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# American Institute of Mathematical Sciences

August  2018, 38(8): 4133-4162. doi: 10.3934/dcds.2018180

## Global regularity for the 2D micropolar equations with fractional dissipation

 1 College of Mathematics and Statistics, Shenzhen University, Shenzhen 518060, China 2 School of Mathematical Science, Anhui University, Hefei 230601, China 3 Department of Mathematics, Oklahoma State University, 401 Mathematical Sciences, Stillwater, OK 74078, USA 4 School of Mathematical Sciences, Beijing Normal University, Laboratory of Mathematics and Complex Systems, Ministry of Education, Beijing 100875, China 5 Department of Mathematics and Statistics, Jiangsu Normal University, 101 Shanghai Road, Xuzhou 221116, Jiangsu, China

* Corresponding author: Zhuan Ye

Received  December 2017 Revised  March 2018 Published  May 2018

Fund Project: Dong was partially supported by the National Natural Science Foundation of China (NO. 11571240; No.11271019) and Research Fund of Shenzhen University (No. 2017056). Wu was partially supported by NSF grants DMS 1209153 and DMS 1624146, by the AT & T Foundation at Oklahoma State University (OSU) and by NSFC (No.11471103, a grant awarded to Professor Baoquan Yuan). Xu was partially supported by the National Natural Science Foundation of China (No. 11371059; No. 11471220; No. 11771045). Ye was supported by the Foundation of Jiangsu Normal University (No. 16XLR029), the Natural Science Foundation of Jiangsu Province (No. BK20170224) and the National Natural Science Foundation of China (No. 11701232)

Micropolar equations, modeling micropolar fluid flows, consist of coupled equations obeyed by the evolution of the velocity $u$ and that of the microrotation $w$. This paper focuses on the two-dimensional micropolar equations with the fractional dissipation $(-Δ)^{α} u$ and $(-Δ)^{β}w$, where $0<α, β<1$. The goal here is the global regularity of the fractional micropolar equations with minimal fractional dissipation. Recent efforts have resolved the two borderline cases $α = 1$, $β = 0$ and $α = 0$, $β = 1$. However, the situation for the general critical case $α+β = 1$ with $0<α<1$ is far more complex and the global regularity appears to be out of reach. When the dissipation is split among the equations, the dissipation is no longer as efficient as in the borderline cases and different ranges of $α$ and $β$ require different estimates and tools. We aim at the subcritical case $\alpha+\beta>1$ and divide $\alpha\in (0,1)$ into five sub-intervals to seek the best estimates so that we can impose the minimal requirements on $\alpha$ and $\beta$. The proof of the global regularity relies on the introduction of combined quantities, sharp lower bounds for the fractional dissipation and delicate upper bounds for the nonlinearity and associated commutators.

Citation: Bo-Qing Dong, Jiahong Wu, Xiaojing Xu, Zhuan Ye. Global regularity for the 2D micropolar equations with fractional dissipation. Discrete & Continuous Dynamical Systems - A, 2018, 38 (8) : 4133-4162. doi: 10.3934/dcds.2018180
##### References:
 [1] H. Bahouri, J.-Y. Chemin and R. Danchin, Fourier Analysis and Nonlinear Partial Differential Equations, Springer, 2011. doi: 10.1007/978-3-642-16830-7. [2] J. Bergh and J. Löfström, Interpolation Spaces, An Introduction, Springer-Verlag, Berlin-Heidelberg-New York, 1976. [3] D. Chamorro and P. G. Lemarié-Rieusset, Quasi-geostrophic equation, nonlinear Bernstein inequalities and α-stable processes, Rev. Mat. Iberoam., 28 (2012), 1109-1122. doi: 10.4171/RMI/705. [4] Q. Chen and C. Miao, Global well-posedness for the micropolar fluid system in critical Besov spaces, J. Differential Equations, 252 (2012), 2698-2724. doi: 10.1016/j.jde.2011.09.035. [5] P. Constantin and C. Foias, Navier-Stokes Equations, Chicago Lectures in Mathematics, University of Chicago Press, Chicago, IL, 1988. [6] A. Córdoba and D. Córdoba, A maximum principle applied to quasi-geostrophic equations, Commun. Math. Phys., 249 (2004), 511-528. doi: 10.1007/s00220-004-1055-1. [7] S. C. Cowin, Polar fluids, Phys. Fluids, 11 (1968), 1919-1927. doi: 10.1063/1.1692219. [8] C. Doering and J. Gibbon, Applied Analysis of the Navier-Stokes Equations, Cambridge Texts in Applied Mathematics, Cambridge University Press, Cambridge, 1995. doi: 10.1017/CBO9780511608803. [9] B. Dong and Z. Chen, Asymptotic profiles of solutions to the 2D viscous incompressible micropolar fluid flows, Discrete Cont. Dyn. Sys., 23 (2009), 765-784. doi: 10.3934/dcds.2009.23.765. [10] B. Dong, J. Li and J. Wu, Global well-posedness and large-time decay for the 2D micropolar equations, J. Differential Equations, 262 (2017), 3488-3523. doi: 10.1016/j.jde.2016.11.029. [11] B. Dong and Z. Zhang, Global regularity of the 2D micropolar fluid flows with zero angular viscosity, J. Differential Equations, 249 (2010), 200-213. doi: 10.1016/j.jde.2010.03.016. [12] M. E. Erdogan, Polar effects in the apparent viscosity of suspension, Rheol. Acta, 9 (1970), 434-438. doi: 10.1007/BF01975413. [13] A. C. Eringen, Theory of micropolar fluids, J. Math. Mech., 16 (1966), 1-18. [14] A. C. Eringen, Micropolar fluids with stretch, Int. J. Engng. Eci., 7 (1969), 115-127. doi: 10.1016/0020-7225(69)90026-3. [15] G. Galdi and S. Rionero, A note on the existence and uniqueness of solutions of micropolar fluid equations, Int. J. Engrg. Sci., 15 (1977), 105-108. doi: 10.1016/0020-7225(77)90025-8. [16] L. Grafakos, Loukas Modern Fourier Analysis, Third edition. Graduate Texts in Mathematics, 250. Springer, New York, 2014. doi: 10.1007/978-1-4939-1230-8. [17] Q. Jiu, C. Miao, J. Wu and Z. Zhang, The two-dimensional incompressible Boussinesq equations with general critical dissipation, SIAM J. Math. Anal., 46 (2014), 3426-3454. doi: 10.1137/140958256. [18] T. Kato and G. Ponce, Commutator estimates and the Euler and the Navier-Stokes equations, Comm. Pure Appl. Math., 41 (1988), 891-907. doi: 10.1002/cpa.3160410704. [19] C. E. Kenig, G. Ponce and L. Vega, Well-posedness of the initial value problem for the Korteweg-de Vries equation, J. American Mathematical Society, 4 (1991), 323-347. doi: 10.1090/S0894-0347-1991-1086966-0. [20] C. E. Kenig, G. Ponce and L. Vega, Well-posedness and scattering results for the generalized Korteweg-de Vries equation via the contraction principle, Comm. Pure Appl. Math., 46 (1993), 527-620. doi: 10.1002/cpa.3160460405. [21] P. G. Lemarie-Rieusset, Recent Developments in the Navier-Stokes Problem, Chapman & Hall/CRC Research Notes in Mathematics Series, CRC Press, 2002. doi: 10.1201/9781420035674. [22] G. Lukaszewicz, Micropolar Fluids. Theory and Applications, Modeling and Simulation in Science, Engineering and Technology, Birkhäuser, Boston, 1999. doi: 10.1007/978-1-4612-0641-5. [23] A. J. Majda and A. L. Bertozzi, Vorticity and Incompressible Flow, Cambridge University Press, Cambridge, 2002. [24] C. Miao, J. Wu and Z. Zhang, Littlewood-Paley Theory and its Applications in Partial Differential Equations of Fluid Dynamics, Science Press, Beijing, China, 2012 (in Chinese). [25] S. Popel, A. Regirer and P. Usick, A continuum model of blood flow, Biorheology, 11 (1974), 427-437. [26] T. Runst and W. Sickel, Sobolev Spaces of Fractional Order, Nemytskij operators and Nonlinear Partial Differential Equations, Walter de Gruyter, Berlin, New York, 1996. doi: 10.1515/9783110812411. [27] V. K. Stokes, Theories of Fluids with Microstructure, Springer, New York, 1984. doi: 10.1007/978-3-642-82351-0. [28] R. Temam, Navier-Stokes Equations: Theory and Numerical Analysis, AMS Chelsea Publishing, Providence, RI, 2001. doi: 10.1090/chel/343. [29] H. Triebel, Theory of Function Spaces II, Birkhauser Verlag, 1992. doi: 10.1007/978-3-0346-0419-2. [30] L. Xue, Well posedness and zero microrotation viscosity limit of the 2D micropolar fluid equations, Math. Methods Appl. Sci., 34 (2011), 1760-1777. doi: 10.1002/mma.1491. [31] K. Yamazaki, Global regularity of the two-dimensional magneto-micropolar fluid system with zero angular viscosity, Discrete Contin. Dyn. Syst., 35 (2015), 2193-2207. doi: 10.3934/dcds.2015.35.2193. [32] Z. Ye and X. Xu, Global well-posedness of the 2D Boussinesq equations with fractional Laplacian dissipation, J. Differential Equations, 260 (2016), 6716-6744. doi: 10.1016/j.jde.2016.01.014. [33] B. Yuan, Regularity of weak solutions to magneto-micropolar fluid equations, Acta Math. Sci., 30B (2010), 1469-1480. doi: 10.1016/S0252-9602(10)60139-7.

show all references

##### References:
 [1] H. Bahouri, J.-Y. Chemin and R. Danchin, Fourier Analysis and Nonlinear Partial Differential Equations, Springer, 2011. doi: 10.1007/978-3-642-16830-7. [2] J. Bergh and J. Löfström, Interpolation Spaces, An Introduction, Springer-Verlag, Berlin-Heidelberg-New York, 1976. [3] D. Chamorro and P. G. Lemarié-Rieusset, Quasi-geostrophic equation, nonlinear Bernstein inequalities and α-stable processes, Rev. Mat. Iberoam., 28 (2012), 1109-1122. doi: 10.4171/RMI/705. [4] Q. Chen and C. Miao, Global well-posedness for the micropolar fluid system in critical Besov spaces, J. Differential Equations, 252 (2012), 2698-2724. doi: 10.1016/j.jde.2011.09.035. [5] P. Constantin and C. Foias, Navier-Stokes Equations, Chicago Lectures in Mathematics, University of Chicago Press, Chicago, IL, 1988. [6] A. Córdoba and D. Córdoba, A maximum principle applied to quasi-geostrophic equations, Commun. Math. Phys., 249 (2004), 511-528. doi: 10.1007/s00220-004-1055-1. [7] S. C. Cowin, Polar fluids, Phys. Fluids, 11 (1968), 1919-1927. doi: 10.1063/1.1692219. [8] C. Doering and J. Gibbon, Applied Analysis of the Navier-Stokes Equations, Cambridge Texts in Applied Mathematics, Cambridge University Press, Cambridge, 1995. doi: 10.1017/CBO9780511608803. [9] B. Dong and Z. Chen, Asymptotic profiles of solutions to the 2D viscous incompressible micropolar fluid flows, Discrete Cont. Dyn. Sys., 23 (2009), 765-784. doi: 10.3934/dcds.2009.23.765. [10] B. Dong, J. Li and J. Wu, Global well-posedness and large-time decay for the 2D micropolar equations, J. Differential Equations, 262 (2017), 3488-3523. doi: 10.1016/j.jde.2016.11.029. [11] B. Dong and Z. Zhang, Global regularity of the 2D micropolar fluid flows with zero angular viscosity, J. Differential Equations, 249 (2010), 200-213. doi: 10.1016/j.jde.2010.03.016. [12] M. E. Erdogan, Polar effects in the apparent viscosity of suspension, Rheol. Acta, 9 (1970), 434-438. doi: 10.1007/BF01975413. [13] A. C. Eringen, Theory of micropolar fluids, J. Math. Mech., 16 (1966), 1-18. [14] A. C. Eringen, Micropolar fluids with stretch, Int. J. Engng. Eci., 7 (1969), 115-127. doi: 10.1016/0020-7225(69)90026-3. [15] G. Galdi and S. Rionero, A note on the existence and uniqueness of solutions of micropolar fluid equations, Int. J. Engrg. Sci., 15 (1977), 105-108. doi: 10.1016/0020-7225(77)90025-8. [16] L. Grafakos, Loukas Modern Fourier Analysis, Third edition. Graduate Texts in Mathematics, 250. Springer, New York, 2014. doi: 10.1007/978-1-4939-1230-8. [17] Q. Jiu, C. Miao, J. Wu and Z. Zhang, The two-dimensional incompressible Boussinesq equations with general critical dissipation, SIAM J. Math. Anal., 46 (2014), 3426-3454. doi: 10.1137/140958256. [18] T. Kato and G. Ponce, Commutator estimates and the Euler and the Navier-Stokes equations, Comm. Pure Appl. Math., 41 (1988), 891-907. doi: 10.1002/cpa.3160410704. [19] C. E. Kenig, G. Ponce and L. Vega, Well-posedness of the initial value problem for the Korteweg-de Vries equation, J. American Mathematical Society, 4 (1991), 323-347. doi: 10.1090/S0894-0347-1991-1086966-0. [20] C. E. Kenig, G. Ponce and L. Vega, Well-posedness and scattering results for the generalized Korteweg-de Vries equation via the contraction principle, Comm. Pure Appl. Math., 46 (1993), 527-620. doi: 10.1002/cpa.3160460405. [21] P. G. Lemarie-Rieusset, Recent Developments in the Navier-Stokes Problem, Chapman & Hall/CRC Research Notes in Mathematics Series, CRC Press, 2002. doi: 10.1201/9781420035674. [22] G. Lukaszewicz, Micropolar Fluids. Theory and Applications, Modeling and Simulation in Science, Engineering and Technology, Birkhäuser, Boston, 1999. doi: 10.1007/978-1-4612-0641-5. [23] A. J. Majda and A. L. Bertozzi, Vorticity and Incompressible Flow, Cambridge University Press, Cambridge, 2002. [24] C. Miao, J. Wu and Z. Zhang, Littlewood-Paley Theory and its Applications in Partial Differential Equations of Fluid Dynamics, Science Press, Beijing, China, 2012 (in Chinese). [25] S. Popel, A. Regirer and P. Usick, A continuum model of blood flow, Biorheology, 11 (1974), 427-437. [26] T. Runst and W. Sickel, Sobolev Spaces of Fractional Order, Nemytskij operators and Nonlinear Partial Differential Equations, Walter de Gruyter, Berlin, New York, 1996. doi: 10.1515/9783110812411. [27] V. K. Stokes, Theories of Fluids with Microstructure, Springer, New York, 1984. doi: 10.1007/978-3-642-82351-0. [28] R. Temam, Navier-Stokes Equations: Theory and Numerical Analysis, AMS Chelsea Publishing, Providence, RI, 2001. doi: 10.1090/chel/343. [29] H. Triebel, Theory of Function Spaces II, Birkhauser Verlag, 1992. doi: 10.1007/978-3-0346-0419-2. [30] L. Xue, Well posedness and zero microrotation viscosity limit of the 2D micropolar fluid equations, Math. Methods Appl. Sci., 34 (2011), 1760-1777. doi: 10.1002/mma.1491. [31] K. Yamazaki, Global regularity of the two-dimensional magneto-micropolar fluid system with zero angular viscosity, Discrete Contin. Dyn. Syst., 35 (2015), 2193-2207. doi: 10.3934/dcds.2015.35.2193. [32] Z. Ye and X. Xu, Global well-posedness of the 2D Boussinesq equations with fractional Laplacian dissipation, J. Differential Equations, 260 (2016), 6716-6744. doi: 10.1016/j.jde.2016.01.014. [33] B. Yuan, Regularity of weak solutions to magneto-micropolar fluid equations, Acta Math. Sci., 30B (2010), 1469-1480. doi: 10.1016/S0252-9602(10)60139-7.
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Types of Nuclear Reactions • Alpha Particles are emitted from alpha radiation from the nucleus of an atom . cm cm mμ 0 2 4 6 8 10 12 14 0 20 40 60 80 100 120 140 160 180 proton range in air R air [cm] proton energy [MeV] It contain 2 protons and 2 neutron which is lightly bound together. Proton Emission. Sometimes specific job postings will arrive here; like this one from Proton Canada: Proton Technologies Canada Inc. has current openings, which are funded in part through a federal program which has criteria for successful applicants: An example of positron emission (β + decay) is shown with magnesium-23 decaying into sodium-23: 23 12 Mg → 23 11 Na + e + + ν e. Because positron emission decreases proton number relative to neutron number, positron decay happens typically in large "proton-rich" radionuclides. For a proton to escape a nucleus, the proton separation energy must be negative – the proton is therefore unbound, and tunnels out of the nucleus in a finite time. • Proton radioactivity. Because it's hard for a positively-charged proton to enter a positively-charged nucleus, proton capture only occurs at high temperature. There is an issue between Cloudflare's cache and your origin web server. In 2005 it was experimentally determined (at the same facility) that … An example of a nuclear equation showing positron emission is shown below: $_{6}^{11}C \rightarrow +_{+1}^{0}\beta +_{5}^{11}B$ Boron has one fewer protons in its nucleus than carbon, but the mass is unchanged because the proton has been replaced by a neutron. Example: Proton and Neutron Decay. Be sure to also look at the images related to proton capture. A proton emission follows a positron emission (beta^+ decay).Proton emission is very rare. 3, is not appropriate as they depend on the number of data points (samples) ﬁtted, and thus on the particle energy and the emission … As a result, the web page can not be displayed. • Beta emission gives off beta particles , ß-is ejected from the nucleus or a radionuclide at high speed. Nitrogen-11 also undergoes proton emission: Carbon-9 has an interesting decay scheme. Proton is always accepting resumes to careers@proton.energy for jobs all over the planet. Source: JANIS (Java-based Nuclear Data Information Software); The JEFF-3.1.1 Nuclear Data Library. Please include the Ray ID (which is at the bottom of this error page). Particle-induced X-ray emission (PIXE) is a technique that induces characteristic XRF by bombarding the surface of the sample with photons or helium ions. Your IP: 103.120.177.86 In addition to the above types of radioactivity, there is a special class of rare beta-decay processes that gives rise to heavy-particle emission. 11.10.3 Example of Two-Proton Emission Decay An example of two-proton emission decay is the decay of iron-45 (a highly proton rich radionuclide with Z. This rare type of emission occurs when a proton is converted to a neutron and a positron in the nucleus, with ejection of the positron. To help support the investigation, you can pull the corresponding error log from your web server and submit it our support team. Isotopes that are "above" the band of stability have a neutron to proton ratio that is too large (too many neutrons). In the Sun (as well as other stars) there is a process called the CNO cycle. 1F. During proton emission, a proton is ejected from an atom's nucleus. II. Somewhere around 25 isotopes are known to engage in proton emission. For the case in study and using the binary cascade model, it was found that … Performance & security by Cloudflare. A radionuclide (radioactive nuclide, radioisotope or radioactive isotope) is an atom that has excess nuclear energy, making it unstable. High as in the temperatures in the Sun. Proton emission When an proton is emitted from a nucleus the decay is called proton emission or proton reactivity. Cloudflare Ray ID: 616149b7baa5dd97 There is an unknown connection issue between Cloudflare and the origin web server. Proton Emission. To help support the investigation, you can pull the corresponding error log from your web server and submit it our support team. Example #1: 7 24 N ---> 7 23 N + 0 1 n. The nitrogen-23 that is produced is, itself, unstable and … II. • The present work focuses on the prompt-gamma emission yields by comparing experimental data with the outcomes of the current version of Geant4 using all applicable proton inelastic models. Example #1: A positron decay equation: 12 23 Mg. For example, Thailand’s Honda City with a 3-cylinder 1.0-litre turbocharged engine weighs between 1,150 – 1,165 kg. Therefore these isotopes are candidates for beta - decay as this converts a neutron to a proton. Proton is a Calgary-based energy company that’s bringing hydrogen into the clean-energy discussion after years of having this abundant resource largely ignored. Proton decay is a rare type of radioactive decay of nuclei containing excess protons, in which a … Proton emission can occur from high-lying excited states in a nucleus following a beta decay, in which case the process is known as beta-delayed proton emission, or can occur from the ground state (or a low-lying isomer) of very proton-rich nuclei, in which case the process is very similar to alpha decay. Somewhere around 25 isotopes are known to engage in proton emission. 1) Something inside the nucleus of an atom breaks down, which causes a proton to become a neutron. For example, a car can get its electric charge via a battery. Proton emission: | | ||| | The decay of a proton rich nucleus A populat... World Heritage Encyclopedia, the aggregation of the largest online encyclopedias available, and the … Neutron Emission. From the Wiki article: From the Wiki article, 69-Tm-147 and 71-Lu-151 also decay by proton emission. An example of the simultaneous emission of two protons in a direct two-proton emission according to the sequence illustrated in the right-hand part of Fig. = Example of Two-Proton Emission Decay An example of two-proton emission decay is the decay of iron-45 (a highly proton rich radionuclide with Z = Show abstract. Example #1: 7 10 N ---> 6 9 C + 1 1 p. Nitrogen-11 also undergoes proton emission: 7 11 N ---> 6 10 C + 1 1 p but the quality of the proton-7Li discrimination was not docu-mented; Fig. The study of proton emission has aided the understanding of nuclear deformation, masses and structure, and it is a wonderfully pure example of quantum tunneling. Nuclei below the belt of stability (low neutron-to-proton ratios): These proton-rich nuclei can increase their ratio by either positron emission or electron capture. Proton decay is also a quantum tunneling process. The decay of a free proton has been predicted, but never observed. 61.6% of the time it decays by positron, then another proton emission to produce Be-8 (which then decays) and 38.4% of the time, it decays by positron, then alpha to Li-5, which then decays by proton emission. 3) The atomic number goes DOWN by one and mass number remains unchanged. The atomic number will decrease by one while the atomic weight does not change. Proton emission is not confined to just the lighter elements. Monte Carlo tools have been long used to assist the research and development of solutions for proton therapy monitoring. 20.18 is taken from the work of Mukha and Schneider (2001). As a result of its low detection limits (between 1 and 100 ppm) for samples weights of a few milligrams, and the higher sensitivity obtained compared to XRF, PIXE is used to detect trace elements as well as major and minor elements. 2) It emits a positron and a neutrino which go zooming off into space. The half-life of this proposed decay is at least 1034 years. It's rather short. Example: Proton and Neutron Decay Source: JANIS (Java-based Nuclear Data Information Software); The JEFF-3.1.1 Nuclear Data Library. How hydrogen could shake up Canada's energy sector There is an issue between Cloudflare's cache and your origin web server. Below is an example of beta + decay which is almost always called "positron emission" Changes within the nucleus in alpha, beta and gamma emission - definition Alpha emission: If the nucleus of a radioactive element X of mass number A and atomic number Z emits an α particle, a new element Y (daughter nucleus) is formed which has mass number equal to (A-4) and atomic number equal to (Z-2). Beta emission decreases the number of neutrons and increases the number of protons in a nucleus, as shown in Equation 21.3. Proton emission is the radioactive decay mode that is expected to determine the limit of observable proton-rich nuclei for most elements. The energy spectrum averaged over 166 shots, as well as the statistical uncertainty, is shown in Fig. Or, you could characterize the decay as the He-2 simply splitting apart (called spontaneous fission). 2 of Ref. Certain decay branches appear to proceed mainly directly to continuum states. Since an atom loses a proton during proton emission, it changes from one element to another. This chain of reactions involves several proton captures and is one of the sources of energy that the Sun (and other stars) generates. For example, after undergoing proton emission, an atom of nitrogen (with 7 protons) becomes an atom of carbon (with 6 protons). Proton emission is one process that unstable atoms can use to become more stable. As an example the beta-delayed proton emission from $$^{11}$$Be is discussed in detail. Is Ballard Power stock the next Tesla? Example: range R 208 of 3 MeV protons in Lead (A=208, r=11.35 g/cm3) R R 208 208 ≈⋅ ⋅ ⋅ ≈⋅ = − − 32 10 208 1135 13 529 10 53 4 3.. []. Or it can also get its electric charge using hydrogen and oxygen (fuel cells). There appear to be around 75 isotopes that decay by neutron emission. Using the absolute values of ˜2, as in their Fig. Positron Emission A positron is a positive electron (a form of antimatter). In order to be emitted, the proton must penetrate a potential barrier. 1G, we report the absolute ... all of which are powered by older generation engines that meet only up to Euro 4 emission standards. If you pursue more on this topic, be sure to look up the names of Hans Bethe and Fred Hoyle. [1] shows only two example events of unknown energy. Additional troubleshooting resources. Proton Emission is rare. In 2002, the simultaneous emission of two protons was observed from the nucleus iron-45 in experiments at GSI and GANIL (Grand Accélérateur National d'Ions Lourds at Caen). Proton Emission is rare. Cloudflare monitors for these errors and automatically investigates the cause. Some nuclei decay via double proton emission, such as 45 Fe. For example ""_11^7N rarr p^+ + ""_10^6C The p+ is indicating a free proton The carbon-10 afterwards gets converted to boron-10 as a result of positron emission ""_10^6C rarr … The proton diagnostics in the LD-EDX setup is a time-of-flight (ToF) spectrometer (see Materials and Methods) (2829, ), which is aligned with the sample slit along the target normal direction. Please include the Ray ID (which is at the bottom of this error page). Proton Decay – Proton Emission. In positron emission, a proton inside the radioactive nucleus is converted into a neutron while releasing a positron; in electron capture, a proton-rich nucleus of a neutral atom absorbs an inner shell electron which then converts a proton into a neutron, emitting an electron neutrino. In Fig. Neutron decay is a type of radioactive decay of nuclei containing excess neutrons (especially fission products), in which a neutron is simply ejected from the nucleus. Proton radioactivity, discovered in 1970, is exhibited by an excited isomeric state of cobalt-53, 53m Co, 1.5 percent of which emits protons: Special beta-decay processes. This is considerably longer than the current estimates for the age of the universe (a bit less than 1.4 x 1010 years). In almost every case, these isotopes are products of fission and have extremely short half-lives. View. One of the least expensive methods for doing so, for example, using methane, has drawn scrutiny for its carbon emissions. Cloudflare monitors for these errors and automatically investigates the cause. If you want to delve more into this topic, I recommend you start with the Wikipedia entry on proton decay. The Wiki article also identifies two isotopes that decay by (simultaneous) emission of two protons: 30  54 Zn  --->    28  52 Ni  +  2 1 1 p. Here's the Wikipedia article on proton capture. 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# Comparing “size” of open sets in a topological space?

In metric space, we have the notion of distance, which enables us to have a general idea on the "size" of an open set, say, by measuring its diameter. If we now have 2 open sets $A,B$ in a topological space, we tends have the impression that $A$ is "smaller than" $B$ if $A\subset B$. But what if we are given 2 disjoint open set? Is there a general way for us to compare "size" of open set in a topological space? (Say, $A$ is smaller than $B$, or $A$ and $B$ have the similar size) How can we get the impression of the size of an open set in a topological space? Thanks.

P.s. Cardinality does not count as a solution because it does not rely on the topology anyway.

• Not in general. – Brian M. Scott May 3 '16 at 14:47
• You have a poset, but nothing more. – Michael Burr May 3 '16 at 14:49
• What about Borel measures? – Augustin May 3 '16 at 14:51
• @Augustin : What makes you think every topological space admits a (nontrivial) measure on its Borel algebra? Random example: Say a space has no notrivial compact sets, then (strong) inner regularity would imply all sets have measure zero, so the only Radon measure on such a space is the trivial one, which would contradict strict positivity (measure of a nonempty open set is positive). There are a number of properties one would like a measure to have, but the topology of the underlying space can make such properties fail. See math.stackexchange.com/questions/43601 for more. – Eric Towers May 3 '16 at 15:07

The comments and answer here seem to be a little depressing - topology can still measure interesting properties of open sets! Michael Burr is probably right in that you can't hope for anything more than a poset, but let's at least construct an interesting one!

Let's try the following: $$\text{Define }\;A \le B\text{ if there is an embedding }\;A \stackrel{f}\hookrightarrow B.$$

We can check that this gives us a partial ordering on topological spaces, hence also topological subspaces. In addition, if $A \subset B$ are subspaces of $X$ then $A \le B$. (Note that if $A \le B$ it may not be the case that $A \subset B$.)

This is a purely topological characterization, and it turns out to be interesting, because we can use topological tools to study it!

One immediate question that you might have is, "Can you produce two spaces $X, Y$ such that $X \lneq Y$?" It turns out this is pretty easy: take $X = \mathbb R^2$, and take $Y = \mathbb{RP}^2$. Or if we want spaces that are incomparable elements consider $\mathbb{RP}^2$ and $S^2$. All these results rely on interesting algebraic topology.

So it seems this is an interesting way to characterize "size" in terms of topological complexity.

A few fun exercises for interested readers, in roughly increasing order of difficulty:

1. Given a space $X$, what subspace(s) is/are minimal in this order? What set is a maximal element?
2. Construct a space $X$ and subsets $A_i \subset X$ such that $$A_1 \lneq A_2 \lneq \cdots \lneq X.$$
3. Give a version of the previous construction where $X$ and the $A_i$ are all connected.
4. Prove that $\mathbb{RP}^2$ and $S^2$ are incomparable elements.
5. Prove that $\mathbb{R}^2 \lneq \mathbb{RP}^2$.
6. Construct a topological space $X$ and subspaces $\{A_r\}_{r \in \mathbb{R}_{\ge 0}}$ with $|X| = \mathfrak{c}$ (the cardinality of the continuum) and $A_r \lneq A_{r'}$ for all $r < r'$. (I'm not 100% sure my answer here is correct, or that this is possible.)
• I enjoyed your answer. – Michael Burr May 3 '16 at 15:53
• This has the negative property that any two open intervals in $\Bbb{R}$, $A, B$ satisfy both $A \leq B$ and $B \leq A$, so in the simplest setting these do not capture "size". The same thing happens for analytic disks in $\Bbb{C}$ (i.e., any two non-empty simply connected open subsets are the same "size"). Note also that the question asks for two open set in the same ambient space (which you have weakly ignored)... – Eric Towers May 4 '16 at 16:40
• Of course it has this drawback. But if you have disjoint open sets (as in the OP's question) then the only topological tools you have to study them come from maps between them, and that's what my answer addresses. I can only try to provide a positive answer to the question that's been asked :). – Thurmond May 4 '16 at 17:04

You cannot. One of the consequences of "rubber sheet geometry" is that distance, hence area, is a fluid thing.

M.C. Escher represented hyperbolic geometry many times in his Circle Limit works, for instance Circle Limit IV. The referenced image is a representation of an infinite metric space. Each demon is congruent (same area) and each angel is congruent in the original space. However, topology allows us to (continuously) magnify and demagnify various parts of a space with impunity. Consequently, Escher's work is topologically equivalent to the original tiling of a hyperbolic space. If we were to imagine translating the space (or what is equivalent, translating the point at the center of our magnifying window into the space), the largest figures would shrink as they moved to the side and smaller figures would grow as they moved to the center of the view.

To borrow words of Yoda, in topology "size matters not."


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# Print the last, middle and first character of your code

The challenge is simple: Print the last, middle, and first character of your program's source code, in that order.

The middle character is defined as follows, assuming a source length of n characters, and 1-indexing:

• If n is even, print the n/2-th and n/2 + 1-th character. (abcdef == cd)
• If n is odd, print (n-1)/2 + 1-th character. (abcde == c)

### Rules

• Given no input, print the last, middle, and first character in your source code, in the form [last][middle][first]. This will be 3-4 characters long.
• Output must not contain any trailing whitespace. However, if whitespace is a first, middle, or last character, it must be printed as such.
• Source code must be n >= 3 characters long.
• Code must consist of >= 3 unique characters.
• Standard loopholes are forbidden.
• This is , so shortest solution in characters wins.

### Samples

# Form: [code] --> [output]
xyz --> zyx
abcd --> dbca
1 --> # not allowed: too short
abcde --> eca
aaabb --> # not allowed: not enough unique characters
System.out.print("S;pr"); --> ;prS
this is a test --> ts t
123[newline]45 --> 53[newline]1

Challenge Proposal

Here is a Stack Snippet to generate both a regular leaderboard and an overview of winners by language.

# Language Name, N bytes

where N is the size of your submission. If you improve your score, you can keep old scores in the headline, by striking them through. For instance:

# Ruby, <s>104</s> <s>101</s> 96 bytes

If there you want to include multiple numbers in your header (e.g. because your score is the sum of two files or you want to list interpreter flag penalties separately), make sure that the actual score is the last number in the header:

# Perl, 43 + 2 (-p flag) = 45 bytes

You can also make the language name a link which will then show up in the leaderboard snippet:

# [><>](http://esolangs.org/wiki/Fish), 121 bytes

var QUESTION_ID=188005;
var OVERRIDE_USER=78850;
var ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe",COMMENT_FILTER="!)Q2B_A2kjfAiU78X(md6BoYk",answers=[],answers_hash,answer_ids,answer_page=1,more_answers=!0,comment_page;function answersUrl(d){return"https://api.stackexchange.com/2.2/questions/"+QUESTION_ID+"/answers?page="+d+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}function commentUrl(d,e){return"https://api.stackexchange.com/2.2/answers/"+e.join(";")+"/comments?page="+d+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+COMMENT_FILTER}function getAnswers(){jQuery.ajax({url:answersUrl(answer_page++),method:"get",dataType:"jsonp",crossDomain:!0,success:function(d){answers.push.apply(answers,d.items),answers_hash=[],answer_ids=[],d.items.forEach(function(e){e.comments=[];var f=+e.share_link.match(/\d+/);answer_ids.push(f),answers_hash[f]=e}),d.has_more||(more_answers=!1),comment_page=1,getComments()}})}function getComments(){jQuery.ajax({url:commentUrl(comment_page++,answer_ids),method:"get",dataType:"jsonp",crossDomain:!0,success:function(d){d.items.forEach(function(e){e.owner.user_id===OVERRIDE_USER&&answers_hash[e.post_id].comments.push(e)}),d.has_more?getComments():more_answers?getAnswers():process()}})}getAnswers();var SCORE_REG=function(){var d=String.rawh\d,e=String.raw\-?\d+\.?\d*,f=String.raw[^\n<>]*,g=String.raw<s>${f}</s>|${f}|${f},h=String.raw[^\n\d<>]*,j=String.raw<[^\n<>]+>;return new RegExp(String.raw<${d}>+String.raw\s*([^\n,]*[^\s,]),.*?+String.raw(${e})+String.raw(?=+String.raw${h}+String.raw(?:(?:${g}|${j})${h})*+String.raw</${d}>+String.raw))}(),OVERRIDE_REG=/^Override\s*header:\s*/i;function getAuthorName(d){return d.owner.display_name}function process(){var d=[];answers.forEach(function(n){var o=n.body;n.comments.forEach(function(q){OVERRIDE_REG.test(q.body)&&(o="
"+q.body.replace(OVERRIDE_REG,"")+"
")});var p=o.match(SCORE_REG);p&&d.push({user:getAuthorName(n),size:+p[2],language:p[1],link:n.share_link})}),d.sort(function(n,o){var p=n.size,q=o.size;return p-q});var e={},f=1,g=null,h=1;d.forEach(function(n){n.size!=g&&(h=f),g=n.size,++f;var o=jQuery("#answer-template").html();o=o.replace("{{PLACE}}",h+".").replace("{{NAME}}",n.user).replace("{{LANGUAGE}}",n.language).replace("{{SIZE}}",n.size).replace("{{LINK}}",n.link),o=jQuery(o),jQuery("#answers").append(o);var p=n.language;p=jQuery(""+n.language+"").text().toLowerCase(),e[p]=e[p]||{lang:n.language,user:n.user,size:n.size,link:n.link,uniq:p}});var j=[];for(var k in e)e.hasOwnProperty(k)&&j.push(e[k]);j.sort(function(n,o){return n.uniq>o.uniq?1:n.uniq<o.uniq?-1:0});for(var l=0;l<j.length;++l){var m=jQuery("#language-template").html(),k=j[l];m=m.replace("{{LANGUAGE}}",k.lang).replace("{{NAME}}",k.user).replace("{{SIZE}}",k.size).replace("{{LINK}}",k.link),m=jQuery(m),jQuery("#languages").append(m)}}
body{text-align:left!important}#answer-list{padding:10px;float:left}#language-list{padding:10px;float:left}table thead{font-weight:700}table td{padding:5px}
  <link rel="stylesheet" type="text/css" href="https://cdn.sstatic.net/Sites/codegolf/primary.css?v=f52df912b654">

Winners by Language

Language	User	Score

Leaderboard

	Author	Language	Size

{{PLACE}}	{{NAME}}	{{LANGUAGE}}	{{SIZE}}

{{LANGUAGE}}	{{NAME}}	{{SIZE}}

• Maybe it's only me, but "Code must consist of >= 3 unique characters." suggests all characters of the program should be unique, while you only require to have at least 3 distinct characters. – Grzegorz Oledzki Jul 9 '19 at 19:22
• @Belhenix Fewer than three unique characters allows solutions as simple as 121 for a great many languages. – Khuldraeseth na'Barya Jul 9 '19 at 19:32
• If anyone can find a stack-based language that uses - for negation and implicitly prints with a linefeed, \n1- is a three-byter. My search has so far been fruitless. – Khuldraeseth na'Barya Jul 9 '19 at 19:35
• @Belhenix Whitespace is certainly possible, it has enough unique valid characters (space, linefeed, tab). – bigyihsuan Jul 9 '19 at 19:46
• Can't change it now with 58 answers, but requiring "first letter of code must be different from last letter of code" would have ruled out a lot of trivial answers, including the current top answer. That may have been the intent of ">= 3 unique characters", but that requirement isn't actually very hard. – manassehkatz-Moving 2 Codidact Jul 10 '19 at 14:10

# JavaScript, 9 7 bytes

N=>a=-N

Outputs: NaN

Try it online!

• Very clever. This was the tact I was trying to take, but just couldn't stumble upon a language with curt enough error output. – ouflak Jul 10 '19 at 6:33
• This is one of the best Javascript answers I've ever seen here – Daniel Jul 10 '19 at 7:42

# Grass, 17 bytes

wWWwwwVVwWWwWWWww

Try it online!

-2 bytes thanks to jimmy23013.

Outputs www. Grass ignores all characters apart from w, W and v. The two Vs are thus ignored; they are there to ensure the middle character is a w – I have no idea how to output v or W in Grass… I could have used another character instead of V, but V preserves the aesthetic of the code.

• "Luckily, the middle character ended up being a w and not a v or a W – I have no idea how to output v or W in Grass…" Lol. xD I tried to do something similar in Whitespace having also only three characters (space, tab, newline), but unfortunately I wasn't so lucky.. Luckily I can simply add a bunch of trailing no-ops to fix it, though. – Kevin Cruijssen Jul 9 '19 at 14:56
• @KevinCruijssen If the middle character had been anything else, I would probably have added in the middle a bunch of xs (or any other non-Grass character, which Grass ignores) to move a w into the correct place. – Robin Ryder Jul 9 '19 at 16:51
• Ah, I'm an idiot.. I can insert a non-whitespace character as well which Whitespace will ignore. ;) -2 bytes thanks to you. – Kevin Cruijssen Jul 9 '19 at 17:03
• 17 bytes: wWWwww  wWWwWWWww – jimmy23013 Jul 10 '19 at 7:05
• @jimmy23013 Thanks! And thanks for posting your tip; I needed it to understand what you did! I changed the spaces to Vs because I find it prettier. ;-) – Robin Ryder Jul 10 '19 at 8:29

# brainfuck, 17 bytes

U-[>+<-UU--]>...U

Try it online!

With unprintables this can be reduced to 7 bytes: ␀..␀_.␀

• Within the rules, but BF doesn't use U as an instruction. Functionally,the code is -[>+<---]>., so the expected output would be .--, which it does not output. – bigyihsuan Jul 9 '19 at 15:20
• @bigyihsuan but the source code still contains the Us and this interpreter happily ignores them. Should every answer which has comments or multiple spaces (or anything else that doesn't necessarily matter to the language) be not counted? – dzaima Jul 9 '19 at 15:25
• Shouldn't it be U-[>+<-UU--]>...U? In its current form the code only shows one U – Helena Jul 9 '19 at 17:32
• @Flater Yes, here they're used to make the challenge easier, but isn't that the whole purpose of the code-golf tag? Many answers here alone make their job easier by adding leading/trailing whitespace, and in general in code-golf here anything following the rules is allowed (and if I had to guess, if there was a rule disallowing unused characters (defining which objectively would already be pretty much impossible) would mean that the challenge might be still hovering at 0 score, and probably be closed as well). – dzaima Jul 11 '19 at 14:05
• Why don't you simply use ÿ-.ÿ..ÿ? That's printable and short. – Dorian Jul 25 '19 at 8:09

# Shakespeare Programming Language, 119 bytes

 ,.Ajax,.Page,.Act I:.Scene I:.[Enter Ajax and Page]
Ajax:  You big big big big big cat.Speak thy.Speak thy.Speak thy.

Try it online!

Prints three spaces. Without printing spaces:

# Shakespeare Programming Language, 227 bytes

,.Ajax,.Page,.Act I:.Scene I:.[Enter Ajax and Page]Ajax:You is the sum of a big big big big big big cat a big pig.
You is the sum of you a large huge large big pig.
Speak thy.Speak thy.You is the sum of you a big pig.Speak thy.

Try it online!

Like my newline-free INTERCAL answer, this is constructed to make the last character of the program the same as the middle character (although in this case it is not also the first), by un-golfing the second half of the code until they match. (I'm not sure if I could have constructed 46 in fewer bytes or not, or if I could have golfed other parts of the first half more.)

• I wasn't expecting Shakespeare to show up for this. This is great – bigyihsuan Jul 10 '19 at 2:34
• Wish I could +1 just for learning this language existed! – Stilez Jul 10 '19 at 22:15
• even in the online interpreter's debug output messages, the authors' sense of humor shows: Warning at line 2: 'mind' expected – dlatikay Jul 12 '19 at 12:52
• 163 bytes for the second version by ending with a ! – Robin Ryder Jul 17 '19 at 10:34
• @UnrelatedString OK, thanks! Done. – Robin Ryder Jul 17 '19 at 17:36

# Python 2, 13 11 bytes

print' t'

To make the whitespace clear:

$xxd solution.py 0000000: 0a70 7269 6e74 2720 7427 20 .print' t'$ python2 solution.py | xxd
0000000: 2074 0a                                t.

Try it online!

-2 bytes, thanks to wastl !

• Both output trailing whitespace (newline), which is not allowed. Python 2 alternative: Try it online! – wastl Jul 9 '19 at 13:59
• Also, I don't think this is valid: Python requires you to have an empty newline at the end of your code. – Geza Kerecsenyi Jul 9 '19 at 15:59
• @GezaKerecsenyi As I understand it, a newline at the end of the source code is just a recommendation. See PEP8. – mypetlion Jul 9 '19 at 16:48
• @GezaKerecsenyi Just a guideline – mbomb007 Jul 9 '19 at 19:01
• @wastl Thanks! Fixed. – TFeld Jul 9 '19 at 19:15

# RProgN 2, 3 bytes

1
0

Try it online!

I think this is valid?

• I'm guessing 1 and 0 are pushed onto a stack, then an implicit pop-print at the end of the program? Yes it's valid – bigyihsuan Jul 9 '19 at 15:21
• @bigyihsuan that's what I'm guessing too; I don't know the language, I just recall it does stuff like this :p – dzaima Jul 9 '19 at 15:22
• @dzaima Haha, when you post an answer and you don't even know how it works. xD I've indeed seen RProgN a few times before with old [quine] challenges, before the quine-rules got changed. – Kevin Cruijssen Jul 9 '19 at 15:48
• I believe this is polyglot in tac on Linux. – tsh Jul 10 '19 at 6:38

# 05AB1E, 7 5 bytes

„ RR

Outputs R \n.

-2 bytes by taking inspiration from @Neil's Batch answer, so make sure to upvote him as well!

Try it online.

5 bytes alternative by @Grimy:

12,1

Outputs 12\n.

Try it online.

'?„«'«?

Outputs ?«'.

Try it online.

Explanation:

          # No-op newline
„ R       # Push 2-char string " R"
R      # Reverse this string to "R "
# (output the result implicitly with trailing newline)

# No-op newline
12,       # Print 12 with trailing newline
1      # Push a 1 (no implicit printing, since we already did an explicit print)

'?       '# Push 1-char string "?"
„«'    '# Push 2-char string "«'"
«    # Concatenate them both together
?   # Output without trailing newline

# PHP, 9 bytes

<?=";;<";

Try it online!

• Actually, since PHP outputs code as text until the first <? tag, you could use one of the trivial 4-byte solutions like 1231 - anything without a special character as first and last, any other 2 characters in the middle. Your answer may be the shortest real code PHP answer. – manassehkatz-Moving 2 Codidact Jul 10 '19 at 14:13

# Jelly, 5 bytes

001 0

Try it online!

The output has two distinct characters, but the code also has a third one.

Trivial 4-byte version:

1231

• Isn't 121 also a valid trivial answer? – Teleporting Goat Jul 10 '19 at 9:51
• @TeleportingGoat: No, because it has only two distinct characters – Oliver Hankeln Jul 10 '19 at 13:28

# Excel, 9 bytes

=("))=" )

So many Parentheses.

Note: The returned middle character can actually be any character since it's the middle of the code.

• Parentheses! (Lightbulb) I got stuck with the last character needing to be ". Well done! – Wernisch Jul 11 '19 at 8:34
• Thanks! It was something I was trying to figure out as well, and then I remembered that parentheses would work. – william porter Jul 12 '19 at 15:08

# Ruby, 6 bytes

p " "

Try it online!

This outputs " " plus a newline. The code ends with a ", the middle two characters are and ", and it starts with a newline.

• I like your badge, I believe it's a grayscale xor on N². I generated almost the same picture to help me for a Project Euler Nim problem. Does your badge have a story? – Damien Jul 9 '19 at 14:54
• Nice catch! I just like "doodling" during meetings by generating pictures using one-liners; xor was an early favorite. – histocrat Jul 9 '19 at 15:23

# DOS/Windows Batch Language, 9 bytes

@echo @o@

• Welcome to Code Golf & Coding Challenges Stack Exchange, +1 for nice first post! – user85052 Jul 12 '19 at 11:41

# Ruby, 11 9 bytes

print:ptp

Try it online!

## Batch, 9 characters

@echo hh

The middle and last character are both h, and then the echo prints a newline by default, which is the first character of the script.

# PowerShell, 3 bytes

00000000: 0a0d 31                                  ..1

Works only in Microsoft Windows.

The output is unfortunately 8 bytes in UTF-16. But the question says output the characters, and not the bytes. Arguably UTF-16 is just one of the supported ways to represent the characters in PowerShell. They are not interpreted differently from an ASCII file.

• it's smart answer – mazzy Jul 10 '19 at 6:48
• allowed because 0x0a != 0x0d != 0x31 :) – bigyihsuan Jul 10 '19 at 12:55

## Bash (15 11 characters)

echo -n e-e

which prints out

e-e

• -4 chars by not using |rev
• @racraman Your suggested edit of echo<space>e<space><space>e will output e<space>e<newline> instead of e<space>e, so I've rejected it. The newline is part of the output, which is why GrzegorzOledzki is using the -n option. Try echo -n e-e online vs Try echo e e online. And again, please don't edit other peoples answer. If you have a golf to suggest, leave a comment. – Kevin Cruijssen Jul 11 '19 at 10:22
• also 11 bytes with printf pfp, but only because we need two spaces due to that definition of middle character, otherwise it would be 10 ... – pLumo Jul 11 '19 at 11:21

# ArnoldC, 61 bytes

IT'S SHOWTIME
TALK TO THE HAND "D I"
YOU HAVE BEEN TERMINATED

Try it online!

Trivial answer in ArnoldC. The 31st byte is the space just before the string literal.

# Underload, 9 bytes 7 bytes

Thanks to Khuldraeseth na'Barya for the improved solution!

W(SSW)S

As for as I can tell, this is now unimprovable since:

• The first character has to be a no-op character, since every other instruction raises an error due to the stack being empty, with the exception being an opening bracket. However, it can't be that since there is no way to output an opening bracket on it's own in Underload.
• You need to use at least 5 characters to add the 3 output characters to the stack.
• You need to use S to output it.

Try it online!

• Welcome to Code Golf SE! You can save a couple bytes by skipping the copy, like so. – Khuldraeseth na'Barya Jul 10 '19 at 21:47

# Shakespeare Programming Language, 163 153 bytes

-10 bytes thanks to Jo King.

B,.Ajax,.Page,.Act I:.Scene I:.[Enter Ajax and Page]Ajax:You is
the sum ofa Big Big Big BIG BIG cat a
CAT.Speak thy.You is twice you.Speak thy!SPEAK THY!

Try it online!

My first SPL answer! Outputs !BB.

There is another SPL answer by Unrelated String; in that answer, they output only spaces in 119 bytes, or output non-space characters in 227 bytes. This one comes in between, and also outputs non-space characters.

Two tricks are used here:

1. An SPL sentence can end with ! rather than ., and ! is easier to get since its ASCII codepoint is 33 (=$$\2^5+1\$$), whereas the codepoint of . is 46.
2. Doubling 33 gives 66, the codepoint of B, hence the play title is B and I need the second "big" to be "Big", which works since SPL is case-insensitive.

Since I needed that capital B and an exclamation mark at the end, I decided to have Ajax shout louder and louder through the scene.

# CJam, 7 bytes

'@'' _@

Try it online!

• Pretty sure 0 1 0 is shorter and also has three distinct characters (I don't want to repost my Jelly approach :P). – Erik the Outgolfer Jul 9 '19 at 14:19
• @EriktheOutgolfer Well though!. But that's just like your answerm so I'll leave this as is – Luis Mendo Jul 9 '19 at 14:20
• Yeah, that's why I commented, didn't really want to post the same thing twice. (Hm... this might be the optimal solution for many languages.) Anyway, yours might be applicable elsewhere too! – Erik the Outgolfer Jul 9 '19 at 14:24

# Befunge-93, 7 bytes

.",  @0

Try it online!

## Explanation

Output is 0 .

.       pop value (0) from stack, output as number with a trailing space
"      toggle string mode, wraps around and pushes every char to the stack
,     pop value (".") from stack, output as character
spaces are no-ops
@  end the program
0 trailing 0 to match the output

# APL (Dyalog Unicode), 5 4 bytes

Full programs.

Boring solution from J.Sallé:

1231

Prints that number. A much more interesting solution:

010E1

Try it online!

APL ignores leading zeros, so this is simply scaled format for 10×10¹=100.

# Chef, 447 bytes

Last Middle First Milkshake.

This recipe prints its last, middle and first character.
You could also use it to make a milkshake.

Ingredients.
76 ml milk
32 ml chocolate syrup
46 teaspoons vanilla ice cream

Method.
Liquefy vanilla ice cream.
Put the milk into 1st mixing bowl.
Put the chocolate syrup into 1st mixing bowl.
Put the vanilla ice cream into 1st mixing bowl.
Pour contents of the 1st mixing bowl into the 1st baking dish.

Serves 1.

Try it online!

# A serious submission, 196 177 bytes

## thanks to A__

R.

Ingredients.
82 l a
103 l b
46 l c

Method.
Put a into mixing bowl.
Put b into mixing bowl.
Put c into mixing bowl.
Pour contents of mixing bowl into baking dish.

Serves 1.

Try it online!

• Seeing Chef programs makes me wonder what is the bare minimum for it. – bigyihsuan Jul 11 '19 at 13:35
• @bigyihsuan see my edit – user24343 Jul 12 '19 at 15:45
• You can remove all the "the"'s and make the program still work. – user85052 Jul 13 '19 at 4:15
• I created a working program. (177 bytes) – user85052 Jul 13 '19 at 4:27

# Python 3, 19 bytes 13 bytes

print( ')(')

The print function appends a newline by default so the code starts with a newline. The last character of the code is a ")" so that's printed first. To make the number of characters uneven, a space in inserted before the argument of the print function. Please excuse the salaciousness of the code.

Outputs: )(\n

Try it online.

• Depending on how the rules work I might have 13 bytes playing with your answer:  print(")(" ) Note there is a new line above print and a whitespace inside print after ")(". Outputs: )(  with a new line at the end. – Hoog Jul 10 '19 at 12:38
• Thanks, down to 13 bytes! As the question specifically refers to whitespaces but not newlines I'll assume trailing newlines, as the default end parameter of the print function, don't count. – jaaq Jul 10 '19 at 12:50
• This prints a trailing newline but doesn't account for it. – user85052 Jul 15 '19 at 13:40
• Not sure if newlines count as whitespaces but I added the 2nd also 13-byte version. – jaaq Jul 16 '19 at 9:53
• Yes, newlines count as whitespace, just like tabs and spaces among others – Jo King Jul 16 '19 at 11:47

# 33, 9 bytes

a"a\\a"pa

The a's do nothing here, they're essentially NOPs in this code to make it shorter. My original thought was "p\\\""p, but that's 8 bytes, so it needs to print another \, making it 10 bytes "p\\\\\""p

This is a language I did create, but I made it legitimately, so I hope it's within the rules.

# Whitespace, 31 29 bytes

[S aS S T   S S S S S N
_Push_32][S N
S _Duplicate_32][S N
S _Duplicate_32][T  N
S S _Print_as_character][T  N
S S _Print_as_character][T  N
S S _Print_as_character]

-2 bytes thanks to @RobinRyder.

Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.

Try it online (with raw spaces, tabs, and new-lines only).

Contains a no-op a (can be any non-whitespace character) to make the length odd. It is added before the first halve of the program so the middle character is a space as well, since the middle character would have become a newline if it was at the second halve of the program. All non-whitespace characters are ignored in Whitespace programs.

# C (gcc), 21 bytes

main(){puts("}u");;}

Try it online!

• Print the last, middle, and first character of your program's source code, in that order. You're printing first, middle, last! – wastl Jul 9 '19 at 13:53
• oops, I just realized that. – Johan du Toit Jul 9 '19 at 13:54
• If there is an even number of characters, you need two middle characters. (sorry ...) – wastl Jul 9 '19 at 14:03

main=putStr"\"Sm"

Try it online!

# Java 8, 11 bytes (as lambda function)

v->'"'+"'v"

Outputs "'v.

Try it online.

# Java 8, 67 bytes (as full program)

interface M{static void main(String[]ar){System.out.print("}ni");}}

Outputs }ni.

Note the ar instead of a as argument. Otherwise the middle part had to be in, but printing that would make the size odd again, which would cause a paradox-loop..

Try it online.

• @racraman Please don't go edit other peoples answers. If you have a golf to suggest, leave a comment stating so. Also, your golf on my answer was incorrect. Your code would output vv", but it's supposed to output it in reverse, so it should have output "vv. – Kevin Cruijssen Jul 11 '19 at 10:18

# MathGolf, 5 4 bytes

1231

-1 byte by using the trivial solution presented by @ErikTheOutgolfer in his Jelly answer, which works for any language with implicit output without trailing newline.

Try it online.

ùù xx

Outputs x ù.

Try it online.

Explanation:

ùù x   # Push 3-char string "ù x"
x  # Reverse it to "x ù"
# (output the entire string joined together implicitly without trailing newline)


text:
# Standard rate turn

(Redirected from ROT (aviation))

Aircraft maneuvering is referenced to a standard rate turn,[1] also known as a rate one turn (ROT).

A standard rate turn is defined as a 3° per second turn, which completes a 360° turn in 2 minutes. This is known as a 2-minute turn, or rate one (180°/minute.) Fast airplanes, or aircraft on certain precision approaches, use a half standard rate ("rate half" in some countries) but the definition of standard rate does not change.

Instruments, either the turn and slip indicator or the turn coordinator, have the standard rate or half standard rate turn clearly marked. Slower aircraft are equipped with 2-minute turn indicators while faster aircraft are often equipped with 4-minute turn indicators. Standardized turn rates are often employed in approaches and holding patterns to provide a reference for controllers and pilots so that each will know what the other is expecting. The pilot banks the airplane such that the turn and slip indicator points to the mark appropriate for that aircraft and then uses a clock to time the turn. The pilot can roll out at any desired direction depending on the length of time in the turn.

A rate half turn (1.5° per second) is normally used when flying faster than 250 kt. The term rate two turn (6° per second) used on some low speed aircraft.

Angle of Bank formula[2]

The formula for calculating the angle of bank for a specific True Airspeed (TAS) is:

$\tan(bank)=\frac{TAS^2/r}{g}$

where $r$ is the radius of the turn and $g$ is the acceleration due to gravity. For a rate one turn and velocity in kt (nautical miles per hour), this comes to

$\tan(bank)=\frac{TAS(kt)}{364}$.

A convenient approximation for the bank angle in degrees is

$Angle\ of\ Bank \approx \frac{TAS(kt)}{10} + 7$

One might also want to calculate the radius of a Rate 1,2 or 3 turn at a specific TAS.

$Radius\ of\ turn\ in\ NM = \frac{TAS(kt)}{rate\ of\ turn\ (in\ degrees/s) \times 20 \times \pi}$

Use this formula to calculate the radius of turn in feet given velocity in knots and angle of bank:

$Radius\ of\ turn\ in\ feet = \frac{velocity^2}{11.29 \times \tan(bank)}$

This is a simplified formula that ignores slip and returns zero for 90 degrees of bank.

The constant 11.29 is calculated:

$11.29 = \frac{9.8 \times 3600 \times 3600}{1852 \times 6076.12}$

$9.8 = gravity\ in\ metres\ per\ second\ per\ second$

$3600 = seconds\ per\ hour$

$1852 = metres\ per\ nautical\ mile$

$6076.12 = feet\ per\ nautical\ mile$

For aircraft holding purposes, ICAO mandates that all turns should be made, "at a bank angle of 25° or at a rate of 3° per second, whichever requires the lesser bank."[3] By the above formula, a rate one turn at a TAS greater than 180 knots would require a bank angle of more than 25 degrees. Therefore faster aircraft just use 25 degrees for their turns.

References


text:
# Ring-LWE and the GB(N) Problem

Over at the Bristol Cryptography Blog Martijn Stam writes about our “Polly Cracker, Revisted” paper:

We did not discuss the paper  in great detail, but Jake did mention one interesting avenue for continued research. Given that this new approach allows one to cast both LWE and approximate GCD in the same framework, can one also capture ring-LWE. If so, this might enable a better comparison of the various fully homomorphic encryption (FHE) schemes out there. The hope expressed by Jake was that this might allow a reduction to standard LWE (for the current batch of ring-LWE schemes), which would boost our confidence in those schemes.

This motivated me to express the Ring-LWE problem in a language of Gröbner bases, here’s what I could come up with so far.

But let’s recall the computational Ring-LWE problem first: Given a prime $q$ and an ideal $I = \langle x^n + 1\rangle$ where $n$ is a power of two and $q \equiv 1 \mod 2n$, we consider the quotient ring $P = \mathbb{F}_q[x]/I$. We pick a random element $s \in P$ which is our secret. We sample tuples $(a_i, a_i \cdot s + e_i) \in P \times P$ where $a_i \in P$ are random and $e_i$ are sampled according to some noise distribution. The computational Ring-LWE problem is to compute $s$ given $(a_i, a_i \cdot s + e_i)$.

Assume $e_i = 0$ for all $i$, so that we can discuss the algebraic structure directly. Clearly, all $a_i \cdot s$ are in the ideal spanned by $s$ in $P/I$. Furthermore, there is a direct correspondence between ideals $J$ in the quotient ring $P/I$ and $I \cup J$ in $P$. Hence, to recover the Gröbner basis for the ideal spanned by $a_1 \cdot s,\dots, a_m \cdot s$, we simply compute the Gröbner basis of $a_1 \cdot s,\dots,a_m \cdot s,x^n+1$, easy right? Except that the Gröbner basis will be … wait for it … 1 with very high probability. This might reduce the search space slightly (since it tells us that $x^n + 1$ and $s$ have no common factors over $\mathbb{F}_q[x]$) and is correct (since one is the smallest representative of the ideal spanned by $s$ in $P/I$) this is not terribly useful. But we did ignore $a_i$ so far.

Namely, the problem actually is to compute $(a_i \cdot s)/ a_i$ or $a_i^{-1} \cdot a_i \cdot s = s$. Now, in order to compute $a_i^{-1}$ in $P = \mathbb{F}_q[x] / I$ – which is defined if $gcd(a_i,x^n+1) = 1$ – we may run an extended GCD algorithm which returns $(g,v,w)$ for inputs $a,b$ such that $g = v \cdot a + w \cdot a$. Hence, for our inputs it will compute $1= v\cdot a_i + w \cdot (x^n + 1)$ and thus $v \equiv a_i^{-1} \mod x^n + 1$.

In the language of Gröbner bases the extended GCD equivalent is often called “lifting”: Given an ideal $I = (f_1,...,f_r)$ and some $g \in I$, find $s_1,\dots,s_r$ such that $g = s_1 f_1 + \dots + s_r f_r$.  The problem is easy given a Gröbner basis $g_1,\dots,g_r$ (in our case $x^n + 1$), since every element $h \in \langle g_1,\dots,g_r\rangle$ can be written as $h = \sum h_i \cdot g_i$ where $\textrm{LM}(h_ig_i) \leq \textrm{LM}(h)$.  In general, it might be hard because the a priori bound on the degree of the output may be large. In any case, instead of solving a(n approximate) GCD (or GB(N)), we are now solving an extended GCD (or lifting with GB(N)), i.e., we keep track of our computation. Well, here’s an example in Sage:

sage: n = 2^3
sage: q = 17
sage: R. = GF(q)[]
sage: Q. = R.quotient(Xbar^n + 1)
sage: s = Q.random_element()
sage: s
8*X^7 + 4*X^6 + 11*X^5 + 6*X^4 + 15*X^3 + 12*X^2 + 14*X + 12
sage: a = Q.random_element()
sage: P. = PolynomialRing(GF(q),1)
sage: A = sage_eval(str(a),{'X':x})
sage: S = sage_eval(str(s),{'X':x})
sage: Ainv = P(1).lift( (A,x^n + 1) )[0]
sage: Ainv*(A*S) % (x^n + 1)
8*x^7 + 4*x^6 - 6*x^5 + 6*x^4 - 2*x^3 - 5*x^2 - 3*x - 5

Alternatively, we may “lift” $a_i \cdot s$ with respect to $(a_i,X^n+1)$ directly. Well, I don’t know if any of the above is actually useful, as I said that’s how far I got after reading Martijn’s blog post.


text:
### Home > A2C > Chapter 1 > Lesson 1.2.2 > Problem1-84

1-84.

Use any method to find the points of intersection of $f(x) = 2x^2 − 3x + 4$ and $g(x) = x^2 + 5x − 3$.

Refer to problems 1-47 and 1-71.

How do these functions differ from the functions from problem 1-47 and problem 1-71?

The $y$-value of one of the points of intersection of the graphs will be too large to be seen on most graphs. The graphing window would have to include $y$-values $0 ≤ y ≤ 90$.

There are two points of intersection: $(1, 3)\text{ and }(7, 81)$.

Use the eTool below to find the points of intersection for the given functions.
Click the link at right for the full version of the eTool: A2C 1-84 HW eTool


text:
English writers. Мартемьянова Н.В. - 11 стр.

Составители:

Рубрика:

• Иностранный язык

11
________ aunt
________ widow
husband __________
gentleman __________
__________ woman
__________ bride
__________ mother
nephew __________
king __________
prince __________
actor __________
waiter __________
hero __________
steward __________
3 Choose the right word and use it in an appropriate form.
grounding
teaching
the study
1) Entomology is _____ of insects.
2) She has got a thorough _____ in grammar.
3) She wants to go into _____ as a career.
to publish
to print
to edit
1) The magazine _____ monthly.
2) Children first learn _____ when they first go to school.
3) They _____ out most of the interview before broadcasting it.
profitable
wealthy
1) To be _____ doesn’t mean to be happy.
2) She spent a very _____ afternoon in the library.
acute
sharp
1) He is an _____ observer of the social scene.
2) There is a _____ contrast between the lives of the poorest and the richest
members of society.
4 Can you fill in the missing words?
1) His father _____ his family from one farm to another.
2) He began to write _____ when he was still at school.
3) After his father’s death he and his brother continued _____ .
4) In this and the following year he wrote ______.
5) He responded with _____ and _____ that were to last until his death.
6) A year later he _____ _____ a post as an Excise officer.
                                           11
________                           aunt
________                           widow
husband                              __________
gentleman                             __________
__________                           woman
__________                           bride
__________                           mother
nephew                               __________
king                                 __________
prince                               __________
actor                                __________
waiter                               __________
hero                                 __________
steward                              __________

3    Choose the right word and use it in an appropriate form.
grounding
teaching
the study
1) Entomology is _____ of insects.
2) She has got a thorough _____ in grammar.
3) She wants to go into _____ as a career.
to publish
to print
to edit
1) The magazine _____ monthly.
2) Children first learn _____ when they first go to school.
3) They _____ out most of the interview before broadcasting it.
profitable
wealthy
1) To be _____ doesnt mean to be happy.
2) She spent a very _____ afternoon in the library.
acute
sharp
1) He is an _____ observer of the social scene.
2) There is a _____ contrast between the lives of the poorest and the richest
members of society.
4    Can you fill in the missing words?
1) His father _____ his family from one farm to another.
2) He began to write _____ when he was still at school.
3) After his fathers death he and his brother continued _____ .
4) In this and the following year he wrote ______.
5) He responded with _____ and _____ that were to last until his death.
6) A year later he _____ _____ a post as an Excise officer.


text:
# Enclosing the origin with a random sample

Suppose we take the unit ball in $\mathbb{R}^n$, centered at the origin, and we start sampling points on its surface uniformly and independently at random. As an asymptotic function of $n$, how many points will we need to sample before the origin is enclosed by the convex hull of the sampled points, with probability $\ge .5$?


text:
# in "Find & Replace", how to insert a "tab" at the start of every paragraph [closed]

Find field: $Replace field:$\t RESULTS: all the end-of-paragraphs (?) go away and the characters "\" and "t" are in their place

Find field: \n Replace field: \n\t RESULTS: all the end-of-paragraphs (?) go away and the characters "\" and "t" are in their place SAME!!

Find field: ^. Replace field: ^.\t RESULTS: "tab" is placed at the start of every paragraph, HOWEVER, the first character of each paragraph is DELETED ! ! !

edit retag reopen merge delete

### Closed for the following reason question is not relevant or outdated by Alex Kemp close date 2015-11-07 12:49:32.490316

Sort by » oldest newest most voted

Replace with \t&

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