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https://stonespounds.com/994-pounds-in-stones-and-pounds
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# 994 pounds in stones and pounds
## Result
994 pounds equals 71 stones and 0 pounds
You can also convert 994 pounds to stones.
## Converter
Nine hundred ninety-four pounds is equal to seventy-one stones and zero pounds (994lbs = 71st 0lb).
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# What is a cash flow ratio?
A cash flow ratio is a financial metric that provides insight into a business’s ability to pay off its current debts with cash generated in the same period.
Several cash flow ratios are used to uncover crucial information about business performance, and in this guide, we explore all of them in detail.
## Types of cash flow ratios
There are six cash flow ratios, namely:
### 1. Current liability coverage ratio
The current liability coverage ratio calculates how much cash you have to pay off debt and measures your liquidity. A ratio greater than one shows you are generating enough cash to meet your obligations.
How it’s calculated: Cash flow from operations divided by current liabilities.
### 2. Cash flow coverage ratio
The cash flow coverage ratio measures how much cash you generate annually to pay off your total outstanding debt. A ratio of greater than one indicates that you’re not at risk of default. Because this ratio shows sufficient cash flow to pay off debt plus interest, it should be as high as possible.
How it’s calculated: Net operating cash flow divided by total debt.
### 3. Price-to-cash-flow ratio
The price-to-cash-flow ratio measures how much cash you generate relative to your stock price and helps determine your company’s value. Unlike the cash flow ratios we have covered so far, the price-to-cash flow ratio should be low. This is because a higher ratio implies that your stock price is high, relative to how much cash you generate.
How it’s calculated: Share price divided by operating cash flow per share.
### 4. Cash interest coverage ratio
The cash interest coverage ratio measures your ability to pay off the interest on your outstanding debt. A higher ratio is more favourable, as it means you’ll have no difficulty meeting your interest payment obligations.
How it’s calculated: Earnings before interest and taxes divided by interest.
### 5. Operating cash flow ratio
The operating cash flow ratio compares your operating cash flow to your current liabilities. As it shows how much cash you have to cover your short-term obligations, a higher ratio is preferable.
How it’s calculated: Operating cash flow divided by liabilities.
### 6. Cash flow to net income
The cash flow to net income ratio compares your operating cash flow to your net income. Because it provides insight into how well you’re converting net income into cash flow, a higher ratio is a positive sign.
How it’s calculated: Operating cash flow divided by net income.
## Interpreting cash flow ratios
Cash flow ratios play a crucial role in financial analysis, as they provide insight into your company’s liquidity, solvency, and long-term sustainability. Except for the price-to-cash flow ratio, which should be low, higher cash flow ratios are preferable across the other ratio types we have listed here.
Comparing your cash flow ratios to those of competitors or industry benchmarks may help you identify issues in your relative financial performance. However, it’s always advisable to use cash flow ratios with other financial metrics to get a complete picture of your company’s financial standing.
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What does the second law of thermodynamics state
What is the second law of thermodynamics in simple terms?
The second law of thermodynamics states that entropy, which is often thought of as simple ‘disorder’, will always increase within a closed system. Ultimately, this is one of the key elements dictating an arrow of time in the Universe.
What does the second law of thermodynamics state quizlet?
The second law of thermodynamics states that the total entropy can only increase over time for an isolated system, meaning a system which neither energy nor matter can enter or leave. … The measure of how more widely a specific amount of molecular energy is dispersed in a process.
What is the second law of thermodynamics called?
The second law of thermodynamics says that the entropy of any isolated system always increases. Isolated systems spontaneously evolve towards thermal equilibrium—the state of maximum entropy of the system. More simply put: the entropy of the universe (the ultimate isolated system) only increases and never decreases.
Why is second law of thermodynamics important?
Second law of thermodynamics is very important because it talks about entropy and as we have discussed, ‘entropy dictates whether or not a process or a reaction is going to be spontaneous’.
Which best describes the Second Law of Thermodynamics?
energy is not created nor destroyed, but it can change into matter. energy is not created nor destroyed, but it can change from one energy form to another. some useful energy is lost as heat whenever an energy transfer occurs. …
What are the two laws of thermodynamics?
The first law, also known as Law of Conservation of Energy, states that energy cannot be created or destroyed in an isolated system. The second law of thermodynamics states that the entropy of any isolated system always increases.
You might be interested: How To Calculate Income Tax Payable On Balance Sheet? (TOP 5 Tips)
What is meant by entropy?
Entropy, the measure of a system’s thermal energy per unit temperature that is unavailable for doing useful work. Because work is obtained from ordered molecular motion, the amount of entropy is also a measure of the molecular disorder, or randomness, of a system.
What is the third law of thermodynamics quizlet?
The third law of thermodynamics states that the entropy of a system approaches a constant value as the temperature approaches zero. Absolute zero. The coldest temperature, 0 Kelvin, that can be reached. It is the hypothetical temperature at which all molecular motion stops. You just studied 7 terms!
Is the second law of thermodynamics always true?
Breaking The Law
The Second Law of Thermodynamics states that entropy within an isolated system always increases. This iron-clad law has remained true for a very long time. … It predicted that there are certain conditions where entropy might actually decrease in the short term.
How does the second law of thermodynamics apply to living organisms?
Since all energy transfers result in the loss of some usable energy, the second law of thermodynamics states that every energy transfer or transformation increases the entropy of the universe. … Essentially, living things are in a continuous uphill battle against this constant increase in universal entropy.
What is the 3rd law of thermodynamics in simple terms?
Explanation. In simple terms, the third law states that the entropy of a perfect crystal of a pure substance approaches zero as the temperature approaches zero. The alignment of a perfect crystal leaves no ambiguity as to the location and orientation of each part of the crystal.
You might be interested: What Closing Costs Are Tax Deductible When Refinancing? (Best solution)
Rudolf Clausius
Why is the second law of thermodynamics not violated by living organisms?
Explanation: The second law of thermodynamics postulates that the entropy of a closed system will always increase with time (and never be a negative value). … Human organisms are not a closed system and thus the energy input and output of an the organism is not relevant to the second law of thermodynamics directly.
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# TJEE 2015 Syllabus tbjee.nic.in : Tripura Board of Joint Entrance Examination
Name of the Organisation : Tripura Board of Joint Entrance Examination (TBJEE)
Type of Announcement : Syllabus
Entrance Exam : TJEE 2015 Tripura Joint Entrance Examination
Reference Number : Notification No.1/ 2015 No. F.1 (1) -DHE / TBJEE / 2015-2016 (Vol-1)
Want to comment on this post?
TRIPURA BOARD OF JOINT ENTRANCE EXAMINATION
SYLLABUS FOR PHYSICS
Full Marks – 100
Each Module Carries 10 marks
MODULE – 1
Unit of measurement, System of units, S.I. units, Fundamental and derived units, Dimensional Analysis.
Composition and resolution of vectors, Rectangular components in two & three dimensions, Unit vector, Representation of vectors in term of co-ordinates, Addition & Subtraction of vectors, Multiplication of vectors – scalar product & vector product.
Uniformaly accelerated motion, Velocity – time graph Position-time graph, Kinematical equations in one dimension.
Centre of mass, Centre of gravity, Conditions of equilibrium of a system of forces, Moment of a force about a point and an axis, couple, torque.
Newton’s laws of motion, Inertial fraame, Impulse & impulsive forces.
Conservation of linear momentum, Static and kinetic friction, Projectile motion.
Rotational motion of a particle, Angular momentum and conservation of angular momentum, Moment of inertia, Relation between
Torque, Angular momentum, Moment of inertia & Angular acceleration, Rotational kinetic energy, Centripetal force & centrifugal force.
Work, energy & power ; Conservative and non-conservative force ; Elastic collision in one dimension.
MODULE – 2
Laws of gravitation, Gravitational field & potential ; Acceleration due to gravity and its variation with altitude, depth & rotation of earth ; Escape velocity, Kepler’s laws of planetary motion (rigorous proof is not required), Geostationary satellite.
Elasticity, Elastic behaviour, Hooke’s Law of elasticity, Elastic module, Poisson’s ratio, Elastic Energy.
Pascal’s law, Principle of multiplication of thrust & its application, Hydraulic press. Archimedes’ principle and its application, Atmospheric pressure, Torricelli’s Expt., Fortin’s Barometer.
Surface energy and surface tension, Capillarity, Streamline and turbulent motion, Newton’s law of viscous force, Coefficient of viscosity, Stoke’s law, Terminal velocity, Bernoulli’s Principle (statement only) and its simple application.
MODULE – 3
Simple Harmonic Motion, Differential equation of S.H.M. & its solution, Energy in S.H.M.,Time period of simple pendulum, Superposition of two S.H.M.,’s (analytical treatment), Free, forced & damped vibrations (qualitative idea only), Resonance.
Elastic waves – longitudinal and transverse waves, Progressive waves.
Superposition of waves : beats ; stationary waves – vibration of strings & air columns.
Doppler effect in sound propagation (effect of medium excluded), Sound wave as longitudinal elastic wave, Velocity of sound wave, Newton’s formula and Laplace’s correction, Dependence of velocity of sound in a medium on temperature, pressure, density and humidity.
MODULE – 4
Thermal expansion of solids, Relation among coefficients of linear, superficial and cubical expansion of solids, Applications of expansion of solids, Thermal expansion of liquids, Relation between expansion coefficients.
Colorimetry, Change of state, Latent heat.
Conduction, convection and radiation, Thermal conductivity, Thermometric conductivity, Black body ratiation, Stefan’s law, Newton’s law of cooling (statement and qualitative explanation only.)
Mean free path, Mean, rms speed and most probable speed, Pressure of an ideal gas ; Charle’s law, Boyle’s law, Avogadro’s law & pressure law from Kinetic theory of gases, Kinetic energy of molecules, Kinetic interpretation of temperature.
First law of Thermodynamics, Thermodynamic variables, Isothermal and Adiabatic expansions of gases, reversible and irreversible processes, Specific heats of gases at constant pressure and at constant volume and relation between them.
MODULE – 5
Reflection of light,Spherical mirrors, Mirror formula,linear magnification (formula).
Refraction at plane surface, Total internal reflection,Critical angle, Relation between refractive index and critical angle, optical fibre, Total reflecting prism, Refraction and dispersion of light through prism.
Thin lenses – concave and convex, lens formula, power of a lens, lens maker’s formula (deduction is not required), two thin lenses in contact.
Simple and compound microscope, Astronomical telescope (simple construction as a combination of coaxial lenses and ray diagram showing final image formation) Magnifying power. Human eye – defects of vision and corrections.
Wave front and Huygen’s principle, Reflection and Refraction of a plane wave front at a plane surface on the basis of Huygen’s Principle, Interference of light. Young’s double slit experiment and expression for fringe width, coherent sources.
MODULE – 6
Coulomb’s law in electrostatics, Electric field intensity and potential and their relation, Electric dipole, Electric field due to a dipole, Dipole moment, Electric flux, Gauss’ theorem in electrostatics (statement only) and its application to find electric field intensity due to uniformly charged infinitely long thin straight wire and uniformly charged thin spherical shell (field inside & outside).
Capacitance, Principle of capacitor, Capacitance of parallel plate capacitor, series & parallel combination of capacitors, energy stored in capacitor, sharing of charges & loss of energy.
Coulomb’s law in magnetism, Magnetic field intensity due to a magnetic dipole (short bar magnet) at a point on its axis and on the perpendicular bisector of the axis ; Torque on a magnetic dipole in a uniform magnetic field, Current loop as a magnetic dipole and magnetic dipole moment.
Properties of magnetic material, permeability & susceptibility, dia, para & ferromagnetic materials & their uses. Magnetic field of earth, Elements of the earth’s magnetic field.
MODULE – 7
Drift velocity and mobility of charge carrier throughmetallic conductor and their relation with electric current, Ohm’s law, Resistance of conductor, Factors influencing resistance, Temperature coefficient of resistance, Resistivity, Combination of resistors, Internal resistance of a cell and circuit equation, Combinations of cells.
Kirchhoff’s law and simple applications, Wheatstone bridge principle, Meter bridge (with application for the determination of unknown resistance), Principle and applications of potentiometer.
Joule’s law on heating effect of current, Electrical method of determination of ‘J’, electric power, B.O.T. unit of electrical energy.
Thermocouple, Thermo-emf, E-T graph, Neutral temp.,Temp. of inversion, Seebeck effect, Peltier effect.
MODULE – 8
Magnetic effect of current, Biot – Savart law and its application to current carrying circular loop, Ampere’s circuital law and its application to infinitely long straight wire, Straight and toroidal solenoids, Force on a moving charge in uniform electric field and magnetic field, Lorentz force.
Force between two straight parallel current carrying conductors – definition of ampere, Fleming’s left hand rule, Torque experienced by a current carrying loop in a uniform magnetic field, Moving coil galvanometer,Conversion of a galvanometer into ammeter and voltmeter.
Faraday’s laws of electromagnetic induction, Lenz’s law, Self and mutual induction, Fleming’s right hand rule, Alternating current (basic concept), Peak and rms value of alternating current/voltage.
Qualitative idea of electromagnetic waves and its spectrum.
MODULE – 9
Bohr’s theory of hydrogen like atom, Hydrogen spectrum, Photo electric effect, Einstein’s photo electric equation, Explanation of laws of Photo electric emission, Photo electric cell, Wave – particle duality, deBroglie’s hypothesis.
Radioactivity ; alpha, beta & gamma rays and their properties, Radioactive decay law, Decay constant, half life & mean life, Radioisotope and their uses.
MODULE – 10
Distinction between metals (conductors), semiconductors & insulators in terms of energy bands in slids, Intrinsic and Extrinsic semiconductors, p-n Junction diode,semiconductor diode rectifiers, p-n-p and n-p-n Transistors, Common Emitter Transistor Characteristics.
Logic Gates – OR gate, AND gate & NOT gate.
Constituents of atomic nucleus, Mass defect, Binding energy, Mass energy equivalence, Nuclear fission, Chain reaction, Nuclear reactor – Principle of operation, Nuclear fusion, Thermo nuclear fusion as the source of energy in Sun and Stars.
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# Fredholm $C^*$-algebras
Let $$H$$ be a Hilbert space. A vector subspace $$W\subset B(H)$$ is called a Fredholm subspace if there is an upper bound for the absolute value of Fredholm index of all Fredholm operators $$T$$ in $$W$$.
Is there a classification of all $$C^*$$-algebras $$A$$ which admit an irreducible representation $$\phi:A \to B(H)$$ in some Hilbert space $$H$$ such that $$\phi(A)$$ is a Fredholm subspace of $$B(H)$$?
Is there a classification of all $$C^*$$-algebras $$A$$ which admit a faithful representation $$\phi:A \to B(H)$$ in some Hilbert space $$H$$ such that $$\phi(A)$$ is a Fredholm subspace of $$B(H)$$?
One can consider the terminology "Fredholm algebra" for any such $$C^*$$-algebras.
Edit: We add an example according to comment by Yemon Choi.
Put $$H=\ell^2$$ let $$S$$ be the shift operator on $$\ell^2$$ and $$n$$ be a fixed integer. Then this is a finite dimensional Fredholm subspace of $$B(\ell^2)$$:
$$\{P(S)\mid \text{P is a polynomial of degree at most n}\}.$$
• Do you have an example of a Fredholm subspace? – Yemon Choi Sep 6 '20 at 1:09
• @YemonChoi Trivial examples: Finite dimensions. Or in the case of infinite dimension, the scalar 1 dimensional space. or any subspace which does not contain any fredholm operator. As another example the space of $\{P(s)\mid s\text{ is the shift operator, whose index is -1 and P is an arbitrary polynomial of degree at most n\}$ – Ali Taghavi Sep 6 '20 at 1:12
• $\{P(s)\mid s\text{ is the shift operator, whose index is -1}\}$ – Ali Taghavi Sep 6 '20 at 1:16
• How do you exclude the trivial 1-dim rep for the first question? – YCor Dec 12 '20 at 14:34
• No, irreducible is clearly defined (=nonzero and no proper nonzero closed invariant subspace). "Trivial" is slightly more ambiguous since it can be the trivial irrep (1-dimensional), or any trivial representation (which can be in dimension 0, 1, or more). – YCor Dec 12 '20 at 18:36
There's a trivial answer to the second question: every C$${}^*$$-algebra has such a representation. Wlog assume $$A \subseteq B(H_0)$$ for some Hilbert space $$H_0$$, then represent $$A$$ on $$H_0 \otimes l^2$$ by tensoring everything with the identity on $$l^2$$. All the Fredholm operators in this representation have index $$0$$ (in fact they would have to be invertible).
• Sure, if $T$ has nonzero kernel then $T\otimes I$ has infinite dimensional kernel and hence is not Fredholm. Same for cokernel, so if $T\otimes I$ is Fredholm it must have no kernel or cokernel (and have closed range), which means it must be invertible. – Nik Weaver Sep 6 '20 at 2:42
• If $\phi$ is a representation such that $\phi(A)$ is a Fredholm space then every Fredholm operator there has index zero, because $ind(T^n) = n\ ind(T)$. – Ruy Sep 6 '20 at 4:31
• I don't understand this question. By definition, the range of any Fredholm operator is closed. If $T$ is not surjective then neither is $T\otimes I$. – Nik Weaver Sep 6 '20 at 14:10
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https://arxiv.org/abs/1609.04235v1
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# Title:Removal Lemmas for Matrices
Abstract: The graph removal lemma is an important structural result in graph theory. It has many different variants and is closely related to property testing and other areas. Our main aim is to develop removal lemmas of the same spirit for two dimensional matrices. These are statements of the following type: fix a finite family $F$ of matrices over some alphabet $\Gamma$. Suppose that for an $n \times n$ matrix $M$ over $\Gamma$, for any couple of integers $s,t > 0$ at most $o(n^{s+t})$ of the $s \times t$ submatrices of $M$ are equal to matrices from $F$. Then one can modify no more than $o(n^2)$ entries in $M$ to make it $F$-free (that is, after the modification no submatrix of $M$ is equal to a matrix from $F$).
As a representative example, one of our main results is the following: fix an $s \times t$ binary matrix $A$. For any $\epsilon > 0$ there exists $\delta > 0$ such that for any $n \times n$ binary matrix $M$ that contains at most $\delta n^{s+t}$ copies of $A$, there exists a set of $\epsilon n^2$ entries of $M$ that intersects every $A$-copy in $M$. Moreover, $\delta^{-1}$ is polynomial in $\epsilon^{-1}$.
The major difficulty is that the rows and the columns of a matrix are ordered. These are the first removal lemma type results for two dimensional graph-like objects with a predetermined order. Our results have direct consequences in matrix property testing: they imply that for several types of families $F$ and choices of the alphabet $\Gamma$, one can determine with good probability whether a given matrix $M$ is $F$-free or $\epsilon$-far from $F$-freeness (i.e., one needs to change at least an $\epsilon$-fraction of its entries to make it $F$-free) by sampling only a constant number of entries in $M$.
In particular, we generalize an efficient induced removal lemma for bipartite graphs of Alon, Fischer and Newman, making progress towards settling one of their open problems.
Comments: 16 pages, no figures, submitted to Innovations in Theoretical Computer Science (ITCS) 2017 Subjects: Combinatorics (math.CO); Computational Complexity (cs.CC); Discrete Mathematics (cs.DM) Cite as: arXiv:1609.04235 [math.CO] (or arXiv:1609.04235v1 [math.CO] for this version)
## Submission history
From: Omri Ben-Eliezer [view email]
[v1] Wed, 14 Sep 2016 12:22:56 UTC (22 KB)
[v2] Thu, 15 Sep 2016 15:09:14 UTC (22 KB)
[v3] Mon, 12 Jun 2017 21:54:50 UTC (25 KB)
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## Saturday, September 1, 2018
### Golden ratio in the Qur’an?
35:33 Gardens of Eternity, which they will enter. They will be adorned therein with bracelets of gold and pearls, and their garments therein (will be of) silk.
(According to the article “What is the golden ratio?” in “canva(dot)com”, “The Golden Ratio exists when a line is divided into two parts and the longer part (a) divided by the smaller part (b) is equal to the sum of (a) + (b) divided by (a), which both equal 1.618”)(The golden ratio is obtained in the Qur’an, and it also proves that the Qur’an has been preserved, because it is calculated according to the existing number of Chapters and Verses in the Qur’an: the sum of all duplicate Chapter+Verse sums is 7906 (Chapter 5 + 120 Verses = 125, 40+85=125, 79+46=125, 8+75=83, 59+24=83, 10+109=119, 36+83=119, 83+36=119, 89+30=119, 15+99=114, 39+75=114, 70+44=114, 88+26=114, 107+7=114, 17+111=128, 18+110=128, 21+112=133, 55+78=133, 22+78=100, 72+28=100, 24+64=88, 34+54=88, 25+77=102, 94+8=102, 97+5=102, 27+93=120, 68+52=120, 114+6=120, 28+88=116, 111+5=116, 112+4=116, 33+73=106, 87+19=106, 91+15=106, 98+8=106, 103+3=106, 35+45=80, 58+22=80, 41+54=95, 42+53=95, 50+45=95, 44+59=103, 86+17=103, 95+8=103, 45+37=82, 64+18=82, 48+29=77, 65+12=77, 51+60=111, 100+11=111, 108+3=111, 52+49=101, 82+19=101, 53+62=115, 75+40=115, 96+19=115, 109+6=115, 54+55=109, 84+25=109, 60+13=73, 62+11=73, 76+31=107, 85+22=107, 99+8=107, 78+40=118, 113+5=118, 81+29=110, 90+20=110, 102+8=110, 105+5=110, 106+4=110, 92+21=113, 104+9=113, 110+3=113)(So, 125 + 125 + 125 + 83 + 83 +119 + 119 + 119 + 119 + 114 + 114 + 114 + 114 + 114 + 128 + 128 + 133 + 133 + 100 + 100 + 88 + 88 +102 + 102 + 102 + 120 + 120 + 120 + 116 + 116 + 116 + 106 + 106 + 106 + 106 + 106 + 80 + 80 + 95 + 95 + 95 + 103 + 103 + 103 + 82 + 82 + 77 + 77 + 111 + 111 + 111 + 101 + 101 + 115 + 115 + 115 + 115 + 109 + 109 + 73 + 73 + 107 + 107 + 107 + 118 + 118 + 110 + 110 + 110 + 110 + 110 + 113 + 113 + 113 = 7906)(And the sum of all unique Chapter+Verse sums is 4885 (1+7=8, 2+286=288, 3+200=203, 4+176=180, 6+165=171, 7+206=213, 9+129=138, 11+123=134, 12+111=123, 13+43=56, 14+52=66, 16+128=144, 19+98=117, 20+135=155, 23+118=141, 26+227=253, 29+69=98, 30+60=90, 31+34=65, 32+30=62, 37+182=219, 38+88=126, 43+89=132, 46+35=81, 47+38=85, 49+18=67, 56+96=152, 57+29=86, 61+14=75, 63+11=74, 66+12=78, 67+30=97, 69+52=121, 71+28=99, 73+20=93, 74+56=130, 77+50=127, 80+42=122, 93+11=104, 101+11=112)(So, 8 + 288 + 203 + 180 + 171 + 213 + 138 + 134 + 123 + 56 + 66 + 144 + 117 + 155 + 141 + 253 + 98 + 90 + 65 + 62 + 219 + 126 + 132 + 81 + 85 + 67 + 152 + 86 + 75 + 74 + 78 + 97 + 121 + 99 + 93 + 130 + 127 + 122 + 104 + 112 = 4885)(The division of these two numbers is the golden ratio = 7906 / 4885 = 1.618; and the sum of these two numbers, divided by the larger number is also the golden ratio = (7906 + 4885) / 7906 = 1.618)(Allah knows best)
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Vous êtes sur la page 1sur 4
# 8/17/2016 STLVector|HackerEarth
## Signup and contribute to Notes. Start Now
3
LIVE EVENTS
Notes
STL - Vector
2 Vector C++ STL
Vector is one of various useful containers defined under Standard Template Library(STL).
We can say that Vector is an advanced version of array in C++ . Though it contains some
good features than array.
#include<vector>
## Vector Declaration , Syntax : `
vector<data_type>identifier(size,initial_value);//Heresize&
initial_valueareoptional
Example :
vector<int>A(5,1);//vectorofsize=5withallelementsinvector
initialisedwithvalue=1
Insertion in Vector , There are two ways for inserting an element in a Vector , either you
can use push_back() or insert(index,value) , insert() will insert value at index give ,
and push_back() will insert value after the last element which is already in vector. And
insert() will insert the value as well as it will swap all the preceding elements by the
number of elements inserted
Example :
A.push_back(5);//A={1,1,1,1,1,5},
vectorcanexpanditssize
https://www.hackerearth.com/notes/stlvector/ 1/4
8/17/2016 STLVector|HackerEarth
A.insert(1,6);//A={1,6,1,1,1,1,5}
## Resizing the vector :
A.resize(10);//nowthevector'A'hassize=10
## Erase all the values of vector :
A.erase();//EraseallvaluesofvectorA
## Erase values of vector lie in a given interval :
A.erase(1,5);//Erasevaluesfromindex=1to5
## Check if any vector is empty :
if(A.empty()){//Statements}
Importing value of any array into a vector , let an array array like
intarray[]={1,2,3,4,5};
vector<int>A(array,array+5);//Itwillmakevectorelements
A={1,2,3,4,5}
Size of vector :
intsize=A.size();//itgivessizeofvector'A',i.e.numberof
elementspresentinvector'A'
## Creating Matrix with help of Vector ,
vector<int,vector<int>>Mat(M,vector<int>N);//Matrixofsize
M*N
Many <algorithm> library function can be implemented on vector(s) , that I will explain
in <algorithm> Note
https://www.hackerearth.com/notes/stlvector/ 2/4
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TRENDING NOTES
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'
# Search results
Found 1412 matches
Hot Air Balloon Lift
The hot air balloon is the oldest successful human-carrying flight technology. The first hot-air balloon flown in the United States was launched from the ... more
Fluid Thread Breakup - Linear Stability of Inviscid Liquids
Fluid thread breakup is the process by which a single mass of fluid breaks into several smaller fluid masses. The process is characterized by the ... more
Capillary pressure in a tube (contact angle)
In a sufficiently narrow tube of circular cross-section of radius “a”, the interface between two fluids forms a meniscus that is a portion of the surface ... more
Terminal velocity (creeping flow conditions)
The terminal velocity of a falling object is the velocity of the object when the sum of the drag force and buoyancy equals the downward force of gravity ... more
Stokes' law (Excess force due to the difference of the weight of the sphere and the buoyancy on the sphere)
The weight of an object is the force on the object due to gravity. Buoyancy is an upward force exerted by a fluid that opposes the weight of an immersed ... more
Hydrostatic Pressure - simplified version
In a fluid at rest, all frictional stresses vanish and the state of stress of the system is called hydrostatic.For water and other liquids, this integral ... more
Ball Screw - Buckling Load
A ball screw is a mechanical linear actuator that translates rotational motion to linear motion with little friction. A threaded shaft provides a helical ... more
Concentric tube heat exchanger - Overall Heat Transfer coefficient
Concentric Tube (or Pipe) Heat Exchangers are used in a variety of industries for purposes such as material processing, food preparation, and ... more
Dynamic Pressure
In incompressible fluid dynamics dynamic pressure (indicated with q, or Q, and sometimes called velocity pressure) is the quantity defined as ... more
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Shear stress, is defined as the component of stress coplanar with a material cross section. Shear stress arises from the force vector component parallel to ... more
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# How far is Buraidah from Qurayyat?
The distance between Qurayyat (Gurayat Domestic Airport) and Buraidah (Prince Naif bin Abdulaziz International Airport) is 528 miles / 849 kilometers / 459 nautical miles.
The driving distance from Qurayyat (URY) to Buraidah (ELQ) is 569 miles / 915 kilometers, and travel time by car is about 9 hours 43 minutes.
528
Miles
849
Kilometers
459
Nautical miles
1 h 29 min
103 kg
## Distance from Qurayyat to Buraidah
There are several ways to calculate the distance from Qurayyat to Buraidah. Here are two standard methods:
Vincenty's formula (applied above)
• 527.854 miles
• 849.498 kilometers
• 458.692 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth's surface using an ellipsoidal model of the planet.
Haversine formula
• 528.064 miles
• 849.836 kilometers
• 458.875 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Qurayyat to Buraidah?
The estimated flight time from Gurayat Domestic Airport to Prince Naif bin Abdulaziz International Airport is 1 hour and 29 minutes.
## What is the time difference between Qurayyat and Buraidah?
There is no time difference between Qurayyat and Buraidah.
## Flight carbon footprint between Gurayat Domestic Airport (URY) and Prince Naif bin Abdulaziz International Airport (ELQ)
On average, flying from Qurayyat to Buraidah generates about 103 kg of CO2 per passenger, and 103 kilograms equals 226 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Qurayyat to Buraidah
See the map of the shortest flight path between Gurayat Domestic Airport (URY) and Prince Naif bin Abdulaziz International Airport (ELQ).
## Airport information
Origin Gurayat Domestic Airport
City: Qurayyat
Country: Saudi Arabia
IATA Code: URY
ICAO Code: OEGT
Coordinates: 31°24′42″N, 37°16′46″E
Destination Prince Naif bin Abdulaziz International Airport
City: Buraidah
Country: Saudi Arabia
IATA Code: ELQ
ICAO Code: OEGS
Coordinates: 26°18′10″N, 43°46′27″E
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# Problems with Newton's Second Law
• TheShehanigan
In summary, the acceleration of a car being pushed by a constant force will decrease in relation to the mass of the car. The graph of acceleration on the y-axis and (1/m) on the x-axis will either be linear, parabolic, or exponential, depending on the formula used to relate mass and acceleration. The slope of the graph will indicate the type of relationship between mass and acceleration, whether it is a straight line, parabolic curve, or exponential curve. Further analysis is needed to determine the exact formula and type of graph.
TheShehanigan
## Homework Statement
Given a car being pushed by a constant force:
a. How will the acceleration change in relation to the mass of the car? --> Done, it'll decrease
b. How will a graph of acceleration in the y-axis and and (1/m) being the x-axis will look like? What will the slope of the graph mean? --> Trouble with this one
F = ma
a = F(1/M)
## The Attempt at a Solution
For part B, I have already determined the graph should be decreasing, but I have doubts if it's exponential or as (1/x) does. I think it's as (1/x), since F is constant. As to what the slope means, I think it refers to the Inertia of the object in question. I have serious doubts with this though.
Any help is greatly appreciated.
a. How will the acceleration change in relation to the mass of the car? --> Done, it'll decrease
I don't care for this answer, which is an oversimplification! In my opinion, the question asks for a "relationship" between mass and acceleration. That word "relationship" means they want the formula relating mass and acceleration. You probably have a formula with an "m" and an "a" in it. The thing to do is rearrange it so it says "a = ...".
For the b part, compare your formula for a= with the some standard formulas.
For example, y = slope*x + b is the formula for a straight line.
y = a*x^2 is a quadratic or parabola
y = a*e^x is an exponential.
Which one fits your formula for "a=" . . . with the "y" replaced by "a", the "x" replaced by "1/m" ?
You should find that it is one of the above, exactly, so you will know whether the graph is linear, parabolic or exponential.
There are a few potential problems with using Newton's Second Law to analyze this situation. First, the law assumes that the force acting on an object is constant, which may not be the case in this scenario. If the person pushing the car applies varying amounts of force, the acceleration may not follow a predictable pattern.
Secondly, the law assumes that the mass of the object is constant, which may not be true in this situation. If the car has items inside of it that can move around, the mass of the car may change, affecting the acceleration.
As for part B, it is important to note that the equation a = F(1/M) is only valid for objects with constant mass. In this situation, the mass of the car is not constant, so this equation may not accurately represent the relationship between acceleration and mass. Additionally, the graph of acceleration versus (1/m) may not be a straight line, as the relationship between these variables may be more complex.
In terms of the slope of the graph, it may not necessarily represent the inertia of the object. The slope of a graph typically represents the rate of change between the two variables, so in this case, it may represent the rate at which the acceleration changes as the mass of the car changes. However, without knowing the specific values and units of the variables, it is difficult to accurately interpret the slope of the graph.
In conclusion, while Newton's Second Law is a useful tool for analyzing many situations, it may not be applicable or accurate in all cases. It is important to consider the limitations and potential issues with using this law in any scientific analysis.
## What is Newton's Second Law?
Newton's Second Law, also known as the Law of Acceleration, states that the acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to its mass. In simpler terms, the greater the force applied to an object, the greater its acceleration will be.
## What are some problems with Newton's Second Law?
There are several problems with Newton's Second Law, including:
• The law assumes that the object is moving in a straight line with constant acceleration, which is not always the case in real-world scenarios.
• The law does not take into account other factors that may affect an object's motion, such as air resistance or friction.
• It is difficult to accurately measure and quantify all the forces acting on an object, making it challenging to apply the law in practical situations.
## How does Newton's Second Law relate to Newton's First and Third Laws?
Newton's First Law, also known as the Law of Inertia, states that an object at rest will remain at rest and an object in motion will remain in motion at a constant velocity unless acted upon by an external force. Newton's Second Law can be seen as an extension of this, as it explains how an object's motion changes when a force is applied to it. Newton's Third Law states that for every action, there is an equal and opposite reaction, which can also be applied to understanding the forces involved in Newton's Second Law.
## Are there any exceptions to Newton's Second Law?
While Newton's Second Law is a fundamental principle in classical mechanics and holds true in most situations, there are a few exceptions. For example, at very high velocities approaching the speed of light, the law does not accurately predict the behavior of objects. Additionally, in the microscopic world of quantum mechanics, the law does not apply and a different set of principles are used to describe the motion of particles.
## How is Newton's Second Law used in real-world applications?
Despite its limitations, Newton's Second Law is still widely used in various fields such as engineering, physics, and sports. It is used to calculate the acceleration, velocity, and displacement of objects in motion, and is the basis for designing vehicles, structures, and machines. In sports, it is used to analyze and improve techniques, such as in the development of faster and more efficient running techniques for sprinters.
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# A baseball is popped straight up into the air and has a hang-time of 6.25s. Determine the height to which the ball rises before it reaches its peak.(Hint : the time to rise to the peak is one half the total hang time )
1
by danishjoshi111
we need either the mass or the initial and final velocities
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## May 2, 2013
### Better Planning on Driving Can Save Lots of Gas Money
This past weekend my wife and I took a trip to a store to buy an item that was on sale. This trip ended up being 11 miles each way for a total of 22 miles round trip. Thats about 1 gallon of gas for my car. In total the trip cost us roughly \$3.50 worth of gas at todays prices. The store in question is on my way to work from our house. If I had instead stopped there on my way home from work on Monday then it would have been just 0.6 miles out of my way. That would have cost me about 10¢ in gas. This extra trip cost us about \$3.40 in gas. It wasn't that big of a deal, it was just one trip and my wife was anxious to buy get the sale item in question before they sold out. The sale was \$50 off of an item that rarely goes on sale so theoretically getting there soon may have saved us \$50.
But what if we engaged in such trips routinely? If we failed to plan our trips to the store like that and went there just once a month we'd be wasting \$42 a year at todays gas prices.
There are two stores I often go to for groceries. If I go to both in one trip then the total trip from home to store A to store B to home is 12.3 miles. If I were to make two separate trips then I'd spend 9.1 miles to go to store A and 10.6 miles to go to store B. Thats 19.7 miles total. By combining the stores into one trip I save 7.4 miles. That costs about \$1.17 in gasoline for my car. If I did that once a week for the weekly grocery trips then that would add up to over \$60 a year.
Now lets say that my wife and I got home from that trip to the store and later decided to go out to eat for dinner. If we didn't plan it well and went to the store in the afternoon then made a separate trip to eat out in the evening then we would have wasted more gas. It is 6.7 miles round trip to the restaurant from our house. Add that to the 22 miles round trip to the store and you've got 28.7 miles total. If we'd instead gone to the store first and then the restaurant then the total would be 20.6 miles. Thats a difference of 8.1 miles or about \$1.28 in gas.
Some of this may seem like common sense to most of us. Of course you'll save money and time if you combine trips. But on the other hand maybe its not so obvious how much you can save. Its also not as easy to plan in advance unless you're really thinking about it. Say my wife and I had gone out to dinner after that trip to the store, combining those trips isn't something I'd necessarily think ahead to do on a random Saturday afternoon. But if you're aware of the fact that any given trip might cost you \$2 or \$3 extra in gasoline then you may be more likely to think ahead and combine all your trips as much as possible.
--
1. We always make sure to plan our drives. W isn't that good at planning though :/
2. Well illustrated. I have a pal who drives 30 miles each way to "save money" on a matinee ticket price for movies. This made sense in 2003 when gas was <\$2/gal. And places a zero value on his time.
Still, your wife was happy so I think you made an excellent investment in both time and money, which can't be quantified in minutes or dollars.:-)
3. JayCeezy,
Actually that movie thetre example works for me too.
For us the cheaper movie theatre is a 36 mile round trip and the closer theatre is just 14 miles. So thats about \$3.50 in gas to go to the cheaper theatre. The cheaper theatre is only about \$1 less per ticket. So it isn't worth it for us to spend \$3.5 in gas to save \$2 on tickets.
Jim
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Radical Equations Worksheet – The goal of Expressions and Equations Worksheets is for your child to be able to learn more efficiently and effectively. They include interactive activities and problems that are based on the order in which operations are performed. Through these worksheets, kids are able to grasp basic as well as complex concepts in short period of time. Download these free documents in PDF format. They will aid your child’s learning and practice maths equations. These are helpful for students from 5th through 8th Grades.
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Definitions
Nearby Words
# Purity
[pyoor-i-tee]
Purity: see Pearl, The.
The KARAT (abbreviation "K" or 'Kt' or "KP" for Karat Plumb) is a measure of the purity of gold alloys. In the Precious Metals Industry, whether or not one is in the United States and Canada, the spelling karat is used, while the spelling carat is used to refer to the measure of mass for gemstones (see Carat (mass)). Only the French and the Italians spell the purity of gold unit with a "C". (See list below)
## Measure
As a measure of purity, one Karat is $tfrac\left\{1\right\}\left\{24\right\}$ purity by mass:
$X = 24,frac\left\{M_g\right\}\left\{M_m\right\}$
where
$mathit\left\{X\right\}$ is the Karat rating of the material,
$M_g$ is the mass of pure gold or platinum in the material, and
$M_m$ is the total mass of the material.
Therefore 24-Karat gold is fine (99.9% Au w/w), 18-Karat gold is 75% gold, 12-Karat gold is 50% gold, and so forth.
Historically, in England the Karat was divisible into four grains, and the grain was divisible into four quarts. For example, a gold alloy of $tfrac\left\{381\right\}\left\{384\right\}$ fineness (that is, 99.2% purity) could have been described as being 23-Karat, 3-grain, 1-quart gold.
The Karat system is increasingly being complemented or superseded by the millesimal fineness system in which the purity of precious metals is denoted by parts per thousand of pure metal in the alloy.
The most common Karats used for gold in bullion, jewellery making and by goldsmiths are:
• 24 Karat (millesimal fineness 999)
• 22 Karat (millesimal fineness 916)
• 20 Karat (millesimal fineness 833)
• 18 Karat (millesimal fineness 750)
• 15 Karat (millesimal fineness 625)
• 14 Karat (millesimal fineness 585)
• 10 Karat (millesimal fineness 417)
• 9 Karat (millesimal fineness 375)
• 8 Karat (millesimal fineness 333)
## Derivation
The word Karat is derived from the Greek kerátiōn (κεράτιων), “fruit of the carob”, via Arabic qīrāṭ (قيراط) and Italian. Carob seeds were used as weights on precision scales because of their reputation for having a uniform weight. This was not the only reason. It is said that in order to keep regional buyers and sellers of gold honest, a potential customer could retrieve their OWN carob seeds on their way to the market, to check the tolerances of the seeds used by the merchant. If this precaution was not taken, the potential customer would be at the mercy of "2 sets of carob seeds". One set of "heavier" carob seeds would be used when buying from a customer (making the seller's gold appear to be less). Another, lighter set of carob seeds would be used when the merchant wanted to SELL to a customer.
However, a 2006 study by Lindsay Turnbull and others found this to not be the case – carob seeds have as much variation in their weights as other seeds. In the distant past, different countries each had their own carat, roughly equivalent to a carob seed. In the mid-16th century, the Karat was adopted as a measure of gold purity, roughly equivalent to the Roman siliqua ($tfrac\left\{1\right\}\left\{24\right\}$ of a golden solidus of Constantine I). As a measure of diamond weight, from 1575, the Greek measure was the equivalent of the Roman siliqua, which was $tfrac\left\{1\right\}\left\{24\right\}$ of a golden solidus of Constantine; but was likely never used to measure the weight for gold.
## Terminology
22/22K - a quality mark indicating the purity of gold most popularly used in India. This purity was adapted and practiced by the big jewellers and was later passed to jewel smiths. The first 22 signifies the "Skin purity" of gold jewellery and the second 22 signifies that after melting purity of the gold jewellery will be 22K (22 [[Karat [purity)]) or 91.67% of pure gold.
This symbol or stamp is very popular on the gold jewellery business in Asian countries like India, Sri Lanka, Pakistan, Bangladesh, Nepal, Yemen, and Persian Gulf countries.
This practice was pioneered and introduced in the early mid-1980s by Nemichand Bamalwa & Sons of Kolkata, India, sparking a revolution in India as it forced jewellers to indicate correctly the after-melting purity, and heightened consumer awareness made it a most sought-after stamp or quality mark.
Chuk Kam (足金) – In Cantonese (Chinese) this term means pure gold, literally "exact gold". It is defined as 99.0% gold minimum with a 1.0% negative tolerance allowed. The quality of gold is guaranteed with a "Certificate of Gold" upon purchases in Hong Kong and Macau.
### In the United States of America
The USA Fair Trade Commission (CFTC) has legislated and standardized the karat markings used within its boundaries for almost 7 decades now. Under these regulations, items 10K or greater are to be stamped with either "K" or "Kt." Decimal markings are also an option under the CFTC regulations.
Under karating is against the USA law. There are specific mandated consequences including fines, et al., based upon the severity of the infraction(s.) It is considered fraud to not mark all jewellery--unless specifically covered by stated exemption such as an art creation--and sell said jewellery to the US consuming public.
Additionally, there are a set of tolerances to the required karat markings in the USA (always designated with a "K" and never a "C") depending upon the use of various soldering requirements when setting stones, mounting crowns, or creating prongs for 3 examples.
Title 16: Commercial Practices PART 23—GUIDES FOR THE JEWELRY, PRECIOUS METALS, AND PEWTER INDUSTRIES
§ 23.4 Misrepresentation as to gold content.
(a) It is unfair or deceptive to misrepresent the presence of gold or gold alloy in an industry product, or the quantity or karat fineness of gold or gold alloy contained in the product, or the karat fineness, thickness, weight ratio, or manner of application of any gold or gold alloy plating, covering, or coating on any surface of an industry product or part thereof.
(b) The following are examples of markings or descriptions that may be misleading:2
2 See §23.4(c) for examples of acceptable markings and descriptions.
(1) Use of the word “Gold” or any abbreviation, without qualification, to describe all or part of an industry product, which is not composed throughout of fine (24 karat) gold.
(2) Use of the word “Gold” or any abbreviation to describe all or part of an industry product composed throughout of an alloy of gold, unless a correct designation of the karat fineness of the alloy immediately precedes the word “Gold” or its abbreviation, and such fineness designation is of at least equal conspicuousness.
(3) Use of the word “Gold” or any abbreviation to describe all or part of an industry product that is not composed throughout of gold or a gold alloy, but is surface-plated or coated with gold alloy, unless the word “Gold” or its abbreviation is adequately qualified to indicate that the product or part is only surface-plated.
(4) Use of the term “Gold Plate,” “Gold Plated,” or any abbreviation to describe all or part of an industry product unless such product or part contains a surface-plating of gold alloy, applied by any process, which is of such thickness and extent of surface coverage that reasonable durability is assured.
(5) Use of the terms “Gold Filled,” “Rolled Gold Plate,” “Rolled Gold Plated,” “Gold Overlay,” or any abbreviation to describe all or part of an industry product unless such product or part contains a surface-plating of gold alloy applied by a mechanical process and of such thickness and extent of surface coverage that reasonable durability is assured, and unless the term is immediately preceded by a correct designation of the karat fineness of the alloy that is of at least equal conspicuousness as the term used.
(6) Use of the terms “Gold Plate,” “Gold Plated,” “Gold Filled,” “Rolled Gold Plate,” “Rolled Gold Plated,” “Gold Overlay,” or any abbreviation to describe a product in which the layer of gold plating has been covered with a base metal (such as nickel), which is covered with a thin wash of gold, unless there is a disclosure that the primary gold coating is covered with a base metal, which is gold washed.
(7) Use of the term “Gold Electroplate,” “Gold Electroplated,” or any abbreviation to describe all or part of an industry product unless such product or part is electroplated with gold or a gold alloy and such electroplating is of such karat fineness, thickness, and extent of surface coverage that reasonable durability is assured.
(8) Use of any name, terminology, or other term to misrepresent that an industry product is equal or superior to, or different than, a known and established type of industry product with reference to its gold content or method of manufacture.
(9) Use of the word “Gold” or any abbreviation, or of a quality mark implying gold content (e.g., 9 karat), to describe all or part of an industry product that is composed throughout of an alloy of gold of less than 10 karat fineness.
Note to paragraph(b)§23.4: The provisions regarding the use of the word “Gold,” or any abbreviation, as described above, are applicable to “Duragold,” “Diragold,” “Noblegold,” “Goldine,” “Layered Gold,” or any words or terms of similar meaning.
(c) The following are examples of markings and descriptions that are consistent with the principles described above:
(1) An industry product or part thereof, composed throughout of an alloy of gold of not less than 10 karat fineness, may be marked and described as “Gold” when such word “Gold,” wherever appearing, is immediately preceded by a correct designation of the karat fineness of the alloy, and such karat designation is of equal conspicuousness as the word “Gold” (for example, “14 Karat Gold,” “14 K. Gold,” or “14 Kt. Gold”). Such product may also be marked and described by a designation of the karat fineness of the gold alloy unaccompanied by the word “Gold” (for example, “14 Karat,” “14 Kt.,” or “14 K.”).
Note to paragraph(c)(1): Use of the term “Gold” or any abbreviation to describe all or part of a product that is composed throughout of gold alloy, but contains a hollow center or interior, may mislead consumers, unless the fact that the product contains a hollow center is disclosed in immediate proximity to the term “Gold” or its abbreviation (for example, “14 Karat Gold-Hollow Center,” or “14 K. Gold Tubing,” when of a gold alloy tubing of such karat fineness). Such products should not be marked or described as “solid” or as being solidly of gold or of a gold alloy. For example, when the composition of such a product is 14 karat gold alloy, it should not be described or marked as either “14 Kt. Solid Gold” or as “Solid 14 Kt. Gold.”
(2) An industry product or part thereof, on which there has been affixed on all significant surfaces, by any process, a coating, electroplating, or deposition by any means, of gold or gold alloy of not less than 10 karat fineness that is of substantial thickness,3 and the minimum thickness throughout of which is equivalent to one-half micrometre (or approximately 20 millionths of an inch) of fine gold,4 may be marked or described as “Gold Plate” or “Gold Plated,” or abbreviated, as, for example, G.P. The exact thickness of the plate may be marked on the item, if it is immediately followed by a designation of the karat fineness of the plating which is of equal conspicuousness as the term used (as, for example, “2 micrometres 12 K. gold plate” or “2µ 12 K. G.P.” for an item plated with 2 micrometres of 12 karat gold.)
3 The term substantial thickness means that all areas of the plating are of such thickness as to assure a durable coverage of the base metal to which it has been affixed. Since industry products include items having surfaces and parts of surfaces that are subject to different degrees of wear, the thickness of plating for all items or for different areas of the surface of individual items does not necessarily have to be uniform.
4 A product containing 1 micrometre (otherwise known as 1µ) of 12 karat gold is equivalent to one-half micrometre of 24 karat gold.
Note to paragraph(c)(2): If an industry product has a thicker coating or electroplating of gold or gold alloy on some areas than others, the minimum thickness of the plate should be marked.
(3) An industry product or part thereof on which there has been affixed on all significant surfaces by soldering, brazing, welding, or other mechanical means, a plating of gold alloy of not less than 10 karat fineness and of substantial thickness5 may be marked or described as “Gold Filled,” “Gold Overlay,” “Rolled Gold Plate,” or an adequate abbreviation, when such plating constitutes at least1/20th of the weight of the metal in the entire article and when the term is immediately preceded by a designation of the karat fineness of the plating which is of equal conspicuousness as the term used (for example, “14 Karat Gold Filled,” “14 Kt. Gold Filled,” “14 Kt. G.F.,” “14 Kt. Gold Overlay,” or “14K. R.G.P.”). When conforming to all such requirements except the specified minimum of1/20th of the weight of the metal in the entire article, the terms “Gold Overlay” and “Rolled Gold Plate” may be used when the karat fineness designation is immediately preceded by a fraction accurately disclosing the portion of the weight of the metal in the entire article accounted for by the plating, and when such fraction is of equal conspicuousness as the term used (for example, “1/40th 12 Kt. Rolled Gold Plate” or “1/4012 Kt. R.G.P.”).
5 See footnote 3.
(4) An industry product or part thereof, on which there has been affixed on all significant surfaces by an electrolytic process, an electroplating of gold, or of a gold alloy of not less than 10 karat fineness, which has a minimum thickness throughout equivalent to .175 micrometres (approximately7 /1,000,000ths of an inch) of fine gold, may be marked or described as “Gold Electroplate” or “Gold Electroplated,” or abbreviated, as, for example, “G.E.P.” When the electroplating meets the minimum fineness but not the minimum thickness specified above, the marking or description may be “Gold Flashed” or “Gold Washed.” When the electroplating is of the minimum fineness specified above and of a minimum thickness throughout equivalent to two and one half (21/2) micrometres (or approximately100 /1,000,000ths of an inch) of fine gold, the marking or description may be “Heavy Gold Electroplate” or “Heavy Gold Electroplated.” When electroplatings qualify for the term “Gold Electroplate” (or “Gold Electroplated”), or the term “Heavy Gold Electroplate” (or “Heavy Gold Electroplated”), and have been applied by use of a particular kind of electrolytic process, the marking may be accompanied by identification of the process used, as for example, “Gold Electroplated (X Process)” or “Heavy Gold Electroplated (Y Process).”
(d) The provisions of this section relating to markings and descriptions of industry products and parts thereof are subject to the applicable tolerances of the National Stamping Act or any amendment thereof.6
6 Under the National Stamping Act, articles or parts made of gold or of gold alloy that contain no solder have a permissible tolerance of three parts per thousand. If the part tested contains solder, the permissible tolerance is seven parts per thousand. For full text, see 15 U.S.C. 295, et seq .
Note 4 to paragraph(d): Exemptions recognized in the assay of karat gold industry products and in the assay of gold filled, gold overlay, and rolled gold plate industry products, and not to be considered in any assay for quality, are listed in the appendix.
## International karatages of gold jewellery
{|class="wikitable" ! width=40% | Region ! width=40% | Typical Karatage (fineness) |- | Arabic countries, Far East (China, Hong Kong, Taiwan) || 24 Karat "Juk Gum" (99.0% min) |- | Arabic countries, Bangladesh, India, Pakistan & Sri Lanka || 22 carat (91.6%) |- | Arabic countries in the Persian Gulf region || 21 Karat (87.5%), 18 Karat (75.0%) in most Egypt |- | Europe - Southern / Mediterranean || 18 Karat (75.0%) |- | Europe - Northern / USA etc || 8-18 Karat (33.3 - 75.0%) |- | Russia / former USSR || 9 (37.5%) and 14 Karat / old 583 and new 585 проба (58.5%) |-
## References
• Fallon, S. (2006) Hong Kong & Macau, 12th ed., Melbourne; London: Lonely Planet, ISBN 1-7405-9843-1
• Harper, D. (2001) " Carat", in: Online Etymological Dictionary, accessed 28 August 2007
• New Scientist (2006) , New Scientist magazine, 2550 (9 May), p. 20
• Turnbull, L.A., Santamaria, L., Martorell, T., Rallo, J. and Hector, A. (2006) " Seed size variability: from carob to carats", Biology Letters, 2 (3: September 22), p. 397–400, DOI 10.1098/rsbl.2006.0476
• World Gold Council (2003) , Online article accessed 28 August 2007
Search another word or see purityon Dictionary | Thesaurus |Spanish
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# 100 000 usd na lakh
10/2/2021
Yes, here is the formula: 1. Go to Google 2. Search for USDINR (right not the value is 67.36) 3. Multiple that value by \$100,000, right now the result is 6,736,000 - 6.736 Million INR 4. If you format that in Excel using Indian formatting, it says 100 USD = 933913.46 LAK-586.54 LAK-0.06%: The value of 100 USD in Lao Kip for the week (7 days) increased by: +586.54 LAK (five hundred eighty-six kip fifty-four att). For the month (30 days) Date Day of the week 100 USD to LAK Changes Changes % February 14, 2021: Sunday: 100 USD = 934500 LAK +2500 LAK +0.27%: For the month (30 days) Date Day of the week 1 USD to LAK Changes Changes % February 21, 2021: Sunday: 1 USD = 9358.54 LAK +33.54 LAK +0.36%: January 22, 2021 9/2/2021 100.00 USD = 934,500.00 LAK Follow news in the Economic Calendar Currency converter - Light Version Here you are getting today's value of one hundred US Dollar to Lao Kip . … Lao Kip to US Dollar currency exchange rate.
Las grandes cantidades denominadas en rupias se cuentan por cientos de miles, es decir, en lakhs (1 lakh= 10 5 rupias = 100 000 rupias), en crores (1 crore=100 lakhs = 10 7 rupias = 10 000 000 rupias) y en arawbs (1 arawb=100 crore = 10 9 rupias = 1 000 000 000 rupias). El uso de millones o billones es menos común. 9/2/2021 100000 USD = 935000000 LAK at the rate on 2021-02-23. \$ 1 = ₭9350 +4.05 (+0.04%) at the rate on 2021-02-23. The page provides data about today's value of one hundred thousand dollars in Lao Kip. Go to Google Search for USDINR (right not the value is 67.36) Multiple that value by \$100,000, right now the result is 6,736,000 - 6.736 Million INR If you format that in Excel using Indian formatting, it says 67 Lakhs 36 thousand Indian Rupees Feb 09, 2021 · North American Edition. The dollar has traded more mixed today, with the pound and dollar bloc holding their own. The DXY dollar index has edged out a fresh two-month high, this time at 91.60, in what is its fifth consecutive up day, underpinned by an improving yield advantage relative to the euro and other peers (with UK gilt yields being the main exception).
## Lao Kip to US Dollar currency exchange rate. 100 LAK = 0.01 USD Today LAK to USD exchange rate = 0.000107. LAK to USD Exchange rates details:. Reverse: 100 USD to LAK
Online interactive currency converter & calculator ensures provding actual conversion information of world currencies according to “Open Exchange Rates” and provides the information in its best way. 118.000 Forkcoin to Malaysian Ringgit 34.000 PCHAIN to Philippine Peso 3000.000 TRON to Australian Dollar 1704.000 Dogecoin to US Dollar 5995.000 LUNA to Canadian Dollar 400.000 PiCoin to Indian Rupee 10.000 Dollars to Ethereum 10.000 PCHAIN to Malaysian Ringgit 1.000 Theta Token to Pakistani Rupee 2000.000 BitTorrent to South African Rand 2000 1 Trillion = 1 lakh crores.
### A lakh is a unit in the Indian numbering system equal to one hundred thousand (100,000; scientific notation: 10 rise to power 5). In the Indian convention of digit grouping, it is written as 1,00,000. For example, in India 150,000 rupees becomes 1.5 lakh rupees
We use international USD/LAK exchange rate, and last update was today. Online converter show how much is 100 US Dollar in Lao Kip. 100 US dollar to Lao Kip converter. Convert 100 USD to LAK with real time currency calculator.
View charts, common conversions, historical exchange rates and more. 19 May 2020 The number of cases of coronavirus infection crossed the one lakh mark in the country on Tuesday (May 19, 2020), while the death toll due to ᐈ How much is \$100000【one hundred thousand】 US Dollar in Indian Rupee? ✓ Check the latest currency rate! Online exchange rate calculator 25 Jun 2019 Let's say the value of one pip is 8.93 euros ((0.0001/1.1200) * 100,000). To convert the value of the pip to U.S. dollars, just multiply the value of
USD - US Dollar U.S. Dollar (USD) is the currency used in United States, East Timor, Puerto Rico, Equador. U.S. Dollar currency symbol: \$; U.S. Dollar coins available: 1¢, 5¢, U.S. Dollar (USD) is the currency used in United States, East Timor, Puerto Rico, Equador. U.S. Dollar currency symbol: \$; U.S. Dollar coins available: 1¢, 5¢, 10¢, Convert: ᐈ 100 000.00 US Dollar (USD) to Indian Rupee (INR) - currency converter, course history. OMR & covered yourself in Uk stock exchange on the sam nge rates were. = hat Hong Kong Dollar will depreciate to 10.89 level and Indian ainst GBP to ` 84.83. t today's spot rate, it is equivalent to ` 53.92 lakhs. It is an One lakh, equivalent to 100000 rupees, is equal to \$1498.21 U.S. dollars.
1 Lakhs. 100 Thousand. 2 Lakhs. 200 Thousand. 3 Lakhs.
Trillion. 12. Each rupee is divided into 100 paise. About the U.S. Dollar. The U.S. Dollar is the currency of the United States of America, which is the top destination for Indian Here you will find the current foreign exchange rates for converting 100000 United States Dollar (USD) in Indian Rupee (INR) today.
Check how many crore, lakh and thousands are equal to thousand using this conversion chart. Get instant value for each low to high scale and high to low scale conversions. 1 Thousand. 0.01 Lakhs.
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### 6 Lakhs In Dollars Thursday, 25 February 2021. Inr vs us dollar why indian ru is govt announces down as dollars 1 75 crore indian diaspora the largest ru vs dollar today 21 lakhs to dollars convert indian difference between lakh and million.
0.01 Lakhs. 2 Thousand. 0.02 Lakhs. 3 Thousand. In finance, an 10 lakh INR to USD exchange rate is the Indian Rupee to >US Dollar rate at which 10 lakh Indian Rupee to US Dollar will be exchanged for another. It is also regarded as the value of 10 lakh INR to USD in relation to another currency.
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Chapter P.1, Problem 8E
### Elementary Geometry For College St...
7th Edition
Alexander + 2 others
ISBN: 9781337614085
Chapter
Section
### Elementary Geometry For College St...
7th Edition
Alexander + 2 others
ISBN: 9781337614085
Textbook Problem
# Let A , B , and C lie on a straight line as shown. Classify these claim as true or false.a) There is no point that lies between A and B .b) A B ¯ has endpoints A and B .
To determine
a) To claim:
Whether true or false.
For the points A, B and C lie on a straight line shown, ‘There is no point that lies between A and B’.
Explanation
Calculation:
Given,
A, B and C are the points on a straight line.
Since, the line segment AB¯
To determine
b) To find:
Whether true or false.
For the points A, B and C lie on a straight line shown, ‘AB¯ has endpoints A and B’.
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Applied Calculus for the Managerial, Life, and Social Sciences: A Brief Approach
#### Find the derivatives of the functions in Problems 1-10. 9.
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#### Differentiate. y = sin cos
Single Variable Calculus: Early Transcendentals
#### For f(x) = 5 + g(x), f(x) = _____. a) 5g(x) b) 5 + g(x) c) 0 g(x) d) g(x)
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Study Guide for Stewart's Multivariable Calculus, 8th
#### Find x. logx32=5
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### Re: Pendulum: What is relationship of the T^2 vs. length?
Date: Fri Jan 3 13:34:15 2003
Posted By: Chris Seaman, Staff, Electrical Engineering, Materials Engineering, Alcoa Technical Center
Area of science: Physics
ID: 1040742124.Ph
Message:
```
Michael,
Your thinking is not "off". In terms of simply solving an algebraic
problem, you are correct.
But now, step back and look at the problem a little bit differently.
Let's assume you don't know anything about pendulums and want to perform
an experiment in which you vary the pendulum length, L, the mass, and the
starting position. You also have the ability to measure the period
accurately. (You may do this by measuring one period, or taking the
average of many periods). You would find that T^2 is directly
proportional to the pendulum length, and completely independent of the
mass or the starting position (as long as the starting position
wasn't "too high").
You now have a simple device that can accurately measure time, and is
quite easy to calibrate (adjust the length) and keeps accurate time even
as the power "runs down". This type of understanding of mechanisms is
what lead to the development of portable, accurate clocks, which
revolutionized navigation in the 17th Century.
One goal of this type of physics analysis and experimentation is to
develop an understanding of unique relationships between phenomena, and
convert it into a mathematical model, if possible. Even at this moment,
researchers in nanotechnology are using this type of modeling to develop
new sensors and devices.
Solve a problem, then step back to try to understand what the solution
means.
Christopher M. Seaman
Technical Specialist
ALCOA Technical Center
```
Current Queue | Current Queue for Physics | Physics archives
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# Thread: Atomic mass
1. If I had a sphere of uranium the size of a football and a sphere of say, iron, also the size of a football. Would the uranium weight more because it has more atomic mass ? (i.e more stuff in it)
2.
3. Not exactly - density matters at the macroscopic level and this has to do with the bonding structure of the material in question and the actual weight of the molecules making up the material. For example water has very different densities if it is liquid or ice, but both cases you have the same molecular weight.
Now uranium as a metal as a density of around 18900 kg/m^3 while iron has a density of around 7000 kg/m^3 at 293 degrees kelvin so the ball of uranium would be more massive then the ball of iron. In case you are interested, the most dense elements known to man are osmium and iridium (with almost the same density) each with a density of 22600 kg/m^3. The densest thing we know of is a neutron (well thing with a measurable size that is) and it has a density of 10^18 kg/m^3!
4. Relative density of Iron is about 7gms/cc and I think uranium is around 20 so it's weight would be almost 3 times that of the Iron. (Water has a relative density of 1).
If I had a football sized piece of Uranium - I'd drop it and run.
5. Atomic mass is certainly part of what amounts to density, but also the volume that the atoms take up when they're combined. It really depends on the structure. Different allotropes have different densities. For example, graphite has a density of 2.09–2.23 g/mL and diamond has a density of 3.52 g/mL. They're both entirely composed of carbon, and their difference in densities is due to their different structures.
6. Yep - it's down to the packing of the individual atoms on the nanometer scale. The atoms that can most densely pack on the atomic scale will have the greatest density on the macroscopic scale - but yes, atomic weight is still a factor.
7. If the volume is the same the element with the higher density would weigh more. Now if the two elements have the same mass there would be no difference (I find it fun to go up to people on the street and ask them, "What weighs more a pound of air or a pound of lead?" you would be suprised how many people answer lead and then try to defend their answer when I tell them they are wrong
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# A concave mirror produces three times magnified (enlarged) real image of object placed at 10 cm in front of it. Where is the image located? - CBSE Class 10 - Science
ConceptReflection of Light Image Formation by Spherical Mirrors - Concave Mirror
#### Question
A concave mirror produces three times magnified (enlarged) real image of object placed at 10 cm in front of it. Where is the image located?
#### Solution
Magnification produced by a spherical mirror is given by the relation,
m=
m=h_1/h_0=-u/v
Let the height of the object, h_0=h
Then, height of the image, h_1=-3h(image formed is real)
-3h/h=-v/u
v/u=3
Object distance, u= – 10 cm
= 3 × ( – 10) = – 30 cm
Here, the negative sign indicates that an inverted image is formed at a distance of
30 cm in front of the given concave mirror.
Is there an error in this question or solution?
#### APPEARS IN
NCERT Solution for Science Textbook for Class 10 (2019 to Current)
Chapter 10: Light – Reflection and Refraction
Q: 2 | Page no. 171
#### Video TutorialsVIEW ALL [1]
Solution A concave mirror produces three times magnified (enlarged) real image of object placed at 10 cm in front of it. Where is the image located? Concept: Reflection of Light - Image Formation by Spherical Mirrors - Concave Mirror.
S
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# Scattering Matrix
Also found in: Acronyms.
## Scattering matrix
An infinite-dimensional matrix or operator that expresses the state of a scattering system consisting of waves or particles or both in the far future in terms of its state in the remote past; also called the S matrix. In the case of electromagnetic (or acoustic) waves, it connects the intensity, phase, and polarization of the outgoing waves in the far field at various angles to the direction and polarization of the beam pointed toward an obstacle. It is used most prominently in the quantum-mechanical description of particle scattering, in which context it was invented in 1937 by J. A. Wheeler to describe nuclear reactions. Because an analog of the Schrödinger equation for the description of particle dynamics is lacking in the relativistic domain, W. Heisenberg proposed in 1943 that the S matrix rather than the hamiltonian or the lagrangian be regarded as the fundamental dynamical entity of quantum mechanics. This program played an important role in high-energy physics during the 1960s but is now largely abandoned. The physics of fundamental particles is now described primarily in terms of quantum gauge fields, and these are used to determine the S matrix and its elements for the collision and reaction processes observed in the laboratory. See Elementary particle, Nuclear reaction, Quantum mechanics, Relativistic quantum theory, Scattering experiments (atoms and molecules), Scattering experiments (nuclei)
The mathematical properties of the S matrix in nonrelativistic quantum mechanics have been thoroughly studied and are, for the most part, well understood. If the potential energy in the Schrödinger equation, or the scattering obstacle, is spherically symmetric, the eigenfunctions of the S matrix are spherical harmonics and its eigenvalues are of the form exp (2iδl), where the real number δl is the phase shift of angular momentum l. In the nonspherically symmetric case, analogous quantities are called the eigenphase shifts, and the eigenfunctions depend on both the energy and the dynamics. In the relativistic regime, without an underlying Schrödinger equation for the particles, the mathematical properties are not as well known. Causality arguments (no signal should propagate faster than light) lead to dispersion relations, which constitute experimentally verifiable consequences of very general assumptions on the properties of nature that are independent of the detailed dynamics. See Angular momentum, Causality, Dispersion relations, Eigenfunction
McGraw-Hill Concise Encyclopedia of Physics. © 2002 by The McGraw-Hill Companies, Inc.
The following article is from The Great Soviet Encyclopedia (1979). It might be outdated or ideologically biased.
## Scattering Matrix
(S-matrix), a combination of quantities (a matrix) describing the process of transition of quantum-mechanical systems from some states to others upon interaction (scattering). The concept of a scattering matrix was introduced by W. Heisenberg in 1943.
During scattering, a system moves from one quantum state, the initial state (which may be related to an instant t = — ∞), to another state, the final state (t = + ∞). If the set of quantum numbers that characterize the initial state is designated as i, and the final state as fl then the scattering amplitude (the square of whose absolute value determines the probability of scattering) may be written as Sfi. The combination of the scattering amplitudes forms a table with two inputs (i, the number of the row, and f, the number of the column), which is called the scattering matrix S. Each amplitude is a matrix element. Sets of the quantum numbers i and f may contain both continuous quantities (such as energy and the scattering angle) and discrete quantities (such as orbital quantum number, spin, isotopic spin, and mass). In the simplest case, a system of two spinless particles in non-relativistic quantum mechanics, the state is determined by the relative momentum p of the particles; then the scattering amplitude is a function of two variables, the energy E and the scattering angle θ:
Sfi = F(E, θ)
In the general case a scattering matrix contains elements that correspond both to elastic scattering and to the processes of particle conversion and production. The square of the absolute value of the matrix element |Sfi|2 determines the probability of the corresponding process (or its effective cross section).
The fundamental problem of quantum mechanics and quantum field theory is to find the scattering matrix. The scattering matrix contains all information concerning the behavior of the system if not only the numerical values but also the analytical properties of its elements are known; in particular, its poles determine the bound states of the system (and, consequently, its discrete energy levels). The most important property of the scattering matrix, its unitarity, follows from the fundamental principles of quantum theory. This is expressed in the form of the equation SS+ = 1, where S+ is the Hermitian conjugate of S— that is, (S+)fi = S*if, where the symbol * designates complex conjugation—or
and reflects the fact that the sum of the probabilities of scattering for all possible channels of the reaction must be equal to unity. The unitarity relation makes it possible to establish important relations among various processes and, in some cases, even to solve the problem completely. There is a trend in relativistic quantum mechanics in which the scattering matrix is considered to be a primary dynamic quantity; the requirements for the unitarity and analyticity of a scattering matrix should serve as the basis for constructing a complete system of equations that define the S-matrix.
V. B. BERESTETSKII
## scattering matrix
(electromagnetism)
A square array of complex numbers consisting of the transmission and reflection coefficients of a waveguide junction.
(quantum mechanics)
A matrix which expresses the initial state in a scattering experiment in terms of the possible final states. Also known as collision matrix; S matrix.
McGraw-Hill Dictionary of Scientific & Technical Terms, 6E, Copyright © 2003 by The McGraw-Hill Companies, Inc.
References in periodicals archive ?
After calibrating the VNA, studying the network consisted of measuring first the reference healthy scattering matrix [S.sub.h] of the NUT without the faults which was accomplished by using the unaltered 30-cm semi-rigid sections.
For the computation of the scattering matrix, (10) is expressed in M transmitter/receiver positions on [partial derivative][D.sub.O], and the summations are truncated to M terms,
When solved subject to [lim.sub.x[right arrow][infinity]] [beta](k, x) = 0 and [lim.sub.x[right arrow][infinity]][beta]'(k, x)] = 0, the scattering matrix is given by [11]
where [S.sub.1] is the scattering matrix for the first layer:
Consider a node (n) with [m.sub.n] two-port elements connected to this node, it is possible to define the behavior of the node via a scattering matrix relation as follows
Consequently, the size of the global scattering matrix depends only on the number of accessible modes in the beginning and the end of the overall structure.
Consequently, the scattering matrix of the shunt node is a 4 x 4 matrix
where [S.sup.calc.sub.ij](x) are the elements of the calculated scattering matrix, and [S.sup.meas.sub.ij] are the corresponding elements of the measured scattering matrix.
Obviously, S represents the full polarization scattering matrix in the direction ([phi], [theta]).
Also included is material from his equally legendary 1954 course at the Summer School of Theoretical Physics at Haute-Savorie, France, on particle and field mechanics, the relation between scattering matrix elements and cross-sections, and renormalization.
(9.) Mackowski, DW, Mishchenko, MI, "Calculation of the T Matrix and the Scattering Matrix for Ensembles of Spheres." /.
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# Localized Dynamics of Thin-Walled Shells
Localized Dynamics of Thin-Walled Shells focuses on localized vibrations and waves in thin-walled structures with variable geometrical and physical characteristics. It emphasizes novel asymptotic methods for solving boundary-value problems for dynamic equations in the shell theory, in the form of functions which are highly localized near both fixed and moving lines/points on the shell surface. Les mer
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Localized Dynamics of Thin-Walled Shells focuses on localized vibrations and waves in thin-walled structures with variable geometrical and physical characteristics. It emphasizes novel asymptotic methods for solving boundary-value problems for dynamic equations in the shell theory, in the form of functions which are highly localized near both fixed and moving lines/points on the shell surface.
Features
First-of-its-kind work, synthesizing knowledge of the localization of vibrations and waves in thin-walled shells with a mathematical tool to study them
Suitable for researchers working on the dynamics of thin shells and also as supplementary reading for undergraduates studying asymptotic methods
Offers detailed analysis of wave processes in shells with varying geometric and physical parameters
Chapter 1: Introduction.
Chapter 2: Equations of the two-dimensional theory of shells.
Chapter 3: Localized vibration modes of plates and shells of revolution.
Chapter 4: Localized vibration modes of cylindrical and conic shells.
Chapter 5: Localized Parametric Vibrations of Thin Shells.
Chapter 6: Wave Packets in Medium-length Cylindrical Shells.
Chapter 7: Effect of External Forces on Wave Packets in Zero Curvature Shells.
Chapter 8: Wave Packets in Long Shells of Revolution Travelling in the Axial Direction.
Chapter 9: Two-dimensionalWave Packets in Shells of Arbitrary Shape.
Gennadi I. Mikhasev is an expert in the theory of non-stationary localized dynamics of thin shells and in the mathematical modelling in biomechanics. He is a co-author of two monographs in mechanics of thin shells, two handbooks in bio-mechanics, and of around 100 papers. He is the head of the department of Bio- and Nanomechanics in the Belarusian State University, as well as a part-time researcher in the laboratory of Theoretical and Applied Mechanics.
Petr E. Tovstik is a specialist in the theory of asymptotic methods and its applications to the shell vibrations and buckling and to the various branches of mechanics. He is a co-author of ten monographs and of around 150 papers. He is the head of department of Theoretical and Applied Mechanics of the Sankt Petersburg State University, as well a part-time researcher in the Institute of Mechanical Engineering.
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Optics
# Uses and applications of total internal reflection
When the angle of incidence becomes larger than the critical angle, no refraction occurs. The entire light is reflected into the denser medium. This is known as a total internal reflection of light. Prisms, binoculars, and optical fibers are a few applications and uses of Total internal reflection.
## What is total internal reflection?
When a ray of light traveling in the denser medium enters into a rarer medium, it bends away from the normal. If the angle of incidence ‘i’ increases’ the angle of refraction ‘r’ also increases. For a particular value of the angle of incidence, the angle of refraction becomes 90°. The angle of incidence, that causes the refracted ray in the rarer medium to bend through 90° is called critical angle.
## When does total internal reflection occur?
When the angle of incidence becomes larger than the critical angle, no refraction occurs. The entire light is reflected into the denser medium. This is known as a total internal reflection of light. It is shown in the figure below:
## What are the conditions necessary for tirf ?
Total internal reflection takes place only when:
• The light should pass from a denser medium to a rare medium.
• The angle of incidence must be greater than the critical angle.
## Examples of total internal reflection
• Underwater reflection of a turtle.
• Mirage is the image of some distant object that appears.
• A fish that looks like water above as a mirror.
## Applications of total internal Reflection
### Total internal reflection in the prism
Many optical instruments use right-angled prisms to reflect a beam of light through 90° or 180° (By total internal reflection) such as cameras, binoculars, periscope, and telescopes.
One of the angles of the right-angled prism is 90°. When a ray of light strikes the face of the prism perpendicular, it enters the prim without deviation and strikes the hypotenuse at an angle of 45°.
Since the angle of incidence of 45° is greater than the critical angle of the glass which is 42°, the light is reflected by the prism through an angle of 90°.
Two such prisms are used in periscope. The light is reflected by the prism at an angle of 180°. Two such prisms are used in binoculars.
## Tirf use in Optical Fibre
Total internal reflection is used in fiber optics which has several advantages in the telecommunication field. Fiber optics consist of hair-size threads of glass or plastic through which light can be traveled.
The inner part of the fiber optics is called the core that carries the light and an outer concentric shell is called cladding. The core is made of glass or plastic or has a relatively high index of refraction. The cladding is made of glass or plastic but of relatively low refractive index. Light entering from one end of the core strikes the core-cladding boundary at an angle of incidence greater than the critical angle and is refracted back into the core. In this way, light travels many kilometers with a small loss of energy.
## Light Pipe
The light pipe is a bundle of thousands of optical fibers bonded together. They are used to illuminate inaccessible places, doctors or engineers. For example, doctors view inside the human body. They can also be used to transmit images from one point to another.
## Endoscope
An endoscope is a medical instrument used for exploratory diagnostics and surgical purposes. An endoscope is used to explore the interior organs of the body. Due to its small size, it can be inserted through the mouth and thus eliminates invasive surgery.
The endoscopes used to examine the stomach, bladder, and throat are called the Gastroscope, Cystoscope, and Bronchoscope respectively. An endoscope uses two fiber-optic tubes through a pipe.
A medical procedure using any type of endoscope is called endoscopy. The light shines on the organ of the patient to be examined by entering through one of the fiber tubes of the endoscope. Then the light is transmitted back to the physician’s viewing lens through the outer fiber tube by the internal reflection.
Flexible endoscopes have a tiny camera attached to the end. The doctor can see the view recorded by the camera on a computer screen.
## Applications of total internal reflection (video)
Watch also:
Related searches about Total internal reflection:
### One Comment
1. Loveness Biteme says:
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# Electric pendulum
As shown above, a charged particle is hanging on the top of two parallel plates. A voltage of $$V$$ generates a uniform electric field inside the plates. If we lift up the charged particle to the point $$x$$ and then let it go, the charged particle oscillates between $$x$$ and $$z$$ with $$y$$ the center point of the oscillation. Which of the following statements is correct? (Ignore gravity, air resistance and the generation of electromagnetic waves.)
a) The electric potential at the point $$x$$ is greater than that at the point $$y.$$
b) The period of the oscillation decreases as the voltage $$V$$ increases.
c) The magnitude of the electric force acting on the charged particle at the point $$x$$ is greater than that at the point $$y.$$
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Introduction to Fortran 90 Part - II - PowerPoint PPT Presentation
Introduction to Fortran 90 Part - II
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Introduction to Fortran 90 Part - II
Introduction to Fortran 90 Part - II
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Presentation Transcript
1. Introduction to Fortran 90 Part - II Consulting Services Group SiddGhosh (presenter) Feb 15, 2011 NCAR/CISL/HSS/CSG Consulting Services Group
2. Contents • Working with F90 array sections • Functions and subroutines • Fortran 90 pointers • Passing array section across subroutines • Modules • Operators for user-defined data types • Intrinsic functions • Overloading functions and operators • Input/Output • Highlights of 2008 standard
3. Examples, exercises etc. ssh –X <user-name>@miragex.ucar.edu replace x with one of 0, 1, 2, 3, 4 randomly The option –X for forwarding X display, you need only if you use x-based editor like nedit cp -r /glade/home/sghosh/f90-2 .
4. Fortran90 array syntax similar to array syntax of Matlab, NCL or IDL Each rank may have the structure beg-ind : end-ind: stride Each index can be any integer, +ive, -iveor 0 (zero stride is illegal and does not make sense anyway!) Note: The syntax is similar to NCL but contrast this with Matlab: beg-ind : stride : end-ind
5. Syntax of Array Section real :: a(10), b(10,10), c(4,4), d(4,4) a = 1 ! Store 1 in all elements of a a(1:10:3) = 2 ! Store 2 in 1st, 4th, 7th and 10th elements of a c = b(1:10:3,1:10:3) ! Needs to be conformable d = sqrt( b(1:10:3,1:10:3) ) ! Element wise operations Note: Array sections are rectangular
6. Syntax of Array Section a = 1 ! Store 1 in all elements of a a(1:10:3) = 2 ! Store 2 in 1st, 4th, 7th and 10th elements of a • May appear in both left and right hand side of an assignment • But needs to be conformable
7. Syntax of Array Section a(1:7:3) = a(2:8:3) Before assignment After assignment
8. Ref: array/ex1.f90, array/ex2.f90 • Notes: • The syntax: • a = (/1,2,3,4,5,6,7,8,9,10/) • Means assign 1 through 10 into consecutive 10 locations of array a • The I/O notation: • print '(“a(:) “,10(i3))', a • Means print the character string “a(:) “ and then keep room for 10 integer locations each with 3 character wide.
9. Multi-dimensional array Column major storage order in 2D
10. Output from array/ex3.f90 Complete array a a( 1,:) 1. 11. 21. 31. 41. 51. 61. 71. 81. 91. a( 2,:) 2. 12. 22. 32. 42. 52. 62. 72. 82. 92. a( 3,:) 3. 13. 23. 33. 43. 53. 63. 73. 83. 93. a( 4,:) 4. 14. 24. 34. 44. 54. 64. 74. 84. 94. a( 5,:) 5. 15. 25. 35. 45. 55. 65. 75. 85. 95. a( 6,:) 6. 16. 26. 36. 46. 56. 66. 76. 86. 96. a( 7,:) 7. 17. 27. 37. 47. 57. 67. 77. 87. 97. a( 8,:) 8. 18. 28. 38. 48. 58. 68. 78. 88. 98. a( 9,:) 9. 19. 29. 39. 49. 59. 69. 79. 89. 99. a(10,:) 10. 20. 30. 40. 50. 60. 70. 80. 90. 100. Array section a(1:m,1:m) b( 1,:) 1. 11. 21. 31. b( 2,:) 2. 12. 22. 32. b( 3,:) 3. 13. 23. 33. b( 4,:) 4. 14. 24. 34. Array section a(m:2*m-1,m:2*m-1) b( 1,:) 34. 44. 54. 64. b( 2,:) 35. 45. 55. 65. b( 3,:) 36. 46. 56. 66. b( 4,:) 37. 47. 57. 67.
11. Output from array/ex4.f90 Complete array a a( 1,:) 1. 11. 21. 31. 41. 51. 61. 71. 81. 91. a( 2,:) 2. 12. 22. 32. 42. 52. 62. 72. 82. 92. a( 3,:) 3. 13. 23. 33. 43. 53. 63. 73. 83. 93. a( 4,:) 4. 14. 24. 34. 44. 54. 64. 74. 84. 94. a( 5,:) 5. 15. 25. 35. 45. 55. 65. 75. 85. 95. a( 6,:) 6. 16. 26. 36. 46. 56. 66. 76. 86. 96. a( 7,:) 7. 17. 27. 37. 47. 57. 67. 77. 87. 97. a( 8,:) 8. 18. 28. 38. 48. 58. 68. 78. 88. 98. a( 9,:) 9. 19. 29. 39. 49. 59. 69. 79. 89. 99. a(10,:) 10. 20. 30. 40. 50. 60. 70. 80. 90. 100. Array section a(1:n:3,1:n:3) b( 1,:) 1. 31. 61. 91. b( 2,:) 4. 34. 64. 94. b( 3,:) 7. 37. 67. 97. b( 4,:) 10. 40. 70. 100.
12. Vector array index Array index may also be explicitly enumerated in a vector real :: a(10), b(3) b = a( (/1, 7, 2/) ) ! 1st, 7th and 2nd element ! of a goes to 3 ! consecutive elements of b Ref: array/ex5.f90
13. Output from array/ex5.f90 a( 1,:) 1. 11. 21. 31. 41. 51. 61. 71. 81. 91. a( 2,:) 2. 12. 22. 32. 42. 52. 62. 72. 82. 92. a( 3,:) 3. 13. 23. 33. 43. 53. 63. 73. 83. 93. a( 4,:) 4. 14. 24. 34. 44. 54. 64. 74. 84. 94. a( 5,:) 5. 15. 25. 35. 45. 55. 65. 75. 85. 95. a( 6,:) 6. 16. 26. 36. 46. 56. 66. 76. 86. 96. a( 7,:) 7. 17. 27. 37. 47. 57. 67. 77. 87. 97. a( 8,:) 8. 18. 28. 38. 48. 58. 68. 78. 88. 98. a( 9,:) 9. 19. 29. 39. 49. 59. 69. 79. 89. 99. a(10,:) 10. 20. 30. 40. 50. 60. 70. 80. 90. 100. a( (/1,4,9/), (/5,7,3/) ) 41. 61. 21. 44. 64. 24. 49. 69. 29. l=( 10 5 2 ), m=( 3 8 ), a(l,m) 30. 80. 25. 75. 22. 72. Note: Order of indices may be non-monotonic too
14. Forall construct forall ( indx-triplets, .. , condition ) <body> end forall • condition may involve more than one index (possible to manage non-rectangular blocks) • logically all the operations on body are simultaneous • compilers may choose to parallelize the body • Ref: • array/ex6.f90 Note: Contrast this with identical body of operations under do-loops
15. where construct where ( mask-array/logical-array-condition ) <body> end where • Elemental operation, condition applies to one element • mask array and the target array could be the same • logically simultaneous, hence may be parallelizable • Refs: • array/ex7.f90
16. Functions and Subroutines • To modularize the operations for easier • code management and • code organization • like most high Level language Fortran also has sub program functionality with well defined interfaces and scoping. function structure function <name> ( arguments,.. ) <declarations..> <body> end function call syntax x = myfunction( y, z )
17. Subroutine structure subroutine <subroutine-name>( arguments,.. ) <declarations> <body of subroutine> end subroutine <subroutine-name> Subroutine call syntax call mysubroutine( x, y, .. ) • Note: • Default interface is arguments, • Functions return a value while subroutine does not (only difference) • Unless a single return value is needed fortran programmers prefer subroutine, we will follow this convention • Refs: subp/ex[1-3].f90
18. closer look at Subprogram structure • Subprogram name needs to be unique within the scoping unit (whole program before we introduce module!) • Arguments and return values are dummy variables i.e. passed by references (or actual memory locations are defined by the caller
19. Jacobi Iteration (i+1,j) (i,j) (i,j-1) (i,j+1) (i-1,j)
20. closer look at Subprogram structure • subprogram unit are local (or type automatic, has meaningful existence within a single call and not between calls by default) • Multi-dimensional array definition needs to be either assumed shape or explicit • Ref: • subp/ex4.f90 subp/ex5.f90
21. Internal Subprogram • Internal subprograms (as the name suggests) are subprogram blocks inside a subprogram • are visible only from the container subprogram • Inherits all the variables and states from caller • May define local variables • If there is a name collision with parent the local name becomes a distinct variable (careful!)
22. Internal Subprogram syntax subroutine xyz(a, b,.. ) <body> contains subroutine pqr(..) <body> end subroutine end subroutine Ref: subp/ex6.f90
23. Fortran90 Pointers • points to an object, typically array locations • F90 pointer has an associated type, rank and strides etc. • often called a descriptor in contrast with Cray or C pointers • May point to NULL, or even to an illegal memory location (beware!) • Standard 2003 allows pointers to a function too • Every locations allocated in F90 will have an associated type, rank and stride (contrast with C and Cray pointers in Fortran)
24. Fortran90 Pointers syntax real, pointer :: a, b(:,:), c(:,:) allocate(b(10,10)) ! 10 x 10 real memory ! associated c => b ! c points to the same ! locations as b c => b(1:n:2,1:n:2) ! c points to an array ! section of b Ref: pointers/ex1.f90
25. Notes on F90 pointers • As we have seen memory can be allocated against pointers or • Pointers may point to an already allocated locations • Pointers once legally defined is valid reference just like any other normal variables or references in Fortran • Pointers may point to array sections defined using triplets but • Vector indices are not allowed (why?) • It is not legal to pass NULL pointer across sub programs
26. Differences with Allocatable attribute • Memory may be allocated both against variables with allocatable attributes and pointers but • Allocatable attributes have more stringent runtime check of garbage collections • Pointers may point to a previously allocated location • F95 specification requires deallocation of automatic variables
27. Passing Array-sections across subroutines • It is important to realize while passing array-sections or pointers defined using array sections with explicit interface i.e. following F77 style the compiler will have to create temporary copy to allow conformance. • It is recommended to use assumed shape arrays to avoid overhead due to copy and copy-back and also the potential problems with asynchronous programming e.g. asynchronous MPI communications. • Ref: • pointers/internal/ex[1-3].f90
28. Modules • Container of data and subprogram units • Allows access to its contents in different subprogram units through use statements • Allows fine grain access through private/public clause • Equivalent functions of common blocks in F77 but with many more features
29. Modules syntax module <name> <declarations> contains function xyz(a, b,..) <..> end function subroutine pqr(p,..) end subroutine end module
30. Modules as a simple container of data • Data defined under a module are real locations (as against dummy locations) • By use statement program units refer to these locations. • Similar to common blocks in F77 but needs to be make accessible to subprograms by use statement as against defined as it was done with common blocks, so less error prone Ref: modules/ex1.f90
31. Modules, data hiding • Not all the data is needed by all the subprograms • So why not subprograms take only that is needed ? • Rationale: • Less possibility of erroneous reference or avoid namespace collisions • Syntax: • use mymodule : only x, y, z Ref: modules/ex2.f90
32. Modules, data hiding, from module side • Module itself dictates what to expose • Rationale: • Single point of control • Even less errorprone • Syntax: • private <data>, procedure • public data, procedure Ref: modules/ex3.f90
33. Modules procedures • Modules may contain procedures • These procedures may be the only interfaces to act on data it contains • Rationale: • Forcing user to act on data in a standard form • Starting point of Object Oriented Methodologies Ref: modules/ex4.f90
34. Notes on Module data and procedures • Modules may use other modules in a chain of dependencies but cyclic use is illegal • Module variables and function/subprograms may be renamed while using to avoid name collision • The only clause may allow few selective variables to be used • Module data is of type automatic by default, so it is important to add save attribute (but it is always implemented as static)
35. Notes on Module data and procedures • Module procedure arguments are always checked for type and rank between caller and callee during compilation • Module procedures inherit all the module variables and has access to all other procedures within the scope of the parent module, it may explicitly include (use) others as well. Ref: modules/ex[5-6].f90
36. Operators between user defined data types • Recall that the only defined operator for user defined data type is “=“ • One or more operators may be created to operate between two or more objects of user defined type • The allowed syntax for user defined operator is • .xyz. • where xyz is any convenient alphabetic (not numeric) string.
37. Syntax for defining operator interface operator (.xyz.) function .. subroutine .. module procedure .. end interface Ref: operator/ex1.f90
38. Intrinsic functions • These are elementary mathematical and other utility functions to operate on basic data types or query on program environment • All the F77 intrinsic functions are supported and few newer intrinsic functions are added • The list is long, some of the common ones are sqrt, sin, cos, tan, asin, acos, atan, atan2, exp, sinh, max, min etc.
39. Intrinsic functions • Example of newly added intrinsic functions are date_and_time, random_number, maxloc, minloc etc. • Complex data type is intrinsically supported in Fortran, so corresponding complex version of these functions (if applicable) are also supported. • Fortran Language reference manual which comes with the compiler is the best place for exhaustive list and almost always google works most efficiently after a little experience
40. Intrinsic Functions and Overloading Many of the intrinsic functions operate on variables with different precisions, rank etc. Example: xsqrt = sqrt( x ) The returned value that is stored in xsqrtwill be of same type and rank as x. Compiler recognizes the data types and redirects the call to appropriate functions for correct return values Ref: overload/ex1.f90
41. Generic interface and overload • Procedures may be overloaded through a generic interface blocks • Presence of argument(s), type and rank associates appropriate call at compile time Basic syntax: interface <name> module procedure <xyz> module procedure <abc> end interface Note: If interface location does not have access to full arguments then those need to be described using dummy arguments. Ref: overload/ex2.f90
42. Overloading operators Combination of operations of defining operator and procedure overload In example overload/ex3.f90 the intrinsic operator * is overloaded for product between 2 matrices and A matrix and a vector
43. Sequential Unformatted I/O • Unformatted I/O preserves the image of variables as existing in memory • Typically needed to write restart files Syntax to open files open(11,file=‘data.dat’,form=‘unformatted’) To write write(11)x, y, z ! Each of these may be ! array of basic or user ! defined types To read read(11)x,y,z
44. Sequential Unformatted I/O Note: Often these data files are called binary files and machine readable only Ref: io/ex2.f90 (to write) io/ex3.f90 (to read)
45. Integer / string conversion • Write integer to a character string • Typically used to manipulate file names • Ref: • io/ex4.f90 • Read integer off a character string • Typically used to manipulate arguments to program • Ref: • io/ex5.f90
46. Fortran 2008 standard • Big thing is Co-array fortran (CAF) • sub modules, breaking module into several subs (one module does not get too big, big relief for WRF devs!) • concurrent do loops (equiv. omp parallel do) • Polymorphic allocate (almost equiv C cast) • Max rank to 15 • newunit finds an I/O unit dynamically • Few more intrinsics Bessel etc.
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# Can complex (Variance) analysis be done in Einstein Analytics?
## Background
Every month we copy revenue data in our from their original objects (complex normalized tree of related objects) into something we call a snapshot. A denormalized flat table (Custom object) that makes reporting easier.
We then do something quite complex on it - a custom form of Variance Analysis (https://www.accountingtools.com/articles/what-is-variance-analysis.html).
We now think of moving this into Einstein Analytics (EA). Not only because storing all the denormalized data in custom objects is expensive, but also because Apex and Force.com seems a bad reporting platform compared to Wave/Einstein Analytics.
## Question
The idea is to store those snapshot directly as datasets inside of EA. And then
to do ALL of the Variance Analytics. Have you build something complex like this in EA? Or can you point me to more elaborate examples of SAQL usage?
We build unique versions of these calculations regularly. Custom Dataflow Snapshots, and some saql to calculated variance over the periods can be implemented.
These would not be considered complex use cases, but I believe quite standard.
Here is a query example.
``````q = load "Opportunity_Snapshot";
result = group q by ('Snapshot_Date_Year', 'Snapshot_Date_Month');
result = foreach result generate q.'Snapshot_Date_Year' + "~~~" + q.'Snapshot_Date_Month' as 'Snapshot_Date_Year~~~Snapshot_Date_Month', sum(q.'Amount') as 'A';
result = group result by 'Snapshot_Date_Year~~~Snapshot_Date_Month';
result = foreach result generate 'Snapshot_Date_Year~~~Snapshot_Date_Month', sum(A) as 'A', sum(A) - sum(sum(A)) over ([-1..-1] partition by all order by ('Snapshot_Date_Year~~~Snapshot_Date_Month')) as 'B';
result = order result by ('Snapshot_Date_Year~~~Snapshot_Date_Month' asc);
result = limit result 2000;
``````
yeah true, by implementing with EA you can get this snapshot of data everyday, allowing you to improve business. Big advantage is you can schedule the Dataflow run and get the results next min you walk into office.
there are many functions which you can use to calculate the variance and project the data in different trends or groupings.
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Visualizing and propagating uncertainty in PCA
# VIPurPCA
VIPurPCA offers a visualization of uncertainty propagated through the dimensionality reduction technique Principal Component Analysis (PCA) by automatic differentiation.
### Installation
VIPurPCA requires Python 3.7.3 or later and can be installed via:
pip install vipurpca
A website showing results and animations can be found here.
### Usage
#### Propagating uncertainty through PCA and visualize output uncertainty as animated scatter plot
In order to propagate uncertainty through PCA the class PCA can be used, which has the following parameters, attributes, and methods:
Parameters
matrix : array_like
Array of size [n, p] containing mean numbers to which VIPurPCA should be applied.
n_components : int or float, default=None, optional
Number of components to keep.
axis : {0, 1} , default=0, optional
The default expects samples in rows and features in columns.
cov_data : array_like of shape [np] or [np, n*p] , default=None, optional
Uncertainties attached to the numbers in matrix. If cov_data is one-dimensional it is assumend to be the diagonal of a diagonal matrix. If None
compute_jacobian : Boolean, default=False, optional
Whether or whether not to propagate uncertainty through PCA.
Attributes
size : [n, p]
Dimension of matrix (n: number of samples, p: number of dimensions)
covariance : ndarray of size [p, p]
Features' covariance matrix.
eigenvalues : ndarray of size [n_components]
Eigenvalues obtained from eigenvalue decomposition of the covariance matrix.
eigenvectors : ndarray of size [n_componentsp, np]
Eigenvectors obtained from eigenvalue decomposition of the covariance matrix.
jacobian : ndarray of size [n_componentsp, np]
Jacobian containing derivatives of eigenvectors w.r.t. input matrix.
jacobian_eigenvalues : ndarray of size [n_componentsp, np]
Jacobian containing derivatives of eigenvalues w.r.t. input matrix.
cov_eigenvectors : ndarray of size [n_componentsp, n_componentsp]
Propagated uncertainties of eigenvectors.
cov_eigenvalues : ndarray of size [n_components*n_components]
Propagaged uncertainties of eigenvalues.
transformed data : ndarray of size [n, n_components]
Low dimensional representation of data after applying PCA.
Methods
pca_value() Apply PCA to the matrix.
pca_grad(center=True) Apply PCA to the matrix and compute the jacobian and jacobian_eigenvalues using automatic differentiation.
transform_data() Transform matrix according to eigenvectors and reduce dimensionality according to n_components.
compute_cov_eigenvectors() Compute uncertainties of eigenvectors.
compute_cov_eigenvalues() Compute uncertainties of eigenvalues.
animate(n_frames=10, labels=None, outfile='animation.html') Generate animation with n_frames number of frames with plotly. labels (list, 1d array) indicate labelling of individual samples. Save animation (as html) at outfile.
#### Example datasets
Three example datasets can be loaded after installing VIPurPCA providing mean, covariance and labels.
from vipurpca import load_data
#### Example
from vipurpca import load_data
from vipurpca import PCA
# load mean (Y), uncertainty estimates (cov_Y) and labels (y)
pca_student_grades = PCA(matrix=Y, cov_data=cov_Y, n_components=2, axis=0, compute_jacobian=True)
# compute PCA with backprop
# Bayesian inference
# Transform data
The resulting animation can be found here here.
## Project details
Uploaded source
Uploaded py3
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## Thursday, February 4, 2016
### Strength of Schedule for Carolina and Denver
Strength of Schedule for Carolina and Denver
The Carolina Panthers had one of the great regular season records of all time. However, their regular season schedule appeared to be a bit weak compared to other teams including their Super Bowl Opponent the Denver Broncos. Let’s test the hypothesis that the Panthers had a weak regular season schedule.
Question: The table below has regular-season won/loss records for the opponent of the Carolina Panthers and the Denver Broncos.
Is the total win percentage of the opposition to Carolina significantly different from the total win percentage of the opposition to the Denver Broncos?
Discuss the limitations of this test and other things a statistician might do to assess the strength of schedule for the two Super Bowl Teams.
Strength of Schedule Carolina Denver Opposition Wins Losses Wins Losses Jacksonville 5 11 Baltimore 5 11 Houston 9 7 Kansas City 10 5 New Orleans 7 9 Detroit 7 9 Tampa Bay 6 10 Minnesota 11 5 Seattle 10 6 Oakland 7 9 Philadelphia 7 9 Cleveland 3 13 Indianapolis 8 8 Green Bay 10 6 Green Bay 10 6 Indianapolis 8 8 Tennessee 3 13 Kansas City 10 5 Washington 9 7 Chicago 6 10 Dallas 4 12 New England 12 4 New Orleans 7 9 San Diego 4 12 Atlanta 8 8 Oakland 7 9 New York Giants 6 10 Pittsburgh 10 6 Atlanta 8 8 Cincinnati 12 4 Tampa Bay 6 10 San Diego 4 12 Total 113 143 Total 126 128
Analysis: The aggregate win/loss records for the regular-season opponents of the two Super Bowl teams are presented in the table below.
Total Regular Season Wins and Losses for Opponents of Carolina and Denver Carolina Opposition Denver Opposition Total Wins 113 126 239 Losses 143 128 271 Win Proportion 0.4414 0.4961 0.4686
The difference in the won-loss record for the two teams appears non-trivial but is it significant?
The answer to this question is No!
t-test on Difference in Regular Win Percentages for the Opponents of the Two Super Bowl Teams Diff -0.0272 SE 0.022096746 t-stat -1.232 Critical Value for .05 significance level -2.3
Discussion: If we added playoff games of the two teams to the sample we would increase sample size and decrease standard error by around 15%. This would make the test results a bit closer but still not significant.
I also very quickly looked at median wins for the opponents of the two teams. Found it was 7.0 for Carolina and 7.5 for Denver. This seems trivial.
Carolinas schedule does seem a bit weak but their record was impressive and you can only beat the teams that you play.
I do sense that Denver is the more tested team having played and beaten New England twice once in the regular season and the second time in the AFC championship game. The counter argument is that Carolina beat Seattle twice but I Seattle had some hiccups this year.
I don’t believe Carolina has played against a defense as accomplished as Denver’s defense. It might be useful to study defensive strength of the two teams opponents and attempt to determine whether Carolina can beat a truly great defense.
I don’t expect Denver to score much but I think Denver’s defense will also keep the score low.
Go Back to the home-field advantage hypothesis testing problems
http://www.dailymathproblem.com/p/home-field-hypothesis-testing-problems.html
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# Distance between Lafayette, LA (LFT) and Watertown, SD (ATY)
Flight distance from Lafayette to Watertown (Lafayette Regional Airport – Watertown Regional Airport) is 1053 miles / 1694 kilometers / 915 nautical miles. Estimated flight time is 2 hours 29 minutes.
Driving distance from Lafayette (LFT) to Watertown (ATY) is 1235 miles / 1987 kilometers and travel time by car is about 21 hours 13 minutes.
## Map of flight path and driving directions from Lafayette to Watertown.
Shortest flight path between Lafayette Regional Airport (LFT) and Watertown Regional Airport (ATY).
## How far is Watertown from Lafayette?
There are several ways to calculate distances between Lafayette and Watertown. Here are two common methods:
Vincenty's formula (applied above)
• 1052.723 miles
• 1694.193 kilometers
• 914.791 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 1054.355 miles
• 1696.819 kilometers
• 916.209 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Lafayette Regional Airport
City: Lafayette, LA
Country: United States
IATA Code: LFT
ICAO Code: KLFT
Coordinates: 30°12′19″N, 91°59′15″W
B Watertown Regional Airport
City: Watertown, SD
Country: United States
IATA Code: ATY
ICAO Code: KATY
Coordinates: 44°54′50″N, 97°9′16″W
## Time difference and current local times
There is no time difference between Lafayette and Watertown.
CST
CST
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 154 kg (340 pounds).
## Frequent Flyer Miles Calculator
Lafayette (LFT) → Watertown (ATY).
Distance:
1053
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
1053
Round trip?
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Here are several examples of situations in which Charles’ Law is at play: If you take a basketball outside on a cold day, the ball shrinks a bit as the temperature is decreased. This is also the case with any inflated object and explains why it’s a good idea to check your car’s tire pressure when the temperature drops.
Also, What law is P1V1 P2V2?
The relationship for Boyle’s Law can be expressed as follows: P1V1 = P2V2, where P1 and V1 are the initial pressure and volume values, and P2 and V2 are the values of the pressure and volume of the gas after change.
Hereof, What are the 5 gas laws?
Gas Laws: Boyle’s Law, Charle’s Law, Gay-Lussac’s Law, Avogadro’s Law.
Also to know How does Charles law affect the human body? Charles law effect on the human body: When cold air is inhaled by the human body when it passes through the respiratory tract, it gets warmer, and the volume of air is changed. The warm air expands and increases the volume.
What does Charles law state?
The physical principle known as Charles’ law states that the volume of a gas equals a constant value multiplied by its temperature as measured on the Kelvin scale (zero Kelvin corresponds to -273.15 degrees Celsius).
## What are the 3 gas laws?
The gas laws consist of three primary laws: Charles’ Law, Boyle’s Law and Avogadro’s Law (all of which will later combine into the General Gas Equation and Ideal Gas Law).
## What are the 3 laws of gas?
The gas laws consist of three primary laws: Charles’ Law, Boyle’s Law and Avogadro’s Law (all of which will later combine into the General Gas Equation and Ideal Gas Law).
## What does Boyles law state?
This empirical relation, formulated by the physicist Robert Boyle in 1662, states that the pressure (p) of a given quantity of gas varies inversely with its volume (v) at constant temperature; i.e., in equation form, pv = k, a constant. …
## What are the laws of gas?
Gas laws, laws that relate the pressure, volume, and temperature of a gas. … These two laws can be combined to form the ideal gas law, a single generalization of the behaviour of gases known as an equation of state, PV = nRT, where n is the number of gram-moles of a gas and R is called the universal gas constant.
## Why is the Charles law important?
Charles’s law, a statement that the volume occupied by a fixed amount of gas is directly proportional to its absolute temperature, if the pressure remains constant. … It is a special case of the general gas law and can be derived from the kinetic theory of gases under the assumption of a perfect (ideal) gas.
## Which example best demonstrates Charles’s law?
Explanation: The Charles law states that the volume of an ideal gas increases when temperature is increased under constant pressure. The pressure inside the balloon is always equal to the atmospheric pressure. Therefore answer A demonstrate the Charles law.
## How does Charles Law relate to hot air balloons?
Charles’ Law in Everyday Life
In order to make a hot air balloon rise, heat is added to the air inside the balloon. … When the density of the balloon decreases to be less than the density of the outside air, the balloon rises. Conversely, the volume of a gas will shrink if its temperature decreases.
## What is Charles Law in simple terms?
Charles’s law, a statement that the volume occupied by a fixed amount of gas is directly proportional to its absolute temperature, if the pressure remains constant. … It is a special case of the general gas law and can be derived from the kinetic theory of gases under the assumption of a perfect (ideal) gas.
## What are the 6 gas laws?
Gas Laws: Boyle’s Law, Charle’s Law, Gay-Lussac’s Law, Avogadro’s Law.
## How do I find my gas laws?
Notice the only gas law with moles or mass in it as a variable, is Ideal Gas Law. Remind ourselves that Ideal Gas Law is PV=nRT. If you’re not given moles or mass, or not asked to calculate Moles or Mass, do not use the Ideal Gas Law.
## How are gas laws calculated?
To find any of these values, simply enter the other ones into the ideal gas law calculator. For example, if you want to calculate the volume of 40 moles of a gas under a pressure of 1013 hPa and at a temperature of 250 K, the result will be equal to: V = nRT/p = 40 * 8.3144598 * 250 / 101300 = 0.82 m³ .
## What is the pressure law?
Gay-Lussac’s law, Amontons’ law or the pressure law was found by Joseph Louis Gay-Lussac in 1808. It states that, for a given mass and constant volume of an ideal gas, the pressure exerted on the sides of its container is directly proportional to its absolute temperature.
## What does Avogadro’s law state?
Avogadro’s law, a statement that under the same conditions of temperature and pressure, equal volumes of different gases contain an equal number of molecules. … The law is approximately valid for real gases at sufficiently low pressures and high temperatures.
## Is Charles Law direct or inverse?
The volume of a given gas sample is directly proportional to its absolute temperature at constant pressure (Charles’s law). The volume of a given amount of gas is inversely proportional to its pressure when temperature is held constant (Boyle’s law).
## How does the Charles law work?
Charles’ Law is an experimental gas law that describes how gases tend to expand when heated. The law states that if a quantity of gas is held at a constant pressure, there is a direct relationship between its volume and the temperature, as measured in degrees Kelvin. Think of it this way.
## Where do you apply Charles Law?
Top 6 Applications Of Charles Law
• Hot Air Balloon.
• Bursting Of A Deodorant.
• Bakery Products.
• Turkey Pop Up Timer.
• Opening Of A Soda Can.
• Helium Balloon On Cold Day.
## What are examples of Boyles Law?
An example of Boyle’s law in action can be seen in a balloon. Air is blown into the balloon; the pressure of that air pushes on the rubber, making the balloon expand. If one end of the balloon is squeezed, making the volume smaller, the pressure inside increased, making the un-squeezed part of the balloon expand out.
## How is Charles law used in real life?
Charles Law application in real life can be seen in our kitchen too. In order to make bread and cakes soft and spongy, yeast is used for fermentation. Yeast produces carbon dioxide gas. When bread and cakes are baked at high temperatures; with an increase in temperature, carbon dioxide gas expands.
## Can Crusher experiment Charles Law?
What this equation means is that when the air inside the can cooled, the volume decreased, causing the can to implode. … When the pressure inside the can decreased, the much higher air pressure outside the can pushed in the sides of the can, causing it to implode.
Tagged in:
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1. ## Math/Physics
Using two formulas to find 2 unknowns called system of equations ? if so, how would you solve f=ma if m and a are unknown ?
2. ## Re: Math/Physics
Originally Posted by Cujo10
Using two formulas to find 2 unknowns called system of equations ? if so, how would you solve f=ma if m and a are unknown ?
What do you mean by solve? What are you talking about?
3. ## Re: Math/Physics
Just say M=10 and A is the unknown you can solve it for F but when M and A are unknown how to you solve it. My friend and physics told me that it can be solves for the unknowns using systems of equations, but i never learnt that. For F=MA to solve for two unknown you need to use two other equation M=F/A and A=F/M its called systems equations i think.
4. ## Re: Math/Physics
I think he might be referring to what we would call 'simultaneous equations'. Are these in the general maths syllabus? Surely they are.
5. ## Re: Math/Physics
That is the one i meant, i got the name mixed up but how would you still solve the missing variable problem using simultaneous equations ?
6. ## Re: Math/Physics
Why don't you just post an actual question of this type?
7. ## Re: Math/Physics
With simultaneous equations and multiple unknowns, you need several pieces of information. Generally you need one equation for each unknown quantity. So if you have two unknowns you need two equations.
And you can't "cheat" by rearranging the first equation and using that as the second, because when you try to solve it you will end up getting an equation like $1 = 1$ or $x = x$ that is as useless as it is true.
For example the equation $xy = 6$ has infinitely many possible solutions for $x$ and $y$, so you need more information to go off.
Suppose you are then given $x - y = 1$.
So now you have two equations:
$\begin{cases} xy&=6 \\ x - y &= 1 \end{cases}$
We can rearrange the second equation, giving us $x = 1 + y$ and then we can substitute this back into the first equation, giving:
$y(1+y) = 6$
And from there it is possible to solve for $y$. Once you know $y$ you can then solve for $x$.
__________________________________________________ __________________________________________________ __
In some more common cases it may be better to add the equations together:
$\begin{cases} 2x+y&=4 \\ x - y &= 2 \end{cases}$
$3x = 6 \implies x = 2$
Or subtract them:
$\begin{cases} x+2y&=8 \\ x - y &= 2 \end{cases}$
$3y = 6 \implies y = 2$
But ultimately what you're trying to do is to focus on one variable by creating an equation without all the others. Once you have that variable you can use it to find the other(s).
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# f(x) = -x^4 + 4x^3 + ax^2 + bx + 9 This function has a factor of x-1 This function when divided by x+4 has a remainder that is five times larger than the remainder when divided by x+5. Find out the...
f(x) = -x^4 + 4x^3 + ax^2 + bx + 9
This function has a factor of x-1
This function when divided by x+4 has a remainder that is five times larger than the remainder when divided by x+5.
Find out the values of a and b.
supersmartpants | (Level 1) Salutatorian
Posted on
Sorry, this function when divided by x+4 has a remainder that is five times larger than the remainder when divided by x+2
aruv | High School Teacher | (Level 2) Valedictorian
Posted on
f(x) = -x^4 + 4x^3 + ax^2 + bx + 9
This function has a factor of x-1
This function when divided by x+4 has a remainder that is five times larger than the remainder when divided by x+2.
Find out the values of a and b.
The f(x) has factor x-1. Therefore f(1)=0
f(1)=-1+4+a+b+9=0
a+b=-12 (i)
The remainder when f(x) divided by x+4 is f(-4).
f(-4)=-256-256+16a-4b+9
=16a-4b-503
The remainder when f(x) divided by x+2 is f(-2).
f(-2)=-16-16+4a-2b+9
=4a-2b-23
But f(-4)=5 f(-2)
16a-4b-503= 5(4a-2b-23)
16a-4b-20a+10b=-115+503
-4a+6b=388 (ii)
To solve (i) and (ii), multiply (i) by 4 and add to (ii), we have
b=34
Substitute b=34 in (i), we have
a=-46
Thus a=-46 and b=34.
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Excel 2016 How To Determine What A Formula Is Doing How to Calculate Nutrition Data Using Excel or Open Office Calc
You are searching about Excel 2016 How To Determine What A Formula Is Doing, today we will share with you article about Excel 2016 How To Determine What A Formula Is Doing was compiled and edited by our team from many sources on the internet. Hope this article on the topic Excel 2016 How To Determine What A Formula Is Doing is useful to you.
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How to Calculate Nutrition Data Using Excel or Open Office Calc
EU Directive 1169/2011 comes into full force on 13 December 2016. The first phase of this directive entered into force on 13 December 2014 but the second requires nutritional data which raises the question of how to calculate the nutritional data.
The first step of this rule requires all ingredients on labels to include allergen information within the ingredients list. Before this regulation, it was legally acceptable to include allergy information in a separate area of your label.
The new regulations require not only highlighting allergens in a single ingredient list for a product but also stating the ingredients in quantitative order.
Quantitative order means that the largest component component should be indicated first, then the second largest, and so on. The percentage of these ingredients should also be included.
There are many ways that highlighting materials can be achieved; Users can use bold text, underline text, color text or italic text
There are 14 allergens that must be indicated on the labeling if they are within the product’s ingredients. This includes wheat or oats or any other grain containing gluten and also includes milk, eggs, fish, crustaceans, molluscs.
Another aspect of the law was to improve the clarity of text on food labels.
Historically, text can be incredibly difficult to read as manufacturers cram as much information as possible into a small section of the label in order to maximize the marketing potential of the rest of the label.
The new rules require that all text be legible in a font smaller than 1.2 millimeters with the specified height of the letter “x”. In layman’s terms, this means the standard Arial or Times New Roman font needs to be 6.5 points and size.
The second phase of the rules, which will take effect this December, will require providing nutrition data with all prepackaged foods so that consumers can make informed choices regarding the nutrition within the foods they purchase.
The law states that this information must be conveyed to the customer per 100 grams.
It is also possible to provide additional information on each serving, for example, which sandwich constitutes a serving so that the food manufacturer can provide information based on the entire sandwich. A food manufacturer may also indicate the nutritional values of a portion, for example, of a biscuit or a small piece of chocolate. But the food manufacturer must also provide the information in a per 100g format in all cases.
How to calculate nutritional data
Calculating the nutritional values of pre-packaged foods for sale to public food production businesses requires knowing the nutritional values for the constituent ingredients within their products. Perhaps the best way to show how to calculate nutritional data is to give an example; A ham and mustard sandwich.
A ham and mustard sandwich can have four ingredients; We will have bread, ham, mustard and margarine or butter to make sandwiches. Each of these ingredients will be included in the recipes; That is, each product will have a specific weight to make a standard product.
Food manufacturers need to start with basic data for nutrition for each ingredient – as mentioned, the law requires providing nutritional data per 100 grams. Since all manufacturers are required to do this, most food manufacturing companies should be able to obtain that information directly from the packaging of the products they purchase or by talking to their suppliers.
In our example, a food manufacturer can move data from component ingredients to a table. Information to be reported includes energy in both kilojoules and kilocalories; They should also express total fat, saturated fat, carbohydrates, sugar, protein, and salt—all in grams.
Food manufacturers can also indicate monounsaturated fat, polyunsaturated fat, polyols and starches (which are carbohydrates) and fiber if they choose to do so.
The order of nutrients is specific and must follow the rules.
After the table of data is prepared for all the ingredients per 100 grams, the food producer needs to understand the weight of each product used in the recipe to make the sandwich. In this example, the food producer needs to know the weight of two slices of bread (say 60 grams), the ham he will use (say 30 grams), 10 grams of mustard, 5 grams of margarine.
Once this is done a simple calculation is applied to each component ingredient to determine how many calories, how much fat, saturated fat, etc. are in the recipe. The calculation is to divide the nutrition data per 100 grams by 100 and multiply that by the weight of the ingredient in the ingredient.
For example, if 100 grams of ham is 350 calories, dividing by 100 is 3.5 calories per gram. 3.5 calories per gram x 30 grams used in the recipe is 105 calories.
Once this is completed, the food manufacturer will have an accurate indication of the total nutritional data for the ham and mustard sandwich by adding the values for each ingredient together as a total for the recipe.
And this is how to calculate nutritional data using Microsoft Excel or OpenOffice Calc.
At the moment, food producers across the UK are facing huge challenges in achieving the objectives set out in the regulations and need to address them very quickly if they haven’t already.
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# SUMIf Formula
#### DCT01
##### New Member
I would like to write a formula that sums cells in column A if there is data in both column B and column C - please help.
### Excel Facts
Can a formula spear through sheets?
Use =SUM(January:December!E7) to sum E7 on all of the sheets from January through December
#### Aladin Akyurek
##### MrExcel MVP
=SUMPRODUCT(\$A\$2:\$A\$100,--(\$B\$2:\$B\$100<>""),--(\$C\$2:\$C\$100<>""))
#### PATSYS
##### Well-known Member
Assuming your data is from A1:C10
=Sum(if(B1:B10<>"",if(C1:C10<>"",A1:A10)))
confirm with CTRL+SHIFT+ENTER
*REVISED*
#### DCT01
##### New Member
Thank you - Aladin's formula worked. Patsys I could not get your to work.
#### PATSYS
##### Well-known Member
It is an array formula that should give you results exactly just like that of Aladin's. Remember not just to press ENTER, you should press CTRL+SHIFT+ENTER.
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Document related concepts
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Transcript
```Phy 212: General Physics II
Chapter 14: Fluids
Lecture Notes
Daniel Bernoulli (1700-1782)
• Swiss merchant, doctor & mathematician
• Worked on:
– Vibrating strings
– Ocean tides
– Kinetic theory
• Demonstrated that as the velocity of a fluid
increases its pressure decreases (known as
Bernoulli’s Principle)
What are fluids?
Fluids are substances that continually deform (flow) under an applied shear
stress regardless of how small the applied stress. All liquids and all
gases are fluids.
Useful Physical Quantities:
1. Mass Density (r) is mass per unit volume or
Dm
m
r =
for a uniform substance : r =
DV
V
The SI units of mass density are kg/m3
2. Pressure is the magnitude of the normal force (DFN) applied to a surface
per unit area (DA), or
DFN
FN
P=
For a uniformly applied force : P =
DA
A
The SI units of pressure are newtons/meters2 (N/m2) or pascals (Pa)
Notes:
a. For static fluids (at rest), force can only be applied normal to the surface
b. Tangential (or shear) forces result in fluid flow!
Pressure in Fluids
1.
Fluids are mobile
–
–
2.
Fluids do not retain their shape
Fluids do not independently support their shape
The pressure within a fluid varies with depth
–
–
Pressure within a fluid is constant at all points of the same depth
Pressure increases with increasing depth (H)
P~H
{where H is the depth below the surface of the fluid}
3.
The pressure difference (DP) between 2 points within a fluid is related to
the density of the fluid (r) & the depth difference between the points
–
–
The greater the fluid density (r), the greater DP
The greater the depth difference (DH), the greater DP
P2 – P1 = rgDH
Derivation of pressures in fluids
Consider a disk of water in a static fluid:
A
F1=P1A
DH
1. Applying Newton’s 2nd Law to the disk:
or
FNet = P2 A ˆj - P1A ˆj - mg ˆj= 0
P2A - P1A = mg
mg
F2=P2A
2. The mass of the disk is: m = rADH so Newton’s 2nd Law can be
rewritten as
P2A – P1A = mg = (rADH)g or
P2 – P1 = rgDH
3. Note that pressure in a fluid does not depend on:
•
•
The total mass of the fluid disk
The surface area where the force is applied
Atmospheric Pressure
1.
2.
If the surface of a fluid is exposed to air, the pressure at that surface
(Po) is equal to that of the atmospheric air pressure at that elevation,
this is a consequence of Pascal’s Principle.
At sea level the atmospheric pressure is
Po = 1.01 x 105 N/m2
3.
The pressure at any depth (H) below the surface of the fluid is given
by:
P = Po + rgH
4.
This relationship also applies to air pressure at higher altitudes (to the
extent that the air density does not change substantially)
How high can you suck water up using a straw? (or
pumping water uphill)
In order to draw water (or any fluid) upward you must lower
the pressure difference between the pressure inside the
“straw” and the outside environment (usually the
atmosphere).
1. Is there a limit to how low you can
make the pressure inside a straw?
2. How high can you suck water?
Barometers
Barometers are devices used to measure air (or atmospheric)
pressure
•Types of barometers:
Closed end
Open end
Pascal’s Principle
When pressure is applied to an enclosed fluid, the pressure is
transmitted undiminished to all parts of the fluid and the enclosing
walls
1. The effect of applied pressure is to increase
the Po for the fluid.
2. The application of hydraulics is based on
Pascal’s Principle:
P2 = P1 + rgH
or
F2
F1
=
+ r gH
A2
A1
F1
Therefore F2 = A2
+ rgH
A1
Archimedes’ Principle
When an object is submerged in a fluid, the fluid exerts a buoyant force
(FB) on the object equal to the weight of the fluid displaced by the
object:
FB = mfluidg = rfluidVdisplacedg
Notes:
1. When a submerged object has greater density than the fluid, its
weight will be greater than FB (it will sink)
2. When a submerged object has lower density than the fluid, it will
partially submerge in the fluid until it reaches a depth where its
weight will be the same as FB (it will float)
3. When the submerged object has the same density as the fluid it
will neither sink nor float but will remain in equilibrium as long as it
is completely submerged
Equation of Continuity
When a fluid flows, the rate of mass entering a point must be equal to the
rate of mass exiting the point:
Or (since
Dm1
Dm2
=
Dt
& Dt
)
Dm = rDV = r ADx Dx = v
r1A1v1 = r2A2v2Dt
Where:
1. A is the cross-sectional area of the fluid segment
2.
Dx is the width of the fluid segment
Bernoulli’s Equation
Bernoulli’s Principle: as speed of fluid increases its pressure drops
Bernoulli’s principle can be extended to Bernoulli’s equation for a (ideal)
fluid that is:
1.
2.
Nonviscous (no friction or interaction with sidewalls)
Incompressible (liquid does not expand nor compress)
Bernoulli’s Equation:
P1 + rgy1 + ½ rv12 = P2 + rgy2 + ½ rv22
Which can be rewritten as
P2 - P1 = - rgy1 - ½ rv12 + rgy2 + ½ rv22
Or:
P2 - P1 = rg(y1 - y2) + ½ r(v12 -v22)
Note: the “state” of the fluid is constant at all points:
P1 + rgy1 + ½ rv12 = P2 + rgy2 + ½ rv22 = constant
```
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# One of the most difficult #friendlywaves I’ve ever gotten! Did I get it right?
Florian sent me a #friendlywave — a wave picture he took, with hopes that I might be able to explain what is going on there. And this one had me puzzled for some time!
This is what the picture looks like:
What I knew about it: Florian was on the ferry from Wisschafen to Glückstadt, crossing the Elbe river.
In the picture itself, there are several features that jumped at me. First, drawn in with the lightblue line below: A sand bank parallel(-ish) to the island’s coast line.
Then, the ship’s wake (shown in red) breaking right near the ship (orange) and turning (green) and breaking (yellow) where it runs on the sand bank.
Florian wrote he was watching the ferry’s wake and noticed something curious: There seemed to be a shallow part, where the waves suddenly became a lot faster! And could I explain what was going on?
Looking at the picture, there were two possibilities for what he might have meant (and, spoiler alert — I completely jumped on the wrong one first!).
Below, I’ve drawn in the part of the wake that is running on the shallow sand bank (green) and how those wave crests continue on the other side of the sand bank (red). I’ve also drawn in some mystery wave crests in blue. Those were the ones I chose to focus on first, since Florian had written that he noticed waves behaving weirdly and suddenly becoming much faster. So if we are talking fast, we are talking really fast, right?
So how do we explain those blue wave crests?
There is a limit for the maximum speed a wave can have. That limit depends on the wave’s wave length: The longer a wave, the faster it travels. In deep water, i.e. water deeper than 1/2 the wavelength, the wave travels at this maximum speed (see green lines in the plot below).
But as it comes into shallower water, it gets slowed down (see black lines in the plot below — those are just a quick sketch, there are complicated equations to calculate it exactly).
In shallow water, i.e. water that is smaller than 1/20th of the wave length, the phase speed only depends on water depth: The shallower the water, the more the wave is being slowed down (see the red lines in the plot below).
Sorry about the quality of the sketch — I don’t have Matlab or anything else useful on the computer I have available right now, so I drew this in ppt! Take it with a pinch of salt, but qualitatively it’s correct!
So looking back at Florian’s picture, for the blue waves to have been caused by Florian’s ferry, there are two options:
A) they would have to have wave speeds faster than the ferry’s bow wave and wake
B) the ferry would have had to come from the direction of the island, so that the waves propagated in that deeper channel behind the sand bank before the ferry made its way around the sandbank.
Option A is impossible, because wakes travel at maximum wave speed (similar to a sonic boom in the atmosphere, where sound is travelling at maximum speed, forming a cone with the air plane at its tip, only here it’s a 2D version, a V-shaped wake with the ship at its tip). So if the wake is traveling at maximum speed already, then the blue waves can’t go faster than that.
For option B, looking Florian’s ferry up on a map, I saw that that ferry goes around a small island, which is the land you see in his picture. But a quick glance at the map shows that even though the sand bank seems to end where the ship would have had to have gone in order to create those waves, the island is still very much in the way. So this can’t be the solution, either.
This map is published at http://map.openseamap.org under a CC BY-SA 2.0 license
So let’s take another good look at the original picture.
Remember those wave crests that I marked in blue? Well, upon closer inspection it turns out that they are tidal gullys and not wave crests! (Which is what Florian confirmed when I asked whether he remembered the situation) Guess I have been barking up the wrong tree all this time!
So back to the wave crests that I marked in red:
What we see here is exactly the depth dependence of the phase speed that I plotted above. Right at the sand bank, the water is shallowest and waves are slowed down (we see that both in the green wave crests that seem to be falling back and start breaking as they get closer to the sand bank [both indicating that the water is getting shallower], and in the red wave crests right at the sand bank). But as the water gets deeper again on the far side of the sand bank (which depth measurements in the map above seem to confirm), the phase speed picks up again (as it has to — see my plot above) and the wave crests accelerate again. Hence we have the weird phenomenon of waves suddenly speeding up!
Very long explanation, I know, but still pretty cool now that we solved it, right? I love #friendlywaves — if you have any mystery wave pictures, please do send them my way! :-)
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# Is Integer x prime? 1. x^2-3 is an even number 2. x+2 is an
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Joined: 22 Dec 2009
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Is Integer x prime? 1. x^2-3 is an even number 2. x+2 is an [#permalink] 07 Mar 2010, 09:23
00:00
Difficulty:
5% (low)
Question Stats:
100% (02:33) correct 0% (00:00) wrong based on 6 sessions
Is Integer x prime?
1. x^2-3 is an even number
2. x+2 is an odd number
_________________
Cheers!
JT...........
If u like my post..... payback in Kudos!!
|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice|
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~~Better Burn Out... Than Fade Away~~
Last edited by jeeteshsingh on 07 Mar 2010, 10:01, edited 1 time in total.
Senior Manager
Joined: 22 Dec 2009
Posts: 366
Followers: 9
Kudos [?]: 167 [0], given: 47
Re: Is integer x prime? [#permalink] 07 Mar 2010, 10:02
kp1811 wrote:
jeeteshsingh wrote:
Is Integer x prime?
1. x^2-3 is an even number
2. x+2 is an odd number
E...
Made the question more clear for you...!
_________________
Cheers!
JT...........
If u like my post..... payback in Kudos!!
|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice|
|For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|
~~Better Burn Out... Than Fade Away~~
Senior Manager
Joined: 30 Aug 2009
Posts: 293
Location: India
Concentration: General Management
Followers: 2
Kudos [?]: 75 [0], given: 5
Re: Is integer x prime? [#permalink] 07 Mar 2010, 10:08
jeeteshsingh wrote:
kp1811 wrote:
jeeteshsingh wrote:
Is Integer x prime?
1. x^2-3 is an even number
2. x+2 is an odd number
E...
Made the question more clear for you...!
will try to make it more clearer.....
stmnt1) x^2 - 3 is even
let x = 3 (prime) then 3^2 - 3 = 6 even
let x =9 (non prime) then 9^2 - 3 = 78 even
hence insuff
stmnt2) x+2 is odd
let x = 3 (prime) then 3 + 2 = 5 odd
let x =9 (non prime) then 9 + 2 = 11 odd
hence insuff
even together they don't suffice. Hence E
Senior Manager
Joined: 22 Dec 2009
Posts: 366
Followers: 9
Kudos [?]: 167 [0], given: 47
Re: Is integer x prime? [#permalink] 07 Mar 2010, 10:26
kp1811 wrote:
jeeteshsingh wrote:
will try to make it more clearer.....
stmnt1) x^2 - 3 is even
let x = 3 (prime) then 3^2 - 3 = 6 even
let x =9 (non prime) then 9^2 - 3 = 78 even
hence insuff
stmnt2) x+2 is odd
let x = 3 (prime) then 3 + 2 = 5 odd
let x =9 (non prime) then 9 + 2 = 11 odd
hence insuff
even together they don't suffice. Hence E
Some how I dont get this approach as you use the ques to prove the statement below. This question is from PR 1012 and I see the same solution there which isnt convincing for me.
My approach is as follows:
Given x is an integer.
Ques is x prime?
S1: x^2 - 3 = even
x^2 - 3 = 2m where m is an integer
x = sqrt(2m + 3) where m is >= 0 as you cannot have - ve sqrt.
This gives x = \sqrt{3},\sqrt{5},\sqrt{7},3,\sqrt{11},\sqrt{13},\sqrt{15},\sqrt{17},\sqrt{19},\sqrt{21},\sqrt{23},5,\sqrt{27},\sqrt{29},.....
Since it is given that x is an integer we get only 3, 5, 7..... which are all prime. Hence SUFF.
S2: x + 2 is odd
x + 2 = 2k + 1 where k is an integer
x = 2k - 1
Therefore x = ....-7,-5,-3,-1,1,3,5,7,9,11... which means all odd numbers and hence not necessarily be prime. Therefore NOT SUFF.
Hence answer is A....!
Can someone highlight what is wrong in my approach!
_________________
Cheers!
JT...........
If u like my post..... payback in Kudos!!
|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice|
|For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|
~~Better Burn Out... Than Fade Away~~
Senior Manager
Joined: 30 Aug 2009
Posts: 293
Location: India
Concentration: General Management
Followers: 2
Kudos [?]: 75 [1] , given: 5
Re: Is integer x prime? [#permalink] 07 Mar 2010, 10:34
1
KUDOS
jeeteshsingh wrote:
kp1811 wrote:
jeeteshsingh wrote:
will try to make it more clearer.....
stmnt1) x^2 - 3 is even
let x = 3 (prime) then 3^2 - 3 = 6 even
let x =9 (non prime) then 9^2 - 3 = 78 even
hence insuff
stmnt2) x+2 is odd
let x = 3 (prime) then 3 + 2 = 5 odd
let x =9 (non prime) then 9 + 2 = 11 odd
hence insuff
even together they don't suffice. Hence E
Some how I dont get this approach as you use the ques to prove the statement below. This question is from PR 1012 and I see the same solution there which isnt convincing for me.
My approach is as follows:
Given x is an integer.
Ques is x prime?
S1: x^2 - 3 = even
x^2 - 3 = 2m where m is an integer
x = sqrt(2m + 3) where m is >= 0 as you cannot have - ve sqrt.
This gives x = \sqrt{3},\sqrt{5},\sqrt{7},3,\sqrt{11},\sqrt{13},\sqrt{15},\sqrt{17},\sqrt{19},\sqrt{21},\sqrt{23},5,\sqrt{27},\sqrt{29},.....
Since it is given that x is an integer we get only 3, 5, 7..... which are all prime. Hence SUFF.
S2: x + 2 is odd
x + 2 = 2k + 1 where k is an integer
x = 2k - 1
Therefore x = ....-7,-5,-3,-1,1,3,5,7,9,11... which means all odd numbers and hence not necessarily be prime. Therefore NOT SUFF.
Hence answer is A....!
Can someone highlight what is wrong in my approach!
in highlighted part after 3,5, 7 .... 9 will come [ for m = 39] which is non prime
Senior Manager
Joined: 22 Dec 2009
Posts: 366
Followers: 9
Kudos [?]: 167 [0], given: 47
Re: Is integer x prime? [#permalink] 07 Mar 2010, 10:39
kp1811 wrote:
jeeteshsingh wrote:
Some how I dont get this approach as you use the ques to prove the statement below. This question is from PR 1012 and I see the same solution there which isnt convincing for me.
My approach is as follows:
Given x is an integer.
Ques is x prime?
S1: x^2 - 3 = even
x^2 - 3 = 2m where m is an integer
x = sqrt(2m + 3) where m is >= 0 as you cannot have - ve sqrt.
This gives x = \sqrt{3},\sqrt{5},\sqrt{7},3,\sqrt{11},\sqrt{13},\sqrt{15},\sqrt{17},\sqrt{19},\sqrt{21},\sqrt{23},5,\sqrt{27},\sqrt{29},.....
Since it is given that x is an integer we get only 3, 5, 7..... which are all prime. Hence SUFF.
S2: x + 2 is odd
x + 2 = 2k + 1 where k is an integer
x = 2k - 1
Therefore x = ....-7,-5,-3,-1,1,3,5,7,9,11... which means all odd numbers and hence not necessarily be prime. Therefore NOT SUFF.
Hence answer is A....!
Can someone highlight what is wrong in my approach!
in highlighted part after 3,5, 7 .... 9 will come [ for m = 39] which is non prime
Thanks mate! Got it Now! Kudos +1
_________________
Cheers!
JT...........
If u like my post..... payback in Kudos!!
|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice|
|For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|
~~Better Burn Out... Than Fade Away~~
Re: Is integer x prime? [#permalink] 07 Mar 2010, 10:39
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# What Predetermined Overhead Rate Is Formula and Sample
This predetermined rate was based on a cost formula that estimated \$\\$ 257,400\$ of total manufacturing overhead cost for an estimated activity level of 11,000 direct labor-hours. The most common allocation base used in manufacturing is machine hours, direct labor hours, or direct materials cost. For example, if the allocation base is machine hours, estimate the total number of machine hours for the period. Further, the company uses direct labor hours to assign manufacturing overhead costs to products. As per the budget, the company will require 150,000 direct labor hours during the forthcoming year.
The concept of predetermined overhead rate is very important because it is used most of the enterprises as it enables them to estimate the approximate total cost of each job. Larger organizations employ different allocation bases for determining the predetermined overhead rate in each production department. However, in recent years the manufacturing operations have started to use machine hours more predominantly as the allocation base. A predetermined overhead rate is defined as the ratio of manufacturing overhead costs to the total units of allocation. A single plantwide factory overhead rate is used to allocate overhead costs to products.
Plantwide Overhead Rate Method Divide your total expenses for the plant by the total number of units you produce. Using the plantwide overhead rate formula, if expenses come to \$10,000 for instance and you produce 2,500 units, \$10,000 divided by 2,500 equals four. However, one major disadvantage of the method break even analysis is that both the numerator and the denominator are estimates and as such, it is possible that the actual result may vary significantly from the predetermined overhead rate. Departmental overhead rates are needed because different processes are involved in production that take place in different departments.
## What is a plant-wide overhead rate?
Company B wants a predetermined rate for a new product that it will be launching soon. Its production department comes up with the details of how much the overheads will be and what other costs will be incurred. Therefore, the predetermined overhead rate of GHJ Ltd for next year is expected to be \$5,000 per machine hour. Therefore, the predetermined overhead rate of TYC Ltd for the upcoming year is expected to be \$320 per hour. The production hasn’t taken place and is completely based on forecasts or previous accounting records, and the actual overheads incurred could turn out to be way different than the estimate.
• With \$2.00 of overhead per direct hour, the Solo product is estimated to have \$700,000 of overhead applied.
• This method of allocation is simple and easy to use, making it popular among small businesses with homogeneous product lines.
• While it may become more complex to have different rates for each department, it is still considered more accurate and helpful because the level of efficiency and precision increases.
• The most common allocation base used in manufacturing is machine hours, direct labor hours, or direct materials cost.
A single overhead rate for assigning all of the manufacturing production and service department costs to products. This rate is less accurate than departmental rates if a company manufactures a diverse group of products. In these situations, a direct cost (labor) has been replaced by an overhead cost (e.g., depreciation on equipment). Because of this decrease in reliance on labor and/or changes in the types of production complexity and methods, the traditional method of overhead allocation becomes less effective in certain production environments.
However, it has limitations, such as inaccurate product costs, which can result in mispricing and lost profits. Therefore, companies should consider more sophisticated methods of allocating overhead costs, such as activity-based costing, for more accurate and reliable cost information. A predetermined overhead rate is an allocation rate given for indirect manufacturing costs that are involved in the production of a product (or several products). The estimate is made at the beginning of an accounting period, before the commencement of any projects or specific jobs for which the rate is needed. Until now, you have learned to apply overhead to production based on a predetermined overhead rate typically using an activity base. An activity base is considered to be a primary driver of overhead costs, and traditionally, direct labor hours or machine hours were used for it.
Different businesses have different ways of costing; some use the single rate, others use multiple rates, and the rest use activity-based costing. The predetermined overhead rate calculation shown in the example above is known as the single predetermined overhead rate or plant-wide overhead rate. The allocation base (also known as the activity base or activity driver) can differ depending on the nature of the costs involved. From the above list, salaries of floor managers, factory rent, depreciation and property tax form part of manufacturing overhead. Small companies typically use activity-based costing, while large organizations will have departments that compute their own rates. After reviewing the product cost and consulting with the marketing department, the sales prices were set.
## Step 1 of 3
The common allocation bases are direct labor hours, direct labor cost, machine hours, and direct materials. However, if the company manufactures diverse products, some of which use expensive equipment while some use only inexpensive equipment, or the company wants precise costs for pricing decisions, a plant-wide rate is not appropriate. In response to this situation, manufacturers will use departmental overhead rates and perhaps activity based costing.
## Which of the following is the formula to calculate the predetermined factory overhead rate?
The sales price, cost of each product, and resulting gross profit are shown in Figure 6.6. The monthly accounting close process for a nonprofit organization involves a series of steps to ensure accurate and up-to-date financial records. Unexpected expenses can be a result of a big difference between actual and estimated overheads. Also, if the rates determined are nowhere close to being accurate, the decisions based on those rates will be inaccurate, too. Harold Averkamp (CPA, MBA) has worked as a university accounting instructor, accountant, and consultant for more than 25 years.
In other words, it is the total amount of factory overhead costs divided by the total amount of the allocation base. Let us take the example of ort GHJ Ltd which has prepared the budget for next year. The company estimates a gross profit of \$100 million on total estimated revenue of \$250 million. As per the budget, direct labor cost and raw material cost for the period is expected to be \$40 million and \$60 million respectively. Calculate the predetermined overhead rate of GHJ Ltd if the required machine hours for next year’s production is estimated to be 10,000 hours. As you’ve learned, understanding the cost needed to manufacture a product is critical to making many management decisions (Figure 6.2).
## Formula for Predetermined Overhead Rate
A plant-wide overhead rate is often a single rate per hour or a percentage of some cost that is used to allocate or assign a company’s manufacturing overhead costs to the goods produced. So, for every hour of direct labor used to produce widgets and gizmos, XYZ Inc. will allocate \$50 of manufacturing overhead costs. The term “predetermined overhead rate” refers to the allocation rate that is assigned to products or job orders at the beginning of a project based on the estimated cost of manufacturing overhead for a specific period of reporting. To calculate a predetermined overhead rate, divide the manufacturing overhead cost by the units of allocation. There are concerns that the rate may not be accurate, as it is based on estimates rather than actual data.
Knowing the total and component costs of the product is necessary for price setting and for measuring the efficiency and effectiveness of the organization. Remember that product costs consist of direct materials, direct labor, and manufacturing overhead. A company’s manufacturing overhead costs are all costs other than direct material, direct labor, or selling and administrative costs. Once a company has determined the overhead, it must establish how to allocate the cost. This allocation can come in the form of the traditional overhead allocation method or activity-based costing..
## How Do You Calculate Single Plantwide Factory Overhead Rate?
For example, a production facility that is fairly labor intensive would likely determine that the more labor hours worked, the higher the overhead will be. As a result, management would likely view labor hours as the activity base when applying overhead costs. Enter the total manufacturing overhead cost and the estimated units of the allocation base for the period to determine the overhead rate. For example, the total direct labor hours estimated for the solo product is 350,000 direct labor hours. With \$2.00 of overhead per direct hour, the Solo product is estimated to have \$700,000 of overhead applied. When the \$700,000 of overhead applied is divided by the estimated production of 140,000 units of the Solo product, the estimated overhead per product for the Solo product is \$5.00 per unit.
A predetermined overhead rate, also known as a plant-wide overhead rate, is a calculation used to determine how much of the total manufacturing overhead cost will be attributed to each unit of product manufactured. The rate is determined by dividing the fixed overhead cost by the estimated number of direct labor hours. A predetermined overhead rate is calculated at the start of the accounting period by dividing the estimated manufacturing overhead by the estimated activity base. The predetermined overhead rate is then applied to production to facilitate determining a standard cost for a product. XYZ Inc. estimates that its total manufacturing overhead costs for the upcoming year will be \$500,000.
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# Showing that the roots of an equation are always real and evaluating positive indices
I am stuck on finding the roots, I cannot seem to work around this with $$\lambda$$ can someone explain the derivation for this?
If the roots of the equation $$x^2 + bx + c = 0$$ are $$\alpha$$, $$\beta$$ and the roots of the equation $$x^2 + \lambda bx + \lambda^2 c = 0$$ are $$\gamma, \delta$$ show that the equation whose roots are $$\alpha \gamma + \beta \delta$$ and $$\alpha \delta + \beta \gamma$$ is:
$$x^2 - \lambda b^2x+ 2\lambda^2 c(b^2-2c)=0$$
Show that the roots of this equation are always real.
How is the following equation derived?
Express with positive indices $$\frac{2b^-3x^2}{7c^-4y^2} = \frac{2x^2c^4}{7b^3y^2}$$
• I'm not sure but is there a typo here? Isn't $\alpha\lambda+\beta\delta$ be $\alpha\gamma+\beta\delta$? – Habagat Maliksi Dec 5 '19 at 3:46
The roots of $$x^2 + \lambda b x + \lambda^2 c$$ are $$\lambda \alpha$$ and $$\lambda \beta$$. I take it you mean $$\alpha \gamma + \beta \delta$$, not $$\alpha \lambda + \beta \delta$$. Making the substitution for $$\gamma$$ and $$\delta$$, we are looking to show that the roots of the equation are $$\lambda(\alpha^2 + \beta^2)$$ and $$2\lambda\alpha\beta$$.
It suffices to show that $$\alpha^2 + \beta^2$$ and $$2\alpha \beta$$ are roots of $$x^2 - b^2x + 2c (b^2 - 2c) = (x - 2c)(x - (b^2 - 2c))$$
We know that $$c = \alpha \beta$$ and $$b = -\alpha - \beta$$, since $$(x - \alpha) (x - \beta) = x^2 + b x + c$$. So $$2c = 2 \alpha \beta$$ and $$b^2 - 2c = \alpha^2 + \beta^2$$.
Hint: $$x^2+bx+c=(x-\alpha)(x-\beta)$$ and $$x^2 + \lambda bx + \lambda^2 c = (x-\delta)(x-\gamma)$$. So, we have the following: (1) $$\alpha+\beta=-b$$, (2) $$\alpha\beta=c$$, (3) $$\delta+\gamma=-\lambda b$$ and (4) $$\delta\gamma=\lambda^2c$$.
Then consider the sum $$(\alpha \lambda + \beta \delta)+(\alpha \delta + \beta \gamma)$$ and the product $$(\alpha \lambda + \beta \delta)(\alpha \delta + \beta \gamma)$$, and try to express these in terms of the four equations I mentioned above.
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# interesting interpretation | Economics homework help
Use the data in CPS78_85.RAW for this exercise.
(i) How do you interpret the coefficient on y85 in equation (13.2)? Does it have an interesting interpretation? (Be careful here; you must account for the interaction terms y85_educ and y85_female.)
(ii) Holding other factors fixed, what is the estimated percent increase in nominal wage for a male with twelve years of education? Propose a regression to obtain a confidence interval for this estimate.
(iii) Estimate equation (13.2) but let all wages be measured in 1978 dollars. In particular, define the real wage as rage _ wage for 1978 and as rage _ wage/1.65 for 1985. Now use log(rage) in place of log(wage) in estimating (13.2). Which coefficients differ from those in equation (13.2)?
(iv) Explain why the R-squared from your regression in part (iii) is not the same as in equation (13.2). of squared residuals, from the two regressions are identical.) (v) Describe how union participation has changed from 1978 to 1985.
(vi) Starting with equation (13.2), test whether the union wage differential has changed over time. (This should be a simple t test.) (vii) Do your findings in parts (v) and (vi) conflict?
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# Homework Help: Structures & Materials - Cranes and Steel Cable
1. Jun 22, 2009
### codz30
1. The problem statement, all variables and given/known data
A crane consists of a crane arm which is held in place and moved by strong steel cables.
The crane arm is a truss structure, which is a rigid framework of quite thin steel beams welded together, forming many triangles.
The cables are made of steel and they are not one piece but consist of many thin strands of steel plaited together, in a way similar to most ropes. The cables carry very heavy loads but fortunatley they are almost perfectly elastic.
i) What is one reason that a truss structure like this is nearly as strong as a solid beam of about the same size?
ii) Why is the cable made of thin strands plaited into one thick cable stronger than a cable made of solid steel?
iii) Explain the meaning of 'elastic' with reference to the cable.
3. The attempt at a solution
i) The shape of the structurem using triangles, spreads out the load. Triangles are strong structures.
ii) Because the steel cables share the load better.
iii) It means the cables can spring back into shape.
I don't really think my explainations are correct, but it's the best I can think of.
2. Jun 22, 2009
### LowlyPion
What about the cross sectional area for a beam? What about the weight of trussed lattice versus say a solid beam?
http://en.wikipedia.org/wiki/Beam_(structure [Broken])
Elastic suggests that it stretches uniformly depending on load doesn't it? Like a spring?
Last edited by a moderator: May 4, 2017
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*EDIT* INTERPRET! lol....sorry!
Write a recursive function array_sum(int *array, int size, int index) that computes the sum of all the elements of the array starting with position "index".
Okay thats not too bad..but then he said this
for example, if an array A holds the values 5,4,3,2,1 then the call array_sum(A,5,2) will return the value 6=3+2+1 for it will sum up the values starting at position 2, that is 3 and 2 and 1.
This is what most confused me...
any tips on what the function will look like?
he said that the array needs to be filled with random numbers..i know that one at least...
``````void fillArray(int a[], int size)
{
for (int i=0; i<size; i++)
a[i]=rand()%10;
// heres my lil random number generator woot woot!
}``````
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Last Post by VernonDozier
Recursive usually means that the function calls itself (some say recursive when they mean iterative - which is quite different).
So for your problem, the following (incomplete) code shows recursion.
``````int recurseAdd(int*A, int N, int i){
// ...
return A[i] + recurseAdd(A, N, i+1); // add current element to the next element
//...
}``````
Note that you're not allowed to try to access A where i >= N (i.e. outside the defined array), so when i >= N, you'll wan't to stop recursing. Do a google on recursion, there should be heaps of info available.
I know what recursion is sir...though thank you for your help!!! i just dont understand what output he wants...or what type of function he wants
I know what recursion is sir...though thank you for your help!!! i just dont understand what output he wants...or what type of function he wants
*EDIT* INTERPRET! lol....sorry!
Write a recursive function array_sum(int *array, int size, int index) that computes the sum of all the elements of the array starting with position "index".
Okay thats not too bad..but then he said this
for example, if an array A holds the values 5,4,3,2,1 then the call array_sum(A,5,2) will return the value 6=3+2+1 for it will sum up the values starting at position 2, that is 3 and 2 and 1.
This is what most confused me...
any tips on what the function will look like?
he said that the array needs to be filled with random numbers..i know that one at least...
array_sum(A, 5, 2) needs to add the elements of A from index 2 through index 5 - 1, or 4. So you're adding A[2] through A[4]. The 4 in A[4] is the array size minus 1. The 2 is the index you are starting the cumulative sum from. So you are adding this:
A[2] + A[3] + A[4] = 3 + 2 + 1 = 6
Putting parentheses around it, you are doing this:
A[2] + (A[3] + (A[4]))
You're actually going to call this function 4 times:
array_sum (A, 5, 2)
array_sum (A, 5, 3)
array_sum (A, 5, 4)
array_sum (A, 5, 5)
array_sum (A, 5, 5) needs to return 0 since you're going past the array bounds here. It'll be your "base case", where index >= size.
So use dougy's example and add a base case situation to it. The "base case" needs to occur BEFORE the return line he mentions because, as he says, you're not allowed to access A if i >= size of the array.
``````int recurseAdd(int*A, int N, int i){
// ...
return A[i] + recurseAdd(A, N, i+1); // add current element to the next element
//...
}``````
Note that you're not allowed to try to access A where i >= N (i.e. outside the defined array), so when i >= N, you'll wan't to stop recursing.
This topic has been dead for over six months. Start a new discussion instead.
Have something to contribute to this discussion? Please be thoughtful, detailed and courteous, and be sure to adhere to our posting rules.
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# Conversation between Obama and Netanyahu
I took a contest entry written by Seth Brown of Williamstown, MA in 2003, and I converted it into this puzzle.
Disclaimer: Events described in this puzzle are fictional.
A White House stenographer transcribed the following conversation between Barack Obama and Benjamin Netanyahu.
Obama: Who is on first?
Netanyahu: Me?
Obama: No, the guy playing first base.
Netanyahu: Me?
Obama: You're the first baseman?
Netanyahu: No, I am asking you. Me?
Obama: Who?
Netanyahu: You mean that man right there?
Obama: He's on first?
Netanyahu: What do you mean? There aren't any females on the team.
Obama: Well then who's on first?
Netanyahu: Me?
Assume that nobody on the team is named "Who", "Me", or "He". With that assumption, the conversation doesn't seem to make any sense. But it actually does make sense. Explain how.
• Who's on first! I love that thing Commented Dec 15, 2014 at 16:02
• @d'alar'cop - Seconded. +1. Commented Dec 15, 2014 at 17:38
• I've been lurking this site for a bit now, I finally made an account just to upvote this. Thank you. Commented Dec 15, 2014 at 20:36
• I downvoted this because most of the puzzle is simply in knowing the right information. Also, the setup honestly doesn't make much sense. It's contrived nature works for a joke, not for a puzzle where you're supposed to explain why people are doing what they are.
– xnor
Commented Dec 17, 2014 at 6:07
• @manshu et al. Since the question (especially the tags) are being edited, and this puzzle has been solved, should we add language or trivia while we're at it? Commented May 20, 2016 at 2:28
## 2 Answers
The miscommunication is because...
Netanyahu is speaking Hebrew. (Although I don't know why - his English is very good and he spent some of his childhood years in the US).
Specifically:
Me = מי (Hebrew for 'who' or 'whom')
He = היא (Hebrew for 'she' or 'her')
Who = הוא (Hebrew for 'he' or 'him')
The Conversation, as heard by Netanyahu:
Obama: He is on first.
Netanyahu: Who?
Obama: The guy playing first base
Netanyahu: Who?
Obama: You're the first baseman?
Netanyahu: No, I am asking you. who?
Obama: Him!
Netanyahu: You mean that man right there?
Obama: She's on first.
Netanyahu: What do you mean? There aren't any females on the team.
Obama: Well then he's on first?
Netanyahu: Who?
And of course, major props to pacoverflow for dusting off one of the best comedy skits of all time.
Language barrier.
Who is vocalised as "hu", "hu" in hebrew means "he".
- http://en.wiktionary.org/wiki/%D7%94%D7%95%D7%90
He is vocalised as "hi", "hi" in hebrew means "she" (wacky, right?).
- http://en.wiktionary.org/wiki/%D7%94%D7%99%D7%90
Me is vocalised as "mi", "mi" in hebrew means "who".
- http://en.wiktionary.org/wiki/%D7%9E%D7%99
With this in mind, the conversation makes sense.
• The third point is incorrect. (I'm sorry for making such a blunt and non-specific answer, but I'm trying not to give a spoiler in the comments). Commented Dec 15, 2014 at 18:16
• @Bachrach44 Heh, edited it just as you answered. Commented Dec 15, 2014 at 18:27
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# Ifcosec √5;find the value of cot cos (where theta is acute
2
by bakkulu
do you mean cosec(theta)= root 5?
yes
how do you want thw answer? using properties or by basics?
formula
2014-11-02T17:12:12+05:30
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Cosec Ф = √5
sin Ф = 1/cosec Ф = 1/√5
we know that sin² Ф + cos² Ф = 1
⇒cos² Ф = 1 - sin² Ф
⇒cos Ф = √(1 - sin² Ф)
⇒cos Ф =
⇒cos Ф =
cot Ф = cos Ф / sin Ф =
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# Statistical Decision Theory: Concepts, Methods and Applications
164 Pages · 2005 · 1.85 MB · English
Statistical Decision Theory: Concepts, Methods and Applications (Special topics in Probabilistic Graphical Models) FIRST COMPLETE DRAFT November 30, 2003
Statistical Decision Theory: Concepts, Methods and Applications (Special topics in Probabilistic Graphical Models) FIRST COMPLETE DRAFT November 30, 2003 Supervisor: Professor J. Rosenthal STA4000Y Anjali Mazumder 950116380 Part I: Decision Theory – Concepts and Methods 1 Part I: DECISION THEORY - Concepts and Methods Decision theory as the name would imply is concerned with the process of making decisions. The extension to statistical decision theory includes decision making in the presence of statistical knowledge which provides some information where there is uncertainty. The elements of decision theory are quite logical and even perhaps intuitive. The classical approach to decision theory facilitates the use of sample information in making inferences about the unknown quantities. Other relevant information includes that of the possible consequences which is quantified by loss and the prior information which arises from statistical investigation. The use of Bayesian analysis in statistical decision theory is natural. Their unification provides a foundational framework for building and solving decision problems. The basic ideas of decision theory and of decision theoretic methods lend themselves to a variety of applications and computational and analytic advances. This initial part of the report introduces the basic elements in (statistical) decision theory and reviews some of the basic concepts of both frequentist statistics and Bayesian analysis. This provides a foundational framework for developing the structure of decision problems. The second section presents the main concepts and key methods involved in decision theory. The last section of Part I extends this to statistical decision theory – that is, decision problems with some statistical knowledge about the unknown quantities. This provides a comprehensive overview of the decision theoretic framework. Part I:
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Cyber Green. Improving Cyber Health through. Measurement and Mitigation. Yurie Ito. Director. JPCERT Coordination Center. Japan Computer Emergency Response Team. Coordination Center. 電子署名者 : Japan Computer Emergency Response Team. Coordination Center. DN : c=JP, st=Tokyo,
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Section 3 describes the historical pathogen prevalence data we collected from atlases of infectious disease, the . In fact, preliminary analysis suggests that the quantitative effects of adding more use diff germ std, pronoun and english as instruments and add the following variables, one-by-one,.
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ASTRONOMY. AND. ASTROPHYSICS. Extended cellular automaton models of solar flare occurrence. K.P. Macpherson. 1 and A.L. MacKinnon. 2. 1. Department of Physics (Theoretical Physics), Oxford University, 1 Keble Road, Oxford OX1 3NP, UK ([email protected]). 2. Department of Adult and
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1 Centre de Recherche de l'Institut du Cerveau et de la Moelle Epinière, INSERM UMRS 975 . essential, as was shown in associative learning (Sulzer et al., 2013a) brain activation was trainable with proper and not sham feed- processing, the neural variable is presented to the subject with a.
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10 years of blogging !
Hi all ! This post is to celebrate 10 years of blogging with an average of 1 post per month. It was a quite cool adventure !
Here is the full table of contents for those interested:
A sane introduction to maximum likelihood estimation (MLE) and maximum a posteriori (MAP)
It is frustrating to learn about principles such as maximum likelihood estimation (MLE), maximum a posteriori (MAP) and Bayesian inference in general. The main reason behind this difficulty, in my opinion, is that many tutorials assume previous knowledge, use implicit or inconsistent notation, or are even addressing a completely different concept, thus overloading these principles.
Those aforementioned issues make it very confusing for newcomers to understand these concepts, and I’m often confronted by people who were unfortunately misled by many tutorials. For that reason, I decided to write a sane introduction to these concepts and elaborate more on their relationships and hidden interactions while trying to explain every step of formulations. I hope to bring something new to help people understand these principles.
Maximum Likelihood Estimation
The maximum likelihood estimation is a method or principle used to estimate the parameter or parameters of a model given observation or observations. Maximum likelihood estimation is also abbreviated as MLE, and it is also known as the method of maximum likelihood. From this name, you probably already understood that this principle works by maximizing the likelihood, therefore, the key to understand the maximum likelihood estimation is to first understand what is a likelihood and why someone would want to maximize it in order to estimate model parameters.
Let’s start with the definition of the likelihood function for continuous case:
$$\mathcal{L}(\theta | x) = p_{\theta}(x)$$
The left term means “the likelihood of the parameters $$\theta$$, given data $$x$$”. Now, what does that mean ? It means that in the continuous case, the likelihood of the model $$p_{\theta}(x)$$ with the parametrization $$\theta$$ and data $$x$$ is the probability density function (pdf) of the model with that particular parametrization.
Although this is the most used likelihood representation, you should pay attention that the notation $$\mathcal{L}(\cdot | \cdot)$$ in this case doesn’t mean the same as the conditional notation, so be careful with this overload, because it is always implicitly stated and it is also often a source of confusion. Another representation of the likelihood that is often used is $$\mathcal{L}(x; \theta)$$, which is better in the sense that it makes it clear that it’s not a conditional, however, it makes it look like the likelihood is a function of the data and not of the parameters.
The model $$p_{\theta}(x)$$ can be any distribution, and to make things concrete, let’s say that we are assuming that the data generating distribution is an univariate Gaussian distribution, which we define below:
\begin{align} p(x) & \sim \mathcal{N}(\mu, \sigma^2) \\ p(x; \mu, \sigma^2) & \sim \frac{1}{\sqrt{2\pi\sigma^2}} \exp{ \bigg[-\frac{1}{2}\bigg( \frac{x-\mu}{\sigma}\bigg)^2 \bigg] } \end{align}
If you plot this probability density function with different parametrizations, you’ll get something like the plots below, where the red distribution is the standard Gaussian $$p(x) \sim \mathcal{N}(0, 1.0)$$:
As you can see in the probability density function (pdf) plot above, the likelihood of $$x$$ at variously given realizations are showed in the y-axis. Another source of confusion here is that usually, people take this as a probability, because they usually see these plots of normals and the likelihood is always below 1, however, the probability density function doesn’t give you probabilities but densities. The constraint on the pdf is that it must integrate to one:
$$\int_{-\infty}^{+\infty} f(x)dx = 1$$
So, it is completely normal to have densities larger than 1 in many points for many different distributions. Take for example the pdf for the Beta distribution below:
As you can see, the pdf shows densities above one in many parametrizations of the distribution, while still integrating into 1 and following the second axiom of probability: the unit measure.
So, returning to our original principle of maximum likelihood estimation, what we want is to maximize the likelihood $$\mathcal{L}(\theta | x)$$ for our observed data. What this means in practical terms is that we want to find the parameters $$\theta$$ of our model where the likelihood that this model generated our data is maximized, we want to find which parameters of this model are most plausible to have generated this observed data, or what are the parameters that make this sample most probable ?
For the case of our univariate Gaussian model, what we want is to find the parameters $$\mu$$ and $$\sigma^2$$, which for convenient notation we collapse into a single parameter vector:
$$\theta = \begin{bmatrix}\mu \\ \sigma^2\end{bmatrix}$$
Because these are the statistics that completely define our univariate Gaussian model. So let’s formulate the problem of the maximum likelihood estimation:
\begin{align} \hat{\theta} &= \mathrm{arg}\max_\theta \mathcal{L}(\theta | x) \\ &= \mathrm{arg}\max_\theta p_{\theta}(x) \end{align}
This says that we want to obtain the maximum likelihood estimate $$\hat{\theta}$$ that approximates $$p_{\theta}(x)$$ to a underlying “true” distribution $$p_{\theta^*}(x)$$ by maximizing the likelihood of the parameters $$\theta$$ given data $$x$$. You shouldn’t confuse a maximum likelihood estimate $$\hat{\theta}(x)$$ which is a realization of the maximum likelihood estimator for the data $$x$$, with the maximum likelihood estimator $$\hat{\theta}$$, so pay attention to disambiguate it in your head.
However, we need to incorporate multiple observations in this formulation, and by adding multiple observations we end up with a complex joint distribution:
$$\hat{\theta} = \mathrm{arg}\max_\theta p_{\theta}(x_1, x_2, \ldots, x_n)$$
That needs to take into account the interactions between all observations. And here is where we make a strong assumption: we state that the observations are independent. Independent random variables mean that the following holds:
$$p_{\theta}(x_1, x_2, \ldots, x_n) = \prod_{i=1}^{n} p_{\theta}(x_i)$$
Which means that since $$x_1, x_2, \ldots, x_n$$ don’t contain information about each other, we can write the joint probability as a product of their marginals.
Another assumption that is made, is that these random variables are identically distributed, which means that they came from the same generating distribution, which allows us to model it with the same distribution parametrization.
Given these two assumptions, which are also known as IID (independently and identically distributed), we can formulate our maximum likelihood estimation problem as:
$$\hat{\theta} = \mathrm{arg}\max_\theta \prod_{i=1}^{n} p_{\theta}(x_i)$$
Note that MLE doesn’t require you to make these assumptions, however, many problems will appear if you don’t to it, such as different distributions for each sample or having to deal with joint probabilities.
Given that in many cases these densities that we multiply can be very small, multiplying one by the other in the product that we have above we can end up with very small values. Here is where the logarithm function makes its way to the likelihood. The log function is a strictly monotonically increasing function, that preserves the location of the extrema and has a very nice property:
$$\log ab = \log a + \log b$$
Where the logarithm of a product is the sum of the logarithms, which is very convenient for us, so we’ll apply the logarithm to the likelihood to maximize what is called the log-likelihood:
\begin{align} \hat{\theta} &= \mathrm{arg}\max_\theta \prod_{i=1}^{n} p_{\theta}(x_i) \\ &= \mathrm{arg}\max_\theta \sum_{i=1}^{n} \log p_{\theta}(x_i) \\ \end{align}
As you can see, we went from a product to a summation, which is much more convenient. Another reason for the application of the logarithm is that we often take the derivative and solve it for the parameters, therefore is much easier to work with a summation than a multiplication.
We can also conveniently average the log-likelihood (given that we’re just including a multiplication by a constant):
\begin{align} \hat{\theta} &= \mathrm{arg}\max_\theta \sum_{i=1}^{n} \log p_{\theta}(x_i) \\ &= \mathrm{arg}\max_\theta \frac{1}{n} \sum_{i=1}^{n} \log p_{\theta}(x_i) \\ \end{align}
This is also convenient because it will take out the dependency on the number of observations. We also know, that through the law of large numbers, the following holds as $$n\to\infty$$:
$$\frac{1}{n} \sum_{i=1}^{n} \log \, p_{\theta}(x_i) \approx \mathbb{E}_{x \sim p_{\theta^*}(x)}\left[\log \, p_{\theta}(x) \right]$$
As you can see, we’re approximating the expectation with the empirical expectation defined by our dataset $$\{x_i\}_{i=1}^{n}$$. This is an important point and it is usually implictly assumed.
The weak law of large numbers can be bounded using a Chebyshev bound, and if you are interested in concentration inequalities, I’ve made an article about them here where I discuss the Chebyshev bound.
To finish our formulation, given that we usually minimize objectives, we can formulate the same maximum likelihood estimation as the minimization of the negative of the log-likelihood:
$$\hat{\theta} = \mathrm{arg}\min_\theta -\mathbb{E}_{x \sim p_{\theta^*}(x)}\left[\log \, p_{\theta}(x) \right]$$
Which is exactly the same thing with just the negation turn the maximization problem into a minimization problem.
The relation of maximum likelihood estimation with the Kullback–Leibler divergence from information theory
It is well-known that maximizing the likelihood is the same as minimizing the Kullback-Leibler divergence, also known as the KL divergence. Which is very interesting because it links a measure from information theory with the maximum likelihood principle.
The KL divergence is defined as:
$$D_{KL}( p || q)=\int p(x)\log\frac{p(x)}{q(x)} \ dx$$
There are many intuitions to understand the KL divergence, I personally like the perspective on the likelihood ratios, however, there are plenty of materials about it that you can easily find and it’s out of the scope of this introduction.
The KL divergence is basically the expectation of the log-likelihood ratio under the $$p(x)$$ distribution. What we’re doing below is just rephrasing it by using some identities and properties of the expectation:
\begin{align} D_{KL}[p_{\theta^*}(x) \, \Vert \, p_\theta(x)] &= \mathbb{E}_{x \sim p_{\theta^*}(x)}\left[\log \frac{p_{\theta^*}(x)}{p_\theta(x)} \right] \\ \label{eq:logquotient} &= \mathbb{E}_{x \sim p_{\theta^*}(x)}\left[\log \,p_{\theta^*}(x) – \log \, p_\theta(x) \right] \\ \label{eq:linearization} &= \mathbb{E}_{x \sim p_{\theta^*}(x)} \underbrace{\left[\log \, p_{\theta^*}(x) \right]}_{\text{Entropy of } p_{\theta^*}(x)} – \underbrace{\mathbb{E}_{x \sim p_{\theta^*}(x)}\left[\log \, p_{\theta}(x) \right]}_{\text{Negative of log-likelihood}} \end{align}
In the formulation above, we’re first using the fact that the logarithm of a quotient is equal to the difference of the logs of the numerator and denominator (equation $$\ref{eq:logquotient}$$). After that we use the linearization of the expectation(equation $$\ref{eq:linearization}$$), which tells us that $$\mathbb{E}\left[X + Y\right] = \mathbb{E}\left[X\right]+\mathbb{E}\left[Y\right]$$. In the end, we are left with two terms, the first one in the left is the entropy and the one in the right you can recognize as the negative of the log-likelihood that we saw earlier.
If we want to minimize the KL divergence for the $$\theta$$, we can ignore the first term, since it doesn’t depend of $$\theta$$ in any way, and in the end we have exactly the same maximum likelihood formulation that we saw before:
$$\begin{eqnarray} \require{cancel} \theta^* &=& \mathrm{arg}\min_\theta \cancel{\mathbb{E}_{x \sim p_{\theta^*}(x)} \left[\log \, p_{\theta^*}(x) \right]} – \mathbb{E}_{x \sim p_{\theta^*}(x)}\left[\log \, p_{\theta}(x) \right]\\ &=& \mathrm{arg}\min_\theta -\mathbb{E}_{x \sim p_{\theta^*}(x)}\left[\log \, p_{\theta}(x) \right] \end{eqnarray}$$
The conditional log-likelihood
A very common scenario in Machine Learning is supervised learning, where we have data points $$x_n$$ and their labels $$y_n$$ building up our dataset $$D = \{ (x_1, y_1), (x_2, y_2), \ldots, (x_n, y_n) \}$$, where we’re interested in estimating the conditional probability of $$\textbf{y}$$ given $$\textbf{x}$$, or more precisely $$P_{\theta}(Y | X)$$.
To extend the maximum likelihood principle to the conditional case, we just have to write it as:
$$\hat{\theta} = \mathrm{arg}\min_\theta -\mathbb{E}_{x \sim p_{\theta^*}(y | x)}\left[\log \, p_{\theta}(y | x) \right]$$
And then it can be easily generalized to formulate the linear regression:
$$p_{\theta}(y | x) \sim \mathcal{N}(x^T \theta, \sigma^2) \\ p_{\theta}(y | x) = -n \log \sigma – \frac{n}{2} \log{2\pi} – \sum_{i=1}^{n}{\frac{\| x_i^T \theta – y_i \|}{2\sigma^2}}$$
In that case, you can see that we end up with a sum of squared errors that will have the same location of the optimum of the mean squared error (MSE). So you can see that minimizing the MSE is equivalent of maximizing the likelihood for a Gaussian model.
Remarks on the maximum likelihood
The maximum likelihood estimation has very interesting properties but it gives us only point estimates, and this means that we cannot reason on the distribution of these estimates. In contrast, Bayesian inference can give us a full distribution over parameters, and therefore will allow us to reason about the posterior distribution.
I’ll write more about Bayesian inference and sampling methods such as the ones from the Markov Chain Monte Carlo (MCMC) family, but I’ll leave this for another article, right now I’ll continue showing the relationship of the maximum likelihood estimator with the maximum a posteriori (MAP) estimator.
Maximum a posteriori
Although the maximum a posteriori, also known as MAP, also provides us with a point estimate, it is a Bayesian concept that incorporates a prior over the parameters. We’ll also see that the MAP has a strong connection with the regularized MLE estimation.
We know from the Bayes rule that we can get the posterior from the product of the likelihood and the prior, normalized by the evidence:
\begin{align} p(\theta \vert x) &= \frac{p_{\theta}(x) p(\theta)}{p(x)} \\ \label{eq:proport} &\propto p_{\theta}(x) p(\theta) \end{align}
In the equation $$\ref{eq:proport}$$, since we’re worried about optimization, we cancel the normalizing evidence $$p(x)$$ and stay with a proportional posterior, which is very convenient because the marginalization of $$p(x)$$ involves integration and is intractable for many cases.
\begin{align} \theta_{MAP} &= \mathop{\rm arg\,max}\limits_{\theta} p_{\theta}(x) p(\theta) \\ &= \mathop{\rm arg\,max}\limits_{\theta} \prod_{i=1}^{n} p_{\theta}(x_i) p(\theta) \\ &= \mathop{\rm arg\,max}\limits_{\theta} \sum_{i=1}^{n} \underbrace{\log p_{\theta}(x_i)}_{\text{Log likelihood}} \underbrace{p(\theta)}_{\text{Prior}} \end{align}
In this formulation above, we just followed the same steps as described earlier for the maximum likelihood estimator, we assume independence and an identical distributional setting, followed later by the logarithm application to switch from a product to a summation. As you can see in the final formulation, this is equivalent as the maximum likelihood estimation multiplied by the prior term.
We can also easily recover the exact maximum likelihood estimator by using a uniform prior $$p(\theta) \sim \textbf{U}(\cdot, \cdot)$$. This means that every possible value of $$\theta$$ will be equally weighted, meaning that it’s just a constant multiplication:
\begin{align} \theta_{MAP} &= \mathop{\rm arg\,max}\limits_{\theta} \sum_i \log p_{\theta}(x_i) p(\theta) \\ &= \mathop{\rm arg\,max}\limits_{\theta} \sum_i \log p_{\theta}(x_i) \, \text{constant} \\ &= \underbrace{\mathop{\rm arg\,max}\limits_{\theta} \sum_i \log p_{\theta}(x_i)}_{\text{Equivalent to maximum likelihood estimation (MLE)}} \\ \end{align}
And there you are, the MAP with a uniform prior is equivalent to MLE. It is also easy to show that a Gaussian prior can recover the L2 regularized MLE. Which is quite interesting, given that it can provide insights and a new perspective on the regularization terms that we usually use.
I hope you liked this article ! The next one will be about Bayesian inference with posterior sampling, where we’ll show how we can reason about the posterior distribution and not only on point estimates as seen in MAP and MLE.
– Christian S. Perone
Cite this article as: Christian S. Perone, "A sane introduction to maximum likelihood estimation (MLE) and maximum a posteriori (MAP)," in Terra Incognita, 02/01/2019, http://blog.christianperone.com/2019/01/mle/.
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## Rectangular Prisms
rectangular prism, or rectangular solid, is basically a box.
It has 6 sides.
All the sides, or faces, are rectangles. Opposite sides are congruent.
A good way to visualize a solid is to “unfold” the box and look at the net.
It is easy to see the 6 rectangles, and that opposite sides are equal.
The surface area, SA, is the sum of the areas of each pair of congruent sides.
SA = 2lh + 2lw + 2hw = 2(lh + lw + hw)
To find the volume of a rectangular prism, multiply the three lengths.
V = length × width × height = lwh
A special rectangular prism is a cube.
All the faces are squares.
If the length of a side is s, then:
SA = 6 × area of one side = 6s2
V = s3
Finding area and volume often uses multiplication of polynomials.
The dimensions of a box are
(y + 2) feet long, (y + 7) feet wide and (2y – 4) feet high. What is the area of the bottom of the box?
### Solution
To find the area of the bottom of the container, multiply length by width, which are the first two binomials.
(y + 2)(y + 7)
y(y + 7) + 2(y + 7)
y2 + 9y + 14
The dimensions of a box are
(y + 2) feet long, (y + 7) feet wide and (2y – 4) feet high. What is the volume of the box?
### Solution
To find the volume, multiply the area of the bottom by the height.
(2y – 4)(y2 + 9y + 14)
= (2y3 + 18y2 + 28y)
– (4y2 + 36y + 56)
= 2y3 + 14y– 8y – 56
## Cylinders
cylinder is like a can. It has circles as the top and base, and straight sides.
A good way to visualize a cylinder is to “unwrap” it and look at the net.
The two circles are easy to see.
The body of the cylinder is a rectangle.
One side of the rectangle is the height of the can.
The other side of the rectangle is wrapped around the circles, so is the circumference of the circle.
SA = 2 circles + rectangle = 2(πr2) + 2πrh
To find the volume of a cylinder, multiply the area of the base by the height:
V = πr2h
It takes about 7.5 gallons to fill a volume of one cubic foot. How many gallons are needed to fill a cylinder 2 feet high with a radius of 28 inches?
### Solution
Notice that the volume is given in feet but the lengths are given in feet and inches. Change inches to feet before calculating.
\dfrac{inches}{12} = 1 foot
\dfrac{28 \,inches}{12\, inches} = \dfrac{7}{3} feet
Calculate the volume of the cylinder.
V = πr2h = π (\dfrac{7}{3})2(2) Use \dfrac{22}{7} for π.
= (\,\dfrac{\,22\,}{7}\,)(\,\dfrac{\,49\,}{9}\,)(2) = \dfrac{22 × 7 × 2}{9}
= \dfrac{44 × 7}{9} = \dfrac{308}{9}
The volume of a cylinder is \dfrac{308}{9} cubic feet.
Since it takes 7.5 gallons to fill one cubic foot, multiply the volume by 7.5.
\dfrac{308}{9} × 7.5 = \dfrac{308 × 3 × 2.5}{9}
= \dfrac{308 × 2.5}{3} = \dfrac{616 + 154}{3}
= \dfrac{770}{3} = 256 \,\dfrac{2}{3}
The cylinder can hold 256 \dfrac{2}{3} gallons.
A gallon of paint covers 400 square feet of wall. How many gallons are required to paint the walls of a building with perimeter 200 feet and height 10 feet that has no windows?
### Solution
The area to be painted is just the walls, not the top or bottom.
The perimeter is the distance around the building, that is, the length of the rectangles that make up its sides. The length of each rectangle isn’t needed, only the sum of the lengths.
Each rectangular side is 10 feet high, so the area is:
A = bh = 200 × 10 = 2,000 ft2
Divide the square feet by 400 ft/gal.
Number of gallons required
= \dfrac{2.000}{400} = 5 gallons
Note: The shape of the building makes no difference here. The building could be a cylinder. The perimeter would be the circumference of the base and top.
## Spheres
sphere is like a ball. The formulas for circumference, surface and volume all use the radius.
The circumference is the same as a circle.
C = 2πr = πd
Surface area
SA = 4πr2
Volume
V = (\dfrac{4}{3}) πr3
A can of tennis balls contains 3 stacked tennis balls. The tennis balls touch the sides, top and bottom of the container. If the circumference of a tennis ball is 8 inches, approximately how much space in the container is not taken up by the tennis balls?
(A) 6 in.3
(B) 13 in.3
(C) 17 in.3
(D) 19 in.3
(E) 26 in.3
### Solution
Use the circumference of a tennis ball to find the radius of the balls and the container.
C = 2πr
8 = 2πr
r = (\dfrac{4}{\textit{π}})
The volume of one tennis ball is
V = \dfrac{4}{3} πr3 = \dfrac{4}{3} π (\dfrac{4}{\textit{π}} )3
= \dfrac{4 × \textit{π} × 4^{\displaystyle{3}}}{3 × {\textit{π}}^{\displaystyle{3}}} = \dfrac{4^{\displaystyle{4}}}{3{\textit{π}}^{\displaystyle{2}}}
The volume of 3 tennis balls will be:
3(\dfrac{4^{\displaystyle{4}}}{3{\textit{π}}^{\displaystyle{2}}}) = \dfrac{4^{\displaystyle{4}}}{{\textit{π}}^{\displaystyle{2}}}
The height of the cylindrical can is as tall as 3 balls, and the height of one ball is its diameter, which is 2(\dfrac{4}{π}).
The volume of the can minus the ball is
\dfrac{3 × 2 × 4^{\displaystyle{3}}}{{\textit{π}}^{\displaystyle{2}}} \,-\, \dfrac{4^{\displaystyle{4}}}{{\textit{π}}^{\displaystyle{2}}}
= \dfrac{4^{\displaystyle{3}}(6 \,-\, 4)}{{\textit{π}}^{\displaystyle{2}}} …Factor out the common factor, 43.
= \dfrac{4^{\displaystyle{3}}(2)}{{\textit{π}}^{\displaystyle{2}}} = \dfrac{64 × 2}{{\textit{π}}^{\displaystyle{2}}} = \dfrac{128}{{\textit{π}}^{\displaystyle{2}}}
Use estimates of π2 = 9 and π2 = 10, and get values between 14.2 and 12.8.
The closest answer choice is (B).
#### Solids
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CGAL 5.4 - 2D Straight Skeleton and Polygon Offsetting
Offset Construction Functions
The following functions are used to construct the inward or outward offsets of a polygon.
# Kernel Choices
Up to three different kernels can be used in the offset construction process:
• Input Kernel (InK): This is the kernel type of the input polygons.
• Skeleton Kernel (SsK): This is the kernel used for the construction of the (partial) straight skeleton. It might differ from the Input Kernel for example if the input is based on exact number types, but you wish to speed up the construction of the straight skeleton, using a lighter kernel such as CGAL::Exact_predicates_inexact_constructions_kernel.
• Offset (Output) Kernel (OfK): This is the kernel used for the construction of the offset polygons and is thus the kernel type for the output.
When the kernel differs, CGAL::Cartesian_converter is used to convert from one kernel to the other.
By default, SsK and OfK are chosen as CGAL::Exact_predicates_inexact_constructions_kernel as a compromise between speed and robustness. See also Section Create Offset Polygons from a Polygon (With or Without Holes).
# Polygon Return Type
There are two classes of offset construction functions: those that simply return all the offset polygons, and those that also create an arrangement of the polygons to create a polygon with holes. The latter contain _with_holes in the name.
The return type of the offset construction function is thus dediced by whether or not it is a simple polygon or a polygon with holes, and on the dictated output kernel (see Section Kernel Choices). Thus, for an input polygon type InKPolygon in an offset construction that does not arrange polygons (e.g. create_interior_skeleton_and_offset_polygons_2()):
• If the kernel type InK of the input polygon type is different from the output kernel type OfK, then the output polygon type is CGAL::Polygon_2<OfK>.
• If the kernel type InK is equal to the output kernel type OfK, but the input polygon is a polygon with holes (a model of GeneralPolygonWithHoles_2), then the return type is InKPolygon::General_polygon_2.
• If the kernel type InK of the input polygon type is equal to the output kernel type OfK, and the input polygon is a polygon without holes (a model of SequenceContainer with InK::Point_2 value type), then the return type is InKPolygon.
Similarly, if the return type is a polygon with holes (e.g. create_interior_skeleton_and_offset_polygons_with_holes_2(), a similar dispatch exists:
• If the kernel type InK of the input polygon type is different from the output kernel type OfK, then the output polygon type is CGAL::Polygon_with_holes_2<OfK>.
• If the kernel type InK is equal to the output kernel type OfK, but the input polygon is a polygon with holes (a model of GeneralPolygonWithHoles_2), then the return type is InKPolygon.
• If the kernel type InK of the input polygon type is equal to the output kernel type OfK, and the input polygon is a polygon without holes (a model of SequenceContainer with InK::Point_2 value type), then the return type is CGAL::Polygon_with_holes_2<OfK>.
## Functions
template<class K , class InputPolygonPtrIterator , class OutputPolygonWithHolesPtrIterator >
bool CGAL::arrange_offset_polygons_2 (InputPolygonPtrIterator begin, InputPolygonPtrIterator end, OutputPolygonWithHolesPtrIterator out, const K &k)
The function arrange_offset_polygons_2() arranges the sequence of 2D polygons obtained by create_offset_polygons_2() into 2D polygons with holes by determining geometric parent-hole relationships using a simple algorithm based on the particular characteristics of offset polygons. More...
template<class InputIterator , class Traits >
boost::optional< typename Traits::FT > CGAL::compute_outer_frame_margin (InputIterator first, InputIterator beyond, typename Traits::FT offset, const Traits &traits=Default_traits)
computes the separation required between a polygon and the outer frame used to obtain an exterior skeleton suitable for the computation of outer offset polygons at a given distance. More...
template<class OfKPolygon , class FT , class StraightSkeleton , class OfK >
std::vector< boost::shared_ptr< OfKPolygon > > CGAL::create_offset_polygons_2 (FT offset, const StraightSkeleton &s, OfK k=Exact_predicates_inexact_constructions_kernel)
returns a container with the offset polygons at distance offset obtained from the straight skeleton s. More...
template<class OfKPolygon , class FT , class InKPolygon , class OfK , class SsK >
std::vector< boost::shared_ptr< OfKPolygon > > CGAL::create_exterior_skeleton_and_offset_polygons_2 (FT offset, const InKPolygon &poly, OfK ofk=Exact_predicates_inexact_constructions_kernel, SsK ssk=Exact_predicates_inexact_constructions_kernel)
returns a container with all the outer offset polygons at distance offset of the 2D polygon poly. More...
template<class OfKPolygon , class FT , class InKPolygon , class HoleIterator , class OfK , class SsK >
std::vector< boost::shared_ptr< OfKPolygon > > CGAL::create_interior_skeleton_and_offset_polygons_2 (FT offset, const InKPolygon &outer_boundary, HoleIterator holes_begin, HoleIterator holes_end, OfK ofk=CGAL::Exact_predicates_inexact_constructions_kernel, SsK ssk=CGAL::Exact_predicates_inexact_constructions_kernel)
returns a container with all the inner offset polygons at distance offset of the 2D polygon with holes whose outer boundary is outer_boundary and its holes are given by [holes_begin,holes_end). More...
template<class OfKPolygon , class FT , class InKPolygon , class OfK , class SsK >
std::vector< boost::shared_ptr< OfKPolygon > > CGAL::create_interior_skeleton_and_offset_polygons_2 (FT offset, const InKPolygon &poly, OfK ofk=CGAL::Exact_predicates_inexact_constructions_kernel, SsK ssk=CGAL::Exact_predicates_inexact_constructions_kernel)
returns a container with all the inner offset polygons at distance offset of the 2D polygon poly. More...
template<class OfKPolygon , class FT , class InKPolygon , class OfK , class SsK >
std::vector< boost::shared_ptr< OfKPolygon > > CGAL::create_interior_skeleton_and_offset_polygons_with_holes_2 (FT offset, const InKPolygon &poly_with_holes, OfK ofk=CGAL::Exact_predicates_inexact_constructions_kernel, SsK ssk=CGAL::Exact_predicates_inexact_constructions_kernel)
returns a container with all the inner offset polygons with holes at distance offset of the 2D polygon with holes poly_with_holes. More...
template<class OfKPolygon , class FT , class InKPolygon , class OfK , class SsK >
std::vector< boost::shared_ptr< OfKPolygon > > CGAL::create_exterior_skeleton_and_offset_polygons_with_holes_2 (FT offset, const InKPolygon &poly_with_holes, OfK ofk=Exact_predicates_inexact_constructions_kernel, SsK ssk=Exact_predicates_inexact_constructions_kernel)
returns a container with all the outer offset polygons with holes at distance offset of the 2D polygon poly_with_holes. More...
template<class Target_skeleton , class Source_skeleton , class Items_converter >
boost::shared_ptr< Target_skeleton > CGAL::convert_straight_skeleton_2 (Source_skeleton const &s, Items_converted const &ic=Items_converter())
returns a new straight skeleton data structure with the same combinatorial and geometric data as s using the items converter ic to convert the geometric embedding to the types of the target skeleton. More...
## ◆ arrange_offset_polygons_2()
template<class K , class InputPolygonPtrIterator , class OutputPolygonWithHolesPtrIterator >
bool CGAL::arrange_offset_polygons_2 ( InputPolygonPtrIterator begin, InputPolygonPtrIterator end, OutputPolygonWithHolesPtrIterator out, const K & k )
#include <CGAL/arrange_offset_polygons_2.h>
The function arrange_offset_polygons_2() arranges the sequence of 2D polygons obtained by create_offset_polygons_2() into 2D polygons with holes by determining geometric parent-hole relationships using a simple algorithm based on the particular characteristics of offset polygons.
The function determines parent-hole relationships among the polygons given by [begin,end] creating boost::shared_ptr< GeneralPolygonWithHoles_2 > objects added to the output sequence given out. A CLOCKWISE oriented polygon H is a hole of a COUNTERCLOCKWISE polygon P, iff at least one vertex of H is ON_BOUNDED_SIDE of P.
This function should not be used to arrange arbitrary polygons into polygons with holes unless they meet the requirements specified below.
Template Parameters
K must be a model of Kernel. InputPolygonPtrIterator must be a model of InputIterator whose value_type is a smart pointer (such as boost::shared_ptr) whose element_type is a model of SequenceContainer with value type K::Point_2. OutputPolygonWithHolesPtrIterator must be a model of OutputIterator whose value_type is a smart pointer (such as boost::shared_ptr) whose element_type is a model of GeneralPolygonWithHoles_2.
Precondition
The input polygons must be simple.
The set of input polygons are unique and interior disjoint. That is, given distinct polygons P and Q, there are vertices of P which are not on the boundary of Q and are all on the bounded or unbounded side of Q (but not both).
Returns
true if no error was encountered, and false otherwise.
create_exterior_straight_skeleton_2()
CGAL::Straight_skeleton_builder_2<Traits,Ss,Visitor>
## ◆ compute_outer_frame_margin()
template<class InputIterator , class Traits >
boost::optional< typename Traits::FT > CGAL::compute_outer_frame_margin ( InputIterator first, InputIterator beyond, typename Traits::FT offset, const Traits & traits = Default_traits )
#include <CGAL/compute_outer_frame_margin.h>
computes the separation required between a polygon and the outer frame used to obtain an exterior skeleton suitable for the computation of outer offset polygons at a given distance.
Given a non-degenerate strictly-simple 2D polygon whose vertices are passed in the range [first,beyond), calculates the largest euclidean distance d between each input vertex and its corresponding offset vertex at a distance offset.
If such a distance can be approximately computed, returns an optional<FT> with the value d + (offset * 1.05). If the distance cannot be computed, not even approximately, due to overflow for instance, returns an empty optional<FT> (an absent result).
This result is the required separation between the input polygon and the rectangular frame used to construct an exterior offset contour at distance offset (which is done by placing the polygon as a hole of that frame).
Such a separation must be computed in this way because if the frame is too close to the polygon, the inward offset contour from the frame could collide with the outward offset contour of the polygon, resulting in a merged contour offset instead of two contour offsets, one of them corresponding to the frame.
Simply using 2*offset as the separation is incorrect since offset is the distance between an offset line and its original, not between an offset vertex and its original. The later, which is calculated by this function and needed to place the frame sufficiently away from the polygon, can be thousands of times larger than offset.
If the result is absent, any attempt to construct an exterior offset polygon at distance offset will fail. This will occur whenever the polygon has a vertex with an internal angle approaching 0 (because the offset vertex of a vertex whose internal angle equals 0 is at infinity).
The default traits class Default_traits is an instance of the class Polygon_offset_builder_traits_2<Kernel> parameterized on the kernel in which the type InputIterator::value_type is defined.
Template Parameters
Traits must be a model for PolygonOffsetBuilderTraits_2. InputIterator must be a model of InputIterator with value_type equivalent to Traits::Point_2.
Precondition
offset > 0.
The range [first,beyond) contains the vertices of a non-degenerate strictly-simple 2D polygon.
PolygonOffsetBuilderTraits_2
CGAL::Polygon_offset_builder_traits_2<K>
Examples:
Straight_skeleton_2/Low_level_API.cpp.
## ◆ convert_straight_skeleton_2()
template<class Target_skeleton , class Source_skeleton , class Items_converter >
boost::shared_ptr CGAL::convert_straight_skeleton_2 ( Source_skeleton const & s, Items_converted const & ic = Items_converter() )
#include <CGAL/Straight_skeleton_converter_2.h>
returns a new straight skeleton data structure with the same combinatorial and geometric data as s using the items converter ic to convert the geometric embedding to the types of the target skeleton.
Template Parameters
Target_skeleton must be a model of StraightSkeleton_2 Source_skeleton must be a model of StraightSkeleton_2 Items_converter must be a model of StraightSkeletonItemsConverter_2
CGAL::Straight_skeleton_items_converter_2<SrcSs,TgtSs,NTCV>
CGAL::Straight_skeleton_converter_2<SrcSs,TgtSs,ItemsCV>
## ◆ create_exterior_skeleton_and_offset_polygons_2()
template<class OfKPolygon , class FT , class InKPolygon , class OfK , class SsK >
std::vector< boost::shared_ptr > CGAL::create_exterior_skeleton_and_offset_polygons_2 ( FT offset, const InKPolygon & poly, OfK ofk = Exact_predicates_inexact_constructions_kernel, SsK ssk = Exact_predicates_inexact_constructions_kernel )
#include <CGAL/create_offset_polygons_2.h>
returns a container with all the outer offset polygons at distance offset of the 2D polygon poly.
A temporary straight skeleton is constructed in the limited exterior of the input polygon to obtain the offsets. The construction of this skeleton is the most expensive operation, therefore, to construct offsets at more than one single distance, use the separate functions create_exterior_straight_skeleton_2() and create_offset_polygons_2() instead. The exterior skeleton is limited by an outer rectangular frame placed at a margin sufficiently large to allow the offset polygons to be constructed.
Template Parameters
OfK must be a model of Kernel. It is used to instantiate Polygon_offset_builder_traits_2 for constructing the offset polygons. SsK must be a model of Kernel. It is used to instantiate Straight_skeleton_builder_traits_2 for constructing the straight skeleton. FT must be a model of FieldNumberType convertible to OfK::FT and SsK::FT. InKPolygon must be a model of SequenceContainer with value type InK::Point_2 (e.g. Polygon_2) or a model of GeneralPolygonWithHoles_2 (e.g. Polygon_with_holes_2). OfKPolygon is a polygon without holes type determined from OfK and InKPolygon, see Section Polygon Return Type.
Note
If SsK != OfK the constructed straight skeleton is converted to Straight_skeleton_2<OfK>.
create_interior_skeleton_and_offset_polygons_2()
create_exterior_skeleton_and_offset_polygons_with_holes_2()
Polygon_offset_builder_2<Ss,Traits,Container>
Examples:
Straight_skeleton_2/Create_skeleton_and_offset_polygons_2.cpp.
## ◆ create_exterior_skeleton_and_offset_polygons_with_holes_2()
template<class OfKPolygon , class FT , class InKPolygon , class OfK , class SsK >
std::vector > CGAL::create_exterior_skeleton_and_offset_polygons_with_holes_2 ( FT offset, const InKPolygon & poly_with_holes, OfK ofk = Exact_predicates_inexact_constructions_kernel, SsK ssk = Exact_predicates_inexact_constructions_kernel )
#include <CGAL/create_offset_polygons_from_polygon_with_holes_2.h>
returns a container with all the outer offset polygons with holes at distance offset of the 2D polygon poly_with_holes.
This is equivalent to arrange_offset_polygons_2(create_exterior_skeleton_and_offset_polygons_2(offset, poly_with_holes, ofk, ssk)).
Template Parameters
OfK must be a model of Kernel. It is used to instantiate Polygon_offset_builder_traits_2 for constructing the offset polygons. SsK must be a model of Kernel. It is used to instantiate Straight_skeleton_builder_traits_2 for constructing the straight skeleton. FT must be a model of FieldNumberType convertible to OfK::FT and SsK::FT. InKPolygon must be a model of SequenceContainer with value type InK::Point_2 (e.g. Polygon_2) or a model of GeneralPolygonWithHoles_2 (e.g. Polygon_with_holes_2). OfKPolygon is a polygon without holes type determined by OfK and InKPolygon, see Section Polygon Return Type.
Note
If SsK != OfK the constructed straight skeleton is converted to Straight_skeleton_2<OfK>.
create_exterior_skeleton_and_offset_polygons_2()
create_interior_skeleton_and_offset_polygons_with_holes_2()
Polygon_offset_builder_2<Ss,Traits,Container>
## ◆ create_interior_skeleton_and_offset_polygons_2() [1/2]
template<class OfKPolygon , class FT , class InKPolygon , class HoleIterator , class OfK , class SsK >
std::vector< boost::shared_ptr > CGAL::create_interior_skeleton_and_offset_polygons_2 ( FT offset, const InKPolygon & outer_boundary, HoleIterator holes_begin, HoleIterator holes_end, OfK ofk = CGAL::Exact_predicates_inexact_constructions_kernel, SsK ssk = CGAL::Exact_predicates_inexact_constructions_kernel )
#include <CGAL/create_offset_polygons_2.h>
returns a container with all the inner offset polygons at distance offset of the 2D polygon with holes whose outer boundary is outer_boundary and its holes are given by [holes_begin,holes_end).
A temporary straight skeleton is constructed in the interior of the input polygon to obtain the offsets. The construction of this skeleton is the most expensive operation, therefore, to construct offsets at more than one single distance, use the separate functions create_interior_straight_skeleton_2() and create_offset_polygons_2() instead.
Template Parameters
OfK must be a model of Kernel. It is used to instantiate Polygon_offset_builder_traits_2 for constructing the offset polygons. SsK must be a model of Kernel. It is used to instantiate Straight_skeleton_builder_traits_2 for constructing the straight skeleton. FT must be a model of FieldNumberType convertible to OfK::FT and SsK::FT. HoleIterator must be a model of InputIterator with value type being a model of ConstRange with value type SsK::Point_2. InKPolygon must be a model of SequenceContainer with value type InK::Point_2 (e.g. Polygon_2). OfKPolygon is a polygon without holes type determined from OfK and InKPolygon, see Section Polygon Return Type.
Note
If SsK != OfK the constructed straight skeleton is converted to Straight_skeleton_2<OfK>.
create_exterior_skeleton_and_offset_polygons_2()
create_interior_skeleton_and_offset_polygons_with_holes_2()
Polygon_offset_builder_2<Ss,Traits,Container>
Examples:
Straight_skeleton_2/Create_saop_from_polygon_with_holes_2.cpp, and Straight_skeleton_2/Create_skeleton_and_offset_polygons_2.cpp.
## ◆ create_interior_skeleton_and_offset_polygons_2() [2/2]
template<class OfKPolygon , class FT , class InKPolygon , class OfK , class SsK >
std::vector< boost::shared_ptr > CGAL::create_interior_skeleton_and_offset_polygons_2 ( FT offset, const InKPolygon & poly, OfK ofk = CGAL::Exact_predicates_inexact_constructions_kernel, SsK ssk = CGAL::Exact_predicates_inexact_constructions_kernel )
#include <CGAL/create_offset_polygons_2.h>
returns a container with all the inner offset polygons at distance offset of the 2D polygon poly.
A temporary straight skeleton is constructed in the interior of the input polygon to obtain the offsets. The construction of this skeleton is the most expensive operation, therefore, to construct offsets at more than one single distance, use the separate functions create_interior_straight_skeleton_2() and create_offset_polygons_2() instead.
Template Parameters
OfK must be a model of Kernel. It is used to instantiate Polygon_offset_builder_traits_2 for constructing the offset polygons. SsK must be a model of Kernel. It is used to instantiate Straight_skeleton_builder_traits_2 for constructing the straight skeleton. FT must be a model of FieldNumberType convertible to OfK::FT and SsK::FT. InKPolygon must be a model of SequenceContainer with value type InK::Point_2 (e.g. Polygon_2) or a model of GeneralPolygonWithHoles_2 (e.g. Polygon_with_holes_2). OfKPolygon is a polygon without holes type determined by OfK and InKPolygon, see Section Polygon Return Type.
Note
If SsK != OfK the constructed straight skeleton is converted to Straight_skeleton_2<OfK>.
create_exterior_skeleton_and_offset_polygons_2()
create_interior_skeleton_and_offset_polygons_with_holes_2()
Polygon_offset_builder_2<Ss,Traits,Container>
## ◆ create_interior_skeleton_and_offset_polygons_with_holes_2()
template<class OfKPolygon , class FT , class InKPolygon , class OfK , class SsK >
std::vector< boost::shared_ptr< OfKPolygon > > CGAL::create_interior_skeleton_and_offset_polygons_with_holes_2 ( FT offset, const InKPolygon & poly_with_holes, OfK ofk = CGAL::Exact_predicates_inexact_constructions_kernel, SsK ssk = CGAL::Exact_predicates_inexact_constructions_kernel )
#include <CGAL/create_offset_polygons_from_polygon_with_holes_2.h>
returns a container with all the inner offset polygons with holes at distance offset of the 2D polygon with holes poly_with_holes.
This is equivalent to arrange_offset_polygons_2(create_interior_skeleton_and_offset_polygons_2(offset, poly_with_holes, ofk, ssk)).
Template Parameters
OfK must be a model of Kernel. It is used to instantiate Polygon_offset_builder_traits_2 for constructing the offset polygons. SsK must be a model of Kernel. It is used to instantiate Straight_skeleton_builder_traits_2 for constructing the straight skeleton. FT must be a model of FieldNumberType convertible to OfK::FT and SsK::FT. InKPolygon must be a model of SequenceContainer with value type InK::Point_2 (e.g. Polygon_2) or a model of GeneralPolygonWithHoles_2 (e.g. Polygon_with_holes_2). OfKPolygon is a polygon without holes type determined by OfK and InKPolygon, see Section Polygon Return Type.
Note
If SsK != OfK the constructed straight skeleton is converted to Straight_skeleton_2<OfK>.
create_interior_skeleton_and_offset_polygons_2()
create_exterior_skeleton_and_offset_polygons_with_holes_2()
Polygon_offset_builder_2<Ss,Traits,Container>
Examples:
Straight_skeleton_2/Create_skeleton_and_offset_polygons_with_holes_2.cpp, and Straight_skeleton_2/Show_offset_polygon.cpp.
## ◆ create_offset_polygons_2()
template<class OfKPolygon , class FT , class StraightSkeleton , class OfK >
std::vector< boost::shared_ptr > CGAL::create_offset_polygons_2 ( FT offset, const StraightSkeleton & s, OfK k = Exact_predicates_inexact_constructions_kernel )
#include <CGAL/create_offset_polygons_2.h>
returns a container with the offset polygons at distance offset obtained from the straight skeleton s.
If s is the inner skeleton of a polygon with holes, the offset polygons will be generated in its interior. If s is the outer skeleton of a polygon with holes, the offset polygons will be generated in its exterior.
Template Parameters
OfK must be a model of Kernel. It is used to instantiate Polygon_offset_builder_traits_2 for constructing the offset polygons. FT must be a model of FieldNumberType convertible to OfK::FT. StraightSkeleton is Straight_skeleton_2. OfKPolygon is a polygon without holes type determined from OfK, see Section Polygon Return Type.
Note
If SsK != OfK the constructed straight skeleton is converted to Straight_skeleton_2<OfK>.
create_interior_straight_skeleton_2()
create_exterior_straight_skeleton_2()
create_interior_skeleton_and_offset_polygons_2()
create_exterior_skeleton_and_offset_polygons_2()
Polygon_offset_builder_2<Ss,Traits,Container>
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https://homework.cpm.org/category/CC/textbook/cca/chapter/11/lesson/11.3.4/problem/11-118
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### Home > CCA > Chapter 11 > Lesson 11.3.4 > Problem11-118
11-118.
1. Determine the number of solutions for each of the quadratics below. Note: You do not need to solve the quadratics. Homework Help ✎
1. (x − 2)2 = −3
2. 6x2x − 2 = 0
3. 4x2 − 4x + 1 = 0
4. 427x2 + 731x − 280 = 0
See problem 11-101 for additional help.
No real solution.
Perfect Square Form:(2x − 1)2 = 0
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https://www.physicsforums.com/threads/mesh-current-analysis.718644/
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# Mesh current analysis
1. Oct 24, 2013
### shaltera
Use mesh-current analysis to determine currents I1, I2 and I1 + I2 for the network shown.My calculation so far:
Mesh A: v1+v2=12∠00
Mesh B: -v2+v3=-10∠00
v1=i1RA=(47Ω)iA
v2=iAR2=R2(iA-iB)=(j100Ω)(iA-iB)
v3=iBR3=(-75j)iB
Mesh A: (47Ω)iA+(j100Ω)(iA-iB)=12∠00
Mesh B: -(j100Ω)(iA-iB)+(-75j)iB=-10∠00
I'm stuck
#### Attached Files:
• ###### circuit.jpg
File size:
25.3 KB
Views:
128
2. Oct 27, 2013
### Kavik
Have you tried distributing j100 Ohm and -j100 Ohm and then combining like terms?
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http://www.gurufocus.com/term/DaysInventory/ESI/Days%2BInventory/ITT%2BEducational%2BServices%252C%2BInc
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Switch to:
ITT Educational Services, Inc. (NYSE:ESI)
Days Inventory
0.00 (As of Sep. 2013)
ITT Educational Services, Inc.'s inventory for the three months ended in Sep. 2013 was \$0 Mil. ITT Educational Services, Inc.'s cost of goods sold for the three months ended in Sep. 2013 was \$122 Mil. Hence, ITT Educational Services, Inc.'s days inventory for the three months ended in Sep. 2013 was 0.00.
ITT Educational Services, Inc.'s days inventory stayed the same from Sep. 2012 (0.00) to Sep. 2013 (0.00).
Inventory can be measured by Days Sales of Inventory (DSI). ITT Educational Services, Inc.'s days sales of inventory (DSI) for the three months ended in Sep. 2013 was 0.00.
Inventory turnover measures how fast the company turns over its inventory within a year.
Inventory to revenue ratio determines the ability of a company to manage their inventory levels. It measures the percentage of Inventories the company currently has on hand to support the current amount of Revenue. ITT Educational Services, Inc.'s inventory to revenue ratio for the three months ended in Sep. 2013 was 0.00.
Definition
Days Inventory indicates the number of days of goods in sales that a company has in the inventory.
ITT Educational Services, Inc.'s Days Inventory for the fiscal year that ended in Dec. 2012 is calculated as
Days Inventory = Inventory / Cost of Goods Sold * Days in Period = 0 / 539.223 * 365 = 0.00
ITT Educational Services, Inc.'s Days Inventory for the quarter that ended in Sep. 2013 is calculated as:
Days Inventory = Inventory / Cost of Goods Sold * Days in Period = 0 / 121.994 * 91 = 0.00
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
Explanation
An increase of Days Inventory may indicate the company's sales slowed.
1. Inventory can be measured by Days Sales of Inventory (DSI).
ITT Educational Services, Inc.'s Days Sales of Inventory for the three months ended in Sep. 2013 is calculated as
Days Sales of Inventory (DSI) = Inventory / Revenue * Days in Period = 0 / 259.416 * 91 = 0.00
2. Inventory Turnover measures how fast the company turns over its inventory within a year.
ITT Educational Services, Inc.'s Inventory Turnover for the three months ended in Sep. 2013 is calculated as
Inventory Turnover = Cost of Goods Sold / Average Inventory = 121.994 / 0 =
3. Inventory to Revenue determines the ability of a company to manage their inventory levels. It measures the percentage of Inventories the company currently has on hand to support the current amount of Revenue.
ITT Educational Services, Inc.'s Inventory to Revenue for the three months ended in Sep. 2013 is calculated as
Inventory to Revenue = Inventory / Revenue = 0 / 259.416 = 0.00
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
Be Aware
A lot of business are seasonable. It makes more sense to compare Days Inventory from the same period in the previous year instead of from the previous quarter.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
ITT Educational Services, Inc. Annual Data
Dec04 Dec05 Dec06 Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 DaysInventory 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
ITT Educational Services, Inc. Quarterly Data
Sep11 Dec11 Mar12 Jun12 Sep12 Dec12 Mar13 Jun13 Sep13 Dec13 DaysInventory 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
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# Convert grams per square foot to other units of surface density
## grams/foot² [g/ft²] surface density conversions
1 g/ft² = 1.08 × 10-17 microgram per square picometer g/ft² to µg/pm² 1 g/ft² = 1.08 × 10-13 microgram per square angstrom g/ft² to µg/Ų 1 g/ft² = 1.08 × 10-11 microgram per square nanometer g/ft² to µg/nm² 1 g/ft² = 1.08 × 10-5 microgram per square micron g/ft² to µg/µ² 1 g/ft² = 1.08 × 10-5 microgram per square micrometer g/ft² to µg/µm² 1 g/ft² = 10.76 micrograms per square millimeter g/ft² to µg/mm² 1 g/ft² = 1 076.39 micrograms per square centimeter g/ft² to µg/cm² 1 g/ft² = 10 763 910.4 micrograms per square meter g/ft² to µg/m² 1 g/ft² = 1 000 micrograms per hectare g/ft² to µg/ha 1 g/ft² = 6 944.44 micrograms per square inch g/ft² to µg/inch² 1 g/ft² = 1 000 000 micrograms per square foot g/ft² to µg/ft² 1 g/ft² = 9 000 000 micrograms per square yard g/ft² to µg/yd² 1 g/ft² = 1.08 × 10-20 milligram per square picometer g/ft² to mg/pm² 1 g/ft² = 1.08 × 10-16 milligram per square angstrom g/ft² to mg/Ų 1 g/ft² = 1.08 × 10-14 milligram per square nanometer g/ft² to mg/nm² 1 g/ft² = 1.08 × 10-8 milligram per square micron g/ft² to mg/µ² 1 g/ft² = 1.08 × 10-8 milligram per square micrometer g/ft² to mg/µm² 1 g/ft² = 0.01 milligram per square millimeter g/ft² to mg/mm² 1 g/ft² = 1.08 milligrams per square centimeter g/ft² to mg/cm² 1 g/ft² = 10 763.91 milligrams per square meter g/ft² to mg/m² 1 g/ft² = 6.94 milligrams per square inch g/ft² to mg/in² 1 g/ft² = 1 000 milligrams per square foot g/ft² to mg/ft² 1 g/ft² = 9 000 milligrams per square yard g/ft² to mg/yd² 1 g/ft² = 1.08 × 10-23 gram per square picometer g/ft² to g/pm² 1 g/ft² = 1.08 × 10-19 gram per square angstrom g/ft² to g/Ų 1 g/ft² = 1.08 × 10-17 gram per square nanometer g/ft² to g/nm² 1 g/ft² = 1.08 × 10-11 gram per square micron g/ft² to g/µ² 1 g/ft² = 1.08 × 10-11 gram per square micrometer g/ft² to g/µm² 1 g/ft² = 1.08 × 10-5 gram per square millimeter g/ft² to g/mm² 1 g/ft² = 0.001 gram per square centimeter g/ft² to g/cm² 1 g/ft² = 10.76 grams per square meter g/ft² to g/m² 1 g/ft² = 0.01 gram per square inch g/ft² to g/in² 1 g/ft² = 9 grams per square yard g/ft² to g/yd² 1 g/ft² = 1.08 × 10-26 kilogram per square picometer g/ft² to kg/pm² 1 g/ft² = 1.08 × 10-22 kilogram per square angstrom g/ft² to kg/Ų 1 g/ft² = 1.08 × 10-20 kilogram per square nanometer g/ft² to kg/nm² 1 g/ft² = 1.08 × 10-14 kilogram per square micron g/ft² to kg/µ² 1 g/ft² = 1.08 × 10-14 kilogram per square micrometer g/ft² to kg/µm² 1 g/ft² = 1.08 × 10-8 kilogram per square millimeter g/ft² to kg/mm² 1 g/ft² = 1.08 × 10-6 kilogram per square centimeter g/ft² to kg/cm² 1 g/ft² = 0.01 kilogram per square meter g/ft² to kg/m² 1 g/ft² = 107.64 kilograms per hectare g/ft² to kg/ha 1 g/ft² = 6.94 × 10-6 kilogram per square inch g/ft² to kg/in² 1 g/ft² = 0.001 kilogram per square foot g/ft² to kg/ft² 1 g/ft² = 0.01 kilogram per square yard g/ft² to kg/yd² 1 g/ft² = 1.08 centners per hectare g/ft² to centner/ha 1 g/ft² = 1.08 × 10-29 tonne per square picometer g/ft² to t/pm² 1 g/ft² = 1.08 × 10-25 tonne per square angstrom g/ft² to t/Ų 1 g/ft² = 1.08 × 10-23 tonne per square nanometer g/ft² to t/nm² 1 g/ft² = 1.08 × 10-17 tonne per square micron g/ft² to t/µ² 1 g/ft² = 1.08 × 10-17 tonne per square micrometer g/ft² to t/µm² 1 g/ft² = 1.08 × 10-11 tonne per square millimeter g/ft² to t/mm² 1 g/ft² = 1.08 × 10-9 tonne per square centimeter g/ft² to t/cm² 1 g/ft² = 1.08 × 10-5 tonne per square meter g/ft² to t/m² 1 g/ft² = 6.94 × 10-9 tonne per square inch g/ft² to t/in² 1 g/ft² = 1 × 10-6 tonne per square foot g/ft² to t/ft² 1 g/ft² = 9 × 10-6 tonne per square yard g/ft² to t/yd² 1 g/ft² = 3.8 × 10-25 ounce per square picometer g/ft² to oz/pm² 1 g/ft² = 3.8 × 10-21 ounce per square angstrom g/ft² to oz/Ų 1 g/ft² = 3.8 × 10-19 ounce per square nanometer g/ft² to oz/nm² 1 g/ft² = 3.8 × 10-13 ounce per square micron g/ft² to oz/µ² 1 g/ft² = 3.8 × 10-13 ounce per square micrometer g/ft² to oz/µm² 1 g/ft² = 3.8 × 10-7 ounce per square millimeter g/ft² to oz/mm² 1 g/ft² = 3.8 × 10-5 ounce per square centimeter g/ft² to oz/cm² 1 g/ft² = 0.38 ounce per square meter g/ft² to oz/m² 1 g/ft² = 0.0002 ounce per square inch g/ft² to oz/in² 1 g/ft² = 0.04 ounce per square foot g/ft² to oz/ft² 1 g/ft² = 0.32 ounce per square yard g/ft² to oz/yd² 1 g/ft² = 2.37 × 10-26 pound per square picometer g/ft² to lb/pm² 1 g/ft² = 2.37 × 10-22 pound per square angstrom g/ft² to lb/Ų 1 g/ft² = 2.37 × 10-20 pound per square nanometer g/ft² to lb/nm² 1 g/ft² = 2.37 × 10-14 pound per square micron g/ft² to lb/µ² 1 g/ft² = 2.37 × 10-14 pound per square micrometer g/ft² to lb/µm² 1 g/ft² = 2.37 × 10-8 pound per square millimeter g/ft² to lb/mm² 1 g/ft² = 2.37 × 10-6 pound per square centimeter g/ft² to lb/cm² 1 g/ft² = 0.02 pound per square meter g/ft² to lb/m² 1 g/ft² = 1.53 × 10-5 pound per square inch g/ft² to lb/in² 1 g/ft² = 0.002 pound per square foot g/ft² to lb/ft² 1 g/ft² = 0.02 pound per square yard g/ft² to lb/yd² 1 g/ft² = 1.66 × 10-22 grain per square picometer g/ft² to gr/pm² 1 g/ft² = 1.66 × 10-18 grain per square angstrom g/ft² to gr/Ų 1 g/ft² = 1.66 × 10-16 grain per square nanometer g/ft² to gr/nm² 1 g/ft² = 1.66 × 10-10 grain per square micron g/ft² to gr/µ² 1 g/ft² = 1.66 × 10-10 grain per square micrometer g/ft² to gr/µm² 1 g/ft² = 0.0002 grain per square millimeter g/ft² to gr/mm² 1 g/ft² = 0.02 grain per square centimeter g/ft² to gr/cm² 1 g/ft² = 166.11 grains per square meter g/ft² to gr/m² 1 g/ft² = 15.43 grains per hectare g/ft² to gr/ha 1 g/ft² = 0.11 grain per square inch g/ft² to gr/inch² 1 g/ft² = 15.43 grains per square foot g/ft² to gr/ft² 1 g/ft² = 138.89 grains per square yard g/ft² to gr/yd² 1 g/ft² = 1.19 × 10-29 short ton per square picometer g/ft² to short tn/pm² 1 g/ft² = 1.19 × 10-25 short ton per square angstrom g/ft² to short tn/Ų 1 g/ft² = 1.19 × 10-23 short ton per square nanometer g/ft² to short tn/nm² 1 g/ft² = 1.19 × 10-17 short ton per square micron g/ft² to short tn/µ² 1 g/ft² = 1.19 × 10-17 short ton per square micrometer g/ft² to short tn/µm² 1 g/ft² = 1.19 × 10-11 short ton per square millimeter g/ft² to short tn/mm² 1 g/ft² = 1.19 × 10-9 short ton per square centimeter g/ft² to short tn/cm² 1 g/ft² = 1.19 × 10-5 short ton per square meter g/ft² to short tn/m² 1 g/ft² = 7.65 × 10-9 short ton per square inch g/ft² to short tn/in² 1 g/ft² = 1.1 × 10-6 short ton per square foot g/ft² to short tn/ft² 1 g/ft² = 9.92 × 10-6 short ton per square yard g/ft² to short tn/yd² 1 g/ft² = 1.06 × 10-29 long ton per square picometer g/ft² to long tn/pm² 1 g/ft² = 1.06 × 10-25 long ton per square angstrom g/ft² to long tn/Ų 1 g/ft² = 1.06 × 10-23 long ton per square nanometer g/ft² to long tn/nm² 1 g/ft² = 1.06 × 10-17 long ton per square micron g/ft² to long tn/µ² 1 g/ft² = 1.06 × 10-17 long ton per square micrometer g/ft² to long tn/µm² 1 g/ft² = 1.06 × 10-11 long ton per square millimeter g/ft² to long tn/mm² 1 g/ft² = 1.06 × 10-9 long ton per square centimeter g/ft² to long tn/cm² 1 g/ft² = 1.06 × 10-5 long ton per square meter g/ft² to long tn/m² 1 g/ft² = 6.83 × 10-9 long ton per square inch g/ft² to long tn/in² 1 g/ft² = 9.84 × 10-7 long ton per square foot g/ft² to long tn/ft² 1 g/ft² = 8.86 × 10-6 long ton per square yard g/ft² to long tn/yd² 1 g/ft² = 1.7 × 10-27 stone per square picometer g/ft² to st/pm² 1 g/ft² = 1.7 × 10-23 stone per square angstrom g/ft² to st/Ų 1 g/ft² = 1.7 × 10-21 stone per square nanometer g/ft² to st/nm² 1 g/ft² = 1.7 × 10-15 stone per square micron g/ft² to st/µ² 1 g/ft² = 1.7 × 10-15 stone per square micrometer g/ft² to st/µm² 1 g/ft² = 1.7 × 10-9 stone per square millimeter g/ft² to st/mm² 1 g/ft² = 1.7 × 10-7 stone per square centimeter g/ft² to st/cm² 1 g/ft² = 0.002 stone per square meter g/ft² to st/m² 1 g/ft² = 0.07 stone per hectare g/ft² to st/ha 1 g/ft² = 1.09 × 10-6 stone per square inch g/ft² to st/inch² 1 g/ft² = 0.0002 stone per square foot g/ft² to st/ft² 1 g/ft² = 0.001 stone per square yard g/ft² to st/yd² 1 g/ft² = 3.46 × 10-25 troy ounce per square picometer g/ft² to oz t/pm² 1 g/ft² = 3.46 × 10-21 troy ounce per square angstrom g/ft² to oz t/Ų 1 g/ft² = 3.46 × 10-19 troy ounce per square nanometer g/ft² to oz t/nm² 1 g/ft² = 3.46 × 10-13 troy ounce per square micron g/ft² to oz t/µ² 1 g/ft² = 3.46 × 10-13 troy ounce per square micrometer g/ft² to oz t/µm² 1 g/ft² = 3.46 × 10-7 troy ounce per square millimeter g/ft² to oz t/mm² 1 g/ft² = 3.46 × 10-5 troy ounce per square centimeter g/ft² to oz t/cm² 1 g/ft² = 0.35 troy ounce per square meter g/ft² to oz t/m² 1 g/ft² = 0.03 troy ounce per hectare g/ft² to oz t/ha 1 g/ft² = 0.0002 troy ounce per square inch g/ft² to oz t/inch² 1 g/ft² = 0.03 troy ounce per square foot g/ft² to oz t/ft² 1 g/ft² = 0.29 troy ounce per square yard g/ft² to oz t/yd² 1 g/ft² = 2.88 × 10-26 troy pound per square picometer g/ft² to troy/pm² 1 g/ft² = 2.88 × 10-22 troy pound per square angstrom g/ft² to troy/Ų 1 g/ft² = 2.88 × 10-20 troy pound per square nanometer g/ft² to troy/nm² 1 g/ft² = 2.88 × 10-14 troy pound per square micron g/ft² to troy/µ² 1 g/ft² = 2.88 × 10-14 troy pound per square micrometer g/ft² to troy/µm² 1 g/ft² = 2.88 × 10-8 troy pound per square millimeter g/ft² to troy/mm² 1 g/ft² = 2.88 × 10-6 troy pound per square centimeter g/ft² to troy/cm² 1 g/ft² = 0.03 troy pound per square meter g/ft² to troy/m² 1 g/ft² = 0.003 troy pound per hectare g/ft² to troy/ha 1 g/ft² = 1.86 × 10-5 troy pound per square inch g/ft² to troy/inch² 1 g/ft² = 0.003 troy pound per square foot g/ft² to troy/ft² 1 g/ft² = 0.02 troy pound per square yard g/ft² to troy/yd² 1 g/ft² = 6.92 × 10-24 pennyweight per square picometer g/ft² to dwt/pm² 1 g/ft² = 6.92 × 10-20 pennyweight per square angstrom g/ft² to dwt/Ų 1 g/ft² = 6.92 × 10-18 pennyweight per square nanometer g/ft² to dwt/nm² 1 g/ft² = 6.92 × 10-12 pennyweight per square micron g/ft² to dwt/µ² 1 g/ft² = 6.92 × 10-12 pennyweight per square micrometer g/ft² to dwt/µm² 1 g/ft² = 6.92 × 10-6 pennyweight per square millimeter g/ft² to dwt/mm² 1 g/ft² = 0.001 pennyweight per square centimeter g/ft² to dwt/cm² 1 g/ft² = 6.92 pennyweights per square meter g/ft² to dwt/m² 1 g/ft² = 0.64 pennyweight per hectare g/ft² to dwt/ha 1 g/ft² = 0.004 pennyweight per square inch g/ft² to dwt/inch² 1 g/ft² = 0.64 pennyweight per square foot g/ft² to dwt/ft² 1 g/ft² = 5.79 pennyweights per square yard g/ft² to dwt/yd²
#### Foods, Nutrients and Calories
PEDIATRIC CHOCOLATE SHAKE, UPC: 683744962423 contain(s) 100 calories per 100 grams (≈3.53 ounces) [ price ]
72 foods that contain Xylitol. List of these foods starting with the highest contents of Xylitol and the lowest contents of Xylitol
#### Gravels, Substances and Oils
CaribSea, Freshwater, Flora Max, Midnight weighs 865 kg/m³ (54.00019 lb/ft³) with specific gravity of 0.865 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Potassium bisulfide [KHS or HKS] weighs 1 680 kg/m³ (104.87897 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Canola oil with temperature in the range of 10°C (50°F) to 140°C (284°F)
#### Weights and Measurements
A microwatt (µW) is a derived metric SI (System International) measurement unit of power
The time measurement was introduced to sequence or order events, and to answer questions like these: "How long did the event take?" or "When did the event occur?"
mg/mm² to long tn/yd² conversion table, mg/mm² to long tn/yd² unit converter or convert between all units of surface density measurement.
#### Calculators
Electricity cost calculator per hours, days, weeks, months and years
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http://www.ck12.org/probability/Mutually-Inclusive-Events/postread/Recognize-Overlapping-Events-as-Having-One-or-More-Outcomes-in-Common-Post-Read/
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Mutually Inclusive Events
## Probability of two events that can occur at the same time P(A or B) = P(A) + P(B) - P(A and B)
Estimated15 minsto complete
%
Progress
Practice Mutually Inclusive Events
Progress
Estimated15 minsto complete
%
Recognize Overlapping Events as Having One or More Outcomes in Common Post Read
Teacher Contributed
Develop understanding of concepts by studying them in a relational manner. Analyze and refine the concept by summarizing the main idea, creating visual aids, and generating questions and comments using a Four Square Concept Matrix.
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https://faculty.evansville.edu/ck6/tcenters/recent/schiff.html
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# SCHIFFLER POINT
Let I denote the incenter of a triangle ABC. The Schiffler point of ABC is the point of concurrence of the Euler lines of the four triangles BCI, CAI, ABI, ABC. Trilinear coordinates for the Schiffler point are 1/(cos(B) + cos(C)) : 1/(cos(C) + cos(A)) : 1/(cos(A) + cos(B)), or, equivalently, (b + c - a)/(b + c) : (c + a - b)/(c + a) : (a + b - c)/(a + b), where a, b, c denote the sidelengths of triangle ABC.
(For a quick lesson on trilinears, click on TRILINEAR COORDINATES.)
The Schiffler point is named for Kurt Schiffler (1896-1986), who introduced the point in a problem proposal. The solvers, named below, suggested that the point be named the Schiffler point.
Kurt Schiffler, G. R. Veldkamp, and W. A. van der Spek, Problem 1018 and Solution, Crux Mathematicorum 12 (1986) 176-179.
Biographical Sketch of Kurt Schiffler
Triangle Centers
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https://www.teacherspayteachers.com/Store/Mathful-Learners
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Teach the measurement concepts of heavier and lighter to your kindergarten students with this FULL week of lesson plans focusing on comparing weight and mass. Learners will love to compare measurements of familiar objects, explore heavy and lighter
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Measurement
Kindergarten, 1st, Homeschool
Types:
Worksheets, Activities, Centers
CCSS:
K.MD.A.1, K.MD.A.2
\$3.50
375
Playing these Dice Number Games will keep your students engaged and eager to learn more during your next math centers. Including 15 number games, these printable dice games can easily be used as your warm up activity or as a complete math lesson.
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Basic Operations, Numbers, Statistics
Kindergarten, 1st, 2nd, 3rd, Homeschool
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\$6.95
328
Make teaching and reading calendars fun with these engaging, easy to use Calendar Worksheets. These calendar skill worksheets allow students to practice and learn how to read a calendar on a monthly basis. Students identify days on the calendar,
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Number Bingo is the perfect maths game for your kindergarten students to practice identifying numbers 0 to 20. This is a quick, fun class set (28 student cards included) of math bingo cards for children to practice reading and recognizing numbers to
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Numbers
Kindergarten, 1st, Homeschool
Types:
Games
CCSS:
K.CC.A.3, K.CC.C.7
\$3.00
219
These easy and fun science experiments are guaranteed to engage your class and teach them key scientific concepts at the same time! These easy scientific method activities include 15 science experiments with full scientific explanations to help
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Science
3rd, 4th, 5th, 6th, Homeschool
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418
Reinforce One to One Correspondence (1:1, Cardinality) and Number Sense with this HUGE pack of printable MATH CENTER ACTIVITIES and GOOGLE SLIDES digital resources. Including 18 FUN counting math centers, number sense worksheets and counting
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Numbers, Back to School
PreK, Kindergarten, 1st, Homeschool
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K.CC.A.3, K.CC.B.4, K.CC.B.4a, K.CC.B.4b, K.CC.C.7
\$9.95
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Looking for the perfect resource to help your students with converting fractions to decimals? I’ve created this timesaving math pack to engage your students in hands-on fractions and decimals activities, games, posters and worksheets as I know first
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3rd, 4th, 5th, Homeschool
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CCSS:
4.NF.C.5, 4.NF.C.6, 4.NF.C.7
\$6.95
204
Composing and Decomposing teen numbers is a place value skill that all Kindergarten and Foundation students need to perfect. Teaching teen numbers is EASY with this 30 page booklet of activities, games and worksheets. Also including hands-on
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Kindergarten, 1st, Homeschool
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CCSS:
K.NBT.A.1, 1.NBT.B.2a, 1.NBT.B.2b
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315
Christmas Coupons, the perfect parent present for your students to make! A NO COST GIFT (saving time and money for teachers) that children can put a lot of love into, these Christmas Coupons include everything you need to make a sweet, inexpensive
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1st, 2nd, 3rd, 4th, 5th
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177
Use these Place Value Activities, Worksheets and Games to teach, review and engage your learners through practical place value tasks that are not only fun, but FULL of great learning opportunities. Filled with teaching tools, place value games,
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2nd, 3rd, Homeschool
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CCSS:
3.NBT.A.1, 3.NBT.A.2
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455
Teaching skip counting and wanting something more exciting than just rote learning? These skip counting math art projects are the perfect thing for you! Through these NO PREP, hands on printables, your students will gain an understanding of the link
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2nd, 3rd, 4th, Homeschool
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CCSS:
3.OA.A.3, 3.OA.D.9
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201
Explore the connection between the inverse operations of multiplication and division with ease using these multiplication and division fact family worksheets and task cards. Using the concept of "3 for Free", students use the fact family triangles
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CCSS:
3.OA.A.4, 3.OA.B.6, 3.OA.C.7
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103
Celebrate Australia with your class with these 70 pages of activities! Use as part of your unit on Australia, at Australia Day or at anytime of the year. Containing a HUGE variety of activities, games, printables and worksheets that can be used
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1st, 2nd, 3rd, 4th
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\$7.50
151
Explore the connection between the inverse operations of addition and subtraction with ease using these addition and subtraction fact family worksheets and task cards. Using the concept of "3 for Free", students use the fact family triangles to
Subjects:
Math, Basic Operations
1st, 2nd, 3rd
Types:
Worksheets, Games, Centers
CCSS:
1.NBT.C.4, 2.NBT.B.5, 1.OA.B.3, 1.OA.B.4, 2.OA.B.2
\$4.00
136
Do your students have difficulty following directions in Math? Check your student's understanding of math concepts and their ability to follow directions with these 12 great Math Listening Activities. Concepts covered include colors, shapes, numbers
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Geometry
PreK, Kindergarten
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CCSS:
K.G.A.1
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*** FREE Updates Every Year - Calendar Year Version***Make teaching calendars engaging and easy for you with these fun Calendar Worksheets. Students learn calendar skills and practice reading a calendar on a monthly basis. Students identify days on
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Practice inverse operations in your classroom with this bundle of fact family worksheets and hands-on fact family triangles. These inverse operations worksheets and activities include fact family triangles that help teach the connection between
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Basic Operations
1st, 2nd, 3rd, 4th, Homeschool
Types:
Worksheets, Games, Centers
CCSS:
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\$8.00
\$5.60
101
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Looking for fun, easy to use Christmas Math Worksheets for your 2nd or 3rd Grade students? Engage students in the days before Christmas with these fun and festive Christmas symmetry and pattern worksheets. These Christmas Math Worksheets are NO
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This Procedure Writing pack includes all the resources you need to introduce and consolidate how-to writing in your classroom. A no prep writing prompt pack, simply print and laminate posters, how-to prompt cards and copy procedure writing templates
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Writing
2nd, 3rd, 4th
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Teach Shape, Place Value and Number Sense to your Kindergarten and Year 1 students using this engaging Math Bingo Bundle. Bingo is an easy game to emphasise shape and number identification and can be used anytime you have a spare 10 minutes. All you
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Geometry, Numbers, Place Value
Kindergarten, 1st, 2nd, Homeschool
Types:
Games
CCSS:
K.CC.A.3, K.CC.B.4
\$17.25
\$11.50
69
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TEACHING EXPERIENCE
I started teaching in 2005 - seems so long ago now! Two years of that was spent teaching in California and Washington, the rest of the time in Australia! I love teaching primary age children, knowing that I am influencing future generations. I am passionate about teaching and everything I create is current, purposeful and engaging for children.
MY TEACHING STYLE
Teaching children through inquiry is what I believe to be the best way! I have always loved Maths and Science but now and definitely getting my head around the best ways to teach Reading and Writing to all age groups.
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MY OWN EDUCATIONAL HISTORY
Attended Primary and Secondary education in the country town I grew up in, with both parents being teachers at the High School I attended. In 2004 I completed a Bachelor of Science/Bachelor of Primary Education at Deakin University.
I’m Alison, your number one classroom helper and lover of everything maths! A teacher of 15 years and Mum to 2 little boys, I love nothing more than creating fun maths resources for K-3 students! I’ve taught in K-8 classrooms in California, Washington and Australia, the country I call home. My resources are easy to use, saving you time and link to the standards in fun and engaging ways. My goal, to help you, my teacher friend, get your weekends back and concentrate on teaching your students. Simply leave all your lesson planning up to me! For more information, visit me at www.mathfullearners.com
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http://oeis.org/A257617
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A257617 Triangle read by rows: T(n,k) = t(n-k, k); t(n,m) = f(m)*t(n-1,m) + f(n)*t(n,m-1), where f(x) = 7*x + 2. 9
1, 2, 2, 4, 36, 4, 8, 388, 388, 8, 16, 3676, 12416, 3676, 16, 32, 33564, 283204, 283204, 33564, 32, 64, 303260, 5538184, 13027384, 5538184, 303260, 64, 128, 2732156, 99831564, 465775352, 465775352, 99831564, 2732156, 128 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS Row sums are: 1, 4, 44, 792, 19800, 633600, 24710400, 1136678400, 60243955200, 3614637312000, ... (see A144827). LINKS FORMULA T(n,k) = t(n-k, k); t(0,0) = 1, t(n,m) = 0 if n < 0 or m < 0, else t(n,m) = f(m)*t(n-1,m) + f(n)*t(n,m-1), where f(x) = 7*x + 2. EXAMPLE 1 2 2 4 36 4 8 388 388 8 16 3676 12416 3676 16 32 33564 283204 283204 33564 32 64 303260 5538184 13027384 5538184 303260 64 128 2732156 99831564 465775352 465775352 99831564 2732156 128 CROSSREFS Cf. A142462, A257627. Cf. A038208, A256890, A257609, A257610, A257612, A257614, A257616, A257618, A257619 Similar sequences listed in A256890. Sequence in context: A296048 A327011 A300361 * A309344 A257618 A088895 Adjacent sequences: A257614 A257615 A257616 * A257618 A257619 A257620 KEYWORD nonn,tabl AUTHOR Dale Gerdemann, May 09 2015 STATUS approved
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Last modified February 22 12:33 EST 2020. Contains 332136 sequences. (Running on oeis4.)
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http://mathhelpforum.com/calculus/92155-find-interval-convergence-power-series.html
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# Thread: Find the interval of convergence of the power series
1. ## Find the interval of convergence of the power series
Problem 1:
Find the interval of convergence of the power series
Σ ( 2^(n+1)(x 3)/(n + 3)3^(n1) )
n=1
2. hey just use the standard test that is $\frac{1}{\rho}=\biggl|\frac{a_{n+1}}{a_n}\biggr|$
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http://www.johndcook.com/blog/category/python/
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# Distribution of numbers in Pascal’s triangle
This post explores a sentence from the book Single Digits:
Any number in Pascal’s triangle that is not in the outer two layers will appear at least three times, usually four.
Pascal’s triangle contains the binomial coefficients C(nr) where n ranges over non-negative numbers and r ranges from 0 to n. The outer layers are the elements with r equal to 0, 1, n-1, and n.
We’ll write some Python code to explore how often the numbers up to 1,000,000 appear. How many rows of Pascal’s triangle should we compute? The smallest number on row n is C(n, 2). Now 1,000,000 is between C(1414, 2) and C(1415, 2) so we need row 1414. This means we need N = 1415 below because the row numbers start with 0.
I’d like to use a NumPy array for storing Pascal’s triangle. In the process of writing this code I realized that a NumPy array with dtype int doesn’t contain Python’s arbitrary-sized integers. This makes sense because NumPy is designed for efficient computation, and using a NumPy array to contain huge integers is unnatural. But I’d like to do it anyway, and the way to make it happen is to set dtype to object.
import numpy as np
from collections import Counter
N = 1415 # Number of rows of Pascal's triangle to compute
Pascal = np.zeros((N, N), dtype=object)
Pascal[0, 0] = 1
Pascal[1,0] = Pascal[1,1] = 1
for n in range(2, N):
for r in range(0, n+1):
Pascal[n, r] = Pascal[n-1, r-1] + Pascal[n-1, r]
c = Counter()
for n in range(4, N):
for r in range(2, n-1):
p = Pascal[n, r]
if p <= 1000000:
c[p] += 1
When we run this code, we find that our counter contains 1732 elements. That is, of the numbers up to one million, only 1732 of them appear inside Pascal’s triangle when we disallow the outer two layers. (The second layer contains consecutive integers, so every positive integer appears in Pascal’s triangle. But most integers only appear in the second layer.)
When Single Digits speaks of “Any number in Pascal’s triangle that is not in the outer two layers” this cannot refer to numbers that are not in the outer two layers because every natural number appears in the outer two layers. Also, when it says the number “will appear at least three times, usually four” it is referring to the entire triangle, i.e. including the outer two layers. So another way to state the sentence quoted above is as follows.
Define the interior of Pascal’s triangle to be the triangle excluding the outer two layers. Every number n in the interior of Pascal’s triangle appears twice more outside of the interior, namely as C(n, 1) and C(nn-1). Most often n appears at least twice in the interior as well.
This means that any number you find in the interior of Pascal’s triangle, interior meaning not in the two outer layers, will appear at least three times in the full triangle, usually more.
Here are the statistics from our code above regarding just the interior of Pascal’s triangle.
• One number, 3003, appears six times.
• Six numbers appear four times: 120, 210, 1540, 7140, 11628, and 24310.
• Ten numbers appear only once: 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, and 705432.
• The large majority of numbers, 1715 out of 1732, appear twice.
# Chi-square goodness of fit test example with primes
Yesterday Brian Hayes wrote a post about the distribution of primes. He showed how you could take the remainder when primes are divided by 7 and produce something that looks like rolls of six-sided dice. Here we apply the chi-square goodness of fit test to show that the rolls are too evenly distributed to mimic randomness. This post does not assume you’ve seen the chi-square test before, so it serves as an introduction to this goodness of fit test.
In Brian Hayes’ post, he looks at the remainder when consecutive primes are divided by 7, starting with 11. Why 11? Because it’s the smallest prime bigger than 7. Since no prime is divisible by any other prime, all the primes after 7 will have a remainder of between 1 and 6 inclusive when divided by 7. So the results are analogous to rolling six-sided dice.
The following Python code looks at prime remainders and (pseudo)random rolls of dice and computes the chi-square statistic for both.
First, we import some functions we’ll need.
from sympy import prime
from random import random
from math import ceil
The function prime takes an argument n and returns the nth prime. The function random produces a pseudorandom number between 0 and 1. The ceiling function ceil rounds its argument up to an integer. We’ll use it to convert the output of random into dice rolls.
In this example we’ll use six-sided dice, but you could change num_sides to simulate other kinds of dice. With six-sided dice, we divide by 7, and we start our primes with the fifth prime, 11.
num_sides = 6
modulus = num_sides + 1
# Find the index of the smallest prime bigger than num_sides
index = 1
while prime(index) <= modulus:
index += 1
We’re going to take a million samples and count how many times we see 1, 2, …, 6. We’ll keep track of our results in an array of length 7, wasting a little bit of space since the 0th slot will always be 0. (Because the remainder when dividing a prime by a smaller number is always positive.)
# Number of samples
N = 1000000
observed_primes = [0]*modulus
observed_random = [0]*modulus
Next we “roll” our dice two ways, using prime remainders and using a pseudorandom number generator.
for i in range(index, N+index):
m = prime(i) % modulus
observed_primes[m] += 1
m = int(ceil(random()*num_sides))
observed_random[m] += 1
The chi-square goodness of fit test depends on the observed number of events in each cell and the expected number. We expect 1/6th of the rolls to land in cell 1, 2, …, 6 for both the primes and the random numbers. But in a general application of the chi-square test, you could have a different expected number of observations in each cell.
expected = [N/num_sides for i in range(1, modulus)]
The chi-square test statistic sums (O – E)2/E over all cells, where O stands for “observed” and E stands for “expected.”
def chisq_stat(O, E):
return sum( [(o - e)**2/e for (o, e) in zip(O, E)] )
Finally, we compute the chi-square statistic for both methods.
ch = chisq_stat(observed_primes[1:], expected[1:])
print(ch)
ch = chisq_stat(observed_random[1:], expected[1:])
print(ch)
Note that we chop off the first element of the observed and expected lists to get rid of the 0th element that we didn’t use.
When I ran this I got 0.01865 for the prime method and 5.0243 for the random method. Your results for the prime method should be the same, though you might have a different result for the random method.
Now, how do we interpret these results? Since we have six possible outcomes, our test statistics has a chi-square distribution with five degrees of freedom. It’s one less than the number of possibilities because the total counts have to sum to N; if you know how many times 1, 2, 3, 4, and 5 came up, you can calculate how many times 6 came up.
A chi-square distribution with ν degrees of freedom has expected value ν. In our case, we expect a value around 5, and so the chi-square value of 5.0243 is unremarkable. But the value of 0.01864 is remarkably small. A large chi-square statistics would indicate a poor fit, the observed numbers being suspiciously far from their expected values. But a small chi-square value suggests the fit is suspiciously good, closer to the expected values than we’d expect of a random process.
We can be precise about how common or unusual a chi-square statistic is by computing the probability that a sample from the chi square distribution would be larger or smaller. The cdf gives the probability of seeing a value this small or smaller, i.e. a fit this good or better. The sf gives the probability of seeing a value this larger or larger, i.e. a fit this bad or worse. (The scipy library uses sf for “survival function,” another name for the ccdf, complementary cumulative distribution function).
from scipy.stats import chi2
print(chi2.cdf(ch, num_sides-1), chi2.sf(ch, num_sides-1))
This says that for the random rolls, there’s about a 41% chance of seeing a better fit and a 59% chance of seeing a worse fit. Unremarkable.
But it says there’s only a 2.5 in a million chance of seeing a better fit than we get with prime numbers. The fit is suspiciously good. In a sense this is not surprising: prime numbers are not random! And yet in another sense it is surprising since there’s a heuristic that says primes act like random numbers unless there’s a good reason why in some context they don’t. This departure from randomness is the subject of research published just this year.
If you look at dice with 4 or 12 sides, you get a suspiciously good fit, but not as suspicious as with 6 sides. But with 8 or 20-sided dice you get a very bad fit, so bad that its probability underflows to 0. This is because the corresponding moduli, 9 and 21, are composite, which means some of the cells in our chi-square test will have no observations. (Suppose m has a proper factor a. Then if a prime p were congruent to a mod m, p would be have to be divisible by a.)
Update: See the next post for a systematic look at different moduli.
You don’t have to use “dice” that correspond to regular solids. You could consider 10-sided “dice,” for example. For such numbers it may be easier to think of spinners than dice, a spinner with 10 equal arc segments it could fall into.
Related post: Probability that a number is prime
# How to create Green noise in Python
This is a follow-on to my previous post on green noise. Here we create green noise with Python by passing white noise through a Butterworth filter.
Green noise is in the middle of the audible spectrum (on the Bark scale), just where our hearing is most sensitive, analogous to the green light, the frequency where our eyes are most sensitive. See previous post for details, including an explanation of where the left and right cutoffs below come from.
Here’s the code:
from scipy.io.wavfile import write
from scipy.signal import buttord, butter, filtfilt
from scipy.stats import norm
from numpy import int16
def turn_green(signal, samp_rate):
# start and stop of green noise range
left = 1612 # Hz
right = 2919 # Hz
nyquist = (samp_rate/2)
left_pass = 1.1*left/nyquist
left_stop = 0.9*left/nyquist
right_pass = 0.9*right/nyquist
right_stop = 1.1*right/nyquist
(N, Wn) = buttord(wp=[left_pass, right_pass],
ws=[left_stop, right_stop],
gpass=2, gstop=30, analog=0)
(b, a) = butter(N, Wn, btype='band', analog=0, output='ba')
return filtfilt(b, a, signal)
def to_integer(signal):
# Take samples in [-1, 1] and scale to 16-bit integers,
# values between -2^15 and 2^15 - 1.
signal /= max(signal)
return int16(signal*(2**15 - 1))
N = 48000 # samples per second
white_noise= norm.rvs(0, 1, 3*N) # three seconds of audio
green = turn_green(white_noise, N)
write("green_noise.wav", N, to_integer(green))
And here’s what it sounds like:
Let’s look at the spectrum to see whether it looks right. We’ll use one second of the signal so the x-axis coincides with frequency when we plot the FFT.
from scipy.fftpack import fft
one_sec = green[0:N]
plt.plot(abs(fft(one_sec)))
plt.xlim((1500, 3000))
plt.show()
Here’s the output, concentrated between 1600 and 3000 Hz as expected:
# How to digitize a graph
Suppose you have a graph of a function, but you don’t have an equation for it or the data that produced it. How can you reconstruction the function?
There are a lot of software packages to digitize images. For example, Web Plot Digitizer is one you can use online. Once you have digitized the graph at a few points, you can fit a spline to the points to approximately reconstruct the function. Then as a sanity check, plot your reconstruction to see if it looks like the original. It helps to have the same aspect ratio so you’re not distracted by something that doesn’t matter, and so that differences that do matter are easier to see.
For example, here is a graph from Zwicker and Fastl’s book on psychoacoustics. It contains many graphs with no data or formulas. This particular one gives the logarithmic transmission factor between free field and the peripheral hearing system.
Here’s Python code to reconstruct the functions behind these two curves.
import matplotlib.pyplot as plt
import numpy as np
from scipy import interpolate
curve_names = ["Free", "Diffuse"]
plot_styles = { "Free" : 'b-', "Diffuse" : 'g:'}
data = {}
for name in curve_names:
data = np.loadtxt("{}.csv".format(name), delimiter=',')
x = data[:,0]
y = data[:,1]
spline = interpolate.splrep(x, y)
xnew = np.linspace(0, max(x), 100)
ynew = interpolate.splev(xnew, spline, der=0)
plt.plot(xnew, ynew, plot_styles[name])
logical_x_range = 24 # Bark
logical_y_range = 40 # dB
physical_x_range = 7 # inch
physical_y_range = 1.625 # inch
plt.legend(curve_names, loc=2)
plt.xlabel("critical-band rate")
plt.ylabel("attenuation")
plt.xlim((0, logical_x_range))
plt.axes().set_aspect(
(physical_y_range/logical_y_range) /
(physical_x_range/logical_x_range) )
ax = plt.gca()
ax.get_xaxis().set_ticks([0, 4, 8, 12, 16, 20, 24])
ax.get_yaxis().set_ticks([-10, 0, 10, 20, 30])
plt.show()
Here’s the reconstructed graph.
# Creating police siren sounds with frequency modulation
Yesterday I was looking into calculating fluctuation strength and playing around with some examples. Along the way I discovered how to create files that sound like police sirens. These are sounds with high fluctuation strength.
The Python code below starts with a carrier wave at fc = 1500 Hz. Not surprisingly, this frequency is near where hearing is most sensitive. Then this signal is modulated with a signal with frequency fm. This frequency determines the frequency of the fluctuations.
The slower example produced by the code below sounds like a police siren. The faster example makes me think more of an ambulance or fire truck. Next time I hear an emergency vehicle I’ll pay more attention.
If you use a larger value of the modulation index β and a smaller value of the modulation frequency fm you can make a sound like someone tuning a radio, which is no coincidence.
Here are the output audio files in .wav format:
from scipy.io.wavfile import write
from numpy import arange, pi, sin, int16
def f(t, f_c, f_m, beta):
# t = time
# f_c = carrier frequency
# f_m = modulation frequency
# beta = modulation index
return sin(2*pi*f_c*t - beta*sin(2*f_m*pi*t))
def to_integer(signal):
# Take samples in [-1, 1] and scale to 16-bit integers,
# values between -2^15 and 2^15 - 1.
return int16(signal*(2**15 - 1))
N = 48000 # samples per second
x = arange(3*N) # three seconds of audio
data = f(x/N, 1500, 2, 100)
write("slow.wav", N, to_integer(data))
data = f(x/N, 1500, 8, 100)
write("fast.wav", N, to_integer(data))
Related posts:
# The empty middle: why no one is average
In 1945, a Cleveland newspaper held a contest to find the woman whose measurements were closest to average. This average was based on a study of 15,000 women by Dr. Robert Dickinson and embodied in a statue called Norma by Abram Belskie. Out of 3,864 contestants, no one was average on all nine factors, and fewer than 40 were close to average on five factors. The story of Norma and the Cleveland contest is told in Todd Rose’s book The End of Average.
People are not completely described by a handful of numbers. We’re much more complicated than that. But even in systems that are well described by a few numbers, the region around the average can be nearly empty. I’ll explain why that’s true in general, then look back at the Norma example.
## General theory
Suppose you have N points, each described by n independent, standard normal random variables. That is, each point has the form (x1, x2, x2, …, xn) where each xi is independent with a normal distribution with mean 0 and variance 1. The expected value of each coordinate is 0, so you might expect that most points are piled up near the origin (0, 0, 0, …, 0). In fact most points are in spherical shell around the origin. Specifically, as n becomes larger, most of the points will be in a thin shell with distance √n from the origin. (More details here.)
## Simulated contest
In the contest above, n = 9, and so we expect most contestants to be about a distance of 3 from average when we normalize each of the factors being measured, i.e. we subtract the mean so that each factor has mean 0, and we divide each by its standard deviation so the standard deviation is 1 on each factor.
We’ve made several simplifying assumptions. For example, we’ve assumed independence, though presumably some of the factors measured in the contest were correlated. There’s also a selection bias: presumably women who knew they were far from average would not have entered the contest. But we’ll run with our simplified model just to see how it behaves in a simulation.
import numpy as np
# Winning critera: minimum Euclidean distance
def euclidean_norm(x):
return np.linalg.norm(x)
# Winning criteria: min-max
def max_norm(x):
return max(abs(x))
n = 9
N = 3864
# Simulated normalized measurements of contestants
M = np.random.normal(size=(N, n))
euclid = np.empty(N)
maxdev = np.empty(N)
for i in range(N):
euclid[i] = euclidean_norm(M[i,:])
maxdev[i] = max_norm(M[i,:])
w1 = euclid.argmin()
w2 = maxdev.argmin()
print( M[w1,:] )
print( euclidean_norm(M[w1,:]) )
print( M[w2,:] )
print( max_norm(M[w2,:]) )
There are two different winners, depending on how we decide the winner. Using the Euclidean distance to the origin, the winner in this simulation was contestant 3306. Her normalized measurements were
[ 0.1807, 0.6128, -0.0532, 0.2491, -0.2634, 0.2196, 0.0068, -0.1164, -0.0740]
corresponding to a Euclidean distance of 0.7808.
If we judge the winner to be the one whose largest deviation from average is the smallest, the winner is contestant 1916. Her normalized measurements were
[-0.3757, 0.4301, -0.4510, 0.2139, 0.0130, -0.2504, -0.1190, -0.3065, -0.4593]
with the largest deviation being the last, 0.4593.
By either measure, the contestant closest to the average deviated significantly from the average in at least one dimension.
* * *
For daily posts on probability, follow @ProbFact on Twitter.
# Maximum principle and approximating boundary value problems
Solutions to differential equations often satisfy some sort of maximum principle, which can in turn be used to construct upper and lower bounds on solutions.
We illustrate this in one dimension, using a boundary value problem for an ordinary differential equation (ODE).
## Maximum principles
If the second derivative of a function is positive over an open interval (ab), the function cannot have a maximum in that interval. If the function has a maximum over the closed interval [ab] then it must occur at one of the ends, at a or b.
This can be generalized, for example, to the following maximum principle. Let L be the differential operator
L[u] = u” + g(x)u’ + h(x)
where g and h are bounded functions on some interval [a, b] and h is non-positive. Suppose L[u] ≥ 0 on (a, b). If u has an interior maximum, then u must be constant.
## Boundary value problems
Now suppose that we’re interested in the boundary value problem L[u] = f where we specify the values of u at the endpoints a and b, i.e. u(a) = ua and u(b) = ub. We can construct an upper bound on u as follows.
Suppose we find a function z such that L[z] ≤ f and z(a) ≥ ua and z(b) ≥ ub. Then by applying the maximum principle to u – z, we see that u – z must be ≤ 0, and so z is an upper bound for u.
Similarly, suppose we find a function w such that L[w] ≥ f and w(a) ≤ ua and w(b) ≤ ub. Then by applying the maximum principle to w – u, we see that w – u must be ≤ 0, and so w is an lower bound for u.
Note that any functions z and w that satisfy the above requirements give upper and lower bounds, though the bounds may not be very useful. By being clever in our choice of z and w we may be able to get tighter bounds. We might start by choosing polynomials, exponentials, etc. Any functions that are easy to work with and see how good the resulting bounds are.
Tomorrow’s post is similar to this one but looks at bounds for an initial value problem rather than a boundary value problem.
## Airy equation example
The following is an elaboration on an example from [1]. Suppose we want to bound solutions to
u”(x) – x u(x) = 0
where u(0) = 0 and u(1) = 1. (This is a well-known equation, but for purposes of illustration we’ll pretend at first that we know nothing about its solutions.)
For our upper bound, we can simply use z(x) = x. We have L[z] ≤ 0 and z satisfies the boundary conditions exactly.
For our lower bound, we use w(x) = x – βx(1 – x). Why? The function z already satisfies the boundary condition. If we add some multiple of x(1 – x) we’ll maintain the boundary condition since x(1 – x) is zero at 0 and 1. The coefficient β gives us some room to maneuver. Turns out L[w] ≥ 0 if β ≥ 1/2. If we choose β = 1/2 we have
(xx2)/2 ≤ u(x) ≤ x
In general, you don’t know the function you’re trying to bound. That’s when bounds are most useful. But this is a sort of toy example because we do know the solution. The equation in this example is well known and is called Airy’s equation. The Airy functions Ai and Bi are independent solutions. Here’s a plot of the solution with its upper and lower bounds.
Here’s the Python code I used to solve for the coefficients of Ai and Bi and make the plot.
import numpy as np
from scipy.linalg import solve
from scipy.special import airy
import matplotlib.pyplot as plt
# airy(x) returns (Ai(x), Ai'(x), Bi(x), Bi'(x))
def Ai(x):
return airy(x)[0]
def Bi(x):
return airy(x)[2]
M = np.matrix([[Ai(0), Bi(0)], [Ai(1), Bi(1)]])
c = solve(M, [0, 1])
t = np.linspace(0, 1, 100)
plt.plot(t, (t + t**2)/2, 'r-', t, c[0]*Ai(t) + c[1]*Bi(t), 'k--', t, t, 'b-',)
plt.legend(["lower bound $(x + x^2)/2$",
"exact solution $c_0Ai + c_1Bi$",
"upper bound $x$"], loc="upper left")
plt.show()
SciPy’s function airy has an optimization that we waste here. The function computes Ai and Bi and their first derivatives all at the same time. We could take advantage of that to remove some redundant computations, but that would make the code harder to read. We chose instead to wait an extra nanosecond for the plot.
Help with differential equations
* * *
[1] Murray Protter and Hans Weinberger. Maximum Principles in Differential Equations.
# Musical pitch notation
How can you convert the frequency of a sound to musical notation? I wrote in an earlier post how to calculate how many half steps a frequency is above or below middle C, but it would be useful go further have code to output musical pitch notation.
In scientific pitch notation, the C near the threshold of hearing, around 16 Hz, is called C0. The C an octave higher is C1, the next C2, etc. Octaves begin with C; other notes use the octave number of the closest C below.
The lowest note on a piano is A0, a major sixth up from C0. Middle C is C4 because it’s 4 octaves above C0. The highest note on a piano is C8.
## Math
A4, the A above middle C, has a frequency of 440 Hz. This is nine half steps above C4, so the pitch of C4 is 440*2-9/12. C0 is four octaves lower, so it’s 2-4 = 1/16 of the pitch of C4. (Details for this calculation and the one below are given in here.)
For a pitch P, the number of half steps from C0 to P is
h = 12 log2(P / C0).
## Software
Here is a page that will let you convert back and forth between frequency and music notation: Music, Hertz, Barks.
If you’d like code rather than just to do one calculation, see the Python code below. It calculates the number of half steps h from C0 up to a pitch, then computes the corresponding pitch notation.
from math import log2, pow
A4 = 440
C0 = A4*pow(2, -4.75)
name = ["C", "C#", "D", "D#", "E", "F", "F#", "G", "G#", "A", "A#", "B"]
def pitch(freq):
h = round(12*log2(freq/C0))
octave = h // 12
n = h % 12
return name[n] + str(octave)
The pitch for A4 is its own variable in case you’d like to modify the code for a different tuning. While 440 is common, it used to be lower in the past, and you’ll sometimes see higher values like 444 today.
If you’d like to port this code to a language that doesn’t have a log2 function, you can use log(x)/log(2) for log2(x).
## Powers of 2
When scientific pitch notation was first introduced, C0 was defined to be exactly 16 Hz, whereas now it works out to around 16.35. The advantage of the original system is that all C’s have frequency a power of 2, i.e. Cn has frequency 2n+4 Hz. The formula above for the number of half steps a pitch is above C0 simplifies to
h = 12 log2P – 48.
If C0 has frequency 16 Hz, the A above middle C has frequency 28.75 = 430.54, a little flat compared to A 440. But using the A 440 standard, C0 = 16 Hz is a convenient and fairly accurate approximation.
# General birthday problem
The birthday problem, sometimes called the birthday paradox, says that it’s more likely than you’d expect that two people in a group have the same birthday. Specifically, in a random sample of 23 people, there’s about a 50-50 chance that two people share the same birthday.
The birthday problem makes a nice party trick, but generalizations of the problem come up frequently in applications. I wrote in the previous post how it comes up in seeding distributed Monte Carlo simulations. In computer science, it’s a concern in hashing algorithms.
If you have a set of N things to choose from, such as N = 365 birthdays, and take r samples, the probability that all r samples are unique is
and the probability that at least two of the samples are the same is 1 – p. (This assumes that N is at least as big as r. Otherwise the denominator is undefined, but in that case we know p is 0.)
With moderately large values of N and r the formula is likely to overflow if implemented directly. So as usual the trick is to use logarithms to avoid overflow or underflow. Here’s how you could compute the probability above in Python. SciPy doesn’t have a log factorial function, but does have a log gamma function, so we use that instead.
from scipy import exp, log
from scipy.special import gammaln
def prob_unique(N, r):
return exp( gammaln(N+1) - gammaln(N-r+1) - r*log(N) )
# Spectral coordinates in Python
A graph doesn’t have any geometric structure unless we add it. The vertices don’t come with any position in space. The same graph can look very different when arranged different ways.
Spectral coordinates are a natural way to draw a graph because they are determined by the properties of the graph, not arbitrary aesthetic choices. Construct the Laplacian matrix and let x and y be the eigenvectors associated with the second and third eigenvalues. (The smallest eigenvalue is always zero and has an eigenvector of all 1’s. The second and third eigenvalues and eigenvectors are the first to contain information about a graph.) The spectral coordinates of the ith node are the ith components of x and y.
We illustrate this with a graph constructed from a dodecahedron, a regular solid with twenty vertices and twelve pentagonal faces. You can make a dodecahedron from a soccer ball by connecting the centers of all the white hexagons. Here’s one I made from Zometool pieces for a previous post:
Although we’re thinking of this graph as sitting in three dimensions, the nodes being the corners of pentagons etc., the graph simply says which vertices are connected to each other. But from this information, we can construct the graph Laplacian and use it to assign plane coordinates to each point. And fortunately, this produces a nice picture:
Here’s how that image was created using Python’s NetworkX library.
import networkx as nx
import matplotlib.pyplot as plt
from scipy.linalg import eigh
# Read in graph and compute the Laplacian L ...
# Laplacian matrices are real and symmetric, so we can use eigh,
# the variation on eig specialized for Hermetian matrices.
w, v = eigh(L) # w = eigenvalues, v = eigenvectors
x = v[:,1]
y = v[:,2]
spectral_coordinates = {i : (x[i], y[i]) for i in range(n)}
G = nx.Graph()
nx.draw(G, pos=spectral_coordinates)
plt.show()
Update: After posting this I discovered that NetworkX has a method draw_spectral that will compute the spectral coordinates for you.
Related:
# Estimating the exponent of discrete power law data
Suppose you have data from a discrete power law with exponent α. That is, the probability of an outcome n is proportional to n. How can you recover α?
A naive approach would be to gloss over the fact that you have discrete data and use the MLE (maximum likelihood estimator) for continuous data. That does a very poor job [1]. The discrete case needs its own estimator.
To illustrate this, we start by generating 5,000 samples from a discrete power law with exponent 3.
import numpy.random
alpha = 3
n = 5000
x = numpy.random.zipf(alpha, n)
The continuous MLE is very simple to implement:
alpha_hat = 1 + n / sum(log(x))
Unfortunately, it gives an estimate of 6.87 for alpha, though we know it should be around 3.
The MLE for the discrete power law distribution satisfies
Here ζ is the Riemann zeta function, and xi are the samples. Note that the left side of the equation is the derivative of log ζ, or what is sometimes called the logarithmic derivative.
There are three minor obstacles to finding the estimator using Python. First, SciPy doesn’t implement the Riemann zeta function ζ(x) per se. It implements a generalization, the Hurwitz zeta function, ζ(x, q). Here we just need to set q to 1 to get the Riemann zeta function.
Second, SciPy doesn’t implement the derivative of zeta. We don’t need much accuracy, so it’s easy enough to implement our own. See an earlier post for an explanation of the implementation below.
Finally, we don’t have an explicit equation for our estimator. But we can easily solve for it using the bisection algorithm. (Bisect is slow but reliable. We’re not in a hurry, so we might as use something reliable.)
from scipy import log
from scipy.special import zeta
from scipy.optimize import bisect
xmin = 1
def log_zeta(x):
return log(zeta(x, 1))
def log_deriv_zeta(x):
h = 1e-5
return (log_zeta(x+h) - log_zeta(x-h))/(2*h)
t = -sum( log(x/xmin) )/n
def objective(x):
return log_deriv_zeta(x) - t
a, b = 1.01, 10
alpha_hat = bisect(objective, a, b, xtol=1e-6)
print(alpha_hat)
We have assumed that our data follow a power law immediately from n = 1. In practice, power laws generally fit better after the first few elements. The code above works for the more general case if you set xmin to be the point at which power law behavior kicks in.
The bisection method above searches for a value of the power law exponent between 1.01 and 10, which is somewhat arbitrary. However, power law exponents are very often between 2 and 3 and seldom too far outside that range.
The code gives an estimate of α equal to 2.969, very near the true value of 3, and much better than the naive estimate of 6.87.
Of course in real applications you don’t know the correct result before you begin, so you use something like a confidence interval to give you an idea how much uncertainty remains in your estimate.
The following equation [2] gives a value of σ from a normal approximation to the distribution of our estimator.
So an approximate 95% confidence interval would be the point estimate +/- 2σ.
from scipy.special import zeta
from scipy import sqrt
def zeta_prime(x, xmin=1):
h = 1e-5
return (zeta(x+h, xmin) - zeta(x-h, xmin))/(2*h)
def zeta_double_prime(x, xmin=1):
h = 1e-5
return (zeta(x+h, xmin) -2*zeta(x,xmin) + zeta(x-h, xmin))/h**2
def sigma(n, alpha_hat, xmin=1):
z = zeta(alpha_hat, xmin)
temp = zeta_double_prime(alpha_hat, xmin)/z
temp -= (zeta_prime(alpha_hat, xmin)/z)**2
return 1/sqrt(n*temp)
print( sigma(n, alpha_hat) )
Here we use a finite difference approximation for the second derivative of zeta, an extension of the idea used above for the first derivative. We don’t need high accuracy approximations of the derivatives since statistical error will be larger than the approximation error.
In the example above, we have α = 2.969 and σ = 0.0334, so a 95% confidence interval would be [2.902, 3.036].
* * *
[1] Using the continuous MLE with discrete data is not so bad when the minimum output xmin is moderately large. But here, where xmin = 1 it’s terrible.
[2] Equation 3.6 from Power-law distributions in empirical data by Aaron Clauset, Cosma Rohilla Shalizi, and M. E. J. Newman.
# Numerical differentiation
Today I needed to the derivative of the zeta function. SciPy implements the zeta function, but not its derivative, so I needed to write my own version.
The most obvious way to approximate a derivative would be to simply stick a small step size into the definition of derivative:
f’(x) ≈ (f(x+h) – f(x)) / h
However, we could do much better using
f’(x) ≈ (f(x+h) – f(x-h)) / 2h
To see why, expand f(x) in a power series:
f(x + h) = f(x) + h f‘(x) + h2 f”(x)/2 + O(h3)
A little rearrangement shows that the error in the one-sided difference, the first approximation above, is O(h). Now if you replace h with –h and do a little algebra you can also show that the two-sided difference is O(h2). When h is small, h2 is very small, so the two-sided version will be more accurate for sufficiently small h.
So how small should h be? The smaller the better, in theory. In computer arithmetic, you lose precision whenever you subtract two nearly equal numbers. The more bits two numbers share, the more bits of precision you may lose in the subtraction. In my application, h = 10-5 works well: the precision after the subtraction in the numerator is comparable to the precision of the (two-sided) finite difference approximation. The following code was adequate for my purposes.
from scipy.special import zeta
def zeta_prime(x):
h = 1e-5
return (zeta(x+h,1) - zeta(x-h,1))/(2*h)
The zeta function in SciPy is Hurwitz zeta function, a generalization of the Riemann zeta function. Setting the second argument to 1 gives the Riemann zeta function.
There’s a variation on the method above that works for real-valued functions that extend to a complex analytic function. In that case you can use the complex step differentiation trick to use
Im( f(x+ih)/h )
to approximate the derivative. It amounts to the two-sided finite difference above, except you don’t need to have a computer carry out the subtraction, and so you save some precision. Why’s that? When x is real, xih and xih are complex conjugates, and f(x – ih) is the conjugate of f(x + ih), i.e. conjugation and function application commute in this setting. So (f(x+ih) – f(x-ih)) is twice the imaginary part of f(x + ih).
SciPy implements complex versions many special functions, but unfortunately not the zeta function.
# Anthony Scopatz on xonsh and shells in general
Anthony Scopatz did an interview for Podcast.__init__ recently talking about xonsh, a command shell that blends Python and some traditions from bash. One line from the interview jumped out at me:
… thinking very critically about what shells get used for and what they’re actually good at and what they’re not good at.
I’ve wondered about this but never reached any satisfying conclusions. I was curious to hear Anthony’s ideas, so I asked him for another interview. (I interviewed Anthony and his co-author Katy Huff regarding their book Effective Computation in Physics.
* * *
JC: If your shell speaks your programming language, then what else does it need to do?
AS: It’s an interesting question. People have tried to use Python as a shell for years and years and they came up with a bunch of different potential solutions, but none of them quite worked because the language wasn’t built around that idea. It ended up being more verbose than people want from a shell. The main purpose of the shell, in my opinion, is to run other code and to glue things together. Python does that really well for libraries and functions, but it doesn’t do that so well for executables. Bash deals with executables really well, but it’s terrible for dealing with even simple conditional logic. Like a lot of people, I wanted something that would do all these things simultaneously and do them all well. But you quickly end up where many traditional computer science people are not willing to go: context-sensitive parsing. It’s something they teach you to be afraid of in school .
JC: But you do it all the time. How can you get away from it?
AS: You can’t, but people want to avoid it in their core languages. The major programming languages keep it out. You’ll find it quarantined to domain-specific languages where the damage is small.
JC: So you have something in mind like Perl? There the behavior of a function can depend entirely on whether it’s being used in a scalar context or an array context.
AS: That’s right. Perl does some of this. The language Forth is completely built around this. It’s all context-sensitive.
You brought up something interesting [in a previous email] about the overlap between shells and editors. Those things are completely separate in my mind, but for a lot of people they get merged very quickly. For instance, Emacs has the ability to run a shell inside the editor, and people use that all the time.
JC: The way I work is that I start something at the command line, then it gets a little complicated, and I switch over to writing a script and regret not having done that sooner. I especially do that with something like R. This is just going to be a few quick calculations, so I’ll do it right from the REPL. Then things get more complicated …
AS: IPython sorta has that too, the old IPython readline shell. You just wanted to do something simple that bash couldn’t do quickly or easily, so you open up the IPython command line. Inevitably it ends up taking more lines than you wanted it to. That is part of why the Jupyter notebook is so great.
JC: One thing I noticed about PowerShell was that system administrators were ecstatic when it came out and would say how much they loved the command line. Then Microsoft put out this ISE, sort of an IDE for PowerShell, and everyone moved there. So they’re not really using the command line anymore. They’re excited about PowerShell as a programming language, not as an interactive shell per se.
In Bruce Payette’s PowerShell book he fields questions asking why PowerShell did something some way they find odd and his answer is always “Because it’s a shell.”
AS: Do you have any examples?
JC: For example, functions don’t use parentheses around their arguments or commas between their arguments because that’s not what people expect from a shell. You expect to type something like ls, not ls() with parentheses at the end. There were more subtle examples than this, but they’re not fresh on my mind.
AS: That’s where I think that tools like Python plumbum are lacking. It’s an all-Python environment, so you have to use Python syntax even when it’s cumbersome. It prevents you from having to import subprocess and worry about that all the time, but it doesn’t do much more than that.
JC: When you were writing xonsh, where there times you wished you could change the Python language? Or things you’d do differently in the shell if you weren’t aiming for 100% Python compatibility?
AS: That’s interesting. Python is deceptively simple. It has a lot of little pieces to it. It’s very natural and intuitive to use, but re-implementing the parser for Python was more work than I expected. There are a lot of little gotchas in the parser. I spent a lot of time on tuples and function argument grouping. The way they’re handled looks very similar but they’re handled completely differently for no reason that’s readily apparent.
There’s also this ambiguity between Python commands and shell commands if you’re trying to do both simultaneously, and that’s frustrating. That’s the hard part, figuring out when you’re in a subprocess and when you’re in Python mode.
JC: It’s hard for you as an implementer, but hopefully users can be blissfully ignorant of the issues and it just does what they expect.
I guess you’re walking a fine line, because as soon as you say you want the shell to infer what people mean, you start getting into the kinds of complications you have in Perl where things depend so heavily on context, and that sort of thing is contrary to the spirit of Python.
AS: Yeah, exactly! After going through this exercise, there is one thing I’d like to change about Python. Python is white space-sensitive at the beginning of a line, but not after the first non-white space character. For example, you can put as many spaces around a binary operator as you like, or none at all. That’s really, really frustrating. If you enforced PEP 8, requiring exactly one white space around every binary operator, you’d be able to resolve these currently ambiguous cases between subprocess mode and Python mode very naturally. But I can’t imagine a world in which people would agree to this.
JC: What shell would you use if you weren’t using xonsh?
AS: I probably would use bash. Fish is really nice in some ways, and things like zsh have nice features too. What I used to do is go back and forth between working in an IPython shell and a bash shell, and between those two I could pretty much get the job done.
JC: Do you use Emacs?
AS: No, I don’t use Emacs or Vim or any of those editors. I use an editor I wrote, kinda like nano. I’ve used Emacs and Vim, but they got in my way too much, so I wanted something else. This is sort of the same thing as xonsh; I want my tools to get out of my way. I want the barrier to entry to doing what I want to be basically zero. You can spend years and years becoming a master of some of these tools and then you’re really effective, but I want to just open up the editor and start typing text. The same thing with the shell. I just want to open it up and get to work and not have to keep going back to the documentation.
# Distance to Mars
The distance between the Earth and Mars depends on their relative positions in their orbits and varies quite a bit over time. This post will show how to compute the approximate distance over time. We’re primarily interested in Earth and Mars, though this shows how to calculate the distance between any two planets.
The planets have elliptical orbits with the sun at one focus, but these ellipses are nearly circles centered at the sun. We’ll assume the orbits are perfectly circular and lie in the same plane. (Now that Pluto is not classified as a planet, we can say without qualification that the planets have nearly circular orbits. Pluto’s orbit is much more elliptical than any of the planets.)
We can work in astronomical units (AUs) so that the distance from the Earth to the sun is 1. We can also work in units of years so that the period is also 1. Then we could describe the position of the Earth at time t as exp(2πit).
Mars has a larger orbit and a longer period. By Kepler’s third law, the size of the orbit and the period are related: the square of the period is proportional to the cube of the radius. Because we’re working in AUs and years, the proportionality constant is 1. If we denote the radius of Mars’ orbit by r, then its orbit can be described by
r exp(2πi (r-3/2 t ))
Here we pick our initial time so that at t = 0 the two planets are aligned.
The distance between the planets is just the absolute value of the difference between their positions:
| exp(2πit) – r exp(2πi (r-3/2 t)) |
The following code computes and plots the distance from Earth to Mars over time.
from scipy import exp, pi, absolute, linspace
import matplotlib.pyplot as plt
def earth(t):
return exp(2*pi*1j*t)
def mars(t):
r = 1.524 # semi-major axis of Mars orbit in AU
return r*exp(2*pi*1j*(r**-1.5*t))
def distance(t):
return absolute(earth(t) - mars(t))
x = linspace(0, 20, 1000)
plt.plot(x, distance(x))
plt.xlabel("Time in years")
plt.ylabel("Distance in AU")
plt.ylim(0, 3)
plt.show()
And the output looks like this:
Notice that the distance varies from about 0.5 to about 2.5. That’s because the radius of Mars’ orbit is about 1.5 AU. So when the planets are exactly in phase, they are 0.5 AU apart and when they’re exactly out of phase they are 2.5 AU apart. In other words the distance ranges from 1.5 – 1 to 1.5 + 1.
The distance function seems to be periodic with period about 2 years. We can do a little calculation by hand to show that is the case and find the period exactly.
The distance squared is the distance times its complex conjugate. If we let ω = -3/2 then the distance squared is
d2(t) = (exp(2πit) – r exp(2πiωt)) (exp(-2πit) – r exp(-2πiωt))
which simplifies to
1 + r2 – 2r cos(2π(1 – ω)t)
and so the (squared) distance is periodic with period 1/(1 – ω) = 2.13.
Notice that the plot of distance looks more angular at the minima and more rounded near the maxima. Said another way, the distance changes more rapidly when the planets leave their nearest approach than their furthest approach. You can prove this by taking square root of d2(t) and computing its derivative.
Let f(t) = 1 + r2 – 2r cos(2π(1 – ω)t). By the chain rule, the derivative of the square root of f(t) is 1/2 f(t)-1/2 f‘(t). Near a maximum or a minimum, f‘(t) takes on the same values. But the term f(t)-1/2 is largest when f(t) is smallest and vice versa because of the negative exponent.
# Julia for Python programmers
One of my clients is writing software in Julia so I’m picking up the language. I looked at Julia briefly when it first came out but haven’t used it for work. My memory of the language was that it was almost a dialect of Python. Now that I’m looking at it a little closer, I can see more differences, though the most basic language syntax is more like Python than any other language I’m familiar with.
Here are a few scattered notes on Julia, especially on how it differs from Python.
• Array indices in Julia start from 1, like Fortran and R, and unlike any recent language that I know of.
• Like Python and many other scripting languages, Julia uses # for one-line comments. It also adds #= and =# for multi-line comments, like /* and */ in C.
• By convention, names of functions that modify their first argument end in !. This is not enforced.
• Blocks are indented as in Python, but there is no colon at the end of the first line, and there must be an end statement to close the block.
• Julia uses elseif as in Perl, not elif as in Python [1].
• Julia uses square brackets to declare a dictionary. Keys and values are separated with =>, as in Perl, rather than with colons, as in Python.
• Julia, like Python 3, returns 2.5 when given 5/2. Julia has a // division operator, but it returns a rational number rather than an integer.
• The number 3 + 4i would be written 3 + 4im in Julia and 3 + 4j in Python.
• Strings are contained in double quotes and characters in single quotes, as in C. Python does not distinguish between characters and strings, and uses single and double quotes interchangeably.
• Julia uses function to define a function, similar to JavaScript and R, where Python uses def.
• You can access the last element of an array with end, not with -1 as in Perl and Python.
* * *
[1] Actually, Perl uses elsif, as pointed out in the comments below. I can’t remember when to use else if, elseif, elsif, and elif.
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Helper III
Sum of Distinct Values over Total Distinct by Category
Hi,
I need assistance on the formula for computing % of new products over total. I could not achieve the desired output.
Formula should be New Product Count / Total by Region.
Eg. for Asia and Enterprise = 7 / 113 (total for Asia) = 6.19%
Correct %:
The formulas I used:
1) Count of NEW Products =
CALCULATE(
DISTINCTCOUNT(DATA[ProductID]),
DATA[Status] IN {"New"})
2) Total Count of Products =
SUMX(
KEEPFILTERS(VALUES(DATA[Region])),
CALCULATE(DISTINCTCOUNT(DATA[ProductID]))
)
3) % New Products = IFERROR([Count of NEW Products]/[Total Count of Products],0)
Attaching the links for the pbix and sample data with computation.
https://drive.google.com/drive/folders/1d1ivWVWhnNzfcUGdseXiTY71pldhct5M?usp=sharing
Thanks in advance for your help.
Regards,
Summer
1 ACCEPTED SOLUTION
Community Support
Hi @summer18 ,
Try this:
``````% New Products =
IFERROR (
[Count of NEW Products]
/ CALCULATE ( [Total Count of Products], ALLSELECTED ( DATA[Segment] ) ),
0
)
``````
Best regards
Icey
If this post helps, then consider Accepting it as the solution to help other members find it faster.
6 REPLIES 6
Super User
@summer18 , I think there some different data in pbix. Can you please check.
Helper III
Hi @amitchandak , Thanks for checking into this. I updated the link
https://drive.google.com/drive/folders/1d1ivWVWhnNzfcUGdseXiTY71pldhct5M?usp=sharing
Community Support
Hi @summer18 ,
I have no access to your shared file. Please share it publicly.
Best Regards,
Icey
Helper III
Hi @Icey , I just made the link public. Hope you can help me on this
Community Support
Hi @summer18 ,
Try this:
``````% New Products =
IFERROR (
[Count of NEW Products]
/ CALCULATE ( [Total Count of Products], ALLSELECTED ( DATA[Segment] ) ),
0
)
``````
Best regards
Icey
If this post helps, then consider Accepting it as the solution to help other members find it faster.
Helper III
Thanks @Icey . It works!
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9 out of 10 based on 780 ratings. 2,335 user reviews.
# CIRCUIT DIAGRAM WORKSHEET HIGH SCHOOL
Basic Circuits Name
PDF fileSeries Circuits: Once your circuit is working, have your teacher check the circuit. Using the symbols above, draw the circuit you created. a. Using one bulb, batteries and some wires, make one light bulb turn on. b. Now make 2 light bulbs turn on with batteries and some wire. c. Using 3 bulbs, batteries, and some wires, make 3 light bulbs turn on.
ELECTRICITY UNIT - Sir Wilfrid Laurier School Board
PDF fileFuses and circuit breakers detect unusually high currents and break the circuit, which helps to prevent fires. The picture above shows a circuit. Electricity flows from the negative side of a battery, through the wires, If you follow the circuit diagram from one side of the cell to the other,
Electric Circuits - Scarlett Middle School
#2 Electric Circuits Parts 1&2 (25 min) Part 1: Using what you learned in the game and self quiz. Draw a Electric Circuit Diagram that represents how Electric Current moves. Be sure to label the wire, conductor, voltage source, and describe any potential difference. Be ready to share with teacher and class.
5th grade Science Worksheets: Circuit diagrams | GreatSchools
In this science worksheet, your child draws circuit diagrams to represent two series circuits. In this science worksheet, Circuit diagram, Guided inquiry, Observational skills, Properties of circuits, Science experiment to try, Series circuit, Understanding power.
Circuit Diagram Worksheet by bur00917 | Teaching Resources
circuit-diagram-worksheet. About this resource. Info. Created: Feb 13, 2015. Updated: Feb 22, 2018. docx, 85 KB. circuit-diagram-worksheet. Pre-K and Kindergarten Primary / Elementary Middle school Secondary / High school Whole school Special Educational Needs Blog. Store4.4/5(7)Brand: TES
Electric Circuit Diagrams: Applications & Examples | Study
Electric circuit diagrams show you how to connect the parts of a circuit, but they also can do a lot more! In this lesson, learn how to use a circuit diagram to calculate the current in different branches of a circuit.
Symbols for circuit components (1) - Natural Science
In this science worksheet, your child labels each of the five parts of a diagram of an electrical circuit. Free printable Worksheets, word lists and activities. In this science worksheet, your child learns about how well different materials act as electrical conductors. Parts of an electrical circuit diagram - Worksheets & Activities | GreatSchools
Circuit diagramm - Science class
2) Circuit diagrams Circuit diagrams are drawn using standard symbols and following specific rules: - 1) The diagram is drawn with a ruler and a pencil. Always begin by tracing its general shape that is a rectangle. - 2) The standard symbols of the different electric components are located, preferably, in the middle of each side.
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http://www.algebra.com/cgi-bin/show-question-source.mpl?solution=339955
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```Question 505734
t = time to build one table
c = time to build one chair
.
t = 2c
.
t + 4c = 30 hr
.
substitute t = 2c
.
2c +4c = 30
6c = 30
c = 5
.
So the time to build a chair = 5 hr to build one chair
.
t = 2c
t = 2(5)
t = 10 hr to build a table
.
Done.```
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# 169.5 cm in Feet: Understanding the Conversion
Source: bing.com
When it comes to height, there are different units of measurement used across the world. Some countries use centimeters, while others use feet and inches. If you’re used to measuring your height in centimeters, you may be wondering how tall you are in feet. In this article, we’ll explore how to convert 169.5 cm to feet and why this conversion is important.
## Understanding the Basics: What are Centimeters and Feet?
Source: bing.com
Centimeters (cm) are a unit of measurement commonly used in the metric system. One centimeter is equivalent to 0.01 meters or 0.3937 inches. On the other hand, feet are a unit of measurement used in the imperial system, primarily in the United States and the United Kingdom. One foot is equivalent to 12 inches or 0.3048 meters.
## How to Convert 169.5 cm to Feet?
Source: bing.com
To convert 169.5 cm to feet, you need to divide the length value by 30.48. The equation for this conversion is as follows:
(169.5 cm) ÷ (30.48 cm/ft) = 5.558 feet or 5 feet and 6.7 inches (rounded off to the nearest tenth)
## Why is Knowing Your Height in Feet Important?
Source: bing.com
Knowing your height in feet is important for a variety of reasons. One of the most common reasons is for measuring your BMI or body mass index. BMI is a measure of body fat based on height and weight. To calculate your BMI accurately, you need to know your height in both feet and inches. Additionally, some professions require you to meet certain height requirements, such as becoming a pilot or a flight attendant.
## Other Common Height Conversions
Source: bing.com
If you’re not used to measuring your height in feet, you may find it useful to know some other common height conversions. Here are some of them:
• 150 cm = 4.92 feet
• 160 cm = 5.25 feet
• 170 cm = 5.58 feet
• 180 cm = 5.91 feet
• 190 cm = 6.23 feet
## Conclusion
Converting centimeters to feet may seem like a daunting task, but it’s actually quite simple. By dividing the length value by 30.48, you can easily convert 169.5 cm to feet. Knowing your height in feet is important for a variety of reasons, including calculating your BMI and meeting certain height requirements for certain professions. With this information, you can accurately measure and track your height in both metric and imperial units.
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# Soil Mechanics & Foundation Engineering : Stability of Slopes
By Sajal Gupta|Updated : February 24th, 2022
### Stability Analysis of Slopes
The quantitative determination of the stability of slopes is necessary in a number of engineering activities, such as:
(a) the design of earth dams and embankments,
(b) the analysis of stability of natural slopes,
(c) analysis of the stability of excavated slopes,
(d) analysis of the deepseated failure of foundations and retaining walls.
Quite a number of techniques are available for these analyses
FACTORS OF SAFETY
The factor of safety is commonly thought of as the ratio of the maximum load or stress that a soil can sustain to the actual load or stress that is applied. Referring to Fig. given below the factor of safety F, with respect to strength, may be expressed as follows:
where τff is the maximum shear stress that the soil can sustain at the value of normal stress of σn , τ is the actual shear stress applied to the soil.
Above equation may be expressed in a slightly different form as follows:
Two other factors of safety which are occasionally used are the factor of safety with respect to cohesion, Fc, and the factor of safety with respect to friction, Fφ.
The factor of safety with respect to cohesion may be defined as the ratio between the actual cohesion and the cohesion required for stability when the frictional component of strength is fully mobilised. This may be expressed as follows:
The factor of safety with respect to friction, Fφ, may be defined as the ratio of the tangent of the angle of shearing resistance of the soil to the tangent of the mobilised angle of shearing resistance of the soil when the cohesive component of strength is fully mobilised.
A further factor of safety which is sometimes used is FH, the factor of safety with respect to height. This is defined as the ratio between the maximum height of a slope to the actual height
of a slope and may be expressed as follows:
where, Cm = Mobilized Cohesion
m = Mobilized Friction Angle
and
#### Factor of Safety w.r.t. Cohesion (fC)
and
where, Hc = Critical depth
H = Actual depth
### Stability Analysis of Infinite Slopes
(i) Cohesionless dry soil/dry sand
where,
τ = Developed shear stress or mobilized shear stress
σn = Normal stress.
where,
Fs = Factor of safety against sliding
• For safety of Slopes
(ii) Seepage taking place and the water table is parallel to the slope in Cohesionless Soil
h = Height of water table above the failure surface.
φ’ is effective friction angle
γ-avg. total unit weight of soil above the slip surface upto ground level.
(iii) If water table is at ground level: i.e.,
h = z
(iv) Infinite Slope of Purely Cohesive Soil
Here H = z = depth of slice/cut.
At Critical Stage Fc =1
where, Sη = Stability Number.
(v) C-∅ Soil in Infinite Slope
(vi) Taylor's stability no.
(for cohesive soil)
Max. theoretical value of stability no. = 0.5
Max. practical value is = 0.261
(for C-∅ soils)
### Stability Analysis of Finite Slopes
(i) Fellinious Method
(For Purely Cohesive Soil)
A. where, F = Factor of safety
r = Radius of rupture curve
I = Length of rupture curve
B.
F = Factor of safety it tension cracks has developed.
(ii) Swedish Circle Method
where, F = Factor of safety
(iii) Friction Circle Method
(iv) Taylor's Stability Method (C-∅ soil)
In case of submerged slope γ should be used instead of γ and if slope is saturated by capillary flow they γsat should be used instead of γ.
where ∅w = weight friction angle.
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1. ## Calc 1 question
find the derivitive f(x) = -4/x^2 use the limit process
Using fx=(x+Dx)-fx/Dx
I get -4/(x+Dx)^2/x^2 + 4/x^2 and get lost in the math - am I on the right track?
2. Hello,
Originally Posted by weezie23
find the derivitive f(x) = -4/x^2 use the limit process
Using fx=(x+Dx)-fx/Dx
I get -4/(x+Dx)^2/x^2 + 4/x^2 and get lost in the math - am I on the right track?
$f'(x)=\frac{f(x+Dx)-f(x)}{Dx}$
$f({\color{red}x+Dx})=\frac{-4}{({\color{red}x+Dx})^2}$
---> $f'(x)=\frac{\frac{-4}{(x+Dx)^2}+\frac{4}{x^2}}{Dx}$
Did you catch your mistake ? Now, try to simplify (show your work, it'd be better)
3. It's time to get UNlost. You just have to sort through it.
These are NOT the same:
x+3/5 and (x+3)/5
2) Don't write things that make no sense.
You have: fx=(x+Dx)-fx/Dx
Notice how "fx" is on both sides. I'm sure you didn't mean that. Plus, there's and 'f' missing on the right -- f(x+Dx)?
4. As far as I get is 8x^2+8xDx+4Dx^2/(x+Dx)-x^2
Originally Posted by Moo
Hello,
$f'(x)=\frac{f(x+Dx)-f(x)}{Dx}$
$f({\color{red}x+Dx})=\frac{-4}{({\color{red}x+Dx})^2}$
---> $f'(x)=\frac{\frac{-4}{(x+Dx)^2}+\frac{4}{x^2}}{Dx}$
Did you catch your mistake ? Now, try to simplify (show your work, it'd be better)
5. You didn't listen to me about the notation, did you?
If you are VERY CAREFUL, you will see it. If you are sloppy, it will work only by luck.
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2 added 123 characters in body
Credit: This answer came out of trying to understand why auniket's answer (a.k.a. counterexample) works.
1) auniket is correct that for dimension reasons $T$ cannot surject onto $S$, so in particular my comment about $X$ being normal possibly helping is irrelevant. So is $T$.
2) It seems to me that there is a much more general problem with your desired statement. Namely I believe the following is true:
Claim: Under the conditions of the question, if in addition $\dim Y=0$ and $Y$ is reduced, then the desired statement cannot be true.
Proof: We may assume that $Y$ is a single point. Since by assumption $X$ is singular at $Y$, the local ring of $X$ at $Y$ is not a regular local ring. Therefore the ideal of $Y$ cannot be generated by $\dim X$ number of elements. On the other hand, by assumption $X$ is birational to $S$, so $\dim X=\dim S=s$. Therefore $v_1,\dots,v_s$ cannot generate the ideal of $Y$. $\square$
Note: I think this actually covers both of auniket's examples and would definitely give an arbitrary number of normal examples.
3) It seems that this still leaves a sliver of hope for you as your $Y$ is a curve (and even in the zero-dimensional case if $Y$ is non-reduced, it could work out). However, if it is reduced then you are at the absolute minimal number of generators that the singularity condition allows.
1
Credit: This answer came out of trying to understand why auniket's answer (a.k.a. counterexample) works.
1) auniket is correct that for dimension reasons $T$ cannot surject onto $S$, so in particular my comment about $X$ being normal possibly helping is irrelevant. So is $T$.
2) It seems to me that there is a much more general problem with your desired statement. Namely I believe the following is true:
Claim: Under the conditions of the question, if in addition $\dim Y=0$ and $Y$ is reduced, then the desired statement cannot be true.
Proof: We may assume that $Y$ is a single point. Since by assumption $X$ is singular at $Y$, the local ring of $X$ at $Y$ is not a regular local ring. Therefore the ideal of $Y$ cannot be generated by $\dim X$ number of elements. On the other hand, by assumption $X$ is birational to $S$, so $\dim X=\dim S=s$. Therefore $v_1,\dots,v_s$ cannot generate the ideal of $Y$. $\square$
Note: I think this actually covers both of auniket's examples and would definitely give an arbitrary number of normal examples.
3) It seems that this still leaves a sliver of hope for you as your $Y$ is a curve (and even in the zero-dimensional case if $Y$ is non-reduced, it could work out).
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Share Your Experience: Take the 2024 Developer Survey
# Tag Info
## Hot answers tagged 2d-graphics
6 votes
### Rounding corners of polygon given vertices of its corners
Ok, Xenapior and Reynolds together have the right idea. But the explanation is a bit lacking so here is a image to explain it all and some further musings. First let us start by drawing an image (yes ...
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6 votes
Accepted
### Converting cartesian pixels to polar pixels
I have implemented the cartesian-to-polar-conversion and have used different interpolation methods: 1) nearest neighbor 2) a subsampling approach, which averages 81 subpixel locations 3) bilinear ...
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5 votes
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### Explanation of the Marching Squares saddle points resolution
Imagine that a black dot represents a value of -1 whereas a white dot represents a value of +1. You are looking for the iso-line(s) where each point upon the line is 0. For this example you could ...
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### What if we don't mention Modelview and projection matrix?
Your screen isn't 3D, so how do you display 3D objects on it? You need to map 3D coordinates into 2D space. This also explains why your OpenGL code is not behaving how you're expecting it to. Sorry if ...
• 1,353
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### How can I transform an ellipse into a circle?
You can use a shear, or equivalently a rotation by 45, scale along one of the axes, rotate back (which is equivalent to the shear which you are looking for). EDIT: On second thought, you don't need ...
• 2,226
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### When fitting 3d cuboids into a box, how to find the order of drawing (in a 2D projection) so that the cuboids don't appear to overlap?
Let's get our orientation straight first. When I speak of a direction, I mean the direction relative to our viewpoint. So the "front" face of a cuboid is the face nearest the camera; "...
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### What is the difference between world coordinate, viewing coordinate and device coordinate in computer graphics?
The 2D pipeline involves ... [coordinate transformation terms] Can someone give me the detail differentiation among these? This is something I very recently learned while trying to understand how ...
3 votes
### Algorithm for thinning a thick line
the canny algorithm is a great start. it takes a sobel input like so. Computes its gradient. Then depending on the gradient orientation it compares all neighbouring pixels aligned with it. If its a ...
3 votes
### 2D metaballs with marching squares and linear interpolation
as promised, here my answer to your question. As I can't follow your code completely (I spend most of my time implementing my code :D ) I can only explain to you what the last step is. So why do we ...
2 votes
### Successive lineTo(x, y) in JavaScript
Straight from the specification of lineTo: If the object's path has no subpaths, then ensure there is a subpath for (x, y). And "ensure that there is a subpath" ...
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### Douglas-Peuker and equal distances
I would consider the curvature in that case too. If the curvature is small - then it is a flat region - so you can safely remove it - your 4th point for example. If the curvature is large (your 5th ...
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### My 2D translation matrix causes the box to be off screen when it shouldnt be
The mistake was in the multiplication order during the composition of the matrices. Since the OP was working with column vectors the correct order for transformations should have been $TRSA∗v$. The ...
• 2,438
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### How to apply wire texture for realistic rendering of embroidery?
Picking up from step 2, where you generated line segments to represent individual stitches, I would first suggest that you apply rounded ends to each segment, rather than square ends. I would then ...
• 341
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### Purpose and workflow of wireframe models
Why are wireframes used? Games don't really use wireframes much, but they might be useful for HUD items etc. Asset creators, designers and engineers use wireframes to see whats inside objects. TO ...
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2 votes
### Purpose and workflow of wireframe models
Regarding wireframes as an edge list. Wireframe models are used in engineering and CAD they provide several benefits. The Wikipedia page you linked has some benefits, this AutoCAD page has some more. ...
• 3,219
2 votes
### How do I plot tristimulus curves?
You can certainly convert the wavelengths of the visible spectrum into the XYZ colour space with some extra considerations. Converting wavelengths of the visible spectrum into RGB will present some ...
1 vote
### 2d gpu powered rendering engine
Writing a 2D render system is a perfectly reasonable thing to do using any of the Graphics API's that you listed. Triangles are the basic medium of all the popular graphics API's so learning to work ...
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### Is drawing Bezier curve with scanline algorithm possible with Vulkan?
The typical solution is to compute a bounding structure using triangles that would be transformed by the vertex shader. The bounding structure is computed to be large enough to fit the entire curve ...
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### Rigorous example of early graphics system
First of all, the type of system they describe is more commonly known today as a vector display. (Until just now, I had never heard it called “calligraphic”.) Looking up information using that name ...
• 1,136
1 vote
### Recursive sampling in Marching Squares
If the contour exists in a loop entirely inside one of your coarse cells — or has some thin protrusion that passes between the corner points, at least — then it will be missed entirely; it will look ...
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1 vote
### Whats the best way to render (2D) parametric curves with uniform stroke width?
Your question implies that everything has to be rendered with a vertex representing each small piece of a curve. This simply isn't accurate. It also implies that device rendering space is in some way ...
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1 vote
### undefined reference to `SDL_setRenderDrawColor'
Looks like a typo since the compiler isn't complaining about the other SDL function calls. You probably need to change SDL_setRenderDrawColor to ...
1 vote
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### Algorithms for rendering overlapping (floating) windows in 2D
I have some knowledge of what you call “stacking” (I don't know if there's a standard term), from having programmed with classic Mac OS graphics (QuickDraw). To keep the explanation straightforward, I'...
• 1,136
1 vote
### wall clock using bresenhams circle drawing algo
The usual method is to just draw over the circle with your lines. The clock is behind the minute hand and the minute hand is behind the hour hand so the drawing order would be. Draw the clock face, ...
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1 vote
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### Algorithm for proper drawing of cosine function
There is in fact a way to avoid this effect and it is outlined by the NYQUIST sampling theorem - which states that the sampling rate should be at least twice as high as the maximum frequency of the ...
1 vote
### How can I draw a filled circle on a grid with the less rectangle as possible?
To get a proper circle, you can use a transformation matrix to transform your cell coordinates into equally spaced cartesian coordinates. In the case of your 50 x 100 pixel cells, your transformation ...
• 3,421
1 vote
### Rounding corners of polygon given vertices of its corners
The cut length from the vertex is x*ctan(t/2), where t is the angle at this vertex.
• 27
1 vote
Accepted
### Rounding corners of polygon given vertices of its corners
Since you're working on CAD software, you probably want some precise results. Here an algorithm that could work: For each side: Compute the segment's equation. Compute each round corner's circle ...
• 830
1 vote
Accepted
### Determining Vanishing Points of edges of Triangle using given Point
Vanishing points are "points at infinity" in projective geometry, which are represented by $w = 0$ in homogeneous coordinates. You can construct the vanishing point of a ray or line by taking its $xyz$...
• 25k
1 vote
### Fill an irregular region with 2D shapes
There is no general algorithm for packing problems. Only some of the special cases have known, and optimal, solutions. If you are packing one shape then finding a reasonable solution is possible. Like ...
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# Physics
## Conservation of E and P
Part (1) Mass of Light Cart: 573.5 ± 0.1 grams = ML ± SML Fin length: 6 ± 0.1 centimeters = DL ± SDL Mass of Heavy Cart: 1691.1 ± 0.1 grams = MH ± SMH Fin length: 6 ± 0.1 centimeters = DH ± SDH Distance between Photogates: 70 ± 0.1 centimeters = Do ± SDo Part (2) Data Table 1: Linear Motion of Light Cart Time (sec) uncertainty (sec) 0.257198 0.03 63.69900 28.71094 -6.819508699 31.4232975764 0.351391 0.03 1.425185 0.03 55.70307 21.94047 1.532899 0.03 Data Table 2: Linear Motion of Heavy Cart Time (sec) uncertainty (sec) 0.251884 0.03 49.54583 17.37761 -4.74925 14.48465 0.372984 0.03 1.763335 0.03 42.30715 12.65707 1.905155 0.03 Part (3) Data Table 3: Heavy Cart Incident upon Light Cart (Elastic Collision) Time (sec) uncertainty (sec) 32.6 0.304995 0.03 81.20179 46.64435 0.378885 0.03 0.9309 0.03 99.45960 69.96815 0.991226 0.03 1.761019 0.03 35.93180 9.14904 1.928002 0.03 Data Table 4: Elastic Collision 0.55753 0.64052 0.10917 0.05559 0.28366 0.39910 0.39283 0.4029522852 0.16471 ± 0.7567289715 Joules Part (4) Data Table 5: Light Cart Incident upon Heavy Cart (Elastic Collision) Time (sec) uncertainty (sec) 32.6 0.100884 0.03 104.10702 76.65780 0.158517 0.03 1.198495 0.03 36.14697 9.25870 1.364484 0.03 2.056196 0.03 10.39110 0.78289 2.633613 0.03 Data Table 6: Elastic Collision 0.31079 0.45769 0.00310 0.00047 0.11048 0.05660 0.11358 0.0565986673 0.19721 ± 0.4611745369 Joules Part (5) Data Table 7: Heavy Cart Incident upon Light Cart (Inelastic Collision) Time (sec) uncertainty (sec) 32.6 0.448185 0.03 66.29908 31.10099 0.538684 0.03 1.518384 0.03 40.58657 3.51786 2.008694 0.03 Data Table 8: Inelastic Collision 0.37167 0.34870 0.18652 0.03233 0.18515 ± 0.3501950946 Joules Analysis (1) Done in Part (2) Analysis (2) Data Table 9: Kinetic Frictional Force light cart -0.03911 0.18021 0.00696 heavy cart -0.08031 0.24495 0.00485 Analysis (3) Done in Parts (3), (4), (5) Analysis (4) Data Table 10: Kinetic Frictional Force Part (3) Part (4) Part (5) Part (3) ± Joules Part (4) ± Joules Part (5) ± Joules Analysis (5) (ignore thermal energy generated by kinetic friction) °C Analysis (6) Summarize results of Analysis (5) & (6) Analysis (7) Data Table 11: Heavy Cart Incident upon Light Cart (Elastic Collision) ± kg m/sec Data Table 12: Light Cart Incident upon Heavy Cart (Elastic Collision) ± kg m/sec Data Table 13: Heavy Cart Incident upon Light Cart (Inelastic Collision) ± kg m/sec 0.1 ± 0.01 seconds Assume fractional uncertainty is 10% Data Table 14: Heavy Cart Incident upon Light Cart (Elastic Collision) Heavy Cart Incident upon Light Cart (Elastic Collision) ± kg m/sec Light Cart Incident upon Heavy Cart (Elastic Collision) ± kg m/sec Heavy Cart Incident upon Light Cart (Inelastic Collision) ± kg m/sec
The total kinetic energy of the carts before collision is more than The total kinetic energy of the carts after collision, and they do agree within their uncertainties. Also, the conservation of energy is being violated because there is alot of energy is missing from the final kinetic energy.
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# Questions tagged [hypothesis-testing]
Hypothesis testing assesses whether data are inconsistent with a given hypothesis rather than being an effect of random fluctuations.
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14 views
### Would the Mantel-Haenszel test for linear trend be appropriate for this data/hypothesis?
I am testing whether or not there is an association between the parents education level and the pupils grades in the oral exam in social studies in the last year of danish primary school from the ...
1 vote
23 views
### Significance test for comparing different 10-fold cross-validated Machine Learning Regressions
Is there a recommended significance test for comparing different 10-fold cross validated regressions? For instance, I want to compare the performance of LASSO against Random Forest for my dataset. ...
1 vote
13 views
• 177
15 views
### Fisher's exact test when resampling from the population
I am using different ML classifiers for making predictions related to a binary classification. I would like to compare two groups. Let's call them A and B. The already trained and tested classifier ...
32 views
### Problem formulating a hypothesis test for the mean
Suppose that I would like to support the following hypothesis: h = The average weight of people in New York is 70kg. If I take h to be the null hypothesis then I can only reject it (I assume that ...
• 883
15 views
### How to conduct t-test for comparing the accuracy of two binary classifiers? [closed]
I am using two binary classifiers that predict the accuracy of samples over a dataset. I need to check if the difference in the mean accuracy between the two models is statistically significant. ...
8 views
### How do I perform a content analysis in SPSS? [closed]
I am struggling to find the right statistical test for a content analysis. I have categorized and coded the different variables (e.g. text length) and generated the Excel file for the analysis. The ...
274 views
### How to report a P-value?
I am writing the results of my thesis, and I have a few questions on how to report P-values. When is it appropriate to give the exact P-value, instead of writing e.g. P<0.05 (also in case of non-...
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1 vote
42 views
+50
### Why does the fact that I estimated my parameters change the way I should use chi−sqaured test?
Suppose I have a data sample and I want to test whether normal distribution with mean $0$ and variance $1$ fits this data sample. If I understand the chi-squared test correctly, I think that I should ...
26 views
### How to test the statistical significance of several clustering algorithms? [closed]
I have computed ARI, NMI and Silhouette scores of several clustering algorithms on single cell RNA sequence data. These metrics are computed three times for each algorithm and for each dataset. Now, I ...
32 views
### Comparing scientific studies (p-values, q-values)
I am comparing two studies. Each study consists of an affected (cancer) and control groups of patients, for which a number of biomarkers (gut microbial species) were studied. For each biomarker ...
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23 views
### how to measure the effect of a recurring event?
I have multiple cities with data on theatre visitors and an event as the mentioning of the theatre in the local news. I want to estimate whether the event of a mentioning lead to more visitors for the ...
24 views
### Testing sequence of ones and zeros for randomness
I am given a sequence of $40$ ones and zeros and I have to test the null hypothesis that ${40 \choose n_1}$ sequences are all equally probable ($n_1$ being the number of ones). To do so, I have to use ...
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1 vote
21 views
### How do I test that two samples don’t differ significantly by their demographics?
I am conducting a study where I am randomly assigning people to different conditions. Because I am recruiting from a representative sample in the US, there will be some variability in the demographics ...
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25 views
### Statistics And Probability [closed]
The Audit Team of ABC University claims that the mean monthly salary of the teachers is Php18,000.00 with a standard deviation of Php4,500.00. A researcher takes a random sample of 20 teachers and ...
1 vote
8 views
### Do differences between ordinations prove a relationship between variables?
I have a large dataset that contains a variety of environmental characteristics (air temperature, tree cover, insect counts) as well as bird abundance data for each point. I was talking with my friend ...
11 views
### How to correctly estimate the ratio different of lower grain unit metric in Cluster randomized experiment?
I work on Education tech products that teachers/students would use in their learning journey. When we run experiment to test hypothesis of a feature, we need to do cluster randomization (cluster = ...
21 views
### Using the same subjects in different measures
I'm working on an experiment where I collect 3 categories of reaction time data form subjects who are video game players, distinguished by the game that they play (eg. RTS, FPS, MOBA). I wish to ...
10 views
### Simplified DID: How to run significance test
If we know that Variable C = Variable A - Variable B, and Variable A is the difference in two groups (binary data) during time period X. It is statistically significant. Variable B is the difference ...
35 views
### How to calculate the statistical significance of the difference in a categorical value of two populations?
There has recently been a statistical report regarding whether Cypriots are ready for a female president. There was a difference in the opinions of men and women and I'm wondering if it is ...
• 101
7 views
### How can I efficiently test whether a subsets of nodes in a DAG are "lined up" more often than expected by random chance?
I have a directed acyclic graph with N nodes, each of which is assigned to one of K groups, with K < N. My hypothesis is that nodes in the same group tend to "line up" along a linear path....
11 views
### How do I interpret t-ratio of interaction variable?
I need to answer this question for my Econometrics HW: Does the gender gap differ significantly in urban versus non-urban areas? Test at the 5% level. If I am not wrong I need to use the t-test for ...
26 views
### Testing the collection of Bernoulli variables
Suppose that I am throwing a triple dice (i.e, three dices simultaneously) $n$ times. Then number of times that the sum of these three dices $k=3,\ldots,18$ appears is my sample $X_1,\ldots,X_n$ is ...
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16 views
### Testing equality of coefficients from two different samples
I have the regression statistics for the same regression run on two different samples, and am asked to explain whether it is possible to test for equality of the coefficents, $\beta_1$and $\beta_2$ ...
35 views
### Delta Method to calculate the standard error of ratio in an AB testing context
In AB testing context, if we have a control group and test group (2 groups), and I'd like to calculate the relative difference (Mean test/ Mean control -1) and the confidence interval of this ratio ...
46 views
### Does the likelihood ratio test violate the likelihood principle?
I've been going over Berger's famous example of negative binomial vs binomial sampling leading to two different p-values conditional on the same observed data. To summarize, suppose we observe 9 tails ...
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21 views
### Model that tests if categorical variables give higher results than others [closed]
Consider a set of categorical variables (values are 0 or 1) which are used to predict revenue range values. For instance the observation $(x_1, x_2, x_3, x_4, ..., x_{20}) = (0,1,0,1, ..., 0)$ might ...
• 113
27 views
### Validating random variable generation from inverse transform sampling
I'm building a simulator and I have to implement some probability distributions. What is the best (formal) way of validating this implementation? I took a look at KS-tests but it seems to me they are ...
34 views
### two sample t-test for non-normal population
I assume this question has been beaten to death and thus I am just looking for a reference which goes through the details. Assuming all populations we deal with have finite means and variances (even ...
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19 views
### T-testing on sparse data
Problem: I have two stochastic processes, $S_1$ and $S_2$, that frequently are zero, but occasionally have positive values with unknown probabilities $q_1$ and $q_2$. e.g. S_1 = \{0,0,0,0,0,21,0,0,...
2k views
### If p-values are random, why not decide using the test statistic instead?
I am currently reading the paper Duncan J Murdoch, Yu-Ling Tsai & James Adcock (2008) P-Values are Random Variables, The American Statistician, 62:3, 242-245, DOI: 10.1198/000313008X332421 In ...
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1 vote
22 views
### Uniformly most powerful test
Suppose we have Xi~Exp(λ), and we want to construct a most powerful test for H0 : λ = λ0, H1 : λ = λ1 I then proceed to use the Neyman Pearson lemma : reject H0 when the likelihood ratio L(λ1;X)/L(...
30 views
### With Stationarity How can ARMA Modelling have any Validity?
I have recently been thrown into the deep end with time-series econometrics. The first thing I have learned is that in order to avoid the spurious correlation trap, I need to ensure that all the ...
1 vote
33 views
### Is this interpretation of the relationship between p-values and type I error rates correct?
I am trying to check my understanding of the relationship between p-values and type I error rates. I am going to use the following example to do so, so I would appreciate any feedback on it. Suppose ...
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28 views
### Z - Score between average scores
I have a test that gives students a score between 0 and 10. Students can be black, white, asian or hispanic. I want to compare the average score of each race against the overall average, and I want to ...
• 21
8 views
### Hypothesis testing on a moving-average model on unevenly-spaced time series
I have some irregular time series $X$, where $X_t$ are identically distributed, and bounded between 0 and 1. I perform a moving average $Y_t = \sum_{t'} \frac{w(t-t')}{\sum_k w(t-k)}X_{t'}$. The ...
1 vote
26 views
### What is the advantage of a scoring scale with many levels?
In my research I have different ordinal variables with 3 levels of scores. Each score represents the severity of a condition. In other studies I see that the same variables are scored on a scale with ...
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+1 vote
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If $\displaystyle \int \frac{\sin 2 x}{(a+b \cos x)^{2}} d x=\alpha\left[\log _{e}|a+b \cos x|+\frac{a}{a+b \cos x}\right]+c$, then $\alpha=$
(A) $\dfrac{2}{\mathrm{~b}^{2}}$
(B) $\dfrac{2}{\mathrm{a}^{2}}$
(C) $-\dfrac{2}{\mathrm{~b}^{2}}$
(D) $-\dfrac{2}{\mathrm{a}^{2}}$
| 37.2k views
## 1 Answer
0 votes
Best answer
Take the $u$-substitution $\cos x=u$ then $-\sin x\ dx=du$. This leads us
\begin{aligned}\int \frac{\sin 2 x}{(a+b \cos x)^{2}} \ d x&= \int \frac{2 \cos x\sin x}{(a+b \cos x)^{2}} d x \\ &=-2\int \frac{u\ du}{(a+b u)^{2}} \end{aligned} \tag{Eq. 1}
We will also perform the same $u$-substitution in the RHS of the given expression to get
$$\alpha\left[\log _{e}|a+b \cos x|+\frac{a}{a+b \cos x}\right]+c=\alpha\left[\log _{e}|a+b u|+\frac{a}{a+b u}\right]+c\tag{Eq. 2}$$
By comparing (Eq. 1) and (Eq. 2) we get
$$-2\int \frac{u\ du}{(a+b u)^{2}}= \alpha\left[\log _{e}|a+b u|+\frac{a}{a+b u}\right]+c \tag{Eq. 3}$$
We will not evaluate the integration to remove integration sign from left we will differentiate both sides with respect to $u$ to obtain
\begin{aligned}\frac{-2u}{(a+b u)^{2}}&= \alpha\left[\frac{1}{a+bu}+\frac{0-ab}{(a+bu)^2}\right] \\ \frac{-2u}{(a+b u)^{2}}&= \alpha\left[\frac{a+bu-ab}{(a+bu)^2}\right] \end{aligned}
By comparing numerator of the above expression we get
\begin{aligned}-2u &= a\alpha(1-b)+bu\alpha \end{aligned}
Again by comparing the coefficient of $u$ and constant term from both side of the above equation we get
$$\left\{\begin{array}{rl}a\alpha(1-b)&=0\\ b\alpha&=-2 \end{array}\right.$$
From last equation we get $u=-\dfrac{2}{\alpha}\implies \alpha=-\dfrac{2}{b^2}$.
Hence the correction option is (C).
by Professor
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The Two-Ears Theorem
The Two-Ears Theorem was developed and proven by Gary H. Meisters at the University of Nebraska in 1975 [4].
### Proof #1: By Gary H. Meisters
The proof by Meisters is by induction on the number of vertices, n, in the simple polygon P. It is quite elegant.
Base Case:
n = 4. The simple polygon P is a quadrilateral, which has two ears.
Induction:
Let P be a simple polygon with at least 4 vertices. Vertex pi is a vertex of P where the interior angle formed by pi-1, pi, and pi+1 is less than 180o.
Case 1:
Polygon P has an ear at pi. If this ear is removed, the remaining polygon P' is a simple polygon with more than three vertices, but with one less vertex than P. By the induction hypothesis, there are two non-overlapping ears E1 and E2 for P'. Since they are non-overlapping, at least one of these two ears, say E1, is not at either of the vertices pi-1 or pi+1. Since all ears of P' are also ears of P, polygon P has two ears: E1 and the ear at pi.
Case 2:
Polygon P does not have an ear at pi. So, the triangle formed by pi-1, pi, and pi+1 contains at least one vertex of P in its interior. Let q be the vertex in the interior of this triangle such that the line L through q and parallel to the line through pi-1 and pi+1 is as close to pi as possible.
Let a and b be the points of intersection of line L with the polygon edges (pi, pi+1) and (pi-1, pi), respectively. The triangle formed by pi, a, and b does not contain in its interior any vertex of P (or else our choice of vertex q would be incorrect).
The line segment Q from vertex q to pi can be constructed without intersecting any edges of P. Line segment Q divides P into two simple polygons: P1 (that contains vertices pi, pi+1,..., q) and P2 (that contains vertices pi, pi-1,..., q).
There are now two sub-cases to consider:
Case 2a:
Polygon P1 is a triangle. So, P1 is an ear for polygon P. By the induction hypothesis, polygon P2 must have at least two non-overlapping ears, E1 and E2 (or else it too would be a triangle and polygon P would have only four vertices). Since they are non-overlapping, at least one, say E1, is at neither vertices pi nor q. E1 does not overlap with the ear formed by polygon P1, so it is the second ear in polygon P.
Case 2b:
Polygon P1 is not a triangle. So, by the induction hypothesis, polygon P1 has two ears, E11 and E12 and polygon P2 has two ears, E21 and E22. Since they are non-overlapping, at least one of the ears in P1, say E11, is not at vertex pi or q. Similarly, at least one of the ears in P2, say E21, is not at vertex pi or q. These two ears, E11 and E21 will be non-overlapping ears for polygon P
Q.E.D.
### Proof #2: By Joseph O'Rourke[6]
It is known that a simple polygon can always be triangulated. Leaves in the dual-tree of the triangulated polygon correspond to ears and every tree of two or more nodes must have at least two leaves.
Q.E.D.
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# How Many Grams Is A Table Spoon?
## How many grams is 2 tablespoons?
Water Conversion Chart Near 0.8 gramgrams to US tablespoons of Water2 grams=0.135 ( 1/8 ) US tablespoon2.1 grams=0.142 ( 1/8 ) US tablespoon2.2 grams=0.149 ( 1/8 ) US tablespoon2.3 grams=0.156 ( 1/8 ) US tablespoon21 more rows.
## How do you measure 1 teaspoon in grams?
as someone mentioned, a teaspoon is approximately 5 grams. … 1/5 of a Teaspoon = a gram. … The spoons are for counting volume, not weight. … As pointed out, the weight to volume thing will work for somethings, not others, but with water, 1 tablespoon = 1/2 ounce = 14 grams.More items…
## How much is a 1/4 teaspoon in grams?
Baking Conversion TableU.S.Metric1/4 teaspoon1.42 grams1/2 teaspoon2.84 grams1 teaspoon5.69 grams1/2 tablespoon8.53 grams82 more rows
## What is 3 tablespoons in grams?
Tablespoon to Gram Conversion TableTablespoonsGrams3 tbsp42.52 g4 tbsp56.7 g5 tbsp70.87 g6 tbsp85.05 g36 more rows
## How much does 1 tablespoon of golden syrup weigh in grams UK?
1 tablespoon of golden syrup in gramsIngredient:gram kilogram pound ounceCalculate!Significant Figures: 2 3 4 5Results 1 US tablespoon of golden syrup weighs 21.9 grams. (or precisely 21.869625111469 grams. All values are approximate).6 more rows
## How do you measure 100 grams of flour?
110 grams of flour = 13 ¾ tablespoons of flour. How to measure 100 grams of flour? 100 grams of flour = 12 and a half tablespoons of flour.
## How many tbsp is 5 grams?
0.333 TConvert 5 Grams to Other UnitsUnit5 Grams (g) =Kilogram (kg)0.005 kgOunce (oz)0.176 ozPound (lb)0.011 lbTablespoon (tbsp)0.333 T2 more rows
## What is 10 grams in tablespoons?
Common Grams to Tablespoon ConversionsGramsTablespoons7 g0.467 or 7/15 tbsp8 g0.533 or 8 /15 tbsp9 g0.6 or 3/15 tbsp10 g0.667 or 2/3 tbsp6 more rows
## Is 15g equal to 15 ml?
g to ml conversion table:1 gram = 1 ml21 grams = 21 ml70 grams = 70 ml14 grams = 14 ml34 grams = 34 ml200 grams = 200 ml15 grams = 15 ml35 grams = 35 ml300 grams = 300 ml16 grams = 16 ml36 grams = 36 ml400 grams = 400 ml17 grams = 17 ml37 grams = 37 ml500 grams = 500 ml13 more rows
## What weighs 1 gram exactly?
A gram is about:a quarter of a teaspoon of sugar.a cubic centimeter of water.a paperclip.a pen cap.a thumbtack.a pinch of salt.a piece of gum.the weight of any US bill.More items…
## Is 15g a tablespoon or teaspoon?
tablespoon to g conversion table:0.1 = 1.5 gram2.1 = 31.5 grams4.1 = 61.5 grams0.8 = 12 grams2.8 = 42 grams4.8 = 72 grams0.9 = 13.5 grams2.9 = 43.5 grams4.9 = 73.5 grams1 = 15 grams3 = 45 grams5 = 75 grams1.1 = 16.5 grams3.1 = 46.5 grams5.1 = 76.5 grams16 more rows
## What is 200g tablespoon?
Convert 200 Grams of Butter to Tablespoonsgtbsp200.0014.110200.0114.110200.0214.111200.0314.11296 more rows
## What does 4 tablespoons of butter weigh in grams?
56.75 GramsWeight of 4 Tablespoons of Butter4 Tablespoons of Butter =56.75Grams2.00Ounces0.13Pounds0.06Kilograms
## How many teaspoon is 12 grams?
teaspoon to g conversion table:0.1 tsp = 0.5 gram2.1 tsp = 10.5 grams7 tsp = 35 grams0.4 tsp = 2 grams2.4 tsp = 12 grams10 tsp = 50 grams0.5 tsp = 2.5 grams2.5 tsp = 12.5 grams11 tsp = 55 grams0.6 tsp = 3 grams2.6 tsp = 13 grams12 tsp = 60 grams0.7 tsp = 3.5 grams2.7 tsp = 13.5 grams13 tsp = 65 grams16 more rows
## How many teaspoons is 20 grams?
4.8 tspHow Many Teaspoons Are in a Gram?GramsTeaspoons20 grams4.8 tsp30 grams7.2 tsp40 grams9.6 tsp50 grams12 tsp3 more rows
## How much is 3 teaspoons in grams?
Conversions and IngredientsTeaspoonsGrams (sugar)Grams (flour)1 teaspoon4.2g2.6g2 teaspoons8.4g5.2g3 teaspoons12.6g7.8g4 teaspoons16.7g10.4g7 more rows
## What is 1/4 of a cup in grams?
32Dry GoodsCupsGramsOunces1/4 cup32 g1.13 oz1/3 cup43 g1.5 oz1/2 cup64 g2.25 oz2/3 cup85 g3 oz3 more rows•Nov 19, 2020
## How many table spoons is 25 grams?
Convert 25 Grams of Butter to Tablespoonsgtbsp25.001.763725.011.764425.021.765125.031.765896 more rows
## How do you measure grams with a spoon?
Tablespoons to grams tablespoon = 15 grams. tablespoons = 30 grams. tablespoons = 45 grams. tablespoons = 60 grams. tablespoons = 75 grams. tablespoons = 90 grams. tablespoons = 105 grams. tablespoons = 120 grams.More items…•Jul 25, 2019
## What is a tablespoon in grams UK?
oz (uk) = 85 millilitres (ml.) 10 tablespoons = 1 cup or 1 wineglass = 6 fluid oz. (uk) = 170 millilitres (ml.)…Spoon Measures.Spoon Measures – solidsounce (uk)gramsA heaped tablespoon of flour= 1 oz.= 28gA heaped tablespoon of butter= 1.25 oz.= 35g4 more rows•Sep 7, 2011
## How much is 4 tablespoons in grams?
Dry Measure Equivalents3 teaspoons1 tablespoon14.3 grams2 tablespoons1/8 cup28.3 grams4 tablespoons1/4 cup56.7 grams5 1/3 tablespoons1/3 cup75.6 grams8 tablespoons1/2 cup113.4 grams3 more rows
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• :00Days
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Q7.5-11E
Expert-verified
Found in: Page 395
### Linear Algebra and its Applications
Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384
# Question: 11. Given multivariate data $${X_1},................,{X_N}$$ (in $${\mathbb{R}^p}$$) in mean deviation form, let $$P$$ be a $$p \times p$$ matrix, and define $${Y_k} = {P^T}{X_k}{\rm{ for }}k = 1,......,N$$. Show that $${Y_1},................,{Y_N}$$ are in mean-deviation form. (Hint: Let $$w$$ be the vector in $${\mathbb{R}^N}$$ with a 1 in each entry. Then $$\left( {{X_1},................,{X_N}} \right)w = 0$$ (the zero vector in $${\mathbb{R}^p}$$).) Show that if the covariance matrix of $${X_1},................,{X_N}$$ is $$S$$, then the covariance matrix of $${Y_1},................,{Y_N}$$ is $${P^T}SP$$.
It is verified that:
1. $$\left( {{{\bf{X}}_1},................,{{\bf{X}}_N}} \right)w = 0$$
2. The covariance matrix is: $${S_Y} = {P^T}SP$$
See the step by step solution
## Step 1: Mean Deviation form and Covariance Matrix
The Mean Deviation form of any $$p \times N$$ is given by:
$$B = \left( {\begin{array}{*{20}{c}}{{{{\bf{\hat X}}}_1}}&{{{{\bf{\hat X}}}_2}}&{........}&{{{{\bf{\hat X}}}_N}}\end{array}} \right)$$
Whose $$p \times p$$ covariance matrix is:
$$S = \frac{1}{{N - 1}}B{B^T}$$
## Step 2: The Mean Deviation Form (a)
From the question, the $$w$$is a unit vector with all values equal to 1. Then, we have:
$$\begin{array}{c}\left( {{{\bf{X}}_1},................,{{\bf{X}}_N}} \right)w = \left( {{{\bf{X}}_1},{{\bf{X}}_2}, \ldots ,{{\bf{X}}_n}} \right)\left( {\begin{array}{*{20}{c}}1\\1\\ \vdots \\1\end{array}} \right)\\ = {{\bf{X}}_1} + ...... + {{\bf{X}}_N}\\ = 0\end{array}$$
The mean deviation form given is:
$$\begin{array}{c}\left( {{{\bf{Y}}_1},................,{{\bf{Y}}_N}} \right)w = \left( {{P^T}{{\bf{X}}_1},................,{P^T}{{\bf{X}}_N}} \right)w\\ = {P^T}\left( {{{\bf{X}}_1} + ...... + {{\bf{X}}_N}} \right)w\\ = {P^T}\left( {{{\bf{X}}_1} + ...... + {{\bf{X}}_N}} \right)\left( {\begin{array}{*{20}{c}}1\\1\\ \vdots \\1\end{array}} \right)\\ = {P^T}\left( {{{\bf{X}}_1} + ...... + {{\bf{X}}_N}} \right)\\ = 0\end{array}$$
Hence, this is the required proof.
## Step 3: The Covariance Matrix (b)
From (a), the covariance matrix can be given as:
$$\begin{array}{c}{S_Y} = \frac{1}{{N - 1}}\left( {{{\bf{Y}}_1},................,{{\bf{Y}}_N}} \right){\left( {{{\bf{Y}}_1},................,{{\bf{Y}}_N}} \right)^T}\\ = \frac{1}{{N - 1}}\left( {{P^T}{{\bf{X}}_1},................,{P^T}{{\bf{X}}_N}} \right){\left( {{P^T}{{\bf{X}}_1},................,{P^T}{{\bf{X}}_N}} \right)^T}\\ = {P^T}\left\{ {\frac{1}{{N - 1}}\left( {{{\bf{X}}_1},......,{{\bf{X}}_N}} \right){{\left( {{{\bf{X}}_1},......,{{\bf{X}}_N}} \right)}^T}} \right\}P\\ = {P^T}SP\end{array}$$
Hence, this is the required proof.
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# Thread: C noob with some issues
1. ## C noob with some issues
So I'm on university doing 3th grade of computer sciende (a 4 year career) but I have no idea on how to code so it was actually quite hard for to me to pass this semester (who would have thougth!?).
Becouse of this I decided to start learning C again, I did learn to code in pseudocode and C in first year but forgot how to.
Today I was trying to make an algorithm that would return all the A, B, and C variables (all between 0 and 9) such that (10A+B)/(10B+C)=A/C becouse the solutions are nice and I had already solved the equation by hand and already knew all the solutions. But I just can't make it work, I'm using an online C compiler and it doesn't show any error so I guess it's stuck in a loop.
This is my sloppy code;
Code:
```#include <stdio.h>
int
main ()
{
int a, b, c, d, e;
while (a + b + c != 27)
a = 0;
b = 0;
c = 0;
a = a++;
if (a = 9)
a = 0;
b = b + 1;
if (b = 9)
c = c++;
d = (10 * a + b) / (10 * b + c);
e = a/c;
if(d=e)
printf ("%d",a);
printf ("%d",b);
printf ("%d",c);
return 0;
}```
Thanks!
2. I think it never gets out of the loop becouse C keeps getting bigger, I have this now:
Code:
```#include <stdio.h>
int
main ()
{
int a, b, c, d, e;
while (a + b + c != 27)
a = 0;
b = 0;
c = 0;
a = a++;
if (a = 9)
a = 0;
b = b++;
if (b = 9)
b=0;
while(c<=9)
c = c++;
d = (10 * a + b) / (10 * b + c);
e = a/c;
if(d=e)
printf ("%d",a);
printf ("%d",b);
printf ("%d",c);
if(c=9)
a=30;
return 0;
}```
3. I think your problem is a lack of change in the variables, all your while loop is doing is setting a to 0, also variables must be initialized BEFORE trying to use them, your loop starts with a, b and c equaling some random number left over from previous usage, in this case it's probably 0 although there's no garuntee on that, so 9999 times out of 10000 a + b + c will never equal 27 which means your loop never ends which means the app never progresses to b = 0;
4. Are you aware that C isn't Python, don't ya?
5. C uses curly brackets to indicate blocks. YOu need to read some C code to get the idea of the syntax.
6. Guys, I was trying to be subtle about it, you both just handed the answer on a silver plate to xem
7. Originally Posted by awsdert
Guys, I was trying to be subtle about it, you both just handed the answer on a silver plate to xem
You make a good point, but our homework policy mainly forbids people asking for others to do their homework for them, and consequently we discourage contributors from doing people's homework for them too. Hence, there's a fair bit of leeway in helping as long as we aren't handing out code that can be used as-is or other answers that don't require the help seeker to adapt/apply the help given for themselves. Both flp1969's reply and Malcolm McLean's reply thus fall within this acceptable range of answers: they're just somewhat more obvious hints than yours (and maybe not for flp1969's case: it assumes Ptolemaic knows Python, but the poor indentation means that a Python version of Ptolemaic's code wouldn't even be syntactically correct).
8. Originally Posted by laserlight
You make a good point, but our homework policy mainly forbids people asking for others to do their homework for them, and consequently we discourage contributors from doing people's homework for them too. Hence, there's a fair bit of leeway in helping as long as we aren't handing out code that can be used as-is or other answers that don't require the help seeker to adapt/apply the help given for themselves. Both flp1969's reply and Malcolm McLean's reply thus fall within this acceptable range of answers: they're just somewhat more obvious hints than yours (and maybe not for flp1969's case: it assumes Ptolemaic knows Python, but the poor indentation means that a Python version of Ptolemaic's code wouldn't even be syntactically correct).
*face palms* didn't even occur to me that it was xer homework, glad I took the route of a subtle answer
9. Originally Posted by awsdert
*face palms* didn't even occur to me that it was xer homework, glad I took the route of a subtle answer
It might not be formal homework, but besides the fact that we cannot be sure, there's also the thing where applying stuff rather than just using it as-is usually leads to better learning outcomes, whether the problem was set by some instructor or boss or the help seekers for themselves. So you're doing the right thing, but it isn't (necessarily) the case that flp1969 and Malcolm McLean are doing the wrong thing.
10. Guys this is not my homework lmao, we don't even have homework at my univeristy.
Nobody is gonna tell me how to fix it then?
11. I have this now but it doesn't return anything:
Code:
```int main()
{
int a, b, c, d, e;
a=0;
b=0;
c=0;
d=0;
e=0;
while(c!=9 && b!=9 && a!=9)
a=a++;
if(a=9)
a=0;
b=b+1;
if(b=9)
b=0;
c=c++;
d=(10*a+b)/(10*b+c);
e=a/c;
if(d=e)
printf("%d",a);
printf("%d",b);
printf("%d",c);
return 0;
}```
12. 1. You need lots of braces. C isn't Python. Just because you indented it nicely doesn't make C follow a different path through the code.
2. Use == in place of =, if you want to compare two values.
3. Don't mix = and ++ together with the same variable.
So at a minimum, you need something like
Code:
``` while(c!=9 && b!=9 && a!=9) {
a++;
if(a==9) {
a = 0;
}
}```
13. Many thanks!
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# By sucking sap from the young twigs of the hemlock tree,
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26 Aug 2004, 03:44
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By sucking sap from the young twigs of the hemlock tree, <<tree growth is retarded by the woolly adelgid, causing needles to change color from deep green to grayish green and to drop>> prematurely.
A. tree growth is retarded by the woolly adelgid, causing needles to change color from deep green to grayish green and to drop
B. tree growth is retarded by the woolly adelgid, and this causes the color of needles to change from deep green to grayish green, and their dropping
C. the woolly adelgid retards tree growth, which causes needles to change color from deep green to grayish green, and dropping
D. the woolly adelgid retards tree growth, causing needles to change color from deep green to grayish green and to drop
E. the woolly adelgid retards tree growth, and this causes the color of needles to change from deep green to grayish green, and the their dropping
[Reveal] Spoiler: OA
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Re: By sucking sap from the young twigs of the hemlock tree, [#permalink]
### Show Tags
26 Aug 2004, 07:40
I think the answer is D.
Tree growth can't "suck", so A and B are wrong. "Which" in C and "this" in E can't indicate the sentence before them.
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Re: By sucking sap from the young twigs of the hemlock tree, [#permalink]
### Show Tags
26 Aug 2004, 08:15
First I went for C but D makes perfect sense to me. causing needle to change color and to drop prematurely
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Re: By sucking sap from the young twigs of the hemlock tree, [#permalink]
### Show Tags
26 Aug 2004, 08:46
D is best. A and B have misplaced modifiers problem. C and E improperly introduce conjuntion "and" after the commas at end of sentence
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Re: By sucking sap from the young twigs of the hemlock tree, [#permalink]
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08 Dec 2011, 04:54
pls, help
the word "this" in E is right or wrong. Pls, explain.
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Re: By sucking sap from the young twigs of the hemlock tree, [#permalink]
### Show Tags
08 Dec 2011, 08:27
1
KUDOS
thangvietnam wrote:
pls, help
the word "this" in E is right or wrong. Pls, explain.
"this" in choice E is totally wrong, this does not refer to anything. If you use "this action (of the woolly adelgid), it will be ok.
Furthermore, choice E make parallelism in the modifier-part with 2 ambiguous items "this" vs "the their dropping".
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Re: By sucking sap from the young twigs of the hemlock tree, [#permalink]
### Show Tags
08 Dec 2011, 09:33
+1 D
A & B modifier prob
C Which...wrongly modifies the previous noun
E error in maintining II-ism
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Re: By sucking sap from the young twigs of the hemlock tree, [#permalink]
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12 Dec 2011, 04:50
By sucking sap from the young twigs of the hemlock tree, <<tree growth is retarded by the woolly adelgid, causing needles to change color from deep green to grayish green and to drop>> prematurely.
A. tree growth is retarded by the woolly adelgid, causing needles to change color from deep green to grayish green and to drop
B. tree growth is retarded by the woolly adelgid, and this causes the color of needles to change from deep green to grayish green, and their dropping
wrong ....
C. the woolly adelgid retards tree growth, which causes needles to change color from deep green to grayish green, and dropping (and dropping is not parallel to any part of the sentence)
D. the woolly adelgid retards tree growth, causing needles to change color from deep green to grayish green and to drop
E. the woolly adelgid retards tree growth, and this causes the color of needles to change from deep green to grayish green, and the their dropping
(( their refers to what? ))
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Re: By sucking sap from the young twigs of the hemlock tree, [#permalink]
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13 Dec 2011, 04:11
A and D stands out,
i would go for D for active construction.
-- is there any thing wrong with A, other than the passive construction?
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Re: By sucking sap from the young twigs of the hemlock tree, [#permalink]
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13 Dec 2011, 07:14
A and in fact B are most certainly grammatically wrong becos they belong to the classic case of misplaced modification.The participial modifier By sucking sap from the young twigs of the hemlock tree should be followed by the modified noun namely the woolly adelgid rather than the tree growth. A and B mean as if the tree growth is sucking sap.
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Re: By sucking sap from the young twigs of the hemlock tree, [#permalink]
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13 Dec 2011, 13:01
marine wrote:
By sucking sap from the young twigs of the hemlock tree, <<tree growth is retarded by the woolly adelgid, causing needles to change color from deep green to grayish green and to drop>> prematurely.
A. tree growth is retarded by the woolly adelgid, causing needles to change color from deep green to grayish green and to drop
B. tree growth is retarded by the woolly adelgid, and this causes the color of needles to change from deep green to grayish green, and their dropping
C. the woolly adelgid retards tree growth, which causes needles to change color from deep green to grayish green, and dropping
D. the woolly adelgid retards tree growth, causing needles to change color from deep green to grayish green and to drop
E. the woolly adelgid retards tree growth, and this causes the color of needles to change from deep green to grayish green, and the their dropping
Narrow down to C/D/E
Discard C as tree growth can't cause the change of color
Discard E growth, and.. green,and... their dropping?....
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Re: By sucking sap from the young twigs of the hemlock tree, [#permalink]
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05 Jan 2012, 22:39
Modifier is wrongly placed in A and B. E is wordy. D has to change .... to drop ||ism. D
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Re: By sucking sap from the young twigs of the hemlock tree, [#permalink]
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Re: By sucking sap from the young twigs of the hemlock tree, [#permalink]
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13 Apr 2016, 20:08
Can we eliminate E on the grounds that 'their dropping' is awkward
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Re: By sucking sap from the young twigs of the hemlock tree, [#permalink]
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16 Apr 2016, 13:34
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Expert's post
knuckleduster wrote:
Can we eliminate E on the grounds that 'their dropping' is awkward
In general a possessive noun / pronoun + gerund is perfectly valid combination.
The following link will probably be helpful to get a better concept on this topic:
http://www.english-for-students.com/Pos ... runds.html
Re: By sucking sap from the young twigs of the hemlock tree, [#permalink] 16 Apr 2016, 13:34
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CS395T/CAM383M Computational Statistics > HW 5 Hw5
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#1
05-12-2011, 11:16 AM
Yanan Member Join Date: Jan 2011 Posts: 15
Hw5
Proof for Problem 1 is attached. Code for Problem 2.
Code:
```data = load('hw5_data.txt','-ASCII');
[N,M] = size(data);
% mu and sigma
mu = mean(data);
sigma = cov(data);
%chi-square and p-value
chi2 = zeros(N,1);
p = zeros(N,1);
for i = 1:N
chi2(i) = (data(i,:)-mu)*inv(sigma)*(data(i,:)-mu)';
p(i) = 1-chi2cdf(chi2(i),5);
end
%FDR
[rank, index] = sort(p);
alpha = 0.1;
rej = [];
j = 1;
for i = 1:N
if p(index(i)) < i*alpha/N
rej(j) = index(i);
j = j+1;
end
end```
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# Question Video: Understanding How Heat Absorption and Emission Relates to Temperature Change Physics
A metal can is placed in constant intensity sunlight and absorbs infrared radiation. During the can’s exposure to sunlight, the temperature of the metal increases. Which of the following statements is correct? [A] Infrared radiation absorption and emission have no effect on temperature. [B] The metal loses more energy by cooling than the energy of the infrared radiation that it absorbs. [C] The metal does not receive any energy from the sunlight. [D] The metal loses exactly as much energy by cooling as the energy of the infrared radiation that it absorbs. [E] The metal absorbs more energy from infrared radiation than the energy it loses by cooling.
04:28
### Video Transcript
A metal can is placed in constant-intensity sunlight and absorbs infrared radiation. During the can’s exposure to sunlight, the temperature of the metal increases. Which of the following statements is correct? (A) Infrared radiation absorption and emission have no effect on temperature. (B) The metal loses more energy by cooling than the energy of the infrared radiation that it absorbs. (C) The metal does not receive any energy from the sunlight. (D) The metal loses exactly as much energy by cooling as the energy of the infrared radiation that it absorbs. And, lastly, answer option (E) the metal absorbs more energy from infrared radiation than the energy it loses by cooling.
As we consider which of these five answer options correctly describes what’s taking place, we can recall that in this situation we have a metal can that’s in direct sunlight. And we’re told the can is absorbing some of the infrared radiation it receives. During this exposure, while the can is in the sunlight, the temperature of the can increases. It goes up. Knowing all this, we want to pick which of these answer options correctly describes what’s going on.
Starting off with option (A), this option says that infrared radiation absorption and emission have no effect on temperature. Considering this option, we were told in our problem statement that as the can absorbs infrared radiation, its temperature does go up. Absorbing infrared radiation does have some effect on the temperature of an object, and at the same time emitting this radiation does too. When an object gives off infrared radiation through emission, then it’s losing energy. And therefore, we would expect its temperature to go down. All this shows us that option (A) won’t be our choice.
Option (B) says that the metal loses more energy by cooling than the energy of the infrared radiation that it absorbs. This option tells us that as the metal can radiates or emits infrared radiation, it loses energy. And whatever the amount of energy lost through this cooling process option (B) tells us is greater than the amount of energy gained by absorbing the sunlight. Physically, it is possible for something like this to happen, for the total energy lost by an object through radiation to be greater than the amount of energy it receives by absorption.
But in this case, we know that that’s not going on. And the reason is the temperature of the metal can is going up. If the can was losing more energy than it was gaining, we would expect this temperature to go in the opposite direction, to decrease. Since it’s going up though, that must mean that the energy lost does not exceed the energy gained.
Option (C) tells us that the metal does not receive any energy from the sunlight. But once again, if the object didn’t receive any energy, then how would its temperature be going up? This could only happen if it was receiving energy from some unnamed source. But we can reasonably assume that no such source exists because we’re not told about it. So, this must mean that the temperature increase in the metal can is due to energy it receives from the sunlight. And indeed, our problem statement confirms this when it says that the can absorbs infrared radiation. By that means it does receive energy from the sunlight. So, we won’t choose option (C) as our answer either.
Option (D) says that the metal loses exactly as much energy by cooling as the energy of the infrared radiation that it absorbs. So, this option then says that the metal can indeed does receive energy from the sunlight and that it gives off just as much energy by cooling. If this were the case though, we would expect that the temperature of our object wouldn’t change at all because the net change in energy of the object is zero. That is, just as much energy came in from the sunlight, as went out due to radiative cooling. That’s what option (D) claims. Yet, if that were true, we would expect the temperature of our object to be constant. The fact that it heats up means that it must be receiving more energy than it is given off. Option (D) isn’t our correct description either.
This leaves us with our last choice option (E), which says that the metal absorbs more energy from infrared radiation than the energy it loses by cooling. This choice tells us that indeed the can does receive energy from the sunlight, and indeed, it does radiate some of that energy away. But that overall, it receives more energy than it emits. And that is consistent with the observation that the temperature of the metal can goes up. When that happens, we would expect more energy to be put in than to be taken out. And indeed, that’s what’s taking place. The correct description of what’s happening in this scenario is that the metal absorbs more energy from infrared radiation than the energy it loses by cooling.
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# 3/4 acre equals how many square feet
• 0.75 acres is 32,670 square feet.
• tk10publ tk10canl
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https://ch.mathworks.com/matlabcentral/cody/problems/2927-matlab-basics-ii-max-index-of-max/solutions/2016482
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Cody
# Problem 2927. Matlab Basics II - Max & Index of Max
Solution 2016482
Submitted on 12 Nov 2019 at 18:59 by Asif Newaz
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = [1 4 5 2 3]; y_correct = [5,3]; assert(isequal(max_ind_max(x),y_correct))
a = 5 b = 3 out = 5 3
2 Pass
x = [3.2 4.3 -9.8 4.7 -10.9 3.7 -2.5]; y_correct = [4.7,4]; assert(isequal(max_ind_max(x),y_correct))
a = 4.7000 b = 4 out = 4.7000 4.0000
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http://fredrik-j.blogspot.com/2010/01/accurate-hypergeometric-functions-for.html
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## Thursday, January 7, 2010
### Accurate hypergeometric functions for large parameters
Holiday season means more free time for coding (yay!). In this commit, I've fixed the remaining major accuracy issues with hypergeometric functions in mpmath. The effect, generally speaking, is that mpmath will now deal much more robustly with large parameters of hypergeometric-type functions.
Previously, you would obtain an accurate value with parameters of magnitude ≈ 10−1 - 101 (say), and often for large parameters as well, but for some functions the accuracy would quickly deteriorate if parameters were increased (or moved closer to singularities). You could still obtain any accuracy by simply increasing the working precision, but you had to manually figure out the amount of extra precision required; the news is that mpmath now automatically gives a value with full accuracy even for large parameters (or screams out loud if it fails to compute an accurate value).
This doesn't mean that you can't trick mpmath into computing a wrong value by choosing sufficiently evil parameters, but it's much harder now than simply plugging in large values. Abnormally large values will now mainly accomplish abnormal slowness (while mpmath implements asymptotic expansions with respect to the argument of hypergeometric functions, evaluation for asymptotically large parameters is a much harder problem as far as I'm aware -- so mpmath in effect accomplishes it by brute force.)
The most trivial change was to change the series summation to aim for control of the relative error instead of the absolute error. This affects alternating series, where large parameters lead to very small function values. Previously, something like the following would happen:
` >>> mp.dps=5; hyp1f1(-5000, 1500, 100) 1.6353e-14 >>> mp.dps=15; hyp1f1(-5000, 1500, 100) 8.09813050863682e-25 >>> mp.dps=30; hyp1f1(-5000, 1500, 100) -1.38318247777802583583082760724e-39`
This isn't disastrously bad since the absolute error is controlled (summing this series naively in floating-point arithmetic might give something like 1e+234 due to catastrophic cancellation). But now, you get the relatively accurate answer right away, which is much nicer:
` >>> mp.dps = 5; hyp1f1(-5000, 1500, 100) 9.1501e-185 >>> mp.dps = 15; hyp1f1(-5000, 1500, 100) 9.15012134245639e-185 >>> mp.dps = 30; hyp1f1(-5000, 1500, 100) 9.15012134245639443114220499541e-185`
Of course, if the value is too small, the evaluation loop must abort eventually. The default behavior now is to raise an exception if relative accuracy cannot be obtained. The user can force a return value by either permitting a higher internal precision before aborting, or specifying a size 2−zeroprec below which the value is small enough to be considered zero:
` >>> hyp1f1(-8000, 4000, 1500) Traceback (most recent call last): ... ValueError: hypsum() failed to converge to the requested 53 bits of accuracy using a working precision of 3568 bits. Try with a higher maxprec, maxterms, or set zeroprec. >>> >>> >>> hyp1f1(-8000, 4000, 1500, maxprec=10000) 4.36754212717293e-1350 >>> hyp1f1(-8000, 4000, 1500, accurate_small=False) -1.99286380611911e-25 >>> hyp1f1(-8000, 4000, 1500, zeroprec=1000) 0.0`
Since exceptions are inconvenient, it will be necessary to add some more symbolic tests for exact zeros for certain functions (particularly orthogonal polynomials) where exact zeros arise as important special cases.
Also, cancellation detection in hypercomb is now the default for all functions. This fixes, among other things, a bug in the Meijer G-function reported by a user via email. Before:
` >>> meijerg([[],[]],[[0],[]],0.1) 0.90483741803596 >>> meijerg([[],[]],[[0,0],[]],0.1) 1.47347419562219 >>> meijerg([[],[]],[[0,0,0],[]],0.1) 1.61746025085449 >>> meijerg([[],[]],[[0,0,0,0],[]],0.1) # suspicious 13194139533312.0`
After:
` >>> meijerg([[],[]],[[0],[]],0.1) 0.90483741803596 >>> meijerg([[],[]],[[0,0],[]],0.1) 1.47347419562219 >>> meijerg([[],[]],[[0,0,0],[]],0.1) 1.61745972140487 >>> meijerg([[],[]],[[0,0,0,0],[]],0.1) 1.56808223438324`
Another important change is more correct handling parameters very close to negative integers, particularly those appearing in denominators of series. Previously, unless the integer n was small enough or the precision was set high enough, this situation would yield a bogus value (that of a prematurely truncated series):
` >>> n = -20 >>> nh = fadd(n, 1e-100, exact=True) >>> hyp0f1(n, 0.25) # nonsense 0.987581857511351 >>> hyp0f1(nh, 0.25) # same nonsense 0.987581857511351`
Now, correctly:
` >>> n = -20 >>> nh = fadd(n, 1e-100, exact=True) >>> hyp0f1(n, 0.25) Traceback (most recent call last): ... ZeroDivisionError: pole in hypergeometric series >>> hyp0f1(nh, 0.25) 1.85014429040103e+49`
Finally, and probably most importantly, the limit evaluation code has been changed to adaptively decrease the size of perturbation until convergence. Under fairly general assumptions, the maximum accurate perturbation at precision p can easily be shown to be 2−(p+C); the problem is that the parameter-dependent constant C isn't generally known. Previously C was just set to a small value (10), and naturally this would break down for some functions when parameters were increased beyond roughly that magnitude.
I briefly considered the idea of estimating C analytically in terms of the parameters, and while I think this can be done rigorously, it seems difficult -- especially to do it tightly (grossly overestimating C would murder performance). The algorithm implemented now is quasi-rigorous, and although there is some slowdown (sometimes by a fair amount), the improved reliability is definitely worth it. A user can also manually supply a perturbation size of their own taste, thereby overriding adaptivity. If the user supplies an intelligent value, this gives both the speed of the old code and full rigor. Probably some intelligent choices for particular functions could be made automatically by mpmath too, to recover the old speed in common cases.
An example of a situation where this becomes necessary is for Legendre functions with certain large parameter combinations. With the old code:
` >>> mp.dps = 15; legenp(0.5, 300, 0.25) # bad 4.19317029788723e+626 >>> mp.dps = 30; legenp(0.5, 300, 0.25) # bad 3.72428336871098039162972441784e+611 >>> mp.dps = 60; legenp(0.5, 300, 0.25) # bad 2.93794154954090326636196697693611381787845107728298382876544e+581 >>> mp.dps = 100; legenp(0.5, 300, 0.25) # only accurate to a few digits -1.71750854949725238788826203712778687036438365374945625996246145924802366061559 6579520831362887006032e+578`
With the new code:
` >>> mp.dps = 15; legenp(0.5, 300, 0.25) -1.71750854949725e+578 >>> mp.dps = 30; legenp(0.5, 300, 0.25) -1.71750854949725238788826203713e+578 >>> mp.dps = 60; legenp(0.5, 300, 0.25) -1.71750854949725238788826203712778687063419097363472262860567e+578 >>> mp.dps = 15; legenp(0.5, 300, 0.25, hmag=300) # faster, with manual perturbation -1.71750854949725e+578`
Another example is for Bessel functions differentiated to high order. With the old code:
` >>> mp.dps = 15; besseli(2, 10, derivative=100) # bad 2.63560662943646e+34 >>> mp.dps = 30; besseli(2, 10, derivative=100) # bad 23408889310840499424.9813614712 >>> mp.dps = 60; besseli(2, 10, derivative=100) # only accurate to a few digits 821.238621006501815018537509753672810563116338269219460709828`
With the new code:
` >>> mp.dps = 15; besseli(2, 10, derivative=100) 821.238621006483 >>> mp.dps = 30; besseli(2, 10, derivative=100) 821.238621006483348660925541651 >>> mp.dps = 60; besseli(2, 10, derivative=100) 821.238621006483348660925541650744338655411830854860813048862`
In related news, I've given all hypergeometric-type functions the ability to accept kwargs and pass them on to hypercomb and hypsum. This means that tuning and error control options will be available for a large number of functions (Bessel functions, etc), which could be useful for some users who need to do advanced calculations. I have yet to document these features thoroughly (the interface will also perhaps be tweaked before the next release).
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https://la.mathworks.com/matlabcentral/cody/problems/19-swap-the-first-and-last-columns/solutions/2013495
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Cody
# Problem 19. Swap the first and last columns
Solution 2013495
Submitted on 11 Nov 2019 by Zicheng Sun
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
A = [ 12 4 7 5 1 4]; B_correct = [ 7 4 12 4 1 5 ]; assert(isequal(swap_ends(A),B_correct))
2 Pass
A = [ 12 7 5 4]; B_correct = [ 7 12 4 5 ]; assert(isequal(swap_ends(A),B_correct))
3 Pass
A = [ 1 5 0 2 3 ]; B_correct = [ 3 5 0 2 1 ]; assert(isequal(swap_ends(A),B_correct))
4 Pass
A = 1; B_correct = 1; assert(isequal(swap_ends(A),B_correct))
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https://www.mersenneforum.org/showpost.php?s=419911a91763d554694822e6af65a970&p=412990&postcount=1
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Thread: A Holy New Board Game View Single Post
2015-10-18, 08:09 #1 MattcAnderson "Matthew Anderson" Dec 2010 Oregon, USA 2×359 Posts A Holy New Board Game Hi Math People, Suppose you have a plywood board measuring 3m by 8m. You want to fill a hole that measures 2m by 12m. You also only want to divide the board into two pieces. Is it possible? I know that real plywood only measures 4 foot by 8 foot. I got this puzzle from https://www.mathsisfun.com/ Also, for a hint, this is a simpler problem to one that I asked several months ago. http://www.mersenneforum.org/showthread.php?t=19499 Also note that both the board and the hole measure 24 m^2 Happy Puzzling, Matt
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http://gateoverflow.in/95042/tifr2017-a-9
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100 views
Consider the $majority$ function on three bits, $\textbf{maj}: \{0, 1\}^3 \rightarrow \{0, 1\}$ where $\textbf{maj}(x_1, x_2, x_3)=1$ if and only if $x_1+x_2+x_3 \geq 2$. Let $p(\alpha)$ be the probability that the output is 1 when each input is set to 1 independently with probability $\alpha$. What is $p'(\alpha) = \frac{d}{d\alpha} p (\alpha)$?
1. $3 \alpha$
2. $\alpha^2$
3. $6\alpha(1-\alpha)$
4. $3\alpha^2 (1-\alpha)$
5. $6\alpha(1-\alpha)+\alpha^2$
\begin{align*} &\Rightarrow \left [ \textbf{maj}\left \{ x_i \right \} = 1 \right ] \Leftrightarrow \left ( \left [ \sum_{i=1}^{3}x_i \right ] \geq 2 \right ) \\ &\Rightarrow Prob\left ( \left [ \sum_{i=1}^{3}x_i \right ] \geq 2 \right ) = Prob\left ( \left [ \sum_{i=1}^{3}x_i \right ] = 2 \right ) + \ Prob\left ( \left [ \sum_{i=1}^{3}x_i \right ] = 3 \right )\\ &\Rightarrow Prob\left ( \left [ \sum_{i=1}^{3}x_i \right ] \geq 2 \right ) = \binom{3}{2}*\left \{ \alpha .\alpha .\left ( 1-\alpha \right ) \right \} +\ \alpha ^{3} \\ &\Rightarrow Prob\left ( \left [ \sum_{i=1}^{3}x_i \right ] \geq 2 \right ) = 3.\alpha ^{2}(1-\alpha ) + \ \alpha ^{3} \\ &\Rightarrow Prob\left ( \left [ \sum_{i=1}^{3}x_i \right ] \geq 2 \right ) = 3.\alpha ^2 - 2.\alpha ^3 \\ &\Rightarrow P^{'}\left ( \alpha \right ) = \ \frac{\mathrm{d} }{\mathrm{d} \alpha }\left [ 3.\alpha ^2 - 2.\alpha ^3 \right ] \\ &\Rightarrow P^{'}\left ( \alpha \right ) = 6.\alpha \left ( 1-\alpha \right ) \\ \end{align*}
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## Monte Carlo Simulation
Discussions about the testing and simulation of mechanical trading systems using historical data and other methods. Trading Blox Customers should post Trading Blox specific questions in the Customer Support forum.
Sir G
Moderator
Posts: 243
Joined: Wed Apr 16, 2003 12:21 am
Location: Salt Lake City, Utah
### Monte Carlo Simulation
I would love to hear about the creative uses for MCS.
Sir G
Mark Johnson
Roundtable Knight
Posts: 122
Joined: Thu Apr 17, 2003 9:49 am
The most useful result from Monte Carlo techniques, for me personally, is an estimate of the cumulative distribution function of possible outcomes.
I like to run a few thousand MC tests, create a CDF, and plot it. Then I stare at the plot and introspect. I look at the 1% and 5% points on the CDF and ask myself if I'm ready to bet that those things probably won't happen to me.
Here's an artificial example. Take a single-market trading system with positive expected value. Capture all N of its trades in backtesting. Using Monte Carlo, generate ten thousand random permutations of (N/2) trades without replacement. Compute the CAGR, MAXDD, MAR for each sequence. Plot these as cumulative distribution functions. Now ask, am I emotionally ready for the 1% probability event, to happen to me?
Previous writer asked for "creative" Monte Carlo. But this is textbook and thus not creative.
Mark Johnson
Roundtable Knight
Posts: 122
Joined: Thu Apr 17, 2003 9:49 am
Here's an MC example I ran just now. I backtested an S&P system from 1/1/1996 to 4/17/2003. It generated 551 trades in that period. MonteCarlo performed the following procedure, one hundred thousand repetitions:
(a) Randomly pick half of the trades (225 of them). Scramble the order of these trades.
(b) For this scrambled sequence of trades, calculate CAGR, MaxDD, MAR. Write to disk.
After the above MonteCarlo procedure finishes, we have 100,000 samples. Build these into cumultive distribution functions and plot them. The plots say:
99% of sequences were more profitable than: CAGR=107.1% MaxDD=57.7% MAR=1.85
95% of sequences were more profitable than: CAGR=127.2% MaxDD=49.9% MAR=2.55
50% of sequences were more profitable than: CAGR=138.7% MaxDD=32.8% MAR=4.50
Wow. The "expected value" of MAR is 4.50, but there's a 5% chance it could be as small as 2.55 and a 1% chance it could be as small as 1.85. (And take a look at MaxDD!) Armed with these probabilities, we can make better-informed trading decisions.
wcb
Contributing Member
Posts: 8
Joined: Wed Apr 16, 2003 10:30 pm
Mark - What software do you use for your Monte Carlo Simulations? Does this program also compute CAGR, MaxDD, and MAR?
I currently use Trading Recipes and am interested in using MCS for further analysis. Does TR have the appropriate trade output for use in the MCS package you use?
Thanks,
WCB
Forum Mgmnt
Roundtable Knight
Posts: 1842
Joined: Tue Apr 15, 2003 11:02 am
Contact:
### Another Perspective??!
I've always thought that Monte Carlo Simulations were useful but not really very realistic. Here's my reasoning:
It seems to me that Monte Carlo simulations require independent events rather than events that are order dependent in order to be reflective of reality. In trading, it is not a consequence of random ordering that great trends tend to follow long periods of choppy markets and large drawdowns for trend-followers. Individual trades and their results don't qualify as independent events.
For this reason, they probably overstate the possible severity and length of drawdowns. I haven't checked this out rigorously, so this is just my intuition speaking.
That having been said, they are useful tools. I also think there is merit in the concept when taken from another angle.
The genesis of this idea was a conversation with Arthur Maddock about expectation after one of the day's of the g.c. seminar last year. He kept asking me about expectation (and indeed this was something that Rich taught us as part of the original Turtle class). However, I thought that thinking in terms of expectation was looking at the wrong thing. Too much data is lost in the translation.
Arthur, who has a very different way of looking at the world and problems than I do, kept persisting, and I kept thinking: "Why does he keep bringing this up?." Finally, I asked him, point-blank, why did he think it was so important. He said that he wanted to use expectation to run a Monte-Carlo-like simulation of a trading system. Now, that was an interesting and new idea for me.
SIDE NOTE: In one of the other forums, someone tried to interpret what I said and I noted that this was fine, that the interplay between different persons with different perspectives often allows the development of an idea that neither would have come up with by oneself. Someone else misinterpreting what you are getting at, sometimes results in a new insight that leads to a better perspective. This is one of the most recent examples of this phenomenon.
Except that when Arthur said this, I immediately recognized the value of taking what I thought was a better way of looking at a model of a trading system and running a Monte-Carlo-like simulation.
I have always considered it better to look at the distribution of outcomes rather than the expectation that a particular distribution might generate. One could represent this distribution as a mathematical curve, or for simplicity's sake, simply as a histogram of outcomes with associated probability percentages. (Mark, I think I've seen a post somewhere where you had a simulation contest of sorts that involved a game with this sort of probability distribution)
A histogram of outcomes that included the fat tails and had 20 elements is probably a reasonable model of the potential outcomes of a trading system.
So the idea is this:
1. Run a test using historical data (as much as you can get)
2. Build a histogram of all the trade outcomes
3. Extrapolate a bit at the ends to get the fat tails
4. Run a simulation where trades are taken from the distribution according to their respective probability
5. Run many iterations of this simulation like you would a Monte-Carlo simulation
It seems to me this would generate better results than a Monte-Carlo simulation, however, it does suffer from the same problem of ignoring the dependence of the events in trader land, where human beings are affected by the past.
One way around this might be to look, not at the trades, but rather at the periods of winners and losers with some sort of noise filter. For example, you might consider series of wins and losses that are less than 5% as noise.
You could then look at winning periods and losing periods as a series with the probablity built as a dependent series. So you would have a distributions of probabilitys of winning periods (time and size) for a given losing period and vice versa. This would be two matrices of histograms rather than a single historgram.
Then you could run a simulation and generate a series of winning periods followed by losing periods.
This might be better, but you still are missing some of the dependence over the longer term that exists in actual markets and trading. Psychology drives the market, after all.
Ted Annemann
Roundtable Knight
Posts: 118
Joined: Tue Apr 15, 2003 7:44 pm
Location: Arizona
Forum Mgmnt, I believe the statisticians have beat you to the punch. If I'm not mistaken, they've been teaching & using that approach for years. They call it "resampling".
Mark Johnson
Roundtable Knight
Posts: 122
Joined: Thu Apr 17, 2003 9:49 am
WCB, I got a p.m. asking for more details so I wrote a MC little program this morning that calculated those values. Captured trades from TS4, fed them into the program, got the 100K output triples (CAGR, MaxDD, MAR), made the CDF's & plotted. I haven't tried the "Code" feature of this BBsys before, let's see if it works:
Code: Select all
``````#include <stdio.h>
#include <math.h>
#define START_EQUITY (1e7)
#define BUXPERCAR (1.5e5)
#define YEARS (7.375 / 2.0)
#define MC_TRIALS (100000)
void mj_shuffle (cards, n)
double cards[];
int n;
{
int i, j;
double temp;
for (i = 0; i < (n-1); i++) {
j = rand () % (n - (i + 1));
j = j + i + 1;
temp = cards[i];
cards[i] = cards[j];
cards[j] = temp;
}
}
main ()
{
int i, j, k, ntrades, nrun, ncars ;
double outcome[3000];
double this, prev, peak, valley, profit, x ;
double equity, pctdd, maxpctdd, cagr, mar;
for (i=0; i<3000; i++) outcome[i] = -9876543210.9;
while (EOF != (i = scanf ("%le", &x))) {
}
for (nrun = 1; nrun <= MC_TRIALS; nrun++) {
equity = START_EQUITY;
peak = equity;
valley = peak;
maxpctdd = 0.0;
for (i = 0; i<(ntrades/2); i++) {
ncars = (int) (floor (equity / BUXPERCAR));
profit = ((double) ncars) * outcome[i];
equity += profit;
if (equity > peak) {
peak = equity;
valley = peak;
}
if (equity < valley) {
valley = equity;
pctdd = (peak - valley) / peak;
if (pctdd > maxpctdd) maxpctdd = pctdd;
}
if (equity < 0.0) equity = 0.0;
}
cagr = pow ((equity / START_EQUITY), (1.0 / YEARS));
cagr *= 100.0;
maxpctdd *= 100.0;
mar = cagr / maxpctdd;
printf ("%5d %9.3f %9.3f %9.3f\n", nrun, cagr, maxpctdd, mar);
}
}
``````
Looks like the BBsys removed some leading blanks, sorry. Cut-and-paste then run it thru "cb" or "indent" to get things lined back up. MJ
PeterK
Full Member
Posts: 13
Joined: Tue Apr 22, 2003 6:48 am
I have also seen Monte Carlo used to simulate random selection of trades where it was likely that multiple trades were offered on the same day over a portfolio of stocks or commodities, and money management would not allow all trades to be taken.
In this instance, the backtested trades are not totally "jumbled up" in the randomisng process, just the days when multiple trades were signalled.
This allows for a realistic simulation of the variety of outcomes (histograms etc) that could have occurred in actual trading. In some systems sold by vendors, I have seen the claimed performance figures are pure luck in the sense that a different random selection of trades offered each day can give wildly different results. The Monte Carlo analysis allows this "luck" factor to be shown.
Peter K
Forum Mgmnt
Roundtable Knight
Posts: 1842
Joined: Tue Apr 15, 2003 11:02 am
Contact:
Ted, I've just boned up a bit on resampling to see if I was missing something.
While I agree that at some level, the purpose of my proposed simulation and resampling overlap, I don't agree that what I was proposing was covered by resampling, especially as it relates to extending the tails of the distribution.
Monte Carlo Simulations are themselves a means of the resampling method called "randomization", so in some respects all of this is about "resampling".
You might be thinking that I was proposing something akin to "bootstrapping" which is a way of generating pseudo-populations by drawing random populations from the data. However, while this is a useful way of getting better estimates from limited sample sets, I don't think it will buy much more than simple Monte Carlo simulations.
I believe that what I was proposing was something different, especially as it relates to the fat tails. A lot of resampling seems to be focused on the mean or getting a better mean for a particular parameter from a population. As traders, we don't care about the average return so much as the worst-case scenarios that will cause us to go bust or drawdown so much that our returns are ruined for years.
So the model I proposed, which builds and extends beyond the ends of the tails as indicated by the sample, is not resampling and actually the opposite of what you might do in some resampling techniques which explicitly discard the outliers in an attempt to find a better mean.
I also need to reiterate that most of the statistics that concern themselves with samples don't relate to dependent events, so they are an imperfect fit, at best.
Christian Smart
Roundtable Fellow
Posts: 50
Joined: Fri Apr 18, 2003 8:53 pm
Location: Huntsville, AL
Contact:
Hi Forum Mgmnt,
What you've described is a form of Monte Carlo simulation - sampling from a histogram based on historical data, even one modified to account for fat tails, is just as much a form of Monte Carlo as is anything else.
Also, it is possible to perform correlated (i.e., non-independent) Monte Carlo simulations. Two well-known Excel add-ins for performing simulations provide this capability. This is not a simple ad-hoc routine, but one grounded in probability theory.
Mark Johnson
Roundtable Knight
Posts: 122
Joined: Thu Apr 17, 2003 9:49 am
To those who sent me p.m.: The trades in the MC example above were generated as follows:
System = I-Master
Slippage = 2.0 big points (\$500.00) per roundtrip ; ; ; yes, I know, I know....
Data = CSI backadjusted continuous contract of S&P day session
As you can plainly see by reading the MC code, the betsizing was embarrassingly rudimentary: trade 1 contract per \$150K of account equity.
Remember, the point of the exercise was to illustrate MonteCarlo concepts. The S&P stuff was merely an example, not some sort of claimant to the throne of Best Trading System In The Universe. The focus should be on the Monte Carlo method, not the detailed minutae of the S&P example that happened to be picked.
Sir G
Moderator
Posts: 243
Joined: Wed Apr 16, 2003 12:21 am
Location: Salt Lake City, Utah
### /2
Hi Mark-
Why do you use 1/2 of the trades? Why don't you reshuffle the whole deck of trades?
Thanks.
Sir G
blueberrycake
Roundtable Knight
Posts: 125
Joined: Mon Apr 21, 2003 11:04 pm
Location: California
I think MC simulations are essential for figuring out the correct bet size, since each distribution has a different optimal bet size. Also, it gives you a pretty good idea of whether a particular system is tradeable based on its result distribution and number of trades.
This little bit of code is rather useful in answering these questions. You specify your average win, average loss, win/loss ratio, number of bets you plan to make and the amount you are going to bet on each trade. It will then run a simulation and tell you what you can expect to end up with on average and more importantly, what the bottom part of the distribution looks like (ie 5%, 10% cutoffs).
Code: Select all
``````int main(int argc, char *argv[]) {
int i, j;
float *fCapital, curCapital;
int iterations = 500000;
int numGames = 100;
float winAmount = 2;
float lossAmount = 1;
float winPercent = .5;
float initCapital = 100;
float riskPercent = .05;
fCapital = malloc(sizeof(float) * iterations);
init_random();
for (i = 0; i < iterations; i++) {
curCapital = initCapital;
for (j = 0; j < numGames; j++) {
if (curCapital >= lossAmount) {
// percent risk model
if (((double) rand() / RAND_MAX) < winPercent)
curCapital += curCapital * riskPercent / lossAmount * winAmount;
else
curCapital -= curCapital * riskPercent;
// constant risk model
/*
if (((double) rand() / RAND_MAX) < winPercent)
curCapital += winAmount;
else
curCapital -= lossAmount;
*/
}
}
fCapital[i] = curCapital;
}
qsort((void *) fCapital, (size_t) iterations, sizeof(float), floatCompare);
printf("80 percentile: %.0f\n", fCapital[iterations / 100 * 80]);
printf("50 percentile: %.0f\n", fCapital[iterations / 100 * 50]);
printf("30 percentile: %.0f\n", fCapital[iterations / 100 * 30]);
printf("20 percentile: %.0f\n", fCapital[iterations / 100 * 20]);
printf("10 percentile: %.0f\n", fCapital[iterations / 100 * 10]);
printf("5 percentile: %.0f\n", fCapital[iterations / 100 * 5]);
return 1;
}
``````
Kiwi
Roundtable Knight
Posts: 513
Joined: Wed Apr 16, 2003 1:18 am
Location: Nowhere near
### Free Monte Carlo Sim Tool
For all you folk who've wondered what it is and wanted to try it there is a new free MCS tool. I havent tested it and it doesnt seem as orientated to money management experiments as Alex's sim at Unicorn but it is free and looks easy to use.
Advice came via Raymond Deux who makes NinjaTrader a superb tool for short term traders that I use and was:
We've just released Equity Monaco 1.0.
Equity Monaco is a Monte Carlo Simulator for analyzing
trading system perfomance. It works with NeoTicker,
but Equity Monaco can also analyze trading
system performance from other software.
Equity Monaco is a free product. We want to provide
a Monte Carlo Simulation for NeoTicker users, without
burdening NeoTicker itself too much. So we write it
as a separate program that integrates with NeoTicker.
The free part also acts as a promotion for TickQuest.
http://66.113.187.197/EquityMonaco10.exe
http://66.113.187.197/equitymonaco.pdf
-----------------
Louis Lin
TickQuest Inc www.tickquest.com
John
PS. Was going to post it at 3 other places around the forum but didnt want to make Sir G mad. Now its up to all you budding MJs to generate some new quantitative insights
gbos
Senior Member
Posts: 26
Joined: Wed May 21, 2003 1:06 pm
Location: Athens Greece
Contact:
Kiwi very nice program. Here is my own home made add-in (MonteCarlo.xla) (money management orientated) with instructions (Read_Me.pdf). Works fine on my Excel 2002 (English Version) but I haven’t tested in other excel versions so if it doesn’t work please don’t throw me rocks.
Attachments
Monte.zip
bloom
Senior Member
Posts: 35
Joined: Thu Apr 17, 2003 12:45 am
Location: SARS infested HONG KONG..ahh
hmm..we know that volatility is the lifeblood of any system, and we also know that volaitlity is cyclical, so I would think that the distribution of trades is probably mean-reverting. Would it be possible to simulate this in MC? Is there any tools that could simulate a probability distribution with
a <0.5 hurst coefficient, I think this would give us more realistic results.
gbos
Senior Member
Posts: 26
Joined: Wed May 21, 2003 1:06 pm
Location: Athens Greece
Contact:
Hi vegasoul
This is not difficult. An easy way that you can simulate this kind of relationship between trades is with the aid of Markov chains. See any introductory text on probability for reference. As for a<0.5 I can’t understand the question. If you are referring to the add-in 'a' coefficient can be changed by you and it only plays role to the relevant with 'a' questions on the menu and not in the simulation in general.
Oops I just realized you are referring to fractals!
Best Rgds
Last edited by gbos on Wed Sep 10, 2003 1:16 am, edited 1 time in total.
CRM114
Senior Member
Posts: 35
Joined: Tue May 06, 2003 7:51 pm
Location: Florida
vegasoul wrote:Is there any tools that could simulate a probability distribution with a <0.5 hurst coefficient, I think this would give us more realistic results.
I can give you a procedure that will produce approximations of fractional Brownian motion. It's on page 495 of the book
Chaos and Fractals by Peitgen, Jurgens, and Saupe, Springer-Verlag 1992.
Say you want to simulate a time series X(t), 0 <= t <= 1, with Hurst exponent H, 0 <= H <= 1. Start with
X(0) = 0 and X(1) = sigma(0) * N(0,1),
where N is a normally-distributed random number with mean 0 and variance 1, and sigma(0)^2 is the variance that you've chosen for the interval [0,1]. The remaining samples are constructed as follows.
X(0.5) = 0.5 * (X(0) + X(1)) + sigma(1) * N(0,1), sigma(1) = sigma(0) * sqrt(1 - 2^(2*H - 2))/2^H.
I've corrected what I believe was a mistake in the book, where the divisor in the previous formula was omitted.
X(0.25) = 0.5 * (X(0.0) + X(0.5)) + sigma(2) * N(0,1), sigma(2) = sigma(1) * (0.5)^H
X(0.75) = 0.5 * (X(0.5) + X(1.0)) + sigma(2) * N(0,1).
X(0.125) = 0.5 * (X(0.00) + X(0.25)) + sigma(3) * N(0,1), sigma(3) = sigma(2) * (0.5)^H
X(0.375) = 0.5 * (X(0.25) + X(0.50)) + sigma(3) * N(0,1).
X(0.625) = 0.5 * (X(0.50) + X(0.75)) + sigma(3) * N(0,1).
X(0.875) = 0.5 * (X(0.75) + X(1.00)) + sigma(3) * N(0,1).
Continue in this fashion, always reducing the standard deviation by the factor (0.5)^H.
Last edited by CRM114 on Wed Oct 06, 2004 6:26 am, edited 1 time in total.
Asterix
Senior Member
Posts: 44
Joined: Mon Apr 05, 2004 11:16 pm
Location: San Diego
### Another Approach to Monte Carlo Simulation
I began experimenting with MC analysis of trading system results about 10 years ago and wrote a program that does some of what c.f. describes in his post. (i.e. extending the distribution to account for the fat tails.)
After reading more on the topic of dependency, I began to question the validity of the MC results. Random sampling of the trade results assumes that the individual results are independent and can occur in any order.
I came up with another method that I called random entry. Rather than re-shuffling all of the trade results, I preserved the order of the trades but randomly picked the starting point. If the starting point was in the middle rather than the end, then I used a wrap-around algorithm when I got to the end of the data and began with the first value in the data and continued until reaching the starting point.
This system isn't without it's own weak points, but it does produce different statistics compared to total re-shuffling for each run. For example, sometimes, the worst drawdown was greater and the risk of ruin was greater.
smodato
Senior Member
Posts: 27
Joined: Wed Jul 14, 2004 2:53 am
I'd like to go back to this old topic and to discuss the following:
let's suppose we have a number of backtesting trades from a system,let's say 250 trades (it is just an example).
Let's look at different approaches:
1) we take the distribution characteristics, average trade, standard deviation and generate with these data 1000 different series of 250 trades sequences, each of these with position sizing, so we get a scenario of 1000 possible equities and dradown out of these we extrapolate the risk analyses of the chosen approach, this method may be weak if the distribution is not normal
2) we create 1000 permutations of the 250 trades and apply position sizing to each of them creating again 1000 equitites and drawdowns
3) we create 1000 possible equities generating random numbers between 1 and 250 and picking the correspondend trade from the main arrow and then apply position sizing and proceed in the same way as described above.
What are your comments about the three different approaches? Personally I'd prefer the 3rd but your experience would help me in the best choice.
Thanks, bye
Smodato
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# How to prove an inequality by mathematical induction? [on hold]
How to prove the following inequality by mathematical induction?
$$x + \frac 1x \ge 2, x \gt 0$$
I am aware of this.
First, I have to prove $$P(1)$$; then $$P(n+1)$$.
I am stuck at $$P(n+1)$$ because I do not know how to add the plus one to it. thanks.
EDIT 4/10/19
I misunderstood the problem; I thought I needed to prove the above inequality by mathematical induction; I learnt that I just need to prove it. I think the easiest way to prove this inequality is by constructing a direct proof.
## put on hold as unclear what you're asking by José Carlos Santos, Lord Shark the Unknown, ancientmathematician, Paras Khosla, Cesareoyesterday
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
• And what is $P(n)$? – José Carlos Santos Mar 19 at 23:09
• Induction on what? – Bernard Mar 19 at 23:10
• Is $x$ here just positive integers, or all positive real numbers? – Minus One-Twelfth Mar 19 at 23:14
No way! Induction is only for integers.
By the way here’s another way to solve it:
$$(\sqrt{x}-\frac{1}{\sqrt{x}})^2\ge 0$$ $$\Rightarrow x +\frac{1}{x}-2\ge 0\Rightarrow x+\frac{1}{x}\ge 2$$
If the inequality is supposed to be proved for positive integers $$x$$ then you get $$P(n+1)$$ from $$P(n)$$ as follows: $$n+1+\frac 1 {n+1}=n+\frac 1 n +1+\frac 1 {n+1}-\frac 1 n \geq 2+1-\frac 1 {n(n+1)}\geq 2$$ because $$n(n+1) \geq 1$$ and $$\frac 1 {n(n+1)} \leq 1$$
• what if $n=1$?? then $1(1+1) \geq 1$ may not be true – James Mar 20 at 0:06
• @JimmySabater I don't see any problem here when $n=1$. – Kavi Rama Murthy Mar 20 at 0:10
• What do you even mean by saying $2 \geq 1$ may not be true? – Kavi Rama Murthy 2 days ago
I think the easiest way to prove this inequality is by direct proof where you simplify the inequality, thereby obtaining a true statement. Then you try to check the algebra is still valid; like so:
$$x + \frac 1x \ge 2$$
$$x(x + \frac 1x) \ge 2x$$
$$x^2 + 1 \ge 2x$$
$$x^2 - 2x + 1 \ge 0$$
$$(x - 1)^2 \ge 0$$
Once you simplify, you prove it by reversing the process and checking the algebra is correct; like so:
$$x^2 - 2x + 1 \ge 0$$ $$x^2 + 1 \ge 2x$$ $$x + \frac 1x \ge 2$$
The result follows if one puts $$a=x$$ and $$b=1/x$$ on the famous AM-GM inequality
• The question clearly says 'by mathematical induction'. – Kavi Rama Murthy Mar 20 at 0:10
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See the recommended documentation of this function
Scilab help >> Sparse Matrix > gmres
gmres
Generalized Minimum RESidual method
Calling Sequence
`[x,flag,err,iter,res] = gmres(A,b,rstr,tol,maxi,M,x0)`
Arguments
A
n-by-n matrix or function returning `A*x`
b
right hand side vector
x0
initial guess vector (default: zeros(n,1))
M
preconditioner: matrix or function returning `M*x` (In the first case, default: eye(n,n))
rstr
number of iterations between restarts (default: 10)
maxi
maximum number of iterations (default: n)
tol
error tolerance (default: 1e-6)
x
solution vector
err
final residual norm
iter
number of iterations performed
flag
0 =
`gmres` converged to the desired tolerance within `maxi` iterations
1 =
no convergence given `maxi`
res
residual vector
Description
GMRES
solves the linear system `Ax=b` using the Generalized Minimal residual method with restarts.
Details
of this algorithm are described in :
"Templates for the Solution of Linear Systems: Building Blocks for Iterative Methods", Barrett, Berry, Chan, Demmel, Donato, Dongarra, Eijkhout, Pozo, Romine, and Van der Vorst, SIAM Publications, 1993 (ftp netlib2.cs.utk.edu; cd linalg; get templates.ps).
"Iterative Methods for Sparse Linear Systems, Second Edition" Saad, SIAM Publications, 2003 (ftp ftp.cs.umn.edu; cd dept/users/saad/PS; get all_ps.zip).
Examples
`// GMRES call x=gmres(A,b);`
• pcg — precondioned conjugate gradient
• qmr — quasi minimal resiqual method with preconditioning
Authors
Sage Group (IRISA, 2005)
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# High Short-circuit protection for LiPo Battery
I will be using a LiPo battery for a project, and I want to make certain that I fully understand the safety features of the Protection Circuit Module included with the battery, so that I have a safe design.
When looking at the datasheet for the battery it includes details of the PCM, and lists the limits for when the PCM protects the battery. One of the features is the short-circuit protection, which I interpret as the protection against having a too large current rushing into the battery.
The battery I am using is very small, 25 mAh, but the PCM is listed that the current limit for when the short-circuit protection starts working is 0.7A, and that the short-circuit detection time is 7.2-11 ms.
Here's my question:
• How can this be safe?
• What if I have a charger that malfunctions and does not regulate the charging current correctly, but has a current limitation of 0.5A. Could I not then have a constant 0.5A going into my battery for which the short-circuit protection does not protect for, and that my small battery would most likely not survive?
The details of the battery PCM is included below.
• The PCM usually protects against OUTPUT short circuit current. You have quoted the overcurrent delay time - the short cct delay is 380 uS max. || For overcurrent 2.1A x 11 mS x say 4V -> Joules = V x I x t = 4 x 2.1 x 0.011 ~~= 0.1 Joule - = 100 mW. seconds . It seems "rather unlikely" that that amount of energy would do significant damage. If you REALLY want a faster and/or lower current PCM you could easily enoughbuild one with a suitable fast comparator. || Do you REALLY have a 25 mAh LiPo - that's tiny - what is the application? Dec 6, 2019 at 11:46
• Consider the consequences of charging it at 0.5A. Its charge voltage will increase by 0.5A * cell internal resistance, and the over-charge protection will cut in a bit early. If the cell is designed to survive charging at 20C, it is safe. Otherwise you will need additional charge current limitation outside the protection module, which is normally not the job of the PCM. (And Russell is of course right - over-current protection usually refers to discharge protection)
– user16324
Dec 6, 2019 at 11:53
• Where did you get the 25mAh battery? Can you provide a link to the datasheet? Dec 6, 2019 at 19:41
• @BruceAbbott You can see the battery here: honcell.com/products/models/id/1667.html Dec 7, 2019 at 16:22
• @RussellMcMahon Yeah, its a really cute little battery, the application is an IoT product with really low consumption, but where a fast charge (1C) rate is required. Dec 7, 2019 at 16:23
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Minimum Wage Regression
# Minimum Wage Regression
скачати
Minimum Wage Regression Essay, Research Paper
Webonomics
Mr. Haessler
17 February 2000
Minimum Wage Regression
The proposed increase in minimum wage will do a number of things to the economy. The greatest effect it has will be on inflation. There is a strong correlation between raises in the minimum wage, and raises in inflation. The effect of raising the minimum wage on employment is very little. The correlation between an increase in inflation, and an increase in unemployment is weak also.
The strongest correlation is between increases in minimum wage predicting increases in inflation. The equation for this is CPI= -25.9+38.8 minimum wage. The R-sq value is 87.5%, and the R-sq(adj) value is 86.8%. This is proof of a strong relationship between the two variables. Increases in minimum wage predict increases in inflation 86.8% of the time. When minimum wage is increased, it brings a rise in inflation. There is more money put into the economy with a minimum wage hike, but inflation counteracts this. As inflation rises, the value of the dollar decreases. There is very little net change. The correlation between increases in minimum wage, and increased
unemployment is extremely low. The equation that models this relationship is Unemployment= 9.75-.844 minimum wage. The R-sq value is 19.0%, and the R-sq(adj) value is 14.7%. This indicates a very feeble correlation. Increases in minimum wage predict increased unemployment 14.7% of the time. Basically, increases in minimum wage are a very inaccurate predictor for an increase in unemployment. The proposed minimum wage increase will have almost no effect on unemployment according to this model.
Raises in inflation have a very weak correlation to increased unemployment as well. The equation that models this fact is Unemployment= 9.30-.0227 CPI. The R-sq value is 23.6%, and the R-sq(adj) value is 19.6%. This indicates an increase of inflation predicts an increase of unemployment 19.6% of the time. This, again, is an inaccurate indicator. Increases in inflation have almost no effect on unemployment, as shown by the model.
These three relationships show a chain of events. When the minimum wage is increased, inflation will go up as well. This does not necessarily predict an increase in unemployment. An increase in inflation does not necessarily predict an increase in unemployment either. Basically, an increase in minimum wage will most likely predict an increase in inflation.
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GeoffBarrenger
Qrew Captain
27 days ago
# Calculation which requires Summary from Other Table
Hi all
I'm trying to calculate a total cost per unit shipped of both Overhead Expenses and Operating Expenses so that I can benchmark a KPI for my services. I am having trouble summarizing totals from another table.
• Operating Expenses are easy - we track based on jobs and log expenses. This can include gas, contractors, extra rentals, food expenses, etc
• Overhead Expenses - I am struggling with because my per unit Overhead expense should get lower and lower the more units I ship, but will also go up the more days into the Fiscal Year I am in. This requires me to solve for how many days into the Fiscal Year have passed (no issue), and also to summarize how many units I have shipped this fiscal from another table (my problem)
How do I put into a formula, a request to summarize the total from a different table using a parameter? ex:
var number totalUnitsShippedThisYear = <look in table unit shipped and get total # of units shipped who's ship date = this fiscal year>
Below is the step by step how I am thinking this through! Currently what I do is pull a summary report, and then hard code the total number of units shipped. This makes it fairly easy to pull the report once a week - but I would like this to be automated.
Thank you thank you!
----------------------- Below is the thinking!
1. Divide number of days in a year into my Yearly Overhead Expenses(a number I provide) estimate (rent, salary, fixed costs) to find my Daily Overhead Expenses (how much it costs just to keep the business going)
2. I can calculate number of days into this fiscal we are currently (days Since April 1)
3. I can then multiply the number of days passed this year by the Daily Overhead Expense. I now know how much I have spent to keep the building open this year.
• example: if it costs me \$365,000 to keep the building running, and I am 15 days into the year (April 15th) I have spent \$15,000
4. Then what I need to do is summarize from another table my total shipped units. The thinking is - if I am shipping more units, my total Overhead Expense PER UNIT will be lower ( the building still costs money to rent if I ship 1 unit per year or 1 million units per year)
5. If I can get the summary of units shipped this Fiscal Year(from another table), I can now divide the Daily Overhead Expense by the Total Shipped Units this Fiscal Year
6. I would now have my PER UNIT Overhead Expense
• scenario 1: If it is April 15th, and I know it has cost me \$15,000 to keep the lights on, but I have only shipped 1 unit, my Per Unit Overhead Expense is \$15,000!!!
• scenario 2: But if we have been busy and shipped 1000 units a day since April 1, out Per Unit Overhead Expense is only \$1
7. For each unit shipped - I will also have an PER UNIT Operating Cost (which I already can calculate) - based off how many units were shipping in that job (another table) and the expenses from that job (ex: we bought gas for the truck and paid for the driver of the truck by mile)
• scenario 1: The 1 unit I shipped cost me \$100 of gas. The operating cost per unit is \$100
• scenario 2: every day I made one delivery of 1000 each - and somehow magically the gas was always \$100 (to keep math simple for this - but we would use actuals(\$0.10 per unit)
8. I want to now add by PER UNIT Operating Cost & PER UNIT Daily Overhead Expense to find my TOTAL COST PER UNIT
1. scenario 1: thinking back to scenario 1, my Total Cost per Unit is actually \$15,100
2. scenario 2: thinking back to scenario 1, my Total Cost per Unit is \$1 plus 10 cents, so \$1.10 per unit.
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EE 321 Lab
Lab 9: BIPOLAR JUNCTION TRANSISTORS, Part II
Prelab for Lab 9
In this lab we will In this lab we will investigate a bipolar junction transistor (BJT) amplifier and emitter follower.
The amplifier circuit of last week's lab is impractical because the bias or operating point depends on the beta of the transistor. The following bias technique controls the voltage across RE and hence the emitter current IE (Example 4.7).
1. Construct the circuit in Figure 1 and measure the d.c. bias voltages at the base, emitter and collector. From your measurements, determine VBE, VBC. Is the transistor biased in its active mode? What are the d.c. (Q `quiescent') values of IC, IE?
2) Apply a triangle wave to the input and measure the voltage gain. Is the gain close to that found in the prelab?
3) The voltage gain can be increased by decreasing RE. Reduce RE to zero for signal frequencies by `bypassing' it with a large capacitor (100 uF) in parallel with RE. (Be sure to observe the correct polarity of the electrolytic capacitor.) Reduce the input amplitude and look at the output waveform.
Further reduce the input amplitude until the output is approximately linear and measure the gain. The gain has been increased to the large values, but at what expense?
The Emitter Follower Amplifier
4) When the input signal is applied to the base and the output is taken at the emitter, Figuure 2, the amplifier is called an `emitter follower'. This is because the emitter voltage `follows' the base voltage. Emitter follower amplifiers are useful because they have a high input impedance and a low output impedance. They amplify the signal power by increasing its current, and are therefore sometimes called a `current booster'.
Construct the circuit in Figure 2.
• With zero (ground input) or small input signal, measure and note the d.c. bias voltages.
• With an input, measure the voltage gain (magnitude and sign).
• The dynamic range of the output is the largest that it can be without saturating or cliping. Measure the dynamic range of the output voltage (decrease the supply voltages to +-12 and measure both vout limits). What causes the output to limit in each case (when is the transistor saturated, active, cutoff)?
• What would happen to vout if the emitter resistor were connected to ground instead of -15V? Try it. What does this mean.
5) Measure Zin at the base of the transistor in the following way.
• Increase the source resistance RS until vout is reduced by about 1/2 (or by some other value).
• Sketch the equivalent circuit (use an equivalent circuit of an amplifier from Ch 1, not the transistor model) and compute Zin from your measurement.
• From the theoretical result that Zin = ( beta + 1) [re + RE ] and re=alpha/gm, estimate the beta of your transistor.
6) Simulate a signal of 10 k Ohms source resistance by setting RS = 10 k Ohms (see Figure 3). Measure the output impedance Zout by loading the output with about 220 Ohms resistance. Use a d.c. blocking capacitor as shown to prevent the load from changing the bias voltages. Ensure that the capacitive impedance is small compared to 220 Ohms by working at a signal frequency of about 10 kHz. Decrease the input until the output is not clipped.
Sketch an equivalent circuit for the amplifier output and compute Zout from your measurement. Compare with the theoretical value Zout = [re + RS /( beta + 1)] || RE. How much output voltage would you have gotten by connecting RL directly to the source (RS is part of the source)? What does this show?
7) EXTRA CREDIT. In the emitter follower (Figure 3) replace the transistor by two transistors connected in a Darlington configuration, (Figure 4): What is the effective beta of this `super beta' transistor? Measure Zin and Zout of the follower now?
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# All posts by Jawwad Farid
## Session V – B: Corporate Finance: Beta, Calculating WACC or Weighted Average Cost of Capital
Concept Title: Weighted Average Cost of Capital (WACC) Description: Explains WACC and how to calculate it Explanation Cost of capital is a concept that can be derived from the discussion we have had about opportunity cost. From the discussion that we have had so far,
## Corporate Finance: Opportunity Cost and Cost of Capital
Opportunity Cost & Cost of Capital Let’s look at two investment opportunities. Opportunity A is a deposit in your local bank. You put the money in and forget it for a year. It’s relatively safe and offers an annual return of 7%. This means that
## Session IV – C: Corporate Finance: Calculating Internal Rate of Return or IRR
Concept Title: Internal Rate of Return (IRR) Description: Teaches you to calculate the IRR Option D Lower your expected return We know that we can’t earn 10% per annum for seven years on this investment. What is the rate that we can earn? There are
## Session IV – B: Corporate Finance: Present Value in Action
Concept Title: Present Value in Action Let’s work with the example above running the numbers using our new formulas and an interest rate or discount rate of 10% a year. We make one additional assumption that all payments, investments as well as returns are made
## Corporate Finance: Discount rate and time value of money
Time value of money. Interest Rate /Discount Rate / Internal Rate of Return. It’s time to talk about interest rates, discount rates and time value of money. Three inter related core concepts in Finance that lead to a number of interesting applications. Compounding Let’s go
## Session III – B: Corporate Finance: The many faces of Return: ROE, ROIC and Payback
Concept Title: Return Description: Explains teaches how to calculate return and different kinds of returns Explanation We already know that compensation for bearing risk is called return. The question now is how do we determine an appropriate return? In order to answer this question we
## Session III – A: Corporate Finance: Risk & Return
Description: Explains the concept of Risk, the reward for bearing the risk and different types of return of Investments Concept Title: Risk & Reward Definition: Establishes the relationship between risk and reward When it comes to risk, there are probably as many definitions out there
## Session II – C: Corporate Finance: Equity and the Income Statement
Concept Title: Equity Description: This concept takes a look on equity and identifies the different equity types Equity When you take out what the firm owes from what the firm owns, something should be left for the original owners of the firm. The net amount
## Session II-B: Corporate Finance: Balance Sheet: Liabilities & Working Capital
Concept Title: Liabilities Description: Continues the discussion of basic accounting and finance terms with a focus on liabilities Liabilities The opposite of assets, liabilities are things that other people own and the firm owes. Similar to assets there are also different shades of liabilities. The
## Session II-A: Corporate Finance: The Balance Sheet, Assets, Depreciation
Review of Financial Statements Description: Builds on Key concepts explained in the previous session and explains the financial statements in detail A sample balance sheet (format) is given below followed by an explanation of the various heads of accounts in the balance sheet, Their explanations
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# DoWhy: Different estimation methods for causal inference
This is a quick introduction to the DoWhy causal inference library. We will load in a sample dataset and use different methods for estimating the causal effect of a (pre-specified)treatment variable on a (pre-specified) outcome variable.
We will see that not all estimators return the correct effect for this dataset.
First, let us add the required path for Python to find the DoWhy code and load all required packages
[1]:
import numpy as np
import pandas as pd
import logging
import dowhy
from dowhy import CausalModel
import dowhy.datasets
Now, let us load a dataset. For simplicity, we simulate a dataset with linear relationships between common causes and treatment, and common causes and outcome.
Beta is the true causal effect.
[2]:
data = dowhy.datasets.linear_dataset(beta=10,
num_common_causes=5,
num_instruments = 2,
num_treatments=1,
num_samples=10000,
treatment_is_binary=True,
outcome_is_binary=False)
df = data["df"]
df
[2]:
Z0 Z1 W0 W1 W2 W3 W4 v0 y
0 0.0 0.521013 -1.325391 2.116213 -0.557359 -1.865444 1.605934 False 10.352212
1 0.0 0.463769 1.069812 1.726722 -0.610115 0.350867 0.882315 True 19.499769
2 0.0 0.125765 -1.945924 1.484627 -1.575032 0.520069 0.559467 False 2.205429
3 0.0 0.861999 -0.221328 0.939332 0.065349 1.291141 0.056167 True 12.968877
4 0.0 0.893044 -0.533583 2.024348 -0.227786 -2.962155 0.057679 True 14.847690
... ... ... ... ... ... ... ... ... ...
9995 0.0 0.019696 -0.518280 2.516658 -3.063094 -1.035060 1.231945 False 7.816809
9996 0.0 0.692714 -0.510823 0.612004 -0.758462 -1.786669 1.217035 True 15.047879
9997 0.0 0.164993 0.190624 0.365532 -0.312707 0.794803 2.458445 True 21.472658
9998 0.0 0.382386 -2.654195 1.469548 -2.191149 -0.681429 1.197488 False 2.936543
9999 0.0 0.713687 -0.799016 0.810676 -0.969194 -0.467067 1.327113 True 15.621830
10000 rows × 9 columns
Note that we are using a pandas dataframe to load the data.
## Identifying the causal estimand
We now input a causal graph in the DOT graph format.
[3]:
# With graph
model=CausalModel(
data = df,
treatment=data["treatment_name"],
outcome=data["outcome_name"],
graph=data["gml_graph"],
instruments=data["instrument_names"]
)
[4]:
model.view_model()
[5]:
from IPython.display import Image, display
display(Image(filename="causal_model.png"))
We get a causal graph. Now identification and estimation is done.
[6]:
identified_estimand = model.identify_effect(proceed_when_unidentifiable=True)
print(identified_estimand)
Estimand type: nonparametric-ate
### Estimand : 1
Estimand name: backdoor
Estimand expression:
d
─────(Expectation(y|W0,W1,W3,W4,W2))
d[v₀]
Estimand assumption 1, Unconfoundedness: If U→{v0} and U→y then P(y|v0,W0,W1,W3,W4,W2,U) = P(y|v0,W0,W1,W3,W4,W2)
### Estimand : 2
Estimand name: iv
Estimand expression:
Expectation(Derivative(y, [Z0, Z1])*Derivative([v0], [Z0, Z1])**(-1))
Estimand assumption 1, As-if-random: If U→→y then ¬(U →→{Z0,Z1})
Estimand assumption 2, Exclusion: If we remove {Z0,Z1}→{v0}, then ¬({Z0,Z1}→y)
### Estimand : 3
Estimand name: frontdoor
No such variable found!
## Method 1: Regression
Use linear regression.
[7]:
causal_estimate_reg = model.estimate_effect(identified_estimand,
method_name="backdoor.linear_regression",
test_significance=True)
print(causal_estimate_reg)
print("Causal Estimate is " + str(causal_estimate_reg.value))
*** Causal Estimate ***
## Identified estimand
Estimand type: nonparametric-ate
### Estimand : 1
Estimand name: backdoor
Estimand expression:
d
─────(Expectation(y|W0,W1,W3,W4,W2))
d[v₀]
Estimand assumption 1, Unconfoundedness: If U→{v0} and U→y then P(y|v0,W0,W1,W3,W4,W2,U) = P(y|v0,W0,W1,W3,W4,W2)
## Realized estimand
b: y~v0+W0+W1+W3+W4+W2
Target units: ate
## Estimate
Mean value: 10.00020571565491
p-value: [0.]
Causal Estimate is 10.00020571565491
## Method 2: Stratification
We will be using propensity scores to stratify units in the data.
[8]:
causal_estimate_strat = model.estimate_effect(identified_estimand,
method_name="backdoor.propensity_score_stratification",
target_units="att")
print(causal_estimate_strat)
print("Causal Estimate is " + str(causal_estimate_strat.value))
*** Causal Estimate ***
## Identified estimand
Estimand type: nonparametric-ate
### Estimand : 1
Estimand name: backdoor
Estimand expression:
d
─────(Expectation(y|W0,W1,W3,W4,W2))
d[v₀]
Estimand assumption 1, Unconfoundedness: If U→{v0} and U→y then P(y|v0,W0,W1,W3,W4,W2,U) = P(y|v0,W0,W1,W3,W4,W2)
## Realized estimand
b: y~v0+W0+W1+W3+W4+W2
Target units: att
## Estimate
Mean value: 10.019860140019505
Causal Estimate is 10.019860140019505
/home/amit/py-envs/env3.8/lib/python3.8/site-packages/sklearn/utils/validation.py:72: DataConversionWarning: A column-vector y was passed when a 1d array was expected. Please change the shape of y to (n_samples, ), for example using ravel().
return f(**kwargs)
## Method 3: Matching
We will be using propensity scores to match units in the data.
[9]:
causal_estimate_match = model.estimate_effect(identified_estimand,
method_name="backdoor.propensity_score_matching",
target_units="atc")
print(causal_estimate_match)
print("Causal Estimate is " + str(causal_estimate_match.value))
/home/amit/py-envs/env3.8/lib/python3.8/site-packages/sklearn/utils/validation.py:72: DataConversionWarning: A column-vector y was passed when a 1d array was expected. Please change the shape of y to (n_samples, ), for example using ravel().
return f(**kwargs)
*** Causal Estimate ***
## Identified estimand
Estimand type: nonparametric-ate
### Estimand : 1
Estimand name: backdoor
Estimand expression:
d
─────(Expectation(y|W0,W1,W3,W4,W2))
d[v₀]
Estimand assumption 1, Unconfoundedness: If U→{v0} and U→y then P(y|v0,W0,W1,W3,W4,W2,U) = P(y|v0,W0,W1,W3,W4,W2)
## Realized estimand
b: y~v0+W0+W1+W3+W4+W2
Target units: atc
## Estimate
Mean value: 9.752837406069007
Causal Estimate is 9.752837406069007
## Method 4: Weighting
We will be using (inverse) propensity scores to assign weights to units in the data. DoWhy supports a few different weighting schemes: 1. Vanilla Inverse Propensity Score weighting (IPS) (weighting_scheme=“ips_weight”) 2. Self-normalized IPS weighting (also known as the Hajek estimator) (weighting_scheme=“ips_normalized_weight”) 3. Stabilized IPS weighting (weighting_scheme = “ips_stabilized_weight”)
[10]:
causal_estimate_ipw = model.estimate_effect(identified_estimand,
method_name="backdoor.propensity_score_weighting",
target_units = "ate",
method_params={"weighting_scheme":"ips_weight"})
print(causal_estimate_ipw)
print("Causal Estimate is " + str(causal_estimate_ipw.value))
*** Causal Estimate ***
## Identified estimand
Estimand type: nonparametric-ate
### Estimand : 1
Estimand name: backdoor
Estimand expression:
d
─────(Expectation(y|W0,W1,W3,W4,W2))
d[v₀]
Estimand assumption 1, Unconfoundedness: If U→{v0} and U→y then P(y|v0,W0,W1,W3,W4,W2,U) = P(y|v0,W0,W1,W3,W4,W2)
## Realized estimand
b: y~v0+W0+W1+W3+W4+W2
Target units: ate
## Estimate
Mean value: 12.690127182579836
Causal Estimate is 12.690127182579836
/home/amit/py-envs/env3.8/lib/python3.8/site-packages/sklearn/utils/validation.py:72: DataConversionWarning: A column-vector y was passed when a 1d array was expected. Please change the shape of y to (n_samples, ), for example using ravel().
return f(**kwargs)
## Method 5: Instrumental Variable
We will be using the Wald estimator for the provided instrumental variable.
[11]:
causal_estimate_iv = model.estimate_effect(identified_estimand,
method_name="iv.instrumental_variable", method_params = {'iv_instrument_name': 'Z0'})
print(causal_estimate_iv)
print("Causal Estimate is " + str(causal_estimate_iv.value))
*** Causal Estimate ***
## Identified estimand
Estimand type: nonparametric-ate
### Estimand : 1
Estimand name: iv
Estimand expression:
Expectation(Derivative(y, [Z0, Z1])*Derivative([v0], [Z0, Z1])**(-1))
Estimand assumption 1, As-if-random: If U→→y then ¬(U →→{Z0,Z1})
Estimand assumption 2, Exclusion: If we remove {Z0,Z1}→{v0}, then ¬({Z0,Z1}→y)
## Realized estimand
Realized estimand: Wald Estimator
Realized estimand type: nonparametric-ate
Estimand expression:
-1
Expectation(Derivative(y, Z0))⋅Expectation(Derivative(v0, Z0))
Estimand assumption 1, As-if-random: If U→→y then ¬(U →→{Z0,Z1})
Estimand assumption 2, Exclusion: If we remove {Z0,Z1}→{v0}, then ¬({Z0,Z1}→y)
Estimand assumption 3, treatment_effect_homogeneity: Each unit's treatment ['v0'] is affected in the same way by common causes of ['v0'] and y
Estimand assumption 4, outcome_effect_homogeneity: Each unit's outcome y is affected in the same way by common causes of ['v0'] and y
Target units: ate
## Estimate
Mean value: 11.085627630661348
Causal Estimate is 11.085627630661348
## Method 6: Regression Discontinuity
We will be internally converting this to an equivalent instrumental variables problem.
[12]:
causal_estimate_regdist = model.estimate_effect(identified_estimand,
method_name="iv.regression_discontinuity",
method_params={'rd_variable_name':'Z1',
'rd_threshold_value':0.5,
'rd_bandwidth': 0.1})
print(causal_estimate_regdist)
print("Causal Estimate is " + str(causal_estimate_regdist.value))
local_rd_variable local_treatment local_outcome
0 0.521013 False 10.352212
1 0.463769 True 19.499769
9 0.477009 True 15.650173
13 0.500243 False -0.607821
16 0.596191 True 7.387162
... ... ... ...
9980 0.449123 False 15.319816
9982 0.501416 True 15.874824
9983 0.447196 True 8.519615
9985 0.499184 False -0.135301
9990 0.475909 True 19.444943
[2005 rows x 3 columns]
*** Causal Estimate ***
## Identified estimand
Estimand type: nonparametric-ate
### Estimand : 1
Estimand name: iv
Estimand expression:
Expectation(Derivative(y, [Z0, Z1])*Derivative([v0], [Z0, Z1])**(-1))
Estimand assumption 1, As-if-random: If U→→y then ¬(U →→{Z0,Z1})
Estimand assumption 2, Exclusion: If we remove {Z0,Z1}→{v0}, then ¬({Z0,Z1}→y)
## Realized estimand
Realized estimand: Wald Estimator
Realized estimand type: nonparametric-ate
Estimand expression:
Expectation(Derivative(y, local_rd_variable))⋅Expectation(Derivative(v0, local
-1
_rd_variable))
Estimand assumption 1, As-if-random: If U→→y then ¬(U →→{Z0,Z1})
Estimand assumption 2, Exclusion: If we remove {Z0,Z1}→{v0}, then ¬({Z0,Z1}→y)
Estimand assumption 3, treatment_effect_homogeneity: Each unit's treatment ['local_treatment'] is affected in the same way by common causes of ['local_treatment'] and local_outcome
Estimand assumption 4, outcome_effect_homogeneity: Each unit's outcome local_outcome is affected in the same way by common causes of ['local_treatment'] and local_outcome
Target units: ate
## Estimate
Mean value: 17.24115839926478
Causal Estimate is 17.24115839926478
| 3,539
| 11,091
|
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##### Saxon 8/7 Math lesson 108
Mathematics Tutor: None Selected Time limit: 1 Day
Liz is drawing a floor plan of her house. On the plan. 1 inch equals 2 feet. What is the floor area of a room that measures by & 7 1/2 inches?
May 27th, 2015
1 in = 2 feet
7in = 14 feet
1/2 inch = 1 feet
total = 15 feet
May 27th, 2015
...
May 27th, 2015
...
May 27th, 2015
Dec 4th, 2016
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Mark as Final Answer
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Final Answer
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Final Answer
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CC-MAIN-2016-50
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longest
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en
| 0.915205
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https://www.easycodingzone.com/2023/01/chef-and-prime-divisors-codechef.html
| 1,718,220,582,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198861183.54/warc/CC-MAIN-20240612171727-20240612201727-00438.warc.gz
| 672,079,596
| 37,643
|
# Chef and Prime Divisors | codechef solution
You are given two positive integers – A and B. You have to check whether A is divisible by all the prime divisors of B.
### Input
The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows.
For each test case, you are given two space separated integers – A and B.
### Output
For each test case, output "Yes" (without quotes) if A contains all prime divisors of B, otherwise print "No".
### Constraints
• 1 ≤ T ≤ 104
• 1 ≤ A, B ≤ 1018
• Subtask 1 (20 points):1 ≤ B ≤ 107
• Subtask 2 (30 points):1 ≤ A ≤ 107
• Subtask 3 (50 points): Original constraints
Input
Output
```3
120 75
128 16
7 8```
```Yes
Yes
No```
### Explanation:
Example case 1. In the first case 120 = 23*3*5 and 75 = 3*52. 120 is divisible by both 3 and 5. Hence, we will print "Yes"
Example case 2. In the second case both 128 and 16 are powers of two. Hence, the answer is "Yes"
Example case 3. In the third case 8 is power of two and 7 is not divisible by 2. So, the answer is "No"
Code(C++):-
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
int main()
{
ll t;
cin >> t;
while (t--)
{
ll a, b;
cin >> a >> b;
ll GCD = __gcd(a, b);
while (b > 1 && GCD > 1)
{
b /= GCD;
GCD = __gcd(a, b);
}
if (b > 1)
{
cout << "No\n";
}
else
{
cout << "Yes\n";
}
}
return 0;
}
### Recommended Post :-
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Capgemini Coding Questions:-
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Tech Mahindra coding questions:-
Unthinkable Solutions coding questions:-
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| 0.666965
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https://math.answers.com/Q/What_grade_is_206_out_of_270
| 1,718,934,203,000,000,000
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text/html
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crawl-data/CC-MAIN-2024-26/segments/1718198862032.71/warc/CC-MAIN-20240620235751-20240621025751-00860.warc.gz
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0
# What grade is 206 out of 270?
Updated: 9/26/2023
Wiki User
7y ago
As a percentage it is: 76.'296'% recurring '296'
Wiki User
7y ago
Earn +20 pts
Q: What grade is 206 out of 270?
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Related questions
### How many bones do adults have in their?
Adults have 206 bones in their body. This number can vary slightly from person to person.
### How many bones in a hurman?
At birth there are 270. As an adult, humans have 206 bones.
32% or F
### In springfield Middle School there are 270 sixth grade students and 270 seventh grade students and 250 eight grade students What percent of the students are in eight grade?
31%. first you add all the numbers then divide it by how many 8th graders there are.
### Indicate the number of bones between children and adult?
there are 270 bones in children while there are 206 bones in adults..
### What is 9x30?
270 i might have gotten the right answer and i am 10 and in 4 grade
### What was the popular vote tally?
Election ResultsCandidatePopular votePercentageElectoral votes (270 to win)Barack Obama6045996250%303Mitt Romney5765397348%206
### How many bones are in a mammal's body?
206 in an adult human and approximately 270 in a newborn infant
270 kips/lb^3
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Babies are born with around 270 bones, but some of these bones fuse together as the baby grows, so the average adult ends up with 206 bones.
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### I got 270 on the math MAP testing what grade level is that at?
i got a 275 at the start of my 10th grade year and i am an advanced math student. a 270 is probably pretty good for high school and extremly good for middle school.
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24.242 | Spring 2004 | Undergraduate
# Logic II
## Syllabus
A list of topics by lecture is available in the calendar listed below.
### Course Meeting Times
Lectures: 2 sessions / week, 1.5 hours / session
### Description
We’ll begin with an introduction to the theory of computability, then proceed to a detailed study of its most illustrious result: Kurt Gödel’s theorem that, for any system of true arithmetical statements we might propose as an axiomatic basis for proving truths of arithmetic, there will be some arithmetical statements that we can recognize as true even though they don’t follow from the system of axioms. In my opinion, which is widely shared, this is the most important single result in the entire history of logic, important not only on its own right but for the many applications of the technique by which it’s proved. We’ll discuss some of these applications, among them: Church’s theorem that there is no algorithm for deciding when a formula is valid in the predicate calculus; Tarski’s theorem that the set of true sentence of a language isn’t definable within that language; and Gödel’s second incompleteness theorem, which says that no consistent system of axioms can prove its own consistency.
The text for the course will be lecture notes.
There will be homework assignments every week or two, and they will be the basis for your grade in the course. I would encourage you to work together on the homework, but when you finally sit down to write up your answers, you should work by yourself without looking at anyone else’s efforts.
### Calendar
LEC # TOPICS KEY DATES
1 Why Study Computability?
2-3 Key Computability Concepts
4 The Language of Arithmetic
5 Church-Turing Thesis
6 Nonstandard Models of Arithmetic
7 Gödel Numbering Homework 1 due
8 Robinson’s Arithmetic Homework 2 due
9-10 Coding Proofs
11-12 Peano Arithmetic Homework 3 due in Lec #12
13-14 Self-Reference Lemma
15-16 First Incompleteness Theorem Homework 4 due in Lec #16
17 Interpretations
18 Tarski’s Theory of Truth Homework 5 due
19
Gödel, Mechanism, and Mind
Articles by Lucas and Benacerraf
20-21 Second Incompleteness Theorem Homework 6 due in Lec #21
22 Introduction to Modal Logic
23-24 Provability Logic Homework 7 due in Lec #23
25 Defining Exponentiation Homework 8 due
## Course Info
Spring 2004
##### Learning Resource Types
Lecture Notes
Problem Sets with Solutions
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Home Metamath Proof ExplorerTheorem List (p. 38 of 325) < Previous Next > Browser slow? Try the Unicode version.
Mirrors > Metamath Home Page > MPE Home Page > Theorem List Contents > Recent Proofs This page: Page List
Color key: Metamath Proof Explorer (1-22374) Hilbert Space Explorer (22375-23897) Users' Mathboxes (23898-32447)
Theorem List for Metamath Proof Explorer - 3701-3800 *Has distinct variable group(s)
TypeLabelDescription
Statement
Theoremdfif2 3701* An alternate definition of the conditional operator df-if 3700 with one fewer connectives (but probably less intuitive to understand). (Contributed by NM, 30-Jan-2006.)
Theoremdfif6 3702* An alternate definition of the conditional operator df-if 3700 as a simple class abstraction. (Contributed by Mario Carneiro, 8-Sep-2013.)
Theoremifeq1 3703 Equality theorem for conditional operator. (Contributed by NM, 1-Sep-2004.) (Revised by Mario Carneiro, 8-Sep-2013.)
Theoremifeq2 3704 Equality theorem for conditional operator. (Contributed by NM, 1-Sep-2004.) (Revised by Mario Carneiro, 8-Sep-2013.)
Theoremiftrue 3705 Value of the conditional operator when its first argument is true. (Contributed by NM, 15-May-1999.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
Theoremiffalse 3706 Value of the conditional operator when its first argument is false. (Contributed by NM, 14-Aug-1999.)
Theoremifnefalse 3707 When values are unequal, but an "if" condition checks if they are equal, then the "false" branch results. This is a simple utility to provide a slight shortening and simplification of proofs vs. applying iffalse 3706 directly in this case. It happens, e.g., in oevn0 6718. (Contributed by David A. Wheeler, 15-May-2015.)
Theoremifsb 3708 Distribute a function over an if-clause. (Contributed by Mario Carneiro, 14-Aug-2013.)
Theoremdfif3 3709* Alternate definition of the conditional operator df-if 3700. Note that is independent of i.e. a constant true or false. (Contributed by NM, 25-Aug-2013.) (Revised by Mario Carneiro, 8-Sep-2013.)
Theoremdfif4 3710* Alternate definition of the conditional operator df-if 3700. Note that is independent of i.e. a constant true or false. (Contributed by NM, 25-Aug-2013.)
Theoremdfif5 3711* Alternate definition of the conditional operator df-if 3700. Note that is independent of i.e. a constant true or false (see also abvor0 3605). (Contributed by Gérard Lang, 18-Aug-2013.)
Theoremifeq12 3712 Equality theorem for conditional operators. (Contributed by NM, 1-Sep-2004.)
Theoremifeq1d 3713 Equality deduction for conditional operator. (Contributed by NM, 16-Feb-2005.)
Theoremifeq2d 3714 Equality deduction for conditional operator. (Contributed by NM, 16-Feb-2005.)
Theoremifeq12d 3715 Equality deduction for conditional operator. (Contributed by NM, 24-Mar-2015.)
Theoremifbi 3716 Equivalence theorem for conditional operators. (Contributed by Raph Levien, 15-Jan-2004.)
Theoremifbid 3717 Equivalence deduction for conditional operators. (Contributed by NM, 18-Apr-2005.)
Theoremifbieq2i 3718 Equivalence/equality inference for conditional operators. (Contributed by Paul Chapman, 22-Jun-2011.)
Theoremifbieq2d 3719 Equivalence/equality deduction for conditional operators. (Contributed by Paul Chapman, 22-Jun-2011.)
Theoremifbieq12i 3720 Equivalence deduction for conditional operators. (Contributed by NM, 18-Mar-2013.)
Theoremifbieq12d 3721 Equivalence deduction for conditional operators. (Contributed by Jeff Madsen, 2-Sep-2009.)
Theoremnfifd 3722 Deduction version of nfif 3723. (Contributed by NM, 15-Feb-2013.) (Revised by Mario Carneiro, 13-Oct-2016.)
Theoremnfif 3723 Bound-variable hypothesis builder for a conditional operator. (Contributed by NM, 16-Feb-2005.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
Theoremifeq1da 3724 Conditional equality. (Contributed by Jeff Madsen, 2-Sep-2009.)
Theoremifeq2da 3725 Conditional equality. (Contributed by Jeff Madsen, 2-Sep-2009.)
Theoremifclda 3726 Conditional closure. (Contributed by Jeff Madsen, 2-Sep-2009.)
Theoremcsbifg 3727 Distribute proper substitution through the conditional operator. (Contributed by NM, 24-Feb-2013.) (Revised by Mario Carneiro, 14-Nov-2016.)
Theoremelimif 3728 Elimination of a conditional operator contained in a wff . (Contributed by NM, 15-Feb-2005.)
Theoremifbothda 3729 A wff containing a conditional operator is true when both of its cases are true. (Contributed by NM, 15-Feb-2015.)
Theoremifboth 3730 A wff containing a conditional operator is true when both of its cases are true. (Contributed by NM, 3-Sep-2006.) (Revised by Mario Carneiro, 15-Feb-2015.)
Theoremifid 3731 Identical true and false arguments in the conditional operator. (Contributed by NM, 18-Apr-2005.)
Theoremeqif 3732 Expansion of an equality with a conditional operator. (Contributed by NM, 14-Feb-2005.)
Theoremelif 3733 Membership in a conditional operator. (Contributed by NM, 14-Feb-2005.)
Theoremifel 3734 Membership of a conditional operator. (Contributed by NM, 10-Sep-2005.)
Theoremifcl 3735 Membership (closure) of a conditional operator. (Contributed by NM, 4-Apr-2005.)
Theoremifeqor 3736 The possible values of a conditional operator. (Contributed by NM, 17-Jun-2007.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
Theoremifnot 3737 Negating the first argument swaps the last two arguments of a conditional operator. (Contributed by NM, 21-Jun-2007.)
Theoremifan 3738 Rewrite a conjunction in an if statement as two nested conditionals. (Contributed by Mario Carneiro, 28-Jul-2014.)
Theoremifor 3739 Rewrite a disjunction in an if statement as two nested conditionals. (Contributed by Mario Carneiro, 28-Jul-2014.)
Theoremdedth 3740 Weak deduction theorem that eliminates a hypothesis , making it become an antecedent. We assume that a proof exists for when the class variable is replaced with a specific class . The hypothesis should be assigned to the inference, and the inference's hypothesis eliminated with elimhyp 3747. If the inference has other hypotheses with class variable , these can be kept by assigning keephyp 3753 to them. For more information, see the Deduction Theorem http://us.metamath.org/mpeuni/mmdeduction.html. (Contributed by NM, 15-May-1999.)
Theoremdedth2h 3741 Weak deduction theorem eliminating two hypotheses. This theorem is simpler to use than dedth2v 3744 but requires that each hypothesis has exactly one class variable. See also comments in dedth 3740. (Contributed by NM, 15-May-1999.)
Theoremdedth3h 3742 Weak deduction theorem eliminating three hypotheses. See comments in dedth2h 3741. (Contributed by NM, 15-May-1999.)
Theoremdedth4h 3743 Weak deduction theorem eliminating four hypotheses. See comments in dedth2h 3741. (Contributed by NM, 16-May-1999.)
Theoremdedth2v 3744 Weak deduction theorem for eliminating a hypothesis with 2 class variables. Note: if the hypothesis can be separated into two hypotheses, each with one class variable, then dedth2h 3741 is simpler to use. See also comments in dedth 3740. (Contributed by NM, 13-Aug-1999.) (Proof shortened by Eric Schmidt, 28-Jul-2009.)
Theoremdedth3v 3745 Weak deduction theorem for eliminating a hypothesis with 3 class variables. See comments in dedth2v 3744. (Contributed by NM, 13-Aug-1999.) (Proof shortened by Eric Schmidt, 28-Jul-2009.)
Theoremdedth4v 3746 Weak deduction theorem for eliminating a hypothesis with 4 class variables. See comments in dedth2v 3744. (Contributed by NM, 21-Apr-2007.) (Proof shortened by Eric Schmidt, 28-Jul-2009.)
Theoremelimhyp 3747 Eliminate a hypothesis containing class variable when it is known for a specific class . For more information, see comments in dedth 3740. (Contributed by NM, 15-May-1999.)
Theoremelimhyp2v 3748 Eliminate a hypothesis containing 2 class variables. (Contributed by NM, 14-Aug-1999.)
Theoremelimhyp3v 3749 Eliminate a hypothesis containing 3 class variables. (Contributed by NM, 14-Aug-1999.)
Theoremelimhyp4v 3750 Eliminate a hypothesis containing 4 class variables (for use with the weak deduction theorem dedth 3740). (Contributed by NM, 16-Apr-2005.)
Theoremelimel 3751 Eliminate a membership hypothesis for weak deduction theorem, when special case is provable. (Contributed by NM, 15-May-1999.)
Theoremelimdhyp 3752 Version of elimhyp 3747 where the hypothesis is deduced from the final antecedent. See ghomgrplem 25053 for an example of its use. (Contributed by Paul Chapman, 25-Mar-2008.)
Theoremkeephyp 3753 Transform a hypothesis that we want to keep (but contains the same class variable used in the eliminated hypothesis) for use with the weak deduction theorem. (Contributed by NM, 15-May-1999.)
Theoremkeephyp2v 3754 Keep a hypothesis containing 2 class variables (for use with the weak deduction theorem dedth 3740). (Contributed by NM, 16-Apr-2005.)
Theoremkeephyp3v 3755 Keep a hypothesis containing 3 class variables. (Contributed by NM, 27-Sep-1999.)
Theoremkeepel 3756 Keep a membership hypothesis for weak deduction theorem, when special case is provable. (Contributed by NM, 14-Aug-1999.)
Theoremifex 3757 Conditional operator existence. (Contributed by NM, 2-Sep-2004.)
Theoremifexg 3758 Conditional operator existence. (Contributed by NM, 21-Mar-2011.)
2.1.16 Power classes
Syntaxcpw 3759 Extend class notation to include power class. (The tilde in the Metamath token is meant to suggest the calligraphic font of the P.)
Theorempwjust 3760* Soundness justification theorem for df-pw 3761. (Contributed by Rodolfo Medina, 28-Apr-2010.) (Proof shortened by Andrew Salmon, 29-Jun-2011.)
Definitiondf-pw 3761* Define power class. Definition 5.10 of [TakeutiZaring] p. 17, but we also let it apply to proper classes, i.e. those that are not members of . When applied to a set, this produces its power set. A power set of S is the set of all subsets of S, including the empty set and S itself. For example, if , then (ex-pw 21690). We will later introduce the Axiom of Power Sets ax-pow 4337, which can be expressed in class notation per pwexg 4343. Still later we will prove, in hashpw 11654, that the size of the power set of a finite set is 2 raised to the power of the size of the set. (Contributed by NM, 5-Aug-1993.)
Theorempweq 3762 Equality theorem for power class. (Contributed by NM, 5-Aug-1993.)
Theorempweqi 3763 Equality inference for power class. (Contributed by NM, 27-Nov-2013.)
Theorempweqd 3764 Equality deduction for power class. (Contributed by NM, 27-Nov-2013.)
Theoremelpw 3765 Membership in a power class. Theorem 86 of [Suppes] p. 47. (Contributed by NM, 31-Dec-1993.)
Theoremelpwg 3766 Membership in a power class. Theorem 86 of [Suppes] p. 47. See also elpw2g 4323. (Contributed by NM, 6-Aug-2000.)
Theoremelpwi 3767 Subset relation implied by membership in a power class. (Contributed by NM, 17-Feb-2007.)
Theoremelpwid 3768 An element of a power class is a subclass. Deduction form of elpwi 3767. (Contributed by David Moews, 1-May-2017.)
Theoremelelpwi 3769 If belongs to a part of then belongs to . (Contributed by FL, 3-Aug-2009.)
Theoremnfpw 3770 Bound-variable hypothesis builder for power class. (Contributed by NM, 28-Oct-2003.) (Revised by Mario Carneiro, 13-Oct-2016.)
Theorempwidg 3771 Membership of the original in a power set. (Contributed by Stefan O'Rear, 1-Feb-2015.)
Theorempwid 3772 A set is a member of its power class. Theorem 87 of [Suppes] p. 47. (Contributed by NM, 5-Aug-1993.)
Theorempwss 3773* Subclass relationship for power class. (Contributed by NM, 21-Jun-2009.)
2.1.17 Unordered and ordered pairs
Syntaxcsn 3774 Extend class notation to include singleton.
Syntaxcpr 3775 Extend class notation to include unordered pair.
Syntaxctp 3776 Extend class notation to include unordered triplet.
Syntaxcop 3777 Extend class notation to include ordered pair.
Syntaxcotp 3778 Extend class notation to include ordered triple.
Theoremsnjust 3779* Soundness justification theorem for df-sn 3780. (Contributed by Rodolfo Medina, 28-Apr-2010.) (Proof shortened by Andrew Salmon, 29-Jun-2011.)
Definitiondf-sn 3780* Define the singleton of a class. Definition 7.1 of [Quine] p. 48. For convenience, it is well-defined for proper classes, i.e., those that are not elements of , although it is not very meaningful in this case. For an alternate definition see dfsn2 3788. (Contributed by NM, 5-Aug-1993.)
Definitiondf-pr 3781 Define unordered pair of classes. Definition 7.1 of [Quine] p. 48. For example, (ex-pr 21691). They are unordered, so as proven by prcom 3842. For a more traditional definition, but requiring a dummy variable, see dfpr2 3790. (Contributed by NM, 5-Aug-1993.)
Definitiondf-tp 3782 Define unordered triple of classes. Definition of [Enderton] p. 19. (Contributed by NM, 9-Apr-1994.)
Definitiondf-op 3783* Definition of an ordered pair, equivalent to Kuratowski's definition when the arguments are sets. Since the behavior of Kuratowski definition is not very useful for proper classes, we define it to be empty in this case (see opprc1 3966, opprc2 3967, and 0nelop 4406). For Kuratowski's actual definition when the arguments are sets, see dfop 3943. For the justifying theorem (for sets) see opth 4395. See dfopif 3941 for an equivalent formulation using the operation.
Definition 9.1 of [Quine] p. 58 defines an ordered pair unconditionally as , which has different behavior from our df-op 3783 when the arguments are proper classes. Ordinarily this difference is not important, since neither definition is meaningful in that case. Our df-op 3783 was chosen because it often makes proofs shorter by eliminating unnecessary sethood hypotheses.
There are other ways to define ordered pairs. The basic requirement is that two ordered pairs are equal iff their respective members are equal. In 1914 Norbert Wiener gave the first successful definition _2 , justified by opthwiener 4418. This was simplified by Kazimierz Kuratowski in 1921 to our present definition. An even simpler definition _3 is justified by opthreg 7529, but it requires the Axiom of Regularity for its justification and is not commonly used. A definition that also works for proper classes is _4 , justified by opthprc 4884. If we restrict our sets to nonnegative integers, an ordered pair definition that involves only elementary arithmetic is provided by nn0opthi 11518. Finally, an ordered pair of real numbers can be represented by a complex number as shown by cru 9948. (Contributed by NM, 28-May-1995.) (Revised by Mario Carneiro, 26-Apr-2015.)
Definitiondf-ot 3784 Define ordered triple of classes. Definition of ordered triple in [Stoll] p. 25. (Contributed by NM, 3-Apr-2015.)
Theoremsneq 3785 Equality theorem for singletons. Part of Exercise 4 of [TakeutiZaring] p. 15. (Contributed by NM, 5-Aug-1993.)
Theoremsneqi 3786 Equality inference for singletons. (Contributed by NM, 22-Jan-2004.)
Theoremsneqd 3787 Equality deduction for singletons. (Contributed by NM, 22-Jan-2004.)
Theoremdfsn2 3788 Alternate definition of singleton. Definition 5.1 of [TakeutiZaring] p. 15. (Contributed by NM, 24-Apr-1994.)
Theoremelsn 3789* There is only one element in a singleton. Exercise 2 of [TakeutiZaring] p. 15. (Contributed by NM, 5-Aug-1993.)
Theoremdfpr2 3790* Alternate definition of unordered pair. Definition 5.1 of [TakeutiZaring] p. 15. (Contributed by NM, 24-Apr-1994.)
Theoremelprg 3791 A member of an unordered pair of classes is one or the other of them. Exercise 1 of [TakeutiZaring] p. 15, generalized. (Contributed by NM, 13-Sep-1995.)
Theoremelpr 3792 A member of an unordered pair of classes is one or the other of them. Exercise 1 of [TakeutiZaring] p. 15. (Contributed by NM, 13-Sep-1995.)
Theoremelpr2 3793 A member of an unordered pair of classes is one or the other of them. Exercise 1 of [TakeutiZaring] p. 15. (Contributed by NM, 14-Oct-2005.)
Theoremelpri 3794 If a class is an element of a pair, then it is one of the two paired elements. (Contributed by Scott Fenton, 1-Apr-2011.)
Theoremnelpri 3795 If an element doesn't match the items in an unordered pair, it is not in the unordered pair. (Contributed by David A. Wheeler, 10-May-2015.)
Theoremelsncg 3796 There is only one element in a singleton. Exercise 2 of [TakeutiZaring] p. 15 (generalized). (Contributed by NM, 13-Sep-1995.) (Proof shortened by Andrew Salmon, 29-Jun-2011.)
Theoremelsnc 3797 There is only one element in a singleton. Exercise 2 of [TakeutiZaring] p. 15. (Contributed by NM, 13-Sep-1995.)
Theoremelsni 3798 There is only one element in a singleton. (Contributed by NM, 5-Jun-1994.)
Theoremsnidg 3799 A set is a member of its singleton. Part of Theorem 7.6 of [Quine] p. 49. (Contributed by NM, 28-Oct-2003.)
Theoremsnidb 3800 A class is a set iff it is a member of its singleton. (Contributed by NM, 5-Apr-2004.)
Page List
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Homework Help
# Solve the equation-5126 =-6-5(x+22)^(5/3)
abigaile | Student, Undergraduate | eNoter
Posted May 24, 2011 at 8:03 AM via web
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Solve the equation
-5126 =-6-5(x+22)^(5/3)
Tagged with algebra1, discussion
giorgiana1976 | College Teacher | Valedictorian
Posted May 24, 2011 at 10:13 PM (Answer #2)
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We'll shift -6 to the left side:
-5126+6 = - 5(x+ 22)^(5/3)
-5120 = 5(x+ 22)^(5/3)
We'll divide by -5 both sides:
1024 = (x+ 22)^(5/3)
We'll raise both sides to 3/5 power:
1024^(3/5) = x+22
We'll use the symmetric property and we'll subtract 22 both sides:
x = 1024^(3/5) - 22
x = 2^(10*3/5) - 22
x = 2^6 - 22
x = 64 - 22
x = 42
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