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Passivity Indices - MATLAB & Simulink - MathWorks Switzerland
Frequency-Domain Characterization
This example shows how to compute various measures of passivity for linear time-invariant systems.
A linear system G(s) is passive when all I/O trajectories
\left(u\left(t\right),y\left(t\right)\right)
{\int }_{0}^{T}{y}^{T}\left(t\right)u\left(t\right)dt>0,\phantom{\rule{1em}{0ex}}\forall T>0
{y}^{T}\left(t\right)
y\left(t\right)
To measure "how passive" a system is, we use passivity indices.
The input passivity index is defined as the largest
\nu
{\int }_{0}^{T}{y}^{T}\left(t\right)u\left(t\right)dt>\nu {\int }_{0}^{T}{u}^{T}\left(t\right)u\left(t\right)dt
The system G is "input strictly passive" (ISP) when
\nu >0
\nu
is also called the "input feedforward passivity" (IFP) index and corresponds to the minimum feedforward action needed to make the system passive.
The output passivity index is defined as the largest
\rho
{\int }_{0}^{T}{y}^{T}\left(t\right)u\left(t\right)dt>\rho {\int }_{0}^{T}{y}^{T}\left(t\right)y\left(t\right)dt
The system G is "output strictly passive" (OSP) when
\rho >0
\rho
is also called the "output feedback passivity" (OFP) index and corresponds to the minimum feedback action needed to make the system passive.
The I/O passivity index is defined as the largest
\tau
{\int }_{0}^{T}{y}^{T}\left(t\right)u\left(t\right)dt>\tau {\int }_{0}^{T}\left({u}^{T}\left(t\right)u\left(t\right)+{y}^{T}\left(t\right)y\left(t\right)\right)dt
The system is "very strictly passive" (VSP) if
\tau >0
Consider the following example. We take the current
I
as the input and the voltage
V
as the output. Based on Kirchhoff's current and voltage law, we obtain the transfer function for
G\left(s\right)
G\left(s\right)=\frac{V\left(s\right)}{I\left(s\right)}=\frac{\left(Ls+R\right)\left(Rs+\frac{1}{C}\right)}{L{s}^{2}+2Rs+\frac{1}{C}}.
R=2
L=1
C=0.1
R = 2; L = 1; C = 0.1;
G = (L*s+R)*(R*s+1/C)/(L*s^2 + 2*R*s+1/C);
Use isPassive to check whether
G\left(s\right)
is passive.
Since PF = true,
G\left(s\right)
is passive. Use getPassiveIndex to compute the passivity indices of
G\left(s\right)
% Input passivity index
nu = getPassiveIndex(G,'in')
% Output passivity index
rho = getPassiveIndex(G,'out')
% I/O passivity index
\tau >0
G\left(s\right)
is very strictly passive.
A linear system is passive if and only if it is "positive real":
G\left(j\omega \right)+{G}^{H}\left(j\omega \right)>0\phantom{\rule{1em}{0ex}}\forall \omega \in R.
The smallest eigenvalue of the left-hand-side is related to the input passivity index
\nu
\nu =\frac{1}{2}\underset{\omega }{\mathrm{min}}{\lambda }_{\mathrm{min}}\left(G\left(j\omega \right)+{G}^{H}\left(j\omega \right)\right)
{\lambda }_{\mathrm{min}}
denotes the smallest eigenvalue. Similarly, when
G\left(s\right)
is minimum-phase, the output passivity index is given by:
\rho =\frac{1}{2}\underset{\omega }{\mathrm{min}}{\lambda }_{\mathrm{min}}\left({G}^{-1}\left(j\omega \right)+{G}^{-H}\left(j\omega \right)\right).
Verify this for the circuit example. Plot the Nyquist plot of the circuit transfer function.
The entire Nyquist plot lies in the right-half plane so
G\left(s\right)
is positive real. The leftmost point on the Nyquist curve is
\left(x,y\right)=\left(2,0\right)
so the input passivity index is
\nu =2
, the same value we obtained earlier. Similarly, the leftmost point on the Nyquist curve for
{G}^{-1}\left(s\right)
gives the output passivity index value
\rho =0.286
It can be shown that the "positive real" condition
G\left(j\omega \right)+{G}^{H}\left(j\omega \right)>0\phantom{\rule{1em}{0ex}}\forall \omega \in R
is equivalent to the small gain condition
||\left(I-G\left(j\omega \right)\right)\left(I+G\left(j\omega \right){\right)}^{-1}||<1\phantom{\rule{1em}{0ex}}\forall \omega \in R.
The relative passivity index (R-index) is the peak gain over frequency of
\left(I-G\right)\left(I+G{\right)}^{-1}
I+G
is minimum phase, and
+\infty
R={‖\left(I-G\right)\left(I+G{\right)}^{-1}‖}_{\infty }.
In the time domain, the R-index is the smallest
r>0
{\int }_{0}^{T}||y-u|{|}^{2}dt<{r}^{2}{\int }_{0}^{T}||y+u|{|}^{2}dt
G\left(s\right)
R<1
, and the smaller
R
is, the more passive the system is. Use getPassiveIndex to compute the R-index for the circuit example.
The resulting
R
value indicates that the circuit is a very passive system.
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ADALINE - Wikipedia
This article is about the neural network. For other uses, see Adaline.
Early single-layer artificial neural network
Learning inside a single layer ADALINE
ADALINE (Adaptive Linear Neuron or later Adaptive Linear Element) is an early single-layer artificial neural network and the name of the physical device that implemented this network.[1][2][3][4][5] The network uses memistors. It was developed by Professor Bernard Widrow and his doctorate student Ted Hoff at Stanford University in 1960. It is based on the McCulloch–Pitts neuron. It consists of a weight, a bias and a summation function.
The difference between Adaline and the standard (McCulloch–Pitts) perceptron is that in the learning phase, the weights are adjusted according to the weighted sum of the inputs (the net). In the standard perceptron, the net is passed to the activation (transfer) function and the function's output is used for adjusting the weights.
A multilayer network of ADALINE units is known as a MADALINE.
3 MADALINE
Adaline is a single layer neural network with multiple nodes where each node accepts multiple inputs and generates one output. Given the following variables as:
{\displaystyle x}
is the input vector
{\displaystyle w}
is the weight vector
{\displaystyle n}
is the number of inputs
{\displaystyle \theta }
some constant
{\displaystyle y}
is the output of the model
then we find that the output is
{\displaystyle y=\sum _{j=1}^{n}x_{j}w_{j}+\theta }
. If we further assume that
{\displaystyle x_{0}=1}
{\displaystyle w_{0}=\theta }
then the output further reduces to:
{\displaystyle y=\sum _{j=0}^{n}x_{j}w_{j}}
{\displaystyle \eta }
is the learning rate (some positive constant)
{\displaystyle y}
{\displaystyle o}
is the target (desired) output
then the weights are updated as follows
{\displaystyle w\leftarrow w+\eta (o-y)x}
. The ADALINE converges to the least squares error which is
{\displaystyle E=(o-y)^{2}}
.[6] This update rule is in fact the stochastic gradient descent update for linear regression.[7]
MADALINE[edit]
MADALINE (Many ADALINE[8]) is a three-layer (input, hidden, output), fully connected, feed-forward artificial neural network architecture for classification that uses ADALINE units in its hidden and output layers, i.e. its activation function is the sign function.[9] The three-layer network uses memistors. Three different training algorithms for MADALINE networks, which cannot be learned using backpropagation because the sign function is not differentiable, have been suggested, called Rule I, Rule II and Rule III.
MADALINE Rule 1 (MRI) - The first of these dates back to 1962 and cannot adapt the weights of the hidden-output connection.[10]
MADALINE Rule 2 (MRII) - The second training algorithm improved on Rule I and was described in 1988.[8] The Rule II training algorithm is based on a principle called "minimal disturbance". It proceeds by looping over training examples, then for each example, it:
finds the hidden layer unit (ADALINE classifier) with the lowest confidence in its prediction,
tentatively flips the sign of the unit,
accepts or rejects the change based on whether the network's error is reduced,
stops when the error is zero.
MADALINE Rule 3 - The third "Rule" applied to a modified network with sigmoid activations instead of signum; it was later found to be equivalent to backpropagation.[10]
Additionally, when flipping single units' signs does not drive the error to zero for a particular example, the training algorithm starts flipping pairs of units' signs, then triples of units, etc.[8]
^ Anderson, James A.; Rosenfeld, Edward (2000). Talking Nets: An Oral History of Neural Networks. ISBN 9780262511117.
^ Youtube: widrowlms: Science in Action
^ 1960: An adaptive "ADALINE" neuron using chemical "memistors"
^ Youtube: widrowlms: The LMS algorithm and ADALINE. Part I - The LMS algorithm
^ Youtube: widrowlms: The LMS algorithm and ADALINE. Part II - ADALINE and memistor ADALINE
^ "Adaline (Adaptive Linear)" (PDF). CS 4793: Introduction to Artificial Neural Networks. Department of Computer Science, University of Texas at San Antonio.
^ Avi Pfeffer. "CS181 Lecture 5 — Perceptrons" (PDF). Harvard University. [permanent dead link]
^ a b c Rodney Winter; Bernard Widrow (1988). MADALINE RULE II: A training algorithm for neural networks (PDF). IEEE International Conference on Neural Networks. pp. 401–408. doi:10.1109/ICNN.1988.23872.
^ Youtube: widrowlms: Science in Action (Madaline is mentioned at the start and at 8:46)
^ a b Widrow, Bernard; Lehr, Michael A. (1990). "30 years of adaptive neural networks: perceptron, madaline, and backpropagation". Proceedings of the IEEE. 78 (9): 1415–1442. doi:10.1109/5.58323.
"Delta Learning Rule: ADALINE". Artificial Neural Networks. Universidad Politécnica de Madrid. Archived from the original on 2002-06-15.
"Memristor-Based Multilayer Neural Networks With Online Gradient Descent Training". Implementation of the ADALINE algorithm with memristors in analog computing.
Retrieved from "https://en.wikipedia.org/w/index.php?title=ADALINE&oldid=1051089253"
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Beta_Cephei_variable Knowpia
Beta Cephei variables, also known as Beta Canis Majoris stars, are variable stars that exhibit small rapid variations in their brightness due to pulsations of the stars' surfaces, thought due to the unusual properties of iron at temperatures of 200,000 K in their interiors. These stars are usually hot blue-white stars of spectral class B and should not be confused with Cepheid variables, which are named after Delta Cephei and are luminous supergiant stars.
Beta Cephei variables are main-sequence stars of masses between about 7 and 20 M
{\displaystyle _{\odot }}
(that is, 7–20 times as massive as the Sun). Among their number are some of the brightest stars in the sky, such as Beta Crucis and Beta Centauri; Spica is also classified as a Beta Cephei variable but mysteriously stopped pulsating in 1970.[1] Typically, they change in brightness by 0.01 to 0.3 magnitudes with periods of 0.1 to 0.3 days (2.4–7.2 hours).[1] The prototype of these variable stars, Beta Cephei, shows variation in apparent magnitude from +3.16 to +3.27 with a period of 4.57 hours. The point of maximum brightness occurs when the star is smallest and hottest. Their variation in brightness is much greater—up to 1 magnitude—in ultraviolet wavelengths.[2] A small number of stars have been identified with periods shorter than one hour, corresponding to 1/4 of the fundamental radial pulsation period and 3/8 of the fundamental period. They also have relatively low amplitudes and a very narrow range of spectral types B2-3 IV-V. They are known as the short period group and the GCVS acronym BCEPS.[3][4]
The pulsations of Beta Cephei variables are driven by the kappa mechanism and p-mode pulsations. At a depth within the star where the temperature reaches 200,000 K, there is an abundance of iron. Iron at these temperatures will increase (rather than decrease) in opacity, resulting in the buildup of energy within the layer. This results in increased pressure that pushes the layer back out again, the cycle repeating itself in a matter of hours. This is known as the Fe bump or Z bump (Z standing for the star's metallicity).[5] The similar slowly pulsating B stars show g-mode pulsations driven by the same iron opacity changes, but in less massive stars and with longer periods.[6]
History of observationsEdit
American astronomer Edwin Brant Frost discovered the variation in radial velocity of Beta Cephei in 1902, initially concluding it was a spectroscopic binary. Paul Guthnick was the first to detect a variation in brightness, in 1913.[7] Beta Canis Majoris and Sigma Scorpii were found to be variable not long afterwards,[2] Vesto Slipher noted in 1904 that Sigma Scorpii's radial velocity was variable, and R.D. Levee and Otto Struve concluded this was due to the star's pulsations in 1952 and 1955 respectively.[8] These variables were often called Beta Canis Majoris variables because Beta Canis Majoris was the most closely studied example in the first half of the 20th century, though its location in the southern sky meant that its lowness in the sky hampered observations.[9] However, Beta Cephei was the first member of the class to be discovered and so they are generally called Beta Cephei variables—despite the similarity of name (and risk of confusion) with Cepheid variables.[2]
Cecilia Payne-Gaposchkin and Sergei Gaposchkin catalogued 17 probable members of the class in their 1938 Variable Stars, though classified them with Delta Scuti variables.[10] 16 Lacertae was another star extensively studied before 1952.[9] The number known jumped from 18 to 41 in 1966.[11] Otto Struve studied these stars extensively in the 1950s, however research declined after his death.[2]
Christiaan L. Sterken and Mikolaj Jerzykiewicz classed 59 stars as definite and 79 more as suspected Beta Cephei variables in 1993.[12] Stankov listed 93 members of the class in a 2005 catalogue, plus 77 candidates and 61 poor or rejected stars.[13] Six stars, namely Iota Herculis, 53 Piscium, Nu Eridani, Gamma Pegasi, HD 13745 (V354 Persei) and 53 Arietis had been found to exhibit both Beta Cephei and SPB variability.[14]
In 2021 β Cru became the first star of any kind to have its pulsation modes identified using polarimetric asteroseismology.[15]
List of Beta Cephei variablesEdit
Maximum Apparent magnitude (mV)[16]
Minimum Apparent magnitude (mV)[16]
Period (hours)[13]
Spectral class[13]
β CMa Canis Major 1909 (William Wallace Campbell[17]) 1m.93 2m.00 6.031 B1II-III Pulsations of 6.03, 6.00, and 4.74 hours.[18]
ξ1 CMa Canis Major [19] 4m.33 4m.36 5.030 B0.5IV
15 CMa Canis Major [19] 4m.79 4m.84 4.429 B1III-IV
V376 Car[20] Carina 4m.91 4m.96 0.4992 B2IV-V BCEPS star
V372 Car Carina [21] 5m.70 2.78 B2III
β Cen Centaurus 0m.61 3.768[12] B1II
ε Cen Centaurus 2m.29 2m.31 4.070 B1V
κ Cen Centaurus 3m.13 3m.14 2.288 B2IV
χ Cen[20] Centaurus 4m.40[13] 0.84 B2V BCEPS star
β Cep Cepheus 1902 (Edwin Brant Frost)[22] 3m.16 3m.27 4.572 B2IIIe Prototype
δ Cet Cetus [19] 4m.05 4m.1 3.867 B2IV
β Cru Crux 1m.23 1m.31 4.589 B0.5IV
δ Cru[16][23] Crux 2m.78 2m.84 3.625 B2IV
ω1 Cyg Cygnus 4m.94 B2.5IV confirmed on hi res spectroscopy.[23]
ν Eri Eridanus 3m.87 4m.01 4.164 B2III Multiperiodic; also a slowly pulsating B star
12 Lac Lacerta 5m.16 5m.28 4.634 B1.5III Also a slowly pulsating B star
16 Lac Lacerta 5m.30 (B) 5m.52 (B) 4.109 B2IV
α Lup Lupus 1956 (Bernard Pagel)[24] 2m.29 2m.34 6.235 B1.5III
δ Lup[13] Lupus 3m.20 3m.24 3.972 B2IV
ε Lup[25] Lupus 3m.36 3m.38 2.316 B2IV + B3V Triple star system; primary is a spectroscopic binary
ι Lup[26] Lupus 3m.54 3m.3.55 B2.5IV not recorded as BCEP since 1997
τ1 Lup[13] Lupus 4m.54 4m.58 4.257 B2IV
19 Mon Monoceros 4m.96 5m.01 4.589 B1IV-Vea
α Mus[16] Musca 2m.68 2m.73 2.167 B2IV-V initially questionable, confirmed on hi res spectroscopy.[23]
θ Oph Ophiuchus 3m.25 3m.31 3.373 B2IV
η Ori Orion 3m.31 3m.35 7.247 B0.5Vea + B3V Quadruple star; also an Algol variable; component Ab is the pulsating star
γ Peg Pegasus 1953 (D. Harold McNamara) 2m.78 2m.89 3.643 B2IV Also a slowly pulsating B star
ε Per Perseus 2m.88 3m.00 3.847 B0.5V
PT Pup Puppis [13] 5m.72 5m.74 3.908 B2III
λ Sco Scorpius 1m.59 1m.65 5.129 B1.5IV + PMS + B2IV Triple system; also an Algol variable
κ Sco Scorpius 2m.41 2m.42 4.795 B1.5III
σ Sco Scorpius 1904 (Vesto Slipher) 2m.86 2m.94 5.923 B1III Quadruple system
Spica Virgo 0m.85 1m.05 6.520 B1IV Brightness variations stopped in 1970[27]
BW Vul Vulpecula 6m.44 6m.68 4.8 B2IIIv Beta Cephei variable with largest change in radial velocity
List of former, excluded or candidate Beta Cephei variablesEdit
ι CMa Canis Major 4m.36 4m.40 33.6[16] B3Ib/II Not considered a β Cep variable[13][28]
FN CMa[29] Canis Major 5m.38 5m.42 36.7[30] B0.5IV No longer considered a β Cep variable[13]
χ Car[31] Carina 3m.46 2.42 B2IV Not considered a β Cep variable[13]
V343 Car Carina 4m.30[13] 57.11 B1.5III Not considered a β Cep variable[13]
ζ Cha[26] Chamaeleon 5m.06 5m.17 25.91[26] B5V considered as a SBP as of 2011[21]
λ Cru Crux 4m.60 4m.64 9.482[16] B4Vne Not considered a β Cep variable[13]
θ2 Cru Crux 4m.70 4m.74 2.134[16] B2IV Not considered a β Cep variable[13]
25 Cyg Cygnus 5m.09[32] 5m.21[32] 5.04[33] B3IVe γ Cas variable, not considered a β Cep variable[13]
ι Her Hercules 2m.93 B3IV No longer classed as Beta Cephei type[13]
η Hya Hydra 4m.27 4m.33 ~4[31] B3V No longer classed as Beta Cephei type[13]
NW Pup Puppis 5m.04 5m.18 3.00 B3Vea Also a rotating ellipsoidal variable, not considered a β Cep variable[13]
α Pyx[19] Pyxis 3m.67 3m.70 B1.5III Candidate β Cephei variable
Merope Taurus 4m.17 4m.19 B6IVe B(e) star, not Beta Cephei type[13]
IS Vel[16] Vela 5m.23 2.592 B1IVn Candidate β Cephei variable[13]
(HW Vel)[16] Vela 5m.46 5m.52 6.275 B6V Candidate β Cephei variable[13]
2 Vul Vulpecula 5m.36 5m.48 14.63 O8IV-B0.5IVeV B(e) star, not Beta Cephei type[13]
^ a b BSJ (16 July 2010). "The Beta Cephei Stars and Their Relatives". Variable Star of the Season. American Association of Variable Star Observers. Retrieved 2 August 2015.
^ a b c d Lesh, Janet Roundtree; Aizenman, Morris L. (1978). "The Observational Status of the β Cephei Stars". Annual Review of Astronomy and Astrophysics. 16: 215–240. Bibcode:1978ARA&A..16..215L. doi:10.1146/annurev.aa.16.090178.001243.
^ Bezdenezhnyi, V. P. (2001). "On the Periods of the β Cephei Stars". Odessa Astronomical Publications. 14: 118. Bibcode:2001OAP....14..118B.
^ Good, Gerry A. (2003). "Pulsating Variable Stars". Observing Variable Stars. pp. 57–95. doi:10.1007/978-1-4471-0055-3_4. ISBN 978-1-85233-498-7.
^ LeBlanc, Francis (2010). An Introduction to Stellar Astrophysics. John Wiley and Sons. p. 196. ISBN 978-0-470-69957-7.
^ Miglio, A. (2007). "Revised instability domains of SPB and β Cephei stars". Communications in Asteroseismology. 151: 48–56. arXiv:0706.3632. Bibcode:2007CoAst.151...48M. doi:10.1553/cia151s48. ISSN 1021-2043. S2CID 16623201.
^ Guthnick, P. (1913). "Nachweis der Veränderlichkeit des kurzperiodischen spektroskopischen Doppelsternsβ Cephei mittels photoelektrischer Messungen". Astronomische Nachrichten. 196 (26): 357–364. Bibcode:1913AN....196..357G. doi:10.1002/asna.19131962602. ISSN 0004-6337.
^ Tkachenko, A.; Aerts, C.; Pavlovski, K.; Degroote, P.; Papics, P. I.; Moravveji, E.; Lehmann, H.; Kolbas, V.; Clemer, K. (2014). "Modelling of Scorpii, a high-mass binary with a Cep variable primary component". Monthly Notices of the Royal Astronomical Society. 442 (1): 616–628. arXiv:1405.0924. Bibcode:2014MNRAS.442..616T. doi:10.1093/mnras/stu885.
^ a b Struve, Otto (1952). "The present state of our knowledge of the β Canis Majoris or β Cephei Stars". Annales d'Astrophysique. 15: 157. Bibcode:1952AnAp...15..157S.
^ Payne-Gaposchkin, Cecilia; Gaposchkin, Sergei (1938). "Variable stars". Cambridge. 5. Bibcode:1938HarMo...5.....P. Reprinted in Payne-Gaposchkin, Cecilia; Gaposchkin, Sergei (1942). "On the Dimensions and Constitution of Variable Stars". Proc. Natl. Acad. Sci. U.S.A. 28 (11): 490–5. Bibcode:1942PNAS...28..490P. doi:10.1073/pnas.28.11.490. PMC 1078523. PMID 16578062.
^ Percy, John R. (1967). "The Beta Cephei Stars". Journal of the Royal Astronomical Society of Canada. 61: 117. Bibcode:1967JRASC..61..117P.
^ a b Sterken, Christiaan; Jerzykiewicz, Mikolaj (1993). "Beta Cephei stars from a photometric point of view". Space Science Reviews. 62 (1–2): 95–171. Bibcode:1993SSRv...62...95S. doi:10.1007/bf00208707. ISSN 0038-6308. S2CID 122300362.
^ a b c d e f g h i j k l m n o p q r s t u v w x Stankov, Anamarija; Handler, Gerald (2005). "Catalog of Galactic β Cephei Stars". The Astrophysical Journal Supplement Series. 158 (2): 193–216. arXiv:astro-ph/0506495. Bibcode:2005ApJS..158..193S. doi:10.1086/429408. ISSN 0067-0049. S2CID 119526948.
^ de Cat, P. (2007). "Observational Asteroseismology of slowly pulsating B stars". Communications in Asteroseismology. 150: 167–74. Bibcode:2007CoAst.150..167D. doi:10.1553/cia150s167.
^ Cotton, D. V.; Buzasi, D. B; Aerts, C.; Bailey, J.; et al. (2021). "Polarimetric detection of non-radial oscillation modes in the β Cephei star β Crucis". Nature Astronomy. doi:10.1038/s41550-021-01531-9.
^ a b c d e f g h i j k Samus, N. N.; Durlevich, O. V.; et al. (2009). "VizieR Online Data Catalog: General Catalogue of Variable Stars (Samus+ 2007–2013)". VizieR On-line Data Catalog: B/GCVS. Originally Published in: 2009yCat....102025S. 1: 02025. Bibcode:2009yCat....102025S.
^ Campbell, W. W. (1909). "Eleven stars having variable radial velocities". The Astrophysical Journal. 29: 224. Bibcode:1909ApJ....29..224C. doi:10.1086/141644. ISSN 0004-637X.
^ Mazumdar, A.; et al. (November 2006), "An asteroseismic study of the β Cephei star β Canis Majoris", Astronomy and Astrophysics, 459 (2): 589–596, arXiv:astro-ph/0607261, Bibcode:2006A&A...459..589M, doi:10.1051/0004-6361:20064980, S2CID 11807580
^ a b c d Hubrig, S.; Briquet, M.; De Cat, P.; Schöller, M.; Morel, T.; Ilyin, I. (April 2009). "New magnetic field measurements of β Cephei stars and slowly pulsating B stars". Astronomische Nachrichten. 330 (4): 317. arXiv:0902.1314. Bibcode:2009AN....330..317H. doi:10.1002/asna.200811187.
^ a b Jakate, S. M. (1979). "A new class of early-type ultra-short-period variables". The Astronomical Journal. 84: 1042. Bibcode:1979AJ.....84.1042J. doi:10.1086/112510. ISSN 0004-6256.
^ a b Dubath, P.; Rimoldini, L.; Süveges, M.; Blomme, J.; López, M.; Sarro, L. M.; De Ridder, J.; Cuypers, J.; Guy, L.; Lecoeur, I.; Nienartowicz, K.; Jan, A.; Beck, M.; Mowlavi, N.; De Cat, P.; Lebzelter, T.; Eyer, L. (2011). "Random forest automated supervised classification of Hipparcos periodic variable stars". Monthly Notices of the Royal Astronomical Society. 414 (3): 2602–17. arXiv:1101.2406. Bibcode:2011MNRAS.414.2602D. doi:10.1111/j.1365-2966.2011.18575.x. S2CID 118560311.
^ Lesh, Janet Roundtree; Aizenman, Morris L. (1978). "The Observational Status of the β Cephei Stars". Annual Review of Astronomy and Astrophysics. 16 (1): 215–240. Bibcode:1978ARA&A..16..215L. doi:10.1146/annurev.aa.16.090178.001243. ISSN 0066-4146.
^ a b c Telting, J. H.; Schrijvers, C.; Ilyin, I. V.; Uytterhoeven, K.; De Ridder, J.; Aerts, C.; Henrichs, H. F. (2006). "A high-resolution spectroscopy survey of β Cephei pulsations in bright stars". Astronomy and Astrophysics. 452 (3): 945–953. Bibcode:2006A&A...452..945T. doi:10.1051/0004-6361:20054730. ISSN 0004-6361.
^ Pagel, B. E. J. (1956), "Results of a search for bright β Cephei variables in the southern sky", Monthly Notices of the Royal Astronomical Society, 116: 10–24, Bibcode:1956MNRAS.116...10P, doi:10.1093/mnras/116.1.10
^ Uytterhoeven, K.; et al. (September 2005), "The orbit of the close spectroscopic binary epsilon Lup and the intrinsic variability of its early B-type components", Astronomy and Astrophysics, 440 (1): 249–260, arXiv:astro-ph/0507376, Bibcode:2005A&A...440..249U, doi:10.1051/0004-6361:20053009, S2CID 16644667
^ a b c ESA (1997). The HIPPARCOS and TYCHO catalogues. Astrometric and photometric star catalogues derived from the ESA HIPPARCOS Space Astrometry Mission. Esa Special Publication. Vol. 1200. Bibcode:1997ESASP1200.....E. ISBN 978-9290923992.
^ Palate, M.; Koenigsberger, G.; Rauw, G.; Harrington, D.; Moreno, E. (2013). "Spectral modelling of theαVirginis (Spica) binary system". Astronomy & Astrophysics. 556: A49. arXiv:1307.1970. Bibcode:2013A&A...556A..49P. doi:10.1051/0004-6361/201321909. ISSN 0004-6361. S2CID 53519054.
^ Balona, L. A.; Engelbrecht, C. A. (1985). "Photometry and frequency analysis of line profile variables". Monthly Notices of the Royal Astronomical Society. 214 (4): 559–574. Bibcode:1985MNRAS.214..559B. doi:10.1093/mnras/214.4.559.
^ Hill, Graham (1967). "On Beta Cephei Stars: a Search for Beta Cephei Stars". The Astrophysical Journal Supplement Series. 14: 263. Bibcode:1967ApJS...14..263H. doi:10.1086/190156. ISSN 0067-0049.
^ Lefèvre, L.; Marchenko, S. V.; Moffat, A. F. J.; Acker, A. (2009). "A systematic study of variability among OB-stars based on HIPPARCOS photometry". Astronomy and Astrophysics. 507 (2): 1141–1201. Bibcode:2009A&A...507.1141L. doi:10.1051/0004-6361/200912304. ISSN 0004-6361.
^ a b Elst, E. W. (1979). "Short Periodic Variation of One Northern and Five Southern Bright Early B-Type Stars". Information Bulletin on Variable Stars. 1562: 1. Bibcode:1979IBVS.1562....1E.
^ a b Percy, J. R.; Harlow, J.; Hayhoe, K. A. W.; Ivans, I. I.; Lister, M.; Plume, R.; Rosebery, T.; Thompson, S.; Yeung, D. (1997). "Photometric Monitoring of Bright Be Stars. III. 1988–89 and 1992–95". Publications of the Astronomical Society of the Pacific. 109: 1215. Bibcode:1997PASP..109.1215P. doi:10.1086/133998. ISSN 0004-6280.
^ Percy, J. R.; Jakate, S. M.; Matthews, J. M. (1981). "Short-period light variations in Be stars". The Astronomical Journal. 86: 53. Bibcode:1981AJ.....86...53P. doi:10.1086/112855. ISSN 0004-6256.
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Magnetic_pressure Knowpia
In physics, magnetic pressure is an energy density associated with a magnetic field. In SI units, the energy density
{\displaystyle P_{B}}
of a magnetic field with strength
{\displaystyle B}
{\displaystyle P_{B}={\frac {B^{2}}{2\mu _{0}}}}
{\displaystyle \mu _{0}}
Any magnetic field has an associated magnetic pressure contained by the boundary conditions on the field. It is identical to any other physical pressure except that it is carried by the magnetic field rather than (in the case of a gas) by the kinetic energy of gas molecules. A gradient in field strength causes a force due to the magnetic pressure gradient called the magnetic pressure force.
Mathematical statementEdit
In SI units, the magnetic pressure is the energy density
{\displaystyle P_{B}}
in a magnetic field of strength
{\displaystyle B}
{\displaystyle P_{B}={\frac {B^{2}}{2\mu _{0}}}}
{\displaystyle \mu _{0}}
Wire loopsEdit
The magnetic pressure force is readily observed in an unsupported loop of wire. If an electric current passes through the loop, the wire serves as an electromagnet, such that the magnetic field strength inside the loop is much greater than the field strength just outside the loop. This gradient in field strength gives rise to a magnetic pressure force that tends to stretch the wire uniformly outward. If enough current travels through the wire, the loop of wire will form a circle. At even higher currents, the magnetic pressure can create tensile stress that exceeds the tensile strength of the wire, causing it to fracture, or even explosively fragment. Thus, management of magnetic pressure is a significant challenge in the design of ultrastrong electromagnets.
The force (in cgs) F exerted on a coil by its own current is[1]
{\displaystyle \mathbf {F} ={\dfrac {I^{2}}{c^{2}R}}\left[\ln \left({\dfrac {8R}{a}}\right)-1+Y\right]}
where Y is the internal inductance of the coil, defined by the distribution of current. Y is 0 for high frequency currents carried mostly by the outer surface of the conductor, and 0.25 for DC currents distributed evenly throughout the conductor. See inductance for more information.
Interplay between magnetic pressure and ordinary gas pressure is important to magnetohydrodynamics and plasma physics. Magnetic pressure can also be used to propel projectiles; this is the operating principle of a railgun.
Force-free fieldsEdit
If any currents present are parallel to a magnetic field, the field lines follow shapes in which the magnetic pressure gradient is balanced by the magnetic tension force. Such a field configuration is called force-free because there is no Lorentz force (
{\displaystyle j\times B=0}
). The familiar potential magnetic field is a special case of a force-free field: potential field configurations occupy space that contains no electric current at all.
^ Garren 1994, p. 3425 harvnb error: no target: CITEREFGarren1994 (help)
Garren & Chen (1994). "Lorentz Self Forces on Curved Current Loops". Physics of Plasmas. 1 (10): 3425–3436. Bibcode:1994PhPl....1.3425G. doi:10.1063/1.870491.
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Extended Euclidean Algorithm | Brilliant Math & Science Wiki
Thaddeus Abiy, Alan Enrique Ontiveros Salazar, Anuj Shikarkhane, and
Vaibhav Nitnaware
Abdullah Mohammad Daihan
The Euclidean algorithm is arguably one of the oldest and most widely known algorithms. It is a method of computing the greatest common divisor (GCD) of two integers
and
b
. It allows computers to do a variety of simple number-theoretic tasks, and also serves as a foundation for more complicated algorithms in number theory.
Recursive Implementation of Euclid's Algorithm
Pseudo-code of the Algorithm
The Euclidean algorithm is basically a continual repetition of the division algorithm for integers. The point is to repeatedly divide the divisor by the remainder until the remainder is 0. The GCD is the last non-zero remainder in this algorithm. The example below demonstrates the algorithm to find the GCD of 102 and 38:
\begin{aligned} 102 &= 2 \times 38 + 26 \\ 38 & = 1 \times 26 + 12\\ 26 & = 2 \times 12 + 2 \\ 12 &= 6 \times 2 + 0. \end{aligned}
The GCD is 2 because it is the last non-zero remainder that appears before the algorithm terminates.
Use Euclid's algorithm to find the GCD of 42823 and 6409.
\begin{aligned} 42823 &= 6409 \times 6 + 4369 \\ 6409 &= 4369 \times 1 + 2040 \\ 4369 &= 2040 \times 2 + 289\\ 2040 &= 289 \times 7 + 17 \\ 289 &= 17 \times 17 + 0. \end{aligned}
The last non-zero remainder is 17, and thus the GCD is 17.
_\square
This algorithm can be beautifully implemented using recursion as shown below:
int greatestCommonDivisor(int m, int n)
return greatestCommonDivisor(n, m % n);
The extended Euclidean algorithm is an algorithm to compute integers
x
y
ax + by = \gcd(a,b)
and
b
The existence of such integers is guaranteed by Bézout's lemma.
The extended Euclidean algorithm can be viewed as the reciprocal of modular exponentiation.
By reversing the steps in the Euclidean algorithm, it is possible to find these integers
x
y
. The whole idea is to start with the GCD and recursively work our way backwards. This can be done by treating the numbers as variables until we end up with an expression that is a linear combination of our initial numbers. We shall do this with the example we used above.
We start with our GCD. We rewrite it in terms of the previous two terms:
2 = 26 - 2 \times 12 .
We replace for
12
by taking our previous line
(38 = 1 \times 26 + 12)
and writing it in terms of 12:
2 = 26 - 2 \times (38 - 1\times 26).
Collect like terms, the
26
's, and we have
2 = 3 \times 26 - 2 \times 38.
2 = 3 \times (102 - 2\times 38) - 2\times 38.
The final result is our answer:
2 = 3 \times 102 - 8 \times 38.
x
y
3
-8
Find two integers
and
b
1914a + 899b = \gcd(1914,899).
First use Euclid's algorithm to find the GCD:
\begin{aligned} 1914 &= 2\times 899 + 116 \\ 899 &= 7 \times 116 + 87 \\ 116 &= 1 \times 87 + 29 \\ 87 &= 3 \times 29 + 0. \end{aligned}
From this, the last non-zero remainder (GCD) is
29
. Now we use the extended algorithm:
\begin{aligned} 29 &= 116 + (-1)\times 87\\ 87 &= 899 + (-7)\times 116. \end{aligned}
87
in the first equation, we have
\begin{aligned} 29 &= 116 + (-1)\times (899 + (-7)\times 116) \\ &= (-1)\times 899 + 8\times 116 \\ &= (-1)\times 899 + 8\times ( 1914 + (-2)\times 899 )\\ &= 8\times 1914 + (-17) \times 899 \\ &= 8\times 1914 - 17 \times 899. \end{aligned}
Since we now wrote the GCD as a linear combination of two integers, we terminate the algorithm and conclude
a = 8, b =-17. \ _\square
Without loss of generality we can assume that
and
b
are non-negative integers, because we can always do this:
\gcd(a,b)=\gcd\big(\lvert a \rvert, \lvert b \rvert\big)
Let's define the sequences
\{q_i\},\{r_i\},\{s_i\},\{t_i\}
r_0=a,r_1=b
i \gets 2
, and increase it at the end of every iteration. We're going to find in every iteration
q_i, r_i, s_i, t_i
r_{i-2}=r_{i-1}q_i+r_i
0 \leq r_i < r_{i-1}
using the division algorithm. We also want to write
r_i
and
b
r_i=s_i a+t_i b
r_i=r_{i-2}-r_{i-1}q_i
r_i=s_{i-2}a+t_{i-2}b-(s_{i-1}a+t_{i-1}b)q_i=(s_{i-2}-s_{i-1}q_i)a+(t_{i-2}-t_{i-1}q_i)b.
s_i=s_{i-2}-s_{i-1}q_i
t_i=t_{i-2}-t_{i-1}q_i
Now, we have to find the initial values of the sequences
\{s_i\}
\{t_i\}
r_i,
\begin{aligned} a=r_0=s_0 a+t_0 b &\implies s_0=1, t_0=0\\ b=r_1=s_1 a+t_1 b &\implies s_1=0, t_1=1. \end{aligned}
Finally, we stop at the iteration in which we have
r_{i-1}=0
. Let's call this the
n^\text{th}
iteration, so
r_{n-1}=0
. That means that
\gcd(a,b)=\gcd(r_0,r_1)=\gcd(r_1,r_2)=\cdots=\gcd(r_{n-2},r_{n-1})=\gcd(r_{n-2},0)=r_{n-2}
, so we found our desired linear combination:
\gcd(a,b)=r_{n-2}=s_{n-2} a + t_{n-2} b.
This algorithm is always finite, because the sequence
\{r_i\}
is decreasing, since
0 \leq r_i < r_{i-1}
2 \leq i < n-1
r_2 > r_3 > \cdots > r_{n-2} > r_{n-1} = 0
. In some moment we reach the value of zero, because all of the
r_i
To implement the algorithm, note that we only need to save the last two values of the sequences
\{r_i\}
\{s_i\}
\{t_i\}
s <- 0; old_s <- 1
t <- 1; old_t <- 0
r <- b; old_r <- a
while r ≠ 0
quotient <- old_r div r
(old_r, r) <- (r, old_r - quotient * r)
(old_s, s) <- (s, old_s - quotient * s)
(old_t, t) <- (t, old_t - quotient * t)
output "Bézout coefficients:", (old_s, old_t)
output "quotients by the gcd:", (t, s)
x
y
for the following equation:
1432x + 123211y = \gcd(1432,123211).
We can write Python code that implements the pseudo-code to solve the problem. (See the code in the next section.)
This gives -22973 and 267 for
x
y,
_\square
print egcd(1432,123211)
Cite as: Extended Euclidean Algorithm. Brilliant.org. Retrieved from https://brilliant.org/wiki/extended-euclidean-algorithm/
|
sweetrainyday8s2 2022-02-28 Answered
Consider the following permutations in
{S}_{7}
\alpha =\left(\begin{array}{ccccccc}1& 2& 3& 4& 5& 6& 7\\ 6& 1& 7& 4& 2& 5& 3\end{array}\right)
\beta =\left(\begin{array}{ccccccc}1& 2& 3& 4& 5& 6& 7\\ 4& 3& 1& 7& 2& 5& 6\end{array}\right)
\beta
Scoopedepalj 2022-02-28 Answered
Let G be a cyclic group of order 6. How many of its elements generate G?
Beverley Rahman 2022-02-28 Answered
Consider the incomplete character table for a group given below:
\begin{array}{|cccccc|}\hline & \left(1\right)& \left(1\right)& \left(2\right)& \left(2\right)& \left(2\right)\\ & 1& a& b& c& d\\ {x}_{1}& 1& 1& 1& 1& 1\\ {x}_{2}& 1& 1& -1& -1& 1\\ {x}_{3}& 1& 1& -1& -1& -1\\ {x}_{4}& 2& -2& 0& 0& 0\\ \hline\end{array}
All the conjugacy classes are there.
1) What is the order of the group?
2) How many characters are missing?
3) Find the missing character and complete the table.
4) Find the order of the Kemel of the missing character.
sacateundisco8i3 2022-02-28 Answered
Express as the product of disjoint cycles
a) (123)(45)(16789)(15)
b) (12)(123)(12)
asistioacer 2022-02-24
Kamari Simon 2022-02-15 Answered
{a}_{1},\cdots ,{a}_{r}\in G
|{a}_{1}\cdots {a}_{r}|
{a}_{1}\cdots {a}_{r}
1cm\left(|{a}_{1}|,\dots ,|{a}_{r}|\right)
{\left({a}_{1},\dots ,{a}_{r}\right)}^{1cm\left(|{a}_{1}|,\dots ,|{a}_{r}|\right)}=1
Bryant Miranda 2022-02-15 Answered
f:\mathbb{R}\to \mathbb{R}
is a homomorphism and p is an integer-coefficient polynomial, then:
f\left(p\left(r\right)\right)=p\left(f\left(r\right)\right)\mathrm{\forall }r\in \mathbb{R}
kaliitcri 2022-02-15 Answered
Northcott Multilinear Algebra Universal Property Proof
Northcott Multilinear Algebra poses a problem. Consider R-modules
{M}_{1},\dots ,{M}_{p}
, M and N. Consider multilinear mapping
\psi :{M}_{1}×\cdots ×{M}_{p}\to N
Northcott calls the universal problem as the problem to find M and multilinear mapping
\varphi :{M}_{1}×\cdots ×{M}_{p}\to M
such that there is exactly one R-module homomorphism
h:M\to N
h\circ \varphi =\psi
\lambda \text{ }\text{and}\text{ }{\lambda }^{\prime }
exist I understand why the equalities at the end of the sentence follow, based on the satisfaction of the universal problem. I can't see however why homomorphisms
\lambda \text{ }\text{and}\text{ }{\lambda }^{\prime }
should exist.
I did more group theory many years ago and this is my first serious foray into "modules" so I wouldn't be surprised if there is something obvious I'm missing.
my thoughts: Clearly M and M′ are both homomorphic to N through h and h′, I'm not sure if this says anything about a relationship between M and M′ though.
If h′ were injective I could say something like
\lambda \left(m\right)={h}^{{}^{\prime }-1}\left(h\left(m\right)\right)
but I don't know if there is any guarantee that h′ is injective..
\varphi
were injective I could define
\lambda \left(m\right)={\varphi }^{\prime }\left({\varphi }^{-1}\left(m\right)\right)
but again I don't know why this would be the case...
I've tried replacing M and N with more familiar vector spaces and R-module homomorphisms by multilinear maps for better intuition but no luck.. I do know that if M and M′ are vector spaces with the same dimension then there is an isomorphism between them. I guess more generally if M and M′ have different dimensions (say
\mathrm{dim}\left({M}^{\prime }\right)>\mathrm{dim}\left(M\right)\right)
then there is a homomorphism from M into a subspace of M′ and another homormophism from M′ onto M. Maybe this carries over to modules and is in the right direction for what I need...?
Alaina Ortiz 2022-02-15 Answered
3×3
A:=\left(\begin{array}{ccc}1& 4& 1\\ 0& 1& 0\\ 0& 1& 2\end{array}\right)
\mathbb{Q}\left[X\right]
V={\mathbb{Q}}^{3}
\mathbb{Q}\left[X\right]×V\to V,\left(P,v\right)\to P\left(A\right)\cdot v
P\left(A\right)\in {\mathbb{Q}}^{3×3}
P\left(A\right)\cdot v
{V}_{A}
{V}_{A}=\mathbb{Q}\left[X\right]\left(\begin{array}{c}1\\ 0\\ 1\end{array}\right)\oplus \mathbb{Q}\left[X\right]\left(\begin{array}{c}0\\ 1\\ -1\end{array}\right)
Jacob Stein 2022-02-15 Answered
Convoluted definition of a set:
H\left(S\right)=\left\{x\in G\mid \mathrm{\exists }n\in N,\mathrm{\exists }\left\{{x}_{1},{x}_{2},\cdots ,{x}_{n}\right\}\subseteq S\cup {S}^{-1},x={x}_{1}\cdots {x}_{n}\right\}
Krystal Villanueva 2022-02-15 Answered
Let A be a ring, S a multiplicatively closed subset. Is it true that
\frac{a}{1}\in {S}^{-1}A
is a non zero-divisor if and only if
a\in A
Quinten Crawford 2022-02-15 Answered
G=a
. The smallest subgroup of G containing
{a}^{8}\text{ }\text{and}\text{ }{a}^{12}
Arithmetical subtraction (-) is binary relation on
2) Z+
4) Z-
pimpinan15t 2022-02-14 Answered
|H|=20\text{ }\text{and}\text{ }|K|=32
|H\cap K|
Aryan Phillips 2022-02-14 Answered
Fundamental question about field extensions
Let k be an algebraically closed field. Let
K\subset k
be a subfield, and suppose that it has an algebraic extension
K\subset L
. Is it true (or at least well defined, because I'm not even sure about this) that
L\subseteq k
? One slightly different point of view is: given injections
i:K\to k,j:K\to L
, the last one being algebraic, is there an injection
l:L\to k
l\circ j=i
(i.e. l is a homomorphism of K-algebras)?
Amiya Arellano 2022-02-14 Answered
Is there a name for the set
E\left(x,n\right)=\left\{{x}^{p}\mid p\in N\wedge \le p\le n-1\right\}\text{ }\text{with}\text{ }x\in G,G
a group?
Seelakant6vr 2022-02-14 Answered
Simple question about finitely generated algebras
I have two finitely generated algebras A and B over a field K such that
B\subseteq A
. Is it true that
A=B\left[{a}_{1},\cdots ,{a}_{n}\right]
{a}_{1},\cdots ,{a}_{n}\in A
uheaeb56e 2022-02-14 Answered
{S}_{9}
\sigma =\left(\begin{array}{ccccccccc}1& 2& 3& 4& 5& 6& 7& 8& 9\\ 3& 5& 2& 1& 4& 6& 9& 7& 8\end{array}\right)
\tau =\left(\begin{array}{ccccccccc}1& 2& 3& 4& 5& 6& 7& 8& 9\\ 5& 2& 6& 7& 1& 3& 8& 9& 4\end{array}\right)
\alpha
\sigma \alpha =\tau
. Write your answer in cycle notation.
squaloideyjj 2022-02-14 Answered
Julio is studying for his final exams in chemistry and algebra. He knows he only has 12 hours to study and it will take him at least 2 more hours to study for algebra than chemistry. The following system of equations represents the situation, where x represents algebra and y represents chemistry:
x+y\le 12
x\ge y+2
Which of the following can Julio do?
Study algebra for 7 hours and chemistry for 5 hours
Study algebra for 7 or 8 hours and chemistry for 4 or 5 hours
Randy Mejia 2022-02-14 Answered
The number of elements in
{S}_{6}
|
Sign function (signum function) - MATLAB sign - MathWorks Deutschland
Plot Sign Function
Plot Real and Imaginary Parts of Sign Function
Sign function (signum function)
Y = sign(x) returns an array Y the same size as x, where each element of Y is:
1 if the corresponding element of x is greater than 0.
0 if the corresponding element of x equals 0.
-1 if the corresponding element of x is less than 0.
x./abs(x) if x is complex.
Find the sign function of a number.
Find the sign function of the values of a vector.
Find the sign function of the values of a matrix.
Find the sign function of a complex number.
Plot the sign function and show its behavior at the zero-crossing. Use eps to represent values just above and below 0.
Plot real and imaginary parts of the sign function over
-3<x<-3
-3<y<3
First, create a mesh of values over -3 < x < 3 and -3 < y < 3 using meshgrid. Then create complex numbers from these values using z = x + 1i*y.
Find the real and imaginary parts of the sign function of z.
Plot the real and imaginary parts.
Input, specified as a scalar, vector, matrix, or multidimensional array.
If an element of x is NaN, then sign returns NaN in the corresponding element of the output.
|
Find a 3×33×3 matrix AA such that Ax⃗ =9x⃗ Ax→=9x→ for all x⃗ x→ in R
Ax=9x
3×3
=\left[\begin{array}{ccc}9& 0& 0\\ 0& 9& 0\\ 0& 0& 9\end{array}\right]
\left(A-9I\right)x =0
x
A
A-9I =0
0
x
A=\left[\begin{array}{ccccc}1& 5& -4& -3& 1\\ 0& 1& -2& 1& 0\\ 0& 0& 0& 0& 0\end{array}\right]
A=\left[\begin{array}{cccc}2& 3& 5& -9\\ -8& -9& -11& 21\\ 4& -3& -17& 27\end{array}\right]
A=\left[\begin{array}{c}-3\\ 2\\ 0\\ 1\end{array}\right]
A=\left[\begin{array}{ccccc}1& 2& -5& 11& -3\\ 2& 4& -5& 15& 2\\ 1& 2& 0& 4& 5\\ 3& 6& -5& 19& -2\end{array}\right]
B=\left[\begin{array}{ccccc}1& 2& 0& 4& 5\\ 0& 0& 5& -7& 8\\ 0& 0& 0& 0& -9\\ 0& 0& 0& 0& 0\end{array}\right]
Find matrix of linear transformation
T:{\mathbb{R}}^{2}\to {\mathbb{R}}^{2}
T\left(i\right)=i+j
T\left(j\right)=2i-j
I have a 3d object, to which I sequentially apply 3
4×4
transformation matrices, A, B, and C. To generalize, each transformation matrix is determined by the multiplication of a rotation matrix by a translation matrix.
How can I calculate the final transformation matrix t, which defines how to get from the original 3d object to the final transformed object?
Do elementary row operations give a similar matrix transformation?
So we define two matrices A, B to be similar if there exists an invertible square matrix P such that
AP=PB
8{x}_{1}-{x}_{2}=4
5{x}_{1}+4{x}_{2}=1
{x}_{1}-3{x}_{2}=2
|
JEE Multinomial Theorem | Brilliant Math & Science Wiki
JEE Multinomial Theorem
Sandeep Bhardwaj, Satyabrata Dash, and Jimin Khim contributed
This page will teach you how to master JEE Multinomial Theorem. We highlight the main concepts, provide a list of examples with solutions, and include problems for you to try. Once you are confident, you can take the quiz to establish your mastery.
As per JEE syllabus, the main concepts under Multinomial Theorem are multinomial theorem and its expansion, number of terms in the expansion of multinomial theorem.
Multinomial theorem and its expansion:
n
(x_1+x_2+x_3+...+x_k)^n=\sum \frac{n!}{n_1!n_2!n_3!...n_k!} x_1^{n_1} x_2^{n_2} x_3^{n_3}...x_k^{n_k},
n_1,n_2,n_3,...,n_k
are all non-negative integers such that
n_1+n_2+n_3+\cdots+n_k=n.
Finding coefficients in the multinomial expansion
Number of terms in the expansion of multinomial theorem:
Number of terms in the expansion of
(x_1+x_2+x_3+\cdots+x_k)^n
, which is equal to the number of non-negative integral solutions of
n_1+n_2+n_3+...+n_k=n,
^{n+k-1}C_{k-1}.
\begin{array} { l l } A) \, & \quad \quad \quad \quad \quad & B) \, \\ C) \, & & D) \, \\ \end{array}
\begin{array} { l l } A) \, & \quad \quad \quad \quad \quad & B) \, \\ C) \, & & D) \, \\ \end{array}
\begin{array} { l l } A) \, & \quad \quad \quad \quad \quad & B) \, \\ C) \, & & D) \, \\ \end{array}
\begin{array} { l l } A) \, & \quad \quad \quad \quad \quad & B) \, \\ C) \, & & D) \, \\ \end{array}
Once you are confident of JEE Multinomial Theorem, move on to JEE Sequence and Series.
Cite as: JEE Multinomial Theorem. Brilliant.org. Retrieved from https://brilliant.org/wiki/jee-multinomial-theorem/
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Riemann SumEdit
{\displaystyle \int \limits _{a}^{b}f(x)dx\approx \sum _{i=1}^{n}f(x_{i}^{*})\Delta x}
{\displaystyle x_{i}^{*}}
is any point in the i-th sub-interval
{\displaystyle [x_{i-1},x_{i}]}
{\displaystyle [a,b]}
Right RectangleEdit
A special case of the Riemann sum, where we let
{\displaystyle x_{i}^{*}=x_{i}}
, in other words the point on the far right-side of each sub-interval on,
{\displaystyle [a,b]}
. Again if we pick n to be finite, then we have:
{\displaystyle \int \limits _{a}^{b}f(x)dx\approx \sum _{i=1}^{n}f(x_{i})\Delta x}
Left RectangleEdit
Another special case of the Riemann sum, this time we let
{\displaystyle x_{i}^{*}=x_{i-1}}
, which is the point on the far left side of each sub-interval on
{\displaystyle [a,b]}
. As always, this is an approximation whe{\displaystyle n}
is finite. Thus, we have:
{\displaystyle \int \limits _{a}^{b}f(x)dx\approx \sum _{i=1}^{n}f(x_{i-1})\Delta x}
Trapezoidal RuleEdit
{\displaystyle \int \limits _{a}^{b}f(x)dx\approx {\frac {b-a}{2n}}\left[f(x_{0})+2\sum _{i=1}^{n-1}{\bigl (}f(x_{i}){\bigr )}+f(x_{n})\right]={\frac {b-a}{2n}}{\bigg (}{f(x_{0})+2f(x_{1})+2f(x_{2})+\cdots +2f(x_{n-1})+f(x_{n})}{\bigg )}}
Simpson's RuleEdit
{\displaystyle \int \limits _{a}^{b}f(x)dx}
{\displaystyle \approx {\frac {b-a}{6n}}\left[f(x_{0})+\sum _{i=1}^{n-1}\left((3-(-1)^{i})f(x_{i})\right)+f(x_{n})\right]}
{\displaystyle ={\frac {b-a}{6n}}{\bigg [}f(x_{0})+4f{\bigl (}{\tfrac {x_{1}}{2}}{\bigr )}+2f(x_{1})+4f{\bigl (}{\tfrac {x_{3}}{2}}{\bigr )}+\cdots +4f{\bigl (}{\tfrac {x_{n-1}}{2}}{\bigr )}+f(x_{n}){\bigg ]}}
|
I have to prove that if P is a R-module
redupticslaz 2022-04-27 Answered
I have to prove that if P is a R-module , P is projective right there is a family
\left\{{x}_{i}\right\}
{f}_{i}:P⇒R
x\in P
x=\sum _{i\in I}{f}_{i}\left(x\right){x}_{i}
x\in P,\text{ }{f}_{i}\left(x\right)=0
i\in I
Put the
{f}_{i}
together to form one giant f from P to
{R}^{\left(I\right)}
, the direct sum of I copies of the ring R. The condition that
x\sum {f}_{i}\left(x\right){x}_{i}
just means that there is some
g:{R}^{\left(I\right)}⇒P
g\left(f\left(x\right)\right)=x
g\left(\begin{array}{ccc}{r}_{1}& \text{ }{r}_{2}& \cdots \end{array}\right)={r}_{1}{x}_{1}+{r}_{2}{x}_{2}+\cdots
. In other words, P is a direct summand of the free module
{R}^{\left(I\right)}
\left({x}_{1},{y}_{1}\right),\left({x}_{2},{y}_{2}\right)
\left({x}_{3},{y}_{3}\right)
\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\ne \frac{{y}_{3}-{y}_{2}}{{x}_{3}-{x}_{2}}\ne \frac{{y}_{1}-{y}_{3}}{{x}_{1}-{x}_{3}},
then there will be a circle passing through them. The general form of the circle is
{x}^{2}+{y}^{2}+ dx +ey+f=0.
x={x}_{i}\text{ }\text{and}\text{ }y={y}_{i}
, there will be a system of equation in three variables, that is:
\begin{array}{rl}\left(\begin{array}{ccc}{x}_{1}& {y}_{1}& 1\\ {x}_{2}& {y}_{2}& 1\\ {x}_{3}& {y}_{3}& 1\end{array}\right)\left(\begin{array}{c}d\\ e\\ f\end{array}\right)& =\left(\begin{array}{c}-\left({x}_{1}^{2}+{y}_{1}^{2}\right)\\ -\left({x}_{2}^{2}+{y}_{2}^{2}\right)\\ -\left({x}_{3}^{2}+{y}_{3}^{2}\right)\end{array}\right).\end{array}
As there are a lot of things going around, the solution is prone to errors. Maybe this solution also has an error.
Is there a better way to solve for the equation of the circle?
Reduce the system
(D2 + 1)[x] − 2D[y] = 2t
(2D − 1)[x] + (D − 2)[y] = 7.
to an equivalent triangular system of the form
P1(D)[y] = f1(t)
P2(D)[x] + P3(D)[y] = f2(t)
U=\left\{\left(x,y,x+y,x-y,2x\right)\in {F}^{5},x,y\in F\right\}
. Find a subspace
W
{F}^{5}
{F}^{5}=U\oplus W
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Counting Shapes | Brilliant Math & Science Wiki
Lawrence Chiou, Christopher Williams, Yash Jain, and
When approaching shape-hunting puzzles, you can think of yourself as a mathematical explorer and zoologist, cataloging and counting the populations of each species of shape in the figure. When scientific adventurers explored the rain forests of East Malaysia, they classified thousands of species, some common, others extremely rare. For example, they encountered approximately 150 species of frogs alone!
Geometric hunting puzzles can be a very difficult kind of puzzle, even for those well practiced at other kinds of mathematics. Most people don't see at least one kind of triangle in a picture like the one below—either the biggest, smallest, or the ones that are 'upside-down' perhaps.
You have to be both creative and meticulous when looking for these rare species that may be very hard to find.
Organizing the Census
The classic example is that of the triangular "matchstick" configuration as shown below.
The usual first step is simply to enumerate all of the distinct shape types of interest. Oftentimes, finding all of the shape types is the most tricky and potentially treacherous step—it can take some experience and meticulousness not to miss anything. Luckily, for this problem, it is relatively simple to enumerate all of the cases. Clearly, there are only three possible sizes of triangles: those having side length 1, side length 2, and side length 3.
The next step is to take a "census" of the different shape types. In this problem, it is fairly straightforward to do the census directly.
9
triangles of side length 1:
3
Finally, there is only
1
triangle of side length 3:
Thus, in total, there are
13
The two steps often work in tandem. Sometimes, the choice of what constitutes a "shape type" (for instance, counting a shape and its reflection as the same type) can make the census of shapes easier or harder. If the census turns out to be difficult, it may be worth making different choices for shape type. Many figures display some degree of symmetry that makes the census straightforward.
Throughout the process, it helps to stay as organized as possible. Position, size, and orientation are three common ways to organize cases. For example, one might count all of the "side-length-1 equilateral triangles," then all of the "side-length-2 equilateral triangles," and so on. Alternatively, one might count the triangles by location, first starting with "all of the triangles that have a corner at the very top of the figure," then "all of the triangles that have their highest corner at the second level of the figure," and so on.
To illustrate these principles in action, consider a slightly trickier example consisting of a star inside a pentagon, as shown below.
The first step is to go through the figure and consider the distinct types of triangles formed. Instead of doing this in a hodgepodge manner, however, it behooves us to use the natural structure that exists in the shape. Labeling the vertices that form the inner pentagon red and outer pentagon blue, one can consider all possible triangles that contain one, two, or three (blue) vertices from the outer pentagon alone (no triangles can be formed with only inner pentagon vertices).
Actually, by organizing the counting in this way, certain parts of the census become automatic. In fact, it is not even necessary to enumerate explicitly all of the distinct triangles that consist of three outer vertices. One can verify that any three outer vertices form a valid triangle. So, using combinations the total number of such triangles is
\binom{5}{3} = 10
The situation is a little more complicated when considering two outer vertices and one inner vertex, as some combinations of such vertices are collinear (i.e., lie on the same line). Furthermore, each inner vertex is only connected to four of the outer vertices. One can count all such triangles as shown:
Alternatively, one can argue that for each inner vertex, there must be
\binom{4}{2} = 6
ways to choose two connected outer vertices, of which
2
are collinear (and thus do not form triangles). Since there are five inner vertices, there must be
4 \cdot 5 = 20
triangles that consist of two outer vertices and one inner vertex.
Finally, it holds that each pair of two inner vertices forms only one triangle with one of the outer vertices, of which there are a total of
5
Overall, there are
10 + 20 + 5 = 35
triangles formed in the figure.
Although there were certainly few enough triangles that a crude attempt to enumerate all distinct shapes would have worked, relying on systematic counting of some of the cases (e.g., the triangles formed by three outer vertices) and symmetry arguments to reduce the number of cases that needed to be counted simplified the problem greatly. For geometric counting problems, efficiency and organization are often the keys to accuracy.
Cite as: Counting Shapes. Brilliant.org. Retrieved from https://brilliant.org/wiki/triangle-triage/
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I'm studying the Cartesian product, which is bound to the idea of a binary relation. Even with Carte
3-2-1
3\oplus 2\oplus 1
Giancarlo Brooks
How about a 3-number average? (Or more numbers if you want an operator that takes more than 3 arguments.) You don't get the correct value by averaging the first two numbers and then averaging that result with the third.
P\left(2\right)+P\left(5\right)
P\left(2\right)=1/6,P\left(5\right)=1/6
P\left(A\cup \left(B\cap C\right)\right)
P\left(A\cap \left(B\cup C\right)\right)=P\left(A\cap B\right)+P\left(A\cap C\right).
P\left(A\cup \left(B\cap C\right)\right)=P\left(A\cup B\right)+P\left(A\cup C\right)?
|
APSIDAL PRECESSION - Encyclopedia Information
Apsidal precession Information
Rotation of a celestial body's orbital line of apsides
In celestial mechanics, apsidal precession (or apsidal advance) [1] is the precession (gradual rotation) of the line connecting the apsides (line of apsides) of an astronomical body's orbit. The apsides are the orbital points closest (periapsis) and farthest (apoapsis) from its primary body. The apsidal precession is the first time derivative of the argument of periapsis, one of the six main orbital elements of an orbit. Apsidal precession is considered positive when the orbit's axis rotates in the same direction as the orbital motion. An apsidal period is the time interval required for an orbit to precess through 360°. [2]
The ancient Greek astronomer Hipparchos noted the apsidal precession of the Moon's orbit; [3] it is corrected for in the Antikythera Mechanism (circa 80 BCE) with the very accurate value of 8.88 years per full cycle, correct to within 0.34% of current measurements. [4] The precession of the solar apsides was discovered in the eleventh century by al-Zarqālī. [5] The lunar apsidal precession was not accounted for in Claudius Ptolemy's Almagest, and as a group these precessions, the result of a plethora of phenomena, remained difficult to account for until the 20th century when the last unidentified part of Mercury's precession was precisely explained.
A variety of factors can lead to periastron precession such as general relativity, stellar quadrupole moments, mutual star–planet tidal deformations, and perturbations from other planets. [6]
For Mercury, the perihelion precession rate due to general relativistic effects is 43″ ( arcseconds) per century. By comparison, the precession due to perturbations from the other planets in the Solar System is 532″ per century, whereas the oblateness of the Sun (quadrupole moment) causes a negligible contribution of 0.025″ per century. [7] [8]
with planetary tidal bulge being the dominant term, exceeding the effects of general relativity and the stellar quadrupole by more than an order of magnitude. The good resulting approximation of the tidal bulge is useful for understanding the interiors of such planets. For the shortest-period planets, the planetary interior induces precession of a few degrees per year. It is up to 19.9° per year for WASP-12b. [9] [10]
Newton derived an early theorem which attempted to explain apsidal precession. This theorem is historically notable, but it was never widely used and it proposed forces which have been found not to exist, making the theorem invalid. This theorem of revolving orbits remained largely unknown and undeveloped for over three centuries until 1995. [11] Newton proposed that variations in the angular motion of a particle can be accounted for by the addition of a force that varies as the inverse cube of distance, without affecting the radial motion of a particle.[ citation needed] Using a forerunner of the Taylor series, Newton generalized his theorem to all force laws provided that the deviations from circular orbits are small, which is valid for most planets in the Solar System.[ citation needed]. However, his theorem did not account for the apsidal precession of the Moon without giving up the inverse-square law of Newton's law of universal gravitation. Additionally, the rate of apsidal precession calculated via Newton's theorem of revolving orbits is not as accurate as it is for newer methods such as by perturbation theory.[ citation needed]
{\displaystyle \varepsilon =24\pi ^{3}{\frac {a^{2}}{T^{2}c^{2}\left(1-e^{2}\right)}}}
where c is the speed of light. [12] In the case of Mercury, half of the greater axis is about 5.79×1010 m, the eccentricity of its orbit is 0.206 and the period of revolution 87.97 days or 7.6×106 s. From these and the speed of light (which is ~3×108 m/s), it can be calculated that the apsidal precession during one period of revolution is ε = 5.028×10−7 radians (2.88×10−5 degrees or 0.104″). In one hundred years, Mercury makes approximately 415 revolutions around the Sun, and thus in that time, the apsidal perihelion due to relativistic effects is approximately 43″, which corresponds almost exactly to the previously unexplained part of the measured value.
Earth's apsidal precession slowly increases its argument of periapsis; it takes about 112,000 years for the ellipse to revolve once relative to the fixed stars. [13] Earth's polar axis, and hence the solstices and equinoxes, precess with a period of about 26,000 years in relation to the fixed stars. These two forms of 'precession' combine so that it takes between 20,800 and 29,000 years (and on average 23,000 years) for the ellipse to revolve once relative to the vernal equinox, that is, for the perihelion to return to the same date (given a calendar that tracks the seasons perfectly). [14]
^ Bowler, M. G. (2010). "Apsidal advance in SS 433?". Astronomy and Astrophysics. 510 (1): A28. arXiv: 0910.3536. Bibcode: 2010A&A...510A..28B. doi: 10.1051/0004-6361/200913471. S2CID 119289498.
^ Jones, A., Alexander (September 1991). "The Adaptation of Babylonian Methods in Greek Numerical Astronomy" (PDF). Isis. 82 (3): 440–453. Bibcode: 1991Isis...82..441J. doi: 10.1086/355836. S2CID 92988054.
^ Freeth, Tony; Bitsakis, Yanis; Moussas, Xenophon; Seiradakis, John. H.; Tselikas, A.; Mangou, H.; Zafeiropoulou, M.; Hadland, R.; et al. (30 November 2006). "Decoding the ancient Greek astronomical calculator known as the Antikythera Mechanism" (PDF). Nature. 444 Supplement (7119): 587–91. Bibcode: 2006Natur.444..587F. doi: 10.1038/nature05357. PMID 17136087. S2CID 4424998. Archived from the original (PDF) on 20 July 2015. Retrieved 20 May 2014.
^ Toomer, G. J. (1969), "The Solar Theory of az-Zarqāl: A History of Errors", Centaurus, 14 (1): 306–336, Bibcode: 1969Cent...14..306T, doi: 10.1111/j.1600-0498.1969.tb00146.x , at pp. 314–317.
^ Kane, S. R.; Horner, J.; von Braun, K. (2012). "Cyclic Transit Probabilities of Long-period Eccentric Planets due to Periastron Precession". The Astrophysical Journal. 757 (1): 105. arXiv: 1208.4115. Bibcode: 2012ApJ...757..105K. doi: 10.1088/0004-637x/757/1/105. S2CID 54193207.
^ Ragozzine, D.; Wolf, A. S. (2009). "Probing the interiors of very hot Jupiters using transit light curves". The Astrophysical Journal. 698 (2): 1778–1794. arXiv: 0807.2856. Bibcode: 2009ApJ...698.1778R. doi: 10.1088/0004-637x/698/2/1778. S2CID 29915528.
^ van den Heuvel, E. P. J. (1966). "On the Precession as a Cause of Pleistocene Variations of the Atlantic Ocean Water Temperatures". Geophysical Journal International. 11 (3): 323–336. Bibcode: 1966GeoJ...11..323V. doi: 10.1111/j.1365-246X.1966.tb03086.x.
Retrieved from " https://en.wikipedia.org/?title=Apsidal_precession&oldid=1082719728"
Apsidal Precession Videos
Apsidal Precession Websites
Apsidal Precession Encyclopedia Articles
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The average of the numbers 1, 2, 3, - Maths - Introduction to Graphs - 9429383 | Meritnation.com
\mathrm{We} \mathrm{have},\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}} \mathrm{average} \mathrm{of} 1, 2, 3, ......., 99, \mathrm{x} = 100\mathrm{x}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\frac{1 + 2 + 3 + .......+99 + \mathrm{x}}{100} = 100\mathrm{x}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒4950 + \mathrm{x} = 10000\mathrm{x} \left[\mathrm{as}, {\mathrm{S}}_{\mathrm{n}} = \frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}, \mathrm{where} \mathrm{n} \mathrm{is} \mathrm{the} \mathrm{natural} \mathrm{numbers}, {\mathrm{S}}_{99} =4950\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒9999\mathrm{x} = 4950\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{x} = \frac{4950}{9999} = \frac{50}{101}
Devasurya answered this
WHY DOVE NEED PARLIMENT
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Linear Algebra/Topic: Input-Output Analysis/Solutions - Wikibooks, open books for an open world
Linear Algebra/Topic: Input-Output Analysis/Solutions
< Linear Algebra | Topic: Input-Output Analysis
Hint: these systems are easiest to solve on a computer.
With the steel-auto system given above, estimate next year's total productions in these cases.
Next year's external demands are: up
{\displaystyle 200}
from this year for steel, and unchanged for autos.
{\displaystyle 100}
for steel, and up
{\displaystyle 200}
for autos.
{\displaystyle 200}
{\displaystyle 200}
These answers were given by Octave.
With the external use of steel as
{\displaystyle 17\,789}
and the external use of autos as
{\displaystyle 21\,243}
{\displaystyle s=25\,952}
{\displaystyle a=30\,312}
{\displaystyle s=25\,857}
{\displaystyle a=30\,596}
{\displaystyle s=25\,984}
{\displaystyle a=30\,597}
In the steel-auto system, the ratio for the use of steel by the auto industry is
{\displaystyle 2\,664/30\,346}
, about
{\displaystyle 0.0878}
. Imagine that a new process for making autos reduces this ratio to
{\displaystyle .0500}
How will the predictions for next year's total productions change compared to the first example discussed above (i.e., taking next year's external demands to be
{\displaystyle 17,589}
for steel and
{\displaystyle 21,243}
for autos)?
Predict next year's totals if, in addition, the external demand for autos rises to be
{\displaystyle 21,500}
because the new cars are cheaper.
Octave gives these answers.
{\displaystyle s=24\,244}
{\displaystyle a=30\,307}
{\displaystyle s=24\,267}
{\displaystyle a=30\,673}
This table gives the numbers for the auto-steel system from a different year, 1947 (see Leontief 1951). The units here are billions of 1947 dollars.
steel used by
auto used by
auto 0 4.40 14.27
Solve for total output if next year's external demands are: steel's demand up 10% and auto's demand up 15%.
How do the ratios compare to those given above in the discussion for the 1958 economy?
Solve the 1947 equations with the 1958 external demands (note the difference in units; a 1947 dollar buys about what $1.30 in 1958 dollars buys). How far off are the predictions for total output?
These are the equations.
{\displaystyle {\begin{array}{*{2}{rc}r}(11.79/18.69)s&-&(1.28/14.27)a&=&11.56\\-(0/18.69)s&+&(9.87/14.27)a&=&11.35\end{array}}}
Octave gives
{\displaystyle s=20.66}
{\displaystyle a=16.41}
These are the ratios.
1947 by steel by autos
use of steel 0.63 0.09
use of autos 0.00 0.69
Octave gives (in billions of 1947 dollars)
{\displaystyle s=24.82}
{\displaystyle a=23.63}
. In billions of 1958 dollars that is
{\displaystyle s=32.26}
{\displaystyle a=30.71}
Predict next year's total productions of each of the three sectors of the hypothetical economy shown below
farm used by
rail used by
shipping used by
farm 25 50 100 500
rail 25 50 50 300
shipping 15 10 0 500
if next year's external demands are as stated.
{\displaystyle 625}
for farm,
{\displaystyle 200}
for rail,
{\displaystyle 475}
{\displaystyle 650}
{\displaystyle 150}
{\displaystyle 450}
This table gives the interrelationships among three segments of an economy (see Clark & Coupe 1967).
food used by
wholesale used by
retail used by
food 0 2 318 4 679 11 869
wholesale 393 1 089 22 459 122 242
retail 3 53 75 116 041
We will do an Input-Output analysis on this system.
Fill in the numbers for this year's external demands.
Set up the linear system, leaving next year's external demands blank.
Solve the system where next year's external demands are calculated by taking this year's external demands and inflating them 10%. Do all three sectors increase their total business by 10%? Do they all even increase at the same rate?
Solve the system where next year's external demands are calculated by taking this year's external demands and reducing them 7%. (The study from which these numbers are taken concluded that because of the closing of a local military facility, overall personal income in the area would fall 7%, so this might be a first guess at what would actually happen.)
Leontief, Wassily W. (Oct. 1951), "Input-Output Economics", Scientific American 185 (4): 15 .
Clark, David H.; Coupe, John D. (Mar. 1967), "The Bangor Area Economy Its Present and Future", Reprot to the City of Bangor, ME .
Retrieved from "https://en.wikibooks.org/w/index.php?title=Linear_Algebra/Topic:_Input-Output_Analysis/Solutions&oldid=3651930"
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\mathrm{End}_{\mathbb R[x]}(M) where M = \frac{\mathbb R[x]}{(x^2 + 1)} is
{\mathrm{End}}_{\mathrm{ℝ}\left[x\right]}\left(M\right)
M=\frac{\mathbb{R}\left[x\right]}{\left({x}^{2}+1\right)}
\mathbb{R}\left[x\right]
For any R-module N, it can be shown that
{\text{End}}_{R}\left(N\right)
is a R-module, hence the result is certainly not gonna be a group like
G{L}_{2}
{\text{End}}_{\mathbb{R}\left[x\right]}\frac{\mathbb{R}\left[x\right]}{\left({x}^{2}+1\right)}\stackrel{\sim }{=}{\text{End}}_{\frac{\mathbb{R}\left[x\right]}{\left({x}^{2}+1\right)}}\frac{\mathbb{R}\left[x\right]}{\left({x}^{2}+1\right)}\stackrel{\sim }{=}\frac{\mathbb{R}\left[x\right]}{\left({x}^{2}+1\right)}\stackrel{\sim }{=}\mathbb{C}
The first isomorphism follows from the general fact
{\text{Hom}}_{R}\left(\frac{M}{IM},\frac{N}{IN}\right)\stackrel{\sim }{=}{\text{Hom}}_{\frac{R}{I}}\left(\frac{M}{IM},\frac{N}{IN}\right)
It's clear that any
\frac{R}{I}
-hom is also R-hom. On the other hand, for any R-homomorphism
\rho :\frac{M}{IM}⇒\frac{N}{IN}
\rho \left(am\right)=a\rho \left(m\right)=0
a\in I,\text{ }m\in \frac{M}{IM}
\rho
\frac{R}{I}
y\le 4
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Coordinates in 3D Practice Problems Online | Brilliant
Understanding graphs and surfaces requires us to delve a little deeper into the place where they live:
\mathbb{R}^3.
This quiz covers the essentials of three-dimensional coordinate systems. We'll need some right triangle trigonometry in order to construct the polar, cylindrical, and spherical coordinate systems.
We'll start with the 3D Cartesian system, which extends the familiar 2D
xy
-coordinate system into the full three dimensions of our day-to-day experience by adding a new axis. (Check out the animation below!)
The new axis is labeled
z,
and we have three coordinates
( x,y,z)
for every point in space
\mathbb{R}^3.
To get to the point
(1, 2 , 2)
pictured below (red dot), we start at the origin where all axes meet, move 1 unit in the positive
x
direction, 2 units in the positive
y
direction, and then 2 units up in the positive
z
What are the coordinates of the green triangle?
Note that you are able to toggle and zoom into the diagram above.
(-1,0,1)
(0,4,-2)
(0,0,-1)
(0,-1,1)
Now, what are the coordinates of the blue rectangle?
(-1,0,1)
(0,4,-2)
(0,0,-1)
(0,-1,-2)
Not all creatures prefer Cartesian coordinates.
Bees do a waggle-dance to communicate distance and direction information, in effect using a coordinate system based on circles instead of perpendicular lines.
If we're to be at least as clever as bees, we should develop alternative coordinate systems, too!
Cylindrical and spherical coordinate systems for
\mathbb{R}^3
are just such coordinate systems, and both are built up from polar coordinates
(r,\theta)
\mathbb{R}^2,
the system of choice for talkative bees.
The rest of the quiz will develop all three coordinate systems from scratch.
Let's start with polar coordinates.
We arrive at the planar point
P =(x,y)
by following a ray starting at the origin and making an angle of
\theta
x
-axis for a distance of
r.
Trigonometry then tells us that
x = r \cos(\theta ).
What option best represents the relationship between
y
and the polar coordinates
(r,\theta) ?
y = \sin( \theta)
y = r \sin( \theta)
y = r \tan( \theta)
y = r^2 \sin( \theta)
Polar coordinates and Cartesian coordinates are related through
x = r \cos(\theta), \quad y = r \sin(\theta) .
Suppose we're given the Cartesian coordinates for
P
but want to know the polar form. Since
\sin^2(\theta)+ \cos^2(\theta) =1 ,
\begin{aligned} \sqrt{x^2+y^2} &= \sqrt{ \big[ r \cos(\theta) \big]^2 + \big[r \sin(\theta)\big]^2} \\ & = \sqrt{ r^2 \big[ \cos^2(\theta)+ \sin^2(\theta)\big]} \\ &= \sqrt{r^2 } \\ &= r. \end{aligned}
What formula lets us calculate
\theta
given the Cartesian coordinates
(x,y)?
\frac{x}{y} = \tan(\theta)
-\frac{x}{y} = \tan(\theta)
\frac{y}{x} = \tan(\theta)
\frac{x}{y} = \sin(\theta)
For 3D problems, we can build a hybrid of the 2D polar system and Cartesian coordinates called polar cylindrical (or just cylindrical) coordinates.
P = (x,y,z) \in \mathbb{R}^3.
The first two numbers represent a point in the plane, which we can describe using
( r, \theta).
The cylindrical coordinates of
P
are then
(r,\theta,z).
Compute the cylindrical coordinates of
(-2,-2, 5).
The interactive below may help.
Adjust the controls until the point sits at the top of the vertical line....
\left( 2 \sqrt{2} , \frac{5 \pi}{4} , 3\right)
\left( -2 \sqrt{2} , \frac{5 \pi}{4} , 5\right)
\left( 2 \sqrt{2} , -\frac{ \pi}{4} , 5\right)
\left( 2 \sqrt{2} , \frac{5 \pi}{4} , 5\right)
Cylindrical coordinates are ideally suited for problems in
\mathbb{R}^3
symmetric about the
z
-axis, like describing
C_{R},
the cylinder of radius
R
z
-axis, which is made up of all points of distance
R
from this line.
Of the equations presented, which one best describes
C_{R}
in cylindrical coordinates?
Note: We use dashed lines to indicate portions of the picture that continue out to infinity. The cylinder above continues parallel to the
z
-axis in both the positive and negative directions.
r = R
\theta = \frac{\pi}{2}
z = R
When a problem has complete symmetry around the origin, spherical coordinates are usually better than cylindrical coordinates.
P = (x,y,z) \in \mathbb{R}^3.
x = r \cos(\theta),\quad y = r \sin(\theta) ,
r
is the distance between the point
( 0,0)
(x,y)
\theta
is taken to be one of the new spherical coordinates; the other two are
\rho
\big(
P
(0,0,0)\big)
\phi,
the angle between the positive
z
-axis and the ray joining
P
with the origin.
Use the diagram to relate
r
\rho.
Hint: It's a geometric fact that the angle made by the solid and dashed lines meeting at
P
form an angle of size
\phi
r = \rho \sin(\phi)
r = \rho \cos(\phi)
r = \rho \tan(\phi)
r = \rho \sin(\phi),
x = \rho \sin(\phi) \cos(\theta), \quad y = \rho \sin(\phi) \sin(\theta).
We need to find a formula for
z
to complete the relationship between Cartesian and spherical coordinates.
Using the picture above, what is
z
( \rho, \theta, \phi) ?
Hint: Remember from the last problem that the angle made by the lines meeting at
P
\phi,
z = \rho \sin(\phi)
z = \rho \cos(\phi)
z = \rho \tan(\phi)
Finally, let's understand why the coordinates
( \rho, \theta, \phi)
x = \rho \sin(\phi) \cos(\theta),\quad y = \rho \sin(\phi) \sin(\theta),\quad z = \rho \cos( \phi)
are called spherical.
Of the options presented, which one correctly describes the sphere
S_{R}
R
centered at the origin in spherical coordinates?
\rho = R
\theta = \frac{\pi}{2}
\phi = \pi
Visualizing mathematical objects like surfaces in
\mathbb{R}^3
can be very helpful in solving many multivariable calculus problems.
We saw one example of this already when we visualized the graph of the depth function at the end of the optimization quiz. At a glance we were able to see where the minimum and maximum values of the depth function occur.
The Cartesian, spherical, and cylindrical systems provide the means for visualizing a large variety of useful objects in multivariable calculus.
The final quiz of this intro chapter shows one particular and very important example.
There, we'll use the 3D coordinate system to understand what place integrals have in the world of multivariable calculus.
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Isotonic regression - Wikipedia
Type of numerical analysis
Isotonic regression for the simply ordered case with univariate
{\displaystyle x,y}
has been applied to estimating continuous dose-response relationships in fields such as anesthesiology and toxicology. Narrowly speaking, isotonic regression only provides point estimates at observed values of
{\displaystyle x.}
Estimation of the complete dose-response curve without any additional assumptions is usually done via linear interpolation between the point estimates. [3]
Problem Statement and Algorithms[edit]
{\displaystyle (x_{1},y_{1}),\ldots ,(x_{n},y_{n})}
be a given set of observations, where the
{\displaystyle y_{i}\in \mathbb {R} }
{\displaystyle x_{i}}
fall in some partially ordered set. For generality, each observation
{\displaystyle (x_{i},y_{i})}
may be given a weight
{\displaystyle w_{i}\geq 0}
, although commonly
{\displaystyle w_{i}=1}
{\displaystyle i}
Isotonic regression seeks a weighted least-squares fit
{\displaystyle {\hat {y}}_{i}\approx y_{i}}
{\displaystyle i}
, subject to the constraint that
{\displaystyle {\hat {y}}_{i}\leq {\hat {y}}_{j}}
{\displaystyle x_{i}\leq x_{j}}
. This gives the following quadratic program (QP) in the variables
{\displaystyle {\hat {y}}_{1},\ldots ,{\hat {y}}_{n}}
{\displaystyle \min \sum _{i=1}^{n}w_{i}({\hat {y}}_{i}-y_{i})^{2}}
{\displaystyle {\hat {y}}_{i}\leq {\hat {y}}_{j}{\text{ for all }}(i,j)\in E}
{\displaystyle E=\{(i,j):x_{i}\leq x_{j}\}}
specifies the partial ordering of the observed inputs
{\displaystyle x_{i}}
(and may be regarded as the set of edges of some directed acyclic graph (dag) with vertices
{\displaystyle 1,2,\ldots n}
). Problems of this form may be solved by generic quadratic programming techniques.
In the usual setting where the
{\displaystyle x_{i}}
values fall in a totally ordered set such as
{\displaystyle \mathbb {R} }
, we may assume WLOG that the observations have been sorted so that
{\displaystyle x_{1}\leq x_{2}\leq \cdots \leq x_{n}}
{\displaystyle E=\{(i,i+1):1\leq i<n\}}
. In this case, a simple iterative algorithm for solving the quadratic program is the pool adjacent violators algorithm. Conversely, Best and Chakravarti[8] studied the problem as an active set identification problem, and proposed a primal algorithm. These two algorithms can be seen as each other's dual, and both have a computational complexity of
{\displaystyle O(n)}
on already sorted data.[8]
To complete the isotonic regression task, we may then choose any non-decreasing function
{\displaystyle f(x)}
{\displaystyle f(x_{i})={\hat {y}}_{i}}
for all i. Any such function obviously solves
{\displaystyle \min _{f}\sum _{i=1}^{n}w_{i}(f(x_{i})-y_{i})^{2}}
{\displaystyle f}
being nondecreasing
and can be used to predict the
{\displaystyle y}
values for new values of
{\displaystyle x}
. A common choice when
{\displaystyle x_{i}\in \mathbb {R} }
would be to interpolate linearly between the points
{\displaystyle (x_{i},{\hat {y}}_{i})}
, as illustrated in the figure, yielding a continuous piecewise linear function:
{\displaystyle f(x)={\begin{cases}{\hat {y}}_{1}&{\text{if }}x\leq x_{1}\\{\hat {y}}_{i}+{\frac {x-x_{i}}{x_{i+1}-x_{i}}}({\hat {y}}_{i+1}-{\hat {y}}_{i})&{\text{if }}x_{i}\leq x\leq x_{i+1}\\{\hat {y}}_{n}&{\text{if }}x\geq x_{n}\end{cases}}}
Centered Isotonic Regression[edit]
As this article's first figure shows, in the presence of monotonicity violations the resulting interpolated curve will have flat (constant) intervals. In dose-response applications it is usually known that
{\displaystyle f(x)}
is not only monotone but also smooth. The flat intervals are incompatible with
{\displaystyle f(x)}
's assumed shape, and can be shown to be biased. A simple improvement for such applications, named centered isotonic regression (CIR), was developed by Oron and Flournoy and shown to substantially reduce estimation error for both dose-response and dose-finding applications.[9] Both CIR and the standard isotonic regression for the univariate, simply ordered case, are implemented in the R package "cir".[4] This package also provides analytical confidence-interval estimates.
^ Kruskal, J. B. (1964). "Nonmetric Multidimensional Scaling: A numerical method". Psychometrika. 29 (2): 115–129. doi:10.1007/BF02289694.
^ "Predicting good probabilities with supervised learning | Proceedings of the 22nd international conference on Machine learning". dl.acm.org. doi:10.1145/1102351.1102430. Retrieved 2020-07-07.
^ Stylianou, MP; Flournoy, N (2002). "Dose finding using the biased coin up-and-down design and isotonic regression". Biometrics. 58 (1): 171–177. doi:10.1111/j.0006-341x.2002.00171.x. PMID 11890313.
^ a b Oron, Assaf. "Package 'cir'". CRAN. R Foundation for Statistical Computing. Retrieved 26 December 2020.
^ Leeuw, Jan de; Hornik, Kurt; Mair, Patrick (2009). "Isotone Optimization in R: Pool-Adjacent-Violators Algorithm (PAVA) and Active Set Methods". Journal of Statistical Software. 32 (5): 1–24. doi:10.18637/jss.v032.i05. ISSN 1548-7660.
^ Xu, Zhipeng; Sun, Chenkai; Karunakaran, Aman. "Package UniIsoRegression" (PDF). CRAN. R Foundation for Statistical Computing. Retrieved 29 October 2021.
^ Pedregosa, Fabian; et al. (2011). "Scikit-learn:Machine learning in Python". Journal of Machine Learning Research. 12: 2825–2830. arXiv:1201.0490. Bibcode:2012arXiv1201.0490P.
^ a b Best, Michael J.; Chakravarti, Nilotpal (1990). "Active set algorithms for isotonic regression; A unifying framework". Mathematical Programming. 47 (1–3): 425–439. doi:10.1007/bf01580873. ISSN 0025-5610.
^ Oron, AP; Flournoy, N (2017). "Centered Isotonic Regression: Point and Interval Estimation for Dose-Response Studies". Statistics in Biopharmaceutical Research. 9 (3): 258–267. arXiv:1701.05964. doi:10.1080/19466315.2017.1286256.
Shively, T.S., Sager, T.W., Walker, S.G. (2009). "A Bayesian approach to non-parametric monotone function estimation". Journal of the Royal Statistical Society, Series B. 71 (1): 159–175. CiteSeerX 10.1.1.338.3846. doi:10.1111/j.1467-9868.2008.00677.x. {{cite journal}}: CS1 maint: multiple names: authors list (link)
Wu, W. B.; Woodroofe, M.; Mentz, G. (2001). "Isotonic regression: Another look at the changepoint problem". Biometrika. 88 (3): 793–804. doi:10.1093/biomet/88.3.793.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Isotonic_regression&oldid=1073267758"
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The streamlines of the PIV and the LES data agree well according to Fig. 6. The plots are superimposed by the normalized magnitude of the velocity field in the symmetry plane, thus <math> ||\vec{U}|| = \sqrt{\langle u^2\rangle + \langle w^2\rangle}/u_{\mathrm{b}}</math>. The approaching turbulent boundary layer is redirected downwards at the flow facing edge of the cylinder caused by a vertical pressure gradient. This downflow reaches the bottom plate of the flume at the stagnation point S3. Here, it is redirected (i) in the out-of-plane direction bending around the cylinder; (ii) towards the cylinder rolling up and forming the corner vortex V3; and (iii) in the upstream direction accelerating and forming a wall-parallel jet. The jet accelerates and exerts a large wall-shear stress on the bottom plate (see Fig. 22). Parts of the downflow form the horseshoe vortex V1. Upstream of this vortex system, the approaching flow is blocked and causes a saddle point S1 with zero velocity magnitude.
[[File:UFR3-35_PIV_streamlines_mag.png|centre|frame|Fig. 6 a) Streamlines of time-averaged flow field superimposed by the in-plane velocity magnitude <math> ||\vec{U}_{\mathrm{PIV}}|| = \sqrt{\langle u^2\rangle + \langle w^2\rangle}/u_{\mathrm{b}}</math> ]]
{\displaystyle c_{\mathrm {p} }(x)}
{\displaystyle c_{\mathrm {f} }(x)}
{\displaystyle \langle u\rangle }
{\displaystyle \langle w\rangle }
{\displaystyle \langle u'_{i}u'_{j}\rangle }
{\displaystyle \langle k\rangle }
{\displaystyle c_{\mathrm {p} }(x)}
{\displaystyle c_{\mathrm {f} }(x)}
{\displaystyle ||{\vec {U}}||={\sqrt {\langle u^{2}\rangle +\langle w^{2}\rangle }}/u_{\mathrm {b} }}
{\displaystyle ||{\vec {U}}_{\mathrm {PIV} }||={\sqrt {\langle u^{2}\rangle +\langle w^{2}\rangle }}/u_{\mathrm {b} }}
{\displaystyle ||{\vec {U}}_{\mathrm {LES} }||={\sqrt {\langle u^{2}\rangle +\langle w^{2}\rangle }}/u_{\mathrm {b} }}
{\displaystyle x/D}
{\displaystyle z/D}
{\displaystyle x/D}
{\displaystyle z/D}
{\displaystyle -0.788}
{\displaystyle 0.03}
{\displaystyle -0.843}
{\displaystyle 0.037}
{\displaystyle -0.918}
{\displaystyle 0}
{\displaystyle -1.1}
{\displaystyle 0}
{\displaystyle -0.533}
{\displaystyle 0}
{\displaystyle -0.534}
{\displaystyle 0}
{\displaystyle -0.507}
{\displaystyle 0.036}
{\displaystyle -0.50}
{\displaystyle 0.04}
{\displaystyle -0.697}
{\displaystyle 0.051}
{\displaystyle -0.735}
{\displaystyle 0.06}
{\displaystyle -0.513}
{\displaystyle 0.017}
{\displaystyle -0.513}
{\displaystyle 0.02}
{\displaystyle x-}
{\displaystyle x_{\mathrm {adj} }={\frac {x-x_{\mathrm {Cyl} }}{x_{\mathrm {Cyl} }-x_{\mathrm {V1} }}}}
{\displaystyle x_{\mathrm {Cyl} }=-0.5D}
{\displaystyle x_{\mathrm {adj} }=-1.0}
{\displaystyle x_{\mathrm {V1} }}
{\displaystyle \langle u(z)\rangle /u_{\mathrm {b} }}
{\displaystyle u(z)}
{\displaystyle x_{\mathrm {adj} }=-0.25}
{\displaystyle x_{\mathrm {adj} }=-0.5}
{\displaystyle {\frac {\partial \langle u\rangle }{\partial z}}}
{\displaystyle {\frac {\partial \langle u\rangle }{\partial z}}}
{\displaystyle \langle u_{i}'u_{j}'(z)\rangle /u_{\mathrm {b} }^{2}}
{\displaystyle \langle k(z)\rangle =0.5(\langle u'^{2}\rangle +\langle w'^{2}\rangle )/u_{\mathrm {b} }^{2}}
{\displaystyle x_{\mathrm {adj} }=-1.5}
{\displaystyle x_{\mathrm {adj} }=-1.0}
{\displaystyle x_{\mathrm {adj} }=-0.5}
{\displaystyle \langle u'u'\rangle }
{\displaystyle \langle u'u'\rangle }
{\displaystyle \langle w'w'\rangle }
{\displaystyle z_{\mathrm {V1} }/D}
{\displaystyle \langle u'w'\rangle }
{\displaystyle \langle w(x)\rangle /u_{\mathrm {b} }}
{\displaystyle z_{\mathrm {V1} }/D}
{\displaystyle \langle w(x)\rangle }
{\displaystyle x-}
{\displaystyle x_{\mathrm {adj} }\approx -0.1}
{\displaystyle x_{\mathrm {adj} }=-0.65}
{\displaystyle \langle u_{i}'u_{j}'(x)\rangle /u_{\mathrm {b} }^{2}}
{\displaystyle \langle k(x)\rangle =0.5(\langle u'^{2}\rangle +\langle w'^{2}\rangle )/u_{\mathrm {b} }^{2}}
{\displaystyle z_{\mathrm {V1} }/D}
{\displaystyle \langle u_{i}'u_{j}'\rangle }
{\displaystyle \langle k\rangle }
{\displaystyle \langle k_{\mathrm {PIV,inplane} }\rangle =0.5(\langle u'^{2}\rangle +\langle w'^{2}\rangle )/u_{\mathrm {b} }^{2}}
{\displaystyle \langle k_{\mathrm {LES,inplane} }\rangle =0.5(\langle u'^{2}\rangle +\langle w'^{2}\rangle )/u_{\mathrm {b} }^{2}}
{\displaystyle \langle k_{\mathrm {LES,total} }\rangle =0.5(\langle u'^{2}\rangle +\langle v'^{2}\rangle +\langle w'^{2}\rangle )/u_{\mathrm {b} }^{2}}
{\displaystyle \langle k\rangle =0.5(\langle u'^{2}\rangle +\langle w'^{2}\rangle )/u_{\mathrm {b} }^{2}}
{\displaystyle \langle k_{\mathrm {PIV,inplane} }\rangle =0.074u_{\mathrm {b} }^{2}}
{\displaystyle \langle k_{\mathrm {LES,inplane} }\rangle =0.079u_{\mathrm {b} }^{2}}
{\displaystyle \langle k_{\mathrm {LES,total} }\rangle =0.5(\langle u'^{2}\rangle +\langle v'^{2}\rangle +\langle w'^{2}\rangle )/u_{\mathrm {b} }^{2}}
{\displaystyle \langle k_{\mathrm {LES,total} }\rangle =0.09u_{\mathrm {b} }^{2}}
{\displaystyle 0=P+\nabla T-\epsilon +C}
{\displaystyle P}
{\displaystyle \nabla T}
{\displaystyle \epsilon }
{\displaystyle C}
{\displaystyle v}
{\displaystyle P=-\langle u_{i}'u_{j}'\rangle {\frac {\partial \langle u_{i}\rangle }{\partial x_{j}}}}
{\displaystyle T=\underbrace {-{\frac {1}{2}}\langle u_{i}'u_{j}'u_{j}'\rangle } _{\text{turbulent fluctuations}}\underbrace {-{\frac {1}{\rho }}\langle u_{i}'p'\rangle } _{\text{pressure transport}}\underbrace {+2\nu \langle u_{j}'s_{ij}\rangle } _{\text{viscous diffusion}}}
{\displaystyle \epsilon =2\nu \langle s_{ij}s_{ij}\rangle }
{\displaystyle s_{ij}={\frac {1}{2}}\left({\frac {\partial u_{i}'}{\partial x_{j}}}+{\frac {\partial u_{j}'}{\partial x_{i}}}\right)}
{\displaystyle \epsilon _{\mathrm {total} }=\epsilon _{\mathrm {res} }+\epsilon _{\mathrm {SGS} }=2\nu \langle s_{ij}s_{ij}\rangle +2\langle \nu _{\mathrm {t} }s_{ij}s_{ij}\rangle }
{\displaystyle C=-\langle u_{i}\rangle {\frac {\partial k}{\partial x_{i}}}}
{\displaystyle D/u_{\mathrm {b} }^{3}}
{\displaystyle P_{\mathrm {PIV} }=-\langle u_{i}'u_{j}'\rangle {\frac {\partial \langle u_{i}\rangle }{\partial x_{j}}}\cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle P_{\mathrm {LES} }=-\langle u_{i}'u_{j}'\rangle {\frac {\partial \langle u_{i}\rangle }{\partial x_{j}}}\cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle 0.3u_{\mathrm {b} }^{3}/D}
{\displaystyle P_{\mathrm {LES} }\approx 0.4u_{\mathrm {b} }^{3}/D}
{\displaystyle P_{\mathrm {PIV} }\approx 0.2u_{\mathrm {b} }^{3}/D}
{\displaystyle x=-0.7D}
{\displaystyle P}
{\displaystyle \nabla T_{\mathrm {turb,PIV} }=-{\frac {1}{2}}{\frac {\partial \langle u_{i}'u_{j}'u_{j}'\rangle }{\partial x_{i}}}\cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle \nabla T_{\mathrm {turb,LES} }=-{\frac {1}{2}}{\frac {\partial \langle u_{i}'u_{j}'u_{j}'\rangle }{\partial x_{i}}}\cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle x=-0.75D}
{\displaystyle 0.4u_{\mathrm {b} }^{3}/D}
{\displaystyle T_{\mathrm {turb,LES} }\approx 0.35u_{\mathrm {b} }^{3}/D}
{\displaystyle \nabla T_{\mathrm {press,LES} }=-{\frac {1}{\rho }}{\frac {\partial \langle u_{i}'p'\rangle }{\partial x_{i}}}\cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle \nabla T_{\mathrm {visc,LES} }=2\nu {\frac {\partial \langle u_{j}'s_{ij}\rangle }{\partial x_{i}}}\cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle \nabla T_{\mathrm {turb} }}
{\displaystyle \nabla T_{\mathrm {press} }}
{\displaystyle \langle w\rangle <0}
{\displaystyle w-}
{\displaystyle w'}
{\displaystyle p'<0}
{\displaystyle \nabla T_{\mathrm {visc} }}
{\displaystyle |0.05|u_{\mathrm {b} }^{3}/D}
{\displaystyle P}
{\displaystyle \nabla T}
{\displaystyle \epsilon }
{\displaystyle \epsilon _{\mathrm {PIV} }=2\nu \langle s_{ij}s_{ij}\rangle \cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle \epsilon _{\mathrm {LES,total} }=(2\nu \langle s_{ij}s_{ij}\rangle +2\langle \nu _{\mathrm {t} }s_{ij}s_{ij}\rangle )\cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle P}
{\displaystyle \epsilon _{\mathrm {LES} }=0.066u_{\mathrm {b} }^{3}/D}
{\displaystyle P_{\mathrm {max} }}
{\displaystyle \epsilon _{\mathrm {max} }}
{\displaystyle C_{\mathrm {PIV} }=-\langle u_{i}\rangle {\frac {\partial k}{\partial x_{i}}}\cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle C_{\mathrm {LES} }=-\langle u_{i}\rangle {\frac {\partial k}{\partial x_{i}}}\cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle x\approx -0.63D}
{\displaystyle C}
{\displaystyle R_{\mathrm {PIV} }=P+\nabla T_{\mathrm {turb} }-\epsilon +C\cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle -\nabla T_{\mathrm {press,LES} }}
{\displaystyle R_{\mathrm {LES} }=P+\nabla T-\epsilon _{\mathrm {total} }+C\cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle <|0.01|u_{\mathrm {b} }^{3}/D}
{\displaystyle T_{\mathrm {turb} }=-{\frac {1}{2}}\langle u_{i}'u_{j}'u_{j}'\rangle }
{\displaystyle c_{\mathrm {p} }(x)}
{\displaystyle c_{\mathrm {f} }(x)}
{\displaystyle c_{\mathrm {p} }}
{\displaystyle c_{\mathrm {p} }={\frac {\langle p\rangle }{{\frac {\rho }{2}}u_{\mathrm {b} }^{2}}}}
{\displaystyle c_{\mathrm {f} }}
{\displaystyle c_{\mathrm {f} }={\frac {\langle \tau _{\mathrm {w} }\rangle }{{\frac {\rho }{2}}u_{\mathrm {b} }^{2}}}}
{\displaystyle z_{1}\approx 0.0036D\approx 10\mathrm {px} }
{\displaystyle z_{1}\approx 0.0005D}
{\displaystyle z-}
{\displaystyle c_{\mathrm {p} }}
{\displaystyle c_{\mathrm {p} }}
{\displaystyle c_{\mathrm {f} }}
{\displaystyle x_{\mathrm {adj} }=-1.0}
{\displaystyle c_{\mathrm {f} }}
{\displaystyle |c_{\mathrm {f} }|=0.01}
{\displaystyle c_{\mathrm {f} }}
{\displaystyle c_{\mathrm {f} }}
{\displaystyle c_{\mathrm {f} }}
{\displaystyle 50\times 171(n\times m)}
{\displaystyle 143\times 131(n\times m)}
{\displaystyle n\cdot m}
{\displaystyle x_{\mathrm {adj} }}
{\displaystyle {\frac {x}{D}}}
{\displaystyle {\frac {z}{D}}}
{\displaystyle {\frac {\langle u\rangle }{u_{\mathrm {b} }}}}
{\displaystyle -}
{\displaystyle {\frac {\langle w\rangle }{u_{\mathrm {b} }}}}
{\displaystyle {\frac {\langle u'u'\rangle }{u_{\mathrm {b} }^{2}}}}
{\displaystyle -}
{\displaystyle {\frac {\langle w'w'\rangle }{u_{\mathrm {b} }^{2}}}}
{\displaystyle -}
{\displaystyle {\frac {\langle u'w'\rangle }{u_{\mathrm {b} }^{2}}}}
{\displaystyle -}
{\displaystyle P{\frac {D}{u_{\mathrm {b} }^{3}}}}
{\displaystyle C{\frac {D}{u_{\mathrm {b} }^{3}}}}
{\displaystyle \nabla T_{\mathrm {turb} }{\frac {D}{u_{\mathrm {b} }^{3}}}}
{\displaystyle -}
{\displaystyle -}
{\displaystyle \epsilon {\frac {D}{u_{\mathrm {b} }^{3}}}}
{\displaystyle -}
{\displaystyle x_{\mathrm {adj} }}
{\displaystyle {\frac {x}{D}}}
{\displaystyle {\frac {z}{D}}}
{\displaystyle {\frac {\langle u\rangle }{u_{\mathrm {b} }}}}
{\displaystyle {\frac {\langle v\rangle }{u_{\mathrm {b} }}}}
{\displaystyle {\frac {\langle w\rangle }{u_{\mathrm {b} }}}}
{\displaystyle {\frac {\langle u'u'\rangle }{u_{\mathrm {b} }^{2}}}}
{\displaystyle {\frac {\langle v'v'\rangle }{u_{\mathrm {b} }^{2}}}}
{\displaystyle {\frac {\langle w'w'\rangle }{u_{\mathrm {b} }^{2}}}}
{\displaystyle {\frac {\langle u'v'\rangle }{u_{\mathrm {b} }^{2}}}}
{\displaystyle {\frac {\langle u'w'\rangle }{u_{\mathrm {b} }^{2}}}}
{\displaystyle {\frac {\langle v'w'\rangle }{u_{\mathrm {b} }^{2}}}}
{\displaystyle P{\frac {D}{u_{\mathrm {b} }^{3}}}}
{\displaystyle C{\frac {D}{u_{\mathrm {b} }^{3}}}}
{\displaystyle \nabla T_{\mathrm {turb} }{\frac {D}{u_{\mathrm {b} }^{3}}}}
{\displaystyle \nabla T_{\mathrm {press} }{\frac {D}{u_{\mathrm {b} }^{3}}}}
{\displaystyle \nabla T_{\mathrm {visc} }{\frac {D}{u_{\mathrm {b} }^{3}}}}
{\displaystyle \epsilon _{\mathrm {total} }{\frac {D}{u_{\mathrm {b} }^{3}}}}
{\displaystyle c_{\mathrm {p} }}
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f\left(x,y\right)=\frac{1}{3}{x}^{3}+{y}^{2}-2xy-6x-3y+4
a) Calculate
{f}_{xx},{f}_{yx},{f}_{xy}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{f}_{yy}
b) Calculate coordinates of stationary points.
c) Classify all stationary points.
F\left(x,y\right)=⟨x{y}^{2}+8x,{x}^{2}y-8y⟩
. Compute the flux
\oint F\cdot nds
of F across a simple closed curve that is the boundary of the half-disk given by
{x}^{2}+{y}^{2}\le 7,y\ge 0
using the vector form Greens
f=\left[{x}^{2}{y}^{2},-\frac{x}{{y}^{2}}\right]
R:1\le {x}^{2}+{y}^{2},+4,x\ge 0,y\ge x
{\int }_{C}F\left(r\right)\cdot dr
counterclockwise around the boundary C of the region R by Green's theorem.
Use the Divergence Theorem to calculate the surface integral
\int {\int }_{S}F·dS
, that is, calculate the flux of F across S.
F\left(x,y,z\right)=\left(\mathrm{cos}\left(z\right)+x{y}^{2}\right)i+x{e}^{-z}j+\left(\mathrm{sin}\left(y\right)+{x}^{2}z\right)k
S is the surface of the solid bounded by the paraboloid
z={x}^{2}+{y}^{2}
and the plane z = 9.
Green’s Theorem, flux form Consider the following regions R and vector fields F.
a. Compute the two-dimensional divergence of the vector field.
b. Evaluate both integrals in Green’s Theorem and check for consistency.
F=⟨x,y⟩,R=\left\{\left(x,y\right):{x}^{2}+{y}^{2}\le 4\right\}
{\oint }_{C}xydx+{x}^{2}dy
, where C is the path going counterclockwise around the boundary of the rectangle with corners (0,0),(2,0),(2,3), and (0,3). You can evaluate directly or use Greens
Comsider this, multivariable fucntion
f\left(x,y\right)=3xy-{x}^{2}+5{y}^{2}-25
a)What is the value of f(-1,3)?
b)Find all x-values such that f(x,x)=0
k\left(a,b\right)=3a{b}^{4}+8\left({1.4}^{b}\right)
\frac{\partial k}{\partial a}
\frac{\partial k}{\partial b}
\frac{\partial k}{\partial b}{\mid }_{a=3}
Consider this multivariable function.
f\left(x,y\right)=y{e}^{3x}+{y}^{2}
a) Find
{f}_{y}\left(x,y\right)
b) What is value of
{f}_{×}\left(0,3\right)
{\int }_{C}\stackrel{\to }{F}\cdot d\stackrel{\to }{r}
\stackrel{\to }{F}=⟨{y}^{3},-{x}^{3}⟩
{x}^{2}+{y}^{2}=3
Divergence Theorem for more general regions Use the Divergence Theorem to compute the net outward flux of the following vector fields across the boundary of the given regions D.
F=⟨z-x,x-y,2y-z⟩
, D is the region between the spheres of radius 2 and 4 centered at the origin.
Suppose that the plane region D, its boundary curve C, and the functions P and Q satisfy the hypothesis of Greens
a) Find the functions
use Green’s Theorem to find the counterclockwise circulation and outward flux for the field F and the curve C.
F=\left({y}^{2}-{x}^{2}\right)i+\left({x}^{2}+{y}^{2}\right)j
C: The triangle bounded by y = 0, x = 3, and y = x
What tis the complete domain D and range R of the following multivariable functions:
w\left(x,y\right)=\sqrt{y-4{x}^{2}}
Use Stokes' theorem to evaluate the line integral
{\oint }_{C}F\cdot dr
where A = -yi + xj and C is the boundary of the ellipse
\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1,z=0
f\left(x,y\right)=\frac{xy}{7}
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Plucks slices from an NDF at arbitrary positions
This application’s function is to extract data at scientifically relevant points such as the spatial location of a source or wavelength of a spectral feature, rather than at data sampling points (for which NDFCOPY is appropriate). This is achieved by the extraction of interpolated slices from an NDF . A slice is located at a supplied set of co-ordinates in the current WCS Frame for some but not all axes, and it possesses one fewer significant dimension per supplied co-ordinate. The slices run parallel to pixel axes of the NDF.
The interpolation uses one of a selection of resampling methods to effect the non-integer shifts along the fixed axes, applied to each output element along the retained axes (see the METHOD, PARAMS, and TOL parameters).
Three routes are available for obtaining the fixed positions, selected using Parameter MODE:
from the parameter system (see Parameter POS);
In the first mode the application loops, asking for new extraction co-ordinates until it is told to quit or encounters an error. However there is no looping if the position has been supplied on the command line.
Each extracted dataset is written to a new NDF, which however, may reside in a single container file (see the CONTAINER parameter).
pluck in axes out method [mode]
AXES( ) = _INTEGER (Read)
The WCS axis or axes to remain in the output NDF. The slice will therefore contain an array comprising all the elements along these axes. The maximum number of axes is one fewer than the number of WCS axes in the NDF.
Each axis can be specified using one of the following options.
A list of acceptable values is displayed if an illegal value is supplied. If the axes of the current Frame are not parallel to the NDF pixel axes, then the pixel axis which is most nearly parallel to the specified current Frame axis will be used.
Name of a text file containing the co-ordinates of slices to be plucked. It is only accessed if Parameter MODE is given the value "File". Each line should contain the formatted axis values for a single position, in the current Frame of the NDF. Axis values can be separated by spaces, tabs or commas. The file may contain comment lines with the first character # or !.
CONTAINER = _LOGICAL (Read)
If TRUE, each slice extracted is written as an NDF component of the HDS container file specified by the OUT parameter. The
n
th component will be named PLUCK_
n
. If set FALSE, each extraction is written to a separate file. On-the-fly format conversion to foreign formats is not possible when CONTAINER=TRUE. [FALSE]
If TRUE, a detailed description of the co-ordinate Frame in which the fixed co-ordinates are to be supplied is displayed before the positions themselves. It is ignored if MODE="Catalogue". [current value]
A catalogue containing a positions list giving the co-ordinates of the fixed positions, such as produced by applications CURSOR, LISTMAKE, etc. It is only accessed if Parameter MODE is given the value "Catalogue". The catalogue should have a WCS Frame common with the NDF, so that the NDF and catalogue FrameSets can be aligned.
"Interface" –- positions are obtained using Parameter POS.
"File" –- positions are obtained from a text file using Parameter COIN.
The NDF structure containing the data to be extracted. It must have at least two dimensions.
The method to use when sampling the input pixel values. For details of these schemes, see the descriptions of routine AST_RESAMPLEx in SUN/210. METHOD can take the following values.
"Linear" –- When resampling, the output pixel values are calculated by linear interpolation in the input NDF among the two nearest pixel values along each axis chosen by AXES. This method produces smoother output NDFs than the nearest-neighbour scheme, but is marginally slower.
\mathrm{\text{sinc}}\left(\pi x\right)
x
is the pixel offset from the interpolation point and
\mathrm{\text{sinc}}\left(z\right)=sin\left(z\right)/z
\mathrm{\text{sinc}}\left(\pi x\right)\mathrm{\text{sinc}}\left(k\pi x\right)
kernel. A valuable general-purpose scheme, intermediate in its visual effect on NDFs between the linear option and using the nearest neighbour.
\mathrm{\text{sinc}}\left(\pi x\right)cos\left(k\pi x\right)
\mathrm{\text{sinc}}\left(\pi x\right){e}^{-k{x}^{2}}
\mathrm{\text{somb}}\left(\pi x\right)
x
\mathrm{\text{somb}}\left(z\right)=2\ast {J}_{1}\left(z\right)/z
{J}_{1}
\mathrm{\text{somb}}\left(\pi x\right)cos\left(k\pi x\right)
N
-dimensional cube.
All methods propagate variances from input to output, but the variance estimates produced by interpolation schemes need to be treated with care since the spatial smoothing produced by these methods introduces correlations variance estimates. The initial default is "SincSinc". [current value]
The name for the output NDF, or the name of the single container file if CONTAINER=TRUE.
PARAMS(1) is required by all the above schemes. It is used to specify how many pixels are to contribute to the interpolated result on either side of the interpolation in each dimension. Typically, a value of 2 is appropriate and the minimum allowed value is 1 (i.e. one pixel on each side). A value of zero or fewer indicates that a suitable number of pixels should be calculated automatically. [0]
PARAMS(2) is required only by the SombCos, Gauss, SincSinc, SincCos, and SincGauss schemes. For the SombCos, SincSinc, and SincCos schemes, it specifies the number of pixels at which the envelope of the function goes to zero. The minimum value is 1.0, and the run-time default value is 2.0. For the Gauss and SincGauss schemes, it specifies the full-width at half-maximum (FWHM) of the Gaussian envelope. The minimum value is 0.1, and the run-time default is 1.0. On astronomical images and spectra, good results are often obtained by approximately matching the FWHM of the envelope function, given by PARAMS(2), to the point-spread function of the input data. []
POS( ) = LITERAL (Read)
An the co-ordinates of the next slice to be extracted, in the current co-ordinate Frame of the NDF (supplying a colon ":" will display details of the current co-ordinate Frame). The position should be supplied as a list of formatted axis values separated by spaces or commas. POS is only accessed if Parameter MODE is given the value "Interface". If the co-ordinates are supplied on the command line only one slice will be extracted; otherwise the application will ask for further positions which may be terminated by supplying the null value (!).
A Title for every output NDF structure. A null value (!) propagates the title from the input NDF to all output NDFs. [!]
The maximum tolerable geometrical distortion that may be introduced as a result of approximating non-linear Mappings by a set of piece-wise linear transforms. Both algorithms approximate non-linear co-ordinate transformations in order to improve performance, and this parameter controls how inaccurate the resulting approximation is allowed to be, as a displacement in pixels of the input NDF. A value of zero will ensure that no such approximation is done, at the expense of increasing execution time. [0.05]
pluck omc1 pos="5:35:13.7,-5:22:13.6" axes=FREQ method=sincgauss params=[3,5] out=omc1_trap
The NDF omc1 is a spectral-imaging cube with (Right ascension, declination, frequency) World Co-ordinate axes. This example extracts a spectrum at RA=5h35m13.7s , Dec=−5°22′13.6′′ using the SincGauss interpolation method. Three pixels either side of the point are used to interpolate, the full-width half-maximum of the Gaussian is five pixels. The resultant spectrum called omc1_trap, is still a cube, but its spatial dimensions each only have one element.
pluck omc1 mode=cat incat=a axes=FREQ container out=omc1_spectra
This example reads the fixed positions from the positions list in file a.FIT. The selected spectra are stored in an HDS container file called omc1_spectra.sdf.
pluck omc1 mode=cat incat=a axes=SPEC container out=omc1_spectra
As the previous example, plucking spectra, this time by selecting the generic spectral axis.
pluck omc1 pos=3.45732E11 axes="RA,Dec" method=lin out=peakplane
This example extracts a plane from omc1 at frequency 3.45732E11 Hz using linear interpolation and stores it in NDF peakplane.
In Interface or File modes all positions should be supplied in the current co-ordinate Frame of the NDF. A description of the co-ordinate Frame being used is given if Parameter DESCRIBE is set to a TRUE value. Application WCSFRAME can be used to change the current co-ordinate Frame of the NDF before running this application if required.
The output NDF has the same dimensionality as the input NDF, although the axes with fixed co-ordinates (those not specified by the AXES parameter) are degenerate, having bounds of 1:1. The retention of these insignificant axes enables the co-ordinates of where the slice originated to be recorded. Such fixed co-ordinates may be examined with say NDFTRACE. NDFCOPY may be used to trim the degenerate axes if their presence prevents some old non-KAPPA tasks from operating.
In Catalogue or File modes the table file need only contain columns supplying the fixed positions. In this case the co-ordinates along the retained axes are deemed to be independent, that is they do not affect the shifts required of the other axes. In practice this assumption only affects File mode, as catalogues made with CURSOR or LISTMAKE will contain WCS information.
In Interface mode representaive co-ordinates along retained axes are the midpoints of the bounds of an array that would contain the resampled copy of the whole input array.
KAPPA: NDFCOPY, REGRID.
The LABEL, UNITS, and HISTORY components, and all extensions are propagated. TITLE is controlled by the TITLE parameter. DATA, VARIANCE, AXIS, and WCS are propagated after appropriate modification. The QUALITY component is not propagated.
The processing of bad pixels and automatic quality masking are supported.
The minimum number of dimensions in the input NDF is two.
Processing a group of input NDFs is not supported unless CONTAINER=TRUE or when only one output NDF is created per input file.
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Use the Laplace transform to solve the following
Erika Bernard 2022-04-08 Answered
Use the Laplace transform to solve the following initial value problem:
y{}^{″}+y=2t
y\left(\frac{\pi }{4}\right)=\frac{\pi }{2}
{y}^{\prime }\left(\frac{\pi }{4}\right)=2-\sqrt{2}
llevochalecoiozq
Laplace-transforming both sides of
y{}^{″}+y=2t
{s}^{2}Y\left(s\right)-{y}_{0}s-{v}_{0}+Y\left(s\right)=\frac{2}{{s}^{2}}
Y\left(s\right)={y}_{0}\left(\frac{s}{{s}^{2}+1}\right)+\left({v}_{0}-2\right)\left(\frac{1}{{s}^{2}+1}\right)+\frac{2}{{s}^{2}}
Taking the inverse Laplace transform,
y\left(t\right)={y}_{0}\mathrm{cos}\left(t\right)+\left({v}_{0}-2\right)\mathrm{sin}\left(t\right)+2t
From the conditions
y\left(\frac{\pi }{4}\right)=\frac{\pi }{2}
{y}^{\prime }\left(\frac{\pi }{4}\right)=2-\sqrt{2}
{y}_{0}+{v}_{0}=2\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}{y}_{0}-{v}_{0}=2
{y}_{0}=2
{v}_{0}=0
y\left(t\right)=2\mathrm{cos}\left(t\right)-2\mathrm{sin}\left(t\right)+2t
Find the orthogonal trajectory.
{y}^{2}=k{x}^{3}
Find a general solution
y\left(x\right)
to the differential equation
{y}^{\prime }=\frac{y}{x}+2{x}^{2}
y\left(t\right)={\int }_{0}^{t}f\left(t\right)dt
If the Laplace transform of y(t) is given
Y\left(s\right)=\frac{19}{\left({s}^{2}+25\right)}
, find f(t)
f\left(t\right)=19\mathrm{sin}\left(5t\right)
f\left(t\right)=6\mathrm{sin}\left(2t\right)
f\left(t\right)=20\mathrm{cos}\left(6t\right)
f\left(t\right)=19\mathrm{cos}\left(5t\right)
A question asks us to solve the differential equation
-u\left(x\right)=\delta \left(x\right)
u\left(-2\right)=0\text{ }\text{and}\text{ }u\left(3\right)=0
\delta \left(x\right)
is the Dirac delta function. But inside the same question, teacher gives the solution in two pieces as
u=A\left(x+2\right)\text{ }\text{for}\text{ }x\le 0\text{ }\text{and}\text{ }u=B\left(x-3\right)\text{ }\text{for}\text{ }x\ge 0
. I understand when we integrate the delta function twice the result is the ramp function R(x). However elsewhere in his lecture the teacher had given the general solution of that DE as
u\left(x\right)=-R\left(x\right)+C+Dx
Y\left(s\right)=\frac{{e}^{-s}}{s\left(2s-1\right)}
use the inverse form of the First translation theorem to find the following laplace transform
{L}^{-1}\left\{\frac{s}{{s}^{2}+4s+5}\right\}
Solve the question with laplace transformation
{y}^{″}+2{y}^{\prime }+y=4
y\left(0\right)=3
{y}^{\prime }\left(0\right)=0
|
\left\{\begin{array}{l}y={x}^{2}\\ {z}^{2}=16-y\end{array}\right\
\left\{\begin{array}{l}4x=y\\ z=0\end{array}\right\
\left\{\begin{array}{l}x=4\\ y=16\end{array}\right\
\left\{\begin{array}{l}y-x=12\\ z=0\end{array}\right\
\left\{\begin{array}{l}x+y=20\\ z=0\end{array}\right\
Definition of polynomial ring
Given a ring R, the polynomial ring is defined as
R\left[x\right]\phantom{\rule{0.222em}{0ex}}=\left\{\sum _{k=0}^{n}{a}_{k}{x}^{k}:n\ge 0,\text{ }{a}_{k}\in R\text{ }\text{for }\text{ }k\in \left\{0,1,\cdots ,n\right\}\right\}.
However, it is not usually specified what x is. In order for multiplication to make sense, I guess it has to be an element in R at least. But is R[x] the set of all functions
P:R\to R,\text{given by }\text{ }x↦\sum _{k=0}^{n}{a}_{k}{x}^{k}
, or the set of those functions evaluated at x?
Sometimes R is required to be commutative. Does that make any difference for R[x]?
A partly filled barrel 300 lb water and 100 lb of ice at 32ºF. How many pounds of steam at 212ºF must be run into the barrel to bring its contents up to 80ºF?
A population of values has a normal distribution with \mu =121.3
and \sigma =5.8
. You intend to draw a random sample of size n=149
P(119.8 < X < 121.2)
Find the probability that a sample of size n=149
is randomly selected with a mean between 119.8 and 121.2.
P(119.8 < M < 121.2)
Enter your answers as numbers accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.
Let L : P2
\to
P3 be a linear transformation for which we
know that L(1) = 1, L(t) = t 2, L(t 2) = t 3 + t.
(a) Find L(2t 2 - 5t + 3). (b) Find L(at 2 + bt + c).
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 121.1-cm and a standard deviation of 2.2-cm. For shipment, 26 steel rods are bundled together.
Find P91, which is the average length separating the smallest 91% bundles from the largest 9% bundles.
P91 = -cm
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Albert Byrd 2022-04-10 Answered
G{L}_{2}\frac{\mathbb{R}}{S}{L}_{2}\left(\mathbb{R}\right)
G{L}_{2}\left(\mathbb{R}\right)
2×2
S{L}_{2}\left(\mathbb{R}\right)
2×2
S{L}_{2}\left(\mathbb{R}\right)
G{L}_{2}\left(\mathbb{R}\right)
|G{L}_{2}\frac{\mathbb{R}}{S}{L}_{2}\left(\mathbb{R}\right)|=\frac{|G{L}_{2}\left(\mathbb{R}\right)|}{|S{L}_{2}\left(\mathbb{R}\right)|},
Ramiro Grant 2022-04-09 Answered
{M}_{2×2}\left(\mathbb{C}\right)
Deangelo Hardy 2022-04-08 Answered
I was doing some practice abstract algebra questions off the internet since I have a quiz coming up soon. However, I am not very skilled at abstract algebra. In fact, I did very average in my group theory class, so I am struggling in my ring theory one. Can someone please help explain what is happening in this proof? I'm very sorry if it's extremely straightforward, I just think I need some time to get used to the way of thinking that's required to solve these questions.
{R}_{1}\text{ }\text{and}\text{ }{R}_{2}
be commutative rings with identities and let
R={R}_{1}×{R}_{2}
. The question asks to show that every ideal I of R is of the form
I={I}_{1}×{I}_{2}
{I}_{1}
an ideal of
{R}_{1}
{I}_{2}
{R}_{2}
Breanna Fisher 2022-04-08 Answered
Understanding a proof of: In a finite group, the number of elements of ' order p is divisible by
p-1
Fields with involution whose fixed field is ordered?
Let K be a field with an involution
\ast
\cdot :K\to K
is an automorphism and
\left(x\ast \right)\ast =x
\xi nK
. Suppose further that the fixed field of
\ast
is ordered (i.e., it can be given an ordering that is compatible with the field structure).
Makayla Stevens 2022-04-04 Answered
Euler's remarkable '-producing polynomial and quadratic UFDs
Example of a polynomial which produces a finite number of 's is:
{x}^{2}+x+41
which produces 's for every integer
0\le x\le 39
Janiyah Hays 2022-04-03 Answered
Examples for when the quotient ring is necessarily / not necessarily an extension of the residue field.
Let R be an integral domain with unique maximal ideal
\mathfrak{m}
F:=Frac\left(R\right)
be the field of fraction of R.
burubukuamaw 2022-04-02 Answered
Show if M is free of rank n as R-module, then
\frac{M}{IM}
is free of rank n as
\frac{R}{I}
Let R be a ring and
I\subset R
a two-sided ideal and M an R-module with
IM=\left\{\sum {r}_{i}{x}_{i}\mid {r}_{i}\in I,{x}_{i}\in M\right\}.
Oxinailelpels3t14 2022-04-02 Answered
Residue of x in polynomial ring
\mathbb{Z}\frac{x}{f}
When we say that α is the residue of x in
\mathbb{Z}\frac{x}{f}
f={x}^{4}+{x}^{3}+{x}^{2}+x
, wouldn't
\alpha
just be x? Because if we divide with the remainder, we would get
x=0\text{ }f+x
Miley Caldwell 2022-04-01 Answered
\frac{{z}^{11}-1}{z-1}={z}^{10}+{z}^{9}+{z}^{8}+{z}^{7}+{z}^{6}+{z}^{5}+{z}^{4}+{z}^{3}+{z}^{2}+z+1
{\zeta }^{1},{\zeta }^{2},\dots ,{\zeta }^{10}
\begin{array}{rl}{A}_{0}& ={x}_{1}+{x}_{2}+{x}_{3}+{x}_{4}+{x}_{5}\\ {A}_{1}& ={x}_{1}+\zeta {x}_{2}+{\zeta }^{2}{x}_{3}+{\zeta }^{3}{x}_{4}+{\zeta }^{4}{x}_{5}\\ {A}_{2}& ={x}_{1}+{\zeta }^{2}{x}_{2}+{\zeta }^{4}{x}_{3}+\zeta {x}_{4}+{\zeta }^{3}{x}_{5}\\ {A}_{3}& ={x}_{1}+{\zeta }^{3}{x}_{2}+\zeta {x}_{3}+{\zeta }^{4}{x}_{4}+{\zeta }^{2}{x}_{5}\\ {A}_{4}& ={x}_{1}+{\zeta }^{4}{x}_{2}+{\zeta }^{3}{x}_{3}+{\zeta }^{2}{x}_{4}+\zeta {x}_{5}\end{array}
{A}_{0},\dots ,{A}_{4}
{x}_{1},\dots ,{x}_{5}
{A}_{0}
\tau
{A}_{j}
{\zeta }^{-j}{A}_{j}
{A}_{j}^{5}
{A}_{j}^{5}
{A}_{j}^{5}
\zeta
\zeta
{A}_{1}^{5}
{A}_{1}
{A}_{j}
{A}_{1}
Octavio Chen 2022-03-31 Answered
{\mathrm{End}}_{\mathrm{ℝ}\left[x\right]}\left(M\right)
M=\frac{\mathbb{R}\left[x\right]}{\left({x}^{2}+1\right)}
\mathbb{R}\left[x\right]
Prove if
F\left(\sqrt[n]{a}\right)
is unramified or totally ramified in certain conditions
Rex Maxwell 2022-03-30 Answered
\frac{{\mathbb{F}}_{2}\left[X,Y\right]}{\left({Y}^{2}+Y+1,{X}^{2}+X+Y\right)}
\frac{\left({\mathbb{F}}_{2}\left[Y\right]\right\}\left\{\left({Y}^{2}+Y+1\right)\right\}\frac{\left\{\right)}{X}}{\left({X}^{2}+X+\stackrel{―}{Y}\right)}
are isomorphic
Aleah Choi 2022-03-29 Answered
I have to prove that if P is a R-module , P is projective
⇔
there is a family
\left\{{x}_{i}\right\}
in P and morphisms
{f}_{i}:P\to R
x\in P
x=\sum _{i\in I}{f}_{i}\left(x\right){x}_{i}
where for each
x\in P,\text{ }{f}_{i}\left(x\right)=0
i\in I
Jasper Dougherty 2022-03-29 Answered
\frac{\mathbb{Z}}{10}\mathbb{Z}
\left\{\stackrel{―}{1},\stackrel{―}{3},\stackrel{―}{7},\stackrel{―}{9}\right\}
{7}^{2}=49\equiv 9±\mathrm{mod}10
r1fa8dy5 2022-03-28 Answered
Proving the generator of
A=\left\{154a+210b:a,b\in \mathbb{Z}\right\}
\left(154,\text{ }210\right)
Hugh Soto 2022-03-28 Answered
n\mid \varphi \left({a}^{n}-1\right)
Aut\left(G\right)
Marzadri9lyy 2022-03-28 Answered
\mathbb{Z}+\left(3x\right)
\mathbb{Z}\left[x\right]
and there is no surjective homomorphism from
\mathbb{Z}\left[x\right]\to \mathbb{Z}+\left(3x\right)
anadyrskia0g5 2022-03-27 Answered
R\left[x\right]\phantom{\rule{0.222em}{0ex}}=\left\{\sum _{k=0}^{n}{a}_{k}{x}^{k}:n\ge 0,\text{ }{a}_{k}\in R\text{ }\text{for }\text{ }k\in \left\{0,1,\cdots ,n\right\}\right\}.
P:R\to R,\text{given by }\text{ }x↦\sum _{k=0}^{n}{a}_{k}{x}^{k}
Asher Olsen 2022-03-25 Answered
{x}^{6}-1
{\left(x+1\right)}^{3}{\left(x+2\right)}^{3}
|
20 blue balls and 11 yellow, drawing 6 times with no replacement, what is the chance that at least o
Giancarlo Shah
OR in mathematics is the inclusive or, so if either condition or both is satisfied the sentence is satisfied. OR in English is ambiguous, it can be inclusive or exclusive. Given the problem, I would take it as inclusive here. As you say, that makes the probability 1. Either the first two balls are blue and hence the same color or at least one of them is yellow. We win either way.
P\left(2\right)+P\left(5\right)
P\left(2\right)=1/6,P\left(5\right)=1/6
\frac{1}{2450}
\frac{1}{50}×\frac{1}{49}
{A}_{14}
{A}_{23}
Pr\left(\text{ (}{A}_{14}\text{ first and }{C}_{23}\text{ second) or (}{C}_{23}\text{ first and }{A}_{14}\text{ second}\right)\right)\phantom{\rule{0ex}{0ex}}=Pr\left({A}_{14}\right)×Pr\left({C}_{23}\mid {A}_{14}\right)+Pr\left({C}_{23}\right)×Pr\left({A}_{14}\mid {C}_{23}\right)=\frac{1}{50}×\frac{1}{49}×2.
\left\{{A}_{14}{C}_{23}\text{ },\text{ }{C}_{23}{A}_{14}\right\}
3-2-1
3\oplus 2\oplus 1
P\left(A\cup \left(B\cap C\right)\right)
P\left(A\cap \left(B\cup C\right)\right)=P\left(A\cap B\right)+P\left(A\cap C\right).
P\left(A\cup \left(B\cap C\right)\right)=P\left(A\cup B\right)+P\left(A\cup C\right)?
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Continuous Wavelet Transform as a Bandpass Filter - MATLAB & Simulink - MathWorks India
CWT as a Filtering Technique
DFT-Based Continuous Wavelet Transform
The continuous wavelet transform (CWT) computes the inner product of a signal,
f\left(t\right)
, with translated and dilated versions of an analyzing wavelet,
\psi \left(t\right).
The definition of the CWT is:
C\left(a,b;f\left(t\right),\psi \left(t\right)\right)={\int }_{-\infty }^{\infty }f\left(t\right)\frac{1}{a}{\psi }^{*}\left(\frac{t-b}{a}\right)dt
You can also interpret the CWT as a frequency-based filtering of the signal by rewriting the CWT as an inverse Fourier transform.
C\left(a,b;f\left(t\right),\psi \left(t\right)\right)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }\stackrel{^}{f}\left(\omega \right)\overline{\stackrel{^}{\psi }}\left(a\omega \right){e}^{i\omega b}d\omega
\stackrel{^}{f}\left(\omega \right)
\stackrel{^}{\psi }\left(\omega \right)
are the Fourier transforms of the signal and the wavelet.
From the preceding equations, you can see that stretching a wavelet in time causes its support in the frequency domain to shrink. In addition to shrinking the frequency support, the center frequency of the wavelet shifts toward lower frequencies. The following figure demonstrates this effect for a hypothetical wavelet and scale (dilation) factors of 1,2, and 4.
This depicts the CWT as a bandpass filtering of the input signal. CWT coefficients at lower scales represent energy in the input signal at higher frequencies, while CWT coefficients at higher scales represent energy in the input signal at lower frequencies. However, unlike Fourier bandpass filtering, the width of the bandpass filter in the CWT is inversely proportional to scale. The width of the CWT filters decreases with increasing scale. This follows from the uncertainty relationships between the time and frequency support of a signal: the broader the support of a signal in time, the narrower its support in frequency. The converse relationship also holds.
In the wavelet transform, the scale, or dilation operation is defined to preserve energy. To preserve energy while shrinking the frequency support requires that the peak energy level increases. The implementation of cwt in Wavelet Toolbox™ uses L1 normalization. The quality factor, or Q factor of a filter is the ratio of its peak energy to bandwidth. Because shrinking or stretching the frequency support of a wavelet results in commensurate increases or decreases in its peak energy, wavelets are often referred to as constant-Q filters.
The equation in the preceding section defined the CWT as the inverse Fourier transform of a product of Fourier transforms.
C\left(a,b;f\left(t\right),\psi \left(t\right)\right)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }\stackrel{\wedge }{f}\left(\omega \right){\stackrel{^}{\psi }}^{*}\left(a\omega \right){e}^{j\omega b}d\omega
The time variable in the inverse Fourier transform is the translation parameter, b.
This suggests that you can compute the CWT with the inverse Fourier transform. Because there are efficient algorithms for the computation of the discrete Fourier transform and its inverse, you can often achieve considerable savings by using fft and ifft when possible.
To obtain a picture of the CWT in the Fourier domain, start with the definition of the wavelet transform:
<f\left(t\right),{\psi }_{a,b}\left(t\right)>=\frac{1}{a}{\int }_{-\infty }^{\infty }f\left(t\right){\psi }^{*}\left(\frac{t-b}{a}\right)dt
If you define:
{\stackrel{˜}{\psi }}_{a}\left(t\right)=\frac{1}{a}{\psi }^{*}\left(-t/a\right)
you can rewrite the wavelet transform as
\left(f\ast {\stackrel{˜}{\psi }}_{a}\right)\left(b\right)={\int }_{-\infty }^{\infty }f\left(t\right){\stackrel{˜}{\psi }}_{a}\left(b-t\right)dt
which explicitly expresses the CWT as a convolution.
To implement the discretized version of the CWT, assume that the input sequence is a length N vector, x[n]. The discrete version of the preceding convolution is:
{W}_{a}\left[b\right]=\sum _{n=0}^{N-1}x\left[n\right]\text{ }\text{ }{\stackrel{˜}{\psi }}_{a}\left[b-n\right]
To obtain the CWT, it appears you have to compute the convolution for each value of the shift parameter, b, and repeat this process for each scale, a.
However, if the two sequences are circularly-extended (periodized to length N), you can express the circular convolution as a product of discrete Fourier transforms. The CWT is the inverse Fourier transform of the product
{W}_{a}\left(b\right)=\frac{1}{N}\sqrt{\frac{2\pi }{\Delta t}}\sum _{k=0}^{N-1}\stackrel{\wedge }{X}\left(2\pi k/N\Delta t\right)\stackrel{\wedge }{\psi }*\left(a2\pi k/N\Delta t\right){e}^{j2\pi kb/N}
where Δt is the sampling interval (period).
Expressing the CWT as an inverse Fourier transform enables you to use the computationally-efficient fft and ifft algorithms to reduce the cost of computing convolutions.
The cwt function implements the CWT.
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Sets (ADT) | Brilliant Math & Science Wiki
Alex Chumbley, Christopher Williams, and Raül Pérez contributed
Sets are a type of abstract data type that allows you to store a list of non-repeated values. Their name derives from the mathematical concept of finite sets.
Unlike an array, sets are unordered and unindexed. You can think about sets as a room full of people you know. They can move around the room, changing order, without altering the set of people in that room. Plus, there are no duplicate people (unless you know someone who has cloned themselves). These are the two properties of a set: the data is unordered and it is not duplicated.
Sets have the most impact in mathematical set theory. These theories are used in many kinds of proofs, structures, and abstract algebra. Creating relations from different sets and codomains are also an important applications of sets.
In computer science, set theory is useful if you need to collect data and do not care about their multiplcity or their order. As we've seen on this page, hash tables and sets are very related. In databases, especially for relational databases, sets are very useful. There are many commands that finds unions, intersections, and differences of different tables and sets of data.
Sample Python Implementation Using a Dictionary
Example Problems with Sets
Built-in Set Functionality in Python
The set has four basic operations.
insert(i) Adds i to the set
remove(i) Removes i from the set
size() Returns the size of the set
contains(i) Returns whether or not the set contains i
Sometimes, operations are implemented that allow interactions between two sets
union(S, T) Returns the union of set S and set T
intersection(S, T) Returns the intersection of set S and set T
difference(S, T) Returns the difference of set S and set T
subset(S, T) Returns whether or not set S is a subset of set T
The only way to adequately implement a set in python is by using a dictionary, or a hash table. This is because the dictionary is the only primitive data structure whose elements are unordered. This require a few more lines of code, but it keeps the sets' principles intact.
Note: Python also has its own Set primitive, but we want to implement our own to show how it works
if len(data) != len(set(data)):
data = set(data)
self.data[d] = d
if i in self.data.keys():
return 'Already in set'
self.data[i] = i
if i not in self.data.keys():
return 'Not in set'
def contains(self, i):
def union(self, otherSet):
setData = list(set(self.data.keys()) | set(otherSet.data.keys()))
unionSet = Set(setData)
def intersection(self, otherSet):
setData = list(set(self.data.keys()) & set(otherSet.data.keys()))
intersectionSet = Set(setData)
def difference(self, otherSet):
setData = list(set(self.data.keys()) ^ set(otherSet.data.keys()))
differenceSet = Set(setData)
def subset(self, otherSet):
if set(self.data.keys()) < set(otherSet.data.keys()):
return '['+', '.join(str(x) for x in self.data.keys())+']'
This code is a little more complex than other data types like the associative array, so let's break it down.
On line 2, we instantiate our class using a python dictionary as our data structure. For more information on dictionaries, check out this page. We use a dictionary to keep the set unordered (dictionaries are unordered because they are implemented using a hash table ).
The functions insert, remove, size, and contains all simply search through the dictionary's list of keys to determine how to proceed with the operation.
The functions union, intersection, difference, and subset all utilize difference pythonic set operators. For example, in the union function, we first turn both Set's data into a set(). Then we say we'll take data values in either set and put them in a new set. We do that using the OR operator. This looks like a |. Then, we change the set() into a list() because that is what our class takes as an argument. Then we create a new instance of our Set, and we return it. There are many other operators such as & and <. If you'd like more information on these, you can check out this page.
We created our own version of the __repr__ function on line 39 so that we can have easy readability for our class. Remember that this class uses a dictionary as its data structure, so when we print it our normally, it would look something like this: {1: 1, 2: 2, 3: 3}. Instead, we just it to print something that looks like a list, so we tell it to only print the keys of this dictonary. This results in something that looks like this [1, 2, 3].
Let's take the same implementation of a Set we had above. Try answering the following questions:
1) What will be the output of the following?
>>> set1 = Set([1, 2, 3, 4, 1, 4])
This set was instantiated with multiple values of 1 and 4. So, those will be removed
>>> set1.union(set2)
We're taking the union of both sets, so we want each element from both sets, but we exclude duplicate because we're using sets.
To take the intersection from both sets, we want to return one copy of each element that exist in both sets.
>>> set1.subset(set2)
We're asking if set1 is a subset of set2. Set1 has the value 1 in it, and set2 doesn't. So, it is false.
Now let's look at the built-in set primitive in Python. This is not the same implementation that we used above. This is data structure provided to you by Python itself. It is very similar to our implementation, but it is far more efficient and has many more functions.
Here are the basic functions:
>>> set1 = set(['Alex', 'Brittany', 'Joyce'])
>>> set1.add('John')
set(['John', 'Alex', 'Brittany', 'Joyce'])
>>> set1.remove('Alex')
set(['John', 'Brittany', 'Joyce'])
Here are functions that relate sets to one another:
>>> odds = set([1, 3, 5, 7, 9])
>>> evens = set([0, 2, 4, 6, 8])
>>> odds | evens #Performs a union
>>> odds & evens #Performs an intersection
>>> evens - odds #Performs a difference
There are many more operations that Python's set can use. For more, check out this page.
As with all abstract data types, there are many ways to implement a Set. A hash table is the most common data structure, so we'll examine its time complexity below:
\begin{array}{c}&&\text{Hash Table Insert- O(1),} &\text{Hash Table Remove - O(1),} \\ &\text{Hash Table Contains - O(1)}\\ \end{array}
Unlike the associative array, we don't have to worry about how we handle collisions in this hash table implementation. This is because we don't handle them ever! If there is a collision, all that means is that value is already there and we don't have to do anything! This makes the hash table implementation of the set the typical choice of programmers.
Cite as: Sets (ADT). Brilliant.org. Retrieved from https://brilliant.org/wiki/sets-adt/
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\frac{\mathbb{F}_2[X,Y]}{(Y^2 + Y + 1,X^2 + X + Y )}
Benjamin Hampton 2022-05-02 Answered
\frac{{\mathbb{F}}_{2}\left[X,Y\right]}{\left({Y}^{2}+Y+1,{X}^{2}+X+Y\right)}
\frac{\left({\mathbb{F}}_{2}\left[Y\right]}{\left({Y}^{2}+Y+1\right)}\frac{\right)\left[X\right]}{\left({X}^{2}+X+\overline{Y}\right)}
narratz5dz
\varphi :{\mathbb{F}}_{2}\left[X,Y\right]\to \left(\frac{{\mathbb{F}}_{2}\left[Y\right]}{\left({Y}^{2}+Y+1\right)}\right)\frac{\left[X\right]}{\left({X}^{2}+X+\stackrel{―}{Y}\right)}
be the morphism of rings which send X and
\stackrel{―}{X}
and Y to
\stackrel{―}{Y}
. It is surjective and, since
\varphi \left({Y}^{2}+Y+1\right)=\varphi \left({X}^{2}+X+Y\right)=0
it induces a surjective ring morphism :
\stackrel{―}{\varphi }:{\mathbb{F}}_{2}\frac{X,Y}{{Y}^{2}+Y+1,{X}^{2}+X+Y}
\to \left({\mathbb{F}}_{2}\frac{Y}{{Y}^{2}+Y+1}\right)\frac{X}{{X}^{2}+X+\stackrel{―}{Y}}
\stackrel{―}{\varphi }\left(P\right)=0
P\left(\stackrel{―}{X},\stackrel{―}{Y}\right)=0
P\left(X,\stackrel{―}{Y}\right)\in \left\{\left({X}^{2}+X+\stackrel{―}{Y}\right)
P\left(X,Y\right)\in \left({X}^{2}+X+Y,{Y}^{2}+Y+1\right)
\frac{{\mathbb{F}}_{2}\left[X,Y\right]}{\left({Y}^{2}+Y+1,{X}^{2}+X+Y\right)}
\stackrel{―}{\varphi }
Fitting a ballistic trajectory to noisy data where both spacial and temporal domains observations are noisy
Fitting a curve to noisy data is somewhat trivial. However it generally assumes that data abscissa is fixed, and the error is computed on the ordinate.
In my setup, I have 3D spacial observations of ballistic trajectories (that I model with a simple parabola), but the observations time are also noisy.
Therefore, I have to estimate the initial position
{y}_{0},{y}_{0},{z}_{0}
{v}_{{x}_{0}},{v}_{{y}_{0}},{v}_{{z}_{0}}
, based on 4D (noisy) observations
\left({X}_{i},{Y}_{i},{Z}_{i},{T}_{i}\right),i\in \left[0,N\right]
, such that they fit the following model:
\left\{\begin{array}{rl}x\left(t\right)& ={x}_{0}+{v}_{{x}_{0}}t\\ y\left(t\right)& ={y}_{0}+{v}_{{y}_{0}}t\\ z\left(t\right)& ={z}_{0}+{v}_{{z}_{0}}t-\frac{g}{2}{t}^{2}\end{array}
with t monotonically increasing with i.
I'm not sure how to formulate such optimization problem because I have 6 parameters to estimate, but also 4N variables with only 3N equations… My intuition tells me there's only one single parabola that minimizes the error (MSE for example), but I can't formulate the problem.
. The average credit card debt for a recent year was $9205. Five years earlier the average credit card debt was $6618. Assume sample sizes of 35 were used and the population standard deviations of both samples were $1928. Find the 95% confidence interval of the difference in means
\left(0,0\right)
f\left(0.01,-0.02\right)
\forall x \in \mathrm{ℝ}, x<2 ⇒{x}^{2}<4
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Simulation of Turbulent Lifted Flames Using a Partially Premixed Coherent Flame Model | J. Eng. Gas Turbines Power | ASME Digital Collection
Yongzhe Zhang,
, 60 Broadhollow Road, Melville, NY 11747
Zhang, Y., and Rawat, R. (February 10, 2009). "Simulation of Turbulent Lifted Flames Using a Partially Premixed Coherent Flame Model." ASME. J. Eng. Gas Turbines Power. May 2009; 131(3): 031505. https://doi.org/10.1115/1.3026559
Partially premixed combustion occurs in many combustion devices of practical interest, such as gas-turbine combustors. Development of corresponding turbulent combustion models is important to improve the design of these systems in efforts to reduce fuel consumption and pollutant emissions. Turbulent lifted flames have been a canonical problem for testing models designed for partially premixed turbulent combustion. In this paper we propose modifications to the coherent flame model so that it can be brought to the simulation of partially premixed combustion. For the primary premixed flame, a transport equation for flame area density is solved in which the wrinkling effects of the flame stretch and flame annihilation are considered. For the subsequent nonpremixed zone, a laminar flamelet presumed probability density function (PPDF) methodology, which accounts for the nonequilibrium and finite-rate chemistry effects, is adopted. The model is validated against the experimental data on a lifted
H2∕N2
jet flame issuing into a vitiated coflow. In general there is fairly good agreement between the calculations and measurements both in profile shapes and peak values. Based on the simulation results, the flame stabilization mechanism for lifted flames is investigated.
combustion, flames, jets, laminar flow, turbulence
Flames, Turbulence, Combustion, Simulation, Fuels
The Stabilization Mechanism of Lifted Diffusion Flames
Experimental Investigation on the Stabilization Mechanism of Jet Diffusion Flames
Liftoff Characteristics of Turbulent Jet Diffusion Flames
Turbulent Flame Propagation in Partially Premixed Ames
Numerical Simulation of Turbulent PropaneÂcair Combustion With Nonhomogeneous Reactants
A Combustion Model for Premixed Flames With Varying Stoichiometry
The Evolution of Surfaces in Turbulence
The Evolution Equation for the Flame Surface Density
,” Center for Turbulence Research Annual Research Briefs, pp.
A Comparison of Flamelet Models for Premixed Turbulent Combustion
Development and Validation of a Coherent Flamelet Model for a Spark-Ignited Turbulent Premixed Flame in a Closed Vessel
, http://me.berkeley.edu/cal/VCB/Data/http://me.berkeley.edu/cal/VCB/Data/.
Simultaneous Laser Raman-Rayleigh-Lif Measurements and Numerical Modeling Results of a Lifted Turbulent H2∕N2 Jet Flame in a Vitiated Coflow
CD-adapco, http://www.cd-adapco.com/http://www.cd-adapco.com/
DARS, http://www.DigAnaRS.com/http://www.DigAnaRS.com/
Entwicklung eines kinetischen Modells der Russbildung mit schneller Polymerisation
PDF Calculations of Turbulent Lifted Flames of H2∕N2 Issuing Into a Vitiated Co-Flow
An Evaluation of Flame Surface Density Models for Turbulent Premixed Jet Flames
Modeling of Inhomogeneously Premixed Combustion With an Extended TFC Model
|
(1-x^2)y''-xy'+m^2y=0, m is a constant. I have to show that
Sydney Stanley 2022-05-02 Answered
\left(1-{x}^{2}\right)y{}^{″}-x{y}^{\prime }+{m}^{2}y=0
, m is a constant. I have to show that
m\in \mathbb{Z}⇒deg\left(y\right)=m
such that y is a solution of the ode.
Note that you have the following, which is valid for
n\ge 0
{a}_{n+2}=\frac{\left(n+m\right)\left(n-m\right)}{\left(n+2\right)\left(n+1\right)}{a}_{n}
m\in \mathbb{Z}
be any. If m is even, take
y\left(0\right)=1\text{ }\text{and}\text{ }{y}^{\prime }\left(0\right)=0
{a}_{0}=1\text{ }\text{and}\text{ }{a}_{1}=0
). Therefore all the coefficients of the form
{a}_{2k+1}
are zero (because of the formula above). Also, as m is an integer,
{a}_{m+2}=0
, and as it is even for all k,
{a}_{m+2k}=0
\text{deg}\text{ }y=m
Otherwise, if m is odd, let
y\left(0\right)=0\text{ }\text{and}\text{ }{y}^{\prime }\left(0\right)=1
. Arguing in a similar way, you conclude that deg
y=m
\frac{dw}{dt}
\frac{dw}{dt}
w=x\mathrm{sin}y,\text{ }x={e}^{t},\text{ }y=\pi -t
t=0
I am interested in the following differential equation:
y{}^{″}-qy
q:{\mathbb{R}}_{+}\to {\mathbb{R}}_{+}^{\cdot }
is a continuous, positive function.
\left(\frac{dy}{dx}\right)=-\frac{{y}^{2}+{x}^{2}}{2xy}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y\left(1\right)=4
Please, solve the differential equation. Write the method you used and solve for the dependent variable it it is possible.
{\left({y}^{\prime }\right)}^{4}x-2y{\left({y}^{\prime }\right)}^{3}+12{x}^{3}=0
\frac{dy}{dt}\frac{{d}^{2}x}{{dt}^{2}}=\frac{dx}{dt}\frac{{d}^{2}y}{{dt}^{2}}
{x}^{\prime }=x\mathrm{sin}\left(\frac{\pi }{x}\right)
is unique
I want to prove that the only solution to the ODE
{x}^{\prime }=\left\{\begin{array}{l}x\mathrm{sin}\left(\frac{\pi }{x}\right)\text{ if }x\ne 0\\ 0\text{ else}\end{array}
Show that the change of variables
x=2u+v,y=3u+v
transforms the linear system
\left\{\begin{array}{rl}{x}^{\prime }\left(t\right)& =\phantom{3}x-2y\\ {y}^{\prime }\left(t\right)& =3x-4y\end{array}
\left\{\begin{array}{rl}{u}^{\prime }\left(t\right)& =-2u\\ {v}^{\prime }\left(t\right)& =-v\end{array}
Verify that the u-axis maps to the line
y=\frac{3}{2}x
and the v-axis maps to the line
y=x
I thought for this I would just have to plug the substitutions into x'(t) and y'(t) but when doing this I receive
\left\{\begin{array}{rlr}{u}^{\prime }\left(t\right)& =2u+v-2\left(3u+v\right)\phantom{\rule{-15pt}{0ex}}& =-4u-v\\ {v}^{\prime }\left(t\right)& =3\left(2u+v\right)-4\left(3u+v\right)\phantom{\rule{-15pt}{0ex}}& =-6u-v\end{array}
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Pantoate—beta-alanine ligase - Wikipedia
Pantoate—beta-alanine ligase
In enzymology, a pantoate-beta-alanine ligase (EC 6.3.2.1) is an enzyme that catalyzes the chemical reaction
ATP + (R)-pantoate + beta-alanine
{\displaystyle \rightleftharpoons }
AMP + diphosphate + (R)-pantothenate
The 3 substrates of this enzyme are ATP, (R)-pantoate, and beta-alanine, whereas its 3 products are AMP, diphosphate, and (R)-pantothenate.
This enzyme belongs to the family of ligases, specifically those forming carbon-nitrogen bonds as acid-D-amino-acid ligases (peptide synthases). The systematic name of this enzyme class is (R)-pantoate:beta-alanine ligase (AMP-forming). Other names in common use include pantothenate synthetase, pantoate activating enzyme, pantoic-activating enzyme, and D-pantoate:beta-alanine ligase (AMP-forming). This enzyme participates in beta-alanine metabolism and pantothenate and CoA biosynthesis.
As of late 2007, 15 structures have been solved for this class of enzymes, with PDB accession codes 1IHO, 1MOP, 1N2B, 1N2E, 1N2G, 1N2H, 1N2I, 1N2J, 1N2O, 1UFV, 1V8F, 2A7X, 2A84, 2A86, and 2A88.
GINOZA HS, ALTENBERN RA (1955). "The pantothenate-synthesizing enzyme in cell-free extracts of Brucella abortus, strain 19". Arch. Biochem. 56 (2): 537–41. doi:10.1016/0003-9861(55)90273-3. PMID 14377603.
MAAS WK (1952). "Pantothenate studies. III. Description of the extracted pantothenate-synthesizing enzyme of Escherichia coli". J. Biol. Chem. 198 (1): 23–32. PMID 12999714.
Maas WK (1956). "Mechanism of the enzymatic synthesis of pantothenate from beta-alanine and pantoate". Fed. Proc. 15: 305–306.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Pantoate—beta-alanine_ligase&oldid=1079676415"
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Magnetic Flux and Faraday's Law | Brilliant Math & Science Wiki
Lawrence Chiou, Dale Gray, July Thomas, and
Faraday's law forms the basis for describing how much of our modern world operates, including how electric current is generated and how modern electronic devices and components function.
Consider a crude device called a crystal radio, which essentially consists of a coil of wire and an earpiece. Why is it that the radio runs perfectly (although perhaps playing on the quiet side) without any external batteries or other power? As it turns out, the air is filled with electromagnetic waves of various spectra, including radio waves. Such waves contain oscillating magnetic (and electric) fields, and changing magnetic fields can lead to electric fields in coils of wire in the form of electric current.
As one of the fundamental relationships of electricity and magnetism, Faraday's law, describes how time-varying magnetic fields produce electric fields.
Much work had been done on electrostatics by the beginning of the nineteenth century, but comparatively little had been done on magnetic fields. It wasn't until around 1820, for instance, that Oested first discovered that electric current produces a magnetic field, which is arguably the beginning of magnetostatics (apart from crude studies of naturally occuring magnets in antiquity).
Thus, it is perhaps surprising that less than two decades later, Michael Faraday had already begun to describe not only changing magnetic fields (thus beginning electrodynamics) but also, as a consequence, the interaction of electric and magnetic fields. Faraday discovered that, in general, a changing magnetic field leads to an electric field, a phenomenon called electromagnetic induction (the electric field is "induced"). The quantitative description of induction is thus named Faraday's law in honor of its discover.
The simplest example of induced electric field is the one generated inside a small circular conducting loop due to a changing magnetic field and responsible for the consequent current. Generally speaking, the induced electric field depends, not only on how the magnetic field,
\mathbf{B}
, changes with time, but also on how the geometric relation between the loop and magnetic field may change as well. For example, even in a uniform and constant magnetic field, changing the shape of a conducting loop or its orientation relative to the magnetic field lines will produce an electric field and hence a current. The appropriate combination of geometry and magnetic field needed to describe the induced electric field, when either changes, is called magnetic flux. The most basic definition is the magnetic flux through a plane figure due to a uniform magnetic field. Consider a plane region of area,
A
, and choose a unit vector,
\mathbf{n}
, perpendicular to the surface. For convenience an area vector,
\mathbf{A}
\mathbf{A}
=
A
\mathbf{n}
\mathbf{A}
will be inclined at some angle
\theta
with the magnetic field lines, i.e.,
\theta
will be the smaller angle between
\mathbf{A}
\mathbf{B}
. (See the figure below.)
The vector labeled 'normal' is our unit vector
\mathbf{n}
, and the magnetic flux through the plane area
A
\Phi = \mathbf{B} \cdot \mathbf{A}.
In essence the dot product projects our original area
A
onto a plane perpendicular to
\mathbf{B}
Magnetic flux through a surface patched together with small planes (think of something like a geodesic dome), is just the sum of that through each 'patch.' To a patch of surface, one assigns a vector
\mathbf{a}_i
that indicates the normal (perpendicular) to the surface. In addition, the magnitude of each
\mathbf{a}_i
is defined to be the area of the corresponding patch.
The magnetic flux,
\Phi_{i}
, through a patch is given by the dot product, which calculates the component of
\mathbf{B}_i
\mathbf{a}_i
\Phi_{i} = \mathbf{B}_i \cdot \mathbf{a}_i
Therefore, the total magnetic flux through a surface composed of many small patches,
\mathbf{a}_i
, is the sum over all patches:
\Phi = \sum_i \mathbf{B}_i \cdot \mathbf{a}_i .
\mathbf{a}_i
become vanishingly small, as in the case of a smooth surface, the sum is replaced with a surface integral:
\Phi = \int_S \mathbf{B} \cdot d\mathbf{a}.
Fortunately, the magnetic flux can often be computed without resorting to computing the integral explicitly. Generally, the calculation for magnetic flux is quite simple because one mostly considers flat loops. However, after the next topic, Faraday's Law, is discussed, it will be necessary to allow the smooth surface
S
to be arbitrary to determine the differential form of Faraday's Law in a subsequent section.
A small, flat loop of wire of area
A
is oriented perpendicular to the uniform magnetic field inside a cylindrical solenoid with turns per unit length
n
I
. What is the flux through the loop?
\mathbf{B} \cdot d\mathbf{a}
is constant, so the flux is simply
\Phi = BA = \mu_0 n I A
Faraday found that the induced emf
\mathcal{E}
(electromotive force) through a current loop was given by
\mathcal{E} = - \frac{d \Phi}{d t},
\Phi
is the magnetic flux through the loop (recall that emf points in the opposite direction as the voltage). In general, one determines
\Phi
t
, which allows for the calculation of
\mathcal{E}
Embedded in Faraday's law is the direction of the induced current in the loop. As with the Biot-Savart law and Ampère's law, the orientation of the loop is defined such that a counterclockwise traversal of the loop (relative to
\mathbf{a}
) is positive. Thus, the negative sign on
d \Phi/dt
indicates that an increase in flux through the loop leads to a clockwise flow of current in the loop. Likewise, a decrease in flux through the loop (positive increase relative to the opposite side of the loop) leads to a counterclockwise flow of current.
It is easy to remember this sign convention using the so-called right-hand rule. If one's thumb points in the direction of magnetic flux increase, then the current flows in the opposite direction to the curl of the fingers (opposite due to the negative sign in Faraday's law).
Note that the loop itself produces a magnetic field. One can verify that the magnetic field is always produced in the direction of decreasing flux (again, due to the negative sign). In some sense, one can think of the induced magnetic field as "opposing" the change in the external magnetic field, a result often referred to as Lenz's law.
A
R
is oriented perpendicular to the uniform, upward-pointing magnetic field inside a cylindrical solenoid with turns per unit length
n
I(t) = I_0 + Bt
I_0
B
are constants. What is the size and direction of the current
I_\text{loop}(t)
induced in the loop as a function of
t
As in the previous example, for a given
t
\mathbf{B} \cdot d\mathbf{a}
is constant, so the flux through the loop is just
\Phi(t) = BA = \mu_0 n (I_0 + Bt) A.
Therefore, Faraday's law yields
\mathcal{E} = - \frac{d\Phi}{dt} = -\mu_0 n A B.
It follows that the current induced in the loop is of magnitude
\mu_0 n A B / R
and points clockwise (due to the negative sign), which can also be verified using the right-hand rule.
A
is initially oriented parallel to a uniform magnetic field of magnitude
B
. If the loop rotates with angular velocity
\omega
, what is the magnitude of the induced emf
\mathcal{E}
in the loop?
Although the area of the loop remains constant, the angle of the loop with respect to the field is constantly changing. The angle with respect to the field is given by
\cos{\omega t}
\Phi = BA \cos{\omega t}
\mathcal{E} = - \frac{d \Phi}{dt} = \omega A B \sin{\omega t}.
In effect, one can generate electricity by mechanical rotation of a wire loop in a magnetic field.
Recall that that the emf
\mathcal{E}
, around any closed path,
C
, is defined in the opposite direction as the voltage, namely
\mathcal{E} = \oint_\text{C} \mathbf{E} \cdot d\mathbf{s}.
C
be the boundary curve of an arbitrary smooth, non closed surface
S
. It follows, from Faraday's law, that the magnetic flux through
S
\Phi = \int_S \mathbf{B} \cdot d\mathbf{a},
and the emf around
C
\oint_\text{C} \mathbf{E} \cdot d\mathbf{s} = - \frac{d}{dt} \int_S \mathbf{B} \cdot d\mathbf{a}.
\int_S \nabla \times \mathbf{E} \cdot d\mathbf{a} = - \frac{d}{dt} \int_S \mathbf{B} \cdot d\mathbf{a}.
Again, one can argue that since the relationship must hold true for any arbitrary surface,
S
, it must be the case that the two integrands are equal, and therefore
\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}.
This is the so-called differential form of Faraday's law. Whereas in electrostatics, the electric field is conservative and
\nabla \times \mathbf{E} = 0
, in the world of electrodynamics, the electric field is not curl-free.
Cite as: Magnetic Flux and Faraday's Law. Brilliant.org. Retrieved from https://brilliant.org/wiki/magnetic-flux-and-faradays-law-quantitative/
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Laws of Thermodynamics - Course Hero
General Chemistry/Thermodynamics/Laws of Thermodynamics
The zeroth law of thermodynamics states that when two systems are in thermal equilibrium with a third system, they are in thermal equilibrium with each other. The first law of thermodynamics states that the total energy of a system is constant.
Four laws of thermodynamics exist. The zeroth law of thermodynamics, developed after the other laws, states that when two thermodynamic systems are in thermal equilibrium with a third system, they are in thermal equilibrium with each other. A thermometer is considered a third system. This is helpful for comparing systems to one another. For example, consider three systems: a glass of ice water, the surrounding room, and a thermometer. The temperature of water with ice in it is measured using the thermometer. The temperature rises on the thermometer as the glass of ice water obtains thermal equilibrium with the surrounding room.
The first law of thermodynamics, also called the law of conservation of energy, states that energy cannot be created or destroyed, only transformed from one type of energy to another type of energy. In any change in the state of a system, some energy produces work, the energy transferred when a force acts on an object, and some is transferred to the surroundings or put into the system as heat. Energy is not created or destroyed; it is merely dispersed. In other words, the total energy of the universe is constant. This law of thermodynamics forbids perpetual motion machines, in which the work done by a system is greater than or equal to the energy input into the system. However, no isolated system can be 100% efficient. Some heat energy will always be lost to the surroundings.
The second law of thermodynamics states that total entropy of an isolated system only increases over time.
The second law of thermodynamics states that the total entropy of an isolated system only increases over time. All isolated systems tend toward an equilibrium state in which entropy is at a maximum value. Thus,
{\Delta S_{\rm{univ}}}={\Delta S_{\rm{sys}}}+{\Delta S_{\rm{surr}}}
\Delta S_{\rm{univ}}>0,
for spontaneous processes. At this point, no energy is left to do work. In terms of thermal energy, this law can be thought of as stating that heat does not flow from a colder region to a hotter one spontaneously.
Most systems are only isolated when considering them theoretically because even insulated systems still exist within surroundings. However, the universe can be considered an isolated system, and it has no surroundings with which to interact. Thus, the universe should be tending toward a maximum state of entropy in which no more energy to do work can exist. The ability for enthalpy to increase is greater if heat is absorbed by the system and if the system is more ordered (low temperature). Therefore the change in enthalpy is proportional to
\Delta{H}_{\rm{sys}}
and inversely proportional to T. This leads to
\Delta S_{\rm{surr}}=-\Delta {H}_{\rm{sys}}/T.
In chemistry a system is considered isolated if zero energy is entering or leaving the system, which means no mass, heat, or work transfer. The system is considered closed as long as no mass enters or leaves the system and the system is not subject to forces from outside the system. Thus, a beaker in which a chemical reaction between two liquids takes place can be considered a closed system because no mass is being transferred into or out of the system and the only work is being done within the system itself. It would not be considered an isolated system because heat is gained or lost in the chemical reaction.
The third law of thermodynamics states that the total entropy of a system approaches a constant value as the temperature of the system approaches absolute zero.
The third law of thermodynamics states that the total entropy of a system approaches a constant value as the temperature of the system approaches absolute zero. Absolute zero is the minimum possible temperature theoretically achievable, equal to 0 K (–273.15°C), at which there is no particle motion. A crystal, a highly ordered microscopic structure, of an element in its most stable form can be considered a system and can be used as a reference for the motion of matter in different states and at different temperature. Thus the third law of thermodynamics allows for the creation of an absolute scale of entropy.
The entropy of one mole of a substance under standard state conditions, expressed in units J/(K mol), is its standard entropy (S°). A standard state is a set of specific conditions under which enthalpies and entropies are measured, typically 0°C and 1 atm pressure. A standard entropy is typically described as a standard molar entropy, which is the standard entropy per mole of a substance. The standard molar entropies of most common substances have been calculated.
Standard entropy can be used to calculate the change in entropy of a chemical reaction. The change in entropy for a reaction (
\Delta S_{\rm{rxn}}
) is the difference between the entropies of the products (
\Sigma S_{\rm{prod}}
) and reactants (
\Sigma S_{\rm{reac}}
\Delta S_{\rm{rxn}}=\Sigma S_{\rm{prod}}-\Sigma S_{\rm{reac}}
For example, the decomposition of hydrogen peroxide forms water and oxygen gas:
2{\rm {H}_{2}{O}_{2}}(l)\rightarrow 2{\rm {H}_{2}{O}}(l)+{\rm O}_{2}(g)
. The change in molar entropy for this reaction can be calculated. Standard molar entropies are as follows:
{\rm {H}_{2}}{\rm {O}_{2}}=232.95\;{\rm J/K}{\cdot}{\rm{mol}}
{\rm {H}_{2}{O}}=188.84\;{\rm J/\rm K{\cdot}\rm{mol}}
{\rm {O}_{2}}=120.15\;{\rm J/\rm K{\cdot{mol}}}
\begin{aligned}\Delta{S}&={(2(188.84\;\rm{J}/{K{\cdot}mol}\;+120.15\;{\rm J/K{\cdot}mol})\;-2(232.95\;\rm{J/K{\cdot}mol}))}\\&=31.93\;\rm{J/K\;mol}\end{aligned}
<What is Thermodynamics?>Free Energy
|
How do you find \sin(\sin^{-1}(\frac{1}{4})) ?
Nataly Best 2022-01-23 Answered
\mathrm{sin}\left({\mathrm{sin}}^{-1}\left(\frac{1}{4}\right)\right)
\mathrm{sin}\left[\mathrm{arcsin}\left(x\right)\right]=x
for x in the domain of
\mathrm{arcsin}\left(x\right)
(The domain of
\mathrm{arcsin}\left(x\right)
-1\le x\le 1
\mathrm{sin}\left[\mathrm{arcsin}\left(\frac{1}{4}\right)\right]=\frac{1}{4}
waijazar1
y=\mathrm{sin}x
\mathrm{arcsin}\left(y\right)=x
\mathrm{arcsin}\left(\mathrm{sin}x\right)=x
\mathrm{arcsin}\left(\mathrm{sin}\left(\frac{1}{4}\right)\right)=\frac{1}{4}
\frac{\mathrm{sec}x}{\mathrm{tan}x}
f\left(x\right)={x}^{2},\phantom{\rule{1em}{0ex}}\text{if}\phantom{\rule{1em}{0ex}}x\le 4\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}f\left(x\right)=m+b,\phantom{\rule{1em}{0ex}}\text{if}\phantom{\rule{1em}{0ex}}x>4
\mathrm{arctan}\left(2\right)+\mathrm{arctan}\left(3\right)
x\mathrm{sin}x
=\sqrt{3i}+\pi +ck,v=4i-j-k
‖u‖,‖v‖
u×v
\mathrm{tan}\frac{3\pi }{2}
{f}^{-1}
f\left(x\right)=\frac{2x}{x+5}
|
{\displaystyle \scriptstyle p=0.9^{4}}
{\displaystyle \scriptstyle 6p^{1.2}}
ScrewTorxRhombusScrew threadSelf-tapping screwCaptive fastenerMechanical engineeringList of screw drivesHexagonScrew conveyor
This article uses material from the Wikipedia article "Cap screw", which is released under the Creative Commons Attribution-Share-Alike License 3.0. There is a list of all authors in Wikipedia
|
(→Budget of turbulent kinetic energy)
(→Distribution of turbulent kinetic energy and its budgets terms: production, diffusive transport, dissipation and convection)
* dissipation: <math>\epsilon = 2\nu\langle s_{ij}s_{ij}\rangle </math>, and <math> s_{ij} = \frac{1}{2}\left(\frac{\partial u_i'}{\partial x_j} + \frac{\partial u_j'}{\partial x_i}\right)</math> as the fluctuating rate-of-strain tensor
** in the LES, the dissipation consists of a ''resolved'' and a ''subgrid scale (SGS)'' part:
** <math>\epsilon = \epsilon_{\mathrm{res}} + \epsilon_{\mathrm{SGS}} = 2\nu\langle s_{ij}s_{ij}\rangle + 2\langle \nu_{\mathrm{t}} s_{ij}s_{ij}\rangle </math>
** <math>\epsilon_{\mathrm{total}} = \epsilon_{\mathrm{res}} + \epsilon_{\mathrm{SGS}} = 2\nu\langle s_{ij}s_{ij}\rangle + 2\langle \nu_{\mathrm{t}} s_{ij}s_{ij}\rangle </math>
* mean convection: <math> C = - \langle u_i\rangle \frac{\partial k}{\partial x_i} </math> (steady state)
Nevertheless, when analysing the distribution and amplitude of the residual of the TKE budget obtained from PIV, we observe a similar structure to the one of <math> -\nabla T_{\mathrm{press,LES}}</math>. Therefore, the missing piece in the experimental data of this flow configuration is the contribution of the pressure fluctuations to the transport mechanisms of the TKE (Jenssen 2019).
[[File:UFR3-35_LES_Budget.png|centre|frame|Fig. 19 b) Residual of turbulent kinetic energy budget <math> R_{\mathrm{LES}} = P + \nabla T - \epsilon{\mathrm{total}} + C \cdot D/u_{\mathrm{b}}^3 </math>]]
[[File:UFR3-35_LES_Budget.png|centre|frame|Fig. 19 b) Residual of turbulent kinetic energy budget <math> R_{\mathrm{LES}} = P + \nabla T - \epsilon_{\mathrm{total}} + C \cdot D/u_{\mathrm{b}}^3 </math>]]
The residual of the LES data is small in wide regions <math> <|0.01|u_{\mathrm{b}}^3/D</math>. In particular around the horseshoe vortex, the residual is close to zero. However, along the cylinder surface and the bottom wall the budget does not fully balance. On the one hand, we assign the large errors to the spatial resolution of the grid in the horizontal direction, which was obviously too coarse to fully resolve the developing boundary layer at the cylinder surface. On the other hand, the sensitivity with respect to the number of samples of the term <math> T_{\mathrm{turb}} = -\frac{1}{2}\langle u_i'u_j'u_j' \rangle </math> containing triple correlations of the velocity can be responsible
{\displaystyle c_{\mathrm {p} }(x)}
{\displaystyle c_{\mathrm {f} }(x)}
{\displaystyle \langle u\rangle }
{\displaystyle \langle w\rangle }
{\displaystyle \langle u'_{i}u'_{j}\rangle }
{\displaystyle \langle k\rangle }
{\displaystyle c_{\mathrm {p} }(x)}
{\displaystyle c_{\mathrm {f} }(x)}
{\displaystyle ||{\vec {U}}||={\sqrt {\langle u^{2}\rangle +\langle w^{2}\rangle }}/u_{\mathrm {b} }}
{\displaystyle ||{\vec {U}}_{\mathrm {PIV} }||={\sqrt {\langle u^{2}\rangle +\langle w^{2}\rangle }}/u_{\mathrm {b} }}
{\displaystyle ||{\vec {U}}_{\mathrm {LES} }||={\sqrt {\langle u^{2}\rangle +\langle w^{2}\rangle }}/u_{\mathrm {b} }}
{\displaystyle x/D}
{\displaystyle z/D}
{\displaystyle x/D}
{\displaystyle z/D}
{\displaystyle -0.788}
{\displaystyle 0.03}
{\displaystyle -0.843}
{\displaystyle 0.037}
{\displaystyle -0.918}
{\displaystyle 0}
{\displaystyle -1.1}
{\displaystyle 0}
{\displaystyle -0.533}
{\displaystyle 0}
{\displaystyle -0.534}
{\displaystyle 0}
{\displaystyle -0.507}
{\displaystyle 0.036}
{\displaystyle -0.50}
{\displaystyle 0.04}
{\displaystyle -0.697}
{\displaystyle 0.051}
{\displaystyle -0.735}
{\displaystyle 0.06}
{\displaystyle -0.513}
{\displaystyle 0.017}
{\displaystyle -0.513}
{\displaystyle 0.02}
{\displaystyle x-}
{\displaystyle x_{\mathrm {adj} }={\frac {x-x_{\mathrm {Cyl} }}{x_{\mathrm {Cyl} }-x_{\mathrm {V1} }}}}
{\displaystyle x_{\mathrm {Cyl} }=-0.5D}
{\displaystyle x_{\mathrm {adj} }=-1.0}
{\displaystyle x_{\mathrm {V1} }}
{\displaystyle \langle u(z)\rangle /u_{\mathrm {b} }}
{\displaystyle u(z)}
{\displaystyle x_{\mathrm {adj} }=-0.25}
{\displaystyle x_{\mathrm {adj} }=-0.5}
{\displaystyle {\frac {\partial \langle u\rangle }{\partial z}}}
{\displaystyle {\frac {\partial \langle u\rangle }{\partial z}}}
{\displaystyle \langle u_{i}'u_{j}'(z)\rangle /u_{\mathrm {b} }^{2}}
{\displaystyle \langle k(z)\rangle =0.5(\langle u'^{2}\rangle +\langle w'^{2}\rangle )/u_{\mathrm {b} }^{2}}
{\displaystyle x_{\mathrm {adj} }=-1.5}
{\displaystyle x_{\mathrm {adj} }=-1.0}
{\displaystyle x_{\mathrm {adj} }=-0.5}
{\displaystyle \langle u'u'\rangle }
{\displaystyle \langle u'u'\rangle }
{\displaystyle \langle w'w'\rangle }
{\displaystyle z_{\mathrm {V1} }/D}
{\displaystyle \langle u'w'\rangle }
{\displaystyle \langle w(x)\rangle /u_{\mathrm {b} }}
{\displaystyle z_{\mathrm {V1} }/D}
{\displaystyle \langle w(x)\rangle }
{\displaystyle x-}
{\displaystyle x_{\mathrm {adj} }\approx -0.1}
{\displaystyle x_{\mathrm {adj} }=-0.65}
{\displaystyle \langle u_{i}'u_{j}'(x)\rangle /u_{\mathrm {b} }^{2}}
{\displaystyle \langle k(x)\rangle =0.5(\langle u'^{2}\rangle +\langle w'^{2}\rangle )/u_{\mathrm {b} }^{2}}
{\displaystyle z_{\mathrm {V1} }/D}
{\displaystyle \langle u_{i}'u_{j}'\rangle }
{\displaystyle \langle k\rangle }
{\displaystyle \langle k_{\mathrm {PIV,inplane} }\rangle =0.5(\langle u'^{2}\rangle +\langle w'^{2}\rangle )/u_{\mathrm {b} }^{2}}
{\displaystyle \langle k_{\mathrm {LES,inplane} }\rangle =0.5(\langle u'^{2}\rangle +\langle w'^{2}\rangle )/u_{\mathrm {b} }^{2}}
{\displaystyle \langle k_{\mathrm {LES,total} }\rangle =0.5(\langle u'^{2}\rangle +\langle v'^{2}\rangle +\langle w'^{2}\rangle )/u_{\mathrm {b} }^{2}}
{\displaystyle \langle k\rangle =0.5(\langle u'^{2}\rangle +\langle w'^{2}\rangle )/u_{\mathrm {b} }^{2}}
{\displaystyle \langle k_{\mathrm {PIV,inplane} }\rangle =0.074u_{\mathrm {b} }^{2}}
{\displaystyle \langle k_{\mathrm {LES,inplane} }\rangle =0.079u_{\mathrm {b} }^{2}}
{\displaystyle \langle k_{\mathrm {LES,total} }\rangle =0.5(\langle u'^{2}\rangle +\langle v'^{2}\rangle +\langle w'^{2}\rangle )/u_{\mathrm {b} }^{2}}
{\displaystyle \langle k_{\mathrm {LES,total} }\rangle =0.09u_{\mathrm {b} }^{2}}
{\displaystyle 0=P+\nabla T-\epsilon +C}
{\displaystyle P}
{\displaystyle \nabla T}
{\displaystyle \epsilon }
{\displaystyle C}
{\displaystyle v}
{\displaystyle P=-\langle u_{i}'u_{j}'\rangle {\frac {\partial \langle u_{i}\rangle }{\partial x_{j}}}}
{\displaystyle T=\underbrace {-{\frac {1}{2}}\langle u_{i}'u_{j}'u_{j}'\rangle } _{\text{turbulent fluctuations}}\underbrace {-{\frac {1}{\rho }}\langle u_{i}'p'\rangle } _{\text{pressure transport}}\underbrace {+2\nu \langle u_{j}'s_{ij}\rangle } _{\text{viscous diffusion}}}
{\displaystyle \epsilon =2\nu \langle s_{ij}s_{ij}\rangle }
{\displaystyle s_{ij}={\frac {1}{2}}\left({\frac {\partial u_{i}'}{\partial x_{j}}}+{\frac {\partial u_{j}'}{\partial x_{i}}}\right)}
{\displaystyle \epsilon _{\mathrm {total} }=\epsilon _{\mathrm {res} }+\epsilon _{\mathrm {SGS} }=2\nu \langle s_{ij}s_{ij}\rangle +2\langle \nu _{\mathrm {t} }s_{ij}s_{ij}\rangle }
{\displaystyle C=-\langle u_{i}\rangle {\frac {\partial k}{\partial x_{i}}}}
{\displaystyle D/u_{\mathrm {b} }^{3}}
{\displaystyle P_{\mathrm {PIV} }=-\langle u_{i}'u_{j}'\rangle {\frac {\partial \langle u_{i}\rangle }{\partial x_{j}}}\cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle P_{\mathrm {LES} }=-\langle u_{i}'u_{j}'\rangle {\frac {\partial \langle u_{i}\rangle }{\partial x_{j}}}\cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle 0.3u_{\mathrm {b} }^{3}/D}
{\displaystyle P_{\mathrm {LES} }\approx 0.4u_{\mathrm {b} }^{3}/D}
{\displaystyle P_{\mathrm {PIV} }\approx 0.2u_{\mathrm {b} }^{3}/D}
{\displaystyle x=-0.7D}
{\displaystyle P}
{\displaystyle \nabla T_{\mathrm {turb,PIV} }=-{\frac {1}{2}}{\frac {\partial \langle u_{i}'u_{j}'u_{j}'\rangle }{\partial x_{i}}}\cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle \nabla T_{\mathrm {turb,LES} }=-{\frac {1}{2}}{\frac {\partial \langle u_{i}'u_{j}'u_{j}'\rangle }{\partial x_{i}}}\cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle x=-0.75D}
{\displaystyle 0.4u_{\mathrm {b} }^{3}/D}
{\displaystyle T_{\mathrm {turb,LES} }\approx 0.35u_{\mathrm {b} }^{3}/D}
{\displaystyle \nabla T_{\mathrm {press,LES} }=-{\frac {1}{\rho }}{\frac {\partial \langle u_{i}'p'\rangle }{\partial x_{i}}}\cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle \nabla T_{\mathrm {visc,LES} }=2\nu {\frac {\partial \langle u_{j}'s_{ij}\rangle }{\partial x_{i}}}\cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle \nabla T_{\mathrm {turb} }}
{\displaystyle \nabla T_{\mathrm {press} }}
{\displaystyle \langle w\rangle <0}
{\displaystyle w-}
{\displaystyle w'}
{\displaystyle p'<0}
{\displaystyle \nabla T_{\mathrm {visc} }}
{\displaystyle |0.05|u_{\mathrm {b} }^{3}/D}
{\displaystyle P}
{\displaystyle \nabla T}
{\displaystyle \epsilon }
{\displaystyle \epsilon _{\mathrm {PIV} }=2\nu \langle s_{ij}s_{ij}\rangle \cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle \epsilon _{\mathrm {LES,total} }=(2\nu \langle s_{ij}s_{ij}\rangle +2\langle \nu _{\mathrm {t} }s_{ij}s_{ij}\rangle )\cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle P}
{\displaystyle \epsilon _{\mathrm {LES} }=0.066u_{\mathrm {b} }^{3}/D}
{\displaystyle P_{\mathrm {max} }}
{\displaystyle \epsilon _{\mathrm {max} }}
Since we analyse a steady state flow, the time derivative becomes irrelevant and the mean convection reduces to
{\displaystyle C=-\langle u_{i}\rangle {\frac {\partial k}{\partial x_{i}}}}
, and is presented in Fig. 18.
{\displaystyle C_{\mathrm {PIV} }=-\langle u_{i}\rangle {\frac {\partial k}{\partial x_{i}}}\cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle C_{\mathrm {LES} }=-\langle u_{i}\rangle {\frac {\partial k}{\partial x_{i}}}\cdot D/u_{\mathrm {b} }^{3}}
The approaching flow separates from the bottom wall of the flume, and consequently, becomes unstable with increasing fluctuations. Therefore, the TKE increases along the streamlines, which is indicated by the negative mean convection upstream of the horseshoe vortex. The same applies for the wall-parallel jet. After the deflection at S3, the jet accelerates, which decreases TKE in the first place. When the jet starts to decelerate (
{\displaystyle x\approx -0.63D}
{\displaystyle C}
{\displaystyle R_{\mathrm {PIV} }=P+\nabla T_{\mathrm {turb} }-\epsilon +C\cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle -\nabla T_{\mathrm {press,LES} }}
{\displaystyle R_{\mathrm {LES} }=P+\nabla T-\epsilon _{\mathrm {total} }+C\cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle <|0.01|u_{\mathrm {b} }^{3}/D}
{\displaystyle T_{\mathrm {turb} }=-{\frac {1}{2}}\langle u_{i}'u_{j}'u_{j}'\rangle }
{\displaystyle c_{\mathrm {p} }(x)}
{\displaystyle c_{\mathrm {f} }(x)}
{\displaystyle c_{\mathrm {p} }}
{\displaystyle c_{\mathrm {p} }={\frac {\langle p\rangle }{{\frac {\rho }{2}}u_{\mathrm {b} }^{2}}}}
{\displaystyle c_{\mathrm {f} }}
{\displaystyle c_{\mathrm {f} }={\frac {\langle \tau _{\mathrm {w} }\rangle }{{\frac {\rho }{2}}u_{\mathrm {b} }^{2}}}}
{\displaystyle z_{1}\approx 0.0036D\approx 10\mathrm {px} }
{\displaystyle z_{1}\approx 0.0005D}
{\displaystyle z-}
{\displaystyle c_{\mathrm {p} }}
{\displaystyle c_{\mathrm {p} }}
{\displaystyle c_{\mathrm {f} }}
{\displaystyle x_{\mathrm {adj} }=-1.0}
{\displaystyle c_{\mathrm {f} }}
{\displaystyle |c_{\mathrm {f} }|=0.01}
{\displaystyle c_{\mathrm {f} }}
{\displaystyle c_{\mathrm {f} }}
{\displaystyle c_{\mathrm {f} }}
{\displaystyle 50\times 171(n\times m)}
{\displaystyle 143\times 131(n\times m)}
{\displaystyle n\cdot m}
{\displaystyle x_{\mathrm {adj} }}
{\displaystyle {\frac {x}{D}}}
{\displaystyle {\frac {z}{D}}}
{\displaystyle {\frac {\langle u\rangle }{u_{\mathrm {b} }}}}
{\displaystyle -}
{\displaystyle {\frac {\langle w\rangle }{u_{\mathrm {b} }}}}
{\displaystyle {\frac {\langle u'u'\rangle }{u_{\mathrm {b} }^{2}}}}
{\displaystyle -}
{\displaystyle {\frac {\langle w'w'\rangle }{u_{\mathrm {b} }^{2}}}}
{\displaystyle -}
{\displaystyle {\frac {\langle u'w'\rangle }{u_{\mathrm {b} }^{2}}}}
{\displaystyle -}
{\displaystyle P{\frac {D}{u_{\mathrm {b} }^{3}}}}
{\displaystyle C{\frac {D}{u_{\mathrm {b} }^{3}}}}
{\displaystyle \nabla T_{\mathrm {turb} }{\frac {D}{u_{\mathrm {b} }^{3}}}}
{\displaystyle -}
{\displaystyle -}
{\displaystyle \epsilon {\frac {D}{u_{\mathrm {b} }^{3}}}}
{\displaystyle -}
{\displaystyle x_{\mathrm {adj} }}
{\displaystyle {\frac {x}{D}}}
{\displaystyle {\frac {z}{D}}}
{\displaystyle {\frac {\langle u\rangle }{u_{\mathrm {b} }}}}
{\displaystyle {\frac {\langle v\rangle }{u_{\mathrm {b} }}}}
{\displaystyle {\frac {\langle w\rangle }{u_{\mathrm {b} }}}}
{\displaystyle {\frac {\langle u'u'\rangle }{u_{\mathrm {b} }^{2}}}}
{\displaystyle {\frac {\langle v'v'\rangle }{u_{\mathrm {b} }^{2}}}}
{\displaystyle {\frac {\langle w'w'\rangle }{u_{\mathrm {b} }^{2}}}}
{\displaystyle {\frac {\langle u'v'\rangle }{u_{\mathrm {b} }^{2}}}}
{\displaystyle {\frac {\langle u'w'\rangle }{u_{\mathrm {b} }^{2}}}}
{\displaystyle {\frac {\langle v'w'\rangle }{u_{\mathrm {b} }^{2}}}}
{\displaystyle P{\frac {D}{u_{\mathrm {b} }^{3}}}}
{\displaystyle C{\frac {D}{u_{\mathrm {b} }^{3}}}}
{\displaystyle \nabla T_{\mathrm {turb} }{\frac {D}{u_{\mathrm {b} }^{3}}}}
{\displaystyle \nabla T_{\mathrm {press} }{\frac {D}{u_{\mathrm {b} }^{3}}}}
{\displaystyle \nabla T_{\mathrm {visc} }{\frac {D}{u_{\mathrm {b} }^{3}}}}
{\displaystyle \epsilon {\frac {D}{u_{\mathrm {b} }^{3}}}}
{\displaystyle c_{\mathrm {p} }}
|
Draw a graph for the linear equation x-2y=4 - Maths - Linear Equations in Two Variables - 10784703 | Meritnation.com
Draw a graph for the linear equation x-2y=4
\mathrm{The} \mathrm{given} \mathrm{equation} \mathrm{is},\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{x} - 2\mathrm{y} = 4\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{When} \mathrm{x} = 0; \mathrm{y} = -2\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{When} \mathrm{x} = 4; \mathrm{y} = 0
Asma Narmawala answered this
I have an answer but app is showing it unappropriate because of some symbols
x= 4+2(1)
Now plot (4,0) and (6,1) on graph
|
Spo-tfidf-y · Matt's Blog
Spo-tfidf-y
26 Dec 2018 • Data-Analytics
> Click here if you just care about the pretty graphs
Vector embeddings are a really powerful method in data analytics and “machine learning”. If you can convert some non-quantitative data, like words, into vectors, you can use traditional data analytics methods to extract meaning from your input.
One of the most notable vector embeddings is Google’s word2vec. By mapping words 2 vectors, the smart folks at Google allowed us lowly non-Google employees to do really cool stuff with the English language! If you have some word
w_i
and want to find a word with a similar meaning
w_s
, you can just look at your vectors and find the word with the closest distance to
w_i
– this vector is your similar word,
w_s
! Who needs thesaurus.com when you have word2vec?
Another neat feature is the “analogy” solving feature. A common analogy (I have no idea if that’s actually the correct term for this) is “man is to woman as king is to ___”. If you didn’t get an A in English, you can use word2vec to figure this out for you! The answer
w_{answer}
is calculated by computing
w_{king} - w_{man} + w_{woman}
. The closest vector to this combination of word vectors turns out to be the vector for queen!
What’s with the Title
Ok it was a pretty bad joke. A common “classical” vector embedding method is called tf-idf, which stands for “term frequency - inverse document frequency”. I’ll be analyzing Spotify data in this blog post, so I thought the pun would be cute. Let’s talk through what tf-idf is before we get to the music part.
With tf-idf, we have some set of documents with words in them, and make the assumption that words in the same document are similar, or at least somewhat related to one another. There are two parts to tf-idf: tf, which stands for term frequency, and idf, which stands for inverse document frequency. The first one is the easiest. In a document
D_i
, for a given word
w_j
, the term frequency of
w_j
D_i
is the number of occurrences of
w_j
D_i
, divided by the total number of words in
D_i
. Note: tf is computed for a word and a document. The same word may have a different tf score for different documents.
If our document is “I really like Matt’s blog, it’s really cool”, the tf of “really” is
2 / 8
, because “really” occurs twice, and there are eight total words!
With idf, you want to lower the score of common words. Every document probably has words like “a” and “the” in it, so their term frequency is probably going to be super high compared to words that might be more representative of the document but occur less frequently. If we have
N
documents and the word
w_j
N_{w_j}
documents, the formula for idf is
log(N / N_{w_j})
. As we can see, if the word appears in a lot of the documents, the denominator will be larger, resulting in a smaller fraction; we take the log of this number, and voila! Note, unlike tf, the idf for a given word is the same across all documents.
If our first document,
D_1
is “I want a hat” and our second document
D_2
is “Matt would like a snowboard”, the idf of “a” is
log(2 / 2)
, while the idf of “Matt” is
log(2 / 1)
Now, like the name implies, we compute
tf * idf
to get our tf-idf score for the current word!
If we look at the previous example, where
D_1
is “I want a hat” and
D_2
is “Matt would like a snowboard”, we can easily compute the tf-idf of “want”. In the first document, its tf is
1/4
, while in the second, its tf is
0
. The idf of “want” in the context of all of our documents is
log(2 / 1)
. Since the tf-idf score is different for each document, we can represent the tf-idf of the word “want” as a vector [
tf-idf
(“want”,
D_1
tf-idf
D_2
)] = [
log(2) / 4
0
]. Nice! We just found a vector encoding of “want”!
Spo-tfidf-y, Revisited
Ok, now that we’re tf-idf experts, we can apply this technique to Spotify songs! Let’s define “words” to be songs, and “documents” to be playlists. We make the assumption here that songs in the same playlist are similar.
I listen to a lot of Kanye, and am pretty familiar with his discography. I wanted to vectorize his songs and then cluster them. This clustering would tell me how his users group his songs, and one could define populist pseudo-albums based on these clusters.
To scrape the playlists, I used Spotify’s super generous API to query for things I felt would be related to Kanye’s music. I ended up scraping
1263
playlists that contained the words “rap”, “hip hop”, and “pop” in the title; I only considered playlists from this query that had at least two Kanye songs in them. I figured I could get better results with more specific keywords, but they would bias the playlists they’re pulling from more – a query like “early 2010s hip hop” is Bound 2 lean towards Yeezus – so I tried to keep it as vague as possible.
Ignore this paragraph if you don’t care about my technical difficulties. Some other issues I came in contact with were re-released albums. There are four instances of The College Dropout on Spotify, along with many more duplicate albums, so I had to manually collect these albums and merge the duplicate songs. Another issue was the endpoint to retrieve albums by Kanye, which is how I defined “Kanye songs”, would not get me all of the duplicate albums. For instance, there are five releases of Kanye West Presents Good Music Cruel Summer, but only one where “Kanye West” is listed as the main artist. Because of this, I just tossed all results for this album, because there might be playlists that do have songs from KWPGMCS but are not technically “Kanye songs”, and I had already finished the super painful scraping process. Anyways, onto the cool stuff!
So finally, I got my really big tf-idf matrix – it had a dimension of 126x1263, because there were 126 Kanye songs and 1263 playlists I looked at. Each row was the vector of a song! Let’s call this whole matrix
M_{playlist}
. I also computed a sibling-matrix of size 126x9. Spotify has an endpoint to compute “audio features” of a given song, which includes information about the song’s acousticness, danceability, energy, instrumentalness, liveness, loudness, speechiness, valence, and tempo. You can read more about what these features mean here – there’s even pretty little charts on how these features are distributed. We’ll call this sibling matrix
M_{features}
I used a couple of things for my data vis. One of the tools I used is called PCA, which looks at some fancy math to find a lower dimension linear combination of your data – PCA is guaranteed to have the most explained variance, which means that if you reduce an input of
n
vectors to
m
vectors, those
m
vectors are the way to encode your data while retaining the most information!
I also used a pretty simple clustering method called KMeans that manages to group your data into
k
distinct groups, or clusters. Here, I set
k
to be 9, as Kanye has 9 albums (give or take). KMeans takes its
k
cluster centers, finds the points that are closer to a given center than to any other center, and moves that center towards the average of those near points. This runs until the centers converge and stop moving.
I’ll explain each of the following graphs below. Before clustering anything, I normalized the input matrix.
For my first experiment, I simply clustered the rows of my
M_{playlist}
matrix and plotted the results in this cool d3 thing I Frankensteined
For my second experiment, I appended
M_{features}
to the end of
M_{playlist}
, clustered that, and plotted it again
I computed a matrix
M_{PCA playlist}
, which is the result of running PCA on
M_{playlist}
, keeping 9 components. Then I clustered
M_{PCA playlist}
and plotted it
I appended
M_{features}
M_{PCA playlist}
, clustered that, and plotted it one last time
I actually had two other experiments where I ran sklearn’s feature agglomeration instead of PCA, but the results, specifically after clustering, were consistently similar.
Please take an intermission from this long af blog post by looking at these Gorgeous graphs.
To reiterate, these are clusterings of Kanye’s albums based, more or less, on how often songs are grouped together in the same playlist!
One interesting but expected grouping is the clustering of skits. Cluster #5 from the first graph is made out of a lot of skits, along with some lesser known tracks, and cluster #5 in the second graph is purely skits, with no actual songs. The reason this second graph might have been better at finding the skit pseudo-album is because it considers factors like “speechiness” and “instrumentalness” – skits typically have a high score in the former and low score in the latter, which would help separate them from their non-skit counterparts. The PCA results also follow the trend of having a skit-based grouping.
Another common trend is clusterings of more mainstream Kanye songs – in the first and last graphs, Stronger, POWER, Heartless, and Gold Digger end up in the same clusters. This makes sense intuitively – people that listen to more mainstream Kanye probably have playlists with the mainstream songs, which leads to closer tf-idf vectors.
There’s a really heavy grouping based on album in all of the graphs. This is expected; people make playlists in the moment and don’t always update them years later when new albums come out.
In general, I think it’s worth noting that the album clusters are pretty sequential – usually The College Dropout isn’t getting clustered with Kids See Ghosts, but rather with Late Registration, while MBDTF and Yeezus are often grouped together. Poetically, 808s and Heartbreaks finds itself alone a lot of the time, which is pretty Bad News.
Scroll through the graphs and see what connections you come up with!
As a side note, I also sorted songs by their tf-idf vector magnitude – theoretically, songs with higher tf-idf magnitudes are more “distinctive”. If we cared about popularity, we can just look at how many playlists each song is in, but that’s less interesting. Here are Kanye’s top 10 “distinctive” songs!
Interestingly, 4/10 of the songs are from MBDTF, which left me So Appalled. Old Kanye was mostly absent – The College Dropout had one song, while Late Registration and Graduation were missing. There was no Yeezus or TLOP either, which is strange since those albums were pretty Famous.
If we only rank songs based on how many playlists they’re in, we get the following results.
Stronger - 336 occurrences
Gold Digger - 333 occurrences
POWER - 296 occurrences
All Of The Lights - 189 occurrences
Can’t Tell Me Nothing - 186 occurrences
Good Life - 160 occurrences
I Love It (& Lil Pump) - 159 occurrences
Black Skinhead - 159 occurrences
Father Stretch My Hands Pt. 1 - 157 occurrences
Flashing Lights - 143 occurrences
Unlike the previous playlist, we have a lot of Graduation in this playlist, and Late Registration, Yeezus, and TLOP are present as well. However, there is no 808s & Heartbreak, which is Amazing.
Many people consider Stronger, Gold Digger, and POWER to be a lot more mainstream, which this data confirms. These songs occur in around 300 playlists, while the 4th place song, All Of The Lights, appears in 189 playlists, a big drop.
“Transorthogonal Linguistics”
I used to go to these meetups in DC called Hack and Tell where people showed off cool things they made – one person I met, Travis, worked on a project called Transorthogonal Linguistics, which took two unit vector embeddings of words in n-dimensional space, drew a “line” between them, traversed that line by a fixed distance at a time, and projected the current point on that line back into the n-dimensional unit sphere. This approximation of traversing the perimeter of a sphere is a lot easier than actually tracing it – while pretty easy in 2D, actually tracing the path between two points on a sphere gets pretty nontrivial pretty quickly as dimension increases.
So what? What’s the use here? In a typical word embedding, we can use this technique to find a “path” between words, where the path represents words that slowly morph in meaning from the start word to the end word. Travis made a repo to let you experiment with this. Using sun as a start and moon as an end, you might get something like sun -> sunlight -> glow -> skies -> shadows -> horizon -> earth -> constellations -> Jupiter -> moons -> moon
I tried applying the same technique to Kanye songs! When using the original vectors alone, the results were pretty bad, mostly due to how sparse the matrices were. When I put in a start word and an end word, the projections were often closest to the start and end word, giving us a trivial path that simply contained our start and end songs. However, after applying some PCA to make the feature matrix a lot more dense, there were some interesting results!
When doing Love Lockdown to Lost In The World, the two intermediate songs were See You In My Nightmares and Paranoid, which showed a big skew to 808s.
I also wanted to know what happened if I took one of Kanye’s oldest songs, All Falls Down, and walked to one of his most recent singles, I Love It. This path went through The Glory, Dark Fantasy, Black Skinhead, I Thought About Killing You, and No Mistakes in that order. I think it’s neat that the walk from old Kanye to new Kanye went chronologically through his discography.
Correlating Playlist Presence and Audio Features
I had one last experiment I wanted to try, which was to see if these vectors, constructed from user playlists, correlated to Spotify’s audio features. I actually found some high correlations! I reduced the dimensionality of
M_{playlist}
using PCA and feature agglomeration, with reduced dimensionalities of 81, 27, 3, and 1. Feature agglomeration produced higher maximum correlations, except for in the case where the data was reduced to a single column.
I look at the correlations between a certain column and a given audio feature. That means that, in the case of a single feature vector, the correlations for the audio features will be unique – there will be only one correlation for speechiness. However, with higher dimensional components, we can have repeats. One column might have a high correlation for speechiness, and another column might also have a high speechiness correlation.
Also note: finding the correlations between the original matrix is not particularly interesting, as each column represents a playlist. Thus, a high correlation between a playlist column and an audio feature column doesn’t tell us much about the trends of Spotify users as a whole – it just tells us that there’s some playlist out there that correlates to high tempo music, which happens all the time. Also, before being reduced in dimensionality, the playlist columns are pretty sparse.
Last note: the absolute value of the correlation represents how predictive the feature is, and the sign of the number encodes whether the correlation is positive or negative.
Evaluating dimensionality reduction of size 1 using PCA
r = 0.69 for speechiness
r = -0.56 for loudness
r = 0.43 for valence
r = 0.4 for acousticness
r = -0.38 for energy
Evaluating dimensionality reduction of size 1 using Feature Agglomeration
r = 0.41 for energy
r = 0.4 for loudness
r = -0.39 for acousticness
r = -0.29 for speechiness
r = -0.14 for valence
Evaluating dimensionality reduction of size 27 using PCA
Evaluating dimensionality reduction of size 27 using Feature Agglomeration
r = 0.4 for energy
r = 0.58 for instrumentalness
I think this is neat! We can see in some of these scores that humans choose songs in their playlists – at least, Kanye songs – based on speechiness, loudness, and instrumentalness, among other things.
As a side note, I also tried calculating these same correlations for the pure frequency matrix, instead of the tf-idf matrix, and it turns out that the tf-idf matrix consistently had higher correlation values, although they were only marginally larger. That means that, at least for estimating audio features of songs given the song’s vectorized form, the tf-idf vector is marginally more predictive.
Hopefully this has been a little enlightening! I have the code for these experiments in a very messy jupyter notebook if you wanna dig through it. Hopefully there were No Mistakes
Data Analytics Spotify Vectors Clustering D3 Visualization
Hello World 15 Dec 2018
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Seminar talk, 16 March 2022 - Geometry of Differential Equations
Speaker: Dmitri Alekseevsky
Title: Special Vinberg cones and their applications
A short survey of the Vinberg theory of convex cones (including its informational geometric interpretation) and homogeneous convex cones will be presented. Then we concentrate on the theory of rank 3 special Vinberg cones, associated to metric Clifford
{\displaystyle Cl({\mathbb {R} }^{n})}
modules.
A generalization of the theory to the indefinite special Vinberg cones, associated to indefinite metric Clifford
{\displaystyle Cl({\mathbb {R} }^{p,q})}
modules is indicated. An application of special Vinberg cones to
{\displaystyle N=2,\,d=5,4,3}
Supergravity will be considered.
Retrieved from "https://gdeq.org/w/index.php?title=Seminar_talk,_16_March_2022&oldid=6639"
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Please help me with this question: - Physics - Motion in one dimension - 16806621 | Meritnation.com
Energy of the bullet while hitting the wooden block is
E=\frac{1}{2}m{v}^{2}=\frac{1}{2}×10×{10}^{-3}×100=0.5 J
This energy is used to do the work (W) in penetrating the wood
W=F.d=Fd\phantom{\rule{0ex}{0ex}}where,\phantom{\rule{0ex}{0ex}}F=average force exerted by the bullet on the block\phantom{\rule{0ex}{0ex}}d=10 cm=0.1 m\phantom{\rule{0ex}{0ex}}So,\phantom{\rule{0ex}{0ex}}F×0.1=0.5\phantom{\rule{0ex}{0ex}}F=\frac{0.5}{0.1}=5 N
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On the Mechanism of Pool Boiling Critical Heat Flux Enhancement in Nanofluids | J. Heat Transfer | ASME Digital Collection
Hyungdae Kim,
Hyungdae Kim
, Yongin-city 446-701, Republic of Korea
Ho Seon Ahn,
, Pohang, Gyungbuk 790-784, Republic of Korea
e-mail: mhkim@postech.ac.kr
Moo Hwan Kim Professor
Kim, H., Ahn, H. S., and Kim, M. H. (March 25, 2010). "On the Mechanism of Pool Boiling Critical Heat Flux Enhancement in Nanofluids." ASME. J. Heat Transfer. June 2010; 132(6): 061501. https://doi.org/10.1115/1.4000746
The pool boiling characteristics of water-based nanofluids with alumina and titania nanoparticles of
0.01 vol %
were investigated on a thermally heated disk heater at saturated temperature and atmospheric pressure. The results confirmed the findings of previous studies that nanofluids can significantly enhance the critical heat flux (CHF), resulting in a large increase in the wall superheat. It was found that some nanoparticles deposit on the heater surface during nucleate boiling, and the surface modification due to the deposition results in the same magnitude of CHF enhancement in pure water as for nanofluids. Subsequent to the boiling experiments, the interfacial properties of the heater surfaces were examined using dynamic wetting of an evaporating water droplet. As the surface temperature increased, the evaporating meniscus on the clean surface suddenly receded toward the liquid due to the evaporation recoil force on the liquid-vapor interface, but the nanoparticle-fouled surface exhibited stable wetting of the liquid meniscus even at a remarkably higher wall superheat. The heat flux gain attainable due to the improved wetting of the evaporating meniscus on the fouled surface showed good agreement with the CHF enhancement during nanofluid boiling. It is supposed that the nanoparticle layer increases the stability of the evaporating microlayer underneath a bubble growing on a heated surface and thus the irreversible growth of a hot/dry spot is inhibited even at a high wall superheat, resulting in the CHF enhancement observed when boiling nanofluids.
alumina, boiling, evaporation, heat transfer, multiphase flow, nanofluidics, nanoparticles, titanium compounds, wetting, critical heat flux, evaporating meniscus, nanofluids, nanoparticle deposition, wetting
Boiling, Critical heat flux, Evaporation, Nanofluids, Nanoparticles, Pool boiling, Water, Wetting, Heat transfer
Understanding, Predicting, and Enhancing Critical Heat Flux
Proceedings of the Tenth International Topical Meeting on Nuclear Reactor Thermal Hydraulics
, Seoul, Republic of Korea, pp.
Effect of Nanoparticles on Critical Heat Flux of Water in Pool Boiling Heat Transfer
Pool Boiling Heat Transfer Experiments in Silica-Water Nano-fluids
Role of Ions in Pool Boiling Heat Transfer of Pure and Silica Nanofluids
Dispersion and Surface Characteristics of Nanosilica Suspensions
, 2006, “Experimental Study on CHF Characteristics of Water-TiO2 Nanofluids,” Nuclear Engineering Technology, 38, pp. 61–68.
Effect of Nanoparticles on CHF Enhancement in Pool Boiling of Nano-fluids
Nanofluids and Critical Heat Flux
,” ASME Paper No. MNHT2008-52204.
Surface Wettability Change During Pool Boiling of Nanofluids and Its Effect on Critical Heat Flux
Effect of Nanoparticle Deposition on Capillary Wicking That Influences the Critical Heat Flux in Nanofluids
Sorption and Agglutination Phenomenon of Nanofluids on a Plane Heating Surface During Pool Boiling
Nanofluid Boiling: The Effect of Surface Wettability
On the Role of Structural Disjoining Pressure and Contact Line Pinning in Critical Heat Flux Enhancement During Boiling of Nanofluids
Mechanisms of Thermal Nanofluids on Enhanced Critical Heat Flux (CHF)
Characterization of TiO2 Nanoparticle Suspensions for Electrophoretic Deposition
Bubble Growth in Saturated Pool Boiling in Water and Surfactant Solution
Correlation of Maximum Heat Flux Data for Boiling of Saturated Liquids
,” Ph.D. thesis, University of California, Los Angeles, CA.
On the Quenching of Steel and Zircaloy Spheres in Water-Based Nanofluids With Alumina, Silica and Diamond Nanoparticles
Nucleate Pool Boiling on Porous Metallic Coatings
Experimental Investigations of Pool Boiling CHF Enhancement in Nanofluids
,” Ph.D. thesis, POSTECH, Pohang, Republic of Korea.
Wettability of Heated Surfaces Under Pool Boiling Using Surfactant Solutions and Nano-fluids
Contact Angles and Interface Behavior During Rapid Evaporation of Liquid on a Heated Surface
Evaporating Menisci of Wetting Fluids
A New Mechanism for Pool Boiling Crisis, Recoil Instability and Contact Angle Influence
High Heat Flux Boiling and Burnout as Microphysical Phenomena: Mounting Evidence and Opportunities
An Analytical Solution for the Total Heat Transfer in the Thin-Film Region of an Evaporating Meniscus
The Boiling Crisis Phenomenon. Part II: Dryout Dynamics and Burnout
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Highly Subcooled Boiling in Crossflow | J. Heat Transfer | ASME Digital Collection
LiDong Huang,
Heat Transfer and Phase Change Laboratory, Department of Mechanical Engineering, University of Houston, Houston, TX 77204-4792
Contributed by the Heat Transfer Division for publication in the JOURNAL OF HEAT TRANSFER. Manuscript received by the Heat Transfer Division August 28, 2000; revision received May 4, 2001. Associate Editor: V. P. Carey.
J. Heat Transfer. Dec 2001, 123(6): 1080-1085 (6 pages)
Huang, L., and Witte, L. C. (May 4, 2001). "Highly Subcooled Boiling in Crossflow ." ASME. J. Heat Transfer. December 2001; 123(6): 1080–1085. https://doi.org/10.1115/1.1413762
Experiments were carried out to determine the influence of fluid flow and liquid subcooling on flow boiling heat transfer of Freon-113 across horizontal tubes. The data cover wide ranges of velocity (1.5 to 6.9
m/s)
and extremely high levels of liquid subcooling (29 to 100°C) at pressures ranging from 122 to 509 kPa. Thin-walled cylindrical electric resistance heaters made of Hastelloy-C with diameter of 6.35 mm were used. The azimuthal wall temperature distributions were measured with five thermocouples around the heaters. The data were compared with Chen’s two-mechanism model with modification for subcooled flow boiling. A new nucleate boiling suppression factor for cross flow was developed. The improved model could predict the present data and Yilmaz and Westwater’s (1980) data well with a mean error ratio of 1.02 and standard deviation of 0.17.
boiling, undercooling, heat transfer, external flows, temperature distribution, Boiling, Heat Transfer, Tubes
Boiling, Flow (Dynamics), Heat transfer, Subcooling, Nucleate boiling, Heat flux
Vaporization Inside Horizontal Tubes
Studies of Heat Transmission Through Boiler Tubing at Pressure From 500 to 3000 Pounds
Heat Transfer to Water at High Flux Densities With and Without Surface Boiling
Jens, W. H., and Lottes, P. A., 1951, “Analysis of Heat Transfer, Burnout, Pressure Drop, and Density Data for High Pressure Water,” USAEC Rep. ANL-4627, Argonne Natl. Lab., Argonne, IL.
Bergles, A. E., and Rohsenow, W. M., 1963, “The Determination of Forced Convection Surface Boiling Heat Transfer,” 6th National Heat Transfer Conference of the ASME-AIChE, Boston, MA.
Huang, L., 1994, “Subcooled Flow Boiling Across Horizontal Cylinders,” Ph.D. dissertation, University of Houston, Houston, TX.
Huang, L., and Witte, L. C., 1995, “Forced Convective Film Boiling Heat Transfer around Horizontal Cylinders in Highly Subcooled Freon-113,” Proceedings of the 4th ASME/JSME Thermal Engineering Joint Conference, Vol. 2, pp. 315–322.
An Experimental Investigation of the Effects of Subcooling and Velocity on Boiling of Freon-113
Correlation of Forced Convection Boiling Heat Transfer Data
A Critical Review of Correlations for Convective Vaporization in Tubes and Tube Banks
The Suppression of Saturated Nucleate Boiling by Forced Convective Flow
Heat Transfer Correlations for Natural Convective Boiling
Simplified General Correlation for Saturated Flow Boiling and Comparisons of Correlation with Data
Discussion: “Heat Transfer and Wall Heat Flux Partitioning During Subcooled Flow Nucleate Boiling–A Review” (Warrier, G.R., and Dhir, V.K., 2006, Journal of Heat Transfer, 128, pp. 1243–1256)
Flow Boiling Heat Transfer From Plain and Microporous Coated Surfaces in Subcooled FC-72
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Investigation of Temperature Effects on Cavitation Erosion Behavior Based on Analysis of Erosion Particles | J. Tribol. | ASME Digital Collection
B. Saleh,
, Assiut 71516, Egypt
e-mail: b_saleh@yahoo.com
e-mail: abouelkasem@yahoo.com
A. Ezz El-Deen,
A. Ezz El-Deen
e-mail: ahmed_ezz@yahoo.com
e-mail: shemy2007@yahoo.com
Saleh, B., Abouel-Kasem, A., El-Deen, A. E., and Ahmed, S. M. (September 10, 2010). "Investigation of Temperature Effects on Cavitation Erosion Behavior Based on Analysis of Erosion Particles." ASME. J. Tribol. October 2010; 132(4): 041601. https://doi.org/10.1115/1.4002069
The effect of temperature on the wear particles produced by vibratory cavitation erosion tests on Al-99.92 in distilled water was analyzed. Scanning electron microscope images of wear particles were obtained, forming a database for further analysis. This study showed that the variation of average particle size with temperature was very much similar to the variation of weight loss with temperature. The average particle size was maximum at
40°C
. It was also observed that the average particle size was time dependent. The particle’s morphology features revealed that the predominant erosion mechanism was fatigue failure, irrespective of the temperature.
aluminium, fatigue, particle size, scanning electron microscopy, wear, wear testing, cavitation erosion, temperature, erosion particles, characterization of particles, erosion mechanism, Al-99.92
Cavitation erosion, Erosion, Particle size, Particulate matter, Temperature, Wear, Damage, Cavitation, Weight (Mass)
Standard Test Method of Vibratory Cavitation Erosion Test
,” Paper No. G32-92, Annual Book of ASTM Standards, Philadelphia, PA, p.
Effect of Temperature, pH and Sulphide on the Cavitation Erosion Behaviour of Super Duplex Stainless Steel
Vibratory Cavitation at Elevated Temperature
Effect of Temperature on the Cavitation Erosion of Cast Iron
Gas Content and Temperature Effects in Vibratory Cavitation Tests
Temperature Effects in Cavitation Damage
Effects of Pressure and Temperature Variation in Vibratory Cavitation Damage Test
STCLOPS—A Qualitative Debris Classification System Developed for RAF Early Failure Detection Centers
Likhterov
Mechanism of Action of Oil Additives in Cavitation Erosion of Steel (Discussion)
Stress Produced in a Solid by Cavitation
A Peculiar Behavior of Cavitation—Nuclei Distributions in Sample Water Under Vibratory Erosion Tests
Bulletin of Faculty of Engineering, Assiut University-Egypt
Marked Surface—Roughness Effects on the Development of Microfracture During the Incubation Period of Vibratory Cavitation Erosion
Proceedings of the Third Japan-China Joint Conference
, Osaka, Japan, Vol.
Application of Ultrasonic Cavitation to Metal Working and Surface Treatment of Mild Steel
Effect of Horn-Tip Shape on Cavitation Erosion of Stationary Specimen in a Vibratory Facility
Journal of Engineering Sciences, Assiut University, Egypt
An Indirect Vibratory Method Capable of Simulating Several Cavitating States
Cavitation Bubbles and Damage of an Indirect Vibratory Method
Erosion Characteristic in Ultrasonic Cavitation
Role of Wear Particles in Severe-Mild Wear Transition
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A-5: Evaluation
The volume of a right circular cone is
V=\mathrm{π} {r}^{2} h/3
. If the radius is proportional to the height, and the height is 2, evaluate V at
r=\mathrm{α} h
, then set h = 2 . (This is known as sequential substitution.)
Control-drag (or type) the equation
V=\dots
V=\mathrm{π} {r}^{2} h/3
\stackrel{\text{assign}}{\to }
V
r=\mathrm{α} h
h=2
\genfrac{}{}{0}{}{V}{\phantom{x=a}}|\genfrac{}{}{0}{}{\phantom{\mathrm{f\left(x\right)}}}{r=\mathrm{α} h}
\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{\mathrm{π}}{\textcolor[rgb]{0,0,1}{\mathrm{α}}}^{\textcolor[rgb]{0,0,1}{2}}{\textcolor[rgb]{0,0,1}{h}}^{\textcolor[rgb]{0,0,1}{3}}
\stackrel{\text{evaluate at point}}{\to }
\frac{\textcolor[rgb]{0,0,1}{8}}{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{\mathrm{π}}{\textcolor[rgb]{0,0,1}{\mathrm{α}}}^{\textcolor[rgb]{0,0,1}{2}}
The eval command (implemented in the evaluation template) does not support sequential substitution. Sequential substitutions have to be made as separate evaluations.
Assign the volume expression to V.
V≔\mathrm{π} {r}^{2} h/3
Use two eval commands.
\mathrm{eval}\left(\mathrm{eval}\left(V,r=\mathrm{α} h\right),h=2\right)
The subs command supports sequential replacement from left to right.
\mathrm{subs}\left(r=\mathrm{α} h,h=2,V\right)
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Let AB in M_n (R). Then which of TF are frac? a) if A or b is non-singular then AB is similar b) If A is similar to B then A^k is similar to B^k for every K in N
If I have the following Hypothese:
H1: Women tend to watch romantic films more than men
Apparently it's a chi-square because I'm measuring relationships between expected frequency and Observed frequency and also because Gender is on the nominal level. But what if this is designed for a survey in which questions are on the nominal, Ordinal and interval level (Likert scale - differential sementic) will the outcome still be a Chi-square?
If u, v, w ∈ R n , then span(u, v + w) = span(u + v, w)
A quadratic function has its vertex at the point (−7,2)
\left(-7,2\right)
. The function passes through the point \left(8,3\right)
. When written in vertex form, the function is f(x)=a(x−h)2+k
f\left(x\right)=a{\left(x-h\right)}^{2}+k
In the relation in the table below, write a value that will make the relation not represent a function.
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Q A block of mass m = 1/3 kg is kept on a rough horizontal plane Friction coefficient - Physics - Work Energy And Power - 12469267 | Meritnation.com
Q. A block of mass m = 1/3 kg is kept on a rough horizontal plane. Friction coefficient is
\mu
= 0.75. The work done by minimum force required to drag the block along the plane by a distance 5 m, is W joule, then find the value of W.
The minimum force required to pull the object upwards is ,
So here the work done will be ,
W = Fs cos
\theta
Here the given values are , s=5 m ,\mu =0.75 and m=\frac{1}{3}kg\phantom{\rule{0ex}{0ex}}but the value of inclination that is \theta is not given .\phantom{\rule{0ex}{0ex}}So kindly put the value of \theta in the equation and get the result .
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Seven integers are arranged from least to greatest If the median is 9 and the only mode is 7, - Maths - Statistics - 10817159 | Meritnation.com
Seven integers are arranged from least to greatest. If
the median is 9 and the only mode is 7, what is the
least possible range for the 7 numbers?
\mathrm{I} \mathrm{assume} \mathrm{numbers} \mathrm{should} \mathrm{be} \mathrm{positive}.\phantom{\rule{0ex}{0ex}}\mathrm{Let} \mathrm{numbers} \mathrm{be} {\mathrm{x}}_{1},{\mathrm{x}}_{2},{\mathrm{x}}_{3},.....,{\mathrm{x}}_{7}\phantom{\rule{0ex}{0ex}}\mathrm{According} \mathrm{to} \mathrm{question}\phantom{\rule{0ex}{0ex}}\frac{{\mathrm{x}}_{1}+{\mathrm{x}}_{2}+{\mathrm{x}}_{3}+.....+{\mathrm{x}}_{7}}{7}=9\phantom{\rule{0ex}{0ex}}⇒{\mathrm{x}}_{1}+{\mathrm{x}}_{2}+{\mathrm{x}}_{3}+.....+{\mathrm{x}}_{7}=63\phantom{\rule{0ex}{0ex}}\mathrm{Mode} \mathrm{is} 7, \mathrm{hence} \mathrm{most} \mathrm{frequently} \mathrm{occured} \mathrm{number} \mathrm{is} 7.\phantom{\rule{0ex}{0ex}}\mathrm{Hence} \mathrm{must} \mathrm{atleast} \mathrm{occur} 2 \mathrm{times}.\phantom{\rule{0ex}{0ex}}\mathrm{Sum} \mathrm{of} \mathrm{reamining} \mathrm{five} \mathrm{numbers}=63-7-7=49\phantom{\rule{0ex}{0ex}}\mathrm{Smallest} \mathrm{possible} \mathrm{integers} \mathrm{are} 1,2,3,4,39\phantom{\rule{0ex}{0ex}}\mathrm{Hence} \mathrm{integer} \mathrm{can} \mathrm{range} \mathrm{from} 1 \mathrm{to} 39.\phantom{\rule{0ex}{0ex}}
Tigerdog answered this
0 to 7 :)
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Damping factor - Wikipedia
The term damping factor can also refer to the damping ratio in any damped oscillatory system.
In an audio system, the damping factor gives the ratio of the rated impedance of the loudspeaker (usually assumed to 8 Ω) to the source impedance. Only the magnitude of the loudspeaker impedance is used. The amplifier output impedance is also assumed to be totally resistive.
Comparison of damping factors for a solid state amplifier (Luxman L-509u) and a tube amplifier (Rogue Atlas).
In typical solid state and tube amplifiers, the damping factor varies as a function of frequency. In solid state amplifiers, the damping factor usually has a maximum value at low frequencies, and it reduces progressively at higher frequencies. The figure to the right shows the damping factor of two amplifiers. One is a solid state amplifier (Luxman L-509u) and the other is a tube amplifier (Rogue Atlas). These results are fairly typical of these two types of amplifiers, and they serve to illustrate the fact that tube amplifiers usually have much lower damping factors than modern solid state amplifiers, which is an undesirable characteristic.
3 The damping circuit
3.1 Effect of voice coil resistance
3.2 Effect of cable resistance
3.3 Amplifier output impedance
The source impedance (that is seen by the loudspeaker) includes the connecting cable impedance. The load impedance
{\displaystyle Z_{\mathrm {L} }}
(input impedance) and the source impedance
{\displaystyle Z_{\mathrm {S} }}
(output impedance) are shown in the circuit diagram.
The damping factor
{\displaystyle DF}
{\displaystyle DF={\frac {Z_{\mathrm {L} }}{Z_{\mathrm {S} }}}\,}
{\displaystyle Z_{\mathrm {S} }}
{\displaystyle Z_{\mathrm {S} }={\frac {Z_{\mathrm {L} }}{DF}}\,}
Pierce[1] undertook an analysis of the effects of amplifier damping factor on the decay time and frequency-dependent response variations of a closed-box, acoustic suspension loudspeaker system. The results indicated that any damping factor over 10 is going to result in inaudible differences between that and a damping factor equal to infinity. However, it was also determined that the frequency-dependent variation in the response of the loudspeaker due to the output resistance of the amplifier is much more significant than the effects on system damping. It is also important to not confuse these effects with damping effects, as they are caused by two quite different mechanisms. The calculations suggested that a damping factor in excess of 50 will not lead to audible improvements, all other things being equal.
For audio power amplifiers, this source impedance
{\displaystyle Z_{\mathrm {S} }}
(also: output impedance) is generally smaller than 0.1 Ω, and from the point of view of the driver voice coil, is a near short circuit.
The loudspeaker's nominal load impedance (input impedance) of
{\displaystyle Z_{\mathrm {L} }}
is usually around 4 to 8 Ω, although other impedance speakers are available, sometimes as low as 1 Ω. However, the impedance rating of a loudspeaker is a number that indicates the nominal minimum impedance of that loudspeaker over a representative portion of its operating frequency range. It needs to be kept in mind that most loudspeakers have an impedance that varies considerably with frequency. For a dynamic loudspeaker driver, a peak in the impedance is present at the free-air resonance frequency of the driver, which can be significantly greater in magnitude than the nominal rated impedance. In addition, the inductance of the voice-coil winding leads to a rising impedance at high frequencies, and crossover networks introduce further impedance variations in multi-way loudspeaker systems. This variation in impedance results in the value of the damping factor of the amplifier varying with frequency when it is connected to a loudspeaker impedance load.
In loudspeaker systems, the value of the damping factor between a particular loudspeaker and a particular amplifier describes the ability of the amplifier to control undesirable movement of the speaker cone near the resonant frequency of the speaker system. It is usually used in the context of low-frequency driver behavior, and especially so in the case of electrodynamic drivers, which use a magnetic motor to generate the forces which move the diaphragm.
Speaker diaphragms have mass, and their surroundings have stiffness. Together, these form a resonant system, and the mechanical cone resonance may be excited by electrical signals (e.g., pulses) at audio frequencies. But a driver with a voice coil is also a current generator, since it has a coil attached to the cone and suspension, and that coil is immersed in a magnetic field. For every motion the coil makes, it will generate a current that will be seen by any electrically attached equipment, such as an amplifier. In fact, the output circuitry of the amplifier will be the main electrical load on the "voice coil current generator". If that load has low resistance, the current will be larger and the voice coil will be more strongly forced to decelerate. A high damping factor (which requires low output impedance at the amplifier output) very rapidly damps unwanted cone movements induced by the mechanical resonance of the speaker, acting as the equivalent of a "brake" on the voice coil motion (just as a short circuit across the terminals of a rotary electrical generator will make it very hard to turn). It is generally (though not universally) thought that tighter control of voice coil motion is desirable, as it is believed to contribute to better-quality sound.
A high damping factor in an amplifier is sometimes considered to result in the amplifier having greater control over the movement of the speaker cone, particularly in the bass region near the resonant frequency of the driver's mechanical resonance. However, the damping factor at any particular frequency will vary, since driver voice coils are complex impedances whose values vary with frequency. In addition, the electrical characteristics of every voice coil will change with temperature; high power levels will increase voice coil temperature, and thus resistance. And finally, passive crossovers (made of relatively large inductors, capacitors, and resistors) are between the amplifier and speaker drivers and also affect the damping factor, again in a way that varies with frequency.
The damping circuit[edit]
The voltage generated by the moving voice coil forces current through three resistances:
the resistance of the voice coil itself;
the resistance of the interconnecting cable; and
the output resistance of the amplifier.
Effect of voice coil resistance[edit]
This is key factor in limiting the amount of damping that can be achieved electrically, because its value is larger (say between 4 and 8 Ω typically) than any other resistance in the output circuitry of an amplifier that does not use an output transformer (nearly every solid-state amplifier on the mass market).
A loudspeaker's flyback current is not only dissipated through the amplifier output circuit, but also through the internal resistance of the loudspeaker itself. Therefore the choice of different loudspeakers will lead to different damping factors when coupled with the same amplifier.
Effect of cable resistance[edit]
The damping factor is affected to some extent by the resistance of the speaker cables. The higher the resistance of the speaker cables, the lower the damping factor. When the effect is small, it is called voltage bridging.
{\displaystyle Z_{\mathrm {L} }}
{\displaystyle Z_{\mathrm {S} }}
Amplifier output impedance[edit]
Modern solid state amplifiers, which use relatively high levels of negative feedback to control distortion, have very low output impedances—one of the many consequences of using feedback—and small changes in an already low value change overall damping factor by only a small, and therefore negligible, amount.
Thus, high damping factor values do not, by themselves, say very much about the quality of a system; most modern amplifiers have them, but vary in quality nonetheless.
Tube amplifiers typically have much lower feedback ratios, and in any case almost always have output transformers that limit how low the output impedance can be. Their lower damping factors are one of the reasons many audiophiles prefer tube amplifiers. Taken even further, some tube amplifiers are designed to have no negative feedback at all.
Typical modern solid-state amplifiers with negative feedback tend to have high damping factors, usually above 50 and sometimes even greater than 150. High damping factors tend to reduce the extent to which a loudspeaker "rings" (undergoes unwanted short-term oscillation after an impulse of power is applied), but the extent to which damping factors higher than about 20 help in this respect is easily overstated;[2] there will be significant effective internal resistance, as well as some resistance and reactance in cross-over networks and speaker cables.[3] Older amplifiers, plus modern triode and even solid-state amplifiers with low negative feedback, will tend to have damping factors closer to unity, or even less than 1 (very low damping factor/high output impedance amplifiers approximate current sources).
Effects of amplifier damping factor on the frequency response when connected to a simulated impedance load typical of a two-way closed box loudspeaker system.
Although extremely high values of damping factor in an amplifier will not necessarily make the loudspeaker–amplifier combination sound better,[4] a high damping factor can serve to reduce the intensity of added frequency response variations that are undesirable. The figure on the right shows the effect of damping factor on the frequency response of an amplifier when that amplifier is connected to a simulated loudspeaker impedance load. This load is moderately demanding but not untypical of high-fidelity loudspeakers that are on the market, and it is based on the circuit proposed by Atkinson.[5] Stereophile have recognised the importance of amplifier damping factor, and have made the use of the simulated loudspeaker load a routine part of their measurements of amplifiers. At around 4 kHz, the real-life difference between an amplifier with a moderate (100) damping factor and one with a low (20) damping factor is about 0.37 dB. Keep in mind though, that the amplifier with the low damping factor is acting more like a graphic equaliser than is the amplifier with the moderate damping factor.
It is clear from the various amplifier frequency response curves that low damping factor values result in significant changes in the frequency response of the amplifier in a number of frequency bands. This will result in broad levels of sound coloration that are highly likely to be audible. In addition, the frequency response changes will depend on the frequency-dependent impedance of whichever loudspeaker happens to be connected to the amplifier. Hence, in high-fidelity sound reproduction systems, amplifiers with moderate to high damping factors are the preferred option if accurate sound reproduction is desired when those amplifiers are connected to typical multi-way loudspeaker impedance loads.
Some amplifier designers, such as Nelson Pass, claim that loudspeakers can sound better with lower electrical damping,[6] although this may be attributed to listener preference rather than technical merit. A lower damping factor helps to enhance the bass response of the loudspeaker by several decibels (where the impedance of the speaker would be at its maximum), which is useful if only a single speaker is used for the entire audio range. Therefore, some amplifiers, in particular vintage amplifiers from the 1950s, 1960s and 1970s, feature controls for varying the damping factor. While such bass "enhancement" may be pleasing to some listeners, it nonetheless represents a distortion of the input signal.
One example of a vintage amplifier with a damping control is the Accuphase E-202, which has a three-position switch described by the following excerpt from its owner's manual:[7]
"Speaker Damping Control enhances characteristic tonal qualities of speakers. The damping factor of solid state amplifiers is generally very large and ideal for damping the speakers. However, some speakers require an amplifier with a low damping factor to reproduce rich, full-bodied sound. The E-202 has a Speaker Damping Control which permits choice of three damping factors and induces maximum potential performance from any speaker. Damping factor with an 8 ohm load becomes more than 50 when this control is set to NORMAL. Likewise, it is 5 at MEDIUM position, and 1 at SOFT position. It enables choosing the speaker sound that one prefers."
In contrast, in modern high-fidelity amplification, the trend is to separate the bass signal and amplify it with a dedicated amplifier. Often, amplifiers for bass reproduction are integrated with the speaker cabinet, a configuration known as the powered subwoofer. In a topology that includes a dedicated amplifier for bass, the damping factor of the main amplifier is not as relevant, and that of the bass amplifier is also irrelevant if that amplifier is integrated with the speaker and cabinet as a unit, since all those components are designed together and optimized for the reproduction of bass.
Damping is also a concern in guitar amplifiers (an application in which controlled distortion is desirable) and low damping can be better. Numerous guitar amplifiers have damping controls, and the trend to include this feature has been increasing since the 1990s. For instance the Marshall Valvestate 8008 rack-mounted stereo amplifier[8] has a switch between "linear" and "Valvestate" mode:
"Linear/Vstate selector. Slide to select linear or Valvestate performance. The Valvestate mode gives extra warm harmonics plus the richness of tone, which is unique to the Valvestate power stage. Linear mode produces a highly defined hi-fi tone that gives a totally different character to the sound and suits certain modern "metal" styles, or PA applications."
This is actually a damping control based on negative current feedback, which is evident from the schematic,[9] where the same switch is labeled as "Output Power Mode: Current/Voltage". The "Valvestate" mode introduces negative current feedback which raises the output impedance, lowers the damping factor, and alters the frequency response, similarly to what occurs in a tube amplifier. (Contrary to the claim in the handbook, this circuit topology has appeared in numerous solid-state guitar amplifiers since the 1970s.)
^ Pierce, Dick (2002). "Damping Factor: Effects On System Response" (PDF). Professional Audio Development. Retrieved 2021-06-12.
^ Toole, Floyd E. (1975). "Damping, Damping Factor, and Damn Nonsense" (PDF). AudioScene Canada (February): 16–17. Retrieved 2021-06-12.
^ Augspurger, George L. (1967). "The Damping Factor Debate" (PDF). Electronics World. Ziff-Davis Publishing Company (January). Retrieved 2021-06-12.
^ Elliott, Rod (2010). "Impedance, and how it affects audio equipment". Elliot Sound Products. Retrieved 2021-06-12.
^ Atkinson, John (1995). "Real-Life Measurements". Stereophile. Retrieved 2021-06-12.
^ Pass, Nelson. "Current Source Amplifiers and Sensitive / Full-Range Drivers" (PDF). Retrieved 2021-06-12.
^ "Accuphase E-202 Integrated Stereo Amplifier" (PDF). Retrieved 2021-06-12.
^ "Marshall Valvestate Power Amplifiers Handbook" (PDF). Retrieved 2021-09-01.
^ "Marshall 8008 Valvestate Schematic" (PDF). Retrieved 22 August 2011.
Julian L Bernstein, Audio Systems, p. 364, John Wiley, 1966.
Radio Electronics - December 1954 and January 1955, D. J. Tomcik, Missing Link in Speaker Operation
Marantz's Legendary Audio Classics. Ben Blish, Damping Factor
Audioholics. AV University. Amplifier Technology. Dick Pierce, Damping factor: Effects on System Response
ProSoundWeb. Studyhall. Chuck McGregor, What is Loudspeaker Damping?
8 Ohm Output and 150 Ohm Input - What is that?
Retrieved from "https://en.wikipedia.org/w/index.php?title=Damping_factor&oldid=1041761664"
Audio amplifier specifications
|
Prove that \mathbb{Z}+(3x) is a subring of \mathbb{Z}[x] and there
\mathbb{Z}+\left(3x\right)
\mathbb{Z}\left[x\right]
\mathbb{Z}\left[x\right]⇒\mathbb{Z}+\left(3x\right)
There is no surjective homomorphism. Suppose that
\phi :\mathbb{Z}\left[x\right]\to \mathbb{Z}+\left(3x\right)
deg\phi \left(f\left(x\right)\right)=deg\left(f\left(\phi \left(x\right)\right)=deg\left(f\left(x\right)\right)deg\left(\phi \left(x\right)\right)
deg\phi \left(x\right)>1
then nothing in the image could be of degree 1, and the map could not be surjective. So
\phi \left(x\right)=3ax+b
a,b\in \mathbb{Z}
\phi
were surjective, something would have to map to 3x, and this forces
a=±1
By composing with the isomorphisms of
\mathbb{Z}\left[x\right]
given by maps
x↦x-c
x↦-x
, we may WLOG assume that
\phi \left(x\right)=3x
The question we ask is, can we find g(x) such that
\phi \left(g\left(x\right)\right)=g\left(3x\right)=3{x}^{2}
Working over
\mathbb{Q}\left[x\right]
, we see if
g\left(3x\right)=3{x}^{2}
g\left(x\right)={x}^{2}/3\overline{)\in }\mathrm{ℤ}\left[x\right]
. And because
\phi
extends to an isomorphism of
\mathbb{Q}\left[x\right]
, this is the only polynomial in
\mathbb{Q}\left[x\right]
that would satisfy the condition. So the map cannot be surjective.
{\mathrm{End}}_{\mathrm{ℝ}\left[x\right]}\left(M\right)
M=\frac{\mathbb{R}\left[x\right]}{\left({x}^{2}+1\right)}
\mathbb{R}\left[x\right]
Let a and b belong to a ring R and let m be an integer. Prove that m(ab) = (ma)b = a(mb)
\varphi
\varphi :\mathbb{Z}\to {\mathbb{Z}}_{8}
phI\left(1\right)=6
Let A be a DVR in its field of fractions F, and
{F}^{\prime }\subset F
a subfield. Then is it true that
A\cap {F}^{\prime }
is a DVR in
{F}^{\prime }?
I can see that it is a valuation ring of
{F}^{\prime }
, but how do I show, for example, that
A\cap {F}^{\prime }
is Noetherian?
If a, b are elements of a ring and m, n ∈ Z, show that (na) (mb) = (mn) (ab)
|
Proving single solution to an initial value problem y' = \left|\frac{1}{1+x^2}
Proving single solution to an initial value problem
y\text{'}=|\frac{1}{1+{x}^{2}}+\mathrm{sin}|{x}^{2}+\mathrm{arctan}{y}^{2}||, y\left({x}_{0}\right)={y}_{0}
or each
\left({x}_{0},{y}_{0}\right)\in \mathbb{R}×\mathbb{R}
I need to prove that there is a single solution defined on
\mathbb{R}
Zemmiq34
g\left(x,y\right)=\mathrm{sin}|{x}^{2}+\mathrm{arctan}\left({y}^{2}\right)|=\mathrm{sin}\left({x}^{2}+\mathrm{arctan}\left({y}^{2}\right)\right)
This function is derivable, hence
{g}_{y}\left(x,y\right)=\frac{2y}{{y}^{4}+1}\mathrm{cos}\left\{\left({x}^{2}+\mathrm{arctan}\left\{\left({y}^{2}\right)\right\}\right)\right\}
{g}_{y}
is continuous thus for each J there is a constant
{L}_{J}
|g\left(x,{y}_{1}\right)-g\left(x,{y}_{2}\right)|\le {L}_{J}|{y}_{1}-{y}_{2}|
|f\left(x,{y}_{1}\right)-f\left(x,{y}_{2}\right)|
=||\frac{1}{1+{x}^{2}}+g\left(x,{y}_{1}\right)|-|\frac{1}{1+{x}^{2}}+g\left(x,{y}_{2}\right)\mid \mid
\le |\frac{1}{1+{x}^{2}}+g\left(x,{y}_{1}\right)-\frac{1}{1+{x}^{2}}-g\left(x,{y}_{2}\right)|
=|g\left(x,{y}_{1}\right)-g\left(x,{y}_{2}\right)|\le {L}_{J}|{y}_{1}-{y}_{2}|
f\left(x,y\right)
is Lipschitz continuous on the 𝑦 variable, and this is true for each box
J×\left(-\mathrm{\infty },\mathrm{\infty }\right)
and therefore there is a single solution on
\mathbb{R}
\left({x}_{0},{y}_{0}\right)\in \mathbb{R}×\mathbb{R}
Find the area of the parallelogram with vertices A(0,1), B(3,0), C(5,-2), and D(2,-1).
Manipulation of equation into quadratic form
\frac{c\sqrt{{d}_{1}+{d}_{2}}-\sqrt{{d}_{1}}{z}_{1}-\delta \left({d}_{2}\right)\sqrt{{P}_{0}{P}_{1}}}{\sqrt{{d}_{1}+{d}_{2}}}=-{z}_{\beta }
\left({\delta }^{2}{p}_{1}{p}_{0}\right){d}_{2}^{2}-{d}_{2}\left[{\left(c+{z}_{\beta }\right)}^{2}-2\sqrt{{d}_{1}{p}_{1}{p}_{0}}{z}_{1}\delta \right]+{d}_{1}\left[{z}_{1}^{2}-{\left(c+{z}_{\beta }\right)}^{2}\right]=0
How do you multiply this radical equation,
\sqrt[3]{9}\left(\sqrt[3]{3}+\sqrt[3]{9}\right)
\mathbf{\text{x}}\text{'}\text{'}\left(t\right)=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]\mathbf{\text{x}}\left(t\right)
\mathbf{\text{x}}\text{'}\left(t\right)=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]\mathbf{\text{x}}\left(t\right)
\mathbf{\text{x}}\left(t\right)=\mathbf{\text{c}}\left[\begin{array}{c}{e}^{-t}\\ {e}^{-t}\end{array}\right]
Function to grab specific digits from a number?
Recently I’ve been thinking a lot about grabbing digits from numbers, for things like calculating multiplication persistence, i.e. turning
1234
1×2×2×3×4
. I’ve been able to come up with
\left[\mathrm{log}x\right]+1
to output the number of digits of the given number x, as well as
⌊\frac{x}{{10}^{⌊\mathrm{log}x⌋-n+1}}⌋
which outputs the first n digits of any number x. But that’s about as far as I’ve been able to think up. Is there any way to construct a function that retrieves the nth digit of a given number? Forgive me if my notation or comprehension is poor, I have no formal education in this field of maths.
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Reliable Dissipative Sampled-Data Control for Uncertain Systems With Nonlinear Fault Input | J. Comput. Nonlinear Dynam. | ASME Digital Collection
Suwon 440-746, South Korea
e-mail: krsakthivel@yahoo.com
S. Vimal Kumar,
S. Vimal Kumar
RVS Technical Campus-Coimbatore,
e-mail: svimalkumar16@gmail.com
D. Aravindh,
e-mail: aravindhjkk@gmail.com
Anna University-Regional Campus,
e-mail: selvamath89@gmail.com
Contributed by the Design Engineering Division of ASME for publication in the JOURNAL OF COMPUTATIONAL AND NONLINEAR DYNAMICS. Manuscript received July 6, 2015; final manuscript received October 29, 2015; published online December 4, 2015. Assoc. Editor: Haiyan Hu.
Sakthivel, R., Vimal Kumar, S., Aravindh, D., and Selvaraj, P. (December 4, 2015). "Reliable Dissipative Sampled-Data Control for Uncertain Systems With Nonlinear Fault Input." ASME. J. Comput. Nonlinear Dynam. July 2016; 11(4): 041008. https://doi.org/10.1115/1.4031980
This paper investigates the robust reliable β-dissipative control for uncertain dynamical systems with mixed actuator faults via sampled-data approach. In particular, a more general reliable controller containing both linear and nonlinear parts is constructed for the considered system. Then, by applying the input delay approach, the sampling measurement of the digital control signal is transformed into time-varying delayed one. The aim of this paper is to design state feedback sampled-data controller to guarantee that the resulting closed-loop system to be strictly (Q, S, R)-β-dissipative. By constructing appropriate Lyapunov function and employing a delay decomposition approach, a new set of delay-dependent sufficient stabilization criteria is obtained in terms of linear matrix inequalities (LMIs). Moreover, the obtained LMIs are dependent, not only upon upper bound of time delay but also depend on the dissipative margin β and the actuator fault matrix. As special cases,
H∞
and passivity control performances can be deduced from the proposed dissipative control result. Finally, numerical simulation is provided based on a flight control model to verify the effectiveness and applicability of the proposed control scheme.
Design optimization , Stability, Dynamical analysis, Dynamical method
Actuators, Closed loop systems, Control equipment, Control systems, Delays, Design, Flight, Simulation results, Stability, Theorems (Mathematics), Uncertain systems, State feedback, Dynamic systems, Trajectories (Physics), Computer simulation, Signals
Further Results on Delay-Range-Dependent Stability With Additive Time-Varying Delay Systems
Robust Adaptive Control Scheme for Uncertain Non-Linear Model Reference Adaptive Control Systems With Time-Varying Delays
Robust Output Feedback Control of Uncertain Time-Delay Systems With Actuator Saturation and Disturbances
Adaptive H∞ Control in Finite Frequency Domain for Uncertain Linear Systems
Reliable Control of a Class of Switched Cascade Nonlinear Systems With Its Application to Flight Control
Non-Fragile H2 Reliable Control for Switched Linear Systems With Actuator Faults
Sig. Process.
Control Design of Interval Type-2 Fuzzy Systems With Actuator Fault: Sampled-Data Control Approach
Reliable H∞ Control for Discrete-Time Fuzzy Systems With Infinite-Distributed Delay
Design of Passive Fault-Tolerant Flight Controller Against Actuator Failure
Robust Sampled-Data Cruise Control Scheduling of High Speed Train
Transp. Res. Part C
Sampled-Data Cooperative Adaptive Cruise Control of Vehicles With Sensor Failures
Robust Sampled-Data H∞ Control for Mechanical Systems
Mixed H∞/ Passive Synchronization for Complex Dynamical Networks With Sampled-Data Control
Sampled-Data Observer-Based Output-Feedback Fuzzy Stabilization of Nonlinear Systems: Exact Discrete-Time Design
Improved Stability Criteria for Synchronization of Chaotic Lur'e Systems Using Sampled-Data Control
Fault-Tolerant Sampled-Data Control of Flexible Spacecraft With Probabilistic Time Delays
Improved Robust Stabilization Method for Linear Systems With Interval Time-Varying Input Delays by Using Wirtinger Inequality
Further Improvement of Free Weighting Matrices Technique for Systems With Time-Varying Delay
Delay-Dependent Exponential Stability Criteria for Neutral Systems With Interval Time-Varying Delays and Nonlinear Perturbations
New Stability Criteria for Singular Systems With Time-Varying Delay and Nonlinear Perturbations
Further Improvement on Delay-Range-Dependent Stability Results for Linear Systems With Interval Time-Varying Delays
Observer Based (Q, V, R)-α-Dissipative Control for T-S Fuzzy Descriptor Systems With Time Delay
Delay-Dependent Robust Dissipativity Conditions for Delayed Neural Networks With Random Uncertainties
Dissipative Reliable Controller Design for Uncertain Systems and Its Application
, “H∞
Output Tracking Control for Flight Control Systems With Time-Varying Delay
Robust Exponential Stability for Uncertain Time-Varying Delay Systems With Delay Dependence
|
Distribution of Primes | Brilliant Math & Science Wiki
Patrick Corn, Sameer Kailasa, Ethan W, and
Trumpets Technologies
The prime number theorem describes the asymptotic distribution of prime numbers. It gives us a general view of how primes are distributed amongst positive integers and also states that the primes become less common as they become larger. Informally, the theorem states that if any random positive integer is selected in the range of zero to a large number
N
, the probability that the selected integer is a prime is about
\frac{1}{\ln N},
\ln N
N
One application of the theorem is that it gives a sense of how long it will take to find a prime of a certain size by a random search. Many cryptosystems (e.g. RSA) require primes
p \approx 2^{512}
; the theorem says that the probability that a randomly chosen number of that size is prime is roughly
\frac{1}{\ln\big(2^{512}\big)} \approx \frac1{355},
\frac1{177}
if the search is restricted to odd numbers. So the expectation is that roughly
177
numbers will have to be tested for primality, which can be computationally expensive.
Counting Primes; Statement of the Theorem
\text{Li}(x)
and Tables of Values
n^\text{th}
Connection with the Riemann Hypothesis
Distribution of Primes Modulo
n
To understand the precise statement of the theorem, define
π(x)
to be the prime-counting function that gives the number of primes less than or equal to
x
x
π(10) = 4
because there are four prime numbers (2, 3, 5, and 7) less than or equal to 10. It is a well-known result that
\lim_{x\to\infty} \pi(x) = \infty
. The next natural question to ask is, "How fast does
\pi(x)
go to infinity?" The prime number theorem provides the answer: at the same rate as
\frac{x}{\ln(x)},
\lim_{x\to\infty}\frac{\pi(x)}{\hspace{2mm} \frac{x}{\ln(x)}\hspace{2mm} }=1,
or written with the asymptotic notation,
\pi(x)\sim\frac{x}{\ln x}.
Note that this notation (and the theorem) does not say anything about the limit of the difference of the two functions as
x \to \infty
. Indeed, this difference tends to infinity as
x
increases. The theorem does say that this difference tends to infinity more slowly than
\frac{x}{\ln x}
\text{Li}(x)
\text{Li}(x) = \int_2^x\frac{dt}{\ln(t)}
is actually a closer approximation to
\pi(x)
. It is an easy calculus exercise that
\text{Li}(x) \sim \frac{x}{\ln(x)}
\text{Li}(x) \sim \pi(x)
as well. Many more precise statements about the size of
\pi(x)
\text{Li}(x)
. Here is a table comparing values of these functions, where the relative error is computed as
\frac{\left\lfloor \text{Li}(x) \right\rfloor -\pi(x)}{\pi(x)}:
x \hspace{25mm}
\pi(x)
\hspace{25mm}
\big\lfloor \text{Li}(x) \big\rfloor
\hspace{25mm}
10
4
5
25\%
10^2
25
29
16\%
10^3
168
177
5.4\%
10^4
1,229
1,245
1.3\%
10^5
9,592
9,630
0.4\%
10^6
78,498
78,628
0.2\%
10^7
664,579
664,917
0.05\%
10^8
5,761,455
5,762,208
0.01\%
10^9
50,847,534
50,849,234
0.003\%
n^\text{th}
The prime number theorem is equivalent to the statement that the
n^\text{th}
p_n
p_n \sim n\ln(n),
the asymptotic notation meaning again that the relative error of this approximation approaches
0
n
The Riemann zeta function has a deep connection with the distribution of primes. The famous Riemann hypothesis, about the zeroes of the zeta function, is equivalent to many statements involving prime numbers. In particular, it is equivalent to a much tighter bound than can currently be proven on the error in the estimate for
\pi(x)
\pi(x) = \operatorname{Li}(x) + O\big(\sqrt{x} \log{x}\big)
(see big O notation for an explanation of the latter term). More explicitly, the Riemann hypothesis implies
\big|\pi(x) - \operatorname{li}(x)\big| < \frac{1}{8\pi} \sqrt{x} \, \log(x) \quad \text{for all } x \ge 2657,
\operatorname{li}(x) = \int_0^x \frac{dt}{\ln(t)} = \operatorname{Li}(x) + \operatorname{li}(2)
c > 1
be a real number. Show that for sufficiently large
x
, there is always a prime between
x
cx
\pi(cx) \sim \frac{cx}{\ln(cx)} \sim \frac{cx}{\ln(x)} \sim c\pi(x).
\lim_{x\to\infty} \frac{\pi(cx)}{\pi(x)} = c,
\frac{\pi(cx)}{\pi(x)} > 1
x
_\square
c=2,
this statement is known as Bertrand's postulate; using more precise estimates for
\pi(x)
, one can show that there is always a prime between
x
2x
x > 3
1 2 3 A finite number greater than 3
\infty
How many numbers in this infinite sequence are perfect squares?
0!, \ \ 1!, \ \ 2!, \ \ 3!,\, \ldots
Bonus: Try to prove your answer!
n
From the classical proof of Dirichlet's theorem on primes in arithmetic progressions, it is known that for any positive integer
n
, the prime numbers are approximately evenly distributed among the reduced residue classes modulo
n
(i.e., the residue classes that are relatively prime to
n
). For example, the reduced residue classes modulo
10
are 1, 3, 5, and 7, so one would expect that
\frac{1}{4}
of all prime numbers end in 1,
\frac{1}{4}
end in 3, and so on.
Yes, definitely Probably not
In general, about
\frac{1}{4}
of all primes less than
end in a 1, 3, 7, and 9, respectively. Would the same symmetry hold if we considered the last digits in pairs of consecutive primes less than, say,
n=100000?
Surprisingly, in 2016, mathematicians Robert Lemke Oliver and Kannan Soundarajan discovered that this equidistribution does not persist, at least for relatively small
x,
when one considers the distribution of pairs of consecutive primes, modulo
n
. Consider, for example, the distribution of prime pairs modulo
3
\pi\big(x; (i,j)\big) := \# \text{ of pairs of consecutive primes } (p, q) \text{ such that } p \equiv i \bmod{3} \text{ and } q \equiv j \bmod{3}.
If the primes were truly random, one might expect the limiting proportions
\lim_{x\to\infty} \frac{\pi\big(x; (1,1)\big)}{\pi(x)} = \lim_{x\to\infty} \frac{\pi\big(x; (1,2)\big)}{\pi(x)} = \lim_{x\to\infty} \frac{\pi\big(x; (2,1)\big)}{\pi(x)} = \lim_{x\to\infty} \frac{\pi\big(x; (2,2)\big)}{\pi(x)} = \frac{1}{4}.
But the numerical evidence suggests otherwise, at least when considering relatively small
x
! Among the first million primes (excluding 2), one sees the erratic distribution
\begin{aligned} \pi \big(10^6; (1,1)\big) &= 215873\\ \pi \big(10^6; (1,2)\big) &=283957\\ \pi \big(10^6; (2,1)\big) &= 283957\\ \pi \big(10^6; (2,2)\big) &= 216213. \end{aligned}
These numbers differ substantially from the expected value
250000
for each one.
Lemke Oliver and Soundarajan have also observed that this discrepancy holds in other bases, such as base 10. They have conjectured an explanation for the biases, but the conjecture has not yet been proven.
2 \le k \le n
, choose a divisor
d_k
k
, uniformly at random from the set of divisors of
k.
P(n)
d_2 + d_3 + \cdots + d_n
Amazingly, there exists a positive integer
N
n\ge N
P(n)
\frac1{32}
. What is the smallest
N
for which this is true?
Cite as: Distribution of Primes. Brilliant.org. Retrieved from https://brilliant.org/wiki/distribution-of-primes/
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Panic Section
10 Panic Section
10.1 Getting help in ICL
10.2 BEGPLOT or DEVICE does not work in ICL
10.3 READF cannot find a column label
10.4 Strange behaviour of DLIMITS
10.5 Error messages from SHOWPONGO in ICL
10.6 AGI problems
10.7 RESET peculiarity (ICL only)
10.8 Plotting large numbers of positions
When in ICL, on-line help on PONGO may be examined using the command:
ICL> help pongo
This will provide a brief description of the package and how to begin and end a PONGO plotting session. Once the PONGO commands have been made available within ICL, i.e. by typing
% pongo
at the C-shell prompt or:
ICL> pongo
at the ICL> prompt, on-line help on any PONGO command may be examined using the command:
ICL> help <pongo command>
will produce a brief introduction to PONGO, and the command:
ICL> help introduction
will produce a more detailed introduction to PONGO and how to get started. These help commands are also available at the PONGO> prompt. A classified list of PONGO applications (HELP CLASSIFIED) is provided within the help system and help is also provided on running the PONGO examples (HELP EXAMPLES).
If difficulties are encountered with PONGO and the on-line help system does not reveal the cause, it is recommended that the relevant sections of this document are read, or re-read. It may be that this document does not provide enough detail concerning the behaviour of the PGPLOT graphics package: for detailed information regarding PGPLOT, the PGPLOT manual should be consulted (available as a Miscellaneous User Document, MUD, at all Starlink sites). As a last resort, the problem may be regarded as a software bug and reported to the Starlink Software support mailing list (starlink@jiscmail.ac.uk).
If the error message:
TOOFEWPARS Not enough procedure parameters
is returned after the BEGPLOT or ENDPLOT commands have been invoked, the reason is that another application package or ICL procedure has switched parameter checking on within ICL (SG/5). The problem is simply overcome by typing the command:
ICL> set nocheckpars
The PONGO commands will then work correctly.
If a column label is used with the READF command instead of the column number (see §5) and READF fails, e.g. :
PONGO> readf myfile.dat xcol=1 ycol=3 zcol="ZCOL"
File: /disk/scratch/jbloggs/plots/myfile.dat
XCOL - 1 is column number 1.
YCOL - 3 is column number 3.
!! ZCOL incorrectly specified as ZCOL.
! READF: Data could not be read.
then either the data file does not have column labels or the incorrect column label has been used. It is of particular significance here that the column labels are case-sensitive. In fact, JBLOGGS found that the column label for his ZCOL data was Zcol:
ZCOL - Zcol is column number 5.
EXCOL - 0 is column number 0.
EYCOL - 0 is column number 0.
LABCOL - 0 is column number 0.
SYMCOL - 0 is column number 0.
42 data points read.
PONGO>
If PONGO does not seem to be doing something correctly, it is often because there are unwanted data in the error columns. This can come about when a file that does not contain errors is read (the appropriate parameters to READF have been set to zero), following one where there have been errors. Here, it should be noted that setting a column parameter in READF to zero means that no data will be read from the file into that particular area, it does not clear the data values. This is a feature rather than a bug, because it allows X and Y data from two different files to be plotted together, as long as each file contains the same number of points (the number of points read from the second file is assumed to be the number required). Although advantageous, it is possible for the internal data areas to get into a mess with this arrangement. If this seems to have happened then the command:
PONGO> clear data
should be used to clear all the data areas.
The commands SHOWPONGO may sometimes deliver the error messages
!! Object ’PONGO_<param>’ not found.
!! DAT_FIND: Error finding a named component in an HDS structure.
when invoked (<param> refers to a PONGO parameter name in this context). The reason is simply that no global parameter of that name currently exists. This can either be because PONGO has not been used before (or not a lot), or because the ADAM global parameter file ($HOME/adam/GLOBAL.sdf) has been deleted since the last PONGO session. Although the error messages look alarming, they are harmless. To stop these error messages being output, the command RESETPONGO should be executed. This command will assign a default value to each of the PONGO global parameters except PONGO_DATA, the data file name used by READF, and SHOWPONGO will be subsequently more informative.
The AGI database is normally kept in a separate HDS file (SG/4) for each machine in your $HOME directory. The command BEGPLOT opens the AGI database and reads database information relevant to the graphics device being used. During a PONGO plotting session, the database file is held open to update the database as the plotting session proceeds. At the end of plotting, when ENDPLOT is executed, the update of the database is completed and the database file is closed. If another non-PONGO application which uses AGI is run before the current PONGO plotting session has been ended (i.e. using ENDPLOT), it will be unable to access the AGI database and will subsequently fail. In the event of this happening, using the PONGO ENDPLOT command will restore the correct behaviour of the non-PONGO application.
What is more important, in order to use AGI successfully, is the use of ENDPLOT before exiting ICL or CL. Because PONGO keeps the AGI database file open throughout a plotting session it is possible to exit ICL or CL without having first executed ENDPLOT. If this is done, it is likely that the AGI database will not have been fully updated before being closed, with the result that the next time AGI is used it will behave inconsistently. PONGO crashing during a plotting session (hopefully, a rare event) can also result in a corrupted AGI database. There is no solution to this problem other than to delete the database file and start again. The AGI database files are kept in the directory $HOME and have names of the form agi_<machine>.sdf, where <machine> is the name of the machine on which AGI is being used. This file may be deleted at any time before a BEGPLOT command or after an ENDPLOT command. The BEGPLOT command may then be used to begin a new PONGO plot, creating a new AGI database as a result.
When using the ADAM parameter system (SG/4) RESET qualifier to set parameters back to their default values (this facility would be particularly useful for BOXFRAME, for example), there will be no effect on parameters which get their values from global parameters. Use the RESETPONGO command to reset global parameters, or alternatively (somewhat drastically) delete the GLOBAL.sdf file.
PONGO can store
5000
positions at any time. This can very occasionally cause problems for specific types of use. One way in which you can actually plot more positions than this, is by reading in the data in chunks (of 5000 points) and plotting these. The following ICL procedure shows one way in which you might do this:
proc superplot file, x1, x2, y1, y2, n
begplot xwindows xmin=(x1) xmax=(x2) ymin=(y1) ymax=(y2)
limits (x1) (x2) (y1) (y2)
loop for i = 0 to (n) step 1
print "Plotting points from" (i*5000) " to " (j*5000-1)
readf data=(file) xcol=1 ycol=2 from=(i*5000) to=(j*5000-1) all accept
This would then be invoked as in (assuming the procedure was kept in a file superplot.icl):
ICL > load superplot.icl
ICL > superplot mygalaxies.dat -3 3 -3 3 10
Which would plot
50000
positions stored in the file mygalaxies.dat.
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Operating Requirements for LNG and LPG gas carriers
Every LNG or LPG tanker should meet Operating Requirements like cargo info and transfer operations, system and controls, personal training and etc.
Entry into spaces
Carriage of cargo at low temperature
Additional operating requirements
Information shall be on board and available to all concerned, giving the necessary data for the safe carriage of the cargo. Such information shall include but not limited for each product carried:
a full description of the physical and chemical properties necessary for the safe containment of the cargo;
action to be taken in the event of spills or leaks;
counter measures against accidental personal contact;
fire fighting procedures and fire fighting media;
procedures for cargo transfer, gas freeing, ballasting, tank cleaning and changing cargoes;
special equipment needed for the safe handling of the particular cargo;
minimum allowable inner hull steel temperatures;
Products required to be inhibited should be refused if the certificate required by “Special Requirements for LNG and LPG gas carriers”Inhibition is not supplied.
A copy of the IGC-Code or national regulations incorporating the provisions of the IGC-Code should be on board every ship covered by the IGC-Code.
The Master should ascertain that the quantity and characteristics of each product to be loaded are within the limits indicated in the International Certificate of Fitness for the Carriage of Liquefied Gases in Bulk provided for in Section 1, A.5 and the Loading and Stability booklet provided for in Section 2, 2.5 and that the products are listed in the International Certificate of Fitness for the Carriage of Liquefied Gases in Bulk as required under Section 3 of the Certificate.
Care should be taken to avoid dangerous chemical reactions if cargoes are mixed. This is of particular significance in respect of:
tank cleaning procedures required between successive cargoes in the same tank;
and simultaneous carriage of cargoes which react when mixed. This should be permitted only if the complete cargo systems including, but not limited only if the complete cargo system including, but not limited to, cargo pipework, tanks, vent systems and refrigeration systems are separate as defined in Section 1,C.32.
Personnel Reference is made to the provisions of the International Convention on Standards of Training, Certification and Watchkeeping for seafarers, 1978 and in particular to the “Mandatory Minimum Requirements for the Training and Qualifications of Masters, Officers and Ratings of Chemical Tankers” – Regulation V/2, Chapter V of the Annex to that Convention and to Resolution 11 of the International Conference on Training and Certification of Seafarers, 1978.x involved in cargo operations should be adequately trained in handling procedures. All personnel should be adequately trained in the use of protective equipment provided on board and have basic training in the procedures, appropriate to their duties, necessary under emergency conditions.
Officers should be trained in emergency procedures to deal with conditions of leakage, spillage or fire involving the cargo Reference is made to the Medical First Aid Guide for use in Accidents Involving Dangerous Goods (MFAG), which provides advice on the treatment of casualties in accordance with the symptoms exhibited as well as equipment and antidotes that may be appropriate for treating the casualty, and to the relevant provisions of STCW Code, parts A and Bx. A sufficient number of them should be instructed and trained in essential first aid for the cargoes carried.
Personnel should not enter cargo tanks, hold spaces, void spaces, cargo handling spaces or other enclosed spaces where gas may accumulate unless:
the gas content of the atmosphere in that space is determined by means of fixed or portable equipment to ensure oxygen sufficiency and the absence of toxic atmosphere;
or personnel wear breathing apparatus and other necessary protective equipment and the entire operation is under the close supervision of a responsible officer.
Personnel entering any space designated as gas dangerous on a ship carrying flammable products should not introduce any potential source of ignition into the space unless it has been certified gas free and is maintained in that condition.
For internal insulation tanks, special fire precautions should be taken in the event of hot work carried out in the vicinity of the tanks. The gas absorbing and de-absorbing characteristics of the insulation material should thereby be taken into account.
For internal insulation tanks, repairs should be carried out in accordance with the procedures provided for in “Cargo containment system of gas vessel”Internal insulation tanks.
When carrying cargoes at low temperatures:
if provided the heating arrangements associated with cargo containment systems should be operated in such a manner as to avoid the temperature falling below that for which the material of the hull structure is designed;
loading should be carried out in such a manner as to ensure that unsatisfactory temperature gradients do not occur in any cargo tank, piping, or other ancillary equipment;
and when cooling down tanks from temperatures at or near ambient the cool down procedure laid down for that particular tank, piping and ancillary equipment should be followed closely.
Personnel should be made aware of the hazards associated with the cargo being handled and should be instructed to act with care and use the appropriate protective equipment as mentioned in “Personal protection of crew on Gas Carriers”Protective equipment during cargo handling.
Cargo emergency shut-down and alarm systems involved in cargo transfer should be tested and/or checked before cargo handling operations begin. Essential cargo handling controls should also be tested and checked prior to transfer operations.
Transfer operations including emergency procedures should be discussed between ship personnel and the persons responsible at the shore facility prior to commencement and communications maintained throughout the transfer operations.
LNG tanker Valletta
The closing time of the valve referred to in “Cargo Tank Instrumentation on Gas Tankers”Overflow control (i.e. time from shut-down signal initiation to complete valve closure) should not be greater than:
\frac{3600·U}{{L}_{R}} \left[s\right],
U – ullage volume at operating signal level [m3];
LR – maximum loading rate agreed between ship and shore facility [m3/h].
The loading rate should be adjusted to limit surge pressure on valve closure to an acceptable level taking into account the loading hose or arm, the ship and the shore piping systems where relevant.
Additional operating requirements will be found in the following paragraphs of the Rules:
Accommodation, service and machinery spaces and control stations Accommodation, service and machinery spaces and control stations Pressure and Temperature Control Pressure relief systems Pressure relief systems Inerting
Inert gas production on board Spaces required to be entered during normal cargo handling operations Spaces required to be entered during normal cargo handling operations Cargo Tank Instrumentation Cargo Tank Instrumentation Protective equipment
Protective equipment First aid equipment Filling limits Information to be provided to the master Arrangement of machinery spaces of Category A Refrigeration systems
Refrigeration systems Exclusion of air from vapour spaces Moisture control Maximum allowable quantity of cargo per tank Submerged electric cargo pumps Ammonia
Chlorine Diethyl ether and vinyl ethyl ether Ethylene oxide Isopropylamine, monoethylamine Methyl acetylene-propadiene mixtures Propylene oxide and mixtures
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Section 60.12 (07IW): Sheaf of differentials—The Stacks project
Section 60.12: Sheaf of differentials (cite)
60.12 Sheaf of differentials
In this section we will stick with the (small) crystalline site as it seems more natural. We globalize Definition 60.6.1 as follows.
Definition 60.12.1. In Situation 60.7.5 let $\mathcal{F}$ be a sheaf of $\mathcal{O}_{X/S}$-modules on $\text{Cris}(X/S)$. An $S$-derivation $D : \mathcal{O}_{X/S} \to \mathcal{F}$ is a map of sheaves such that for every object $(U, T, \delta )$ of $\text{Cris}(X/S)$ the map
\[ D : \Gamma (T, \mathcal{O}_ T) \longrightarrow \Gamma (T, \mathcal{F}) \]
is a divided power $\Gamma (V, \mathcal{O}_ V)$-derivation where $V \subset S$ is any open such that $T \to S$ factors through $V$.
This means that $D$ is additive, satisfies the Leibniz rule, annihilates functions coming from $S$, and satisfies $D(f^{[n]}) = f^{[n - 1]}D(f)$ for a local section $f$ of the divided power ideal $\mathcal{J}_{X/S}$. This is a special case of a very general notion which we now describe.
Please compare the following discussion with Modules on Sites, Section 18.33. Let $\mathcal{C}$ be a site, let $\mathcal{A} \to \mathcal{B}$ be a map of sheaves of rings on $\mathcal{C}$, let $\mathcal{J} \subset \mathcal{B}$ be a sheaf of ideals, let $\delta $ be a divided power structure on $\mathcal{J}$, and let $\mathcal{F}$ be a sheaf of $\mathcal{B}$-modules. Then there is a notion of a divided power $\mathcal{A}$-derivation $D : \mathcal{B} \to \mathcal{F}$. This means that $D$ is $\mathcal{A}$-linear, satisfies the Leibniz rule, and satisfies $D(\delta _ n(x)) = \delta _{n - 1}(x)D(x)$ for local sections $x$ of $\mathcal{J}$. In this situation there exists a universal divided power $\mathcal{A}$-derivation
\[ \text{d}_{\mathcal{B}/\mathcal{A}, \delta } : \mathcal{B} \longrightarrow \Omega _{\mathcal{B}/\mathcal{A}, \delta } \]
Moreover, $\text{d}_{\mathcal{B}/\mathcal{A}, \delta }$ is the composition
\[ \mathcal{B} \longrightarrow \Omega _{\mathcal{B}/\mathcal{A}} \longrightarrow \Omega _{\mathcal{B}/\mathcal{A}, \delta } \]
where the first map is the universal derivation constructed in the proof of Modules on Sites, Lemma 18.33.2 and the second arrow is the quotient by the submodule generated by the local sections $\text{d}_{\mathcal{B}/\mathcal{A}}(\delta _ n(x)) - \delta _{n - 1}(x)\text{d}_{\mathcal{B}/\mathcal{A}}(x)$.
We translate this into a relative notion as follows. Suppose $(f, f^\sharp ) : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \mathcal{O}')$ is a morphism of ringed topoi, $\mathcal{J} \subset \mathcal{O}$ a sheaf of ideals, $\delta $ a divided power structure on $\mathcal{J}$, and $\mathcal{F}$ a sheaf of $\mathcal{O}$-modules. In this situation we say $D : \mathcal{O} \to \mathcal{F}$ is a divided power $\mathcal{O}'$-derivation if $D$ is a divided power $f^{-1}\mathcal{O}'$-derivation as defined above. Moreover, we write
\[ \Omega _{\mathcal{O}/\mathcal{O}', \delta } = \Omega _{\mathcal{O}/f^{-1}\mathcal{O}', \delta } \]
which is the receptacle of the universal divided power $\mathcal{O}'$-derivation.
Applying this to the structure morphism
\[ (X/S)_{\text{Cris}} \longrightarrow \mathop{\mathit{Sh}}\nolimits (S_{Zar}) \]
(see Remark 60.9.6) we recover the notion of Definition 60.12.1 above. In particular, there is a universal divided power derivation
\[ d_{X/S} : \mathcal{O}_{X/S} \to \Omega _{X/S} \]
Note that we omit from the notation the decoration indicating the module of differentials is compatible with divided powers (it seems unlikely anybody would ever consider the usual module of differentials of the structure sheaf on the crystalline site).
Lemma 60.12.2. Let $(T, \mathcal{J}, \delta )$ be a divided power scheme. Let $T \to S$ be a morphism of schemes. The quotient $\Omega _{T/S} \to \Omega _{T/S, \delta }$ described above is a quasi-coherent $\mathcal{O}_ T$-module. For $W \subset T$ affine open mapping into $V \subset S$ affine open we have
\[ \Gamma (W, \Omega _{T/S, \delta }) = \Omega _{\Gamma (W, \mathcal{O}_ W)/\Gamma (V, \mathcal{O}_ V), \delta } \]
where the right hand side is as constructed in Section 60.6.
Lemma 60.12.3. In Situation 60.7.5. For $(U, T, \delta )$ in $\text{Cris}(X/S)$ the restriction $(\Omega _{X/S})_ T$ to $T$ is $\Omega _{T/S, \delta }$ and the restriction $\text{d}_{X/S}|_ T$ is equal to $\text{d}_{T/S, \delta }$.
Lemma 60.12.4. In Situation 60.7.5. For any affine object $(U, T, \delta )$ of $\text{Cris}(X/S)$ mapping into an affine open $V \subset S$ we have
\[ \Gamma ((U, T, \delta ), \Omega _{X/S}) = \Omega _{\Gamma (T, \mathcal{O}_ T)/\Gamma (V, \mathcal{O}_ V), \delta } \]
Lemma 60.12.5. In Situation 60.7.5. Let $(U, T, \delta )$ be an object of $\text{Cris}(X/S)$. Let
\[ (U(1), T(1), \delta (1)) = (U, T, \delta ) \times (U, T, \delta ) \]
in $\text{Cris}(X/S)$. Let $\mathcal{K} \subset \mathcal{O}_{T(1)}$ be the quasi-coherent sheaf of ideals corresponding to the closed immersion $\Delta : T \to T(1)$. Then $\mathcal{K} \subset \mathcal{J}_{T(1)}$ is preserved by the divided structure on $\mathcal{J}_{T(1)}$ and we have
\[ (\Omega _{X/S})_ T = \mathcal{K}/\mathcal{K}^{[2]} \]
Proof. Note that $U = U(1)$ as $U \to X$ is an open immersion and as (60.9.1.1) commutes with products. Hence we see that $\mathcal{K} \subset \mathcal{J}_{T(1)}$. Given this fact the lemma follows by working affine locally on $T$ and using Lemmas 60.12.4 and 60.6.5. $\square$
It turns out that $\Omega _{X/S}$ is not a crystal in quasi-coherent $\mathcal{O}_{X/S}$-modules. But it does satisfy two closely related properties (compare with Lemma 60.11.2).
Lemma 60.12.6. In Situation 60.7.5. The sheaf of differentials $\Omega _{X/S}$ has the following two properties:
$\Omega _{X/S}$ is locally quasi-coherent, and
for any morphism $(U, T, \delta ) \to (U', T', \delta ')$ of $\text{Cris}(X/S)$ where $f : T \to T'$ is a closed immersion the map $c_ f : f^*(\Omega _{X/S})_{T'} \to (\Omega _{X/S})_ T$ is surjective.
Proof. Part (1) follows from a combination of Lemmas 60.12.2 and 60.12.3. Part (2) follows from the fact that $(\Omega _{X/S})_ T = \Omega _{T/S, \delta }$ is a quotient of $\Omega _{T/S}$ and that $f^*\Omega _{T'/S} \to \Omega _{T/S}$ is surjective. $\square$
Comment #1608 by Rakesh Pawar on August 26, 2015 at 09:49
In the lemma 49.12.2., in the central formula the subscript of
\Omega
on the right should be replaced with
\Gamma(W/\mathcal{O}_W)/\Gamma(V/\mathcal{O}_V),\delta
\Omega
\Gamma(W/\mathcal{O}_W)/\Gamma(V/\mathcal{O}_V),\delta
Similar correction in lemma 49.12.4.
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Compute consistent initial conditions for ode15i - MATLAB decic - MathWorks Australia
Compute Consistent Initial Conditions for Implicit Equations
fixed_y0
fixed_yp0
y0_new
yp0_new
resnrm
Compute consistent initial conditions for ode15i
[y0_new,yp0_new] = decic(odefun,t0,y0,fixed_y0,yp0,fixed_yp0)
[y0_new,yp0_new] = decic(odefun,t0,y0,fixed_y0,yp0,fixed_yp0,options)
[y0_new,yp0_new,resnrm] = decic(___)
[y0_new,yp0_new] = decic(odefun,t0,y0,fixed_y0,yp0,fixed_yp0) uses y0 and yp0 as guesses for the initial conditions of the fully implicit function odefun, holds the components specified by fixed_y0 and fixed_yp0 as fixed, then computes values for the nonfixed components. The result is a complete set of consistent initial conditions. The new values yo_new and yp0_new satisfy odefun(t0,y0_new,yp0_new) = 0 and are suitable to be used as initial conditions with ode15i.
[y0_new,yp0_new] = decic(odefun,t0,y0,fixed_y0,yp0,fixed_yp0,options) also uses the options structure options to specify values for AbsTol and RelTol. Create the options structure using odeset.
[y0_new,yp0_new,resnrm] = decic(___) returns the norm of odefun(t0,y0_new,yp0_new) as resnrm. If the norm seems unduly large, then use options to decrease the relative error tolerance RelTol, which has a default value of 1e-3.
Consider the implicit system of equations
\begin{array}{cc}0& =2{y}_{1}^{\prime }-{y}_{2}\\ 0& ={y}_{1}+{y}_{2}\end{array}
These equations are straightforward enough that it is simple to read off consistent initial conditions for the variables. For example, if you fix
{y}_{1}=1
{y}_{2}=-1
according to the second equation and
{y}_{1}^{\prime }=-1/2
according to the first equation. Since these values of
{y}_{1}
{y}_{1}^{\prime }
{y}_{2}
satisfy the equations, they are consistent.
Confirm these values by using decic to compute consistent initial conditions for the equations, fixing the value
{y}_{1}=1
. Use guesses of y0 = [1 0] and yp0 = [0 0], which do not satisfy the equations and are thus inconsistent.
odefun = @(t,y,yp) [2*yp(1)-y(2); y(1)+y(2)];
yp0 = [0 0];
[y0,yp0] = decic(odefun,t0,y0,[1 0],yp0,[])
{\mathrm{ty}}^{2}{\left({\mathit{y}}^{\prime }\right)}^{3}-{\mathit{y}}^{3}{\left({\mathit{y}}^{\prime }\right)}^{2}+\mathit{t}\left({\mathit{t}}^{2}+1\right){\mathit{y}}^{\prime }-{\mathit{t}}^{2}\mathit{y}=0
\mathit{f}\left(\mathit{t},\mathit{y},{\mathit{y}}^{\prime }\right)=0
\mathit{t}
\mathit{y}
{\mathit{y}}^{\prime }
\mathit{f}\left({\mathit{t}}_{0},\mathit{y},{\mathit{y}}^{\prime }\right)=0
\mathit{y}\left({\mathit{t}}_{0}\right)=\sqrt{\frac{3}{2}}
{\mathit{y}}^{\prime }\left({\mathit{t}}_{0}\right)
{\mathit{y}}^{\prime }\left({\mathit{t}}_{0}\right)=0
\left[1\text{\hspace{0.17em}}10\right]
\mathit{y}\left(\mathit{t}\right)=\sqrt{{\mathit{t}}^{2}+\frac{1}{2}}
Functions to solve, specified as a function handle that defines the functions to be integrated. odefun represents the system of implicit differential equations that you want to solve using ode15i.
f\left(t,y,y\text{'}\right)
. odefun must accept all three input arguments, t, y, and yp even if one of the arguments is not used in the function.
y\text{'}-y=0
\begin{array}{l}y{\text{'}}_{1}-{y}_{2}=0\\ y{\text{'}}_{2}+1=0\text{\hspace{0.17em}},\end{array}
t0 — Initial time
Initial time, specified as a scalar. decic uses the initial time to compute consistent initial conditions that satisfy odefun(t0,y0_new,yp0_new) = 0.
y0 — Initial guesses for y-components
Initial guesses for y-components, specified as a vector. Each element in y0 specifies an initial condition for one dependent variable
{y}_{n}
in the system of equations defined by odefun.
fixed_y0 — y-components to hold fixed
vector of 1s and 0s | []
y-components to hold fixed, specified as a vector of 1s and 0s, or as [].
Set fixed_y0(i) = 1 if no change is permitted in the guess for y0(i).
Set fixed_y0 = [] if any entry can be changed.
You cannot fix more than length(yp0) components. Depending on the specific problem, it is not always possible to fix certain components of y0 or yp0. It is a best practice not to fix more components than is necessary.
yp0 — Initial guesses for y'-components
Initial guesses for y'-components, specified as a vector. Each element in yp0 specifies an initial condition for one differentiated dependent variable
y{\text{'}}_{n}
fixed_yp0 — y'-components to hold fixed
y'-components to hold fixed, specified as a vector of 1s and 0s, or as [].
Set fixed_yp0(i) = 1 if no change is permitted in the guess for yp0(i).
Set fixed_yp0 = [] if any entry can be changed.
Options structure, specified as a structure array. Use the odeset function to create or modify the options structure. The relevant options for use with the decic function are RelTol and AbsTol, which control the error thresholds used to compute the initial conditions.
Example: options = odeset('RelTol',1e-5)
y0_new — Consistent initial conditions for y0
Consistent initial conditions for y0, returned as a vector. If the value of resnrm is small, then yo_new and yp0_new satisfy odefun(t0,y0_new,yp0_new) = 0 and are suitable to be used as initial conditions with ode15i.
yp0_new — Consistent initial conditions for yp0
Consistent initial conditions for yp0, returned as a vector. If the value of resnrm is small, then yo_new and yp0_new satisfy odefun(t0,y0_new,yp0_new) = 0 and are suitable to be used as initial conditions with ode15i.
resnrm — Norm of residual
Norm of residual, returned as a vector. resnrm is the norm of odefun(t0,y0_new,yp0_new).
A small value of resnrm indicates that decic successfully computed consistent initial conditions that satisfy odefun(t0,y0_new,yp0_new) = 0.
If the value of resnrm is large, try adjusting the error thresholds RelTol and AbsTol using the options input.
The ihb1dae and iburgersode example files use decic to compute consistent initial conditions before solving with ode15i. Type edit ihb1dae or edit iburgersode to view the code.
You can additionally use decic to compute consistent initial conditions for DAEs solved by ode15s or ode23t. To do this, follow these steps.
Rewrite the system of equations in fully implicit form f(t,y,y') = 0.
Call decic to compute consistent initial conditions for the equations.
Specify y0_new as the initial condition in the call to the solver, and specify yp_new as the value of the InitialSlope option of odeset.
ode15i | odeget | odeset
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Arylsulfatase - Wikipedia
Arylsulfatase monomer, Pseudomonas aeruginosa
Arylsulfatase (EC 3.1.6.1, sulfatase, nitrocatechol sulfatase, phenolsulfatase, phenylsulfatase, p-nitrophenyl sulfatase, arylsulfohydrolase, 4-methylumbelliferyl sulfatase, estrogen sulfatase) is a type of sulfatase enzyme with systematic name aryl-sulfate sulfohydrolase.[1][2][3][4] This enzyme catalyses the following chemical reaction
a phenol sulfate + H2O
{\displaystyle \rightleftharpoons }
a phenol + sulfate
This group of enzymes has similar specificities.
Arylsulfatase A (also known as "cerebroside-sulfatase")
Arylsulfatase B (also known as "N-Acetylgalactosamine-4-Sulfatase")
Steroid sulfatase (formerly known as "Arylsulfatase C")
^ Dodgson KS, Spencer B, Williams K (October 1956). "Studies on sulphatases. 13. The hydrolysis of substituted phenyl sulphates by the arylsulphatase of Alcaligenes metalcaligenes". The Biochemical Journal. 64 (2): 216–21. PMC 1199721. PMID 13363831.
^ Roy AB (1960). "The synthesis and hydrolysis of sulfate esters". Advances in Enzymology and Related Subjects of Biochemistry. 22: 205–35. PMID 13744184.
^ Roy AB (April 1976). "Sulphatases, lysosomes and disease". The Australian Journal of Experimental Biology and Medical Science. 54 (2): 111–35. doi:10.1038/icb.1976.13. PMID 13772.
^ Webb EC, Morrow PF (September 1959). "The activation of an arysulphatase from ox liver by chloride and other anions". The Biochemical Journal. 73 (1): 7–15. PMC 1197004. PMID 13843260.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Arylsulfatase&oldid=1045005539"
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High school probability problems, solved and explained
High school probability questions and answers
Recent questions in High school probability
sibuzwaW 2020-11-27 Answered
X denotes a binomial random variable with parameters n and p. For each exercise, indicate which area under the appropriate normal curve would be determined to approximate the specified binomial probability.
P\left(7<X<10\right)
Assume a Poisson distribution with lambda
=5.0
X=1
Find the number of possible outcomes. A die is rolled 8 times.
Assume that when adults with smartphones are randomly selected, 46% use them in meetings or classes. If 20 adult smartphone users are randomly selected, find the probability that exactly 15 of them use their smartphones in meetings or classes.
A bag contains 6 red, 4 blue and 8 green marbles. How many marbles of each color should be added so that the total number of marbles is 27, but the probability of randomly selecting one marble of each color remains unchanged.
The spinner below is divided into eight equal parts. Find the theoretical probability described below as a fraction.
If x is a binomial random variable, find the probabilities below by using the binomial probability table.
P\left(x<11\right)\text{ }for\text{ }n=15,p=0.1
P\left(x\ge 11\right)\text{ }for\text{ }n=25,p=0.4
P\left(x=2\right)\text{ }for\text{ }n=20,p=0.6
What is the row of Pascal’s triangle containing the binomial coefficients (nk),
0\le k\le 9
The score of adults on an IQ test are approximately Normal with mean 100 and standart deviation 15. Alysha scores 135 on such a test. She scores highter than what percent of all adults?
What is the probability that a randomly chosen location will have a snow depth between 2.25 and 2.75 inches
If two events A and B are independent and you know that P(A)=0.3, what is the value of P(A|B)?
Suppose that when the weather is rainy the probability that Lily goes to her outdoor baseball practice is 0.1, but when it's not raining the probability is 0.9. The probability it will rain tomorrow is 0.3. Determine the probability that She will go to her baseball practice tomorrow.
Eighty percent of the employees at the General Mills plant on Laskey Road have their bimonthly wages sent directly to their bank electronic funds transfers. This is called direct deposit . Suppose we select a random sample of seven recipients. What is the probability that all seven employees use direct deposit?
According to Exercise 16, the probability that a U.S. resident has traveled to Canada is 0.18, to Mexico is 0.09, and to bith countries is 0.04.
A binomial probability is given: P(x <= 64) Find the answer that corresponds to the binomial probability statement. a) P(63.5< x < 64.5) b) P(x > 63.5) c) P(x < 63.5) d) P(x > 64.5) e) P(x < 64.5)
A croissant shop has plain croissants, cherry croissants, chocolate croissants, almond croissants, apple croissants, and broccoli croissants. How many ways are there to choose 6 dozen croissants?
A business traveler has five shirts, four pairs of pants, and two jackets in his closet. How many possible outfits, consisting of a shirt, a pair of pants, and a jacket, could he wear?
How many 3-letter combinations of the letters in the word EIGHT are possible?
State the criteria for a binomial probability experiment.
High school probability is one of those interesting tasks that young students receive as part of their statistics and probability assignments. You should also find the answers to probability exercises for high school along with the high school probability questions that will help you come up with the most efficient solutions. If you would like to focus on statistical challenges, you should also use high school probability problems as well as it is based on our examples. Be it related to equations or graphs that will talk about probability, follow our examples and things will become clearer!
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1. Sailing ships used to send messages with signal flags flown from their masts. How many different
For question 1, we must understand that with a minimum of 2 flags and a maximum of 4 flags being used for each signal configuration, the following configurations are possible:
1. 2 Flags Used
In the case of only 2 flags being used, any 4 of the flags can be picked initially as the possible first flag, leaving 3 flags remaining as the possible second flag. Thus, the number of different signals that can be made here is:
4\cdot \left(4-1\right)=4\cdot 3=12
In the case of using 3 flags, any 4 of the flags can be picked initially as the possible first flag, leaving 3 flags remaining as the possible second flag, leaving 2 flags remaining as the possible third flag. Thus, the number of different signals that can be made here is:
4\cdot \left(4-1\right)\cdot \left(4-2\right)=4\cdot 3\cdot 2=24
In the final case of using all 4 flags, any 4 of the flags can be picked initially as the possible first flag, leaving 3 flags remaining as the possible second flag, leaving 2 flags remaining as the possible third flag, leaving only 1 flag as the final flag to be chosen. Thus the number of different signals that can be made here is:
4\cdot \left(4-1\right)\cdot \left(4-2\right)\cdot \left(4-3\right)=4\cdot 3\cdot 2\cdot 1=24
Adding all of these possible signals will give us the answer:
12+24+24=60
For question 2, we can think of a tree-diagram to represent this. Starting with 3 separate trees: Science, Geography, and Phys. Ed at the top. From here, there are 4 possible classes to branch off of the first class: Art, Music, French, or business. We then branch off of these classes to Math or English, and the final period will be Math if the third period was English, or English if the third period was Math, in order to ensure no duplicates.
From here,
3\cdot 4\cdot 2=24
possible timetables can be chosen.
{26}^{2}×{10}^{4}
{}^{6}{P}_{2}
A
B
S
S=A\cup B
A\cap B=\mathrm{\varnothing }
\mathcal{P}\left(X\right)
X
|Y|
Y
|\mathcal{P}\left(A\right)|+|\mathcal{P}\left(B\right)|=|\mathcal{P}\left(A\right)\cup \mathcal{P}\left(B\right)|
\mathcal{P}\left(A\right)
\mathcal{P}\left(B\right)
\mathcal{P}\left(X\right)
|Y|
N\left(A\cup B\cup C\right)=N\left(A\right)+N\left(B\right)+N\left(C\right)-N\left(A\cap B\right)-N\left(A\cap C\right)-N\left(B\cap C\right)+N\left(A\cap B\cap C\right)
21=9+10+7-N\left(A\cap B\right)-N\left(A\cap C\right)-N\left(B\cap C\right)+5
N\left(A\cap B\right)+N\left(A\cap C\right)+N\left(B\cap C\right)=10
N\left(A\cap B\cap C\right)
=N\left(A\cap B+B\cap C+A\cap C\right)-2\ast N\left(A\cap B\cap C\right)=10-2\ast 5=0
X=\left\{\left(123\right),\left(132\right),\left(124\right),\left(142\right),\left(134\right),\left(143\right),\left(234\right),\left(243\right)\right\}
{A}_{4}
X
x=\left(123\right)
4=|\mathcal{O}\left(x\right)|=|G|/|{G}_{x}|=12/|{G}_{x}|
{G}_{x}=\left\{1\right\}
f
A\to B
|A|=4,|B|=3
{3}^{4}-\left(\genfrac{}{}{0}{}{3}{1}\right){2}^{4}+\left(\genfrac{}{}{0}{}{3}{2}\right){1}^{4}
\left(\genfrac{}{}{0}{}{3}{2}\right){1}^{4}
\left(\genfrac{}{}{0}{}{3}{1}\right){2}^{4}
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Show if M is free of rank n as R-module,
\frac{M}{IM}
\frac{R}{I}
I\subset R
IM=\left\{\sum {r}_{i}{x}_{i}\mid {r}_{i}\in I,{x}_{i}\in M\right\}.
Your proof that
\frac{M}{IM}
is a submodule of M is surely wrong. Take
R=M=\mathbb{Z}
I=2\mathbb{Z}
\frac{M}{IM}=\frac{\mathbb{Z}}{2\mathbb{Z}}
that doesn't even embed in M.
Saying that M is a free module of rank n is the same as saying that
M\stackrel{\sim }{=}{R}^{n\mid }
and it's not restrictive to take
M={R}^{n}
I{R}^{n}={I}^{n}
\frac{M}{IM}=\frac{{R}^{n}}{{I}^{n}}\stackrel{\sim }{=}{\left(\frac{R}{I}\right)}^{n}
as R-modules via the map
\left({x}_{1},{x}_{2},\cdots ,{x}_{n}\right)+{I}^{n}↦\left({x}_{1}+I,{x}_{2}+I,\cdots ,{x}_{n}+I\right)
This is also an isomorphism of
\frac{R}{I}
-modules, as you can verify.
Multiples of 4 as sum or difference of 2 squares
Is it true that for any
n\in \mathbb{N}
4n={x}^{2}+{y}^{2}
4n={x}^{2}-{y}^{2}
x,y\in \mathbb{N}\cup \left(0\right)
I was just working out a proof and this turns out to be true from
n=1\text{ }\text{to}\text{ }n=20
. After that I didn't try, but I would like to see if a counterexample exists for a greater value of n.
K⇝K\left(X\right)
f\left(a\right)=Ua{U}^{-1}
{Z}_{12}
|a|,|b|
|a+b|
a=5,b=4
\frac{{\mathbb{F}}_{2}\left[X,Y\right]}{\left({Y}^{2}+Y+1,{X}^{2}+X+Y\right)}
\frac{\left({\mathbb{F}}_{2}\left[Y\right]\right\}\left\{\left({Y}^{2}+Y+1\right)\right\}\frac{\left\{\right)}{X}}{\left({X}^{2}+X+\stackrel{―}{Y}\right)}
Abelian group G such that the infinite-order elements form a subgroup with the identity.
\left\{g\in G\mid g=e\text{ }\text{or g has infinite order}\right\}
is a subgroup of G, what can we say about the order of the elements of G?
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Custom Nonlinear ENSO Data Analysis - MATLAB & Simulink - MathWorks Australia
Load Data and Fit Library and Custom Fourier Models
Use Fit Options to Constrain a Coefficient
Create Second Custom Fit with Additional Terms and Constraints
Create a Third Custom Fit with Additional Terms and Constraints
This example fits the ENSO data using several custom nonlinear equations. The ENSO data consists of monthly averaged atmospheric pressure differences between Easter Island and Darwin, Australia. This difference drives the trade winds in the southern hemisphere.
The ENSO data is clearly periodic, which suggests it can be described by a Fourier series:
y\left(x\right)={a}_{0}+\sum _{i=1}^{\infty }{a}_{i}\mathrm{cos}\left(2\pi \frac{x}{{c}_{i}}\right)+{b}_{i}\mathrm{sin}\left(2\pi \frac{x}{{c}_{i}}\right)
where ai and bi are the amplitudes, and ci are the periods (cycles) of the data. Determine how many cycles exist.
As a first attempt, assume a single cycle and fit the data using one cosine term and one sine term.
{y}_{1}\left(x\right)={a}_{0}+{a}_{1}\mathrm{cos}\left(2\pi \frac{x}{{c}_{1}}\right)+{b}_{1}\mathrm{sin}\left(2\pi \frac{x}{{c}_{1}}\right)
If the fit does not describe the data well, add additional cosine and sine terms with unique period coefficients until a good fit is obtained.
The equation is nonlinear because an unknown coefficient c1 is included as part of the trigonometric function arguments.
Load the data and open the Curve Fitter app.
The app includes the Fourier series as a nonlinear library equation. However, the library equation does not meet the needs of this example because its terms are defined as fixed multiples of the fundamental frequency w. Refer to Fourier Series for more information. Create the built-in library Fourier fit to compare with your custom equations:
In the app, on the Curve Fitter tab, in the Data section, click Select Data. In the Select Fitting Data dialog box, select month as the X Data value and pressure as the Y Data value.
On the Curve Fitter tab, in the Fit Type section, click the arrow to open the gallery. In the fit gallery, click Fourier in the Regression Models group.
In the Table Of Fits pane, double-click the Fit name value and enter Fourier.
In the Fit Options pane, change the number of terms to 8.
Observe the library model fit. In the next steps, you will create custom equations to compare.
Duplicate your fit. Right-click your fit in the Table Of Fits pane and select Duplicate "Fourier".
Name the new fit Enso1Period.
On the Curve Fitter tab, in the Fit Type section, open the fit type gallery and click Custom Equation in the Custom group.
In the Fit Options pane, replace the example text in the equation edit box with the following:
a0 + a1*cos(2*pi*x/c1) + b1*sin(2*pi*x/c1)
The app applies the fit to the enso data.
The graphical and numerical results shown here indicate that the fit does not describe the data well. In particular, the fitted value for c1 is unreasonably small. Your initial fit results might differ from these results because the starting points are randomly selected.
By default, the coefficients are unbounded and have random starting values from 0 to 1. The data include a periodic component with a period of about 12 months. However, with c1 unconstrained and with a random starting point, this fit failed to find that cycle.
To assist the fitting procedure, constrain c1 to a value from 10 to 14. In the Fit Options pane, click Advanced Options to expand the section and view the constraints for the coefficients. Observe that by default the coefficients are unbounded (bounds of -Inf and Inf).
In the Coefficient Constraints table, change the Lower and Upper bounds for c1 to constrain the cycle from 10 to 14 months, as shown next.
The Curve Fitter app updates the fit.
Observe the new fit and the residuals plot. If necessary, click Residuals Plot in the Visualization section of the Curve Fitter tab.
The fit appears to be reasonable for some data points but clearly does not describe the entire data set very well. As predicted, the numerical results in the Results pane (c1=11.94) indicate a cycle of approximately 12 months. However, the residuals show a systematic periodic distribution, indicating that at least one more cycle exists. There are additional cycles that you should include in the fit equation.
To refine your fit, you need to add an additional sine and cosine term to y1(x) as follows:
{y}_{2}\left(x\right)={y}_{1}\left(x\right)+{a}_{2}\mathrm{cos}\left(2\pi \frac{x}{{c}_{2}}\right)+{b}_{2}\mathrm{sin}\left(2\pi \frac{x}{{c}_{2}}\right)
and constrain the upper and lower bounds of c2 to be roughly twice the bounds used for c1.
Duplicate your fit by right-clicking it in the Table Of Fits pane and selecting Duplicate "Enso1Period".
In the Fit Options pane, add two terms to the end of the previous equation so that the equation box displays the following terms:
a0 + a1*cos(2*pi*x/c1) + b1*sin(2*pi*x/c1) +
a2*cos(2*pi*x/c2) + b2*sin(2*pi*x/c2)
Click Advanced Options to expand the section. In the Coefficient Constraints table, observe the Lower and Upper bounds for c1, which constrain the cycle from 10 to 14 months. Add more coefficient constraints.
Change the Lower and Upper bounds for c2 to be roughly twice the bounds used for c1 (20<c2<30).
Change the StartPoint value for a0 to 5.
As you change each setting, the Curve Fitter app updates the fit. You can observe the fit plot and the residuals plot.
The fit appears reasonable for most data points. However, the residuals indicate that you should include another cycle to the fit equation.
As a third attempt, add an additional sine and cosine term to y2(x)
{y}_{3}\left(x\right)={y}_{2}\left(x\right)+{a}_{3}\mathrm{cos}\left(2\pi \frac{x}{{c}_{3}}\right)+{b}_{3}\mathrm{sin}\left(2\pi \frac{x}{{c}_{3}}\right)
and constrain the lower bound of c3 to be roughly triple the value of c1.
a2*cos(2*pi*x/c2) + b2*sin(2*pi*x/c2) +
Click Advanced Options to expand the section. Observe that your previous fit options are still present.
In the Coefficient Constraints table, change the Lower bound for c3 to 36, which is roughly triple the value of c1.
The fit is an improvement over the previous two fits, and appears to account for most of the cycles in the ENSO data set. The residuals appear random for most of the data, although a pattern is still visible indicating that additional cycles might be present, or you can improve the fitted amplitudes.
In conclusion, Fourier analysis of the data reveals three significant cycles. The annual cycle is the strongest, but cycles with periods of approximately 44 and 22 months are also present. These cycles correspond to El Nino and the Southern Oscillation (ENSO).
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I have just recently covered the Intermediate Value Theorem, and I wanted to practice solving proble
Dashawn Robbins 2022-05-03 Answered
a>0
f\left(x\right)=f\left(x+2a\right)
x\in R
c\in \left[0,a\right]
f\left(c\right)=f\left(c+a\right)
hadnya1qd
Good question! Try the function
g\left(x\right)=f\left(x+a\right)-f\left(x\right)
since the sum/difference of continuous functions is continuous,
g
is continuous. If you still can't get it, leave a comment for me. note this may not work, but on first look I am pretty sure it will
{x}^{3}+2x-\frac{1}{x}=0
\left(\frac{1}{4},1\right)
f:\left[a,b\right]\to \mathbb{R}
is continuous and that
f\left(a\right)<0
f\left(b\right)>0
. By the intermediate-value theorem, the set
S=\left\{x\in \left[a,b\right]:f\left(x\right)=0\right\}
is nonempty. If
c=supS
c\in S
My first thought was to show that
S
is finite therefore
c\in S
, but there is no guarantee that
f
doesn't have infinitely many zeros.
A thought I have now is to show that
c>max\left(S\right)
can not be true, but I do not know how to show this, is this even the correct thing to show? Thank you in advance for any input.
I=\left[a,b\right]
a<b
u:I\to \mathbb{R}
be a function with bounded pointwise variation, i.e.
Va{r}_{I}u=sup\left\{\sum _{i=1}^{n}|u\left({x}_{i}\right)-u\left({x}_{i-1}\right)|\right\}<\mathrm{\infty }
where the supremum is taken over all partition
P=\left\{a={x}_{0}<{x}_{1}<...<{x}_{n-1}<b={x}_{n}\right\}
. How can I prove that if
u
satisfies the intermediate value theorem (IVT), then
u
is continuous?
My try:
u
can be written as a difference of two increasing functions
{f}_{1},{f}_{2}
. I know that a increasing function that satisfies the (ITV) is continuous, hence, if I prove that
{f}_{1},{f}_{2}
satisfies the (ITV) the assertion follows. But, is this true? I mean,
{f}_{1},{f}_{2}
satisfies (ITV)?
\left(X,M,\mu \right)
f
a\le f\left(x\right)\le b
\mu
x\in X
g
c\in \left[a,b\right]
{\int }_{X}f|g|d\mu =c{\int }_{X}|g|d\mu
a,b\in \mathbb{R}
a<b
f
be a differentiable real-valued function on an open subset of
\mathbb{R}
that contains [a,b]. Show that if
\gamma
is any real number between
{f}^{\prime }\left(a\right)
{f}^{\prime }\left(b\right)
c\in \left(a,b\right)
\gamma ={f}^{\prime }\left(c\right)
Hint: Combine mean value theorem with the intermediate value theorem for the function
\frac{\left(f\left({x}_{1}\right)-f\left({x}_{2}\right)\right)}{{x}_{1}-{x}_{2}}
\left\{\left({x}_{1},{x}_{2}\right)\in {E}^{2}:a\le {x}_{1}<{x}_{2}\le b\right\}
I am having a lot of trouble trying to start on this problem.
f\left(x\right)={x}^{2}+10\mathrm{sin}\left(x\right)
c
f\left(c\right)=1000
f\left(0\right)=0
f\left(90\right)={90}^{2}+10\mathrm{sin}\left(90\right)=8100+10\ast 1=8110
f\left(c\right)=1000
f:\mathbb{R}\to \mathbb{R}
k
{x}_{1},{x}_{2},...,{x}_{k}
\left[a,b\right]
z
\left[a,b\right]
f\left(z\right)=\left(f\left({x}_{1}\right)+f\left({x}_{2}\right)+...+f\left({x}_{k}\right)\right)/k
a<{x}_{1}<b,a<{x}_{2}<b,...,a<{x}_{k}<b
f\left(a\right)<f\left({x}_{1}\right)<f\left(b\right),...,f\left(a\right)<f\left({x}_{k}\right)<f\left(b\right)
k.f\left(a\right)<f\left({x}_{1}\right)+...+f\left({x}_{k}\right)<k.f\left(b\right)
f\left(a\right)<\left(f\left({x}_{1}\right)+f\left({x}_{2}\right)+...+f\left({x}_{k}\right)\right)/k<f\left(b\right)
f\left(a\right)<f\left({x}_{k}\right)<f\left(b\right)
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Fractals/Iterations in the complex plane/Mandelbrot set/mandelbrot - Wikibooks, open books for an open world
Fractals/Iterations in the complex plane/Mandelbrot set/mandelbrot
1 Speed improvements - optimisation
1.2 Bailout test
1.3 interior detection
1.4 Period detection
1.4.1 Cardioid and period-2 checking
1.4.2 Periodicity checking
1.5.1.1 Newton-Raphson zooming
1.5.1.2 Glitches
Speed improvements - optimisationEdit
Speed improvements by Robert Munafo
fractalforums : help-optimising-mandelbrot
Koebe 1/4 theorem by claudiusmaximus
The Mandelbrot set is symmetric with respect to the x-axis in the plane :
colour(x,y) = colour(x,-y)
its intersection with the x-axis ( real slice of Mandelbrot set ) is an interval :
<-2 ; 1/4>
It can be used to speed up computations ( up to 2-times )[1]
Bailout testEdit
Instead of checking if magnitude ( radius = abs(z) ) is greater then escape radius ( ER):
{\displaystyle {\sqrt {Z_{x}^{2}+Z_{y}^{2}}}<ER}
compute square of ER:
ER2 = ER*ER
once and check :
{\displaystyle Z_{x}^{2}+Z_{y}^{2}<ER^{2}}
It gives the same result and is faster.
interior detectionEdit
Mandelbrot set with Interior detection method
Interior detection method[2]
time with detection versus without detection is : 0.383s versus 8.692s so it is 23 times faster !!!!
// gives last iterate = escape time
// output 0< i < iMax
int iterate(double complex C , int iMax)
double complex Z= C; // initial value for iteration Z0
complex double D = 1.0; // derivative with respect to z
{ if(cabs(Z)>EscapeRadius) break; // exterior
if(cabs(D)<eps) break; // interior
D = 2.0*D*Z;
Z=Z*Z+C; // complex quadratic polynomial
Period detectionEdit
"When rendering a Mandelbrot or Julia set, the most time-consuming parts of the image are the “black areas”. In these areas, the iterations never escape to infinity, so every pixel must be iterated to the maximum limit. Therefore, much time can be saved by using an algorithm to detect these areas in advance, so that they can be immediately coloured black, rather than rendering them in the normal way, pixel by pixel. FractalNet uses a recursive algorithm to split the image up into blocks, and tests each block to see whether it lies inside a “black area”. In this way, large areas of the image can be quickly coloured black, often saving a lot of rendering time. ... (some) blocks were detected as “black areas” and coloured black immediately, without having to be rendered. (Other) blocks were rendered in the normal way, pixel by pixel." Michael Hogg [3]
// cpp code by Geek3
// http://commons.wikimedia.org/wiki/File:Mandelbrot_set_rainbow_colors.png
bool outcircle(double center_x, double center_y, double r, double x, double y)
{ // checks if (x,y) is outside the circle around (center_x,center_y) with radius r
x -= center_x;
y -= center_y;
if (x * x + y * y > r * r)
// skip values we know they are inside
if ((outcircle(-0.11, 0.0, 0.63, x0, y0) || x0 > 0.1)
&& outcircle(-1.0, 0.0, 0.25, x0, y0)
&& outcircle(-0.125, 0.744, 0.092, x0, y0)
&& outcircle(-1.308, 0.0, 0.058, x0, y0)
&& outcircle(0.0, 0.25, 0.35, x0, y0))
// code for iteration
Cardioid and period-2 checkingEdit
One way to improve calculations is to find out beforehand whether the given point lies within the cardioid or in the period-2 bulb. Before passing the complex value through the escape time algorithm, first check if:
{\displaystyle (x+1)^{2}+y^{2}<{\frac {1}{16}}}
to determine if the point lies within the period-2 bulb and
{\displaystyle q=\left(x-{\frac {1}{4}}\right)^{2}+y^{2}}
{\displaystyle q\left(q+\left(x-{\frac {1}{4}}\right)\right)<{\frac {1}{4}}y^{2}.}
to determine if the point lies inside the main cardioid.
another description [4]
// http://www.fractalforums.com/new-theories-and-research/quick-(non-iterative)-rejection-filter-for-mandelbrotbuddhabrot-with-benchmark/
public static void quickRejectionFilter(BlockingCollection<Complex> input, BlockingCollection<Complex> output)
foreach(var item in input.GetConsumingEnumerable())
if ((Complex.Abs(1.0 - Complex.Sqrt(Complex.One - (4 * item))) < 1.0)) continue;
if (((Complex.Abs(item - new Complex(-1, 0))) < 0.25)) continue;
if ((((item.Real + 1.309) * (item.Real + 1.309)) + item.Imaginary * item.Imaginary) < 0.00345) continue;
if ((((item.Real + 0.125) * (item.Real + 0.125)) + (item.Imaginary - 0.744) * (item.Imaginary - 0.744)) < 0.0088) continue;
if ((((item.Real + 0.125) * (item.Real + 0.125)) + (item.Imaginary + 0.744) * (item.Imaginary + 0.744)) < 0.0088) continue;
//We tried every known quick filter and didn't reject the item, adding it to next queue.
Periodicity checkingEdit
Most points inside the Mandelbrot set oscillate within a fixed orbit. There could be anything from ten to tens of thousands of points in between, but if an orbit ever reaches a point where it has been before then it will continually follow this path, will never tend towards infinity and hence is in the Mandelbrot set. This Mandelbrot processor includes periodicity checking (and p-2 bulb/cardioid checking) for a great speed up during deep zoom animations with a high maximum iteration value.
Perturbation theoryEdit
"The thing we call perturbation consist of 2 things: Calculate one pixel with high precision and use it as a reference for all other pixels.
This method will fail though, however thanks to Pauldelbrot we have a trustable method of detecting the pixels where the reference fails to compute the pixel with hardware precision. These pixels can be rendered with a reference closer to these pixels, so a typical perturbation render use several references. This method gives a speedup at about 10 times on depths of 10^100 Use a truncated Taylor series to approximate a starting value for a specific iteration, which make you able to skip all previous iterations. This method gives a speedup of typically Another 10 times on depths of 10^100, and together the speed up is typically 100 times. This, which we call Series Approximation, is where we have issues since we do not have any solid theoretically way of finding when too many iterations are skipped - for all pixels in the view. The more terms you include in the Taylor series, the more iterations you are able to skip. So if you stay below say 50 terms, it is not likely that you ever encounter any issues. Because some views can be correctly rendered 1000 or 100,000 times faster than full precision for each pixel with many terms - can you imagine a month turned into seconds! shocked K.I.Martin originally used only 3 terms " - Kalles Fraktaler [5]
Very highly magnified images require more than the standard 64-128 or so bits of precision most hardware floating-point units provide, requiring renderers use slow "bignum" or "arbitrary precision"[6] math libraries to calculate. However, this can be sped up by the exploitation of perturbation theory.[7] Given
{\displaystyle z_{n+1}=z_{n}^{2}+c}
as the iteration, and a small epsilon, it is the case that
{\displaystyle (z_{n}+\epsilon )^{2}+c=z_{n}^{2}+2z_{n}\epsilon +\epsilon ^{2}+c}
{\displaystyle z_{n+1}+2z_{n}\epsilon +\epsilon ^{2}}
so if one defines
{\displaystyle \epsilon _{n+1}=2z_{n}\epsilon _{n}+\epsilon _{n}^{2}}
one can calculate a single point (e.g. the center of an image) using normal, high-precision arithmetic (z), giving a reference orbit, and then compute many points around it in terms of various initial offsets epsilon-zero plus the above iteration for epsilon. For most iterations, epsilon does not need more than 16 significant figures, and consequently hardware floating-point may be used to get a mostly accurate image.[8] There will often be some areas where the orbits of points diverge enough from the reference orbit that extra precision is needed on those points, or else additional local high-precision-calculated reference orbits are needed. This rendering method, and particularly the automated detection of the need for additional reference orbits and automated optimal selection of same, is an area of ongoing, active research. Renderers implementing the technique are publicly available and offer speedups for highly magnified images in the multiple orders of magnitude range.[9][10][11]
Newton-Raphson zoomingEdit
One can "use newton's method to find and progressively refine the precision of the location of the minibrot at the center of a pattern. This allows them to arbitrarily select a magnification between the location they started at and the final minibrot they calculate to be at the center of that location."[12][13]
Glitches[14][15]
2 independent descriptions :
Fast Mandelbrot set with infinite resolution by Sergey Khashin (July 9, 2011)[16][17]
perturbation method by K.I. Martin (2013-05-18)[18] ( only use 3 Series Approximation terms )
Pauldelbrot's glitch detection method [19]
extending Martin's description to handle interior distance estimation too by Claude Heiland-Allen[20]
SuperFractalThing in Java by K.I. Martin[21]
mightymandel by Claude Heiland-Allen[22]
mandelbrot-perturbator by Claude Heiland-Allen : http://code.mathr.co.uk/mandelbrot-perturbator/
Kalles Fraktaler for windows
perturbation_algebra and et
interactive SageMath worksheet explaining how it works, try changing the formula
T Myers: Introduction to perturbation ...
↑ How to use symetry of set
↑ A Cheritat wiki: Mandelbrot_set#Interior_detection_methods
↑ FractalNet by Michael Hogg
↑ Fractal forums: quick-(non-iterative)-rejection-filter-for-mandelbrotbuddhabrot-with-benchmark/
↑ Fractal Forums > Fractal Software > Help & Support > (C++) How to deep zoom in mandelbrot set?
↑ arbitrary precision at wikipedia
↑ perturbation theory in wikipedia
↑ "Superfractalthing - Arbitrary Precision Mandelbrot Set Rendering in Java by K.I. Martin 2013-05-18". http://www.fractalforums.com/announcements-and-news/superfractalthing-arbitrary-precision-mandelbrot-set-rendering-in-java/.
↑ "Kalles Fraktaler 2". http://www.chillheimer.de/kallesfraktaler/.
↑ Fast Mandelbrot set with infinite resolution, ver. 2 by Sergey Khashin, July 9, 2011
↑ Perturbation techniques applied to the Mandelbrot set by Claude Heiland-Allen October 21, 2013
↑ newton-raphson-zooming by quaz0r
↑ Newton-Raphson zooming and Evolution zoom method by Dinkydau
↑ pertubation-theory-glitches-improvement - fractal forum
↑ fractalNotes by Gerrit
↑ Fast Mandelbrot set with infinite resolution, ver. 2
↑ Fast calculation of the Mandelbrot set with infinite resolution by Sergey Khashin, October 12, 2016
↑ fractalforums : superfractalthing-arbitrary-precision-mandelbrot-set-rendering-in-java
↑ fractalforums - pertubation-theory-glitches-improvement
↑ Perturbation glitches by Claude Heiland-Allen
↑ superfractalthing By K.I. Martin
↑ mightymandel by Claude Heiland-Allen
Retrieved from "https://en.wikibooks.org/w/index.php?title=Fractals/Iterations_in_the_complex_plane/Mandelbrot_set/mandelbrot&oldid=4008636"
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Use suitable linear approximation to find the approximate values for given functions at the points i
Use suitable linear approximation to find the approximate values for given functions at the points indicated:
f\left(x,y\right)=x{e}^{y+{x}^{2}}
bailaretzy33
L\left(x,y\right)
f\left({x}_{0},{y}_{0}\right)+{f}_{x}\left({x}_{0},{y}_{0}\right)\left(x-{x}_{0}\right)+{f}_{y}\left({x}_{0},{y}_{0}\right)\left(y-{y}_{0}\right)
L\left(x,y\right)\approx f\left(x,y\right)
\left({x}_{0},{y}_{0}\right)=\left(2,-4\right)
L\left(x,y\right)=2+9\left(x-2\right)+2\left(y+4\right)\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}f\left(2.05,-3.92\right)\approx L\left(2.05,-3.92\right)=2.61
Notice, from a calculator,
f\left(2.05,-3.92\right)=2.7192
Denoting the partial derivatives by
{f}_{x}^{\prime }
{f}_{y}^{\prime }
, the formula is:
f\left({x}_{0}+h,{y}_{0}+k\right)=f\left({x}_{0},{y}_{0}\right)+{f}_{x}^{\prime }\left({x}_{0},{y}_{0}\right)h+{f}_{y}^{\prime }\left({x}_{0},{y}_{0}\right)k+o\left(‖\left(h,k\right)‖\right).
Use the linear approximation of Use the linear approximation of f(x,y)=e2x2+3y at (0,0) to estimate f(0.01,−0.02). at
\left(0,0\right)
f\left(0.01,-0.02\right)
Linear approximation to find
\frac{1}{4.002}
Determine how accurate should we measure the side of a cube so that the calculated surface area of the cube lies within
3
% of its true value, using Linear Approximation.
A\left(x\right)=TSA
x=side
A\left(x\right)=6{x}^{2}
Find the linear approximation
Y
f\left(x\right)
x=a
f\left(x\right)=x+{x}^{4},\phantom{\rule{1em}{0ex}}a=0
Use linear approximation, i.e. the tangent line, to approximate
{11.2}^{2}
as follows :
f\left(x\right)={x}^{2}
and find the equation of the tangent line to
f\left(x\right)
x=11
. Using this, find your approximation for
{11.2}^{2}
l\left(x\right)
be the linear approximation of
f\left(x\right)={x}^{2/5}
a=32
. Approximation?
Use linear approximation to estimate
\mathrm{tan}\left(\frac{\pi }{4}+0.05\right)
. Identify the differentials
dy
and
in the situation.
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Binary, Decimal, and Hexadecimal Practice Problems Online | Brilliant
This quiz is a quick overview and refresher of the concepts from discrete math needed to understand memory. In particular, the number bases 2 and 16, and converting into and out of various bases are important concepts for working with memory.
Binary, Decimal, and Hexadecimal
If we have a device that can be in one of two states, we can store a single digit of a base 2, or binary, number on that device. If we have many of these devices, we can store larger numbers, one binary digit per device.
In the picture below, we have a machine with buckets on a conveyor belt. To store a number, the machine fills specific buckets with water. We can consider this machine as representing a base 2 number, where filled buckets represent a 1 and empty buckets represent a 0.
What number is stored on the machine? Answer in base 10.
Whether it’s stored in RAM, in a bank of vacuum tubes, or on a conveyor belt of water buckets, digital data in its most basic form is just a series of individual binary digits, also known as bits. One bit represents one digit in base 2, which is either 0 or 1.
In this course, we’ll adopt the convention used in some programming languages to prefix binary numbers with 0b. What is the base 10 number 173 in binary? Prefix the binary with 0b.
0b10010 0b173 0b10101101 0b10101110
How many bits are required to represent 128 in binary?
We can determine the largest base 10 value that can be represented by a certain number of bits in binary by setting all of the bits to 1 and converting to base 10. What’s the largest base 10 number we can represent with 8 bits?
8 bits are referred to as a byte. As we saw in the last problem, the largest number that can be stored in a single byte is 255. Clearly, we require three digits to represent 255 in base 10. What is the smallest base that can represent 255 using only 2 digits?
Base 16 is also referred to as hexadecimal, or hex for short. In this course, we’ll adopt the convention used in some programming languages to prefix hexadecimal numbers with 0x. For example, since 32 is
2 \times 16^1 + 0 \times 16^0
, it will be represented in hexadecimal as 0x20.
Since one hexadecimal digit needs to represent up to the value 15 in decimal, we will use letters for some of the digits:
\begin{array} { | c | c| } \hline \text{dec} & \text{hex} \\ \hline 0 & 0 \\ 1 & 1 \\ 2 & 2 \\ \vdots & \vdots \\ 9 & 9 \\ 10 & \text{A} \\ 11 & \text{B} \\ 12 & \text{C} \\ 13 & \text{D} \\ 14 & \text{E} \\ 15 & \text{F} \\ \hline \end{array}
\begin{array} { | c c r c | } \hline \text{dec} & & & & \text{hex} \\ \hline 0 &= & 0 \times 16^0 & = & 0\text{x}0 \\ 1 &= & 1 \times 16^0 & = & 0\text{x}1 \\ 9 &= & 9 \times 16^0 & = & 0\text{x}9 \\ 10 &= & 10 \times 16^0 & = & 0\text{xA} \\ 11 &= & 11 \times 16^0 & = & 0\text{xB} \\ 12 &= & 12 \times 16^0 & = & 0\text{xC} \\ 15 &= & 15 \times 16^0 & = & 0\text{xF} \\ 16 &= & 1\times 16^1 + 0 \times 16^0 & = & 0\text{x}10 \\ 17 &= & 1\times 16^1 + 1 \times 16^0 & = & 0\text{x}11 \\ 25 &= & 1\times 16^1 + 9 \times 16^0 & = & 0\text{x}19 \\ 26 &= & 1\times 16^1 + 10 \times 16^0 & = & 0\text{x}1\text{A} \\ 31 &= & 1\times 16^1 + 15 \times 16^0 & = & 0\text{x}1\text{F} \\ 32 &= & 2\times 16^1 + 0 \times 16^0 & = & 0\text{x}20 \\ 159 &= & 9\times 16^1 + 15 \times 16^0 & = & 0\text{x}9\text{F} \\ 160 &= & 10\times 16^1 + 0 \times 16^0 & = & 0\text{xA}0 & \\ 255 &= & 15\times 16^1 + 15 \times 16^0 & = & 0\text{xFF} \\ \hline \end{array}
The decimal number 47872 is represented as 0xBB00 in hexadecimal. What is the decimal number 48000 in hexadecimal?
0xBB37 0xBB80 0x48000 0xBB4912
As we saw in the previous question, 255 in hexadecimal is 0xFF. This happens to be the maximum possible value that can be represented by 2 digits in hexadecimal. Recall that 255 in binary is 0b11111111, which is also the maximum possible value that can be represented by 8 digits in binary.
This property is important because it means that 2 digits of hexadecimal are exactly equivalent to 8 digits of binary. One implication is that concatenating bytes in either hexadecimal or binary form will give the same result. For example:
\begin{array} { | c | c| c | } \hline \text{Binary} & \text{Hexadecimal} & \text{Decimal} \\ \hline \text{0b11101111} & \text{0xEF} & 239 \\ \text{0b00111001} & \text{0x39} & 057 \\ \hline \end{array}
\text{0b1110111100111001} = 61241
\text{0xEF39} = 61241
We can see that concatenating the hexadecimal forms of 239 and 057 is equivalent to concatenating the binary forms. However, if we concatenate the two decimal values, we get something completely different: 239057.
We normally deal with integers of fixed sizes in multiples of bytes. Recall that a byte is 8 bits. It’s common to see 8-bit, 16-bit, 32-bit, and 64-bit integers in contemporary code. The largest value of a 32-bit integer is
2^{32} - 1
which is 4294967295 in decimal. If we store a much smaller number, the remaining unused bits must be set to 0, since sizes are fixed.
What is the decimal value 763 stored in a 32-bit integer, padding the unused bits with 0s? Answer in hexadecimal.
0x2FB 0x000002FB 0x763 0x00000763
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(10R)-dioxygenase Wikipedia
Linoleate 10R-lipoxygenase (EC 1.13.11.62, 10R-DOX, (10R)-dioxygenase, 10R-dioxygenase) is an enzyme with systematic name linoleate:oxygen (10R)-oxidoreductase.[1][2] This enzyme catalyses the following chemical reaction
linoleate + O2
{\displaystyle \rightleftharpoons }
(8E,10R,12Z)-10-hydroperoxy-8,12-octadecadienoate
Linoleate 10R-lipoxygenase is involved in biosynthesis of oxylipins.
^ Garscha U, Oliw EH (May 2009). "Leucine/valine residues direct oxygenation of linoleic acid by (10R)- and (8R)-dioxygenases: expression and site-directed mutagenesis oF (10R)-dioxygenase with epoxyalcohol synthase activity". The Journal of Biological Chemistry. 284 (20): 13755–65. doi:10.1074/jbc.M808665200. PMC 2679477. PMID 19289462.
^ Jernerén F, Garscha U, Hoffmann I, Hamberg M, Oliw EH (April 2010). "Reaction mechanism of 5,8-linoleate diol synthase, 10R-dioxygenase, and 8,11-hydroperoxide isomerase of Aspergillus clavatus". Biochimica et Biophysica Acta (BBA) - Molecular and Cell Biology of Lipids. 1801 (4): 503–7. doi:10.1016/j.bbalip.2009.12.012. PMID 20045744.
Linoleate+10R-lipoxygenase at the US National Library of Medicine Medical Subject Headings (MeSH)
|
How many different permutations are there of the letters in
How many different permutations are there of the letters in the word MISSISSIPPI?
1M+4i+4S+2P
\frac{11!}{1!4!4!2!}=\frac{11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5}{4\cdot 3\cdot 2\cdot 2}
=11\cdot 10\cdot 9\cdot 7\cdot 5
The standard deviation of monthly rents paid by students in a particular city is assumed to be $ 40.
If a random sample of 100 students is taken, the sample mean must be at least $ 5 higher than the population average what is the probability?
A radar center consists of two units operating independently. The probability that one of the units detects an incoming missile is 0.99, and the probability that the other unit detects it is 0.95. What is the probability that
(i) both units will detect?
(ii) at least one will detect?
(iii) neither will detect?
From Theoretical and Experimental Probability, which of the two method is more appropriate and valid to use in determining probabilities? Why?
Consider two products, A and B, both of which have the same purchase probability. Product A has a population size of 1000 and a 50 percent awareness probability. Product B has a population size of 10,000 and a 5 percent awareness probability. Which product do you expect to sell more? Product B
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Sheryl Awuor 2022-03-28
Find k
such that the following matrix M
Let AB = Im, where A is an m×n matrix and B is an n×m matrix. If y ∼ MVN(μ,In) and BA is symmetric, find the distribution of yT BAy.
I need solution of Q2 of given assignment
describe the motion of the particle if normal component of acceleration is 0
Find a 3x3 matrix A such that Ax⃗ =9x⃗ for all x⃗ in R
Determine which of the following transformations are linear transformations.
A. The transformation T defined by T(x1,x2,x3)=(1,x2,x3)T(x1,x2,x3)=(1,x2,x3)
B. The transformation T defined by T(x1,x2)=(2x1−3x2,x1+4,5x2)T(x1,x2)=(2x1−3x2,x1+4,5x2).
C. The transformation T defined by T(x1,x2)=(4x1−2x2,3|x2|)T(x1,x2)=(4x1−2x2,3|x2|).
D. The transformation T defined by T(x1,x2,x3)=(x1,x2,−x3)T(x1,x2,x3)=(x1,x2,−x3)
E. The transformation T defined by T(x1,x2,x3)=(x1,0,x3)
A second hand car dealer has ten cars for sale. He decides to investigate the link between the
ages of the cars, x years and the mileage, y thousand miles. The data collected from the cars
Age, x years
Mileage, y thousand miles
(a) Construct a well labeled scatter diagram in the graph paper provided to present the
information. (3 marks)
(b) Find Sxx and Sxy. (3 marks)
(c) Find the regression equation in the form y = α + βx. (4 marks)
(d) Fit the regression equation obtained in (c) to the scatter plot in (a) and label the line
appropriately. (2 marks)
(e) Give a practical interpretation of the slope β. (1 mark)
(f) Using your answer in (c), fifind the mileage for a 6 year old car.
(f) Using your answer in (c), fifind the mileage for a 6 year old car
x + y - z = -22x - y + 3z = 9x - 4y - 2z = 1find the solution
Identify each of the following functions as exponential growth or decay y=39(0.98)^t
Find the expected count and the contribution to the chi-square statistic for the (Treatment, Disagree) cell in the two-way table below.
The Smithson family is going to visit family in California for spring break. They plan to spend 1/3 of their time fishing, 1/4 of their time swimming. 1/4 of their time just relaxing, and the rest of their time at an amusement park. Approximately what fraction of time will the Smithson family spend at the amusement park? What percentage of time will the Smithson family spend doing water sports? Round to the nearest whole number. If the Smithson family wanted to spend 50% of their time relaxing, which activity or activities should they give up? List all possible choices and/or combinations, including those that would allow for nearly 50% relaxation.
a 5kg block rest on the 30degree incline. the coefficient of static friction between the block and the incline is 0.2 . how large a holizontal force must push on the block if the block is to be on the verge of sliding (a) up the incline ,(b) down the incline
Find the Laplace transform of the following function using the properties and table of laplace f(t) = 2δ(t − 1)
Convert the following readings of pressure to kPa assuming that barometer reads 760 mm of Hg. Assuming density of Hg, ρHg= 13.596 × 1000 kg/ m3 1. 80 cm of Hg 2. 30 cm Hg vacuum 3. 1.35 m H2O gauge 4. 4.2 bar.
In Exercises 3–12, determine whether each 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, y), where , with the standard operations on .
Kristina Mcclain 2022-02-02 Answered
Given Matrix transformation matrix
A=\left(\begin{array}{ccc}1& 1& 1\\ 0& 2& 2\\ 0& 0& 3\end{array}\right)
Are there vectors in
{\mathbb{R}}^{3}
that do not change under this transformation, i.e. such that
T\left(u\right)=u
? Explain why yes or why not.
Lets say we have the following matrix
M=\left[\begin{array}{cc}4& 3\\ 4& 3\end{array}\right]
Then I want to transform all numbers in the matrix with the following function:
f\left(x\right)={\left(x-4\right)}^{2}+2x-4
What would be the correct notation for this?
From my working, P would equate to
P=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]
Now to find the transformation shown by
{P}^{3}
, would I have to cube the matrix or is there something that I'm missing here?
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if 0 5 mol of BaCl2 is mixed with 0 2 mol of Na3PO4 How many maximum moles of Ba3(PO4)2 - Chemistry - Solutions - 6875005 | Meritnation.com
Balanced equation for reaction between BaCl2 and Na3PO4 is as follows:
3BaC{l}_{2}\quad +\quad 2N{a}_{3}P{O}_{4}\quad \quad \to \quad B{a}_{3}{\left(P{O}_{4}\right)}_{2}\quad +\quad 6NaCl
3 moles of BaCl2 react with 2 moles of Na3PO4 to give 1 mole of Ba3(PO4)2
0.5 moles of BaCl2 will react with (2/3) x 0.5 = 0.33 moles of Na3PO4
Available moles of Na3PO4 = 0.2
So, Na3PO4 is the limiting reagent
Now, 2 moles of Na3PO4 give 1 mole of Ba3(PO4)2
So, 0.2 moles of Na3PO4 will give 1/2) x 0.2 = 0.1 mole of Ba3(PO4)2
Hence, maximum number of moles of Ba3(PO4)2 formed = 0.1
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Stable Math - Balancer
Weighted Math
Stable Math is designed to allow for swaps between any assets that have the same price, or are "pegged" to the same asset. The most common examples are stablecoins that track US Dollars (DAI, USDT, USDC), and assets that track the price of Bitcoin (WBTC, renBTC, sBTC). Prices are determined by the pool balances, the amplification parameter, and amounts of the tokens that are being swapped.
In an ideal scenario, it would make sense to simply allow 1-to-1 trades for these assets; this would be a Constant Sum curve. In a worst case scenario where one or more of these assets loses their peg and their value diverges, it would make sense to enforce trade rules for uncorrelated assets; this would be a Constant Product curve, such as the one in Weighted Math.
Since most cases are neither ideal nor disasters, the Stable Math curve combines the Constant Sum and Constant Product curves and is designed to facilitate approximately 1-to-1 trades that incur large price changes only when token balances differ greatly. The amplification parameter,
A
, defines the degree to which the Stable Math curve approximates the Constant Product curve (when
A=0
), or the Constant Sum curve (when
A\rightarrow \infty
For more stable math formulas, check out the Developer Docs.
StableSwap approaches Constant Product as A->0 and Constant Sum as A->∞
Since the Stable Math equation is quite complex, determining the invariant,
D
, is typically done iteratively. For an example of how to do this, please refer to this function.
A \cdot n^n \cdot \sum{x_i} +D = A \cdot D \cdot n^n + { \frac{D^{n+1}}{{n}^{n}\cdot \prod{x_i} } }
n
is the number of tokens
x_i
is is balance of token
i
A
is the amplification parameter
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Special Issue in honour of Joseph Krasil'shchik 70th birthday - Geometry of Differential Equations
Special Issue in honour of Joseph Krasil'shchik 70th birthday
This is a special collection of articles arising from the Conference "Local and Nonlocal Geometry of PDEs and Integrability" (8-12 October 2018, SISSA, Trieste, Italy) dedicated to the 70th birthday of Joseph Krasil'shchik
by Volodya Rubtsov, JGP Editor
All papers for the Special Issue in honour of Joseph Krasil'shchik 70th birthday which you can see on this page are published in the "Journal of Geometry and Physics" in 2019-2020. Usually, the Journal never publishes any Proceedings of Conferences and Workshops and accepts a publication of Special Issues in a form of "regular" articles along with the dates of its submissions. This papers pass the usual peer-reviewing process under the standards of the Journal. Special invited Guest Editors (together with one of Managing Editors of the JGP) are participate in the evaluation and handling process.
This Special Issue were prepared by the Guest Editors Valentin Lychagin, Alik Verbovetsky, Rafaelle Vitolo, and myself. The subject of it ("Geometry and Integrability of PDE's") is one of the most important and popular in the wide spectrum of the Journal optic.
My dear friend professor Joseph Krasil'shchik invested a lot both in this scientific area and, in particular, in the popularization of this direction in the JGP. He is not only an active author of many interesting and valuable articles published in the JGP, but also he was the Guest Editor of three Special Issues. Joseph is always careful and reliable referee. His serious and detailed reports helped a lot to the Editors in their decision, but in the same time are very helpful to the authors, especially, to young and novices.
I wish you, my dear Joseph, long life, excellent health and new wonderful results and beautiful papers (which, I hope, you will publish in our Journal!)
Shortest and Straightest geodesics in Sub-Riemannian Geometry
J. Geom. Phys. 155 (2020), 103713, arXiv:1909.08275
On symmetries of the Gibbons-Tsarev equation
Hynek Baran, Petr Blaschke, Joseph Krasil'shchik, and Michal Marvan
J. Geom. Phys. 144 (2019), 54-80, arXiv:1811.08199
On the Geometry of Twisted Symmetries: Gauging and Coverings
Diego Catalano Ferraioli and Giuseppe Gaeta
Invertible linear ordinary differential operators and their generalizations
J. Geom. Phys. 151 (2020), 103617
Einstein Metrics, Projective Structures and the
{\displaystyle SU(\infty )}
Toda equation
Maciej Dunajski and Alice Waterhouse
Lie remarkable partial differential equations characterized by Lie algebras of point symmetries
Matteo Gorgone and Francesco Oliveri
J. Geom. Phys. 144 (2019), 314-323, arXiv:2108.02171
Integrable hierarchies in the
{\displaystyle \mathbb {N} \times \mathbb {N} }
-matrices related to powers of the shift operator
Gerard Helminck and Jeffrey Weenink
On Lie algebras responsible for integrability of (1+1)-dimensional scalar evolution PDEs
Sergei Igonin and Gianni Manno
Hovhannes Khudaverdian and Theodore Voronov
Joseph Krasil'shchik, Oleg Morozov, and Petr Vojčák
Exact solutions of the Burgers-Huxley equation via dynamics
Alexei Kushner and Ruslan Matviichuk
Paolo Lorenzoni and Raffaele Vitolo
Phase transitions in filtration of Redlich-Kwong gases
Valentin Lychagin and Mikhail Roop
On equivalence of third order linear differential operators on two-dimensional manifolds
Valentin Lychagin and Valeriy Yumaguzhin
(Super-)integrable systems associated to 2-dimensional projective connections with one projective symmetry
Gianni Manno and Andreas Vollmer
Classification of integrable vector equations of geometric type
Anatoly Meshkov and Vladimir Sokolov
Soliton transmutations in KdV-Burgers layered media
Haantjes algebras and diagonalization
Piergiulio Tempesta and GiorgioTond
On inverse variational problem for one class of quasilinear equations
Symmetries and conservation laws for a generalization of Kawahara equation
Jakub Vašícek
Retrieved from "https://gdeq.org/w/index.php?title=Special_Issue_in_honour_of_Joseph_Krasil%27shchik_70th_birthday&oldid=6281"
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Given the probability p = 0.7
Given the probability
p=0.7
that an event will happen, how do you find the probability that the event will not happen?
Probability of an event happening
p=0.7
Probability of the event not happening
=?
\text{Probability of an event happening}+\text{Probability of the event not happening}=1
=1-
=1-0.7=0.3
The probability that a man will be alive in 25 years is 3/5, and the probability that his wife will be alive in 25 years is 2/3.
Determine the probability that both will be alive.
The probability that a shopper will buy a PS3 is 0.40.
The probability that a shopper will buy an X-Box 360 is 0.54.
The probability that a shopper will buy a Nintendo DS is 0.30.
The probability that a shopper will buy both a PS3 and an X-Box 360 is 0.13. The probability that a shopper will buy both a PS-3 and a DS is 0.12.
The probability that a shopper will buy both an X-Box 360 and a DS is 0.07. There is no chance that a shopper will buy all three systems.
Find the probability that a random shopper will:
a) Buy only a PS-3
b) Buy exactly one type of system
c) Buy exactly two types of systems d) Not buy any of the three systems
A. The odds in favor of an event are given. Compute the probability of the event. (Enter the probability as a fraction.) 2 to 9
B. The probability of an event is given. Find the odds in favor of the event. 0.3
C. The probability of an event is given. Find the odds in favor of the event. 0.35
In a rational world, husband and wife are arguing over a matter. The probability of the husband being Right is 0.8 and the probability of the wife being right is 0.78. And the probability of husband being Right provided that wife is Right is 0.83. What is the probability that the wife is Right provided that the husband is Right?
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Convert radiation pattern from azimuth/elevation form to u/v form - MATLAB azel2uvpat - MathWorks France
u/v
u
v
u/v
u
v
u/v
u
v
u
v
\begin{array}{l}u=\mathrm{sin}\theta \mathrm{cos}\varphi \\ v=\mathrm{sin}\theta \mathrm{sin}\varphi \end{array}
\begin{array}{l}u=\mathrm{cos}el\mathrm{sin}az\\ v=\mathrm{sin}el\end{array}
\begin{array}{l}-1\le u\le 1\\ -1\le v\le 1\\ {u}^{2}+{v}^{2}\le 1\end{array}
\begin{array}{l}\mathrm{tan}\varphi =v/u\\ \mathrm{sin}\theta =\sqrt{{u}^{2}+{v}^{2}}\end{array}
\begin{array}{l}\mathrm{sin}el=v\\ \mathrm{tan}az=\frac{u}{\sqrt{1-{u}^{2}-{v}^{2}}}\end{array}
\begin{array}{l}\mathrm{sin}el=\mathrm{sin}\varphi \mathrm{sin}\theta \\ \mathrm{tan}az=\mathrm{cos}\varphi \mathrm{tan}\theta \\ \mathrm{cos}\theta =\mathrm{cos}el\mathrm{cos}az\\ \mathrm{tan}\varphi =\mathrm{tan}el/\mathrm{sin}az\end{array}
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Please answer with solution.
asistioacer 2022-02-24 Answered
2. Since
y={x}^{2}
y=4x-{x}^{2}
{x}^{2}=4x-{x}^{2}
2{x}^{2}-4x=0⇒2x\cdot \left(x-2\right)=0
2x=0⇒x=0
x=0⇒y={x}^{2}⇒y=0
A=\left(0,0\right)
x-2=0⇒x=2
x=2⇒y={x}^{2}⇒y=4
B=\left(2,4\right)
dx
slice (that mean the Area revolving about x-axis)
The interval of the integral become
x\in \left[0,2\right]
now let set up the integral of Area
A={\int }_{a}^{b}{y}_{2}-{y}_{1}\ast dx
A={\int }_{0}^{2}\left(4x-{x}^{2}\right)-\left({x}^{2}\right)\cdot dx
A={\int }_{0}^{2}4x-2{x}^{2}\cdot dx={\left[2{x}^{2}-\frac{2}{3}{x}^{3}\right]}_{0}^{2}
\left[8-\frac{16}{3}\right]=\frac{24-16}{3}=\frac{8}{3}=2.667{\left(unite\right)}^{2}
An object moves in simple harmonic motion with period 7 minutes and amplitude 17m. At time =t0 minutes, its displacement d from rest is 0m, and initially it moves in a positive direction
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ca foundation mock test in ratio and proportion of Business Mathematics | Online Exam
Home Quantitative Aptitude ca foundation mock test-ratio and proportion
ca foundation mock test in ratio and proportion of Business Mathematics
The integral part of a logarithm is called___ and the decimal part of a logarithm is called____
mantissa,characteristic
characteristic ,mantissa
whole,decimal
The integral part of a logarithm is called characteristic and the decimal part of a logarithm is called mantissa
X,Y,Z together start a business.if X invests 3 times as much as Y invests and Y invests 2/3rd of what Z invests,then the ratio of X,Y,Z is
given,X=3Y and Y=2/3*Z
X/Y=3/1 and Y/Z=2/3
X:Y=3:1 and Y:Z=2:3
=3*2:1*2=6:2
if log 2 = 0.3010 and log 3 =0.4771,then the value of log 24 is
log 24=log(2x2x2x3)
=log2+log2+log2+log3
=3 x 0.3010 + 0.4771
=0.9030+0.4771 = 1.3801
There are 23 coins of Rs 1, Rs 2 and Rs 5 in a bag.if their value is Rs 43 the ratio of coins of Rs 1 and Rs 2 is 3:2.then the number of coins of Rs 1 is
total no of coins = 23
Rs 1 :Rs 2 =3:2
let no of Rs1 =3x
no of Rs2 =2x
no of Rs5 23-3x-2x=23-5x
total value of all coins is 43
3x*1+2x*2+(23-5x)5=43
3x+4x+115-25x=42
-18x=43-115
no of Rs 1 coins=3x=3*4=12
If a:b=2:3,b:c=4:5 and c:d=6:7,then a:d is
a:b=2:3=a/b=2/3………(i)
b:c=4:5=b/c=4/5……….(ii)
c:d=6:7=c/d=6/7……….(iii)
multiply equation i,ii and iii
a/b*b/c*c/d=2/3*4/5*6/7=16/35
The ratio compounded of 4:5 and sub-duplicate of a:9 is 8:15.then value of ‘a’ is
if {\mathrm{log}}_{2}\left(x\right)+{\mathrm{log}}_{4}\left(x\right)=6,then the value of x is
if X varies inversely as square of y and given that Y=2 for X=1, then the value of X for Y =6 will be
find 2 numbers such that the mean proportional between them is 18 and 3rd proportional between them is 144
in a film shooting , A and B received money in a certain ratio , B and C also received the money in the same ratio.if A gets rs 160000 and C gets rs 250000.find the amount received by B?
The mean proportional between 24 and 54 is
mean proportional=\sqrt{24×54}\phantom{\rule{0ex}{0ex}} =\sqrt{1296}\phantom{\rule{0ex}{0ex}} =36
mean proportional=\sqrt{24×54}\phantom{\rule{0ex}{0ex}} =\sqrt{1296}\phantom{\rule{0ex}{0ex}} =36
The triplicate ratio of 4:5 is
Triplicate ratio of 4:5={4}^{3}:{5}^{3}=64:125
Triplicate ratio of 4:5={4}^{3}:{5}^{3}=64:125
Find the three numbers in the ratio 1:2:3,to that the sum of their squares is equal to 504
The value of {\mathrm{log}}_{4}\left(9\right).{\mathrm{log}}_{3}\left(2\right) is
{\mathrm{log}}_{4}\left(9\right).{\mathrm{log}}_{3}\left(2\right)=\frac{\mathrm{log}9}{\mathrm{log}4}.\frac{\mathrm{log}2}{\mathrm{log}3}\phantom{\rule{0ex}{0ex}} =\frac{\mathrm{log}{3}^{2}}{\mathrm{log}{2}^{2}}.\frac{\mathrm{log}2}{\mathrm{log}3}\phantom{\rule{0ex}{0ex}} =\frac{2\mathrm{log}3}{2\mathrm{log}2}.\frac{\mathrm{log}2}{\mathrm{log}3}\phantom{\rule{0ex}{0ex}} =1
{\mathrm{log}}_{4}\left(9\right).{\mathrm{log}}_{3}\left(2\right)=\frac{\mathrm{log}9}{\mathrm{log}4}.\frac{\mathrm{log}2}{\mathrm{log}3}\phantom{\rule{0ex}{0ex}} =\frac{\mathrm{log}{3}^{2}}{\mathrm{log}{2}^{2}}.\frac{\mathrm{log}2}{\mathrm{log}3}\phantom{\rule{0ex}{0ex}} =\frac{2\mathrm{log}3}{2\mathrm{log}2}.\frac{\mathrm{log}2}{\mathrm{log}3}\phantom{\rule{0ex}{0ex}} =1
The value of \left({\mathrm{log}}_{y}\left(x\right).{\mathrm{log}}_{z}\left(y\right).{\mathrm{log}}_{x}\left(z\right){\right)}^{3} is
\left({\mathrm{log}}_{y}\left(x\right).{\mathrm{log}}_{z}\left(y\right).{\mathrm{log}}_{x}\left(z\right){\right)}^{3}=\left[\frac{\mathrm{log}x}{\mathrm{log}y}.\frac{\mathrm{log}y}{\mathrm{log}z}.\frac{\mathrm{log}z}{\mathrm{log}x}{\right]}^{3}={1}^{3}=1
\left({\mathrm{log}}_{y}\left(x\right).{\mathrm{log}}_{z}\left(y\right).{\mathrm{log}}_{x}\left(z\right){\right)}^{3}=\left[\frac{\mathrm{log}x}{\mathrm{log}y}.\frac{\mathrm{log}y}{\mathrm{log}z}.\frac{\mathrm{log}z}{\mathrm{log}x}{\right]}^{3}={1}^{3}=1
Divided 80 into two parts so that their product is maximum,then the numbers are
If x:y=2:3,then (5x+2y):(3x-y)=_____
given,x:y=2:3
let x=2k,y=3k
(5x+2y):(3x-y)=5x+2y/3x-y
=5*2k+2*3k/3*2k-3k
=10k+6k/6k-3k
=16k/3k=16:3
if the salary of P is 25% lower than that of Q and the salary of R is 20% higher than that of Q , the ratio of the salary of R and P will be
let salary of Q = 100
salary of P = 100-25% = 75
salary of R = 100 + 20% = 120
ratio of salary R and P are 120:75=8:5
A person has assets worth rs 148200. he wishes to divide if among his wife , son and daughter in the ratio 3:2:1. from this assets , the share of his son will be
persons assets worth =rs 148200
wife:son:daughter=3:2:1
sum of ratio =3+2+1=6
share of son = 2/6 x 148200 =49400
for 3 months , the salary of a person are in the ratio 2:4:5.if the difference between the product of salaries of the 1st 2 months and last 2 months is rs 48000000, then the salary of the person for the 2nd month will be
A dealer mixes rice costing rs 13.85 per kg with rice costing rs 15.54 and sell the mixture at rs 17.60 per kg. so , he earned a profit of 14.6% on his sale price. the proportion in which he mixes the two qualities of rice is
the ratios of 3rd proportion of 12, 30 to the mean proportion of 9,25 is
what number must be added to each of the numbers 10,18,22,38 to make the number is proportion?
7log(16/15)+5log(25/24)+3log(81/80)=
if log(a+b/4)=1/2(loga+logb) then a/b+b/a=
Next articleca foundation mock test-logarithm
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Design and Testing of a Unitized Regenerative Fuel Cell | J. Electrochem. En. Conv. Stor | ASME Digital Collection
Jeremy Fall,
Drew Humphreys,
Fall, J., Humphreys, D., and Guo, S. M. (May 12, 2009). "Design and Testing of a Unitized Regenerative Fuel Cell." ASME. J. Fuel Cell Sci. Technol. August 2009; 6(3): 031003. https://doi.org/10.1115/1.3005575
A unitized regenerative fuel cell (URFC) is designed and tested for energy conversion and storage under the support of a NASA funded student design project. The URFC is of the proton exchange membrane type with an active cell area of
25cm2
. In the URFC design, liquid water is stored internally to the fuel cell within graphite bipolar plates while hydrogen and oxygen gases, electrolyzed from water, are stored in containers external to the fuel cell. A spraying technique is used to produce a functional membrane electrode assembly. Catalyst ink is prepared using E-TEK Inc. platinum and iridium catalysts loaded on Vulcan XC-72. Platinum catalyst is used for the hydrogen electrode.
50wt%
platinum∕
50wt%
iridium catalyst is used for the oxygen electrode. The metal weight on carbon is 30% for both the platinum and iridium catalysts. Water management within the fuel cell is handled by treatment of the gas diffusion layer with a Teflon emulsion to create the proper balance of hydrophobic and hydrophilic pores. The single cell unit is tested in either fuel cell mode or electrolysis mode for different catalyst loadings. Polarization curves for the URFC are generated to evaluate system performance.
catalysts, design engineering, electrochemical electrodes, electrolysis, hydrogen, iridium, oxygen, platinum, proton exchange membrane fuel cells, spraying, PEMFC, unitized regenerative fuel cell, design
Catalysts, Design, Electrolysis, Fuel cells, Gas diffusion layers, Hydrogen, Oxygen, Water, Membranes, Electrodes, Inks, Membrane electrode assemblies, Proton exchange membranes, Platinum, Testing
Kitwanzawa
Iridium Oxide∕Platinum Eletrocayalysts for Unitized Regenerative Polymer Electrolyte Fuel Cells
Nonmetallic Gasket Materials and Forms
The Seals Book: A Basic Reference Manual for Design Engineers. Seals, Packings, Gaskets, Sealants
Proton-Exchange Membrane Regenerative Fuel Cells
A Review of the Latest Developments in Electrodes for Unitized Regenerative Fuel Cells
Bifunctional Electrodes With a Thin Catalyst Layer for ‘Unitized’ Proton Exchange Membrane Regenerative Fuel Cell
Influence of PTFE Coating on Gas Diffusion Backing for Unitized Regenerative Polymer Electrolyte Fuel Cells
Thin Film Electrocatalyst Layer for Unitized Regenerative Polymer Electrolyte Fuel Cells
Bifunctional Oxygen Electrode With Corrosion-Resistive Gas Diffusion Layer for Unitized Regenerative Fuel Cell
Development of Supported Bifunctional Electrocatalysts for Unitized Regenerative Fuel Cells
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Weighted Math - Balancer
Weighted Math is designed to allow for swaps between any assets whether or not they have any price correlation. Prices are determined by the pool balances, pool weights, and amounts of the tokens that are being swapped.
Balancer's Weighted Math equation is a generalization of the
x*y=k
constant product formula recommended for Automated Market Makers (AMMs) in a post by Vitalik Buterin. Balancer's generalization accounts for cases with
n \geq2
tokens as well as weightings that are not an even 50/50 split.
As the price of each token changes, traders and arbitrageurs rebalance the pool by making swaps. This maintains the desired weighting of the value held by each token whilst collecting trading fees from the traders.
For more weighted math formulas, check out the Developer Docs.
V
V= \prod_t B_t^{W_t}
t
ranges over the tokens in the pool
B_t
is the balance of the token in the pool
W_t
is the normalized weight of the tokens, such that the sum of all normalized weights is 1.
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GraphTheory/IsBiregular - Maple Help
Home : Support : Online Help : GraphTheory/IsBiregular
test if graph is biregular
IsBiregular(G)
IsBiregular(G, P)
partition=truefalse
If partition=true and G is biregular, two lists of vertices comprising a biregular partition of G are returned. Otherwise a simple Boolean value is returned indicating whether the graph is biregular.
IsBiregular returns true if the graph G is biregular and false otherwise. If a variable name P is specified, then this name is assigned a bipartition of the vertices as a list of lists.
A graph G is biregular if its set of vertices can be partitioned into two sets,
{V}_{1}
{V}_{2}
{V}_{1}
to a vertex in
{V}_{2}
and if there exist nonnegative integers
{\mathrm{D}}_{1}
{\mathrm{D}}_{2}
such that every vertex in
{V}_{1}
{\mathrm{D}}_{1}
and every vertex in
{V}_{2}
{\mathrm{D}}_{2}
\mathrm{with}\left(\mathrm{GraphTheory}\right):
\mathrm{K32}≔\mathrm{CompleteGraph}\left(3,2\right)
\textcolor[rgb]{0,0,1}{\mathrm{K32}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{Graph 1: an undirected unweighted graph with 5 vertices and 6 edge\left(s\right)}}
\mathrm{IsBiregular}\left(\mathrm{K32},'\mathrm{partition}'\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}\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}]
\mathrm{DrawGraph}\left(\mathrm{K32},\mathrm{style}=\mathrm{bipartite}\right)
\mathrm{AdjacencyMatrix}\left(\mathrm{K32}\right)
[\begin{array}{ccccc}\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{1}\\ \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{1}\\ \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{1}\\ \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}\\ \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}\end{array}]
G≔\mathrm{CycleGraph}\left(5\right)
\textcolor[rgb]{0,0,1}{G}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{Graph 2: an undirected unweighted graph with 5 vertices and 5 edge\left(s\right)}}
\mathrm{IsBiregular}\left(G\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}}
The GraphTheory[IsBiregular] command was introduced in Maple 2019.
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Sheftel M. Recursion operators and bi-Hamiltonian representations of cubic evolutionary (2+1)-dimensional systems (abstract) - Geometry of Differential Equations
Sheftel M. Recursion operators and bi-Hamiltonian representations of cubic evolutionary (2+1)-dimensional systems (abstract)
Speaker: Mikhail Sheftel
We construct all (2+1)-dimensional PDEs which have the Euler-Lagrange form and determine the corresponding Lagrangians. We convert these equations and their Lagrangians to two-component forms and find Hamiltonian representations of all these systems using Dirac's theory of constraints. Integrability properties of one-parameter equations that are cubic in partial derivatives of the unknown are derived by our method of skew factorization of the symmetry condition. Lax pairs and recursion relations for symmetries are determined both for one-component and two-component form. For the integrable cubic one-parameter equations in the two-component form we obtain recursion operators in
{\displaystyle 2\times 2}
matrix form and bi-Hamiltonian representations, thus discovering new bi-Hamiltonian (2+1) systems.
Slides: SheftelAMVconf2021slides.pdf
Retrieved from "https://gdeq.org/w/index.php?title=Sheftel_M._Recursion_operators_and_bi-Hamiltonian_representations_of_cubic_evolutionary_(2%2B1)-dimensional_systems_(abstract)&oldid=6473"
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use squeeze theorem find the following lim x,y-0,0
x+3 if x≤ -1
2^x if x>-1
The length of a rectangular swimming pool is exactly three times as long as its width. If the pool has a perimeter of 288m find the width of the pool.
set-builder form {y I x is a natural number and y >2}
A large cooler contains the following drinks: 6 lemonade, 8 Sprite, 15 Coke, and 7 root beer. You randomly pick two cans, one at a time (without replacement).
(a) What is the probability that you get 2 cans of Sprite?
(b) What is the probability that you do not get 2 cans of Coke?
(c) What is the probability that you get either 2 root beer or 2 lemonade?
(d) What is the probability that you get one can of Coke and one can of Sprite?
(e) What is the probability that you get two drinks of the same type?
If the minute hand is 4 inches. Find the area of the clock
functions pigeonhole principle
\left\{1,2,3,\dots ,346\right\}
are divisible by 5 or 7 ?
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Dawn of arithmetic[edit]
The earliest historical find of an arithmetical nature is a fragment of a table: the broken clay tablet Plimpton 322 (Larsa, Mesopotamia, ca. 1800 BC) contains a list of "Pythagorean triples", that is, integers
{\displaystyle (a,b,c)}
{\displaystyle a^{2}+b^{2}=c^{2}}
. The triples are too many and too large to have been obtained by brute force. The heading over the first column reads: "The takiltum of the diagonal which has been subtracted such that the width..."[2]
{\displaystyle \left({\frac {1}{2}}\left(x-{\frac {1}{x}}\right)\right)^{2}+1=\left({\frac {1}{2}}\left(x+{\frac {1}{x}}\right)\right)^{2},}
which is implicit in routine Old Babylonian exercises.[4] If some other method was used,[5] the triples were first constructed and then reordered by
{\displaystyle c/a}
Euclid IX 21–34 is very probably Pythagorean;[10] it is very simple material ("odd times even is even", "if an odd number measures [= divides] an even number, then it also measures [= divides] half of it"), but it is all that is needed to prove that
{\displaystyle {\sqrt {2}}}
is irrational.[11] Pythagorean mystics gave great importance to the odd and the even.[12] The discovery that
{\displaystyle {\sqrt {2}}}
is irrational is credited to the early Pythagoreans (pre-Theodorus).[13] By revealing (in modern terms) that numbers could be irrational, this discovery seems to have provoked the first foundational crisis in mathematical history; its proof or its divulgation are sometimes credited to Hippasus, who was expelled or split from the Pythagorean sect.[14] This forced a distinction between numbers (integers and the rationals—the subjects of arithmetic), on the one hand, and lengths and proportions (which we would identify with real numbers, whether rational or not), on the other hand.
Classical Greece and the early Hellenistic period[edit]
{\displaystyle {\sqrt {3}},{\sqrt {5}},\dots ,{\sqrt {17}}}
{\displaystyle f(x,y)=z^{2}}
{\displaystyle f(x,y,z)=w^{2}}
{\displaystyle f(x_{1},x_{2},x_{3})=0}
{\displaystyle g_{1},g_{2},g_{3}}
{\displaystyle r}
{\displaystyle s}
{\displaystyle x_{i}=g_{i}(r,s)}
{\displaystyle i=1,2,3}
{\displaystyle f(x_{1},x_{2},x_{3})=0.}
Āryabhaṭa, Brahmagupta, Bhāskara[edit]
{\displaystyle n\equiv a_{1}{\bmod {m}}_{1}}
{\displaystyle n\equiv a_{2}{\bmod {m}}_{2}}
could be solved by a method he called kuṭṭaka, or pulveriser;[27] this is a procedure close to (a generalisation of) the Euclidean algorithm, which was probably discovered independently in India.[28] Āryabhaṭa seems to have had in mind applications to astronomical calculations.[24]
Arithmetic in the Islamic golden age[edit]
Al-Haytham as seen by the West: on the frontispiece of Selenographia Alhasen [sic] represents knowledge through reason and Galileo knowledge through the senses.
Early modern number theory[edit]
One of Fermat's first interests was perfect numbers (which appear in Euclid, Elements IX) and amicable numbers;[note 7] these topics led him to work on integer divisors, which were from the beginning among the subjects of the correspondence (1636 onwards) that put him in touch with the mathematical community of the day.[38]
In 1638, Fermat claimed, without proof, that all whole numbers can be expressed as the sum of four squares or fewer.[39]
Fermat's little theorem (1640):[40] if a is not divisible by a prime p, then
{\displaystyle a^{p-1}\equiv 1{\bmod {p}}.}
{\displaystyle a^{2}+b^{2}}
is not divisible by any prime congruent to −1 modulo 4;[41] and every prime congruent to 1 modulo 4 can be written in the form
{\displaystyle a^{2}+b^{2}}
.[42] These two statements also date from 1640; in 1659, Fermat stated to Huygens that he had proven the latter statement by the method of infinite descent.[43]
{\displaystyle x^{2}-Ny^{2}=1}
as a challenge to English mathematicians. The problem was solved in a few months by Wallis and Brouncker.[44] Fermat considered their solution valid, but pointed out they had provided an algorithm without a proof (as had Jayadeva and Bhaskara, though Fermat wasn't aware of this). He stated that a proof could be found by infinite descent.
Fermat stated and proved (by infinite descent) in the appendix to Observations on Diophantus (Obs. XLV)[45] that
{\displaystyle x^{4}+y^{4}=z^{4}}
{\displaystyle x^{3}+y^{3}=z^{3}}
has no non-trivial solutions, and that this could also be proven by infinite descent.[46] The first known proof is due to Euler (1753; indeed by infinite descent).[47]
Fermat claimed (Fermat's Last Theorem) to have shown there are no solutions to
{\displaystyle x^{n}+y^{n}=z^{n}}
{\displaystyle n\geq 3}
Euler[edit]
{\displaystyle p=x^{2}+y^{2}}
{\displaystyle p\equiv 1{\bmod {4}}}
; initial work towards a proof that every integer is the sum of four squares (the first complete proof is by Joseph-Louis Lagrange (1770), soon improved by Euler himself[53]); the lack of non-zero integer solutions to
{\displaystyle x^{4}+y^{4}=z^{2}}
{\displaystyle x^{2}+Ny^{2}}
, some of it prefiguring quadratic reciprocity.[57] [58][59]
Diophantine equations. Euler worked on some Diophantine equations of genus 0 and 1.[60][61] In particular, he studied Diophantus's work; he tried to systematise it, but the time was not yet ripe for such an endeavour—algebraic geometry was still in its infancy.[62] He did notice there was a connection between Diophantine problems and elliptic integrals,[62] whose study he had himself initiated.
"Here was a problem, that I, a 10 year old, could understand, and I knew from that moment that I would never let it go. I had to solve it."[63] – Sir Andrew Wiles about his proof of Fermat's Last Theorem.
Lagrange, Legendre, and Gauss[edit]
{\displaystyle mX^{2}+nY^{2}}
{\displaystyle ax^{2}+by^{2}+cz^{2}=0}
[64] and worked on quadratic forms along the lines later developed fully by Gauss.[65] In his old age, he was the first to prove Fermat's Last Theorem for
{\displaystyle n=5}
(completing work by Peter Gustav Lejeune Dirichlet, and crediting both him and Sophie Germain).[66]
Maturity and division into subfields[edit]
Main subdivisions[edit]
Elementary number theory[edit]
Analytic number theory[edit]
Algebraic number theory[edit]
{\displaystyle f(x)=0}
{\displaystyle x}
{\displaystyle x^{5}+(11/2)x^{3}-7x^{2}+9=0}
(say) is an algebraic number. Fields of algebraic numbers are also called algebraic number fields, or shortly number fields. Algebraic number theory studies algebraic number fields.[83] Thus, analytic and algebraic number theory can and do overlap: the former is defined by its methods, the latter by its objects of study.
{\displaystyle a+b{\sqrt {d}}}
{\displaystyle a}nd
{\displaystyle b}
are rational numbers an{\displaystyle d}
{\displaystyle {\sqrt {-5}}}
{\displaystyle 6}
{\displaystyle 6=2\cdot 3}
{\displaystyle 6=(1+{\sqrt {-5}})(1-{\sqrt {-5}})}
{\displaystyle 2}
{\displaystyle 3}
{\displaystyle 1+{\sqrt {-5}}}
{\displaystyle 1-{\sqrt {-5}}}
are irreducible, and thus, in a naïve sense, analogous to primes among the integers.) The initial impetus for the development of ideal numbers (by Kummer) seems to have come from the study of higher reciprocity laws,[84] that is, generalisations of quadratic reciprocity.
{\displaystyle x^{2}+y^{2}=1,}
{\displaystyle (x,y)}
{\displaystyle a^{2}+b^{2}=c^{2}}
{\displaystyle x=a/c}
{\displaystyle y=b/c}
{\displaystyle x^{2}+y^{2}=1}
{\displaystyle f(x,y)=0}
{\displaystyle f}
is a polynomial in two variables—turns out to depend crucially on the genus of the curve. The genus can be defined as follows:[note 13] allow the variables in
{\displaystyle f(x,y)=0}
{\displaystyle f(x,y)=0}
{\displaystyle f(x,y)=0}
{\displaystyle x}
{\displaystyle a/q}
{\displaystyle \gcd(a,q)=1}
{\displaystyle x}
{\displaystyle |x-a/q|<{\frac {1}{q^{c}}}}
{\displaystyle c}
{\displaystyle x}
{\displaystyle x}
Other subfields[edit]
Probabilistic number theory[edit]
{\displaystyle 0}
Arithmetic combinatorics[edit]
{\displaystyle A}
{\displaystyle a}
{\displaystyle a+b,a+2b,a+3b,\ldots ,a+10b}
{\displaystyle A}
{\displaystyle A}
{\displaystyle A}
{\displaystyle A+A}
{\displaystyle A}
{\displaystyle A}
Computational number theory[edit]
{\displaystyle ax+by=c}
^ Already in 1921, T. L. Heath had to explain: "By arithmetic, Plato meant, not arithmetic in our sense, but the science which considers numbers in themselves, in other words, what we mean by the Theory of Numbers." (Heath 1921, p. 13)
^ Take, for example, Serre 1973 harvnb error: no target: CITEREFSerre1973 (help). In 1952, Davenport still had to specify that he meant The Higher Arithmetic. Hardy and Wright wrote in the introduction to An Introduction to the Theory of Numbers (1938): "We proposed at one time to change [the title] to An introduction to arithmetic, a more novel and in some ways a more appropriate title; but it was pointed out that this might lead to misunderstandings about the content of the book." (Hardy & Wright 2008)
^ Robson 2001, p. 201. This is controversial. See Plimpton 322. Robson's article is written polemically (Robson 2001, p. 202) with a view to "perhaps [...] knocking [Plimpton 322] off its pedestal" (Robson 2001, p. 167); at the same time, it settles to the conclusion that
^ Perfect and especially amicable numbers are of little or no interest nowadays. The same was not true in medieval times—whether in the West or the Arab-speaking world—due in part to the importance given to them by the Neopythagorean (and hence mystical) Nicomachus (ca. 100 CE), who wrote a primitive but influential "Introduction to Arithmetic". See van der Waerden 1961, Ch. IV.
{\displaystyle a\equiv b{\bmod {m}}}
{\displaystyle xa\equiv 1{\bmod {p}}}
); this fact (which, in modern language, makes the residues mod p into a group, and which was already known to Āryabhaṭa; see above) was familiar to Fermat thanks to its rediscovery by Bachet (Weil 1984, p. 7). Weil goes on to say that Fermat would have recognised that Bachet's argument is essentially Euclid's algorithm.
^ Up to the second half of the seventeenth century, academic positions were very rare, and most mathematicians and scientists earned their living in some other way (Weil 1984, pp. 159, 161). (There were already some recognisable features of professional practice, viz., seeking correspondents, visiting foreign colleagues, building private libraries (Weil 1984, pp. 160–61). Matters started to shift in the late 17th century (Weil 1984, p. 161); scientific academies were founded in England (the Royal Society, 1662) and France (the Académie des sciences, 1666) and Russia (1724). Euler was offered a position at this last one in 1726; he accepted, arriving in St. Petersburg in 1727 (Weil 1984, p. 163 and Varadarajan 2006, p. 7). In this context, the term amateur usually applied to Goldbach is well-defined and makes some sense: he has been described as a man of letters who earned a living as a spy (Truesdell 1984, p. xv); cited in Varadarajan 2006, p. 9). Notice, however, that Goldbach published some works on mathematics and sometimes held academic positions.
^ The Galois group of an extension L/K consists of the operations (isomorphisms) that send elements of L to other elements of L while leaving all elements of K fixed. Thus, for instance, Gal(C/R) consists of two elements: the identity element (taking every element x + iy of C to itself) and complex conjugation (the map taking each element x + iy to x − iy). The Galois group of an extension tells us many of its crucial properties. The study of Galois groups started with Évariste Galois; in modern language, the main outcome of his work is that an equation f(x) = 0 can be solved by radicals (that is, x can be expressed in terms of the four basic operations together with square roots, cubic roots, etc.) if and only if the extension of the rationals by the roots of the equation f(x) = 0 has a Galois group that is solvable in the sense of group theory. ("Solvable", in the sense of group theory, is a simple property that can be checked easily for finite groups.)
{\displaystyle y^{2}=x^{3}+7}
{\displaystyle (a+bi)^{2}=(c+di)^{3}+7}
^ Neugebauer & Sachs 1945, p. 40. The term takiltum is problematic. Robson prefers the rendering "The holding-square of the diagonal from which 1 is torn out, so that the short side comes up...".Robson 2001, p. 192
{\displaystyle (p^{2}-q^{2},2pq,p^{2}+q^{2})}
. Van der Waerden gives both the modern formula and what amounts to the form preferred by Robson.(van der Waerden 1961, p. 79)
^ Neugebauer (Neugebauer 1969, pp. 36–40) discusses the table in detail and mentions in passing Euclid's method in modern notation (Neugebauer 1969, p. 39).
^ Iamblichus, Life of Pythagoras,(trans., for example, Guthrie 1987) cited in van der Waerden 1961, p. 108. See also Porphyry, Life of Pythagoras, paragraph 6, in Guthrie 1987 Van der Waerden (van der Waerden 1961, pp. 87–90) sustains the view that Thales knew Babylonian mathematics.
^ Any early contact between Babylonian and Indian mathematics remains conjectural (Plofker 2008, p. 42).
^ Colebrooke 1817, p. lxv, cited in Hopkins 1990, p. 302. See also the preface in Sachau 1888 harvnb error: no target: CITEREFSachau1888 (help) cited in Smith 1958, pp. 168
^ Weil 1984, p. 118. This was more so in number theory than in other areas (remark in Mahoney 1994, p. 284). Bachet's own proofs were "ludicrously clumsy" (Weil 1984, p. 33).
^ Tannery & Henry 1891, Vol. II, p. 204, cited in Weil 1984, p. 63. All of the following citations from Fermat's Varia Opera are taken from Weil 1984, Chap. II. The standard Tannery & Henry work includes a revision of Fermat's posthumous Varia Opera Mathematica originally prepared by his son (Fermat 1679).
^ Weil 1984, p. 174. Euler was generous in giving credit to others (Varadarajan 2006, p. 14), not always correctly.
Apostol, Tom M. (n.d.). "An Introduction to the Theory of Numbers". (Review of Hardy & Wright.) Mathematical Reviews (MathSciNet). American Mathematical Society. MR 0568909. Retrieved 2016-02-28. {{cite journal}}: Cite journal requires |journal= (help) (Subscription needed)
Friberg, Jöran (August 1981). "Methods and Traditions of Babylonian Mathematics: Plimpton 322, Pythagorean Triples and the Babylonian Triangle Parameter Equations". Historia Mathematica. 8 (3): 277–318. doi:10.1016/0315-0860(81)90069-0.
Rashed, Roshdi (1980). "Ibn al-Haytham et le théorème de Wilson". Archive for History of Exact Sciences. 22 (4): 305–21. doi:10.1007/BF00717654. S2CID 120885025.
Robson, Eleanor (2001). "Neither Sherlock Holmes nor Babylon: a Reassessment of Plimpton 322" (PDF). Historia Mathematica. 28 (3): 167–206. doi:10.1006/hmat.2001.2317. Archived from the original (PDF) on 2014-10-21.
Tannery, Paul; Henry, Charles (eds.); Fermat, Pierre de (1891). Oeuvres de Fermat. (4 Vols.) (in French and Latin). Paris: Imprimerie Gauthier-Villars et Fils. {{cite book}}: |first2= has generic name (help) Volume 1 Volume 2 Volume 3 Volume 4 (1912)
Iamblichus; Taylor, Thomas (trans.) (1818). Life of Pythagoras or, Pythagoric Life. London: J.M. Watkins. Archived from the original on 2011-07-21. {{cite book}}: CS1 maint: bot: original URL status unknown (link) For other editions, see Iamblichus#List of editions and translations
Vardi, Ilan (April 1998). "Archimedes' Cattle Problem" (PDF). American Mathematical Monthly. 105 (4): 305–19. CiteSeerX 10.1.1.383.545. doi:10.2307/2589706. JSTOR 2589706.
Algebraic number theory (class field theory, non-abelian class field theory, Iwasawa theory, Kummer theory)
Analytic number theory (probabilistic number theory, sieve theory)
Arithmetic combinatorics (additive number theory)
Arithmetic geometry (anabelian geometry, P-adic Hodge theory)
Retrieved from "https://en.wikipedia.org/w/index.php?title=Number_theory&oldid=1086374771"
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Emerson Barnes 2022-02-01 Answered
Find the matrix representation of the linear transformation T that maps the input vector
{\stackrel{\to }{x}}_{1}={\left(1,1\right)}^{T}
to the output vector
{\stackrel{\to }{y}}_{1}={\left(2,-3\right)}^{T}
and maps the input vector
{\stackrel{\to }{x}}_{2}={\left(1,2\right)}^{T}
{\stackrel{\to }{y}}_{2}={\left(5,1\right)}^{T}
Danitat6 2022-02-01 Answered
Consider the rectangle formed by the points (2,7),(2,6),(4,7) and (4,6). Is it still a rectangle after transformation by?
A=\left(\begin{array}{cc}3& 1\\ 2& \frac{1}{2}\end{array}\right)
By what factor has its area changed ?
Determine transformation matrix
f:{P}_{2}\left(\mathbb{R}\right)\to {P}_{2}\left(\mathbb{R}\right)
f\left(p\left(x\right)\right)=3\cdot p\left(1\right)-{x}^{2}\cdot p\left(0\right)+\left(x-1\right)\cdot {p}^{\prime }\left(1\right)
Writing a composite transformation as a matrix multiplication
I have two matrices, P and Q as follows:
P=\left(\begin{array}{cc}\frac{1}{2}& \frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}& -\frac{1}{2}\end{array}\right)
Q=\left(\begin{array}{cc}-\frac{1}{2}& \frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}& \frac{1}{2}\end{array}\right)
- A⃗ = (4, 3) , B⃗= (2, 5)
Excellent website by the way, but my only issue, is when i try calculating with the square root its give me a different value. And it is physic .
Celia Horne 2022-01-31 Answered
How can you find a reflection matrix about a given line, using matrix multiplication and the idea of composition of transformations?
The line of:
y=-\frac{2x}{3}
, all in
{\mathbb{R}}^{2}
How to find the linear transformation associated with a given matrix?
It is well known that given two bases (or even one if we consider the canonical basis) of a vector space, every linear transformation
T:V\to W
can be represented as a matrix, but since this is an isomorphism between
L\left(V,W\right)
{M}_{m×n}
where the latter represents the space
m×n
matrices on the same field in which are defined respectively vector spaces.
a) Find the standard matrix of the linear operator
T:{R}^{2}\to {R}_{2}
given by the orthogonal projection onto the vector (1,-2).
b) Given the linear transformation
T:{R}^{2}\to {R}_{3}
Find T(2,3).
Can a triangle be represented as a matrix?
Triangle1 (1, 1), (1, 2), (3, 1)
Find the matrix which represents the stretch that maps triangle T1 onto triangle T2.
Seamus Kent 2022-01-31 Answered
1) Calculate the transformation matrix
2) Calculate the dimension of the kernel of the transformation, justify.
T:{\mathbb{R}}^{\mathbb{3}}\to {\mathbb{R}}^{\mathbb{3}}
is a linear transformation such that:
T\left({\stackrel{\to }{e}}_{1}\right)={\stackrel{\to }{e}}_{1}-{\stackrel{\to }{e}}_{2}+2{\stackrel{\to }{e}}_{3}
T\left({\stackrel{\to }{e}}_{1}+{\stackrel{\to }{e}}_{2}\right)=2{\stackrel{\to }{e}}_{3}
T\left({\stackrel{\to }{e}}_{1}+{\stackrel{\to }{e}}_{2}+{\stackrel{\to }{e}}_{3}\right)=-{\stackrel{\to }{e}}_{2}+{\stackrel{\to }{e}}_{3}
T:{R}^{3}\to {R}^{3}
be the linear transformation such that
T\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=\left[\begin{array}{c}-2\\ 5\\ -2\end{array}\right],\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}T\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]=\left[\begin{array}{c}4\\ 1\\ 4\end{array}\right],\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}T\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]=\left[\begin{array}{c}1\\ -1\\ 1\end{array}\right]
a) Find a matrix A such that
T\left(x\right)=Ax
x\in {R}^{3}
b) Find a linearly independent set of vectors in
{R}^{3}
that spans the range of T
Berasiniz1 2022-01-31 Answered
Show that the given transformation is linear, by showing it is a matrix transformation:
F\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}x=y\end{array}\right]
Gabriela Duarte 2022-01-31 Answered
I have a square matrix A over
\mathbb{C}
then is its rank invariant under a congruence transformation
A\to {P}^{t}AP
? What's the easiest way to see this?
Haialarmz6 2022-01-31 Answered
A standard matrix is given:
A=\left[\begin{array}{ccc}0& -1& 3\\ 1& 1& -3\\ 2& 2& -5\end{array}\right]
representing the linear transformation
L:{\mathbb{R}}^{3}\to {\mathbb{R}}^{3}
L\left(2,\text{ }-3,\text{ }1\right)?
Tessa Leach 2022-01-31 Answered
How do we know that S must be induced by some matrix B?
T:{\mathbb{R}}^{n}\to {\mathbb{R}}^{m}
are called transformations from
{\mathbb{R}}^{n}
{\mathbb{R}}^{m}
T:{\mathbb{R}}^{n}\to {\mathbb{R}}^{n}
has an inverse if there is some transformation S such that
T\circ S=S\circ T={1}_{{\mathbb{R}}^{n}}
T={T}_{A}:{\mathbb{R}}^{n}\to {\mathbb{R}}^{n}
denote the matrix transformation induced by the
n×n
matrix A, that is
T\left(x\right)=Ax
BA\mathbf{x}=S\left[T\left(\mathbf{x}\right)\right]=\left(S\circ T\right)\mathbf{x}={1}_{{\mathbb{R}}^{\mathbb{n}}}\left(\mathbf{x}\right)=\mathbf{x}={I}_{n}\mathbf{x}
Is there a set of matrix transformations that convert a 1D vector into a 2D matrix?
I have a vector:
\left[\begin{array}{cccccc}a& b& c& d& e& f\end{array}\right]
to convert into a 2D matrix.
\left[\begin{array}{cc}a& b\\ c& d\\ e& f\end{array}\right]
Is there a series of multiplicative matrix transformations that performs this reshaping, and if so, what is the general name for this operation?
Can a matrix transformation ever make a linearly dependent matrix linearly independent?
For example, if A is a linearly dependent matrix, and B any matrix, could BA ever come out to be linearly independent?
Miguel Davenport 2022-01-30 Answered
Find invariant points under matrix transformation
Q=\left[\begin{array}{cc}-1& 2\\ 0& 1\end{array}\right]
urlacainnxp 2022-01-30 Answered
Finding an expression for x and y variables in terms of the corresponding image X and Y variables, find the image of the line
y=2x-7
under the transformation given by
M=\left(\begin{array}{cc}1& -3\\ -3& 2\end{array}\right)
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Rotational Form of Newton's Second Law | Brilliant Math & Science Wiki
Rohit Gupta, July Thomas, Jimin Khim, and
The rotational form of Newton's second law states the relation between net external torque and the angular acceleration of a body about a fixed axis. The result looks similar to Newton's second law in linear motion with a few modifications.
Translational quantity Rotational analogue Symbol
Force Torque
\tau
I
Translational acceleration Angular acceleration
\alpha
{\overrightarrow \tau _\text{net,external}} = I\overrightarrow \alpha
Rotational Form of Newton's Second Law for Point Mass
Rotational Form of Newton's Second Law for Rigid Body
First consider a case where all the mass is in one place.
Light rod having a small object attached at its end, rotating about an axis perpendicular to its length and passing through its other end
Suppose a point object of mass
m
attached to a light rigid rod of length
l
is rotating about an axis perpendicular to the rod and passing through its end. A force acts on the particle to increase the angular velocity of rotation. Break the force into its components. One component is towards the axis and called the radial component of force
{F_r}
and the other component is in the tangential direction
{F_t}
The torque of the radial component about the axis is zero as the line of action of the force is passing through the axis itself. The torque of the tangential component will try to increase the object's angular velocity and produce angular acceleration.
The net torque of the force along the axis is
\begin{aligned} \overrightarrow \tau &= \overrightarrow r \times \overrightarrow F \\ &= \overrightarrow r \times \Big({\overrightarrow F _r} + {\overrightarrow F _t}\Big) \\ &= \overrightarrow r \times {\overrightarrow F _t} &\qquad \Big(\text{since } r \times {\overrightarrow F _r} = 0 \text{ with } r \text{ parallel to } {F_r}\Big)\\ &= l{F_t}. \qquad (1) \end{aligned}
The acceleration of the particle along the tangential direction will be
{a_t} = \frac{{{F_t}}}{m}
. For motion along a circular path, the object's angular acceleration will be
\alpha = \frac{{{a_t}}}{R}
R
is the radius of the circular path of the particle, which in this case will be the length of the rod. Thus,
\begin{aligned} \alpha &= \frac{{{F_t}}}{{ml}}\\ {F_t} &= ml\alpha \\ {F_t}l &= m{l^2}\alpha. \qquad (2) \end{aligned}
Comparing equations (1) and (2),
\tau = m{l^2}\alpha.
Now, since the moment of inertia of a particle about the axis of rotation is
I = m{l^2},
\tau = I\alpha. \ _\square
If the relative distance between any two particles on a body remains the same throughout the motion, then the body is said to be rigid. Such a body maintains its shape and size irrespective of the forces acting on it.
A rigid body rotating about a fixed axis under the influence of multiple forces and torques
If a rigid body is rotating about a fixed axis and multiple forces are acting on it to change its angular velocity, then the body can be considered to be made up of many small point masses attached at the end of massless rods and rotating about the same axis. As the body is rigid, all the particles complete their circular motion together and the angular acceleration for all the particles is the same.
Applying the rotational form of Newton's second law for individual particles,
{\tau _1} = {I_1}\alpha ,\,{\tau _2} = {I_2}\alpha ,\ldots, {\tau _n} = {I_n}\alpha.
\begin{aligned} {\tau _1} + {\tau _2} + \cdots + {\tau _n} &= ({I_1} + {I_2} +\cdots + {I_n})\alpha \\\\ {\tau _\text{net}} &= {I_\text{rigidbody}}\alpha. \end{aligned}
Which one rolls faster down a hill, a hollow cylinder or a solid cylinder? Consider both are of equal mass and radius.
If the two objects are of the same size and mass, then the external torques will be the same. The hollow object has a larger moment of inertia as the mass is spread away from the axis. As
\tau = I\alpha
, for a given torque, its angular acceleration will be less. This remains true even if the objects have different masses. This is because the mass cancels out of the equation: the torque due to gravitational force is proportional to the mass, as is the moment of inertia.
A rod of mass M and length L hinged at one end without friction and held horizontally is suddenly released under gravity. Find the angular acceleration of the rod just after the release.
A rod hinged at its left end is released under gravity.
To analyze the motion of the rod just after its release, first draw its free body diagram. The free body diagram shows two forces acting on the rod:
\quad \text{1) } Mg
, gravitational force acting at the center of gravity of the rod
\quad \text{2) } {F_\text{hinge}}
, force due to the hinge on the rod.
About the hinge point, the net torque
{\tau _\text{net}}
equals the vector sum of torque due to the hinge force and the gravitational force.
Torque due to the hinge force about the hinge is zero because the hinge force passes through the hinge. Torque due to the gravitational force about hinge is
\overrightarrow \tau = \overrightarrow r \times \overrightarrow F
{\tau _{Mg/o}} = Mg\frac{L}{2}.
{\tau _\text{net}} = \frac{{MgL}}{2}
{\tau _\text{net}} = I\alpha .
\frac{{MgL}}{2} = I\alpha.
The moment of inertia (
I
) of the rod about an axis passing through the hinge and perpendicular to the rod is
\frac{{M{L^2}}}{3}.
\begin{aligned} \frac{{MgL}}{2} &= \frac{{M{L^2}}}{3}\alpha \\ \alpha &= \frac{{3g}}{{2L}}. \end{aligned}
\frac{{2mgt}}{{R(M + 2m)}}
\frac{{2mgt}}{{R(M - 2m)}}
\frac{{2Mgt}}{{(M + 2m)}}
\frac{{mgRt}}{{(M + m)}}
In the figure above, a body of mass
m
is tied at one end of a light string and this string is wrapped around the solid cylinder of mass
M
R
. At the moment
t=0
the system starts moving. If the friction is negligible, then what is the angular velocity at time
t?
Cite as: Rotational Form of Newton's Second Law. Brilliant.org. Retrieved from https://brilliant.org/wiki/rotational-form-of-newtons-second-law/
|
Free answers to linear algebra questions
Linear algebra questions and answers
Recent questions in Linear algebra
Given the linear transformation:
T\left(M\right)=\left[\begin{array}{cc}1& 0\\ 0& 2\end{array}\right]M-M\left[\begin{array}{cc}1& 0\\ 0& 2\end{array}\right]
a) Find the matrix B of T w respect to the standard basis B of
{\mathbb{R}}^{2×2}
b) Find bases of the image and kernel of B?
Cassarrim1 2022-01-29 Answered
e=\left(a,b,c\right)
be a unit vector in
{\mathbb{R}}^{3}
and let T be the linear transformation on
{\mathbb{R}}^{3}
of rotation by
{180}^{\circ }
about e. Find the matrix for T with respect to the standard basis
{e}_{1}=\left(1,0,0\right),\text{ }{e}_{2}=\left(0,1,0\right)
{e}_{3}=\left(0,0,1\right)
The rotation matrix in
{\mathbb{R}}^{3}
{180}^{\circ }
\left[\begin{array}{ccc}-1& 0& 0\\ 0& -1& 0\\ 0& 0& 1\end{array}\right]
So rotating e by
{180}^{\circ }
gives :
\left[\begin{array}{c}-a\\ -b\\ c\end{array}\right]
After that how to get the transformation matrix w.r.t the standard basis?
Emmy Combs 2022-01-29 Answered
f:{\mathbb{R}}^{3}\to {\mathbb{R}}^{3}
\left(\begin{array}{ccc}4& 1& 3\\ 2& -1& 3\\ 2& 2& 0\end{array}\right)
Establish x,y and z such that,
f\left(\begin{array}{c}x\\ y\\ z\end{array}\right)
Do I just need to multiply the values of f for the
3×3
matrix? What does this mean overall?
FiessyFrimatsd0 2022-01-29 Answered
Find the basis for kernel of a matrix transformation
\psi \text{ }:{\left\{Mat\right\}}_{2×2}\left(\mathbb{R}\right)\text{ }\to \text{ }{\left\{Mat\right\}}_{2×2}\left(\mathbb{R}\right)
\psi :\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\to \left(\begin{array}{cc}a+b& a-c\\ a+c& b-c\end{array}\right)
Find basis for ker
\psi
Reese Munoz 2022-01-29 Answered
Let (x, y) be the co-ordinates of a point P referred to a set of rectangular axes OX, OY. Then its co-ordinates
\left(x\prime ,y\prime \right)
referred to OX′, OY′, obtained by rotating the former axes through an angle
\theta
{x}^{\prime }=x\mathrm{cos}\left\{\theta \right\}+y\mathrm{sin}\left\{\theta \right\}
{y}^{\prime }=-x\mathrm{sin}\left\{\theta \right\}+y\mathrm{cos}\left\{\theta \right\}
How do we derive the result for
\left(x\prime ,\text{ }y\prime \right)
from (x, y) ?
Suppose that U and W are subspaces of a vector space V . Prove that if W is subset of U, then U +W = U.
A local university football team has ordera national power to next yerar’s schedule.the order team has agreed to play the game guaranteed fee of $100000 .plus 25 percent of the gate receipts .assume the ticket price is $ 12 . (a) determine the number od tickets which must be sold to recover the $ 100000 guarantee. (b) if college official hope to net a profit of $2400000 from the game how many tickets must be sold .(c) if a sellout of 50000 fans is assured , what ticket price would allow the university to earn the desired profit of $ 240000. ( d) assuming a total sell out , what would total profit equal if the the $ 12price is charged.
Branden Valentine 2022-01-24 Answered
m4tx45w 2022-01-24 Answered
Suppose there was a basis for and a certain number of dimensions for subspace W in
{\mathbb{R}}^{4}
.Why is the number of dimensions 2?
W=\left\{⟨4s-t,s,t,s⟩\mid s,t\in \mathbb{R}\right\}
For instance, apparently,
\left\{⟨0,1,4,1⟩,⟨1,1,3,1⟩\right\}
is a valid set, and it happens to be of dimension 2 in
{\mathbb{R}}^{4}
. Does a basis for
{\mathbb{R}}^{n}
have to have n vectors?
V={\mathbb{R}}^{3}
W=\left\{\left(x,y,z\right)\mid x,y,z\in \mathbb{Q}\right\}
W\le V
\left(0,0,0\right)\in W
\alpha ,\beta \in W
\alpha =\left(x,y,z\right),\beta =\left({x}^{\prime },{y}^{\prime },{z}^{\prime }\right)
\alpha ,\beta =\left(x+{x}^{\prime },y+{y}^{\prime },z+{z}^{\prime }
\alpha +\beta \in W
c\in \mathbb{R},\alpha \in W
\alpha =\left(x,y,z\right)
c\alpha =\left(cx,cy,cz\right)
c\alpha \in W
W\le V
maliaseth0 2022-01-24 Answered
How could I determine whether vectors
P<-2,7,4>,Q<-4,8,1>,\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}R<0,6,7>
are all in the same plane?
licencegpopc 2022-01-24 Answered
V=Span\left\{{f}_{1},{f}_{2},{f}_{3}\right\}
{f}_{1}=1,{f}_{2}={e}^{x},{f}_{3}=x{e}^{x}
a) Prove that
S=\left\{{f}_{1},{f}_{2},{f}_{3}\right\}
is a basis of V. b) Find the coordinates of
g=3+\left(1+2x\right){e}^{x}
with respect to S. c) Is
\left\{{f}_{1},{f}_{2},{f}_{3}\right\}
linearly independent?
jelentetvq 2022-01-24 Answered
How do I find the unit vector for
v=<2,-5,6>
What is the magnitude of vector
AB
A=\left(4,2,-6\right)
B=\left(9,-1,3\right)
Lainey Goodwin 2022-01-24 Answered
Geometrically, the span of two non-parallel vectors in
{R}^{3}
1) one octant
2) a line
4) the whole 3-space
5) a plane
\stackrel{\to }{{v}_{1}}=\left[\begin{array}{c}2\\ 3\end{array}\right]
\stackrel{\to }{{v}_{1}}=\left[\begin{array}{c}4\\ 6\end{array}\right]
\stackrel{\to }{{v}_{1}}
\stackrel{\to }{{v}_{1}}
shangokm 2022-01-24 Answered
Prove that in a real vector space
V\cdot c\left({}^{\prime }\alpha -\beta \right)=c\cdot \alpha -c\cdot \beta
c\in \mathbb{R};\alpha ,\beta \in V
Let say K and L are two different subspace real vector space V. If given
\mathrm{dim}\left(K\right)=\mathrm{dim}\left(L\right)=4
, how to determine minimal dimensions are possible for V?
Anika Klein 2022-01-23 Answered
Solve the system by inverting the coefficient matrix
{x}_{1}+3{x}_{2}+{x}_{3}=4
2{x}_{1}+2{x}_{2}+{x}_{3}=-1
2{x}_{1}+3{x}_{2}+{x}_{3}=3
trefoniu1 2022-01-23 Answered
Linear transformation has unique standard matrx?
\left[T\right]=-I
Finding detailed linear algebra problems and solutions has always been difficult because the textbooks would never provide anything that would be sufficient. Since it is used not only by engineering students but by anyone who has to work with specific calculations, we have provided you with a plethora of questions and answers in their original form. It will help you to see some logic as you are solving complex numbers and understand the basic concepts of linear Algebra in a clearer way. If you need additional help or would like to connect several solutions, compare more than one solution as you approach your task.
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Area of a Sector of a Circle | Formulas, Arc Length, & Radians
How to Find Area of a Sector
Find the Radius of a Circle
Area of a Sector of a Circle Examples
Anytime you cut a slice out of a pumpkin pie, a round birthday cake, or a circular pizza, you are removing a sector. A sector is created by the central angle formed with two radii, and it includes the area inside the circle from that center point to the circle itself. The portion of the circle's circumference bounded by the radii, the arc, is part of the sector.
Acute central angles will always produce minor arcs and small sectors. When the central angle formed by the two radii is
90°
, the sector is called a quadrant (because the total circle comprises four quadrants, or fourths). When the two radii form a
180°
, or half the circle, the sector is called a semicircle and has a major arc.
Unlike triangles, the boundaries of sectors are not established by line segments. True, you have two radii forming the central angle, but the portion of the circumference that makes up the third "side" is curved, so finding the area of the sector is a bit trickier than finding area of a triangle. The distance along that curved "side" is the arc length.
You cannot find the area of a sector if you do not know the radius of the circle. Be careful, though; you may be able to find the radius if you have either the diameter or the circumference. You may have to do a little preliminary mathematics to get to the radius.
Given the diameter,
d
, of a circle, the radius,
r
r=\frac{d}{2}
Given the circumference,
C
of a circle, the radius,
r
r=\frac{C}{\left(2\pi \right)}
Once you know the radius, you have the lengths of two of the parts of the sector. You only need to know arc length or the central angle, in degrees or radians.
The central angle lets you know what portion or percentage of the entire circle your sector is. A quadrant has a
90°
central angle and is one-fourth of the whole circle. A
45°
central angle is one-eighth of a circle.
Those are easy fractions, but what if your central angle of a 9-inch pumpkin pie is, say,
31°
[insert drawing of pumpkin pie with sector cut at +/- 31°]
This formula helps you find the area,
A
, of the sector if you know the central angle in degrees,
n°
, and the radius,
r
, of the circle:
A = \left(\frac{n°}{360°}\right) × \pi × {r}^{2}
For your pumpkin pie, plug in
31°
and 9 inches:
A = \left(\frac{31}{360}\right) × \pi × {9}^{2}
A = 0.086111 × \pi × 81
\mathbf{A = }\mathbf{21.9126}\mathbf{ }{\mathbf{in}}^{\mathbf{2}}
If, instead of a central angle in degrees, you are given the radians, you use an even easier formula.
To find Area,
A
, of a sector with a central angle
\theta
radians and a radius,
r
A = \left(\frac{\theta }{2}\right) × {r}^{2}
Our beloved
\pi
seems to have disappeared! It hasn't, really. Radians are based on
\pi
(a circle is
2\pi
radians), so what you really did was replace
\frac{n°}{360°}
\frac{\theta }{2}\pi
\frac{\theta }{2}\pi
is used in our original formula, it simplifies to the elegant
\left(\frac{\theta }{2}\right) × {r}^{2}
You have a personal pan pizza with a diameter of
30 cm
. You have it cut into six equal slices, so each piece has a central angle of
60°
. What is the area, in square centimeters, of each slice?
A = \left(\frac{n°}{360°}\right) × \pi × {r}^{2}
Try it yourself first, before you look ahead!
A = \left(\frac{60°}{360°}\right) × \pi × {15}^{2}
A = \left(\frac{1}{6}\right) × \pi × 225
A = 117.8097 c{m}^{2}
Did you remember to take half the diameter to find the radius?
Area of a Sector Example Using Radians
Suppose you have a sector with a central angle of
0.8
radians and a radius of
1.3
meters. Your formula is:
A = \left(\frac{\theta }{2}\right) × {r}^{2}
Try it yourself before you look ahead!
A = \left(\frac{0.8}{2}\right) × {1.3}^{2}
A = 0.676 {m}^{2}
You can also find the area of a sector from its radius and its arc length. The formula for area,
A
, of a circle with radius, r, and arc length,
L
A = \frac{\left(r × L\right)}{2}
Here is a three-tier birthday cake
6
inches tall with a diameter of
10
[insert cartoon drawing, or animate a birthday cake and show it getting cut up]
You cut it into
16
even slices; ignoring the volume of the cake for now, how many square inches of the top of the cake does each person get?
Each slice has a given arc length of
1.963
inches. The radius is
5
inches, so:
A = \frac{\left(5 × 1.963\right)}{2}
A = 4.9075 i{n}^{2}
Since the cake has volume, you might as well calculate that, too:
V = \frac{\left(\pi × {5}^{2} × 6 × 22.5°\right)}{360°} = 29.452 i{n}^{3}
, or cubic inches.
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The Dynamic Analysis of the Cash Flows on ATM
The Dynamic Analysis of the Cash Flows on ATM
Zhengyou Wang
Department of Film and Television Arts, Shanghai Publishing and Printing College, Shanghai, China
Based on the time series of cash flows on ATM, the varying rule of withdrawal is analyzed. The model of autoregression and moving average is established by Matlab and the reliability is checked. According to the model, the cash flows on ATM are forecasted in the coming 10 days. It is important for banks to prepare the cash.
Cash Flows, Autoregression Model, Dynamic Analysis, Forecast
As the pace of life is accelerating, the application of ATM (Automated Teller Machine) becomes wider and wider. Every day, there is considerable amount of cash withdrew through the means of ATMs. Therefore, how to accurately predict the daily ATM cash flows becomes a significant issue for banks in their preparation and allocation of cash. This article, based on the known time series of ATM cash flows, analyzes the varying rules of cash withdrawal amount of ATMs. It builds an autoregressive and moving average model (ARMA) (p, q) via model recognition and parameter estimation. Based on the model, ATM cash flows for the coming 10 days are forecasted.
The main steps for predicting ATM cash flows by using time series analysis are as shown in Figure 1.
Since the observed time series data often contains random interference components, the building of mathematic model for time series will offset the interference to the maximum, so as to focus on the essential characteristics of time series.
Figure 1. ATM cash flows predicting procedures.
Time series analysis is a fast developing discipline due to its wide range of application, which leaves us enormous mathematic models illustrating time series. Main models include ARIMA (autoregressive integrated moving average model), wavelet model, and neural network model [1] [2] [3] . However, this article adopts the most typical model and a model applied most widely―the autoregressive moving average model for the illustration of the varying rules of ATM cash flows.
Time series is featured by that the data values vary with time, which means, there is great randomness in values or locations of points for different moments. It is not possible to accurately predict values by historical values. However, the values before and after the moment or the location of data points have certain correlations [4] [5] [6] . This article adopts Matlab as the tool for modeling and analysis. It imports original data of daily withdrawal of an ATM in the year of 2016, which constitutes the time series to be analyzed and is stored in the system as output column vector
\left\{X\left(t\right)\right\}
2.2. Stationarity Evaluation
Use Matlab toolbox to analyze the stationarity of time series and employ default ARMA models and relevant parameters provided by Matlab for time series modeling. As shown in Figure 2, the stationarity of original data is weak.
By observing the fit of the model, we can find that: the fit of the model is 3.3381. Since the stationarity of the model is relatively weak, the fit of the model is not strong. Therefore, the original data requires stationarity processing.
In order to better model and predict, data should be firstly subject to stationarity processing. The method adopted is to differencing the data by EXCEL and get a more stationary group of series. Import the processed data into Matlab and we will see Figure 3 in the Matlab toolbox.
The stationarity of the series curve after stationarity processing is better than the curve of original data, and the fit of the model is also better than the original ones. As shown in Figure 4, compared with the original time series, the time series after preprocessing offset the trend components of the original time series, which means, the processed one filters out the low frequency part.
2.4. Order Determination
After stationarity processing, it is the stage of order determination, which means, to determine the values of p and q in ARMA (p, q). Since the order determination is different, the consequent fit values of models are different. Hence, the choosing of a proper order is vital for determining the fit between the model and original data and also for the accuracy of data prediction after modeling.
There are a number of methods for order determination, such as FPE, AIC, BIC and etc. Test in accordance with the following orders which are arranged from low to high orders: ARMA (2, 1) ARMA (1, 1), ARMA (3, 2), ARMA (3, 1), ARMA (2, 2), ARMA (4, 3), ARMA (4, 2), ARMA (4, 1), ARMA (3, 3) and ARMA (4, 4) models. The order is determined through the fit values between the series and models of different orders.
Comparison of model fit is shown in Figure 5.
Figure 2. Time series plot of original data.
Figure 3. Series after stationarity processing.
Figure 4. Comparison of model fit.
It is concluded that the model fit of (p = 4, q = 3) is better than the fit chart of other selected orders. Therefore, ARMA (4, 3) is selected.
And the standard form of the ARMA model is:
\left\{1-\sum _{i=1}^{p}{a}_{i}{B}^{i}\right\}x\left(t\right)=\left\{1-\sum _{j=1}^{q}{c}_{j}{B}^{j}\right\}\epsilon \left(t\right)
B
is the backward shift operator and
\epsilon \left(t\right)
is white noise. This model may be simplified as:
A\left(B\right)x\left(t\right)=C\left(B\right)x\left(t\right)
The model after order determination is outputted as:
C\left(B\right)=1-2.444B+2.07{B}^{2}-0.6207{B}^{3}
C\left(B\right)=1-2.444B+2.07{B}^{2}-0.6207{B}^{3}
First evaluate the independency of the model residual, that is, to evaluate whether analyzing residuals have autocorrelation, so as to get the corresponding residual analysis graph Figure 6.
In Figure 6, we can see clearly that the residuals are strictly distributed between −0.1 and 0.1, by which we can conclude that the model residuals have no autocorrelation and the independency of the model residuals are strong. Therefore, this ARMA model is suitable for simulating this time series.
In addition, difference analysis is conducted and the residual plot is as shown in Figure 7.
Besides, by calculation, we can get the average error of 14.12% (Average accuracy is 85.88%), which proves that this model has sound original data for fit and that this model may be employed to predict future values.
After the model is built, we proceed to the stage of prediction. The adopted principle for prediction is to obtain corresponding estimates through simulation of data by the model and the estimates are the predicted values. The prediction may be easily completed by using Matlab toolbox. This is the predicted cash
Figure 6. Residual analysis graph.
Figure 7. Residual plot.
flows of this ATM for the next 10 days. The predicted values for the next 10 days are:
71.850, 40.116, 52.453, 56.345, 57.699, 62.215, 65.810, 52.470, 44.929, 41.179.
Data analysis is always a popular topic for research, creating various new cross-disciplines. It gathers research findings of various disciplines including machine learning, database, pattern recognition, statistics, AI, management information system and etc., and attracts experts and scholars of different backgrounds to engage in research and development in this area. It is deemed as an important tool bringing huge returns by the business circles. Especially, the utilization of times series research methods in solution and prediction of actual issues are applied in a wide range of areas. It has been transferred from pure theoretical studies to more practical connection, integration and application in the real world and engineering practices. By using the varied types of times series and through analysis and choosing of proper mathematic models, we can predict future occurrence and realize our purpose of understanding of things.
This article predicts cash flows of an ATM by analyzing the ATM cash flows data and building ARMA models. As we all know, the application of ATM is an integral part of a society of which the life pace is speeding up. Every day, there is a considerable amount of cash withdrew by means of ATMs. Therefore, the accurate prediction of daily cash flow is of great importance for banks and corresponding organizations in preparation and allocation of cash.
Wang, Z.Y. (2018) The Dynamic Analysis of the Cash Flows on ATM. Journal of Computer and Communications, 6, 38-43. https://doi.org/10.4236/jcc.2018.64003
1. Tian, Z. (2016) Theories and Methods for Dynamic Data Processing—Time Series Analysis. Northwest University Press, Xi’an.
2. Chen, Z.G. (2016) Time Series and Spectrum Analysis. Science Press, Beijing.
3. Guo, G.D. (2015) Data Simplification and Categorization Based on Space Division. Journal of Chinese Computer Systems, 23, 456-459.
4. Ganti, V., Gehrke, J. and Ramakrishnan, R. (2013) DEMO: Mining and Monitoring Evolving Data. IEEE Transactions on Knowledge and Data Engineering, 13, 50-63. https://doi.org/10.1109/69.908980
5. Zakirr, J. (2014) Scalable Algorithms for Association Mining. IEEE Transactions on Knowledge and Data Engineering, 12, 372-390. https://doi.org/10.1109/69.846291
6. Agrawal, R., Faloutscs, C. and Swami, A. (1993) Efficient Similarity Search in Sequence Databases. Proceedings of the 4th International Conference on Foundations of Data Organization and Algorithms, 13-15 October 1993, Chicago, Illinois, 69-84. https://doi.org/10.1007/3-540-57301-1_5
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Addition rule of probability questions, solved and explained
Get help with addition rule of probability questions
Recent questions in Addition Rule of Probability
P\left(A\cup \left(B\cap C\right)\right)
P\left(A\cap \left(B\cup C\right)\right)=P\left(A\cap B\right)+P\left(A\cap C\right).
P\left(A\cup \left(B\cap C\right)\right)=P\left(A\cup B\right)+P\left(A\cup C\right)?
\frac{1}{2450}
\frac{1}{50}×\frac{1}{49}
{A}_{14}
{A}_{23}
Pr\left(\text{ (}{A}_{14}\text{ first and }{C}_{23}\text{ second) or (}{C}_{23}\text{ first and }{A}_{14}\text{ second}\right)\right)\phantom{\rule{0ex}{0ex}}=Pr\left({A}_{14}\right)×Pr\left({C}_{23}\mid {A}_{14}\right)+Pr\left({C}_{23}\right)×Pr\left({A}_{14}\mid {C}_{23}\right)=\frac{1}{50}×\frac{1}{49}×2.
\left\{{A}_{14}{C}_{23}\text{ },\text{ }{C}_{23}{A}_{14}\right\}
P\left(2\right)+P\left(5\right)
P\left(2\right)=1/6,P\left(5\right)=1/6
yopopolin10d 2022-05-02 Answered
3-2-1
3\oplus 2\oplus 1
Esther Hoffman 2022-04-30 Answered
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Why Quantum? Practice Problems Online | Brilliant
In everyday life, the objects we encounter and interact with tend to follow certain rules. As a result, our brains are wired with an internal logic for how we expect objects in the world around us to act: a tennis ball thrown against a wall will tend to rebound in a predictable fashion, a laundry hamper full of socks, when paired and sorted by color, will remain paired and sorted forever.
But we’ve all heard of quantum weirdness: the unpredictable laws of quantum behavior that caused Einstein himself to famously reject the conclusions of his own theory. Over the astoundingly productive first 27 years of the twentieth century, a group of physicists outlined laws describing microscopic phenomena by performing experiments with atomic and subatomic particles. Quantum objects don't just rebound off walls, and no matter how carefully you sort them, they always seem to get mixed up.
When it comes to quantum objects, it rapidly became clear that when things get small, things get weird.
This quote is often misattributed to various preeminent physicists including William Thomson, the
1^\text{st}
Baron of Kelvin in the closing years of the
19^\text{th}
century. It’s usually interpreted with a knowing chuckle: little did these clever physicists know that the world of classical mechanics was about to be turned on its head and shoved out the door by the coming revolutions of relativity and quantum mechanics.
In fact, the true quote is a bit different, and reads:
“The future truths of physical science are to be looked for in the sixth place of decimals.”
The preeminent physicists in the closing years of the
19^\text{th}
century knew exactly where the secrets of physics were hiding: in increasingly accurate measurements of microscopic systems. Only when studying the very small with high precision would the shortcomings of classical mechanics become obvious.
The most common way to begin study in quantum mechanics, the physical laws which describe microscopic phenomena, is to follow its real-world historical development.
The story often starts as early as the double slit experiment showing the diffraction of light in 1801, or the spectral lines of hydrogen atoms first observed in 1859 that puzzled the physicists of the day. Each of these experiments connects to microscopic phenomena that would later be explained with quantum mechanics.
Every quantum mechanics textbook is sure to describe step by step how Max Planck and his radiation law upended the classical view of light in 1900, and how Niels Bohr developed a radical, new model of the atom in 1913. All of these breakthroughs came from precise measurement of microscopic quantum objects like photons and electrons.
The number of experiments that contradicted classical law began to pile up in the first decades of the twentieth century, and by studying them chronologically learners can come to empathize with how the physicists felt at the time: forced to abandon firmly held tenets of classical physics, and compelled to build up the laws of quantum mechanics as we know them today.
But we’re not going to follow this approach here. We’re not physicists with one foot still in the nineteenth century, so we can start with a clear and open mind.
This course will fast-forward through the quantum revolution and instead begin with an experiment from 1922 that exemplifies the behavior of quantum objects better than any other.
We will start by experimenting with a purely quantum property: spin.
The spinning subatomic particles we’ll be investigating are the least classical and most “quantum” objects that we could think of. With only simple two-state spin systems, the entire physical and mathematical framework of quantum mechanics will fall out as logical conclusions of the experiments, which is exactly what we'll do in Chapters 1 and 2.
But two-state spinning quantum objects are far from just a useful rhetorical tool. Gaining a firm understanding of spin systems and the curious case of quantum measurement is immensely rewarding to anyone interested in the cutting edge of quantum information, photonics, and statistical physics. Quantum computers are little more than carefully assembled arrays of two-state quantum objects, which you’ll be able to fully understand and appreciate in Chapter 2 of this course.
In Chapter 3, equipped with a well-developed intuition and all the mathematical tools required to explain the behavior and measurement of quantum objects, we’ll circle back to the rest of the quantum world to describe all of light and matter with quantum mechanics.
The quantum theory of light introduces the concept of a photon and leads to the radical idea that all of matter has a wave-nature associated with its motion. Along the way, an equation of motion that describes the time-evolution of quantum objects emerges: the Schrodinger equation. Together with spin, these ideas can explain many of the wonders of the modern world: stimulated emission with lasers, semiconductor transistors, and magnetic resonance imaging will be explored in the later chapters.
Of course, quantum mechanics isn’t limited to such complex modern phenomena: all of classical physics including Newton’s laws are buried somewhere in the Schrodinger equation too, as we’ll soon see.
We certainly have our work cut out for us. Understanding quantum objects is challenging, but gaining an intuition of quantum behavior is incredibly fun and rewarding. The rest of this chapter will walk through a series of experiments into particle spin, and ease in to the fundamental and strange aspects of quantum measurement.
Our exploration will begin largely conceptually in an effort to build intuition, which we can then match with mathematics in the next chapter.
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vv A second hand car dealer has ten cars
Equation of regression line is
\stackrel{^}{Y}=a+bX
b=\frac{n\sum _{ }XY-\sum _{ }X\sum _{ }Y}{n\sum _{ }{X}^{2}-{\left(\sum _{ }X\right)}^{2}}
b=\frac{10*1818.5-41*406}{10*188-{\left(41\right)}^{2}}
b = 7.734
a=\frac{\sum _{ }Y-\left(b*\sum _{ }X\right)}{n}
a =\frac{406 - \left( 7.7337 * 41 \right)}{10}
a = 8.892
Equation of regression line becomes
\stackrel{^}{Y}=8.892+7.7337X
X = 5
\stackrel{^}{Y}= 8.892 + 7.734 X
\stackrel{^}{Y}= 8.892 + 7.734 * 5
\stackrel{^}{Y} = 47.56
\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)
r\left(t\right)=<{t}^{2},\frac{2}{3}{t}^{3},t>
<4,-\frac{16}{3},-2>
\begin{array}{cc}& \text{Age}\\ \text{Facebook user?}& \begin{array}{lccc}& \begin{array}{c}\text{ Younger }\\ \left(18-22\right)\end{array}& \begin{array}{c}\text{ Middle }\\ \left(23-27\right)\end{array}& \begin{array}{c}\text{ Older }\\ \left(28\text{ and up)}\end{array}\\ \text{ Yes }& 78& 49& 21\\ \text{ No }& 4& 21& 46\end{array}\end{array}
Find the coordinates of T given that S is the midpoint of RT, R(2,6) and S(-2, 0).
Find a basis for the space spanned by the given vectors,
{v}_{1}\dots ,{v}_{5}
\left[\begin{array}{ccccc}1& -2& 6& 5& 0\\ 0& 1& -1& -3& 3\\ 0& -1& 2& 3& -1\\ 1& 1& -1& -4& 1\end{array}\right]\sim \left[\begin{array}{ccccc}1& 0& 0& -1& -2\\ 0& 1& 0& -3& 5\\ 0& 0& 1& 0& 2\\ 0& 0& 0& 0& 0\end{array}\right]
Determine the eigenvalues, eigenveltI and eigenspace of the follewing matix
\left[\begin{array}{cc}1& 3\\ 5& 3\end{array}\right]
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Bond issuers - Treasury | Brilliant Math & Science Wiki
Amir Shamsubarov and Calvin Lin contributed
The United States Department of Treasury issues several types of government debt instruments to finance the national debt of the United States, which are called Treasuries. There are 4 types of marketable treasuries:
* Treasury bills
* Treasury notes
* Treasury bonds
* Treasury Inflation Protection Securities
These are extremely liquid and are heavily traded on the secondary market.
Treasury Notes (T-notes) and Treasury Bonds (T-bonds)
A Treasury bill is a discount bond which has a maturity of less than 270 days. They pay no interest to maturity, and are sold at a discount to par value to create a positive yield. They are sold in a Dutch auction weekly, and all participants of the issue have the same purchase price.
In the secondary market, Treasury bills are quoted by the yield, based on a 360-day year. Recall that the yield is inversely proportional to the price. Because the bills are quoted by their respective yield, rather than the price, the bid will be higher than the ask, which is different from normal markets. To determine the price of a Treasury bill from the yield, we use the formula
\text{Price} = \left( 1 - \frac{ \text{yield}} { 100} \times \frac{ \text{Days to maturity} } { 360 } \right) \times (\text{Face value})
The ask yield is the bond equivalent yield when calculated on the basis of a 365-day year. We calculate this by assuming that the bond is purchased for the ask price / offer and annualize it over it's life. hence, the formula is
r_{BEY} = \frac{ \text{Face value} - P } { \text{Face value}} \times \frac{365} { \text{Days to maturity}} = \frac{ \text{Yield} } { 100} \times \frac{ 365}{360}.
Buy 1.02 and sell 1.01 Need more information about interest rates Sell 1.02 and buy 1.01 Buy 1.01 Buy 1.02
A broker calls you up with a market in the 80 day to maturity Treasury bill. His market is 1.02 bid at 1.01. What trade do you want to do?
Treasury notes have a maturity of 2 to 10 years.
Treasury bonds have a maturity of greater than 10 years. Nowadays, they are usually issued only in 30 year maturities.
Both T-notes and T-bonds:
* Have a semi-annual coupon payment.
* Are sold in a Dutch auction.
* Are issued at face values of $1000.
In the secondary market, T-notes and T-bonds are quoted at percentage of par in thirty-seconds of a point. A quote of
98:06
98'06
98-06
) indicates that it is trading at a discount, namely
\left( 96 + \frac{ 6}{ 32 } \right) \%
\$ 1000 = \$961.875
. Sometimes, a
+
is ended to the end, to indicate an additional
\frac{1}{64}
point. Prices are quoted
$221.00 $10265.625 $10221.00 $265.625
If you bought a $10,000 Treasury note with a quote of 102:21, what is the cost of the trade?
Ignore accrued interest
The Treasury note that matures in 18 months and pays a coupon payment of 5% semi-annually has a quote of 108:19 bid at 108:20. If you buy this note, what is the yield to maturity?
Assume that an interest payment was just made yesterday.
Assume that the face value of the note is $1000. Then, it can be bought for a price of
\left(108 + \frac{20}{32} \right) \% \times \$1000 = 1086.25
To find the yield to maturity, we need the value of
y
1086.25 = \frac{ 50 } { ( 1 + y \%) ^ { 0.5 } }+ \frac{ 50 } { ( 1 + y \% ) ^ 2 } + \frac{ 1050} { ( 1 + y \%) ^ 3 }
Solving this, we obtain
y = 4.06
Treasury Inflation-Protection Securities are inflation-indexed bonds in which the principal is adjusted to the Consumer Price Index, which is a measure of inflation. When CPI rises, indicating more inflation, then the principal would increase further. The coupon rate is constant, but there are different amounts of interest generated in each period as the principal has been adjusted. This protects the holder against inflation, hence is useful for retirees or pensioners.
Cite as: Bond issuers - Treasury. Brilliant.org. Retrieved from https://brilliant.org/wiki/bond-issuers-treasury/
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James A. D. W. Anderson - Wikipedia
James A. D. W. Anderson
(Redirected from James Anderson (computer scientist))
For other people with the same name, see James Anderson (disambiguation).
Transreal arithmetic
Geoffrey Daniel Sullivan
James Arthur Dean Wallace Anderson Known as James Anderson is a retired member of academic staff member in the School of Systems Engineering at the University of Reading, England, where he used to teach compilers, algorithms, fundamentals of computer science and computer algebra, and in the past he has taught programming and computer graphics.[1]
Anderson quickly gained publicity in December 2006 in the United Kingdom when the regional BBC South Today reported his claim of "having solved a 1200 year old problem", namely that of division by zero. However, commentators quickly responded that his ideas are just a variation of the standard IEEE 754 concept of NaN (Not a Number), which has been commonly employed on computers in floating point arithmetic for many years.[2]
Dr Anderson defended against the criticism of his claims on BBC Berkshire on 12 December 2006, saying, "If anyone doubts me I can hit them over the head with a computer that does it."[3]
Anderson was banned from teaching Transreal Arithmetic at the University of Reading in 2019 when he was reported to have been teaching it during a class "Fundamentals of Computer Science". Andersons nullity and transreal arithmetic are unaccepted by mathematicians and computer scientists alike, and is not a fundamental part of computer science. Shortly after he quit around the end of 2019.[citation needed]
1 Research and background
2 Transreal arithmetic
2.1 Transreal arithmetic and other arithmetics
Research and background[edit]
Anderson is a member of the British Computer Society, the British Machine Vision Association, Eurographics, and the British Society for the Philosophy of Science.[4] He is also a teacher in the Computer Science department (School of Systems Engineering) at the University of Reading.[1] He was a psychology graduate who worked in the Electrical and Electronic Engineering departments at the University of Sussex and Plymouth Polytechnic (now the University of Plymouth). His doctorate is from the University of Reading for (in Anderson's words) "developing a canonical description of the perspective transformations in whole numbered dimensions".
He has written two papers on division by zero[5][6] and has invented what he calls the "Perspex machine".
Anderson claims that "mathematical arithmetic is sociologically invalid" and that IEEE floating-point arithmetic, with NaN, is also faulty.[7]
Transreal arithmetic[edit]
In mathematical analysis, the following limits can be found:
{\displaystyle \lim _{x\to 0}{\frac {0}{x}}=0}
{\displaystyle \lim _{x\to 0^{+}}{\frac {1}{x}}=+\infty }
{\displaystyle \lim _{x\to 0^{-}}{\frac {1}{x}}=-\infty }
{\displaystyle \lim _{x\to 0}{\frac {\sin x}{x}}=1}
{\displaystyle \lim _{x\to 0}{\frac {1-\cos x}{x}}=0}
{\displaystyle 0^{0}}
is also an indeterminate form. See exponentiation.
In IEEE floating-point arithmetic:
{\displaystyle {\frac {0}{0}}=NaN}
In several computer programming languages, including C's pow function,
{\displaystyle 0^{0}}
{\displaystyle 1}
, as that is the most convenient value for numerical analysis programs, since it makes
{\displaystyle f(x)=x^{0}}
(and many other functions) continuous at zero, with the notable exception of
{\displaystyle f(x)=0^{x}}
In transreal arithmetic:
{\displaystyle {\frac {0}{0}}=\Phi }
{\displaystyle 0^{0}=\Phi \,}
by Anderson's proof, reported on by the BBC, that:
{\displaystyle 0^{0}={\frac {0}{0}}}
Anderson's transreal numbers were first mentioned in a 1997 publication,[9] and made well known on the Internet in 2006, but not accepted as useful by the mathematics community. These numbers are used in his concept of transreal arithmetic and the Perspex machine. According to Anderson, transreal numbers include all of the real numbers, plus three others: infinity (
{\displaystyle \infty }
), negative infinity (
{\displaystyle -\infty }
) and "nullity" (
{\displaystyle \Phi }
), a numerical representation of a non-number that lies outside of the affinely extended real number line. (Nullity, confusingly, has an existing mathematical meaning.)
Anderson intends the axioms of transreal arithmetic to complement the axioms of standard arithmetic; they are supposed to produce the same result as standard arithmetic for all calculations where standard arithmetic defines a result. In addition, they are intended to define a consistent numeric result for the calculations which are undefined in standard arithmetic, such as division by zero.[10]
Transreal arithmetic and other arithmetics[edit]
"Transreal arithmetic" is entirely plagiarized from IEEE floating point arithmetic, a floating point arithmetic commonly used on computers. IEEE floating point arithmetic, like transreal arithmetic, uses affine infinity (two separate infinities, one positive and one negative) rather than projective infinity (a single unsigned infinity, turning the number line into a loop).
Here are some identities in transreal arithmetic with the IEEE equivalents:
IEEE standard floating point arithmetic
{\displaystyle 0\div 0=\Phi }
{\displaystyle 0\div 0=NaN}
{\displaystyle \infty \times 0=\Phi }
{\displaystyle \infty \times 0=NaN}
{\displaystyle \infty -\infty =\Phi }
{\displaystyle \infty -\infty =NaN}
{\displaystyle \Phi +a=\Phi \ }
{\displaystyle NaN+a=NaN}
{\displaystyle \Phi \times a=\Phi }
{\displaystyle NaN\times a=NaN}
{\displaystyle -\Phi =\Phi \ }
{\displaystyle -NaN=NaN}
(i.e. applying unary negation to NaN yields NaN)
{\displaystyle +1\div 0=+\infty }
{\displaystyle 1\div +0=-1\div -0=+\infty }
{\displaystyle -1\div 0=-\infty }
{\displaystyle 1\div -0=-1\div +0=-\infty }
{\displaystyle \Phi =\Phi \Rightarrow True\ }
{\displaystyle NaN=NaN\Rightarrow False}
The main difference is that IEEE arithmetic replaces the real (and transreal) number zero with positive and negative zero. (This is so that it can preserve the sign of a nonzero real number whose absolute value has been rounded down to zero. See also infinitesimal.) Division of any non-zero finite number by zero results in either positive or negative infinity.
Another difference between transreal and IEEE floating-point operations that nullity compares equal to nullity, whereas NaN does not compare equal to NaN. This is due to conflicting semantics of the equality operator rather than a computational difference. In both cases, NaN and nullity are special error values assigned to an indeterminate. In IEEE, the inequality is because two expressions which both fail to have a numerical value cannot be numerically equivalent. In transreals, the equality allows certain identities which, although the result isn't numerical per se, both sides are either equal or both nullity (e.g. (-x)/0 = -(x/0))
Anderson's analysis of the properties of transreal algebra is given in his paper on "perspex machines".[11]
Due to the more expansive definition of numbers in transreal arithmetic, several identities and theorems which apply to all numbers in standard arithmetic are not universal in transreal arithmetic. For instance, in transreal arithmetic,
{\displaystyle a-a=0}
is not true for all
{\displaystyle a}
{\displaystyle \Phi -\Phi =\Phi }
. That problem is addressed in ref.[11] pg. 7. Similarly, it is not always the case in transreal arithmetic that a number can be cancelled with its reciprocal to yield
{\displaystyle 1}
. Cancelling zero with its reciprocal in fact yields nullity.
Examining the axioms provided by Anderson,[10] it is easy to see that any term which contains an occurrence of the constant
{\displaystyle \Phi }
is provably equivalent to
{\displaystyle \Phi }
. Formally, le{\displaystyle t}
be any term with a sub-term
{\displaystyle \Phi }
{\displaystyle t=\Phi }
is a theorem of the theory proposed by Anderson.
Anderson's transreal arithmetic, and concept of "nullity" in particular, were introduced to the public by the BBC with its report in December 2006[5] where Anderson was featured on a BBC television segment teaching schoolchildren about his concept of "nullity". The report implied that Anderson had discovered the solution to division by zero, rather than simply attempting to formalize it. The report also suggested that Anderson was the first to solve this problem, when in fact the result of zero divided by zero has been expressed formally in a number of different ways (for example, NaN).
The BBC was criticized for irresponsible journalism, but the producers of the segment defended the BBC, stating that the report was a light-hearted look at a mathematical problem aimed at a mainstream, regional audience for BBC South Today rather than at a global audience of mathematicians. The BBC later posted a follow-up giving Anderson's response to many claims that the theory is flawed.[3]
Anderson has been trying to market his ideas for transreal arithmetic and "Perspex machines" to investors. He claims that his work can produce computers which run "orders of magnitude faster than today's computers".[7][12] He has also claimed that it can help solve such problems as quantum gravity,[7] the mind-body connection,[13] consciousness[13] and free will.[13]
British computer scientist's new "nullity" idea provokes reaction from mathematicians
^ a b "Computer Science at Reading - Dr. James Anderson". University of Reading. Retrieved 28 February 2011.
^ Mark C. Chu-Carroll (7 December 2006). "Nullity: The Nonsense Number". Good Math, Bad Math. Archived from the original on 9 December 2006. Retrieved 7 December 2006.
^ a b "Nullity is a number, and that makes a difference". BBC News. 12 December 2006. Retrieved 12 December 2006.
^ "About the Ambient & Pervasive Intelligence Research Group". University of Reading. Retrieved 16 January 2007.
^ a b Ben Moore; Ollie Williams (7 December 2006). "1200-year-old problem "easy"". BBC News. Schoolchildren from Caversham have become the first to learn a brand new theory that dividing by zero is possible using a new number—"nullity". But the suggestion has left many mathematicians cold. .
^ "Professor Comes Up With a Way to Divide by Zero". Slashdot. Retrieved 7 December 2006.
^ a b c Dr James A.D.W. Anderson. "Transreal Computing Research and Portfolio – Company Showcase" (PDF). Archived from the original (PDF) on 23 January 2007. Retrieved 11 December 2006.
^ John Benito (April 2003). "Rationale for International Standard – Programming Languages – C" (PDF). Revision 5.10: 182. {{cite journal}}: Cite journal requires |journal= (help)
^ Anderson, James A. D. W. (1997). "Representing Geometrical Knowledge". Philosophical Transactions of the Royal Society of London, Series B. 352 (1358): 1129–39. Bibcode:1997RSPTB.352.1129A. doi:10.1098/rstb.1997.0096. PMC 1692011. PMID 9304680.
^ a b J. A. D. W. Anderson (2006). "Perspex Machine VIII: Axioms of Transreal Arithmetic" (PDF). In Longin Jan Latecki; David M. Mount; Angela Y. Wu (eds.). Vision Geometry XV: Proceedings of SPIE. Vol. 6499.
^ a b J. A. D. W. Anderson (2006). "Perspex Machine IX: Transreal Analysis" (PDF). In Longin Jan Latecki; David M. Mount; Angela Y. Wu. (eds.). Vision Geometry XV: Proceedings of SPIE. Vol. 6499.
^ "Transreal Computing Ltd". Archived from the original on 8 January 2007. Retrieved 12 December 2006.
^ a b c http://www.bookofparagon.com/
John Graham-Cumming (11 December 2006). "The Midas Number (or why divide by zero?)".
Philip Dorrell (16 December 2006). "Zero Divided By Zero: Application to Spherical Coordinates". Archived from the original on 18 January 2007. Retrieved 31 December 2006.
Reading University Profile page
Book of Paragon — personal homepage
Retrieved from "https://en.wikipedia.org/w/index.php?title=James_A._D._W._Anderson&oldid=1080886807"
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Superposition of Electric Fields | Brilliant Math & Science Wiki
A Former Brilliant Member, July Thomas, Andrew Ellinor, and
Every charged particle in the universe creates an electric field in the space surrounding it. This field can be calculated with the help of Coulomb's law. The principle of superposition allows for the combination of two or more electric fields.
In 1-dimension, electric fields can be added according to the relationship between the directions of the electric field vectors.
Same direction: Add the magnitudes together to find the net field.
Opposite directions: Subtract the smaller magnitude from the larger magnitude to find the net field. The net field will point in the direction of the greater field.
q_{1}
q_{2}
AB
L = 2\text{ m}
in vacuum. What is the magnitude of electric field at the center of the rod due to these 2 charges?
q_{1} = +10^{-4} C
q_{2} = +10^{-5} C
\dfrac{1}{4\pi\epsilon_{0}} = 9 \times 10^{9} \dfrac{\text{Nm}^{2}}{\text{C}^{2}}
q_{1}
q_{2}
E_{1}
E_{2},
\begin{aligned} |E_{1}| &= \dfrac{q_{1}}{4\pi\epsilon_{0}(1)^{2}} = 9 \times 10^{9} \times 10^{-4} = 9 \times 10^{5} \left(\dfrac{\text{N}}{\text{C}}\right)\\ |E_{2}| &= \dfrac{q_{2}}{4\pi\epsilon_{0}(1)^{2}} = 9 \times 10^{9} \times 10^{-5} = 9 \times 10^{4} \left(\dfrac{\text{N}}{\text{C}}\right). \end{aligned}
E_{1}
E_{2}
are oppositely directed and the angle between them is
\theta = \pi
radians, we have
\begin{aligned} |E_{net}| &= \sqrt{E_{1}^{2} + E_{2}^{2} + 2E_{1}E_{2}\cos\theta} \\ &= \sqrt{E_{1}^{2} + E_{2}^{2} - 2E_{1}E_{2}} \\ & = \Big| |E_{1}| - |E_{2}| \Big|\\ \Rightarrow E_{net} &= 9 \times 10^{5} - 9 \times 10^{4} \\ &= 86 \times 10^{4} \left(\frac{\text{N}}{\text{C}}\right).\ _\square \end{aligned}
+16
C is fixed at each of the points
x=3, 9, 15, \ldots, \infty
x
-axis, and a charge
-16
x=6, 12, 18, \ldots, \infty
x
-axis. Then find the potential at the origin due to these charges.
If the potential is of the form
\frac { A\ln B }{ C\pi { \varepsilon }_{ 0 } }
A+B+C
A
C
are co-prime and
B
Try more here!
In 2-dimensions, considerations need to be made for the relative directions of the electric fields.
The two electric fields at a point in space are
\vec{E}_1 = 3\hat{x} \frac{\text{N}}{\text{C}}
\vec{E}_2 = 4\hat{y} \frac{\text{N}}{\text{C}}.
What is the net electric field?
The net electric field is
\vec{E}_{\text{net}} = \vec{E}_1 + \vec{E}_2 = (3\hat{x} + 4\hat{y})\frac{\text{N}}{\text{C}}.
The magnitude of this electric field is
E = \sqrt{3^2 + 4^2} = 5 \left(\frac{\text{N}}{\text{C}}\right).
The direction is
\theta = \tan^{-1}\frac{y}{x} = \tan^{-1}\frac{4}{3} = 53.13^\circ.\ _\square
3\text{ cm}, 4\text{ cm}, 5\text{ cm}.
2\times 10^{-6}\text{C}
\dfrac 1{4\pi\epsilon_0} = 9\times 10^9\text{ Nm}^2/\text{C}^2
Cite as: Superposition of Electric Fields. Brilliant.org. Retrieved from https://brilliant.org/wiki/superposition-electric-fields/
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TriakisTetrahedon - Maple Help
Home : Support : Online Help : Mathematics : Geometry : 3-D Euclidean : Duality : TriakisTetrahedon
Reciprocal Polyhedra of the Thirteen Archimedean Solids
TriakisTetrahedron(gon, o, r)
TetrakisHexahedron(gon, o, r)
TriakisOctahedron(gon, o, r)
PentakisDodecahedron(gon, o, r)
TriakisIcosahedron(gon, o, r)
RhombicDodecahedron(gon, o, r)
RhombicTriacontahedron(gon, o, r)
TrapezoidalIcositetrahedron(gon, o, r)
TrapezoidalHexecontahedron(gon, o, r)
HexakisOctahedron(gon, o, r)
HexakisIcosahedron(gon, o, r)
PentagonalIcositetrahedron(gon, o, r)
PentagonalHexacontahedron(gon, o, r)
The functions are to define the reciprocal polyhedra of the thirteen Archimedean solids where o is the center of the polyhedron, and r the mid-radius.
To access the information relating to these particular type of polyhedra, use the following function calls:
returns the center of gon.
returns the mid-radius of gon.
\mathrm{with}\left(\mathrm{geom3d}\right):
Define a trapezoidal icositetrahedron with center (1,2,3), radius of the mid-sphere 3
\mathrm{HexakisIcosahedron}\left(t,\mathrm{point}\left(o,1,2,3\right),3\right)
\textcolor[rgb]{0,0,1}{t}
\mathrm{coordinates}\left(\mathrm{center}\left(t\right)\right)
[\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}]
\mathrm{form}\left(t\right)
\textcolor[rgb]{0,0,1}{\mathrm{HexakisIcosahedron3d}}
\mathrm{radius}\left(t\right)
\textcolor[rgb]{0,0,1}{3}
\mathrm{schlafli}\left(t\right)
\textcolor[rgb]{0,0,1}{\mathrm{dual}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{\mathrm{_t}}\textcolor[rgb]{0,0,1}{}\left([[\textcolor[rgb]{0,0,1}{3}]\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{5}]]\right)\right)
\mathrm{draw}\left(t,\mathrm{style}=\mathrm{patch},\mathrm{lightmodel}=\mathrm{light3}\right)
geom3d[duality]
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type/SERIES - Maple Help
Home : Support : Online Help : type/SERIES
The SERIES data structure represents an expression as a truncated series in one specified indeterminate, expanded at a particular point. A call to the MultiSeries[multiseries] function will always return an object of this type, or 0.
The SERIES structure has nine entries:
the scale (a table, see MultiSeries,scale);
the list of coefficients, which can be SERIES themselves;
\mathrm{O}\left(...\right)
term, which can be a function of the variable varying more slowly than the expansion variable. It is 0 if the series is exact with respect to its expansion variable (see below);
the common type of its coefficients;
the list of exponents;
the exponent of the
\mathrm{O}\left(...\right)
the type of the exponents;
the expansion variable, i.e. the element of the asymptotic basis being used;
the expression being expanded.
\mathrm{with}\left(\mathrm{MultiSeries},\mathrm{multiseries}\right):
s≔\mathrm{multiseries}\left(\mathrm{sin}\left(x\right),x,3\right)
\textcolor[rgb]{0,0,1}{s}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{-}\frac{{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{3}}}{\textcolor[rgb]{0,0,1}{6}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\mathrm{O}}\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{4}}\right)
\mathrm{lprint}\left(s\right)
SERIES(Scale,[1, -1/6],1,rational,[1, 3],4,integer,_var[x],sin(_var[x]))
s≔\mathrm{multiseries}\left(\frac{1}{x}+5+6{x}^{2},x,3\right)
\textcolor[rgb]{0,0,1}{s}\textcolor[rgb]{0,0,1}{≔}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{5}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{6}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}
\mathrm{lprint}\left(s\right)
SERIES(Scale,[1, 5, 6],0,integer,[-1, 0, 2],infinity,integer,_var[x],1/_var[x]+5+6*_var[x]^2)
f≔\frac{1}{1-\frac{\mathrm{exp}\left(-\frac{1}{x}\right)}{1-x}}-1
\textcolor[rgb]{0,0,1}{f}\textcolor[rgb]{0,0,1}{≔}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{-}\frac{{\textcolor[rgb]{0,0,1}{ⅇ}}^{\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{x}}}}{\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}{1}
s≔\mathrm{multiseries}\left(f,x,1\right):
s≔\mathrm{multiseries}\left(f,\mathrm{op}\left(1,s\right),3,\mathrm{op}\left(1,s\right)[\mathrm{list}][2..2]\right)
\textcolor[rgb]{0,0,1}{s}\textcolor[rgb]{0,0,1}{≔}\frac{\textcolor[rgb]{0,0,1}{1}}{\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{x}\right)\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{x}}}}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{1}}{{\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{x}\right)}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}{\left({\textcolor[rgb]{0,0,1}{ⅇ}}^{\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{x}}}\right)}^{\textcolor[rgb]{0,0,1}{2}}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\mathrm{O}}\textcolor[rgb]{0,0,1}{}\left(\frac{\textcolor[rgb]{0,0,1}{1}}{{\left({\textcolor[rgb]{0,0,1}{ⅇ}}^{\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{x}}}\right)}^{\textcolor[rgb]{0,0,1}{3}}}\right)
\mathrm{lprint}\left(s\right)
SERIES(Scale,[1/(1-_var[x]), 1/(1-_var[x])^2],1,algebraic,[1, 2],3,integer,_var[1/exp(1/x)],-1+1/(1-_var[1/exp(1/x)]/(1-_var[x])))
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Pendulum (mechanics) - Wikipedia @ WordDisk
Arbitrary-amplitude period
Approximate formulae for the nonlinear pendulum period
Arbitrary-amplitude angular displacement Fourier series
Physical interpretation of the imaginary period
{\displaystyle {\textbf {F}}={\frac {d}{dt}}(m{\textbf {v}})}
This article uses material from the Wikipedia article Pendulum (mechanics), and is written by contributors. Text is available under a CC BY-SA 4.0 International License; additional terms may apply. Images, videos and audio are available under their respective licenses.
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I'm missing something here. Let X = <mo fence="false" stretchy="false">{ <mo stretchy="
Perla Galloway 2022-04-30 Answered
X=\left\{\left(123\right),\left(132\right),\left(124\right),\left(142\right),\left(134\right),\left(143\right),\left(234\right),\left(243\right)\right\}
{A}_{4}
X
x=\left(123\right)
4=|\mathcal{O}\left(x\right)|=|G|/|{G}_{x}|=12/|{G}_{x}|
{G}_{x}=\left\{1\right\}
Litzy Fuentes
The stabilizer is not the identity, because your action in conjugation, not multiplication. Every group element stabilizes itself under conjugation, and every power of a group element stabilizes the original element, too.
1. Sailing ships used to send messages with signal flags flown from their masts. How many different signals are possible with a set of four distinct flags if a minimum of two flags is used for each signals?
2. A Gr. 9 students may build a timetable by selecting one course for each period, with no duplication of courses. Period 1 must be science, geography, or physical education. Period 2 must be art, music, French, os business. Period 3 and 4 must be math or English. How many different timetables could a student choose?
N\left(A\cup B\cup C\right)=N\left(A\right)+N\left(B\right)+N\left(C\right)-N\left(A\cap B\right)-N\left(A\cap C\right)-N\left(B\cap C\right)+N\left(A\cap B\cap C\right)
21=9+10+7-N\left(A\cap B\right)-N\left(A\cap C\right)-N\left(B\cap C\right)+5
N\left(A\cap B\right)+N\left(A\cap C\right)+N\left(B\cap C\right)=10
N\left(A\cap B\cap C\right)
=N\left(A\cap B+B\cap C+A\cap C\right)-2\ast N\left(A\cap B\cap C\right)=10-2\ast 5=0
A
B
be subsets of the finite set
S
S=A\cup B
A\cap B=\mathrm{\varnothing }
\mathcal{P}\left(X\right)
the power set of
X
|Y|
the number of elements in the set
Y
Given a statement
|\mathcal{P}\left(A\right)|+|\mathcal{P}\left(B\right)|=|\mathcal{P}\left(A\right)\cup \mathcal{P}\left(B\right)|
Use the Addition Counting Principle to prove or disprove the statement.
I understand that its asking me to find the elements of
\mathcal{P}\left(A\right)
\mathcal{P}\left(B\right)
, but where does
\mathcal{P}\left(X\right)
|Y|
fit in to solve this question?
\left(5+4\right)+\left(4+4\right)+\left(3+3\right)+\left(7+4\right)=9+8+6+11=34
I am to create a six character password that consists of 2 lowercase letters and 4 numbers. The letters and numbers can be mixed up in any order and I can also repeat the same number and letter as well. How many possible passwords are there?
What I have pieced together so far:
Well, from the fundamental counting principle, we would definitely need
{26}^{2}×{10}^{4}
but obviously this is not all the possibilities since I can rearrange letters and numbers. Since it is a password the order matters so would I try and do a permuation of some sort like
{}^{6}{P}_{2}
since there are 6 slots to try to rearrange 2 objects (letters)?
\frac{24\cdot 1}{24\cdot 23}=\frac{1}{23}
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Truncation - WikiMili, The Best Wikipedia Reader
In mathematics, limiting the number of digits right of the decimal point
For other uses, see Truncation (disambiguation).
In mathematics and computer science, truncation is limiting the number of digits right of the decimal point.
Truncation and floor function
Causes of truncation
Main article: Floor and ceiling functions
Truncation of positive real numbers can be done using the floor function. Given a number
{\displaystyle x\in \mathbb {R} _{+}}
to be truncated and
{\displaystyle n\in \mathbb {N} _{0}}
, the number of elements to be kept behind the decimal point, the truncated value of x is
{\displaystyle \operatorname {trunc} (x,n)={\frac {\lfloor 10^{n}\cdot x\rfloor }{10^{n}}}.}
However, for negative numbers truncation does not round in the same direction as the floor function: truncation always rounds toward zero, the floor function rounds towards negative infinity. For a given number
{\displaystyle x\in \mathbb {R} _{-}}
, the function ceil is used instead.
{\displaystyle \operatorname {trunc} (x,n)={\frac {\lceil 10^{n}\cdot x\rceil }{10^{n}}}}
In some cases trunc(x,0) is written as [x].[ citation needed ] See Notation of floor and ceiling functions.
With computers, truncation can occur when a decimal number is typecast as an integer; it is truncated to zero decimal digits because integers cannot store non-integer real numbers.
An analogue of truncation can be applied to polynomials. In this case, the truncation of a polynomial P to degree n can be defined as the sum of all terms of P of degree n or less. Polynomial truncations arise in the study of Taylor polynomials, for example. [1]
In mathematics, a complex number is a number that can be expressed in the form a + bi, where a and b are real numbers, and i is a symbol, called the imaginary unit, that satisfies the equation i2 = −1. Because no real number satisfies this equation, i was called an imaginary number by René Descartes. For the complex number a + bi, a is called the real part and b is called the imaginary part. The set of complex numbers is denoted by either of the symbols or C. Despite the historical nomenclature "imaginary", complex numbers are regarded in the mathematical sciences as just as "real" as the real numbers and are fundamental in many aspects of the scientific description of the natural world.
In mathematics, the factorial of a non-negative integer n, denoted by n!, is the product of all positive integers less than or equal to n:
In computing, floating-point arithmetic (FP) is arithmetic using formulaic representation of real numbers as an approximation to support a trade-off between range and precision. For this reason, floating-point computation is often used in systems with very small and very large real numbers that require fast processing times. In general, a floating-point number is represented approximately with a fixed number of significant digits and scaled using an exponent in some fixed base; the base for the scaling is normally two, ten, or sixteen. A number that can be represented exactly is of the following form:
An integer is colloquially defined as a number that can be written without a fractional component. For example, 21, 4, 0, and −2048 are integers, while 9.75, 5+1/2, and √2 are not.
A number is a mathematical object used to count, measure, and label. The original examples are the natural numbers 1, 2, 3, 4, and so forth. Numbers can be represented in language with number words. More universally, individual numbers can be represented by symbols, called numerals; for example, "5" is a numeral that represents the number five. As only a relatively small number of symbols can be memorized, basic numerals are commonly organized in a numeral system, which is an organized way to represent any number. The most common numeral system is the Hindu–Arabic numeral system, which allows for the representation of any number using a combination of ten fundamental numeric symbols, called digits. In addition to their use in counting and measuring, numerals are often used for labels, for ordering, and for codes. In common usage, a numeral is not clearly distinguished from the number that it represents.
In mathematics, a square root of a number x is a number y such that y2 = x; in other words, a number y whose square (the result of multiplying the number by itself, or y ⋅ y) is x. For example, 4 and −4 are square roots of 16, because 42 = (−4)2 = 16. Every nonnegative real number x has a unique nonnegative square root, called the principal square root, which is denoted by where the symbol is called the radical sign or radix. For example, the principal square root of 9 is 3, which is denoted by because 32 = 3 ⋅ 3 = 9 and 3 is nonnegative. The term (or number) whose square root is being considered is known as the radicand. The radicand is the number or expression underneath the radical sign, in this case 9.
In mathematics, the p-adic number system for any prime number p extends the ordinary arithmetic of the rational numbers in a different way from the extension of the rational number system to the real and complex number systems. The extension is achieved by an alternative interpretation of the concept of "closeness" or absolute value. In particular, two p-adic numbers are considered to be close when their difference is divisible by a high power of p: the higher the power, the closer they are. This property enables p-adic numbers to encode congruence information in a way that turns out to have powerful applications in number theory – including, for example, in the famous proof of Fermat's Last Theorem by Andrew Wiles.
Balanced ternary is a ternary numeral system that uses a balanced signed-digit representation of the integers in which the digits have the values −1, 0, and 1. This stands in contrast to the standard (unbalanced) ternary system, in which digits have values 0, 1 and 2. The balanced ternary system can represent all integers without using a separate minus sign; the value of the leading non-zero digit of a number has the sign of the number itself. While binary numerals with digits 0 and 1 provide the simplest positional numeral system for natural numbers, balanced ternary provides the simplest self-contained positional numeral system for integers. The balanced ternary system is an example of a non-standard positional numeral system. It was used in some early computers and also in some solutions of balance puzzles.
The IEEE Standard for Floating-Point Arithmetic is a technical standard for floating-point arithmetic established in 1985 by the Institute of Electrical and Electronics Engineers (IEEE). The standard addressed many problems found in the diverse floating-point implementations that made them difficult to use reliably and portably. Many hardware floating-point units use the IEEE 754 standard.
Positional notation usually denotes the extension to any base of the Hindu–Arabic numeral system. More generally, a positional system is a numeral system in which the contribution of a digit to the value of a number is the value of the digit multiplied by a factor determined by the position of the digit. In early numeral systems, such as Roman numerals, a digit has only one value: I means one, X means ten and C a hundred. In modern positional systems, such as the decimal system, the position of the digit means that its value must be multiplied by some value: in 555, the three identical symbols represent five hundreds, five tens, and five units, respectively, due to their different positions in the digit string.
The fractional part or decimal part of a non‐negative real number is the excess beyond that number's integer part. If the latter is defined as the largest integer not greater than x, called floor of x or , its fractional part can be written as:
In computing, the modulo operation returns the remainder or signed remainder of a division, after one number is divided by another.
Methods of computing square roots are numerical analysis algorithms for finding the principal, or non-negative, square root of a real number. Arithmetically, it means given , a procedure for finding a number which when multiplied by itself, yields ; algebraically, it means a procedure for finding the non-negative root of the equation ; geometrically, it means given the area of a square, a procedure for constructing a side of the square.
In mathematics, a negligible function is a function :\mathbb {N} \to \mathbb {R} } such that for every positive integer c there exists an integer Nc such that for all x > Nc,
A negative base may be used to construct a non-standard positional numeral system. Like other place-value systems, each position holds multiples of the appropriate power of the system's base; but that base is negative—that is to say, the base b is equal to −r for some natural number r.
A repeating decimal or recurring decimal is decimal representation of a number whose digits are periodic and the infinitely repeated portion is not zero. It can be shown that a number is rational if and only if its decimal representation is repeating or terminating. For example, the decimal representation of 1/3 becomes periodic just after the decimal point, repeating the single digit "3" forever, i.e. 0.333.... A more complicated example is 3227/555, whose decimal becomes periodic at the second digit following the decimal point and then repeats the sequence "144" forever, i.e. 5.8144144144.... At present, there is no single universally accepted notation or phrasing for repeating decimals.
In mathematics, a real number is a value of a continuous quantity that can represent a distance along a line. The adjective real in this context was introduced in the 17th century by René Descartes, who distinguished between real and imaginary roots of polynomials. The real numbers include all the rational numbers, such as the integer −5 and the fraction 4/3, and all the irrational numbers, such as . Included within the irrationals are the real transcendental numbers, such as π (3.14159265...). In addition to measuring distance, real numbers can be used to measure quantities such as time, mass, energy, velocity, and many more. The set of real numbers is denoted using the symbol R or and is sometimes called "the reals".
↑ Spivak, Michael (2008). Calculus (4th ed.). p. 434. ISBN 978-0-914098-91-1.
Wall paper applet that visualizes errors due to finite precision
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Laboratoire de Recherche en Energétique et Météorologie de l’Espace (LAREME), Université Norbert Zongo (anciennement Université de Koudougou), Koudougou, Burkina Faso.
Abstract: The coronal mass ejections (CMEs) produce by Sun poloidal magnetic fields contribute to geomagnetic storms. The geomagnetic storm effects produced by one-day-shock, two-days-shock and three-days-shock activities on Ouagadougou station F2 layer critical frequency time variation are analyzed. It is found that during the solar minimum and the increasing phases, the shock activity produces both positive and negative storms. The positive storm is observed during daytime. At the solar maximum and the decreasing phases only the positive storm is produced. At the solar minimum there is no three-days-shock activity. During the solar increasing phase the highest amplitude of the storm effect is due to the one-day-shock activity and the lowest is produced by the two-days-shock activity. At the solar maximum phase the ionosphere electric current system is not affected by the shock activity. Nevertheless, the highest amplitude of the storm effect is caused by the two-days-shock activity and the lowest by the one-day-shock activity. During the solar decreasing phase, the highest amplitude provoked by the storm is due to the three-days-shock activity and the lowest by the one-day-shock activity.
Keywords: CMEs Activity, Shock Activity, F2 Layer Critical Frequency, Geomagnetic Activity, Ionosphere Time Variation
\sigma =\sqrt{V}
\frac{1}{N}{\displaystyle {\sum }_{i=1}^{N}{\left({x}_{i}-\stackrel{¯}{x}\right)}^{2}}
\stackrel{¯}{x}
{\text{foF2}}_{\text{two-days-shock}}\text{value}>{\text{foF2}}_{\text{quietest-days}}\text{value}
{\text{foF2}}_{\text{one-day-shock}}\text{value}>{\text{foF2}}_{\text{quietest-days}}\text{value}
\text{foF}{2}_{\text{one-day-shock}}\text{value}>\text{foF}{2}_{\text{three-days-shock}}\text{value}>\text{foF}{2}_{\text{two-days-shock}}\text{value}
\text{foF}{2}_{\text{two-days-shock}}>\text{foF}{2}_{\text{three-days-shock}}>\text{foF}{2}_{\text{one-day-shock}}
\text{foF}{2}_{\text{two-days-shock}}\cong \text{foF}{2}_{\text{three-days-shock}}>\text{foF}{2}_{\text{one-day-shock}}
\text{foF}{2}_{\text{three-days-shock}}>\text{foF}{2}_{\text{two-days-shock}}>\text{foF}{2}_{\text{one-day-shock}}
\text{foF}{2}_{\text{one-day-shock}}>\text{foF}{2}_{\text{three-days-shock}}\cong \text{foF}{2}_{\text{two-days-shock}}
{\text{foF2}}_{\text{three-days-shock}}\text{value}>{\text{foF2}}_{\text{two-days-shock}}\text{value}>{\text{foF2}}_{\text{one-day-shock}}\text{value}
{\text{foF2}}_{\text{three-days-shock}}\text{value}>{\text{foF2}}_{\text{one-day-shock}}\text{value}>{\text{foF2}}_{\text{two-days-shock}}\text{value}
{\text{foF2}}_{\text{three-days-shock}}\text{value}\cong {\text{foF2}}_{\text{one-day-shock}}\text{value}>{\text{foF2}}_{\text{two-days-shock}}\text{value}
\text{foF}{2}_{\text{one-day-shock}}\text{value}>\text{foF}{2}_{\text{three-days-shock}}\text{value}>\text{foF}{2}_{\text{two-days-shock}}\text{value}
\text{foF}{2}_{\text{two-days-shock}}>\text{foF}{2}_{\text{three-days-shock}}>\text{foF}{2}_{\text{one-day-shock}}
\text{foF}{2}_{\text{two-days-shock}}>\text{foF}{2}_{\text{three-days-shock}}>\text{foF}{2}_{\text{one-day-shock}}
Cite this paper: Gyébré, A. , Gnabahou, D. and Ouattara, F. (2018) The Geomagnetic Effects of Solar Activity as Measured at Ouagadougou Station. International Journal of Astronomy and Astrophysics, 8, 178-190. doi: 10.4236/ijaa.2018.82013.
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3 Ways to Order Decimals from Least to Greatest - wikiHow
1 Ordering Decimals Using Place Value
2 Ordering Decimals Using a Number Line
3 Understanding Place Value
You can order decimals just as you can order other numbers. With decimals, it is especially important to understand place value. You can use a place value chart to compare the value of comparable digits in each number, or you can use a number line and assess each number’s relative position.
Ordering Decimals Using Place Value Download Article
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Look at the whole number. The quickest way to determine the least or greatest number is to compare their whole numbers. If one number has a larger whole number than the others, it is automatically the greatest number. If one number has a smaller whole number than the others, it is automatically the least number.
For example, if you are comparing the numbers 12.45, 12.457, and 11.47, compare the whole numbers: 12, 12, and 11. Since 11 is less than 12, you know that 11.47 is going to be the smallest, or least, number.
Set up a table for the remaining numbers. The table should have one row for each number, and a column for each digit in the numbers. You should also add a column to include the decimal point.[1] X Research source Label the place values above the table.
For example, since you are comparing 12.45 and 12.457, you will make a table with two rows and six columns--one column for each place value in the longest number, plus a column for the decimal point.
From left to right, the columns would be labeled tens, ones, decimal, tenths, hundredths, thousandths.
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Fill in the numbers on the table. Make sure the decimal points line up. If the numbers are different lengths, fill in zeros for any open columns.[2] X Research source
For example, since 12.45 has four digits, and 12.457 has five digits, you will need to add a 0 in the thousandths place for 12.45.
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Compare the tenths column. If either number has a larger digit in the tenths column, it is the larger number. If the numbers have the same digit in the tenths column, you need to move on to compare the hundredths column.
For example, 12.45 and 12.457 both have a 4 in the tenths place, so you cannot tell yet which is greater.
Compare the hundredths column. Again, compare the digits in this place value. If either number has a larger digit here, it is the larger number. If not, you need to move on to the thousandths column.
For example, 12.45 and 12.457 both have a 5 in the hundredths place, so you cannot tell yet which is greater.
Compare the smaller fractional place values. Keep comparing the digits in the columns of your table, until you find a number that has a larger digit. This will be the largest number.[3] X Research source
For example, 12.45 has a 0 in the thousandths place, and 12.457 has a 7 in the thousandths place. Thus, 12.457 is greater than 12.45.
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Order the numbers from least to greatest. The smallest number should come first in your list, and the largest number should come last. The other numbers should be listed in between in ascending order.
For example, the numbers listed from least to greatest are 11.47, 12.45, 12.457. You can also write them using the less than symbol: 11.47 < 12.45 < 12.457.
Ordering Decimals Using a Number Line Download Article
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Compare the whole numbers. The quickest way to order decimals is to look at their whole numbers. If one whole number is greater than or less than the others, you know the decimal is greater than or less than the others.
For example, if you are comparing 14.36, and 13.458, and 14.369, you would compare 14, 13, and 14. Since 13 is less than 14, you know that 13.458 is the smallest number.
Determine the first place value where the digits differ. Moving from left-to-right, compare the digits in each place value. Noting where the numbers differ will help you set up your number line.
For example, 14.36 and 14.369 have the same digits until the thousandths place.
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Determine the two fractional numbers the decimals fall between. These will be the first and last numbers you label on your number line. To find these numbers, look to the last place value where the numbers shared the same digit. This will be the first number on your number line. To find the last number, add 1 to the last place value the numbers shared.
For example, since the last place value where 14.36 and 14.369 share a digit is the hundredths place, the first number on the number line will be 14.36. Determine the last number by adding 1 to the hundredths place. So, the last number on the number line is 14.37.
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Divide the number line into tenths. Since each new place value increases or decreases by a factor of 10, you should divide the line into tenths to represent the 10 smaller numbers that come between the first and last numbers on the number line.[4] X Research source You do not need to label each number, but doing so can make it easier to plot the numbers you are comparing.
For example, between 14.36 and 14.37, you should label 14.361, 14.362, 14.363, 14.364, 14.365, 14.366, 14.367, 14.368, and 14.369.
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Plot the numbers you are comparing on the number line. If you labeled all the numbers on your number line, simply find the corresponding number and draw a dot. If you did not label, you will need to count the hash marks. If a number you are comparing does not fall directly on a labeled point of the number line, plot the number between two other numbers or hash marks.[5] X Research source
For example, if you are comparing 14.36 and 14.369, draw a dot on the first number of the number line to show 14.36. Draw a dot on the second-to-last number to show 14.369.
Compare the location of each number on the number line. On a number line, numbers grow larger left-to-right. So, the greatest number will be the farthest right on the number line, and the least number will be the farthest left on the number line.[6] X Research source
For example, since 14.369 is to the right of 14.36, then 14.369 is the larger number.
Write the numbers in order from least to greatest. The smallest number should be listed first, and the largest number should be listed last. The other numbers should be listed between these two, in ascending order.
For example, 13.458, 14.36, 14,369. You can also use the less than sign to show their relationship: 13.458 < 14.36 < 14,369.
Understanding Place Value Download Article
Locate the decimal symbol. A decimal symbol looks like a dot or period. (In Europe, the decimal symbol looks like a comma.) It separates the number into its whole and fractional parts.[7] X Research source The digits to the left of a decimal symbol represent whole numbers. The digits to the right of the decimal symbol represent fractional numbers.
For example, if you have the decimal 12.38, the digits 1 and 2 represent whole numbers (the number 12), and the digits 3 and 8 represent fractional numbers (.38 is less than 1, or a fraction of 1).
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Learn place values for whole numbers. Place value indicates the value of a digit. A digit can have a different value, depending on where it is located in a number.[8] X Research source From right to left, the place values are ones, tens, hundreds, thousands, ten-thousands, hundred-thousands, and millions.[9] X Research source
For example, In the number 51, the digit 5 is in the tens place. It has a value of 5 tens, or 50. But in the number 50,001, the digit 5 is in the ten-thousands place. It has a value of 5 ten-thousands, or 50,000.
The smallest place value for whole numbers is the ones place. Once you have 10 ones, you exchange them for 1 ten. Thus, you have a 1 in the tens place, and a 0 in the ones place. Once you have 10 tens, you exchange them for 1 hundred. Thus, you have a 1 in the hundreds place, and a 0 in the tens and ones place. In other words, each new place value you add on increases by a factor of 10.[10] X Research source This pattern continues for the higher place values.
Note that place value gets larger from right-to-left. In a whole number, a digit has more value the farther away it is from the decimal point.
Learn the place value for fractional numbers. Just like place value indicates the value of digit in a whole number, it also indicates the value of a digit in a fractional number. From left-to-right from the decimal symbol, the place values are tenths, hundredths, thousandths, ten-thousandths, hundred-thousandths, and millionths.[11] X Research source
For example, in the number 1.5, the digit 5 is in the tenths place. It has the value of 5 tenths, or
{\displaystyle {\frac {5}{10}}}
. But in the number 1.0005, the digit 5 has a value of 5 ten-thousandths, or
{\displaystyle {\frac {5}{10,000}}}
The largest place value for fractional numbers is the tenths place. You need 10 hundredths to make 1 tenth. You need 10 thousandths to make 1 hundredth. In other words, each new place value you add on decreases by a factor of 10.[12] X Research source This pattern continues for the smaller place values.
Note that, similar to place value in whole numbers, place value in fractional numbers gets bigger moving from right-to-left. However, in a fractional number, a digit has less value the farther away it is from the decimal point. It is a common misconception that 1 tenth is less than 1 hundredth, since 1 ten is less than 1 hundred. However, 1 tenth is greater than 1 hundredth. When comparing fractional place value, it can help to think of the digits expressed as fractions:
{\displaystyle {\frac {1}{10}}>{\frac {1}{100}}}
Which is least: 0.105, 0.501, 0.015, or 0.15?
All four numbers have nothing but a zero to the left of the decimal point, so the next thing to look at is the first place to the right of the decimal point. The numbers in that place are 1, 5, 0 and 1. So the number with the 0 in that place is the least: 0.015.
↑ http://www.mathsisfun.com/ordering_decimals.html
↑ http://www.virtualnerd.com/middle-math/decimals/comparing-ordering/compare-decimals-example
↑ http://mathworld.wolfram.com/DecimalPoint.html
↑ http://www.virtualnerd.com/middle-math/decimals/representing/place-value-definition
↑ http://www.coolmath.com/prealgebra/02-decimals/01-decimals-place-value-01
↑ https://www.mathsisfun.com/definitions/decimal-point.html
Español:ordenar decimales de menor a mayor
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