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Bit shift right arithmetic - MATLAB bitsra - MathWorks Benelux
Shift Right a Built-in int8 Input
Shift Right a Floating-Point Input
Bit shift right arithmetic
c=bitsra(a,k)
c=bitsra(a,k) returns the result of an arithmetic right shift by k bits on input a for fixed-point operations. For floating-point operations, it performs a multiply by 2-k.
If the input is unsigned, bitsra shifts zeros into the positions of bits that it shifts right. If the input is signed, bitsra shifts the most significant bit (MSB) into the positions of bits that it shifts right.
bitsra ignores fimath properties such as RoundingMode and OverflowAction.
Create a signed fixed-point fi object with a value of –8, word length 4, and fraction length 0. Then display the binary value of the object.
Shift a right by 1 bit.
disp(bin(bitsra(a,1)))
Use bitsra to shift an int8 input right by 2 bits.
bitsra(a,2)
bitsra(a,k)
Scale a floating-point double input by
{2}^{-3}
a = double(128);
Data that you want to shift, specified as a scalar, vector, matrix, or multidimensional array of fi objects or built-in numeric types.
Data Types: fi |single | double | int8 | int16 | int32 | int64 | uint8 | uint16 | uint32 | uint64
Number of bits to shift, specified as a non-negative integer-valued scalar fi object or built-in numeric type.
Generates sra operator in VHDL® code.
Generates >>> operator in Verilog® code.
bitsll | bitsrl | bitshift | pow2 |
Energy Loss of a Heavy Particle Near 3D Rotating Hairy Black Hole
Jalil Naji, Hassan Saadat, "Energy Loss of a Heavy Particle Near 3D Rotating Hairy Black Hole", Advances in High Energy Physics, vol. 2014, Article ID 720713, 5 pages, 2014. https://doi.org/10.1155/2014/720713
Jalil Naji1 and Hassan Saadat2
1Physics Department, Ilam University, P.O. Box 69315-51, Ilam, Iran
2Department of Physics, Shiraz Branch, Islamic Azad University, P.O. Box 71555-477, Shiraz, Iran
We consider rotating black hole in 3 dimensions with a scalar charge and discuss energy loss of heavy particle moving near the black hole horizon. We find that drag force was increased by scalar charge while it was decreased due to the rotation of black hole. We also study quasnormal modes.
The lower dimensional theories may be used as toy models to study some fundamental ideas which yield to better understanding of higher dimensional theories, because they are easier to study [1]. Moreover, these are useful for application of AdS/CFT correspondence [2–5]. This paper is indeed an application of AdS/CFT correspondence to probe moving charged particle near the three-dimensional black holes which are recently introduced by Xu et al. [6, 7] where charged black holes with a scalar hair in (2 + 1) dimensions and rotating hairy black hole in (2 + 1) dimensions are constructed, respectively. Here, we are interested in the case of rotating black hole with a scalar hair in (2 + 1) dimensions. Recently, a charged rotating hairy black hole in 3 dimensions corresponding to infinitesimal black hole parameters was constructed [8]. Also, thermodynamics of such systems is recently studied in [9, 10]. We consider this background in AdS side as a dual picture of a QCD model as CFT side.
In this paper, we would like to study the motion of a heavy charged particle near the black hole horizon and calculate the energy loss. The energy loss of moving heavy charged particle through a thermal medium is known as the drag force. One can consider a moving heavy particle (such as charm and bottom quarks) near the black hole horizon with the momentum , mass , and constant velocity , which is influenced by an external force . So, one can write the equation of motion as , where in the nonrelativistic motion , and in the relativistic motion ; also is called the friction coefficient. In order to obtain drag force, one can consider two special cases. The first case is the constant momentum which yields to obtain for the norelativistic case. In this case, the drag force coefficient will be obtained. In the second case, the external force is zero, so one can find . In other words, by measuring the ratio or , one can determine friction coefficient without any dependence on mass . These methods lead us to obtain the drag force for a moving heavy particle. The moving heavy particle in context of QCD has dual picture in the string theory in which an open string is attached to the D-brane and stretched to the horizon of the black hole. Therefore, we can apply AdS/CFT correspondence to probe a charged particle (such as a quark) moving through 3D hairy black hole background.
Similar studies are already performed in several backgrounds [11–22]. Most of them considered and super Yang-Mills plasma with asymptotically AdS geometries. Also [20] considered 4D Kerr-AdS black holes. All of the mentioned studies used AdS5/CFT4 correspondence. Now, we are going to consider the same problem in a rotating hairy 3D background and use AdS3/CFT2 correspondence [23–25].
This paper is organized as follows. In the next section, we review rotating hairy black hole in (2 + 1) dimensions. In Section 3, we obtain equation of motion and in Section 4, we try to obtain solution and discuss about drag force. In Section 5, we give linear analysis and discuss quasinormal modes. Finally, in Section 6, we summarized our results.
2. Rotating Hairy Black Hole in (2 + 1) Dimensions
Rotating hairy black hole in (2 + 1) dimensions is described by the following action: which yields to the following line element [1]: where where is a rotation parameter related to the angular momentum of the solution and is related to the cosmological constant via . is integration constants depending on the black hole mass: and scalar charge is related to the scalar field as Also, one can obtain Ricci scalar of this model is given by We can see that Ricci scalar is singular at the origin.
Black hole horizon, which is obtained by , may be written as follows: where we defined
3. The Equations of Motion
The moving heavy particle near the black hole may be described by the following Nambu-Goto action: where is the string tension. The coordinates and are corresponding to the string world-sheet. Also, is the induced metric on the string world-sheet with determinant obtained as follows: where we used static gauge in which , , and the string only extends in one direction . Then, the equation of motion is obtained as follows: We should obtain canonical momentum densities associated with the string as follows: The simplest solution of the equation of motion is static string described by = constant with total energy of the form, where is an arbitrary location of D-brane. As we expected, the energy of static particle is interpreted as the remaining mass.
4. Time Dependent Solution
In the general case, we can assume that the particle moves with constant speed ; in that case, the equation of motion (12) reduces to where Equation (15) gives the following expression: where is an integration constant which will be determined by using reality condition of . Therefore, we yield to the following canonical momentum densities: These give us loosing energy and momentum through an endpoint of string: As we mentioned before, reality condition of gives us constant . The expression is real for . In the case of small , one can obtain which yields to Therefore, we can write drag force as follows: We draw drag force in terms of velocity and in agreement with the previous works such as [11–22]; the value of drag force increased by . In Figure 1, we can see behavior of drag force with rotation parameter and scalar charge. It is shown that the scalar charge increases the value of drag force but the increasing rotational parameter decreases the value of the drag force.
Drag force in terms of for , , and ; (dotted line), (solid line), and (dashed line).
Motion of string yields to small perturbation after late time due to the drag force. In that case, the speed of particle is infinitesimal and one can write . Also, we assume that , where is the friction coefficient. Therefore, one can rewrite the equation of motion as follows: We assume outgoing boundary conditions near the black hole horizon and use the following approximation: which suggests the following solutions: where is the black hole temperature. In the case of infinitesimal , we can use the following expansion: Inserting this equation in the relation (24) gives = constant, and where is a constant. Assuming near horizon limit enables us to obtain the following solution: Comparing (25) and (27) gives the following quasinormal mode condition: It is interesting to note that these results recover drag force (22) for infinitesimal speed. In Figure 2, we can see behavior of with rotational parameter and scalar charge. We find that scalar charge increases the value of friction coefficient, but the effect of rotation decreases .
in terms of : and (blue dashed line), and (blue solid line), and (blue dotted line), and and (green dashed line), and and (green solid line).
5.1. Low Mass Limit
Low mass limit means that , and we use the following assumptions: so, by using relation (23) we can write Then, we can obtain constant as follows: It tells that yields to divergence; therefore we called this a critical behavior of the friction coefficient and found that Figure 3 shows behavior of critical friction coefficient with the black hole parameters.
in terms of for and ; (blue line), (black line), and (red line).
In this paper, we considered rotating 3D black hole together with a scalar charge as a background where a charged particle moves with speed and then calculated drag force. We used motivation of AdS/CFT correspondence and string theory method to study motion of charged particle. This is indeed in the context of AdS3/CFT2 where drag force on moving heavy particle is calculated. Numerically, we found that the scalar charge increases the value of drag force but rotational parameter decreases the value of the drag force. Therefore, in order to have the most free motion we need to increase and decrease . It means that and may cancel the effect of each other on the drag force. We can find critical values of scalar charge and rotational parameters in which the value of drag force will be infinite as Then, we studied quasinormal modes and obtained friction coefficient which was enhanced by the black hole charge and reduced by rotation. Quasinormal mode analysis also reproduced drag force at slow velocities. It is also possible to study dispersion relations which again reproduce the drag force which was obtained in (22). For the future work, we will consider charged rotating 3D hairy black hole and study drag force.
M. Hortaçsu, H. T. Özçelik, and B. Yapışkan, “Properties of solutions in
2+1
dimensions,” General Relativity and Gravitation, vol. 35, no. 7, pp. 1209–1221, 2003. View at: Publisher Site | Google Scholar | Zentralblatt MATH | MathSciNet
M. Henneaux, C. Martínez, R. Troncoso, and J. Zanelli, “Black holes and asymptotics of
2+1
gravity coupled to a scalar field,” Physical Review D, vol. 65, Article ID 104007, 4 pages, 2002. View at: Publisher Site | Google Scholar | MathSciNet
M. Hasanpour, F. Loran, and H. Razaghian, “Gravity/CFT correspondence for three-dimensional Einstein gravity with a conformal scalar field,” Nuclear Physics B, vol. 867, no. 2, pp. 483–505, 2013. View at: Publisher Site | Google Scholar | Zentralblatt MATH | MathSciNet
D. F. Zeng, “An exact hairy black hole solution for AdS/CFT superconductors,” In press, http://arxiv.org/abs/0903.2620. View at: Google Scholar
B. Chen, Z. Xue, and J.-j. Zhang, “Note on thermodynamics method of black hole/CFT correspondence,” Journal of High Energy Physics, vol. 2013, article 102, 2013. View at: Publisher Site | Google Scholar | MathSciNet
W. Xu and L. Zhao, “Charged black hole with a scalar hair in (2+1) dimensions,” Physical Review D, vol. 87, Article ID 124008, 11 pages, 2013. View at: Publisher Site | Google Scholar
L. Zhao, W. Xu, and B. Zhu, “Novel rotating hairy black hole in (2+1)-dimensions,” In press, http://arxiv.org/abs/1305.6001. View at: Google Scholar
J. Sadeghi, B. Pourhassan, and H. Farahani, “Rotating charged hairy black hole in (2+1) dimensions and particle acceleration,” In press, http://arxiv.org/abs/1310.7142. View at: Google Scholar
A. Belhaj, M. Chabab, H. EL Moumni, and M. B. Sedra, “Critical behaviors of 3D black holes with a scalar hair,” In press, http://arxiv.org/abs/1306.2518. View at: Google Scholar
J. Sadeghi and H. Farahani, “Thermodynamics of a charged hairy black hole in (2+1) dimensions,” In press, http://arxiv.org/abs/1308.1054. View at: Google Scholar
C. Hoyos-Badajoz, “Drag and jet quenching of heavy quarks in a strongly coupled Script
𝒩
= 2* plasma,” Journal of High Energy Physics, vol. 2009, no. 9, article 068, 2009. View at: Publisher Site | Google Scholar
J. Sadeghi and B. Pourhassan, “Drag force of moving quark at
𝒩
= 2 supergravity,” Journal of High Energy Physics A, vol. 2008, no. 12, article 026, 2008. View at: Publisher Site | Google Scholar | MathSciNet
C. P. Herzog, A. Karch, P. Kovtun, C. Kozcaz, and L. G. Yaffe, “Energy loss of a heavy quark moving through
𝒩
= 4 supersymmetric Yang-Mills plasma,” Journal of High Energy Physics A, vol. 2006, no. 7, article 013, 2006. View at: Publisher Site | Google Scholar | MathSciNet
J. Sadeghi, B. Pourhassan, and S. Heshmatian, “Application of AdS/CFT in quark-gluon plasma,” Advances in High Energy Physics, vol. 2013, Article ID 759804, 13 pages, 2013. View at: Publisher Site | Google Scholar | MathSciNet
C. P. Herzog, “Energy loss of a heavy quark from asymptotically AdS geometries,” Journal of High Energy Physics A, vol. 2006, no. 9, article 032, 2006. View at: Publisher Site | Google Scholar | MathSciNet
S. S. Gubser, “Drag force in AdS/CFT,” Physical Review D, vol. 74, no. 12, Article ID 126005, 4 pages, 2006. View at: Publisher Site | Google Scholar | MathSciNet
E. Nakano, S. Teraguchi, and W. Y. Wen, “Drag force, jet quenching, and AdS/QCD,” Physical Review D, vol. 75, no. 8, Article ID 085016, 6 pages, 2007. View at: Publisher Site | Google Scholar
E. Cáceres and A. Güijosa, “Drag force in a charged
𝒩
= 4 SYM plasma,” Journal of High Energy Physics A, vol. 2006, no. 11, article 077, 2006. View at: Publisher Site | Google Scholar | MathSciNet
J. F. Vazquez-Poritz, “Drag force at finite 't Hooft coupling from AdS/CFT,” In press, http://arxiv.org/abs/0803.2890. View at: Google Scholar
A. N. Atmaja and K. Schalm, “Anisotropic drag force from 4D Kerr-AdS black holes,” http://arxiv.org/abs/1012.3800. View at: Google Scholar
B. Pourhassan and J. Sadeghi, “STU-QCD correspondence,” Canadian Journal of Physics, vol. 91, no. 12, pp. 995–1019, 2013. View at: Publisher Site | Google Scholar
E. Caceres and A. Guijosa, “On drag forces and jet quenching in strongly coupled plasmas,” Journal of High Energy Physics, vol. 2006, no. 12, article 068, 2006. View at: Publisher Site | Google Scholar
P. Kraus, “Lectures on black holes and the
{\mathrm{AdS}}_{3}/{\mathrm{CFT}}_{2}
correspondence,” in Supersymmetric Mechanics. Vol. 3, vol. 755 of Lecture Notes in Physics, pp. 193–247, Springer, Berlin, Germany, 2008. View at: Google Scholar | Zentralblatt MATH | MathSciNet
R. Borsato, O. Ohlsson Sax, and A. Sfondrini, “All-loop Bethe ansatz equations for
{\mathrm{AdS}}_{3}/{\mathrm{CFT}}_{2}
,” Journal of High Energy Physics, vol. 4, article 116, 2013. View at: Google Scholar | MathSciNet
D. Momeni, M. Raza, M. R. Setare, and R. Myrzakulov, “Analytical holographic superconductor with backreaction using
Ad{S}_{3}/CF{T}_{2}
,” International Journal of Theoretical Physics, vol. 52, no. 8, pp. 2773–2783, 2013. View at: Publisher Site | Google Scholar | Zentralblatt MATH | MathSciNet
Copyright © 2014 Jalil Naji and Hassan Saadat. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. The publication of this article was funded by SCOAP3. |
Screw thread profile with a square cross-section
Find sources: "Square thread form" – news · newspapers · books · scholar · JSTOR (June 2021)
Dimensions of a square thread form
The square thread form is a common screw thread profile, used in high load applications such as leadscrews and jackscrews. It gets its name from the square cross-section of the thread.[1] It is the lowest friction and most efficient thread form, but it is difficult to fabricate.
The greatest advantage of square threads is that they have a much higher intrinsic efficiency than trapezoidal threads (Acme or metric trapezoidal). Due to the lack of a thread angle there is no radial pressure, or bursting pressure, on the nut. This also increases the nut life.[1]
The greatest disadvantage is the difficulty in machining such a thread. The single-point cutting tools or taps and dies used to cut the thread cannot have efficient rake and relief angles (because of the square form), which makes the cutting slow and difficult. Square threads also cannot carry as much load as a trapezoidal thread, because the root of the square thread is smaller. Also, there is no way to compensate for wear on the nut, so it must be replaced when worn out.[1]
Square threads are defined as follows by ISO standards:
{\displaystyle Sq\,60\times 9}
where Sq designates a square thread, 60 is the nominal diameter in millimeters, and 9 is the pitch in millimeters. When there is no suffix it is a single start thread. If there is a suffix then the value after the multiplication sign is the lead and the value in the parentheses is the pitch. For example:
{\displaystyle Sq\,60\times 18(P9)LH}
would denoted two starts, as the lead divided by the pitch is two. The "LH" denotes a left hand thread.[2]
Standard pitches for metric diameters[2]
Main article: Leadscrew mechanics
^ a b c Bhandari, p. 203.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Square_thread_form&oldid=1088514775" |
Adenosine triphosphate - New World Encyclopedia
Previous (Adenine)
Next (Adhesive)
[[[5-(6-aminopurin-9-yl)-3,4-dihydroxy-oxolan-2-yl]
methoxy-hydroxy-phosphoryl]
oxy-hydroxy-phosphoryl] oxyphosphonic acid
Abbreviations ATP
Molecular mass 507.181 g mol-1
Adenosine triphosphate (ATP) is the chemical compound known in biochemistry as the "molecular currency" of intracellular energy transfer; that is, ATP is able to store and transport chemical energy within cells. All cells—both prokaryotic, such as bacteria, and eukaryotic, such as with amoeba, fungi, plants, and animals—use ATP as the main molecule for carrying energy, and as the principal energy source for endergonic, or energy-requiring, reactions.
Living cells require energy to survive and function, and most of this energy comes either via radiant energy or from chemical energy tied up in interatomic bonds of nutrient molecules. When nutrient molecules, such as those derived from carbohydrates and fats, are oxidized by cells, a portion of the free energy released can be captured in the chemical bonds of ATP. ATP allows cells to store energy as chemical potential and to circulate and use this energy. Cells are constantly creating and circulating ATP, and when cells need energy, they "spend ATP," leading it to be commonly referred to as the energy currency of life.
In addition to its energy-related function, ATP also plays an important role in the synthesis of nucleic acids and further in signal transduction pathways in which it provides the phosphate for the protein-kinase reactions.
4 ATP in the human body
The ubiquitous presence of ATP in the cells of all living organisms provides support for the view that newer creations are built on the foundation of earlier creations, with ATP having appeared very early in the history of cellular life. The universal use of ATP likewise reflects the conservative nature of creation, where the same or similar metabolic processes and chemical compounds repeatedly occur, and it reflects a connectedness from the simplest organisms to humans. The intricate manner in which ATP is integrated in fundamental metabolic pathways also reveals the complex coordination required between the parts of living systems.
ATP consists of adenosine and three attached phosphate groups (triphosphate). Adenosine itself is composed of two major molecular entities, adenine (a nitrogen-containing molecule) and ribose (a five-carbon sugar). Adenosine monophosphate (AMP) has one phosphate group attached to adenosine, and adenosine diphosphate (ADP) has two attached phosphate groups.
The three linked phosphoryl groups, starting with that on AMP, are referred to as the alpha (α), beta (β), and gamma (γ) phosphates. These linked phosphate groups are the "business end" of the molecule, as ATP stores energy in the bonds between the phosphate groups. A molecule of ATP is sometimes written as A~P~P~P, with the "~" representing a bond that contains potential chemical energy.
ATP is extremely rich in chemical energy, in particular between the second and third phosphate groups. As these chemical bonds are broken (as ATP is converted into ADP and an inorganic phosphate) the energy release is -12 kCal / mole in vivo (inside a living cell), and -7.3 kCal / mole in vitro (in laboratory conditions). Such a relatively massive release of energy from a single chemical change with the whole cycle of charging and discharging the molecule integrated perfectly into the regular cellular metabolism is what makes ATP so valuable to all forms of life. The molecules can be charged up at one site and transported to another site for discharge, somewhat like a dry cell battery.
Space filling image of ATP
ATP can be produced by various cellular processes. Under aerobic conditions, the synthesis occurs in mitochondria during oxidative phosphorylation, which is catalyzed by ATP synthase; to a lesser degree, under anaerobic conditions, this is done through substrate phosphorylation catalyzed by two enzymes: phosphoglycerate kinase (PGK) and pyruvate kinase.
ATP is also synthesized through several so-called "replenishment" reactions catalyzed by the enzyme families of NDKs (nucleoside diphosphate kinases), which use other nucleoside triphosphates as a high-energy phosphate donor, and the ATP guanido-phosphotransferase family, which uses creatine.
{\displaystyle \to }
In plants, ATP is synthesized in chloroplasts by photosynthesis during the light reactions of photosynthesis. However, this ATP is then used to power the Calvin cycle step of photosynthesis and so photosynthesis does not result in an overall production of ATP.
The main fuels for ATP synthesis are glucose and fatty acids. First, glucose is broken down into pyruvate in the cytosol yielding two molecules of ATP for each glucose molecule. Further breakdown of the glucose molecule for synthesizing ATP is carried out in the mitochondria in a process that yields about 30 molecules of ATP for each molecule of glucose that is oxidized. (See citric acid cycle.)
ATP energy is released through hydrolysis (breakdown through reaction with water) of the high-energy phosphate-phosphate bonds. An enzyme, ATPase, aids in the breaking of the bond between the second and third phosphate groups, as ATP is converted to ADP. The hydrolysis yields free inorganic phosphate (Pi) and ADP. Although this may result in free phosphate ions, usually the phosphate group is transferred to another molecule in a process called phosphorylation.
Energy is also released when the bond between the first and second phosphate groups is broken, as ADP is converted to AMP. That is, ADP can be broken down further to another Pi and AMP. ATP can also be broken down to AMP directly, with the formation of pyrophosphate (PPi). This last reaction has the advantage of being an effectively irreversible process in aqueous solution.
This energy can be used by a variety of enzymes, motor proteins, and transport proteins to carry out the work of the cell.
ATP in the human body
The total quantity of ATP in the human body at any one time is about 0.1 mole. Yet, adults convert daily a quantity of ATP corresponding to at least half their body weight, and nearly a ton during a day of hard work. That is, the energy used by human cells requires the hydrolysis of 200 to 300 moles of ATP daily. This means that each ATP molecule is recycled 2,000 to 3,000 times during a single day. There is limited capacity to store ATP in a cell, and it is depleted in seconds, hence its consumption must closely follow its synthesis. That is, cells need to continually replenish or re-synthesize ATP.
Abrahams, J.P., A. G. Leslie, R. Lutter, and J. E. Walker. 1994. Structure at 2.8 Å resolution of F 1 -ATPase from bovine heart mitochondria. Nature 370:621–628.
Boyer, P. D. 1993. The binding change mechanism for ATP synthase: Some probabilities and possibilities. Biochimica et Biophysica Acta 1140:215–250.
Boyer, P. D. 1997. The ATP synthase - a splendid molecular machine. Annual Review in Biochemistry 66:717–749.
Lutsenko, S., and J. H. Kaplan. 1996. Organization of P-type ATPases: Significance of structural diversity. Biochemistry 34:15607–15613.
Möller, J. V., B. Juul, and M. le Maire. 1996. Structural organization, ion transport, and energy transduction of P-type ATPases. Biochimica et Biophysica Acta 1286:1–51.
Skou, J. C. 1957. The influence of some cations on an adenosine triphosphatase from peripheral nerves. Biochimica et Biophysica Acta 23:394–401.
Skou, J. C., and M. Esmann. 1992. The Na, K-ATPase. Journal of Bioenergetics and Biomembranes 24:249–261.
Lingrel, J. B. 1992. Na-K-ATPase: Isoform structure, function, and expression. Journal of Bioenergetics and Biomembranes 24:263–270.
Adenosine triphosphate history
History of "Adenosine triphosphate"
Retrieved from https://www.newworldencyclopedia.org/p/index.php?title=Adenosine_triphosphate&oldid=994007 |
Arithmetic Progression Sum Practice Problems Online | Brilliant
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19.
The number of red circles in the above diagram forms an arithmetic progression. What is the sum of the first 10 terms of this sequence?
You're the captain of your cricket team which is taking part in a series of matches. In the first match you score 40 runs and in the last match you score 100 runs. Provided that your score increases by 12 runs in each match, find your total score throughout the series.
If the sum of first
n
terms of an arithmetic progression is given by
S_n=3n^2+2n
, what is the 24th term of the AP?
\begin{array} { l l l } 2 & = 2 & = 1 \times 2 \\ 2 + 4 & = 6 & = 2 \times 3 \\ 2 + 4 + 6 & = 12 & = 3 \times 4 \\ \vdots & \vdots & \vdots \\ \end{array}
According to the above pattern, what is
2 + 4 + 6 + \ldots + 30 ?
14 \times 15
15 \times 16
29 \times 30
30 \times 31 |
Spin-weighted spherical harmonics - Wikipedia
Not to be confused with spinor spherical harmonics.
In special functions, a topic in mathematics, spin-weighted spherical harmonics are generalizations of the standard spherical harmonics and—like the usual spherical harmonics—are functions on the sphere. Unlike ordinary spherical harmonics, the spin-weighted harmonics are U(1) gauge fields rather than scalar fields: mathematically, they take values in a complex line bundle. The spin-weighted harmonics are organized by degree l, just like ordinary spherical harmonics, but have an additional spin weight s that reflects the additional U(1) symmetry. A special basis of harmonics can be derived from the Laplace spherical harmonics Ylm, and are typically denoted by sYlm, where l and m are the usual parameters familiar from the standard Laplace spherical harmonics. In this special basis, the spin-weighted spherical harmonics appear as actual functions, because the choice of a polar axis fixes the U(1) gauge ambiguity. The spin-weighted spherical harmonics can be obtained from the standard spherical harmonics by application of spin raising and lowering operators. In particular, the spin-weighted spherical harmonics of spin weight s = 0 are simply the standard spherical harmonics:
{\displaystyle {}_{0}Y_{lm}=Y_{lm}\ .}
Spaces of spin-weighted spherical harmonics were first identified in connection with the representation theory of the Lorentz group (Gelfand, Minlos & Shapiro 1958). They were subsequently and independently rediscovered by Newman & Penrose (1966) and applied to describe gravitational radiation, and again by Wu & Yang (1976) as so-called "monopole harmonics" in the study of Dirac monopoles.
1 Spin-weighted functions
1.1 The operator ð
2 Spin-weighted harmonics
3 Representation as functions
4 Orthogonality and completeness
6 First few spin-weighted spherical harmonics
6.1 Spin-weight s = 1, degree l = 1
7 Relation to Wigner rotation matrices
8 Triple integral
Spin-weighted functions[edit]
Regard the sphere S2 as embedded into the three-dimensional Euclidean space R3. At a point x on the sphere, a positively oriented orthonormal basis of tangent vectors at x is a pair a, b of vectors such that
{\displaystyle {\begin{aligned}\mathbf {x} \cdot \mathbf {a} =\mathbf {x} \cdot \mathbf {b} &=0\\\mathbf {a} \cdot \mathbf {a} =\mathbf {b} \cdot \mathbf {b} &=1\\\mathbf {a} \cdot \mathbf {b} &=0\\\mathbf {x} \cdot (\mathbf {a} \times \mathbf {b} )&>0,\end{aligned}}}
where the first pair of equations states that a and b are tangent at x, the second pair states that a and b are unit vectors, the penultimate equation that a and b are orthogonal, and the final equation that (x, a, b) is a right-handed basis of R3.
A spin-weight s function f is a function accepting as input a point x of S2 and a positively oriented orthonormal basis of tangent vectors at x, such that
{\displaystyle f{\bigl (}\mathbf {x} ,(\cos \theta )\mathbf {a} -(\sin \theta )\mathbf {b} ,(\sin \theta )\mathbf {a} +(\cos \theta )\mathbf {b} {\bigr )}=e^{is\theta }f(\mathbf {x} ,\mathbf {a} ,\mathbf {b} )}
for every rotation angle θ.
Following Eastwood & Tod (1982), denote the collection of all spin-weight s functions by B(s). Concretely, these are understood as functions f on C2\{0} satisfying the following homogeneity law under complex scaling
{\displaystyle f\left(\lambda z,{\overline {\lambda }}{\bar {z}}\right)=\left({\frac {\overline {\lambda }}{\lambda }}\right)^{s}f\left(z,{\bar {z}}\right).}
This makes sense provided s is a half-integer.
Abstractly, B(s) is isomorphic to the smooth vector bundle underlying the antiholomorphic vector bundle O(2s) of the Serre twist on the complex projective line CP1. A section of the latter bundle is a function g on C2\{0} satisfying
{\displaystyle g\left(\lambda z,{\overline {\lambda }}{\bar {z}}\right)={\overline {\lambda }}^{2s}g\left(z,{\bar {z}}\right).}
Given such a g, we may produce a spin-weight s function by multiplying by a suitable power of the hermitian form
{\displaystyle P\left(z,{\bar {z}}\right)=z\cdot {\bar {z}}.}
Specifically, f = P−sg is a spin-weight s function. The association of a spin-weighted function to an ordinary homogeneous function is an isomorphism.
The operator ð[edit]
The spin weight bundles B(s) are equipped with a differential operator ð (eth). This operator is essentially the Dolbeault operator, after suitable identifications have been made,
{\displaystyle \partial :{\overline {\mathbf {O} (2s)}}\to {\mathcal {E}}^{1,0}\otimes {\overline {\mathbf {O} (2s)}}\cong {\overline {\mathbf {O} (2s)}}\otimes \mathbf {O} (-2).}
Thus for f ∈ B(s),
{\displaystyle \eth f\ {\stackrel {\text{def}}{=}}\ P^{-s+1}\partial \left(P^{s}f\right)}
defines a function of spin-weight s + 1.
Spin-weighted harmonics[edit]
Just as conventional spherical harmonics are the eigenfunctions of the Laplace-Beltrami operator on the sphere, the spin-weight s harmonics are the eigensections for the Laplace-Beltrami operator acting on the bundles E(s) of spin-weight s functions.
Representation as functions[edit]
The spin-weighted harmonics can be represented as functions on a sphere once a point on the sphere has been selected to serve as the North pole. By definition, a function η with spin weight s transforms under rotation about the pole via
{\displaystyle \eta \rightarrow e^{is\psi }\eta .}
Working in standard spherical coordinates, we can define a particular operator ð acting on a function η as:
{\displaystyle \eth \eta =-\left(\sin {\theta }\right)^{s}\left\{{\frac {\partial }{\partial \theta }}+{\frac {i}{\sin {\theta }}}{\frac {\partial }{\partial \phi }}\right\}\left[\left(\sin {\theta }\right)^{-s}\eta \right].}
This gives us another function of θ and φ. (The operator ð is effectively a covariant derivative operator in the sphere.)
An important property of the new function ðη is that if η had spin weight s, ðη has spin weight s + 1. Thus, the operator raises the spin weight of a function by 1. Similarly, we can define an operator ð which will lower the spin weight of a function by 1:
{\displaystyle {\bar {\eth }}\eta =-\left(\sin {\theta }\right)^{-s}\left\{{\frac {\partial }{\partial \theta }}-{\frac {i}{\sin {\theta }}}{\frac {\partial }{\partial \phi }}\right\}\left[\left(\sin {\theta }\right)^{s}\eta \right].}
The spin-weighted spherical harmonics are then defined in terms of the usual spherical harmonics as:
{\displaystyle {}_{s}Y_{lm}={\begin{cases}{\sqrt {\frac {(l-s)!}{(l+s)!}}}\ \eth ^{s}Y_{lm},&&0\leq s\leq l;\\{\sqrt {\frac {(l+s)!}{(l-s)!}}}\ \left(-1\right)^{s}{\bar {\eth }}^{-s}Y_{lm},&&-l\leq s\leq 0;\\0,&&l<|s|.\end{cases}}}
The functions sYlm then have the property of transforming with spin weight s.
Other important properties include the following:
{\displaystyle {\begin{aligned}\eth \left({}_{s}Y_{lm}\right)&=+{\sqrt {(l-s)(l+s+1)}}\,{}_{s+1}Y_{lm};\\{\bar {\eth }}\left({}_{s}Y_{lm}\right)&=-{\sqrt {(l+s)(l-s+1)}}\,{}_{s-1}Y_{lm};\end{aligned}}}
The harmonics are orthogonal over the entire sphere:
{\displaystyle \int _{S^{2}}{}_{s}Y_{lm}\,{}_{s}{\bar {Y}}_{l'm'}\,dS=\delta _{ll'}\delta _{mm'},}
and satisfy the completeness relation
{\displaystyle \sum _{lm}{}_{s}{\bar {Y}}_{lm}\left(\theta ',\phi '\right){}_{s}Y_{lm}(\theta ,\phi )=\delta \left(\phi '-\phi \right)\delta \left(\cos \theta '-\cos \theta \right)}
Calculating[edit]
These harmonics can be explicitly calculated by several methods. The obvious recursion relation results from repeatedly applying the raising or lowering operators. Formulae for direct calculation were derived by Goldberg et al. (1967) harvtxt error: no target: CITEREFGoldbergMacfarlaneNewmanRohlich1967 (help). Note that their formulae use an old choice for the Condon–Shortley phase. The convention chosen below is in agreement with Mathematica, for instance.
The more useful of the Goldberg, et al., formulae is the following:
{\displaystyle {}_{s}Y_{lm}(\theta ,\phi )=\left(-1\right)^{m}{\sqrt {\frac {(l+m)!(l-m)!(2l+1)}{4\pi (l+s)!(l-s)!}}}\sin ^{2l}\left({\frac {\theta }{2}}\right)\times \sum _{r=0}^{l-s}{l-s \choose r}{l+s \choose r+s-m}\left(-1\right)^{l-r-s}e^{im\phi }\cot ^{2r+s-m}\left({\frac {\theta }{2}}\right)\,.}
A Mathematica notebook using this formula to calculate arbitrary spin-weighted spherical harmonics can be found here.
With the phase convention here:
{\displaystyle {\begin{aligned}{}_{s}{\bar {Y}}_{lm}&=\left(-1\right)^{s+m}{}_{-s}Y_{l(-m)}\\{}_{s}Y_{lm}(\pi -\theta ,\phi +\pi )&=\left(-1\right)^{l}{}_{-s}Y_{lm}(\theta ,\phi ).\end{aligned}}}
First few spin-weighted spherical harmonics[edit]
Analytic expressions for the first few orthonormalized spin-weighted spherical harmonics:
Spin-weight s = 1, degree l = 1[edit]
{\displaystyle {\begin{aligned}{}_{1}Y_{10}(\theta ,\phi )&={\sqrt {\frac {3}{8\pi }}}\,\sin \theta \\{}_{1}Y_{1\pm 1}(\theta ,\phi )&=-{\sqrt {\frac {3}{16\pi }}}(1\mp \cos \theta )\,e^{\pm i\phi }\end{aligned}}}
Relation to Wigner rotation matrices[edit]
{\displaystyle D_{-ms}^{l}(\phi ,\theta ,-\psi )=\left(-1\right)^{m}{\sqrt {\frac {4\pi }{2l+1}}}{}_{s}Y_{lm}(\theta ,\phi )e^{is\psi }}
This relation allows the spin harmonics to be calculated using recursion relations for the D-matrices.
Triple integral[edit]
The triple integral in the case that s1 + s2 + s3 = 0 is given in terms of the 3-j symbol:
{\displaystyle \int _{S^{2}}\,{}_{s_{1}}Y_{j_{1}m_{1}}\,{}_{s_{2}}Y_{j_{2}m_{2}}\,{}_{s_{3}}Y_{j_{3}m_{3}}={\sqrt {\frac {\left(2j_{1}+1\right)\left(2j_{2}+1\right)\left(2j_{3}+1\right)}{4\pi }}}{\begin{pmatrix}j_{1}&j_{2}&j_{3}\\m_{1}&m_{2}&m_{3}\end{pmatrix}}{\begin{pmatrix}j_{1}&j_{2}&j_{3}\\-s_{1}&-s_{2}&-s_{3}\end{pmatrix}}}
Dray, Tevian (May 1985), "The relationship between monopole harmonics and spin-weighted spherical harmonics", J. Math. Phys., American Institute of Physics, 26 (5): 1030–1033, Bibcode:1985JMP....26.1030D, doi:10.1063/1.526533 .
Eastwood, Michael; Tod, Paul (1982), "Edth-a differential operator on the sphere", Mathematical Proceedings of the Cambridge Philosophical Society, 92 (2): 317–330, Bibcode:1982MPCPS..92..317E, doi:10.1017/S0305004100059971 .
Gelfand, I. M.; Minlos, Robert A.; Shapiro, Z. Ja. (1958), Predstavleniya gruppy vrashcheni i gruppy Lorentsa, ikh primeneniya, Gosudarstv. Izdat. Fiz.-Mat. Lit., Moscow, MR 0114876 ; (1963) Representations of the rotation and Lorentz groups and their applications (translation). Macmillan Publishers.
Goldberg, J. N.; Macfarlane, A. J.; Newman, E. T.; Rohrlich, F.; Sudarshan, E. C. G. (November 1967), "Spin-s Spherical Harmonics and ð", J. Math. Phys., American Institute of Physics, 8 (11): 2155–2161, Bibcode:1967JMP.....8.2155G, doi:10.1063/1.1705135 (Note: As mentioned above, this paper uses a choice for the Condon-Shortley phase that is no longer standard.)
Newman, E. T.; Penrose, R. (May 1966), "Note on the Bondi-Metzner-Sachs Group", J. Math. Phys., American Institute of Physics, 7 (5): 863–870, Bibcode:1966JMP.....7..863N, doi:10.1063/1.1931221 .
Wu, Tai Tsun; Yang, Chen Ning (1976), "Dirac monopole without strings: monopole harmonics", Nuclear Physics B, 107 (3): 365–380, Bibcode:1976NuPhB.107..365W, doi:10.1016/0550-3213(76)90143-7, MR 0471791 .
Retrieved from "https://en.wikipedia.org/w/index.php?title=Spin-weighted_spherical_harmonics&oldid=983280421" |
Empty set - Simple English Wikipedia, the free encyclopedia
In mathematics, the empty set is the set that has nothing in it. It is often written as
{\displaystyle \varnothing }
{\displaystyle \emptyset }
{\displaystyle \{\}}
.[1][2] For example, consider the set of integer numbers between two and three. Since there is no integer between two and three, the set of integer numbers between them is empty.
Any statement about all elements of the empty set is automatically true. For example, all integers between two and three are greater than seven. This only makes sense because there are no integers between two and three. This kind of truth is called vacuous truth.
Note that the symbol
{\displaystyle \varnothing }
comes from the Latin letter Ø, and not the Greek letter φ. The empty set is also sometimes called the null set.[3]
↑ Weisstein, Eric W. "Empty Set". mathworld.wolfram.com. Retrieved 2020-10-10.
↑ Taylor, Courtney (March 23, 2018). "What Is the Empty Set in Set Theory?". ThoughtCo. Retrieved 2020-10-11. {{cite web}}: CS1 maint: url-status (link)
Retrieved from "https://simple.wikipedia.org/w/index.php?title=Empty_set&oldid=7141033" |
Janae Pritchett, Zandra Vinegar, and Jason Dyer contributed
Angles, Surface Area, and Volume
Angles and the Pythagorean Theorem
Exponents and Exponential Growth
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1-2-\sqrt{5}
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General Chemistry/Reactions of Acids and Bases - Wikibooks, open books for an open world
General Chemistry/Reactions of Acids and Bases
← Buffer Systems ·Phases of Matter →
← Buffer Systems · General Chemistry · Phases of Matter →
3 Anhydrides
5 Lewis Acids/Bases
To summarize the properties and behaviors of acids and bases, this chapter lists and explains the various chemical reactions that they undergo. You may wish to review chemical equations and types of reactions before attempting this chapter.
The following reactions are net ionic equations. In other words, spectator ions are not written. If an ion does not partake in the reaction, it is simply excluded. The spectator ions can be found because they occur on both the reactant and the product side of the equation. Cross them out and rewrite the equation without them. Of course, the coefficients must be equal.
Canceling out the spectator ions explains the net of net ionic equations. The ionic part means that dissolved compounds are written as ions instead of compounds. Acids, bases, and salts are all ionic, so they are written as separate ions if they have dissociated.
Soluble salts are written as ions.
e.g.: Na+ + Cl-
Solids, liquids, and gases are written as compounds.
e.g.: NaCl(s), H2O(l), HCl(g)
Strong acids and strong bases are written as ions (because they dissociate almost completely).
e.g.: H+ + NO3-
Weak acids and weak bases are written as compounds (because they barely dissociate).
e.g.: HNO2
As an example, sodium bicarbonate (NaHCO3) would be written as Na+ and HCO3- because the salt will dissociate, but the bicarbonate will not dissociate (it's a weak acid).
When an acid and a base react, they form a neutral substance, often water and a salt.
First, let's examine the neutralization of a strong acid with a strong base.
{\displaystyle {\hbox{KOH}}+{\hbox{H}}^{+}+{\hbox{Cl}}^{-}}
Solid potassium hydroxide is added to an aqueous solution of hydrochloric acid. Notice how the solid is written as a compound, but the acid is written as ions because it dissociates.
{\displaystyle {\hbox{K}}^{+}+{\hbox{OH}}^{-}+{\hbox{H}}^{+}+{\hbox{Cl}}^{-}\to {\hbox{K}}^{+}+{\hbox{Cl}}^{-}+{\hbox{H}}_{2}{\hbox{O}}}
The hydrogen ions will react with hydroxide ions to form water.
{\displaystyle {\hbox{H}}^{+}+{\hbox{OH}}^{-}\to {\hbox{H}}_{2}{\hbox{O}}}
Ignoring spectator ions, this is the net ionic equation.
Whenever a strong acid neutralizes a strong base, the net ionic equation is always H+ + OH- → H2O.
Now, let's see some examples involving weak acids and weak bases.
{\displaystyle {\hbox{PO}}_{4}^{3-}+3{\hbox{H}}^{+}\to {\hbox{H}}_{3}{\hbox{PO}}_{4}}
Excess hydrochloric acid is added to a solution of sodium phosphate. Phosphoric acid is weak, so the phosphate ions will react with hydrogen ions. The result is a solution with some, but much less, hydrogen ions, so it is much closer to neutral than either of the original reactants.
{\displaystyle {\hbox{PO}}_{4}^{3-}+{\hbox{H}}^{+}\to {\hbox{H}}{\hbox{PO}}_{4}^{-2}}
Equimolar amounts of sodium phosphate and hydrochloric acid are mixed. Notice the difference between this reaction and the previous one.
{\displaystyle {\hbox{HCO}}_{3}^{-}+{\hbox{OH}}^{-}\to {\hbox{H}}_{2}{\hbox{O}}+{\hbox{CO}}_{3}^{-2}}
A strong base is added to a solution of calcium bicarbonate. (Bicarbonate is a weak acid.)
{\displaystyle {\hbox{HCO}}_{3}^{-}+{\hbox{H}}^{+}\to {\hbox{H}}_{2}{\hbox{O}}+{\hbox{CO}}_{2}}
A strong acid is added to a solution of calcium bicarbonate. Gas bubbles appear.
Many reactions result in the formation of gas bubbles or a solid precipitate that will make the solution cloudy. The last equation brings up an interesting application. Many rocks and minerals contain calcium carbonate or calcium bicarbonate. To identify these rocks, geologists can perform the "acid test". A drop of acid is applied, and the presence of gas bubbles indicates carbonate.
Many ions and compounds are amphoteric. They can react with (or behave as) acids and bases. Bicarbonate is a good example, as you can see by comparing the last two equations above.
Here are more examples of neutralization reactions.
{\displaystyle {\hbox{NH}}_{4}{\hbox{Cl}}+{\hbox{OH}}^{-}\to {\hbox{H}}_{2}{\hbox{O}}+{\hbox{NH}}_{3}+{\hbox{Cl}}^{-}}
Solid ammonium chloride crystals are dissolved into a solution of sodium hydroxide. The smell of ammonia is detected.
{\displaystyle {\hbox{NH}}_{3}+{\hbox{H}}^{+}\to {\hbox{NH}}_{4}^{+}}
Ammonia gas is bubbled through a solution of hydrochloric acid. This reaction is essentially the opposite of the previous. In that reaction, ammonium ions react with base to form ammonia gas. In this reaction, ammonia gas reacts with acid to form ammonium ions.
{\displaystyle {\hbox{NH}}_{3}+{\hbox{CH}}_{3}{\hbox{COOH}}\to {\hbox{NH}}_{4}^{+}+{\hbox{CH}}_{3}{\hbox{COO}}^{-}}
Ammonia (a weak base) reacts with acetic acid (also weak). The resulting solution is nearly neutral, but it will be slightly basic because ammonia is stronger than acetic acid.
{\displaystyle {\hbox{H}}_{2}{\hbox{S}}+2{\hbox{OH}}^{-}\to 2{\hbox{H}}_{2}{\hbox{O}}+{\hbox{S}}^{2-}}
Hydrogen sulfide gas is bubbled into a strong base.
{\displaystyle 2{\hbox{H}}^{+}+{\hbox{S}}^{2-}\to {\hbox{H}}_{2}{\hbox{S}}}
A strong acid is added to the above result, and hydrogen sulfide gas is released.
An anhydride is a substance that does not contain water. More specifically, it is a substance that reacts with water to form an acid or base. Anhydrides are usually in the form of a gas that dissolves into water and reacts to form an acid or base. They can also be solids that will react with water.
{\displaystyle {\hbox{N}}_{2}{\hbox{O}}_{5}+{\hbox{H}}_{2}{\hbox{O}}\to 2{\hbox{H}}^{+}+2{\hbox{NO}}_{3}^{-}}
Gaseous dinitrogen pentoxide is bubbled through water to form nitric acid.
{\displaystyle {\hbox{N}}_{2}{\hbox{O}}_{3}+{\hbox{H}}_{2}{\hbox{O}}\to 2{\hbox{HNO}}_{2}}
Dinitrogen trioxide is mixed with water to form nitrous acid.
The main difference between those two equations is the fact that nitrous acid is weak and thus does not dissociate, whereas nitric acid is strong and dissociates into ions.
Here are a few more examples of anhydride reactions.
{\displaystyle {\hbox{K}}_{2}{\hbox{O}}+{\hbox{H}}_{2}{\hbox{O}}\to 2{\hbox{K}}^{+}+2{\hbox{OH}}^{-}}
Solid potassium oxide is added to water to form a strong base.
{\displaystyle {\hbox{P}}_{2}{\hbox{O}}_{5}+3{\hbox{H}}_{2}{\hbox{O}}\to 2{\hbox{H}}_{3}{\hbox{PO}}_{4}}
Phosphorus(V) oxide powder is mixed into water to form a weak acid.
It is important to remember which acids are strong and which are weak. Review this if necessary.
Anhydrides can also react with acidic or basic solutions. In that case, you may find it helpful to first determine the anhydride's reaction with water, then determine that reaction with the acid or base.
For example, sulfur dioxide gas (acidic anhydride) is bubbled through a solution of calcium hydroxide (basic).
{\displaystyle {\hbox{SO}}_{2}+{\hbox{H}}_{2}{\hbox{O}}\to {\hbox{H}}_{2}{\hbox{SO}}_{3}}
First, determine the reaction of the anhydride with water.
{\displaystyle {\hbox{H}}_{2}{\hbox{SO}}_{3}+{\hbox{Ca}}({\hbox{OH}})_{2}\to {\hbox{CaSO}}_{3}+2{\hbox{H}}_{2}{\hbox{O}}}
Then, determine the reaction of the acid and base. This is a double replacement reaction.
{\displaystyle {\hbox{SO}}_{2}+{\hbox{H}}_{2}{\hbox{O}}+{\hbox{H}}_{2}{\hbox{SO}}_{3}+{\hbox{Ca}}({\hbox{OH}})_{2}\to {\hbox{H}}_{2}{\hbox{SO}}_{3}+{\hbox{CaSO}}_{3}+2{\hbox{H}}_{2}{\hbox{O}}}
Add the two reactions together.
{\displaystyle {\hbox{SO}}_{2}+{\hbox{Ca}}^{2-}+2{\hbox{OH}}^{-}\to {\hbox{CaSO}}_{3}+{\hbox{H}}_{2}{\hbox{O}}}
Cancel out spectators. Also, calcium hydroxide should be ionized (but calcium sulfite is a solid precipitate). This is the final net ionic equation.
{\displaystyle {\hbox{CaO}}+2{\hbox{H}}^{+}\to {\hbox{H}}_{2}{\hbox{O}}+{\hbox{Ca}}^{2+}}
Calcium oxide crystals (basic anhydrides) are added to a strong acid. Notice that it does not matter what the acid is (nitric, sulfuric, etc.) because it is strong and this reaction only requires the hydrogen ions. In other words, the anions of the strong acid are spectators and are not written.
{\displaystyle {\hbox{SO}}_{2}+{\hbox{OH}}^{-}\to {\hbox{HSO}}_{3}^{-}}
Excess sulfur dioxide gas is bubbled into a dilute solution of strong base. The base is the limiting reactant.
{\displaystyle {\hbox{SO}}_{2}+2{\hbox{OH}}^{-}\to {\hbox{H}}_{2}{\hbox{O}}+{\hbox{SO}}_{3}^{-}}
Sulfur dioxide gas is bubbled into an excess of basic solution.
Remember that water is involved in these reactions, but it is not written if it occurs on both sides of the equation.
Anhydrides can undergo neutralization reactions, even without the presence of water.
{\displaystyle {\hbox{CaO}}+{\hbox{CO}}_{2}\to {\hbox{CaCO}}_{3}}
Solid calcium oxide (basic anhydride) is exposed to dry ice gas (acidic anhydride). The resulting solid is a salt.
{\displaystyle {\hbox{CaO}}+{\hbox{SO}}_{3}\to {\hbox{CaSO}}_{4}}
Solid calcium oxide is exposed to a stream of sulfur trioxide gas. The resulting solid is a neutral salt.
A salt of a weak acid and strong base dissociates and reacts in water to form OH-. A salt of a strong acid and weak base dissociates and reacts in water to form H+. This process is called hydrolysis.
In this first example, aluminum nitrate is dissolved in water.
{\displaystyle {\hbox{Al}}({\hbox{NO}}_{3})_{3}+{\hbox{H}}_{2}{\hbox{O}}\to {\hbox{Al}}^{3+}+3{\hbox{NO}}_{3}^{-}+{\hbox{H}}_{2}{\hbox{O}}}
First, the salt dissociates in the water. It isn't necessary to write H2O in this reaction.
{\displaystyle {\hbox{Al}}^{3+}+{\hbox{H}}_{2}{\hbox{O}}\to {\hbox{Al}}({\hbox{OH}})^{2+}+{\hbox{H}}^{+}}
Now, at least one of the ions will react with water. You know that nitric acid is strong, so the nitrate ion will not take an H+ ion from water. Instead, the aluminum ion will react with water, releasing a hydrogen ion.
{\displaystyle {\hbox{Al}}({\hbox{NO}}_{3})_{3}+{\hbox{H}}_{2}{\hbox{O}}\to {\hbox{Al}}({\hbox{OH}})^{2+}+{\hbox{H}}^{+}+3{\hbox{NO}}_{3}^{-}}
This is the net ionic equation. The resulting solution is acidic.
The solution is acidic not because nitric acid is strong, but because aluminum is a weak base.
{\displaystyle {\hbox{NaNO}}_{2}+{\hbox{H}}_{2}{\hbox{O}}\to {\hbox{Na}}^{+}+{\hbox{NO}}_{2}^{-}+{\hbox{H}}_{2}{\hbox{O}}}
First, the salt dissociates. Again, the H2O need not be written.
{\displaystyle {\hbox{NO}}_{2}^{-}+{\hbox{H}}_{2}{\hbox{O}}\to {\hbox{HNO}}_{2}+{\hbox{OH}}^{-}}
Sodium ions will not react with water. Even if they did, they would form NaOH, which is a strong base, so it would immediately dissociate. Instead, the NO2 reacts with water. Being the conjugate of a weak acid, the nitrite ions will accept a proton from water to form nitrous acid (weak) and hydroxide ions (basic).
{\displaystyle {\hbox{NaNO}}_{2}+{\hbox{H}}_{2}{\hbox{O}}\to {\hbox{Na}}^{+}+{\hbox{HNO}}_{2}+{\hbox{OH}}^{-}}
This is the net ionic equation for the hydrolysis of sodium nitrite. The resulting solution is basic.
There is no hydrolysis reaction for neutral salts. Salts like NaCl and K2SO4 (and any other composed of the conjugates of both a strong acid and strong base) will not react with water. They create neutral solutions when dissolved.
Lewis Acids/BasesEdit
Lewis acids accept an electron pair. Lewis bases donate an electron pair. Together they react and bond to form an adduct.
Lewis acids/bases do not require the presence of water. However, H+ can be thought of as a Lewis acid because it accepts electron pairs. OH- can donate an electron pair, making it a Lewis base.
{\displaystyle {\hbox{BF}}_{3}+{\ddot {\hbox{N}}}{\hbox{H}}_{3}\to {\hbox{F}}_{3}{\hbox{B}}-{\hbox{NH}}_{3}}
Boron trifluoride (Lewis acid) is exposed to ammonia (a Lewis base, as shown by the electron pair over the N). The electron pair is shared between the nitrogen and boron, creating a bond. (The boron trifluoride is written backwards as F3B only to demonstrate the B-N bond. Its structure has not changed.)
{\displaystyle {\hbox{B}}_{2}{\hbox{H}}_{6}+2{\hbox{H}}^{-}\to 2{\hbox{BH}}_{4}^{-}}
Diborane accepts the two electrons from H- and forms a Lewis adduct.
Practice ProblemsEdit
Write the net ionic equations for the following. Make note of any solid precipitates or gas bubbles that would form.
Equimolar solutions of sodium biphosphate and potassium hydroxide are mixed.
Equimolar solutions of sodium biphosphate and hydrochloric acid are mixed.
Excess sulfur dioxide gas is bubbled into a dilute solution of sodium hydroxide. Acid is then added.
Aluminum chloride is dissolved into water.
Sodium fluoride is dissolved into water. Strong acid is then added.
Solid calcium oxide is exposed to a stream of sulfur trioxide gas. If the resulting compound is dissolved, will the solution be acidic, basic, or neutral?
Gaseous hydrogen chloride is bubbled into a solution of silver nitrate.
Ammonium chloride crystals are dissolved in water. Sodium hydroxide is then added.
Calcium hydroxide crystals are dissolved into a solution of sodium bicarbonate.
Phosphine gas is sprayed onto pebbles of aluminum trichloride. (Hint: these are Lewis acids/bases.)
Retrieved from "https://en.wikibooks.org/w/index.php?title=General_Chemistry/Reactions_of_Acids_and_Bases&oldid=3293127" |
Determine whether the following statement is true or false. A system of linear equations A
Determine whether the following statement is true or false. A system of linear equations Ax=b has the same solutions as the system of linear equations
Determine whether the following statement is true or false. A system of linear equations Ax=b has the same solutions as the system of linear equations Rx=c, where [R c] is the reduced row echelon form of [A b].
True, because we solve the system by finding the reduced row echelon form.
r\left(t\right)=<{t}^{2},\frac{2}{3}{t}^{3},t>
<4,-\frac{16}{3},-2>
\left(1,3,0\right),\left(-2,0,2\right),\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left(-1,3,-1\right)
The curvature given by the vector function r is
k\left(t\right)=\frac{|{r}^{\prime }\left(t\right)×r{}^{″}\left(t\right)|}{{|{r}^{\prime }\left(t\right)|}^{3}}
Use the formula to find the curvature of
r\left(t\right)=⟨\sqrt{15t},\text{ }{e}^{t},\text{ }{e}^{-t}⟩
\left(0,\text{ }1,\text{ }1\right)
Determine whether the set equipped with the given operations is a vector space.
For those that are not vector spaces identify the vector space axioms that fail.
The set of all pairs of real numbers of the form (x,0) with the standard operations on
{\mathbb{R}}^{2}
\circ
V is a vector space.
\circ
V is not a vector space, and Axiom 7,8, 9 fails to hold.
\circ
V is not a vector space, and Axioms 4 and 5 fail to hold.
\circ
\circ
V is not a vector space, and Axiom 10 fails to hold.
3/4x+2=9/10 No decimals
the failure of certain electronic device is suspected to increase linearly with its temp .Fit atleast regression line throgh the following data Temp 55 65 75 85 95 105Failure Rate 1.90 1.93 1.97 2.00 2.01. 2.01 |
Chi-Square tests are nonparametric tests. Discuss the following: What are the d
Chi-Square tests are nonparametric tests. Discuss the following: What are the differences between parametric tests and nonparametric tests? What are the requirements for Chi-Square tests? What are the limitations of Chi-Square tests?
Chi-Square tests are nonparametric tests. Discuss the following:
What are the differences between parametric tests and nonparametric tests?
What are the requirements for Chi-Square tests?
What are the limitations of Chi-Square tests?
Clelioo
Chi-square tests are non parametric tests
Difference between parametric and nonparametric tests:
Parametric tests make assumptions about the parameters of the population distribution from which the sample is drawn. This is often the assumption that the population data are normally distributed. Nonparametric tests don’t require that your data follow the normal distribution. They’re known as distribution-free tests and, as such, can be used for non-Normal variables.
Requirements for Chi-Square tests:
The sampling method should be simple random sampling. The variable under study should be categorical. The expected value of the number of sample observations in each level of the variable should be at least 5.
Limitations of Chi-Square tests:
All participants measured must be independent, means that an individual cannot fit in more than one category. The data must be frequency data The expected value of the number of sample observations in each level of the variable should be at least 5. Chi-square not be used if the sample size is less than 50
A chi-square value of 6.15 is calculated from data in a
4x5
contingency table. Assuming
\alpha =0.05
, identify the critical value.
A study is designed to investigate whether there is a difference in response to various treatments in patients with Rheumatoid arthritis. The outcome is the patient’s self-reported effect of treatment. The data are shown below.
\begin{array}{|ccccc|}\hline & \text{ Symptoms Worsened }& \text{ No Effect }& \text{ Symptoms Improved }& \text{ Total }\\ \text{ Treatment 1 }& 22& 14& 14& 50\\ \text{ Treatment 2 }& 14& 15& 21& 50\\ \text{ Treatment 3 }& 9& 12& 29& 50\\ \hline\end{array}
Is there a significant difference in effect of treatment? Run the test at a 5% level of significance.
The chi-square test is appropriate for use when:
you have one categorical independent variable and one numerical scaled dependent variable.
you have the dependent and independent variables need to be measured on an interval or ratio scale.
you have data from two numerical variables.
you have data from two categorial variables.
The Bitter Bottling Company has developed "Featherweight," the cola with fewer calories and less taste. To evaluate this new product, the marketing manager gives a taste test to a random sample of 300 people. Each person in the sample tastes Featherweight and 4 other diet cola brands. To avoid bias, the actual brand labels are replaced by the letters A, B, C, D, and E. The results of the sample are shown in the following table:
\begin{array}{|cc|}\hline \text{ Brand }& \text{ Number of Preferring Brand }\\ \text{ A }& 50\\ \text{ B }& 65\\ \text{ C }& 45\\ \text{ D }& 70\\ \text{ E }& 70\\ \hline\end{array}
Do the customers prefer a particular brand more than the others?
Use an alpha of 0.05.
1.What type of hypothesis testing should be used?
a. Chi-square Goodness of Fit Test
b. Chi-square Test on Independence
2. What are the null and alternative hypothesis?
{H}_{0}
: The market does not have preferences;
{H}_{1}
: The market does have preference
{H}_{0}
: The market does have preferences;
{H}_{1}
: The market does not have preference
3. What are the significance level and the type of test?
\alpha =0.05
, Right-tailed
\alpha =0.05
, left-tailed
\alpha =0.05
, two-tailed
A random sample of 175 students was selected from a large high school. They were each asked their current grade level and whether they had a driver's license. The responses are displayed in the table.
\begin{array}{cc}& \text{Grade Level}\\ \text{Driver's License?}& \begin{array}{|ccccc|}\hline & \text{ 9th }& \text{ 10th }& \text{ 11th }& \text{ 12th}\\ \text{ Yes }& 1& 5& 21& 48\\ \text{ No }& 28& 42& 20& 10\\ \hline\end{array}\text{ }\end{array}
The principal would like to know if these data provide convincing evidence of a relationship between grade level and having a driver's license in the population of all students in her school. What is the appropriate inference procedure?
1. chi-square test for goodness of fit because the data come from one random sample
2. chi-square test for homogeneity because the data come from independent random samples
3. chi-square test for association/independence because the data come from one random sample
4. chi-square test for association/independence because the data come from independent random samples
Table 1 gives the distribution of the number of acceptances of 100 students into 3 departments in College of Commerce and Business Administration. Test at the 5% level of significance that the distribution of acceptances is approximately binomial if the probability of a student's being accepted into college is 0.40.
What is the computed value of chi-square? Blank 1
What is the tabular value of chi-square? Blank 2
Choose a or b Blank 3
a. Accept the null hypothesis because the tabular value of chi-square is larger than the computed value of chi-square.
b. Accept the alternative hypothesis because the tabular value of chi-square is smaller than the computed value of chi-square.
\begin{array}{|cc|}\hline \text{ Number of Acceptances }& \text{ Number of Students }\\ 0& 25\\ 1& 34\\ 2& 31\\ 3& 10\\ \hline\end{array}
How to interpret the result according to the table below?
\begin{array}{cc}& \text{Chi-square Tests}\\ & \begin{array}{|cccc|}\hline & \text{ Value }& \text{ df }& \text{ Asymptotic Significance (2-sided) }\\ \text{ Pearson Chi-Square }& & 2& .034\\ \text{ Likelihood Ratio }& & 2& .033\\ \text{ Linear-by-Linear Association }& & 1& .380\\ \text{ N of Valid Cases }& & & \\ \hline\end{array}\text{ }\end{array}
a) Since 0.034<0.05, the Ho hypothesis is rejected
b) Since 0.033<0.05, the Ho hypothesis is rejected
c) Since 0.034<0.5, the Ho hypothesis is rejected
d) Since 0.380>0.05, the Ho hypothesis is accepted
e) Since 0.033<0.5, the Ho hypothesis is rejected |
Solve the given system of differential equations. [Dx+Dy+(D+1)z=0 Dx+y=e^{t} Dx+y-2z=50sin(2t)
Solve the given system of differential equations.[Dx+Dy+(D+1)z=0Dx+y=e^{t}Dx+y-2z=50sin(2t)
Solve the given system of differential equations.
Dx+Dy+\left(D+1\right)z=0
Dx+y={e}^{t}
Dx+y-2z=50\mathrm{sin}\left(2t\right)
The given system of differential equations is,
Dx+Dy+\left(D+1\right)z=0
Dx+y={e}^{t}
Dx+y-2z=50\mathrm{sin}\left(2t\right)
Dx+y
in the third equation, we get,
Dx+y-2z=50\mathrm{sin}\left(2t\right)
⇒{e}^{t}-2z=50\mathrm{sin}\left(2t\right)
⇒z=\frac{1}{2}{e}^{t}-25\mathrm{sin}\left(2t\right)
Now differentiating both sides with respect to t, we get,
Dz=\frac{1}{2}{e}^{t}-50\mathrm{cos}\left(2t\right)
⇒\left(D+1\right)z=Dz+z=\left(\frac{1}{2}{e}^{t}-50\mathrm{cos}\left(2t\right)\right)+\frac{1}{2}{e}^{t}-25\mathrm{sin}\left(2t\right)
⇒\left(D+1\right)z={e}^{t}-25\mathrm{sin}\left(2t\right)-50\mathrm{cos}\left(2t\right)
Dx={e}^{t}-y
\left(D+1\right)z
in first equation, we get,
{e}^{t}-y+Dy+{e}^{t}-25\mathrm{sin}2t-50\mathrm{cos}2t=0
⇒Dy-y=25\mathrm{sin}2t+50\mathrm{cos}2t-2{e}^{t}
So the integrating factor of this differential equation is,
I.F={e}^{\int \left(-1\right)dt}
={e}^{-t}
Hence the solution is,
y{e}^{-t}=\int \left(25\mathrm{sin}2t+50\mathrm{cos}2t-2{e}^{t}\right){e}^{-t}dt
=\int 25\mathrm{sin}2t{e}^{-t}dt+\int 50\mathrm{cos}2t{e}^{-t}dt-2\int {e}^{t}{e}^{-t}dt
=-10{e}^{-t}\mathrm{cos}2t-5{e}^{-t}\mathrm{sin}2t+20{e}^{-t}\mathrm{sin}2t-10{e}^{-t}\mathrm{cos}2t-2t+C
⇒y=-10\mathrm{cos}2t-5\mathrm{sin}2t+20\mathrm{sin}2t-10\mathrm{cos}2t-2t{e}^{t}+C{e}^{t}
⇒y=-20\mathrm{cos}2t+15\mathrm{sin}2t-2t{e}^{t}+C{e}^{t}
Now putting this value of
y
in the given second differential equation, we get:
Dx={e}^{t}-y
={e}^{t}-\left(-20\mathrm{cos}2t+15\mathrm{sin}2t-2t{e}^{t}+C{e}^{t}\right)
={e}^{t}+20\mathrm{cos}2t-15\mathrm{sin}2t+2t{e}^{t}-C{e}^{t}
x=\int \left({e}^{t}+20\mathrm{cos}2t-15\mathrm{sin}2t+2t{e}^{t}-C{e}^{t}\right)dt
{y}^{\prime }+y=x,\text{ }y\left(0\right)=1
xydy-{y}^{2}dx=\left(x+y{\right)}^{2}{e}^{\left(}-y/x\right)
Make use of vectors to re-write the following second order differential equation into a 1st order differential equation. (note: I do not need to solve it).
2.1{x}^{-2}\frac{{d}^{2}y}{{dx}^{2}}-3{e}^{2x}{y}^{4}\frac{dy}{dx}=-6y
y=4.2
\frac{dy}{dx}=-3.1
x=0.5
Im not sure were to begin any help is greatly appreciated.
Solve the first-order differential equation State which method you are using!
{x}^{2}{y}^{2}dx-\left({x}^{3}+1\right)dy=0
Explain why or why not determine whether the following statements are true and give an explanation or counterexample.
a. The differential equation y+2y
The marginal productive output of workers in a small manufacturing firm is given by
MP=8L-{L}^{2}+20
where L is the number of workers hired by the firm. Rewrite the equation above as a first order differential equation and find the general solution for the total productive output function P(L).
xdy/dx-y={x}^{2}\mathrm{ln}x |
Physics Study Guide/Fluids - Wikibooks, open books for an open world
Physics Study Guide/Fluids
BuoyancyEdit
Buoyancy is the force due to pressure differences on the top and bottom of an object under a fluid (gas or liquid).
Net force = buoyant force - force due to gravity on the object
Fluid flow is a complex phenomenon. An ideal fluid may be described as:
The fluid flow is steady i.e its velocity at each point is constant with time.
The fluid is incompressible. This condition applies well to liquids and in certain circumstances to gases.
The fluid flow is non-viscous. Internal friction is neglected. An object moving through this fluid does not experience a retarding force. We relax this condition in the discussion of Stokes' Law.
The fluid flow is irrotational. There is no angular momentum of the fluid about any point. A very small wheel placed at an arbitrary point in the fluid does not rotate about its center. Note that if turbulence is present, the wheel would most likely rotate and its flow is then not irrotational.
As the fluid moves through a pipe of varying cross-section and elevation, the pressure will change along the pipe. The Swiss physicist Daniel Bernoulli (1700-1782) first derived an expression relating the pressure to fluid speed and height. This result is a consequence of conservation of energy and applies to ideal fluids as described above.
Consider an ideal fluid flowing in a pipe of varying cross-section. A fluid in a section of length
{\displaystyle \Delta x_{1}}
moves to the section of length
{\displaystyle \Delta x_{2}}
{\displaystyle \Delta t}
. The relation given by Bernoulli is:
{\displaystyle P+{\tfrac {1}{2}}\rho v^{2}+\rho gh={\text{constant}}}
{\displaystyle P}
is pressure at cross-section
{\displaystyle h}
is height of cross-section
{\displaystyle \rho }
is density
{\displaystyle v}
is velocity of fluid at cross-section
In words, the Bernoulli relation may be stated as: As we move along a streamline the sum of the pressure (
{\displaystyle P}
), the kinetic energy per unit volume and the potential energy per unit volume remains a constant.
Retrieved from "https://en.wikibooks.org/w/index.php?title=Physics_Study_Guide/Fluids&oldid=3658949" |
Prime element - Wikipedia
Analogue of a prime number in a commutative ring
2 Connection with prime ideals
3 Irreducible elements
Connection with prime ideals[edit]
Irreducible elements[edit]
Main article: Irreducible element
The following are examples of prime elements in rings:
The integers ±2, ±3, ±5, ±7, ±11, ... in the ring of integers Z
the complex numbers (1 + i), 19, and (2 + 3i) in the ring of Gaussian integers Z[i]
the polynomials x2 − 2 and x2 + 1 in Z[x], the ring of polynomials over Z.
2 in the quotient ring Z/6Z
x2 + (x2 + x) is prime but not irreducible in the ring Q[x]/(x2 + x)
In the ring Z2 of pairs of integers, (1, 0) is prime but not irreducible (one has (1, 0)2 = (1, 0)).
In the ring of algebraic integers
{\displaystyle \mathbf {Z} [{\sqrt {-5}}],}
the element 3 is irreducible but not prime (as 3 divides
{\displaystyle 9=(2+{\sqrt {-5}})(2-{\sqrt {-5}})}
and 3 does not divide any factor on the right).
^ Hungerford 1980, Theorem III.3.4(i), as indicated in the remark below the theorem and the proof, the result holds in full generality.
^ Hungerford 1980, Theorem III.3.4(iii)
^ Hungerford 1980, Remark after Definition III.3.5
Section III.3 of Hungerford, Thomas W. (1980), Algebra, Graduate Texts in Mathematics, vol. 73 (Reprint of 1974 ed.), New York: Springer-Verlag, ISBN 978-0-387-90518-1, MR 0600654
Retrieved from "https://en.wikipedia.org/w/index.php?title=Prime_element&oldid=1077599386" |
The spinner is spun twice. Two possible outcomes art AC and CA. a. Create a sam
The spinner is spun twice. Two possible outcomes art AC and CA. a. Create a sample space that lists all of the equally likely outcomes. b. Refer to the sample space to find the probability of spinning A one or more times in two spins.
The spinner is spun twice. Two possible outcomes art AC and CA.
a. Create a sample space that lists all of the equally likely outcomes.
b. Refer to the sample space to find the probability of spinning A one or more times in two spins.
2k1enyvp
a.Any of the three letters can be an outcome in each spin so the sample space is:
AA,AB,AC,BA,BB,BC,CA,CB,CC
b.Check for outcomes that have one or two A’s. There are 5 outcomes that have one or two A’s out of the sample space of 9. So, the probability is:
\frac{5}{9}
Seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Find the probability that (a) 3 red, 2 blue, and 2 green balls are withdrawn; (b) at least 2 red balls are withdrawn; (c) all withdrawn balls are the same color; (d) either exactly 3 red balls or exactly 3 blue balls are withdrawn.
You are expecting two sets of exams- “easy” and “hard”. The probability of getting a hard exam is 0.80. The probability that the first question on the exam will be difficult is 0.90 if the set is “hard” and is 0.15 if the set is “easy.
a) What is the probability that the first question on your exam is marked as difficult.
b) What is the probability that your exam is hard given that the first question on the exam is marked as difficult?
A process yields 10% defective items. For testing purposes, 100 items are randomly selected from the process. A normal distribution may be approximated in order to get the probability of any event.
- What is the probability that the number of defectives exceeds 13?
- What is the probability that the number of defectives is less than 8? |
specification - Maple Help
Home : Support : Online Help : Mathematics : Discrete Mathematics : Combinatorics : Combinatorial Structures : specification
Grammar Specification of a Combinatorial Class
A combinatorial class is either an elementary class, or is built from simpler classes with "constructors." The elementary classes are Epsilon, which represents an object of size zero, and Atom, which represents an object of size one. The available constructors are listed in the following table.
object of size 0
object of size 1 (Z is a predefined atom)
Union(A,B,...)
disjointunion of the classes A,B, ...
Prod(A,B,...)
partitional product of the classes A, B, ...
all sets with repetitions whose elements are in A
all sets without repetitions whose elements are in A
Sequence(A)
all sequences of elements of A
Cycle(A)
all directed cycles of elements of A
Subst(A,B)
B-objects whose atoms are replaced by A-objects
For the constructors Set, PowerSet, Sequence, and Cycle, it is possible to add restrictions on the cardinality. For example,
\mathrm{Set}\left(A,1\le \mathrm{card}\right)
means all nonempty sets whose elements are in
A
\mathrm{Sequence}\left(A,\mathrm{card}\le 3\right)
means all sequences of at most three elements of
A
\mathrm{Cycle}\left(A,\mathrm{card}=5\right)
means all directed cycles of five elements from
A
None of these constructors accept an object that generates Epsilon as an argument. In some cases, no cardinality restriction or
k\le \mathrm{card}
with any constructor but PowerSet, such a grammar generates an infinite number of objects of size 0, whereas this system is for classes with a finite number of members of each size.
In the others, PowerSet or
\mathrm{card}\le k
\mathrm{card}=k
, while there are only a finite number of objects of size 0, the current system does not handle grammars with such Epsilon productions.
\mathrm{Subst}\left(A,B\right)
A
B
may produce objects of size 0.
A specification is a set of productions of the form
A=\mathrm{rhs}
A
is the name of the class being defined, and
\mathrm{rhs}
is an expression involving elementary classes, constructors, and other classes. For example, the following table lists specifications of some well-known combinatorial classes.
When the labeling type is 'labeled'.
A=\mathrm{Prod}\left(Z,\mathrm{Set}\left(A\right)\right)
nonplane trees
B=\mathrm{Union}\left(Z,\mathrm{Prod}\left(B,B\right)\right)
plane binary trees
C=\mathrm{Prod}\left(Z,\mathrm{Sequence}\left(C\right)\right)
plane general trees
\mathrm{D}=\mathrm{Set}\left(\mathrm{Cycle}\left(Z\right)\right)
F=\mathrm{Set}\left(\mathrm{Set}\left(Z,1\le \mathrm{card}\right)\right)
G=\mathrm{Union}\left(Z,\mathrm{Prod}\left(Z,\mathrm{Set}\left(G,\mathrm{card}=3\right)\right)\right)
nonplane ternary trees
H=\mathrm{Union}\left(Z,\mathrm{Set}\left(H,2\le \mathrm{card}\right)\right)
L=\mathrm{Set}\left(\mathrm{Set}\left(\mathrm{Set}\left(Z,1\le \mathrm{card}\right),1\le \mathrm{card}\right)\right)
3-balanced hierarchies
M=\mathrm{Sequence}\left(\mathrm{Set}\left(Z,1\le \mathrm{card}\right)\right)
N=\mathrm{Set}\left(\mathrm{Cycle}\left(A\right)\right),A=\mathrm{Prod}\left(Z,\mathrm{Set}\left(A\right)\right)
When the labeling type is 'unlabeled'.
A=\mathrm{Set}\left(\mathrm{Sequence}\left(Z,1\le \mathrm{card}\right)\right)
B=\mathrm{Sequence}\left(\mathrm{Union}\left(Z,Z\right)\right)
C=\mathrm{Cycle}\left(\mathrm{Set}\left(Z,1\le \mathrm{card}\right)\right)
\mathrm{D}=\mathrm{Prod}\left(Z,\mathrm{Set}\left(\mathrm{D}\right)\right)
rooted unlabeled trees
F=\mathrm{Union}\left(Z,\mathrm{Set}\left(F,\mathrm{card}=2\right)\right)
nonplane binary trees
G=\mathrm{Union}\left(Z,\mathrm{Set}\left(G,\mathrm{card}=3\right)\right)
H=\mathrm{Union}\left(Z,\mathrm{Set}\left(H,2\le \mathrm{card}\right)\right)
unlabeled hierarchies
J=\mathrm{Set}\left(\mathrm{Cycle}\left(\mathrm{D}\right)\right),\mathrm{D}=\mathrm{Prod}\left(Z,\mathrm{Set}\left(\mathrm{D}\right)\right)
random mappings patterns
K=\mathrm{Union}\left(Z,\mathrm{Subst}\left(\mathrm{Union}\left(\mathrm{Prod}\left(Z,Z\right),\mathrm{Prod}\left(Z,Z,Z\right)\right),K\right)\right)
L=\mathrm{PowerSet}\left(\mathrm{Sequence}\left(Z,1\le \mathrm{card}\right)\right)
integer partitions without
It is possible to use Epsilon as a way of marking certain objects without affecting their size.
There are also predefined structures, for example, combinations, built into the system. For more information, see combstruct[structures].
\mathrm{with}\left(\mathrm{combstruct}\right):
Generate words on two letters of the form
C=\left(aCb\right)^``
\mathrm{sys}≔{C=\mathrm{Sequence}\left(\mathrm{Prod}\left(a,C,b\right)\right),a=\mathrm{Atom},b=\mathrm{Atom}}:
\mathrm{word}≔\mathrm{draw}\left([C,\mathrm{sys}],\mathrm{size}=6\right)
\textcolor[rgb]{0,0,1}{\mathrm{word}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{Sequence}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{\mathrm{Prod}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Sequence}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{\mathrm{Prod}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Ε}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{b}\right)\right)\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{b}\right)\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Prod}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Ε}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{b}\right)\right)
If you do not need the derivation structure, remove it.
\mathrm{eval}\left(\mathrm{subs}\left(\mathrm{`=`}\left(\mathrm{Prod},\left(\right)↦\mathrm{args}\right),\mathrm{`=`}\left(\mathrm{Sequence},\left(\right)↦\mathrm{args}\right),\mathrm{Ε}=\mathrm{NULL},\mathrm{word}\right)\right)
\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{b}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{b}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{b}
To model series and parallel circuits of resistors, a parallel circuit is made up of two or more resistors in series, and a series circuit is made up of two or more parallel circuits.
\mathrm{circuit}≔{C=\mathrm{Union}\left(P,S,R\right),P=\mathrm{Set}\left(\mathrm{Union}\left(S,R\right),2\le \mathrm{card}\right),R=\mathrm{Atom},S=\mathrm{Set}\left(\mathrm{Union}\left(P,R\right),2\le \mathrm{card}\right)}:
\mathrm{draw}\left([C,\mathrm{circuit},\mathrm{labeled}],\mathrm{size}=5\right)
\textcolor[rgb]{0,0,1}{\mathrm{Set}}\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{R}}_{\textcolor[rgb]{0,0,1}{4}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Set}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{\mathrm{Set}}\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{R}}_{\textcolor[rgb]{0,0,1}{5}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Set}}\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{R}}_{\textcolor[rgb]{0,0,1}{1}}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{R}}_{\textcolor[rgb]{0,0,1}{2}}\right)\right)\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{R}}_{\textcolor[rgb]{0,0,1}{3}}\right)\right)
However, you cannot tell from this answer which resistors are in series and which are in parallel. In this case, you obtain more information about the derivation by using Epsilon tags.
\mathrm{circuit2}≔{C=\mathrm{Union}\left(P,S,R\right),P=\mathrm{Prod}\left(\mathrm{par},\mathrm{Set}\left(\mathrm{Union}\left(S,R\right),2\le \mathrm{card}\right)\right),R=\mathrm{Atom},S=\mathrm{Prod}\left(\mathrm{ser},\mathrm{Set}\left(\mathrm{Union}\left(P,R\right),2\le \mathrm{card}\right)\right),\mathrm{par}=\mathrm{Ε},\mathrm{ser}=\mathrm{Ε}}:
\mathrm{draw}\left([C,\mathrm{circuit2},\mathrm{labeled}],\mathrm{size}=5\right)
\textcolor[rgb]{0,0,1}{\mathrm{Prod}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{\mathrm{ser}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Set}}\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{R}}_{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Prod}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{\mathrm{par}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Set}}\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{R}}_{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{R}}_{\textcolor[rgb]{0,0,1}{4}}\right)\right)\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Prod}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{\mathrm{par}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Set}}\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{R}}_{\textcolor[rgb]{0,0,1}{5}}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{R}}_{\textcolor[rgb]{0,0,1}{1}}\right)\right)\right)\right)
Since Epsilon has size 0, taking the product with Epsilon does not change the number of objects of each size. |
Film frame - Wikipedia
(Redirected from Animation frame)
One of the many still images which compose the complete moving picture
"still frame" redirects here. For the song, see Still Frame.
Find sources: "Film frame" – news · newspapers · books · scholar · JSTOR (May 2013) (Learn how and when to remove this template message)
2 Physical film frames
3 Video frames
3.1 Line and resolution
3.3 Moving picture
3.4 Example (System B)
5 Fourth wall usage
When the moving picture is displayed, each frame is flashed on a screen for a short time (nowadays, usually 1/24, 1/25 or 1/30 of a second) and then immediately replaced by the next one. Persistence of vision blends the frames together, producing the illusion of a moving image.
The frame is also sometimes used as a unit of time, so that a momentary event might be said to last six frames, the actual duration of which depends on the frame rate of the system, which varies according to the video or film standard in use. In North America and Japan, 30 frames per second (fps) is the broadcast standard, with 24 frames/s now common in production for high-definition video shot to look like film. In much of the rest of the world, 25 frames/s is standard.
In systems historically based on NTSC standards, for reasons originally related to the Chromilog NTSC TV systems, the exact frame rate is actually (3579545 / 227.5) / 525 = 29.97002616 fps.[a] This leads to many synchronization problems which are unknown outside the NTSC world, and also brings about hacks such as drop-frame timecode.
In film projection, 24 fps is the normal, except in some special venue systems, such as IMAX, Showscan and Iwerks 70, where 30, 48 or even 60 frame/s have been used. Silent films and 8 mm amateur movies used 16 or 18 frame/s.
Physical film frames[edit]
In a strip of movie film, individual frames are separated by frame lines. Normally, 24 frames are needed for one second of film. In ordinary filming, the frames are photographed automatically, one after the other, in a movie camera. In special effects or animation filming, the frames are often shot one at a time.
The most common film format, 35 mm, has a frame size of 36 by 24 mm when used in a still 35 mm camera where the film moves horizontally, but the frame size varies when used for motion picture where the film moves vertically (with the exception of VistaVision and Technirama where the film moves horizontally). Using a 4-perf pulldown, there are exactly 16 frames in one foot of 35 mm film, leading to film frames sometimes being counted in terms of "feet and frames". The maximum frame size is 18 by 24 mm, (silent/full aperture), but this is significantly reduced by the application of sound track(s). A system called KeyKode is often used to identify specific physical film frames in a production.
Video frames[edit]
Historically, video frames were represented as analog waveforms in which varying voltages represented the intensity of light in an analog raster scan across the screen. Analog blanking intervals separated video frames in the same way that frame lines did in film. For historical reasons, most systems used an interlaced scan system in which the frame typically consisted of two video fields sampled over two slightly different periods of time. This meant that a single video frame was usually not a good still picture of the scene, unless the scene being shot was completely still.
With the dominance of digital technology, modern video systems now represent the video frame as a rectangular raster of pixels, either in an RGB color space or a color space such as YCbCr, and the analog waveform is typically found nowhere other than in legacy I/O[clarification needed] devices.
Standards for the digital video frame raster include Rec. 601 for standard-definition television and Rec. 709 for high-definition television.
Video frames are typically identified using SMPTE time code.
Line and resolution[edit]
The frame is composed of picture elements just like a chess board. Each horizontal set of picture elements is known as a line. The picture elements in a line are transmitted as sine signals where a pair of dots, one dark and one light can be represented by a single sine. The product of the number of lines and the number of maximum sine signals per line is known as the total resolution of the frame. The higher the resolution the more faithful the displayed image is to the original image. But higher resolution introduces technical problems and extra cost. So a compromise should be reached in system designs both for satisfactory image quality and affordable price.
Viewing distance[edit]
The key parameter to determine the lowest resolution still satisfactory to viewers is the viewing distance, i.e. the distance between the eyes and the monitor. The total resolution is inversely proportional to the square of the distance. If d is the distance, r is the required minimum resolution and k is the proportionality constant which depends on the size of the monitor;
{\displaystyle r=k^{2}\cdot {\frac {1}{d^{2}}}}
Since the number of lines is approximately proportional to the resolution per line, the above relation can also be written as
{\displaystyle n=k\cdot {\frac {1}{d}}}
where n is the number of lines. That means that the required resolution is proportional to the height of the monitor and inversely proportional to the viewing distance.
Moving picture[edit]
In moving picture (TV) the number of frames scanned per second is known as the frame rate. The higher the frame rate, the better the sense of motion. But again, increasing the frame rate introduces technical difficulties. So the frame rate is fixed at 25 (System B/G) or 29.97 (System M). To increase the sense of motion it is customary to scan the very same frame in two consecutive phases. In each phase only half of the lines are scanned; only the lines with odd numbers in the first phase and only the lines with even numbers in the second phase. Each scan is known as a field. So the field rate is two times the frame rate.
Example (System B)[edit]
Main article: CCIR System B
In system B the number of lines is 625 and the frame rate is 25. The maximum video bandwidth is 5 MHz.[1] The maximum number of sine signals the system is theorically capable of transmitting is given as follows:
The system is able to transmit 5 000 000 sine signals in a second. Since the frame rate is 25, the maximum number of sine signals per frame is 200 000. Dividing this number by the number of lines gives the maximum number of sine signals in a line which is 320. (Actually about 19% of each line is devoted to auxiliary services. So the number of maximum useful sine signals is about 260.)
Still frame[edit]
A badly chosen still can give a misleading impression.
This still may imply that the content concerns the letter W (thumbtime=1).
A better preview, which implies an interview, for the same video (thumbtime=58).
A still frame is a single static image taken from a film or video, which are kinetic (moving) images. Still frames are also called freeze frame, video prompt, preview or misleadingly thumbnail, keyframe, poster frame,[2][3] or screen shot/grab/capture/dump. Freeze frames are widely used on video platforms and in video galleries, to show viewers a preview or a teaser. Many video platforms have a standard to display a frame from mid-time of the video. Some platforms offer the option to choose a different frame individually.[4][5]
Video and film artists sometimes use still frames within the video/film to achieve special effects, like freeze-frame shots or still motion.[6]
For criminal investigations it has become a frequent use to publish still frames from surveillance videos in order to identify suspect persons and to find more witnesses.[7] Videos of the J.F. Kennedy assassination have been often discussed frame-by-frame for various interpretations.[8] For medical diagnostics it is very useful to watch still frames of Magnetic resonance imaging videos.[9]
Fourth wall usage[edit]
Some humor in animation is based on the fourth wall aspect of the film frame itself, with some animation showing characters leaving what is assumed to be the edge of the film or the film malfunctioning. This latter one is used often in films as well. This hearkens back to some early cartoons, where characters were aware that they were in a cartoon, specifically that they could look at the credits and be aware of something that isn't part of the story as presented. These jokes include:
Split frames – Where the fourth wall is broken by two frames, the lower half of the previous frame and the upper part of the next frame, showing at once, usually showing frame lines, with jokes involving them including a character crossing the frame itself.
Film break – A famous form of the joke, where the film either snaps or is deliberately broken, with often the fourth wall coming into play during this period when, rightfully, there should be nothing on screen.
Gate hair – A famous form of joke where the animator intentionally places fake "gate hairs" within the frame, which one of the animated characters plucks and removes from the frame.
Editorial marks – Where those marks which an editor would normally employ on a "work print" to indicate the intended presence of a fade or a dissolve or a "wipe" to the SFX department are animated, and the film follows suit, or doesn't, depending upon the intended effect.
Cue marks – Where those marks, usually circular for non-Technicolor titles and "serrated" for Technicolor titles to indicate a reel changeover are animated for a humorous effect. This could also be employed for the famous "false ending" effect, employed even today in popular songs. For Inglourious Basterds, the cue marks for the reel changes of the Nation's Pride pseudo-documentary employed exceptionally large scribed circles with a large "X" scribed within it—marks which would never be utilized in actual editorial practice (motor and changeover cue marks are supposed to be clearly visible to the projectionist, but not obvious to the audience).
Exiting the frame – This joke, an extension of the split frames joke, has characters depart from the sides of the frame, sometimes finding themselves falling out of the cartoon entirely.
Freeze frame (disambiguation)
Freeze frame television
^ In actual practice, the master oscillator is 14.31818 MHz, which is divided by 4 to give the 3.579545 MHz color "burst" frequency, which is further divided by 455 to give the 31468.5275 KHz "equalizing pulse" frequency, this is further divided by 2 toorizontal line rate), the "equalizing pulse" frequency is divided by 525 to give the 59.9401 Hz "vertical drive" frequency, and this is further divided by 2 to give the 29.9700 vertical frame rate. "Equalizing pulses" perform two essential functions: 1) their use during the vertical retrace interval allows for the vertical synch to be more effectively separated from the horizontal synch, as these, along with the video itself, are an example of "in band" signaling, and 2) by alternately including or excluding one "equalizing pulse", the required half-line offset necessary for interlaced video may be accommodated.
^ Reference Data for Radio Engineers, ITT Howard W.Sams Co., New York, 1977, section 30
^ Microsoft: Add a poster frame to your video, retrieved 29 June 2014
^ Indezine: Poster Frames for Videos in PowerPoint 2010 for Windows, retrieved 29 June 2014
^ Vimeo: How do I change the thumbnail of my video?, retrieved 29 June 2014
^ MyVideo: Editing my video, retrieved 29 June 2014
^ Willie Witte: SCREENGRAB, retrieved 29 June 2014
^ Wistv: Assaults, shooting in Five Points under investigation, retrieved 29 June 2014
^ "Archived copy". motherboard.vice.com. Archived from the original on 30 November 2012. Retrieved 11 January 2022. {{cite web}}: CS1 maint: archived copy as title (link)
^ Lister Hill National Center for Biomedical Communications: A classic diagnosis with a new ‘spin’, retrieved 29 June 2014
The image areas on a 35 mm film frame
Retrieved from "https://en.wikipedia.org/w/index.php?title=Film_frame&oldid=1086551751" |
Answer the given question with proper steps - Maths - Statistics - 12548519 | Meritnation.com
To solve this question , we first need to create a cumulative frequency table.
Salary No.of persons Cumulative frequency
Now, the formula for finding the median is:
Median = {x}_{l} + \frac{\frac{N}{2}-{F}_{l}}{{f}_{m}} × c\phantom{\rule{0ex}{0ex}}{x}_{l} = Lower class boundary of the median class\phantom{\rule{0ex}{0ex}}N = Total frequency\phantom{\rule{0ex}{0ex}}{F}_{l} = Less than type cumulative frequency corresponding to the {x}_{l}\phantom{\rule{0ex}{0ex}}{f}_{m} =frequency of the median class\phantom{\rule{0ex}{0ex}}c = Class width\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}From the given data, we first need to select the median class.\phantom{\rule{0ex}{0ex}}N = 280\phantom{\rule{0ex}{0ex}}\frac{N}{2} = \frac{280}{2} = 140\phantom{\rule{0ex}{0ex}}The cumulative frequency more than 140 is 182, corresponding to class 10-15.\phantom{\rule{0ex}{0ex}}Thus, the salary range 10-15 is the median class.\phantom{\rule{0ex}{0ex}}Thus,\phantom{\rule{0ex}{0ex}}{x}_{l} = 10\phantom{\rule{0ex}{0ex}}N = 280\phantom{\rule{0ex}{0ex}}{F}_{l} = less than type c.f corresponding to lower boundary 10 = 49\phantom{\rule{0ex}{0ex}}{f}_{m} = 133\phantom{\rule{0ex}{0ex}}c = 5\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}So, Median = {x}_{l} + \frac{\frac{N}{2}-{F}_{l}}{{f}_{m}} × c = 10 + \frac{140 - 49}{133}×5 = 13.42
Thus, the median salary is 13.42 thousand. |
(Redirected from Rewriting logic)
Replacing subterm in a formula with another term
Example cases[edit]
{\displaystyle \neg \neg A\to A}
{\displaystyle \neg (A\land B)\to \neg A\lor \neg B}
{\displaystyle \neg (A\lor B)\to \neg A\land \neg B}
{\displaystyle (A\land B)\lor C\to (A\lor C)\land (B\lor C)}
{\displaystyle A\lor (B\land C)\to (A\lor B)\land (A\lor C),}
{\displaystyle \to }
{\displaystyle {\begin{aligned}A+0&\to A&{\textrm {(1)}},\\A+S(B)&\to S(A+B)&{\textrm {(2)}},\\A\cdot 0&\to 0&{\textrm {(3)}},\\A\cdot S(B)&\to A+(A\cdot B)&{\textrm {(4)}}.\end{aligned}}}
{\displaystyle S(S(0))+S(S(0))}
{\displaystyle \;\;{\stackrel {(2)}{\to }}\;\;}
{\displaystyle S(\;S(S(0))+S(0)\;)}
{\displaystyle \;\;{\stackrel {(2)}{\to }}\;\;}
{\displaystyle S(S(\;S(S(0))+0\;))}
{\displaystyle \;\;{\stackrel {(1)}{\to }}\;\;}
{\displaystyle S(S(S(S(0)))),}
{\displaystyle S(S(0))\cdot S(S(0))}
{\displaystyle \;\;{\stackrel {(4)}{\to }}\;\;}
{\displaystyle S(S(0))+S(S(0))\cdot S(0)}
{\displaystyle \;\;{\stackrel {(4)}{\to }}\;\;}
{\displaystyle S(S(0))+S(S(0))+S(S(0))\cdot 0}
{\displaystyle \;\;{\stackrel {(3)}{\to }}\;\;}
{\displaystyle S(S(0))+S(S(0))+0}
{\displaystyle \;\;{\stackrel {(1)}{\to }}\;\;}
{\displaystyle S(S(0))+S(S(0))}
{\displaystyle \;\;{\stackrel {\textrm {s.a.}}{\to }}\;\;}
{\displaystyle S(S(S(S(0)))),}
{\displaystyle {\rm {A\rightarrow X}}}
{\displaystyle {\rm {S\rightarrow NP\ VP}}}
Abstract rewriting systems[edit]
{\displaystyle {\overset {*}{\rightarrow }}}
{\displaystyle \rightarrow }
{\displaystyle \leftrightarrow }
{\displaystyle \rightarrow }
{\displaystyle {\overset {*}{\leftrightarrow }}}
{\displaystyle \rightarrow }
{\displaystyle x{\overset {*}{\leftrightarrow }}y}
{\displaystyle x\rightarrow y}
{\displaystyle x{\stackrel {*}{\rightarrow }}y}
{\displaystyle x{\downarrow }}
{\displaystyle x\downarrow y}
{\displaystyle x{\overset {*}{\rightarrow }}z{\overset {*}{\leftarrow }}y}
{\displaystyle x{\overset {*}{\leftrightarrow }}y}
{\displaystyle x\downarrow y}
{\displaystyle x{\overset {*}{\leftarrow }}w{\overset {*}{\rightarrow }}y}
{\displaystyle x\downarrow y}
{\displaystyle x\leftarrow w\rightarrow y}
{\displaystyle x{\mathbin {\downarrow }}y}
{\displaystyle x_{0}\rightarrow x_{1}\rightarrow x_{2}\rightarrow \cdots }
String rewriting systems[edit]
{\displaystyle R}
{\displaystyle (\Sigma ,R)}
{\displaystyle \Sigma }
{\displaystyle R}
{\displaystyle {\underset {R}{\rightarrow }}}
{\displaystyle R}
{\displaystyle \Sigma ^{*}}
{\displaystyle s,t\in \Sigma ^{*}}
{\displaystyle s{\underset {R}{\rightarrow }}t}
{\displaystyle x,y,u,v\in \Sigma ^{*}}
{\displaystyle s=xuy}
{\displaystyle t=xvy}
{\displaystyle uRv}
{\displaystyle {\underset {R}{\rightarrow }}}
{\displaystyle \Sigma ^{*}}
{\displaystyle (\Sigma ^{*},{\underset {R}{\rightarrow }})}
{\displaystyle R}
{\displaystyle {\underset {R}{\rightarrow }}}
{\displaystyle R}
{\displaystyle {\overset {*}{\underset {R}{\rightarrow }}}}
{\displaystyle x{\overset {*}{\underset {R}{\rightarrow }}}y}
{\displaystyle uxv{\overset {*}{\underset {R}{\rightarrow }}}uyv}
{\displaystyle x,y,u,v\in \Sigma ^{*}}
{\displaystyle {\underset {R}{\rightarrow }}}
{\displaystyle {\overset {*}{\underset {R}{\leftrightarrow }}}}
{\displaystyle {\overset {*}{\underset {R}{\leftrightarrow }}}}
{\displaystyle R}
{\displaystyle R}
{\displaystyle {\overset {*}{\underset {R}{\rightarrow }}}}
{\displaystyle {\overset {*}{\underset {R}{\leftrightarrow }}}}
{\displaystyle {\overset {*}{\underset {R}{\leftrightarrow }}}}
{\displaystyle {\mathcal {M}}_{R}=\Sigma ^{*}/{\overset {*}{\underset {R}{\leftrightarrow }}}}
{\displaystyle \Sigma ^{*}}
{\displaystyle {\mathcal {M}}}
{\displaystyle {\mathcal {M}}_{R}}
{\displaystyle (\Sigma ,R)}
{\displaystyle {\mathcal {M}}}
{\displaystyle \{a,b\}}
{\displaystyle \{ab\rightarrow \varepsilon ,ba\rightarrow \varepsilon \}}
{\displaystyle \varepsilon }
{\displaystyle \{ab\rightarrow \varepsilon \}}
{\displaystyle (\Sigma ,R)}
Term rewriting systems[edit]
{\displaystyle l\longrightarrow r}
{\displaystyle p}
{\displaystyle \sigma }
{\displaystyle x*(y*z)}
{\displaystyle {\frac {a*((a+1)*(a+2))}{1*(2*3)}}}
{\displaystyle (\vee )}
{\displaystyle (\wedge )}
{\displaystyle (\neg )}
Formal definition [edit]
{\displaystyle l\rightarrow r}
{\displaystyle l\rightarrow r}
{\displaystyle \sigma }
{\displaystyle s}
{\displaystyle \sigma }
{\displaystyle r}
{\displaystyle \sigma }
{\displaystyle s}
{\displaystyle t}
{\displaystyle R}
{\displaystyle s\rightarrow _{R}t}
{\displaystyle s{\underset {R}{\rightarrow }}t}
{\displaystyle s{\overset {R}{\rightarrow }}t}
{\displaystyle t_{1}}
{\displaystyle t_{n}}
{\displaystyle t_{1}{\underset {R}{\rightarrow }}t_{2}{\underset {R}{\rightarrow }}\cdots {\underset {R}{\rightarrow }}t_{n}}
{\displaystyle t_{1}}
{\displaystyle t_{n}}
{\displaystyle t_{1}{\overset {+}{\underset {R}{\rightarrow }}}t_{n}}
{\displaystyle {\overset {+}{\underset {R}{\rightarrow }}}}
{\displaystyle {\underset {R}{\rightarrow }}}
{\displaystyle {\overset {*}{\underset {R}{\rightarrow }}}}
{\displaystyle {\underset {R}{\rightarrow }}}
{\displaystyle s{\overset {*}{\underset {R}{\rightarrow }}}t}
{\displaystyle s=t}
{\displaystyle s{\overset {*}{\underset {R}{\rightarrow }}}t}
{\displaystyle R}
{\displaystyle {\underset {R}{\rightarrow }}}
{\displaystyle x*(y*z)\rightarrow (x*y)*z}
{\displaystyle *}
{\displaystyle {\frac {a*((a+1)*(a+2))}{1*(2*3)}}}
{\displaystyle \{x\mapsto a,\;y\mapsto a+1,\;z\mapsto a+2\}}
{\displaystyle (a*(a+1))*(a+2)}
{\displaystyle {\frac {(a*(a+1))*(a+2)}{1*(2*3)}}}
{\displaystyle *}
{\displaystyle {\frac {a*((a+1)*(a+2))}{1*(2*3)}}}
{\displaystyle {\frac {a*((a+1)*(a+2))}{(1*2)*3}}}
{\displaystyle {\begin{aligned}f(x,x)&\rightarrow g(x),\\f(x,g(x))&\rightarrow b,\\h(c,x)&\rightarrow f(h(x,c),h(x,x)).\\\end{aligned}}}
{\displaystyle f(0,1,x)\rightarrow f(x,x,x)}
{\displaystyle g(x,y)\rightarrow x,}
{\displaystyle g(x,y)\rightarrow y.}
{\displaystyle {\begin{aligned}&f(g(0,1),g(0,1),g(0,1))\\\rightarrow &f(0,g(0,1),g(0,1))\\\rightarrow &f(0,1,g(0,1))\\\rightarrow &f(g(0,1),g(0,1),g(0,1))\\\rightarrow &\cdots \end{aligned}}}
{\displaystyle R_{1}}
{\displaystyle R_{2}}
{\displaystyle R_{1}}
{\displaystyle R_{2}}
{\displaystyle R_{1}}
{\displaystyle R_{2}}
Higher-order rewriting systems[edit]
Graph rewriting systems[edit]
Trace rewriting systems[edit]
{\displaystyle x*(y*z)}
{\displaystyle a*((a+1)*(a+2))} |
B Expression syntax
B.1 Details of expressions
B.2 Mathematical functions provided
B.3 Rules for expressions
B.4 Operator precedence
This appendix discusses the syntax permitted for expressions passed to the expression parser. Expressions are input to CAT as argument EXPR of subroutine CAT_EIDNT (see Section 7).
Expressions have an algebraic format, and comprise: columns, vector column elements, parameters and constants linked by arithmetic operators and mathematical functions. For example, suppose that a catalogue contained scalar columns called x, y and z and parameters called p and q. Some valid expressions are:
\begin{array}{c}\text{x}\hfill \\ \text{p}\hfill \\ \text{x + p}\hfill \\ \text{(x + y + 2) / (p + q)}\hfill \\ \text{(2.0*x + y + 3.75*p) + 13.0) / (z + 1.8*q)}\hfill \end{array}
Remember that in CAT column and parameter names are not case-sensitive. Thus the following column or parameter names would all be considered equivalent:
HD_NUMBER
Vector column elements occur in expressions with their usual syntax: the name of the base column followed by the element number enclosed in square brackets. The first element in a vector is numbered one. For example, an expression to add two to the fourth element of vector FLUX would be ‘FLUX[4] + 2.0’.
brackets (‘(’ and ‘)’) may be used as required.
Table 16 lists the mathematical functions which are provided. The letters denote data types permitted, coded as follows: B = BYTE, H = half INTEGER, I = INTEGER, R = REAL, D = DOUBLE PRECISION, C = CHARACTER, L = LOGICAL. The appearance of N as an argument means that any numeric type (BHIRD) is permitted, as a result it means that the type is the widest type of any of the arguments. R/D means that the result is REAL unless one or more arguments is of DOUBLE PRECISION type in which case D is the result.
B = BYTE(N) convert to BYTE data type
H = HALF(N) convert to INTEGER*2 data type
I = INT(N) convert to INTEGER data type
R = REAL(N) convert to REAL data type
D = DBLE(N) convert to DOUBLE PRECISION data type
I = NINT(N) convert to nearest INTEGER
N = MIN(N,N) the function must have precisely two arguments
N = MAX(N,N) the function must have precisely two arguments
N = MOD(N,N) remainder
N = ABS(N) absolute value
R/D = SQRT(N) square root
R/D = LOG(N) natural logarithm
R/D = LOG10(N) logarithm to the base 10
R/D = EXP(N) exponential
R/D = SIN(N) sine; argument in radians
R/D = COS(N) cosine; argument in radians
R/D = TAN(N) tangent; argument in radians
R/D = ASIN(N) arc-sine; result in radians
R/D = ACOS(N) arc-cosine; result in radians
R/D = ATAN(N) arc-tangent; result in radians
R/D = ATAN2(N,N) arc-tangent (two arguments) result in radians
I = IAND(I,I) bitwise logical AND
I = IOR(I,I) bitwise logical OR
I = XOR(I,I) bitwise logical exclusive OR
R/D = DTOR(N) degrees to radians conversion
R/D = RTOD(N) radians to degrees conversion
C = UPCASE(C) convert character string to upper case
C = STRIP(C) leading and trailing spaces are removed
C = SUBSTR(C,N,N) returns characters from positions argument 2
to argument 3 inclusive, with the positions
L = NULL(*) .TRUE. if argument is NULL
D = HMSRAD(N,N,N) converts 3 arguments hours, minutes and
D = DMSRAD(C,N,N,N) first argument is the sign (‘
+
’ or ‘-’ ),
converts degrees, minutes and seconds to radians
D = GREAT(N,N,N,N) great circle distance between two spherical
coordinates. All the input arguments and the
return argument are in radians. The input arguments
are in the order: ,
D = PANGLE(N,N,N,N) Position angle of point
\left({\alpha }_{2},{\delta }_{2}\right)
\left({\alpha }_{1},{\delta }_{1}\right)
. All the input arguments and the return
argument are in radians. The input arguments are in
the order: ,
Table 16: Mathematical functions which may be used in expressions
The expression string can contain constants, column and parameter names, operators, functions, and parentheses. In general the usual rules of algebra and Fortran should be followed, with some minor exceptions, as noted below.
Spaces are permitted between items, except that a function-name must be followed immediately by a left parenthesis. Spaces are not permitted within items such as names and numerical constants, but can be used within character strings and date/time values in curly braces.
Lower-case letters are treated everywhere as identical to the corresponding upper-case letter.
Column and parameter names can be up to CAT__SZCMP characters long, and may consist of letters, digits, and underscores, except that the first character must not be a digit.
Vector elements are supported but with a restricted syntax: they may consist of a name followed by an unsigned integer constant subscript enclosed in square brackets, for example FLUX[4] or MAGNITUDE[13]. The first element of the vector is numbered one.
CHARACTER constants may be enclosed in a pair of single or double quotes; embedded quotes of the same type may be denoted by doubling up on the quote character within the string, for example ’DON’’T’ or "DON""T".
LOGICAL constants may be .TRUE. or .FALSE. but abbreviations of these words are allowed down to .T. and .F.
Numerical constants may appear in any valid form for Fortran 77 (except that embedded spaces are not allowed). Some additional forms are also permitted, as shown below.
%Xstring %Ostring %Bstring for hexadecimal, octal and binary INTEGER constants respectively.
Angles in sexagesimal notation: colons must be used to separate items, for example hours:minutes:seconds (or degrees:minutes:seconds). If there is a leading sign then the value will be taken as degrees:minutes:seconds, otherwise hours:minutes:seconds. In either case the value is converted to RADIANS.
A date/time value may be given as a string enclosed in curly braces; a range of common formats are permitted, with order year-month-day or day-month-year, and the month as a number or three-character abbreviation. The time may follow with colons separating hours:minutes:seconds. Examples of some valid dates:
26/7/92T3:45
Relational operators are supported in both Fortran 77 form (for example .GE. .NE.) as well as in the Fortran 90 forms (for example,
\text{>=}
\text{/=}
Single-symbol forms for .AND. .OR. and .NOT. are provided as an alternative: &
\text{|}
# respectively.
The dots may be left off the Fortran 77 forms of the relational operators and the logical operators .AND. and .OR. where spaces or parentheses separate them from names or constants, but the logical constants and the .NOT. operator need the enclosing dots to distinguish them from other lexical items in all cases.
INTEGER division does not result in truncation (as in Fortran) but produces a floating-point result. The NINT or INT function should be used (as appropriate) if an INTEGER result is required.
The functions MAX and MIN must have exactly two arguments.
All arithmetic is carried out internally in DOUBLE PRECISION (but the compiler works out the effective data type of the result using the normal expression rules).
Exponentiation is performed by log/exp functions, with use of ABS to avoid taking logs of negative arguments, thus -2**3 will come out as ‘+8’ not ‘-8’.
The operator precedence rules are show in Table 17. The rules of Fortran 90 are used as far as possible; in this table the larger numbers denote higher precedence (tighter binding).
Precedence Function/operator
2 start/end of expression
8 .EQV. .NEQV.
10 .OR.
\text{|}
12 .AND. &
14 .NOT. #
16 .EQ. .GE. .GT. .LE. .LT. .NE.
\text{== >= > <= < /=}
18 FROM TO
22 + - (binary operators)
24 + - (unary operators)
30 all functions
Table 17: Operator precedence rules
Note that all operators except ** associate from left to right, but ** and functions associate from right to left. |
Staking, Liquidity, and Inflation Model - Astar & Shiden documentation
This includes advanced content.
Astar Network's token ecosystem is built into Polkadot. Therefore, this document includes the same formula and values as Polkadot.
Astar tokens has four main roles:
Staking for consensus, rewards for validators and nominators
Transaction fees used to prevent harmful behaviors
Block rewards for dApps operators; sustainable reward designed for applications
Good/bad voting, dApps operators
Astar tokens are intended to be used as a liquidity token. Tokens are issued through multiple Lockdrops to prevent zero-value collateral and increase the number of token holders. Astar tokens are expected to be operated at the ratio of
1:1 = Staking:Liquidity
1:1 = Staking: Liquidity
In the previous chapter, we defined the algorithm that determines the issue amount and distribution method when issuing new Astar Network tokens. The Astar Network is structured that the new token issuance fee is shared with dApps Rewards and a reward for securing the chain. The consensus algorithm of the Astar Network is expected to be NPoS. Thereby, there are two types of Staking actions: Staking (NPoS) for Validator and Staking (dApps Rewards) for smart contracts. Both rewards from each staking are equally proportional to the amount of staking. Users who stake on validators / smart contracts are collectively called nominators. The ideal ratio of Staking for validators and Staking for smart contracts is:
Staking_{validators}
represents the action of staking on validators.
Staking_{contracts}
represents the action of staking on smart contracts.
Then, the below formula is the expected ratio between Staking for validators and Staking for smart contracts.
5:1 = Staking_{validators} : Staking_{contracts}
Rewards paid to Operators are dApps Rewards. Operator rewards increase in proportion to the inflation rate due to Staking. dApps Rewards rewards 50% of the total reward when meeting the ideal
q
from quote the dApps Rewards chapter. The rewards given to the Operator at that time is maximized. To show the specific reward distribution, we introduce the following variables:
Rewards_{operators}
is the total amount of reward got by the Operator.
Rewards_{stakers_{validators}}
is the total amount of rewards got by staking a validator.
Reards_{stakers_{contracts}}
is the total amount of reward got by staking smart contracts.
t
is a coefficient that represents how many times the total amount of rewards earned by the operator is greater than the rewards earned by taking a smart contract.
t= 4
from quote the dApps Rewards chapter and 50% of the total reward for meeting the ideal
q
will go to the dApps Rewards reward. Therefore, the ideal distribution ratio of remuneration is determined as follows.
\begin{aligned} Rewards_{stakers_{validators}}:Rewards_{stakers_{contracts}}:Rewards_{operators} & = 5:1:4 \end{aligned}
Also, the percentage of Staking and the percentage of reward are equal as follows:
Staking_{validators}:Staking_{contracts}=Rewards_{stakers_{validators}}:Rewards_{stakers_{contracts}}
Astar tokens use the same NPoS as Polkadot. This nominator and validator can operate the token at a certain annual interest rate for Staking. Also, token rewards will be paid to the Astar dApps operator's Nominator and Operator as well. Astar Network's inflation model is defined as follows: First, follow the Polkadot inflation model and define the following variables:
x
is the total amount of staking divided by the total amount of tokens issued.
X_{ideal}
is the ideal value of
x
Staking: Liquidity = 1: 1
X_ideal = 0.5
q
is the amount of staking to the validator divided by the total amount of staking.
q=\frac{Staking_{validators}}{Staking_{validators}+Staking_{contracts}}
Q_{ideal}
q
5:1 = Staking_{validators} : Staking_{contracts}
, the ideal value of
q
Q_{ideal} = 5/6
i(x,q)
is the average annual interest getting by Staker. It is a monotonically decreasing function of
x
| Q_{ideal} -q |
(difference from the ideal ratio). To make both
x
q
close to the ideal value, when
x
| Q_{ideal} -q |
is low, raise the interest rate as an incentive to increase the amount of stake. When
x
| Q_{ideal} -q |
is high, lower interest rates as an incentive to reduce stake.
i_{ideal}
is the average annual interest rate of Staker
i (x, q)
x
q
are ideal values. in other words,
i_{ideal} = i (X_{ideal }, Q_{ideal})
I_{Staking}
is the inflation rate by Staking.
I_{Staking}
, a bivariate function involving
x
q
, draws a three-dimensional convex function. Expressing Staking total amount x interest rate = inflation rate and expressing it as
x * i (x, q) = I_{Staking}
Also, this value is maximized when
x
q
is the ideal value from the reward design of
i (x, q)
. The ideal state equation can be expressed as
X_{ideal} * i (X_{ideal}, Q_{ideal}) = Maxmium I_{Staking}
I_0
is the lower limit of inflation rate. When
x = 1 or x = 0
, converge to the lower limit.
I_0
is equivalent to the operating cost of the validator. The reason is that if you do not secure at least the incentive to operate the validator, the chain will break, so I_0 = 0.025 is recommended here.
d
is an adjustable decay rate for each
x
x
d
X_{ideal}
I_{Staking}
is reduced by 50%. In other words,
I_{Staking} (X_{ideal} + d, Q_{ideal}) \ge I_{Staking} /2
.We recommend d = 0.02.
g
is an adjustable decay rate on
q
q
g
Q_{ideal}
I_{Staking}
I_{Staking} (X_{ideal}, Q_{ideal} \ pm e) \ge I_{Staking} / 2
. We recommend g = 0.15.
i_{staking}
is the average annual interest earned by the nominator through Staking. This can be determined by dividing inflation by the Staking ratio. In other words,
i_{staking} = \frac{I_{Staking}}{x}
I_{operators}
is the inflation rate due to the rewards that the Operator can get. This is t times the ratio (1-q) of Staking to Operator in I{Staking}.
Staking_{validators}:Staking_{contracts} = Rewards_{stakers_{validators}}:Rewards_{stakers_{contracts}}
Rewards_{stakers_{validators}}:Rewards_{stakers_{contracts}}:Rewards_{operators} = y:1:t
Rewards_{staking_{validators}}(y+1) = (Rewards_{staking_{contract}}+Rewards_{staking_{validators}})y
q =\frac{Staking_{validators}}{Staking_{validators}+Staking_{contracts}}
=\frac{Rewards_{stakers_{validators}}}{Rewards_{stakers_{validators}}+Rewards_{stakers_{contracts}}}
=y/(y+1)
y =q/(1-q)
Rewards_{stakers_{validators}}:Rewards_{stakers_{contracts}}:Rewards_{operators} =q/(1-q):1:t
Rewards_{stakers}:Rewards_{operators} =q/(1-q)+1:t
Here, the ratio of the amount of reward and the ratio of the inflation rate are equal.
\begin{aligned} I_{Staking}:I_{operators}& =q/(1-q)+1:t \\\\ I_{operstors}(q/(1-q)+1)& =I_{Staking}t \\\\ I_{operators}& =\frac{tI_{Staking}}{\frac{q}{1-q}+1} \\\\ & = \frac{tI_{Staking}}{\frac{q}{1-q}+\frac{1-q}{1-q}} \\\\ \frac{tI_{Staking}}{\frac{1}{1-q}}=t(1-q)I_{Staking} \end{aligned}
represents the average (based on the amount staked) interest rate of the operator's reward. From the auxiliary formula,
i_{operators}=\frac{I_{operators}}{x(1-q)}
I
is the overall inflation rate. This is
I = I_{Staking} + I_{operators}
, which is the sum of the reward for Staking and the inflation rate due to the reward for Operator.
I_{Staking}
is followed the below formula.
I_{Staking} = \begin{cases} I_0 + x(i_{ideal} - \frac{I_0}{X_{ideal}})\cdot2^{-|q-Q_{ideal}|/g} & (0 \lt x \le X_{ideal}) \\ I_0 + (i_{ideal} \cdot X_{ideal} - I_0) \cdot 2^{(X_{ideal}-x)/d-|q-Q_{ideal}|/g} & (X_{ideal} \lt x \le 1) \end{cases}
The below figure is a graph simulating the inflation rate when each parameter is set as follows.
\begin{aligned} i_{ideal}=0.2 \\\\ X_{ideal}=0.5 \\\\ Q_{ideal}=5/6 \\\\ I_0=0.025 \\\\ d=0.02 \\\\ g=0.15 \\\\ t=4 \\\\ \end{aligned}
The above graph is fixed at
q = Q_{ideal}
. Here, the upper green line is the average annual interest rate (
i_{operators}
) for the operator's staking amount, the lower green line is the average annual interest rate of the staking (
i_{staking}
), and the red line is the overall inflation rate (
I
), the upper blue line indicates the inflation rate due to Staking reward (
I_{Staking}
), and the lower blue line indicates the inflation rate due to Operator reward (
I_{Operator}
). The inflation rate when both
x
q
are ideal values is 0.166 ... (1/6) at maximum. Next, the graph when
q = 0.2
is shown in the below figure. When
q = 0.2
, the Staking percentage is
1: 5 = Staking_{validators}: Staking_{contracts}
, and the reward percentage is as follows:
Rewards_{stakers_{validators}}:Rewards_{stakers_{contracts}}:Rewards_{operators}=1:5:20
Then, although the ratio of the operator's reward is increasing, the upper green line representing the average annual interest rate for the operator's staking amount is low because
q
is far from the ideal value. As a result, even if the ratio of Staking to smart contracts increases, the reward paid to Operators is not much different from the ideal state. Also, the lower green line, which represents the average annual interest rate of Staking rewards, has been reduced, giving the incentive for Staker to take validators to Stake in order to maintain balance. As an extreme example, The below figure shows a graph when
q = 1.0
. At this time, no one has taken Staking for the smart contract, and the reward will be as follows.
Rewards_{stakers_{validators}}:Rewards_{stakers_{contracts}}:Rewards_{operators}=1:0:0
The green line that represents the average annual rate of reward for Staking is lower than ideal because the reward that the Operator gets is zero and
q
is far from ideal. In this case, too, Staker creates an incentive to take Stakes on smart contracts to maintain balance. Note that the red line that represents the overall inflation rate and the blue line that represents the inflation rate due to the Staking reward overlap, making the latter invisible. Also, note that these graphs meet the following closings:
The average annual interest functions
i_{staking}, i_{operator}
are monotonic with respect to x.
i_{staking}, i_{operator}
maximize when q is an ideal value.
I_{Staking}, I_{Operator}, I
maximize when
x and q
are both ideal values.
I_0
is the lower limit of the inflation rate.
Rewards_{staking}: Rewards_{operator} = 5 + 1: 4 = 6: 4 = 3: 2
q
is the ideal value. In other words, when
q
is the ideal value, it satisfies
I_{staking}: I_{operator} = 3: 2
By adding the above-inflation model, we will adjust the incentives of Astar users and encourage the actions expected of Astar Network. |
Tell whether the function represents exponential growth or exponential decay. Th
Tell whether the function represents exponential growth or exponential decay. Then graph the function. f(x)=(0.25)^x
Tell whether the function represents exponential growth or exponential decay. Then graph the function.
f\left(x\right)={\left(0.25\right)}^{x}
Want to know more about Exponential growth and decay?
Jaylen Fountain
Video answer of function
The count in a bacteria culture was 200 after 15 minutes and 1600 after 30 minutes. Assuming the count grows exponentially, what was the initial size of the culture? What was the doubling period?
Identify each of the following functions as exponential growth or decay. Then give the rate of growth or decay as a percent.
y=a{\left(\frac{3}{2}\right)}^{t}
An unknown radioactive element decays into non-radioactive substances. In 720 days the radioactivity of a sample decreases by 33 percent.
(a) What is the half-life of the element? half-life:_____
(6) How long will it take for a sample of 100 mg to decay to 47 mg? time needed:____
Using the concept of exponential growth and exponential decay. Solve the given problem.
1. A small locality has a population of 52,365 in 2012. If its population increases 3% every 2 years,
a. Derive a function P that determines the population t years after 2012.
b. What is the expected population in 2020?
Identify the function as exponential growth or exponential decay.
y=0.3\cdot {0.5}^{x}
Consider the following case of exponential growth. Complete parts a through c below.
The population of a town with an initial population of 75,000 grows at a rate of 5.5% per year.
a. Create an exponential function of the form
Q={Q}_{0}×\left(1+r\right)t
, (where r>0 for growth and r<0 for decay) to model the situation described
State whether the equation represents exponential growth, exponential decay, or neither.
y=0.82\cdot {3}^{x} |
A characterization of maps in $H^1 (B^3, S^2)$ which can be approximated by smooth maps
{H}^{1}\left({B}^{3},{S}^{2}\right)
Bethuel, F.
author = {B\'ethuel, Fabrice},
title = {A characterization of maps in $H^1 (B^3, S^2)$ which can be approximated by smooth maps},
TI - A characterization of maps in $H^1 (B^3, S^2)$ which can be approximated by smooth maps
Bethuel, F. A characterization of maps in $H^1 (B^3, S^2)$ which can be approximated by smooth maps. Annales de l'I.H.P. Analyse non linéaire, Tome 7 (1990) no. 4, pp. 269-286. http://www.numdam.org/item/AIHPC_1990__7_4_269_0/
[B] H. Brezis, private communication.
[BCL] H. Brezis, J.M. Coron and E.H. Lieb, Harmonic maps with defects.Comm. Math. Phys., t. 107, 1986, p. 649-705. | MR 868739 | Zbl 0608.58016
[Bel] F. Bethuel, The approximation problem for Sobolev maps between two manifolds, to appear. | MR 1120602 | Zbl 0756.46017
[BZ] F. Bethuel and X. Zheng, Density of smooth functions between two manifolds in Sobolev spaces. J. Func. Anal., t. 80, 1988, p. 60-75. | MR 960223 | Zbl 0657.46027
[CG] J.M. Coron and R. Gulliver, Minimizing p-harmonic maps into spheres, preprint. | MR 1018054
[H] F. Helein, Approximations of Sobolev maps between an open set and an euclidean sphere, boundary data, and singularities, preprint. | MR 1010196
[SU] R. Schoen and K. Uhlenbeck, A regularity theory for harmonic maps. J. Diff. Geom., t. 17, 1982, p. 307-335. | MR 664498 | Zbl 0521.58021
[W] B. White, Infima of energy functionals in homotopy classes. J. Diff. Geom, t. 23, 1986, p. 127-142. | MR 845702 | Zbl 0588.58017 |
solve this The table below shows the salaries of 280 persons Salary (In thousand)No of persons5-104910-1513315-206320-251525-306 - Maths - Statistics - 12548379 | Meritnation.com
The table below shows the salaries of 280 persons
\begin{array}{cc}Salary \left(In thousand\right)& No. of persons\\ 5-10& 49\\ 10-15& 133\\ 15-20& 63\\ 20-25& 15\\ 25-30& 6\end{array}\phantom{\rule{0ex}{0ex}} \begin{array}{cc}30-35 & 7\\ 35-40 & 4\\ 40-45 & 2\\ 45-50 & 1\end{array}
Calculate the median salary of the data,
Hope this will helps you..!!
Anmol Gupta answered this
Arre ab koi fayda ni h iska
Waise iska ans 13.42 tha
Thankyou....just checking out how was my maths prepration ......Hehehe
Ayush Yeole answered this
The answer is 13.42 And do you know that CBSE is going to reconduct class 10th maths paper held today
Hey are we ---- to answer your asked question
Yes i know ayush!!
Tanish Gupta answered this
Its not 13.42 rather 13420 |
Analysis of Heat Transfer Enhancement in Minichannel Heat Sinks With Turbulent Flow Using H2O–Al2O3 Nanofluids | J. Electron. Packag. | ASME Digital Collection
Analysis of Heat Transfer Enhancement in Minichannel Heat Sinks With Turbulent Flow Using
H2O–Al2O3
Bergman, T. L. (April 2, 2009). "Analysis of Heat Transfer Enhancement in Minichannel Heat Sinks With Turbulent Flow Using
H2O–Al2O3
Nanofluids." ASME. J. Electron. Packag. June 2009; 131(2): 021008. https://doi.org/10.1115/1.3103949
Heat transfer enhancement associated with use of a nanofluid coolant is analyzed for small electronic heat sinks. The analysis is based on the
ε
-NTU heat exchanger methodology, and is used to examine enhancement associated with use of
H2O–Al2O3
nanofluids in a heat sink experiencing turbulent flow. Predictive correlations are generated to ascertain the degree of enhancement based on the fluid’s thermophysical properties. The enhancement is quite small, suggesting the limited usefulness of nanofluids in this particular application.
alumina, coolants, heat exchangers, heat sinks, heat transfer, nanofluidics, turbulence, nanofluid, heat sink, enhancement
Heat sinks, Heat transfer, Nanofluids, Turbulence, Water, Coolants, Fluids
Challenges in Electronic Cooling—Opportunities for Enhanced Thermal Management Techniques—Microprocessor Liquid Cooled Minichannel Heat Sink
Heat Transfer in Nanofluids–A Review
Mean-Field Versus Microconvection Effects in Nanofluid Thermal Conduction
Beyond the Maxwell Limit: Thermal Conduction in Nanofluids With Percolating Fluid Structures
Thermal Conductance of Nanofluids: Is the Controversy Over?
Cauthier
Heat Transfer Enhancement Using Al2O3—Water Nanofluid for an Electronic Liquid Cooling System
Assessment of the Effectiveness of Nanofluids for Single-Phase and Two-Phase Heat Transfer in Micro-Channels
Effect of Reduced Specific Heats of Nanofluids on Single Phase, Laminar Internal Convection
Heat Sink Optimization With Applications to Microchannels
Numerical Study of Turbulent Heat Transfer and Pressure Drop Characteristics in a Water-Cooled Microchannel Heat Sink
Forced Convection, Internal Flow in Ducts
Measurement of the Specific Heat Capacity of Water-Based Al2O3 Nanofluid
Enhancement of Molar Heat Capacity of Nanostructured Al2O3
Effective Viscosities and Thermal Conductivities of Aqueous Nanofluids Containing Low Volume Concentrations of Al2O3 Nanoparticles
Effects of Al 2 O 3 -Water Nanofluid and Angular Orientation on Entropy Generation and Convective Heat Transfer of an Elliptical Micro-Pin-Fin Heat Sink
Single-Side Heated Monoblock, High Heat Flux Removal Using Water Subcooled Flow Boiling
Effect of High Coolant Velocity on Heat Transfer in Compact Heat Exchangers |
Roots of Unity Geometry Practice Problems Online | Brilliant
A "hop" is a movement of 4 sides at a time counterclockwise around the regular nonagon above. From the starting point, What is the minimum numbers of "hops" it will take to end up back at the starting point?
A regular hexagon has a vertex at
(1,0)
and has its centroid at the origin. Which of these is another coordinate of the hexagon?
\left(-\frac{1}{2},\frac{7}{8}\right)
\left(-\frac{1}{2},\frac{17}{20}\right)
\left(-\frac{1}{2},\frac{\sqrt3}{2}\right)
\left(-\frac{1}{2},\frac{\sqrt{6}}{3}\right)
Regular polygons are placed on the coordinate plane such that they each have a vertex at
(1,0)
, and centroid at the origin. There is one regular
n
-gon for each value of
n
3
50
n
are there such that the
n
-gon does not share any vertex other than
(1,0)
with any of the other
n
-gons?
(5,-3)
\dfrac{\pi}{4}
radians counterclockwise about the point
(2,4)
. What is the resulting image?
\left(9,3\sqrt{2}-3\right)
\left(2+5\sqrt{2},4-2\sqrt{2}\right)
\left(16-4\sqrt{3},3\sqrt{3}-4\right)
\left(\frac{127}{14},\frac{6}{5}\right)
Each of the partial sums
\sum\limits_{k=1}^{n}{e^{k\pi i/3}}
is graphed on the complex plane for
n\in\{1,2,3,4,5,6\}
6^\text{th}
roots of unity coincides with one of these partial sums?
-1
0
1
\frac{1}{2}+i\frac{\sqrt{3}}{2} |
Observer-Based Feedback with Matlab | Brilliant Math & Science Wiki
Observer-Based Feedback with Matlab
Samuel Hansen contributed
This wiki covers the aspects of a typical control problem which is solved using Matlab Suppose we are given the plant,
P(s)= \frac{1}{s(s+1)}
which we would like to stabilize it and perform tracking using pole placement and observer design techniques in matlab.
a. Place the poles according to the time domain specifications: overshoot is less than 20% and the rise time is less than 0.5 seconds.
We can use the transfer function and pick it in controllable canonical form to obtain the A,B matrices. First we’ll want to convert the system to state space. Recall that A state-space model can be extracted from the system variable G with the following command:
[A,B,C,D] = ssdata(G) % or with [A,B,C,D] = tf2ss(num,den)
Which return the matrices in controllable canonical form. Alternatively, for this simple system we can write the system as a set of differential equations,
y''+y'=u
which is trivial to put into state space formulation. In either case, the command
* place(A,B,[-1.8+3.177j,-1.8-3.177j]) *
yields K values of 2.6 and 13.333.
Now we close the loop and simulate the time response. Acl=A-B*K; sys=ss(Acl, B, C, D); step(sys); %to simulate response and verify time response
But what’s this? The step response is off by a constant factor! Well this is because the controller we have designed so far meets our transient requirements, but now we must address the steady-state error. This is done with the use of a constant precompensator. Looking at our block diagram we can derive that we must achieve unity DC gain, thus let
G=(C(-A+BK)^{-1})^{-1}
b.Next, we want to design an observer and incorporate it into the system. Note that the step response is exactly the same. Show we this is the case.
So we add an observer with dynamics,
\hat{\dot{x}} = (A-LC)\hat{x}+By+Ly
Again we can use place to choose the location of the poles, say twice as far (it doesn’t matter in this case.)
Cite as: Observer-Based Feedback with Matlab. Brilliant.org. Retrieved from https://brilliant.org/wiki/observer-based-feedback-with-matlab/ |
Form (a) the coefficient matrix and (b) the augmented matrix for the system of l
Form (a) the coefficient matrix and (b) the augmented matrix for the system of linear equations. displaystyle{leftlbracebegin{matrix}{x}+{y}={0}{5}{x}-{2}{y}-{2}{z}={12}{2}{x}+{4}{y}+{z}={5}end{matrix}right.}
Form (a) the coefficient matrix and (b) the augmented matrix for the system of linear equations.
\left\{\begin{array}{c}x+y=0\\ 5x-2y-2z=12\\ 2x+4y+z=5\end{array}
For a system of equations
\left\{\begin{array}{c}ax+by+cz=j\\ dx+ey+fz=k\\ gx+hy+iz=l\end{array}
, the coefficient matrix is
\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]
and the augmented matrix is
\left[\begin{array}{ccccc}a& b& c& |& j\\ d& e& f& |& k\\ g& h& i& |& l\end{array}\right]
a) For the system
\left\{\begin{array}{c}x+y=0\\ 5x-2y-2z=12\\ 2x+4y+z=5\end{array}
a=1 , b=1 , c=0,d=5,e=-2, f=-2, g=2 , h=4 and i=1 so the coefficient matrix is
\left[\begin{array}{ccc}5& 1& 0\\ 5& -2& -2\\ 2& 4& 1\end{array}\right]
b) For the system
\left\{\begin{array}{c}x+y=0\\ 5x-2y-2z=12\\ 2x+4y+z=5\end{array}
a=1 , b=1 , c=0,d=5,e=-2, f=-2, g=2 , h=4 , i=1 , j=0 , k=12 and l=5 so the augmented matrix is
\left[\begin{array}{ccccc}5& 1& 0& |& 0\\ 5& -2& -2& |& 12\\ 2& 4& 1& |& 5\end{array}\right]
\left[\begin{array}{ccc}5& 1& 0\\ 5& -2& -2\\ 2& 4& 1\end{array}\right]
\left[\begin{array}{ccccc}5& 1& 0& |& 0\\ 5& -2& -2& |& 12\\ 2& 4& 1& |& 5\end{array}\right]
f\left(x\right)=\sqrt{4-x}
a=0
\sqrt{3.9}
\sqrt{3.99}
Write the following linear differential equations with constant coefficients in the form of the linear system
\stackrel{˙}{x}=Ax
\left(a\right)\stackrel{¨}{x}+\stackrel{˙}{x}-2x=0
\left(b\right)\stackrel{¨}{x}+x=0
\left(c\right)d\stackrel{¨}{x}-2\stackrel{¨}{x}-\stackrel{˙}{x}+2x=0
{x}_{1}=x,{x}_{2}=\stackrel{˙}{x}
I have tried to do this in the following way but I do not know if I am doing well:
{x}_{1}=x,{x}_{2}=\stackrel{˙}{x},{x}_{3}=\stackrel{¨}{x},{x}_{4}={x}^{\dots }
, then the previous system becomes
\left(a\right){x}_{3}+{x}_{2}-2{x}_{1}=0,\left(b\right){x}_{3}+{x}_{1}=0,\left(c\right){x}_{4}-{x}_{3}-{x}_{2}+2{x}_{1}=0
\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\\ {x}_{4}\end{array}\right]=\left[\begin{array}{c}\frac{{x}_{2}}{3}\\ {x}_{2}\\ \frac{-{x}_{2}}{3}\\ 0\end{array}\right]
. Is this what it should be or is it different? Thank you very much.
\left\{\begin{array}{l}\stackrel{˙}{x}=-4y\\ \stackrel{˙}{y}=x\end{array}
x\left(t\right)
y\left(t\right)
are unknown real functions.
\left(\begin{array}{c}x\left(t\right)\\ y\left(t\right)\end{array}\right)={c}_{1}\left(\begin{array}{c}\mathrm{cos}\left(2t\right)\\ \frac{1}{2}\mathrm{sin}\left(2t\right)\end{array}\right)+{c}_{2}\left(\begin{array}{c}-2\mathrm{sin}\left(2t\right)\\ \mathrm{cos}\left(2t\right)\end{array}\right)
{c}_{1}
{c}_{2}
are real parameters.
\left(\begin{array}{c}x\left(t\right)\\ y\left(t\right)\end{array}\right)={k}_{1}{e}^{i2t}\left(\begin{array}{c}{P}_{11}\\ {P}_{21}\end{array}\right)+{k}_{2}{e}^{-i2t}\left(\begin{array}{c}{P}_{12}\\ {P}_{22}\end{array}\right)
{k}_{1}
{k}_{2}
should (?) be complex parameters.
Obtain just a particular solution (the general solution can alwaysbe obtained easily by adding an arbitrary multiple of a solution of the associated homogeneous equation). In these exercises, the forcing function is an elementary sinusoidal function. If the forcing function is
\mathrm{cos}5t
, then a particular solution of the form A
\mathrm{cos}5t
will not work. However, a linear combination of
\mathrm{cos}5t
\mathrm{sin}5t
will work. Thus, the form for xp is
A\mathrm{cos}5t+B\mathrm{sin}5t
, where the constants A and B are determined by substituting the assumed form of the particular solution into the differential equation for x. In Exercises 47–52, the forcing function has several different terms. Use a form for xp consisting of the sum of the forms for each term.
\frac{dx}{dt}+x=2{e}^{3t}+\mathrm{sin}t
I mean, that whenever we talk about linear equations, we say that they always form a straight line on a graph. We make this sure by making a graph of some possible solutions to the equation and plotting the points on the graph. But can we make sure that all the solutions of the linear equation will always form a straight line because we cannot plot each and every solution of the equation on the graph because there are infinitely many solutions to a linear equation in 2 variables?
{x}^{\prime }=Ax+b
The linear equation is the equation that has a from:
{a}_{1}{x}_{1}+{a}_{2}{x}_{2}+\dots +{a}_{n}{x}_{n}=b
But is this really the definition of a linear equation? I thought that the definition should be the form of linear mapping. Like,
f\left(x+y\right)=f\left(x\right)+f\left(y\right)
f\left(ax\right)=af\left(x\right)
What is the definition of a linear equation? |
Into the Mystic: Hypothesis Testing Practice Problems Online | Brilliant
Many of the decisions and judgments we make are based on limited or incomplete information.
Amidst all of this uncertainty, statistics plays a special role:
Statistics provides the tools to make the best possible judgments with limited data.
We have a long way to go before we can fully appreciate how statistics accomplishes this, but this chapter starts us on our journey by introducing the most essential ideas underlying statistics.
In this first quiz, we'll explore how probability is used in statistics to make informed decisions.
Into the Mystic: Hypothesis Testing
Despite being a magician by trade, Rupert is a skeptic.
Sure, he's built a lucrative career convincing his audience that he can defy the very laws of nature, but it's all sleight-of-hand and well-crafted illusions designed to fool the senses.
So when he hears about Alberta The Oracle, who claims that she draws the answer to any yes or no question from supernatural sources, he sets out to determine the truth for himself.
Rupert thinks Alberta is just a good guesser, so he gathers a stockpile of yes/no questions that only he can answer and then heads out from Magicians Alliance HQ one cold and foggy night to put The Oracle to the test...
In The Oracle's parlor, Rupert takes a seat, thanks his host for the opportunity, and then poses his first question.
Suppose for the moment that Alberta really does just guess the answer to every question she's given by responding "yes" or "no" with equal probability.
What's the probability Alberta answers Rupert's first question correctly just by guessing?
0
\frac{1}{2}
\frac{2}{3}
1
The probabilities in the last problem obey a uniform distribution, which is common in statistics.
To get a sense of what "uniform" means, let's say the set of all possible outcomes of our experiment — called the sample space — is finite.
For example, the sample space in the last problem is
\{ \text{yes}, \text{no} \},
the only possible responses from The Oracle. Another example is
\{ \tt{HH, TH, HT, TT}\},
the outcomes for two back-to-back flips of a coin with one side labeled
\tt{H}
\tt{T}.
It's always true that the total probability is
1
no matter how it's distributed, so if we list the outcomes in the sample space in no particular order, then
\begin{aligned} & \text{(Probability of first outcome)} \\ &+ \text{(Probability of second outcome)} \\ & + \dots + \text{(Probability of last outcome)} \\ &=1. \end{aligned}
In the uniform distribution, all of these probabilities are equal, so
\begin{aligned} & \text{(Probability of an outcome in the uniform distribution)} \\ & \times \text{(Size of the sample space)} = 1, \end{aligned}
\text{(Probability of an outcome in the uniform distribution)} \\[0.8em] = \frac{1}{\text{(Size of the sample space)}}.
Suppose Rupert poses three questions to The Oracle.
In that case, her possible responses can be encoded as
3
-letter strings of
\tt{Y}
\tt{N}
\tt{YNY}
represents Alberta answering "yes" to the first question, "no" to the second, and "yes" to the third. These three-letter strings make up the sample space for Rupert's experiment.
If Alberta simply guesses her answers indiscriminately as Rupert believes, then every outcome in the sample space has the same probability of being observed.
In other words, the probability is distributed uniformly over the sample space.
So what's the probability that Alberta answers Rupert's three questions correctly if the magician is right about the way The Oracle responds?
\frac{1}{8}
\frac{1}{4}
\frac{1}{3}
\frac{2}{5}
To test The Oracle, Rupert needs to know the probability Alberta will answer
n
consecutive questions correctly if he's right about her method of guessing "yes" and "no" with equal likelihood.
Use what we discovered in the last problem to find this probability.
\frac{1}{2n}
\frac{1}{n^2}
\frac{1}{2^{n}}
\frac{1}{2(n + 1)}
It's helpful at this point to introduce some terminology that we'll use throughout the course:
A hypothesis is an assumption made before an experiment is done.
Rupert, being a skeptic, hypothesizes that Alberta doesn't in fact commune with otherworldly powers to get her answers: rather, he thinks that she simply guesses "yes" or "no."
Quizzing The Oracle with his list of questions gives Rupert data, and that data either supports his hypothesis or gives him reason to reject it. But data can never "prove" or "disprove" his hypothesis!
For instance, Alberta could answer correctly for a very long stretch just by random chance, so this doesn't prove she has supernatural knowledge.
On the other hand, she also could receive messages from beyond this world, but her source may not be right all the time: providing one wrong answer doesn't disprove her oracular powers.
In short, nothing is ever "proven" in statistics! Hypotheses are either ruled likely true or likely untrue based on data gathered.
So how exactly does Rupert test his hypothesis that Alberta guesses every answer she gives?
Well, the probability that Alberta gets
n
consecutive questions correct is
\frac{1}{2^{n}}
if Rupert's hypothesis is true.
This decays exponentially, so if she did give a long string of correct answers, Rupert's hypothesis looks pretty unlikely: he'd think "Maybe there's something to this oracle business after all!"
Since Rupert is so skeptical, he'll only reject his hypothesis and admit Alberta is an oracle if she gives so many correct answers that the probability she's guessing is less than
0.001.
How many consecutive correct answers must Alberta give in order for Rupert to change his mind about her abilities?
Use the interactive below if you don't have a calculator handy:
3
questions At least
5
7
10
Rupert's skepticism led him to choose
0.001
since he needs an "extreme" (i.e. highly unlikely) outcome in order to give up the hypothesis that Alberta simply guesses.
Given what we found in the last problem, Rupert considers
10
questions answered correctly and consecutively enough to abandon his initial assumption that Alberta guesses her responses.
As we'll see later in the course, there are many practical considerations that go into choosing the threshold at which we give up on a hypothesis, but a probability of
5\%
is commonly used.
So what makes Rupert so skeptical of Alberta The Oracle?
Well, truth be told, Rupert often "communes with spirits" in his magic act, too, but the source of his supposed knowledge isn't supernatural: it's all about playing the odds.
Rupert could be drummed out of the Magicians Alliance for revealing his methods, but he's willing to talk to us about one of the simplest tricks that most psychics use:
A quick internet search shows that about
5.5
million of the
330
million people living in the US is named James, so a randomly selected American has a probability
p = \frac{1}{60}
of being a "James."
If a psychic plays to a crowd of size
n
and says "The spirits tell me there's a James here," how big does
n
have to be for the psychic to have at least an
80 \%
chance of being right?
Assume the names of the
n
audience members are all independent
Hint: Calculate the probability there are no Jameses in the audience, and then use the rule of complement to find the probability there's at least one James. The plot below will be of help.
n
has to be at least
23.
n
42.
n
71.
n
96.
Of course, just being
80 \%
confident there's at least one James in the audience doesn't mean there actually will be one present every time, so a good psychic needs a way to backpedal.
That's why psychics start their acts by explaining that the messages they get can be vague, and then go on to tell the audience that they'll need help understanding what the messages mean.
It's worth taking a moment to think about the probabilistic idea at the center of this magic trick:
No matter how unlikely an event is, the probability of observing it in some experiment goes up with the number of trials you perform.
The probability of randomly selecting a James from the US population is only
1.7 \%,
but sample
96
US citizens and the probability one of them is a James is over
80 \%.
We'll see this idea again later in the course when we look at sources of false positives, which is a type of error made when a hypothesis is rejected even though it's actually true.
We mentioned this earlier, but it's worth repeating: nothing is ever "proven" with statistics.
Instead, a statistician rejects or doesn't reject a hypothesis based on the "extremeness" of data.
Most statistical analysis boils down to the following steps:
\\[-1em]
Come up with a hypothesis, which is an assumption made before experimentation.
Decide on a criterion that rejects the hypothesis if the experimental results are too "extreme."
Gather and analyze data, and then reject or don't reject the hypothesis based on the results.
\\[-1em]
We saw all of these ideas play out in this quiz:
\\[-1em]
First, Rupert hypothesized "The Oracle guesses 'yes' or 'no'."
He then came up with a criterion for rejecting it: "If she guesses
10
correct questions in a row, I think my hypothesis probably isn't true."
He gathered data by asking Alberta a series of questions.
\\[-1em]
Yet this quiz only scratches the surface: our journey to mastering the basic tools and concepts used in modern hypothesis testing truly begins in Chapter
3.
In the next quiz, we'll introduce two other important aspects of statistics, estimation and sampling, topics we'll take up in full in Chapter
2. |
Describe in words the region of R^{3} represented by the equation(s) or inequality. x = 5
nicekikah 2021-02-22 Answered
{R}^{3}
x=5
The equation of the plane parallel to yz- plane is
x=a
x=5
x=5
x=5
{R}^{3}
whose x coordinate is 5.
\left(x,y,z\right)|\text{ }x=5,y\in R,z\in R
This is a plane which is parellel to yz plane and five units in front of it.
x=5
represents a plane in
{R}^{3}
parallel to yz plane and five units in front of it.
x=5
{R}^{3}
{x}^{2}+{y}^{2}+{z}^{2}+8x-6y+2z+17=0
{\mathrm{sin}}^{3}
{\mathrm{sin}}^{3}
\left(\theta ,\varphi \right)
\mathrm{sin}\left\{3\right\}\left(\theta \right)
{v}_{1}=100k\frac{m}{s}
{v}_{2}=200k\frac{m}{s}
=2
{M}_{1}=5.74e33g
{M}_{2}=2.87e33g
Give a full answer for the given question: The covalent backbone of DNA and RNA consists of: ?
\left(9,\pi ,\frac{\pi }{2}\right)
\left(x,y,z\right)=?
Let C be a circle, and let P be a point not on the circle. Prove that the maximum and minimum distances from P to a point X on C occur when the line X P goes through the center of C. [Hint: Choose coordinate systems so that C is defined by
x2+y2=r2
and P is a point (a,0)
on the x-axis with a
\ne ±r,
use calculus to find the maximum and minimum for the square of the distance. Don’t forget to pay attention to endpoints and places where a derivative might not exist.]
{R}^{3}
{x}^{2}+{y}^{2}=4
Below are various vectors in cartesian, cylindrical and spherical coordinates. Express the given vectors in two other coordinate systems outside the coordinate system in which they are expressed
a\right)\stackrel{\to }{A}\left(x,y,z\right)={\stackrel{\to }{e}}_{x}
d\right)\stackrel{\to }{A}\left(\rho ,\varphi ,z\right)={\stackrel{\to }{e}}_{\rho }
g\right)\stackrel{\to }{A}\left(r,\theta ,\varphi \right)={\stackrel{\to }{e}}_{\theta }
j\right)\stackrel{\to }{A}\left(x,y,z\right)=\frac{-y{\stackrel{\to }{e}}_{x}+x{\stackrel{\to }{e}}_{y}}{{x}^{2}+{y}^{2}} |
Huanhe Dong, Yanfeng Zhang, Yongfeng Zhang, Baoshu Yin, "Generalized Bilinear Differential Operators, Binary Bell Polynomials, and Exact Periodic Wave Solution of Boiti-Leon-Manna-Pempinelli Equation", Abstract and Applied Analysis, vol. 2014, Article ID 738609, 6 pages, 2014. https://doi.org/10.1155/2014/738609
Huanhe Dong,1 Yanfeng Zhang,1 Yongfeng Zhang,1 and Baoshu Yin2,3
1College of Mathematics and System Science, Shandong University of Science and Technology, Qingdao 266590, China
2Institute of Oceanology, China Academy of Sciences, Qingdao 266071, China
3Key Laboratory of Ocean Circulation and Wave, Chinese Academy of Sciences, Qingdao 266071, China
We introduce how to obtain the bilinear form and the exact periodic wave solutions of a class of ( )-dimensional nonlinear integrable differential equations directly and quickly with the help of the generalized -operators, binary Bell polynomials, and a general Riemann theta function in terms of the Hirota method. As applications, we solve the periodic wave solution of BLMP equation and it can be reduced to soliton solution via asymptotic analysis when the value of is 5.
It is significantly important to research nonlinear evolution equations in exploring physical phenomena in depth [1, 2]. Since the soliton theory has been proposed, the research on seeking the exact solutions of the soliton equations has attracted great attention and made great progress. A series of methods have been proposed, such as Painléve test [3], Bäcklund transformation method [4, 5], Darboux transformation [6], inverse scattering transformation method [7], Lie group method [8, 9], and Hamiltonian method [10, 11]. Particularly, Hirota direct method [12, 13] provides a direct approach to solve a kind of specific bilinear differential equations among the exciting methods. As we all know, once the bilinear forms of nonlinear differential equations are obtained, we can construct the multisoliton solutions, the bilinear Bäcklund transformation, and Lax pairs easily. It is clear that the key of Hirota direct method is to find the bilinear forms of the given differential equations by the Hirota differential -operators. Recently, Ma put forward generalized bilinear differential operators named -operators in [14] which are used to create bilinear differential equations. Furthermore, different symbols are also used to furnish relations with Bell polynomials in [15], and even for trilinear equations in [16].
In this paper, we would like to explore the relations between multivariate binary Bell polynomials [17–19] and the -operators and to find the bilinear form of Boiti-Leon-Manna-Pempinelli (BLMP) equation [20, 21]. Then, we can obtain the exact periodic wave solution [22–25] of the BLMP equation with the help of a general Riemann theta function in terms of Hirota method.
The paper is structured as follows. In Section 2, we will give a brief introduction about the difference between the Hirota differential -operators and the generalized -operators. In Section 3, we will explore the relations between multivariate binary Bell polynomials and the -operators. In Section 4, we will use the relation in Section 2 to seek the differential form of the BLMP equation and then take advantage of the Riemann theta function [26, 27] and Hirota method to obtain its exact periodic wave solution which can be reduced to the soliton solution via asymptotic analysis.
2. Hirota Bilinear -Operators and the Generalized -Operators
It is known to us that Hirota bilinear -operators play a significant role in Hirota direct method. The -operators are defined in [14] as the following: where . Generally, we have where .
For instance, for the Boussinesq equation under , we have we can get its bilinear form with -operators
However, based on the Hirota -operators, Professor Ma put forward a kind of generalized bilinear -operators in [14]: where, for an integer , the th power of is defined by with .
For example, if , the -operators are Hirota operators.
By (6) and (8), it is clear to see that
Now, under , the generalized bilinear Boussinesq equation can be expressed as
Then, we would like to discuss how to use the -operator to seek the bilinear differential form of other nonlinear integrable differential equations with the help of binary Bell polynomial.
3. Binary Bell Polynomial
As we all know, Bell proposed three kinds of exponent-form polynomials. Later, Lambert, Gilson, and their partners generalized the third type of Bell polynomials in [28, 29] which is used mainly in this paper.
The multidimensional binary Bell polynomials which we will use are defined as follows:
in which .
For convenience, we assume that and read that
We find that the link between -polynomials and the -operator can be given as the following through the above deduction:
Particularly, when , we define -polynomials by
When , we can obtain that
Let us now utilize the -polynomials given above to seek the bilinear form of BLMP equation with the -operators.
4. Boiti-Leon-Manna-Pempinelli Equation
In this section, firstly, we will give the bilinear form of BLMP equation with the help of -polynomials and the -operators. And then, we construct the exact periodic wave solution of BLMP equation with the aid of the Riemann theta function, Hirota direct method, and the special property of the -operators when acting on exponential functions.
4.1. Bilinear Form
BLMP equation can be written as
Setting , inserting it into (18), and integrating with respect to yields where is an integral constant.
Based on (17), (19) can be expressed as
From the above, we can get the bilinear form of (18): with .
4.2. Periodic Wave Solutions
When acting on exponential functions, we find that -operators have a good property if we assume that
As a result of the property above, we consider Riemann’s theta function solution of (18): where , with , , and being constants to be determined.
Then, we have where .
To the bilinear form of BLMP equation, satisfies the period characters when . The powers of obey rule (7), noting that where .
From (26) we can infer that
For (21), we may let
Also, the powers of obey rule (7). For the sake of computational convenience, we denote that
By (28), (29), and (30), we can get that
In view of (30), it is easy to see that
Thus, we obtain the periodic wave solution of BLMP equation: where is given by (24) and are satisfied with (31).
Then, assuming , based on (29), we may obtain that which lead to So, we have and , as .
It is interesting that if we set which can infer that
From all the above, it can be proved that the periodic wave solution (32) just goes to the soliton solution
Thus, if we assume that , , and to the solution solution of (18) can be shown in Figure 1.
Solution of (18).
In this paper, we obtain the bilinear form of bilinear differential equations by applying the -operators and binary Bell polynomials, which has proved to be a quick and direct method. Furthermore, together with Riemann theta function and Hirota method, we successfully get the exact periodic wave solution and figure of BLMP equation when .
There are many other interesting questions on bilinear differential equations, for example, how to apply the generalized operators into the discrete equations; it is known that researches on the discrete and differential equations are also significant. Besides, we will try to explore other operators to construct more nonlinear evolution equations simply and directly in the near future.
This work was supported by National Natural Science Foundation of China (no. 11271007), Nature Science Foundation of Shandong Province of China (no. ZR2013AQ017), the Strategic Pioneering Program of Chinese Academy of Sciences (no. XDA 10020104), SDUST Research Fund (no. 2012 KYTD105), Open Fund of the Key Laboratory of Ocean Circulation and Waves, Chinese Academy of Science (no. KLOCAW1401), and Graduate Innovation Foundation from Shandong University of Science and Technology (no. YC130321).
H. W. Yang, B. S. Yin, and Y. L. Shi, “Forced dissipative Boussinesq equation for solitary waves excited by unstable topography,” Nonlinear Dynamics, vol. 70, no. 2, pp. 1389–1396, 2012. View at: Publisher Site | Google Scholar | MathSciNet
H. W. Yang, X. R. Wang, and B. S. Yin, “A kind of new algebraic Rossby solitary waves generated by periodic external source,” Nonlinear Dynamics, vol. 76, no. 3, pp. 1725–1735, 2014. View at: Publisher Site | Google Scholar | MathSciNet
G. Wei, Y. Gao, W. Hu, and C. Zhang, “Painlev\'e analysis, auto-B\"acklund transformation and new analytic solutions for a generalized variable-coefficient Korteweg-de Vries (KDV) equation,” The European Physical Journal B. Condensed Matter and Complex Systems, vol. 53, no. 3, pp. 343–350, 2006. View at: Publisher Site | Google Scholar | MathSciNet
Y. Yang and Y. Chen, “Pseudopotentials, lax pairs and bäcklund transformations for generalized fifth-order KdV equation,” Communications in Theoretical Physics, vol. 55, no. 1, pp. 25–28, 2011. View at: Publisher Site | Google Scholar | MathSciNet
E. Fan, “New bilinear Bäcklund transformation and Lax pair for the supersymmetric two-BOSon equation,” Studies in Applied Mathematics, vol. 127, no. 3, pp. 284–301, 2011. View at: Publisher Site | Google Scholar | MathSciNet
S. Lou and X. Hu, “Broer-Kaup systems from Darboux transformation related symmetry constraints of KADomtsev-Petviashvili equation,” Communications in Theoretical Physics, vol. 29, no. 1, pp. 145–148, 1998. View at: Google Scholar | MathSciNet
H. Dong and X. Wang, “A Lie algebra containing four parameters for the generalized Dirac hierarchy,” Applied Mathematics and Computation, vol. 215, no. 2, pp. 459–463, 2009. View at: Publisher Site | Google Scholar | Zentralblatt MATH | MathSciNet
Y. Zhang, “A generalized Boite–Pempinelli–Tu (BPT) hierarchy and its bi-Hamiltonian structure,” Physics Letters A, vol. 317, no. 3-4, pp. 280–286, 2003. View at: Publisher Site | Google Scholar | MathSciNet
W. Ma, “Nonlinear continuous integrable Hamiltonian couplings,” Applied Mathematics and Computation, vol. 217, no. 17, pp. 7238–7244, 2011. View at: Publisher Site | Google Scholar | Zentralblatt MATH | MathSciNet
W. Ma, Y. Zhang, Y. Tang, and J. Tu, “Hirota bilinear equations with linear subspaces of solutions,” Applied Mathematics and Computation, vol. 218, no. 13, pp. 7174–7183, 2012. View at: Publisher Site | Google Scholar | MathSciNet
W. Ma, “Bilinear equations, Bell polynomials and linear superposition principle,” Journal of Physics: Conference Series, vol. 411, no. 1, Article ID 012021, 2013. View at: Publisher Site | Google Scholar
W. X. Ma, “Bilinear equations and resonant solutions characterized by Bell polynomials,” Reports on Mathematical Physics, vol. 72, no. 1, pp. 41–56, 2013. View at: Publisher Site | Google Scholar | MathSciNet
W. Ma, “Trilinear equations, Bell polynomials, and resonant solutions,” Frontiers of Mathematics in China, vol. 8, no. 5, pp. 1139–1156, 2013. View at: Publisher Site | Google Scholar | Zentralblatt MATH | MathSciNet
Y. Wang and Y. Chen, “Integrability of an extended (2+1)-dimensional shallow water wave equation with Bell polynomials,” Chinese Physics B, vol. 22, no. 5, Article ID 050509, 2013. View at: Publisher Site | Google Scholar
Y. Wang and Y. Chen, “Binary Bell polynomials, bilinear approach to exact periodic wave solutions of
\left(2+1\right)
-dimensional nonlinear evolution equations,” Communications in Theoretical Physics, vol. 56, no. 4, pp. 672–678, 2011. View at: Publisher Site | Google Scholar | MathSciNet
C. R. Gilson, J. J. C. Nimmo, and R. Willox, “A
\left(2+1\right)
-dimensional generalization of the {AKNS} shallow water wave equation,” Physics Letters A, vol. 180, no. 4-5, pp. 337–345, 1993. View at: Publisher Site | Google Scholar | MathSciNet
Q. U. Chang- Zheng, “Symmetry algebras of generalized
\left(2+1\right)
-dimensional KdV equation,” Communications in Theoretical Physics, vol. 25, no. 3, pp. 369–372, 1996. View at: Google Scholar | MathSciNet
T. Xia, B. Li, and H. Zhang, “New explicit and exact solutions for the Nizhnik-NOVikov-Vesselov equation,” Applied Mathematics E-Notes, vol. 1, pp. 139–142, 2001. View at: Google Scholar | MathSciNet
Z. Zhao, Y. Zhang, and T. Xia, “Double periodic wave solutions of the
\left(2+1\right)
-dimensional Sawada-Kotera equation,” Abstract and Applied Analysis, vol. 2014, Article ID 534017, 6 pages, 2014. View at: Publisher Site | Google Scholar | MathSciNet
T. C. Xia, F. J. Yu, and D. Y. Chen, “Multi-component C-KdV hierarchy of soliton equations and its multi-component integrable coupling system,” Communications in Theoretical Physics, vol. 42, no. 4, pp. 494–496, 2004. View at: Google Scholar | MathSciNet
J. D. Fay, “Riemann’s theta function,” in Theta Functions on Riemann Surfaces, vol. 352 of Lecture Notes in Mathematics, pp. 1–15, Springer, Berlin, Germany, 1973. View at: Publisher Site | Google Scholar
F. Lambert, I. Loris, J. Springael, and R. Willox, “On a direct bilinearization method: Kaup’s higher-order water wave equation as a modified nonlocal Boussinesq equation,” Journal of Physics A: Mathematical and General, vol. 27, no. 15, pp. 5325–5334, 1994. View at: Publisher Site | Google Scholar | MathSciNet
C. Gilson, F. Lambert, J. Nimmo, and R. Willox, “On the combinatorics of the Hirota
D
-operators,” Proceedings of the Royal Society. London. Series A. Mathematical, Physical and Engineering Sciences, vol. 452, no. 1945, pp. 223–234, 1996. View at: Publisher Site | Google Scholar | Zentralblatt MATH | MathSciNet
Copyright © 2014 Huanhe Dong et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. |
IsRightDistributive - Maple Help
Home : Support : Online Help : Mathematics : Algebra : Magma : IsRightDistributive
test whether a finite magma is right distributive
IsRightDistributive( m )
A magma is right distributive if it satisfies the right distributive law (XY)Z = (XZ)(YZ).
The IsRightDistributive command returns true if the given magma is right distributive. It returns false otherwise.
\mathrm{with}\left(\mathrm{Magma}\right):
m≔〈〈〈1|1|1〉,〈2|1|2〉,〈3|3|3〉〉〉
\textcolor[rgb]{0,0,1}{m}\textcolor[rgb]{0,0,1}{≔}[\begin{array}{ccc}\textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{1}\\ \textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{2}\\ \textcolor[rgb]{0,0,1}{3}& \textcolor[rgb]{0,0,1}{3}& \textcolor[rgb]{0,0,1}{3}\end{array}]
\mathrm{IsRightDistributive}\left(m\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
m≔〈〈〈1|2|3〉,〈2|3|3〉,〈3|1|2〉〉〉
\textcolor[rgb]{0,0,1}{m}\textcolor[rgb]{0,0,1}{≔}[\begin{array}{ccc}\textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{3}\\ \textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{3}& \textcolor[rgb]{0,0,1}{3}\\ \textcolor[rgb]{0,0,1}{3}& \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{2}\end{array}]
\mathrm{IsRightDistributive}\left(m\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}}
The Magma[IsRightDistributive] command was introduced in Maple 15. |
Home : Support : Online Help : Programming : Document Tools : Components : Meter Component
generate XML for a Meter Component
Meter( rng, opts )
action : string; A string which parses to one or more valid statements in 1-D Maple Notation. These statements form the Value Changed Action Component Code that executes when the Meter position is manually adjusted.
The Meter command in the Component Constructors package returns an XML function call which represents a Meter Component.
\mathrm{with}\left(\mathrm{DocumentTools}\right):
\mathrm{with}\left(\mathrm{DocumentTools}:-\mathrm{Components}\right):
\mathrm{with}\left(\mathrm{DocumentTools}:-\mathrm{Layout}\right):
Executing the Meter command produces a function call.
S≔\mathrm{Meter}\left(\mathrm{identity}="Meter0"\right)
\textcolor[rgb]{0,0,1}{S}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{_XML_EC-Meter}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{"id"}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{"Meter0"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"lower-bound"}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{"0"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"upper-bound"}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{"100"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"control-position"}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{"0"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"major-tick-spacing"}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{"20"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"minor-ticks"}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{"10"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"show-labels"}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{"true"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"show-ticks"}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{"true"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"continuous-update"}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{"true"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"inputenabled"}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{"true"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"visible"}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{"true"}\right)
\mathrm{xml}≔\mathrm{Worksheet}\left(\mathrm{Group}\left(\mathrm{Input}\left(\mathrm{Textfield}\left(S\right)\right)\right)\right):
\mathrm{InsertContent}\left(\mathrm{xml}\right):
\mathrm{codestring}≔"\ns := true;\n\nt := false;"
\textcolor[rgb]{0,0,1}{\mathrm{codestring}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{"s := true; t := false;"}
S≔\mathrm{Meter}\left(0...9.0,\mathrm{identity}="Meter0",\mathrm{tooltip}="My example Meter",\mathrm{showlabels},\mathrm{position}=3.27,\mathrm{width}=150,\mathrm{height}=75,\mathrm{action}=\mathrm{codestring}\right):
\mathrm{xml}≔\mathrm{Worksheet}\left(\mathrm{Group}\left(\mathrm{Input}\left(\mathrm{Textfield}\left(S\right)\right)\right)\right):
The previous example's call to the InsertContent command inserted a component with identity "Meter0", which still exists in this worksheet. Inserting additional content whose input contains another component with that same identity "Meter0" incurs a substitution of the input identity in order to avoid a conflict with the identity of the existing component.
\mathrm{lookup}≔\mathrm{InsertContent}\left(\mathrm{xml},\mathrm{output}=\mathrm{table}\right)
\textcolor[rgb]{0,0,1}{\mathrm{lookup}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{table}\textcolor[rgb]{0,0,1}{}\left([\textcolor[rgb]{0,0,1}{"Meter0"}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{"Meter1"}]\right)
\mathrm{lookup}["Meter0"]
\textcolor[rgb]{0,0,1}{"Meter1"}
\mathrm{GetProperty}\left(\mathrm{lookup}["Meter0"],\mathrm{value}\right)
\textcolor[rgb]{0,0,1}{3.27}
The DocumentTools:-Components:-Meter command was introduced in Maple 2015. |
A paraglider is flying horizontally at a constant speed.Assume that only two for
A paraglider is flying horizontally at a constant speed.Assume that only two forces act on it in the vertical direction,its weight and a vertical lift
A paraglider is flying horizontally at a constant speed.Assume that only two forces act on it in the vertical direction,its weight and a vertical lift force exerted on its wings by theair. The lift force has a magnitude of 1800 N.
(a) What is the magnitude and direction of the force that theparaglider exerts on the earth ?
(b)If the lift force should suddenly decrease to 1200 N, whatwould be the vertical acceleration of the glider ? For bothquestions, take the upward direction to be the + y direction.
Let us assume that the skier is falling in the - y direction
The Lift force
{F}_{l\phantom{\rule{1em}{0ex}}\text{if}\phantom{\rule{1em}{0ex}}t}=1800N
There are two forces acting on him, 1) the Lift forceacting upwards
2) his wight acting downwards
The Net Force
\sum {F}_{y}=mg-{F}_{l\phantom{\rule{1em}{0ex}}\text{if}\phantom{\rule{1em}{0ex}}t}
\sum {F}_{y}=m{a}_{y}
{a}_{y}=\frac{mg-{F}_{l\phantom{\rule{1em}{0ex}}\text{if}\phantom{\rule{1em}{0ex}}t}}{m}
=\frac{mg-1800}{m}
{a}_{y}=g-\frac{1800}{m}
b) If the vertical lift is reduced to 1200N,
{a}_{y}=g-\frac{1200}{m}
As part of your work out, you lie on your back and push with your feet against a platform attached to two stiff springs arranged side by side so that they are parallel to each other. when u push the platform, you compress the springs. You do 80.0G of work when u compress the springs 0.200m from their uncompressed length.
a) What magnitude of force must u apply to hold the platform in this position?
b) How much additional work must you do to move the platform 0.200m farther, and what maximum force must you apply?
For tests using a ballistocardiograph, a patient lies on a horizontal platform that is supported on jets of air. Because of the air jets, the friction impeding the horizontal motion of the platform is negligible. Each time the heart beats, blood is pushed out of the heart in a direction that is nearly parallel to the platform. Since momentum must be conserved, the body and the platform recoil, and this recoil can be detected to provide information about the heart. For each beat, suppose that 0.050 kg of blood is pushed out of the heart with a velocity of +0.30 m/s and that the mass of the patient and the platform is 85 kg. Assuming that the patient does not slip with respect to the platform, and that the patient and the platform start from rest, determine the recoil velocity.
Force F = (2.0 N) i - (3.0 N) k acts on a pebble with positionvector r = (O.50 m) j - (2.0 m) k, relative to the origin. What isthe resulting torque acting on the pebble about
a.) the origin and
b.) a point with coordinates (2.0 m, 0, - 3.0 m)?
What are the bla?
Often new technology spreads exponentially. Between 1995 and 2005, each year the number of internet domain hosts was 1.43 times the number of hosts in the preceding year. In 1995, the number of hosts was 8.2 million.
a) Explain why the number of hosts is an exponential function of time.
The number of hosts grows by a factor of _____ each year, this is an exponential function because the number is growing by constant multiples.
b) Find a formula for the exponential function that gives the number N of hosts, in millions, as a function of the time t in years since 1995.
N=
c) According to this model, in what year did the number of hosts reach 35 million? (Round your answer to the nearest year.)
A placekicker must kick a football from a point 36.0 m (about 40 yards) from the goal. Half the crowd hopes the ball will clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 20.0 m/s at an angle of 53.08 to the horizontal. (a) By how much does the ball clear or fall short of clearing the crossbar?
A string of length L = 2.9 m and mass m = 0.095 kg is fixed between two stationary points, and when the string is plucked a transverse wave of frequency f = 84 Hz is generated.
Part A: What is the strings linear density,
\rho
, in kilograms per meter?
Part B: If the wavelength is 10.0 cm, which harmonic is this, counting the fundamental as 1?
Part C: For the case described in Part (b), what is the tension in N? |
A Black Box Puzzle Practice Problems Online | Brilliant
This course will teach you how to compute using the principles of quantum mechanics. We’ll start with a simple puzzle and solve it using qubits in two ways:
First, we’ll limit our qubits to their two computational states:
1
0,
just like in a classical circuit.
Next, we’ll try it using qubits that are in a superposition state. To do this you'll build your first quantum circuit and run it to see the quantum advantage.
Let's get started. But first, let's meet the classical version of the puzzle.
A Black Box Puzzle
You’re given four coins. Each one looks real, but you think that one or more of them might be counterfeit, so that it weighs slightly more or less than a real coin.
Thankfully, you’re also given a coin checker: a physical black box that has a mechanism, very similar to a vending machine, that can accept one coin at a time.
Put in a real coin, and it will pass through with no problem. But put in a counterfeit coin and it will set off an alarm, alerting you that the coin was counterfeit.
How many times do you have to use this coin checker to find out which (if any) of the coins are counterfeit?
1
2
3
4
Looking at this puzzle more mathematically, we can think of the coin checker as a black box that computes a function of the four inputs that gets output as the alarm:
1
if the alarm goes off, and
0
if not.
Rather than physically inputting the coins, the black box checker knows which coins are counterfeit and uses this knowledge to output the function
f(x)
and set off the alarm. Black box circuits like this are also known as oracles.
x
is always a string of length
4
that describes which coin is being weighed. A
1
in a given position means that the corresponding coin is being checked.
For example, if the second coin is counterfeit, then
f(0100) = 1
f(1000) = f(0010) = f(0001) = 0.
At this point we're ready to attack the counterfeit coin problem with quantum computing. But first, let's get some basics out of the way.
Reading a quantum circuit left to right, we see the qubits in their initial state, followed by a sequence of quantum gates, followed by a column of measurement devices. The gates operate on qubits to change their quantum state.
When the qubits are measured, their quantum state collapses and the computation is over.
Quantum computing then is about processing information with the quantum state and extracting the final form of that information through the measurements.
Here's the coin puzzle again except, instead of inserting physical coins, each row now corresponds to a qubit.
The first four qubits represent the coins. Each qubit starts in the
\ket{0}
state so, if we want to "insert" a coin, we have to flip the corresponding qubit from
\ket{0}
\ket{1}.
The blue meters on the right are measurement devices, if they display
0\%
it means that the given qubit was measured in the
\ket{0}
state, and
100\%
means that the qubit was measured in the
\ket{1}
The final qubit is the alarm. If it's measured in the
\ket{1}
state, it means that the inserted coin is counterfeit.
If we were operating on classical bits, we would use the
\mathbf{NOT}
operation to flip a
0
1
(or vice versa), but when we operate on qubits we call it an
\mathbf{X}
operation, for reasons we'll see later.
\begin{aligned} \ket{0} &\ce{->[\mathbf{X}]} \ket{1}\\ \ket{1} &\ce{->[\mathbf{X}]} \ket{0}. \end{aligned}
You can "insert coins" one at a time by dragging the
\mathbf{X}
gate onto the corresponding qubit.
Submit the circuit above once you've found the counterfeit coin and set off the alarm qubit.
In the quantum circuit, we do away with strings in favor of kets to describe the state of our qubits.
If we want to "check" the first coin, we can set the state of the first qubit to
\ket{1}
\mathbf{X}
operation, and leave the second, third, and fourth qubits in the
\ket{0}
state. The collective state of the four qubits would then be represented by
\ket{1000}.
\mathbf{X}
gates in the circuit above to check the coins one at a time. When the alarm goes off, the qubits will have some measured state.
\ket{0010}
\ket{0011}
\ket{1111}
We’ve learned that while bits can only be either
0
1,
the state of a qubit can be a combination of the states
\ket{0}
\ket{1}
at the same time: a superposition, like
\dfrac{\ket{0} + \ket{1}}{\sqrt{2}}
This raises a natural question. Instead of checking the first coin, then the second, and so on, can we use a superposition to check every coin at once?
In fact, many quantum algorithms exploit this as part of their operation. When a qubit that's in superposition passes through a circuit, the circuit can operate on all of the states of the superposition simultaneously, so it doesn't require separate queries.
This approach is so commonplace that there's a quantum gate, called
\mathbf{H},
designed to prepare this sort of combination:
\ket{0} \ce{->[\mathbf{H}] \ket{0}} + \ket{1}.
This is one of the two most important quantum gates that you'll use in this course, and it's often the first step in transforming a classical algorithm into its quantum counterpart.
While there are
16
different possible input strings to our checker, using
\mathbf{H}
on each qubit can prepare a superposition state that contains them all.
To store the classical states on the left, we would need a set of
4
physical classical bits to represent each one, or
64
physical classical bits altogether. The quantum state on the right can be stored using just
4
physical qubits.
But when we pass the superposition state to the checker circuit, it produces a muddled mess
(
all of the qubit measurements yield
50\%).
By putting in a combination of every possible input, we’ve produced an incomprehensible output which seems to say that the alarm both does and does not go off.
In a way this isn't surprising — we wouldn't expect a binary light to be able to deliver nuanced feedback like "one of the coins you gave me was counterfeit but not the other".
From here on out, we have to give up on the idea of the fifth qubit as an "alarm" and figure out another way to extract information from the quantum state.
But all is not lost, the quantum state coming out of the checker has changed. Every combination of inputs containing a single counterfeit coin has been marked with an alarm; we just can't see it yet.
Using an advanced feature of the circuit simulator, which you'll use later in the course, we can see the exact quantum superposition state of the output:
How has the black box changed the quantum state? Is there any information we could use to figure out which coins are counterfeit from the quantum state alone?
Yes, the checker circuit has stored information in the signs (
+
-
) of the kets No, there is no information about the coin check in this state
The superposition state that's output from the black box checker contains all the information we need to see which coins are counterfeit.
By comparing some of the input terms to the output ones, we have the following:
\begin{aligned} + \ket{1000} & \rightarrow +\ket{1000} \quad \textrm{No alarm}\\ + \ket{0100} & \rightarrow +\ket{0100} \quad \textrm{No alarm}\\ + \ket{0010} & \rightarrow -\ket{0010} \quad \color{#D61F06} \textrm{Alarm!} \color{#333333}\\ + \ket{0001} & \rightarrow -\ket{0001}\quad \color{#D61F06} \textrm{Alarm!} \color{#333333}. \end{aligned}
In only one query, all the possible coin inputs that set off the alarm have been tagged with a minus sign (
-
There's a big problem here, however. As we learned in the last quiz, it isn't possible to extract the coefficients of a superposition through measurement — you'll only be able to measure a random selection of one of the superposed states (e.g.
\ket{0010}
\ket{1011}
The only time we can be sure that the measurement of a qubit corresponds to the state it was in prior to measurement is when it's in one of the computational states:
\ket{0}
\ket{1}.
Thankfully, the
\mathbf{H}
operation has another trick up its sleeve: not only can it flip a computational state into a superposition, it can also do the opposite to transform different superpositions back into defined computational states:
\begin{aligned} \ket{0} + \ket{1} &\ce{->[\mathbf{H}]} \ket{0} \\ \ket{0} - \ket{1} &\ce{->[\mathbf{H}]} \ket{1}. \end{aligned}
\mathbf{H}
gate to find the counterfeit coins in a single query, then submit the circuit above when you think you've found it.
Hint: all the qubits need to be taken out of the superposition state, including the fifth one.
Preparing a superposition state for the oracle, followed by decoding the mixed jumble of a state it yields back into one of the computational states is a common approach in quantum algorithms. In problems like the counterfeit coin puzzle, a brute force search that could take many queries can be sped up significantly. This approach is called quantum parallelism, and we'll become well-versed in the careful choreography of quantum problem-solving in this course.
The construction of this quantum circuit is not meant to be obvious to you now, and we’ll spend much of this course unpacking how superposition and other quantum properties can be leveraged to speed up computation. Throughout the course, drag-and-drop quantum circuits like this one will allow you to try out the computation for yourself, and get a feel for the operations you’re performing as equations never could. |
Numbers and Operations - Vocabulary - Course Hero
College Algebra/Numbers and Operations/Vocabulary
number that indicates how many times a base is multiplied by itself; shown by a raised number, such as:
4^3=4\cdot4\cdot4
symbols, such as parentheses
(\; )
[ \;]
\lbrace\;\rbrace
, that are used to separate a part of an expression. Some operation symbols, such as fraction bars and radicals, may also function as grouping symbols.
n
in the radical
\sqrt[n]{x}
n
th root of
x
. For example, the index of
\sqrt[3]{x}
is 3. In the radical
\sqrt{x}
, the index is not shown, but it is understood to be 2.
Radical and Power Functions
number in the set of whole numbers and their opposites:
...-\!3,-2,-1,0,1,2,3 ,...
number in the set of nonterminating, nonrepeating decimals, such as
\sqrt{2}
\pi
number in the set of counting numbers:
1,2,3,4,...
set of rules indicating which calculations to perform first to simplify a mathematical expression
\sqrt{\phantom{0}}
used to represent
n
th roots, where
n
is a natural number greater than or equal to 2
expression under a radical
exponent that is a rational number
number in the set that can be written as a fraction of two integers where the denominator is not zero, such as
\frac{1}{2}
-\frac{2}{3}
rewriting a fraction to remove radical expressions from the denominator
number in the set of all rational and irrational numbers
notation used to express a number as the product of two factors. The first is a number greater than or equal to 1 and less than 10, and the second is a power of 10.
number in the set that includes zero and the counting numbers
1, 2, 3, 4,...
<Overview>Sets of Numbers |
Fermat's Last Theorem - Simple English Wikipedia, the free encyclopedia
Fermat's Last Theorem or FLT is a very famous idea in mathematics. It says that:
{\displaystyle n}
is a whole number larger than 2, then the equation
{\displaystyle x^{n}+y^{n}=z^{n}}
has no solutions when x, y and z are natural numbers.
It is impossible to express in whole numbers two cubes, which added equal a third cube. Furthermore, it is impossible with anything higher than squares.
This means that there are no examples where
{\displaystyle x}
{\displaystyle y}
{\displaystyle z}
are natural numbers, i.e. whole numbers larger than zero, and where
{\displaystyle n}
is a whole number bigger than 2. Pierre de Fermat wrote about it in 1637 inside his copy of a book called Arithmetica. He said "I have a proof of this theorem, but there is not enough space in this margin". However, no correct proof was found for 357 years. It was finally proven in 1995. Most mathematicians do not think that Fermat, in fact, ever had a margin proof of this theorem.
In its original the problem is as follows:
Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos & generaliter nullam in infinitum ultra quadratum potestatem in duos eiusdem nominis fas est dividere cuius rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caperet.
1.1 On the equation's quadratic
2.1 Criticism of proof
2.2 Proof without elliptic
Fermat's Last Theorem is a more general form of the Pythagorean theorem,[1] which is an equation that says:
{\displaystyle a^{2}+b^{2}=c^{2}}
{\displaystyle a}
{\displaystyle b}
{\displaystyle c}
are whole numbers this is called a "Pythagorean triple". For example,
{\displaystyle 3^{2}+4^{2}=9+16=25}
{\displaystyle {\sqrt[{2}]{25}}=5}
{\displaystyle 3^{2}+4^{2}=5^{2}}
is a Pythagorean triple. Fermat's Last Theorem rewrites this as
{\displaystyle x^{n}+y^{n}=z^{n}}
and claims that, if you make the
{\displaystyle n}
a larger whole number than 2, then
{\displaystyle a}
{\displaystyle b}
{\displaystyle c}annot all be natural numbers. For example,
{\displaystyle 3^{3}+4^{3}=27+64=91}
{\displaystyle {\sqrt[{3}]{91}}=4.49794144528}
{\displaystyle 3^{3}+4^{3}=4.49794144528^{3}}
is an example that confirms this.
On the equation's quadratic[change | change source]
The x and y are two unknown sums, summing imaginary third sum z. Despite there being 4 terms: n, x, y & z, the n is a function summing the total of unknown sums. Zero is missing from this equation by the rule of "1 plus 1 is 2 and no more", written 1+1=2+0.
To give clarification, the n is known to be a sum.
Proof[change | change source]
British mathematician Andrew Wiles
The proof was made for some values of
{\displaystyle n}
{\displaystyle n=3}
{\displaystyle n=4}
{\displaystyle n=5}
{\displaystyle n=7}
, which was managed by many mathematicians including Fermat, Euler, Sophie Germain. However, since there are an infinite number of Pythagorean triples,[2] as numbers count upwards forever, this made Fermat's Last Theorem hard to prove or disprove; the full proof must show that the equation has no solution for all values of
{\displaystyle n}
(whe{\displaystyle n}
is a whole number bigger than 2) but it is not possible to simply check every combination of numbers if they continue forever.
An English mathematician named Andrew Wiles found a solution in 1995, 358 years after Fermat wrote about it.[3][4][5] Richard Taylor helped him find the solution[source?]. The proof took eight years of research. He proved the theorem by first proving the modularity theorem, which was then called the Taniyama–Shimura conjecture. Using Ribet's Theorem, he was able to give a proof for Fermat's Last Theorem. He received the Wolfskehl Prize from Göttingen Academy in June 1997: it amounted to about $50,000 U.S. dollars.[6]
After a few years of debate, people agreed that Andrew Wiles had solved the problem. Andrew Wiles used a lot of modern mathematics and even created new maths when he made his solution. This mathematics was unknown when Fermat wrote his famous note, so de Fermat could not have used it. This leads one to believe that de Fermat did not in fact have a complete solution of the problem.
Criticism of proof[change | change source]
Vos Savant wrote in 1995, that Wiles' proof should be rejected for its use of non-Euclidean geometry. She said, "the chain of proof is based in hyperbolic (Lobachevskian) geometry", and because this geometry allows things like squaring the circle, a "famous impossibility" despite being possible in hyperbolic geometry, then "if we reject a hyperbolic method of squaring the circle, we should also reject a hyperbolic proof of Fermat's last theorem."
Proof without elliptic[change | change source]
Where n is known to sum two ordinal values, it cannot exceed the counted value 2 if the larger is taken as 1 unit.
Generalization[change | change source]
Beal's Generalization Conjecture, or the Beal Conjecture, posed by investor Andrew Beal, asks why there are always common factors (like cells in batteries), in equations like this, of the general form aˣ+bʸ=cᶻ.
↑ Stillwell J (2003). Elements of Number Theory. New York: Springer-Verlag. pp. 110–112. ISBN 0-387-95587-9.
↑ Wiles, Andrew (1995). "Modular elliptic curves and Fermat's Last Theorem" (PDF). Annals of Mathematics. 141 (3): 443–551. doi:10.2307/2118559. JSTOR 2118559. OCLC 37032255. Archived from the original (PDF) on 2011-05-10. Retrieved 2013-04-20.
↑ Taylor R, Wiles A (1995). "Ring theoretic properties of certain Hecke algebras". Annals of Mathematics. 141 (3): 553–572. doi:10.2307/2118560. JSTOR 2118560. OCLC 37032255. Archived from the original on 2001-11-27. Retrieved 2011-10-18.
↑ Neil Pieprzak. "Fermat's last theorem and Andrew Wiles". Plus Magazine. Retrieved 2012-04-30.
↑ Singh S 1998. Fermat's Enigma, p284. New York: Anchor Books. ISBN 978-0-385-49362-8
Aczel, Amir (30 September 1996). Fermat's Last Theorem: unlocking the secret of an ancient mathematical problem. Four Walls Eight Windows. ISBN 978-1-568-58077-7.
Friberg, Joran (2007). Amazing traces of a Babylonian origin in Greek mathematics. World Scientific Publishing Company. ISBN 978-9812704528.
Kleiner I (2000). "From Fermat to Wiles: Fermat's Last Theorem becomes a theorem" (PDF). Elem. Math. 55: 19–37. doi:10.1007/PL00000079. S2CID 53319514. Archived from the original (PDF) on 2012-02-19. Retrieved 2011-08-17.
Mordell L.J (1921). Three Lectures on Fermat's Last Theorem. Cambridge: Cambridge University Press.
Ribenboim P (2000). Fermat's Last Theorem for amateurs. New York: Springer-Verlag. ISBN 978-0387985084.
Singh, Simon (October 1998). Fermat's Enigma. New York: Anchor Books. ISBN 978-0-385-49362-8.
Daney, Charles (1997-10-29). "The Mathematics of Fermat's Last Theorem". Archived from the original on 2011-08-13. Retrieved 2011-08-17.
Kleiner, Israel (2000). "From Fermat to Wiles: Fermat's Last Theorem becomes a theorem" (PDF). Elemente der Mathematik. Archived from the original (PDF) on 2012-02-19. Retrieved 2011-08-17.
Shay, David (2003). "Fermat's Last Theorem". Archived from the original on 2012-02-27. Retrieved 2011-08-17.
Weisstein, Eric W. "Fermat's Last Theorem". MathWorld. Retrieved 2011-08-17.
Retrieved from "https://simple.wikipedia.org/w/index.php?title=Fermat%27s_Last_Theorem&oldid=8160717" |
In Bohr's Orbit,
\frac{\mathrm{nh}}{2\mathrm{\pi }}
Angular momentum =
\mathrm{m\nu r}=\quad \frac{\mathrm{nh}}{2\mathrm{\pi }}
\frac{{\mathrm{K}}_{\mathrm{p}}}{{\mathrm{K}}_{\mathrm{c}}}
for following reaction will be
{\mathrm{CO}}_{\left(\mathrm{g}\right)}+\frac{1}{2}\quad {\mathrm{O}}_{2\left(\mathrm{g}\right)}\to {\mathrm{CO}}_{2\left(\mathrm{g}\right)}
\frac{1}{\mathrm{RT}}
\frac{1}{\sqrt{\mathrm{RT}}}
\frac{\mathrm{RT}}{2}
\frac{1}{\sqrt{\mathrm{RT}}}
{∆}_{{\mathrm{n}}_{\mathrm{g}}}={\mathrm{n}}_{\mathrm{p}}-{\mathrm{n}}_{\mathrm{r}\quad }=\quad 1-\frac{3}{2}
{∆}_{{\mathrm{n}}_{\mathrm{g}}}=\quad \frac{-1}{2}
. Hence KP = Kc(RT)-1/2
\frac{{\mathrm{K}}_{\mathrm{p}}}{{\mathrm{K}}_{\mathrm{c}}}=\frac{1}{\left(\mathrm{RT}{\right)}^{1/2}}=\frac{1}{\sqrt{\mathrm{RT}}}
Arrange the following gases in order of their critical temperature.
NH3 > H2O > CO2 > O2
O2 > CO2 > H2O > NH3
H2O > NH3 > CO2 > O2
CO2 >O2 >H2O > NH3
Greater are the intermolecular forces of attraction, higher is the critical temperature.
Threshold frequency of a metal is 5
\text{×}
1013 s-1 upon which 1 x 1014s-1 frequency light is focused. Then the maximum kinetic energy of emitted electron is
\text{×}
\text{×}
\text{×}
\text{×}
\text{×}
Following the conservation of energy principle,
\left[\frac{1}{2}{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}^{2}\right]
= h(
\text{v}
{\mathrm{v}}_{\circ }
=(6.626
\text{×}
10-34) (1014s-1 - 5
\text{×}
1013 s-1)
=(6.626 x 10-34J s) (5 x 1013 s-1)
=3.313 x 10-20J.
The dipole moment is minimum in
BF3 has zero dipole moment
Bond dissociation energy of CH4 is 360 kJ/mol and C2H6 is 620 kJ/mol. Then bond dissociation energy of C-C bond is
Dissociation energy of methane = 360 kJ mol-1
Bond energy of C-H bond =
\frac{360}{4}=90\quad \mathrm{kJ}
Bond energy of ethane,
1(C-C) 6 (C-H) =620 kJ/mol
(C-C) + 6
\times
(C-C) + 540 = 62
C-C = 80 kJ mo1-1
Bond dissociation of C-C bond= 80 kJ moJ-1.
At 60° and 1 atm, N2O4 is 50% dissociated into NO2 then KP is
{\mathrm{N}}_{2}{\mathrm{O}}_{4}\quad \rightleftharpoons \quad 2{\mathrm{NO}}_{2}
At equili. 1-
\mathrm{\alpha }
\mathrm{\alpha }
N2O4 is 50% dissociated, so
\mathrm{\alpha }
\frac{1}{2}
{\mathrm{K}}_{\mathrm{p}}=\frac{{\mathrm{p}}_{{\mathrm{NO}}_{2}}^{2}}{{\mathrm{p}}_{{\mathrm{N}}_{2}{\mathrm{O}}_{4}}}=\frac{(2\times {\displaystyle \frac{1}{2}}{)}^{2}}{(1-{\displaystyle \frac{1}{2}})}=2\quad \mathrm{atm}
pKa increases in benzoic acid when substituent "x" is bonded at para-position, then "x" is
Larger the value of pKa smaller will be its acidity. Out of the four groups, -COOH, -NO2 and -CN are e- withdrawing which makes benzoic acid more acidic whereas -OCH3 is e- donating which reduces the acidity (makes H less easily available). pKa value increases if OCH3 is present at para-position of benzoic acid.
At equilibrium which is correct?
\text{∆}
\text{∆}
\text{∆}
\text{∆}
{\text{G}}^{°}
\text{∆}
In unfavorable reaction have Delta G values that are positive (also called endergonic reaction). When Delta G for reaction is zero, a reaction is said to be at equilibrium. Equilibrium does not mean equal concentrations. If the Delt G is zero, there is no net change in A and B, as the system is at equilibrium.
Which has the highest pH?
NH4Cl solution is acidic, its pH < 7. NaNO3 solution is neutral, its pH = 7. CH3COOK and Na2CO4solutions are basic their pH > 7. But Na2CO3 solution is more basic, its pH > pH of CH3COOK solution. |
Inverse gas chromatography - Wikipedia
Analytical gas chromatography A (top) compared with inverse gas chromatography B (bottom). While in gas chromatography a sample containing multiple species is separated into its components on a stationary phase, Inverse gas chromatography uses injection of a single species to probe the characteristics of a stationary phase sample.
Inverse gas chromatography is a physical characterization analytical technique that is used in the analysis of the surfaces of solids.[1]
Inverse gas chromatography or IGC is a highly sensitive and versatile gas phase technique developed over 40 years ago to study the surface and bulk properties of particulate and fibrous materials. In IGC the roles of the stationary (solid) and mobile (gas or vapor) phases are inverted from traditional analytical gas chromatography (GC). In GC, a standard column is used to separate and characterize several gases and/or vapors. In IGC, a single gas or vapor (probe molecule) is injected into a column packed with the solid sample under investigation. Instead of an analytical technique, IGC is considered a materials characterization technique.
During an IGC experiment a pulse or constant concentration of a known gas or vapor (probe molecule) is injected down the column at a fixed carrier gas flow rate. The retention time of the probe molecule is then measured by traditional GC detectors (i.e. flame ionization detector or thermal conductivity detector). Measuring how the retention time changes as a function of probe molecule chemistry, probe molecule size, probe molecule concentration, column temperature, or carrier gas flow rate can elucidate a wide range of physico-chemical properties of the solid under investigation. Several in depth reviews of IGC have been published previously.[2][3]
IGC experiments are typically carried out at infinite dilution where only small amounts of probe molecule are injected. This region is also called Henry's law region or linear region of the sorption isotherm. At infinite dilution probe-probe interactions are assumed negligible and any retention is only due to probe-solid interactions. The resulting retention volume, VRo, is given by the following equation:
{\displaystyle V_{R}^{\circ }={\frac {j}{m}}F(t_{R}-t_{o}){\frac {T}{273.15}}}
where j is the James–Martin pressure drop correction, m is the sample mass, F is the carrier gas flow rate at standard temperature and pressure, tR is the gross retention time for the injected probe, to is the retention time for a non-interaction probe (i.e. dead-time), and T is the absolute temperature.
1 Surface energy determination
1.2 Polymers and coatings
1.6 Metakaolins
Surface energy determination[edit]
The main application of IGC is to measure the surface energy of solids (fibers, particulates, and films). Surface energy is defined as the amount of energy required to create a unit area of a solid surface; analogous to surface tension of a liquid. Also, the surface energy can be defined as the excess energy at the surface of a material compared to the bulk. The surface energy (γ) is directly related to the thermodynamic work of adhesion (Wadh) between two materials as given by the following equation:
{\displaystyle W_{\mathrm {adh} }=2(\gamma _{1}\gamma _{2})^{1/2}}
where 1 and 2 represent the two components in the composite or blend. When determining if two materials will adhere it is common to compare the work of adhesion with the work of cohesion, Wcoh = 2γ. If the work of adhesion is greater than the work of cohesion, then the two materials are thermodynamically favored to adhere.
Surface energies are commonly measured by contact angle methods. However, these methods are ideally designed for flat, uniform surfaces. For contact angle measurements on powders, they are typically compressed or adhered to a substrate which can effectively change the surface characteristics of the powder. Alternatively, the Washburn method can be used, but this has been shown to be affected by column packing, particle size, and pore geometry.[4] IGC is a gas phase technique, thus is not subject to the above limitations of the liquid phase techniques.
To measure the solid surface energy by IGC a series of injections using different probe molecules is performed at defined column conditions. It is possible to ascertain both the dispersive component of the surface energy and acid-base properties via IGC. For the dispersive surface energy, the retention volumes for a series of n-alkane vapors (i.e. decane, nonane, octane, heptanes, etc.) are measured. The Dorris and Gray.[5] or Schultz [6] methods can then be used to calculate the dispersive surface energy. Retention volumes for polar probes (i.e. toluene, ethyl acetate, acetone, ethanol, acetonitrile, chloroform, dichloromethane, etc.) can then be used to determine the acid-base characteristics of the solid using either the Gutmann,[7] or Good-van Oss theory.[8]
Other parameters accessible by IGC include: heats of sorption [1], adsorption isotherms,[9] energetic heterogeneity profiles,[10][11] diffusion coefficients,[12] glass transition temperatures [1],[13] Hildebrand [14][15] and Hansen [16] solubility parameters, and crosslink densities.[17]
IGC experiments have applications over a wide range of industries. Both surface and bulk properties obtained from IGC can yield vital information for materials ranging from pharmaceuticals to carbon nanotubes. Although surface energy experiments are most common, there are a wide range of experimental parameters that can be controlled in IGC, thus allowing the determination of a variety of sample parameters. The below sections highlight how IGC experiments are utilized in several industries.
Polymers and coatings[edit]
IGC has been used extensively for the characterization of polymer films, beads, and powders. For instance, IGC was used to study surface properties and interactions amongst components in paint formulations.[18] Also, IGC has been used to investigate the degree of crosslinking for ethylene propylene rubber using the Flory–Rehner equation [17]. Additionally, IGC is a sensitive technique for the detection and determination of first and second order phase transitions like melting and glass transition temperatures of polymers.[19] Although other techniques like differential scanning calorimetry are capable of measuring these transition temperatures, IGC has the capability of glass transition temperatures as a function of relative humidity.[20]
The increasing sophistication of pharmaceutical materials has necessitated the use for more sensitive, thermodynamic based techniques for materials characterization. For these reasons, IGC, has seen increased use throughout the pharmaceutical industry. Applications include polymorph characterization,[21] effect of processing steps like milling,[22] and drug-carrier interactions for dry powder formulations.[23] In other studies, IGC was used to relate surface energy and acid-base values with triboelectric charging[24] and differentiate the crystalline and amorphous phases [23].
Surface energy values obtained by IGC have been used extensively on fibrous materials including textiles,[25] natural fibers,[26] glass fibers,[27] and carbon fibers.[28] Most of these and other related studies investigating the surface energy of fibers are focusing on the use of these fibers in composites. Ultimately, the changes in surface energy can be related to composite performance via the works of adhesion and cohesion discussed previously.
Similar to fibers, nanomaterials like carbon nanotubes, nanoclays, and nanosilicas are being used as composite reinforcement agents. Therefore, the surface energy and surface treatment of these materials has been actively studied by IGC. For instance, IGC has been used to study the surface activity of nanosilica, nanohematite, and nanogeoethite.[29] Further, IGC was used to characterize the surface of as received and modified carbon nanotubes.[30]
Metakaolins[edit]
IGC was used to characterize the adsorption surface properties of calcined kaolin (metakaolin) and the grinding effect on this material.[31]
Other applications for IGC include paper-toner adhesion,[32] wood composites,[33] porous materials [3], and food materials.[34]
^ Mohammadi-Jam, S.; Waters, K.E. (2014). "Inverse gas chromatography applications: A review". Advances in Colloid and Interface Science. 212: 21–44. doi:10.1016/j.cis.2014.07.002. ISSN 0001-8686.
^ J. Condor and C. Young, Physicochemical measurement by gas chromatography, John Wiley and Sons, Chichester, UK (1979)
^ F. Thielmann, Journal of Chromatography A. 1037 (2004) 115.
^ J. Dove, G. Buckton, and C. Doherty, International Journal of Pharmaceutics. 138 (1996) 199–206.
^ G.M. Doris and D.G. Gray, Journal of Colloids and Interfacial Science. 56 (1964) 353.
^ J. Schultz, L. Lavielle, and C. Martin, Journal of Adhesion. 77 (1980) 353–362.
^ V. Gutmann, Coordination Chemistry Reviews. 2 (1966) 239–256.
^ C.J. van Oss, R.J. Good, and M.K. Chaudhury, Langmuir. 4 (1988) 884–891.
^ E. Cremer and H. Huber, in Gas Chromatography., ed. N. Brenner, et al., Academic Press, New York (1962) p 169.
^ P.P. Yla-Maihaniemi, J.Y.Y. Heng, F. Thielmann, and D.R. Williams, Langmuir. 24 (2008) 9551–9557.
^ F. Thielmann, D.J. Burnett, and J.Y.Y. Heng, Drug Development and Industrial Pharmacy. 33 (2007) 1240–1253.
^ J. van Deemter, F.J. Zuiderweg, and A. Klinkenberg, Chemical Engineering Science. 5 (1965) 271.
^ G. Buckton, A. Ambarkhane, and K. Pincott, Pharmaceutical Research. 21 (2004) 1554–1557.
^ 14 D. Benczedi, D, I. Tomka, and F. Escher, Macromolecules. 31 (1998) 3055.
^ G. DiPaola and J.E. Guillet, Macromolecules. 11 (1978) 228.
^ K. Adamska and A. Voelkel, International Journal of Pharmaceutics. 304 (2005) 11–17.
^ G.J. Price, K.S. Siow, and J.E. Guillet, Macromolecules. 22 (1989) 3116–3119.
^ A. Ziani, R. Xu, H.P. Schreiber, and T. Kobayashi, Journal of Coatings Technology. 71 (1999) 53–60.
^ A. Lavole and J.E. Guillet, Macromolecules. 2 (1969) 443.
^ F. Thielmann and D.R. Williams, Deutsche Lebensmittel-Rundschau. 96 (2000) 255–257.
^ H.H.Y. Tong, B.Y. Shekunov, P. York, and A.H.L Chow, Pharmaceutical Research. 19 (2002) 640–648.
^ J.Y.Y. Heng, F. Thielmann, and D.R. Williams, Pharmaceutical Research, 23 (2006) 1918–1927.
^ J. Feeley, P. York, B. Sumby, and H. Dicks, International Journal of Pharmaceutics. 172 (1998) 89–96.
^ N. Ahfat, G. Buckton, R. Burrows, and M. Ticehurst, European Journal of Pharmaceutical Sciences. 9 (2000) 271–276.
^ E. Cantergiani and D. Benczedi, Journal of Chromatography A. 969 (2002) 103–110.
^ J.Y.Y. Heng, D.F. Pearse, F. Thielmann, T. Lampke, and A. Bismarck, Composite Interfaces. 14 (2007) 581–604.
^ K. Tsutsumi and T. Ohsuga, Colloid and Polymer Science. 268 (1990) 38–44.
^ L. Lavielle and J. Schultz, Langmuir. 7 (1991) 978–981.
^ K. Batko and A. Voelkel, Journal of Colloid and Interface Science. 315 (2007) 768–771.
^ R. Menzel, A. Lee, A. Bismarck, and M.S.P. Shaffer, Langmuir. (2009) in press.
^ Gamelas, J.; Ferraz, E.; Rocha, F. (2014). "An insight into the surface properties of calcined kaolinitic clays: the grinding effect". Colloids and Surfaces A: Physicochemical and Engineering Aspects. 455: 49–57. doi:10.1016/j.colsurfa.2014.04.038.
^ J. Borch, Journal of Adhesion Science and Technology. 5 (1991) 523–541.
^ R.H. Mills, D.J. Gardner, and R. Wimmer. Journal of Applied Polymer Science. 110 (2008) 3880–3888.
^ Q. Zhou and K.R. Cadwallader. Journal of Agricultural and Food Chemistry. 54 (2006) 1838–1843.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Inverse_gas_chromatography&oldid=1031350794" |
Deception Practice Problems Online | Brilliant
Statistics is such a powerful tool because it uses data to convey important information. This information may be visualized as a graph, chart, or table. It might be given numerically as the mean, median, mode, or percent change. There are so many different ways to frame information.
Because people can choose how to show statistics, sometimes it can be deceiving. Let's look at some examples.
The median is a measure of central tendency. It is the middle value in a set of numbers arranged from least to greatest.
The mode is the value that occurs most in a data set.
The percent change is how much something increases or decreases in terms of its initial amount. The formula is:
\frac{new\text{ }value - old\text{ }value}{old\text{ }value}
From the data above, what can we conclude about home prices in the United States? Select all that apply.
The median home price tripled from March 2009 to January 2011. The median home price increased by $50,000 from March 2009 to January 2011. The median home price increased at a constant rate from March 2009 to January 2011.
A study is done on 250 people who claim they have psychic abilities.
A regular deck of cards has an equal number of red and black cards. The deck is shuffled, and each person predicts the color of the top card. The top card is then flipped face up to see if they are correct. This is repeated 8 times in a row.
One participant gets all 8 predictions correct! The researcher claims that this person has exhibited psychic abilities.
Based on just the way the experiment is designed, is the researcher's claim a viable hypothesis? (Note: we're not judging the plausibility of psychic ability in itself, just this particular experiment's design.)
Deception in statistics can be quite intentional. In the first problem, the choice of what months to display did not happen by accident! The researcher probably wanted to convey the data in a certain way.
Deception can also be unintentional. In the second problem, the researcher examining psychic phenomena may have been earnest about their hypothesis, but they still interpreted the results of their experiment poorly.
Alternatively, sometimes the statistics themselves can deceive. It depends who gathers the data and how they do it. The questions that follow will demonstrate this issue.
The Transportation Association of Canada wanted to study the effects of wearing a helmet on motorcycle accidents. The data below was collected by Canadian hospitals between 2002 and 2014. Each motorcyclist who was in a traffic accident and arrived at a hospital had the information below recorded.
Wearing helmet? Yes No
Arrived in ambulance 18% 9%
Had concussion 9% 3%
Needed blood transfusion 0.8% 0.4%
Based on the table, what is true?
A. Motorcyclists who came to a Canadian hospital that were wearing helmets in traffic accidents were twice as likely to arrive in an ambulance than motorcyclists who were not wearing helmets.
B. Motorcyclists who came to a Canadian hospital that were wearing helmets in traffic accidents were three times as likely to have a concussion than motorcyclists who were not wearing helmets.
C. 0.8% of all Canadian motorcycle riders who wear helmets will need a blood transfusion.
(Data source: Journal of Internal Medicine.)
Only A Only B A and B A, B, and C
The previous problem showed how it is important to remember the data source. Not everyone who gets in an accident goes to the hospital!
The table below shows all data regarding motorcycle accidents in Canada between 2002 and 2014.
Which quadrants would be missing from the hospital's data?
A and C B and D A and B C and D
The data from the previous two questions demonstrates what's known as Berkson's bias. In taking what appears to be a representative sample, a specific group may still be excluded. Hospitals won't see those who are healthy enough to not need a hospital, so the data won't include this part of the population.
Recall that the table below shows all motorcycle accidents in Canada between 2002 and 2014.
Look at all accidents involving motorcyclists wearing helmets. What percentage of these accidents did the hospital see?
Statistics cannot be done in a mathematical vacuum. Usually it involves real-life knowledge about the circumstances. This can be as important as any sort of calculation.
When looking at graphs or charts, it is important to inspect the axes and labels. When analyzing data, it is important to remember the data source and how the data was collected. This can help you recognize deceptive statistics.
It may seem, at this point, that statistics can only serve to confound. However, you'll find statistics can be used to pierce deception rather than just create it — continue the course to find out how! |
Modifier | PoE Wiki
3.3 Scourge mods
Non-unique Synthesised items were exclusive to the Synthesis league. They can be found in other challenge leagues content like Heist and Harvest.
Regal OrbRegal OrbStack Size: 10Upgrades a magic item to a rare itemRight click this item then left click a magic item to apply it. Current modifiers are retained and a new one is added.
Shift click to unstack. upgraded it to rare item and enchanted an additional suffix modifier.
Eldritch currency - Eldritch Ichors and Eldritch Embers from Eater of Worlds or Searing Exarch-influenced monsters, respectively, will replace the item's implicit modifier with an Eldritch mod, but will retain the Eldritch mod of the other type.
Labyrinth Enchantments are obtained from the Divine Font at the end of the Labyrinth. Gloves, Boots, and Helmets can receive Labyrinth enchantments. Belts can also be given enchantments in the Eternal Labyrinth of Potential.
Certain Harvest crafting options allows you to add enchantments to weapons, body armours, maps and flasks. Read Wild Brambleback, Primal Cystcaller, Vivid Nestback and Vivid Striketail for details.
Weapons and body armour with enchantments are possible rewards in Grand Heists. The enchantments can be added by using Tailoring OrbTailoring OrbStack Size: 20Adds or replaces an enchantment on a body armour. This may reforge the body armour's sockets.Right click this item then left click on the item you wish to modify.
Scourge mods
If an item becomes Scourged using the Blood Crucible, it gains two Scourge implicit modifiers, one positive and one negative. Scourge mods are separate from implicit mods and enchantments. Blood Crucible upgrades can cause an item to gain additional Scourge implicit mods.
Monsters have different means of receiving modifiers:
Magic packs of monsters spawn with one Archnemesis modifier
Rare monsters spawn with at least two Archnemesis modifiers. At Act 6, they can have up to three, and in Maps, they can have up to four.
Magic and Rare monsters receive a Delirium modifier during a Delirium encounter; they are removed when the encounter is over
Picking up a Talisman gives monsters a Talisman modifier (Talisman league only as well as specific Delve node that contains a Talisman-possessed monster)
Shift click to unstack. , Regal OrbRegal OrbStack Size: 10Upgrades a magic item to a rare itemRight click this item then left click a magic item to apply it. Current modifiers are retained and a new one is added.
Shift click to unstack. , and Ancient OrbAncient OrbStack Size: 20Reforges a unique item as another of the same item classRight click this item then left click a unique item to apply it.
Shift click to unstack. and Exalted OrbExalted OrbStack Size: 10Augments a rare item with a new random modifierRight click this item then left click a rare item to apply it. Rare items can have up to six random modifiers.
Shift click to unstack. may be used to add a random modifier on Magic and Rare items respectively
Orb of AnnulmentOrb of AnnulmentStack Size: 20Removes a random modifier from an itemRight click this item then left click on a magic or rare item to apply it.
Item The mod is generated on items (but not jewels or flasks, etc)
Atlas The mod is generated by Sextants.
Leaguestone The mod is generated by leaguestones. Whole leaguestone mechanic is not added to the core gameplay
Synthesis Area The mod is generated on Synthesis unique map, due to the map item mod Map has # additional random Modifiers; used to be part of Synthesis instance building mechanic
Synthesis Area (Global) The mod is generated on Synthesis unique map, due to the map item mod Map has # additional Synthesis Global Modifiers; used to be part of Synthesis instance building mechanic
Heist Gear The mod is generated on rogue's equipment
veiled modifier The mod is generated on item that drops from Immortal Syndicate
Expedition Remnants
Eldritch Altar monster augmentation mod from altar in Searing Exarch or Eater of Worlds-influenced map
Synthesis unknown
Synthesis Globals Synthesis global area mod. Now only used on Synthesis unique maps
Synthesis Bonus
Blight Tower
Monster Affliction
Scourge Benefit
Scourge Detriment
Scourge Gimmick
Eater of Worlds generated from Eldritch Ichor currencies , etc
- Icius Perandus, Antiquities Collection, Item 202 normally has a level requirement of 1, however putting the 0.5% of Damage Leeched as Life if you've Killed Recently enchantment from the 3rd labyrinth on it, will increase the level requirement to
\color [rgb]{0.6392156862745098,0.5529411764705883,0.42745098039215684}\left\lceil {66\times 0.8}\right\rceil =52
Glorious PlateGlorious PlateArmour: (776-892)Requires Level 68, 191 Str has the tags str_armour • body_armour • armour • default.
Assassin's GarbAssassin's GarbEvasion: (737-811)Requires Level 68, 183 Dex3% increased Movement Speed has the tags dex_armour • body_armour • armour • default.
Result for Glorious PlateGlorious PlateArmour: (776-892)Requires Level 68, 191 Str : The mod can spawn because it has the tag str_armour and a positive spawn weight.
Result for Assassin's GarbAssassin's GarbEvasion: (737-811)Requires Level 68, 183 Dex3% increased Movement Speed : The mod can not spawn because only default matches and its spawn weight is 0.
Currency such as Exalted OrbExalted OrbStack Size: 10Augments a rare item with a new random modifierRight click this item then left click a rare item to apply it. Rare items can have up to six random modifiers.
Shift click to unstack. , as well as beastcrafting recipes "add affix, remove affix", interact with item which has metamod of Spellcraft - Cannot roll Attack Modifiers (Suffix) or of Weaponcraft - Cannot roll Caster Modifiers (Suffix). Which Exalted Orb would not add mod to the item with attack mod tag (of Spellcraftof Spellcraft
de:Modifikatoren ru:Модификатор
Retrieved from ‘https://www.poewiki.net/w/index.php?title=Modifier&oldid=1168536’ |
A 1.3 kg toaster is not plugged in. The coefficient ofstatic friction between th
A 1.3 kg toaster is not plugged in. The coefficient ofstatic friction between the toaster and a horizontal countertop is 0.35. To make the toaster sta
A 1.3 kg toaster is not plugged in. The coefficient ofstatic friction between the toaster and a horizontal countertop is 0.35. To make the toaster start moving, you carelesslypull on its electric cord.
PART A: For the cord tension to beas small as possible, you should pull at what angle above thehorizontal?
PART B: With this angle, how largemust the tension be?
Viktor Wiley
\mathrm{cos}
x = u.m.g then
F=um\frac{g}{\mathrm{cos}x}=\left(0.35\right)\left(1.3\right)\frac{9.8}{\mathrm{cos}x}
F=\frac{4.46}{\mathrm{cos}x}
a. taking differential of this we have :
{F}^{\prime }=0\left(\mathrm{cos}x\right)-\left(4.46\right)\frac{-\mathrm{sin}x}{{\mathrm{cos}}^{2}x}=4.46\frac{\mathrm{sin}x}{{\mathrm{cos}}^{2}x}
taken its stationary we'll have
{F}^{\prime }=4.46\frac{\mathrm{sin}x}{{\mathrm{cos}}^{2}x}=0\to \mathrm{sin}x=0\to x={0}^{o}
F=\frac{4.46}{\mathrm{cos}0}=\frac{4.46}{1}=4.46N
An airplane propeller is 2.08 m in length (from tip to tip) and has a mass of 117 kg. When the airpline's engine is first started, it applies a constant torque of
1950\text{ }N\cdot m
to the propeller, which starts from rest.
a) What is the angular acceleration of the propeller? Model the propeller as a slender rod.
b) What is the propeller's angular speed after making 5.00 revolutions?
c) How much work is done by the engine during the first 5.00 revolutions?
e) What is the instantaneous power output of the motor at the instant that the propeller has turne through 5.00 revolutions?
A 17.8 kg flood light in a park is supported at the end of a horizontal beam of negligible mass that is hinged to a pole. A cable connected to the beam helps to support the light. (In thefigure,
\theta
= 34.2 degrees. Find (A) The tension in the cable and (B) the horizontal and (C) vertical forces exerted on the beamby the pole.
An exceptional standing jump would raise a person 0.80 m off the ground. To do this, what force must a 66-kg person exert against the ground? Assume the person crouches a distance of 0.20 m prior to jumping, and thus the upward force has this distance to act over before he leaves the ground.
A radar station, located at the origin of xz plane, as shown in the figure , detects an airplane coming straight at the station from the east. At first observation (point A), the position of the airplane relative to the origin is
\stackrel{\to }{{R}_{A}}
. The position vector
\stackrel{\to }{{R}_{A}}
has a magnitude of
{360}^{m}
and is located at exactly
{40}^{\circ }
above the horizon. The airplane is tracked for another
{123}^{\circ }
} in the vertical east-west plane for
{5.0}^{s}
, until it has passed directly over the station and reached point B. The position of point B relative to the origin is
\stackrel{\to }{{R}_{B}}
(the magnitude of
\stackrel{\to }{{R}_{B}}
{880}^{m}
). The contact points are shown in the diagram, where the x axis represents the ground and the positive z direction is upward.
Define the displacement of the airplane while the radar was tracking it:
{\stackrel{\to }{R}}_{BA}={\stackrel{\to }{R}}_{B}-{\stackrel{\to }{R}}_{A}
. What are the components of
{\stackrel{\to }{R}}_{BA}
{\stackrel{\to }{R}}_{BA}
in meters as an ordered pair, separating the x and z components with a comma, to two significant figures.
In the following, simplify using absolute values as necessary.
\sqrt[3]{8{a}^{51}{b}^{6}}
A mass sitting on a horizontal, frictionless surface is attached to one end of a spring; the other end is fixed to a wall.3.0 J of work is required to compress the spring by 0.12 m. If the mass is released s from rest with the spring compressed, it experiences a maximum acceleration of 15
\frac{m}{{s}^{2}}
. Find the value of (a) the spring constant and(b) the mass.
A bank teller has a total of 70 bills in five-, ten-, andtwenty-dollar denominations. the number of fives is threetimes the number of tens, while the total value of the money is$960. Find the number of each type of bill. Solve usngthe Echelon method. Can someone help me setting up this problem |
Rotational Kinetic Energy - Conservation of Energy | Brilliant Math & Science Wiki
Rohit Gupta, Tom Verhoeff, and Jimin Khim contributed
The total energy of the universe is always conserved. The same statement can be stated in other words: the energy can never be created nor be destroyed, but it can only be transferred from one form to another.
There are different types of energies: for example, kinetic energy, heat energy, light energy, sound energy, etc. The sum of all these energies of the universe is always constant. The universe can be broken down into two segments:
1) system: that part of the universe which is under observation
2) surrounding: everything else other than system.
Surroundings may exchange energies with the system, and thus the total energy of a system alone may not always be conserved.
Energies can also be categorized as
1) Mechanical energy: It consists of kinetic and potential energies. Kinetic energy consists of energy due to translational motion of the center of mass of the body and due to rotation of the body about center of mass.
2) Non-mechanical energy: All other energies except kinetic and potential energies fall under non-mechanical energies.
For an isolated system, in the absence of non-conservative forces, the mechanical energy of the system remains conserved.
The mechanical energy of a system can be changed by mechanical work. Work is done by force or torque of a force. Force can be categorized into two:
Conservative forces: Those forces whose work done does not depend on the path taken and the work done in all closed loops is zero are called conservative forces. Because the work done in a closed loop is zero, these forces and their torques can never convert the mechanical energy of the system into non-mechanical forms. Examples of such forces are gravitational forces, electrostatic forces, etc.
Non-conservative forces: These forces convert mechanical energy of the system into a non-mechanical form and generally work done in closed loops is negative.
Kinetic energy can be changed by the work of both types of forces, and thus conservative forces must change the potential energy of the system as well, to ensure the mechanical energy remains conserved.
It can be concluded that in the absence of external forces and internal non-conservative forces, the total mechanical energy of the system remains conserved.
m
is connected by a string, which is wound on a solid cylinder of mass
M
R.
The block is released to fall under gravity. The axle of the cylinder is fixed and smooth. Find the speed of the block as a function of the height descended
h
by the block.
As the block descends, its potential energy decreases and the kinetic energy of the block and the cylinder increases. According to conservation of energy, (loss in potential energy) = (gain in kinetic energy). If the block descends by distance
h,
the block gains speed
v
and the cylinder rotates by angular velocity
\omega
mgh = \frac{1}{2}m{v^2} + \frac{1}{2}I{\omega ^2}.
As the points on the circumference of the cylinder are rotating in circular motion, for them it can be written that
v = R\omega.
Moment of inertia of the cylinder about its axis is
\frac{{M{R^2}}}{2}
\begin{aligned} mgh &= \frac{1}{2}\left( {m + \frac{M}{2}} \right){v^2}\\ v &= \sqrt {\frac{{4mgh}}{{2m + M}}}. \ _\square \end{aligned}
Cite as: Rotational Kinetic Energy - Conservation of Energy. Brilliant.org. Retrieved from https://brilliant.org/wiki/rotational-kinetic-energy-conservation-of-energy/ |
Chimney - Turkcewiki.org
Not to be confused with cooling tower.
"Smokestack" redirects here. For the 1963 avant-garde jazz album, see Smokestack (album). For smokestacks on ships, see Funnel (ship).
Find sources: "Chimney" – news · newspapers · books · scholar · JSTOR (March 2013) (Learn how and when to remove this template message)
Residential flue liners[edit]
Chimney pots, caps and tops[edit]
Chimney draught or draft[edit]
{\displaystyle Q=C\,A\,{\sqrt {2\,g\,H\,{\frac {T_{i}-T_{e}}{T_{e}}}}}}
Draft hood[edit]
Maintenance and problems[edit]
Dual-use chimneys[edit]
Cooling tower used as an industrial chimney[edit]
Retrieved from "http://en.turkcewiki.org/w/index.php?title=Chimney&oldid=1067747328" |
Use proof by Contradiction to prove that the sum of an irrational number and a r
A real number r is rational if and only if there exists two integer a and b such that
r=a/b
( b is not equal to zero) Let us assume that x is an irrational number and y is a rational number . Suppose for the sake of contradiction that
x+y
is not irrational
y-a/b
since s is irrational , x cannot be written as the ratio of two integers Since
x+y
in not irrational,
x+y
is rational . By the definition of rational, there exists integers c and d not equal to zero such that
x+y=c/d
\text{since}\text{ }y=a/b
x+a/b=c/d
x=c/d-a/b
x=\left(bc-ad\right)/bd
Since a,b,c,d are integers bc - ad and bd are also integers. Moreover bd is zero as b and d are both non zero. However , this then implies that x is a rational number (as it is written as the ratio of tow integers). which is in contradiction with the fact that with the fact that x is an irrational number. Thus our supposition that
x+y
is not irrational is false. Which means that
x+y
is irrational .
\left(A-B\right)-C=\left(A-C\right)-\left(B-C\right)
\cap \cup
\left(a,b\right)\in R
{\left(x+y+z\right)}^{30}
Determine the truth value using truth table (Attach your answer, any image file format):
p\vee \left(q\wedge r\right)\equiv \left(p\vee q\right)\wedge \left(urc\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\ne rp\vee r\right)
Consider a Poisson process on
\left[0,\mathrm{\infty }\right)
\lambda
and let T be a random variable independent of the process. Assume T has an exponential distribution with parameter v. Let
{N}_{T}
denote the number of particles in the interval
\left[0,T\right]
. Compute the discrete density of
{N}_{T} |
What is the range in the set of data below
What is the range in the set of data below 35, 39, 25, 57 62, 46, 53, 41?
rotgelb7kjxw
The range of a data set is the difference between the largest and the smallest value in the set.
Here the range is:r=62-25=37
Explain how to will the mean and the median affected after correcting the score.
The instructor entered the test scores from statistics class of 25 students and calculated the mean and median of the scores. After checking it was discovered that the top score was entered wrong as 46 instead of 56.
How do you find the geometric mean for the pair of numbers: -18, -36?
Given the set:
\left\{2,-\frac{1}{2},-\frac{3}{4},x,\frac{7}{2}\right\}
, for what x would the mean of the set be 1/2?
Given a set {-3,2, -1,x}, for what x would the mean of the set be -5? |
Let left{v_{1}, v_{2}, dots, v_{n}right} be a basis for a vector space V. Prove
Let left{v_{1}, v_{2}, dots, v_{n}right} be a basis for a vector space V.
\left\{{v}_{1},\text{ }{v}_{2},\dots ,\text{ }{v}_{n}\right\}
be a basis for a vector space V. Prove that if a linear transformation
T\text{ }:\text{ }V\to \text{ }V
T\left({v}_{1}\right)=0\text{ }\text{for}\text{ }i=1,\text{ }2,\stackrel{˙}{s},\text{ }n,
then T is the zero transformation.
Getting Started: To prove that T is the zero transformation, you need to show that
T\left(v\right)=0
for every vector v in V.
(i) Let v be an arbitrary vector in V such that
v={c}_{1}\text{ }{v}_{1}\text{ }+\text{ }{c}_{2}\text{ }{v}_{2}\text{ }+\text{ }\stackrel{˙}{s}\text{ }+\text{ }{c}_{n}\text{ }{v}_{n}.
(ii) Use the definition and properties of linear transformations to rewrite
T\text{ }\left(v\right)
T\text{ }\left({v}_{1}\right)
(iii) Use the fact that
T\text{ }\left({v}_{i}\right)=0
to conclude that
T\text{ }\left(v\right)=0
, making T the zero tranformation.
\left\{{v}_{1},\text{ }{v}_{2},\dots ,\text{ }{v}_{n}\right\}
be the basis for a vector space.
Define the map,
T\text{ }:\text{ }V\to V\text{ }\text{such that}\text{ }T\text{ }\left({v}_{i}\right)=0\text{ }\text{for}\text{ }i=1,\text{ }2,\stackrel{˙}{s},\text{ }n.
To prove, the above defined map is a linear transformation.
\text{Let}\text{ }{v}_{1},\text{ }{v}_{2}\text{ }\in \text{ }V\text{ }\text{and}\text{ }r\text{ }\in \mathbb{F}\text{ }\text{a scalar.}
⇒\text{ }T\text{ }\left({v}_{1}\right)=0=T\left({v}_{2}\right)
\text{As}\text{ }{v}_{1}\text{ }{v}_{2}\in \text{ }V⇒\text{ }{v}_{1}\text{ }+\text{ }{v}_{2}\text{ }\in \text{ }V,\text{ }r{v}_{1}\text{ }\in \text{ }V
⇒\text{ }T\text{ }\left({v}_{1}\text{ }+\text{ }{v}_{2}\right)=0
=0\text{ }+\text{ }0
=T\text{ }\left({v}_{1}\right)\text{ }+\text{ }T\text{ }\left({v}_{2}\right)
\text{Also}
T\text{ }\left(r{v}_{1}\right)=0=r.0
⇒\text{ }T\text{ }\left(r{v}_{1}\right)=rT\text{ }\left({v}_{1}\right)
Therefore, the above defined map is a linear transformation and this is true for every vector in V.
Let v be an arbitrary vector in V such that
v={c}_{1}\text{ }{v}_{1}\text{ }+\text{ }{c}_{2}\text{ }{v}_{2}\text{ }+\text{ }\stackrel{˙}{s}\text{ }+\text{ }{c}_{n}\text{ }{v}_{n}
{C}_{i}
's are scalars.
The vector v can be written as such linear combination since
\left\{{v}_{1},\text{ }{v}_{2},\dots ,\text{ }{v}_{n}\right\}
is the basis for a vector space V.
Apply the transformation T on both sides and use the fact that T is linear transformation.
⇒\text{ }T\text{ }\left(v\right)=T\text{ }\left({c}_{1}{v}_{1}\text{ }+\text{ }{c}_{2}{v}_{2}\text{ }+\text{ }\stackrel{˙}{s}\text{ }+\text{ }{c}_{n}{v}_{n}\right)
=T\text{ }\left({c}_{1}{v}_{1}\right)\text{ }+\text{ }T\left({c}_{2}{v}_{2}\right)\text{ }+\text{ }\stackrel{˙}{s}\text{ }+\text{ }T\left({c}_{n}{v}_{n}\right)
={c}_{1}T\text{ }\left({v}_{1}\right)\text{ }+\text{ }{c}_{2}T\text{ }\left({v}_{2}\right)\text{ }+\text{ }\stackrel{˙}{s}\text{ }+\text{ }{c}_{n}T\text{ }\left({v}_{n}\right)
\mathrm{△}\text{ }CBF\text{ }\stackrel{\sim }{=}\mathrm{△}EDF
\mathrm{△}\text{ }CBF\text{ }\stackrel{\sim }{=}\text{ }\mathrm{△}\text{ }EDF
Triangle ABC has vertices
A\left(1,3\right),B\left(-2,-1\right)\text{ }\text{and}\text{ }C\left(3,-2\right)
. Graph
\mathrm{△}ABC
and its image after the indicated composite transformation.
First transformation: Translation: along
\left(x,y\right)\to \left(x+2,y\right)
Second Transformation: Reflection across y-axis.
Coordinate of B after translation
B'( ?, ?).
Coordinate of B after Reflection
B''( ?, ?)
Whether T is a linear transformation, that is
T:C\left[0,1\right]\to C\left[0,1\right]
T\left(f\right)=g
g\left(x\right)={e}^{x}f\left(x\right)
{k}^{2}\text{ }+\text{ }{m}^{2}
Whether the function is a linear transformation or not.
T\text{ }:\text{ }{R}^{2}\to {R}^{3},T\left(x,y\right)=\left(\sqrt{x},xy,\sqrt{y}\right)
\left({T}_{2}{T}_{1}\right)\left(v\right)
for an arbitrary vector v in V.
\left\{{v}_{1},{v}_{2}\right\}
is vasis for the vector space V.
The linear transformation with satisfying equations
{T}_{1}\left({v}_{1}\right)=3{v}_{1}+{v}_{2},
{T}_{1}\left({v}_{1}\right)=-3{v}_{1}+{v}_{2},\text{ }{T}_{2}\left({v}_{1}\right)=-5{v}_{2},
{T}_{2}\left({v}_{2}\right)=-{v}_{1}+6{v}_{2}
{T}_{1}\text{ }:\text{ }V\text{ }\to V
{T}_{2}\text{ }:\text{ }V\to V.
a\text{ }<\text{ }\frac{a+b}{2}\text{ }<\text{ }b
a and b are real numbers. |
Euler scheme for SDEs with non-Lipschitz diffusion coefficient : strong convergence
Berkaoui, Abdel ; Bossy, Mireille ; Diop, Awa
ESAIM: Probability and Statistics, Tome 12 (2008), pp. 1-11.
We consider one-dimensional stochastic differential equations in the particular case of diffusion coefficient functions of the form
{|x|}^{\alpha }
\alpha \in \left[1/2,1\right)
. In that case, we study the rate of convergence of a symmetrized version of the Euler scheme. This symmetrized version is easy to simulate on a computer. We prove its strong convergence and obtain the same rate of convergence as when the coefficients are Lipschitz.
Classification : 65C30, 60H35, 65C20
Mots clés : discretization scheme, strong convergence, CIR process
@article{PS_2008__12__1_0,
author = {Berkaoui, Abdel and Bossy, Mireille and Diop, Awa},
title = {Euler scheme for {SDEs} with {non-Lipschitz} diffusion coefficient : strong convergence},
AU - Berkaoui, Abdel
AU - Bossy, Mireille
AU - Diop, Awa
TI - Euler scheme for SDEs with non-Lipschitz diffusion coefficient : strong convergence
ID - PS_2008__12__1_0
Berkaoui, Abdel; Bossy, Mireille; Diop, Awa. Euler scheme for SDEs with non-Lipschitz diffusion coefficient : strong convergence. ESAIM: Probability and Statistics, Tome 12 (2008), pp. 1-11. doi : 10.1051/ps:2007030. http://www.numdam.org/articles/10.1051/ps:2007030/
[1] A. Alfonsi, On the discretization schemes for the CIR (and Bessel squared) processes. Monte Carlo Methods Appl. 11 (2005) 355-384. | MR 2186814 | Zbl 1100.65007
[2] A. Berkaoui, Euler scheme for solutions of stochastic differential equations. Potugalia Mathematica Journal 61 (2004) 461-478. | MR 2113559 | Zbl 1065.60061
[3] M. Bossy and A. Diop, Euler scheme for one dimensional SDEs with a diffusion coefficient function of the form
{|x|}^{a}
a
in [1/2,1). Annals Appl. Prob. (Submitted).
[4] M. Bossy, E. Gobet and D. Talay, A symmetrized Euler scheme for an efficient approximation of reflected diffusions. J. Appl. Probab. 41 (2004) 877-889. | MR 2074829 | Zbl 1076.65009
[5] J. Cox, J.E. Ingersoll and S.A. Ross, A theory of the term structure of the interest rates. Econometrica 53 (1985) 385-407. | MR 785475
[6] G. Deelstra and F. Delbaen, Convergence of discretized stochastic (interest rate) processes with stochastic drift term. Appl. Stochastic Models Data Anal. 14 (1998) 77-84. | MR 1641781 | Zbl 0915.60064
[7] O. Faure, Simulation du Mouvement Brownien et des Diffusions. Ph.D. thesis, École nationale des ponts et chaussées (1992).
[8] P.S. Hagan, D. Kumar, A.S. Lesniewski and D.E. Woodward, Managing smile risk. WILMOTT Magazine (September, 2002).
[9] J.C. Hull and A. White, Pricing interest-rate derivative securities. Rev. Finan. Stud. 3 (1990) 573-592.
[10] I. Karatzas and S.E. Shreve, Brownian Motion and Stochastic Calculus. Springer-Verlag, New York (1988). | MR 917065 | Zbl 0638.60065 |
For each polynomial function, one zero is given. Find all rational zeros and fac
For each polynomial function, one zero is given. Find all rational zeros and factor the polynomial. Then graph the function. f(x)=3x^{3}+x^{2}-10x-8, zero:2
For each polynomial function, one zero is given. Find all rational zeros and factor the polynomial. Then graph the function.
f\left(x\right)=3{x}^{3}+{x}^{2}-10x-8
, zero:2
sovienesY
Step 1 Since 2 is a zero of
f\left(x\right)=3{x}^{3}+{x}^{2}-10x-8
3{x}^{2}+7x+4
f\left(x\right)=\left(x-2\right)\left(3{x}^{2}+7x+4\right)
Factor the trinomial
3{x}^{2}+7x+4=\left(3x+4\right)\left(x+1\right)
So, the function in factored form is
f\left(x\right)=\left(x-2\right)\left(3x+4\right)\left(x+1\right)
Set each factor equal to zero
x-2=0
3x+4=0
x+1=0
So, all rational zeros are
2,-\frac{4}{2},-1
Step 2 Since, all rational zeros are
2,-\frac{4}{3},-1
So, the graph of the function crosses the x-axis at
\left(2,0\right),\left(-\frac{4}{3},0\right),\left(-1,0\right)
To find the y-intercept evaluate f(0)
f\left(x\right)=3{x}^{3}+{x}^{2}-10x-8
f\left(0\right)=3{\left(0\right)}^{3}+{\left(0\right)}^{2}-10\left(0\right)-8
Substitute 0 for x
=-8
The leading coefficient is 3 (positive), and the function of degree 3 (odd) So, the end behavior is
x\to \mathrm{\infty },f\left(x\right)\to \mathrm{\infty }
x\to -\mathrm{\infty },f\left(x\right)\to -\mathrm{\infty }
See the graph of f(x)
For the following exercise, for each polynomial, a. find the degree; b. find the zeros, if any; c. find the y-intercept(s), if any; d. use the leading coefficient to determine the graph’s end behavior; and e. determine algebraically whether the polynomial is even, odd, or neither.
f\left(x\right)=-3{x}^{2}+6x
f\left(x\right)={x}^{5}+2{x}^{4}+3{x}^{3}-2{x}^{6}-9{x}^{2}-6x+4
Is it true or false the degree of this polynomial function is 5? If not, why? What is the degree?
The area A of the region S that lies under the graph of the continuous function is the limit of the sum of the areas of approximating rectangles.
A=limn\to \infty Rn=limn\to \infty \left[f\left({x}_{1}\right)\Delta x+f\left({x}_{2}\right)\Delta x+...+f\left(xn\right)\Delta x\right]
Use this definition to find an expression for the area under the graph of f as a limit. Do not evaluate the limit.
f\left(x\right)=7x\mathrm{cos}\left(7x\right),0\le x\le \pi 2
Use the intermediate value theorem to determine whether the function
f\left(x\right)={x}^{3}+2x-4
has a root or not. between
x=1\text{ }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{ }x=2
. Ifyes, then find the root to five decimal places.
How does one find the lovely asymptotes of a polynomial graph?
Solve the polynomial inequality graphically.
-{x}^{3}-3{x}^{2}-9x+4<0
Find the cubic polynomial whose graph passes through the points
\left(-1,-1\right),\left(0,1\right),\left(1,3\right),\left(4,-1\right). |
Octal Knowpia
The octal numeral system, or oct for short, is the base-8 number system, and uses the digits 0 to 7, that is to say 10octal represents eight and 100octal represents sixty-four. However, English, like most languages, uses a base-10 number system, hence a true octal system might use different vocabulary.
In the decimal system, each place is a power of ten. For example:
{\displaystyle \mathbf {74} _{10}=\mathbf {7} \times 10^{1}+\mathbf {4} \times 10^{0}}
In the octal system, each place is a power of eight. For example:
{\displaystyle \mathbf {112} _{8}=\mathbf {1} \times 8^{2}+\mathbf {1} \times 8^{1}+\mathbf {2} \times 8^{0}}
Octal numerals can be easily converted from binary representations (similar to a quaternary numeral system) by grouping consecutive binary digits into groups of three (starting from the right, for integers). For example, the binary representation for decimal 74 is 1001010. Two zeroes can be added at the left: (00)1 001 010, corresponding to the octal digits 1 1 2, yielding the octal representation 112.
0 at the base, 7 at the top, 1 to 3 on the right, 4 to 6 on the left
The eight bagua or trigrams of the I Ching correspond to octal digits:
0 = ☷, 1 = ☳, 2 = ☵, 3 = ☱,
4 = ☶, 5 = ☲, 6 = ☴, 7 = ☰.
Gottfried Wilhelm Leibniz made the connection between trigrams, hexagrams and binary numbers in 1703.[1]
By Native AmericansEdit
By EuropeansEdit
It has been suggested that the reconstructed Proto-Indo-European (PIE) word for "nine" might be related to the PIE word for "new". Based on this, some have speculated that proto-Indo-Europeans used an octal number system, though the evidence supporting this is slim.[4]
In 1716, King Charles XII of Sweden asked Emanuel Swedenborg to elaborate a number system based on 64 instead of 10. Swedenborg argued, however, that for people with less intelligence than the king such a big base would be too difficult and instead proposed 8 as the base. In 1718 Swedenborg wrote (but did not publish) a manuscript: "En ny rekenkonst som om vexlas wid Thalet 8 i stelle then wanliga wid Thalet 10" ("A new arithmetic (or art of counting) which changes at the Number 8 instead of the usual at the Number 10"). The numbers 1-7 are there denoted by the consonants l, s, n, m, t, f, u (v) and zero by the vowel o. Thus 8 = "lo", 16 = "so", 24 = "no", 64 = "loo", 512 = "looo" etc. Numbers with consecutive consonants are pronounced with vowel sounds between in accordance with a special rule.[6]
In the mid-19th century, Alfred B. Taylor concluded that "Our octonary [base 8] radix is, therefore, beyond all comparison the "best possible one" for an arithmetical system." The proposal included a graphical notation for the digits and new names for the numbers, suggesting that we should count "un, du, the, fo, pa, se, ki, unty, unty-un, unty-du" and so on, with successive multiples of eight named "unty, duty, thety, foty, paty, sety, kity and under." So, for example, the number 65 (101 in octal) would be spoken in octonary as under-un.[9][10] Taylor also republished some of Swedenborg's work on octal as an appendix to the above-cited publications.
In computersEdit
Newer languages have been abandoning the prefix 0, as decimal numbers are often represented with leading zeroes. The prefix q was introduced to avoid the prefix o being mistaken for a zero, while the prefix 0o was introduced to avoid starting a numerical literal with an alphabetic character (like o or q), since these might cause the literal to be confused with a variable name. The prefix 0o also follows the model set by the prefix 0x used for hexadecimal literals in the C language; it is supported by Haskell,[19] OCaml,[20] Python as of version 3.0,[21] Raku,[22] Ruby,[23] Tcl as of version 9,[24] PHP as of version 8.1,[25] Rust[26] and it is intended to be supported by ECMAScript 6[27] (the prefix 0 originally stood for base 8 in JavaScript but could cause confusion,[28] therefore it has been discouraged in ECMAScript 3 and dropped in ECMAScript 5[29]).
Octal numbers that are used in some programming languages (C, Perl, PostScript...) for textual/graphical representations of byte strings when some byte values (unrepresented in a code page, non-graphical, having special meaning in current context or otherwise undesired) have to be to escaped as \nnn. Octal representation may be particularly handy with non-ASCII bytes of UTF-8, which encodes groups of 6 bits, and where any start byte has octal value \3nn and any continuation byte has octal value \2nn.
In aviationEdit
Transponders in aircraft transmit a "squawk" code, expressed as a four-octal-digit number, when interrogated by ground radar. This code is used to distinguish different aircraft on the radar screen.
Conversion between basesEdit
Decimal to octal conversionEdit
Method of successive Euclidean division by 8Edit
Method of successive multiplication by 8Edit
Method of successive duplicationEdit
Octal to decimal conversionEdit
{\displaystyle k=\sum _{i=0}^{n}\left(a_{i}\times 8^{i}\right)}
Octal to binary conversionEdit
Binary to octal conversionEdit
Octal to hexadecimal conversionEdit
Hexadecimal to octal conversionEdit
Computer number format – Internal representation of numeric values in a digital computer
^ Leibniz, Gottfried Wilhelm (1703). "Explanation of binary arithmetic". leibniz-translations.com. Retrieved 2022-03-02.
^ Winter, Werner (1991). "Some thoughts about Indo-European numerals". In Gvozdanović, Jadranka (ed.). Indo-European numerals. Trends in Linguistics. Vol. 57. Berlin: Mouton de Gruyter. pp. 13–14. ISBN 3-11-011322-8. Retrieved 2013-06-09.
^ Küveler, Gerd; Schwoch, Dietrich (2007-10-04). Informatik für Ingenieure und Naturwissenschaftler: PC- und Mikrocomputertechnik, Rechnernetze (in German). Vol. 2 (5 ed.). Vieweg, reprint: Springer-Verlag. ISBN 978-3-83489191-4. 978-3-83489191-4. Retrieved 2015-08-05.
^ "Haskell 98 Lexical Structure".
^ Rust literals and operators: https://doc.rust-lang.org/rust-by-example/primitives/literals.html |
Consider the following two bases for R^3 : alpha := left{ begin{bmatrix} 2 1
Consider the following two bases for R^3
Consider the following two bases for
{R}^{3}
\alpha :=\left\{\left[\begin{array}{c}2\\ 1\\ 3\end{array}\right],\left[\begin{array}{c}-1\\ 0\\ 1\end{array}\right],\left[\begin{array}{c}3\\ 1\\ -1\end{array}\right]\right\}and\text{ }\beta :=\left\{\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right],\left[\begin{array}{c}-2\\ 3\\ 1\end{array}\right],\left[\begin{array}{c}2\\ 3\\ -1\end{array}\right]\right\}
\left[x{\right]}_{\alpha }={\left[\begin{array}{c}1\\ 2\\ -1\end{array}\right]}_{\alpha }then\text{ }find\text{ }\left[x{\right]}_{\beta }
(that is, express x in the
\beta
\alpha =\left\{\left[\begin{array}{c}2\\ 1\\ 3\end{array}\right],\left[\begin{array}{c}-1\\ 0\\ 1\end{array}\right],\left[\begin{array}{c}3\\ 1\\ -1\end{array}\right]\right\}\beta =\left\{\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right],\left[\begin{array}{c}-2\\ 3\\ 1\end{array}\right],\left[\begin{array}{c}2\\ 3\\ -1\end{array}\right]\right\}
\left[x{\right]}_{\alpha }={\left[\begin{array}{c}1\\ 2\\ -1\end{array}\right]}_{\alpha }\to x=1.\left[\begin{array}{c}2\\ 1\\ 3\end{array}\right]+2\left[\begin{array}{c}-1\\ 0\\ 1\end{array}\right]-1\left[\begin{array}{c}3\\ 1\\ 1\end{array}\right]=\left[\begin{array}{c}-3\\ 0\\ 6\end{array}\right]
x={C}_{1}\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]+{C}_{2}\left[\begin{array}{c}-2\\ 3\\ 1\end{array}\right]+{C}_{3}\left[\begin{array}{c}2\\ 3\\ -1\end{array}\right]⇒\begin{array}{c}{C}_{1}-2{C}_{2}+2{C}_{3}=-3-\left(1\right)\\ {C}_{1}+3C+2+3{C}_{3}=0-\left(2\right)\\ {C}_{1}+{C}_{2}-{C}_{3}=6-\left(3\right)\end{array}
{C}_{3}={C}_{1}+{C}_{2}-6
{C}_{1}-2{C}_{2}+2{C}_{1}+2{C}_{2}-12=-3⇒3{C}_{1}=+9⇒{C}_{1}=3
{C}_{1}+3{C}_{2}+3{C}_{1}+3{C}_{2}-18=0⇒4{C}_{1}+6{C}_{2}=18⇒6{C}_{2}={C}_{2}
{C}_{3}=3+1-6=-2
\left[\begin{array}{c}-3\\ 0\\ 6\end{array}\right]=+3\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]+1\left[\begin{array}{c}-2\\ 3\\ 1\end{array}\right]-2\left[\begin{array}{c}2\\ 3\\ -1\end{array}\right]
\left[x{\right]}_{\beta }=\left[\begin{array}{c}3\\ 1\\ -2\end{array}\right]
To compare and contrast: the rectangular, cylindrical and spherical coordinates systems.
(a) Find the bases and dimension for the subspace
H=\left\{\left[\begin{array}{c}3a+6b-c\\ 6a-2b-2c\\ -9a+5b+3c\\ -3a+b+c\end{array}\right];a,b,c\in R\right\}
(b) Let be bases for a vector space V,and suppose (i) Find the change of coordinate matrix from B toD. (ii) Find
{\left[x\right]}_{D}f\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}x=3{b}_{1}-2{b}_{2}+{b}_{3}
To find: the distance between the lakes to the nearest mile.
(26,15) and (9,20)
The system of equation
\left\{\begin{array}{l}2x+y=1\\ 4x+2y=3\end{array}
by graphing method and if the system has no solution then the solution is inconsistent.
Given: The linear equations is
\left\{\begin{array}{l}2x+y=1\\ 4x+2y=3\end{array}
We can describe the location of a point in the plane using different ________ systems. The point P shown in the figure has rectangular coordinates ( _ , _ ) and polar coordinates ( _, _).
\text{ }x=\text{ }r\mathrm{cos}\theta
\text{ }y=\text{ }r\mathrm{sin}\theta
b. From rectangular to polar:
r=±\sqrt{{x}^{2}\text{ }+\text{ }{y}^{2}}
\mathrm{cos}\theta =\frac{x}{r},\mathrm{sin}\theta =\frac{y}{r},\mathrm{tan}\theta =\frac{x}{y}
Given: equation in rectangular-coordinate is
y=x
Converting into equivalent polar equation -
y=x
x=r\mathrm{cos}\theta ,\text{ }y=r\mathrm{sin}\theta ,
⇒\text{ }r\mathrm{sin}\theta =r\mathrm{cos}\theta
⇒\text{ }\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }=1
⇒\text{ }\mathrm{tan}\theta =1
Thus, desired equivalent polar equation would be
\theta =1
The equivalent polar equation for the given rectangular-coordinate equation:
{x}^{2}\text{ }+\text{ }{y}^{2}\text{ }+\text{ }8x=0 |
Advanced Syllogistic Logic Practice Problems Online | Brilliant
Which of the following statements is
\color{#D61F06}{\text{not}}
consistent with the above diagram?
All Teees are not Viiis, but some Doubleuus are Viiis No Teees are Viiis, but some Viiis are Doubleuus No Youus are Doubleuus, and no Teees are Doubleuus Some Doubleuus are Viiis, but no Youuus are Viiis
All Wannts are Tuuus.
Some Tuuus are Triiis.
All Triiis are Forees.
Is the following statement necessarily true?
All Forees are Wannts.
It is necessarily true It is possibly but not necessarily true It is necessarily false
All bookstores sell stationary items.
All convenience stores sell stationary items.
Which of the following conclusions might be true?
\text{(1)}
All bookstores are convenience stores.
\text{(2)}
All convenience stores are bookstores.
(2) might be true (1) might be true Both (1) and (2) might be true Both are definitely false
A friend of Will's is not a friend of Jack's.
A friend of Hannibal's is not a friend of Jack's.
Which of the following conclusions must be true?
\text{(1)}
Some friend of Will's is a friend of Hannibal's.
\text{(2)}
Some friend of Hannibal's a friend of Will's.
(2) only (1) only (1) and (2) None
All Faivvs are Sivkks.
All Faivvs are Sevvens.
Given the above two statements (and the fact at least one Faivv exists), is the following statement true?
No Sivkks are Sevvens.
Yes, always Sometimes, but not always No, never |
Alkanes, Alkenes, Alkynes | Brilliant Math & Science Wiki
Ashish Menon, David Huang, Skanda Prasad, and
Stuart Ferrie
Ashrit Ramadurgam
Organic chemistry is the study of carbon compounds, so the study of organic chemistry is important because all living things are based on carbon compounds. Carbon is unique in that it can form up to four bonds in a compound, so they can easily bond with other carbon atoms, forming long chains or rings. In addition, the type of bonding in organic compounds is almost always covalent. Organic compounds don't dissociate in solutions because there are no ionic bonds, and therefore organic compounds are poor conductors of electricity and do not behave as electrolytes in solution. Generally, the larger and more complicated the organic substance, the higher its boiling and melting points.
Why is silicon, another element in group 14 of the periodic table, unable to make the great variety of molecules that carbon atoms can?
Silicon can make large molecules called silicones; however, the silicon-silicon bond is much weaker than the carbon-carbon bond, especially with respect to the silicon-oxygen bond.
_\square
So carbon can have four bonds, but the number of bonds carbon makes per atom can also vary. Just to get the terminology out of the way, we'll be looking at what's known as hydrocarbons, which essentially covers molecules that have only carbon and hydrogen atoms. Even more particularly, we'll study aliphatic compounds, which means the compounds will be open-chained. The details of which will be explained later.
When it comes to Alkanes, Alkenes, and Alkynes, the acidity is in the order of Alkynes > Alkenes > Alkanes. The acidity is mainly due to the increase in the s-character which causes an increase in acidity.
A large and structurally simple class of hydrocarbons includes those substances in which all the carbon-carbon bonds are single bonds. These are called saturated hydrocarbons or alkanes. The simplest alkanes are methane
(\ce{CH_4}),
(\ce{C_2H_6}),
and propane
(\ce{C_3H_8}).
The alkanes are also called as paraffins. In an alkane, all
4
valencies of the carbon atom are satisfied with other hydrogen atoms.
This gives them a general formula :
\ce{C_{n}H}_{2n+2}.
Methane gas is the first member of the homologous series of alkanes. The valency of a single carbon atom is satisfied by four hydrogen atoms which form single covalent bonds.
In nature, methane is formed by the microbial activity of organic matter.
Due to decomposition of organic matter in marshy areas (an area of low wetland).
Natural gas which is a byproduct of petroleum mining contains 70% methane.
Coal gas too contains 25-40% methane.
Laboratory preparation of methane gas
\color{#3D99F6}{\text{Chemicals required:}}
You would need the following chemicals:
1) Sodium acetate
(\ce{CH_3COONa})
2) Soda lime
(\ce{NaOH + CaO})
\color{#3D99F6}{\text{Chemical reaction:}}
\ce{CH_3COONa + NaOH ->[\ce{CaO}] {Na}_2CO_3 + CH_4}
\color{#3D99F6}{\text{Collection of methane gas:}}
Since methane gas is insoluble in water, it is collected by downward displacement of water.
Another way of preparation of methane:
\ce{{Al}_4C_3 + 12H_2O -> 4Al{(OH)}_3 + 3CH_4}
Another method of preparation of methane:
\ce{CH_3I + 2[H]}\,(\text{from Zn/Cu couple})\ce{-> CH_4 + HI}
In an environment of excess oxygen, methane burns with a pale blue flame. No soot is formed.
\ce{CH_4 + 2O_2 -> CO_2 + 2H_2O}
In an environment of insufficient oxygen, air pollution can be caused due to release of excess carbon monoxide into the atmosphere.
\ce{2CH_4 + 3O_2 -> 2CO + 4H_2O}
In this reaction, the hydrogen atom or atoms in the hydrocarbon are substituted by more reactive atoms such as chlorine, bromine, etc. or group of atoms such as
\ce{OH, SO_4}
, etc. When hydrogen atoms of an alkane are substituted by chlorine, the reaction is called chlorination.
\begin{aligned} \ce {CH_4 + {Cl}_2 &->[\text{Sunlight}] HCl + CH_3Cl} &&(\text{Monochloromethane})\\\\ \ce {CH_4 + 2{Cl}_2 &->[\text{Sunlight}] 2HCl + CH_2{Cl}_2} &&(\text{Dichloromethane})\\\\ \ce {CH_4 + 3{Cl}_2 &->[\text{Sunlight}] 3HCl + CH{Cl}_3} &&(\text{Trichloromethane})\\\\ \ce {CH_4 + 4{Cl}_2 &->[\text{Sunlight}] 4HCl + C{Cl}_4} &&(\text{Tetrachloromethane or carbon tetrachloride}) \end{aligned}
The process of decomposition of a hydrocarbon into elements on heating in the absence of air is called pyrolysis.
\ce{CH_4 ->[1000°\text{C}] C + 2H_2}
Catalytic or controlled oxidation
\begin{aligned} \ce{2CH_4 + O_2 ->[\text {Copper - 200°C - 100Atms}] &2CH_3OH}\\\\ \ce{2CH_3OH + O_2 ->[\text {Copper - 200°C - 100Atms}] &2HCHO + 2H_2O}\\\\ \ce{2HCHO + O_2 ->[\text {Copper - 200°C - 100Atms}] &2HCOOH} \end{aligned}
\ce{CH_4 + O_2 ->[\text{MoO in 350°C-500°C}] HCHO + H_2O}
Examples of reactions with methane
\begin{aligned} \ce {CH_4 + {Cl}_2 &-> HCl + CH_3Cl}\\\\ \ce {CH_3Cl + KOH &-> KCl + CH_3OH} \end{aligned}
\begin{aligned} \ce {CH_4 + {Cl}_2 -> &HCl + CH_3Cl}\\\\ \ce {CH_3Cl + KOH -> &KCl + CH_3OH}\\\\ \ce {CH_3OH + CuO ->[300°\text{C}] &Cu + H_2O + HCHO} \end{aligned}
\begin{aligned} \ce {CH_4 + {Cl}_2 -> &HCl + CH_3Cl}\\\\ \ce {CH_3Cl + KOH -> &KCl + CH_3OH}\\\\ \ce {CH_3OH + 2[O] ->[\ce{K_2{Cr}_2O_7 + H_2SO_4}] &HCOOH + H_2O} \end{aligned}
Laboratory preparation of ethane gas
\color{#3D99F6}{\text{Chemicals required:}}
(\ce{C_2H_5COONa})
(\ce{NaOH + CaO})
\color{#3D99F6}{\text{Chemical reaction:}}
\ce{C_2H_5COONa + NaOH ->[\ce{CaO}] {Na}_2CO_3 + C_2H_6}
\color{#3D99F6}{\text{Collection of ethane gas:}}
Since ethane gas is insoluble in water, it is collected by downward displacement of water.
Another method of preparation of ethane:
\ce{C_2H_5I + 2[H]}\,(\text{from Zn/Cu couple})\ce{-> C_2H_6 + HI}
In an environment of excess oxygen, ethane burns with a pale blue flame. No soot is formed.
\ce{C_2H_6 + 7O_2 -> 4CO_2 + 6H_2O}
\ce{2CH_4 + 5O_2 -> 4CO + 6H_2O}
\ce{OH, SO_4}
\begin{aligned} \ce {C_2H_6 + {Cl}_2 &->[\text{Sunlight}] HCl + C_2H_5Cl} &&(\text{Monochloroethane})\\\\ \ce {C_2H_6 + 2{Cl}_2 &->[\text{Sunlight}] 2HCl + C_2H_4{Cl}_2} &&(\text{Dichloroethane})\\\\ \ce {C_2H_6 + 3{Cl}_2 &->[\text{Sunlight}] 3HCl + C_2H_3{Cl}_3} &&(\text{Trichloroethane})\\\\ \ce {C_2H_6 + 4{Cl}_2 &->[\text{Sunlight}] 4HCl + C_2H_2{Cl}_4} &&(\text{Tetrachloroethane})\\\\ \ce {C_2H_6 + 5{Cl}_2 &->[\text{Sunlight}] 5HCl + C_2H{Cl}_5} &&(\text{Pentachloroethane})\\\\ \ce {C_2H_6 + 6{Cl}_2 &->[\text{Sunlight}] 6HCl + C_2{Cl}_6} &&(\text{Hexachloroethane}) \end{aligned}
Catalytic or controlled oxidation:
\begin{aligned} \ce{2C_2H_6 + O_2 &->[\text {Copper - 200°C - 100Atms}] 2C_2H_5OH}\\\\ \ce{2C_2H_5OH + O_2 &->[\text {Copper - 200°C - 100Atms}] 2CH_3CHO + 2H_2O}\\\\ \ce{2CH_3CHO + O_2 &->[\text {Copper - 200°C - 100Atms}] 2CH_3COOH} \end{aligned}
\ce{C_2H_6 + O_2 ->[\text{MoO in 350°C-500°C}] CH_3CHO + H_2O}
Examples of reactions with ethane
\begin{aligned} \ce {C_2H_6 + {Cl}_2 &-> HCl + C_2H_5Cl}\\\\ \ce {C_2H_5Cl + KOH &-> KCl + C_2H_5OH} \end{aligned}
\begin{aligned} \ce {C_2H_6 + {Cl}_2 -> &HCl + C_2H_5Cl}\\\\ \ce {C_2H_5Cl + KOH -> &KCl + C_2H_5OH}\\\\ \ce {C_2H_5OH + CuO ->[300°\text{C}] &Cu + H_2O + CH_3CHO} \end{aligned}
\begin{aligned} \ce {C_2H_6 + {Cl}_2 -> &HCl + C_2H_5Cl}\\\\ \ce {C_2H_5Cl + KOH -> &KCl + C_2H_5OH}\\\\ \ce {C_2H_5OH + 2[O] ->[\ce{K_2{Cr}_2O_7 + H_2SO_4}] &CH_3COOH + H_2O} \end{aligned}
Alkenes are the unsaturated hydrocarbons in which there is a double bond between two carbon atoms. In these compounds, unsaturation is due to the presence of the double bond.
Alkenes have the general formula
\ce{C_nH_{2n}},
n
is the number of carbon atoms in the molecule. For example,
n=2,
the alkene is called as ethene or also as ethylene with the formula
\ce{C_2H_4};
n=3,
the alkene would be called as propene or propylene with the formula
\ce{C_3H_6}.
Note: There is no alkene with only
one
carbon atom.
Laboratory preparation of ethene gas
\color{#3D99F6}{\text{Chemicals required:}}
We would require the following chemicals:
1) Ethyl alcohol or ethanol
(\ce{C_2H_5OH})
2) Concentrated sulphuric acid
(\ce{H_2SO_4})
\color{#3D99F6}{\text{Chemical reaction:}}
\begin{aligned} \ce{C_2H_5OH + H_2SO_4 ->[>160°\text{C}]& C_2H_5HSO_4 + H_2O}\\\\ \ce{C_2H_5HSO_4 ->[<160°\text{C} \text{ in excess of }\ce{H_2SO_4}] &C_2H_4 + H_2SO_4} \end{aligned}
\color{#3D99F6}{\text{Collection of ethene gas:}}
Ethene is insoluble in water and hence it is collected by downward displacement of water.
Another method of preparation of ethene:
\ce{CH_2{Br}_2 + Zn ->[\text{Ethanol} + \text{Heat}] C_2H_4 + Zn{Br}_2}
\ce{C_2H_6 ->[\ce{Al_2O_3 - SiO_2} - 500°\text{C}] C_2H_4 + H_2}
Chemical properties of ethene
In excess of air, ethene burns with a pale blue flame.
\ce{C_2H_4 + 3O_2 ->[\text{Excess of air}] 2CO_2 + 2H_2O}
Ethene is a highly reactive unsaturated hydrocarbon because of the presence of a double bond between two carbon atoms. In the double bond between the carbon atoms, one single bond is weaker. This bond breaks freeing one valence electron of each carbon atom. The free valence electrons then bind to other atoms to form additive compounds in which all valencies of carbon atoms are fully satisfied by single covalent bonds.
\ce{C_2H_4 + H_2 ->[\ce{Ni-300°C}] C_2H_6}
\ce{C_2H_4 + {Cl}_2 ->[\ce{C{Cl}_4}] C_2H_4{Cl}_2}
\ce{C_2H_4 + {Br}_2 ->[\ce{C{Cl}_4}] C_2H_4{Br}_2}
\ce{C_2H_4 + HCl -> C_2H_5Cl}
\ce{C_2H_4 + HBr -> C_2H_5Br}
\ce{C_2H_4 + H_2SO_4 -> C_2H_5HSO_4}
\ce{C_2H_4 + H_2O + [O] ->[\text{Alkaline }\ce{MnO_4}] C_2H_4{(OH)}_2}
The process of forming molecules of higher molecular mass by combining similar molecules of lower molecular mass is termed as polymerization.
\ce{n(H_2C = CH_2) -> -{(H_2C - CH_2)}_n-}
Examples of reactions with ethene
\begin{aligned} \ce {C_2H_4 + HCl &-> C_2H_5Cl}\\\\ \ce {C_2H_5Cl + KOH &-> KCl + C_2H_5OH} \end{aligned}
\begin{aligned} \ce {C_2H_4 + HCl -> &C_2H_5Cl}\\\\ \ce {C_2H_5Cl + KOH -> &KCl + C_2H_5OH}\\\\ \ce {C_2H_5OH + CuO ->[300°\text{C}] &Cu + H_2O + CH_3CHO} \end{aligned}
\begin{aligned} \ce {C_2H_4 + HCl -> &C_2H_5Cl}\\\\ \ce {C_2H_5Cl + KOH -> &KCl + C_2H_5OH}\\\\ \ce {C_2H_5OH + 2[O] ->[\ce{K_2{Cr}_2O_7 + H_2SO_4}] &CH_3COOH + H_2O} \end{aligned}
Alkynes contain only one triple bond between two adjacent carbon atoms. But if the number of triple bonds is more than one in any compound, the standard IUPAC nomenclature is used. They are commonly called as acetylenes.
They have the general formula
\ce{C_nH_{2n-2}},
n
is the number of carbon atoms in the molecule.
For example, if an alkyne has 2 carbon atoms, then it would be called as ethyne with molecular formula
\ce{C_2H_2}.
If an alkyne has 3 carbon atoms, then it would be propyne with formula
\ce{C_3H_4}.
\color{#3D99F6}{\text{Chemicals required:}}
We would require the following chemicals:-
1) Calcium carbide
(\ce{CaH_2})
2) Distilled water
(\ce{H_2O})
\color{#3D99F6}{\text{Chemical reaction:}}
\ce{CaC_2 + 2H_2O -> C_2H_2 + Ca{(OH)}_2}
\color{#3D99F6}{\text{Collection of ethyne gas:}}
Ethyne is insoluble in water and hence is collected by downward displacement of water.
Another method of preparation of ethyne:
\ce{C_2H_2{Br}_4 + 2Zn ->[\text{Ethanol} + \text{Heat}] C_2H_2 + 2Zn{Br}_2}
\ce{C_2H_2{Br}_2 + 2KOH -> C_2H_2 + 2KBr + 2H_2O}
Chemical properties of ethyne
In excess of air, ethene burns with a brilliant white flame.
\ce{2C_2H_2 + 5O_2 ->[\text{Excess of air}] 4CO_2 + 2H_2O}
\ce{C_2H_2 + H_2 ->[\ce{Ni-300°C}] C_2H_4 + H_2 ->[\ce{Ni-300°C}] CH_4 }
\ce{C_2H_2 + {Cl}_2 ->[\ce{C{Cl}_4}] C_2H_2{Cl}_2 + {Cl}_2 ->[\ce{C{Cl}_4}] C_2H_2{Cl}_4}
\ce{C_2H_2 + {Br}_2 ->[\ce{C{Cl}_4}] C_2H_2{Br}_2 + {Br}_2 ->[\ce{C{Cl}_4}] C_2H_2{Br}_4}
\ce{C_2H_2 + HCl -> C_2H_3Cl + HCl -> C_2H_4{Cl}_2 }
\ce{C_2H_2 + HBr -> C_2H_3Br + HBr -> C_2H_4{Br}_2 }
\ce{C_2H_2 + 2CuCl + 2NH_4OH -> C_2{Cu}_2 + 2NH_4Cl + 2H_2O}
\ce{C_2H_2 + 2AgNO_3 + 2NH_4OH -> C_2{Ag}_2 + 2NH_4NO_3 + 2H_2O}
\ce{3(H_2C_2 = C_2H_2) ->[\text{Red hot silica tube}] C_6H_6}
Cite as: Alkanes, Alkenes, Alkynes. Brilliant.org. Retrieved from https://brilliant.org/wiki/alkanes-alkenes-alkynes/ |
To determine: The smallest nonnegative integer x that satisfies the given system of congruences. x\equiv 6\pmod 8 x\equiv 17\pmod {25}
x\equiv 6±od8
x\equiv 17±od\left\{25\right\}
hajavaF
x\equiv 6±od8
x\equiv 17±od\left\{25\right\}
We see that the solution x is unique modulo 25.8=200.
Now, 25(1)-8(3)=1.
x=6.25(1)-17.8(3)
x=-58±od\left\{200\right\}
x=142±od\left\{40\right\}
Therefore, x=142.
\stackrel{―}{UQ}\stackrel{\sim }{=}\stackrel{―}{VQ}
\stackrel{―}{SU}\stackrel{\sim }{=}\stackrel{―}{SV},\mathrm{\angle }1\stackrel{\sim }{=}\mathrm{\angle }2
x\equiv 4\left(mod\text{ }5\right)
x\equiv 3\left(mod\text{ }8\right)
x\equiv 2\left(mod\text{ }3\right)
Write a paragraph proof.
\stackrel{―}{AD}\text{ }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{ }\stackrel{―}{EC}
intersect at B.
\mathrm{△}ABE\sim \mathrm{△}DBC
To find: Whether the given triangles are congruent or not, to write the corresponding.
To prove : Diagonals of a rectangle are congruent.
Prove directly from the definition of congruence modulo n that if a,c, and n are integers,
n>1
a\equiv c\left(\text{mod}\text{ }n\right),then\text{ }{a}^{3}\equiv {c}^{3}\left(\text{mod}\text{ }n\right)
Solve the congruence
{x}^{20}-1\equiv 0\left(\text{mod}61\right) |
Will Abramson - Getting Started With MIRACL Core
Getting Started With MIRACL Core
Over the last few weeks I have been getting to grips with the MIRACL core cryptography library - which describes itself as:
MIRACL Core is a multi-lingual and architecturally agnostic Cryptographic library that supports elliptic curve cryptography, pairing-friendly curve cryptography, RSA, AES symmetric encryption and hash functions. All code is entirely in the supported high-level languages. No assembly language and no third party code is required - these libraries are completely self-contained. MIRACL Core was designed from the ground up with side-channel attack resistance in mind. Multiple curves can be supported in a single application.
It is a fantastic library, which opens up a lot of opportunities to implement cryptography without worrying about managing some of the basement level challenges that arise when implementing the cryptographic primitives this library provides. For example, implementing and handling BIG integers. I have been particularly interested in learning how to use the pairing friendly curves contained in this library, as many of the newest cryptographic protocols require this type of curve.
Here is my attempt to synthesise my knowledge, so that you can have an easier on-boarding journey and more people can start harnessing some of the magical capabilities that cryptography research has discovered. Note: I focused on the Rust programming language, so some of this will be specific to that although by in large the API’s are consistent across languages so this hopefully will be useful to everyone.
The first challenge is to actually get the library installed and built on your machine so you can use the library when developing your own protocols. There are some pretty good instructions in the repository for this already, but I will walk through it for completeness.
First thing to do is clone the core repo, giving you the code for the library implemented in all the different languages supported, each under its own folder.
git clone https://github.com/miracl/core.git
Then choose your desired language and cd into it’s directory within the repository.
Now the next step, was a little strange to me, you build the library by running a python script, either config32.py or config64.py, then selecting the specific curves you wish the library to support. I spoke to Mike Scott about this and he explained this was for consistency and simplicity across the different platforms build processes, which makes sense.
It is worth reading through the readme in your specific languages folder for any language dependent details.
The readme for the rust section suggests using the libcore.rlib file with the compiler rustc to compile any code files you create. However, it is possible to use the library with cargo for an easier, more typical rust developer experience. Once you have built the library by running the python script, the contents of the rust folder will have changed, specifically all code files will now be under the folder core. This folder is just a rust library and can be included in any project.
First create a new rust project somewhere.
cargo new mymiraclproject
Then copy the core folder that was just created by the build into your new project.
Finally, update the cargo.toml file in your new project to include the core library as a dependency. Here path must equal the path to the core folder, which if copied into the project should be just under the root.
core={path="core", version="0.1.0"}
You should now be able to run cargo build on this project and see the dependencies compile, if so you are ready to start developing some crypto.
The structure of the library is fairly straightforward. Each folder under the src folder represents all the code for that curvev (apart from rsa2048). The rest of the files contain implementations for other useful functions such as encryption (aes.rs), hashing, and a random number generator (rand.rs).
Random Number Generator (rand.rs)
Whenever developing cryptography, you need a source of randomness that you use to instantiate a random number generator which is then used when creating any random numbers required for the protocol. Cryptography relies heavily on randomness, however when experimenting a simple way to instantiate the random number generator in MIRACL is as follows:
let mut raw: [u8; 50] = [0; 50];
let mut rng = RAND::new();
rng.clean();
raw[i] = i as u8
rng.seed(50, &raw);
This simply seeds the random number generator, rng, with a byte array of [0, … , 50]. It should not be used for production.
This library supports many different elliptic curves, which can be used to implement many different protocols. My focus has been on digital signatures using pairing friendly curves.
A curve, like the bls12383 shown above is made of the following files. As a consumer of this library, a lot of the code in these files are abstracted away such that, it is not necessary to fully understand their contents. The important files containing functions that you will use regularly are ecp.rs, ecp2.rs, big.rs, pair.rs (if a pairing curve), and rom.rs.
Static Curve Constants (rom.rs)
Each curve has their own set of constants which the library uses for it’s calculations. This file might be a bit confusing, but just think it is a set of predefined constant values using byte arrays. There are a lot of constants defined in this file, however there are two that are important to understand and that you will need to use regularly:
MODULUS - the large prime that all numbers are modulus of, also known as p is some papers.
CURVE_ORDER - This is the order of the group defined by the elliptic curve bound by the field described by the MODULUS. This is the most commonly used static constant.
Now I got, and still get sometimes, confused by these two numbers. A way I like to think of it is relating it to generators in group theory. For the group defined by an elliptic curve it has an order - the number of unique curve points of CURVEORDER. That means that nG for n = 1 up to n = CURVEORDER will always produce a point on the elliptic curve. The points - (x,y) values will be constrained by the MODULUS => x or y can never be larger than the MODULUS.
BIG Numbers (big.rs)
Cryptography works with large numbers, typically 256 bits, whereas our computers default maximum is only 64-bit or 32-bit depending on the operating system. So the ability to manage these numbers needs to be implemented, although some programming languages come with their own implementations. For the interested, this paper describes how the MIRACL library handles BIG numbers.
The great thing about this library is you don’t need to handle this complexity, you just need to understand how to use the big number structure defined. Here are a few of the functions I used regularly although I recommend familiarising yourself with the code within big.rs:
Creating a new BIG from a byte array:
let mut q = BIG::new_ints(&rom::CURVE_ORDER);
This creates a new big int which represents the order of the elliptic curve
Creating random numbers, you will use this alot
pub fn randomnum(q: &BIG, rng: &mut RAND) -> BIG
Here you pass in a random number generator and a modulus which the number cannot be larger than. This is typically the CURVE_ORDER, q.
Creating a BIG integer representing 1.
let mut one = BIG::new();
Specifically you have to create the integer before you set it to one.
Convert to string - tostring()
Modular Math - typically this will always be modulo q
Addition - modadd
/* return a*b mod m */
pub fn modmul(a1: &BIG, b1: &BIG, m: &BIG) -> BIG
/* return this^e mod m */
pub fn powmod(&mut self, e1: &BIG, m: &BIG) -> BIG
Reduce modulo n - sometimes there is a need to reduce big integers so they fit back within a certain modulo. Will be needed whenever hashing a string to a BIG.
/* reduce self mod n */
pub fn rmod(&mut self, n: &BIG)
Curve Points (ecp.rs and ecp2.rs)
The ecp.rs file (and ecp2.rs for pairing friendly curves) contains all the functions for creating and manipulating curve points. Note, there is no curve object in the MIRACL library. Also, remember that all curves have a generator which when multiplied by any number less than the CURVE_ORDER creates a valid elliptic curve point.
With that in mind there are a number of functions that you will need to use a lot
Creating a generator point
ECP::generator() in rust, but all libraries should have a generator function
This create a generator for the curve that is always the same point on the curve.
Creating a point on the curve with a known modulus.
Typically used for creating a public and private key pair
let g1 = ECP::generator();
// must be modulus the curve order q
let x = BIG::randomnum(q, rng);
// note pair.rs gives a more efficient way to do this
let X = g1.mul(&x);
Creating a random point on the curve with unknown modulus
Some protocols need random points on the curve, never knowing the modulus is a safer way to program this
let rn=BIG::randomnum(&q, & mut rng2);
// Hashes the random number to an point on the curve with unknown modulus
let mut A=ECP::hashit(&rn);
/* self+=Q */
pub fn add(&mut self, Q: &ECP)
Pairing (pair.rs)
All pairing friendly curves have a file containing the pairing operations available. You generally want to be able to compute some variation of the pairing function:
e(g_1^x, g_2^y) \rightarrow e(g_1, g_2)^{xy}
A lot of cryptographic protocols are based on this transformation, taking two points on separate elliptic curves and mapping them to a point on a third curve. This is a good paper on pairing for cryptography if you aren’t too familiar. This is computed in the library using two functions, first computing the ate pairing then the final exponentiation.
let mut lhs = pair::ate(&g2, &A);
lhs = pair::fexp(&lhs);
Note: Both functions must be used for a correct pairing.
Additionally, as mentioned, this file includes functions for multiplication of elliptic curve points by big numbers which Mike suggested is faster. These are g1mul and g2mul respectively depending on the curve your point is from.
The last thing that I had to learn to implement some of the signature schemes I have been experimenting with was how to hash a message string, the thing being signed, into a big number so that it could be used within the protocol.
The MIRACL library comes with 4 hash implementations: hash256.rs, hash384.rs, hash512.rs and sha3.rs. They all follow the same API to produce hashes of byte arrays, the difference is the output size which is either a 256, 284 or 512 bit number. When choosing the hashing algorithm to use you need to make sure you are producing numbers big enough for the curve you are using. For example the BLS381 curve will throw an error if you use a 256 bit hash.
The API is as follows.
First, initialise the hasher.
let mut hasher = HASH384::new();
For each message, process it into the hasher in byte form
hasher.process_array(message.as_bytes());
Return the hashed bytes from the hasher and convert into a BIG integer
let mut m = BIG::frombytes(&hasher.hash());
Finally, remember to reduce the BIG integer modulo the curve order
m.rmod(&q);
There is also a number of general purpose hashing functions available in hmac.rs which I believe abstract a lot of this away including choosing the appropriate hasher. However, I have not yet got to grips with the code yet.
Converting Math to Code
To close off I will just give an example of some of the conversions from a mathematical formula into code that you generally come across when reading a cryptography paper.
\tilde{g} ~{\leftarrow}{\$}~ \mathbb{G_2}
This refers to selecting a generator the the second pairing curve (ECP2). So in miracl core you could represent this as:
let g2 = ECP2::generator()
Selecting random BIG integers
Typically used when selecting a private key
(x, y)~{\leftarrow}{\$}~ \mathbb{Z}_p^2
This means select two big integers
x
y
randomly from the set
\mathbb{Z}_p
. A couple of confusing points here - the small 2 is because you are selecting 2 elements and the
p
often used to represent the modulus actually should mean the curve order. Particularly if
x
y
form a private key.
let x = BIG::randomnum(&q, rng)
Note: I have not initialised rng above which would need to happen at the start of your code.
Creating public keys
(\tilde{X}, \tilde{Y}) \leftarrow (\tilde{g}^x, \tilde{g^y})
Here again, perhaps a little confusing because
g^x
is not possible in elliptic curves and this notation actually represents
x \times g
. Here if available, I would use the functions in pair.rs to calculate these values.
let x_tilde = pair::g2mul(&g2, &x)
I have attempted to give a very basic getting started guide to the MIRACL library and cryptography programming. This is the guide I would have liked when I was beginning to dive into this last year and will work best for those already exploring the field of cryptography.
Cryptography is both challenging and fascinating. I hope this article can help a few more people start experimenting with these cryptographic primitives in a practical way. |
Bending (metalworking) - 3D CAD Models & 2D Drawings
Bending (metalworking) (7171 views - Mechanical Engineering)
A disadvantage of air bending is that, because the sheet does not stay in full contact with the dies, it is not as precise as some other methods, and stroke depth must be kept very accurate. Variations in the thickness of the material and wear on the tools can result in defects in parts produced.[2]. Thus, the use of adequate process models is important[3].
In bottoming, the sheet is forced against the V opening in the bottom tool. U-shaped openings cannot be used. Space is left between the sheet and the bottom of the V opening. The optimum width of the V opening is 6 T (T stands for material thickness) for sheets about 3 mm thick, up to about 12 T for 12 mm thick sheets. The bending radius must be at least 0.8 T to 2 T for sheet steel. Larger bend radius require about the same force as larger radii in air bending, however, smaller radii require greater force—up to five times as much—than air bending. Advantages of bottoming include greater accuracy and less springback. A disadvantage is that a different tool set is needed for each bend angle, sheet thickness, and material. In general, air bending is the preferred technique.[2]
Joggling,[5] also known as joggle bending, is an offset bending process in which the two opposite bends are each less than 90° (see following section for how bend angle is measured), and are separated by a neutral web so that the offset (in the usual case where the opposite bends are equal in angle) is less than 5 workpiece thicknesses.[6] Often the offset will be one workpiece thickness, in order to allow a lap joint which is smooth on the 'show-face'.
The neutral line (also called the neutral axis) is an imaginary line that can be drawn through the cross-section of the workpiece that represents the locus where no tensile nor compressive stresses are present on the work. Its location in the material is a function of the forces used to form the part and the material yield and tensile strengths. In the bend region, the material between the neutral line and the inside radius will be under compression during the bend. The material between the neutral line and the outside radius will be under tension during the bend. For small radii, the line with no tension/compression does no longer coincide with the line with zero strain. Furthermore, the bend allowance (see below) in air bending depends on the shoulder distance of the die [8]. As a result, the bending process is more complicated than it appears to be at first sight.
There is sheet metal software available, that performs all calculations automatically, like SolidWorks and LITIO.[9]
{\displaystyle BA=A\left({\frac {\pi }{180}}\right)\left(R+(K\times T\right))}
{\displaystyle BD=2\left(R+T\right)\tan {\frac {A}{2}}-BA}
{\displaystyle BD=R\left(2-A\right)+T\left(2-KA\right)}
In sheet metal design, the K-factor is used to calculate how much sheet metal one needs to leave for the bend in order to achieve particular final dimensions, especially for between the straight sides next the bend. Use the known K-factor and the known inner bending radius to calculate the bending radius of the neutral line. Then use the neutral bending radius to calculate the arc length of the neutral line ("circumference of circle" multiplied by the "bend angle as fraction of 360deg"). The arc length of the neutral line is the length of the sheet metal you have to leave for the bend.
{\displaystyle K={\frac {-R+{\frac {BA}{\pi A/180}}}{T}}}
Thickness to 3 x thickness 0.40 0.43 0.45
Greater than 3 x thickness 0.50 0.50 0.50
{\displaystyle K={\frac {\log(\min(100,{\frac {\max(20R,T)}{T}}))}{2\log(100)}}}
Bending is a cost effective near net shape process when used for low to medium quantities. Parts usually are lightweight with good mechanical properties. A disadvantage is that some process variants are sensitive to variations in material properties. For instance, differences in spring-back have a direct influence on the resulting bend angle. To mitigate this, various methods for in-process control have been developed[13]. Other approaches include combining brakeforming with incremental forming[14].
Broadly speaking, each bend corresponds with a set-up (although sometimes, multiple bends can be formed simultaneously). The relatively large number of set-ups and the geometrical changes during bending make it difficult to address tolerances and bending errors a priori during set-up planning, although some attempts have been made [15]
Brake (sheet metal bending)Hydraulic brake水理学板金Honing (metalworking)熱処理金属加工高周波炉誘導加熱Induction heaterInduction forgingSteel mill鉄鉱石スクラップ高炉金属工学製錬BloomerySemi-finished casting productsBar stockShear (sheet metal)ダイヤルゲージSine barSineマッフル炉Industrial furnace製鉄所型抜きファインブランキングプレスWeb (manufacturing)Roll slitting金型カムシャフト搗鉱機 |
Calculus/Hyperbolic functions - Wikibooks, open books for an open world
Calculus/Hyperbolic functions
3 Some simple identities
4 Derivatives of hyperbolic functions
5 Principal values of the main hyperbolic functions
6 Inverse hyperbolic functions
7 Derivatives of inverse hyperbolic functions
The unit hyperbola has a sector with an area half of the hyperbolic angle
The independent variable of a hyperbolic function is called a hyperbolic angle. Just as the circular functions sine and cosine can be seen as projections from the unit circle to the axes, so the hyperbolic functions sinh and cosh are projections from a unit hyperbola to the axes.
The hyperbolic functions are defined in analogy with the trigonometric functions:
{\displaystyle \sinh(x)={\frac {e^{x}-e^{-x}}{2}}}
{\displaystyle \cosh(x)={\frac {e^{x}+e^{-x}}{2}}}
{\displaystyle \tanh(x)={\frac {\sinh(x)}{\cosh(x)}}={\frac {e^{x}-e^{-x}}{e^{x}+e^{-x}}}}
The reciprocal functions csch, sech, coth are defined from these functions:
{\displaystyle {\rm {csch}}(x)={\frac {1}{\sinh(x)}}}
{\displaystyle {\rm {sech}}(x)={\frac {1}{\cosh(x)}}}
{\displaystyle \coth(x)={\frac {1}{\tanh(x)}}}
Some simple identitiesEdit
{\displaystyle \cosh ^{2}(x)-\sinh ^{2}(x)=1}
{\displaystyle 1-\tanh ^{2}(x)={\rm {sech}}^{2}(x)}
{\displaystyle \sinh(2x)=2\sinh(x)\cosh(x)}
{\displaystyle \cosh(2x)=\cosh ^{2}(x)+\sinh ^{2}(x)}
Derivatives of hyperbolic functionsEdit
{\displaystyle {\frac {d}{dx}}\sinh(x)=\cosh(x)}
{\displaystyle {\frac {d}{dx}}\cosh(x)=\sinh(x)}
{\displaystyle {\frac {d}{dx}}\tanh(x)={\rm {sech}}^{2}(x)}
{\displaystyle {\frac {d}{dx}}{\rm {csch}}(x)=-{\rm {csch}}(x){\rm {coth}}(x)}
{\displaystyle {\frac {d}{dx}}{\rm {sech}}(x)=-{\rm {sech}}(x)\tanh(x)}
{\displaystyle {\frac {d}{dx}}{\rm {coth}}(x)=-{\rm {csch}}^{2}(x)}
Principal values of the main hyperbolic functionsEdit
There is no problem in defining principal braches for sinh and tanh because they are injective. We choose one of the principal branches for cosh.
{\displaystyle \sinh :\mathbb {R} \to \mathbb {R} }
{\displaystyle \cosh :[0,\infty ]\to [1,\infty ]}
{\displaystyle \tanh :\mathbb {R} \to (-1,1)}
With the principal values defined above, the definition of the inverse functions is immediate:
{\displaystyle {\rm {arsinh:\mathbb {R} \to \mathbb {R} }}}
{\displaystyle {\rm {arcosh:[1,\infty ]\to [0,\infty ]}}}
{\displaystyle {\rm {artanh:(-1,1)\to \mathbb {R} }}}
{\displaystyle {\rm {arcsch}}}
{\displaystyle {\rm {arsech}}}
{\displaystyle {\rm {arcoth}}}
We can also write these inverses using the logarithm function,
{\displaystyle {\rm {arsinh}}(x)=\ln {\big (}x+{\sqrt {x^{2}+1}}{\big )}}
{\displaystyle {\rm {arcosh}}(x)=\ln {\big (}x+{\sqrt {x^{2}-1}}{\big )}}
{\displaystyle {\rm {artanh}}(x)=\ln \left({\sqrt {\frac {1+x}{1-x}}}\right)}
These identities can simplify some integrals.
Derivatives of inverse hyperbolic functionsEdit
{\displaystyle {\frac {d}{dx}}{\rm {arsinh}}(x)={\frac {1}{\sqrt {1+x^{2}}}}}
{\displaystyle {\frac {d}{dx}}{\rm {arcosh}}(x)={\frac {1}{\sqrt {x^{2}-1}}}\ ,\ x>1}
{\displaystyle {\frac {d}{dx}}{\rm {artanh}}(x)={\frac {1}{1-x^{2}}}\ ,\ |x|<1}
{\displaystyle {\frac {d}{dx}}{\rm {arcsch}}(x)=-{\frac {1}{|x|{\sqrt {1+x^{2}}}}}\ ,\ x\neq 0}
{\displaystyle {\frac {d}{dx}}{\rm {arsech}}(x)=-{\frac {1}{x{\sqrt {1-x^{2}}}}}\ ,\ 0<x<1}
{\displaystyle {\frac {d}{dx}}{\rm {arcoth}}(x)={\frac {1}{1-x^{2}}}\ ,\ |x|>1}
Transcendental FunctionsEdit
Hyperbolic functions are examples of transcendental functions -- they are not algebraic functions. They include trigonometric, inverse trigonometric, logarithmic and exponential functions.
Retrieved from "https://en.wikibooks.org/w/index.php?title=Calculus/Hyperbolic_functions&oldid=3535549" |
Supervised Learning | Brilliant Math & Science Wiki
Akshay Padmanabha, Karleigh Moore, Christopher Williams, and
Supervised learning is the machine learning task of determining a function from labeled data. For example, in a machine learning algorithm that detects if a post is spam or not, the training set would include posts labeled as "spam" and posts labeled as "not spam" to help teach the algorithm how to recognize the difference. Supervised learning algorithms infer a function from labeled data and use this function on new examples. Supervised learning is a core concept of machine learning and is used in areas such as bioinformatics, computer vision, and pattern recognition.
An example of k-nearest neighbors, a supervised learning algorithm. The algorithm determines the classification of a data point by looking at its k nearest neighbors.[1]
Supervised learning begins by operating on a training dataset, data points that are labeled with their appropriate outputs. For example, in the image above, the training set would be the location of the blue squares and the red triangles, and the labels for each data point would be whether the point is a blue square or a red triangle. The learning algorithm looks for a function
g(x)
that can map the input data
x
to their appropriate labels
y
well. The overall goal of the algorithm is to generalize this function so that it performs well on unknown examples.
In determining how good a specific supervised learning algorithm is, two types of risks can be minimized:
Empirical risk - Empirical risk is the expected loss of the function
g
that the supervised learning algorithm infers from the training dataset. For example, if
g
correctly maps all training data points
x_i
to their respective labels
y_i
, the empirical risk is 0. Mathematically, the empirical risk function for
N
training data points is
R(g) = \frac{1}{N}\sum\limits_{i=1}^{N} L(y_i, g(x_i))
L
is the user-defined loss function that determines the penalty of incorrectly labeling a specific data point.
In minimizing empirical risk, the supervised learning algorithm is taught to match the training data as well as possible. However, as shown in the image below, a solution can minimize empirical risk without being a good candidate function for unknown data points. This is called overfitting, which occurs when the proposed function focuses more on noise rather than the actual data, as seen below with the blue line.
For the given set of red input points, both the green and blue lines minimize error to 0. However, the green line may be more successful at predicting the coordinates of unknown data points, since it seems to generalize the data better.[2]
Structural risk - Structural risk is used to prevent the supervised learning algorithm from overfitting the training data. Structural risk minimization introduces a regularization penalty that can prefer certain solutions over others. Mathematically, the regularization penalty is a function
C(g)
that is used along with empirical risk to determine a solution. Specifically, structural risk minimizes
R(g) + \lambda C(g)
R(g)
is empirical risk and
\lambda
is a user-defined parameter that controls the regularization penalty. For example, if
\lambda=0
, the optimization problem minimizes empirical risk as before.
A good way of determining an appropriate value for
\lambda
is by using cross validation, a method that trains the supervised algorithm on training data and tests its performance on a validation dataset (data points where the correct label is known). The algorithm is then updated to minimize its error on the validation set, while still being trained on the training dataset. An example of this in action is described in ridge regression, and is widely used to determine functions that perform well on unknown data.
There are many challenges in constructing supervised learning algorithms, with four important ones described below.
Bias-variance tradeoff - Suppose a supervised learning algorithm is trained on multiple datasets. If the algorithm is unable to correctly label a specific data point, it is said to be biased for that input. Additionally, if the algorithm produces different output values when trained on different datasets, it is said to have high variance. Empirical risk focuses on bias, whereas structural risk focuses on variance. There is usually a tradeoff between bias and variance, where a low bias implies high variance and vice-versa. A problem for supervised algorithms is finding the balance between these two concepts that works best with unknown data points. The blue curve minimizes the error of the data points (low bias) but has high variance. Conversely, the black line does not minimize errors (has a higher bias), but fits the data well (low variance).[3]
Complexity - The function that the supervised learning algorithm is trying to mimic could be either simple or complex. If the expected function is simple, the algorithm should have low variance to well fit the data. However, if the expected function is complex, the algorithm should have high variance to adapt to unknown data points. Supervised learning algorithms should be able to determine variance appropriately based on the amount of data and the type of function to be expecting.
Many dimensions - When a supervised learning algorithm is given a dataset consisting of many dimensions, it may attempt to identify trends between irrelevant factors. This increases the variance of the inferred function and can decrease the accuracy of the algorithm. Two ways of combating this issue include running a different algorithm to discard irrelevant variables and reducing the input data to a lower number of dimensions.
Format of Data - If the training data has errors in its labeling or in its data value, the supervised learning algorithm should not attempt to exactly match the training examples. This can lead to overfitting and will not perform well for unknown values. Additionally, if the training data contains redundant information, supervised learning algorithms may perform poorly due to over-relying on specific examples. Filtering the data or regularizing the algorithm properly can mitigate these two issues.
Ajanki, A. Example of k-nearest neighbour classificationnb. Retrieved May 28, 2016, from https://en.wikipedia.org/wiki/K-nearest_neighbors_algorithm#/media/File:KnnClassification.svg
Cite as: Supervised Learning. Brilliant.org. Retrieved from https://brilliant.org/wiki/supervised-learning/ |
Bashicu matrix calculator shows calculation process of Bashicu matrix. Set initial variable in the form of BM[n], where BM is a sequence expression of Bashicu matrix, n is a natural number, and press "Go" button. See some examples below. You can also download and use offline version of this program with milder limit on input variables.
(Japanese) バシク行列計算機は、バシク行列の計算過程を表示します。Initial variable に、初期値 BM[n] (BM はバシク行列の数列表記) を入力して Go ボタンを押して下さい。Maximum length は計算を終了する数列の長さ（行列の列数）、n increment は n の値をどのように変化させるかを指定します。下に例を示します。オフラインバージョンをインストールすることもできます。
Bashicu matrix
Bashicu matrix system is a notation designed to produce large numbers. Bashicu matrix is a matrix such as
% <![CDATA[ \begin{pmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\end{pmatrix} %]]>
where all elements are nonnegative integers. The matrix can be written in the form of
(a_{11},a_{21})(a_{12},a_{22})(a_{13},a_{23})
; sequence of transpose of each column. With an algorithm invented by Bashicu in 2014 and updated as shown below in version history, Bashich matrix BM works as a function from a natural number n to a natural number BM[n] (provided that the calculation ends), and written as (0,0)(1,1)(1,1)[3]. It is known that 2-row matrix, pair sequence, can be approximated with Hardy function. When the function is approximated with Hardy function, the matrix itself represents the ordinal of the Hardy function, and therefore can be written as:
Official definition of the algorithm of Bashicu matrix is in the source code of C program of this site, Bashicu matrix calculator. Human readable definition and analysis of Bashicu matrix system is available at the entry at googology wiki.
Versions of BM
Version 1 (BM1) NT Bashicu 2014
Version 2.1 (BM2.1) NT koteitan 2018
Version 2.2 (BM2.2) B koteitan 2018
Version 3.1 (BM3.1) B Nish 2018
Version 3.3 (BM3.3) B rpakr, Ecl1psed 2019
Version 4 (BM4) B Bashicu 2018
Status column shows if (0,0,0,...,0)(1,1,1,...,1)[n] always terminates.
T: It is proved that it always terminates. There is no such version yet.
B: Believed to terminate but there is no proof.
NT: It is proved that it does not always terminate.
See original definitions of each version for detail.
(0)(1)[3] (Hardy)
(0)(1)(1)[3] (Hardy)
(0)(1)(1)(1)[3] (Hardy)
(0,0)(1,1)[3] (Hardy)
(0,0)(1,1)[2]
(0,0)(1,1)(1,1)[2]
(0,0,0)(1,1,1)[2]
(0,0,0)(1,1,1)(2,0,0)(1,1,0)(2,1,0)(3,1,0)[2]
(0,0,0)(1,1,1)(2,0,0)(1,1,1)[2]
(0,0,0,0)(1,1,1,1)(2,2,1,1)(3,3,1,1)(4,2,0,0)(5,1,1,1)(6,2,1,1)(7,3,1,1)[1] |
Frequency response of digital filter - MATLAB freqz - MathWorks Korea
H\left(z\right)=\frac{0.05634\left(1+{z}^{-1}\right)\left(1-1.0166{z}^{-1}+{z}^{-2}\right)}{\left(1-0.683{z}^{-1}\right)\left(1-1.4461{z}^{-1}+0.7957{z}^{-2}\right)}.
H\left(z\right)=\frac{0.05634\left(1+{z}^{-1}\right)\left(1-1.0166{z}^{-1}+{z}^{-2}\right)}{\left(1-0.683{z}^{-1}\right)\left(1-1.4461{z}^{-1}+0.7957{z}^{-2}\right)}.
\beta =8
0.5\pi
0.35\pi
0.8\pi
0
0.1\pi
0.9\pi
H\left({e}^{j\omega }\right)=\frac{B\left({e}^{j\omega }\right)}{A\left({e}^{j\omega }\right)}=\frac{\text{b(1)}+\text{b(2)}\text{\hspace{0.17em}}{e}^{-j\omega }+\text{b(3)}\text{\hspace{0.17em}}{e}^{-j2\omega }+\cdots +\text{b(M)}\text{\hspace{0.17em}}{e}^{-j\left(M-1\right)\omega }}{\text{a(1)}+\text{a(2)}\text{\hspace{0.17em}}{e}^{-j\omega }+\text{a(3)}\text{\hspace{0.17em}}{e}^{-j2\omega }+\cdots +\text{a(N)}\text{\hspace{0.17em}}{e}^{-j\left(N-1\right)\omega }}. |
Update performance metrics in linear incremental learning model given new data and train model - MATLAB updateMetricsAndFit - MathWorks Switzerland
Update Performance Metrics and Train Model on Data Stream
Update performance metrics in linear incremental learning model given new data and train model
Given streaming data, updateMetricsAndFit first evaluates the performance of a configured incremental learning model for linear regression (incrementalRegressionLinear object) or linear binary classification (incrementalClassificationLinear object) by calling updateMetrics on incoming data. Then updateMetricsAndFit fits the model to that data by calling fit. In other words, updateMetricsAndFit performs prequential evaluation because it treats each incoming chunk of data as a test set, and tracks performance metrics measured cumulatively and over a specified window [1].
updateMetricsAndFit provides a simple way to update model performance metrics and train the model on each chunk of data. Alternatively, you can perform the operations separately by calling updateMetrics and then fit, which allows for more flexibility (for example, you can decide whether you need to train the model based on its performance on a chunk of data).
Mdl = updateMetricsAndFit(Mdl,X,Y) returns an incremental learning model Mdl, which is the input incremental learning model Mdl with the following modifications:
updateMetricsAndFit measures the model performance on the incoming predictor and response data, X and Y respectively. When the input model is warm (Mdl.IsWarm is true), updateMetricsAndFit overwrites previously computed metrics, stored in the Metrics property, with the new values. Otherwise, updateMetricsAndFit stores NaN values in Metrics instead.
updateMetricsAndFit fits the modified model to the incoming data by following this procedure:
Initialize the solver with the configurations and linear model coefficient and bias estimates of the input model Mdl.
Mdl = updateMetricsAndFit(Mdl,X,Y,Name,Value) uses additional options specified by one or more name-value pair arguments. For example, you can specify that the columns of the predictor data matrix correspond to observations, and set observation weights.
Fit the incremental model to the training data by using the updateMetricsAndFit function. At each iteration:
{\beta }_{1}
{\beta }_{1}
{\beta }_{1}
Track Performance Metrics and Fit Model
Perform incremental learning on the rest of the data by using the updateMetricsAndFit function. At each iteration:
Call updateMetricsAndFit to update the cumulative and window epsilon insensitive loss of the model given the incoming chunk of observations, and then fit the model to the data. Overwrite the previous incremental model with a new one. Specify that the observations are oriented in columns, and specify the observation weights.
{\beta }_{313}
IncrementalMdl = updateMetricsAndFit(IncrementalMdl,Xil(:,idx),Yil(idx), ...
beta313(j+1) = IncrementalMdl.Beta(end);
{\beta }_{313}
The cumulative loss gradually changes with each iteration (chunk of 500 observations), whereas the window loss jumps. Because the metrics window is 200 by default, updateMetricsAndFit measures the performance based on the latest 200 observations in each 500 observation chunk.
{\beta }_{313}
changes, but levels off quickly, as fit processes chunks of observations.
Incremental learning model whose performance is measured and then the model is fit to data, specified as an incrementalClassificationLinear or incrementalRegressionLinear model object. You can create Mdl directly or by converting a supported, traditionally trained machine learning model using the incrementalLearner function. For more details, see the corresponding reference page.
If Mdl.IsWarm is false, updateMetricsAndFit does not track the performance of the model. For more details, see Performance Metrics.
Chunk of predictor data with which to measure the model performance and then to fit the model to, specified as a floating-point matrix of n observations and Mdl.NumPredictors predictor variables. The value of the ObservationsIn name-value argument determines the orientation of the variables and observations. The default ObservationsIn value is "rows", which indicates that observations in the predictor data are oriented along the rows of X.
If Mdl.NumPredictors = 0, updateMetricsAndFit infers the number of predictors from X, and sets the corresponding property of the output model. Otherwise, if the number of predictor variables in the streaming data changes from Mdl.NumPredictors, updateMetricsAndFit issues an error.
updateMetricsAndFit supports only floating-point input predictor data. If your input data includes categorical data, you must prepare an encoded version of the categorical data. Use dummyvar to convert each categorical variable to a numeric matrix of dummy variables. Then, concatenate all dummy variable matrices and any other numeric predictors. For more details, see Dummy Variables.
Y — Chunk of responses (or labels)
Chunk of responses (or labels) with which to measure the model performance and then fit the model to, specified as a categorical, character, or string array, logical or floating-point vector, or cell array of character vectors for classification problems; or a floating-point vector for regression problems.
updateMetricsAndFit supports binary classification only.
If Y contains a label that is not a member of Mdl.ClassNames, updateMetricsAndFit issues an error.
If an observation (predictor or label) or weight contains at least one missing (NaN) value, updateMetricsAndFit ignores the observation. Consequently, updateMetricsAndFit uses fewer than n observations to compute the model performance and create an updated model, where n is the number of observations in X.
The chunk size n and the stochastic gradient descent (SGD) hyperparameter mini-batch size (Mdl.BatchSize) can be different values, and n does not have to be an exact multiple of the mini-batch size. If n < Mdl.BatchSize, updateMetricsAndFit uses the n available observations when it applies SGD. If n > Mdl.BatchSize, the function updates the model with a mini-batch of the specified size multiple times, and then uses the rest of the observations for the last mini-batch. The number of observations for the last mini-batch can be smaller than Mdl.BatchSize.
Chunk of observation weights, specified as the comma-separated pair consisting of 'Weights' and a floating-point vector of positive values. updateMetricsAndFit weighs the observations in X with the corresponding values in Weights. The size of Weights must equal n, which is the number of observations in X.
When you call updateMetricsAndFit, the following conditions apply:
If the model is not warm, updateMetricsAndFit does not compute performance metrics. As a result, the Metrics property of Mdl remains completely composed of NaN values. For more details, see Performance Metrics.
If Mdl.EstimationPeriod > 0, updateMetricsAndFit estimates hyperparameters using the first Mdl.EstimationPeriod observations passed to it; the function does not train the input model using that data. However, if an incoming chunk of n observations is greater than or equal to the number of observations remaining in the estimation period m, updateMetricsAndFit estimates hyperparameters using the first n – m observations, and fits the input model to the remaining m observations. Consequently, the software updates the Beta and Bias properties, hyperparameter properties, and recordkeeping properties such as NumTrainingObservations.
For classification problems, if the ClassNames property of the input model Mdl is an empty array, updateMetricsAndFit sets the ClassNames property of the output model Mdl to unique(Y).
updateMetricsAndFit tracks model performance metrics, specified by the row labels of the table in Mdl.Metrics, from new data when the incremental model is warm (IsWarm property is true). An incremental model is warm after an incremental fitting, like updateMetricsAndFit, fits the incremental model to Mdl.MetricsWarmupPeriod observations, which is the metrics warm-up period.
If Mdl.EstimationPeriod > 0, updateMetricsAndFit estimates hyperparameters before fitting the model to data. Therefore, the functions must process an additional EstimationPeriod observations before the model starts the metrics warm-up period.
The Metrics property of the incremental model stores two forms of each performance metric as variables (columns) of a table, Cumulative and Window, with individual metrics in rows. When the incremental model is warm, updateMetricsAndFit updates the metrics at the following frequencies:
Cumulative — The function computes cumulative metrics since the start of model performance tracking. The function updates metrics every time you call the function and bases the calculation on the entire supplied data set.
For classification problems, if the prior class probability distribution is known (in other words, the prior distribution is not empirical), updateMetricsAndFit normalizes observation weights to sum to the prior class probabilities in the respective classes. This action implies that observation weights are the respective prior class probabilities by default.
For regression problems or if the prior class probability distribution is empirical, the software normalizes the specified observation weights to sum to 1 each time you call updateMetricsAndFit.
Use saveLearnerForCoder, loadLearnerForCoder, and codegen (MATLAB Coder) to generate code for the updateMetricsAndFit function. Save a trained model by using saveLearnerForCoder. Define an entry-point function that loads the saved model by using loadLearnerForCoder and calls the updateMetricsAndFit function. Then use codegen to generate code for the entry-point function.
To generate single-precision C/C++ code for updateMetricsAndFit, specify the name-value argument "DataType","single" when you call the loadLearnerForCoder function.
This table contains notes about the arguments of updateMetricsAndFit. Arguments not included in this table are fully supported.
If you configure Mdl to shuffle data (Mdl.Shuffle is true, or Mdl.Solver is 'sgd' or 'asgd'), the updateMetricsAndFit function randomly shuffles each incoming batch of observations before it fits the model to the batch. The order of the shuffled observations might not match the order generated by MATLAB®. Therefore, the fitted coefficients computed in MATLAB and by the generated code might not be equal. |
Vanish at infinity - Wikipedia
In mathematics, a function is said to vanish at infinity if its values approach 0 as the input grows without bounds. There are two different ways to define this with one definition applying to functions defined on normed vector spaces and the other applying to functions defined on locally compact spaces. Aside from this difference, both of these notions correspond to the intuitive notion of adding a point at infinity, and requiring the values of the function to get arbitrarily close to zero as one approaches it. This definition can be formalized in many cases by adding an (actual) point at infinity.
2 Rapidly decreasing
A function on a normed vector space is said to vanish at infinity if the function approaches
{\displaystyle 0}
as the input grows without bounds (that is,
{\displaystyle f(x)\to 0}
{\displaystyle \|x\|\to \infty }
{\displaystyle \lim _{x\to -\infty }f(x)=\lim _{x\to +\infty }f(x)=0.}
in the specific case of functions on the real line.
{\displaystyle f(x)={\frac {1}{x^{2}+1}}}
defined on the real line vanishes at infinity.
Alternatively, a function
{\displaystyle f}
on a locally compact space vanishes at infinity, if given any positive number ε, there exists a compact subset
{\displaystyle K}
{\displaystyle \|f(x)\|<\varepsilon }
whenever the point
{\displaystyle x}
lies outside of
{\displaystyle K.}
[1][2][3] In other words, for each positive number ε the set
{\displaystyle \left\{x\in X:\|f(x)\|\geq \varepsilon \right\}}
has compact closure. For a given locally compact space
{\displaystyle \Omega }
the set of such functions
{\displaystyle f:\Omega \to \mathbb {K} }
valued in
{\displaystyle \mathbb {K} ,}
which is either
{\displaystyle \mathbb {R} }
{\displaystyle \mathbb {C} ,}
forms a
{\displaystyle \mathbb {K} }
-vector space with respect to pointwise scalar multiplication and addition, which is often denoted
{\displaystyle C_{0}(\Omega ).}
As an example, the function
{\displaystyle h(x,y)={\frac {1}{x+y}}}
{\displaystyle x}
{\displaystyle y}
are reals greater or equal 1 and correspond to the point
{\displaystyle (x,y)}
{\displaystyle \mathbb {R} _{\geq 1}^{2}}
vanishes at infinity.
A normed space is locally compact if and only if it is finite-dimensional so in this particular case, there are two different definitions of a function "vanishing at infinity". The two definitions could be inconsistent with each other: if
{\displaystyle f(x)=\|x\|^{-1}}
in an infinite dimensional Banach space, then
{\displaystyle f}
vanishes at infinity by the
{\displaystyle \|f(x)\|\to 0}
definition, but not by the compact set definition.
Rapidly decreasing[edit]
Main article: Schwartz space
Refining the concept, one can look more closely to the rate of vanishing of functions at infinity. One of the basic intuitions of mathematical analysis is that the Fourier transform interchanges smoothness conditions with rate conditions on vanishing at infinity. The rapidly decreasing test functions of tempered distribution theory are smooth functions that are
{\displaystyle O\left(|x|^{-N}\right)}
{\displaystyle N}
{\displaystyle |x|\to \infty }
, and such that all their partial derivatives satisfy the same condition too. This condition is set up so as to be self-dual under Fourier transform, so that the corresponding distribution theory of tempered distributions will have the same property.
Infinity – Mathematical concept
Projectively extended real line – Real numbers with an added point at infinity
Zero of a function – Point where function's value is zero
^ "Function vanishing at infinity - Encyclopedia of Mathematics". www.encyclopediaofmath.org. Retrieved 2019-12-15.
^ "vanishing at infinity in nLab". ncatlab.org. Retrieved 2019-12-15.
^ "vanish at infinity". planetmath.org. Retrieved 2019-12-15.
Hewitt, E and Stromberg, K (1963). Real and abstract analysis. Springer-Verlag. {{cite book}}: CS1 maint: multiple names: authors list (link)
Retrieved from "https://en.wikipedia.org/w/index.php?title=Vanish_at_infinity&oldid=1083869630" |
Moment of Inertia | Brilliant Math & Science Wiki
Rohit Gupta, Sudeep Salgia, Nihar Mahajan, and
The moment of inertia is a physical quantity which describes how easily a body can be rotated about a given axis. It is a rotational analogue of mass, which describes an object's resistance to translational motion. Inertia is the property of matter which resists change in its state of motion. Inertia is a measure of the force that keeps a stationary object stationary, or a moving object moving at its current speed. The larger the inertia, the greater the force that is required to bring some change in its velocity in a given amount of time. Suppose a heavy truck and a light car are both at rest, then intuitively we know that more force will be required to push the truck to a certain speed in a given amount of time than will be needed to push the car to that same speed in the same amount of time.
Similarly, moment of inertia is that property where matter resists change in its state of rotatory motion. The larger the moment of inertia, the greater the amount of torque that will be required to bring the same change in its angular velocity in a given amount of time. Here, torque and angular velocity are the angular analogues of force and velocity, relating to moment of inertia in the exact same way that force and velocity relate to mass.
Unlike inertia, moment of inertia depends not only on the mass but also the distribution of mass around the axis about which the moment of inertia is to be calculated. An object can have different moments of inertia about different axes. That is, to rotate an object about different axes with an equal angular acceleration, different torque (or effort) is required. This concept is relevant and highly necessary throughout mechanics. While life would be simple if nothing rotated, realistically we need to have a way to deal with both translation and rotation (often at the same time). This is a necessary piece in analyzing more complex motion.
Moment of inertia, how intuitive is it?
General Properties and Ideas
Relation of Moment of Inertia with Kinetic Energy of the System
Center of Mass - Acceleration
Center of Mass Distribution
Moment of inertia shows the tendency of an object to stay in its state of rotatory motion. If an object has more mass, then it is more difficult to rotate it. Consider two spheres of the same radius: one made up of wood, and the other of iron. Both are at the same distance from the axis of rotation.
Which one is easier to rotate? The one which is made up of wood will be easier to rotate as it is of lesser mass. It requires more efforts for larger mass to set up into the rotation. Thus, moment of inertia depends upon mass.
Consider the two identical objects of the same mass at different distances from the axis of rotation:
Two identical objects at different distances from axis of rotation
It is not equally easy to rotate both of them about the same axis of rotation. More efforts are required for the object at a greater distance to accelerate to the same angular velocity. Thus, it can be calculated that moment of inertia is dependent on the distance from the axis. If the mass is farther away from the axis, its moment of inertia is greater.
Consider a cricket bat as shown in the diagram. There are two axes about which the bat can be rotated. About which axis is it easier to rotate? Do we require the same or different amount of torque to produce the same angular acceleration about the two axes? If different, then about which axis is less torque required?
Well, the answer is that it is easier to rotate about axis 2. As the mass moves away from the axis, it becomes more difficult to rotate. Thus, the bat has a different moment of inertia about the axes. The moment of inertia of the bat is less about axis 2 compared to that about axis 1. Therefore, we can say that as the mass moves away from the axis, its moment of inertia increases and it becomes more difficult to rotate.
Moment of inertia increases Moment of inertia remains the same as the total mass remains the same Moment of inertia decreases due to the decrease in the distance of the redistributed mass Not enough information
A metal ring is melted and a solid sphere is made out of it. What happens to the moment of inertia about the vertical axis through the center?
Moment of inertia is a tensor quantity. It has different values for different axes.
It depends upon the mass as well as the mass's distribution around its axis.
A body can have different moments of inertia about different axes.
It is an inherent property of matter by which it tries to maintain its state of angular motion unless and until it is compelled by external torques.
It is an extensive (additive) property: the moment of inertia of a composite system is the sum of the moments of inertia of its components' subsystems (all taken about the same axis).
The moment of inertia of a point mass
m
about an axis at a perpendicular distance of
r
mr^2
Therefore, if the distance of a point mass from the axis is doubled, then the moment of inertia will be quadrupled. If the mass is doubled, then the moment of inertia will also be doubled.
The moment of inertia of an
-point mass system
\{ m_i \}_{i=1}^n
at perpendicular distances
\{ r_i \}_{i=1}^n
from the axis of rotation is given by
\displaystyle \sum_{i=1}^n m_i r_i^2
The moment of inertia of the masses adds up just as a scalar quantity would.
Avoid this pitfall:
It should also be noted that the moment of inertia of a system of particles about an axis is not the same as the moment of inertia of the center of mass of the system of particles about the same axis.
Find the moment of inertia of a point mass system consisting of six equal masses, each of mass
m
placed at the corners of a regular hexagon of side length
about an axis passing through the center of the hexagon and perpendicular to its plane. Find the moment of inertia of mass of
6m
placed at the center of mass of the above system.
According to the formula, the moment of inertia is
\displaystyle I = \sum_{i=1}^6 m_i r_i^2
. Here all the masses are the same, so
m_i = m
i = 1,2,3,4,5,6
. Also, in a regular hexagon, the distances of the all the corners from the center are the same and equal to the side length of the hexagon. Thus,
r_i = a
i = 1,2,3,4,5,6
I = \sum_{i=1}^6 m a^2 = 6ma^2 .
Moment of inertia of continuous mass distribution:
A continuous mass system can be thought of as a collection of infinite mass particles. A bigger object can be broken down into infinitely small elemental point masses. The total moment of inertia will be the sum of all these particles. An integration helps to add the moments of inertia of all these particles.
Thus, the moment of inertia of a distributed mass system can be written as
I = \int dm\, r^2,
I
dm
is the mass of a small element considered on the object, and
r
is the distance of the elemental mass from the axis.
Calculate the moment of inertia of a uniform, straight rod of mass
M
L
about an axis passing through one of its ends.
Consider an elemental particle present at a distance of
x
, having a length of
dx
from the axis. Now we know that the total mass
M
is present in a length of
L
and hence the mass of a particle of width
dx
\frac{M\cdot dx}{L}=dm
. We then can simply integrate this as follows:
\begin{aligned} I&=\int_0^Ldm \cdot x^2\\ &=\int_0^L\frac{Mdx}L\times x^2\\ &=\frac ML\times \left[\frac{x^3}{3}\right]_0^L\\ &=\frac{ML^2}{3}. \end{aligned}
65\text{ kg m}^2
55\text{ kg m}^2
60\text{ kg m}^2
50\text{ kg m}^2
A mass of
1\text{ kg}
is placed at
(1\text{ m}, 2\text{ m}, 0).
Another mass of
2\text{ kg}
(3\text{ m}, 4\text{ m}, 0).
Find the moment of inertia of the system about the
z
As we now know, a body can have different moments of inertia about different axes. Then, is there any relation among these moments of inertia? Yes, the moments of inertia about a few axes can be found using two theorems:
For the particle of mass
m_i
r_i
from the axis, the linear velocity is
v_i=r_i\omega
. Thus the kinetic energy
(k_i)
of the motion of the particle is
k_i=\dfrac{1}{2}m_iv_i^2=\dfrac{1}{2} m_ir_i^2\omega^2.
The total kinetic energy
K
of the body is given by the sum of kinetic energies of individual particles (assuming there are
n
particles). Hence, we have
K=\sum_{i=1}^{n} k_i=\dfrac{1}{2} \sum_{i=1}^{n} \left(m_ir_i^2\omega^2\right).
\omega
is the angular velocity constant for all particles, we can take it out of the summation to obtain
K=\dfrac{1}{2}\omega^2 \sum_{i=1}^{n} \left(m_ir_i^2\right).
Since we have defined moment of inertia as
I=\displaystyle\sum_{i=1}^n m_ir_i^2,
we have the relation
K=\dfrac{1}{2}I\omega^2.
For a planar object, the moment of inertia about an axis perpendicular to the plane is the sum of the moments of inertia of two perpendicular axes through the same point in the plane of the object:
{I_x} + {I_y} = {I_z}.
Perpendicular axis theorem is valid only for planar objects like a thin disc, rings, triangular plates, etc. For objects like spheres, cylinders, cones, etc., this theorem cannot be applied.
Consider a lamina lying in the
xy
-plane. The
z
-axis passes through the intersection of the
x
y
-axes and is perpendicular to the plane of lamina:
If we take any point on the lamina of mass
dm
at coordinate
(x,y,z),
then the moment of inertia about the
x
-axis can be calculated as
{I_x} = \int_{}^{} {dm\,{y^2}}. \qquad (1)
{I_y} = \int_{}^{} {dm\,{x^2}}. \qquad (2)
For the moment of inertia about the
z
-axis, let the distance from the
z
-axis be
r = \sqrt {{x^2} + {y^2}},
{I_z} = \int_{}^{} {dm\,{r^2}}= \int_{}^{} {dm\big({x^2}+{y^2}\big)}. \qquad (3)
(1)
(2)
(3),
\begin{aligned} {I_x} + {I_y} &= \int_{}^{} {dm\,{x^2}} + \int_{}^{} {dm\,{y^2}} \\ &= \int_{}^{} {dm\big({x^2} + {y^2}\big)} \\ &= {I_z}. \end{aligned}
To learn about the second theorem, i.e. parallel axis theorem, we first need to learn about the concept of center of mass.
The center of mass of a system is a very special point. It is that point whose acceleration depends only on the external forces. For a system of particles, the acceleration of the center of mass equals net external force divided by the total mass of the system:
\overrightarrow {{a_\text{cm}}} = \frac{{{{\overrightarrow F }_\text{net,external}}}}{{{m_\text{total}}}}.
Consider the diagram shown below. A block of mass
{m_1}
is acted upon by a force of
F
in the horizontal direction. If the ground is smooth, then what is the acceleration of the center of mass?
{a_\text{cm}} = \frac{{{F_\text{net,external}}}}{{{m_\text{total}}}}.
{F_\text{net,external}} = F.
{a_\text{cm}} = \frac{F}{{{m_1} + {m_2}}}.
It's also possible to find the actual center of mass, the center of mass of a collection of points.
Suppose that we have
N
particles of masses
{m_1}, {m_2}, \ldots, {m_N}
at their respective coordinates
({x_1},{y_1},{z_1}),\,({x_2},{y_2},{z_2}),\ldots,({x_N},{y_N},{z_N}).
x
-coordinate of the center of mass is
\displaystyle {x_\text{cm}} = \frac{{{m_1}{x_1} + {m_2}{x_2} +\cdots+ {m_N}{x_N}}}{{{m_1} + {m_2} + \cdots+ {m_N}}}.
y
\displaystyle {y_\text{cm}} = \frac{{{m_1}{y_1} + {m_2}{y_2} + \cdots+ {m_N}{y_N}}}{{{m_1} + {m_2} +\cdots+ {m_N}}}.
Likewise, the
z
\displaystyle {z_\text{cm}} = \frac{{{m_1}{z_1} + {m_2}{z_2} + \cdots+ {m_N}{z_N}}}{{{m_1} + {m_2} + \cdots + {m_N}}}.
Find the distance of the center of mass of a two-particle system of masses
{m_1}
{m_2}
from mass
{m_1}
, given that the distance between the two masses is
r.
In the diagram below, let mass
{m_1}
be at the origin, and mass
{m_2}
at the coordinate
(r,0).
{x_\text{cm}} = \frac{{{m_1}0 + {m_2}r}}{{{m_1} + {m_2}}}
{y_\text{cm}} = \frac{{{m_1}0 + {m_2}0}}{{{m_1} + {m_2}}} = 0.
Thus the distance of the center of mass from
{m_1}
\displaystyle \frac{{{m_2}r}}{{{m_1} + {m_2}}}.
Note: The distance of the center of mass from
{m_2}
\displaystyle \frac{{{m_1}r}}{{{m_1} + {m_2}}}.
A continuous mass distribution contains infinite point mass particles. A center of mass can be defined for such a system of particles with the help of integration. First, break the system into infinite small point masses and then integrate to get the location of the center of mass.
The location of the center of mass of a continuous mass distribution can be calculated as
X_\text{cm} = \frac{\int x\, dm}{\int \, dm},
dm
is the mass of an elementary particle,
x
x
-coordinate of the elementary particle, and
{X_\text{cm}}
x
-coordinate of the center of mass.
Y_\text{cm} = \frac{\int y\, dm\,} {\int \, dm}, \quad Z_\text{cm} = \frac{\int z\, dm\,} {\int \,dm} .
The theorem of parallel axes states that the moment of inertia of a rigid body about any axis is equal to its moment of inertia about a parallel axis through its center of mass plus the product of the mass of the body and the square of the perpendicular distance between the two parallel axes.
If an object has the moment of inertia
{I_\text{cm}}
about an axis passing through its center of mass, then the moment of inertia
I
about another parallel axis at a distance of
d
I = {I_\text{cm}} + m d^2 .
Cite as: Moment of Inertia. Brilliant.org. Retrieved from https://brilliant.org/wiki/calculating-moment-of-inertia-of-point-masses/ |
A shape has ________ if it can be folded about a line so that its two parts matc
A shape has ________ if it can be folded about a line so that its two parts match exactly.
A shape has Symmetry if it can be folded about a line so that its two parts match exactly. Indeed this is the definition of Symmetries of a Graph and the line is called as axis oF summetry of the graph.
\left(A-B\right)-C=\left(A-C\right)-\left(B-C\right)
\left(a,b\right)\in R
Prove the following fact by induction: Fol all
n\ge 1,\text{ }{4}^{n}\ge {3}^{n}+{n}^{2}
Consider a relation R on Z — {0} defined by the rule that (z,y)
\in
R if and only if xy > 0. a) Prove that R is an equivalence relation. b) Determine all elements in the equivalence class containing 1. How many distinct equivalence classes are there?
Given discrete math
A=\left[\begin{array}{ccc}1& 1& 1\\ 1& 0& 1\\ 1& 1& \end{array}\right]B=\left[\begin{array}{ccc}1& 1& 0\\ 1& 1& 1\\ 1& 0& 0\end{array}\right] |
Improper Integrals | Brilliant Math & Science Wiki
Henry Maltby, Brandon Monsen, Nihar Mahajan, and
An improper integral is a type of definite integral in which the integrand is undefined at one or both of the endpoints. Strictly speaking, it is the limit of the definite integral as the interval approaches its desired size.
Improper integrals may be evaluated by finding a limit of the indefinite integral of the integrand. However, such a value is meaningful only if the improper integral converges in the first place.
Improper integrals appear frequently in the study of probability distributions, asymptotic behavior, and calculus in general. In order to evaluate them properly, it is crucial to understand precisely what is meant by these integrals.
One type of improper integral is an integral where one of the endpoints is (approaching) infinity. For instance,
\int_0^\infty \frac{1}{1 + x^2} \, dx = \lim_{b \to \infty} \int_0^b \frac{1}{1 + x^2} \, dx.
Another type of improper integral is an integral where both of the endpoints are (approaching) infinity. For instance,
\int_{-\infty}^\infty e^{-x^2} \, dx = \int_{-\infty}^0 e^{-x^2} \, dx + \int_0^\infty e^{-x^2} \, dx
is an integral used in Gaussian distributions.
A third type of improper integral is an integral for which an asymptote appears at one (or both) of the endpoints. For instance,
\int_0^1 \frac{1}{\sqrt{x}} \, dx = \lim_{a \to 0^+} \int_a^1 \frac{1}{\sqrt{x}} \, dx.
A fourth type of improper integral is one that is not properly defined: an integral whose integrand is not defined at all the values within the interval of integration. For instance,
\int_0^2 \frac{1}{\sqrt[3]{(x - 1)^2}} \, dx
is undefined because the function
f(x) = \tfrac{1}{\sqrt[3]{(x-1)^2}}
is not defined throughout the interval of integration, even though its antiderivative
F(x) = 3\sqrt[3]{x-1}
is defined throughout the interval of integration. A concept known as the Cauchy principal value permits a reinterpretation of the integral as
\int_0^2 \frac{1}{\sqrt[3]{(x - 1)^2}} \, dx = \lim_{b \to 1^-} \int_0^b \frac{1}{\sqrt[3]{(x - 1)^2}} \, dx + \lim_{a \to 1^+} \int_a^2 \frac{1}{\sqrt[3]{(x - 1)^2}} \, dx
in order to assign a value to determine a value for it, provided each of the integrals converges. In general, the Cauchy principal value splits the interval of integration of an improper integral into closed intervals, on whose interiors the integrand is defined.
An improper integral may be interpreted unambiguously if its integrand is defined at all points in the interior of the interval of integration (as opposed to the fourth possibility above) and one of the endpoints. For such improper integrals, the value may either exist and be finite, diverge to
\pm \infty
, or diverge to no particular value.
\int_{-\infty}^\infty \cos x \, dx
diverges to no particular value, since the cosine function oscillates.
Note that it may be impossible to define clearly an improper integral like
\int_{-\infty}^\infty x \, dx
where the integrand is undefined at both endpoints. In such a case, the Cauchy principal value would be
\lim_{a \to \infty} \int_{-a}^a x \, dx = 0
, but the expression itself is undefined.
An improper integral converges if its limit exists (and is finite). This may be determined by considering the limiting behavior of an antiderivative of the integrand, but it is not necessary to consider the antiderivative. It is possible to prove existence by showing that, taking the integral as a geometric statement, the area in question is bounded and the limiting behavior is monotonic (strictly increasing or strictly decreasing).
Determine whether or not the improper integral
\int_0^\infty \frac{1}{1 + x^2} \, dx
The function in question is defined for all real numbers. Observe that an antiderivative
\int \tfrac{1}{1 + x^2} \, dx = \arctan x
is finite:
\lim_{b \to \infty} \arctan b = \frac{\pi}{2},
so the limit of the antiderivative exists, thus making the improper integral exist.
On the other hand, the integrand is positive for all real numbers, so the limiting behavior is strictly increasing. Furthermore, comparison to a left Riemann sum yields the inequality
\begin{aligned} \int_0^\infty \frac{1}{1 + x^2} \, dx &< 1 + \int_1^\infty \frac{1}{1 + x^2} \, dx \\ &< 1 + \int_1^\infty \frac{1}{x^2} \, dx \\ &< 1 + \sum_{n=1}^\infty \frac{1}{n^2} \\ &= 1 + \frac{\pi^2}{6}. \end{aligned}
Since the limit is bounded and increasing, it must exist, and the improper integral must also exist.
_\square
I I and II I and III I, II, and III II
Which of the following integrals converge?
\displaystyle{\int_0^\infty \frac{\ln(x)}{x^2 + 1} \, dx}
\displaystyle{\int_1^\infty \frac{1}{(x-1)^2} \, dx}
\displaystyle{\int_0^{\pi/2} \sqrt{\tan(x)} \, dx}
If an improper integral is clearly defined, its value may be determined by evaluating the limit.
\int_0^\infty \frac{1}{1 + x^2} \, dx.
As proved above, this value exists. Then
\int_0^\infty \frac{1}{1 + x^2} \, dx = \arctan(\infty) - \arctan(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}. \ _\square
\int_0^1 \frac{1}{\sqrt{x}} \, dx.
\begin{aligned} \int_0^1 \frac{1}{\sqrt{x}} \, dx &= \lim_{a \to 0^+} \int_a^1 \frac{1}{\sqrt{x}} \, dx \\ &= \lim_{a \to 0^+} \left[ 2\sqrt{x} \right]_a^1 \\ &= \lim_{a \to 0^+} 2\sqrt{1} - 2\sqrt{a} \\ &= 2. \ _\square \end{aligned}
\frac{\pi}{4}
\frac{\pi}{2}
\pi
The integral diverges.
\int _{ 0 }^{ { \pi } / { 2 } }{ \dfrac { \sqrt { \cot{x} } }{ \sqrt{\cot{x}}+ \sqrt{\tan{x}}}} \, dx = \, ?
and
b
are positive coprime integers in the following integral:
\int_0^\infty \frac{2x^2-1}{x^4+5x^2+4}dx=\frac{a\pi}{b},
a+b
Cite as: Improper Integrals. Brilliant.org. Retrieved from https://brilliant.org/wiki/improper-integrals/ |
Interpolate signal using cascaded integrator-comb filter - Simulink - MathWorks 日本
CIC interpolated output, returned as a vector or a matrix. The data type of the output is determined by the settings in the block dialog. The complexity of the output matches that of the input. The number of output rows is R✕Num, where R is the interpolation factor and Num is the number of input rows.
\text{WL}=\mathrm{ceil}\left({\mathrm{log}}_{2}\left(\frac{{\left(RM\right)}^{N}}{R}\right)\right)+I
\begin{array}{l}H\left(z\right)={\left[\underset{k=0}{\overset{RMâ1}{â}}{z}^{âk}\right]}^{N}=\frac{{\left(1â{z}^{âRM}\right)}^{N}}{{\left(1â{z}^{â1}\right)}^{N}}=\frac{{\left(1â{z}^{âRM}\right)}^{N}}{1}·\frac{1}{{\left(1â{z}^{â1}\right)}^{N}}={H}_{\text{C}}{}^{N}\left(z\right)·{H}_{\text{I}}{}^{N}\left(z\right)\\ \end{array}
H\left(z\right)=\frac{{\left(1â{z}^{âM}\right)}^{N}}{{\left(1â{z}^{â1}\right)}^{N}}.
[1] Hogenauer, E.B. “An Economical Class of Digital Filters for Decimation and Interpolation†IEEE Transactions on Acoustics, Speech and Signal Processing. Vol. 29, Number 2, 1981, pp. 155–162, 1981. |
Variation of information - Wikipedia
In probability theory and information theory, the variation of information or shared information distance is a measure of the distance between two clusterings (partitions of elements). It is closely related to mutual information; indeed, it is a simple linear expression involving the mutual information. Unlike the mutual information, however, the variation of information is a true metric, in that it obeys the triangle inequality.[1][2][3]
Information diagram illustrating the relation between information entropies, mutual information and variation of information.
1.1 Explicit information content
Suppose we have two partitions
{\displaystyle X}
{\displaystyle Y}
{\displaystyle A}
into disjoint subsets, namely
{\displaystyle X=\{X_{1},X_{2},\ldots ,X_{k}\}}
{\displaystyle Y=\{Y_{1},Y_{2},\ldots ,Y_{l}\}}
{\displaystyle n=\sum _{i}|X_{i}|=\sum _{j}|Y_{j}|=|A|}
{\displaystyle p_{i}=|X_{i}|/n}
{\displaystyle q_{j}=|Y_{j}|/n}
{\displaystyle r_{ij}=|X_{i}\cap Y_{j}|/n}
Then the variation of information between the two partitions is:
{\displaystyle \mathrm {VI} (X;Y)=-\sum _{i,j}r_{ij}\left[\log(r_{ij}/p_{i})+\log(r_{ij}/q_{j})\right]}
This is equivalent to the shared information distance between the random variables i and j with respect to the uniform probability measure on
{\displaystyle A}
{\displaystyle \mu (B):=|B|/n}
{\displaystyle B\subseteq A}
Explicit information content[edit]
We can rewrite this definition in terms that explicitly highlight the information content of this metric.
The set of all partitions of a set form a compact Lattice where the partial order induces two operations, the meet
{\displaystyle \wedge }
and the join
{\displaystyle \vee }
, where the maximum
{\displaystyle {\overline {\mathrm {1} }}}
is the partition with only one block, i.e., all elements grouped together, and the minimum is
{\displaystyle {\overline {\mathrm {0} }}}
, the partition consisting of all elements as singletons. The meet of two partitions
{\displaystyle X}
{\displaystyle Y}
is easy to understand as that partition formed by all pair intersections of one block of,
{\displaystyle X_{i}}
{\displaystyle X}
and one,
{\displaystyle Y_{i}}
{\displaystyle Y}
{\displaystyle X\wedge Y\subseteq X}
{\displaystyle X\wedge Y\subseteq Y}
Let's define the entropy of a partition
{\displaystyle X}
{\displaystyle H\left(X\right)\,=\,-\sum _{i}\,p_{i}\log p_{i}}
{\displaystyle p_{i}=|X_{i}|/n}
{\displaystyle H({\overline {\mathrm {1} }})=0}
{\displaystyle H({\overline {\mathrm {0} }})=\log \,n}
. The entropy of a partition is a monotonous function on the lattice of partitions in the sense that
{\displaystyle X\subseteq Y\Rightarrow H(X)\geq H(Y)}
Then the VI distance between
{\displaystyle X}
{\displaystyle Y}
{\displaystyle \mathrm {VI} (X,Y)\,=\,2H(X\wedge Y)\,-\,H(X)\,-\,H(Y)}
{\displaystyle d(X,Y)\equiv |H\left(X\right)-H\left(Y\right)|}
is a pseudo-metric as
{\displaystyle d(X,Y)=0}
doesn't necessarily imply that
{\displaystyle X=Y}
{\displaystyle {\overline {\mathrm {1} }}}
{\displaystyle \mathrm {VI} (X,\mathrm {1} )\,=\,H\left(X\right)}
If in the Hasse diagram we draw an edge from every partition to the maximum
{\displaystyle {\overline {\mathrm {1} }}}
and assign it a weight equal to the VI distance between the given partition and
{\displaystyle {\overline {\mathrm {1} }}}
, we can interpret the VI distance as basically an average of differences of edge weights to the maximum
{\displaystyle \mathrm {VI} (X,Y)\,=\,|\mathrm {VI} (X,{\overline {\mathrm {1} }})\,-\,\mathrm {VI} (X\wedge Y,{\overline {\mathrm {1} }})|\,+\,|\mathrm {VI} (Y,{\overline {\mathrm {1} }})\,-\,\mathrm {VI} (X\wedge Y,{\overline {\mathrm {1} }})|\,=\,d(X,X\wedge Y)\,+\,d(Y,X\wedge Y)}
{\displaystyle H(X)}
as defined above, it holds that the joint information of two partitions coincides with the entropy of the meet
{\displaystyle H(X,Y)\,=\,H(X\wedge Y)}
and we also have that
{\displaystyle d(X,X\wedge Y)\,=\,H(X\wedge Y|X)}
coincides with the conditional entropy of the meet (intersection)
{\displaystyle X\wedge Y}
{\displaystyle X}
The variation of information satisfies
{\displaystyle \mathrm {VI} (X;Y)=H(X)+H(Y)-2I(X,Y)}
{\displaystyle H(X)}
{\displaystyle X}
{\displaystyle I(X,Y)}
is mutual information between
{\displaystyle X}
{\displaystyle Y}
with respect to the uniform probability measure on
{\displaystyle A}
{\displaystyle \mathrm {VI} (X;Y)=H(X,Y)-I(X,Y)}
{\displaystyle H(X,Y)}
is the joint entropy of
{\displaystyle X}
{\displaystyle Y}
{\displaystyle \mathrm {VI} (X;Y)=H(X|Y)+H(Y|X)}
{\displaystyle H(X|Y)}
{\displaystyle H(Y|X)}
are the respective conditional entropies.
The variation of information can also be bounded, either in terms of the number of elements:
{\displaystyle \mathrm {VI} (X;Y)\leq \log(n)}
Or with respect to a maximum number of clusters,
{\displaystyle K^{*}}
{\displaystyle \mathrm {VI} (X;Y)\leq 2\log(K^{*})}
^ P. Arabie, S.A. Boorman, S. A., "Multidimensional scaling of measures of distance between partitions", Journal of Mathematical Psychology (1973) , vol. 10, 2, pp. 148–203, doi: 10.1016/0022-2496(73)90012-6
^ W.H. Zurek, Nature, vol 341, p119 (1989); W.H. Zurek, Physics Review A, vol 40, p. 4731 (1989)
^ Marina Meila, "Comparing Clusterings by the Variation of Information", Learning Theory and Kernel Machines (2003), vol. 2777, pp. 173–187, doi:10.1007/978-3-540-45167-9_14, Lecture Notes in Computer Science, ISBN 978-3-540-40720-1
Arabie, P.; Boorman, S. A. (1973). "Multidimensional scaling of measures of distance between partitions". Journal of Mathematical Psychology. 10 (2): 148–203. doi:10.1016/0022-2496(73)90012-6.
Meila, Marina (2003). "Comparing Clusterings by the Variation of Information". Learning Theory and Kernel Machines. Lecture Notes in Computer Science. 2777: 173–187. doi:10.1007/978-3-540-45167-9_14. ISBN 978-3-540-40720-1.
Meila, M. (2007). "Comparing clusterings—an information based distance". Journal of Multivariate Analysis. 98 (5): 873–895. doi:10.1016/j.jmva.2006.11.013.
Kingsford, Carl (2009). "Information Theory Notes" (PDF). Retrieved 22 September 2009.
Kraskov, Alexander; Harald Stögbauer; Ralph G. Andrzejak; Peter Grassberger (2003). "Hierarchical Clustering Based on Mutual Information". arXiv:q-bio/0311039.
Partanalyzer includes a C++ implementation of VI and other metrics and indices for analyzing partitions and clusterings
C++ implementation with MATLAB mex files
Retrieved from "https://en.wikipedia.org/w/index.php?title=Variation_of_information&oldid=1085275463" |
Find the augmented matrix for the following system of linear equations: 3x+7y-
Find the augmented matrix for the following system of linear equations: 3x+7y-20z=-45x+12y-34z=-7
Find the augmented matrix for the following system of linear equations:
3x+7y-20z=-4
5x+12y-34z=-7
f\left(x\right)=\sqrt{4-x}
a=0
\sqrt{3.9}
\sqrt{3.99}
Determine the equation of the line in slope-intercept form which passes through the point (6. 6) and has the slope
-\frac{1}{3}
. Simplify the answer, please.
Let X be a smooth hypersurface in Projective space
{\mathbb{P}}^{n}
of degree d defined by the equation f=0. Given that we have a vector bundle E of rank
r\ge 1
on X such that we have the following exact sequence on
{\mathbb{P}}^{n}
0\to O{\left(-1\right)}^{rd}\to {O}^{rd}\to E\to 0
My question is as follows. What is the morphism from
O{\left(-1\right)}^{rd}\to {O}^{rd}
? A paper indicated that it is given by a
rd×rd
matrix of linear forms. Why is this? I am not able to see it. If this is so, can we say where in
{\mathbb{P}}^{n}
the determinant of that matrix vanishes?
I have no idea how to solve it. Should be linear equation of order one since I am passing through this chapter, but I can't put into the form of
{y}^{\prime }+P\left(x\right)y=Q\left(x\right)
\left(2xy+{x}^{2}+{x}^{4}\right)dx-\left(1+{x}^{2}\right)dy=0
It is not exact since partial derivatives are not equal.
A\in {M}_{m×n}\left(\mathbb{Q}\right)
b\in {\mathbb{Q}}^{m}
. Suppose that the system of linear equations Ax=b has a solution in
{\mathbb{R}}^{n}
. Does it necessarily have a solution in
{\mathbb{Q}}^{n}
and I thought I'd give an interesting, possibly wrong, approach to solving it. I'm not sure if such things can be done, if not maybe you can help me refine.
I considered the form of the equality as
{A}^{\left(1\right)}{x}_{1}+\cdots +{A}^{\left(n\right)}{x}_{n}=b
{A}^{\left(i\right)}
is a column vector of A. I then noticed that for
{x}_{i}\in \mathbb{R}\mathrm{\setminus }\mathbb{Q}=\mathbb{T}
then, and this is where I think I'm doing something forbidden, each x has the represenation
{x}_{1}={k}_{11}{\tau }_{1}+\cdots +{k}_{1p}{\tau }_{p}
{x}_{2}={k}_{21}{\tau }_{1}+\cdots +{k}_{2p}{\tau }_{p}
⋮
{x}_{n}={k}_{n1}{\tau }_{1}+\cdots +{k}_{np}{\tau }_{p}
{\tau }_{i}
is a distinct irrational number,
{k}_{ij}\in \mathbb{R}
, and p is the number of such distinct irrational numbers. I wound this out, but there may be a discrepancy with p and m. I feel this method can lead me to the answer, but I'm not sure where to go from here.
I end up getting something like this, I believe, after substitution:
A\left({k}^{\left(1\right)}{\tau }_{1}+\cdots +{k}^{\left(p\right)}{\tau }_{p}\right)=b
{k}^{\left(i\right)}
\left(\left({k}_{1i}\right),\left(\dots \right),\left({k}_{ni}\right)\right)
I think there is no discrepancy with p and m because
A\in {M}_{m×n}\left(\mathbb{Q}\right),K\in {M}_{n×p}\left(\mathbb{R}\right)
\tau \in {M}_{p×1}\left(\mathbb{T}\right)
\left(m×n\right)\cdot \left(n×p\right)\cdot \left(p×1\right)=m×1
Define Absolute system of Linear Equations?
During some computations I came up with the following system of linear recurrences:
{B}_{n+2}=3{B}_{n}+{A}_{n}
{A}_{n}={A}_{n-1}+{B}_{n-1}
Here I am trying to find the solution for B (hoping to get some sort of homogeneous equation, find it roots and get the closed form formula).
But I can't solve it. The only thing I can do is
{B}_{n+2}=3{B}_{n}+\sum _{i=0}^{n-1}{B}_{i}
which will not help me to solve recurrence. So is there a way I can find B without the summation through n? |
Trigonometric Even-odd Functions | Brilliant Math & Science Wiki
Hemang Agarwal, Mei Li, Jonas Menzi, and
Even and odd functions are functions satisfying certain symmetries: even functions satisfy
f(x)=f(-x)
x
, while odd functions satisfy
f(x)=-f(-x)
. Trigonometric functions are examples of non-polynomial even (in the case of cosine) and odd (in the case of sine and tangent) functions.
The properties of even and odd functions are useful in analyzing trigonometric functions, particularly in the sum and difference formulas.
Trigonometric Even and Odd Functions
The cosine and sine functions satisfy the following properties:
\begin{aligned} \cos(-\theta) &= \cos \theta \\ \sin(-\theta) &= -\sin \theta. \end{aligned}
x= \cos \theta \quad \text{ and } \quad y= \sin \theta.
\theta
-\theta
x
\cos \theta=\cos (-\theta)
y
\sin (-\theta)=-\sin\theta
_\square
Now that we have the above identities, we can prove several other identities, as shown in the following example:
\begin{aligned} \tan(-\theta) &= -\tan \theta\\ \cot(-\theta) &= -\cot \theta\\ \csc(-\theta) &= -\csc \theta\\ \sec(-\theta) &= \sec \theta. \end{aligned}
\begin{aligned} \tan(-\theta) &=\frac{\sin(-\theta)}{\cos(-\theta)}=\frac{-\sin \theta}{\cos \theta}=-\tan \theta\\ \cot(-\theta) &=\frac{1}{\tan(-\theta)}=\frac{1}{-\tan \theta}=-\cot \theta\\ \csc(-\theta) &=\frac{1}{\sin (-\theta)}=\frac{1}{-\sin \theta}=-\csc \theta\\ \sec(-\theta) &=\frac{1}{\cos(-\theta)}=\frac{1}{\cos \theta}=\sec \theta. \ _\square \end{aligned}
\sin(-60^\circ) .
From trigonometric even-odd functions, we have
\begin{aligned} \sin (-60^\circ) &= -\sin 60^\circ \\ &= -\frac{\sqrt{3}}{2}. \ _\square \end{aligned}
\tan(-45^\circ).
\begin{aligned} \tan (-45^\circ) &= -\tan 45^\circ \\ &= -1. \ _\square \end{aligned}
\cos(-30^\circ)+\sec(-60^\circ).
\begin{aligned} \cos(-30^\circ)+\sec(-60^\circ) &= \cos 30^\circ + \sec 60^\circ \\ &= \frac{\sqrt{3}}{2} + \frac{1}{\cos 60^\circ} \\ &= \frac{\sqrt{3}}{2} + 2 \\ &= \frac{\sqrt{3} + 4}{2}. \ _\square \end{aligned}
\tan x \times \cot(-x).
\begin{aligned} \tan x \times \cot(-x) &= \tan x \times (-\cot x ) \\ &= \tan x \times \left(-\frac{1}{\tan x}\right) \\ &= -1. \ _\square \end{aligned}
ABCD
O
with radius 1 in the above diagram. Let
\theta
be the angle formed by
\overline {AO}
x
-axis. Then which vertex has
2\sin\left(-\frac{\theta}{2}\right)\cos\left(-\frac{\theta}{2}\right)
y
Observe that the expression can be rewritten as follows:
\begin{aligned} 2\sin\left(-\frac{\theta}{2}\right)\cos\left(-\frac{\theta}{2}\right) &= -2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} &&\quad \big(\text{since } \sin (-x) = -\sin x, \cos(-x) = \cos x\big) \\ &= -\sin \theta. &&\quad (\text{since } \sin 2\theta = 2\sin x \cos x ) \end{aligned}
Also observe that the
y
A
\sin \theta
A
C
are symmetric with respect to the origin. Then the
y
C
-\sin \theta.
Therefore, our answer is vertex
C.
_\square
Cite as: Trigonometric Even-odd Functions. Brilliant.org. Retrieved from https://brilliant.org/wiki/trigonometric-even-odd-functions/ |
Define Hermitian Matrices.
Hermitian matrix is a complex square matrix that is equal to its own conjugate transpose, that is for which
A={A}^{H}
{A}^{H}
denotes the conjugate transpose.
In other words, we can say the matrix whose matrix whose element in the i-th row and j-th column is equal to the complex conjugate of the element in the j-th row and i-th column is called Hermitian matrix.
{a}_{ij}={\overline{a}}_{ji}
A=\left[\begin{array}{cc}1& -i\\ i& 1\end{array}\right]
\overline{A}=\left[\begin{array}{cc}1& i\\ -i& 1\end{array}\right]
{\overline{A}}^{T}=\left[\begin{array}{cc}1& -i\\ i& 1\end{array}\right]
{\overline{A}}^{T}=A
Therefore, A is hermitian matrix.
2×2
\text{Basis }=\left\{\left[\begin{array}{cc}& \\ & \end{array}\right],\left[\begin{array}{cc}& \\ & \end{array}\right]\right\}
If A and B are
n×n
matrices. Find the cost of calculating
{A}^{2}+BAB
\left[\begin{array}{cc}3& 0\\ -3& 1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}6\\ -7\end{array}\right]
3×3
matrix A that has eigenvalues
\lambda =0
, 4 ,-4 with corresponding eigenvectors
\left[\begin{array}{c}0\\ 1\\ -1\end{array}\right],\left[\begin{array}{c}1\\ -1\\ 1\end{array}\right],\left[\begin{array}{c}0\\ 1\\ 1\end{array}\right]
Write each of the following linear combinations of columns as a linear system of the form in (4):
a\right){x}_{1}\left[\begin{array}{c}2\\ 0\end{array}\right]+{x}_{2}\left[\begin{array}{c}1\\ 3\end{array}\right]=\left[\begin{array}{c}4\\ 2\end{array}\right]
b\right){x}_{1}\left[\begin{array}{c}1\\ 2\\ -1\end{array}\right]+{x}_{2}\left[\begin{array}{c}0\\ 1\\ 2\end{array}\right]+{x}_{3}\left[\begin{array}{c}3\\ 4\\ 5\end{array}\right]+{x}_{4}\left[\begin{array}{c}1\\ 3\\ 4\end{array}\right]=\left[\begin{array}{c}2\\ 5\\ 8\end{array}\right]
Sketch the graph of the inequality.(x + 2)2 + y2 > 1 |
Hexadecimal Numbers | Brilliant Math & Science Wiki
Sharky Kesa, Aloysius Ng, Yash Singhal, and
Hexadecimal numbers (also known as base-16) are a system of numbers which have 16 digits, instead of 10. These digits are
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F,
A
is 10,
B
is 11, etc.
In hexadecimal, place value is determined by powers of 16, instead of 10. For example,
2A3_{16} = 2 \times 16^2 + 10 \times 16^1 + 3 \times 16^0 = 675_{10}.
8C91_{16}
in base 10.
8C91_{16}
8 \times 16^3 + 12 \times 16^2 + 9 \times 16^1 + 1 \times 16^0 = 32768 + 3072 + 144 + 1 = 35985_{10}.
35985.
_ \square
Now that you have learned about hexadecimal numbers, give this problem a try:
FACE_{16} ?
3A5_{16}
2D1_{16}
We will first convert these numbers to decimal (base 10) and work out the sum, and at the end we will convert back:
\begin{aligned} 3A5_{16} &= 3 \times 16^2 +10 \times 16^1 + 5 \times 16^0 = 768 + 160 + 5 = 933_{10}\\ 2D1_{16} &= 2 \times 256 + 13 \times 16^1 + 1 \times 16^0 = 512 + 208 + 1 = 721_{10}. \end{aligned}
Now we need to find the sum of
933
721
1654
Finally we convert back. Since
1654 = 6 \times 16^2 + 7 \times 16^1 + 6 \times 16^0 = 676_{16},
676_{16}.
_ \square
However, with a lot of practice, the sum can also be done without changing the base back to 10. For instance,
3A5_{16}
2D1_{16}
\begin{aligned} 3A5+2D1 &=3A0+2D0+6\\ &=300+200+176\\ &=676. \end{aligned}
Note all numbers in this segment are in base 16.
_\square
Multiplication will be complicated as instead of a
10 \times 10
multiplication chart, we have a
16 \times 16
one. Much harder!
3A5_{16}
2D1_{16}
3A5 \times 2D1= 74A00+2F610+3A5=A43B5.
In base 10, we have just calculated
721 \times 933 = 672693. \ _\square
Cite as: Hexadecimal Numbers. Brilliant.org. Retrieved from https://brilliant.org/wiki/hexadecimal-numbers/ |
Percentages In Calculation Practice Problems Online | Brilliant
Mike can find 3% of any number by finding 1% of the number and multiplying that value by 3.
The percentage wage increase of $20 per hour to $24 per hour is greater than the percentage wage increase of $15 per hour to $18 per hour.
If a positive number is doubled, its value increases by which percent?
50 \%
100\%
200\%
A rectangle's width is 40% of its length. If the perimeter of the rectangle is 56 units, what is the area of the rectangle?
The number of people owning memberships at a gym increased by 20% to 60. Did the gym have 40 or 50 members before the increase? |
lexorder - Maple Help
Home : Support : Online Help : Programming : Operations : Ordering : lexorder
test for lexicographical order
lexorder(s1, s2)
strings or unevaluated symbols
The procedure lexorder returns true if s1 occurs before s2 in lexicographical order, or if s1 is equal to s2. Otherwise, it returns false.
The lexicographical order depends in part on the ordering of the underlying character set, which is system-dependent.
For strings and symbols consisting of ordinary letters, lexicographical order is the standard alphabetical order.
The most common use of lexorder is as the second (optional) argument to the sort command. This allows you to sort a list of strings (or symbols) lexicographically. Note, however, that it is the symbol lexorder rather than the lexorder procedure that should be used as an option to sort. Both will work, with identical results, but the symbol lexorder is recognized by sort, causing it to use a builtin algorithm that is faster than the one that calls the lexorder procedure. (See the example below.)
The lexorder command is thread-safe as of Maple 15.
\mathrm{lexorder}\left(a,b\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
\mathrm{lexorder}\left(A,a\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
\mathrm{lexorder}\left("a",a\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
\mathrm{lexorder}\left(\mathrm{greatest},\mathrm{great}\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}}
\mathrm{lexorder}\left(\mathrm{`*`},\mathrm{`^`}\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
\mathrm{lexorder}\left("first","second"\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
\mathrm{sort}\left(["first","second","third","fourth","fifth"],'\mathrm{lexorder}'\right)
[\textcolor[rgb]{0,0,1}{"fifth"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"first"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"fourth"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"second"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"third"}]
Calling the builtin procedure lexorder() directly is slower than using the 'lexorder' algorithm that is built in to sort().
L≔[\mathrm{seq}\left(\mathrm{StringTools}:-\mathrm{Random}\left(1000,'\mathrm{alnum}'\right),i=1..10000\right)]:
\mathrm{time}\left(\mathrm{sort}\left(L,'\mathrm{lexorder}'\right)\right)
\textcolor[rgb]{0,0,1}{0.007}
\mathrm{time}\left(\mathrm{sort}\left(L,\mathrm{eval}\left(\mathrm{lexorder}\right)\right)\right)
\textcolor[rgb]{0,0,1}{0.198}
Reversing the sense of the comparison is more costly still, because a full Maple procedure call is incurred for each comparison.
\mathrm{time}\left(\mathrm{sort}\left(L,\left(a,b\right)↦\mathrm{lexorder}\left(b,a\right)\right)\right)
\textcolor[rgb]{0,0,1}{0.339}
A better way to do this is to use a linear reversal algorithm after sorting the list with the builtin algorithm. Since sorting dominates at O(n*ln(n)), using the sorting algorithm with the smaller constant factor delivers better performance.
revsort := proc( los::list(string) )::list(string);
L := sort( los, 'lexorder' );
[ seq( L[ -i ], i = 1 .. nops( L ) ) ]
\mathrm{time}\left(\mathrm{revsort}\left(L\right)\right)
\textcolor[rgb]{0,0,1}{0.013}
\mathrm{evalb}\left(\mathrm{revsort}\left(L\right)=\mathrm{sort}\left(L,\left(a,b\right)↦\mathrm{lexorder}\left(b,a\right)\right)\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
StringTools,Random |
Parabolic reflector - Wikipedia
Reflector that has the shape of a paraboloid
A parabolic (or paraboloid or paraboloidal) reflector (or dish or mirror) is a reflective surface used to collect or project energy such as light, sound, or radio waves. Its shape is part of a circular paraboloid, that is, the surface generated by a parabola revolving around its axis. The parabolic reflector transforms an incoming plane wave travelling along the axis into a spherical wave converging toward the focus. Conversely, a spherical wave generated by a point source placed in the focus is reflected into a plane wave propagating as a collimated beam along the axis.
Parabolic reflectors are used to collect energy from a distant source (for example sound waves or incoming star light). Since the principles of reflection are reversible, parabolic reflectors can also be used to collimate radiation from an isotropic source into a parallel beam.[1] In optics, parabolic mirrors are used to gather light in reflecting telescopes and solar furnaces, and project a beam of light in flashlights, searchlights, stage spotlights, and car headlights. In radio, parabolic antennas are used to radiate a narrow beam of radio waves for point-to-point communications in satellite dishes and microwave relay stations, and to locate aircraft, ships, and vehicles in radar sets. In acoustics, parabolic microphones are used to record faraway sounds such as bird calls, in sports reporting, and to eavesdrop on private conversations in espionage and law enforcement.
2.1 Focus-balanced reflector
2.2 Scheffler reflector
2.3 Off-axis reflectors
If a parabola is positioned in Cartesian coordinates with its vertex at the origin and its axis of symmetry along the y-axis, so the parabola opens upward, its equation is
{\displaystyle \scriptstyle 4fy=x^{2}}
{\displaystyle \scriptstyle f}
is its focal length. (See "Parabola#In a cartesian coordinate system".) Correspondingly, the dimensions of a symmetrical paraboloidal dish are related by the equation:
{\displaystyle \scriptstyle 4FD=R^{2},}
{\displaystyle \scriptstyle F}
is the focal length,
{\displaystyle \scriptstyle D}
is the depth of the dish (measured along the axis of symmetry from the vertex to the plane of the rim), and
{\displaystyle \scriptstyle R}
is the radius of the dish from the center. All units used for the radius, focal point and depth must be the same. If two of these three quantities are known, this equation can be used to calculate the third.
A more complex calculation is needed to find the diameter of the dish measured along its surface. This is sometimes called the "linear diameter", and equals the diameter of a flat, circular sheet of material, usually metal, which is the right size to be cut and bent to make the dish. Two intermediate results are useful in the calculation:
{\displaystyle \scriptstyle P=2F}
(or the equivalent:
{\displaystyle \scriptstyle P={\frac {R^{2}}{2D}})}
{\displaystyle \scriptstyle Q={\sqrt {P^{2}+R^{2}}},}
{\displaystyle \scriptstyle F,}
{\displaystyle \scriptstyle D,}
{\displaystyle \scriptstyle R}
are defined as above. The diameter of the dish, measured along the surface, is then given by:
{\displaystyle \scriptstyle {\frac {RQ}{P}}+P\ln \left({\frac {R+Q}{P}}\right),}
{\displaystyle \scriptstyle \ln(x)}
means the natural logarithm of
{\displaystyle \scriptstyle x}
, i.e. its logarithm to base "e".
The volume of the dish is given by
{\displaystyle \scriptstyle {\frac {1}{2}}\pi R^{2}D,}
where the symbols are defined as above. This can be compared with the formulae for the volumes of a cylinder
{\displaystyle \scriptstyle (\pi R^{2}D),}
a hemisphere
{\displaystyle \scriptstyle ({\frac {2}{3}}\pi R^{2}D,}
{\displaystyle \scriptstyle D=R),}
and a cone
{\displaystyle \scriptstyle ({\frac {1}{3}}\pi R^{2}D).}
{\displaystyle \scriptstyle \pi R^{2}}
is the aperture area of the dish, the area enclosed by the rim, which is proportional to the amount of sunlight the reflector dish can intercept. The area of the concave surface of the dish can be found using the area formula for a surface of revolution which gives
{\displaystyle \scriptstyle A={\frac {\pi R}{6D^{2}}}\left((R^{2}+4D^{2})^{3/2}-R^{3}\right)}
. providing
{\displaystyle \scriptstyle D\neq 0}
. The fraction of light reflected by the dish, from a light source in the focus, is given by
{\displaystyle \scriptstyle 1-{\frac {\arctan {\frac {R}{D-F}}}{\pi }}}
{\displaystyle F,}
{\displaystyle D,}
{\displaystyle R}
are defined as above.
Parallel rays coming into a parabolic mirror are focused at a point F. The vertex is V, and the axis of symmetry passes through V and F. For off-axis reflectors (with just the part of the paraboloid between the points P1 and P3), the receiver is still placed at the focus of the paraboloid, but it does not cast a shadow onto the reflector.
Focus-balanced reflector[edit]
It is sometimes useful if the centre of mass of a reflector dish coincides with its focus. This allows it to be easily turned so it can be aimed at a moving source of light, such as the Sun in the sky, while its focus, where the target is located, is stationary. The dish is rotated around axes that pass through the focus and around which it is balanced. If the dish is symmetrical and made of uniform material of constant thickness, and if F represents the focal length of the paraboloid, this "focus-balanced" condition occurs if the depth of the dish, measured along the axis of the paraboloid from the vertex to the plane of the rim of the dish, is 1.8478 times F. The radius of the rim is 2.7187 F.[a] The angular radius of the rim as seen from the focal point is 72.68 degrees.
Scheffler reflector[edit]
Off-axis reflectors[edit]
A circular paraboloid is theoretically unlimited in size. Any practical reflector uses just a segment of it. Often, the segment includes the vertex of the paraboloid, where its curvature is greatest, and where the axis of symmetry intersects the paraboloid. However, if the reflector is used to focus incoming energy onto a receiver, the shadow of the receiver falls onto the vertex of the paraboloid, which is part of the reflector, so part of the reflector is wasted. This can be avoided by making the reflector from a segment of the paraboloid which is offset from the vertex and the axis of symmetry. For example, in the above diagram the reflector could be just the part of the paraboloid between the points P1 and P3. The receiver is still placed at the focus of the paraboloid, but it does not cast a shadow onto the reflector. The whole reflector receives energy, which is then focused onto the receiver. This is frequently done, for example, in satellite-TV receiving dishes, and also in some types of astronomical telescope (e.g., the Green Bank Telescope, the James Webb Space Telescope).
Accurate off-axis reflectors, for use in solar furnaces and other non-critical applications, can be made quite simply by using a rotating furnace, in which the container of molten glass is offset from the axis of rotation. To make less accurate ones, suitable as satellite dishes, the shape is designed by a computer, then multiple dishes are stamped out of sheet metal.
Off-axis-reflectors heading from medium latitudes to a geostationary TV satellite somewhere above the equator stand steeper than a coaxial reflector. The effect is, that the arm to hold the dish can be shorter and snow tends less to accumulate in (the lower part of) the dish.
The principle of parabolic reflectors has been known since classical antiquity, when the mathematician Diocles described them in his book On Burning Mirrors and proved that they focus a parallel beam to a point.[4] Archimedes in the third century BCE studied paraboloids as part of his study of hydrostatic equilibrium,[5] and it has been claimed that he used reflectors to set the Roman fleet alight during the Siege of Syracuse.[6] This seems unlikely to be true, however, as the claim does not appear in sources before the 2nd century CE, and Diocles does not mention it in his book.[7] Parabolic mirrors were also studied by the physicist Ibn Sahl in the 10th century.[8] James Gregory, in his 1663 book Optica Promota (1663), pointed out that a reflecting telescope with a mirror that was parabolic would correct spherical aberration as well as the chromatic aberration seen in refracting telescopes. The design he came up with bears his name: the "Gregorian telescope"; but according to his own confession, Gregory had no practical skill and he could find no optician capable of actually constructing one.[9] Isaac Newton knew about the properties of parabolic mirrors but chose a spherical shape for his Newtonian telescope mirror to simplify construction.[10] Lighthouses also commonly used parabolic mirrors to collimate a point of light from a lantern into a beam, before being replaced by more efficient Fresnel lenses in the 19th century. In 1888, Heinrich Hertz, a German physicist, constructed the world's first parabolic reflector antenna.[11]
Antennas of the Atacama Large Millimeter Array on the Chajnantor Plateau[12]
Lighting the Olympic Flame with a parabolic reflector
The Olympic Flame is traditionally lit at Olympia, Greece, using a parabolic reflector concentrating sunlight, and is then transported to the venue of the Games. Parabolic mirrors are one of many shapes for a burning glass.
A parabolic reflector pointing upward can be formed by rotating a reflective liquid, like mercury, around a vertical axis. This makes the liquid-mirror telescope possible. The same technique is used in rotating furnaces to make solid reflectors.
Liquid-mirror telescope, paraboloids produced by rotation
^ The closeness of this number to the value of "e", the base of natural logarithms, is just an accidental coincidence, but it does make a useful mnemonic.
^ Fitzpatrick, Richard (2007-07-14). "Spherical Mirrors". Farside.ph.utexas.edu. Retrieved 2012-11-08.
^ "Servicing Mission 1". NASA. Archived from the original on April 20, 2008. Retrieved April 26, 2008.
^ Administrator. "The Scheffler-Reflector". www.solare-bruecke.org.
^ pp. 162–164, Apollonius of Perga's Conica: text, context, subtext, Michael N. Fried and Sabetai Unguru, Brill, 2001, ISBN 90-04-11977-9.
^ pp. 73–74, The forgotten revolution: how science was born in 300 BC and why it had to be reborn, Lucio Russo, Birkhäuser, 2004, ISBN 3-540-20068-1.
^ "Archimedes' Weapon". Time Magazine. November 26, 1973. Archived from the original on October 12, 2007. Retrieved 2007-08-12.
^ p. 72, The Geometry of Burning-Mirrors in Antiquity, Wilbur Knorr, Isis 74 #1 (March 1983), pp. 53–73, doi:10.1086/353176.
^ pp. 465, 468, 469, A Pioneer in Anaclastics: Ibn Sahl on Burning Mirrors and Lenses, Roshdi Rashed, Isis, 81, #3 (September 1990), pp. 464–491, doi:10.1086/355456.
^ Chambers, Robert (1875). A biographical dictionary of eminent Scotsmen. Oxford University. p. 175.
^ McLean, Ian S (2008-07-29). Electronic Imaging in Astronomy: Detectors and Instrumentation. Google Books. ISBN 9783540765820. Retrieved 2012-11-08.
^ "Prehistory of Radio Astronomy". www.nrao.edu.
^ "ALMA Doubles its Power in New Phase of More Advanced Observations". ESO Announcement. Retrieved 11 January 2013.
^ "Parabolic Reflector Free WiFi Booster". Do-It-Yourself Wireless Antennas Update and Wi-Fi Resource Center | WiFi Wireless Q & A. Binarywolf.com. 2009-08-26. Archived from the original on 2019-06-09. Retrieved 2012-11-08.
^ "Slideshow: Wi-Fi Shootout in the Desert". Wired. 2004-08-03. Retrieved 2012-11-08.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Parabolic_reflector&oldid=1079209618" |
Engineering Acoustics/Electro-acoustic analogies - Wikibooks, open books for an open world
Engineering Acoustics/Electro-acoustic analogies
1 Electro-acoustical Analogies
1.1 Acoustical Mass
1.1.1 Acoustical Impedance
1.1.2 Acoustical Mobility
1.1.3 Impedance Analog vs. Mobility Analog
1.1.4 Acoustical Resistance
1.2 Acoustical Generators
1.3 Acoustical Compliance
1.4 Examples of Electro-Acoustical Analogies
Electro-acoustical AnalogiesEdit
Acoustical MassEdit
Consider a rigid tube-piston system as following figure.
Piston is moving back and forth sinusoidally with frequency of f. Assuming
{\displaystyle f<<{\frac {c}{l\ or\ {\sqrt {S}}}}}
(where c is sound velocity
{\displaystyle c={\sqrt {\gamma RT_{0}}}}
), volume of fluid in tube is,
{\displaystyle \Pi _{v}=S\ l}
Then mass (mechanical mass) of fluid in tube is given as,
{\displaystyle M_{M}=\Pi _{v}\rho _{0}=\rho _{0}S\ l}
For sinusoidal motion of piston, fluid move as rigid body at same velocity as piston. Namely, every point in tube moves with the same velocity.
Applying the Newton's second law to the following free body diagram,
{\displaystyle SP'=(\rho _{0}Sl){\frac {du}{dt}}}
{\displaystyle {\hat {P}}=\rho _{0}l(j\omega ){\hat {u}}=j\omega ({\frac {\rho _{0}l}{S}}){\hat {U}}}
Where, plug flow assumption is used.
"Plug flow" assumption:
Frequently in acoustics, the velocity distribution along the normal surface of
fluid flow is assumed uniform. Under this assumption, the acoustic volume velocity U is
simply product of velocity and entire surface.
{\displaystyle U=Su}
Acoustical ImpedanceEdit
Recalling mechanical impedance,
{\displaystyle {\hat {Z}}_{M}={\frac {\hat {F}}{\hat {u}}}=j\omega (\rho _{0}Sl)}
acoustical impedance (often termed an acoustic ohm) is defined as,
{\displaystyle {\hat {Z}}_{A}={\frac {\hat {P}}{\hat {U}}}={\frac {Z_{M}}{S^{2}}}=j\omega ({\frac {\rho _{0}l}{S}})\quad \left[{\frac {Ns}{m^{5}}}\right]}
where, acoustical mass is defined.
{\displaystyle M_{A}={\frac {\rho _{0}l}{S}}}
Acoustical MobilityEdit
Acoustical mobility is defined as,
{\displaystyle {\hat {\xi }}_{A}={\frac {1}{{\hat {Z}}_{A}}}={\frac {\hat {U}}{\hat {P}}}}
Impedance Analog vs. Mobility AnalogEdit
Acoustical ResistanceEdit
Acoustical resistance models loss due to viscous effects (friction) and flow resistance (represented by a screen).
File:Ra analogs.png rA is the reciprocal of RA and is referred to as responsiveness.
Acoustical GeneratorsEdit
The acoustical generator components are pressure, P and volume velocity, U, which are analogus to force, F and velocity, u of electro-mechanical analogy respectively. Namely, for impedance analog, pressure is analogus to voltage and volume velocity is analogus to current, and vice versa for mobility analog. These are arranged in the following table.
Impedance and Mobility analogs for acoustical generators of constant pressure and constant volume velocity are as follows:
File:Acoustic gen.png
Acoustical ComplianceEdit
Consider a piston in an enclosure.
File:Enclosed Piston.png
When the piston moves, it displaces the fluid inside the enclosure. Acoustic compliance is the measurement of how "easy" it is to displace the fluid.
Here the volume of the enclosure should be assumed to be small enough that the fluid pressure remains uniform.
Assume no heat exchange 1.adiabatic 2.gas compressed uniformly , p prime in cavity everywhere the same.
from thermo equitation File:Equ1.jpg it is easy to get the relation between disturbing pressure and displacement of the piston File:Equ3.gif where U is volume rate, P is pressure according to the definition of the impendance and mobility, we can getFile:Equ4.gif
Mobility Analog VS Impedance Analog
File:Comp.gif
Examples of Electro-Acoustical AnalogiesEdit
Example 1: Helmholtz Resonator
Assumptions - (1) Completely sealed cavity with no leaks. (2) Cavity acts like a rigid body inducing no vibrations.
- Impedance Analog - File:Example2holm1sol.JPG
Example 2: Combination of Side-Branch Cavities
File:Exam2prob.JPG
- Impedance Analog - File:Exam2sol.JPG
Retrieved from "https://en.wikibooks.org/w/index.php?title=Engineering_Acoustics/Electro-acoustic_analogies&oldid=3232718" |
Least-squares reverse time migration in the presence of velocity errorsLSRTM with velocity errors | Geophysics | GeoScienceWorld
. E-mail: ceeyaj@nus.edu.sg; ceeliyy@nus.edu.sg (corresponding author); ceeccha@nus.edu.sg.
. E-mail: liuyuzhu@tongji.edu.cn; dlg@tongji.edu.cn.
Jizhong Yang, Yunyue Elita Li, Arthur Cheng, Yuzhu Liu, Liangguo Dong; Least-squares reverse time migration in the presence of velocity errors. Geophysics 2019;; 84 (6): S567–S580. doi: https://doi.org/10.1190/geo2019-0005.1
Least-squares reverse time migration (LSRTM), which aims to match the modeled data with the observed data in an iterative inversion procedure, is very sensitive to the accuracy of the migration velocity model. If the migration velocity model contains errors, the final migration image may be defocused and incoherent. We have used an LSRTM scheme based on the subsurface offset extended imaging condition, least-squares extended reverse time migration (LSERTM), to provide a better solution when large velocity errors exist. By introducing an extra dimension in the image space, LSERTM can fit the observed data even when significant errors are present in the migration velocity model. We further investigate this property and find that after stacking the extended migration images along the subsurface offset axis within the theoretical lateral resolution limit, we can obtain an image with better coherency and fewer migration artifacts. Using multiple numerical examples, we to demonstrate that our method provides superior inversion results compared to conventional LSRTM when the bulk velocity errors are as large as 10%.
Elastic least-squares reverse time migration using the energy norm
A wavefield-separation-based elastic least-squares reverse time migration
Q
-least-squares reverse time migration with viscoacoustic deblurring filters |
InteriorPoint - Maple Help
Home : Support : Online Help : Mathematics : Geometry : Polyhedral Sets : Properties of a Set : InteriorPoint
get an interior point of a polyhedral set
InteriorPoint(polyset)
This command computes an interior point of the highest dimensional face for the given polyhedral set polyset. If polyset is of lower dimension than its ambient space, then the point lies on the boundary of polyset, but the point is not an extreme point of the set.
The command returns NULL if there are no interior points in polyset. In other words, if polyset is a vertex or an empty set. Otherwise, the command returns a list of rationals representing a point in polyset's coordinate space.
\mathrm{with}\left(\mathrm{PolyhedralSets}\right):
Get a point inside the cube
c≔\mathrm{ExampleSets}:-\mathrm{Cube}\left(\right):
p≔\mathrm{InteriorPoint}\left(c\right)
\textcolor[rgb]{0,0,1}{p}\textcolor[rgb]{0,0,1}{≔}[\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}]
For a 2-D triangular set in 3-D space, the command returns a point which is interior to the triangle, but lies on the boundary of the set strictly speaking.
\mathrm{triangle}≔\mathrm{PolyhedralSet}\left([z=1,-1\le y,-1\le x,x\le -y+1],[x,y,z]\right)
\textcolor[rgb]{0,0,1}{\mathrm{triangle}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{{}\begin{array}{lll}\textcolor[rgb]{0,0,1}{\mathrm{Coordinates}}& \textcolor[rgb]{0,0,1}{:}& [\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{z}]\\ \textcolor[rgb]{0,0,1}{\mathrm{Relations}}& \textcolor[rgb]{0,0,1}{:}& [\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{\le }\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{\le }\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{\le }\textcolor[rgb]{0,0,1}{1}]\end{array}
p≔\mathrm{InteriorPoint}\left(\mathrm{triangle}\right)
\textcolor[rgb]{0,0,1}{p}\textcolor[rgb]{0,0,1}{≔}[\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}]
The point lies on a 2 dimensional set, while the set is in a 3 dimensional space.
\mathrm{face_with_p}≔\mathrm{LocatePoint}\left(p,\mathrm{triangle}\right)
\textcolor[rgb]{0,0,1}{\mathrm{face_with_p}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{{}\begin{array}{lll}\textcolor[rgb]{0,0,1}{\mathrm{Coordinates}}& \textcolor[rgb]{0,0,1}{:}& [\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{z}]\\ \textcolor[rgb]{0,0,1}{\mathrm{Relations}}& \textcolor[rgb]{0,0,1}{:}& [\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{\le }\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{\le }\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{\le }\textcolor[rgb]{0,0,1}{1}]\end{array}
\mathrm{Dimension}\left(\mathrm{face_with_p}\right)
\textcolor[rgb]{0,0,1}{2}
\mathrm{nops}\left(\mathrm{Coordinates}\left(\mathrm{face_with_p}\right)\right)
\textcolor[rgb]{0,0,1}{3}
The PolyhedralSets[InteriorPoint] command was introduced in Maple 2015. |
\sqrt{x^2+y^2}=x+y?
\large \sqrt{a^2 + b^2} = a + b
Why some people say it's true: It's an example of the distributive property which works since exponents are just repeated multiplication. So, just like
5(a+b) = 5a + 5b, \sqrt{a^2 + b^2} = \sqrt{a^2} + \sqrt{b^2} = a + b.
Why some people say it's false: The distributive property doesn't work for the square root functions, so you can't say that
\sqrt{a^2 + b^2} = \sqrt{a^2} + \sqrt{b^2}
\color{#20A900}{\text{Reveal the Correct Answer:}}
\color{#D61F06}{\textbf{false}}
a = 3 , b = 4
LHS = \sqrt{ 3^2 + 4^2 } = \sqrt{ 25 } = 5
RHS = 3 + 4 = 7
a^2 + b^2 = (a+b)^2 = a^2 + 2ab + b^2.
This is clearly not an identity since
2ab
is not always 0. Thus, we can conclude that this is an identity if and only if
2ab = 0
The distributive property is an identity that relates addition and multiplication:
m(a+b) = ma + mb
. However, it does not apply to the square root function. To understand why, consider the shape of the square-root function:
Consider any two points on this curve,
(a, \sqrt{a})
(a+b, \sqrt{a+b})
b > 0
. What is the effect on the difference between
\sqrt{a}
\sqrt{a+b}
as the two points shift together, along the function and to the right (in other words, as
a
increases, but
b
remains fixed)?
As the region shifts to the right, the
\sqrt{x}
curve flattens out, which, algebraically, means that the difference between
\sqrt{a}
\sqrt{a+b}
drops to 0 as
a
b
stays constant. On the other hand, the difference between
\sqrt{a}
\sqrt{a} + \sqrt{b}
\sqrt{b}
, which doesn't change of course, as the region shifts right, since
b
doesn't change.
\color{#3D99F6}{\text{See Common Rebuttals:}}
Rebuttal: For the most part, I agree, but
\sqrt{x^2} = \pm x,
|x|
Reply: That's another misconception. See the "
\text{Is }\sqrt{x^2} = \pm x?
" misconception page for details.
Rebuttal: Consider a linear function,
f(x) = mx +b
. As a region defined by two points a fixed width apart moves right, it stays the same height; does that mean that all linear functions obey the distributive property? (That is, if
f(x) = mx + c
c,
f(a+b) = f(a) + f(b).
Reply: No. There's actually a second condition in order for the distributive property to be applicable to a function:
f(0)
must equal 0. Otherwise,
f(a) = f(a+0) = f(a) + f(0)
could not possibly be true. Therefore, functions of the form
f(x) = mx + b
only obey the distributive property on the domain of real numbers if
b=0
It's true for all real numbers It's true if and only if
a=0
b=0
It's false for all real numbers None of the other options are correct
\large \sqrt{a^2 + b^2} = a + b
and
b
\sqrt{a^2+b^2} = \sqrt{a^2} + \sqrt{b^2} .
(a, b)
-10 \leq a \leq 10, -10 \leq b \leq 10
\sqrt{ a^2 + b^2} = \sqrt{a^2} + \sqrt{b^2} ?
\text{Is }(a+b)^2 = a^2 + b^2?
" Misconception Page
\text{Is }\sqrt{x^2} = \pm x?
Cite as: Does
\sqrt{x^2+y^2}=x+y?
. Brilliant.org. Retrieved from https://brilliant.org/wiki/is-square-root-of-a2b2-equal-a-b/ |
PolynomialTutor - Maple Help
Home : Support : Online Help : Education : Student Packages : Precalculus : Interactive : PolynomialTutor
Student[Precalculus][PolynomialTutor] - demonstrate graphing of polynomial
PolynomialTutor(p)
(optional) polynomial in at most one variable
The PolynomialTutor(p) command launches a tutor interface that demonstrates the graphing of the polynomial p. If the polynomial has any zeroes, they are also displayed.
If p is not specified, PolynomialTutor uses a default function.
\mathrm{with}\left(\mathrm{Student}[\mathrm{Precalculus}]\right):
\mathrm{PolynomialTutor}\left(\right)
\mathrm{PolynomialTutor}\left(2{x}^{3}+{x}^{2}-2\right) |
Determine whether a linear, quadratic, or exponential function best models the data. Then,
Determine whether a linear, quadratic, or exponential function best models the data. Then, use regression to find the function that models the data. x
Determine whether a linear, quadratic, or exponential function best models the data. Then, use regression to find the function that models the data.
\begin{array}{|cccccc|}\hline x& 0& 1& 2& 3& 4\\ y& 100& 89.5& 78.9& 68.4& 57.8\\ \hline\end{array}
a) Check the 1st differences of the consecutive y-values. Since they are nearly constant, we can use a linear model.
\begin{array}{|cc|}\hline y& \text{1st differences}\\ 100& \\ 89.5& 89.5-100=-10.5\\ 78.9& 78.9-89.5=-10.6\\ 68.4& 68.4-78.9=-10.5\\ 57.8& 57.8-68.4=-10.6\\ \hline\end{array}
b)At the WolframAlpha web site you can enter: fit linear
\left\{\left\{0,100\right\},\left\{1,89.5\right\},\left\{2,78.9\right\},\left\{3,68.4\right\},\left\{4,57.8\right\}\right\}
Which results i
y=100.02-10.55x
6\in U\left(700\right)
Find the product of the complex numbers
{z}_{1}=1+i
{z}_{2}=2+2i
. Leave answers in polar form.
Find the product of the complex number and its conjugate.
I want to find this limit for complex variable z
\underset{z\to -1}{lim}\left(z+1\right)\mathrm{sin}\left(\frac{1}{z+1}\right)
In the real case I know sin(z) is bounded by -1,1, and the limit is 0. But in the complex case
\mathrm{sin}\left(z\right)
is not bounded. How can I find the limit or prove that it doesn't exist?
What is 2+2i minus 7+6i?
1+{\omega }^{m}+{\omega }^{2m}+\cdots +{\omega }^{\left(n-1\right)m}=0
{z}^{n}=1
\omega =\mathrm{cos}\frac{2\pi }{n}+i\mathrm{sin}\frac{2\pi }{n}
1+{\omega }^{m}+{\omega }^{2m}+\cdots +{\omega }^{\left(n-1\right)m}=0 |
(→Atmospheric correction: note on using MOD08)
m (→Atmospheric correction: On using MOD08 as an AOD input for i.atcorr (thanks to YannC))
'''Sources for AOD estimations'''
* MODIS MOD08 product [http://modis.gsfc.nasa.gov/data/dataprod/dataproducts.php?MOD_NUMBER=08 MOD08 - Gridded Atmospheric Product].
** Daily MOD08 products can be downloaded directly from NASA's FTP server: [ftp://ladsweb.nascom.nasa.gov/allData/51/MOD08_D3/ '''ftp://ladsweb.nascom.nasa.gov/allData/51/MOD08_D3/'''].
** File and layer specifications are available, for example, at [http://www.atmos.washington.edu/~robwood/modis/filespecs/MOD08_D3.CDL.fs MODIS HDF File Specification MOD08_D3: MODIS Level 3 Daily Atmosphere Gridded Product]
** The layer '''Optical_Depth_Land_And_Ocean_Mean''', described as ''Aerosol Optical Thickness at 0.55 microns for both Ocean (best) and Land (corrected): Mean'', can be probably used as an input for {{cmd|i.atcorr}}. '''Note,''' the layer's values have to be scaled by using the scaling factor ''Optical_Depth_Land_And_Ocean_Mean:scale_factor ='' '''0.001d''' as reported in the specifications.
'''TοDο:'''
{\displaystyle QCAL}
{\displaystyle DN}
{\displaystyle QCALMIN}
{\displaystyle QCALMAX}
{\displaystyle QCALMINM}
{\displaystyle DN}
{\displaystyle QCALMAX}
{\displaystyle DN=255}
{\displaystyle DN}
{\displaystyle DNM/math>value(<math>QCALMIN}
{\displaystyle DN}
{\displaystyle QCALMAX}
{\displaystyle QCAL}
{\displaystyle DN}
{\displaystyle 42*100/60}
{\displaystyle 6S}
0.09745 # e.g. rayleigh optical depth at 0km altitude for Midlatitude summer, estimated by Anthony Buchholz
explain how to use the MODIS Gridded Atmospheric Product (MOD08). Note, MOD08 products can be downloaded directly from [1].
explain how to use Rayleigh Optical Depths estimated at 0km altitude for six different atmosphere models from the paper Rayleigh-scattering calculations for the terrestrial atmosphere by Anthony Buchholz, table 4, page 2270. |
What Makes A Good Problem? | Brilliant Math & Science Wiki
Pi Han Goh, Sandeep Bhardwaj, Josiah Kiok, and
When you publish a problem on Brilliant, it's only natural to hope that the community falls in love with it and wants to work on your problem. It feels good when thousands of people have viewed your problem, and you bask in their admiration. However, only a few problems have that special distinction. This wiki explains how you can greatly improve the quality of your problems by
identifying your target audience and
choosing the best presentation.
Applying these Ideas to Improve a Problem
1) What makes the problem stand out?
Represents a daily life scenario: people love seeing how math and science are applied to the real world.
Clarifies a misconception: people are curious when they come across a counter-intuitive result, and want to understand how to think about it.
Thought-provoking: it encourages people to apply their knowledge instead of just relying on memorization of formulas.
Involves multiple concepts: a problem becomes more challenging if it involves multiple concepts.
2) What level is the problem?
Easy: It should neither involve a lot of tedious calculations nor contain unexplained technical jargon.
Medium: It may involve a single application of a difficult concept but doesn't require a lot of knowledge.
Hard: Showcase a complex situation which requires the use of multiple concepts. It requires a lot of thought processes to completely solve the problem.
This gives you insight into the ability of your audience and determines the amount of information that you should provide in the problem.
Everything on the internet is competing for your limited attention. If the presentation and content do not feel outstanding, then we have been conditioned to hit the backspace and move on. To avoid losing your audience, be aware of the following:
1) Phrasing: People can only read what we have written down, and not what we were thinking or intending.
Keep it short and uncomplicated. In general, remove unnecessary information that is not relevant to the problem.
Organize the information. Keep similar information together so that it is quicker to understand.
Identify the crux of the problem. Make it the focus of the problem.
2) Theme/Motivation: People will not engage if they find the problem boring.
Make the problem engaging with a relevant theme. Real-life applications tend to work well, but do not overdo it.
Consider providing a hint to encourage answering the problem.
Keep your target audience in mind as you write up the problem.
3) Contains imagery: People will only engage with the problem if they can easily comprehend its meaning.
A picture says a thousand words. When relevant, it helps to draw the reader in and quickly provides the necessary context.
For geometry problems, having a picture often makes it easier to understand what is being described.
4) Directive/Answer options: Meaningful options makes a sensible question.
Keep it simple. Avoid making the reader do unnecessary work in order to submit an answer.
Consider using multiple-choice options to encourage people to give the problem a try, even if they are not fully certain.
How can we improve the following problem?
My 2 favorite positive integers satisfy the property that if I take any one of these numbers and multiply by itself by a total number of times, where this number is numerically equal to the other number, then the resultant product is 16. What is the sum of my 2 favorite positive integers?
First, let's identify our target audience.
What makes the problem stand out? It involves both multiplication and indices.
What level is the problem? This is likely of easy difficulty because the reader is required to know how multiplication and indices are related to each other.
Second, let's choose the best presentation.
Phrasing: The problem seems convoluted. For example, to verbally describe the second sentence without any variables will make it extremely difficult to digest. It's much easier to just use mathematical notations to describe it. Plus, it isn't necessary to state that they are my 2 favorite numbers, simply stating that there are 2 specific positive integers in question is sufficient.
Theme / Motivation: Readers can easily see the symmetry behind the second sentence once we've written out the math expressions:
x^y = 16, y^x = 16
Imagery: In this case, the math expressions speak for themselves.
Options: A numerical answer would work best in this case. There's little to no benefit by adding multiple choices because it doesn't show how one can arrive at these numbers.
Based on the above, we can improve the problem by
rewriting the question by providing math notations and
removing the context of "favorite numbers."
As such, this leads us to create the following problem:
x
y
x^y = 16
y^x = 16
. What is the sum
x+y?
There is a point that is 5 away from the edge of a circle that is towards the center of the circle. If you draw all of the chords through this point, you will find that the shortest chord has length 30. What is the radius of the circle?
What makes the problem stand out? It is thought-provoking because there doesn't seem to be enough information in the problem to proceed.
What level is the problem? This is likely of medium difficulty because it is a simple application of a concept.
Phrasing: The problem seems convoluted. For example, the second sentence can be made explicit/immediate by phrasing it as "The shortest chord of the circle that passes through this point has length 30."
Theme/Motivation: This is actually a really interesting geometry problem, but that has not been expressed as yet. Too much time is spent trying to figure out what the problem is, instead of appreciating it. In addition, a hint might be helpful, because there doesn't seem to be enough information at the start to proceed.
Imagery: This will strongly benefit from having an attached image.
Options: A numerical answer would work best in this case. Having multiple-choice options doesn't make it more tempting to answer, because it is not clear how we can arrive at those values.
providing a pictorial image for people to understand what the description is and
providing a hint for people to get started.
t=0
, a particle
A
is located at the origin
(0,0)
and a particle
B
is located on the
y
(0,-d)
A
starts traveling along the
x
-axis at a constant velocity of
u.
B
, on the other hand, travels with a constant speed of
u
such that, at every instant, its velocity vector is oriented towards
A
's current location. Let
r(t)
denote the distance between the particles at time
t
\lim_{t \to \infty} r(t)
\frac{d}{2}
d
2d
What makes the problem stand out? It represents a daily life scenario where we want to chase an item directly. As they work through the problem, it might showcase a misconception.
What level is the problem? This is likely of hard difficulty because we have to relate several concepts and solve a differential equation.
Phrasing: Let's be honest, most people would not make it through the entire paragraph, because it appears boring.
Theme/Motivation: Being a classic "chasing problem," we could add some characters to make the problem more interesting.
Imagery: Since it can be a real-world application, having an image that illustrates all the information would be very helpful.
Options: While we could convert this into a numerical answer by setting
d = 1
, having the options not only makes it easier for people to guess, it also showcases that there is a "nice" answer to this seemingly complicated problem.
figuring out how best to present the problem as a real-world/familiar scenario and
providing a pictorial image for people to immediately grasp what is happening.
Tom and Jerry both have equal top running speeds
u
and are initially at points
A
B,
respectively, separated by a distance of
d
. They both spot each other and immediately start running at their top speeds. Jerry runs on a straight line perpendicular to the line
AB
and Tom runs in such a way that its velocity always points towards the current location of Jerry. Let
r(t)
denote the distance between Tom and Jerry at time
t
\displaystyle \lim_{t \to \infty} r(t)
\frac{d}{2}
d
2d
Now that you've seen these ideas in play, take a problem of yours and see how you can make it great. If you would like to help going through a problem, comment in the feedback box below!
Cite as: What Makes A Good Problem?. Brilliant.org. Retrieved from https://brilliant.org/wiki/what-makes-a-good-problem/ |
To explain:Why the test for homogeneity follows the same procedures as the test
To explain:Why the test for homogeneity follows the same procedures as the test for independence.
The procedures for the chi-square test for homogeneity are identical to those for a chi-square test for independence. The data differs for the two tests. Tests for independence are used to determine whether there is a significant relationship between two categorical variables from a single population. A single population is segmented based on the value of two variables. Hence, there will be a column variable and a row variable.
A chi-square test for the homogeneity of proportions can be used to compare the population proportions from two or more independent samples, it determines whether the frequency counts are distributed identically across different populations.
Thus, the test for homogeneity follows the same procedures as the test for independence because the assumptions for performing the chi-square test for independence and chi-square test for homogeneity are the same.
Assume that you have invited back to your high school to be a guest teacher for a day. You task is to explain the importance of Chi square test in real life applications of Statistics. Your target audience is students in Grade 10 (or Ordinary Level students). short speech with examples.
A bullying prevention program is underway in the phoenix school districts. The researchers ran multiple analyses with the baseline data. The researchers want to determine if there is a relationship between age (13, 14, 15, 16, etc.) and victimization score (measured on a multi-item, Likert scale). What type of statistical test should be used?
Is age group related to the device people prefer to use to watch television? A research firm sought to answer this question by surveying a random sample of 260 U.S. adults age 18 and over. The results are shown below.
\begin{array}{|ccccc|}\hline & & \text{Device}& & \\ & & \text{Television}& \text{Computer}& \text{Mobile device}\\ & 18-24& 19& 6& 12\\ \text{Age Group}& 25-34& 23& 7& 39\\ & 35-54& 34& 8& 28\\ & 55+& 33& 8& 13\\ \hline\end{array}
Which of the following tests should be used?
Either (a) or (c).
A chi-square goodness of fit test.
A z-test for the difference of two proportions.
A chi-square test for homogeneity.
A chi-square test for independence.
In Unit 3 we studied several hypothesis tests: 1-Prop z-Test, 2-Prop z-Test, 1-Sample t-Test, 2- Sample t-Test, Paired t-Test, Chi-Square Goodness-of-Fit Test, and Chi-Square Test for Independence. For each scenario, identify the hypothesis test that should be applied.
1. c) A researcher wants to test a claim that students perform better on math problems when not listening to music as compared to when they do listen to music.
2. d) A researcher wants to test a claim that the average amount of time that kids spend reading books has decreased.
3. e) A researcher wants to test a claim that there is an association between socoal media an mood.
4x5
\alpha =0.05
1. When the null hypothesis for a chi square test for independence is true.
a. a large difference between the observed frequencies and the expected frequencies.
b.little or no difference between the observed frequencies and the expected frequencies.
c. no difference between the observed frequencies and the marginals
d. no differences between the row and the column marginal
2. In the Chi Square test, degrees of freedom are calculated as :
{N}_{1}+{N}_{2}-2
c. (r+1)(c+1)
d. (r-1)(c-1)
What kinds of research questions are appropriate to use with a chi-square analysis? |
Calculate geodetic latitude, longitude, and altitude above planetary ellipsoid from Earth-centered Earth-fixed (ECEF) position - Simulink - MathWorks Switzerland
\left(\overline{p}\right)
\left(\overline{\mu }\right)
\left(\overline{\iota }\right)
\left(\overline{h}\right)
\overline{p}=\left[\begin{array}{c}{\overline{p}}_{x}\\ {\overline{p}}_{y}\\ {\overline{p}}_{z}\end{array}\right].
\iota =\text{atan}\left(\frac{{p}_{y}}{{p}_{x}}\right).
\left(\overline{\mu }\right)
\left(\overline{\mu }\right)
\left(\overline{\beta }\right)
\begin{array}{c}\overline{\beta }=\text{atan}\left(\frac{{p}_{z}}{\left(1-f\right)s}\right)\\ \\ \overline{\mu }=\text{atan}\left(\frac{{p}_{z}+\frac{{e}^{2}\left(1-f\right)}{\left(1-{e}^{2}\right)}R{\left(\mathrm{sin}\beta \right)}^{3}}{s-{e}^{2}R{\left(\mathrm{cos}\beta \right)}^{3}}\right)\end{array}
s=\sqrt{{p}_{x}^{2}+{p}_{y}^{2}}.
\left(\overline{\beta }\right)
\beta =\text{atan}\left(\frac{\left(1-f\right)\mathrm{sin}\mu }{\mathrm{cos}\mu }\right)
\left(\overline{\mu }\right)
\overline{\mu }
\left(\overline{h}\right)
h=s\mathrm{cos}\mu +\left({p}_{z}+{e}^{2}N\mathrm{sin}\mu \right)\mathrm{sin}\mu -N,
\left(\overline{N}\right)
N=\frac{R}{\sqrt{1-{e}^{2}{\left(\mathrm{sin}\mu \right)}^{2}}}. |
WorkbookData - Maple Help
Home : Support : Online Help : Programming : Input and Output : File Manipulation : ExcelTools : WorkbookData
Retrieve sheet and named range data from an Excel spreadsheet
WorkbookData(file)
string ; the name of the Excel file from which workbook data will be retrieved.
The WorkbookData(file) function retrieves the sheet and named range data from the workbook saved in the Excel file. This data is returned in the form of a list consisting of two sublists:
The first sublist gives the names of the sheets in the workbook, as strings.
The second sublist gives all the named ranges in the workbook, as strings.
\mathrm{with}\left(\mathrm{ExcelTools}\right):
L≔\mathrm{WorkbookData}\left("Employees.xls"\right)
The ExcelTools[WorkbookData] command was introduced in Maple 15. |
The Comparison of Area under the Curve of Serum Estradiol during Controlled Ovarian Stimulation between Long Gonadotrophin Releasing Hormone Agonist Protocol and Short Gonadotrophin Antagonist Protocol in Invitro Fertilization
Hartanto Bayuaji1*, Wiryawan Permadi1, Herri S. Sastramihardja2, Heda Melinda Nazaruddin Nataprawira3
1Department of Obstetrics & Gynecology, Faculty of Medicine, Universitas Padjadjaran/Hasan Sadikin Hospital, West Java, Indonesia.
2Department of Pharmacology and Therapy, Faculty of Medicine, Universitas Padjadjaran/Hasan Sadikin Hospital, West Java, Indonesia.
3Department of Child Health, Faculty of Medicine, Universitas Padjadjaran/Hasan Sadikin Hospital, West Java, Indonesia.
Background: during stimulation, estradiol levels represent the dynamic hormonal interaction of follicle development. Some studies have been done to evaluate the influence of day-hCG estradiol assay and IVF outcome. However, this assay does not reflect the total exposure of estradiol throughout stimulation. Hence the concept of calculating the area under the curve (AUC) of estradiol during ovarian stimulation is thought to be a better parameter to evaluate total exposure of estradiol to the entire process of IVF. Methods: a retrospective analysis of data from a fertility center in a tertiary hospital. Between 2011-2016, there were long GnRH agonist cycles and short GnRH antagonist cycles eligible for analysis. Data including subject’s characteristics, the level of estradiol on day-2, day-6, and day-hCG during controlled ovarian stimulation were obtained. Those data were plotted, and area under the curve of each subject was calculated. Statistical analysis was done using student t-test while p < 0.05 was considered significant. Results: there were no significant differences in subject’s age (33.4 vs. 34.2 years, p = 0.12), day-3 of menstrual cycle FSH level (7.1 vs. 7.3 mIU/mL, p = 0.64), and day-3 estradiol level (35.1 vs. 36.5 pg/mL, p = 0.61). The estradiol level on day-5 of COS was 324.9 (273.1) pg/mL and 363.3 (293) pg/mL (p = 0.36), and at the day of hCG administration was 2461.8 (2283.7) ng/mL and 2558.2 (2043) ng/mL (p = 0.76). Estradiol area under the curve for long protocol and short protocol was 9205.2 (7683.7) pg/mL and 9732.9 (6893.9) pg/mL respectively (p = 0.63). Conclusion: our results indicate that there are no significant differences in the area under the curve of estradiol between two protocols.
Area under the Curve, Controlled Ovarian Stimulation, Estradiol, Long Agonist Protocol, Short Antagonist Protocol
Bayuaji, H. , Permadi, W. , Sastramihardja, H. and Nataprawira, H. (2017) The Comparison of Area under the Curve of Serum Estradiol during Controlled Ovarian Stimulation between Long Gonadotrophin Releasing Hormone Agonist Protocol and Short Gonadotrophin Antagonist Protocol in Invitro Fertilization. Open Access Library Journal, 4, 1-7. doi: 10.4236/oalib.1103832.
Controlled ovarian stimulation (COS) is an important and integral part of in vitro fertilization (IVF) program, with the goal to obtain more oocytes in supraphysiological conditions. This may lead to higher pregnancy chance, and the spare embryos can be cryo-stored for future transfer [1] [2] . The COS protocol mainly consists of exogenous follicle-stimulating hormone (FSH) administration using long-agonist protocol, and short-antagonist protocol. The long-agonist protocol induces hypophyseal desensitization to prevent premature luteinization, and significantly reduce cycle cancellation. The oocyte yield was considered to be higher compared to the short antagonist protocol. However, in a specific patient group such as polycystic ovary syndrome (PCOS), on the other hand, the short antagonist protocol completely blocks pituitary GnRH receptors and induces reversible and rapid gonadotrophin secretion. This may be achieved by administration of GnRH antagonist drugs such as cetrorelix and ganirelix [3] [4] [5] .
The ovarian response during COS should be monitored adequately [2] [6] [7] . During COS, the ovarian produces E2 in supraphysiological level, and this may reflect ovarian response. The goal of COS monitoring was to ensure adequate ovarian response while avoiding complication potentials i.e., ovarian hypers- timulation syndrome or inadequate response due to inaccurate FSH dosage [2] [5] [8] . However, the serum E2 assay is mostly done by serial measurement during COS and may not reflect the actual exposure of E2. The area under the curve (AUC) concept, therefore proposed in several studies to calculate the amount of E2 exposure during COS [9] [10] .
This study was aimed to compare the AUC during COS between long agonist- and short antagonist protocol. We hypothesized that there was no significant difference in AUC between two protocols, which may reflect the same efficacy of both protocols.
This retrospective study was done by reviewing the records of female infertility patients in The Assisted Reproductive Technology Unit, Hasan Sadikin Hospital Bandung, Indonesia during the 2011-2016 period. Subjects included in this study were 20 - 40 years old infertile female who underwent IVF program. The inclusion criteria were a normal ovarian reserve (serum follicle stimulating hormone/FSH level < 10 mIU/mL and estradiol level < 70 pg/mL on day-3 of the menstrual cycle, or anti-Mullerian hormone/AMH > 1.4 ng/mL) [11] , which was evaluated before the subject started the IVF program. Subjects with endome- triosis, polycystic ovary syndrome, poor ovarian stimulation response, or ovarian hyperstimulation syndrome were excluded from this study. This study was approved by the Ethical Committee for Health Research, Hasan Sadikin Hospital Bandung. All patients have given their general consent before receiving management from the hospital.
The data obtained from medical records included subject’s characteristics, serum estradiol level on the beginning of COS, day-5 of COS, and at the day of hCG trigger administration during ovarian stimulation. Data then plotted in the curve with the x-axis represents the day of menstrual cycle, and the y-axis represents the estradiol level. The area under the curve was calculated by adding the area of rectangles and triangles for each time interval. The formula for calculating the AUC was proposed by Pruessner et al. as
AUC=\frac{\left\{\left(m2+m1\right)t1\right\}}{2}+\frac{\left\{\left(m3+m2\right)t2\right\}}{2}
where m = measurement of estradiol level, and t = time interval [12] . The mean of AUC of both groups was analyzed using student’s t-test. Statistical analysis was performed using IBM SPSS version 24 (IBM Analytics, Chicago, IL, USA). The p value < 0.05 was considered significant.
During the study period, there were 185 subjects enrolled in the IVF program. There were 96 subjects using long agonist protocol and 89 subjects using short antagonist protocol. The characteristics of study subjects were presented in Table 1.
The serum E2 level during COS was presented in Table 2.
The AUC of E2 in long agonist protocol group was 9205.2 (7683.7) pg/mL, while in short antagonist protocol group was 9732.9 (6893.9) pg/mL. Using the student t-test, there was no statistical difference of the AUC in both groups (p = 0.63) (Figure 1).
The age and ovarian reserve parameters were not significantly different between
*student t-test.
Table 2. The comparison of E2 level.
Figure 1. The comparison of area under the curve of serum estradiol from both study groups.
both study groups, reflecting the similar characteristics. Long agonist protocol was related to more severe suppression of pituitary, resulting downregulated condition before ovarian stimulation commenced. Therefore, this protocol may not be suitable for patients with diminished ovarian reserve or advanced age. Our subjects’ characteristics in long agonist group the mean of age were still below 35 years old, the age of decreased ovarian reserve would be anticipated. The day-3 FSH and E2 level, which used for evaluating the ovarian reserve, did not significantly differ as well. Due to its simplicity and convenience, the short antagonist protocol now became much applied in our hospital. However, the questions may arise from the patients whether the short antagonist protocol may yield the same results as the already established long agonist protocol. To answer the question, we evaluate the dynamics of serum E2 level, the marker that may reflect the ovarian response during COS. There was no significant difference of E2 level throughout COS. After calculating the AUC of E2, there was no significant difference of AUC in both groups. This may reflect the similar efficacy of both protocols.
The E2 dynamics during COS was evaluated during COS for several reasons: to ensure the adequate ovarian response which may yield the optimum results, and to prevent the probability of the complication such as ovarian hyperstimulation syndrome [13] . The final response of COS usually evaluated by measuring E2 level before the administration of hCG which served as a trigger to induce oocyte maturation. However, this might not reflect the total exposure, or total estradiol secreted by the developing follicles [9] . Hence the concept of area under the curve calculation was then introduced. This was done by calculating the sum of the area in estradiol assay result in day to day basis. In our hospital, the serum estradiol assay was done in the commencement, on the day-5 of stimulation, and at the end of stimulation. This might lead to the limited data obtained and hence became the limitation of our study. Mitwally et al. conducted the study using the graph plotted based on the estradiol level on the day-to-day basis in long agonist protocol [9] . The AUC of serum E2 was 9357 (4555) pg/mL. They conclude that AUC of estradiol reflected the amount of E2 produced during COS more accurately. Kutlu et al. conducted a study to evaluate whether the AUC of E2 might be used as a predictor of clinical pregnancy or implantation failure. They found that there was a significant difference between AUC of E2 in women with successful implantation and those whose implantation was failed [10] . They concluded that the AUC of E2 might be used for predicting implantation success.
The serum E2 resulted from COS was considered in supraphysiologic level, that in several circumstances may interfere the receptivity of endometrium, leading to implantation failure. This was observed in several studies, who found the lower implantation rate in high estradiol level condition during IVF ET cycle [10] [14] [15] . However, the difference in pregnancy results was not the primary goal of this study.
The limitation of this study was a relatively small subject to analyze and less serum estradiol assay checkpoint compared with other studies. The day-to-day estradiol assay will produce smoother curve, therefore the area under the curve calculation will reveal more accurate results. However, our results was in similar range with other study [9] .
There was no significant difference of area under the curve of serum estradiol between long agonist protocol and short antagonist protocol in controlled ovarian stimulation.
AUC: area under the curve;
E2: estradiol;
ET: embryo transfer;
FSH: follicle-stimulating hormone;
GnRH: gonadotrophin-releasing hormone;
[1] Farquhar, C., Marjoribanks, J., Brown, J., Fauser, B., Lethaby, A., Mourad, S., et al. (2017) Management of Ovarian Stimulation for IVF: Narrative Review of Evidence Provided for World Health Organization Guidance. Reproductive Biomedicine Online, 35, 3-16.
[2] Kwan, I., Bhattacharya, S., Kang, A. and Woolner, A. (2014) Monitoring of Stimulated Cycles in Assisted Reproduction (IVF and ICSI). Cochrane Database of Systematic Reviews, 8, Article ID: CD005289.
[3] Macklon, N.S., Stouffer, R.L., Giudice, L.C. and Fauser, B.C. (2006) The Science behind 25 Years of Ovarian Stimulation for in Vitro Fertilization. Endocrine Reviews, 27, 170-207.
[4] Santos, M.A., Kuijk, E.W. and Macklon, N.S. (2010) The Impact of Ovarian Stimulation for IVF on the Developing Embryo. Reproduction, 139, 23-34.
[5] Vandekerckhove, F., Gerris, J., Vansteelandt, S. and De Sutter, P. (2014) Adding Serum Estradiol Measurements to Ultrasound Monitoring Does Not Change the Yield of Mature Oocytes in IVF/ICSI. Gynecological Endocrinology, 30, 649-652.
[6] Hendriks, D.J., Klinkert, E.R., Bancsi, L.F., Looman, C.W., Habbema, J.D., Velde, E.R., et al. (2004) Use of Stimulated Serum Estradiol Measurements for the Prediction of Hyperresponse to Ovarian Stimulation in in Vitro Fertilization (IVF). Journal of Assisted Reproduction and Genetics, 21, 65-72.
https://doi.org/10.1023/B:JARG.0000027016.65749.ad
[7] Macklon, N.S. and Fauser, B.C. (2005) Gonadotrophins in Ovulation Induction. Reproductive Biomedicine Online, 10, 25-31.
[8] Martins, W.P., Vieira, C.V., Teixeira, D.M., Barbosa, M.A., Dassuncao, L.A. and Nastri, C.O. (2014) Ultrasound for Monitoring Controlled Ovarian Stimulation: A Systematic Review and Meta-Analysis of Randomized Controlled Trials. Ultrasound in Obstetrics Gynecology, 43, 25-33.
[9] Mitwally, M.F., Bhakoo, H.S., Crickard, K., Sullivan, M.W., Batt, R.E. and Yehl, J. (2005) Area under the Curve for Estradiol Levels Do Not Consistently Reflect Estradiol Levels on the Day of HCG Administration in Patients Undergoing Controlled Ovarian Hyperstimulation for IVF-ET. Journal of Assisted Reproduction and Genetics, 22, 57-63.
[10] Kutlu, T., Ozkaya, E., Ayvaci, H., Devranoglu, B., Sanverdi, I., Sahin, Y., et al. (2016) Area under Curve of Temporal Estradiol Measurements for Prediction of the Detrimental Effect of Estrogen Exposure on Implantation. International Journal of Gynaecology and Obstetrics, 135, 168-171.
[11] Homburg, R. (2014) Ovulation Induction and Controlled Ovarian Stimulation. Springer International Publishing, Gewerbestrasse.
[12] Pruessner, J.C., Kirschbaum, C., Meinlschmid, G. and Hellhammer, D.H. (2003) Two Formulas for Computation of the Area under the Curve Represent Measures of Total Hormone Concentration versus Time-Dependent Change. Psychoneuroendocrinology, 28, 916-931.
[13] Kwan, I., Bhattacharya, S., McNeil, A. and Van Rumste, M.M. (2008) Monitoring of Stimulated Cycles in Assisted Reproduction (IVF and ICSI). Cochrane Database of Systematic Reviews, 2, Article ID: CD005289.
[14] Valbuena, D., Martin, J., de Pablo, J.L., Remohi, J., Pellicer, A. and Simon, C. (2001) Increasing Levels of Estradiol Are Deleterious to Embryonic Implantation because They Directly Affect the Embryo. Fertility and Sterility, 76, 962-968.
[15] Blumenfeld, Z. (2015) Why More Is Less and Less Is More When It Comes to Ovarian Stimulation. Journal of Assisted Reproduction and Genetics, 32, 1713-1719. |
Template for Package Overview Help Page - Maple Help
Home : Support : Online Help : Applications and Example Worksheets : Language and System : Template for Package Overview Help Page
<Package Overview Template>
<For more information on using this template refer to the ?Templates/PackageOverview help page. Delete this text and the title.>
Overview of the <Package_Name> Package
<Package_Name>[command](arguments)
<The first bullet item should be an overview of the package functionality. Include any details, for example, specific scope or generality of package.>
Each command in the <Package_Name> package can be accessed by using either the long form or the short form of the command name in the command calling sequence.
As the underlying implementation of the <Package_Name> package is a module, it is also possible to use the form <Package_Name>:-command to access a command from the package. For more information, see Module Members.
<Insert hyperlinks to package command help pages.>
help1 help2 help3 help4
To display the help page for a particular <Package_Name> command, see Getting Help with a Command in a Package.
\mathrm{example1}
\mathrm{example2}
<related help topic>, <related help topic>, <related help topic>, UsingPackages, with |
Intro to Computation Practice Problems Online | Brilliant
In math, a function takes an input and returns an output. For example
f(x)=x^2
returns the square of the input. Another example - common in computer science - is
\text{sort(list)},
which returns a list with the same elements as the input list except in increasing order.
For how many distinct inputs is
\text{sort(list)}=[1,2,3,4]?
That is, how many lists, when sorted, give
[1,2,3,4]?
There are a lot of things we can say about functions in math, but we don't usually concern ourselves with how "difficult" a function is to evaluate; it just has an input and an output.
However, how "difficult" a function is to evaluate is a central question we ask when working with algorithms. It determines which computations are feasible in a reasonable amount of time with a computer, and which are not.
Computers know how to directly evaluate some "native" functions (like comparisons, addition, multiplication, etc.), and can only compute higher-order functions using these native operations.
Thus, a central question is how many native operations it takes to complete a higher-order function for different inputs. Let's take a look at concrete example of this in the next question.
Let's take an ElementList object that has two native functions:
You can compare any two values in the list to see which one is bigger.
You can swap any two values in the list.
Below is a routine that uses some number of these native operations to sort the list. Across all four-element input lists, what is the maximum number of element comparisons made?
Feel free to modify the input list in line 4 and re-run the code!
from brilliant.algorithms.lists import *
# Try changing the order of these elements and running the code!
lst = ElementList('c', 'b', 'd', 'a')
assert n==4
if lst[i] > lst[j]:
lst.switch(i, j)
print('Sorted list: ' + str(lst))
print('Number of comparisons: '+str(lst.num_comparisons))
print('Number of switches: ' +str(lst.num_switches))
from brilliant.algorithms.lists import * # Try changing the order of these elements and running the code! lst = ElementList('c', 'b', 'd', 'a') n = len(lst) assert n==4 for i in range(n-1): for j in range(i+1,n): if lst[i] > lst[j]: lst.switch(i, j) print('Sorted list: ' + str(lst)) print('Number of comparisons: '+str(lst.num_comparisons)) print('Number of switches: ' +str(lst.num_switches))
Across all four-element input lists, what is the maximum number of switches made?
We have seen the worst-case number of switches and comparisons for one particular way of sorting the list. But what if we want to make a statement about any method of sorting that uses these two native operations (comparisons and switches)?
To start, we will explicitly define a comparison as an operation that takes in two numbers and has one of two results depending on their values. Using this definition, one way to get a lower bound for the number of comparisons needed is the Pigeonhole principle. For instance, we use it below to show that a sorting algorithm needs more than two comparisons to sort a list of four elements.
Suppose that a sorting algorithm uses at most two comparisons on any four-element list. Each comparison has two possible outcomes, so there are only
2^2=4
possible outcomes from the combination of the two. This means that with two comparisons, we can only reorder the list in four different ways.
However, there are 24 different possible initial orderings for a four element list, so to sort every possible input, we need to be able to reorder the list in 24 different ways. As a result, in order for us to sort the entire list with two comparisons, we would need to contain 24 different actions in our four potential outcomes.
Unfortunately, the pigeonhole principle, and common sense, tell us that this is impossible, meaning that we can't sort a 4 element list with just two comparisons.
Using this argument, what is the largest
N
for which we can definitively say that any algorithm that makes at most
N
comparisons cannot sort a four-element list?
In this quiz, we've seen how we can measure algorithms by the number of times they call a primitive function. By studying algorithms, we can learn efficient ways to solve common problems, as well as get a rich toolkit with which to reason about how well a given algorithm performs. |
\textcolor[rgb]{0.407843137254902,0.250980392156863,0.36078431372549}{\mathrm{ω}}
\mathrm{ω}
\mathrm{π}:E→M
be a fiber bundle, with base dimension
m
{\mathrm{π}}^{\mathrm{∞}}:{J}^{\mathrm{∞}}\left(E\right) → M
E
({x}^{i}, {u}^{\mathrm{α}}, {u}_{{i}_{}}^{\mathrm{α}}, {u}_{{i}_{}j}^{\mathrm{α}}
{u}_{\mathrm{ij} \cdot \cdot \cdot k}^{\mathrm{α}}, ....)
{\mathrm{Θ}}^{\mathrm{α}} = {\mathrm{du}}^{\mathrm{α}}-{u}_{\mathrm{ℓ}}^{\mathrm{α}}{\mathrm{dx}}^{\mathrm{ℓ}}
{\mathrm{\Omega }}^{\left(n,s\right)}\left({J}^{\infty }\left(E\right)\right)
n
s.
\mathrm{ω} ∈{\mathrm{\Omega }}^{\left(n,s\right)}\left({J}^{\infty }\left(E\right)\right)
{E}_{\mathrm{α}}\left(\mathrm{ω}\right) ∈ {\mathrm{Ω}}^{\left(n-1,s\right)}\left({J}^{\infty }\left(E\right)\right)
\mathrm{ω}
I: {\mathrm{\Omega }}^{\left(n,s\right)}\left({J}^{\infty }\left(E\right)\right)→{\mathrm{\Omega }}^{\left(n,s\right)}\left({J}^{\infty }\left(E\right)\right)
I\left(\mathrm{ω}\right) = \frac{1}{s}{\mathrm{Θ}}^{\mathrm{α} }∧{E}_{\mathrm{α}}\left(\mathrm{ω}\right).
I
\mathrm{η}
\left(n-1, s\right),
I\left({d}_{H}\mathrm{η}\right) = 0
{d}_{H }\mathrm{η}
\mathrm{η}
\mathrm{ω}
\left(n,s\right)
I\left(\mathrm{ω}\right) =0,
\left(n-1, s\right)
\mathrm{ω} = {d}_{H }\mathrm{η}
I
I∘I = I
\textcolor[rgb]{0.407843137254902,0.250980392156863,0.36078431372549}{\mathrm{\omega }}
\left(n, s\right)
I\left(\mathrm{ω}\right)
{J}^{3}\left(E\right)
E
\left(x,u\right)→ x.
{\mathrm{ω}}_{1}
\textcolor[rgb]{0,0,1}{\mathrm{ω1}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{_DG}}\textcolor[rgb]{0,0,1}{}\left([[\textcolor[rgb]{0,0,1}{"biform"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{E}\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}]]\textcolor[rgb]{0,0,1}{,}[[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{a}]\textcolor[rgb]{0,0,1}{,}[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{b}]\textcolor[rgb]{0,0,1}{,}[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{c}]\textcolor[rgb]{0,0,1}{,}[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{d}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{x}\right)]]]\right)
\textcolor[rgb]{0,0,1}{\mathrm{_DG}}\textcolor[rgb]{0,0,1}{}\left([[\textcolor[rgb]{0,0,1}{"biform"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{E}\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}]]\textcolor[rgb]{0,0,1}{,}[[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{d}}_{\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{c}}_{\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{b}}_{\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{a}]]]\right)
{\mathrm{ω}}_{2}
\textcolor[rgb]{0,0,1}{\mathrm{ω2}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{_DG}}\textcolor[rgb]{0,0,1}{}\left([[\textcolor[rgb]{0,0,1}{"biform"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{E}\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}]]\textcolor[rgb]{0,0,1}{,}[[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{a}]\textcolor[rgb]{0,0,1}{,}[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{b}]\textcolor[rgb]{0,0,1}{,}[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{c}]]]\right)
\textcolor[rgb]{0,0,1}{\mathrm{ω3}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{_DG}}\textcolor[rgb]{0,0,1}{}\left([[\textcolor[rgb]{0,0,1}{"biform"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{E}\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}]]\textcolor[rgb]{0,0,1}{,}[[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}\frac{{\textcolor[rgb]{0,0,1}{c}}_{\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}}}{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{b}}_{\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{a}]\textcolor[rgb]{0,0,1}{,}[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{c}}_{\textcolor[rgb]{0,0,1}{x}}}{\textcolor[rgb]{0,0,1}{2}}]\textcolor[rgb]{0,0,1}{,}[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{c}]]]\right)
{\mathrm{ω}}_{3}
{\mathrm{ω}}_{3}
\textcolor[rgb]{0,0,1}{\mathrm{_DG}}\textcolor[rgb]{0,0,1}{}\left([[\textcolor[rgb]{0,0,1}{"biform"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{E}\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}]]\textcolor[rgb]{0,0,1}{,}[[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}\frac{{\textcolor[rgb]{0,0,1}{c}}_{\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}}}{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{b}}_{\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{a}]\textcolor[rgb]{0,0,1}{,}[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{c}}_{\textcolor[rgb]{0,0,1}{x}}}{\textcolor[rgb]{0,0,1}{2}}]\textcolor[rgb]{0,0,1}{,}[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{c}]]]\right)
{J}^{3}\left(E\right)
E
\left(x,y, u, v\right)→ \left(x,y\right)
{\mathrm{ω}}_{4}.
\textcolor[rgb]{0,0,1}{\mathrm{ω4}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{_DG}}\textcolor[rgb]{0,0,1}{}\left([[\textcolor[rgb]{0,0,1}{"biform"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{E}\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}]]\textcolor[rgb]{0,0,1}{,}[[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{a}]\textcolor[rgb]{0,0,1}{,}[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{b}]\textcolor[rgb]{0,0,1}{,}[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{c}]\textcolor[rgb]{0,0,1}{,}[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{6}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{d}]\textcolor[rgb]{0,0,1}{,}[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{7}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{e}]\textcolor[rgb]{0,0,1}{,}[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{8}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{f}]]]\right)
\textcolor[rgb]{0,0,1}{\mathrm{_DG}}\textcolor[rgb]{0,0,1}{}\left([[\textcolor[rgb]{0,0,1}{"biform"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{E}\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}]]\textcolor[rgb]{0,0,1}{,}[[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{d}}_{\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{c}}_{\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{a}]\textcolor[rgb]{0,0,1}{,}[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{f}}_{\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{e}}_{\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{b}]]]\right)
{\mathrm{ω}}_{5}.
\textcolor[rgb]{0,0,1}{\mathrm{ω5}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{_DG}}\textcolor[rgb]{0,0,1}{}\left([[\textcolor[rgb]{0,0,1}{"biform"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{E}\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}]]\textcolor[rgb]{0,0,1}{,}[[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{7}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{a}]]]\right)
\textcolor[rgb]{0,0,1}{\mathrm{_DG}}\textcolor[rgb]{0,0,1}{}\left([[\textcolor[rgb]{0,0,1}{"biform"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{E}\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}]]\textcolor[rgb]{0,0,1}{,}[[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{7}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}\frac{{\textcolor[rgb]{0,0,1}{a}}_{\textcolor[rgb]{0,0,1}{x}}}{\textcolor[rgb]{0,0,1}{2}}]\textcolor[rgb]{0,0,1}{,}[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{12}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{a}}{\textcolor[rgb]{0,0,1}{2}}]\textcolor[rgb]{0,0,1}{,}[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}]\textcolor[rgb]{0,0,1}{,}\frac{{\textcolor[rgb]{0,0,1}{a}}_{\textcolor[rgb]{0,0,1}{x}}}{\textcolor[rgb]{0,0,1}{2}}]\textcolor[rgb]{0,0,1}{,}[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{9}]\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{a}}{\textcolor[rgb]{0,0,1}{2}}]]]\right)
{\mathrm{ω}}_{6}
\mathrm{η}.
\textcolor[rgb]{0,0,1}{\mathrm{\eta }}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{_DG}}\textcolor[rgb]{0,0,1}{}\left([[\textcolor[rgb]{0,0,1}{"biform"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{E}\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}]]\textcolor[rgb]{0,0,1}{,}[[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{6}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{7}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{9}]\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}}]]]\right)
\textcolor[rgb]{0,0,1}{\mathrm{ω6}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{_DG}}\textcolor[rgb]{0,0,1}{}\left([[\textcolor[rgb]{0,0,1}{"biform"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{E}\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}]]\textcolor[rgb]{0,0,1}{,}[[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{6}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{7}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{9}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}}]\textcolor[rgb]{0,0,1}{,}[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{6}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{7}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{16}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}}]\textcolor[rgb]{0,0,1}{,}[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{6}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{9}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{13}]\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}}]\textcolor[rgb]{0,0,1}{,}[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{7}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{9}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{11}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}}]]]\right)
\textcolor[rgb]{0,0,1}{\mathrm{_DG}}\textcolor[rgb]{0,0,1}{}\left([[\textcolor[rgb]{0,0,1}{"biform"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{E}\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}]]\textcolor[rgb]{0,0,1}{,}[[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}]]]\right) |
Compute the Inverse Laplace Transform of F(s) =frac{s}{R^2s^2+16pi^2} take R=70
Compute the Inverse Laplace Transform of F(s) =frac{s}{R^2s^2+16pi^2}take R=70
Compute the Inverse Laplace Transform of
F\left(s\right)=\frac{s}{{R}^{2}{s}^{2}+16{\pi }^{2}}
R=70
R=70
The given function is,
F\left(s\right)=\frac{s}{{R}^{2}{s}^{2}+16{\pi }^{2}}
F\left(s\right)=\frac{s}{{70}^{2}{s}^{2}+16{\pi }^{2}}
F\left(s\right)=\frac{s}{4900{s}^{2}+16{\pi }^{2}}
f\left(t\right)={L}^{-1}\left(\frac{s}{4900{s}^{2}+16{\pi }^{2}}\right)
{L}^{-1}\left(sF\left(s\right)\right)=\frac{d}{\left(dt\right)}f\left(t\right)+f\left(0\right)
Apply the above property and obtain the inverse Laplace transform of the given function as shown below.
f\left(t\right)={L}^{-1}\left(\frac{s}{4900{s}^{2}+16{\pi }^{2}}\right)
f\left(t\right)={L}^{-1}\left(\frac{1}{4900{s}^{2}+16{\pi }^{2}}\right)+{L}^{-1}\left(\frac{1}{4900{s}^{2}+16{\pi }^{2}}\right)\left(0\right)
f\left(t\right)=\frac{d}{\left(dt\right)}\left(\frac{1}{280}\mathrm{sin}\left(\frac{\left(2\pi t\right)}{35}\right)\right)+0
f\left(t\right)=\frac{1}{4900}\cdot \mathrm{cos}\left(\frac{2\pi t}{35}\right)
Thus, the required inverse Laplace transform is obtained.
One property of Laplace transform can be expressed in terms of the inverse Laplace transform as
{L}^{-1}\left\{\frac{{d}^{n}F}{d{s}^{n}}\right\}\left(t\right)=\left(-t{\right)}^{n}f\left(t\right)
f={L}^{-1}\left\{F\right\}
. Use this equation to compute
{L}^{-1}\left\{F\right\}
F\left(s\right)=\mathrm{arctan}\frac{23}{s}
Solution of I.V.P for harmonic oscillator with driving force is given by Inverse Laplace transform
y"+{\omega }^{2}y=\mathrm{sin}\gamma t,y\left(0\right)=0,{y}^{\prime }\left(0\right)=0
y\left(t\right)={L}^{-1}\left(\frac{\gamma }{\left({s}^{2}+{\omega }^{2}{\right)}^{2}}\right)
y\left(t\right)={L}^{-1}\left(\frac{\gamma }{{s}^{2}+{\omega }^{2}}\right)
y\left(t\right)={L}^{-1}\left(\frac{\gamma }{\left({s}^{2}+{\gamma }^{2}{\right)}^{2}}\right)
y\left(t\right)={L}^{-1}\left(\frac{\gamma }{\left({s}^{2}+{\gamma }^{2}\right)\left({s}^{2}+{\omega }^{2}\right)}\right)
Solve the initial value problem below using the method of Laplace transforms
y"-35y=144t-{36}^{-6t}
y\left(0\right)=0
{y}^{\prime }\left(0\right)=47
\frac{dy}{dx}+5y=15
f\left(t\right)=500{e}^{0.05t}
Use Laplace Transform to solve the given equation:
{y}^{″}+6{y}^{\prime }+5y=t\cdot U\left(t-2\right)
y\left(0\right)=1
{y}^{\prime }\left(0\right)=0
Solve the following differential equations using the indicated method.
SEPARATION OF VARIABLES:
{x}^{2}dx+y\left(x-1\right)dy=0 |
Lock-in amplifier - Wikipedia
Type of amplifier that emphasizes a specific frequency in a noisy signal
A lock in amplifier uses a multiplier and a low pass filter to compare a reference signal against a noisy signal
A lock-in amplifier is a type of amplifier that can extract a signal with a known carrier wave from an extremely noisy environment. Depending on the dynamic reserve of the instrument, signals up to a million times smaller than noise components, potentially fairly close by in frequency, can still be reliably detected. It is essentially a homodyne detector followed by low-pass filter that is often adjustable in cut-off frequency and filter order.
The device is often used to measure phase shift, even when the signals are large, have a high signal-to-noise ratio and do not need further improvement.
Recovering signals at low signal-to-noise ratios requires a strong, clean reference signal with the same frequency as the received signal. This is not the case in many experiments, so the instrument can recover signals buried in the noise only in a limited set of circumstances.
Example of a lock-in amplifier
The lock-in amplifier is commonly believed to have been invented by Princeton University physicist Robert H. Dicke who founded the company Princeton Applied Research (PAR) to market the product. However, in an interview with Martin Harwit, Dicke claims that even though he is often credited with the invention of the device, he believes that he read about it in a review of scientific equipment written by Walter C. Michels, a professor at Bryn Mawr College.[1] This could have been a 1941 article by Michels and Curtis,[2] which in turn cites a 1934 article by C. R. Cosens,[3] while another timeless article was written by C. A. Stutt in 1949.[4]
Whereas traditional lock-in amplifiers use analog frequency mixers and RC filters for the demodulation, state-of-the-art instruments have both steps implemented by fast digital signal processing, for example, on an FPGA. Usually sine and cosine demodulation is performed simultaneously, which is sometimes also referred to as dual-phase demodulation. This allows the extraction of the in-phase and the quadrature component that can then be transferred into polar coordinates, i.e. amplitude and phase, or further processed as real and imaginary part of a complex number (e.g. for complex FFT analysis).
2 Digital lock-in amplifiers
3 Signal measurement in noisy environments
The operation of a lock-in amplifier relies on the orthogonality of sinusoidal functions. Specifically, when a sinusoidal function of frequency f1 is multiplied by another sinusoidal function of frequency f2 not equal to f1 and integrated over a time much longer than the period of the two functions, the result is zero. Instead, when f1 is equal to f2 and the two functions are in phase, the average value is equal to half of the product of the amplitudes.
In essence, a lock-in amplifier takes the input signal, multiplies it by the reference signal (either provided from the internal oscillator or an external source, and can be sinusoidal or square wave[5]), and integrates it over a specified time, usually on the order of milliseconds to a few seconds. The resulting signal is a DC signal, where the contribution from any signal that is not at the same frequency as the reference signal is attenuated close to zero. The out-of-phase component of the signal that has the same frequency as the reference signal is also attenuated (because sine functions are orthogonal to the cosine functions of the same frequency), making a lock-in a phase-sensitive detector.
For a sine reference signal and an input waveform
{\displaystyle U_{\text{in}}(t)}
, the DC output signal
{\displaystyle U_{\text{out}}(t)}
can be calculated for an analog lock-in amplifier as
{\displaystyle U_{\text{out}}(t)={\frac {1}{T}}\int _{t-T}^{t}\sin \left[2\pi f_{\text{ref}}\cdot s+\varphi \right]U_{\text{in}}(s)\,ds,}
where φ is a phase that can be set on the lock-in (set to zero by default).
If the averaging time T is large enough (e.g. much larger than the signal period) to suppress all unwanted parts like noise and the variations at twice the reference frequency, the output is
{\displaystyle U_{\text{out}}={\frac {1}{2}}V_{\text{sig}}\cos \theta ,}
{\displaystyle V_{\text{sig}}}
is the signal amplitude at the reference frequency, and
{\displaystyle \theta }
is the phase difference between the signal and reference.
Many applications of the lock-in amplifier only require recovering the signal amplitude rather than relative phase to the reference signal. For a simple so called single-phase lock-in-amplifier the phase difference is adjusted (usually manually) to zero to get the full signal.
More advanced, so called two-phase lock-in-amplifiers have a second detector, doing the same calculation as before, but with an additional 90° phase shift. Thus one has two outputs:
{\displaystyle X=V_{\text{sig}}\cos \theta }
is called the "in-phase" component, and
{\displaystyle Y=V_{\text{sig}}\sin \theta }
the "quadrature" component. These two quantities represent the signal as a vector relative to the lock-in reference oscillator. By computing the magnitude (R) of the signal vector, the phase dependency is removed:
{\displaystyle R={\sqrt {X^{2}+Y^{2}}}=V_{\text{sig}}.}
The phase can be calculated from
{\displaystyle \theta =\arctan \left({\frac {Y}{X}}\right).}
Digital lock-in amplifiers[edit]
The majority of today's lock-in amplifiers are based on high-performance digital signal processing (DSP). Over the last 20 years, digital lock-in amplifiers have been replacing analog models across the entire frequency range, allowing users to perform measurements up to a frequency of 600 MHz. Initial problems of the first digital lock-in amplifiers, e.g. the presence of digital clock noise on the input connectors, could be completely eliminated by use of improved electronic components and better instrument design. Today's digital lock-in amplifiers outperform analog models in all relevant performance parameters, such as frequency range, input noise, stability and dynamic reserve. In addition to better performance, digital lock-in amplifiers can include multiple demodulators, which allows analyzing a signal with different filter settings or at multiple different frequencies simultaneously. Moreover, experimental data can be analyzed with additional tools such as an oscilloscope, FFT spectrum analyzers, boxcar averager or used to provide feedback by using internal PID controllers. Some models of the digital lock-in amplifiers are computer-controlled and feature a graphical user interface (can be a platform-independent browser user interface) and a choice of programming interfaces.
Signal measurement in noisy environments[edit]
Typical experimental setup
Signal recovery takes advantage of the fact that noise is often spread over a much wider range of frequencies than the signal. In the simplest case of white noise, even if the root mean square of noise is 103 times as large as the signal to be recovered, if the bandwidth of the measurement instrument can be reduced by a factor much greater than 106 around the signal frequency, then the equipment can be relatively insensitive to the noise. In a typical 100 MHz bandwidth (e.g. an oscilloscope), a bandpass filter with width much narrower than 100 Hz would accomplish this. The averaging time of the lock-in amplifier determines the bandwidth and allows very narrow filters, less than 1 Hz if needed. However, this comes at the price of a slow response to changes in the signal.
In summary, even when noise and signal are indistinguishable in the time domain, if the signal has a definite frequency band and there is no large noise peak within that band, noise and signal can be separated sufficiently in the frequency domain.
If the signal is either slowly varying or otherwise constant (essentially a DC signal), then 1/f noise typically overwhelms the signal. It may then be necessary to use external means to modulate the signal. For example, when detecting a small light signal against a bright background, the signal can be modulated either by a chopper wheel, acousto-optical modulator, photoelastic modulator at a large enough frequency so that 1/f noise drops off significantly, and the lock-in amplifier is referenced to the operating frequency of the modulator. In the case of an atomic-force microscope, to achieve nanometer and piconewton resolution, the cantilever position is modulated at a high frequency, to which the lock-in amplifier is again referenced.
When the lock-in technique is applied, care must be taken to calibrate the signal, because lock-in amplifiers generally detect only the root-mean-square signal of the operating frequency. For a sinusoidal modulation, this would introduce a factor of
{\displaystyle {\sqrt {2}}}
between the lock-in amplifier output and the peak amplitude of the signal, and a different factor for non-sinusoidal modulation.
In the case of nonlinear systems, higher harmonics of the modulation frequency appear. A simple example is the light of a conventional light bulb being modulated at twice the line frequency. Some lock-in amplifiers also allow separate measurements of these higher harmonics.
Furthermore, the response width (effective bandwidth) of detected signal depends on the amplitude of the modulation. Generally, linewidth/modulation function has a monotonically increasing, non-linear behavior.
^ Oral History Transcript — Dr. Robert Dicke.
^ Michels, W. C.; Curtis, N. L. (1941). "A Pentode Lock-In Amplifier of High Frequency Selectivity". Review of Scientific Instruments. 12 (9): 444. Bibcode:1941RScI...12..444M. doi:10.1063/1.1769919.
^ Cosens, C. R. (1934). "A balance-detector for alternating-current bridges". Proceedings of the Physical Society. 46 (6): 818–823. Bibcode:1934PPS....46..818C. doi:10.1088/0959-5309/46/6/310.
^ Stutt, C. A. (1949). "Low-frequency spectrum of lock-in amplifiers". MIT Technical Report (MIT) (105): 1–18.
^ Horowitz and Hill, 1985, the Art of electronics figure 14.35
Scofield, John H. (February 1994). "Frequency-domain description of a lock-in amplifier". American Journal of Physics. AAPT. 62 (2): 129–133. Bibcode:1994AmJPh..62..129S. doi:10.1119/1.17629.
Jaquier, Pierre-Alain; Jaquier, Alain (March 1994). "Multiple-channel digital lock-in amplifier with PPM resolution". Review of Scientific Instruments. AIP. 65 (3): 747. Bibcode:1994RScI...65..747P. doi:10.1063/1.1145096.
Wang, Xiaoyi (1990). "Sensitive digital lock-in amplifier using a personal computer". Review of Scientific Instruments. AIP. 61 (70): 1999–2001. Bibcode:1990RScI...61.1999W. doi:10.1063/1.1141413.
Wolfson, Richard (June 1991). "The lock-in amplifier: A student experiment". American Journal of Physics. AAPT. 59 (6): 569–572. Bibcode:1991AmJPh..59..569W. doi:10.1119/1.16824.
Dixon, Paul K.; Wu, Lei (October 1989). "Broadband digital lock-in amplifier techniques". Review of Scientific Instruments. AIP. 60 (10): 3329. Bibcode:1989RScI...60.3329D. doi:10.1063/1.1140523.
van Exter, Martin; Lagendijk, Ad (March 1986). "Converting an AM radio into a high-frequency lock-in amplifier in a stimulated Raman experiment". Review of Scientific Instruments. AIP. 57 (3): 390. Bibcode:1986RScI...57..390V. doi:10.1063/1.1138952.
Probst, P. A.; Collet, B. (March 1985). "Low-frequency digital lock-in amplifier". Review of Scientific Instruments. AIP. 56 (3): 466. Bibcode:1985RScI...56..466P. doi:10.1063/1.1138324.
Temple, Paul A. (1975). "An introduction to phase-sensitive amplifiers: An inexpensive student instrument". American Journal of Physics. AAPT. 43 (9): 801–807. Bibcode:1975AmJPh..43..801T. doi:10.1119/1.9690.
Burdett, Richard (2005). "Amplitude Modulated Signals - The Lock-in Amplifier". Handbook of Measuring System Design. Wiley. doi:10.1002/0471497398.mm588. ISBN 978-0-470-02143-9.
About LIAs from Stanford Research Systems. Application note detailing how lock-in amplifiers work.
Lock-in amplifier tutorial from Bentham Instruments. Comprehensive tutorial about the why and how of lock-in amplifiers.
Lock-in Technical Notes Range of Technical and Applications notes describing the design of digital and analog lock-ins, and guide to their specifications from SIGNAL RECOVERY.
PCSC-Lock-in Tool for data acquisition on acoustic chopping frequency using a computer sound card.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Lock-in_amplifier&oldid=1088748541" |
Find parametric equations for the tangent line to the curve with the given param
Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x=e^{-8t} cos(8t), y=e^{-8t} sin(8t), z=e^{-8t}, (1, 0, 1)
Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point.
x={e}^{-8t}\text{ }\mathrm{cos}\left(8t\right),\text{ }y={e}^{-8t}\text{ }\mathrm{sin}\left(8t\right),\text{ }z={e}^{-8t},\text{ }\left(1,\text{ }0,\text{ }1\right)
Bertha Stark
Step 1 Write the expression to find the parametric equations for a line through the point
\left({x}_{0},\text{ }{y}_{0},\text{ }{z}_{0}\right)\text{ }\text{are parallel to the vector}\text{ }v=\text{ }<\text{ }a,\text{ }b,\text{ }c\text{ }>.
x={x}_{0}\text{ }+\text{ }at,\text{ }y={y}_{0}\text{ }+\text{ }bt,\text{ }z={z}_{0}\text{ }+\text{ }ct
Step 2 Write the parametric equations of the curveas follows.
x={e}^{-8t}\text{ }\mathrm{cos}\left(8t\right),\text{ }y={e}^{-8t}\text{ }\mathrm{sin}\left(8t\right),\text{ }z={e}^{-8t}
Step 3 Write the vector equation from the parametric equations of the curve as follows.
r\left(t\right)=⟨\text{ }{e}^{-8t}\text{ }\mathrm{cos}\left(8t\right),\text{ }{e}^{-8t}\text{ }\mathrm{sin}\left(8t\right),\text{ }{e}^{-8t}⟩
Step 4 The tangent vector of the curve is the derivative of the vector function r(t). To find the derivative of the vector function, differentiate each component of the vector function.
{r}^{\prime }\left(t\right)=⟨\text{ }\frac{d}{dt}\left({e}^{-8t}\text{ }\mathrm{cos}\left(8t\right)\right),\text{ }\frac{d}{dt}\left({e}^{-8t}\text{ }\mathrm{sin}\left(8t\right)\right),\text{ }\frac{d}{dt}\left({e}^{8t}\right)⟩
=\text{ }⟨\text{ }{e}^{-8t}\left(-8\text{ }\mathrm{sin}\left(8t\right)\text{ }+\text{ }\mathrm{cos}\left(8t\right)\left(-8{e}^{-8t}\right),\text{ }{e}^{-8t}\left(8\text{ }\mathrm{cos}\left(8t\right)\right)\text{ }+\text{ }\mathrm{sin}\left(8t\right)\left(-8{e}^{-8t},\text{ }-8{e}^{-8t}\right)⟩
=\text{ }⟨\text{ }-8{e}^{-8t}\text{ }\mathrm{sin}\left(8t\right)\text{ }-\text{ }8{e}^{-8t}\text{ }\mathrm{cos}\left(8t\right),\text{ }8{e}^{-8t}\text{ }\mathrm{cos}\text{ }-\text{ }8{e}^{-8t}\text{ }\mathrm{sin}\left(8t\right),\text{ }-8{e}^{-8t}\right)⟩
Step 5 Given that the point
\left(1,\text{ }0,\text{ }1\right)
1={e}^{-8t}\text{ }\mathrm{cos}\left(8t\right)\text{ }⇒\text{ }{e}^{-8t}=1\text{ }or\text{ }\mathrm{cos}\left(8t\right)=1\text{ }⇒\text{ }t=0
0={e}^{-8t}\text{ }\mathrm{sin}\left(8t\right)\text{ }⇒\text{ }{e}^{-8t}\text{ }\ne \text{ }0\text{ }so\text{ }\mathrm{sin}\left(8t\right)=0\text{ }⇒\text{ }t=0
1={e}^{-8t}\text{ }⇒\text{ }t=0
As the specified point
\left(1,0,1\right)\text{ }\text{corresponds to}\text{ }t=0
, consider the value of scalar parameter t as 0 and substitute in the parametric equations of the curve to obtain the point, which is on the required line. Substitute 0 for t in equation (2),
x={e}^{-8\left(0\right)}\text{ }\mathrm{cos}\left(8\left(0\right)\right),\text{ }y={e}^{-8\left(0\right)}\text{ }\mathrm{sin}\left(8\left(0\right)\right),\text{ }z={e}^{-8\left(0\right)}
x={e}^{0}\text{ }\mathrm{cos}\left(0\right),\text{ }y={e}^{0}\text{ }\mathrm{sin}\left(0\right),\text{ }z={e}^{-8\left(0\right)}
x=1,\text{ }y=0,\text{ }z=1
The point on the required line is
\left(1,\text{ }0,\text{ }1\right).
As the point on the required line is same as the specified point
\left(1,\text{ }0,\text{ }1\right),\text{ }\text{the tangent vector is}\text{ }{r}^{\prime }\left(t\right)\text{ }at\text{ }t=0.
Step 6 Substitute 0 for t in
Use set theoretic or vector notation or both to describe the points that lie in the given configurations. The line passing through (0,2,1) in the direction of
2i-k
How do you write the parametric equations represent the ellipse given by
\frac{{x}^{2}}{9}+\frac{{y}^{2}}{81}=1
Consider the helix represented investigation by the vector-valued function
r\left(t\right)=\text{ }<\text{ }2\text{ }\mathrm{cos}\text{ }t,\text{ }2\text{ }\mathrm{sin}\text{ }t,\text{ }t\text{ }>
. Solve for t in the relationship derived in part (a), and substitute the result into the original set of parametric equations. This yields a parametrization of the curve in terms of the arc length parameter s.
x=\left(80{\mathrm{cos}45}^{o}\right)t,y=6+\left(80{\mathrm{sin}45}^{o}\right)t-16{t}^{2}
. t = 2.
Determine the area of the region below the parametric curve given by the set of parametric equations. For each problem you may assume that each curve traces out exactly once from right to left for the given range of t. For these problems you should only use the given parametric equations to determine the answer. 1.
x={t}^{2}+5t-1y=40-{t}^{2}-2\le t\le 5
x=3co{s}^{2}\left(t\right)—si{n}^{2}\left(t\right)y=6+cos\left(t\right)-\frac{\pi }{2}\ne t\le 0
x={e}^{\frac{1}{4}t}—2y=4+{e}^{\frac{1}{4t}}—{e}^{\frac{1}{4}t}-6\le t\le 1
Find the volume of the solid generated by revolving the standed region about the x-axis.
Te volume of the solid is ? cubic units.
4x+3y=24 |
Circulant matrix - Wikipedia
For the symmetric graphs, see Circulant graph.
In linear algebra, a circulant matrix is a square matrix in which all row vectors are composed of the same elements and each row vector is rotated one element to the right relative to the preceding row vector. It is a particular kind of Toeplitz matrix.
In numerical analysis, circulant matrices are important because they are diagonalized by a discrete Fourier transform, and hence linear equations that contain them may be quickly solved using a fast Fourier transform.[1] They can be interpreted analytically as the integral kernel of a convolution operator on the cyclic group
{\displaystyle C_{n}}
and hence frequently appear in formal descriptions of spatially invariant linear operations. This property is also critical in modern software defined radios, which utilize Orthogonal Frequency Division Multiplexing to spread the symbols (bits) using a cyclic prefix. This enables the channel to be represented by a circulant matrix, simplifying channel equalization in the frequency domain.
In cryptography, a circulant matrix is used in the MixColumns step of the Advanced Encryption Standard.
3 Analytic interpretation
4 Symmetric circulant matrices
5 Hermitian circulant matrices
6.1 In linear equations
6.2 In graph theory
{\displaystyle n\times n}
{\displaystyle C}
{\displaystyle C={\begin{bmatrix}c_{0}&c_{n-1}&\cdots &c_{2}&c_{1}\\c_{1}&c_{0}&c_{n-1}&&c_{2}\\\vdots &c_{1}&c_{0}&\ddots &\vdots \\c_{n-2}&&\ddots &\ddots &c_{n-1}\\c_{n-1}&c_{n-2}&\cdots &c_{1}&c_{0}\\\end{bmatrix}}}
or the transpose of this form (by choice of notation). When the term
{\displaystyle c_{i}}
{\displaystyle p\times p}
square matrix, then the
{\displaystyle np\times np}
{\displaystyle C}
is called a block-circulant matrix.
A circulant matrix is fully specified by one vector,
{\displaystyle c}
, which appears as the first column (or row) of
{\displaystyle C}
. The remaining columns (and rows, resp.) of
{\displaystyle C}
are each cyclic permutations of the vector
{\displaystyle c}
with offset equal to the column (or row, resp.) index, if lines are indexed from 0 to
{\displaystyle n-1}
. (Cyclic permutation of rows has the same effect as cyclic permutation of columns.) The last row of
{\displaystyle C}
{\displaystyle c}
shifted by one in reverse.
Different sources define the circulant matrix in different ways, for example as above, or with the vector
{\displaystyle c}orresponding to the first row rather than the first column of the matrix; and possibly with a different direction of shift (which is sometimes called an anti-circulant matrix).
{\displaystyle f(x)=c_{0}+c_{1}x+\dots +c_{n-1}x^{n-1}}
is called the associated polynomial of matrix
{\displaystyle C}
Eigenvectors and eigenvalues[edit]
The normalized eigenvectors of a circulant matrix are the Fourier modes, namely,
{\displaystyle v_{j}={\frac {1}{\sqrt {n}}}\left(1,\omega ^{j},\omega ^{2j},\ldots ,\omega ^{(n-1)j}\right),\quad j=0,1,\ldots ,n-1,}
{\displaystyle \omega =\exp \left({\tfrac {2\pi i}{n}}\right)}
{\displaystyle n}
-th root of unity and
{\displaystyle i}s the imaginary unit.
(This can be understood by realizing that multiplication with a circulant matrix implements a convolution. In Fourier space, convolutions become multiplication. Hence the product of a circulant matrix with a Fourier mode yields a multiple of that Fourier mode, i.e. it is an eigenvector.)
The corresponding eigenvalues are given by
{\displaystyle \lambda _{j}=c_{0}+c_{n-1}\omega ^{j}+c_{n-2}\omega ^{2j}+\dots +c_{1}\omega ^{(n-1)j},\quad j=0,1,\dots ,n-1.}
As a consequence of the explicit formula for the eigenvalues above, the determinant of a circulant matrix can be computed as:
{\displaystyle \det(C)=\prod _{j=0}^{n-1}(c_{0}+c_{n-1}\omega ^{j}+c_{n-2}\omega ^{2j}+\dots +c_{1}\omega ^{(n-1)j}).}
Since taking the transpose does not change the eigenvalues of a matrix, an equivalent formulation is
{\displaystyle \det(C)=\prod _{j=0}^{n-1}(c_{0}+c_{1}\omega ^{j}+c_{2}\omega ^{2j}+\dots +c_{n-1}\omega ^{(n-1)j})=\prod _{j=0}^{n-1}f(\omega ^{j}).}
The rank of a circulant matrix
{\displaystyle C}
{\displaystyle n-d}
{\displaystyle d}
is the degree of the polynomial
{\displaystyle \gcd(f(x),x^{n}-1)}
Any circulant is a matrix polynomial (namely, the associated polynomial) in the cyclic permutation matrix
{\displaystyle P}
{\displaystyle C=c_{0}I+c_{1}P+c_{2}P^{2}+\dots +c_{n-1}P^{n-1}=f(P),}
{\displaystyle P}
{\displaystyle P={\begin{bmatrix}0&0&\cdots &0&1\\1&0&\cdots &0&0\\0&\ddots &\ddots &\vdots &\vdots \\\vdots &\ddots &\ddots &0&0\\0&\cdots &0&1&0\end{bmatrix}}.}
{\displaystyle n\times n}
circulant matrices forms a{\displaystyle n}
-dimensional vector space with respect to addition and scalar multiplication. This space can be interpreted as the space of functions on the cyclic group of order
{\displaystyle n}
{\displaystyle C_{n}}
, or equivalently as the group ring of
{\displaystyle C_{n}}
Circulant matrices form a commutative algebra, since for any two given circulant matrices
{\displaystyle A}
{\displaystyle B}
{\displaystyle A+B}
is circulant, the product
{\displaystyle AB}
is circulant, and
{\displaystyle AB=BA}
{\displaystyle U}
that is composed of the eigenvectors of a circulant matrix is related to the discrete Fourier transform and its inverse transform:
{\displaystyle U_{n}^{*}={\frac {1}{\sqrt {n}}}F_{n},\quad {\text{and}}\quad U_{n}={\frac {1}{\sqrt {n}}}F_{n}^{-1},{\text{ where }}F_{n}=(f_{jk}){\text{ with }}f_{jk}=e^{-2jk\pi i/n},\,{\text{for }}0\leq j,k<n.}
Consequently the matrix
{\displaystyle U_{n}}
diagonalizes
{\displaystyle C}
. In fact, we have
{\displaystyle C=U_{n}\operatorname {diag} (F_{n}c)U_{n}^{*}={\frac {1}{n}}F_{n}^{-1}\operatorname {diag} (F_{n}c)F_{n},}
{\displaystyle c}
is the first column of
{\displaystyle C}
. The eigenvalues of
{\displaystyle C}
are given by the product
{\displaystyle F_{n}c}
. This product can be readily calculated by a fast Fourier transform.[3]
{\displaystyle p(x)}
be the (monic) characteristic polynomial of an
{\displaystyle n\times n}
{\displaystyle C}
{\displaystyle p'(x)}
be the derivative of
{\displaystyle p(x)}
. Then the polynomial
{\textstyle {\frac {1}{n}}p'(x)}
is the characteristic polynomial of the following
{\displaystyle (n-1)\times (n-1)}
{\displaystyle C}
{\displaystyle C_{n-1}={\begin{bmatrix}c_{0}&c_{n-1}&\cdots &c_{3}&c_{2}\\c_{1}&c_{0}&c_{n-1}&&c_{3}\\\vdots &c_{1}&c_{0}&\ddots &\vdots \\c_{n-3}&&\ddots &\ddots &c_{n-1}\\c_{n-2}&c_{n-3}&\cdots &c_{1}&c_{0}\\\end{bmatrix}}}
(see [4] for the proof).
Analytic interpretation[edit]
Circulant matrices can be interpreted geometrically, which explains the connection with the discrete Fourier transform.
Consider vectors in
{\displaystyle \mathbb {R} ^{n}}
as functions on the integers with period
{\displaystyle n}
, (i.e., as periodic bi-infinite sequences:
{\displaystyle \dots ,a_{0},a_{1},\dots ,a_{n-1},a_{0},a_{1},\dots }
) or equivalently, as functions on the cyclic group of order
{\displaystyle n}
{\displaystyle C_{n}}
{\displaystyle \mathbb {Z} /n\mathbb {Z} }
) geometrically, on (the vertices of) the regular
{\displaystyle n}
-gon: this is a discrete analog to periodic functions on the real line or circle.
Then, from the perspective of operator theory, a circulant matrix is the kernel of a discrete integral transform, namely the convolution operator for the function
{\displaystyle (c_{0},c_{1},\dots ,c_{n-1})}
; this is a discrete circular convolution. The formula for the convolution of the functions
{\displaystyle (b_{i}):=(c_{i})*(a_{i})}
{\displaystyle b_{k}=\sum _{i=0}^{n-1}a_{i}c_{k-i}}
(recall that the sequences are periodic) which is the product of the vector
{\displaystyle (a_{i})}
by the circulant matrix for
{\displaystyle (c_{i})}
The discrete Fourier transform then converts convolution into multiplication, which in the matrix setting corresponds to diagonalization.
{\displaystyle C^{*}}
-algebra of all circulant matrices with complex entries is isomorphic to the group
{\displaystyle C^{*}}
-algebra of
{\displaystyle \mathbb {Z} /n\mathbb {Z} }
Symmetric circulant matrices[edit]
For a symmetric circulant matrix
{\displaystyle C}
one has the extra condition that
{\displaystyle c_{n-i}=c_{i}}
. Thus it is determined by
{\displaystyle \lfloor n/2\rfloor +1}
{\displaystyle C={\begin{bmatrix}c_{0}&c_{1}&\cdots &c_{2}&c_{1}\\c_{1}&c_{0}&c_{1}&&c_{2}\\\vdots &c_{1}&c_{0}&\ddots &\vdots \\c_{2}&&\ddots &\ddots &c_{1}\\c_{1}&c_{2}&\cdots &c_{1}&c_{0}\\\end{bmatrix}}.}
The eigenvalues of any real symmetric matrix are real. The corresponding eigenvalues become:
{\displaystyle \lambda _{j}=c_{0}+2c_{1}\Re \omega _{j}+2c_{2}\Re \omega _{j}^{2}+\dots +2c_{n/2-1}\Re \omega _{j}^{n/2-1}+c_{n/2}\omega _{j}^{n/2}}
{\displaystyle n}
even, and
{\displaystyle \lambda _{j}=c_{0}+2c_{1}\Re \omega _{j}+2c_{2}\Re \omega _{j}^{2}+\dots +2c_{(n-1)/2}\Re \omega _{j}^{(n-1)/2}}
{\displaystyle n}
odd, where
{\displaystyle \Re z}
{\displaystyle z}
. This can be further simplified by using the fact that
{\displaystyle \Re \omega _{j}^{k}=\cos(2\pi jk/n)}
Symmetric circulant matrices belong to the class of bisymmetric matrices.
Hermitian circulant matrices[edit]
The complex version of the circulant matrix, ubiquitous in communications theory, is usually Hermitian. In this case
{\displaystyle c_{n-i}=c_{i}^{*},\;i\leq n/2}
and its determinant and all eigenvalues are real.
If n is even the first two rows necessarily takes the form
{\displaystyle {\begin{bmatrix}r_{0}&z_{1}&z_{2}&r_{3}&z_{2}^{*}&z_{1}^{*}\\z_{1}^{*}&r_{0}&z_{1}&z_{2}&r_{3}&z_{2}^{*}\\\dots \\\end{bmatrix}}.}
in which the first element
{\displaystyle r_{3}}
in the top second half-row is real.
If n is odd we get
{\displaystyle {\begin{bmatrix}r_{0}&z_{1}&z_{2}&z_{2}^{*}&z_{1}^{*}\\z_{1}^{*}&r_{0}&z_{1}&z_{2}&z_{2}^{*}\\\dots \\\end{bmatrix}}.}
Tee[5] has discussed constraints on the eigenvalues for the Hermitian condition.
In linear equations[edit]
{\displaystyle C\mathbf {x} =\mathbf {b} ,}
{\displaystyle C}
is a circulant square matrix of size
{\displaystyle n}
we can write the equation as the circular convolution
{\displaystyle \mathbf {c} \star \mathbf {x} =\mathbf {b} ,}
{\displaystyle \mathbf {c} }
{\displaystyle C}
, and the vectors
{\displaystyle \mathbf {c} }
{\displaystyle \mathbf {x} }
{\displaystyle \mathbf {b} }
are cyclically extended in each direction. Using the circular convolution theorem, we can use the discrete Fourier transform to transform the cyclic convolution into component-wise multiplication
{\displaystyle {\mathcal {F}}_{n}(\mathbf {c} \star \mathbf {x} )={\mathcal {F}}_{n}(\mathbf {c} ){\mathcal {F}}_{n}(\mathbf {x} )={\mathcal {F}}_{n}(\mathbf {b} )}
{\displaystyle \mathbf {x} ={\mathcal {F}}_{n}^{-1}\left[\left({\frac {({\mathcal {F}}_{n}(\mathbf {b} ))_{\nu }}{({\mathcal {F}}_{n}(\mathbf {c} ))_{\nu }}}\right)_{\!\nu \in \mathbb {Z} }\right]^{\rm {T}}.}
This algorithm is much faster than the standard Gaussian elimination, especially if a fast Fourier transform is used.
In graph theory[edit]
In graph theory, a graph or digraph whose adjacency matrix is circulant is called a circulant graph (or digraph). Equivalently, a graph is circulant if its automorphism group contains a full-length cycle. The Möbius ladders are examples of circulant graphs, as are the Paley graphs for fields of prime order.
^ Davis, Philip J., Circulant Matrices, Wiley, New York, 1970 ISBN 0471057711
^ A. W. Ingleton (1956). "The Rank of Circulant Matrices". J. London Math. Soc. s1-31 (4): 445–460. doi:10.1112/jlms/s1-31.4.445.
^ Golub, Gene H.; Van Loan, Charles F. (1996), "§4.7.7 Circulant Systems", Matrix Computations (3rd ed.), Johns Hopkins, ISBN 978-0-8018-5414-9
^ Kushel, Olga; Tyaglov, Mikhail (July 15, 2016), "Circulants and critical points of polynomials", Journal of Mathematical Analysis and Applications, 439 (2): 634–650, arXiv:1512.07983, doi:10.1016/j.jmaa.2016.03.005, ISSN 0022-247X
^ Tee, G J (2007). "Eigenvectors of Block Circulant and Alternating Circulant Matrices". New Zealand Journal of Mathematics. 36: 195–211.
R. M. Gray, Toeplitz and Circulant Matrices: A Review doi:10.1561/0100000006
Weisstein, Eric W. "Circulant Matrix". MathWorld.
IPython Notebook demonstrating properties of circulant matrices
Retrieved from "https://en.wikipedia.org/w/index.php?title=Circulant_matrix&oldid=1070986566" |
The mean marks of a group of students is 60 If their marks are 85, 62, 36, 48,72, x - Maths - Data Handling - 10598901 | Meritnation.com
The mean marks of a group of students is 60. If their marks are 85, 62, 36, 48,72, x , 75, and 39 then find the value of x ?
\mathrm{We} \mathrm{know} \mathrm{that} :\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{mean} = \frac{\mathrm{sum} \mathrm{of} \mathrm{observations}}{\mathrm{number} \mathrm{of} \mathrm{observations}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒60 = \frac{85+62+36+48+72+\mathrm{x}+75+39}{8}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒480 = 417 + \mathrm{x}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{x} = 480 - 417\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{x} = 63
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Entropy (Information Theory) | Brilliant Math & Science Wiki
Alexander Katz, Tejas Suresh, David Holcer, and
In information theory, the major goal is for one person (a transmitter) to convey some message (over a channel) to another person (the receiver). To do so, the transmitter sends a series (possibly just one) partial messages that give clues towards the original message. The information content of one of these partial messages is a measure of how much uncertainty this resolves for the receiver. For instance,
A partial message that cuts the number of possibilities in half transmits one bit of information about the message. For instance, if the transmitter were trying to transmit the result of a randomly chosen digit to the receiver, the partial message "the number is odd" would transmit one bit of information.
A partial message that doesn't reduce the number of possibilities at all transmits no information at all; for instance, the partial message "the number is less than 10" transmits zero bits of information.
In essence, the "information content" can be viewed as how much useful information the message actually contains.
The entropy, in this context, is the expected number of bits of information contained in each message, taken over all possibilities for the transmitted message. For example, suppose the transmitter wanted to inform the receiver of the result of a 4-person tournament, where some of the players are better than others. The entropy of this message would be a weighted average of the amount of information each of the possible messages (i.e. "player 1 won", "player 2 won", "player 3 won", or "player 4 won") provides.
In essence, the "entropy" can be viewed as how much useful information a message is expected to contain. It also provides a lower bound for the "size" of an encoding scheme, in the sense that the expected number of bits to be transmitted under some encoding scheme is at least the entropy of the message. An encoding scheme that manages to achieve this lower bound is called lossless.
Formal Definition (Information)
Formal Definition (Entropy)
Application to Encoding
Application to Data Compression
Before we define
H
formally, let us see the properties of
H:
H(X)
2) Conditioning reduces entropy, i.e.
H(X/Y)<H(X).
3) Invariance under relabeling:
H(X)=H(f(X))
for any bijective
f.
A situation with several possible outcomes can be modeled by a random variable
X
, where for any possible outcome
x_i
that has probability
p_i
of occurring:
\text{Pr}(X=x_i) = p_i.
Communication can be viewed as giving information about this random variable
X
The simplest type of message is one that specifies the value of
X
. The information content of a message specifying
X
resulted in
x_i
I(x_i)=\log_2\left(\frac{1}{p_i}\right).
\log_25
Alice randomly chooses an 8-bit integer
X
, and gives Bob another 8-bit integer
Y
. Alice then tells Bob that the Hamming distance between
X
Y
is 1. How much information has Bob been given about
X
This makes an intuitive sense, as specifying a high-probability event occurred is not transmitting much information (as it doesn't resolve much uncertainty--the result was "expected"), while specifying a low-probability event occurred is transmitting a significant amount of information. The reason that
I
is logarithmic stems from the simple reason that
I
is additive over independent events; i.e.
I(x_ix_j)=I(x_i)+I(x_j)
meaning that, in essence, if the situation were run twice and the information about both results were transmitted simultaneously, the information content would be the same as if information about the results were separately transmitted. It is clear that the above condition is satisfied by the logarithmic function, hence the definition.
It is also worth noting that the definition is satisfying for two reasons:
Transmitting a result that had probability
\frac{1}{2}
transmits
\log_2\left(\frac{1}{\frac{1}{2}}\right)=1
bit of information, which agrees with the introduction (that cutting the number of possibilities by half is transmitting 1 bit of information).
Transmitting a result that had probability 1 transmits
\log_2\left(\frac{1}{1}\right)=0
bits of information, which also agrees with the introduction (that transmitting a message that doesn't reduce the number of possibilities transmits no information).
Furthermore, this definition continues to hold even when more complex messages are received. For instance, in the previous example of transmitting the result of a random digit, the message "the digit is at least 3" had a
\frac{7}{10}
chance of being received; hence, the information content of such a message is
\log_2\left(\frac{1}{\frac{7}{10}}\right)=\log_2\left(\frac{10}{7}\right).
It is also worth examining what happens upon the further message "the digit is at most 3": this has a
\frac{1}{7}
chance of being received, so the information content of this message is
\log_2\left(\frac{1}{\frac{1}{7}}\right)=\log_2 7.
Therefore, the information content of these messages taken together is
\log_2\left(\frac{10}{7}\right)+\log_2 7=\log_2 10,
which is exactly the information content of the message "the digit is 3." This agrees with the fact that information is additive over independent events, and affirms the choice of a logarithmic function.
The entropy of a message is defined as the expected amount of information to be transmitted about the random variable
X
defined in the previous section. More formally, if
X
takes on the states
x_1, x_2, \ldots, x_n
, the entropy is defined as
\sum_{i=1}^n p_i\log_2\left(\frac{1}{p_i}\right).
Alice rolls two dice, and wants to communicate their sum to Bob. What is the entropy of this message?
This comes from repeating the scenario involving
X
a large number
N
times. The expected number of times
x_i
is transmitted is
n_i=Np_i
, and the amount of information received from that message is
I(x_i)=\log_2\left(\frac{1}{p_i}\right)
. Hence the total amount of information received over the
N
messages (recall that information is additive) is
\sum_{i=1}^n n_iI(x_i)=\sum_{i=1}^nNp_iI(x_i)=N\sum_{i=1}^np_iI(x_i),
so the expected amount of information per message is
\sum_{i=1}^np_iI(x_i)
A small issue worth pointing out immediately: in the case where
p_i=0
, it can be assumed that
p_i\log(p_i)=0
, which is justified as
\lim_{x \rightarrow 0}x\log x=0
. This also makes intuitive sense: if an event never occurs, it cannot contribute to the entropy, as it is never expected to occur.
A major goal of information theory is to encode messages as binary strings in such a way that
a string of transmitted bits (binary digits) can be reconstructed into a sequence of messages, and
the number of bits necessary to transmit message is as small as possible.
For instance, the encoding scheme
is efficient but extremely flawed, as the message "AB" and the message "C" are indistinguishable (both would be transmitted as 01). In contrast, the encoding scheme
is correct (any stream of bits can be reconstructed into the original message), but extremely inefficient as significantly less bits are necessary to reliably transmit messages.
It is natural to ask what the "best" encoding scheme is, and while the answer strongly depends on application (for example, transmissions over noisy channels would do well to employ error correcting codes), one natural answer is "the correct encoding scheme that minimizes the expected length of a message." That theoretical minimum is given by the entropy of the message.
For example, suppose a message details the value of a random variable
X
x
\text{Pr}(X = x)
The entropy of this message is
\begin{aligned} \sum_{i=1}^{n} p_i\log_2\left(\frac{1}{p_i}\right) &= 0.5\log_2\left(\frac{1}{0.5}\right) + 0.25\log_2\left(\frac{1}{0.25}\right) + 0.125\log_2\left(\frac{1}{0.125}\right) + 0.125\log_2\left(\frac{1}{0.125}\right) \\ &=0.5 \cdot 1 + 0.25 \cdot 2 + 0.125 \cdot 3 + 0.125 \cdot 3 \\ &= 1.75, \end{aligned}
so the expected length of any correct encoding scheme is at least
1.75
. In this case, this can be realized with a relatively simple scheme:
Letter Encoding Length
It can be verified that the above encoding is correct (it is an example of Huffman coding), and it has expected length
1 \cdot 0.5 + 2 \cdot 0.25 + 3 \cdot 0.125 + 3 \cdot 0.125 = 1.75,
which is equal to the entropy of the message.
A ubiquitous application of encoding schemes, and thus entropy, is to data compression: the act of transferring a large file into a smaller, equivalent file for storage (but usually not human readability). One simple example of such a scheme is a run-length code, which replaces each sequence of repeated bits with two numbers: the bit and the number of times it is to appear. For instance, the message "0011100011" could become "02130312".
In essence, this kind of data compression is doing the same thing entropy was meant to measure: events that occur with higher likelihood give less information content, and thus should be encoded with fewer bits. Analogously, data compression identifies inherent structure in the large file, and replaces the more commonly appearing structures with smaller compressions. For instance[1],
can be compressed to
2=Sam-I-Am
3=I do not like
4=green eggs and ham
5=you like
6=here or there
7=I would not like them
T1 2!
3 t1 2!
Do 5 4?
Would 5 them 6?
3 them, 2.
which is a shortening from 322 to 186 characters (and the entire story would experience even better compression).
Garcia, E. How does file compression work?. Retrieved March 16th, 2016, from http://qr.ae/RS2Y2m
Cite as: Entropy (Information Theory). Brilliant.org. Retrieved from https://brilliant.org/wiki/entropy-information-theory/ |
Moving Average Model - MATLAB & Simulink - MathWorks Switzerland
The moving average (MA) model captures serial autocorrelation in a time series yt by expressing the conditional mean of yt as a function of past innovations,
{\epsilon }_{t-1},{\epsilon }_{t-2},\dots ,{\epsilon }_{t-q}
. An MA model that depends on q past innovations is called an MA model of degree q, denoted by MA(q).
{y}_{t}=c+{\epsilon }_{t}+{\theta }_{1}{\epsilon }_{t-1}+\dots +{\theta }_{q}{\epsilon }_{t-q},
{\epsilon }_{t}
is an uncorrelated innovation process with mean zero. For an MA process, the unconditional mean of yt is μ = c.
{L}^{i}{y}_{t}={y}_{t-i}
. Define the degree q MA lag operator polynomial
\theta \left(L\right)=\left(1+{\theta }_{1}L+\dots +{\theta }_{q}{L}^{q}\right).
You can write the MA(q) model as
{y}_{t}=\mu +\theta \left(L\right){\epsilon }_{t}.
By Wold’s decomposition [2], an MA(q) process is always stationary because
\theta \left(L\right)
is a finite-degree polynomial.
For a given process, however, there is no unique MA polynomial—there is always a noninvertible and invertible solution [1]. For uniqueness, it is conventional to impose invertibility constraints on the MA polynomial. Practically speaking, choosing the invertible solution implies the process is causal. An invertible MA process can be expressed as an infinite-degree AR process, meaning only past events (not future events) predict current events. The MA operator polynomial
\theta \left(L\right)
is invertible if all its roots lie outside the unit circle. |
Atomic number of an element is equal to the number of :
Atomic number of an element is always equal to the number of protons in its atom while mass number is equal to the sum of the number of protons and neutrons .
Number of electrons surrounding Kr in KrF2 is :
Total no. of electron which are found around , Kr in KrF2 are 10 .
Which is the correct relation for diffusion of gases :
\propto
\propto
\sqrt{\mathrm{d}}
\propto
\frac{1}{\sqrt{\mathrm{d}}}
\propto
\frac{1}{\sqrt{\mathrm{d}}}
According to Graham for an ideal gas ,
\propto
\frac{1}{\sqrt{\mathrm{d}}}
Hydrolysis of sodium acetate gives :
amphoteric solution
Sodium acetate is a salt of weak acid and strong base hence its aquous solution is basic .
C-H3CONa
\stackrel{{\mathrm{H}}_{2}\mathrm{O}}{⇌}
\underset{\mathrm{Weak}}{{\mathrm{CH}}_{3}\mathrm{COOH}}
\underset{\mathrm{Strong}}{\mathrm{NaOH}}
Which has smallest size ?
Among isoelectronic species , size decreases with increase in nuclear charge .Hence , the order of size is _
Mg2+ < Na+ < O2-
Isotopic pair is :
Isotopes are the atoms of an element which have same atomic number but different mass number .
H - I - O bond angle in H2O2 :
The bond angle of H2O2 is 94.8o .
Cationic hydrolysis gives the following solution :
Cationic hydrolysis always forms basic solution .
Hybridization present in ClF3 is :
In ClF3 molecule , Cl atom is sp3d hybridised and due to the presence of two lone pairs of electrons , it has T-shaped geometry .
Electron deficient molecule is :
Structure of BF3 is ;
In this example octet is incomplete . |
Solution of Second-Order Ordinary Differential Equations via Simulated Annealing ()
1Department of General Studies, Petroleum Training Institute, Delta, Nigeria.
DOI: 10.4236/ojop.2019.81003 PDF HTML XML 604 Downloads 1,241 Views Citations
Bilesanmi, A. , Wusu, A. and Olutimo, A. (2019) Solution of Second-Order Ordinary Differential Equations via Simulated Annealing. Open Journal of Optimization, 8, 32-37. doi: 10.4236/ojop.2019.81003.
{y}^{″}=f\left(x,y\right)
{y}^{″}+p\left(t\right){y}^{\prime }+q\left(t\right)y=r\left(t\right)
{u}^{″}=f\left(t,u\right);u\left(a\right)={\eta }_{1};u\left(b\right)={\eta }_{2}
{y}^{″}=f\left(t,y\right);\text{ }y\left({t}_{0}\right)={y}_{0},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{y}^{\prime }\left({t}_{0}\right)={{y}^{\prime }}_{0},\text{ }t\in \left[a,b\right].
{x}_{new}
{x}_{best}
f\left({x}_{new}\right)\le f\left({x}_{best}\right)
{x}_{new}
{x}_{best}
{x}_{new}
\mathrm{exp}\left(b\left(i,\Delta f,{f}_{0}\right)\right)
\Delta f
{f}_{0}
b\left(i,\Delta f,{f}_{0}\right):=\frac{-\Delta f\mathrm{log}\left(i+1\right)}{10}
\mathrm{min}\left(2d,50\right)
d=1
y\left(t\right)=\underset{i=0}{\overset{k}{\sum }}\text{ }{\psi }_{i}{t}^{i},\text{ }k\in {ℤ}^{+}
{\psi }_{i}
\underset{i=2}{\overset{k}{\sum }}\text{ }\text{ }i\left(i-1\right){\psi }_{i}{t}^{i-2}=f\left(t,y\left(t\right)\right)
{\left[\underset{i=0}{\overset{k}{\sum }}\text{ }\text{ }{\psi }_{i}{t}^{i}\right]}_{t={t}_{0}}={y}_{0}\text{ }\text{and}\text{ }{\left[\underset{i=1}{\overset{k}{\sum }}\text{ }i{\psi }_{i}{t}^{i-1}\right]}_{t={t}_{0}}={{y}^{\prime }}_{0}
{t}_{n}
{\mathcal{E}}_{n}\left(t\right)={\left[\underset{i=2}{\overset{k}{\sum }}\text{ }\text{ }i\left(i-1\right){\psi }_{i}{t}^{i-2}-f\left(t,y\left(t\right)\right)\right]}_{t={t}_{n}}\simeq 0
\left\{{\psi }_{i}|i=0\left(1\right)k\right\}
\underset{n=1}{\overset{\frac{b-a}{h}}{\sum }}\text{ }{\mathcal{E}}_{n}^{2}\left(t\right)
\text{Minimize:}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\underset{n=1}{\overset{\frac{b-a}{h}}{\sum }}\text{ }{\mathcal{E}}_{n}^{2}\left(t\right)
\text{Subject}\text{\hspace{0.17em}}\text{to:}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\left[\underset{i=0}{\overset{k}{\sum }}\text{ }{\psi }_{i}{t}^{i}\right]}_{t={t}_{0}}={y}_{0}\text{ }\text{and}\text{ }{\left[\underset{i=1}{\overset{k}{\sum }}\text{ }\text{ }i{\psi }_{i}{t}^{i-1}\right]}_{t={t}_{0}}={{y}^{\prime }}_{0}
\left\{{\psi }_{i}|i=0,1,\cdots ,k\right\}
{\sum }_{n=1}^{\frac{b-a}{h}}{\mathcal{E}}_{n}^{2}\left(t\right)
{y}^{″}\left(t\right)-y\left(t\right)=t-1;\text{ }y\left(0\right)=2,\text{ }{y}^{\prime }\left(0\right)=-2
y\left(t\right)=1-t+\mathrm{exp}\left(-t\right)
k=10
\left\{{\psi }_{i}|i=0\left(1\right)10\right\}
\left\{2,-2,\frac{416995243}{834315644},-\frac{105164777}{636059757},\frac{69800031}{1811256752},-\frac{6535788}{1275358859}\right\}.
h=0.01
{y}^{″}\left(t\right)=\left(1+{t}^{2}\right)y\left(t\right);\text{ }y\left(0\right)=1,\text{ }{y}^{\prime }\left(0\right)=0
y\left(t\right)=\mathrm{exp}\left(\frac{{t}^{2}}{2}\right)
k=11
\left\{{\psi }_{i}|i=0\left(1\right)11\right\}
\begin{array}{l}\left\{1,0,\frac{1306409430}{2612828131},\frac{29397245}{1713857114397},\frac{3187969586}{25524507753},\frac{172091099}{436085951591},\\ \frac{313833621}{15857243966},\frac{382909153}{201794651238},\frac{117010789}{496614906383},\\ \frac{766119929}{389265107664},-\frac{130287162}{172534329575},\frac{125796527}{456658410146}\right\}\end{array}
[1] George, D.M. (2006) On the Application of Genetic Algorithms to Differential Equations. Romanian Journal of Economic Forecasting, 3, 5-9.
[2] Mastorakis, N.E. (2005) Numerical Solution of Non-Linear Ordinary Differential Equations via Collocation Method (Finite Elements) and Genetic Algorithms. Proceedings of the 6th WSEAS International Conference on Evolutionary Computing, Lisbon, 16-18 June 2005, 36-42.
[3] Mastorakis, N.E. (2006) Unstable Ordinary Differential Equations: Solution via Genetic Algorithms and the Method of Nelder-Mead. Proceedings of the 6th WSEAS International Conference on Systems Theory & Scientific Computation, Elounda, 21-23 August 2006, 1-6.
[4] Junaid, A., Raja, A.Z. and Qureshi, I.M. (2009) Evolutionary Computing Approach for the Solution of Initial Value Problems in Ordinary Differential Equations. World Academic of Science, Engineering and Technology, 55, 578-581.
[5] Omar, A.A., Zaer, A., Shaher, M. and Nabil, S. (2012) Solving Singular Two-Point Boundary Value Problems Using Continuous Genetic Algorithm. Abstract and Applied Analysis, 2012, Article ID: 205391.
[6] Bakre, O.F., Wusu, A.S. and Akanbi, M.A. (2015) Solving Ordinary Differential Equations with Evolutionary Algorithms. Open Journal of Optimization, 4, 69-73.
[7] Wusu, A.S. and Akanbi, M.A. (2016) Solving Oscillatory/Periodic Ordinary Differential Equations with Differential Evolution Algorithms. Communications in Optimization Theory, 2016, Article ID: 7. |
Yi-Chou Chen, Wei-Shih Du, "New Optimality Conditions for a Nondifferentiable Fractional Semipreinvex Programming Problem", Journal of Applied Mathematics, vol. 2013, Article ID 527183, 5 pages, 2013. https://doi.org/10.1155/2013/527183
Yi-Chou Chen1 and Wei-Shih Du2
1Department of General Education, National Army Academy, Taoyuan 320, Taiwan
We study a nondifferentiable fractional programming problem as follows: subject to , where is a semiconnected subset in a locally convex topological vector space , , and , . If , , and , , are arc-directionally differentiable, semipreinvex maps with respect to a continuous map satisfying and , then the necessary and sufficient conditions for optimality of are established.
In recent years, there has been an increasing interest in studying the develpoment of optimality conditions for nondifferentiable multiobjective programming problems. Many authors established and employed some different Kuhn and Tucker type necessary conditions or other type necessary conditions to research optimal solutions; see [1–27] and references therein. In [7], Lai and Ho used the Pareto optimality condition to investigate multiobjective programming problems for semipreinvex functions. Lai [6] had obtained the necessary and sufficient conditions for optimality programming problems with semipreinvex assumptions. Some Pareto optimality conditions are established by Lai and Lin in [8]. Lai and Szilágyi [9] studied the programming with convex set functions and proved that the alternative theorem is valid for convex set functions defined on convex subfamily of measurable subsets in and showed that if the system has on solution, where stands for zero vector in a topological vector space, then there exists a nonzero continuous linear function such that In this paper, we study the following optimization problem: where is a semiconnected subset in a locally convex topological vector space , , and , , are functions satisfying some suitable conditions. The purpose of this study is dealt with such constrained fractional semipreinvex programming problem. Finally, we established the Fritz John type necessary and sufficient conditions for the optimality of a fractional semipreinvex programming problem.
Throughout this paper, we let be a locally convex topological vector space over the real field . Denote by the space of all linear operators from into .
Let be a nonempty convex subset of . Let be differentiable at . Then there is a linear operator , such that Recall that a function is called convex on , if or If is convex and differentiable at , then by (3) and (5), we have In 1981, Hanson [13, 14] introduced a generalized convexity on , so-called invexity; that is, is replaced by a vector in (6), or So an invex function is indeed a generalization of a convex differentiable function.
Definition 1 (see [6]). (1) A set is said to be semiconnected with respect to a given if
(2) A map is said to be semipreinvex on a semiconnected subset if each corresponds a vector such that where stands for the zero vector of .
The following is an example of a bounded semiconnected set in , which is semiconnected with respect to a nontrivial .
Example 2. Let , and be bounded sets. Let be defined by Then is a bound semiconnected set with respect to .
Theorem 3 (see [6, Theorem 2.2]). Let be a semiconnected subset and a semipreinvex map. Then any local minimum of is also a global minimum of over .
From the assumption in problem (9), there exists a positive number such that Consequently, we can reduce the problem (9) to an equivalent nonfractional parametric problem: where is a parameter.
We will prove that the problem is equivalent to the problem ( ) for the optimal value . The following result is our main technique to derive the necessary and sufficient optimality conditions for problem .
Theorem 4. Problem has an optimal solution with optimal value if and only if and is an optimal solution of .
Proof. If is an optimal solution of with optimal value , that is, It follows from (12) that Thus, we have Then, by (14), we get Therefore, is an optimal solution of ( ) and .
Conversely, if is an optimal solution of ( ) with optimal value , then So It follows from (17) that and hence Therefore, and we know is an optimal solution of with optimal value .
3. The Existence of the Necessary and Sufficient Conditions for Semipreinvex Functions
Definition 5 (see [6]). A mapping is said to be arcwise directionally (in short, arc-directionally) differentiable at with respect to a continuous arc if for with that is, the continuous function is differentiable from right at , and the limit
Note that the arc directional derivative is a mapping from into . Moreover, how can we make to be a semiconnected set? Indeed, we can construct a function concerned with defined as follows.
For any and , we choose a vector then Let , and , , be semipreinvex maps on a semiconnected subset in . Consider a constrained programming problem as .
The following Fritz John type theorem is essential in this section for programming problem .
Theorem 6 (Necessary Optimality Condition). Suppose that , and , are arc-directionally differentiable at and semipreinvex on with respect to a continuous arc defined as in Definition 5. If minimizes locally for the semipreinvex programming problem , then there exist and such that where and
Proof. By Theorem 4, the minimum solution to is also a minimum to ( ). Then is the local minimal solution to ( ). By Theorem 3, we have is the global minimal solution to ( ). It follows that the system has no solution in , then we have has no solution in for any . Thus for any , for some . Putting in (29), we get Since and , it follows that So (26) is proved.
As is a semiconnected set, for any and , we have For , the point does not solve the system (27). So substituting in (29) and using the result (26), we obtain Since and are arc-directionally differentiable with respect to , choose a vector as (23), so that (24) hold. It follows that if we divide (33) by and take the limit as , then we have which proves (25) and the proof of theorem is completed.
Theorem 7 (Sufficient Optimality Condition). Let , and , be arc-directionally differentiable at and semipreinvex on with respect to a continuous arc defined as in Definition 5. If there exist and satisfying with and then is an optimal solution for problem .
Proof. Suppose to the contrary that is not optimal for problem and . Then . Therefore, thus .
By Theorem 4, was not optimal for problem . Then there is an such that for . Moreover, we have for any . Thus Since the semi-preinvex maps , and , are arc-directionally differentiable, it follows that for there corresponds a vector such that and so Letting , we have and the last inequalities imply Consequently, from (41) and (44), we obtain which contradicts the fact of (35). Therefore is an optimal solution of problem .
Since any global minimal is a local minimal, applying Theorems 6 and 7, we can obtain the necessary and sufficient conditions for problem .
Theorem 8. Suppose that , and , are arc-directionally differentiable at at and semi-preinvex on with respect to a continuous arc defined as in Definition 5. If minimizes globally for the semi-preinvex programming problem if and only if there exists , , such that where and
Remark 9. Our results also hold for preinvex functions.
The research of Wei-Shih Du was supported partially under Grant no. NSC 101-2115-M-017-001 by the National Science Council of the Republic of China.
B. D. Craven, “Invex functions and constrained local minima,” Bulletin of the Australian Mathematical Society, vol. 24, no. 3, pp. 357–366, 1981. View at: Publisher Site | Google Scholar | Zentralblatt MATH | MathSciNet
B. D. Craven and B. M. Glover, “Invex functions and duality,” Australian Mathematical Society, vol. 39, no. 1, pp. 1–20, 1985. View at: Publisher Site | Google Scholar | Zentralblatt MATH | MathSciNet
F. H. Clarke, Optimization and Nonsmooth Analysis, Canadian Mathematical Society Series of Monographs and Advanced Texts, John Wiley & Sons, New York, NY, USA, 1983. View at: MathSciNet
F. H. Clarke, Optimization and Nonsmooth Analysis, John Wiley & Sons, New York, NY, USA, 2nd edition, 1983.
H. Dietrich, “On the convexification procedure for nonconvex and nonsmooth infinite-dimensional optimization problems,” Journal of Mathematical Analysis and Applications, vol. 161, no. 1, pp. 28–34, 1991. View at: Publisher Site | Google Scholar | Zentralblatt MATH | MathSciNet
H.-C. Lai, “Optimality conditions for semi-preinvex programming,” Taiwanese Journal of Mathematics, vol. 1, no. 4, pp. 389–404, 1997. View at: Google Scholar | Zentralblatt MATH | MathSciNet
H. C. Lai and C. P. Ho, “Duality theorem of nondifferentiable convex multiobjective programming,” Journal of Optimization Theory and Applications, vol. 50, no. 3, pp. 407–420, 1986. View at: Publisher Site | Google Scholar | Zentralblatt MATH | MathSciNet
H.-C. Lai and L.-J. Lin, “Moreau-Rockafellar type theorem for convex set functions,” Journal of Mathematical Analysis and Applications, vol. 132, no. 2, pp. 558–571, 1988. View at: Publisher Site | Google Scholar | Zentralblatt MATH | MathSciNet
H. C. Lai and P. Szilágyi, “Alternative theorems and saddlepoint results for convex programming problems of set functions with values in ordered vector spaces,” Acta Mathematica Hungarica, vol. 63, no. 3, pp. 231–241, 1994. View at: Publisher Site | Google Scholar | Zentralblatt MATH | MathSciNet
H. C. Lai and J. C. Liu, “Minimax fractional programming involving generalised invex functions,” The ANZIAM Journal, vol. 44, no. 3, pp. 339–354, 2003. View at: Publisher Site | Google Scholar | Zentralblatt MATH | MathSciNet
J.-C. Chen and H.-C. Lai, “Fractional programming for variational problem with
F,\rho ,\theta
-invexity,” Journal of Nonlinear and Convex Analysis, vol. 4, no. 1, pp. 25–41, 2003. View at: Google Scholar | MathSciNet
J. C. Liu, “Optimality and duality for generalized fractional programming involving nonsmooth (
F,\rho
)-convex functions,” Computers and Mathematics with Applications, vol. 42, pp. 437–446, 1990. View at: Google Scholar
M. A. Hanson, “On sufficiency of the Kuhn-Tucker conditions,” Journal of Mathematical Analysis and Applications, vol. 80, no. 2, pp. 545–550, 1981. View at: Publisher Site | Google Scholar | Zentralblatt MATH | MathSciNet
M. A. Hanson and B. Mond, “Necessary and sufficient conditions in constrained optimization,” Mathematical Programming, vol. 37, no. 1, pp. 51–58, 1987. View at: Publisher Site | Google Scholar | Zentralblatt MATH | MathSciNet
S. K. Mishra and G. Giorgi, Invexity and Optimization, vol. 88 of Nonconvex Optimization and Its Applications, Springer, Berlin, Germnay, 2008. View at: Publisher Site | MathSciNet
S. K. Mishra, S. Wang, and K. K. Lai, V-Invex Functions and Vector Optimization, vol. 14 of Springer Optimization and Its Applications, Springer, New York, NY, USA, 2008. View at: MathSciNet
S. K. Mishra, S.-Y. Wang, and K. K. Lai, Generalized Convexity and Vector Optimization, vol. 90 of Nonconvex Optimization and Its Applications, Springer, Berlin, Germnay, 2009. View at: MathSciNet
S. K. Mishra and V. Laha, “On approximately star-shaped functions and approximate vector variational inequalities,” Journal of Optimization Theory and Applications. In press. View at: Publisher Site | Google Scholar
S. K. Mishra and V. Laha, “On V-r-invexity and vector variational Inequalities,” Filomat, vol. 26, no. 5, pp. 1065–1073, 2012. View at: Google Scholar
S. K. Mishra and B. B. Upadhyay, “Nonsmooth minimax fractional programming involving
\eta
-pseudolinear functions,” Optimization. In press. View at: Publisher Site | Google Scholar
S. K. Mishra and M. Jaiswal, “Optimality conditions and duality for nondifferentiable multiobjective semi-infinite programming,” Vietnam Journal of Mathematics, vol. 40, pp. 331–343, 2012. View at: Google Scholar
N. G. Rueda and M. A. Hanson, “Optimality criteria in mathematical programming involving generalized invexity,” Journal of Mathematical Analysis and Applications, vol. 130, no. 2, pp. 375–385, 1988. View at: Publisher Site | Google Scholar | Zentralblatt MATH | MathSciNet
T. W. Reiland, “Nonsmooth invexity,” Bulletin of the Australian Mathematical Society, vol. 12, pp. 437–446, 1990. View at: Google Scholar
T. Weir, “Programming with semilocally convex functions,” Journal of Mathematical Analysis and Applications, vol. 168, no. 1, pp. 1–12, 1992. View at: Publisher Site | Google Scholar | Zentralblatt MATH | MathSciNet
T. Weir and V. Jeyakumar, “A class of nonconvex functions and mathematical programming,” Bulletin of the Australian Mathematical Society, vol. 38, no. 2, pp. 177–189, 1988. View at: Publisher Site | Google Scholar | Zentralblatt MATH | MathSciNet
T. Antczak, “(p, r)-invex sets and functions,” Journal of Mathematical Analysis and Applications, vol. 263, no. 2, pp. 355–379, 2001. View at: Publisher Site | Google Scholar | Zentralblatt MATH | MathSciNet
Y. L. Ye, “D-invexity and optimality conditions,” Journal of Mathematical Analysis and Applications, vol. 162, no. 1, pp. 242–249, 1991. View at: Publisher Site | Google Scholar | Zentralblatt MATH | MathSciNet
Copyright © 2013 Yi-Chou Chen and Wei-Shih Du. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. |
Trigonometric Equations - Triple Angle Formula | Brilliant Math & Science Wiki
The trigonometric triple-angle identities give a relationship between the basic trigonometric functions applied to three times an angle in terms of trigonometric functions of the angle itself.
\begin{aligned} \sin 3 \theta &= 3 \sin \theta - 4 \sin ^3 \theta \\\\ \cos 3\theta &= 4 \cos ^ 3 \theta - 3 \cos \theta \end{aligned}
Proof: To prove the triple-angle identities, we can write
\sin 3 \theta
\sin(2 \theta + \theta)
. Then we can use the sum formula and the double-angle identities to get the desired form:
\begin{aligned} \sin 3 \theta & = \sin (2 \theta + \theta)\\ & = \sin 2 \theta \cos \theta + \cos 2 \theta \sin \theta \\ & = (2 \sin \theta \cos \theta) \cos \theta + \big(1 - 2 \sin^2 \theta \big) \sin \theta \\ & = 2 \sin \theta \cos^2 \theta + \sin \theta - 2 \sin^3 \theta \\ & = 2 \sin \theta \big(1 - \sin^2 \theta\big) + \sin \theta - 2 \sin^3 \theta \\ & = 2 \sin \theta - 2 \sin^3 \theta + \sin \theta - 2 \sin^3 \theta \\ & = 3 \sin \theta - 4 \sin^3 \theta.\ _\square \end{aligned}
The triple-angle identity of
\cos 3 \theta
can be proved in a very similar manner.
From these formulas, we also have the following identities for
\sin^3 \theta
\cos^3 \theta
in terms of lower powers:
\begin{aligned} \sin^3 \theta &= \frac{3 \sin \theta - \sin 3\theta }{4}\\\\ \cos^3 \theta & = \frac{\cos 3\theta + 3 \cos \theta}{4}. \end{aligned}
Cite as: Trigonometric Equations - Triple Angle Formula. Brilliant.org. Retrieved from https://brilliant.org/wiki/trigonometric-equations-triple-angle-formula/ |
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