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# Do random functions have synchronous, alternating circuits with non-injective first layers?

After discussing in the comments, I think a clearer definition of the question is as follows: for a random function $$f : \{0, 1\}^n \rightarrow \{0, 1\}$$, what is the probability that there exists a non-injective function $$g : \{0, 1\}^n \rightarrow \{0, 1\}^m$$ defined by $$m$$ conjunctions on $$2$$ variables, with arbitrary negations, such that $$\forall x, y \in \{0, 1\}^n, g(x) = g(y) \Rightarrow f(x) = f(y)$$ with $$g$$?

Here is the former description for context:

Here, consider circuits to have $$AND$$ and $$OR$$ gates with fan-in $$2$$, with negations optionally present at any input to a gate, and to be alternating and synchronous (Harper 1977), the latter meaning that there are layers such that layer $$1$$ only accesses the inputs, layer $$2$$ only accesses layer $$1$$, and so on, and the former meaning that each layer has exclusively $$AND$$s or $$OR$$s, which one alternating layer by layer. For any function $$f : \{0, 1\}^n \rightarrow \{0, 1\}$$, we can then consider an arbitrary circuit $$C$$ which computes it, and "chop off" everything above the first layer. Doing so, if that layer had $$w$$ gates (width $$w$$), we get new, multi-output circuit which computes a function $$g_C : \{0, 1\}^n \rightarrow \{0, 1\}^w$$.

My question is: for uniformly random $$f$$, what is the probability that there exists such a $$C$$ which compute $$f$$ such that $$g_C$$ is not injective?

• Your description of the circuit model appears incomplete. As written, the circuits are monotone, hence a random Boolean function cannot be computed in this model at all, with overwhelming probability. Nov 18 '21 at 8:51
• I made it explicit, though maybe there is a way to more concisely state it with a common name for the basis with ANDs, ORs with arbitrary negations? Nov 18 '21 at 18:58
• If you have negations for free anywhere, then what’s the point of making the layers alternating? You can turn any AND into an OR by negating all its input and output wires, and vice versa. Nov 18 '21 at 19:04
• Yes, good point. Is the typical way to push the negations to the bottom and make it alternating? Nov 18 '21 at 19:05
• I just care about the existence of $m$ width 2 conjunctions (possibly containing negations) computing a non-injective function $g : \{0, 1\}^n \rightarrow \{0, 1\}^m$ such that $\forall x, y, g(x) = g(y) \Rightarrow f(x) = f(y)$. Nov 18 '21 at 19:07


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# MCQ Questions for Class 9 Science Chapter 9 Force and Laws of Motion

Students are advised to practice the NCERT MCQ Questions for Class 9 Science Chapter 9 Force and Laws of Motion with Answers Pdf free download is available here. MCQ Questions for Class 9 Science with Answers are prepared as per the Latest Exam Pattern. Students can solve these Force and Laws of Motion Class 9 MCQs Questions with Answers and assess their preparation level.

## Force and Laws of Motion Class 9 MCQs Questions with Answers

Solving the Force and Laws of Motion Multiple Choice Questions of Class 9 Science Chapter 9 MCQ can be of extreme help as you will be aware of all the concepts. These MCQ Questions on Force and Laws of Motion Class 9 with answers pave for a quick revision of the Chapter thereby helping you to enhance subject knowledge. Have a glance at the MCQ of Chapter 9 Science Class 9 and cross-check your answers during preparation.

Multiple Choice Questions

Question 1.
Rocket works on the principle of conservation of:
(a) mass
(b) energy
(c) momentum
(d) velocity

Question 2.
Among the equal-sized stone and a football, the inertia will be higher of:
(a) football
(b) stone
(c) both
(d) none of them

Question 3.
A batsman hits a cricket ball which then rolls on the ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because:
(a) the batsman did not hit the ball hard enough.
(b) velocity is proportional to the force exerted on the ball.
(c) there is a force on the ball opposing the motion.
(d) there is no unbalanced force on the ball so the ball would want to come to rest.

Answer: (c) there is a force on the ball opposing the motion.

Question 4.
What is the momentum of an object of mass m, moving with a velocity v?
(a) (mv)2
(b) mv2
(c) $$\frac{1}{2}$$ mv2
(d) mv

Question 5.
Friction is:
(a) useful to us
(b) harmful to us
(c) both useful as well as harmful to us
(d) none of them.

Answer: (c) both useful as well as harmful to us

Fill in the Blanks.

Question 6.
The SI unit of momentum is _________.

Question 7.
The natural tendency of objects to resist a change in their state of rest or of uniform motion is called ______

Question 8.
To every action, there is an ______ and opposite reaction.

Question 9.
The resultant force of balanced forces is ________

Question 10.
The force can change the motion, direction, or _______ of an object.

Question 11.
The value of inertia depends on the _______ of an object.

Question 12.
The rate of change of momentum of an object is ______ to the applied unbalanced force in the direction of the force.

True/False.

Question 13.
In 1586, Galileo Galilei wrote his first scientific book ‘The Little Balance’.

Question 14.
In practical situations, it is difficult to achieve a zero balanced force.

Question 15.
Force is a scalar quantity.

Question 16.
The mass of an object is a measure of its inertia.

Question 17.
In an isolated system, the total momentum remains conserved.

Match the Column.

Question 18.

 A B 1. Unit of force (i) Mass × acceleration 2. Unit of pressure (ii) kgms-1 3. Test of purity of milk (iii) Pascal 4. Force (iv) Lactometer 5. S.I unit of momentum (v) Newton

 A B 1. Unit of force (v) Newton 2. Unit of pressure (iii) Pascal 3. Test of purity of milk (iv) Lactometer 4. Force (i) Mass × acceleration 5. SI unit of momentum (ii) kgms-1

Question 19.
Write the unit of pressure.

Question 20.
A body is moving with constant velocity, then what will be the acceleration of that body?

Question 21.
Write the S.I unit of force.

Question 22.
On what factor the inertia of an object depends?

Question 23.
Express the second law of motion of newton in the context of the equation of momentum.

Answer: Force = $$\frac{\text { Change in momentum } \Delta P}{\text { time }}=\frac{m(v-u)}{t}$$

format: plain


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# Custom Metrics with Keras

How to define a custom performance metric in Keras?

I am trying to use it but I can not see the metrics values on each epoch.

clf51.compile(optimizer=sgd51, loss='binary_crossentropy', metrics=["accuracy"])

clf51.fit(X_train, Y_train, batch_size=384, epochs=5, callbacks=[metrics], validation_split=0.30, verbose=2)

As you can see:

Train on 139554 samples, validate on 59810 samples

Epoch 1/5 - 4s - loss: 0.3576 - acc: 0.9885 - val_loss: 0.0531 - val_acc: 0.9989

Epoch 2/5 - 3s - loss: 0.0261 - acc: 0.9987 - val_loss: 0.0135 - val_acc: 0.9987

Am I doing something wrong? I would like to make a earlystop using f1s.

## 1 Answer

The callback you are using isn't for displaying the desired metrics, just recording them. For example if you want to access the F1-score you need to type: metrics.f1s. This is useful, let's say, if you want to make a graph on how the F1-score reduced during training.

To use the EarlyStopping callback, however, f1-score needs to be a metric not a callback like you have it!

You need to write (or find) a function that calculates the F1-score through keras' backend functions. You might want to check if this works for you.


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• ### The Astropy Project: Building an inclusive, open-science project and status of the v2.0 core package(1801.02634)

Jan. 16, 2018 astro-ph.IM
The Astropy project supports and fosters the development of open-source and openly-developed Python packages that provide commonly-needed functionality to the astronomical community. A key element of the Astropy project is the core package Astropy, which serves as the foundation for more specialized projects and packages. In this article, we provide an overview of the organization of the Astropy project and summarize key features in the core package as of the recent major release, version 2.0. We then describe the project infrastructure designed to facilitate and support development for a broader ecosystem of inter-operable packages. We conclude with a future outlook of planned new features and directions for the broader Astropy project.
• ### Earliest phases of star formation (EPoS): Dust temperature distributions in isolated starless cores(1606.04318)

June 14, 2016 astro-ph.GA
Constraining the temperature and density structure of dense molecular cloud cores is fundamental for understanding the initial conditions of star formation. We use Herschel observations of the thermal FIR dust emission from nearby isolated molecular cloud cores and combine them with ground-based submillimeter continuum data to derive observational constraints on their temperature and density structure. The aim of this study is to verify the validity of a ray-tracing inversion technique developed to derive the dust temperature and density structure of isolated starless cores directly from the dust emission maps and to test if the resulting temperature and density profiles are consistent with physical models. Using this ray-tracing inversion technique, we derive the dust temperature and density structure of six isolated starless cloud cores. We employ self-consistent radiative transfer modeling to the derived density profiles, treating the ISRF as the only heating source. The best-fit values of local strength of the ISRF and the extinction by the outer envelope are derived by comparing the self-consistently calculated temperature profiles with those derived by the ray-tracing method. We find that all starless cores are significantly colder inside than outside, with the core temperatures showing a strong negative correlation with peak column density. This suggests that their thermal structure is dominated by external heating from the ISRF and shielding by dusty envelopes. The temperature profiles derived with the ray-tracing inversion method can be well-reproduced with self-consistent radiative transfer models.
• ### Three-Dimensional Radiation Transfer in Young Stellar Objects(1307.0561)

July 1, 2013 astro-ph.GA, astro-ph.SR
We have updated our publicly available dust radiative transfer code (HOCHUNK3D) to include new emission processes and various 3-D geometries appropriate for forming stars. The 3-D geometries include warps and spirals in disks, accretion hotspots on the central star, fractal clumping density enhancements, and misaligned inner disks. Additional axisymmetric (2-D) features include gaps in disks and envelopes, "puffed-up inner rims" in disks, multiple bipolar cavity walls, and iteration of disk vertical structure assuming hydrostatic equilibrium. We include the option for simple power-law envelope geometry, which combined with fractal clumping, and bipolar cavities, can be used to model evolved stars as well as protostars. We include non-thermal emission from PAHs and very small grains, and external illumination from the interstellar radiation field. The grid structure was modified to allow multiple dust species in each cell; based on this, a simple prescription is implemented to model dust stratification. We describe these features in detail, and show example calculations of each. Some of the more interesting results include the following: 1) Outflow cavities may be more clumpy than infalling envelopes. 2) PAH emission in high-mass stars may be a better indicator of evolutionary stage than the broadband SED slope; and related to this, 3) externally illuminated clumps and high-mass stars in optically thin clouds can masquerade as YSOs. 4) Our hydrostatic equilibrium models suggest that dust settling is likely ubiquitous in T Tauri disks, in agreement with previous observations.
• ### The standard model of low-mass star formation applied to massive stars: a multi-wavelength picture of AFGL 2591(1212.1719)

Dec. 7, 2012 astro-ph.GA, astro-ph.SR
This paper aims to investigate the hypothesis that the embedded luminous star AFGL2591-VLA3 (2.3E5Lsun at 3.33kpc) is forming according to a scaled-up version of a low-mass star formation scenario. We present multi-configuration VLA 3.6cm and 7mm, as well as CARMA C18O and 3mm continuum observations to investigate the morphology and kinematics of the ionized gas, dust, and molecular gas around AFGL2591. We also compare our results to ancillary near-IR images, and model the SED and 2MASS image profiles of AFGL2591 using a dust continuum radiative transfer code. The observed 3.6cm images uncover for the first time that the central powering source AFGL2591-VLA3 has a compact core plus collimated jet morphology, extending 4000AU eastward from the central source with an opening angle of <10deg at this radius. However, at 7mm VLA3 does not show a jet morphology, but instead compact (<500AU) emission, some of which (<0.57 mJy of 2.9mJy) is estimated to be from dust. We determine that the momentum rate of the jet is not sufficient to ionize itself via only shocks, and thus a significant portion of the emission is instead likely created in a photoionized wind. The C18O emission uncovers dense entrained material in the outflow(s) from the young stars in the region. The main features of the SED and 2MASS images of AFGL2591-VLA3 are also reproduced by our model dust geometry of a rotationally flattened envelope with and without a disk. The above results are consistent with a picture of massive star formation similar to that seen for low-mass protostars. However, within its envelope, AFGL2591-VLA3 contains at least four other young stars, constituting a small cluster. Therefore it appears that AFGL2591-VLA3 may be able to source its accreting material from a shared gas reservoir while still exhibiting the phenomena expected during the formation of low-mass stars. (Abridged)
• ### A Water Maser and Ammonia Survey of GLIMPSE Extended Green Objects (EGOs)(1210.5528)

Oct. 19, 2012 astro-ph.GA, astro-ph.SR
We present the results of a Nobeyama 45-m water maser and ammonia survey of all 94 northern GLIMPSE Extended Green Objects (EGOs), a sample of massive young stellar objects (MYSOs) identified based on their extended 4.5 micron emission. We observed the ammonia (1,1), (2,2), and (3,3) inversion lines, and detect emission towards 97%, 63%, and 46% of our sample, respectively (median rms ~50 mK). The water maser detection rate is 68% (median rms ~0.11 Jy). The derived water maser and clump-scale gas properties are consistent with the identification of EGOs as young MYSOs. To explore the degree of variation among EGOs, we analyze subsamples defined based on MIR properties or maser associations. Water masers and warm dense gas, as indicated by emission in the higher-excitation ammonia transitions, are most frequently detected towards EGOs also associated with both Class I and II methanol masers. 95% (81%) of such EGOs are detected in water (ammonia(3,3)), compared to only 33% (7%) of EGOs without either methanol maser type. As populations, EGOs associated with Class I and/or II methanol masers have significantly higher ammonia linewidths, column densities, and kinetic temperatures than EGOs undetected in methanol maser surveys. However, we find no evidence for statistically significant differences in water maser properties (such as maser luminosity) among any EGO subsamples. Combining our data with the 1.1 mm continuum Bolocam Galactic Plane Survey, we find no correlation between isotropic water maser luminosity and clump number density. Water maser luminosity is weakly correlated with clump (gas) temperature and clump mass.
• ### Galactic structure based on the ATLASGAL 870mum survey(1112.4609)

Dec. 20, 2011 astro-ph.SR
The ATLASGAL 870mum continuum survey conducted with the APEX telescope is the first survey covering the whole inner Galactic plane (60deg>l>-60deg & b<1.5deg) in submm continuum emission tracing the cold dust of dense and young star-forming regions. Here, we present the overall distribution of sources within our Galactic disk. The submm continuum emission is confined to a narrow range around the galactic plane, but shifted on average by ~0.07deg below the plane. Source number counts show strong enhancements toward the Galactic center, the spiral arms and toward prominent star-forming regions. Comparing the distribution of ATLASGAL dust continuum emission to that of young intermediate- to high-mass young stellar objects (YSOs) derived from Spitzer data, we find similarities as well as differences. In particular, the distribution of submm dust continuum emission is significantly more confined to the plane than the YSO distribution (FWHM of 0.7 and 1.1deg, corresponding to mean physical scale heights of approximately 46 and 80pc, respectively). While this difference may partly be caused by the large extinction from the dense submm cores, gradual dispersal of stellar distributions after their birth could also contribute to this effect. Compared to other tracers of Galactic structure, the ATLASGAL data are strongly confined to a narrow latitude strip around the Galactic plane.
• ### Spectral Energy Distributions of High Mass Proto Stellar Objects - Evidence for High Accretion Rates(0711.2261)

Sept. 24, 2008 astro-ph
The spectral energy distributions (SEDs), spanning the mid-infrared to millimeter wavelengths, of a sample of 13 high-mass protostellar objects (HMPOs) were studied using a large archive of 2-D axisymmetric radiative transfer models. Measurements from the Spitzer GLIMPSE and MIPSGAL surveys and the MSX survey were used in addition to our own surveys at millimeter and submillimeter wavelengths to construct the SEDs, which were then fit to the archive of models. These models assumed that stars of all masses form via accretion and allowed us to make estimates for the masses, luminosities and envelope accretion rates for the HMPOs. The models fit the observed SEDs well. The implied envelope accretion rates are high, $\approx 10^{-2.5} \msun/yr$, consistent with the accretion-based scenario of massive star formation. With the fitted accretion rates and with mass estimates of up to $\sim 20 \msun$ for these objects, it appears plausible that stars with stellar masses $M_{\ast} > 20 \msun$ can form via accretion.
• ### Molecular Outflows and a Mid-Infrared Census of the Massive Star-Formation Region Associated with IRAS 18507+0121(0706.4330)

June 28, 2007 astro-ph
We have observed the central region of the IR-dark cloud filament associated with IRAS 18507+0121 at millimeter wavelengths in CO(1-0), 13CO(1-0), and C18O(1-0) line emission and with the Spitzer Space Telescope at mid-IR wavelengths. Five massive outflows from two cloud cores were discovered. Three outflows are centered on or near an Ultracompact HII region (G34.4+0.23) while the remaining two outflows originate from the millimeter core G34.4+0.23 MM. Modeling of the SEDs of the mid-IR sources identified 31 young stellar objects in the filament with a combined stellar mass of ~127 +/- 27 Msun. An additional 22 sources were identified as probable cluster members based on the presence of strong 24 micron emission. The total star formation efficiency in the G34.4 cloud filament is estimated to be ~7% while the massive and intermediate mass star formation efficiency in the entire cloud filament is estimated to be roughly 2%. A comparison of the gravitational binding energy with the outflow kinetic energy suggests that the compact core containing G34.4+0.23 MM is being destroyed by its molecular outflows whereas the outflows associated with more massive core surrounding the G34.4 UC HII region are not likely to totally disrupt the cloud. Additionally, a qualitative evaluation of the region appears to suggest that stars in this region may have formed in two stages: first lower mass stars formed and then, a few Myrs later, the more massive stars began to form.
• ### The morphology of minor axis gaseous outflows in edge-on Seyfert galaxies(astro-ph/0612741)

Dec. 27, 2006 astro-ph
(Abridged) Spiral galaxies often have extended outflows that permeate beyond the region of the disk. Such outflows have been seen both in starburst galaxies, actively star forming galaxies and galaxies with an AGN. In the latter galaxies it is unknown whether the large-scale outflows are driven by star formation activity or purely by the active nucleus. The aim of our investigation is to study the frequency of extended minor-axis outflows in edge-on Seyfert galaxies to investigate the role of the AGN, the circumnuclear environment and star formation activity within the disk regions, and their importance for IGM enrichment on large scales. Narrowband imaging in two different ionizational stages (H-alpha and [OIII]) was performed to attempt a discrimination between processes associated with the active nucleus and those connected to star forming activity within the disk. The H-alpha morphology of the Seyfert galaxies is usually complex, but only in three out of 14 galaxies did we find evidence for minor axis disk outflows. At the sensitivity of our observations [OIII] emission is generally detected only in the nuclear region. Overall, our results show that extraplanar emission of similar brightness and extent as in the previously known cases of NGC3079 and NGC4388 is not common in Seyfert galaxies of otherwise similar properties. Comparison with our previous results shows that for nearby edge-on spiral galaxies star formation may be a more powerful mechanism for producing DIG than AGN activity. While in general AGN activity undoubtedly plays some role in driving minor-axis outflows, this probably requires higher AGN luminosities than are encountered in our small distance-limited sample.
• ### The Frequency of Mid-Infrared Excess Sources in Galactic Surveys(astro-ph/0612235)

Dec. 8, 2006 astro-ph
We have identified 230 Tycho-2 Spectral Catalog stars that exhibit 8 micron mid-infrared extraphotospheric excesses in the MidCourse Space Experiment (MSX) and Spitzer Space Telescope Galactic Legacy MidPlane Survey Extraordinaire (GLIMPSE) surveys. Of these, 183 are either OB stars earlier than B8 in which the excess plausibly arises from a thermal bremsstrahlung component or evolved stars in which the excess may be explained by an atmospheric dust component. The remaining 47 stars have spectral classifications B8 or later and appear to be main sequence or late pre-main-sequence objects harboring circumstellar disks. Six of the 47 stars exhibit multiple signatures characteristic of pre-main-sequence circumstellar disks, including emission lines, near-infrared K-band excesses, and X-ray emission. Approximately one-third of the remaining 41 sources have emission lines suggesting relative youth. Of the 25 GLIMPSE stars with SST data at >24 microns, 20 also show an excess at 24 microns. Three additional objects have 24 micron upper limits consistent with possible excesses, and two objects have photospheric measurements at 24 microns. Six MSX sources had a measurement at wavelengths >8 microns. We modeled the excesses in 26 stars having two or more measurements in excess of the expected photospheres as single-component blackbodies. We determine probable disk temperatures and fractional infrared luminosities in the range 191 < T < 787 and 3.9x10^-4 < L_IR/L_* < 2.7x10^-1. We estimate a lower limit on the fraction of Tycho-2 Spectral Catalog main-sequence stars having mid-IR, but not near-IR, excesses to be 1.0+-0.3%.


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Q)

# A compound on analysis,was found to have the following composition :(i) Na=14.31% (ii) S=9.97% (iii) O=69.50% (iv) H=6.22%.Calculate the molecular formula of the compound assuming that whole of hydrogen in the compound is present as water of crystallization .Molecular mass of the compound is 322.

$\begin{array}{1 1}(a)\;Na_2SO_4.10H_2 O&(b)\;Na_2SO_4\\(c)\;NaSO_49H_2O&(d)\;Na_2SO_3 10H_2O\end{array}$

Comment
A)
The empirical formula $=Na_2SH_{20}O_{14}$
Empirical formula mass =$(2\times 23)+32+(20\times 1)$
$\qquad\qquad\qquad\qquad\;=322$
$n=\large\frac{Mol.mass}{Emp.mass}$
$\;\;\;=\large\frac{322}{322}$
$\;\;\;=1$
$\therefore$ Molecular formula =$Na_2SH_{20}O_{14}$
So molecular formula =$Na_2SO_4.10H_2 O$


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# Fill colour for ring (pstricks)

I have a drawing for a cylinder and have filled in the ring section with light gray.

However, because my background is light orange, the image turns out to be a bit weird.

Here is the image:

Here is my code:

    \begin{figure}[H]
\centering
\psset{xunit=1.0cm,yunit=1.0cm,algebraic=true,dimen=middle,dotstyle=o,dotsize=3pt 0,linewidth=0.8pt,arrowsize=3pt 2,arrowinset=0.25}
\begin{pspicture*}(-5.03999786579,-2.60435804065)(9.52396200612,6.62258178636)
\pscircle[fillcolor=lightgray,fillstyle=solid](-2.17742644269,0.466153228618){2.53961608687}
\pscircle[fillcolor=white,fillstyle=solid](-2.17742644269,0.466153228618){2.05904260258}
\psline(-3.15956373091,2.80817291593)(4.45168632659,5.71420083763)
\psline(-1.20065102466,-1.87810777464)(6.41205030135,1.02847425142)
\parametricplot{-1.177820125593466}{1.9703016360052805}{1.*2.53948165422*cos(t)+0.*2.53948165422*sin(t)+5.43958261193|0.*2.53948165422*cos(t)+1.*2.53948165422*sin(t)+3.37437997525}
\parametricplot[linestyle=dotted]{1.9703016360052805}{5.10536518158612}{1.*2.53982297387*cos(t)+0.*2.53982297387*sin(t)+5.43958261193|0.*2.53982297387*cos(t)+1.*2.53982297387*sin(t)+3.37437997525}
\pscircle[linestyle=dotted](5.43958261193,3.37437997525){2.05992773607}
\psline[linestyle=dotted](-2.97422018908,2.36477803751)(4.6369046976,5.27148578244)
\psline[linestyle=dotted](-1.3859945665,-1.43471289623)(6.2269636645,1.4708744313)
\psline[linestyle=dashed,dash=3pt 3pt](-2.17742644269,0.466153228618)(-1.56471932744,2.43192189002)
\psline[linestyle=dashed,dash=3pt 3pt](-2.17742644269,0.466153228618)(0.202564742151,-0.420013701907)
\rput[tl](-1.92632368628,-0.0862728354885){8 cm}
\rput[tl](-2.87963195551,1.7969978376){5 cm}
\rput[tl](-0.168604391391,4.93578229275){14 cm}
\psdots[dotstyle=*](-2.17742644269,0.466153228618)
\rput[bl](-2.67698534012,0.316814882465){{$O$}}
\end{pspicture*}
\end{figure}

Another PSTricks solution:

\documentclass[pstricks]{standalone}
\usepackage{pst-solides3d}
\begin{document}

\begin{pspicture}(-4,-4)(4,4)
\psset{lightsrc=viewpoint,viewpoint=100 20 20 rtp2xyz,Decran=50,shortput=nab}
\psSolid[object=anneau,h=12,ngrid=72 4,grid,fillcolor={[hsb]{0.118 1 1}},R=5,r=4,
incolor=green!50,RotY=90]
\psSolid[object=anneau,h=0.1,ngrid=72 4,grid,fillcolor=red,R=5,r=4,
incolor=green!50,RotY=90](6,0,0)
\psPoint(6,0,0){O}\psdot(O)
\psPoint(6,5 120 cos mul,5 120 sin mul){Ext}
\pcline{->}(O)(Ext)_[nrot=:D]{R1}
\psPoint(6,4 30 cos mul,4 30 sin mul){Int}
\pcline{->}(O)(Int)^[nrot=:U]{R2}
\end{pspicture}

\end{document}

It is the choice of the perspective that is bad. Here is an example with pst-solides3d.

\documentclass{article}
\usepackage{pst-solides3d}
\begin{document}
\begin{center}
\begin{pspicture}(-4,-2)(4,10)
\psset{lightsrc=viewpoint,viewpoint=100 20 20 rtp2xyz,Decran=50,solidmemory}
\psSolid[object=prisme,h=12,grid,
fillcolor={[hsb]{0.118 1 1}},
base=0 5 360 { % 72 faces extérieurs 360/5
/angle ED
5 angle cos mul
5 angle sin mul
} for
360 -5 0 { % 72 faces intérieures
/angle ED
4 angle cos mul
4 angle sin mul
} for,
fcol=0 (red) 72 1 144 {/i ED i (cyan)} for,
RotY=90]
\defFunction[algebraic]{cercle1}(t){5*cos(t)}{5*sin(t)}{}
\defFunction[algebraic]{cercle2}(t){4*cos(t)}{4*sin(t)}{}
\psSolid[object=plan,
definition=equation,
args={[1 0 0 -12]},
base=-10 10 -10 10,
action=none,
name=monplan]
\psset{plan=monplan}
\psProjection[object=courbeR2,
range=0 2 Pi mul,resolution=360,
function=cercle2]
\psProjection[object=courbeR2,
range=0 2 Pi mul,resolution=360,
function=cercle1]
\composeSolid
\psPoint(12,0,0){O}\psdot(O)
\psPoint(12,5 120 cos mul,5 120 sin mul){Ext}
\psline{->}(O)(Ext)
\pcline(Ext)(O)
\aput{:U}{R1}
\psPoint(12,4 30 cos mul,4 30 sin mul){Int}
\pcline{->}(O)(Int)
\aput{:U}{R2}
\end{pspicture}
\end{center}
\end{document}

• Just a little thing: the dot in the middle should be 3D, it would be less confusing for the human brain when it tries to place it in the correct place in the 3D space. – yo' Mar 7 '15 at 16:19


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# The vaccine distribution conundrum

The context is that of a pandemic that is spreading wildly and requiring global vaccination of the population.

You are working in a distribution center for the vaccines.

One day you have been requested to prepare and deliver 40 boxes of vaccines to 3 vaccination centers. You have a minivan and will do a roundtrip visiting each center once.

Unfortunately you haven't been told how many boxes each vaccination center needs. You will learn the need of each center only when you arrive there. It could even be that a center didn't order any vaccine for that day. You just know the total is 40 boxes.

The problem is that the boxes need to be delivered in refrigerated sealed containers. You can't reopen any container when delivering the boxes. You must deliver the containers still sealed.

You don't have to deliver exactly one container per center. You could just put each box in its own container. But it is a lot of work to refrigerate and seal a container. And you are lazy. So you want to figure out a better way to arrange boxes in containers in such a way that you are sure, regardless of the needs of each vaccination center, that you can deliver to each one the exact number of boxes they need.

Of course, you want to minimize the number of containers you need to prepare.

So, how many containers do you prepare and how many boxes do you put in each container?

Disclaimer: The events portayed in this puzzle are fictitious. Any resemblence to real events or situations is purely coincidential.

PS: clarification: The unit is the box, which is indivisible. The number of doses per box is irrelevant. All boxes are equal. The vaccination centers ordered an integer number of boxes. You cannot return boxes. You must pack exactly 40 boxes.

• oeis.org/A005428 – loopy walt May 1 at 2:17
• @loopywalt I think this OEIS sequence considers the relaxed problem where you know all demands at once (my initial misinterpretation). – RobPratt May 1 at 2:23
• For numbers of boxes that are not necessarily partial sums of A005428, the minimum number of containers appears to match oeis.org/A061420. – RobPratt May 1 at 15:58

You are looking for a partition of $$40$$ with the minimum number of parts that is a common refinement of all $$154$$ partitions of $$40$$ into at most $$3$$ parts.

You can satisfy all $$154$$ scenarios with

8 containers:

14, 9, 6, 4, 3, 2, 1, 1

Note that no container can contain more than 14 boxes because the centers might demand 14, 13, and 13, respectively.

Update: I had initially assumed that the demands for all three centers were known upon reaching the first one, but I see now that the problem wording implies uncertainty about the remaining two centers upon reaching the first center. It turns out that enforcing "nonanticipativity" over the resulting $$861$$ scenarios yields the same solution as above.

I used integer linear programming, and one of the decision variables is how many containers to deliver to the first center, independent of the remaining two centers. The resulting solution for the first center is as follows:

0 = 0
1 = 1
2 = 2
3 = 3
4 = 4
5 = 1 + 4
6 = 6
7 = 1 + 6
8 = 2 + 6
9 = 9
10 = 1 + 9
11 = 2 + 9
12 = 3 + 9
13 = 4 + 9
14 = 14
15 = 1 + 14
16 = 2 + 14
17 = 3 + 14
18 = 4 + 14
19 = 1 + 4 + 14
20 = 6 + 14
21 = 1 + 6 + 14
22 = 2 + 6 + 14
23 = 9 + 14
24 = 1 + 9 + 14
25 = 2 + 9 + 14
26 = 3 + 9 + 14
27 = 4 + 9 + 14
28 = 1 + 4 + 9 + 14
29 = 6 + 9 + 14
30 = 1 + 6 + 9 + 14
31 = 2 + 6 + 9 + 14
32 = 3 + 6 + 9 + 14
33 = 4 + 6 + 9 + 14
34 = 1 + 4 + 6 + 9 + 14
35 = 2 + 4 + 6 + 9 + 14
36 = 3 + 4 + 6 + 9 + 14
37 = 1 + 3 + 4 + 6 + 9 + 14
38 = 2 + 3 + 4 + 6 + 9 + 14
39 = 1 + 2 + 3 + 4 + 6 + 9 + 14
40 = 1 + 1 + 2 + 3 + 4 + 6 + 9 + 14

In hindsight, this is just a greedy allocation. Use the largest available container and repeat until the demand is satisfied. But there are also non-greedy optimal allocations to the first center.

The optimization model enforces that the demands for the remaining two centers can always be satisfied after removing these containers.

• That is correct. But can you show that this combination will always work and explain how you decide which container to deliver in the first center you visit? – Florian F May 1 at 8:53
• You mean 14, 13, 13 instead of 14, 14, 13? – justhalf May 2 at 3:15
• Yes, corrected just now. – RobPratt May 2 at 4:32

Let us consider the series https://oeis.org/A005428

which starts with $$a_1 = 1$$ and proceeds with (0) $$a_{n+1} = \left \lceil \frac {1 + \sum_{i=1}^n a_i} 2 \right \rceil$$. First eight terms: 1,1,2,3,4,6,9,14

Let us write $$A_n = \sum_{i=1}^n a_i$$. First eight terms: 1,2,4,7,11,17,26,40

We will now show by induction that the following three statements are all true:

Note that wherever I say "set" below I should really have said "multi set" but that would have been tedious so I didn't.

1. For any unordered 3-partition $$A_n = a^{(n)}_1 + a^{(n)}_2 + a^{(n)}_3$$ there exists a 3-way (set) partition of $$\{a_1,...,a_n\}$$ such that the sums of the three subsets are $$a^{(n)}_1,a^{(n)}_2,a^{(n)}_3$$

2. For any $$B there exists a subset S of $$\{a_1,...,a_n\}$$ such that for every 2-partition $$B=b_1+b_2$$ there exists a 2-way (set) partition of $$S$$ such that the sums of the two subsets are $$b_1,b_2$$

3. Any ordered sequence $$\{b_1\le b_2\le ...\le b_n\}$$ with property (1) has $$b_i\le a_i$$ for all $$i$$.

All three statements are easily verified for small $$n$$.

Now observe that by construction of $$a_n$$ the following holds:

4. Every 3-partition of $$A_n$$ has at least one term, $$a^{(n)}_1$$ say, which is $$\ge a_n$$.
A representation in terms of $$\{a_1,...,a_{n-1}\}$$ of the 3-partition $$a^{(n)}_1-a_n,a^{(n)}_2,a^{(n)}_3$$ of $$A_{n-1}$$ can therefore be made into a representation in terms of $$\{a_1,...,a_n\}$$ of the 3-partition $$a^{(n)}_1,a^{(n)}_2,a^{(n)}_3$$ of $$A_n$$ simply by adding $$a_n$$ to the first subset. This proves (1).

Similarly:

5. For every $$b_1+b_2=B < A_n$$ we either have $$B \le A_{n-1}$$ or at least one of $$b_1,b_2$$ is $$\ge a_n$$. In the second case we can proceeed essentially as above, the first case is trivial. This proves (2).

Finally,

6. Because we must always be able to cover the even split ($$\pm \frac 2 3$$) of $$A_n$$ the last and largest term $$a_n$$ cannot be larger than what it is assigned by the formation law (0). Assertion (3) if satisfied for $$a_i,b_i$$ for $$i \le n$$ thus implies $$b_{n+1} \le \left \lceil \frac {1 + \sum_{i=1}^n b_i} 2 \right \rceil \le \left \lceil \frac {1 + \sum_{i=1}^n a_i} 2 \right \rceil = a_{n+1}$$. This proves (3).

To summarize:

The series we have cited is the best solution for both the two-stage and the simultaneous partioning problems. From the proof of statement (2) we can also derive how to find the right subset in case of ambiguity.

• Does this prove the optimality? – justhalf May 2 at 3:21
• @justhalf statement 3 does. – loopy walt May 2 at 7:04
• @loopywalt , I couldn't understand anything after the line : "We will now show by induction that the following three statements are all true:" It is too technical a language for me. Can you please explain in simpler language ? – Hemant Agarwal May 2 at 20:45

When I wrote the question I had a more concrete solution in mind. The other answers given are very computational or mathematical. They don't really show how the solution is constructed imho.

So here is what I had in mind.

Since there are 40 boxes to distribute and only 3 centers, there must be at least one center that ordered 14 boxes or more. That is because if no center ordered more than 13, then the total could not exceed 39.
So we can prepare one container with 14 boxes and reserve it for the first center that requires that many boxes.

When you are done, you can see that you face the same problem again, but smaller:
You need to distribute the 26 remaining boxes without knowing in advance how many each center requires, (after deduction of the first container).
With 26 boxes total, one center needs 9 boxes or more (9 is 26/3 rounded up). So you can prepare a container with 9 boxes to be delivered to the first center that needs so many (possibly additionally to the 14-container).

An now you have 17 boxes to deliver under the same condidions...

When you repeat this process you get the sequence 14, 9, 6, 4, 3, 2, 1, 1.

This shows a solution that works and explains why it needs to be these numbers. It works without knowing the quantities in advance. When you reach a center, you fill the order by choosing first the largest containers possible. This ensures that the largest container gets delivered and the smaller ones are delivered according to the remaining needs.

Now, unfortunately, this doesn't prove optimality.
But you can see that in each iteration the container you prepare is the largest possible. For example the first container cannot have more than 14 boxes because the required amounts could be 14, 13, 13. A larger container couldn't be delivered. And there is the intuitive idea that larger containers are better, because it makes the the remaining problem smaller. This makes it plausible that the method is optimal. Or not too bad. It is not a proof, though.

• This is very nice solution! – justhalf May 2 at 3:22
• @justhalf , can you help us in proving that this will give us an optimal solution ? – Hemant Agarwal May 2 at 12:35
• loopy already did – justhalf May 2 at 14:31
• @justhalf , I asked loopy for an explanation by commenting below his answer but he did not respond. Could you please explain the proof. This was my comment : Loopy, I couldn't understand anything after the line : "We will now show by induction that the following three statements are all true:" It is too technical a language for me. Can you please explain in simpler language ? – Hemant Agarwal Jul 5 at 15:01
• Statement 1 is: "We can always use the OEIS 5428 sequence of number of boxes in each container to handle any distribution to 3 centers", Statement 3 is: "Any solution must have container sizes smaller than OEIS 5428", which means they need to use at least as many containers as OEIS 5428. So OEIS 5428 gives the minimum number of containers. I'm not sure of the significance of Statement 2. – justhalf Jul 5 at 18:04


text:
# Proof of “generic curve of genus at least 2 has no nontrivial maps to a positive genus curve”

I searched for it for a long time, but it seems that everybody is taking this for granted and does not bother to point out a proof. Would it be possible that someone points me to a proof or makes me see the obvious.

Thank you very much.

• What is your definition of "generic"? I expect that there are countably many subvarieties in the moduli space parameterizing curves that admit a positive degree morphism to some genus 1 curve. – Jason Starr Feb 23 '15 at 16:26
• Of course, Jason is correct. However, a very general point of $M_g$ (by which I mean, exclude a countable union of proper sub varieties) corresponds to a curve with a simple Jacobian. This curve can't map onto anything of smaller genus. (This is over $\mathbb{C}$.) – Donu Arapura Feb 23 '15 at 16:47
• @DonuArapura Could you give me a reference published somewhere? Thanks a lot. – Syed Feb 23 '15 at 17:19
• @DonuArapura because the set of morphism is a map between this union of sub varieties to $\cup M_{g_i<g}$ ? Is this the classical proof? – Syed Feb 23 '15 at 19:13
• For your first question, it follows from from Cor. 17.5.2 of Birkenhake and Lange's book on Complex Abelian Varieties (2nd ed), but I'm sure there are more suitable references. – Donu Arapura Feb 23 '15 at 19:42

OK, here's an answer. As I said in the comments, if the $C$ is a (smooth projective) curve with a simple Jacobian, then it can't map onto a curve of smaller genus. Since I was a bit curious myself, I found a reference for the next step: Koizumi "The ring of correspondences on a generic curve of genus g" Nagoya (1976). The result is better than I expected! Koizumi proves that if $C$ is the geometric generic curve of $M_g$ over the algebraic closure of the prime field, then $End(J( C) =\mathbb{Z}$; so in particular, it is simple.


text:
# The field of Laurent series over $\mathbb{C}$ is quasi-finite

How can I prove that the field of Laurent series over $\mathbb{C}$ is quasi-finite, which means that it has a unique extension of degree $n$ for all $n \geqslant 1$ ?

The article http://en.wikipedia.org/wiki/Quasi-finite_field says that the extension of degree $n$ is $\mathbb{C}((T^{1/n}))$. I think that I understand why it is an extension of degree $n$ (the polynomial $X^n – T$ is irreducible by Eisenstein’s criterion ?), but how can I prove that it is unique please ?

#### Solutions Collecting From Web of "The field of Laurent series over $\mathbb{C}$ is quasi-finite"

The assertion is a consequence of the ramification theory of valuations: the field of Laurent series $\mathbb{C}((t))$ carries the discrete valuation $v(a_kt^k+a_{k+1}t^{k+1}+\ldots ):=k$, $a_k\neq 0$. It is complete with respect to this valuation; consequently $v$ has a unique extension $w$ to every finite extension field $K$ of $\mathbb{C}((t))$.

Assume $[K:\mathbb{C}((t))]=n$, then by the fundamental equality of valuation theory one has $(\Gamma :\mathbb{Z})[\overline{K}:\mathbb{C}]=n$, where $\Gamma$ is the value group of $w$ and $\overline{K}$ is the residue field of $w$. Since $\mathbb{C}$ is algebraically closed, one gets $\overline{K}=\mathbb{C}$ and thus $(\Gamma :\mathbb{Z})=n$.

Now assume for the moment that $K/\mathbb{C}((t))$ is Galois. Ramification theory then yields that the Galois group $G$ is isomorphic to $\Gamma /\mathbb{Z}$, which is cyclic of order $n$. In particular every intermediate field of $K/\mathbb{C}((t))$ is Galois with cyclic Galois group.

Since every finite extension $L/\mathbb{C}((t))$ is contained in a finite Galois extension $K/\mathbb{C}((t))$, this shows that every finite extension $L/\mathbb{C}((t))$ is Galois with cyclic Galois group.

Since in a finite cyclic group for every divisor $d$ of the group order there exists exactly one subgroup of index $d$, for every degree $d$ there exists exactly one extension $K/\mathbb{C}((t))$ of degree $d$.


text:
# Contents

## Introduction

Much of what is discussed below is either referred to as non-equilibrium thermodynamics or stochastic thermodynamics.

## Markov processes and the master equation

A continuous-time, finite state Markov process is a pair $(V,H)$ where $V$ is a finite set of states and $H : \mathbb{R}^V \to \mathbb{R}^V$ is an infinitesimal stochastic,

(1)$H_{ij} \geq 0 \; \: i \neq j \quad \text{and} \quad \; \sum_i H_{ij} = 0,$

Hamiltonian generating the time evolution of a population distribution $p(t) \in \mathbb{R}^V$ via the master equation

(2)$\frac{dp}{dt} = Hp .$

The infinitesimal stochastic condition requires that the off-diagonal components of $H$ are non-negative and that the columns sum to zero, meaning that $H_{ii} = -\sum_{j \neq i} H_{ji}$, or that the diagonal elements are equal to minus the sum of all the other elements in that column. The entry $H_{ij}$ is interpreted as the transition rate from state $j$ to state $i$. Thus the diagonal entries are equal to minus the sum of the outgoing rates from a certain state.

An equilibrium distribution $q \in \mathbb{R}^V$ is a population distribution that does not change with time,

(3)$\frac{dq}{dt} = Hq = 0.$

### Currents

In terms of the indices, the master equation reads:

(4)$\frac{dp_i}{dt} = \sum_{j} H_{ij}p_j.$

We can use the infinitesimal stochastic property of $H$ to rewrite this as:

(5)$\frac{dp_i}{dt} = \sum_{j} H_{ij}p_j - H_{ji}p_i.$

If we define the current from $j$ to $i$ as $J_{ij} = H_{ij}p_j - H_{ji}p_i$, we can write the master equation in the simple form:

(6)$\frac{dp_i}{dt} = \sum_j J_{ij}.$

Note that this definition of current requires an implicit choice that a positive current $J_{ij}$ corresponds to a net flow population into the $i^{\text{th}}$ state.

### Detailed Balance

We say an equilibrium distribution $q$ satisfies detailed balance if

(7)$H_{ij}q_j = H_{ji}q_i.$

Equilibrium distributions always satisfy $\frac{dq_i}{dt} = \sum_j J_{ij} = 0$ for all $i$, but for an equilibrium satisfying detailed balance each term in this sum vanishes identically.

For an equilibrium not satisfying detailed balance there can be non-zero individual currents flowing $J_{ij} \neq 0$ in such a way that $\sum_j J_{ij} = 0$ is still satisfied. Such an equilibrium is typically referred to as a non-equilibrium steady state.

### Representation of Markov processes using labelled graphs

One can represent a Markov process as a directed graph $(V,E,s,t)$, where $V$ is a finite set of states, $E$ is a finite set of edges, and $s,t : E \to V$ are source and target maps respectively, along with a map $r : E \to [0,\infty)$ giving the labels or the transition rates associated to each edge, i.e. for an edge $e \in E$, $r(e) = H_{ t(e) s(e) }$.

### The cycle basis

Given a labelled, directed graph $G=(V,E,s,t,r)$ one can forget the directedness of the edges and consider the undirected graph $(V,E)$. A spanning tree is then a connected subgraph containing all the vertices $V$ of $G$, but with no cycles. A connected undirected graph with no cycles is called a tree. Given a spanning tree $T = (V,E_T )$ the leftover edges $E-E_T$ form the set of chords of the graph. Adding any chord to the spanning tree produces a cycle. The set of cycles obtained when adding each of the chords individually to a given spanning tree provides one choice of a cycle basis for the underlying graph $G$.

For Markov processes one cannot discard the information regarding the direction of the edges. Similarly we will need a directed cycle basis. To achieve this we choose an arbitrary orientation on all of the cycles in our basis, such as clockwise. It is easy to see that for a graph $G$ with $n$-vertices $|V|=n$ a spanning tree $T=(V,E_T)$ on those n-vertices will have $n-1$ edges, $|E_T| = n-1$. Therefore we have $\nu = |E|-|E_T| = |E| - n + 1$ chords and hence $\nu$ elements in our cycle basis. Therefore let us denote the cycle basis by $\{C_1,C_2,....C_{\nu} \}$.

Given an arbitrary directed cycle $C$ on the graph $G$ we can write $C = \sum_{\alpha=1}^{\nu} (C,C_{\alpha})C_{\alpha}$. Here we have introduced an inner product on cycle space $(C,C_{\alpha}) = S_{\alpha}(C)S_{\alpha}(C_{\alpha}) \quad 1 \leq \alpha \leq \nu$, where

(8)$S_{\alpha}(C) = \left\{ \begin{array}{ccc} +1 & \text{if} & \alpha || C \\ -1 & \text{if} & -\alpha || C \\ 0 & \text{if} & \alpha \notin C. \end{array} \right.$


text:
# zbMATH — the first resource for mathematics

Strong approximation for the total space of certain quadric fibrations. (English) Zbl 1328.11060
Let $$X$$ be a variety over a number field $$F$$ such that the set $$X(F)$$ of rational points is nonempty. Let $$S$$ be a finite set of places in $$F$$. The strong approximation for $$X$$ off the set $$S$$ means that the diagonal image of $$X(F)$$ is dense in the space of $$S$$-adéles $$X({\mathbb A}_{F}^{S}).$$ The $$S$$-adéles are the adéles with places in $$S$$ omitted. If the property of strong approximation holds for $$X$$ then it implies a local-global principle for the existence of integral points on models over the ring of $$S$$-integers of $$F.$$ There is an obstruction called the Brauer-Manin obstruction to strong approximation off $$S.$$
In this paper, the authors investigate strong approximation for varieties $$X/F$$ given by an equation $q(x_{1},\dots ,x_{n}) = p(t),$ where $$q$$ is a quadratic form of rank $$n$$ in $$n\geq 3$$ variables and $$p(t)$$ is a nonzero polynomial.

##### MSC:
 11G05 Elliptic curves over global fields 14G05 Rational points 14F22 Brauer groups of schemes
Full Text:


text:
## Discarding correlation among inputs in a neural network

3

I am working on a problem with 4 inputs and 1 continuous output variable. The sum of all values of the 4 input variables is always 1.

a1+a2+a3+a4=1

So, they are correlated.

My question is: should I use all 4 variables for neural network training? Or, should I use any 3 of them to get rid of correlation? Is there any problem if I use all 4?

– TwinPenguins – 2018-05-18T07:18:24.243

So, according to this answer, I can use all 4 inputs without any problem. That is ok. But do I get the same answer if I use only 3 of them? – Saptarshi Roy – 2018-05-18T07:51:14.073

Well, how are you going to choose three of four variables? Something like PCA? If yes do try, but you need some sort of way of selecting the subset of features, or maybe random subsampling of features? I would first take all and closely observe the validation curve and do some regularizations first! – TwinPenguins – 2018-05-18T08:46:08.597

That said, have you tested the data for multicollinearity? Maybe you have but I don't think that a1+a2+a3+a4=1 implies a high degree of correlation.


text:
# Why one cell in a table have different height?

I am trying to make a table having equal height and width to all the cells. I tried this code:

\documentclass[conference]{IEEEtran}
\setlength{\extrarowheight}{1pt}
\begin{table}
\centering
\caption{My caption}
\label{my-label}
\begin{tabular}{|*{12}{p{1.0cm}|}}
\hline
labels     & airplane & automobile & bird & cat & deer & dog & frog & horse & ship & truck & accuracy \\ \hline
airplane   & 915      & 4          & 17   & 19  & 3    & 1   & 0    & 2     & 27   & 12    & 91.50\%  \\ \hline
automobile & 8        & 934        & 3    & 4   & 0    & 0   & 3    & 0     & 10   & 38    & 93.40\%  \\ \hline
bird       & 60       & 1          & 813  & 37  & 19   & 23  & 30   & 10    & 7    & 0     & 81.30\%  \\ \hline
cat        & 18       & 1          & 34   & 746 & 25   & 113 & 37   & 18    & 8    & 0     & 74.60\%  \\ \hline
deer       & 24       & 1          & 38   & 33  & 809  & 19  & 44   & 29    & 2    & 1     & 80.90\%  \\ \hline
dog        & 4        & 0          & 37   & 106 & 23   & 792 & 9    & 26    & 2    & 1     & 79.20\%  \\ \hline
frog       & 2        & 5          & 19   & 35  & 1    & 20  & 912  & 2     & 3    & 1     & 91.20\%  \\ \hline
horse      & 14       & 0          & 26   & 20  & 18   & 28  & 4    & 886   & 3    & 1     & 88.60\%  \\ \hline
ship       & 35       & 10         & 3    & 2   & 0    & 2   & 1    & 0     & 936  & 11    & 93.60\%  \\ \hline
truck      & 23       & 37         & 4    & 10  & 1    & 2   & 2    & 0     & 15   & 906   & 90.60\%  \\ \hline
\end{tabular}
\end{table}

and got output like below:

I couldn't figure out why the third row i.e automobile row is having different height.. Why is that??

• You need to provide more information (a minimum working example, starting with \documentclass). But I suspect that the word automobile is running up against a column margin, perhaps. – Steven B. Segletes Apr 20 '16 at 18:57
• I edited by giving the document class.. and i tried by increasing the width of the columns and it works – Setu Kumar Basak Apr 20 '16 at 19:01
• ...and the preamble content, too. What packages are loaded and in what order could affect things. You really need to post a compileable document that produces the image you show. – Steven B. Segletes Apr 20 '16 at 19:02
• you have defined all the cells as paragraphs, so anything that exceeds the specified width will force an additional line. "automobile" does this. (you can tell because the space to its right is wider than the space to its left; there will also be a warning in the log file.) – barbara beeton Apr 20 '16 at 19:22
• how can i make the text inside cell to left aligned?? – Setu Kumar Basak Apr 20 '16 at 19:28

I propose two solutions, with tabularx, siunitx and eqparbox, for different alignments in columns:

\documentclass[conference]{IEEEtran}
\usepackage{showframe, eqparbox, siunitx} \usepackage{tabularx}
\setlength{\extrarowheight}{2pt}
\newlength{\colwd}

\begin{document}

\begin{table*}
\centering
\setlength\tabcolsep{3.5pt}
\caption{My caption}
\label{my-label}
\begin{tabularx}{\linewidth}{|*{12}{ >{\centering\arraybackslash}X|}}%{|*{12}{p{1.1cm}|}}%
\hline
labels & airplane & automobile & bird & cat & deer & dog & frog & horse & ship & truck & accuracy \\ \hline
airplane & 915 & 4 & 17 & 19 & 3 & 1 & 0 & 2 & 27 & 12 & 91.50\,\% \\ \hline
automobile & 8 & 934 & 3 & 4 & 0 & 0 & 3 & 0 & 10 & 38 & 93.40\,\% \\ \hline
bird & 60 & 1 & 813 & 37 & 19 & 23 & 30 & 10 & 7 & 0 & 81.30\,\% \\ \hline
cat & 18 & 1 & 34 & 746 & 25 & 113 & 37 & 18 & 8 & 0 & 74.60\,\% \\ \hline
deer & 24 & 1 & 38 & 33 & 809 & 19 & 44 & 29 & 2 & 1 & 80.90\,\% \\ \hline
dog & 4 & 0 & 37 & 106 & 23 & 792 & 9 & 26 & 2 & 1 & 79.20\,\% \\ \hline
frog & 2 & 5 & 19 & 35 & 1 & 20 & 912 & 2 & 3 & 1 & 91.20\,\% \\ \hline
horse & 14 & 0 & 26 & 20 & 18 & 28 & 4 & 886 & 3 & 1 & 88.60\,\% \\ \hline
ship & 35 & 10 & 3 & 2 & 0 & 2 & 1 & 0 & 936 & 11 & 93.60\,\% \\ \hline
truck & 23 & 37 & 4 & 10 & 1 & 2 & 2 & 0 & 15 & 906 & 90.60\,\% \\ \hline
\end{tabularx}
\end{table*}
%
\begin{table*}
\centering
\setlength\tabcolsep{3.5pt}
\sisetup{table-format=3.0, table-number-alignment=center}
\caption{My caption}
\label{thy-label}
\begin{tabularx}{\linewidth}{|l|*{10}{S[]|}S[table-format =2.2,table-space-text-post = \,\%]}%{|*{12}{p{1.1cm}|}}%
\hline
airplane & 915 & 4 & 17 & 19 & 3 & 1 & 0 & 2 & 27 & 12 & 91.50\,\% \\ \hline
automobile & 8 & 934 & 3 & 4 & 0 & 0 & 3 & 0 & 10 & 38 & 93.40\,\% \\ \hline
bird & 60 & 1 & 813 & 37 & 19 & 23 & 30 & 10 & 7 & 0 & 81.30\,\% \\ \hline
cat & 18 & 1 & 34 & 746 & 25 & 113 & 37 & 18 & 8 & 0 & 74.60\,\% \\ \hline
deer & 24 & 1 & 38 & 33 & 809 & 19 & 44 & 29 & 2 & 1 & 80.90\,\% \\ \hline
dog & 4 & 0 & 37 & 106 & 23 & 792 & 9 & 26 & 2 & 1 & 79.20\,\% \\ \hline
frog & 2 & 5 & 19 & 35 & 1 & 20 & 912 & 2 & 3 & 1 & 91.20\,\% \\ \hline
horse & 14 & 0 & 26 & 20 & 18 & 28 & 4 & 886 & 3 & 1 & 88.60\,\% \\ \hline
ship & 35 & 10 & 3 & 2 & 0 & 2 & 1 & 0 & 936 & 11 & 93.60\,\% \\ \hline
truck & 23 & 37 & 4 & 10 & 1 & 2 & 2 & 0 & 15 & 906 & 90.60\,\% \\ \hline
\end{tabularx}
\end{table*}

\end{document}


text:
# Cat program in x86 NASM

I wrote cat program in x86 NASM and I would like to know if there is anything that can be improved.

This implementation of cat supports:

• Multiple arguments.

I checked few cases and program behaves exactly like original cat command.

I was thinking if I should add labels like write even if I wouldn't jmp to them, so code would be cleaner.

Code:

section .text
global _start

_start:
;esi=1
mov esi, 1

;if(argc==1)
;goto open
cmp dword [esp], 1
jnz open

;else
;fd=STDIN_FILENO
mov dword [fd], 0

open:
;fd=open(path,oflag)
mov eax, 5
mov ebx, [esp+esi*4+4]
mov ecx, 0
int 0x80
mov [fd], eax

mov eax, 3
mov ebx, [fd]
mov ecx, buf
mov edx, BUFSIZE
int 0x80

mov edx, eax
mov eax, 4
mov ebx, 1
mov ecx, buf
int 0x80

cmp edx, 0

;close(fd)
mov eax, 6
mov ebx, [fd]
int 0x80

;esi++
inc esi

;if(esi<argc)
cmp dword esi, [esp]
jl open

exit:
mov eax, 1
mov ebx, 0
int 0x80

section .data
BUFSIZE equ 1024

section .bss
buf resb BUFSIZE
fd resd 1

I was thinking if I should add labels like write even if I wouldn't jmp to them, so code would be cleaner.

Writing labels that aren't jumped to is fine. Provided they have meaningful names, it can help to understand the program.

## My comments (not in a particular order)

;esi=1
mov esi, 1

This is a redundant comment! I already can see what the instruction does. What I can't know is what the ESI register will be used for.

;if(argc==1)
;goto open
cmp dword [esp], 1
jnz open

Here the comment does not correspond to the code. The code actually does if(argc!=1).

cmp edx, 0

The more usual way to test for zero would be test edx, edx. It has a smaller encoding and is generally a bit faster.

mov ebx, 0

Similarly, zeroing a register is best done using the xor instruction (xor ebx, ebx). This is both faster and shorter.

mov dword [fd], 0

Your program would not need this instruction at all, if you would move the fd variable from the .bss section to the .data section using fd dd 0.

This in turn opens up an opportunity to shorter the code from:

    cmp dword [esp], 1
jnz open
mov dword [fd], 0
open:

to:

    cmp dword [esp], 1
open:

Also note that in conjunction with the cmp instruction, it's better to use je and jne.
In conjunction with the test instruction, it's preferable to use jz and jnz.

je and jz have an identical encoding but choosing the right instruction mnemonic better expresses what the intention of the code is.

;if(esi<argc)
cmp dword esi, [esp]
jl open

The argc is by its very nature an unsigned number. After all it's just a count. You should use the conditional jumps that are provided to deal with the unsigned conditions. So better use jb open (JumpIfBelow).
Also there's no point in writing the dword size-tag. The mention of the dword register ESI already dictates the size.

mov edx, eax
mov eax, 4
mov ebx, 1
mov ecx, buf
int 0x80

I always prefer to write the function number directly above the int 0x80 instruction. That way it's immediately clear what function is getting invoked. And for perfection I even assign the registers in a sorted manner:

mov edx, eax
mov ecx, buf
mov ebx, 1
mov eax, 4
int 0x80

## Putting it all together

• not being afraid to write redundant labels
• shaving off one more byte by using the fact that ESI==1 when cmp dword [esp], 1 is executed.
section .text
global _start

_start:
mov  esi, 1
cmp  [esp], esi        ;if(argc==1) goto read
open:
xor  ecx, ecx          ;fd=open(path,oflag)
mov  ebx, [esp+esi*4+4]
mov  eax, 5
int  0x80
mov  [fd], eax
mov  ecx, buf
mov  ebx, [fd]
mov  eax, 3
int  0x80
write:
mov  ecx, buf
mov  ebx, 1
mov  eax, 4
int  0x80
more:
close:
mov  ebx, [fd]         ;close(fd)
mov  eax, 6
int  0x80
next:
inc  esi
cmp  esi, [esp]        ;if(esi<argc)
jb   open
exit:
xor  ebx, ebx
mov  eax, 1
int  0x80

section .data
fd   dd 0              ;fd=STDIN_FILENO

section .bss
BUFSIZE equ 1024
buf  resb BUFSIZE

## Concern

Nowhere in your code do you check for any errors from api calls like open or read. Imagine what might happen in these cases?

• Thanks for pointing out these comments. Can you tell me how do you know that test is faster (same with xor)? Or how can I check myself is something is faster in NASM? I forgot I can define this variable, so it will be 0 at start and assign other value later. I was always assigning registers "for perfection" starting with eax, so it looks like in most programming languages i.e. <function name>(<args>). I'm not checking for errors, cause writing this code took me bit of time and I wanted to know if I'm on right track. I will add error handling later. Thanks for that answer. – DeBos Oct 31 '19 at 17:05
• By the way, is there any difference if constant is defined in .bss section, not in .data? – DeBos Oct 31 '19 at 17:07
• Can you also tell me why are you comparing [esp] with esi, not just 1? Is it faster when you are using register for that? – DeBos Oct 31 '19 at 17:10
• I did everything you described in answer and comments and code looks much cleaner, but for some reason executable file that is created is 2 times bigger. Is it cause I used some instruction I didn't use before or I did something wrong? – DeBos Oct 31 '19 at 17:30
• @DeBos99 I could not suggest to you a tutorial mainly because I've never used one. When writing code for the x86 architecture, you should download the relevant Intel manuals from their website. You could also peek at the Intel Optimization Manual. And where NASM is concerned, I think the accompanying manual should suffice. Good hunting! – Sep Roland Oct 31 '19 at 17:58


text:
# Enumerate all lists of $m$ non-negative integers that add to $n$ [duplicate]

Let $$x_1,\dots,x_m$$ be non-negative integers such that $$\sum_{i=1}^m x_i=n$$, where $$m,n$$ are given. How can I enumerate all such lists of $$m$$ integers that add to $$n$$?

Note that IntegerPartitions[n,{m}] counts two such lists as one if they are a permutation of each other, but I would count them as distinct.

Thus the list I want must have $$\binom{m+n-1}{m-1}$$ elements.

## marked as duplicate by Mr.Wizard♦Sep 24 '18 at 15:53

• Apply[Sequence]@*Permutations /@ IntegerPartitions[n, {k}] – AccidentalFourierTransform Sep 24 '18 at 15:37
• Try FrobeniusSolve[ConstantArray[1, m], n]. (I am sure this is a dupe.) – J. M. is away Sep 24 '18 at 15:38
• @AccidentalFourierTransform Thanks, that works. I didn't know that Permutations considered repeated elements as identical. Feel free to post an answer. – becko Sep 24 '18 at 15:42


text:
# Sandbox for drafts of long, complex posts

This sandbox is intended for saving drafts of long, complex posts, especially posts whose composition takes a long time. It serves to localize to one thread the front-page "bumps" caused by edits to drafts of such posts, so that they may be easily ignored. Also, it helps to guard against losing longly-composed posts due to system crashes.

When you are happy with your draft here, you may simply copy the code and paste it to the desired location.

## Proper Use of the Sandbox

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## locked by user642796Jan 12 '17 at 6:11

This question’s answers are a collaborative effort. If you see something that can be improved, just edit the answer to improve it! No additional answers can be added here.

• I have added a [sandbox] tag to allow people ignore it more easily (via software support of ignoring tags), and since it seems that we have two sandboxes now, a tag may seem a bit more in place here. – Asaf Karagila Jul 18 '12 at 8:35
• (+1) For thinking outside the (sand)box. – cardinal Jul 18 '12 at 19:40
• At the suggestion of the moderators, I have gone and changed the associated owners of all the answers here to the Community user. This way, the original owners will not receive excess pings for each time another user uses the draft space for their work. Enjoy! – Grace Note Oct 5 '12 at 14:45
• To prevent crashes I've found the "Bookmarks to disable/enable MathJax", provided in here, pretty useful. – leo Dec 17 '12 at 18:03
• PSA: Between the creation of this sandbox (in July 2012) and today (December 2015), technology has advanced. Something like StackEdit (or others, it's simply the only one I know) essentially solves all the limitations of this sandbox. You can have multiple concurrent drafts, you don't have to worry about polluting meta's front page, you can leave your draft untouched for days and expect it to still be there, you don't have to explicitly clear up your draft when you're done... Maybe someday we can get rid of this outdated crutch. – Najib Idrissi Dec 2 '15 at 14:07

$$\begin{array}{r} &&&&&&&U&X\\ &&&&-&-&-&-&- \\ L&M&N&)&R&S&T&U&N \\ &&&&R&T&Y&X \\ &&&&-&-&-&- \\ &&&&&T&Y&Y&N \\ &&&&&T&Y&Y&J \\ &&&&&-&-&-&- \\ &&&&&&&&Y \end{array}$$

Out of the ten letters, $$\{L,M,N,R,S,T,U,X,Y\}$$

• L, R, T, U can't be $$0$$ because they are the first digit of a number.
• X, Y can't be zero or because of the column $$\quad \begin{array}{c}U\\X\\-\\Y\end{array}\quad$$
• Similarly J (and Y) can't be $$0$$.
• N can't be $$0$$ because UN mod $$10$$ equals X and X $$\ne 0$$.
• S can't be $$0$$ because of $$\quad \begin{array}{cC}R&S\\ R&T\\ -&-\\ &T \end{array}$$

Thus $$M = 0$$.

Since the domain of our variables is the set of digits base ten, we will use ABC (for example) to mean $$100A+10B+C$$ and we will use $$A \cdot B \cdot C$$ to mean $$A$$ times $$B$$ times $$C$$.

Then, because $$M=0$$,

• $$U \cdot N = YX$$
• $$X \cdot N = YJ$$
• $$U \cdot L = RT$$
• $$X \cdot L = TY$$

As @RossMillikan said, $$T \in \{1,2,3,4\}$$. Note that, if $$Y \ge 5$$, then $$S$$ has to be odd. And if $$Y \le 4$$, then $$S$$ has to ber even. That implies these possibilities.

$$\begin{array}{c} T & S & Y \\ \hline 1 & 3 & 5 \\ 2 & 4 & 1 \\ 2 & 5 & 6 & *\\ 3 & 6 & 1 & *\\ 3 & 7 & 6 \\ 4 & 8 & 2 \\ 4 & 9 & 7 & * \end{array}$$

Because $$M=0$$, then $$XL = 10T+Y$$ needs to be a product of two digits. So we can remove the starred items.

$$\begin{array}{c} T & S & Y \\ \hline 1 & 3 & 5 \\ 2 & 4 & 1 \\ 3 & 7 & 6 \\ 4 & 8 & 2 \end{array}$$

Next we look at $$X \cdot LMN = TYYJ$$

We see that $$X \ge 2$$, $$X \cdot L = TY$$, and $$X \cdot N = YJ$$

When $$TY=15$$, then $$X \in \{3,5\}$$. But $$X \ne 5$$ since $$Y=5$$, so we must have $$X=3$$. But then $$L=5$$. So $$(T,S,Y) \ne (1,3,5)$$.

With that sort of reasoning, I found the only solution is

$$\begin{array}{c} T & S & Y & X & L & M & N & J \\ \hline 4 & 8 & 2 & 7 & 6 & 0 & 4 & 1 \end{array}$$

Hence the solution is

$$\begin{array}{r} &&&&&&&9&7\\ &&&&-&-&-&-&- \\ 6&0&3&)&5&8&4&9&3 \\ &&&&5&4&2&7 \\ &&&&-&-&-&- \\ &&&&&4&2&2&3 \\ &&&&&4&2&2&1 \\ &&&&&-&-&-&- \\ &&&&&&&&2 \end{array}$$

The integral corresponding to this entry is: $$$$I= \int_{0}^{\infty}\frac{x\: d{x}}{(ax^2+2bx+c)^{n}}$$$$ The value of this integral, as has been mentioned in the tables, is- \begin{align} \int_{0}^{\infty}\frac{x\: d{x}}{(ax^2+2bx+c)^{n}} & = \frac{(-1)^n}{(n-1){!}{!}}\frac{\partial^{n-2}}{\partial c^{n-2}}\Big(\frac{1}{2(ac-b^2)}-\frac{b}{2(ac-b^2)^{\frac{3}{2}}} cot^{-1}(\frac{b}{\sqrt{ac-b^2}})\Big)\, [ac>b^2] \\ & = \frac{(-1)^n}{(n-1){!}{!}}\frac{\partial^{n-2}}{\partial c^{n-2}}\Big(\frac{1}{2(ac-b^2)}+\frac{b}{4(b^2-ac)^{\frac{3}{2}}} \ln\Big(\frac{b+\sqrt{b^2-ac}}{b-\sqrt{b^2-ac}}\Big)\Big)\ [b^2>ac>0] \\ & = \frac{a^{n-2}}{2(n-1)(2n-1)b^{2n-2}}\qquad [ac=b^2] \end{align} For the subsequent discussion, please refer to the rules in this paper.

We use rule $$P_2$$, to expand the denominator as a bracket-series: $$$$(ax^2+2bx+c)^{-n}=\sum_{n_1,n_2,n_3=0}^{\infty}\phi_{n_1n_2n_3}a^{n_1}(2b)^{n_2}c^{n_3}x^{2n_1+n_2}\frac{}{\Gamma(n)}$$$$ Substituting the above expansion in the first expression for I, and employing the definition that $$ \mapsto \int_{0}^{\infty}x^{a-1}dx$$ , we have: $$$$\label{eq:14} I=\sum_{n_1,n_2,n_3=0}^{\infty}\phi_{n_1n_2n_3}a^{n_1}(2b)^{n_2}c^{n_3}\frac{<2n_1+n_2+2>}{\Gamma(n)}$$$$ Following is the system of linear equations we are supposed to solve: \begin{align*} 2n_1 + n_2 + 2 &= 0 \\ n + n_1 + n_2 + n_3 &= 0 \end{align*} There are 3 variables and 2 equations, thus, there are $${{no. of sums}\choose {index}}={3\choose 1}=3$$ different solutions, taking one variable to be free at a time.

Following are the solutions obtained:

• Region $$|\frac{ac}{b^2}|<1$$:$$I_1=\frac{c^{2-n} \Gamma (1-n) \, _2\tilde{F}_1\left(1,\frac{3}{2};3-n;\frac{a c}{b^2}\right)}{4 b^2}$$
• Region $$|\frac{b^2}{ac}|<1$$:$$I_2=\frac{\sqrt{a} c^{1-n} \Gamma (n-1) \, _2F_1\left(1,n-1;\frac{1}{2};\frac{b^2}{a c}\right)-\sqrt{\pi } b c^{\frac{1}{2}-n} \Gamma \left(n-\frac{1}{2}\right) \left(1-\frac{b^2}{a c}\right)^{\frac{1}{2}-n}}{2 a^{3/2} \Gamma (n)}$$
• Region $$|\frac{ac}{b^2}|<1$$:$$I_3=\frac{4^{1-n} a^{n-2} b^{2-2 n} \Gamma (2-n) \Gamma (2 n-2) \left(1-\frac{a c}{b^2}\right)^{\frac{1}{2}-n}}{\Gamma (n)}$$

It could be seen that both $$I_1$$ and $$I_3$$ converge in the same region, so we should add them up, as per rule $$E_3$$: $$I_4=-\frac{\Gamma (1-n) \left(2 a^{n-2} b^{4-2 n} \Gamma \left(n-\frac{1}{2}\right) \left(1-\frac{a c}{b^2}\right)^{\frac{1}{2}-n}-\sqrt{\pi } c^{2-n} \, _2\tilde{F}_1\left(1,\frac{3}{2};3-n;\frac{a c}{b^2}\right)\right)}{4 \sqrt{\pi } b^2}$$

The solution for this region, $$|\frac{ac}{b^2}|<1$$, thus obtained, must correspond to $$I$$, for a suitable choice of the parametric-triad (a,b,c). Also, it's known from the book of Gradshteyn and Ryzhik that for $$(0, the integral could be expressed in the following closed form: $$\int_{0}^{\infty}\frac{x\: d{x}}{(ax^2+2bx+c)^{n}} = \frac{(-1)^n}{(n-1){!}{!}}\frac{\partial^{n-2}}{\partial c^{n-2}}\Big(\frac{1}{2(ac-b^2)}+\frac{b}{4(b^2-ac)^{\frac{3}{2}}} \ln\Big(\frac{b+\sqrt{b^2-ac}}{b-\sqrt{b^2-ac}}\Big)\Big)\: [b^2>ac>0]$$

However, a derivation of this closed-form starting with the solution obtained using the rules of the Method of Brackets: $$I_4$$, turns out to be too non-trivial a task to accomplish, so a numerical consistency check is called for. However, this method has its own set of nasty complications.

It is known that the gamma function has poles of order 1 at all non-positive integers. Owing to this, $$I_1$$ (and so, $$I_4$$) fails to provide a suitable expression, $$\forall$$ n $$\in$$ $$\mathbb{N}$$.

Also, the hypergeometric $$_2{F}_1$$ happens to have the following standard expansion: \begin{align} _2{F}_1(a,b;c;z) &= \sum_{k=0}^{\infty}\frac{(a)_k(b)_k}{(c)_kk{!}}z^k \\ &= \sum_{k=0}^{\infty}\frac{\Gamma(a+k)\Gamma(b+k)\Gamma(c)}{\Gamma(a)\Gamma(b)\Gamma(c+k)\Gamma(k+1)}z^n \end{align} which is convergent in $$|z|<1$$, as long as Re(c$$-$$a$$-$$b)$$>$$0.

Here, a$$=$$1, b$$=\frac{3}{2}$$, c$$=$$ 3$$-$$n, and z $$=$$ $$\frac{ac}{b^2}$$. Now, $$\Gamma$$(3$$-$$n) is a simple pole $$\forall$$ n $$\in$$ $$\mathbb{N}\setminus\{1,2\}$$. Thus, a numerical check for $$I_1$$ turns out to be impossible.

For $$|\frac{ac}{b^2}|<1$$, the other contributing term is $$I_3$$. For reasons explained before, this expression suffers from an indeterminate form at n$$=$$0, and happens to have simple poles $$\forall$$ n $$\in$$ $$\mathbb{N}$$. To elucidate this point, note that the factor $$\Gamma$$(2-n) has a simple pole $$\forall$$ n $$\in$$ $$\mathbb{N}\setminus\{1\}$$. However, $$\Gamma$$(2n-2) has a simple pole at n$$=$$1, thus, there's a simple pole for $$I_3$$ $$\forall$$ n $$\in$$ $$\mathbb{N}$$.

A numerical consistency check for relevant expressions, and the integral itself, seems a hopeless endeavor for $$|\frac{ac}{b^2}|<1$$. So, let's proceed to the region $$|\frac{b^2}{ac}|<1$$, and see for ourselves what fate awaits us.

$$I_2$$ is a sum of two terms, the first of which happens to have a simple pole at n$$=$$1, and overall has an indeterminate form at n$$=$$0.

Let's show explicitly that the first term of $$I_2$$ indeed has a simple pole at n$$=$$1. Ignoring the constant factors, we can write: \begin{align} \frac{\Gamma(n-1)\; _2{F}_1(1,n-1;\frac{1}{2};\frac{b^2}{ac})}{\Gamma(n)} &= \sum_{k=0}^{\infty}\frac{\Gamma(k+1)\Gamma(k+n-1)\Gamma(\frac{1}{2})}{\Gamma(1)\Gamma(n-1)\Gamma(k+\frac{1}{2})\Gamma(k+1)}\Big(\frac{b^2}{ac}\Big)^k \\ &= \sqrt{\pi}\Big(\frac{1}{\Gamma(n)}\sum_{k=0}^{\infty}\frac{\Gamma(k+n-1)}{\Gamma(k+\frac{1}{2})}\Big(\frac{b^2}{ac}\Big)^k\Big)_{n=1} \\ &= \sqrt{\pi}\sum_{k=0}^{\infty}\frac{\Gamma(k)}{\Gamma(k+\frac{1}{2})}\Big(\frac{b^2}{ac}\Big)^k \end{align} which blows up at k=0.

For sake of numerical consistency check, we set n$$=$$5, a$$=$$3, b$$=$$2, c$$=$$2. The differences between the numerical values obtained via employing different expressions is of the minuscule order of $$10^{-16}$$.

Now, let's look at the case $$b^2=ac$$. Since we see that for n$$=$$5, $$I_2$$ yields a nice solution as long as $$|\frac{b^2}{ac}|<1$$, we can take a dig at $$|\frac{b^2}{ac}|=1$$, using Gauss's hypergeometric theorem, which states the following: $$_2{F}_1(a,b;c;1)=\frac{(c-b)_{-a}}{(c)_{-a}}=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}\qquad [Re(c-a-b)>0]$$ Hence, we may write the following:

\begin{align} _2{F}_1(1,n-1;\frac{1}{2};1) &= \frac{\Gamma(\frac{1}{2})\Gamma(\frac{1}{2}-n)}{\Gamma(\frac{1}{2}-1)\Gamma(\frac{1}{2}-n+1)} \\ &= \frac{\Gamma(\frac{1}{2})\Gamma(\frac{1}{2}-n)}{\Gamma(-\frac{1}{2})\Gamma(\frac{1}{2}-n+1)} \\ &= \frac{\Gamma(-\frac{1}{2}+1)\Gamma(\frac{1}{2}-n)}{\Gamma(-\frac{1}{2})(\frac{1}{2}-n)\Gamma(\frac{1}{2}-n)} \\ &= \frac{-\frac{1}{2}\Gamma(-\frac{1}{2})}{(\frac{1}{2}-n)\Gamma(-\frac{1}{2})} \\ &= \frac{1}{2n-1} \end{align} For $$b^2=ac$$, the second term in $$I_2$$ blows up, while the first one evaluates to a finite value, for given values of a, c, and n. The blowing up of the second term is actually due to one of the roots of b in $$b^2=ac$$, namely, $$b=-\sqrt{ac}$$.

That $$b=\sqrt{ac}$$ is indeed responsible for the finite part, and that $$b=-\sqrt{ac}$$ is responsible for the divergence, could be ascertained as follows:

• $$b=\sqrt{ac}$$:$$\int_{0}^{\infty}\frac{x\: d{x}}{(ax^2+2bx+c)^{n}}=\int_{0}^{\infty}\frac{x\: d{x}}{(ax^2+2\sqrt{ac}x+c)^{n}}=\int_{0}^{\infty}\frac{x\: d{x}}{(\sqrt{a}x+\sqrt{c})^{2n}}\qquad$$ which yields a finite result for given values of a, c, and n. It is seen that this result is in exact numerical agreement with that of the first term in $$I_2$$, for any given set of parameters (a,c,n).
• $$b=-\sqrt{ac}$$:$$\int_{0}^{\infty}\frac{x\: d{x}}{(ax^2+2bx+c)^{n}}=\int_{0}^{\infty}\frac{x\: d{x}}{(ax^2-2\sqrt{ac}x+c)^{n}}=\int_{0}^{\infty}\frac{x\: d{x}}{(\sqrt{a}x-\sqrt{c})^{2n}}\qquad$$ which rapidly diverges for any given set of parameters (a,c,n).

Now that I am done with all the details of the calculation (please, feel free to ask me to show some step explicitly if the corresponding calculation seems blurry), let me put forth my queries:

• In the book by Gradshteyn and Ryzhik (7th ed., pg. 1005, section 9.10), it is said that for non-troublesome values of the triad of parameters $$(\alpha,\beta,\gamma)$$, a hypergeometric series $$F(\alpha,\beta;\gamma;z)$$ diverges on the entire unit circle, if $$Re(\alpha+\beta-\gamma) \geq 1$$, and converges throughout the entire unit circle (except at $$z=1$$), for $$1>Re(\alpha+\beta-\gamma) \geq 0$$.

Now looking at $$I_1$$, it could be seen that $$Re(\alpha+\beta-\gamma)=n-\frac{1}{2}$$, which is $$<1$$ for $$n=1$$, but $$>1$$ $$\forall$$ n $$\in$$ $$\mathbb{N}\setminus\{1\}$$.

Putting aside the calculations for awhile, let's now take a look at the following table: $$\begin{array}{c|c|c|c|c|c|c|c|} \text{(a,b,c)} & \text{n} & \text{Int} & \text{RHS} & \text{MoB} & \text{|Int-RHS|} & \text{|Int-MoB|} & \text{|RHS-MoB|} \\ \hline \text{(1,2,1)} & 5 & 0.0045 & 0.0045,0.0045 & 0.0045 & 0.,0. & O(10^{-18}) & O(10^{-18}),O(10^{-18}) \\ \hline \text{(3,7,5)} & 5 & 2.87*10^{-6} & 2.87*10^{-6},2.87*10^{-6} & 2.87*10^{-6} & O(10^{-22}),O(10^{-21}) & O(10^{-21}) & O(10^{-21}),O(10^{-20}) \\ \hline \text{(3,1,5)} & 5 & 0.000031 & 0.000031,0.000031 & 0.000031 & O(10^{-21}),O(10^{-20}) & O(10^{-21}) & O(10^{-20}),O(10^{-20}) \end{array}$$

(this question is not complete yet---I am currently editing it, as it is too long to write at one go)

My post needs more work, and I don't have the time for it today. Leaving this to be used by anyone.

This entry is available for anyone to commandeer since Oct. 17th, 2018.

This space free for whoever wants it

This answer is available for anyone to use.

This slot is open for anyone to use.

My first ansatz, which gives at best an asymptotic power series (with convergence radius zero!), seemed to be rough and likely much suboptimal. But for completeness I'll give the method here, also because it provides that nice initial guess $$a_\text{init}$$ which is very near at the true value.

The ansatz employs essentially the Bernoulli-numbers, but not as in the classic Bernoulli-polynomials but in some "transposed" manner.

The individual $$\log(1+1/a_k)$$ - epressions have a mercator series, such that we can reformulate the original sum $$S_d(a,N)$$ as a double sum $$\begin{array}{rll} S_d(a,N) &=& 1/a - 1/2/a^2 + 1/3/ a^3 - 1/4/ a^4 + \cdots - \cdots \\ &+& 1/(a+d) - 1/2/(a+d)^2 + 1/3/ (a+d)^3 - 1/4/ (a+d)^4 + \cdots - \cdots \\ &+& 1/(a+2d) - 1/2/(a+2d)^2 + 1/3/ (a+2d)^3 - 1/4/ (a+2d)^4 + \cdots - \cdots \\ &...& \\ &+& 1/(a+(N-1)d) - 1/2/(a+(N-1)d)^2 + 1/3/ (a+(N-1)d)^3 - 1/4/ (a+(N-1)d)^4 + \cdots - \cdots \\ \end{array}$$ Summing this vertically gives the column-sums $$\begin{array}{rll} C_{d,1}(a,N) &=& 1/a + 1/(a+d)+ 1/(a+2d) + ... + 1/(a+(N-1)d) \\ C_{d,2}(a,N) &=&\frac12 ( 1/a^2 + 1/(a+d)^2+ 1/(a+2d)^2 + ... + 1/(a+(N-1)d)^2 )\\ C_{d,3}(a,N) &=&\frac13 ( 1/a^3 + 1/(a+d)^3+ 1/(a+2d)^3 + ... + 1/(a+(N-1)d)^3 )\\ &...& \end{array}$$ Such sums are essentially Hurwitz-Zeta sums, and can be expressed as power-series with zeta at negative integers as coefficients.
For easiness of handling and memorizing I've the "ZETA"-matrix in my toolbox, which provides that coefficients for the application to power series.
The ZETA-matrix looks like

and has the composition of Zeta-values
Reading it horizontally one observes the coefficients for the integrals of the Bernoulli-polynomials which are configured to allow the summing-of-like-powers.
However, reading vertically, they give the coeffients for power series to determine the sums-of-reciprocals-of-like-powers. However, that power series have convergence-radius zero, but because of alternating signs we can use them as asymptotic power series, truncated at some meaningful index.

For instance, the power series taken from the second,third, fourth and so on column, and calculated as $$f_1(a) = a z_{0,1} + a^2 z_{1,1} + a^3 z_{2,1} + ... \to \zeta(2,a) \\ f_2(a) = a z_{0,2} + a^2 z_{1,2} + a^3 z_{2,2} + ... \to \zeta(3,a) \\ ...$$ where $$z_{r,c}$$ indicates the element of the ZETA-matrix from row $$r$$ and column $$c$$ (index based at $$0$$).
That allows immediately to determine the sums of reciprocals from $$a$$ to $$a+(N-1)$$ by $$s_1(a) = f_1(a) - f_1(a+N) = 1/a^2 + 1/(a+1)^2+... + 1/(a+(N-1))^2 \\ s_2(a) = f_2(a) - f_2(a+N) = 1/a^3 + 1/(a+1)^3+... + 1/(a+(N-1))^3 \\ ...$$

For the first column we need a correction factor which comes out to be $$f_0(a) = a z_{0,0} + a^2 z_{1,0} + a^3 z_{2,0} + ... \to \\ \log(1/a)+\left(\sum_{k=1}^{a-1} \frac1k \right) -\gamma$$
such that $$s_0(a) = f_0(a) - f_0(a+N) = 1/a + 1/(a+1)+... + 1/(a+(N-1)) \\ + \log(1/a) - \log(1/(a+(N-1))$$

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Consider the following initial value problem (IVP) to the first order ODE:

$$\tag{1} \dot x = f(t, x), \ \ \ x(t_0) = x_0.$$

The Existence and Uniqueness Theorem describes when one has exactly one solutions. This is true (e.g.) when $$f$$ is continuously differentiable.

There are examples where uniqueness fails and existence fails (here or here):

In all of the examples where I am aware of, whenever one has more than one solutions, one actually has infinitely many. Hence my question:

Can an IVP has more than one solutions, but only finitely many solutions?

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The claim is made that $$\frac NS \ge \frac {\ln 2}{ \ln3}$$ which also means $$\text{conj: } \liminf _{k \to \infty} N^*(k) \ln3 - k \ln2 \ge 0$$ This would be satisfied if in our construction of infinitely many partial trajectories on $$\{x_1,x_2,...,x_{N_x},y_1,...\}$$ only finitely many exceptions for the inequality $$N_x \ln3 - S_x \ln2 \ge 0 \qquad \qquad \qquad \text{or} \qquad \qquad \qquad \qquad \qquad \qquad \\ {2^{S_x} \over 3^{N_x}} \le 1 \qquad \text{ for some partial trajectory x_1,...,x_{N_x} }$$ Now we rewrite our formula to adapt $${2^S\over 3^N} = \frac{a_1}{b_1} \cdot (1+{1\over 3a_1 })(1+{1\over 3a_2 })\cdots(1+{1\over 3a_N}) \le 1$$ That means also $$a_1 \cdot (3a_1+1)(3a_2+1)\cdots(3a_N +1) \le b_1 \cdot 3a_1 \cdot 3a_2 \cdots 3a_N \\ 3^N a_1 a_2 \cdots a_N + ... + (3a_1+3a_2+\cdots +3a_N) + N \le b_1/a_1 \cdot 3^N a_1 a_2 \cdots a_N$$

Let's shorten the formula by the first parenthese on $$a_1$$: $$2^{S-1} = 3^{N-1} \cdot \frac{1}{b_1} \cdot ({3a_1+1\over 2 })(1+{1\over 3a_2 })\cdots(1+{1\over 3a_N})$$

A well known tool developed for the discussion of cycles is useful also here: the productformula of all elements of a (partial) trajectory equalling the product of its iterates. Of course we have equality $$a_2 a_3 \cdots a_N a_{N+1} = {3a_1+1\over 2^{A_1}}{3a_2+1\over 2^{A_N}}\cdots{3a_N+1\over 2^{A_N}}$$ Writing $$S$$ for the sum of all exponents $$A_k$$, rearranging and cancelling $$a_2,...a_{N+1}$$ gives $$2^S = ({3a_1+1\over 1 })(3+{1\over a_2 })\cdots(3+{1\over a_N}) \cdot \frac1{a_{N+1}}$$ A smoothing occurs if we extract $$a_1$$ from the first parenthese and rewrite $$2^S = \frac{a_1}{a_{N+1}} \cdot (3+{1\over a_1 })(3+{1\over a_2 })\cdots(3+{1\over a_N})$$ We finally extract the $$N$$ fold occuring $$3$$ in the parentheses and also introduce that $$a_{N+1} = b_1$$ by our definitions to write the basic formula $$2^S = 3^N \cdot \frac{a_1}{b_1} \cdot (1+{1\over 3a_1 })(1+{1\over 3a_2 })\cdots(1+{1\over 3a_N})$$ From our construction of the sequence of partial trajectories we have the following properties: $$a_2= {3a_1+1\over2 } \qquad \text{ because a_1 \to a_2 must be increasing } \\ a_1< b_1 < a_2 = {3a_1+1\over2 } \qquad \text{ and even } \\ a_1< b_1 < \min(a_2,a_3,...,a_N)\qquad \text{ by construction of partial trajectories }$$

In the following I use the "syracuse" (or "compacted") notation for the Collatz-transformation $$Y:= a_{k+1} = {3a_k+1\over 2^{A_k}}$$ with $$a_k$$ odd and the exponent $$A_k\ge1$$ such that also $$a_{k+1}$$ is odd
I use $$N$$ for the number of subsequent iterations and $$S(N)$$ or simply $$S$$ for the sum of exponents of the subsequent iterations and always capital letters for exponents and small letters for the elements (also using indexes) or simple temporary variables.

Let's define an assumed divergent trajectory from its minimal element $$a_1$$ on $$\{a_1,a_2,...,a_{N_a}, b_1,b_2,...,b_{N_b},c_1,... \}$$ where we assume $$a_1$$ the minimal element of the whole trajectory, $$b_1$$ the minimal element of the remaining trajectory after $$a_1$$, then $$c_1$$ the minimal element of the remaining trajectory after $$b_1$$ and so on. We construct with this a monotonic increasing sequence of elements, separating the full trajectory into infinitely many partial trajectories. There cannot be equal elements because then we would have a cycle whose trajectory obviously cannot diverge infinitely and which is thus not in the scope of this discussion.

We have always $$a_1 and $$b_1 and so on except if $$b_1=a_2$$ but then we would reduce the first partial trajectory to the element $$a_1$$ only and $$N_a=1$$ and use then $$a_2$$ only to keep index-notations in formulae consistent/easy.

Function $$f$$ is defined as $$f(x)=\frac{1}{(x-1)^3(x^3+2x^2+x+1)}\tag 1$$ Note that $$x^3+2x^2+x+1=(x+1)(x^2+x+1)\tag 2$$ and so $$f(x)=\frac{1}{(x-1)^3(x+1)(x^2+x+1)}\tag 3$$ and also $$f(x)=\frac{1}{(x-1)}\cdot\frac{1}{(x^2-1)}\cdot\frac{1}{(x^3-1)}=\sum_{i=0}^\infty x^i \cdot\sum_{k=0}^\infty x^{2k} \cdot\sum_{j=0}^\infty x^{3j}$$

$$\frac{1}{(x-1)^3(x+1)(x^2+x+1}=\frac{A_1}{(x-1)}+\frac{A_2}{(x-1)^1}+\frac{A_3}{(x-1)^3}+\frac{B}{(x+1)}+\frac{Cx+D}{(x^2+x+1)}$$ $$1=A_1(x-1)^2(x+1)(x^2+x+1)+A_2(x-1)(x+1)(x^2+x+1)+A_3(x+1)(x^2+x+1)+B(x-1)^3(x^2+x+1)+(Cx+D)(x-1)^3(x+1)$$

• Oh god, I still remember having tests that required computing these king of generating functions by hand. If result was correct, the method was verified, otherwise no points. – enedil Feb 11 at 21:53

Edit: This answer got more popular than I expected, so let me add some details about what the structure here actually looks like.

Before talking about manifolds, which we'll get to in a bit, let's discuss some algebra. If you're not comfortable with algebra you can skip down to where I start talking about topology again. There are lots of examples in algebra of times where we are talking about the same underlying sets, but considering different structures, and the best way to think about those structures is through their morphisms.

For example, if we have the category of $$R$$-modules, over some ring $$R$$, we are considering the (homo)morphisms which preserve the structure of being a module over $$R$$. What does that mean, precisely? Given a pair of modules $$M,M' \in \textbf{R-Mod}$$ we say that a map $$f:M \to M'$$ is a homomorphism iff $$f$$ is a homomorphism of abelian groups and $$f$$ preserves the $$R$$ action (i.e. $$r \cdot f(a) = f(r\cdot a)$$).

We have a whole rich set of other possible morphisms we could care about in this context, though, depending on what we are doing. The classical example is that if there is some subring $$S \subset R$$, then we can consider morphisms that only preserve the action by $$S$$. This actually gives us a new category, $$S-\textbf{Mod}$$, the category of $$S$$ modules. Another example is that we could just drop the $$R$$ action all together and just think about homomorphisms of abelian groups. We could also take the approach where we don't care about the group structure at all and instead think only about $$f$$ preserving the group action by $$R$$. These all lead to different and important areas of math.

My point in this diversion is simply to say that the richness of mathematical objects actually doesn't come from their inherit structure, like their topology or their geometry, instead it comes from studying those structures paired with the morphisms that preserve certain aspects of them. It is from this pairing that we start to see the bigger picture about what is and isn't true for a whole class of objects, and even how those classes relate to each other. So that is the approach I'll take in describing some more detail.

Alright, lets get to manifolds now. Before introducing anything about Riemannian or Smooth manifolds let me start out by contrasting a topological space and a topological manifold. The reason I want to briefly focus on this contrast is that the morphisms of a topological space behaves a lot like those of other objects you learn about in math, but the morphisms of a topological manifold, although preserving a very similar structure, behave much more like those of other kinds of manifolds than anything.

A pair of topological spaces $$X, Y$$ with topology $$\mathscr{T,T'}$$ has a morphism $$f: X \to Y$$, called a continuous map, iff $$\forall A \in \mathscr{T'}, f^{-1}(A) \in \mathscr{T}$$. This is an isomorphism, called a homeomorphism, iff $$\forall A \in \mathscr{T}, f(A) \in \mathscr{T'}$$. This is what we see all over math, so it should look familiar.

A topological manifold is a topological space $$X$$ with the local property of being locally homeomorphic to $$\mathbb R^n$$. Here we have introduced two things that will make all manifolds behave "differently" than topological spaces: a local property, and a relationship to $$\mathbb R^n$$.

It's been a long time since I last did this stuff, but nobody better seems to have turned up, so I'll have a go.

I'll give the box $$\square \phi$$ its usual meaning here: "when I translate $$\phi$$ into the formal system using Gödel numbers, then there is a collection of mechanical transformations inside the formal system that together form an internal-system-proof of $$\phi$$'s quotation". I will also refer to the formal system in question in the first person, with the pronoun "I".

Then the formula $$\square (\square \phi \to \phi) \to \square \phi$$ should be read as:

If there is a number which encodes a proof that "having a number which encodes a proof of $$\phi$$ means I can actually prove $$\phi$$", then there is a number which encodes a proof of $$\phi$$.

Or, a bit more loosely,

If I, internally, can prove that my proof of $$\phi$$ respects the logic in which I operate (so my proof of $$\phi$$ means that in my surrounding logic, $$\phi$$ is actually true), then in fact there is a proof of $$\phi$$ that I will be satisfied with.

This is more concretely understood through its universal quantification, Löb's theorem (which is true in any system rich enough to support Peano arithmetic):

If I can prove within myself that my own proof-checking process respects the semantics of the logic in which I operate, then the logic realises that actually I can satisfy myself of any statement at all.

Specifically, $$\square (\square \phi \to \phi)$$ means "I contain a proof that my proofs are semantically valid in the surrounding logic", while $$\square \square \phi \to \square \phi$$ means "it's true in the surrounding logic that if I want to prove $$\phi$$, it's enough to prove provability-of-$$\phi$$".

The two statements were very unlikely to be the same: one is talking about proof within the system, and one is talking about facts of the surrounding logic.


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# Experiences with OpenFOAM for water turbine applications Paper in proceedings, 2007

OpenFOAM has been used successfully at Chalmers for water turbine applications since the beginning of 2005. OpenFOAM has been validated for the flow in a Kaplan water turbine runner and draft tube, and for the swirling unsteady flow in a combustor. The flow in the combustor resembles the flow in a water turbine draft tube (which is a diffuser), with its adverse pressure gradient and unsteady flow features. The results compare well with the experimental results. For the flow in the draft tube the OpenFOAM results are almost identical to those of the CFX-5 CFD code. For the flow in the combustor the OpenFOAM results show the same trend as the results using the Fluent CFD code. OpenFOAM gives similar results as the CALC-PMB in-house CFD code that was developed specifically for water turbine applications (however, no comparisons are shown here). Many interesting flow features in water turbines are unsteady. There is an interaction between steady and rotating parts of the machine, which should be properly resolved. There are also flow instabilities like the break-up of the vortex after the runner, yielding a highly unsteady flow (here referred to as the vortex rope). A CFD code used for water turbine applications should thus be able to include rotor-stator interaction and allow unsteady flow features to appear without numerical damping. OpenFOAM is not yet fully developed for full rotor-stator interaction in water turbines. In the present work a rotating inlet boundary condition is shown, which makes it possible to include guide vane wakes or non-axisymmetry from the spiral casing from a previous simulation of the distributor. The present work evaluates a newly developed filtering technique for the $k-\omega$ SST turbulence model. It is shown that the filter is necessary to get unsteady and accurate time averaged results of the flow in the combustor. The Reynolds numbers in water turbines are high, the geometries are complicated, and the computational grids are usually coarse and skew. It is thus necessary to use models and methods that can give accurate results also under non-ideal conditions.

## Author

### Håkan Nilsson

Chalmers, Applied Mechanics, Fluid Dynamics

### Subject Categories

Fluid Mechanics and Acoustics


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# How do you use the distributive property to factor 21b-15a?

Mar 25, 2017

See the solution process below:

#### Explanation:

First, factor a $\textcolor{red}{3}$ from each term and rewrite the expression as:

$\left(\textcolor{red}{3} \cdot 7 b\right) - \left(\textcolor{red}{3} \cdot 5 a\right)$

Now, factor the $\textcolor{red}{3}$ out of both terms:

$\textcolor{red}{3} \left(7 b - 5 a\right)$


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# Math Help - Ellipse

1. ## Ellipse

[LEFT]Determine the length of the ray focal of the point $M(3,\frac{7}{4})$who is on the ellipse $7x^2+16y^2=112$

Answer:
$\frac{25}{4}$ and $\frac{7}{4}$

2. Originally Posted by Apprentice123
[LEFT]Determine the length of the ray focal of the point $M(3,\frac{7}{4})$who is on the ellipse $7x^2+16y^2=112$

Answer:
$\frac{25}{4}$ and $\frac{7}{4}$
That ellipse can also be written (divide both sides by 112) $\frac{x^2}{112}+ \frac{y^2}{7}= \frac{x^2}{(4\sqrt{7})^2}+ \frac{y^2}{(\sqrt{7})^2}= 1$.

That has semi-axes of length $\sqrt{7}$ and $4\sqrt{7}$. Can you calculate the coordinates of the foci from that? Then it should be easy to find the distance from that point to each point.


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# Why do irrelevant operators require infinitely many counterterms?

As far as I understand it, in the Wilsonian picture of renormalization, we view a theory as having some fixed cutoff and bare couplings, and integrate out high-momentum modes to understand what happens at low momentum. We say that an operator is relevant if its coupling constant grows when we go to low momentum scales, and irrelevant if it shrinks.

Now, in the "usual" picture of renormalization, we have a QFT which we want to define as a continuum limit of a theory with a cutoff, i.e. the limit where the cutoff goes to infinity. We want to take this limit holding the physical coupling constants at some energy scale to be fixed; to do this, we add cutoff-dependent counterterms to the bare Lagrangian. We say that an interaction is renormalizable if we only need to add a finite number of counterterms, and non-renormalizable if we need an infinite number of counterterms.

However, I don't understand how these two pictures fit together. In particular, it is usually stated that irrelevant operators are non-renormalizable, and relevant operators are renormalizable, but this doesn't seem obvious to me. Can someone explain why this is true?

-


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• ### Detection of thermal neutrons with the PRISMA-YBJ array in Extensive Air Showers selected by the ARGO-YBJ experiment(1512.01326)

We report on a measurement of thermal neutrons, generated by the hadronic component of extensive air showers (EAS), by means of a small array of EN-detectors developed for the PRISMA project (PRImary Spectrum Measurement Array), novel devices based on a compound alloy of ZnS(Ag) and $^{6}$LiF. This array has been operated within the ARGO-YBJ experiment at the high altitude Cosmic Ray Observatory in Yangbajing (Tibet, 4300 m a.s.l.). Due to the tight correlation between the air shower hadrons and thermal neutrons, this technique can be envisaged as a simple way to estimate the number of high energy hadrons in EAS. Coincident events generated by primary cosmic rays of energies greater than 100 TeV have been selected and analyzed. The EN-detectors have been used to record simultaneously thermal neutrons and the air shower electromagnetic component. The density distributions of both components and the total number of thermal neutrons have been measured. The correlation of these data with the measurements carried out by ARGO-YBJ confirms the excellent performance of the EN-detector.
• ### The transverse momentum dependence of charged kaon Bose-Einstein correlations in the SELEX experiment(1501.04316)

Nov. 14, 2015 hep-ex, nucl-ex
We report on the measurement of the one-dimensional charged kaon correlation functions using 600~GeV/{\it c} $\Sigma^-$, $\pi^-$ and 540~GeV/{\it c} $p$ beams from the SELEX~(E781) experiment at the Fermilab Tevatron. $K^{\pm}K^{\pm}$ correlation functions are studied for three transverse pair momentum, $k_T$, ranges and parameterized by a Gaussian form. The emission source radii, $R$, and the correlation strength, $\lambda$, are extracted. The analysis shows a decrease of the source radii with increasing kaon transverse pair momentum for all beam types.