodies - whose gravitational field is appreciable and whose motions are subject of great interest - are usually spherical. Our prime interest, therefore, is to derive expression for field strength of solid sphere. Conceptually, a solid sphere can be considered being composed of infinite numbers of closely packed spherical shells. In turn, a spherical shell can be conceptualized to be aggregation of thin circular rings of different diameters.

The process of finding the net effect of these elements fits perfectly well with integration process. Our major task, therefore, is to suitably set up an integral expression for elemental mass and then integrate the elemental integral between appropriate limits. It is clear from the discussion here that we need to begin the process in the sequence starting from ring -->spherical shell -->solid sphere.

## Gravitational field due to a uniform circular ring

We need to find gravitational field at a point “P” lying on the central axis of the ring of mass “M” and radius “a”. The arrangement is shown in the figure. We consider a small mass “dm” on the circular ring. The gravitational field due to this elemental mass is along PA. Its magnitude is given by :

$dE=\frac{Gdm}{{PA}^{2}}=\frac{Gdm}{\left({a}^{2}+{r}^{2}\right)}$

We resolve this gravitational field in the direction parallel and perpendicular to the axis in the plane of OAP.

$d{E}_{||}=dE\mathrm{cos}\theta$

$d{E}_{\perp }=dE\mathrm{sin}\theta$

We note two important things. First, we can see from the figure that measures of “y” and “θ” are same for all elemental mass. Further, we are considering equal elemental masses. Therefore, the magnitude of gravitational field due any of the elements of mass “dm” is same, because they are equidistant from point “P”.

Second, perpendicular components of elemental field intensity for pair of elemental masses on diametrically opposite sides of the ring are oppositely directed. On integration, these perpendicular components will add up to zero for the whole of ring. It is clear that we can assume zero field strength perpendicular to axial line, if mass distribution on the ring is uniform. For uniform ring, the net gravitational intensity will be obtained by integrating axial components of elemental field strength only. Hence,

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# Changes

## Conservation of Angular Momentum

27 bytes removed, 20:47, 13 December 2016
Maths formatting
Solving the above four equations yields:
:$\cos{\theta } = 1 - \frac{I \omega^2}{gl(2m+M)}$
Plugging in for angular velocity from the initial equations above yields:
:$\cos{\theta } = 1 - \frac{m^2 v^2 l}{Ig(2m+M)}$
Calculating the moment of inertia $I$ now becomes necessary for a rod of length ''$l''$ and mass ''$M''$, with a small block of mass ''$m''$ at its end.
An ordinary rod of length ''$l''$ has the following moment of inertia relative to an axis of rotation at one end:
:$\mathbf{I}= \frac{M}{3}\times{l^2}$
A The [[moment of inertia]] is additive, and defined as follows:
::$I \ \stackrel{\mathrm{def}}{=}\ \sum_{i=1}^{N} {m_{i} r_{i}^2}\,\!$
where ''$m''$ is the mass at each (perpendicular) distance ''$r''$ from the axis of rotation.
Thus the moment of inertia ''$I''$ for a rod of length ''$l''$ and mass ''$M''$, with a small block of mass ''$m''$ at its end is simply this:
:$\mathbf{I}= \frac{M}{3}\times{l^2} + m\times{l^2}$
which is:
:$\mathbf{I}= \frac{1}{3}\times(3m+M)\times{l^2}$
Plugging this back into the unsolved equation above yields:
:$\mathbf{cos{\theta} = 1 - \frac{{3m^2}{v^2}}{l\times{g}\times{lg (2m+M)(3m+M)}}$
If we complicate the problem further by assuming the small block began with velocity zero from an incline of height h, then applying conservation of energy to the moment in time just prior to its collision with the rod yields the following velocity of impact:
:$\mathbf{\frac{m\times{v^2}}{2}}=m\times{g}\times{h}mgh$
and hence
:$\mathbf{v^2}=2gh$
and thus the solution is:
:$\cos{\theta } = 1 - \frac{{6m^2}{h}}{l(2m+M)(3m+M)}$
or
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# What is the vertex form of y=x^4+2x^2-2?

Feb 4, 2016

I think the best we can do is
$\textcolor{w h i t e}{\text{XXX}} y = {\left({x}^{2} + 1\right)}^{2} - 3$

#### Explanation:

A reference to "the vertex form" implies that we are dealing with a parabolic equation; $y = {x}^{4} + 2 {x}^{2} - 2$ is not a parabolic equation.

We could replace ${x}^{2}$ with $z$, so
$\textcolor{w h i t e}{\text{XXX}} y = {z}^{2} + 2 z - 2$

color(white)("XXX")y=(z^2+2zcolor(red)(+1)-2color(red)(-1)
$\textcolor{w h i t e}{\text{XXX}} y = {\left(z + 1\right)}^{2} - 3$
which is a parabolic equation with the vertex at $\left(\textcolor{b l u e}{z} , y\right) = \left(- 1 , - 3\right)$

Replacing $z$ with ${x}^{2}$ gives
$\textcolor{w h i t e}{\text{XXX}} y = \left({x}^{2} + 1\right) - 3$
but as noted, this is not a parabola (see graph below)
graph{x^4-2x^2-2 [-6.783, 7.26, -4.515, 2.505]}


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Proof or counterexample. Here $v, w, z$ are vectors in a real inner product space $H$.
a) Let $v, w, z$ be vectors in a real inner product space. If $\langle v, w\rangle=0$ and $\langle v, z\rangle=0$, then $\langle w, z\rangle=0$.
b) If $\langle v, z\rangle=\langle w, z\rangle$ for all $z \in H$, then $v=w .$
c) If $A$ is an $n \times n$ symmetric matrix then $A$ is invertible.
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a. Counterexample: Let $H=\mathbb{R}^2$, $v=(1, 0)$, $w=(0, 1)$ and $z=(1, 1)$. Then $\langle v, w\rangle=0$ and $\langle v, z\rangle=0$, but $\langle w, z\rangle=1 \neq 0$.

b. Counterexample: Let $H=\mathbb{R}^2$, $v=(1, 0)$ and $w=(1, 1)$. Then $\langle v, z\rangle=\langle w, z\rangle$ for all $z \in H$, but $v \neq w$.

c. True: Since $A$ is symmetric, it is a self-adjoint operator and thus has an orthonormal basis of eigenvectors. Since the eigenvalues are non-zero, the matrix is invertible.

by SIlver Status (12,174 points)

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# An example of a regular function over an open set

Look at the example 2.1.7 at page 19 of these notes (which is the same example present in Mumford's red book at page 21). The author shows that a regular function isn't a ratio of two polynomial, and in particular in the above example inside the open set $U$ we have the following regular function: $$\phi(p)=\left\{\begin{array} {lll} \frac{X_1}{X_2} & \textrm{if p\in X_2\neq 0}\\\\ \frac{X_3}{X_4} & \textrm{if p\in X_4\neq 0} \end{array}\right.$$

I agree with the fact that $\frac{X_1}{X_2}=\frac{X_3}{X_4}$ in $K(X)$ when $X_2\neq0$ and $X_4\neq0$, but the point is that such two functions shoud be the same in the whole $V$ to make sensible the example. Infact in in the intersection $\{X_2\neq0\}\cap\{X_4\neq0\}$ we have the regular function described by $\frac{X_1}{X_2}$ or $\frac{X_3}{X_4}$, but what about the rest of $U$?

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But the point is that a single formula cannot work everywhere, so you shouldn't expect them to agree everywhere because they are not even defined everywhere on the open set! – Dylan Moreland May 25 '12 at 16:12

The situation is that we have a cone $Y\subset \mathbb A^4_k$ of equation $X_1X_4=X_2X_3$, which contains the closed plane $P\subset Y$ given by $X_2=X_4=0$.
We are interested in the complementary open subset $U=Y\setminus P\subset Y$ and its ring of regular functions $\mathcal O_Y(U)$.
The function $\phi\in \mathcal O_Y(U)$ that you describe, which has $P$ as its polar set , cannot be written $\phi=\frac{f\mid U}{g\mid U}$ with $f,g\in \mathcal O_Y(Y)$ (where $\mathcal O_Y(Y)$ consists of restrictions to $Y$ of genuine polynomials of $k[X_1,X_2,X_3,X_4]$).
The reason is that if we had such a description of $\phi$, then the plane $P$ would be the zero locus of the single regular function $g\in \mathcal O_Y(Y)$ i.e. would be a Cartier divisor, and this is actually not the case.


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#### NEO: Non Equilibrium Sampling on the Orbits of a Deterministic Transform

Sampling from a complex distribution $\pi$ and approximating its intractable normalizing constant $\mathrm{Z}$ are challenging problems. In this paper, a novel family of importance samplers (IS) and Markov chain Monte Carlo (MCMC) samplers is derived. Given an invertible map $\mathrm{T}$, these schemes combine (with weights) elements from the forward and backward Orbits through points sampled from a proposal distribution $\rho$. The map $\mathrm{T}$ does not leave the target $\pi$ invariant, hence the name NEO, standing for Non-Equilibrium Orbits. NEO-IS provides unbiased estimators of the normalizing constant and self-normalized IS estimators of expectations under $\pi$ while NEO-MCMC combines multiple NEO-IS estimates of the normalizing constant and an iterated sampling-importance resampling mechanism to sample from $\pi$. For $\mathrm{T}$ chosen as a discrete-time integrator of a conformal Hamiltonian system, NEO-IS achieves state-of-the art performance on difficult benchmarks and NEO-MCMC is able to explore highly multimodal targets. Additionally, we provide detailed theoretical results for both methods. In particular, we show that NEO-MCMC is uniformly geometrically ergodic and establish explicit mixing time estimates under mild conditions.


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# David Marr's tri-levels of computational neuroscience

• Computational level

• what does the system do (e.g.: what problems does it solve or overcome) and similarly, why does it do these things
• 最適化の評価関数を決める
• Algorithmic/representational level

• how does the system do what it does, specifically, what representations does it use and what processes does it employ to build and manipulate the representations
• 事象をどう表現（数式化、数値化）するか決める
• Implementational/physical level

• how is the system physically realized (in the case of biological vision, what neural structures and neuronal activities implement the visual system)
• 脳のどの部位がどの計算をしているかを決める


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Thus meV stands for milli-electron volt. Modern English Version (Bible edition) MEV. The symbol for volt is V. There are 0.000000001 volts in a nanovolt. Ultraviolet (UV) Extreme UV: EUV: 1240 - 12.4: 1 - 100: 1e7 - 1e5: 3e5 - 3e3: 0.00334 - 0.334: Vacuum UV: VUV, UV-C: 12.4 - 6.53: 100 - 190: 100000 - 52600: 3000 - 1580 V = eV / e . What does millielectronvolt mean? A photon has an energy of 1.9 meV (milli electron Volt). They come from many sources and are not checked. Mega (Million) Electron-Volts. metres squared, grams, moles, feet per second, and many more! millielectronvolt to femtojoule. 100 millielectronvolt to electronvolt = 0.1 electronvolt. Milli-electronvolt monochromatization of hard X-rays with a sapphire backscattering monochromator Users experience is very important, that's why I use non-intrusive ads. meV means Milli electron volt. Ostatní významy MEV Kromě Milli-Elektronvolt má MEV jiné významy. Miscellaneous » Unclassified. 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It is commonly used with the metric prefixes milli-, kilo-, mega-, giga-, tera-, peta- or exa- (meV, keV, MeV, GeV, TeV, PeV and EeV respectively). Always check the results; rounding errors may occur. electronvolt Note that rounding errors may occur, so always check the results. 1 Electron volt (eV) = 1.602176565 x 10-19 joules (J). Acronym. Have you found the page useful? Type in your own numbers in the form to convert the units! 50 millielectronvolt to electronvolt = 0.05 electronvolt. millielectronvolt or Be warned. Show transcribed image text. Translation memories are created by human, but computer aligned, which might cause mistakes. the rest mass of an electron [1]: me = 9.1093897(54) 10 31 kg me = (9.1093897 0.0000054) 10 31 kg $m_{\mathrm{e}} = \SI{9.1093897(54)e-31}{\kg}$\\ $m_{\mathrm{e}} = \SI[seperr]{9.1093897(54)e-31}{\kg}$ A number of effects are available as options. 1 Millielectron Volt to common energy units; 1 meV = 1.60218E-22 joules (J) 1 meV = 1.60218E-25 kilojoules (kJ) 1 meV = 3.8293021032505E-23 calories (cal) 1 meV = 3.8293021032505E-26 kilocalories (kcal) 1 meV = 0.001 electron volt (eV) 1 meV = 4.4505E-26 watt hour (Wh) 1 meV = 3.6749405469679E-5 atomic unit of energy (au) 1 meV = 3.8293021032505E-32 tons of TNT (tTNT) In some older documents, and in the name Bevatron, the symbol BeV is used, which stands for billion electron volts; it … The most common shorthand of "Milli-Electron Volt" is mEV. Conversions Table. 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Atomic properties like the ionization energy are often quoted in electron volts. to use the unit converter. Thus meV stands for milli-electron volt. Definition of millielectronvolt in the Definitions.net dictionary. An Electron Volt is the amount of charge one electron gains or loses when it crosses an electrical potential difference of 1 volt. Cosmology sets limits on how large neutrino masses can be. Information and translations of Electronvolt in the most comprehensive dictionary definitions resource on the web. One electron-volt is equal to 1.602176565⋅10-19 joules: 1eV = 1.602176565e-19 J = 1.602176565⋅10 -19 J So the energy in joules E (J) is equal to the energy in electron-volts E (eV) times 1.602176565⋅10 -19 : Have a definition for milli-electron-volt ? Thus meV stands for milli-electron volt. Convert 3 millivolts to volts: Contact | Please note that Milli-Electron Volt is not the only meaning of MEV. Hard x-ray monochromator with milli-electron volt bandwidth for high-resolution diffraction studies of diamond crystals. millielectronvolt to terajoule Exchange reading in millijoules unit mJ into electron volts unit eV as in an equivalent measurement result (two different units but the same identical physical total value, which is also equal to their proportional parts when divided or multiplied). kiloelectron-volt to kilogram (keV—kg) measurement units conversion. I've spent over 10 trillion microseconds (and counting), on this project. These are fully explained in Section 11. In chemistry, it is often useful to have the molar equivalent, that is the energy that would be produced by one mole of charge (6.022 141 29 (27) × 1023) passing through a potential difference of … By definition, it is equal to the amount of kinetic energy gained by a single unbound electron when it accelerates through an electrostatic potential difference of one volt. An electron volt is a unit of measurement that can be used to describe the energy of vibrating particles. milli-+ electronvolt. Meaning of millielectronvolt. the Milli-Electron-Volt Scale∗ Eduard Mass´o Grup de F´ısica Teorica and Institut de F´ısica d’Altes Energies Universitat Autonoma de Barcelona, 08193 Bellaterra, Spain PreprintUAB-FT-662 Abstract A short review about vacuum energy and the cosmological constant is presented. Price of commodities: gold, silver, oil, live cattles and more. A short review about vacuum energy and the cosmological constant is presented. Like the elementary charge on which it is based, it is not an independent quantity but is equal to (1 J/C)(2 h α / μ0 c0)0.5 It is a common unit of energy within physics, widely used … Most people chose this as the best definition of millielectronvolt: (physics) A unit of energ... See the dictionary meaning, pronunciation, and sentence examples. Showing page 1. The Note that rounding errors may occur, so always check the results. Share this. We are proud to list acronym of MEV in the largest database of abbreviations and acronyms. Meaning of Electronvolt. Varying the gap of the undulator alters the strength of the magnetic field that the electron bunches experience and changes the radiation wavelength according to Eq. Milli Electron Volt. The answer is 1000. area, mass, pressure, and other types. It is based on resonant two-photon ionization of meta- stable Ar*(4s3P,) atoms in a collimated beam and pulsed production and extraction of negative … They come from many sources and are not checked. Please use the following to spread the word: About | Contact Us iOS app | Android Suggest | Recent Last Searched Popular Abbreviations Popular Categories Statistics. Expert Answer . 3 Millijoules to Electron Volts = 1.87245280298E+16. or. Expert Answer . A photon is a quantum of EM radiation. Keep four significant digits. 1 Millijoules to Electron Volts = 6.24150934326E+15. 70 Millijoules to Electron Volts = 4.36905654028E+17. Stoupin S(1), Shvyd'ko Y, Shu D, Khachatryan R, Xiao X, DeCarlo F, Goetze K, Roberts T, Roehrig C, Deriy A. 'Integrated definitions for 'millielectronvolt' and related words: 'unit', 'energy', 'thousandth', 'electronvolt', 'MeV', 'meV' This page is all about the acronym of MEV and its meanings as Milli-Electron Volt. 1 millielectron volt is equal to 1.6021773E-22 joule. Any feedback is appreciated. Convert Electric Resistance from MilliVolt/milliAmp to MilliOhm or to different units such as MicroOhm, MilliOhm, Ohm, KiloOhm, MegaOhm, AbOhm, Volt/Ampere, Kilovolt/Ampere, Kilovolt/KiloAmp, Millivolt/milliAmp and more MilliVolt / milliAmp and MilliOhm are the units to measure Electric Resistance, where 1 MilliVolt / milliAmp = 1,000 MilliOhm Add to My List Edit this Entry Rate it: (1.00 / 1 vote) Translation Find a translation for Milli Electron Volt in other languages: Select another language: - Select - 简体中文 (Chinese - Simplified) 繁體中文 (Chinese - Traditional) We needed access to a better microscope,” Klie said. The SI derived unit for energy is the joule. 1 Electron Volt: Change in energy by moving a single electron across an electric potential difference of one volt. 80 Millijoules to Electron Volts = 4.99320747461E+17. Unit Descriptions; 1 Millijoule: One millijoule is exactly 0.001 Joules (Joule is the SI unit for enery). 200 millielectronvolt to electronvolt = 0.2 electronvolt. Add to My List Edit this Entry Rate it: (1.00 / 1 vote) Translation Find a translation for Milli Electron Volt in other languages: Select another language: - Select - 简体中文 (Chinese - Simplified) 繁體中文 (Chinese - Traditional) How is Milli-Electron Volt abbreviated? ConvertUnits.com provides an online millielectron volt or joule. Screencast on voltage and the Electron Volt (eV). Radiative charge transfer in He + + H 2 collisions in the milli- to nano-electron-volt range: A theoretical study within state-to-state and optical potential approaches Previous question Next question Transcribed Image Text from this Question. Find My Location: IP address, city, latitude & longitude. mEV is defined as Milli-Electron Volt somewhat frequently. In chemistry , it is often useful to have the molar equivalent , that is the energy that would be produced by one mole of charge ( 6.022 1 41 2 9(27) × 10 23 ) passing through a potential difference of one volt. Expert Answer . … We assume you are converting between millielectronvolt and electronvolt. You can find metric conversion tables for SI units, as well Justin XoXo :), Gigagrams of TNT -> british thermal unit Mean. Write it here to share it with the entire community. Please support this site by disabling or whitelisting the Adblock for "justintools.com". conversion calculator for all types of measurement units. For example, when a tennis ball is hit by a racket and stops momentarily, the forces acting upon it (like gravity and the resistance of the racket), make it stay still in that position. One of the definitions of mEV is "Milli-Electron Volt". Aluminizing the early solar system. An electron-volt (or eV) is the kinetic energy that an electron gains in passing through one volt of potential, and a milli-electron-volt (or meV) is one thousandth of an eV. Thank you. Have a fact about milli-electron-volt ? volt = electronvolt / elementary charge. The following image shows one of the definitions of MEV in English: Milli-Electron Volt. Be warned. Exchange reading in millijoules unit mJ into electron volts unit eV as in an equivalent measurement result (two different units but the same identical physical total value, which is also equal to their proportional parts when divided or multiplied). In some older documents, and in the name Bevatron, the symbol BeV is used, which stands for billion electron volts; it … Keep four significant digits. A photon has an energy of 1.9 meV (milli electron Volt). Thanks to Steve Uren. Note that rounding errors may occur, so always check the results. (Métrologie) (Physique) Unité de mesure d’énergie en dehors du Système international (mais dont l’usage est accepté avec le SI), valant 10−3 électronvolt, et dont le symbole est meV. related. MEV. The problem is that, theoretically, the predicted vacuum energy is many orders of magnitude larger than 10-3 eV. The Rydberg constant to electron-volt [eV] conversion table and conversion steps are also listed. 1 millielectronvolt to electronvolt = 0.001 electronvolt, 10 millielectronvolt to electronvolt = 0.01 electronvolt, 50 millielectronvolt to electronvolt = 0.05 electronvolt, 100 millielectronvolt to electronvolt = 0.1 electronvolt, 200 millielectronvolt to electronvolt = 0.2 electronvolt, 500 millielectronvolt to electronvolt = 0.5 electronvolt, 1000 millielectronvolt to electronvolt = 1 electronvolt. Information and translations of Electronvolt in the most comprehensive dictionary definitions resource on the web. Please enable Javascript noun Physics. The Niowave FEL design uses a 5 mega-electron volt (MeV) electron beam in conjunction with a ten period variable gap undulator to generate the radiation. Abbreviations: MeV, Mev, mev. millielectronvolt to quad What does Electronvolt mean? Add fact ! Convert millielectronvolt to electronvolt, Quick conversion chart of millielectronvolt to electronvolt, millielectronvolt to kilogram-force meter. Found 1 sentences matching phrase "milli electron volt".Found in 3 ms. Found 1 sentences matching phrase "milli-electron-volt".Found in 2 ms. An infrared photon has a frequency of 7.8 E12 Hz. Mission Extension Vehicle (spacecraft technique) MEV. Free online force conversion. Translation memories are created by human, but computer aligned, which might cause mistakes. millielectronvolt (plural millielectronvolts) A unit of energy equal to one thousandth of an electronvolt, symbol meV mEV stands for Milli-Electron Volt. 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## Section4.6Chapter exercises

### Exercises4.6.1Exercises

In the game of roulette, a wheel is spun and you place bets on where it will stop. One popular bet is that it will stop on a red slot; such a bet has an $18/38$ chance of winning. If it stops on red, you double the money you bet. If not, you lose the money you bet. Suppose you play 3 times, each time with a $1 bet. Let $Y$ represent the total amount won or lost. Write a probability model for $Y\text{.}$ Solution 0 wins (-$3): 0.1458. 1 win (-1$): 0.3936. 2 wins (+1$): 0.3543. 3 wins (+$3): 0.1063. ###### 2.Speeding on the I-5, Part I. The distribution of passenger vehicle speeds traveling on the Interstate 5 Freeway (I-5) in California is nearly normal with a mean of 72.6 miles/hour and a standard deviation of 4.78 miles/hour. 1 1. What percent of passenger vehicles travel slower than 80 miles/hour? 2. What percent of passenger vehicles travel between 60 and 80 miles/hour? 3. How fast do the fastest 5% of passenger vehicles travel? 4. The speed limit on this stretch of the I-5 is 70 miles/hour. Approximate what percentage of the passenger vehicles travel above the speed limit on this stretch of the I-5. S. Johnson and D. Murray. “Empirical Analysis of Truck and Automobile Speeds on Rural Interstates: Impact of Posted Speed Limits”. In: Transportation Research Board 89th Annual Meeting. 2010. ###### 3.University admissions. Suppose a university announced that it admitted 2,500 students for the following year's freshman class. However, the university has dorm room spots for only 1,786 freshman students. If there is a 70% chance that an admitted student will decide to accept the offer and attend this university, what is the approximate probability that the university will not have enough dormitory room spots for the freshman class? Solution Want to find the probability that there will be 1,786 or more enrollees. Using the normal approximation, with $\mu= np = 2,500 \times 0.7 = 1750$ and $\sigma = \sqrt{np(1-p} = \sqrt{2,500 \times 0.7 \times 0.3} \approx 23\text{,}$ $Z = 1.61\text{,}$ and $P(Z > 1.61) = 0.0537\text{.}$ With a 0.5 correction: 0.0559. ###### 4.Speeding on the I-5, Part II. Exercise 4.6.1.2 states that the distribution of speeds of cars traveling on the Interstate 5 Freeway (I-5) in California is nearly normal with a mean of 72.6 miles/hour and a standard deviation of 4.78 miles/hour. The speed limit on this stretch of the I-5 is 70 miles/hour. 1. A highway patrol officer is hidden on the side of the freeway. What is the probability that 5 cars pass and none are speeding? Assume that the speeds of the cars are independent of each other. 2. On average, how many cars would the highway patrol officer expect to watch until the first car that is speeding? What is the standard deviation of the number of cars he would expect to watch? ###### 5.Auto insurance premiums. Suppose a newspaper article states that the distribution of auto insurance premiums for residents of California is approximately normal with a mean of$1,650. The article also states that 25% of California residents pay more than $1,800. 1. What is the $Z$-score that corresponds to the top 25% (or the $75^{\text{th}}$ percentile) of the standard normal distribution? 2. What is the mean insurance cost? What is the cutoff for the $75^{\text{th}}$ percentile? 3. Identify the standard deviation of insurance premiums in California. Solution (a) $Z=0.67$ (b) $\mu=1650\text{,}$ $x=1800\text{.}$ $0.67=\frac{1800-1650}{\sigma} \rightarrow \sigma = 233.88\text{.}$ ###### 6.SAT scores. SAT scores (out of 1600) are distributed normally with a mean of 1100 and a standard deviation of 200. Suppose a school council awards a certificate of excellence to all students who score at least 1350 on the SAT, and suppose we pick one of the recognized students at random. What is the probability this student's score will be at least 1500? (The material covered in Section 3.2 would be useful for this question.) ###### 7.Married women. The American Community Survey estimates that 47.1% of women ages 15 years and over are married. 2 1. We randomly select three women between these ages. What is the probability that the third woman selected is the only one who is married? 2. What is the probability that all three randomly selected women are married? 3. On average, how many women would you expect to sample before selecting a married woman? What is the standard deviation? 4. If the proportion of married women was actually 30%, how many women would you expect to sample before selecting a married woman? What is the standard deviation? 5. Based on your answers to parts (c) and (d), how does decreasing the probability of an event affect the mean and standard deviation of the wait time until success? U.S. Census Bureau, 2010 American Community Survey, Marital Status. Solution (a) $(1-0.471)^2 \times 0.471 = 0.1318\text{.}$ (b) $0.471^3 = 0.1045\text{.}$ (c) $\mu = 1/0.471 = 2.12\text{,}$ $\sigma = \sqrt{2.38} = 1.54\text{.}$ (d) $\mu = 1/.30 = 3.33\text{,}$ $\sigma = 2.79\text{.}$ (e) When p is smaller, the event is rarer, meaning the expected number of trials before a success and the standard deviation of the waiting time are higher. ###### 8.Survey response rate. Pew Research reported that the typical response rate to their surveys is only 9%. If for a particular survey 15,000 households are contacted, what is the probability that at least 1,500 will agree to respond? 3 ###### 9.Overweight bags. Suppose weights of the checked baggage of airline passengers follow a nearly normal distribution with mean 45 pounds and standard deviation 3.2 pounds. Most airlines charge a fee for baggage that weigh in excess of 50 pounds. Determine what percent of airline passengers incur this fee. Solution $Z=1.56\text{,}$ $P(Z > 1.56)= 0.0594\text{,}$ i.e. 6%. ###### 10.Heights of 10 year olds, Part I. Heights of 10 year olds, regardless of gender, closely follow a normal distribution with mean 55 inches and standard deviation 6 inches. 1. What is the probability that a randomly chosen 10 year old is shorter than 48 inches? 2. What is the probability that a randomly chosen 10 year old is between 60 and 65 inches? 3. If the tallest 10% of the class is considered “very tall”, what is the height cutoff for “very tall”? ###### 11.Buying books on Ebay. The textbook you need to buy for your chemistry class is expensive at the college bookstore, so you consider buying it on Ebay instead. A look at past auctions suggest that the prices of that chemistry textbook have an approximately normal distribution with mean$89 and standard deviation $15. 1. What is the probability that a randomly selected auction for this book closes at more than$100?

2. Ebay allows you to set your maximum bid price so that if someone outbids you on an auction you can automatically outbid them, up to the maximum bid price you set. If you are only bidding on one auction, what are the advantages and disadvantages of setting a bid price too high or too low? What if you are bidding on multiple auctions?

3. If you watched 10 auctions, roughly what percentile might you use for a maximum bid cutoff to be somewhat sure that you will win one of these ten auctions? Is it possible to find a cutoff point that will ensure that you win an auction?

4. If you are willing to track up to ten auctions closely, about what price might you use as your maximum bid price if you want to be somewhat sure that you will buy one of these ten books?

Solution

(a) $Z = 0.73\text{,}$ $P(Z > 0.73) = 0.2327\text{.}$

(b) If you are bidding on only one auction and set a low maximum bid price, someone will probably outbid you. If you set a high maximum bid price, you may win the auction but pay more than is necessary. If bidding on more than one auction, and you set your maximum bid price very low, you probably won't win any of the auctions. However, if the maximum bid price is even modestly high, you are likely to win multiple auctions.

(c) An answer roughly equal to the 10th percentile would be reasonable. Regrettably, no percentile cutoff point guarantees beyond any possible event that you win at least one auction. However, you may pick a higher percentile if you want to be more sure of winning an auction.

(d) Answers will vary a little but should correspond to the answer in part (c). We use the $10^{th}$ percentile: $Z = -1.28 \rightarrow 69.80\text{.}$

###### 12.Heights of 10 year olds, Part II.

Heights of 10 year olds, regardless of gender, closely follow a normal distribution with mean 55 inches and standard deviation 6 inches.

1. The height requirement for Batman the Ride at Six Flags Magic Mountain is 54 inches. What percent of 10 year olds cannot go on this ride?

2. Suppose there are four 10 year olds. What is the chance that at least two of them will be able to ride Batman the Ride?

3. Suppose you work at the park to help them better understand their customers' demographics, and you are counting people as they enter the park. What is the chance that the first 10 year old you see who can ride Batman the Ride is the 3rd 10 year old who enters the park?

4. What is the chance that the fifth 10 year old you see who can ride Batman the Ride is the 12th 10 year old who enters the park?

###### 13.Heights of 10 year olds, Part III.

Heights of 10 year olds, regardless of gender, closely follow a normal distribution with mean 55 inches and standard deviation 6 inches.

1. What fraction of 10 year olds are taller than 76 inches?

2. If there are 2,000 10 year olds entering Six Flags Magic Mountain in a single day, then compute the expected number of 10 year olds who are at least 76 inches tall. (You may assume the heights of the 10-year olds are independent.)

3. Using the binomial distribution, compute the probability that 0 of the 2,000 10 year olds will be at least 76 inches tall.

4. The number of 10 year olds who enter Six Flags Magic Mountain and are at least 76 inches tall in a given day follows a Poisson distribution with mean equal to the value found in part (b). Use the Poisson distribution to identify the probability no 10 year old will enter the park who is 76 inches or taller.

Solution

(a) $Z = 3.5\text{,}$ upper tail is 0.0002. (More precise value: 0.000233, but we'll use 0.0002 for the calculations here.)

(b) $0.0002 \times 2000 = 0.4\text{.}$ We would expect about 0.4 10 year olds who are 76 inches or taller to show up.

(c) ${2000 \choose 0}(0.0002)^0(1-0.0002)^2000 = 0.67029\text{.}$

(d) $\frac{0.4^0 \times e^{-0.4}}{0!} = \frac{1\times e^{-0.4}}{1} = 0.67032\text{.}$

###### 14.Multiple choice quiz.

In a multiple choice quiz there are 5 questions and 4 choices for each question (a, b, c, d). Robin has not studied for the quiz at all, and decides to randomly guess the answers. What is the probability that

1. the first question she gets right is the $3^{\text{rd}}$ question?

2. she gets exactly 3 or exactly 4 questions right?

3. she gets the majority of the questions right?

###### 15.Overweight baggage, Part II.

Suppose weights of the checked baggage of airline passengers follow a nearly normal distribution with mean 45 pounds and standard deviation 3.2 pounds. What is the probability that the total weight of 10 bags is greater than 460 lbs?

Solution

This is the same as checking that the average bag weight of the 10 bags is greater than 46 lbs. $SD_{\bar{x}}= \frac{3.2}{\sqrt{10}} = 1.012\text{;}$ $z=\frac{46-45}{1.012} = 0.988\text{;}$ $P(z > 0.988) = 0.162 = 16.2%\text{.}$

Students are asked to count the number of chocolate chips in 22 cookies for a class activity. The packaging for these cookies claims that there are an average of 20 chocolate chips per cookie with a standard deviation of 4.37 chocolate chips.

1. Based on this information, about how much variability should they expect to see in the mean number of chocolate chips in random samples of 22 chocolate chip cookies?

2. What is the probability that a random sample of 22 cookies will have an average less than 14.77 chocolate chips if the company's claim on the packaging is true? Assume that the distribution of chocolate chips in these cookies is approximately normal.

3. Assume the students got 14.77 as the average in their sample of 22 cookies. Do you have confidence or not in the company's claim that the true average is 20? Explain your reasoning.

###### 17.Young Hispanics in the US.

The 2012 Current Population Survey (CPS) estimates that 38.9% of the people of Hispanic origin in the Unites States are under 21 years old. 4  Calculate the probability that at least 35 people among a random sample of 100 Hispanic people living in the United States are under 21 years old.

United States Census Bureau. 2012 Current Population Survey. The Hispanic Population in the United States: 2012. Web.
Solution

First we need to check that the necessary conditions are met. There are $100 \times 0.389 = 38.9$ expected successes and $100 \times (1 -0.389) = 61.1$ expected failures, therefore the success-failure condition is met. Calculate using either (1) the normal approximation to the binomial distribution or (2) the sampling distribution of $\hat{p}\text{.}$ (1) The binomial distribution can be approximated by $N(\mu = 0.389, \sigma = 4.88)\text{.}$ $P( \gt 35) = P(Z > -0.80 = 1-0.2119 = 0.7881\text{.}$ (2) The sampling distribution of $\hat{p} ~ N(\mu = 0.389, \sigma = 0.0488\text{.}$ $P(\hat{p}>0.35)=P(Z>-0.8) = 0.7881\text{.}$

###### 18.Poverty in the US.

The 2013 Current Population Survey (CPS) estimates that 22.5% of Mississippians live in poverty, which makes Mississippi the state with the highest poverty rate in the United States. 5  We are interested in finding out the probability that at least 250 people among a random sample of 1,000 Mississippians live in poverty.

1. Estimate this probability using the normal approximation to the binomial distribution.

2. Estimate this probability using the distribution of the sample proportion.


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# The 24th International Spin Symposium

18-22 October 2021
Matsue, Shimane Prefecture, Japan
Asia/Tokyo timezone

## Azimuthal asymmetries in unpolarised semi-inclusive DIS at COMPASS

18 Oct 2021, 19:26
18m
Room 601 (Kunibiki Messe)

### Room 601

#### Kunibiki Messe

Parallel Session Presentation Transverse momentum structure (TMD)

### Speaker

Jan Matousek (Charles University (Prague, CZ))

### Description

In QCD the momentum and spin structure of the nucleon explored in hard-scattering reactions is described in terms of Transverse Momentum Dependent Parton Distribution Functions (TMD PDFs). The PDFs were studied primarily in the collinear approximation, until it was demonstrated that the intrinsic transverse motion of partons plays an important role.
In Semi-Inclusive Deep Inelastic Scattering (SIDIS) the non-zero transverse momentum of partons $k_T$ induces azimuthal dependence of the cross-section and affects the transverse momentum distribution of the produced hadrons. The SIDIS cross-section contains three nucleon polarization independent azimuthal modulations that are related to different combinations of twist-2 or higher twist PDFs and fragmentation functions. Those are the so-called Cahn effect reflected in $cos(\phi_{h})$ modulation, the $cos(2\phi_{h})$ term related to the Boer-Mulders PDF and $sin(\phi_{h})$ twist-3 effect known as beam-spin asymmetry.
In 2016 and 2017, the COMPASS experiment at CERN collected a large sample of DIS events using a longitudinally polarised 160 GeV/c muon beam scattering on a liquid hydrogen target. The azimuthal asymmetries $A_{UU}^{\cos(\phi_{h})}$, $A_{UU}^{\cos(2\phi_{h})}$ and $A_{UU}^{\sin(\phi_{h})}$ have been extracted from part of the data. A new procedure has been developed to subtract a background coming from the decay of diffractively produced vector mesons. The results presented in this talk qualitatively agree with earlier COMPASS results obtained with an isoscalar target.

### Primary author

Jan Matousek (Charles University (Prague, CZ))


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# Need help graphing an absolute value function

The function I have is $$|x-1|+|y+2|\leq2$$

So I tried graphing it the same way I would $$|x|+|y|\leq1$$

I know that $$|x-1|$$ is $$x-1$$ when $$x>0$$ and $$-x+1$$ when $$x<0$$

Similarly, I know that $$|y+2|$$ is $$y+2$$ when $$y>0$$ and $$-y-2$$ when $$y<0$$

So, I have four cases:

I. $$x>0,y>0$$ (Quadrant I)

II. $$x<0,y>0$$ (Quadrant II)

III. $$x<0,y<0$$ (Quadrant III)

IV. $$x>0,y<0$$ (Quadrant IV)

So for Quadrant I, we need to solve $$x-1+y+2\leq2$$ and get $$y\leq-x+1$$

Quadrant II, we solve $$-x+1+y+2\leq2$$ and get $$y\leq x-1$$

Quadrant III, we solve $$-x+1-y-2\leq2$$ and get $$y\geq -x-3$$

Quadrant IV, we solve $$x-1-y-2\leq2$$ and get $$y\geq x-5$$

When I graph these lines in their respective quadrants, I do not get the correct answer. Can someone please tell me where I went wrong in my method?

• "I know that $|x - 1| = x - 1$ when $x > 0$..." This is wrong. It should be $x - 1> 0$. You just need to remember the one inside the absolute value is positive, rather than $x$ itself. Sep 6, 2021 at 2:15
• A better way to do this would be to plot $|x|+|y|\le2$ first and then shifting the graph along $x$ and $y$ by $1$ and $-2$ units respectively Sep 6, 2021 at 2:49
• A more intuitive way to see this might be to note that |𝑥 − 1| is the distance from the line 𝑥 = 1, and |𝑦 + 2| is the distance from the line 𝑦 = -2.  So you want points where the sum of those distances is ≤ 2.  (This will be a diamond, centred on the point where those lines cross.) Sep 6, 2021 at 11:00

You know that $$|x|\geq 0\iff x\geq0$$

Replace $$x$$ with $$x-1$$ and you get $$|x-1|\geq 0\iff x-1\geq0$$ which is equivalent to

$$|x-1|\geq 0\iff x\geq1$$

So you will need to make $$4$$ cases-

$$x\leq1, y\geq2$$

$$x\leq1, y\leq2$$

$$x\geq1, y\leq2$$

$$x\geq1, y\geq2$$

If you know simple graph transformation methods then from simple equation , you can reach this equation

$$|x|+|y|\leq2$$

Replace $$x$$ with $$x-1$$ and shift whole graph $$1$$ units right

Now you got graph of $$|x-1|+|y|\leq2$$

Now replace $$y$$ with $$y+2$$ and shift graph $$2$$ units down and you will get graph of $$|x-1|+|y+2|\leq 2$$

• thank you. your method of just doing $|x|+|y|\leq2$ and shifting it 1 unit to the right and 2 units down was much easier Sep 6, 2021 at 4:30
• @user8358234 If my answer helped you , please accept it :-) Sep 6, 2021 at 4:32

Hint: Let $$L_1, L_2$$ be two non parallel lines lines, then $$a|L_1|+b|L_2|=c$$ represents a parallelogram with diagonals as $$L_1=0=L_2$$. If $$L_1,L_2$$ are perpendicular then it is asquare or rhombus, otherwise rectangle/parallelogram. With sises as $$aL_1+bL_2=c, -aL_!+bL_2=c, aL_1-bL_2=c,-aL_1+bL_2=c.$$ If adjacent sides are perpendicular then it is square or rectangle. So in this case we have a square centered at $$(1,-2)$$.

$$|x-1|=x-1$$ when $$x \geqq 1$$ and not 0. Same is for $$y$$. You may check this by putting $$x=\frac{1}{2}$$. You get $$-0.5$$ from your analysis but it should have been $$+0.5$$.


text:
# A relation for the state of a scatteing process

I am studying the article "A Formal Optical Model", of Bell and Squires (the article is here https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.3.96) where they proof that the self-energy of the Green function is equal to the optical potential. They consider the scattring of a particle by a target (both being identical fermions) and they say that the scattering state can be write as

$$| \alpha \rangle = \int_{-\infty}^{t} dt' e^{-iEt'} \hat{\psi} (\textbf{r}, t') | 0 \rangle$$

where $$E$$ is the energy of the incident particle and $$| 0 \rangle$$ is the state of the target. What is the justification or deduction for this relation? I have never seen it before... and it's weird, because the ket is depending on the position $$\textbf{r}$$. Can someone enlighten me what they are actually doing?


text:
# No IE

By Murray Bourne, 23 Dec 2005

I like this logo:

I am developing a new site for work. We are using WordPress and will be using it basically as a content management system.

Everything was going well (rearranging pages, fixing links, adding favicon, messing with colours, etc) and I then checked it in IE. Wak - things were all over the place - DIVs were out of alignment, or too big. Images were not showing in the excerpts. Also, the links were not behaving - the URL showing in the address bar was always the root of the site - not where we actually were. Grrrr... 🙁

After quite a bit of trial and error, I found it was some floating div problems. Only an issue in IE - all other browsers were okay.

Be the first to comment below.

### Comment Preview

HTML: You can use simple tags like <b>, <a href="...">, etc.

To enter math, you can can either:

1. Use simple calculator-like input in the following format (surround your math in backticks, or qq on tablet or phone):
a^2 = sqrt(b^2 + c^2)
(See more on ASCIIMath syntax); or
2. Use simple LaTeX in the following format. Surround your math with $$ and $$.
$$\int g dx = \sqrt{\frac{a}{b}}$$
(This is standard simple LaTeX.)

NOTE: You can't mix both types of math entry in your comment.


text:
# Vincent gathered groups of 1 to 8 people and gave every group the same positive salt. He recorded the number of minutes it took each you to solve the

Vincent gathered groups of 1 to 8 people and gave every group the same positive salt. He recorded the number of minutes it took each you to solve the puzzle in Porter resort. What is the best description of this relationship￼

$Vincent gathered groups of 1 to 8 people and gave every group the same positive salt. He recorded t$

## This Post Has 4 Comments

1. Expert says:

5 x 14?

im new, sorry if this is wrong.

step-by-step explanation:

2. Expert says:

ratio is 2 : 9

the answer is three

13.5 : 3

$Can someone plz answer this for me? whoever answers it first, i will give !$

3. Expert says:

idk

step-by-step explanation:

4. Expert says:

use this formula:

p(a\b) = p(a and b)/p(b).

take it from here.


text:
## Tag Archives: quotient group

### Two consequences of Frattini’s argument regarding normalizers and centralizers

Let $G$ be a finite group, let $p$ be a prime, let $P \leq G$ be a Sylow $p$-subgroup, and let $N \leq G$ be a normal subgroup such that $p \not| |N|$. Prove the following.

1. $N_{G/N}(PN/N) = N_G(P)N/N$
2. $C_{G/N}(PN/N) = C_G(P)N/N$

Note that $P \cap N = 1$, so that $PN \cong P \times N$. In particular, $P \leq C_G(N)$ and $N \leq C_G(P)$.

Note that $P \leq PN \cong P \times N$ is a Sylow subgroup since $p$ does not divide $|N|$, and that $PN \leq N_G(PN)$ is normal. By Frattini’s Argument, we have $N_G(PN) = PN N_G(P) = N_G(P)N$. Thus $N_G(P)N/N = N_G(PN)/N = N_{G/N}(PN/N)$, as desired.

Now note that $C_G(PN) \leq C_G(P)$, so that $C_{G/N}(PN/N) = C_G(PN)/N \leq C_G(P)N/N$. Moreover, we have $[C_G(P)/N, PN/N] = [C_G(P),PN]/N$ $= [C_G(P),N]/N$ $= 1$. Thus $C_G(P)/N \leq C_{G/N}(PN/N)$.

### If the first two lower central quotients of a group are cyclic, then the second derived subgroup is trivial

Show that if $G^\prime/G^{\prime\prime}$ and $G^{\prime\prime} / G^{\prime\prime\prime}$ are both cyclic then $G^{\prime\prime} = 1$.

We begin by proving a lemma.

Lemma: Let $G$ be a group and let $N \leq G$ be normal. Then $(G/N)^\prime = G^\prime N /N$. Proof: $(\subseteq)$ If $[xN,yN] \in G/N$ is a commutator, then $[xN,yN] = [x,y]N \in G^\prime N/N$. $(\supseteq)$ $G^\prime \leq G^\prime N$, so that $G/G^\prime N \cong (G/N)/(G^\prime N/N)$ is abelian. By Theorem 5.7 in the text, $(G/N)^\prime \leq G^\prime N/N$. $\square$

Now for the main result.

Note that $G^{\prime\prime\prime} \leq G$ is characteristic, hence normal, so that $G^{\prime\prime}/G^{\prime\prime\prime} \leq G/G^{\prime\prime\prime}$ is a cyclic, hence abelian, normal subgroup.

By this previous exercise, $(G/G^{\prime\prime\prime})/(G^{\prime\prime}/G^{\prime\prime\prime}) \cong G/G^{\prime\prime}$ (via the Third Isomorphism Theorem) acts on $G^{\prime\prime}/G^{\prime\prime\prime}$ by conjugation on the left as follows: $(gG^{\prime\prime}) \cdot (a G^{\prime\prime\prime}) = (gag^{-1}) G^{\prime\prime\prime}$. Now conjugation is an automorphism of $G^{\prime\prime}/G^{\prime\prime\prime}$, and since this quotient is cyclic, its automorphism group is abelian. Thus, for all $g,h \in G$ and $a \in G^{\prime\prime}$, we have $(gh)a(gh)^{-1} G^{\prime\prime\prime} = (hg)a(hg)^{-1} G^{\prime\prime\prime}$. Via some arithmetic, we see that $[g,h] a [g,h]^{-1} a^{-1} \in G^{\prime\prime\prime}$; in particular, each element of $G^{\prime\prime}/G^{\prime\prime\prime}$ commutes with each generator of $G^{\prime}/G^{\prime\prime\prime}$, so that $G^{\prime\prime}/G^{\prime\prime\prime} \leq Z(G^{\prime}/G^{\prime\prime\prime})$.

Using the Third Isomorphism Theorem, $(G^\prime/G^{\prime\prime\prime})/Z(G^\prime/G^{\prime\prime\prime}) \cong ((G^\prime/G^{\prime\prime\prime})/(G^{\prime\prime}/G^{\prime\prime\prime}))/(Z(G^\prime/G^{\prime\prime\prime})/(G^{\prime\prime}/G^{\prime\prime\prime}))$ $\cong (G^\prime/G^{\prime\prime})/(Z(G^\prime/G^{\prime\prime\prime})/(G^{\prime\prime}/G^{\prime\prime\prime}))$ is cyclic; hence $G^\prime/G^{\prime\prime\prime}$ is abelian, and using the lemma we have $1 = (G^\prime/G^{\prime\prime\prime})^\prime = G^{\prime\prime}/G^{\prime\prime\prime}$. Thus $G^{\prime\prime} = G^{\prime\prime\prime}$.

### The class of nilpotent groups is closed under subgroups and quotients, but not extensions

Prove that subgroups and quotient groups of nilpotent groups are nilpotent. (Your proof should work for infinite groups.) Give an explicit example of a group $G$ which possesses a normal subgroup $H$ such that $H$ and $G/H$ are nilpotent but $G$ is not nilpotent.

We showed that subgroups and quotients of nilpotent groups are nilpotent in the lemmas to this previous exercise.

Consider now $D_6$. $\langle r \rangle \leq D_6$ is a normal subgroup of order 3, and thus $\langle r \rangle \cong Z_3$ is abelian, hence nilpotent. Moreover, $D_6/\langle r \rangle \cong Z_2$ is abelian, hence nilpotent. However, $D_6$ has distinct Sylow 2-subgroups $\langle s \rangle$ and $\langle sr \rangle$ (for example). By Theorem 3, $D_6$ is not nilpotent.

### If a group mod its center is nilpotent, then the group is nilpotent

Prove that if $G/Z(G)$ is nilpotent, then $G$ is nilpotent.

We proved, as a lemma to the previous exercise, that $Z_t(G/Z(G)) = Z_{t+1}(G)/Z(G)$ for all $t$. Now suppose $G/Z(G)$ is nilpotent of nilpotence class $k$. Then $G/Z(G) = Z_k(G/Z(G)) = Z_{k+1}(G)/Z(G)$, and we have $G = Z_{k+1}(G)$. Thus $G$ is nilpotent of nilpotence class at most $k+1$.

### If the quotients of a group by two subgroups are abelian, then the quotient by their intersection is abelian

Let $A,B \leq G$ be normal subgroups such that $G/A$ and $G/B$ are abelian. Prove that $G/(A \cap B)$ is abelian.

Since $G/A$ is abelian we have $[G,G] \leq A$. Likewise, $[G,G] \leq B$. Thus $[G,G] \leq A \cap B$, so that $G/(A \cap B)$ is abelian.

### Compute a given quotient of a free abelian group

Let $n$ and $k$ be positive integers and let $A$ be the free abelian group of rank $n$ written additively. Prove that $A/kA$ is isomorphic to $(\mathbb{Z}/k\mathbb{Z})^n$, where $kA = \{ ka \ |\ a \in A \}$.

We have $A = \mathbb{Z}^n$. Note that $x \in k \mathbb{Z}^n$, then $x = k(a_i)$, so that $x = (ka_i)$, hence $x \in (k\mathbb{Z})^n$.

By this previous exercise, $\mathbb{Z}^n / k\mathbb{Z}^n \cong (\mathbb{Z}/(k))^n$.

### A quotient by a product is isomorphic to the product of quotients

Let $G = A_1 \times A_2 \times \cdots \times A_n$ and for each $i$ let $B_i \leq A_i$ be a normal subgroup. Prove that $B_1 \times \cdots \times B_n \leq G$ is normal and that $(A_1 \times \cdots \times A_n)/(B_1 \times \cdots \times B_n) \cong (A_1/B_1) \times \cdots \times (A_n/B_n)$.

We begin with some lemmas.

Lemma 1: Let $\varphi_1 : G_1 \rightarrow H_1$ and $\varphi_2 : G_2 \rightarrow H_2$ be group homomorphisms. Then $\mathsf{ker}(\varphi_1 \times \varphi_2) = (\mathsf{ker}\ \varphi_1) \times (\mathsf{ker}\ \varphi_2)$. Proof: $(\subseteq)$ Let $(a,b) \in \mathsf{ker}(\varphi_1 \times \varphi_2)$. Then $(1,1) = (\varphi_1 \times \varphi_2)(a,b) = (\varphi_1(a), \varphi_2(b))$, hence $\varphi_1(a) = 1$ and $\varphi_2(b) = 1$. Thus $a \in \mathsf{ker}\ \varphi_1$ and $b \in \mathsf{ker}\ \varphi_2$, and $(a,b) \in (\mathsf{ker}\ \varphi_1) \times (\mathsf{ker}\ \varphi_2)$. $(\supseteq)$ If $(a,b) \in (\mathsf{ker}\ \varphi_1) \times (\mathsf{ker}\ \varphi_2)$, then $(\varphi_1 \times \varphi_2)(a,b) = (1,1)$; hence $(a,b) \in \mathsf{ker}(\varphi_1 \times \varphi_2)$. $\square$

Lemma 2: Let $A_1$ and $A_2$ be groups with normal subgroups $B_1 \leq A_1$ and $B_2 \leq A_2$. Then $B_1 \times B_2$ is normal in $A_1 \times A_2$ and $(A_1 \times A_2)/(B_1 \times B_2) \cong (A_1/B_1) \times (A_2/B_2)$. Proof: Let $\pi_1 : A_1 \rightarrow A_1/B_1$ and $\pi_2 : A_2 \rightarrow A_2/B_2$ denote the natural projections. These mappings are surjective, so that $\pi_1 \times \pi_2$ is a surjective homomorphism $A_1 \times A_2 \rightarrow (A_1/B_1) \times (A_2/B_2)$. By Lemma 1, we have $\mathsf{ker}(\pi_1 \times \pi_2) = B_1 \times B_2$. The conclusion follows by the First Isomorphism Theorem. $\square$

The main result now follows by induction on $n$.

### An upper bound on the number of Sylow subgroups of a finite quotient

Let $G$ be a finite group and suppose $N \leq G$ is normal. Prove that $n_p(G/N) \leq n_p(G)$.

We begin with a lemma.

Lemma: Every Sylow $p$-subgroup of $G/N$ has the form $PN/N$ for some Sylow $p$-subgroup $P$ of $G$. Proof: Let $P$ be a Sylow $p$-subgroup of $G$. In this previous exercise, we saw that $PN/N$ is a Sylow $p$-subgroup of $G/N$. Now let $Q \leq G/N$ be any Sylow $p$-subgroup; then $Q = (gN)(PN/N)(g^{-1}N) = (gPg^{-1})N/N$ for some $g \in G$ by Sylow’s Theorem, and $gPg^{-1}$ is a Sylow $p$-subgroup of $G$. $\square$

Now to the main result.

Note that the mapping $\Phi : \mathsf{Syl}_p(G) \rightarrow \mathsf{Syl}_p(G/N)$ given by $P \mapsto PN/N$ is a surjection of finite sets. Thus $n_p(G) \geq n_p(G/N)$.

### In a finite group, Sylow subgroups induce Sylow subgroups in normal subgroups and their corresponding quotients

Let $G$ be a finite group, $P \in \mathsf{Syl}_p(G)$, and $N \leq G$ a normal subgroup. Use the conjugacy part of Sylow’s Theorem to prove that $P \cap N$ is a Sylow $p$-subgroup of $N$. Deduce that $PN/N$ is a Sylow $p$-subgroup of $G/N$.

Let $Q$ be a Sylow $p$-subgroup of $N$. $Q$ is a $p$-subgroup of $G$, so that $Q \leq xPx^{-1}$ for some $x \in G$. Then $Q \leq xPx^{-1} \cap N = xPx^{-1} \cap xNx^{-1}$ since $N$ is normal, hence $Q \leq x(P \cap N)x^{-1}$, thus $x^{-1}Qx \leq P \cap N$. Now $x^{-1}Qx$ is a subgroup of $N$, and has maximal $p$-power order in $N$, since $|x^{-1}Qx| = |Q|$. Since $P \cap N$ is a $p$-subgroup of $N$, $P \cap N$ is a maximal $p$-power subgroup of $N$, thus a Sylow $p$-subgroup.

Note that $PN/N \cong P/(P \cap N)$ by the Second Isomorphism Theorem, so that $PN/N$ is a $p$-subgroup of $G/N$. By the Third Isomorphism Theorem, $[G/N : PN/N] = [G : PN]$; since $P$ is a maximal $p$-power order subgroup of $G$, $p$ does not divide $[G : PN]$, so $p$ does not divide $[G/N : PN/N]$. Hence $PN/N$ is a Sylow $p$-subgroup of $G/N$.

### An action of a quotient group on an abelian normal subgroup

Let $G$ be a group and let $A$ be an abelian, normal subgroup of $G$. Show that $G/A$ acts (on the left) by conjugation on $A$ by $[g] \cdot a = gag^{-1}$. (In particular, show that this action is well defined.) Give an explicit example to show that this action is not well defined if $A$ is nonabelian.

To see well definedness, suppose $gh^{-1} = b \in A$. That is, $g = bh$. Now $gag^{-1} = bhah^{-1}b^{-1}$; note that $hah^{-1} \in A$ since $A$ is normal, hence $gag^{-1} = hah^{-1}$ since $A$ is abelian. Thus $[g] \cdot a = [h] \cdot a$, and the action is well defined.

Clearly we have $[1] \cdot a = a$ and $[gh] \cdot a = ghah^{-1}g^{-1} = [g] \cdot ([h] \cdot a) = ([g][h]) \cdot a$. So $G/A$ acts on $A$.

Now identify $D_6 = \langle r^2, s \rangle$ in $D_{12}$. Note that $D_6 \leq D_{12}$ has index 2, hence is normal, and that $D_6$ is nonabelian. Moreover, $r$ and $rs$ are distinct representatives of $rD_6$. However, $rr^2r^{-1} = r^2$ and $rsr^2rs = r^4$, so that the action described above is not well defined.


text:
# Malba Tahan: The exciting conclusion!

Last Wednesday, I continued my series on People Who Are Not What They Seem with Malba Tahan, the Islamic intellectual and writer who was actually the fictional alter ego of a Brazilian math teacher, created to help his students learn word problems.

That post ended on a mathematical cliffhanger (bwahahahaha!). Our heroes, Tahan and his friend Beremiz Samir, happened upon an argument among three brothers about how to divide up their father’s inheritance. Their father left them a herd of 35 camels, with the following instructions (as reported by the eldest brother):

According to the express wishes of my father half of them belong to me, one-third to my brother Hamed, and one-ninth to Harim, the youngest.

Samir’s solution was clever, but it required some risk – he added into the herd the camel that Tahan was riding, making a new herd of 36. He then divided the new herd according to the father’s instructions: one-half (18) to the eldest, one-third (12) to the middle, and one-ninth (4) to the youngest. All three brothers were satisfied with this arrangement, which left two camels remaining. One, of course, was Tahan’s that had been added at the beginning. Samir requested the other as his payment for arranging this solution – and since all three brothers were satisfied, they agreed. Samir grabbed the strongest, most beautiful member of the herd, and the pair rode off together into the sunset.

It’s a happy ending. Everyone is satisfied, especially our heroes. And you have to admire Samir’s Raven-level trickeration in getting something for nothing. But how did he solve the problem?

When faced with a word problem, often the best first step is to write down what you know and what you want to find out. Before Tahan and Samir arrive, here is the situation the brothers face:

• What we know
• Total camels: 35
• Fraction to each brother: eldest: $\frac{1}{2}$, middle: $\frac{1}{3}$, youngest: $\frac{1}{9}$
• What we want to find out
• How many camels should each brother get?

In theory, this should be an easy problem: for the eldest brother, divide 35 by 2, and repeat for the others. Thus, the eldest brother should get 17 $\frac{1}{2}$ camels — not too pleasant for the camel! And besides, half a camel is not that useful anyway. Clearly a better solution is needed.

Tahan and Samir arrive, Samir offers Tahan’s camel for the herd, and the problem changes. Now we have:

• What we know
• Total camels: 36
• Fraction to each brother: eldest: $\frac{1}{2}$, middle: $\frac{1}{3}$, youngest: $\frac{1}{9}$
• What we want to find out
• How many camels should each brother get?

Now we’re getting somewhere.

36 divided by 2 is 18, 36 divided by 3 is 12, and 36 divided by 9 is 4. Thus, the three brothers get eighteen, twelve, and four camels, all of which give them full camels instead of useless fractional camels.*

Adding up all three brothers’ camelshare gives 18 + 12 + 4 = 34 camels, with two remaining from the herd. One was Tahan’s, one is now Samir’s. Everything is A-OK.

But where did that extra camel come from?

Re-read the father’s instructions again, carefully:

According to the express wishes of my father half of them belong to me, one-third to my brother Hamed, and one-ninth to Harim, the youngest.

At this stage, there are two ways to approach the problem. The slightly easier way is to convert the fractional shares. You can always multiply the top (numerator) of a fraction by any number, and the bottom (denominator) by the same number, and the fraction will be the same. One-half ($\frac{1}{2}$) is the same as two-fourths ($\frac{2}{4}$). So, let’s multiply each fractional camelshare by the number of camels, which is now 36. Thus, the father’s instructions now read:

According to the express wishes of my father $\frac{18}{36}$ of them belong to me, $\frac{12}{36}$ to my brother Hamed, and $\frac{4}{36}$ to Harim, the youngest.

Or, if you prefer, you can convert the fractions to percentages (rounded to the nearest tenth of a percent):

According to the express wishes of my father 50% of them belong to me, 33.3% to my brother Hamed, and 11.1% to Harim, the youngest.

Either way, it quickly becomes clear: the father’s will was incomplete! The percentages don’t add up to 100%, so no matter how many camels were in the herd, some would be left over after the division.

Cool, huh?

*Useless Fractional Camels is the name of my They Might Be Giants cover band.


text:
A. Tanya and Stairways
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

Little girl Tanya climbs the stairs inside a multi-storey building. Every time Tanya climbs a stairway, she starts counting steps from $1$ to the number of steps in this stairway. She speaks every number aloud. For example, if she climbs two stairways, the first of which contains $3$ steps, and the second contains $4$ steps, she will pronounce the numbers $1, 2, 3, 1, 2, 3, 4$.

You are given all the numbers pronounced by Tanya. How many stairways did she climb? Also, output the number of steps in each stairway.

The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways.

Input

The first line contains $n$ ($1 \le n \le 1000$) — the total number of numbers pronounced by Tanya.

The second line contains integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — all the numbers Tanya pronounced while climbing the stairs, in order from the first to the last pronounced number. Passing a stairway with $x$ steps, she will pronounce the numbers $1, 2, \dots, x$ in that order.

The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways.

Output

In the first line, output $t$ — the number of stairways that Tanya climbed. In the second line, output $t$ numbers — the number of steps in each stairway she climbed. Write the numbers in the correct order of passage of the stairways.

Examples
Input
71 2 3 1 2 3 4
Output
23 4
Input
41 1 1 1
Output
41 1 1 1
Input
51 2 3 4 5
Output
15
Input
51 2 1 2 1
Output
32 2 1


text:
avsnitt 4, Ämne 10
Pågår
Fredrik Svensson maj 31, 2020
avsnitt Progress
0% färdig

One strength of Python is its relative ease in handling and manipulating string data. Pandas builds on this and provides a comprehensive set of vectorized string operations that become an essential piece of the type of munging required when working with (read: cleaning up) real-world data. In this section, we’ll walk through some of the Pandas string operations, and then take a look at using them to partially clean up a very messy dataset of recipes collected from the Internet.

Introducing Pandas String Operations

We saw in previous sections how tools like NumPy and Pandas generalize arithmetic operations so that we can easily and quickly perform the same operation on many array elements. For example:In [1]:

import numpy as np
x = np.array([2, 3, 5, 7, 11, 13])
x * 2

Out[1]:

array([ 4,  6, 10, 14, 22, 26])

This vectorization of operations simplifies the syntax of operating on arrays of data: we no longer have to worry about the size or shape of the array, but just about what operation we want done. For arrays of strings, NumPy does not provide such simple access, and thus you’re stuck using a more verbose loop syntax:In [2]:

data = ['peter', 'Paul', 'MARY', 'gUIDO']
[s.capitalize() for s in data]

Out[2]:

['Peter', 'Paul', 'Mary', 'Guido']

This is perhaps sufficient to work with some data, but it will break if there are any missing values. For example:In [3]:

data = ['peter', 'Paul', None, 'MARY', 'gUIDO']
[s.capitalize() for s in data]

---------------------------------------------------------------------------
AttributeError                            Traceback (most recent call last)
<ipython-input-3-fc1d891ab539> in <module>()
1 data = ['peter', 'Paul', None, 'MARY', 'gUIDO']
----> 2 [s.capitalize() for s in data]

<ipython-input-3-fc1d891ab539> in <listcomp>(.0)
1 data = ['peter', 'Paul', None, 'MARY', 'gUIDO']
----> 2 [s.capitalize() for s in data]

AttributeError: 'NoneType' object has no attribute 'capitalize'

Pandas includes features to address both this need for vectorized string operations and for correctly handling missing data via the str attribute of Pandas Series and Index objects containing strings. So, for example, suppose we create a Pandas Series with this data:In [4]:

import pandas as pd
names = pd.Series(data)
names

Out[4]:

0    peter
1     Paul
2     None
3     MARY
4    gUIDO
dtype: object

We can now call a single method that will capitalize all the entries, while skipping over any missing values:In [5]:

names.str.capitalize()

Out[5]:

0    Peter
1     Paul
2     None
3     Mary
4    Guido
dtype: object

Using tab completion on this str attribute will list all the vectorized string methods available to Pandas.

Tables of Pandas String Methods

If you have a good understanding of string manipulation in Python, most of Pandas string syntax is intuitive enough that it’s probably sufficient to just list a table of available methods; we will start with that here, before diving deeper into a few of the subtleties. The examples in this section use the following series of names:In [6]:

monte = pd.Series(['Graham Chapman', 'John Cleese', 'Terry Gilliam',
'Eric Idle', 'Terry Jones', 'Michael Palin'])

Methods similar to Python string methods

Nearly all Python’s built-in string methods are mirrored by a Pandas vectorized string method. Here is a list of Pandas str methods that mirror Python string methods:

len()lower()translate()islower()
ljust()upper()startswith()isupper()
rjust()find()endswith()isnumeric()
center()rfind()isalnum()isdecimal()
zfill()index()isalpha()split()
strip()rindex()isdigit()rsplit()
rstrip()capitalize()isspace()partition()
lstrip()swapcase()istitle()rpartition()

Notice that these have various return values. Some, like lower(), return a series of strings:In [7]:

monte.str.lower()

Out[7]:

0    graham chapman
1       john cleese
2     terry gilliam
3         eric idle
4       terry jones
5     michael palin
dtype: object

But some others return numbers:In [8]:

monte.str.len()

Out[8]:

0    14
1    11
2    13
3     9
4    11
5    13
dtype: int64

Or Boolean values:In [9]:

monte.str.startswith('T')

Out[9]:

0    False
1    False
2     True
3    False
4     True
5    False
dtype: bool

Still others return lists or other compound values for each element:In [10]:

monte.str.split()

Out[10]:

0    [Graham, Chapman]
1       [John, Cleese]
2     [Terry, Gilliam]
3         [Eric, Idle]
4       [Terry, Jones]
5     [Michael, Palin]
dtype: object

We’ll see further manipulations of this kind of series-of-lists object as we continue our discussion.

Methods using regular expressions

In addition, there are several methods that accept regular expressions to examine the content of each string element, and follow some of the API conventions of Python’s built-in re module:

MethodDescription
match()Call re.match() on each element, returning a boolean.
extract()Call re.match() on each element, returning matched groups as strings.
findall()Call re.findall() on each element
replace()Replace occurrences of pattern with some other string
contains()Call re.search() on each element, returning a boolean
count()Count occurrences of pattern
split()Equivalent to str.split(), but accepts regexps
rsplit()Equivalent to str.rsplit(), but accepts regexps

With these, you can do a wide range of interesting operations. For example, we can extract the first name from each by asking for a contiguous group of characters at the beginning of each element:In [11]:

monte.str.extract('([A-Za-z]+)', expand=False)

Out[11]:

0     Graham
1       John
2      Terry
3       Eric
4      Terry
5    Michael
dtype: object

Or we can do something more complicated, like finding all names that start and end with a consonant, making use of the start-of-string (^) and end-of-string ($) regular expression characters:In [12]: monte.str.findall(r'^[^AEIOU].*[^aeiou]$')

Out[12]:

0    [Graham Chapman]
1                  []
2     [Terry Gilliam]
3                  []
4       [Terry Jones]
5     [Michael Palin]
dtype: object

The ability to concisely apply regular expressions across Series or Dataframeentries opens up many possibilities for analysis and cleaning of data.

Miscellaneous methods

Finally, there are some miscellaneous methods that enable other convenient operations:

MethodDescription
get()Index each element
slice()Slice each element
slice_replace()Replace slice in each element with passed value
cat()Concatenate strings
repeat()Repeat values
normalize()Return Unicode form of string
pad()Add whitespace to left, right, or both sides of strings
wrap()Split long strings into lines with length less than a given width
join()Join strings in each element of the Series with passed separator
get_dummies()extract dummy variables as a dataframe

Vectorized item access and slicing

The get() and slice() operations, in particular, enable vectorized element access from each array. For example, we can get a slice of the first three characters of each array using str.slice(0, 3). Note that this behavior is also available through Python’s normal indexing syntax–for example, df.str.slice(0, 3) is equivalent to df.str[0:3]:In [13]:

monte.str[0:3]

Out[13]:

0    Gra
1    Joh
2    Ter
3    Eri
4    Ter
5    Mic
dtype: object

Indexing via df.str.get(i) and df.str[i] is likewise similar.

These get() and slice() methods also let you access elements of arrays returned by split(). For example, to extract the last name of each entry, we can combine split() and get():In [14]:

monte.str.split().str.get(-1)

Out[14]:

0    Chapman
1     Cleese
2    Gilliam
3       Idle
4      Jones
5      Palin
dtype: object

Indicator variables

Another method that requires a bit of extra explanation is the get_dummies()method. This is useful when your data has a column containing some sort of coded indicator. For example, we might have a dataset that contains information in the form of codes, such as A=”born in America,” B=”born in the United Kingdom,” C=”likes cheese,” D=”likes spam”:In [15]:

full_monte = pd.DataFrame({'name': monte,
'info': ['B|C|D', 'B|D', 'A|C',
'B|D', 'B|C', 'B|C|D']})
full_monte

Out[15]:

infoname
0B|C|DGraham Chapman
1B|DJohn Cleese
2A|CTerry Gilliam
3B|DEric Idle
4B|CTerry Jones
5B|C|DMichael Palin

The get_dummies() routine lets you quickly split-out these indicator variables into a DataFrame:In [16]:

full_monte['info'].str.get_dummies('|')

Out[16]:

ABCD
00111
10101
21010
30101
40110
50111

With these operations as building blocks, you can construct an endless range of string processing procedures when cleaning your data.

We won’t dive further into these methods here, but I encourage you to read through ”Working with Text Data” in the Pandas online documentation, or to refer to the resources listed in Further Resources.

Example: Recipe Database

These vectorized string operations become most useful in the process of cleaning up messy, real-world data. Here I’ll walk through an example of that, using an open recipe database compiled from various sources on the Web. Our goal will be to parse the recipe data into ingredient lists, so we can quickly find a recipe based on some ingredients we have on hand.

The scripts used to compile this can be found at https://github.com/fictivekin/openrecipes, and the link to the current version of the database is found there as well.

As of Spring 2016, this database is about 30 MB, and can be downloaded and unzipped with these commands:In [17]:

# !curl -O http://openrecipes.s3.amazonaws.com/recipeitems-latest.json.gz
# !gunzip recipeitems-latest.json.gz

The database is in JSON format, so we will try pd.read_json to read it:In [18]:

try:
except ValueError as e:
print("ValueError:", e)

ValueError: Trailing data

Oops! We get a ValueError mentioning that there is ”trailing data.” Searching for the text of this error on the Internet, it seems that it’s due to using a file in which each line is itself a valid JSON, but the full file is not. Let’s check if this interpretation is true:In [19]:

with open('recipeitems-latest.json') as f:

Out[19]:

(2, 12)

Yes, apparently each line is a valid JSON, so we’ll need to string them together. One way we can do this is to actually construct a string representation containing all these JSON entries, and then load the whole thing with pd.read_json:In [20]:

# read the entire file into a Python array
with open('recipeitems-latest.json', 'r') as f:
# Extract each line
data = (line.strip() for line in f)
# Reformat so each line is the element of a list
data_json = "[{0}]".format(','.join(data))
# read the result as a JSON

In [21]:

recipes.shape

Out[21]:

(173278, 17)

We see there are nearly 200,000 recipes, and 17 columns. Let’s take a look at one row to see what we have:In [22]:

recipes.iloc[0]

Out[22]:

_id                                {'$oid': '5160756b96cc62079cc2db15'} cookTime PT30M creator NaN dateModified NaN datePublished 2013-03-11 description Late Saturday afternoon, after Marlboro Man ha... image http://static.thepioneerwoman.com/cooking/file... ingredients Biscuits\n3 cups All-purpose Flour\n2 Tablespo... name Drop Biscuits and Sausage Gravy prepTime PT10M recipeCategory NaN recipeInstructions NaN recipeYield 12 source thepioneerwoman totalTime NaN ts {'$date': 1365276011104}
url                   http://thepioneerwoman.com/cooking/2013/03/dro...
Name: 0, dtype: object

There is a lot of information there, but much of it is in a very messy form, as is typical of data scraped from the Web. In particular, the ingredient list is in string format; we’re going to have to carefully extract the information we’re interested in. Let’s start by taking a closer look at the ingredients:In [23]:

recipes.ingredients.str.len().describe()

Out[23]:

count    173278.000000
mean        244.617926
std         146.705285
min           0.000000
25%         147.000000
50%         221.000000
75%         314.000000
max        9067.000000
Name: ingredients, dtype: float64

The ingredient lists average 250 characters long, with a minimum of 0 and a maximum of nearly 10,000 characters!

Just out of curiousity, let’s see which recipe has the longest ingredient list:In [24]:

recipes.name[np.argmax(recipes.ingredients.str.len())]

Out[24]:

'Carrot Pineapple Spice & Brownie Layer Cake with Whipped Cream & Cream Cheese Frosting and Marzipan Carrots'

That certainly looks like an involved recipe.

We can do other aggregate explorations; for example, let’s see how many of the recipes are for breakfast food:In [25]:

recipes.description.str.contains('[Bb]reakfast').sum()

Out[25]:

3524

Or how many of the recipes list cinnamon as an ingredient:In [26]:

recipes.ingredients.str.contains('[Cc]innamon').sum()

Out[26]:

10526

We could even look to see whether any recipes misspell the ingredient as ”cinamon”:In [27]:

recipes.ingredients.str.contains('[Cc]inamon').sum()

Out[27]:

11

This is the type of essential data exploration that is possible with Pandas string tools. It is data munging like this that Python really excels at.

A simple recipe recommender

Let’s go a bit further, and start working on a simple recipe recommendation system: given a list of ingredients, find a recipe that uses all those ingredients. While conceptually straightforward, the task is complicated by the heterogeneity of the data: there is no easy operation, for example, to extract a clean list of ingredients from each row. So we will cheat a bit: we’ll start with a list of common ingredients, and simply search to see whether they are in each recipe’s ingredient list. For simplicity, let’s just stick with herbs and spices for the time being:In [28]:

spice_list = ['salt', 'pepper', 'oregano', 'sage', 'parsley',
'rosemary', 'tarragon', 'thyme', 'paprika', 'cumin']

We can then build a Boolean DataFrame consisting of True and False values, indicating whether this ingredient appears in the list:In [29]:

import re
spice_df = pd.DataFrame(dict((spice, recipes.ingredients.str.contains(spice, re.IGNORECASE))
for spice in spice_list))

Out[29]:

cuminoreganopaprikaparsleypepperrosemarysagesalttarragonthyme
0FalseFalseFalseFalseFalseFalseTrueFalseFalseFalse
1FalseFalseFalseFalseFalseFalseFalseFalseFalseFalse
2TrueFalseFalseFalseTrueFalseFalseTrueFalseFalse
3FalseFalseFalseFalseFalseFalseFalseFalseFalseFalse
4FalseFalseFalseFalseFalseFalseFalseFalseFalseFalse

Now, as an example, let’s say we’d like to find a recipe that uses parsley, paprika, and tarragon. We can compute this very quickly using the query() method of DataFrames, discussed in High-Performance Pandas: eval() and query():In [30]:

selection = spice_df.query('parsley & paprika & tarragon')
len(selection)

Out[30]:

10

We find only 10 recipes with this combination; let’s use the index returned by this selection to discover the names of the recipes that have this combination:In [31]:

recipes.name[selection.index]

Out[31]:

2069      All cremat with a Little Gem, dandelion and wa...
74964                         Lobster with Thermidor butter
93768      Burton's Southern Fried Chicken with White Gravy
113926                     Mijo's Slow Cooker Shredded Beef
137686                     Asparagus Soup with Poached Eggs
140530                                 Fried Oyster Po’boys
158475                Lamb shank tagine with herb tabbouleh
158486                 Southern fried chicken in buttermilk
163175            Fried Chicken Sliders with Pickles + Slaw
Name: name, dtype: object