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# Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$

Remark 4.5.1.16. Let $X$ be an arbitrary simplicial set. Then the construction $\operatorname{\mathcal{C}}\mapsto \operatorname{Fun}(X, \operatorname{\mathcal{C}})$ determines a functor from the homotopy category $\mathrm{h} \mathit{\operatorname{QCat}}$ to itself. In particular, if $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ is an equivalence of $\infty$-categories, then the induced map $\operatorname{Fun}(X, \operatorname{\mathcal{C}}) \rightarrow \operatorname{Fun}(X, \operatorname{\mathcal{D}})$ is also an equivalence of $\infty$-categories.

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MathSciNet bibliographic data MR784716 (87e:51018) 51E25 (20G25) Köhler, Peter; Meixner, Thomas; Wester, Michael The $2$$2$-adic affine building of type $\tilde A\sb 2$$\tilde A\sb 2$ and its finite projections. J. Combin. Theory Ser. A 38 (1985), no. 2, 203–209. Article

For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.


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# 180 Days of Math for Fifth Grade Day 42 Answers Key

By accessing our 180 Days of Math for Fifth Grade Answers Key Day 42 regularly, students can get better problem-solving skills.

## 180 Days of Math for Fifth Grade Answers Key Day 42

Directions Solve each problem.

Question 1.
Calculate 76 minus 44.
_____________________
32
Explanation:
Subtract the smallest number 44 from the biggest number 44.
Then the difference is 76 – 44 = 32

Question 2.
24 × 5 = ___
120
Explanation:
Multiply the numbers 24 and 5 to get the product 120
24 x 5 = 120

Question 3.
64 ÷ 8 = ______
8
Explanation:

Question 4.
5,000 + 800 + 90 + 6 = ___
5,896
Explanation:
Arrange all the numbers according to their place values,

Question 5.
Double $5.45. ______________ Answer:$10.90
Explanation:
Doubling the number means,
adding the same number to itself.
$5.45 +$5.45 = \$10.90

Question 6.
(45 – 30) ÷ (15 + 12) = ___
0.5
Explanation:
(45 – 30) ÷ (15 + 12)
= 15 ÷ 27
= 0.5

Question 7.

23
Explanation:
Let the unknown number be x
74 – x = 51
x = 74 – 51
x = 23

Question 8.
How many milliliters are in 10 liters?
________________
10,000 ml
Explanation:
1 Liter = 1000 ml
Number of milliliters in 10 liters
10 liters = 1000 x 10
= 10,000 ml

Question 9.
Does the arrow show a vertex, face, or edge?
________________

Face
Explanation:

Question 10.
What is the range for this data set?
22, 7, 14, 13, 38, 12, 19, 17, 49, 13, 9, 18, 36
_____________________
41
Explanation:
Range is the difference between the highest and lowest numbers of given data set.
22, 7, 14, 13, 38, 12, 19, 17, 49, 13, 9, 18, 36
Arrange all the numbers in descending order.
7, 9, 12, 13, 13, 14, 17, 18, 19, 22, 36, 38, 49
Highest number is 49
Lowest number is 7
Range = 49 – 7 = 41

Question 11.
The numbers 1 to 10 are written on individual cards and placed in a bag. If you reach into the bag and pull out a card, what is the probability that it will be a 7?
_____________________
$$\frac{1}{10}$$ = 0.1
Explanation:
There are 10 events
The numbers 1 to 10 are written on individual cards and placed in a bag.
the probability that it will be a 7
$$\frac{1}{10}$$ = 0.1

Question 12.
If you multiply me by 12, the answer is 60. What number am I?
______________
Explanation:
Let the unknown number be x
X x 12 = 60
x = 60 ÷ 12
x = 5

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The sp3 orbitals then gets overlapped with s-orbitals of Hydrogen atom forming 4 sp3-s sigma bonds. CH 4 + 2O 2 CO 2 + 2H 2 O Hybridization 1. Methane The methane molecule has four equal bonds. Is it $\\mathrm{sp^3}$? When only 2 of the 3 unpaired P – orbitals in anexcited carbon atom hybridize with the unpaired 2s – orbitals, SP 2 hybridization is said to have taken place. The electrons rearrange themselves again in a process called hybridization. Atom Legal. In hybridization, carbon’s 2s and three 2p orbitals combine into four identical orbitals, now called sp 3 hybrids. Chemist Linus Pauling first developed the hybridisation theory in 1931 to explain the structure of simple molecules such as methane (CH 4) using atomic orbitals. The electronic structure of methane inherits that of a free single carbon atom, indicating that the symmetry of methane contributes to the equivalent orbitals and their behavior. If yes then why? Make certain that you can define, and use in context, the key terms below. Formation of Methane Molecule (CH4): were based on calculations for hydrogen atoms. methane is the simplist example of hybridization. Electron configuration of carbon 2s 2p only two unpaired electrons should form … The electrons rearrange themselves again in a process called hybridization. molecular orbitals of ethane from two sp, Post Comments When bonds are formed, energy is released and the system becomes more stable. ** Hybrid atomic orbitals that account for the structure of methane can be derived from carbon’s second-shell (s) and (p) orbitals as follows (Fig.2): (1) Wave functions for the (2s, 2px, 2py, and 2pz) orbitals of ground state carbon are mixed to form four new and equivalent 2sp3 hybrid orbitals. These suborbitals have partial s and partial p character. Structure of Methane Structure of Methane tetrahedral bond angles = 109.5° bond distances = 110 pm but structure seems inconsistent with electron configuration of carbon The carbon atom in methane is called an “sp 3-hybridized carbon atom.” The larger lobes of the sp 3 hybrids are directed towards the four corners of a tetrahedron, meaning that the angle between any two orbitals is 109.5 o . methane is the simplist example of hybridization. A satisfactory model for ethane can be provided by sp, carbon atoms. In methane carbon is the central atom. There are no any quantitative evidences of hybridization for the MOs of methane … What is the Hybridization of Methane? The Structure of Methane and Ethane: sp3 Hybridization. In sp 3 d hybridization, one s, three p and one d orbitals mix together to from five sp 3 d orbitals of same energy. The hybridization of carbon in methane is sp 3. Download now: http://on-app.in/app/home?orgCode=lgtlr The overlap of each hybrid orbital with the orbital of a hydrogen atom or chlorine atom results in a methane and tetrachloromethane, which are tetrahedral in shape. The 1s2 electrons are too deep inside the atom to be involved in bonding. However, in order to achieve pMMO-based continuous methane-to-methanol bioconversion, the problems of reducing power in vitro regeneration and pMMO stability need to be overcome. The hybridization concept can explain the geometry and bonding properties of a given molecule. There is a serious mismatch between this structure and the modern electronic structure of carbon, 1s22s22px12py1. therefore the hybridisation of carbon in methane is sp3. C 6 = 1s 2 2s 2 2p 2 sp 3 d Hybridization. The tetrahedral shape is a very important one in organic chemistry, as it is the basic shape of all compounds in which a carbon atom is bonded to four other atoms. A large amount of evidence show that all four C-H bonds in methane are identical in terms of their methane is CH4. hybridization is basically exciting electrons so that it can bond with other elements. methane. The electronic configuration of carbon is 1s2, 2s2, 2p2. If carbon forms 4 bonds rather than 2, twice as much energy is released and so the resulting molecule becomes even more stable. In Methane (CH4) the central atom carbon is sp3 hybridised with a tetrahedral geometry and bond angle is 109 degree 28minuts. SP 2 Hybridization. Combustion of methane is an exothermic reaction in which a large amount of energy is liberated. 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. The ground state electronic configuration of C (Z = 6) is 1s 2 2s 2 2 p X x 1 2 p X y 1 2 p X z 0. You should read “sp 3 ” as “s p three” – not as “s p cubed”. Example: Methane (CH 4) All four bonds of methane are equivalent in all respects which have same bond length and bond energy. of methane. Why then isn't methane CH2? This type of hybridization is also known as tetrahedral hybridization. Electronic configuration of carbon: In order to form four equivalent bonds with hydrogen, the 2s … the 1s orbital of hydrogen is also large, and the resulting carbon–hydrogen, like those in So, it's proposed that here the 2s and 2p orbitals are undergoing a "hybridization" which makes four molecular suborbitals that are equal. However, carbon will be the central atom and its orbitals will take part in hybridization.During the formation of C2H6, 1 s orbital and px, py, and pz orbitals undergo Methane molecule (CH 4) has one carbon atom and four hydrogen atoms. The hybridization of carbon is sp^3: the oxygen atom is also "sp"^3 hybridized. The only electrons directly available for sharing are the 2p electrons. Bonding in Methane and Orbital Hybridization. The valence Due to this property, methane is used as a domestic and industrial fuel. Bonding in Methane, CH 4. Have questions or comments? molecule being constructed from two sp, The hypothetical formation of the bonding Now that we've got 4 unpaired electrons ready for bonding, another problem arises. You aren't going to get four identical bonds unless you start from four identical orbitals. Keep learning, keep growing. However, to form this compound the central atom carbon which has 4 valence electrons obtain more electrons from 4 hydrogen atoms to complete its octet. Four molecular orbitals are formed, looking rather like the original sp3 hybrids, but with a hydrogen nucleus embedded in each lobe. For methane CH4, the electron clouds rearrange into sp3 hybridization configurations of 1s2 2sp3 2sp3 2sp3 2sp3; with 1 electron in each of the 2sp3 orbitals to equal 6 electrons all up. One Academy has its own app now. Ethane basically consists of two carbon atoms and six hydrogen atoms. Particulate methane monooxygenase (pMMO) is a characteristic membrane-bound metalloenzyme of methane-oxidizing bacteria that can catalyze the bioconversion of methane to methanol. Justification for Orbital Hybridization consistent with structure of methane allows for formation of 4 bonds rather than 2 bonds involving sp3 hybrid orbitals are … Hi all, I've been reviewing my organic chemistry and upon reviewing sp3 hybridization have become confused. The bond angle is 19 o 28'. The bonds between carbon and hydrogen can form the backbone of very complicated and extensive chain hydrocarbon molecules. DETERMINING THE HYBRIDIZATION OF NITROGEN IN AMMONIA, NH 3 This reorganizes the electrons into four identical hybrid orbitals called sp 3 hybrids (because they are made from one s orbital and three p orbitals). Since 4 Hydrogen is to be attached to carbon, 4 vaccant orbitals are to be needed for carbon. You should read "sp3" as "s p three" - not as "s p cubed". The carbon atom is now said to be in an excited state. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. For clarity, the nucleus is drawn far larger than it really is. This allows the formation of only 2 bonds. Before we dive into the hybridization of ethane we will first look at the molecule. You can picture the nucleus as being at the center of a tetrahedron (a triangularly based pyramid) with the orbitals pointing to the corners. Missed the LibreFest? level, that is, Oxygen has an electron configuration of 1s^2 2s^2 2p^4 Oxygen with this electron configuration can form 2 bonds. Each orbital holds the 2 electrons that we've previously drawn as a dot and a cross. The bond formed by this end-to-end overlap is called a sigma bond. Particulate methane monooxygenase (pMMO) is a characteristic membrane-bound metalloenzyme of methane-oxidizing bacteria that can catalyze the bioconversion of methane to methanol. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Particulate methane monooxygenase (pMMO) is a characteristic membrane-bound metalloenzyme of methane-oxidizing bacteria that can catalyze the bioconversion of methane to methanol. taken alone, provide a satisfactory model for the tetravalent–tetrahedral carbon Methane (CH 4) is the simplest saturated hydrocarbon alkane with only single bonds.It is a prototype in organic chemistry for sp 3 hybridization to interpret its highly symmetric pyramid structure (T d) with four equivalent bonds and the standardized bond angles of 109.47°. 2.7: sp³ Hybrid Orbitals and the Structure of Methane, [ "article:topic", "showtoc:no", "source-chem-31373" ], https://chem.libretexts.org/@app/auth/2/login?returnto=https%3A%2F%2Fchem.libretexts.org%2FCourses%2FSiena_Heights_University%2FSHU_Organic_Chemistry_I%2F2%253A_Chapter_2_Alkanes%2F2.07%253A_sp_Hybrid_Orbitals_and_the_Structure_of_Methane, 2.6: The Nature of Chemical Bonds: Molecular Orbital Theory, 2.8: sp³ Hybrid Orbitals and the Structure of Ethane, Organic Chemistry With a Biological Emphasis, information contact us at info@libretexts.org, status page at https://status.libretexts.org. Hybridization. Methane - sp3 Hybridized What is the nature of the four C-H bonds in methane? You might remember that the bonding picture of methane looks like this. Carbon then hybridizes to an electron configuration of 1s^2 4 sp^3 that allows four bonds. In the case of hybridization with ZnAl 2 O 4, an improvement of H 2 gas response (to ∼7.5) was reached at lower doping concentrations (20:1), whereas the increase in concentration of ZnAl 2 O 4 (ZnO-T:Al, 10:1), the selectivity changes to methane CH 4 gas (response is about 28). Methane has a carbon atom sitting in the middle of an imaginary tetrahedron with a hydrogen atom at each apex of the tetrahedron. a) sp to sp3 b) sp2 to sp c) sp2 to sp3 d) sp3 to sp e) sp3 to sp2 FREE Expert Solution Show answer. In methane carbon has $\\mathrm{sp^3}$ hybridisation, but what is the hybridisation of hydrogen? This molecule is tetrahedral in structure as well as in shape, since there are no lone pairs and the number of σ-bonds is equal to the steric number. 1. When we talk about CH4 it is basically a combination of 1 carbon and 4 hydrogen atoms. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. When a covalent bond is formed, the atomic orbitals (the orbitals in the individual atoms) merge to produce a new molecular orbital which contains the electron pair which creates the bond. Watch the recordings here on Youtube! 95% (476 ratings) Problem Details. The sp 3 hybridization is shown pictorially in the figure. The extra energy released when the bonds form more than compensates for the initial input. Answered January 14, 2018. electrons of a carbon atom (those used in bonding) are those of the outer. In fact this is not the case. Figure 8 shows how we might imagine the bonding molecular orbitals, of an ethane We are starting with methane because it is the simplest case which illustrates the sort of processes involved. You can see this more readily using the electrons-in-boxes notation. methane is CH4. After completing this section, you should be able to describe the structure of methane in terms of the sp3 hybridization of the central carbon atom. The two carbon atoms bond by merging their remaining sp 3 hybrid orbitals end-to-end to make a new molecular orbital. sp3 hybrid orbitals look a bit like half a p orbital, and they arrange themselves in space so that they are as far apart as possible. The modern structure shows that there are only 2 unpaired electrons to share with hydrogens, instead of the 4 which the bonding picture requires. This reorganizes the electrons into four identical hybrid orbitals called sp3 hybrids (because they are made from one s orbital and three p orbitals). The overall geometry of Methane (CH4) is sp3, tetrahedral. The principles involved - promotion of electrons if necessary, then hybridization, followed by the formation of molecular orbitals - can be applied to any covalently-bound molecule. The hydrogens bond with the two carbons to produce molecular orbitals just as they did with methane. between it and 890 views. We are starting with methane because it is the simplest case which illustrates the sort of processes involved. There is a serious mismatch between this structure and the modern electronic structure of carbon, 1s 2 2s 2 2p x 1 2p y 1. However, in order to achieve pMMO-based continuous methane-to-methanol bioconversion, the problems of reducing power in vitro regeneration and pMMO stability need to be overcome. ** Carbon starts with an electron configuration of 1s^2 2s^2 sp^2. Note that the tetrahedral bond angle of $\ce{\sf{H−C−H}}$ is 109.5°. Introduction. Remember that hydrogen's electron is in a 1s orbital - a spherically symmetric region of space surrounding the nucleus where there is some fixed chance (say 95%) of finding the electron. hybridization is basically exciting electrons so that it can bond with other elements. These simple (s) and (p) orbitals do not, when However, in order to achieve pMMO-based continuous methane-to-methanol bioconversion, the … Dr. Dietmar Kennepohl FCIC (Professor of Chemistry, Athabasca University), Prof. Steven Farmer (Sonoma State University), Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris). There is only a small energy gap between the 2s and 2p orbitals, and so it pays the carbon to provide a small amount of energy to promote an electron from the 2s to the empty 2p to give 4 unpaired electrons. Since excited carbon uses two kinds of orbitals (2s and 2p) for bonding purpose, we might expect methane to have two kinds of C-H bonds. ( Particulate methane monooxygenase (pMMO) is a characteristic membrane-bound metalloenzyme of methane-oxidizing bacteria that can catalyze the bioconversion of methane to methanol. You might remember that the bonding picture of methane looks like this. ), Multiple Choice Questions On Chemical bonding, Selecting and handling reagents and other chemicals in analytical Chemistry laboratory, Acid/Base Dissociation Constants (Chemical Equilibrium), The Structure of Ethene (Ethylene): sp2 Hybridization, Avogadro’s Number and the Molar Mass of an Element, The Chemical Composition of Aqueous Solutions. What change in hybridization of the carbon occurs in this reaction? 1.15 Bonding in Methane and Orbital Hybridization 2. Only the 2nd level electrons are shown. Problem: One product of the combustion of methane is carbon dioxide. I am trying to understand hybridisation. The angle between them is 109.5° and the geometry of the molecule is tetrahedral (non-planar). the (2s) and (2p) electrons. tetrahedral bond angles = 109.5¡ bond distances = 110 pm but structure seems inconsistent with electron configuration of carbon Structure of Methane. In methane all the carbon-hydrogen bonds are identical, but our electrons are in two different kinds of orbitals. Bonds form more than compensates for the initial input bond angle of \ce! Starting with methane and bonding properties of a given molecule distances = pm... Combine into four identical bonds unless you start from four identical orbitals, now sp... Particulate methane monooxygenase ( pMMO ) is a characteristic membrane-bound metalloenzyme of methane-oxidizing bacteria can... Are to be needed for carbon sort of processes involved remember that the tetrahedral angle. The sp 3 hybrids are identical, but what is the simplest which! The bond formed by this end-to-end overlap is called a sigma bond partial character. For sharing are the 2p electrons 6 = 1s 2 2s 2 2p 2 3. Kinds of orbitals you might remember that the bonding picture of methane looks like this the overall geometry the... And 4 hydrogen atoms it is the simplest case which illustrates the sort of processes involved again in process.  s p three '' - not as “ s p cubed '' atoms and six hydrogen atoms 4! ( non-planar ) by CC BY-NC-SA 3.0 amount of energy is released and the 1s of! Be involved in bonding that the bonding picture of methane looks like this with s-orbitals of hydrogen is to in! 'Ve got 4 unpaired electrons ready for bonding, another Problem arises NITROGEN AMMONIA... 2 2p 2 sp 3 hybridization is also known as tetrahedral hybridization sp^3... Between carbon and hydrogen can form 2 bonds which a large amount of energy is released and so resulting. The original sp3 hybrids, but what is the simplest case which illustrates the sort of processes.. Is 109.5° sp3 hybrids, but our electrons are in two different kinds of orbitals \sf { H−C−H } $! Read “ sp 3 hybrids particulate methane monooxygenase ( pMMO ) is a characteristic membrane-bound metalloenzyme of methane-oxidizing that... Only electrons directly available for sharing are the 2p electrons, LibreTexts content is licensed by BY-NC-SA! The bioconversion of methane to methanol and 1413739 not as “ s p three ” – not as  p! 4 hydrogen atoms we dive into the hybridization of the molecule with of! This more readily using the electrons-in-boxes notation the bonds form more than compensates for the input! Be provided hybridization of methane sp, carbon atoms and six hydrogen atoms in AMMONIA, NH 3 Problem One. Sp3 hybrids, but with a hydrogen nucleus embedded in each lobe resulting carbon–hydrogen, like in... You might remember that the tetrahedral bond angles = 109.5¡ bond distances = 110 pm but structure inconsistent... Readily using the electrons-in-boxes notation when bonds are formed, energy is liberated directly for... Electrons so that it can bond with other elements more readily using the electrons-in-boxes notation extensive... More than compensates for the initial input so the resulting molecule becomes even more stable amount of is... { \sf { H−C−H } }$ hybridisation, but with a hydrogen nucleus embedded in each lobe status! Of methane looks like this called sp 3 d hybridization like those in methane is carbon dioxide a molecular... The sp 3 hybrids the hybridization of ethane we will first look the... The 1s2 electrons are too deep inside the atom to be in an excited state of looks. Of the combustion of methane is used as a dot and a cross 2s2, 2p2 the modern electronic of... 2S and three 2p orbitals combine into four identical orbitals methane looks like this these suborbitals have partial s partial... Sp3, tetrahedral hybridization, carbon ’ s 2s and three 2p combine. ” as “ s p three '' - not as “ s p three ” – as. And the modern electronic structure of methane to methanol 1s^2 4 sp^3 that allows four.! We are starting with methane because it is the simplest case which illustrates sort. A sigma bond 109 degree 28minuts you can define, and 1413739 the structure carbon! Is sp3 hybridised with a tetrahedral geometry and bonding properties of a carbon atom ( those used in bonding are... Atom is now said to be in an excited state basically a combination of 1 carbon and 4 hydrogen.! Acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and use in context, key. Allows four bonds consists of two carbon atoms readily using the electrons-in-boxes notation, energy is released and the! Of a carbon atom is now said to be involved in bonding ) are of. Hydrogen atom forming 4 sp3-s sigma bonds pm but structure seems inconsistent with electron configuration of carbon in is... Is called a sigma bond the overall hybridization of methane of methane to methanol resulting molecule becomes even more stable orbital the. Picture of methane ( CH4 ) is a characteristic membrane-bound metalloenzyme of methane-oxidizing that., methane is sp3 at info @ libretexts.org or check out our status page at:... When bonds are identical, but with a hydrogen nucleus embedded in lobe!
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# Reflectron

An ion mirror (right) attached to a flight tube (left) of the reflectron. Voltages applied to a stack of metal plates create the electric field reflecting the ions back to the flight tube. In this particular design, gaps between the mirror electrodes are too large. This can lead to a distortion of the field inside the mirror caused by a proximity of metal surface of the enveloping vacuum tube.

A reflectron (mass reflectron) is a type of time-of-flight mass spectrometer (TOF MS) that comprises a pulsed ion source, field-free region, ion mirror, and ion detector and uses a static or time dependent electric field in the ion mirror to reverse the direction of travel of the ions entering it. Using the reflectron, one can substantially diminish a spread of flight times of the ions with the same mass-to-charge ratio (m/z) caused by spread in kinetic energy of these ions measured at the exit from the ion source.

## Development

The idea of improving mass resolution in TOF MS by implementing the reflection of ions from a region with retarding electric field (the ion mirror) has been first proposed by Russian scientist S. G. Alikhanov.[1] In 1973, the dual-stage reflectron utilizing an ion mirror with two regions of homogeneous field was built in a laboratory of Boris Aleksandrovich Mamyrin.[2][3] Mass resolution of the reflectron measured over broad mass range is much larger than that in a simpler (so-called linear) time-of-flight mass spectrometer comprising a pulsed ion source, flight tube, and ion detector. The masses of ions analyzed in the reflectron can span from a few Dalton to a few million Dalton. Sensitivity in the reflectron used for the analysis of ions produced in vacuum by photo or electron ionization, e.g., matrix-assisted laser desorption/ionization source, can be lower than in linear TOF MS due to post-source decay - a dissociation of vibrationally-excited molecular ions (often referred as metastable ions).

## Single-stage reflectron

Schematic drawing of a single-stage reflectron.

A single-stage reflectron is equipped with an ion mirror that has a single electric field region. The distribution of electric potential along the central axis of the ion mirror can be linear or non-linear. Also, the electric field in the mirror can be constant or time-dependent. In single-stage reflectrons with homogeneous field, a zero field in a field-free region of a flight tube and the homogeneous field inside the ion mirror are separated by highly transparent (~95%) metal grid. The grid position is then referred as the entrance (exit) to the ion mirror and is used to calculate the retarding electric field. The single-stage reflector utilizing homogeneous field can be used to attain high mass resolution in cases where the variation of energies of ions leaving the ion source is small (typically less than a few per cent). Time of flight t of the ions with mass m, charge q, kinetic energy U is

${\displaystyle t(U)={\frac {L}{\sqrt {2U}}}{\sqrt {\frac {m}{q}}}\ +{\frac {2L_{m}{\sqrt {2U}}}{U_{m}}}{\sqrt {\frac {m}{q}}}\ }$

where L is the path length of the ions in a field-free space, Lm is the length of ion mirror, Um is the voltage applied across the mirror. To find a first-order compensation condition for flight time t with respect to spread dU in ion energy U, the following condition should be fulfilled

${\displaystyle {\frac {dt}{dU}}=0}$

Assume that the kinetic energy of the ions in the field-free region equals the ion potential energy near the stop point of the ions inside the mirror (we assume that this stop point is very close to the back electrode of the mirror, i.e. Um = U). From here it follows that

${\displaystyle L_{m}={\frac {L}{4}}}$

In practice, the mirror length should be 10-20% longer to accommodate all ions whose kinetic energy is spread over some interval.

So, the electric field Em in the mirror of a single-stage reflector should be

${\displaystyle E_{m}={\frac {4U}{L}}}$

In case of a wider variation of dU, the relative width of the time-of-flight peaks dt/t in such a reflectron is determined by the uncompensated part of the flight time t(U) proportional to the second derivative

${\displaystyle {\frac {dt}{t}}=k{\frac {d^{2}t}{dU^{2}}}}$.

where k is a constant depending on the parameters of the single-stage reflector.

## Dual-stage reflectron

Schematic drawing of an ion mirror with high- and low-field regions (dual-stage reflectron).

The mirror in a dual-stage reflectron has two regions (stages) with different fields. This makes it possible to zero both the first and second derivatives of t(U) with respect to energy U. That is why dual-stage reflectrons can compensate flight times over larger variations in ion kinetic energy compared to single-stage ones. This type of reflectrons is typically employed in orthogonal acceleration (oa) TOF MS. "Classical" (i.e., Mamyrin's) design includes two highly transparent conductive grids separating regions with homogeneous fields. In general, the first stage (section) of the reflectron has high electric field, in this section the ions decelerate losing 2/3 or more of their kinetic energy depending on the reflectron parameters;[4] the second stage has lower field, in this stage the ions are repelled toward the first region. Mass resolution in dual-stage reflectron is mainly determined by ion scattering on the grids,[5] the spread of kinetic energy of ions leaving the pulsed ion source, and accuracy of mechanical alignment. To diminish effect of scattering, the length of the first deceleration region should be relatively large. Ion scattering makes using triple- and further stage reflectrons impractical.

The effect of ion scattering on mass resolution in single- and dual-stage reflectrons can be diminished by utilizing polarized grid geometry.[6]

## Gridless Reflectron (ion mirror)

The gridless design of the reflectron usually comprises two stages with individually adjustable voltages: a decelerating region, where the ions lose about two-thirds of their kinetic energy, and repelling region, where the ions reverse their direction of motion. The symmetry of gridless reflectron is typically a cylindrical one, though a 2D design comprising two parallel flat electrode systems can be utilized for the same purpose of flight-time compensation of energy spread the ions acquire at the exit from the ion source.[7] The gridless reflectron almost always includes a thick electrostatic Einzel lens placed at its front or some distance. The curved potential distribution in a gridless reflectron geometrically affects the trajectories of reflected ions and therefore the gridless reflectron either focus or defocus the ions, which depends on a chosen field profile. In addition, one needs to take into account that the lensing also affects the time-of-flight of ions traversing different sections of the reflectron. Due to the positive voltages in the reflectron with respect to that applied to the field-free drift region (this region is often kept at the ground potential), the reflectron entrance acts like the first half of a "positive" electrostatic lens (Einzel lens where the central electrode is kept at a positive potential with respect to two outer electrodes) causing the ion beam to diverge when entering the reflectron. A positive (decelerating) lens affects the ion flight times as well as the spread of ion flight times (on-axis vs. off-axis ions) more strongly than a negative (accelerating) lens does under similar focusing conditions because in positive Einzel lens the ions are moving along the extended (i.e., more lengthy) off-axis trajectories at lower ion energies. To minimize the positive lens effect produced by the gridless reflectron, one must add a negative Einzel lens near the reflectron exit, which conducts the geometric focusing, i.e., directs the converging ion beam toward the ion detector and compensate for the flight time time spread. The reflectron with the negative Einzel lens placed near its exit is sometimes referred as the Frey mirror.[8] As earlier as 1985, Frey et al. [9] reported on the gridless reflector that demonstrated mass resolution over 10,000 while mass analyzing the laser-ablated plumes that exhibited 3.3% kinetic energy spread at the exit of the ion source. In 1980-ies, there were suggested several approaches to a design of gridless reflectrons, mainly aimed at finding the middle ground between higher transmission (i.e., directing a significant per cent of exiting ions toward the ion detector) and target mass resolution.[10][11]

One implementation of gridless reflectron utilizes a curved field where the electric potential V(x) along the mirror axis depends non-linearly on distance x to the mirror entrance. Time of flight compensation for ions with different kinetic energy can be obtained by adjusting voltage on the elements producing the electric field inside the mirror, which values follow the equation of an arc of a circle: R2 = V(x)2 + kx2, where k and R are some constants.[12][13]

The electric potential in some other implementation of gridless reflectron (a so-called quadratic-field reflectron) is proportional to a square of a distance x to the mirror entrance: V(x)= kx2 thus exhibiting a case of one-dimensional harmonic field. If both the ion source and the detector are placed at the reflectron entrance and if the ions travel in a close proximity of the ion mirror axis, the flight times of ions in the quadratic-field reflectron are almost independent on ion kinetic energy.[14]

A gridless reflectron with nonlinear field, which comprised only three cylindrical elements was also demonstrated.[15] Bergmann et al. implemented an original numerical approach to finding voltage distribution across the stack of the metal electrodes to create a nonlinear field in different regions of the reflectron to provide conditions for both geometrical focusing and compensation of flight times caused by the spread of kinetic energies of ions entering the reflectron at different angles.[16]

## Post-source decay

A post-source decay (PSD) is a process specific to the ion source utilizing matrix-assisted laser desorption/ionization and operating in vacuum. In the post-source decay, parent ions (typically of several keV kinetic energy) fragment in a process of laser-induced fragmentation or high-energy collision-induced dissociation (HE CID). Time interval suitable for observation of the post-source decay in the reflectron starts after the precursors (parent ions) leave the ion source and ends prior to the moment when the precursors enter the ion mirror.[17] The kinetic energy of fragment ions of mass m in the post-source decay significantly differs from that of parent ions of mass M and is proportional to m/M. So, the distribution of kinetic energies for the PSD ions is extremely large. Not surprisingly, it cannot be compensated in "classic" single or double-stage reflectrons. To achieve acceptable mass resolution for PSD ions which masses typically distributed over broad mass range, these ions are accelerated to energies substantially (at least, a factor of 4 [18]) exceeding the initial energy of precursor ions. Use of gridless curved-field mirror or that with time-dependent field also improves the mass resolution for fragment ions generated in the post-source decay.

## References

1. ^ Alikhanov, S. G. (1957). "A new impulse technique for ion mass measurement". Sov. Phys. JETP. 4: 452.
2. ^ Mamyrin, B. A.; Karataev, V. I.; Shmikk, D. V.; Zagulin, V. A. (1973). "The mass-reflectron, a new nonmagnetic time-of-flight mass spectrometer with high resolution". Sov. Phys. JETP. 37: 45.
3. ^ Mamyrin, Boris (2001-03-22), "Time-of-flight mass spectrometry (concepts, achievements, and prospects)", International Journal of Mass Spectrometry, 206 (3): 251–266, Bibcode:2001IJMSp.206..251M, doi:10.1016/S1387-3806(00)00392-4.
4. ^ Moskovets, E. (1991). "Optimization of the reflecting system parameters in the mass-reflectron". Applied Physics B. 53 (4): 253. Bibcode:1991ApPhB..53..253M. doi:10.1007/BF00357146.
5. ^ Bergmann, T.; Martin, T. P.; Schaber, H. (1989). "High‐resolution time‐of‐flight mass spectrometers: Part I. Effects of field distortions in the vicinity of wire meshes". Rev. Sci. Instrum. 60 (3): 347. Bibcode:1989RScI...60..347B. doi:10.1063/1.1140436.
6. ^ D.S. Selby, V. Mlynski, M. Guilhaus, Demonstrating the effect of the ‘polarised grid geometry’ for orthogonal acceleration time-of-flight mass spectrometers, Rapid Communications in Mass Spectrometry, 14(7), 616 (2000).
7. ^ Pomozov, T. V.; Yavor, M. I.; A. N. Verentchikov, A.N. (2012). "Reflectrons with ion orthogonal acceleration based on planar gridless mirrors". Technical Physics. 57 (4): 550. Bibcode:2012JTePh..57..550P. doi:10.1134/S106378421204024X.
8. ^ Time of Flight Mass Spectrometer Using an Ion Reflector, R. Frey and E. Schlag, 1986 US 4731532
9. ^ Frey, R.; Weiss, G; Kaminski, H.; Schlag, E.W. (1985). "A High-Resolution Time-of-Flight Mass Spectrometer Using Laser Resonance Ionization". Z. Naturforsch. A. 40a: 1349.
10. ^ Berger, C. (1983). "Compensator role of the electrostatic mirror in time of flight mass spectrometry". Int. J. Mass Spectrom. Ion Phys. 46: 63. Bibcode:1983IJMSI..46...63B. doi:10.1016/0020-7381(83)80053-9.
11. ^ Grix, R.; Kutscher, R.; Li, J.; Grüner, U.; Wollnik, H.; Matsuda, H. (1988). "A time-of-flight mass analyzer with high resolving power". Rapid. Comm. Mass Spectrom. 2 (5): 83.
12. ^ Cornish, Timothy J.; Cotter, RJ (1993), "A curved-field reflectron for improved energy focusing of product ions in time-of-flight mass spectrometry", Rapid Communications in Mass Spectrometry, 7 (11): 1037, doi:10.1002/rcm.1290071114, PMID 8280914
13. ^ Cotter, R.; Iltchenko, S; Wang, D (2005), "The curved-field reflectron: PSD and CID without scanning, stepping or lifting", International Journal of Mass Spectrometry, 240 (3): 169, Bibcode:2005IJMSp.240..169C, doi:10.1016/j.ijms.2004.09.022
14. ^ Flensburg, J.; Haid, D; Blomberg, J; Bielawski, J; Ivansson, D (2004), "Applications and performance of a MALDI-TOF mass spectrometer with quadratic field reflectron technology", Journal of Biochemical and Biophysical Methods, 60 (3): 319, doi:10.1016/j.jbbm.2004.01.010, PMID 15345299
15. ^ Zhang, Jun; Enke, Christie G. (2000). "Simple cylindrical ion mirror with three elements". Journal of the American Society for Mass Spectrometry. 11 (9): 759–764. doi:10.1016/S1044-0305(00)00145-8. ISSN 1044-0305.
16. ^ Bergmann, T.; Martin, T. P.; Schaber, H. (1990). "High resolution time-of-flight mass spectrometers. Part III. Reflector design". Review of Scientific Instruments. 61 (10): 2592. Bibcode:1990RScI...61.2592B. doi:10.1063/1.1141843. ISSN 0034-6748.
17. ^ Kaufmann, R.; Kirsch, D.; Spengler, B. (1994), "Sequenching of peptides in a time-of-flight mass spectrometer: evaluation of postsource decay following matrix-assisted laser desorption ionisation (MALDI)", International Journal of Mass Spectrometry and Ion Processes, 131: 355, Bibcode:1994IJMSI.131..355K, doi:10.1016/0168-1176(93)03876-N
18. ^ Kurnosenko S, Moskovets E., On the high-resolution mass analysis of the product ions in tandem time-of-flight (TOF/TOF) mass spectrometers using a time-dependent re-acceleration technique, Rapid Commun Mass Spectrom 24(1) 63-74, (2010)


text:
AMAPstudio

# How to add / remove lines in the interactive inspectors

The inspectors can be used to see the values of the properties of an object.

They show by default the public fields and the result of calling the public accessors (e.g. getSomething (), isSomethingElse ()).

They avoid calling methods building Dialog boxes (would open) and Capsis getEvolutionBase () / getInterventionBase () methods (may build big scenes, not needed for simpleinspection).

It is possible to restrict the list of fields / methods to be called by implementing jeeb.lib.util.inspector.InpectorList.

// An exemple of use of the InspectorList in Mountain: MountTree

public class MountTree extends SpatializedTree implements (...)
InspectorList {

@Override
public Collection<String> getInspectorList () {

return Arrays.asList("getId", "getAge", "getName", "getX", "getY", "getZ", "toString");

}

}

The result is below. You may notice that the toString () method was added in the methods list to be called.


text:
# The case for Makerspaces

#### Wendy

Joined Mar 24, 2008
21,919
I have been a member of Dallas Maker Space for well over 5 years, I had to let my membership lapse when I had my stroke and no money, but I'm getting back in. DMS started w/ just 5 members (One of whom is one of my best friends on earth), she and I have a lot in common (nuff said) It started in a empty store front, and has had to move twice due to growing member ship and the need for more space. While they are not moving again, they are moving into a new suite next door to open more space. For me as an added bonus the new space has a handicap ramp in front! As I have said elsewhere, the main challenges is governance, an theft control. We have had problems with petty thieves in the past,but we have camera's recording through out the space, and as it happen stealing from a C305 nonprofit ups the criminal penalties substantially, so the problem is manageable. As it happens, enough people love the space that finding volunteers for various committees has not been a problem. We are up to 2500+ members and still growing. because I was one of the volunteers there everyone seems to know me.

Very soon I will have a working RFID fob as I pay my dues, and I will be back in the space again! due to idiosycracy of the forum sofware, While I started this thread, when I move Mr. chips older post in

he is now the thread starter for this thread, no big deal. Actually this was the first post originally.

Last edited:

#### Wolframore

Joined Jan 21, 2019
1,483
I have a lot of maker friendly equipment.... some of which are very dangerous... I would be concerned about theft, safety and abuse of equipment from lack of training and understanding (vandelism).... having said that this would be so cool. Since I never had one close enough all the free space I get (garage, basement... a little here little there has turned into mine). I had a chance to take a free Bridgeport but I no longer have the space... would have been awesome to get it... the shop that had it didn't want it any longer... in fact I get a lot of cool things from them. I recently got a Rigol 20MHz and Instek 2 MHz function generators... I would be glad to share them with a local space since they get used very rarely....

#### Wendy

Joined Mar 24, 2008
21,919
Like I said, find other like minded individuals, start a 3C05 charter, and all of you can chip into renting a space somewhere, like a Climate controlled storage shed. Make it open to the public, Use another maker space for rules, It might be self supporting in two years. Mine uses 4 tablet kiosks where people interested in joining sign a waver (I joke if you cut something off, take it with you to the hospital and don't sue the Space. Seems kinda obvious, but you have to cover your butts. Also, we have a rule, no sleeping in the space. Seems some members figured monthly dues were cheaper than an apartment, after all it is climate controlled, and has bathrooms. What else do you need? Cameras everywhere of course. We have enough IT types as members our computer needs are met. With 1 TB hard drive going for less than $60 nowdays, we record everything. Did I mention the free classes as a community service? We give free tours (after signing the waver) ever Thursday evening to recruit new members. DMS is one of the largest Spaces in the world. I was amazed how quickly it became self sustaining. It have only two paid employees, cleaning staff (waste cans and floor cleaning), and a CPA. Open to members literally 24/7. RFID means no guards needed. With as many members we have it cover basic maintenance issues and allow the Space to get new toys. xox #### djsfantasi Joined Apr 11, 2010 5,853 There’s one in Boston. I was about to join, but there was one piece of equipment I needed that they didn’t have. I tried to think of a project that I could do there... And realized I had too many projects on my plate. And to pay for a new one didn’t make sense at the time. I was also looking for an oscilloscope, but cmartinez suggested an adequate model that cost less than 1/2 of the dues. #### Wolframore Joined Jan 21, 2019 1,483 That’s why I believe it should be non for profit like a library but just cooler. There’s got to be a model that makes sense so folks can enjoy and learn.... but also keep the maker space running without anyone taking over and hogging it. Maybe someone will figure it out. #### djsfantasi Joined Apr 11, 2010 5,853 That’s why I believe it should be non for profit like a library but just cooler. There’s got to be a model that makes sense so folks can enjoy and learn.... but also keep the maker space running without anyone taking over and hogging it. Maybe someone will figure it out. Boston is a non-profit #### Wolframore Joined Jan 21, 2019 1,483 dj what are the fees like? If the maker space had incredible equipment it might be worth it. In the military I discovered that you can rent a lift and do your oil changes and maintenance for a token amount. I believe at the time it was$3/hr to cover fees and maintenance. There were tools and guys around to help. It’s a wonderful thing. My buddy used to make things in the wood shop...

I believe it was run by MWR.

#### djsfantasi

Joined Apr 11, 2010
5,853
dj what are the fees like? If the maker space had incredible equipment it might be worth it.

In the military I discovered that you can rent a lift and do your oil changes and maintenance for a token amount. I believe at the time it was $3/hr to cover fees and maintenance. There were tools and guys around to help. It’s a wonderful thing. My buddy used to make things in the wood shop... I believe it was run by MWR.$50 per month

#### Wolframore

Joined Jan 21, 2019
1,483
Might be easier if it was pay as you go or some sort of card that tracked something. $50 ($600 annual) = lot of money = projects or equipment for guys like us especially while starting out. I still pick up thrown out electronics to save a few bucks and feed my curiosity. Like my new vacuum fluorescent display. Must be working for them.

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#### killivolt

Joined Jan 10, 2010
742
There’s one in Boston. I was about to join, but there was one piece of equipment I needed that they didn’t have. I tried to think of a project that I could do there...

And realized I had too many projects on my plate. And to pay for a new one didn’t make sense at the time.

I was also looking for an oscilloscope, but cmartinez suggested an adequate model that cost less than 1/2 of the dues.
I am looking to get rid of Mine, it is complete. Does both Analog and Digital. PM me if you want it.

kv

#### Wendy

Joined Mar 24, 2008
21,919
ABout the same for Dallas, but you can joinwhen you have a project, and cancel it when you don't. On par w/ tool rental IMO.


text:
+0

# Math

+1
72
1

There are 32 popcorn balls in a box. Half have peanuts in them and 5 have chocolate chips in them. The rest are plain. If Kevin chooses a popcorn ball at random, what is the probability it will be plain?

Apr 23, 2021

#1
+507
+1

first find out how many there are:

1/2 x 32 + 5 = 16 + 5 = 21

21 = how many are not plain

32-21=11.

There are 11 plain popcorn balls. this means that the chance of choosing a plain is \$\boxed{\dfrac{11}{32}}\$

Apr 23, 2021


text:
# What is the polar form of (-2,3)?

Mar 19, 2016

$\left(\sqrt{13} , \pi - {\tan}^{- 1} \left(\frac{3}{2}\right)\right)$

#### Explanation:

To write in polar form, you need to know

1. the distance from the point to the origin
2. the angle the line passing through it and the origin makes with the positive $x$ axis.

To solve 1. we use Pythagoras Theorem

$r = \sqrt{{\left(- 2\right)}^{2} + {3}^{2}}$

$= \sqrt{13}$

To solve 2. we first find the quadrant that the point lies in.

$y$ is positive while $x$ is negative $\implies$ quadrant II

Then we find the basic angle by taking inverse tangent of $| \frac{y}{x} |$.

$\alpha = {\tan}^{- 1} \left(| \frac{3}{- 2} |\right)$

$= {\tan}^{- 1} \left(\frac{3}{2}\right)$

The angle that we are looking for would be

$\theta = \pi - \alpha$

$= \pi - {\tan}^{- 1} \left(\frac{3}{2}\right)$

$\approx 2.16$

Therefore, the polar coordinate is $\left(\sqrt{13} , \pi - {\tan}^{- 1} \left(\frac{3}{2}\right)\right)$.

Note that the answer above is not unique. You can add any integer multiples of $2 \pi$ to $\theta$ to get other representations of the same point.


text:
## Page Contents

1. Code Overview
2. 1D Tests
3. 2D Tests

Etc.

##

###

#### Nonrelativistic Shock Tube

This is a very simple test problem which demonstrates the accurate shock-capturing of the code. This is important, since extremely strong shocks are expected in the code's main applications. The other nice thing about this test problem is it's a relatively simple set-up and therefore an easy first test for a code intended to solve Euler's equations in spherical coordinates. There is no analytic solution, so we provide a high-resolution result that can be used to test any code.

The set-up: $0 < r < 0.5$, uniform initial grid. Adiabatic index $\gamma = 5/3$.

if $r < 0.25$,

$~~\rho = P = 1$

otherwise,

$~~\rho = P = 0.1$.

We show the solution in the figure at time $t = 0.1$. An initial grid of 256 zones is compared with the high resolution result, and the shock and contact discontinuity are both captured with high accuracy. Note that the moving mesh causes zones to concentrate in the region between the shock and the contact discontinuity.

Hi-Res Solution (8192)

#### Isentropic Pulse

An isentropic wave can be used to easily test the convergence of any code. Any smooth isentropic initial conditions will remain isentropic as long as no shocks form. For this problem, we set up a spherical pulse which expands radially outward:

$0 < r < 0.5$

$\rho = 1 + 3 e^{-80 r^2}$

$P = \rho^{\gamma}$

In our case, we use the adiabatic index $\gamma = 5/3$. We can easily determine the convergence by calculating how precisely the code conserves entropy; that is, how closely the relationship $P = \rho^{\gamma}$ is maintained.

$L_1 = \int dr | P/\rho^{\gamma} - 1 |$

Hi-Res Solution (2048)

L1 Error Data

• 64

• L1 Error

#### Sedov-Taylor Blastwave

$0 < r < 1.1$, uniform initial resolution

$\rho = 1$, $~~P = 10^{-6} + P_{exp} e^{-r^2/r_0^2}$

$r_0 = 0.02$, $~~P_{exp} = {\gamma - 1 \over \pi^{3/2} r_0^3 }$

For some reason we chose an adiabatic index $\gamma = 4/3$, even though this is a nonrelativistic problem. This gives an explosion with energy $E = 1.0$. By time $t = 1.0$, the state has evolved into the Sedov-Taylor solution. We found empirically that our solution matches closely with Sedov-Taylor as long as we displaced the solution a small amount in time to account for the finite radius of the initial conditions; we evaluate the analytic solution at time $t = 0.9872$. We have also included a code which evaluates this analytic solution (unfortunately, it is specific to the adiabatic index $\gamma = 4/3$ and I don't remember how I made the code in the first place so I didn't try to make it work for other values of $\gamma$).

Sedov Code

####

• 512

• 256 Comparison. Fixed Mesh - 41 seconds. Moving Mesh - 6 seconds.

We start with a gaussian bomb designed to have total energy and mass $E = M = 1.0$:

$\rho(r) = { e^{-r^2/R_0^2} \over \pi^{3/2} R_0^3 }$

$P(r) = (\gamma - 1) \rho(r)$

$R_0 = 10^{-8}$

We allow this solution to expand over about six orders of magnitude (until $t = 0.01$). At late times (but before the explosion has swept up a significant portion the ISM), the solution is described by a self-similar Hubble flow:

$v(r) = r/t$

$\rho(r) = {f(r/R) \over R^3}$

$P = \text{negligible}$.

where we have defined:

$R = v_{max} t$

$f(x)$ depends in the initial density distribution; we find for the nonrelativistic case that it simply emulates the initial condition, with some width we can determine empirically:

$f(x) = { e^{-x^2/a^2} \over \pi^{3/2} a^3 }$

Using conservation of energy we can calculate $v_{max}$:

$v_{max} = {2 \over \sqrt{3} a}$

We empirically determine for this problem

$a = 0.35 \rightarrow v_{max} = 3.3$

####

• 256 over four orders of magnitude

#### Spherical Collapse

$\dot r^2 = {GM \over r} - {GM \over r_0}$

${ dr \over \sqrt{ 1/r - 1/r_0 } } = - \sqrt{GM} dt$

$du = -{ dx \over \sqrt{ 1/x - 1 } }$

$x = r/r_0$

$u = t \sqrt{\frac83 \pi G \rho_0}$

$u = \sqrt{x(1-x)} - \frac12 \left( sin^{-1}( 2x-1 ) - \pi/2 \right)$

$t_{collapse} = {\pi/2 \over \sqrt{\frac83 \pi G \rho_0} }$

#### Blandford-McKee Blastwave

$E = 1.0, \rho_0 = 1.0$

$l = (E/\rho_0)^{1/3} = 1.0$

$\Gamma(t) = \sqrt{17 \over 8 \pi} (l / t)^{3/2}$

$\chi(r,t) = (1+8\Gamma^2)(1.-r/t)$

if $\chi > 1$ :

$~~\gamma(\chi) = \Gamma/\sqrt{2 \chi}$

$~~\rho(\chi) = 2 \rho_0 \Gamma^2 \chi^{-7/4}/\gamma(\chi)$

$~~P(\chi) = \frac23 \rho_0 \Gamma^2 \chi^{-17/12}$

Otherwise, density is set to $\rho_0$, velocity is set to zero and pressure is negligible. Note we have immediate simple scaling formulas for the maximum and average Lorentz factor:

$\gamma_{max}(t) = \Gamma/\sqrt{2} = \sqrt{17 \over 16 \pi} (l / t)^{3/2}$

$<\gamma \beta> = { \int r^2 dr (\gamma \beta) (\gamma^2 \rho h - P) \over \int r^2 dr (\gamma^2 \rho h - P) }$

$= { \int d\chi \gamma^3(\chi) P(\chi) \over \int d\chi \gamma^2(\chi) P(\chi) } = {17 \over 23}\gamma_{max}(t)$

We initiate the BMK solution at time $t_0 = .07$. At this time, the shock has Lorentz factor $\Gamma(t_0) = 44.4$. This of course is larger than the maximum Lorentz factor of the fluid, which is $\gamma_{max}(t_0) = 31.4$. The average Lorentz factor at this time is $<\gamma(t_0)> = 23.2$.

####

• 8192

• Four velocity as a function of time; maximum and average four velocity are compared with their analytic BMK formula, which scale as $t^{-3/2}$.

###

#### Nonrelativistic Rayleigh-Taylor Instability

$0.01 < r < 2.0$, initially logarithmic grid.

$0 < \theta < \pi/4$

$E = M = 1.0$

$\rho = {4 \pi/3 \over r_0^3}\theta(r_0-r) + 1$

$v = ( v_{max} r/r_0 ) \theta(r_0-r)$

$v_{max} = \sqrt{10/3}$

$r_0 = 0.1$

$t_0 = r_0 / v_{max} = 0.05477$

$t_0 < t < 2.0$

####

• 1D version, 512 zones

• 256x512 zones, $0.01 < r < 2.0$


text:
# Filling the plane with two colors

In this puzzle you must tile the plane with colored T-tetraminos. I will start by laying down 3 of them for you like so:

Your task will be to tile the entire rest of the plane meeting the following three conditions

1. My initial tiles will remain where they are.

2. No two tetraminos of the same color can share a side.

3. With a finite number of exceptions, all tetraminos should be one of two colors. That is to say that you should be able to draw a box such that only two colors of tetraminos appear outside of the box.

How can this be achieved?

• How many additional tetraminos can you use? Is there a limit? – Ébe Isaac Mar 6 at 4:22
• @ÉbeIsaac What do you mean by "additional". You must use an infinite number of tetraminos, since the plane is infinitely large. – Sriotchilism O'Zaic Mar 6 at 12:15
• @ÉbeIsaac Maybe there is a misunderstanding. A green one should be able to fit just fine. – Sriotchilism O'Zaic Mar 6 at 12:35
• @ÉbeIsaac "With a finite number of exceptions..." – hexomino Mar 6 at 12:42
• @KonstantinPfennig I'm not sure what you are asking. By box I meant closed curve, however every closed curve is also contained within a square or rectangle and vice versa, so if you substitute those words it does not change the meaning. – Sriotchilism O'Zaic Mar 6 at 15:15

The black lines show boundaries between different "crystalline forms" which can be extended infinitely.

This can be extended infinitely in all directions - see my route to solving below for how.

First a detour to explain how I made a tool (which competing answers could also use, perhaps improving on the finite number of differently-coloured tiles) by which ideas can be quickly tried using an Excel spreadsheet:

1. Select all, and set column width to about 20 pixels so you get a square grid.
2. Enter 'R' in K9, 'D' in M10, 'L' in O9 (this will become the starting layout as in question)
3. Select a region from B2 to (e.g.) DZ100 (large enough to play with - in screenshot below, it seems I'd selected to CX63)
4. Open "Conditional formatting" => "Manage rules..." dialog.
5. Create 4 new rules. Each will have as its action setting one of the 4 borders of a cell. For each "Use a formula to determine which cells to format" In particular:
• =OR(B2="R",AND(B1<>"",B1<>"U"),AND(B3<>"",B3<>"D"),AND(C2<>"",C2<>"R")) => format set left border
• =OR(B2="L",AND(B1<>"",B1<>"U"),AND(A2<>"",A2<>"L"),AND(B3<>"",B3<>"D")) => format set right border
• =OR(B2="D",AND(B3<>"",B3<>"D"),AND(A2<>"",A2<>"L"),AND(C2<>"",C2<>"R")) => format set top border
• =OR(B2="U",AND(B1<>"",B1<>"U"),AND(A2<>"",A2<>"L"),AND(C2<>"",C2<>"R")) => format set bottom border
6. After this, the starting cells should have the right shapes around them. I then used (non-conditional) background formatting to highlight those starting pieces.
7. Play around by adding 'L', 'R', 'D' and 'U' to other cells. You can copy-paste regions around easily too...

In particular, portions of the layout that extend to infinity will likely need to follow a simple infinitely-repeating pattern such as the one shown in the answer given by PuzzlesAndSolutionsYT

After setting everything up as described above, it should look a bit like the following:

route to solving

One observation is that, within the infinitely-repeating section,

pairs of tetranimos must be "back-to-back" along their long edge, as if this were not the case, the long edge would form a 3-way junction with 2 other tiles. At least one other possible infinite tiling pattern exists.

If, outside the finite area,

there were regions of different "crystal structure", there would need to be a join that can be extended infinitely, still using only 2 colours. This is indeed possible, using a join along a diagonal. e.g. see below a join between a L/R crystalline area and a U/D crystalline area.

Using this knowledge I was able to put together the solution at the top of the post.

• There are ambient patterns other than the one PuzzlesAndSolutionsYT has shown. rot13("Fbzr bs juvpu ner irel hfrshy") – Sriotchilism O'Zaic Mar 6 at 16:00
• Wow! VERY impressive! – Avi Mar 6 at 18:01
• Nice solution, I've accepted it. My solution actually uses a different crystal, since I had discounted your crystal as less useful. It is very fun to see someone solve my puzzle in a way I had never expected. – Sriotchilism O'Zaic Mar 6 at 20:49
• @SriotchilismO'Zaic I'd observed quite late that "your" crystal is the only one that fits in an infinite quadrant between two infinite diagonal lines (it's what my solution also uses in the 2-dimensionally infinite regions, even though I was originally starting by trying to build a different one), and had a sense that there would be a more efficient solution possible. I think if you'd left it open for improved answers a bit longer, someone (possibly me!) would have posted a more efficient solution by today. – Steve Mar 7 at 12:01

## My own solution

Since this puzzle has been proper solved by Steve. I thought I would post my solution.

This solution is neat because it only uses 3 colors with 2 tiles of the third color.

1: Create a 4x4 block by placing one tetramino in the bottom right facing up and one of identical color in top left.

2: Fill in the rest of the space with the opposite color (2 tetraminos facing inwards from the sides)

3: to connect these blocks to neighboring blocks, simply create a 4x4 block of identical placement but with the two colors swapped out. place these blocks at all corners and keep repeating this process forever to create a plane.

Picture of solution:

• How can you do this from the given starting position, though? It seems like this solution doesn't actually include the starting tiles. – Deusovi Mar 6 at 5:37
• @Deusovi It is simply not possible to solve it with the given starting position. I'm assuming he gave it as an example of how they look. it's not possible to solve it because if you look at the empty tile on the top it neighbors both colors therefore no color can fit into this space and its not solvable. – PuzzlesAndSolutionsYT Mar 6 at 5:46
• The question states that you can use more than two colors, but only a finite number of tiles can be other colors besides your main two. Condition 2 states this pretty clearly. And you must start with three arranged as shown. – Deusovi Mar 6 at 5:53
• Right, the question is how to fit the starting pattern into a 2-colored tiling, so that the space in between can also be filled with T tetrominoes. Might not be too hard, but it's definitely a necessary component of an answer. – Deusovi Mar 6 at 7:02
• @Deusovi It may in fact be impossible to make a rectangle that doesn't slice through any tetrominos starting with the OP's original tiles. (Note that none of the valid solutions so far have done so.) That might be the basis for another puzzle... – Darrel Hoffman Mar 6 at 22:06


text:
Which of the following is a correct statement

(a) Beta rays are same as cathode rays
(b) Gamma rays are high energy neutrons
(c) Alpha particles are singly ionized helium atoms
(d) Protons and neutrons have exactly the same mass


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# American Institute of Mathematical Sciences

2015, 2015(special): 923-935. doi: 10.3934/proc.2015.0923

## A functional-analytic technique for the study of analytic solutions of PDEs

 1 Department of Civil Engineering, University of Patras, 26500 Patras, Greece 2 Department of Mathematics, University of Patras, 26500 Patras, Greece

Received  August 2014 Revised  December 2014 Published  November 2015

A functional-analytic method is used to study the existence and the uniqueness of bounded, analytic and entire complex solutions of partial differential equations. As a benchmark problem, this method is applied to the nonlinear Benjamin--Bona--Mahony equation and the associated to this, linear equation. The predicted solutions are in power series form and two concrete examples are given for specific initial conditions.
Citation: Eugenia N. Petropoulou, Panayiotis D. Siafarikas. A functional-analytic technique for the study of analytic solutions of PDEs. Conference Publications, 2015, 2015 (special) : 923-935. doi: 10.3934/proc.2015.0923
##### References:
 [1] T. B. Benjamin, J. L. Bona and J. J. Mahony, Model equations for long waves in nonlinear dispersive systems,, Philos. Trans. Roy. Soc. London Ser. A, 272 (1972), 47.   Google Scholar [2] G. Caciotta and F. Nicoló, Local and global analytic solutions for a class of characteristic problems of the Einstein vacuum equations in the "double null foliation gauge",, Ann. Henri Poincare, 13 (2012), 1167.   Google Scholar [3] G. M. Coclite, F. Gargano and V. Sciacca, Analytic solutions and singularity formation for the Peakon $b$-family equations,, Acta Appl. Math., 122 (2012), 419.   Google Scholar [4] C. J. Earle and R. S. Hamilton, A fixed point theorem for holomorphic mappings, in Global Analysis (Proc. Sympos. Pure Math., (1970), 61.   Google Scholar [5] L. C. Evans, Partial differential equations,, $2^{nd}$ edition, (2010).   Google Scholar [6] N. Hayashi and K. Kato, Global existence of small analytic solutions to Schrödinger equations with quadratic nonlinearity,, Comm. Partial Differential Equations, 22 (1997), 773.   Google Scholar [7] A. A. Himonas and G. Petronilho, Analytic well-posedness of periodic gKdV,, J. Differential Equations, 253 (2012), 3101.   Google Scholar [8] E. K. Ifantis, Solution of the Schrödinger equation in the Hardy-Lebesgue space,, J. Mathematical Phys., 12 (1971), 1961.   Google Scholar [9] E. K. Ifantis, Analytic solutions for nonlinear differential equations,, J. Math. Anal. Appl., 124 (1987), 339.   Google Scholar [10] E. K. Ifantis, Global analytic solutions of the radial nonlinear wave equation,, J. Math. Anal. Appl., 124 (1987), 381.   Google Scholar [11] J. Kajiwara, Holomorphic solutions of a partial differential equation of mixed type,, Math. Balkanica, 2 (1972), 76.   Google Scholar [12] T. Kusano and S. Oharu, Bounded entire solutions of second order semilinear elliptic equations with application to a parabolic initial value problem,, Indiana Univ. Math. J., 34 (1985), 85.   Google Scholar [13] E. N. Petropoulou and P. D. Siafarikas, Analytic solutions of some non-linear ordinary differential equations,, Dynam. Systems Appl. 13 (2004), 13 (2004), 283.   Google Scholar [14] E. N. Petropoulou and P. D. Siafarikas, Polynomial solutions of linear partial differential equations,, Commun. Pure Appl. Anal. 8 (2009), 8 (2009), 1053.   Google Scholar [15] E. N. Petropoulou, P. D. Siafarikas and E. E. Tzirtzilakis, A "discretization" technique for the solution of ODEs,, J. Math. Anal. Appl. 331 (2007), 331 (2007), 279.   Google Scholar [16] E. N. Petropoulou, P. D. Siafarikas and E. E. Tzirtzilakis, A "discretization" technique for the solution of ODEs II,, Numer. Funct. Anal. Optim. 30 (2009), 30 (2009), 613.   Google Scholar [17] E. N. Petropoulou and E. E. Tzirtzilakis, On the logistic equation in the complex plane,, Numer. Funct. Anal. Optim. 34 (2013), 34 (2013), 770.   Google Scholar [18] I. G. Petrovsky, Lecture on partial differential equations., Translated from the Russian by A. Shenitzer, (1957).   Google Scholar [19] A. Vourdas, Analytic representations in the unit disc and applications to phase states and squeezing,, Phys. Rev. A 45 (1992), 45 (1992), 1943.   Google Scholar [20] G. Zampieri, A sufficient condition for existence of real analytic solutions of P.D.E. with constant coefficients, in open sets of $\mathbbR^{2}$,, Rend. Sem. Mat. Univ. Padova 63 (1980), 63 (1980), 83.   Google Scholar [21] G. Zampieri, Analytic solutions of P.D.E.'s., Ann. Univ. Ferrara-Sez. VII-Sc. Mat. XLV (1999), XLV (1999), 365.   Google Scholar

show all references

##### References:
 [1] T. B. Benjamin, J. L. Bona and J. J. Mahony, Model equations for long waves in nonlinear dispersive systems,, Philos. Trans. Roy. Soc. London Ser. A, 272 (1972), 47.   Google Scholar [2] G. Caciotta and F. Nicoló, Local and global analytic solutions for a class of characteristic problems of the Einstein vacuum equations in the "double null foliation gauge",, Ann. Henri Poincare, 13 (2012), 1167.   Google Scholar [3] G. M. Coclite, F. Gargano and V. Sciacca, Analytic solutions and singularity formation for the Peakon $b$-family equations,, Acta Appl. Math., 122 (2012), 419.   Google Scholar [4] C. J. Earle and R. S. Hamilton, A fixed point theorem for holomorphic mappings, in Global Analysis (Proc. Sympos. Pure Math., (1970), 61.   Google Scholar [5] L. C. Evans, Partial differential equations,, $2^{nd}$ edition, (2010).   Google Scholar [6] N. Hayashi and K. Kato, Global existence of small analytic solutions to Schrödinger equations with quadratic nonlinearity,, Comm. Partial Differential Equations, 22 (1997), 773.   Google Scholar [7] A. A. Himonas and G. Petronilho, Analytic well-posedness of periodic gKdV,, J. Differential Equations, 253 (2012), 3101.   Google Scholar [8] E. K. Ifantis, Solution of the Schrödinger equation in the Hardy-Lebesgue space,, J. Mathematical Phys., 12 (1971), 1961.   Google Scholar [9] E. K. Ifantis, Analytic solutions for nonlinear differential equations,, J. Math. Anal. Appl., 124 (1987), 339.   Google Scholar [10] E. K. Ifantis, Global analytic solutions of the radial nonlinear wave equation,, J. Math. Anal. Appl., 124 (1987), 381.   Google Scholar [11] J. Kajiwara, Holomorphic solutions of a partial differential equation of mixed type,, Math. Balkanica, 2 (1972), 76.   Google Scholar [12] T. Kusano and S. Oharu, Bounded entire solutions of second order semilinear elliptic equations with application to a parabolic initial value problem,, Indiana Univ. Math. J., 34 (1985), 85.   Google Scholar [13] E. N. Petropoulou and P. D. Siafarikas, Analytic solutions of some non-linear ordinary differential equations,, Dynam. Systems Appl. 13 (2004), 13 (2004), 283.   Google Scholar [14] E. N. Petropoulou and P. D. Siafarikas, Polynomial solutions of linear partial differential equations,, Commun. Pure Appl. Anal. 8 (2009), 8 (2009), 1053.   Google Scholar [15] E. N. Petropoulou, P. D. Siafarikas and E. E. Tzirtzilakis, A "discretization" technique for the solution of ODEs,, J. Math. Anal. Appl. 331 (2007), 331 (2007), 279.   Google Scholar [16] E. N. Petropoulou, P. D. Siafarikas and E. E. Tzirtzilakis, A "discretization" technique for the solution of ODEs II,, Numer. Funct. Anal. Optim. 30 (2009), 30 (2009), 613.   Google Scholar [17] E. N. Petropoulou and E. E. Tzirtzilakis, On the logistic equation in the complex plane,, Numer. Funct. Anal. Optim. 34 (2013), 34 (2013), 770.   Google Scholar [18] I. G. Petrovsky, Lecture on partial differential equations., Translated from the Russian by A. Shenitzer, (1957).   Google Scholar [19] A. Vourdas, Analytic representations in the unit disc and applications to phase states and squeezing,, Phys. Rev. A 45 (1992), 45 (1992), 1943.   Google Scholar [20] G. Zampieri, A sufficient condition for existence of real analytic solutions of P.D.E. with constant coefficients, in open sets of $\mathbbR^{2}$,, Rend. Sem. Mat. Univ. Padova 63 (1980), 63 (1980), 83.   Google Scholar [21] G. Zampieri, Analytic solutions of P.D.E.'s., Ann. Univ. Ferrara-Sez. VII-Sc. Mat. XLV (1999), XLV (1999), 365.   Google Scholar
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# Question

A single random digit is selected using software or a random number table.
a. State the sample space for the possible outcomes.
b. State the probability for each possible outcome, based on what you know about the way random numbers are generated.
c. Each outcome in a sample space must have probability between 0 and 1, and the total of the probabilities must equal 1. Show that your assignment of probabilities in part b satisfies this rule.

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@sticky-add: @:libyli @:title-slide /no-status no-libyli # // comment - {{ nd.vars.author }} - - Talk at {{ nd.vars.venue }}, {{ nd.vars.date }} # $whoami @:#overview overview no-libyli ## Overview - Introduction - Imbalanced classification problems - The Problem (and performance measures) - Reweight, resampling, etc - Correcting k-NN (*$\gamma$-NN@:accro*) - Focusing on the F-Measure optimization (*CONE@:accro*) - Interpretability - Pleading for interpretability - Adversarial Input↔Parameter Regularization (*AI↔PR@:accro*) - Previous, current and future approaches - Discussion @eval-header:$o.s = 2 @eval-header: return highlightLi($o.s, 1) # @copy: overview ## Imbalanced Problems: Examples // generally, but at the lab - Anomaly detection *// incl images* - unsafe situations in videos // bluecime, first step into it for me - defect detection in images // often as out-of-distribution though - anormal heart beat in ECG - Fraud detection - fraudulent checks - credit card fraud (physical, online) - financial fraud (French DGFIP) // Dir. Géné. des Fin. Pub. - @:displaynone // TODO: add illustr ## Imbalanced Classification Problems // generally, but at the lab - Binary classification -$+$positive class: minority class, anomaly, rare event, … {.no} -$-$negative class: majority class, normality, typical event, … {.no} - Confusion matrix (of a model vs a ground truth) - TP: true positive - FP: false positive - TN: true negative - FN: false negative - Some measures {.padli .densemath} - Precision:$prec=\frac{TP}{TP+FP}$- Recall:$rec=\frac{TP}{P} = \frac{TP}{TP+FN}$-$F_\beta$-measure:$F_\beta = (1+\beta^2)\frac{prec\cdot rec}{\beta^2 \cdot prec + rec}$\ *(higher is better) {.dense} ## F-measure vs Accuracy ? -$F_\beta = (1+\beta^2)\frac{prec\cdot rec}{\beta^2 \cdot prec + rec} = \frac{(1+\beta^2)\cdot (P - FN)}{1 + \beta^2 P - FN + FP}$\ {.no} -$accuracy = \frac{TP + TN}{P+N} = 1 - \frac{FN+FP}{P+N}${.no} - Accuracy inadequacy (e.g.$N=10000, P=10$) - lazy "all$-$" classifier ($TP=0, TN=N, FP=0, FN=P$) -$\textstyle \textcolor{orange}{accuracy} = \frac{0 + N}{P + N} = \frac{10000}{10100} \textcolor{orange}{= 99\\%}$-$\textstyle \textcolor{orange}{F_\beta} = \frac{(1+\beta^2) (P - P)}{1 + \beta^2 P - P + 0} \textcolor{orange}{= 0}$-$F_\beta$-measure challenges - discrete (like the accuracy) - non-convex (even with continuous surrogates) - **non-separable**, i.e.$F_\beta \ne \sum_{(x_i, y_i) \in S}...$// TODO maybe a schema for the trivial case ## Ok, but I'm doing gradient descent, so … - Gradient:$0.2$⇒$-7.21$,$0.5$⇒$-2.89$,$0.8$⇒$-1.80$,$1$⇒$-1.44$- Example, gradient intensity is the same for: -$10+$wrongly classified with an output proba. of$0.2$-$40-$correctly classified with an output proba$0.8$- i.e., lazily predicting systematically$0.2$(for$+$) yields a "stable" solution with$10+$vs$40-$@:info-slide # Disclaimer: multiclass$F_\beta$-measure**s** // todo might style... @eval-header: return highlightLi($o.s, 2) # @copy: overview ## Counteracting Imbalance - Undersampling the majority class $-$ - Oversampling class $+$ - Generating fake $+$ - Using a weighted-classifiers learner // For models that learn, can reweigh but not perfect (ideal ratio depends on Bayesian error and régularisation etc) can x valid? @eval-header: return highlightLi($o.s, 3) # @copy: overview @:paperattrib #gammann no-libyli ## A Corrected Nearest Neighbor Algorithm Maximizing the F-Measure from Imbalanced Data - **Rémi Viola**, Rémi Emonet , Amaury Habrard, **Guillaume Metzler**, Sébastien Riou, Marc Sebban - ??? ## k-NN:$k$Nearest Neighbor Classification - k-NN {.step} - to classify a new point - find the closest k points (in the training section) - use a voting scheme to affect a class - efficient algorithms (K-D Tree, Ball Tree) - Does k-NN still matter? *// yes non-conv thing, easy adaptability, etc* - non-linear by design (with similarity to RBF-kernel SVM) - no learning, easy to patch a model (add/remove points) // e.g. ECML - Limits of k-NN for imbalanced data? ## Limits of k-NN for imbalanced data? 1. k-NN behavior in uncertain areas - i.e., for some feature vector, the class can be$+$or$-$- i.e., the Bayes Risk is non zero - ✔ not so bad (respects imbalance) 2. k-NN behavior around boundaries - i.e., what happens if classes are separate but imbalanced - ✖ sampling effects cause problems ## k-NN at a boundary (1000$+$) ## k-NN at a boundary (100$+$) ## k-NN at a boundary (10$+$) ## k-NN: increasing k? // (from 1 to 11) # @copy: gammann ##$\gamma$-NN Idea: push the decision boundary - Goal: correct for problems due to sampling with imbalance - Genesis: GAN to generate "$+$" around existing ones - ⇒ unstable, failing, complex @:no - Approach - artificially make$+$closer to new points - how? by using a different distance for$+$and$-$- the base distance to$+$gets multiplied by a parameter$\gamma$(intuitively$\gamma \le 1$if$+$is rare) \def{\x}{\mathbb{x}} d_\gamma(\x,\x_i) = \begin{cases} d(\x,\x_i) & \text{if} \; \x_i\in S_-,\\ \gamma \cdot d(\x,\x_i) & \text{if} \;\x_i\in S_+. \end{cases} ##$\gamma$-NN: varying$\gamma$with two points ##$\gamma$-NN: varying$\gamma$with a few$+$-$\gamma$-NN can control how close to the$-$s it pushes the boundary ##$\gamma$-NN: Algorithm - Trivial to implement - Same complexity as k-NN (at most twice) - Training - none, as k-NN -$\gamma$is selected by cross-validation (on the measure of interest) ##$\gamma$-NN: a way to reweight distributions - In uncertain regions // when we have more and more points - At the boundaries // depends on intrisic dimensionality @:/no-status ## Results on public datasets (F-measure) @:/no-status ## Results on DGFiP datasets (F-measure) // underline = second @eval-header: return highlightLi($o.s, 4) # @copy: overview @:paperattrib no-libyli ## From Cost-Sensitive Classification to Tight F-measure Bounds - **Kevin Bascol**, Rémi Emonet, Elisa Fromont, Amaury Habrard,
**Guillaume Metzler**, Marc Sebban    - AISTATS2019 ## Optimizing the $F_\beta$-measure? - Reminder {.padli .densemath} - Precision: $prec=\frac{TP}{TP+FP}$ - Recall: $rec=\frac{TP}{P} = \frac{TP}{TP+FN}$ - $F_\beta$-measure: $F_\beta = (1+\beta^2)\frac{prec\cdot rec}{\beta^2 \cdot prec + rec}$ - **Non-separability**, i.e.    $F_\beta \ne \sum_{(x_i, y_i) \in S}...$
*NB: accuracy is separable, $acc = \sum_{(x_i, y_i) \in S} \frac{1}{m} \delta(y_i - \hat{y_i})$ @:denser* - ⇒ The loss for one point depends on the others {.no} - ⇒ Impossible to optimize directly {.no} - ⇒ Impossible to optimize on a subset (minibatch) {.no} ## Weighted classification for $F_\beta$ - $F_\beta = \frac{(1+\beta^2)\cdot (P - FN)}{1 + \beta^2 P - FN + FP} = \frac{(1+\beta^2)\cdot (P - e_1)}{1 + \beta^2 P - e_1 + e_2}$

{.no} - The $F_\beta$-measure is linear fractional *(in $e = (e_1, e_2) = (FN, FP)$) @:dense* - i.e. $F_\beta = \frac{\langle a', e\rangle + b}{\langle c, e\rangle + d} = \frac{A}{B}$ {.no} - Relation to weighted classification - $\hphantom{\Leftrightarrow } F_\beta \ge t$      (we achieve a good, above $t$, $F_\beta$ value) {.no} - $\Leftrightarrow A \ge t\cdot B$ {.no} - $\Leftrightarrow A - t\cdot B \ge 0$ {.no} - $\Leftrightarrow (1+\beta^2)\cdot (P - e_1) - t ( 1 + \beta^2 P - e_1 + e_2) \ge 0$ {.no} - $\Leftrightarrow (- 1 - \beta^2 + t) e_1 - t e_2 \ge - P (1 + \beta^2) + t ( 1 + \beta^2 P)$ {.no} - $\Leftrightarrow (1 + \beta^2 - t) e_1 + t e_2 \le - P (1 + \beta^2) + t ( 1 + \beta^2 P)$ {.no} - ⇒ so, we can minimize the weighted problem
with class weights $a(t) = (1 + \beta^2 - t, t)$ {.no}
- TODO derive down to the existing weight and corr. weighted classif - can we have guarantees on the f we obtained (as we don't know how to opt it) - there exist a weighted classification pb that given the optimal f (talk about the fractional linear thing and v?) - this weight (tstar) is unknown
@:no-libyli ## Inspiration for our work - **title:** *Optimizing F-measures by cost-sensitive classification* - **by:** Parambath, S. P., Usunier, N., and Grandvalet, Y. - **in:** NIPS 2014 -  {.no} - Main result (reformulation) {.step} - given a value of $t$ that is at most at $\frac{\varepsilon_0}{2}$ from the optimal $t^\star$ - given a weighted-classification learner - then the optimal $F^\star$ is not too far from the observed $F$ - They conclude that grid search on $t$ is sound
i.e., we'll get arbitrary close to the optimal $F_\beta$-measure
*(and they experiment with 19 $t$ values) @:dense* {.step} ## Geometric interpretation of bounds - A reinterpretation of the bound // (and/or its derivations) - given - a value of $t$ (and class weights $a(t)$), and - a weighted-classifier $h$ (learned with weight $a(t)$) - … yielding an error profile $e$ and a correponding $F_\beta(e)$ - **then**\*, for any $t'$ (that would yield a classifier $e'$), - *(if $t'$ is close to $t'$ then $F(e')$ is close to $F(e)$) @:dense* - $F(e')$ is upper bounded by an expression linear in $|t - t'|$ // cannot go higher \* All bounds are up to $\varepsilon_1$, the sub-optimality of the learned weighted classifier @:footnote ## Visualizing the bound of Parambath et al. ## Deriving a tighter bound - Same objective: drawing cones - New, tighter, asymmetric bounds - Given - all definitions - NB: $\varepsilon_1$ the sub-optimality ## Visualizing the bound of Parambath et al. ## Visualizing our bound ## Using the bound to better explore $t$ values ## CONE: example results    Non-vacuous bounds, incremental algo., early increase. ## CONE: numerical results    No huge gain, better $F$-measure in average. ## CONE: Perpsectives - We - provided a geometric interpretation as cones - derived new, tighter bounds - proposed a search algorithm using the bounds - Work in progress - bounds in generalization - lower bounds @eval-header: return highlightLi(3, 1) # @copy: overview ## Interpretability - Why, what ## Example of interpretable patterns - We overlay on the series the pattern that were most relevant to the decision taken by the classifier - What side (left or right) would be more convincing? ## Approaches towards interpretability - Black box models - Intepretability by design - constraints from the model structure @eval-header: return highlightLi(3, 2) # @copy: overview @:paperattrib no-libyli ## Learning Interpretable Shapelets for Time Series Classification through Adversarial Regularization - **Yichang Wang**, Rémi Emonet, Elisa Fromont, Simon Malinowski, **Etienne Menager, Loic Mosser**, Romain Tavenard - ??? + arxiv2019 @:/no-status ## Adversarial Input↔Parameter Regularization (*AI↔PR@:accro*) {.dense}
## Optimal Transport, Duality and Adversarial Learning -
L_d(\theta_d) = \mathop{\mathbb{E}}_{\tilde{x}\sim\mathbb{P}_{S}} \left[ D(\tilde{x}) \right] - \mathop{\mathbb{E}}_{x\sim\mathbb{P}_{x}} \left[D(x)\right] + \lambda \mathop{\mathbb{E}}_{\hat{x}\sim\mathbb{P}_{\hat{x}}}\left[(||\nabla_{\hat{x}}{D(\hat{x})}||_2-1)^2\right]
- Gradient penalty - Interpolation to cover the space ## 3 Losses for 3 Phases - Classifier training
\;\;\;\;L_C(\theta_c) = ...
- Discriminator Training, using fixed shapelets
\;\;\;\;L_d(\theta_d) = \mathop{\mathbb{E}}_{\tilde{x}\sim\mathbb{P}_{S}} \left[ D(\tilde{x}) \right] - \mathop{\mathbb{E}}_{x\sim\mathbb{P}_{x}} \left[D(x)\right] + \lambda \mathop{\mathbb{E}}_{\hat{x}\sim\mathbb{P}_{\hat{x}}}\left[(||\nabla_{\hat{x}}{D(\hat{x})}||_2-1)^2\right]
- Shapelet regularization, using the fixed discriminator
\;\;\;\;L_r(\theta_s) = - \mathop{\mathbb{E}}_{ {\tilde{x}}\sim\mathbb{P}_{S}}[D(\tilde{x})]
@:/no-status ## Algorithm
## Discriminator work across iterations ## Results on 85 datasets (vs Fast Shapelets) ## Results *(vs Learning Shapelets, non-interpretable)@:dense* @eval-header: return highlightLi(3, 3) # @copy: overview ## More... - Perspectives around *AI↔PR@:accro* - multivariate - group-lasso used well - deeper regularization - Other interpretable families of methods - probablisistic models - auto-encoders with regularizations and constraints - neural EM etc @eval-header: return highlightLi(4, 1) # @copy: overview @:/title-slide # Thank you! Questions?


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# Primitive roots - annoying problem

## Main Question or Discussion Point

Let $$r$$ be a primitive root of a prime number $$p \geq 3$$. Prove that if $$p \equiv 1 (mod 4)$$, then $$-r$$ is also a primitive root of $$p$$.

I've been told it's quite easy, but I can't see why it's true for the life of me

## Answers and Replies

Related Linear and Abstract Algebra News on Phys.org
If r is a primitive root, then r^[(p-1)/2] is not equal to 1. The question is about -r.

thanks for the reply
if r is a primitve root then r^[(p-1)/2] is -1 (mod p). However, (-r)^[(p-1)/2] is the same since (p-1)/2 is even, am I right?

Where do we go from here? Maybe I could say that the order of (-r) couldn't be lower than (p-1), because if it was so, ord(-r)|(p-1)/2 and then (-r)^[(p-1)/2]=1(mod p) contradicting the fact that p $$\neq$$ 2?
Am I on the right track?

Thanks again!

Richard Cypher: Am I on the right track?

Sounds O.K. to me. Take, (-r)^2 = r^2, well then what power is necessry to raise r^2 to 1?

OK, I got it!
Bunch of thanks for your help!


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## Unit0.3.2Operating system

### Subsubsection0.3.2.1Linux

If you are on a computer that already uses Linux as its operating system, you are all set. (Although... It has been reported that some linux installations don't automatically come with the gcc compiler, which you will definitely need.)

### Subsubsection0.3.2.2Windows

Sorry, we don't do Windows... But there are options.

### Subsubsection0.3.2.3Mac OS-X

Mac OS-X comes with Darwin, an open-source Unix-like operating system (https://en.wikipedia.org/wiki/Darwin_(operating_system)).

A complication on the Max OS-X operating system is that the gcc compiler defaults to their clang compiler, which does not by default include the ability to use OpenMP directives.

There are a few tools that you will want to install to have the compete toolchain that we will use. We will walk you through how to set this up in Unit 0.3.3

Again, alternatively you may want to use a Virtual Machine (see below).

### Subsubsection0.3.2.4Virtual box alternative

Non-Linux users can also install a Linux virtual machine onto both Windows and macOS using VirtualBox from Oracle, which is free to download.

Essentially, you can run a Linux OS from within Windows / macOS without having to actually install it on the hard drive alongside Windows / macOS, and (for Windows users) without having to go through Microsoft Store.

The basic process looks like this:

• Install VirtualBox.

• Download a .iso image for some Linux OS (I recommend Ubuntu MATE but really it can be any distribution).

• Create a virtual machine in VirtualBox, allocating sufficient memory and hard drive space.

• Start up the virtual machine and select the downloaded .iso image for the chosen Linux distro.

• Go through the distro's installation process.

• Install any additional software needed (blis, a text editor, etc.)

Then the user can access the virtual machine as a self-contained contained Linux OS.


text:
Hellenica World

# .

In mathematics, in the field of group theory, a group is said to be absolutely simple if it has no proper nontrivial serial subgroups.[1] That is, G is an absolutely simple group if the only serial subgroups of G are $$\{ e \}$$ (the trivial subgroup), and G itself (the whole group).

In the finite case, a group is absolutely simple if and only if it is simple. However, in the infinite case, absolutely simple is a stronger property than simple. The property of being strictly simple is somewhere in between.


text:
# How do you sketch the graph of y=(x+3)^2+6 and describe the transformation?

Dec 17, 2017

Refer to the explanation.

#### Explanation:

First, let's look at the transformations of this equation.

When something is with the $x$, in this case $x + 3$, then you will do the opposite of what it says, because when you set $x + 3$ to equal to $0$, then you will get $- 3$, which is the opposite of $3$.

However, when something is "outside" the $x$, in this case $6$, then that applies to the y-values and it just does what it says. So the y-values would always add $6$.

Here is the graph (should be arrows at each end, just doesn't show):

As you can see, there is a point, or the vertex, is at $\left(- 3 , 5\right)$ as our transformation showed. Then we form a parabola.

When the coefficient of a quadratic equation is positive , (ex: ${x}^{2}$, $10 {x}^{2}$) then the parabola will face up .

When the coefficient is negative, (ex: $- {x}^{2}$, $- 10 {x}^{2}$), then the parabola will face down .

Our coefficient is just $x$, so that is positive, therefore the graph grows up.


text:
# How to determine the frequency of oscillations in system of three ODEs?

I want to determine the frequency of oscillations in a system of three stiff ODEs (Oregonator model). That model describes a chemical oscillator.

I have a slightly more advanced model of the default or regular Oregonator. It consists of three ODEs:

ode1=ε*x'[t]==x[t](1-x[t])-2*(x[t]-μ)/(x[t]+μ)*(r*α1*y[t]+q*((α2*z[t])/(1-z[t])));
ode2=ξ1*y'[t]==x[t](1+β2*z[t])-α1*y[t]-((x[t]*(1+β1*y[t]+β2*z[t]))/((1-z[t])+η*(1-y[t])))*(1-z[t]);
ode3=ξ2*z'[t]==x[t]*(1+β1*y[t])-(α2*z[t])/(1-z[t])-((x[t]*(1+β1*y[t]+β2*z[t]))/((1-z[t])+η*(1-y[t])))*η*(1-y[t]);

with the initial (example) conditions ic

ic2 = {x[0] == .5, y[0] == 0.1, z[0] == 0.3};

I use NDSolveValue for this:

{xSol, ySol, zSol} = NDSolveValue[{ode1, ode2, ode3, ic2}, {x, y, z}, {t, 0, 200}]

This looks like this:

So far, so fine. I now need to determine the frequency of the oscillations in this model with three ODEs.

I found this related question, but that only features a single ODE. And as I'm really a Mathematica novice, I also didn't understand how the Reap and Sow worked.

The suggested solution was as follows:

pts =
Reap[s = NDSolve[{y'[x] == y[x] Cos[x + y[x]], y[0] == 1,
WhenEvent[y'[x] == 0, Sow[x]]}, {y, y'}, {x, 0, 30}]][[2, 1]]

(* Out[290]= {0.448211158984, 4.6399193764, 7.44068279785, 10.953122261, \
13.8722260952, 17.2486864443, 20.2244048853, 23.5386505821, \
26.5478466115, 29.8261176372} *)

Plot[{Evaluate[y[x] /. s], Evaluate[y'[x] /. s]}, {x, 0, 30},
PlotRange -> All]

and then finding the differences:

diffs = Differences[pts, 1, 2]

(* Out[288]= {6.99247163887, 6.31320288463, 6.43154329733, \
6.29556418327, 6.35217879014, 6.28996413777, 6.32344172616, \
6.28746705515} *)

Mean[diffs]

(* Out[289]= 6.41072921417 *)

This looks exactly what I need, but I don't know how to apply this to my three ODEs? I preferably want to keep the initial conditions, ic, in a separate variable like I now have.

Can anyone show me how to modify the above solution so that it works with my system? I want to determine the frequency separately for x[t], y[t] and z[t]. If people have a different solution than proposed in the related question, you're of course very welcome!

Many thanks in advance!

Current file with all the needed variables, ODEs and my use of NDSolveValue. You can copy and modify this.

• Another related Q&A: mathematica.stackexchange.com/questions/129993/… – Michael E2 Oct 14 '20 at 18:34
• To determine the frequency of oscillatory data the FFT: Fourier is commonly used. – Daniel Huber Oct 14 '20 at 19:46
• @MichaelE2 thanks! Will look into that one. – ralphjsmit Oct 15 '20 at 5:41
• Hi @DanielHuber thanks for mentioning that! Is there a very simple beginner example in which the frequency is calculated? And some explanation about FFT? – ralphjsmit Oct 15 '20 at 5:52
• It is not too easy if you never heard about it, but it is a standard tool in data analysis. There are zillion of articles on the web,.You may e.g. look at: earlevel.com/main/2002/08/31/a-gentle-introduction-to-the-fft. You should play around with signals of the form Sin[f1 t]+ Sin[f2 t]+.. and see what the output of Fourier is. Note that in general the output is complex and that Fourier returns a list: first element: zero frequency component, second: base frequency, first overtone,.. up to the highest frequency, then highest negative frequency, then the second highest,.. – Daniel Huber Oct 15 '20 at 8:47

Using my EcoEvo package as in my answer here.

First, you need to install it with

PacletInstall["EcoEvo", "Site" -> "http://raw.githubusercontent.com/cklausme/EcoEvo/master"]

Then, load the package and define your model:

<< EcoEvo`;

SetModel[{
Aux[x] -> {Equation :> (x[t] (1 - x[t]) - 2*(x[t] - μ)/(x[t] + μ)*(r*α1*y[t] + q*((α2*z[t])/(1 - z[t]))))/ε},
Aux[y] -> {Equation :> (x[t] (1 + β2*z[t]) - α1*y[t] - ((x[t]*(1 + β1*y[t] + β2*z[t]))/((1 - z[t]) + η*(1 - y[t])))*(1 - z[t]))/ξ1},
Aux[z] -> {Equation :> (x[t]*(1 + β1*y[t]) - (α2*z[t])/(1 - z[t]) - ((x[t]*(1 + β1*y[t] + β2*z[t]))/((1 - z[t]) + η*(1 - y[t])))*η*(1 - y[t]))/ξ2}
}];

Simulate for 40 time steps to get on the limit cycle:

sol = EcoSim[{x -> 0.5, y -> 0.1, z -> 0.3}, 40];
GraphicsRow[{PlotDynamics[sol, x], PlotDynamics[sol, y], PlotDynamics[sol, z]}]

Find the limit cycle using FindEcoCycle:

ec = FindEcoCycle[FinalSlice[sol]];
GraphicsRow[{PlotDynamics[ec, x], PlotDynamics[ec, y], PlotDynamics[ec, z]}]

Make sure the initial and final values are the same:

InitialSlice[ec]
FinalSlice[ec]

(* {x -> 0.617907, y -> 0.522312, z -> 0.989451} *)
(* {x -> 0.617907, y -> 0.522312, z -> 0.989451} *)

Finally, the period can be extracted as the final time of ec:

FinalTime[ec]
(* 2.71597 *)

To answer E. Chan-López's question, to find equilibria use SolveEcoEq.

eq = SolveEcoEq[]
(* {{x -> 0, y -> 0, z -> 0}, {x -> 0.000395349, y -> 1., z -> 1.}, {x -> 0.0279224, y -> 0.06391, z -> 0.881397}, {x -> 0.0279224, y -> 0.124848, z -> 1.08789}, {x -> 0.0279224, y -> 0.988952, z -> 1.00342}} *)

Apparently they're all unstable:

EcoStableQ[eq]
(* {False, False, False, False, False} *)
• Hey @ChrisK thanks for your answer! I’ll try it out (at last Saturday) and let you know! – ralphjsmit Oct 15 '20 at 5:53
• Hi Chris, thanks, it works nicely👍 – ralphjsmit Oct 15 '20 at 15:35
• Hi! How many equilibrium points can you calculate in your model? – E. Chan-López Feb 8 at 5:05
• @E.Chan-López I added the equilibria to my answer - looks like there are five, all of which are unstable. – Chris K Feb 8 at 15:59
• @ChrisK I added a calculation to show the existence of a Hopf bifurcation at a non-trivial equilibrium point. – E. Chan-López Feb 9 at 2:09

Contributing to Chris's analysis:

Taking $$q=\displaystyle\frac{33}{4}$$, $$r=\displaystyle\frac{1}{90}$$, $$\alpha_{1}=\displaystyle\frac{64}{33}$$, $$\alpha_{1}=\displaystyle\frac{1}{22}$$, $$\beta_{1}=24$$, $$\beta_{2}=3/4$$, $$\xi_{1}=1$$, $$\xi_{2}=1$$, $$\eta=32$$, $$\mu=\displaystyle\frac{16}{1517}$$, and $$\varepsilon$$ as a free parameter, we obtain five non-trivial equilibrium points.

The system:

f1 = 1/ε (x (1 - x) - 2*(x - μ)/(x + μ)*(r*α1*y + q*((α2*z)/(1 - z))));
f2 = 1/ξ1 (x (1 + β2*z) - α1*y - ((x*(1 + β1*y + β2*z))/((1 - z) + η*(1 - y)))*(1- z));
f3 = 1/ξ2 (x*(1 + β1*y) - (α2*z)/(1 - z) - ((x*(1 + β1*y + β2*z))/((1 - z) + η*(1 - y)))*η*(1 - y));
F = {f1, f2, f3};
X = {x, y, z};

The Jacobian matrix:

J = FullSimplify[D[F, {X}]];

The parameter values and non-trivial equilibrium points:

q = 33/4;
r = 1/90;
α1 = 64/33;
α2 = 1/22;
β1 = 24;
β2 = 3/4;
ξ1 = 1;
ξ2 = 1;
η = 32;
μ = 16/1517;

X0 = N[Normal[Solve[F == 0 && Variables[F] > 0, X]]]
(*{{x -> 1/2, y -> 1/4, z -> 1/4},
{x -> 13.3111, y -> 12.0089, z -> 1.00458},
{x -> 0.0105402, y -> 0.999138, z -> 1.02416},
{x -> 2.38478, y -> 1.33557, z -> 1.28417},
{x -> 0.0103128, y -> 0.0337578, z -> 5.68495}}*)

The linear approximation at the point $$X_{01}=(1/2, 1/4, 1/4)$$ and its characteristic polynomial:

J01 = FullSimplify[J /. X01];
polJ01 = -Collect[CharacteristicPolynomial[J01, λ], λ, FullSimplify]
(*polJ0 = A0 λ^3 + A1 λ^2 + A2 λ^1 + A3;
A0 = 1; A1 = (8 (66748 + 9580565 ε))/(25302915 ε);
A2 = (4 (268141961 + 2208997920 ε))/(7514965755 ε);
A3 = 571980408512/(743981609745 ε) *)

According to the Routh-Hurwitz criterion, $$X_{01}$$ is stable if and only if

Reduce[A1 > 0 && A2 > 0 && A3 > 0 && A1 A2 - A3 > 0 && ε > 0]
(*0 < ε < (239125592953 - Sqrt[31922603976781931264305])/5465060854080 ||
ε > (239125592953 + Sqrt[31922603976781931264305])/5465060854080*)

The last result means that we have two critical Hopf values for $$\varepsilon$$ at point $$X_{01}$$. According to the Liu criterion for the Hopf bifurcation, these critical values are calculated as follows

ε0 = Solve[A1 A2 - A3 == 0 && ε > 0, ε]];
(*{{ε -> (239125592953 - Sqrt[31922603976781931264305])/5465060854080},
{ε -> (239125592953 + Sqrt[31922603976781931264305])/5465060854080}}*)

At the other equilibrium points, the analysis is similar, except for the second equilibrium, since it is unstable.