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## Found 283 Documents (Results 1–100)

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### Methods for studying the stability of linear periodic systems depending on a small parameter. (English. Russian original)Zbl 1484.34060

J. Math. Sci., New York 258, No. 1, 115-127 (2021); translation from Itogi Nauki Tekh., Ser. Sovrem. Mat. Prilozh., Temat. Obz. 163, 113-126 (2019).
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### Construction of $$2\times 2$$ Fuchsian system with five singular points. (English)Zbl 07355616

MSC:  65-XX 34-XX
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### On the deformation of linear Hamiltonian systems. (English)Zbl 1460.35073

MSC:  35F20 34L05
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### Periodic solutions and Hyers-Ulam stability of atmospheric Ekman flows. (English)Zbl 1467.34043

MSC:  34C25 34D10
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### Stability boundary approximation of periodic dynamics. (English)Zbl 1479.70051

Lacarbonara, Walter (ed.) et al., Nonlinear dynamics of structures, systems and devices. Proceedings of the first international nonlinear dynamics conference, NODYCON 2019, Rome, Italy, February 17–20, 2019. Volume I. Cham: Springer. 13-23 (2020).
MSC:  70K20
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### Bifurcations of periodic solutions of a Hamiltonian system with a discrete symmetry group. (English. Russian original)Zbl 1459.37042

Program. Comput. Softw. 46, No. 2, 84-97 (2020); translation from Programmirovanie 46, No. 2, 14-29 (2020).
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### Hurwitz numbers from Feynman diagrams. (English. Russian original)Zbl 1453.81052

Theor. Math. Phys. 204, No. 3, 1166-1194 (2020); translation from Teor. Mat. Fiz. 204, No. 3, 396-429 (2020).
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### Generalized Picard-Fuchs operators from Whitham hierarchy in $$\mathcal{N} = 2$$ supersymmetric gauge theory with massless hypermultiplets. (English. Russian original)Zbl 1445.81038

Theor. Math. Phys. 202, No. 2, 150-164 (2020); translation from Teor. Mat. Fiz. 202, No. 2, 170-186 (2020).
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### Solution space monodromy of a special double confluent Heun equation and its applications. (English. Russian original)Zbl 1437.34087

Theor. Math. Phys. 201, No. 1, 1426-1441 (2019); translation from Teor. Mat. Fiz. 201, No. 1, 17-36 (2019).
MSC:  34M03 34M35 82D55
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### Solving scattering problems in the half-line using methods developed for scattering in the full line. (English)Zbl 1429.81095

MSC:  81U05 34M35 78A60
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MSC:  92D30
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MSC:  35B10

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### A monodromy matrix for the almost Mathieu equation with small coupling constant. (English. Russian original)Zbl 1416.39012

Funct. Anal. Appl. 52, No. 4, 311-315 (2018); translation from Funkts. Anal. Prilozh. 52, No. 4, 89-93 (2018).
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### The inverse monodromy problem. (English)Zbl 1459.46035

Alpay, Daniel (ed.) et al., Indefinite inner product spaces, Schur analysis, and differential equations. A volume dedicated to Heinz Langer. Cham: Birkhäuser. Oper. Theory: Adv. Appl. 263, 73-105 (2018).
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### Monodromization and difference equations with meromorphic periodic coefficients. (English. Russian original)Zbl 1390.39053

Funct. Anal. Appl. 52, No. 1, 77-81 (2018); translation from Funkts. Anal. Prilozh. 52, No. 1, 92-97 (2018).
MSC:  39A45 39A23
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### Pseudospectral functions of various dimensions for symmetric systems with the maximal deficiency index. (English)Zbl 1443.34094

J. Math. Sci., New York 229, No. 1, 51-84 (2018) and Ukr. Mat. Visn. 14, No. 2, xx (2017).
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### The fluctuations of the ring supported with threads. (Russian. English summary)Zbl 1446.74153

MSC:  74K10 74H55 74H50

### Current presentation for the super-Yangian double $$DY(\mathfrak{gl}(m|n))$$ and Bethe vectors. (English. Russian original)Zbl 1377.82026

Russ. Math. Surv. 72, No. 1, 33-99 (2017); translation from Usp. Mat. Nauk 72, No. 1, 37-106 (2017).
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### Limit cycles for regularized discontinuous dynamical systems with a hyperplane of discontinuity. (English)Zbl 1377.34025

MSC:  34A36 34C05 34D20
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### Multiple commutation relations in the models with $$\mathfrak{gl}(2|1)$$ symmetry. (English. Russian original)Zbl 1361.81050

Theor. Math. Phys. 189, No. 2, 1624-1644 (2016); translation from Teor. Mat. Fiz. 189, No. 2, 256-278 (2016).
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### $$\mathrm{SU}(2)/\mathrm{SL}(2)$$ knot invariants and Kontsevich-Soibelman monodromies. (English. Russian original)Zbl 1346.81073

Theor. Math. Phys. 187, No. 2, 678-694 (2016); translation from Teor. Mat. Fiz. 187, No. 2, 263-282 (2016).
MSC:  81T13 58J28 14D05
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### Divergent series, summability and resurgence I. Monodromy and resurgence. (English)Zbl 1355.34003

Lecture Notes in Mathematics 2153. Cham: Springer (ISBN 978-3-319-28735-5/pbk; 978-3-319-28736-2/ebook). xxi, 298 p. (2016).
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### Unique determination of a system by a part of the monodromy matrix. (English. Russian original)Zbl 1350.34025

Funct. Anal. Appl. 49, No. 4, 264-278 (2015); translation from Funkts. Anal. Prilozh. 49, No. 4, 33-49 (2015).
MSC:  34A55
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### One-dimensional two-component Bose gas and the algebraic Bethe ansatz. (English. Russian original)Zbl 1327.82025

Theor. Math. Phys. 183, No. 3, 800-821 (2015); translation from Teor. Mat. Fiz. 183, No. 3, 409-433 (2015).
MSC:  82B23 82B20
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### Trace formula for linear Hamiltonian systems with its applications to elliptic Lagrangian solutions. (English)Zbl 1316.35276

MSC:  35Q70 70F10 35B10
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### Numerical computation of periodic responses of nonlinear large-scale systems by shooting method. (English)Zbl 1368.74076

MSC:  74S30 74H45 65M20 34C25 65L20
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MSC:  65Q10
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### Notes on thin matrix groups. (English)Zbl 1365.11039

Breuillard, Emmanuel (ed.) et al., Thin groups and superstrong approximation. Selected papers based on the presentations at the workshop held at MSRI, Berkeley, CA, USA, February 6–10, 2012. Cambridge: Cambridge University Press (ISBN 978-1-107-03685-7/hbk). Mathematical Sciences Research Institute Publications 61, 343-362 (2014).
MSC:  11F06 20H25 22E40
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### The ubiquity of thin groups. (English)Zbl 1355.11029

Breuillard, Emmanuel (ed.) et al., Thin groups and superstrong approximation. Selected papers based on the presentations at the workshop held at MSRI, Berkeley, CA, USA, February 6–10, 2012. Cambridge: Cambridge University Press (ISBN 978-1-107-03685-7/hbk). Mathematical Sciences Research Institute Publications 61, 73-92 (2014).
MSC:  11F11 20H25

### $$H_\infty$$ norm computation of linear continuous-time periodic systems by a structure-preserving algorithm. (English)Zbl 1317.93097

MSC:  93B36 93C15 34C25
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### Asymptotics of the spectrum and eigenfunctions of the magnetic induction operator on a compact two-dimensional surface of revolution. (English. Russian original)Zbl 1315.35162

Math. Notes 95, No. 3, 374-387 (2014); translation from Mat. Zametki 95, No. 3, 417-432 (2014).
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### Andronov-Hopf bifurcation theorem for relay systems. (English. Russian original)Zbl 1314.34080

J. Math. Sci., New York 200, No. 1, 134-142 (2014); translation from Sovrem. Mat. Prilozh. 85 (2012).
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### Monodromization method in the theory of almost-periodic equations. (English. Russian original)Zbl 1326.39011

St. Petersbg. Math. J. 25, No. 2, 303-325 (2014); translation from Algebra Anal. 25, No. 2, 203-235 (2013).
MSC:  39A24
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MSC:  74-XX
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### A single compartment model of pacemaking in dissasociated substantia nigra neurons. Stability and energy analysis. (English)Zbl 1382.92054

MSC:  92C20 92B25
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### Symplectic difference systems with periodic coefficients: Krein’s traffic rules for multipliers. (English)Zbl 1380.39009

MSC:  39A12 15A18
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### On some constructive methods for the matrix Riemann-Hilbert boundary-value problem. (English. Russian original)Zbl 1288.34078

J. Math. Sci., New York 195, No. 2, 146-174 (2013); translation from Sovrem. Mat. Prilozh. 81 (2012).
MSC:  34M35 34M50
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### Classification of Fuchsian systems and their connection problem. (English)Zbl 1307.34134

MSC:  34M35 34M40 34M15
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### On a trivial monodromy criterion for the Sturm-Liouville equation. (English. Russian original)Zbl 1288.34080

Math. Notes 94, No. 4, 508-523 (2013); translation from Mat. Zametki 94, No. 4, 552-568 (2013).
MSC:  34M35
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### On the degenerate spectral gaps of the 1D Schrödinger operators with 4-term periodic delta potentials. (English)Zbl 1292.34084

MSC:  34L40 34L15
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### Converging bounds for the effective shear speed in 2D phononic crystals. (English)Zbl 1329.74121

MSC:  74J05 74Q20 74E15
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### Construction of a Pfaff system of Fuchs type with three singular surfaces on $$CP^n$$. (English)Zbl 1312.34136

Rogosin, S. V. (ed.), Analytic methods of analysis and differential equations. AMADE 2009. Selected papers of the 5th international conference, Minsk, Belarus, September 14–19, 2009. Dedicated to the memory of Anatoly Kilbas. Cambridge: Cambridge Scientific Publishers (ISBN 978-1-908106-13-1/pbk). 1-8 (2012).
MSC:  34M50 34M35 58A17

### Integral symmetry and the deformed hypergeometric equation. (English)Zbl 1291.34150

Bruno, Alexander D. (ed.) et al., Painlevé equations and related topics. Proceedings of the international conference, Saint Petersburg, Russia, June 17–23, 2011. Berlin: de Gruyter (ISBN 978-3-11-027558-2/hbk; 978-3-11-027566-7/ebook). De Gruyter Proceedings in Mathematics, 231-235 (2012).

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### Teichmüller spaces as degenerated symplectic leaves in Dubrovin-Ugaglia Poisson manifolds. (English)Zbl 1257.30041

MSC:  30F10 30F60
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### Elementary models of unbounded Jacobi matrices with a few bounded gaps in the essential spectrum. (English)Zbl 1282.47043

MSC:  47B36 47A10 47B25
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### The factorized $$F$$-matrices for arbitrary $$U(1)^{N-1}$$ integrable vertex models. (English)Zbl 1246.82021

MSC:  82B20 16T25 14D05
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### A Fuchsian matrix differential equation for Selberg correlation integrals. (English)Zbl 1245.82030

MSC:  82B44 15B52 34M35
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### The monodromy matrix in the F-basis for arbitrary six-vertex models. (English)Zbl 1229.82043

MSC:  82B20 16T25 14D05
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### Invariants of the homoclinic trajectories of a two-dimensional diffeomorphism. (English. Russian original)Zbl 1244.37016

Math. Notes 89, No. 5, 712-725 (2011); translation from Mat. Zametki 89, No. 5, 755-770 (2011).
MSC:  37C29 37C05
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### Sensitivity of Schur stability of monodromy matrix. (English)Zbl 1225.39019

MSC:  39A30 39A06
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### Quantization conditions on Riemannian surfaces and the semiclassical spectrum of the Schrödinger operator with complex potential. (English. Russian original)Zbl 1234.81077

Math. Notes 88, No. 2, 209-227 (2010); translation from Mat. Zametki 88, No. 2, 229-248 (2010).
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### Additive Deligne-Simpson problem for non-Fuchsian systems. (English)Zbl 1246.15018

MSC:  15A24 34M50 34M35
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### Two classes of solvable subgroups of $$\text{SL}(3,\mathbb C)$$ and application for differential equations in $$\mathbb C$$. (English)Zbl 1224.20074

MSC:  20H20 34M15 20E07

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### Bifurcation of cycles of automatic control systems with ideal relay. (English. Russian original)Zbl 1223.34058

Autom. Remote Control 71, No. 11, 2291-2306 (2010); translation from Avtom. Telemekh. 2010, No. 11, 38-54 (2010).
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### Two classes of solvable subgroups of $$\text{SL}(2k,\mathbb C)$$ and application in Fuchsian systems. (Chinese. English summary)Zbl 1224.20073

MSC:  20H20 34M15 20E07

### Bifurcations of periodic orbits and integrability of dynamical systems. (English)Zbl 1204.37064

MSC:  37J20 37J30 37J45
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MSC:  83E30
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### On the algebraic aspects of Saxon-Hutner theorem. (English)Zbl 1168.81328

Kielanowski, Piotr (ed.) et al., XXVI international workshop on geometrical methods in physics, Białowieża, Poland, 1-7 July 2007. Melville, NY: American Institute of Physics (AIP) (ISBN 978-0-7354-0470-0/hbk). AIP Conference Proceedings 956, 169-174 (2007).
MSC:  81Q05 81T25

### Discontinuity correction in piecewise-linear models of oscillators for phase noise characterization. (English)Zbl 1134.93390

MSC:  93C99 37D99 94C99
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### Extremal unitary local systems on $$\mathbb P^1-{p_1,...,p_s}$$. (English)Zbl 1133.14048

Mehta, V. B. (ed.), Algebraic groups and homogeneous spaces. Proceedings of the international colloquium, Mumbai, India, January 6–14, 2004. New Delhi: Narosa Publishing House/Published for the Tata Institute of Fundamental Research (ISBN 978-81-7319-802-1/hbk). Studies in Mathematics. Tata Institute of Fundamental Research 19, 37-64 (2007).

### On the influence of random perturbations on the propagation of waves described by a periodic Schrödinger operator. (English)Zbl 1135.35102

Dawson, Donald A. (ed.) et al., Probability and mathematical physics. A volume in honor of Stanislav Molchanov. Proceedings of a conference in honor of Stanislav Molchanov’s 65th birthday, Montreal, Canada, June 27–July 1, 2005. Providence, RI: American Mathematical Society (AMS) (ISBN 978-0-8218-4089-4/pbk). CRM Proceedings and Lecture Notes 42, 167-179 (2007).

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# Topological Quantum Field Theory Seminar

### Topological Field Theory in dimension $3$

I will give an overview of some aspects of $3d$ TFT, from the Turaev-Viro and Reshetikin-Turaev invariants of oriented $3$-manifolds, to the more recent classifications of fully extended theories in terms of fusion categories and once extended theories in terms of Modular Tensor Categories.

Current organizers: Roger Picken, Marko Stošić.

FCT Projects PTDC/MAT-GEO/3319/2014, Quantization and Kähler Geometry, PTDC/MAT-PUR/31089/2017, Higher Structures and Applications.

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TH String Theory Seminar

# Holographic models for QCD in the Veneziano limit

## by Matti Jarvinen (Heraklion U.)

Tuesday, 7 January 2014 from to (Europe/Zurich)
at CERN ( 4-3-006 - TH Conference Room )
 Description I discuss holographic bottom-up models for QCD in the Veneziano limit (large N_f and N_c with fixed x=N_f/N_c). After a brief introduction, I will go on discussing a specific class of models, named V-QCD. The models are constructed by putting together holographic models of Yang-Mills theory with Sen-type tachyon actions, and by considering full backreaction. The resulting phase diagrams and mass spectra meet expectations from QCD at qualitative level.


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# Calculate $\sin\frac{3\pi}{14}-\sin\frac{\pi}{14}-\sin\frac{5\pi}{14}$

I have interesting trigonometric expression for professionals in mathematical science. So, here it is: $$\sin\dfrac{3\pi}{14}-\sin\dfrac{\pi}{14}-\sin\dfrac{5\pi}{14};$$ Okay! I attempt calculate it: \begin{gather} \sin\dfrac{3\pi}{14}-\left(\sin\dfrac{\pi}{14}+\sin\dfrac{5\pi}{14}\right)=\\ =\sin\dfrac{3\pi}{14}-\left(2\sin\dfrac{3\pi}{14}\cdot\cos\dfrac{\pi}{7}\right)=\\ =\sin\dfrac{3\pi}{14}\left[1-2\cdot\cos\dfrac{\pi}{7}\right]=... \end{gather} Tried everything... Here deadlock. I really do not know what to do next. Help somebody, please.

• if you are satsified with the value alone, it is -1/2, according to wolframalpha – Bananach Apr 19 '16 at 8:28
• Hint: $\sum\limits_{k=1}^7 \sin\left(\frac{(4k-1)\pi}{14}\right) = 0$ – achille hui Apr 19 '16 at 8:34

For simplicity, let $x = \frac{\pi}{14}$, then we want to simplify: $$\sin 3x-\sin x -\sin 5x$$ Multiply by $\cos x$ to get: $$\color{blue}{\sin 3x\cos x}-\color{green}{\sin x\cos x} -\color{red}{\sin 5x\cos x} \quad (*)$$ With $\sin\alpha\cos\beta = \tfrac{1}{2}\left( \sin(\alpha+\beta)+\sin(\alpha-\beta) \right)$, you have: $$\color{blue}{\sin 3x\cos x = \tfrac{1}{2}\left( \sin 4x +\sin 2x \right)} \quad \mbox{and} \quad \color{red}{\sin 5x\cos x = \tfrac{1}{2}\left( \sin 6x +\sin 4x \right)}$$ and $\color{green}{\sin x\cos x = \tfrac{1}{2}\sin 2x}$; so: $$\require{cancel} (*) \quad \tfrac{1}{2}\left( \cancel{\sin 4x} +\bcancel{\sin 2x} \right) - \bcancel{\tfrac{1}{2}\sin 2x} - \tfrac{1}{2}\left( \sin 6x +\cancel{\sin 4x} \right) = - \tfrac{1}{2}\sin 6x$$ Divide again by $\cos x$: $$- \tfrac{1}{2}\frac{\sin 6x}{\cos x} = - \tfrac{1}{2}\frac{\sin \frac{6\pi}{14}}{\cos \frac{\pi}{14}}= - \tfrac{1}{2}\frac{\cos\left( \frac{\pi}{2}-\frac{6\pi}{14}\right)}{\cos \frac{\pi}{14}} =- \tfrac{1}{2}\frac{\cos\frac{\pi}{14}}{\cos \frac{\pi}{14}} = -\frac{1}{2}$$

Let $14x=\pi$

$$S=\sin3x-\sin x-\sin5x=\sin3x+\sin(-x)+\sin(-5x)$$

$$\sin(-2x)\cdot S=\cos5x-\cos7x$$

$$-\sin2x\cdot S=\cos5x=\sin2x$$

As $\cos7x=0,\cos5x=\sin2x$ as $5x+2x=\dfrac\pi2$

Can you take it from here?


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QUESTION: WHICH CLOSED HYPERSURFACES OF A FOUR-SPHERE HAVE ERGODIC GEODESIC FLOW?

Peter Sarnak and coworkers have extensively studies the ergodicity of geodesic flow on negatively curved manifolds but completely integrable systems such as spheres do not have this ergodicity with the standard metric.  Now Zelditch had proved that using the degeneracy of the eigenspaces of the Laplacian on $S^2$ and using the Haar measure on unitary matrices preserving the eigenspaces, there is generic ergodicity almost surely.  We happen to believe that the actual universe is deterministic and a four-sphere where the empirical material universe is an evolving subspace.  So another example of a positively curved manifold with ergodicity in geodesic flow is addressed according a 2014 paper referring to Burns and Gedeon, “Metrics with ergodic geodesic flows on spheres”

We are interested in the question of whether the evolution of a hypersurface along orthogonal geodesics for example where horizontal dynamics is the Hamiltonian dynamics has ergodic geodesic flow. Intuitively this we expect to be true, and if true we have some better sense of determinism and uncertainty in nature.


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# Cylinder with a horizontal tube (Fluid dynamics problem)

#### ValeForce46

Homework Statement
A container is made up of a vertical cylinder, diameter $D=9.0 cm$, on which is placed an horizontal tube, diameter $d=3.0 cm$, with axis at a distance $l=5.0 cm$ from the bottom of the cylinder. The other end of the horizontal tube is plugged and the container is filled with water until the height of $h=50 cm$ (see picture). Assuming that the surface is perfectly smooth (no friction), determine the value of the necessary force to keep the container still when the plug has been removed. [Advice: the lowering speed of the column of water should NOT be neglected.]
Homework Equations
Bernoulli's Principle: $p+ρgz+\frac{1}{2}ρv^2=constant$

I'll call $v_2$ the flow speed on the horizontal tube and $v_1$ the lowering speed of the column.
Even if the lowering speed isn't negligible, the problem says "...when the plug has been removed", so can I consider the height unchanged?
In order to find the force, I need the pressure when the plug is off. Is it $p'=p_{atm}+ρg(h-l)+\frac{1}{2}ρv_2^2$ ?
First thing I know is $v_1*A_1=v_2*A_2$ ($A_1$ and $A_2$ are the two section).
Second thing I know is $p_{atm}+\frac{1}{2}ρv_2^2=p_{atm}+ρg(h-l)+\frac{1}{2}ρv_1^2$ thanks to Bernoulli's principle.
So I can find $v_2$.
The force should be
$$F=p'*A=[p_{atm}+ρg(h-l)+\frac{1}{2}ρv_2^2]*(\frac{D}{2})^2*π$$
Is something wrong?

#### Attachments

• 24.2 KB Views: 19
Related Introductory Physics Homework Help News on Phys.org

#### haruspex

Homework Helper
Gold Member
2018 Award
Is it $p'=p_{atm}+ρg(h-l)+\frac{1}{2}ρv_2^2$ ?
I see no basis for such an equation. Can you justify it?

Having found v2, think about rate of change of momentum.

#### MrsZee

I am not sure about the accuracy of the EQ haruspex is pointing to.
But for sure you need to start with the pressure when the plug is ON.
After unplugging, everything will decrease with the time ( the pressure and both velocities ) , until there is no liquid going through pipe.

For the stability of the system, velocity out of the pipe in the un[lugging moment can be the only reason for container to NOT stay stable.

#### ValeForce46

I see no basis for such an equation. Can you justify it?
I thought that was the pressure to a certain height $(h-l)$ and flow speed $v_2$.
How do I find the force, having found v2? $F=\frac{mΔv}{Δt}=ρ*v_2*A_2*Δv$ because flow rate is constant?

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#### Chestermiller

Mentor
You can't consider the height unchanged. If the flow out the pipe is v2, what is the downward flow rate in the vertical tube v1 (given that the fluid can be considered incompressible)? From this, what does Bernoulli equation give you?

#### ValeForce46

You can't consider the height unchanged. If the flow out the pipe is v2, what is the downward flow rate in the vertical tube v1 (given that the fluid can be considered incompressible)? From this, what does Bernoulli equation give you?
The downward flow rate should be $v_1=v_2*\frac{A_2}{A_1}$
Bernoulli's principle gives me $p_{atm}+ρg(h-l-x)+\frac{1}{2}ρv_1^2=p_{atm}+\frac{1}{2}ρv_2^2$ if I can't consider the height unchanged...

#### Chestermiller

Mentor
The downward flow rate should be $v_1=v_2*\frac{A_2}{A_1}$
Bernoulli's principle gives me $p_{atm}+ρg(h-l-x)+\frac{1}{2}ρv_1^2=p_{atm}+\frac{1}{2}ρv_2^2$ if I can't consider the height unchanged...
They're only (tacitly) asking for the force F at short time when h has not changed significantly yet.

Also, in this problem, to get the force, you would find it easier to work in terms of gauge pressures and not absolute pressures.

#### MrsZee

I thought that was the pressure to a certain height $(h-l)$ and flow speed $v_2$.
How do I find the force, having found v2? $F=\frac{mΔv}{Δt}=ρ*v_2*A_2*Δv$ because flow rate is constant?
the flow is not constant; it will decrease as the fluid drains

#### haruspex

Homework Helper
Gold Member
2018 Award
I thought that was the pressure to a certain height $(h-l)$ and flow speed $v_2$.
How do I find the force, having found v2? $F=\frac{mΔv}{Δt}=ρ*v_2*A_2*Δv$ because flow rate is constant?
Sorry, in my cutting and pasting I picked out the wrong equation. The one I see no justification for is
$$F=p'*A=[p_{atm}+ρg(h-l)+\frac{1}{2}ρv_2^2]*(\frac{D}{2})^2*π$$
As I posted, when you have found v2, think about horizontal momentum.

But as @Chestermiller points out, your Bernoulli equation needs to allow for the downward velocity at the top of the tank.
It is a bit of a fudge. On the one hand they want you to treat the level of water in the tank as unchanged, on the other, not to treat it as unchanging. I.e. pretend that it instantly transits from static to steady state flow.

#### haruspex

Homework Helper
Gold Member
2018 Award
the flow is not constant; it will decrease as the fluid drains
Yes, but the question asks about the force after the level has dropped very little.

#### ValeForce46

As I posted, when you have found v2, think about horizontal momentum.
Why is this wrong? $F=\frac{mΔv}{Δt}=ρ∗v_2∗A_2∗Δv=ρ*A_2*v_2^2$
v1 is vertical so it doesn't affect horizontal momentum

#### haruspex

Homework Helper
Gold Member
2018 Award
Why is this wrong? $F=\frac{mΔv}{Δt}=ρ∗v_2∗A_2∗Δv=ρ*A_2*v_2^2$
v1 is vertical so it doesn't affect horizontal momentum
I didn't say that was wrong, but I omitted to say it was right.
You have all the pieces.

#### MrsZee

Yes, but the question asks about the force after the level has dropped very little.
which is the force when the plug is released; which will be the product of the cross sectional area of the pipe multiplied with the pressure at the pipe level

#### Chestermiller

Mentor
which is the force when the plug is released; which will be the product of the cross sectional area of the pipe multiplied with the pressure at the pipe level
This is not correct. The exit pressure from the pipe is zero gauge. Moreover, the fluid is being accelerated horizontally, so the tank is exerting a force to accelerate the fluid, and the fluid is exerting a corresponding reaction force on the tank. The OP's result is spot on, and captures this force correctly.

#### haruspex

Homework Helper
Gold Member
2018 Award
which is the force when the plug is released; which will be the product of the cross sectional area of the pipe multiplied with the pressure at the pipe level
No. That would be the force before the plug is removed.
The questioner wants us to consider the force when a more-or-less steady state has been achieved, on the assumption that the level has not declined much.
In practice, of course, the flow increases swiftly from 0, reaches a maximum, then declines as the level in the tank drops.
Usually, with this question, we are not even required to worry about the rate at which the level drops, but here we are advised to do so. Using Bernoulli’s equation and the incompressibility of water we get that the force is $\frac{2gh\rho a}{1-(\frac aA)^2}$.

#### MrsZee

potato-potato
the pressure before the plug is released is the one to create the force of the flow as the plug is released; split seconds; after the plug is released everything starts moving because of this force

we can agree to disagree
btw, the question is will the system trip over, or will it get to the steady state you are waiting to happen to start calculations

#### haruspex

Homework Helper
Gold Member
2018 Award
potato-potato
the pressure before the plug is released is the one to create the force of the flow as the plug is released; split seconds; after the plug is released everything starts moving because of this force

we can agree to disagree
btw, the question is will the system trip over, or will it get to the steady state you are waiting to happen to start calculations
Note the wording: "the necessary force to keep the container still when the plug has been removed."
The most reasonable interpretation is to find the maximum force that occurs subsequent to the unplugging. Your approach yields $\rho gha$. As I stated in post #15, Bernoulli's equation gives a little more than double that. However, I regard the denominator there as bogus. A full analysis (if I remember correctly) gives a peak value a bit less than $2\rho gha$.

#### Chestermiller

Mentor
potato-potato
the pressure before the plug is released is the one to create the force of the flow as the plug is released; split seconds; after the plug is released everything starts moving because of this force

we can agree to disagree
btw, the question is will the system trip over, or will it get to the steady state you are waiting to happen to start calculations
You omitted the part of the force necessary for accelerating the fluid horizontally, from zero horizontal velocity at the top of the tank to the horizontal velocity in the exit pipe. Before the plug is removed, the fluid is not being accelerated horizontally. This is the reason the correct answer is twice as high as your (incorrect) answer. The OP and haruspex analyzed the problem correctly. For more details on how to properly do this kind of calculation, see Bird, Stewart, and Lightfoot, Transport Phenomena, section on Macroscopic Momentum Balances.

"Cylinder with a horizontal tube (Fluid dynamics problem)"

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text:
# If license plates have three numbers and three letters in any order, how many unique license plates can there be?

Oct 13, 2015

$351520000$

#### Explanation:

There are 10 possible digits for the numbers (0,1,2,...,9), and 26 possible letters (A,B,C,.....,Z).

Since repetitions are allowed, we have that for each letter used, 26 still remain for the next choice, and for each digit used, 10 still remain for the next choice.

Hence, by the multiplication principle, the total different number of letter permutations is $26 \times 26 \times 26 = 17576$ and the total number of digit permutations is $10 \times 10 \times 10 = 1000$

But now since they can be in any order, we basically require in how many different ways can 3 be chosen from 6 which is the combination
""^6C_3=(6!)/((6-3)!3!)=20

So for each of the original permutations of numbers and letters, we must consider the 20 different combinations of positions they can be in.

Therefore by the multiplication principle, the total possible different number plates is
$17576 \times 1000 \times 20 = 351520000$


text:
# Zermelo's Theorem (Set Theory)

## Theorem

Every set of cardinals is well-ordered with respect to $\le$.

## Proof

Let $S_1$ and $S_2$ be sets which are not empty.

Suppose there exists an injection $f: S_1 \to S_2$ and another injection $g: S_2 \to S_1$.

Then by the Cantor-Bernstein-Schröder Theorem there exists a bijection between $S_1$ and $S_2$ and by definition $S_1$ is equivalent to $S_2$.

Let $\AA$ be the set of invertible mappings $\phi: A \to B$ where $A \subseteq S_1$ and $B \subseteq S_2$.

Since $S_1$ and $S_2$ are not empty, $\exists s_1 \in S_1$ and $\exists s_2 \in S_2$.

Thus we can construct the mapping $\alpha: \set {s_1} \to \set {s_2}$ such that $\map \alpha {s_1} = s_2$.

This is trivially an invertible mapping, so $\AA$ is not empty.

We can impose an ordering $\le$ on $\AA$ by letting $\phi_1 \le \phi_2 \iff \phi_1 \subseteq \phi_2$, that is, if $\phi_2$ is an extension of $\phi_1$.

Let $\CC$ be a chain in $\struct {\AA, \le}$.

Then $\ds \bigcup \set {\phi \in \CC}$ is an upper bound of every $\phi \in \CC$, and it lies in $\AA$.

The conditions of Zorn's Lemma are satisfied, so we can find a maximal element $M$ in $\AA$.

Let:

$M_1 = \set {s_1: \tuple {s_1, s_2} \in M}$
$M_2 = \set {s_2: \tuple {s_1, s_2} \in M}$

We have that $M_1 \subsetneq S_1$ and $M_2 \subsetneq S_2$ both together contradict the fact that $M$ is maximal element.

Thus either $M_1 = S_1$ or $M_2 = S_2$, and possibly both.

Thus either:

$M$ is an injection of $S_1$ into $S_2$

or:

$M^{-1}$ is an injection of $S_2$ into $S_1$.

Thus either $S_1 \le S_2$ or $S_2 \le S_1$, and the result follows.

$\blacksquare$

#### Axiom of Choice

This proof depends on the Axiom of Choice.

Because of some of its bewilderingly paradoxical implications, the Axiom of Choice is considered in some mathematical circles to be controversial.

Most mathematicians are convinced of its truth and insist that it should nowadays be generally accepted.

However, others consider its implications so counter-intuitive and nonsensical that they adopt the philosophical position that it cannot be true.

## Also known as

This is called by some authors the Trichotomy Problem.

## Source of Name

This entry was named for Ernst Friedrich Ferdinand Zermelo.


text:
# What does “supersingular” mean?

Are supersingular primes and supersingular elliptic curves related?

(this was essentially a subquestion in my earlier question, but still looks sufficiently different to me to deserve a separate post)

-

Let F be the finite field with p elements.

A supersingular elliptic curve is an elliptic curve E/F with the property that the endomorphism ring (ring of homomorphisms from E to E) of E over the algebraic closure of F_p is has rank 4 as a Z-module.

It is a theorem that End(E/F) has rank 2 or rank 4 (and that in the former case, End(E/F) is isomorphic to an order in an imaginary quadratic number ring whereas in the latter case, End(E/F) is isomorphic to an order in a quaternion algebra over Q). Hence, a supersingular elliptic curve over F_p can be thought of as an elliptic curve over F with a "big" endomorphism ring.

It is a theorem that another characterization supersingular elliptic curves is that E/F is supersingular is if and only if the number of points on E/F is exactly p + 1 (edit: for p > 3, see Voloch's comment below).

I believe that the etymology of the term "supersingular" is as follows: if you start with an elliptic curve E/Q over Q, for all but finitely many primes p, reduction (mod p) gives an elliptic curve E/F. For a generic E/Q (specifically, one without "complex multiplication") then the set of primes such that reduction (mod p) turns E/Q into a supersingular E/F has asymptotic density 0. Such primes are called "supersingular primes for E/Q" - supersingular refers to "really unusual." The reductions for such primes are then called supersingular elliptic curves over F. I'm pretty sure that every elliptic curve over F is a (mod p) reduction of an elliptic curve over Q so that all supersingular elliptic curves arise in this way.

I'll remark (following Silverman) that a supersingular elliptic curve over F is not "singular" in the sense of algebraic geometry - by definition all elliptic curves are nonsingular.

I do not know much about supersingular primes in the context of monstrous moonshine. According to Wikipedia, a supersingular prime is a prime that divides the order of the monster group; and there are 15 such primes. Given E/Q, by a theorem of Elkies there will be infinitely many super singular primes p for E/Q. So it's difficult to imagine how the list of 15 supersingular (with respect to moonshine) primes could emerge from the notion of "supersingular elliptic curve." I imagine that the etymology of the term "supersingular" in the context of moonshine is again that that supersingular primes are special - but that they special in a completely different way from the supersingular primes for a elliptic curve over Q.

-
"E/F is supersingular is if and only if the number of points on E/F is exactly p + 1." This is only true for p prime, p > 3. The correct statement is that the curve is supersingular if and only if the number of points is p+1-t and t is divisible by p. Since $|t| \le 2\sqrt{p}$, it follows that t = 0, when p is prime, p > 3. – Felipe Voloch Feb 15 '10 at 15:32
I don't know the actual etymology but I have thought it is as follows: In the (very) old days a singular (in the sense of special) modulus was the $j$-invariant of a lattice in $\mathbb C$ with complex multiplication. From the point of view of elliptic curves this corresponds to the $j$-invariants of complex elliptic curves with endomorphism ring larger than $\mathbb Z$. In positive characteristic there are some elliptic curves with larger endomorphism rings than complex curves can have, they are clearly even more special, hence supersingular. – Torsten Ekedahl Mar 21 '10 at 9:57

There are many equivalent ways to define supersingularity for an elliptic curve over a characteristic $p$ field. One of them is that the $p$-torsion of the curve is connected, i.e., it is a purely infinitesimal group scheme of order $p^2$. As Jonah mentioned, supersingular means very special, and is not a statement about smoothness. There is a theorem of Deuring that implies the j-invariant of a supersingular elliptic curve always lies in $\mathbb{F}_{p^2}$, and as a consequence, all such curves are defined over a finite degree extension of $\mathbb{F}_p$.

There are two notions of supersingular prime: one is relative to a fixed elliptic curve over $\mathbb{Q}$, and one is absolute. For any elliptic curve $E/\mathbb{Q}$, a prime $p$ is supersingular for $E$ if $E$ has good supersingular reduction at $p$. Such primes are known to be asymptotically density zero, but infinite in number (by a theorem of Elkies). Lang and Trotter conjectured that the number of supersingular primes less than $N$ limits to a constant times $\sqrt{N}/\log(N)$ as $N$ gets large.

Supersingular primes in the absolute sense are those primes $p$ for which all supersingular elliptic curves over an algebraic closure of $\mathbb{F}_p$ have $j$-invariant in $\mathbb{F}_p$ instead of just $\mathbb{F}_{p^2}$. These happen to be the primes that divide the order of the monster simple group, and they are also the primes for which the normalizer of $\Gamma_0(p) \in SL_2(\mathbb{R})$ acts on the complex upper half plane with a genus zero quotient. For general $p$, this normalizer contains $\Gamma_0(p)$ as an index 2 subgroup, with the nontrivial coset called the "Fricke involution" (a special case of Atkin-Lehner involultion). There is a standard order 2 representative, taking $z \mapsto -1/pz$. The quotient curve classifies unordered pairs of elliptic curves with dual degree p isogenies between them. I do not know any canonical relations between these characterizations of supersingularity.

Edit: Thanks to Emerton for pointing out the connection. I'll try to expand on it a bit. The moduli problem of generalized elliptic curves with $\Gamma_0(p)$-structure has a coarse moduli space that is a smooth irreducible curve away from p, but has mod p fiber given by taking a disjoint union of two copies of $X(1)$ (a genus zero curve) and gluing along supersingular points (this description is more or less in Katz-Mazur, chapter 13). A geometric point describing an elliptic curve with j-invariant $\alpha \in \mathbb{F}_{p^2}$ is glued to a geometric point on the other irreducible component describing an elliptic curve with j-invariant $\alpha^p$. The Fricke involution switches the components, so the quotient of $X_0(p)$ by this involution is a genus zero curve glued to itself at finitely many supersingular points. The quotient has arithmetic genus zero if and only if all supersingular geometric points are glued to themselves - otherwise, the flat modular deformation to characteristic zero yields a smooth curve of higher genus. In other words, it is necessary and sufficient that all supersingular geometric points have no Frobenius conjugates, i.e., that the j-invariants of all supersingular curves lie in $\mathbb{F}_p$.

More Edit: I should give a more honest reply to Mariano's question, which was originally raised by Ogg in the mid 1970s (and he famously offered a bottle of Jack Daniels to anyone who could solve it). Half of the question has an answer. If we combine the results of Borcherds's paper Monstrous moonshine and monstrous Lie superalgebras with the results of the paper Modular equations and the genus zero property of moonshine functions by Cummins and Gannon, we get the following fact:

Let G be a finite group acting faithfully on a conformal vertex algebra V by conformal symmetries, and suppose V has central charge 24 and character $\operatorname{Tr}(q^{L_0-1}|V) = j(\tau)-744$. Then for any element $g \in G$, the series $\operatorname{Tr}(gq^{L_0-1}|V)$ is the q-expansion of a modular function that is holomorphic on the upper half plane, invariant under a discrete group $\Gamma \subset PSL_2(\mathbb{R})$ satisfying $\Gamma \supset \Gamma_0(N)$ for some $N$, and generates the function field of the quotient curve $\Gamma \backslash \mathbf{H}$. In particular, the quotient curve is genus zero.

The monster simple group arises in this context because I. Frenkel, Lepowsky, and Meurman constructed a conformal vertex algebra satisfying the above hypotheses, whose group of conformal automorphisms is the monster. The fact given above implies that for each prime p dividing the order of the monster, the quotient $X_0^+(p) = X_0(p)/\langle w_p \rangle$ is genus zero, where $w_p$ is the Fricke involution. In particular, each prime dividing the order of the monster is necessarily supersingular.

I know some people who would like there to be a conceptual (read: non-enumerative) explanation for why all supersingular primes divide the order of the monster. So far, the best I've heard is that the monster is really big, while there aren't that many supersingular primes.

-
What is the reason the primes which divide the order of the monster simple group show up here? – Mariano Suárez-Alvarez Dec 9 '09 at 12:55
I don't think a good answer is known. – S. Carnahan Dec 9 '09 at 18:18
Dear Scott, I'm not sure if this is what you referred to in the last sentence of your answer, but it is not too hard to prove that $p$ is a supersingular prime if and only if $X_0(p)$ mod the A-L involution has genus zero directly from the description of the bad fibre of $X_0(p)$ mod $p$. (The point being that this description directly involves the s.s. elliptic curves mod p, and the action of Frobenius on their j-invariants.) – Emerton Jan 9 '10 at 20:55
Thank you for pointing that out. Somehow the phrases "AL involution" and "moduli problem at p" never made it into the same part of my brain at the same time. I will edit the answer when I understand the bad fiber a little better. – S. Carnahan Jan 12 '10 at 17:31


text:
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# Volume of the cylinder segment

Calculates volume of the cylinder segment

Our user asked to make a cylinder segment volume calculator. I imagine the segment like this - see picture

Accordingly, we set four parameters and obtain the volume segment. If the inner radius is set 0 and the number of degrees is 360, we will calculate the volume of cylinder and so on.

The formula is simple - the area difference of the segments obtained by the first and the second radius, multiplied by the height.
Segment area$S=(\frac{\alpha}{360})\pi R^2$

#### Volume of the cylinder segment

Digits after the decimal point: 2
Segment volume

#### Similar calculators

PLANETCALC, Volume of the cylinder segment


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# Magnetic domains and domain wall pinning in atomically thin CrBr3 revealed by nanoscale imaging

## Abstract

The emergence of atomically thin van der Waals magnets provides a new platform for the studies of two-dimensional magnetism and its applications. However, the widely used measurement methods in recent studies cannot provide quantitative information of the magnetization nor achieve nanoscale spatial resolution. These capabilities are essential to explore the rich properties of magnetic domains and spin textures. Here, we employ cryogenic scanning magnetometry using a single-electron spin of a nitrogen-vacancy center in a diamond probe to unambiguously prove the existence of magnetic domains and study their dynamics in atomically thin CrBr3. By controlling the magnetic domain evolution as a function of magnetic field, we find that the pinning effect is a dominant coercivity mechanism and determine the magnetization of a CrBr3 bilayer to be about 26 Bohr magnetons per square nanometer. The high spatial resolution of this technique enables imaging of magnetic domains and allows to locate the sites of defects that pin the domain walls and nucleate the reverse domains. Our work highlights scanning nitrogen-vacancy center magnetometry as a quantitative probe to explore nanoscale features in two-dimensional magnets.

## Introduction

The discovery of atomically thin van der Walls (vdW) magnetic materials enables fundamental studies of magnetism in various spin systems in the two-dimensional (2D) limit1,2. Advantages, such as easy fabrication and a wide variety of control mechanisms3,4,5,6,7,8,9, make the vdW magnets and their heterostructures promising candidates for next-generation spintronic devices. This combination of fundamental and technological interest has motivated both the search for new room-temperature vdW magnets and the investigation of the mechanisms determining magnetic properties of already discovered materials. The vdW magnets have been intensively studied at micrometer scale by several probe techniques, such as magneto-optical Kerr effect microscopy3,4, magnetic circular dichroism microscopy6,10, anomalous hall effect11,12, and electron tunneling8,13,14. Despite many significant results, these methods have limited spatial resolution due to the laser diffraction limit and electrode size. The nanoscale features of the atomically thin vdW magnets, such as magnetic domains15,16,17 and topological spin textures18,19,20,21, are largely unexplored. For example, magnetic domains in layered CrBr3 have been predicted from its anomalous hysteresis loop in magneto-photoluminescence and micro-magnetometry measurements15,16, but the magnetic domain structure and its evolution has not been detected in real space. As a ferromagnetic insulator, CrBr3 provides a unique spin system to study the ferromagnetism and spin fluctuation in the 2D limit10,15,16, and is a building block of vdW heterostructures22,23,24,25. Quantitatively studying the magnetic properties at the nanoscale would allow a better understanding of the material and benefit device design.

Many high-spatial-resolution magnetic imaging techniques, such as magnetic force microscopy and Lorentz transmission electron microscopy, have been successfully used to study magnetic thin film materials. However, recent works show that it remains challenging to probe atomically thin vdW magnets with them due to the weak signal level17,20. Scanning superconducting quantum interference device can achieve very high magnetic field sensitivity even with a probe diameter of ~50 nm (refs. 26,27). The method works well at temperatures below a few Kelvin, but the work temperature range is limited by the low-temperature superconducting material of the probe. In contrast, the negatively charged nitrogen-vacancy (NV) center has been demonstrated as a high sensitivity magnetometer with operational temperatures from below one to several hundreds of Kelvin28, which is suitable to probe most of the discovered vdW magnets. Scanning magnetometry combining atomic force microscopy and NV center magnetometers allows for quantitative nanoscale imaging of magnetic fields, and has been well established in room-temperature measurements29,30,31,32,33,34. Recently, its cryogenic implementation has been demonstrated35,36 and has already been employed to image the magnetization in layered CrI3 (ref. 37). In these works, the magnetic field is measured using the continuous wave (cw) optically detected magnetic resonance (ODMR) scheme, which is robust but still has some limitations. On the one hand, the simultaneously applied laser and microwave results in power broadening for the electronic spin resonance linewidth, which degrade the sensitivity of the NV magnetometer38. On the other hand, the heating effect of cw-microwave can increase the sample temperature by a few Kelvin37, which is unfavorable for probing magnetic phenomena sensitively depending on the temperature19,21,39.

In this work, we demonstrate cryogenic scanning magnetometry using a single NV center in a diamond probe. We show that by using a pulsed measurement scheme, the optimal magnetic field sensitivity that can be achieved with a commercial diamond probe is ~0.3 $$\upmu {{\rm{THz}}}^{-\frac{1}{2}}$$, and the microwave heating is significantly reduced. With this setup, the magnetization of few-layer CrBr3 samples is quantitatively studied, and the domain structure in a CrBr3 bilayer is imaged in real space. We also study the magnetic domain evolution, and observe the domain wall pinning and reverse domain nucleation at defect sites.

## Results

### Cryogenic scanning NV magnetometry

The schematics of the scanning NV magnetometry setup is depicted in Fig. 1a. The NV center is implanted in the apex of a pillar etched from a diamond cantilever, which is attached to the tuning fork of an atomic force microscope (AFM). We optically detect the NV spin state through the diamond cantilever supporting the pillar (see “Methods”). To efficiently apply microwave-driven field to the NV center, we deposit a coplanar waveguide on a SiO2/Si substrate and transfer the sample under test to one gap of it. In this work, the CrBr3 samples are encapsulated with hexagonal boron nitride (hBN) on both sides, and transferred in a glovebox filled with pure nitrogen (see “Methods” for details of sample preparation). A heater and resistive thermometer is placed near the sample to control/monitor the sample temperature. The microscope head is suspended in an insertion tube filled with helium buffer gas, which is dipped in the liquid helium cryostat equipped with a set of vector superconducting coils. In the measurement, the AFM operates in a frequency modulation mode with a tip oscillation amplitude of ~1.5 nm. The topography of the sample is obtained from the AFM readout with a lateral spatial resolution of ~200 nm, which is limited by the diameter of the pillar apex. A typical in situ AFM image of part of the CrBr3 sample is shown in Fig. 1b. Figure 1c shows the height of the step along the vertical direction by an average of the data in the dashed box in Fig. 1b. The 2 nm step height indicates a bilayer sample.

The stray magnetic field is mapped by measuring the electronic spin resonance spectrum via the pulsed ODMR scheme38 at each pixel. The measurement sequence is shown in Fig. 1d. Instead of the cw-laser and -microwave, we apply short laser pulses and π pulse microwave to the NV center. The ODMR curve is obtained by sweeping the microwave frequency, which is achieved by modulating a microwave signal with pairs of sinusoidal signals in quadrature via an IQ-mixer. All the control signals of the measurement sequences are generated using an arbitrary waveform generator so that the measurement sequence is well synchronized, and fast microwave frequency sweeping is realized by altering the frequency of the sinusoidal signals in each unit segment. In our experiment, we use commercial diamond probes (Qnami) with NV center formed from implanted 14N atom. Each spin transition from the $$\left|{m}_{\mathrm{s}}=0\right\rangle$$ to $$\left|{m}_{\mathrm{s}}=\pm 1\right\rangle$$ states includes three sublevels, resulting from the hyperfine interaction with the nuclear spin of the 14N atom. The optimal sensitivity can be achieved by generating three pairs of sinusoidal signals to simultaneously excite the three hyperfine split transitions and detecting at the steepest slope of the ODMR resonance line. With a simple optimization, the sensitivity estimated from the ODMR curve is ~0.3 $$\upmu {{\rm{THz}}}^{-\frac{1}{2}}$$ (see Supplementary Information for detail). This pulsed scheme also significantly decreases microwave heating compared with cw-ODMR37. In our experiment, the sample temperature only increases by a few hundred milli-Kelvin from the base temperature of ~4.2 K during measurements. Moreover, the magnetic field is measured via microwave pulses, which are applied 600 ns after the laser beam has been switched off. Thus, the measured stray magnetic field is not disturbed by laser-induced excitations, such as spin waves40,41.

### Magnetization and magnetic domains of bilayer CrBr3

The stray magnetic field is obtained by fitting the ODMR curve with a Lorentzian lineshape. Figure 2a shows a typical stray magnetic field image of a CrBr3 bilayer under a 2 mT external magnetic field after being cooled down under zero field. The external magnetic field is used to split the energy levels $$\left|{m}_{\mathrm{s}}=\pm 1\right\rangle$$ so that the direction of the stray magnetic field can be determined. The external magnetic field direction is set parallel to the NV axis in all measurements in this work to avoid the mixing of spin states due to an off-axis magnetic field component42. The axis of all the NV centers in the (100)-oriented diamond cantilevers we used here is ~54.7 with respect to the vertical direction, as shown in Fig. 1a. The pixel size in this work is set to 30 nm and the data accumulation time is 2 s at each pixel. The resulting stray magnetic field map clearly shows magnetic domains with prominent positive and negative values of the magnetic field and domain walls with nearly zero fields. To reveal further details, we reconstruct the magnetization from the stray magnetic field using reverse-propagation protocol34,37,43,44.

As discussed in more detail in the Supplementary Information, to uniquely determine the magnetization, we need some initial knowledge of the sample, such as the direction of spin polarization. It has been reported that few-layer CrBr3 has an out-of-plane easy axis and can be polarized by a small external magnetic field of ~4 mT, while a relatively high external field ($${B}_{\parallel }^{c}\approx 0.44$$ T) is required to polarize the spins in the in-plane direction45. The in-plane external field component in this work is much lower than the critical field $${B}_{\parallel }^{c}$$, allowing us to use the assumption of out-of-plane magnetization in the magnetization reconstruction. In addition, we neglect the finite thickness of the domain walls. Figure 2b presents the magnetization image reconstructed from the stray magnetic field image in Fig. 2a. It clearly shows the magnetic domain structure, with positive (negative) values, indicating the magnetization direction parallel (antiparallel) to the external magnetic field. The sample can be polarized by increasing the external magnetic field. Figure 2c shows a magnetization image taken at 11 mT external magnetic field. The common areas in Fig. 2b, c are marked with dashed boxes. The saturation magnetization can be estimated using the magnetization statistics of the two magnetization images, as shown by the histograms in Fig. 2d, e. Due to the sample imperfection, measurement error, and truncation error in the reconstruction, the reconstructed magnetization is distributed in a range around the zero-magnetization and the saturation magnetization values. The near-zero-magnetization pixels are mostly in domain walls, defects, and the non-sample area on the left part of the images. The saturation magnetization values are ~26(−28) and ~26 μBnm−2, respectively, with μB the Bohr magneton. These values are close to the 3 μB saturation moment per Cr3+ ion in CrBr3 at 0 K, i.e., ~ 32 μBnm−2 for a CrBr3 bilayer46.

### Domain wall pinning effect

In addition to elucidating the magnetic domain structure of 2D magnets, our scanning magnetometry measurements enable a more detailed study of coercivity mechanisms in these systems. A multi-domain ferromagnet typically reverses its magnetization direction through processes, such as nucleation of reverse domains and their growth through domain wall motion47,48. Defects in the material alter the energy of the magnetic domain walls and hence affect domain wall motion. This behavior can be demonstrated by taking magnetization images, while varying the external magnetic field. Figure 3a–g shows magnetization images obtained while increasing the field from 2 to 6 mT after the sample is thermally demagnetized and cooled down under zero field. The area of positive (negative) domains grows (shrinks) with increasing field, as the domain walls move toward the negative domains. Before entirely disappearing, the negative domain size becomes very small, and only near-zero-magnetization spots of several tens of nanometer diameter are revealed in the magnetization images. As we are limited by the spatial resolution of the NV center (~80 nm in this work, determined by the distance between the NV center and the sample for the diamond probe), we cannot obtain the detailed magnetization pattern inside the spots. These spots are usually associated with defects, which alter the local switching field. Domain walls are pinned by these defects (see Fig. 3a–i, solid and dashed yellow circles). To confirm the pinning sites, we compare the magnetization images at 4 and 6 mT for different thermal cycles (dashed box in Fig. 3e, g and h, i). Though the magnetic domain structures are different upon successive thermal cycles, the positions of pinning sites are reproducible.

To verify that the pinning effect is a dominant coercivity mechanism, we extract the initial magnetization curve of the thermally demagnetized sample by estimating its average magnetization as $$M/{M}_{\mathrm{sat}}=\frac{{N}_{+}-{N}_{-}}{{N}_{+}+{N}_{-}}$$, where Msat is the saturation magnetization and N+ (N) is the number of pixels with evident positive (negative) magnetization (absolute value >10 μBnm−2, according to the Fig. 2e). When pinning effects are negligible, the magnetic domain walls can move freely, resulting in a high initial magnetization even with a small external magnetic field. Defects, however, increase the energy barrier for the displacement of domain walls. In samples with a large defect density, the magnetization increases slowly until the external magnetic field is large enough to overcome the pinning energy. The initial magnetization curve shown in Fig. 3j is measured at external magnetic fields >2 mT. We extend the curve to the origin using B-spline interpolation, assuming zero magnetization in a thermally demagnetized sample. The average permeability is very low when the field is <2 mT and it significantly increases when the field is >2 mT (see the blue bars in Fig. 3j), which is consistent with the behavior of a pinning effect dominated initial magnetization.

We use a similar approach to measure the magnetic hysteresis loop. Figure 4a shows the magnetization image at an external magnetic field of 2 mT after thermal demagnetization. We choose the area marked by the dashed box (#1) to analyse the magnetization, as we cycle the external magnetic field to saturate the magnetization in both the positive and negative z direction. Representative magnetization images are shown in Fig. 4b–i (see Supplementary Information for all images). Remarkably, a magnetic domain with the same irregular shape, but opposite magnetization direction is observed in Fig. 4e, i. The border of the magnetic domain (#2) is denoted by the dashed line. The hysteresis loop of the domain #2 is nearly rectangular, as shown by the red curve in Fig. 4j. There are several defects around the domain #2. Some of these defects act as nucleation centers of the reversed domains that appear immediately, when the external field is reduced from the saturation field. The domain walls move as the reversed domain grows and some of domain walls are finally pinned at the border of the magnetic domain (#2), which might be a grain boundary. Therefore, the hysteresis loop averaged over area #1, which is represented by the black curve in Fig. 4j, indicates a lower coercive field.

## Discussion

To verify the observations, we reproduce the demagnetized ground state using a micromagnetic simulation with parameters within the ranges estimated from measurement of CrBr3 bulk49 (see “Methods” and Supplementary Information). This shows that the domain sizes and parameters and so on are all plausible. However, the micromagnetic simulation does not factor in the pinning effect, and therefore does not capture the reversal process itself. From this we can deduce that the magnetization (reversal) process is strongly dependent on the defects of the sample. In turn, this makes nanoscale magnetic imaging even more crucial, because the behavior cannot be predicted form simple simulation.

We also observe similar magnetic domain structures and domain wall pinning in other few-layer CrBr3 samples, as presented in the Supplementary Information. The domain evolution measurement in a three-layer and four-layer sample indicates nearly rectangular hysteresis loops with higher coercive fields than domain #2 in the bilayer sample. Indeed, the influence of laser heating need to be determined as in previous works using NV center31,50. We show that the laser heating effect is negligible in this work by measuring both the domain structure and magnetization at various laser power in a three-layer CrBr3 sample.

In conclusion, we demonstrate a cryogenic scanning magnetometer based on a single NV center in a diamond probe with a pulsed measurement scheme. We show that our setup can achieve a high sensitivity and significantly reduce the microwave heating. Using this setup, we have studied magnetic domains in few-layer samples of CrBr3 by quantitatively mapping the stray magnetic field. The magnetization of bilayer CrBr3 is determined, and the magnetic domain evolution is observed in real space. We show that domain wall pinning is the dominant coercivity mechanism by observing the evolution of both the individual magnetic domains and the average magnetization, with changing external magnetic field. Our approach is also compatible with other pulsed measurement sequences that can be used to detect electronic spin resonance51, nuclear magnetic resonance32, and spin waves52 in the 2D magnetic materials.

## Methods

### Sample fabrication

The hBN flakes of 10–30 nm were mechanically exfoliated onto 90 nm SiO2/Si substrates, and examined by optical and atomic force microscopy under ambient conditions. Only atomically clean and smooth flakes were used for making samples. A V/Au (10/200 nm) microwave coplanar waveguide was deposited onto an 285 nm SiO2/Si substrate, using standard electron beam lithography with a bilayer resist (A4 495 and A4 950 polymethyl methacrylate) and electron beam evaporation. CrBr3 crystals were exfoliated onto 90 nm SiO2/Si substrates in an inert gas glovebox with water and oxygen concentration <0.1 p.p.m. The CrBr3 flake thickness was identified by optical contrast and atomic force microscopy. The layer assembly was performed in the glovebox using a polymer-based dry transfer technique. The flakes were picked up sequentially: top hBN, CrBr3, and bottom hBN. The resulting stacks were then transferred and released in a gap of the pre-patterned coplanar waveguide. In the resulting heterostructure, the CrBr3 flake is fully encapsulated on both sides. Finally, the polymer was dissolved in chloroform for <5 min to minimize the exposure to ambient conditions.

### Confocal microscope

The optics of the confocal microscope consists of the low-temperature objective (Attocube LT-APO/VISIR/0.82) with 0.82 numerical aperture and home-built optics head. The 515 nm excitation laser generated by an electrically driven laser diode is transmitted to the optics head through a polarization maintaining single-mode fiber and collimated by an objective lens. A pair of steering mirrors are used to align the beam for perpendicular incidence to the center of the objective. The NV center’s fluorescence photons are also collected by the objective and transmitted through the same free-beam path to the optics head. The collected fluorescence photons are separated from the green laser beam via a dichroic mirror and then passed through a band-pass filter to further decrease background photons. Finally, the photons are coupled to a single-mode fiber and detected by a fiber-coupled single-photon detector.

### Stray magnetic field measurement

With an external magnetic field applied along the NV axis, B, the spin states $$\left|{m}_{\mathrm{s}}=0\right\rangle$$ and $$\left|{m}_{\mathrm{s}}=\pm 1\right\rangle$$ exhibit Zeeman splitting of f = Ds ± γeB, where Ds ≈ 2.87 GHz is the zero field splitting between levels $$\left|{m}_{\mathrm{s}}=\pm 1\right\rangle$$ and $$\left|{m}_{\mathrm{s}}=0\right\rangle$$, and γe = 28 GHzT−1 is the electronic spin gyromagnetic ratio. In the pulsed ODMR measurement, the NV center is optically initialized in spin state $$\left|{m}_{\mathrm{s}}=0\right\rangle$$. After a delay of τ = 600 ns, a π-pulse (~80 ns) of microwave radiation is applied. If the microwave frequency is in resonance with one of the transitions, the NV center is driven to spin state $$\left|{m}_{\mathrm{s}}=\pm 1\right\rangle$$. The population difference between the spin states $$\left|{m}_{\mathrm{s}}=0\right\rangle$$ and $$\left|{m}_{\mathrm{s}}=\pm 1\right\rangle$$ can then be optically readout via fluorescence contrast. In our experiment, the laser pulse duration is 600 ns, and only the first 400 ns of the fluorescence photon signal is used in the data analysis. Details on the NV center ground-state energy levels and sensitivity can be found in Supplementary Information.

### Micromagnetic simulation

To complement the experimental results, micromagnetic simulations of the systems magnetic ground state were conducted using MuMax3 (ref. 53). Saturation magnetization was set to Ms = 270 kAm−1, which is in accordance to the values measured here and reported in ref. 49. Uniaxial magnetic anisotropy constant was assumed to be Ku = 86 kJm−3 along the normal axis, which has been reported for bulk material49. A global exchange stiffness constant Aex was first roughly estimated from Curie temperature and experimentally observed domain wall width to lie within 10−12–10−14 Jm−1. Magnetization was initialized in a random configuration and then relaxed to the minimum energy state at zero external field. This was done for varying values of Aex from the interval estimated above.

The results for the normal magnetization component are shown in the Supplementary Information for a system trying to locally approximate the irregular shape of the real sample. This simulation assumed Aex = 3 × 10−13 Jm−1, which resulted in the closest match for the experiment. The simulation shows a domain structure that is qualitatively very similar to the experimental observations, which are further supported by this result. However, due to the limited capacity of the simulation to account for structural defects and pinning effects, the hysteresis behavior observed experimentally could not accurately be recreated by it. This further supports the conclusion that pinning is a major factor in this materials hysteresis.

## Data availability

The data that support the findings of this study are available from the corresponding author upon reasonable request

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## Acknowledgements

The authors thank Dr. Patrick Maletinsky, Dr. Felipe Favaro de Oliveira, and Dr. Durga Dasari for the fruitful discussion, and Dr. Thomas Oeckinghaus and Dr. Roman Kolesov for the help with the experiment. J.W. acknowledges the Baden-Württemberg Foundation, the European Research Council (ERC; SMel grant agreement no. 742610). R.S. thanks the EU ASTERIQS. The work at U. Washington is mainly supported by DOE BES DE-SC0018171. Device fabrication is partially supported by AFOSR MURI program, grant no. FA9550-19-1-0390. The authors also acknowledge the use of the facilities and instrumentation supported by NSF MRSEC DMR-1719797.

## Funding

Open Access funding enabled and organized by Projekt DEAL.

## Author information

Authors

### Contributions

J.W. and X.X. supervised the project. Q.-C.S., A.B. and R.S. built the experimental setup. Q.-C.S. performed the experiment. T. Song and E.A. fabricated and characterized the samples. T. Shalomayeva provided experimental assistance. J.F. and J.G. performed the micromagnetic simulation. T.T. and K.W. provided and characterized bulk hBN crystals. Q.-C.S., T. Song, E.A., R.S., J.W. and X.X. wrote the paper with input from all authors. All authors discussed the results.

### Corresponding authors

Correspondence to Qi-Chao Sun or Rainer Stöhr.

## Ethics declarations

### Competing interests

The authors declare no competing interests.

Peer review informationNature Communications thanks the anonymous reviewer(s) for their contribution to the peer review of this work. Peer reviewer reports are available.

Publisher’s note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

## Rights and permissions

Reprints and Permissions

Sun, QC., Song, T., Anderson, E. et al. Magnetic domains and domain wall pinning in atomically thin CrBr3 revealed by nanoscale imaging. Nat Commun 12, 1989 (2021). https://doi.org/10.1038/s41467-021-22239-4

• Accepted:

• Published:

• DOI: https://doi.org/10.1038/s41467-021-22239-4

• ### Nanoscale magnetic field imaging for 2D materials

• Estefani Marchiori
• Lorenzo Ceccarelli
• Martino Poggio

Nature Reviews Physics (2022)

• ### Magnetization dependent tunneling conductance of ferromagnetic barriers

• Zhe Wang
• Ignacio Gutiérrez-Lezama
• Alberto F. Morpurgo

Nature Communications (2021)


text:
# $\int\sec^3x\,dx$; the integral of a function raised to some power is equal to a fraction of the sum of its integral and its derivative.

While looking at the solution

$$\int\sec^3x\,dx \,=\, \frac 12\ln|\sec x + \tan x|\,+\,\frac 12 \sec x \tan x+C\,,$$ from integration by parts, I see that $\frac d{dx}\sec x$ is formed by both $u=\sec x$ and $v=\tan x$ when we let $\int\sec^3x\,dx=\int u\,dv$ so $$\int\sec^3x\,dx \,=\, \frac 12\left(\int\sec x\,dx\,+\,\frac d{dx}\sec x\right)+C\,.$$

I find that this is a simple way to remember the result from now on (and avoid having to evaluate the integral to convince myself) but I wonder why such a nice result happens with this integral and

is it trivial that the integral of a function raised to some power is equal to a fraction of the sum of its integral and its derivative?

Why does this happen with this particular integral and are there other interesting examples of similar behavior?

Thanks for the input!

• Just for clarity, you're asking if/when $uv = u^\prime$? – Will Craig Apr 11 '17 at 19:05
• @WillCraig I mean that the integral of the function cubed is equal to half the integral of the function plus half the derivative of the function. Now that I wrote it this way I see that it sound much better, I will edit the question I agree that it is not quite clear right now, thank you. – user409521 Apr 11 '17 at 19:53
• Oh I see that now. That's much more interesting. – Will Craig Apr 11 '17 at 20:26
• This a product of the over all method known as the reduction formula for secant, you can actually work out the more general case. There is in fact a whole Wikipedia article written specifically for this particular integral en.m.wikipedia.org/wiki/Integral_of_secant_cubed – Triatticus Apr 11 '17 at 20:39

Following a similar pattern, $$\int\csc^3x dx=\frac 12\left(\int csc x dx+\frac{d}{dx}\csc x\right)$$

On a similar theme, $$\int \operatorname{sech}^3x dx=\frac 12\left(\int\operatorname{sech} xdx\color{red}{-}\frac {d}{dx}\operatorname{sech} x\right)$$

This question appears to be asking when $uv = u^\prime$. We can solve this as follows (we let $u = u(t), v = v(t)$);

$$uv = \dfrac{du}{dt} \implies v dt = \dfrac{du}{u} \implies \int vdt = \int \dfrac{du}{u} \implies \ln{|u|} + C = \int vdt$$

This suggests that such convenient properties occur whenever the antiderivative of $v$ is expressible as a logarithm. In the case given, we have that $\int \tan{t}dt = -\ln{|cos{x}|} + C$. I am not aware of any other integrals that work this way that are not either trivial ($v = 1/t$) or essentially the same as this ($u = \csc{t}, v = \cot{t}$) but if anyone else is aware of some I would love to see!

Taking the derivative of both sides of $$\int\sec^3(x)\ dx = \frac{1}{2}\left(\int\sec(x)\ dx+\frac{d}{dx}\sec(x) \right)$$ shows that $\sec(x)$ satisfies the differential equation: $$y^3 = \frac{1}{2}(y+y'').$$ Now, we can are in a position to generalize this; we can search for solutions to the differential equation $$y^n=\frac{1}{k}(y+y'')\quad\Leftrightarrow\quad y'' = ky^n-y = f(y)$$ for $k>0$ and $n\in\mathbb{Z}_{\geq 0}$. This is a second order autonomous ODE, and has the following implicit solution: $$K\pm x = \int\frac{1}{\sqrt{\displaystyle C+2\int f(y)\ dy}}\ dy = \int\frac{1}{\sqrt{\frac{2k}{n+1}y^{n+1}-y^2+C}}\ dy$$ for $C$ and $K$ arbitrary constants. In the case where $n=3$, $k=2$, $C=K=0$, and we take positive $x$ on the left we have the solution $$x = \int\frac{1}{\sqrt{y^4-y^2}}\ dy = \int\frac{1}{|y|\sqrt{y^2-1}}\ dy = \operatorname{arcsec}(y)+A$$ and therefore $y=\sec(x-A)$. When $K=0$, the left hand side is $-x$, and $n=3$, $k=2$, and $C=0$, we have $$x = \int\frac{-1}{\sqrt{y^4-y^2}}\ dy = \int\frac{-1}{|y|\sqrt{y^2-1}}\ dy = \operatorname{arccsc}(y)+A$$ and so $y=\csc(x-A)$. Taking $f(y) = y-ky^n$ will give you things like David Quinn's second example. Maybe the masters of integration around here will have more to say about integrals of the form: $$\int\frac{1}{\sqrt{\frac{2k}{n+1}y^{n+1}-y^2+C}}\ dy.$$

• Very interesting, thanks for the insight! – user409521 Apr 12 '17 at 15:46
• @InfiniteMonkey Happy to help! – user171308 Apr 12 '17 at 16:08

We can consider the reduction formula:

$$I_n=\int sec^n(x)dx \ s.t. \ n>2$$

Expressing $sec^nx=sec^{n-2}x\cdot sec^2x$ we have:

$$I_n=\int (sec^{n-2}(x)sec^2(x))dx$$

Integrating by parts, and setting $f=sec^{n-2}(x)$ and $g'=sec^2(x)$,

$$I_n=sec^{n-2}(x)tan^2(x)-(n-2)\int (sec^{n-2}(x)tan^2(x))dx$$

$$=sec^{n-2}(x)tan^2(x)-(n-2)\int (sec^{n-2}(x)(sec^2(x)-1))dx$$

$$=sec^{n-2}(x)tan^2(x)-(n-2)(I_n-I_{n-2})$$

Solving for $I_n$, we get:

$$I_n=\frac {sec^{n-1}(x)sin(x)}{n-1}+\frac {n-2}{n-1}I_{n-2}$$

So as $n$ increases, we get a summation of terms including only $sec^{n-2}(x)tan(x)$ and integrals of $sec$, two degrees below the previous integration.


text:
Challenging Distance, Speed, and Time Word Problem

Two people, $$A$$ and $$B$$, walk from P to Q and back. $$A$$ starts 1 hour after $$B$$, overtakes him 2 miles from Q, meets him 32 minutes later, and arrives at P when $$B$$ is 4 miles off. Find the distance from P to Q.

I really cannot think of anything to solve this problem. I tried drawing a diagram to model the situation, but that didn't seem to help very much. The problem I am having is translating the English to mathematical language; which is really what makes me struggle with word problems in general.

Even if I could just have the equations to describe the problem, that would help tremendously; my problem is forming the equations, but solving them is easy for me. If I could also get some tips for how to form equations based on the word problems, that would be much help. Thanks.

• Please say more about your attempts at solving the problem. What you said is too vague to really see what you are thinking. Also: Is one to assume that each walker goes at a constant rate? (If not, the question seems unanswerable.) Mar 23 at 1:28
• @paw88789 I think the question is assuming they are going at a constant rate. Mar 23 at 1:49
• @boojum would B's rate be $\frac{x}{92}$ ? Mar 23 at 1:56
• I guess I had thought A and B met at Q due to the "meets him 32 minutes later" line; since the only way they can meet after A has overtaken B is if A stops for B. Mar 23 at 2:08
• So the total time B went from P to Q is constituted by 60 minutes, the time it took for A to overtake B, and the other 32 minutes? Mar 23 at 2:11

Let $$d$$ be the distance from P to Q. Let $$r_a$$ and $$r_b$$ be the walking rates of A and B respectively (in miles per hour).

The key facts are:

(1) B takes one hour longer than A to walk $$d-2$$ miles.

(2) In 32 minutes, the sum total distance walked by A and B is $$4$$ miles.

(3) B takes one hour longer to walk $$2d-4$$ miles than it takes A to walk $$2d$$ miles.

From (1) you get (4): $$\frac{d-2}{r_b}=\frac{d-2}{r_a}+1$$.

From (2) you get (5): $$\frac{32}{60}\cdot(r_a+r_b)=4$$

From (3) you get (6): $$\frac{2d-4}{r_b}=\frac{2d}{r_a}+1$$

This gives you three equations in three unknowns for you to solve, which you said would be the easy part for you. [It may be helpful to note that the left side of equation (6) is exactly twice the left side of equation (4).]


text:
# Bivariate transformation using CDF method

1. Dec 3, 2015

### rabbed

If I have the following relations:
X = sqrt(1-V^2)*cos(U)
Y = sqrt(1-V^2)*sin(U)
Z = V
where (-pi < U < pi) and (-1 < V < 1) are independent random variables, both with uniform distributions.
How do I use the CDF method to find X_pdf(x)?
X_pdf(x) =
X_cdf'(x) =
( P( X < x ) )' =
( P( sqrt(1-V^2)*cos(U) < x ) )' = ?

2. Dec 4, 2015

### haruspex

Can you see how to write that as a condition on V, and hence as an integral wrt u?

3. Dec 5, 2015

### rabbed

As a first step, is this correct?

( P( sqrt(1-V^2) < x/cos(U) ) )' =
( P( 1-V^2 < (x/cos(U))^2 ) )' =
( P( -1+V^2 > -(x/cos(U))^2 ) )' =
( P( V^2 > 1-(x/cos(U))^2 ) )' =
( P( |V| > sqrt(1-(x/cos(U))^2) ) )' =
( P( -V > sqrt(1-(x/cos(U))^2) OR V > sqrt(1-(x/cos(U))^2) ))' =
( P( V < -sqrt(1-(x/cos(U))^2) OR V > sqrt(1-(x/cos(U))^2) ))' = ?

Then i think i should make a 2D graph with V as Y-axis and U as X-axis, drawing:

V1 = sqrt(1-(x/cos(U))^2)
V2 = -sqrt(1-(x/cos(U))^2)

By solving 1-(x/cos(U))^2 < 0
i get that the functions are imaginary when
pi*n + arccos(x) < U < pi*n + arccos(-x) for some n
and real when
pi*(n-1) + arccos(-x) <= U <= pi*n + arccos(x) for some n
For there to be at least one real solution i need to limit (-1 <= x <= 1)

I think the volume i should be calculating is a double integral of the joint distribution
of U and V, which is 1/(1-(-1)) * 1/(pi-(-pi)) = 1/(4*pi)?
And i should get the integration limits by solving some inequality using the 2D graph?

Sorry for my bad skills, my only source is the internet :)

Last edited: Dec 5, 2015
4. Dec 5, 2015

### haruspex

fine up to there, went a bit wrong after that.
But now use that V has a uniform distribution on (-1,1).
You will need to do something with the U, though, Bringing in an integral involving u, the value of U.

5. Dec 5, 2015

### rabbed

Sorry, i'm lost :(
Do you mean that the expression 1-(x/cos(U))^2 also has uniform distribution on (-1,1) and if I evaluate the integral of the expression on U/u within those limits it would equal 1?
Is the graphing i described above another way to do this or have i misunderstood that?

6. Dec 5, 2015

### haruspex

No.
Suppose the value of U is in (u, u+du). You can write down the probability of that. Now 1-(x/cos(u))^2 is just a value, and you know from the uniform distribution of V the probability that |V| exceeds that.
You need to consider, in a sense, all the possible values of U, the probability (density) of that value, and the probability of the value of V meeting that condition. Then you sum these up (integrate).
For any two r.v.s X, Y, $P(Y>X)=\int P(Y>x).dF_X(x)$.

7. Dec 7, 2015

### rabbed

Ok, my interpretation of what you said:

P(u < U < u+du) = U_cdf(u+du)-U_cdf(u) = U_pdf(u)*du

P(|V| > sqrt(1-(x/cos(U))^2)) = 1-V_cdf(sqrt(1-(x/cos(U))^2)) =
1 - integral wrt v from -1 to sqrt(1-(x/cos(U))^2) of V_pdf(v) =
1 - (1/2*(sqrt(1-(x/cos(U))^2)) - 1/2*(-1)) =
1/2*(1 - sqrt(1-(x/cos(U))^2))

integral wrt u from -pi to pi of
1/2*(1-sqrt(1-(x/cos(U))^2)) * U_pdf(u)*du =
1/(4*pi)*(1-sqrt(1-(x/cos(U))^2)) * du

that integral gave me a nasty result.. :)

what am i calculating? the joint probability that U=u and V=v in the relation X = sqrt(1-V^2)*cos(U)?

8. Dec 11, 2015

### rabbed

Hope someone can advice me how to proceed with this?

9. Dec 11, 2015

### haruspex

Sorry, I forgot to come back to this. Thanks for the nudge.
You have a few Us that should be u, but other than that you are on the right lines.
The integral is correct, but gives the CDF of X. You want the PDF. If you could solve the integral then you could find the PDF from the CDF, but there is another route. What can you do instead?

10. Dec 12, 2015

### rabbed

Thanks for coming back :)
I think you are referring to directly derivate the integral without evaluating the integral?

But I guess in this case it's different since I need to derivate a double integral?
I read something about Leibniz integral rule, is that what i should use?

Another thing is bugging me:
P(|V| > sqrt(1-(x/cos(U))^2)) = 1-V_cdf(sqrt(1-(x/cos(U))^2))
I think it's only valid to say that if we have V instead of |V| in the left hand side?
Graphing |y|>sqrt(1-(c/cos(x))^2) with c variable between -1 and 1 on this page: https://www.desmos.com/calculator
the area is both above and below the "circle", so I'm wondering if this is correct:

P(|V| > sqrt(1-(x/cos(U))^2)) =
P(V < -sqrt(1-(x/cos(U))^2) OR V > sqrt(1-(x/cos(U))^2)) =
P(V < -sqrt(1-(x/cos(U))^2)) + P(V > sqrt(1-(x/cos(U))^2)) =
V_cdf(-sqrt(1-(x/cos(U))^2)) + (1-V_cdf(sqrt(1-(x/cos(U))^2)))

with V_pdf(v) = 1/2 (-1 < v < 1) and by symmetry we get
V_cdf(-sqrt(1-(x/cos(U))^2)) = (1-V_cdf(sqrt(1-(x/cos(U))^2)))
so that
P(|V| > sqrt(1-(x/cos(U))^2)) = 2*(1-V_cdf(sqrt(1-(x/cos(U))^2))) ?

11. Dec 12, 2015

### haruspex

It's not a double integral, but yes, it's the Leibniz integral rule you need. The link you posted does not cover it.
So what integral do you get after differentiating? Can you solve it?
Wrt V and |V|, the easiest is to realise that you only care about |V|, and that is uniform on (0,1).

12. Dec 12, 2015

### rabbed

I was thinking it was double generally, including the integral wrt v also.

Hm, seems I failed somewhere:

I(x) = integral wrt u from -pi to pi of 1/(4*pi)*(1-sqrt(1-(x/cos(u))^2)) * du

I'(x) = d[I(x)]/dx =
integral wrt u from -pi to pi of d[1/(4*pi)*(1-sqrt(1-(x/cos(u))^2))]/dx * du =
integral wrt u from -pi to pi of x/(4*Pi*cos^2(u)*sqrt(1-(x/cos(u))^2)) * du =
[ -sqrt(1-(x/cos(u))^2)/(4*Pi) ] wrt u from -pi to pi =
-sqrt(1-(x/cos(pi))^2)/(4*Pi) + sqrt(1-(x/cos(-pi))^2)/(4*Pi) =
0

13. Dec 12, 2015

### haruspex

First, there's a step I missed out. We need to fix up the range of integration. When cos(u)<x the integrand (before differentiation) makes no sense. The probability of |V| being negative is zero, so the integration range should be truncated. Also, by symmetry, we only need to consider positive u. Cos(-u)=cos(u). This leads to an integration for the cdf from 0 to arccos(x) (which I'll call y below).
As a result, when differentiating, you get another term courtesy of the fundamental theorem of calculus. But that term is not an integral, so presents no difficulty.
This leaves us with the core issue of integrating $\int_0^y\frac{\sec^2(u).du}{\sqrt{1-x^2\sec^2(u)}}$.
sec2 plays nicely with tan, both directly and via calculus. So try t=tan(u).

14. Dec 14, 2015

### rabbed

Just to check that we're on the same page.

So first you made a change in the integral range wrt v:
P(|V| > sqrt(1-(x/cos(U))^2)) = 1-V_cdf(sqrt(1-(x/cos(u))^2)) =
1 - integral wrt v from 0 to sqrt(1-(x/cos(u))^2) of V_pdf(v) =
1 - [x/2] wrt v from 0 to sqrt(1-(x/cos(u))^2) =
1 - (sqrt(1-(x/cos(u))^2/2 - 0) = 1-1/2*sqrt(1-(x/cos(u))^2

And then you made a change in the integral range wrt u:
I(x) = integral wrt u from 0 to arccos(x) of 1/(4*pi)*(1-1/2*sqrt(1-(x/cos(u))^2) * du

Which gives us:
I'(x) = d[I(x)]/dx =
integral wrt u from 0 to arccos(x) of d[1/(4*pi)*(1-1/2*sqrt(1-(x/cos(u))^2)]/dx * du =
integral wrt u from 0 to arccos(x) of x*sec^2(u)/(8*pi*sqrt(1-(x/cos(u))^2)) * du

There is a mismatch between our integrals where I have x*sec^2(u) in the nominator, you only have sec^2(u).
Also I assume you moved the constant 8*Pi from the denominator outside the integral sign?

Meanwhile, I'm brushing up my integration skills..

15. Dec 14, 2015

### haruspex

You really will have to get into LaTeX.
Yes, I have an x in the numerator, but it's just a constant as far as the integral is concerned. Same with the pi etc.

16. Dec 15, 2015

### rabbed

17. Dec 15, 2015

### haruspex

I started again from the top and realised we need to be more careful about ranges.
By symmetry, we can restrict U to the range (0,pi/2) and V to the range (0,1). This will give the distribution for |X|.
Doing this, I found that the term that comes from the fundamental theorem of calculus cancels with another term, leaving only the integral:
$\frac 2{\pi}\int_{u=0}^{arccos(x)}\frac{x\sec^2(u).du}{\sqrt{1-x^2\sec^2(u)}}$
t = tan(u):
$\frac 2{\pi}\int_{t=0}^{\frac 1x\sqrt{1-x^2}}\frac{xdt}{\sqrt{1-x^2-x^2t^2}}$
s=xt/sqrt(1-x2):
$\frac 2{\pi}\int_{s=0}^1\frac{ds}{\sqrt{1-s^2}}$
s=sin(theta):
$\frac 2{\pi}\int_{\theta=0}^{\pi/2}1.d\theta = 1$
Sure enough, when I simulate it in a spreadsheet I see a uniform distribution.

18. Dec 15, 2015

### haruspex

I felt there should be a much simpler geometric path to the answer, and there is.
Using the method Archimedes employed to show that the surface area of a sphere is the same as that of the enclosing cylinder, we can see that the given distributions of U and V are equivalent to a uniform distribution over the surface of a sphere with coordinates X, Y and Z=V. Rotating the cylinder's axis around, it follows immediately that X and Y also have uniform distributions.

19. Dec 17, 2015

### rabbed

So you're saying that X_pdf(x) = Y_pdf(y) = Z_pdf(z) = 1/2?
That's awesome, thank you so much for your help!
I will try to calculate Y_pdf(y) by myself, just for the practice. This is one of the trickiest topics i've come across in math so far.
Thanks again!

20. Dec 17, 2015

### haruspex

The pdf of X would normally be written fX(x), etc., but yes. Did you understand the geometry argument? There should be some way to turn it into algebra, but I've not tried.


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What is the formula of finding the average (mean) of a binomial distribution?
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The formula for the mean of binomial distribution is: μ = n *p Where “n” is the number of trials and “p” is the probability of success.
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text:
# $$56 \div \left[\frac{1}{3}\left\{ {15 + 12 - \left( {9 + 6 - \overline {5 + 7} } \right)} \right\} \right]= ?$$

Free Practice With Testbook Mock Tests

## Options:

1. 12

2. 4032

3. 7

4. 448

### Correct Answer: Option 3 (Solution Below)

This question was previously asked in

RRC Group D Previous Paper 1 (Held On: 17 Sep 2018 Shift 1)

## Solution:

Concept used:

Calculation:

$$56 \div \left[\frac{1}{3}\left\{ {15 + 12 - \left( {9 + 6 - \overline {5 + 7} } \right)} \right\} \right]= ?$$

⇒ 56 ÷ [(1/3){15 + 12 - (9 + 6 - 12)}]

⇒ 56 ÷ [(1/3){15 + 12 - (15 - 12)}]

⇒ 56 ÷ [(1/3){15 + 12 - 3}]

⇒ 56 ÷ [(1/3){27 - 3}]

⇒ 56 ÷ [(1/3)(24)]

⇒ 56 ÷ 8

⇒ 7

∴ ? = 7

If in any simplification problem there is (  ̅  ) bar sign above any operation then we should perform the operation at the beginning, whatever the operation is. Then we will follow the chart given above.