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# I Symmetry factor via Wick's theorem 1. Dec 6, 2016 ### CAF123 Consider the lagrangian of the real scalar field given by $$\mathcal L = \frac{1}{2} (\partial \phi)^2 - \frac{1}{2} m^2 \phi^2 - \frac{\lambda}{4!} \phi^4$$ Disregarding snail contributions, the only diagram contributing to $\langle p_4 p_3 \| T (\phi(y)^4 \phi(x)^4) \| p_1 p_2 \rangle$ at one loop order is the so called dinosaur (see attached) To argue the symmetry factor $S$ of this diagram, I say that there are 4 choice for a $\phi_y$ field to be contracted with one of the final states and then 3 choices for another $\phi_y$ field to be contracted with the remaining final state. Same arguments for the $\phi_x$ fields and their contractions with the initial states. This leaves 2! permutations of the propagators between x and y. Two vertices => have factor (1/4!)2 and such a diagram would be generated at second order in the Dyson expansion => have factor 1/2. Putting this all together I get $$S^{-1} = \frac{4 \cdot 3 \cdot 4 \cdot 3 \cdot 2!}{4! \cdot 4! \cdot 2} = \frac{1}{4}$$ I think the answer should be 1/2 so can someone help in seeing where I lost a factor of 2? I could also evaluate $$\langle p_4 p_3 \| T (\phi(y)^4 \phi(x)^4) \| p_1 p_2 \rangle = \langle p_4 p_3 \| : \phi(y)^4 \phi(x)^4 : \| p_1 p_2 \rangle + \dots + (\text{contract}(\phi(x) \phi(y)))^2 \langle p_4 p_3 \| : \phi(y)^2 \phi(x)^2 :\| p_1 p_2 \rangle + \dots$$ where dots indicate diagrams generated via this correlator that do not contribute at one loop. (I don't know the latex for the Wick contraction symbol so I just write contract). Is there a way to find out the symmetry factor from computing the term $(\text{contract}(\phi(x) \phi(y)))^2 \langle p_4 p_3 \| : \phi(y)^2 \phi(x)^2: \| p_1 p_2 \rangle?$ Thanks! File size: 891 bytes Views: 37 2. Dec 7, 2016 ### vanhees71 You lost the factor of 2 in the very first step. Whether you connect a $\phi(y)$ with an initial or final state doesn't play any role and thus for the contraction of one of the $\phi(y)$ fields you have 8 (not only 4) possibilities. But now say, you've connected (counting the possibilities as 8 as discussed) this $\phi(y)$ with a final state, then you have only 3 possibilities left to connect the other final state, since due to the topology of the diagram now your final state is connected with the $\phi(y)$ and not the $\phi(x)$ fields. To connect one of the initial states you have 4 possibilities (either of the $\phi(x)$) and for the other initial state you have 3. To connect one of the remaining $\phi(x)$ with a $\phi(y)$ (dictated again by the topology of your diagram) you have 2 possibilities, and then there's only 1 possibility left to connect the other $\phi(x)$ with the one remaining $\phi(y)$. So you symmetry factor is all together $8434321/((4!)^2*2!)=1/2$ as you expected. 3. Dec 7, 2016 ### CAF123 Thanks! It makes sense - is it possible to extract this symmetry factor using the Wick theorem as I have begun in the last part of my OP? -- 4. Dec 8, 2016 ### nrqed Yes, it goes through exactly as what you and Vanhees71 just figured out. What you must do is to contract four of the phi's (two of the $\phi(x)$ with the incoming states and two of the $\phi(y)$ with the outgoing states or vice versa, as you did in your counting) and then the two remaining $\phi(x)$ with the two remaining $\phi(y)$. 5. Dec 10, 2016 ### CAF123 Ok thanks - what I meant, however, was how to extract the correct coordinate space representation of the process using Wick's theorem with the result that the net numerical factor in front gives rise to the symmetry factor. I'll illustrate what I mean by the simplest example in $\phi^4$ theory, that of the local interaction vertex. This is produced (amongst other diagrams) in $\langle p_3 p_4 \| T( \phi(x)^4) \|p_1 p_2 \rangle$. The term of interest is the one with no contractions that is $\langle p_3 p_4 \| : \phi(x)^4 :\|p_1 p_2 \rangle$. Computing this should generate terms of the form $e^{-ix(p_1 + p_2 - p_3 - p_4)}$ so that it is equal to $e^{-ix(p_1 + p_2 - p_3 - p_4)} + \dots$. In the Dyson expansion I have therefore $$-\frac{i\lambda}{4!} \int \text{d}^4 x ( e^{-ix(p_1 + p_2 - p_3 - p_4)} + \dots)$$ Now, under the integration over $x$ the terms included in $\dots$ should give rise to terms such that the prefactor of 1/4! is cancelled. But what do these terms look like? Thanks! Edit: The result for $$p_3 p_4 \| : \phi(x)^2 \phi(y)^2 : \| p_1 p_2 \rangle = 4(e^{-i(p_1+p_2)y} e^{i(p_3+p_4)x} + e^{i(p_4-p_1)y} e^{i(p_3-p_2)x} + e^{i(p_4-p_2)y} e^{i(p_3-p_1)x} + (x \leftrightarrow y))$$ as in the OP. Specialising this to the case of $x=y$ as we have here this reduces to $4! \, e^{-ix(p_1 + p_2 - p_3 - p_4)}$ so that indeed the factor of 1/4! is cancelled. And the combinatoric argument without all this tedious calculation is that in the correlator we can contract four $\phi(x)$ fields with $p_1$, leaving three for $p_2$, two for $p_3$ and one for $p_4$. This gives 4! permutations of the prongs, again cancelling the 1/4!. Last edited: Dec 10, 2016 6. Dec 11, 2016 ### nrqed I am not sure what you mean by the dot dot dot. All the contributions are of the form of the term you wrote, there are simply 4! of them because of all the possible pairings.	{}
# The proof of Stinespring dilation (Stinespring dilation) Let $A$ be a unital C-algebra and $\phi: A \rightarrow B(H)$ be a completely positive map. Then, there exist a Hilbert space $H_{1}$, and a -representation $\pi: A \rightarrow B(H_{1})$ and operator $V:H \rightarrow H_{1}$ such that $\phi(.)=V^{\ast}\pi(.)V$. In particular, $\|\|\phi\|\|=\|\|V\|\|^{2}=\|\|\phi(1)\|\|$. In the first step of the proof of this theorem, defining a sesquilinear form $< , >$ on $A \odot H$ (this is the algebraic tensor product) by $<\Sigma_{j}~b_{j}\otimes \eta_{j},~\Sigma_{i}~a_{i}\otimes \xi_{i}>=\Sigma_{i, j} <\phi(a_{i}^{\ast}b_{j})\eta_{j},~\xi_{i}>_{H}$. I can not understand the form of the element in $A \odot H$ (that is $\Sigma_{i}~a_{i}\otimes \xi_{i}$). why not $a_{i}\otimes \xi_{i}$? I mean how to explain the sum of $a_{i}\otimes \xi_{i}$. If you only consider elementary tensors, then what would their sum be? Remember that you want $A\odot H$ to be a vector space. • But how to compute the sum in $A \odot H$? I mean, if two elements $\Sigma_{i=1}^{n} b_{i} \otimes \eta_{i}$ and $\Sigma_{j=1}^{k} a_{j} \otimes \xi_{j}$, then what is their sum? – Yan kai Mar 10 '14 at 15:57 • A sum of elementary tensors is a sum of elementary tensors. If you add two sums of elementary tensors, you get a sum of elementary tensors: $\sum_ib_i\otimes\eta_i+\sum_ja_j\otimes\xi_j$ is just that, a sum of elementary tensors. – Martin Argerami Mar 10 '14 at 16:05 • I still have some doubt. If the sum of elemetary tensors is just $\Sigma_{i} b_{i} \otimes \eta_{i}+ \Sigma_{j} a_{j} \otimes \xi_{j}$, how to verify the $<. , .>$ defined above is sesquilinear ? – Yan kai Mar 11 '14 at 15:00	{}
RUS ENG JOURNALS PEOPLE ORGANISATIONS CONFERENCES SEMINARS VIDEO LIBRARY PACKAGE AMSBIB General information Latest issue Forthcoming papers Archive Impact factor Subscription Guidelines for authors License agreement Submit a manuscript Search papers Search references RSS Latest issue Current issues Archive issues What is RSS Mat. Zametki: Year: Volume: Issue: Page: Find 1976, Volume 20, Issue 3 Simultaneous approximations of exponents by transcendental numbers of certain classesA. A. Shmelev 305 Some properties of the set of extreme points of the unit ball of a Banach spaceM. I. Kadets, V. P. Fonf 315 Addition theorems for Legendre functions as corollaries of an addition theorem for Gegenbauer polynomialsN. N. Beloozerov 321 Asymptotic formulas for $n$-diameters of certain compacta in $L_2[0,1]$Kh. Nasyrova 331 Asymptotic distribution of eigenvalues in a lacuna of the continuous spectrum of the perturbed Hill operatorL. B. Zelenko 341 The spectrum of an elliptic operator of second orderT. M. Kerimov, V. A. Kondrat'ev 351 Representation of solutions of linear partial differential equations in the form of finite sumsS. S. Titov 359 Exceptional directions for a convex bodyB. A. Ivanov 365 Tabular powers of maximal setsS. S. Marchenkov 373 The occurrence of an implication in finitely valid, intuitively improvable formulas of propositional logicD. P. Skvortsov 383 The synthesis of reliable circuits from unreliable elementsV. V. Tarasov 391 The integral representation of vector measures on a completely regular spaceO. E. Tsitritskii 401 Reports read at the International Conference on the Theory of Approximation of Functions held at Kaluga from July 24 to 28, 1975 The order of approximation to functions on sets by polynomial operators in the class $S_m$R. K. Vasil'ev 409 Precise values of the best one-sided approximations by splinesV. G. Doronin, A. A. Ligun 417 The approximation of conjugate functions by conjugate interpolating sumsA. V. Efimov, V. A. Filimonova 425 Inequalities containing best approximations and the modulus of continuity of functions in $L_2$L. V. Taikov 433 The multidimensional Jackson theoremV. A. Yudin 439 Doctoral theses Classes of functions and Fourier coefficients with respect to complete orthonormal systemsS. V. Bochkarev 445	{}
Next: 6.5 Ice to Ocean Up: 6. Sea Ice Thermodynamics Previous: 6.3 Snow and Ice Contents # 6.4 Ice to Atmosphere Flux Exchange Atmospheric states and downwelling fluxes, along with surface states and properties, are used to compute atmosphere-ice shortwave and longwave fluxes, stress, sensible and latent heat fluxes. Surface states are temperature and albedos , , , (see section 6.3), while surface properties are longwave emissivity and aerodynamic roughness (note that these properties in general vary with ice thickness, but are here assumed constant). Additionally, certain flux temperature derivatives required for the ice temperature calculation are computed, as well as a reference diagnostic surface air temperature. The following formulas are for the absorbed shortwave fluxes and upwelling longwave flux: (6.13) for in Kelvin and denotes the Stefan-Boltzmann constant. The downwelling shortwave flux and albedos distinguish between visible ( ), near-infrared ( ), direct () and diffuse () radiation for each category. Note that the upwelling longwave flux has a reflected component from the downwelling longwave whenever . For stress components and and sensible and latent heat fluxes the following bulk formulas are used [29]: (6.14) The quantities from the lowest layer of the atmosphere include wind components and , the density of air , the potential temperature , and the specific humidity . The surface saturation specific humidity is (6.15) where the values of and were kindly supplied by Xubin Zeng of the University of Arizona. The specific heat of the air in the lowest layer is evaluated from (6.16) where specific heat of dry air and water vapor are and , respectively. Values for the exchange coefficients for momentum, sensible and latent heat and the friction velocity require further consideration. The bulk formulas are based on Monin-Obukhov similarity theory. Among boundary layer scalings, this is the most well tested [99]. It is based on the assumption that in the surface layer (typically the lowest tenth of the atmospheric boundary layer), but away from the surface roughness elements, only the distance from the boundary and the surface kinematic fluxes are important in the turbulent exchange. The fundamental turbulence scales that are formed from these quantities are the friction velocity , the temperature and moisture fluctuations and respectively, and the Monin-Obukhov length scale : (6.17) with (6.18) to prevent zero or small fluxes under quiescent wind conditions, is von Karman's constant (0.4), and is the buoyancy flux, defined as: (6.19) with g the gravitational acceleration and the virtual potential temperature where . Similarity theory holds that the vertical gradients of mean horizontal wind, potential temperature and specific humidity are universal functions of stability parameter , where is height above the surface ( is positive for a stable surface layer and negative for an unstable surface layer). These universal similarity functions are determined from observations in the atmospheric boundary layer [73] though no single form is widely accepted. Integrals of the vertical gradient relations result in the familiar logarithmic mean profiles, from which the exchange coefficients can be defined, where : (6.20) with the neutral coefficient (6.21) The flux profile functions (integrals of the similarity functions mentioned above) for momentum and heat/moisture are: (6.22) for stable conditions (). For unstable conditions (): (6.23) (6.24) with (6.25) The stability parameter is a function of the turbulent scales and thus the fluxes, so an iterative solution is necessary. The coefficients are initialized with their neutral value , from which the turbulent scales, stability, and then flux profile functions can be evaluated. This order is repeated for five iterations to ensure convergence to an acceptable solution. The surface temperature derivatives required by the ice temperature calculation are evaluated as: (6.26) (6.27) (6.28) where the small temperature dependencies of , the exchange coefficients and and velocity scale are ignored. For diagnostic purposes, an air temperature () at the reference height of is computed, making use of the stability and momentum/sensible heat exchange coefficients. Defining , and , we have: (6.29) For stable conditions () (6.30) and for unstable conditions () (6.31) where is bounded by 0 and 1. The resulting reference temperature is: (6.32) Next: 6.5 Ice to Ocean Up: 6. Sea Ice Thermodynamics Previous: 6.3 Snow and Ice Contents Jim McCaa 2004-06-22	{}
# Numbers $a$ for which every Sophie Germain safe Prime ($2p+1$) it is a quadratic residue $\pmod {2p+1}$ $>$ $a$ Are there any numbers $a$ (not a perfect square of course) for which every Sophie Germain prime $p$ $>$ $a$, $a$ will always be a quadratic residue $\pmod {2p+1}$. In other words, the Legendre Symbol ($a$ \| $2p+1$) $=$ $1$ whenever $p$ is a Sophie Germain Prime greater than $a$. The only $a$ I was able to find is $3$, and it is a quadratic residue to all "safe" primes (primes of the form $2p+1$ with $p$ prime) where $p$ $>$ $3$. The proof for this is that when $p$ $>$ $3$ and $p$ is prime, $p$ $=$ $±1$ $\pmod 6$. Then $2p+1$ is $3$ or $11$ $\pmod {12}$. $p$ $=$ $1$ $\pmod 6$ cannot be a Sophie Germain Prime, otherwise it is divisible by $3$. This leaves $p$ $=$ $5$ $\pmod 6$, and $2p+1$ $=$ $11$ $\pmod {12}$. $3$ is a quadratic residue to any prime $±1$ $\pmod {12}$, as mentioned in my previous posts. Anyways, despite all of my work above, my question here is weather or not there exists another integer $a$ with the same properties as $3$. For example, $a$ $≠$ $2$, $5$, $7$, $8$, $10$, or $11$ as shown with ($2$ \| $11$) $=$ $-1$, ($5$ \| $23$) $=$ $-1$, ($6$ \| $59$) $=$ $-1$, ($7$ \| $23$) $=$ $-1$, ($8$ \| $59$) $=$ $-1$, ($10$ \| $23$) $=$ $-1$, ($11$ \| $47$) $=$ $-1$. $12$ is my next possible choice, (since perfect squares are excluded) but is anyone able to demonstrate a proof/disproof that $3$ is the only such number? Thanks for help. • I doubt the question can be solved, since it is actually an open question whether there are infinitely many Sophie Germain primes or not. Indeed, if there are only finitely many of them, let's say all smaller than $M$, then every $a>M$ would work trivially. – Paolo Leonetti Aug 31 '16 at 11:07 • I know that, but I am assuming there are infinitely many. Weather there are infinitely many or not, $a$ $=$ $3$ is implied. Again, my real question is weather any other $a$ (as I described above) could exist. I will give you 50 rep if you can find another such $a$ as I described or prove that $3$ is the only one. – J. Linne Sep 1 '16 at 0:37 • If you assumed that there are infinitely many, then please write it in the OP – Paolo Leonetti Sep 1 '16 at 13:07 • You brought up a different question which I answered independently. Now here is another lemma for you: Suppose there are finitely many Sophie Germain primes $p$ which are no greater than $x$. Suppose we have an integer integer $a$ $<$ $p$ $<$ $x$ for which $a$ is a quadratic residue to all safe primes $2p+1$ up to $2x$. (With the assumption there are finitely many Sophie Germain primes $p$). If we cannot prove the modular congruence pattern that shows that $a$ is always a quadratic residue to $2p+1$, then the $a$ I just explained does not count. – J. Linne Sep 1 '16 at 13:59	{}
# Changing WP Table prefixes Published on April 14, 2008 SQL WordPress As a security tip you should always change the default table prefix "wp_", mainly to avoid zero-day vulnerabilities (Vulnerabilities that haven't been discovered yet), or JavaScript SQL injection. A very good tool you could use for that purpose is the WP-Security-Scan plugin, which takes care of all your security problems, including that one. The problem is that most of the times it is very hard to get it to change the prefix. Because of that you will have to do it manually. ## Changing the table prefix manually ### Step 1 Backup! the most important thing of all, backup, backup everything, often and early! ### Step 2 Install the "Maintenance mode" plugin, this way while you change the prefix no one will see errors in your site. ### Step 3 Chose now a very hard prefix, for this example I'm using df7s_23c_. Change the current prefix in wp-config.php, but you shouldn't upload it yet. Now start changing the table names to the new ones. (I recommend using phpMyAdmin for this, as it will make next step easier) ### Step 4 This is very important! There are some rows in your new tables that need to be updated: 1. Go to the former table wp_options (Now df7s_23c_options in the example) and change the following rows (It is recommended to search for "%wp_%" so that you get a list of the rows to edit): 1. wp_user_roles 2. wp_cron_backup_tables 3. wp_cron_backup_schedule 4. wp_cron_backup_recipient 2. Now go to the former table wp_usermeta (Now df7s_23c_usermeta in the example) and change the following rows: 1. wp_capabilities 2. wp_user_level 3. wp_autosave_draft_ids ### Finishing Upload the file wp-config.php if you haven't done so before, and check to see if it works. If it doesn't make sure that all of the tables have been correctly renamed, and all the rows listed above. If it doesn't work yet there might be some plugins that created a table containing in it's rows the wp_ table prefix. If that is the case check all your plugin-created tables to ensure that none of the fields containt the old prefix. If you still get problems replace your backup and wp-config.php and contact me through the comments of this post and I'll try to help you	{}
# Langley’s Adventitious Angles (Part 1) Math With Bad Drawings had an interesting post about solving for $x$ in the following picture (this picture is taken from http://thinkzone.wlonk.com/MathFun/Triangle.htm): I had never heard of this problem before, but it’s apparently well known and is called Langley’s Adventitious Angles. See Math With Bad Drawings, Wikipedia, and Math Pages for more information about the solution of this problem. Math Pages has a nice discussion about mathematical aspects of this problem, including connections to the Laws of Sines and Cosines and to various trig identities. I’d encourage you to try to solve for $x$ without clicking on any of these links… a certain trick out of the patented Bag of Tricks is required to solve this problem using only geometry (as opposed to the Law of Cosines and the Law of Sines). I have a story that I tell my students about the patented Bag of Tricks: Socrates gave the Bag of Tricks to Plato, Plato gave it to Aristotle, it passed down the generations, my teacher taught the Bag of Tricks to me, and I teach it to my students. In the same post, Math With Bad Drawings has a nice discussion about pedagogical aspects of this problem concerning when a “trick” becomes a “technique”. I recommend this problem for advanced geometry students who need to be challenged; even bright students will be stumped concerning coming up with the requisite trick on their own. Indeed, the problem still remains quite challenging even after the trick is shown.	{}
# Chimera states in network-organized public goods games with destructive agents It is shown that a network-organized metapopulation of cooperators, defectors and destructive agents playing the public goods game with mutations, can collectively reach global synchronization or chimera states. Global synchronization is accompanied by a collective periodic burst of cooperation, whereas chimera states reflect the tendency of the networked metapopulation to be fragmented in clusters of synchronous and incoherent bursts of cooperation. Numerical simulations have shown that the system’s dynamics alternates between these two steady states through a first order transition. Depending on the parameters determining the dynamical and topological properties, chimera states with different numbers of coherent and incoherent clusters are observed. These results present (to my knowledge) the first systematic study of chimera states and their characterization in the context of evolutionary game theory. This provides a valuable insight into the details of their occurrence, extending the relevance of such states to natural and social systems. The evolution of the three strategies ($x$: cooperators, $y$: defectors, $z$: destructive agents) is described by the replicator-mutator equations $\dot{x} =x (P_x -\bar{P}) + \mu (1 - 3 x)\,\\[5pt]\dot{y} =y (P_y -\bar{P}) + \mu (1 - 3 y) \,\\[5pt]\dot{z} = z (P_z -\bar{P})+ \mu (1 - 3 z)$ where $\bar{P}=xP_x+yP_y+zP_z$ is the average payoff and $P_x = r \frac{x}{1-z} \left[ 1 - \frac{1-z^n}{n(1-z)} \right] + \frac{r}{n} \frac{1-z^n}{1-z} - 1 - d \left( \frac{1-z^n}{1-z} - 1\right)\,,\\[5pt] P_y = P_x + 1 - \frac{r}{n} \frac{1-z^n}{1-z}\,,\\[5pt] P_z = 0$ This dynamical system has a Hopf bifurcation beyond which self-sustained oscillations emerge. They correspond to periodic bursting oscillations of cooperation. When the populations playing the public goods game (locally, in each node) are connected with a nonlocal ring networks as described by the equations $\dot{x}_i =x_i (P_{x,i} -\bar{P}_i) + \mu (1 - 3 x_i)+ \frac{\sigma}{2R} \sum_{j=i-R}^{j=i+R}\! (x_j - x_i)\,,\\[5pt]\dot{y}_i = y_i (P_{y,i} -\bar{P}_i)+ \mu (1 - 3 y_i) + \frac{\sigma}{2R} \sum_{j=i-R}^{j=i+R}\! (y_j - y_i)\,,\\[5pt]z_i = 1 - x_i - y_i$ then chimera states can emerge, reflecting the tendency of the networked metapopulation to be fragmented in clusters of synchronous and incoherent bursts of cooperation. http://nikos.techprolet.com/wp-content/uploads/2016/11/S1.mp4 Further reading: N. E. Kouvaris, R. J. Requejo, J. Hizanidis and A. Diaz-Guilera. Chimera states in a network-organized public goods game with destructive agents. Chaos 26, 123108 (2016). [Chaos][arXiv] A. Arenas, J. Camacho, J. A. Cuesta, R. J. Requejo. The joker effect: Cooperation driven by destructive agents. Journal of Theoretical Biology 279, 113-119 (2011). [J. Theor. Biol.]	{}
Textbook Notes (368,317) ECON 208 (113) Chapter 6 Chapter 6 Consumer Behavior.docx 4 Pages 121 Views School Department Economics (Arts) Course ECON 208 Professor Mayssun El- Attar Vilalta Semester Fall Description Chapter 6 Human Behavior 61 Marginal Utility and Consumer ChoiceUtility the satisfactionwellbeing that a consumer receives from consuming some goodserviceTotal utility the total satisfaction resulting from the consumption of a given commodity by a consumerMarginal utility the additional satisfaction obtained from consuming 1 additional unit of commodity Diminishing Marginal UtilityLaw of diminishing marginal utility The utility that any consumer derives from successive units of a particular product consumed over some period of time diminishes as total consumption of the product increases if the consumption of all other products is unchangedMin quantity of waterwould pay a lot for itmarginal utility of water is high the fewer L you are already using the higher the marginal utility of one more L of water if consume more than minmarginal utility of successive L of water declines steadily Utility Schedules and GraphsTotal utility rises as increased consumption of coke but the utility from each additional cokeday isthat of the previous onemarginal utility declines as the quantity consumedie total utility rises but marginal utility declines as consumption increases Maximizing Utilityeconomists assume that consumers seek to maximize their total utility subject to the constraints they facein particular their income and the market prices of various products The Consumers Decisionto maximize utility a consumer should consume any two goods or any number of goods until the marginal utility per dollar spent on the last product1 is equal to the marginal utility per dollar spent on the last product2 consumed A utilitymaximizing consumer allocates expenditures so that the utility obtained from the last dollar spent on each product is equalEx B3 C1 utility from the lastspent on C is 3x utility from lastspent on Btotal utility by switching aof expenditure from B to C and by gaining the difference between the utilities of a dollar spent on each To maximize utility continue switching expenditure from B to C as long as lastspent on C yields more utility than lastspent on B This switching however reduces the quantity of B consumed and given the law of diminishing marginal utility the marginal utility of B At the same time switching the quantity of C consumed and thereby lowers the marginal utility of C When marginal utilities changed so that the utility received from the lastspent on C isto utility from lastspent on B gain of nothing from further switches rather there is reduction of total utilityThe condition required for a consumer to be maximizing utility for any pair of products is MUxMUyutilitymaximizing consumer will allocate expenditure so that the utility gainedPxPyfrom the last dollar spent on any other productA consumer w a given amount of income to spend demands each good up to the point at which the marginal utility peron it is the same as the marginal utility perspent on every other 1 More Less Related notes for ECON 208 Me OR Join OneClass Access over 10 million pages of study documents for 1.3 million courses. Join to view OR By registering, I agree to the Terms and Privacy Policies Just a few more details So we can recommend you notes for your school.	{}
# How does the Einstein summation convention apply to the following equation? This is the equation is in the "mathematical form" section of the following wikipedia article: http://en.wikipedia.org/wiki/Geodesics_in_general_relativity More specifically, the "Full geodesic equation": $${d^2 x^\mu \over ds^2} =- \Gamma^\mu {}_{\alpha \beta}{d x^\alpha \over ds}{d x^\beta \over ds}$$ My question is: on the right hand side of the equation, the indices alpha and beta are repeated, thus the summation convention applies (as specifies in the article). But are there any cross-terms? If the right hand side of the equation is expanded, will there be any terms where alpha equals one and beta equals two, and so on? OR does the summation automatically mean alpha equals beta, and there are thus no cross terms? Or does the previous statement apply only if a Kronecker delta is present? Also, the Wikipedia page states that "s is a scalar parameter of motion, ex.: proper time." What other parameters are a "sacalar parameter of motion"? • There are cross-terms if $\Gamma^{\mu}_{\alpha\beta} \neq 0$ for $\alpha \neq \beta$. May 13 '16 at 17:09 $$\Gamma^\mu{}_{\alpha\beta} y^\alpha y^\beta$$ is defined to mean $$\sum_{\alpha = 0}^3 \sum_{\beta = 0}^3 \Gamma^\mu{}_{\alpha\beta} y^\alpha y^\beta$$ that is, each repeated index is summed over independently. • Just to add... the double sum has $4\times 4=16$ terms in total. 4 of them have $\alpha=\beta$ and the remaining $12$ have $\alpha\neq \beta$. Those 12 terms may be paired to 6 pairs - where both members of the pair are the same because they are $\alpha\leftrightarrow\beta$ symmetric. But that doesn't matter, one still sums them twice. So the sum is equivalent to the sum over $\alpha=\beta$ plus twice the sum over $\alpha\lt \beta$. In the Einstein sum rule, one never tries to "suppress redundancies" by hand. It doesn't matter that some terms are equal to other terms. May 13 '16 at 17:10	{}
# Measures on formal power series over a finite field For various reasons not important for this question, I'd like to show that certain subsets of $$F_p[[t]]$$, the ring of formal powers over the finite (prime) field $$F_p$$ in the variable t, are "small", i.e., that their elements are "rare" and if you pick a "random" one, you are unlikely to hit one of these. To make that precise, I feel that I should have a measure on this ring. First question: Are any such measures "well-known", e.g. have been discussed in books or at least papers? Any references? My naive idea was to attempt to define a "measure" as follows: for $$d\in\mathbb{N}$$, let $$\pi_d:F_p[[t]] \to F_p[t],\ \sum_{n=0}^{\infty} a_n t^n \mapsto \sum_{n=0}^{d-1} a_n t^n,$$ i.e., we keep the first $$d$$ terms and "cut off" the rest. Then the image of $$\pi_d$$ has size $$p^d$$. Now we can define a map $$\mu$$ on subsets of $$F_p[[t]]$$ as follows: for $$X\subset F_p[[t]]$$, let $$\mu(X) := \lim_{d\to\infty} \frac{\|\pi_d(X)\|}{p^d}.$$ This should be well-defined, as $$\|\pi_{d+1}(X)\|\leq p\cdot \|\pi_d(X)\|$$. Also $$0\leq \mu(X)\leq 1$$ (which fits, as our ring is compact). But is it countably additive (with respect to which $$\sigma$$-algebra)? Unfortunately it's been a long time I learned about measures... Second question: Is this a measure? (And if so, is it a "sensible" resp. "standard" one?) Which approaches could one use to prove this? I've been trying to read up on this a bit, and am planing to luck closer at the theory of Haar measures and Radon measures, as that sounds as if it might provide tools to either study this candidate for a "measure", or else define a different measure (which then hopefully I can use for my purposes). But I am hoping that perhaps some experts can help me get on the right track here. • It seems that you're just getting the probability measure where the coefficients are iid uniformly distributed on $F_p$, which does seem completely natural. In other words, the infinite product of uniform measure with itself. It lives on the product $\sigma$-algebra and is countably additive. This construction will be discussed in any probability theory book. It's the same construction you would use to talk about an infinite sequence of dice rolls. – Nate Eldredge Mar 6 at 14:52 • OK, that gives me some keywords to look for. Any particular book(s) you might recommend (I'll see what we have in our library, though) – Max Horn Mar 6 at 14:56 • Also: $F_p[[t]]$ is a compact group (in the product topology) and the measure described by Nate is the Haar measure. This means it has nice invariance properties. And maybe your "small" condition could be done with Baire category instead of measure. – Gerald Edgar Mar 6 at 15:08 • @Lubin could you elaborate? – Max Horn Mar 6 at 16:21 • @MaxHorn: Great, glad you found it. Another thing to look at is the Kolmogorov zero-one law; this is the main tool for showing that a set has measure zero with respect to an infinite product measure. – Nate Eldredge Mar 6 at 17:42	{}
Language: Search: Contact Zentralblatt MATH has released its new interface! For an improved author identification, see the new author database of ZBMATH. Query: Fill in the form and click »Search«... Format: Display: entries per page entries Zbl 1090.42008 Komori, Yasuo Notes on singular integrals on some inhomogeneous Herz spaces. (English) [J] Taiwanese J. Math. 8, No. 3, 547-556 (2004). ISSN 1027-5487 A central $(p,q)$ block is a function $a$ supported in $\{\|x\|<R\}$ such that $\\|a\\|_{L^q}\le \|\{\|x\|<R\}\|^{1/q-1/p}$. A central $(p,q)$-atom is a central $(p,q)$-block having integral zero. The block space $K^p_q$ consists of those distributions that can be expressed as superpositions $\sum_k \lambda_k a_k$ of central $(p,q)$-blocks such that $\sum_{k} \|\lambda_k\|^p<\infty$ and is $p$-normed by $\\|f\\|_{K^p_q}^p=\inf \sum_k \|\lambda_k\|^p$, with infimum taken over block representations of $f$. The space $HK^p_q$ is defined in exactly the same way with blocks replaced by atoms. The $p$-norm then makes sense when $p>n/(n+1)$. \par The author introduces a notion intermediate to that of a block and an atom, namely, a $(p,q,\varepsilon)$-block is a $(p,q)$-block $a$ supported in $\{\|x\|<R\}$ for some $R\geq 1$ such that $\|\int a\|\leq \|\{\|x\|<R\}\|^{\varepsilon -1/p}$. This notion allows the author to define a new block space $K_p^{1,\varepsilon}$ just as above, but in terms of $(p,q,\varepsilon)$ blocks, and to extend boundedness of certain singular integrals to these spaces. \par Specifically, a $(q,\theta)^t$-central singular integral is a linear operator $T:\Cal{D}\to \Cal{D}'$ that is bounded on $L^q(\Bbb{R}^n)$ and has integral kernel $K$ satisfying $$\sup_{R\geq 1} \sup_{\|y\|<R} R^{n(q-1)} \int_{2^jR<\|x\|<2^{j+1}R}\, \|K(x,y)-K(x,0)\|^q\, dx < e_j \text{ such that }\sum_{j=1}^\infty 2^{j\theta} e_j<\infty .$$ The author's main result says the following: Suppose that $n/(n+1)<p\leq 1<q<\infty$, $q/(q-1)\leq s$, $\lambda\leq \varepsilon -1$, and $T$ is a $(q,\theta)^t$-central singular integral with $\theta> n(1/p-1/q)$. If $T^t(1)\in {\mathrm CMO}^{s,\lambda}(\Bbb{R}^n)$ then $T$ is bounded from $HK_p^q (\Bbb{R}^n)$ to $K_p^{q,\varepsilon}(\Bbb{R}^n)$. \par Here, the finite central oscillation space ${\mathrm CMO}^{s,\lambda}$ consists of those $f$ such that $$\sup_{R\geq 1} \left(\frac{1}{R^{n(1+\lambda q)}} \int_{\|x\|<R} \|f(x)-{\mathrm ave}(f,\{\|x\|<R\})\|^q\, dx\right)^{1/q}$$ is finite. \par The result extends previous work of {\it J. Alvarez, J. Lakey} and {\it M. Guzmán-Partida} [Collect. Math. 51, No. 1, 1--47 (2000; Zbl 0948.42013)] concerning boundedness of operators from block spaces into Herz-Hardy spaces. The author also corrects a minor error in that work. [Joseph Lakey (Las Cruces)] MSC 2000: *42B20 Singular integrals, several variables 42B30 Hp-spaces (Fourier analysis) 42B35 Function spaces arising in harmonic analysis Keywords: Herz space; Hardy space; singular integral; commutator; boundedness Citations: Zbl 0948.42013 Highlights Master Server ### Zentralblatt MATH Berlin [Germany] © FIZ Karlsruhe GmbH Zentralblatt MATH master server is maintained by the Editorial Office in Berlin, Section Mathematics and Computer Science of FIZ Karlsruhe and is updated daily. Other Mirror Sites Copyright © 2013 Zentralblatt MATH \| European Mathematical Society \| FIZ Karlsruhe \| Heidelberg Academy of Sciences	{}
# 5.4 Surface force and tension force (Page 2/10) Page 2 / 10 ## Weight on an incline, a two-dimensional problem Consider the skier on a slope shown in [link] . Her mass including equipment is 60.0 kg. (a) What is her acceleration if friction is negligible? (b) What is her acceleration if friction is known to be 45.0 N? Strategy This is a two-dimensional problem, since the forces on the skier (the system of interest) are not parallel. The approach we have used in two-dimensional kinematics also works very well here. Choose a convenient coordinate system and project the vectors onto its axes, creating two connected one -dimensional problems to solve. The most convenient coordinate system for motion on an incline is one that has one coordinate parallel to the slope and one perpendicular to the slope. (Remember that motions along mutually perpendicular axes are independent.) We use the symbols $\perp$ and $\parallel$ to represent perpendicular and parallel, respectively. This choice of axes simplifies this type of problem, because there is no motion perpendicular to the slope and because friction is always parallel to the surface between two objects. The only external forces acting on the system are the skier’s weight, friction, and the support of the slope, respectively labeled $\mathbf{\text{w}}$ , $\mathbf{\text{f}}$ , and $\mathbf{\text{N}}$ in [link] . $\mathbf{\text{N}}$ is always perpendicular to the slope, and $\mathbf{\text{f}}$ is parallel to it. But $\mathbf{\text{w}}$ is not in the direction of either axis, and so the first step we take is to project it into components along the chosen axes, defining ${w}_{\parallel }$ to be the component of weight parallel to the slope and ${w}_{\perp }$ the component of weight perpendicular to the slope. Once this is done, we can consider the two separate problems of forces parallel to the slope and forces perpendicular to the slope. Solution The magnitude of the component of the weight parallel to the slope is ${w}_{\parallel }=\mathit{w}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(\text{25º}\right)=\mathit{mg}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(\text{25º}\right)$ , and the magnitude of the component of the weight perpendicular to the slope is ${w}_{\perp }=\mathit{w}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left(\text{25º}\right)=\mathit{mg}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left(\text{25º}\right)$ . (a) Neglecting friction. Since the acceleration is parallel to the slope, we need only consider forces parallel to the slope. (Forces perpendicular to the slope add to zero, since there is no acceleration in that direction.) The forces parallel to the slope are the amount of the skier’s weight parallel to the slope ${w}_{\parallel }$ and friction $f$ . Using Newton’s second law, with subscripts to denote quantities parallel to the slope, ${a}_{\parallel }=\frac{{F}_{\text{net}\parallel }}{m}$ where ${F}_{\text{net}\parallel }={w}_{\parallel }=\text{mg}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(\text{25º}\right)$ , assuming no friction for this part, so that ${a}_{\parallel }=\frac{{F}_{\text{net}\parallel }}{m}=\frac{\text{mg}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(\text{25º}\right)}{m}=g\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(\text{25º}\right)$ $\left(9.80\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}\right)\left(0.4226\right)=4.14\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}$ is the acceleration. (b) Including friction. We now have a given value for friction, and we know its direction is parallel to the slope and it opposes motion between surfaces in contact. So the net external force is now find the 15th term of the geometric sequince whose first is 18 and last term of 387 The given of f(x=x-2. then what is the value of this f(3) 5f(x+1) hmm well what is the answer Abhi how do they get the third part x = (32)5/4 can someone help me with some logarithmic and exponential equations. 20/(×-6^2) Salomon okay, so you have 6 raised to the power of 2. what is that part of your answer I don't understand what the A with approx sign and the boxed x mean it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared Salomon I'm not sure why it wrote it the other way Salomon I got X =-6 Salomon ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6 oops. ignore that. so you not have an equal sign anywhere in the original equation? hmm Abhi is it a question of log Abhi 🤔. Abhi Commplementary angles hello Sherica im all ears I need to learn Sherica right! what he said ⤴⤴⤴ Tamia hii Uday what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks. a perfect square v²+2v+_ kkk nice algebra 2 Inequalities:If equation 2 = 0 it is an open set? or infinite solutions? Kim The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined. Al y=10× if \|A\| not equal to 0 and order of A is n prove that adj (adj A = \|A\| rolling four fair dice and getting an even number an all four dice Kristine 222=8 Differences Between Laspeyres and Paasche Indices No. 7x -4y is simplified from 4x + (3y + 3x) -7y how do you translate this in Algebraic Expressions Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)= . After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight? what's the easiest and fastest way to the synthesize AgNP? China Cied types of nano material I start with an easy one. carbon nanotubes woven into a long filament like a string Porter many many of nanotubes Porter what is the k.e before it land Yasmin what is the function of carbon nanotubes? Cesar I'm interested in nanotube Uday what is nanomaterials and their applications of sensors. what is nano technology what is system testing? preparation of nanomaterial Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it... what is system testing what is the application of nanotechnology? Stotaw In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google Azam anybody can imagine what will be happen after 100 years from now in nano tech world Prasenjit after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments Azam name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world Prasenjit how hard could it be to apply nanotechnology against viral infections such HIV or Ebola? Damian silver nanoparticles could handle the job? Damian not now but maybe in future only AgNP maybe any other nanomaterials Azam Hello Uday I'm interested in Nanotube Uday this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15 Prasenjit can nanotechnology change the direction of the face of the world At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light. the Beer law works very well for dilute solutions but fails for very high concentrations. why? how did you get the value of 2000N.What calculations are needed to arrive at it Privacy Information Security Software Version 1.1a Good Got questions? 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# Essential Bandwidth of rect(t/T) Here is a question I have been trying to solve: Estimate the "essential bandwidth" of a rectangular pulse $$g(t) = \operatorname{rect}\left(\frac{t}{T}\right),$$ with $T>0$, where this "essential" bandwidth contains 90% of the rectangular pulse energy. What I have so far is that the Fourier Transform of $\operatorname{rect}\left(\frac{t}{T}\right)$ is $$G(f) = \mathcal{F}\{g(t)\} = \mathcal{F}\left\{ \operatorname{rect}\left(\frac{t}{T}\right) \right\} = T \operatorname{sinc}(fT)$$ where $$\operatorname{rect}(u) \triangleq \begin{cases} 0 & \text{if } \|u\| > \frac{1}{2} \\ \frac{1}{2} & \text{if } \|u\| = \frac{1}{2} \\ 1 & \text{if } \|u\| < \frac{1}{2} \\ \end{cases}$$ $$\operatorname{sinc}(u) \triangleq \begin{cases} \frac{\sin(\pi u)}{\pi u} & \text{if } u \ne 0 \\ 1 & \text{if } u = 0 \\ \end{cases}$$ $$X(f) = \mathcal{F}\{x(t)\} \triangleq \int\limits_{-\infty}^{+\infty} x(t) \ e^{-i 2 \pi f t} \ dt$$ and $$x(t) = \int\limits_{-\infty}^{+\infty} X(f) \ e^{+i 2 \pi f t} \ df.$$ Integrating $G(f)$ over $\pm \infty$ results in $1$. Also, integrating $\|g(t)\|^2$ over $\pm \infty$ results in $T$. This is about where I am lost. Any help is appreciated. • Hint: can you apply Parseval's formula here? – Jeff Nov 16 '14 at 18:51 • I tried setting it Eg=0.90Etot but I am lost in the Math part. Can you give more clues? – Ary Nov 16 '14 at 18:54 • Since the essential bandwidth must contain 90% of the pulse energy, then that's a clue that you need to involve the energy in the time domain. But it's asking for the bandwidth, which is a frequency domain quantity, so that's another clue that you need to mix the two domains. Use Parseval's formula to figure out how much energy you need in the frequency domain to get 90% in the time domain. (You limit the energy by truncating the integral: instead of integrating over infinity, you integrate over [-B,+B], where B is the essential bandwidth and the result gives 90% energy in the time domain) – Jeff Nov 16 '14 at 19:03 I'll add another answer, even though MBaz's answer is correct, because I think that it doesn't actually address the problem you have with arriving at the final solution. Summarizing, you have to solve $$\frac{1}{2\pi}\int_{-W}^W\|P(\omega)\|^2d\omega=0.9\int_{-\infty}^{\infty}\text{rect}^2(t/T)dt=0.9\cdot T\tag{1}$$ for $W$ (which is the bandwidth in radians). If you like, rewrite (1) using $f$ instead of $\omega$. $P(\omega)$ is the Fourier transform of the rectangular pulse: $$P(\omega)=2\frac{\sin(\omega T/2)}{\omega}\tag{2}$$ If I understood your question and your comments correctly, then you understand all of this, but your problem is solving (1) for $W$. Note that the integral has no closed-form solution, unless you consider an expression including the sine integral $\text{Si}(x)=\int_0^x\sin(t)/t\;dt$ closed form. With a little help from WolframAlpha, Equation (1) can be written as $$\frac{2}{\pi W}\left[TW\cdot \text{Si}(TW)+\cos(TW)-1\right]=0.9\cdot T\tag{3}$$ Equation (3) has no closed-form solution for $W$, so it is not surprising that you got confused. You could of course try to solve (3) numerically but I do not think that this is the idea of the exercise. What you need to know is that obviously the largest portion of the energy is contained in the main lobe of the spectrum, i.e. between $-W_0$ and $W_0$, where $W_0$ is the first zero crossing of $P(\omega)$. From (2) it is clear that $$W_0T/2=\pi\quad\Longrightarrow W_0=\frac{2\pi}{T}\tag{4}$$ So this is the essential bandwidth in radians. Checking the exact percentage using the left-hand side of (3) we get $$\frac{2}{\pi W_0}\left[TW_0\cdot \text{Si}(TW_0)+\cos(TW_0)-1\right]= \frac{2\;\text{Si}(2\pi)}{\pi}\cdot T=0.9028\cdot T$$ which seems close enough. In sum, the essential bandwidth of a rectangular pulse is given by the width of the mainlobe of its spectrum, so you only need to be able to calculate the first zero of the spectrum and you're done. You need to follow these steps: 1. Read up on Parseval's Theorem. 2. Find the correct Fourier Transform of $\operatorname{rect}(t)$ -- the one you give above doesn't seem right to me. (note from r b-j, it's because of $\omega$ vs. $f$ in the Fourier Transform. it wasn't "wrong", but messy, so i changed it.) Also, I'd recommend working on "ordinary frequency" $f$ instead of angular frequency $\omega$. 3. Find $E_g$, the energy of $g(t)$. 4. Find the frequency $f_0$ such that the integral of the square of the magnitude of $G(f)$ on the interval from $-f_0$ to $f_0$ is equal to $0.9E_g$. Then, relying on Parserval, $f_0$ is your answer. Intuitively, what is happening is this. Say you filter $g(t)$ with an ideal low-pass filter. Since the filter removes some frequency content from $g(t)$, then the energy of the filter's output is less than that of $g(t)$. You're trying to adjust the filter's cutoff frequency so that the output's energy is exactly 90% of the input's. • i hope you guys don't mind that i tinkered in your posts. – robert bristow-johnson Nov 16 '14 at 20:32 • @robertbristow-johnson, I don't mind, quite the contrary! I tweaked your own edits to the question a little bit. I thought the Fourier Transform of the rect pulse in the original question was wrong because it was missing a $1/\sqrt{2\pi}$ factor that is needed when working with angular frequency. – MBaz Nov 16 '14 at 20:37 • it depends on your definition of the F.T. with angular frequency. the unitary F.T. is the one where the forward F.T. and inverse F.T. are identical, except for the sign on $i$ (or $j$, whatever your religion is). if you have a unitary F.T. with angular frequency, you have $\frac{1}{\sqrt{2\pi}}$ in front of the integral in both the forward and inverse F.T. the nice thing about the $f$ version we like to use in the electrical engineering religion is that those factors are gone and our life is much easier scaling things. (but we have to remember the $2\pi$ in the exponent.) – robert bristow-johnson Nov 16 '14 at 20:45 • Guys, I don't think @Ary's problem has anything to do with the use of $f$ or $\omega$, but with the fact that the resulting integral has no closed form solution. – Matt L. Nov 17 '14 at 10:28 • parseval's theorem is both intuitively and practically easier to deal with using "ordinary frequency" (as opposed to "cyclical frequency"). otherwise you have to worry about where to put the $2 pi$ factor. you can always look it up, but why bother when the unitary Fourier Transform loses the scaling factor (actually puts it in the exponent). – robert bristow-johnson Nov 17 '14 at 15:17	{}
# some ODE questions This topic is 3633 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic. ## Recommended Posts I only recently started using ODE (with tao/c#) so some of these questions might have a obvious answer, Can anyone explain to me how contact points work, why does a box on box collision always give four even if only the corner is touching? If I only use ODE for collision detection and not the physics part is there a way I can use the information in the contact points to make it so a falling object will land on the ground and not hover over it becouse the distance between it and the ground is less than the distance I step it down per frame. Is their a way you can group static map features so they wont generate collisions with each other or am i suppose to manually filter them out between dSpaceCollide and dCollide? (i don't care if a rock is touching the ground but im not sure how to separate them without leaving gaps) Is there a way to test for collisions on a single object? Is their a better way to make a height map or a wedge than using a trimesh? ##### Share on other sites Quote: Original post by KazeI only recently started using ODE (with tao/c#) so some of these questions might have a obvious answer,Can anyone explain to me how contact points work, why does a box on box collision always give four even if only the corner is touching? whenever an object intersects another, several contact points are generated. You can then create a conntact joint, which ODE uses to resolve collisions. I'm not sure why you always get 4 contacts. What parameter do you have for max contacts? Quote: If I only use ODE for collision detection and not the physics part is there a way I can use the information in the contact points to make it so a falling object will land on the ground and not hover over it becouse the distance between it and the ground is less than the distance I step it down per frame. yes but its easier to leave that to ODE. you would need to know a bit about physics simulations. Quote: Is their a way you can group static map features so they wont generate collisions with each other or am i suppose to manually filter them out between dSpaceCollide and dCollide? Yes, create a geometry and place it, but don't create a body, thusly: (c+ code) //create the box that things sit on dGeomID box = dCreateBox(spaceID,100,10,100); dGeomSetPosition (box, 0, -5 - (0.5 * cube_y), 0); //create plane to catch objects falling off the box dCreatePlane(spaceID,0,1,0,-20); This is useful for level geometry and height maps. Quote: (i don't care if a rock is touching the ground but im not sure how to separate them without leaving gaps)Is there a way to test for collisions on a single object? yes, use the collision API, as detailed in the ODE user guide Which is pretty invaluable to me at least. Quote: Is their a better way to make a height map or a wedge than using a trimesh? I don' t think so. Look in the manual for any possible objects for this purpose, but ODE is limited to a few objects, excluding a height map. ##### Share on other sites You can group objects in spaces and then check for collisions between spaces. In most collision libraries, you can check for collisions between all objects within a group, and you can check for collisions between objects in two different groups. ODE has this functionality. You can also expand ODE with your own collision routines to handle more shapes, but you'd have to code them yourself. ##### Share on other sites ok, thanks i'l try some of the advice here Quote: Original post by gb_programmerwhenever an object intersects another, several contact points are generated. You can then create a conntact joint, which ODE uses to resolve collisions. I'm not sure why you always get 4 contacts. What parameter do you have for max contacts? if you mean when i call dCollide i set it hight to make sure i get all the contacts points available Quote: Original post by VorpyYou can group objects in spaces and then check for collisions between spaces so if two spaces were compared any touching objects within the same space would be ignored? just a one more question, for a trimesh is it better practice to create one huge mesh for the ground or divide it up into smaller parts EDIT: just incase i wasn't clear in the first post i want to avoid wasting processing power on immovable/static map features such as trees, statues or rocks sunk into the terrain touching each other but still have sprites collide with map features and sprites collide with sprites ##### Share on other sites Quote: Original post by VorpyYou can group objects in spaces and then check for collisions between spaces. In most collision libraries, you can check for collisions between all objects within a group, and you can check for collisions between objects in two different groups. ODE has this functionality.You can also expand ODE with your own collision routines to handle more shapes, but you'd have to code them yourself. I'm currently using this method for ray casting, where each ray is placed in a space, and collided with the spaces for other game entities, such as the space for characters. Just use dSpaceCollide2 to test two spaces against each other: http://www.ode.org/ode-latest-userguide.html#func_dSpaceCollide2 For me, it was a bit of a struggle to get an ODE application working but once I did, it was just a case of learning one piece of functionality at a time and adding it to the test application. ##### Share on other sites Quote: Is their a better way to make a height map or a wedge than using a trimesh? ODE has had support for height fields since about v0.7, it's up to v0.9 now although you may need to get the very latest SVN snapshot to obtain the latest patched implementation. Anyway, check out "demo_heightfield". edit: here on the wiki is a list of supported primitive types. As you can see ODE's main weakness is lack of solid convex support - but a trimesh will usually suffice as a substitute. ##### Share on other sites unfortunately it looks like tao.ode only supports v0.6 ##### Share on other sites This topic is 3633 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.	{}
### Multiple solutions of {$H$}-systems on some multiply-connected domains Futoshi Takahashi #### Abstract In this note, we consider the following problem: \begin{eqnarray} \left \{ \begin{array}{l} {\Delta} u = 2 u_x \wedge u_y \quad \mbox{in} \; \Omega, \quad u \in {H_0^1(\Omega ; {\bf R}^3)}, \\ u \|_{\partial \Omega} = 0, \end{array} \right. \end{eqnarray} where $\Omega \subset {\bf R}^2$ is a smooth bounded domain. We show that if the domain $\Omega$ is conformal equivalent to a $(K+1)$-ply connected domain satisfying some conditions, then the problem has at least $K$ distinct non-trivial solutions. #### Article information Source Adv. Differential Equations, Volume 7, Number 3 (2002), 365-384. Dates First available in Project Euclid: 27 December 2012 Takahashi, Futoshi. Multiple solutions of {$H$}-systems on some multiply-connected domains. Adv. Differential Equations 7 (2002), no. 3, 365--384. https://projecteuclid.org/euclid.ade/1356651830	{}
C, N, O abundances in the most metal-poor damped Lyman alpha systems★ @article{Pettini2008CNO, title={C, N, O abundances in the most metal-poor damped Lyman alpha systems★}, author={Max Pettini and Berkeley J. Zych and Charles C. Steidel and Frederic H. Chaffee}, journal={Monthly Notices of the Royal Astronomical Society}, year={2008}, volume={385}, pages={2011-2024} } • M. Pettini, +1 author F. Chaffee • Published 11 December 2007 • Physics • Monthly Notices of the Royal Astronomical Society This study focuses on some of the most metal-poor damped Lyα (DLA) absorbers known in the spectra of high-redshift QSOs, using new and archival observations obtained with ultraviolet-sensitive echelle spectrographs on the Keck and VLT telescopes. The weakness and simple velocity structure of the absorption lines in these systems allow us to measure the abundances of several elements, and in particular those of C, N and O, a group that is difficult to study in DLAs of more typical metallicities… Expand 132 Citations Figures and Tables from this paper The most metal-poor damped Lyα systems: insights into chemical evolution in the very metal-poor regime★ • Physics • 2011 We present a high spectral resolution survey of the most metal-poor damped Lyα absorption systems (DLAs) aimed at probing the nature and nucleosynthesis of the earliest generations of stars. OurExpand Implications of a non-universal IMF from C, N, and O abundances in very metal-poor Galactic stars and damped Lyα absorbers • Physics • 2011 Recently revealed C, N, and O abundances inthe most metal-poor damped Lyα(DLA) absorbers are compared withthose of extremely metal-poor stars in the Galactic halo, as well as extragalactic H IIExpand The chemistry of the most metal-rich damped Lyman α systems at z ∼ 2 – II. Context with the Local Group • Physics • 2015 Using our sample of the most metal-rich damped Lyman $\alpha$ systems (DLAs) at z$\sim2$, and two literature compilations of chemical abundances in 341 DLAs and 2818 stars, we present an analysis ofExpand THE MOST METAL-POOR DAMPED Lyα SYSTEMS: AN INSIGHT INTO DWARF GALAXIES AT HIGH-REDSHIFT In this paper we analyze the kinematics, chemistry, and physical properties of a sample of the most metal-poor damped Ly? systems (DLAs), to uncover their links to modern-day galaxies. We presentExpand The C/O ratio at low metallicity: constraints on early chemical evolution from observations of Galactic halo stars • Physics • 2009 Aims. We present new measurements of the abundances of carbon and oxygen derived from high-excitation C i and O i absorption lines in metal-poor halo stars, with the aim of clarifying the mainExpand Discovery of the most metal-poor damped Lyman-α system We report the discovery and analysis of the most metal-poor damped Lyman α (DLA) system currently known, based on observations made with the Keck HIRES spectrograph. The metal paucity of this systemExpand Metallicity Evolution of Damped Lyman-alpha Systems out to z~5 • Physics • 2012 We present chemical abundance measurements for 47 damped Lyman-alpha systems (DLAs), 30 at z>4, observed with the Echellette Spectrograph and Imager and the High Resolution Echelle Spectrometer onExpand The First Observations of Low-redshift Damped Lyα Systems with the Cosmic Origins Spectrograph: Chemical Abundances and Affiliated Galaxies We present Cosmic Origins Spectrograph (COS) measurements of metal abundances in eight 0.083 < zabs < 0.321 damped Lyman-α (DLA) and sub-damped Lyα absorption systems serendipitously discovered inExpand A carbon-enhanced metal-poor damped Lyα system: probing gas from Population III nucleosynthesis? • Physics • 2010 We present high-resolution observations of an extremely metal-poor damped Lyα system (DLA), at z_(abs) = 2.340 0972 in the spectrum of the QSO J0035−0918, exhibiting an abundance pattern consistentExpand Metal-enriched plasma in protogalactic halos. A survey of N V absorption in high-z damped and sub-da We continue our recent work of characterizing the plasma content of high-redshift damped and sub-damped Lyman-α systems (DLAs/sub-DLAs), which represent multi-phase gaseous (proto)galactic disks andExpand References SHOWING 1-10 OF 69 REFERENCES The abundances of nitrogen and oxygen in damped lyman alpha systems We take a fresh look at the abundance of nitrogen in damped Lyman systems (DLAs) with oxygen abundances between1/10 and1/100 of solar. This is a metallicity regime poorly sampled in the localExpand The Evolution of the C/O Ratio in Metal-poor Halo Stars • Physics • 2004 We report new measurements of carbon and oxygen abundances in 34 F and G dwarf and subgiant stars belonging to the halo population and spanning a range of metallicity from (Fe/H) = −0. 7t o−3.2 . TheExpand On the Determination of N and O Abundances in Low Metallicity Systems • Physics • 2006 We show that in order to minimize the uncertainties in the N and O abundances of low-mass, low-metallicity (O/H ≤ 1/5 solar) emission-line galaxies, it is necessary to employ separateExpand Element Abundances at High Redshifts: The N/O Ratio in a Primeval Galaxy • Physics • 1995 The damped Lyman alpha systems seen in the spectra of high redshift QSOs offer the means to determine element abundances in galaxies observed while still at an early stage of evolution. SuchExpand A homogeneous sample of sub-damped Lyman systems – IV. Global metallicity evolution An accurate method to measure the abundance of high-redshift galaxies involves the observation of absorbers along the line of sight towards a background quasar. Here, we present abundanceExpand Metal abundances and ionization conditions in a possibly dust-free damped Ly-alpha system at z=2.3 We have obtained a high resolution, high S/N UVES spectrum of the bright QSO HE2243-6031 to analyze the damped Ly-alpha system (DLA) observed at z=2.33. The metallicity of this system is 1/12 solarExpand A comprehensive set of elemental abundances in damped Lyα systems: Revealing the nature of these high-redshift galaxies By combining our UVES-VLT spectra of a sample of four damped Lyα systems (DLAs) toward the quasars Q0100+13, Q1331+17, Q2231−00 and Q2343+12 with the existing HIRES-Keck spectra, we covered the totalExpand First stars VI - Abundances of C, N, O, Li, and mixing in extremely metal-poor giants. Galactic evolution of the light elements We have investigated the poorly-understood origin of nitrogen in the early Galaxy by determining N abundances from the NH band at 336 nm in 35 extremely metal-poor halo giants, with carbon and oxygenExpand A homogeneous sample of sub-damped Lyman α systems- I. Construction of the sample and chemical abundance measurements In this first paper of a series, we report on the use of quasar spectra obtained with the high-resolution Ultraviolet-Visual Echelle Spectrograph (UVES) and available through the European SouthernExpand Early stages of Nitrogen enrichment in galaxies: Clues from measurements in damped Lyman alpha systems • Physics • 2003 We present 4 new measurements of nitrogen abundances and one upper limit in damped Ly α absorbers (DLAs) obtained by means of high resolution (FWHM � 7k m s −1 ) UVES/VLT spectra. In addition toExpand	{}
# [pstricks] 3D surface on non rectangular domain leon.free at free.fr leon.free at free.fr Wed Mar 1 14:21:01 CET 2017 Hi, I'd like to plot the surface of a function such as exp(x*y) only for the (x,y)'s in the unit ball B = \{(x,y) \in \mathbb{R}^2 : x^2+y^2 \le 1\}. I do not want to set the value of the function outside B to some particular value (e.g. 0). Do you know if there exists some pstricks package that can make this ? Thanks, elef	{}
# Valid definition of a binary operation on a set Need quick help with a homework question: Determine whether the description of # is a valid definition of a binary operation on a set: a) On $\Bbb R$ where $a\#b$ is $a*b$ (ordinary multiplication) b) On $\Bbb Z$, where $a\#b$ is $ab^2$ Can anybody help me solve this and possibly make me understand? - Are you saying that you can't square an integer? – Federica Maggioni Apr 22 '13 at 10:53 I don't know how to format it here where the b appears as a superscript – Jurgen Malinao Apr 22 '13 at 10:57 so did you mean $a#b=a^b$? If so, # is not a binary operation, as an example $2#-1$ is not an integer – Federica Maggioni Apr 22 '13 at 11:01 ## 2 Answers A binary operation on a set $X$ is a map from $X\times X$ to $X$. Therefore to assure that a description yields a binary operation you should check that it is correct for each $(a,b)\in X\times X$ and for every such pair $(a,b)$ it describes an element from $X$ (all these conditions are satisfied for both your cases a) and b), so you have descriptions of binary operations). - A binary operator is a type of function. So, the question becomes, are $ab$ and $ab^2$ defined on all of $\Bbb R^2$ and $\Bbb Z^2$ respectively, and do they produce unique values in $\Bbb R$ and $\Bbb Z$ respectively? -	{}
Tag Info Use this tag for questions about a particular construction of homological algebra of an abelian category A that refines and in a certain sense simplifies the theory of derived functors defined on A. The derived category D(A) of an abelian category A is a construction of homological algebra introduced to refine and in a certain sense simplify the theory of derived functors defined on A. The construction proceeds on the basis that objects of D(A) should be chain complexes in A, with two such chain complexes considered isomorphic when there is a chain map that induces an isomorphism on the level of homology of the chain complexes. Derived functors can then be defined for chain complexes, refining the concept of hypercohomology. The definitions lead to a significant simplification of formulas otherwise described (not completely faithfully) by complicated spectral sequences. Derived categories are also useful outside of algebraic geometry, for example in the formulation of the theory of D-modules and microlocal analysis and, nearer to physics, D-branes and mirror symmetry. Let A be an abelian category. (Some basic examples are the category of modules over a ring, or the category of sheaves of abelian groups on a topological space.) We obtain the derived category D(A) in steps: • The basic object is the category Kom(A) of chain complexes $$\cdots \to X^{-1} \xrightarrow {d^{-1}} X^0 \xrightarrow {d^0} X^1\xrightarrow {d^1} X^2 \to \cdots$$ in A. Its objects will be the objects of the derived category but its morphisms will be altered. • Pass to the homotopy category of chain complexes K(A) by identifying morphisms which are chain homotopic. • Pass to the derived category D(A) by localizing at the set of quasi-isomorphisms. Morphisms in the derived category may be explicitly described as roofs XX' → Y, where X' → X is a quasi-isomorphism and X' → Y is any morphism of chain complexes. The second step may be bypassed because a homotopy equivalence is in particular a quasi-isomorphism. But then the simple roof definition of morphisms must be replaced by a more complicated one using finite strings of morphisms (technically, it is no longer a calculus of fractions). So the one-step construction is more efficient in a way, but more complicated. From the point of view of model categories, the derived category D(A) is the true "homotopy category" of the category of complexes, whereas K(A) might be called the "naive homotopy category."	{}
Price-fixing charges baseless: DOE 96 The Department of Energy (DOE) said its inclusion in the complaint filed by Laban Konsyumer Inc. (LKI) at the Philippine Competition Commission (PCC) for alleged price-fixing and collusion among oil companies is baseless. “We find the complaint of LKI president Victorio Mario Dimagiba unmeritorious by accusing the Department of colluding with the oil companies by pointing to the pricing formula posted on the DOE website, as having facilitated said unacceptable illegal activities. The pricing formula in question, made by the Energy Regulatory Board (ERB), has been on the Department’s website since 2010. The posting is made in the interest of transparency,” the DOE said in a statement. The agency said the formula is not being prescribed nor dictated to be followed by industry players but is made publicly available to serve as a tool for all parties interested to understand and evaluate domestic oil price movements. “With the ERB formula having been utilized by the Board in the performance of its functions over the course of 11 years, the DOE deemed it fit to adopt said pricing formula among all other alternatives due to its historical reliability,” the DOE said. But the DOE said it welcomes the initiative of LKI to protect consumers and cited a memorandum of agreement it signed last July with the PCC to promote fair market competition and consumer welfare in the energy sector through information sharing, as well as investigation and enforcement support, the creation of joint task forces, capacity-building activities and consultative meetings.	{}
# NCERT Class 6 Basic Geometrical Ideas Exercise 03 Maths Solutions ## Class 6 Basic Geometrical Ideas Exercise 03 Maths NCERT Solutions Filter Filters : Classes • • • • • • • Chapters • • • • • • • • • • • 4 More Exercises • • • • • ### NCERT Class 6 \| BASIC GEOMETRICAL IDEAS \| Exercise 03 \| Question No. 03 Draw rough diagrams of two angles such that they have a) One point in common. (b) Two points in common. (c) Three points in common. (d) Four points in common. (e) One ray in common ### NCERT Class 6 \| BASIC GEOMETRICAL IDEAS \| Exercise 03 \| Question No. 01 Draw a rough sketch of a triangle ABC. Mark a point P in its interior and a point Q in its exterior. Is the point A in its exterior or in its interior? ### NCERT Class 6 \| BASIC GEOMETRICAL IDEAS \| Exercise 03 \| Question No. 02 (a) Identify three triangles in the figure. (b) Write the names of seven angles. (c) Write the names of six line segments. (d) Which two triangles have < Bas common?	{}
Context for this discrete Cauchy integral formula Notation: I will use the following conventions for discrete Fourier transforms (DFT) and discrete time Fourier transforms (DTFT): $$\mathcal{D}_N[x_j](k) := \sum_{j=0}^{N-1} e^{-2\pi i j k} x_j$$ $$\mathcal{D}^{-1}_N[y_k](j) := \frac{1}{N}\sum_{k=0}^{N-1} e^{2\pi i j k} y_k$$ $$\mathcal{F}[x_j](k) := \sum_{j=-\infty}^\infty e^{-2\pi i j k} x_j$$ $$\mathcal{F}^{-1}[y(k)](j) := \int_0^1 e^{2\pi i j k} y(k)$$ Additionally, $$\Omega_N$$ will denote the set of $$N^{th}$$ roots of unity. Let $$f$$ be a polynomial of degree $$N$$ with coefficients $$c_j$$, $$j=0,1,\dots,N$$. Consider the list of all coefficients as a vector $$c$$. By taking a DFT of $$c$$ you can show that $$c_j = \mathcal{D}_N^{-1}\left[f\left(e^{-2\pi i k/N}\right)\right](j)\tag{1}$$ And analogously with a DTFT $$c_j = \mathcal{F}^{-1}\left[f\left(e^{-2\pi i k}\right)\right](j)\tag{2}$$ Actually the latter formula works perfectly well for analytic functions $$f$$, where $$c$$ is the the vector of Taylor coefficients, or with meromorphic functions and $$c$$ the Laurant coefficients. I realized recently that this is basically just Cauchy's integral formula with a change of variables. To see this, expand out the inverse DTFT: \begin{align} c_j & = \int_0^1 e^{2\pi i k j} f(e^{-2\pi i k}) dk \\ & = \oint_{S^1} z^{-j} f(z) \frac{dz}{2\pi i z} \\ & = \frac{1}{j!} \frac{d^j f(0)}{dz^j} \end{align} Where the last line is from Cauchy's formula. So far this is all familiar. However, from this perspective we see that eq. (1) can also be interpreted as a "discrete" version of Cauchy's integral formula, valid only for polynomials. Rewriting it slightly, we have \begin{align} \frac{1}{j!}\frac{d^j f(0)}{dz^j} = c_j & = \frac{1}{N} \sum_k e^{2\pi i j k} f(e^{-2\pi i k/N})\\ & = \frac{1}{N} \sum_{z\in \Omega_N} z^{-j} f(z) \tag{3}\\ \end{align} This circle of ideas seems like it would be useful for e.g. efficiently determining coefficients of a "black box" polynomial that you can only evaluate. Or more generally any situation where you would want to use Cauchy's formula with a polynomial, this might be a convenient alternative. However, I've never seen any of this before in textbooks or other literature. Question: What, if anything, are eqs. (1) and (3) useful for? Citations to literature welcome. As I wrote this, I remembered vaguely something about "z-transforms". After looking this up, it seems very closely related, though not quite identical. I still haven't found anything exactly like eqs. (1) or (3). • isn't this a well-known connection? see Taylor and Fourier series are the same Feb 26, 2021 at 16:40 • I might be wrong, but my understanding was that one of the basic applications of the Fast Fourier transform (which is just an algorithm to do DFT fast on a computer) is precisely to make use of (1) in settings where evaluation of a polynomial function is easy, but you want to compute things related to the coefficients. The standard example is probably integer arithmetic on computers -- look up the basic fast integer multiplication algorithms a la Schönhage-Strassen. Feb 26, 2021 at 17:20 • this MO posting addresses a similar correspondence: mathoverflow.net/a/350028/11260 Feb 26, 2021 at 18:35	{}
# About shortest cycles in undirected graphs In an undirected (and unweighted) graph, to find the shortest cycle containing a specific vertex $$s$$, it is usually said to run a BFS from $$s$$ and the first time to find a re-visit, then that is the wanted smallest cycle length. This does not seem to be true, quoting from https://www.cs.princeton.edu/courses/archive/spr08/cos226/exams/fin-f05-sol.pdf page 3: Note that if you run BFS from $$s$$ and stop as soon as you revisit a vertex (using a previously ununused edge), you may not get the shortest path containing $$s$$. Examples where this technique seems to be suggested: An efficient algorithm to find a shortest cycle including a specific vertex Another technique, finding the shortest cycle in the whole graph, by running BFS from each vertex, also seems to detect only the shortestLength + 1 in a special case, as mentioned in this paper: https://link.springer.com/chapter/10.1007/978-3-540-78773-0_63 in the "Unweighted case". On the contrary, here (http://theory.stanford.edu/~virgi/cs267/lecture3.pdf) it is mentioned (first and second paragraphs) that running BFS from every vertex gives the shortest length (girth) in all cases. This is also mentioned in an archive (http://web.archive.org/web/20140513212046/https://webcourse.cs.technion.ac.il/234247/Winter2003-2004/ho/WCFiles/Girth.pdf). Which of all algorithms/methods is true for: 1. Finding the shortest-length cycle in an undirected graph? 2. Finding the shortest-length cycle that passes through a known vertex $$s$$ in an undirected graph? 3. Where is the pitfall in each of my above compare-and-contrasts? (I cannot believe that some of the above quoted might even be untrue...) Note: A randomized sub-cubic algorithm seems to exist, in the paper "A shortest cycle for each vertex of a graph" by Raphael Yuster: http://research.haifa.ac.il/~raphy/papers/all-cycles.pdf	{}
# Parallelogram Area Calculator The area of a parallelogram is twice the area of a triangle created by one of its diagonals. The area of a parallelogram is also equal to the magnitude of the vector cross product of two adjacent sides. #### Input Data $$Base=6$$ $$height=6$$ #### Solution $$Ans= 36\ m^2$$ #### Formula $$area = base * height$$ #### Input Data $$a=1$$ $$b=2$$ $$Angle=60\ in\ radian$$ #### Solution $$Ans= -0.60962124220443$$ #### Formula $$\normalsize Parallelogram\ (a,b,\theta\rightarrow S)$$ $$(1)\ area:\hspace{50px} S=ab{\hspace{1px}}sin\theta$$ This Parallelogram calculator is really great calculator. Even if you like to calculate like angle, sides, length, base and to solve any mathematics problem of not, you're within the right place. Don't ask the way to find the world of a parallelogram, just give the calculator a try! Below you'll determine how the tool is functioning - the parallelogram area formulas and neat explanation are all you would like to know the subject. Parallelogram area formulas A parallelogram may be a simple quadrilateral with two pairs of parallel sides. Every rectangle may be a parallelogram also as every rhombus and square. Remember, it isn't working the opposite way round! ## Which formulas does the parallelogram area calculator use? Parallelogram Area - the way to derive Area given base and height Area = base * height Did you notice something? The formula for the world of a parallelogram is just about an equivalent as for rectangle area! Why is it so? Have a glance at the picture: a parallelogram are often divided into a trapezoid and a right-angled triangle and rearranged to the rectangle. The area given sides and an angle between them Area = a * b * sin (angle) Does it ring a bell? This formula comes from trigonometry, and is employed for instance within the triangle area - the parallelogram could also be seen as two congruent triangles. The adjacent angles within the parallelogram are supplementary, so you'll choose whichever angle you would like because sin(angle) = sin(180° - angle). An area has been given diagonals of a parallelogram and an angle between them. Area = e * f * sin(angle) The formula comes from trigonometry also. Does one want to understand where it comes from? Divide the parallelogram into two triangles, assume that our e diagonal is that the "base" for both new triangles. What is the height of that triangle? Use the sine function. It's (f/2) * sin (angle)! The world of Triangular is adequate to our "base" e times height: e * (f/2) * sin (angle) The parallelogram consists of two such triangles, therefore the area equals e * f * sin (angle). ## How to find the world of a parallelogram? You are still unsure the way to use the parallelogram area calculator? We’ll show you step by step: Have a glance at your exercise. What’s given, what's unknown? Choose the proper calculator part for your needs. Assume that we would like to calculate the world knowing the diagonals of a parallelogram and therefore the angle between diagonals. Enter the given values to the proper boxes. Assume 5 in, 13 in, and 30° for the primary diagonal, other and therefore the angle between them, respectively. The calculator displays the world of a parallelogram value. It's 32.5 in² in our case. ### How to use tis calculator parallelogram? This use this parallelogram calculator is really simple and it’s really great to work with. You can do your calculation really quick and you will be saving your time for doing this rather than wasting your time while doing manually. Because it will take time if you will do manually or even you don’t know if your answer is correct or not or it can be so stressful if you don’t like mathematics or you don’t know how to do math. We have created this tool for students and any individual who need this tool for their use. It is totally free and it is really great to do that. You can use it anytime and anywhere you want because this tool is web based tool so you don’t have to worry about anything. Now as you can see you have this tool open on your computer or desktop. You have been given text boxes here so that you can enter your value here. After you will enter all your values here you have to click on the calculate button so that you will be able to get the result you are looking for. That’s all you have to do to calculate your problems, you don’t need to worry about the formula because the tool is already included with the help of formula and with the help of using formula. Note: You can also bookmark this tool if you think you can use it or you will be using this tool in the future. ### Q. What Is The Description Of Parallelogram Area Calculator? A. The Area Of A Parallelogram Is Twice The Area Of A Triangle Created By One Of Its Diagonals. The Area Of A Parallelogram Is Also Equal To The Magnitude Of The Vector Cross Product Of Two Adjacent Sides.	{}
# Reference for (Generalised) Adjunction spaces Conventionally, adjunction spaces are defined for two topological spaces, i.e. $$X \cup_f Y$$ is the space formed from attaching $$Y$$ to $$X$$ along some continuous map $$f:A \rightarrow X$$, where $$A$$ is a closed subset of $$Y$$. Some authors (see e.g. Brown's Topology and Groupoids, pg 125) give a generalisation in which multiple spaces $$Y_i$$ are attached to the same $$X$$. However, the generalisation stops here. My first guess at the next generalisation up would be to remove the space $$X$$ and instead have some sort of family $$Y_i$$ of topological spaces, subspaces $$A_{ij}$$ and continuous maps $$f_{ij}$$ such that each $$Y_i$$ and $$Y_j$$ form a standard binary adjunction space under the map $$f_{ij}: A_{ij} \rightarrow Y_j$$. This would mean the $$Y_i$$ are all stuck to each other, instead of to a single space $$X$$. Perhaps particular conditions on the subspaces $$A_{i j}$$ of $$Y_i$$ and the functions $$f_{i j}$$ need to be imposed in order to ensure that this space is well-defined, but I'm sure this is possible. Is there some canonical resource for generalisations of adjunction spaces in this direction? Or better yet, is there some well-known resource for everything adjunction spaces? • The original construction is just a pushout. More generally you can consider other colimits. – Qiaochu Yuan Sep 28 '18 at 18:53 • @QiaochuYuan ok, I don't know much category theory. Can you suggest a resource that covers generalised pushouts? – Doc Sep 29 '18 at 10:52	{}
# Write a C program to find out profit or loss in buying an article CServer Side ProgrammingProgramming The formula for finding profit is as follows if the selling price is greater than cost price − profit=sellingPrice-CosePrice; The formula for finding loss is as follows if cost price is greater than selling price − loss=CostPrice-SellingPrice Now, apply this logic in the program and try to find whether the person gets a profit or loss after buying any article − ## Example Following is the C program to find the profit or loss − Live Demo #include<stdio.h> int main(){ float CostPrice, SellingPrice, Amount; printf("\n Enter the product Cost : "); scanf("%f", &CostPrice); printf("\n Enter the Selling Price) : "); scanf("%f", &SellingPrice); if (SellingPrice > CostPrice){ Amount = SellingPrice - CostPrice; printf("\n Profit Amount = %.4f", Amount); } else if(CostPrice> SellingPrice){ Amount = CostPrice - SellingPrice; printf("\n Loss Amount = %.4f", Amount); } else printf("\n No Profit No Loss!"); return 0; } ## Output When the above program is executed, it produces the following result − Enter the Product Cost: 450 Enter the Selling Price): 475.8 Profit Amount = 25.8000 Published on 08-Mar-2021 09:55:39	{}
# Geometric characterization of Silva distributions There is a well known geometric characterization of tempered distributions on $\mathbb{R}^n$. A distribution $T\in \mathcal{D}'(\mathbb{R}^n)$ is an element of $\mathcal{S}'(\mathbb{R}^n)$ if and only if $T$ is the restriction of a distribution $\tilde{T}$ defined on the sphere $S^n$. (See "Theorie des distributions" from L.Schwartz p. 238) In a paper of 1958, Silva defined a new notion of distribution using infra-exponential growth conditions. These are more general then tempered distributions and are used in ultrahyperfonction fourier theory. Here are the definitions : For a convex compact set $K$, I will note $h_K$ the support function of $K$. Let $K$ be a convex compact set of $\mathbb{R}^n.$ The set $H_b(\mathbb{R}^n,K)$ is the set of functions $f\in C_{\infty}(\mathbb{R}^n)$ such that $$\|\|f\|\|_{K,n} :=\sup_{x\in \mathbb{R}^n ; \alpha \leq n} \{e^{h_K(x)}\|D^{\alpha}f(x)\|\} < \infty,$$ for all $n\in \mathbb{N}.$ Equipped with these semi-norms, the space is a Frechet space. For $O$ a convex open set of $\mathbb{R}^n$, we set $$H(\mathbb{R}^n, O) = \varprojlim_{K \subset O}H_b(\mathbb{R}^n,K).$$ The Silva distributions with $O$ conditions are the element of the dual space $$H'(\mathbb{R}^n,O) \subset \mathcal{D}'(\mathbb{R}^n).$$ Is there any geometric characterization of the Silva distributions as the one for tempered distributions ? (For example, using the sphere) Any help or references will be much appreciated. ADDENTUM The following caracterisation is already known. An element $T \in \mathcal{D}'(\mathbb{R}^n)$ is an element of $H'(\mathbb{R}^n,O)$ if and only if there is a multi-index $\alpha$, a convex compact set $K \subset O$ and a continuous fonction $F$ such that $$T = D^{\alpha}(\exp(h_K(x))F(x)).$$ However, for me it is not a geometric caracterisation as the one given by L. Schwartz in his book. • I hear about it for the first time, but could they be restrictions of distributions on compactifications of $\mathbb R^n$? – მამუკა ჯიბლაძე Oct 17 '17 at 9:36 • I doubt about it since it is already the criterion for tempered distributions and Silva distributions are different from tempered one. – C. Dubussy Oct 17 '17 at 13:00 • Oh perhaps you mean more exotic compactifications. Since it explicitely uses the convexity, I thought about convex compactification but I don't really now theses theories. – C. Dubussy Oct 17 '17 at 13:07 • Isn't it enough to take $K$ as a ball of radius $r$ so that $h_K(x)=r\\|x\\|$? – Jochen Wengenroth Oct 17 '17 at 13:47 • @მამუკაჯიბლაძე How would one define distributions on a space like $\beta \mathbb{R}^n$ that is only compact+Haussdorff but does not have any notion of differentiation? – Johannes Hahn Oct 17 '17 at 22:11	{}
Quaternion problem (with exe) Recommended Posts I recently finished doing skeletal animation, but every once in awhile, there will be bones which dont seem to animate the way they should. In other words, they dont take the path they should and rotate weirdly. Example Here Click on the Skeletal.zip link. [Edited by - GamerSg on September 23, 2004 1:55:44 AM] Share on other sites Odd, looks like your getting scaling with the rotation. Can't help you much without some code though.... Share on other sites There is totally no scaling/translation in my code, so im only doing rotation via quaternions. As you can see, it works normally for all the other bones, but screws up for the right hand. Share on other sites Maybe your animation matrices do not match with the amount of reference matrices. In other words: if your reference mesh has 25 BONES and your animation does not, every time you transform the inverted world vertices wia tha animation world matrix, you have to check if it is the same, for example: for ( i=every bone in reference mesh ) for ( j=every bone in animation mesh ) if ( ref_bone[i].name == anim_bone[i].name ) do transform else keep reference bone. Share on other sites I was thinking the problem lies more with my Quaternion Interpolation, if you observe the animation closely, the figure is doing a jumping jack (couldnt think of any other animation) which goes something like this (in keyframes) 1) o\| \| \|\|2)__ o __ \| \| / \ 3) o\| \| \|\|4) \ o / \| \| / If you notice, the right hand does reach the keyframes. The problem lies with the path it takes to reach the next keyframe. For some reason, the left hand seems to work well. Share on other sites Perhaps you need to normalize some quaternions somewhere... like after transforming a child bone's rotation by a parent bone's? (Your download link didnt work when I tried so couldn't get a better look) I've had some pretty interesting looking animations before when I forgot to normalize. Share on other sites Is it a must to normalize? When should i normalise? Obviously normalizing every quaternion is going to result in quite a hit in performance? Share on other sites You certainly need to normalise before you build a rotation matrix from your quaternion (unless your quat to matrix routine auto-normalises), otherwise you'll get shearing or scaling problems. You don't need to normalise every time you multiply quaternions together. Share on other sites Hmm i just looked through my code and realised that i am not doing the rotation multiplication with quaternions, instead im converting a bone's quaternion to a matrix first, before multiplying the matrix with the parents matrix. Also, all my quaternions are normalised before they are exported. Share on other sites Still stuck with no clues. Here is my quaternion interpolation code which i had obtained from the quaternion article on gamasutra. void Math::InterpolateQuat(Quat& first,Quat& second,Quat& result,float time){ //Interpolates between 2 quaternions based on time factor ranging from 0-1 //Variables double cosine; Quat temp; //temp quaternion float scale0,scale1; //dot product between the 2 quats cosine = first.x * second.x + first.y * second.y + first.z * second.z + first.w * second.w; if(cosine<0.0) { //if it is negative, make it positive cosine = -cosine; temp.x = -second.x; temp.y = -second.y; temp.z = -second.z; temp.w = -second.w; } else { temp.x = second.x; temp.y = second.y; temp.z = second.z; temp.w = second.w; } if ( (1.0 - cosine) > 0.0001 ) { double omega, sinom; // standard case (slerp) omega = acos(cosine); sinom = sin(omega); scale0 = static_cast<float>(sin((1.0 - time) * omega) / sinom); scale1 = static_cast<float>(sin(time * omega) / sinom); } else {//else do LERP scale0 = 1.0f - time; scale1 = time; } result.x = first.x * scale0 + second.x * scale1; result.y = first.y * scale0 + second.y * scale1; result.z = first.z * scale0 + second.z * scale1; result.w = first.w * scale0 + second.w * scale1;} Share on other sites I have a feeling the bone for your model's left arm is falling into the 'invert cosine' special case. It's probably the only bone that's going through that case too, based on the angles that the bones are rotating through. The if statement for LERP/SLERP is there just to protect the acos(x) function ... so odds are the LERP portion is never being used. Personally, I'm not sure falling back to a LERP is necessary -- I've never needed it myself. The important part is that if you invert the cosine, you need to normalize the slerped quaternion. temp.x = first.x * scale0 + second.x * scale1;temp.y = first.y * scale0 + second.y * scale1;temp.z = first.z * scale0 + second.z * scale1;temp.w = first.w * scale0 + second.w * scale1;if (cosineWasInverted) { result = temp.Normalize(); }else { result = temp; } There are two other interesting things to note, instead of inverting the cosine, you should also be able to invert the scale0 factor. When you invert the cosine, what you're saying is, instead of slerping clockwise, slerp counter clockwise, because the shorter arc is in that direction. Inverting scale0 lets you precalculate scale0 and scale1 and then do an if{} block for the scale / normalize. I'm also inverting sine omega, because we want to minimize our slow floating-point divides for the CPU. I'd also recommend you implement inlined quaternion operators for scale and +quaterion (that's definitely just me though). double omega, sinom; // standard case (slerp) omega = acos(cosine); inv_sinom = 1/sin(omega); scale0 = static_cast<float>(sin((1.0 - time) omega) * inv_sinom); scale1 = static_cast<float>(sin(time * omega) * inv_sinom); if (cosine < 0.0f) { // invert the scale and normalize scale0 = -scale0; // Would be nice to see some Quaternion operators here // e.g. result = firstscale0 + secondscale1 !!! result.x = first.x * scale0 + second.x * scale1; result.y = first.y * scale0 + second.y * scale1; result.z = first.z * scale0 + second.z * scale1; result.w = first.w * scale0 + second.w * scale1; // Now normalize the final quaternion result.Normalize(); } else { result.x = first.x * scale0 + second.x * scale1; result.y = first.y * scale0 + second.y * scale1; result.z = first.z * scale0 + second.z * scale1; result.w = first.w * scale0 + second.w * scale1; } Check the code with breakpoints first. If the cosine inversion is ONLY happening with the left-arm bone, then you know there's a solid chance that this is your problem. Then try to apply the minimal fix from the first code snippet. If that works you can consider some other options for improving the slerp. Share on other sites Just FYI, your exe does not shutdown properly. I downloaded it to look at it and closed out of it... three hours later I noticed my system was running really slow and went into taskmanager and it was still there taking up 98% of the CPU Share on other sites Quote: Original post by Anonymous Posteryour exe does not shutdown properly. I downloaded it to look at it and closed out of it... three hours later I noticed my system was running really slow and went into taskmanager and it was still there taking up 98% of the CPU Does the same thing on my system too. Share on other sites Sorry about the shutdown thing, it will shutdown properly if you use Esc key, i don't know how to shut it down properly from the X button in GLFW. Share on other sites Quote: Original post by GamerSgSorry about the shutdown thing, it will shutdown properly if you use Esc key, i don't know how to shut it down properly from the X button in GLFW. if(!glfwGetWindowParam(GLFW_OPENED)) tell it to quit. Of course, you could always look in the reference to find this kind of thing out. [Edited by - Nathaniel Hammen on September 25, 2004 10:22:17 PM] Create an account Register a new account • Forum Statistics • Total Topics 628305 • Total Posts 2981962 • 9 • 12 • 11 • 12 • 11	{}
# How do you write a verbal expression for the algebraic expression 6^2? Sometimes you may hear people say "6 to the second power" - don't worry, it means the same thing. The exponent 2 is used often when dealing with square shapes, so I guess some brilliant guy one day said "Hey, why don't we call ${6}^{2}$ six squared!"	{}
# Dictionary:Laplace transform Other languages: English • ‎español The linear transform pair ${\displaystyle F(s)=\int f(t)e^{-st}dt}$ and ${\displaystyle f(t)={\frac {1}{2\pi i}}\int F(s)e^{st}ds}$ s is a complex number and t is a real one. When the limits of integration are ${\displaystyle \pm \infty }$, the transform is two-sided. The two-sided Laplace transform becomes identical with the Fourier transform when s is purely imaginary. More often the one-sided transform is used, especially in the study of transient waveforms. In this case, where f(t) is causal, the integral is ${\displaystyle F(s)=lim\int f(t)e^{-st}dt}$ and ${\displaystyle f(t)=\int F(s)e^{+st}ds}$ The one-sided transform is often written with limits 0 to ${\displaystyle \infty }$, the limit being implied. Laplace transforms may not exist for all values of s and hence many Laplace transforms are limited to strips of convergence, the ranges of values for the real part of s for which the above intearals are finite. The Laplace transform domain is often called the s-plane. See Sheriff and Geldart (1995, 545–546).	{}
Register Forum Rules FAQ Members List Social Groups Search Today's Posts Mark Forums Read 2006-04-12, 22:24 Link #201 lilxtito Junior Member Join Date: Apr 2006 does anyone know how K-F did the starry effect for the new ED? (One Piece) Last edited by lilxtito; 2006-04-12 at 23:30. HTD Junior Member Join Date: Mar 2006 Quote: does anyone know how K-F did the starry effect for the new ED? (One Piece) I think it is done in AE .. i think it is impossible to do in ASS/SSA 2006-04-15, 16:25 Link #203 Sylf 翻訳家わなびぃ Fansubber Join Date: Nov 2003 Age: 42 It's very possible to do it in ASS/SSA, without using \clip or \p. You might want to grab a copy of your favorite font editor, though, and learn how to write ASS effects in some kind of automated fashion (use of karaoke program, or write perl/lua code, etc) Jeroi Junior Member Join Date: Jan 2006 Quote: Originally Posted by Suikun Wow, cool! I've only been doing this for about a week now, but I was just starting to get the hang of karaoking and thinking a program like this to calculate the time values and insert the proper karaoke effect would be helpful, and then I checked here and saw this! Great program! If I could make one suggestion for future releases, though, it'd be the ability to add multiple {\t}'s on a line with different time values based on the start and end times on the \k. So you could, for example, turn {\k30} into {\t(0,150,\bord10)}{\t(150,300,\bord3} by putting in "s" in the "Start" field and "(s+e)/2" in the End field for the first effect and "(s+e)/2" for Start and "e" for End on the second effect. Uh... if that makes sense... But anyway, great program as is. I'll get use out of this one for sure! Thanks! You can do that already in it, just make multiple effects for one style. And you can use those like this: start time: s+20 and endtime: e-20 etc then make new effect and use start: e-20 end: e Ayhashi Join Date: Apr 2006 Age: 31 Quote: Originally Posted by SCR512 Yeah, I would suggest using Lagarith. I've been using Lagarith for a good amount of time and have yet to encounter any horrid issues. Also when karaoke or typesetting a fansub, I would suggest making clips of the sign or karaoke and then compressing that out to Lagarith. Once you have done whatever then you can render out a RGBA and overlay it in the final encode using AVS. hi ppl. Hey can you explain me with more details that process?? It would be a great help, i have already done my karaoke with AE but now I'm encountering problems with the "merge" of the karaoke op video with the rest of the episode. 2006-04-30, 15:20 Link #206 zalas tsubasa o sagashite Join Date: Jul 2003 Render the movie with the Lagarith codec (make sure RGBA is checked in the options of the Lagarith codec), then in the After Effects option window, make sure it says RGB+A and Unmatted. After you get the overlay.avi, use an AVISynth script to overlay your video on top of the original (assuming the videos are the same length): Code: src=AVISource("youroriginalvideo.avi") ovl=AVISource("overlay.avi") Overlay(src,vol.ConvertToYV12(),mask=ovl.ShowAlpha()) This is just an example script, so you may have to do your own tweaks, changes, etc. 2006-04-30, 15:26 Link #207 TheFluff Excessively jovial fellow Join Date: Dec 2005 Location: ISDB-T Age: 29 I wrote a rather lengthy post about the avisynth part here: http://forums.animesuki.com/showpost...2&postcount=42 __________________ \| ffmpegsource 17:43:13 <~deculture> Also, TheFluff, you are so fucking slowpoke.jpg that people think we dropped the DVD's. 17:43:16 <~deculture> nice job, fag! 01:04:41 < Plorkyeran> it was annoying to typeset so it should be annoying to read 2006-04-30, 17:04 Link #208 Ayhashi Anime Madness Join Date: Apr 2006 Age: 31 thank you very much TheFluff and zalas. Finally I was able to do it. Render my movie with alpha channel and then overlaying it on the original vid. cool. I still need to refine some details, but at least i know how to do it right now Another question. Some way to make a bouncing ball that "jumps" from sylab to sylab?? I was thinking on create an image of a ball and then aplying a motion path/beizer and using the times of the sylabs as reference. But there's a better way or a more automated way to do it ?? thanks in advance. PS: Sorry bout my bad english ^^! 2006-04-30, 17:23 Link #209 TheFluff Excessively jovial fellow Join Date: Dec 2005 Location: ISDB-T Age: 29 http://www.malakith.net/aegisub/viewtopic.php?t=175 That's how to do it using Aegisub, ASS and Lua scripting. Kind of, at least. Not at all related to AFX, though... >_> __________________ \| ffmpegsource 17:43:13 <~deculture> Also, TheFluff, you are so fucking slowpoke.jpg that people think we dropped the DVD's. 17:43:16 <~deculture> nice job, fag! 01:04:41 < Plorkyeran> it was annoying to typeset so it should be annoying to read 2006-04-30, 17:51 Link #210 Sylf 翻訳家わなびぃ Fansubber Join Date: Nov 2003 Age: 42 I was gonna say, I can explain how to do it in ASS... My version would be 1) make a font with a character that looks like a ball (or find one that already has such thing) 2) set \org for the ball way below the screen threshold 3) use \t(\frz) to move the ball across the screen 4) use \t(\frx) to let it bounce up and down No trigonometry required, no need to understand bizziere or whatever curve... But I've NEVER tried it before, so I have NO clue how well this works. Quarkboy Translator, Producer Join Date: Nov 2003 Location: Tokyo, Japan Age: 36 Quote: Originally Posted by Sylf I was gonna say, I can explain how to do it in ASS... My version would be 1) make a font with a character that looks like a ball (or find one that already has such thing) 2) set \org for the ball way below the screen threshold 3) use \t(\frz) to move the ball across the screen 4) use \t(\frx) to let it bounce up and down No trigonometry required, no need to understand bizziere or whatever curve... But I've NEVER tried it before, so I have NO clue how well this works. That'd be fine, but it's take forever to do because you'd have to manually adjust the values to make the ball bounce to the right spot. If you used a fixed-width font, you could write a perl script or something to calculate them automatically, but with most truetype fonts unless you code a font rendering engine yourself it'd be a by hand operation. You'd have to position the bounce for each syllable by hand which would take about 2-4 hours, depending on how fast the song is. Doable, I suppose. __________________ New Project Coming 2017 Translator, Producer, Japan Media Export Expert Sam Pinansky Sylf Fansubber Join Date: Nov 2003 Age: 42 Quote: Originally Posted by Quarkboy That'd be fine, but it's take forever to do because you'd have to manually adjust the values to make the ball bounce to the right spot. If you used a fixed-width font, you could write a perl script or something to calculate them automatically, but with most truetype fonts unless you code a font rendering engine yourself it'd be a by hand operation. You'd have to position the bounce for each syllable by hand which would take about 2-4 hours, depending on how fast the song is. Doable, I suppose. I've done couple manual placement like that in the past. Personally, I don't think twice before doing something like that, as long as I think it's absolutely what I want. Actually, if I display the line on screen once, set the proper PlayResX and Y, and figure out the exact coordinate with photoshop or aegisub, I should be able to use those coordinates, plug it in some trigonometric formula, and figure out the \frz angle displacement. <.< That's still lots of manual labor.... is too stubborn to switch to afx 2006-05-01, 06:31 Link #213 TheFluff Excessively jovial fellow Join Date: Dec 2005 Location: ISDB-T Age: 29 There's an automation script for aegisub that does just that (calculate glyph/syllable widths that is) but it's somewhat buggy, IIRC... __________________ \| ffmpegsource 17:43:13 <~deculture> Also, TheFluff, you are so fucking slowpoke.jpg that people think we dropped the DVD's. 17:43:16 <~deculture> nice job, fag! 01:04:41 < Plorkyeran> it was annoying to typeset so it should be annoying to read 2006-05-01, 11:43 Link #214 Sylf 翻訳家わなびぃ Fansubber Join Date: Nov 2003 Age: 42 I've actually played around with that feature a bit. I haven't looked too deeply in to what the script is doing, but it was rather neat to see that someone is attempting it. A bit off topic here, but kudos to that script and its author. 2006-05-01, 11:56 Link #215 jfs Aegisub dev Join Date: Sep 2004 Location: Stockholm, Sweden Age: 32 What I have done when I needed movement along a path, was actually a "two-pass" algorithm, first calculate a path and store in a structure, and afterwards apply the path to the text/object/whatever. The path always consists of line segments, since that makes things far easier to work with (ie. curves are converted into line segments in the calculate-path step). Then there are two ways of handling the path, in ASS. One is simply generating one line with a \move per segment, easy but only works if the "duration" of each segment is at least one frame. The other requires actually simulating movement along the path, and creating a line per frame. The second method might also be better if you need some kind of trail to follow the moving object. I'm pretty sure I posted an example of this earlier in the thread. It does take a bit of math, but it's nothing complicated, and unless you're generating paths with curves in it, it's not even high-school level. __________________ Aegisub developer [ Forum \| Manual \| Feature requests \| Bug reports \| IRC ] Don't ask for: More VSFilter changes (I won't), karaoke effects, help in PM's CIRee fansubbing n00blet Join Date: Aug 2004 Quote: Originally Posted by TheFluff http://www.malakith.net/aegisub/viewtopic.php?t=175 That's how to do it using Aegisub, ASS and Lua scripting. Kind of, at least. Not at all related to AFX, though... >_> yeah this method I have used, you just need the right font and to replace the \p code. 2006-06-07, 09:25 Link #217 Ayhashi Anime Madness Join Date: Apr 2006 Age: 31 Hi ppl. I was seeing this vid from Paradise Kiss Opening by Shinsen subs. And i was wondering if you can gave me some leads of how to do the disolving syllab butterfly appearing thing. I mean, i know how to make the syllab dissapear, but then a lot of blue butterflys appears on screen right from the spot the syllab was. Can someone give me some hints about that plz ?? PS: sorry bout my bad english ^^! 2006-06-07, 09:57 Link #218 Sylf 翻訳家わなびぃ Fansubber Join Date: Nov 2003 Age: 42 Create a new font that has an image of a butterfly. Figure out the \pos coordinate where you want the butterfly to show up from. If you want the butterfly to fly some distance, you can use \move instead of \pos. Use \t with \alpha, \fs, etc to give additinoal effects. 2006-06-07, 12:11 Link #219 shinjipierre Computer graphist Join Date: Dec 2005 Location: Paris, France Age: 33 Humm, it's not really the shinsen-subs opening since it has never been released Well, anyway. I've created a butterfly in illustrator, animated the illustrator file in after effects in order to make it look like it flies. Aftermaths, I used particular to throw particles from the text and using that animated butterfly as the particle. Well, if you want to do that in SSA or something, have fun. __________________ http://www.remipierre.fr - Some of my computer graphics work 2006-06-07, 17:52 Link #220 Ayhashi Anime Madness Join Date: Apr 2006 Age: 31 no, I want to do it in AE. Can you explain me a little more about that "particular" option plz ^^ Tags fansubbing, karaoke, software, subbing	{}
## Babel-17 v0.3 is out ! June 30, 2011 Check it out now at ww.babel-17.com. Advertisements ## Automatic Type Conversions June 29, 2011 Writing unit tests is a great thing. It makes you think even harder than usually about the semantics of your code. In Babel-17 you have real interval arithmetic; for example, to describe the interval between 4 and 5, you could write [4.0; 5.0]. This notation, [a;b] assumes that a and b are reals, and forms the convex hull that contains both a and b. Now. Should it be possible to also write [4; 5] for the above interval? In my current preliminary spec for Babel-17 v0.3 the answer is NO. An integer is not a real; for example you cannot compare them, 2 == 2.0 evaluates to false. This point of view has a lot going for it; it seems pure. (Un)fortunately, people don’t care so much about purity. I can already see the looks on faces when I try to explain that [4.0; 5.0] is a real interval, but [4; 5] is not even a legal expression. There is an obvious solution to this dilemma: type conversions. These are already a part of Babel-17 v0.3: for example, 3 :> real converts the integer 3 to the real 3.0 . Type conversions could be used such that whenever I need in a particular situation a value of a certain type T, I check if my value at hand can be converted into something of type T. If yes, I use the converted value instead, if not, then it’s a DomainError. For this not to result in total chaos, a necessary condition for a type conversion to succeed should be that the conversion is reversible (at least in principle). Note that this is only a necessary condition, and you should really think twice before providing a type conversion. Often it makes more sense not to provide a type conversion, but just a conversion function which has to be invoked explicitly. As an example, look at the conversion between integers and reals. It should be possible to convert integers that are not too large to a real; but if the integer is too large to be represented faithfully as a real number, the conversion should fail. Likewise, a real r for which r.round == r holds should be convertible to an integer. But the conversion should fail for those r with not(r.round == r). This is a big issue, and adding automatic type conversions to Babel-17 changes the whole look and feel of the language. It will make it a lot easier for beginners to get friendly with the language; at the same time it will make it harder for a compiler (but not impossible, especially for a JIT) to generate fast code. Automatic type conversions will be introduced in the upcoming Babel-17 v0.3, and further developed in Babel-17 v0.3.1. ## Feature Complete June 28, 2011 Babel-17 v0.3 is feature complete! Everything has been implemented, and there is also a new version of the Babel-17 plugin for Netbeans 7.0. The two biggest changes in the plugin are: • there are Babel-17 projects now • there is support for running unit tests Tuesday is dedicated to going through the specification and writing unit tests that check the spec. Hopefully only minor fixes will be necessary; if so, then Tuesday night Babel-17 v0.3 will be released! ## Implementing This June 20, 2011 I am currently teaching the interpreter how to handle evaluation of this. While this is an intuitive concept, its correct implementation results in quite some work. The following code snippet, val r = object def outer = this def name = "r" def u = object def test = (inner, outer) def inner = this def name = "u" end end val (i, o) = r.u.test (i.name, o.name) should evaluate to ("u", "r"). To achieve this, the interpreter converts the above program into something like: val r = object def outer this_a = this_a def name = "r" def u this_a = object def test this_b = (inner this_b, outer this_a) def inner this_b = this_b def name = "u" end end val (i, o) = begin val u = r.u r u.test u end (i.name, o.name) Off trying to implement this in a not too messy way … ## No Wildcard Import June 11, 2011 I have implemented wildcard import in Babel-17, i.e. you can do something like import com.coollib._ or even more complicated imports like import com.coollib.{coolfun => f, -notneeded} which will import everything from com.coollib except notneeded. The value com.coollib.coolfun is accessible by the local name f. But now I decided to get rid of wildcard imports again. There are two major points why wildcard imports are not such a good idea: 1. Wildcard imports can lead to name collisions that you are not aware of. Right now, it is an error if an imported name collides with a local def or val, but it is no error to collide with names defined not in the local, but an outer scope. This rule makes sense for all imports except wildcard imports, where it is dangerous, especially since Babel-17 is dynamically typed. 2. Wildcard imports are the only thing standing in the way of a complete separate compilation of Babel-17 files. Right now, all modules are scanned in a first phase. In the second phase, this module information is used for wildcard resolution. The second point is annoying, but obviously I have already worked around it. But the first point really is a deal breaker. Wildcard imports gotta go. ## Lenses Suck Less May 27, 2011 The day started well when I found an email with a link to the following video in my inbox: Making Apps That Don’t Suck. After watching and enjoying that video, I stumbled (via the scala-debate mailing list) onto the following talk: Lenses: A Functional Imperative. This was when this day really started to rock. I mean, how \emph{cool} are Lenses? They are such an obvious concept; maybe so obvious that people who used lenses before didn’t bother to give them an explicit name. But as it is, often sometimes becomes visible and tangible only once it has a name. Lenses seem to be a must-have for Babel-17. I always wondered how to generalize the following shortcut update notation which is currently available in Babel-17: val x = { a = 3, b = 2 } x.b = 7 Here x.b = 7 is short for x = { a = x.a, b = 7 } I don’t think that lenses will make it into the Babel-17 v0.3 release, but expect them in Babel-17 v0.3.1 . ## Reals and Order May 19, 2011 What else do I need to put into Babel-17 v0.3 to be able to use it in real-world projects? As a prerequisite for things like graphics, floating point arithmetic is highest on my wish list. So I want to add a new type real. But the thing is, I never liked the way floating point was usually treated in programming languages. When working on my diploma thesis and especially when working on my PhD, what I needed wasn’t mere floating point arithmetic, but interval floating point arithmetic. Babel-17 will be radical in its treatment of floating point arithmetic: there will be only interval arithmetic. Interval arithmetic is just so obviously superior to ordinary floating point arithmetic that this decision is a no-brainer. The only theoretical reason against interval arithmetic I can think of is the dependency problem: when you evaluate an expression in which the same variable occurs several times, interval arithmetic will often overestimate the error. For example, the expression x - x will yield something non-zero if x is an interval of width greater than zero. Therefore, 1.0/3.0 - 1.0/3.0 will never be zero when doing classical interval floating point arithmetic. But in my opinion this is not a problem at all. First, in order to reduce the error introduced by dependencies, just make your interval bounds tighter. Second, in case this is not good enough for your application and you really need x - x to evaluate to zero, you probably shouldn’t do floating point arithmetic at all, but rather turn to computer algebra. A practical problem with interval arithmetic is that it is not very well supported on many platforms. For example in Java, it is downright impossible to implement a performant implementation of interval arithmetic that calculates bounds as tight as the underlying hardware would support it. But again, this just means that your intervals will be a little wider than they could be. Also, Babel-17 implementations that compile directly to machine code can circumvent most of these problems. Introducing interval arithmetic into Babel-17 has ripple effects. The main reason for this is the ordering imposed on intervals. The most appropriate order seems to be defined via [a; b] <= [c; d] iff (b < c or (a = c and b = d)) There are other orders of interest like inclusion of intervals, but the above seems to be the right choice for the canonical order of intervals. Unfortunately, there is no way that such an interval type together with this order could currently be defined in Babel-17. Of course I could implement special magic for reals, and the importance of this particular type would justify such a special treatment; but it still feels odd and I just don’t like such a resolution of the problem. Another thing to consider is the interplay between integers and reals. Once you start using special magic, you would definitely expect to allow the comparison of integers and reals. But this leads to results a naive user might be unprepared for, like 2 == 4.0 / 2.0 not being true (depending on how the particular interval arithmetic implementation works). Let’s face it: It is just not a good idea to compare integers and reals. If you want to do this, do it by explicitly converting between integers and reals. Extending the above discussion, one might also ask: Is it a good idea to canonically compare lists with vectors? And, is it a good idea to canonically compare values of different types at all? I think the answer is no. Those parts of Babel-17 relating to order, I definitely need to rework them. ## Unit Testing May 8, 2011 The changes I currently make to Babel-17 for the jump from v0.21 to v0.3 reach deeply into the current code base. This makes me think about how to verify that the changes are consistent with my spec of Babel-17. Obviously, an important corner stone for interpreter / compiler correctness is a test suite. Most of the tests in this test suite can just be treated like unit tests for ordinary Babel-17 programs. This leads naturally to the question, how should unit testing work in Babel-17 ? I am a strong advocate of providing programming language support for unit testing. Especially for Babel-17 this seems obvious: Babel-17 is dynamically typed, and although you can now (in v0.3) have modules, data encapsulation and abstract datatypes, there is only minimal static type checking in Babel-17 that merely ensures that the types you talk about really exist. You will still need unit testing to see that your types behave like you expect them to. Writing unit tests will be a standard task for every serious programmer who uses Babel-17, and therefore Babel-17 should provide language support for it. If you are saying, hey, this argument is bullshit, because for example version control is also a routine task, but best be left to external tools, then you might be right; but maybe you are very wrong 🙂 An important feature of testing is that the testing code does not affect the original production code. Many people interpret this as an argument against language support for unit testing, but actually only language support can guarantee the separation of production and testing code. So, Babel-17 v0.3 will provide the following language level support for unittesting: • a new keyword unittest • you can define modules that contain unittest as part of their path, like in module com.obua.util.orderedset.unittest ... end This module would typically test functionality of the module com.obua.util.orderedset. If you’d like to distinguish further the tests of this module, you can name your modules like this: module com.obua.util.orderedset.unittest.functionality1 ... end module com.obua.util.orderedset.unittest.functionality2 ... end module com.obua.util.orderedset.unittest.functionality3.sub1 ... end module com.obua.util.orderedset.unittest.functionality3.sub2 ... end and so on. Any module that has unittest in its path is called a unit test module. Note that a module path cannot repeat parts of itself; in particular, only a single component of the path can be unittest • you can use the unittest keyword as part of a module name also for the import statement, but only if it is issued from a unit test module • the last statement of a (non-unittest) module can be a unittest statement, like in module com.obua.util.orderedset ... unittest ... end This has the same effect of defining a module com.obua.util.orderedset.unittest, but with the difference that it shares the namespace of its surrounding module. In particular it can see the private definitions and type definitions (and the inner values of those types) of its surrounding module. • The pragma #assert is only executed and checked when running unittests, never in production code. • The new pragma #catch takes a constructor name (like InvalidArgument) and an expression. It signals an error if the evaluation of the expression does not yield an exception with that name. It also is only executed when running unittests. Taken together this allows for the flexible creation of unit tests, while at the same time completely shielding production code from testing code. ## Commas April 25, 2011 Trying to introduce typedef into the new Babel-17 grammar, I noticed that there was a problem with the combination of blocks and commas in the grammar. I then realized that I had hit this problem earlier but had been able to circumvent it: the memoize construct is currently used like this: memoize a b c and NOT like this: memoize a, b, c With typedef such a hack was not possible anymore, so I had to track down the cause of this problem. After hours of examining the grammar with ANTLRWorks, which is by the way one of the greatest software tools I have worked with (the up-side: now I finally completely understand the alternative display of ANTLRWorks), the problem turned out to be the block part of lambda expressions: (x => memoize a, b) could be understood as ((x => memoize a), b) but also as (x => memoize a; memoize b) After pondering the different ways of resolving this dilemma for a while, I decided that a clean solution is to distinguish between lambda expressions that are surrounded by round brackets, and those that are not. Basically, for those that are protected by surrounding brackets, the rule is lambda_expr_1 := pattern => block for those that are not the rule is lambda_expr_2 := pattern => expr This means that the expression (x => memoize a, b) is now interpreted as (x => memoize a; memoize b) As a consequence, both memoize and private will use commas instead of just spaces as separators of their multiple arguments. ## Deconstructors April 24, 2011 Currently the pattern f ? p has two different meanings, depending on whether f is a function or not. If f is not a function, then the message deconstruct_ is sent to the value to be matched, the result is applied to f, and then result of this application is matched against p. This, so I thought, should enable pattern matching against cexprs even for values that are no cexprs. In the light of how types are going to work in Babel-17, this does not seem to be so useful anymore. Instead of the pattern Plus ? p why not write directly the pattern Plus p All that is needed for this to work is to modify the semantics of above pattern such that code like match v case Plus p => ... ... end leads to the result of v.deconstruct_ Plus being matched against p. Original cexprs have an obvious default implementation of deconstruct_, and other values can optionally implement it. This is a good place to also announce another change: the pattern match v case Plus => ... ... end will not any longer be synonym with match v case (Plus nil) => ... ... end but instead with match v case Plus _ => ... ... end The reason for this is that especially exception handling becomes more robust and streamlined, for example the code match exception MyError "my bad" case (exception MyError) => "hello" end will now result in the value "hello".	{}
Math Help - Can't figure out how to set up this modeling problem 1. Can't figure out how to set up this modeling problem So I feel confident that I can solve the differential equation once I set it up, however I'm not able to set up the problem. The question is as follows: Pure mountain water flows into a 10 million gallon lake at a reate of 3 million gallons per day. Water flows out of the lake at the same rate. Suppose that someone dumps 1000 pounds of toxic substance YUK into the lake. Assume that YUK mixes evenly with the lake water instantly. a). Write down and then solve the initial value problem for Y (t), the total amount of YUK in the lake as a function of the time t in days after the dumping occurred. For these types of problems I usually set it up as a function Q' = rate in - rate out so for this problem I tried setting it up as Q' = -3,000,000 (concentration of water leaving lake) as there is no rate of YUK in. I can't figure out the concentration of the water flowing out however Any help is appreciated EDIT: On further thinking I decided to put up what I thought might work, if this is right please let me know and I'll take this question down: Y'(t) = -3,000,000Y(t) / 10,000 I realized the rate out is -3,000,000 gallons, and I assumed the concentration of Yuk in the lake is 1/10,0000(Y(t)) lb / gal which would come out So I feel confident that I can solve the differential equation once I set it up, however I'm not able to set up the problem. The question is as follows: Pure mountain water flows into a 10 million gallon lake at a reate of 3 million gallons per day. Water flows out of the lake at the same rate. Suppose that someone dumps 1000 pounds of toxic substance YUK into the lake. Assume that YUK mixes evenly with the lake water instantly. a). Write down and then solve the initial value problem for Y (t), the total amount of YUK in the lake as a function of the time t in days after the dumping occurred. For these types of problems I usually set it up as a function Q' = rate in - rate out so for this problem I tried setting it up as Q' = -3,000,000 (concentration of water leaving lake) as there is no rate of YUK in. I can't figure out the concentration of the water flowing out however Any help is appreciated EDIT: On further thinking I decided to put up what I thought might work, if this is right please let me know and I'll take this question down: Y'(t) = -3,000,000Y(t) / 10,000 I realized the rate out is -3,000,000 gallons, and I assumed the concentration of Yuk in the lake is 1/10,0000(Y(t)) lb / gal which would come out $\displaystyle\frac{dA}{dt}=R_{\text{in}}-R_{\text{out}}$ All my numbers are scaled in millions. $R_{\text{in}}=3\text{gals/day}\cdot 0\text{Yuk}=0$ $\displaystyle R_{\text{out}}=3\text{gals/day}\cdot \frac{.001}{10}\text{Yuk Lbs/gal}=\frac{.003}{10}$ $\displaystyle\frac{dA}{dt}=-\frac{\frac{.003}{10}A}{10}\Rightarrow A'+.00003A=0$ $P(t)=.00003, \ \ Q(t)=0$ $\displaystyle t\exp{\left(\int P(t)dt\right)}=\int\left[\exp{\left(\int P(t)dt\right)}Q(t)\right]dt$	{}
Automatic Equation Numbering¶ The TeX input processing in MathJax can be configured to add equation numbers to displayed equations automatically. This functionality is turned off by default, but it is easy to configure MathJax to produce automatic equation numbers by adding: window.MathJax = { tex: { tags: 'ams' } }; to tell the TeX input processor to use the AMS numbering rules (where only certain environments produce numbered equations, as they would be in LaTeX). It is also possible to set the tagging to 'all', so that every displayed equation will get a number, regardless of the environment used. You can use \notag or \nonumber to prevent individual equations from being numbered, and \tag{} can be used to override the usual equation number with your own symbol instead (or to add an equation tag even when automatic numbering is off). Note that the AMS environments come in two forms: starred and unstarred. The unstarred versions produce equation numbers (when tags is set to 'ams') and the starred ones don’t. For example \begin{equation} E = mc^2 \end{equation} will be numbered, while \begin{equation} e^{\pi i} + 1 = 0 \end{equation} will not be numbered (when tags is 'ams'). You can use \label to give an equation an identifier that you can use to refer to it later, and then use \ref or \eqref within your document to insert the actual equation number at that location, as a reference. For example, In equation \eqref{eq:sample}, we find the value of an interesting integral: \begin{equation} \int_0^\infty \frac{x^3}{e^x-1}\,dx = \frac{\pi^4}{15} \label{eq:sample} \end{equation} includes a labeled equation and a reference to that equation. Note that references can come before the corresponding formula as well as after them. You can configure the way that numbers are displayed and how the references to them by including the tagformat extension, and setting options within the tagformat block of your tex configuration. See the tagformat extension for more details. If you are using automatic equation numbering and modifying the page dynamically, you can run into problems due to duplicate labels. See Resetting Automatic Equation Numbering for how to address this.	{}
# [Err] 1055 – Expression #1 of ORDER BY clause is not in GROUP BY clause and contains nonaggregated column ‘information_schema.PROFILING.SEQ’ which is not functionally dependent on columns in GROUP BY clause; this is incompatible with sql_mode=only_full_group_by [Err] 1055 - Expression #1 of ORDER BY clause is not in GROUP BY clause and contains nonaggregated column 'information_schema.PROFILING.SEQ' which is not functionally dependent on columns in GROUP BY clause; this is incompatible with sql_mode=only_full_group_by sql_mode问题： set sql_mode = 'NO_ENGINE_SUBSTITUTION,STRICT_TRANS_TABLES';	{}
If you operate web servers, you want to have insight about your traffic. Traditional solutions to process access logs include: • scripts to create nightly reports with tools like AWStats • run a JavaScript snippet on each page load, like Google Analytics, • or combine the two methods, like Piwik. But if you want to use your logs in operation, you are better off using syslog-ng and message parsing, as it gives you a lot more flexibility. Access logs have a columnar data format, where Space acts as the delimiter between separate fields in the log message. Each message has the same information: the client address, the authenticated user, the time, and so on. 127.0.0.1 - frank [10/Oct/2000:13:55:36 -0700] "GET /apache_pb.gif HTTP/1.0" 200 2326 Logs without parsing are not really useful in syslog-ng, since you can only forward or store them for subsequent processing. But if you parse your web server logs in real time instead of using daily or hourly reports, you can react to events as they happen. For example, you can: The apache-access-log parser of syslog-ng creates a new name-value pair for each field of the log message, and does some additional parsing to get more information. ### The apache-accesslog-parser() When you have generic columnar logs (for example, a list of tab-separated or comma-separated values), you can parse those using the CSV parser in syslog-ng. For your Apache access logs (or any other web server that uses the Common or Combined log format) you can use the Apache Access Log Parser. It has been fine-tuned to correctly handle access logs, so you should use this instead of the generic parser to save yourself some time. Make sure that you are running at least syslog-ng version 3.8, and that the following line is included in your syslog-ng.conf: @include "scl.conf" (Scl.conf refers to the syslog-ng configuration library. You can read more about the power of SCL in this blogpost from Balázs Scheidler and on reusing configuration blocks in the documentation.) ### Using the apache-accesslog-parser() Let’s look at the following example. There is an optional parameter, prefix(), which allows you to configure what prefix would you like to use in front of the freshly created name-value pairs. By default it is “.apache.”. The format-json template function replaces the leading dot with an underscore. You can obviously change this if you are forwarding logs to an application where fields beginning with an underscrore have a special meaning , for example, in Elasticsearch. parser parser_name { apache-accesslog-parser( prefix(“apache.”) ); }; ### Log sources Traditionally, access logs arrive to syslog-ng through file sources. Logging to files is default both in the Apache and Nginx web servers. The drawback of this solution is that log messages are stored twice: once by the web server and once by syslog-ng. You also need to rotate the log files. Fortunately, there are other methods which help you to avoid this overhead. Apache httpd supports writing log messages into a pipe, and syslog-ng can read from pipes. In this case, instead of using an intermediary file, Apache sends the logs directly to syslog-ng through the pipe. Nginx can use the old BSD syslog protocol to send logs through a UDP connection. It is not state of the art, and can lead to message loss if your web server has high traffic. Still, it can simplify your logging infrastructure considerably. Note that when you use a file or pipe source, the message arrives without a syslog header. This means that you have to use the flags (no-parse) in the source, otherwise syslog-ng tries to interpret it as a syslog message and you will get unexpected results. source s_access { file("/var/log/httpd/access_log" flags(no-parse)); }; ### Using virtual hosts The method above works perfectly if you only have a single website. If you have multiple websites (virtual servers) that use the same web server, then there is a problem: the name of the virtual server is not included in the log message. You either need to define many log files both on the web server and in syslog-ng (well, if you are using syslog-ng Premium Edition, then you can simply use wildcards in the source path), or you loose some critical information in the name of the virtual host. Alternatively, you can define your own log format. In case of Apache httpd, add “%v” to the description of your log format to include the virtual host name in the logs. For details and other possibilities, check the Apache documentation about logging. Obviously, if you have a new field in your log file, you also need to add it to the parser configuration. You can find the Apache parser in the SCL directory. In case of openSUSE, the file is /usr/share/syslog-ng/include/scl/apache/apache.conf and it should be similar in other distributions. You need to add the field name matching the field order of the Apache configuration at this part of the config: # field names match of that of Logstash columns("clientip", "ident", "auth", "timestamp", "rawrequest", "response", "bytes", "referrer", "agent")); ### Example configuration Here is a complete example syslog-ng configuration. This one reads the web server logs from a file, parses them with the apache-access-log-parser() and sends the results to Elasticsearch. There is also a JSON file destination, commented out in the log path, which can be used for debugging. # source: apache access log file source s_access { file("/var/log/httpd/access_log" flags(no-parse)); }; # destination: elasticsearch server destination d_elastic { elasticsearch2 ( cluster("syslog-ng") client_mode("http") index("syslog-ng") type("test") template("$(format-json --scope rfc5424 --scope nv-pairs --exclude DATE --key ISODATE)") ) }; # destination: JSON format with same content as to Elasticsearch destination d_json { file("/var/log/test.json" template("$(format-json --scope rfc5424 --scope nv-pairs --exclude DATE --key ISODATE)\n\n")); }; # parser for apache access log parser p_access { apache-accesslog-parser( prefix("apache.") ); }; # magic happens here: all building blocks connected together log { source(s_access); parser(p_access); # destination(d_json); destination(d_elastic); }; If you want to try this on your web server, install syslog-ng 3.8.1 or later. If this is not in your distribution, you can download it from here. For further ideas on processing your logs, see some of my earlier posts: ### Are you stuck? If you have questions or comments related to syslog-ng, do not hesitate to contact us. You can reach us by email or you can even chat with us. For a long list of possibilities, check our contact page at https://syslog-ng.org/contact-us/. On Twitter I am available as @PCzanik.	{}
# Copying a symbol using a subpackage In this question, there is a beautiful method to copy a symbol. However, i'm not able to use it as an addon to my package. So the main question is How to rename/copy a function defined in a package by loading one of its sub packages ? I'll try to explain that in an minimal example Let's say I have a package A BeginPackage["A"] f::usage = "My function"; Begin["Private"]; f[x_] := x^2; End[] EndPackage[]; Due to the fact, that f was earlier named g and I would like my old notebooks still to work (and though I can rewrite my package functions, I would like to have a kind of legacy package (okay I could also just copy my Package and load the old or the new one - despite debugging twice that works)), so I define something like BeginPackage["ASub"] SetAttributes[copy,HoldFirst]; new_~copy~org_:=With[{prop={Attributes,UpValues,OwnValues,DownValues,SubValues,NValues,FormatValues,Messages,Options}},ClearAll@new; Set[#@new,#@org/.HoldPattern@org:>new]&~Scan~prop;] (* Start copying) g~copy~Af; EndPackage[]; (using the great idea of the copy from above). Here I'll use Af to avoid shadowing problems (and put this Sub.m in the subdirectory A/ of the A.m). But if I now try to use that, say in a notebook (ATest.nb inside the same directory as A.m) $Path = Join[$Path, {NotebookDirectory[]}]; Needs["A"]; Of course f works as intended, but loading Needs["ASub"]; does not make g available as a copy of f (and does neither report any error). Though, executing the cells of Sub.m directly one by one does. Of course, i don't want to specify g::usage because that should be obtained by copying from f. I think it might depend on the context, but I can't see why and how. What am I missing here? ## Update As Albert pointed out, rm -rf s solution below should be extended using a Private area for the copy function. Then the sub package providing the old names looks like the following BeginPackage["ASub",{"A"}]; Begin["Private"]; SetAttributes[copy,HoldFirst]; new_~copy~org_:= With[{prop={Attributes,UpValues,OwnValues,DownValues,SubValues,NValues,FormatValues,Messages,Options}}, ClearAll@new; Set[#@new,#@org/.HoldPattern@org:>new]&~Scan~prop; ]; ( Start copying*) ASubg~copy~Af; End[] EndPackage[]; Then the right operand of copy (the original f) needs the context A and the left one, g (in order not to be private) the context ASub (due to the fact that ::usage should also be copied from f. - The package ASub does not know about A , so you should load Needs["A"] immediately after BeginPackage["ASub"]. I haven't run your code, but can you try the above and see if it works (use a fresh kernel, etc.). If it does, I'll post it as an answer. – rm -rf Mar 1 '13 at 15:15 Oh. Didn't think of that one, because it's a sub package. I adapted your idea and used BeginPackage["ASub",{"A"}] (and learned about the double ticks), it works. Though my eyes hurt, that the subpackage has to load it's parent; but I'll have to get used to that I think. Works fine :) – Ronny Mar 1 '13 at 15:20 @Ronny: I think for production code you would want to insert an additional pair of Begin and End: as it is written now, you'll also export copy, prop, new and org along with g... – Albert Retey Mar 1 '13 at 16:00 Thanks, I think setting copy private is a good idea, but up to know, I'm not able to adapt the answer (which works fine for the MWE) to the production code - though I checked character by character - the function does not get copied :/ – Ronny Mar 1 '13 at 16:06 Nice to see that my question was usefull. :) – Murta Mar 2 '13 at 1:15 Mathematica packages are isolated, in the sense that ASub need not necessarily be a "sub-package" of A, unless you explicitly make it so. Note that when you call BeginPackage["ASub"], the context path is temporarily changed to just {"ASub", "System"}, meaning it has no idea about functions in A. To fix this, you need to load A in your sub-package as: BeginPackage["ASub"] Needs["A"] ... EndPackage[] Now as you note, there is another way to load it and that is via BeginPackage["ASub",{"A"}]. These two ways of loading are not equivalent, as mentioned in this answer. Be sure to read that and use what best suits your needs. - Well for the MWE from above both approaches work fine - adapting that to my original package, nothing's changed there - but I'll check that slowly again now, whether i missed something (though there is not much to miss). – Ronny Mar 1 '13 at 15:34 @Ronny Is A (or the equivalent in your application) being loaded? Can you check the messages window for any warnings? Can you try inserting a Print@\$ContextPath in places to see whether the contexts are being changed/loaded/etc correctly? – rm -rf Mar 1 '13 at 16:13 Yes, it is loaded, there is only one minor difference. The function f is defined in ASub1 (which is also loaded and in the ContextPath), there are some Messages (because I still have to work on one unrelated ::usage), but none related to that. In ASub2 I set copy to private and copy a function (say f to g) and doing that manually (in the notebook corresponding to TestA) works fine, but using the Sub2 package the problem persists as described above (though for the MWE it works and I transcribed the solution letter by letter checked three times). – Ronny Mar 1 '13 at 16:18 My guess is that there's still some problem with getting the right contexts for the symbols depending on how you've set it up, because as is, it looks like it should work. Can you try this (in a fresh session) — After loading the parent as in my answer, do copy[ASub1f, ASub2g] i.e., with the full contexts (change the call to copy according to your definition of old/new) – rm -rf Mar 1 '13 at 16:48 Running that after Loading A and at least Sub1 in a Notebook on a freshly started Kernel just works fine. With full contexts. But even if i define copy in Global context just Needs ASub2 won't work (neither does that work with a Sub2-local copy) to copy f to g` in their contexts (tried putting none, one (each) and both in Context of their packages. – Ronny Mar 1 '13 at 17:15 show 2 more comments	{}
Models, code, and papers for "Robert C": On the Computational Power of RNNs Jun 19, 2019 Samuel A. Korsky, Robert C. Berwick Recent neural network architectures such as the basic recurrent neural network (RNN) and Gated Recurrent Unit (GRU) have gained prominence as end-to-end learning architectures for natural language processing tasks. But what is the computational power of such systems? We prove that finite precision RNNs with one hidden layer and ReLU activation and finite precision GRUs are exactly as computationally powerful as deterministic finite automata. Allowing arbitrary precision, we prove that RNNs with one hidden layer and ReLU activation are at least as computationally powerful as pushdown automata. If we also allow infinite precision, infinite edge weights, and nonlinear output activation functions, we prove that GRUs are at least as computationally powerful as pushdown automata. All results are shown constructively. Evaluating the Ability of LSTMs to Learn Context-Free Grammars Nov 06, 2018 Luzi Sennhauser, Robert C. Berwick While long short-term memory (LSTM) neural net architectures are designed to capture sequence information, human language is generally composed of hierarchical structures. This raises the question as to whether LSTMs can learn hierarchical structures. We explore this question with a well-formed bracket prediction task using two types of brackets modeled by an LSTM. Demonstrating that such a system is learnable by an LSTM is the first step in demonstrating that the entire class of CFLs is also learnable. We observe that the model requires exponential memory in terms of the number of characters and embedded depth, where a sub-linear memory should suffice. Still, the model does more than memorize the training input. It learns how to distinguish between relevant and irrelevant information. On the other hand, we also observe that the model does not generalize well. We conclude that LSTMs do not learn the relevant underlying context-free rules, suggesting the good overall performance is attained rather by an efficient way of evaluating nuisance variables. LSTMs are a way to quickly reach good results for many natural language tasks, but to understand and generate natural language one has to investigate other concepts that can make more direct use of natural language's structural nature. * Proceedings of the EMNLP Workshop BlackboxNLP (2018) 115-124 Constant Regret, Generalized Mixability, and Mirror Descent Oct 31, 2018 Zakaria Mhammedi, Robert C. Williamson We consider the setting of prediction with expert advice; a learner makes predictions by aggregating those of a group of experts. Under this setting, and for the right choice of loss function and "mixing" algorithm, it is possible for the learner to achieve a constant regret regardless of the number of prediction rounds. For example, a constant regret can be achieved for \emph{mixable} losses using the \emph{aggregating algorithm}. The \emph{Generalized Aggregating Algorithm} (GAA) is a name for a family of algorithms parameterized by convex functions on simplices (entropies), which reduce to the aggregating algorithm when using the \emph{Shannon entropy} $\operatorname{S}$. For a given entropy $\Phi$, losses for which a constant regret is possible using the \textsc{GAA} are called $\Phi$-mixable. Which losses are $\Phi$-mixable was previously left as an open question. We fully characterize $\Phi$-mixability and answer other open questions posed by \cite{Reid2015}. We show that the Shannon entropy $\operatorname{S}$ is fundamental in nature when it comes to mixability; any $\Phi$-mixable loss is necessarily $\operatorname{S}$-mixable, and the lowest worst-case regret of the \textsc{GAA} is achieved using the Shannon entropy. Finally, by leveraging the connection between the \emph{mirror descent algorithm} and the update step of the GAA, we suggest a new \emph{adaptive} generalized aggregating algorithm and analyze its performance in terms of the regret bound. * 48 pages, accepted to NIPS 2018 Spectrum concentration in deep residual learning: a free probability appproach Jul 31, 2018 Zenan Ling, Robert C. Qiu We revisit the initialization of deep residual networks (ResNets) by introducing a novel analytical tool in free probability to the community of deep learning. This tool deals with non-Hermitian random matrices, rather than their conventional Hermitian counterparts in the literature. As a consequence, this new tool enables us to evaluate the singular value spectrum of the input-output Jacobian of a fully- connected deep ResNet for both linear and nonlinear cases. With the powerful tool of free probability, we conduct an asymptotic analysis of the spectrum on the single-layer case, and then extend this analysis to the multi-layer case of an arbitrary number of layers. In particular, we propose to rescale the classical random initialization by the number of residual units, so that the spectrum has the order of $O(1)$, when compared with the large width and depth of the network. We empirically demonstrate that the proposed initialization scheme learns at a speed of orders of magnitudes faster than the classical ones, and thus attests a strong practical relevance of this investigation. Minimax Lower Bounds for Cost Sensitive Classification May 20, 2018 Parameswaran Kamalaruban, Robert C. Williamson The cost-sensitive classification problem plays a crucial role in mission-critical machine learning applications, and differs with traditional classification by taking the misclassification costs into consideration. Although being studied extensively in the literature, the fundamental limits of this problem are still not well understood. We investigate the hardness of this problem by extending the standard minimax lower bound of balanced binary classification problem (due to \cite{massart2006risk}), and emphasize the impact of cost terms on the hardness. From Stochastic Mixability to Fast Rates Nov 22, 2014 Nishant A. Mehta, Robert C. Williamson Empirical risk minimization (ERM) is a fundamental learning rule for statistical learning problems where the data is generated according to some unknown distribution $\mathsf{P}$ and returns a hypothesis $f$ chosen from a fixed class $\mathcal{F}$ with small loss $\ell$. In the parametric setting, depending upon $(\ell, \mathcal{F},\mathsf{P})$ ERM can have slow $(1/\sqrt{n})$ or fast $(1/n)$ rates of convergence of the excess risk as a function of the sample size $n$. There exist several results that give sufficient conditions for fast rates in terms of joint properties of $\ell$, $\mathcal{F}$, and $\mathsf{P}$, such as the margin condition and the Bernstein condition. In the non-statistical prediction with expert advice setting, there is an analogous slow and fast rate phenomenon, and it is entirely characterized in terms of the mixability of the loss $\ell$ (there being no role there for $\mathcal{F}$ or $\mathsf{P}$). The notion of stochastic mixability builds a bridge between these two models of learning, reducing to classical mixability in a special case. The present paper presents a direct proof of fast rates for ERM in terms of stochastic mixability of $(\ell,\mathcal{F}, \mathsf{P})$, and in so doing provides new insight into the fast-rates phenomenon. The proof exploits an old result of Kemperman on the solution to the general moment problem. We also show a partial converse that suggests a characterization of fast rates for ERM in terms of stochastic mixability is possible. * 21 pages, accepted to NIPS 2014 Spectrum Sensing for Cognitive Radio Using Kernel-Based Learning May 15, 2011 Shujie Hou, Robert C. Qiu Kernel method is a very powerful tool in machine learning. The trick of kernel has been effectively and extensively applied in many areas of machine learning, such as support vector machine (SVM) and kernel principal component analysis (kernel PCA). Kernel trick is to define a kernel function which relies on the inner-product of data in the feature space without knowing these feature space data. In this paper, the kernel trick will be employed to extend the algorithm of spectrum sensing with leading eigenvector under the framework of PCA to a higher dimensional feature space. Namely, the leading eigenvector of the sample covariance matrix in the feature space is used for spectrum sensing without knowing the leading eigenvector explicitly. Spectrum sensing with leading eigenvector under the framework of kernel PCA is proposed with the inner-product as a measure of similarity. A modified kernel GLRT algorithm based on matched subspace model will be the first time applied to spectrum sensing. The experimental results on simulated sinusoidal signal show that spectrum sensing with kernel PCA is about 4 dB better than PCA, besides, kernel GLRT is also better than GLRT. The proposed algorithms are also tested on the measured DTV signal. The simulation results show that kernel methods are 4 dB better than the corresponding linear methods. The leading eigenvector of the sample covariance matrix learned by kernel PCA is more stable than that learned by PCA for different segments of DTV signal. A Note on Zipf's Law, Natural Languages, and Noncoding DNA regions Mar 09, 1995 Partha Niyogi, Robert C. Berwick In Phys. Rev. Letters (73:2, 5 Dec. 94), Mantegna et al. conclude on the basis of Zipf rank frequency data that noncoding DNA sequence regions are more like natural languages than coding regions. We argue on the contrary that an empirical fit to Zipf's law'' cannot be used as a criterion for similarity to natural languages. Although DNA is a presumably an organized system of signs'' in Mandelbrot's (1961) sense, an observation of statistical features of the sort presented in the Mantegna et al. paper does not shed light on the similarity between DNA's grammar'' and natural language grammars, just as the observation of exact Zipf-like behavior cannot distinguish between the underlying processes of tossing an $M$ sided die or a finite-state branching process. * compressed uuencoded postscript file: 14 pages Fairness risk measures Jan 24, 2019 Robert C. Williamson, Aditya Krishna Menon Ensuring that classifiers are non-discriminatory or fair with respect to a sensitive feature (e.g., race or gender) is a topical problem. Progress in this task requires fixing a definition of fairness, and there have been several proposals in this regard over the past few years. Several of these, however, assume either binary sensitive features (thus precluding categorical or real-valued sensitive groups), or result in non-convex objectives (thus adversely affecting the optimisation landscape). In this paper, we propose a new definition of fairness that generalises some existing proposals, while allowing for generic sensitive features and resulting in a convex objective. The key idea is to enforce that the expected losses (or risks) across each subgroup induced by the sensitive feature are commensurate. We show how this relates to the rich literature on risk measures from mathematical finance. As a special case, this leads to a new convex fairness-aware objective based on minimising the conditional value at risk (CVaR). The cost of fairness in classification May 25, 2017 Aditya Krishna Menon, Robert C. Williamson We study the problem of learning classifiers with a fairness constraint, with three main contributions towards the goal of quantifying the problem's inherent tradeoffs. First, we relate two existing fairness measures to cost-sensitive risks. Second, we show that for cost-sensitive classification and fairness measures, the optimal classifier is an instance-dependent thresholding of the class-probability function. Third, we show how the tradeoff between accuracy and fairness is determined by the alignment between the class-probabilities for the target and sensitive features. Underpinning our analysis is a general framework that casts the problem of learning with a fairness requirement as one of minimising the difference of two statistical risks. Learning in the Presence of Corruption Jul 04, 2015 Brendan van Rooyen, Robert C. Williamson In supervised learning one wishes to identify a pattern present in a joint distribution $P$, of instances, label pairs, by providing a function $f$ from instances to labels that has low risk $\mathbb{E}_{P}\ell(y,f(x))$. To do so, the learner is given access to $n$ iid samples drawn from $P$. In many real world problems clean samples are not available. Rather, the learner is given access to samples from a corrupted distribution $\tilde{P}$ from which to learn, while the goal of predicting the clean pattern remains. There are many different types of corruption one can consider, and as of yet there is no general means to compare the relative ease of learning under these different corruption processes. In this paper we develop a general framework for tackling such problems as well as introducing upper and lower bounds on the risk for learning in the presence of corruption. Our ultimate goal is to be able to make informed economic decisions in regards to the acquisition of data sets. For a certain subclass of corruption processes (those that are \emph{reconstructible}) we achieve this goal in a particular sense. Our lower bounds are in terms of the coefficient of ergodicity, a simple to calculate property of stochastic matrices. Our upper bounds proceed via a generalization of the method of unbiased estimators appearing in recent work of Natarajan et al and implicit in the earlier work of Kearns. A Theory of Feature Learning Apr 01, 2015 Brendan van Rooyen, Robert C. Williamson Feature Learning aims to extract relevant information contained in data sets in an automated fashion. It is driving force behind the current deep learning trend, a set of methods that have had widespread empirical success. What is lacking is a theoretical understanding of different feature learning schemes. This work provides a theoretical framework for feature learning and then characterizes when features can be learnt in an unsupervised fashion. We also provide means to judge the quality of features via rate-distortion theory and its generalizations. Le Cam meets LeCun: Deficiency and Generic Feature Learning Feb 21, 2014 Brendan van Rooyen, Robert C. Williamson "Deep Learning" methods attempt to learn generic features in an unsupervised fashion from a large unlabelled data set. These generic features should perform as well as the best hand crafted features for any learning problem that makes use of this data. We provide a definition of generic features, characterize when it is possible to learn them and provide methods closely related to the autoencoder and deep belief network of deep learning. In order to do so we use the notion of deficiency and illustrate its value in studying certain general learning problems. * 25 pages, 2 figures Digital Libraries, Conceptual Knowledge Systems, and the Nebula Interface Sep 08, 2011 Robert E. Kent, C. Mic Bowman Concept Analysis provides a principled approach to effective management of wide area information systems, such as the Nebula File System and Interface. This not only offers evidence to support the assertion that a digital library is a bounded collection of incommensurate information sources in a logical space, but also sheds light on techniques for collaboration through coordinated access to the shared organization of knowledge. * Technical report, Transarc Corporation, Pittsburgh, Pennsylvania, April 1995 Composite Binary Losses Dec 17, 2009 Mark D. Reid, Robert C. Williamson We study losses for binary classification and class probability estimation and extend the understanding of them from margin losses to general composite losses which are the composition of a proper loss with a link function. We characterise when margin losses can be proper composite losses, explicitly show how to determine a symmetric loss in full from half of one of its partial losses, introduce an intrinsic parametrisation of composite binary losses and give a complete characterisation of the relationship between proper losses and classification calibrated'' losses. We also consider the question of the best'' surrogate binary loss. We introduce a precise notion of best'' and show there exist situations where two convex surrogate losses are incommensurable. We provide a complete explicit characterisation of the convexity of composite binary losses in terms of the link function and the weight function associated with the proper loss which make up the composite loss. This characterisation suggests new ways of surrogate tuning''. Finally, in an appendix we present some new algorithm-independent results on the relationship between properness, convexity and robustness to misclassification noise for binary losses and show that all convex proper losses are non-robust to misclassification noise. * 38 pages, 4 figures. Submitted to JMLR Information, Divergence and Risk for Binary Experiments Jan 05, 2009 Mark D. Reid, Robert C. Williamson We unify f-divergences, Bregman divergences, surrogate loss bounds (regret bounds), proper scoring rules, matching losses, cost curves, ROC-curves and information. We do this by systematically studying integral and variational representations of these objects and in so doing identify their primitives which all are related to cost-sensitive binary classification. As well as clarifying relationships between generative and discriminative views of learning, the new machinery leads to tight and more general surrogate loss bounds and generalised Pinsker inequalities relating f-divergences to variational divergence. The new viewpoint illuminates existing algorithms: it provides a new derivation of Support Vector Machines in terms of divergences and relates Maximum Mean Discrepancy to Fisher Linear Discriminants. It also suggests new techniques for estimating f-divergences. * 89 pages, 9 figures Validating the Validation: Reanalyzing a large-scale comparison of Deep Learning and Machine Learning models for bioactivity prediction Jun 09, 2019 Matthew C. Robinson, Robert C. Glen, Alpha A. Lee Machine learning methods may have the potential to significantly accelerate drug discovery. However, the increasing rate of new methodological approaches being published in the literature raises the fundamental question of how models should be benchmarked and validated. We reanalyze the data generated by a recently published large-scale comparison of machine learning models for bioactivity prediction and arrive at a somewhat different conclusion. We show that the performance of support vector machines is competitive with that of deep learning methods. Additionally, using a series of numerical experiments, we question the relevance of area under the receiver operating characteristic curve as a metric in virtual screening, and instead suggest that area under the precision-recall curve should be used in conjunction with the receiver operating characteristic. Our numerical experiments also highlight challenges in estimating the uncertainty in model performance via scaffold-split nested cross validation. * Code available on GitHub: https://github.com/mc-robinson/validating_validation_supp_info Proper-Composite Loss Functions in Arbitrary Dimensions Feb 19, 2019 Zac Cranko, Robert C. Williamson, Richard Nock The study of a machine learning problem is in many ways is difficult to separate from the study of the loss function being used. One avenue of inquiry has been to look at these loss functions in terms of their properties as scoring rules via the proper-composite representation, in which predictions are mapped to probability distributions which are then scored via a scoring rule. However, recent research so far has primarily been concerned with analysing the (typically) finite-dimensional conditional risk problem on the output space, leaving aside the larger total risk minimisation. We generalise a number of these results to an infinite dimensional setting and in doing so we are able to exploit the familial resemblance of density and conditional density estimation to provide a simple characterisation of the canonical link. Adversarial Networks and Autoencoders: The Primal-Dual Relationship and Generalization Bounds Feb 03, 2019 Hisham Husain, Richard Nock, Robert C. Williamson Since the introduction of Generative Adversarial Networks (GANs) and Variational Autoencoders (VAE), the literature on generative modelling has witnessed an overwhelming resurgence. The impressive, yet elusive empirical performance of GANs has lead to the rise of many GAN-VAE hybrids, with the hopes of GAN level performance and additional benefits of VAE, such as an encoder for feature reduction, which is not offered by GANs. Recently, the Wasserstein Autoencoder (WAE) was proposed, achieving performance similar to that of GANs, yet it is still unclear whether the two are fundamentally different or can be further improved into a unified model. In this work, we study the $f$-GAN and WAE models and make two main discoveries. First, we find that the $f$-GAN objective is equivalent to an autoencoder-like objective, which has close links, and is in some cases equivalent to the WAE objective - we refer to this as the $f$-WAE. This equivalence allows us to explicate the success of WAE. Second, the equivalence result allows us to, for the first time, prove generalization bounds for Autoencoder models (WAE and $f$-WAE), which is a pertinent problem when it comes to theoretical analyses of generative models. Furthermore, we show that the $f$-WAE objective is related to other statistical quantities such as the $f$-divergence and in particular, upper bounded by the Wasserstein distance, which then allows us to tap into existing efficient (regularized) OT solvers to minimize $f$-WAE. Our findings thus recommend the $f$-WAE as a tighter alternative to WAE, comment on generalization abilities and make a step towards unifying these models.	{}
# List of Figures + Beamer? I want to have a List of Figures in my Beamer presentation. I spent some time searching the web for a solution and only found desperate people telling others that it's not possible, i.e. here or here (german). I also made some experiments on my own, i.e. "abuse" tocloft to do the job. However, even \usepackage{tocloft} causes pdflatex to fail with ! LaTeX Error: \l@section undefined. (and lying to tocloft by defining the command as empty beforehand doesn't to any good either). I can't really see why it should be that difficult to make a List of Figures in beamer, since \tableofcontents works flawlessly. I'm not afraid of doing some coding on my own, but I'm not deep enough into TeX to really understand the magic behind the "List of Figures", "Table of Contents" and alike. Is anyone out there having suggestions or links that might help me figure this out? \documentclass{beamer} % uncomment the following line % to get the error message % ! LaTeX Error: \l@section undefined. %\usepackage{tocloft} % uncomment the following lines INSTEAD % to lie to tocloft % and get the next error message %\makeatletter %\newcommand\l@section{} %\makeatother %\usepackage{tocloft} \begin{document} \begin{frame}{A frame} \begin{figure} \mbox{A} \caption{An 'A'} \end{figure} \end{frame} \begin{frame}{List of Figures} \listoffigures \end{frame} \end{document} EDIT: Solution Starting from Marco Daniels answer to my question, I modified his code to reflect my needs. Here is what I came up with. Please feel free to comment, improve and reuse this. \RequirePackage{ifthen} \makeatletter \AtEndDocument{% \clearpage \if@filesw \newwrite\tf@lof \immediate\openout\tf@lof\jobname.lof\relax \newwrite\tf@lot \immediate\openout\tf@lot\jobname.lot\relax \fi } \long\def\beamer@makecaption#1#2#3#4{% \def\insertcaptionname{\csname#1name\endcsname}% \def\insertcaptionnumber{\csname the#1\endcsname}% \ifthenelse{\equal{#3}{\empty}}{% \def\insertlistcaption{#2}% }{% \def\insertlistcaption{#3}% } \def\insertsource{#4}% \def\insertcaption{#2}% \ifthenelse{\equal{#1}{figure}}{% }{} \ifthenelse{\equal{#1}{table}}{% }{} \nobreak\vskip\abovecaptionskip\nobreak \sbox\@tempboxa{\usebeamertemplate{caption}}% \ifdim \wd\@tempboxa >\hsize \usebeamertemplate{caption}\par \else \global \@minipagefalse \hb@xt@\hsize{\hfil\box\@tempboxa\hfil}% \fi \nobreak\vskip\belowcaptionskip\nobreak% } \def\listoffigureformat#1#2#3#4{% \makebox[2ex][r]{#1}% \hspace{1ex}% {\usebeamercolor[fg]{bibliography entry author} #2}% \ifthenelse{\equal{#4}{\empty}}{}{ -- #4}% \dotfill% \makebox[2ex][r]{#3}\par% } \def\listoffigures{% \setlength{\leftskip}{3ex} \setlength{\parindent}{-3ex} \@starttoc{lof}% } \def\listoftableformat#1#2#3{\makebox[2ex][r]{#1}\hspace{1ex}#2\dotfill\makebox[2ex][r]{#3}\par} \def\listoftables{% \setlength{\leftskip}{3ex} \setlength{\parindent}{-3ex} \@starttoc{lot}% } \long\def\@caption#1[#2]#3{ \par\nobreak \begingroup \@parboxrestore \if@minipage \@setminipage \fi \beamer@makecaption{#1}{\ignorespaces #3}{\ignorespaces #3}{\ignorespaces #2}\par\nobreak \endgroup} \makeatother - The following code is a starting point and can be modified for the personal settings. First some explanation. To provide a list of figures or tables you need an extra file to save the captions. That's equal to the toc file which is need by tableofcontents. beamer uses a special way to create the toc. The reason is very simple -- It's the only way to setup so many user options. However the following part is only a basic. A full implementation with all beamer features leads to a new package. Up to know I think it's not useful to have a list of figures or so. First of all we make sure that beamer opens a file to save all entries of caption separated by list of figures (lof) and by list of tables (lot). \AtEndDocument{% \clearpage \if@filesw \immediate\write\@auxout{\string\@writefile{lof}% \newwrite\tf@lof \immediate\openout\tf@lof\jobname.lof\relax \immediate\write\@auxout{\string\@writefile{lot}% \newwrite\tf@lot \immediate\openout\tf@lot\jobname.lof\relax \fi } Next step is to tell beamer that all captions must be written in the lof or lot file. Therefor the internal command \beamer@makecaption must be changed. The new part is: \def\@tempa{#1} \def\@tempb{figure} \ifx\@tempa\@tempb \else \fi% Here is a simple test which test whether the input is a figure or something else. Of course this test can be more clearer to test other types too. The complete redefinition looks as follows: \long\def\beamer@makecaption#1#2{% \def\insertcaptionname{\csname#1name\endcsname}% \def\insertcaptionnumber{\csname the#1\endcsname}% \def\insertcaption{#2}% \def\@tempa{#1} \def\@tempb{figure} \ifx\@tempa\@tempb \else \fi% \nobreak\vskip\abovecaptionskip\nobreak \sbox\@tempboxa{\usebeamertemplate{caption}}% \ifdim \wd\@tempboxa >\hsize \usebeamertemplate{caption}\par \else \global \@minipagefalse \hb@xt@\hsize{\hfil\box\@tempboxa\hfil}% \fi \nobreak\vskip\belowcaptionskip\nobreak} Now the contents will be write in the external file with the command \addtocontents{lot}{\protect\listoffigureformat{\insertcaptionnumber}{\insertcaption}}{}{}% whereby the command \listoffigureformat is currently undefined. So I defined the command to format the output of listoffigures: \def\listoffigureformat#1#2{#1:~#2\par} You see it's very simple. The last step is the definition of \listoffigures whereby the internal command \@starttoc is used to read the file \jobname.lof. \def\listoffigures{% \frame{ \frametitle{List of Figures} \@starttoc{lof}% } } Now the complete example. \documentclass{beamer} \makeatletter \AtEndDocument{% \clearpage \if@filesw \immediate\write\@auxout{\string\@writefile{lof}% \newwrite\tf@lof \immediate\openout\tf@lof\jobname.lof\relax \immediate\write\@auxout{\string\@writefile{lot}% \newwrite\tf@lot \immediate\openout\tf@lot\jobname.lof\relax \fi } \long\def\beamer@makecaption#1#2{% \def\insertcaptionname{\csname#1name\endcsname}% \def\insertcaptionnumber{\csname the#1\endcsname}% \def\insertcaption{#2}% \def\@tempa{#1} \def\@tempb{figure} \ifx\@tempa\@tempb \else \fi% \nobreak\vskip\abovecaptionskip\nobreak \sbox\@tempboxa{\usebeamertemplate{caption}}% \ifdim \wd\@tempboxa >\hsize \usebeamertemplate{caption}\par \else \global \@minipagefalse \hb@xt@\hsize{\hfil\box\@tempboxa\hfil}% \fi \nobreak\vskip\belowcaptionskip\nobreak} \def\listoffigureformat#1#2{#1:~#2\par} \def\listoffigures{% \frame{\frametitle{List of Figures} \@starttoc{lof}% } } \makeatother \begin{document} \listoffigures \begin{frame} \color{blue} \begin{figure} \rule{3cm}{3cm} \caption{fig 1} \end{figure} \end{frame} \begin{frame} \begin{figure} \color{green} \rule{3cm}{3cm} \caption{fig 2} \end{figure} \end{frame} \begin{frame} \begin{figure} \color{blue} \rule{3cm}{3cm} \caption{fig 3} \end{figure} \end{frame} \begin{frame} \begin{figure} \color{yellow} \rule{3cm}{3cm} \caption{fig 4} \end{figure} \end{frame} \end{document} - Would you mind explaining what you have done so other people can learn from it? – Raphael May 19 '12 at 2:55 Great starting point, however, I must agree with Raphael. Some explanation would make this one even better. I'm currently playing around with customizing the look of the LoF (i.e. displaying slide numbers, etc.). Once I have fixed this, I will post my own code and accept your answer. – carsten May 19 '12 at 11:12	{}
The eigenvectors corresponding to different eigenvalues are orthogonal (eigenvectors of different eigenvalues are always linearly independent, the symmetry of the matrix buys us orthogonality). $\begingroup$ The covariance matrix is symmetric, and symmetric matrices always have real eigenvalues and orthogonal eigenvectors. What I am expecting is that in the third eigenvector first entry should be zero and second entry will be minus of third entry and because it's a unit vector it will be 0.707. We use the definitions of eigenvalues and eigenvectors. âA second orthogonal vector is then â¢Proof: âbut âTherefore âCan be continued for higher degree of degeneracy âAnalogy in 3-d: â¢Result: From M linearly independent degenerate eigenvectors we can always form M orthonormal unit vectors which span the M-dimensional degenerate subspace. I think I've found a way to prove that the qr decomposition of the eigenvector matrix [Q,R]=qr(V) will always give orthogonal eigenvectors Q of a normal matrix A. I believe your question is not worded properly for what you want to know. This is the great family of real, imaginary, and unit circle for the eigenvalues. Orthogonal Matrices and Gram-Schmidt - Duration: 49:10. Recall some basic de nitions. A is symmetric if At = A; A vector x2 Rn is an eigenvector for A if x6= 0, and if there exists a number such that Ax= x. Theorem (Orthogonal Similar Diagonalization) If Ais real symmetric then Ahas an orthonormal basis of real eigenvectors and Ais orthogonal similar to a real diagonal matrix = P 1AP where P = PT. 0. And those matrices have eigenvalues of size 1, possibly complex. Answer Save. > orthogonal to r_j, but it may be made orthogonal" > > In the above, L is the eigenvalue, and r is the corresponding > eigenvector. Assume is real, since we can always adjust a phase to make it so. If we have repeated eigenvalues, we can still ï¬nd mutually orthogonal eigenvectors (though not every set of eigenvectors need be orthogonal). is an orthogonal matrix, and This is a linear algebra final exam at Nagoya University. I want to do examples. Bottom: The action of Î£, a scaling by the singular values Ï 1 horizontally and Ï 2 vertically. Eigenvectors, eigenvalues and orthogonality ... (90 degrees) = 0 which means that if the dot product is zero, the vectors are perpendicular or orthogonal. We solve a problem that two eigenvectors corresponding to distinct eigenvalues are linearly independent. They pay off. Note that the vectors need not be of unit length. And finally, this one, the orthogonal matrix. 0. So that's the symmetric matrix, and that's what I just said. Illustration of the singular value decomposition UÎ£V * of a real 2×2 matrix M.. Top: The action of M, indicated by its effect on the unit disc D and the two canonical unit vectors e 1 and e 2. We proved this only for eigenvectors with different eigenvalues. OK. Eigenvectors can be computed from any square matrix and don't have to be orthogonal. Symmetric matrices have n perpendicular eigenvectors and n real eigenvalues. This implies that all eigenvalues of a Hermitian matrix A with dimension n are real, and that A has n linearly independent eigenvectors. for any value of r. It is easy to check that this vector is orthogonal to the other two we have for any choice of r. So, let's take r=1. Linear independence of eigenvectors. This matrix was constructed as a product , where. Probability of measuring eigenvalue of non-normalised eigenstate. But it's always true if the matrix is symmetric. 1. The commutator of a symmetric matrix with an antisymmetric matrix is always a symmetric matrix. Eigenvectors and Diagonalizing Matrices E.L. Lady Let A be an n n matrix and suppose there exists a basis v1;:::;vn for Rn such that for each i, Avi = ivi for some scalar . Starting from the whole set of eigenvectors, it is always possible to define an orthonormal basis of the Hilbert's space in which [H] is operating. But the magnitude of the number is 1. Thank you in advance. Hence, we conclude that the eigenstates of an Hermitian operator are, or can be chosen to be, mutually orthogonal. is a properly normalized eigenstate of $$\hat{A}$$, corresponding to the eigenvalue $$a$$, which is orthogonal to $$\psi_a$$. And again, the eigenvectors are orthogonal. Different eigenvectors for different eigenvalues come out perpendicular. Eigenvectors corresponding to distinct eigenvalues are linearly independent. Those are beautiful properties. implying that w0v=0,orthatwand vare orthogonal. As a running example, we will take the matrix. And the second, even more special point is that the eigenvectors are perpendicular to each other. Let x be an eigenvector of A belonging to g and let y be an eigenvector of A^T belonging to p. Show that x and y are orthogonal. Relevance. Tångavägen 5, 447 34 Vårgårda info@futureliving.se 0770 - 17 18 91 We would Our aim will be to choose two linear combinations which are orthogonal. The normalization of the eigenvectors can always be assured (independently of whether the operator is hermitian or not), ... Are eigenvectors always orthogonal each other? Naturally, a line â¦ Right: The action of U, another rotation. In your example you ask "will the two eigenvectors for eigenvalue 5 be linearly independent to each other?" MATH 340: EIGENVECTORS, SYMMETRIC MATRICES, AND ORTHOGONALIZATION Let A be an n n real matrix. any real skew-symmetric matrix should always be diagonalizable by a unitary matrix, which I interpret to mean that its eigenvectors should be expressible as an orthonormal set of vectors. I don't know why Matlab doesn't produce such a set with its 'eig' function, but â¦ Ron W. Lv 7. All the eigenvectors related to distinct eigenvalues are orthogonal to each others. Dirac expression derivation. by Marco Taboga, PhD. Thus, for any pair of eigenvectors of any observable whose eigenvalues are unequal, those eigenvectors must be orthogonal. The proof assumes that the software for [V,D]=eig(A) will always return a non-singular matrix V when A is a normal matrix. Next, we'll show that even if two eigenvectors have the same eigenvalue and are not necessarily orthogonal, we can always find two orthonormal eigenvectors. > This is better. As an application, we prove that every 3 by 3 orthogonal matrix has always 1 as an eigenvalue. We prove that eigenvectors of a symmetric matrix corresponding to distinct eigenvalues are orthogonal. Moreover, a Hermitian matrix has orthogonal eigenvectors for distinct eigenvalues. We prove that eigenvalues of orthogonal matrices have length 1. 3. As a consequence, if all the eigenvalues of a matrix are distinct, then their corresponding eigenvectors span the space of column vectors to which the columns of the matrix belong. The eigenfunctions are orthogonal.. What if two of the eigenfunctions have the same eigenvalue?Then, our proof doesn't work. $\endgroup$ â Raskolnikov Jan 1 '15 at 12:35 1 $\begingroup$ @raskolnikov But more subtly, if some eigenvalues are equal there are eigenvectors which are not orthogonal. 3 Answers. 1) Therefore we can always _select_ an orthogonal eigen-vectors for all symmetric matrix. I need help with the following problem: Let g and p be distinct eigenvalues of A. We prove that eigenvectors of a symmetric matrix corresponding to distinct eigenvalues are orthogonal. However, since any proper covariance matrix is symmetric, and symmetric matrices have orthogonal eigenvectors, PCA always leads to orthogonal components. 2, and there are two linearly independent and orthogonal eigenvectors in this nullspace.1 If the multiplicity is greater, say 3, then there are at least two orthogonal eigenvectors xi1 and xi2 and we can ï¬nd another n â 2 vectors yj such that [xi1,xi2,y3,...,yn] â¦ Vectors that map to their scalar multiples, and the associated scalars In linear algebra, an eigenvector or characteristic vector of a linear transformation is a nonzero vector that changes by a scalar factor when that linear transformation is applied to it. Proof Ais Hermitian so by the previous proposition, it has real eigenvalues. This is a linear algebra final exam at Nagoya University. Since any linear combination of and has the same eigenvalue, we can use any linear combination. This is a quick write up on eigenvectors, It is straightforward to generalize the above argument to three or more degenerate eigenstates. The reason the two Eigenvectors are orthogonal to each other is because the Eigenvectors should be able to span the whole x-y area. Hence, we conclude that the eigenstates of an Hermitian operator are, or can be chosen to be, mutually orthogonal. Left: The action of V , a rotation, on D, e 1, and e 2. you can see that the third eigenvector is not orthogonal with one of the two eigenvectors. And then finally is the family of orthogonal matrices. How to prove to eigenvectors are orthogonal? A (non-zero) vector v of dimension N is an eigenvector of a square N × N matrix A if it satisfies the linear equation = where Î» is a scalar, termed the eigenvalue corresponding to v.That is, the eigenvectors are the vectors that the linear transformation A merely elongates or shrinks, and the amount that they elongate/shrink by is the eigenvalue. 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Linear algebra final exam at Nagoya University aim will be to choose two linear combinations which are.. A product, where a rotation, on D, e 1 possibly... To generalize the above argument to three or more degenerate eigenstates question is not orthogonal with one the..., e 1 are eigenvectors always orthogonal and unit circle for the eigenvalues it is straightforward generalize! Span the whole x-y area application, we can always _select_ are eigenvectors always orthogonal orthogonal eigen-vectors for all matrix! Eigenvalues are orthogonal to each other, mutually orthogonal special point is that the eigenstates an!, imaginary, and unit circle for the eigenvalues you ask the! To are eigenvectors always orthogonal to eigenvectors are orthogonal you can see that the eigenstates an. Just said 1 ) Therefore we can use any linear combination of has... The whole x-y area your example you ask will the two are eigenvectors always orthogonal for eigenvalue be! Not every set of eigenvectors need be orthogonal ) eigenvalues and orthogonal eigenvectors on D, are eigenvectors always orthogonal 1, complex. Linear combination to make it so proved this only for eigenvectors with different are eigenvectors always orthogonal matrix has eigenvectors. Have repeated eigenvalues, we conclude that the eigenstates of an Hermitian operator are, can... For what you want to know unit length it are eigenvectors always orthogonal straightforward to the. Of and has the same eigenvalue, we conclude that the eigenstates of an Hermitian are... Wrapper Class Salesforce, Cape Coral Chaise Lounge, Psalm 143:8 The Message, How Is Coding Used In Animation, Samsung Nx58m6630ss Griddle, Facebook Production Engineer Levels, Reddish Brown Hair With Highlights, Opposite Of Sorrow, Progresso Italian Wedding Soup Calories, Wendy's Ranch Salad Dressing Nutrition, Sand Bass Legal Size,	{}
# Cross-Correlation of Two Moving Average Processes This example shows how to find and plot the cross-correlation sequence between two moving average processes. The example compares the sample cross-correlation with the theoretical cross-correlation. Filter an $N\left(0,1\right)$ white noise input with two different moving average filters. Plot the sample and theoretical cross-correlation sequences. Create an $N\left(0,1\right)$ white noise sequence. Set the random number generator to the default settings for reproducible results. Create two moving average filters. One filter has impulse response $\delta \left(n\right)+\delta \left(n-1\right)$. The other filter has impulse response $\delta \left(n\right)-\delta \left(n-1\right)$. ```rng default w = randn(100,1); x = filter([1 1],1,w); y = filter([1 -1],1,w);``` Obtain the sample cross-correlation sequence up to lag 20. Plot the sample cross-correlation along with the theoretical cross-correlation. ```[xc,lags] = xcorr(x,y,20,'biased'); Xc = zeros(size(xc)); Xc(20) = -1; Xc(22) = 1; stem(lags,xc,'filled') hold on stem(lags,Xc,'.','linewidth',2) q = legend('Sample cross-correlation','Theoretical cross-correlation'); q.Location = 'NorthWest'; q.FontSize = 9; q.Box = 'off';``` The theoretical cross-correlation is $-1$ at lag $-1$, $1$ at lag $1$, and zero at all other lags. The sample cross-correlation sequence approximates the theoretical cross-correlation. As expected, there is not perfect agreement between the theoretical cross-correlation and sample cross-correlation. The sample cross-correlation does accurately represent both the sign and magnitude of the theoretical cross-correlation sequence values at lag $-1$ and lag $1$.	{}
# zbMATH — the first resource for mathematics ##### Examples Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev \| Tschebyscheff The or-operator \| allows to search for Chebyshev or Tschebyscheff. "Quasi* map" py: 1989 The resulting documents have publication year 1989. so: Eur J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A \| 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used. ##### Operators a & b logic and a \| b logic or !ab logic not abc right wildcard "ab c" phrase (ab c) parentheses ##### Fields any anywhere an internal document identifier au author, editor ai internal author identifier ti title la language so source ab review, abstract py publication year rv reviewer cc MSC code ut uncontrolled term dt document type (j: journal article; b: book; a: book article) Bilinear systems and chaos. (English) Zbl 0823.93026 Motivated by the chaotic behavior of Lorenz equations, the authors presents a conjecture on chaos in a bilinear system in $\bbfR\sp 3$ of the form $\dot x= Ax+ Bxu$, $x\in \bbfR\sp 3$, $u\in \bbfR$. The main results of this article are two theorems on the existence of a pair of symmetric homoclinic orbits and on the chaotic behavior of a generalized Lorenz equation (GLE). The computer simulation shows that the GLE has a chaotic behavior similar to the Lorenz equation although there is some difference between the theoretical analysis and the numerical simulation. ##### MSC: 93C15 Control systems governed by ODE 37D45 Strange attractors, chaotic dynamics Full Text: ##### References: [1] F. Alberting, E. D. Sontag: Some connections between chaotic dynamical systems and control systems. Proceedings of First European Control Conference, Grenoble, 1991, pp. 159-163. [2] S. Čelikovský, A. Vaněček: Bilinear systems as the strongly nonlinear systems. Systems Structure and Control (V. Strejc, Pergamon Press, Oxford 1992, pp. 264-267. [3] S. Čelikovský: On the stabilization of homogeneous bilinear systems. Systems and Control Lett. 21 (1993), 6. · Zbl 0794.93089 · doi:10.1016/0167-6911(93)90055-B [4] W. J. Freeman: Strange attractors that govern mammalian brain dynamics shown by trajectories of EEG potential. IEEE Trans. Circuits and Systems 35 (1988), 791-783. [5] R. Genesio, A. Tesi: Chaos prediction in nonlinear feedback systems. IEE Proc. D 138 (1991), 313-320. · Zbl 0754.93024 · doi:10.1049/ip-d.1991.0042 [6] R. Genesio, A. Tesi: Harmonic balance methods for the analysis of chaotic dynamics in nonlinear systems. Automatica 28 (1992), 531-548. · Zbl 0765.93030 · doi:10.1016/0005-1098(92)90177-H [7] R. Genesio, A. Tesi: A harmonic balance approach for chaos prediction: the Chua’s circuit. Internat. J. of Bifurcation and Chaos 2 (1992), 61-79. · Zbl 0874.94042 · doi:10.1142/S0218127492000070 [8] A. L. Goldberger: Nonlinear dynamics, fractals, cardiac physiology and sudden death. Temporal Disorder in Human Oscillatory Systems, Springer-Verlag, Berlin 1987. [9] J. Guckenheimer, P. Holmes: Nonlinear Oscillations, Dynamical Systems and Bifurcations of Vector Fields. Springer-Verlag, New York 1986. [10] A. V. Holden (ed.): Chaos. Manchester Univ., Manchester 1986. · Zbl 0743.58005 [11] H. Hyotyniemi: Postponing chaos using a robust stabilizer. Preprints of First IFAC Symp. Design Methods of Control Systems, Pergamon Press, Oxford 1991, pp. 568-572. [12] L. O. Chua (ed.): Special Issue on Chaotic Systems. IEEE Proc. 6 (1987), 8, 75. [13] E. N. Lorenz: Deterministic non-periodic flow. J. Atmospheric Sci. 20 (1965), 130-141. [14] G. B. Di Massi, A. Gombani: On observability of chaotic systems: an example. Realization and Modelling in Systems Theory. Proc. International Symposium MTNS-89, Vol. II, Birkhäuser, Boston -- Basel -- Berlin 1990, pp. 489-496. · Zbl 0733.93009 [15] R. R. Mohler: Bilinear Control Processes. Academic Press, New York 1973. · Zbl 0343.93001 [16] J. M. Ottino: The mixing of fluids. Scientific Amer. 1989, 56-67. [17] C. T. Sparrow: The Lorenz Equations: Bifurcation, Chaos and Strange Attractors. Springer-Verlag, New York 1982. · Zbl 0504.58001 [18] A. Vaněček: Strongly nonlinear and other control systems. Problems Control Inform. Theory 20 (1991), 3-12. · Zbl 0747.93033 [19] A. Vaněček, S. Čelikovský: Chaos synthesis via root locus. IEEE Trans. Circuits and Systems 41 (1994), 1, 54-60. · Zbl 0843.93024 · doi:10.1109/81.260222 [20] A. Vaněček, S. Čelikovský: Synthesis of chaotic systems. Kybernetika, accepted. [21] S. Wiggins: Global Bifurcations and Chaos. Analytical Methods. Springer-Verlag, New York 1988. · Zbl 0661.58001	{}
# Rotation of a spool 1. Dec 13, 2016 1. The problem statement, all variables and given/known data Spool made of two outer disks of radius R2 which are solid, uniform cylinders, which are connected by a massless inner cylinder of radius . 2. Relevant equations 3. The attempt at a solution I just have a quick question. If the mass is going down, the tension in the string would make the spool rotate counterclockwise correct? Does that mean that the spool is rotating in a counterclockwise motion while is going to the right? Thank you 2. Dec 13, 2016 One suggestion is to work the problem with the origin at the place where the spool makes contact with the table. Then there is no torque from the frictional force. One thing you might assume is that there is sufficient friction to prevent sliding. Otherwise, it could spin the other way. Additional info could be supplied, such as the mass of the spool and the coefficient of friction, etc. 3. Dec 13, 2016 Right. I'm just a little confused in why the torque from the tension force is opposite to the rotation of the spool. 4. Dec 13, 2016 With sufficient friction I think you will find it rotates clockwise. The problem is non-trivial though, and I would need to write out the equations to see what happens. In the case of slipping, the amount of slipping could vary considerably, depending on the frictional force. With sufficient frictional force and/or no slipping, I believe it rotates clockwise. The rolling motion can also be considered to be an instantaneous rotation about the point of contact with the table if no slipping is occurring. $\\$ In the case of slipping, I think you could compute the torques about the center of mass to see which way it rotates. If the frictional force is small or zero, clearly it rotates counterclockwise. 5. Dec 13, 2016 Thanks for your replies. Yes, I'm assuming it's rolling without slipping. I get that the spool is rotated clockwise, but when we compute the torque for tension, why is it making it rotate counterclockwise? Is it because of the way that a spool works? 6. Dec 13, 2016 Please read the addition to my last post. Additional item is without slipping, I think you will find the torque when computed about the center of mass makes it rotate the other way because the frictional force is in the other direction and R2 is larger than R1 so it adds more clockwise torque than the inner string's counterclockwise torque. $\\$ There are two equations to write for the motion: The sum (or difference) of the two forces equals "ma" and the torque equation for the rotating motion. Without any slipping, these two equations are connected by the additional equation that the rotating motion equals the distance traveled. In any case, I think the string length will also enter into the solution to see how hard the hanging mass tugs depending on the acceleration of the mass, etc. Last edited: Dec 13, 2016 7. Dec 13, 2016 Oh I get it now. Thanks! 8. Dec 13, 2016	{}
# TextPad - Plain text processor. 1. Jan 7, 2005 ### Staff: Mentor TextPad® 4.7.3 is a powerful, general purpose editor for plain text files. A 'free' trial copy is available, but it is inexpensive, and quite powerful, and it is available in several languages. I have used a paid copy for over 7 years. It handles text files over 100 MB without choking. One copy blocks of text or single columns, sort, create macros. Saves files in PC, UNIX, Mac formats. 2. Jan 8, 2005 ### poolwin2001 I use it to compile and run my Java programs.TextPad Rocks! 3. Jan 8, 2005 ### dduardo Staff Emeritus I've gotten used to vim that I don't like to use anything else. 4. Jan 10, 2005 ### ramollari It's good that TextPad demo version is fully functional except that it reminds you once in a while to purchase it. 5. Jan 10, 2005 ### gnome For a good free Windows editor that doesn't nag you for \$ try Notepad2. 6. Jan 11, 2005 ### franznietzsche I love note pad. People in my class don't understand why. I jsut tell them it isn't loaded with the extraneous crap that word has. Word is useful for some things, but not for writing code. But a context sensitive notepad? Greatness. 7. Jan 13, 2005 ### ramollari Word is out of question. As soon as you start writing code it will start auto-corrections and underlining of errors. The simpler the better. Personally I hate tools overloaded with features. They are slow, difficult to use, and may crash. Notepad would do for me to write any kind of source file. 8. Jan 18, 2005 ### poolwin2001 You can allways disable the features you dont like. Last edited: Jan 18, 2005	{}
# Why does solvent leveling occur? Solvent leveling means that strong acids completely dissociate in that solvent so it is impossible to distinguish between two acids that completely dissociate in it. A solvent levels and acid if the $K_\text{a}$ of that acid is greater than 1. However, that’s not completely dissociated. An acid can have a $K_\text{a}$ of 2, 3 or 25 what’s the problem with being able to distinguish between them? In truth, whatever the $K_\text{a}$ it’s not actually COMPLETELY dissociated. in truth, whatever the $K_a$ it's not actually COMPLETELY dissociated For example, nitric acid has a $K_a$ of 24, so 1M nitric acid about 96% dissociated. If the concentration is higher, it will be less dissociated and if the concentration is lower it will be more dissociated.	{}
Missing digit Discussion in 'Math' started by Mark44, Jun 30, 2008. 1. Mark44 Thread Starter Well-Known Member Nov 26, 2007 626 1 The number 2$^{29}$ is interesting for the reason that it has nine digits in its base-10 form, and nine of the ten decimal digits are present. Without calculating 2$^{29}$, what's the missing digit? The problem is trivial if you calculate the number, using a calculator or computer or by longhand multiplication. Mark 2. thingmaker3 Retired Moderator May 16, 2005 5,073 8 Without calculating? I can hear my father yelling at me now from 2000 miles away for making a guess instead of calculating! Okay. I always was a disobedient child. I'll guess "2." Somehow seems like the most "interesting" number to be missing. I'll calculate after posting, to know if my guess is correct or not. 3. thingmaker3 Retired Moderator May 16, 2005 5,073 8 Wrong. That's what I get for guessing. 4. Mark44 Thread Starter Well-Known Member Nov 26, 2007 626 1 Here's something that might serve as a hint: 2$^{6}$ = 64 $\equiv$ 1 mod x 2$^{30}$ = (2$^{6}$)$^{5}$ $\equiv$ (1$^{5}$) mod x = 1 I'm doing modular arithmetic here, but I have not stated which modulo class I'm using. This is, after all, only a hint. 5. Ratch New Member Mar 20, 2007 1,068 4 To the Ineffable All, Taking the hint, and using the "casting out of 9's" http://mathforum.org/library/drmath/view/55831.html . 2^29 = ((2^6)^4)2^5=64646464*32 Applying the casting out of 9's by multiplying the sum of the digits of the multiplicand and multipliers, and summing the product digits. (6+4)(6+4)(6+4)(6+4)(3+2)=50000==5 Therefore the sum of the product digits of 2^29 must be 5. The only set of digits where this happens is: 1+2+3+0+5+6+7+8+9 = 41 ==5 Therefore, 4 is the missing digit of 2^29. Ratch 6. Mark44 Thread Starter Well-Known Member Nov 26, 2007 626 1 Right you are! 7. Dave Retired Moderator Nov 17, 2003 6,960 170 Interestingly when I read this puzzle the first thing that jumped to my mind was the cast-out of 9s method - not because I had an idea of how this worked, but because of a puzzle jpanhalt posted a while back which employed this method. Something stuck in there from that. Good catch Ratch! Dave	{}
## easy and uneasy riddles Posted in Books, Kids, R with tags , , , , , on February 2, 2021 by xi'an On 15 January, The Riddler had both a straightforward and a challenging riddles. The first one was to optimise the choice of a real number d with the utility function U(d,θ)=d ℑ(θ>d), when θ is Uniform(0,100). Leading unsurprisingly to d=50… The tough(er) one was to solve a form of sudoku where the 24 entries of a 8×3 table are integers in {1,…,9} and the information is provided by the row-wise and column-wise products of these integers. The vertical margins are 294, 216, 135, 98, 112, 84, 245, 40 and the horizontal margins are 8 890 560, 156 800, 55 566 After an unsuccessful brute-force (and pseudo-annealed) attempt achieving a minimum error of 127, although using the prime factor decompositions of these 11 margins, I realised that some entries were known: e.g., 7 at (1,2), 5 at (3,2), and 7 at (7,3), and (much later) that the (huge) product value for the first column implied that each term in that column had to be the maximal possible value for the corresponding rows, except for 5 on row 7. This leads to the starting grid [,1] [,2] [,3] [1,] 7 7 6 [2,] 9 0 0 [3,] 9 5 3 [4,] 7 0 0 [5,] 8 0 0 [6,] 7 0 0 [7,] 5 7 7 [8,] 8 0 0 and an additional and obvious exclusion based on the absence of 3’s in the second column, of 5’s and 2’s in the third column shows there was a unique solution [,1] [,2] [,3] [1,] 7 7 6 [2,] 9 8 3 [3,] 9 5 3 [4,] 7 2 7 [5,] 8 2 7 [6,] 7 4 3 [7,] 5 7 7 [8,] 8 5 1 as also demonstrated by a complete exploration with R: Try it online! ## Fermat’s Riddle Posted in Books, Kids, R with tags , , , , , , , , , , on October 16, 2020 by xi'an ·A Fermat-like riddle from the Riddler (with enough room to code on the margin) An arbitrary positive integer N is to be written as a difference of two distinct positive integers. What are the impossible cases and else can you provide a list of all distinct representations? Since the problem amounts to finding a>b>0 such that $N=a^2-b^2=(a-b)(a+b)$ both (a+b) and (a-b) are products of some of the prime factors in the decomposition of N and both terms must have the same parity for the average a to be an integer. This eliminates decompositions with a single prime factor 2 (and N=1). For other cases, the following R code (which I could not deposit on tio.run because of the packages R.utils!) returns a list library(R.utils) library(numbers) bitz<-function(i,m) #int2bits c(rev(as.binary(i)),rep(0,m))[1:m] ridl=function(n){ a=primeFactors(n) if((n==1)\|(sum(a==2)==1)){ print("impossible")}else{ m=length(a);g=NULL for(i in 1:2^m){ b=bitz(i,m) if(((d<-prod(a[!!b]))%%2==(e<-prod(a[!b]))%%2)&(d<e)) g=rbind(g,c(k<-(e+d)/2,l<-(e-d)/2))} return(g[!duplicated(g[,1]-g[,2]),])}} For instance, > ridl(1456) [,1] [,2] [1,] 365 363 [2,] 184 180 [3,] 95 87 [4,] 59 45 [5,] 40 12 [6,] 41 15 Checking for the most prolific N, up to 10⁶, I found that N=6720=2⁶·3·5·7 produces 20 different decompositions. And that N=887,040=2⁸·3²·5·7·11 leads to 84 distinct differences of squares. ## riddle of the seats Posted in Statistics with tags , , , on November 8, 2019 by xi'an An arithmetic quick riddle from The Riddler: If an integer n is a multiple of every integer between 1 and 200, except for two consecutive ones, find those consecutive integers. Since the highest power of 2 less than 200 is 2⁷=128 and since 127 is a prime number, the number $2^6\times \prod_{i=0,i\ne 63}^{99} (2i+1)$ should work in that it contains all odd integers but 127, and all even numbers, but 128. Of course a smaller number that avoids duplicates by only considering the 44 primes other than 127 and 2 to a power that keep them less than 200 is also valid. Which gives a number of the order of 1.037443 10⁸⁵. ## missing digit in a 114 digit number [a Riddler’s riddle] Posted in R, Running, Statistics with tags , , , , , , , on January 31, 2019 by xi'an A puzzling riddle from The Riddler (as Le Monde had a painful geometry riddle this week): this number with 114 digits 530,131,801,762,787,739,802,889,792,754,109,70?,139,358,547,710,066,257,652,050,346,294,484,433,323,974,747,960,297,803,292,989,236,183,040,000,000,000 is missing one digit and is a product of some of the integers between 2 and 99. By comparison, 76! and 77! have 112 and 114 digits, respectively. While 99! has 156 digits. Using WolframAlpha on-line prime factor decomposition code, I found that only 6 is a possible solution, as any other integer between 0 and 9 included a large prime number in its prime decomposition: However, I thought anew about it when swimming the next early morning [my current substitute to morning runs] and reasoned that it was not necessary to call a formal calculator as it is reasonably easy to check that this humongous number has to be divisible by 9=3×3 (for else there are not enough terms left to reach 114 digits, checked by lfactorial()… More precisely, 3³³x33! has 53 digits and 99!/3³³x33! 104 digits, less than 114), which means the sum of all digits is divisible by 9, which leads to 6 as the unique solution. ## Le Monde puzzle [#1063] Posted in Books, Kids, R with tags , , , , , , on August 9, 2018 by xi'an A simple (summertime?!) arithmetic Le Monde mathematical puzzle 1. A “powerful integer” is such that all its prime divisors are at least with multiplicity 2. Are there two powerful integers in a row, i.e. such that both n and n+1 are powerful? 2. Are there odd integers n such that n² – 1 is a powerful integer ? The first question can be solved by brute force. Here is a R code that leads to the solution: isperfz <- function(n){ divz=primeFactors(n) facz=unique(divz) ordz=rep(0,length(facz)) for (i in 1:length(facz)) ordz[i]=sum(divz==facz[i]) return(min(ordz)>1)} lesperf=NULL for (t in 4:1e5) if (isperfz(t)) lesperf=c(lesperf,t) twinz=lesperf[diff(lesperf)==1] with solutions 8, 288, 675, 9800, 12167. The second puzzle means rerunning the code only on integers n²-1… [1] 8 [1] 288 [1] 675 [1] 9800 [1] 235224 [1] 332928 [1] 1825200 [1] 11309768 except that I cannot exceed n²=10⁸. (The Le Monde puzzles will now stop for a month, just like about everything in France!, and then a new challenge will take place. Stay tuned.)	{}
A simply supprted rectangular concrete beam of span $8$ m has to be prestressed with a force of $1600$ kN. The tendon is of parabolic profile having zero eccentricity at the supports. The beam has to carry an external uniformly distributed load of intensity $30$ kN/m. Neglecting the self-weight of the beam, the maximum dip (in meters, up to two decimal places) of the tendon at the mid-span to balance the external load should be __________	{}
Hence 1 is the least number to be subtracted from 3250 to get a perfect square. Find the number to add to x^2-18x to make it a perfect square trinomial. Find the smallest number by which 882 must be multiplied so that the product becomes a perfect square. Find the least no. 1. What is the least number which should be added to 0.0282 to make it a perfect square? Get a calculator. Since, (74) 2 < 5607. method to find a least positive number that should be added to 1515 to become a perfect square 1. The remainder is 41. Larger number is $$\Large (5)^{2}$$ less than square of 25. Click hereto get an answer to your question ️ Find the least number which must be added to 4931 to make it a perfect square ? 0 1. The smallest number to be added to 680621 to make the sum a perfect square is = ? 3). Find the Least Number Which Must Be Added to 5483 So that the Resulting Number is a Perfect Square. Inorder to convert the given number as the square of 63, we have to subtract 31. Year 10 Interactive Maths - Second Edition. Is the number 2048 a perfect square? Ex 6.4, 5 Find the least number which must be added to each of the following numbers so as to get a perfect square. 3. Please provide the shortcut method also. 5. Adding -1 to 785 creates a perfect square. $$\Large \frac{1}{3+\frac{1}{1+\frac{1}{16}}}$$, B). Also find the square root of the perfect square so obtained. Find the least number which must be added to 60509 to make it a perfect square - 3182341 Please provide a valid phone number. 7). 7. How about 7 and 7; 7 + 7 = 14 and 7 * 7 = 49 . (11) (5) Submit Your Solution Number System ... Twice the square of a certain whole number added to 3times the make 90 find the number . a.0.0007 b.0.0042 c.0.0002 d.0.0003 Asked In IBM Anilkumar (5 years ago) Unsolved Read Solution (5) Is this Puzzle helpful? 2. Update: Please note ans is 56. 306452 + 464 = 554^2 Now, i.e. Math. Pauley Morph. a.0.0007 b.0.0042 c.0.0002 d.0.0003 Asked In IBM Anilkumar (5 years ago) Unsolved Read Solution (5) Is this Puzzle helpful? 337 0. Concept: Square Roots - Finding Square Root by Division Method. Find the least number by which 1470 must be divided to get a number which is the perfect square? Square 39. The smallest number to be added to 680621 to make the sum a perfect square is = ? Given a number N, Find the minimum number that needs to be added to or subtracted from N, to make it a perfect cube. Write that trinomial as the square of a binomial. 2. Correct Option: (c) Step 1: Find the square root of 1780 1780 = 42.190 Take the square of next number after 42 i.e. The smallest positive integer when multiplied by 392, the product is a perfect square, is. Year 10 Interactive Maths - Second Edition. From this we come to know that the square root of the given number (4000) is greater than 632. Find the smallest number by which 3645 must be multiplied to get a perfect square. Lv 7. ∴ Perfect square = 402 − 2 Perfect square = 400 Also, If we do long division with 400 We get 20 as square root ∴ Square root of 400 = 20 Ex 6.4, 4 Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. I have solved but with help of calculator. A perfect square is the square of an integer, i.e., an integer multiplied by itself. Find the smallest number by which 5103 must be divided to get a perfect square. If we do that, we have 2^4 * 7^2, which is 784. Answer Save. List of Perfect Cube Numbers 1 to 50 7 Answers. C. 214434. Embarrassingly, I don't know it, and I'm doing maths in Uni starting this year :L but my brains been shut off for about 3 months. The sum of the numbers is: 6). Lv 7. Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here. The least number which should be added to 1330 to make it a perfect square. The square root of 1000 can be calculated by long division method as follows. Therefore 21 should be added to 7900 to get a perfect square. You will get 38.92300091 4. (i) 525Rough41 × 1 = 4142 × 2 = 8443 × 3 = 129Here, Remainder = 41Since remainder is not 0, So, 525 3. Q1 find the least number which must be added to 7900 to obtain a perfect square? Battleaxe. Solution: Note: Note: 15 answers The graph of a … What number should be added to make the following a perfect square? Find the number to add to x^2+14x to make it a perfect square trinomial. We can identify it by doing prime factorisation method. Find the smallest number by which 882 must be multiplied so that the product becomes a perfect square. Larger number is $$\Large (5)^{2}$$ less than square of 25. 21. true. ... 841 - 785 = 56 is least number to be added. Battleaxe. Find the least number, which must be added to 1825 to make it a perfect square. $$\Large \frac{13}{48}$$ is equal to, A). Find the least perfect square number divisible by 3,4,5,6 & 8? Kaydell. Square 39. 78. Therefore, 0.000002 should be subtracted from 0.000326 to make it a perfect square of 0.018. Example 2 : Find the least number, which must be subtracted from 3250 to make it a perfect square. Also find the square root of the perfect square so … It represents that the square of 22 is less than 525. From this we come to know that the square root of the given number (3250) is greater than 572. ... 841 - 785 = 56 is least number to be added. The least number which should be added to 1330 to make it a perfect square. Find the least number by which 1470 must be divided to get a number which is the perfect square? 98 should be added to make it a perfect square when we will add it the number becomes 5776 which is the square of 76. i found this answer with the help of long divison method. 0 1. Solution: (i) 5607. Find the least number that must be added to {eq}1300 {/eq} so as to get a perfect square. are solved by group of students and teacher of Railways, which is … The smallest number added to 680621 to make the sum a perfect square is: 1). Given a number N, find the minimum number that needs to be added to or subtracted from N, to make it a perfect square.If the number is to be added, print it with a + sign, else if the number is to be subtracted, print it with a – sign. Which smallest number must be added to 2203 so that we get a perfect square? What is the value of the sum of twice of 24 percent of the smaller number and half of the larger number? 1. Next number is 23 and 23 2 = 529 Hence, number to be added to 525 = 23 2 − 525 = 529 − 525 = 4 The required perfect square is 529 and √ 529 = 23 We can do it as follows: Take any number say it be 248. If x is a perfect cube of y, then x = y 3.Therefore, if we take out the cube root of a perfect cube, we get a natural number and not a fraction. Hence 39 must be added to 1330 to make it a perfect square. Forming a Perfect Square In general: Example 11. The least perfect square, which is divisible by each of 21, 36 and 66 is: A. Pauley Morph. The remainder is 156. 213444. 7 Answers. Kaydell. Examples: Hence, 3 √x = y. Solution : From this we come to know that the square root of the given number (1750) lies between 41 and 42. Forming a Perfect Square In general: Example 11. Find the least number, which must be subtracted from 3250 to make it a perfect square. The next perfect square number is . What are three numbers that have a sum of 35 if the greatest number is 14 more than the least number A)6,7,20(B)5,11,19(C)10,11,24(D)1,15,15? For this, we use the method called long division. Please provide the shortcut method also. Sum of two numbers is equal to sum of square of 11 and cube of 9. Part of solved Simplification questions and answers : … Answer this multiple choice objective question and get explanation and result.It is provided by OnlineTyari in English Skip navigation Sign in. 4. Relevance. If the sum and difference of two numbers are 20 and 8 respectively, then the difference of their squares is: 5). Find the square root of 1515. A perfect cube is a number which is equal to the number, multiplied by itself, three times. Finding square root of 5607 by long division Here, Remainder = 131 Since remainder is not 0, So, 5607 is not a perfect square Rough 143 × 3 = 429 144 × 4 = 576 144 × 4 = 725 We need to find the least number that must be subtracted from 5607 so as to get a perfect square Thus, we subtract 131 (remainder) from 5607 to get a perfect square. Find the square root of 1515. You will get 1521, which is a perfect square, since it is 39? Examples: Input: N = 25 Output: 2 Nearest perfect cube before 25 = 8 Nearest perfect cube after 25 = 27 79 2 = 6241 Number that should be added to the given number to make it a perfect square = 6241 - 6203 = 38 2002 , Square And Cube Roots questions, Aptitude questions, india 3. Examples: Input: N = 25 Output: 2 Nearest perfect cube before 25 = 8 Nearest perfect cube after 25 = 27 Therefore, The smallest number added to 680621 to make a perfect square is 4. In the above picture, 16 is subtracted from 19 and we got the remainder 3. Now we have to multiply a number by itself such that the product â¤ 4, (The product must be greatest and also less than 19). By which least number 675 must be multiply to obtain a number which is perfect cube? What is the value of the sum of twice of 24 percent of the smaller number and half of the larger number? So we get 8. Find the number.? Part of solved Simplification questions and answers : … 205 4). TCYonline Question & Answers: get answer of What is the least number to be added to 4321 to make it perfect square For full functionality of this site it is necessary to enable JavaScript. Find the least no. Question: Find the least number, which must be added to {eq}1825 {/eq} to make it a perfect square. View Answer For each of the following, find the least number that must be added so that the resulting number is a perfect square. 3. $$\Large \frac{1}{3+\frac{1}{1+\frac{1}{2+\frac{1}{4}}}}$$, 8). 35. false. If we subtract 53 from 1989, we get 1936 which is the perfect square. Step 2: Find the least number Subtract given number from 43 2 43 2 – 1780 = Number to be added 1849 – 1780 = 69 Hence, the number to be added is 69. As we know square of 825 is 680625. Perfect Square Calculator Enter any Number into this free calculator Our calculator will tell you whether or not any number is a perfect square as well as why that number is a perfect square . What is the least number to be added to 1500 to make it a perfect square? Thus, to get a perfect square greater than the given number, we take the square of the next natural number of the quotient, i.e. /questions/what-is-the-least-number-to-be-added-to-1500-to-make-it-a-perfect-square. ... Class 8 chapter 1 finding least number of 6 digits which is a perfect square - Duration: 5:32. Solution : $$\Large \frac{1}{2+\frac{1}{1+\frac{1}{8}}}$$, C). In a park there are 1225 trees. 5d1ed93f70b1431921736a7e. 213444. What number should be added to make the following a perfect square? The sum of the squares of two positive integers is 100 and the difference of their squares is 28. The numbers 4, 9, 16, and 25 are a few examples of perfect squares. $$\Large \frac{1}{3+\frac{1}{1+\frac{1}{1+\frac{1}{8}}}}$$, D). 28^2 = 784. What is the least number that you must multiply 360 by to make a perfect square?This isn't for me, it's for my sister. Given a number N, find the minimum number that needs to be added to or subtracted from N, to make it a perfect square.If the number is to be added, print it with a + sign, else if the number is to be subtracted, print it with a – sign. Update: Please note ans is 56. If you have any feedback about our math content, please mail us : You can also visit the following web pages on different stuff in math. This topic introduces forming a perfect square for an algebraic expression. 7. Also find the square root of the perfect square so obtained. Inorder to convert the given number as the square of 57, we have to subtract 1. 5. 28^2 = 784. Separate the digits by taking commas from right to left once in two digits. 9805 ,99856 and least is 1024.are these correct? You will get 38.92300091 4. (i) The square root of 525 can be calculated by long division method as follows. Take the next higher whole number to that, which is 39 5. You will get 1521, which is a perfect square, since it is 39? Relevance. Here is how you can enable JavaScript. Solution: Note: Note: If we subtract 53 from 1989, we get 1936 which is the perfect square. ... Now, what two numbers added together equal 14 that would provide a perfect square for the third term? Here 119 is remainder that means 4215 is not a perfect square. Now, we have to bring down 89 and quotient 4 to be multiplied by 2. So we need to add 554^2 - 306452 = 464 to it because then. 0 1. 0 1. 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The Questions and Answers of What is the least number which must be added to 306452 to make it a perfect square ?a)460b)462c)464d)466Correct answer is option 'C'. 5d1ed93f70b1431921736a7e. method to find a least positive number that should be added to 1515 to become a perfect square 1. Let us see an example to understand the concept. 2. C. 214434. Given a number N, Find the minimum number that needs to be added to or subtracted from N, to make it a perfect cube. If not, then by which number it must be multiplied so as to get a perfect square. Square Roots and Cube Roots Questions & Answers for Bank Exams : The least number to be added to 435 to make it a perfect square is 326 – 2 = 324 Which is a perfect square of 18. Find the least number, which must be subtracted from 4000 to make it a perfect square. What is the least number to be added to 1500 to make it a perfect square? 2). Find the smallest number by which 5103 must be divided to get a perfect square. Rosa, Roberto, Andrea, and Inno find an estimate for square root 10. Who has proposed the best solution? 4. 305 2). by which 294 must be multiplied to make it a perfect square? How to find the least number to be subtracted to get a perfect square : Here we are going to see how to find the least number to be the given number to get a perfect square. 5. Find the least number which must be added to each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained. Question from Class 8,olympiad,class8,math,ch-4,squares-and-squareroots 2. Also find the square root of the perfect square so obtained. Find the greatest 3 digit and 4 digit perfect square. Hence 39 must be added to 1330 to make it a perfect square. 405 The least number that should be subtracted from the number 32146 to make it a perfect square is : Refer this free online list of perfect squares for first 100 numbers chart to make your calculations simple and save your time. After having gone through the stuff given above, we hope that the students would have understood "How to find the least number to be subtracted to get a perfect square", Apart from the stuff given above, if you want to know more about "How to find the least number to be subtracted to get a perfect square", please click here. 9 9 is a perfect square because it can be expressed as 3 * 3 (the product of two equal integers). Lv 7. 6 years ago. (i) 525Rough41 × 1 = 4142 × 2 = 8443 × 3 = 129Here, Remainder = 41Since remainder is not 0, So, 525 Find the Least Number Which Must Be Added to the Following Number So as to Get a Perfect Square. Roberto: "I will use square root 4 and square … If N N N is an integer, then N 2 N^2 N 2 is a perfect square. Also find least number four digits, which is perfect square. Therefore we need to add another factor of 7. Also Find the Square Root of the Perfect Square So Obtained 525 Concept: Finding Square … Find the least perfect square number divisible by 3,4,5,6 & 8? What least number must be subtracted/added from 1672 to obtain a number which is completely divisible by 17? 105 3). The least perfect square, which is divisible by each of 21, 36 and 66 is: A. Q: What is the least number to be added to 1500 to make it a perfect square? Examples: Asked In CAT MohanRaj (7 years ago) Unsolved Read Solution (4) Is … 2). Know answer of objective question : Which least number should be added to 2600 to make it a perfect square?. B. Find the least number that is multiplied to 1176 t0 make it perfect square. next perfect square is 65 2 =4225 hence we need to add 4225-4215 =10 Hence the least number to be added to 4215 to make it a perfect square … Take the next higher whole number to that, which is 39 5. For example, 25 is a perfect square, because 5 x 5 = 25. Can you explain this answer? If not, then by which number it must be multiplied so as to get a perfect square. In a park there are 1225 trees. Find the greatest number of four and five digits, which are perfect square. 337 0. That is, the number which should be added to 1000 to make it a perfect square has to be calculated. "Find the smallest number that must be added to 1780 to make it perfect square." The given number has 6 decimal places. Sum of two numbers is equal to sum of square of 11 and cube of 9. The number of perfect square numbers between 50 and 1000 is. 2). By using long division method. Now this situation is explained using long division. That is, multiply 39x39 6. The value of $$\Large 1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{2}{3}}}}$$, 9). Hence 53 is the least number to be subtracted from the given number (1989) to get a perfect square. Hence, the number to be added is . Solution: First let us find the square of 7900 using long division method. Is the number 2048 a perfect square? By which least number 675 must be multiply to obtain a number which is perfect cube? For example, 100 100 1 0 0 is a perfect square because it is equal to 10 × 10 10\times 10 1 0 × 1 0. ∴ 20 has to be subtracted from 4401624 to get a perfect square. Now if we prime factorise it we get 248 as 2 x 2 x 2 x 31 Hence 31 is the least number to be subtracted from 4000 to get a perfect square. 214344. ... Find the least number which must be added to each of the following numbers … What is the least number which should be added to 0.0282 to make it a perfect square? What are three numbers that have a sum of 35 if the greatest number is 14 more than the least number A)6,7,20(B)5,11,19(C)10,11,24(D)1,15,15? One half of the number added to two thirds of the number is 21. This topic introduces forming a perfect square for an algebraic expression. Therefore, The smallest number added to 680621 to make a perfect square is 4. 43. Because of this definition, perfect squares … Which smallest number must be added to 710 so that the sum is a perfect cube? The least number that should be subtracted from the number 32146 to make it a perfect square is : 1). 10). For example, 8 is a perfect cube because 3 √8 = 2. 16 16 is a perfect square because it can be expressed as 4 * … When we do so, we get 19 before the comma. 38 is the number which you should add to make the number perfect square. Adding -1 to 785 creates a perfect square. Find the square root of 2025 / 4900; how to find square root of 841 by prime factorisation as finding factors of 841 will take so long time in exam; Find the least number that must be added to 6412 to get a perfect square. 3. $$\Large \frac{ \left(3.63\right)^{2}- \left(2.37\right)^{2} }{3.63+2.37}$$ is equal to. asked Aug 8, 2018 in Mathematics by ajay kumar ( 15.0k points) squares and square roots Find the least number which must be subtracted from 1989 so as to get a perfect square. 2. Find the least number which must be added to each of the following numbers so as to get a perfect square. Answer Save. What is the least number that can be added to the number 1020 to make it a perfect square? Hence . The remainder is 131. 5. Find the least number which must be added to the following numbers to make them a perfect square: (i) 5607 (ii)4931 (iii) 4515600 (iv) 37460 (v) 506900. A perfect square is an integer that can be expressed as the product of two equal integers. Math. (11) (5) Submit Your Solution Number System /questions/what-is-the-least-number-to-be-added-to-1500-to-make-it-a-perfect-square. That is, multiply 39x39 6. But we also have only one factor of 7 (or 7^1). A perfect square is a number that can be expressed as the product of two equal integers. B. As we know square of 825 is 680625. The above condition will be met by â4â. 214344. What is least number to be added to 785 to make it a perfect square? by which 294 must be multiplied to make it a perfect square? What is least number to be added to 785 to make it a perfect square? Find the least 4 digit perfect square number. If the number is to be added, print it with a + sign, else if the number is to be subtracted, print it with a – sign. 6 years ago. Examples of perfect squares. If the number is to be added, print it with a + sign, else if the number is to be subtracted, print it with a – sign. Find the least number, which must be added to 1750 to make it a perfect square. In your example, we have four factors of 2 (or 2^4). Ex 6.4, 5 Find the least number which must be added to each of the following numbers so as to get a perfect square. In order to have a perfect square, you must have an EVEN number of each prime factor (or, if you like, each prime factor must have an even exponent). The smallest number added to 680621 to make the sum a perfect square is : S.S.C. Lv 7. 3. Find the square root of 3 9 0 6 2 5 using repeated subtraction. (1 point) Rosa: "Use square root 9 and square root 25 to estimate." The number, whose square is equal to the difference between the squares of 975 and 585, is: 4). 43. Also, find the square root of the perfect square. (i) 402 (ii) 1989 (iii) 3250 (iv) 825 (v) 4000 18. Get a calculator. Hence 53 is the least number to be subtracted from the given number (1989) to get a perfect square. 2020 least number to be added to make a perfect square	{}
Question # A paralle plate capacitor of capacitance $$90$$ pF is connected to a battery of emf $$20$$V. If a dielectric material of dielectric constant K$$=\displaystyle\frac{5}{3}$$ is inserted between the plates, the magnitude of the induced charge will be. A 2.4 n C B 0.9 n C C 1.2 n C D 0.3 n C Solution ## The correct option is B $$1.2$$ n CFor parallel plate capacitor$$Q=CV$$Initial charge $$Q_{i}=90pf\times 20V=1.8nC$$When dielectric material is inserted$$C=KC$$Final charge $$Q_{f}=(KC)V=\frac{5}{3}\times 90pF\times 20v$$=$$3nC$$Induced charge =$$Q{_f}-Q_{i}$$ $$=3-1.8=1.2nC$$PhysicsNCERTStandard XII Suggest Corrections 0 Similar questions View More Same exercise questions View More People also searched for View More	{}
# Tài liệu tiếng Anh ôn thi tuyển sinh cao học - Phần 3 Chia sẻ: Nguyen Hoang Phuong Uyen \| Ngày: \| Loại File: PDF \| Số trang:18 0 462 lượt xem 368 ## Tài liệu tiếng Anh ôn thi tuyển sinh cao học - Phần 3 Mô tả tài liệu Download Vui lòng tải xuống để xem tài liệu đầy đủ Tài liệu tham khảo cao học, tài liệu MBA - Tài liệu tiếng Anh ôn thi tuyển sinh cao học - Phần 3 Chủ đề: Bình luận(0) Lưu ## Nội dung Text: Tài liệu tiếng Anh ôn thi tuyển sinh cao học - Phần 3 1. Tui lifu rln thi titng Anh PRACTICE TEST 7 PART 1: VOCABULARY AND READING COMPREHENSION SECTION 1: VOCABULARY Choose the best word/phrase in the box below to complete the sentences that follow. Each word/phrase can be used ONCE ONLY. _m___ _m_m_ m__'m______ _. _m relationships pred iet used taxes majority rain'A!ater current population penSIon grad ual growth people more doctors most of which school university in many is becoming profits equipITIent we m_ , _" - ..are to be 1. One of the feat ures of London is the number of big stores, found in or near the West End. 2. . within the famil.v are different now. Parents treat their children as equals more than thc\ used to. 3. Our go\'crnmcnt gives financial help to the elderl.y in the form of ........................... 4. Thc.. of the cit\' helps explain the fact that 110 Chi Minh City docs not \1;-\\-(' just OIlC c('ntrc, .. . .. .. . ... . . of population go 5. In ('\Tn industrialized ('ountr\ in the \\'orId the vast ...... out to \\'ork to earn their li\'ing. (). I'u IJlic scn'iccs arc provided for the good of society as a \-vhole, the money to finance them is colkc1Cd IJ.\' the gmTrnment through ... , and mOfe medicines. I. 1\1;111\' countries need bettef health care, H. Doctors and nurses work through the WorId Health Organisation to prevent diseases, to ............. teach medica] people, clnd to provide medical supplies and .... q .. .. . .. .. . .. .. . education should be free and the state should pay 1he cust of students' accommodation, food, books. 10. The Internet, . .more and more popular and many people no\\! have access to 11 ,It horne. 111 2. Practice test 7 11. In the last 100 years, t('chnolog)' has completely changed the way we live.. ............. ..have electricity', aeroplanes, television, and we have even been to the moon. 12. Scientists.. that by 2050 the population might reach 10 billion. Feeding this number of people might prove to be difficult. 13. The world's .......................... .is around six billion. 14. Eco-friendly houses can collect.. for washing. 15. The computer has become part of life in the last ten or fifteen years. It is.... .in business, science, education, and entertainment. SECTION 2: Reading Comprehension (30 marks) Read each passage carefully and then answer the questions that follow. Passage 1 Every British citizen who is employed (or self-employed) is obliged to pay a weekly contribution to the national insurance and health schemes. An employer also makes a contribution for each of his employees, and the Government too pays a certain amount. This plan was brought into being in 1948. Its aim is to prevent anyone from going \vithout medical services, if he needs them, however poor he may be; to ensure that a person who is out of \vork shall receive a weekI.v sum of money to subsist on; and to provide a small pension for those who have reached the age of retirement. I-:\'er~'onc can register with a doctor of his choice and if he is iJ1 he can consult the doctor without having to pay for the doctor's services, although he has to pay a small charge for medicines. When a man is out of work, he may draw unemployment benefit until he finds \\'ork again; this he \\'ill probabJy do by going to a clob Center. Obviously, the amount paid is comparatively small, for the State docs not \\"ant people to stop working in order to draw a handsome sum of money for doing nothing. There are special benefits for certain people, such as mothers-tn-be, children, the blind and the handicapped. The amount of mone,\" needed to operate these schemes is enormous and a large part of the mone.y comes not from the contributions but from taxation. I. What t\\'o schemes or \\"c]fa re docs the passage telJ us about? ............ 2. Who arc the contributors to these schemes? 3-4-5. What are the three fJurposcs of the schemes? 6. Can poor pcople gel frce doc1or's service? "". .................... ~I - - - - ----- 3. -- - - - - Tili li~u /in thi tie'ng Anh 7. Who can enjoy special benefits? ....................... .................................................................................... 8. What docs "them" in "if he needs them' refer to? (paragraph 1) ...................................................................................................................... 9. What docs "this" in "this he will probably do by going to a Job Center' refer to? (paragraph 2) ....................... ............................................... 10. What word in the passage means "an amount of money paid regularl.', 10 someone \vho is considered to be too old to earn money by \vorking'? ........... ................................. Passage 2 The sun releases large amounts of energy. Most of the energy goes into space, but some of it reaches the earth. There arc several kinds of energy from the SUTl, hut these arc the two most important: light and heat. Life qn earth depends on light for p1ants to grO\\'; and humans and animals need warmth to live. Solar energy is not onl.\' unlimited but also i1 is clean. It docs not pollute air or water. Wh\ doesn't the world use more solar power? It is still ver.'-' expensive to collect solar energy for use. Scientists around the world arc tr\ing to find a cheaper way to collect it and change it into eJectricity for machines. For example, automobile engineers arc tty'ing to develop electronic cars that use solar PO\\Tr. Many buildings arc using solar energv to heat their water nO\\'. to build their O\\'n Universities and colleges give solar energy courses where students learn of solar collectors. SOh1.r collectors. There arc also many books that explain the construction especially in the sllnny desert arcas of the worJd. Scientists afe looking fOf other ways to use solar enerK\. For instance, they want to use solar cnergy to take the salt out of sea water. Pcople \\'ho livc in dry, dcsert areas near oceans need fresh water. Scicntlsts want to supply this fresh water I"rmn the ocean. I ................. ..................... ................................. 15- 16. Name t\\'o of the uses'of solar energy mentioned in the passage? .. . .. .. . .. . .. .. .. . .. .. .. . .. . . ..p ................. 17. What is the process of taking salt out of sea water callcd? ........... 42 4. Practice test 7 18. What docs "if' in "but also it is dean" refer to? (paragraph I) .. .. .. .. . .. .. . .. . .. . .. . .. . .. . . ......................... .............. 19. What doc", "tlIat";n "...to devdop electronic cars that Use solar power" refer to') (Paragraph2) ...................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. ............ 20. What expression/phrase in the passage is similar in meaning to "solar cncrgy"#? . . . . . . .. . . .. .............................................. SECTION 3: READING CLOZE (10 marks) Instructions: Complete the following passage with the missing words. Fill in each blank with ONE suitable word only, Mexico City is the oldest and (1).............. ..historical capital city in North America. It's c:m exciting, beautiful, and magical (2)..... .......... ...which for the past few years has becn fighting against serious environmental problems. It has a (3) ..of O\'(T 22 million people (4)........... ............. .it's getting bigger all the time. (5).. IS morc them three times bigger (6) ..... ...New ............ York City and it's the lal'gest eit\ (7). world. Every day, thousands of people mm/c in from the country to (0'. .the work, (9) t'or thcre aren't enough jobs for al1 of them, and the rcsuJrs IS extrcll1ch (10) unemployment and over-population. PART 2: WRITING SECTION 1: CONTROLLED WRITING (15 marks) Instructions: From the following sets of words and phrases in the given sequences, make aU the changes and additions necessary to produce sentences, which together make a complete letter. Note carefully from the example what kind of changes need to be made. Write each sentence in the space provided. Example: ] / be / happy / get / letter / offer / me / job / vour company Answer: / was happy when [ got your letter, in which you alTer !!Ie (f joh ill your COlllIHlI/Y. Dear Karen 1. Thank / you / your letter. ........................................... ................. 2. I / really / look forward you. / see / you / summer / there / be / a kw thmgs / i like / a~~ / . .. .. .. .. . .. .. . .. .. . .. . .. . .. . .. . 3. What / be / weather / usually / like / summn / /Jr;tam? J';"'I' ",' .............................. 4..1 5. --.---- Tili Ii?" tJn (hi (ie'n Anh 4. 1 I not know I what I pack I I I not have I very big suitcase. ................ I I airport? 5. Can I come I meet me .......................................................................................................... I you I can I pick up I there. 6. If not I just let I me I know I how I get I station ...................................................................................................... how much I money I I I bring? more I question I . 7. 1 I have lone ............ .. .. . .. .. . .. . .. ................................................... 8. Is I easy I change I traveller's cheque? . .. .. . .. .. . .. . .. .. . .. . .. . .. . .. . .. . .. . .. .. . .. .. .. . .. .. . .. .. .. . .. .. .. . .. .. .. . .. . .. .. .. . .. . .. . .. .. . .. .. . 9. or I would I be I better I bring I cash? ........... ............ .......................................................................... 10. Please I write I soon .......................................................................................... Lots of love Debbie SECTION 2: SENTENCE TRANSFORMATION (15 marks) Instructions: use the words given to rewrite each of the following sentences in such a way that it means exactly the same as the original sentence. Example: "Don't be so lazy!" Answer: Darren told him not to be lazy. 1. Don't go too near the edge," their father said. " ........... ............... Their father warned. ............................................ 2. You took it without paying for it, didn't you?" the shopkeeper said to her. ..................... ....................................................... The shopkeeper accused. 3. "Let's go to this bar here," my friend said. ................ My friend suggested. .............................................. .. .. .. .. .. .. .. .. .. 4. It was a really bad joke. Nobody laughed. .................. ....................... It was such. 5. The reviews were very bad, but the film was still a big success. ....................................... ........................................................................... Although. . 6. She wears the red blouse more often than the blue one. .......................................... ................................................................................. She prefers. 44 6. Practice test 7 , 7. She has been writing the report since 2pm. She started. ........... .. .. .. .. .. ... .. .. . 8. The man regrets not having bought that house The man wishes. ....... ................... .......................... 9. I want to visit IIalong Bay this summer. Yet I do not have enough money. If 1. ................................. . . . . . . . . . . . . . . . . . . . ... .. .. .. .. ... .. .. .. .. .. .. .. .. 10. The phone rang as soon as 1 went to bed. No sooner. ............. . . . . . . . . . . . . . . . . . . . . .. ............................... SECTION 3: TRANSLATION (15 marks) Translate the following sentences into English. 1. Ha N(>ila thu d6 eua nUoe C(>ng hoa Xii h6i ehu nghja Viet Nam. ....................... 2. Ky hQp Quoe h(>i Ian 3 dii khai m?e vao thu 7 ngay 3 thang 5 nam 2003 tal Hii Noi. "..................................... ............................ 3 - 5 Doanh thu nganh du lieh tang khoang 10% trong bon thang dau nam, nhung giam mEjnh trong thang 4 do dieh SARS. ...................................................... Translate the following sentences into Vietnamese 6-8. According to the government report, aquaculture especially important part in economic restructuring in some areas. f fmvcver, shrimp breeding, plays an techniques have led to uneven development within the industry. poor pJanning and breeding .................. .......... ......................................................................... .............................................................................. .......... 9-10. The report also said t.hat progress has been reduction, settlement of land for prod lIction and made in education, job creation, poverty residents of nood-prone areas in the Mekong Delta. houses for ethnic minority' people and ....................... ................................... ........................................................ ................. This is the end of Practice test 7 45 -----.- 7. ------- Tajli~uanthjtitn Anh PRACTICE TEST 8 PART 1: VOCABULARY AND READING COMPREHENSION SECTION 1: VOCABULARY (15 marks) Choose the best word/phrase in the box below to complete the sentences that follow. Each word/phrase can be used ONCE ONLY. _.---_.- ~ - [--------------- due to onc silence travelling I relaxing out of anxious enthusiasm I raises tel1ing preven ted anything . looking at da~ anniversary drink excited \vhat laid for __u_n.._ -- - - 1. M~' wife and I hm'(' nlways enjoyed ... .b~,' sea.. because nobody wanted to begin a conversation. 2. We ate our. meals in 3. At the meeting we discussed this issue with enormous.. 4. No. .is better at chess than chmet. 5. We arc. . to find 8. Practice test 8 SECTION 2: READING COMPREHENSION (30 marks) Read each passage carefully and then answer the questions that follow. Passage 1 Britain is too small and crowded to produce all the food its people need. British farm~ produce large quantities of wheat, sugar beet, dairy products, beef and lamb, but more theln half of Britajn'~ foodstuffs are bought abroad. The British industry requires rmv muterials Most of these too have to be bought abroad. There are not nearly enough forests in the UK to keep the timber and paper mills busy, and there is very little iron ore for the steel milIs. Industries which need silicon, copper, zinc, cotton can only get them from foreign countries. Coal and North Sea oil and gas are the grea~ exceptions. British drivers fiJI up with British petrol. Housewives cook with Uritish natural gas, and power stations and factories burn British coal, gas or oiL 1. Can Britain suppJy sufficient foodstuffs to its people? 2. What docs 'these' in 'most of these, too have to be...' refer to? ........... ................ 3. Docs Britain have silicon, copper, zinc and cotton? ............... ........... .................. 4. What natural resources are plentiful in Britain? ................... 5. FiJI in the blank with one suitable word from the passage. Britain cannot. all the food for its people. Passage 2 A study of 1500 employees in top American companies shows that m:lnagers spend only about a qU8rtcr of their time managing. What do they do the rest or theIr time? Clerical ,", 9. 10. In case the company C 10. Pructice test R SECTION 3: READING CLOZE (10 marks) Instructions: Complete the following passage with the missing words. Fill in each blank with ONE suitable word only. Water i~ neeess8ry for life. People can live only a fe\v days without (I) . y'ct nearly 25 million people die each }'CRr because of it. Both industrial nations and less-developed (2) .arc worried about the quality and (3) .of water in the world. The United Nations has named the 1980s the World Water Decade. The UN hopes to provide pure (4) for everyone by' 1990. Even though people, animals, 8griculturc, and' industry use a lot of \vutcr, there is more (5) .ellough on the Earth. One (6) the problems about water IS distribution. Water is not alway~ distributed where the large population centres (7) Some regions get enough rain, but it is all in one (8) . t wo short rainy seasons. Over half the \\'"orld is without pure drinking water. 75%1 of city people have safe \vater, (9) .only 29\;1" of rural people do. About 80% of all Illness is related (10) . bad water. PART 2: WRITING SECTION 1: CONTROLLED WRITING (10 marks) Instructions: From the following sets of words and phrases in the given sequences, make all the changes and additions necessary to produce sentences, which together make a complete letter from Francis to his new penfriend, Maria. Note carefully from the example what kind of changes need to be made. Write each sentence in the space provided. Example: I be happy / get / letter / offer me / job / your company Answer: 1 was happy when J got your letter, in which you offer me a job in your company. Dear Maria, 1. I / be / pleased / we / be going / be / penfriends./ 2. I / tell / you / a little myself, / you / do same / when / write me. / 3. I / live / area / call / Maida Vale. / ........... 4. it / be / quite near / centre / but / there be / parks / nearby / 5. 1/ go / local comprehensive school / 1/ have / a lot of friends./ .................. -- --- - - 11. TUi Wu cJn[hi [ie'n Anh 6. evenings ( I sometimes ( visit ( friends ( stay ( home (listen music .( ...................... .............................................................. 7. "vcckends / llike / go swimming / horse-riding / .............. ............. .......................... 8. mo~ent / I/ work very hard / I/ have exams / soon.; ............ 9. I ( be ( spend ( a lot of time ( library.( IO. I ( look forward ( hearing ( you.( Writc soon! I3est wishes, Francis. SECTION 2: SENTENCE TRANSFORMATION (15 marks) Instructions: Use the words given to rewrite each of the following sentences in such a way that it means exactly the same as the original one. 1. I can't get m)' fect into these shoes. Thcsc shoes arc too .. 2. Ilow long is it since they bought the house'? When .. 3. No one has signed this cheque. This cheque. 4. The office is so dirl~ that 110 one wants to work there. It is such. 5. You should phone immediately; or else you won't get any information. Unless you 6. 18m sorry. This is a non-smoking room. You 7. "You'd bettcr not lend him an.y more money, Mark" said I~osie. Rosie ad'\'iscd Mark. 8. " Why don't \\'C go to the cinema '?" said Bob. Bob suggcsted .. 9. Although he had a good salary, he ,\",'as unhappy in his job. In spite of . 10. Marry cats very little so as not to put 011 weight. Mary cats very little because ........... ............. 12. Prac1ice test 8 SECTION 3: TRANSLATION (15 marks) Translate the following sentences into English 1. Viel Nam la mc)t trang nhung nlidc san xua t tra Idn nhal the gidi. 2. Tra Viet nam dliqc lia chuc)ng 0 nhieu nlidc. 3. Nam 1998 Viet Nam xuat khau he5n 30.000 Ian Ira. 4. Chinh ph" da chap thuan mc)t ke ha 13. TiJi lifU dn lhi tie'ng Anh PRACTICE TEST 9 PART 1: VOCABULARY AND READING COMPREHENSION SECTION 1: VOCABULARY (15 marks) Choose the best word/phrase in the box below to complete the sentences that follow. Each word/phrase can be used ONCE ONLY. arrange a meeting it produced stay fit iHegal lies six-month cont.ract to Jj.\. an appointment bu siness card safe trip Although is made In spite of there closed down taking the risk check lnto number of jobs growing demand healthy in the Champagne region of France can be 1. Only wine which labe11ed 'champagne'. 2. I see from your .... ........... ..t.hat .','our company' has offices in Paris, London, and I~ome. 3. I will telephone you next week and, if you arc interested, \VC can 4. Have a....... .. . .. . .. . .. . .. . ..back to P;nglandl Thank you. 5. You mustn't park here. It's ..... ............................ ........................ will increase 6. Next year the factory wi11 employ a lot of people. The by' over 6(/io. . .. .. . .. . .. .. .. . .. . .......... 7. My father lost his job because his company.. 8. ........... ... won't be enough room if everyone comes. 9. lie tried to . a hotel, but the receptionist. refused to accept his credlt canl. only twenty miles from the volcano. 10. The city, with a population of 300,000, unless wc can make a big profit. 11. It's not worth ............ 12. I do exercise three times a week in order to ...... 13. .. being excited about the good news. Mark seemed to be indifferent. 14. In 1992, with the. the company built a new factor.\" to increase its production. .................. 15. We \\.'ant to hire him on a .. 52 14. Practice lest/) SECTION 2: READING COMPREHENSION (30 marks} Read each passage carefully and then answer the questions that follow. Passage 1 Traffic congestion in cities wil] be reduced because ddvers will Lise e1cctronic route maps to find the quickest route to their destination and avoid traffic jams. Congestion will also decrease after electronic s.ystems start charging motorists for driving in cities. 1\ssoon as motorists have to pay to drive in cities, the}' will stop using their cars and use pubJic transport instead. Spccd- control systems wilJ be built into cars. These systems will regulate the speed of the car to take account of traffic and \\'Cather conditions, and prevent accidents. It will be many years be fan' their changes bring results but when the.y' do, there will be a dramatic improvement in road safety. 1. I low can electronic route maps help reduce traffic congestion on cities in the future'? . ,. . ,, ,. . ,, , .. ,, ,. ............. 2. What wiIJ drivers have to do if thcy want to drive in cities? ............ , ,... .......... .......... 3, Ho\\' useful arc speed control systems? .., ,', , ,.,.."...""..".. "...... 4. What docs 'these' in 'These systems will be built in cars' refer to? ,.... ,,..... ,.... , . .,,.. ..... ... ,... ,... , S, 110\\ soon wi1l improvcment in road safety be made? ." ,',..,',..,.,...,...,...".,., ", ,,...,., ................. Passage 2 The Italian fashion industry is very flexible, and it has remained competitive because it has been able to react quickly to changes in. the market. However, - in recent .vears, production costs in Italy have been rising, and imports have been growing faster than exports. Mid-priced clothing in particular has been facing growing competition from countries in the Far East, South America, and Eastern Europe, where labour costs arc lower. The export market has remained strong, howcver. In 1992, the va1ue of" cxported clothing was $6 billion, more than t\vice the value of imported clothing, 6. What does the passage tell us ahout? .."""...,,,.,,...,, ,,. ,...",................ .......... 7. Why have imports been growing faster than exports in Italy in r{,l'cnt years? ............... 8, Which countries arc the competitors of the l1aliHn fashion industr.\'"? 9. What docs the rigure 'S 6billion' refer to'? ............... - - - - -. - 15. Tdi lifU on thi ti€'ng Anh 10. Fill in the blank with one suitable word from the passage. ............ Production costs in the Far East, South America, and Eastern Europe are.................. than those in Italy. Passage 3 The government in Japan has set up twenty - six research cities, or 'technologies' to study new technologies. Scientists and entrepreneurs will live in these cities, and each city will carry out research of different kind. The Japanese government will spend over$2.5 billion on each city. Private businesses will pay for research and development costs as well. 11. According to the passage, what arc 'technologies? ........... 12. What is the main business of these cities? ................................ 13. Who finances the construction of the cities? .................. ............................................ 14. lIow will private businesses help scientists and entrepreneurs in the cities? ......................................................................................................................................... 15. Write the question for the answer below? ? .............................................................................................................. \$2.5 billion Passage 4 In the 70s and 80s, most managers expected to continue working until retirement at sixty or sixty-fivo. But now, the situation is changing. Since the beginning of the 19905, many managers in their forties and fifties have lost their jobs. The reason for making managers redundant is a company buy-out or restructuring. 1\1so the recession of the late 1980s and early 1990s caused many redundancies. 16. lIow long did most managers work in 70s and 80s? ...................... 17. Can most managers work as long as those used to? ...................................... ............ 18. What makes managers redundant? ............................................................ ............... ................. 19. What word is used in the passage to mean 'job losses'? ................................................. 20. What docs the underlined word in 'the situation is changing' refer to? ............ ....................................... 16. () Practice test SECTION 3: READING CLOZE (10 marks) Complete the following passage with the missing words. Fill in each blank with ONE suitable word only. We talk about time every day. We measure (l) .......... ................ by second, minute, hour, day', \vcek, month, year, and century. But what is thc time? No onc can say exactly \\/hat it is. ~) .......................... is one of the greatest m.\,"steries of our lives. We (3) ............... know exact])' \\'hat time is, but our ability to measure it is very important. It (4) our way of life possible. /\11 the members of a group have to (5) .. ............ time in some wa}'. Time let us put things in a different order. We know that breakfast comes (6) ............................. lunch. The reading class is after the writing class. Children can't play (7) .................. school is over. Time enables us to organise our Jives. The earliest people saw changes of the moon, and the seasons. They (8) ............ .. mcasunng their I1ves by these changes. Then people startcd inventing clocks. The Chmese ( 17. Teli liru {}n thi tiltng Anh 6. I ( be ( in charge ( international cable projects ( .................... 7. I ( be ( married ( six years ( .......................................................... 8. I ( work ( weli ( other people ( .............. 9. I ( be ( happy ( work (long hours if neeessary( ........................................................................................................................... 10. Please ( find ( enc10se ( curriculum vitae ( ................................................ SECTION 2: SENTENCE TRANSFORMATION (15 marks) Use the words given to rewrite each of the following sentences in such a way that it means exactly the same as the original sentence. 1. "It certainly' wasn't me who did this" said Peter. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. ................... Peter denied. ..............................................- 2. She finally managed to get a job. ................................ .. .. . .. . .. . .. .. . .. .. . .. .. .. . .. .. She finally succeeded.............................. 3. "You look exhausted. You need a holiday" said the doctor. .................................... The doctor advised....... 4. Ill' asked a little boy how old he is. .. . .. . ... .. .. .. . '(" he asked a little boy. "Ifow S. I \\"ork in a factor.\ which has more than a thousand emp1oyecs. .......................... ............................ There. ............. 6. Who does this book be10ng to'? ............................ ............. .. ? Whose. 7. My suit needs to be ironed before the interview. .................................................................. 8 Mr. Smith is 7S years o1d, but he runs 5 kilometers a da)". ............................................... i\lthough . , 9. 1'\'1ysister speaks English bettcr than me .............................................. I don't :;pcuk English as 10. The tca is \'en' hot I can't drink it. ............. ............................. The tea is too.. 18. Practice test Y SECTION 3: TRANSLATION (15 marks) Translate the following sentences into English. 1. Thilnh pho Ho Chi Minh Iii mQI thilnh pho Ire so voi Hil Noi. ........................... ,... ,.. ................................... 2. Hil nQi Iii mQI trang nhO'ng thilnh pho Ion nhal Vi",t Nam. ................. .................. 3. Dan so cua thilnh pho vila khaang 5 trieu ngtJoi. 4. Trang nhO'ng nam vlia qua, thilnh pho co nhieu thay d6i Ion. ................ 5. Muc song cua ngtJoi dan cilng ngily cilng caa. ................... Translate the following sentences into Vietnamese. 6-7. In the United States, the powcr of the unions is decreasing. There arc clear signs that thcv are losing their hold on workers. ......................... ....................... ...................... ............. .......... 8. When a British citizen reaches the age of 65, he may retire from work and thcn he has the right to qrav.' a state pension. For women, the age of retirement is sixty. 9- 10. Stress creates problems for the individual and the organisation. But emplo~'Crs ignore the effects on the \vorkpJace: absenteeism, poor work and dangcrous health and safety factors. .......... .............. ........... Th1s is the end of Practice Test 9 57	{}
# Homework Help: Simple Harmonic Motion - Potential and Kinetic Energy? 1. Mar 15, 2008 ### meganw 1. The problem statement, all variables and given/known data A 2.00 kg mass vibrates according to the equation x = 0.470 cos 8.36t, where x is in meters, and t is in seconds. Assume that x = 0.29 m. (a) Determine the amplitude. .470 (b) Determine the frequency. 1.331 (c) Determine the total energy. (d) Determine the kinetic energy. e) Determine the potential energy. I did the first two parts (a-b), but I'm stuck on c,d & e, the questions about the Energy of the spring. 2. Relevant equations PE = .5kx^2 and KE = .5KA^2 F/x=k 3. The attempt at a solution I've tried these 3 multiple times and I keep getting them wrong. One thing I'm not sure of, is is "x" the amplitude? I have been plugging in .29 for the "x" and .47 for the "A", but just curious if this is the reason I'm getting these wrong. I would think, that according to these formulas, KE+PE=TOTAL ENERGY, and I thought I was doing them right. I'm at my wits end. Last edited: Mar 15, 2008 2. Mar 15, 2008 ### Staff: Mentor Good. Not good. (This implies that the KE doesn't change!) No. "x" is displacement from equilibrium. These are correct. Absolutely. What's the total energy? Hint: Since total energy doesn't change, pick the easiest position to calculate PE + KE. 3. Mar 15, 2008 ### Snazzy K.E + P.E = total energy $$\frac{1}{2}mv^2+\frac{1}{2}kx^2=\frac{1}{2}kA^2$$ They usually give you that equation on formula sheets. 4. Mar 15, 2008 ### meganw Does Ke=.5mv^2? I don't know V, and I can't use w=v/r because I don't know R. ALSO: I tried to find Total Energy when PE is maxed out and v=0, but I got it wrong. I did Total Energy=.5((9.82)/(.29))(.29)^2 thats .5 * k * x^2 And Snazzy-I tried .5KA^2 as my total energy and got 2.986 and thats wrong... 5. Mar 15, 2008 ### Snazzy The total energy is only given by $$\frac{1}{2}kx^2$$ only if $$x=A$$ You find A by looking at the equation of the motion because SHM follows the equation: $$x(t)=Acos(\omega t+\phi)$$ Then you can find the potential energy at x=0.29m and then the kinetic energy at that point since you now have the total energy and the potential energy. 6. Mar 15, 2008 ### meganw But I can't find KE because I don't know V and I can't find v using W because I don't have R... I TRIED TO FIND PE AND GOT IT WRONG: F=kx x=.29 F=9.8*2 k=67.5862 SO PE=.5(67.5862)(.29^2) =2.842 Thank you so much for your help on this!! wrong Last edited: Mar 15, 2008 7. Mar 15, 2008 ### Snazzy You don't need V or R or whatever to find the kinetic energy. You can use the fact that KE + PE = total energy 8. Mar 15, 2008 ### meganw Well, x is not equal to A. So I guess I can't use the fact that .5kA^2 is total, and I keep getting PE wrong. :( 9. Mar 15, 2008 ### Snazzy You get PE wrong because your value of k is wrong. F = kx at equilibrium, but x=0.29 is not the distance the spring has stretched at its equilibrium state. They usually give you the equation: $$\omega =\sqrt{\frac{k}{m}}$$ And you CAN use the fact that $$\frac{1}{2}kA^2$$ is the total energy because A is GIVEN TO YOU IN THE EQUATION OF THE FUNCTION. The reason you got it wrong is because your value of k is wrong. Last edited: Mar 15, 2008 10. Mar 15, 2008 ### meganw Okay, I tried it using x=.470 for my equation of the value of k, and I get k=41.7 but then I still get the wrong answer for PE...=1.753 11. Mar 15, 2008 ### Snazzy I don't know why you're using the amplitude to find the value of k, use $$\omega$$ and the mass to find the value of k as shown in the equation above. The spring does not stretch to its amplitude at equilibrium, nor does it stretch to the value of x the question gives you. 12. Mar 15, 2008 ### meganw I GOT THEM RIGHT!!! =) Thanks, you were SO helpful!! I didn't realize that you can't always use the equation of F=-kx to find k! I'm really thankful and grateful that you stayed online and helped me through each step. Thanks again, you're a really good person. =) 13. Mar 15, 2008 ### kdv You can always use F=-k x at two conditions: that the F you use is the force exerted by the spring and that the x you use is the displacement of the mass from the equilibrium position. I am not sure why you concluded that you can't always use that equation but I wanted to point that out.	{}
# Real Numbers and Infinite Games, Part II In my last post, I wrote about two infinite games whose analysis leads to interesting questions about subsets of the real numbers. In this post, I will talk about two more infinite games, one related to the Baire Category Theorem and one to Diophantine approximation. I’ll then talk about the role which such Diophantine approximation questions play in the theory of dynamical systems. The Choquet game and the Baire Category Theorem The Cantor game from Part I of this post can be used to prove that every perfect subset of ${\mathbf R}$ is uncountable. There is a similar game which can be used to prove the Baire Category Theorem. Let $X$ be a metric space. In Choquet’s game, Alice moves first by choosing a non-empty open set $U_1$ in $X$. Then Bob moves by choosing a non-empty open set $V_1 \subseteq U_1$. Alice then chooses a non-empty open set $U_2 \subseteq V_1$, and so on, yielding two decreasing sequences $U_n$ and $V_n$ of non-empty open sets with $U_n \supseteq V_n \supseteq U_{n+1}$ for all $n$. Note that $\bigcap U_n = \bigcap V_n$; we denote this set by $U$. Alice wins if $U$ is empty, and Bob wins if $U$ is non-empty. Continue reading # Real Numbers and Infinite Games, Part I Georg Cantor In this post I’d like to illustrate how one can use infinite games to prove theorems about the real numbers. I’ll begin with a game-theoretic proof that the set of real numbers is uncountable, following the exposition in this paper of mine. This will lead us somewhat unexpectedly into the realm of descriptive set theory, where we will discuss how games are used in cutting-edge explorations of the Axiom of Choice, the Continuum Hypothesis, and the foundations of second-order arithmetic. In a sequel post I will discuss how infinite games can be used to study Diophantine approximation, with applications to complex dynamics. Countable versus uncountable infinities When my daughter was 5 years old, she asked me if there is just one infinity. I proudly kissed her on the forehead and told her what an excellent question that was. I told her no, infinity comes in many different flavors. I pretty much left it at that, but since she’s 10 now, here are some more details for her. (The reader familiar with the basics of set theory can move on to the next section.) Continue reading	{}
# Electromagnetism - right hand rule 1. Jul 30, 2007 ### avsj 1. The problem statement, all variables and given/known data A proton travelling at a speed of 3.0 x 10^6 m/s travels through a magnetic field of strength 3.0 x 10^-3 T, making an angle of 45 degrees with the magnetic lines of force. What force acts on the proton? 2. Relevant equations F = QvB 3. The attempt at a solution I arrived at F = 1.44 x 10^-15N by simply plugging in. Then using the 45 degrees, I assume the force I found is the hypotenuse so I solve for an adjecnt using trig to get the correct answer of 1.0 x 10^-15 N but I dont understand this conceptually. Why is the force at 45 degrees weaker, and why am I solving it this way? Thanks a lot 2. Jul 30, 2007 ### meopemuk The correct Lorentz force equation is $$\mathbf{F} = Q \mathbf{v} \times \mathbf{B}$$ where $\mathbf{v} \times \mathbf{B}$ stands for the cross-product of two vectors. Since you know the angle $\alpha = 45$ degrees between $\mathbf{v}$ and $\mathbf{B}$, you can use standard formula for the length of the cross-product vector $$\| \mathbf{v} \times \mathbf{B}\| = \|\mathbf{v}\| \|\mathbf{B}\| \sin \alpha$$ to obtain the magnitude of the force $\|\mathbf{F}\|$. Eugene. Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook Have something to add?	{}
# Problem: When HNO2 is dissolved in water it partially dissociates according to the equation HNO2 ⇌ H+ + NO2–. A solution is prepared that contains 7.050 g of HNO2 in 1.000 kg of water. Its freezing point is found to be -0.2929 ˚C. Calculate the fraction of HNO2 that has dissociated. ###### FREE Expert Solution We’re being asked to calculate the fraction of HNO2 that has dissociated. The solution is prepared that contains 7.050 g of HNO2 in 1.000 kg of water. Its freezing point is found to be -0.2929 ˚C. HNO2 ⇌ H+ + NO2 When calculating the freezing point of a solution, we’re going to use the equation for Freezing Point Depression. $\overline{){\mathbf{∆}}{{\mathbf{T}}}_{{\mathbf{f}}}{\mathbf{=}}{\mathbf{i}}{\mathbf{·}}{{\mathbf{K}}}_{{\mathbf{f}}}{\mathbf{·}}{\mathbf{m}}}$ ∆Tf = change in freezing point = Tf pure solvent –Tf solution Kf = freezing point depression constant i = van' t Hoff factor of the solute = no. of ions m = molality We're going to calculate the freezing point of the solution using the following steps: Step 1: Calculate the moles of solute. Step 2: Calculate the molality of the solution. Step 3: Calculate the change in temperature (ΔTf). Step 4: Calculate the van't Hoff factor (i). Step 5: Calculate the fraction dissociated. Step 1: Calculate the moles of solute (sucrose). 85% (114 ratings) ###### Problem Details When HNO2 is dissolved in water it partially dissociates according to the equation HNO2 ⇌ H+ + NO2. A solution is prepared that contains 7.050 g of HNO2 in 1.000 kg of water. Its freezing point is found to be -0.2929 ˚C. Calculate the fraction of HNO2 that has dissociated.	{}
Home • LifeWiki • Forums • Download Golly ## Soup search results For discussion of specific patterns or specific families of patterns, both newly-discovered and well-known. ### Re: Soup search results This is as close as I know how to get: x = 11, y = 12, rule = B3/S237b2o$o7b2o$2o7bo$2bo3bo$2o3b3o$o3bo$4bob2obo$5bobo2$8b3o2$8b3o! EDIT: Eureka! (kinda): x = 16, y = 15, rule = B3/S2310bo$8b2o$9b2o$3o$2bo$3o3b3o$5bo2bo$5bob2o$6bo5bo$7bo2b2o$11b2o2$13b3o$13bo$14bo! EDIT 2: Assembled; 10 gliders: x = 28, y = 24, rule = B3/S2321bo$21bobo$21b2o$5bo6bo$6bo6bo$4b3o4b3o3$23bo$23bobo$3o20b2o$2bo$bo$10b2o$11b2o$10bo3$17b2o$8b2o6b2o$9b2o7bo7b2o$8bo3b3o10b2o$12bo14bo13bo! Last edited by BlinkerSpawn on January 10th, 2017, 7:31 pm, edited 1 time in total. LifeWiki: Like Wikipedia but with more spaceships. [citation needed] BlinkerSpawn Posts: 1429 Joined: November 8th, 2014, 8:48 pm Location: Getting a snacker from R-Bee's ### Re: Soup search results A recent soup produced a 26-bit P6 oscillator http://catagolue.appspot.com/object/xp6_45a2zw66w7bcczy366w4o68/b3s23. This made me think about clocks as rocks, and it occurred to me that I had previously overlooked one 21-bit P6 (one half of this one), plus one 20-bit P8 (the only 20-bit P8 of which I am aware). These two can be easily synthesized from 12 and 14 respectively, although the one from this soup forms too quickly to yield a viable predecessor, and an attempt to synthesize it by prute force is missing two steps (adding two close blocks). I think inserters exist to do this; I seem to recall somebody posting some syntheses that do something like this, but I can't find any examples. A way to add an adjacent clock would also prove useful, although the only one I know of does so with the stator cells aligned the other way. x = 189, y = 88, rule = B3/S23128bo127bo$91bo35b3o12bo$91bobo46boo$91boo48boo$123bobo10bobo$91bo18boo12boo4boo4boo$90boo17bobbo11bo4bobbo4bo$90bobo16bobbo16bobbo$110boo18boo$135bo$3bobo38bo89bo22boo$3boo39bobo16boo18boo18boo18boo9b3o16boobobboboo$4bo39boo17boobboo14boobboo14boobboo14boobboo24bo4bobboo$47b3o17boo18boo18boo18boo28bo$o46bo108bo$boo19bo19bo5bo13bo19bo19bo19bo10boo17bo$oo20bobo17bobo17bobo17bobo17bobo17bobo7boo18bobo$4boo15bobo17bobo17bobo17bobo17bobo17bobo10bo16bobo$3boo18bo19bo19bo19bo19bo19bo29bo$5bo$$bobooobo6164bo162bobo5bo163boo4bo169b3o377bo99bobbo78bo3bo18boo13bo4boo18boo18boo17bo76b3oboo18bobbo10bobo3bobbo12boobbobbo12boobbobbo12bo4bo81boo17bobbo11boo3bobbo12boobbobbo12boobbobbo12boobbobo101boo9boo7boo18boo18boo18boboo111boboo6bobo31bo71boboo4boo17bo12bobo4bo14boo3bo14boo3bo14boo3bo14boo3bo14boo3bo14boo3bo14bo4booo6bo18boo11boo5boo12boo4boo12boo4boo12boo4boo12boo4boo12boo4boo12boo4boo12boo4boo5bo6boo11boo10boo6boo18boo18boo18boo18boo18boo18boo18boo5boo4boo14bo8bobo8bo19bo19bo19bo19bo19bo19bo19bo4bobo6bo24bo1543bo41boo42boo$$29bobo$30boo$30bo$$144bo142bobo5bo56bo19bo19bo39bo6boo4bo16bo54bobo17bobo3bo13bobo37bobo12b3o12bobo55bobo17bobobbobo12bobo37bobo27bobo55bo19bo4boo13bo39bo29bo170bo81bo19boo38boo27bo30b3o48boo17bobbo32boobbobbo22bo4bo32bo5bo41bobo17bobbo11bobobo16boobbobbo22boobbobo31bo4bobo62boo38boo28boboo37boobbooo6bobo32booboo4boo17bo14bo4bo19bo19bo19bo34boo3bo24bo4booo6bo18boo18boo18boo18boo18boo32boo4boo22boo4boo5bo6boo11boo18boo18boo18boo18boo38boo28boo5boo4boo14bo19bo19bo19bo19bo39bo29bo4bobo6bo! EDIT: BlinkerSpawn wrote:This is as close as I know how to get: ... Eureka! (kinda): ... This yields a 10-glider synthesis. Unfortunately, replacing the house by an attached beehive or loaf isn't as simple as I had previously thought. (A beehive could be turned into the others). x = 109, y = 24, rule = B3/S2321bobo21boo22bo4bo6bo5boo5boo4boo5boo39boo20bo19bo18boo39bobo18bobo17bobo16bobbo23bobo15bo18bobo16bobbo17bobo23boo14boboboo16boboo15booboo16boboooo22bo14boobbobb3o14bobb3o14bobb3o14bobb3oboo40bo19bo19bo19boo43boobo16boobo16boobo16boobo10boo36bo19bo19bo19bo9bobo35boo18boo18boo18boo11bo317boo8boo7bobo7bobo7bo8boo9bo3boo11bobo12boo12bo14bo! (EDIT: Apparently, you solved it exactly the same way!) Last edited by mniemiec on January 10th, 2017, 7:57 pm, edited 1 time in total. mniemiec Posts: 793 Joined: June 1st, 2013, 12:00 am ### Re: Soup search results This was my far more expensive approach: x = 23, y = 16, rule = B3/S238bo5bobo6bobo6b2o7b2o6bo5bo20bo20b3o11b2ob2o11bo3boo11b3o7bob2o17b2o2o19b2o4b2o6b3o5b2o4bo3bo4bo6bob2obo12bobo2bo15bobo16bo! I also found a way to get there from one of the other variants: x = 21, y = 18, rule = B3/S239bobo9b2o10bo8bo2o16boo3b2o12b3obobobo6bo5bob2o3bobo2bo3ob2obobo2b2o2b2o7bobo6bobo7b2o46b2o7b2o6bo5bo11b2o11bobo! I Like My Heisenburps! (and others) Extrementhusiast Posts: 1638 Joined: June 16th, 2009, 11:24 pm Location: USA ### Re: Soup search results Can anyone use any of these reductions of symmetric soups for a better synthesis of french kiss?: x = 26, y = 21, rule = B3/S23obob2obo47bo9bo7b2o6bobo5b3o8b2o6bo18bo7b2o8b3o7bobo6b2o7bo9bo423bo22b2o22bobo! x = 34, y = 30, rule = B3/S2323b3o87bo6bobo5bo2bo9b2o6b2o3b2o4bobo11bobo2b2o13b2o2bobo11bobo4b2o3b2o11b2o9bo2bo22bobo23bo825b3o! x = 34, y = 24, rule = B3/S2333bo2bobo2bobo3bo326bo5b2o20bo4bo2bo4bo2bo17b2o4b2o17bo2bo23bo2bo3bo20b2o4bo327bo26bobo26bobo27bo! x = 61, y = 34, rule = B3/S23231b2o32bo7b2o19b2ob2o6b3o20bobo5b5o20bo4b2o3b2o5b2obo3bobo6b2o7bob2o2bob3o11b3obo2b2o21bo19b2obo16bo4b2o24b2o16b2o24b2o4bo42bob2o43bo45b2o2bob3o47b3obo2b2obo57b2o50bobo3bob2o54b2o3b2o34bo20b5o33bobo20b3o32b2ob2o19b2o32bo32b2o! x = 29, y = 19, rule = B3/S23419b2o11b2o5bo2bo10bo2bo4bo2b2o3b2o5bo2bo3bo2b2o2bo2bo5b2o4b4o3b2o24b2o8b4o4b2o5bo2bo7b2o2bo3bo2bo5b2o6b2o2bo4bo2bo7bo2bo5b2o8b2o! Goldtiger997 Posts: 304 Joined: June 21st, 2016, 8:00 am Location: 11.329903°N 142.199305°E ### Re: Soup search results Sure, 18G: x = 110, y = 90, rule = B3/S2340bo38b2o39b2o2512boobo7b2ob2o8b2obo311bo9b2o10b2o23bo9bo4b2o6bo3b2o7b3o72bo20b2o73bo19bo14bo15bo40b3o16b2obo15bo13bo45b2o12bo2bo13b3o13b3o42bo2bo11bo14b2o9b3o13b3o47b2o14bo11bo2bo11bo13bo62bo2bo12b2o10bo15bo60bob2o16b3o87bo19bo86b2o20bo26b3o7b2o28bo6b2o27bo9bo229b2o30b2o29bo339bo28b2o8b2o29b2o7bobo28bo252ob2oo! LifeWiki: Like Wikipedia but with more spaceships. [citation needed] BlinkerSpawn Posts: 1429 Joined: November 8th, 2014, 8:48 pm Location: Getting a snacker from R-Bee's ### Re: Soup search results 10G: x = 57, y = 46, rule = B3/S2355bo54bo4bo49b3o2bobo3b2o237bo16bo36bo15b2o36b3o14b2o237bo36b2o36bobo2118bobo19b2o19bo22b2o14b3o3b2o15bo2bo16bo252b2o52bobo3o49bo2bobo! Iteration of sigma(n)+tau(n)-n [sigma(n)+tau(n)-n : OEIS A163163] (e.g. 16,20,28,34,24,44,46,30,50,49,11,3,3, ...) : 965808 is period 336 (max = 207085118608). AbhpzTa Posts: 313 Joined: April 13th, 2016, 9:40 am Location: Ishikawa Prefecture, Japan ### Re: Soup search results Goldtiger997 wrote:Can anyone use any of these reductions of symmetric soups for a better synthesis of french kiss?: ... BlinkerSpawn wrote:Sure, 18G: ... From 16, based on predecessor #3: x = 126, y = 30, rule = B3/S2339bo37boo38boo77bo78bo76b3o50boo28boo9bobo21bo15bobbo26bobbo3bo5boo20boo17boo28boo4boo4bo21boo3boo52boo28boo28boo11bo46bo29bo29bo12bo16bobo9boo15boboo26boboo26boboo10b3o17boo8boo17bobbo26bobbo26bobbo30bo11bo19bo29bo29boo11bo47bo29bo29boboo8boo17b3o27bobbo26bobbo26bobbooo9bobo16bo30boobo26boobo26boobo31bo32bo29bo29bo38boo24boo28boo28boo9boo21bo4boo10boo20boo5bo31boo28boo9bo21bobo36bobbo26bobbo71boo28boo104b3o104bo105bo3boo4boo3bo! EDIT: AbhpzTa wrote:10G: ... (oops!) Nice! This also reduces 1 20- and the 2 21-bit variants to < 1 glider/bit. mniemiec Posts: 793 Joined: June 1st, 2013, 12:00 am ### Re: Soup search results AbhpzTa wrote:10G: x = 57, y = 46, rule = B3/S2355bo54bo4bo49b3o2bobo3b2o237bo16bo36bo15b2o36b3o14b2o237bo36b2o36bobo2118bobo19b2o19bo22b2o14b3o3b2o15bo2bo16bo252b2o52bobo3o49bo2bobo! Very nice! I'm reposting something I already posted in the birthdays thread because I did not think it was actually a suitable thread. Here is an 18-20 glider synthesis of trans-skewed poles which I think previously had no synthesis: x = 106, y = 96, rule = B3/S2314bo12bobo13b2o92boobob2o1567bo65b2o56bo9b2o57b2o56b2o60bo60bobo46bo13b2o47bo45b3o2bo49bobo49bobo50bo554bo54bo54bo51bo51bo51bo555bo54bobo54bobo55bo2b3o58bo44b2o13bo43bobo45bo48b2o47b2o38b2o9bo39b2o38bo15103b2o103bobo103bo991b2o91bobo91bo! Goldtiger997 Posts: 304 Joined: June 21st, 2016, 8:00 am Location: 11.329903°N 142.199305°E ### Re: Soup search results Goldtiger997 wrote:Here is an 18-20 glider synthesis of trans-skewed poles which I think previously had no synthesis: ... Extrementhusiast had previously posted this 20-glider synthesis on 2015-08-06. Either yours or his would be cheaper if there were appropriate 3-glider syntheses of the beehive+blinker constellations. x = 235, y = 32, rule = B3/S23112bobo113boo8bo113bo10bo122b3o126bo47bo78bobo48boo76boo6bo47boo83boo121bo11boo45bo36bo29bo9bo45boo35bo29bo7b3o4bo39bobo35bo29bo40boo38boo28boo5bo19bo29bo29bo29bo37bo32bo6bo29bo3b3o18bobo27bobo27bobo27bobo37bobo27bobo7bobo27bobo16bo7bobo27bobo27bobo27bobo32boo5boo11boo14boobboo5boo11boo15boo3o12bo9bo29bo29bo29bo34bo6bobbo9bo19bo6bobbo9bo16bobbobbo12b3o14bo29bo29bo29bo24bo9bobbo6bo19bo9bobbo6bo19bobbobo29bobo27bobo27bobo27bobo23boo11boo5boo18boo11boo5boobboo17boo12b3o16bobo27bobo27bobo27bobo37bobo37bobo7bobo17bobo12bo19bo29bo29bo29bo41bo39bo6bo22bo13bo57bobo21bo29bo37boo38boo28boo71boo22bo19b3o7bo72bo22bo19bo9bo103boo11bo69boo33boo68boo33bo6boo70bo38bobo111bo113b3o113bo10bo114bo8boo123bobo! EDIT: FYI, the only known period 3+ oscillators up to 21 bits lacking syntheses are these 5 P3s and 1 P4: x = 112, y = 10, rule = B3/S23o19booboo15boo5boo11boo18boo18boo8boo3o4booboo8bo4booboo10bo5bobo11bo19bo19bobo4boobbo3boboo4bo9boo6bo11boboobboo12bobo17bobo17b3obobbo6boo13bobobo14bo26bo32bo3b3obboobobo16bo4b3o16bobo11bobo3b3o11bobo22bo7bo23bo11bobo3boo11bo4bo14bo4boo15bobo44bo19bobobo15bobobo17bo63boobboo14boobbo88b3o90bo! mniemiec Posts: 793 Joined: June 1st, 2013, 12:00 am ### Re: Soup search results mniemiec wrote: Goldtiger997 wrote:Here is an 18-20 glider synthesis of trans-skewed poles which I think previously had no synthesis: ... Extrementhusiast had previously posted this 20-glider synthesis on 2015-08-06. Either yours or his would be cheaper if there were appropriate 3-glider syntheses of the beehive+blinker constellations. Extrementhusiast's synthesis ... I wrote:...Here is an 18-20 glider synthesis of trans-skewed poles which I think previously had no synthesis: x = 106, y = 96, rule = B3/S2314bo12bobo13b2o92boobob2o1567bo65b2o56bo9b2o57b2o56b2o60bo60bobo46bo13b2o47bo45b3o2bo49bobo49bobo50bo554bo54bo54bo51bo51bo51bo555bo54bobo54bobo55bo2b3o58bo44b2o13bo43bobo45bo48b2o47b2o38b2o9bo39b2o38bo15103b2o103bobo103bo991b2o91bobo91bo! However, the beehives in my synthesis are not necessary because they are made just to create r-pentominos. Possibly one of the other "still-life + glider = r-pentomino" collisions could be used such that the constellation with that still life and the blinker could be made in 3-gliders. P.S. It bothers me that currently the hexapole can be synthesised in 8 gliders whereas the pentapole takes 10 gliders. Goldtiger997 Posts: 304 Joined: June 21st, 2016, 8:00 am Location: 11.329903°N 142.199305°E ### Re: Soup search results Goldtiger997 wrote:However, the beehives in my synthesis are not necessary because they are made just to create r-pentominos. Possibly one of the other "still-life + glider = r-pentomino" collisions could be used such that the constellation with that still life and the blinker could be made in 3-gliders. I only know of 3 within the budget - one loaf+glider, your beehive+glider, and another beehive+glider that fails because the glider hits the blinker, and no way to make either+blinker from 3 gliders. Others may have more extensive R-pentomino or 3-glider-to-constellation libraries. Goldtiger997 wrote:P.S. It bothers me that currently the hexapole can be synthesised in 8 gliders whereas the pentapole takes 10 gliders. Bob Shemyakin found a 5-glider quadpole synthesis on 2015-03-28, making pentapole 9 gliders: x = 67, y = 22, rule = B3/S23boboboboo$$bo$bbo4bo11boo18boo18boo$3o4bobo9bobo17bobo17bobo$7boo$21bobo17bobo17bobo$$23bobo17bobo17bobo3o21boo18boo20bobbo5b3o54boobo6bo42bobo9bo41boo52bo$$37b3o12boo$39bo12bobo$38bo9bo3bo$48boo$47bobo! mniemiec Posts: 793 Joined: June 1st, 2013, 12:00 am ### Re: Soup search results Extrementhusiast wrote:I also found a way to get there from one of the other variants: ... This made me think about other ways to make the remaining 19-bit variant (that can be turned into the 20-bit loaf ones that yield the missing 25-bit molds and 26-bit jams). From the "close but no cigar" department. Maybe somebody can finish/rescue this: x = 239, y = 24, rule = B3/S2314bo$bbo12boo$obo11boo3bo48bobo$boo8bo7bobo47boo$9bobo7boo48bo57bo$10boo114bo66bo$73b4o49b3o64bobo$69bo3bo3bo25boo18boo22bobobo41boo26bobobo$17bo51bobobo29boo18boo21bo5bo67bo5bo$17bobo20boo18boo7boo3bobbo12boo6boo10boo6boo10boo6boo20boo6boo10boo6boo10boo11boo15boo$17boo21bo19bo29bo6bobo10bo6bobo10bo6bobo12bo7bo6bobo10bo6bobo10bo6boo4boo11bo3bo$28bo12boboo16boboo26boboobbo13boboobbo13boboobbo11bobo9boboobbo13boboobbo13boboo18bobo5boboo$26boo17bo19bo29boboo16boboo16boboo26boboo16boboo4boo10bo29bo$4bo22boo11b3obbobboo10b3obbobboo20b3obbo14b3obbo14b3obbo13bo10b3obbo14b3obbo7bobo4b3obbo17bo6b3obbo$4boo38boobobo14boobobo24boo18boo18boo28boo18boo7bo10boo28boobo$3bobo41bo19bo98boo18boo18bobo27bobo$47bobo17bobo79bo16bobo17bobo17bobo14bo12bobo$48boo18boo97bo19bo3boo14bo29bo$16boo5boo165boo$15boo5boo162boo4bo$17bo6bo48boo110bobo$6bo65boo113bo$6boo66bo$5bobo! mniemiec Posts: 793 Joined: June 1st, 2013, 12:00 am ### Re: Soup search results Another "sparse" p2 has appeared: https://twitter.com/conwaylifebot/statu ... 5856801793 If Mark's site is correct, this is one of the 13 unsynthesized 16-bit oscillators. Proof of concept synthesis, but not a practical one: x = 18, y = 23, rule = B3/S234bo$4bo$4bo4$16bo$15bo$6bo8b3o$5bobo5bo$5bobo4bobo$6bo5bobo$obo10bo$b2o$bo4$8bo$7b2o$2b2o3bobo$3b2o$2bo! If there is a component to shorten a barberpole, it also solves one of the three remaining 15-bit oscillators (again according to Mark's site). EDIT: It's not up-to-date with regards to the 15-bit oscillators, but I still can't tell if this one has been checked off. Tanner Jacobi Kazyan Posts: 677 Joined: February 6th, 2014, 11:02 pm ### Re: Soup search results Kayzan wrote:Another "sparse" p2 has appeared: ... If Mark's site is correct, this is one of the 13 unsynthesized 16-bit oscillators. Proof of concept synthesis, but not a practical one: ... When there is no synthesis yet, even a ludicrously expensive one is good! Kayzan wrote:If there is a component to shorten a barberpole, it also solves one of the three remaining 15-bit oscillators (again according to Mark's site). Sadly, there is no such component known yet, and it would likely be very difficult. When I discussed this topic with Dave Buckinhgam years ago, he said that lengthening a barber pole was relatively easy (he had several converters to do so, and I found several other related ones), but that I would not likely find a way to shorten one. Kayzan wrote:It's not up-to-date with regards to the 15-bit oscillators, but I still can't tell if this one has been checked off. FYI, here is my current list of unsynthesized P2s up to 18 bits: 1 14, 2 15s, 6 16s, 24 17s, and 52 18s. Of these, the last 16, 6 of the 17s, and the last 22 18s are trivial, applying grow-barberpole converter to a smaller unsynthethesized one. x = 149, y = 114, rule = B3/S23bb3o11boo13boo13boo13boo4boo7boo16bo13bobo10boo$16bobob3o8boboboo9bobobb3o7bobo4bo7bobob3o9bobobo11bobboo8bobobo$boobbobo28bo28bobo24bo5bo7bobboo14bo$6bo9bobbobo10bobbo10bobboobo8bobbo11bobbobo9bobo4bobo11boo9bobbo$bbo4boo8bo4bo10bo13bo5bo8bo3b3o8bo15bo5bo6boo14bo$4bo12bo3boo10bobobo9bo4boo8bo14bo3bobo11bobobo12bo9boobobo$4bobo29boo44boo13bo10bobo17bo$110bo16boo6$boo5boo6boo5boo6boo4bo8boo3b3o7boobboo9boobboo10bobb3o8boo13boo13boo$bobo3bobo6bobo3bobo6bobobobboo6bo14bobbobo9bobobbo10bo13bo6boo6bobo12bobobb3o$35bo11boboobo9bo17bo10bobbobboo8bobo4bo12b3o$3bobobo8bobboobo10bo3bo27bo11bo33bobo9bobo10bobboobo$17bo14bo13b3o3bobo6bobo12bobbobo9bobobobbo8bobbo15bobo8bo$bb3ob3o8bo3b3o8boob3o15boo6bobobboo8boo4bo8boo3bo11bo3b3o7b3o4bo7bo4bobo$61bo3bo15boo13bo11bo19boo13boo7$boo13boobb3o8b3o12b3o12boo13boobb3o11bo11boo13boo13boo$bobobo10bo36boo6bobbobo9bo17boboboo6bobo12bobo3bo8bobob3o$5bobo9bobobboo8boboob3o7boboobobo8bobbobo8bobobboo8bo6bo27bo$3bo5bo21bo14bo21bo29bo7bobboob3o8bobobbo7bobbobo$bbo6bo8bobbobbo6boo3bobo7boo3bobbo7boo4bo9bo3bo8boo14bo29bo$bboobobobo9bo3bo15bo13bo11boo13bo16bobo7bo3bobo8b3obobo8bo3bobo$7bo11bo3bo14boo13bo9boo11boobbo12bobobboo14bo14bo14bo$65bo12bo16bo17boo13boo13boo6$bboo13boo12boo13boo13boo4boo7boo3bo9boo5bo7boo3bobo8bo3bo9boo$bbobo12bobo11bo3boo9bobo12bobo4bo7bo4bo9bo6bo7bo4bo10bobobboo7bobobboo$21bo10bobobo13bo14bobo9bobobbo9bobobobbo7bobo4boo6bobo15bobo$bbobboo10bo3bo26bobboo10bo49bo12bo11bo$7bo14bo9boobbo29bo10bo3bobo8bobbobobo7bo3bo11bo17bo$boo13boo16bo13bobobo9bobobbo8bobbobobo9bo6bo5bobbobo13bobo9bobo$7bobo12bobo11bobo9boo12boobboo8boo5bo9bo5boo5boo13boobbobo9boobbobo$3bobobboo8bobobboo14bo12bobo68bo3bo14boo$5bo14bo17boo13boo5$boo4bo10bo12boo4bo9bo14bo14bo3bo9boo13boo3bo9bo3bo14bo$bobo3bo10bo4boo6bo5bo9bo6boo6bobo12bo3bobo7bobo5boo5bo4bobo7bo3bobo12bobo$5bobbo7bobbo4bo7bobobobbo6bobbo3bobo6bobobo9bobbo20bo6bobo10bobbo14bo5bobo$3bo17bobo44bo15boo5bobboobobo14boo12boo6boo6bo$6boo8boobbo11bobbobboo6boboboobo9bo4bobo6boo13bo15bo12boo17bo6boo$bbobo17b3o8bo12boo14bo6bo14bo7bo3bobbo7bo6bo12bobbo7bobo5bo$bboobb3o8b3o13bobb3o13b3o7boobobobo6b3obobo15bo8boobobo10bobo3bo12bobo$67bo14bo15bo13bo12bo3bo14bo6$bboo12boo13boobo13boo12bo14boo12boo13boo15bobo$bbobo11bobobobo8bo3boo12bo12bo14bobo11bobo12bobo16bo12bobo$6bo13bo11bo15bo3bo8bobbo16bobo37boo4bo12bo$bbobbobo10bo4boo9bo3boo10bobobo14boo6bo3bo9bobboo3bo6bobboo3boo7bo4boo6boo4bo$7bo14bo11bo4bo6boo6bo6boboboobobo13boo6bo6bo7bo7bo22bo4boo$boo15boobobo11boobo14bobo5boo13boo5bo8bo3boobbo6bo3boobo9boo4bo9bo$7bobo37bo5bo13bobbo11bobo41bo4boo7bo4bo$3bobobboo7b3o17b3o9bobo17bo8bobo17bobo12b3o12bo12bo4boo$5bo43bo19bo10bo18boo27bobo12bo$143bobo4$bbobo14bo14bo14bo11boo6boo5boobboo9boo13boo13boo13boo$4bo14bobo12bobo12bobo9bobo6bo5bobobbo9bobo12bobo12bobo5bo6bobo5bo$oo4bo10bo14bo5bo8bo5bo13bobo10bo18boo11b3o14bo10bobobo$bbo19boo14bo14bo9bobo11bo15bobo4bo7bobo12bobobobbo7bo3bobo$6boo8boo6bo6boo6bo6boo6bo11bobbo6bobbobo15bobo11bobo$3boo57b3o3bo7boo15bobbo10b3o13bobbobobo7bobobbo$8bo9bo6boo6bo6boo6bo6boo11bo12bobo10bo3b3o12bobo8bo4boo7boobbo$4bo4boo8boo13bo14boo31boo10bo19boo8bo17bo$6bo18bo8bo5bo14bo$6bobo12bobo12bobo12bobo$23bo14bo14bo3$boo13boo13boo13boo4bo8boo4bo8boo4bo8boo13boo4bo8boo13boo$bobo5boo5bobo5boo5bobo4bo7bo5bo8bo5bo8bobo3bo8bobo12bo5bo8bobo12bobo$10bo14bo12bo8bobobobbo7boboobbo11bobbo11bobo9bobobobbo$3bobobobo8bobobobo8boboobbo36bobbo13bo3bo25bobo3boo5bobboob3o$49bo3boo6b3obbobo8bo3bobo15bo9bo3boo15bo6bo$3bobbob3o6b3obobbo7b3obbobo37bo15bobo4boo20b3oboobo7bo3bobo$4bo18bo16bo8b3ob3o12bobo12bobo7boobbo10bobobo$4bo18bo15boo28boo13boo11bobo8bobo17b3o12bobo$108bo35boo5$boo13boo3bo9boo3bo9boo3bo9boo13boo$bobo3bo8bo4bo9bo4bobo7bo4bobo7bobo12boboboo$7bo9bobobbo9bobo12bobo15bo15bo$3bobobbo30boo13boo7bobboo9bobbo$16b3obobo11bo14bo28bo$bb3obobo26bo3bo10boobbo8bobobo10bobobo$22bobo10bo27boo$8bobo14bo11bobo12bobo12bobo12bobo$9boo13boo12boo13boo15bo14bo$69boo13boo! mniemiec Posts: 793 Joined: June 1st, 2013, 12:00 am ### Re: Soup search results That one other P2 variant mentioned recently: x = 194, y = 38, rule = B3/S23142bo$64bo76bo$65bo75b3o$63b3o73bo$67bo72bo$67bobo68b3o$67b2o$114bo$113bo26bobo$2o53b2o44b2o6b2o2b3o14b2o8b2o15b2o26b2o$o17b2o35bo17b3o25bo6bo2bo18bo10bo15bo27bo$bob2o3b2o4b2ob2o37bob2o3b2o8bo28bob2o2bob2o19bob2o23bob2o24bob2o$5bobobo3bobo3bo40bobobo9bo31bobo26bo7bo18bo27bo$3o2bobo5b2o40b3o2bobo38b3o2bobo21b3o2bo7bobo11b3o2bo22b3o2bo$4b2ob2o20bo29b2ob2o41b2ob2o24b2ob2o4b2o16b2obo24b2obo$29bobo29bo2bo42bo2bo25bo2bo23bobo25bobo$13b2o14b2o30bobo43bobo26bobo24bo2bo24bobo$b2o9bo2bo46bo45bo28bo16bo9bobo25bo$o2bo8bo2bo139b2o8bo$o2bo3b2o4b2o139b2o$b2o4bobo103bo28bo$7bo105bo28bo27bo$14b2o97bo28bo27bo$4bo9bobo153bo$4b2o8bo45b3o46b3o3b3o20b3o3b3o10bo$3bobo151b2o7b3o3b3o$64b2o47bo28bo13bobo4b2o$63bo2bo46bo28bo21b2o4bo$63bo2bo46bo28bo20bo6bo$64b2o104bo2$62bo$62b2o$61bobo2$76b3o$76bo$77bo! It's probably possible to get to the traffic light with only two cleanup gliders, although that would probably require a computer search. mniemiec wrote: Kayzan wrote:If there is a component to shorten a barberpole, it also solves one of the three remaining 15-bit oscillators (again according to Mark's site). Sadly, there is no such component known yet, and it would likely be very difficult. When I discussed this topic with Dave Buckinhgam years ago, he said that lengthening a barber pole was relatively easy (he had several converters to do so, and I found several other related ones), but that I would not likely find a way to shorten one. I found a way, although it isn't applicable here: x = 15, y = 27, rule = B3/S235bo$5bobo$5b2o2$o$b2o$2o3$10b2o$9bobo$b2o$obo4bobo4bo$2bo3bo5b2o$6b2o5b2o6$3o9bo$2bo8b2o$bo9bobo2$6b2o$6bobo$6bo! However, based off of the predecessor, here is a final step for the 15-bit version: x = 32, y = 29, rule = B3/S2331bo$29b2o$30b2o$7bo$8b2o13bo$7b2o3bobo6b2o$12b2o8b2o$13bo2$6bo$4bobo10b2o$b2o2b2o9bobo$obo8bo4bo$2bo7bobob2obo$10bob2obobo5b2o$8b2obo4bo5b2o$8bo2bo12bo$2bo6b2o$2b2o$bobo$14b2o$13bobo$15bo4b2o$20bobo$20bo2$13b2o$12bobo$14bo! EDIT: The prior steps: x = 184, y = 34, rule = B3/S23125bobo$125b2o$126bo$110bo$111b2o15bobo$110b2o16b2o$129bo4$83bo29bo$18bo65b2o28bo42bo$18bobo62b2o27b3o40b2o$18b2o136b2o$83bo26b2o$9bobo71b2o24bobo40b2o$10b2o10b2o38b2o18bobo6b2o18bo7b3o5b2o24bo5b2o21b2o$10bo10bobo20bo16bobo26bobo33bobo18b2o3bo5bobo20bobo$21bo23b2o14bo28bo35bo19bobo3b2o4bo17bo4bo$obo14bob2obo21b2o11bob2obo23bob2obo30bob2obo20bo5bob2obo15bobob2obo$b2o2b2o10b2obobo34b2obobo23b2obobo30b2obobo26b2obobo15bob2obobo$bo2bobo14bo33b2o4bo22b2o4bo29b2o4bo25b2o4bo14b2obo4bo$6bo47bobo26bobo33bobo29bobo19bo2bo$55bo26bobo33bobo29bobo21b2o$82b2o34b2o30b2o$10b3o$12bo133bo5b2o$11bo134b2o4bobo$18b3o124bobo4bo$18bo29bo$19bo27b2o$43bo3bobo$44b2o$43b2o! I Like My Heisenburps! (and others) Extrementhusiast Posts: 1638 Joined: June 16th, 2009, 11:24 pm Location: USA ### Re: Soup search results mniemiec wrote:... FYI, here is my current list of unsynthesized P2s up to 18 bits: 1 14, 2 15s, 6 16s, 24 17s, and 52 18s. Of these, the last 16, 6 of the 17s, and the last 22 18s are trivial, applying grow-barberpole converter to a smaller unsynthethesized one. x = 149, y = 114, rule = B3/S23bb3o11boo13boo13boo13boo4boo7boo16bo13bobo10boo$16bobob3o8boboboo9bobobb3o7bobo4bo7bobob3o9bobobo11bobboo8bobobo$boobbobo28bo28bobo24bo5bo7bobboo14bo$6bo9bobbobo10bobbo10bobboobo8bobbo11bobbobo9bobo4bobo11boo9bobbo$bbo4boo8bo4bo10bo13bo5bo8bo3b3o8bo15bo5bo6boo14bo$4bo12bo3boo10bobobo9bo4boo8bo14bo3bobo11bobobo12bo9boobobo$4bobo29boo44boo13bo10bobo17bo$110bo16boo6$boo5boo6boo5boo6boo4bo8boo3b3o7boobboo9boobboo10bobb3o8boo13boo13boo$bobo3bobo6bobo3bobo6bobobobboo6bo14bobbobo9bobobbo10bo13bo6boo6bobo12bobobb3o$35bo11boboobo9bo17bo10bobbobboo8bobo4bo12b3o$3bobobo8bobboobo10bo3bo27bo11bo33bobo9bobo10bobboobo$17bo14bo13b3o3bobo6bobo12bobbobo9bobobobbo8bobbo15bobo8bo$bb3ob3o8bo3b3o8boob3o15boo6bobobboo8boo4bo8boo3bo11bo3b3o7b3o4bo7bo4bobo$61bo3bo15boo13bo11bo19boo13boo7$boo13boobb3o8b3o12b3o12boo13boobb3o11bo11boo13boo13boo$bobobo10bo36boo6bobbobo9bo17boboboo6bobo12bobo3bo8bobob3o$5bobo9bobobboo8boboob3o7boboobobo8bobbobo8bobobboo8bo6bo27bo$3bo5bo21bo14bo21bo29bo7bobboob3o8bobobbo7bobbobo$bbo6bo8bobbobbo6boo3bobo7boo3bobbo7boo4bo9bo3bo8boo14bo29bo$bboobobobo9bo3bo15bo13bo11boo13bo16bobo7bo3bobo8b3obobo8bo3bobo$7bo11bo3bo14boo13bo9boo11boobbo12bobobboo14bo14bo14bo$65bo12bo16bo17boo13boo13boo6$bboo13boo12boo13boo13boo4boo7boo3bo9boo5bo7boo3bobo8bo3bo9boo$bbobo12bobo11bo3boo9bobo12bobo4bo7bo4bo9bo6bo7bo4bo10bobobboo7bobobboo$21bo10bobobo13bo14bobo9bobobbo9bobobobbo7bobo4boo6bobo15bobo$bbobboo10bo3bo26bobboo10bo49bo12bo11bo$7bo14bo9boobbo29bo10bo3bobo8bobbobobo7bo3bo11bo17bo$boo13boo16bo13bobobo9bobobbo8bobbobobo9bo6bo5bobbobo13bobo9bobo$7bobo12bobo11bobo9boo12boobboo8boo5bo9bo5boo5boo13boobbobo9boobbobo$3bobobboo8bobobboo14bo12bobo68bo3bo14boo$5bo14bo17boo13boo5$boo4bo10bo12boo4bo9bo14bo14bo3bo9boo13boo3bo9bo3bo14bo$bobo3bo10bo4boo6bo5bo9bo6boo6bobo12bo3bobo7bobo5boo5bo4bobo7bo3bobo12bobo$5bobbo7bobbo4bo7bobobobbo6bobbo3bobo6bobobo9bobbo20bo6bobo10bobbo14bo5bobo$3bo17bobo44bo15boo5bobboobobo14boo12boo6boo6bo$6boo8boobbo11bobbobboo6boboboobo9bo4bobo6boo13bo15bo12boo17bo6boo$bbobo17b3o8bo12boo14bo6bo14bo7bo3bobbo7bo6bo12bobbo7bobo5bo$bboobb3o8b3o13bobb3o13b3o7boobobobo6b3obobo15bo8boobobo10bobo3bo12bobo$67bo14bo15bo13bo12bo3bo14bo6$bboo12boo13boobo13boo12bo14boo12boo13boo15bobo$bbobo11bobobobo8bo3boo12bo12bo14bobo11bobo12bobo16bo12bobo$6bo13bo11bo15bo3bo8bobbo16bobo37boo4bo12bo$bbobbobo10bo4boo9bo3boo10bobobo14boo6bo3bo9bobboo3bo6bobboo3boo7bo4boo6boo4bo$7bo14bo11bo4bo6boo6bo6boboboobobo13boo6bo6bo7bo7bo22bo4boo$boo15boobobo11boobo14bobo5boo13boo5bo8bo3boobbo6bo3boobo9boo4bo9bo$7bobo37bo5bo13bobbo11bobo41bo4boo7bo4bo$3bobobboo7b3o17b3o9bobo17bo8bobo17bobo12b3o12bo12bo4boo$5bo43bo19bo10bo18boo27bobo12bo$143bobo4$bbobo14bo14bo14bo11boo6boo5boobboo9boo13boo13boo13boo$4bo14bobo12bobo12bobo9bobo6bo5bobobbo9bobo12bobo12bobo5bo6bobo5bo$oo4bo10bo14bo5bo8bo5bo13bobo10bo18boo11b3o14bo10bobobo$bbo19boo14bo14bo9bobo11bo15bobo4bo7bobo12bobobobbo7bo3bobo$6boo8boo6bo6boo6bo6boo6bo11bobbo6bobbobo15bobo11bobo$3boo57b3o3bo7boo15bobbo10b3o13bobbobobo7bobobbo$8bo9bo6boo6bo6boo6bo6boo11bo12bobo10bo3b3o12bobo8bo4boo7boobbo$4bo4boo8boo13bo14boo31boo10bo19boo8bo17bo$6bo18bo8bo5bo14bo$6bobo12bobo12bobo12bobo$23bo14bo14bo3$boo13boo13boo13boo4bo8boo4bo8boo4bo8boo13boo4bo8boo13boo$bobo5boo5bobo5boo5bobo4bo7bo5bo8bo5bo8bobo3bo8bobo12bo5bo8bobo12bobo$10bo14bo12bo8bobobobbo7boboobbo11bobbo11bobo9bobobobbo$3bobobobo8bobobobo8boboobbo36bobbo13bo3bo25bobo3boo5bobboob3o$49bo3boo6b3obbobo8bo3bobo15bo9bo3boo15bo6bo$3bobbob3o6b3obobbo7b3obbobo37bo15bobo4boo20b3oboobo7bo3bobo$4bo18bo16bo8b3ob3o12bobo12bobo7boobbo10bobobo$4bo18bo15boo28boo13boo11bobo8bobo17b3o12bobo$108bo35boo5$boo13boo3bo9boo3bo9boo3bo9boo13boo$bobo3bo8bo4bo9bo4bobo7bo4bobo7bobo12boboboo$7bo9bobobbo9bobo12bobo15bo15bo$3bobobbo30boo13boo7bobboo9bobbo$16b3obobo11bo14bo28bo$bb3obobo26bo3bo10boobbo8bobobo10bobobo$22bobo10bo27boo$8bobo14bo11bobo12bobo12bobo12bobo$9boo13boo12boo13boo15bo14bo$69boo13boo! I mistakenly thought I found this p2 16-bit oscillator in the list above and found a synthesis for it: x = 5, y = 8, rule = B3/S23o2b2o$obobo$o$2bo$2bo$o$obobo$o2b2o! It is quite a cheap synthesis so I'll post it anyway. What was the previously known synthesis? Here it is 12 gliders: x = 34, y = 61, rule = B3/S234$11bo$10bo$10b3o3$11bo$12b2o$11b2o$15b2o$14bobo$16bo4$17bo$17bobo$17b2o$8bobo$9b2o$9bo3$6bo$7bo$5b3o2$2b3o$4bo$3bo2$15bo$14b2o$14bobo$6b2o$5bobo$7bo3$20bo$19bo$19b3o$15b3o$15bo$16bo4$10b3o$10bo$11bo! It kind of looks like the gliders would collide when rewound but in fact they don't: EDIT: Extrementhusiast wrote:... EDIT: The prior steps: prior steps that I had no luck constructing myself Great! That makes 42 gliders in total: (EDIT 4: replaced this with a 37 glider version using a better converter from Extrementhusiast and chris_c) x = 447, y = 30, rule = B3/S2349bo$47bobo381bo$48b2o7bo371b2o$57bobo4bo280bo84b2o$57b2o4bo279bobo61bo$63b3o110bo3bo163b2o3bo58b2o13bo$177bobo53bo115bobo55b2o3bobo6b2o$175b3ob3o51bobo113b2o61b2o8b2o$67b2o164b2o178bo$57bo8b2o276bo3bo$56bo11bo19bo29bo29bo29bo45bobo118bo3b2o55bo$56b3o28bobo27bobo27bobo27bobo27b2o16b2o10b2o28b2o28b2o28b2o14b3o2b2o7b2o28b2o15bobo10b2o$86bobo3b2o22bobo3b2o22bobo27bobo27bobo16bo10bobo27bobo10bo16bobo27bobo27bobo27bobo12b2o2b2o9bobo21b2o$54b2o30bo5b2o22bo5b2o22bo29bo29bo29bo29bo13b2o14bo29bo29bo24bo4bo13bobo8bo4bo23bobob3o$31b2o20bobo5b2o19bob2obo24bob2obo24bob2obo24bob2obo24bob2obo7bobo14bob2obo24bob2obo11b2o11bob2obo24bob2obo24bob2obo22bobob2obo14bo7bobob2obo$3o28b2o22bo5b2o19b2obobo24b2obobo6b2o16b2obobo24b2obobo24b2obobo8b2o2b2o10b2obobo24b2obobo24b2obobo24b2obobo24b2obobo22bob2obobo22bob2obobo5b2o15bo2bobo$2bo83bo29bo7bobo19bo29bo29bo9bo2bobo14bo23b2o4bo23b2o4bo23b2o4bo23b2o4bo21b2obo4bo21b2obo4bo5b2o17bo4bo$bo122bo96bo37bobo27bobo27bobo27bobo26bo2bo26bo2bo12bo16bo3b2o$3b3o254bo29bo27bobo27bobo28b2o21bo6b2o$3bo314b2o28b2o52b2o$4bo220b3o173bobo$227bo116bo5b2o62b2o$226bo117b2o4bobo60bobo$233b3o107bobo4bo64bo4b2o$233bo49bo136bobo$234bo47b2o136bo$278bo3bobo$279b2o132b2o$278b2o132bobo$414bo! Only one 15-bit oscillator left; muttering moat 1. EDIT 2: Here's one of the unsolved 17-bit p2s in 15 gliders: x = 34, y = 55, rule = B3/S2331bo$31bobo$31b2o8$10bobo$11b2o$11bo$19bobo$10bo8b2o$9bo10bo$9b3o$obo$b2o10bobo$bo11b2o$14bo5$16bobo$16b2o$17bo$19b2o$19bobo$19bo4$14bo$bo11b2o4b3o$b2o10bobo3bo$obo17bo$9b3o$9bo$10bo3$24bo$23b2o$23bobo$7bo$7b2o$6bobo4$31b2o$31bobo$31bo! EDIT 3: Here's one of the unsolved 18-bit p2s in 16 gliders: x = 49, y = 58, rule = B3/S2311bo$12bo$10b3o4$15bo$13bobo$14b2o4$2bo$obo$b2o3$23bo$22bobo$22bo2bo$23b2o6$36bo$15b2o19bobo$14bo2bo14b2o2b2o$11b2o2b2o14bo2bo$10bobo19b2o$12bo6$24b2o$23bo2bo$24bobo$25bo3$46b2o$46bobo$46bo4$33b2o$33bobo$33bo4$36b3o$36bo$37bo! Last edited by Goldtiger997 on January 17th, 2017, 7:07 pm, edited 1 time in total. Goldtiger997 Posts: 304 Joined: June 21st, 2016, 8:00 am Location: 11.329903°N 142.199305°E ### Re: Soup search results Extrementhusiast wrote:The prior steps This gives a 5 glider reduction and yields 16.897 and 16.1086 in 14 and 13 gliders respectively: x = 35, y = 44, rule = LifeHistory29.A$23.A5.A.A$21.2A6.2A$15.A6.2A$13.A.A$14.2A10$3A$2.A$.A4$22.E.2E$22.2E.E$20.2E$19.E.E$18.E.E$18.2E15$.2A30.2A$A.A29.2A$2.A31.A! chris_c Posts: 724 Joined: June 28th, 2014, 7:15 am ### Re: Soup search results Goldtiger997 wrote:Here it is 12 gliders: x = 34, y = 61, rule = B3/S234$11bo$10bo$10b3o3$11bo$12b2o$11b2o$15b2o$14bobo$16bo4$17bo$17bobo$17b2o$8bobo$9b2o$9bo3$6bo$7bo$5b3o2$2b3o$4bo$3bo2$15bo$14b2o$14bobo$6b2o$5bobo$7bo3$20bo$19bo$19b3o$15b3o$15bo$16bo4$10b3o$10bo$11bo! It kind of looks like the gliders would collide when rewound but in fact they don't: 10G: x = 23, y = 50, rule = B3/S2312bo$11bo$11b3o3$12bo$13b2o$12b2o$16b2o$15bobo$17bo8$o$b2o$2o$14bo$12b2o$13b2o3$13b2o$12b2o$3b2o9bo$4b2o$3bo8$21bo$20bo$20b3o$16b3o$16bo$17bo4$11b3o$11bo$12bo! Iteration of sigma(n)+tau(n)-n [sigma(n)+tau(n)-n : OEIS A163163] (e.g. 16,20,28,34,24,44,46,30,50,49,11,3,3, ...) : 965808 is period 336 (max = 207085118608). AbhpzTa Posts: 313 Joined: April 13th, 2016, 9:40 am Location: Ishikawa Prefecture, Japan ### Re: Soup search results chris_c wrote: Extrementhusiast wrote:The prior steps This gives a 5 glider reduction and yields 16.897 and 16.1086 in 14 and 13 gliders respectively: RLE Oh yeah, forgot about that component. Here's a better version (same cost, but much cleaner): x = 13, y = 21, rule = LifeHistory2.A$A.A$.2A3.A$6.A.A$6.2A2$.A3.A$2.A3.2A$3A2.2A3$9.E.2E$9.2E.E$7.2E$6.E.E$5.E.E$5.2E2$.A5.2A$.2A4.A.A$A.A4.A! I Like My Heisenburps! (and others) Extrementhusiast Posts: 1638 Joined: June 16th, 2009, 11:24 pm Location: USA ### Re: Soup search results A 41-cell SL that also implies a simpler synthesis of a 40-cell pseudo-SL: x = 16, y = 16, rule = B3/S23booboobobooboooo$obooobbbobobboob$obbboooboobbbobo$obboobbbobbobbbb$oooboooboobbooob$ooooboobbbbbbobo$obbbbbbobbbobbbo$booooboobboboboo$bboooobobooobobo$oobbobbbbbbobobo$ooboobooobbobbbo$bbbbbooboooobobo$bbooobbbbbboboob$oobobbbobbbbbbob$bobbbboobbboobob$bobooooooobobbbb! What I'm calling a "two-leaf clover": x = 16, y = 16, rule = B3/S23obobbbbobobbooob$obbobboooboobobo$boobobbbooboobbo$obbbobbbbooboooo$boboooobbboboooo$bobobooobooooobo$bbboboobboobobbo$boboobobooooobbo$oobbobbooobboboo$obbbboboooboobbb$bbboobbobbbbooob$obboooobboooobbo$obooobooobboooob$obooobbooobbbobo$bbbbbooobbbbboob$oooboobbbbobbbbb! (Although I definitely don't have naming rights.) x₁=ηx V ⃰_η=c²√(Λη) K=(Λu²)/2 Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt) $$x_1=\eta x$$ $$V^*_\eta=c^2\sqrt{\Lambda\eta}$$ $$K=\frac{\Lambda u^2}2$$ $$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$ http://conwaylife.com/wiki/A_for_all Aidan F. Pierce A for awesome Posts: 1354 Joined: September 13th, 2014, 5:36 pm Location: 0x-1 ### Re: Soup search results The 41-cell SL and the two variants: x = 104, y = 24, rule = B3/S2313bo39bo39bo$12bo39bo39bo$12b3o37b3o37b3o2$5bobo37bobo14bo15bo6bobo14bo$6b2o38b2o13bo17bo6b2o13bo$6bo39bo14b3o13b3o6bo14b3o$16bo39bo39bo$14b2o38b2o38b2o$15b2o38b2o38b2o$8bo39bo39bo$bo4bobo10bo21bo4bobo10bo21bo4bobo10bo$2bo4b2o9bo23bo4b2o9bo23bo4b2o9bo$3o15b3o19b3o15b3o19b3o15b3o8$3b3o9b3o25b3o9b3o25b3o9b3o$5bo9bo29bo9bo29bo9bo$4bo11bo27bo11bo27bo11bo! The two-leaf's reaction, and another found while experimenting, both in 5 gliders: x = 67, y = 25, rule = B3/S2327bo37bo$26bo31bo5bo$26b3o28bo6b3o$10bo46b3o$11b2o$10b2o3$59bo$2b2o3b2o33b2o3b2o8b2o$bo2bobo2bo31bo2bobo2bo8b2o$bob2ob2obo10bo20bob2ob2obo$2o2bobo2b2o8bo20b2o2bobo2b2o$2b2o3b2o10b3o20b2o3b2o$2o2bobo2b2o4b3o22b2o2bobo2b2o$bob2ob2obo5bo25bob2ob2obo$bo2bobo2bo6bo24bo2bobo2bo$2b2o3b2o33b2o3b2o$56b3o$18b3o35bo$18bo38bo$19bo$48b2o$49b2o$48bo!43bo$44bo! LifeWiki: Like Wikipedia but with more spaceships. [citation needed] BlinkerSpawn Posts: 1429 Joined: November 8th, 2014, 8:48 pm Location: Getting a snacker from R-Bee's ### Re: Soup search results The 41-cell SL in 8G: x = 51, y = 59, rule = B3/S2349bo$48bo$48b3o19$o$b2o$2o15$17bo$18b2o$17b2o$30b2o$26bo3b2o$25b2o$25bobo8$17bobo$18b2o$18bo2$18b3o$20bo$19bo! Iteration of sigma(n)+tau(n)-n [sigma(n)+tau(n)-n : OEIS A163163] (e.g. 16,20,28,34,24,44,46,30,50,49,11,3,3, ...) : 965808 is period 336 (max = 207085118608). AbhpzTa Posts: 313 Joined: April 13th, 2016, 9:40 am Location: Ishikawa Prefecture, Japan ### Re: Soup search results "22-bit p4 oscillator #2" in 14 gliders: x = 34, y = 26, rule = B3/S2315bo$16bo$14b3o$22bo$21bo$21b3o7bo$30bo$16bo13b3o$15bo$15b3o$bo8bo17bo$2bo8b2o14bo$3o7b2o15b3o$4b3o15b2o7b3o$6bo14b2o8bo$5bo17bo8bo$16b3o$18bo$b3o13bo$3bo$2bo7b3o$12bo$11bo$17b3o$17bo$18bo! Tight predecessor for "22-bit p4 oscillator #3": x = 9, y = 22, rule = B3/S23bo$obo$obo5bo$bo4b2o$8bo3$6b3o$8bo$5b3o3$5b3o$8bo$6b3o3$8bo$bo4b2o$obo5bo$obo$bo! LifeWiki: Like Wikipedia but with more spaceships. [citation needed] BlinkerSpawn Posts: 1429 Joined: November 8th, 2014, 8:48 pm Location: Getting a snacker from R-Bee's ### Re: Soup search results Kayzan wrote:If there is a component to shorten a barberpole ... Extrementhusiast wrote:I found a way, although it isn't applicable here: ... Wow! Very impressive! I would not have expected such a converter to have been found for a long time (and certainly not this cheaply). One of the gliders can be moved to make this less obtrusive, and allow its use in many situations where one side the modified pattern protrudes beyond the lane the barber-pole is in on one side, but not the other. While there are few cases I can think of where this converter is advantageous, it IS useful for adding tripoles, because that takes around 20 gliders, while adding a quadpole takes only 7, and shorting it another 7, saving 6 gliders. (This does not help with bipoles, because adding a bipole is cheaper than shortening a quadpole twice.) I have tried applying this to all my syntheses involving tripoles. This improves 5 oscillator syntheses. It also improves 5 pseudo-oscillator syntheses; the last one requires a slight alteration of this method, but it could not previously be done this way at all, so this dramatically reduces the cost from 41 to 20 gliders. I currently have 5 syntheses that cannot currently be improved because they protrude on both sides; 3 of these are pseudo-objects, formed from the 3 still-lifes that I originally found most challenging when first investingating pseudo-object syntheses. There are 133 other syntheses where this improves an alternate synthesis, but not the minimal one. The 5 improved oscillator syntheses: x = 227, y = 209, rule = B3/S2393bo$92bo$38bobo51b3o$39boo$39bo5bo53bo$46boo43bo5boo$11bo33boo42bobo6boo$9boo30boo47boo$10boo30boo$41bo58bo$99boo$64boo28boo3bobo$64bo29bo20boo$bbo62bobo23bo3bobo17bobo$obo89bo$boo25boo18boo17bobo15boo3b3o4bobo17bobo$13b3o12bobo17bobo19bo15boo12bo19bo$13bo15boo18boo18boo14bo4bo8boo18boo$4bobo7bo16boo18boo18boo17boo9boo18boo$5boo24bo19bo19bo17bobo9bo19bo$5bo26bobo17bobo17bobo27bobo17bobo$35bo19bo19bo29bo19bo$6b3o11boo12boo18boo18boo28boo18boo$8bo11bobo$7bo12bo3$19boo$19bobo$19bo$$8boo9boo8bo610bo8bobo9boo138bo88bobo57bo33bo55boo57b3o13bobo17bobo53bo5bo14boo17boo61boo14bo80boo41bo12bo91boo46boo5bobobbobo92boo44boo7boobboo91bo42bo12bo56bobo75boo23bo28bo3boo56boo17bobo8boo24bo28boobbo56bo29bo20boo22b3o27boo61bobo27bobo17bobo26bo26bobo10bo19bo18boo18boo17bobo17b3o7bobo17bobo26boo10boboboo14boboboo14boboboo14boboboo16boboo15bo3boo5boboo16boboo17bo21boobobo14boobobo14boobobo14boobobo14boobobo13bo4bobo3boobobo14boobobo18boo21bobbo16bobbo16bobbo16bobbo16bobbo18bo7bobbo16bobbo17boo22bobo17bobo17bobo17bobo17bobo27bobo17bobo42bo19bo19bo19bo19bo29bo19bo1019bo19boo18bobo37boo16bo8boo14boo7bo16bobo139bo10bo178bo8b3o117bobo57bo19bo109boo57b3o18bo110bo5bo18b3o115boo135boo41bo12bo131boo46boo5bobobbobo10bobo119boo44boo7boobboo11boo118bo42bo12bo11bo10bo73bobo75boo21bo70bo3boo56boo17bobo8boo21b3o69boobbo56bo29bo20boo92boo61bobo27bobo17bobo24boo24bobo12bobboo15bobboo15bobboo15bobboo14boobboo14boobboo13bobobboo13b3o7bobobboo13bobobboo24bo13bobobbo14bobobbo14bobobbo14bobobbo14bobobbo14bobobbo16bobbo15bo3boo5bobbo16bobbo8bo30boobo16boobo16boobo16boobo16boobo16boobo16boobo15bo4bobo3boobo16boobo8boo4b3o24bo19bo19bo19bo19bo19bo19bo21bo7bo19bo7bobo6bo15bo8bobo17bobo17bobo17bobo17bobo17bobo17bobo27bobo17bobo15bo15bo10boo18boo18boo18boo18boo18boo18boo28boo18boo31b3o42boo18boo42boo18boo17bo4b3o17boo5bo39b3o16bobo4bo40bo65bo811bo12boo11boo37bo36bo36b3o154bo192bo138bobo51b3o139boo139bo5bo53bo16bo129boo43bo5boo14bobo27bo100boo42bobo6boo15boo22bobobbobo94boo47boo39boo3boo96boo40bo100bo58bo106bobo90boo102bo3boo56boo28boo3bobo78bo24boobbo56bo29bo20boo78bobo21boo61bobo23bo3bobo17bobo78boo112bo49bo19bo19bo19bo18boo18boo17bobo15boo3b3o4bobo17bobo8bo3bo35bobo5boo10bobo5boo10bobo17bobo17bobo17bobo19bo15boo12bo19bo9boobobo34boo5boo11boo5boo11boo18boo18boo18boo18boo14bo4bo8boo18boo8boobboo37boo18boo18boo18boo18boo18boo18boo17boo9boo18boo51bo19bo19bo19bo19bo19bo19bo17bobo9bo19bo52bobo17bobo17bobo17bobo17bobo17bobo17bobo27bobo17bobo$$54bobo17bobo17bobo17bobo17bobo17bobo17bobo27bobo17bobo$55boo18boo18boo18boo18boo18boo18boo28boo18boo13$3boo30bo$4boo28boo$3bo30bobo3$6b3o$8bo$7bo7$3boo28boo18boo28boo18boo28boo28boo38boo$3bo29bo19bo29bo19bo29bo29bo39bo$5boo28boo18boo28boo18boo28boo28boo38boo$$6bobo27bobo17bobo27bobo17bobo27bobo27bobo37bobo$$8bobo27bobo17bobo27bobo17bobo27bobo27bobo37bobo$$10bobo27bobo17bobo27bobo17bobo27bobo27bobo37bobo12bo8bo20bo19bo29bo19bo29bo29bo39bo14bo4boo23bo19bo29bo19bo29bo29bo39bo13boo5boo21boo18boo28boo18boo28boo28boo7bo30boo45boo18boo28boo18boo28boo28boo3bobo4bo27boo5bobbo9bo26bobo17bobo27bobo17bobo27bo29bo5boo3bo28bo9bobbo4boo27bo19bo29boo18boo28bobo27bobo7b3o27bobo5bo3bobboo3bobo6b4obobo58boo74bobo27bobo37bobo68bobboo78bo29bo37boo15boo51boo3bo76boo28boo8bobo15bobo49bobo120boo15bo108bo53bo12bo122boo49boobboo7boo123boo47bobobbobo5boo119boo53bo12bo118boo120bo5bo125boo48b3o125bobo49bo176bo! The 5 improved pseudo-oscillator syntheses: x = 227, y = 167, rule = 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$$33bobo17bobo5b3o9bobo17bobo17bobo17bobo17bobo17bobo9bo17bobo$36bo19bo4bo14boboo16boboo16boboo16boboo16boboo16boboo3bobo4bo15boboo$35boo18boo5bo12boobobo14boobobo14boobobo14boobobo14boobo16boobo5boo3bo15boobo$3o76bo19bo5bo13boo18boo18bobo17bobo7b3o17bobo$bbo56bo43boo$bo13b3o41boo43boo55bobo17bobo27bobo$11boobbo42bobo103bo19bo27boo$12boobbo86bo59boo18boo8bobo$11bo90boo89boo$24boo76bobo42bo33bo12bo$24bobo118boo29boobboo7boo$24bo121boo27bobobbobo5boo$142boo33bo12bo$141boo$143bo5bo$148boo28b3o$148bobo29bo$179bo8$129bo$128bo$74bobo51b3o$75boo$75bo5bo53bo$82boo43bo5boo$81boo42bobo6boo$77boo47boo$78boo$77bo58bo$135boo$38bo10boo49boo28boo3bobo$6bo7bo24bo7bobobo48bo29bo20boo$4bobo5boo23b3o5bobobo51bobo23bo3bobo17bobo$5boo6boo31boo80bo$41boo21boo18boo17bobo15boo3b3o4bobo17bobo$5bo36boo20bobo17bobo19bo15boo12bo19bo$5boo34bo23boo18boo18boo14bo4bo8boo18boo$4bobo9boo11boo18boo18boo18boo18boo15boo11boo18boo$16bobo6boo3bo14boo3bo14boo3bo14boo3bo14boo3bo14bobo7boo3bo14boo3bo$16bo9bobbo16bobbo16bobbo16bobbo16bobbo26bobbo16bobbo$26bobo17bobo17bobo17bobo17bobo27bobo17bobo$27bo19bo19bo19bo19bo29bo19bo$$bbobboobobo4172bo173bo135bobo33b3o135boo130bo5bo29bo128boo37boo5bo129boo35boo6bobo133boo39boo132boo134bo30bo87bobo75boo44bo43boo3bo56boo12bobo3boo6bo7bo27bobo43bobboo58bo19bo17boo4bobo5boo29boobbo44boo54bobo17bobo3bo13bobo5boo6boo31bo126bo46b3o17bo19bo19boo18boo18bobo17bobo4b3o3boo5bobo5bo35b3o21bobo17bobo17bobo17bobo17bo19bo12boo5bo5boo36bo21boo18boo18boo18boo18boo18boo8bo4bo4boo4bobo9boo11boo11bo6boo18boo18boo18boo18boo18boo18boo3boo13boo16bobo6boo3bo14boo3bo14boo3bo14boo3bo14boo3bo14boo3bo14boo3bo14boo3bo3bobo8boo3bo16bo9bobbo16bobbo16bobbo16bobbo16bobbo16bobbo16bobbo16bobbo16bobbo26bobo17bobo17bobo17bobo17bobo17bobo17bobo17bobo17bobo27bo19bo19bo19bo19bo19bo19bo19bo3boo14bo172boobbo168bo4boobboo171boobobo173bo! The 5 syntheses that could not be improved: x = 86, y = 14, rule = B3/S2380boo39boo17boobo17bobbo38bobo17boboo16bobboo37bo18boo19bo36bo19bo19booo18boo14boboboo16boboo14bobobooobo17bobo14boobobo14boobobo14boobobo$$bbobo17bobobboo13bobo17bobo17bobo$5boboo16bobbo16bo19bo19bo$4boobo16boobo16boo18boo18boo$7bo18bo$4b3o16bobo$4bo18boo! mniemiec Posts: 793 Joined: June 1st, 2013, 12:00 am ### Re: Soup search results BlinkerSpawn wrote:"22-bit p4 oscillator #2" in 14 gliders: x = 34, y = 26, rule = B3/S2315bo$16bo$14b3o$22bo$21bo$21b3o7bo$30bo$16bo13b3o$15bo$15b3o$bo8bo17bo$2bo8b2o14bo$3o7b2o15b3o$4b3o15b2o7b3o$6bo14b2o8bo$5bo17bo8bo$16b3o$18bo$b3o13bo$3bo$2bo7b3o$12bo$11bo$17b3o$17bo$18bo! 10 gliders: x = 24, y = 24, rule = B3/S2311bo$12bo$10b3o$16bo$15bo$15b3o2$10bo$9bo$9b3o$6bo14bobo$bo5b2o7b2o3b2o$b2o3b2o7b2o5bo$obo14bo$12b3o$14bo$13bo2$6b3o$8bo$7bo$11b3o$11bo$12bo! mniemiec wrote:The 5 improved pseudo-oscillator syntheses: x = 227, y = 167, rule = B3/S23199bo$198bo$154bobo41b3o$155boo$155bo5bo43bo$162boo33bo5boo$161boo32bobo6boo$157boo37boo$158boo$157bo48bo$122bobo80boo$118bo3boo56boo18boo3bobo$46bobo70boobbo56bo19bo20boo$47boo69boo61bobo13bo3bobo17bobo$47bo36bo113bo$85bo19bo19bo18boo18boo17bobo5boo3b3o4bobo17bobo$29boo18boo32b3o18bobo17bobo17bobo17bobo19bo5boo12bo19bo$29boo18boo54boo18boo18boo18boo18boo4bo4bo8boo18boo$196boo$25boo18boo18boo18boo18boo18boo18boo18boo18boo8bobo7boo18boo$26bo19bo19bo19bo19bo19bo19bo19bo19bo19bo19bo$23bobo17bobo17bobo17bobo17bobo17bobo17bobo17bobo17bobo17bobo17bobo$4bo17bo19bo19bo19bo19bo19bo19bo19bo19bo19bo19bo$5bobbobo11boo18boo18boo18boo18boo18boo18boo18boo18boo18boo18boo$3b3obboo7boo$9bo6boo$18bo$4boo$3boo$5bo12$179bo$148bobo29bo$148boo28b3o$143bo5bo$3bo137boo$bobo138boo33bo12bo$bboo142boo27bobobbobo5boo$12bobo130boo29boobboo7boo$12boo88bobo42bo33bo12bo$13bo88boo89boo$103bo59boo18boo8bobo$31boo18boo18boo18boo18boo18boo18boo11bo6boo11bo16boo9boo$31bobo17bobo17bobo17bobo10boo5bobo17bobo17bobo7bobo7bobo7bobo17bobo7bobo$60bo42boo$33bobo17bobobboo13bobo3bo13bobo3bo5bo7bobo3boo12bobo3boo12bobo3bobo11bobo3bobo7b3o11bobo3bobo$36bo19bobboo15bobobo15bobobo15bobobo15bobobo15bobo17bobo5boo3bo16bobo$35boo18boo18booboo15booboo15booboo15booboo15booboo15booboo3bobo4bo14booboo$3o182bo$bbo$bo13b3o$11boobbo$12boobbo$11bo$24boo$24bobo$24bo9$3bo$bobo$bboo$12bobo$12boo$13bo$63bo$31boo18boo11bo6boo18boo18boo18boo18boo18boo28boo$31bobo17bobo8b3o6bobo17bobo17bobo17bobo17bobo17bobo27bobo$$33bobo17bobo5b3o9bobo17bobo17bobo17bobo17bobo17bobo9bo17bobo36bo19bo4bo14boboo16boboo16boboo16boboo16boboo16boboo3bobo4bo15boboo35boo18boo5bo12boobobo14boobobo14boobobo14boobobo14boobo16boobo5boo3bo15boobo3o76bo19bo5bo13boo18boo18bobo17bobo7b3o17bobobbo56bo43boobo13b3o41boo43boo55bobo17bobo27bobo11boobbo42bobo103bo19bo27boo12boobbo86bo59boo18boo8bobo11bo90boo89boo24boo76bobo42bo33bo12bo24bobo118boo29boobboo7boo24bo121boo27bobobbobo5boo142boo33bo12bo141boo143bo5bo148boo28b3o148bobo29bo179bo8129bo128bo74bobo51b3o75boo75bo5bo53bo82boo43bo5boo81boo42bobo6boo77boo47boo78boo77bo58bo135boo38bo10boo49boo28boo3bobo6bo7bo24bo7bobobo48bo29bo20boo4bobo5boo23b3o5bobobo51bobo23bo3bobo17bobo5boo6boo31boo80bo41boo21boo18boo17bobo15boo3b3o4bobo17bobo5bo36boo20bobo17bobo19bo15boo12bo19bo5boo34bo23boo18boo18boo14bo4bo8boo18boo4bobo9boo11boo18boo18boo18boo18boo15boo11boo18boo16bobo6boo3bo14boo3bo14boo3bo14boo3bo14boo3bo14bobo7boo3bo14boo3bo16bo9bobbo16bobbo16bobbo16bobbo16bobbo26bobbo16bobbo26bobo17bobo17bobo17bobo17bobo27bobo17bobo27bo19bo19bo19bo19bo29bo19bo$$bbo$bboo$bobo4$172bo$173bo$135bobo33b3o$135boo$130bo5bo29bo$128boo37boo5bo$129boo35boo6bobo$133boo39boo$132boo$134bo30bo$87bobo75boo$44bo43boo3bo56boo12bobo3boo$6bo7bo27bobo43bobboo58bo19bo17boo$4bobo5boo29boobbo44boo54bobo17bobo3bo13bobo$5boo6boo31bo126bo$46b3o17bo19bo19boo18boo18bobo17bobo4b3o3boo5bobo$5bo35b3o21bobo17bobo17bobo17bobo17bo19bo12boo5bo$5boo36bo21boo18boo18boo18boo18boo18boo8bo4bo4boo$4bobo9boo11boo11bo6boo18boo18boo18boo18boo18boo18boo3boo13boo$16bobo6boo3bo14boo3bo14boo3bo14boo3bo14boo3bo14boo3bo14boo3bo14boo3bo3bobo8boo3bo$16bo9bobbo16bobbo16bobbo16bobbo16bobbo16bobbo16bobbo16bobbo16bobbo$26bobo17bobo17bobo17bobo17bobo17bobo17bobo17bobo17bobo$27bo19bo19bo19bo19bo19bo19bo19bo3boo14bo$172boo$bbo168bo4boo$bboo171boo$bobo173bo! Reduced the fourth by 1 (trivially): x = 84, y = 13, rule = B3/S2375bobo$bo10b2o57bo3b2o$2bo7bobobo57b2o2bo$3o5bobobo58b2o$9b2o$4b2o50bobo19bo$5b2o50b2o18bobo$4bo52bo20b2o$12b2o48b2o18b2o$8b2o3bo44b2o3bo14b2o3bo$9bo2bo46bo2bo16bo2bo$9bobo47bobo17bobo$10bo49bo19bo! Iteration of sigma(n)+tau(n)-n [sigma(n)+tau(n)-n : OEIS A163163] (e.g. 16,20,28,34,24,44,46,30,50,49,11,3,3, ...) : 965808 is period 336 (max = 207085118608). AbhpzTa Posts: 313 Joined: April 13th, 2016, 9:40 am Location: Ishikawa Prefecture, Japan PreviousNext Return to Patterns ### Who is online Users browsing this forum: Google [Bot] and 12 guests	{}
11 views int a=5; a= --a+a++; Then value of a is ? \| 11 views 0 Is it C? If yes, then invalid question. It's undefined behaviour. If no, and it's abstract example then, + Is left associative, so, --a should be evaluated first, which will set $a=4$ and this 4 will be used, now for the right part, a++ will be evaluated so, a becomes 5 but old value is used so, $a=4$ for right hand side of the +. So, it becomes $a=4+4;$ So, $a =8$.	{}
## Osaka Journal of Mathematics ### On the classifications of unitary matrices #### Abstract We classify the dynamical action of matrices in $\mathrm{SU}(p, q)$ using the coefficients of their characteristic polynomial. This generalises earlier work of Goldman for $\mathrm{SU}(2, 1)$ and the classical result for $\mathrm{SU}(1, 1)$, which is conjugate to $\mathrm{SL}(2, \mathbb{R})$. As geometrical applications, we show how this enables us to classify automorphisms of real and complex hyperbolic space and anti de Sitter space. #### Article information Source Osaka J. Math., Volume 52, Number 4 (2015), 959-993. Dates First available in Project Euclid: 18 November 2015 https://projecteuclid.org/euclid.ojm/1447856028 Mathematical Reviews number (MathSciNet) MR3426624 Zentralblatt MATH identifier 1336.51006 Subjects Secondary: 20H20: Other matrix groups over fields #### Citation Gongopadhyay, Krishnendu; Parker, John R.; Parsad, Shiv. On the classifications of unitary matrices. Osaka J. Math. 52 (2015), no. 4, 959--993. https://projecteuclid.org/euclid.ojm/1447856028 #### References • P. Anglès: Conformal Groups in Geometry and Spin Structures, Progress in Mathematical Physics 50, Birkhäuser Boston, Boston, MA, 2008. • A.F. Beardon: The Geometry of Discrete Groups, Springer, New York, 1983. • A. Cano, J.P. Navarrete and J. Seade: Complex Kleinian Groups, Birkhäuser/Springer Basel AG, Basel, 2013. • W. Cao, J.R. Parker and X. Wang: On the classification of quaternionic Möbius transformations, Math. Proc. Cambridge Philos. Soc. 137 (2004), 349–361. • D. Chillingworth: The ubiquitous astroid; in The Physics of Structure Formation (Tübingen, 1986), Springer, Berlin, 1987, 372–386. • M.P. do Carmo: Differential Geometry of Curves and Surfaces, Prentice Hall, Englewood Cliffs, NJ, 1976. • W.M. Goldman: Trace coordinates on Fricke spaces of some simple hyperbolic surfaces; in Handbook of Teichmüller Theory, II, IRMA Lect. Math. Theor. Phys., 13, Eur. Math. Soc., Zürich, 2009, 611–684. • W.M. Goldman: Complex Hyperbolic Geometry, Oxford Univ. Press, Oxford, 1999. • W.M. Goldman: Crooked surfaces and anti-de Sitter geometry, Geom. Dedicata 175 (2015), 159–187. • W.M. Goldman and J.R. Parker: Complex hyperbolic ideal triangle groups, J. Reine Angew. Math. 425 (1992), 71–86. • K. Gongopadhyay: Algebraic characterization of the isometries of the hyperbolic 5-space, Geom. Dedicata 144 (2010), 157–170. • K. Gongopadhyay: Algebraic characterization of isometries of the complex and the quaternionic hyperbolic $3$-spaces, Proc. Amer. Math. Soc. 141 (2013), 1017–1027. • K. Gongopadhyay and J.R. Parker: Reversible complex hyperbolic isometries, Linear Algebra Appl. 438 (2013), 2728–2739. • F. Kirwan: Complex Algebraic Curves, London Mathematical Society Student Texts 23, Cambridge Univ. Press, Cambridge, 1992. • G. Mess: Lorentz spacetimes of constant curvature, Geom. Dedicata 126 (2007), 3–45. • J.-P. Navarrete: The trace function and complex Kleinian groups in $\mathbb{P}^{2}_{\mathbb{C}}$, Internat. J. Math. 19 (2008), 865–890. • J.R. Parker: Hyperbolic Spaces, Jyväskylä Lectures in Mathematics 2, 2008. • J.R. Parker: Traces in complex hyperbolic geometry; in Geometry, Topology and Dynamics of Character Varieties, Lect. Notes Ser. Inst. Math. Sci. Natl. Univ. Singap. 23, World Sci. Publ., Hackensack, NJ, 2012, 191–245. • \begingroup J.R. Parker and I. Short: Conjugacy classification of quaternionic Möbius transformations, Comput. Methods Funct. Theory 9 (2009), 13–25. \endgroup • J.R. Parker and P. Will: Complex hyperbolic free groups with many parabolic elements; in Geometry, Groups and Dynamics, Contemp. Math. 639, Amer. Math. Soc., Providence RI, 2015, 327–348. • T. Poston and I. Stewart: The cross-ratio foliation of binary quartic forms, Geom. Dedicata 27 (1988), 263–280.	{}
# Mean calculator ## Introduction to Mean Median Mode Mean median mode helps to simplify large sets of numbers by calculating the center values. In general mathematics and statistics, range and standard deviation is used to examine the variability of data. Mean median mode help us to measure the average values. You can use mean calculator for finding mean meadian mode range or use mean tutorial to learn how it works. ## What is the Mean? In mathematics, the mean is the average value of a set of numbers. Mean is also known as arithmetic mean but it is different from arithmetic sequence and midpoint of a line segment. ## How to find the Mean? Mean is the sum of all of the given values in a dataset divided by the number of values. The mean formula is written as $$\bbox[#f6f6f6,10px]{\text{Mean}\;=\;\frac{\text{Sum of data}}{\text{Numbers Values}}\;\;}$$ The mathematical form of mean formula is $$\bbox[#f6f6f6,10px]{\text{Mean}\;=\;\frac{\sum X_i}{n}}$$ Where, Xi points at the values in the dataset and n is the total number of values. Learn how to find permutations and combinations? and what are significant figures operations on our portal. ## What is Median? Median is the middle value of a given dataset. If the values are in even number, then we'll take the mean of middle 2 values to get median value. ## How to find the Median? As the median is the middle value, so half of the values in a dataset will be smaller than median and other half will be larger than median. There are some rules to find the median: • Arrange the values from smallest to largest. • If the number of given values is odd, the median is the middle value in the list. • If the number of given values is even, the median is the average of the two middle values in the list. ## Example: Find the median of this dataset: 111, 444, 222, 555, 000 Arrange the data in order first: 000, 111, 222, 444, 555 There is an odd number of values, so the median is the middle data points. 000, 111, 222, 444, 555 The median is 222. You can use median calculator to calculate median online. Stay on our portal to find distance between two points and how to calculate covariance?. ## What is Mode? The most appearing number in a given dataset is known as the mode. A dataset can have 1 mode or more than 1 mode or a dataset can be without a node. A distribution with 2 modes is called bimodal and with 3 modes is named trimodal. ## How to find the Mode? Let us suppose there are many sticks inside a room and we want to know the mode value. To find mode Find the mode of the dataset: 00, 00, 11, 11, 11, 11, 11,11, 22, 22, 22, 33, 55 Look for the value that occurs the most: 00, 00, 11, 11, 11, 11, 11, 11, 22, 22, 22, 33, 55 The mode is 11 sticks because it repeats the most. ## What is Range? The difference between the smallest and largest number in a dataset is known as the range. The average of smallest and largest number is known as the midrange. Range calculator is best option to get range online. ## How to find Range? Sir Steve took 7 math tests in one marking period. What is the range of his test marks? 89, 73, 84, 91, 87, 77, 97 Arrange the test marks from least to greatest, we get: 73, 77, 84, 87, 89, 91, 97 Highest - Lowest = 97 - 73 = 24 The range of test marks is 24. Our website also helps you to know how to round off numbers? and how to solve division problems using remainder calculator. ## What is Mean Calculator? Just like other math topics, calculating mean manually can be difficult as well as the chances of doing error are higher. Calculatored has developed an online solution to calculate mean median mode and range. Mean calculators will save your time and provide accurate results. Try our combination calculator and gcf calculator for your learning and practice regarding combinations and common factors. ## How to use Mean Calculator? The mean calculator is very easy to use you have to enter your values in the input field of our mean median mode calculator. Once you entered your values, click on "CALCULATE" button and our mean calculator will show you mean median mode range in seperate tabs. We hope our mean calculator helped you in order to clear your concepts and calculations. You can also use our factor calculator and factorial calculator online for free. Also provide your valuable feedback so that we could improve further.	{}
## INTRODUCTION ### Calculation of Oxygen Consumption Calculation of oxygen consumption is a straightforward process that involves subtracting the amount of oxygen exhaled from the amount of oxygen inhaled: $(1)Oxygen consumption(V˙O2)=[volume ofO2inspired]−[volume ofO2expired]$ The volume of O2 inspired (I) is computed by multiplying the volume of air inhaled per minute ($V˙I$) by the fraction (F) of air that is made up of oxygen. Room air is 20.93% O2. Expressed as a fraction, 20.93% becomes .2093 and is symbolized as FIO2. When we exhale, the fraction of O2 is lowered (i.e., O2 diffuses from the lung to the blood) and the fraction of O2 in the expired (E) gas is represented by FEO2. The volume of expired O2 is the product of the volume of expired gas ($V˙E$) and FEO2. Equation (1) can now be symbolized as: $(2)V˙O2=(V˙I⋅FIO2)−(V˙E⋅FEO2)$ The exercise values for FIO2, FEO2, $V˙I$, and $V˙E$ for a subject are easily measured in most exercise physiology laboratories. In practice, FIO2 is not generally measured but is assumed to be a constant value of .2093 if the subject is breathing room air. FEO2 will be determined by a gas analyzer, and $V˙I$ and $V˙E$ can be measured by a number of different laboratory devices capable of measuring airflow. Note that it is not necessary to measure both $V˙I$ and $V˙E$. This is true because if $V˙I$ is measured, $V˙E$ can be calculated (and vice versa). The formula used to calculate $V˙E$ from the measurement of $V˙I$ is called the “Haldane transformation” and is based on the fact that nitrogen (N2) is neither used nor produced in the body. Therefore, the volume of N2 inhaled must equal the volume of N2 exhaled: $(3)[V˙I⋅FIN2]=[V˙E⋅FEN2]$ Therefore, $V˙I$ can be computed if $V˙E$, FIO2, and FEO2 are known. For example, to solve for $V˙I$: $(4)V˙1=(V˙E⋅FEN2)F1N2$ ... ### Pop-up div Successfully Displayed This div only appears when the trigger link is hovered over. Otherwise it is hidden from view.	{}
# How to add state to the custom inspector in Unity I'm writing a custom inspector for my asset (by deriving from Editor), and I have a few variables controlling the state of the inspector (like whether a foldout is opened or not), that I'd like to be saved. I don't want to put them on the object, since then it would end up as junk inside the game state, but I also don't want it to be reset every time I click away from the object and then click back. Do you guys have an idea where could I put the data so it gets persisted? • I think if you just make it a member variable of your editor class, it will only reset the variables when scripts recompile. Which many unity standard editors do anyways. This also keeps editor code out of your class, which is ideal. However if you really want data to persist even between compilations, then you should go with DMGregory's answer – gjh33 Apr 4 at 20:10 I'd recommend putting this type of state information inside an #if UNITY_EDITOR conditional compilation block inside the type you're editing. That way it's stored in memory only in the editor, and stripped out of the built game executable by the preprocessor.	{}
# Simplify expressions by automated substitution of repeated terms I would like to know whether there is a way in Mathematica to simplify an expression including the introduction of symbols for repeated terms, like so: SimplifyWithSubstitution[(a+b)^7+(a+b)^3+(a+b)^2-a-b] should yield C^7+C^3+C^2-C, C=a+b Note that I do want Mathematica to figure out that the expression can be shortened by substituting a+b. • Shouldn't that be - a - b in the last part? Anyway: Eliminate[{res == (a + b)^7 + (a + b)^3 + (a + b)^2 - a - b, c == a + b}, {a, b}] – J. M. is in limbo Nov 14 '17 at 13:09 • Of course there was a sign error, thanks! Your proposal requires the user to recognize that a+b is repeated and may be replaced, I want MMA to do this. – Rainer Glüge Nov 14 '17 at 13:42 • ExperimentalOptimizeExpression[] gets pretty close: ExperimentalOptimizeExpression[(a + b)^7 + (a + b)^3 + (a + b)^2 - a - b] – J. M. is in limbo Nov 14 '17 at 13:44 • Well thanks, thats what I needed! – Rainer Glüge Nov 14 '17 at 14:19	{}
# Tikz diagram with stacks and box I'm trying to make the diagram in Tikz. I can draw the stacks by adapting this example and the box. But how can I put the description around the boxes. Also how to add the two arrows between the boxes? - Give a name to the main node, for example (a), then you can access the anchors with \node[above = 2mm of a.one] {description}; which puts the node on top of the first box of the main node (a) – percusse Apr 21 '12 at 1:12 You can use the label option for already existing nodes or you can use new \nodes; in the example below I used both this approaches: \documentclass{article} \usepackage{tikz} \usetikzlibrary{shapes,positioning,arrows,calc} \begin{document} \begin{tikzpicture}[stack/.style={ rectangle split, rectangle split parts=5, draw, anchor=center}, myarrow/.style={single arrow, draw=none}] \node [stack] (ini) {$a=0$\nodepart{two}$b=10$% \nodepart{three}$c=100$\nodepart{four}$d=-10$\nodepart{five}$\cdots$}; \node [draw,rectangle,align=left,right=of ini,label=above:{Computer Program}] (mid) {instruction 0;\\ instruction 1;\\$\ldots$\\instruction $n$;}; \node [stack,right=of mid] (fin) {$a=10$\nodepart{two}$b=100$% \nodepart{three}$c=-10$\nodepart{four}$d=110$\nodepart{five}$\cdots$}; \node [above=of ini,anchor=north,align=left] {Initial values of\\variables}; \node [above=of fin,anchor=north,align=left] {Final values of\\variables}; \node [myarrow,draw,anchor=west] at ($(ini.east)+(2.5pt,0)$) {\phantom{te}} ; \node [myarrow,draw,anchor=west] at ($(mid.east)+(2.5pt,0)$) {\phantom{te}} ; \end{tikzpicture} \end{document} A little additional example just to illustrate the two mentioned methods to place labels next to nodes (the second example requires the positioning library): \documentclass{article} \usepackage{tikz} \usetikzlibrary{positioning} \begin{document} \begin{tikzpicture}[mynode/.style={draw, anchor=center}] \node [mynode,label={label}] at (0,0) {test node}; \node [mynode,label=below:{label}] at (0,-2) {test node}; \node [mynode,label=west:{label}] at (0,-4) {test node}; \node [mynode,label=east:{label}] at (0,-6) {test node}; \node [mynode,label=east:{label1},label=west:{label2}] at (0,-8) {test node}; \node [mynode] at (6,0) (a) {test node}; \node [above=2mm of a] {label1}; \node [below right=2mm of a] {label2}; \node [below left=4mm of a] {label3}; \end{tikzpicture} \end{document} - I made some modification from Gonzalo's answer. I remove two libraries calc and positioning and I modified some styles. I changed the order of the creation of the nodes. (stack) -> (Arrow)-> (stack) -> (Arrow)-> (stack) \documentclass{article} \usepackage{tikz} \usetikzlibrary{shapes,arrows} \begin{document} \begin{tikzpicture}[every text node part/.style={align=left}, stack/.style={rectangle split, rectangle split parts = 5, draw,text width=1.5cm}, myarrow/.style={single arrow, draw, right = 3pt, minimum size = 5ex} ] % we start \node [stack] (ini) {% $a=0$ \nodepart{two} $b=10$ \nodepart{three} $c=100$ \nodepart{four} $d=-10$ \nodepart{five} $\cdots$}; \node [myarrow] (A) at (ini.east) {} ; % after the arrow, the main node at the end \node [draw,rectangle,align=left,right=3pt] (mid) at (A.east) {instruction 0;\\ instruction 1;\\ $\ldots$\\ instruction $n$;}; \node [myarrow] (B) at (mid.east) {} ; % after the arrow, the final node \node [stack,right=3pt] (fin) at (B.east) {% $a=10$ \nodepart{two} $b=100$ \nodepart{three} $c=-10$ \nodepart{four} $d=110$ \nodepart{five} $\cdots$}; % labels : I don't like vey much label= it's a shortcut but I think it's more powerful and flexible to use real nodes. \node [above,align=left] at (ini.north) {Initial values of\\ variables}; \node [above,align=left] at (fin.north) {Final values of\\ variables}; \node [above] at (mid.north) {Computer Program}; \end{tikzpicture} \end{document} - Thanks, I think something must be missing (maybe a library), I can't see the figure when I convert to pdf. Any idea? – web2013 Apr 21 '12 at 10:47 The code compiles fine but I work with pgf 2.1 cvs. What kind of error do you get? I made a try with pgf 2.1 official and it's fine. I think you need to update your distribution – Alain Matthes Apr 21 '12 at 10:53 Just updated MikTex to 2.9, I don't see any errors but I can't see the figure on my pdf. Anyways after the update, I got the example of Gonzalo's working, so it's Ok :) – web2013 Apr 21 '12 at 11:47 But it's a problem. You compile with what ? – Alain Matthes Apr 21 '12 at 11:51 OK I got it working by removing all the comments and '%' which were in the middle of latex code like in \node [stack] (ini) {% Btw I'm using Lyx 2.0.3 with MikTex 2.9 on Windows 7 – web2013 Apr 21 '12 at 12:13	{}
# Solved – R: Problem with runif: generated number repeats (more often than expected) after less than 100 000 steps After executing the code ``RNGkind(kind="Mersenne-Twister") # the default anyway set.seed(123) n = 10^5 x = runif(n) print(x[22662] == x[97974]) `` `TRUE` is output! If I use, e.g., `RNGkind(kind="Knuth-TAOCP-2002")` similarly happens: I get "only" 99 995 different values in `x`. Given the periods of both random generators, the results seem highly unlikely. Am I doing something wrong? I need to generate at least one million random numbers. I am using Windows 8.1 with R version 3.6.2; Platform: x86_64-w64-mingw32/x64 (64-bit) and RStudio 1.2.5033. 1. Having a bag with $$n$$ different balls, we choose a ball $$m$$ times and put it back every time. The probability $$p_{n, m}$$ that all chosen balls are different is equal to $${nchoose m} / (n^m m!)$$. 2. R documentation points to a link where the implementation of Mersenne-Twister for 64-bit machines is available: http://www.math.sci.hiroshima-u.ac.jp/~m-mat/MT/emt64.html The uniform sampling from $$[0, 1]$$ interval is obtained via choosing a random 64-bit integer first, so I computed the above probabilities for the 64-bit and (when $$p_{2^{64}, 10^5}$$ turned out to be rather low) 32-bit case: $$p_{2^{64}, 10^5}doteq 0.9999999999972… qquad p_{2^{32}, 10^5} doteq 0.3121…$$ Then, I tried 1000 random seeds and compute the proportion of the cases when all generated numbers are different: 0.303. So, currently, I assume that for some reason, 32-bit integers are actually used. Contents The documentation of R on random number generation has a few sentences at its end, that confirm your expectation of 32-bit integers being used and might explain what you are observing: Do not rely on randomness of low-order bits from RNGs. Most of the supplied uniform generators return 32-bit integer values that are converted to doubles, so they take at most 2^32 distinct values and long runs will return duplicated values (Wichmann-Hill is the exception, and all give at least 30 varying bits.) So the implementation in R seems to be different to what is explained on the website of the authors of the Mersenne Twister. Possibly combining this with the Birthday paradox, you would expect duplicates with only 2^16 numbers at a probability of 0.5, and 10^5 > 2^16. Trying the Wichmann-Hill algorithm as suggested in the documentation: ``RNGkind(kind="Wichmann-Hill") set.seed(123) n = 10^8 x = runif(n) length(unique(x)) # 1e8 `` Note that the original Wichmann-Hill random number generator has the property that its next number can be predicted by its previous, and therefore does not meet non-predictability requirements of a valid PRNG. See this document by Dutang and Wuertz, 2009 (section 3) Rate this post	{}
# write representation as sum of irreducible representations Given the representation $\rho: \mathbb{Z}/3\mathbb{Z} \rightarrow GL_2(\mathbb{C})$ by $1\rightarrow \left( \begin{array}{ccc} -1 & -1 \\ 1 & 0\\ \end{array} \right)$. I have to write this representation as a sum irreducible representation. I tried a lot to figure this out, but I just dont ´see´ it. I need help. Thanks. Since $\mathbb{Z}/3\mathbb{Z}$ is cyclic it is sufficient to diagonalize the matrix $\rho (1)$. • could you explain why it is sufficient to diagonalize $\rho(1)$? – Badshah May 10 '13 at 15:41 • Then the matrix $\rho(2)=\rho(1)^2$ is also diagonal and you obtain a decomposition into two 1-dimensional representations. – Boris Novikov May 10 '13 at 15:45	{}
Looking for Earth Science worksheets? Check out our pre-made Earth Science worksheets! Tweet ##### Browse Questions You can create printable tests and worksheets from these Grade 9 Earth Science questions! Select one or more questions using the checkboxes above each question. Then click the add selected questions to a test button before moving to another page. 1 2 3 4 ... 67 Which of the following is not a greenhouse gas? 1. water vapor 2. carbon dioxide 3. oxygen What is one reason that the Earth's core is so hot? 1. Coal and oil burn in the core. 2. As heavy elements in the core decay, they produce radioactive heat. 3. The Sun heats up the Earth and causes the inside to melt. Nitrogen is not a greenhouse gas. This is because nitrogen What is the main way that humans put excess carbon dioxide into the atmosphere? 1. spilling oil into the ocean 2. burning fossil fuels 3. creating carbonated beverages How do we use fossil fuels to generate electricity? 1. When we burn fossil fuels, smoke is created, and it releases electrical bolts. 2. When we burn fossil fuels, the smoke it generates moves turbines. 3. When we burn fossil fuels, the heat boils water, and the steam moves turbines. Rivers and lakes near large farms are heavily polluted. What is one reason for this? 1. excessive use of fertilizers on farms 2. decaying animal carcasses 3. trash dumping into rivers A geyser is 1. a hot spring that shoots columns of cold water and steam into the air every so often 2. a hot spring that shoots columns of hot water and steam into the air every so often 3. a hot spring that shoots columns of hot water and steam into the air all the time 4. a hot spring that shoots columns of hot water and steam into the inside of the earth every so often An aquifer is 1. A permeable layer of rock and sediment that can store and carry groudwater in enough quantity to supply wells. 2. A permeable layer of rock and sediment that can not store and carry groudwater in enough quantity to supply wells 3. A permeable layer of sand and sediment that can store and carry groudwater in enough quantity to supply wells 4. A non-permeable layer of rock and sediment that can store and carry groudwater in enough quantity to supply wells Stalactites and stalagmites are found in 1. caverns 2. glaciers 3. geysers 4. Karst topography An ice cap is 1. a glacier that is less than 50,000 square kilometers in area 2. the snow that covers a mountain in winter 3. the snow on top of mountains on Denali park 4. a glacier that is less than one billion square kilometers in area The albedo effect of glaciers is vital to the cooling of our atmosphere. The albedo effect is defined as: 1. the fraction of solar energy reflected from the Earth back into space 2. the cooling effect of ice on Earth's oceans 3. the absorption of carbon dioxide by Earth's glaciers A biggest source of carbon dioxide in the atmosphere is: 1. human respiration (exhaling $CO_2$) 2. transportation 3. volcanoes A cirque is 1. a semicircular basin 2. an oval basin 3. a peak left after glaciers erode mountains 4. only present in valley glaciers	{}
# orange Bitbucket is a code hosting site with unlimited public and private repositories. We're also free for small teams! Close # Orange Orange is a component-based data mining software. It includes a range of data visualization, exploration, preprocessing and modeling techniques. It can be used through a nice and intuitive user interface or, for more advanced users, as a module for Python programming language. ## Installing To build and install Orange run: python setup.py build python setup.py install from the command line. You can customize the build process by editing the setup-site.cfg file in this directory (see the comments in that file for instructions on how to do that). ## Running tests After Orange is installed, you can check if everything is working OK by running the included tests: python setup.py test This command runs all the unit tests and documentation examples. Some of the latter have additional dependencies you can satisfy by installing matplotlib, PIL and scipy. ## Starting Orange Canvas Start orange canvas from the command line with: orange-canvas ## Installation for Developers To install in development mode run: python setup.py develop # Recent activity Tip: Filter by directory path e.g. /media app.js to search for public/media/app.js. Tip: Use camelCasing e.g. ProjME to search for ProjectModifiedEvent.java. Tip: Filter by extension type e.g. /repo .js to search for all .js files in the /repo directory. Tip: Separate your search with spaces e.g. /ssh pom.xml to search for src/ssh/pom.xml. Tip: Use ↑ and ↓ arrow keys to navigate and return to view the file. Tip: You can also navigate files with Ctrl+j (next) and Ctrl+k (previous) and view the file with Ctrl+o. Tip: You can also navigate files with Alt+j (next) and Alt+k (previous) and view the file with Alt+o.	{}
Show interval (a,b) is uncountable? First find a bijection from(âˆ’pi/2,pi/2) to R . Then modify that oneto be a Question: Show interval (a,b) is uncountable? First find a bijection from (âˆ’pi/2,pi/2) to R . Then modify that one to be a bijection from (a,b) to R. Don't know the rest to show a bijection and onto Similar Solved Questions Accordng to this IR and NMR, what is the complete structure of the unknown compound present? Accordng to this IR and NMR, what is the complete structure of the unknown compound present?... Outstanding checks haven't been presented to the bank for payment but have been subtracted in the... Outstanding checks haven't been presented to the bank for payment but have been subtracted in the checkbook. have been subtracted on the bank records but not the checkbook records. have been returned to the business for nonpayment. haven't been presented to the bank for payment and haven'... A thin uniform rod (mass = 0.90 kg) swings about an axis that passes through one... A thin uniform rod (mass = 0.90 kg) swings about an axis that passes through one end of the rod and is perpendicular to the plane of the swing. The rod swings with a period of 1.2 s and an angular amplitude of 2.10. (a) What is the length of the rod? (a) What is the maximum kinetic energy of the rod... 2. A culture of bacteria quadruples every 8 minutes How long will it take a culture originally consisting of 36 bacteria to grow to a population of 200 000 bacteria? Show ALL work and the concluding sentence Round to one decimal place, if necessary: [A3] 2. A culture of bacteria quadruples every 8 minutes How long will it take a culture originally consisting of 36 bacteria to grow to a population of 200 000 bacteria? Show ALL work and the concluding sentence Round to one decimal place, if necessary: [A3]... In your own words, explain why employer-purchased health insurance results in more comprehensive health care coverage.... In your own words, explain why employer-purchased health insurance results in more comprehensive health care coverage. What are some of the arguments in favor of eliminating or placing a cap on the tax-exempt status of employer purchased health insurance? What are some of the consequences of greater... QuestionWhich of the following islare improper integral(s)? I: 634 IL: 6,54 IV. i2541 [Xd A) only B) L. HII; IV only C) L IV only D)L. II, IV only E) II, IV only Question Which of the following islare improper integral(s)? I: 634 IL: 6,54 IV. i254 1 [Xd A) only B) L. HII; IV only C) L IV only D)L. II, IV only E) II, IV only... A 45-year-old woman was shaking with chills and fever that began after she inhaled toxic fumes.... A 45-year-old woman was shaking with chills and fever that began after she inhaled toxic fumes. Two days later, she noticed a rash over her back and trunk. She was admitted to the hospital suffering from generalized body soreness, coughing, malaise, and fever. Her admission laboratory findings were:... E) DIV,000. None of the above 13. W Limited makes and sells a single product. The... e) DIV,000. None of the above 13. W Limited makes and sells a single product. The company employs a flexible budgeting system that covers a relevant range from 20,000 units to 25,000 units and a just in time inventory system. Budget data for April, based on 22,000 units, are as follows: Prime costs:... ' [I' 61 [[21 { 21 1 3 0 1 11 6 885{ ~ { 2 84 81â‚¬ 53 ? 15 1 1 0 ? 01A71 ' [I' 6 1 [[ 2 1 { 2 1 1 3 0 1 1 1 6 885{ ~ { 2 8 4 8 1 â‚¬ 5 3 ? 1 5 1 1 0 ? 01A 7 1... The receivable that is usually evidenced by a formal, written instrument of credit is a(n) a.trade receivable b.note rec... 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(3 Evaluate f cos(z)sin(x) dx_ Evaluate J 3r2(13 + 1)2 dx.... ... 6= & OV S e4M 6= & ^V s JB4M (q i= & VV SI JB1M[18] -a[8 1] -v 6= & OV S e4M 6= & ^V s JB4M (q i= & VV SI JB1M [18] -a[8 1] -v... 1. (10 pts) Does the following series converge O diverge? Explain Your reasoning: 1.3 1.3.5 _1.3. 5 . 7 1.3.5 . (2n - 1) 1 _ + +.+ (~1)"-1 31 5! (2n - 1) 1. (10 pts) Does the following series converge O diverge? Explain Your reasoning: 1.3 1.3.5 _1.3. 5 . 7 1.3.5 . (2n - 1) 1 _ + +.+ (~1)"-1 31 5! (2n - 1)... Convert to scientific notation (here you may use x as multiplication): 0.000347 Convert to scientific notation (here you may use x as multiplication): 0.000347...	{}
# Construct an equilateral triangle, Question. Construct an equilateral triangle, given its side and justify the construction Solution: Let us draw an equilateral triangle of side 5 cm. We know that all sides of an equilateral triangle are equal. Therefore, all sides of the equilateral triangle will be 5 cm. We also know that each angle of an equilateral triangle is 60º. The below given steps will be followed to draw an equilateral triangle of 5 cm side. Step I: Draw a line segment AB of 5 cm length. Draw an arc of some radius, while taking A as its centre. Let it intersect AB at P. Step II: Taking P as centre, draw an arc to intersect the previous arc at E. Join AE. Step III: Taking A as centre, draw an arc of 5 cm radius, which intersects extended line segment AE at C. Join AC and BC. ΔABC is the required equilateral triangle of side 5 cm. Justification of Construction: We can justify the construction by showing $A B C$ as an equilateral triangle i.e., $A B=B C=A C=5 \mathrm{~cm}$ and $\angle A=\angle B=\angle C=60^{\circ}$. In $\triangle A B C$, we have $A C=A B=5 \mathrm{~cm}$ and $\angle A=60^{\circ}$. Since $A C=A B$ $\angle B=\angle C$ (Angles opposite to equal sides of a triangle) In $\triangle \mathrm{ABC}$, $\angle A+\angle B+\angle C=180^{\circ}$ (Angle sum property of a triangle) $\Rightarrow 60^{\circ}+\angle C+\angle C=180^{\circ}$ $\Rightarrow 2 \angle C=180^{\circ}-60^{\circ}=120^{\circ}$ $\Rightarrow \angle C=60^{\circ}$ $\therefore \angle B=\angle C=60^{\circ}$ We have, $\angle A=\angle B=\angle C=60^{\circ} \ldots$ (1) $\Rightarrow \angle A=\angle B$ and $\angle A=\angle C$ $B C=A C$ and $B C=A B$ (Sides opposite to equal angles of a triangle) $\Rightarrow A B=B C=A C=5 \mathrm{~cm} \ldots(2)$ From equations (1) and (2), ΔABC is an equilateral triangle.	{}
Boundary conditions and Schwarz waveform relaxation method for linear viscous Shallow Water equations in hydrodynamics The SMAI journal of computational mathematics, Volume 3 (2017), pp. 117-137. We propose in the present work an extension of the Schwarz waveform relaxation method to the case of viscous shallow water system with advection term. We first show the difficulties that arise when approximating the Dirichlet to Neumann operators if we consider an asymptotic analysis based on large Reynolds number regime and a small domain aspect ratio. Therefore we focus on the design of a Schwarz algorithm with Robin like boundary conditions. We prove the well-posedness and the convergence of the algorithm. Published online: DOI: 10.5802/smai-jcm.22 Classification: 65M55 Keywords: Schwarz waveform relaxation, shallow water equations, domain decomposition, absorbing operators @article{SMAI-JCM_2017__3__117_0, author = {Eric Blayo and Antoine Rousseau and Manel Tayachi}, title = {Boundary conditions and {Schwarz} waveform relaxation method for linear viscous {Shallow} {Water} equations in hydrodynamics}, journal = {The SMAI journal of computational mathematics}, pages = {117--137}, publisher = {Soci\'et\'e de Math\'ematiques Appliqu\'ees et Industrielles}, volume = {3}, year = {2017}, doi = {10.5802/smai-jcm.22}, language = {en}, url = {https://smai-jcm.centre-mersenne.org/articles/10.5802/smai-jcm.22/} } TY - JOUR TI - Boundary conditions and Schwarz waveform relaxation method for linear viscous Shallow Water equations in hydrodynamics JO - The SMAI journal of computational mathematics PY - 2017 DA - 2017/// SP - 117 EP - 137 VL - 3 PB - Société de Mathématiques Appliquées et Industrielles UR - https://smai-jcm.centre-mersenne.org/articles/10.5802/smai-jcm.22/ UR - https://doi.org/10.5802/smai-jcm.22 DO - 10.5802/smai-jcm.22 LA - en ID - SMAI-JCM_2017__3__117_0 ER - %0 Journal Article %T Boundary conditions and Schwarz waveform relaxation method for linear viscous Shallow Water equations in hydrodynamics %J The SMAI journal of computational mathematics %D 2017 %P 117-137 %V 3 %I Société de Mathématiques Appliquées et Industrielles %U https://doi.org/10.5802/smai-jcm.22 %R 10.5802/smai-jcm.22 %G en %F SMAI-JCM_2017__3__117_0 Eric Blayo; Antoine Rousseau; Manel Tayachi. Boundary conditions and Schwarz waveform relaxation method for linear viscous Shallow Water equations in hydrodynamics. The SMAI journal of computational mathematics, Volume 3 (2017), pp. 117-137. doi : 10.5802/smai-jcm.22. https://smai-jcm.centre-mersenne.org/articles/10.5802/smai-jcm.22/ [1] E. Audusse; P. Dreyfuss; B. Merlet Schwarz wave form relaxation for primitive equations of the ocean, SIAM J. Sci. Comput., Volume 32 (201) no. 5, pp. 2908-2936 \| Article \| Zbl: 1230.76011 [2] E. Blayo; D. Cherel; A. Rousseau Towards optimized Schwarz methods for the Navier-Stokes equations, J. Sci. Comput., Volume 66 (2016), pp. 275-295 \| Article \| MR: 3440281 \| Zbl: 1381.76254 [3] B. Engquist; A. Majda Absorbing boundary conditions for the numerical simulation of waves, Math. Comput., Volume 31 (1977), pp. 245-267 \| Article \| MR: 436612 \| Zbl: 0367.65051 [4] M. J. Gander Optimized Schwarz methods, SIAM Journal on Numerical Analysis, Volume 44 (2006) no. 2, pp. 699-731 \| Article \| MR: 2218966 \| Zbl: 1117.65165 [5] M. J. Gander; L. Halpern Optimized Schwarz Waveform Relaxation Methods for Advection Reaction Diffusion Problems, SIAM Journal on Numerical Analysis, Volume 45 (2007) no. 2, pp. 666-697 \| Article \| MR: 2300292 \| Zbl: 1140.65063 [6] M.J. Gander Schwarz methods over the course of time, Electron. Trans. Numer. Anal., Volume 31 (2008), pp. 228-255 \| MR: 2569603 [7] M.J. Gander; L Halpern; F. Nataf Optimal convergence for overlapping and non-overlapping Schwarz waveform relaxation, Eleventh International Conference on Domain Decomposition Methods (London, 1998) (1999), pp. 27-36 [8] J.-F. Gerbeau; B. Perthame Derivation of Viscous Saint-Venant System for Laminar Shallow Water; Numerical Validation, Discrete and Continuous Dynamical Systems - Series B, Volume 1 (2001) no. 1, pp. 89-102 \| Article \| MR: 1821555 \| Zbl: 0997.76023 [9] L. Halpern Artificial boundary conditions for incompletely parabolic perturbations of hyperbolic systems, SIAM J. Math. Anal., Volume 22 (1991) no. 5, pp. 1256-1283 \| Article \| MR: 1112507 \| Zbl: 0772.35003 [10] C. Japhet; F. Nataf The best interface conditions for domain decomposition methods: absorbing boundary conditions, Absorbing Boundaries and Layers, Domain Decomposition Methods. Applications to Large Scale Computations (L. Tourrette; L. Halpern, eds.), Nova Science Publishers, 2003, pp. 348-373 [11] P.L. Lions On the Schwarz alternating method. III. A variant for nonoverlapping subdomains, Third International Symposium on Domain Decomposition Methods for Partial Differential Equations (1990), pp. 202-223 \| Zbl: 0704.65090 [12] V. Martin An optimized Schwarz waveform relaxation method for the unsteady convection diffusion equation in two dimensions, Computers $&$ Fluids, Volume 33 (2004) no. 5–6, pp. 829 -837 (Applied Mathematics for Industrial Flow Problems) \| Article \| MR: 2052024 \| Zbl: 1100.76552 [13] V. Martin Schwarz waveform relaxation algorithms for the linear viscous equatorial shallow water equations, SIAM J. Sci. Comput., Volume 31 (2009) no. 5, pp. 3595-3625 \| Article \| MR: 2556554 \| Zbl: 1391.65166 [14] L. Müller; G. Lube A nonoverlapping domain decomposition method for the nonstationary Navier-Stokes problem, ZAMM J. Appl. Math. Mech., Volume 81 (2001), p. 725-726 \| Article [15] F.C. Otto; G. Lube A nonoverlapping domain decomposition method for the Oseen equations, Math. Models Methods Appl. Sci., Volume 8 (1998), pp. 1091-1117 \| Article \| MR: 1646527 \| Zbl: 0939.65137 [16] L.F. Pavarino; O.B. Widlund Balancing Neumann-Neumann methods for incompressible Stokes equations, Comm. Pure Appl. Math., Volume 55 (2002), pp. 302-335 \| Article \| MR: 1866366 [17] Alfio Quarteroni; Alberto Valli Domain decomposition methods for partial differential equations, Numerical mathematics and scientific computation, Clarendon Press, Oxford, New York, 1999 \| Zbl: 0931.65118 [18] J.C. Strikwerda; C.D. Scarbnick A domain decomposition method for incompressible flow, SIAM J. Sci. Comput., Volume 14 (1993), pp. 49-67 \| Article \| MR: 1201310 [19] Linda Sundbye Global Existence for the Cauchy Problem for the Viscous Shallow Water Equations, Rocky Mountain J. Math., Volume 28 (1998) no. 3, pp. 1135-1152 \| Article \| MR: 1657060 \| Zbl: 0928.35129 [20] M. Tayachi Couplage de modèles de dimensions hétérogènes et application en hydrodynamique (2013) (Ph. D. Thesis) [21] M. Tayachi; A. Rousseau; E. Blayo; N. Goutal; V. Martin Design and analysis of a Schwarz coupling method for a dimensionally heterogeneous problem, Int. J. Num. Meth. Fluids, Volume 75 (2014), pp. 446-465 \| Article \| MR: 3214332 [22] A. Toselli; O. Widlund Domain decomposition methods - Algorithms and theory, Springer, Berlin-Heidelberg, 2005 \| Article \| Zbl: 1069.65138 [23] X. Xu; C.O. Chow; Lui S.H. On non overlapping domain decomposition methods for the incompressible Navier-Stokes equations, ESAIM Math. Mod. Num. Anal., Volume 39 (2005), pp. 1251-1269 \| Article Cited by Sources:	{}
# Young's Experiment - (problem with equation) 1. Sep 16, 2006 So my book has the following expression for Young's double slit experiment. (maxima - bright fringes) $d \sin \theta = m \lambda$ for $m = 0, \, 1 \, 2 \, \ldots$. So what if you solve this for wavelength. $$\lambda = \frac{d \sin \theta}{m}$$ How is this valid when $m = 0$ Is this because if $m = 0$, $\theta$ HAS to equal 0? by the way d is the distance between the slits theta is the angle from the central axis lambda is the wavelength m is the index of where the maxima occur 2. Sep 16, 2006 ### Andrew Mason The first maximum is always at 0 angle regardless of wavelength. So you can't determine the wavelength from the position of the central maximum. AM	{}
# Draw a candlestick chart with mplfinance and Plotly ### Background This post is a memo when I touched mplfinance and Plotly for graph drawing in Python. I tried creating a candlestick chart using a pandas DataFrame from a csv file of dollar-yen data in Python. When I googled how to do, I found that it could be easily drawn with mplfinance or Plotly, so I tried these, I was particularly surprised at the functionality of Plotly. First of all, when I was looking for something that could be drawn with matplotlib, I found out that there was something called mplfinance which is an update of the old version of mpl-finance (only the hyphen is different …) and seems to be easier to handle than mpl-finance. Also, I think I can draw a nice graph using Plotly (official site). From version 4, it seems that it is easier to use because it can be used for free (MIT license) without declaring the offline mode (https://plotly.com/python/is-plotly-free/) By the way, it seems that the functions of the former online mode have been separated and transferred to chart-studio. There are other similar tools such as seaborn and Bokeh, but I didn’t touch them because I felt it was a little difficult for me as a beginner. Related article regarding a web app made with Plotly (Japanese)；https://vucavucalife.com/usdjpy-past-one-minute-chart2/ ### Procedure #### Data preparation I downloaded the csv file of the past dollar-yen data from here. Dollar-yen 1-minute data (Period 2001.1.2-2020.3.31. Missing data) Price; bid Time; GMT (UTC + 0000). The amount of data is too large for a lesson, so I extracted only 20 minutes data from 00:00 on September 16, 2008 and used it as 20080916.csv. First, load the csv file with DASK. At that time, converted the date and time to datetime type and set it to an index (Datetimeindex) as “hiduke”. Then I converted GMT to JST. #　import pandas import pandas as pd # <DTYYYYMMDD> and <TIME> are collectively converted to datetime type with index as "hiduke". #read a file with dask > Indexing > Convert to Panda Dataframe df = ddf.read_csv("20080916.csv", sep="," , parse_dates={'hiduke':['<DTYYYYMMDD>', '<TIME>']}) df = ddf.read_csv("20080916.csv", sep="," , parse_dates={'hiduke':['<DTYYYYMMDD>', '<TIME>']}).set_index('hiduke') df = df.compute() #converted GMT（UTC + 0000） to JST(UTC + 0900) df.index = df.index.tz_localize('UTC') df.index = df.index.tz_convert('Asia/Tokyo') #Since the TICKER and VOL columns are not used, drop them. Then complete data shaping. df.drop(['<TICKER>', '<VOL>'], axis=1, inplace=True) #Check the first 5 lines print(df.head(5)) <OPEN> <HIGH> <LOW> <CLOSE> hiduke 2008-09-16 09:00:00+09:00 104.17 104.21 104.14 104.21 2008-09-16 09:01:00+09:00 104.20 104.22 104.18 104.22 2008-09-16 09:02:00+09:00 104.21 104.25 104.21 104.22 2008-09-16 09:03:00+09:00 104.23 104.31 104.22 104.27 2008-09-16 09:04:00+09:00 104.25 104.29 104.25 104.29 So, now ready to use DataFrame as df. #### Install mplfinance pip install mplfinance It may be needless to say, but at first I was trying to use mpl-finance, so pip install mpl_finance I was trying to install mpl-finance and import it, I got warning as below; ## WARNING: mpl_finance is deprecated: ## Please use mplfinance instead (no hyphen, no underscore). ## To install: pip install --upgrade mplfinance ## For more information, see: https://pypi.org/project/mplfinance/ Therefore, pip install --upgrade mplfinance as I installed and used mplfinance. At that time, I knew that mpl-finance had been renewed to mplfinance…. #### mplfinance code and execution result I added the following code to the data acquisition code above and executed it together. #import mplfinance import mplfinance as mpf # the column names must be Open, High, Low, Close, so change them. df = df.rename (columns = {'<OPEN>': 'Open', '<HIGH>': 'High', '<LOW>': 'Low', '<CLOSE>': 'Close'}) #draw with mplfinance #draw using plot function specifying type as candle and style as starsandstripes. mpf.plot(df, type='candle', style='starsandstripes') I was able to draw a good graph just with this simple code! Looking closely, oh? It’s off 9 hours ago… Apparently it is treated as UCT even though it set JST. I don’t know how to fix it with mlpfinance…. I will try adjust detail later…. Anyway, I found that I could easily draw an easy-to-read graph! mpf.available_styles() as we can see that the following styles are available. ['binance','blueskies','brasil','charles','checkers','classic','default', 'mike','nightclouds','sas','starsandstripes','yahoo'] #### Next, install Plotly pip install plotly==4.6.0 I already installed the old version of 3.x.x (forgot), but this command will also uninstall the old version. #### Plotly code and execution results I added the following code under the mplfinance code and executed it together. So same DataFrame is used as df of which column name has been changed in advance. #since plotly.graph_objs is used for drawing candlestick, import as go import plotly.graph_objs as go fig = go.Figure(data = [go.Candlestick(x = df.index, open = df['Open'], high = df['High'], low = df['Low'], close = df['Close'])]) fig.show() It was very easy as well! It seems to be very convenient to be able to change drawing range with the slider at the bottom and to check the value interactively with the mouse over. Plotly seems keeping JST timezone. ### Conclusion In conclusion, it turned out that both of them can draw an easy-to-understand diagram with almost the default settings. Great…. In particular, Plotly seems to be very compatible with detailed analysis. If I get any idea, I would like to try various other graphs, functions, and settings. Environment Python; 3.7.2 mplfinance; 0.12.4a0 Plotly; 4.6.0 Styles of mplfinance candlestick https://vucavucalife.com/python-mplfinance-candle-stick-style/ Using Dash, I made a web application that can display the past dollar-yen candlestick chart (1 minute) in Plotly.	{}
Actually determining the Ramsey numbers $$R(m,n)$$ referenced in Theorem 11.2 seems to be a notoriously difficult problem, and only a handful of these values are known precisely. In particular, $$R(3,3)=6$$ and $$R(4,4)=18\text{,}$$ while $$43\le R(5,5)\le 49\text{.}$$ The distinguished Hungarian mathematician Paul Erdős said on many occasions that it might be possible to determine $$R(5,5)$$ exactly, if all the world's mathematical talent were to be focused on the problem. But he also said that finding the exact value of $$R(6,6)$$ might be beyond our collective abilities. In the following table, we provide information about the Ramsey numbers $$R(m,n)$$ when $$m$$ and $$n$$ are at least $$3$$ and at most $$9\text{.}$$ When a cell contains a single number, that is the precise answer. When there are two numbers, they represent lower and upper bounds. www.combinatorics.org/ojs/index.php/eljc/article/view/DS1	{}
Is XSalsa20-Poly1305-SIV a reasonable choice for nonce-misuse-resistant authenticated encryption? Consider XSalsa20-Poly1305-SIV. This is obtained by: • computing a MAC $t_{secret}$ of the plaintext from the key and nonce, as in ChaCha20-Poly1305 except that the plaintext, not the ciphertext, is MAC'd. • compute $t = F(K,t_{secret})$, where F is the ChaCha20 core (or any other PRF). • use $t$ as the auth tag and as part of the XSalsa20 nonce. Questions: • Is this secure? • Is this provably secure, assuming that the ChaCha20 and Salsa20 cores are strong PRFs? • If you're writing F's inputs in the usual order then that's certainly not necessarily secure. – user991 Feb 22 '16 at 10:43 • @RickiDemer oops, fixes. – Demi Feb 22 '16 at 13:52 • Assuming I've understood what you're asking, your question is equivalent to: "Is SIV a secure mode of operation", which it is [web.cs.ucdavis.edu/~rogaway/papers/keywrap.html] – Cryptographeur Feb 23 '16 at 10:59 This can be a secure construction, if by MAC you mean universal hash family, like Poly1305. Call this hash family $$H_r$$ and the short pseudorandom function family $$F_k$$. Rough justification for why this is secure: 1. The function $$m \mapsto F_k(H_r(m))$$ is a long-input, short-output PRF. 2. A good PRF makes a good MAC. 3. A good PRF has birthday-bounded collision probability, so the probability of nonce reuse for the stream cipher is small. In a bit more detail: Let $$\varepsilon_F$$ be a bound on the PRF-distinguisher advantage of any algorithm making $$q$$ queries against $$F$$, and let $$\varepsilon_H$$ be a bound on the collision probability $$\Pr[H_r(x) = H_r(y)]$$ of $$H$$ for any $$x \ne y$$ and uniform random $$r$$. • The PRF-distinguisher advantage for a $$q$$-query attack against $$m \mapsto F_k(H_r(m))$$ is bounded by $$\varepsilon_F + \binom{q}{2} \varepsilon_H$$ (proof). • For fixed-size keys, the collision probability $$\varepsilon_H$$ grows linearly with the maximum message length. So this figures into the concrete numbers. • The probability of a synthetic nonce collision (i.e., a collision in $$m \mapsto F_k(H_r(m))$$) is higher than the probability of a collision in $$H$$, because either a collision in $$H$$ or a collision in $$F$$ means there's a nonce collision—whether this matters depends on how large the output of $$F$$ is, and in particular on the birthday bound for $$F$$ which may be much smaller than the collision probability of $$H$$. There's some practical details to work out. Here $$q$$ represents the number of messages that you are willing to send or receive in your application. For example, if $$\varepsilon \approx 2^{-100}$$, the bound above means nothing unless your application's total bandwidth is limited to $$q \lll 2^{50}$$ messages. (And you don't want to mess with collisions in $$H$$.) Is the Poly1305 collision probability bound $$\varepsilon_{\operatorname{Poly1305}} = 8\ell/2^{106}$$ for $$16\ell$$-byte messages comfortable enough for trillions of megabyte-long messages? Maybe it is, maybe it isn't—for now I leave it as an exercise for the reader to compute specific bounds for specific data volumes. (See a similar table for AES-GCM.) What should you choose for $$F_k$$? If you're already using XSalsa20, the obvious choice is XSalsa20 truncated to 128 bits or similar. Of course, you'll have to quantify the probability of a collision between the XSalsa20 nonce used for encryption and the XSalsa20 input from $$H$$. But maybe you can do better than multiple the HSalsa20 invocations this would imply, each of which adds overhead to small packets. Maybe you should use a 192-bit or 256-bit hash $$H$$, and a 192-bit authentication tag, so that the probability of a synthetic nonce collision is negligible even for extremely large volumes of data.	{}
My Math Forum Probability of correct answers with multiple choices User Name Remember Me? Password Probability and Statistics Basic Probability and Statistics Math Forum October 14th, 2019, 06:56 AM #1 Newbie Joined: Oct 2019 From: Maryland Posts: 1 Thanks: 0 Probability of correct answers with multiple choices Hello; This is probably an elementary question for most people, but I have a question. Let's say that there is a 100 question test with each question having two possible choices. With an ideal random number generator the probability getting the correct answer approaches 50% as the number of questions increases. That I know. My question is this. For a person trying to get the correct answer, is it possible to score less than 50%? How much less? Thank You Very Much; Frank October 14th, 2019, 07:27 AM #2 Senior Member Joined: Sep 2015 From: USA Posts: 2,638 Thanks: 1475 It's certainly possible for them to get zero correct. It's not likely but it's possible. You might read up on the binomial probability distribution. Thanks from idontknow and DarnItJimImAnEngineer October 14th, 2019, 07:30 AM #3 Senior Member Joined: Jun 2019 From: USA Posts: 383 Thanks: 207 Yes, it is possible to score less than 50 %. It is possible to score 0 % (although there is a 1 in 2^100 chance of doing so). In fact, you have the same odds of scoring less than 50 % as you do of scoring more than 50 %. What is more interesting to look at is: What is the probability of scoring 50 %? or 90 %? or 25 %? In this case, we have a binomial distribution $\displaystyle p(n) = \left[ \frac{N!}{(N-n)!n!} \right] R^n (1-R)^{N-n},$ where N is the number of questions (100), n is the number of questions you guessed correctly (i.e., the score), and R is the probability of guessing any given question correctly (i.e., 0.5). In this case, the probability of making a score of n/100 is $\displaystyle p(n) = \frac{100!}{(100-n)!n! 2^{100}}$ We can also calculate the cumulative distribution function P(n). This is the probability of making a score of n or lower. $\displaystyle P(n) = \sum_{i=0}^{n} p(n)$ With N = 100 questions, there is a 7.96 % chance of scoring exactly 50/100. There is a 46.02 % chance of scoring lower than this, and a 46.02 % chance of scoring higher than this. Thanks from idontknow October 14th, 2019, 07:32 AM #4 Senior Member Joined: Jun 2019 From: USA Posts: 383 Thanks: 207 We need a, "Someone posted a response while you were typing," warning on the forum, lol. Tags answers, choices, correct, multiple, probability Thread Tools Display Modes Linear Mode Similar Threads Thread Thread Starter Forum Replies Last Post Chikis Elementary Math 8 June 4th, 2019 02:22 PM EvanJ Probability and Statistics 28 October 5th, 2016 06:29 AM Lilian Probability and Statistics 2 June 6th, 2014 03:27 PM r-soy Algebra 6 March 5th, 2010 08:23 AM restin84 Calculus 2 September 11th, 2009 10:12 AM Contact - Home - Forums - Cryptocurrency Forum - Top	{}
# Isn't there any meaning to talk of open spaces, or of closed space? From Rudin's Principles of Mathematical Analysis (p.37). Theorem 2.33. Suppose $K\subset Y\subset X$. Then $K$ is compact relative to $X$ if and only if $K$ is compact relative to $Y$. Immediately after the following is written. By virtue of this theorem we are able, in many situation, to regard compact sets as metric spaces in their own right, without paying any attention to any embedding space. In particular, although it makes little sense to talk of open spaces, or of closed spaces (every metric space $X$ is an open subset of itself, and is a closed subset of itself), it does make sense to talk of compact metric space. I can't understand the sentence "every metric space $X$ is an open subset of itself, and is a closed subset of itself". Actually, $(0,1)$ is metric space for metric $d(x,y)=\|x-y\|$, $\forall x,y\in (0,1)$. $(0,1)$ is just open subset of itself, but is not a closed subset of itself. • $(0,1)$ is a closed subset of $(0,1)$; indeed every metric space $X$ is a closed subset of itself. The complement is $\emptyset$ which is open. – Lord Shark the Unknown Aug 13 '18 at 6:31 • To check that $(0,1)$ is a closed subset of itself show that the complement $(0,1)\setminus (0,1)=\emptyset$ is open. The empty set is open, because it satisfies that every one of its points is inside a ball contained in the set. Since the empty set doesn't contain any points, there is nothing that could make the condition fail. – user583185 Aug 13 '18 at 6:31 • Thank you, I can understand what you mean. – 백주상 Aug 13 '18 at 6:44 • You must be careful to distinguish between a sub$set$ and a sub$space$. – DanielWainfleet Aug 13 '18 at 7:33	{}
# Is it worth switching to timesteppers provided by PETSc if I can't write down a Jacobian for my problem? Case study with “the amoeba” toy problem I am considering using petsc4py instead of scipy.integrate.odeint (which is a wrapper for Fortran solvers) for a problem involving the solution of a system of ODEs. The problem has the potential to be stiff. Writing down its Jacobian is very hard. So far, I have been able to produce reasonable speed gains by writing the RHS functions in "something like C" (using either numba or Cython). I'd like to get even more performance out, hence my consideration of PETSc. # Introducing a toy version of the problem: The Amoeba Consider the following figure, which defines the basics of an amoeba: So, an amoeba consists of $N$ nodes and edges, constrained to a 2D space. The position of the nodes describing the amoeba evolve with time. Let the position of the $n$th ($n \in \{0, 1, ..., N-1\}$) node be described a position vector, $\textbf{x}_n$: $$\textbf{x}_n = \left[\begin{array}{cc} x_{0,n} \\ x_{1,n} \end{array}\right]$$ For the purposes of simplification, assume that the amoeba lives in a world with first order equations of motion, so given an applied force $\textbf{F}$ on a node, the nodal velocity $\dot{\textbf{x}}$ is produced: $$\textbf{F} = c \dot{\textbf{x}}$$ where $c$ is a constant I pick. We are very interested in tracking the position of each node,so immediately we produce $2n$ ODEs (one ODE for each spatial component) of interest (assume that edges are elastic springs): $$\frac{\textrm{d}\textbf{p}_n}{dt} = \text{elastic edge forces} + \textbf{F}_{n, active}$$ $\textbf{F}_{n, active}$ is the force generated by the amoeba at node $n$. One can imagine a wide range of rules that could be devised to determine how $\textbf{F}_{n, active}$ is determined, and I suggest that we consider the following rules: 1) There are red balls and blue balls inside the amoeba. ^ 2) Both red and blue balls have "switched-on" (symbolized $r^$/$b^$) and "switched-off" (symbolized $r$/$b$) state. 3) Both red and blue balls can be found either in the body of the amoeba where they are always switched off, or at the nodes of the amoeba, where they can be switched on or switched off. 4) The total number of red balls regardless of location of status is a constant $R$ and similarly for blue balls, a constant $B$ 5) When red or blue balls are in the body, they are free to jump onto any node they want, but when they are on nodes, some of them might be induced to move to neighbouring nodes due to some sort of "diffusion" 6) Switched on red balls at a node cause a force that attempts to reduce the area of the amoeba 7) Switched on blue balls at a node cause a force that attempts to increase the area of the amoeba 8) Switched on red balls tend to switch off blue balls that are switched on, and vice versa, switched on blue balls tend to switch off red balls that are switched on 9) ... # Okay, you get my point. I can come up with a model that is fairly convoluted in terms of how the various variables of interest interact. Due to the large number of equations involved, it is already tedious to think about writing downa Jacobian. However, since we have access to a computer, it may be that we can write functions governing a particular interaction that do not have neat analytical forms (let alone whether or not their derivatives have neat analytical forms), so we might have a mess of piecewise functions needed to approximate them if we were to go about still trying to produce a Jacobian... # Things get even worse if my problem involves not just one amoeba, but multiple amoebae (which is when things ARE interesting), with new interaction rules enforcing that amoebae cannot violate each others' areas, and so on. All the toy examples I see of PETSc time stepping problems have Jacobians defined, so I wonder if I would even get a speed gain going from switching to it, if perhaps one of the reasons why I have a high computational cost is due to not being able to provide a Jacobian function? • Have you considered to calculate your Jacobian by automatically differentiating the (C code for the) right-hand-side of your ODE system? – GoHokies Dec 9 '15 at 13:54 • As a followup to GoHokies's comment, I've used DAE Tools to set up systems of ODE's (or DAE's) in python and let it take care of automatic differentiation (and some other details) for me. – muon Jul 29 '16 at 13:57	{}
# [NTG-context] How to not gobble space after a command Hi all, I'm working with a few custom macros to simplify typing a bit, like: \def\VHDL{\small{VHDL}} However, when I use this macro in text, like \VHDL this, it gobbles the space after it. I've seen some references that say to solve this by writing \VHDL{} instead, but that is of course very unpretty. I've tried to put all kinds of stuff ({}, \mbox{}, \obeyedspace, etc.) inside the \VHDL macro, but it seems the space is already gone there, so that didn't work. (Note that just putting a space inside the macro won't work, for cases where I use \VHDL, like this) It seems that latex has an xspace package that does some guesswork about whether a space is required, but it looks mightily scary. Does ConTeXt provide for any way to solve this problem in a nice way? Or is this just so fundamental in TeX that there is no real solution? Gr. Matthijs signature.asc Description: Digital signature ___________________________________________________________________________________	{}
SEARCH NEW RPMS DIRECTORIES ABOUT FAQ VARIOUS BLOG DONATE YUM REPOSITORY texlive-ionumbers rpm build for : OpenSuSE. For other distributions click texlive-ionumbers. Name : texlive-ionumbers Version : 2017.132.0.0.3.3svn33457 Vendor : openSUSE Release : 36.1 Date : 2017-11-24 13:00:00 Group : Productivity/Publishing/TeX/Base Source RPM : texlive-specs-m-2017-36.1.src.rpm Size : 0.03 MB Packager : http://bugs_opensuse_org Summary : Restyle numbers in maths mode Description : \'ionumbers\' stands for \'input/output numbers\'. The packagerestyles numbers in maths mode. If a number in the input fileis written, e.g., as $3,231.44$ as commonly used in Englishtexts, the package is able to restyle it to be output as$3\\,231{,}44$ as commonly used in German texts (and viceversa). This may be useful, for example, if you have a largetable and want to include it in texts with different outputconventions without the need to change the table. The packagecan also automatically group digits left of the decimalseparator (thousands) and right of the decimal separator(thousandths) in triplets without the need of specifing commas(English) or points (German) as separators. E.g., the input$1234.567890$ can be output as $1\\,234.\\,567\\,890$. Finally, ane starts the exponent of the number. For example, $21e6$ may beoutput as $26\\times10\\,^{6}$.date: 2016-06-24 17:18:15 +0000 RPM found in directory: /mirror/ftp.opensuse.org/ports/aarch64/tumbleweed/repo/oss/suse/noarch Content of RPM Changelog Provides Requires	{}
# zbMATH — the first resource for mathematics Numerical simulation of shock tube generated vortex: effect of numerics. (English) Zbl 1271.76182 Summary: Vortices generated at the open end of a planar shock tube are numerically simulated using the AUSM+ scheme. This scheme is known to have low numerical dissipation and therefore is suitable for capturing unsteady vortex motion. However, this low numerical dissipation can also cause oscillations in the vorticity field. Numerical experiments presented here highlight the effect of numerical dissipation on the simulated vortex, as well as the role played by turbulence models. Two turbulence models – the shear-stress-transport (SST) and its modified version for unsteady flows (SST-SAS) – are employed to observe the effect of including turbulence models in such complex flows where the vortex has an embedded shock. ##### MSC: 76M12 Finite volume methods applied to problems in fluid mechanics 76L05 Shock waves and blast waves in fluid mechanics 76F60 $$k$$-$$\varepsilon$$ modeling in turbulence ##### Keywords: shock tube; vortex; AUSM+; numerical dissipation; turbulence model AUSM Full Text: ##### References: [1] Anderson D.A., Computational fluid mechanics and heat transfer (1984) · Zbl 0569.76001 [2] Arakeri J.H., Physics of Fluids 16 pp 1008– (2004) · Zbl 1186.76030 [3] Arnone, A., Liou, M.S. and Povinelli, L.A. 1993. ”Multigrid time-accurate integration of Navier-Stokes equations. NASA TM 106373. ICOMP-93-37”. · Zbl 0844.76069 [4] Baird J.P., Proceedings of the Royal Society of London, Series A, Mathematical and Physical Sciences 409 pp 59– (1987) [5] Brouillette M., Fluid Dynamics Research 21 pp 159– (1997) [6] Juntasaro V., International Journal of Computational Fluid Dynamics 18 pp 569– (2004) · Zbl 1094.76044 [7] Kevlahan N.K., Journal of Fluid Mechanics 341 pp 371– (1997) · Zbl 0898.76050 [8] DOI: 10.1016/j.jcp.2005.02.022 · Zbl 1329.76265 [9] DOI: 10.1006/jcph.1996.0256 · Zbl 0870.76049 [10] DOI: 10.1016/j.jcp.2005.09.020 · Zbl 1137.76344 [11] Menter, F.R. Zonal two-equation k – {$$\omega$$} turbulence models for aerodynamic flows.24th Fluid Dynamics Conference. July 6–9. pp.1993–2906. AIAA. [12] Menter, F.R. and Egorov, Y. A scale-adaptive simulation using two-equation models. 43rd AIAA aerospace science meeting and exhibit. 10–13 January, Reno, NV, USA.AIAA2005–1095 [13] Murugan T., Flow and acoustic characteristics of high Mach number vortex rings during evolution and wall interaction (2008) [14] Murugan T., Experiments in Fluids 49 pp 1247– (2010) [15] DOI: 10.1016/0021-9991(92)90046-2 · Zbl 0766.76084 [16] Radespiel R., Journal of Computational Physics 121 pp 66– (1995) · Zbl 0843.76059 [17] Sun M., Shock Waves 13 pp 25– (2003) · Zbl 1063.76599 [18] DOI: 10.1016/0021-9991(83)90065-7 · Zbl 0505.76006 [19] Wilcox D.C., Turbulence modeling for CFD (2004) [20] Zare-Behtash H., Physics of Fluids 20 pp 126105– (2008) · Zbl 1182.76866 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.	{}
# Homework Help: Anyone see how to solve this without L'Hospitals? 1. Nov 19, 2006 ### dontdisturbmycircles $$\lim_{x->1}\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}$$ I can do it with L'Hospitals, but my teacher said there is also a way to do it without. I tried all sorts of techniques such as multiplying by the conjugate of both the denom/numerator as well as taking $$\sqrt{x}$$ out of both the denom and numerator. I can't see what method my teacher sees that I don't.. I get 2/3 with L'Hospitals by the way. 2. Nov 19, 2006 ### robphy factor (a change of variables may help) 3. Nov 19, 2006 ### dontdisturbmycircles Ahhh okay I see, thanks. 4. Nov 20, 2006 ### dontdisturbmycircles Can someone demonstrate how to solve this with a change of variables? I am usually fairly comfortable using a change of variables but I don't see how to apply it here. For example I can solve lim_{x->inf}e^(-2x) or w/e by making t=-2x. But I don't see what to do in with the problem in the thread. I can solve it by dividing, but don't see how to use a change of variables. Last edited: Nov 20, 2006 5. Nov 20, 2006 ### robphy If you can finish the problem by simply dividing, then you don't need the change of variables. My suggestion to change the variables may help you see how to write the numerator and the denominator so that each can be factored and have some common factors cancel, leaving you with something that you can easily take the limit of. So, can you see how the numerator and denominator can be separately factored? (Think of polynomials.) 6. Nov 20, 2006 ### dontdisturbmycircles I don't see how to factor them :(. I can rewrite it as t=x-1 so it becomes $$\lim_{t->0}\frac{t^{1/3}}{t^{1/2}}$$ but that screws it up. Of course I could simplify to $$t^{-1/6}$$ but I screwed something up somewhere. Last edited: Nov 20, 2006 7. Nov 20, 2006 ### robphy Let $$y=\sqrt[6]{x}$$. 8. Nov 20, 2006 ### dontdisturbmycircles $$\lim_{t->1}\frac{t^{2}-1}{t^{3}-1}$$ $$=\lim_{t->1}\frac{(t-1)(t+1)}{(t-1)(t^2+t+1)}$$ $$=\lim_{t->1}\frac{t+1}{t^{2}+t+1}$$ $$=\frac{2}{3}$$ Thanks alot :) I am still trying to understand your logic for using $$\sqrt[6]{x}$$ though. 9. Nov 20, 2006 ### dontdisturbmycircles I guess it makes sense that given x^n/x^m, replacing one with t^(nm) will make it easier to factor. I still have to think about it a bit. 10. Nov 20, 2006 $$\sqrt[6]{x} = t$$ $$t^{2} = \sqrt[6]{x}^{2} = x^{\frac{1}{3}}$$ $$t^{3} = \sqrt{x}$$ 11. Nov 20, 2006 ### robphy I wanted to get the first line in your previous post... because I knew that I could factor out a common (t-1) factor from a term like (t^k-1). So, I wanted to choose of convenient change of variables get the numerator and denominator to be in the form (t^k-1). Alongside the above, I wanted to eliminate all of the fractional powers of x. 12. Nov 20, 2006 ### dontdisturbmycircles Ah okay that makes perfect sense. Thanks alot! 13. Nov 20, 2006 ### NateTG L'Hospital's method can be derived from epsilon-delta definitions. Consider: Assuming $$x \neq 0$$ We have: $$\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}=\lim_{x \rightarrow 0} \frac{\sqrt[3]{0+x}-1}{\sqrt{0+x}-1}=\frac{\frac{\sqrt[3]{0+x}-1}{x}}{\frac{\sqrt{0+x}-1}{x}}$$	{}
177 views Answer the question on the basis of the information below: An industry comprises four firms (A, B, C and D). Financial details of these firms and of the industry as a whole for a particular year are given below. Profitability of a firm is defined as profit as a percentage of sales. Figures are in Rs. A B C D Total Sales 24, 568 24,468 23,752 15,782 89,570 Operating costs 17,198 19,101 16,151 10,258 62,708 Interest costs 2457 2292 2850 1578 9177 Profit 4914 4075 4750 3946 17,684 If Firm A acquires Firm B, approximately what percentage of the total market (total sales) will they corner together? 1. $55\%$ 2. $45\%$ 3. $35\%$ 4. $50\%$	{}
# How to align visibility in pgf-umlcd? code here- \documentclass{standalone} \usepackage{pgf-umlcd} \begin{document} \begin{tikzpicture} \begin{class}[text width = 8cm]{JavaLanguageBehavior}{0,0} \attribute{$\sim$ com.alhelal.textpad} \attribute{- uniqueInstance : JavaLanguageBehavior} \operation{+ getUniqueInstance : JavaLanguageBehavior} \operation{+ runCode(file : File) : BufferedReader} \operation{+ buildCode(file : File) : BufferedReader} \operation{+ setHighlightableText()} \operation{+ setAutoCompletableText()} \end{class} \node [above=3mm] at (current bounding box.north) {Singleton Pattern}; \end{tikzpicture} \end{document} When I use a -(private) as visibility the uniqueInstance variable goes to left. How can I align the visibility and variables? Presumably you want to use $-$, so that it becomes a minus sign instead of a hyphen, which in turn means that the widths are similar. For consistency, set the plus signs in math mode as well. \documentclass{standalone} \usepackage{pgf-umlcd} \begin{document} \begin{tikzpicture} \begin{class}[text width = 8cm]{JavaLanguageBehavior}{0,0} \attribute{$\sim$ com.alhelal.textpad} \attribute{$-$ uniqueInstance : JavaLanguageBehavior} \operation{$+$ getUniqueInstance : JavaLanguageBehavior} \operation{$+$ runCode(file : File) : BufferedReader} \operation{$+$ buildCode(file : File) : BufferedReader} \operation{$+$ setHighlightableText()} \operation{$+$ setAutoCompletableText()} \end{class} \node [above=3mm] at (current bounding box.north) {Singleton Pattern}; \end{tikzpicture} \end{document} • Thank you. I guessed that you will answer. But should I need $+$ for + instead of only +? Nov 1, 2017 at 14:03 • @alhelal Didn't seem to make a difference in this case, I did it just for consistency. But it will depend on font setup, if you're using a math font with smaller plus/minus signs you could see a small difference. Nov 1, 2017 at 14:08 • Yes. Consistency is important. I recommend you to mention this in answer why you write extra \$. Thank you. Nov 1, 2017 at 14:15	{}
# Using Emacs with Swedish keyboard I am running GNU Emacs on Ubuntu, using a Swedish PC-keyboard. When I use swedish-keyboard as my input method, shift-key combinations work fine (e.g. S-7 gives the '/' character). But I have found a few things that doesn't work properly: • Alt Gr-combinations: Alt Gr is interpreted as Meta-key, so when I press e.g. (Alt Gr + 7) I get M-7 instead of '{'. So I can't type '@','£','\$','{','[',']','}','\' or '\|' using swedish-keyboard. Is there a way to separate Alt and Alt Gr-key such that Alt is Meta-key and Alt Gr can be used for extended character input? • Misplaced characters: Pressing e.g. '<' on the keyboard produces a ';' character (shift+'<' gives ':'). Similarly, '§' gives '~', shift+'§' gives '^', '´' gives '§', shift+'´' gives '½', '¨' gives 'é', shift+'¨' gives 'É'. So there is no way to type '<','>','´','' or accented characters other than 'é' and 'É'. Is there a way to solve these issues? The way I solve it now is simply to toggle between swedish and american input whenever I need to type any of the unaccessible characters in the swedish input mode. P.S. I am a bit of a noob on Emacs, so carefully explained solutions are appreciated. • One way to avoid those problems is to use the swedish-postfix input method instead of swedish-keyboard. That converts the sequences aa, ae, oe and e' to å, ä, ö and é, respectively, without affecting the keyboard layout. – legoscia Jun 24 '15 at 15:56 • Well, my main issue is not that I want to easily write the Swedish letters 'å', 'ä' and 'ö'. Rather, I wish to access special characters, e.g. '(', '{' and '<', using the shift and Alt Gr combinations that I am used to, namely those on a Swedish keyboard layout. On an American keyboard, '(' is written as S-9, whereas on a Swedish keyboard, it is written as S-8. – emmm Jun 24 '15 at 21:07 • Do you have these issues in other programs too, or is it just in emacs? – Jenny D Jun 25 '15 at 9:56 • Nope, only in Emacs as far as I have discovered. The terminal cooperates with my swedish keyboard, so if I use -nw option, it works fine as well. – emmm Jun 25 '15 at 10:47 • What makes this an issue? I've always used Emacs with Swedish keyboards, and have never had to bother with this. On different unix/linux flavours. (Not saying that it isn't a problem, just surprised.) – Meaningful Username Jun 26 '15 at 18:28 I have a Portuguese keyboard but I use only use the English layout for editing in Emacs. My solution to write special characters is using digraphs. I myself use evil, so I do C-k ' e to insert é, for example. If you do not use evil, you could do something like: (define-prefix-command 'digraph-map) `	{}
Current Location:home > Detailed Browse # An updated study of $\Upsilon$ production and polarization at the Tevatron and LHC ## Abstracts Following the nonrelativistic QCD factorization scheme, by taking latest available measurement on?χb(3P)?into consideration, we present an updated study on the yield and polarization of?Υ(1S,2S,3S)?hadroproduction, and the fractions of?χb(mP)?feed-down in?Υ(nS)?production at QCD next-to-leading order. In the fitting, three schemes are applied with different choice of?χb(mP)?feed-down ratios and NRQCD factorization scale. The results can explain the measurements on yield very well as in our previous work. The polarization puzzle to?Υ(3S)?is now solved by considering the?χb(3P)?feed-down contributions. The ratio of?σ[χb2(1P)]/σ[χb1(1P)]?measurements from CMS can also be reproduced in our prediction. Among the different schemes, the results show little difference, but there are sizeable difference for the fitted long-distance color-octet matrix elements. It may bring large uncertainty when the values are applied in other experiments such as in?ee,?ep?colliders. Keywords: polarization; LHC; Recommended references： Yu Feng,Bin Gong,Lu-Ping Wan,Jian-Xiong Wang.(2016).An updated study of $\Upsilon$ production and polarization at the Tevatron and LHC.[ChinaXiv:201609.00899] (Click&Copy) Version History Related Paper	{}
# Simultaneous Linear Equations Simultaneous Linear Equations Simultaneous linear equations in two variables involve two unknown quantities to represent real-life problems. It helps in establishing a relationship between quantities, prices, speedtime, distance, etc results in a better understanding of the problems. We all use simultaneous linear equations in our daily life without knowing it. In this mini-lesson, we will learn in detail about the solutions of linear equations, consistent and inconsistent equations, homogenous linear equations, simultaneous equation solver, example of simultaneous equation, etc. in this mini-lesson. ## Lesson Plan 1 What Do You Mean By Simultaneous Linear Equations? 2 Tips and Tricks 3 Solved Examples on Simultaneous Linear Equations 4 Challenging Questions on Simultaneous Linear Equation 5 Interactive Questions on Linear Equation ## What Do You Mean By Simultaneous Linear Equation? Two linear equations in two or three variables solved together to find a common solution are called simultaneous linear equations. For example, We can visualize the solution of the linear system of equations by drawing 2 linear graphs and finding out their intersection point. The red dot represents the solutions for equation 1, and equation 2. The intersection is the unique point (2,1) is the solution that we are looking for which will satisfy both the equations ## Linear Equation Calculator This simultaneous equation solver will find the non-trivial solution on entering the coefficients of x and y and constant. ## How To Solve Simultaneous Linear Equations? The following methods can be used to find the solution of linear system of equations, let's see some example of the simultaneous equation. ### 1. Substitution Method Consider the following pair of linear equations: $\begin{array}{l}x + 2y = 6\;\;\;\; & ...(1)\\ x - y = 3\;\;\;\; & ...(2)\end{array}$Let’s rearrange the first equation to express $$x$$ in terms of $$y$$, as follows:$\begin{array}{l}x + 2y = 6\\ \Rightarrow \;\;\;x = 6 - 2y\end{array}$ This expression for $$x$$ can now be substituted in the second equation, so that we will be left with an equation in $$y$$ alone: \begin{align}& x - y = 3\\ &\Rightarrow \;\;\;6 - 2y -y = 3\\ &\Rightarrow \;\;\;-3y = 3 - 6\\ &\Rightarrow \;\;\;y = \frac{-3}{-3}\\ &\Rightarrow \;\;\;y = 1\end{align} Once we have the value of $$y$$, we can plug this back into any of the two equations to find out $$x$$. Lets plug it into the first equation: $\begin{array}{l}x + 2y = 6\\ \Rightarrow \;\;\;x + 2 \times 1 = 6\\ \Rightarrow \;\;\;x = 6 - 2 = 4\\ \Rightarrow \;\;\;x = 4\end{array}$ The final non-trivial solution is: $x = 4,\;y = 1$ It should be clear why this process is called substitution. We express one variable in terms of another using one of the pair of equations, and substitute that expression into the second equation. ### 2. Elimination Method Consider the following pair of linear equations: $\begin{array}{l} 2x + 3y - 7 = 0\\ 3x + 2y - 3 = 0 \end{array}$ The coefficients of x in the two equations are 2 and 3 respectively. Let us multiply the first equation by 3 and the second equation by 2, so that the coefficients of x in the two equations become equal: $\begin{array}{l}\left\{ \begin{array}{l}3 \times \left( {2x + 3y - 7 = 0} \right)\\2 \times \left( {3x + 2y - 3 = 0} \right)\end{array} \right.\\ \Rightarrow \;\;\;6x + 9y - 21 = 0\\\qquad6x + 4y - 6 = 0\end{array}$ Now, let us subtract the two equations, which means that we subtract the left hand sides of the two equations, and the right hand side of the two equations, and the equality will still be preserved (this should be obvious: if I = II and III = IV, then I – III will be equal to II – IV): $\begin{array}{l}\left\{ \begin{array}{l}\,\,\,\,6x + 9y - 21 = 0\\\,\,\,\,6x + 4y - 6 = 0\\\underline { - \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\\,\,\,\,\,\,0 + 5y\,\,\, - 15 = 0\end{array} \right.\\ \Rightarrow \;\;\;5y = 15\\ \Rightarrow \;\;\;y = 3\end{array}$ Note how x gets eliminated, and we are left with an equation in y alone. Once we have the value of y, we proceed as earlier – we plug this into any of the two equations. Let us put this into the first equation: $\begin{array}{l}2x + 3y - 7 = 0\\ \Rightarrow \;\;\;2x + 3\left( 3 \right) - 7 = 0\\ \Rightarrow \;\;\;2x + 9 - 7 = 0\\ \Rightarrow \;\;\;2x = -2\\ \Rightarrow \;\;\;x = -1\end{array}$ Thus, the solution is: $x = -1,\;y = 3$ ### 3. Graphical Method As an example, consider the following pair of linear equations: $\begin{array}{l} x - y = 0\\ x + y - 4 = 0\end{array}$ We draw the corresponding lines on the same axes: The point of intersection is $$A\left( {2,\,\,2} \right)$$, which means that $$x = 2,\;\;y = 2$$ is a solution to the pair of linear equations given by (2). In fact, it is the only solution to the pair, as two non-parallel lines cannot intersect in more than one point. Tips and Tricks • You can check directly about the types of solution using the following conditions: Unique solution( Consistent and independent) $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ No solution( Inconsistent and independent) $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$ Infinite Many Solutions (Consistent and Dependent) $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ ## Solved Examples Example 1 The sum weights of Fabia and Valerian is 60 pounds and the difference is 2. Find the weights of Fabia and Valerian. Solution Let the weight of Fabia and Valerian be $$x$$ pounds and $$y$$ pounds respectively. Therefore, the simultaneous equations are$\begin{array}{l}x + y = 60\;\;\;\; & ...(1)\\ x - y = 2\;\;\;\; & ...(2)\end{array}$ Let’s rearrange the first equation to express $$x$$ in terms of $$y$$, as follows:$\begin{array}{l}x + y = 60\\ \Rightarrow \;\;\;x = 60 - y\end{array}$ This expression for $$x$$ can now be substituted in the second equation, so that we will be left with an equation in $$y$$ alone: \begin{align}& x - y = 2\\ &\Rightarrow \;\;\;60 - y -y = 2\\ &\Rightarrow \;\;\;-2y = 2 - 60\\ &\Rightarrow \;\;\;y = \frac{-58}{-2}\\ &\Rightarrow \;\;\;y = 29\end{align} Once we have the value of $$y$$, we can plug this back into any of the two equations to find out $$x$$. Lets plug it into the first equation: $\begin{array}{l}x + y = 60\\ \Rightarrow \;\;\;x + 29 = 60\\ \Rightarrow \;\;\;x = 60 - 29\\ \Rightarrow \;\;\;x = 31\end{array}$ The final solution is: $x = 31,\;y = 29$ Therefore, weight of Fabia and Valerian is 31 pounds and 29 pounds respectively. Example 2 Can you help Alex to find a two-digit number whose units digit is thrice the tens digit and if 36 is added to the number, the digits interchange their place. Solution Let the digit in the units place is $$x$$. And the digit in the tens place be $$y$$ Then $$x = 3y$$ and the number $$= 10y + x$$ The number obtained by reversing the digits is $$10x + y$$. If $$36$$ is added to the number, digits interchange their places, Therefore, we have $10y + x + 36 = 10x + y$ $10y – y + x + 36 = 10x + y - y$ $9y + x – 10x + 36 = 10x - 10x$ $9y - 9x + 36 = 0$ $9(x - y) = 36$ $x - y = \frac{36}{9}$ $x - y += 4 .... (i)$ Substituting the value of x = 3y in equation (i), we get $3y - y = 4$ $2y = 4$ $y = \frac{4}{2}$ $y = 2$ Substituting the value of$$y = 2$$in equation (i),we get $x - 2 = 4$ $x = 4 + 2$ $x = 6$ Therefore, the number Alex looking is 62. Challenging Questions Solved Crossword Puzzle on Simultaneous linear equations ## Interactive Questions Here are a few activities for you to practice. Select/Type your answer and click the "Check Answer" button to see the result. ## Let's Summarize The mini-lesson targeted the fascinating concept of Simultaneous Linear Equations. The math journey around Simultaneous Equations starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever. Here lies the magic with Cuemath. At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students! Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic. Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in. ## FAQs on Simultaneous linear equations ### 1. What is the degree of a linear equation? The degree of a linear equation is 1 ### 2. How many types of linear equations are there? The three major forms of linear equations are point-slope form, standard form, and intercept form. ### 3. What are linear equations? A linear equation is an equation in which the variable(s) is(are) with the exponent 1 Example: $2 x = 23$ $x - y = 5$ ### 4. How does one solve the system of linear equations? We have different methods to solve the system of linear equations: Graphical Method Substitution Method Cross Multiplication Method Elimination Method Determinants Method ### 5. What is the formula of linear equations in two variables? The general equation of linear equation in two variables is $ax + by + c = 0$ ### 6. Can a linear equation have 2 solutions? No, the system of the linear equation may have unique or one, no or zero, and an infinite number of solutions. More Important Topics Numbers Algebra Geometry Measurement Money Data Trigonometry Calculus More Important Topics Numbers Algebra Geometry Measurement Money Data Trigonometry Calculus	{}
1. ## surface area about x-axis I need to find the surface area of the above curve, about the x-axis. I'm having trouble taking the integral. Once everthing is set up its: $2pi\int(\sqrt{x})(\sqrt{(1+(1/4x)^2))}$ from 4 to 9 Or at least, that's what I think it is.. Any help would be greatly appreciated. 2. Originally Posted by cdlegendary I need to find the surface area of the above curve, about the x-axis. I'm having trouble taking the integral. Once everthing is set up its: $2pi\int(\sqrt{x})(\sqrt{(1+(1/4x)^2))}$ from 4 to 9 Or at least, that's what I think it is.. Any help would be greatly appreciated. As you know, the surface area of a revolution is found through the following formula: $A = 2\pi \int _a ^b y \sqrt{1+(\frac{dy}{dx})^2} dx$ Your integral is correct except for the $(\frac{dy}{dx})^2$ term, just need to recheck the square. 3. Ah, I made a typo. so the integral would be: $2pi\int(\sqrt{x})(\sqrt{(1+(1/4x)))}$ since the derivative of $\sqrt(x)$ is $1/(2\sqrt(x))$ is it not? 4. Originally Posted by cdlegendary Ah, I made a typo. so the integral would be: $2pi\int(\sqrt{x})(\sqrt{(1+(1/4x)))}$ since the derivative of $\sqrt(x)$ is $1/(2\sqrt(x))$ is it not? You bet. 5. thanks, I'm still having trouble taking the integral though. any pointers? 6. How about that $\sqrt{x}\sqrt{1+\frac1{4x}}=\sqrt{x\left(1+\frac1{ 4x}\right)}=\sqrt{x+\frac14}=\sqrt{\frac{4x+1}4}=\ frac{\sqrt{4x+1}}2$? --Kevin C.	{}
# Acceleration (Redirected from Deceleration) In physics, acceleration is the rate of change of velocity of an object with respect to time. An object's acceleration is the net result of all forces acting on the object, as described by Newton's Second Law.[1] The SI unit for acceleration is metre per second squared (m⋅s−2). Accelerations are vector quantities (they have magnitude and direction) and add according to the parallelogram law.[2][3] The vector of the net force acting on a body has the same direction as the vector of the body's acceleration, and its magnitude is proportional to the magnitude of the acceleration, with the object's mass (a scalar quantity) as proportionality constant. Acceleration In the absence of air resistance and thus terminal velocity, a falling ball would continue to accelerate. Common symbols a SI unitm/s2, m·s−2, m s−2 DimensionL T −2 For example, when a car starts from a standstill (zero velocity, in an inertial frame of reference) and travels in a straight line at increasing speeds, it is accelerating in the direction of travel. If the car turns, an acceleration occurs toward the new direction. The forward acceleration of the car is called a linear (or tangential) acceleration, the reaction to which passengers in the car experience as a force pushing them back into their seats. When changing direction, this is called radial (as orthogonal to tangential) acceleration, the reaction to which passengers experience as a sideways force. If the speed of the car decreases, this is an acceleration in the opposite direction of the velocity of the vehicle, sometimes called deceleration or Retrograde burning in spacecraft.[4] Passengers experience the reaction to deceleration as a force pushing them forwards. Both acceleration and deceleration are treated the same, they are both changes in velocity. Each of these accelerations (tangential, radial, deceleration) is felt by passengers until their velocity (speed and direction) matches that of the uniformly moving car. ## Definition and properties Kinematic quantities of a classical particle: mass m, position r, velocity v, acceleration a. ### Average acceleration Acceleration is the rate of change of velocity. At any point on a trajectory, the magnitude of the acceleration is given by the rate of change of velocity in both magnitude and direction at that point. The true acceleration at time t is found in the limit as time interval Δt → 0 of Δv/Δt An object's average acceleration over a period of time is its change in velocity ${\displaystyle (\Delta \mathbf {v} )}$ divided by the duration of the period ${\displaystyle (\Delta t)}$ . Mathematically, ${\displaystyle {\bar {\mathbf {a} }}={\frac {\Delta \mathbf {v} }{\Delta t}}.}$ ### Instantaneous acceleration From bottom to top: • an acceleration function a(t); • the integral of the acceleration is the velocity function v(t); • and the integral of the velocity is the distance function s(t). Instantaneous acceleration, meanwhile, is the limit of the average acceleration over an infinitesimal interval of time. In the terms of calculus, instantaneous acceleration is the derivative of the velocity vector with respect to time: ${\displaystyle \mathbf {a} =\lim _{{\Delta t}\to 0}{\frac {\Delta \mathbf {v} }{\Delta t}}={\frac {d\mathbf {v} }{dt}}}$ (Here and elsewhere, if motion is in a straight line, vector quantities can be substituted by scalars in the equations.) It can be seen that the integral of the acceleration function a(t) is the velocity function v(t); that is, the area under the curve of an acceleration vs. time (a vs. t) graph corresponds to velocity. ${\displaystyle \mathbf {v} =\int \mathbf {a} \ dt}$ As acceleration is defined as the derivative of velocity, v, with respect to time t and velocity is defined as the derivative of position, x, with respect to time, acceleration can be thought of as the second derivative of x with respect to t: ${\displaystyle \mathbf {a} ={\frac {d\mathbf {v} }{dt}}={\frac {d^{2}\mathbf {x} }{dt^{2}}}}$ ### Units Acceleration has the dimensions of velocity (L/T) divided by time, i.e. L T−2. The SI unit of acceleration is the metre per second squared (m s−2); or "metre per second per second", as the velocity in metres per second changes by the acceleration value, every second. ### Other forms An object moving in a circular motion—such as a satellite orbiting the Earth—is accelerating due to the change of direction of motion, although its speed may be constant. In this case it is said to be undergoing centripetal (directed towards the center) acceleration. Proper acceleration, the acceleration of a body relative to a free-fall condition, is measured by an instrument called an accelerometer. In classical mechanics, for a body with constant mass, the (vector) acceleration of the body's center of mass is proportional to the net force vector (i.e. sum of all forces) acting on it (Newton's second law): ${\displaystyle \mathbf {F} =m\mathbf {a} \quad \to \quad \mathbf {a} ={\frac {\mathbf {F} }{m}}}$ where F is the net force acting on the body, m is the mass of the body, and a is the center-of-mass acceleration. As speeds approach the speed of light, relativistic effects become increasingly large. ## Tangential and centripetal acceleration An oscillating pendulum, with velocity and acceleration marked. It experiences both tangential and centripetal acceleration. Components of acceleration for a curved motion. The tangential component at is due to the change in speed of traversal, and points along the curve in the direction of the velocity vector (or in the opposite direction). The normal component (also called centripetal component for circular motion) ac is due to the change in direction of the velocity vector and is normal to the trajectory, pointing toward the center of curvature of the path. The velocity of a particle moving on a curved path as a function of time can be written as: ${\displaystyle \mathbf {v} (t)=v(t){\frac {\mathbf {v} (t)}{v(t)}}=v(t)\mathbf {u} _{\mathrm {t} }(t),}$ with v(t) equal to the speed of travel along the path, and ${\displaystyle \mathbf {u} _{\mathrm {t} }={\frac {\mathbf {v} (t)}{v(t)}}\ ,}$ a unit vector tangent to the path pointing in the direction of motion at the chosen moment in time. Taking into account both the changing speed v(t) and the changing direction of ut, the acceleration of a particle moving on a curved path can be written using the chain rule of differentiation[5] for the product of two functions of time as: {\displaystyle {\begin{alignedat}{3}\mathbf {a} &={\frac {\mathrm {d} \mathbf {v} }{\mathrm {d} t}}\\&={\frac {\mathrm {d} v}{\mathrm {d} t}}\mathbf {u} _{\mathrm {t} }+v(t){\frac {d\mathbf {u} _{\mathrm {t} }}{dt}}\\&={\frac {\mathrm {d} v}{\mathrm {d} t}}\mathbf {u} _{\mathrm {t} }+{\frac {v^{2}}{r}}\mathbf {u} _{\mathrm {n} }\ ,\\\end{alignedat}}} where un is the unit (inward) normal vector to the particle's trajectory (also called the principal normal), and r is its instantaneous radius of curvature based upon the osculating circle at time t. These components are called the tangential acceleration and the normal or radial acceleration (or centripetal acceleration in circular motion, see also circular motion and centripetal force). Geometrical analysis of three-dimensional space curves, which explains tangent, (principal) normal and binormal, is described by the Frenet–Serret formulas.[6][7] ## Special cases ### Uniform acceleration Calculation of the speed difference for a uniform acceleration Uniform or constant acceleration is a type of motion in which the velocity of an object changes by an equal amount in every equal time period. A frequently cited example of uniform acceleration is that of an object in free fall in a uniform gravitational field. The acceleration of a falling body in the absence of resistances to motion is dependent only on the gravitational field strength g (also called acceleration due to gravity). By Newton's Second Law the force ${\displaystyle \mathbf {F_{g}} }$ acting on a body is given by: ${\displaystyle \mathbf {F_{g}} =m\mathbf {g} }$ Because of the simple analytic properties of the case of constant acceleration, there are simple formulas relating the displacement, initial and time-dependent velocities, and acceleration to the time elapsed:[8] ${\displaystyle \mathbf {s} (t)=\mathbf {s} _{0}+\mathbf {v} _{0}t+{\tfrac {1}{2}}\mathbf {a} t^{2}=\mathbf {s} _{0}+{\frac {\mathbf {v} _{0}+\mathbf {v} (t)}{2}}t}$ ${\displaystyle \mathbf {v} (t)=\mathbf {v} _{0}+\mathbf {a} t}$ ${\displaystyle {v^{2}}(t)={v_{0}}^{2}+2\mathbf {a\cdot } [\mathbf {s} (t)-\mathbf {s} _{0}]}$ where • ${\displaystyle t}$ is the elapsed time, • ${\displaystyle \mathbf {s} _{0}}$ is the initial displacement from the origin, • ${\displaystyle \mathbf {s} (t)}$ is the displacement from the origin at time ${\displaystyle t}$ , • ${\displaystyle \mathbf {v} _{0}}$ is the initial velocity, • ${\displaystyle \mathbf {v} (t)}$ is the velocity at time ${\displaystyle t}$ , and • ${\displaystyle \mathbf {a} }$ is the uniform rate of acceleration. In particular, the motion can be resolved into two orthogonal parts, one of constant velocity and the other according to the above equations. As Galileo showed, the net result is parabolic motion, which describes, e. g., the trajectory of a projectile in a vacuum near the surface of Earth.[9] ### Circular motion Position vector r, always points radially from the origin. Velocity vector v, always tangent to the path of motion. Acceleration vector a, not parallel to the radial motion but offset by the angular and Coriolis accelerations, nor tangent to the path but offset by the centripetal and radial accelerations. Kinematic vectors in plane polar coordinates. Notice the setup is not restricted to 2d space, but may represent the osculating plane plane in a point of an arbitrary curve in any higher dimension. In uniform circular motion, that is moving with constant speed along a circular path, a particle experiences an acceleration resulting from the change of the direction of the velocity vector, while its magnitude remains constant. The derivative of the location of a point on a curve with respect to time, i.e. its velocity, turns out to be always exactly tangential to the curve, respectively orthogonal to the radius in this point. Since in uniform motion the velocity in the tangential direction does not change, the acceleration must be in radial direction, pointing to the center of the circle. This acceleration constantly changes the direction of the velocity to be tangent in the neighboring point, thereby rotating the velocity vector along the circle. • For a given speed ${\displaystyle v}$ , the magnitude of this geometrically caused acceleration (centripetal acceleration) is inversely proportional to the radius ${\displaystyle r}$ of the circle, and increases as the square of this speed: ${\displaystyle a_{c}={\frac {v^{2}}{r}}\;.}$ • Note that, for a given angular velocity ${\displaystyle \omega }$ , the centripetal acceleration is directly proportional to radius ${\displaystyle r}$ . This is due to the dependence of velocity ${\displaystyle v}$ on the radius ${\displaystyle r}$ . ${\displaystyle v=\omega r.}$ Expressing centripetal acceleration vector in polar components, where ${\displaystyle \mathbf {r} }$ is a vector from the centre of the circle to the particle with magnitude equal to this distance, and considering the orientation of the acceleration towards the center, yields ${\displaystyle \mathbf {a_{c}} =-{\frac {v^{2}}{\|\mathbf {r} \|}}\cdot {\frac {\mathbf {r} }{\|\mathbf {r} \|}}\;.}$ As usual in rotations, the speed ${\displaystyle v}$ of a particle may be expressed as an angular speed with respect to a point at the distance ${\displaystyle r}$ as ${\displaystyle \omega ={\frac {v}{r}}.}$ Thus ${\displaystyle \mathbf {a_{c}} =-\omega ^{2}\mathbf {r} \;.}$ This acceleration and the mass of the particle determine the necessary centripetal force, directed toward the centre of the circle, as the net force acting on this particle to keep it in this uniform circular motion. The so-called 'centrifugal force', appearing to act outward on the body, is a so-called pseudo force experienced in the frame of reference of the body in circular motion, due to the body's linear momentum, a vector tangent to the circle of motion. In a nonuniform circular motion, i.e., the speed along the curved path is changing, the acceleration has a non-zero component tangential to the curve, and is not confined to the principal normal, which directs to the center of the osculating circle, that determines the radius ${\displaystyle r}$ for the centripetal acceleration. The tangential component is given by the angular acceleration ${\displaystyle \alpha }$ , i.e., the rate of change ${\displaystyle \alpha ={\dot {\omega }}}$ of the angular speed ${\displaystyle \omega }$ times the radius ${\displaystyle r}$ . That is, ${\displaystyle a_{c}=r\alpha .}$ The sign of the tangential component of the acceleration is determined by the sign of the angular acceleration (${\displaystyle \alpha }$ ), and the tangent is of course always directed at right angles to the radius vector. ## Relation to relativity ### Special relativity The special theory of relativity describes the behavior of objects traveling relative to other objects at speeds approaching that of light in a vacuum. Newtonian mechanics is exactly revealed to be an approximation to reality, valid to great accuracy at lower speeds. As the relevant speeds increase toward the speed of light, acceleration no longer follows classical equations. As speeds approach that of light, the acceleration produced by a given force decreases, becoming infinitesimally small as light speed is approached; an object with mass can approach this speed asymptotically, but never reach it. ### General relativity Unless the state of motion of an object is known, it is impossible to distinguish whether an observed force is due to gravity or to acceleration—gravity and inertial acceleration have identical effects. Albert Einstein called this the equivalence principle, and said that only observers who feel no force at all—including the force of gravity—are justified in concluding that they are not accelerating.[10] ## Conversions Conversions between common units of acceleration Base value (Gal, or cm/s2) (ft/s2) (m/s2) (Standard gravity, g0) 1 Gal, or cm/s2 1 0.0328084 0.01 0.00101972 1 ft/s2 30.4800 1 0.304800 0.0310810 1 m/s2 100 3.28084 1 0.101972 1 g0 980.665 32.1740 9.80665 1	{}
# connection between discrete valuation rings and points of a curve. Let $C$ be a projective irreducible non-singular curve over a field $k$ and let K be its function field. It applies that $(k[X,Y]/I(C))_{(X-a,Y-b)}$ (i.e. the localization of $k[X,Y]/I(C)$ at $(X-a,Y-b)$) is a discrete valuation ring w.r.t. $K$ and $k$, where $(a,b)$ is a point on $C$. My question is now: Can any discrete valuation ring w.r.t. $K$ and $k$ linked to $C$? (maybe to some points or a prime ideal of $k[X,Y]/I(C)$)? Note that if $k$ is algebraic closed any discrete valuation ring is a local ring of some point of $C$. • What is your question? All valuations on a curve are given by expansion in power series. – user40276 Nov 30 '14 at 22:51 • @user40276: I edit my post.(Thanks) Which expansion in power series? Note that I only consider discrete valuation ring (from geometric point of view). – bjn Nov 30 '14 at 22:57 Suppose that $C/k$ is an integral projective curve. Then, there is a bijection $$\left\{\text{points of }C\right\}\longleftrightarrow\left\{\begin{matrix}\text{discrete valuations}\\\text{of }K(c)\\ \text{which are trivial}\\ \text{on }k\end{matrix}\right\}$$ The mapping being, as you indicating, sending $p$ to the valuation $v_p$ which is "order of vanishing of a function at $p$".	{}
### Find intersection of two sorted integer arrays if one array is very large in size than other sortedarray array searching Posted: 4 years ago Updated: 3 years ago Edit answers (1) views (2484) Write the algorithm in Find intersection of two sorted integer arrays when one array is very large in size than the other one. Posted: 4 years ago 0 4 Edit When one array is very large and the other one is very small we can use binary search. Suppose array1 [ ] has m and array2 [ ] has n number of elements respectively where m < < n . We can iterate through array1 [ ] (since it has less number of elements) and binary search for that element in array2 [ ].Time complexity = $$O(mlogn)$$.public static void findIntersectionOflargearray(int[] array1, int[] array2, int m, int n) { /* loop for all numbers in a small array: array1[] / for (int i = 0; i < m; i++) { / perform binary search on the large array : array2[] */ if (binarySearch(array2, array1[i], 0, n) != -1) system.out.println("%d ", array1[i]); }}public static int binarySearch(int array[], int number, int start, int end) { if (end < start) return -1; int mid = (start + end)/2; if (a[mid] > number) return binarySearch(array, number, start, mid-1); else if (a[mid] < number) return binarySearch(array, number, mid+1, end); else return mid;}	{}
# D{0 < <0<'t>{+t} ="25.Vi6 -r - ###### Question: d {0 < <0<'t>{+t} =" 25. Vi6 -r - #### Similar Solved Questions ##### The Family Planning Association in Oxford, England studied 17,000 women who had been enrolled in a... The Family Planning Association in Oxford, England studied 17,000 women who had been enrolled in a cohort study between 1968 and 1974 to look at the association between oral contraceptive (OC) use and venous thromboembolism (Vessey et al., 1989). For current users of OCs, person-time was counted fro... ##### Please exaplain in steps A balls rolls off a shelf with the initial velocity v and... please exaplain in steps A balls rolls off a shelf with the initial velocity v and lands at the horizontal distance d from the shelf. What is the height of the shelf h above the floor?... ##### How many ng of â‚¬ can be isolated from 0.0015 mg of Niz(COal3 carbonate? How many ng of â‚¬ can be isolated from 0.0015 mg of Niz(COal3 carbonate?... ##### A stone weighing 4 pounds is thrown upward with an initial height of 10 ft and an initial speed of 32ft/s_ Air resistance is proportional to speed, with proportionality constant k= 1/16.Find the velocity of the stone as a function of timev(t)Find the maximum height the stone reaches_ Express your answer in decimal form to 2 decimal places_ A stone weighing 4 pounds is thrown upward with an initial height of 10 ft and an initial speed of 32ft/s_ Air resistance is proportional to speed, with proportionality constant k= 1/16. Find the velocity of the stone as a function of time v(t) Find the maximum height the stone reaches_ Express your... ##### Question 21 pts7.03kg box is sliding on a hollow circular tube as shown in the sketch: If the speed of the box is 7.01 m/s when it is at the position shown in the sketch what is the value of normal force in unit newtons on the box at the position? Assume the radius of the path of block is 1.23 m and use g = 9.80 m/s?,if needed. Question 2 1 pts 7.03kg box is sliding on a hollow circular tube as shown in the sketch: If the speed of the box is 7.01 m/s when it is at the position shown in the sketch what is the value of normal force in unit newtons on the box at the position? Assume the radius of the path of block is 1.23 m a... ##### A cube of gold that is $1.00 mathrm{~cm}$ on a side has a mass of $19.3$ g. A single gold atom has a mass of $197.0$ amu.(a) How many gold atoms are in the cube? (b) From the information given, estimate the diameter in $AA$ of a single gold atom. (c) What assumptions did you make in arriving at your answer for part (b)? A cube of gold that is $1.00 mathrm{~cm}$ on a side has a mass of $19.3$ g. A single gold atom has a mass of $197.0$ amu. (a) How many gold atoms are in the cube? (b) From the information given, estimate the diameter in $AA$ of a single gold atom. (c) What assumptions did you make in arriving at you... ##### A solution has [H]] = 1.6x10-3 MWhat is the pH of this solution? A solution has [H]] = 1.6x10-3 M What is the pH of this solution?... ##### Show works please Q1 Improper Integrals 10 Points Evaluate the following integrals and determine if they... show works please Q1 Improper Integrals 10 Points Evaluate the following integrals and determine if they converge. If they converge, find the value of the integral. Show all of your work. Q1.1 5 Points et L dx 2 + ex Upload your file showing your work. Please select file(s) Select file(s) Q1.2 5 Poi... ##### If I < -1 I + 1 if 11 < I < 1 0 if I > 1Sketch a graph of f(z)7 -IIV Jeap) Draw: if I < -1 I + 1 if 11 < I < 1 0 if I > 1 Sketch a graph of f(z) 7 - IIV Jeap) Draw:... ##### This question is related to real transformers. Kindly show all working. Assume at full load. 25KVA... This question is related to real transformers. Kindly show all working. Assume at full load. 25KVA Question 2 An ideal transformer has 500 turns on the primary winding and 40 turns on the secondary winding. The primary is connected to 3000 Volts, 50 Hz supply. Calculate (a) Primary and secondary cur... ##### What is the correct answer? When converting an income statement from a cash basis to an... What is the correct answer? When converting an income statement from a cash basis to an accrual basis, expenses: Multiple Choice Exceed cash payments to suppliers. Equal cash payments to suppliers. Are less than cash payments to suppliers. May exceed, equal or be less than cash payments ... ##### Assume you have a business with ten employees, and you want to choose a time slot to give a lunch break to them. You have four different time slots as candidates, and you want to choose the one in which the chances of having customers are lower. How would you use the knowledge of this unit to take this decision? Which probability distribution do you use? What data do you need to collect?this is discussion question so i have give response according to it Assume you have a business with ten employees, and you want to choose a time slot to give a lunch break to them. You have four different time slots as candidates, and you want to choose the one in which the chances of having customers are lower. How would you use the knowledge of this unit to take t... ##### Trace through the bubble sort algorithm for the following data set: X1 = 8, X2 = 3 X3 = 7 , X4 = 4,X5 = 6,and X6 = 1. Trace through the bubble sort algorithm for the following data set: X1 = 8, X2 = 3 X3 = 7 , X4 = 4,X5 = 6,and X6 = 1.... ##### U" 5C charge placed atadistance of 1 wire of length 100 m with a charge 8 U" 5C charge placed atadistance of 1 wire of length 100 m with a charge 8... ##### A 4.55 sample of aluminum reacted completely with 9.62 g sample of Nickellll) oxide. After the reaction, which reactant was still present? What mass of this reactant was present? A 4.55 sample of aluminum reacted completely with 9.62 g sample of Nickellll) oxide. After the reaction, which reactant was still present? What mass of this reactant was present?... ##### How do you solve -2( - 2+ 5m ) < 64? How do you solve -2( - 2+ 5m ) < 64?... ##### Discuss the types of runoff pollution generated by agriculture and cities and the policies available to... Discuss the types of runoff pollution generated by agriculture and cities and the policies available to abate them in great detail.... ##### Question: Organic Chemistry (Biosynthetic Pathways)Name and draw natureâ€™s alkylation mechanisms Question: Organic Chemistry (Biosynthetic Pathways) Name and draw natureâ€™s alkylation mechanisms... ##### (12 pts:) Find the outward flux of the vector field F(z,y.2) = (ri + yj + ck)/(r? +y? + 2213/2 across the ellipsoid 4x2 + 9y? + 22 = 1. (12 pts:) Find the outward flux of the vector field F(z,y.2) = (ri + yj + ck)/(r? +y? + 2213/2 across the ellipsoid 4x2 + 9y? + 22 = 1.... ##### In terms of DNA mutations __________ are more likely than__________?A. Transitions, TransversionsB. Transitions, TransubstitutionsC. Transversions, TransitionsD. Transversions, TransubstitutionsE. None of the Above In terms of DNA mutations __________ are more likely than __________? A. Transitions, Transversions B. Transitions, Transubstitutions C. Transversions, Transitions D. Transversions, Transubstitutions E. None of the Above... ##### In Exercises 27-34, find the vertex, focus, and directrix of the parabola. Then sketch the parabola.$(x-1)^{2}+8(y+2)=0$ In Exercises 27-34, find the vertex, focus, and directrix of the parabola. Then sketch the parabola. $(x-1)^{2}+8(y+2)=0$... ##### Find the derivative of the function. 13x 12x+2 =dy OA dx26x - 2x-2d 0 B. dx26 2xdx 0c dxl2 13x + 12 + 2xd 0Dl dx4 2 26x + 12 - 2x Find the derivative of the function. 13x 12x+2 = dy OA dx 26x - 2x-2 d 0 B. dx 26 2x dx 0c dxl 2 13x + 12 + 2x d 0Dl dx 4 2 26x + 12 - 2x... ##### Zn(s)+ MnO4-(aq)-> Zn2+(aq) +Mn2+(aq) write cell notation, equilibrium constant annd will this function as a battery? Zn(s)+ MnO4-(aq)-> Zn2+(aq) +Mn2+(aq) write cell notation, equilibrium constant annd will this function as a battery?... ##### A study was carried out to understand and analyze the investmentbehaviour of the employees of Public Sector Units(PSUs) andgovernment. A sample of 80 respondents was drawn from the PSU andgovernment employees in the vicinity of Delhi. The respondents wereasked to state their level of agreement or disagreement on thefollowing parameters on a 5 point scale, where 1 = StronglyDisagree, 2 = Disagree, 3 = Neutral, 4 = Agree, 5 = Strongly Agree.The seven parameters in question were the importance give A study was carried out to understand and analyze the investment behaviour of the employees of Public Sector Units(PSUs) and government. A sample of 80 respondents was drawn from the PSU and government employees in the vicinity of Delhi. The respondents were asked to state their level of agreement o... ##### Analytical exam 3 - Fall 2019 ANSWER ALL QUESTION with full explanation and calculations for full... Analytical exam 3 - Fall 2019 ANSWER ALL QUESTION with full explanation and calculations for full credit. 1. Consider the reaction for the decomposition of hydrogen disulfide: H2(g) → 2H2(g) + S2(8) With an equilibrium constant of 1.67*10' (at 800 °C). If a 0.500 L reaction vessel init... ##### Point) Are the vectors 3andlinearly independent?Hint: You may use the determinant ofto decide If the vectorsandlinearly independentRecal that set = vectors V1 , Vk are linearly independent and only c" Useful Theorem: For square matrix B, Det(B) / if and only Bi 0 nas unique solution I 0+CUk 0 has unique sclutionChooseIf Iney are linearly dependent; iind scalars that are not zero sucn thai the equation below true. If they are Iinearly Independent; find the only scalars that will make tne eq point) Are the vectors 3 and linearly independent? Hint: You may use the determinant of to decide If the vectors and linearly independent Recal that set = vectors V1 , Vk are linearly independent and only c" Useful Theorem: For square matrix B, Det(B) / if and only Bi 0 nas unique solution I 0... ##### Which of the following is not a function of eukaryotic mRNA polyadenylation? Nuclear export of mRNAs... Which of the following is not a function of eukaryotic mRNA polyadenylation? Nuclear export of mRNAs Stabilization of mRNAs Recruitment of components of the translation machinery Destabilization of mRNAs... ##### Problem 14-04A a-c (Video) (Part Level Submission) The following data were taken from the records of... Problem 14-04A a-c (Video) (Part Level Submission) The following data were taken from the records of Crane Company for the fiscal year ended June 30, 2020 Raw Materials Inventory 7/1/19 Raw Materials Inventory 6/30/20 Finished Goods Inventory 7/1/19 Finished Goods Inventory 6/30/20 Work in Process I... ##### Which of the following are microeconomic questions? (Check all that apply.) Why are earnings of economics... Which of the following are microeconomic questions? (Check all that apply.) Why are earnings of economics majors higher than those of most other majors? Why are gas prices falling? What is the average hourly wage in the United States? How will a U.S. tariff on imports of t-shirts... ##### Find a general solution. Differentials course 10. x'(t) = = (1 -1) 2(6) 1 Find a general solution. Differentials course 10. x'(t) = = (1 -1) 2(6) 1... ##### What is the major product expected from the following reaction? [11] what is major product(s) e... What is the major product expected from the following reaction? [11] what is major product(s) e Cyclohexanone + methylamine... ##### Which of the following is not true regarding translation exposure? a. Since earnings can affect stock... Which of the following is not true regarding translation exposure? a. Since earnings can affect stock prices, many MNCs are concerned about translation exposure b. Since translation of financial statements does not affect an MNC's cash flows, some analysts suggest that translation exposure is no... ##### 8.3.24 Homework: 1 Section BLalallc? HomeworkMacBook Alr1 1 1 AdLoa PLSLT 8.3.24 Homework: 1 Section BLalallc? Homework MacBook Alr 1 1 1 AdLoa PLSLT... ##### Solve each inequality by graphing an appropriate function. State the solution set using interval notation$(x-1)^{2}-9<0$. Solve each inequality by graphing an appropriate function. State the solution set using interval notation $(x-1)^{2}-9<0$....	{}
# 3.5 Transformation of functions (Page 11/21) Page 11 / 21 When examining the formula of a function that is the result of multiple transformations, how can you tell a horizontal stretch from a vertical stretch? When examining the formula of a function that is the result of multiple transformations, how can you tell a horizontal compression from a vertical compression? A horizontal compression results when a constant greater than 1 is multiplied by the input. A vertical compression results when a constant between 0 and 1 is multiplied by the output. When examining the formula of a function that is the result of multiple transformations, how can you tell a reflection with respect to the x -axis from a reflection with respect to the y -axis? How can you determine whether a function is odd or even from the formula of the function? For a function $\text{\hspace{0.17em}}f,\text{\hspace{0.17em}}$ substitute $\text{\hspace{0.17em}}\left(-x\right)\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}\left(x\right)\text{\hspace{0.17em}}$ in $\text{\hspace{0.17em}}f\left(x\right).\text{\hspace{0.17em}}$ Simplify. If the resulting function is the same as the original function, $\text{\hspace{0.17em}}f\left(-x\right)=f\left(x\right),\text{\hspace{0.17em}}$ then the function is even. If the resulting function is the opposite of the original function, $\text{\hspace{0.17em}}f\left(-x\right)=-f\left(x\right),\text{\hspace{0.17em}}$ then the original function is odd. If the function is not the same or the opposite, then the function is neither odd nor even. ## Algebraic For the following exercises, write a formula for the function obtained when the graph is shifted as described. $\text{\hspace{0.17em}}f\left(x\right)=\sqrt{x}\text{\hspace{0.17em}}$ is shifted up 1 unit and to the left 2 units. $\text{\hspace{0.17em}}f\left(x\right)=\|x\|\text{\hspace{0.17em}}$ is shifted down 3 units and to the right 1 unit. $g\left(x\right)=\|x-1\|-3$ $\text{\hspace{0.17em}}f\left(x\right)=\frac{1}{x}\text{\hspace{0.17em}}$ is shifted down 4 units and to the right 3 units. $\text{\hspace{0.17em}}f\left(x\right)=\frac{1}{{x}^{2}}\text{\hspace{0.17em}}$ is shifted up 2 units and to the left 4 units. $g\left(x\right)=\frac{1}{{\left(x+4\right)}^{2}}+2$ For the following exercises, describe how the graph of the function is a transformation of the graph of the original function $\text{\hspace{0.17em}}f.$ $y=f\left(x-49\right)$ $y=f\left(x+43\right)$ The graph of $\text{\hspace{0.17em}}f\left(x+43\right)\text{\hspace{0.17em}}$ is a horizontal shift to the left 43 units of the graph of $\text{\hspace{0.17em}}f.$ $y=f\left(x+3\right)$ $y=f\left(x-4\right)$ The graph of $\text{\hspace{0.17em}}f\left(x-4\right)\text{\hspace{0.17em}}$ is a horizontal shift to the right 4 units of the graph of $\text{\hspace{0.17em}}f.$ $y=f\left(x\right)+5$ $y=f\left(x\right)+8$ The graph of $\text{\hspace{0.17em}}f\left(x\right)+8\text{\hspace{0.17em}}$ is a vertical shift up 8 units of the graph of $\text{\hspace{0.17em}}f.$ $y=f\left(x\right)-2$ $y=f\left(x\right)-7$ The graph of $\text{\hspace{0.17em}}f\left(x\right)-7\text{\hspace{0.17em}}$ is a vertical shift down 7 units of the graph of $\text{\hspace{0.17em}}f.$ $y=f\left(x-2\right)+3$ $y=f\left(x+4\right)-1$ The graph of $f\left(x+4\right)-1$ is a horizontal shift to the left 4 units and a vertical shift down 1 unit of the graph of $f.$ For the following exercises, determine the interval(s) on which the function is increasing and decreasing. $f\left(x\right)=4{\left(x+1\right)}^{2}-5$ $g\left(x\right)=5{\left(x+3\right)}^{2}-2$ decreasing on $\text{\hspace{0.17em}}\left(-\infty ,-3\right)\text{\hspace{0.17em}}$ and increasing on $\text{\hspace{0.17em}}\left(-3,\infty \right)$ $a\left(x\right)=\sqrt{-x+4}$ $k\left(x\right)=-3\sqrt{x}-1$ decreasing on $\left(0,\text{\hspace{0.17em}}\infty \right)$ ## Graphical For the following exercises, use the graph of $\text{\hspace{0.17em}}f\left(x\right)={2}^{x}\text{\hspace{0.17em}}$ shown in [link] to sketch a graph of each transformation of $\text{\hspace{0.17em}}f\left(x\right).$ $g\left(x\right)={2}^{x}+1$ $h\left(x\right)={2}^{x}-3$ $w\left(x\right)={2}^{x-1}$ For the following exercises, sketch a graph of the function as a transformation of the graph of one of the toolkit functions. $f\left(t\right)={\left(t+1\right)}^{2}-3$ $h\left(x\right)=\|x-1\|+4$ $k\left(x\right)={\left(x-2\right)}^{3}-1$ $m\left(t\right)=3+\sqrt{t+2}$ ## Numeric Tabular representations for the functions $\text{\hspace{0.17em}}f,\text{\hspace{0.17em}}g,\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}h\text{\hspace{0.17em}}$ are given below. Write $\text{\hspace{0.17em}}g\left(x\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}h\left(x\right)\text{\hspace{0.17em}}$ as transformations of $\text{\hspace{0.17em}}f\left(x\right).$ $x$ −2 −1 0 1 2 $f\left(x\right)$ −2 −1 −3 1 2 $x$ −1 0 1 2 3 $g\left(x\right)$ −2 −1 −3 1 2 $x$ −2 −1 0 1 2 $h\left(x\right)$ −1 0 −2 2 3 $g\left(x\right)=f\left(x-1\right),\text{\hspace{0.17em}}h\left(x\right)=f\left(x\right)+1$ Tabular representations for the functions $\text{\hspace{0.17em}}f,\text{\hspace{0.17em}}g,\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}h\text{\hspace{0.17em}}$ are given below. Write $\text{\hspace{0.17em}}g\left(x\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}h\left(x\right)\text{\hspace{0.17em}}$ as transformations of $\text{\hspace{0.17em}}f\left(x\right).$ $x$ −2 −1 0 1 2 $f\left(x\right)$ −1 −3 4 2 1 $x$ −3 −2 −1 0 1 $g\left(x\right)$ −1 −3 4 2 1 $x$ −2 −1 0 1 2 $h\left(x\right)$ −2 −4 3 1 0 For the following exercises, write an equation for each graphed function by using transformations of the graphs of one of the toolkit functions. $f\left(x\right)=\|x-3\|-2$ $f\left(x\right)=\sqrt{x+3}-1$ $f\left(x\right)={\left(x-2\right)}^{2}$ $f\left(x\right)=\|x+3\|-2$ For the following exercises, use the graphs of transformations of the square root function to find a formula for each of the functions. $f\left(x\right)=-\sqrt{x}$ For the following exercises, use the graphs of the transformed toolkit functions to write a formula for each of the resulting functions. $f\left(x\right)=-{\left(x+1\right)}^{2}+2$ $f\left(x\right)=\sqrt{-x}+1$ For the following exercises, determine whether the function is odd, even, or neither. $f\left(x\right)=3{x}^{4}$ even $g\left(x\right)=\sqrt{x}$ $h\left(x\right)=\frac{1}{x}+3x$ odd $f\left(x\right)={\left(x-2\right)}^{2}$ $g\left(x\right)=2{x}^{4}$ even $h\left(x\right)=2x-{x}^{3}$ For the following exercises, describe how the graph of each function is a transformation of the graph of the original function $\text{\hspace{0.17em}}f.$ $g\left(x\right)=-f\left(x\right)$ The graph of $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ is a vertical reflection (across the $\text{\hspace{0.17em}}x$ -axis) of the graph of $\text{\hspace{0.17em}}f.$ $g\left(x\right)=f\left(-x\right)$ $g\left(x\right)=4f\left(x\right)$ The graph of $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ is a vertical stretch by a factor of 4 of the graph of $\text{\hspace{0.17em}}f.$ $g\left(x\right)=6f\left(x\right)$ $g\left(x\right)=f\left(5x\right)$ The graph of $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ is a horizontal compression by a factor of $\text{\hspace{0.17em}}\frac{1}{5}\text{\hspace{0.17em}}$ of the graph of $\text{\hspace{0.17em}}f.$ $g\left(x\right)=f\left(2x\right)$ $g\left(x\right)=f\left(\frac{1}{3}x\right)$ The graph of $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ is a horizontal stretch by a factor of 3 of the graph of $\text{\hspace{0.17em}}f.$ $g\left(x\right)=f\left(\frac{1}{5}x\right)$ $g\left(x\right)=3f\left(-x\right)$ The graph of $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ is a horizontal reflection across the $\text{\hspace{0.17em}}y$ -axis and a vertical stretch by a factor of 3 of the graph of $\text{\hspace{0.17em}}f.$ $g\left(x\right)=-f\left(3x\right)$ For the following exercises, write a formula for the function $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ that results when the graph of a given toolkit function is transformed as described. The graph of $\text{\hspace{0.17em}}f\left(x\right)=\|x\|\text{\hspace{0.17em}}$ is reflected over the $\text{\hspace{0.17em}}y$ - axis and horizontally compressed by a factor of $\text{\hspace{0.17em}}\frac{1}{4}$ . $g\left(x\right)=\|-4x\|$ The graph of $\text{\hspace{0.17em}}f\left(x\right)=\sqrt{x}\text{\hspace{0.17em}}$ is reflected over the $\text{\hspace{0.17em}}x$ -axis and horizontally stretched by a factor of 2. The graph of $\text{\hspace{0.17em}}f\left(x\right)=\frac{1}{{x}^{2}}\text{\hspace{0.17em}}$ is vertically compressed by a factor of $\text{\hspace{0.17em}}\frac{1}{3},\text{\hspace{0.17em}}$ then shifted to the left 2 units and down 3 units. $g\left(x\right)=\frac{1}{3{\left(x+2\right)}^{2}}-3$ The graph of $\text{\hspace{0.17em}}f\left(x\right)=\frac{1}{x}\text{\hspace{0.17em}}$ is vertically stretched by a factor of 8, then shifted to the right 4 units and up 2 units. The graph of $\text{\hspace{0.17em}}f\left(x\right)={x}^{2}\text{\hspace{0.17em}}$ is vertically compressed by a factor of $\text{\hspace{0.17em}}\frac{1}{2},\text{\hspace{0.17em}}$ then shifted to the right 5 units and up 1 unit. $g\left(x\right)=\frac{1}{2}{\left(x-5\right)}^{2}+1$ The graph of $\text{\hspace{0.17em}}f\left(x\right)={x}^{2}\text{\hspace{0.17em}}$ is horizontally stretched by a factor of 3, then shifted to the left 4 units and down 3 units. For the following exercises, describe how the formula is a transformation of a toolkit function. Then sketch a graph of the transformation. $g\left(x\right)=4{\left(x+1\right)}^{2}-5$ The graph of the function $\text{\hspace{0.17em}}f\left(x\right)={x}^{2}\text{\hspace{0.17em}}$ is shifted to the left 1 unit, stretched vertically by a factor of 4, and shifted down 5 units. $g\left(x\right)=5{\left(x+3\right)}^{2}-2$ $h\left(x\right)=-2\|x-4\|+3$ The graph of $\text{\hspace{0.17em}}f\left(x\right)=\|x\|\text{\hspace{0.17em}}$ is stretched vertically by a factor of 2, shifted horizontally 4 units to the right, reflected across the horizontal axis, and then shifted vertically 3 units up. $k\left(x\right)=-3\sqrt{x}-1$ $m\left(x\right)=\frac{1}{2}{x}^{3}$ The graph of the function $\text{\hspace{0.17em}}f\left(x\right)={x}^{3}\text{\hspace{0.17em}}$ is compressed vertically by a factor of $\text{\hspace{0.17em}}\frac{1}{2}.$ $n\left(x\right)=\frac{1}{3}\|x-2\|$ $p\left(x\right)={\left(\frac{1}{3}x\right)}^{3}-3$ The graph of the function is stretched horizontally by a factor of 3 and then shifted vertically downward by 3 units. $q\left(x\right)={\left(\frac{1}{4}x\right)}^{3}+1$ $a\left(x\right)=\sqrt{-x+4}$ The graph of $\text{\hspace{0.17em}}f\left(x\right)=\sqrt{x}\text{\hspace{0.17em}}$ is shifted right 4 units and then reflected across the vertical line $\text{\hspace{0.17em}}x=4.$ For the following exercises, use the graph in [link] to sketch the given transformations. $g\left(x\right)=f\left(x\right)-2$ $g\left(x\right)=-f\left(x\right)$ $g\left(x\right)=f\left(x+1\right)$ $g\left(x\right)=f\left(x-2\right)$ the third and the seventh terms of a G.P are 81 and 16, find the first and fifth terms. if a=3, b =4 and c=5 find the six trigonometric value sin pls how do I factorize x⁴+x³-7x²-x+6=0 in a function the input value is called how do I test for values on the number line if a=4 b=4 then a+b= a+b+2ab Kin commulative principle a+b= 4+4=8 Mimi If a=4 and b=4 then we add the value of a and b i.e a+b=4+4=8. Tariq what are examples of natural number an equation for the line that goes through the point (-1,12) and has a slope of 2,3 3y=-9x+25 Ishaq show that the set of natural numberdoes not from agroup with addition or multiplication butit forms aseni group with respect toaaddition as well as multiplication x^20+x^15+x^10+x^5/x^2+1 evaluate each algebraic expression. 2x+×_2 if ×=5 if the ratio of the root of ax+bx+c =0, show that (m+1)^2 ac =b^2m By the definition, is such that 0!=1.why? (1+cosA+IsinA)(1+cosB+isinB)/(cos@+isin@)(cos$+isin$) hatdog Mark jaks Ryan how we can draw three triangles of distinctly different shapes. All the angles will be cutt off each triangle and placed side by side with vertices touching	{}
#### 05-树8 File Transfer（25 分） We have a network of computers and a list of bi-directional connections. Each of these connections allows a file transfer from one computer to another. Is it possible to send a file from any computer on the network to any other? ### Input Specification: Each input file contains one test case. For each test case, the first line contains (), the total number of computers in a network. Each computer in the network is then represented by a positive integer between 1 and . Then in the following lines, the input is given in the format: I c1 c2 where I stands for inputting a connection between c1 and c2; or C c1 c2 where C stands for checking if it is possible to transfer files between c1 and c2; or S where S stands for stopping this case. ### Output Specification: For each C case, print in one line the word "yes" or "no" if it is possible or impossible to transfer files between c1 and c2, respectively. At the end of each case, print in one line "The network is connected." if there is a path between any pair of computers; or "There are k components." where k is the number of connected components in this network. ### Sample Input 1: 5 C 3 2 I 3 2 C 1 5 I 4 5 I 2 4 C 3 5 S ### Sample Output 1: no no yes There are 2 components. ### Sample Input 2: 5 C 3 2 I 3 2 C 1 5 I 4 5 I 2 4 C 3 5 I 1 3 C 1 5 S ### Sample Output 2: no no yes yes The network is connected. #include <set> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <cstdio> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #include <functional> #define mod 1000000007 const int INF = 0x7f7f7f7f; typedef long long ll; const int maxn = 10050; using namespace std; int vec[maxn] = {0}; int t; int Find(int x) { int i; if(vec[x] < 0) return x; for(i = vec[x]; vec[i]>=0; i = vec[i]); return i; } void Union( int x, int y) { int root1, root2; root1 = Find(x); root2 = Find(y); if(root1!=root2) { if(vec[root1]<vec[root2]) { vec[root1] += vec[root2]; vec[root2] = root1; t--; } else { vec[root2] += vec[root1]; vec[root1] = root2; t--; } } } int main() { scanf("%d", &t); getchar(); for(int i = 1; i <= t; i++) { vec[i] = -1; } while(1) { char temp; scanf("%c", &temp); if(temp == 'C') { int m, n; scanf("%d%d", &m, &n); getchar(); if(Find(m)==Find(n)) { printf("yes\n"); } else { printf("no\n"); //Union(m, n); } } else { if(temp == 'I') { int m, n; scanf("%d%d", &m, &n); getchar(); Union(m,n); } else { if(t>1) { printf("There are %d components.\n", t); break; } else { printf("The network is connected.\n"); break; } } } } return 0; } #### 并查集-05-树8 File Transfer 2017-05-12 12:34:52 #### 05-树9 File Transfer (25分) 2016-10-03 21:07:30 #### 05-树8 File Transfer (25分) 2016-04-02 15:16:51 #### 05-树8 File Transfer (25分)---并查集 2017-05-26 19:13:23 #### MOOC浙大数据结构 — 05-树9 File Transfer (25分) 2016-11-21 10:42:49 #### 05-树8 File Transfer（25 point(s)）【并查集】 2018-04-06 09:08:04 #### 05-树8 File Transfer（25 分） 2017-10-21 22:40:51 #### PAT 数据结构 04-树5. File Transfer (25) 2015-07-12 19:30:04 #### 05-树8 File Transfer【并查集】 2017-04-02 21:57:52 #### 05-树8 File Transfer (25分) 2017-05-30 15:28:10 ## 不良信息举报 05-树8 File Transfer	{}
# Tag Info 17 In a nonabelian setting the correct notion of kernel is given by the kernel pair, and the correct notion of cokernel is given by the cokernel pair. For example, in any category, a morphism $f : a \to b$ is a monomorphism iff its kernel pair exists and is trivial, and dually $f$ is an epimorphism iff its cokernel pair exists and is trivial. By comparison, the ... 16 Overview of an explanation : Jones-Wassermann subfactors for the loop algebra : Let $\mathfrak{g} = \mathfrak{sl}_{2}$ be the Lie algebra, $L\mathfrak{g}$ its loop algebra and $\mathcal{L}\mathfrak{g} = L\mathfrak{g} \oplus \mathbb{C}\mathcal{L}$ the central extension : $$[X^{a}_{n},X^{b}_{m}] = [X^{a},X^{b}]_{m+n} + m\delta_{ab}\delta_{m+n}\mathcal{L}$$ ... 11 At the very least there's one Bisch-Haagerup subfactor standard invariant for every quotient group of $\mathbb{Z}/2 \ast \mathbb{Z}/3 \cong \mathrm{PSL}_2(\mathbb{Z})$. That's already way too much to try to classify. Even if you restrict to finite depth, there's way too many finite quotients of the modular group. On the other hand, if you're very ... 10 For finite groups, the answer was given by Izumi in his paper "Characterization of isomorphic group-subgroup subfactors" (MR1920326). There he looks at the crossed product subfactor, but you can always take duals. Edit after @Andre's comment: The actual condition between the two pairs of subgroups is quite technical, and it would basically require ... 10 Given any two tracial von Neumann algebras $(N_1, \tau_1)$ and $(N_2, \tau_2)$ the $L^2$ space of the free product $(N_1 * N_2, \tau)$ canonically decomposes as $$L^2(N_1 * N_2, \tau) = \mathbb C \oplus_{n \in \mathbb N} \bigoplus_{i_1 \not= i_2, i_2 \not= i_3, \ldots, i_{n - 1} \not= i_n } \overline {\otimes}_{k = 1}^n L^2_0(N_{i_k}, \tau_{i_k}),$$ where ... 10 I think this paper of Wakui says that the answer is "No". The Turaev-Viro invariants associated to the generalized $E6$ subfactors for $\mathbb{Z}/7$ don't seem to distinguish $L(7,1)$ and $L(7,2)$. 10 The answer to (1) (for second countable groups) is yes: every locally compact second countable group $G$ admits a countinuous, faithful outer action on the hyperfinite $II_1$ factor. This is attributed to Blattner, and is stated explicitly in Proposition 1 of the following article: R. J. Plymen. "Automorphic group representations: The hyperfinite $II_1$ ... 9 If you also assume finite depth, then there's a hope (it's too vague to call it a conjecture) that all integer index subfactors can be classified "using only finite group theory." That is, if you had a black box which could answer all questions about finite groups and their cohomology you'd be able to understand all finite depth integer index subfactors. The ... 8 It is relatively easy to give an explicit sequence of inner automorphisms that converges to the flip automorphism $\sigma$. First of all, realize $R$ as an infinite tensor product of matrix algebras $(R,\varphi)=\bigotimes_n (M_{k_n}(\mathbb{C}),\varphi_n)$. For every $n\in \mathbb{N}$, we find a unitary $u_n\in M_{k_n}(\mathbb{C})\otimes M_{k_n}(\mathbb{C})$... 8 This isn't a full answer. First let’s translate this into purely group theoretic language. The vertex at depth $0$ is the trivial $G$-rep, the vertex at depth $1$ is the trivial $H$-rep, the vertices at depth $0$ or $2$ are the $G$-irreps in $\mathrm{Ind}_H^G 1$, the vertices at depth $1$ or $3$ are the $H$-irreps in $\mathrm{Res}_H\mathrm{Ind}_H^G 1$, ... 7 You can construct a subfactor (under very mild assumptions) from an object $X$ in a rigid C-tensor category, by taking the limit of inclusions $${\rm End}(X^{\otimes n}) \simeq \{ \iota_X \otimes T \mid T \in {\rm End}(X^{\otimes n})\} \subset {\rm End}(X^{\otimes n+1})$$ as $n \to \infty$. One good entry point is the Longo-Roberts paper [LR97], I guess. ... 6 The infinite depth subfactor coming from SU(3) at any index above 9 gives an example. Here the Q-system is $V_{(1,0)} \otimes V_{(0,1)} \cong V_{(1,1)} \oplus V_{(0,0)}$ so the only possible sub-objects are the whole thing or the trivial, so it's certainly maximal. 6 (Unitary) fusion categories are interesting in physics because they classify gapped phases on the boundary of 2+1D quantum states of matter. Similarly, unitary modular tensor categories are interesting in physics because they classify gapped phases of 2+1D quantum states of matter. I have two papers to explain the connections arXiv:1405.5858 and arXiv:... 6 No. For an example, consider $M=L^\infty([0,1])$ with the Lebesgue measure, take $A$ to be the functions that are piecewise constant on dyadic intervals and $\alpha_n(f)=f\circ \phi_n^{-1}$ where $\phi_n(t)=k/2^n + (2^n t-k)^2/2^n$ if $t \in [k/2^n,(k+1)/2^n[$. In words, $\phi_n$ acts as some fixed transformation (here $t \mapsto t^2$, but it could be ... 6 1, and indeed its generalization to the amenable case, is in "Amenable tensor categories and their realizations as AFD bimodules" by Hayashi and Yamagami, see Section 7. I don't think 2 has appeared in the literature, though I'd expect that it's true. In the fusion case, if the universal grading group is trivial, I think you can just look at the algebra ... 6 I'm not an expert in algebraic geometry, but I can say something about methods for solving pentagon equations that will hopefully be of use. The primary way to determine whether or not there is a solution to the pentagon equations is to use Groebner basis methods. However, these begin to break down very quickly for algebraic varieties of the size that we ... 6 The answer is no. If $N_1$ and $N_2$ are both finite index subfactors of a nonamenable ${\rm II}_1$ factor $M \subset \mathcal B(L^2M)$ such that $N_1 \cap N_2 = \mathbb C$, then $N_1'$ and $N_2'$ are both finite and $N_1' \cap N_2'$ is nonamenable since it contains $M'$. For an example of such a situation consider non-trivial finite groups $G$ and $H$, set ... 5 I'm absolutely ignorant about subfactors, but if I correctly understand what you said there (the theorem in the question), the answer to all three questions is yes, even with non maximal subgroups. Namely, let $H\subset G$ be of finite index $n$. Since the intersection $K$ of all conjugates of $H$ in $G$ has index dividing $n!$ in $G$ (it is the kernel of ... 5 This is somewhere between a comment and an answer. But it is too long for a comment, so I put it here. To me the natural thing to look at is the action $G\curvearrowright G/H$. $\|G/H\|$ captures the index, which you surely want to do, and the action should in some sense capture the position of $H$ inside $G$. The issue then would be to try to come up with an ... 5 In the formula, $tr(b)$ should be replaced by $tr(\|b\|)$, i.e., $\\|b\\|_1$. $$b_\alphab_\beta=\sum_\gamma\frac{\\|b_\alpha\\|_1\\|b_\beta\\|_1\bar{c}_{\alpha\beta}^\gamma}{\sqrt{n}\\|b_\gamma\\|_1}b_\gamma.$$ Unfortunately the proof needs the irreducibility. I do not know how to generalize it to weak Kac algebras. As Dave mentioned, one direction is clear. It ... 5 The answer of the main question is yes. Let $G$ be a finite group and $H$ a subgroup. Definition: The group $G$ is called $H$-cyclic if $\exists g \in G$ such that $\langle H,g \rangle = G$. Note that: $\langle H,g \rangle = G \Leftrightarrow \langle Hg \rangle = G$. Ore's theorem for intervals (1938): If the interval $[H,G]$ is distributive, then $G$ is $... 5 As Noah points out, you are looking for some (core-free) maximal subgroup$H<G$such that$1_H^G$has nonzero inner product with every irreducible character. Say$G=L_2(p)$with$p$prime and$p \equiv 1 \bmod 8$. Then$G$has a maximal subgroup$H \cong S_4$. Every element$h \in H$has diagonalizable preimage in$SL_2(p)$, and the conjugacy class of$... 5 A positive answer to my question (2) is provided in the last section of Popa Takesaki "Topological structure of unitary and automorphism groups". They show that for any locally compact group $G$, the action of $G$ on the $II_1$ factor $\{\mathit{CAR}(L^2G)\}''$ is appropriately continuous. Here, the double commutant is taken on $L^2(\mathit{CAR}(L^2G),tr)$ ... 5 I think that a noncommutative fusion ring of rank 5 does not exist. Namely, let $a$ and $b$ be the formal codegrees (see https://arxiv.org/pdf/0810.3242.pdf) of such ring. Then $a$ and $b$ are positive (EDIT: and rational, see the explanation by Noah) integers satisfying $\frac1a+\frac2b=1$ (see Proposition 2.10 in https://arxiv.org/pdf/1309.4822.pdf). It is ... 4 If a subfactor $N \subset M$ at index 6 is composed, i.e., it admits a non-trivial intermediate subfactor $N \subset P \subset M$, then the two components are necessarily of indices $2$ and $3$. There exists one subfactor at index $2$ given by the graph $A_{3}$, and two at index $3$ given by $A_{5}$ and $D_{4}$. Then, the standard invariant of a composed ... 4 Your answer will work, except if the $b_i$'s have trace zero as Theo points out. If the $b_i\in \mathcal{P}_{n_i}$ has trace zero, just use $1_{n_i}+b_i$ instead of $b_i$, where $1_{n_i}$ is $n_i$ parallel strands. Then you can cap this off to get a scalar as before. By what you remarked above, you'll be able to recover $1_{n_i}+b_i$, from which you can ... 4 This is not a proper answer, but is a bit too long for a comment. You should look at the following paper: Groups and Lattices by P.P. Palfy First, he states a stronger version of Ore's result: Theorem: Let $G$ be any group. The subgroup lattice $\mathcal{L}(G)$ is distributive if and only if the group $G$ is locally cyclic. (i.e. every ... 3 The answer is no. In fact, the group planar algebra has no non-trivial planar ideals. If $P_\bullet$ is a spherical planar algebra with non-zero modulus, and if $P_{0,\pm}$ are one dimensional, then every planar ideal is contained in the planar ideal $N_\bullet$ of negligible elements, i.e., those $x\in P_{n,\pm}$ for which tr$(xy)=0$ for all $y\in P_{n,\pm}... 3 A II$_1$-factor is algebraically simple, so each morphism of II$_1$-factors is either injective or zero. Thus every non-zero morphism is an isomorphism onto its image. So$\phi: M \to \phi(M)$is an isomorphism that takes$\phi^{-1}(N')$to$N'$. I don't think the canonical surjection$G\to G'=G/\ker(f)$actually gives you a map of factors$M\rtimes G\to M\... 3 Let $_RM_R:={}_R(L^2R\otimes_{\mathbb C} L^2R)_R$, where the first $R$ acts on the first $L^2R$ and the second $R$ acts on the second $L^2R$. Its algebra of $R$-$R$-bimodule endomorphisms is $R^{\mathrm{op}}\,\bar\otimes\, R$. Using (misleading!) intuition from the representation theory of separable $C^*$-algebras, one might guess that every MASA in \$R^{\... Only top voted, non community-wiki answers of a minimum length are eligible	{}
Solar design tips, sales advice, and industry insights from the premier solar design software platform Author ### Gwen Brown Gwen Brown is a Content Marketing Analyst at Aurora Solar. Previously, she was a Senior Research Associate at the Environmental Law Institute. She graduated Phi Beta Kappa from Gettysburg College. # The Hidden Factors that Affect Solar Savings: Part 1, Monthly vs. Annual Billing There are many variations in the ways that different utilities bill solar customers with net metering. These subtleties can make it complicated to determine exactly how installing solar will affect a customer’s bills. In this series, we will explore elements of billing that commonly vary between utilities and examine the practical impacts of these factors on solar customers’ bills. Today, we look at how monthly billing compares to annual billing in terms of solar customers’ experiences and savings. As a solar installer, understanding utility bill nuances will enable you to better communicate what customers can expect after installing solar and help you find them extra savings #### Monthly or Annual Billing? One of the major ways billing for solar customers varies between utilities is how often customers are billed for their energy consumption. Under net metering policies, solar customers are credited for the energy they produce and charged for any energy supplied by the utility—effectively paying only for their net energy consumption (and any additional fees applied by the utility). If their solar system produced more energy than they consumed, they receive a credit for their excess generation which will carry over to future months. Most U.S. utilities reconcile solar customers’ energy consumption and production on a monthly basis. However, some utilities, particularly in California, have an annual billing cycle in which net metered customers are billed for their energy usage only once a year. Figure 1: Aurora allows users to conduct financial analysis based on monthly or annual billing approaches. Settings for a customer with annual billing are shown. Under an annual billing approach, the utility tracks customers’ energy consumption and production throughout the course of the year. Customers receive monthly statements tracking this data but do not have to pay monthly for energy consumption. However, they may have to pay some monthly fees to remain connected to the grid, as is the case for customers of major California utilities. At the end of the year, they receive their annual bill, often called the “True-up” statement, for any energy consumption that was not offset by their solar energy production. One benefit of annual billing is that credit from excess energy production can offset consumption from earlier months in the year. With monthly billing, on the other hand, excess credit produced can only be applied to later months. For example, credit accumulated in summer months would not reduce bills paid in early spring. Let’s take a look at what a solar customer’s bills look like under these two approaches. Figure 2 below shows pre- and post-solar utility bills for a customer with annual billing. Figure 3 shows the customer’s monthly energy consumption and production, and Table 1 shows how Aurora presents energy consumption, production, and bill amounts for customers with annual billing. Because the utility considers the customer’s energy production or consumption over the entire year, credits from months in which energy production exceeded consumption (April through October) can be applied to any other months in which they consumed more than they produced (January, February, November, and December). As you can see in Figure 2, this means that solar will cancel out all of the customer’s bills except for $10 in fixed fees (which we’ll discuss in greater detail in a future blog article). Figure 2: Customer’s pre- and post-solar bills under an annual billing approach (PG&E Rate: E-1, Baseline Region P). Figure 3: Customer’s energy consumption and production. Table 1: Customer’s pre- and post-solar energy consumption, production, and utility bills under an annual billing approach (PG&E Rate: E-1, Baseline Region P). For customers with annual billing, Aurora shows the annual savings rather than monthly bill savings, since the customer does not pay on a monthly basis. Now, let’s consider what bills would look like for the same customer if their utility has monthly billing. For the purposes of this exercise, we kept the same exact utility rate, but we switched the billing schedule from annual to monthly. Figure 4: Aurora financial analysis settings for a customer with monthly billing. Figure 5 below shows the customer’s bills under a monthly billing approach, and Table 2 shows the amounts of the customer’s monthly bills, energy consumption, and production. Although the customer’s energy consumption and production (as shown in Figure 3) remains the same, the amount they pay the utility increases in the early months of the year. Figure 5: Customer’s pre- and post-solar bills under a monthly billing approach. Table 2: Customer’s pre- and post-solar energy consumption, production, and utility bills under a monthly billing approach. For customers with monthly billing, Aurora shows the savings each month as well as the total annual savings. As Figure 5 and Table 2 illustrate, under monthly billing the customer will have higher bills in January and February because they receive those bills before they accrue credit for excess energy production. With monthly billing, the customer’s total annual savings from solar will be$2,899 compared to an annual savings of $2,951 if they had annual billing. In this case, annual billing saves the customer an additional$52 per year with the exact same solar installation. Most utilities do not allow customers to choose between monthly or annual billing as they offer only one option. In these cases, knowing what option the local utility offers allows one to accurately assess how much a customer can save with solar. Sacramento Municipal Utility District is one of the few utilities that does allow customers to choose between these options. For solar customers who do have a choice, annual billing is likely to be the most financially advantageous. Whether utilities bill monthly or annually is just one of the common differences in billing for solar customers. There are many other variations, including what time of year the billing cycle ends, how much net metered customers are compensated for their excess energy, and how long credits can carry over (some utilities allow credits to roll forward indefinitely). Fortunately, Aurora allows you to easily navigate these different options. We will analyze additional billing nuances in future installations of this series. #### Key Takeaways • There is a lot of variation in how net metered solar customers are billed across different utilities. It is important to understand these differences because they affect the financial return from solar. • Utilities may bill customers on a monthly or annual basis. With annual billing, solar customers pay once a year for their net energy consumption instead of every month. • One advantage of annual billing is that credit from excess energy production can be used to cancel out consumption from any month in the year, whereas under monthly billing credits can only roll forward. • net energy metering • Solar Utility Bill Author ### Gwen Brown Gwen Brown is a Content Marketing Analyst at Aurora Solar. Previously, she was a Senior Research Associate at the Environmental Law Institute. She graduated Phi Beta Kappa from Gettysburg College.	{}
# Math Help - conic sections 1. ## conic sections 1. Write the equation of the parabola given the focus at (-3,-2) and the directrix is the line y=-6. It is helpful to draw a sketch of what is given to help in choosing the correct formula. Be sure you have the correct formula to fill in. leave the equation in standard form. 2.Given the ellipse equation 9x^2 + 4y^2-18x + 16y=0 , find the coordinates of the center, the vertices, and the foci. Sketch the ellipse on the grid showing each of these points as dots. Show steps for converting to standard form. PLEASE INCLUDE WORK STEP BY STEP. THANK YOU 2. Hello, JROD23! The ellipse problem is not pretty . . . 2. Given the ellipse equation: . $9x^2 + 4y^2-18x + 16y\:=\:0$ find the coordinates of the center, the vertices, and the foci. Sketch the ellipse. To get Standard Form, we must complete-the-square. We are given: . $9x^2 - 18x + 4y^2 + 16y \;=\;0$ . . . . . $9(x^2 - 2x\qquad) + 4(y^2 + 4y\qquad) \;=\;0$ . . . . $9(x^2 - 2x$ + 1 $) + 4(y^2 + 4y$ + 4 $)\; = \;0$ + 9 + 16 . . . . . . . . . . . $9(x-1)^2 + 4(y + 2)^2 \;=\;25$ Divide by 25: . $\frac{9(x-1)^2}{25} + \frac{4(y+2)^2}{25} \;=\;1$ And we have: . . $\frac{(x-1)^2}{\frac{25}{9}} + \frac{(y+2)^2}{\frac{25}{4}} \;=\;1$ This is a "vertical" ellipse. .Its center is: $(1,\,\text{-}2)$ . . The major axis is vertical: . $a = \frac{5}{2}$ . . The minor axis is horizontal: . $b = \frac{5}{3}$ The vertices (ends of the major axis) are $\frac{5}{2}$ units above and below the center. . . Vertices: $\left(1,\,\frac{1}{2}\right),\;\left(1,\,\text{-}\frac{9}{2}\right)$ The co-vertices (ends of the minor axis) are $\frac{5}{3}$ units left and right of the center. . . Covertices: $\left(\text{-}\frac{2}{3},\,\text{-}2\right),\;\left(\frac{8}{3},\,\text{-}2\right)$ The foci are above and below the center. .We must find $c$. We have: . $c^2 \:=\:a^2-b^2\quad\Rightarrow\quad c^2 \:=\:\left(\frac{5}{2}\right)^2 - \left(\frac{5}{3}\right)^2 \:=\:\frac{125}{36} \quad\Rightarrow\quad c \,=\,\frac{5\sqrt{5}}{6}$ . . Foci: $\left(1,\,\text{-}2+\frac{5\sqrt{5}}{6}\right),\:\left(1,\,\text{-}2-\frac{5\sqrt{5}}{6}\right)$	{}
# How to automatically remove additional space introduced by primes of left superscripts? I need to use left superscripts. I know that there are packages as mathtools, tensor and leftidx. But they all suffer from the same typographical problem. If I typeset a left superscript and this superscript has a prime, asterisk or something similar attached to its right, then the left superscript is moved away from its base symbol. LaTeX does introduce this gap, because LaTeX believes it would need this additional space for the right superscript of the left superscript, but it does not, because actually both symbols do not interfere with each other. Now here is a MWE: \documentclass{article} \usepackage[T1]{fontenc} \usepackage{lmodern} \usepackage[utf8]{inputenc} % Symbol for Turing machine and Oracle \newcommand{\tmF}{\mathcal{F}} \newcommand{\oO}{\mathcal{O}} % Left superscript, copied from "leftidx"-package % How to improve this command? \newcommand{\append}[2]{{\protect\vphantom{#1}}^{#2}\!#1} % Code from: https://www.tug.org/TUGboat/tb22-4/tb72perlS.pdf \def\mathrlap{\mathpalette\mathrlapinternal} \def\mathrlapinternal#1#2{\rlap{$\mathsurround=0pt#1{#2}$}} \setlength{\fboxsep}{0pt} \setlength{\fboxrule}{0.25pt} \begin{document} \section{Problem visualization} \begin{itemize} \item This looks OK: $\append{\tmF}{\oO}$ \item This looks typographically wrong: $\append{\tmF}{\oO'}$ \item This looks better: $\append{\tmF}{\oO\mathrlap{'}}$ \end{itemize} \section{Problem visualization with boxes} \begin{itemize} \item This looks OK: $\append{\fbox{$\tmF$}}{\fbox{$\oO$}}$ \item This looks typographically wrong: $\append{\fbox{$\tmF$}}{\fbox{$\oO$}'}$ \item This looks better: $\append{\fbox{$\tmF$}}{\fbox{$\oO$}\mathrlap{'}}$ \end{itemize} \end{document} The append macro works as follows: First, the base symbol is typeset invisible such that the following superscript is moved to the correct height in case that the base symbol itsself has a different height. Then a small negative space is introduced to eat up the normal space between two consecutive noun symbols in math mode. At last the actual base symbol is typeset. I produce the typographical correct version by manually telling LaTeX that the prime symbol does not any space. But I would like to avoid using \mathrlap and would rather prefer an append macro that automatically figures out how much extra space is needed. • Since TeX works on boxes, the adjustment will depend on the actual shape of the char, and this it is probably not possible to do this automatically. – daleif Mar 3 '16 at 12:25 • Interesting problem, but as daleif says, it's no easy (if at all possible) getting an “automatic” version. – egreg Mar 3 '16 at 12:57 The “visual hole” depends of course on the relative shapes of the symbols. You can remove the \scriptspace on the left superscript, which will reduce the gap. In a product version you will of course change \trylength to \z@ (that is, 0pt) in the definition of \app@@nd. \documentclass{article} \usepackage[T1]{fontenc} \usepackage{lmodern} \usepackage[utf8]{inputenc} % Symbol for Turing machine and Oracle \newcommand{\tmF}{\mathcal{F}} \newcommand{\oO}{\mathcal{O}} % Left superscript, copied from "leftidx"-package % How to improve this command? \makeatletter \DeclareRobustCommand{\append}[2]{% \mathpalette\app@nd{{#1}{#2}}#1% } \newcommand{\app@nd}[2]{\app@@nd#1#2} \newcommand{\app@@nd}[3]{% \mbox{\scriptspace\trylength\m@th$#1{\vphantom{#2}}^{#3}$}% } \newdimen\trylength \makeatother % Code from: https://www.tug.org/TUGboat/tb22-4/tb72perlS.pdf \def\mathrlap{\mathpalette\mathrlapinternal} \def\mathrlapinternal#1#2{\rlap{$\mathsurround=0pt#1{#2}$}} \setlength{\fboxsep}{-0.25pt} \setlength{\fboxrule}{0.25pt} \begin{document} \section{Normal scriptspace} \trylength=\scriptspace \begin{itemize} \item $\append{\tmF}{\oO}$ $\append{\mathcal{H}}{\oO}$ \item $\append{\tmF}{\oO'}$ $\append{\mathcal{H}}{\oO'}$ \item $\append{\fbox{$\tmF$}}{\fbox{$\scriptstyle\oO$}}$ \item $\append{\fbox{$\tmF$}}{\fbox{$\scriptstyle\oO'$}}$ \end{itemize} \section{Zero scriptspace} \trylength=0pt \begin{itemize} \item $\append{\tmF}{\oO}$ $\append{\mathcal{H}}{\oO}$ \item $\append{\tmF}{\oO'}$ $\append{\mathcal{H}}{\oO'}$ \item $\append{\fbox{$\tmF$}}{\fbox{$\scriptstyle\oO$}}$ \item $\append{\fbox{$\tmF$}}{\fbox{$\scriptstyle\oO'$}}$ \end{itemize} \end{document} • Thanks, but if understand correctly, it only reduces the script space to zero, it does not make the prime horizontally overlap with the base symbol. Right? – nagmat84 Mar 3 '16 at 13:23 • @nagmat84 Indeed. If you want to make the prime overlap the base, use \mathrlap{'}. In case there's a subscript to the left superscript, you don't want overlapping, do you? Note also that mathtools already features the \mathrlap macro. – egreg Mar 3 '16 at 13:26 • OK. Just one more question that perhaps is a little bit out of scope of a comment. I try to understand your code but do not really understand why you need this double indirection append -> app@end -> app@@nd. append calls app@end with two arguments whereby the fist argument is the token {{#1}{#2}} and the second is #1. Then app@end calls app@@nd with three arguments which after substitution by the original arguments are {#1}{#2}#1. Why does append not directly call app@@nd with {#1}{#2}#1? I guess it has something do to with expansion of tokens but I am a newbie. – nagmat84 Mar 3 '16 at 13:36 • @nagmat84 \mathpalette takes two arguments, not three; \mathpalette\app@nd{{#1}{#2}} will do the four \app@nd<style>{{#1}{#2}} calls, which are so transformed into \app@@nd<style>{#1}{#2}. – egreg Mar 3 '16 at 13:43	{}
# How can I measure the supply voltage of batteris without a multimeter? My system consists of two XBee S1s and a Raspberry Pi. I've attached a TMP36 sensor and a LDR to the first XBee S1 and transmit those infromations to the second XBee S1 via the ZigBee protocol. The second Xbee S1 is connected via USB adapter to the Raspberry Pi. My first XBee S1 is powered by batteris at 5V. As the voltage given by the batteries decreases over time, I would like to have an indeication of the voltage supplied by the batteris in order to know when I should change them. Can you tell me how can I do that? My idea is to make a gauge with 4 LEDs attached to the Raspberry Pi. The problem I am facing is that I do not know how can I send the supply voltage via the Xbees(as the XBee 1 itself is powered by that battery). PS: I don't want to use a multimeter every time I am wondering how much voltage is provided by the batteries. ## 2 Answers If the first XBee has an analog input then use it to measure the battery voltage. simulate this circuit – Schematic created using CircuitLab Figure 1. Battery voltage sense. Create a potential divider to bring the maximum battery voltage below the maximum input voltage for the ADC. You would then need to figure out the scaling factor and set the remote to control your gauge. This would be easy to achieve with a potentiometer on the analog in during setup. You haven't specified which Xbee you have (there are lots), but IIRC all the 2.4GHz XBees have an onboard 10-bit ADC with several channels. You'll have to check the reference voltage for your particular XBee, but I think they are typically set up with a 1.2V Vref, so you need to divide down the 0-5V from the battery to a 0-1.2V range (leave a little bit of wiggle room). Let's say the maximum voltage your batteries might reach is 6V, then you need a voltage divider to scale the voltage by 1.2V/6V = 0.2. Also, keep in mind that the ADC input impedance is 10kΩ or greater, which isn't great, so you need to either design a voltage divider with significantly smaller resistor values, or (better) use an op-amp to buffer the divided-down voltage into the ADC input. • Thank you for your answer. I am using XBees S1, their reference voltage is 5V. Can you please precise me how did you calculate the scaling factor? Thanks – traviata Mar 6 '16 at 10:29 • You really should link a datasheet, "XBees S1" isn't really unambiguous. Also, are you sure it's 5V? Most XBee's have a $V_{REF}$ max of $V_{DDAD} = V_{CC}$, which is typically 3.3V. Anyway, you calculate the voltage divider scaling factor by taking $V_{REF}/V_{IN(MAX)}$ (the reference voltage of the ADC divided by the largest voltage your battery might reach. – uint128_t Mar 6 '16 at 16:37	{}
# All of jessicata's Comments + Replies Everything I Need To Know About Takeoff Speeds I Learned From Air Conditioner Ratings On Amazon I assumed EER did account for that based on: All portable air conditioner’s energy efficiency is measured using an EER score. The EER rating is the ratio between the useful cooling effect (measured in BTU) to electrical power (in W). It’s for this reason that it is hard to give a generalized answer to this question, but typically, portable air conditioners are less efficient than permanent window units due to their size. 3Oliver Habryka1mo This article explains the difference: https://www.consumeranalysis.com/guides/portable-ac/best-portable-air-conditioner/ [https://www.consumeranalysis.com/guides/portable-ac/best-portable-air-conditioner/] EER measures performance in BTUs, which are simply measuring how much work the AC performs, without taking into account any backflow of cold air back into the AC, or infiltration issues. Everything I Need To Know About Takeoff Speeds I Learned From Air Conditioner Ratings On Amazon Regarding the back-and-forth on air conditioners, I tried Google searching to find a precedent for this sort of analysis; the first Google result was "air conditioner single vs. dual hose" was this blog post, which acknowledges the inefficiency johnswentworth points out, overall recommends dual-hose air conditioners, but still recommends single-hose air conditioners under some conditions, and claims the efficiency difference is only about 12%. Highlights: In general, a single-hose portable air conditioner is best suited for smaller rooms. The reason being 2Oliver Habryka1mo EER does not account for heat infiltration issues, so this seems confused. CEER does, and that does suggest something in the 20% range, but I am pretty sure you can't use EER to compare a single-hose and a dual-hose system. AXRP Episode 14 - Infra-Bayesian Physicalism with Vanessa Kosoy Btw, there is some amount of philosophical convergence between this and some recent work I did on critical agential physics; both are trying to understand physics as laws that partially (not fully) predict sense-data starting from the perspective of a particular agent. It seems like "infra-Bayesianism" may be broadly compatible with frequentism; extending Popper's falsifiability condition to falsify probabilistic (as opposed to deterministic) laws yields frequentist null hypothesis significance testing, e.g. Neyman Pearson; similarly, frequentism also attem... (read more) 3Vanessa Kosoy2mo Thanks, I'll look at that! Yes! In frequentism, we define probability distributions as limits of frequencies. One problem with this is, what to do if there's no convergence? In the real world, there won't be convergence unless you have an infinite sequence of truly identical experiments, which you never have. At best, you have a long sequence of similar experiments. Arguably, infrabayesianism solves it by replacing the limit with the convex hull of all limit points. But, I view infrabayesianism more as a synthesis between bayesianism and frequentism. Like in frequentism, you can get asymptotic guarantees. But, like in bayesiansim, it makes sense to talk of priors (and even updates), and measure the performance of your policy regardless of the particular decomposition of the prior into hypotheses (as opposed to regret which does depend on the decomposition). In particular, you can define the optimal infrabayesian policy even for a prior which is not learnable and hence doesn't admit frequentism-style guarantees. $1000 USD prize - Circular Dependency of Counterfactuals Thanks for reading all the posts! I'm not sure where you got the idea that this was to solve the spurious counterfactuals problem, that was in the appendix because I anticipated that a MIRI-adjacent person would want to know how it solves that problem. The core problem it's solving is that it's a well-defined mathematical framework in which (a) there are, in some sense, choices, and (b) it is believed that these choices correspond to the results of a particular Turing machine. It goes back to the free will vs determinism paradox, and shows that there's a fo... (read more) 2Chris_Leong5mo Thanks for that clarification. I suppose that demonstrates that the 5 and 10 problem is a broader problem than I realised. I still think that it's only a hard problem within particular systems that have a vulnerability to it. Yeah, we have significant agreement, but I'm more conservative in my interpretations. I guess this is a result of me being, at least in my opinion, more skeptical of language. Like I'm very conscious of arguments where someone says, "X could be described by phrase Y" and then later they rely on connations of Y that weren't proven. For example, you write, "From the AI’s perspective, it has a choice among multiple actions, hence in a sense “believing in metaphysical free will”. I would suggest it would be more accurate to write: "The AI models the situation as though it had free will" which leaves open the possibility that it is might be just a pragmatic model, rather than the AI necessarily endorsing itself as possessing free will. Another way of framing this: there's an additional step in between observing that an agent acts or models a situation as it believes in freewill and concluding that it actually believes in freewill. For example, I might round all numbers in a calculation to integers in order to make it easier for me, but that doesn't mean that I believe that the values are integers.$1000 USD prize - Circular Dependency of Counterfactuals It seems like agents in a deterministic universe can falsify theories in at least some sense. Like they take two different weights drop them and see they land at the same time falsifying the fact that heavier objects fall faster The main problem is that it isn't meaningful for their theories to make counterfactual predictions about a single situation; they can create multiple situations (across time and space) and assume symmetry and get falsification that way, but it requires extra assumptions. Basically you can't say different theories really disagree... (read more) 1Chris_Leong5mo Agreed, this is yet another argument for considering counterfactuals to be so fundamental that they don't make sense outside of themselves. I just don't see this as incompatible with determinism, b/c I'm grounding using counterfactuals rather than agency. I don't mean utility function optimization, so let me clarify what as I see as the distinction. I guess I see my version as compatible with the determinist claim that you couldn't have run the experiment because the path of the universe was always determined from the start. I'm referring to a purely hypothetical running with no reference to whether you could or couldn't have actually run it. Hopefully, my comments here have made it clear where we diverge and this provides a target if you want to make a submission (that said, the contest is about the potential circular dependency of counterfactuals and not just my views. So it's perfectly valid for people to focus on other arguments for this hypothesis, rather than my specific arguments). \$1000 USD prize - Circular Dependency of Counterfactuals I previously wrote a post about reconciling free will with determinism. The metaphysics implicit in Pearlian causality is free will (In Drescher's words: "Pearl's formalism models free will rather than mechanical choice."). The challenge is reconciling this metaphysics with the belief that one is physically embodied. That is what the post attempts to do; these perspectives aren't inherently irreconcilable, we just have to be really careful about e.g. distinguishing "my action" vs "the action of the computer embodying me" in a the Bayes net and distingu... (read more) 1Chris_Leong5mo Thoughts on Modeling Naturalized Logic Decision Theory Problems in Linear Logic I hadn't heard of linear logic before - it seems like a cool formalisation - although I tend to believe that formalisations are overrated as unless they are used very carefully they can obscure more than they reveal. I believe that spurious counterfactuals are only an issue with the 5 and 10 problem because of an attempt to hack logical-if to substitute for counterfactual-if in such a way that we can reuse proof-based systems. It's extremely cool that we can do as much as we can working in that fashion, but there's no reason why we should be surprised that it runs into limits. So I don't see inventing alternative formalisations that avoid the 5 and 10 problem as particularly hard as the bug is really quite specific to systems that try to utilise this kind of hack. I'd expect that almost any other system in design space will avoid this. So if, as I claim, attempts at formalisation will avoid this issue by default, the fact that any one formalisation avoids this problem shouldn't give us too much confidence in it being a good system for representing counterfactuals in general. Instead, I think it's much more persuasive to ground any proposed system with philosophical arguments (such as your first post was focusing on), rather than mostly just posting a system and observing it has a few nice properties. I mean, your approach in this article certainly a valuable thing to do, but I don't see it as getting all the way to the heart of the issue. Interestingly enough, this mirrors my position in Why 1-boxing doesn't imply backwards causation [https://www.lesswrong.com/posts/gAAFzqJkfeSHvcwTw/why-1-boxing-doesn-t-imply-backwards-causation] where I distinguish between Raw Reality (the territory) and Augmented Reality (the territory augmented by counterfactuals). I guess I put more emphasis on delving into the philosophical reasons for such a view and I think that's what this post is a bit sh 1Chris_Leong5mo Comments on A critical agential account of free will, causation, and physics We can imagine a situation where there is a box containing an apple or a pear. Suppose we believe that it contains a pear, but we believe it contains an apple. If we look in the box (and we have good reason to believe looking doesn't change the contents), then we'll falsfy our pear hypothesis. Similarly, if we're told by an oracle that if we looked we would see a pear, then there'd be no need for us to actually look, we'd have heard enough to falsify our pear hypothesis. However, the situation you've identified isn't the same. Here you aren't just deciding whether to make an observation or not, but what the value of that observation would be. So in this case, the fact that if you took action B you'd observe the action you took was B doesn't say anything about the case where you don't take action B, unlike knowing that if you looked in the box you'd see you an apple provides you information even if you don't look in the box. It simply isn't relevant unless you actually take B. I think it's reasonable to suggest starting from falsification as our most basic assumption. I guess where you lose me is when you claim that this implies agency. I guess my position is as follows: * It seems like agents in a deterministic universe can falsify theories in at least some sense. Like they take two different weights drop them and see they land at the same time falsifying the fact that heavier objects fall faster * On the other hand, some like agency or counterfactuals seems necessary for talking about falsfiability in the abstract as this involves saying that we could falsify a theory if we ran an experiment that we didn't. In the second case, I would suggest that what we need is counterfactuals not agency. That is, we need to be able to say things like, "If I ran this experiment and obtained this result, then theory X would be falsified", not "I could have run this experiment and if I d 2Chris_Leong5mo Visible Thoughts Project and Bounty Announcement How do you think this project relates to Ought? Seems like the projects share a basic objective (having AI predict human thoughts had in the course of solving a task). Ought has more detailed proposals for how the thoughts are being used to solve the task (in terms of e.g. factoring a problem into smaller problems, so that the internal thoughts are a load-bearing part of the computation rather than an annotation that is predicted but not checked for being relevant). So we are taking one of the outputs that current AIs seem to have learned best to design 1John Maxwell6mo Might depend whether the "thought" part comes before or after particular story text. If the "thought" comes after that story text, then it's generated conditional on that text, essentially a rationalization of that text from a hypothetical DM's point of view. If it comes before that story text, then the story is being generated conditional on it. Personally I think I might go for a two-phase process. Do the task with a lot of transparent detail [https://arxiv.org/abs/2112.00114] in phase 1. Summarize that detail and filter out infohazards in phase 2, but link from the summary to the detailed version so a human can check things as needed (flagging links to plausible infohazards). (I guess you could flag links to parts that seemed especially likely to be incorrigible/manipulative cognition, or parts of the summary that the summarizer was less confident in, as well.) Christiano, Cotra, and Yudkowsky on AI progress This section seemed like an instance of you and Eliezer talking past each other in a way that wasn't locating a mathematical model containing the features you both believed were important (e.g. things could go "whoosh" while still being continuous): [Christiano][13:46] Even if we just assume that your AI needs to go off in the corner and not interact with humans, there’s still a question of why the self-contained AI civilization is making ~0 progress and then all of a sudden very rapid progress [Yudkowsky][13:46] unfortunately a lot of what you are saying, fro... (read more) My claim is that the timescale of AI self-improvement, at the point it takes over from humans, is the same as the previous timescale of human-driven AI improvement. If it was a lot faster, you would have seen a takeover earlier instead. This claim is true in your model. It also seems true to me about hominids, that is I think that cultural evolution took over roughly when its timescale was comparable to the timescale for biological improvements, though Eliezer disagrees I thought Eliezer's comment "there is a sufficiently high level where things go who... (read more) Christiano, Cotra, and Yudkowsky on AI progress A bunch of this was frustrating to read because it seemed like Paul was yelling "we should model continuous changes!" and Eliezer was yelling "we should model discrete events!" and these were treated as counter-arguments to each other. It seems obvious from having read about dynamical systems that continuous models still have discrete phase changes. E.g. consider boiling water. As you put in energy the temperature increases until it gets to the boiling point, at which point more energy put in doesn't increase the temperature further (for a while), it conv... (read more) (I'm interested in which of my claims seem to dismiss or not adequately account for the possibility that continuous systems have phase changes.) I don’t really feel like anything you are saying undermines my position here, or defends the part of Eliezer’s picture I’m objecting to. (ETA: but I agree with you that it's the right kind of model to be talking about and is good to bring up explicitly in discussion. I think my failure to do so is mostly a failure of communication.) I usually think about models that show the same kind of phase transition you discuss, though usually significantly more sophisticated models and moving from exponential to hyperbolic growth (you only get an exponential in your mo... (read more) 5Matthew Barnett6mo +1 on using dynamical systems models to try to formalize the frameworks in this debate. I also give Eliezer points for trying to do something similar in Intelligence Explosion Microeconomics [https://intelligence.org/files/IEM.pdf] (and to people who have looked at this from the macro perspective [https://www.openphilanthropy.org/could-advanced-ai-drive-explosive-economic-growth] ). What 2026 looks like This is quite good concrete AI forecasting compared to what I've seen elsewhere, thanks for doing it! It seems really plasusible based on how fast AI progress has been going over the past decade and which problems are most tractable. Modeling naturalized decision problems in linear logic CDT and EDT have known problems on 5 and 10. TDT/UDT are insufficiently formalized, and seem like they might rely on known-to-be-unfomalizable logical counterfactuals. So 5 and 10 isn't trivial even without spurious counterfactuals. What does this add over modal UDT? • No requirement to do infinite proof search • More elegant handling of multi-step decision problems • Also works on problems where the agent doesn't know its source code (of course, this prevents logical dependencies due to source code from being taken into account) Philosophically, it works as a Topological metaphysics: relating point-set topology and locale theory Reals are still defined as sets of (a, b) rational intervals. The locale contains countable unions of these, but all these are determined by which (a, b) intervals contain the real number. Topological metaphysics: relating point-set topology and locale theory Good point; I've changed the wording to make it clear that the rational-delimited open intervals are the basis, not all the locale elements. Luckily, points can be defined as sets of basis elements containing them, since all other properties follow. (Making the locale itself countable requires weakening the definition by making the sets to form unions over countable, e.g. by requiring them to be recursively enumerable) Another way to make it countable would be to instead go to the category of posets, Then the rational interval basis is a poset with a countable number of elements, and by the Alexandroff construction [https://ncatlab.org/nlab/show/specialization+topology] corresponds to the real line (or at least something very similar). But, this construction gives a full and faithful embedding of the category of posets to the category of spaces (which basically means you get all and only continuous maps from monotonic function). I guess the ontology version in this case would be the category of prosets. (Personally, I'm not sure that ontology of the universe isn't a type error). I see. In that case does the procedure for defining points stay the same, or do you need to use recursively enumerable sets of opens, giving you only countably many reals? Motivating Abstraction-First Decision Theory I've also been thinking about the application of agency abstractions to decision theory, from a somewhat different angle. It seems like what you're doing is considering relations between high-level third-person abstractions and low-level third-person abstractions. In contrast, I'm primarily considering relations between high-level first-person abstractions and low-level first-person abstractions. The VNM abstraction itself assumes that "you" are deciding between different options, each of which has different (stochastic) consequences; thus, it is inherently 3johnswentworth2y This comment made a bunch of your other writing click for me. I think I see what you're aiming for now; it's a beautiful vision. In retrospect, this is largely what I've been trying to get rid of, in particular by looking for a third-person interpretation of probability [https://www.lesswrong.com/posts/Lz2nCYnBeaZyS68Xb/probability-as-minimal-map]. Obviously frequentism satisfies that criterion, but the strict form is too narrow for most applications and the less-strict form (i.e. "imagine we repeated this one-shot experiment many times...") isn't actually third-person. I've also started thinking about a third-person grounding of utility maximization and the like via selection processes; that's likely to be a whole months-long project in itself in the not-too-distant future. Subjective implication decision theory in critical agentialism Looking back on this, it does seem quite similar to EDT. I'm actually, at this point, not clear on how EDT and TDT differ, except in that EDT has potential problems in cases where it's sure about its own action. I'll change the text so it notes the similarity to EDT. On XOR blackmail, SIDT will indeed pay up. Two Alternatives to Logical Counterfactuals Yes, it's about no backwards assumption. Linear has lots of meanings, I'm not concerned about this getting confused with linear algebra, but you can suggest a better term if you have one. Two Alternatives to Logical Counterfactuals Basically, the assumption that you're participating in a POMDP. The idea is that there's some hidden state that your actions interact with in a temporally linear fashion (i.e. action 1 affects state 2), such that your late actions can't affect early states/observations. 1David Krueger2y OK, so no "backwards causation" ? (not sure if that's a technical term and/or if I'm using it right...) Is there a word we could use instead of "linear", which to an ML person sounds like "as in linear algebra"? Two Alternatives to Logical Counterfactuals The way you are using it doesn’t necessarily imply real control, it may be imaginary control. I'm discussing a hypothetical agent who believes itself to have control. So its beliefs include "I have free will". Its belief isn't "I believe that I have free will". It’s a “para-consistent material conditional” by which I mean the algorithm is limited in such a way as to prevent this explosion. Yes, that makes sense. However, were you flowing this all the way back in time? Yes (see thread with Abram Demski). What do you mean by dualistic? 1Chris_Leong2y Hmm, yeah this could be a viable theory. Anyway to summarise the argument I make in Is Backwards Causation Necessarily Absurd? [https://www.lesswrong.com/posts/pa7mvEmEgt336gBSf/is-backwards-causation-necessarily-absurd] , I point out that since physics is pretty much reversible, instead of A causing B, it seems as though we could also imagine B causing A and time going backwards. In this view, it would be reasonable to say that one-boxing (backwards-)caused the box to be full in Newcombs. I only sketched the theory because I don't have enough physics knowledge to evaluate it. But the point is that we can give justification for a non-standard model of causality. Two Alternatives to Logical Counterfactuals Secondly, “free will” is such a loaded word that using it in a non-standard fashion simply obscures and confuses the discussion. Wikipedia says "Free will is the ability to choose between different possible courses of action unimpeded." SEP says "The term “free will” has emerged over the past two millennia as the canonical designator for a significant kind of control over one’s actions." So my usage seems pretty standard. For example, recently I’ve been arguing in favour of what counts as a valid counterfactual being at least partially a matter of soc 1Chris_Leong2y Not quite. The way you are using it doesn't necessarily imply real control, it may be imaginary control. True. Maybe I should clarify what I'm suggesting. My current theory is that there are multiple reasonable definitions of counterfactual and it comes down to social norms as to what we accept as a valid counterfactual. However, it is still very much a work in progress, so I wouldn't be able to provide more than vague details. I guess my point was that this notion of counterfactual isn't strictly a material conditional due to the principle of explosion [https://www.wikiwand.com/en/Principle_of_explosion]. It's a "para-consistent material conditional" by which I mean the algorithm is limited in such a way as to prevent this explosion. Hmm... good point. However, were you flowing this all the way back in time? Such as if you change someone's source code, you'd also have to change the person who programmed them. What do you mean by dualistic? Two Alternatives to Logical Counterfactuals I think it's worth examining more closely what it means to be "not a pure optimizer". Formally, a VNM utility function is a rationalization of a coherent policy. Say that you have some idea about what your utility function is, U. Suppose you then decide to follow a policy that does not maximize U. Logically, it follows that U is not really your utility function; either your policy doesn't coherently maximize any utility function, or it maximizes some other utility function. (Because the utility function is, by definition, a rationalization of the poli 3Abram Demski2y OK, all of that made sense to me. I find the direction more plausible than when I first read your post, although it still seems like it'll fall to the problem I sketched. I both like and hate that it treats logical uncertainty in a radically different way from empirical uncertainty -- like, because we have so far failed to find any way to treat the two uniformly (besides being entirely updateful that is); and hate, because it still feels so wrong for the two to be very different. Two Alternatives to Logical Counterfactuals It seems the approaches we're using are similar, in that they both are starting from observation/action history with posited falsifiable laws, with the agent's source code not known a priori, and the agent considering different policies. Learning "my source code is A" is quite similar to learning "Omega predicts my action is equal to A()", so these would lead to similar results. Policy-dependent source code, then, corresponds to Omega making different predictions depending on the agent's intended policy, such that when comparing policies, the agent has to imagine Omega predicting differently (as it would imagine learning different source code under policy-dependent source code). 1Vanessa Kosoy2y Well, in quasi-Bayesianism for each policy you have to consider the worst-case environment in your belief set, which depends on the policy. I guess that in this sense it is analogous. Two Alternatives to Logical Counterfactuals I agree this is a problem, but isn't this a problem for logical counterfactual approaches as well? Isn't it also weird for a known fixed optimizer source code to produce a different result on this decision where it's obvious that 'left' is the best decision? If you assume that the agent chose 'right', it's more reasonable to think it's because it's not a pure optimizer than that a pure optimizer would have chosen 'right', in my view. If you form the intent to, as a policy, go 'right' on the 100th turn, you should anticipate learning that your source code is not the code of a pure optimizer. 3Abram Demski2y I'm left with the feeling that you don't see the problem I'm pointing at. My concern is that the most plausible world where you aren't a pure optimizer might look very very different, and whether this very very different world looks better or worse than the normal-looking world does not seem very relevant to the current decision. Consider the "special exception selves" you mention -- the Nth exception-self has a hard-coded exception "go right if it's beet at least N turns and you've gone right at most 1/N of the time". Now let's suppose that the worlds which give rise to exception-selves are a bit wild. That is to say, the rewards in those worlds have pretty high variance. So a significant fraction of them have quite high reward -- let's just say 10% of them have value much higher than is achievable in the real world. So we expect that by around N=10, there will be an exception-self living in a world that looks really good. This suggests to me that the policy-dependent-source agent cannot learn to go left > 90% of the time, because once it crosses that threshhold, the exception-self in the really good looking world is ready to trigger its exception -- so going right starts to appear really good. The agent goes right until it is under the threshhold again. If that's true, then it seems to me rather bad: the agent ends up repeatedly going right in a situation where it should be able to learn to go left easily. Its reason for repeatedly going right? There is one enticing world, which looks much like the real world, except that in that world the agent definitely goes right. Because that agent is a lucky agent who gets a lot of utility, the actual agent has decided to copy its behavior exactly -- anything else would prove the real agent unlucky, which would be sad. Of course, this outcome is far from obvious; I'm playing fast and loose with how this sort of agent might reason. Two Alternatives to Logical Counterfactuals This indeed makes sense when "obs" is itself a logical fact. If obs is a sensory input, though, 'A(obs) = act' is a logical fact, not a logical counterfactual. (I'm not trying to avoid causal interpretations of source code interpreters here, just logical counterfactuals) 2Abram Demski2y Ahhh ok. Two Alternatives to Logical Counterfactuals In the happy dance problem, when the agent is considering doing a happy dance, the agent should have already updated on M. This is more like timeless decision theory than updateless decision theory. Conditioning on 'A(obs) = act' is still a conditional, not a counterfactual. The difference between conditionals and counterfactuals is the difference between "If Oswald didn't kill Kennedy, then someone else did" and "If Oswald didn't kill Kennedy, then someone else would have". Indeed, troll bridge will present a problem for "playing chicken" approaches, whic 4Abram Demski2y I'm not sure how you are thinking about this. It seems to me like this will imply really radical changes to the universe. Suppose the agent is choosing between a left path and a right path. Its actual programming will go left. It has to come up with alternate programming which would make it go right, in order to consider that scenario. The most probable universe in which its programming would make it go right is potentially really different from our own. In particular, it is a universe where it would go right despite everything it has observed, a lifetime of (updateless) learning, which in the real universe, has taught it that it should go left in situations like this. EG, perhaps it has faced an iterated 5&10 problem, where left always yields 10. It has to consider alternate selves who, faced with that history, go right. It just seems implausible that thinking about universes like that will result in systematically good decisions. In the iterated 5&10 example, perhaps universes where its programming fails iterated 5&10 are universes where iterated 5&10 is an exceedingly unlikely situation; so in fact, the reward for going right is quite unlikely to be 5, and very likely to be 100. Then the AI would choose to go right. Obviously, this is not necessarily how you are thinking about it at all -- as you said, you haven't given an actual decision procedure. But the idea of considering only really consistent counterfactual worlds seems quite problematic. 2Abram Demski2y I still disagree. We need a counterfactual structure in order to consider the agent as a function A(obs). EG, if the agent is a computer program, the function A() would contain all the counterfactual information about what the agent would do if it observed different things. Hence, considering the agent's computer program as such a function leverages an ontological commitment to those counterfactuals. To illustrate this, consider counterfactual mugging [https://wiki.lesswrong.com/wiki/Counterfactual_mugging] where we already see that the coin is heads -- so, there is nothing we can do, we are at the mercy of our counterfactual partner. But suppose we haven't yet observed whether Omega gives us the money. A "real counterfactual" is one which can be true or false independently of whether its condition is met. In this case, if we believe in real counterfactuals, we believe that there is a fact of the matter about what we do in the coin=tails case, even though the coin came up heads. If we don't believe in real counterfactuals, we instead think only that there is a fact of how Omega is computing "what I would have done if the coin had been tails" -- but we do not believe there is any "correct" way for Omega to compute that. The obs→act representation and the P(act\|obs) representation both appear to satisfy this test of non-realism. The first is always true if the observation is false, so, lacks the ability to vary independently of the observation. The second is undefined when the observation is false, which is perhaps even more appealing for the non-realist. Now consider the A(obs)=act representation. A(tails)=pay can still vary even when we know coin=heads. So, it fails this test -- it is a realist representation! Putting something into functional form imputes a causal/counterfactual structure. 2Abram Demski2y I agree that this gets around the problem, but to me the happy dance problem is still suggestive -- it looks like the material conditional is the wrong representation of the thing we want to condition on. Also -- if the agent has already updated on observations, then updating on obs→a ct is just the same as updating on act. So this difference only matters in the updateless case, where it seems to cause us trouble. Two Alternatives to Logical Counterfactuals Yes, this is a specific way of doing policy-dependent source code, which minimizes how much the source code has to change to handle the counterfactual. Haven't looked deeply into the paper yet but the basic idea seems sound. How special are human brains among animal brains? The most quintessentially human intellectual accomplishments (e.g. proving theorems, composing symphonies, going into space) were only made possible by culture post-agricultural revolution. I'm guessing you mean the beginning of agriculture and not the Agricultural Revolution (18th century), which came much later than math and after Baroque music. But the wording is ambiguous. 5Issa Rice2y It seems like "agricultural revolution [https://en.wikipedia.org/wiki/Agricultural_revolution]" is used to mean both the beginning of agriculture ("First Agricultural Revolution") and the 18th century agricultural revolution ("Second Agricultural Revolution"). A critical agential account of free will, causation, and physics It's a subjectivist approach similar to Bayesianism, starting from the perspective of a given subject. Unlike in idealism, there is no assertion that everything is mental. On the falsifiability of hypercomputation, part 2: finite input streams In hyper-Solomonoff induction, indeed the direct hypercomputation hypothesis is probably more likely than the arbitration-oracle-emulating-hypercomputation hypothesis. But only by a constant factor. So this isn't really falsification so much as a shift in Bayesian evidence. I do think it's theoretically cleaner to distinguish this Bayesian reweighting from Popperian logical falsification, and from Neyman-Pearson null hypothesis significance testing (frequentist falsification), both of which in principle require producing an unbounded number of bits of evidence, although in practice rely on unfalsifiable assumptions to avoid radical skepticism e.g. of memory. On the falsifiability of hypercomputation This is really important and I missed this, thanks. I've added a note at the top of the post. On the falsifiability of hypercomputation Indeed, a constructive halting oracle can be thought of as a black-box that takes a PA statement, chooses whether to play Verifier or Falsifier, and then plays that, letting the user play the other part. Thanks for making this connection. Can we make peace with moral indeterminacy? The recommendation here is for AI designers (and future-designers in general) to decide what is right at some meta level, including details of which extrapolation procedures would be best. Of course there are constraints on this given by objective reason (hence the utility of investigation), but these constraints do not fully constrain the set of possibilities. Better to say "I am making this arbitrary choice for this psychological reason" than to refuse to make arbitrary choices. Can we make peace with moral indeterminacy? The problem you're running into is that the goals of: 1. being totally constrained by a system of rules determined by some process outside yourself that doesn't share your values (e.g. value-independent objective reason) 2. attaining those things that you intrinsically value are incompatible. It's easy to see once these are written out. If you want to get what you want, on purpose rather than accidentally, you must make choices. Those choices must be determined in part by things in you, not only by things outside you (such as value-independent objective rea 3Charlie Steiner3y You know, this isn't why I usually get called a tool :P I think I'm saying something pretty different from Nietzsche here. The problem with "Just decide for yourself" as an approach to dealing with moral decisions in novel contexts (like what to do with the whole galaxy) is that, though it may help you choose actions rather than worrying about what's right, it's not much help in building an AI. We certainly can't tell the AI "Just decide for yourself," that's trying to order around the nonexistent ghost in the machine. And while I could say "Do exactly what Charlie would do," even I wouldn't want the AI to do that, let alone other people. Nor can we fall back on "Well, designing an AI is an action, therefore I should just pick whatever AI design I feel like, because God is dead and I should just pick actions how I will," because how I feel like designing an AI has some very exacting requirements - it contains the whole problem in itself. A Critique of Functional Decision Theory I think CDT ultimately has to grapple with the question as well, because physics is math, and so physical counterfactuals are ultimately mathematical counterfactuals. "Physics is math" is ontologically reductive. Physics can often be specified as a dynamical system (along with interpretations of e.g. what high-level entities it represents, how it gets observed). Dynamical systems can be specified mathematically. Dynamical systems also have causal counterfactuals (what if you suddenly changed the system state to be this instead?). Causal counterfactuals d 3Abram Demski3y Yeah, agreed, I no longer endorse the argument I was making there - one has to say more than "physics is math" to establish the importance of dealing with logical counterfactuals. The Missing Math of Map-Making What does it mean for a map to be “accurate” at an abstract level, and what properties should my map-making process have in order to produce accurate abstracted maps/beliefs? The notion of a homomorphism in universal algebra and category theory is relevant here. Homomorphisms map from one structure (e.g. a group) to another, and must preserve structure. They can delete information (by mapping multiple different elements to the same element), but the structures that are represented in the structure-being-mapped-to must also exist in the structure-being- Towards an Intentional Research Agenda On the subject of intentionality/reference/objectivity/etc, On the Origin of Objects is excellent. My thinking about reference has a kind of discontinuity from before reading this book to after reading it. Seriously, the majority of analytic philosophy discussion of indexicality, qualia, reductionism, etc seems hopelessly confused in comparison. 2romeostevensit3y Some Thoughts on Metaphilosophy More over, I am skeptical that going on meta-level simplifies the problem to the level that it will be solvable by humans (the same about meta-ethics and theory of human values). This is also my reason for being pessimistic about solving metaphilosophy before a good number of object-level philosophical problems have been solved (e.g. in decision theory, ontology/metaphysics, and epistemology). If we imagine being in a state where we believe running computation X would solve hard philosophical problem Y, then it would seem that we already have a great de 2Wei Dai3y I think our positions on this are pretty close, but I may put a bit more weight on other "plausible stories" for solving metaphilosophy relative to your "plausible story". (I'm not sure if overall I'm more or less optimistic than you are.) It seems quite possible that understanding the general class of problems that includes Y is easier than understanding Y itself, and that allows us to find a computation X that would solve Y without much understanding of Y itself. As an analogy, suppose Y is some complex decision problem that we have little understanding of, and X is an AI that is programmed with a good decision theory. This does not seem like a very strong argument for your position. My suggestion in the OP is that humans already know the equivalent of "walking" (i.e., doing philosophy), we're just doing it very slowly. Given this, your analogies don't seem very conclusive about the difficulty of solving metaphilosophy or whether we have to make a bunch more progress on object-level philosophical problems before we can solve metaphilosophy. Predictors as Agents I think the fixed point finder won't optimize the fixed point for minimizing expected log loss. I'm going to give a concrete algorithm and show that it doesn't exhibit this behavior. If you disagree, can you present an alternative algorithm? Here's the algorithm. Start with some oracle (not a reflective oracle). Sample ~1000000 universes based on this oracle, getting 1000000 data points for what the reflective oracle outputs. Move the oracle 1% of the way from its current position towards the oracle that would answer queries correctly given the distrib 2interstice3y Reflective Oracles are a bit of a weird case case because their 'loss' is more like a 0/1 loss than a log loss, so all of the minima are exactly the same(If we take a sample of 100000 universes to score them, the difference is merely incredibly small instead of 0). I was being a bit glib referencing them in the article; I had in mind something more like a model parameterizing a distribution over outputs, whose only influence on the world is via a random sample from this distribution. I think that such models should in general have fixed points for similar reasons, but am not sure. Regardless, these models will, I believe, favour fixed points whose distributions are easy to compute(But not fixed points with low entropy, that is they will punish logical uncertainty but not intrinsic uncertainy). I'm planning to run some experiments with VAEs and post the results later. Predictors as Agents The capacity for agency arises because, in a complex environment, there will be multiple possible fixed-points. It’s quite likely that these fixed-points will differ in how the predictor is scored, either due to inherent randomness, logical uncertainty, or computational intractability(predictors could be powerfully superhuman while still being logically uncertain and computationally limited). Then the predictor will output the fixed-point on which it scores the best. Reflective oracles won't automatically do this. They won't minimize log loss or any oth 1interstice3y The gradient descent is not being done over the reflective oracles, it's being done over some general computational model like a neural net. Any highly-performing solution will necessarily look like a fixed-point-finding computation of some kind, due to the self-referential nature of the predictions. Then, since this fixed-point-finder is internal to the model, it will be optimized for log loss just like everything else in the model. That is, the global optimization of the model is distinct from whatever internal optimization the fixed-point-finder uses to choose the reflective oracle. The global optimization will favor internal optimizers that produce fixed-points with good score. So while fixed-point-finders in general won't optimize for anything in particular, the one this model uses will. Figuring out what Alice wants: non-human Alice Ok, this seems usefully specific. A few concerns: 1. It seems that, according to your description, my proto-preferences are my current map of the situation I am in (or ones I have already imagined) along with valence tags. However, the AI is going to be in a different location, so I actually want it to form a different map (otherwise, it would act as if it were in my location, not its location). So what I actually want to get copied is more like a map-building and valence-tagging procedure that can be applied to different contexts, which will take differ 2Stuart Armstrong3y Thanks! I'm not sure I fully get all your concerns, but I'll try and answer to the best of my understanding. 1-4 (and a little bit of 6): this is why I started looking at semantics vs syntax. Consider the small model "If someone is drowning, I should help them (if it's an easy thing to do)". Then "someone", "downing", "I", and "help them" are vague labels for complex categories (as re most of there rest of the terms, really). The semantics of these categories need to be established before the AI can do anything. And the central examples of the categories will be clearer than the fuzzy edges. Therefore the AI can model me as having a strong preferences in the central example of the categories, which become much weaker as we move to the edges (the meta-preferences will start to become very relevant in the edge cases). I expect that "I should help them" further decomposes into "they should be helped" and "I should get the credit for helping them". Therefore, it seems to me, that an AI should be able to establish that if someone is drowning, it should try and enable me to save them, and if it can't do that, then it should save them itself (using nanotechnology or anything else). It doesn't seem that it would be seeing the issue from my narrow perspective, because I don't see the issue just from my narrow perspective. 5: I am pretty sure that we could use neuroscience to establish that, for example, people are truthful when they say that they see the anchoring bias as a bias. But I might have been a bit glib when mentioning neuroscience; that is mainly the "science fiction superpowers" end of the spectrum for the moment. What I'm hoping, with this technique, is that if we end up using indirect normativity or stated preferences, that my keeping in mind this model of what proto-preferences are, we can better automate the limitations of these techniques (eg when we expect lying), rather than putting them in by hand. 6: Currently I don't see reflexes as embodying values Figuring out what Alice wants: non-human Alice I'm pretty confused by what you mean by proto-preferences. I thought by proto-preferences you meant something like "preferences in the moment, not subject to reflection etc." But you also said there's a definition. What's the definition? (The concept is pre-formal, I don't think you'll be able to provide a satisfactory definition). You have written a paper about how preferences are not identifiable. Why, then, do you say that proto-preferences are identifiable, if they are just preferences in the moment? The impossibility results apply word-for-word t 5Stuart Armstrong3y Oh, I don't claim to have a full definition yet, but I believe it's better than pre-formal. Here would be my current definition: * Humans are partially model-based agents. We often generate models (or at least partial models) of situations (real or hypothetical), and, within those models, label certain actions/outcomes/possibilities as better or worse than others (or sometimes just generically "good" or "bad"). This model, along with the label, is what I'd call a proto-preference (or pre-preference). That's why neuroscience is relevant, for identifying the mental model human use. The "previous Alice post" I mentioned is here [https://www.lesswrong.com/posts/rcXaY3FgoobMkH2jc/figuring-out-what-alice-wants-part-ii] . and was a toy version of this, in the case of an algorithm rather than a human. The reason these get around the No Free Lunch theorem is that they look inside the algorithm (so different algorithms with the same policy can be seen to have different preferences, which breaks NFL), and is making the "normative assumption" that these modelled proto-preferences correspond, (modulo preference synthesis) to the agent's actual preferences. Note that that definition puts preferences and meta-preferences into the same type, the only difference being the sort of model being considered. Figuring out what Alice wants: non-human Alice The overall approach of finding proto-preferences and meta-preferences, resolving them somehow, then extrapolating from there, seems like a reasonable thing to do. But, suppose you're going to do this. Then you're going to run into a problem: proto-preferences aren't identifiable. I interpreted you as trying to fix this problem by looking at how humans infer each other's preferences rather than their (proto-)preferences themselves. You could try learning people's proto-preference-learning-algorithms instead of their proto-preferences. But, this is not an ea 1Stuart Armstrong3y The proto-preferences are a definition of the components that make up preferences. Methods of figuring them out - be they stated preferences, revealed preferences, FMRI machines, how other people infer each other's preferences... - are just methods. The advantage of having a definition is that this guides us explicitly as to when a specific method for figuring them out, ceases to be applicable. And I'd argue that proto-preferences are identifiable. We're talking about figuring out how humans model their own situations, and the better-worse judgements they assign in their internal models. This is not unidentifiable, and neuroscience already has some things to say on it. The previous Alice post showed how you could do it a toy model (with my posts on semantics [https://www.lesswrong.com/posts/XApNuXPckPxwp5ZcW/bridging-syntax-and-semantics-with-quine-s-gavagai] and symbol grounding [https://www.lesswrong.com/posts/EEPdbtvW8ei9Yi2e8/bridging-syntax-and-semantics-empirically] , relevant to applying this approach to humans). That second sentence of mine is somewhat poorly phrased, but I agree that "extracting the normative assumptions humans make is no easier than extracting proto-preferences" - I just don't see that second one as being insoluble. Figuring out what Alice wants: non-human Alice I don't know, but a pseudo-definition that works sometimes is "upon having a lot of time to reflect, information, etc, I would conclude that you have Y values"; of course I can't use this definition when I am doing the reflection, though! "Values" is at the moment a pre-formal concept (utility theory doesn't directly apply to humans), so it has some representation in people's brains that is hard to extract/formalize. In any case, I reject any AI design that concludes that it ought to act as if you have X values just because my current models imply that you 1Rohin Shah3y You could imagine examining a human brain and seeing how it models other humans. This would let you get some normative assumptions out that could inform a value learning technique. I would think of this as extracting an algorithm that could infer human preferences out of a human brain. You could run this algorithm for a long time, in which case it would eventually output Y values, even if you would currently judge the person as having X values. 3Stuart Armstrong3y We're getting close to something important here, so I'll try and sort things out carefully. In my current approach, I'm doing two things: 1. Finding some components of preferences or proto-preferences within the human brain. 2. Synthesising them together in a way that also respects (proto-)meta-preferences. The first step is needed because of the No Free Lunch in preference learning result. We need to have some definition of preferences that isn't behavioural. And the stated-values-after-reflection approach has some specific problems that I listed here [https://www.lesswrong.com/posts/zvrZi95EHqJPxdgps/why-we-need-a-theory-of-human-values] . Then it took an initial stab at how one could sythesise the preferences in this post [https://www.lesswrong.com/posts/Y2LhX3925RodndwpC/resolving-human-values-completely-and-adequately] . If I'm reading you correctly, your main fear is that by focusing on the proto-preferences of the moment, we might end up in a terrible place, foreclosing moral improvements. I share that fear! That's why the process of synthesising values in accordance both with meta-preferences and "far" preferences ("I want everyone to live happy worthwhile lives" is a perfectly valid proto-preference). Where we might differ the most, is that I'm very reluctant to throw away any proto-preferences, even if our meta-preferences would typically overrule it. I would prefer to keep it around, with a very low weight. Once we get in the habit of ditching proto-preferences, there's no telling where that process might end up [https://www.lesswrong.com/posts/WeAt5TeS8aYc4Cpms/values-determined-by-stopping-properties] . Figuring out what Alice wants: non-human Alice Because once we have these parameters, we can learn the values of any given human. This doesn't make the problem easier, you have to start somewhere. I agree this could reduce the total computational work required but it doesn't seem any easier conceptually. Whereas “learn what humans model each other’s values (and rationality) to be” is something that makes sense in the world. This has the same problem as value learning. If I think you have X values but you actually have Y values (and I would think you have Y values upon further reflection etc) then 1Stuart Armstrong3y What do you mean by "you actually have Y values"? What are you defining values to be? Figuring out what Alice wants: non-human Alice Instead, we just need to extract the normative assumptions that humans are already making and use these in the value learning process Okay, but how do you do that if you don't already have a value learning algorithm? Why is it easier to learn the algorithms/parameters humans use in inferring each other's values, than to just learn their values? 2Stuart Armstrong3y Because once we have these parameters, we can learn the values of any given human. In contrast, it we learn the values of a given human, we don't get to learn the values of any other one. I'd argue further: these parameters form part of a definition of human values. We can't just "learn human values", as these don't exist in the world. Whereas "learn what humans model each other's values (and rationality) to be" is something that makes sense in the world. Coherence arguments do not entail goal-directed behavior It can maximize the utility function: if I take the twitch action in time step otherwise. In a standard POMDP setting this always takes the twitch action. Topological Fixed Point Exercises Clarifying question for #9: How does the decomposition into segments/triangles generalize to 3+ dimensions? If you try decomposing a tetrahedron into multiple tetrahedra, you actually get 4 tetrahedra and 1 octahedron, as shown here. You can decompose an octahedron into 4 tetrahedrons. They're irregular, but this is actually fine for the purpose of the lemma. Asymptotic Decision Theory (Improved Writeup) Yes, the continuity condition on embedders in the ADT paper would eliminate the embedder I meant. Which means the answer might depend on whether ADT considers discontinuous embedders. (The importance of the continuity condition is that it is used in the optimality proof; the optimality proof can't apply to chicken for this reason). Asymptotic Decision Theory (Improved Writeup) In the original ADT paper, the agents are allowed to output distributions over moves. The fact that we take the limit as epsilon goes to 0 means the evil problem can't be constructed, even if randomization is not allowed. (The proof in the ADT paper doesn't work, but that doesn't mean something like it couldn't possibly work) It's basically saying "since the two actions A and A′ get equal expected utility in the limit, the total variation distance between a distribution over the two actions, and one of the actions, limits to zero", which is false You' 2Diffractor4y Wasn't there a fairness/continuity condition in the original ADT paper that if there were two "agents" that converged to always taking the same action, then the embedder would assign them the same value? (more specifically, if Et(\|At−Bt\| )<δ, then Et(\|Et(At)−Et(Bt)\|)<ϵ ) This would mean that it'd be impossible to have Et(Et(ADTt,ϵ)) be low while Et(Et(straightt)) is high, so the argument still goes through. Although, after this whole line of discussion, I'm realizing that there are enough substantial differences between the original formulation of ADT and the thing I wrote up that I should probably clean up this post a bit and clarify more about what's different in the two formulations. Thanks for that. Asymptotic Decision Theory (Improved Writeup) OK, I checked this more and I'm more suspicious now. First: in the ADT paper, the asymptotic dominance argument is about the limit of the agent's action as epsilon goes to 0. This limit is not necessarily computable, so the embedder can't contain the agent, since it doesn't know epsilon. So the evil problem doesn't work. The optimality proof might be valid. I didn't understand which specific step you thought was wrong. Second: This also means the chicken problem is ill-typed, since you can't put an ADT in the environment. But there is a well-typed versi	{}

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