text:
Definition of Infrared Spectroscopy

Infrared spectroscopy (IR) is a characterization tool chemists use to help determine the molecular structure. IR capitalizes on the concept that functional groups absorb specific frequencies of energy based on their structure. When a functional group absorbs energy, it can vibrate in a bending or stretching mode and the characteristic energy for this vibrational mode is reported in wavenumbers. A change in molecular dipole is required for a bending or stretching mode to be visible in by IR. In general, stronger bonds vibrate at higher wavenumbers, as do bonds between atoms of very different size. IR is a reliable and sensitive technique that has important applications in crime-scene investigation, for example in solving a case of arson, IR can identify traces of fuel or accelerant.

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# Britt

Newest questions and responses by Britt
1. ## U.S History

Which countries did Simon Bolivar help to gain independence from Spain? (Select all that apply.) A. Haiti B. Mexico C. Peru D. Venezuela E. Ecuador C,D, E?

2. ## U.S. History

Which best describes a significant event that occurred during the American Revolutionary War? The Emancipation Proclamation declared the freedom of 3.1 million slaves. James Madison wrote the first 10 amendments of the Constitution, known as the Bill of

3. ## U.S History

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4. ## Chemistry

At room temperature, 20° C, water (H2O) is a liquid and oxygen (O2) is a gas. Based on kinetic molecular theory, which answer best explains the states of the two fluids? aThe particles in the liquid have greater kinetic energy than the particles in the

40. ## history

How was the civil rights movement of the 1950s and 1960s a continuation of the Reconstruction-era struggle for blacks to achieve equality?

41. ## history

How was the civil rights movement of the 1950s and 1960s a continuation of the Reconstruction-era struggle for blacks to achieve equality?

42. ## chemistry

Based on electronegativity values, which bond has the highest degree of ionic character: H--S Si--Cl Cs--Br?

43. ## science

you are riding your bike at 9km/h. ten minutes later; your speed is 6km/hr.

44. ## science

What is the average speed of a car that travels 320km in 3.5 hours?

45. ## chemistry

Ammonia, NH3, can be made by reacting nitrogen with hydrogen. N2 + 3H2 = 2NH3 How many moles of NH3 can be made if 7.5 moles of H2 react with enough N2? Please help!

46. ## University Physics

Building A and Building B are 30m away from each other, on the 4th floor; 35m high, someone has set up a slingshot from the window on the 4th floor of building A and is trying to sling a balloon into the window directly across from them in building B. the

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Discuss the importance of landscape in crafting battle strategy for both the French and the English. Help please! talking in general.. no one specific battle

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Solve for x: 2x=ex+1

49. ## calculus

Solve the equation 14=65(19)x for x using logs.

50. ## trigonometry

Use De Moivre’s Theorem to simplify each expression. Write the answer in the form a + bi. {1 - i(/3)}^4

51. ## accounting

Christy Burch and Myung Lim, partners, compared their current income statement with their income statement of a year ago. They noted that sales were 12.0% higher than a year ago. They also noted that the total expenses were 20.0% higher than a year ago.

52. ## trigonometry

Find a complex number a + bi such that a^2 + b^2 is irrational. Justify and explain your reasoning.

53. ## trigonometry

Find a complex number a + bi such that a^2 + b^2 is irrational. Justify and explain your reasoning.

54. ## trigonometry

{1/(sec x - 1)} - {1/(sec x +1)} = 2 cot^2 x solve using identities?

55. ## trigonometry

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56. ## trigonometry

1) (sin2t) (cos5t) + (cos2t) (sin5t) 2) {1/(sec x - 1)} - {1/(sec x +1)} = 2 cot^2 x

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A 0.15 kg ball of dough is thrown straight up into the air with an initial velocity of 13 m/s. (a) Find the momentum of the of dough at its maximum height. kg·m/s upward (b) Find the momentum of the ball of dough halfway to its maximum height on the way

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for two consecutive numbers, the sum of the smaller number plus twice the lage number is 38. what are the numbers?

85. ## statistics

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86. ## Chem

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87. ## MATH

Which division sentence will give an answer that is not in equal groups? a.) 26/4 b.) 35/7 c.) 42/6 d.) 45/5

88. ## language

what sentence could you use please in an appaositive sentence

89. ## MATH

Coin box Question: Assume that there are 25 pennies, 10 nickels, 10 dimes, and 5 quarters in a box. after shuffeling the box, you pick a coin at random and note down its face value. let x represent the face value of the coin in cents. then construct the

90. ## Statistics

Coin box Question: Assume that there are 25 pennies, 10 nickels, 10 dimes, and 5 quarters in a box. after shuffeling the box, you pick a coin at random and note down its face value. let x represent the face value of the coin in cents. then construct the

91. ## Stat

Coin box Question: Assume that there are 25 pennies, 10 nickels, 10 dimes, and 5 quarters in a box. after shuffeling the box, you pick a coin at random and note down its face value. let x represent the face value of the coin in cents. then construct the

92. ## physics

The distance from an observer on the plain to the top of a nearby mountain is 5.3km, and the angle between this line and the horizontal is 8.4 degrees. How tall is the mountain?

93. ## Algebra

69: 16x^2+8x-3. I inserted this into the quadratic formula and got -8+/- the square root of (8^2-4(16)(-3)/(2 times 16). This reduced to -8+/- the square root of (64+120)/32, which further reduced to -8+/-the square root of 184/32. I reduced the top part

94. ## math

identifying domain and range

95. ## Physics

A fire helicopter carries a 679 kg bucket of water at the end of a 15.2 m long cable. Flying back from a fire at a constant speed of 42.5 m/s, the cable makes an angle of 49.0o with respect to the vertical. Determine the force of air resistance on the

96. ## Physics

Your groceries are in a bag with paper handles. The handles will tear off if a force greater than 50.8 N is applied to them. What is the greatest mass of groceries that can be lifted safely with this bag, given that the bag is raised with constant speed.

97. ## Chem

You've just prepared a solution by transferring 20.0 g of glucose into a 1.0 - L volumetric flask, and adding water to the mark. Calculate the concentration of the glucose solution in terms of millimoles/Liter, mg/mL, and % (m/v). Ok, I think i'm

98. ## Chem

You've just prepared a solution by transferring 20.0 g of glucose into a 1.0 - L volumetric flask, and adding water to the mark. Calculate the concentration of the glucose solution in terms of millimoles/Liter, mg/mL, and % (m/v).

99. ## math

I need help with finding function that appears in newspaper, and what is the domain and range of each. My question is what i'm i suppose to look for in the newspaper a article?

format: plain


text:
# AX

The AX function computes the relative rotational displacement of marker I with respect to marker J about the X-axis of marker J.

AX(I, J)

## Description

The computed angle takes into account the total number of revolutions between the two markers. While computing this angle, it is assumed that rotations about the other two axes of marker J (Y and Z) are zero. The AX angle is the angle between the two Y-axes measured counterclockwise from the Y-axis of marker J.

## Arguments

I
The marker whose rotational displacement is to be computed. Must be specified.
J
The marker with respect to which the rotational displacement is to be computed. This argument is optional. If omitted, it defaults to the ground coordinate system.

## Example

<Force_Vector_TwoBody
id                   = "104004"
type                 = "TorqueOnly"
i_marker_id          = "10415720"
j_floating_marker_id = "10416721"
ref_marker_id        = "10416720"
tx_expression        = "-(1000)*(AX(10415720,10416720))+(0)-(10)*WX(10415720,10416720,10416720)"
ty_expression        = "-(1000)*(AY(10415720,10416720))+(0)-(10)*WY10415720,10416720,10416720)"
tz_expression        = "-(100)*(AZ(10415720,10416720))+(0)-(1)  *WZ(10415720,10416720,10416720)"
/>
OR
<Post_Request
comment              = "Left Caster/Kingpin Inclination"
id                   = "71000030"
type                 = "EXPRESSION"
expr1                = "NULL"
expr2                = "(AY(10803060,30101010))*RTOD"
expr3                = "-(AX(10803060))*RTOD"
expr4                = "NULL"
expr5                = "NULL"
expr6                = "NULL"
expr7                = "NULL"
expr8                = "NULL"
/>


text:
Document Type : Review paper

Authors

Faculty of Mathematics, K. N. Toosi University of Technology, P. O. Box: 16765-3381, Tehran, Iran

Abstract

The theory of derivatives and integrals of fractional in fractional calculus have found enormous
applications in mathematics, physics and engineering so for that reason we need an efficient and accurate computational method for the solution of fractional differential equations. This paper presents a numerical method for solving a class of linear and nonlinear multi-order fractional differential equations with constant coefficients subject to initial conditions based on the fractional order Chebyshev functions that this function is defined as follows:
\begin{equation*}
\overline{T}_{i+1}^{\alpha}(x)=(4x^{\alpha}-2)\overline{T}_{i}^{\alpha}(x)\overline{T}_{i-1}^{\alpha}(x),\,i=0,1,2,\ldots,
\end{equation*}
where $\overline{T}_{i+1}^{\alpha}(x)$ can be defined by introducing the change of variable $x^{\alpha},\,\alpha>0$, on the shifted Chebyshev polynomials of the first kind. This new method is an adaptation of collocation method in terms of truncated fractional order Chebyshev Series. To do this method, a new operational matrix of fractional order differential in the Hilfer sense for the fractional order Chebyshev functions is derived. By using this method we reduces such problems to those of solving a system of algebraic equations thus greatly simplifying the problem. At the end of this paper, several numerical experiments are given to demonstrate the efficiency and accuracy of the proposed method.

Keywords

Main Subjects

###### ##### References

[1] M. Ahmadi Darani, A. Saadatmandi, The operational matrix of fractional derivative of the fractional-order Chebyshev functions and its applications, Comput. Methods. Differ. Equ. 5 (1) (2017), 67-87.

[2] A. Abedian, A. Düster, Equivalent Legendre polynomials: Numerical integration of discontinuous functions in the finite element methods, Comput. Methods. Appl. Mech. Eng. 343 (2019), 690-720.

[3] M. A. Abdelkawy, A collocation method based on Jacobi and fractional order Jacobi basis functions for multi-dimensional distributed-order diffusion equations, Int. J. Nonlinear. Sci. Numer. Simul. 19 (7-8) (2018), 781-792.

[4] A. Ansari, A. Refahi Sheikhani, H. Saberi Najafi, Solution to system of partial fractional differential equations using the fractional exponential operators, Math. Methods. Appl. Sci. 35 (1) (2012), 119-123.

[5] D. Baleanu, B. Shiri, H. M. Srivastava, M. Al Qurashi, A Chebyshev spectral method based on operational matrix for fractional differential equations involving non-singular Mittag-Leffler kernel, Adv. Differ. Equ. 2018, 2018:353.

[6] M. Behroozifar, N. Habibi, A numerical approach for solving a class of fractional optimal control problems via operational matrix Bernoulli polynomials, J. Vib. Control. 24 (12) (2018), 2494-2511.

[7] A. H. Bhrawy, M. A. Zaky, J. A. T. Machado, Numerical solution of the two-sided space-time fractional
telegraph equation via Chebyshev tau approximation, J. Optim. Theory. Appl. 174 (1) (2017), 321-341.

[8] C. Canuto, M. Y. Hussaini, A. Quarteroni, T. A. Zang, Spectral Methods, Springer-Verlag, Berlin, 2006.

[9] E. H. Doha, A. H. Bhrawy, S. S. Ezz-Eldien, A Chebyshev spectral method based on operational matrix for initial and boundary value problems of fractional order, Comput. Math. Appl. 62 (5) (2011), 2364-2373.

[10] M. R. Eslahchi, M. Dehghan, S. Amani, The third and fourth kinds of Chebyshev polynomials and best uniform approximation, Math. Comput. Model. 55 (5-6) (2012), 1746-1762.

[11] C. M. Fan, C. N. Chu, B. Sarler, T. H. Li, Numerical solutions of waves-current interactions by generalized finite difference method, Eng. Anal. Bound. Elem. 100 (2019), 150-163.

[12] R. Garra, R. Gorenflo, F. Polito, Z. Tomovski, Hilfer-Prabhakar derivatives and some applications, Appl. Math. Momput. 242 (2014), 576-589.

[13] Y. Gu, W. Qu, W. Chen, L. Song, C. Zhang, The generalized finite difference method for long-time dynamic modeling of three-dimensional coupled thermoelasticity problems, J. Comput. Phys. 384 (2019), 42-59.

[14] O. E. Hepson, A. Korkmaz, I. Dag, Exponential B-spline collocation solutions to the Gardner equation, Int. J. Comput. Math. 97 (4) (2020), 837-850.

[15] M. H. Heydari, Z. Avazzadeh, A new wavelet method for variable-order fractional optimal control problems, Asian. J. Control. 20 (5) (2018), 1804-1817.

[16] E. Keshavarz, Y. Ordokhani, M. Razzaghi, Numerical solution of nonlinear mixed Fredholm-Volterra integro-differential equations of fractional order by Bernoulli wavelets, Comput. Methods. Differ. Equ. 7 (2) (2019), 163-176.

[17] A. A. Kilbas, H. M. Srivastava, J. J. Trujillo, Theory and applications of fractional differential equations (Vol. 204), Elsevier Science Limited, North-Holland Mathematics Studies, 2006.

[18] D. Mahapatra, S. Sanyal, S. Bhowmick, An approximate solution of functionally Graded Timoshenko beam using B-spline collocation method, J. Solid. Mech. 11 (2) (2019), 297-310.

[19] V. A. Manrique, M. Ramirez, Simple Approach to Special Polynomials: Laguerre, Hermite, Legendre, Tchebycheff and Gegenbauer, Appl. Math. IntechOpen. (2019), 1-18.

[20] M. Mashoof, A. Refahi Sheikhani, H. S. Naja, Stability analysis of distributed order Hilfer-Prabhakar differential equations, Hacettepe. J. Math. Stat. 47 (2) (2018), 299-315.

[21] I. Podlubny, Fractional Differential Equations, Academic Press, New York, 1999.

[22] P. Rana, N. Shukla, Y. Gupta, I. Pop, Homotopy analysis method for predicting multiple solutions in the channel flow with stability analysis, Commun. Nonlinear. Sci. Numer. Simul. 66 (2019), 183-193.

[23] K. Rabiei, Y. Ordokhani, Solving fractional pantograph delay differential equations via fractional order Boubaker polynomials, Eng. Comput. 35 (4) (2019), 1431-1441.

[24] H. Rezazadeh, H. Aminikhah, A. Refahi Sheikhani, Stability analysis of Hilfer fractional differential systems, Math. Commun. 21 (1) (2016), 45-64.

[25] B. Sahiner, M. Sezer, Determining constant breadth curve mate of a curve on a surface via Taylor collocation method, New Trends. Math. Sci. 6 (3) (2018), 103-115.

[26] M. R. M. Shabestari, R. Ezzati, T. Allahviranloo, Numerical solution of fuzzy fractional integro differential equation via two-dimensional Legendre wavelet method, J. Intell. Fuzzy. Syst. 34 (4) (2018), 2453-2465.

[27] A. Shojaei, U. Galvanetto, T. Rabczuk, A. Jenabi, M. Zaccariotto, A generalized finite difference method based on the Peridynamic differential operator for the solution of problems in bounded and unbounded domains, Comput. Methods. Appl. Mech. Eng. 343 (2019), 100-126.

[28] E. V. P. Spreafico, M. Rachidi, New approach of Bernoulli and Genocchi Numbers and their associated polynomials via generalized Fibonacci sequences of order, Proceeding Series. Braz. Soc. Comput. Appl. Math. 6 (2) (2018), 010436-(1-7).

[29] K. Yadav, J. P. Jaiswal, On the operational matrix for fractional integration and its application for solving Abel integral equation using Bernoulli wavelets, Glob. J. Pure. Appl. Math. 15 (1) (2019), 81-101.

[30] F. Yousefi, A. Rivaz, W. Chen, The construction of operational matrix of fractional integration for solving fractional differential and integro-differential equations, Neural. Comput. Appl. 31 (6) (2019), 1867-1878.


text:
# Metric entropy

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

of a dynamical system

One of the most important invariants in ergodic theory. Basic is the concept of the entropy $h ( S)$ of an endomorphism $S$( see Metric isomorphism) of a Lebesgue space $( X , \mu )$. For any finite measurable decomposition (measurable partition) $\xi$ the limit

$$h ( S , \xi ) = \lim\limits _ {n \rightarrow \infty } \ \frac{1}{n} H ( \xi _ {S} ^ {n} ) ,$$

$$\xi _ {S} ^ {n} = \xi \lor S ^ {-1} \xi \lor \dots \lor S ^ {- n+ 1} \xi$$

(the entropy of $\xi$ in unit time relative to $S$) exists, where $H ( \xi )$ is the entropy (cf. Entropy of a measurable decomposition) of $\xi$, and $\xi \lor \eta$ is the partition whose elements are the intersections of the elements of $\xi$ and $\eta$. (This definition carries over verbatim to $\xi$ with $H ( \xi ) < \infty$; by another method $h ( S , \xi )$ can be defined for any measurable $\xi$.) The entropy $h ( S)$ is defined as the least upper bound of the $h ( S , \xi )$ over all possible finite measurable $\xi$. (It may be $\infty$; the use of all $\xi$ with $H ( \xi ) < \infty$ or of all measurable $\xi$ yields the same entropy.)

Originally the entropy was defined by A.N. Kolmogorov somewhat differently (see ); the version given above came later (see [2]). In the basic case of an aperiodic automorphism of a Lebesgue space the definitions are ultimately equivalent [3].

It turns out that $h ( S ^ {n} ) = n h ( S)$, and if $S$ is an automorphism, then $h ( S ^ {-1} ) = h ( S)$. Therefore, the entropy of a cascade $\{ S ^ {n} \}$ is naturally taken to be $h ( S)$. For a measurable flow $\{ S _ {t} \}$ it turns out that $h ( S _ {t} ) = | t | h( S _ {1} )$. Therefore the entropy of a flow is naturally taken to be $h ( S _ {1} )$. The definition of the entropy for other transformation groups with an invariant measure is somewhat different. (It does not reduce to the entropy of a single transformation in the group; see [5], [6].) There are modifications of the entropy for the case of an infinite invariant measure [7]; another modification is the $A$- entropy (where $A = \{ k _ {n} \}$ is an ascending sequence of natural numbers), which is obtained when $\xi _ {S} ^ {n}$ is replaced by

$$S ^ {- k _ {1} } \xi \lor \dots \lor S ^ {- k _ {n} } \xi$$

and $\lim\limits$ by $\overline{\lim\limits}\;$( see [8]).

The entropy is a metric isomorphism invariant of dynamical systems and is fundamentally different from the earlier-known invariants, which are basically connected with the spectrum of a dynamical system. In particular, by means of the entropy of Bernoulli automorphisms (cf. Bernoulli automorphism; see ) it was first established that there exist non-isomorphic ergodic systems with the same continuous spectrum (which contrasts with the situation for a discrete spectrum). In a wider setting the role of the entropy is related to the fact that a new trend arose in ergodic theory: the entropy theory of dynamical systems (see [3], [4], and Ergodic theory).

The entropy provides a tool for characterizing the rate of mixing of sets of small measure (more accurately, the collection of those that form the partition). Side-by-side with this "global" role, the entropy also plays a "local" role, which is established by Breiman's ergodic theorem (an individual ergodic theorem of information theory): For ergodic $S$ and almost-all $x$,

$$\frac{1}{n} | \mathop{\rm log} \mu ( C _ {\xi _ {X} ^ {s} } ( x) ) | \rightarrow h ( S , \xi ) \ \textrm{ for } n \rightarrow \infty ,$$

where $C _ \eta ( x)$ is the element of the partition $\eta$ containing $x$ and the logarithm is taken to the same base as in the definition of $H$( see , [4]). (Breiman's theorem is true for $\xi$ with $H ( \xi ) < \infty$[10], but, generally speaking, not for countable $\xi$ with $H ( \xi ) = \infty$[11]; there are variants for non-ergodic $S$( see [4], [12]) and an infinite $\mu$[13]. A weaker assertion on the convergence in the sense of $l _ {1}$ has been proved for a certain general class of transformation groups [6].)

For smooth dynamical systems with a smooth invariant measure a connection has been established between the entropy and the Lyapunov characteristic exponent of the equations in variations (see [14][16]).

The name "entropy" is explained by the analogy between the entropy of dynamical systems and that in information theory and statistical physics, right up to the fact that in certain examples these entropies are the same (see, for example, [4], [17]). The analogy with statistical physics was one of the stimuli for introducing in ergodic theory (even in a not-purely metric context and for topological dynamical systems, cf. Topological dynamical system) new concepts such as "Gibbsian measures" , the "topological pressure" (an analogue to the free energy) and the "variational principle" for the latter (see the references to $Y$- system; Topological entropy).

#### References

 [1a] A.N. Kolmogorov, "A new metric invariant of transitive dynamical systems, and Lebesgue space automorphisms" Dokl. Akad. Nauk SSSR , 119 : 5 (1958) pp. 861–864 (In Russian) [1b] A.N. Kolmogorov, "On entropy per unit time as a metric invariant of automorphisms" Dokl. Akad. Nauk SSSR , 124 : 4 (1959) pp. 754–755 (In Russian) [2] Ya.G. Sinai, "On the notion of entropy of dynamical systems" Dokl. Akad. Nauk SSSR , 124 : 4 (1959) pp. 768–771 (In Russian) [3] V.A. Rokhlin, "Lectures on the entropy theory of transformations with invariant measure" Russian Math. Surveys , 22 : 5 (1967) pp. 1–52 Uspekhi Mat. Nauk , 22 : 5 (1967) pp. 3–56 [4] P. Billingsley, "Ergodic theory and information" , Wiley (1965) [5] A.V. Safonov, "Information parts in groups" Math. USSR. Izv. , 22 (1984) pp. 393–398 Izv. Akad. Nauk SSSR Ser. Mat. , 47 : 2 (1983) pp. 421–426 [6] J.C. Kieffer, "A generalized Shannon–McMillan theorem for the action of an amenable group on a probability space" Ann. of Probab. , 3 : 6 (1975) pp. 1031–1037 [7] V. Krengel, "Entropy of conservative transformations" Z. Wahrscheinlichkeitstheor. Verw. Geb. , 7 : 3 (1967) pp. 161–181 [8] A.G. Kushnirenko, "Metric invariants of entropy type" Russian Math. Surveys , 22 : 5 (1967) pp. 53–61 Uspekhi Mat. Nauk , 22 : 5 (1967) pp. 37–65 [9a] L. Breiman, "The individual ergodic theorem of information theory" Ann. Math. Stat. , 28 : 3 (1957) pp. 809–811 [9b] L. Breiman, "Correction to "The individual ergodic theorem of information theory" " Ann. Math. Stat. , 31 : 3 (1960) pp. 809–810 [10] K.L. Chung, "A note on the ergodic theorem of information theory" Ann. Math. Stat. , 32 : 3 (1961) pp. 612–614 [11] B.S. Pitskeĺ, "Nonuniform distribution of entropy for processes with a countable set of states" Probl. Peredatsi Inform. , 12 : 2 (1976) pp. 98–103 (In Russian) [12] A. Ionesco-Tulcea, "Contributions to information theory for abstract alphabets" Arkiv for Mat. , 4 : 2–3 (1961) pp. 235–247 [13] E.M. Klimko, L. Sucheston, "On convergence of information in spaces with infinite invariant measure" Z. Wahrscheinlichkeitstheor. Verw. Geb. , 10 : 3 (1968) pp. 226–235 [14] V.M. Millionshchikov, "A formula for the entropy of smooth dynamical systems" Differential Eq. , 12 (1976) pp. 1527–1530 Differents. Uravnen. , 12 : 12 (1976) pp. 2188–2192 [15] Ya.B. Pesin, "Characteristic Lyapunov exponents, and smooth ergodic theory" Russian Math. Surveys , 32 : 4 (1977) pp. 55–114 Uspekhi Mat. Nauk , 32 : 4 (1977) pp. 55–112 [16] R. Mañé, "A proof of Pesin's formula" Ergod. Th. and Dynam. Syst. , 1 : 1 (1981) pp. 95–102 [17] D.W. Robinson, D. Ruelle, "Mean entropy of states in classical statistical mechanics" Comm. Math. Phys. , 5 : 4 (1967) pp. 288–300

Instead of $A$- entropy the term sequence entropy is used in the English literature. See e.g. [a1], § 4.11. For several useful recent references concerning the computation of entropy, see [a2].


text:
How to cat fastq files but keep different samples separate
3
0
Entering edit mode
2.5 years ago

I'm trying to find a fast way to cat multiple fastq files in a directory, but keep the output merged files separate for each samples. For example, if I have 2 samples split across multiple lanes in a directory:

Sample1_C_S4_L001_R1_001.fastq.gz Sample1_C_S4_L001_R2_001.fastq.gz Sample1_C_S4_L002_R1_001.fastq.gz Sample1_C_S4_L002_R2_001.fastq.gz Sample1_L_S4_L001_R1_001.fastq.gz Sample1_L_S4_L001_R2_001.fastq.gz Sample1_L_S4_L002_R1_001.fastq.gz Sample1_L_S4_L002_R2_001.fastq.gz

I've tried something like: for f in Sample1_*.fastq.gz; do cat $f >$f-merged.fastq.gz; done but I can't get it to work, the output I want is two files named Sample_1_C-merged.fastq.gz and Sample_1_L-merged.fastq.gz. I don't need to keep the R1 and R2 reads separate

next-gen sequencing dna sequence • 1.3k views
0
Entering edit mode

Can you rerun demultiplexer and not split by lane.?

2
Entering edit mode
2.5 years ago
Prakash ★ 2.2k

$f used in output will create output for each line. you need to try : for f in Sample1_C*.fastq.gz; do cat$f > Sample1_C_merged.fastq.gz; done
for f in Sample1_L*.fastq.gz; do cat f > Sample1_L_merged.fastq.gz; done  ADD COMMENT 0 Entering edit mode Can I do that in a single command? Or does it have to be done separately? ADD REPLY 0 Entering edit mode Can be in one loop, see my answer below and the concerns I have with both your strategy. ADD REPLY 1 Entering edit mode 2.5 years ago ATpoint 65k Careful, the solutions from both Prakash and swbarnes2 merge the paired-end data into a single file which is probably not what should be happening even though OP does not explicitely says so. I am reasonaby sure you want forward and reverse read being separated for proper alignment. @OP, this is paired end data so combining both into one file would double-count each sequenced fragment since every of the two paired-end reads comes from the same piece of DNA. Are you aware of that? Better do: for f in Sample*.fastq.gz do Basename={f%_L00*}
## merge R1
ls ${Basename}_L00*_R1_001.fastq.gz | xargs cat >${Basename}_merged_R1.fastq.gz
## merge R2
ls ${Basename}_L00*_R2_001.fastq.gz | xargs cat >${Basename}_merged_R2.fastq.gz
done

0
Entering edit mode

I thought that was what I wanted. But I confirmed with the sequencing centre that there is little to no overlap between the forward and reverse reads as the bulk of each library has fragments with inserts >300bp. Also I will be using these reads for metagenomic analysis (metaphlan2) which doesn't require merged reads.

Or am I going about this the completely wrong way, should I be keeping the forward and reverse reads separate during the trimming and filtering step?

0
Entering edit mode

I am not into metagenomics so I do not kow for sure but for most other analysis like RNA-seq, ChIP, ATAC etc you always want to keep them separated. In your specific case it might also work that way but as said I do not know how the metagenomics field works. Others might comment on that. What could be a problem is that when you do any aalysis with tools such as Picard (e.g. to get rate of duplicates) it might be messed up since you have two reads with the same name not being flagged as paired. I'd keep them separated as this is how paired-end sequencing is typically interpreted by most tools.

0
Entering edit mode

If you don't need to merge your reads to run metaphlan (I can't remember what form of data it expects to ingest), then don't.

If you need all of the input reads in a single file, then cat them together with the suggestions above. It's not clear to me what the difference between your L and C samples are, so its hard to say whether it makes sense to 'pool' them or not, since apparently everything is Sample1.

0
Entering edit mode

Or am I going about this the completely wrong way, should I be keeping the forward and reverse reads separate during the trimming and filtering step?

If your reads don't overlap (sounds like they mostly don't) then you should keep them separate. Some programs may expect interleaved data (Read1_R1, Read1_R2, Read2_R1, Read2_R2 etc). In that case you can interleave the data using reformat.sh from BBMap suite.

0
Entering edit mode

I agree with you ATpoint, paired-end data R1 and R2 should be kept separate. Thanks for highlighting this point. Few things are still no clear to me such as

no overlap between forward and reverse reads

also as Joe highlighted

difference between L and C samples

I hope @m.radz would be able to choose better.

0
Entering edit mode
2.5 years ago

You don't want a loop.

cat Sample1_C*fastq.gz > Sample1_C.fastq.gz

0
Entering edit mode

That will cat all the "C" files but not the "L". I know how to do them separately, I was hoping I could write a command that would cat the "C" files then move on to the "L" files automatically.


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# How to get started with CubeSats

I am new both to CubeSats and to Stack Exchange.

I have recently started studying and planning on how to build a CubeSat, as cheaply as possible. I have read some blog posts and watched some videos, but I don't know exactly where to look for more technical knowledge, or what software/things should I learn.

I am studying Electronics and Telecommunication Engineering (2nd year). I don't want to perform a copy-paste project where you read a few things and build a replica. I want to have a more thorough knowledge.

But where can I start to get a better understanding of what the main issues are and the major tradeoffs that I'll have to look at in order to keep the cost down but to still have a functional CubeSat that can be launched?

– uhoh
Nov 10, 2020 at 3:42
• @uhoh Thanks a lot. The question seems much better now. And thanks for the welcome and guidelines, I will keep them in mind. Nov 10, 2020 at 4:32
• Assuming you're wanting to build a cubesat, the biggest question is "Why?" Realistically, building a cubesat isn't too difficult and anyone who has a degree in something like electrical engineering or adjacent fields should be able to pull it off. There are even places you can buy an almost-finished, just-add-your-payload cubesat systems. In fact, the entire cubesat architecture is intended to make bringing an experiment to space as easy as possible. I'd start with a function, and then design the rest of the satellite around it. Nov 10, 2020 at 21:35

Personal Suggestion - It's always best to learn by getting your hands dirty and since you are a student, do make an effort to start a Student Satellite Project in your college if it does not already have one.

With that being said, having been a part of one such endeavor in my college, I can disseminate whatever exposure I have gotten so far. I was a member of the ADCS (Attitude Determination and Control System) subsystem where I worked on the feasibility study of RCS thrusters, reaction wheels, magnetorquers, etc. to control the orientation of 1U CubeSat in space.

First, you need to be thorough with all the major subsystems -

• Power and propulsion
• Structure and thermal considerations
• Avionics - electronics and software
• Telemetry, tracking, and command
• OBDH (On-Board Data Handling),
• Payload - basically the purpose for which you want a CubeSat, which is a whole different ballgame and a critical one

The next thing would be to pick up the related skills. It is hard for a person to specialize in all of them, hence the need for a multi-disciplinary team. But as a leader (or a single person army, if you prefer), you need to have a working knowledge of each of them. On a broader sense that includes -

• Mechanical Engineering skills - CAD modeling, structural analysis and simulation, heat transfer, material science, dynamics and control
• Electrical/Electronics Engineering skills - power, propulsion, avionics (including electronic design automation), communication, signal processing, tracking station
• Computer Science skills - on-board software development, simulations, mission control software, data analysis

Of course, the above list is not exhaustive. Here's a PDF for a brief introduction to the individual subsystems - Satellite Subsystems and here are some of the books from my own collection which you may find useful (read them when you have worked sufficiently on the aforementioned skills) - Resources for CubeSat.

There are a lot of Student Satellite Projects around the world, so I would suggest you look for them and read their reports. I highly appreciate this IIT Bombay Student Satellite Project wiki where they have documented their knowledge and experience from launching their first CubeSat 'Pratham'.

Hope this helps you to get started. If ever you run across some problem, Stack Exchange is always there to help. It's a wonderful community! Welcome :)

• You see I never understood why do we need telemetry. And somewhere I read about 'bus'. I have no idea what does that mean either. Nov 10, 2020 at 13:54
• @Gurkirat If you launch a CubeSat and then don't know what's happening with it or getting the data for which it is intended for, what then would be the use? It would be like launching a stone in space and then forgetting you ever launched anything. Nov 10, 2020 at 13:57
• Thanks for your answer. It is really helpful. I don't know if you know this, but the drive links which you have provided are all private. Hence I cannot access them. Nov 10, 2020 at 13:58
• @Gurkirat Ah sorry! I have changed the share setting for books folder and corrected the permission for subsystem PDF. It should work now. Let me know if you face any problem or need further help. Nov 11, 2020 at 10:25
• Thank you so much Nov 11, 2020 at 16:16

(This was going to be a comment rather than an "Answer", but it got long...)

I'll assume you have or can get the engineering skills, and give a few CubeSat-specific resources.

If you haven't seen it yet, NASA's CubeSat 101 has a lot of information about design, licensing, integration, and such. They don't give exhaustive detail there, but it maps out the territory so you know what to look for, and they have more specific resources, too.

You'll need a ground station to talk to your satellite. There are commercial ground station services like Leaf and Amazon Ground Station, or you can build your own and link it to the SatNOGS network. These options give you global coverage, as opposed to the more limited coverage if you build your own standalone station.

Off-the-shelf components are available, like solar cells, reaction wheels, structural parts, etc., so you can concentrate on mission-specific components. I can't evaluate the suppliers, but I have a soft spot for the name Pumpkin.

To get your cubesat up, you might get a free ride from NASA if your cubesat helps them to fulfill their own mission, as they describe in their CubeSats 101 book. United Launch Alliance has a competition to schools for free launches, that's from 2016. I don't know if there's anything more recent, but you can always contact them. Otherwise I think it's around \$250,000 to ride up in a dispenser on a Centaur. SpaceX and Rocket Lab are also in the business, Wikipedia suggests as low as \$100,000, but I haven't found specific links on short notice.

It's actually amazing how much support there is for you. It's still expensive, but space has never been cheaper or more accessible. I've never built or launched a satellite myself, I trust you'll soon know more about it than I do.

• Thanks, @Greg all of this helps me very much. But I was wondering, wouldn't commercial products be more expensive than in-house production? Because I would be on a very tight budget with this project, so I want it to be as cheap as possible. And I had one more question, will SatNOGS also provide a global network like the others you have mentioned? Thank you once again for helping me. Nov 11, 2020 at 0:58
• @Gurkirat Whether to use commercial products depends on your situation. You might easily make your own structure but want to buy solar panels, for instance. And I would trust commercial reaction wheels to be more reliable than something I could build myself. And it's not enough that it just work on the bench, it has to survive environmental testing. But it totally depends on what you can and are willing to do. And SatNOGS basically offers free worldwide access for making your own contribution to the network. You may have to schedule your time in advance.
– Greg
Nov 11, 2020 at 19:07
• @Gurkirat Speaking of scheduling, You would normally have your launch scheduled way in advance, maybe more than a year in advance, and build to that launcher's requirements. Nothing about space is not meticulously planned out ahead of time.
– Greg
Nov 11, 2020 at 19:15
• Sir, yes sir (@Greg). Thank you for your support. You all are nicer than... You know who. Nov 11, 2020 at 19:26


text:
## anonymous one year ago Is acceleration a dimensionless number?

1. anonymous

@amistre64

2. amistre64

what is the definition of acceleration?

3. amistre64

or, if you know how the acceleration due to gravity is defined ... is it dimensionless?

4. anonymous

if the acceleration is equal to the gravity sin theta, then is it dimensionless?$a=g \sin \Theta$

5. anonymous

and we know the angle is like 40 degrees...

6. amistre64

sin(theta) is dimensionless, gravity is not

7. anonymous

well, thank you

8. amistre64

a/g = sin(theta) ... the dimensions cancel out to 1 and we are left with sin(theta) as a dimensionless value whose only purpose is to offer some sense of scale to the problem.

9. amistre64

at least thats my view of it

10. anonymous

Another approach. Acceleration is the rate of change of velocity, i.e.$a=\frac{ \Delta v }{ \Delta t }$If you know the dimensions (units) of velocity and time, you can work out the dimension of acceleration.


text:
# The Legendary Question Six IMO 1988

Published:

The final problem of the International Mathematics Olympiad (IMO) 1988 is considered to be the most difficult problem on the contest.

Arthur Engel wrote the following about the problem’s difficulty:

Nobody of the six members of the Australian problem committee could solve it. Two of the members were husband and wife George and Esther Szekeres, both famous problem solvers and problem creators. Since it was a number theoretic problem it was sent to the four most renowned Australian number theorists. They were asked to work on it for six hours. None of them could solve it in this time. The problem committee submitted it to the jury of the XXIX IMO marked with a double asterisk, which meant a superhard problem, possibly too hard to pose. After a long discussion, the jury finally had the courage to choose it as the last problem of the competition. Eleven students gave perfect solutions.

## Problem

#### Let a and b be positive integers such that (ab + 1) divides (a^2 + b^2). Show that (a^2 + b^2) / (ab + 1) is the square of an integer.

Alright, let’s try to solve the above problem.

## Intro

First, let’s denote that `k = (a^2 + b^2) / (ab + 1) ( * )`.

We know that `a` and `b` are positive integers (`a > 0` and `b > 0`). Consequently, if `k` is an integer, then it must be positive (`k > 0`).

From the problem statement, we know that `k` is not always an integer (e.g. `a` = 3 and `b` = 1). If so, we can ignore this case.

## Proof

We’re going to prove the statement with proof by contradiction approach.

This means that we need to make an assumption that the statement we’re going to prove is false. Recall that the original statement is k is the square of an integer.

Therefore, our assumption would be k is NOT the square of an integer.

In other words, we’re trying to see what would happen if we work under such an assumption.

### a & b

Since we assume that the solutions to `( * )` where `k` is not the square of an integer are exists, we may take the smallest value for each of `a` and `b` from the list of solutions to `( * )`.

Let’s assume that `a1` and `b1` are the smallest solution to `( * )` where `k` is not the square of an integer. This also means that `a1 + b1` should yield the minimum value compared with other pairs of `a` and `b`.

### Rearrange the Equation

Let’s rearrange `( * )` into a more usual form.

``````a^2 - kba + b^2 - k = 0
``````

Well, the above is a quadratic function.

We know that a polynomial equation can be expressed in terms of its root products such as the following.

``````f(x) = a.x^n + b.x^(n-1) + c.x^(n-2) + ... + constant = a(x-p)(x-q)(x-r)...

where p, q, r, etc. are the roots (f(x) = 0).
``````

Same as the above, if we consider `a` as the variable and the others fixed, then our quadratic can be expressed as `(a - p)(a - q)`.

According to the rule of sum and product of polynomial roots, we know the followings.

``````Adding the roots yields -b/a.

Multiplying the roots yields:

* z/a (for even degree polynomials)
* -z/a (for odd degree polynomials)

Where:
* "a" is the coefficient of x^n.
* "b" is the coefficient of x^(n-1)
* "z" is the constant at the end
``````

So, we can apply the above rules into our quadratic (even degree polynomial).

Recall that our quadratics is factored into `(a - p)(a - q)` which means that the roots are `p` and `q`.

Previously we also assume that `( * )` has `a1` and `b1` as the smallest solution. Since the quadratic is formed by considering `a` as the variable (the others are fixed), we may say that `a1` is one of the roots of the quadratic. Consequently, the other root `a2` can be calculated by leveraging the above rules.

``````f(a) = a^2 - (k.b1)a + (b1)^2 - k

The roots of f(a) are "a1" and "a2".

From the above rules, we know that:

a1 . a2 = ((b1)^2 - k)
a1 + a2 = (k.b1)

Therefore,

a2 = ((b1)^2 - k) / a1    ( ** )

OR

a2 = (k.b1) - a1    ( *** )
``````

Now, let’s check several characteristics of `a2`.

From `( ** )` we know that `a2` will never be zero because `k` is not the square of an integer. In other words, `k` will never be the same as `b^2`.

From `( *** )` we know that `a2` is an integer because `k, b1,` and `a1` are all integers.

Up til now we know that `a2` is an integer and not zero. However, is it a positive or negative integer?

To find out the sign of integer, let’s back to our original formula `( * )`.

``````Since a2 is also the solution to (*), we'll get the following.

k = ((a2)^2 + (b1)^2) / ((a2.b1) + 1)
``````

In the Intro section, we know that `k` is a positive integer (`k > 0`).

We also know that the numerator (`( (a2)^2 + (b1)^2 )`) will always be positive.

In addition, `b1` in the denominator is a positive integer. To be consistent with the fact that `k > 0`, we know for sure that `a2` must always be positive so that the denominator yields positive value as well.

In conclusion, we finally know that `a2` is a positive integer.

### Meet the Contradiction

Alright, the next question would be what’s the relationship between `a1` and `a2`?

Let’s assume that `a1 >= b1`.

If we square both sides, the inequality will still hold. So we got `(a1)^2 >= (b1)^2`.

Now, observe that `(b1)^2 > (b1)^2 - k` since `k > 0`. Consequently, `(a1)^2 > (b1)^2 - k`.

If we divide both sides by `a1`, the inequality will still hold. So we got `a1 > [(b1)^2 - k] / a1`.

Recall that from `( ** )`, `((b1)^2 - k) / a1` equals `a2`.

Therefore, we know that `a1 > a2 ( **** )`.

It’s weird here.

Initially we assumed that `a1` and `b1` are the smallest solution to `( * )`. This consequently makes `a1 + b1` as the minimum value.

From `( **** )` we know that `a2 < a1`. This obviously should make `a2 + b1` as the minimum value and NOT `a1 + b1`. In other words, this states that `a2` and `b1` are the smallest solution to `( * )`. This information clearly contradicts our assumption that `a1` and `b1` are the smallest solution to `( * )`.

Well, in this step, we might say that our assumption that `a1 >= b1` doesn’t hold true.

How about if we assume that `a1 < b1` instead?

Doing the same trick, if we square both sides, the inequality will still hold. So we got `(a1)^2 < (b1)^2`.

Also, observe that `(a1)^2 < (b1)^2 - k`. If we divide both sides by `a1` we’ll get `a1 < [(b1)^2 - k] / a1`.

However, the inequality in `a1 < [(b1)^2 - k] / a1` will stay the same for all `k` that makes `[(b1)^2 - k] / a1` always greater than `a1`. In other words, this inequality will stay the same for some values of `k` only.

Eventually, we’d end up the same as before in which `a1 > a2`, therefore, contradicts our assumption that `a1` and `b1` are the smallest solution to `( * )`.

This information states that our assumption that `a1 < b1` doesn’t hold true.

## What does it all mean then?

Well, we might say that this what would happen if we assume that `a1` and `b1` are the smallest solution to `( * )` where `k` is not a perfect square.

However, the assumption of the smallest solution can’t be falsified because since we assume that there exists solutions to `( * )` where `k` is not a perfect square, we can always pick the smallest value of `a` and `b` from the pool of solutions to `( * )` where `k` is not the square of an integer. And in this case, we decided to choose `a1` and `b1` as the smallest positive integers.

If on the other hand, we assumed that `a1` and `b1` are not the smallest solutions to `( * )`, we might not end up with such a contradiction. However, this doesn’t merely show that `( * )` where `k` is not a perfect square holds true since a contradiction comes up when we work with the smallest solutions.

Therefore, we conclude that there might be no solutions to `( * )` where `k` is not a perfect square. In other words, our first assumption that there exists solutions to `( * )` where `k` is not a perfect square is false.

In similar terms, we say that any value of `k` where it is not the square of an integer should not be exist since it doesn’t have any solution.

Finally, we conclude that `k` must be the square of an integer.

Q.E.D

Tags:


text:
Fri 21 Jan 2022 11:10 - 11:35 at Salon III - Semantics 1 Chair(s): Alan Jeffrey

We present a construction which, under suitable assumptions, takes a model of Moggi’s computational $\lambda$-calculus with sum types, effect operations and primitives, and yields a model that is adequate and fully abstract. The construction, which uses the theory of fibrations, categorical glueing, $\top\top$-lifting, and $\top\top$-closure, takes inspiration from O’Hearn & Riecke’s fully abstract model for PCF. Our construction can be applied in the category of sets and functions, as well as the category of diffeological spaces and smooth maps and the category of quasi-Borel spaces, which have been studied as semantics for differentiable and probabilistic programming.

#### Fri 21 JanDisplayed time zone: Eastern Time (US & Canada) change

 10:20 - 12:00 Semantics 1POPL at Salon III Chair(s): Alan Jeffrey Roblox 10:2025mResearch paper From Enhanced Coinduction towards Enhanced InductionRemotePOPLDavide Sangiorgi University of Bologna; Inria DOI Media Attached 10:4525mResearch paper A Fine-Grained Computational Interpretation of Girard’s Intuitionistic Proof-NetsInPersonPOPLDelia Kesner Université de Paris; CNRS; IRIF; Institut Universitaire de France DOI Media Attached 11:1025mResearch paper Fully Abstract Models for Effectful λ-Calculi via Category-Theoretic Logical RelationsRemotePOPLOhad Kammar University of Edinburgh, Shin-ya Katsumata National Institute of Informatics, Philip Saville University of Oxford DOI Media Attached 11:3525mResearch paper Layered and Object-Based Game SemanticsInPersonPOPLArthur Oliveira Vale Yale University, Paul-André Melliès CNRS; Université de Paris, Zhong Shao Yale University, Jérémie Koenig Yale University, Leo Stefanesco IRIF, University Paris Diderot & CNRS DOI Media Attached


text:
# Managing music using mpd/mpc

Music Player Daemon (MPD) is a lightweight application. It hardly takes any cpu. Moreover it runs in background and does not clutter and workspace. If you dont use mouse frequently or wants to avoid it (programmers use keyboard), then mpd is for you. It is one of those applications which does one job at a time and does it really well (well, if you get is working once which is not very hard). The job of MPD is to read/decode audio files available in its music directory on demand (one can specify it in config file) and create sound. It them make sound available to user using ALSA (or some other) soundcard driver. How to setup MPD is explained nicely on arch-linux wiki (gentoo wiki should also have something) otherwise there is some help always available on http://unix.stackexchange.com or http://superuser.stackexchange.com. Here is my config file stored as ~/.mpd/mpdconf. All of my music is kept in/download to ~/Bhandar/Music .

A minimal example for mpdconf file which works on my system.

On ubuntu, if you install mpd, a file is automatically created in /etc/mpd.conf. You can edit this file also. You have to create some paths before you can continue, if you create a config file like mine.

 $mkdir -p ~/.mpd/playlists$ touch ~/.mpd/playlists

Now type mpd in terminal, if it starts without any messages then we are good to go. Sometimes, you might get this warning. Failed to bind to ‘127.0.0.1:6600‘: Address already in use. This means mpd is already running. Check the running process

 $ps -e | grep mpd 14073 ? 00:00:28 mpd  Terminate it, $ sudo pkill mpd
$sudo pkill mpd  or $ sudo kill -9 14073
$sudo kill -9 14073 Check again. $ ps -e | grep mpd
$ps -e | grep mpd  This time we should have an empty line. Run mpd again. If there is a problem, then time to ask people or debug by yourself. ## Music player client (MPC) Now we need a client to send commands to mpd. MPC is a command line tool (perl scripts) while gmpc is a gui-based client. Mind you, mpd does not have those advance features of rythmbox and bashee. If you like to use amplifier and other fancy feature to enhance the sound, there is no point reading further. The very first thing we need to do, it to update database. $ mpc -h gabbar@localhost update
Updating DB (#1) ...

Password ‘gabbar’ was given in conf file. Without it, we will not have access to update the database. Check what password you have given and see the results. This much we need to do only once.

## Creating playlists and playing them

mpc help command will show you how to search for songs and add them to playlist. I woluld suggest to learn it the hard way. I have written few scripts to make my life slightly easier (short term memory losses). Script [1] does search for a given word in an of songs and add them all matched songs.

 $./mpc_search_add.sh "bhimsen joshi"$./mpc_search_add.sh "bhimsen joshi"

It will add songs having ‘bhimsen joshi’ in their metadata, if they are not already added. Another common problem is to play the fav song quickly. There is script [3] for this also. For instance if I want to listen to "Yellow, Cold play" then I do the following.

 $./mpc_play "yellow" [INFO] Searching for yellow [INFO] No song found for your query.  Ok, yellow is not found that means it is not added to playlist. Use script 2 to add "Coldplay" to playlist. Menwhile, lets play a song which I am pretty sure of having in my playlist: ‘Seene main jalan’. It will play the first match.  #./mpc_play "seene" [INFO] A matching song is found in playlist at 1911 Seene main jalan - Gaman.mp3 [playing] #1911/1911 0:00/4:30 (0%) volume:100% repeat: on random: on single: off consume: off  Depending on size of playlist, this might take 1-3 seconds. Next, previous, ## Play/pause and volume MPC comes with standard command to do them mpc next, mpc previous, mpc toggle etc etc. Each windown manager allows key mapping. Map these commands to some key combination. I use Alt+Ctlr+<some_key> for user defined commands. ## Deleting currently playing song or adding it to favs A lot of songs gets downloaded and once in a while suddenly some random guy starting singing about somebodys butt and breasts. Who wants such sort of non-sense in ones playlist :-p. Sometimes a great songs appears and you wish to add it to your a folder. Script [2] does both of these functions. This acts on currently playing song. If you path to music_dir is different than mine, then you have to change the paths accordingly. To delete currently playing song $./manage_mpc.sh -d

To add the song to a folder of gems ~/Bhandar/Music/MyCollection

 \$./manage_mpc.sh -a

You can bind these two commands to some keys. The important thing is that these scripts must be executable (chmod +x scipt_name) and available in system path (copy them to /usr/local/bin ). Script [4] provides color-support in terminal which is needed for other scripts to work. It should be in the same folder as other scripts. One can use it for other purposes. Everyone loves colors.

Scroll to Top


text:
## Calculus 8th Edition

Published by Cengage

# Chapter 11 - Infinite Sequences and Series - 11.1 Sequences - 11.1 Exercises - Page 744: 43

#### Answer

converges to $0$

#### Work Step by Step

Given: $a_n=\frac{cos^{2}n}{2^{n}}$ Note that we can write $\frac{0}{2^{n}} \leq \frac{cos^{2}n}{2^{n}}\leq \frac{1}{2^{n}}$ Also, $0 \leq \frac{cos^{2}n}{2^{n}}\leq \frac{1}{2^{n}}$ Now, $\lim\limits_{n \to \infty}\frac{1}{2^{n}}=\frac{1}{\infty}=0$ Therefore, by Sandwich Theorem $\lim\limits_{n \to \infty}\frac{cos^{2}n}{2^{n}}=0$ Hence, the given sequence converges to $0$.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
# 03. Predict with pre-trained YOLO models¶

This article shows how to play with pre-trained YOLO models with only a few lines of code.

First let’s import some necessary libraries:

from gluoncv import model_zoo, data, utils
from matplotlib import pyplot as plt

Let’s get an YOLOv3 model trained with on Pascal VOC dataset with Darknet53 as the base model. By specifying pretrained=True, it will automatically download the model from the model zoo if necessary. For more pretrained models, please refer to Model Zoo.

net = model_zoo.get_model('yolo3_darknet53_voc', pretrained=True)

## Pre-process an image¶

Next we download an image, and pre-process with preset data transforms. Here we specify that we resize the short edge of the image to 512 px. You can feed an arbitrarily sized image. Once constraint for YOLO is that input height and width can be divided by 32.

You can provide a list of image file names, such as [im_fname1, im_fname2, ...] to gluoncv.data.transforms.presets.yolo.load_test() if you want to load multiple image together.

This function returns two results. The first is a NDArray with shape (batch_size, RGB_channels, height, width). It can be fed into the model directly. The second one contains the images in numpy format to easy to be plotted. Since we only loaded a single image, the first dimension of x is 1.

im_fname = utils.download('https://raw.githubusercontent.com/zhreshold/' +
'mxnet-ssd/master/data/demo/dog.jpg',
path='dog.jpg')
print('Shape of pre-processed image:', x.shape)

Out:

Downloading dog.jpg from https://raw.githubusercontent.com/zhreshold/mxnet-ssd/master/data/demo/dog.jpg...
Shape of pre-processed image: (1, 3, 512, 683)

## Inference and display¶

The forward function will return all detected bounding boxes, and the corresponding predicted class IDs and confidence scores. Their shapes are (batch_size, num_bboxes, 1), (batch_size, num_bboxes, 1), and (batch_size, num_bboxes, 4), respectively.

We can use gluoncv.utils.viz.plot_bbox() to visualize the results. We slice the results for the first image and feed them into plot_bbox:

class_IDs, scores, bounding_boxs = net(x)

ax = utils.viz.plot_bbox(img, bounding_boxs[0], scores[0],
class_IDs[0], class_names=net.classes)
plt.show()

Total running time of the script: ( 0 minutes 1.467 seconds)

Gallery generated by Sphinx-Gallery


text:
# Which computational method/basis set should I use for small molecular ions?

I apologize for a perhaps newbie question.

I would like to compute the potential energy surface for several small molecular ions, like $\ce{H2+}$ and $\ce{H3+}$, using quantum chemistry software. What are the proper method and basis set to do that and why? I am thinking about ROHF/STO-3G or UHF.

$\ce{H2+}$ or $\ce{H3+}$ have unpaired electrons because they are ions. Due to this, you should be using an unrestricted shell theoretical model which doesn't restrict electrons to paired (closed shell) orbitals. You should start with unrestricted Hartree-Fock (UHF) and can increase the level of theory to include electron correlation with something like 2nd order Møller Plesset (MP2) or 4th order (MP4).

The electron correlation models acknowledge the fact that electron locations are not totally random, repulsion between electrons is a real effect and can give more realistic results.

With regard to basis set, STO-3G would be considered a starting point. If time and resources allow, you could build up to something like a 6-31G(d,p) which adds p-orbitals to the hydrogen atoms allowing more optimization of molecular orbitals. Increasing the basis set quality approximates the radial components of the atomic orbitals with more Gaussian curves, thereby increasing the possible accuracy of the calculation.

Exploring Chemistry with Electronic Structure Methods - Foresman and Frisch

• For the people who are not familiar with the terminology, could you add references? – rcollyer Aug 21 '13 at 20:16
• Thanks for the suggestion, I've added some additional detail and a reference. From the OP's terminology it appears that Gaussian might be the software used so I stuck with their nomenclature. – scs217 Aug 21 '13 at 20:32
• $\ce{H3+}$ does not have unpaired electrons. – Martin - マーチン Dec 29 '15 at 15:51

if u accurately want to make ur computations a bigger basis set and high level are necessery (CI(SDQ) or CAS), however so bigger basis set so less human readable are results. If the only aim is to see nice orbitals, minimal basis and DFT or even semiemprical methods should be enough. As stated before unrestricted methods are necessary unless u want to play with half-occupations.


text:
## The Annals of Statistics

### Confidence bands in generalized linear models

#### Abstract

Generalized linear models (GLM) include many useful models. This paper studies simultaneous confidence regions for the mean response function in these models. The coverage probabilities of these regions are related to tail probabilities of maxima of Gaussian random fields, asymptotically, and hence, the so-called tube formula is applicable without any modification. However, in the generalized linear models, the errors are often nonadditive and non-Gaussian and may be discrete. This poses a challenge to the accuracy of the approximation by the tube formula in the moderate sample situation. Here two alternative approaches are considered. These approaches are based on an Edgeworth expansion for the distribution of a maximum likelihood estimator and a version of Skorohod’s representation theorem, which are used to convert an error term (which is of order $n^{-1 /2}$ in one-sided confidence regions and of \$n^{-1} in two-sided confidence regions) from the Edgeworth expansion to a “bias” term. The bias is then estimated and corrected in two ways to adjust the approximation formula. Examples and simulations show that our methods are viable and complementary to existing methods. An application to insect data is provided. Code for implementing our procedures is available via the software parfit

#### Article information

Source
Ann. Statist., Volume 28, Number 2 (2000), 429-460.

Dates
First available in Project Euclid: 15 March 2002

Permanent link to this document
https://projecteuclid.org/euclid.aos/1016218225

Digital Object Identifier
doi:10.1214/aos/1016218225

Mathematical Reviews number (MathSciNet)
MR1790004

Zentralblatt MATH identifier
1106.62343

#### Citation

Sun, Jiayang; Loader, Catherine; McCormick, William P. Confidence bands in generalized linear models. Ann. Statist. 28 (2000), no. 2, 429--460. doi:10.1214/aos/1016218225. https://projecteuclid.org/euclid.aos/1016218225

#### References

• Bliss, C. L. (1935). The calculation ofthe dosage-mortality curve. Ann. Appl. Biology 22 134-167.
• Brown, L. D. (1986). Fundamentals of Statistical Exponential Families: with Applications in Statistical Decision Theory. IMS, Hayward, CA.
• Faraway, J. and Sun, J (1995). Simultaneous confidence bands for linear regression with heteroscedastic errors. J. Amer. Statist. Assoc. 90 1094-1098.
• Hall, P. and Titterington, D. M. (1988). On confidence bands in nonparametric density estimation and regression. J. Multivariate Anal. 27 228-254.
• Hall, P. (1992). The Bootstrap and Edgeworth Expansion. Springer, New York.
• H¨ardle, W. and Marron, J. S. (1991). Bootstrap simultaneous error bars for nonparametric regression. Ann. Statist. 19 778-796.
• Johansen, S. and Johnstone, I. (1990). Hotelling's theorem on the volume oftubes: some illustrations in simultaneous inference and data analysis. Ann. Statist. 18 652-684.
• Knafl, G., Sacks, J. and Ylvisaker, D. (1985). Confidence bands for regression functions. J. Amer. Statist. Assoc. 80 683-691.
• Knowles, M. and Siegmund, D. (1989). On Hotelling's geometric approach to testing for a nonlinear parameter in regression. Internat. Statist. Rev. 57 205-220.
• Kreyszig, E. (1968). Introduction to Differential Geometry and Riemannian Geometry. Univ. Toronto Press.
• Loader, C. (1993). Nonparametric regression, confidence bands and bias correction. In Computing Science and Statistics. Proceedings of the 25th Symposium on the Interface 131-136. Interface Foundation of North America, Fairfax Station, VA.
• Loader, C. and Sun, J. (1997). Robustness oftube formula. J. Comput. Graph. Statist. 6 242-250.
• Naiman, D. Q. (1987). Simultaneous confidence bounds in multiple regression using predictor variable constraints. J. Amer. Statist. Assoc. 82 214-219.
• Naiman, D. Q. (1990). On volumes oftubular neighborhoods ofspherical polyhedra and statistical inference. Ann. Statist. 18 685-716.
• Naiman, D. Q. and Wynn, H. P. (1997). Abstract tubes, improved inclusion-exclusion identities and inequalities and importance sampling. Ann. Statist. 25 1954-1983.
• Scheff´e, H. (1959). The Analysis of Variance. Wiley, New York. Skovgaard, I. M. (1981a). Transformation ofan Edgeworth expansion by a sequence ofsmooth functions. Scand. J. Statist. 8 207-217. Skovgaard, I. M. (1981b). Edgeworth expansions ofthe distribution ofmaximum likelihood estimators in the general (non i.i.d.) case. Scand. J. Statist. 8 227-236.
• Sun, J. (1993). Tail probabilities ofthe maxima ofGaussian random fields. Ann. Probab. 21 34-71.
• Sun, J. and Loader, C. (1994). Simultaneous confidence bands for linear regression and smoothing. Ann. Statist. 22 1328-1345.
• Sun, J., Loader, C. and McCormick, W. (1998). Confidence bands for generalized confidence bands. Available at http://sun.cwru.edu/ jiayang/glm.ps.
• Sun, J., Raz., J. and Faraway, J. (1999). Simultaneous confidence bands for growth and response curves. Statist. Sinica 9 679-698.
• Weyl, H. (1939). On the volume oftubes. Amer. J. Math. 61 461-472.


text:
Czechoslovak Mathematical Journal, Vol. 62, No. 4, pp. 1033-1053, 2012

# On the dimension of the solution set to the homogeneous linear functional differential equation of the first order

## Alexander Domoshnitsky, Robert Hakl, Bedřich Půža

Alexander Domoshnitsky, Department of Mathematics and Computer Science, The Ariel University Center of Samaria, 44837 Ariel, Israel, e-mail: adom@ariel.ac.il; Robert Hakl, Institute of Mathematics, Academy of Sciences of the Czech Republic, Branch in Brno, Žižkova 22, 616 62 Brno, Czech Republic, e-mail: hakl@ipm.cz; Bedřich Půža, Institute of Mathematics, Academy of Sciences of the Czech Republic, Branch in Brno, Žižkova 22, 616 62 Brno, Czech Republic, e-mail: puza@math.muni.cz

Abstract: Consider the homogeneous equation
$$u'(t)=\ell(u)(t)\qquadfor a.e. t\in[a,b]$$
where $\ell C([a,b];\Bbb R)\to L([a,b];\Bbb R)$ is a linear bounded operator. The efficient conditions guaranteeing that the solution set to the equation considered is one-dimensional, generated by a positive monotone function, are established. The results obtained are applied to get new efficient conditions sufficient for the solvability of a class of boundary value problems for first order linear functional differential equations.

Keywords: functional differential equation, boundary value problem, differential inequality, solution set

Classification (MSC 2010): 34K06, 34K10

Full text available as PDF.


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# Homework Help: Acceleration of a Rolling Cylinder

1. Oct 22, 2011

### Megzzy

1. The problem statement, all variables and given/known data
Prove that a=⅔g sin θ will find the acceleration of a cylinder rolling down an incline of angle θ.

2. Relevant equations
a=⅔g sin θ

3. The attempt at a solution
I don't understand how to do this without numbers. I have a feeling the principle of conservation of mechanical energy should be used somehow?

2. Oct 22, 2011

### I like Serena

Hi Megzzy!

You can basically do this 2 ways.

You can set up an equation for the conservation of mechanical energy and rewrite it to find "a" (using the derivative).

Or you can set up a diagram with the forces.
Set up the equations of movement and find "a".

Either way, you will need a the moment of inertia of a cylinder.
And you will need to set up a relation between the angle the cylinder has rolled and the distance it has rolled.

How much do you already know?

3. Oct 22, 2011

### Megzzy

thanks for the help!

I have no values so that is why I am having problems I think. I know the moment of inertia of a cylinder is I=1/2MR^2. I'm still lost though.

4. Oct 22, 2011

### I like Serena

That's okay.
And yes, that is the right moment of inertia.

Don't worry yet whether you have numbers or not.

Which (relevant) formulas do you know?

Like the one for kinetic energy of a linear movement?
And the one for kinetic energy of a rotating movement?
Energy of gravity?

Or the one for the force of gravity?
The equation for the resulting force on the cylinder?
The equation for the resulting moment on the cylinder?

5. Oct 22, 2011

### Megzzy

I know KE rotational= 1/2Iw^2
KE linear=1/2mv^2
PE=mgh

I'm not sure where to go from there.

6. Oct 22, 2011

### I like Serena

This will get you on the way.

Now suppose the cylinder rolls down a certain distance d.
What will be the change in angle of the cylinder?
And by what amount will its height decrease?

Btw, what I'm aiming for are the relations between v, w, h, and a.
But to get there, we're introducing d and the angle of the cylinder.

7. Oct 22, 2011

### Megzzy

So I am picturing the right triangle?
d is therefore the hypotenuse and the height is opposite the angle of inclination.

I'm sorry this process is taking so long! The question is just really strange to me.

8. Oct 22, 2011

### I like Serena

Exactly. :)

9. Oct 22, 2011

### Megzzy

alright thank you!
I am still unsure of where to go from there and how to connect that with v,w,h and a.

10. Oct 22, 2011

### I like Serena

Well, you pictured a right triangle with d on the hypotenusa and h as the opposite of the angle of inclination theta.

So which formula describes the relation between d, h, and theta?

11. Oct 22, 2011

### Megzzy

all I can think of is sin theta=height/distance

12. Oct 22, 2011

### I like Serena

That's the one!

Can you rewrite that in the form:
height = ?
And substitute it in the total energy formula?

Btw, I'm off to bed now. :zzz:

I'll give you another hint to work on.
Can you find a formula that relates the angular velocity w of the cylinder to the linear velocity v of the cylinder?

13. Oct 23, 2011

### Megzzy

I'd get height=sin theta x distance.
I don't understand how it plugs into the total energy formula though.

14. Oct 23, 2011

### I like Serena

Let's see...

Your total energy (which must be conserved) is:
E = KE rotational + KE linear + PE
$E= {1 \over 2}I \omega^2 + {1 \over 2}m v^2 + mgh$

If the ball rolls a distance d along the slope, the height h decreases by (sin theta x distance).
So the new energy is:
$E= {1 \over 2}I \omega^2 + {1 \over 2}m v^2 + mg(h - \sin(\theta) \times d)$.
Note that w and v will have greater values in this equation to conserve energy.

Now when the cylinder has rolled down a distance d, how many radians did it turn?

15. Oct 24, 2011

### Megzzy

Or am I supposed to be making an actual estimate here?

16. Oct 25, 2011

### I like Serena

Yes, that is the relation between radians and degrees.
For problems like this, it is easiest to work in radians.

Suppose the cylinder rolled over an angle of 1 radian, which distance would it have rolled?

17. Oct 27, 2011

### I like Serena

Let me try to make it clearer.

The circumference of the cylinder is $2\pi R$.
This means that an angle of $2\pi$ radians corresponds to a distance of $2\pi R$.

So if the cylinder rolls over an angle of 1 radian, it covers a distance of $d = {2\pi R \over 2\pi} = R$.

Are you still with me?

18. Oct 28, 2011

### Megzzy

Sorry for taking so long to get back to you! It's been a busy week.
I seem to understand what you have explained so far.

19. Oct 28, 2011

### I like Serena

Okay, so if the cylinder rolls 2 radians the distance is 2R.
More generally the $distance = \phi R$
if $\phi$ is the angle over which the cylinder has rolled.

The speed $v$ of the cylinder is the change in distance per unit of time.
This is the derivative. Did you know that?

So $v = {d \over dt} distance$.
Similarly the angular velocity is the change in angle per unit of time.
That is, $\omega = {d\phi \over dt}$.

Could you apply this to $d = \phi R$ to find the relation between $v$ and $\omega$?


text:
Fall 2011 Strategic Practice 7: Section 1 (Joint, Conditional, and Marginal Distributions) - Question 4
A group of $n\geq 2$ people decide to play an exciting game of Rock-Paper-Scissors. As you may recall, Rock smashes Scissors, Scissors cuts Paper, and Paper covers Rock (despite Bart Simpson saying "Good old rock, nothing beats that!"). Usually this game is played with 2 players, but it can be extended to more players as follows. If exactly 2 of the 3 choices appear when everyone reveals their choice, say a, b $\epsilon$ {Rock, Paper, Scissors} where a beats b, the game is decisive: the players who chose a win, and the players who chose b lose. Otherwise, the game is indecisive and the players play again. For example, with 5 players, if one player picks Rock, two pick Scissors, and two pick Paper, the round is indecisive and they play again. But if 3 pick Rock and 2 pick Scissors, then the Rock players win and the Scissors players lose the game. Assume that the n players independently and randomly choose between Rock, Scissors, and Paper, with equal probabilities. Let X, Y, Z be the number of players who pick Rock, Scissors, Paper, respectively in one game.
(a) Find the joint PMF of X, Y, Z.
(b) Find the probability that the game is decisive. Simplify your answer (it should not involve a sum of many terms).
(c) What is the probability that the game is decisive for n = 5? What is the limiting probability that a game is decisive as n ! 1? Explain briefly why your answer makes sense.
Solution: (a) This is a Multinomial(n, (1/3, 1/3, 1/3)) distribution. (b) The game is decisive if and only if exactly one of X, Y, Z is 0. These cases are disjoint so by symmetry, the probability is 3 times the probability that X is zero and Y and Z are nonzero. Note that if X = 0 and Y = k, then Z = n−k. This gives P (decisive) = ([2^n] - 2) / (3^[n-1]). As a check, when n = 2 this reduces to 2/3, which makes sense since for 2 players, the game is decisive if and only if the two players do not pick the same choice. (c) The probability is (2^5-2)/3^4 = 30/81 = 0.37 approximately for n = 5. The limiting probability is 0 as n goes to infinity, which make sense since if the number of players is very large, it is very likely that there will be at least one of each of Rock, Paper, and Scissors.
"Mathematics is the logic of certainty, but statistics is the logic of uncertainty."


text:
# Dynamic Programming to Maximize Profit

I have this problem to resolve in dynamic programming:

An ice cream shop owner has 4 stores. To meet the demand in the summer he purchased 7 refrigerators for the ice cream. Because the stores are in different places, there are different points for selling the ice cream. It was estimated the profit (in monetary unit) that could be obtained in each store:

$$\begin{array}{c|lcr} N refrigerators & \text{store1} & \text{store2} & \text{store3} & \text{store4} \\ \hline 1 & 4 & 2 & 6 & 1 \\ 2 & 6 & 6 & 12 & 4 \\ 3 & 10 & 8 & 14 & 8 \\ 4 & 12 & 10 & 18 & 16 \\ 5 & 14 & 12 & 20 & 20 \\ 6 & 16 & 16 & 22 & 21 \\ 7 & 18 & 20 & 22 & 24 \end{array}$$

The problem asks to find the working scheme that maximizes the financial return.

As dynamic programming aims to reuse the code I know that it is necessary to use a recursive function, but when analyzing the problem I assumed that my answer field is in a matrix where the lines are referring to the number of refrigerators and the columns the stores.

Then I assumed that $M(i, j)$ is my function that returns a monetary value, assuming that $i$ is the number of refrigerators, ranging from $1 \le i \le 7$, and $j$ is a particular store, ranging from $1 \le j \le 4$.

I also assumed that my function would have 4 calls of this same function $M(i, j)$, because there are 4 stores to divide 7 refrigerators:

The equation would be:

$$MAX \{ M(i, 1) + M (i, 2) + M (i, 3) + M (i, 4)\}$$

Where $\sum_{i=1}^4 i \le 7$

As a base case of recursion I put this:

$$M(0, 1) + M (0, 2) + M (0, 3) + M (0, 4) = 0$$

But my problem is not knowing how to treat this problem recursively. Maybe I'm doing the wrong analysis, too. Every help is welcome.

The dynamic programming approach is to compute recursively the maximal profit that can be obtained from using $x$ refrigerators in the first $y$ stores (and not using any in the other stores). I'll let you fill in the missing details.


text:
# Laurent series

1. Apr 10, 2009

### krindik

Hi,
can u pls help me on this?

1. The problem statement, all variables and given/known data
Find Laurent series that converges for

$$\, 0 < |z - z_0| < R }$$ and determine precise region of convergance

$$\, \frac {1}{z^2 + 1} \,\,$$

2. Relevant equations

3. The attempt at a solution
I tried to spilt this into fractions
i.e
$$f(x) \, = \, \frac{A}{z-i} + \, \frac{B}{z+i}$$

as I would have done for

$$\frac {1}{z^2 - 1} \,\,$$

But in that case I would expand it with a geometrical series.
The problem rises with $$i$$ instead of $$1$$

2. The problem statement, all variables and given/known data
Can u pls explain how can I choose the method of expansion (Laurent, Taylor) given a function f(x) ?

Thanks
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 10, 2009

### latentcorpse

can't you just say $w=z^2$ then you can make

$\frac{1}{w-1}=\frac{1}{w} \frac{1}{1-\frac{1}{w}}$ and expand as a geometric series...

3. Apr 10, 2009

### krindik

Thanks.

In fact the question is
$$\, \frac {1}{z^2 + 1} \,\,$$

not

$$\, \frac {1}{z^2 - 1} \,\,$$

I'll restate the question

Find Laurent series that converges for

$$\, 0 < |z - z_0| < R }$$

and determine precise region of convergance for
$$\, \frac {1}{z^2 + 1} \,\,$$

where
$$z_0 = i$$

Additionally, I would like to know how u use $$\, 0 < |z - z_0| < R }$$
in coming up with a solution

Last edited: Apr 10, 2009
4. Apr 10, 2009

### Dick

The Laurent series will be an expansion in powers of (z-i). Use your partial fractions form. The A/(z-i) is the (-1) power. Write the B/(z+i) part as B/(2i+(z-i)) and do the geometric series expansion with (z-i) as the variable.

5. Apr 10, 2009

### latentcorpse

why can we not just change variable here?

6. Apr 10, 2009

### Dick

You can. Isn't that basically what I did when I wrote z+i=(z-i)+2i?

7. Apr 10, 2009

### krindik

Thank you,

I have few questions
1. Is it that you choose $(z - i)$ terms because the question specifies it should converge in $0 < |z-i| < R$ ?

2. So $\frac{B}{z + i}$ will be expanded like

$\frac{1}{2i} * \frac{1}{1+\frac{z-i}{2i}} = \sum_{n=0}^{\infty}{s^n}$

where
$s = \frac{z-i}{2i}$

so the answer to the solution would be
$\frac{A}{z-i} + \{expansion\, in\, 2\}$ ?

Thanks again

8. Apr 11, 2009

### Dick

I chose (z-i) because you specified that you want to expand around z0=i in post 3. I couldn't tell where you wanted to put the expansion from the first post. And yes, you've got the expansion right.

9. Apr 11, 2009

### krindik

Thank you very much Dick

I shall summarize here, so that anybody else following this thread can get the point.

$\frac{1}{1+z^2} = \frac{A}{z-i} + \frac{B}{z+i}$

expand
$\frac{B}{z+i} = \frac{B}{2i} * \frac{1}{1-(-1 * \frac{z-i}{2i})}$

then by expanding in geometric series
$\frac{1}{1-(-1 * \frac{z-i}{2i})} = \sum_{n=0}^\infty{[(\frac{-1}{2i})^n * (z-i)^n]}$


text:
# Low Dimensional Topology

## July 19, 2010

### Combinatorics of Jacobi diagrams

Filed under: Combinatorics,Knot theory,Quantum topology — dmoskovich @ 2:25 pm

A Jacobi diagram is surely the most tantalizing object in all of quantum topology. Its definition is combinatorial and easy to visualize, to understand, and to manipulate; and the smallest insight into the structure of the graded space of Jacobi diagrams would lead to huge rewards in quantum topology and in the study of Lie algebras. And yet, despite the concerted efforts of some of the finest minds in mathematics (for instance and in no particular order: Greg Kuperberg, Dror Bar-Natan, Alex Stoimenow, Oliver Dasbach, Pierre Vogel, Jerry Levine, David Yetter, Tomotada Ohtsuki, Justin Sawon, David Broadhurst, Maxim Kontsevich, and many others), they remain forever teasing, tempting, but forever maddeningly beyond reach.

A primitive Jacobi diagram is a graph (actually a multigraph) whose vertices have valence 1 or 3 and whose trivalent vertices are oriented, i.e. a cyclic ordering of 3 edges around each trivalent vertex is fixed (clockwise or counterclockwise), and whose legs are labeled by elements of some finite set $S$. The degree of a Jacobi diagram is defined to be half the total number of vertices of the diagram. A Jacobi diagram is a formal sum of primitive Jacobi diagrams over a ring $R$. Here’s an example of a Jacobi diagram over the matrix ring $M_2(\mathbb{Z}/2\mathbb{Z})$, whose legs are labeled by elements of the set $S =\{1,2,3,4,5\}$

Embarassingly, almost nothing is known about Jacobi diagrams when $R$ is not the field of rational numbers $\mathbb{Q}$. The space of Jacobi diagrams is the vector space over $\mathbb{Q}$ spanned by Jacobi diagrams subject to the AS and IHX relations, which are local moves between Jacobi diagrams which differ inside a dotted circle as indicated below. The space of Jacobi diagrams is graded by degree. A Jacobi diagram is said to be $n$-loop if its Euler number is $1-n$, i.e. if its first Betti number is $n$. (A Jacobi diagram of the type we have just defined is sometimes called an open Jacobi diagram, and the space of these Jacobi diagrams is sometimes denoted $\mathcal{B}$ in the literature.)

Jacobi diagrams are important, I suppose, because finite-type invariants can be constructed by evaluating a weight system (a map $W: \mathcal{B}\longrightarrow \mathbb{Q}$ satisfying some relations) on a Jacobi diagram. So information about Jacobi diagrams is information about finite type invariants. For more stuff along these lines on the blogosphere, see this Everything Seminar post.

### Tree-like Jacobi diagrams

Information about Jacobi diagrams is also information about Lie algebras. To see this most clearly, think of Serre’s procedure to write non-associative words as rooted trees. So write

Then, wonder of wonders, the AS and IHX relations are seen to be anti-symmetry and the Jacobi relation of the Lie bracket. So you see that the space of Jacobi diagrams is some sort of a generalization of a free Lie algebra- specifically, a free Lie algebra $\mathcal{L}(n)$ over $n$ letters is a space of Jacobi diagrams over the set $S=\{0,1,\ldots,n\}$ whose underlying graphs are trees, with one special leg chose to be the root whose label is $0$, and all of whose other legs are labeled by elements of $\{1,\ldots,n\}$.

What about the whole space of tree-like Jacobi diagrams with $m$ legs $B^{\text{tree}}_m(S)$? There’s actually a tremendously useful identification, which I think is actually due to Shigeyuki Morita. It tells us that, over $\mathbb{Q}$, we have the following short exact sequence:

$B^{\text{tree}}_m(S)\overset{\alpha}{\longrightarrow} \mathcal{L}_{m-1}(S)\oplus S \overset{\beta}{\longrightarrow} \mathcal{L}_m(S)\longrightarrow 0.$

To go from $B^{\text{tree}}_m(S)$ to $\mathcal{L}_{m-1}(S)\oplus S$, for each primitive tree-like Jacobi diagram choose a leg $l$ to be the root (so now you have an element of $\mathcal{L}_{m-1}(S)$ by the construction of Serre above), and tensor with its label $s$. The map $\alpha$ is the sum over all choices of legs to be the root. You have to sum over all choices, because there is no way of choosing a canonical leg for a general tree-like Jacobi diagram. The map $\mathcal{L}_{m-1}(S)\oplus S \overset{\beta}{\longrightarrow} \mathcal{L}_m(S)$ is the left-bracketing map. The map $\alpha$ is an isomorpism onto its image (its injective)- to see this, construct its inverse by “re-labeling” $l$ by $s$ and forget it was the root. So, starting from a primitive Jacobi diagram, you get a sum of $m$ primitive elements in $\mathcal{L}_{m-1}(S)\oplus S$, which maps back to $m$ copies of the Jacobi diagram which you started with. We are working over $\mathbb{Q}$, so we can divide by $m$, so mapping a Jacobi diagram to $m$ copies of itself is an isomorphism.

Things are more interesting over $\mathbb{Z}$, where you can’t just divide by $m$. This is the subject of the final paper of Jerry Levine, published 4 months after he died. You still have the exact sequence

$B^{\text{tree}}_m(S)\overset{\alpha}{\longrightarrow} \mathcal{L}^\prime_{m-1}(S)\otimes S \overset{\beta}{\longrightarrow} \mathcal{L}^\prime_m(S)\longrightarrow 0,$

but it’s no longer clear that $\alpha$ is an injective (what’s is $\mathcal{L}^\prime_{m}(S)$? That’s the key subtlety- AS is anti-symmetry $[a,b]=-[b,a]$ of the Lie bracket, but $[a,a]=0$, which holds true in a Lie algebra, is a strictly stronger relation in a ring where 2 fails to be invertible. Thus, over $\mathbb{Z}$, we want to map to a quasi-Lie algebra in which, indeed, the Lie bracket is anti-symmetric, but it may be that $[a,a]\neq 0$). It looks like it should be dead easy… I worked on related stuff pretty hard though when I was starting out as a grad student (before Levine’s paper), and I can testify that it’s not, at least not for me. This is a tantalizing conjecture, and it’s maddening not to know whether or not it’s true.

### Planarity

Here’s another maddening conjecture, due to Sergei Duzhin. Is every Jacobi diagram equivalent under the IHX relation to a Jacobi diagram which is a sum of primitive Jacobi diagrams whose underlying graphs are planar? This looks like it should be moronically easy (by Kuratowski’s Theorem all you would have to prove is that any copy of $K_{3,3}$ as a subgraph can be killed), but it’s wide open. I actually suspect it’s false, but I don’t know.

Why is this an conjecture important? Because it leads to estimates for the growth rate of $\mathcal{B}$ as the degree increases.

### Vanishing of Jacobi diagrams with an odd number of legs

I suppose that perhaps one of the more interesting open problems about the space of Jacobi diagrams is that, for $S=\{pt\}$ (all legs share the same label), every Jacobi diagram of odd degree vanishes (over $\mathbb{Q}$, or over $\mathbb{Z}[\frac{1}{2}]$ if you want more fun). This is an important open problem, because it would imply that no finite-type invariant can distinguish between a knot and its inverse, where the inverse of a knot is that knot with the opposite orientation. Consequences of that are discussed in this paper by Greg Kuperberg.

This is easy for 0-loop Jacobi diagrams (tree-like Jacobi diagrams), because they all vanish for $S=\{pt\}$. The AS relation applied to any trivalent vertex gives back the same tree-like Jacobi diagram you started with, with a minus sign, so the Jacobi diagram equals minus itself, and because 2 is invertible this means that the Jacobi diagram vanishes.
Good. On to 1-loop Jacobi diagrams. The AS relation applies to all vertices gives back the same Jacobi diagram, with sign $-1^m$ where $m$ is the degree. The same trick works for 2-loop Jacobi diagrams. The fun starts for 3-loop Jacobi diagrams, where applying an AS relation to all vertices might just give you something else (you could tell the difference between a Jacobi diagram and its reflection).

Not to toot my own horn, but Tomotada Ohtsuki and I managed to prove the 3-loop case in this paper. I suppose that with quite a bit of effort we could do the 4-loop case, but the 5-loop case should already by rather interesting… prove it and I’ll throw you a party!


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## Stream: Is there code for X?

### Topic: Lattice Homomorphisms?

#### Reed Mullanix (Sep 13 2020 at 15:44):

I'm working on some point-free topology, and was struggling to find a notion of a lattice homomorphism anywhere.

#### Aaron Anderson (Sep 13 2020 at 18:52):

The file order/rel_iso.lean has everything relevant that we have

#### Aaron Anderson (Sep 13 2020 at 18:55):

In short, we have order embeddings and order isomorphisms, and we know how those relate to lattices, and a general relation homomorphism that generalizes monotone maps, but we don’t have a dedicated type for maps that may not be embeddings/isos but do preserve all lattice operations.

#### Aaron Anderson (Sep 13 2020 at 19:02):

I don’t know your background, but if you’re looking for a place to start/first PR, studying that file and writing a definition of lattice homomorphism seems to me like a decent place to start

#### Reed Mullanix (Sep 13 2020 at 20:10):

I'll probably pester you all with some more silly questions, but I'll give that a go later today :)

#### Johan Commelin (Sep 14 2020 at 08:38):

Note that there is in fact preorder_hom

#### Johan Commelin (Sep 14 2020 at 08:38):

@Reed Mullanix @Aaron Anderson see order/preorder_hom.lean

#### Bryan Gin-ge Chen (Mar 13 2021 at 15:02):

@Reed Mullanix did you ever make any progress with lattice homomorphisms? I think we'll be wanting them soon, as @Noam Atar has been defining properties of order ideals.

#### Scott Morrison (Mar 13 2021 at 22:21):

Yes please, I'd like them too!

#### Scott Morrison (Mar 13 2021 at 22:21):

Do we have bundled sublattices? (Like subsemiring and friends.) They would be nice too.

#### Bryan Gin-ge Chen (Mar 13 2021 at 22:51):

@Scott Morrison No, we don't have those either yet. Imitating the algebraic hierarchy around subsemiring, ring_hom, etc. would be a good way to start, right?

#### Scott Morrison (Mar 13 2021 at 22:58):

Yes. On the other hand it's starting to look pretty creaky to have to do this by hand again.

#### Eric Wieser (Mar 14 2021 at 00:06):

I think mathlib is in dire need of a "subobject" generator that takes a list of "closure under" axioms and builds the 200 or so lines of boilerplate

#### Eric Wieser (Mar 14 2021 at 00:07):

I know one of @Kevin Buzzard's students was going to start looking at sets closed under +, -, and smul that contain 0 recently - which is basically submodule but extending subgroup instead of submonoid and without a ring assumption that makes the - irrelevant. The number of the subobjects that can exist is exponential in the number of operators, which feels really bad.

#### Aaron Anderson (Mar 14 2021 at 03:56):

A lot of the API can come from the fact that closure is a Galois Insertion representing a finitary closure operator, I think this might get us a lot of the way there

#### Bryan Gin-ge Chen (Mar 14 2021 at 05:52):

Aaron Anderson said:

A lot of the API can come from the fact that closure is a Galois Insertion representing a finitary closure operator, I think this might get us a lot of the way there

I'd love to see how this works, if you wouldn't mind sketching some of this out.

#### Kevin Buzzard (Mar 14 2021 at 08:37):

That's normally the quickest way to get to the sublattice I think. Here this would involve defining the G-invariant-subgroup-closure of either a subgroup or a G-invariant subset (note that the definition of G-invariant subgroup is just to switch on the old structure command and then extend sub_mul_action and subgroup, so the game seems to be that you have to make it back to one of the things you extended). You can either do it with an inductive prop or prove closed under arbitrary intersections first and use this. Then you prove enough about it to be able to use the theorem which pulls back complete lattice structures along Galois insertions. Then I still want "kernel and range of a G-module map are G-submodules" and a map and comap which are also a Galois connection

#### Kevin Buzzard (Mar 14 2021 at 08:49):

It's all a great exercise for a beginner by the way

#### Aaron Anderson (Mar 14 2021 at 16:25):

Or we could define lattices as substructures of lattices in a given first-order language, and just prove that boilerplate once more for substructures

#### Bhavik Mehta (Apr 30 2021 at 14:50):

Is there any progress on this? I'd like to try getting to things like Stone duality for profinite sets, so some notions of lattice homomorphisms would be helpful

Last updated: May 17 2021 at 16:26 UTC