text:
## Hoff, David

Compute Distance To:
 Author ID: hoff.david Published as: Hoff, David; Hoff, D. Homepage: https://math.indiana.edu/about/emeriti/hoff-david.html External Links: MGP
 Documents Indexed: 77 Publications since 1977, including 1 Book Co-Authors: 21 Co-Authors with 39 Joint Publications 474 Co-Co-Authors
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### Co-Authors

 36 single-authored 7 Smoller, Joel Alan 4 Chen, Gui-Qiang G. 4 Trivisa, Konstantina 4 Zumbrun, Kevin R. 3 Tsyganov, Eugene N. 3 Ziane, Mohammed 2 Jenssen, Helge Kristian 2 Liu, Tai-Ping 2 Perepelitsa, Misha 2 Serre, Denis 2 Zarnowski, Roger E. 1 Chadam, John M. 1 Conway, Edward Daire 1 DiBenedetto, Emmanuele 1 Johnson, Charles Royal 1 Khodja, Mohamed R. 1 Lucier, Bradley J. 1 Nasserasr, Shahla 1 Ortoleva, Peter J. 1 Santos, Marcelo M. 1 Suen, Anthony 1 Youngs, Matthias 1 Zhao, Jennifer Jing
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### Serials

 10 Journal of Differential Equations 8 Archive for Rational Mechanics and Analysis 5 Indiana University Mathematics Journal 5 Transactions of the American Mathematical Society 5 SIAM Journal on Mathematical Analysis 4 ZAMP. Zeitschrift für angewandte Mathematik und Physik 4 SIAM Journal on Numerical Analysis 3 Mathematics of Computation 3 SIAM Journal on Applied Mathematics 2 Communications in Mathematical Physics 2 Annales de l’Institut Henri Poincaré. Analyse Non Linéaire 2 Communications in Partial Differential Equations 1 Communications on Pure and Applied Mathematics 1 IMA Journal of Applied Mathematics 1 Journal of Mathematical Analysis and Applications 1 Rocky Mountain Journal of Mathematics 1 Physica D 1 RAIRO. Modélisation Mathématique et Analyse Numérique 1 Acta Mathematicae Applicatae Sinica. English Series 1 Linear Algebra and its Applications 1 Proceedings of the Royal Society of Edinburgh. Section A. Mathematics 1 Comptes Rendus de l’Académie des Sciences. Série I. Mathématique 1 Journal of Mathematical Fluid Mechanics 1 Mathematical Surveys and Monographs
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### Fields

 69 Partial differential equations (35-XX) 57 Fluid mechanics (76-XX) 12 Numerical analysis (65-XX) 6 Classical thermodynamics, heat transfer (80-XX) 4 Dynamical systems and ergodic theory (37-XX) 2 Biology and other natural sciences (92-XX) 1 Linear and multilinear algebra; matrix theory (15-XX) 1 Ordinary differential equations (34-XX) 1 Statistical mechanics, structure of matter (82-XX) 1 Geophysics (86-XX)

### Citations contained in zbMATH Open

63 Publications have been cited 2,066 times in 1,226 Documents Cited by Year
Global solutions of the Navier-Stokes equations for multidimensional compressible flow with discontinuous initial data. Zbl 0836.35120
Hoff, David
1995
Multi-dimensional diffusion waves for the Navier-Stokes equations of compressible flow. Zbl 0842.35076
Hoff, David; Zumbrun, Kevin
1995
Large time behavior of solutions of systems of nonlinear reaction- diffusion equations. Zbl 0383.35035
Conway, Edward; Hoff, David; Smoller, Joel
1978
Strong convergence to global solutions for multidimensional flows of compressible, viscous fluids with polytropic equations of state and discontinuous initial data. Zbl 0836.76082
Hoff, David
1995
Discontinuous solutions of the Navier-Stokes equations for multidimensional flows of heat-conducting fluids. Zbl 0904.76074
Hoff, David
1997
Pointwise decay estimates for multidimensional Navier-Stokes diffusion waves. Zbl 0882.76074
Hoff, David; Zumbrun, Kevin
1997
Global existence for 1D, compressible, isentropic Navier-Stokes equations with large initial data. Zbl 0656.76064
Hoff, David
1987
The failure of continuous dependence on initial data for the Navier- Stokes equations of compressible flow. Zbl 0741.35057
Hoff, David; Serre, Denis
1991
Compressible flow in a half-space with Navier boundary conditions. Zbl 1095.35025
Hoff, David
2005
Dynamics of singularity surfaces for compressible, viscous flows in two space dimensions. Zbl 1020.76046
Hoff, David
2002
The inviscid limit for the Navier-Stokes equations of compressible, isentropic flow with shock data. Zbl 0674.76047
Hoff, David; Liu, Tai-Ping
1989
Global solutions of the compressible Navier-Stokes equations with large discontinuous initial data. Zbl 0977.35104
Chen, Gui-Qiang; Hoff, David; Trivisa, Konstantina
2000
Spherically symmetric solutions of the Navier-Stokes equations for compressible, isothermal flow with large, discontinuous initial data. Zbl 0765.35033
Hoff, David
1992
Uniqueness and continuous dependence of weak solutions in compressible magnetohydrodynamics. Zbl 1077.35071
Hoff, David; Tsyganov, Eugene
2005
Non-formation of vacuum states for compressible Navier-Stokes equations. Zbl 0988.76081
Hoff, David; Smoller, Joel
2001
The zero-Mach limit of compressible flows. Zbl 0907.35098
Hoff, David
1998
Solutions in the large for certain nonlinear parabolic systems. Zbl 0578.35044
Hoff, David; Smoller, Joel
1985
Global solutions of the equations of one-dimensional, compressible flow with large data and forces, and with differing end states. Zbl 0913.35031
Hoff, David
1998
Global well-posedness of the Cauchy problem for the Navier-Stokes equations of nonisentropic flow with discontinuous initial data. Zbl 0762.35085
Hoff, David
1992
Global solutions to a model for exothermically reacting, compressible flows with large discontinuous initial data. Zbl 1022.76056
Chen, Gui-Qiang; Hoff, David; Trivisa, Konstantina
2003
Invariant regions for systems of conservation laws. Zbl 0535.35056
Hoff, David
1985
Stability and convergence of finite difference methods for systems of nonlinear reaction-diffusion equations. Zbl 0411.76062
Hoff, David
1978
Global low-energy weak solutions of the equations of three-dimensional compressible magnetohydrodynamics. Zbl 1284.76414
Suen, Anthony; Hoff, David
2012
Uniqueness of weak solutions of the Navier–Stokes equations of multidimensional, compressible flow. Zbl 1100.76052
Hoff, David
2006
Asymptotic behavior of multidimensional scalar viscous shock fronts. Zbl 0967.76049
Hoff, David; Zumbrun, Kevin
2000
Construction of solutions for compressible, isentropic Navier-Stokes equations in one space dimension with nonsmooth initial data. Zbl 0635.35074
Hoff, David
1986
Symmetric nonbarotropic flows with large data and forces. Zbl 1060.76099
Hoff, David; Jenssen, Helge Kristian
2004
Lagrangian structure and propagation of singularities in multidimensional compressible flow. Zbl 1144.76049
Hoff, David; Santos, Marcelo M.
2008
Global existence for systems of parabolic conservation laws in several space variables. Zbl 0624.35044
Hoff, David; Smoller, Joel
1987
Discontinuous solutions of the Navier-Stokes equations for compressible flow. Zbl 0732.35071
Hoff, David
1991
Reactive infiltration instabilities. Zbl 0619.76116
Chadam, J.; Hoff, D.; Merino, E.; Ortoleva, P.; Sen, A.
1986
An interface tracking algorithm for the porous medium equation. Zbl 0564.76090
DiBenedetto, E.; Hoff, David
1984
The sharp form of Oleinik’s entropy condition in several space variables. Zbl 0528.35062
Hoff, David
1983
Pointwise Green’s function bounds for multidimensional scalar viscous shock fronts. Zbl 1028.35108
Hoff, David; Zumbrun, Kevin
2002
On the Navier-Stokes equations for exothermically reacting compressible fluids. Zbl 1032.76056
Chen, Gui-Qiang; Hoff, David; Trivisa, Konstantina
2002
Time analyticity and backward uniqueness of weak solutions of the Navier-Stokes equations of multidimensional compressible flow. Zbl 1387.35454
Hoff, David; Tsyganov, Eugene
2008
The global attractor and finite determining nodes for the Navier-Stokes equations of compressible flow with singular initial data. Zbl 0977.35105
Hoff, David; Ziane, Mohammed
2000
Globally smooth solutions of quasilinear hyperbolic systems in diagonal form. Zbl 0488.35057
Hoff, David
1982
A finite-difference scheme for the Navier-Stokes equations of one- dimensional, isentropic, compressible flow. Zbl 0727.76094
Zarnowski, Roger; Hoff, David
1991
Local solutions of a compressible flow problem with Navier boundary conditions in general three-dimensional domains. Zbl 1246.35149
Hoff, David
2012
A finite difference scheme for a system of two conservation laws with artificial viscosity. Zbl 0447.65056
Hoff, David
1979
Attracteurs compacts pour les équations de Navier-Stokes d’un fluide compressible monodimensionel. (Compact attractors for the Navier-Stokes equations of one-dimensional, compresible flow.) Zbl 0926.35113
Hoff, David; Ziane, Mohammed
1999
Error bounds for finite-difference approximations for a class of nonlinear parabolic systems. Zbl 0613.65096
Hoff, David; Smoller, Joel
1985
Stability of coexisting phases for compressible van der Waals fluids. Zbl 0787.35065
Hoff, David; Khodja, M.
1993
Continuous dependence on initial data for discontinuous solutions of the Navier-Stokes equations for one-dimensional, compressible flow. Zbl 0860.35094
Hoff, David
1996
Existence of solutions to a model for sparse, one-dimensional fluids. Zbl 1214.35045
Hoff, David
2011
Continuous dependence in $$L^ 2$$ for discontinuous solutions of the viscous $$p$$-system. Zbl 0836.35157
Hoff, David; Zarnowski, Roger
1994
A constructive theory for shock-free, isentropic flow. Zbl 0416.76015
Hoff, David
1980
A linearly implicit finite-difference scheme for the one-dimensional porous medium equation. Zbl 0603.76096
Hoff, David
1985
A convergent finite-difference scheme for the Navier-Stokes equations of one-dimensional, nonisentropic, compressible flow. Zbl 0810.76060
Zhao, Jing; Hoff, David
1994
Locally Lipschitz solutions of a single conservation law in several space variables. Zbl 0443.34005
Hoff, David
1981
Systems of partial differential equations of mixed hyperbolic-parabolic type. Zbl 1055.35007
Tsyganov, Eugene; Hoff, David
2004
Error bounds for Glimm difference approximations for scalar conservation laws. Zbl 0579.65096
Hoff, David; Smoller, Joel
1985
A characterization of the blow-up time for the solution of a conservation law in several space variables. Zbl 0501.35054
Hoff, David
1982
On the smoothing of initial discontinuities and the development of singularities in solutions of certain quasilinear hyperbolic equations. Zbl 0532.35049
Hoff, David
1983
Instantaneous boundary tangency and cusp formation in two-dimensional fluid flow. Zbl 1197.35070
Hoff, David; Perepelitsa, Misha
2009
Convergence and error bound analysis of a finite difference scheme for the one-dimensional Navier-Stokes equations. Zbl 0890.76055
Zhao, Jennifer Jing; Hoff, David
1997
Asymptotic behavior of solutions to a model for the flow of a reacting fluid. Zbl 1193.35154
Hoff, David
2010
Totally positive shapes and $$TP_{k}$$-completable patterns. Zbl 1242.15023
Hoff, D.; Johnson, C. R.; Nasserasr, S.
2012
A scheme for computing solutions and interface curves for a doubly- degenerate parabolic equation. Zbl 0613.65090
Hoff, David
1985
Boundary tangency for density interfaces in compressible viscous fluid flows. Zbl 1364.35272
Hoff, David; Perepelitsa, Misha
2012
Approximation and decay of solutions of systems of nonlinear diffusion equations. Zbl 0379.35037
Hoff, David
1977
Finite-dimensional attractors and exponential attractors for the Navier-Stokes equations of compressible flow. Zbl 1044.35054
Hoff, David; Ziane, Mohammed
2003
Global low-energy weak solutions of the equations of three-dimensional compressible magnetohydrodynamics. Zbl 1284.76414
Suen, Anthony; Hoff, David
2012
Local solutions of a compressible flow problem with Navier boundary conditions in general three-dimensional domains. Zbl 1246.35149
Hoff, David
2012
Totally positive shapes and $$TP_{k}$$-completable patterns. Zbl 1242.15023
Hoff, D.; Johnson, C. R.; Nasserasr, S.
2012
Boundary tangency for density interfaces in compressible viscous fluid flows. Zbl 1364.35272
Hoff, David; Perepelitsa, Misha
2012
Existence of solutions to a model for sparse, one-dimensional fluids. Zbl 1214.35045
Hoff, David
2011
Asymptotic behavior of solutions to a model for the flow of a reacting fluid. Zbl 1193.35154
Hoff, David
2010
Instantaneous boundary tangency and cusp formation in two-dimensional fluid flow. Zbl 1197.35070
Hoff, David; Perepelitsa, Misha
2009
Lagrangian structure and propagation of singularities in multidimensional compressible flow. Zbl 1144.76049
Hoff, David; Santos, Marcelo M.
2008
Time analyticity and backward uniqueness of weak solutions of the Navier-Stokes equations of multidimensional compressible flow. Zbl 1387.35454
Hoff, David; Tsyganov, Eugene
2008
Uniqueness of weak solutions of the Navier–Stokes equations of multidimensional, compressible flow. Zbl 1100.76052
Hoff, David
2006
Compressible flow in a half-space with Navier boundary conditions. Zbl 1095.35025
Hoff, David
2005
Uniqueness and continuous dependence of weak solutions in compressible magnetohydrodynamics. Zbl 1077.35071
Hoff, David; Tsyganov, Eugene
2005
Symmetric nonbarotropic flows with large data and forces. Zbl 1060.76099
Hoff, David; Jenssen, Helge Kristian
2004
Systems of partial differential equations of mixed hyperbolic-parabolic type. Zbl 1055.35007
Tsyganov, Eugene; Hoff, David
2004
Global solutions to a model for exothermically reacting, compressible flows with large discontinuous initial data. Zbl 1022.76056
Chen, Gui-Qiang; Hoff, David; Trivisa, Konstantina
2003
Finite-dimensional attractors and exponential attractors for the Navier-Stokes equations of compressible flow. Zbl 1044.35054
Hoff, David; Ziane, Mohammed
2003
Dynamics of singularity surfaces for compressible, viscous flows in two space dimensions. Zbl 1020.76046
Hoff, David
2002
Pointwise Green’s function bounds for multidimensional scalar viscous shock fronts. Zbl 1028.35108
Hoff, David; Zumbrun, Kevin
2002
On the Navier-Stokes equations for exothermically reacting compressible fluids. Zbl 1032.76056
Chen, Gui-Qiang; Hoff, David; Trivisa, Konstantina
2002
Non-formation of vacuum states for compressible Navier-Stokes equations. Zbl 0988.76081
Hoff, David; Smoller, Joel
2001
Global solutions of the compressible Navier-Stokes equations with large discontinuous initial data. Zbl 0977.35104
Chen, Gui-Qiang; Hoff, David; Trivisa, Konstantina
2000
Asymptotic behavior of multidimensional scalar viscous shock fronts. Zbl 0967.76049
Hoff, David; Zumbrun, Kevin
2000
The global attractor and finite determining nodes for the Navier-Stokes equations of compressible flow with singular initial data. Zbl 0977.35105
Hoff, David; Ziane, Mohammed
2000
Attracteurs compacts pour les équations de Navier-Stokes d’un fluide compressible monodimensionel. (Compact attractors for the Navier-Stokes equations of one-dimensional, compresible flow.) Zbl 0926.35113
Hoff, David; Ziane, Mohammed
1999
The zero-Mach limit of compressible flows. Zbl 0907.35098
Hoff, David
1998
Global solutions of the equations of one-dimensional, compressible flow with large data and forces, and with differing end states. Zbl 0913.35031
Hoff, David
1998
Discontinuous solutions of the Navier-Stokes equations for multidimensional flows of heat-conducting fluids. Zbl 0904.76074
Hoff, David
1997
Pointwise decay estimates for multidimensional Navier-Stokes diffusion waves. Zbl 0882.76074
Hoff, David; Zumbrun, Kevin
1997
Convergence and error bound analysis of a finite difference scheme for the one-dimensional Navier-Stokes equations. Zbl 0890.76055
Zhao, Jennifer Jing; Hoff, David
1997
Continuous dependence on initial data for discontinuous solutions of the Navier-Stokes equations for one-dimensional, compressible flow. Zbl 0860.35094
Hoff, David
1996
Global solutions of the Navier-Stokes equations for multidimensional compressible flow with discontinuous initial data. Zbl 0836.35120
Hoff, David
1995
Multi-dimensional diffusion waves for the Navier-Stokes equations of compressible flow. Zbl 0842.35076
Hoff, David; Zumbrun, Kevin
1995
Strong convergence to global solutions for multidimensional flows of compressible, viscous fluids with polytropic equations of state and discontinuous initial data. Zbl 0836.76082
Hoff, David
1995
Continuous dependence in $$L^ 2$$ for discontinuous solutions of the viscous $$p$$-system. Zbl 0836.35157
Hoff, David; Zarnowski, Roger
1994
A convergent finite-difference scheme for the Navier-Stokes equations of one-dimensional, nonisentropic, compressible flow. Zbl 0810.76060
Zhao, Jing; Hoff, David
1994
Stability of coexisting phases for compressible van der Waals fluids. Zbl 0787.35065
Hoff, David; Khodja, M.
1993
Spherically symmetric solutions of the Navier-Stokes equations for compressible, isothermal flow with large, discontinuous initial data. Zbl 0765.35033
Hoff, David
1992
Global well-posedness of the Cauchy problem for the Navier-Stokes equations of nonisentropic flow with discontinuous initial data. Zbl 0762.35085
Hoff, David
1992
The failure of continuous dependence on initial data for the Navier- Stokes equations of compressible flow. Zbl 0741.35057
Hoff, David; Serre, Denis
1991
Discontinuous solutions of the Navier-Stokes equations for compressible flow. Zbl 0732.35071
Hoff, David
1991
A finite-difference scheme for the Navier-Stokes equations of one- dimensional, isentropic, compressible flow. Zbl 0727.76094
Zarnowski, Roger; Hoff, David
1991
The inviscid limit for the Navier-Stokes equations of compressible, isentropic flow with shock data. Zbl 0674.76047
Hoff, David; Liu, Tai-Ping
1989
Global existence for 1D, compressible, isentropic Navier-Stokes equations with large initial data. Zbl 0656.76064
Hoff, David
1987
Global existence for systems of parabolic conservation laws in several space variables. Zbl 0624.35044
Hoff, David; Smoller, Joel
1987
Construction of solutions for compressible, isentropic Navier-Stokes equations in one space dimension with nonsmooth initial data. Zbl 0635.35074
Hoff, David
1986
Reactive infiltration instabilities. Zbl 0619.76116
Chadam, J.; Hoff, D.; Merino, E.; Ortoleva, P.; Sen, A.
1986
Solutions in the large for certain nonlinear parabolic systems. Zbl 0578.35044
Hoff, David; Smoller, Joel
1985
Invariant regions for systems of conservation laws. Zbl 0535.35056
Hoff, David
1985
Error bounds for finite-difference approximations for a class of nonlinear parabolic systems. Zbl 0613.65096
Hoff, David; Smoller, Joel
1985
A linearly implicit finite-difference scheme for the one-dimensional porous medium equation. Zbl 0603.76096
Hoff, David
1985
Error bounds for Glimm difference approximations for scalar conservation laws. Zbl 0579.65096
Hoff, David; Smoller, Joel
1985
A scheme for computing solutions and interface curves for a doubly- degenerate parabolic equation. Zbl 0613.65090
Hoff, David
1985
An interface tracking algorithm for the porous medium equation. Zbl 0564.76090
DiBenedetto, E.; Hoff, David
1984
The sharp form of Oleinik’s entropy condition in several space variables. Zbl 0528.35062
Hoff, David
1983
On the smoothing of initial discontinuities and the development of singularities in solutions of certain quasilinear hyperbolic equations. Zbl 0532.35049
Hoff, David
1983
Globally smooth solutions of quasilinear hyperbolic systems in diagonal form. Zbl 0488.35057
Hoff, David
1982
A characterization of the blow-up time for the solution of a conservation law in several space variables. Zbl 0501.35054
Hoff, David
1982
Locally Lipschitz solutions of a single conservation law in several space variables. Zbl 0443.34005
Hoff, David
1981
A constructive theory for shock-free, isentropic flow. Zbl 0416.76015
Hoff, David
1980
A finite difference scheme for a system of two conservation laws with artificial viscosity. Zbl 0447.65056
Hoff, David
1979
Large time behavior of solutions of systems of nonlinear reaction- diffusion equations. Zbl 0383.35035
Conway, Edward; Hoff, David; Smoller, Joel
1978
Stability and convergence of finite difference methods for systems of nonlinear reaction-diffusion equations. Zbl 0411.76062
Hoff, David
1978
Approximation and decay of solutions of systems of nonlinear diffusion equations. Zbl 0379.35037
Hoff, David
1977
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### Cited by 1,086 Authors

 37 Zhu, Changjiang 26 Xin, Zhouping 25 Zhao, Huijiang 23 Tan, Zhong 22 Feireisl, Eduard 22 Hoff, David 21 Yao, Zhengan 21 Zumbrun, Kevin R. 18 Jiang, Song 18 Qin, Yuming 18 Wang, Weike 18 Yang, Tong 17 Huang, Xiangdi 17 Zhang, Ting 16 Guo, Zhenhua 16 Wen, Huanyao 16 Xu, Jiang 16 Yao, Lei 15 Chen, Gui-Qiang G. 15 Li, Jing 15 Wang, Dehua 14 Zhang, Jianwen 13 Danchin, Raphaël 13 Fan, Jishan 13 Fang, Daoyuan 13 Huang, Lan 13 Vasseur, Alexis F. 12 Li, Hailiang 12 Novotný, Antonín 12 Ou, Yaobin 12 Wang, Yi 11 Guo, Boling 11 Mucha, Piotr Bogusław 10 Ducomet, Bernard 10 Guermond, Jean-Luc 10 Li, Fucai 10 Peng, Hongyun 10 Popov, Boyan 10 Suen, Anthony 10 Sun, Yongzhong 10 Wang, Wenjun 9 Howard, Peter 9 Jiu, Quansen 9 Kagei, Yoshiyuki 9 Nečasová, Šárka 9 Zhao, Junning 8 Haspot, Boris 8 Kweon, Jae Ryong 8 Li, Yeping 8 Li, Zilai 8 Lian, Ruxu 8 Wu, Zhigang 8 Zhang, Yinghui 8 Zhong, Xin 8 Zi, Ruizhao 7 Frid, Hermano 7 Hou, Xiaofeng 7 Hu, Xianpeng 7 Hu, Yuxi 7 Ju, Qiangchang 7 Kawashima, Shuichi 7 Mimura, Masayasu 7 Pan, Ronghua 7 Serre, Denis 7 Trivisa, Konstantina 7 Ukai, Seiji 7 Wang, Yuzhu 7 Wu, Guochun 7 Zatorska, Ewelina 7 Zhang, Zhifei 7 Zhu, Shengguo 6 Ding, Shijin 6 Dou, Changsheng 6 Duan, Qin 6 Duan, Renjun 6 Huang, Feimin 6 Liang, Zhilei 6 Lü, Boqiang 6 Petzeltová, Hana 6 Qin, Xulong 6 Simsen, Jacson 6 Wang, Yuanming 6 Zhai, Xiaoping 6 Zhang, Jianlin 6 Zhang, Peixin 5 Bresch, Didier 5 Chadam, John M. 5 Colombo, Rinaldo M. 5 Cui, Haibo 5 Hong, GuangYi 5 Jiang, Fei 5 Kang, Moon-Jin 5 Li, Jinkai 5 Liu, Shengquan 5 Liu, Tai-Ping 5 Lyng, Gregory Daniel 5 Ma, Shixiang 5 Perepelitsa, Misha 5 Racke, Reinhard 5 Shi, Weixuan ...and 986 more Authors
all top 5

### Cited in 178 Serials

 187 Journal of Differential Equations 79 Journal of Mathematical Analysis and Applications 68 Archive for Rational Mechanics and Analysis 44 ZAMP. Zeitschrift für angewandte Mathematik und Physik 37 Nonlinear Analysis. Theory, Methods & Applications. Series A: Theory and Methods 31 Journal of Mathematical Fluid Mechanics 30 Journal of Mathematical Physics 30 Discrete and Continuous Dynamical Systems 27 SIAM Journal on Mathematical Analysis 26 Mathematical Methods in the Applied Sciences 26 Nonlinear Analysis. Real World Applications 22 Communications in Partial Differential Equations 21 Journal de Mathématiques Pures et Appliquées. Neuvième Série 20 Communications in Mathematical Physics 20 M$$^3$$AS. Mathematical Models & Methods in Applied Sciences 19 Boundary Value Problems 18 Journal of Hyperbolic Differential Equations 18 Science China. Mathematics 17 Applicable Analysis 16 Journal of Dynamics and Differential Equations 15 Acta Applicandae Mathematicae 12 Computers & Mathematics with Applications 12 Chinese Annals of Mathematics. Series B 12 Acta Mathematicae Applicatae Sinica. English Series 12 Applied Mathematics Letters 12 Proceedings of the Royal Society of Edinburgh. Section A. Mathematics 12 Communications on Pure and Applied Analysis 11 Journal of Mathematical Biology 10 Physica D 10 Nonlinear Analysis. Theory, Methods & Applications 9 Applied Mathematics and Computation 9 Annales de l’Institut Henri Poincaré. Analyse Non Linéaire 9 Acta Mathematica Sinica. English Series 9 Discrete and Continuous Dynamical Systems. Series B 8 Journal of Computational Physics 7 Communications on Pure and Applied Mathematics 7 Nonlinearity 7 Numerische Mathematik 7 Transactions of the American Mathematical Society 6 Annali di Matematica Pura ed Applicata. Serie Quarta 6 Journal of Computational and Applied Mathematics 6 Journal of Functional Analysis 6 Quarterly of Applied Mathematics 6 Journal of Evolution Equations 6 Kinetic and Related Models 5 Mathematics of Computation 5 Bulletin of Mathematical Biology 5 Advances in Mathematics 5 Indiana University Mathematics Journal 5 SIAM Journal on Numerical Analysis 5 Mathematical and Computer Modelling 5 Discrete and Continuous Dynamical Systems. Series S 4 Journal of Fluid Mechanics 4 Publications of the Research Institute for Mathematical Sciences, Kyoto University 4 SIAM Journal on Applied Mathematics 4 Communications in Contemporary Mathematics 4 Acta Mathematica Scientia. Series B. (English Edition) 4 Analysis and Applications (Singapore) 4 European Series in Applied and Industrial Mathematics (ESAIM): Mathematical Modelling and Numerical Analysis 3 Computer Methods in Applied Mechanics and Engineering 3 Mathematical Biosciences 3 Rocky Mountain Journal of Mathematics 3 Annales de l’Institut Fourier 3 Automatica 3 Calcolo 3 Inventiones Mathematicae 3 Mathematische Annalen 3 Osaka Journal of Mathematics 3 Zeitschrift für Analysis und ihre Anwendungen 3 Applied Numerical Mathematics 3 Japan Journal of Applied Mathematics 3 Applications of Mathematics 3 Journal of Nonlinear Science 3 SIAM Journal on Scientific Computing 3 Wuhan University Journal of Natural Sciences (WUJNS) 3 Differential Equations 3 Frontiers of Mathematics in China 2 Applied Mathematics and Optimization 2 Studies in Applied Mathematics 2 Advances in Applied Mathematics 2 Systems & Control Letters 2 RAIRO. Modélisation Mathématique et Analyse Numérique 2 Numerical Methods for Partial Differential Equations 2 Journal of Scientific Computing 2 European Journal of Applied Mathematics 2 Applied Mathematical Modelling 2 Boletim da Sociedade Brasileira de Matemática. Nova Série 2 NoDEA. Nonlinear Differential Equations and Applications 2 Opuscula Mathematica 2 Abstract and Applied Analysis 2 Computational Geosciences 2 Journal of Applied Mathematics 2 Central European Journal of Mathematics 2 Advances in Difference Equations 2 Annali dell’Università di Ferrara. Sezione VII. Scienze Matematiche 2 Evolution Equations and Control Theory 2 Advances in Nonlinear Analysis 2 Journal of Function Spaces 1 Acta Mechanica 1 International Journal for Numerical and Analytical Methods in Geomechanics ...and 78 more Serials
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### Cited in 36 Fields

 1,132 Partial differential equations (35-XX) 795 Fluid mechanics (76-XX) 103 Numerical analysis (65-XX) 98 Biology and other natural sciences (92-XX) 70 Classical thermodynamics, heat transfer (80-XX) 29 Ordinary differential equations (34-XX) 26 Dynamical systems and ergodic theory (37-XX) 24 Mechanics of deformable solids (74-XX) 18 Statistical mechanics, structure of matter (82-XX) 13 Harmonic analysis on Euclidean spaces (42-XX) 12 Optics, electromagnetic theory (78-XX) 11 Operator theory (47-XX) 10 Astronomy and astrophysics (85-XX) 10 Operations research, mathematical programming (90-XX) 8 Integral equations (45-XX) 8 Geophysics (86-XX) 8 Systems theory; control (93-XX) 7 Calculus of variations and optimal control; optimization (49-XX) 5 Functional analysis (46-XX) 5 Global analysis, analysis on manifolds (58-XX) 5 Probability theory and stochastic processes (60-XX) 3 General topology (54-XX) 3 Quantum theory (81-XX) 2 Functions of a complex variable (30-XX) 2 Differential geometry (53-XX) 1 General and overarching topics; collections (00-XX) 1 Linear and multilinear algebra; matrix theory (15-XX) 1 Real functions (26-XX) 1 Measure and integration (28-XX) 1 Difference and functional equations (39-XX) 1 Approximations and expansions (41-XX) 1 Integral transforms, operational calculus (44-XX) 1 Mechanics of particles and systems (70-XX) 1 Relativity and gravitational theory (83-XX) 1 Information and communication theory, circuits (94-XX) 1 Mathematics education (97-XX)


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# When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.

Q.8.    When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.

$R=\frac{12}{0.0025}=4800\Omega =4.8k\Omega$

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# Triangle medians and centroids (2D proof)

Video transcript
- [Voiceover] In the video on triangle medians and centroids I did, essentially, the proof that the centroid is 2/3 along the way of a median. I did it using a two-dimensional triangle in three dimensions, and I mentioned that I thought, at least it made the math a little simpler, but someone mentioned that they'd be interested in seeing the two-dimensional version of the proof, so why not do that? So let's just draw an arbitrary triangle, and I'll make one side of the triangle right along the x-axis just to make, I think it'll make the math a little bit easier, maybe there's easier ways of doing it. So that's one side of the triangle, that's the other side of the triangle, and let's put the height of the triangle, let's put it on the y-axis, so this triangle right here, this is the y-axis that is the y-axis, and then this right here, is our x-axis. Now, we said that this is an arbitrary triangle, so let's just make this distance equal to a, so this point right here, this vertex would be the point a comma zero, so that distance is a. Let's make this distance right here be equal to, I don't know, let's make it equal to b, so this is the point negative b comma zero, and then let's make this distance, this is the point y is equal to c, so this is a point zero comma c, so it's an arbitrary triangle, any triangle can be represented this way. Now, let's think about its medians. Let's think about its medians and the centroid. I'm only going to do it for two of the medians, because we know that the third median will also intersect at the same centroid, so we have a median on this, we have a midpoint, I should say, on this side of the triangle, and its coordinates are just going to be the midpoint of those two points, or zero plus a over two, which would just be a over two, and then c plus zero over two, which would just be c over two. Let's do the same thing over here. The midpoint of this side right over here is going to be negative b plus zero over two, so it's negative b over two, and then zero plus c over two, so it's just going to be c over two. Now that we know all the coordinates for the vertices and the midpoints, at least of these two sides, we can find the equations for the lines that the medians are part of, so we can find the equation for this line, we could find the equation for this line in purple, then we can find the equation for the other line that connects to these two dots, find their intersection, and we'll essentially have the coordinates of our centroid, of the intersection of the medians, and we can do it for this one, too, but that would be redundant, because it's going to intersect in the same point. So what would be the equation for this line right over here? Well, our slope is going to be change and y, so c over two minus zero, so that's just c over two, over change and x, so a over two minus minus b, so minus negative b, I should say, so it's plus b, I could write a plus b over here, but just to makes the maths involved right, plus 2b over two. I just subtracted a negative b, which is the same thing as adding a b, and adding a b is the same thing as adding a 2b over two, and I did that so these have the same denominator. I could add them, or I could just multiply the numerator and the denominator both by two and I'll get the slope is equal to, c over a plus 2b, so that's the slope of this line right over here, and then we know some points here. We could just use the point slope formula for the equation of a line. We know the point negative b comma zero is on the line, so we know that the equation of this line in purple, let me stay in the purple, is going to be y minus zero, is equal to x minus negative b, or I could say x plus b, times our slope, times c over a plus 2b. And obviously, this would just simplify to y is equal to all of this business over here, so this is essentially the equation of this median right here if the line just kept going, or the median is a line segment that is part of this line. Let us do the same thing for this line right over here, this median right over here. So once again, its slope is change and y over change and x, so the change and y, c over two minus zero, so the slope here is equal to c over two minus zero, which is just c over two, over negative b over two, minus a, or I could say minus 2a over two, that's the same thing as minus a, and do the same thing, multiply the numerator and the denominator by two. We get c over negative b minus 2a, so the equation of this line right here, the other median, the equation of this other median right over here, well we have this point over here, so we could use the point slope again. We get y minus zero is equal to x minus a times the slope, times c over negative b minus 2a, so we have the two equations for these lines. We can now use that information to find their intersection, which is going to be the centroid. Well, to do that, we have both of these in terms of y, right? Y minus zero is the same thing as y, so we could just set this as being equal to that, so let's do that. We get x plus b times c, over a plus 2b is equal to x minus a times c over negative b minus 2a. Well, both sides are divisible by c, we can assume that c is non-zero, so this triangle actually exists, it actually exists in two dimensions, so we can divide both sides by c and we get that. Let's see, now we can cross-multiply. We can multiply x plus b times this quantity right here, and that's going to be equal to a plus 2b times this quantity over here. We can just distribute it, so it's going to be x times all of this business, it's going to be x times negative b minus 2a, plus b times all of this, so that's going to be minus b squared minus 2ab, so I just multiply that times that, is going to equal to this times that, so it's going to be x times all of this business. X times a plus 2b, minus a times that, so distribute the a, minus a squared minus 2ab. Let's see if we can simplify this. We have a negative 2ab on both sides, so let's just subtract that out, so we cancelled things out, let's subtract this from both sides of the equation, so we'll have a minus x times a plus 2b, and I'll subtract that from here, but I'll write it a little different, it'll simplify things, so this will become x times, I'll distribute the negative inside the a plus 2b, so negative a minus 2b, and let's add a b squared to both sides, so plus b squared, plus b squared. I want to collect all the x terms on one side and all of the constants on the other. This becomes on the left-hand side, the coefficient on x, I have negative b minus 2b is negative 3b negative 2a minus a is negative 3a, times x is equal to, these guys cancel out, and these guys cancel out, is equal to b squared minus a squared. Let's see if we can factor, this already looks a little suspicious in a good way, something that we should be able to solve. This can be factored. We can factor out a negative three. We can actually factor out a three. We'll get three times b minus a, actually, let's factor out a negative three, so negative three times b plus a times x is equal to, now this is the same thing as b plus a times b minus a. So we can divide both sides by b plus a, we have negative three times x is equal to b minus a, let's just divide both sides by negative three. We get x is equal to b minus a over negative three, which is the same thing as a minus b over three, so we have the x-coordinate so far for our centroid. It is a minus b over three. You kind of see hints of the solution over here. A minus b is this entire distance. It's one third of this entire distance, although I don't want to jump the gun too much because it complicates just the x-coordinate. But let's figure out the y-coordinate. To do that, we can just substitute back into one of these equations up here. This was the equation of this median, the grey median that we had done, so lets substitute back in. We have, I'll do it over here, y is equal to x minus a times c. X is this thing over here, so it's a minus b over three minus a, so I'm going to subtract, instead of just writing a minus a over here, I'm going to write a negative three over three, so minus three a over three. That's this minus a right over here. I just multiplied and divided it by three. It's going to be all of that times c over negative b minus 2a, so this is going to be equal to c and the numerator up here, we have an a minus 3a, so this is equal to negative b, we have a negative b there and then we have a minus 3a is minus 2a, so this becomes negative b minus 2a, we can factor out the 1/3, times c over three, so that's what the numerator is. All of that over negative b minus 2a, well, that's going to cancel with that, and so our y-coordinate is going to be equal to c over three, so our y-coordinate over here is c over three. Now, we could just use the distance formula now. We know all the coordinates, and feel free to do it if you like, but there's a slight simplification, or at least in my mind a slight simplification. We know that the height, we know that the y-coordinate of the centroid is c over three, so this right here, this point over here, is c over three. We could actually use, maybe an easier argument, because remember, our whole point was to show that this point is 2/3 along the median away from the vertex. Let me draw this vertex now here. Our whole point of this video as we said this is the midpoint of this side over here is to show that this right here is 2/3 along the way, or that this length is twice this length, or that this length is 2/3 of the entire length of the median. So how can we do that? Well, the simplest way I can do that, I mean you could actually just use the coordinates and use the distance formula, but a simpler way, just the way that we've arranged this, is just to use an argument of similar triangles. We know that this entire height right here in blue is c. We know that this height right over here is c over three. Or you could do that as 1/3 times c, so we know that this height over here we know that that distance is 2/3 c, so we can use a similar triangle argument. Let's say that this entire thing right here is the length of the median, and let's just call that, I don't know, let's call that l, and let's say we want to figure out how far along the median away from the vertex that is and let's just call that, let me use another, let me just call that distance, let me do this in a different color. Let's call this distance, right over here, let's see, I've already used a, b, and c, let's call that d. So if we can show that the ratio of d to l is 2/3, that this is 2/3 of the distance, then we're done. We can do that by similar triangles. We have two similar triangles. Let me use a new color. We have this triangle, which is kind of embedded inside of this larger triangle right over here. They both share the same vertex, this angle over here, they both have right angles over here, and since they have two of the same angles, their third angle has to be the same, so they're definitely similar triangles. So we can just set up a ratio here. 2/3 c, which is this length right over here, that's just kind of the vertical side of these smaller triangles, we could write 2/3 c over this entire length, over the length of the larger, this corresponding side of the larger similar triangle, over c is equal to the hypotenuse of the smaller similar triangle, d over the hypotenuse of the longer similar triangle, or the bigger similar triangle. D over l. This is clearly just dividing the numerator and denominator by c, this clearly becomes 2/3, so the ratio of this length to this larger length is 2/3, or this is 2/3 along the median away from the vertex. So there you go, that's the two-dimensional proof that the centroid is 2/3 along the way of any median from the vertex, or 1/3 along the median, or 1/3 the distance of the median from the opposite midpoint. Anyway, hopefully you enjoyed that.


text:
Limited access

At a certain large high school, $58\%$ of all students participate in at least one extra-curricular activity. Dora does not know this, however, and is exploring the issue for herself. One day, she stands at the door to the lunchroom and randomly asks students who are entering if they participate in at least one extra-curricular activity.

What is the probability that a student who participates in at least one extra-curricular activity will be either the $2^\text{nd}$ or $3^\text{rd}$ student that Dora questions?

A

$0.1023$

B

$0.2436$

C

$0.3459$

D

$0.58$

E

$0.7941$

Select an assignment template


text:
# Simple harmonic motion of clock

1. Feb 25, 2006

### gandharva_23

let us consider a bob suspended by means of a thread from a fixed point (wrt ground and i m considering ground to be an inertial frame in this problem ) . What will be the time period of small oscillations of this pandulum as seen by a man accelerating with an acceleration a along the horizontal . will it be 2pi(root(l/g)) or will it be 2pi(root(l/root(a^2+g^2)) .
logic1 : if we solve the problem in ground frame we get T=2pi(root(l/g)) . now at small speeds this time is not frame dependent , hence the time period should be same in mans frame also .
logic2 : We want to calculate T as seen by man so lets solve the problem in mans frame . Since its a noninertial frame a pseudo force will act on the bob and the time period will come out to be
2pi(root(l/root(a^2+g^2)) .
which one is correct ?

2. Feb 25, 2006

### Tom Mattson

Staff Emeritus
If the pendulum itself is inertial then it acts as a clock. In Galilean relativity all clocks keep the same time in all frames. In Special Relativity they don't.

Which one are you using?

3. Feb 25, 2006

### gandharva_23

i am not getting into special relativity and i totally agree that the time period should be 2piroot(l/g) but what is the flaw in logic 2 ?

4. Feb 25, 2006

### topsquark

I believe the problem is that the pedulum will have the same period, but look as if it were pushed off to one side, as if it were not hanging straight down. The pseudo-force will act only in the horizontal direction and thus will not affect the period of motion.

-Dan

5. Feb 25, 2006

### gandharva_23

If pseudo force acts in the horizontal direction then it would definitely alter the time period . if we make the fbd and calculate the time period the time period after application of pseudo force comes out to be 2pi(l/((a^2+g^2)^1/2))^1/2

6. Feb 26, 2006

### topsquark

Also:
I was thinking in the shower this morning. (Always a good place to meditate! ) I think we have both been laboring under a particular assumption, me tacitly and you (gandharva) specifically, that the pendulum bob experiences an extra constant acceleration from the pseudo-force. That a freely falling mass will be "shunted" off to one side by the pseudo-force is true, but the pendulum mass is certainly NOT freely falling. It is constrained to move in a circle, and that circle will still be a circle in the (non-relativisitic) accelerating frame.

I am no longer going to blithely state what the motion of the pendulum is because the simple answer I gave before could easily be wrong. Certainly my previous post is incorrect in regards to detail.

The whole solution to the problem rests with Galilean Relativity. In Galilean Relativity time in an invariant. So whatever time we calculate in one reference frame is going to be the same time in any other reference frame. I am not prepared to say that the pendulum experiences nothing due to the action of the pseudo-force, but the period MUST be the same in both reference frames. I suspect this means the pseudo-force is not constant on the pendulum mass. (After all, it's a PSEUDO-force, so F=ma need not apply!)

-Dan

7. Feb 26, 2006

### gandharva_23

you seem to have meditated intensely :-) on i dont know what ;-) but the solution does not seem to be convincing ....... the period has to be 2piroot(l/g) but the logic that you give to defy the second argument that i gave does not seem to be too convincing ..... neways keep meditating on it... and i ll do the same

8. Feb 26, 2006

### topsquark

Convincing or not, the Galilean Relativity transforms leave t alone. Remember we aren't working in Einsteinian Relativity where there are no preferred reference frames. There ARE preferred reference frames in Galilean Relativity, called "inertial frames." Your observer isn't in one, so a calculation of the period of a pendulum (which really isn't a simple problem in general) might not work out to be so easy.

-Dan

9. Feb 26, 2006

### gandharva_23

Galilean Relativity transforms leave t alone ..... thats exactly i want to say . the time period that should appear to the observer should be same as the time period of pandulum in the ground frame which is 2 pi root(l/g) . but try solving the problem in thet non inertial frame and u 'll get a different result . i know it isnt really simple but i did calculated that and that came out to be equal to 2pi(l/((a^2+g^2)^1/2))^1/2 .

10. Feb 27, 2006

### topsquark

That was exactly my point when I said I suspected the pseudo-force was not acting in such a way to provide a constant acceleration to the pendulum mass. In order to figure out what the pseudo-force is doing in that frame we need to specify that the period is $$2 \pi \sqrt{l/g}$$ and work backward to figure out what the pseudo-force is doing. It doesn't sound like a pretty problem doing it that way :yuck: so, frankly, I'm not going to do it.

-Dan

11. Feb 27, 2006

### qtp

why can't u just add the acceleration into the calculation of the EOM for the pendulum and see what you get? i would think that it would change the measured period in the same way that things on the earth experience a coriolis effect.

12. Mar 2, 2006

### topsquark

gandharva_23 I owe you an apology for two reasons. First, I have made two errors in logic on this thread, but more importantly that I had convinced myself that I knew the Physics of the situation when in retrospect I clearly didn't. I can excuse myself for making a mistake (everyone does) but I can't excuse myself for posting an answer when I didn't completely understand the problem. (I'm a Quantum Physicist, not a Relativist, so I don't usually do accelerating frames.)

As to my error, in my defense, no one corrected it including the mentor-teachers on the forum. It's apparently an easy mistake to make.

Let me demonstrate the problem that's been bugging me for a few days since I made my last post. I concluded that the action of the pseudo-force was not constant because the pendulum period was the same in both an accelerating frame and an inertial frame. This is an obvious mistake as the following experiment would show. Consider the pendulum, not in motion, but as hanging at its equilibrium point. The accelerating observer will NOT see the pendulum hanging off to one side. The accelerating observer will be able to measure the point where the string meets the ceiling to support the pendulum and will be able to tell that it is directly over the pendulum.

So if there is a pseudo-force acting on the pendulum, why isn't it hanging off to one side? The answer is quite simple: EVERYTHING in the reference frame is accelerated by the pseudo-force! Our error was to assume that it applied merely to the pendulum mass.

In retrospect this should be obvious. According to the Galilean transformations, any and all objects in the stationary frame will be "accelerated" by the same amount according to an accelerating observer. The Galilean transforms apply to ANY measurement made in the stationary frame. The error is easy to make since we often only look at the object under study when we use Galilean Relativity. An Einsteinian Relativist might not be so prone to the error since is it quite normal to consider the time transformation at several points in the reference frame. (A similar comment should hold true for Cosmologists as well.)

I hope this has cleared up the confusion generated by my previous posts as well as answering what went wrong in logical argument 2 in your original post.

-Dan

13. Mar 3, 2006

### gandharva_23

What we can say is that if we get into the frame of observer the pivot will also be accelerating in the backward direction . thats what i thought about 2 days ago .


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# Energy Modeling of a Supertall Building Using Simulated 600 m Weather File Data

• Published : 2019.06.01
• 192 5

#### Abstract

Assessing the energy performance of supertall buildings often does not consider variations in energy consumption due to the change of environmental conditions such as temperature, pressure, and wind speed associated with differing elevations. Some modelers account for these changing conditions by using a conventional temperature lapse rate, but not many studies confirm to the appropriateness of applying it to tall buildings. This paper presents and discusses simulated annual energy consumption results from a 600 m tall skyscraper floor plate located in Dubai, UAE, assessed using ground level weather data, a conventional temperature lapse rate of $6.5^{\circ}C/km$, and more accurate simulated 600 m weather data. A typical office floorplate, with ASHRAE 90.1-2010 standards and systems applied, was evaluated using the EnergyPlus engine through the OpenStudio graphical user interface. The results presented in this paper indicate that by using ground level weather data, energy consumption at the top of the building can be overestimated by upwards of 4%. Furthermore, by only using a lapse rate, heating energy is overestimated by up to 96% due to local weather phenomenon such as temperature inversion, which can only be conveyed using simulated weather data. In addition, sizing and energy consumption of fans, which are dependent both on wind and atmospheric pressure, are not accurately captured using a temperature lapse rate. These results show that that it is important, with the ever increasing construction of supertall buildings, to be able to account for variations in climatic conditions along the height of the building. Adequately modeling these conditions using simulated weather data will help designers and engineers correctly size mechanical systems, potentially decreasing overall building energy consumption, and ensuring that these systems are able to provide the necessary indoor conditions to maintain occupant comfort levels.

#### Keywords

Energy modeling;Microclimate;Simulated weather;Lapse rate;Supertall buildings

#### File

Figure 1. Three different lapse rates for Dubai summer design conditions.

Figure 2. Three different lapse rates for Dubai winter design conditions.

Figure 3. Wind speed variation by elevation.

Figure 4. Typical floor axonometric view and plan indicating zonal distribution and window to wall ratio.

Table 1. Comparison of Lapse Rate Methods and Simulated Weather Data for Dubai Design Conditions

Table 2. Internal Gains and Occupancy Density

Table 3. Energy End Use Distribution and Totals by Climate Inputs

#### References

1. Phillips, D. (2012 June). Meteorological Data in a Humid Climate at Grade and 1970ft /600m, ASHRAE Annual Conference, Houston, TX.
2. ASHRAE. (2005). Fundamentals Handbook 2005. ASHRAE, USA.
3. Leung, L., & Weismantle, P. (2008). How Supertall Buildings Can Benefit From Height. CTBUH 2008 8th World Congress, Dubai, CTBUH.
4. Ellis, P.G. & Torcellini, P.A. (2005). Simulating Tall Buildings Using Energy Plus, Ninth International Building Performance Simulation Association Conference and Exhibition, Montreal, IBPSA.
5. Saroglou, T., Meir, I.A., & Theodosiou, T. (2017). Quantifying Energy Consumption in Skyscrapers of Various Heights. Procedia Environmental Sciences, 38, pp. 314-321. https://doi.org/10.1016/j.proenv.2017.03.085
6. Lotfabadi, P. (2013). High-rise buildings and environmental factors. Renewable and Sustainable Energy Reviews, 38, pp. 285-295.


text:
# 10.25. list Type Conversion Function¶

Python has a built-in type conversion function called list that tries to turn whatever you give it into a list. For example, try the following:

The string “Crunchy Frog” is turned into a list by taking each character in the string and placing it in a list. In general, any sequence can be turned into a list using this function. The result will be a list containing the elements in the original sequence. It is not legal to use the list conversion function on any argument that is not a sequence.

It is also important to point out that the list conversion function will place each element of the original sequence in the new list. When working with strings, this is very different than the result of the split method. Whereas split will break a string into a list of “words”, list will always break it into a list of characters.

Next Section - 10.26. Tuples and Mutability


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The birthday problem II : three people or more

This is a follow on from the previous post about the birthday problem.  Now we will look at the more general case where we want more than two people to share a birthday.  Incidentally, the reason for my interest in this problem is that we used to use it as a coding/probability exercise for data science interviews at Qubit (not anymore!), and because there seems to be surprisingly little written about it online.

We will be using the falling factorial  notation $(x)_n$ to ease exposition slightly (because writing latex in wordpress is not so fun).

To simplify, we will assume that birthdays are uniformly distributed throughout the year.  Consider the case where we want to know if three or more people from a group of $N$ share a birthday.  There are other discussions of this on the internet, e.g. this post, but what follows is an easier derivation than what I have seen.  Again, we look at the complement probability – let’s try and work out the probability that all days have at most two people with that as their birthday.  We will break into cases based on how many pairs of people share a birthday, which we will label $m$, so that $0 \leq m \leq \lfloor \frac{N}{2} \rfloor$.  The idea is to count how many ways we can split our $N$ people into $m$ pairs, and $N - 2m$ singletons, and count the distinct number of ways to put these into 365 bins.  This is :

$(N)_{2m}\cdot \frac{1}{2 ^m} \cdot (365)_m \cdot (365 - m)_{N-2m} .$

where the four terms are respectively : the number of ways of selecting $m$ pairs from the $N$ people, a correction term, since we can swap two people in a pair around, the number of ways of selecting the $m$ ‘pair-days’ from 365, and finally the number of way of selecting the remaining $n-2m$ people from the remaining $365 - m$ days.  Simplifying this slightly and taking the sum over $m$, we have the probability of $N$ people having at least one triple is

$1- \frac{1}{365^N} \sum \limits_{m = 0}^{ \lfloor \frac{N}{2} \rfloor }\frac{ (N)_{2m} \cdot (365)_{N-m}} {2 ^ m}.$

It seems unlikely that there is a clean closed formula to calculate the general case here (one was claimed in this stackexchange post, but it’s wrong, there’s some erroneous double counting here).  A natural way to approximate it would be to run a simulation (and in interviews we would push the candidate towards this), but for curiosity’s sake, I’m going to show how you can work it out precisely.

If we think about the case where we want there to be no quadruples, it is already getting quite daunting to try and use our previous strategy to break this up in terms of smaller subgroups (we’d have to look at all the ways that different numbers of triples, pairs and singletons which add up to $N$ can be arranged into 365 bins) – you can do it with an induction onto lower terms (left as an exercise!), but it’s messy, slow, and memory intensive to implement.  Instead, we are going to try and reduce the number of days we need to think about.  To this end, let $F_M (N, k)$ be the number of ways of pigeonholing $N$ objects into $k$ bins so that there are no more than $M$ objects in any one bin. Then the probability that amongst $N$ randomly selected people, there is one set of at least $M$ who all share a birthday is

$1 - \frac{F_{M - 1} (N, 365)}{365^N}.$

Consider a binning that is counted by $F_M (N, k)$.  Let’s assume there are $m$ objects in the first bin, so that $0 \leq m \leq M$.  There are ${N \choose m}$ sets of objects this could be.  So considering what remains, and summing over all valid $m$‘s, we have the recurrence

$F_M (N, k) = \sum_{m=0}^M {N \choose m} F_M (N - m, k - 1).$

This together with the base cases $F_M (N, 1) = 1$ for $0 \leq N \leq M$ and zero otherwise, allows us to calculate the probability exactly.  Here is some python code to do this (this is essentially a tail recursion on $k$ up to 365.  Note the use of Decimal, to keep precision.

import math
from decimal import Decimal
from decimal import getcontext

#bit of a trade off between speed/memory and precision here
getcontext().prec = 50

def binom(n, m):
return Decimal(math.factorial(n)) / (Decimal(math.factorial(m)) * Decimal(math.factorial(n-m)))

def get_birthday_prob(N, M):
current_row = [Decimal(0)] * (N + 1)

#base case
for k in range(M):
current_row[k] = Decimal(1)

#recursion
for d in range(1, 365):
new_row = [Decimal(0)] * (N + 1)
for n in range(N + 1):
s = Decimal(0)
for m in range(min(n + 1, M)):
s += binom(n, m) * current_row[n-m]
new_row[n] = s
current_row = new_row

complement_prob = current_row[N] / Decimal(math.pow(365, N))
return float(1 - complement_prob)

print get_birthday_prob(23, 2)


text:
Thread: CRT optimisation? View Single Post
2016-03-30, 02:08   #9
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26·131 Posts

Quote:
 Originally Posted by jasonp The problem I was thinking of was that in general (a mod b) mod c is not the same as (a mod c) mod b. So if your m0 was larger than m1, then (r1 - r0) mod m1 may not be the same as (r1 - (r0 mod m1)) mod m1. I could be overthinking it though.
to further point out this problem I decided to post an example:

(30 mod 5) mod 7 -> 0 mod 7
(30 mod 7) mod 5 -> 2 mod 5

all we did was change the order of the modulo operations and it changes what we get back.

I think jasonp is overthinking it if CRT is chinese remainder theorem as the mod is never done twice and if you subtract the same thing from two things that are congruent they stay congruent. in the two congruences you give ( unless you meant m0 in one of them) you can think of them as polynomials r1-r0 can be (m1*y+r1)-(m1*z+r2). though I still don't see how you can cross values mod the m's like that. if you remember that any value mod another number is always congruent to itself you can rewrite the second as the first.

Last fiddled with by science_man_88 on 2016-03-30 at 02:11


text:
IntMath Home » Forum home » Complex Numbers » Complex conjugates

# Complex conjugates [Solved!]

### My question

In section 2 the part on dividing complex numbers, it is written that the numerator and denominator are multiplied by the conjugate so that
"we would be left with no radical (square root) in the denominator. Dividing with complex numbers is similar."

Does that mean that we divide by the conjugate so that no complex number will be left in the denominator?

Thank you.

### Relevant page

2. Basic Operations in Complex Numbers

### What I've done so far

X

In section 2 the part on dividing complex numbers, it is written that the numerator and denominator are multiplied by the conjugate so that
"we would be left with no radical (square root) in the denominator. Dividing with complex numbers is similar."

Does that mean that we divide by the conjugate so that no complex number will be left in the denominator?

Thank you.
Relevant page

2. Basic Operations in Complex Numbers

What I've done so far

## Re: Complex conjugates

Hello Phinah. Not exactly (but I can understand why you took that meaning from it).

I revised that part just now on that page. I hope it makes more sense now.

X

Hello Phinah. Not exactly (but I can understand why you took that meaning from it).

I revised that part just now on that page. I hope it makes more sense now.

## Re: Complex conjugates

Okay. I read it. Thank you.

X

Okay.  I read it.  Thank you.


text:
# Efficiency of a Heat Engine

Redbelly98
Staff Emeritus
Homework Helper
Looking good.

Just a side note: those p1 v1f terms could just as easily be p3 v3f, since that is a constant quantity along path 31. This might come in handy later, when you have to manipulate expressions to get things to look like the given answer from post #1.

What's left now are Q12 and Q23.

Q23 (the heat out) isn't really necessary for calculating efficiency, since that is really Qc here and doesn't enter into the efficiency calculation.

So we just need Q12 (the heat in, or energy cost). There are a few ways one might do this:

1. Use Q = ∫ T dS along path 12
2. Use Q = ΔU + W along path 12
3. Use ΔU = 0 = (Q12+Q23+Q31) - (W12+W23+W31) around the entire cycle

The key is to pick which one is the easiest way to arrive at the answer.

I'm going to be away from the computer for several hours, until at least 6 pm (Eastern USA time zone). See how far you can get, then I can check back later.

Once you have Q12, you'll need to calculate the final answer:

Efficiency = (W12+W23+W31) / Q12

and that will involve playing with the algebra to get it in the form given in the problem statement,

$$e = 1 - \frac{1}{f}\left(\frac{1 - \frac{p_3}{p_1}}{1 - \frac{v_1}{v_3}}\right)$$

where "f" is really γ, and I have fixed a typo (v1/v3 instead of v3/v1).

Good luck!

Okay, so I have chosen method 2,

Q = U + W along path 12

so

W = (V3 - V1)P1

$$\Delta U = C_V \Delta T$$ (P const.)

Since Ideal gas, use ideal gas law,

$$T = \frac{PV}{nR}$$

Thus:

$$T_1 = \frac{p_1v_1}{nR}$$

$$T_2 = \frac{p_1v_3}{nR}$$

So:

$$\Delta T = T_f - T_i = T_2 - T_1 = \frac{p_1v_3}{nR} - \frac{p_1v_1}{nR}$$

and thus:

$$\Delta U = C_V (\frac{p_1v_3}{nR} - \frac{p_1v_1}{nR})$$

thus:

$$Q_{12} = (C_V (\frac{p_1v_3}{nR} - \frac{p_1v_1}{nR}) + (V_3 - V_1)P_1$$

Now efficiency,

$$e = \frac{W_{12} + W_{23} + W{31}}{Q_{12}}$$

$$e = \frac{\left[\frac{p_1v_1^f}{v_1^{f - 1}(-f + 1)}\right] - \left[\frac{p_1v_1^f}{v_3^{f - 1}(-f + 1)}\right] + (v_3 - v_1)p_1 + 0}{(C_V (\frac{p_1v_3}{nR} - \frac{p_1v_1}{nR}) + (V_3 - V_1)P_1}$$

$$e = \frac{\left[\frac{p_1v_1^f}{v_1^{f - 1}(-f + 1)}\right] - \left[\frac{p_1v_1^f}{v_3^{f - 1}(-f + 1)}\right] + (v_3 - v_1)p_1}{(C_V (\frac{p_1v_3}{nR} - \frac{p_1v_1}{nR}) + (V_3 - V_1)P_1}$$

I am assuming there is a way to simplify this...

I will keep looking, but I am sure I can get the 1 at the neginning from the last part of the fraction, and the 1/gamma isclose, as I have 1/Cv, and gamma is just CP/Cv,

But how can I get rid of the nR's?

Redbelly98
Staff Emeritus
Homework Helper
Good, that answer is a correct expression. And yes, it must be simplified to get the asked-for expression.

It's all algebra from this point onward.

• Anything you know that relates Cv, Cp, gamma, and nR to one another
• p1 v1f = p3 v3f (because path 31 is adiabatic)

Like I said, it's just playing around with the algebra at this point.

Good luck!

Okay, so:

$$\gamma = f = \frac{C_p}{C_v}$$

$$C_p - C_v = nR$$

So:

$$e = \frac{\left[\frac{p_1v_1^f}{v_1^{f - 1}(-f + 1)}\right] - \left[\frac{p_1v_1^f}{v_3^{f - 1}(-f + 1)}\right] + (v_3 - v_1)p_1}{(C_V (\frac{p_1v_3}{nR} - \frac{p_1v_1}{nR}) + (V_3 - V_1)P_1}$$

$$e = \frac{\left[\frac{p_1v_1^f}{v_1^{f - 1}(-f + 1)}\right] - \left[\frac{p_1v_1^f}{v_3^{f - 1}(-f + 1)}\right] + (v_3 - v_1)p_1}{(C_V (\frac{p_1v_3}{C_p - C_v} - \frac{p_1v_1}{C_p - C_v}) + (V_3 - V_1)P_1}$$

Now the equation has a 1/\gamma, so would this be on the right path?

$${C_v} \frac{C_p}{f}$$

$$e = \frac{\left[\frac{p_1v_1^f}{v_1^{f - 1}(-f + 1)}\right] - \left[\frac{p_1v_1^f}{v_3^{f - 1}(-f + 1)}\right] + (v_3 - v_1)p_1}{(\frac{C_p}{f} (\frac{p_1v_3}{C_p - C_v} - \frac{p_1v_1}{C_p - C_v}) + (V_3 - V_1)P_1}$$

and since p1v1 ^f = v3p3^f

$$e = \frac{\frac{p_1v_1^f}{v_1^{f - 1}(-f + 1)} - \frac{p_3v_3^f}{v_3^{f - 1}(-f + 1)} + (v_3 - v_1)p_1}{(\frac{C_p}{f} (\frac{p_1v_3}{C_p - C_v} - \frac{p_1v_1}{C_p - C_v}) + (V_3 - V_1)P_1}$$

Does this look okay?

Andrew Mason
Homework Helper
Those are the right limits, but in the wrong order. To see why, note that path 31 starts at "3" and ends at "1". So the limits are

$$\left[ \text{...} \right] ^{v1}_{v3}$$

EDIT: including figure from 1st post:
There is a simple way to do this. Apply the first law. There is no change in U in one cycle. So the work is simply Qh-Qc = Q12-Q23. You just have to work out the temperatures to determine what those heats are.

AM

Okay so:

$$Work = Q_1_2 - Q_3_1$$

$$Q_{12) = (C_V (\frac{p_1v_3}{nR} - \frac{p_1v_1}{nR}) + (V_3 - V_1)P_1$$

Okay so Q23

Q = U + W along path 23

W = 0

$$\Delta U = C_P \Delta T$$ (V const)

using Ideal gas law, T = PV/nR

$$\Delta T = \frac{P_3V_3}{nR} - \frac{P_1V_3}{nR}$$

$$\Delta U = C_P \frac{P_3V_3}{nR} - \frac{P_1V_3}{nR}$$

$$Q_{23} = C_P \frac{P_3V_3}{nR} - \frac{P_1V_3}{nR}$$

So:

$$e = \frac{Work}{Q_{12}}$$

$$Work = \left(C_V (\frac{p_1v_3}{nR} - \frac{p_1v_1}{nR}) + (V_3 - V_1)P_1\right) - \left(C_P \frac{P_3V_3}{nR} - \frac{P_1V_3}{nR}\right)$$

$$e = \frac{\left(C_V (\frac{p_1v_3}{nR} - \frac{p_1v_1}{nR}) + (V_3 - V_1)P_1\right) - \left(C_P \frac{P_3V_3}{nR} - \frac{P_1V_3}{nR}\right)}{(C_V (\frac{p_1v_3}{nR} - \frac{p_1v_1}{nR}) + (V_3 - V_1)P_1}$$

So does this look okay now?

Andrew Mason
Homework Helper
...
So:

$$e = \frac{Work}{Q_{12}}$$

$$Work = \left(C_V (\frac{p_1v_3}{nR} - \frac{p_1v_1}{nR}) + (V_3 - V_1)P_1\right) - \left(C_P \frac{P_3V_3}{nR} - \frac{P_1V_3}{nR}\right)$$

$$e = \frac{\left(C_V (\frac{p_1v_3}{nR} - \frac{p_1v_1}{nR}) + (V_3 - V_1)P_1\right) - \left(C_P \frac{P_3V_3}{nR} - \frac{P_1V_3}{nR}\right)}{(C_V (\frac{p_1v_3}{nR} - \frac{p_1v_1}{nR}) + (V_3 - V_1)P_1}$$
Try:

$Q_{12} = n\gamma C_v(T_2-T_1)$ and $Q_{23} = nC_v(T_3-T_2)$

$$\eta = W/Q_h = \frac{Q_{12} - Q_{23}}{Q_{12}} = \frac{(\gamma(T_2-T_1) - (T_3 - T_2)}{\gamma(T_2-T_1)}$$

Work out T1 and T3 using:

$$TV^{\gamma-1} = const.$$

ie.

$$T_3 = T_1\left(\frac{V_1}{V_3}\right)^{\gamma-1} = \frac{P_1V_1}{nR}\left(\frac{V_1}{V_3}\right)^{\gamma-1}$$

$$T_2 = P_1V_3/nR$$

AM

I see. So:

$$Q = n C_P \Delta T$$

From the expression for gamma,

$$C_p = f C_V$$ (I am using f because gamma doesn't show well on latex)

so

$$Q = n f C_V \Delta T$$

$$Q_{12} = n f C_V (T_2 - T_1)$$

$$Q_{23} = n C_V (T_3 - T_2)$$

$$e = W/Q_h = \frac{Q_{12} - Q_{23}}{Q_{12}}$$

$$\frac{Q_{12} - Q_{23}}{Q_{12}} = \frac{nf C_V (T_2 - T_1) - n C_V (T_3 - T_2)}{n f C_V (T_2 - T_1)}$$

Cancels to:

$$\frac{Q_{12} - Q_{23}}{Q_{12}} = \frac{f(T_2 - T_1) - (T_3 - T_2)}{f(T_2 - T_1)}$$

$$TV^f = c$$

$$T_{1}V_{1}^{f -1}= T_{3}V_{3}^{f-1}$$

$$T_{1}= T_{3}\frac{V_{3}^{f-1}}{V_{1}^{f -1}}$$

Use ideal gas law:

T = PV/nR

$$\T_1= \frac{P_3V_3}{nR}\frac{V_{3}^{f-1}}{V_{1}^{f -1}}$$

$$T_2 = \frac{P_1V_3}{nR}$$

$$T_3 = \frac{P_1V_1}{nR}\frac{V_{1}^{f-1}}{V_{3}^{f -1}}$$

so:

$$\frac{(f(\frac{P_1V_3} - \frac{P_3V_3}{nR}\frac{V_{3}^{f-1}}{V_{1}^{f -1}}) - (\frac{P_1V_1}{nR}\frac{V_{1}^{f-1}}{V_{3}^{f -1}} - \frac{P_1V_3})}{f(\frac{P_1V_3} - \frac{P_3V_3}{nR}\frac{V_{3}^{f-1}}{V_{1}^{f -1}})}$$

Okay so far?

Andrew Mason
Homework Helper
I
$$\frac{(f(\frac{P_1V_3} - \frac{P_3V_3}{nR}\frac{V_{3}^{f-1}}{V_{1}^{f -1}}) - (\frac{P_1V_1}{nR}\frac{V_{1}^{f-1}}{V_{3}^{f -1}} - \frac{P_1V_3})}{f(\frac{P_1V_3} - \frac{P_3V_3}{nR}\frac{V_{3}^{f-1}}{V_{1}^{f -1}})}$$

Okay so far?
Notice that:

$$\eta = \frac{(\gamma(T_2-T_1) - (T_3 - T_2)}{\gamma(T_2-T_1)}$$

reduces to:

$$\eta = 1 - \frac{1}{\gamma}\frac{(T_3 - T_2)}{(T_2-T_1)}$$

So you just have to show that:

$$\frac{(T_3 - T_2)}{(T_2-T_1)} = \frac{(1-\frac{P_3}{P_1})}{(1-\frac{V_3}{V_1})}$$

AM

Notice that:

$$\eta = \frac{(\gamma(T_2-T_1) - (T_3 - T_2)}{\gamma(T_2-T_1)}$$

reduces to:

$$\eta = 1 - \frac{1}{\gamma}\frac{(T_3 - T_2)}{(T_2-T_1)}$$
I see that now.

$$\eta = \frac{(\gamma(T_2-T_1) - (T_3 - T_2)}{\gamma(T_2-T_1)}$$

is the same as:

$$\eta = \frac{(\gamma(T_2-T_1)}{\gamma(T_2-T_1)} - \frac{(T_3 - T_2)}{\gamma(T_2-T_1)}$$

Which cancels down to:

$$\eta = 1 - \frac{1}{\gamma}\frac{(T_3 - T_2)}{(T_2-T_1)}$$

Okay so:

$$\frac{(T_3 - T_2)}{(T_2-T_1)}$$

$$T_1= \frac{P_3V_3}{nR}\frac{V_{3}^{f-1}}{V_{1}^{f -1}}$$

$$T_2 = \frac{P_1V_3}{nR}$$

$$T_3 = \frac{P_1V_1}{nR}\frac{V_{1}^{f-1}}{V_{3}^{f -1}}$$

So:

$$\frac{(\frac{P_1V_1}{nR}\frac{V_{1}^{f-1}}{V_{3}^{f -1}} - \frac{P_1V_3}{nR})}{(\frac{P_1V_3}{nR} -\frac{P_3V_3}{nR}\frac{V_{3}^{f-1}}{V_{1}^{f -1}})}$$

Factorise out:

$$\frac{(\frac{P_1V_1}{nR}\frac{V_{1}^{f-1}}{V_{3}^{f -1}} - \frac{P_1V_3}{nR})}{\frac{V_3}{nR}(P_1 - P_3 \frac{V_3^{\gamma - 1}}{v_1^{\gamma - 1}})}$$

and:

$$\frac{\frac{p_1}{nR}(v_1\frac{v_1^{\gamma - 1}}{v_3^{\gamma - 1}} - V_3)}{\frac{V_3}{nR}(P_1 - P_3 \frac{V_3^{\gamma - 1}}{v_1^{\gamma - 1}})}$$

Does this look okay so far?

I am also assuming that the nR can cancel, so:

$$P_1(v_1\frac{v_1^{\gamma - 1}}{v_3^{\gamma - 1}} - V_3)}{V_3(P_1 - P_3 \frac{V_3^{\gamma - 1}}{v_1^{\gamma - 1}})}$$

Does this look right?

Redbelly98
Staff Emeritus
Homework Helper
To be honest I have lost track. However, it would be desirable to simplify the term
v1 v1γ-1
The idea is, once you have an expression for the efficiency, for you to manipulate and simplify it algebraically and get

$$e = 1 - \frac{1}{\gamma}\left(\frac{1 - \frac{p_3}{p_1}}{1 - \frac{v_1}{v_3}}\right)$$

which, as I had mentioned earlier, has a typo corrected from what was given in Post #1 (v1/v3 instead of v3/v1).

Oops, my Latex went a bit wrong. it should have been:

$$\frac{P_1(v_1\frac{v_1^{\gamma - 1}}{v_3^{\gamma - 1}} - V_3)}{V_3(P_1 - P_3 \frac{V_3^{\gamma - 1}}{v_1^{\gamma - 1}})}$$


text:
# Recognizing Musical Melodies for Secure Audio Pairing

## Machine Learning for Computer Security

By Pranav Rajpurkar, Brad Girardeau on October 2nd 2016

In this post, we share our work at the intersection of machine learning and security. We look at secure device pairing, a task in which we are trying to establish a secure connection between two devices wirelessly. We establish a protocol in which the devices can be paired securely if each device can recognize a melody played by the other. To enable this application, we build a machine learning system that can perform melody recognition.

We begin this post with a focus on the security application: describing the secure device pairing problem, highlighting drawbacks of existing solutions, and finally demonstrating how an audio-based pairing approach can be used to overcome them. In the second half of the post, we focus on the machine learning task of melody recognition: sharing our experience exploring different approaches to the task, and detailing a solution that performs well even in noisy environments.

Listen to the melody below, which consists of 8-notes played in sequence (in a noisy environment) – can you identify any of them? The system we build later in the post gets them all right.

# Secure Device Pairing

Most of us have used Bluetooth at some point in our lives. We are familiar with the routine of “pairing” two devices, like a phone and car, which finishes with entering a PIN or comparing some numbers. The goal of this last step, of course, is security. You don’t want someone next to you to be able to eavesdrop on you or trick you into connecting to them to steal your data. However this security step is often annoying (typing a PIN) or easily skippable (comparing numbers). This problem promises to become increasingly common as new applications like smart cars and connected homes continue linking things in the physical world together.

We set out to create a better way of securing device pairing in the physical world. Instead of typing a PIN or comparing numbers, a simple audio melody is played and automatically recognized. An attacker would only be able to subvert the pairing by playing their own melody, which would be easily noticed. Otherwise we do not need to do anything.

So how does this work? How can entering a PIN, comparing numbers, or recognizing an audio sequence make things secure? Let’s first learn about secure device pairing in general, before delving into the details of how we use audio to improve the process.

## Background

Like many secure communication discussions, we start with our characters Alice and Bob. In this case, Alice has a device she would like to pair with a device of Bob’s that she hasn’t connected to before. At the end of the pairing, the goal is for Alice’s device and Bob’s device to share a secret key that only they know, even if there are attackers who listen or interfere. The devices would then be securely paired because they can use the secret key to encrypt all of their communication so that no one else will be able to eavesdrop on or tamper with them.

There are two types of attackers that Alice and Bob need to be concerned with: a passive attacker we’ll call Eve and an active attacker Mallory. Eve, as a passive attacker, can only eavesdrop on, not tamper with the communication nor send messages herself. If Alice and Bob only need to defend against Eve, they can use the Diffie-Hellman protocol to safely create a shared secret key. The challenge is when Mallory, an active attacker, is involved. While Mallory can’t subvert the Diffie-Hellman key exchange directly, Mallory can insert herself between Alice’s device and Bob’s device, tricking Alice’s device and Bob’s device into separately pairing with Mallory’s device. Both Alice and Bob will now think they share a secret key with each other, while in reality Alice and Mallory share one key, while Bob and Mallory share a different key. This allows Mallory to eavesdrop on or impersonate Alice and Bob.

To prevent Mallory’s attack, we need to add some way for Alice and Bob to be sure they are connecting to each other, not an attacker, for the Diffie-Hellman key exchange. There are several ways to do this using a third party that Alice and Bob both trust to verify each other’s identity. However, for devices and protocols like Bluetooth, finding a single third party everyone will trust is impractical. This is where Short Authenticated Strings can help.

### Short Authenticated Strings (SAS)

A short authenticated string uses the idea that certain channels are naturally “authenticated” even if they are not secret. This means we know we are communicating with another party because an attacker can’t tamper with messages across this channel or will be detected if they do. For example, if Alice and Bob have a face to face conversation, they know they are communicating to the right person and could tell (and ignore) if Mallory tried to shout a different message at them. Even over the phone, Alice and Bob might be able to recognize each other’s voice, so they can “authenticate” information exchanged during the conversation.

Going back to the Bluetooth pairing example at the beginning, entering a PIN and comparing numbers also form authenticated channels. Alice sees the PIN on Bob’s device and types it into her device or compares it with a PIN on her device; an attacker presumably cannot tamper with Alice’s vision, so these channels allow short strings like the PIN to be authenticated.

Once we have an authenticated channel, we can use it to authenticate Alice and Bob’s Diffie-Hellman key exchange. All they need to do is ensure that they have agreed on the same key, since Mallory’s attacks will cause them to have different keys. One way to do this is by sending a hash, or summary, of the key over the authenticated channel. However, the hash in this case is too long for this to be practical; no one will compare or type 100 characters just to pair their phone with a car or computer. We need short authenticated strings, and it turns out there is a protocol that lets us authenticate a message using much shorter authenticated strings.

### Secure Audio Pairing

However typing a PIN or comparing numbers still require a lot of active human intervention. Instead, we can use audio as a authenticated channel and automate the process. If we encode the short authenticated string as melodic notes, Alice’s device can simply the play the sound and Bob’s device can recognize it. If an attacker tries to interfere with the audio channel, Alice or Bob will hear it, so the audio channel is naturally authenticated.

Putting the steps together, two users can pair devices securely by doing a standard key exchange over a normal communications channel like Bluetooth, then authenticate the shared key for the connection with a Short Authenticated String protocol. The Short Authenticated String protocol involves each device playing a short audio sequence, which the other device must recognize. No intervention beyond passively listening is required by the users, as the only attack they need to detect is if they hear a third unexpected device playing audio. From the user’s perspective, this is an easy and seamless process.

# Exploring Approaches to Melody Recognition

A key part of the secure audio pairing system is recognizing the audio notes played, which can be a challenging task in noisy environments. We use machine learning for melody recognition. To recognize a melody is to recognize the sequence of notes that constitute it. We seek to develop a system which can listen to an audio file and recognize the melody. An example recognition of a four-note melody is $[A, C, E, G]$. If we have sheet music for an audio file, we know exactly when each note is played, and for how long. In this case, the input audio and labellings are said to have alignment, and the data is said to be strongly labelled. The harder case is when we have the sequence of notes that constitute the melody, but no information on when each note is played. In this case, input data and labellings are said to have unknown alignment, the data is said to be weakly labelled, and the task is called temporal classification. We often encounter temporal classification in speech recognition, where we may know from an audio recognition that a person says “hello world”, but we don’t know the precise moments when each of the words (or phonemes/graphemes) is uttered.

## Without Alignment Information

When the alignment between audio data and melodies is not given, a model must map an input sequence of audio data to an output sequence of discrete notes that constitute the melody – a temporal classification task.

In the early 2000s, graphical models such as Hidden Markov Models (HMMs) and Conditional Random Fields (CRFs) were the reigning kings for temporal classification. At that time, Recurrent Neural Networks (RNNs), although able to provide a powerful mechanism for time series modelling, were not directly applicable to temporal classification. RNNs had an API constraint: they required target outputs for each point in the training sequence. Temporal classification with a single neural network was a feat yet unachieved.

In 2006, Alex Graves et al. introduced Connectionist Temporal Classification (CTC), a way to label sequence data with neural networks without need for either alignment data or post-processing. CTC (also called CTC-RNN to make clear that the choice of network is recurrent) defines a differentiable loss function that calculates the probability of a melody conditioned on the input as a sum over all possible alignments of the audio input with the notes in the melody. The CTC network has since evolved, and is used in state-of-the-art speech recognitions systems (see DeepSpeech2) today.

## With Alignment Information

One simple way to use alignment information between audio and notes is to split the audio data into segments such that each audio segment corresponds to one output class (note). A model can then learn to map from an audio segment to a single note, as opposed to a sequence of notes. Rather than recognize the melody all at once from the entire audio sequence, we can thus recognize each note that constitutes it one audio-segment at a time. This idea is more generally called the sliding window approach, is commonly used in pedestrian detection and photo OCR (see Coursera video by Andrew Ng on sliding windows.

A major drawback of having models work on segments, rather than on the entire sequence is that context information is lost. Just as certain sequences of words make much more sense than others in accordance with the rules of syntax and grammar, certain sequences of notes are more likely than others in terms of musicality. Any model that operates at the level of segments (or words) rather than melodies (or sentences) signs off any benefit that it may have acquired from context. A popular workaround is to hybridize segment-level models, responsible for making local predictions, with HMMs, responsible for modelling global behaviors of data.

Models can also learn to recognize the entire melody from the audio sequence without any segmentations. In recent years, working with sequential inputs and outputs for neural networks has become extraordinarily effective, and has led to the popularization of the sequence-to-sequence learning paradigm.

# A System for Melody Recognition

While the CTC / sequence-to-sequence class of approaches to the problem of melody recognition seemed promising, their implementations (at time of writing) were hard to get to work well. Nevertheless, this still seems a promising direction for future work, given their results in speech recognition.

We turn instead to the pre-CTC era approach of sliding windows, where a model detects individual notes in an audio segment (window) that slides through the audio sequence; the notes outputted are then preprocessed to get the melody. We’ll see that sliding windows can work incredibly well with the right choice of segment-level model.

Let’s start understanding what the system we will implement will do. We can think of the system as having three sequential components:

• Preprocessing: Segment the audio sequence by sliding a small window across time.
• Model: Recognize the note that constitutes each segment.
• Postprocessing: Combine the note outputs over all of the windows to get the melody.
Taking sliding time windows over the audio, a model identifies the presence of single notes in each of the segments, the outputs of which are postprocessed to get the melody.

The preprocessing step is straightforward: one simply needs to do is find a window size large enough that it can contain the entirety of one note, but small enough so it doesn’t contain two notes. In the domain object detection, where different objects can be of different scales, it is common to find sliding windows of multiple scales being used. However, in this application, we can get away with setting a constant window size, because the notes are of the same length.

The model and postprocessing step, then, are the meat of the difficulty. The model is responsible for recognizing the note that constitutes each segment, and the postprocessing for combining the outputs of the model to infer the underlying melody.

## A Model for Recognizing Notes in Audio Segments

Our model will take a segment of audio as input and determine which note, if any, was played in that segment. One way to formulate this task is as a multiclass classification problem with $n + 1$ classes, where the extra note class denotes silence or background noise. An alternative, often seen in character recognition, is to break the task into detection and recognition steps such that a detector decides whether or not a segment contains a note and a recognizer then classifying the positive detected note into one of $n$ classes. In our experiments, we found that posing the task as an $n + 1$ class classification worked better.

We haven’t yet defined what an audio segment is. We can think of an audio segment as a sequence of samples $[s_1, s_2, … s_t]$, where each sample is a two-dimensional real-valued vector $[ v_l , v_r ]$. Here, the values represent the instantaneous amplitude of the audio signal in the left and right audio channels. For the purposes of our example, let’s use the following length-3 sample input: $x = [ [1.0, 4.0] , [2.0, 5.0], [3.0, 3.0] ]$. Our model maps this input to a probability distribution over all the possible notes. If the possible outputs were A, B, C (and the noise class), then a sample output might look like $y = [noise: 0.1, A: 0.2, B: 0.6, C: 0.1]$.

### Recurrent Neural Networks to Model Temporal Dependencies

The temporal properties of audio make it inherently sequential. Recurrent Neural Networks (RNNs) are a special class of neural networks that can exploit this sequential structure, using information from the past to guide decision making in the present.

The recurrent network takes in a sample at each time step as input (red), and at the last time step, produces a probability distribution over the possible notes as output (blue). In the hidden (green) layer, the network is propagating information through timesteps.

(See Alex Graves’ thesis and Andrej Karpathy’s post for some really cool RNN work)

To an RNN, we feed in the input one audio sample at a time until we’ve exhausted all of the samples in the audio segment. Typically, when one uses an RNN network, the specific variant of recurrent network used is an LSTM, which is better at handling longer sequences of input. It is also typical to stack multiple hidden (green) LSTM layers on top of each other to form a deep network, so that each subsequent layer learns a slightly more abstract representation of the data, which turns out to be useful in improving the accuracy of the classification. Once all of the samples in the segment are fed into the network, the network outputs scores for each note, which are normalized to translate to probabilities for each class – how likely it was that any of the notes were played in that specific segment.

We thus train an RNN on several hours of synthetically generated audio data. One effective trick used for training the network, also commonly seen in speech recognition, is to mix different kinds of noise on the clean audio data. With this, we are not only able to augment the size of the dataset, but also make the network more robust to noisy audio environmments.

### Postprocessing Sliding Window Outputs To Label Sequences

The model described above is trained to output a probability distribution over the possible classes for each segment of the sliding window. The final step is to combine the outputs of each of these segments to get the melody.

Sliding window outputs over time show changing probabilities of classes across time. Non-Maximum Suppression keeps the local maxima, which can be used to recover the underlying melody [Note A, Note B]

To recover the underlying melody from sliding window outputs, we use a variant of Non-Maximum Suppression (NMS). The basic idea of NMS is to look at each point in the sequence, and suppress neighbouring points that are smaller to zero. At the end of the NMS run, only the local maxima remain. Maxima separated by the silence / noise class can then be read off as the notes constituting the melody.

# In Summary

We have described a simple machine learning approach to recognize musical melodies with a concrete application in secure audio pairing. For the learning system, we use a sliding window to break a large chunk of audio into smaller segments, run a Recurrent Neural Network – which captures temporal properties of audio – over each of those segments to make segment-level predictions, and finally use Non-Maximum Suppression to postprocess the outputs. This system is a part of a security protocol which involves devices establishing a secure connection with each other by playing a short audio sequence, which the other device must recognize.

This is work done at Stanford University, advised by Professor Dan Boneh


text:
Definition 14.3.1. Let $\mathcal{C}$ be a category.

1. A simplicial object $U$ of $\mathcal{C}$ is a contravariant functor $U$ from $\Delta$ to $\mathcal{C}$, in a formula:

$U : \Delta ^{opp} \longrightarrow \mathcal{C}$
2. If $\mathcal{C}$ is the category of sets, then we call $U$ a simplicial set.

3. If $\mathcal{C}$ is the category of abelian groups, then we call $U$ a simplicial abelian group.

4. A morphism of simplicial objects $U \to U'$ is a transformation of functors.

5. The category of simplicial objects of $\mathcal{C}$ is denoted $\text{Simp}(\mathcal{C})$.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


text:
# How to programatically calibrate Turningy esc?

I have a Turnigy ESC and I am controlling it from AVR. Now I need to calibrate it to set the range of the input. With a servo tester I managed to calibrate it without any problems, more or less by following the user guide, but when I try to do the same procedure from code, the ESC starts beeping in some confused pattern and then enters programming mode.

My code looks like this:

void calibrate_turningy_esc()
{
servo_set16(SERVO_RANGE_TICKS);
for (uint16_t i = 0; i < 10000; ++i)
_delay_ms(1);
servo_set16(-SERVO_RANGE_TICKS);
for (uint16_t i = 0; i < 10000; ++i)
_delay_ms(1);
servo_set16(0);
for (uint16_t i = 0; i < 10000; ++i)
_delay_ms(1);
}

where +SERVO_RANGE_TICKS is 2.2ms pulse length, -SERVO_RANGE_TICKS is 0.8ms pulse length and 0 is 1.5ms. The timeouts of 10s were measured during the manual calibration with a stopwatch.

I have checked with an oscilloscope that the output servo signal looks the way I would expect it -- 10 seconds of 2.2ms pulses, 10 seconds of 0.8ms pulses and then 1.5ms pulses. Edit: I made a mistake here, see my answer.

Do you have any idea what to change to calibrate the ESC?


text:
## Growth of solutions of nonhomogeneous linear differential equations.(English)Zbl 1172.34058

Summary: This paper studies the growth of solutions of linear differential equations of type
$f^{(k)}+A_{k - 1}(z)f^{(k - 1)}+ \dots +A_{1}(z)f^{\prime}+A_{0}(z)f=H(z),$
where $$A_{j}$$ ($$j=0,\dots,k - 1)$$ and $$H$$ are entire functions of finite order.

### MSC:

 34M10 Oscillation, growth of solutions to ordinary differential equations in the complex domain
Full Text:

### References:

 [1] W. K. Hayman, Meromorphic Functions, Oxford Mathematical Monographs, Clarendon Press, Oxford, UK, 1964. · Zbl 0115.06203 [2] G. Jank and L. Volkmann, Einführung in die Theorie der ganzen meromorphen Funktionen mit Anwendungen auf Differentialgleichungen, Birkhäuser, Basel, Switzerland, 1985. · Zbl 0682.30001 [3] I. Laine, Nevanlinna Theory and Complex Differential Equations, Walter de Gruyter, Berlin, Germany, 1993. · Zbl 0784.30002 [4] G. G. Gundersen, E. M. Steinbart, and S. Wang, “The possible orders of solutions of linear differential equations with polynomial coefficients,” Transactions of the American Mathematical Society, vol. 350, no. 3, pp. 1225-1247, 1998. · Zbl 0893.34003 [5] S. Hellerstein, J. Miles, and J. Rossi, “On the growth of solutions of certain linear differential equations,” Annales Academiæ Scientiarum Fennicæ. Series A I, vol. 17, no. 2, pp. 343-365, 1992. · Zbl 0759.34005 [6] G. G. Gundersen and E. M. Steinbart, “Finite order solutions of nonhomogeneous linear differential equations,” Annales Academiæ Scientiarum Fennicæ. Mathematica, vol. 17, pp. 327-341, 1992. · Zbl 0765.34004 [7] J. Wang and I. Laine, “Growth of solutions of second order linear differential equations,” Journal of Mathematical Analysis and Applications, vol. 342, no. 1, pp. 39-51, 2008. · Zbl 1151.34069 [8] J. Tu and C.-F. Yi, “On the growth of solutions of a class of higher order linear differential equations with coefficients having the same order,” Journal of Mathematical Analysis and Applications, vol. 340, no. 1, pp. 487-497, 2008. · Zbl 1141.34054 [9] Z.-X. Chen, “On the hyper order of solutions of higher order differential equations,” Chinese Annals of Mathematics. Series B, vol. 24, no. 4, pp. 501-508, 2003. · Zbl 1047.30019 [10] C.-C. Yang and H.-X. Yi, Uniqueness Theory of Meromorphic Functions, vol. 557 of Mathematics and Its Applications, Kluwer Academic Publishers, Dordrecht, The Netherlands, 2003. · Zbl 1070.30011 [11] Z.-X. Chen and C.-C. Yang, “Some further results on the zeros and growths of entire solutions of second order linear differential equations,” Kodai Mathematical Journal, vol. 22, no. 2, pp. 273-285, 1999. · Zbl 0943.34076 [12] A. Markushevich, Theory of Functions of a Complex Variable, vol. 2, Prentice-Hall, Englewood Cliffs, NJ, USA, 1965. · Zbl 0142.32602 [13] G. G. Gundersen, “Estimates for the logarithmic derivative of a meromorphic function, plus similar estimates,” Journal of the London Mathematical Society, vol. s2-37, no. 121, pp. 88-104, 1998. · Zbl 0638.30030 [14] G. G. Gundersen, “Finite order solutions of second order linear differential equations,” Transactions of the American Mathematical Society, vol. 305, no. 1, pp. 415-429, 1988. · Zbl 0669.34010 [15] I. Laine and R. Yang, “Finite order solutions of complex linear differential equations,” Electronic Journal of Differential Equations, vol. 2004, no. 65, pp. 1-8, 2004. · Zbl 1063.30031
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
# PHP Table generation

I have been using the following function (in the kohana html class in a table) to generate tables.

function table($columns,$reverse=false,$fill=false,array$attr=null){
$c = ''; //ugly hack, looking for fix$max=sizeof(max($columns)); for($i=0;$i<sizeof($columns);$i++){$column = $columns[$i];

if($fill==true){$l = ($max-sizeof($column));
for($ii=0;$ii<$l;$ii++){
array_push($column,''); } } if($reverse==true){
$columns[$i] = array_reverse($column); }else{$columns[$i] =$column;
}
}
for($i=0,$l=sizeof($columns);$i<$l;$i++){
$column =$columns[$i];$c.="
";
for($ii=0,$ll=sizeof($column);$ii<$ll;$ii++){
$c.="".$column[$ii]."	"; }$c.="
";
}
return "
".$c."
";
}

And then I have been calling it like so:

echo html::table(array(
array('colunm 1 row 1','colunm 2 row 1'),
));

And a reversed table that auto fills the columns.

echo html::table(array(
array('colunm 1 row 1','colunm 2 row 1'),
array('column 1 row 2','column 2 row 2')
),true,true);

• There's also PEAR's HTML_Table package, if you're into that sort of thing. :-) May 31, 2011 at 0:06

Didn't test, but I would write it something like this:

function table($rows,$reverse = false, $fill = false, array$attr = null) {
$c = '';$max_cols = sizeof(max($columns)); foreach ($rows as $columns) { if ($fill && sizeof($columns) <$max_cols) {
$columns = array_merge($columns, array_fill(0, $max_cols - sizeof($columns), ''));
}
if ($reverse) {$columns = array_reverse($columns); }$c .= '
' . implode('	', $columns) . '
'; } return '
' .$c . '
';
}

• Sweet, I wasn't aware of array_fill or implode. Very convenient and it shortens the function by a lot. There was one bug $fill_to should be $max_cols. I decided to change some of the naming from $columns to $rows. Here is the edited version: codepad.org/UQyynzqW Thanks. May 6, 2011 at 21:36
• that's what i get for not testing. fixed the variable, and naming to make more sense. May 7, 2011 at 7:26
• Some of the variable names could benefit from being slightly less terse.
• Aren't you passing in an array of rows? Each element in the array is then an array of columns.
• What is max? Is it trying to find the row with the most columns? Either rename to something more self-documenting or add a comment.
• array_map may be worth a look and could feasibly allow for reverse to be replaced with a callback function.
• Consider changing the last lines of the function to:

return "
".$c."
";

• I'm not sure what I was thinking for the return statement. Or rather I wasn't thinking %). Also, yes I'm passing in an array of rows which contains an array of columns. May 6, 2011 at 2:39
• Yes max is attempting to find the max number of columns. Basically if you reverse a table with out all the rows having the same number of columns, the table will fill to the left. By passing in true to $fill it adds additional empty cells to the end of table so it doesn't fill to the left. What would you advise for a naming alternative to $max? That's the php function name so I thought it was well suited. May 6, 2011 at 2:42
• Perhaps $max could be called$num_tbl_columns since it's the number of columns in the table. May 6, 2011 at 12:51

replace for loops with foreach

• I would love to use foreach, but I am unsure how to use it in place of the first for loop. I am reassigning itself as a reversed array at the end of the loop. I don't believe php allows you to reassign the looping variable inside a foreach. May 6, 2011 at 2:52
• Actually, you can using foreach($array as$key => &\$value) May 6, 2011 at 17:03


text:
## modphysnoob Group Title A mechanism on Earth used to shoot down geosynchronous satellites that house laser-based weapons is finally perfected and propels golf balls at 0.94c. a) How far will a detector riding with the golf ball initially measure the distance to the satellite? [Geosynchronous satellites are placed 3.58x104 km above the surface of the Earth.] b) How much time will it take the golf ball to make the journey to the satellite in the Earth’s frame? How much time will it take in the golf ball’s frame? one year ago one year ago

1. modphysnoob Group Title

@Jemurray3 do I use length contraction on this? if so , how?

2. Jemurray3 Group Title

Yes, for the first part. As straightforwardly as you possibly could... $L = L_0 \sqrt{1-v^2/c^2}$

3. modphysnoob Group Title

so L0 would be 3.58x10^4 ?

4. Jemurray3 Group Title

Yes, the proper distance is measured in the earth frame because the geosynchronous satellites are at rest relative to the earth.

5. modphysnoob Group Title

for b) do I do just normal calcuation using t= l/v then lorentz transform that t?

6. modphysnoob Group Title

@Jemurray3

7. Jemurray3 Group Title

Yes. Alternatively, you could divide the contracted distance by the velocity.

8. modphysnoob Group Title

oh I see, that works too Thank you for your help