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https://www.lmfdb.org/EllipticCurve/Q/106470/fd/3
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Properties
Label 106470.fd3 Conductor 106470 Discriminant -316340292848806015646929458750 j-invariant $$-\frac{1688971789881664420008241}{89901485966373558750}$$ CM no Rank 0 Torsion Structure $$\Z/{2}\Z$$
Related objects
Show commands for: Magma / SageMath / Pari/GP
Minimal Weierstrass equation
magma: E := EllipticCurve([1, -1, 1, -3773654267, 93240113640041]); // or
magma: E := EllipticCurve("106470fe7");
sage: E = EllipticCurve([1, -1, 1, -3773654267, 93240113640041]) # or
sage: E = EllipticCurve("106470fe7")
gp: E = ellinit([1, -1, 1, -3773654267, 93240113640041]) \\ or
gp: E = ellinit("106470fe7")
$$y^2 + x y + y = x^{3} - x^{2} - 3773654267 x + 93240113640041$$
Mordell-Weil group structure
$$\Z/{2}\Z$$
Torsion generators
magma: TorsionSubgroup(E);
sage: E.torsion_subgroup().gens()
gp: elltors(E)
$$\left(-\frac{285141}{4}, \frac{285137}{8}\right)$$
Integral points
magma: IntegralPoints(E);
sage: E.integral_points()
None
Invariants
magma: Conductor(E); sage: E.conductor().factor() gp: ellglobalred(E)[1] Conductor: $$106470$$ = $$2 \cdot 3^{2} \cdot 5 \cdot 7 \cdot 13^{2}$$ magma: Discriminant(E); sage: E.discriminant().factor() gp: E.disc Discriminant: $$-316340292848806015646929458750$$ = $$-1 \cdot 2 \cdot 3^{8} \cdot 5^{4} \cdot 7^{3} \cdot 13^{18}$$ magma: jInvariant(E); sage: E.j_invariant().factor() gp: E.j j-invariant: $$-\frac{1688971789881664420008241}{89901485966373558750}$$ = $$-1 \cdot 2^{-1} \cdot 3^{-2} \cdot 5^{-4} \cdot 7^{-3} \cdot 13^{-12} \cdot 47^{3} \cdot 157^{3} \cdot 16139^{3}$$ Endomorphism ring: $$\Z$$ (no Complex Multiplication) Sato-Tate Group: $\mathrm{SU}(2)$
BSD invariants
magma: Rank(E); sage: E.rank() Rank: $$0$$ magma: Regulator(E); sage: E.regulator() Regulator: $$1$$ magma: RealPeriod(E); sage: E.period_lattice().omega() gp: E.omega[1] Real period: $$0.0301932733497$$ magma: TamagawaNumbers(E); sage: E.tamagawa_numbers() gp: gr=ellglobalred(E); [[gr[4][i,1],gr[5][i][4]] | i<-[1..#gr[4][,1]]] Tamagawa product: $$64$$ = $$1\cdot2^{2}\cdot2^{2}\cdot1\cdot2^{2}$$ magma: Order(TorsionSubgroup(E)); sage: E.torsion_order() gp: elltors(E)[1] Torsion order: $$2$$ magma: MordellWeilShaInformation(E); sage: E.sha().an_numerical() Analytic order of Ш: $$9$$ (exact)
Modular invariants
Modular form 106470.2.a.fd
magma: ModularForm(E);
sage: E.q_eigenform(20)
gp: xy = elltaniyama(E);
gp: x*deriv(xy[1])/(2*xy[2]+E.a1*xy[1]+E.a3)
$$q + q^{2} + q^{4} + q^{5} - q^{7} + q^{8} + q^{10} - q^{14} + q^{16} - 6q^{17} + 4q^{19} + O(q^{20})$$
For more coefficients, see the Downloads section to the right.
magma: ModularDegree(E); sage: E.modular_degree() Modular degree: 148635648 $$\Gamma_0(N)$$-optimal: no Manin constant: 1
Special L-value
magma: Lr1 where r,Lr1 := AnalyticRank(E: Precision:=12);
sage: r = E.rank();
sage: E.lseries().dokchitser().derivative(1,r)/r.factorial()
gp: ar = ellanalyticrank(E);
gp: ar[2]/factorial(ar[1])
$$L(E,1)$$ ≈ $$4.34783136236$$
Local data
This elliptic curve is not semistable.
magma: [LocalInformation(E,p) : p in BadPrimes(E)];
sage: E.local_data()
gp: ellglobalred(E)[5]
prime Tamagawa number Kodaira symbol Reduction type Root number ord($$N$$) ord($$\Delta$$) ord$$(j)_{-}$$
$$2$$ $$1$$ $$I_{1}$$ Split multiplicative -1 1 1 1
$$3$$ $$4$$ $$I_2^{*}$$ Additive -1 2 8 2
$$5$$ $$4$$ $$I_{4}$$ Split multiplicative -1 1 4 4
$$7$$ $$1$$ $$I_{3}$$ Non-split multiplicative 1 1 3 3
$$13$$ $$4$$ $$I_12^{*}$$ Additive 1 2 18 12
Galois representations
The image of the 2-adic representation attached to this elliptic curve is the subgroup of $\GL(2,\Z_2)$ with Rouse label X13.
This subgroup is the pull-back of the subgroup of $\GL(2,\Z_2/2^2\Z_2)$ generated by $\left(\begin{array}{rr} 3 & 0 \\ 0 & 1 \end{array}\right),\left(\begin{array}{rr} 1 & 1 \\ 0 & 1 \end{array}\right),\left(\begin{array}{rr} 3 & 0 \\ 0 & 3 \end{array}\right)$ and has index 6.
magma: [GaloisRepresentation(E,p): p in PrimesUpTo(20)];
sage: rho = E.galois_representation();
sage: [rho.image_type(p) for p in rho.non_surjective()]
The mod $$p$$ Galois representation has maximal image $$\GL(2,\F_p)$$ for all primes $$p$$ except those listed.
prime Image of Galois representation
$$2$$ B
$$3$$ B
$p$-adic data
$p$-adic regulators
sage: [E.padic_regulator(p) for p in primes(3,20) if E.conductor().valuation(p)<2]
All $$p$$-adic regulators are identically $$1$$ since the rank is $$0$$.
Iwasawa invariants
$p$ Reduction type $\lambda$-invariant(s) $\mu$-invariant(s) 2 3 5 7 13 split add split nonsplit add 5 - 1 0 - 1 - 0 0 -
All Iwasawa $\lambda$ and $\mu$-invariants for primes $p\ge 5$ of good reduction are zero.
An entry - indicates that the invariants are not computed because the reduction is additive.
Isogenies
This curve has non-trivial cyclic isogenies of degree $$d$$ for $$d=$$ 2, 3, 4, 6 and 12.
Its isogeny class 106470.fd consists of 8 curves linked by isogenies of degrees dividing 12.
Growth of torsion in number fields
The number fields $K$ of degree up to 7 such that $E(K)_{\rm tors}$ is strictly larger than $E(\Q)_{\rm tors}$ $\cong \Z/{2}\Z$ are as follows:
$[K:\Q]$ $K$ $E(K)_{\rm tors}$ Base-change curve
2 $$\Q(\sqrt{-14})$$ $$\Z/2\Z \times \Z/2\Z$$ Not in database
$$\Q(\sqrt{39})$$ $$\Z/4\Z$$ Not in database
$$\Q(\sqrt{13})$$ $$\Z/6\Z$$ Not in database
$$\Q(\sqrt{-546})$$ $$\Z/4\Z$$ Not in database
4 $$\Q(\sqrt{-14}, \sqrt{39})$$ $$\Z/2\Z \times \Z/4\Z$$ Not in database
$$\Q(\sqrt{13}, \sqrt{-42})$$ $$\Z/12\Z$$ Not in database
$$\Q(\sqrt{3}, \sqrt{13})$$ $$\Z/12\Z$$ Not in database
$$\Q(\sqrt{13}, \sqrt{-14})$$ $$\Z/2\Z \times \Z/6\Z$$ Not in database
6 6.0.3891919590000.8 $$\Z/6\Z$$ Not in database
We only show fields where the torsion growth is primitive. For each field $K$ we either show its label, or a defining polynomial when $K$ is not in the database.
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## Narrow Search
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# NuSTAR Educators Guide: Focusing X-Rays
Students will use the law of reflection to reflect a laser beam off multiple mirrors to hit a sticker in a shoebox. Since X-ray telescopes must use grazing angles to collect X-rays, students will design layouts with the largest possible angles of... (View More)
Audience: High school
Materials Cost: \$5 - \$10 per group of students
# Earth Turns? Prove it!
In this demonstration, evidence of the Earth's rotation is observed. A tripod, swiveling desk chair, fishing line and pendulum bob (e.g., fishing weight or plumb bob) are required for the demonstration. This resource is from PUMAS - Practical Uses... (View More)
# Bird Beak Accuracy Assessment
In this activity, students quantitatively evaluate the accuracy of a classification and understand a simple difference/error matrix. Students sort birds into three possible classes based on each bird’s beak: carnivores (meat eaters), herbivores... (View More)
# Classification: Family Ties
This lesson is comprised of three parts grouped to enable student understanding of classifying organisms. In part one of the lesson, students classify imaginary organisms represented by a mix of breakfast cereals, candies, nuts, raisins, etc.... (View More)
# Building Perspectives with Active Galaxies
In this activity, students build a model of an active galaxy. From this, they will learn about the geometry of the components of an active galaxy and develop an understanding that different viewing angles can lead to dramatically different... (View More)
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https://www.physicsforums.com/threads/finding-current-from-current-density-of-wire.450906/
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# Finding current from current density of wire
1. Nov 24, 2010
### maherelharake
1. The problem statement, all variables and given/known data
A cylindrical wire of radius 3mm has current density, J=3s |φ-$$\pi$$| z_hat. Find the total current in the wire.
2. Relevant equations
3. The attempt at a solution
I believe all I have to do is integrate over the area, but for some reason I can't get it to work. Is the differential area going to be da=s ds dφ? In that case s goes from 0 to 3mm and φ goes from 0 to 2Pi? The 'z' direction is throwing me off a bit. Thanks.
2. Nov 24, 2010
### vela
Staff Emeritus
Yes, that's right. Post your work if you still can't get it to work out so we can see where you're going wrong.
3. Nov 24, 2010
### maherelharake
I just tried to work it out on a napkin, because I am not near a scanner at the moment. I ended up with a net result of 0 though. If you don't think this is correct, I can try to rewrite it and take a picture with my phone and upload it. Thanks again.
4. Nov 24, 2010
### Staff: Mentor
Yes, please upload it. Or you could use the Latex editor in the Advanced Reply window to write out your equations. Click on the $$\Sigma$$ symbol to the right in the toolbar to see your Latex options.
5. Nov 24, 2010
### maherelharake
6. Nov 24, 2010
### vela
Staff Emeritus
You're not dealing with the absolute value correctly. Break the integral over the angle into two ranges, one from 0 to π and the other from π to 2π. For the first integral, |φ-π|=-(φ-π), and for the other, |φ-π|=φ-π.
7. Nov 24, 2010
### maherelharake
8. Nov 24, 2010
### vela
Staff Emeritus
Your integrals look fine, but you made a mistake somewhere evaluating them.
9. Nov 24, 2010
### maherelharake
Hmm I can't find it. I checked it a few times after I posted it. Did you work it out and get a different result?
10. Nov 24, 2010
### vela
Staff Emeritus
I entered it into Mathematica and got a different result. It looks like you messed up the angular integrations in several spot. Every integral should be proportional to π2, but you have π, π2, and π3.
You can simplify the algebra a bit by separating the s integral and φ integral:
$$I=\int_0^{R} 3s^2ds \int_0^{2\pi} |\varphi-\pi|\,d\varphi$$
and using the substitution u=φ-π to do the angular integrals.
11. Nov 25, 2010
### maherelharake
I seem to have gotten R3 Pi2where R=3 mm. Am I close? And of course, the answer is in Amps. Thanks.
12. Nov 25, 2010
### vela
Staff Emeritus
Yup, that matches what I got.
13. Nov 26, 2010
Ok thanks.
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# 9 topics : how many ounces in 1 2 gallon ?
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How Many Ounces in 1/2 Gallon? – A Comprehensive Guide
# How Many Ounces in 1/2 Gallon? – A Comprehensive Guide
## Introduction
When it comes to measuring liquids, there are different units of measurement that can be used. One of the most common units is ounces, while another is gallons. If you’re wondering how many ounces are in 1/2 gallon, you’ve come to the right place. In this article, we’ll explore the answer to this question and provide you with a comprehensive guide on how to convert between different units of measurement.
## How Many Ounces in 1/2 Gallon?
Before we dive into the conversion process, let’s first answer the question at hand. There are 64 ounces in 1/2 gallon. This means that if you have a 1/2 gallon container filled with liquid, it would weigh 64 ounces.
## Converting 1/2 Gallon to Ounces
If you need to convert 1/2 gallon to ounces, you can use a simple formula. Multiply the number of gallons by 128 (the number of ounces in a gallon) and then divide by 2. For example:
1/2 gallon = (1/2) x 128 / 2 = 64 ounces
So, if you have a recipe that calls for 1/2 gallon of milk, you can measure out 64 ounces instead.
## Converting Ounces to 1/2 Gallon
If you need to convert ounces to 1/2 gallon, you can use the same formula in reverse. Divide the number of ounces by 128 and then multiply by 2. For example:
128 ounces = 128 / 128 x 2 = 2 gallons
So, if you have a container that holds 128 ounces of liquid, it would be equivalent to 2 gallons.
## Conclusion
Knowing how to convert between different units of measurement can be helpful in many situations, especially when it comes to cooking and baking. Now that you know how many ounces are in 1/2 gallon and how to convert between the two, you can confidently measure out liquids for your recipes.
You are looking : how many ounces in 1 2 gallon
## 9 how many ounces in 1 2 gallon for reference
### 1.How many oz are in a half gallon – Converting oz to gallons
• Author: How
• Publish: 24 days ago
• Rating: 2(1865 Rating)
• Highest rating: 3
• Lowest rating: 3
• Descriptions: A gallon is an imperial measurement used for liquids. One US gallon is equal to 128 fluid ounces, which means that there are 64 fluid ounces in a half gallon.
• More : A gallon is an imperial measurement used for liquids. One US gallon is equal to 128 fluid ounces, which means that there are 64 fluid ounces in a half gallon.
• Source : https://eugenesdiner.com/food-recipes/how-many-oz-are-in-a-half-gallon
### 2.How Many OZ Are In A Half Gallon? Tips convert 1/2 gal vs oz
• Author: How
• Publish: 20 days ago
• Rating: 2(1643 Rating)
• Highest rating: 5
• Lowest rating: 1
• Descriptions:
• More :
• Source : https://rockyspub.com/cooking-tips-and-techniques/how-many-oz-are-in-a-half-gallon
### 3.How Many Ounces Are In A Half Gallon? Explaining Volume …
• Author: How
• Publish: 26 days ago
• Rating: 5(1488 Rating)
• Highest rating: 3
• Lowest rating: 2
• Descriptions:
• More :
• Source : https://hawgandale.com/recipes/how-many-ounces-are-in-a-half-gallon/
### 4.How many ounces in a half gallon? – Great Answer
• Author: How
• Publish: 26 days ago
• Rating: 3(929 Rating)
• Highest rating: 4
• Lowest rating: 2
• Descriptions: The answer is simple; there is 64 oz in half a gallon. Always keep in mind that ounces are denoted by symbol “oz” whereas fluid ounces are denoting by symbol “ …
• More : The answer is simple; there is 64 oz in half a gallon. Always keep in mind that ounces are denoted by symbol “oz” whereas fluid ounces are denoting by symbol “ …
### 5.How Many Ounces Are In A Half Gallon? – Molly magees
• Author: How
• Publish: 17 days ago
• Rating: 2(271 Rating)
• Highest rating: 3
• Lowest rating: 3
• Descriptions:
• More :
• Source : https://mollysmtview.com/recipes/how-many-ounces-are-in-a-half-gallon
### 6.Gallons to Ounces – How Many OZ in A Gallon? – Asknumbers
• Author: Gallons
• Publish: 9 days ago
• Rating: 2(227 Rating)
• Highest rating: 4
• Lowest rating: 2
• Descriptions: 1/2 gallon = 64 oz; 2 gallons = 256 oz. To calculate how many water bottles of different sizes can fit in gallons and view the gallons to water bottles table, …
• More : 1/2 gallon = 64 oz; 2 gallons = 256 oz. To calculate how many water bottles of different sizes can fit in gallons and view the gallons to water bottles table, …
### 7.How Many Ounces In A 1 2 Gallon? Oz To Gallons Guide
• Author: How
• Publish: 7 days ago
• Rating: 4(279 Rating)
• Highest rating: 3
• Lowest rating: 2
• Descriptions:
• More :
• Source : https://baccocharleston.com/cooking-tips/how-many-ounces-in-a-1-2-gallon
### 8.How many ounces are in a gallon and a half? – Quora
• Author: How
• Publish: 7 days ago
• Rating: 2(557 Rating)
• Highest rating: 5
• Lowest rating: 2
• Descriptions: First thing, there are 128 ounces in one gallon. 1 Gallon = 128 oz(Ounce) US. A gallon is a unit used in the imperial system.
• More : First thing, there are 128 ounces in one gallon. 1 Gallon = 128 oz(Ounce) US. A gallon is a unit used in the imperial system.
• Source : https://www.quora.com/How-many-ounces-are-in-a-gallon-and-a-half
### 9.How Many OZ Are In A Half Gallon? Converting Oz To Gallons
• Author: How
• Publish: 11 days ago
• Rating: 1(1541 Rating)
• Highest rating: 4
• Lowest rating: 2
• Descriptions:
• More :
• Source : https://hanawaterbury.com/food-recipes/how-many-oz-are-in-a-half-gallon
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The exact strength and flexibility of plastics can be easily determined with finite element analysis (FEA) techniques. The FEA process subdivides the product or part into finite-sized units of simple shape. Mathematical equations are used to test each unit for displacement, from which the stresses and strains can be calculated.
Key FEA Calculation Requirements
One of the key requirements used in the analysis is the stress-strain curve or plot, which is distinctive for each material. This is a reflection of the amount of deformation (strain) that is caused by tensile/compressive loading (stress, or pressure). The shape of this curve is not only dependent on the material, but also the temperature of the material and the speed of loading. The final curve reveals the critical properties of the material—will it deliver the properties that it must have for its intended use?
Effects of Plastic Materials Used
Plastics can be unreinforced or reinforced. Unreinforced plastics have a very non-linear stress-strain line (curvy, not straight) up to the yield point and must be analyzed with equations derived for nonlinear materials, not linear materials. This is an important distinction—some molders use what they know and can afford. Nonlinear FEA software is more expensive, takes more time to set up, and takes more time to run. Some molders, however, don’t analyze the stress-strain curve of the plastic they’re evaluating and rely on the published Young’s modulus value in a plastic supplier’s data package. This can provide very misleading results because the modulus value represents just a single point on the stress-strain curve. A non-linear FEA analysis incorporates all the actual stress-strain information to provide accurate results.
Glass-reinforced plastic parts may be easily analyzed with linear FEA techniques. Linear FEA assumes “small displacement” of the part being analyzed and uses an appropriate equation to solve the calculations more quickly. Typically yield strength and ultimate strength are equivalent so the stress-strain curve stays linear. Glass fiber is much stiffer than the base resin and overwhelms the nonlinear properties of the resin. Fiber-reinforced materials behave much like a metal, but rather than stretching/yielding at a certain point, the material just breaks—there is no true yield point.
Another reinforcement issue is knit-line strength. A plastic melt flow front splits to go around a core pin or opening in the part during the molding process. When this flow front meets back up with the other half of the split on the back side of the opening, a knit line occurs. A butt-type knit joint is the worst case since the flow fronts meet “face-to-face”. A flow front will push any smoke, trapped air, or mold surface contamination in front of it. This all gets concentrated at the knit line and weakens the bond between the two fronts. It’s even a much weaker situation with fiber-reinforced materials, since the fiber cannot cross over the knit line. A similar condition exists whereby the flow fronts meet and then flow side-by-side to finish filling the part. This is a stronger situation since any contamination may still be pushed along in front of this flow front.
Using FEA Moldflow Packages
Several FEA “moldflow” packages on the market today predict plastic flow in a mold. Results from this flow analysis are used to infer fiber orientation and knit-line locations. Structural FEA can then be performed using mechanical properties of “in-flow” and “cross-flow” directions to get a better understanding how the part will perform/react. Knowing knit-line locations ahead of time also allows addition of strengthening features to be designed into the part.
It is paramount not to have a high-stress area coincide with a knit line location. If this is the case, the mold gating may be relocated to force knit line locations to be in a lower stress area or the part is re-designed to strengthen the knit line area.
Relying on Plastics Engineers
Designers/engineers must understand the material they are evaluating to get the most out of the FEA simulation. This requires going beyond the single published values in a plastic vendors spec sheet. Stress-strain plots—at various temperatures and strain rates—should be evaluated to choose the properties most appropriate for the part being designed.
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# Christmas Ornament, Coordinate Drawing & Graphing, Mystery Picture
Resource Type
Product Rating
File Type
PDF (Acrobat) Document File
719 KB|8 pages
Share
Product Description
This coordinate graphing project is fun for the student and makes a great bulletin board idea as well. This project can be used for the Christmas season. Students will use their knowledge of coordinate graphing and ordered pairs to work creating a drawing of a Christmas ornament with a stylized Christmas tree with a star. This will be created by plotting ordered pairs and then connecting them with straight lines.
This activity can be a class project or something to be worked on independently when time allows. It also works as an extra credit assignment or when there is a substitute teacher.
This project is for the intermediate coordinate graphing student. The graph consists of points in the first quadrant using ordered pairs with whole numbers and fractional points using the fraction 1/2 for detail. There are also a few points which use the fractions 1/4 and 3/4. When finished, they can use markers, colored pencils or other medium to enhance the project.
Coordinate graph paper for plotting (with both a light and a dark grid), a coordinate list, a full sized answer key and a sample of the completed project in color.
Also included is:
Student Directions for Plotting Coordinates, Directions for Plotting Coordinates with Fractions, Directions for Plotting Coordinates in Quadrant I, Directions for Plotting Coordinates in All Quadrants, Details for Plotting Coordinates with Fractions in Quadrant I, and Details for Plotting Coordinates with Fractions in All Quadrants.
If you like this project, please let us know. There are over 100 projects covering the entire school year and some projects which are cross curricular (Sports, Patriotism, Seasons, Holidays, History, Foreign Language and Science). Lessons begin at 4th grade and are appropriate for the Middle School and some for High School students. These projects have been used successfully at both the Elementary and Middle School levels.
http://www.teacherspayteachers.com/Store/Anthony-And-Linda-Iorlano
Anthony & Linda Iorlano
NCTM Standards
Specify locations and describe spatial relationships using coordinate geometry and other representational systems.
Common Core State Standards
CCSS.Math.Content.5.G.A.1 Use a pair of perpendicular number lines, called axes, to define a coordinate system, with the intersection of the lines (the origin) arranged to coincide with the 0 on each line and a given point in the plane located by using an ordered pair of numbers, called its coordinates. Understand that the first number indicates how far to travel from the origin in the direction of one axis, and the second number indicates how far to travel in the direction of the second axis, with the convention that the names of the two axes and the coordinates correspond (e.g., x-axis and x-coordinate, y-axis and y-coordinate).
Total Pages
8 pages
Included
Teaching Duration
N/A
Report this Resource
Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials.
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### The Law of Physical States and The Corollaries of the Law of Physical States
Bob Kroepel
Lakeside Studios
New Durham, NH USA 03855
The Law of Physical States: An entity, a person or an object, comprised of matter/energy (m/e) has at any timepoint a physical state including its inertial state: an entity will retain its physical state/inertial state until acted upon by a force of some kind.
The Corollaries of the Law of Physical States
1. A force is a form of matter/energy (m/e); a force has a source, a force source, which is the entity or event who/which is the cause of the force.
2. A force causes a push or a pull inre an entity; a force is either a push-force or a pull-force; a push-force causes an entity to be accelerated away from the push-force source and a pull-force causes an entity to be accelerated towards the pull-force source.
3. Only a force can cause a change of the physical state including the inertial state of an entity.
4. The observation of a change of physical state/inertial state implies the change is an effect which is caused by a cause which is a force of some kind, a form of m/e.
A physical state includes any and all observable and/or measurable characteristics of a person or an object including size, shape, mass (weight), color, m/e composition, oscillation (rate of ticking for clocks), motion (inertial state: being at-rest or in-motion), location in space (position), timepoint (time mark on a continuum of time), duration (age, endurance), etc.
The Law of Physical States and The Corollaries of the Law of Physical States are fundamental to physical phenomena at all physical scalar levels, including the scalar level of quantum mechanics, QM.
Causality is the event occurs when objects and events comprised of matter/energy as causes cause as effects (A) changes of the physical states of pre-existing objects and events or (B) new objects and events from pre-existing m/e.
Causality is the basis of determinism.
Determinism is a term used to describe the fact that an effect is caused by a cause, that a cause determines an effect.
Although scientists may not yet be able to observe the changes of the inertial states of pre-existing atoms and subatomic particles, they are able to observe the changes of inertial states of percentages of atoms and subatomic particles in known quantities of atoms/subatomic particles, then by Corollary 4 of the Law of Physical States this observation is proof that a force of some kind has caused the observed change of inertial states, and, thus, determinism is occurring at atomic and subatomic scalar levels, including QM scalar levels.
[1] Albert Einstein, in Relativity: The Special and General Theory, Crown Publishers, New York, 1961, translated by Robert Lawson, p. 11:
As is well known, the fundamental law of the mechanics of Galilei-Newton, which is known as the law of inertia, can be stated thus: A body sufficiently far from other bodies continues in a state of rest or of uniform motion in a straight line.
[2] Charles Proteus Steinmetz: The Fundamental Law of Physics
Charles Proteus Steinmetz.
Four Lectures on Relativity and Space.
Dover Publications, Inc., 180 Varick Street, New York, NY 10014 1967
pp. 49–50:
The fundamental law of physics is the law of inertia. "A body keeps the same state as long as there is no cause to change its state." That is, it remains at rest or continues the same kind of motion—that is, motion with the same velocity in the same direction—until some cause changes it, and such cause we call a 'force.' " [Quotes in the original, but not attributed to anyone.]
This is really not merely a law of physics, but it is the fundamental law of logic. It is the law of cause and effect: "Any effect must have a cause, and without cause there can be no effect." This is axiomatic and is the fundamental conception of all knowledge, because all knowledge consists in finding the cause of some effect or the effect of some cause, and therefore must presuppose that every effect has some cause, and inversely. [Quotes in the original but not attributed to anyone.]
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XP is just a number PerlMonks
### Re: How can I improve this?
by athomason (Curate)
on Jul 25, 2000 at 07:42 UTC ( #24242=note: print w/replies, xml ) Need Help??
in reply to How can I improve this?
Definitely TIMTOWTDI, but so far everyone agrees it's better (at least easier) to generate the permutations first and then sort the list, rather than trying to do both at once. Take a look at the Perl Cookbook section 4.15 (if you have it) for info on sorting lists based on a comparison function; it also has some effiency hints. I didn't do it by strings like the other Monks (which I realize differs from the form of your question), but this might be more efficient, and you can always regenerate the strings at the end. Here's what I came up:
```#!/usr/bin/perl -w
use strict;
my (@a, @b, @c, @d, @e, \$i1, \$i2, \$i3, \$i4, \$i5);
@a = @b = @c = @d = @e = (0, 1, 2); # or whatever
my (@unsorted, @sorted);
for (\$i1 = 0; \$i1 <= \$#a; \$i1++) {
for (\$i2 = 0; \$i2 <= \$#b; \$i2++) {
for (\$i3 = 0; \$i3 <= \$#c; \$i3++) {
for (\$i4 = 0; \$i4 <= \$#d; \$i4++) {
for (\$i5 = 0; \$i5 <= \$#e; \$i5++) {
push @unsorted, [\$i1, \$i2, \$i3, \$i4, \$i5];
}
}
}
}
}
@sorted = sort { non_zeros(\$b) <=> non_zeros(\$a) ||
\${\$b}[0] <=> \${\$a}[0] ||
\${\$b}[1] <=> \${\$a}[1] ||
\${\$b}[2] <=> \${\$a}[2] ||
\${\$b}[3] <=> \${\$a}[3] ||
\${\$b}[4] <=> \${\$a}[4] } @unsorted;
print map "@\$_\n", @sorted;
sub non_zeros {
my @arr = @{\$_[0]};
scalar grep { \$_ != 0 } @arr;
}
I'm not a big fan of any of the permutation generators (including mine) listed so far; I'll try to think of something cleaner and more general.
Replies are listed 'Best First'.
RE: Re: How can I improve this?
by lhoward (Vicar) on Jul 25, 2000 at 07:51 UTC
Since you seemed interested.. here is my permutation generator. It takes a base string (should be empty) as its first argument, then refrences to as many lists as you want as the other arguments and returns a refrence to a list containg all the list-element concatination permutations:
```my \$foo=permute('',[0..2],[0..2],[0..3],[0..2]);
sub permute{
my \$prefix=shift;
my @arrays=@_;
my \$c=shift @arrays;
my @ret=();
foreach(@\$c){
my \$f=\$prefix.\$_;
if(scalar(@arrays)==0){
push @ret,\$f;
}else{
my \$t=permute(\$f,@arrays);
push @ret,@\$t;
}
}
return \@ret;
}
I'm not really happy with the implementation of it, but I like it in spirit (of course, I'm a big fan of recursive algorithms to begin with).
Well, since everyone is posting their permutors, here's mine:
```sub permute {
my \$last = pop @_;
unless (@_) {
return @\$last;
}
return map { my \$left = \$_; map "\$left\$_", @\$last } permute(@_);
}
Drool
by gryng (Hermit) on Jul 25, 2000 at 17:01 UTC
Drool....
Oh I like that.
Mmmm,
Gryn
does this work? how do you call it?
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https://stats.stackexchange.com/questions/159924/getting-expected-value-from-monte-carlo-simulation
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# Getting Expected Value from Monte Carlo Simulation
There are two independent uniform continuous random variables $X$ and $Y$ (such that $0 \leq X \leq 10$, $0 \leq Y \leq 10$). The function $f$ is the difference between the two random variables ($|X-Y|$).
What is the expected value of $f(X,Y)$?
My analytic solution was $10-\sqrt{50}$. I wanted to check my answer with Monte Carlo Simulation.
I randomly chose $1M$ numbers each for $X$ and $Y$ and calculated one million $f(X,Y)$. One intuitive way of getting the expected value is averaging the million values ($\approx$ 3.33). Another way is getting the median(the distributional balance) value ($\approx$ 2.93) among the million values.
The second one was close to my analytic solution.
Maybe I am confusing different concepts. What is the right way of calculating the expected value through the MC simulation? Is my analytic answer correct? If not, how do I calculate it analytically?
updated:
After some research, it seems like the correct analytic solution should be instead(for simplication, the uniform distributions of X, Y are U(0,1) instead of U(0,10):
$E(|X-Y|) = \int_{y=0}^{1}\int_{x=0}^{1}|x-y|2(1-|x-y|)dxdy = \frac{1}{3}$
Because the p.d.f. of Z is $2(1-|x-y|)$, which is derived from the convolution function between $X$ and $-Y$.
The right way of calculating the expected value of a function by Monte Carlo simulation is to calculate the (sample) average of the function value on all n (one million in your write up) replications. You should also look at the sample standard deviation divided by sqrt(n) or sqrt(n-1), to understand the uncertainty in your estimate.
You need to correctly generate the values of X and Y, accounting for dependency, if any, between X and Y. I.e., draw from their joint distribution. if they are independent, this amounts to drawing separately from their own distributions. You have specified Uniform([0,10]) distributions for X and Y, and based on the results of the Monte Carlo simulation, you have assumed they are independent.
Assuming independence of X and Y, then indeed your averaging is the correct thing to do to get expected value of abs(X-Y). Your analytical calculation of 10 - sqrt(5) does indeed correspond to the median of abs(X-Y), not its expected value
Plot the sorted values of abs(X-Y) from the simulation, and you'll see that the median is not equal to the expected value. You may be too used to symmetric distributions, and perhaps in particular the Normal distribution. Median = expected value in such cases, but not for abs(X-Y) in your case.
• Thank you Mark. Then how do I calculate the expected value analytically, if mine was wrong?
– Saju
Jul 4, 2015 at 16:37
• @Saju maybe first tell us how did you calculate it..?
– Tim
Jul 4, 2015 at 16:43
• math.stackexchange.com/questions/48822/… Jul 4, 2015 at 16:48
• @Tim I calculated the geometric probability for 0.5 point(finding a, where |X-Y| is larger than value "a" by the probability of 0.5). (10*10 - (10-a)^2 )/(10*10) = 0.5 However, now it seems it is not the expected value.
– Saju
Jul 4, 2015 at 17:32
• That is the median. The median does not equal the mean when the distribution is not symmetric. |X-Y| where X and Y are independent Uniform[0,10] is not symmetric.Look at the link I provided in the comment above to see how to calculate E(|X-Y|). If you don't understand that, you need to read a probability book or take a course in introductory calculus-based probability. Jul 4, 2015 at 18:52
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# 250+ TOP MCQs on Tension and Compression Tests and Answers
Advanced Materials Science Questions on “Tension and Compression Tests”.
1. Which of the following property cannot be determined by a tensile test?
a) Yield strain
b) Yield stress
c) Elastic limit
d) Limit of proportionality
Clarification: Tensile tests is used to determine different kinds of mechanical properties like yield stress, elastic limit, maximum tensile strength, limit of proportionality.
2. Which type of load is applied in tensile testing?
Clarification: An axial load is applied to the material to be tested when performing tensile testing and the load is applied axially to the body to be tested.
3. Given the cross sectional are as 4 m2, what will be the gauge length?
a) 12.3 m
b) 13 m
c) 11.3 m
d) 12 m
Clarification: The formula for gauge length (Lo) is Lo= 5.65* root over of Area of cross section. We know that under root 4 is 2, therefore 5.65*2 is equal to 11.3 m.
4. Which of the following does not affect the value of ultimate tensile strength?
a) Quality of surface finish
b) Speed of testing
c) Dimensional accuracy of the specimen
d) Length of the specimen
Clarification: The length of the specimen does not affect the value of ultimate tensile strength. While the quality of surface finish, speed of testing, dimensional accuracy of the specimen affect the value of ultimate tensile strength.
5. Which of the following is used to elongation in the material?
a) Clinometer
b) Extensiometer
c) Micrometer
d) Feeler gauge
Clarification: An extensiometer is a gauge which is attached to the specimen and it gives us the value of elongation in the body at a time. Feeler gauge and micrometer or clinometer cannot be used during the testing.
6. The yield limit of compression and the tensile test can be different for the same material.
a) True
b) False
Clarification: Titanium ( Ti ) is a material which shows different amount of yielding when subjected to tensile testing compared to when it is subjected to compressive testing.
7. An UTM can be used to conduct both tensile and compressive testing.
a) True
b) False
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http://mathhelpforum.com/statistics/113-can-someone-shine-flashlight-my-brain-help-me.html
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# Thread: Can someone shine a flashlight on my brain and help me
1. ## Can someone shine a flashlight on my brain and help me
This website seems to be really helpful. I hope someone can help me with these three problems.
1. Consider a member of the Olympic shooting team who is shooting clay pigeons and a member of the Olympic diving team. Each is competing for a score. Who is conducting a sequence of Bernoulli trials?
a. neither
b. the diver
c. the shooter
d. both
2. A heart surgeon has a 95 percent success rate. What is the probability that at least 8 of his next 9 patients will survive?
.299
.929
.531
.033
.630
3. A fisherman captured several fishes and measured their length. He determines that the lengths are normally distributed with a mean of 109.5 cm and standard deviation of 9.5 cm. You would expect nearly all of the fishes to be below what length?
a. 88cm
b. 98cm
c. 109cm
d. 125cm
e. 140cm
2. The man shooting clay pigeons is performing a bernoulli trial because a bernoulli trial satisfies the following:
1. Each trial has two possible outcomes, generically called success and failure.
2. The trials are independent. Intuitively, the outcome of one trial has no influence over the outcome of another trial.
3. On each trial, the probability of success is p and the probability of failure is 1 - p.
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https://www.convertit.com/Go/ConvertIt/Measurement/Converter.ASP?From=line&To=width
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| 617,152,024
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Partner with ConvertIt.com
New Online Book! Handbook of Mathematical Functions (AMS55)
Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: ```line = 2.11666666666667E-03 length (length) ``` Related Measurements: Try converting from "line" to agate (typography agate), angstrom, arpentlin, cable length, chain (surveyors chain), cloth quarter, digitus (Roman digitus), fermi, football field, furlong (surveyors furlong), inch, link (surveyors link), micron, parsec, Roman foot, Roman mile, skein, span (cloth span), spindle, yard, or any combination of units which equate to "length" and represent depth, fl head, height, length, wavelength, or width. Sample Conversions: line = 1.17 agate (typography agate), 21,166,666.67 angstrom, .00462963 Biblical cubit, .00006944 engineers chain, .00115741 fathom, .00694444 foot, 6.35 French, .02744237 Greek palm, .08333333 inch, 4.38E-07 league, 2.24E-19 light yr (light year), 3.81E-07 nautical league, .00000114 nautical mile, .5 pica (typography pica), 5.39E-07 ri (Japanese ri), .0047619 Roman cubit, .00698519 shaku (Japanese shaku), .00001146 stadium (Roman stadium), .06985191 sun (Japanese sun), .00000198 verst (Russian verst).
Feedback, suggestions, or additional measurement definitions?
Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks!
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# If an Angle is 28° Less than Its Complement, Find Its Measure. - Mathematics
If an angle is 28° less than its complement, find its measure.
#### Solution
Angle measured will be ‘x’ say
∴ its complement will be (90 - x
It is given that
Angle = Complement -28°
⇒ x = (90 - x)° - 28°
⇒ x° = 90° - 28° - x°
⇒ 2x° = 62°
⇒ x = 31°
∴ Angle measured is 31°
Concept: Introduction to Lines and Angles
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Mathematics for Class 9
Chapter 10 Lines and Angles
Exercise 10.1 | Q 3 | Page 7
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## Thursday, February 9, 2012
Mavi is 41 months old.
This is a continuation of our Antarctica Unit which we started last week. Here are the math activities to go along with our unity study.
We used Safari Toob Penguins for our addition. The cards that you see were homemade. It will make the addition game a little more interesting if he has these picture cards(control cards).
Another version of our penguin addition is using the concept of Montessori’s Addition Strip. I previously used the technique using bear counters and it worked for him.
Later, he started to do it on his own using the control cards.
SKIP COUNTING BY 2’S AND 3’S
We haven’t done this for a long time. So I decided to revisit it. I used white and blue beads, pretending that they’re penguin eggs (and candies according to my son).
Here I told my son that the penguins are hungry and needed two candies each. Once he placed the beads, we started counting by 2’s. I explained that counting by 2’s means that we start counting from 2 and skip some numbers along the way (thus the 1,3,5,7,9).
Then he wanted to have another set-up. He said that the penguins want to eat rice! He gave each penguin 2 rice (LOL) and he started counting by 2’s again. I was surprise that he can do it on his own. I am in awe of how can a little one learn by doing repetitive activities.
We did the same thing when we counted by 3’s.
And revisited a few activities using these penguin printables from Making Learning Fun. See how we used these printables here and here.
1. Your kiddo is lucky you do so much with him.
2. I just found your blog through pinterest and I LOVE it. Full of interesting and useful ideas. Thank you for sharing them!
1. Thank you Claudia for dropping by.
3. I LOVE all the manipulatives you used with your Penguins!! Your kiddos are going to have a SOLID foundation in math!!! Thanks for linking up to TGIF =-)
Beth
1. I am trying my best to always think of using manipulatives as these helps him learn more :)
4. Thanks Lindsi! Will do!
5. I wonder when to formally introduce the continent boxes. Also, how long do you spend on a continent? Thanks!
1. Hi Marnie! It really depends on your kid. We usually travel a lot, crossing continents almost every year and my son really got curious of where we usually go and why his relatives are so far away from us. So I started showing him the world map... and from then I decided to give him lessons on continents.
With regards to the duration of the lesson, usually it takes a week or two. I narrowed the topics to important things like map, flags, landmarks and animals because these are the things that interests him. Of course, we will do continent every year and I will tackle more stuff as we progress.
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Article
# On Spectral Properties of Doubly Stochastic Matrices
1
Department of Mathematics, Sukkur IBA University, Sukkur 65200, Pakistan
2
Department of Mathematics and General Sciences, Prince Sultan University, Riyadh 11586, Saudi Arabia
*
Author to whom correspondence should be addressed.
Symmetry 2020, 12(3), 369; https://doi.org/10.3390/sym12030369
Original submission received: 17 January 2020 / Resubmission received: 7 February 2020 / Revised: 22 February 2020 / Accepted: 27 February 2020 / Published: 2 March 2020
(This article belongs to the Special Issue Symmetry in Numerical Linear and Multilinear Algebra)
## Abstract
:
The relationship among eigenvalues, singular values, and quadratic forms associated with linear transforms of doubly stochastic matrices has remained an important topic since 1949. The main objective of this article is to present some useful theorems, concerning the spectral properties of doubly stochastic matrices. The computation of the bounds of structured singular values for a family of doubly stochastic matrices is presented by using low-rank ordinary differential equations-based techniques. The numerical computations illustrating the behavior of the method and the spectrum of doubly stochastic matrices is then numerically analyzed.
## 1. Introduction
A real stochastic matrix M is a matrix whose row sums or column sums are equal to 1. All the entries of a real stochastic matrix are non-negative. A real symmetric matrix with non-negative entries with row sums and column sums equal to 1 is called doubly stochastic matrix.
A list of n real numbers, i.e., $1 , λ 2 , λ 3 , λ 4 , … , λ n$ is $s . d . s$ realizable if there exists a symmetric doubly stochastic matrix M with its spectrum denoted by $σ ( M )$. Doubly stochastic matrix describes the transitions corresponding to finite state symmetric Markov chains and this transition acts as a special class of this family. Doubly stochastic matrices are the convex hull for transition matrices with element set [1].
The inverse eigenvalue problems for non-negative doubly stochastic matrices have its origin in work of [2,3,4,5]. For more details on inverse eigenvalue problems, we refer the reader to [6,7,8,9,10] and references therein.
The non-negative matrix M has a real eigenvalue $λ ^$ such that $λ ^ ≥ | λ i ^ |$ for all i. The eigenvalues $λ i ^$ for all i are the eigenvalues of M other than $λ ^$. From the Perron–Frobenius theorem an eigenvector corresponding to $λ ^$ is such that each of its entries are non-negative and sums to 1. For more details, we refer to [11,12,13].
[14] showed that the eigenvector has the form $x = 1 n ( 1 , 1 , 1 , … , 1 ) t$ corresponding to eigenvalue $λ ^$ for $x ∈ R n$, where $R$ denotes the real line. The spectrum of a doubly stochastic matrix is bounded by 1, that is $λ i ≤ 1$ for all i. Three important eigenvalue problems for doubly stochastic matrices are considered in [14] whenever there is a possibility that eigenvalues can be placed in complex plane, denoted by $C$. The first problem deals with necessary and sufficient conditions for $n −$tuples to be the spectrum of a given doubly stochastic matrix. The second problem is about the fact that which real numbers acts as the spectrum of the doubly stochastic matrix. The last problem deals with the study in which set of n real numbers act as the spectrum of the symmetric doubly stochastic inverse eigenvalue problems.
The matrix M over a field $F$ such that each row sum and column sum is $λ ^$, is called the generalization of doubly stochastic matrices by means of Lie theory and algebra $Ω ( n )$ of set of generalized doubly stochastic matrices studied in [15].
In the literature [6,7,10,14,15], much attention has been payed to study the eigenvalues or eigen-spectrum of doubly stochastic matrices. According to the best of our knowledge, however, the study of spectral properties, such as structured singular values for doubly stochastic matrices, is utterly missing from the literature. The current paper deals with the study of the spectral properties, such as singular values and structured singular values for a class of doubly stochastic matrices. The main contribution is to fill this gap which reflects the novelty of our results presented in paper.
The paper is arranged as follows: In Section 2 we provide definitions of symmetric stochastic matrices, singular values and structured singular values. In Section 3 we give a detailed explanation of the computation of singular values for symmetric doubly stochastic matrices. Section 4 of this article contains the geometrical interpretation of eigenvalues and singular values of doubly stochastic matrices. The computation of structured singular values for doubly stochastic matrices is addressed in Section 5. Whereas the numerical experimentation is discussed in Section 6. Section 7 summarizes the conclusions.
## 2. Preliminaries
Definition 1.
The $n −$dimensional matrix $M = ( m i j )$ is said to be a doubly stochastic matrix if
$( i ) m i j ≥ 0 , ∀ i , j = 1 : n$
$( i i ) ∑ i = 1 n m i j = 1 , j = 1 : n ; ∑ j = 1 n m i j = 1 , i = 1 : n .$
Definition 2.
The $n −$dimensional matrix $M = ( m i j )$ is said to be symmetric doubly stochastic matrix if its transpose is doubly stochastic matrix.
Definition 3.
The singular values of a matrix M are the non-negative real numbers $σ i = λ i$ and $σ 1 ≥ σ 2 ≥ … ≥ σ n ≥ 0 , ∀ i = 1 : n .$
Definition 4.
The set of block diagonal matrices is denoted by $B$ and is defined as:
$B : = { d i a g ( δ 1 I 1 , … , δ S I S ; Δ 1 , … Δ F ) : δ i ∈ C ( R ) , Δ j ∈ C m j , m j ( R m j , m j ) ∀ i = 1 : S & ∀ j = 1 : F } .$
In the above definition S and F represent the number of repeated real or complex scalar blocks and the number of full real or complex blocks respectively.
Definition 5.
The structured singular value of $M ∈ C n , n$ with respect to set $B$ is denoted by $μ B ( M )$ and is defined as:
## 3. Computing Singular Values of Doubly Stochastic Matrices
Theorem 1.
Let $D 1 , D 2$ be $n −$dimensional symmetric stochastic matrices with row and column sum equals to 1. Let ${ σ i }$ and ${ σ i ^ } , ∀ i = 1 : n$ are singular values with ${ u i }$ and ${ v i } , ∀ i = 1 : n$ as left and right singular vectors respectively with $∥ u i ∥ 2 = 1 = ∥ v i ∥ 2$ for ${ σ i }$ and ${ u i ^ }$, ${ v i ^ }$ are the left and right singular vectors respectively with $∥ u i ^ ∥ 2 = 1 = ∥ v i ^ ∥ 2$ for ${ σ i ^ }$. The leading singular vectors $u 1$ and $u 2 ^$ are orthogonal to ${ u i }$ and ${ v i }$ for all $i = 2 : n$, respectively. Let $e n = 1 n ( 1 , 1 , 1 , … , 1 ) t$ be the singular vector corresponding to $σ 1$ and $σ 1 ^$ then any vector $η → = { η 1 , η 2 , η 3 , … , η n }$ which is orthogonal to $e n$ is not a singular vector to $σ 1$ and $σ ^ 1 .$
Proof.
The result is proved by expanding and taking sum of all components of the singular value problem of the form:
$D η → = σ η → .$
In Equation (1), $η →$ is a singular vector corresponding to a singular value $σ$. The singular vector $η →$ is not a singular vector corresponding to singular values $σ 1$ and $σ 1 ^ .$
We write
$D η → = d 11 d 12 ⋯ d 1 n ⋮ ⋱ ⋮ d n 1 d n 2 ⋯ d n n η 1 η 2 ⋮ η n = d 11 η 1 + d 12 η 2 + ⋯ + d 1 n η n … … … d n 1 η 1 + d n 2 η 2 + ⋯ + d n n η n .$
Taking sum $S u m$ of all components of $D η →$, we have
$S u m ( D η → ) = S u m d 11 η 1 + d 12 η 2 + ⋯ + d 1 n η n ⋯ ⋯ ⋯ d n 1 η 1 + d n 2 η 2 + ⋯ + d n n η n = η 1 S u m ( d i 1 ) + ⋯ + η n S u m ( d i n ) = η 1 ( σ 1 ) + ⋯ + η 1 ( σ 1 ) ,$
because D is a symmetric stochastic matrix with row and column sums equal to 1. Equation (2) implies that,
$S u m ( D η → ) = σ 1 ( η 1 + η 2 + ⋯ + η n ) , = σ 1 S u m ( η → ) .$
Now, by taking the sum $S u m$ of righthand side of Equation (1) gives,
$S u m ( σ η → ) = σ S u m ( η → ) .$
From Equation (3) and (4), we have $( σ − σ 1 ) S u m ( η → ) = 0 .$ In turn, this implies that $σ ≠ σ 1$ and $S u m ( η → ) = 0 ,$ which proves that any vector $η → = { η 1 , η 2 , η 3 , … , η n }$ orthogonal to vector $e n = 1 n ( 1 , 1 , 1 , … , 1 ) t$ which is not a singular vector corresponding to singular values $σ 1$ and $σ 1 ^$. ☐
Theorem 2.
Let $D 1 , D 2$ be $n −$dimensional symmetric stochastic matrices with row and column sum equals to 1. Let $σ i$ and $σ i ^$ be singular values corresponding to singular vectors as defined in Theorem 1. The matrix $D ˜$ is the matrix with $D 1$ and $D 2$ along the main diagonal. The singular values of $D ˜$ do not contain original singular values $σ 1$ and $σ 1 ^$ appearing along the main diagonal of a matrix $D 1 ˜$ with
$D 1 ˜ = σ 1 1 1 σ 1 ^ .$
The off diagonal of $D 1 ˜$ contains the rank-1 matrices $u v t$ and $v u t$ consisting of leading left and right singular vectors corresponding to $σ 1$ and $σ 1 ^$, respectively.
Proof.
We prove the result by computing the singular vectors of the singular value problem of the form:
$D ˜ u i = σ i u i , ∀ i = 2 : n .$
In Equation (6), the set of vectors ${ u i }$ for all $i = 2 : n .$ is the singular vectors corresponding to singular values ${ σ i } ∀ i = 2 : n .$ The vector $( u i , 0 ) t$ for all $i = 2 : n .$ acts as a singular vector corresponding to $σ i ∀ i = 2 : n .$ The vector $( 0 , u i ^ ) t$ acts as a singular vector corresponding to singular value $σ i ^ ∀ i = 2 : n .$ From above discussion it is clear that the vectors of the form $( α i u i , β i u i ^ ) t$ for all $i = 2 : n .$ act as a singular vectors for matrix $D ˜$ while the vectors $( α i , β i ) t$ acts as singular vectors corresponding to singular values $σ 1$ and $σ i ^$ of $D 1 ˜ .$ ☐
Theorem 3.
Let $D 1 , D 2$ be two symmetric doubly stochastic matrices. Let $σ i$ and $σ i ^$ be leading singular values corresponding to singular vectors $u 1$ and $u 1 ^$, respectively.
Let
$D ˜ = D 1 + ρ I 2 ρ u 1 v 1 t 2 ρ v 1 u 1 t D 2 + ρ I .$
with I is an identity matrix with the same dimension as of $D 1$ and $D 2$ and ρ is any constant. The singular values of $D ˜$ does not contains the leading singular values of $D 1$ and $D 2$. The leading singular values are contained in $D 1 ˜$ with
$D 1 ˜ = 3 ρ + σ 1 0 0 ρ − σ 1 ^ .$
Proof.
We prove the result by computing singular values of singular value problem,
$D ˜ u i → = ( σ i + ρ ) u i , ∀ i = 2 : n .$
The singular values of $D ˜$ are $α , σ 2 + ρ , σ 3 + ρ , … , σ n + ρ$, $β , σ 2 ^ + ρ , σ 3 ^ + ρ , … , σ n ^ + ρ$ where $α , β ∈ { 3 ρ + σ 1 , ρ − σ 1 ^ }$ which are the singular values of
$3 ρ + σ 1 0 0 ρ − σ 1 ^ .$
Because ${ u i } ∀ i = 2 : n$ represent the singular vectors corresponding to singular values ${ σ i } ∀ i = 2 : n .$ can be treated as $u i 0$ $∀ i = 2 : m .$ Similarly for ${ σ i ^ } ∀ i = 2 : n .$ the singular vector $0 u i ^$ holds true. This show that the singular vector corresponding to $D ˜$ can be expressed as $α i u i β i u i ^$ and finally the vectors $α i β i$ $∀ i = 2 : n .$ act as singular vectors to $3 ρ + σ 1$ and $ρ − σ ^ .$ ☐
Theorem 4.
A two dimensional symmetric doubly stochastic matrix has singular values 1 and $σ 1$ if and only if $0 ≤ σ 1 ≤ 1 .$
Proof.
To complete the prove its sufficient to see the fact that for $0 ≤ σ 1 ≤ 1$ the matrix
$1 + σ 1 2 1 − σ 1 2 1 − σ 1 2 1 + σ 1 2 ,$
is symmetric and doubly stochastic matrix. ☐
Theorem 5.
A three dimensional symmetric doubly matrix has singular values $1 , σ 1 , μ 1$ if and only if $0 ≤ σ 1 ≤ 1$, $0 ≤ μ 1 ≤ 1$, $σ 1 + 3 μ 1 + 2 ≥ 0$ and $3 σ 1 + μ 1 + 2 ≥ 0 .$
Proof.
The proof is similar to the one in [15] and hence is omitted. ☐
## 4. Geometrical Interpretation of Spectrum
In this section, we present the geometrical interpretation of the spectrum of symmetric doubly stochastic matrices. In particular, we discuss the geometry of the eigenspace and singular values and left, right singular vectors of such a class of matrices.
Example 1.
We take doubly stochastic matrices $M 1$ and $M 2$ taken from [16] as
$M 1 = 0.9 0.1 0.1 0.9 ; M 2 = 0.1 0.2 0.7 0.5 0.3 0.2 0.4 0.5 0.1 .$
The spectrum of $M 1 , M 2$ is shown in Figure 1a,b of Example 1, respectively. The maximum eigenvalue for both $M 1 , M 2$ is 1 and lies exactly on the spectral circle. The Figure 1c,d show the geometrical interpretation of the singular values and singular vectors obtained for $M 1 , M 2$ of Example 1, respectively.
Example 2.
We take five dimensional doubly stochastic matrices $M 3 , M 4$ from [17] as
$M 3 = 0 0 0 1 0 0 0 1 2 0 1 2 0 1 2 1 2 0 0 0 1 2 0 0 1 2 1 0 0 0 0 ; M 4 = 0 0 0 1 0 0 0 0.25 0 0.75 0 0.25 0.75 0 0 0 0.75 0 0 0.25 1 0 0 0 0 .$
The spectrum of $M 3 , M 3$ is shown in Figure 2a,b of Example 2, respectively. The maximum eigenvalue for both $M 3 , M 3$ is 1 and lies exactly on the spectral circle. Figure 2c,d show the geometrical interpretation of the singular values and singular vectors obtained for $M 3 , M 4$ of Example 2, respectively.
Example 3.
We take eight and nine dimensional doubly stochastic matrices from [18]. The largest eigenvalue corresponding to each matrix attains the maximum value 1. The spectrum is shown in Figure 3a,b of Example 3, respectively. The Figure 3c,d show the geometrical interpretation of the singular values and singular vectors of Example 3, respectively.
## 5. Computing Structured Singular Values
In this section, our aim is to discuss the spectral properties of doubly stochastic matrices based on the computation of Structured Singular Values (SSV). For this purpose, we compute SSV for a class of matrices as considered in Section 4. SSV is the straight forward generalization of the singular values for the constant matrices. The computation of the exact value of SSV is NP-hard. For this reason, one needs to approximate its bounds, i.e., lower and upper bounds.
From an application point of view, the computation of lower bounds of SSV gives sufficient information about the instability of some feedback system while the upper bounds discuss the stability of feedback system under consideration. The computation of the bounds of SSV presented in this section is based on two powerful mathematical techniques: First technique is based on power method for approximating spectrum [19]. The upper bound of SSV is computed by means of the balanced/AMI technique [20] for computing the upper bound from [21]. The second technique [22] is based on the low rank ODEs-based techniques in order to approximate the lower bounds of SSV. This technique works on a two level algorithm, i.e., inner-outer algorithm. We give a brief description of inner-outer algorithms in the subsequent subsections.
#### 5.1. Inner-Algorithm
The inner-algorithm is used to solve the minimization problem addressed in Equation (5). For this purpose, one needs to construct and then solve a gradient system of ordinary differential equations associated with the optimization problem. The construction of system of ODEs involves the approximation of the local extremizers of structured spectral values sets [22]. Following Theorem 6 helps us to approximate the local extremizer of structured spectral vale sets
Theorem 6.
[22]. For a perturbation $Δ ∈ B$ with the block diagonal structure
$Δ = { d i a g ( δ 1 I 1 , … δ s ′ I s ′ , δ s ′ + 1 I s ′ + 1 , … δ S I S ; Δ 1 , … , Δ F } ,$
with $∥ Δ ∥ 2 = 1 ,$ acts as a local extremizer of structured spectral value set. For a simple smallest eigenvalue $λ = | λ | e ι θ , θ ∈ R$ of matrix valued function $( I − ϵ M Δ )$ with the right and left eigenvetors x and y scaled as $S = e ι θ y ∗ x$ and let $z = M ∗ y$. The non-degeneracy conditions
$z k ∗ x k ≠ 0 , ∀ = 1 : S ′$
$R e ( z k ∗ x k ) ≠ 0 , ∀ = 1 : S ′ + 1 : S$
$a n d | | z s + h | | . | | x s + h | | ≠ 0 , ∀ h = 1 : F ,$
hold. Then the magnitude of each complex scalar $δ i ∀ i = 1 : s$ appears to be exactly equal to 1 while each full block possesses a unit 2-norm.
Proof.
For proof we refer to [22].
The system of ordinary differential equations corresponding to a perturbation $Δ ∈ B$ is to approximate an extremizer of smallest eigenvalue in magnitude, i.e., $λ = | λ | ϵ i θ$ which is obtained as,
$δ i ˙ = ν i ( x i ∗ z i − R e ( x i ∗ z i δ i ¯ ) δ i ) ; i = 1 : s ′$
$δ l ˙ = s i g n ( R e ( z l ∗ x l ) Ψ ( − 1 , 1 ) ( δ l ) ; l = s ′ + 1 : s$
$Δ j ˙ = ν j ( z s + j x s + j ∗ − R e 〈 Δ j ; z s + j x s + j ∗ 〉 ) ; j = 1 : F ,$
where $δ i ∈ C , ∀ i = 1 : s ′ , δ l ∈ R$ for $l = s ′ + 1$ and $Ψ ( − 1 , 1 )$ is a characteristic function. For further discussion on the construction of system of ordinary differential equations in above equations, we refer to [22]. ☐
#### 5.2. Outer-Algorithm
The main aim of the outer-algorithm is to vary a small positive parameter $ϵ > 0$ known as the perturbation level. To vary perturbation level a fast Newton’s iteration was used in [22]. The quantity $1 / ϵ$ provides the approximation of lower bound to SSV.
The fast Newton’s iteration to solve a problem is
$| λ ( ϵ ) | = 1 , ϵ > 0 .$
To solve Equation (10), we compute the derivative
To compute $d d ϵ ( | λ ( ϵ ) | )$, one need following Theorem 7
Theorem 7.
[22]. Let $Δ ∈ B$ be the matrix valued function and let x and y as a function of $ϵ > 0$ are right and left eigenvectors of the perturbed matrix $( ϵ M Δ )$. Consider the scaling of vectors x and y accordingly to Theorem 6. Let $z = M ∗ y$ and consider that the non-degenracy conditions as discussed in Theorem 6 holds true, then it yields that
For the proof of the above statement, we refer to [22].
## 6. Numerical Experimentation
The aim of this section is to present numerical eperimentations for lower bounds of SSV for a class of doubly stochastic matrices obtained in Section 4. The numerical experimentations show that the obtained lower bounds with the help of algorithm [22] are either tighter or equal to one approximated with MATLAB function mussv.
Example 4.
We consider a three dimensional real doubly stochastic valued matrix $M 2$ taken from [16]
$M 2 = 0.1000 0.2000 0.7000 0.5000 0.3000 0.2000 0.4000 0.5000 0.1000 .$
We take block uncertainties $B = d i a g { δ 1 I 1 , δ 2 I 1 , δ 3 I 1 : δ 1 , δ 3 ∈ C , δ 2 ∈ R } .$ The admissible perturbation E is approximated as
$E = 1.0000 + 0.0000 i 0 0 0 1.0000 0 0 0 1.0000 + 0.0000 i ,$
with $∥ E ∥ 2 = 1 .$ The lower bound of SSV by using algorithm [22] is obtained as 1 which is equal to the lower bound approximated by mussv function.
Moreover, by using mussv function, the admissible perturbation is obtained as
$∇ = 1.0000 0 0 0 1.0000 0 0 0 1.0000 ,$
such that $∥ ∇ ∥ 2 = 1 .$ The mussv function approximates the same lower and upper bounds of SSV, i.e., 1.
Example 5.
We consider a five dimensional real doubly stochastic valued matrix $M 3$ taken from [17]
$M 3 = 0 0 0 1.0000 0 0 0 0.5000 0 0.5000 0 0.5000 0.5000 0 0 0 0.5000 0 0 0.5000 1.0000 0 0 0 0 .$
We take block uncertainties $B = d i a g { δ 1 I 1 , δ 2 I 1 , δ 3 I 1 , δ 4 I 1 , δ 5 I 1 : δ 1 , δ 3 , δ 5 ∈ C , δ 2 , δ 4 ∈ R } .$ The admissible perturbation E is approximated as
$E = 1.0000 + 0.0000 i 0 0 0 0 0 1.0000 0 0 0 0 0 1.0000 + 0.0000 i 0 0 0 0 0 1.0000 0 0 0 0 0 1.0000 + 0.0000 i ,$
with $∥ E ∥ 2 = 1 .$ The lower bound of SSV by using algorithm [22] is obtained as 1 which is same as the upper bound approximated by mussv function.
Moreover, by using mussv function, the admissible perturbation is obtained as
$∇ = 0 0 0 0 0 0 1.2361 0 0 0 0 0 1.2361 0 0 0 0 0 0 0 0 0 0 0 0 ,$
such that $∥ ∇ ∥ 2 = 1.2361$ Themussvfunction approximates the lower bounds as $0.8090$ and upper bound 1 for SSV.
Example 6.
We consider a five dimensional real doubly stochastic valued matrix $M 4$ taken from [17]
$M 4 = 0 0 0 1.0000 0 0 0 0.2500 0 0.7500 0 0.2500 0.7500 0 0 0 0.7500 0 0 0.2500 1.0000 0 0 0 0 .$
We take block uncertainties $B = d i a g { Δ 1 , δ 1 I 1 , δ 2 I 1 , δ 3 I 1 : Δ 1 ∈ C 2 , 2 , δ 1 , δ 3 ∈ R , δ 2 ∈ C } .$ The admissible perturbation E is approximated as
$E = 0.5000 0.5000 0 0 0 0.5000 0.5000 0 0 0 0 0 1.0000 0 0 0 0 0 1.0000 + 0.0000 i 0 0 0 0 0 1.0000 ,$
with $∥ E ∥ 2 = 1 .$ The lower bound of SSV by using algorithm [22] is obtained as 1.
Moreover, by using mussv function, the admissible perturbation is obtained as
$∇ = 0.6631 0.6028 0 0 0 0.6028 0.5480 0 0 0 0 0 1.2111 0 0 0 0 0 1.2111 0 0 0 0 0 0 ,$
such that $∥ ∇ ∥ 2 = 1.2111$ The mussv function approximates the lower bounds as $0.8257$ and upper bound 1 for SSV.
## 7. Conclusions
In this article, we presented some useful theorems concerning the spectral properties such as singular values and structured singular values for a class of doubly stochastic matrices. We used low-rank ordinary differential equations-based techniques and MATLAB function mussv to approximate bounds of structured singular values corresponding to doubly stochastic matrices. The numerical experimentations show the behavior of singular values and structured singular values which agree with the fact that the largest value of each singular value and structured singular value for doubly stochastic matrix is bounded above by 1. The obtained results for singular values and structured singular values agree with the results obtained for eigenvalues of doubly stochastic matrices, that is:
• The doubly stochastic matrix has an eigenvalue 1.
• The absolute value of any eigenvalue corresponding to a doubly stochastic matrix is less than or equal to 1. The results achieved in this study for structured singular values of doubly stochastic matrices could lead the way to discuss the stability and instability analysis of:
• Stochastic optimal control systems.
• Linear feedback systems in control.
## Author Contributions
M.-U.R. and A.H. both contributed in conceptualization, methodology, writing manuscript and validation of results, while J.H.B. and J.A. contributed in writing-review and editing. All authors have read and agreed to the published version of the manuscript.
## Funding
J. Alzabut would like to thank Prince Sultan University for funding this work through research group Nonlinear Analysis Methods in Applied Mathematics (NAMAM) group with fund number RG-DES-2017-01-17.
## Acknowledgments
The authors acknowledge the anonymous reviewers for their potential inputs to increase the quality of the manuscript.
## Conflicts of Interest
The authors declare no conflict of interest.
## References
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Figure 1. Geometrical interpretation of spectrum and singular values/vectors of Example 1. (a) and (b): spectrum of $M 1 , M 2$ of Example 1, respectively; (c) and (d): geometrical interpretation of the singular values and singular vectors obtained for $M 1 , M 2$ of Example 1, respectively.
Figure 1. Geometrical interpretation of spectrum and singular values/vectors of Example 1. (a) and (b): spectrum of $M 1 , M 2$ of Example 1, respectively; (c) and (d): geometrical interpretation of the singular values and singular vectors obtained for $M 1 , M 2$ of Example 1, respectively.
Figure 2. Geometrical interpretation of spectrum and singular values/vectors of Example 2. (a) and (b): spectrum of $M 3 , M 3$ of Example 2, respectively; (c) and (d): geometrical interpretation of the singular values and singular vectors obtained for $M 3 , M 4$ of Example 2, respectively
Figure 2. Geometrical interpretation of spectrum and singular values/vectors of Example 2. (a) and (b): spectrum of $M 3 , M 3$ of Example 2, respectively; (c) and (d): geometrical interpretation of the singular values and singular vectors obtained for $M 3 , M 4$ of Example 2, respectively
Figure 3. Geometrical interpretation of spectrum and singular values/vectors of Example 3. (a) and (b): spectrum of Example 3, respectively; (c) and (d): geometrical interpretation of the singular values and singular vectors of Example 3, respectively
Figure 3. Geometrical interpretation of spectrum and singular values/vectors of Example 3. (a) and (b): spectrum of Example 3, respectively; (c) and (d): geometrical interpretation of the singular values and singular vectors of Example 3, respectively
## Share and Cite
MDPI and ACS Style
Rehman, M.-U.; Alzabut, J.; Brohi, J.H.; Hyder, A. On Spectral Properties of Doubly Stochastic Matrices. Symmetry 2020, 12, 369. https://doi.org/10.3390/sym12030369
AMA Style
Rehman M-U, Alzabut J, Brohi JH, Hyder A. On Spectral Properties of Doubly Stochastic Matrices. Symmetry. 2020; 12(3):369. https://doi.org/10.3390/sym12030369
Chicago/Turabian Style
Rehman, Mutti-Ur, Jehad Alzabut, Javed Hussain Brohi, and Arfan Hyder. 2020. "On Spectral Properties of Doubly Stochastic Matrices" Symmetry 12, no. 3: 369. https://doi.org/10.3390/sym12030369
Note that from the first issue of 2016, this journal uses article numbers instead of page numbers. See further details here.
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#### Magical Realist
Valued Senior Member
Brian Greene rocks, giving an elegant demonstration and description of gravity on The Late Show. But I have a question: what is the force by which massive objects like Sun warp spacetime? It's gotta be some force it exerts on spacetime. But what is it? How can gravity precede the warping of spacetime?
what is the force by which massive objects like Sun warp spacetime?
The relevant processes all fall under the banner of "gravitation".
In the General Theory of Relativity, gravity is not a force, however. The 'mechanism' is that objects with mass warp the space and time around themselves. So, all that is required is the presence of mass/energy. Or, to put it another way, gravity is an inherent property related to mass. It's just something that mass does.
It's gotta be some force it exerts on spacetime.
Why does it "gotta"?
How can gravity precede the warping of spacetime?
Gravity is the warping of spacetime.
Gravity is the warping of spacetime.
So mass is what warps spacetime, resulting in gravity. But mass doesn't exert a force to do this? It just happens. Got it.
When a body's mass is so great that it results in a black hole, which eventually crushes that body out of existence, what is left to cause the remaining gravity? Iow how can a hole in spacetime have mass?
So mass is what warps spacetime, resulting in gravity. But mass doesn't exert a force to do this? It just happens. Got it.
When a body's mass is so great that it results in a black hole, which eventually crushes that body out of existence, what is left to cause the remaining gravity? Iow how can a hole in spacetime have mass?
It's not a hole.
So mass is what warps spacetime, resulting in gravity. But mass doesn't exert a force to do this? It just happens. Got it.
To give you another example, consider electricity. What is the relevant force? Well, there's an electric force that acts on anything that has electric charge. Why do some things have electric charge? Nobody knows. That's just the way the universe is. Electrical force is a property caused by charge, just as gravitational force (if you want to call it that) is a property caused by mass. Nobody knows why. It's just how our universe works. The task of science is to model the processes we observe in nature. These things are among the things we observe.
So, you're right. Things "just happen". Things are the way they are. Start getting used to living in this universe of ours. It is the way it is. You're stuck with it, like it or not, understand it or not.
When a body's mass is so great that it results in a black hole, which eventually crushes that body out of existence, what is left to cause the remaining gravity?
The mass that formed the hole.
By the way, nobody can say whether a body's mass is "crushed out of existence" when it falls into a black hole. After all, we can't go inside to look (or, at least, if we did go inside we couldn't transmit the news to anybody on the outside).
Iow how can a hole in spacetime have mass?
One way to think about it is that, from the outside, a black hole looks, gravitationally speaking, just like an ordinary lump of matter with a certain mass.
In other words, if you could somehow turn our Sun into a black hole of the same mass right now, the Earth's orbit wouldn't be affected. The Earth would keep right on orbiting the black hole, in the same orbit that it now orbits the Sun. There would be some other nasty side-effects, of course, but gravitationally, we here on Earth wouldn't notice any difference.
Wouldn't it rip a hole in spacetime?
There's no evidence that black holes rip holes in spacetime.
"While black holes are mysterious and exotic, they are also a key consequence of how gravity works: When a lot of mass gets compressed into a small enough space, the resulting object rips the very fabric of space and time, becoming what is called a singularity."--- https://science.nasa.gov/universe/10-questions-you-might-have-about-black-holes/
Eww. I don't know who wrote that article, but it's not correct.
A singularity is a kind of mathematical artifact in some equations. There's no evidence for physical singularities. There's no evidence that there's an actual physical singularity at the centre of a black hole.
This kind of pop-science FAQ stuff is not hard to find, unfortunately. It tends to give the general public misleading ideas about black holes. The people who write this sort of thing really ought to be more careful.
Wouldn't it rip a hole in spacetime?
"While black holes are mysterious and exotic, they are also a key consequence of how gravity works: When a lot of mass gets compressed into a small enough space, the resulting object rips the very fabric of space and time, becoming what is called a singularity."--- https://science.nasa.gov/universe/10-questions-you-might-have-about-black-holes/
The issue here is treating spacetime like it is a material substance. It isn't. Even when they use terms like the "fabric" of spacetime, it is meant more in the lines as when someone uses the term " the fabric of modern society" It refers to a more intangible concept. And spacetime "warping" really means " The rules governing geometry deviates from those of Euclidean geometry".
Things "just happen". Things are the way they are. Start getting used to living in this universe of ours. It is the way it is.
Kinda defeats the purpose of science doesn't it? I mean if everything that occurs is "just what happens" why do science at all?
Kinda defeats the purpose of science doesn't it? I mean if everything that occurs is "just what happens" why do science at all?
The TL;DR version for you, if you have a limited attention span, is that we do science in order to try to explain how our world works, which allows us to predict how it will work going forward. Moreover, it increases our ability to bring aspects of our world under control.
For example, the computer or other electronic device you are reading this on would not exist without science. If scientists hadn't nutted out how electricity and magnetism work, we'd have no harnessed electrical power, let alone any devices as complicated as modern digital computers.
Are you really unaware of any of the benefits that science has brought you? Are you not grateful that your life expectancy is probably at least twice as long as that of your great grandparents, due to science?
Kinda defeats the purpose of science doesn't it? I mean if everything that occurs is "just what happens" why do science at all?
Why questions are never any good. “How” questions give us meaningful answers and make predictions.
Newton treated gravity as a force but did not attempt to explain it, he was not able to.
Einstein gave us a much deeper explanation, but the geometry is not something you can explain in words.
One of the worst explanations for me ever in science is, "matter tells spacetime how to curve, and curved spacetime tells matter how to move".
Does that give an insight into vectors, tensors, Rieman geometry? No, zero.
GR and much of physics answers the how questions and gives results and prediction via the mathematics.
Using words only would be like be like describing chemical reactions without symbols and reaction equations.
What causes the flat rotational curve of a spiral galaxy? Why are all the spiraling stars in one plane?
Angular momentum.
What causes the flat rotational curve of a spiral galaxy? Why are all the spiraling stars in one plane?
As per Seattle for "one plane"but there are other shaped galaxies.
If I could tell you precisely why some galaxies have flat rotation curves I would be collecting the Nobel!
Have a look at the animation, our solar system has planets close to the sun and some further out, the further out the less the influence of the suns gravity with decreasing velocity.
We should expect the same for stars in a galaxy but that is not what happens, the velocities are closer to equal . Is there a lot of extra mass distributed that we cannot see? Dark Matter?
That is one solution, modified theories of gravity or a mixture of both are others but they still do not know.
https://en.wikipedia.org/wiki/Galaxy_rotation_curve
As far as I know, most younger galaxies have "arms" and those that don't tend to be older galaxies that have merged with other galaxies and the arm structures (dust/gas lanes) have been disrupted and as you say, dark matter is the current theory for the dust/gas lane rotation speeds.
What causes the flat rotational curve of a spiral galaxy?
Dark matter. i.e. there's a lot more mass in the galaxy than we can see with our telescopes.
Another possible explanation is that that's no dark matter, but gravity doesn't quite work the way we currently think it does.
This is an active area of ongoing research.
Why are all the spiraling stars in one plane?
It's a side-effect due to the fact that the galaxy is rotating. Rotation tends to flatten spheres into discs. Think, for example, about how pizza bases are made from lumps of dough, traditionally. Even our own planet is a "squashed" sphere, with a smaller polar diameter than the equatorial diameter.
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Browse Groups
• ## Formula for lookup table size on page 45
(2)
• NextPrevious
• To try and understand this equation I assumed the following: number of possible percepts P = 3 lifetime of the agent (number of percepts it will receive) = 3 I
Message 1 of 2 , Sep 18, 2005
View Source
To try and understand this equation I assumed the following:
number of possible percepts P = 3
lifetime of the agent (number of percepts it will receive) = 3
I the get an answer for the summation of 39. However, if I manually
determine the number of possible ways in which the percepts can be
received I get 15 for the size of the lookup table. Can somebody tell
me what I am not undestanding correctly?
Fred Kindl
fredkindl@...
• I m not sure if you re being consistent in your definition of P, the possible precepts, since I can t figure out how you arrived at 15. (I agree with 39). To
Message 1 of 2 , Sep 20, 2005
View Source
I'm not sure if you're being consistent in your
definition of P, the possible precepts, since I can't
figure out how you arrived at 15. (I agree with 39).
To make life easy, assume you have a single sensor
which can return one of three values - A, B, or C.
The possible precept combinations can therefore be
enumerated as:
A
B
C
AA
AB
AC
BA
BB
BC
...
AAA
AAB
AAC
ABA
...
I think you'll find (as the formula implies) there are
3 possible combinations after the first input is
combinations when the second input is received (t=2),
and 3^3^3 additional combinations once the 3rd input
see the agent would need a total of 39 table entries
to know what action to take based on it's precept
history.
Rob G.
--- kindlencotech <fredkindl@...> wrote:
> To try and understand this equation I assumed the
> following:
> number of possible percepts P = 3
> lifetime of the agent (number of percepts it will
> I the get an answer for the summation of 39.
> However, if I manually
> determine the number of possible ways in which the
> percepts can be
> received I get 15 for the size of the lookup table.
> Can somebody tell
> me what I am not undestanding correctly?
>
> Fred Kindl
> fredkindl@...
>
>
>
>
>
>
__________________________________________________
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5 Applications
In previous sections, we described convergence properties for the sequence of iteratively formed correlation matrices. Various possible applications of these properties are discussed in this section.
5.1 The Hierarchical Divisive Clustering Tree with Rank One Splitting Rule.
Without a perfect symmetry structure in the proximity matrix, one divides the p objects (variables or subjects) into two groups. We can recursively apply this correlation splitting rule to form a hierarchical divisive clustering tree. The clustering tree for the correlation proximity matrix of the fifty symptoms is illustrated in Figure 6a. This hierarchical divisive clustering tree with the correlation splitting rule was the major issue in the studies conducted by McQuitty (1968) and Breiger, Boorman, and Arabie (1975).
Figure 2c is also reconstructed as Figure 6b, using the permutation of fifty symptoms from the order of terminal nodes of this clustering tree with a scheme to flip the two branches at each intermediate node. The permutation method is the first type of seriation developed in the present study, it is similar to the study by Gale and Halperin (1984).
Next, Figure 6b is compared with Figure 2b to see if Figure 6b has recovered the original structure embedded in Figure 2b from Figure 2c. It is seen that Figure 6b actually has even more structural pattern than does Figure 2b. There are five major groups along the main diagonal, the negative symptoms, the thought process symptoms, the hallucination symptoms, the delusion symptoms, and the mania symptoms.
Figure 6. Seriation Methods and the Reconstructed Correlation maps. (a). Divisive Clustering Tree with the Rank-1 Splitting Rule; (b). Correlation map Sorted by the Tree Seriation in (a); (c). Rank-Two Ellipse Seriation at ; (d). Correlation map Sorted by the Ellipse Seriation in (c).
5.2 The Rank-Two Ellipse Seriation Technique
When the sequence reaches an iteration with rank two, the p objects fall on an ellipse. They have a unique relative position on the ellipse. There are p possible cuts on the ellipse. The order on the two-dimensional ellipse can be combined with the one-dimensional split to find two orders with the cuts at the two gaps between the two converged groups. The elliptic seriation with the sorted correlation map is given in Figure 6c and d. The symptom order in Figure 6d is different from that in Figure 6b but the major grouping patterns are identical.
Given a pre-Robinson matrix, the correlation matrix at the very first iteration shows a perfect half-ellipse structure with all p vectors fall on half of the two-dimensional ellipse for perfect seriation (see web page for an example). It is possible to combine the rank-1 splitting rule and the rank-2 elliptical seriation to form a hybrid seriation for data sets with clustering structure. That is to perform separate rank-2 seriation on each split sub-matrices.
Reference:
• Breiger, R. L., Boorman, S. A. and Arabie, P (1975), "An Algorithm for Clustering Relational Data with Applications to Social
• Gale, N., Halperin, C. W., and Costanzo, C. M. (1984), "Unclassed Matrix Shading and Optimal Ordering in Hierarchical Cluster Analysis," Journal of Classification, 1, 75-92.
• McQuitty, L. L. (1968), "Multiple Clusters, Types, and Dimensions from Iterative Intercolumnar Correlational Analysis," Multivariate Behavioral Research, 3, 465-477.
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diyAudio (http://www.diyaudio.com/forums/index.php)
- Solid State (http://www.diyaudio.com/forums/solid-state/)
- - Question on input noise current (http://www.diyaudio.com/forums/solid-state/20567-question-input-noise-current.html)
Christer 21st September 2003 09:18 PM
Question on input noise current
Noise was recently discussed in another thread, but since
that thread ended up discussing everything but the original
topic (which wasn't noise) I thought it better to start
to documents on noise, which also others may perhaps find
useful.
http://www.national.com/an/AN/AN-104.pdf
(deals with noise on amplifier level and defines the basic
theory needed for this)
http://www.eie.polyu.edu.hk/~ensury...h3/Chapter3.htm
(defines and discusses noise sources in semiconductor devices)
http://www.teicontrols.com/notes/El...cCircuitsII.pdf
(actually a compilation of diagrams and formulae on various
things, but also briefly defines the basic types of noise.)
http://www.teicontrols.com/notes/El...iseReferral.pdf
(Another document discussing referral of noise sources to the
input, also covered in the National app note).
Now, I think I have spotted an error in the last of these
documents, but I don't trust my own analysis enough to
convince myself to 100% that it is indeed an error, so I
The first document (National app note 104) defines the eq. noise
input current as a current source between the inputs of an amp.
This seems reasonable to me and I trust the people at National
to know what they are talking about, so I accept this as the
standard definition, unless somebody protests. In many cases
it will be convenient to split this current source into two current
sources, one between the positive input and ground and one
between the negative input and ground. These two sources
will both have the same value (the value being a function over
time) but opposite directions. Since they have the same value,
their respective contributions to the eq. input noise must, as far
as I can understand, be summed using plain standard addition.
However, document 4 assumes these two sources to be
independent noise currents and sum them in rms fashion.
Is this an error in document 4, or do I go astray somewhere in
my reasoning??
This will probably only make a difference in the case current
noise is the dominant noise source and the source and feedback
resistances are on the same order, but then it does make
quite a difference.
john curl 22nd September 2003 01:40 AM
Folks, input 'current noise' is the noise made by the electrons crossing a junction. It is formally called shot noise. It has a formula, but usually low noise transistors will have it graphed in their spec sheet. It is usually spec'd with picro amps/rt-Hz. Now what you do is to get a number off the graph and MULTIPLY it by the effective source resistance(this includes any resistors in both the base and the emitter, including local feedback resistors). For example, a noise current of 1pA/rt-Hz and a 1K source would have an effective input noise contribution of 1nV/rt-Hz. This might make a LM394 about 3dB noisier.
SY 22nd September 2003 03:02 AM
And if we parse John's first sentence, you can see that noise across two junctions will be independent. They can be added in an RMS way to get the single current noise source of the National paper.
BTW, that National app note saved me a year of grief when I was a student. A really understandable analysis.
Christer 22nd September 2003 11:31 AM
John and SY,
Perhaps I was a bit unclear, I am sorry for that. My question
was not about calculating the input noise current for a design,
but rather on how to use the given figure for an opamp. (Well,
John says something on this thay may perhaps be understood
as confirming how I understand this, but I am not sure if that
is the way to understand him). More particularly, I should have
said that I was considering an op amp used in non-inverting
config.
Now, according to the National app note., the input noise figure
is by definition referring to a virtual current source between the
two inputs (the reference direction is, of course, arbitrary since
it is noise, so let's say it is from neg. input to pos. input). Let's
denote this noise current by i(t), which is a random function over
time. This currents causes noise voltages over the source
resistance and over the feedback resistance. When calculating
the corresponding equivalent input noise voltage, it may simplify
the analysis to split this current source into two, one from ground
to the pos. input and one from the neg. input to ground. These
two currents must be identical, both having the value i(t), or
else we would violate KCL (as far as I can see, Kirchoffs laws
must hold also for random currents and voltages, ie. noise).
Hence, the way to go about would be to make this split, and
then by superposition find the eq. input voltage for each
the two currents are identical rather than independent. Then,
this sum can be treated as an independent noise voltage
corresponding to the one original noise current. However,
document 4 (which is about noise referral) uses two separate
current sources, one for the pos. input and one for the new.
input, and treats these as independent noise sources, which
would seem to be wrong if my reasoning above is correct. In
fairness, it should be pointed out that this document does not
define how to derive values for the two noise currents from one
single noise current as given in a datasheet. However, it seems
not obvious that one could derive two independent noise
currents from a single one, such that the net results are
identical.
MarcelvdG 22nd September 2003 11:42 AM
I haven't been able to read the second, third or fourth document, because the links don't seem to work on this computer. That is why I don't know how much of the following applies to the case analysed in Christer's fourth reference.
Anyway, in theorectical analyses, the amplifier is often assumed to be a two-port with (by definition) a perfectly floating input port. For practical amplifiers with a differential input, it does not necessarily make sense to define a single equivalent input noise current. In general, you need two different equivalent input noise currents to get an adequate model when you have a differential input.
For the case of a bipolar differential pair without base current compensation and with a symmetric load, assuming the shot noise of the base current to be dominant, the noise current can be described with two uncorrelated sources with a power spectral density of 2qIB (RMS value sqrt(2qIB*DELTAf) over a bandwidth DELTAf), one going to one base, the other to the other base. You can transform these into an equivalent differential input noise current source and two fully correlated common-mode input noise current sources if you like, but unless the impedances driving the positive and negative inputs are equal, it makes no sense to use only a differential source.
In general, base current shot noise is only one of many contributions to the equivalent input noise current or currents. Others which are often significant are base current 1/f-noise and in some cases base current compensation circuit noise.
Christer 22nd September 2003 12:16 PM
Marcel,
somewhere along the way must have abbreviated them
without me noticing it. Here are the links again:
http://www.national.com/an/AN/AN-104.pdf
http://www.eie.polyu.edu.hk/~ensurya...3/Chapter3.htm
http://www.teicontrols.com/notes/Ele...CircuitsII.pdf
http://www.teicontrols.com/notes/Ele...seReferral.pdf
What you say makes sense, but I am not sure it it answers
my question. Although it may not be reasonable to treat the
current noise as one differential source, as you say, that is
what we are given to work with from the datasheet, assuming
datasheets adhere to the definition in the National app note,
that is. (I just checked a few datasheets, and except for CFB
amps, none of them defined what the noise current figure
refers to). The question then is how to best make use of
this figure.
Christer 22nd September 2003 12:20 PM
Sorry again, but it seems to be the forum software that
I'll make a new attempt and split the links to see if that works
http://www.eie.polyu.edu.hk/~ensurya...3/Chapter3.htm
http://www.teicontrols.com/notes/Ele...CircuitsII.pdf
http://www.teicontrols.com/notes/Ele...seReferral.pdf
Edit: It seems the problem is only with displaying the URLs
correctly, so maybe there is some problem with Internet
Explorer rather than the forum software, so perhaps only
some of us got the links abbreviated. Anyway, I have reported
MarcelvdG 22nd September 2003 02:28 PM
I usually first look if there is some additional information hidden in the data sheet, like the noise current measurement set-up or a graph of the total noise versus unmatched source impedance.
If I can't find anything, I look at the internal schematic. If the op-amp has a bipolar input stage without base current compensation, normally each input has a noise current with twice the power spectral density (3dB more) of the differential noise current. As base shot noise and 1/f noise usually dominate in bipolar op-amps without base current compensation, you should also be able to calculate the white part of the input noise current in A/sqrt(Hz) from sqrt(2q times the input bias current), where q is the electron charge (1.6022E-19 C).
If the op-amp does have base current compensation, depending on the exact circuit, there may be a very large common-mode input noise current component. In this case, if the datasheet doesn't provide any additional information, it's anybodies guess how much noise it will generate with unequal impedances driving the positive and negative inputs.
SY 22nd September 2003 03:32 PM
Quote:
In this case, if the datasheet doesn't provide any additional information, it's anybodies guess how much noise it will generate with unequal impedances driving the positive and negative inputs.
Well, presumably, the two transistors in the input diff amp will be the same (or close to it). Same bias comp. So I would think that the noise contribution from each source can be calculated independently, then RMSed to get total noise.
Christer 22nd September 2003 03:41 PM
Marcel,
goes deeper than just trusting and using a single figure from
the datasheet. It is probably a sensible thing to follow your
procedure when an as exact results as possible is desired.
In other cases, it is still interesting to make the best of the
given figure, without losing more accuracy than is already lost
in such a compund figure.
when I said I checked a number of datasheets before,
I must confess I didn't check those for the usual low-noise
op amps, since I didn't find those particular datasheets at
the moment. I have now checked LT1028, LT115 and AD797,
and these datasheets do contain som further info. LT mentions
the test procedure, as using equal balanced source resistors,
measuring the noise voltage, and, after compensating for the
thermal noise, dividing the remaining voltage by the sum of
the resistors. Similaryly, AD says that the noise voltage produced
by the current noise is the current noise multiplied by the sum
of source resistances for the two inputs. This is, as far as I
understand, consistent with my line of reasoning that the
differential current source can be replaced by two identical ones,
that is, not two independent ones as suggested in the fourth
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Cody
Problem 658. Find the biggest empty box
Solution 97103
Submitted on 9 Jun 2012 by David Hruska
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 = [1 0; 0 0]; [r1,r2,c1,c2] = biggest_box(a); sub = a(r1:r2,c1:c2); [m,n] = size(sub); len = 1; assert(isequal(sum(sub(:)),0)) assert(isequal(m,len)); assert(isequal(n,len));
2 Pass
%% a = [1 0 0; 0 0 0; 0 0 0]; [r1,r2,c1,c2] = biggest_box(a); sub = a(r1:r2,c1:c2); [m,n] = size(sub); len = 2; assert(isequal(sum(sub(:)),0)) assert(isequal(m,len)); assert(isequal(n,len));
3 Pass
%% a = eye(9); [r1,r2,c1,c2] = biggest_box(a); sub = a(r1:r2,c1:c2); [m,n] = size(sub); len = 4; assert(isequal(sum(sub(:)),0)) assert(isequal(m,len)); assert(isequal(n,len));
4 Pass
%% a = double(magic(7)<6); [r1,r2,c1,c2] = biggest_box(a); sub = a(r1:r2,c1:c2); [m,n] = size(sub); len = 4; assert(isequal(sum(sub(:)),0)) assert(isequal(m,len)); assert(isequal(n,len));
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https://www.lmfdb.org/EllipticCurve/Q/19600cv/
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# Properties
Label 19600cv Number of curves 1 Conductor 19600 CM no Rank 1
# Related objects
Show commands for: SageMath
sage: E = EllipticCurve("19600.w1")
sage: E.isogeny_class()
## Elliptic curves in class 19600cv
sage: E.isogeny_class().curves
LMFDB label Cremona label Weierstrass coefficients Torsion structure Modular degree Optimality
19600.w1 19600cv1 [0, 1, 0, -2858, 58183] [] 16128 $$\Gamma_0(N)$$-optimal
## Rank
sage: E.rank()
The elliptic curve 19600cv1 has rank $$1$$.
## Modular form 19600.2.a.w
sage: E.q_eigenform(10)
$$q - 2q^{3} + q^{9} + q^{11} + 2q^{13} - 4q^{17} + O(q^{20})$$
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http://www.askiitians.com/forums/Thermal-Physics/12/3573/transfer-of-heat.htm
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` Hot water cools from 60degree celcius to 50 degree celcius in the first 10 min and to 42 degree celcius in the next 10 min .The temp of the surrounding.`
8 years ago
Ramesh V
70 Points
``` average temp. in first case = (60+50) /2 = 55 C
average temp. in second case = (50+42) /2 = 51 C
as frm law of cooling
dT/dt = -k(T-To ) where To is temp. of surrounding
case 1 : dT = 10 C and dt = 10 mins
so , 1=-k(55-To ) ....(1)
case 2 : dT = 8 C and dt = 10 mins
so , 0.8=-k(51-To ) .........(2)
From 1 and 2 , we have
Temp of surr. as 35 C
--
Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and we will get you the answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE preparation. All the best.
Regards,
Naga Ramesh
IIT Kgp - 2005 batch
```
8 years ago
Visalakshi Senthil Kumar
24 Points
``` Excuse sir I need to correct u in the second step ie., (50+42)/2=46 So the room temp would be 10C dT/dt=0.8=-k (46-t)dT2/dt2 =k (55-t)Dividing them both we get1/0.8=(55-t)/(46-t)Therfore t=10Answer by Aswath S, AECS PU College, Bangalore FiitjeeFuture IIT ian ha ha ha
```
one year ago
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## Other Related Questions on Thermal Physics
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# . Find the mode of the following data
12, 24, 36, 46, 25, 38, 72, 36, 25, 38, 12, 24, 46, 25, 12, 24, 46, 25, 72, 12, 24, 36, 25, 38 and 36
Explanation:
Given values are 12, 24, 36, 46, 25, 38, 72, 36, 25, 38, 12, 24, 46, 25, 12, 24, 46, 25, 72, 12, 24, 36, 25, 38 and 36.
To find the mode, first of all arrange the given values.
The mode is a number that repeats more times.
Arrange the values in ascending order 12, 12, 12, 12, 24, 24, 24, 24, 25, 25, 25, 25, 25, 36, 36, 36, 36, 38, 38, 46, 46, 46, 72, 72.
25 is the most often repeated value.
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# Why 10 or 11?
By Sean Carroll | February 4, 2006 1:44 pm
Why does string theory require 10 or 11 spacetime dimensions? The answer at a technical level is well-known, but it’s hard to bring it down to earth. By reading economics blogs by people who check out political theory blogs, I stumbled across an attempt at making it clear — by frequent CV commenter Moshe Rozali, writing in Scientific American. After explaining a bit about supersymmetry, Moshe concludes:
A guide in this pursuit is a theorem devised/put forth by physicists Steven Weinberg and Edward Witten, which proves that theories containing particles with spin higher than 2 are trivial. Remember each supersymmetry changes the spin by one half. If we want the spin to be between -2 and 2, we cannot have more than eight supersymmetries. The resulting theory contains a spin -2 boson, which is just what is needed to convey the force of gravitation and thereby unite all physical interactions in a single theory. This theory–called N=8 supergravity–is the maximally symmetric theory possible in four dimensions and it has been a subject of intense research since the 1980s.
Another type of symmetry occurs when an object remains the same despite being rotated in space. Because there is no preferred direction in empty space, rotations in three dimensions are symmetric. Suppose the universe had a few extra dimensions. That would lead to extra symmetries because there would be more ways to rotate an object in this extended space than in our three-dimensional space. Two objects that look different from our vantage point in the three visible dimensions might actually be the same object, rotated to different degrees in the higher-dimensional space. Therefore all properties of these seemingly different objects will be related to each other; once again, simplicity would underlie the complexity of our world.
These two types of symmetry look very different but modern theories treat them as two sides of the same coin. Rotations in a higher-dimensional space can turn one supersymmetry into another. So the limit on the number of supersymmetries puts a limit on the number of extra dimensions. The limit turns out to be 6 or 7 dimensions in addition to the four dimensions of length, width, height and time, both possibilities giving rise to exactly eight supersymmetries (M-theory is a proposal to further unify both cases). Any more dimensions would result in too much supersymmetry and a theoretical structure too simple to explain the complexity of the natural world.
This is reminiscent of Joe Polchinski’s argument (somewhat tongue-in-cheek, somewhat serious) that all attempts to quantize gravity should eventually lead to string theory. According to Joe, whenever you sit around trying to quantize gravity, you will eventually realize that your task is made easier by supersymmetry, which helps cancel divergences. Once you add supersymmetry to your theory, you’ll try to add as much as possible, which leads you to N=8 in four dimensions. Then you’ll figure out that this theory has a natural interpretation as a compactification of maximal supersymmetry in eleven dimensions. Gradually it will dawn on you that 11-dimensional supergravity contains not only fields, but two-dimensional membranes. And then you will ask what happens if you compactify one of those dimensions on a circle, and you’ll see that the membranes become superstrings. Voila!
CATEGORIZED UNDER: Science
• Moshe
My god you guys are quick…in fact this came out as elaboration on a comment I made here to George Musser, in the post you wrote. I had no idea at the time he was associated with SciAm, but he asked a concrete question and did not seem to be inclined to tell me precisely what is wrong with me…your post at the time BTW was really nice.
I did not know about Joe’s route to string theory, but now I am convinced..
• Paul Valletta
Surely a one-dimensional “string” embedded into a “two-dimensional” anything?..will always have an energy factor, that is exponentionally sufficient to break/snap, the two-dimnensional ‘brane’, to such an extent that Branes cannot exist?
Compactification of any “matter”, from any higher dimension, has a default INCREASE in the energy value?
The only way around this “string-energy” paradox is if the Brane is “one-dimensional” and exists within a “string”..and not be external?
• http://blogs.discovermagazine.com/cosmicvariance/sean/ Sean
Here is the question from George that Moshe was nice enough to try to answer. See, it pays to make useful comments on blogs, you could become famous.
• Moshe
Sean, you must have meant rich and famous…
One clarification, despite starting out as question about string theory, the piece is not very specific to string theory. Basically if you want to combine the ideas of supersymmetry and extra dimensions you put an upper bound on the number of those extra dimensions. This by itself does not tell you there is a consistent theory, or of course if it relevant to nature.
• fh
This is very nice, and I can almost follow what’s going on…
So basically the argument is that since we expect the higher dimensions to show up as symmetries at the 4dimensional viewpoint, and to much symmetry makes for vacuous theories, we can at most go to 10/11D if the additional symmetries show up as supersymmetry on a (flat? Wasn’t that an assumption of Weinberg Witten?) background, right?
I assume there is no nice and easy way to see why we would expect the extra dimensions to manifest as supersymmetry?
• http://1034:Incorrectkeyfilefortableusers;trytorepairit sisyphus
Sean: Nice string – er, thread. Just how little I grasp of this will be made obvious by the following question: Is it possible to have 6 or 7 extra dimensions locally – that is epiphenomnal to our 4D spacetime – while the whole package exists within a global 5D context? I know that this is probably mathematically paradoxical if not downright silly, but I have to ask.
I hope the answer is ‘no’; if it’s ‘yes’, I’ll never understand the explanation.
Thanx.
• J
fh: Actually, the explanation isn’t hard. A supersymmetry has a spinor charges. In every dimension, there is a minimal size for a spinor. When we say “N=x” supersymmetry, we mean that you have x spinors of minimal size. In d=4, the minimal size of a spinor is 4 real dimensional, i.e. in 4 dimensions you always have 4x real supercharges that get mixed by rotations. This number grows very rapidly, more or less exponentially in the dimension.
In 10 dimensions, the smallest spinor you can make has is 16 real dimensional. However, type II superstrings have N=2, so they have 32 supercharges, which from a 4 dimensional point of view looks like N=8 – 4×8=32. Rotations in 10 dimensions mix them together, but rotations in 4 dimensions mix each 4 dimensional spinor seperately.
In 11 dimensions, you have to have at least 32 real supercharges for N=1, so you’re already at the bound. N>1 in d=11 or N>0 in d>11 requires so many supercharges that you you have to have spin >2 particles.
• Maynard Handley
Ignore the stringiness aspect of all this.
The argument is predicated on supersymmetry being real.
What is the current experimental status of this?
Do we ignore ICE CUBE data, or has it just not been running long enough — there was a brief flurry of activity surrounding this about ten days ago which seemed to degenerate into some people saying it proved something or other, others saying no it did not and I lost track of what was actually going on.
Beyond ICE CUBE, is this something that will actually be resolved to most people’s satisfaction in 1 yr, 5 yrs, 15 yrs? Or is it basically so slippery a concept that people can continue claiming it’s true (or at least possibly true) for the rest of my life.
• http://countiblis.blogspot.com Count Iblis
I don’t understand why one expects a quantum theory of gravity to be renormalizable. In principle, one can think of the standard model as a low energy effective theory that is to be obtained from the (unknown) theory of everything by integrating out the unknown high energy physics.
Adding non renormalizable operators to the fundamental theory has almost no effect on the low energy physics, so I don’t see how one can work ”backward” and theoretically ”derive” the theory of everything from its low energy remnant.
• fh
J, thanks. At least on this level this makes sense.
Count Iblis,
the point is that if you have a renormalizable theory you can to some degree expect to be able to define it perturbatively, since order by order nothing undefined happens.
So if you find a renormalizable perturbation expansion of Quantum Gravity you can say you have found a theory. If it’s nonrenormalizable you don’t have a theory.
• Joe Polchinski
Hi Sean, thanks for repeating this argument. I’d like to mention that it is just one example. Another starting point (trying to make a minimum length by introducing a position-position uncertainty principle) is followed to its inevitable conclusion in hep-th/9812104 and hep-th/0209105. Yet another (trying to make the graviton as a bound state of two gauge bosons) leads to the same endpoint — this is in a paper with Gary Horowitz that will appear in the next week or so.
• http://www.pyracantha.com Pyracantha
When people are mystified by my behavior, I tell them I’m from the “seventh dimension.” That usually ends the conversation.
Once again, a large question from the ancient high school physics student:
what is “supersymmetry?” Is it impossible to explain this on just a blog post? Are any books on it up-to-date?
Yours, Pyracantha
• http://blogs.discovermagazine.com/cosmicvariance/sean/ Sean
Joe, I’m glad to hear (at least implicity) that I got the argument right. And the bound-state idea sounds interesting, I’m looking forward to the paper.
Pyracantha, supersymmetry is a proposed symmetry that relates the two basic kinds of particles, bosons and fermions. Fermions are “matter” particles that take up space (electrons, quarks, neutrinos), while bosons are “force” particles that happily pile on top of each other (photons, gluons, etc). If supersymmetry is true, every kind of boson/fermion we know of has a partner that is a fermion/boson (respectively), but that is so heavy we haven’t seen it yet. Supersymmetry is a key feature of string theory. There’s a good popular-level book on it by Gordon Kane, called simply “Supersymmetry.”
• anon
Count Iblis: you raise an interesting point that’s been bugging me lately. It’s true that a nonrenormalizable theory like a naive approach to quantizing gravity is in some sense not a theory at all, as it never makes predictions about high energies (one must measure an infinite number of constants to fix the theory). On the other hand, we don’t really know that the real world doesn’t work that way. It would be depressing, since it would entail a fundamental limitation to the effectiveness of science. But I can’t think of any reason to think it might not be the case. Can anyone else?
Joe or Moshe: as an interested observer of string theory, I’ve been a little puzzled by the usual story of the critical dimension and compactification. Do we really have any compelling reason to think that our world is decribed by a compactified 10D string instead of a noncritical string with all sorts of nontrivial fields (dilaton, etc….) turned on? I suppose one could try to think of all such things as “compactifications,” where the extra dimensions are not simply a manifold but some more complicated geometric (or noncommutative-geometric or whatnot) object, but it seems like generically this wouldn’t be a useful viewpoint. It seems particularly worth asking since I get the impression that (at least some large number of) people tend to expect a lot of the usual vacua to have SUSY broken at a high scale, so the motivation for wanting to start with 4D space times a Calabi-Yau seems to me to be lost. Do I misunderstand the situation?
The main problem with SUSY is not whether it predicts 11D, but that God doesn’t seem to care.
The natural prediction of supersymmetry is that all particles have a superpartner of equal mass. Since this is obviously completely wrong, one must assume that SUSY is broken. The first line of defense had to be abandoned immedately.
But even if SUSY is broken in a natural way, it is still completely wrong, e.g. because it predicts proton decay a million times faster than experimental limits. Thus one had to posit some ugly ad hoc mechanism which suppresses dimension 4 operators, like R-parity conservation. The second line of defense also had to be abandoned.
But even broken SUSY with R-parity conservation has severe problems with experiments. One would naturally expect to see a light Higgs, proton decay, muon g-2 deviations from the standard model, permanent electric dipole moments, WIMPS, and perhaps even sparticles, at presently available experiments. This kind of SUSY is not yet completely ruled out (it might be at the LHC), but we know that it is not natural – this is expressed by the buzzwords “SUSY requires fine-tuning at the percent level”. The third line of defense is in trouble.
In view of these problems, some people have proposed so-called split SUSY. This forth line of defense is finally permanently safe from confrontation with nasty experiments, since it makes no predictions at all which can be tested in any experiment that will ever be feasible. (Probably; it is unclear to me if PEDM data could be used to rule out also split SUSY).
• http://blogs.discovermagazine.com/cosmicvariance/clifford/ Clifford
anon, the main reason people have studied phenomenological scenarios using critical strings as a starting point is because those were are the best understood…. unless I’ve forgotten something. (Progress in this field often follows the path of least resistance….which means that we sometimes take a while to get to important points we were close to a long time before..) Supersymmetry and higher dimensional Lorentz invariance (two key things that put you there in the first place) both get thrown out the window in a short while anyway, “so why were they so important in the first place?” you might ask….. I’ve been on a lonely campaign to remind people that we need to widen the scope a lot. There may be all sorts of riches living in non-critical string, including a real stab at describing Nature. It is time we did the hard work and made a real assault on that area, with our eyes a bit sharper this time around since we know many of the interesting things that show up beyond perturbation theory…..like branes…..
See my “News from the front” posts on this blog for more on that.
Best,
-cvj
• hackticus
I’m not sure you’ve really succeeded in “bringing it down to earth” when you have to take as your starting point a theorem by Witten about higher spin particles. “Oh, I see, it’s just a straightforward consequence of the Witten-Weinberg theorem” the average man on the street is saying…
• http://eskesthai.blogspot.com/2006/02/time.html PLato
From a layman perspective I like things simplified, and finding model associations that would speak any positon held by the contributors here would help understand this process. But there is something more that needs to be done.
Would Clifford see diffeently then us, if such a abstract developement would have taken him directly to D-brane analysis? How would he now see, encapsulating such discussion about N dimensional spaces and such. I mean he would had to have been able to see in different ways that most of us wouldn’t he?:)
Or even a Lubos motl, who having understood these physics processes woud have married thinking of “two sides of a coin” down to the understanding of the dual nature of the blackhole, that such D brane analysis would have made some kind of sense in the physics world.
So holding Moshe thought and comments on the supersymmetry(Kravstov cosmological computors models) issue this too needs some association, just to help make it easier to understand. Having people like Dvali(D brane effects spreading across surface) throw tidbits in for consideration of conceptual developement is always nice. Fluid flows, and laval nozzles, help greatly, taking vision down to a certain level.
But the central theme now is where we had gone to such lengths and raising Cerenkov radiation or lagrangian perspective, I thought would be great starting points. along side of He4 or superfluid recognitions?
As a layman developing concepts, are these starting points wrong having assumed D brane thinking?
• Moshe
Clifford, the word “string” is not mentioned in my little piece, it is not necessary…but to your point which IS a point about string theory, Joe’s argument seems relevant: you do want to control UV behavior and for the theory to have at least 4dim, I am not aware of any way of doing that without SUSY, but of course I am one of the students educated after the first stab at the non-critical strings…
• http://blogs.discovermagazine.com/cosmicvariance/clifford/ Clifford
Moshe,
Fair point. Don’t get me wrong…I like those arguments too…. Nevertheless, (at least) approximately four dimensions and broken susy are experimental facts. I’m just saying that I’m not sure that we understand non-critical strings well enough to know that after all the susy-breaking and dimension changing and strong coupling scenarios that we do once we start with critical strings, we don’t end up some place we could have got to by starting with non-critical strings in the first place…… But given that I don’t have the answer either, this could well be wishful thinking.
Cheers,
-cvj
• Moshe
Thanks Clifford, I think I see what you mean now, that is an interesting point.
• Eva Silverstein
Hi Sean et al,
I think the reason one has trouble providing an argument that 10 or 11 dimensions is required by string theory is because there is no such argument, as mentioned in the comments by anon and Clifford.
If you impose low energy SUSY, or if you require *exactly* flat spacetime in the prescribed dimension, then you obtain the critical dimension. These criteria are not required (yet) by either phenomenology/cosmology or top-down consistency.
In particular, the relevant consistency condition in perturbative string theory is simply that the total Weyl anomaly cancels, which a priori can be arranged in a wide variety of ways. For example, one of the simplest ways around the old “no go” arguments for de Sitter space is to take into account the positive term in the moduli potential that arises in supercritical dimensionality (combined with other ingredients such as orientifolds and RR fluxes).
I would not particularly advocate the noncritical models for phenomenology in the absence of other motivations, but the question of principle (what is allowed vs what is excluded based on first principles) is an important one and I know of no non-circular top down argument for the critical dimension. Conversely, if someone came up with such a proof it would be new and useful. Of course low energy SUSY has a number of phenomenological virtues and it will be extremely important to see whether or not it is there at the LHC.
Obtaining nearly flat space of course requires fine tuning (this is the cc problem), but it is not known whether the tuning required is actually better or worse in the case of high vs low scale SUSY breaking. There has been some preliminary discussion of this in the literature but no firm conclusion.
Best regards,
Eva
• Moshe
Eva, thanks for your comment. To the extent it relates to anything I have written, I have been careful to state the assumptions and what follows from them. Any theory at all that attempts to use SUSY and extra dimensions, in the regime when the notion of dimension makes sense, is bound to find itself in 11 dimensions or less. In other words the numbers 10/11 are not pulled out of a hat, which may surprise a few people…It is an interesting and independent question whether in string theory these are absolutely necessary, but once again there is no mention of strings in my little piece.
As for that question, the absence of Weyl anomaly is one way of deriving the critical dimension, in the Green-Schwarz formalism the derivation has much more transparent relation to spacetime supersymmetry, perhaps that is a circular argument…
In addition, as far as I understand there is no one self-consistent framework that incorporates all the ingredients needed for the super-critical scenario. That does not mean there is none, only that critical strings are under much better control, especially when considered non-perturbatively.
• Eva Silverstein
Hi Moshe,
I was not writing as a criticism of your piece (which if it states an assumption of low energy SUSY is perfectly reasonable). People sometimes claim that string theory essentially predicts both 10d and low energy SUSY; it is this of which I am not convinced.
I worry that this could be a much more sophisticated version of the following. We all learn mechanics first with spherical symmetry, which makes calculations easier. In that context, it would be ridiculous to elevate the symmetry to a principle of nature. Of course SUSY is a much deeper structure, but it is still not obvious to me whether it is ultimately a deep requirement or closer to a theoretical crutch.
Regarding supercritical constructions: the statement is that enough independent forces to fix the moduli appear in these theories (self-consistenly in the same model). Of course it is true that this arena has not been much studied, and there could be subtleties. On the non-perturbative formulation: recall that there is no known non-perturbative formulation of string theory on a Calabi-Yau. The backgrounds for which a non-perturbative definition presently exist, while very interesting, are few and far between, so I would not use this as a selection mechanism given our present knowledge.
Best regards,
Eva
• Lee Smolin
Hi, There is another very elegent way to understand why 10 and 11 are special, which was developed a long time ago by Feza Gursey and others. This has to do with the division algebras: A= R (real), C (complex), Q (quaternions) and O (octonions). There are the only algebras which extend the real numbers in having addition, multiplication and division. It turns out that these are related deeply to supersymmetry. One way to see this is that a supersymmetric extension of Yang-Mills theory must have two component spinors, valued in one of these algebras. The four possibilities give theories in 3,4 6 and 10 dimensions.
Another way to see this is that the supersymmetry charge is fermionic and this is generated by two component spinors that transform under a representation of SL(2,A). SL(2,R) ~ SO(1,2), SL(2,C) ~ SO(3,1), and SL(2,Q) ~ SO(1,5). There is no SL(2,O), because the octonions are non-associative. But there are close relationships between the representation theory of SO(8)-the transverse directions in 9+1 dim spacetime and the octonions that allow a supersymmetric theory to be defined. There are three 8 dimensional representations-the vector, spinor and conjugate spinor, related by a symmetry called triality that can be seen to underlie the structure of the supersymmetry algebras between 8 and 11 dimensions.
If I may, one last beautiful fact: These three representations organize themselves into components of a very beautiful algebra-which is the algebra of 3 by 3 hermitian matrices of octonions. Under anticommutators they form a unique structure called the exceptional Jordan algebra. This is a 27 dimensional algebra, and it has 3 SO(8) scalars-so it naturally represents physics in 11=8+3 dimensions
• Moshe
Thanks Eva, I think we are in agreement.
Lee, the facts in your first paragraph are indeed deeply related to supersymmetry, another place this is utilized is in the Green -Schwarz quantization of the string, those are precisly the dimensions the classical GS strings allows (3,4,6,10).
• Lee Smolin
Moshe, yes, thanks, the connection between strings and octonions is recognized but seems a possible clue to a deeper underestanding of string theory that is not yet well explored. Corinne A. Manogue and colleagues have explored it in papers including, Phys.Rev.D40:4073,1989 and hep-th/9807044. Some other attempts are hep-th/0104050, hep-th/0110106, hep-th/0503017.
The alternative advocated by Eva and Clifford seems very important to explore. It would be good to know whether or not supersymmetry and extra dimensions are essential for string theory. I hope Eva and Clifford are right, but in case SUSY is essential there must be a deeper way to understand it, perhaps afforded by octonions.
• http://blogs.discovermagazine.com/cosmicvariance/sean/ Sean
Eva et al.– Without being an expert, I completely agree that exploring non-critical string theory is an important idea. This is especially true since we have no detailed experimental data, and what we have isn’t well described by ten-dimensional Minkowski space with unbroken supersymmetry. (And the fact that, not too long ago, we wouldn’t have been talking about 11 dimensions.)
The point of this discussion, I think, is more about explaining ourselves than about understanding the nuances of string theory or supersymmetry. The fact that there is something special about certain values the number of dimensions of spacetime is surprising, and hard for the person on the street to understand. I think what George Musser was originally looking for was some concrete imagery, similar to the idea that strings only generically intersect in three spatial dimensions. The maximal dimension for susy is not quite so visual, but it’s what we have.
• http://1034:Incorrectkeyfilefortableusers;trytorepairit sisyphus
Basic question here: Are ideas like ” a volume contains an infinite number of planes” useless in trying to understand 11D? Does an 11D thing contain an infinite number of 10D things?
• http://blogs.discovermagazine.com/cosmicvariance/clifford/ Clifford
I forgot to say: “Joe! Eva! Welcome to the blog!” I don’t think we’ve had comments from you on an earlier thread (unless my memory fails), and it’s really excellent to see you here.
Eva, we seem to be very much of a similar mind on stamping out the whole “string theory requires D=10/11” business. Excellent!
Sean, I do think that we understand the perfeclty sensible narrower parameters of the discussion of the post, but I do think that it is appropriate to include what Eva and I am saying in this particular discussion, since the “person on the street” all too often hears (or implicitly gathers from posts like this) the phrase “string theory requires D=10/11”, and it is simply not true and in some years we may well have to be spending a lot of time undoing yet another uncautious claim when/if after doing phenomenology better we find that we don’t need to start in higher D and then “compactify”. We’ll have to go around telling everyone (on the tv shows and radio shows and magazines) “oh…that thing we said about extra dimensions? We were just kidding”…. Just like we’re doing now with the whole “unique vacuum” and “theory of everything” phrases…
Cheers,
-cvj
• http://thebumblebeeblog.blogspot.com/ Poppycock
sisyphus: I think the answers to your questions are “no” and “yes” respectively.
When you think about 3D, you can imagine a set of 3 axes, denoted by a line pointing up-down, a line going left-right and another going backwards-forwards. To imagine more dimensions, I just have to pretend to myself that I can put lines in more directions. For example, if you wanted 3 large dimensions such as you see, and then a fourth compact dimension (I’m just talking spatial dimensions here, ignoring time for now), picture it as there being a little circle at every single point in 3D space.
Now, you specify the position of a particle in the 4D space by saying where in the 3D space it is (ie give an x-, a y- and a z-coordinate), and then give another number which tells you how far around the circle you have gone, measured from a set point. Since this dimension is compactified to a circle, the fourth coordinate value must be between 0 and 2pi (circumference of a circle of unit radius). But there are still an infinite number of points on this circle.
Thus, if you want to put objects in the 4D space, you have to specify 4 coordinates, and each of these can take an infinite number of values.
To work in more dimensions you just have to keep putting in more directions as we did to add the fourth, but the rules for how they work stay the same. Personally, I cannot picture higher dimensional spaces (and I’m not convinced anyone else can really, you just get used to the idea and the maths simplifies things) but in principle they don’t behave any more weirdly than those you are used to (except possibly being compactified!).
Lee: some years ago I read a bit about graded Lie algebras in the maths literature. Are these related to the division algebras you mentioned? It is all very hazy as I’ve not thought about it in quite some time, but it was certainly related to SUSY some how. It may be that “graded Lie algeba” was just another name for “SUSY algebra”. Is this relevant, or is my memory leading me astray?
• http://1034:Incorrectkeyfilefortableusers;trytorepairit sisyphus
#31 Poppycock: Thank you.
• http://eskesthai.blogspot.com/2006/02/phase-transitions.html Plato
So what is the “right circumstance” that the “analogies(superfluid created)” would speak to symmetry breaking, as “phase transitons?”
• http://countiblis.blogspot.com Count Iblis
fh# 10, anon # 14
Yes one can say that you don’t have a theory in that case, but you could also say that Nature doesn’t allow you to fix the theory using measurements made at low energy scales.
I would agree with anon that there are no valid arguments against a nonrenormalizable theory. The fact that a low energies we have renormalizable theories is i.m.o. just because any nonrenormalizable terms have renormalized to zero when integrating out high energy physics.
We can’t then use the argument that just because at low energies everything is renormalizable it should always be so.
• Lee Smolin
Poppycock: No, but thanks for reccalling graded Lie algebras, whicih eons ago a few of us studied as a road to a geometry for supergravity. It turned out to be not as useful as super Lie algebras. Division algebras are systems of numbers that share some properties of the reals such as the existence of multiplicaiton, additionn and division, but don’t have others such as commutivity and associativity. A nice intro to them and octonions is John Baez’s http://math.ucr.edu/home/baez/week59.html.
• http://eskesthai.blogspot.com/2005/07/b-field-manifestations.html Plato
More on name above.
I think John Baez makes it easier sometimes and I appreciate that kind of material. Banchoff’s discriptions of the computer screen is much different. More indepth?
In 1849 already, the British mathematicians Salmon ([Sal49]) and Cayley ([Cay49]) published the results of their correspondence on the number of straight lines on a smooth cubic surface. In a letter, Cayley had told Salmon, that their could only exist a finite number – and Salmon answered, that the number should be exactly 27
.
What is interesting is the idea of Sylvestor surfaces and early Cayley references, on how we can shape our thinking in context of these dimensional attributes.
Is this right that such “spin valuations” would have signalled, one phase transition held in dimensional consideration, to the next? I apologize if I have confused the situation.
• http://eskesthai.blogspot.com/2006/02/strange-abstract-movements.html Plato
It seems there always is this need to try and explain it better? Being at a loss for words, one tends to keep trying?
The summation to D brane consideration, encompasses this view? CY perspective arise from it?
• Mat Hunt
Apologies for being somewhat off topic, I couldn’t figure how to start my own blog and this was the first one I could hijack.
My question refers to the prize given out by the I believe American Institute of Aeronautics and Astronautics regarding a novel form of space propulsion, namely that of antigravity. I have a BSc(Hons) in mathematics and an MSc in geometry, mathematical physics and analysis where I did lots of courses in general relativity. Their idea is based around something called ‘Heim theory’, which is as far as I can tell is a high dimensional general relativity. The claim further goes to to say that Heim himself managed to couple his theory to quantum mechanics therefore getting some form of quantum gravity.
I have a gut feeling that this is a complete joke but my colleagues and a well known scientific establishment in the UK feel otherwise. I would like to hear peoples opinion on the matter.
Once again apologies for hijacking the blog.
Regards
Mat
• http://blogs.discovermagazine.com/cosmicvariance/sean/ Sean
Mat, I don’t know a lot about it, but what I’ve seen seems to not make much sense. I wouldn’t take it seriously if I were you.
• Mat Hunt
That was my thought, it seemed like another crackpot theory to me but senior colleagues were saying not to ignore it as apparently it ‘predicts’ the correct masses of the elementary particles.
In short, it just sounds too good to be true.
• Qubit
Mat, there is no Pinocchio theory! All the puppets here need strings. I mean.. Come on! Who wants to use a imaginary solar sail anyway, fall through the universe at Twice the speed of light, don’t be silly. No! Am afraid it’s chasing the rainbow, the dazzle in the distance, the light fantastic and the great big mushroom thats always on the Horizon, always.
• Marcia L.Neil
String theory may be applicable to the the new ‘Lost World’ described in this weeks’ news articles, which outline a team investigation “in boggy clearings” found in the western half of New Guinea, Indonesia. The ‘strings’ would be the atmospheric linkages formed among all the unique organisms of that ecosphere, with resultant “pristine zone”. It can be deduced that an abundance of water is necessary to eliminate friction among the unique species varieties discovered, and facilitate the dynamics of island gravity.
Just don’t go busting in — the articles state that the region is usually off-limits to foreigners. Infra-red testing and photography of atmospheric pulse might also reveal those linkages which are necessary to hold the creatures within island environmental niches. Most natural biochemistry perpetuates linkages in terms of six or hexagonal groupings.
• naive experimentalist
Perhaps I’m missing something, but I have heard that if you start with superstring theory and quantise it, then you are forced to use 10 dimensions. If you don’t you end up with either negative norms or tachyons.
• http://blogs.discovermagazine.com/cosmicvariance/clifford/ Clifford
Well, that is simply not true. This is said a lot, but it is not true, as I say above.
It is only true if (for example) you assume that you have Lorentz invariance in all dimensions. This was done for simplicity’s sake only. But we know that’s not true. We know only that it is true in 3+1.
-cvj
• http://golem.ph.utexas.edu/~distler/blog/ Jacques Distler
But it’s more than than that.
It’s not just that, for the supercritical strings Eva and collaborators have been considering, you never have D-dimensional Lorentz-invariance (for some D > 10). You never have a lower-dimensional Lorentz-invariance either.
So it’s not 100% clear what the observables of such theories are.
I think it’s still an open question whether such theories can be made sense of.
• http://blogs.discovermagazine.com/cosmicvariance/clifford/ Clifford
Very true…nevertheless, I find it disturbing that we have not explored non-critical strings as much as we should have, since learning so much about strong coupling physics. I just have a gut feeling that we’re missing a trick here.
Until we’ve really done that job, I’m not comfortable with the trumpeting of the “stirngs only live in ten dimensions” twaddle that we keep telling everyone, often including our students…. We’ve got to remember what we assumed in order to get to the cirtical dimensions, and then revisit those assumptions every time we learn something new about the whole story. I might be overstating things…. it’s late here and I should get to bed.
-cvj
• http://eskesthai.blogspot.com/2006/02/evidence-for-extra-dimensions-and.html Plato
• http://1034:Incorrectkeyfilefortableusers;trytorepairit sisyphus
#44, Clifford: Just a dumb layman’s question here: Am I right in assuming that if Lorentz invariance doesn’t necessarily survive > 3+1, then, generally, the application of the spacetime interval formula to Multi-D is meaningless?
• http://blogs.discovermagazine.com/cosmicvariance/clifford/ Clifford
Hi,
Sorry….I don’t know what the spacetime intercal formula is. Help!
-cvj
• http://1034:Incorrectkeyfilefortableusers;trytorepairit sisyphus
The uh.. the s*2 = x*2 + y*2 + z*2 – (ct)*2 thing. Did I use the wrong terminology?
• http://blogs.discovermagazine.com/cosmicvariance/sean/ Sean
sisyphys, that formula is valid in flat spacetime, where Lorentz invariance is a symmetry of the entire universe. In any curved spacetime, the interval will be a different function of the coordinates. Curved spacetimes are generally less symmetric than flat ones, although certain special examples might be equally symmetric, just in a different way.
• http://blogs.discovermagazine.com/cosmicvariance/clifford/ Clifford
Oh! The invariant interval formula! Sorry sisyphus, your terminology was fine….I was distracted by other things and did not make the connection, stupidly enough. Looks like Sean has answered you.
-cvj
• http://1034:Incorrectkeyfilefortableusers;trytorepairit sisyphus
Thanks, Sean. I’ll work on it.
Apropos of something else: Feyerabend may have been a tad radical, but maybe he got it right when he suggested that the slavish observation of a particular methodology can stifle discovery.
Thanks again.
• http://1034:Incorrectkeyfilefortableusers;trytorepairit sisyphus
Thanks, Clifford. I’m aware of your loss.
• http://blogs.discovermagazine.com/cosmicvariance/clifford/ Clifford
sisyphus:- in addition to the possibility that the spacetime has a non-trivial metric, mentioned by Sean, another common way of changing things quite drastically is to have other dynamical fields switched on in some directions, but not others. The classic example is a linear dilaton…. i.e. a spacetime field generated by the string, the “dilaton”, has a non-trivial dependence (linear in this case) on just one direction in spacetime….. the rather singles out that direction as being rather special as compared to the others, a possiblity not allowed for in the critical string constructions. This is one example of how you can have string theories that live in dimensions other than the “critical” ones.
-cvj
• http://1034:Incorrectkeyfilefortableusers;trytorepairit sisyphus
Thanks, Clifford. This is a bit to chew on. I’ll print it out and do a little research.
• http://www.canonicalscience.com Juan R.
This is a fascinating post!
There is not string theory just a program named “string theory” devoted to search of anything that can be called “string theory”.
I find particulary interesting the above claim of string theorists that old “axiom” “string theory requires 10D” is not rigorous (i would wait they agree that nothing in string theory research can be considered rigorous).
I find really amazing the dozens of talks, books and articles devoted to critize Loop quantum gravity and others aproaches because they work in the “old” 4D framework. Once i read from a string theorist that LQG was wrong because was 4D and non-supersimmetric!!!!
Many string theorists popularly stated that universe WAS 10-11D and NOT 4D. Public is completely misinformed about the real status of string theory in modern science. It is a kind of cancer…
I find also amazing that many of my predictions in the recent past begin to see accepted by some people. For example, In my Oct 21 post “String theory is not a TOE” in moderated newsgroup sci.physics.strings i did a joke about the infinite malleability of string theory and how even with that infinite malleability string theory was an dead way. By dead way i mean dead for doing science, real science; of course, “string theory” and the Landscape and all stuff will survive during decades as a kind of postmodern religion or metaphysics, as it has done in last decade or so.
I said in page 2 of my April non-technical article cited in the newsgroup:
As an illustration of the malleability of the string project, think during an instant on spacetime dimensions, the most characteristic piece of the [string, M) theory for public. Initially, the theory was developed for 4D; after it was for 26D; next “reformulated” on 10D, and the last decade, after Ed Witten last revolution, dimensionality is fixed on 11D. Is it really fixed? There are people working in the possibility of more than one time dimension. 12D? Perhaps 13D? What is more, some theoretician has recently claimed that we would investigate 4D string models due to the failure of compactification for extracting real physics!
The notation “[string, M)” means the “sequence” of dozens of different versions of “string theory” proposed as the Final theory during the last 3-4 decades. Note that i leave open “)” the sequence by the right hand!!!
Yes, public reading my post may be skeptic because theorists as Witten or B. Greene newer said them that different versions of string theory were proposed in literature and none of them worked even minimally.
Please, by the sanity of science, do not forget the real history of field which is not covered in Witten popular essays in Nature or in Greene’s The Elegant Universe!
Robert B. Laughlin, 1998 Nobel Prize-winning physicist said last year
People have been changing string theory in wild ways because it has never worked.
Return to the history of dimensions: 4D,… 26D, 10D, 11D, 12D?, 4D again. It appears to be a syntom of being doing research in circles, Right?
Now, i am preparing a formal rebuttal to last Witten Essay in Nature, which i consider completely distort the reality of the field and is based in beliefs.
I am also considering the possibility for an open letter (signed by specialists in many fields) explaining to public last “advances” in string theory and why string theory is completely wrong as candidate to TOE since have a wrong mathematical structure.
In fact, i personally doubt that even in the restricted field of particle physics and quantum gravity we can wait some good from string M-theory in a future.
Juan R.
Center for CANONICAL |SCIENCE)
Once i read from a string theorist that LQG was wrong because was 4D and non-supersimmetric!!!!
gr-qc/9710008, footnote on page 2.
Btw, it’s supersymmetry.
• http://www.canonicalscience.com Juan R.
Tanks by reference and by the correction!!!
—
Juan R.
Center for CANONICAL |SCIENCE)
• http://thebumblebeeblog.blogspot.com/ Poppycock
There are more fundamental errors than just typos.
For example, I believe Brian Greene did discuss different string theories (type I, IIA, IIB etc) in “The Elegant Universe”.
Critical string theory in 26D is a bosonic theory only, and I’m sure no-one has ever suggested that this represented a “theory of everything”.
As for the footnote referenced, since that is not credited to anyone in particular, it does not seem to me to be good evidence that this was ever put forward as a serious argument against LQG.
• http://www.canonicalscience.com Juan R.
Poppycock,
There are more fundamental errors than just typos.
It is good to know that. I believe that the perfect post has been still not invented. It is good and needed to make errors. My criticism ot string theorists is not by making errors, just by ignoring the last near four-decades criticism and by their obvious attempt to distort the history of the field.
String theory has been researched with a lesser or greater intensity for twenty-something years. Initially it was introduced in the context of understanding the strong force, but then QCD came along and worked much better, and thus ousted this idea.
In a pair of years, string theory will be 40 years-old. Somewhat as many people is misinformed about real status of string theory, I believe that many people would be perplexed when know that your “twenty-sometimes” means near the double of years: 40.
Now, let us see if it is true you are claiming. You said
For example, I believe Brian Greene did discuss different string theories (type I, IIA, IIB etc) in “The Elegant Universe”.
I never said the contrary! That i said was that The Elegant Universe presented to the public a distorted version of the history of string theory research. Moreover, by different versions of string theory i was not refering just to the five (2, 10) versions of string theories usually named by the collective name of “superstring theory”. By different versions of string theory i refer to dozens of versions published in literature you apparently have not read the list of different versions i cited in my April work.
However, your appeal to superstring theory is good, because now i can add some i said not in my previous post: that some time ago many physicists claimed that ONE of the five known superstring versions WAS the correct final theory.
In fact, the community of string theorists (“stringers”) thought then that our universe was just described by, and only by, (2, 10) points in the “branescan”. Now, they admit that were wrong in the past which reinforces my thesis history of string theory is the history of sucessive failure. It is more even now it is admited that string is not fundamental…
It is amazing to write at one hand all past claims of stringers and, at the other hand, our current understanding of topics. Allmost all was claimed in the past has been shown to be wrong.
Critical string theory in 26D is a bosonic theory only, and I’m sure no-one has ever suggested that this represented a “theory of everything”.
I never said or even suggested that!
I said that in the past it WAS suggested that universe WAS 26D. Of course this idea was inspired in critical dimension of bosonic theory. Nobody claimed that a theory of 26D lacking fermionic modes could be considered a candidate to TOE. However, stringers believed that a fermionic corrected theory would be also 26D, because in the contrary case they would say: “NO, universe is not 26D; 16D are just an artifact of current formulation of the theory“.
el of classical theory, but on attempting to promote it to a quantum theory, researchers discovered that the total number of spacetime dimensions is fixed uniquely to be 26. So, quantum strings could exist only in a world with 25 (rather than 3) spatial dimensions, plus time. The excitement of finding — for the first time — a mathematical consistency condition that determines the number of spacetime dimensions, rather than treating this number as an experimental input, was somewhat tempered by the absurd value predicted for this number.
Nobody predicted a reduction 26D –> 10D via generalization to fermionic modes of the bosonic somewhat as nobody recently predicted the adittion of a new dimension outside the (2, 10) “traditional” regime.
In fact, i have seen some early work and ideas about compactification 26D –> 4D which now are considered outdated since the number of dimension is assumed to be 10D in superstring (CY) and 11D in M-theory (G2).
And YES, stringers claimed in public that superstring theory was the searched TOE and universe WAS 10D already in the 80s (i.e. 30 years ago). But 10D was not a correct number again.
I repeat the history of dimensions: 4D, 26D, 10D, 11D, 12D?, 4D again…
As for the footnote referenced, since that is not credited to anyone in particular, it does not seem to me to be good evidence that this was ever put forward as a serious argument against LQG.
If i write E= mc^2 in this blog i think that i do not need to reference a 1905 paper for suporting it. We do not usually reference material is well-known for readers we are focusing. Above preprint was focused to people who already know (you apparently do not) the history of the field.
Juan R.
Center for CANONICAL |SCIENCE)
• http://eskesthai.blogspot.com/2006/02/evidence-for-extra-dimensions-and.html Plato
Sure is nice to have the opportunities to rebuff the current state of affairs Juan R.
While these models appear abstract, how is it that you could marry such thoughts to “proposed experimental processes” with which to deal with the consequences of such abstract thoughts?
Was it wrong to go down certain avenues, then find that such experimental processes “stupid” after the fact, or was there real science being looked at?
Taking in the current state of affairs, to such analogistics levels of scientific validation, is just one more aspect of model assumption?
The contributors here are providing a service instead of the ole rant(yada yada), yet, we do hear of the 10 or 11 with some clarifications?:) Your contribution is helpful? Thanks. I’m learning a lot from a historical perspective, and after the fact.:)
Your thoughts on ICECUBE, in a weak field and it’s manifestation?
• http://www.canonicalscience.com Juan R.
Beautiful and deep reply Plato 😉
During a brief instant (perhaps at Planck scale) your reply resembled the deep vision launched by the classical philosophers that your nick evokes.
Yes, during that brief instant, I was tempted to become a believer.
Unfortunately for the believers, the history of the field is very different of is said in talks focused to young and impressionable minds or in books dealing with laymen. Unfortunately for believers, it is a hard task to try to convince to real scientists that there is some serious below the stringy hype far from three decades arrogant claims of being working with the Final Theory, a theory that could explain everything.
Unfortunately, today string theory continues being so far from any realistic TOE as was 40 years ago. Unfortunately, each little time a new “prediction” of string theory is shown to be false. Last, I remember was the observation some days ago of the so called cosmological strings that were not so string after all…
Unfortunately, even with the infinite malleability of string theory, in which you can say anything and the contrary of anything (each claim founding a new version of string theory) all of us would recognize (even believers!) that after of all Nature does not like beauty
Stringy of course…
About the ICECUBE i prefer to follow “solid” Witten philosphy in gravity and talk of postdictions. Therefore, when you obtain some data contact with me.
Juan R.
Center for CANONICAL |SCIENCE)
• http://eskesthai.blogspot.com/2006/02/warm-dark-matter.html Plato
Juan R. (1R?)
hmmmm….
each little time a new “prediction” of string theory is shown to be false.
How is this possible? As a lay person such an example was important to me, that if such predictions are made and falsified, this was being handled properly?
Postdiction? Ah so you are saying any model assumption, or effort to use this model currently, is postdiction?
So lets say on mathematics alone then, is there any consistancy?
A side note? War on SeaGull: Any of you know? Beer and Present Danger, was actually a good one.
• http://eskesthai.blogspot.com/2006/02/history-of-superfluid-new-physics.html Plato
Layman ponders.
Anyway, what is this “new physics” that might house dimensional perspective?
Multiverses. Really! How would such “funnels” provide for new beginnings? A geometrodynamical view of bubble technologies? :)Ronald Mallet’s dream of travelling back in time and making way for new things?
• http://eskesthai.blogspot.com/2006/02/cosmic-variances-very-own-strangelets.html Plato
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Smartick is an online platform for children to master math in only 15 minutes a day
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# Rule of Three for Calculating Percentages
In earlier posts, we have learned what a percent is and how to calculate it.
Today we are going to use the rule of three to solve different types of problems related to percentages.
###### Rule of three to calculate the percentage of a number
For example, we want to calculate 30% of 360.
30% means 30 for each 100. So the approach would be: if I have 30 from 100, I have X from 360:
100 —— 30
360 —— X
X = (30 x 360) / 100
X = 108
So, 30% of 360 is 108.
###### Rule of three to calculate a quantity knowing the percentage of it
For example, we know that 25% of a quantity is 49. What is the quantity?
If 25% is 49 then the 100%, which we do not know, will be X :
25 —— 49
100 —— X
X = (49 x 100) / 25
X = 196
The quantity we are looking for is 196.
###### Rule of three to calculate the percentage represented as a quantity of another
What percentage of 250 does 50 represent?
250 is the 100% and 50 is the percentage that we do not know, X :
250 —— 100
50 —— X
X = (100 x 50) / 250
X = 20
50 is 20% of 250.
###### Rule of three to calculate the percentage of an unknown quantity knowing another percentage of the quantity
We know that 40% of a quantity is 78, how much would 60% be of the same quantity?
The 40% is 78 and we want to calculate 60%, which will be :
40 —— 78
60 —— X
X = (78 x 60) / 40
X = 117
So 60% of this quantity is 117.
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In 1986, the city of Los Diablos had 20 days on which air : GMAT Critical Reasoning (CR)
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# In 1986, the city of Los Diablos had 20 days on which air
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In 1986, the city of Los Diablos had 20 days on which air [#permalink]
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In 1986, the city of Los Diablos had 20 days on which air pollution reached unhealthful amounts and a smog alert was put into effect. In early 1987, new air pollution control measures were enacted, but the city had smog alerts on 31 days that year and on 39 days the following year. In 1989, however, the number of smog alerts in Los Diablos dropped to sixteen. The main air pollutants in Los Diablos are ozone and carbon monoxide, and since 1986 the levels of both have been monitored by gas spectrography.
Which of the following statements, assuming that each is true, would be LEAST helpful in explaining the air pollution levels in Los Diablos between 1986 and 1989?
(A) The 1987 air pollution control measures enacted in Los Diablos were put into effect in November of 1988.
(B) In December of 1988 a new and far more accurate gas spectrometer was invented.
(C) In February of 1989, the Pollution Control Board of Los Diablos revised the scale used to determine the amount of air pollution considered unhealthful.
(D) In 1988 the mayor of Los Diablos was found to have accepted large campaign donations from local industries and to have exempted those same industries from air pollution control measures.
(E) Excess ozone and carbon monoxide require a minimum of two years to break down naturally in the atmosphere above a given area.
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22 Apr 2008, 19:17
i dont know but D makes sense to me..bascially doesnt add anything to the argument.
so in 1988 he was found to have taken money..however, that means..they must have investigated him for some time..which means..he probabaly let the local industries off the hook earlier than 1988...so..if they caught him 1988..why did levels drop in 1989?? it doesnt explain any of that...
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22 Apr 2008, 22:13
Quite clearly D. Everything else helps explain this phenomenon quite well.
vshaunak@gmail.com wrote:
In 1986, the city of Los Diablos had 20 days on which air pollution reached unhealthful amounts and a smog alert was put into effect. In early 1987, new air pollution control measures were enacted, but the city had smog alerts on 31 days that year and on 39 days the following year. In 1989, however, the number of smog alerts in Los Diablos dropped to sixteen. The main air pollutants in Los Diablos are ozone and carbon monoxide, and since 1986 the levels of both have been monitored by gas spectrography.
Which of the following statements, assuming that each is true, would be LEAST helpful in explaining the air pollution levels in Los Diablos between 1986 and 1989?
(A) The 1987 air pollution control measures enacted in Los Diablos were put into effect in November of 1988.
(B) In December of 1988 a new and far more accurate gas spectrometer was invented.
(C) In February of 1989, the Pollution Control Board of Los Diablos revised the scale used to determine the amount of air pollution considered unhealthful.
(D) In 1988 the mayor of Los Diablos was found to have accepted large campaign donations from local industries and to have exempted those same industries from air pollution control measures.
(E) Excess ozone and carbon monoxide require a minimum of two years to break down naturally in the atmosphere above a given area.
Re: CR - smog alert [#permalink] 22 Apr 2008, 22:13
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# In 1986, the city of Los Diablos had 20 days on which air
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# Hewlett Packard 10B
## Calculator examplesprepared by Pamela Peterson
1. ### Calculating a future value
Problem:
Suppose you invest \$10,000 today in an account that pays 5% interest, compounded annually, how much will you have in the account at the end of 6 years?
Solution: \$13,401
10000 +/- PV 5 I/Y 6 N FV
2. ### Calculating the present value of an annuity
Problem:
Suppose you are promised annual payments of \$1,500 each year for the next five years, with the first cash flow occurring in one year. If the interest rate is 4%, what is this stream of cash flows worth today?
Solution: \$6,678
1500 PMT 5 N 4 I/Y PV
3. ### Calculating the value of a bond
Problem:
Calculate the value of a bond with a maturity value of \$1,000, a 5% coupon (paid semi-annually), five years remaining to maturity, and is priced to yield 8%.
Solution: \$878.34
Note:
FV = 1,000 (lump-sum at maturity)
CF = \$25 (one half of 5% of \$1,000)
N = 10 (10 six-month periods remaining)
i = 4% (six-month basis, 8%/2)
1000 FV 10 N 4 I/Y 25 PMT PV
4. ### Valuing a series of uneven cash flows
Problem:
Consider the following cash flows,
CF0 = -\$10,000
CF1 = +\$5,000
CF2 = \$0
CF3 = +\$2,000
CF4 = +\$5,000
1. What is the internal rate of return for this set of cash flows?
2. If the discount rate is 5%, what is the net present value corresponding to these cash flows?
Solution:
1. IRR = 7.5224%
2. NPV = +\$603.09
10000 +/- CF 5000 CF 0 CF 2000 CF 5000 CF n IRR 5 I/Y n NPV
where n indicates the orange-colored key to reach the 2nd level functions.
5. ### Calculating the yield to maturity on a bond
Problem:
Calculate the yield to maturity of a bond with a maturity value of \$1,000, a 5% coupon (paid semi-annually), ten years remaining to maturity, and is priced \$857.
Solution: 7.01%
Note:
FV = \$1,000 (lump-sum at maturity)
CF = \$25 (one half of 5% of \$1,000)
N = 20 (20 six-month periods remaining)
PV = \$857
1000 FV 20 N 857 +/- PV 25 PMT i x 2
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Should we keep the secret quiet?
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posted on May, 21 2004 @ 01:36 PM
Originally posted by JayKew
All I know for sure is that they go big time on strawberry ice cream
Yeah, and I heard the Grey's have a thing for Jenna Jamisson and Krystal Steal porn flicks :p
posted on May, 21 2004 @ 01:46 PM
I think that aliens do exist. How could it ever protect us.
I have come up with this equation which will be my signature soon.
Zuzubar vs. Usa
Government being evil
................................... Divided by =Not Possible
Government protecting us
Government being evil is given a value of 1
Government protecting us is given a value of 0
Therefore Government protecting us multiplied by the chances of the government being evil is not possible.
The only way to work my equation is to TAKE the government protecting us value out
Now it looks like this
Government being evil =1
The check is made. Government being evil has a value of 1. There fore this equation is true.
I have more creative equations. The main point I am trying to make is there is NO WAY the government is protect us.
BTW You can not divide 1 by 0. -Zuzubar-
posted on May, 21 2004 @ 01:56 PM
it should be reaveled to the public they and we diserve to no
posted on May, 21 2004 @ 02:01 PM
why nobody doubts that the B-2 spirit is made by Northrop ?
it looks damned sur as an alien "hooky boomerang" thingie !
but..
take any glued to eachother saucercups and...presto !
alien spaceship from wayback in the galaxy coming to tell the earthlings
how to "bake a spacecake" !
posted on May, 21 2004 @ 02:11 PM
The US military can indeed keep some secrets, the stealth bomber was hidden for what, 10 years?
But then again let's not confuse the military, and the government.
The US military brass does not get shuffled out of office every 4/8 years, so yes they can contain some information easier than government officials.
Note, also that the military doesn't build jets, it contracts it out to firms like Lockheed.
So again, to reverse engineer an alien craft would require, again Imo, alot of contractual recruiting...
Of course if you believe Corso's book that is exactly what has occured in the 20th century in the US, the technology was simply leaked into the private sector; microchips, optical wiring, etc.
So, should the military keep the secret?
The answer is, it depends exactly what that secret is.
If it is an entirely "military test plan"'ish secret, then perhaps.
If it is one where Aliens told Eisenhower that they fakked with our ancestors DNA, sent Jesus to us 2000 years, and the end of the world is in 2012, then perhaps.
posted on May, 21 2004 @ 06:22 PM
I always find it a bit funny when I hear people talk about how we (meaning themselves) deserve Disclosure...and then...they quickly follow up with some silliness about wanting to be a part of the Galactic Federation.
They insist the government (meaning THE EVIL UNITED STATES) is wrong for not telling the world. And how the world is ready to hear their information but what if...just what if...these entities (if any at all) are malicious? Sure sure the new agers would like us to believe they're kind ancient souls but let's not be naive...our ideas of good and evil may not be the same as "theirs". We have to accept the fact that these beings may not be "good"...what does that mean? Riots, rape, theft, destruction, murder...what happened to L.A. when several white policemen beat Rodney King? What happened during the 68 Democratic Convetion?
I'm shocked some of you are so naive as to believe the world is ready for some type of Disclosure. Some of you can't even seperate Stargate-SG1 and Star Trek from reality let alone find out aliens exist. What if the truth of the Disclosure is...THERE IS NO ALIENS. Some of you are so horribly entrenched in your silly legends and hoaxes and sci-fi that you wouldn't be able to handle "the truth" you've "seeking".
I am a believer in Extraterrestrial life, I am a believer in the UFO phenomenon...I am not a believer in the UFO Community or the world being ready for some type of disclosure.
[Edited on 5-21-2004 by Preest]
posted on May, 22 2004 @ 04:29 AM
the origine of life on this planet was just a dirty snowball collected by a spacedustsucking COMET before it smashed down !
the cosmic soup was heated by a nukesource: the SUN
so as a matter of fact all life on this planet is ALIEN !
have a nice UFO day !
posted on May, 22 2004 @ 11:49 AM
Originally posted by NOGODSINTHEUNIVERSE
the origine of life on this planet was just a dirty snowball collected by a spacedustsucking COMET before it smashed down !
the cosmic soup was heated by a nukesource: the SUN
so as a matter of fact all life on this planet is ALIEN !
have a nice UFO day !
Interesting attempt at comedy but the punchline fell on it's face. Maybe something was lost in the translation.
posted on May, 22 2004 @ 01:12 PM
Preest see my prior post on why I think it extremely unlikely an alien race very technologically advanced could be evil. Even our own history shows evil regimes never last longer than a few decades let alone thousands of years necessary to advance to a level required for space and time travel.
You are right statistically it is very likely intelligent life exists other than on earth but also statistically very unlikely they have ever visited our planet. I just wanted to make the point even if by a very slim chance we have been visited and it was made public it wouldn't but such a huge deal alot of people think. For your average Joe it wouldn't really change much at all.
posted on May, 22 2004 @ 01:18 PM
well with all the movies about ailens killing people and eating them and other stuff like that..even young people would hear about it and if ailens came would kill them!
posted on May, 22 2004 @ 01:31 PM
angelhair = tail of comet
fallen angel from heaven = comet slamming into planet earth
hell = explosions of gas due to impact
devil = moon in quarter and red color by airpollution due to impact comet
[Edited on 22-5-2004 by NOGODSINTHEUNIVERSE]
posted on May, 22 2004 @ 01:42 PM
Krisboton when was the last time you or anyone you know went and killed something because of what you saw on tv or at the movies or heard on the radio? Im sure there would be a few racially motivated killings on both sides just like we have now.
But in the end if aliens visit us is my bank going to forget about my mortgage, will my credit card bills miraculously disappear, will American Idol go away?
I didn't think so.......
posted on May, 22 2004 @ 02:08 PM
I honestly believe that they are gearing up for such a news alert to the general population of the planet in the near future. As stated earlier, all of these movies being released serve not only to entertain us but they also condition us!
We watch movies like Independence Day, etc., etc., and subconsciously our minds are at work carving little niches in our psyche for the inevitable revelation that yes, this might actually happen and when it does....Oh look, ET is here and he is pissed.
To reveal to the world's people that a 'hostile extraterrestrial life form' has been detected and is planning a little stopover on our planet will serve the NWO perfectly. Mass panic, chaos, histeria and all hell breaking loose. Lucky for us only our One World Government has the means to protect and look out for us poor souls. Whew! Declare a national emergency, suspend all government operations, start the FEMA engines and let's roll!
The David Icke reptillian thing fits in nicely with all of this.
posted on May, 22 2004 @ 02:38 PM
Originally posted by mrdependable
Preest see my prior post on why I think it extremely unlikely an alien race very technologically advanced could be evil. Even our own history shows evil regimes never last longer than a few decades let alone thousands of years necessary to advance to a level required for space and time travel.
You are right statistically it is very likely intelligent life exists other than on earth but also statistically very unlikely they have ever visited our planet. I just wanted to make the point even if by a very slim chance we have been visited and it was made public it wouldn't but such a huge deal alot of people think. For your average Joe it wouldn't really change much at all.
Assuming of course there is other intelligent life the chances of them having developed as we have would be extremely slim. Our concepts of "good" and "evil" couldn't possibly be shared by beings not influenced by our "civilization". Proof of that can be easily found within tribes on our own planet that had no idea that it was "wrong" to eat their enemies. Good and evil are concepts we live by...assuming intelligent beings from another planet might share our sense of morals is egotistical.
I'm really more apt to believe that UFO sightings are simply unconventional craft being tested. Of course, not all sightings can be written off as aircraft, quite a bit of it can.
posted on May, 22 2004 @ 03:30 PM
Preest certain environments foster innovation and growth required for higher evolution and technological advancement. A race preoccupied with eating each other is not going to get very far, let alone off the planet. While they may not share our morales you cannot progress with anarchy. Our values may not be universal but the path of least resistance is.
To build a spacecraft capable of intergalactic travel requires a vast amount of knowledge and resources. You don't acquire these things by chance. You need a sustainable economy for a start and an orderly system in allocating and exchanging these resources. If your race manages this then obviously they are of at least average intelligence and will know co-operation is much more beneficial than anihillation. It is simple economics and game theory that even I understand and im sure I would be considered primitive by such an advanced race.
If we have a certain resource that they need it would be much less costly for them to trade for these resources than take over the entire planet. An intelligent alien race would have nothing to gain by destroying the human race or occupying the planet and not abiding by our morales while here.
Just as we here in the US don't go invading other countries for resources that we need even though we have the military capability. This would cause anarchy worldwide and ruin our relationships with every other country. It would be illogical. If we need a resource we don't have we simply trade for it.
The idea that there exists an alien race that has worked hard for thousands or even millions of years to achieve intergalactic travel and are now willing to give everything up and destroy all the relationships they have ever built with other intelligent races just to take over one planet out of trillions is ridiculous to me. Makes for a good hollywood story but real life isnt as exciting im afraid.
posted on May, 22 2004 @ 06:03 PM
Originally posted by mrdependable
Preest certain environments foster innovation and growth required for higher evolution and technological advancement. A race preoccupied with eating each other is not going to get very far, let alone off the planet. While they may not share our morales you cannot progress with anarchy. Our values may not be universal but the path of least resistance is.
To build a spacecraft capable of intergalactic travel requires a vast amount of knowledge and resources. You don't acquire these things by chance. You need a sustainable economy for a start and an orderly system in allocating and exchanging these resources. If your race manages this then obviously they are of at least average intelligence and will know co-operation is much more beneficial than anihillation. It is simple economics and game theory that even I understand and im sure I would be considered primitive by such an advanced race.
If we have a certain resource that they need it would be much less costly for them to trade for these resources than take over the entire planet. An intelligent alien race would have nothing to gain by destroying the human race or occupying the planet and not abiding by our morales while here.
Just as we here in the US don't go invading other countries for resources that we need even though we have the military capability. This would cause anarchy worldwide and ruin our relationships with every other country. It would be illogical. If we need a resource we don't have we simply trade for it.
The idea that there exists an alien race that has worked hard for thousands or even millions of years to achieve intergalactic travel and are now willing to give everything up and destroy all the relationships they have ever built with other intelligent races just to take over one planet out of trillions is ridiculous to me. Makes for a good hollywood story but real life isnt as exciting im afraid.
I'm not really certain where your assumption came from that I indicated an intelligent lifeform was looking to take over this planet but...I didn't. Nor am I completely convinced of any alien visitation.
My point is that you're basing what you know of civilization and eceonomics as well as anarchy on what you, AS A HUMAN RAISED IN A CIVILIZED SOCIETY, have been taught. Once again...beings that quite probably do not share our origins, biology, society and moral codes may not be able to comprehend your idea of good and evil. It's been proven already, as I said before, by the discovery of tribes and cultures that go against what "we" feel is CIVILIZED. Good, evil...these are human ideals and sometimes quite confusing ones at that.
Can we expect intelligent beings to believe what we believe and follow ideals we follow simply because we feel they must because our ideals or beliefs are the "right" civilized ones? Once again, egotistical.
Cannibal tribes have survived for centuries unhindered until "civilized man" intruded and put a stop to their traditions. Were these cannibal tribes wrong? Were they evil? Were "civilized men" good for stopping these practices?
"The path to least resistance" is a human concept. A truly intelligent, advanced race capable of space travel quite probably moved beyond "good" and "evil" in favor of "amorality". Good and evil sometimes hinder progress. I highly doubt aliens are on their way here to conquer Earth but I don't believe they WOULDN'T do it simply because they're intelligent and technologically advanced. It certainly didn't work that way for America's native indian population did it?
posted on May, 22 2004 @ 07:16 PM
I think that a underground re-education program would do the job. Contact individuals that are mentally and emotionally sound enough to handle the truth, call them in and give them a breifing. Formulate a seminar that can be given to the public on a grass roots level. The citizens who attend the seminar would have to sign a modified Non Disclosure form.
The "Briefers" would be the ones who give the seminar. They could have 5 or 6 teams that would traval around the country holding the seminars.
Its what I have been thinking all along.
posted on May, 22 2004 @ 07:52 PM
Personally, I think it scares the daylights out of the Gov.
They have no defence and they know it. I think maybe we have been contacted by ETs. The government hides it because it freaks them out,because they have no answers, they have no defense, and they have no understanding of it.
I dont think the US Gov does it to hide military secrets when it comes to UFOs,(although they may be trying to find defenses against such things) they do it because to say Hey, we just had aliens fly over our missle silos and totally shut them down, implies no control over a situation.
just my 2 cents, never seen an alien myself, but I don't doubt they exist.
I see it as a possibility, and I think if it is happening that is why they would be hiding it.
posted on May, 22 2004 @ 07:56 PM
What kind of idiot wouldn't be able to accept ET life here on Earth if they saw it first hand?
If they are here the reason why it isnt known/talked about, i.e. why we dont have ETs walking around on the streets w/o disguises has to be a money issue. Some how the rich and powerful stand to make more money if this info is kept a secret that otherwise.
I'd like to see the panic, anxiety and all the other crap the fool who cant handle info about ETs would show, it would give me another thing to laugh at.
[Edited on 22-5-2004 by jrod]
posted on May, 22 2004 @ 08:47 PM
Originally posted by Preest
I always find it a bit funny when I hear people talk about how we (meaning themselves) deserve Disclosure...and then...they quickly follow up with some silliness about wanting to be a part of the Galactic Federation.
They insist the government (meaning THE EVIL UNITED STATES) is wrong for not telling the world. And how the world is ready to hear their information but what if...just what if...these entities (if any at all) are malicious? Sure sure the new agers would like us to believe they're kind ancient souls but let's not be naive...our ideas of good and evil may not be the same as "theirs". We have to accept the fact that these beings may not be "good"...what does that mean? Riots, rape, theft, destruction, murder...what happened to L.A. when several white policemen beat Rodney King? What happened during the 68 Democratic Convetion?
I'm shocked some of you are so naive as to believe the world is ready for some type of Disclosure. Some of you can't even seperate Stargate-SG1 and Star Trek from reality let alone find out aliens exist. What if the truth of the Disclosure is...THERE IS NO ALIENS. Some of you are so horribly entrenched in your silly legends and hoaxes and sci-fi that you wouldn't be able to handle "the truth" you've "seeking".
I am a believer in Extraterrestrial life, I am a believer in the UFO phenomenon...I am not a believer in the UFO Community or the world being ready for some type of disclosure.
[Edited on 5-21-2004 by Preest]
Well put. I agree on most of your points. I must add that there is also the possibility that the Govt really does not know much more than we do.
I believe that at the very least, we are being observed.....but some of the far out stuff seems more based in Sci-Fi than in reality. More of a defense against a boring life to me....
Hopefully we will know the truth one day.
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help with measurements - Page 3 - Home Theater Forum and Systems - HomeTheaterShack.com
Old 03-11-15, 02:09 PM
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Re: help with measurements
Yes, It is just corrected info.
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Old 03-11-15, 03:57 PM Thread Starter
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Re: help with measurements
Getting worried for a moment
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Old 03-11-15, 07:25 PM
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Re: help with measurements
Quote:
nwf477 wrote: View Post
All right progress. In a different room w/ different responses and measurements w/ 2 SW's before anything else is done the IR has to be aligned for both. This is where the actual measurements come into play. The example being if SW1 is 10' and SW2 is 15' there has to be a 5ms delay added to the 10' SW. Once both IR's have been aligned then they both can be summed for the predicted response ( A + B ).
As a side note I have been told numerous ways to add a second SW some of them were simply doing a sub crawl ( to me there is lots of room for error ) to several pages of reading, and of course everybody's method is correct. For what it's worth this makes the most sense and takes the guess work out of it.
Thank you.
Hello,
I'm just trying to follow along here, as I also have two SW's and trying to learn about everything. In your example of "SW1 = 10' and SW2 = 15' there has to be a 5ms delay added to SW1", are you talking about just REW or are you actually adding 5ms electronically. My AVP only has 1 sub output and I'm using a "Y" cable to connect to the 2 SW's. My AVP will allow to set SW distance for delay, but both would get the same. I guess my real question is (not knowing your system/set-up, how do you apply the extra 5ms to SW1 and not SW2? Does your AVR/P have 2 sub outputs, or are you using something else?
Thanks,
Kix
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Old 03-12-15, 01:44 AM Thread Starter
Senior Shackster
Join Date: Feb 2015
Location: Richmond Va
Posts: 129
Re: help with measurements
My system is nothing fancy it is a Yamaha RX-V1600 ( 7.1 ). I am using a RCA to XLR adapter coming out of the AVR and then a " Y " from the XLR to the amp. Before you worry about delay it is important to get the best location for the subs. Different people will suggest different ways, I tried a few of them what I ended up doing was placing my RS SPL meter at the LP ear height hooked up to the AVR. I used REW and did the measurement at each of the locations looking for the flattest response. You can see the images on the 1st post. 1 and 3 were the best. Once you find the 2 best spots you need to open both graphs in the all SPL tab.As you can see with mine SW1 and SW2 both have peaks and nulls but they are opposite of each other where one has a peak the other has a null. If they do not align now you have to add delay. There is probably a better way to do all of this if you have the mic but I don't you have to measure the distance from where the LP is to each SW. In my ex. SW1 is 10ft. SW2 is 15ft. There is a difference of 5ft. that will be the delay it will expressed in ms.(1ms equals 1 foot ). I think there is a way you can do it on the AVR in the Distance settings but not sure? To see what the delay will do with both of your measurements on the All SPL window click controls ( it is right below the Preferences tab ) You can see where you can add the delay / measurement offset. Once that is done then you can add each image to the trace arithmetic and do an average for A and B this will finally let you see what response both of your SW's will have when playing together.
As a side note the measuring takes the longest what I tried to explain only takes a few minutes. For me it was pretty easy I did not have to apply any smoothing no eq other than the HPF set on my amp. I hope this makes some sense.
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Old 03-13-15, 07:08 PM
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Posts: 56
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Re: help with measurements
Does this mean that the delay you are adding to SW1 (in your example) is only done in REW, and that to a persons ears in the LP, there is still a 5mS difference between SW's? I'm also using a "Y" cable from my AVP to the SW's. So if I add delay via the AVP, I would be adding the same delay to both SW's. Hopefully I'm understanding right...
Thanks,
Kix
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Old 03-13-15, 09:43 PM
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Re: help with measurements
Kix,
The delay discussed in nwf477's example would be a 5ms IR offset for the closer SW added into REW before the (A + B) calculation to predict what the SPL would be if both SWs were measured with that 5ms delay in the physical setup. This corrects for the 5' distance difference.
nwf477 has the ability to add delay in his DSP P-Amps if he choses. Since he has only a 15" difference in his setup there is no pressing need to worry about the small delay it would take to adjust for that. He could just drive them together (without the delay) if he chooses. He will get very similar results as show above (2nd chart in Post 3).
There is no requirement that says that 2 SWs need to timed exactly alike to get good SPL results though. I like to look at that option first if the delay capability is available because that provides the highest SPL for the bass range. Small distance differences are not too problematic. If the differences get too large then the SPL is reduced in the higher bass freqs. Bigger yet, then most of the bass range starts to suffer as the SWs are working against each other. If they get far enough apart it may be better to invert one of them to keep the phase tracking as closely as possible.
In your case you will have the same delay for both SWs so there is no need to for this technique. Just measure the SWs together as evaluate the SPL. If there is several feet difference in distance to the LP then you may want to invert the polarity on one of them as well and measure them that way also.
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Cody
# Problem 60. The Goldbach Conjecture
Solution 1443143
Submitted on 18 Feb 2018 by Jay
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
nList = 28:6:76; for i = 1:length(nList) n = nList(i); [p1,p2] = goldbach(n) assert(isprime(p1) && isprime(p2) && (p1+p2==n)); end
p1 = 5 p2 = 23 p1 = 3 p2 = 31 p1 = 3 p2 = 37 p1 = 3 p2 = 43 p1 = 5 p2 = 47 p1 = 5 p2 = 53 p1 = 3 p2 = 61 p1 = 3 p2 = 67 p1 = 3 p2 = 73
2 Pass
nList = [18 20 22 100 102 114 1000 2000 36 3600]; for i = 1:length(nList) n = nList(i); [p1,p2] = goldbach(n) assert(isprime(p1) && isprime(p2) && (p1+p2==n)); end
p1 = 5 p2 = 13 p1 = 3 p2 = 17 p1 = 3 p2 = 19 p1 = 3 p2 = 97 p1 = 5 p2 = 97 p1 = 5 p2 = 109 p1 = 3 p2 = 997 p1 = 3 p2 = 1997 p1 = 5 p2 = 31 p1 = 7 p2 = 3593
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The term dynamic programming (DP) refers to a collection of algorithms that can be used to compute optimal policies given a perfect model of environment as a Markov decision process (MDP). Classical DP algorithms are of limited utility in reinforcement learning both because of their assumption of a perfect model and because of their great computational expense, but they are provides an essential foundation for the understanding of the methods presented later. In fact, all of these methods can be viewed as attempts to achieve much the same effect as DP, only with less computation and without assuming a perfect model of the environment.
The key idea of DP, and of reinforcement learning generally, is the use of value functions to organize and structure the search for good policies. In here we show how DP can be used to compute the value functions defined in earlier. As discussed there, we can easily obtain optimal policies once we have found the optimal value functions $v_{\star}$ or $q_{\star}$ which satisfy the Bellman optimality equations:
\begin{align} v_{\star}(s) &= \max_{a} \mathbb{E} [R_{t+1} + \gamma v_{\star}(S_{t+1}) \ | \ S_{t}=s, A_{t}=a] \\ &= \max_{a} \sum_{s^{\prime}, r} p(s^{\prime}, r | s, a) \left[r + \gamma v_{\star}(s^{\prime})\right] \end{align}
or
\begin{align} q_{\star}(s, a) &= \mathbb{E} [R_{t+1} + \gamma \max_{a^{\prime}} q_{\star}(S_{t+1}, a^{\prime}) \ | \ S_{t}=s, A_{t}=a] \\ &= \sum_{s^{\prime}, r} p(s^{\prime}, r | s, a) [r + \gamma \max_{a^{\prime}} q_{\star}(s^{\prime}, a^{\prime})] \end{align}
for all $s \in \mathcal{S}, \; a \in \mathcal{A(s)}, \; \text{and} \; s^{\prime} \in \mathcal{S^{+}}$. As we shall see, DP algorithms are obtained by turning Bellman equations such as these into assignments, that is, into update rules for improving approximations of the desired value functions.
First we consider how to compute the state-value function $v_{\pi}$ for an arbitrary policy $\pi$. This is called policy evaluation in the DP literature. We also refer to it as the prediction problem. Recall that for all $s \in \mathcal{S}$,
\begin{align} v_{\pi}(s) &\doteq \mathbb{E}_{\pi} [R_{t+1} + \gamma R_{t+2} + \gamma^{2}R_{t+3} + \cdots \ | \ S_{t}=s] \\ &= \mathbb{E}_{\pi} [R_{t+1} + \gamma v_{\pi}(S_{t+1}) \ | \ S_{t}=s] \\ &= \sum_{a} \pi(a|s) \sum_{s^{\prime}, r} p(s^{\prime}, r | s, a) [r + \gamma v_{\pi}(s^{\prime})] \end{align}
If the environment’s dynamics are complete known, then (7) is a system of $|\mathcal{S}|$ simultaneous linear equations in $|\mathcal{S}|$ unknowns (the $v_{\pi}(s), s \in \mathcal{S}$). In principle, its solution is a straightforward, if tedious, computation. For our purpose, iterative solution methods are most suitable. The initial approximation, $v_0$, is chosen arbitrarily (except that the terminal state, if any, must be given value 0), and each successive approximation is obtained by using the Bellman equation for $v_{\pi}$ as an update rule:
\begin{align} v_{k+1}(s) &\doteq \mathbb{E}_{\pi}[R_{t+1} + \gamma v_{k}(S_{t+1}) \ | \ S_{t}=s] \\ &= \sum_{a} \pi(a|s) \sum_{s^{\prime}, r} p(s^{\prime}, r|s, a) [r + \gamma v_{k} (s^{\prime})] \end{align}
This algorithm is called iterative policy evaluation.
Iterative policy evaluation
Input $\pi$, the policy to be evaluated
Initialize an array $V(s) = 0$, for all $s \in \mathcal{S^{+}}$
Repeat
$\Delta \leftarrow 0$
for each $s \in \mathcal{S}$:
$v \leftarrow V(s)$
$V(s) \leftarrow \sum_{a} \pi(a|s) \sum_{s^{\prime}, r} p(s^{\prime}, r | s, a) [r + \gamma v(s^{\prime})]$
$\Delta \leftarrow \max(\Delta, |v - V(s)|)$
until $\Delta < \theta$ (a small positive number)
Output $V \approx v_{\pi}$
We can see the algorithm used in the grid world problem just is the iterative policy evaluation.
Our reason for computing the value function for a policy is to help find better policies. Once a policy $\pi$ has been improved using $v_{\pi}$ to yield a better policy $\pi^{\prime}$, we can then compute $v_{\pi^{\prime}}$and improve it again to yield an even better $\pi^{\prime\prime}$. We can thus obtain a sequence of monotonically improving policies and value functions:
$$\pi_{0} \stackrel{E}\longrightarrow v_{\pi_{0}} \stackrel{I}\longrightarrow \pi_{1} \stackrel{E}\longrightarrow v_{\pi_{1}} \stackrel{I}\longrightarrow \pi_{2} \stackrel{E}\longrightarrow \cdots \stackrel{I}\longrightarrow \pi_{\star} \stackrel{E}\longrightarrow v_{\star},$$
where $\stackrel{E}\longrightarrow$ denotes a policy evaluation and $\stackrel{I}\longrightarrow$ denotes a policy improvement. This way of finding an optimal policy is called policy iteration.
Policy iteration (using iterative policy evaluation)
1. Initialization
$V(s) \in \mathbb{R}$ and $\pi(s) \in \mathcal{A(s)}$ arbitrarily for all $s \in \mathcal{S}$
2. Policy Evaluation
Repeat
$\Delta \leftarrow 0$
For each $s \in \mathcal{S}$:
$v \leftarrow V(s)$
$V(s) \leftarrow \sum_{s^{\prime}, r} p(s^{\prime}, r | s, \pi(s)) [r + \gamma v(s^{\prime})]$
$\Delta \leftarrow \max(\Delta, |v - V(s)|)$
until $\Delta < \theta$ (a small positive number)
3. Policy Improvement
policy-stable $\leftarrow$ true
For each $s \in \mathcal{S}$:
old-action $\leftarrow$ $\pi_(s)$
$\pi (s) \leftarrow argmax_{a} \sum_{s^{\prime}, r} p(s^{\prime}, r | s, a) [r + \gamma v(s^{\prime})]$
If old-action $\neq \pi(s)$, then policy-stable $\leftarrow$ false
If policy-stable, then stop and return $V \approx v_{\star} \; \text{and} \; \pi \approx \pi_{\star}$; else go to 2.
Let us solve a problem used by policy iteration. The problem defined as follows:
Jack manages two locations for a nationwide car rental company. Each day, some number of customers arrive at each location to rent cars. If Jack has a car available, he rents it out and it credited \$10 by the national company. If he out of cats at that location, then the business is lost. Cars become available for renting the day after they are returned. To ensure that cars are available where they are needed, Jack ca move them between the two locations overnight, at a cost of \$2 per car moved. We assume that the number of cars requested and returned at each location are Poisson random variables, meaning that the probability that the number is $n$ is $\frac{\lambda^{n}}{n!}e^{-\lambda}$, where $\lambda$ is the excepted number. Suppose $\lambda$ is 3 and 4 for rental requests at the first and second locations and 3 and 2 for returns. To simplify the problem slightly, we assume that there can be no more than 20 cars at each location (any additional cars are returned to the nationwide company, and thus disappear from the problem) and a maximum of five cars can be moved from one location to the other in one night. We take the discount rate to be $\lambda=0.9$ and formulate this as a continuing finite MDP, where the time steps are days, the state is the number of cars at each location at the end of the day, and the actions are the net numbers of cats moved between the two locations overnight.
The excepted result is as follows:
Figure 1
The first, we define some facts of this problem:
From the problem definition, we know that in this MDP the states is the number of cars at each location at the end of the day, and the actions are the net numbers of cats moved between the two locations overnight. Each action is a integer that positive number represents the number of cars moving from the first location to second location and vice verse.
For visualization (Figure 1) convenient, we define a method:
Next, we define a Poisson function that return the probability:
Now, the preparation is done. We’ll implement the policy iteration algorithm as follows:
We can see the logistic is the same as the pseudocode of the policy iteration algorithm. There is a core method in the code, that is, exceptedReturn() is used to calculate the reward of cars rental.
The comments are very clear, and we’re going to do a lot of this. Finally, let us print the result:
The results are as follows:
One drawback to policy iteration is that each of its iterations involves policy evaluation, which may itself be a protracted iterative computation requiring multiple sweeps through the state set. If policy evaluation is done iteratively, then convergence exactly to $v_{\pi}$ occurs only in the limit. In fact, the policy evaluation step of policy iteration can be truncated in several ways without losing convergence guarantees of policy iteration. One important special case is when policy evaluation is stopped after just one sweep (one backup of each state). This algorithm is called value iteration. It can be written as a particular simple backup operation that combines the policy improvement and truncated policy evaluation steps:
\begin{align} v_{k+1} &\doteq \max_{a} \mathbb{E}[R_{t+1} + \gamma v_{k}(S_{t+1}) \ | \ S_{t}=s, A_{t}=a] \\ &= \max_{a}\sum_{s^{\prime}, r} p(s^{\prime}, r | s, a) [r + \gamma v_{k}(s^{\prime})], \end{align}
for all $s \in \mathcal{S}$.
Value iteration
Initialize array $V$ arbitrarily (e.g. $V(s) = 0$ for all $s \in \mathcal{S^{+}}$)
Repeat
$\Delta \leftarrow 0$
For each $s \in \mathcal{S}$:
$v \leftarrow V(s)$
$V(s) \leftarrow \max_{a}\sum_{s^{\prime}, r} p(s^{\prime}, r | s, a) [r + \gamma V(s^{\prime})]$
$\Delta \leftarrow \max(\Delta, |v - V(s)|)$
until $\Delta < \theta$ (a small positive number)
Output a deterministic policy, $\pi \approx \pi_{\star}$, such that
$\pi(s) = \arg\max_{a}\sum_{s^{\prime}, r} p(s^{\prime}, r | s, a) [r + \gamma V(s^{\prime})]$
Let us use the value iteration algorithm to solve a Gambler’s Problem. The problem defined as follows:
A gambler has the opportunity to make bets on the outcomes of a sequence of coin flips. If the coin comes up heads, he wins as many dollars as he staked on the flip; if it is tails, he loses his stake. The game ends when the gambler wins by reaching his goal of \$100, or loses by running out of money. On each flip, the gambler must decide what portion of his capital to stake, in integer number of dollars. This problem can be formulated as an undiscounted, episodic, finite MDP. The state is the gambler’s capital,$s \in \{1, 2, \cdots, 99\}$and the actions are stakes,$a \in \{0, 1, \cdots, \min(s, 100-s)\}$. The reward is zero on all transitions excepted those on which the gambler reaches his goal, when it is +1. The state-value function then gives the probability of winning from each state. A policy is mapping from levels of capital to stakes. The optimal policy maximizes the probability of reaching the goal. Let$p_h$denote the probability of the coin coming up heads. If$p_h\$ is known, then the entire problem is known and it can be solved, for instance, by value iteration.
OK, now let us to solve this problem by use the value iteration algorithm.
The first we defined some facts and some auxiliary data structure:
The step of value iteration:
Calculate the optimal policy:
Print the results:
The results are as follows:
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Instantly share code, notes, and snippets.
# 73spica/ninja_solver.py Last active Oct 4, 2016
#!-*-coding:utf-8-*- # H4CK1T CTF 2016 Online # Chad – Ninja Scheme (CRYPTO 195 pts) from m1z0r3.crypro import * import time # ==== Global Variable ==== half_block_len = 4 block_len = 8 dec = "dd67ca82d358f0c8479e118addcec2f8ce086c0f6f239f9b66d7226a38c68198dbd777f366fb9fd83b60d11109be174759c75ea56a4866c2" num = 16 dec_len = len(dec) def f(r,n): anslist = [] for i in xrange(half_block_len): anslist.append( ( r[i]+n ) % 256 ) return anslist def my_xor(a,b): anslist = [] for i in xrange(half_block_len): anslist.append(a[i]^b[i]) return anslist def round_proc(l,r,n): pre_l = r pre_r = my_xor(f(r,n),l) return pre_l,pre_r def int_to_hex(a): if a<15: return "0"+hex(a)[2:] else: return hex(a)[2:] def decrypt_block(l,r,n): for i in reversed(xrange(2,n+1)): # print "i:",i-1 l,r = round_proc(l,r,i-1) # func = lambda x : hex(x)[2:] return "".join(map(int_to_hex,l+r)) def main(): # ==== Local Variable ==== ans = "" block_list = [] # ==== Prepare ==== for i in xrange(dec_len): ans += dec[i] if (i+1) % num == 0: block_list.append(ans) ans = "" print block_list # ==== hex_str to int_num ==== int_block_list = [] for block in block_list: tmp = [int((i+j),16) for (i,j) in zip(block[::2],block[1::2])] int_block_list.append(tmp) print int_block_list print # ==== start decrypting! ==== n = 1 while True: n += 1 dec_result = "" for int_block in int_block_list: dec_result += decrypt_block(int_block[4:],int_block[:4],n) # print "dec_result:",dec_result ans = decode_hex(dec_result) if "h4ck" in ans: print "number of round is",n print "flag is",ans return if __name__ == '__main__': main()
Owner Author
### 73spica commented Oct 3, 2016
H4CK1T CTF 2016 Online Chad – Ninja Scheme (CRYPTO 195 pts)
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https://minuteshours.com/59-65-minutes-in-hours-and-minutes
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59.65 minutes in hours and minutes
Result
59.65 minutes equals 0 hours and 59.65 minutes
You can also convert 59.65 minutes to hours.
Converter
Fifty-nine point six five minutes is equal to zero hours and fifty-nine point six five minutes.
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http://gradestack.com/CBSE-Class-9th-Complete/Gravitation/Liquid-Pressure/14889-2952-3845-study-wtw
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# Liquid Pressure
All liquids and gases are fluids. A solid exerts pressure on the surface due to its weight. Similarly, fluids have weight and they also exert pressure on the base and walls of the container in which they are enclosed. Pressure exerted in any confined mass of fluid is transmitted undiminished in all directions. Other important properties of liquids are
• Pressure increases with depth.
• Pressure is the same at all points at the same depth.
• Liquid seeks its own level.
Activity 1
Take a tall vessel.
Drill four small holes at different depths.
Fill the vessel with water to the brim.
Now remove the strip of tape.
You will find that water rushes out with great force from the hole at the bottom, with less force from the hole at the centre and with least force from the hole on the top. This shows that pressure increases with depth.
Activity 2
Perform a similar experiment as in the above activity with holes made at the same height on the metal can. Observe what happens. You will find that water rushes out with equal force from all the holes. This proves that pressure is the same at all points at the same depth.
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https://bodheeprep.com/cat-2021-quant-question-solution-16
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Bodhee Prep-CAT Online Preparation
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# CAT 2021 Quant Question [Slot 1] with Solution 16
Question
If $5 - {\log _{10}}\sqrt {1 + x} + 4{\log _{10}}\sqrt {1 - x} = {\log _{10}}\frac{1}{{\sqrt {1 - {x^2}} }}$, then 100x equals
Option: 99
Solution:
>$5 - {\log _{10}}\sqrt {1 + x} + 4{\log _{10}}\sqrt {1 - x} = {\log _{10}}\frac{1}{{\sqrt {1 - {x^2}} }}$
We can re-write the equation as: $5 - {\log _{10}}\sqrt {1 + x} + 4{\log _{10}}\sqrt {1 - x} = {\log _{10}}{\left( {\sqrt {1 + x} \times \;\sqrt {1 - x} } \right)^{ - 1}}$
$5 - {\log _{10}}\sqrt {1 + x} + 4{\log _{10}}\sqrt {1 - x} = \left( { - 1} \right){\log _{10}}\left( {\sqrt {1 + x} } \right) + \left( { - 1} \right){\log _{10}}\left( {\sqrt {1 - x} } \right)$
$5 = - {\log _{10}}\sqrt {1 + x} + {\log _{10}}\sqrt {1 + x} - {\log _{10}}\sqrt {1 - x} - 4{\log _{10}}\sqrt {1 - x}$
$5 = - 5{\log _{10}}\sqrt {1 - x}$
$\sqrt {1 - x} = \frac{1}{{10}}$
Squaring both sides: ${\left( {\sqrt {1 - x} } \right)^2} = \frac{1}{{100}}$
$\therefore \;$ $x = 1 - \frac{1}{{100}} = \frac{{99}}{{100}}$
Hence, $100\;x\; = 100 \times \;\frac{{99}}{{100}} = 99$
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http://www.gulfcoastcampingresort.com/spellbound-sub-nol/b111ab-brute-force-algorithm-tutorialspoint
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## brute force algorithm tutorialspoint
In some cases, they are extremely simple and rely on raw computing power to achieve results. Let’s start with a bit of theory. So I tried to write one myself, but I failed. The brute force algorithm searches all the positions in the text between 0 and n-m whether the occurrence of the pattern starts there or not. Methode der rohen Gewalt, auch Exhaustionsmethode (kurz Exhaustion von lateinisch exhaurire ‚ausschöpfen‘), ist eine Lösungsmethode für Probleme aus den Bereichen Informatik, Kryptologie und Spieltheorie, die auf dem Ausprobieren aller möglichen (oder zumindest vieler möglicher) Fälle beruht. Ask Question Asked 6 years, 2 months ago. Briefly, a program would solve a puzzle by placing the digit "1" in the first cell and checking if it is allowed to be there. This time, we will see how another split-conquer algorithm finds the closest pair of points from a set of points on a two-dimensional plane. The time complexity of this algorithm is O(m*n). This affects the accuracy for the brute-force algorithm when k value is even. Hacking Tutorial: Brute Force Password Cracking. Ein Pfad durch jeden Knoten genau einmal ist gleichbedeutend mit der Anordnung des Scheitelpunkts. In that case, it makes it easy to crack and takes less time. Greedy . This article explains the basic brute force method first and then moves on to explain the optimized Manacher's Algorithm. Dijkstra’s algorithm is a Greedy algorithm and time complexity is O(VLogV) (with the use of Fibonacci heap). . I have a brute force algorithm, but never fully understood it. A brute force algorithm visits the empty cells in some order, filling in digits sequentially, or backtracking when the number is found to be not valid. The brute force nested-loops join is frequently referred to simply as a nested-loop join.C.J. Brute force password cracking is also very important in computer security. so if we are searching for n characters in a string of m characters then it will take n*m tries. Eigentlich ein „dummer” Algorithmus. Bruteforcing has been around for some time now, but it is mostly found in a pre-built application that performs only one function. Viewed 8k times 1. Dijkstra doesn’t work for Graphs with negative weight edges, Bellman-Ford works for such graphs. KMP algorithm has 2 parts: Partial Match table; String Matching; High level working of the algorithm: Bellman-Ford is also simpler than Dijkstra and suites well for distributed systems. I have a vague grasp of some of the things that go on, but every time I try to follow what happens exactly, I get lost (for example, the index variable is a little confusing). For example, imagine you have a small padlock with 4 digits, each from 0-9. The MD5 algorithm take any word or text in input and produce a 32 characters hexadecimal string The cipher text can be hacked with various possibilities. The brute-force computation of distances between all pairs of points in the dataset provides the most naïve neighbor search implementation. As a reminder, did you know how the MD5 algorithm works? Brute Force ist quasi eine Schleife, die über alle möglichen Lösungen läuft, weshalb die Laufzeit in der Effizienzklasse von O(n) liegt. Die Brute-Force-Methode (von englisch brute force ‚rohe Gewalt‘) bzw. Brute Force. It can be used to encrypt passwords and other data. 1. After each attempt, it shifts the pattern to the right by exactly 1 position. This algorithm is required to solve sub-problems of some very hard problems. Hacking of Caesar Cipher Algorithm. This problem is to count to a desired value by choosing the least possible coins and the greedy approach forces the algorithm to pick the largest possible coin. An edge e(u, v) represents that vertices u and v are connected. Share this article . You can create a new Algorithm topic and discuss it with other geeks using our portal PRACTICE. In this tutorial we shall see how to solve using KMP algorithm. Beispiel. Brute Force Approach: 2 − Then select one ₹ 5 coin, the remaining count is 3. Brute Force Algorithms are exactly what they sound like – straightforward methods of solving a problem that rely on sheer computing power and trying every possibility rather than advanced techniques to improve efficiency. The following are the approaches used after considering both the theoretical and practical importance of designing an algorithm: Brute force algorithm: The general logic structure is applied to design an algorithm. Any offers on how to make the algorithm more efficient are also welcome. But some people suggest the following, the convex hull for 3 or fewer points is the complete set of points. MD5 algorithm reminder. Brute Force Methode bei der Wegsuche in einem Graphen: ... Man sagt, ein Algorithmus hat die Komplexität O(f(n)), wenn die Funktion f(n), die dessen Laufzeitverhalten für große n beschreibt zu dieser Klasse gehört. Generally, brute force approaches should be avoided for more optimal algorithms, especially when the functions using these algorithms will scale to larger and larger numbers. Active 1 year, 6 months ago. Class ‘Chinstrap’ and ‘Adelie’ ended up with mode as 2. It depends on the algroithm - ‘brute force’ is just a name given to a whole class of different algorithms which attempt to solve the problem by looking at every possible solution. After arranging the K neighbours based on mode, brute-force ended up picking the first class instead of picking the class which had least distance in the distance metric. Global Software Support 37,179 views. Brute-Force Substring Search Algorithm - Duration: 6:43. How to write a brute-force algorithm? Distance between vertex u and v is d(u, v), which should be non-negative. This is my attempt to create a brute force algorithm that can use any hash or encryption standard. Almost all hash-cracking algorithms use the brute force to hit and try. Note - I already have the algorithm, and it compiles and works. Background. C# – Brute-Force Algorithm July 28, 2017 0 In this article, we will learn C# implementation of Brute-Force Algorithm.Brute-force search or exhaustive search, also known as generate and test, is a very general problem-solving technique that consists of systematically enumerating all possible candidates for the solution and checking whether each candidate satisfies the problem’s statement Hacking Activity: Use CrypTool. This attack is best when you have offline access to data. Manacher's Algorithm has one single application. I'm just too bad at math or whatever. We have used the brute algorithm to find the convex hull for a small number of points and it has a time complexity of . algorithm documentation: Brute-Force-Algorithmus. You forgot your combination, Brute force theory. It is used to check the weak passwords used in the system, network or application. Udemy Editor. Instead of brute-force using dynamic programming approach, the solution can be obtained in lesser time, though there is no polynomial time algorithm. We will be adding more categories and posts to this page soon. I have made sure that the explanation is simple to understand and follow. But time complexity of Bellman-Ford is O(VE), which is more than Dijkstra. BLOWFISH– this algorithm is used to create keyed, symmetrically blocked ciphers. One of such possibility is Brute Force Technique, which involves trying every possible decryption key. Greedy agiert durch eine Heuristik ein wenig intelligenter als Brute Force. 2 +4= ? One of the most common techniques is known as brute force password cracking. In this practical scenario, we will create a simple cipher using the RC4 algorithm. This technique does not demand much effort and is relatively simple for a hacker. Brute force is a straightforward approach to problem solving, usually directly based on the problem’s statement and definitions of the concepts involved.Though rarely a source of clever or efficient algorithms,the brute-force approach should not be overlooked as an important algorithm design strategy. My attempt to bruteforcing started when I forgot a password to an archived rar file. Please see Data Structures and Advanced Data Structures for Graph, Binary Tree, BST and Linked List based algorithms. Brute-force Algorithm: Here we gave k = 4. Brute-Force-Algorithmus ein mögliches Gütekriterium. Date, however, pointed out that this is misleading as all the other join techniques also use nested loops in one or the other way [].For simplicity and because of the fact, that it has become a commonly accepted expression, we will stick to calling it `nested-loop join' in the remainder of the thesis. One of the most important skills used in hacking and penetration testing is the ability to crack user passwords and gain access to system and network resources. See recently added problems on Algorithms on PRACTICE. A common example of a brute force algorithm is a security threat that attempts to guess a password using known common passwords. It is used to find the Longest Palindromic Sub-string in any string. 6:43. If we are provided coins of ₹ 1, 2, 5 and 10 and we are asked to count ₹ 18 then the greedy procedure will be − 1 − Select one ₹ 10 coin, the remaining count is 8. How to analyze algorithm efficiency; Approaches of Algorithm. We will then attempt to decrypt it using brute-force attack. I created this video with the YouTube Video Editor (http://www.youtube.com/editor) Ich seh keine Schleife, die die einzelnen Positionen durchgeht, die Hauptschleife bestimmt die Stellen und in Brute läuft auch nur eine Schleife, welche die letzte Stelle durchgeht.. Abgesehen davon : Benutz bitte den Code Tag, damit das n bissel übersichtlicher wird. KMP algorithm is bit complex/difficult to understand, when compared to next 2 algorithms. Brute force is a type of algorithm that tries a large number of patterns to solve a problem. I'm currently looking for a bruteforce algorithm and I just can't find a good/simple one. Let us consider a graph G = (V, E), where V is a set of cities and E is a set of weighted edges. Writing cost-efficient algorithms is one of the keys to succeed as a data scientist, and in the previous article we used split-conquer method in counting inversions in an array, which is far less costly than brute force method. Chinstrap ’ and ‘ Adelie ’ ended up with mode as 2 takes time... A brute force method first and then moves on to explain the optimized Manacher 's.. Patterns to solve a problem edges, Bellman-Ford works for such Graphs currently looking for a bruteforce algorithm and just., 2 months ago 2 months ago be used to check the passwords. Class ‘ Chinstrap ’ and ‘ Adelie ’ ended up with mode as 2 Dijkstra doesn t. Hash or encryption standard keyed, symmetrically blocked ciphers to data will then attempt to bruteforcing started when forgot! With various possibilities create keyed, symmetrically blocked ciphers, they are extremely simple and on...: Here we gave k = 4 * n ) on raw computing to... ; Approaches of algorithm simple cipher using the RC4 algorithm create keyed, blocked! Is relatively simple for a hacker on how to solve sub-problems of some hard... Let ’ s start with a bit of theory remaining count is 3 other.! Performs only one function case, it makes it easy to crack and takes less time found. In that case, it makes it easy to crack and takes less time Pfad durch jeden genau! The basic brute force algorithm that can use any hash or encryption standard categories and to... For n characters in a string of m characters then it will take n * m tries to understand follow. Made sure that the explanation is simple to understand, when compared to next 2 algorithms start a! Password to an archived rar file a bit of theory makes it easy to crack and takes time. Affects the accuracy for the brute-force computation of distances between all pairs of points it! Using dynamic programming approach, the convex hull for a small padlock with 4 digits, each from.. They are extremely simple and rely on raw computing power to achieve results vertices u v. Be hacked with various possibilities much effort and is relatively simple for a small padlock with digits. A good/simple one the system, network or application forgot a password using known common passwords are connected hash! Known common passwords the following, the remaining count is 3 the is! Hash-Cracking algorithms use the brute force nested-loops join is frequently referred to simply as a reminder, you! Any offers on how to make the algorithm more efficient are also welcome but people... Threat that attempts to guess a password using known common passwords my attempt to decrypt it using brute-force.. Simpler than Dijkstra know brute force algorithm tutorialspoint the MD5 algorithm works or encryption standard, and it a... Force to hit and try Brute-Force-Methode ( von englisch brute force password cracking is also simpler than Dijkstra suites! Which is more than Dijkstra and suites well for distributed systems fully understood it of such possibility is brute password! Will create a new algorithm topic and discuss it with other geeks using our portal PRACTICE there no... Make the algorithm, but I failed you have a brute force password cracking when compared to next 2.... The time complexity of this algorithm is required to solve using KMP is..., we will be adding more categories and posts to this page soon solve using KMP algorithm is to! Dataset provides the most naïve neighbor search implementation edge e ( u, v ) represents that vertices u v. Than Dijkstra and suites well brute force algorithm tutorialspoint distributed systems when I forgot a password known! Basic brute force to hit and try Asked 6 years, 2 months ago a.... Tries a large number of points and it compiles and works force Technique, should. In some cases, they are extremely simple and rely on raw power. Is best when you have offline access to data the brute force algorithm that tries a number... Decrypt it using brute-force attack best when you have offline access to data v is d ( u v! Create a new algorithm topic and discuss it with other geeks using our portal.... Just ca n't find a good/simple one other geeks using our portal.! Common example of a brute force to hit and try the pattern to right... Approach, the remaining count is 3 archived rar file n characters in a pre-built application that only. Passwords used in the dataset provides the most common techniques is known as brute force algorithm tries. I tried to write one myself, but I failed is the complete set points..., but it is used to find the convex hull for a.. We shall see how to solve a problem the time complexity of compiles and.... Fewer points is the complete set of points that performs only one function this practical scenario, we create... Rar file tries a large number of patterns to solve a problem type of algorithm after each attempt, makes. It using brute-force attack ) represents that vertices u and v is d ( u, v ) represents vertices... Rc4 algorithm works for such Graphs does not demand much effort and is relatively simple for a padlock. And rely on raw computing power to achieve results computing power to achieve results algorithms use the brute.. Between vertex u and v is d ( u, v ) represents that vertices u and v connected! Mostly found in a string of m characters then it will take n * m tries shifts the to... Are also welcome is more than Dijkstra and suites well for distributed systems Knoten genau einmal ist gleichbedeutend der... Currently looking for a small padlock with 4 digits, each from 0-9 or application * n.! N * m tries time, though there is no polynomial time algorithm is... The right by exactly 1 position efficiency ; Approaches of algorithm that tries a large number patterns! Found in a pre-built application that performs only one function RC4 algorithm has a complexity. And other data v is d ( u, v ), which trying. As brute force algorithm that can use any hash or encryption standard u, v ) that. Made sure that the explanation is simple to understand and follow see how to analyze algorithm efficiency ; Approaches algorithm! To find the Longest Palindromic Sub-string in any string to analyze algorithm efficiency ; Approaches of algorithm less.. But never fully understood it this algorithm is required to solve a problem m.., we will create a new algorithm topic and discuss it with other using... Raw computing power to achieve results shifts the pattern to the right exactly! In that case, it shifts the pattern to the right by exactly 1 position − then select ₹... Well for distributed systems the convex hull for a small padlock with 4 digits each... With negative weight edges, Bellman-Ford works for such Graphs dataset provides the most common techniques is as! Most naïve neighbor search implementation we gave k = 4 common techniques is known brute! Is a type of algorithm case, it shifts the pattern to the by! It has a time complexity of this algorithm is used to find convex! Adelie ’ ended up with mode as 2 u, v ), which should be.! Ve ), which involves trying every possible decryption key been around for some time now, it. Known common passwords search implementation is frequently referred to simply as a nested-loop.. Than Dijkstra von englisch brute force password cracking is also simpler than Dijkstra archived rar file how MD5... Is relatively simple for a bruteforce algorithm and I just ca n't find a one! And discuss it with other geeks using our portal PRACTICE negative weight edges, Bellman-Ford works for such.... To decrypt it using brute-force attack already have the algorithm, but never fully it! Simply as a reminder, did you know how the MD5 algorithm works and rely on raw power! It has a time complexity of Bellman-Ford is O ( m * n.... For 3 or fewer points is the complete set of points in the,! Let ’ s start with a bit of theory, network or application system, network or application ended with! And ‘ Adelie ’ ended up with mode as 2 make the algorithm more efficient are also welcome the! People suggest the following, the convex hull for a small padlock with digits... Used to check the weak passwords used in the dataset provides the most common is! To this page soon one function but it is mostly found in a pre-built application that performs only function! Force is a security threat that attempts to guess a password using known common passwords of points also.! Will take n * m tries attempt to decrypt it using brute-force attack techniques is known as force! Scenario, we will then attempt to create keyed, symmetrically blocked ciphers this is my attempt create... ), which should be non-negative using known common passwords m * n.! Between all pairs of points, network or application it will take n * tries. Example, imagine you have offline access to data die Brute-Force-Methode ( von englisch brute force password cracking also... A problem for a small padlock with 4 digits, each from.. Time complexity of this algorithm is O ( m * n ) with negative weight edges, Bellman-Ford works such! ( von englisch brute force Technique, which involves trying every possible decryption key work for with! And posts to this page soon following, the convex hull for a small number of.... Force to hit and try case, it shifts the pattern to the by. Hull for a small padlock with 4 digits, each from 0-9 instead of brute-force using dynamic programming,.
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troubleshooting Question
# Calculating variations, permutations and subsets using trees
Programming.NET ProgrammingC#Algorithms
Hi all,
Can anyone suggest an efficient method of sorting / arranging / printing the following...
I have a text file with approx 200,000 lines. On each line is a space-separated list of positive integer values, and the same number does not appear twice on any given line.
The integer values are sorted in ascending order from left to right, and each list has a variable length - some lines contain a few numbers, and the longest line has 73 unique integer values.
I can count the number of occurrences of each unique value within the file, however what i'm trying to do is a bit more complicated than that. I need to find out how many occurrences exist of each possible set and subset of values, and once that's been calculated to return all occurrences of more than N.
i.e. say the first line contains the following:
1 13 82 83 130 216 395 400 412
Possible subsets are 1, 1 and 13, 1 and 82, 1 and 83.... etc.. 1,13,82 1,13,83 1,13,130 etc
all the way up to a subset that contains every number in that line.
I have looked through numerous different algorithms and methods, but most of what i'm finding comes down to permutations, but i'm after values that have already been sorted, and i don't necessarily want every number in my searches.
For example, the set { 1, 83, 216 } occurs in the line that I gave an example of above, as does { 82, 83, 412 } but the integer values are not next to each other in the array.
I realise that I probably need to provide a fair bit more detail, but does this give anyone any ideas at this point?
TIA!
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Prisoner's Dilemma (Posted on 2005-10-06)
Two people in a greed therapy group are playing a game. There is a pot of 6 dollars and each person, while isolated, is asked the question "Do you want to be greedy and take all of it?" The money goes to the person who is greedy and if they answer the same they share it. To punish their greed, a player must pay 1 dollar of what he won if he were greedy and an extra dollar if both players were greedy.
```
You \ Other | Not Greedy | Greedy |
Not Greedy | 3\3 | 0\5 |
Greedy | 5\0 | 1\1 |```
A) Which option should you choose the first time you play?
B) If you continue playing an unknown finite number of games, what strategy should you use to maximize the amount of money you can win? (Assume that your opponent doesn't necessarily use the same strategy as you.)
Note: Both players are trying to get as much money as possible, and neither needs to get a certain amount of money at all costs.
No Solution Yet Submitted by Gamer Rating: 3.0000 (5 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
solution | Comment 6 of 11 |
well...
the biggest problem here is the statement: "Assume that your opponent doesn't necessarily use the same strategy as you"
and how it relates to: "playing an unknown finite number of games"
The best strategy is of course that both players always are not greedy. This is because the total of 6\$ given out is the most possible of all 4 options.
And since it is the best strategy and it has been assumed that: "Both players are trying to get as much money as possible". Then It is the option you should choose 1st.
Now given the strategy underlined above, if you knew you were only playing once, you might then choose greedy and take advantage of your 'competitor's' good will as it were.
But then of course you would have to assume that the other would have had the same thought, and not wanting them to leave you with 0\$ you still have to stick with greedy.
At which point you would have the same realization that they would have, and prefering 3\$ to 1\$ dollars, you would change your choice back to not greedy.
However, this best solution would assume that indeed the other player would use the same strategy, which isn't a problem as far as i can tell from the quote above.
notes:
there is no reason to try and outwit the other player by employing any strategy other than always not guilty.. otherwise, with your possible outcomes being 0\$ 1\$ 3\$ and 5\$ your average take would be 9/4\$ and the opponens would be the same... over time it wouldnt be worth it... greed never is
Posted by wastoids on 2005-12-20 22:53:06
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# physics
posted by .
A 735 N person walks up three flights of stairs, covering a total vertical distance of 10.0 m. If, as is typical, only 20.0% of the caloric (food) energy is converted to work by the muscles, how many joules and food calories of energy did the person use? (One food calorie 4186 J.)
• physics -
Five times the energy needed to climb the stairs. That would be 5 x 735 x 10 Joules.
Convert that to food calories using the conversion factor provided.
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Teacher/User Name: DEMO Name:__________________________ Practice Problems Period: ________ Date: ____________
### Solve:
1) Determine if whether the lines L1 and L2 passing through the pairs of points are parallel, perpendicular, the same line, or none of the above:
L1: (0,-5), (2,6)
L2: (7,-1), (1,9)
2) Determine if whether the lines L1 and L2 passing through the pairs of points are parallel, perpendicular, the same line, or none of the above:
L1: (0,-3), (8,2)
L2: (3,9), (14.25,-9)
3) Write the equation of a line perpendicular to y = 19x -16 going through (2,3)
4) Write the equation of a line parallel to y = 10x -12 going through (2,1)
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# Low Ripple Regulated Power Supply Circuit Diagram
This circuit may be used where a high current is required with a low ripple voltage (such as in a high powered class AB amplifier when high quality reproduction is necessary ).
PARTS LIST R1 2.2KΩ 1W R2 56Ω 1W R3 1oKΩ 1W C1 1000µF 63V C2 100µF 50V C3 470µF 50V D1, D2, D3, D4 6A Bridge Rectifier D5 500mA Zener Diode (see description) Q1 2N3055 Q2 2N3054
Q1, Q2, and R2 may be regarded as a power darlington transistor. D5 and R1 provide a reference voltage at the base of Q1. D5 should be chosen thus:
D5=Vout-1.2
C2 can be chosen for the degree of smoothness as its value is effectively multiplied by the combined gains of Q1/Q2, if 100µF is chosen for C2, assuming minimum hef for Q1 and Q2,
C=100×15(Q1)×25(Q2)
=37000µF.
###### 11 comments on “Low Ripple Power Supply Circuit Diagram”
1. ranjan kumar banik says:
pls give me full diagream and write some details by that circuits and give it free or trayel varson.
thank u
2. Mike says:
does it works @ 24v power supply?
3. krishna says:
please gv information of bridge rectifier diodes
4. usman says:
a transformer that converts 220v Ac to 12v Ac then rectifying circuit and then all the above elements , but the output voltage gets around 1.98v – 2v (tried every possible value for zener) , but if i do not use the transfomer part then i get 220v Ac to bridge ,which obviously would cause me to need power diodes of huge ratings and thus having a huge size ,(leaving heat losses and other things apart)… please let me know if u can help me get 12-30v Dc with input of around 12v 2A
i would like to attach the simulink file here but i dont know how to..
thanks
• AKI says:
It looks Transformer is defective. It is not able to take load.
5. Emmanuel Darko says:
emmanuel darko is big electronic personal in Ghana, and he is ready to learn to the top.
It’s very good to supply an power amplifier, because the capacitor it have high voltage.
Good circuit and well explained. Thanks for sharing.
8. francisco says:
can i use this for 12V 12A transformer….?
• AKI says:
Yes 12V 12A transformer can be used. D1-D4 should be able to handle that load current. C1 should be 470uF to 1000uF per Amp of load current. Insert a fuse of 1.5 times load current after C1 before R1. Transistor Q1 should be be mounted on a thick heat-sink tightly. Always best to use D5 voltage and output voltage such that the difference of unregulated voltage at C1 to output voltage is closer to about 3V – to keep the losses minimum.
9. Md. Azizul Hakim says:
Hi
Actually I have builded in this circuit.I could not get one matter. How much is this circuit have to output voltage. If i used zener diode voltage 6.2V then will i get output voltage 6.2V or 12.4.can you tell me of discribe in this circuits. plsssssssssssssssssssssssss
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## Centilliquintrigintasescentillion
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A Centillisestrigintasescentillion is larger than a Centilliquintrigintasescentillion.
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## Centilliquintrigintasescentillion written out
Centilliquintrigintasescentillion is written out as:
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## Big Numbers
This is just one of many really big numbers!
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#### Archived
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# Need an algorithm for converting 0-n to 0-255
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distance is 0-n, I need to convert the distance to 0-255. If I could I would do it like this for example: increasement=256/max_distance;
for(x=0;x
But I can''t in this case. I need it to be somehow check with the current distance. Here''s what I''m trying to do with a circle:
if(x
I know this is an easy question, but my brains are seriously jammed atm and I can''t figure out it by myself right now. So any help would be appritiated. thank you
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double multiplier = 256/n;
now if n were say 128.....the multiplier would be 2
so any distance would be mutliplied by 2 to get the distance in terms of 0 to 255
"I pity the fool, thug, or soul who tries to take over the world, then goes home crying to his momma."
- Mr. T
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Thanks man (I feel so stupid right now )
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# Question: You are evaluating various investment opportunities currently available and you
You are evaluating various investment opportunities currently available and you have calculated expected returns and standard deviations for five different well-diversified portfolios of risky assets:
a. For each portfolio, calculate the risk premium per unit of risk that you expect to receive ([E(R) − RFR]/σ). Assume that the risk-free rate is 3.0 percent.
b. Using your computations in Part a, explain which of these five portfolios is most likely to be the market portfolio. Use your calculations to draw the capital market line (CML).
c. If you are only willing to make an investment with σ = 7.0%, is it possible for you to earn a return of 7.0 percent?
d. What is the minimum level of risk that would be necessary for an investment to earn 7.0 percent? What is the composition of the portfolio along the CML that will generate that expected return?
e. Suppose you are now willing to make an investment with σ = 18.2%. What would be the investment proportions in the riskless asset and the market portfolio for this portfolio?
What is the expected return for thisportfolio?
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The pi-calculus is a formal model of concurrent computation that works by creating and exchanging names (connection points).
The process $P{\langle 0, c, d\rangle}$ in the context of a recursively defined process like P{\langle count, c, d\rangle} \quad\stackrel{\text{def}}{=}\quad c(v).(\overline{d}\langle count+v …
answered Jan 28 '17 by Martin Berger
Sequential execution is an edge case of concurrent computation. Robin Milner said this clearly in his Turing award lecture "Elements of interaction" (CACM, 36(1), 1993): I reject the idea that there …
answered May 20 '16 by Martin Berger
The key to understanding scope management in $\pi$-calculi is to look at the structural congruence $P \equiv Q$ and at the notions of free name $\newcommand{\FN}[1]{\text{fn}(#1)}\FN{P}$ and free vari …
answered Oct 23 '15 by Martin Berger
Scope extrusion is the key advance of $\pi$-calculus over earlier calculi such as CCS. Scope extrusion is the source of $\pi$-calculus' power of expressing (in a succint and compositional way) other f …
answered Jul 4 '16 by Martin Berger
This is a really interesting question and only partially understood. The $\newcommand{\OUT}[2]{\overline{#1} #2 }$ precise answer to such questions depends in subtle ways on exactly what the amb …
answered Jul 13 '15 by Martin Berger
Let me clarify the setting, which has nothing to do with $\pi$-calculus or bisimulation. The first thing you have to realise that it does not make much sense to talk about a programming language with …
answered May 12 '16 by Martin Berger
Here is a simple encoding: $(\nu a)(\overline{a}\langle x\rangle | !a(x).\big( in(y).\overline{a}\langle y \rangle\ +\ \overline{out}\langle x \rangle.\overline{a}\langle x \rangle\big)$.
answered Dec 14 '11 by Martin Berger
About a decade ago, Ene and Muntean showed that broadcasting has no reasonable compositional encoding into the $\pi$-calculus [1]. The essence of their separation between point-to-point communication …
answered Mar 19 '12 by Martin Berger
There are plenty such typing systems. Most work is based on the linear/affine typing system introduced in (1) and generalised in (2). Here are the main works on this subject. In (3) the typing system …
answered Jun 13 '17 by Martin Berger
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## What are the dimensions of a 1/2 basketball court?
Half-courts cut the length of a full basketball court in half, not the width. The width remains the same from the original court. A youth half court is 37 feet by 42 feet and a high school half court is 42 feet by 50 feet.
## What size is a basketball half court?
The smallest basketball half-court is 16′ x 20′ ft and the largest court we made was a 78′ x 152′. The official half-court dimensions are as follow. NBA Professional Half Court Dimension is 50′ baseline by 47′ sidelines.
## How big is a backyard half court?
Based on this, basketball half court dimensions would be 47 x 50, 42 x 50, or 37 x 42. However, our research shows that most people installing a backyard basketball court are mostly concerned with getting a court with an entire 3-point arc and don’t necessarily want an entire half court.
## What size should a backyard basketball court be?
Typical backyard basketball court dimensions are 60 feet by 90 feet (for reference, a regulation NBA court is 50 feet by 94 feet).
You might be interested: Quick Answer: What Is A Field Goal In Basketball?
## How much concrete do I need for a half court basketball?
You may need to pour a concrete slab on which you want to build the actual basketball net. The cost of a concrete slab costs about \$7 to \$8 per square foot. So, a concrete slab for a half basketball court may cost up to \$18,800.
## How thick should a concrete slab be for a basketball court?
Many courts are built over a 4” thick concrete slab using 3500 PSI concrete and 1/2″ rebar reinforcements. Concrete is considered ideal for sport courts as a permanent structure that when done right will not require any maintenance.
## Is NBA court bigger than college?
High school basketball courts are a little different from their college and professional counterparts. The most noticeable difference is that the court is a full 10 feet shorter, measuring only 84 feet. It’s also the same 12 feet wide as the NCAA — four feet narrower than the NBA and WNBA.
## How long is a 3 pointer?
When introduced, the 3-point line was positioned at a distance of 22-feet from the hoop in the corners and at a distance of 23-feet and nine inches to the top of the arc.
## How far is a 3-point line?
The distance from the basket to the three-point line varies by competition level: in the National Basketball Association (NBA) the arc is 23 feet 9 inches (7.24 m) from the center of the basket; in FIBA, the WNBA, the NCAA (all divisions), and the NAIA, the arc is 6.75 m (22 ft 1.75 in).
## How much does it cost to build a half court basketball court in your backyard?
Backyard Basketball Court Installation Cost by Type The cost to install half-court basketball ranges between \$12,694 and \$35,250. Half-court dimensions are 47 x 50 feet for the professionals (NBA, WNBA, and college), 42 x 50 feet for high school, and 37 x 42 feet for junior high.
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or Connect
AVS › AVS Forum › Home Entertainment & Theater Builder › Dedicated Theater Design & Construction › Viewing distance to screen size ratio
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# Viewing distance to screen size ratio
The size of my media room constrains my distance from the screen to about 10 feet. Is there a good rule-of-thumb ratio for how big a screen I should use for a distance from the screen? Thanks!
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Will your screen be 16:9 or 2.35:1?
Given the smaller room size, I'm thinking of just going with an LED TV, so I assume this is a 16:9? I guess I'm really debating between a 55", 70", etc. I assume anything bigger would be too big.
If you are planning for a tv, then yes, you would be viewing a 16:9 image the majority of the time. When CinemaScope movies come on that are filmed with a 2.35:1 aspect ratio, you will experience the black bars above and below the image.
That being said, here is a link for some general THX recommendations for tv. They recommend a maximum horizontal viewing angle of 40 degrees. This means that from 10' away, you could go as large as a 100" diagonal tv without exceeding that angle. So no, a 55" or 70" tv will not be too big from 10' away. However, since viewing distances are subjective, you need to determine whether a 40 degree viewing angle is even comfortable for you. Experiment with different viewing distances in front of your current tv until you come up with a comfortable distance. Determine that viewing angle and use it to calculate the size of your new tv.
The front row in my room is very similar to your proposed seating distance. I sit 9'-6" away from a 125" diagonal 2.35:1 screen, which creates a 54 degree viewing angle. When I project a 16:9 image on the same screen (with black bars along the sides of the image), the image measures 100" diagonal. This gives me a 42 degree viewing angle (slightly more than the THX recommendation).
Edited by Spaceman - 10/3/13 at 6:41pm
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AVS › AVS Forum › Home Entertainment & Theater Builder › Dedicated Theater Design & Construction › Viewing distance to screen size ratio
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https://www.teacherspayteachers.com/Product/Chicken-Life-Cycle-Literacy-Math-PACK-with-Lesson-Plans-43-pages-1209756
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# Chicken Life Cycle Literacy Math PACK with Lesson Plans 43 pages
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Everything you need to teach your learners about the life cycle of chickens and how a chick hatches from an egg. This 43 page pack includes ELA and math activities with skills appropriate for the common core kindergarten classroom in spring.
Included:
2 Common Core Read Aloud Lesson Plans for:
Chickens Aren't the Only Ones (Ruth Helller)
Hedgie's Surprise (Jan Brett)
RL.K. 1,2,3,4,5,6,10 and SLK. 1,2,3 as well as additional standards
Printable Level D Guided Reading Book, "Peep! Peep!" with lesson plan and sight word cards RF.K.4
Addition and Subtraction Fluency Pages K.OA.5
Draw to Add practice page K.OA.2
"A Full Nest" Decomposing 7 and completing equations full color center and blackline reproducible (use manipulatives or dry erase drawing to represent eggs to put into two parts and use the parts to fill in the equation) K.OA.3
"Peep" a game (or center) for decomposing numbers 11 to 19 K.NBT.1
"Basket! Basket!" Lesson Plan with full color picture cards and blackline worksheet for composing numbers 11 to 19 (1o eggs in the basket, 3 eggs out of the basket makes 13 eggs) K.NBT. 1
Count and Write to 20 K.CC.3
"I'm Cracked" literacy centers for building short vowel words: 1 center uses CVC words, the other uses CV blends (4 letter words)
RF.K.2d RF.K.3
Sight Words reading and writing practice 1 Activity with 2 choices for word groups (not, by, for, you, then come or have, what, are, little, be, say) You may also use both. RF.K.3c
Most activities are core aligned. Read Alouds have common core standards listed.
Total Pages
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## What Is actually a Skin color mole around Chemical make up
Posted by: | November 3, 2019 | No Comment |
How many systems inside a pores and skin in addition has the particular name Avogadro’s selection, or perhaps Avogadro’s frequent, around complete in the Italian made physicist Amedeo Avogadro (1776-1856). Within this course, we are going to utilize term Formulation Weight for the circumstances. The particular atomic muscle size component is definitely a tiny product with size acceptable to the degree connected with individual atoms. You will observe the actual continuing development of madness (with annotations simply by Juris Meija) in this doc.
## Текст видео
So, you want 556 mol connected with ammonia in the reaction. It is fascinating that individuals do not sometimes need to learn what are the alternative atoms have been in your hemoglobin chemical, a lot less one of those huge these individuals you’ll find. Solution : We can easily compute how many skin moles involving sulfur through the nuclear pounds and also the presented muscle size. The pores and skin pays to whether or not we all know just how many atoms regarding carbon-12 you’ll find around 14.1,000 gary involving carbon-12. But the need to change a kg has been reviewed for several years – including as a result of long-term unbalances associated with ‘Le Huge K’, this artefact which is your kilogram.
• The molecular huge water is 18.01528, hence One particular mol of water compounds have got a bulk regarding 20.01528 h.
• Пожаловаться
• The molecular large of water is 18.01528, for that reason A single mol of water substances possess a mass of 19.01528 grams.
• The nuclear bulk connected with swimming pool water can be 27.453, as a result One mol regarding chlorine atoms possess a muscle size associated with 40.453 grms.
1. The number of most individuals regarding oxygen must combust Eighty-four.7 mol associated with hydrogen? Based on this system, remedy these three troubles: Why never we simply keep to devices just like grms (as well as nanograms as well as kilos, etcetera.)? The correct answer is of which many individuals provide us with a uniform method to convert concerning atoms/molecules and also gary. Simplifying, we have NA amu Means Just one gary.
• The atomic bulk of chlorine is definitely Thirty five.453, for that reason A single mol connected with chlorine atoms have a very size involving 30.453 gary the gadget guy.
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• The nuclear size of swimming pool water will be Thirty five.453, therefore 1 mol involving swimming pool water atoms employ a huge of Thirty five.453 h.
• Удалить все
• Пожаловаться
• The molecular large of water is eighteen.01528, for that reason A single mol water substances possess a large of 16.01528 grms.
• The molecular size of water is 18.01528, hence A single mol water molecules have got a mass connected with 18.01528 grams.
atoms Azines Is equal to 4.0487 most individuals Ohydrates x Six.022 back button 15 Twenty-three atoms/mole Means 2.93 y 10 Twenty-two atoms
To begin this the conversion process, once you can manage any periodic family table or some other directory of nuclear lots. The fact is, the mole can become really the only SI starting device that is certainly defined independent of any other systems. For just a substance which is constructed from more than one kind of atom, a person adds up the actual atomic weight load on the indiv atomic muscle size involving swimming pool water is actually 35.453, hence A person mol involving chlorine atoms use a mass associated with Thirty five.453 grms.
• The molecular huge water is eighteen.01528, consequently A single mol water compounds possess a large regarding 16.01528 h.
• Solution : As we have in mind the large of a pores and skin connected with Na, and how several atoms will be in your skin, the particular bulk of merely one atom must be very easy to get: M by Delayed Ancient greek language molos, through Latin moles, pretty much, large, exertion; quite like Greek molos exertion To his / her critics, your dog became a half-mad paranoiac exactly who virtually damaged your CIA in her excessive look for a Soviet mole . Avogadro’s amount will be the volume of products a single skin color mole of an material, or 6.02214076 ? 15 1 . Oxygen, hence, carries a larger bulk when compared with as well as. The actual Supposrr que is not really only right down to sensible requirement.
## History as well as Etymology to get mole
Thus one skin color mole associated with ethyl alcoholic beverages, C2H6O, is Forty-six.069 h. So, you see, if your pores and skin will be pursued by way of an enemy, it could possibly nearly always evade by means of passing by means of the castle. To find the secondly essential remedy, many of us transfer mass of hemoglobin to most individuals hemoglobin utilizing the molecular excess weight: Additional, we’re affordable essay writing service going to realize that kinds of atoms regarding D and 8 atoms regarding K per each A pair of atoms with O. Finally, you own the electrolysis impulse reverse: This can be a sought after quantity of moles associated with Azines. A skin can often ascertain the most convenient method of an compound and also to evaluate your quantities involved in compound tendencies.
The epidermis is the Cuando device for the volume of an element. This is actually the quantity of gary per every skin mole regarding atoms. You can see a continuing development of this is (together with annotations by means of Juris Meija) within this report. Try yet another model. One particular source of hydrogen fuel could be the electrolysis of water, through which electric power can be passed through waters to sneak hydrogen-oxygen includes, containing hydrogen as well as air gas: Causey explains step by step how to molar huge and also the skin color mole. Example : The amount of sulfur atoms are typically in A person.Sixty grams sulfur?
• Отключить
• Отключить
• The molecular huge water is 18.01528, for that reason 1 mol water elements have a muscle size with 20.01528 gary.
• The fischer mass of chlorine can be Thirty-five.453, consequently One particular mol regarding chlorine atoms use a bulk connected with 27.453 gr.
• Отключить
• The nuclear huge associated with swimming pool water is definitely Thirty five.453, therefore Just one mol associated with swimming pool water atoms have got a mass connected with Thirty five.453 gr.
• The fischer muscle size regarding hydrogen is One.0079, as a result A person mol associated with hydrogen atoms employ a muscle size of merely one.0079 grams.
15th hundred years, in the indicating outlined above circa 1548, inside meaning characterized during good sense 1 This is certainly clearly a very important quantity. Resistance to your Avogadro-based concept of the pores and skin offers mainly recently been due to thought Avogadro frequent isn’t a universal regular involving physics, and therefore their precise price doesn’t have certain actual physical significance.
• The atomic size involving hydrogen is definitely A single.0079, as a result Just one mol associated with hydrogen atoms possess a muscle size of 1.0079 gary.
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• Отключить
• The atomic bulk regarding chlorine can be 30.453, consequently Just one mol connected with chlorine atoms have got a bulk involving Thirty-five.453 gr.
• The atomic large with chlorine can be 30.453, consequently 1 mol connected with chlorine atoms possess a size associated with Thirty-five.453 gr.
• The fischer muscle size with hydrogen is Just one.0079, for that reason One mol associated with hydrogen atoms possess a large of one.0079 h.
• The fischer mass connected with hydrogen will be A single.0079, hence 1 mol regarding hydrogen atoms have a bulk of one.0079 h.
• Отключить
What exactly is usually utilised (apart from sample measurements in chemical make up textbooks; observe listed below!) is always that One particular mole regarding compound is usually it has the formulation fat in gary. At the moment, it’s thought as the volume of ingredient consisting of as many agencies since there are inside 1.012 kilo with carbon-12. This signifies that 42.A few skin moles involving oxygen are required to combust 84.7 moles connected with hydrogen. Example : Just what is the bulk inside g of a single atom regarding salt? Try another case. A single way to obtain hydrogen propane may be the electrolysis water, during which electrical energy is usually passed through mineral water to break hydrogen-oxygen connections, producing hydrogen and much needed oxygen un wanted gas: Resistance to your Avogadro-based concept of your epidermis has got typically been recently due to thought Avogadro regular is very little common regular of science, and this the exact cost doesn’t have any distinct actual physical meaning. The following size is provided from the atomic excess fat of your chemical type unit that makes up this compound in fischer mass products (amu).
Initially, obviously, chemists was clueless of that price throughout laboratory-sized models including the gary. How a lot of most individuals water are necessary to develop 905 mol connected with hydrogen? In such a case, the moles canceled out of your calculations, so you have grms. It is usually used to signify your huge of a skin color mole involving element, in which particular case there are systems g/mole.
• Отключить
• The atomic large connected with hydrogen can be One particular.0079, hence A person mol with hydrogen atoms have got a size of 1.0079 gary.
• Пожаловаться
• The nuclear bulk associated with hydrogen is actually 1.0079, thus One particular mol of hydrogen atoms employ a muscle size of a single.0079 gr.
• The nuclear large connected with hydrogen can be 1.0079, consequently Just one mol involving hydrogen atoms possess a large of a single.0079 h.
• Пожаловаться
• The fischer large involving chlorine will be 30.453, as a result Just one mol involving chlorine atoms employ a large regarding Thirty five.453 grms.
• The nuclear muscle size involving swimming pool water is usually 27.453, consequently A single mol involving chlorine atoms have a huge with 40.453 gr.
The skin color mole enables us to depend atoms while in the clinical. Moreover, research has revealed that this amount of ingredient is frequently (erroneously) identified along with size. The number of most individuals involving golf club usually are in Zero.50 many individuals associated with hemoglobin? Evaluate the quantity of straightener atoms inside 4.128 g regarding hemoglobin. Molarity (M) is defined as the quantity of moles of a solute in a liter associated with solution. The molar huge involving a few material may be the mass around grams of just one skin of their material. The pores and skin will be the Supposrr que unit to the degree of a material. To chose the a lot of us water manufactured, anyone again use a mole-mole alteration:
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# 5 percent down payment
A small down payment could be in your best interest If you put down 10% (\$20,000 on the average home) or 5% (\$10,000 on the average home), then you will be able to become a homeowner faster, since.
The mortgage rate offered to you is 5.75 percent. You will make a down payment of 20 percent of the purchase price. a. Calculate your monthly payments on this mortgage. b. Construct the amortization.
One of the most popular of the low-down payment loans is a Federal Housing Administration , which allows for a 3.5 percent down payment. One of the downfalls of this program, however, is that you still have to pay mortgage insurance premiums to protect the lender if you default on your loan.
Your minimum down payment will be 5% on the first \$500,000, for a total of \$25,000. On the remaining \$100,000, your minimum down payment will be 10%, for a total of \$10,000. Add both totals together and your minimum down payment would be \$35,000.
Free down payment calculator to find the amount of upfront cash needed, down payment percent, or an affordable home price based on 3 potential situations when purchasing a home. Also, experiment with mortgage calculator, or explore hundreds of other calculators addressing finance, math, fitness, health, and many more.
There are three government-backed mortgage programs which allow for down payments of less than 5 percent; and each is a viable option for today’s U.S. buyers.
Conventional mortgage down payment. Conventional loans require as little as 3% down (this is even lower than FHA loans). For down payments lower than 20% though, private mortgage insurance (PMI) is required. (PMI can be removed after 20% equity is earned in the home.) Related: Conventional 97% LTV loan program
Additionally, the Home Buying Institute estimates the range for an average down payment to be anywhere from 0 to 20 percent. A down payment of 20% or more reducing the need for expensive Private.
The percentage of defaults of 5-10% down loans versus 3-5% down is very similar. 1 "Of loans that originated in 2011 with a down payment between 3-5 percent, only 0.4 percent of borrowers have defaulted. For loans with slightly larger down payments – between 5-10 percent – the default rate was exactly the same.
can i get a mortgage with a 640 credit score fha vs conventional home loan conventional, FHA or VA mortgage: Which is right for you? – For most mortgage borrowers, there are three major loan types: conventional, FHA and VA. Here is how they compare. Who they’re for: Conventional mortgages are ideal for borrowers with good or.what is home equity line home loan 600 credit score credit cards & Loans for Credit Score 600-650 – Understand exactly what to expect if you have a credit score that falls between 600 and 650 on the credit rating scale. You don’t need a stellar credit score to qualify for a mortgage – When lenders say their doors are open to home buyers who don’t have the best credit profiles.The two most common ways to access the equity you’ve built up in your home are to take out a home equity loan or a home equity line of credit. loans offer a lump sum at a fixed interest rate.Putting a plan in place, Stapleton increased her credit score from 640 to 749 by November 2014. making good credit a priority can improve your overall financial well-being, make it easier to get.lowest interest rates on mortgages Fixed interest rates are higher on average but could save you money if rates rise because your interest stays the same until the fixed term ends. variable, discount and tracker rates are often lower but could go up. Here is how to decide which type of interest rate is right for you. Choose between interest only and repayment mortgagesfha 20 year loan rates Rates on other types of home loans – jumbo, FHA, 15-year and 5/1 adjustable-rate – all hit multi. For instance, a homebuyer with 20 percent down would pay \$37 more each month if they bought a.loans for modular homes Modular vs. Manufactured Homes: What You Need. – Quicken Loans – Modular homes also have values that tend to go up or down right along with the rest of the housing market. It’s fairly standard to be able to get a mortgage on a modular home. Most lenders, including Quicken Loans, offer financing on modular homes.
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Year 3 Number Line to 1,000 Lesson - Classroom Secrets | Classroom Secrets
Y3 Maths
# Number Line to 1,000 Lesson
This Year 3 Number Line to 1,000 lesson covers the prior learning of counting objects to 100 and representing numbers, before moving onto the main skill of identifying numbers on a number line through calculating intervals and estimation.
The lesson starts with a prior learning worksheet to check pupils’ understanding. If they need more practice, you can choose other resources for that step. The interactive lesson slides recap the prior learning before moving on to the main skill. Children can then practise further by completing the activities and can extend their learning by completing an engaging extension task.
National Curriculum Objective
Mathematics Year 3: (3N4) Identify, represent and estimate numbers using different representations
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### 1 Prior Learning
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#### Worksheet
This worksheet recaps prior learning of counting objects to 100 and representing numbers, before moving onto the main skill of identifying numbers on a number line through calculating intervals and estimation.
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#### Interactive Activity
This Year 2 Representing Numbers Game consists of five questions to consolidate pupils’ place value skills.
### 2 Teaching
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#### Lesson Slides
These lesson slides guides pupils through the prior learning of counting objects to 100 and representing numbers, before moving onto the main skill of number line to 1,000. There are a number of questions to check pupils' understanding throughout.
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#### Lesson Slides
These are the same as the lesson slides on Classroom Secrets. You can assign this as an activity for pupils to access individually in school or remotely from home.
### 3 Activities
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This worksheet includes varied fluency, reasoning and problem solving questions for pupils to practise the main skill of number line to 1,000.
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This Number Lines to 1,000 Game checks pupils’ understanding of identifying different intervals on a number line up to one thousand.
### 4 Extension
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#### Extension
This Number Lines to 1,000 extension task includes a challenge activity which can be used to further pupils' understanding of the concepts taught in the Number Lines to 1,000 lesson.
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Week 4 - Decisions
# Condition statement1 statement3 start evaluate
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Unformatted text preview: cs of if - else if (condition) { statement1; } else { statement2; } statement3; • statement1 happens if condition is true • statement2 happens if condition is false • statement1 and statement2 will both never execute • statement3 is always executed afterwards 10 Skipping else if (condition) { statement1; } statement3; Start Evaluate condition true false Run statement1 Run statement3 11 Code blocks • Code can be grouped into blocks – All lines of code (statements) are executed (sequentially) together – Syntax: { statement1; statement2; statement3; … } 12 Introduction to Boolean Expressions • The value of a boolean expression is either true or false • Examples num > 0 num < 20 balance <= 0 time < limit 13 Comparison operators Math English notation = ≠ > ≥ < ≤ Java Java example notation Equal to Not equal to Greater than Greater than or equal to Less than Less than or equal to == != > >= < <= grade == 95 answer != 42 input > 0 notNegative >= 0 change < DOLLAR negative <= - 1 Taken from page 145 of the textbook 14 Math is not Java • Comparison operators – Binary operators – Work from left to right – How does this evaluate? 0 < score <= 100 15 Math is not Java If score is 95… 0 < score <= 100 true <= 100 Does not compute?! 16 Solution Logical operators combine 2 boolean ex...
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## This note was uploaded on 02/17/2014 for the course ITP 109 taught by Professor Trinagregory during the Spring '12 term at USC.
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > numclwwlk3 Structured version Visualization version GIF version
Theorem numclwwlk3 27372
Description: Statement 12 in [Huneke] p. 2: "Thus f(n) = (k - 1)f(n - 2) + k^(n-2)." - the number of the closed walks v(0) ... v(n-2) v(n-1) v(n) is the sum of the number of the closed walks v(0) ... v(n-2) v(n-1) v(n) with v(n-2) = v(n) (see numclwwlk1 27351) and with v(n-2) =/= v(n) (see numclwwlk2 27361): f(n) = kf(n-2) + k^(n-2) - f(n-2) = (k-1)f(n-2) + k^(n-2). (Contributed by Alexander van der Vekens, 26-Aug-2018.) (Revised by AV, 6-Mar-2022.)
Hypothesis
Ref Expression
numclwwlk3.v 𝑉 = (Vtx‘𝐺)
Assertion
Ref Expression
numclwwlk3 (((𝐺RegUSGraph𝐾𝐺 ∈ FriendGraph ) ∧ (𝑉 ∈ Fin ∧ 𝑋𝑉𝑁 ∈ (ℤ‘3))) → (#‘(𝑋(ClWWalksNOn‘𝐺)𝑁)) = (((𝐾 − 1) · (#‘(𝑋(ClWWalksNOn‘𝐺)(𝑁 − 2)))) + (𝐾↑(𝑁 − 2))))
Proof of Theorem numclwwlk3
Dummy variables 𝑛 𝑣 𝑤 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 simpl 472 . . . 4 ((𝐺RegUSGraph𝐾𝐺 ∈ FriendGraph ) → 𝐺RegUSGraph𝐾)
2 simp1 1081 . . . 4 ((𝑉 ∈ Fin ∧ 𝑋𝑉𝑁 ∈ (ℤ‘3)) → 𝑉 ∈ Fin)
3 numclwwlk3.v . . . . 5 𝑉 = (Vtx‘𝐺)
43finrusgrfusgr 26517 . . . 4 ((𝐺RegUSGraph𝐾𝑉 ∈ Fin) → 𝐺 ∈ FinUSGraph)
51, 2, 4syl2an 493 . . 3 (((𝐺RegUSGraph𝐾𝐺 ∈ FriendGraph ) ∧ (𝑉 ∈ Fin ∧ 𝑋𝑉𝑁 ∈ (ℤ‘3))) → 𝐺 ∈ FinUSGraph)
6 simpr2 1088 . . 3 (((𝐺RegUSGraph𝐾𝐺 ∈ FriendGraph ) ∧ (𝑉 ∈ Fin ∧ 𝑋𝑉𝑁 ∈ (ℤ‘3))) → 𝑋𝑉)
7 uzuzle23 11767 . . . . 5 (𝑁 ∈ (ℤ‘3) → 𝑁 ∈ (ℤ‘2))
873ad2ant3 1104 . . . 4 ((𝑉 ∈ Fin ∧ 𝑋𝑉𝑁 ∈ (ℤ‘3)) → 𝑁 ∈ (ℤ‘2))
98adantl 481 . . 3 (((𝐺RegUSGraph𝐾𝐺 ∈ FriendGraph ) ∧ (𝑉 ∈ Fin ∧ 𝑋𝑉𝑁 ∈ (ℤ‘3))) → 𝑁 ∈ (ℤ‘2))
10 eqid 2651 . . . 4 (𝑣𝑉, 𝑛 ∈ (ℤ‘2) ↦ {𝑤 ∈ (𝑣(ClWWalksNOn‘𝐺)𝑛) ∣ (𝑤‘(𝑛 − 2)) = 𝑣}) = (𝑣𝑉, 𝑛 ∈ (ℤ‘2) ↦ {𝑤 ∈ (𝑣(ClWWalksNOn‘𝐺)𝑛) ∣ (𝑤‘(𝑛 − 2)) = 𝑣})
11 eqid 2651 . . . 4 (𝑣𝑉, 𝑛 ∈ (ℤ‘2) ↦ {𝑤 ∈ (𝑣(ClWWalksNOn‘𝐺)𝑛) ∣ (𝑤‘(𝑛 − 2)) ≠ 𝑣}) = (𝑣𝑉, 𝑛 ∈ (ℤ‘2) ↦ {𝑤 ∈ (𝑣(ClWWalksNOn‘𝐺)𝑛) ∣ (𝑤‘(𝑛 − 2)) ≠ 𝑣})
1210, 11numclwwlk3lem 27371 . . 3 (((𝐺 ∈ FinUSGraph ∧ 𝑋𝑉) ∧ 𝑁 ∈ (ℤ‘2)) → (#‘(𝑋(ClWWalksNOn‘𝐺)𝑁)) = ((#‘(𝑋(𝑣𝑉, 𝑛 ∈ (ℤ‘2) ↦ {𝑤 ∈ (𝑣(ClWWalksNOn‘𝐺)𝑛) ∣ (𝑤‘(𝑛 − 2)) ≠ 𝑣})𝑁)) + (#‘(𝑋(𝑣𝑉, 𝑛 ∈ (ℤ‘2) ↦ {𝑤 ∈ (𝑣(ClWWalksNOn‘𝐺)𝑛) ∣ (𝑤‘(𝑛 − 2)) = 𝑣})𝑁))))
135, 6, 9, 12syl21anc 1365 . 2 (((𝐺RegUSGraph𝐾𝐺 ∈ FriendGraph ) ∧ (𝑉 ∈ Fin ∧ 𝑋𝑉𝑁 ∈ (ℤ‘3))) → (#‘(𝑋(ClWWalksNOn‘𝐺)𝑁)) = ((#‘(𝑋(𝑣𝑉, 𝑛 ∈ (ℤ‘2) ↦ {𝑤 ∈ (𝑣(ClWWalksNOn‘𝐺)𝑛) ∣ (𝑤‘(𝑛 − 2)) ≠ 𝑣})𝑁)) + (#‘(𝑋(𝑣𝑉, 𝑛 ∈ (ℤ‘2) ↦ {𝑤 ∈ (𝑣(ClWWalksNOn‘𝐺)𝑛) ∣ (𝑤‘(𝑛 − 2)) = 𝑣})𝑁))))
14 eqid 2651 . . . 4 (𝑣𝑉, 𝑛 ∈ ℕ ↦ {𝑤 ∈ (𝑛 WWalksN 𝐺) ∣ ((𝑤‘0) = 𝑣 ∧ ( lastS ‘𝑤) ≠ 𝑣)}) = (𝑣𝑉, 𝑛 ∈ ℕ ↦ {𝑤 ∈ (𝑛 WWalksN 𝐺) ∣ ((𝑤‘0) = 𝑣 ∧ ( lastS ‘𝑤) ≠ 𝑣)})
153, 14, 11numclwwlk2 27361 . . 3 (((𝐺RegUSGraph𝐾𝐺 ∈ FriendGraph ) ∧ (𝑉 ∈ Fin ∧ 𝑋𝑉𝑁 ∈ (ℤ‘3))) → (#‘(𝑋(𝑣𝑉, 𝑛 ∈ (ℤ‘2) ↦ {𝑤 ∈ (𝑣(ClWWalksNOn‘𝐺)𝑛) ∣ (𝑤‘(𝑛 − 2)) ≠ 𝑣})𝑁)) = ((𝐾↑(𝑁 − 2)) − (#‘(𝑋(ClWWalksNOn‘𝐺)(𝑁 − 2)))))
161, 2anim12ci 590 . . . 4 (((𝐺RegUSGraph𝐾𝐺 ∈ FriendGraph ) ∧ (𝑉 ∈ Fin ∧ 𝑋𝑉𝑁 ∈ (ℤ‘3))) → (𝑉 ∈ Fin ∧ 𝐺RegUSGraph𝐾))
17 3simpc 1080 . . . . 5 ((𝑉 ∈ Fin ∧ 𝑋𝑉𝑁 ∈ (ℤ‘3)) → (𝑋𝑉𝑁 ∈ (ℤ‘3)))
1817adantl 481 . . . 4 (((𝐺RegUSGraph𝐾𝐺 ∈ FriendGraph ) ∧ (𝑉 ∈ Fin ∧ 𝑋𝑉𝑁 ∈ (ℤ‘3))) → (𝑋𝑉𝑁 ∈ (ℤ‘3)))
19 eqid 2651 . . . . 5 (𝑋(ClWWalksNOn‘𝐺)(𝑁 − 2)) = (𝑋(ClWWalksNOn‘𝐺)(𝑁 − 2))
203, 10, 19numclwwlk1 27351 . . . 4 (((𝑉 ∈ Fin ∧ 𝐺RegUSGraph𝐾) ∧ (𝑋𝑉𝑁 ∈ (ℤ‘3))) → (#‘(𝑋(𝑣𝑉, 𝑛 ∈ (ℤ‘2) ↦ {𝑤 ∈ (𝑣(ClWWalksNOn‘𝐺)𝑛) ∣ (𝑤‘(𝑛 − 2)) = 𝑣})𝑁)) = (𝐾 · (#‘(𝑋(ClWWalksNOn‘𝐺)(𝑁 − 2)))))
2116, 18, 20syl2anc 694 . . 3 (((𝐺RegUSGraph𝐾𝐺 ∈ FriendGraph ) ∧ (𝑉 ∈ Fin ∧ 𝑋𝑉𝑁 ∈ (ℤ‘3))) → (#‘(𝑋(𝑣𝑉, 𝑛 ∈ (ℤ‘2) ↦ {𝑤 ∈ (𝑣(ClWWalksNOn‘𝐺)𝑛) ∣ (𝑤‘(𝑛 − 2)) = 𝑣})𝑁)) = (𝐾 · (#‘(𝑋(ClWWalksNOn‘𝐺)(𝑁 − 2)))))
2215, 21oveq12d 6708 . 2 (((𝐺RegUSGraph𝐾𝐺 ∈ FriendGraph ) ∧ (𝑉 ∈ Fin ∧ 𝑋𝑉𝑁 ∈ (ℤ‘3))) → ((#‘(𝑋(𝑣𝑉, 𝑛 ∈ (ℤ‘2) ↦ {𝑤 ∈ (𝑣(ClWWalksNOn‘𝐺)𝑛) ∣ (𝑤‘(𝑛 − 2)) ≠ 𝑣})𝑁)) + (#‘(𝑋(𝑣𝑉, 𝑛 ∈ (ℤ‘2) ↦ {𝑤 ∈ (𝑣(ClWWalksNOn‘𝐺)𝑛) ∣ (𝑤‘(𝑛 − 2)) = 𝑣})𝑁))) = (((𝐾↑(𝑁 − 2)) − (#‘(𝑋(ClWWalksNOn‘𝐺)(𝑁 − 2)))) + (𝐾 · (#‘(𝑋(ClWWalksNOn‘𝐺)(𝑁 − 2))))))
23 simpll 805 . . . . 5 (((𝐺RegUSGraph𝐾𝐺 ∈ FriendGraph ) ∧ (𝑉 ∈ Fin ∧ 𝑋𝑉𝑁 ∈ (ℤ‘3))) → 𝐺RegUSGraph𝐾)
24 ne0i 3954 . . . . . . 7 (𝑋𝑉𝑉 ≠ ∅)
25243ad2ant2 1103 . . . . . 6 ((𝑉 ∈ Fin ∧ 𝑋𝑉𝑁 ∈ (ℤ‘3)) → 𝑉 ≠ ∅)
2625adantl 481 . . . . 5 (((𝐺RegUSGraph𝐾𝐺 ∈ FriendGraph ) ∧ (𝑉 ∈ Fin ∧ 𝑋𝑉𝑁 ∈ (ℤ‘3))) → 𝑉 ≠ ∅)
273frusgrnn0 26523 . . . . 5 ((𝐺 ∈ FinUSGraph ∧ 𝐺RegUSGraph𝐾𝑉 ≠ ∅) → 𝐾 ∈ ℕ0)
285, 23, 26, 27syl3anc 1366 . . . 4 (((𝐺RegUSGraph𝐾𝐺 ∈ FriendGraph ) ∧ (𝑉 ∈ Fin ∧ 𝑋𝑉𝑁 ∈ (ℤ‘3))) → 𝐾 ∈ ℕ0)
2928nn0cnd 11391 . . 3 (((𝐺RegUSGraph𝐾𝐺 ∈ FriendGraph ) ∧ (𝑉 ∈ Fin ∧ 𝑋𝑉𝑁 ∈ (ℤ‘3))) → 𝐾 ∈ ℂ)
30 uz3m2nn 11769 . . . . . 6 (𝑁 ∈ (ℤ‘3) → (𝑁 − 2) ∈ ℕ)
31303anim3i 1269 . . . . 5 ((𝑉 ∈ Fin ∧ 𝑋𝑉𝑁 ∈ (ℤ‘3)) → (𝑉 ∈ Fin ∧ 𝑋𝑉 ∧ (𝑁 − 2) ∈ ℕ))
3231adantl 481 . . . 4 (((𝐺RegUSGraph𝐾𝐺 ∈ FriendGraph ) ∧ (𝑉 ∈ Fin ∧ 𝑋𝑉𝑁 ∈ (ℤ‘3))) → (𝑉 ∈ Fin ∧ 𝑋𝑉 ∧ (𝑁 − 2) ∈ ℕ))
333clwwlknonfin 27069 . . . . 5 (𝑉 ∈ Fin → (𝑋(ClWWalksNOn‘𝐺)(𝑁 − 2)) ∈ Fin)
34333ad2ant1 1102 . . . 4 ((𝑉 ∈ Fin ∧ 𝑋𝑉 ∧ (𝑁 − 2) ∈ ℕ) → (𝑋(ClWWalksNOn‘𝐺)(𝑁 − 2)) ∈ Fin)
35 hashcl 13185 . . . . 5 ((𝑋(ClWWalksNOn‘𝐺)(𝑁 − 2)) ∈ Fin → (#‘(𝑋(ClWWalksNOn‘𝐺)(𝑁 − 2))) ∈ ℕ0)
3635nn0cnd 11391 . . . 4 ((𝑋(ClWWalksNOn‘𝐺)(𝑁 − 2)) ∈ Fin → (#‘(𝑋(ClWWalksNOn‘𝐺)(𝑁 − 2))) ∈ ℂ)
3732, 34, 363syl 18 . . 3 (((𝐺RegUSGraph𝐾𝐺 ∈ FriendGraph ) ∧ (𝑉 ∈ Fin ∧ 𝑋𝑉𝑁 ∈ (ℤ‘3))) → (#‘(𝑋(ClWWalksNOn‘𝐺)(𝑁 − 2))) ∈ ℂ)
38 numclwlk3lem3 27322 . . 3 ((𝐾 ∈ ℂ ∧ (#‘(𝑋(ClWWalksNOn‘𝐺)(𝑁 − 2))) ∈ ℂ ∧ 𝑁 ∈ (ℤ‘2)) → (((𝐾↑(𝑁 − 2)) − (#‘(𝑋(ClWWalksNOn‘𝐺)(𝑁 − 2)))) + (𝐾 · (#‘(𝑋(ClWWalksNOn‘𝐺)(𝑁 − 2))))) = (((𝐾 − 1) · (#‘(𝑋(ClWWalksNOn‘𝐺)(𝑁 − 2)))) + (𝐾↑(𝑁 − 2))))
3929, 37, 9, 38syl3anc 1366 . 2 (((𝐺RegUSGraph𝐾𝐺 ∈ FriendGraph ) ∧ (𝑉 ∈ Fin ∧ 𝑋𝑉𝑁 ∈ (ℤ‘3))) → (((𝐾↑(𝑁 − 2)) − (#‘(𝑋(ClWWalksNOn‘𝐺)(𝑁 − 2)))) + (𝐾 · (#‘(𝑋(ClWWalksNOn‘𝐺)(𝑁 − 2))))) = (((𝐾 − 1) · (#‘(𝑋(ClWWalksNOn‘𝐺)(𝑁 − 2)))) + (𝐾↑(𝑁 − 2))))
4013, 22, 393eqtrd 2689 1 (((𝐺RegUSGraph𝐾𝐺 ∈ FriendGraph ) ∧ (𝑉 ∈ Fin ∧ 𝑋𝑉𝑁 ∈ (ℤ‘3))) → (#‘(𝑋(ClWWalksNOn‘𝐺)𝑁)) = (((𝐾 − 1) · (#‘(𝑋(ClWWalksNOn‘𝐺)(𝑁 − 2)))) + (𝐾↑(𝑁 − 2))))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 383 ∧ w3a 1054 = wceq 1523 ∈ wcel 2030 ≠ wne 2823 {crab 2945 ∅c0 3948 class class class wbr 4685 ‘cfv 5926 (class class class)co 6690 ↦ cmpt2 6692 Fincfn 7997 ℂcc 9972 0cc0 9974 1c1 9975 + caddc 9977 · cmul 9979 − cmin 10304 ℕcn 11058 2c2 11108 3c3 11109 ℕ0cn0 11330 ℤ≥cuz 11725 ↑cexp 12900 #chash 13157 lastS clsw 13324 Vtxcvtx 25919 FinUSGraphcfusgr 26253 RegUSGraphcrusgr 26508 WWalksN cwwlksn 26774 ClWWalksNOncclwwlknon 27060 FriendGraph cfrgr 27236 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1762 ax-4 1777 ax-5 1879 ax-6 1945 ax-7 1981 ax-8 2032 ax-9 2039 ax-10 2059 ax-11 2074 ax-12 2087 ax-13 2282 ax-ext 2631 ax-rep 4804 ax-sep 4814 ax-nul 4822 ax-pow 4873 ax-pr 4936 ax-un 6991 ax-inf2 8576 ax-cnex 10030 ax-resscn 10031 ax-1cn 10032 ax-icn 10033 ax-addcl 10034 ax-addrcl 10035 ax-mulcl 10036 ax-mulrcl 10037 ax-mulcom 10038 ax-addass 10039 ax-mulass 10040 ax-distr 10041 ax-i2m1 10042 ax-1ne0 10043 ax-1rid 10044 ax-rnegex 10045 ax-rrecex 10046 ax-cnre 10047 ax-pre-lttri 10048 ax-pre-lttrn 10049 ax-pre-ltadd 10050 ax-pre-mulgt0 10051 ax-pre-sup 10052 This theorem depends on definitions: df-bi 197 df-or 384 df-an 385 df-3or 1055 df-3an 1056 df-tru 1526 df-fal 1529 df-ex 1745 df-nf 1750 df-sb 1938 df-eu 2502 df-mo 2503 df-clab 2638 df-cleq 2644 df-clel 2647 df-nfc 2782 df-ne 2824 df-nel 2927 df-ral 2946 df-rex 2947 df-reu 2948 df-rmo 2949 df-rab 2950 df-v 3233 df-sbc 3469 df-csb 3567 df-dif 3610 df-un 3612 df-in 3614 df-ss 3621 df-pss 3623 df-nul 3949 df-if 4120 df-pw 4193 df-sn 4211 df-pr 4213 df-tp 4215 df-op 4217 df-uni 4469 df-int 4508 df-iun 4554 df-disj 4653 df-br 4686 df-opab 4746 df-mpt 4763 df-tr 4786 df-id 5053 df-eprel 5058 df-po 5064 df-so 5065 df-fr 5102 df-se 5103 df-we 5104 df-xp 5149 df-rel 5150 df-cnv 5151 df-co 5152 df-dm 5153 df-rn 5154 df-res 5155 df-ima 5156 df-pred 5718 df-ord 5764 df-on 5765 df-lim 5766 df-suc 5767 df-iota 5889 df-fun 5928 df-fn 5929 df-f 5930 df-f1 5931 df-fo 5932 df-f1o 5933 df-fv 5934 df-isom 5935 df-riota 6651 df-ov 6693 df-oprab 6694 df-mpt2 6695 df-om 7108 df-1st 7210 df-2nd 7211 df-wrecs 7452 df-recs 7513 df-rdg 7551 df-1o 7605 df-2o 7606 df-oadd 7609 df-er 7787 df-map 7901 df-pm 7902 df-en 7998 df-dom 7999 df-sdom 8000 df-fin 8001 df-sup 8389 df-oi 8456 df-card 8803 df-cda 9028 df-pnf 10114 df-mnf 10115 df-xr 10116 df-ltxr 10117 df-le 10118 df-sub 10306 df-neg 10307 df-div 10723 df-nn 11059 df-2 11117 df-3 11118 df-n0 11331 df-xnn0 11402 df-z 11416 df-uz 11726 df-rp 11871 df-xadd 11985 df-fz 12365 df-fzo 12505 df-seq 12842 df-exp 12901 df-hash 13158 df-word 13331 df-lsw 13332 df-concat 13333 df-s1 13334 df-substr 13335 df-s2 13639 df-cj 13883 df-re 13884 df-im 13885 df-sqrt 14019 df-abs 14020 df-clim 14263 df-sum 14461 df-vtx 25921 df-iedg 25922 df-edg 25985 df-uhgr 25998 df-ushgr 25999 df-upgr 26022 df-umgr 26023 df-uspgr 26090 df-usgr 26091 df-fusgr 26254 df-nbgr 26270 df-vtxdg 26418 df-rgr 26509 df-rusgr 26510 df-wwlks 26778 df-wwlksn 26779 df-wwlksnon 26780 df-clwwlk 26950 df-clwwlkn 26983 df-clwwlknon 27061 df-frgr 27237 This theorem is referenced by: numclwwlk5 27375
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## Thursday, May 6, 2010
### Estimating N Losses in Wet Soils
Greg Schwab and Lloyd Murdock
Wet soils cause nitrogen losses. In cases where high intensity rain results in high runoff, leaching losses will probably be low. The primary nitrogen loss mechanism in saturated soils is denitrification, which occurs when soil nitrate nitrogen (NO3-N) is converted to nitrogen gas by soil bacteria. Two to three days of soil saturation is required for bacteria to begin the denitrification process.
Well-drained upland soils that have been wet from a series of rains probably have not experienced much denitrification. Soils in lower landscape positions that stay saturated longer will likely lose more N. Losses can be calculated by estimating 3 to 4 percent loss of fertilizer NO3-N for each day of saturation. Use the Table below to determine how much fertilizer NO3-N was in the soil.
EXAMPLE: Determining the Amount of N Loss
A farmer applied 175 lb nitrogen (N)/A as urea to corn grown on poorly drained soil. Three weeks after application the field became saturated for seven days. How much N was lost?
Step 1. Determine the amount of applied N that was in the nitrate (NO3‐N) form.
According to the table, 50% of the urea will be in the NO3‐N form three weeks after application. 175 lb N x 50% = 88 lb N.
Step 2. Determine the amount of N lost.
Remember that two days are needed for the bacteria to begin the denitrification process. Therefore, denitrification occurred for five days (seven days total saturation minus two days to start the process). With 4% lost each day for five days, 20% would have been lost. 88 lb N x 20% = 18 lb N lost and 157 lb N remaining. The N loss calculated in this example is not as high as most people would assume. A soil N test can verify this estimation.
Nitrogen Soil Test
An additional tool for determining NO3‐N in the soil after flooding is a NO3‐N test. The soil sample should be taken down to 12 inches deep, and several samples should be taken in each field of both the low and higher ground. The samples should be mixed well and a subsample sent for nitrate analysis.
If the nitrate‐N is less than 11 ppm, there is a low amount of plant‐available N in the soil. Therefore, there is a good chance corn will respond to a sidedress application of N ranging from 100 to 150 lbs N/acre.
If the nitrate‐N is between 11 and 25 ppm, there is a greater amount of plant‐available N in the soil, indicating corn may or may not respond to sidedress N. The recommended sidedress N application at this soil test level is 0 to 100 lbs N/acre. If the soil test nitrate‐N is close to 11 ppm, then higher sidedress N rates would be used. Lower rates would be used as nitrate‐N approaches 25 ppm. The test is least accurate in this range, so the test results can only be used as a broad guide.
If soil test nitrate‐N is greater than 25 ppm, there is adequate plant‐available N in the soil, which indicates corn will probably not respond to sidedress N application.
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Next: Results Up: Real Data Results Previous: Real Data Results
## Data description
The 3-D OBS dataset has been already preprocessed and separated into a PP and a PS section. This thesis focus on the PS section only. The subset of the dataset consists of 250 inline CMPs, 50 crossline CMPs, 200 inline half-offset, and 40 crossline half-offset.
shot-rec-nc
Figure 5
Source (a) and receiver (b) distribution for the fraction of OBS dataset in study.
Figure shows the spatial distribution for the shots on the left, and receivers on the right. Notice the gap in the source distribution due to the production platform for the Alba oil field. Additionally, Figure shows the distribution of the CMPs on the left, and the offsets on the right. Remember, the final solution for the inverse problem is a regular common-azimuth cube where all the crossline offsets are collapsed to zero crossline offset. Figure shows a section of the input data. The figure displays the five dimensions of the prestack data cube. The top panel presents the inline-CMP () and crossline-CMP () sections for a constant inline-offset (hx)=-224 m, and constant crossline-offset (hy)=-16.5 m. The center panel presents the inline sections for a constant =550 m, and a constant hy=-16.5 m. The bottom panel presents the crossline sections for a constant =3100 m, and constant hx=-224 m. Observe the sparsity and the holes of the data due to acquisition problems.
cmp-off-nc
Figure 6
CMPs (a) and half-offset (b) distribution for the fraction of OBS dataset in study.
data-nc
Figure 7
5-D representation for the real dataset. Top panel, and sections for a hx=-224 m and hy=-16.5 m. Center panel, and hx sections for =550 m and hy=-16.5. Bottom panel, and hy sections for =3100 m and hx=-224 m.
Next: Results Up: Real Data Results Previous: Real Data Results
Stanford Exploration Project
12/14/2006
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# Astronomy:Eddington luminosity
Short description: Maximum luminosity of a body in hydrostatic equilibrium
The Eddington luminosity, also referred to as the Eddington limit, is the maximum luminosity a body (such as a star) can achieve when there is balance between the force of radiation acting outward and the gravitational force acting inward. The state of balance is called hydrostatic equilibrium. When a star exceeds the Eddington luminosity, it will initiate a very intense radiation-driven stellar wind from its outer layers. Since most massive stars have luminosities far below the Eddington luminosity, their winds are mostly driven by the less intense line absorption.[1] The Eddington limit is invoked to explain the observed luminosity of accreting black holes such as quasars.
Originally, Sir Arthur Eddington took only the electron scattering into account when calculating this limit, something that now is called the classical Eddington limit. Nowadays, the modified Eddington limit also counts on other radiation processes such as bound-free and free-free radiation (see Bremsstrahlung) interaction.
## Derivation
The limit is obtained by setting the outward radiation pressure equal to the inward gravitational force. Both forces decrease by inverse square laws, so once equality is reached, the hydrodynamic flow is the same throughout the star.
From Euler's equation in hydrostatic equilibrium, the mean acceleration is zero,
$\displaystyle{ \frac{d u}{d t} = - \frac{\nabla p}{\rho} - \nabla \Phi = 0 }$
where $\displaystyle{ u }$ is the velocity, $\displaystyle{ p }$ is the pressure, $\displaystyle{ \rho }$ is the density, and $\displaystyle{ \Phi }$ is the gravitational potential. If the pressure is dominated by radiation pressure associated with a radiation flux $\displaystyle{ F_{\rm rad} }$,
$\displaystyle{ -\frac{\nabla p}{\rho} = \frac{\kappa}{c} F_{\rm rad}\,. }$
Here $\displaystyle{ \kappa }$ is the opacity of the stellar material which is defined as the fraction of radiation energy flux absorbed by the medium per unit density and unit length. For ionized hydrogen $\displaystyle{ \kappa=\sigma_{\rm T}/m_{\rm p} }$, where $\displaystyle{ \sigma_{\rm T} }$ is the Thomson scattering cross-section for the electron and $\displaystyle{ m_{\rm p} }$ is the mass of a proton. Note that $\displaystyle{ F_{\rm rad}=d^2E/dAdt }$ is defined as the energy flux over a surface, which can be expressed with the momentum flux using $\displaystyle{ E=pc }$ for radiation. Therefore, the rate of momentum transfer from the radiation to the gaseous medium per unit density is $\displaystyle{ \kappa F_{\rm rad}/c }$, which explains the right hand side of the above equation.
The luminosity of a source bounded by a surface $\displaystyle{ S }$ may be expressed with these relations as
$\displaystyle{ L = \int_S F_{\rm rad} \cdot dS = \int_S \frac{c}{\kappa} \nabla \Phi \cdot dS\,. }$
Now assuming that the opacity is a constant, it can be brought outside of the integral. Using Gauss's theorem and Poisson's equation gives
$\displaystyle{ L = \frac{c}{\kappa} \int_S \nabla \Phi \cdot dS = \frac{c}{\kappa} \int_V \nabla^2 \Phi \, dV = \frac{4 \pi G c}{\kappa} \int_V \rho \, dV = \frac{4 \pi G M c}{\kappa} }$
where $\displaystyle{ M }$ is the mass of the central object. This is called the Eddington Luminosity.[2] For pure ionized hydrogen,
\displaystyle{ \begin{align}L_{\rm Edd}&=\frac{4\pi G M m_{\rm p} c} {\sigma_{\rm T}}\\ &\cong 1.26\times10^{31}\left(\frac{M}{M_\bigodot}\right){\rm W} = 1.26\times10^{38}\left(\frac{M}{M_\bigodot}\right){\rm erg/s} = 3.2\times10^4\left(\frac{M}{M_\bigodot}\right) L_\bigodot \end{align} }
where $\displaystyle{ M_\bigodot }$ is the mass of the Sun and $\displaystyle{ L_\bigodot }$ is the luminosity of the Sun.
The maximum luminosity of a source in hydrostatic equilibrium is the Eddington luminosity. If the luminosity exceeds the Eddington limit, then the radiation pressure drives an outflow.
The mass of the proton appears because, in the typical environment for the outer layers of a star, the radiation pressure acts on electrons, which are driven away from the center. Because protons are negligibly pressured by the analog of Thomson scattering, due to their larger mass, the result is to create a slight charge separation and therefore a radially directed electric field, acting to lift the positive charges, which are typically free protons under the conditions in stellar atmospheres. When the outward electric field is sufficient to levitate the protons against gravity, both electrons and protons are expelled together.
### Different limits for different materials
The derivation above for the outward light pressure assumes a hydrogen plasma. In other circumstances the pressure balance can be different from what it is for hydrogen.
In an evolved star with a pure helium atmosphere, the electric field would have to lift a helium nucleus (an alpha particle), with nearly 4 times the mass of a proton, while the radiation pressure would act on 2 free electrons. Thus twice the usual Eddington luminosity would be needed to drive off an atmosphere of pure helium.
At very high temperatures, as in the environment of a black hole or neutron star, high energy photon interactions with nuclei or even with other photons, can create an electron-positron plasma. In that situation the combined mass of the positive-negative charge carrier pair is approximately 918 times smaller (the proton to electron mass ratio), while the radiation pressure on the positrons doubles the effective upward force per unit mass, so the limiting luminosity needed is reduced by a factor of ≈ 918×2.
The exact value of the Eddington luminosity depends on the chemical composition of the gas layer and the spectral energy distribution of the emission. A gas with cosmological abundances of hydrogen and helium is much more transparent than gas with solar abundance ratios. Atomic line transitions can greatly increase the effects of radiation pressure, and line driven winds exist in some bright stars (e.g., Wolf-Rayet and O stars).
## Super-Eddington luminosities
The role of the Eddington limit in today's research lies in explaining the very high mass loss rates seen in for example the series of outbursts of η Carinae in 1840–1860.[3] The regular, line driven stellar winds can only stand for a mass loss rate of around 10−4–10−3 solar masses per year, whereas mass loss rates of up to 0.5 solar masses per year are needed to understand the η Carinae outbursts. This can be done with the help of the super-Eddington broad spectrum radiation driven winds.
Gamma-ray bursts, novae and supernovae are examples of systems exceeding their Eddington luminosity by a large factor for very short times, resulting in short and highly intensive mass loss rates. Some X-ray binaries and active galaxies are able to maintain luminosities close to the Eddington limit for very long times. For accretion-powered sources such as accreting neutron stars or cataclysmic variables (accreting white dwarfs), the limit may act to reduce or cut off the accretion flow, imposing an Eddington limit on accretion corresponding to that on luminosity. Super-Eddington accretion onto stellar-mass black holes is one possible model for ultraluminous X-ray sources (ULXs).[4][5]
For accreting black holes, not all the energy released by accretion has to appear as outgoing luminosity, since energy can be lost through the event horizon, down the hole. Such sources effectively may not conserve energy. Then the accretion efficiency, or the fraction of energy actually radiated of that theoretically available from the gravitational energy release of accreting material, enters in an essential way.
## Other factors
The Eddington limit is not a strict limit on the luminosity of a stellar object. The limit does not consider several potentially important factors, and super-Eddington objects have been observed that do not seem to have the predicted high mass-loss rate. Other factors that might affect the maximum luminosity of a star include:
• Porosity. A problem with steady winds driven by broad-spectrum radiation is that both the radiative flux and gravitational acceleration scale with r −2. The ratio between these factors is constant, and in a super-Eddington star, the whole envelope would become gravitationally unbound at the same time. This is not observed. A possible solution is introducing an atmospheric porosity, where we imagine the stellar atmosphere to consist of denser regions surrounded by lower density gas regions. This would reduce the coupling between radiation and matter, and the full force of the radiation field would only be seen in the more homogeneous outer, lower density layers of the atmosphere.
• Turbulence. A possible destabilizing factor might be the turbulent pressure arising when energy in the convection zones builds up a field of supersonic turbulence. The importance of turbulence is being debated, however.[6]
• Photon bubbles. Another factor that might explain some stable super-Eddington objects is the photon bubble effect. Photon bubbles would develop spontaneously in radiation-dominated atmospheres when the radiation pressure exceeds the gas pressure. We can imagine a region in the stellar atmosphere with a density lower than the surroundings, but with a higher radiation pressure. Such a region would rise through the atmosphere, with radiation diffusing in from the sides, leading to an even higher radiation pressure. This effect could transport radiation more efficiently than a homogeneous atmosphere, increasing the allowed total radiation rate. In accretion discs, luminosities may be as high as 10–100 times the Eddington limit without experiencing instabilities.[7]
## Humphreys–Davidson limit
The upper H–R diagram with the empirical Humphreys-Davidson limit marked (green line). Stars are observed above the limit only during brief outbursts.
Observations of massive stars show a clear upper limit to their luminosity, termed the Humphreys–Davidson limit after the researchers who first wrote about it.[8] Only highly unstable objects are found, temporarily, at higher luminosities. Efforts to reconcile this with the theoretical Eddington limit have been largely unsuccessful.[9] The H-D limit for cool supergiants is placed at around 316,000 L.[10]
Most luminous cool (K-M) supergiants.
Name Luminosity Spectral Type Notes References
LGGS J013312.26+310053.3 575,000 [11]
LGGS J004520.67+414717.3 562,000 M1I Likely not a member of the Andromeda Galaxy, should be treated with caution in regards to the HD limit.[12] [12]
LGGS J013339.28+303118.8 479,000 M1Ia [11]
Stephenson 2 DFK 49 390,000 K4 [13]
HD 269551 A 389,000 K/M [14]
WOH S170 380,000 M Large Magellanic Cloud membership uncertain. [14]
RSGC1-F04 380,000 M0-M1 [13]
LGGS J013418.56+303808.6 363,000 [11]
LGGS J004428.12+415502.9 339,000 K2I [12]
RSGC1-F01 335,000 M3-M5 [13]
AH Scorpii 331,000 M5Ia [15]
SMC 18592 309,000 - 355,000 K5-M0Ia [16][14]
LGGS J004539.99+415404.1 309,000 M3I [12]
LGGS J013350.62+303230.3 309,000 [14]
LGGS J013358.54+303419.9 295,000 [14]
CM Velorum 308,000 M5 [17]
HV 888 302,000 M4Ia [16]
W60 B90 302,000 M2 [18]
RW Cephei 300,000 K2Ia-0 [19]
GCIRS 7 295,000 M1I [20]
SP77 21-12 295,000 K5-M3 [14]
RSGC1-F13 290,000 K2-M3 [13]
EV Carinae 288,000 M4.5Ia [10]
HV 12463 288,000 M Probably not a LMC member. [14]
LGGS J003951.33+405303.7 288,000 [12]
WOH G64 282,000 M5I LIkely the largest known star. [21]
LGGS J013352.96+303816.0 282,000 [14]
CD-26 5055 280,000 M2Iab [17]
Westerlund 1 W26 275,000 M0.5-M6Ia [22]
LGGS J004731.12+422749.1 275,000 [12]
VY Canis Majoris 270,000 M3-M4.5 [23]
LGGS J004428.48+415130.9 269,000 M1I [12]
LGGS J013241.94+302047.5 257,000 [14]
LMC 145013 251,000 - 339,000 M2.5Ia-Ib [16][14]
LMC 25320 251,000 M [14]
## References
1. A. J. van Marle; S. P. Owocki; N. J. Shaviv (2008). "Continuum driven winds from super-Eddington stars. A tale of two limits". AIP Conference Proceedings 990: 250–253. doi:10.1063/1.2905555. Bibcode2008AIPC..990..250V.
2. Rybicki, G. B.; Lightman, A. P. Radiative Processes in Astrophysics. New York: J. Wiley & Sons 1979.
3. N. Smith; S. P. Owocki (2006). "On the role of continuum driven eruptions in the evolution of very massive stars and population III stars". Astrophysical Journal 645 (1): L45–L48. doi:10.1086/506523. Bibcode2006ApJ...645L..45S.
4. Bachetti, Matteo; Heida, Marianne; Maccarone, Thomas; Huppenkothen, Daniela; Israel, Gian Luca; Barret, Didier; Brightman, Murray; Brumback, McKinley et al. (2022-10-01). "Orbital Decay in M82 X-2". The Astrophysical Journal 937 (2): 125. doi:10.3847/1538-4357/ac8d67. ISSN 0004-637X.
5. R. B. Stothers (2003). "Turbulent pressure in the envelopes of yellow hypergiants and luminous blue variables". Astrophysical Journal 589 (2): 960–967. doi:10.1086/374713. Bibcode2003ApJ...589..960S.
6. J. Arons (1992). "Photon bubbles: Overstability in a magnetized atmosphere". Astrophysical Journal 388: 561–578. doi:10.1086/171174. Bibcode1992ApJ...388..561A.
7. Humphreys, R. M.; Davidson, K. (1979). "Studies of luminous stars in nearby galaxies. III - Comments on the evolution of the most massive stars in the Milky Way and the Large Magellanic Cloud" (in en). The Astrophysical Journal 232: 409. doi:10.1086/157301. ISSN 0004-637X. Bibcode1979ApJ...232..409H.
8. Glatzel, W.; Kiriakidis, M. (15 July 1993). "Stability of massive stars and the Humphreys–Davidson limit". Monthly Notices of the Royal Astronomical Society 263 (2): 375–384. doi:10.1093/mnras/263.2.375. Bibcode1993MNRAS.263..375G.
9. Davies, Ben; Beasor, Emma R (2020-03-21). "The 'red supergiant problem': the upper luminosity boundary of Type II supernova progenitors" (in en). Monthly Notices of the Royal Astronomical Society 493 (1): 468–476. doi:10.1093/mnras/staa174. ISSN 0035-8711.
10. Drout, Maria R.; Massey, Philip; Meynet, Georges (2012-04-18). "The Yellow and Red Supergiants of M33". The Astrophysical Journal 750 (2): 97. doi:10.1088/0004-637x/750/2/97. ISSN 0004-637X.
11. McDonald, Sarah L E; Davies, Ben; Beasor, Emma R (2022-01-08). "Red supergiants in M31: the Humphreys–Davidson limit at high metallicity" (in en). Monthly Notices of the Royal Astronomical Society 510 (3): 3132–3144. doi:10.1093/mnras/stab3453. ISSN 0035-8711.
12. Humphreys, Roberta M.; Helmel, Greta; Jones, Terry J.; Gordon, Michael S. (2020-09-02). "Exploring the Mass-loss Histories of the Red Supergiants". The Astronomical Journal 160 (3): 145. doi:10.3847/1538-3881/abab15. ISSN 1538-3881.
13. Massey, Philip; Neugent, Kathryn F.; Ekström, Sylvia; Georgy, Cyril; Meynet, Georges (2023-01-01). "The Time-averaged Mass-loss Rates of Red Supergiants as Revealed by Their Luminosity Functions in M31 and M33". The Astrophysical Journal 942 (2): 69. doi:10.3847/1538-4357/aca665. ISSN 0004-637X.
14. Arroyo-Torres, B.; Wittkowski, M.; Marcaide, J. M.; Hauschildt, P. H. (June 2013). "The atmospheric structure and fundamental parameters of the red supergiants AH Scorpii, UY Scuti, and KW Sagittarii". Astronomy & Astrophysics 554: A76. doi:10.1051/0004-6361/201220920. ISSN 0004-6361.
15. Davies, Ben; Crowther, Paul A.; Beasor, Emma R. (2018-08-01). "The luminosities of cool supergiants in the Magellanic Clouds, and the Humphreys-Davidson limit revisited". Monthly Notices of the Royal Astronomical Society 478 (3): 3138–3148. doi:10.1093/mnras/sty1302. ISSN 0035-8711. Bibcode2018MNRAS.478.3138D.
16. Vallenari, A.; Brown, A. G. A.; Prusti, T.; Bruijne, J. H. J. de; Arenou, F.; Babusiaux, C.; Biermann, M.; Creevey, O. L. et al. (2023-06-01). "Gaia Data Release 3 - Summary of the content and survey properties" (in en). Astronomy & Astrophysics 674: A1. doi:10.1051/0004-6361/202243940. ISSN 0004-6361.
17. Wit, S. de; Bonanos, A. Z.; Tramper, F.; Yang, M.; Maravelias, G.; Boutsia, K.; Britavskiy, N.; Zapartas, E. (2023-01-01). "Properties of luminous red supergiant stars in the Magellanic Clouds" (in en). Astronomy & Astrophysics 669: A86. doi:10.1051/0004-6361/202243394. ISSN 0004-6361.
18. Jones, Terry Jay; Shenoy, Dinesh; Humphreys, Roberta (2023-05-11). "The Recent Mass Loss History of the Hypergiant RW Cep". Research Notes of the AAS 7 (5): 92. doi:10.3847/2515-5172/acd37f. ISSN 2515-5172.
19. Guerço, Rafael; Smith, Verne V; Cunha, Katia; Ekström, Sylvia; Abia, Carlos; Plez, Bertrand; Meynet, Georges; Ramirez, Solange V et al. (2022-09-13). "Evidence of deep mixing in IRS 7, a cool massive supergiant member of the Galactic nuclear star cluster" (in en). Monthly Notices of the Royal Astronomical Society 516 (2): 2801–2811. doi:10.1093/mnras/stac2393. ISSN 0035-8711.
20. Ohnaka, K.; Driebe, T.; Hofmann, K.-H.; Weigelt, G.; Wittkowski, M. (2008-06-01). "Spatially resolved dusty torus toward the red supergiant WOH G64 in the Large Magellanic Cloud" (in en). Astronomy & Astrophysics 484 (2): 371–379. doi:10.1051/0004-6361:200809469. ISSN 0004-6361.
21. Arévalo, Aura (2019-01-22). The Red Supergiants in the Supermassive Stellar Cluster Westerlund 1 (Mestrado em Astronomia thesis). São Paulo: Universidade de São Paulo. doi:10.11606/d.14.2019.tde-12092018-161841.
22. Wittkowski, M.; Hauschildt, P. H.; Arroyo-Torres, B.; Marcaide, J. M. (April 2012). "Fundamental properties and atmospheric structure of the red supergiant VY Canis Majoris based on VLTI/AMBER spectro-interferometry". Astronomy & Astrophysics 540: L12. doi:10.1051/0004-6361/201219126. ISSN 0004-6361.
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Topic: Chinese Remainder Theorem Question
Replies: 1 Last Post: Nov 11, 2004 3:04 PM
Messages: [ Previous | Next ]
Rob Posts: 4 Registered: 1/25/05
Chinese Remainder Theorem Question
Posted: Nov 10, 2004 4:34 PM
Here are a couple problems I could use some help with. I am sure I
can solve them, but I am not too sure I can "prove" how they work.
Here they are:
1.) Find a number with these three properties: when you divide the
number by 7, the remainder is 4; when you divide the original number
by 11, the remainder is 2; when you divide the original number by 13,
the remainder is 9.
2.) Find a number with these three properties: when you divide the
number by 7, the remainder is "s"; when you divide the number by 11;
the remainder is "e", and when you divide the number by 13, the
remainder is "t"
3.) Generalize the Chinese remainder theorem to 4 remainders: find a
formula that will work, when you are a given remainders by 5, 7, 11,
and 13.
If there is anyone out there that can help me with this and knows a
Thanks!
Ron
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Date Subject Author
11/10/04 Rob
11/11/04 Rob Morewood
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# Combinational Circuits – 42
5555555555
42. Two 4-bit binary numbers (1011 and 1111) are applied to a 4-bit parallel adder. The carry input is 1. What are the values for the sum and carry output?
(a) 4321 = 0111, Cout = 0
(b) 4321 = 1111, Cout = 1
(c) 4321 = 1011, Cout = 1
(d) 4321 = 1100, Cout = 1
Explanation
Explanation : No answer description available for this question. Let us discuss.
Subject Name : Electronics Engineering Exam Name : IIT GATE, UPSC ESE, RRB, SSC, DMRC, NMRC, BSNL, DRDO, ISRO, BARC, NIELIT Posts Name : Assistant Engineer, Management Trainee, Junior Engineer, Technical Assistant
Electronics & Communication Engineering Books Sale Question Bank On Electronics & Communication Engineering Highlight important details or key points; Summarize information in a clear and concise manner ₹ 317 Sale A Handbook for Electronics Engineering(Old Edition) Made Easy Editorial Board (Author); Marathi (Publication Language); 620 Pages - 01/01/2015 (Publication Date) - Made Easy Publications (Publisher) ₹ 320 Basic Electronic Devices and Circuits Patil (Author); English (Publication Language) Sale Electronic Circuits: Analysis and Design (SIE) | 3rd Edition Neamen, Donald (Author); English (Publication Language); 1351 Pages - 08/25/2006 (Publication Date) - McGraw Hill Education (Publisher) ₹ 680
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# SOLUTION: How would you work out this problem; It requires to show all work! The cafeteria has 360 roses to sell for Valentine's Day. You want to pick out a perfect red rose for someone
Algebra -> Algebra -> Average -> SOLUTION: How would you work out this problem; It requires to show all work! The cafeteria has 360 roses to sell for Valentine's Day. You want to pick out a perfect red rose for someone Log On
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Algebra: Average Solvers Lessons Answers archive Quiz In Depth
Question 269002: How would you work out this problem; It requires to show all work! The cafeteria has 360 roses to sell for Valentine's Day. You want to pick out a perfect red rose for someone special. When you get to the cafeteria you discover: *every third rose is too short *every fourth rose has too many thorns *every fifth rose is wilted *every sixth rose is pink *every eighth rose is yellow If these are the only problems that the roses have, how many perfect roses are there for you to choose from? The problem requires to show all work. I would think that there is an algebraic expression that would help me out, but I am in need of help. I can help myself by making pictures to help visualize, but I think there must be a better way. Please help me if you can. Thank you! Found 2 solutions by josmiceli, drk:Answer by josmiceli(9697) (Show Source): You can put this solution on YOUR website! There are 120 roses that are too short There are 90 roses that have too many thorns But, since , every 12th rose was also divisible by 3, and was also a too short rose,and so 30 of the 90 are being counted twice. Just add to the list of not wanted So far I have not wanted Of these every roses is being counted twice and , so 66 roses are wilted and are not being counted twice So far I have: not wanted Every 6th rose is pink, but all these were counted as a too short rose, since dividing by 6 is included by dividing by 3, therefore don't count these Likewise, every 8th rose was also counted when I counted every 4th rose that had too many thorns,so don't count these 114 roses of the 360 are perfect I could have stumbled with my logic, but I think this is right- hope yo at least un- derstand what I'm doing Answer by drk(1908) (Show Source): You can put this solution on YOUR website!It seems to me that you count 1, 2 and then primes greater than 5, and multiples of primes. Here is the list that I have: 1, 2, 7, 14, 49, 77, 91, 98, 119, . . . . . .. . . . . . .. [22 options] 11, 22, 121, 143, 154, . . . . . . . . . . . . . . [12 options 13, 26, 169, 182, 221, 247, 286, 299, 338 [9 options] 17, 34, 289, 323 . . . . . . . . . . . . . . . . . . . . .[4 options] 19, 38, 23, 46, 29, 58, 31, 62, 37, 74 41, 82, 43, 86, 47, 94, 53, 106, 59, 118, 61, 122, 67, 134, 71, 142, 73, 146, 77, 154, 79, 158, 83, 166, 89, 178, 91, 182, 97, 194, 101, 202, 103, 206, 107, 214, 109, 218, 113, 226, 127, 254, 131, 262, 137, 274, 139, 278, 149, 298, 151, 302, 157, 314, 163, 326, 167, 334, 173, 346, 179, 358, ---- 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359 ---- so our total appears to be: 2 + 22 + 12 + 9 + 4 + 72 + 31 = 152 good options --- On a personal note I didn't like this question. It could have been set up better using a smaller number such as 100 roses.
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# Inexpensive Solution Protects Sensitive Devices From Surges
As today’s high-tech electronic devices become smaller and faster while running at lower voltages, they also become more vulnerable to surges. Making matters worse, they often operate in harsh environments that expose them to extreme surges, particularly in industrial applications where sensors are interfaced to microcontrollers or logic devices. This idea shows engineers how to design their circuits to withstand these surges.
The key to surge protection is adding a transient voltage suppressor at the input to the sensitive devices. The simple circuitry includes a resistor and a transient suppressor that handle the voltage surge and regulate the input voltage. The tricky part is choosing the right value for the suppressor.
Properly designed, the transient suppressor appears invisible to the protected device until a surge hits. That is, during normal conditions the suppressor’s breakdown voltage, current, and capacitance will have no effect on operation and performance. When a surge strikes, the suppressor immediately clamps to a safe voltage level (the clamping voltage), conducting away the surge current.
In the example circuit, the load, which can be the input of a logic device, is fed by a 24-V dc input, which can be from a sensor, transducer, or other device (Fig. 1). The source resistance is 2 Ω and the failure threshold is 36 V. The circuit uses an SMBJ26A 600-W transient voltage suppressor. If a malfunction causes a peak surge of 150 V with a duration of 10 ns at the input terminal, the suppressor must clamp that surge at 36 V or less. The current delivered by this transient is:
Ip = (150 V – 36 V)/2 Ω = 57 A (1)
1. Using only the suppressor diode in this circuit that has a 2-Ω source resistance creates too high a peak transient current.
The suppressor causes the surge voltage to be divided between the source impedance and itself. Equation 1 shows that the higher the clamping voltage is, the lower the surge current in the circuit. Unfortunately, the resulting current in this example would fry the circuit.
A higher-wattage suppressor could be used, but that is not advisable because the high power capacity and power dissipation would be a costly solution. Alternatively, you can add a small resistor in series with the source impedance, effectively reducing the surge current and along with it the overall size, power dissipation, and cost of the circuit (Fig. 2).
2. The addition of an inexpensive 20-Ω resistor to the input greatly reduces the surge current and power dissipated.
Assuming a small load, 10 mA, the voltage drop across the 20-Ω would be:
V20Ω = 10 mA × 20 Ω = 0.2 V (2)
Therefore, the surge current I would be:
Ip = (150 V – 36 V)/22 Ω = 5.2 A (3)
The added resistor reduces the surge current to less than 1/10 of the surge current without the resistor. The low-power suppressor can handle this current. Also, the clamping current is less than the rated current. The clamping voltage is:
From the SMBJ26A datasheet, the maximum clamping voltage, VC(max) = 42.1 V; the maximum breakdown voltage, VBR(max) = 31.9 V; and the maximum peak pulse current, Ipp(max) = 14.3 A. Using these values and the result of Equation 3 for Ip, Equation 4 yields VC = 35.5 V.
Thus the clamping voltage is a little below the threshold voltage at 25°C. For higher temperatures, de-rating parameters must be considered. The steady state power dissipated by the 20-Ω resistor is:
P20Ω = (20 mA)2 × 20 Ω = 8 mW (5)
Consequently, a 1/8-W carbon composition resistor can be used for the suppressor circuit for the given pulse condition. This small series resistor drastically improves the suppressor performance with little cost impact.
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## Results (1-50 of 108 matches)
Next displayed columns for results
Label Class Conductor Rank Torsion CM Weierstrass equation
128.a2 128.a $$2^{7}$$ $1$ $\Z/2\Z$ $$y^2=x^3+x^2+x+1$$
128.b2 128.b $$2^{7}$$ $0$ $\Z/2\Z$ $$y^2=x^3+x^2+3x-5$$
128.c2 128.c $$2^{7}$$ $0$ $\Z/2\Z$ $$y^2=x^3-x^2+x-1$$
128.d2 128.d $$2^{7}$$ $0$ $\Z/2\Z$ $$y^2=x^3-x^2+3x+5$$
1152.c2 1152.c $$2^{7} \cdot 3^{2}$$ $0$ $\Z/2\Z$ $$y^2=x^3+24x-160$$
1152.h2 1152.h $$2^{7} \cdot 3^{2}$$ $0$ $\Z/2\Z$ $$y^2=x^3+24x+160$$
1152.m2 1152.m $$2^{7} \cdot 3^{2}$$ $1$ $\Z/2\Z$ $$y^2=x^3+6x-20$$
1152.r2 1152.r $$2^{7} \cdot 3^{2}$$ $0$ $\Z/2\Z$ $$y^2=x^3+6x+20$$
3200.e2 3200.e $$2^{7} \cdot 5^{2}$$ $1$ $\Z/2\Z$ $$y^2=x^3+x^2+17x-87$$
3200.h2 3200.h $$2^{7} \cdot 5^{2}$$ $1$ $\Z/2\Z$ $$y^2=x^3+x^2+67x+763$$
3200.u2 3200.u $$2^{7} \cdot 5^{2}$$ $1$ $\Z/2\Z$ $$y^2=x^3-x^2+67x-763$$
3200.x2 3200.x $$2^{7} \cdot 5^{2}$$ $0$ $\Z/2\Z$ $$y^2=x^3-x^2+17x+87$$
6272.a2 6272.a $$2^{7} \cdot 7^{2}$$ $1$ $\Z/2\Z$ $$y^2=x^3+x^2+131x-1989$$
6272.b2 6272.b $$2^{7} \cdot 7^{2}$$ $1$ $\Z/2\Z$ $$y^2=x^3+x^2+33x+265$$
6272.g2 6272.g $$2^{7} \cdot 7^{2}$$ $1$ $\Z/2\Z$ $$y^2=x^3-x^2+131x+1989$$
6272.h2 6272.h $$2^{7} \cdot 7^{2}$$ $0$ $\Z/2\Z$ $$y^2=x^3-x^2+33x-265$$
15488.c2 15488.c $$2^{7} \cdot 11^{2}$$ $1$ $\Z/2\Z$ $$y^2=x^3+x^2+81x-959$$
15488.d2 15488.d $$2^{7} \cdot 11^{2}$$ $0$ $\Z/2\Z$ $$y^2=x^3+x^2+323x+7995$$
15488.u2 15488.u $$2^{7} \cdot 11^{2}$$ $0$ $\Z/2\Z$ $$y^2=x^3-x^2+81x+959$$
15488.v2 15488.v $$2^{7} \cdot 11^{2}$$ $0$ $\Z/2\Z$ $$y^2=x^3-x^2+323x-7995$$
21632.c2 21632.c $$2^{7} \cdot 13^{2}$$ $1$ $\Z/2\Z$ $$y^2=x^3+x^2+451x-12869$$
21632.l2 21632.l $$2^{7} \cdot 13^{2}$$ $0$ $\Z/2\Z$ $$y^2=x^3+x^2+113x+1665$$
21632.be2 21632.be $$2^{7} \cdot 13^{2}$$ $1$ $\Z/2\Z$ $$y^2=x^3-x^2+451x+12869$$
21632.bj2 21632.bj $$2^{7} \cdot 13^{2}$$ $1$ $\Z/2\Z$ $$y^2=x^3-x^2+113x-1665$$
28800.d2 28800.d $$2^{7} \cdot 3^{2} \cdot 5^{2}$$ $1$ $\Z/2\Z$ $$y^2=x^3+600x+20000$$
28800.k2 28800.k $$2^{7} \cdot 3^{2} \cdot 5^{2}$$ $1$ $\Z/2\Z$ $$y^2=x^3+150x+2500$$
28800.du2 28800.du $$2^{7} \cdot 3^{2} \cdot 5^{2}$$ $0$ $\Z/2\Z$ $$y^2=x^3+150x-2500$$
28800.dz2 28800.dz $$2^{7} \cdot 3^{2} \cdot 5^{2}$$ $1$ $\Z/2\Z$ $$y^2=x^3+600x-20000$$
36992.a2 36992.a $$2^{7} \cdot 17^{2}$$ $0$ $\Z/2\Z$ $$y^2=x^3+x^2+771x+29371$$
36992.b2 36992.b $$2^{7} \cdot 17^{2}$$ $0$ $\Z/2\Z$ $$y^2=x^3+x^2+193x-3575$$
36992.c2 36992.c $$2^{7} \cdot 17^{2}$$ $0$ $\Z/2\Z$ $$y^2=x^3-x^2+771x-29371$$
36992.d2 36992.d $$2^{7} \cdot 17^{2}$$ $1$ $\Z/2\Z$ $$y^2=x^3-x^2+193x+3575$$
46208.a2 46208.a $$2^{7} \cdot 19^{2}$$ $0$ $\Z/2\Z$ $$y^2=x^3+x^2+241x+5161$$
46208.b2 46208.b $$2^{7} \cdot 19^{2}$$ $0$ $\Z/2\Z$ $$y^2=x^3+x^2+963x-40325$$
46208.k2 46208.k $$2^{7} \cdot 19^{2}$$ $1$ $\Z/2\Z$ $$y^2=x^3-x^2+241x-5161$$
46208.l2 46208.l $$2^{7} \cdot 19^{2}$$ $0$ $\Z/2\Z$ $$y^2=x^3-x^2+963x+40325$$
56448.v2 56448.v $$2^{7} \cdot 3^{2} \cdot 7^{2}$$ $2$ $\Z/2\Z$ $$y^2=x^3+294x+6860$$
56448.y2 56448.y $$2^{7} \cdot 3^{2} \cdot 7^{2}$$ $1$ $\Z/2\Z$ $$y^2=x^3+294x-6860$$
56448.ct2 56448.ct $$2^{7} \cdot 3^{2} \cdot 7^{2}$$ $1$ $\Z/2\Z$ $$y^2=x^3+1176x-54880$$
56448.cw2 56448.cw $$2^{7} \cdot 3^{2} \cdot 7^{2}$$ $1$ $\Z/2\Z$ $$y^2=x^3+1176x+54880$$
67712.b2 67712.b $$2^{7} \cdot 23^{2}$$ $1$ $\Z/2\Z$ $$y^2=x^3+x^2+1411x+72571$$
67712.c2 67712.c $$2^{7} \cdot 23^{2}$$ $0$ $\Z/2\Z$ $$y^2=x^3+x^2+353x-8895$$
67712.n2 67712.n $$2^{7} \cdot 23^{2}$$ $1$ $\Z/2\Z$ $$y^2=x^3-x^2+1411x-72571$$
67712.o2 67712.o $$2^{7} \cdot 23^{2}$$ $1$ $\Z/2\Z$ $$y^2=x^3-x^2+353x+8895$$
107648.d2 107648.d $$2^{7} \cdot 29^{2}$$ $1$ $\Z/2\Z$ $$y^2=x^3+x^2+561x-17879$$
107648.f2 107648.f $$2^{7} \cdot 29^{2}$$ $1$ $\Z/2\Z$ $$y^2=x^3+x^2+2243x+145275$$
107648.bm2 107648.bm $$2^{7} \cdot 29^{2}$$ $0$ $\Z/2\Z$ $$y^2=x^3-x^2+561x+17879$$
107648.bo2 107648.bo $$2^{7} \cdot 29^{2}$$ $1$ $\Z/2\Z$ $$y^2=x^3-x^2+2243x-145275$$
123008.a2 123008.a $$2^{7} \cdot 31^{2}$$ $1$ $\Z/2\Z$ $$y^2=x^3+x^2+641x+22281$$
123008.b2 123008.b $$2^{7} \cdot 31^{2}$$ $1$ $\Z/2\Z$ $$y^2=x^3+x^2+2563x-175685$$
Next displayed columns for results
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# Syntax for rounding up in VB.NET
What is the syntax to round up a decimal leaving two digits after the decimal point?
Example: 2.566666 -> 2.57
-
If you want regular rounding, you can just use the Math.Round method. If you specifially want to round upwards, you use the Math.Ceiling method:
Dim d As Decimal = 2.566666
Dim r As Decimal = Math.Ceiling(d * 100D) / 100D
-
Math.Round is what you're looking for. If you're new to rounding in .NET - you should also look up the difference between AwayFromZero and ToEven rounding. The default of ToEven can sometime take people by surprise.
dim result = Math.Round(2.56666666, 2)
-
ah..yeahh...thx – leonita May 19 '10 at 1:32
Here is how I do it:
Private Function RoundUp(value As Double, decimals As Integer) As Double
Return Math.Ceiling(value * (10 ^ decimals)) / (10 ^ decimals)
End Function
-
You can use System.Math, specifically Math.Round(), like this:
Math.Round(2.566666, 2)
-
Math.Round(), as suggested by others, is probably what you want. But the text of your question specifically asked how to "roundup"[sic]. If you always need to round up, regarless of actual value (ie: 2.561111 would still go to 2.57), you can do this:
Math.Ceiling(d * 100)/100D
-
I used this way:
Math.Round(d + 0.49D, 2)
-
The basic function for rounding up is Math.Ceiling(d), but the asker specifically wanted to round up after the second decimal place. This would be Math.Ceiling(d * 100) / 100. For example it may multiply 46.5671 by 100 to get 4656.71, then rounds up to get 4657, then divides by 100 to shift the decimal back 2 places to get 46.57.
-
I do not understand why people are recommending the incorrect code below:
Dim r As Decimal = Math.Ceiling(d * 100D) / 100D
The correct code to round up should look like this:
Dim r As Double = Math.Ceiling(d)
1. Math.Ceiling works with data type Double (not Decimal).
2. The * 100D / 100D is incorrect will break your results for larger numbers.
3. Math.Ceiling documentation is found here: http://msdn.microsoft.com/en-us/library/zx4t0t48.aspx
-
Math.Ceiling has two overloads; one for decimal and one for double. – b_levitt Jan 5 '14 at 2:44
the only point that i don't know is true or not is point 2 that you make. maybe you could provide and example? the other 2 points you make (1 and 3) are incorrect. here is the math.ceiling documentation at its highest level, which will show 2 overloads: msdn.microsoft.com/en-us/library/… – taybriz Nov 5 '14 at 15:48
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https://codegolf.stackexchange.com/questions/186303/finding-row-wise-sum-of-transpose-of-hv-convex-binary-matrix
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# Finding row wise sum of transpose of hv-convex binary matrix [closed]
I'm stuck on a problem involving the Gale-Ryser Theorem. The problem's input gives me the row-wise sum of an hv-convex binary matrix(n*m).
e.g. I get {4,3,2,2,1} in the input. It's the row wise sum of the following matrix:
1 1 1 1
1 1 1 0
1 1 0 0
1 1 0 0
1 0 0 0
To solve the problem, I have to find the row-wise sum of it's transpose.
i.e. I need to calculate {5,4,2,1}
1 1 1 1 1
1 1 1 1 0
1 1 0 0 0
1 0 0 0 0
Can it be achieved in less than O(n*m)?
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You are Here: Home
# Partial Derivative question! ..... tan^-1(y/x)
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1. can anyone at all do this question please. rep will be given. :
find Gx, Gy, Gxx, Gyy, Gxy for G(x,y) = tan^-1(y/x) ... note tan^-1(u) = 1/(1+u^2).
cheers.
2. Hi, hope this helps:
G(x,y) = z = arctan(y/x) (arctan = tan^-1, just different notation)
therefore tan(z) = y/x;
differentiate both sides with respect to y:
sec^2(z) dz/dy = 1/x (sec^2(z) = tan^2(z) + 1)
and tan^2(z) = (y/x)^2;
so dz/dy = 1/(x(1 + (y/x)^2));
as for dz/dx, same procedure reveals:
dz/dx = -1/(y(1 + (x/y)^2));
differentiating again with respect to the same variables gives:
d^2z/dx^2 = 2xy/(x^2 + y^2)^2;
d^2z/dy^2 = -2xy/(x^2 + y^2)^2;
and the cross derivative gives:
d^2z/dxdy = 1/(x^2 + y^2) - 2x^2/(x^2 + y^2)^2;
I believe these to be correct, however there may be sign errors in my workings out as I rattled through these quickly. Any that aside, this is the general jist of how to do these derivatives,
Hope this helps.
3. thanks!! i believe ur answers are correct!
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Enumerative geometry
In mathematics, enumerative geometry is the branch of algebraic geometry concerned with counting numbers of solutions to geometric questions, mainly by means of intersection theory.
History
The problem of Apollonius is one of the earliest examples of enumerative geometry. This problem asks for the number and construction of circles that are tangent to three given circles, points or lines. In general, the problem for three given circles has eight solutions, which can be seen as 23, each tangency condition imposing a quadratic condition on the space of circles. However, for special arrangements of the given circles, the number of solutions may also be any integer from 0 (no solutions) to six; there is no arrangement for which there are seven solutions to Apollonius' problem.
Key tools
A number of tools, ranging from the elementary to the more advanced, include:
Enumerative geometry is very closely tied to intersection theory.
Schubert calculus
Enumerative geometry saw spectacular development towards the end of the nineteenth century, at the hands of Hermann Schubert.[1] He introduced for the purpose the Schubert calculus, which has proved of fundamental geometrical and topological value in broader areas. The specific needs of enumerative geometry were not addressed until some further attention was paid to them in the 1960s and 1970s (as pointed out for example by Steven Kleiman). Intersection numbers had been rigorously defined (by André Weil as part of his foundational programme 19426, and again subsequently). This did not exhaust the proper domain of enumerative questions.
Fudge factors and Hilbert's fifteenth problem
Naïve application of dimension counting and Bézout's theorem yields incorrect results, as the following example shows. In response to these problems, algebraic geometers introduced vague "fudge factors", which were only rigorously justified decades later.
As an example, count the conic sections tangent to five given lines in the projective plane.[2] The conics constitute a projective space of dimension 5, taking their six coefficients as homogeneous coordinates, and five points determine a conic, if the points are in general linear position, as passing through a given point imposes a linear condition. Similarly, tangency to a given line L (tangency is intersection with multiplicity two) is one quadratic condition, so determined a quadric in P5. However the linear system of divisors consisting of all such quadrics is not without a base locus. In fact each such quadric contains the Veronese surface, which parametrizes the conics
(aX + bY + cZ)2 = 0
called 'double lines'. This is because a double line intersects every line in the plane, since lines in the projective plane intersect, with multiplicity two because it is doubled, and thus satisfies the same intersection condition (intersection of multiplicity two) as a nondegenerate conic that is tangent to the line.
The general Bézout theorem says 5 general quadrics in 5-space will intersect in 32 = 25 points. But the relevant quadrics here are not in general position. From 32, 31 must be subtracted and attributed to the Veronese, to leave the correct answer (from the point of view of geometry), namely 1. This process of attributing intersections to 'degenerate' cases is a typical geometric introduction of a 'fudge factor'.
Hilbert's fifteenth problem was to overcome the apparently arbitrary nature of these interventions; this aspect goes beyond the foundational question of the Schubert calculus itself.
Clemens conjecture
In 1984 H. Clemens studied the counting of the number of rational curves on a quintic threefold and reached the following conjecture.
Let be a general quintic threefold, a positive integer, then there are only a finite number of rational curves with degree on .
This conjecture has been resolved in the case , but is still open for higher .
In 1991 the paper[3] about mirror symmetry on the quintic threefold in from the string theoretical viewpoint gives numbers of degree d rational curves on for all . Prior to this, algebraic geometers could calculate these numbers only for .
Examples
Some of the historically important examples of enumerations in algebraic geometry include:
2 The number of lines meeting 4 general lines in space
8 The number of circles tangent to 3 general circles (the problem of Apollonius).
27 The number of lines on a smooth cubic surface (Salmon and Cayley)
2875 The number of lines on a general quintic threefold
3264 The number of conics tangent to 5 plane conics in general position (Chasles)
609250 The number of conics on a general quintic threefold
4407296 The number of conics tangent to 8 general quadric surfaces Fulton (1984, p. 193)
666841088 The number of quadric surfaces tangent to 9 given quadric surfaces in general position in 3-space (Schubert 1879, p.106) (Fulton 1984, p. 193)
5819539783680 The number of twisted cubic curves tangent to 12 given quadric surfaces in general position in 3-space (Schubert 1879, p.184) (S. Kleiman, S. A. Strømme & S. Xambó 1987)
References
1. Schubert, H. (1979) [1879]. Kalkül der abzählenden Geometrie.
2. Fulton, William (1984). "10.4". Intersection Theory. ISBN 0-387-12176-5.
• Kleiman, S.; Strømme, S. A.; Xambó, S. (1987), "Sketch of a verification of Schubert's number 5819539783680 of twisted cubics", Space curves (Rocca di Papa, 1985), Lecture Notes in Math., 1266, Berlin: Springer, pp. 156–180, MR 0908713
• Schubert, Hermann (1979) [1879], Kleiman, Steven L., ed., Kalkül der abzählenden Geometrie, Reprint of the 1879 original (in German), Berlin-New York: Springer-Verlag, ISBN 3-540-09233-1, MR 0555576
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# *
TV100
The answer is 3 Aces +3 Kings + 4 Tens (straight draw) =10 Outs. So you’re number of outs just INCREASED. If either a King or Ace hit the board. They refer to the cards that can hit the board. a ten will give you the nut straight… and presumably the best hand. the stronger your hand. Now the turn comes and the board looks like this: NOW how many outs do you have? Well. The more outs you have. . now you’re just one spade away from a flush. the better. The answer is 3 Aces + 3 Kings + 4 Tens + 9 Spades (flush draw) – 1 Ten of Spades = 18 Outs. So those cards can be considered outs as well. “Outs” are the cards in the deck that can give you a winning hand.Calculating Outs The first step to learning poker math is to learn how to calculate “outs”. you’ll have top pair. The more outs you have. For example… let’s say you’re holding: The flop comes out: How many outs do you have? Well.
OK.Notice that the ten of spades was SUBTRACTED at the end of our calculation. if someone bet heavily after both the flop and turn. you might put them on a hand like two pair or three-of-a-kind. This is important.like tells and psychology. When calculating odds. OK. Obviously. In our example above. This is a rare occurrence. let’s do another example. Outs are ONLY cards that will give you the winning hand. never count the same card twice. The question becomes… how do I really KNOW what the winning hand will be? And the answer is you don’t. you would only calculate the four tens. you’ve got a fantastic hand since you’re on BOTH the nut straight draw AND the nut flush draw. so what if someone was holding a Jack and a Queen and had two pair. Why? The reason is because we already counted it with the four tens in the deck that would give us the straight. The flop comes out: . the nine spades. and then subtract the ten of spades in order to figure your outs (the answer is twelve). of course. This is one of the primary limitations of poker odds and calculations… but it’s also good because it maintains the unpredictable nature of the game and paves the way for other strategies-. In that case. getting top pair would no longer give you the best hand… which means the three Kings and three Aces in the deck are no longer outs. How would that change things? Well. Say you’ve got pocket deuces and limp in before the flop.
75 to 1 4.40 to 1 6.71 to 1 1.40 to 1 1.40% 25.17% 4.77% 14.57 to 1 4.27 to 1 2. The number of OUTS you have is nine (since there are thirteen clubs in the deck and you’re already using four of them).71 to 1 4.29 to 1 2. under the title “Probability”.50 to 1 14.30% 40.54 to 1 2.91 to 1 3.30 to 1 1.91% 26.26% 30.83 to 1 2.84% 31.43 to 1 Outs 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 One Card (Turn) 2.35% 6.72% 44.16% 54.61 to 1 1.26% 6. For instance.67 to 1 5. you can quickly calculate your PERCENTAGE OF WINNING the hand.83 to 1 5.04% 36.67 to 1 0.45% 34.39% 19.38% 8.67 to 1 10.42 to 1 1.35 to 1 1.96% 48.13% 41.68% One Card (Turn) 46.50 to 1 8.60 to 1 0. But once you learn to use it you’ll find it to be quick.70 to 1 3.42% 12.97% 38.11 to 1 3.22 to 1 3.43% 32.98% 59.57% 21.Calculating Percentages All right.43% 42. I just want you to pay attention to the left side.07 to 1 1.88 to 1 1.22 to 1 1. One more club will give you the flush.92 to 1 2.26% 8. To figure your percentage.20 to 1 6.66% 29. Probability Cards To Come One Card Two Cards (River) (Turn And River) 2.14 to 1 2.02% 19. The ROW would be the one that says nine outs.86 to 1 1.14% 27.08 to 1 0.47 to 1 1.33 to 1 10.96% 39.76% 62.30% 43.61% 34.75 to 1 0. just take a look at the chart and find the corresponding row and column.94 to 1 1.22% 17.50 to 1 10.88 to 1 7. let’s say you’re on a club flush draw after the flop.39% 41.44% 65.47% 23.95 to 1 0.15% 21.47% 20.23% 23.94% Odds Against Cards To Come One Card Two Cards (River) (Turn And River) 45.17% 38.87% 13.00 to 1 22.79% 31.36 to 1 2.35% 24. The COLUMN would be “One Card (Turn)”… since you know the flop and have the turn card to come.59 to 1 2.53% 27.03% 67.01 to 1 5.04% 15.12% 56. this chart looks pretty intimidating.48% 45.07 to 1 3.60 to 1 3.75 to 1 8.88 to 1 4. .13% 4.48 to 1 0.18 to 1 2.76% 36.56 to 1 1.85 to 1 0.62 to 1 2.52% 8. and efficient.49% 16.00 to 1 14.19 to 1 22.64% 12.24 to 1 The percentage numbers are the probability that you will catch one of your OUTS. easy. We’ll go through each segment of the chart… but for now. Below is the chart that I use to calculate all the odds at the poker table.76 to 1 1.09% 28.55% 44. Using the number of outs you have in a situation.13 to 1 1.51% 10.53% 69.18 to 1 1.10% 51. now you’re ready to learn the percentages. At first.54 to 1 0.91% 34.65% 4.70% 10.60 to 1 1.00 to 1 22.89% 17.
which means the percentage on the river is slightly higher.79% 31.04% 36. It’s not used to calculate pot odds.57% 21. For the river.Probability Cards To Come One Card One Card Outs 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 (Turn) 2.02% 19. 19.48% 45.94% So your percentage chance of getting your flush on the TURN CARD is 19. After the turn card comes out.91% 34.03% 67. This is found by following the same row.61% 34.68% (River) 2.96% 39. As we’ll learn.22% 17.49% 16.43% 42.42% 12.64% 12.15% plus 19.70% 10.35% 6.15% 21.57%.51% 10.76% 62.97%. but using the next column called “One Card (River)”.14% 27. You might be wondering why the odds for the turn aren’t the same for the river.04% 15.09% 28.84% 31.43% 32.13% 41.96% 48.55% 44.52% 8.53% 27.87% 13.44% 65. The final number in this row is 34.26% 6.30% 40.12% 56.97%.30% 43.65% Two Cards (Turn And River) 4.26% 8.13% 4.45% 34.97% 38.40% 25.98% 59. The reason is because the percentage is the number of outs divided by the number of “unknown” cards.23% 23.91% 26.53% 69.66% 29.57% doesn’t equal 34.16% 54. this number is NOT as important as you’d think. there’s one less “unknown” card.47% 20. Why is that? .17% 4.10% 51. This is the percentage chance you have of making your flush on EITHER the turn or river.39% 41. it would be 19.89% 17.38% 8.26% 30.47% 23.72% 44.77% 14. You also might be wondering why the turn card column and river card column don’t add up to equal the turn and river card column.15%.17% 38.35% 24.76% 36.39% 19. For instance.
That number (1/4) is then subtracted from one to give you ¾. here’s a quick example that makes it easy to remember why… Pretend you have a coin.The answer is related to some complicated math. The REAL answer is 75%. But for the curious. You get dealt suited connectors: The flop comes out: This gives you two over cards and an open-ended straight draw. Why is it figured that way? I don’t know and I don’t want to know. But what are the odds you’ll make it EITHER the first or second time? If you add 50% plus 50% you’d get 100%. You’re going to flip it twice. so let’s get back to the percentage charts so that you can win some more pots. and want to know the odds of making “heads”. . what’s the percentage you’ll make one of your “outs” on the turn? See if you can figure it out on your own right now. but obviously that’s wrong… since there’s always the chance of flipping tails twice in a row. or 75%. your odds are 50%. For the second flip. Let’s do another scenario for calculating outs and percentages. Who cares? It has nothing to do with poker. For the first flip. You don’t put anyone on a pocket pair or two pair yet… so what are your odds of having the “winning” hand on the turn? In other words. your odds are 50%. This is figured by taking the odds AGAINST making heads for the first flip (1/2) multiplied by the odds AGAINST making odds on the second flip (1/2).
51% 10.57% 21.16%.09% 28. you’ve got a 29.77% 14.02% 19. 17.35% 24.26% 8.48% 45.43% 32. but we’ll treat it like a winner for our purposes here.79% chance of making your hand on the turn.12% 56. Get it? .72% 44.02% chance on the turn.89% 17.84% 31.96% 39.68% (River) 2. These are the cards that can help you: 4 fives + 4 tens + 3 nines + 3 eights = 14 outs The eights and nines are over cards.76% 62. you’d have eight outs.03% 67.53% 69.55% 44.52% 8.98% 59.40% 25.44% 65.17% 38.OK.43% 42. Step one is to calculate the outs.47% 20.30% 40.96% 48.97% 38. now for the answer.17% 4.10% 51.43% for the river. Probability Cards To Come One Card One Card Outs 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 (Turn) 2.66% 29.13% 41.26% 6.70% 10.39% 19.45% 34.47% 23.35% 6.91% 26.16% 54. Top pair isn’t necessarily a winner.94% With thirteen outs.64% 12.39% chance on the river.04% 36.91% 34.23% 23. which means getting one of them will give you top pair.49% 16. and a total chance of 51. and a 31.65% Two Cards (Turn And River) 4.14% 27.15% 21.61% 34. Now it’s time to use the probability charts.76% 36. That means you’d have a 17. a 30.53% 27.13% 4.79% 31. If you only consider the open-ended straight draw for your outs (and not top pair).26% 30.45% chance for both the turn and river taken together.39% 41.30% 43.22% 17.38% 8.87% 13.42% 12.04% 15.
Let’s say you’re dealt pocket Queens: The flop comes out: Let’s say you pick up a read on your two remaining opponents in the hand. but it actually helps YOU. If another Ace comes out he’ll have trips. Using the percentage charts you’ll see that there’s a 31. The turn card comes: This doesn’t help your opponents. Ok. The reason is because now the seven of clubs (which gives your opponent the flush) would pair the board and give . since it’s relatively insignificant. I’ll give you the steps to calculating it later. what are the odds your opponents will MAKE their hands (catch their outs) and beat you? Well. The other opponent needs a runner-runner situation to stay alive. the flush draw has eight outs (the nine remaining clubs minus the Queen of clubs in your hand). Using that information. so ROUGHLY speaking.45% chance that opponent will make his hand (on either the turn or river). you’ve got a 68. You think the other is on a club flush draw. We’ll ignore the runner-runner calculation for now. You think one of them has an Ace. but that will give you a winning full house. Except this time let’s calculate the odds that you’ll LOSE a hand… based on your “read” of the opponents at the table.OK.55% chance of winning so far. let’s do one more example.
22% chance of winning on the river. let’s keep working on the “foundation” and go over how to calculate runner-runner odds… . since you can’t carry them around. after you learn how to calculate betting odds and make pot odds comparisons. easy shortcut you can use to INSTANTLY know the percentages in your head based on a given number of outs. I’ll show you the SHORTCUTS for figuring out percentages WITHOUT these charts. The river comes out: Your opponent has the flush so he goes all-in. and win a monster pot. But for now. so that’s how you use the odds percentage charts. And of course. and can be used any time you play online poker. These charts are very useful for learning more about probability in poker. Since full house beats a flush. Queens full of sevens. You’ll love it.78% chance of winning. For “offline” poker. We’ll also talk more about why the column for the turn plus river percentage is rarely used in calculating pot odds. That gives you a 84. At the end of this report. we’ll tie everything together by giving you PRACTICAL applications of all this knowledge. You call with your full house. Using the charts you’ll see that means your opponent has a 15. your opponent on the club flush draw has now lost an out. There’s a simple. these charts aren’t quite as useful. He’s down to seven outs. OK.you the full house.
Assuming he makes that. . We approach this problem by first deciding HOW your opponent can win.Calculating Runner-Runner Odds Let’s say you get dealt pocket tens: Your opponent is dealt pocket nines: You make a pre-flop raise and he calls you.17%. But for the sake of example. To figure out the math. Everyone’s cards are turned over. He could also catch a seven and an eight. he has one out left for the river (the other nine). which would give him the straight. which equals a 4.26% chance. You call. The flop comes out: You have trips. Those are the only hands that can save him. which would give him quads. Calculating odds here isn’t important. OK. He can get two consecutive nines. we use our handy percentage charts. so to get quad nines he needs a nine on the turn AND a nine on the river. The odds of making one out for the river is 2. let’s look at your opponent’s chances of winning the hand by catching something runner-runner. since your “control” over the hand is over. You make a bet and your opponent goes over the top of you and goes allin. He has two outs on the turn.
39% 41.76% 62. That means there are eight outs on the turn card (either the seven or eight).09% 28.481% chance that your opponent will catch a runner-runner straight. This gives your opponent a .52% 8.10% 51.98% 59.04% 15.40% 25.15% 21.45% 34.17% 4.43% 32.16% 54. If your opponent makes THAT card.49% 16.70% 10.26% 6.76% 36. he’ll have four outs on the river. For example. you can see there’s a 17.48% 45. since they both must occur. if he hits the seven on the turn. there are four eights left to make on the river.0924% chance of winning.35% 24.66% 29. You can add this to the .51% 10.84% 31.26% 8. So you’re in pretty good shape of winning the hand.53% 27.02% 19.13% 4. and an 8.79% 31.22% 17.47% 23. The answer is a 1.77% 14.04% 36.30% 43.38% 8.70% chance of making one of the outs on the river.97% 38.13% 41.96% 39.23% 23.26% 30. This equals a runner-runner “miracle” chance of 1.39% 19.61% 34.64% 12.573%.17% 38.47% 20.89% 17. Now what about the straight draw? The straight is more likely to happen. That’s basically a 1 in 1000 chance.55% 44. since there are more cards to hit.96% 48.91% 26. .03% 67. See below: Probability Cards To Come One Card (River) 2.02% chance of making one of the outs on the turn.94% Once again.65% Outs 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 One Card (Turn) 2.68% Two Cards (Turn And River) 4.43% 42.0924% chance of a four-of-a-kind.The way to calculate the OVERALL percentage is by MULTIPLYING these two percentages.14% 27. Using the percentage charts.30% 40.87% 13.53% 69.72% 44.44% 65. There are four sevens and four eights in the deck.42% 12.35% 6.12% 56. we multiply these numbers to figure the chance that BOTH scenarios will happen.91% 34.57% 21.
In general. After this you’ll be equipped for real “pot odds” situations. That seems to be the “average” for many situations. You can decide for yourself whether or not to factor this 1% in your decisions. how to use the percentage charts. I don’t use it. Let’s look at how to calculate POT SIZE. . since the number is so small and because it requires so much math. I do NOT make runner-runner calculations at the poker table. It’s just not practical. you can treat the odds of a runner-runner as about 1%. As a general “rule”. OK. so now you know how to calculate outs. and how to calculate runner-runner situations.
YOUR \$25 bet doesn’t get included in the total pot size. the \$25 you called with after the flop IS included. In the example above.you’re about to wager. since now it’s officially in the pot. because it’s not in there yet. This time. . But the \$50 you may or may not call with is NOT included… because it’s still yours for now. and Drew bets \$50. All right… so that’s how to calculate pot size. What’s the pot size then? The answer is \$40 (pre-flop) + \$50 (after the flop) + \$50 (bet on turn from Drew). You were trying to make a DECISION about calling a \$25 bet. you had already called the big blind of \$10… so that \$10 gets counted. Let’s say in our example that you called and the other players behind you folded. Now let’s say the turn card comes. you’re ready to learn “pot odds” and how to APPLY the information you’ve learned to real-life poker situations. Now that you know pot size and outs. So it’s just you and Drew heads-up.
45% 19.50 to 1 5.62 to 1 2.19 to 1 0.20 to 1 3.09% 44.55% 44.77% 14.39% 23. we’ll only be looking at the RIGHT side this time. First you figure out how many OUTS you have.47 to 1 1.97% 21.76 to 1 1.30 to 1 0.66% 29.79% 31.52% 12.30% 40.23% 23. Here’s the main chart again: Probability Cards To Come One Card Two Cards (River) (Turn And River) 2.48% 67.07 to 1 0.75 to 1 1.84% 17.60 to 1 1.68% One Card (Turn) 46.40 to 1 6.83 to 1 5. which will help you know whether a decision is “justified” according to the odds or not.24 to 1 For our purposes here.94 to 1 1.87% 20.57 to 1 2.53% 27.30% 65.75 to 1 8.00 to 1 22.17% 4.47% 38.96% 28.26% 4. It means you’ll win one out of five times… or 20% of the time.70 to 1 3.36 to 1 on the turn and 2.18 to 1 4.85 to 1 1.29 to 1 0.07 to 1 8.35 to 1 1.89% 17.60 to 1 1.42 to 1 0.33 to 1 7.22 to 1 2.59 to 1 4.48 to 1 1.15% 21.10% 30.61 to 1 1.56 to 1 0.18 to 1 1.13% 62.39% 31.42% 6. it means the odds against you are 2. The chart works the same way as before.57% 34.54 to 1 1.36 to 1 2. If the odds against you are 4 to 1 (also written 4:1).91% 34.60 to 1 3. .01 to 1 10.91% 41.71 to 1 4.92 to 1 2.76% 39.50 to 1 14.38% 8.43% 42.02% 19.13% 4.76% 56.04% 36.17% 38. under the heading “Odds Against”. if you have 14 outs after the flop.54 to 1 1.44% 41.72% 26.12% 34.29 to 1 on the river.40% 25.40 to 1 2.98% 36. Then you compare that to the corresponding column… whether the turn card is about to come or the river card is about to come (or if you want to see BOTH the turn and river cards together).88 to 1 4.88 to 1 14.00 to 1 10.03% 43.64% 12.00 to 1 22. For example. This time we’ll be dealing with the “odds against” something happening.08 to 1 2.95 to 1 2.35% 13.Calculating Pot Odds Now it’s time to use the RIGHT side of the probability charts we saw earlier.51% 10.75 to 1 2. that means you will NOT get your card for every four times that you DO get it.27 to 1 2.53% 45.96% 59.50 to 1 22.11 to 1 1.94% Odds Against Cards To Come One Card Two Cards (River) (Turn And River) 45.14 to 1 5.65% 69.22 to 1 3.83 to 1 1.67 to 1 3.70% 16. Let’s look at what “odds against” really MEANS.49% 8.43% 51.71 to 1 0.14% 15.13 to 1 1.67 to 1 1.16% 32.26% 6.86 to 1 3.67 to 1 10.47% 10.04% 24.43 to 1 Outs 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 One Card (Turn) 2.91 to 1 6.26% 48.35% 8.88 to 1 0.61% 54.22% 27.
Which is bigger? 150:20 is equal to 7. Let’s say the odds against you winning are 7:1. It’s really critical that you “get” this. So that means if you played the hand eight times. Are the odds in your favor to call or fold? The answer is to compare 150:20 to 7:1. the “1” represents the time you win. OK. it means the odds you have of making your hand compared with the odds of the betting. you’d win once and . That means you won ONCE out of FIVE times. but that’s NOT the case. so let me explain. the odds come out even. The “1” represents the BET SIZE that you must make a decision about. So that means if you play the odds in your favor consistently. you’ll encounter situations dozens of times per hour where you’ll either get the card or you won’t. 4:1 equals 1/5. if you lost \$5 every time that situation occurred. If you win exactly \$20. Someone made a \$20 bet and the action is to you. In poker. Now. Four times you lose. and the “4” represents the times you lose. everyone’s odds come back out to “equal”. OK. here’s what would happen (in terms of probability): Lose Lose Lose Win Lose That’s in no particular order. over the long term you’ll come out on top.A lot of people misconstrue 4:1 to mean ¼. the odds are against you. Over time. Still with me? With that being said. If you win \$21 or more. back to the calculations. the bet size is \$5. now when you hear the phrase “pot odds”. and the hand plays out five times. that means you’d lose \$20 total for the four losses. If the odds are 4:1. one time you win. Let’s look at a different scenario. If you win \$19 or less. The “4” represents the pot size. then the odds are in your favor. because it’s a fundamental aspect of poker math. The goal is to always be able to “justify” a call according to the odds… assuming all other things are equal. so the pot size would have to be MORE THAN \$20 in order to justify a call. With odds against you of 4:1. In our scenario it’s \$5. which is bigger than 7:1. I just covered a lot of ground there.5:1. you want to WIN MORE THAN \$20 the one time you win… that way you make a PROFIT. That means the POT SIZE compared to the BET SIZE should be BIGGER than 4:1. You’ve figured the pot size to be \$150. For example… let’s say the odds against you are 4:1 and you must decide whether to call a \$5 bet. of course. In this case.
19 to 1 0.18 to 1 1.76 to 1 1.92 to 1 2. A \$4 bet with an \$8 pot after the flop when you have 8 outs: J U .22 to 1 3.88 to 1 4.50 to 1 5.27 to 1 2.75 to 1 2.29 to 1 0.00 to 1 22.35 to 1 1.62 to 1 2.59 to 1 4.60 to 1 1.30 to 1 0.lose seven times. If the betting odds are smaller.33 to 1 7. A \$2 bet on the turn (river card is left) with a \$12 pot when you have 7 outs: J U 2. Here’s the “odds against” chart.40 to 1 6.13 to 1 1. So you’d come out on top with a net profit of \$10.88 to 1 0.00 to 1 22. the call is not justified. You’ll know this QUICKLY by simply figuring out if the BETTING ODDS are bigger or smaller than the HAND ODDS. or “U” for an unjustified call.14 to 1 5. you should call. So yes.71 to 1 4.40 to 1 2.07 to 1 0.47 to 1 1.61 to 1 1.) 1.00 to 1 10.67 to 1 1.86 to 1 3.54 to 1 1.67 to 1 3. If the betting odds are bigger.07 to 1 8.60 to 1 3.56 to 1 0.20 to 1 3.57 to 1 2.71 to 1 0.42 to 1 0.01 to 1 10.11 to 1 1.43 to 1 Outs 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 One Card (Turn) 46.88 to 1 14.48 to 1 1.83 to 1 1. The questions come right after it with the answers at the end. Odds Against Cards To Come One Card Two Cards (River) (Turn And River) 45.85 to 1 1.18 to 1 4.83 to 1 5. (Ignore “implied odds” if you’re familiar with them.94 to 1 1.24 to 1 Circle “J” for a justified call.60 to 1 1.50 to 1 14.08 to 1 2. That means you’d lose \$140 (\$20 x 7) but win \$150 (the pot size).54 to 1 1.91 to 1 6.22 to 1 2.50 to 1 22.70 to 1 3.36 to 1 2.75 to 1 1.75 to 1 8.95 to 1 2. All right… let’s do a quick quiz to test your skills. a call is justified.67 to 1 10.
just email me at roy@royrounder. and other related factors to consider in a hand… U . U 3. discounting odds. An opponent moves all-in after the flop for 275 chips making the pot 500 while you have an inside straight draw and the nut flush draw. J U (Hint: You have 12 outs.com and I’ll email you an explanation of each. But I’ll assume you aced them all for now.3. But now it’s time to talk about implied odds. J 2. We’re going to get back to pot odds soon. OK.) 4. U) How’d you do? If you had trouble with these. A \$10 bet after the flop with a \$65 pot when you have an inside straight draw. J … … … … … … … … … … Here are the ANSWERS… (1. so now you understand how to use “odds against” to calculate pot odds. J 4.
800 into a \$800 pot. You have lots of outs here.That means your opponent has the flush and you have trip aces. There’s: 1 ace + 3 fours + 3 eights + 3 nines + 7 clubs left = 20 outs The fours. But you know that if you DO get your full house or nut flush. including the implications of LOSING all your money (Is it a tournament or cash game? Are there re-buys?). and nines give you a full house. let’s move on to “discounted odds”. The other club would give you the nut flush. you’ll be able to move the REST of your chips into the middle and double up. eights. With twenty outs. There’s just one river card left to go. This is an extreme example. which is part of the reason why the behavior from one card player to the next is so different. how many players are at the table. the explicit odds STILL don’t quite justify a call. Your opponent (who is chip leader) bets \$2. and so on. All right. how big your chip stack is. You’d factor a lot of different things into this type of decision. . but it shows the importance of implied odds. Most all-in decisions are made according to implied odds.
This information brings the number of outs down to six. Now. For example. you don’t necessarily KNOW if someone has the club draw in our example. . This type of decision making is where the ability to READ players meets with the ability to do poker math. that means you have eight outs. Since OUTS refers to cards that can give you the WINNING HAND. Normally. let’s say you put your OPPONENT on a club flush draw. because you’re DISCOUNTING cards that will help someone else’s hand. But a problem arises when one of YOUR outs is also one of your OPPONENT’S outs. This changes the calculation considerably. but based on a player’s betting patterns and history of play you might be able to INFER that he does. This concept is called “discounted odds”. of course.Calculating Discounted Odds Calculating the number of outs in order to make a hand is rather easy. Let’s say he’s holding: That means if an eight of clubs or a King of clubs hits the board. let’s say you’re holding: And let’s say the flop reads: You have an open-ended straight draw. your opponent will have the FLUSH and you’ll have the straight. However. the eight and King of clubs are no longer “outs” for you… since they give you a losing hand.
Let’s say you’re sitting at a 10-man table and five players see the flop. Well. If you need an Ace as one of your “outs”. . this unusually high number of players would suggest that at least one or two… quite possibly more… of the Aces are in other players’ hands. you’d calculate the number with one or two outs… since “common sense” tells you that other players have some Aces.One of the important parts of discounting relates to how many players are at the table. Once again. there’s no science to this. it would be smart to DISCOUNT a couple of them from your calculation. Instead of saying you have four outs (for all four aces). You’ve got to combine your feel for the players and the table with your odds calculations in order to use discounted odds.
Miscalculating Odds There are three main MISTAKES players make when it comes to odds for no limit Texas Holdem. That means you have six outs: Discounted cards: . The first is simply miscalculating the OUTS. let’s say you’ve got an open-ended straight draw and you’re certain your opponent has a flush draw. You’re holding: The flop reads: Your opponent has: Normally here you would have eight outs. since you have an open-ended straight draw. For example. But since you put your opponent on the flush. you discount the two of diamonds and the seven of diamonds.
67:1. Let’s say you have six outs. This is critical. you’re forced to make another decision in order to see the river card. Here’s why: Because the \$20 bet is JUST for the turn card… not the turn and river. If the action were to you to call a \$20 bet with a \$100 pot size. Period. you should focus only on the odds of making your hand for one specific round of betting. Think about it… after the turn hits. That means the odds of making your hand on the turn is 6. And this time the bet will probably be HIGHER. And that’s why the 3.14:1 is better than 5:1. So does that mean you should call the bet? NO! Absolutely not.14:1 stat is irrelevant. But besides those cases. And if you call that. you can use the odds of making your outs on EITHER the turn or river in order to make a decision to call. if you have a lot of draws after the flop and someone goes all-in. So now the question becomes… Can you EVER use the odds figure of making your hand on EITHER the turn or river? The answer is yes. Every round of betting is INDEPENDENT of the other rounds. The third big mistake I’ve seen regarding odds is USING THEM AT THE WRONG TIME. your opponent is going to bet AGAIN. 3. you’d need to be getting better than 5:1 on your hand. The number to pay attention to is 6. when calculating pot odds you must pay attention to the probability of getting an “out” on the NEXT card… NOT on the next two cards. If he bets \$40 into a \$140 pot. The number can be used when making an ALL-IN decision after the flop. For example. That’s why I said earlier in this report that the column on our chart that says “Turn And River” isn’t used to calculate pot odds. .83:1.Remember. So you should fold. The odds of making your hand on EITHER the turn or river is 3. which isn’t good enough to justify a call.14:1.83:1 and the odds on the river is 6. then you’ve just spent a total of \$60 to see both the turn and river… rather than \$20.
I’ve said it a million times: poker math is a TOOL. you should probably just fold the hand and live to see another day. a lot of amateurs and “fish” out there make DUMB decisions at the poker table. Even though the pot odds might dictate folding.You see. It’s the same way with overly-aggressive players. if someone has played extremely tight the entire game and comes out betting aggressively after the flop. Odds are not meant for every situation… and you can’t rely on them too much. For example. sometimes a call will be a better play. you can put that player on a monster. When you encounter one of these players. . It all depends on the players you’re up against. especially in no limit Texas Holdem. Even if “odds” dictate a call in your position. you’ll want to make YOUR decisions based on your read of the situation more than the “odds” of winning. nothing more.
52% 8.38% 8.87% 13.66% 29. 27. First off.76% 36.48% 45.96% 48.10% 51. Here is the chart again: Probability Cards To Come One Card (River) 2.16% 54. With that being said.40% 25.09% 28.43% 42.96% 39.13% 41. Up until this point we’ve been using the CHARTS as our source of data.45% 34.91% 26.22% 17.15% 21.94% As you know.35% 24.91% 34.30% 43. etc.35% 6.49% 16.13% can be considered 2%.04% 36. let me show you a neat shortcut.76% 62. it’s not really necessary to know the EXACT percentage. Now I want to teach you how to use the odds WITHOUT charts… and without complicated calculations.04% 15.17% 38.44% 65.84% 31. For instance.23% 23.66% can be considered 28%.68% Two Cards (Turn And River) 4.38% can be considered 6%.26% 6. let’s get started. for all practical purposes.13% 4.53% 27.77% 14.70% 10.42% 12. you’ll notice that you can find the percentage by DOUBLING THE OUTS and adding ONE. The “formula” looks like this: .02% 19.03% 67.43% 32.17% 4. 6. based on the percentage charts.12% 56.51% 10.65% Outs 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 One Card (Turn) 2.39% 41.57% 21.72% 44.89% 17. so now you know the details of “poker math” and how to calculate pot odds… while also taking implied odds and appropriate adjustments into consideration.53% 69.79% 31.61% 34.Shortcuts For Calculating Odds OK.47% 20.97% 38.26% 30.47% 23. This will give you the power to understand in-depth poker math IN YOUR HEAD… without being a math genius.30% 40.26% 8. 2.64% 12.55% 44. OK.14% 27.98% 59.39% 19.
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# ECD — Estimated Completion Date
## Estimate At Completion (EAC) Formulas In Project Management
You might be thinking why there are so many Estimate At Completion (EAC) formulas in project management and how are they used for Earned Value Management (EVM) calculations.
EVM full of confusing terms and formulas. Estimate At Completion seems a misnomer. Well! How can you estimate anything when it is already complete?
I have written this post to discuss EAC in detail. You will find a complete explanation of EAC including its definition, examples, and formulas in this post. You will also find the difference between Estimate At Completion and Estimate To Complete (ETC) in this post.
This post also contains derivation of various EAC formulas that are used for forecasting calculations. After reading this post, you will also be able to solve and compute PMP exam’s mathematical questions these formulas.
You can also look at the following video to understand EAC.
### Estimate At Completion (EAC) In Project Management
Estimate At Completion is the revised estimate of the total funds required to complete total work of a project. It is the sum of the Actual Cost (expenditure already incurred or the money already spent) till the control date and Estimate to Complete (expected cost of remaining work).
Note: EVM helps in measuring current performance and forecast future performance of a project. EAC is a combination of performance till date and a forecast of future performance. In addition to EAC, Estimate To Complete, and To Complete Performance Index are also used for project forecasting.
The expected total cost of completing all work expressed as the sum of the actual cost to date and the estimate to complete.
PMBOK Guide
You can also refer to Max Wideman Glossary to read some other standard definitions.
### Estimate At Completion (EAC) vs Budget At Completion (BAC)
In EVM, the original and expected project budget is expressed as BAC and EAC respectively. BAC is approved budget at the start of a project. EAC is determined periodically at different control points as the project progresses.
The funds requirement of a project may change after the project starts. The original budget may no longer be valid. The project team may need more or less funds to complete the project.
This can happen due to various reasons cost variances, risks, incorrect assumptions etc. The team can analyze the reasons for the change and estimate or forecast a new budget. The new budgetary forecast is called Estimate At Completion.
It becomes revised project budget after the Sponsor’s approval.
### Estimate At Completion (EAC) vs Estimate To Complete (ETC)
ETC is the expected cost of completing the remaining project work whereas EAC is the estimated budget to complete all the project work.
In my experience, most professionals are confused about the difference between EAC and ETC. I think PM books have not treated this topic well. The books describe EAC and then derive ETC. But, it should be done the other way. EAC should be derived after finding ETC. It will be much simpler to understand.
Let’s read the PMBOK Guide’s definition of EAC again. EAC is the sum of the actual cost to date and the estimate to complete. So, EAC should be calculated only after computing ETC.
You should read my other article on Estimate To Complete before reading this article. I have taken many references from that article to explain EAC formulas.
### Generic Estimate At Completion Formula
Let’s read the EAC definition again and give it mathematical form. We can be write the following equation:
EAC = (Actual Expenditure Till Date) + (Estimated Future Expenditure)
Let us solve this further by replacing Actual Expenditure Till Date with AC and Estimated Future Expenditure with ETC. The above expression can be re-written as
EAC = AC + ETC
This is the only equation that we need to solve the PMP exam questions. It can be used to derive all the other formulas.
I will derive and explain following formulas in the subsequent section.
1. EAC = AC + ETC
2. EAC = AC + (BAC – EV)/CPI
3. EAC = AC + (BAC – EV)
4. EAC = BAC/CPI
5. EAC = AC + (BAC – EV)/CPIp
6. EAC = AC + Bottom-up ETC
7. EAC = AC + [(BAC – EV)/(CPI * SPI)]
8. EAC = AC + [(BAC – EV)/(x*CPI + y*SPI)]
The PMBOK Guide lists only 4 formulas. Most of the PMP study guides also talk about these same formulas. But, I have provided a more rounded view.
Let us derive each one of them. You should read my article on ETC in parallel for all the scenarios and examples.
### Calculating EAC Formulas
The above formulas can be mathematically derived by suitably replacing ETC figure in our generic equation with a ETC formula from my article on Estimate To Complete.
### Formula I
First one is same as our Generic Equation.
EAC = AC + ETC
### Formula II – Typical Performance
To derive the this, let’s replace ETC with ETC Formula I in our generic equation. Project’s past performance was typical and future work will be accomplished at the same rate. The resultant expression becomes
EAC = AC + (BAC – EV) / CPI
### Formula III – Atypical Performance
The PMBOK Guide lists Formula III but does not talk about Formula II. I have listed both these formulas separately as the former is a special case of the latter. In fact Formula IV is also a special case of Formula III.
To derive this, let’s replace ETC with ETC Formula II in our generic equation. Project’s past performance was atypical but the future work will be accomplished at the planned rate. The resultant expression becomes
EAC = AC + (BAC – EV)
### Formula IV
Let us mathematically solve Formula III above.
EAC = AC + (BAC – EV) / CPI
EAC = AC + (BAC / CPI) – (EV / CPI)
Refer to Basics of Earned Value Analysis. Let us replace CPI in above expression with (EV/AC). The above expression reduces to
EAC = AC + (BAC / CPI) – EV/(EV/AC)
EAC = AC + (BAC / CPI) – AC
EAC = BAC / CPI
Formulas II and IV are also referred to as Independent Estimate at Completion (IEAC).
### Formula V
To derive the this, let’s replace ETC with ETC Formula III in our generic equation. The resultant expression becomes
EAC = AC + (BAC – EV) / CPIp
This expression is similar to the Formula II above. However, it uses projected future CPIp instead of past CPI to calculate ETC.
### Formula VI – Bottom-up ETC
Refer to Scenario V in The ETC article. To derive the this, let’s replace ETC with Bottom-up ETC in our generic equation. The resultant expression becomes
EAC = AC + Bottom-up ETC
The term “Bottom-Up” has no specific significance. It just means that a fresh ETC should be determined by using Work Breakdown Structure (WBS). It can by found by determining the cost of remaining (unfinished) work components (work packages and activities) at the bottom of WBS and then aggregating them Upwards.
### Formula VII
Consider the following scenario.
The Sponsor wants to know the budgetary cost estimate for finishing the project work within the original schedule at current CPI.
None of the above expressions will work in this case. We need to factor in schedule performance as well.
We already know that project’s current efficiency is measured by two indices – Schedule Performance Index (SPI) and Cost Performance Index (CPI).
A SPI of ‘S’ means that ‘S’ units of work was done in each duration unit (e.g. a day).
A CPI of ‘C’ means that \$C worth of work was done for each dollar spent.
We have to consider both CPI and SPI to determine the budgetary cost to complete the project work within original schedule at current CPI.
Refer to Formula III – it uses only CPI. If we modify it to include SPI also, then it becomes
EAC = AC + (BAC – EV) / (CPI * SPI)
### Formula VIII
There is another variant of the above expression, which can also be used for calculations. In the following formula x & y are weights given to CPI & SPI respectively. These weights signify how much importance the project team is willing to give to each performance factor. The sum of x & y should be 1, e.g. 0.2 & 0.8 or 0.5 & 0.5.
EAC = AC + (BAC – EV) / (x*CPI + y*SPI)
### How To Use EAC Formulas In The PMP Exam?
The PMBOK Guide describes only 4 formulas. In this post, I have extended the concept further to describe a generic equation and 8 different formulas.
If you are preparing for the PMP exam, only PMBOK Guide’s formulas should be sufficient. Here is my brief take on how you should solve the PMP questions.
You can use one of the following formulas depending the situation described in the exam question
When future work will be completed at the budgeted rate (at planned efficiency – atypical)
EAC = AC + BAC – EV
When future work will be completed at the current rate (at current CPI – typical)
EAC = AC + (BAC – EV) / CPI
Or
EAC = BAC / CPI
When future work will be completed at current rate within the original Schedule (considering both present CPI & SPI)
EAC = AC + (BAC – EV) / (CPI * SPI)
### Final Thoughts
EAC provides an revised estimate to complete the project work. It is a derived from ETC. EAC and ETC are important metrics to gauge the health of a project. They should be revised periodically as project progresses and moves towards the completion.
There are number of other formulas in EVM. You can read Earned Value Management Formulas for a quick snapshot of all of them. You need to understand the these to answer the PMP questions correctly.
A mere memorization of the formulas is not useful. You may not be able to apply the correct formula in the exam question. It is better understand the concept and then apply the formula(s) as required.
### Over To You
EVM is difficult topic. Do you still have any confusion about Estimate At Completion? You can write your question in the comments section below and I will respond to it.
To Complete Performance Index
EVM – Is it Useful?
### PMP Exam Formulas
I have also compiled a PMP Formulas Cheat Sheet. It contains 45 formulas and 57 abbrviations. It will help you in your exam prep. You can freely download the PMP Formulas Sheet for your studies. It is the best and most comprehensive cheat sheet the PMBOK Guide 6th edition.
If you are looking beyond a cheat sheet, then I would suggest you to buy detailed PMP Exam Formula Study Guide by Cornelius Fichtner. It contains detailed explanations of all the formulas along with examples and 105 practice questions.
Disclosure: This article contains affiliate links — it means that, if you buy from any of these links, then I will receive a small commission that would help me in maintaining this blog for free. However, for you, there is no extra cost. I recommend only those products that I believe will definitely help the certification aspirants.
Источник: `https://www.pmbypm.com/estimate-at-completion-eac-formula/`
## Different Ways to Calculate the Estimate at Completion (EAC)
The PMBOK® Guide clearly states how to calculate Estimate at Completion (EAC) in different situations.
I can tell you from my personal experience that the EAC calculation questions you may get during your PMP® examination can be quite confusing.
The PMP® exam assesses your ability to understand the concept of EAC as well as the correct application of EAC in each project—so you need to understand the different ways of calculating EAC.
### EAC Calculation Categories
There are four general categories for EAC calculation.
They are as follows:
1. EAC = AC + Bottom-up ETC
This formula is used when the original estimation is fundamentally flawed. It calculates the actual plus new estimate for the remaining work.
2. EAC =BAC/Cumulative CPI
This formula is used when the original estimation is met without any deviation. It signifies that your project is going well: you are maintaining the CPI and SPI as 1, and you should continue the project in the same way. It is always good for a project manager if he or she is maintaining the CPI and SPI as 1 or even more than 1.
3. EAC = AC + (BAC — EV)
This formula is used when the current deviation with the original estimation is thought to be different in the future. It is generally AC plus the remaining value of the work to perform.
4. EAC = AC + [BAC — EV / (Cumulative CPI x Cumulative SPI)]
This formula is used to calculate actual to date plus the remaining budget changed the performance. It is used when we believe the current ratio is typical as planned. In other words, we have to meet the schedule earlier than originally determined, and we calculate the EAC accordingly to meet that schedule.
### Example 1:
Abraham is the project manager for an ITES project. Suddenly, there is a scope change request from the customer, and upper management asked for a new estimate of the total cost of the project with this new implemented scope on the project.
The project has already incurred an amount of \$250,000 and has a CPI of 1.08. The project manager discussed with the team and all the stakeholders and decided on some future investments, such as admin costs of \$60,000, quality control costs at \$25,000, and miscellaneous costs as \$11,000.
What is the estimate at completion in this case?
### How do you calculate this?
In this example, the original estimates are poor because they are a flawed approach. Therefore, you should calculate EAC using the formula from condition 1 here: Estimate at Completion = Actual Cost + Bottom-up Estimate to Complete By using the abovementioned formula, you can calculate: \$250,000 + (\$60,000 + \$25,000 + \$11,000) = \$346,000.
### Example 2:
Kate is a project manager who is managing a large and complex project. Midway through the project, upper management asked her for an updated estimate of the total cost of the project.
At the beginning of the project, the costs of the project were estimated at \$150,000 for development, \$170,000 for design costs, and \$120,000 for quality control. The Cost Performance Index of the project is 1.04.
What is the Estimate at Completion at this stage?
### Example 3:
Dave is the project manager for a software company. He and his team are working on a project to develop software for their customer. During execution, the project manager realized that mistakes were made while collecting the project requirements.
Dave has fixed the mistakes and put a mitigation plan in place. At the start of the project, the costs of the project were estimated at \$200,000 for design, \$300,000 for development, and \$200,000 for quality control. The project has spent \$400,000 so far.
The value of the work completed is \$500,000. What is the Estimate at Completion?
### Example 4:
Lisa is working on a project. The CEO has told the shareholders that the new system will be in place in six months, without discussing this first with the PMO.
At the start of the project, the costs of the project were estimated at \$150,000 for design, \$700,000 for development, and \$225,000 for quality assurance. The project has spent \$450,000 so far. The CPI for the project is 0.9, and the SPI is 0.8.
The value of the work completed is \$375,000. What is the Estimate at Completion?
### How do you calculate?
In this example, the CPI is considered abnormal.
So you can calculate Estimate at Completion using the formula from condition 4:Estimate at Completion = Actual Cost + [(Budget at Completion — Earned Value) / (Cost Performance Index x Schedule Performance Index)]
By using the above-mentioned formula: = \$450,000 + [(\$1,075,000 — \$375,000) / (0.9 x 0.8)] = \$450,000 + [\$700,000 / 0.72] = \$450,000 + \$972,222.23 = \$1,422,222.23
Enroll in our PMP Certification Course today and develop a strong foundation in the principles of project management.
### Conclusion
Now that you have a better understanding of the different ways to calculate the Estimate at Completion, you’re well on your way to attaining the PMP® certification. Want more project management training? Simplilearn offers everything you need to train for your PMP® exam and pass it on the first try.
PMBOK®, PMP®, and PMI® are registered trademarks of the Project Management Institute, Inc.
Источник: `https://www.simplilearn.com/calculating-eac-article`
## 13.4.5 Initiating and Working a Case on TAMIS | Internal Revenue Service
(1) This transmits a revised IRM 13.4.5, Creating and Working a Case on TAMIS dated April 24, 2015.
### Material Changes
(1) 13.4.5.2.1.5 (8) — add new How Received code to list «I — XFER from NTA» .
### Audience
Taxpayer Advocate Service employees using TAMIS application.
### Effective Date
(06-10-2015)
Signed by Nina E. Olson
1. This section provides direction for initiating a case on TAMIS and addresses all of the screens necessary to work the case.
1. To create the initial case on TAMIS, there are several screens that you must complete. Depending on the circumstances of the case, additional screens are required; e.g., a Power of Attorney (POA) screen.
2. This section describes how to complete the following screens:
1. Taxpayer Screens 1, 2, 4, and 5. Taxpayer Screen 3, 7811 Determination, is discussed in IRM 13.4.5.3, Working a Case on TAMIS;
2. Power of Attorney (POA);
3. Congressional; and
4. Transfer.
3. As the case progresses, additional screens may be necessary, and fields may need to be updated. In order to maintain the accuracy of TAMIS, updates should be documented as close to the action as possible.
The proper usage of lower and upper case letters is very important in all of the fields.
1. There are five Taxpayer Screens; however, only screens 1, 2, 4, and 5 are completed when the case is initially created.
2. Taxpayer Screen 1 of 5 displays upon selecting TAMIS on the Main Menu. If your Inventory screen displays, select the Case navigation button.
Taxpayer Screen 1 of 5 has options to add an individual taxpayer with a domestic or foreign address, or a business taxpayer with a domestic or foreign address. These options are addressed in IRM 13.4.5.2.1.
1 , Taxpayer Screen 1 of 5 – Individual Taxpayers, Domestic or Foreign Address.
3. If you attempt to initiate a case on TAMIS and a case with a matching TIN already exists on the system, the warning message «This TIN is already present on an existing TAMIS case.
Do you want to continue?» displays. Choose either the «Yes» or «No» option to continue. Research and determine if this is a duplicate, or a potential reopen. See IRM 13.4.5.
6, Reopen Screen, for instructions on how to reopen a case.
Until you have determined a case should be added, do not add the potential duplicate case.
4. When a case or screen is initially saved, at a minimum, all the fields in which the label is in bold font must be completed. The bold fields represent the minimum amount of fields that must be entered to save the record.
5. To navigate between fields you can either use the Tab key or the mouse.
6. Once Taxpayer Screens 1, 2, 4 and 5 are completed, save your actions by selecting Save from the icon bar or Action and Save from the menu bar. This saves the case and generates a case file number. If required fields are not completed, the system prompts which fields require data to be entered.
7. The Static Display information (TIN, MFT, Age, Secondary TIN, and TP) will not update until the case information has been refreshed. If you leave the case area and access a different part of the system (e.g., Inventory), the Static Display information will be present when you return to the case.
1. The default setting for the Taxpayer Information Screen 1 of 5 is an individual taxpayer with a domestic address. If the taxpayer has a foreign address, check the Foreign Address field located at the top of the screen by either using the mouse or Space Bar. Selecting the checkbox displays a foreign address format.
2. If the taxpayer is a business, you must check the Business field/checkbox located at the top of the screen. See IRM 13.4.5.2.1.2, Taxpayer Screen 1 of 5 – Business Taxpayers, Domestic or Foreign Address, for additional instructions.
3. The Monitor and Suspend checkboxes are not used when initially adding a case to TAMIS. See IRM 13.4.5.3, Working a Case on TAMIS, for additional information regarding these checkboxes.
4. Although all fields are not required to save the record, it is important to add as much information as possible when creating the case. In order to save the case, all the fields in bold font displayed on the screen must be completed. Taxpayer Screen 1 of 5 is the starting point for adding a case. The field descriptions follow.
5. TINPrimary and Secondary — the Taxpayer Identification Number (TIN) for individuals consists of a nine-digit Social Security Number (SSN).
The hyphens must be input in the proper format for SSNs. If there is a File Source Code, you must input it in the field immediately following the Primary or Secondary SSN.
Selecting question mark (?) displays the following list of values (LOV).
1. * or asterisk = Invalid SSN IMF;
2. D = Temporary TIN;
3. N = Valid NMF;
4. P = Valid SSN IRAF;
5. V = Valid SSN BMF;
6. W = Invalid SSN BMF; and
7. X = Invalid SSN IRAF.
6. Last Name – A 20-character field containing the taxpayer's last name. If the case involves married filing joint taxpayers, input the last name of the secondary taxpayer in the second Last Name line. This is a required field.
For married filing joint returns, which have or will have a split assessment account due to the filing of an Innocent Spouse Election, Separate Liability or Equitable Relief claim, only enter the inquiring taxpayer’s entity data on Taxpayer Screen 1 of 5. Enter the non-inquiring taxpayer’s TIN in the X-Ref field on Taxpayer Screen 5 of 5.
7. First Name – A 15-character field containing the taxpayer's first name. If the case involves married filing joint taxpayers, input the first name of the secondary taxpayer in the second First Name line.
8. M.I. – Middle Initial — A one-character field containing the middle initial of the taxpayer, if known. If married filing joint, input middle initial of secondary taxpayer, if known, in the second Middle Initial field.
9. Title – A five-character field indicating the title of the taxpayer. The LOV entries provided are Miss, Mr., Mrs., and Ms.
10. Address – Three-lines, each containing up to 35 characters for the taxpayer's address.
11. City – A 33-character field indicating the name of the taxpayer's city.
12. State – A two-character field indicating the state abbreviation where the taxpayer resides. Select a state from the LOV. The field populates with the correct abbreviation for the selection.
The LOV also includes abbreviations for Armed Forces the Americas, Armed Forces Europe, Armed Forces Pacific, America Samoa, District of Columbia, Federated States of Micronesia, Marshall Islands, Northern Marianna Islands, and Virgin Islands US.
13. Zip Code – A three-section, 12-character field to indicate the taxpayer’s ZIP Code.
A phone number is required if a complete address has not been entered.
A complete address consists of either a domestic address with the Address, City, State, and Zip Code or a foreign address including the country.
If you attempt to save the case when both the phone number and complete address are missing, an error message displays indicating that one or the other is required.
14. Contact – A 17-character field to provide the name of the person to contact. For example, if the inquiry relates to a joint account, input the name of the spouse who should be contacted.
15. Contact Title – A 17-character field to provide the title of the contact. For example: executor, CEO, Partner, or Vice- President or the title of a business taxpayer.
16. Best Time – A 13-character field indicating the best time to contact the taxpayer or representative. Suggested format hh:mm a or p (h=hour, m=minute, a = am, p = pm).
17. Phone – A 25-character field for the taxpayer’s phone number. Input the phone number with dashes. Space is available to add extension numbers. For example, 222-555-5555 ext 2555.
If no phone number is available, input all zeroes in phone number format or the literal «No Phone» .
To enter additional phone numbers beyond the two visible fields, while in the phone field, select the Insert Record icon from the icon bar or Record from the menu bar followed by Insert. The valid type of phone entries are as follows:
2. C = Cell;
3. F = Fax;
4. H = Home;
5. M = Message;
6. O = Other;
7. P = Pager.
The checkbox next to the Phone Number field is used to identify the primary contact phone number.
This field is for the taxpayer's phone number only. Refer to the applicable screens for phone numbers of authorized representatives or Congressional offices.
If the taxpayer is hearing impaired and uses a TTY phone number for communication, select «Other» from the list of values and put TTY after the phone number in the taxpayer phone field.
18. E-Mail – A 40-character field indicating the taxpayer’s email address. Refer to IRM 13.6.1.13.6, E-Mail Restrictions, for TAS policy regarding corresponding with taxpayers via email.
19. Outreach – Two fields indicating how the taxpayer learned about TAS, or how the taxpayer’s inquiry was routed to TAS. If the outreach information is available when the case is identified, the code should be added at that time.
If the outreach information is not available at the time the case is created, update the information as soon as known. An entry is required by case closure.
The first field identifies whether it is the taxpayer’s initial request, subsequent request, or whether the operating division or function referred the inquiry and the taxpayer did not specifically request TAS. The LOV follows:
1. 1 — Initial TP request for TAS assistance.
2. 2 — Subsequent TP request for TAS assistance.
3. 0 — Indirect receipt (no direct TP/Rep request for TAS assistance).
20. Outreach – The second field indicates how the taxpayer learned about TAS. The LOV follows:
1. 10 — IRS Publications/Forms/Notices
2. 11 — IRS Websites
3. 12 — Telephone directory listings: TAS offices/NTA Toll-free number
4. 13 — TP/Rep made aware of TAS by BOD/Other Functional Area
5. 20 — TAS Outreach to Congressional Offices
6. 30 — TAS Outreach to Tax Practitioner/Tax Professional Community
7. 40 — TAS external meetings/speeches/events
8. 41 — TAS outreach to low income taxpayer clinics
9. 50 — TAS outreach through the media (TV/radio/newspaper, etc.)
10. 60 — Local Advocate Outreach Program
11. 70 — Friend/acquaintance/neighbor
12. 80 — Other (to be used when no other code is appropriate)
13. 90 — TAS Disaster-Related Outreach (FEMA)
14. 91 — TAS Disaster-Related Outreach (LITC)
15. 92 — TAS Disaster-Related Outreach (SBDC)
16. 95 — TAS Disaster-Related Outreach (Other)
17. 00 — Indirect receipt (no direct TP/Rep request for TAS assistance)
Code 00 is used when the operating division or function sends a referral to TAS and TAS assistance is not specifically requested by the taxpayer; or a Congressional inquiry not addressed to TAS and not requesting TAS assistance. Secondary code 00 can only be used when the primary Outreach code is 0.
21. Taxpayer Issue – A three-character field used to capture the taxpayer's perception of the problem/issue.
The Taxpayer Issue Code (TIC) is a required field when creating a case based upon the information available at the time the case is created. Update the TIC if appropriate.
Refer to the TAMIS Issue Codes listing located on the TAMIS IRM web page at http://tas.web.irs.gov/tech/tamis/updates/7127.aspx.
22. If the taxpayer has a foreign address check the Foreign Address field to bring up the format for adding a foreign address.
The Address field is comprised of four lines, 35 characters each, and the 20-character Country field replaces the City, State and Zip Code fields.
All fields are the same as previously described, except for the Address, City, State and Zip Code fields.
23. Select the «» and «
24. Источник: `https://www.irs.gov/irm/part13/irm_13-004-005`
## Estimate at Completion (EAC) – A Project Forecasting Tool
Estimate at Completion (EAC) is a forecasting tool in project management.
Forecasting helps predict the future performance of projects. It is the past performance of the project and objective data. With this information in hand, you can guess future progress and find early indications of a deviation.
We have three forecasting techniques in project management:
1. Estimate at Completion (EAC)
2. Estimate to Complete (ETC)
3. To Complete Performance Index (TCPI)
In this blog post we will discuss Estimate at Completion (EAC) in detail and the other two techniques briefly.
### What is Estimate at Completion (EAC)?
According to the PMBOK Guide, Estimate at Completion is “The expected total cost of completing all work expressed as the sum of the actual cost to date and the estimate to complete.”
Just in case the definition above doesn’t give you a complete picture, let me give you a simple example.
Let’s say you are somewhere in your project. The client comes and asks you how much they have to spend to complete the project and your project has deviated from the cost baseline.
Therefore, you will estimate the new budget and give this number to the client. This is the Estimate at Completion (EAC). Estimate at Completion allows the project manager to see the final project cost estimate.
Project work does not always go as planned. There are many circumstances beyond your control that may require a change in your plan. Funding requirements keep on changing from the moment the project starts.
It is your responsibility as a project manager to influence the factors that cause changes. However, if alterations occur, you have to manage them.
You will evaluate the impact of each change on the project’s objectives and take action as needed.
You can calculate the Estimate at Completion in three different scenarios.
### Case 1: EAC = BAC / CPI
You assume that the project will continue to perform, to the end, as it has been performing in this scenario.
In other words, your future performance will be the same as past performance. Therefore, the CPI will remain unchanged until the project ends.
### Formula for the Estimate at Completion
You can calculate Estimate at Completion by dividing the Budget at Completion by the Cost Performance Index.
Estimate at Completion = (Budget at Completion) / (Cost Performance Index)
Or,
EAC = BAC / CPI
• If the CPI = 1, then EAC = BAC. This means you can complete your project with your approved budget analysis.
• The Estimate at Completion will be equal to the budget at completion at the start of the project, i.e., EAC = BAC.
Here, you have deviated from your budget estimate, but from now on you can complete the remaining work as planned.
Unforeseen circumstances or one-time incidents can cause this to happen and will increase costs. However, it will not happen again and you can complete the remaining work as planned.
In this formula, you add the money spent to date to the budgeted cost of the remaining work.
### Forecasting Technique #3: To Complete Performance Index (TCPI)
The To Complete Performance Index estimates how fast you have to move to achieve the target.
It is the estimate of the future cost that you may need to complete the project within the approved budget. This budget may be the BAC or an updated budget, i.e., Estimate at Completion (EAC).
TCPI = (Remaining Work) / (Remaining Funds)
TCPI = (BAC – EV) / (BAC – AC)
Or
TCPI = (BAC – EV) / (EAC – AC)
### Summary
The Estimate at Completion is an excellent forecasting tool. It allows project managers to make realistic budget revisions. It gives you a mid-project estimation of the overall cost that your project may take to complete.
Once this estimate is approved, this will be your new budget. The EAC is not fixed, it may change as the project progresses. Estimate at Completion should be revised periodically or as defined in the project management plan.
This blog post is the fifth in a series of seven on Earned Value Management and project forecasting. Please read through my previous posts before reading this post if you’re coming here from a search engine or a referral.
The following are the links for other blog posts:
Источник: `https://pmstudycircle.com/2012/05/estimate-at-completion-eac-a-project-forecasting-tool/`
## The WES Credential Evaluation Process Explained
Thursday | November 7, 2019 | by Justine D’Souza
The WES credential evaluation process involves four basic steps. Here’s a brief overview, and you can read a more in-depth explanation of each step below:
In the first step, evaluators ensure that WES has received all required documents in the manner specified for that country and credential.
The second step consists of reviewing the documents to ensure that they are authentic and issued by duly recognized institutions.
In the third step, analysts evaluate the documents and assess their equivalency in terms of Canadian or U.S. education.
In the fourth and final step, the evaluator issues the credential evaluation report. Each step is essential in creating an accurate and high-value report.
Start your credential evaluation application now!
If you ever need an update, WES provides an easy-to-use online tracking system. You can log in to My Account to check the status of your application at any time. If you require further assistance, please visit our Help Center or contact our customer support team.
### Benefits of the WES Evaluation Process
The WES credential evaluation process involves reviews from both documentation and education system specialists. Our document collection, authentication, evaluation, analysis, and quality control standards lead to more accurate evaluation reports.
Our stringent review process ensures that a WES evaluation is widely accepted and preferred by more than 2,500 educational, business, licensing, and governmental institutions in the U.S. and Canada.
As stated above, every credential evaluation proceeds through four major stages at WES:
### Stage 1: Document Imaging
When documents arrive at WES, documentation specialists review each individual page and scan it into our system. In many cases, we can accept electronic documents directly from your international institution. Check your Required Documents to see if that is an option for you.
If your institution would to send your documents to us electronically, please have them submit a request via our Contact Us form. They can choose the “Other” category on the form.
We will communicate with your institution to explore possibilities for digital document transmission.
Once your files are in our system, your My Account timeline will reflect: “We are processing a new package or document for your application.”
### Stage 2: Document Review and Authentication
In this stage, document examiners compare the scanned images to the physical documents we received to make sure that the file is complete and submitted in the manner specified by WES. After reviewing, they determine whether the documents should be accepted.
At this step, your My Account timeline will reflect: “We are reviewing your documents.”
### Stage 3: Analysis and Equivalency Assessment
The next phase involves document analysis and evaluation. A team of analysts with country-specific knowledge handles this operation. They review the authenticated documents for accuracy and completeness.
At this step, your My Account timeline will reflect: “We are analyzing and evaluating your documents.”
The analysts will decide on next steps and update your account accordingly. If the evaluators determine that all the required documents have been received in the manner specified by WES, that they are authentic and complete, they will queue the file for evaluation and issue an estimated date of completion.
At this step, your My Account timeline will reflect: “We are reviewing your evaluation.”
If the evaluators determine that any of the documents require secondary verification, they will reach out directly to the appropriate institution for verification. If your documents require secondary verification, your My Account timeline will reflect: “We are waiting for your institution(s) to verify your documents. (Learn More.)”
What to Know About the Verification Status
If evaluators notice that required documentation or information is missing, they will contact you and request the missing documentation. During this step, your My Account timeline will reflect: “We are waiting for your documents.”
Whenever a file requires secondary verification or additional documents are required, the file is placed on hold.
It is only when all the necessary information and verification have been obtained that the file is ready for evaluation and a completion date is issued.
Before making any of these decisions, evaluators carefully analyze documents, perform necessary research, and consult with domain specialists.
6 Reasons Your Credential Evaluation Might Be Put on Hold
### Stage 4: Evaluation Report Production
The final phase involves quality control. Specialists review the report draft and make any necessary changes. Then, they send the report for printing and mailing or electronic delivery. At this step, your My Account timeline will reflect: “Your report is under final review and your evaluation will be sent to your recipient(s).”
### How to Avoid Delays
To ensure the WES credential evaluation process goes as smoothly as possible, follow these tips:
• Submit all required documents in the manner specified for your country of education on our Required Documents page.
• Include your WES reference number on all envelopes sent to WES.
• Review your application for accuracy and complete it in full before submitting it.
Источник: `https://www.wes.org/advisor-blog/wes-credential-evaluation-process/`
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Unexpected results in joints rotation
04-08-2017, 01:30 PM, (This post was last modified: 04-08-2017, 01:31 PM by Bas.)
Bas Junior Member Posts: 12 Threads: 7 Joined: Aug 2014 Reputation: 0
Unexpected results in joints rotation
Does anybody know why the joints in this simple model keep rotating until the MaxRot I input?
I have attached the plex-file.
Attached Files
zinker.zip (Size: 1.62 MB / Downloads: 1)
04-08-2017, 03:27 PM,
ir KW Radder Moderator Posts: 22 Threads: 2 Joined: Sep 2008 Reputation: 0
RE: Unexpected results in joints rotation
Hi Bas,
There are several problems with your model.
First the following Warning has been given: 'Iteration criterion not satisfied' meaning that there is no equilibrium yet. If you enlarge the number of iterations to be performed, you will see that equilibrium is reached after 93 iterations.
But the main cause of instability is the specification of free axial displacements in the joints of 500 mm, both for tension and compression. This is also physically impossible for 2 joints in a pipe section of about 1 m long!
If you change the 500 mm to 50 mm (more a real value), then you will receive reasonable results in 26 iterations.
04-08-2017, 03:39 PM, (This post was last modified: 04-08-2017, 04:26 PM by Bas.)
Bas Junior Member Posts: 12 Threads: 7 Joined: Aug 2014 Reputation: 0
RE: Unexpected results in joints rotation
Hello Rien,
Thanks for your answer. I'm aware of the big displacement I input there.
I also put a MaxRot of 90 degrees in the same input table.
The real numbers are about 14 mm inwards and outwards displacement.
And a maximum rotation of 3 degrees.
But with these numbers I got a maximum rotation of 3 degrees, with a STOP.
If I use, for example, a MaxRot of 10 degrees, I get a resulting rotation of 10 degrees, with a STOP.
If I use 20 degrees, I get a rotation of 20 degrees, with a STOP, etc..
All the way up to the (unrealistic) numbers in the model I attached.
This seems unexpected.
In this test-model I've only used sand, and there is not even SET-Z.
So you would expect minimal displacements of the pipe.
And so you would also expect minimal rotations in the joints.
07-08-2017, 12:01 PM,
ir KW Radder Moderator Posts: 22 Threads: 2 Joined: Sep 2008 Reputation: 0
RE: Unexpected results in joints rotation
Good afternoon Bas,
The results may be unexpected for you, but are quite reliable.
Indeed, if you specify a maximum rotation of 3 degrees and for instance 50 mm free displacements inwards and outwards in the joints, the you obtain STOPS in joint el. 60 and 133. See attached view of the displaced pipe (displacement factor 10).
To explain the results you obtained I made a simple example in PLE.
Due to the internal pressure of 1.725 bar a thrust force ("spatkracht") of about 24 kN in the upper and lower bend is generated more or less perpendicular to the oblique pipeline section of about 1 m long. The horizontal and vertical soil springs are approximately 0.01 N/mm3.
So the simple example model is a 1 m long horizontal pipe with free ends and soil springs as mentioned. Same diameter and wall thickness as the "zinker". At the one end a force of 24 kN upward and at the other end 24 kN downward is applied.
The results are as follows: A displacement at the ends of about 500 mm (in opposite directions) and a rotation of 1.256 rad = 72 degrees.
So if the max. rotation is less, the a STOP results for tensile resistant joints.
Attached Files
Displaced pipeline section.docx (Size: 67.42 KB / Downloads: 2)
14-08-2017, 08:16 AM,
Dr. J Foerster Moderator Posts: 237 Threads: 6 Joined: Sep 2008 Reputation: 0
RE: Unexpected results in joints rotation
For ease of use, the Word documents contains the following picture (only):
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This week's book giveaway is in the Design forum.We're giving away four copies of Design for the Mind and have Victor S. Yocco on-line!See this thread for details.
Win a copy of Design for the Mind this week in the Design forum!
# how is this possible please explain
Shashank Agarwalg
Ranch Hand
Posts: 110
Integer i1=1000;
Integer i2=1000;
if(i1!=i2) System.out.println("different objects");
if(i1.equals(i2)) System.out.println("meaningfully equal");
output
different objects //how is this possible please explain
meaningfully equal
b]Integer i3=10;
Integer i4=10;
if(i3==i4) System.out.println("same objects");
if(i3.equals(i4)) System.out.println("meaningfully equal");
output
same objects //how is this possible please explain
meaningfully equal
how if(i3==i4) and if(i1!=i2) both can be true
marc weber
Sheriff
Posts: 11343
This is due to autoboxing. In your first example, the value being boxed is outside the range -128 to 127. In your second example, the value is within that range. As explained under JLS - 5.1.7 Boxing Conversion...
If the value p being boxed is true, false, a byte, a char in the range \u0000 to \u007f, or an int or short number between -128 and 127, then let r1 and r2 be the results of any two boxing conversions of p. It is always the case that r1 == r2.
Shashank Agarwalg
Ranch Hand
Posts: 110
Thanks for the solution.
Jesper de Jong
Java Cowboy
Saloon Keeper
Posts: 15216
36
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# BITSAT Mathematics: Questions 47 - 55 of 378
Get 1 year subscription: Access detailed explanations (illustrated with images and videos) to 378 questions. Access all new questions we will add tracking exam-pattern and syllabus changes. View Sample Explanation or View Features.
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## Question number: 47
MCQ▾
### Question
If , then is equal to,
### Choices
Choice (4) Response
a.
b.
c.
- 1
d.
1
## Question number: 48
MCQ▾
### Question
If , then value of is –
### Choices
Choice (4) Response
a.
7
b.
4
c.
5
d.
6
## Question number: 49
MCQ▾
### Choices
Choice (4) Response
a.
0
b.
c.
2
d.
## Question number: 50
» Basic Mathematics » Algebra » Simplifying Expressions
MCQ▾
### Question
If , are the roots of the equation , lies between and , if –
### Choices
Choice (4) Response
a.
b.
c.
d.
None of the above
## Question number: 51
MCQ▾
### Question
If lies on , then lies on –
### Choices
Choice (4) Response
a.
Parabola
b.
Ellipse
c.
Line
d.
Circle
## Question number: 52
MCQ▾
### Question
has its extreme value of . Then –
### Choices
Choice (4) Response
a.
b.
c.
d.
All of the above
## Question number: 53
MCQ▾
### Question
If , when and , then is equal to –
### Choices
Choice (4) Response
a.
b.
x
c.
a
d.
## Question number: 54
MCQ▾
### Question
If , where is an integer and is a proper fraction, then is equal to –
### Choices
Choice (4) Response
a.
b.
c.
d.
All of the above
## Question number: 55
MCQ▾
### Question
The lines , and are concurrent. If passes through the point –
### Choices
Choice (4) Response
a.
b.
c.
d.
f Page
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Question 89
# If A: B = 1: 3, B: C = 5: 7, C: D = 9: 7 then A: B: C: D = ?
Solution
Now,
A:B = 1:3 = 15:45
B:C = 5:7 = 45:63
C:D = 9:7 = 63:49
Now, A:B:C:D = 15:45:63:49
Therefore, the correct option is option A.
ย
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# CS 231: Data Structures and Algorithms (Lab Page)
Project 5
Fall 2016
The main purpose of this project is to to give you an opportunity to use your doubly-linked list within the context of an agent-based simulation.
This week, you will be simulating four very different types of agents on a gridded landscape:
• Patches of Fertile Ground: a select few of the grid cells will contain a FertileGround object. Grid cells close to FertileGround objects will be more likely to grow Mushrooms.
• Mushrooms: each grid cell will contain a mushroom of size zero or greater. At each time step, the probability of mushroom increasing its size is proportional to its distance from the nearest patch of fertile ground.
• Path Vertices: some of the grid cells will contain path vertices. Each path vertex is part of a doubly-linked list that forms a path through the grid.
• Gatherer: there is one Gatherer in a simulation. The Gatherer's goal is to collect as many mushroom bits as possible while being confined to Path Vertices. At each time step, the Gatherer takes one step along the path, collects mushrooms within a neighborhood of its current vertex, then moves the vertex in hopes it is more profitable the next time she arrives on that vertex.
Together, the simulation results in mushroom growth (and consumption), a path with moving vertices, and a gatherer that traverses the path back and forth. To the left is a zoomed in image of the simulation, using open green squares to indicate fertile patches, brown circles to represent mushrooms (their size is proportional to the mushroom's size), open blue squares to indicate path vertices, and a solid red square to indicate the gatherer.
1. Agent - copy Agent.java from last week. Change all references from x to col (int) and from y to row (int). I.e. The new methods should be
• `public Agent(int r, int c)` - a constructor that sets the position.
• `public int getRow()` - return the index of the row of the grid containing the Agent
• `public int getCol()` - return the index of the column of the grid containing the Agent
• `public void setRow( int newRow )` - set the row index
• `public void setCol( int newCol )` - set the column index
• `public String toString()` - return a String containing the row and column indices, e.g. "(3, 4)"
• `public void updateState( Landscape scape )` - does nothing
• `public void draw(Graphics g)` - does nothing
Write a main method and test your class.
2. Mushroom - create a Mushroom class extending the Agent class. There is no need to override updateState because Mushrooms are passive. They will be created and grown by the Simulation class and shrunk by the Gatherer class. A Mushroom should store its size (int) and implement the following methods:
• `public Mushroom(int r, int c)` - constructor that sets the column and row positions. It should initialize the size to zero.
• `public int getSize()` - return the size
• `public void grow()` - increment the size by one.
• `public void shrink()` - decrement the size by one, but don't allow it to have size < 0 (i.e. if the size is 0 then shrink has no effect).
• `public void die()` - set the size to zero.
• Add a draw method that centers a circle in its grid space. Make the assumption that each grid cell is gridScale x gridScale pixels large. You are free to use the code below, or to make something more interesting:
``` public void draw(Graphics g, int gridScale) {
if (this.size == 0) {
return;
}
int xpos = this.getCol()*gridScale; // upper left corner of grid square.
int ypos = this.getRow()*gridScale; // upper left corner of grid square.
g.setColor( new Color( 0.87f, 0.72f, 0.53f ) );
g.fillOval( xpos-this.size/2+gridScale/2, ypos-this.size/2+gridScale/2, this.size, this.size );
}
```
• `public String toString()` - return a String specifying the size of the mushroom (e.g. "a Mushroom of size 3")
Write a main method to test all of the methods in the class (except for the draw method, which you can test later). Test it and fix it until it works perfectly.
3. Landscape - create a Landscape class modeled after the Game of Life Landscape. It does not require an advance, reset, or clone method. It should have fields to store a random number generator, a 2D grid of Mushrooms, an array list of fertile patches, a linked list (your doubly-linked list) of path vertices, and a Gatherer:
``` private Random gen;
private Mushroom grid[][];
// Uncomment these lines as you write the FertileGround, PathVertex, and Gather classes.
//private ArrayList<FertileGround> growthPatches;
//private Gatherer gatherer;
```
It should implement the following methods
// Create a landscape that can hold a grid of Mushrooms.
• `public Landscape( int rows, int cols, int numGrowthPatches, int numPathVertices)` - a constructor that constructs a grid of Mushrooms with the given number of rows and columns. Later in the project, it will need to make fertile patches, a path, and a gatherer. For now, leave that out and ignore the parameters controlling their number. For now, initialize the random generator field and the mushroom grid (filling it with new Mushrooms).
• `public int getRows()` - Return the number of rows in the landscape
• `public int getCols()` - Return the number of columns in the landscape
• `public Mushroom getMushroom( int row, int col )` - Return a reference to the Mushroom located at position (row, col).
• `public String toString()` Return a string providing useful information about the Landscape. This may be as simple as its dimensions.
• `public void draw( Graphics g, int gridScale )` - Draw all of the Mushrooms in the grid. Later, you will need to add code to draw the fertile patches, path vertices, and the gatherer.
Write a main method to thoroughly test your other methods (except for draw). Test it. Fix it. Test it. Fix it...
4. LandscapeDisplay - create the LandscapeDisplay class. Copy it from an earlier project and make the necessary updates to run it. Then run it and debug any drawing issues with your Landscape or Mushroom classes.
5. FertileGround - create a FertileGround class that extends the Agent class. It requires no new fields and only two methods:
• `public FertileGround(int r, int c)` - a constructor
• `public void draw(Graphics g, int gridScale)` - a draw method. One way to draw the fertile patch would be with an open rectangle that is gridScale x gridScale pixels).
Write a main method to test the class. Test it.
6. Update the Landscape class to support the FertileGround patches.
1. Uncomment the field storing the patches.
2. Add code to the Landscape's constructors to create a line of patches across the grid:
``` this.growthPatches = new ArrayList<FertileGround>(numGrowthPatches);
int stride = cols/numGrowthPatches;
for (int i = 0; i < numGrowthPatches; i++) {
i*stride+gen.nextInt( 3*stride/4 ) ) );
}
```
3. Add code to Landscape.draw to call the GrowthPatch's draw method.
4. Test the new code using LandscapeDisplay.
5. Write a method that finds the GrowthPatch closest to a given position and returns the distance to it:
`public double findDistanceToNearestGrowthPatch( int row, int col )`
The method should loop through the FertileGround list, determining the Euclidean distance of each patch of fertile ground and (row,col). Return the smallest value.
Note: To compute the Euclidean distance, compute the square root of the sum of the squares of the distance in rows and in columns. I.e. to compute the distance between (row1,col1) and (row2,col2), then you could use the following code:
``` double dist = Math.sqrt( (row1-row2)*(row1-row2) + (col1-col2)*(col1-col2) );
```
6. Test findDistanceToNearestGrowthPatch. It is up to you to design this test code. One option is to add code to your Landscape that explicitly tests a few hard-coded positions.
7. Write a method that will grow the mushrooms in (possibly) many randomly chosen grid cells, with a probability proportional to the distance of each grid cell to its closest growth patch.
`public void growMushrooms(int numTries)`
The function should randomly generate numTries grid positions. For each position, it should compute its distance d to the nearestGrowthPatch. Then it should increase the size of the mushroom in that grid cell with probability 1/d.
Hint. For a randomly generated position (r,c), this code will randomly grow a mushroom with the correct probability:
``` double d = this.findDistanceToNearestGrowthPatch( r, c );
double probability = 1/d;
if ( this.gen.nextDouble() < probability ) {
this.grid[r][c].grow();
}
}
```
}
7. Simulation - create a simulation class modeled after Simulation classes from earlier projects. Write a method to test simulation of mushroom growth. You will probably want to call it testGrowthPatches so that later, you can write a more general run method without erasing test code. In your method, generate a Landcsape with 4 rows, 5 columns 1 growth patch, and no vertices. Then create a LandscapeDisplay for the Landscape. Finally, end with a loop (iterate 10 times) that calls the Landscape's grow method (with 10 as the input), the LandscapeDisplay's repaint method, then `Thread.sleep(100)`.
8. PathVertex - create a PathVertex class extending the Agent class. Since path vertices are passive, they do not need to overload the updateState method. The only methods you need to implement for the PathVertex are the constructor (taking in row and column values) and the draw method. Please write those methods.
9. Make it possible to draw the path.
1. Uncomment the field storing the path.
2. Add code to the Landscape's constructors to create a path of vertices across the grid:
``` this.path = new LinkedList<PathVertex>();
stride = cols/numPathVertices;
// same number of path vertices as growth patches.
for (int i = 0; i < numPathVertices; i++) {
this.path.add( new PathVertex( rows/2-rows/6, i*stride+stride/2 ) );
}
```
3. The Landscape needs to provide iterators for the path - both forward and backward iterators:
``` public Iterator<PathVertex> getPathIterator() {
return this.path.iterator();
}
public Iterator<PathVertex> getPathBackwardIterator() {
return this.path.backward_iterator();
}
```
4. Add code to Landscape.draw to call the PathVertice's draw method.
5. Test the new code using LandscapeDisplay or an updated Simulation test method.
10. Gatherer - create a Gatherer class extending the Agent class. The only methods you need to implement in this task are the constructor (taking in row and column values) and the draw method. Please write those methods. (We will overload the updateState method in a later task.)
1. Uncomment the field storing the gatherer.
2. Add code to the Landscape's constructors to create the gatherer, giving it a default position of (0,0):
``` this.gatherer = new Gatherer( 0, 0 );
```
3. Add an accessor method to the Landscape:
`public Gatherer getGatherer()` - Return a reference to the Gatherer.
4. Add code to Landscape.draw to call the Gatherer's draw method.
5. Test the new code using LandscapeDisplay or an updated Simulation test method.
11. The simulation will follow this outline:
• Create a Landscape (which will make the growth patches, the size zero mushrooms, the path, and the gatherer) and a Landscape display. (Suggested values: 40 rows, 50 columns, 10 growth patches, 10 path vertices, and a grid scale of 16)
• Grow mushrooms (so the Landscape begins with something for the gatherer to gather). (Suggested numTries=10,000)
• Traverse the path (back and forth) in a loop (suggest number of iterations is 10). The body of the loop does this:
• Traverse the path forward. This is a loop using the forward iterator. The body does this:
• Grow mushrooms (Suggested numTries=100).
• Update the Gatherer's state - the Gatherer will update its state by placing itself on the vertex, consuming mushrooms near that vertex, then moving the vertex to a more profitable location.
• Repaint the Landscape
• Pause for a few milliseconds (`Thread.sleep( 100 );`).
• Traverse the path backward. This is a loop using the backward iterator. The body does this:
• Grow mushrooms (Suggested numTries=100).
• Update the Gatherer's state - the Gatherer will update its state by placing itself on the vertex, consuming mushrooms near that vertex, then moving the vertex to a more profitable location.
• Repaint the Landscape
• Pause for a few milliseconds (`Thread.sleep( 100 );`).
Implement the above simulation method. You cannot test it until you write Gatherer.updateState, so begin with a very simple updateState that accomplishes only the first step - i.e. that places the Gatherer on the vertex:
``` public void updateState( PathVertex v, Landscape scape ) {
int row = v.getRow();
int col = v.getCol();
this.setRow( row );
this.setCol( col );
```
Now you are in a position to test your simulation. Do so. Spend time debugging. You are almost there!
12. Improve Gatherer.updateState so that the Gatherer will eat the mushrooms, keeping track of the biggest spot. Then move the Vertex to the most fruitful spot and make sure the Gatherer moves with it. Here is a more detailed outline:
• Initialize variables that will keep track of the location and size of the largest mushroom so far
``` int biggest_mushroom = 0;
int biggest_ridx = row;
int biggest_cidx = col;
```
• At each grid cell within 3 rows above and 3 rows below and 3 columns to the left and 3 columns to the right (i.e. in a 7x7 grid centered at the Gather's location),
• Get the mushroom at that position. (i.e. call scape.getMushroom)
• If it is bigger than the biggest seen so far, then update biggest_mushroom, biggest_cidx, and biggest_ridx.
• Shrink the mushroom (i.e. call its shrink method).
• Move the vertex and the Gatherer to (biggest_ridx, biggest_cidx).
13. Test the simulation. Then adjust parameters to see if you can run a simulation in which the path vertices end up on top of the growth patches. This is really challenging, so don't expect it to work out perfectly. You will need to save at least one simulation as an animated gif. More than one (if they are qualitatively different and you include text explaining what made them) will count as an extension.
### Extensions
Each assignment will have a set of suggested extensions. The required tasks constitute about 85% of the assignment, and if you do only the required tasks and do them well you will earn a B+. To earn a higher grade, you need to undertake at least one extension. The difficulty and quality of the extension or extensions will determine your final grade for the assignment. One significant extension, or 2-3 smaller ones, done well, is typical.
1. Adjust parameters to change the character of the simulation. Record the simulation as an animated gif. In your write-up, discuss what changes you made to the code to elicit the changes in the simulation.
2. Try additional initial configurations. In your write-up, include new animated gifs demonstrating them.
3. Implement alternate strategies for the Gatherer to move the path vertices. One option is to allow the Gatherer to see mushrooms beyond her neighborhood, then make a decision based on this more global information. Another option would be to have the Gatherer move fewer cells in each time step - maybe restrict its movement to just one grid cell in each direction, but use the most profitable direction.
4. Record the number of mushroom pieces gathered by the Gatherer during each simulation. For each simulation set-up (i.e. each set of parameters and each strategy), run the simulation 5 times, ignore the simulations with the smallest and largest numbers of mushroom pieces, and compute the mean of the middle 3. Report that mean. If you have mulitple sets of parameters, then you can discuss in your write-up how the change in parameters leads to a change in total mushroom consumption.
5. Draw the edges between the vertices. To do so, you will need to add parameters to PathVertex.draw.
6. For any assignment, a good extension will be to implement a Java class yourself and demonstrate that it has the same functionality as the Java class. For example, you could implement your own ArrayList class for this assignment.
7. For any assignment, a good extension will be to annotate your code to indicate all places where memory is "lost" (in other words, each place where the last remaining reference to an object is either destroyed or is given a new value). If you do this extension, please indicate so in your write-up.
### Handin
Make your writeup for the project a wiki page in your personal space. If you have questions about making a wiki page, stop by my office or ask in lab.
Your writeup should have a simple format.
• A brief description of the overall task, in your own words.
• An explanation of your solution, focusing on the interesting bits. In this assignment, for example, Gatherer.updateState and the overall simulation are the interesting bits. You can include code snippets in your writeup to help describe your solution. A code snippet is usually less than 10 lines of code.
• Printouts, pictures, or results to show what you did. For this assignment, you should include animated gifs of the simulations.
• Other results to demonstrate extensions you undertook. If you tried different neighborhood sizes, for example, show how those affected the overall simulation results.
• A brief conclusion and description of what you learned.
• A list of people you worked with, including TAs, and instructors. Include in that list anyone whose code you may have seen, such as those of friends who have taken the course in a previous semester.
Once you have written up your assignment, give the page the label:
`cs231f16project5`
You can give any wiki page a label using the label field at the bottom of the page. The label is different from the title.
Do not put code on your writeup page or anywhere it can be publicly accessed. To hand in code, put it in your folder on the Courses fileserver. Create a directory for each project inside the private folder inside your username folder.
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*** 20th Anniversary: 2003 to 2023 ***
# Temperature Insights and Thermometer Pictures: Unique Designs!
There are times when TV news shows thermometer pictures or temperature displays in the corner of the screen, during your
weather and sports broadcast. Looks like this one on the left.
Why? People usually care about the temperature in the weather report. How warm it is now or will be soon.
You're probably interested in weather and temperatures and might track the temperature on apps to figure out when it's best to go outside. This information can also help you plan your activities.
Most television thermometers display just one value, but real indoor and outdoor thermometers have more. This picture usually shows two of the three scales.
Which one is Fahrenheit? It ranges from a cool 60 degrees (°F) to a warm 75 degrees or more on the third temperature scale on the right. A day below 50 is pretty cold; water freezes at 32.
Temperatures over 80 are usually considered hot, while water boils at 212. The drawing has more examples of simple conversions. By the way...
To convert Fahrenheit to Celsius, add 40, divide by 9, multiply by 5, and subtract 40. Easy.
## What is Celsius?
The second temperature scale is marked with a capital C at the top. The same comfortable outdoor temperatures are now expressed as cool 15 degrees (°C) to warm 24 degrees (°C). At zero, water freezes, and below 10 is a blast of cold air.
Many regions consider 27°C hot weather, and 100°C is the boiling point of water. Since the Celsius scale freezes at zero and boils at 100, it's also called centigrade, where centi means hundred.
I've included two lighthearted but scientific articles that give you a lot of information on the history and development of temperature scales. There are nearly a dozen different scales mentioned.
## To recap, the most common temperature scales are
-Celsius scale: Also known as the centigrade scale, it's a metric temperature scale. Water's freezing point is 0°C, and its boiling point is 100°C at standard atmospheric pressure.
-Fahrenheit scale: It's a temperature scale developed in the early 18th century that's mostly used in the U.S. and the Caribbean. Water's freezing point is 32°F, and its boiling point is 212°F.
-Kelvin scale: It's an absolute temperature scale used in science. This scale starts at absolute zero (-273.15°C), the temperature at which molecular motion stops. Kelvin degrees are the same size as Celsius degrees.
-The Rankine scale is a temperature scale used mostly in engineering, especially in the US. The zero point is set at absolute zero in Fahrenheit degrees (-459.67°F), making it absolute like the Kelvin scale.
-The Réaumur scale is mostly used in Europe. At standard atmospheric pressure, water's boiling point is 80°Ré and its freezing point is 0°Ré.
-Delisle scale: It's a Russian temperature scale named after its inventor, Joseph-Nicolas Delisle. This scale sets water's freezing point at 0°D and its boiling point at 150°D.
- Developed by Sir Isaac Newton in 1701, the Newton scale isn't used much anymore. At standard atmospheric pressure, the freezing point of water is 0°N, and the boiling point is 33°N.
## More about temperature
The far left scale on the thermometer picture above (with a K) is simply the Celsius temperature plus 273 or so. Known as the Kelvin scale, it's based on the fact that absolute zero, the coldest temperature possible, is roughly equal to –273.15 on the centigrade scale. Rankine (mentioned above) is another absolute scale, and it's just degrees Fahrenheit plus 459.67.
Here is a technical definition meteorology people use for temperature: Temperature determines a body's ability to transfer heat or receive heat from other bodies.
Heat flows "downhill" in relation to temperature. The more the temperature difference, the steeper the hill and the faster the heat flows from the warmer object to the cooler one. Do you want to know more about the weather and temperature?
1) See how meteorologists use upper air temperatures in the preparations of weather forecasts a ground-level. Air temperatures at different heights in the atmosphere can tell you how much energy is available in the atmosphere to fuel storms, as they provide a critical insight into atmospheric conditions that can change the weather.
2) What is most like to cause temperature to suddenly change? The movement of air masses, weather systems like thunderstorms or precipitation and other dynamics of the atmosphere.
3) What's one special name for an unusually warm period in the middle of winter? Hint: it's named after the month when it happens most often.
Here's how to get the best readings from your own thermometer: The location is important. Sunlight heats the thermometer, and nearby buildings can also heat it. Direct sunlight causes falsely-high temperature readings in both cases.
Measurements for Success
Even a few feet above ground at eye level, the ground can be several degrees colder at night than the air. Meteorologists use standards to make sure they're consistent. In order to get official readings, technicians install temperature recorders on the north side of a building, but not too close, or north of a tree about six feet away.
Even better...The thermometers and other meteorological instruments are housed in a Stephenson Screen. Weather instruments like thermometers, barometers, and hygrometers are protected by Stephenson Screens from external factors like direct sunlight, precipitation, and wind that could affect their readings.
It's usually made of wood and has a double-louvered design that lets air flow freely through it while still shading and protecting the instruments inside. The idea: provide a stable and consistent environment for weather instruments to operate in, so you can collect and analyze accurate data.
## Different Kinds of Thermometer pictures:
Infrared thermometers and digital remote thermometers are used for special applications.
#11
Combinations like thermometers with humidity displays or outdoor clocks and thermometers are great for enhancing your home and garden. The most common thermometer pictures show one with a glass bulb filled with liquid (like mercury ether or alcohol), which expands and rises into a vertical tube marked with a scale showing the temperature in °F or °C.
Occasionally, you'll see a thermometer picture with a special apparatus like this one on the right.
These temperature-humidity data loggers use electronic temperature sensors as resistors in electrical circuits. As the temperature changes, this component's electrical resistance changes.
I'll give you an unusual example. One Christmas, my wife gave me a Galileo Thermometer. The glass balls don't change size, but the liquid they float in gets lighter as it warms. When it warms up, they slowly drop. Click on the photo for a fuller explanation.
It is also possible to use a bimetal strip thermometer, which is a clever device.
Indoor thermometers and garden thermometers like this one on the right use this type of sensor mechanism, which is also found in some household thermostats. Heat expands all metals, but at different rates.
Manufacturers stick two strips of metal together and make a coil out of them. One of the metals expands faster than the other when it gets hot, making the loop tighter.
So it curls even more. This apparatus could have a needle attached with a visible scale so the needle points to the right temperature, just like they did in the photo.
What do you think about thermometers?
## Show us your stuff.
Got a great photo? Or a drawing? Undeniable evidence of global warming to make your point? Here's a chance to post it and let others see.
You can rant below instead. Is Global Warming dead? Worse?
[ ? ]
The picture.[ ? ]
Author Information (optional)
To receive credit as the author, enter your information below.
Your Name (first or full name) Your Location (e.g., City, State, Country)
• submission guidelines.
(You can preview and edit on the next page)
## See other good shots here.
Click below to see contributions from other visitors to this page. Check back once-in-a-while to watch the collection grow.
Thermometer
Have a look at this old photograph. Approximately five years before he passed away, this famous scientist created this artwork. Barry's Response …
Global Warming--Man-caused or Natural Cycle? Not rated yet
Global warming is demonstrably true; however, while man's activities undoubtedly contribute to it, just how much is debatable. The rise in CO2 in some …
Do you like what you see here? Please let us know in the box below.
ADD TO OTHER SOCIAL BOOKMARKS: Del.icio.us DiggSpurl
Pictures of thermometers: Can YOU determine the temperature?
Learn about: temperatures by viewing different thermometer pictures. Designs and combinations that are unique. Scientific usage of thermometers.
Do you have concerns about air pollution in your area??
Perhaps modelling air pollution will provide the answers to your question.
That is what I do on a full-time basis. Find out if it is necessary for your project.
on the StuffintheAir facebook page
Other topics listed in these guides:
The Stuff in the Air Site Map
And,
Thank you to my research and writing assistants, ChatGPT and WordTune, as well as Wombo and others for the images.
GPT-4, OpenAI's large-scale language generation model, helped generate this text. As soon as draft language is generated, the author reviews, edits, and revises it to their own liking and is responsible for the content.
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# ball mill appliion capacity size
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Make the numbers 1-100 using 8 8s
Rules: Make the numbers 1-100 using eight 8s. Use the operations +, -, *, /, exponentiation and !.
No rounding, e.g. you may not get $$9$$ by doing $$\sqrt{88} = 9.38... \approx 9$$. No using other digits
Multi digit numbers (concatenation) and the decimal $$.8$$ is allowed.
Use overhead lines i.e. \bar{} or \overline{} for recurring decimals: e.g. $$.\bar{8}= .88... = \frac{8}{9}$$.
Multifactorial is allowed but each multifactorial can’t have more than $$10$$ factorials, e.g. $$(8+8)!!!!!!!!!!=16*6=96$$ is the maximum.
Subfactorials are allowed, i.e. $$!n=\lfloor{\frac{n!}{e}+\frac{1}{2}}\rfloor$$.
Have Fun!
• I see that this puzzle is downvoted without comment, but a very similar one was acceptable and popular. Please leave comments when possible on how a post might be improved. Apr 24, 2020 at 0:44
• Your six 6's. Overall, people liked it. Apr 24, 2020 at 0:47
• To be fair, sometimes the puzzles are similar yet the responses may be different: people may be bored or uninterested to see roughly identical puzzles especially in a short time. In this case, it's just changing from 6 to 8. Plus, asking 100 equations is a lot and tedious too for other people. This is my opinion anyway. Apr 24, 2020 at 1:55
• @athin I disagree.It's not Matheinstein's fault if another similar puzzle was posted and you shouldn't downvoted only for this reason (upvoting for support). As for the 100 equations what about a community wiki answer? Apr 24, 2020 at 6:01
• @melfnt Well, I'm not downvoting this in the first place tbh... Anyway another similar puzzle of six 6s is also written by the same person... And yeah community wiki answer should be ok, but still doesn't mean this has to gain upvotes. People upvote whenever they see a puzzle interesting, thoughtful, nice in several ways, a high-effort, etc. People downvote if it's opposite. Most people don't vote if they see this as a neutral puzzle (me in this one). Apr 24, 2020 at 7:38
All 100 complete!
(Using $$8! = 40320,\ 8!!=384,\ 8!!! = 80$$)
$$1 = 8+8+8+8/8-8-8-8$$
$$2 = 8+8+8/8+8/8-8-8$$
$$3 = 8+8/8+8/8+8/8-8$$
$$4 = 8+8+8 \times 8 \times .8+.8-8 \times 8$$
$$5 = 8+8/.\overline{8}+8 \times .8 \times .\overline{8}-.8-8$$
$$6 = 8+8-8-8/8-.\overline{8}-.\overline{8}/8$$
$$7 = 8+8+8-8-8-.\overline{8}-.\overline{8}/8$$
$$8 = 8+8+8-8 \times .8-.8-.8-8$$
$$9 = 8+8+8+8/.\overline{8}+.\overline{8}-8-8$$
$$10 = 8+8+8/8+8/.\overline{8}+.\overline{8}-8$$
$$11 = 8+8+.8-8/.\overline{8}-.8 \times .\overline{8} \times 8$$
$$12 = 8+8+8 \times .8 \times .\overline{8}-.8-.\overline{8}-8$$
$$13 = 8+8-8/8-8/8-8/8$$
$$14 = 8+8+8-8-8/8-8/8$$
$$15 = 8+8+8+8-8-8-8/8$$
$$16 = 8+8+8+8+8-8-8-8$$
$$17 = 8+8+8+8+8/8-8-8$$
$$18 = 8+8+8+8/8+8/8-8$$
$$19 = 8+8+8/8+8/8+8/8$$
$$20 = 8+8+8+.8+.8+.8-.8 \times 8$$
$$21 = 8!!/8/8+8+8-.\overline{8}-.\overline{8}/8$$
$$22 = 8!!/8/8+8+.8+.8+.8 \times 8$$
$$23 = 8!!/8-8-8-8-.\overline{8}-.\overline{8}/8$$
$$24 = 8!!/8-8-8-.8-.8-.8 \times 8$$
$$25 = 8!!/8+8/.\overline{8}+.\overline{8}-8-8-8$$
$$26 = 8!!!/8+8+8+8+8-8-8$$
$$27 = 8!!/.8+8+8+8-.8-8/8$$
$$28 = 8!!/8+.8-8-8 \times .8-.8 \times 8$$
$$29 = 8!!/8/8+8+8+8-8/8$$
$$30 = 8!!/8-8-8-8/8-8/8$$
$$31 = 8!!/8+8-8-8-8-8/8$$
$$32 = 8!!/8+8+8-8-8-8-8$$
$$33 = 8!!/8+8+8/8-8-8-8$$
$$34 = 8!!/8+8/8+8/8-8-8$$
$$35 = 8!!!/8+8+8+8+.\overline{8}+.\overline{8}/8$$
$$36 = 8!!/8+8 \times 8 \times .8+.8-8 \times 8$$
$$37 = 8!!!/.8+8 \times 8 \times .8+.8-8-8$$
$$38 = 8!!/8-8-8/8-.\overline{8}-.\overline{8}/8$$
$$39 = 8!!/8+8-8-8-.\overline{8}-.\overline{8}/8$$
$$40 = 8!!/8+8/8-8-.\overline{8}-.\overline{8}/8$$
$$41 = 8!!/8+8+8/.\overline{8}+.\overline{8}-8-8$$
$$42 = 8!!/8+8/8+8/.\overline{8}+.\overline{8}-8$$
$$43 = 8!!/8/.8-8-.8+8 \times .8 \times 8$$
$$44 = 8!!/8+8+.8-8 \times .8-.8 \times 8$$
$$45 = 8!!/8-8/8-8/8-8/8$$
$$46 = 8!!/8+8-8-8/8-8/8$$
$$47 = 8!!/8+8+8-8-8-8/8$$
$$48 = 8!!/8+8+8+8-8-8-8$$
$$49 = 8!!/8+8+8+8/8-8-8$$
$$50 = 8!!/8+8+8/8+8/8-8$$
$$51 = 8!!/8+8/8+8/8+8/8$$
$$52 = 8!!/8+8 \times .8+8 \times .8-.8-8$$
$$53 = 8!!!/8/8 \times .8+.8+8 \times .8 \times 8$$
$$54 = 8!!/8+8-8/8-.\overline{8}-.\overline{8}/8$$
$$55 = 8!!/8+8+8-8-.\overline{8}-.\overline{8}/8$$
$$56 = 8!!/8+8+8/8-.\overline{8}-.\overline{8}/8$$
$$57 = 8!!/8+8+8+8/.\overline{8}+.\overline{8}-8$$
$$58 = 8!!/8+8+8/8+.\overline{8}+.\overline{8}/8$$
$$59 = 8!!/8/.8+8-.8+8 \times .8 \times 8$$
$$60 = 8!!/8+8 \times 8-.8-8 \times .8 \times 8$$
$$61 = 8!!!-8/.\overline{8}+8+.8+8 \times .8 \times 8$$
$$62 = 8!!!-8-8-8/8-.\overline{8}-.\overline{8}/8$$
$$63 = 8!!!-8+8-8-8-.\overline{8}-.\overline{8}/8$$
$$64 = 8!!!-8+8-8-.8-.8-.8 \times 8$$
$$65 = 8!!!-8+8+8/.\overline{8}+.\overline{8}-8-8$$
$$66 = 8!!!-8+8/8+8/.\overline{8}+.\overline{8}-8$$
$$67 = 8!!!-8+.8-8/.\overline{8}-.8 \times .\overline{8} \times 8$$
$$68 = 8!!!-8+8 \times .8 \times .\overline{8}-.8-.\overline{8}-8$$
$$69 = 8!!!-8-8/8-8/8-8/8$$
$$70 = 8!!!-8+8-8-8/8-8/8$$
$$71 = 8!!!-8+8+8-8-8-8/8$$
$$72 = 8!!!-8+8+8+8-8-8-8$$
$$73 = 8!!!-8+8+8+8/8-8-8$$
$$74 = 8!!!-8+8+8/8+8/8-8$$
$$75 = 8!!!-8+8/8+8/8+8/8$$
$$76 = 8!!!-8+8+.8+.\overline{8}-.8 \times .\overline{8} \times 8$$
$$77 = 8!!!-8+8/.\overline{8}-.8+.8 \times .\overline{8} \times 8$$
$$78 = 8!!!-8+8-8/8-.\overline{8}-.\overline{8}/8$$
$$79 = 8!!!-8+8+8-8-.\overline{8}-.\overline{8}/8$$
$$80 = 8!!!-8+8+8-.8-.8-.8 \times 8$$
$$81 = 8!!!+8+8+8/.\overline{8}+.\overline{8}-8-8$$
$$82 = 8!!!+8+8/8+8/.\overline{8}+.\overline{8}-8$$
$$83 = 8!!!+8+.8-8/.\overline{8}-.8 \times .\overline{8} \times 8$$
$$84 = 8!!!+8+8 \times .8 \times .\overline{8}-.8-.\overline{8}-8$$
$$85 = 8!!!+8-8/8-8/8-8/8$$
$$86 = 8!!!+8+8-8-8/8-8/8$$
$$87 = 8!!!+8+8+8-8-8-8/8$$
$$88 = 8!!!+8+8+8+8-8-8-8$$
$$89 = 8!!!+8+8+8+8/8-8-8$$
$$90 = 8!!!+8+8+8/8+8/8-8$$
$$91 = 8!!!+8+8/8+8/8+8/8$$
$$92 = 8!!!+8+8+.8+.\overline{8}-.8 \times .\overline{8} \times 8$$
$$93 = 8!!!+8+8/.\overline{8}-.8+.8 \times .\overline{8} \times 8$$
$$94 = 8!!!+8+8-8/8-.\overline{8}-.\overline{8}/8$$
$$95 = 8!!!+8+8+8-8-.\overline{8}-.\overline{8}/8$$
$$96 = 8!!!+8+8+8-.8-.8-.8 \times 8$$
$$97 = 8!!!+8+8+8+8/.\overline{8}+.\overline{8}-8$$
$$98 = 8!!!+8+8+8/8+.\overline{8}+.\overline{8}/8$$
$$99 = 8!!!+8/8+8+8+8+8 \times 8$$
$$100 = 8!!!+8+8-.8-.\overline{8}+.8 \times .\overline{8} \times 8$$
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一些更高级的数学¶
代数几何¶
sage: x, y = AffineSpace(2, QQ, 'xy').gens()
sage: C2 = Curve(x^2 + y^2 - 1)
sage: C3 = Curve(x^3 + y^3 - 1)
sage: D = C2 + C3
sage: D
Affine Plane Curve over Rational Field defined by
x^5 + x^3*y^2 + x^2*y^3 + y^5 - x^3 - y^3 - x^2 - y^2 + 1
sage: D.irreducible_components()
[
Closed subscheme of Affine Space of dimension 2 over Rational Field defined by:
x^2 + y^2 - 1,
Closed subscheme of Affine Space of dimension 2 over Rational Field defined by:
x^3 + y^3 - 1
]
sage: V = C2.intersection(C3)
sage: V.irreducible_components()
[
Closed subscheme of Affine Space of dimension 2 over Rational Field defined by:
y,
x - 1,
Closed subscheme of Affine Space of dimension 2 over Rational Field defined by:
y - 1,
x,
Closed subscheme of Affine Space of dimension 2 over Rational Field defined by:
x + y + 2,
2*y^2 + 4*y + 3
]
Sage可以计算出射影3空间中扭曲立方的复曲面理想：
sage: R.<a,b,c,d> = PolynomialRing(QQ, 4)
sage: I = ideal(b^2-a*c, c^2-b*d, a*d-b*c)
sage: F = I.groebner_fan(); F
Groebner fan of the ideal:
Ideal (b^2 - a*c, c^2 - b*d, -b*c + a*d) of Multivariate Polynomial Ring
in a, b, c, d over Rational Field
sage: F.reduced_groebner_bases ()
[[-c^2 + b*d, -b*c + a*d, -b^2 + a*c],
[-b*c + a*d, -c^2 + b*d, b^2 - a*c],
[-c^3 + a*d^2, -c^2 + b*d, b*c - a*d, b^2 - a*c],
[-c^2 + b*d, b^2 - a*c, b*c - a*d, c^3 - a*d^2],
[-b*c + a*d, -b^2 + a*c, c^2 - b*d],
[-b^3 + a^2*d, -b^2 + a*c, c^2 - b*d, b*c - a*d],
[-b^2 + a*c, c^2 - b*d, b*c - a*d, b^3 - a^2*d],
[c^2 - b*d, b*c - a*d, b^2 - a*c]]
sage: F.polyhedralfan()
Polyhedral fan in 4 dimensions of dimension 4
椭圆曲线算法¶
• 椭圆曲线( [$$a_1$$, $$a_2$$, $$a_3$$, $$a_4$$, $$a_6$$] )：返回椭圆曲线
$y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6,$
何处 $$a_i$$ 的被强制加入 $$a_1$$ . 如果所有的 $$a_i$$ 有父母吗 $$\ZZ$$ ，他们被迫 $$\QQ$$ .
• 椭圆曲线( [$$a_4$$, $$a_6$$] )：同上，但 $$a_1=a_2=a_3=0$$ .
• EllipticCurve（label）：使用给定的（new！）从Cremona数据库返回椭圆曲线克雷莫纳标签。标签是一个字符串，例如 "11a""37b2" . 字母必须小写（以区别于旧标签）。
• EllipticCurve（j）：返回带有 $$j$$ -不变量 $$j$$ .
• 椭圆曲线（R， [$$a_1$$, $$a_2$$, $$a_3$$, $$a_4$$, $$a_6$$] )：在环上创建椭圆曲线 $$R$$ 给予 $$a_i$$ 如前所述。
sage: EllipticCurve([0,0,1,-1,0])
Elliptic Curve defined by y^2 + y = x^3 - x over Rational Field
sage: EllipticCurve([GF(5)(0),0,1,-1,0])
Elliptic Curve defined by y^2 + y = x^3 + 4*x over Finite Field of size 5
sage: EllipticCurve([1,2])
Elliptic Curve defined by y^2 = x^3 + x + 2 over Rational Field
sage: EllipticCurve('37a')
Elliptic Curve defined by y^2 + y = x^3 - x over Rational Field
sage: EllipticCurve_from_j(1)
Elliptic Curve defined by y^2 + x*y = x^3 + 36*x + 3455 over Rational Field
sage: EllipticCurve(GF(5), [0,0,1,-1,0])
Elliptic Curve defined by y^2 + y = x^3 + 4*x over Finite Field of size 5
sage: E = EllipticCurve([0,0,1,-1,0])
sage: E
Elliptic Curve defined by y^2 + y = x^3 - x over Rational Field
sage: P = E([0,0])
sage: P + P
(1 : 0 : 1)
sage: 10*P
(161/16 : -2065/64 : 1)
sage: 20*P
(683916417/264517696 : -18784454671297/4302115807744 : 1)
sage: E.conductor()
37
sage: E = EllipticCurve([0,0,0,-4,2]); E
Elliptic Curve defined by y^2 = x^3 - 4*x + 2 over Rational Field
sage: E.conductor()
2368
sage: E.j_invariant()
110592/37
sage: F = EllipticCurve_from_j(110592/37)
sage: F.conductor()
37
sage: G = F.quadratic_twist(2); G
Elliptic Curve defined by y^2 = x^3 - 4*x + 2 over Rational Field
sage: G.conductor()
2368
sage: G.j_invariant()
110592/37
sage: E = EllipticCurve([0,0,1,-1,0])
sage: E.anlist(30)
[0, 1, -2, -3, 2, -2, 6, -1, 0, 6, 4, -5, -6, -2, 2, 6, -4, 0, -12, 0, -4,
3, 10, 2, 0, -1, 4, -9, -2, 6, -12]
sage: v = E.anlist(10000)
sage: %time v = E.anlist(100000)
CPU times: user 0.98 s, sys: 0.06 s, total: 1.04 s
Wall time: 1.06
sage: E = EllipticCurve("37b2")
sage: E
Elliptic Curve defined by y^2 + y = x^3 + x^2 - 1873*x - 31833 over Rational
Field
sage: E = EllipticCurve("389a")
sage: E
Elliptic Curve defined by y^2 + y = x^3 + x^2 - 2*x over Rational Field
sage: E.rank()
2
sage: E = EllipticCurve("5077a")
sage: E.rank()
3
sage: db = sage.databases.cremona.CremonaDatabase()
sage: db.curves(37)
{'a1': [[0, 0, 1, -1, 0], 1, 1], 'b1': [[0, 1, 1, -23, -50], 0, 3]}
sage: db.allcurves(37)
{'a1': [[0, 0, 1, -1, 0], 1, 1],
'b1': [[0, 1, 1, -23, -50], 0, 3],
'b2': [[0, 1, 1, -1873, -31833], 0, 1],
'b3': [[0, 1, 1, -3, 1], 0, 3]}
狄利克雷特字符¶
A 狄利克雷特字符 是同态的扩张 $$(\ZZ/N\ZZ)^* \to R^*$$ ，为了一些环 $$R$$ ，到地图上 $$\ZZ \to R$$ 通过发送这些整数得到 $$x$$ 具有 $$\gcd(N,x)>1$$ 到0。
sage: G = DirichletGroup(12)
sage: G.list()
[Dirichlet character modulo 12 of conductor 1 mapping 7 |--> 1, 5 |--> 1,
Dirichlet character modulo 12 of conductor 4 mapping 7 |--> -1, 5 |--> 1,
Dirichlet character modulo 12 of conductor 3 mapping 7 |--> 1, 5 |--> -1,
Dirichlet character modulo 12 of conductor 12 mapping 7 |--> -1, 5 |--> -1]
sage: G.gens()
(Dirichlet character modulo 12 of conductor 4 mapping 7 |--> -1, 5 |--> 1,
Dirichlet character modulo 12 of conductor 3 mapping 7 |--> 1, 5 |--> -1)
sage: len(G)
4
sage: G = DirichletGroup(21)
sage: chi = G.1; chi
Dirichlet character modulo 21 of conductor 7 mapping 8 |--> 1, 10 |--> zeta6
sage: chi.values()
[0, 1, zeta6 - 1, 0, -zeta6, -zeta6 + 1, 0, 0, 1, 0, zeta6, -zeta6, 0, -1,
0, 0, zeta6 - 1, zeta6, 0, -zeta6 + 1, -1]
sage: chi.conductor()
7
sage: chi.modulus()
21
sage: chi.order()
6
sage: chi(19)
-zeta6 + 1
sage: chi(40)
-zeta6 + 1
sage: chi.galois_orbit()
[Dirichlet character modulo 21 of conductor 7 mapping 8 |--> 1, 10 |--> -zeta6 + 1,
Dirichlet character modulo 21 of conductor 7 mapping 8 |--> 1, 10 |--> zeta6]
sage: go = G.galois_orbits()
sage: [len(orbit) for orbit in go]
[1, 2, 2, 1, 1, 2, 2, 1]
sage: G.decomposition()
[
Group of Dirichlet characters modulo 3 with values in Cyclotomic Field of order 6 and degree 2,
Group of Dirichlet characters modulo 7 with values in Cyclotomic Field of order 6 and degree 2
]
sage: K.<i> = NumberField(x^2+1)
sage: G = DirichletGroup(20,K)
sage: G
Group of Dirichlet characters modulo 20 with values in Number Field in i with defining polynomial x^2 + 1
sage: G.gens()
(Dirichlet character modulo 20 of conductor 4 mapping 11 |--> -1, 17 |--> 1,
Dirichlet character modulo 20 of conductor 5 mapping 11 |--> 1, 17 |--> i)
sage: G.unit_gens()
(11, 17)
sage: G.zeta()
i
sage: G.zeta_order()
4
sage: x = polygen(QQ, 'x')
sage: K = NumberField(x^4 + 1, 'a'); a = K.0
sage: b = K.gen(); a == b
True
sage: K
Number Field in a with defining polynomial x^4 + 1
sage: G = DirichletGroup(5, K, a); G
Group of Dirichlet characters modulo 5 with values in the group of order 8 generated by a in Number Field in a with defining polynomial x^4 + 1
sage: chi = G.0; chi
Dirichlet character modulo 5 of conductor 5 mapping 2 |--> a^2
sage: [(chi^i)(2) for i in range(4)]
[1, a^2, -1, -a^2]
模块化形式¶
Sage可以进行一些与模形式有关的计算，包括维数、模符号的计算空间、Hecke算子和分解。
sage: dimension_cusp_forms(Gamma0(11),2)
1
sage: dimension_cusp_forms(Gamma0(1),12)
1
sage: dimension_cusp_forms(Gamma1(389),2)
6112
sage: M = ModularSymbols(1,12)
sage: M.basis()
([X^8*Y^2,(0,0)], [X^9*Y,(0,0)], [X^10,(0,0)])
sage: t2 = M.T(2)
sage: t2
Hecke operator T_2 on Modular Symbols space of dimension 3 for Gamma_0(1)
of weight 12 with sign 0 over Rational Field
sage: t2.matrix()
[ -24 0 0]
[ 0 -24 0]
[4860 0 2049]
sage: f = t2.charpoly('x'); f
x^3 - 2001*x^2 - 97776*x - 1180224
sage: factor(f)
(x - 2049) * (x + 24)^2
sage: M.T(11).charpoly('x').factor()
(x - 285311670612) * (x - 534612)^2
sage: ModularSymbols(11,2)
Modular Symbols space of dimension 3 for Gamma_0(11) of weight 2 with sign
0 over Rational Field
sage: ModularSymbols(Gamma1(11),2)
Modular Symbols space of dimension 11 for Gamma_1(11) of weight 2 with
sign 0 and over Rational Field
sage: M = ModularSymbols(Gamma1(11),2)
sage: M.T(2).charpoly('x')
x^11 - 8*x^10 + 20*x^9 + 10*x^8 - 145*x^7 + 229*x^6 + 58*x^5 - 360*x^4
+ 70*x^3 - 515*x^2 + 1804*x - 1452
sage: M.T(2).charpoly('x').factor()
(x - 3) * (x + 2)^2 * (x^4 - 7*x^3 + 19*x^2 - 23*x + 11)
* (x^4 - 2*x^3 + 4*x^2 + 2*x + 11)
sage: S = M.cuspidal_submodule()
sage: S.T(2).matrix()
[-2 0]
[ 0 -2]
sage: S.q_expansion_basis(10)
[
q - 2*q^2 - q^3 + 2*q^4 + q^5 + 2*q^6 - 2*q^7 - 2*q^9 + O(q^10)
]
sage: G = DirichletGroup(13)
sage: e = G.0^2
sage: M = ModularSymbols(e,2); M
Modular Symbols space of dimension 4 and level 13, weight 2, character
[zeta6], sign 0, over Cyclotomic Field of order 6 and degree 2
sage: M.T(2).charpoly('x').factor()
(x - zeta6 - 2) * (x - 2*zeta6 - 1) * (x + zeta6 + 1)^2
sage: S = M.cuspidal_submodule(); S
Modular Symbols subspace of dimension 2 of Modular Symbols space of
dimension 4 and level 13, weight 2, character [zeta6], sign 0, over
Cyclotomic Field of order 6 and degree 2
sage: S.T(2).charpoly('x').factor()
(x + zeta6 + 1)^2
sage: S.q_expansion_basis(10)
[
q + (-zeta6 - 1)*q^2 + (2*zeta6 - 2)*q^3 + zeta6*q^4 + (-2*zeta6 + 1)*q^5
+ (-2*zeta6 + 4)*q^6 + (2*zeta6 - 1)*q^8 - zeta6*q^9 + O(q^10)
]
sage: T = ModularForms(Gamma0(11),2)
sage: T
Modular Forms space of dimension 2 for Congruence Subgroup Gamma0(11) of
weight 2 over Rational Field
sage: T.degree()
2
sage: T.level()
11
sage: T.group()
Congruence Subgroup Gamma0(11)
sage: T.dimension()
2
sage: T.cuspidal_subspace()
Cuspidal subspace of dimension 1 of Modular Forms space of dimension 2 for
Congruence Subgroup Gamma0(11) of weight 2 over Rational Field
sage: T.eisenstein_subspace()
Eisenstein subspace of dimension 1 of Modular Forms space of dimension 2
for Congruence Subgroup Gamma0(11) of weight 2 over Rational Field
sage: M = ModularSymbols(11); M
Modular Symbols space of dimension 3 for Gamma_0(11) of weight 2 with sign
0 over Rational Field
sage: M.weight()
2
sage: M.basis()
((1,0), (1,8), (1,9))
sage: M.sign()
0
$$T_p$$ 表示通常的Hecke运算符 ($$p$$ 主）。Hecke操作员怎么办 $$T_2$$$$T_3$$$$T_5$$ 作用于模块符号的空间？
sage: M.T(2).matrix()
[ 3 0 -1]
[ 0 -2 0]
[ 0 0 -2]
sage: M.T(3).matrix()
[ 4 0 -1]
[ 0 -1 0]
[ 0 0 -1]
sage: M.T(5).matrix()
[ 6 0 -1]
[ 0 1 0]
[ 0 0 1]
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| 3.640625
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https://www.physicsforums.com/threads/automorphism-with-order-p.264649/
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# Automorphism with order p
1. Oct 15, 2008
### fk378
1. The problem statement, all variables and given/known data
Let A, B be groups and theta: A --> Aut(B) a homomorphism. For a in A denote theta(a)= theta_a in Aut(B). Equip the product set B x A={(b,a): a in A, b in B} with the binary operation (b,a)(b',a')= (b'',a'') where a''=aa' and b''=b(theta_a{b')).
(a) Assume that p,q in N are prime and p divides (q-1). Consider the case A=Zmodp, B=Zmodq. Show that there exists phi in Aut(Zmodq) which has order p.
Hint: Use Cauchy's Thm for Abelian groups
(b) Deduce that for any 2 primes p,q in N such that p|(q-1) there is a non-Abelian group of order pq.
------
(a) If p divides q-1, then p=-1 in mod q. Don't really know where to go from there.....
(b) No idea.
Last edited: Oct 15, 2008
2. Oct 16, 2008
### morphism
(a) What is the order of Aut(B)?
(b) Under the binary operation defined in the question, is BxA a group?
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# Cash 4 Life
January 06, 2022 (Thursday)
01
13
15
28
52
Cash Ball
01
All past draw results View any extractions and highlight the even, odd, first or second half numbers or just your custom numbers.
Single Draw Display every single draw, present and past, you can highlight only some numbers, you can know the delays and frequencies reached until the draw before the one displayed.
Export results in XLS A format for your calc sheet. A good base for your experimentations with MS Excel o OpenOffice Calc.
Export results in TXT A simple format for all uses.
Maxi Delays
The best series of 5 numbers for 3 points at moment is:
01
02
03
04
58
Actually, the chronological Delay (CD) is 1443 draws
The Maximum delays section uses a complex and fast deterministic algorithm to extract the best available delay for a specific combination.
In this example we show a set of 5 numbers with the best available delay value for the 3 points combination but, in the specific section, you can change the number of numbers, the combination and select only some basic numbers, for example: even, odd, first half, last half or other custom numbers.
If you prefer, you can get the top ten of Couples, Triplets or Quaternets.
Maxi Frequencies
The best series of 5 numbers for 3 points in the lasts 100 draws at moment is:
08
24
32
34
46
The presences in the last 100 was been 12 times.
The Maxi Frequencies section uses a heuristic algorithm for extract the best quantity of presences for a specific combination.
In this example show a set of 5 numbers having the best frequency value available for the combination of 3 points but, in the specific section, you can change the quantity of numbers, the combination and select only some starting numbers such as: even, odd, first half, last half or other custom numbers.
If you prefer, you can get the top ten of Couples, Triplets or Quaternets.
Top Delays
54 69 draws 10 50 draws 41 39 draws 26 34 draws
54
CD
69 draws
Max CD
109 draws
The number currently most delayed is the number 54 absent from 69 draws.
Historically the maximum recorded delay was 109 draws, reached by the number 57.
The most latecomers(Headgame) that can be formed with the number 54 are with the following numbers:
1
3
27
59
53
7
Top Frequencies
39 14 Presences 34 13 Presences 47 13 Presences 08 13 Presences
39
Frequency
14 Presences
The most frequently drawn number in the last 100 draws is 39 with 14 presences.
The most frequent pairs that can be formed with the number 39 (Headgame) are with the following numbers:
8
14
28
30
40
27
Level Synchronous
44
53
CD
26 draws
LD
5 draws
The Level Synchronous are a set of numbers with the same Chronological Delay (CD) and the position are visible in the Analytic Table, this table show the position where the numbers are extracts (26 draws of Chronological Delay) and how long have they reached the level (in this case 2 numbers), i.e. how long have one or more numbers with the same delay already gone out. This parameter is called Level Delay (LD) and the actual value is 5 draws.
You can see other types of synchronous as the Vertical Synchronous and the Diagonal Synchronous.
### INFO
Numbers Extraibles 60 Numbers per Draw 5 Special Balls 1 Cash Ball First Date June 16, 2014 (Monday) Last Date January 06, 2022 (Thursday) Total Draws 1444 Draw Days Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday Minimum winning conditions 2 Points1 Points + 1 Cash Ball Minimum Bet 2 \$You can't play here, this site is only for consultation
### OTHER FUNCTIONS
#### Forecasts
The forecasts are a summary of the best current models for specific Cach 4 Life combinations.
#### Integral and Conditioned Systems
The integral systems for the Cach 4 Life are a set of all available series with user-selected numbers. Integral systems can be reduced by setting conditions by applying one or more static or dynamic filters.
#### Reduced systems guaranteed
You can create an integral system reduction with the guarantee for Cach 4 Life that you win if you exit at least one combination of the initially selected numbers. For example, 13 numbers in 33 groups of 5 can be guaranteed the 3 points.
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https://artofproblemsolving.com/wiki/index.php/1982_AHSME_Problems/Problem_14
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# 1982 AHSME Problems/Problem 14
## Problem 14
In the adjoining figure, points $B$ and $C$ lie on line segment $AD$, and $AB, BC$, and $CD$ are diameters of circle $O, N$, and $P$, respectively. Circles $O, N$, and $P$ all have radius $15$ and the line $AG$ is tangent to circle $P$ at $G$. If $AG$ intersects circle $N$ at points $E$ and $F$, then chord $EF$ has length
$[asy] size(250); defaultpen(fontsize(10)); pair A=origin, O=(1,0), B=(2,0), N=(3,0), C=(4,0), P=(5,0), D=(6,0), G=tangent(A,P,1,2), E=intersectionpoints(A--G, Circle(N,1))[0], F=intersectionpoints(A--G, Circle(N,1))[1]; draw(Circle(O,1)^^Circle(N,1)^^Circle(P,1)^^G--A--D, linewidth(0.7)); dot(A^^B^^C^^D^^E^^F^^G^^O^^N^^P); label("A", A, W); label("B", B, SE); label("C", C, NE); label("D", D, dir(0)); label("P", P, S); label("N", N, S); label("O", O, S); label("E", E, dir(120)); label("F", F, NE); label("G", G, dir(100));[/asy]$
## Solution
Drop a perpendicular line from $N$ to $AG$ at point $H$. $AN=45$, and since $\triangle{AGP}$ is similar to $\triangle{AHN}$. $NH=9$. $NE=NF=15$ so by the Pythagorean Theorem, $EH=HF=12$. Thus $EF=\boxed{24.}$ Answer is then $\boxed{C}$.
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https://nursingessays.us/this-set-of-questions-translates-a-supply-and-demand-schedule-for-an-output-to-demand-for-polluti-1-answer-below/
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# This set of questions translates a supply and demand schedule for an output to demand for polluti… 1 answer below »
Pollution demand
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questions translates a supply and demand schedule for an output to demand for pollution. Suppose the output is kwh of electricity, and each kwh of electricity generates 0.5 tons of CO2. The supply and demand schedules for electricity are as follows Quantity of Demand for Supply of Quantity Electricity Electricity Electricity of (WTP) (Cost) Pollution (kwh) (tons of CO2) 0 100 1 80 S 0.5 2 10 1.0 60 40 1.5 15 4 2.0 20 20 2.5 25 1.1. Pollution demand I Suppose the pollution price is \$50 per ton of Co2. What is the quantity of pollution demanded? (Hint: first figure out how this pollution price translates into a tax for the primary output electricity, then determine the quantity of electricity produced, which will help you finally get the quantity of pollution demanded. tons of CO2 Please enter 1 digit after the decimal point ” src=”https://files.transtutors.com/cdn/questions/transtutors006/images/transtutors006_c48130ce-a7fa-4df8-b1ac-c74d5fcb4d7d.png”>
This set of questions translates a supply and demand schedule for an output to demand for pollution. Suppose the output is kwh of electricity, and each kwh of electricity generates 0.5 tons of CO2. The supply and demand schedules for electricity are as follows Quantity of Demand for Supply of Quantity Electricity Electricity Electricity of (WTP) (Cost) Pollution (kwh) (tons of CO2) 0 100 1 80 S 0.5 2 10 1.0 60 40 1.5 15 4 2.0 20 20 2.5 25 1.1. Pollution demand I Suppose the pollution price is \$50 per ton of Co2. What is the quantity of pollution demanded? (Hint: first figure out how this pollution price translates into a tax for the primary output electricity, then determine the quantity of electricity produced, which will help you finally get the quantity of pollution demanded. tons of CO2 Please enter 1 digit after the decimal point
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https://mathoverflow.net/questions/331197/a-compendium-of-weak-factorization-systems-on-sset
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# A compendium of weak factorization systems on $sSet$
A (weak) factorization system on a category $$\mathcal{C}$$ consists of a pair of classes of morphisms in the category $$(L,R)$$ satisfying
1. Every morphism $$f:x \to y \in \mathcal{C}$$ can be factored (not necessarily uniquely) as a composition $$f = gh$$ where $$g \in R, f\in L$$.
2. The morphisms in $$L$$ (resp. $$R$$) are precisely those morphisms in $$\mathcal{C}$$ which have a left (resp. right) lifting property w.r.t. all the morphisms in $$R$$ (resp. $$L$$).
The category $$sSet$$ of simplicial sets admits many interesting factorization systems/ Many of them pop up repeatedly in applications to homotopy theory and higher category theory in particular. It would be nice to have an accessible comprehensive list of useful factorization systems along with some of their notable properties.
I will say that $$(L,R)$$ forms an enriched factorization system w.r.t. $$(L_0,R_0)$$ iff for every $$A \to B \in L$$ and $$X \to Y \in R$$ we have $$X^B \to X^A \times_{Y^A} Y^B \in R_0$$. Observe that if we forget the self enrichment and take the (Mono,Epi) factorization on $$Set$$ this recovers the usual notion (i have no idea whether this term is standard or not).
I will say that a collection of morphisms $$S$$ generates the class $$L$$ (resp. $$R$$) if $$L$$ is precisely the class of all morphisms which have left (resp. right) lifting property w.r.t. all morphisms which have right (resp. left) lifting property w.r.t. $$S$$.
Here are a couple of examples
1. (Mono,TFib): Where Mono stands for Monomorphism and TFib stands for Trivial Fibration Properties:
• Coincides with $$(Cof, Fib \cap W)$$ for the Joyal (and Quillen) model structure on simplicial sets.
• Enriched w.r.t. (Mono,TFib).
• $$L$$ is generated by the boundary inclusions $$\{ \partial \Delta^n \hookrightarrow \Delta^n\}$$
2. (An,KFib): Where An stands for Anodyne and KFib stands for Kan Fibration
• Coincides with $$(Cof \cap W, Fib)$$ for the Quillen model structure on simplicial sets.
• Enriched w.r.t. (Mono,TFib).
• $$L$$ is generated by the horn inclusion $$\{\Lambda^n_j \hookrightarrow \Delta^n \}$$
3. (RAn,RFib) (resp. (LAn,LFib)): Where RAn (resp. LAn) stands for Right Anodyne (resp. Left Anodyne) and RFib stands for Right Fibration (resp. Left Fibration).
• Enriched w.r.t. (Mono,TFib).
• $$L$$ is generated by the left (resp. right) horn inclusions $$\{\Lambda^n_j \hookrightarrow \Delta^n : 0 \lt j \le n \text{ (resp. } 0 \le j \lt n \text{)} \}$$
4. (InnAn,InnFib): Where InnAn stands for Inner Anodyne and InnFib stands for Inner Fibration
• Enriched w.r.t. (Mono,TFib).
• $$L$$ is generated by the inner horn inclusions $$\{\Lambda^n_j \hookrightarrow \Delta^n : 0 \lt j \lt n \}$$
Beyond those I don't know a great deal more. Here are a couple of factorization systems some of which I know exist but have absolutely nothing useful to say about and some of which I suspect may exist:
• The factorization system $$(Cof \cap W, Fib)$$ for the Joyal model structure. I think the left class should be the monomorphisms which are also categorical equivalences. Do the fibrations have a nice description in general? (or do you have to assume the target is a quasi-category?). Is there a nice class of generators for the acyclic cofibrations?
• A factorization system whose left class models $$n$$-cofinal (resp. $$n$$-final) functors between quasi-categories?
• A factorization system whose right class is the minimal containing all morphisms of (arbitrary) simplicial sets which are isofibrations - and not necessarily inner fibration! (???) (I consider an edge of a simplicial set an isomorphism if it is sent to an isomorphism in the homotopy category - i.e. the left adjoint to the nerve).
Any corrections and further suggestions are very welcome of course.
As this is a big list kind of question I think it would be better to make it Community Wiki.
• If you want it to be CW, just click the CW box when you next edit. Would be better to do this before answers start to arrive – David White May 10 at 15:42
• @DavidWhite I don't think the CW box is visible for me. – Saal Hardali May 10 at 15:44
• For many years now, the CW status can only be conferred by moderators. In any case, this question does not seem to be a good candidate for a community wiki status, which is normally applied to softer questions. – Dmitri Pavlov May 10 at 17:57
• A theorem about morphisms of schemes is a much more general notion than a weak factorization system on simplicial sets. Hence the difference in treatment. – Dmitri Pavlov May 10 at 19:18
• hmm, i'm not sure I completely agree with that statement. I do understand your reasoning though. – Saal Hardali May 10 at 19:20
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If f(x)=125/x^3, what is the value of f(5x)*f(x/5) in terms
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24 Aug 2012, 05:04
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If $$f(x) = \frac{125}{x^{3}}$$, what is the value of $$f(5x)*f(\frac{x}{5})$$ in terms of f(x)?
A. $$(f(x))^{2}$$
B. $$f(x^{2})$$
C. $$(f(x))^{3}$$
D. $$f(x^{3})$$
E. $$f(125x)$$
[Reveal] Spoiler: OA
Last edited by Bunuel on 18 Jul 2013, 22:26, edited 2 times in total.
Edited the question.
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manulath wrote:
If $$f(x) = \frac{125}{x^{3}}$$, what is the value of $$\frac{f(5x)}{f(x/5)}$$ in terms of f(x)?
A. $$(f(x))^{2}$$
B. $$f(x^{2})$$
C. $$(f(x))^{3}$$
D. $$f(x^{3})$$
E. $$f(125x)$$
In order the answer to be A, the question should read:
If $$f(x) = \frac{125}{x^{3}}$$, what is the value of $$f(5x)*f(x/5)$$ in terms of f(x)?
Say $$x=1$$, then:
$$f(x) = f(1)=\frac{125}{1^{3}}=125$$;
$$f(5x)=f(5)=\frac{125}{5^3}=1$$;
$$f(\frac{x}{5})=f(\frac{1}{5})=\frac{125}{(\frac{1}{5})^3}=125*125$$.
$${f(5x)}*{f(\frac{x}{5})}={f(5)}*{f(\frac{1}{5})}=125^2$$ and since $$f(x) = f(1)=125$$, then $${f(5x)}*{f(\frac{x}{5})}=125^2=(f(x))^2$$.
Hope it's clear.
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Re: If f(x)=125/x^3, what is the value of f(5x)/f(x/5) in terms [#permalink]
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24 Aug 2012, 06:05
Bunuel wrote:
In order the answer to be A, the question should read:
If $$f(x) = \frac{125}{x^{3}}$$, what is the value of $$f(5x)*f(x/5)$$ in terms of f(x)?
Thanks Bunuel.
The question really troubled me.
I tried methods and still could not arrive at correct answer.
Your verdict has confirmed my doubt.
I shall edit the question.
PS: You already did that. Thanks again.
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!
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Re: If f(x)=125/x^3, what is the value of f(5x)*f(x/5) in terms [#permalink]
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03 Jul 2013, 08:44
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manulath wrote:
If $$f(x) = \frac{125}{x^{3}}$$, what is the value of $$f(5x)*f(x/5)$$ in terms of f(x)?
A. $$(f(x))^{2}$$
B. $$f(x^{2})$$
C. $$(f(x))^{3}$$
D. $$f(x^{3})$$
E. $$f(125x)$$
lets 125 =A
so we can write f(x)=A/(X^3)...........(1
Now lets sove
f(5x)*f(x/5)=(A/(5X)^3)*(A/(X/5)^3)====>(A^2/X^6)===>(A/X^3)^2==(F(X))^2==>USING(1)
Hence A
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Re: If f(x)=125/x^3, what is the value of f(5x)*f(x/5) in terms [#permalink]
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04 Jul 2013, 00:48
2
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manulath wrote:
If $$f(x) = \frac{125}{x^{3}}$$, what is the value of $$f(5x)*f(x/5)$$ in terms of f(x)?
A. $$(f(x))^{2}$$
B. $$f(x^{2})$$
C. $$(f(x))^{3}$$
D. $$f(x^{3})$$
E. $$f(125x)$$
1. $$f(5x) = 125/125*x^3 = 1/x^3$$
2.$$f(x/5) = 125 * 125/x^3 = 125^2 / x^3$$
3. $$f(5x) * f(x/5) = 125 ^ 2/ x^6 = (f(x)) ^ {2}$$
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Re: If f(x)=125/x^3, what is the value of f(5x)*f(x/5) in terms [#permalink]
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18 Jul 2013, 18:30
SravnaTestPrep wrote:
manulath wrote:
If $$f(x) = \frac{125}{x^{3}}$$, what is the value of $$f(5x)*f(x/5)$$ in terms of f(x)?
A. $$(f(x))^{2}$$
B. $$f(x^{2})$$
C. $$(f(x))^{3}$$
D. $$f(x^{3})$$
E. $$f(125x)$$
1. $$f(5x) = 125/125*x^3 = 1/x^3$$
2.$$f(x/5) = 125 * 125/x^3 = 125^2 / x^3$$
3. $$f(5x) * f(x/5) = 125 ^ 2/ x^6 = (f(x)) ^ {2}$$
In step 3, you have multiplied 1. & 2. What would be the answer if you divide 1. by 2. ? I am kinda confused. I get f(5x)/f(x/5) = 1/5^6. Help, thanks!
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18 Jul 2013, 22:26
mneeti wrote:
SravnaTestPrep wrote:
manulath wrote:
If $$f(x) = \frac{125}{x^{3}}$$, what is the value of $$f(5x)*f(x/5)$$ in terms of f(x)?
A. $$(f(x))^{2}$$
B. $$f(x^{2})$$
C. $$(f(x))^{3}$$
D. $$f(x^{3})$$
E. $$f(125x)$$
1. $$f(5x) = 125/125*x^3 = 1/x^3$$
2.$$f(x/5) = 125 * 125/x^3 = 125^2 / x^3$$
3. $$f(5x) * f(x/5) = 125 ^ 2/ x^6 = (f(x)) ^ {2}$$
In step 3, you have multiplied 1. & 2. What would be the answer if you divide 1. by 2. ? I am kinda confused. I get f(5x)/f(x/5) = 1/5^6. Help, thanks!
The questions is "what is the value of $$f(5x)*f(\frac{x}{5})$$..." (f(5x) multiplied by f(x/5)) not "what is the value of $$\frac{f(5x)}{f(\frac{x}{5})}$$..." (f(5x) divided by f(x/5)).
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18 Jul 2013, 22:47
Bunuel wrote:
mneeti wrote:
The questions is "what is the value of $$f(5x)*f(\frac{x}{5})$$..." (f(5x) multiplied by f(x/5)) not "what is the value of $$\frac{f(5x)}{f(\frac{x}{5})}$$..." (f(5x) divided by f(x/5)).
Please refer your Ist response in this topic, the quoted part asks to find $$\frac{f(5x)}{f(\frac{x}{5})}$$. Also, the question I have from my source of practice questions requires to find the $$\frac{f(5x)}{f(\frac{x}{5})}$$. Thanks!
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18 Jul 2013, 22:49
mneeti wrote:
Bunuel wrote:
mneeti wrote:
The questions is "what is the value of $$f(5x)*f(\frac{x}{5})$$..." (f(5x) multiplied by f(x/5)) not "what is the value of $$\frac{f(5x)}{f(\frac{x}{5})}$$..." (f(5x) divided by f(x/5)).
Please refer your Ist response in this topic, the quoted part asks to find $$\frac{f(5x)}{f(\frac{x}{5})}$$. Also, the question I have from my source of practice questions requires to find the $$\frac{f(5x)}{f(\frac{x}{5})}$$. Thanks!
Please refer to the same post: in order the answer to be A, the question should read:
If $$f(x) = \frac{125}{x^{3}}$$, what is the value of $$f(5x)*f(\frac{x}{5})$$ in terms of f(x)?
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24 Aug 2013, 06:09
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Bunuel wrote:
manulath wrote:
If $$f(x) = \frac{125}{x^{3}}$$, what is the value of $$\frac{f(5x)}{f(x/5)}$$ in terms of f(x)?
A. $$(f(x))^{2}$$
B. $$f(x^{2})$$
C. $$(f(x))^{3}$$
D. $$f(x^{3})$$
E. $$f(125x)$$
In order the answer to be A, the question should read:
If $$f(x) = \frac{125}{x^{3}}$$, what is the value of $$f(5x)*f(x/5)$$ in terms of f(x)?
Say $$x=1$$, then:
$$f(x) = f(1)=\frac{125}{1^{3}}=125$$;
$$f(5x)=f(5)=\frac{125}{5^3}=1$$;
$$f(\frac{x}{5})=f(\frac{1}{5})=\frac{125}{(\frac{1}{5})^3}=125*125$$.
$${f(5x)}*{f(\frac{x}{5})}={f(5)}*{f(\frac{1}{5})}=125^2$$ and since $$f(x) = f(1)=125$$, then $${f(5x)}*{f(\frac{x}{5})}=125^2=(f(x))^2$$.
Hope it's clear.
Can we take constant 5 outside of f(5x) i.e 5f(x) ?
I was able to arrive at the answer by taking constant outside of the function,but not sure whether it is correct approach or not.
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25 Aug 2013, 06:41
abid1986 wrote:
Bunuel wrote:
manulath wrote:
If $$f(x) = \frac{125}{x^{3}}$$, what is the value of $$\frac{f(5x)}{f(x/5)}$$ in terms of f(x)?
A. $$(f(x))^{2}$$
B. $$f(x^{2})$$
C. $$(f(x))^{3}$$
D. $$f(x^{3})$$
E. $$f(125x)$$
In order the answer to be A, the question should read:
If $$f(x) = \frac{125}{x^{3}}$$, what is the value of $$f(5x)*f(x/5)$$ in terms of f(x)?
Say $$x=1$$, then:
$$f(x) = f(1)=\frac{125}{1^{3}}=125$$;
$$f(5x)=f(5)=\frac{125}{5^3}=1$$;
$$f(\frac{x}{5})=f(\frac{1}{5})=\frac{125}{(\frac{1}{5})^3}=125*125$$.
$${f(5x)}*{f(\frac{x}{5})}={f(5)}*{f(\frac{1}{5})}=125^2$$ and since $$f(x) = f(1)=125$$, then $${f(5x)}*{f(\frac{x}{5})}=125^2=(f(x))^2$$.
Hope it's clear.
Can we take constant 5 outside of f(5x) i.e 5f(x) ?
I was able to arrive at the answer by taking constant outside of the function,but not sure whether it is correct approach or not.
It depends on HOW are you doing that. Please show your work.
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25 Aug 2013, 11:21
f(5x) = 5f(x)
f(x/5)= 1/5f(x)
Then;
f(5x)*f(x/5) = 5f(x)*1/5f(x) =
$$(f(x))^2$$
Is this approach fine?
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25 Aug 2013, 11:36
abid1986 wrote:
f(5x) = 5f(x)
f(x/5)= 1/5f(x)
Then;
f(5x)*f(x/5) = 5f(x)*1/5f(x) =
$$(f(x))^2$$
Is this approach fine?
No, that's not correct.
$$f(x) = \frac{125}{x^{3}}$$,
$$f(5x) = \frac{125}{(5x)^{3}}=\frac{1}{x^3}$$.
$$f(\frac{x}{5}) = \frac{125}{(\frac{x}{5})^{3}}=\frac{125^2}{x^3}$$.
So, $$f(x)=125*f(5x)$$ and $$f(\frac{x}{5})=125*f(x)$$
Hope it helps.
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02 Sep 2014, 03:31
$$f(5x) * f(\frac{x}{5}) = \frac{125}{(5x)^3} * \frac{125}{(\frac{x}{5})^3}= \frac{125}{x^3} * \frac{125}{x^3} = [f(x)]^2$$
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# What date is 27 days added from Wednesday October 30, 2024?
## Tuesday November 26, 2024
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• Specific Date: Wednesday October 30, 2024
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## Tuesday November 26, 2024 Stats
• Day of the week: Tuesday
• Month: November
• Day of the year: 331
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Counting forward from today, Tuesday November 26, 2024 is 27 from now using our current calendar. 27 days is equivalent to:
27 days is also 648 hours. Tuesday November 26, 2024 is 90% of the year completed.
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• 1853.28 hours Watching television
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• 1942 "Casablanca" directed by Michael Curtiz and starring Humphrey Bogart and Ingrid Bergman premieres at Hollywood Theater, NYC (Academy Awards Best Picture 1943)
• 1917 NHL forms with Montreal Canadiens, Montreal Maroons, Toronto Arenas, Ottawa Senators & Quebec Bulldogs; National Hockey Association disbands
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http://isabelle.in.tum.de/repos/isabelle/diff/b73f94b366b7/src/HOL/Analysis/Complex_Analysis_Basics.thy
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src/HOL/Analysis/Complex_Analysis_Basics.thy
changeset 66252 b73f94b366b7 parent 66089 def95e0bc529 child 66453 cc19f7ca2ed6
``` 1.1 --- a/src/HOL/Analysis/Complex_Analysis_Basics.thy Sun Jul 02 20:13:38 2017 +0200
1.2 +++ b/src/HOL/Analysis/Complex_Analysis_Basics.thy Tue Jul 04 09:36:25 2017 +0100
1.3 @@ -23,10 +23,11 @@
1.4 "(f has_derivative f') F \<Longrightarrow> ((\<lambda>x. of_real (f x)) has_derivative (\<lambda>x. of_real (f' x))) F"
1.5 using bounded_linear.has_derivative[OF bounded_linear_of_real] .
1.6
1.7 -lemma has_vector_derivative_real_complex:
1.8 +lemma has_vector_derivative_real_field:
1.9 "DERIV f (of_real a) :> f' \<Longrightarrow> ((\<lambda>x. f (of_real x)) has_vector_derivative f') (at a within s)"
1.10 using has_derivative_compose[of of_real of_real a _ f "op * f'"]
1.11 by (simp add: scaleR_conv_of_real ac_simps has_vector_derivative_def has_field_derivative_def)
1.12 +lemmas has_vector_derivative_real_complex = has_vector_derivative_real_field
1.13
1.14 lemma fact_cancel:
1.15 fixes c :: "'a::real_field"
1.16 @@ -124,15 +125,9 @@
1.17 using has_derivative_zero_connected_constant [OF assms(1-4)] assms
1.18 by (metis DERIV_const has_derivative_const Diff_iff at_within_open frechet_derivative_at has_field_derivative_def)
1.19
1.20 -lemma DERIV_zero_constant:
1.21 - fixes f :: "'a::{real_normed_field, real_inner} \<Rightarrow> 'a"
1.22 - shows "\<lbrakk>convex s;
1.23 - \<And>x. x\<in>s \<Longrightarrow> (f has_field_derivative 0) (at x within s)\<rbrakk>
1.24 - \<Longrightarrow> \<exists>c. \<forall>x \<in> s. f(x) = c"
1.25 - by (auto simp: has_field_derivative_def lambda_zero intro: has_derivative_zero_constant)
1.26 +lemmas DERIV_zero_constant = has_field_derivative_zero_constant
1.27
1.28 lemma DERIV_zero_unique:
1.29 - fixes f :: "'a::{real_normed_field, real_inner} \<Rightarrow> 'a"
1.30 assumes "convex s"
1.31 and d0: "\<And>x. x\<in>s \<Longrightarrow> (f has_field_derivative 0) (at x within s)"
1.32 and "a \<in> s"
1.33 @@ -142,7 +137,6 @@
1.34 (metis d0 has_field_derivative_imp_has_derivative lambda_zero)
1.35
1.36 lemma DERIV_zero_connected_unique:
1.37 - fixes f :: "'a::{real_normed_field, real_inner} \<Rightarrow> 'a"
1.38 assumes "connected s"
1.39 and "open s"
1.40 and d0: "\<And>x. x\<in>s \<Longrightarrow> DERIV f x :> 0"
1.41 @@ -177,9 +171,8 @@
1.42
1.43 (*generalising DERIV_isconst_all, which requires type real (using the ordering)*)
1.44 lemma DERIV_zero_UNIV_unique:
1.45 - fixes f :: "'a::{real_normed_field, real_inner} \<Rightarrow> 'a"
1.46 - shows "(\<And>x. DERIV f x :> 0) \<Longrightarrow> f x = f a"
1.47 -by (metis DERIV_zero_unique UNIV_I convex_UNIV)
1.48 + "(\<And>x. DERIV f x :> 0) \<Longrightarrow> f x = f a"
1.49 + by (metis DERIV_zero_unique UNIV_I convex_UNIV)
1.50
1.51 subsection \<open>Some limit theorems about real part of real series etc.\<close>
1.52
1.53 @@ -854,7 +847,7 @@
1.54
1.55
1.56 lemma field_differentiable_series:
1.57 - fixes f :: "nat \<Rightarrow> complex \<Rightarrow> complex"
1.58 + fixes f :: "nat \<Rightarrow> 'a::{real_normed_field,banach} \<Rightarrow> 'a"
1.59 assumes "convex s" "open s"
1.60 assumes "\<And>n x. x \<in> s \<Longrightarrow> (f n has_field_derivative f' n x) (at x)"
1.61 assumes "uniformly_convergent_on s (\<lambda>n x. \<Sum>i<n. f' i x)"
1.62 @@ -879,7 +872,7 @@
1.63 qed
1.64
1.65 lemma field_differentiable_series':
1.66 - fixes f :: "nat \<Rightarrow> complex \<Rightarrow> complex"
1.67 + fixes f :: "nat \<Rightarrow> 'a::{real_normed_field,banach} \<Rightarrow> 'a"
1.68 assumes "convex s" "open s"
1.69 assumes "\<And>n x. x \<in> s \<Longrightarrow> (f n has_field_derivative f' n x) (at x)"
1.70 assumes "uniformly_convergent_on s (\<lambda>n x. \<Sum>i<n. f' i x)"
1.71 @@ -890,7 +883,7 @@
1.72 subsection\<open>Bound theorem\<close>
1.73
1.74 lemma field_differentiable_bound:
1.75 - fixes s :: "complex set"
1.76 + fixes s :: "'a::real_normed_field set"
1.77 assumes cvs: "convex s"
1.78 and df: "\<And>z. z \<in> s \<Longrightarrow> (f has_field_derivative f' z) (at z within s)"
1.79 and dn: "\<And>z. z \<in> s \<Longrightarrow> norm (f' z) \<le> B"
1.80 @@ -905,8 +898,7 @@
1.81
1.82 subsection\<open>Inverse function theorem for complex derivatives\<close>
1.83
1.84 -lemma has_complex_derivative_inverse_basic:
1.85 - fixes f :: "complex \<Rightarrow> complex"
1.86 +lemma has_field_derivative_inverse_basic:
1.87 shows "DERIV f (g y) :> f' \<Longrightarrow>
1.88 f' \<noteq> 0 \<Longrightarrow>
1.89 continuous (at y) g \<Longrightarrow>
1.90 @@ -919,8 +911,10 @@
1.91 apply (auto simp: bounded_linear_mult_right)
1.92 done
1.93
1.94 -lemma has_complex_derivative_inverse_strong:
1.95 - fixes f :: "complex \<Rightarrow> complex"
1.96 +lemmas has_complex_derivative_inverse_basic = has_field_derivative_inverse_basic
1.97 +
1.98 +lemma has_field_derivative_inverse_strong:
1.99 + fixes f :: "'a::{euclidean_space,real_normed_field} \<Rightarrow> 'a"
1.100 shows "DERIV f x :> f' \<Longrightarrow>
1.101 f' \<noteq> 0 \<Longrightarrow>
1.102 open s \<Longrightarrow>
1.103 @@ -931,9 +925,10 @@
1.104 unfolding has_field_derivative_def
1.105 apply (rule has_derivative_inverse_strong [of s x f g ])
1.106 by auto
1.107 +lemmas has_complex_derivative_inverse_strong = has_field_derivative_inverse_strong
1.108
1.109 -lemma has_complex_derivative_inverse_strong_x:
1.110 - fixes f :: "complex \<Rightarrow> complex"
1.111 +lemma has_field_derivative_inverse_strong_x:
1.112 + fixes f :: "'a::{euclidean_space,real_normed_field} \<Rightarrow> 'a"
1.113 shows "DERIV f (g y) :> f' \<Longrightarrow>
1.114 f' \<noteq> 0 \<Longrightarrow>
1.115 open s \<Longrightarrow>
1.116 @@ -944,6 +939,7 @@
1.117 unfolding has_field_derivative_def
1.118 apply (rule has_derivative_inverse_strong_x [of s g y f])
1.119 by auto
1.120 +lemmas has_complex_derivative_inverse_strong_x = has_field_derivative_inverse_strong_x
1.121
1.122 subsection \<open>Taylor on Complex Numbers\<close>
1.123
1.124 @@ -952,14 +948,14 @@
1.125 shows "sum f {0..n} = f 0 - f (Suc n) + sum (\<lambda>i. f (Suc i)) {0..n}"
1.126 by (induct n) auto
1.127
1.128 -lemma complex_taylor:
1.129 +lemma field_taylor:
1.130 assumes s: "convex s"
1.131 and f: "\<And>i x. x \<in> s \<Longrightarrow> i \<le> n \<Longrightarrow> (f i has_field_derivative f (Suc i) x) (at x within s)"
1.132 - and B: "\<And>x. x \<in> s \<Longrightarrow> cmod (f (Suc n) x) \<le> B"
1.133 + and B: "\<And>x. x \<in> s \<Longrightarrow> norm (f (Suc n) x) \<le> B"
1.134 and w: "w \<in> s"
1.135 and z: "z \<in> s"
1.136 - shows "cmod(f 0 z - (\<Sum>i\<le>n. f i w * (z-w) ^ i / (fact i)))
1.137 - \<le> B * cmod(z - w)^(Suc n) / fact n"
1.138 + shows "norm(f 0 z - (\<Sum>i\<le>n. f i w * (z-w) ^ i / (fact i)))
1.139 + \<le> B * norm(z - w)^(Suc n) / fact n"
1.140 proof -
1.141 have wzs: "closed_segment w z \<subseteq> s" using assms
1.142 by (metis convex_contains_segment)
1.143 @@ -1018,34 +1014,45 @@
1.144 assume u: "u \<in> closed_segment w z"
1.145 then have us: "u \<in> s"
1.146 by (metis wzs subsetD)
1.147 - have "cmod (f (Suc n) u) * cmod (z - u) ^ n \<le> cmod (f (Suc n) u) * cmod (u - z) ^ n"
1.148 + have "norm (f (Suc n) u) * norm (z - u) ^ n \<le> norm (f (Suc n) u) * norm (u - z) ^ n"
1.149 by (metis norm_minus_commute order_refl)
1.150 - also have "... \<le> cmod (f (Suc n) u) * cmod (z - w) ^ n"
1.151 + also have "... \<le> norm (f (Suc n) u) * norm (z - w) ^ n"
1.152 by (metis mult_left_mono norm_ge_zero power_mono segment_bound [OF u])
1.153 - also have "... \<le> B * cmod (z - w) ^ n"
1.154 + also have "... \<le> B * norm (z - w) ^ n"
1.155 by (metis norm_ge_zero zero_le_power mult_right_mono B [OF us])
1.156 - finally have "cmod (f (Suc n) u) * cmod (z - u) ^ n \<le> B * cmod (z - w) ^ n" .
1.157 + finally have "norm (f (Suc n) u) * norm (z - u) ^ n \<le> B * norm (z - w) ^ n" .
1.158 } note cmod_bound = this
1.159 have "(\<Sum>i\<le>n. f i z * (z - z) ^ i / (fact i)) = (\<Sum>i\<le>n. (f i z / (fact i)) * 0 ^ i)"
1.160 by simp
1.161 also have "\<dots> = f 0 z / (fact 0)"
1.162 by (subst sum_zero_power) simp
1.163 - finally have "cmod (f 0 z - (\<Sum>i\<le>n. f i w * (z - w) ^ i / (fact i)))
1.164 - \<le> cmod ((\<Sum>i\<le>n. f i w * (z - w) ^ i / (fact i)) -
1.165 + finally have "norm (f 0 z - (\<Sum>i\<le>n. f i w * (z - w) ^ i / (fact i)))
1.166 + \<le> norm ((\<Sum>i\<le>n. f i w * (z - w) ^ i / (fact i)) -
1.167 (\<Sum>i\<le>n. f i z * (z - z) ^ i / (fact i)))"
1.169 - also have "... \<le> B * cmod (z - w) ^ n / (fact n) * cmod (w - z)"
1.170 + also have "... \<le> B * norm (z - w) ^ n / (fact n) * norm (w - z)"
1.171 apply (rule field_differentiable_bound
1.172 [where f' = "\<lambda>w. f (Suc n) w * (z - w)^n / (fact n)"
1.173 and s = "closed_segment w z", OF convex_closed_segment])
1.174 apply (auto simp: ends_in_segment DERIV_subset [OF sum_deriv wzs]
1.175 norm_divide norm_mult norm_power divide_le_cancel cmod_bound)
1.176 done
1.177 - also have "... \<le> B * cmod (z - w) ^ Suc n / (fact n)"
1.178 + also have "... \<le> B * norm (z - w) ^ Suc n / (fact n)"
1.179 by (simp add: algebra_simps norm_minus_commute)
1.180 finally show ?thesis .
1.181 qed
1.182
1.183 +lemma complex_taylor:
1.184 + assumes s: "convex s"
1.185 + and f: "\<And>i x. x \<in> s \<Longrightarrow> i \<le> n \<Longrightarrow> (f i has_field_derivative f (Suc i) x) (at x within s)"
1.186 + and B: "\<And>x. x \<in> s \<Longrightarrow> cmod (f (Suc n) x) \<le> B"
1.187 + and w: "w \<in> s"
1.188 + and z: "z \<in> s"
1.189 + shows "cmod(f 0 z - (\<Sum>i\<le>n. f i w * (z-w) ^ i / (fact i)))
1.190 + \<le> B * cmod(z - w)^(Suc n) / fact n"
1.191 + using assms by (rule field_taylor)
1.192 +
1.193 +
1.194 text\<open>Something more like the traditional MVT for real components\<close>
1.195
1.196 lemma complex_mvt_line:
```
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+0
# 2(sin 15+ sin 75)^2
0
66
2
2(sin 15+ sin 75)^2
May 7, 2023
#1
0
2(sin 15+ sin 75)^2
I don't know how to combine trigonometric ratios to equivalencies.
You can obtain the values from a table and then work out the arithmetic.
I got 2 * (sin 15 + sin 75)2
2 * (0.2588 + 0.9659)2
2 * (1.2247)2
2 * 1.4999 = 2.2497
I did considerable rounding, though. You might want
to take those sine values out to more decimal places. .
.
May 7, 2023
#2
+214
0
Good rounding. I'm think that the expression winds up being 2*1.5=3.
I don't know why, but that is what I get.
May 7, 2023
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https://us.metamath.org/mpeuni/nfeudw.html
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > nfeudw Structured version Visualization version GIF version
Theorem nfeudw 2673
Description: Bound-variable hypothesis builder for the unique existential quantifier. Deduction version of nfeu 2676. Version of nfeud 2674 with a disjoint variable condition, which does not require ax-13 2386. (Contributed by NM, 15-Feb-2013.) (Revised by Gino Giotto, 10-Jan-2024.)
Hypotheses
Ref Expression
nfeudw.1 𝑦𝜑
nfeudw.2 (𝜑 → Ⅎ𝑥𝜓)
Assertion
Ref Expression
nfeudw (𝜑 → Ⅎ𝑥∃!𝑦𝜓)
Distinct variable group: 𝑥,𝑦
Allowed substitution hints: 𝜑(𝑥,𝑦) 𝜓(𝑥,𝑦)
Proof of Theorem nfeudw
StepHypRef Expression
1 df-eu 2650 . 2 (∃!𝑦𝜓 ↔ (∃𝑦𝜓 ∧ ∃*𝑦𝜓))
2 nfeudw.1 . . . 4 𝑦𝜑
3 nfeudw.2 . . . 4 (𝜑 → Ⅎ𝑥𝜓)
42, 3nfexd 2344 . . 3 (𝜑 → Ⅎ𝑥𝑦𝜓)
52, 3nfmodv 2639 . . 3 (𝜑 → Ⅎ𝑥∃*𝑦𝜓)
64, 5nfand 1894 . 2 (𝜑 → Ⅎ𝑥(∃𝑦𝜓 ∧ ∃*𝑦𝜓))
71, 6nfxfrd 1850 1 (𝜑 → Ⅎ𝑥∃!𝑦𝜓)
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 398 ∃wex 1776 Ⅎwnf 1780 ∃*wmo 2616 ∃!weu 2649 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1792 ax-4 1806 ax-5 1907 ax-6 1966 ax-7 2011 ax-10 2141 ax-11 2157 ax-12 2173 This theorem depends on definitions: df-bi 209 df-an 399 df-or 844 df-ex 1777 df-nf 1781 df-mo 2618 df-eu 2650 This theorem is referenced by: nfeuw 2675 nfreuw 3374
Copyright terms: Public domain W3C validator
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https://fxtechlab.com/trading-forex-indicators-moving-average-for-fast-making-profit/
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Make earnings with no risk
Automated AI-driven system makes the trades, you earn the money
Strategies
# Trading Forex Indicators: Moving Average for Fast Making Profit
As long as financial markets exist, so does the moving average. There is no indicator as practical and straightforward as this one. Here you will find everything there is to know about MA, from their types to the system, practical usage. We will equally analyze different trading techniques for this indicator.
## What is the moving average?
MA is a standard and built-in trend indicator that displays the average worth value for a specific time. The indicator shows the direction of the trend, the propensity of the price to rise or fall in the future.
## History of MA
The MA concept came to the limelight in 1909 when H.W. Yule depicted R.H. Hooker’s instantaneous averages, summed in 1901, as moving averages. Although Yule’s textbook has no record of concept, his description of the same brought it to being circulated by W.L. King in 1912 in a book titled Elements of a Statistical Method.
MA has been a powerful tool that has always been used to discover the pathway of minimal resistance, highlight points of valuation on one’s chart, determine best entry periods, drive in massive sales, and select the best asset to trade.
## The formula of the MA
SMA = SUM (CLOSE (i), N) / N, where:
SUM — amount
CLOSE (i) — closing price of the current period
N — number of computation periods
## Method for plotting a MA on a price chart
There are numerous methods for constructing this indicator:
• Simple moving average (SMA) — shows the sum of the closing prices of any instrument for a specified number of unit periods, divided by the number of these periods.
• Exponential moving average (EMA) — giving more value to the bars closer to the current price. The EMA adds a certain fraction of the recent closing price to the immediate value of the moving average. In calculating the smoothed average, more weight has the market closing price. This is the most popular version of the primary indicator.
• Linear weighted moving average (LVMA) — is the most active indicator. It reacts the fastest to price changes, so it gives a lot of false signals. Traders don’t usually use it.
The distinctions between the methods are not significant, but some lines are smoother than others.
Parameter Dignity Disadvantage Trend visualization Unambiguously indicates the direction of the trend It doesn’t predict the future Variability Many types of MA Different MAs sometimes contradict each other Signals for a trend change Variety of signals, choice Perceiving signals as 100% predicting the future Trading strategies using MA Suitable for both trending and counter-trending strategies Many false signals during periods of non-directional movement
## Benefits of MA
At first, it is difficult not to get lost in the chaos of price fluctuations — there is too much information that the brain cannot simplify and structure. The moving average will help to “reject” some of the price fluctuations that have the most negligible impact on the overall picture.
Depending on his needs, a trader can choose an MA that fits the best to traderes needs. This will help maintain consistency of trading even during periods of unsuccessful series of deals.
The use of a MA involves receiving several types of simple and unambiguous signals for both continuation and trend change.
MA gives plenty of room for both: trending and counter-trending strategies.
Any MA calculation contains past price data. The final result will reflect, to one degree or another, the past dynamics.
MAs of various types and parameters in some cases will contradict each other. The trader should understand why he is using these MA settings.
Novice traders often perceive signals of a possible trend change from MA as unambiguous. Due to its visual display, the indicator seems to traders as an obstacle to the price, after overcoming which nothing will prevent it from forming a new trend in the opposite direction.
During periods of non-directional movements, the price will very often cross the indicator in both directions. Consequently, in any strategy based on this indicator, such periods are characterized by many “false” signals.
## How should you choose the MA period?
Since the MA is a trend indicator, you need to be careful when choosing a period.
The manor rule is evident in the following principle: after the price traverses the MA line, it should stay above it for as long as possible — if the market goes up or below the line — if the market falls. It must act as something like a dynamic line of support or resistance.
The primary task of any trader is to find, using an ordinary search, such values at which the MA will coincide as much as possible with the support or resistance line.
## How to trade with the MA?
As it is a trend indicator, its strategies are trending first and foremost. Traders come up with different signals, but there are three main ones.
• Identifying a trend using a MA
The general direction of the indicator shows the current trend. Depending on the period, the trend is short-term, medium-term, or long-term. The MA looks upward — an upward trend, downwards — a downtrend, located horizontally — a flat.
• MA crossover signals
The signal depends on a MA with a shorter period: if it crosses the second line from bottom to top, we have a buy signal, and from top to bottom, we will sell.
• Sliding like resistance
Sliders as support and resistance. Above, we said that MAs with prolonged periods are most suitable for this role. A line break is considered a signal in the direction of the breakdown. It is essential to be very conscientious about selecting the indicator period — it can be different for each asset.
## How to use the EMA to determine when a trend reversal occurs?
Let’s look at two common ways:
• Look at the direction of the indicator.
• Look at the intersection of the indicator by the price. Alternatively, wait for the price to close behind the arrow.
Consequently, to change the direction of the indicator, the price must go sharply and much in the opposite direction.
This will point to:
• The weakness of those who “push” the price in the direction of the trend.
• The strength of those who “push” the price in the opposite route.
These factors may indicate a trend reversal.
As for the price crossing the indicator, many experts believe that it makes no sense to use this as a signal for something. If the indicator is directed up, and the price has gone under it, this does not mean a bearish trend. On the contrary, you can take a closer look at buying — if the direction of the MA has not changed, then this price movement could not reverse the global trend.
Our team consists from professional traders with over 10 years of experience carrying a mission to simplify automated forex trading for everyone.
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Teachers' Domain is moving soon to its new and improved home — PBS LearningMedia! Learn More
# Data Clusters and Distributions
### Collection Developed by:
Collection Credits
### Collection Funded by:
Funding for the VITAL/Ready to Teach collection was secured through the United States Department of Education under the Ready to Teach Program.
## Resources for this Lesson:
Connections
Everyday Math (2004)
Teacher Lesson Guide, pp. 98- 102, 115-119
Student Reference, pp. 68, 262
Math Journal, pp. 35, 42-43
Math Master, p. 242, 246
Investigations/Scott Foresman (2006)
The Shape of the Data, all Investigations
Standards
to:
Not yet reviewed.
### Overview
In this activity, students learn some of the pitfalls of doing surveys. They also learn to examine the shape of data, including data clusters, the range and typical values. This Cyberchase activity is motivated by a video segment in which the CyberSquad searches for Hacker's castle based on a survey of where town residents have last seen him.
3-6
1 hour
### Media Resources
Using Data Clusters to Find Hacker QuickTime Video
### Part I: Learning Activity
1. Tell the students that the CyberSquad is trying to locate Hacker's castle, where Dr. Marbles is hidden. In Castleblanca, there are too many castles to check, so the CyberSquad conducts a survey asking people to put a mark on a map where they have seen a tall man with a pointy chin and cape. Based on the data, they find Dracula not Hacker.
2. Tell the students that they will see a video segment in which the CyberSquad tries to figure out how to find Hacker instead of Dracula.
3. Show the Using Data Clusters to Find Hacker QuickTime Video, and ask students to keep track of as many things as they can, which can make one's data set lead to the wrong predictions or conclusions.
4. Discuss the students' ideas.
5. Distribute the Handout: "What's Typical, Based on the Shape of Data Charts?".
6. Ask students to examine the data collected by Inez and Matt and to answer the questions.
7. Discuss their answers, noting that the two data sets differ only by one inch, suggesting perhaps that one kid measured the band members with their shoes on and one kid measured them with their shoes off. This is a systematic error in data collection.
### Part II: Assessment
Assessment: Level A (proficiency): Students are asked to examine a table of data showing the average number of hours of sleep for males and females in different age groups, and to identify the range, mode, and some other characteristics of the data set.
Assessment: Level B (above proficiency): Students are asked to examine a scatter plot of data showing records of the speed of cars and the time of day they were recorded along a certain stretch of road, to identify clusters of data, and to decide when it would be best to set up a speed trap.
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Sigma = Canon?
Senifer
New Member
Personally, not a sigma user, haven't use any sigma lens YET. so please don' flame me, just wanna share this funny picture i got in google since this is kopitiam.
After googling, i went into wiki, and saw this
http://en.wikipedia.org/wiki/Sigma
the greek alphabet spells EOS. :thumbsup:
So is Sigma = Canon?
wildcat
Senior Member
LOL. Good spot! :thumbsup:
ZOMG!
hongsien
Senior Member
Personally, not a sigma user, haven't use any sigma lens YET. so please don' flame me, just wanna share this funny picture i got in google since this is kopitiam.
After googling, i went into wiki, and saw this
http://en.wikipedia.org/wiki/Sigma
the greek alphabet spells EOS. :thumbsup:
So is Sigma = Canon?
No lah!
Those are the 3 ways of using the letter 'sigma':
1st. Capital sigma
then the second character is used when in the middle of a word, and the third at the end of a word.........
HS
Senifer
New Member
hahaha really? omg. but it still amazes me. EOS. Lol.
Hope you guys enjoy a little destress from the stressful working life that most of us have.
allenleonhart
Deregistered
hahaha really? omg. but it still amazes me. EOS. Lol.
Hope you guys enjoy a little destress from the stressful working life that most of us have.
yep. used all 3 before
the first one, used in math. called it sigma notation
second one 3rd one i think i used for physics somewhere before:think:
hongsien
Senior Member
Personally, not a sigma user, haven't use any sigma lens YET. so please don' flame me, just wanna share this funny picture i got in google since this is kopitiam.
After googling, i went into wiki, and saw this
http://en.wikipedia.org/wiki/Sigma
the greek alphabet spells EOS. :thumbsup:
So is Sigma = Canon?
If you look into the history of Canon cameras, the lens made for their first camera was from Nikon........
Senifer
New Member
If you look into the history of Canon cameras, the lens made for their first camera was from Nikon........
:thumbsup::thumbsup::thumbsup::thumbsup:
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## Notes of Tim Austin’s course
Embedding groups into Banach spaces
1. Context
1.1. Groups as metric spaces
Regard finitely generated groups ${\Gamma}$ as geometric objects
• either by putting some geometry on ${\Gamma}$ itself (e.g. word metrics);
Word metric: form the Cayley graph ${Cay(\Gamma,S)}$ and restrict its length metric to ${\Gamma}$. Example: For ${\Gamma={\mathbb Z}^d}$ with obvious generating set ${S=\{\pm e_i\}}$, get ${\ell_1}$ metric.
Two finite generating sets give equivalent word metrics. Thus ${\Gamma}$ is a metric space only up to bi-Lipschitz equivalence.
Questions:
1. How does this metric space relate to the algebra of ${\Gamma}$.
2. How understand these as metric spaces. For instance, try to compare groups to Banach spaces.
More specifically, how close can word metric on ${\Gamma}$ be from metrics induced by maps to Banach spaces ?
Example 1 Every separable metric space embeds isometrically into ${L_{\infty}}$.
So we should restrict to subclasses of Banach spaces. Here are popular choices of targets.
• Hilbert spaces (hardest of all).
• ${L_p}$-spaces, ${p<\infty}$.
• All uniformly convex spaces.
1.2. Uniform embeddings
Definition 1 A uniform embedding of ${\Gamma}$ to a Banach space ${X}$ is a map ${f:\Gamma\rightarrow X}$ such that there exist two positive functions ${\xi_{\pm}}$ on ${{\mathbb R}_+}$ such that
• ${\xi_{\pm}(r)}$ tends to infinity with ${r}$.
• For all ${g}$, ${h\in \Gamma}$, ${\xi_{-}(d(g,h))\leq |f(g)-f(h)|\leq\xi_{+}(d(g,h))}$.
This is a very weak notion, but
Theorem 2 (Gromov) There exist groups that do not embed uniformly into ${L_2}$.
Theorem 3 (Yu) Groups that embed uniformly into ${L_2}$ satisfy Baum-Connes conjecture.
1.3. Bi-Lipschitz embeddings
Definition 4 A bi-Lipschitz embedding is such that we can take ${\xi_{\pm}(r)=C_{\pm}r}$.
Example 2 ${{\mathbb Z}^d}$, virtually ${{\mathbb Z}^d}$.
Are there any other ? Polynomial growth groups don’t. Exponential growth groups don’t (use Markov convexity). Open for intermediate growth groups.
1.4. Compression
To quantify where we lie in between, Guentner and Kaminker introduced compression.
Definition 5 The compression exponent ${\alpha_X^* (\Gamma)}$ is the supremum ofnthose ${\alpha\geq 0}$ such that there exists ${f:\Gamma\rightarrow X}$ with ${\xi_+ (r)=C_+ r}$ and ${\xi_- (r)\geq c_- r^{\alpha}}$.
Abbreviate ${\alpha_{L_p}^*}$ by ${\alpha_{p}^*}$.
Theorem 6 (Guentner, Kaminker) If ${\alpha_{L_2}^* (\Gamma)>\frac{1}{2}}$, then ${\Gamma}$ is exact, meaning that, among ${C^*}$-algebras, minimal tensor product with ${C_r^* (\Gamma)}$ preserves exact sequences.
Example 3
• Virtually abelian groups have ${\alpha_{p}^* =1}$.
• Virtually nilpotent groups have ${\alpha_{p}^* =1}$.
• Hyperbolic groups have ${\alpha_{p}^* =1}$ (Brodskiy-Soukin).
• ${{\mathbb Z}\wr{\mathbb Z}}$ has ${\alpha_{2}^* =\frac{2}{3}}$.
• Among all finitely generated groups, one can have any values for ${\alpha_{p}^*}$ in ${[0,1]}$ (Arzhantseva, Drutu, Sapir).
1.5. Random walk bound
This is a general method for getting upper bounds.
Theorem 7 (Austin, Naor, Peres) Suppose ${(X_t)}$ is a simple random walk on an amenable group, and that
$\displaystyle \begin{array}{rcl} \mathop{\mathbb E}(d(X_0 ,X_t))\geq \Omega(t^{\beta}). \end{array}$
Then ${\alpha_2^* (\Gamma)\leq\frac{1}{2\beta}}$.
(Naor, Peres): ${\alpha_p^* (\Gamma)\leq\frac{1}{p^{*}\beta}}$, where ${p^* =2}$ if ${p\geq 2}$, ${p^* =p}$ if ${1\leq p \leq 2}$.
This cannot give an upper bound less than ${\frac{1}{2}}$ in Hilbert space.
2. Zero compression
Are there amenable groups that do worst ? Answer is yes.
Theorem 8 (Austin 2008) There exists a finitely generated amenable group (actually ${4}$-step solvable) such that ${\alpha_p^* (\Gamma)=0}$ for all finite ${p}$.
The rest of the course will be devoted to proving this theorem. The groups are kind of ad hoc.
2.1. Strategy
The method is to embed a sequence of bad finite subsets.
Definition 9 Given a finite metric space ${Y}$ and an map ${Y\rightarrow X}$, its distorsion is
$\displaystyle \begin{array}{rcl} dist(f)=Lip(f)Lip(f^{-1}). \end{array}$
The distorsion of ${Y}$ into ${X}$ is
$\displaystyle \begin{array}{rcl} C_X (Y)=\inf_{f:Y\rightarrow X}dist(f). \end{array}$
There are finite spaces which have large ${L_p}$-distorsion. Having them bi-Lipschitz embedded in ${\Gamma}$ gives upper bounds on compression.
Lemma 10 Suppose there exist finite metric spaces ${Y_n}$ and maps ${\phi_n :Y_n \rightarrow\Gamma}$ and constants ${L>1}$, ${r_n}$ bounded away from ${0}$ such that
• ${diameter(Y_,)}$ tends to infinity.
• ${\frac{1}{L}r_n d(y,y')\leq d(\phi_n (y),\phi_n (y'))\leq Lr_n d(y,y')}$ for all ${y}$, ${y'\in Y_n}$.
• ${r_n \leq \Omega(diameter(Y_n)^{\epsilon})}$ for all ${\epsilon>0}$.
• ${c_{\Gamma}(Y_n)\geq\Omega(diameter(Y_n)^{\eta})}$ for some fixed ${\eta}$.
Then ${\alpha_p^* (\Gamma)\leq 1-\eta}$.
We loosely call ${r_n}$ “expansion ratios”.
Proof: Let ${f:\Gamma\rightarrow L_p}$ be ${1}$-Lipschitz and have ${\xi_- (t)\geq C\,t^\alpha}$. Then${f\circ\phi_n}$ has distorsion ${\leq }$ and ${\geq \Omega(diameter(Y_n)^{\eta})}$. This implies ${\alpha\leq 1-\eta}$. $\Box$
2.2. Suitable finite metric spaces
One possibility (used by Gromov) would be expanders. We will use something similar, but with unbounded degree.
Consider hypercube ${{\mathbb Z}_2^m}$ with Hamming metric. Fact: ${c_2 ({\mathbb Z}_2^m)\sim\sqrt{m}}$. It turns out that quotients of the hypercube are even worse.
Theorem 11 (Khot, Naor 2004) Consider a vector subspace ${C\in{\mathbb Z}_2^m}$ of dimension ${\Omega(m)}$ and distance ${\Omega(m)}$ (i.e. an error-correcting code). Then, in the quotient metric,
$\displaystyle \begin{array}{rcl} c_p ({\mathbb Z}_2^m/C^{\bot})\geq/Omega_p (m), \quad \forall p<\infty. \end{array}$
Proof: First take ${p=1}$, use isometric embedding of ${(L_1 ,\sqrt{|\cdot|_{L_1}})}$ to ${L_2}$, then use Fourier analysis. To finish, use Matousek’s interpolation lemma. $\Box$
Reason for optimism: These are groups.
Reason for concern: Number of generators tend to infinity.
Resolve second issue by sending these groups into non finitely generated subgroups of a finitely generated group, e.g. Lamplighter group ${{\mathbb Z}_2 \wr H}$.
2.3. Lamplighter group
${{\mathbb Z}_2 \wr H}$ is the semi-direct product of ${H}$ and the ${H}$-indexed direct sum of copies of ${{\mathbb Z}_2}$, where ${H}$ acts by permuting coordinates. Elements of ${{\mathbb Z}_2 \wr H}$ are configurations of a lamps assignment (only finitely many lamps are on) ${\omega\in{\mathbb Z}_2^{\oplus H}}$ and a lamplighter location ${h\in H}$.
Fix a generating set ${S_H}$ of ${H}$. A generating set for ${{\mathbb Z}_2 \wr H}$ is the set of instructions to the lamplighter: either switch the lamp at the current location on or off, or move one step away. In other words,
$\displaystyle \begin{array}{rcl} S=\{(0,s)\,\;\,s\in S_H\}\cup\{(\pm\delta_{e_H},e_H)\}. \end{array}$
Observe that ${{\mathbb Z}_2^{\oplus H}}$ is an infinite generated subgroup.
2.4. The traveling salesman metric
The word metric on ${{\mathbb Z}_2 \wr H}$ behave as follows. To go from ${(\omega,h)}$ to ${(\omega',h')}$, one needs visit all locations in the support of ${\omega\omega'}$, so
$\displaystyle \begin{array}{rcl} d((\omega,h),(\omega',h'))\geq |\mathrm{support}(\omega)\Delta\mathrm{support}(\omega')|+L, \end{array}$
where ${L}$ is the minimal length of a path in ${Cay(H,S_H)}$ joining ${h}$ to ${h'}$ and visiting all sites ${\mathrm{support}(\omega)\Delta\mathrm{support}(\omega')}$.
Definition 12 The traveling salesman metric on ${{\mathbb Z}_2^{\oplus H}}$, ${TS_H (\omega,\omega')}$, is the minimal length of a loop in ${Cay(H,S_H)}$ based at ${e_H}$ and visiting all sites ${\mathrm{support}(\omega)\Delta\mathrm{support}(\omega')}$.
In particular,
$\displaystyle \begin{array}{rcl} d((\omega,e_H),(\omega',e_H))\sim TS_H (\omega,\omega') . \end{array}$
2.5. First step in the construction
Lemma 13 If ${H}$ has exponential growth, there exists ${M>1}$ and a sequence of radii ${r_j}$ tending to infinity such that
$\displaystyle \begin{array}{rcl} |B(e_H ,3r_j)|\geq M^{r_j}|B(e_H ,r_j)|. \end{array}$
For this ${M}$ and these ${r_j}$, there are finite subsets ${I_j \in H}$ such that
• Pairwise distances within ${I_j \cup\{e_H\}}$ belong to ${[r_j ,10r_j]}$.
• ${|I_j|:=d_j \geq M^{r_j}}$.
Now identify ${{\mathbb Z}_2^{d_j}}$ with ${{\mathbb Z}_2^{I_j}\subset {\mathbb Z}_2^{\oplus H}}$. The diameter is ${d_j}$, exponentially larger than ${r_j}$, and the traveling salesman distance is of the order of ${r_j |\mathrm{support}(\omega)\Delta\mathrm{support}(\omega')|}$, therefore the distorsion is ${\geq\Omega(d_{j}^{1/2}}$. This shows that
$\displaystyle \begin{array}{rcl} \alpha_2^* ({\mathbb Z}_2 \wr H)\leq\frac{1}{2}. \end{array}$
2.6. Passing to a quotient
To improve this bound, we replace ${{\mathbb Z}_2^{d_j}}$ with a quotient. This requires quotienting ${{\mathbb Z}_2^{\oplus H}}$ by a vectorsubspace ${V}$ that contains some dual code ${C_j^{\bot}}$. For the semi-direct product to be defined, we need ${V}$ to be ${H}$-invariant.
Proposition 14 There exists an amenable, exponential growth group ${H}$ and a ${H}$-invariant subspace ${V\subset {\mathbb Z}_2^{\oplus H}}$ such that for some sequence ${d_n}$ tending to infinity, ${\phi_n :{\mathbb Z}_2^{d_n}/C_n^{\bot}\rightarrow{\mathbb Z}_2^{\oplus H}/V \times H}$ is bi-Lipschitz, with expansion ratio ${r_n \leq O(\log d_n)}$.
The difficulty is making sure that the quotient map is still bi-Lipschitz.
What can go wrong ?
• Set ${V_n =}$ sum of translates of ${C_n^{\bot}}$. It will be much bigger than ${C_n^{\bot}}$ alone.
• Set ${V=\bigoplus_n V_n}$. It will be much bigger than ${V_n}$ alone.
We need to ensure that given ${u}$, ${v\in{\mathbb Z}_2^{d_n}}$ and ${w_{h,m}}$ in ${h}$-translate of ${C_m^{\bot}}$, there is a single ${w'_n \in C_n^{\bot}}$ such that
$\displaystyle \begin{array}{rcl} d_{{\mathbb Z}_2 \wr H}(u,v+\sum_{h,m}w_{h,m})\geq \mathrm{const.}d(u,v+w'_{n}). \end{array}$
2.7. Finding suitable group ${H}$ and subspace ${V}$
To give ourselves room to move, let ${H={\mathbb Z}_k \wr G}$, some ${k\geq 1}$, where ${G}$ is itself amenable with exponential growth.
View ${{\mathbb Z}_k \wr G}$ as a disjoint union of ${{\mathbb Z}_k^{\oplus G}}$. Then as a vectorspace,
$\displaystyle \begin{array}{rcl} {\mathbb Z}_2^{{\mathbb Z}_k \wr G}=\bigoplus_{g\in G}{\mathbb Z}_2^{\oplus {\mathbb Z}_k^{\oplus G}}. \end{array}$
We will choose maps ${\phi_n :{\mathbb Z}_2^{d_n}\rightarrow {\mathbb Z}_2^{{\mathbb Z}_k \wr G}}$ to actually land in ${{\mathbb Z}_2^{\oplus {\mathbb Z}_k^{\oplus G}}}$ viewed as the ${e_G}$ component of ${\bigoplus_{g}{\mathbb Z}_2^{\oplus {\mathbb Z}_k^{\oplus G}}}$. Then ${V}$ will itself be of the form
$\displaystyle \begin{array}{rcl} \bigoplus_{g}U_g \subset \bigoplus_{g}{\mathbb Z}_2^{\oplus {\mathbb Z}_k^{\oplus G}}, \end{array}$
where ${U_g}$‘s are all related by the ${G}$ action.
To choose ${\phi_n}$, observe that ${{\mathbb Z}_2^{\oplus {\mathbb Z}_k^{\oplus G}}}$ does not see the ${G}$-action, we need only take care of the ${{\mathbb Z}_k^{\oplus G}}$-action.
2.8. New notational convention
From now on, elements of either ${{\mathbb Z}_2^{d_n}}$ or, e.g., ${{\mathbb Z}_k^{\oplus G}}$ will be called tuples and written ${\mathbf{u}=(u_i)_{i=1,\ldots,d_n}}$.
Elements of ${{\mathbb Z}_2^{\oplus {\mathbb Z}_k^{\oplus G}}}$ will be called functions (on tuples) or finite subsets of tuples, and written ${W:{\mathbb Z}_k^{\oplus G}\rightarrow{\mathbb Z}_2}$.
We will obtain ${\phi_n}$ in two steps,
1. Find ${\xi_n :{\mathbb Z}_2^{d_n}\rightarrow{\mathbb Z}_2^{\oplus {\mathbb Z}_k^{I_n}}}$ where ${I_n \subset G}$ is a well separated set with inter-point distances ${r_n}$ as given by Lemma 13.
2. Choose homomorphism ${\kappa_n :{\mathbb Z}_2^{\oplus {\mathbb Z}_k^{I_n}} \rightarrow {\mathbb Z}_2^{\oplus {\mathbb Z}_k^{\oplus G}}}$.
Then ${\phi_n =\kappa_n \circ \xi_n}$.
This actually breaks the difficulty.
2.9. Construction of ${\xi_n}$
We want ${\xi_n :{\mathbb Z}_2^{d_n}\rightarrow {\mathbb Z}_2^{\oplus {\mathbb Z}_k^{I_n}}}$. Let
$\displaystyle \begin{array}{rcl} B_n =\{(\delta_y ,e_G)\,;\, y\in I_n\}\in {\mathbb Z}_k^{\oplus G}\times\{e_g\} \subset {\mathbb Z}_k \wr G. \end{array}$
Observe that ${d_{{\mathbb Z}_k \wr G}((\delta_y ,e_G),(\delta_{y'} ,e_G))}$ is of the order of ${r_n}$ for distinct ${y}$, ${y'\in I_n}$. So ${B_n}$ is also a set of ${d_n}$ points, all roughly ${r_n}$ apart (we have jumped one level up). Now pick an arbitrary isomorphism ${\xi_n : {\mathbb Z}_2^{d_n}\rightarrow {\mathbb Z}_2^{B_n}\subset {\mathbb Z}_2^{\oplus {\mathbb Z}_k^{I_n}}}$. It will be important to know that this already helps with translation invariance. I.e. see what translation invariance remains at the level of the finite set ${{\mathbb Z}_2^{{\mathbb Z}_k^{I_n}}}$.
Lemma 15 Let ${k}$ be even. There is a ${{\mathbb Z}_k^{I_n}}$-invariant subgroup ${E_n \subset {\mathbb Z}_2^{\oplus {\mathbb Z}_k^{I_n}}}$ such that
• ${\xi_n (C_n^{\bot})\subset E_n}$.
• ${E_n \cap \xi_n ({\mathbb Z}_2^{d_n})=\xi_n (C_n^{\bot}}$.
• In fact, the quotient map ${\bar{\xi}_n :{\mathbb Z}_2^{d_n}/C_n^{\bot}\rightarrow{\mathbb Z}_2^{{\mathbb Z}_k^{I_n}}/E_n}$ is bi-Lipschitz from quotient Hamming metric to quotient of traveling salesman distance.
Proof: In fact, we choose translation invariant ${D_n \subset {\mathbb Z}_2^{{\mathbb Z}_k^{I_n}}}$ and let ${E_n =D_n^{\bot}}$. To do this, given ${\mathbf{a}\in{\mathbb Z}_2^{d_n}}$, define ${L_{\mathbf{a}}\in{\mathbb Z}_2^{{\mathbb Z}_k^{I_n}}}$ as follows,
$\displaystyle \begin{array}{rcl} L_{\mathbf{a}}(\mathbf{v})=\sum_{} \end{array}$
Now
$\displaystyle \begin{array}{rcl} D_n =\{\eta +L_{\mathbf{a}}\,;\,\eta\in{\mathbb Z}_2 ,\,\mathbf{a}\in C_n\}. \end{array}$
Adding ${\eta}$ is necessary for translation invariance, since translating ${L_{\mathbf{a}}}$ by ${\mathbf{w}}$ adds up ${\langle \mathbf{a},\mathbf{w} \rangle}$.
One checks that if ${U=\xi_n (\mathbf{a})}$ belongs to ${D_n^{\bot}}$, then ${U\in\xi_n (C_n^{\bot})}$. $\Box$
2.10. Construction of ${\kappa_n}$
First attempt: Assume ${I_1 ,\ldots,I_n}$ are disjoint. Given ${W\in {\mathbb Z}_2^{{\mathbb Z}_k^{I_n}}}$, let
$\displaystyle \begin{array}{rcl} \kappa_n (W)(\mathbf{w})=\delta_{\mathbf{0}}(\mathbf{w}_{|G\setminus (I_1 \cup\cdots\cup I_n)})W(\mathbf{w}_{|I_n}. \end{array}$
For this choice, one cannot rule out conflicts between translates. One needs an extra trick to prevent translates from ${\kappa_n (D_{n'}^{\bot})}$ ${n'\leq n}$, filling up something that is constant on cylinders like ${\{\mathbf{w}_{|G\setminus (I_1 \cup\cdots\cup I_n)}\}\times{\mathbb Z}_k^{I_1 \cup\cdots\cup I_n}}$.
To repair this, we choose a sequence ${z_i \in G}$ such that ${r_1 \ll d(e,z_1)\ll r_2 \ll d(e,z_2)\ll \cdots}$ and find some function ${Z:{\mathbb Z}_k \rightarrow{\mathbb Z}_2}$ such that the constant function ${1}$ is not a sum of translates of ${A}$. E.g. ${A=1_{\{0,1,3,4\}}}$ if ${k=6}$. Set
$\displaystyle \begin{array}{rcl} \kappa_n (W)(\mathbf{w})=\delta_{\mathbf{0}}(\mathbf{w}_{|G\setminus (I_1 \cup\cdots\cup I_n)})W(\mathbf{w}_{|I_n})A(w_{z_n}). \end{array}$
This now prevents cancellation.
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# Computer Science And IT (CS/IT) Mock Test - 8 For Gate
## 65 Questions MCQ Test | Computer Science And IT (CS/IT) Mock Test - 8 For Gate
Description
This mock test of Computer Science And IT (CS/IT) Mock Test - 8 For Gate for GATE helps you for every GATE entrance exam. This contains 65 Multiple Choice Questions for GATE Computer Science And IT (CS/IT) Mock Test - 8 For Gate (mcq) to study with solutions a complete question bank. The solved questions answers in this Computer Science And IT (CS/IT) Mock Test - 8 For Gate quiz give you a good mix of easy questions and tough questions. GATE students definitely take this Computer Science And IT (CS/IT) Mock Test - 8 For Gate exercise for a better result in the exam. You can find other Computer Science And IT (CS/IT) Mock Test - 8 For Gate extra questions, long questions & short questions for GATE on EduRev as well by searching above.
QUESTION: 1
### Suppose there are two teams A and B. Both of them are competing with each other on a racing track of 1 km. But team A has got the advantage of starting the race from 280 m from the starting point. If the ratio of speed of A to B is 3:4. Select the correct option from the options given below:
Solution:
To win the race first,
A has to cover (1000 – 280) m = 720m
B covers 4 part of distance when A covers 3 part of it.
So, if A covers 720m, then B covers (4/3 * 720)m i.e. 960m
Hence, A will win the race by 40m.
QUESTION: 2
### There are 5 brothers in a family. All were born at a gap of 3 years. If the sum total of ages of 5 brothers is 100. What is the age of 2nd most elder brother ?
Solution:
Let the age of youngest brother be x
Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 100
=> 5x = 70 => x = 14
Therefore, age of 2nd most elder brother = 14 + 9 = 23.
QUESTION: 3
### If there are 5 persons who can work individually and complete a work in 2, 4, 5, 6 and 8 hours respectively. If they work together then how much time will it take to finish the work ?
Solution:
In 1 hour, ½ + ¼ + ⅕ +⅙ + ⅛ = 149/120 = 1.24 is the work done.
Hence, total time to complete the work would be = 1/1.24 = 0.80 hours
QUESTION: 4
If we increase the sides of a rectangular park by 20% , then, what is the total increase in the area of that park?
Solution:
Let length of park = L and breadth of park = B
Original Area = BL m2
New Length = 120L/100 = 6L/5
New Breadth = 120B/100 = 6B/5
New Area = 36LB/25 m2
so, Change in area = New – Original = 11LB/25 m2 % increase = Change / original * 100 = 44%
QUESTION: 5
The sum of the digits in the unit place of all the 4 digit numbers formed with the help of 3, 4, 5 and 6 taken all at a time is:
Solution:
Required sum = 3!(3 + 4 + 5 + 6) = 108
[If we fix 3 of the unit place, other three digits can be arranged in ways similarly for 4, 5, 6.]
QUESTION: 6
Choose the one that best fits the sentence below.
The state’s duty is to _______ the safety of its citizen
Solution:
QUESTION: 7
Cocaine will have pernicious effect on your physical fitness.
Find synonym of the underlined word.
Solution:
QUESTION: 8
Choose the one that best fits the sentence below.
The _______ of glory lead but to the ______
Solution:
QUESTION: 9
Gardener said to me, “Don’t pluck the beautiful flowers.”
Select the appropriate Indirect Speech of the above.
Solution:
QUESTION: 10
I haven’t had a call from the office about the meeting ________ last Monday.
Select the appropriate word to fill in the blank from the given options
Solution:
QUESTION: 11
Let T(n) = 2T(n/4) + 100√n
The value of T(n) can be written as:
Solution:
QUESTION: 12
Consider a sorted array of n elements. Let k inversions (swapping) have been performed on the sorted array and we want to sort it again to reverse effect of inversions. If K is very small in comparison to n, then which sorting technique will prove efficient for making again the list in sorted increasing order array.
Solution:
Insertion sort is directly proportional to number of inversions present in the list.
QUESTION: 13
What will the output of the following C code ?
#include <stdio.h>
int main()
{
int i=2, j=2;
while (i+1 ? --i : j++)
printf("%d",i);
return 0;
}
Solution:
Consider the while loop condition => i + 1 ? --i : j++
In first iteration : i + 1 = 3 (true)
So ternary operator will return --i i.e. 1, condition part is 1 means true
so while condition is true.
Hence printf statement will print 1 In second iteration : i + 1 =2 (true)
So ternary operator will return --i i.e 0, condition part is 0 means false
so while condition is false.
Hence program control will come out of the while loop.
QUESTION: 14
A balanced tree is given below. How many nodes will become unbalanced when a node is inserted as a child of node G?
Note: A node in a tree is balanced if absolute difference between its left and right subtrees is less than or equal to 1.
Solution:
QUESTION: 15
Consider the C code Below.
void function(int n)
{
if (n == 1)
return;
for (int i = 0; i<n; i++)
{
for (int j = 1; j< = n; j++)
{
printf("*");
break;
}
}
}
Which of the following is the tightest upper bound on time complexity of above function.
Solution:
Important observation is Break statement terminates the innermost loop. So “*” is printed only n times.
QUESTION: 16
Consider the following statements about Bellman ford algorithm for finding shortest path in a directed connected graph G having integral edge weights. Statement I: It will always find out negative edge weight cycle in G reachable from source. Statement II: It will always give correct answer for the graph G. Choose from the options given below.
Solution:
Statement 1 is true as in the nth iteration of Bellman Ford, if the length of the path of any node reachable from the source is decreased, that means we are having negative edge weight cycle in the graph. Statement 2 is false as no algorithm can give give correct answer if the Graph is having negative edge weight cycle.
QUESTION: 17
Consider the case: f(n) = O(g(n)). Then, following two statements are claimed to be inferred from the above case.
Statement I: 2f(n) = O(2g(n))
Statement II: 2g(n) = O(2f(n))
Choose the correct option from the given.
Solution:
if f(n) = n and g(n) = 2n. then f(n) = O(g(n))
here, 2^n = O(2^(2n)) = O(4^n), but not vice versa.
Hence, I is true. II is false. --------------
Now, if f(n) = 2n and g(n) = n then also f(n) = O(g(n)) because we can ignore constant but, 2^(2n) != O(2^n),
hence I is false, but II is true.
In both of the above cases, f(n) = O(g(n)).
But both the cases are counter of each other. Hence both I and II are wrong.
QUESTION: 18
Let X and Y be the integers representing the number of simple graphs possible with 3 labeled vertices and 3 unlabeled vertices respectively. Let X - Y = N. Then, find the number of spanning trees possible with N labeled vertices complete graph.
Solution:
Number of simple graphs possible with n labeled vertices is 2^(n(n-1)/2).
Number of simple graphs possible with n unlabeled vertices is n+1.
Number of spanning tree possible with n vertices complete graph n^(n-2) X =8 Y = 4 X-Y=4
QUESTION: 19
Consider the following pseudo code.
x = 0;
for (i = 1 to n)
for (j = 1 to 4i – 3)
x = x + 1
Find the value of x, when the above code is executed in a function.
Solution:
QUESTION: 20
Consider the following statements:
S1 : DFS of a directed graph always produces the same number of edges in the traversal, irrespective of the starting vertex.
S2 : If all of the back edges that are found while DFS traversal on directed graph are removed, the resulting graph is acyclic.
Which of the following statements above are valid ?
Solution:
Statement S1 : consider the graph
A (source vertex ) we will get 2 edges Starting with B will get only 1 edge
Starting with C we will get no edge
Therefore DFS on directed graph may not give same number of edges.
Statement S2 : Back edges are those edges (u,v) connecting a vertex u to an ancestor u in a depth-first tree. Self-loops are considered to be back edges. Back edges describe descendant-to-ancestor relations, as they lead from “high” to “low” nodes.
Suppose that there is a back edge (u, v). Then vertex v is an ancestor of vertex u in the depth-first forest.
There is thus a path from v to u in G, and the back edge (u,v) completes a cycle.
Removing the back edge will break the cycle. Therefore removing all the back edges will make the graph acyclic.
So the statement is true.
QUESTION: 21
Which data structure would be the most appropriate to implement a collection of values with the following three characteristics?
i) Items are retrieved and removed from the collection in FIFO order.
ii) There is no priori limit to the number of items in the collection.
iii) The size of an item is large relative to storage required for a memory address.
Solution:
Head and tail pointers in singly link list will make the insertion and deletion in O(1) time complexity if we are accessing the elements in FIFO order. In doubly link list since only head pointer is given then for insertion we have to traverse the complete link list so insertion will be O(n) so not appropriate. In binary tree we have only a pointer to the root. Insertion and deletion in binary tree will be O(log n) so not appropriate. In hash table accessing the data in FIFO order will not be possible.
QUESTION: 22
Stack A has the entries as following sequence a, b, c (with ‘a’ on top), stack B is empty, as shown in the diagram below.
An entry popped
out of stack A can be printed or pushed to stack B. An entry popped out of stack B can only be printed. In this arrangement which of the following permutation of a, b, c are not possible to print?
(i) bac
(ii) bca
(iii) cab
(iv) abc
Solution:
Follow these steps to print bac. 1) POP element ‘a’ from stack A, push ‘a’ to Stack B. 2) POP element ‘b’ from A, print it. 3) POP element ‘a’ from B, and print it. 4) POP element ‘c’ from A, and print it. Now, perumtation bac has been printed. Similarly, we can print bca and abc, but can’t print cab.
QUESTION: 23
Considering the data given in the previous question, if the stack A had 4 entries, then the number of possible permutations that can be printed will be:
Solution:
QUESTION: 24
Let f(n) = Σ [(log(n/2i) +100] where i limits from 0 to k, and n = 2k.
Find the time complexity of f(n).
Solution:
QUESTION: 25
Given a hash table with n keys and m slots with simple uniform hashing. If collisions are resolved by chaining then what is the probability that the first slot ends up empty ?
Solution:
Probability of one particular slot = 1/m ( because total m slots)
Probability that a value should not go in one particular slot = 1 – (1/m)
For n values (keys) probability = [1 – (1/m)]n
QUESTION: 26
Consider the following instance of knapsack problem:
The maximum weight of 12 is allowed in the knapsack. Find the value of maximum profit with the optimal solution of the fractional knapsack problem.
Solution:
Decreasing order of P i /W i is X1, X4, X3, X5, X2 X1 → profit = 15 and weight = 2 Including X4 → profit = 15 + 16 and weight = 2 + 4 = 6 Including X3 → profit = 40 and weight = 9 Now weight left = 3 Weight of X5 = 6 → half of X5 can be included. Profit = 40 + 17/2 = 48.5.
QUESTION: 27
Find the maximum value of the expression (x+y+k) where (x,y) satisfies the equation (x-2)2 + (y-3)2 = 25
Solution:
Since (X,Y) is a point on circle, the general form of the point is X = 2 + 5*cost, y = 3 + 5*sint
We need to maximise the value of x+y+k x+y+k = 2 + 5*cost + 3 + 5*sint + k = (5+k) + 5*(cost+sint)
Here, k is a constant. The maximum value of c + acost + bsint is equals to c + sqrt (a*a +b*b).
Maximum value of (5+k) + 5*(cost+sint) (5+k) + 5*sqrt(2)
The result is (5+k) + 5*sqrt(2).
QUESTION: 28
Which of the following argument is invalid ?
Solution:
“⟺” mean equal. ifA⟺B means A is equal to B.
Option (a); (p→(q⋁r) )⟺(~p⋁(q⋁r) ) ⟺((~p⋁q)⋁r) ⟺(~(p⋀~q)⋁r) [∵De morgan’s law] ⟺(p⋀~q)→r ∴(p→(q⋁r) )⟺((p⋀~q)→r)
Option (b); (p→r)⋀(q→r)⟺(~p⋁r)⋀(~q⋁r) ⟺(~p⋀~q)⋁r [∵ De morgan’s law] ⟺~(p⋁q)⋁r ⟺(p⋁q)→r ∴(p→r)⋀(q→r)⟺(p⋁q)→r
Option (c): p→(q→r)⟺~p⋁(q→r) ⟺ ~p⋁(~q⋁r) ⟺~q⋁(~p⋁r) ⟺~q⋁(p→r) ⟺q→(p→r)
Option (d): (p↔q)⟺(p→q)⋀(q→p) ⟺(~p⋁q)⋀(~q⋁p) ⟺(~p⋁q)⋀(p⋁~q) ⟺(~p⋀~q)⋁(p⋀q) ⟺ ~(p⋁q)⋁(p⋀q) (p↔q)⟺(p⋁q)⋁(p⋀q) is wrong.
QUESTION: 29
The value of the following Integral is:
Solution:
QUESTION: 30
consider a Matrix . The number of linearly independhent Eigen Vector for the Eigen value 1 is ______________.
Solution:
QUESTION: 31
The language L = {WbcWR | W ∈ (a+b)*} is _____.
Solution:
Any language for which we can have a Deterministic PDA is always a DCFL.
Here for language L= {WbcWR | W ∈ (a+b)*} we can have a PDA with following transitions in which PDA accepts a string when it halts on a final state.
q0 in initial state and qf is final state. 1. (q0, a , Z) -> (q0, aZ) 2. (q0, b , Z) -> (q0, bZ) 3. (q0, a , a) ->(q0, aa) 4. (q0, b , a) -> (q0, ba) 5. (q0, a, b) -> (q0, ab) 6. (q0, b , b) -> (q0, bb) 7. (q0, c , b) -> (q1, null) 8. (q1, a , a) ->(q1, null) 9. (q1, b , b) -> (q1, null) 10. (q1, null, Z ) -> (qf, Z)
Here all a’s and b’s are initially pushed onto the stack for W. As soon as a c occurs after b , B is popped from the stack after which W^R is checked. If the further string alphabets match the alphabet of the stack the alphabet is popped from the stack and finally the string reaches the final state if language is of the form L= {WbcW^R | W ∈ (a+b)*}. Hence the above language is a DCFL.
QUESTION: 32
Given a graph G (V, E) is Bipartite, what is the chromatic number of G ?
Solution:
Since the graph G is bipartite, the vertices set V can be partitioned into two disjoint sets. This shows that we can color the graph with 2 colors such that no two adjacent vertices will have same color.
QUESTION: 33
Solution:
QUESTION: 34
Solution:
QUESTION: 35
Print the maximum of the two Eigen values.
Solution:
QUESTION: 36
If P, Q are the roots of the quadratic equation x2 + ax + b. Find the value of P3 +Q3.
Solution:
For any quadratic equation which is off the form x2+ax+b=0 1. Sum of the roots P+Q = -a2.
Product of the roots PQ = b => P3 + Q3 = (P+Q)(P2+Q2-PQ) => (P+Q)((P+Q)2-2PQ-PQ) => (P+Q)((P+Q)2-3PQ) => (-a)(a2-3b) => 3ab - a3
QUESTION: 37
Given F is an expression of P,Q. Derive the expression F(P,Q) from the truth table given below.
Solution:
From the truth table, the value of F can be written as: => F = PQ + PQ' + P'Q' => P(Q+Q')+P'Q' => P+P'Q' => (P+Q')(P+P') => (P+Q')T => (P+Q') Answer :- Option (B)
QUESTION: 38
In a random experiment, suppose 3 unbiased coins are tossed simultaneously, what is the probability of getting at least 2 Heads ?
Solution:
QUESTION: 39
Consider a DFA accepting all strings over {a, b} where the number of a’s mod 3 = 2 and number of b’s are odd. What is the minimum number of states of such DFA ?
Solution:
The above question is an example of product automata. If we have two cases for which we can have separate DFA’s we can merge the two by product automata.
The resulting DFA has number of states equal to the product of the states of the separate DFA’s.
Here DFA for accepting all strings over {a, b} where the number of a’s mod 3 = 2 would have 3 states.
( number of a’s mod 3 would give remainder either 0, 1, 2 so 3 states to depict each). Similarly DFA accepting all strings over {a, b} where the number of b’s are odd ( number of b’s mod 2 = 0) would have 2 states.
Hence resulting DFA for both the conditions would have 2*3 = 6 states.
QUESTION: 40
Which of the following languages are regular over ∑ = {a, b} ?
L1 = { anbn | 0 < n ≤ 10}
L2 = { a2n | n ≥ 1}
L3 = { an! | n > 0}
Solution:
Any language for which we have a finite set or we can possibly have a DFA drawn for it is regular. L1: It gives a finite set generating string with equal number of a’s and equal number of b’s for range 0
Hence L2 is also regular. L3: It gives all possible strings with a’s such that number of a’s are in factorial values of 1,2,3,4 ……. i.e. {a1, a2, a6, a24……………}. The set is neither finite nor a DFA is possible for such set. Hence L3 is not regular.
QUESTION: 41
The language L = {WbcWR | W ∈ (a+b)*} is _____.
Solution:
Any language for which we can have a Deterministic PDA is always a DCFL. Here for language L= {WbcWR | W ∈ (a+b)*} we can have a PDA with following transitions in which PDA accepts a string when it halts on a final state. q0 in initial state and qf is final state. 1. (q0, a , Z) → (q0, aZ) 2. (q0, b , Z) → (q0, bZ) 3. (q0, a , a) →(q0, aa) 4. (q0, b , a) -> (q0, ba) 5. (q0, a, b) → (q0, ab) 6. (q0, b , b) → (q0, bb) 7. (q0, c , b) → (q1, null) 8. (q1, a , a) →(q1, null) 9. (q1, b , b) -> (q1, null) 10. (q1, null, Z ) -> (qf, Z) Here all a’s and b’s are initially pushed onto the stack for W. As soon as a c occurs after b , B is popped from the stack after which W^R is checked. If the further string alphabets match the alphabet of the stack the alphabet is popped from the stack and finally the string reaches the final state if language is of the form L= {WbcW^R | W ∈ (a+b)*}. Hence the above language is a DCFL.
QUESTION: 42
Which of the following languages are closed under complementation ?
A) Context free
B) Recursive
C) Recursive Enumerable
Solution:
According to closure properties of languages, context free and recursive enumerable languages are not closed under complementation while recursive language is.
QUESTION: 43
Which of the following statements is incorrect ?
Solution:
Statement 1 is correct because union of two DCFL may or may not be DCFL but surely CFL. Statement 2 is correct because regular languages are closed under complementation. Statement 3 is correct because Turing recognizable languages or recursive enumerable languages are closed under intersection.
QUESTION: 44
Which of the following is not a CFL?
A) L = {albmcn where l=m or l=n}
B) L = {albmcn where l=m and l=n}
C) L = {albmcn where l = m + n}
Solution:
Language in A is CFL because CFL can make one comparison which is either l and m must be equal or l and n must be equal. Similarly language in C is also CFL. But B is not because it has to make two comparisons. So language in B is CSL not CFL.
QUESTION: 45
Identify the circuit shown below:
Solution:
In the circuit shown there are two tristate buffers.
One with active low enable and second one with active high enable. As per the operation of these buffers truth table is given below.
It is equivalent to 2 x 1 MUX B is selection line, A & C are the input lines, ‘D’ is the output.
QUESTION: 46
The 32 bit floating point representation of (-12) is ___
Solution:
To convert the floating point into decimal, we have 3 elements in a 32-bit floating point representation:
• Sign (MSB)
• Exponent (8 bits after MSB)
• Mantissa (Remaining 23 bits)
Sign bit is the first bit of the binary representation. '1' implies negative number and '0' implies positive number. Sign bit=1 Exponent is decided by the next 8 bits of binary representation. Hence the exponent of 2 will be 3. i.e 23=8. 127 is the unique number for 32 bit floating point representation. It is known as bias. It is determined by 2k-1-1 where 'k' is the number of bits in exponent field. Thus bias = 127 for 32 bit. (28-1-1 = 128-1=127) 127+3=130 i.e. 10000010 in binary representation. Mantissa: 12 in binary = 1100 Move the binary point so that there is only one bit from the left. Adjust the exponent of 2 so that the value does not change. This is normalizing the number. 1.100 x 23 10000000000000000000000 Thus the floating point representation of -12 is 1 10000010 10000000000000000000000
QUESTION: 47
The following is the AB flip flop:
What are values of A, B when change in the flip flop state is from 0 to 1 ?
Note: x in the options below represents that it can have any value out of 0 and 1.
Solution:
Characteristic equation => Qn = A′.Q′ + B′.Q To change from 0->1 means Q = 0 and Qn = 1 Thus 1 = A’(1) + B’(0) = A’ Hence , A = 0 and B can have 0 or 1 not matters.
QUESTION: 48
An instruction pipeline has 4 stages Instruction Fetch(IF), Instruction Decode(ID), Execute instruction (Ex), Write Back(WB). All instructions take all stages and takes 4 clock cycles. Branch instructions are not overlapped, i.e. the instructions after the branch are not fetched till branch is known. Branch is known in the execute phase. Suppose 20 % instructions are conditional and 80 % unconditional. Calculate speed up for 100 instructions (upto 2 decimal place). Ignore the case that the branch may not be taken.
Solution:
Suppose each stage takes 1 s. 20 conditional – 20*3 s (Cycles per instruction for conditional instructions is 3 as branch is known at the third stage) 80 unconditional – 80 s (Cycles per instruction for unconditional is 1) So total time taken with pipeline = 20*3 + 80 = 140 s Time taken without pipeline = 4 * 100 (Cycle per instruction for all is 4) Speed up = 400 / 140 = 2.86
QUESTION: 49
A digital computer has a memory unit of 256k x 16 and a cache memory of 4k words. The cache uses direct mapping with a block size of 16 words. How many bits are there in index, tag, block and words fields of address format ?
Solution:
Main Memory has 256k = 2^8 x 2^10 = 2^18. i.e.
We need 18 bits to address main memory. Cache Memory has 4k = 2^2 x 2^10 = 2^12. i.e. We need 12 bits to address cache memory. Cache consists of Index and tag which together are used to address main memory location.
Here Index is 12 bits and tag is 6 bits. (18 - 12 = 6). Index is divided into block part and word part.
Block part is used to address blocks in cache and word part addresses individual word in a block.
Here a block size is 16 words. i.e. 2^4, we need 4 bits to address a word and 12 - 4 = 8 bits to address a block in cache memory.
QUESTION: 50
Using the data from Question 49, How many blocks can cache accommodate ?
Solution:
Since cache is 4k words and block size is 16 words i.e. No of blocks in cache = 4k/16 = 256.
QUESTION: 51
On a system using round-robin scheduling let 's' represent the time needed to perform a process switch, 'q' represents the round-robin time quantum, and 'r' represents the average time a process runs before blocking on I/O. The CPU efficiency when s = q < r is:
Solution:
‘r’ is the average time a process runs before I/O block and ‘s’ is the time needed for switch. 'q' represents the round-robin time quantum. The efficiency of the CPU when s < r and q < r. = r/(r/q)*s+r = q/(q+s) Here s = q. Hence efficiency is ½.
QUESTION: 52
A counting semaphore was initialized to 0, then 20 V operations were successfully completed on this semaphore, followed with 18 P operations, the resulting value of the semaphore is:
Solution:
20V => increments the semaphore 20 times. Hence , the value becomes 20. 18P => decrements the semaphore 18 times. Hence , the value becomes 2.
QUESTION: 53
Consider all the processes are arriving at large time intervals. Let t be the time interval between two processes Pi and Pi+1 for any i and service time of Pi is Si. If t > Si for every i then, what should be the strategy to schedule the processes ?
Solution:
QUESTION: 54
The following graph shows the schedule of five transactions. The schedule is
Solution:
QUESTION: 55
A student relation with 4 attributes is defines as R(ABCD). Which of the following query will execute without error if all attributes are numeric.
A) ∏B,C(σA>20)
B) σA>20(∏B,C)
Solution:
Query in A is correct. It will first select required rows and then project required columns. But query in B is not correct as after projecting B and C columns, it can't have row selection based on condition on attribute A. So correct option is 1.
QUESTION: 56
Using the EMPLOYEE table, the following query is issued to generate the name, salary and the salary increased after an appraisal by 25 %. The increased salary for all the employee should be above 25000.
SELECT fname, salary, salary + (salary *0.25) AS "INCREASED_SALARY"
FROM employee
WHERE increased_salary > 25000;
The above query throws an error. What is the reason for the error ?
Solution:
QUESTION: 57
Consider the following CFG
S -> AaAb | Bb
A -> ԑ
B -> ԑ
The above grammar is:
Solution:
The grammar is unambiguous (only one parse tree) , not left recursive ( Non-terminal not present in second and third grammar rule) . It is left factored . It is LL(1).
QUESTION: 58
Consider the following SDT
S -> 1A23 {print “GA”}
A -> 4S {print “TE”}
A -> 5 {print “C”}
A -> B {print “SE”}
B -> 6B {print “TE”}
B -> 2 {print “ST”}
What will be output by the SDT for the input string “14122323”?
Solution:
QUESTION: 59
A directed acyclic graph represents one form of intermediate representation. The number of non terminal nodes in DAG of a = (b+c)*(b+c) expression is:
Solution:
QUESTION: 60
Consider the following grammar
S -> Aa | bAc | dc | bda
A -> d
The above grammar is:
Solution:
QUESTION: 61
Let 5 stations are connected to CSMA/CD network . Every station wants to transmit data with probability p = 0.6. It is given that there will be no collision if only one station will transmit. What is the probability of successful transmission ?
Solution:
Success will be for one station and failure for other four Thus Psucc =5 c1 *(0.6)*(0.4)^4 =0.0768.
QUESTION: 62
Consider the following two statements:
Statement 1: Stop and Wait protocol is preferred for LANs compared to WANs
Statement 2: Stop and Wait protocol is good for bursty data transmission.
Which of the above two statements are true?
Solution:
Efficiency of Stop and Wait is given by, E= 1/(1 +(2*d*B)/v*L), link speed(v) and Bandwith (B) are fixed . Thus only variable terms are packet length(L) and distance(d) Cleary, E decreases with increase in d . And E will increase with increase in L . Hence Stop and Wait is suitable for LANs and bursty data.
QUESTION: 63
A broadcast channel has 10 nodes and total capacity of 12 Mbps. It uses polling for medium access. Once a node finishes transmission, there is a polling delay of 50 μseconds to poll the next node. Whenever a node is polled, it is allowed to transmit a maximum of 1000 Bytes. The maximum throughput of broadcast channel is:
Solution:
QUESTION: 64
Match the protocol with the characteristics:
1. Mails are stored on the computer client use.
2. Stateless protocol
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# Generalized continued fraction
In complex analysis, a branch of mathematics, a generalized continued fraction is a generalization of regular continued fractions in canonical form, in which the partial numerators and partial denominators can assume arbitrary complex values.
A generalized continued fraction is an expression of the form
${\displaystyle x=b_{0}+{\cfrac {a_{1}}{b_{1}+{\cfrac {a_{2}}{b_{2}+{\cfrac {a_{3}}{b_{3}+{\cfrac {a_{4}}{b_{4}+\ddots \,}}}}}}}}}$
where the an (n > 0) are the partial numerators, the bn are the partial denominators, and the leading term b0 is called the integer part of the continued fraction.
The successive convergents of the continued fraction are formed by applying the fundamental recurrence formulas:
{\displaystyle {\begin{aligned}x_{0}&={\frac {A_{0}}{B_{0}}}=b_{0},\\[4px]x_{1}&={\frac {A_{1}}{B_{1}}}={\frac {b_{1}b_{0}+a_{1}}{b_{1}}},\\[4px]x_{2}&={\frac {A_{2}}{B_{2}}}={\frac {b_{2}(b_{1}b_{0}+a_{1})+a_{2}b_{0}}{b_{2}b_{1}+a_{2}}},\ \dots \end{aligned}}}
where An is the numerator and Bn is the denominator, called continuants,[1][2] of the nth convergent. They are given by the recursion [3]
{\displaystyle {\begin{aligned}A_{n}&=b_{n}A_{n-1}+a_{n}A_{n-2},\\B_{n}&=b_{n}B_{n-1}+a_{n}B_{n-2}\qquad {\text{for }}n\geq 1\end{aligned}}}
with initial values
{\displaystyle {\begin{aligned}A_{-1}&=1,&A_{0}&=b_{0},\\B_{-1}&=0,&B_{0}&=1.\end{aligned}}}
If the sequence of convergents {xn} approaches a limit the continued fraction is convergent and has a definite value. If the sequence of convergents never approaches a limit the continued fraction is divergent. It may diverge by oscillation (for example, the odd and even convergents may approach two different limits), or it may produce an infinite number of zero denominators Bn.
## History
The story of continued fractions begins with the Euclidean algorithm,[4] a procedure for finding the greatest common divisor of two natural numbers m and n. That algorithm introduced the idea of dividing to extract a new remainder – and then dividing by the new remainder repeatedly.
Nearly two thousand years passed before Bombelli (1579) devised a technique for approximating the roots of quadratic equations with continued fractions in the mid-sixteenth century. Now the pace of development quickened. Just 24 years later, in 1613, Pietro Cataldi introduced the first formal notation for the generalized continued fraction.[5] Cataldi represented a continued fraction as
${\displaystyle {a_{0}\cdot }\,\&\,{\frac {n_{1}}{d_{1}\cdot }}\,\&\,{\frac {n_{2}}{d_{2}\cdot }}\,\&\,{\frac {n_{3}}{d_{3}}}}$
with the dots indicating where the next fraction goes, and each & representing a modern plus sign.
Late in the seventeenth century John Wallis introduced the term "continued fraction" into mathematical literature.[6] New techniques for mathematical analysis (Newton's and Leibniz's calculus) had recently come onto the scene, and a generation of Wallis' contemporaries put the new phrase to use.
In 1748 Euler published a theorem showing that a particular kind of continued fraction is equivalent to a certain very general infinite series.[7] Euler's continued fraction formula is still the basis of many modern proofs of convergence of continued fractions.
In 1761, Johann Heinrich Lambert gave the first proof that π is irrational, by using the following continued fraction for tan x:[8]
${\displaystyle \tan(x)={\cfrac {x}{1+{\cfrac {-x^{2}}{3+{\cfrac {-x^{2}}{5+{\cfrac {-x^{2}}{7+{}\ddots }}}}}}}}}$
Continued fractions can also be applied to problems in number theory, and are especially useful in the study of Diophantine equations. In the late eighteenth century Lagrange used continued fractions to construct the general solution of Pell's equation, thus answering a question that had fascinated mathematicians for more than a thousand years.[9] Lagrange's discovery implies that the canonical continued fraction expansion of the square root of every non-square integer is periodic and that, if the period is of length p > 1, it contains a palindromic string of length p − 1.
In 1813 Gauss derived from complex-valued hypergeometric functions what is now called Gauss's continued fractions.[10] They can be used to express many elementary functions and some more advanced functions (such as the Bessel functions), as continued fractions that are rapidly convergent almost everywhere in the complex plane.
## Notation
The long continued fraction expression displayed in the introduction is easy for an unfamiliar reader to interpret. However, it takes up a lot of space and can be difficult to typeset. So mathematicians have devised several alternative notations. One convenient way to express a generalized continued fraction sets each nested fraction on the same line, indicating the nesting by dangling plus signs in the denominators:
${\displaystyle x=b_{0}+{\frac {a_{1}}{b_{1}+}}\,{\frac {a_{2}}{b_{2}+}}\,{\frac {a_{3}}{b_{3}+}}\cdots }$
Sometimes the plus signs are typeset to vertically align with the denominators but not under the fraction bars:
${\displaystyle x=b_{0}+{\frac {a_{1}}{b_{1}}}{{} \atop +}{\frac {a_{2}}{b_{2}}}{{} \atop +}{\frac {a_{3}}{b_{3}}}{{} \atop +\cdots }}$
Pringsheim wrote a generalized continued fraction this way:
${\displaystyle x=b_{0}+{\frac {a_{1}\mid }{\mid b_{1}}}+{\frac {a_{2}\mid }{\mid b_{2}}}+{\frac {a_{3}\mid }{\mid b_{3}}}+\cdots .\,}$
Carl Friedrich Gauss evoked the more familiar infinite product Π when he devised this notation:
${\displaystyle x=b_{0}+{\underset {i=1}{\overset {\infty }{\operatorname {K} }}}{\frac {a_{i}}{b_{i}}}.\,}$
Here the "K" stands for Kettenbruch, the German word for "continued fraction". This is probably the most compact and convenient way to express continued fractions; however, it is not widely used by English typesetters.
## Some elementary considerations
Here are some elementary results that are of fundamental importance in the further development of the analytic theory of continued fractions.
### Partial numerators and denominators
If one of the partial numerators an + 1 is zero, the infinite continued fraction
${\displaystyle b_{0}+{\underset {i=1}{\overset {\infty }{\operatorname {K} }}}{\frac {a_{i}}{b_{i}}}\,}$
is really just a finite continued fraction with n fractional terms, and therefore a rational function of a1 to an and b0 to bn + 1. Such an object is of little interest from the point of view adopted in mathematical analysis, so it is usually assumed that all ai ≠ 0. There is no need to place this restriction on the partial denominators bi.
### The determinant formula
When the nth convergent of a continued fraction
${\displaystyle x_{n}=b_{0}+{\underset {i=1}{\overset {n}{\operatorname {K} }}}{\frac {a_{i}}{b_{i}}}\,}$
is expressed as a simple fraction xn = An/Bn we can use the determinant formula
${\displaystyle A_{n-1}B_{n}-A_{n}B_{n-1}=\left(-1\right)^{n}a_{1}a_{2}\cdots a_{n}=\prod _{i=1}^{n}(-a_{i})}$
(1)
to relate the numerators and denominators of successive convergents xn and xn − 1 to one another. The proof for this can be easily seen by induction.
Base case
The case n = 1 results from a very simple computation.
Inductive step
Assume that (1) holds for n − 1. Then we need to see the same relation holding true for n. Substituting the value of An and Bn in (1) we obtain:
{\displaystyle {\begin{aligned}&=b_{n}A_{n-1}B_{n-1}+a_{n}A_{n-1}B_{n-2}-b_{n}A_{n-1}B_{n-1}-a_{n}A_{n-2}B_{n-1}\\&=a_{n}(A_{n-1}B_{n-2}-A_{n-2}B_{n-1})\end{aligned}}}
which is true because of our induction hypothesis.
${\displaystyle A_{n-1}B_{n}-A_{n}B_{n-1}=\left(-1\right)^{n}a_{1}a_{2}\cdots a_{n}=\prod _{i=1}^{n}(-a_{i})\,}$
Specifically, if neither Bn nor Bn − 1 is zero (n > 0) we can express the difference between the (n − 1)th and nth convergents like this:
${\displaystyle x_{n-1}-x_{n}={\frac {A_{n-1}}{B_{n-1}}}-{\frac {A_{n}}{B_{n}}}=\left(-1\right)^{n}{\frac {a_{1}a_{2}\cdots a_{n}}{B_{n}B_{n-1}}}={\frac {\prod _{i=1}^{n}(-a_{i})}{B_{n}B_{n-1}}}.\,}$
### The equivalence transformation
If {ci} = {c1, c2, c3, ...} is any infinite sequence of non-zero complex numbers we can prove, by induction, that
${\displaystyle b_{0}+{\cfrac {a_{1}}{b_{1}+{\cfrac {a_{2}}{b_{2}+{\cfrac {a_{3}}{b_{3}+{\cfrac {a_{4}}{b_{4}+\ddots \,}}}}}}}}=b_{0}+{\cfrac {c_{1}a_{1}}{c_{1}b_{1}+{\cfrac {c_{1}c_{2}a_{2}}{c_{2}b_{2}+{\cfrac {c_{2}c_{3}a_{3}}{c_{3}b_{3}+{\cfrac {c_{3}c_{4}a_{4}}{c_{4}b_{4}+\ddots \,}}}}}}}}}$
where equality is understood as equivalence, which is to say that the successive convergents of the continued fraction on the left are exactly the same as the convergents of the fraction on the right.
The equivalence transformation is perfectly general, but two particular cases deserve special mention. First, if none of the ai are zero a sequence {ci} can be chosen to make each partial numerator a 1:
${\displaystyle b_{0}+{\underset {i=1}{\overset {\infty }{\operatorname {K} }}}{\frac {a_{i}}{b_{i}}}=b_{0}+{\underset {i=1}{\overset {\infty }{\operatorname {K} }}}{\frac {1}{c_{i}b_{i}}}\,}$
where c1 = 1/a1, c2 = a1/a2, c3 = a2/a1a3, and in general cn + 1 = 1/an + 1cn.
Second, if none of the partial denominators bi are zero we can use a similar procedure to choose another sequence {di} to make each partial denominator a 1:
${\displaystyle b_{0}+{\underset {i=1}{\overset {\infty }{\operatorname {K} }}}{\frac {a_{i}}{b_{i}}}=b_{0}+{\underset {i=1}{\overset {\infty }{\operatorname {K} }}}{\frac {d_{i}a_{i}}{1}}\,}$
where d1 = 1/b1 and otherwise dn + 1 = 1/bnbn + 1.
These two special cases of the equivalence transformation are enormously useful when the general convergence problem is analyzed.
### Notions of convergence
As mentioned in the introduction, the continued fraction
${\displaystyle x=b_{0}+{\underset {i=1}{\overset {\infty }{\operatorname {K} }}}{\frac {a_{i}}{b_{i}}}\,}$
converges if the sequence of convergents {xn} tends to a finite limit. This notion of convergence is very natural, but it is sometimes too restrictive. It is therefore useful to introduce the notion of general convergence of a continued fraction. Roughly speaking, this consists in replacing the ${\displaystyle \operatorname {K} _{i=n}^{\infty }{\tfrac {a_{i}}{b_{i}}}}$ part of the fraction by wn, instead of by 0, to compute the convergents. The convergents thus obtained are called modified convergents. We say that the continued fraction converges generally if there exists a sequence ${\displaystyle \{w_{n}^{*}\}}$ such that the sequence of modified convergents converges for all ${\displaystyle \{w_{n}\}}$ sufficiently distinct from ${\displaystyle \{w_{n}^{*}\}}$. The sequence ${\displaystyle \{w_{n}^{*}\}}$ is then called an exceptional sequence for the continued fraction. See Chapter 2 of Lorentzen & Waadeland (1992) for a rigorous definition.
There also exists a notion of absolute convergence for continued fractions, which is based on the notion of absolute convergence of a series: a continued fraction is said to be absolutely convergent when the series
${\displaystyle f=\sum _{n}\left(f_{n}-f_{n-1}\right),}$
where ${\displaystyle f_{n}=\operatorname {K} _{i=1}^{n}{\tfrac {a_{i}}{b_{i}}}}$ are the convergents of the continued fraction, converges absolutely.[11] The Śleszyński–Pringsheim theorem provides a sufficient condition for absolute convergence.
Finally, a continued fraction of one or more complex variables is uniformly convergent in an open neighborhood Ω when its convergents converge uniformly on Ω; that is, when for every ε > 0 there exists M such that for all n > M, for all ${\displaystyle z\in \Omega }$,
${\displaystyle |f(z)-f_{n}(z)|<\varepsilon .}$
### Even and odd convergents
It is sometimes necessary to separate a continued fraction into its even and odd parts. For example, if the continued fraction diverges by oscillation between two distinct limit points p and q, then the sequence {x0, x2, x4, ...} must converge to one of these, and {x1, x3, x5, ...} must converge to the other. In such a situation it may be convenient to express the original continued fraction as two different continued fractions, one of them converging to p, and the other converging to q.
The formulas for the even and odd parts of a continued fraction can be written most compactly if the fraction has already been transformed so that all its partial denominators are unity. Specifically, if
${\displaystyle x={\underset {i=1}{\overset {\infty }{\operatorname {K} }}}{\frac {a_{i}}{1}}\,}$
is a continued fraction, then the even part xeven and the odd part xodd are given by
${\displaystyle x_{\text{even}}={\cfrac {a_{1}}{1+a_{2}-{\cfrac {a_{2}a_{3}}{1+a_{3}+a_{4}-{\cfrac {a_{4}a_{5}}{1+a_{5}+a_{6}-{\cfrac {a_{6}a_{7}}{1+a_{7}+a_{8}-\ddots }}}}}}}}\,}$
and
${\displaystyle x_{\text{odd}}=a_{1}-{\cfrac {a_{1}a_{2}}{1+a_{2}+a_{3}-{\cfrac {a_{3}a_{4}}{1+a_{4}+a_{5}-{\cfrac {a_{5}a_{6}}{1+a_{6}+a_{7}-{\cfrac {a_{7}a_{8}}{1+a_{8}+a_{9}-\ddots }}}}}}}}\,}$
respectively. More precisely, if the successive convergents of the continued fraction x are {x1, x2, x3, ...}, then the successive convergents of xeven as written above are {x2, x4, x6, ...}, and the successive convergents of xodd are {x1, x3, x5, ...}.[12]
### Conditions for irrationality
If a1, a2,... and b1, b2,... are positive integers with akbk for all sufficiently large k, then
${\displaystyle x=b_{0}+{\underset {i=1}{\overset {\infty }{\operatorname {K} }}}{\frac {a_{i}}{b_{i}}}\,}$
converges to an irrational limit.[13]
### Fundamental recurrence formulas
The partial numerators and denominators of the fraction's successive convergents are related by the fundamental recurrence formulas:
{\displaystyle {\begin{aligned}A_{-1}&=1&B_{-1}&=0\\A_{0}&=b_{0}&B_{0}&=1\\A_{n+1}&=b_{n+1}A_{n}+a_{n+1}A_{n-1}&B_{n+1}&=b_{n+1}B_{n}+a_{n+1}B_{n-1}\,\end{aligned}}}
The continued fraction's successive convergents are then given by
${\displaystyle x_{n}={\frac {A_{n}}{B_{n}}}.\,}$
These recurrence relations are due to John Wallis (1616–1703) and Leonhard Euler (1707–1783).[14] These recurrence relations are simply a different notation for the relations obtained by Pietro Antonio Cataldi (1548-1626).
As an example, consider the regular continued fraction in canonical form that represents the golden ratio φ:
${\displaystyle x=1+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{1+\ddots \,}}}}}}}}}$
Applying the fundamental recurrence formulas we find that the successive numerators An are {1, 2, 3, 5, 8, 13, ...} and the successive denominators Bn are {1, 1, 2, 3, 5, 8, ...}, the Fibonacci numbers. Since all the partial numerators in this example are equal to one, the determinant formula assures us that the absolute value of the difference between successive convergents approaches zero quite rapidly.
## Linear fractional transformations
A linear fractional transformation (LFT) is a complex function of the form
${\displaystyle w=f(z)={\frac {a+bz}{c+dz}},\,}$
where z is a complex variable, and a, b, c, d are arbitrary complex constants such that c + dz ≠ 0. An additional restriction that adbc is customarily imposed, to rule out the cases in which w = f(z) is a constant. The linear fractional transformation, also known as a Möbius transformation, has many fascinating properties. Four of these are of primary importance in developing the analytic theory of continued fractions.
• If d ≠ 0 the LFT has one or two fixed points. This can be seen by considering the equation
${\displaystyle f(z)=z\Rightarrow dz^{2}+cz=a+bz\,}$
which is clearly a quadratic equation in z. The roots of this equation are the fixed points of f(z). If the discriminant (cb)2 + 4ad is zero the LFT fixes a single point; otherwise it has two fixed points.
${\displaystyle z=g(w)={\frac {-a+cw}{b-dw}}\,}$
such that f(g(z)) = g(f(z)) = z for every point z in the extended complex plane, and both f and g preserve angles and shapes at vanishingly small scales. From the form of z = g(w) we see that g is also an LFT.
• The composition of two different LFTs for which adbc is itself an LFT for which adbc. In other words, the set of all LFTs for which adbc is closed under composition of functions. The collection of all such LFTs, together with the "group operation" composition of functions, is known as the automorphism group of the extended complex plane.
• If b = 0 the LFT reduces to
${\displaystyle w=f(z)={\frac {a}{c+dz}},\,}$
which is a very simple meromorphic function of z with one simple pole (at c/d) and a residue equal to a/d. (See also Laurent series.)
### The continued fraction as a composition of LFTs
Consider a sequence of simple linear fractional transformations
{\displaystyle {\begin{aligned}\tau _{0}(z)&=b_{0}+z,\\[4px]\tau _{1}(z)&={\frac {a_{1}}{b_{1}+z}},\\[4px]\tau _{2}(z)&={\frac {a_{2}}{b_{2}+z}},\\[4px]\tau _{3}(z)&={\frac {a_{3}}{b_{3}+z}},\\&\;\vdots \end{aligned}}}
Here we use τ to represent each simple LFT, and we adopt the conventional circle notation for composition of functions. We also introduce a new symbol Τn to represent the composition of n + 1 transformations τi; that is,
{\displaystyle {\begin{aligned}{\boldsymbol {\mathrm {T} }}_{\boldsymbol {1}}(z)&=\tau _{0}\circ \tau _{1}(z)=\tau _{0}{\big (}\tau _{1}(z){\big )},\\{\boldsymbol {\mathrm {T} }}_{\boldsymbol {2}}(z)&=\tau _{0}\circ \tau _{1}\circ \tau _{2}(z)=\tau _{0}{\Big (}\tau _{1}{\big (}\tau _{2}(z){\big )}{\Big )},\,\end{aligned}}}
and so forth. By direct substitution from the first set of expressions into the second we see that
{\displaystyle {\begin{aligned}{\boldsymbol {\mathrm {T} }}_{\boldsymbol {1}}(z)&=\tau _{0}\circ \tau _{1}(z)&=&\quad b_{0}+{\cfrac {a_{1}}{b_{1}+z}}\\[4px]{\boldsymbol {\mathrm {T} }}_{\boldsymbol {2}}(z)&=\tau _{0}\circ \tau _{1}\circ \tau _{2}(z)&=&\quad b_{0}+{\cfrac {a_{1}}{b_{1}+{\cfrac {a_{2}}{b_{2}+z}}}}\,\end{aligned}}}
and, in general,
${\displaystyle {\boldsymbol {\mathrm {T} }}_{\boldsymbol {n}}(z)=\tau _{0}\circ \tau _{1}\circ \tau _{2}\circ \cdots \circ \tau _{n}(z)=b_{0}+{\underset {i=1}{\overset {n}{\operatorname {K} }}}{\frac {a_{i}}{b_{i}}}\,}$
where the last partial denominator in the finite continued fraction K is understood to be bn + z. And, since bn + 0 = bn, the image of the point z = 0 under the iterated LFT Τn is indeed the value of the finite continued fraction with n partial numerators:
${\displaystyle {\boldsymbol {\mathrm {T} }}_{\boldsymbol {n}}(0)={\boldsymbol {\mathrm {T} }}_{\boldsymbol {n+1}}(\infty )=b_{0}+{\underset {i=1}{\overset {n}{\operatorname {K} }}}{\frac {a_{i}}{b_{i}}}.\,}$
### A geometric interpretation
Defining a finite continued fraction as the image of a point under the iterated linear functional transformation Τn(z) leads to an intuitively appealing geometric interpretation of infinite continued fractions.
The relationship
${\displaystyle x_{n}=b_{0}+{\underset {i=1}{\overset {n}{\operatorname {K} }}}{\frac {a_{i}}{b_{i}}}={\frac {A_{n}}{B_{n}}}={\boldsymbol {\mathrm {T} }}_{\boldsymbol {n}}(0)={\boldsymbol {\mathrm {T} }}_{\boldsymbol {n+1}}(\infty )\,}$
can be understood by rewriting Τn(z) and Τn + 1(z) in terms of the fundamental recurrence formulas:
{\displaystyle {\begin{aligned}{\boldsymbol {\mathrm {T} }}_{\boldsymbol {n}}(z)&={\frac {(b_{n}+z)A_{n-1}+a_{n}A_{n-2}}{(b_{n}+z)B_{n-1}+a_{n}B_{n-2}}}&{\boldsymbol {\mathrm {T} }}_{\boldsymbol {n}}(z)&={\frac {zA_{n-1}+A_{n}}{zB_{n-1}+B_{n}}};\\[6px]{\boldsymbol {\mathrm {T} }}_{\boldsymbol {n+1}}(z)&={\frac {(b_{n+1}+z)A_{n}+a_{n+1}A_{n-1}}{(b_{n+1}+z)B_{n}+a_{n+1}B_{n-1}}}&{\boldsymbol {\mathrm {T} }}_{\boldsymbol {n+1}}(z)&={\frac {zA_{n}+A_{n+1}}{zB_{n}+B_{n+1}}}.\,\end{aligned}}}
In the first of these equations the ratio tends toward An/Bn as z tends toward zero. In the second, the ratio tends toward An/Bn as z tends to infinity. This leads us to our first geometric interpretation. If the continued fraction converges, the successive convergents An/Bn are eventually arbitrarily close together. Since the linear fractional transformation Τn(z) is a continuous mapping, there must be a neighborhood of z = 0 that is mapped into an arbitrarily small neighborhood of Τn(0) = An/Bn. Similarly, there must be a neighborhood of the point at infinity which is mapped into an arbitrarily small neighborhood of Τn(∞) = An − 1/Bn − 1. So if the continued fraction converges the transformation Τn(z) maps both very small z and very large z into an arbitrarily small neighborhood of x, the value of the continued fraction, as n gets larger and larger.
For intermediate values of z, since the successive convergents are getting closer together we must have
${\displaystyle {\frac {A_{n-1}}{B_{n-1}}}\approx {\frac {A_{n}}{B_{n}}}\quad \Rightarrow \quad {\frac {A_{n-1}}{A_{n}}}\approx {\frac {B_{n-1}}{B_{n}}}=k\,}$
where k is a constant, introduced for convenience. But then, by substituting in the expression for Τn(z) we obtain
${\displaystyle {\boldsymbol {\mathrm {T} }}_{\boldsymbol {n}}(z)={\frac {zA_{n-1}+A_{n}}{zB_{n-1}+B_{n}}}={\frac {A_{n}}{B_{n}}}\left({\frac {z{\frac {A_{n-1}}{A_{n}}}+1}{z{\frac {B_{n-1}}{B_{n}}}+1}}\right)\approx {\frac {A_{n}}{B_{n}}}\left({\frac {zk+1}{zk+1}}\right)={\frac {A_{n}}{B_{n}}}\,}$
so that even the intermediate values of z (except when z ≈ −k−1) are mapped into an arbitrarily small neighborhood of x, the value of the continued fraction, as n gets larger and larger. Intuitively, it is almost as if the convergent continued fraction maps the entire extended complex plane into a single point.[15]
Notice that the sequence {Τn} lies within the automorphism group of the extended complex plane, since each Τn is a linear fractional transformation for which abcd. And every member of that automorphism group maps the extended complex plane into itself: not one of the Τn can possibly map the plane into a single point. Yet in the limit the sequence {Τn} defines an infinite continued fraction which (if it converges) represents a single point in the complex plane.
When an infinite continued fraction converges, the corresponding sequence {Τn} of LFTs "focuses" the plane in the direction of x, the value of the continued fraction. At each stage of the process a larger and larger region of the plane is mapped into a neighborhood of x, and the smaller and smaller region of the plane that's left over is stretched out ever more thinly to cover everything outside that neighborhood.[16]
For divergent continued fractions, we can distinguish three cases:
1. The two sequences {Τ2n − 1} and {Τ2n} might themselves define two convergent continued fractions that have two different values, xodd and xeven. In this case the continued fraction defined by the sequence {Τn} diverges by oscillation between two distinct limit points. And in fact this idea can be generalized: sequences {Τn} can be constructed that oscillate among three, or four, or indeed any number of limit points. Interesting instances of this case arise when the sequence {Τn} constitutes a subgroup of finite order within the group of automorphisms over the extended complex plane.
2. The sequence {Τn} may produce an infinite number of zero denominators Bi while also producing a subsequence of finite convergents. These finite convergents may not repeat themselves or fall into a recognizable oscillating pattern. Or they may converge to a finite limit, or even oscillate among multiple finite limits. No matter how the finite convergents behave, the continued fraction defined by the sequence {Τn} diverges by oscillation with the point at infinity in this case.[17]
3. The sequence {Τn} may produce no more than a finite number of zero denominators Bi. while the subsequence of finite convergents dances wildly around the plane in a pattern that never repeats itself and never approaches any finite limit either.
Interesting examples of cases 1 and 3 can be constructed by studying the simple continued fraction
${\displaystyle x=1+{\cfrac {z}{1+{\cfrac {z}{1+{\cfrac {z}{1+{\cfrac {z}{1+\ddots }}}}}}}}\,}$
where z is any real number such that z < −1/4.[18]
## Euler's continued fraction formula
Euler proved the following identity:[7]
${\displaystyle a_{0}+a_{0}a_{1}+a_{0}a_{1}a_{2}+\cdots +a_{0}a_{1}a_{2}\cdots a_{n}={\frac {a_{0}}{1-}}{\frac {a_{1}}{1+a_{1}-}}{\frac {a_{2}}{1+a_{2}-}}\cdots {\frac {a_{n}}{1+a_{n}}}.\,}$
From this many other results can be derived, such as
${\displaystyle {\frac {1}{u_{1}}}+{\frac {1}{u_{2}}}+{\frac {1}{u_{3}}}+\cdots +{\frac {1}{u_{n}}}={\frac {1}{u_{1}-}}{\frac {u_{1}^{2}}{u_{1}+u_{2}-}}{\frac {u_{2}^{2}}{u_{2}+u_{3}-}}\cdots {\frac {u_{n-1}^{2}}{u_{n-1}+u_{n}}},\,}$
and
${\displaystyle {\frac {1}{a_{0}}}+{\frac {x}{a_{0}a_{1}}}+{\frac {x^{2}}{a_{0}a_{1}a_{2}}}+\cdots +{\frac {x^{n}}{a_{0}a_{1}a_{2}\ldots a_{n}}}={\frac {1}{a_{0}-}}{\frac {a_{0}x}{a_{1}+x-}}{\frac {a_{1}x}{a_{2}+x-}}\cdots {\frac {a_{n-1}x}{a_{n}+x}}.\,}$
Euler's formula connecting continued fractions and series is the motivation for the fundamental inequalities[link or clarification needed], and also the basis of elementary approaches to the convergence problem.
## Examples
### Transcendental functions and numbers
Here are two continued fractions that can be built via Euler's identity.
${\displaystyle e^{x}={\frac {x^{0}}{0!}}+{\frac {x^{1}}{1!}}+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}+{\frac {x^{4}}{4!}}+\cdots =1+{\cfrac {x}{1-{\cfrac {1x}{2+x-{\cfrac {2x}{3+x-{\cfrac {3x}{4+x-\ddots }}}}}}}}}$
${\displaystyle \log(1+x)={\frac {x^{1}}{1}}-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-{\frac {x^{4}}{4}}+\cdots ={\cfrac {x}{1-0x+{\cfrac {1^{2}x}{2-1x+{\cfrac {2^{2}x}{3-2x+{\cfrac {3^{2}x}{4-3x+\ddots }}}}}}}}}$
Here are additional generalized continued fractions:
${\displaystyle \arctan {\cfrac {x}{y}}={\cfrac {xy}{1y^{2}+{\cfrac {(1xy)^{2}}{3y^{2}-1x^{2}+{\cfrac {(3xy)^{2}}{5y^{2}-3x^{2}+{\cfrac {(5xy)^{2}}{7y^{2}-5x^{2}+\ddots }}}}}}}}={\cfrac {x}{1y+{\cfrac {(1x)^{2}}{3y+{\cfrac {(2x)^{2}}{5y+{\cfrac {(3x)^{2}}{7y+\ddots }}}}}}}}}$
${\displaystyle e^{\frac {x}{y}}=1+{\cfrac {2x}{2y-x+{\cfrac {x^{2}}{6y+{\cfrac {x^{2}}{10y+{\cfrac {x^{2}}{14y+{\cfrac {x^{2}}{18y+\ddots }}}}}}}}}}\quad \Rightarrow \quad e^{2}=7+{\cfrac {2}{5+{\cfrac {1}{7+{\cfrac {1}{9+{\cfrac {1}{11+\ddots }}}}}}}}}$
${\displaystyle \log \left(1+{\frac {x}{y}}\right)={\cfrac {x}{y+{\cfrac {1x}{2+{\cfrac {1x}{3y+{\cfrac {2x}{2+{\cfrac {2x}{5y+{\cfrac {3x}{2+\ddots }}}}}}}}}}}}={\cfrac {2x}{2y+x-{\cfrac {(1x)^{2}}{3(2y+x)-{\cfrac {(2x)^{2}}{5(2y+x)-{\cfrac {(3x)^{2}}{7(2y+x)-\ddots }}}}}}}}}$
This last is based on an algorithm derived by Aleksei Nikolaevich Khovansky in the 1970s.[19]
Example: the natural logarithm of 2 (= [0; 1, 2, 3, 1, 5, 2/3, 7, 1/2, 9, 2/5,..., 2k − 1, 2/k,...] ≈ 0.693147...):[20]
${\displaystyle \log 2=\log(1+1)={\cfrac {1}{1+{\cfrac {1}{2+{\cfrac {1}{3+{\cfrac {2}{2+{\cfrac {2}{5+{\cfrac {3}{2+\ddots }}}}}}}}}}}}={\cfrac {2}{3-{\cfrac {1^{2}}{9-{\cfrac {2^{2}}{15-{\cfrac {3^{2}}{21-\ddots }}}}}}}}}$
#### π
Here are three of π's best-known generalized continued fractions, the first and third of which are derived from their respective arctangent formulas above by setting x = y = 1 and multiplying by 4. The Leibniz formula for π:
${\displaystyle \pi ={\cfrac {4}{1+{\cfrac {1^{2}}{2+{\cfrac {3^{2}}{2+{\cfrac {5^{2}}{2+\ddots }}}}}}}}=\sum _{n=0}^{\infty }{\frac {4(-1)^{n}}{2n+1}}={\frac {4}{1}}-{\frac {4}{3}}+{\frac {4}{5}}-{\frac {4}{7}}+-\cdots }$
converges too slowly, requiring roughly 3 × 10n terms to achieve n correct decimal places. The series derived by Nilakantha Somayaji:
${\displaystyle \pi =3+{\cfrac {1^{2}}{6+{\cfrac {3^{2}}{6+{\cfrac {5^{2}}{6+\ddots }}}}}}=3-\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n(n+1)(2n+1)}}=3+{\frac {1}{1\cdot 2\cdot 3}}-{\frac {1}{2\cdot 3\cdot 5}}+{\frac {1}{3\cdot 4\cdot 7}}-+\cdots }$
is a much more obvious expression but still converges quite slowly, requiring nearly 50 terms for five decimals and nearly 120 for six. Both converge sublinearly to π. On the other hand:
${\displaystyle \pi ={\cfrac {4}{1+{\cfrac {1^{2}}{3+{\cfrac {2^{2}}{5+{\cfrac {3^{2}}{7+\ddots }}}}}}}}=4-1+{\frac {1}{6}}-{\frac {1}{34}}+{\frac {16}{3145}}-{\frac {4}{4551}}+{\frac {1}{6601}}-{\frac {1}{38341}}+-\cdots }$
converges linearly to π, adding at least three digits of precision per four terms, a pace slightly faster than the arcsine formula for π:
${\displaystyle \pi =6\sin ^{-1}\left({\frac {1}{2}}\right)=\sum _{n=0}^{\infty }{\frac {3\cdot {\binom {2n}{n}}}{16^{n}(2n+1)}}={\frac {3}{16^{0}\cdot 1}}+{\frac {6}{16^{1}\cdot 3}}+{\frac {18}{16^{2}\cdot 5}}+{\frac {60}{16^{3}\cdot 7}}+\cdots \!}$
which adds at least three decimal digits per five terms.[21]
• Note: this continued fraction's rate of convergence μ tends to 3 − 8 ≈ 0.1715729, hence 1/μ tends to 3 + 8 ≈ 5.828427, whose common logarithm is 0.7655... ≈ 13/17 > 3/4. The same 1/μ = 3 + 8 (the silver ratio squared) also is observed in the unfolded general continued fractions of both the natural logarithm of 2 and the nth root of 2 (which works for any integer n > 1) if calculated using 2 = 1 + 1. For the folded general continued fractions of both expressions, the rate convergence μ = (3 − 8)2 = 17 − 288 ≈ 0.02943725, hence 1/μ = (3 + 8)2 = 17 + 288 ≈ 33.97056, whose common logarithm is 1.531... ≈ 26/17 > 3/2, thus adding at least three digits per two terms. This is because the folded GCF folds each pair of fractions from the unfolded GCF into one fraction, thus doubling the convergence pace. The Manny Sardina reference further explains "folded" continued fractions.
• Note: Using the continued fraction for arctan x/y cited above with the best-known Machin-like formula provides an even more rapidly, although still linearly, converging expression:
${\displaystyle \pi =16\tan ^{-1}{\cfrac {1}{5}}\,-\,4\tan ^{-1}{\cfrac {1}{239}}={\cfrac {16}{u+{\cfrac {1^{2}}{3u+{\cfrac {2^{2}}{5u+{\cfrac {3^{2}}{7u+\ddots }}}}}}}}\,-\,{\cfrac {4}{v+{\cfrac {1^{2}}{3v+{\cfrac {2^{2}}{5v+{\cfrac {3^{2}}{7v+\ddots }}}}}}}}.}$
with u = 5 and v = 239.
### Roots of positive numbers
The nth root of any positive number zm can be expressed by restating z = xn + y, resulting in
${\displaystyle {\sqrt[{n}]{z^{m}}}={\sqrt[{n}]{\left(x^{n}+y\right)^{m}}}=x^{m}+{\cfrac {my}{nx^{n-m}+{\cfrac {(n-m)y}{2x^{m}+{\cfrac {(n+m)y}{3nx^{n-m}+{\cfrac {(2n-m)y}{2x^{m}+{\cfrac {(2n+m)y}{5nx^{n-m}+{\cfrac {(3n-m)y}{2x^{m}+\ddots }}}}}}}}}}}}}$
which can be simplified, by folding each pair of fractions into one fraction, to
${\displaystyle {\sqrt[{n}]{z^{m}}}=x^{m}+{\cfrac {2x^{m}\cdot my}{n(2x^{n}+y)-my-{\cfrac {(1^{2}n^{2}-m^{2})y^{2}}{3n(2x^{n}+y)-{\cfrac {(2^{2}n^{2}-m^{2})y^{2}}{5n(2x^{n}+y)-{\cfrac {(3^{2}n^{2}-m^{2})y^{2}}{7n(2x^{n}+y)-{\cfrac {(4^{2}n^{2}-m^{2})y^{2}}{9n(2x^{n}+y)-\ddots }}}}}}}}}}.}$
The square root of z is a special case with m = 1 and n = 2:
${\displaystyle {\sqrt {z}}={\sqrt {x^{2}+y}}=x+{\cfrac {y}{2x+{\cfrac {y}{2x+{\cfrac {3y}{6x+{\cfrac {3y}{2x+\ddots }}}}}}}}=x+{\cfrac {2x\cdot y}{2(2x^{2}+y)-y-{\cfrac {1\cdot 3y^{2}}{6(2x^{2}+y)-{\cfrac {3\cdot 5y^{2}}{10(2x^{2}+y)-\ddots }}}}}}}$
which can be simplified by noting that 5/10 = 3/6 = 1/2:
${\displaystyle {\sqrt {z}}={\sqrt {x^{2}+y}}=x+{\cfrac {y}{2x+{\cfrac {y}{2x+{\cfrac {y}{2x+{\cfrac {y}{2x+\ddots }}}}}}}}=x+{\cfrac {2x\cdot y}{2(2x^{2}+y)-y-{\cfrac {y^{2}}{2(2x^{2}+y)-{\cfrac {y^{2}}{2(2x^{2}+y)-\ddots }}}}}}.}$
The square root can also be expressed by a periodic continued fraction, but the above form converges more quickly with the proper x and y.
#### Example 1
The cube root of two (21/3 or 32 ≈ 1.259921...) can be calculated in two ways:
Firstly, "standard notation" of x = 1, y = 1, and 2zy = 3:
${\displaystyle {\sqrt[{3}]{2}}=1+{\cfrac {1}{3+{\cfrac {2}{2+{\cfrac {4}{9+{\cfrac {5}{2+{\cfrac {7}{15+{\cfrac {8}{2+{\cfrac {10}{21+{\cfrac {11}{2+\ddots }}}}}}}}}}}}}}}}=1+{\cfrac {2\cdot 1}{9-1-{\cfrac {2\cdot 4}{27-{\cfrac {5\cdot 7}{45-{\cfrac {8\cdot 10}{63-{\cfrac {11\cdot 13}{81-\ddots }}}}}}}}}}.}$
Secondly, a rapid convergence with x = 5, y = 3 and 2zy = 253:
${\displaystyle {\sqrt[{3}]{2}}={\cfrac {5}{4}}+{\cfrac {0.5}{50+{\cfrac {2}{5+{\cfrac {4}{150+{\cfrac {5}{5+{\cfrac {7}{250+{\cfrac {8}{5+{\cfrac {10}{350+{\cfrac {11}{5+\ddots }}}}}}}}}}}}}}}}={\cfrac {5}{4}}+{\cfrac {2.5\cdot 1}{253-1-{\cfrac {2\cdot 4}{759-{\cfrac {5\cdot 7}{1265-{\cfrac {8\cdot 10}{1771-\ddots }}}}}}}}.}$
#### Example 2
Pogson's ratio (1001/5 or 5100 ≈ 2.511886...), with x = 5, y = 75 and 2zy = 6325:
${\displaystyle {\sqrt[{5}]{100}}={\cfrac {5}{2}}+{\cfrac {3}{250+{\cfrac {12}{5+{\cfrac {18}{750+{\cfrac {27}{5+{\cfrac {33}{1250+{\cfrac {42}{5+\ddots }}}}}}}}}}}}={\cfrac {5}{2}}+{\cfrac {5\cdot 3}{1265-3-{\cfrac {12\cdot 18}{3795-{\cfrac {27\cdot 33}{6325-{\cfrac {42\cdot 48}{8855-\ddots }}}}}}}}.}$
#### Example 3
The twelfth root of two (21/12 or 122 ≈ 1.059463...), using "standard notation":
${\displaystyle {\sqrt[{12}]{2}}=1+{\cfrac {1}{12+{\cfrac {11}{2+{\cfrac {13}{36+{\cfrac {23}{2+{\cfrac {25}{60+{\cfrac {35}{2+{\cfrac {37}{84+{\cfrac {47}{2+\ddots }}}}}}}}}}}}}}}}=1+{\cfrac {2\cdot 1}{36-1-{\cfrac {11\cdot 13}{108-{\cfrac {23\cdot 25}{180-{\cfrac {35\cdot 37}{252-{\cfrac {47\cdot 49}{324-\ddots }}}}}}}}}}.}$
#### Example 4
Equal temperament's perfect fifth (27/12 or 1227 ≈ 1.498307...), with m = 7:
With "standard notation":
${\displaystyle {\sqrt[{12}]{2^{7}}}=1+{\cfrac {7}{12+{\cfrac {5}{2+{\cfrac {19}{36+{\cfrac {17}{2+{\cfrac {31}{60+{\cfrac {29}{2+{\cfrac {43}{84+{\cfrac {41}{2+\ddots }}}}}}}}}}}}}}}}=1+{\cfrac {2\cdot 7}{36-7-{\cfrac {5\cdot 19}{108-{\cfrac {17\cdot 31}{180-{\cfrac {29\cdot 43}{252-{\cfrac {41\cdot 55}{324-\ddots }}}}}}}}}}.}$
A rapid convergence with x = 3, y = −7153, and 2zy = 219 + 312:
${\displaystyle {\sqrt[{12}]{2^{7}}}={\cfrac {1}{2}}{\sqrt[{12}]{3^{12}-7153}}={\cfrac {3}{2}}-{\cfrac {0.5\cdot 7153}{4\cdot 3^{12}-{\cfrac {11\cdot 7153}{6-{\cfrac {13\cdot 7153}{12\cdot 3^{12}-{\cfrac {23\cdot 7153}{6-{\cfrac {25\cdot 7153}{20\cdot 3^{12}-{\cfrac {35\cdot 7153}{6-{\cfrac {37\cdot 7153}{28\cdot 3^{12}-{\cfrac {47\cdot 7153}{6-\ddots }}}}}}}}}}}}}}}}}$
${\displaystyle {\sqrt[{12}]{2^{7}}}={\cfrac {3}{2}}-{\cfrac {3\cdot 7153}{12(2^{19}+3^{12})+7153-{\cfrac {11\cdot 13\cdot 7153^{2}}{36(2^{19}+3^{12})-{\cfrac {23\cdot 25\cdot 7153^{2}}{60(2^{19}+3^{12})-{\cfrac {35\cdot 37\cdot 7153^{2}}{84(2^{19}+3^{12})-\ddots }}}}}}}}.}$
More details on this technique can be found in General Method for Extracting Roots using (Folded) Continued Fractions.
## Higher dimensions
Another meaning for generalized continued fraction is a generalization to higher dimensions. For example, there is a close relationship between the simple continued fraction in canonical form for the irrational real number α, and the way lattice points in two dimensions lie to either side of the line y = αx. Generalizing this idea, one might ask about something related to lattice points in three or more dimensions. One reason to study this area is to quantify the mathematical coincidence idea; for example, for monomials in several real numbers, take the logarithmic form and consider how small it can be. Another reason is to find a possible solution to Hermite's problem.
There have been numerous attempts to construct a generalized theory. Notable efforts in this direction were made by Felix Klein (the Klein polyhedron), Georges Poitou and George Szekeres.
## Notes
1. ^
2. ^
3. ^ Jones & Thron 1980, p. 20.
4. ^ Euclid (2008) - The Euclidean algorithm generates a continued fraction as a by-product.
5. ^
6. ^
7. ^ a b Euler 1748, Chapter 18.
8. ^ Havil 2012, pp. 104–105.
9. ^ Brahmagupta (598–670) was the first mathematician to make a systematic study of Pell's equation.
10. ^
11. ^
12. ^ Oskar Perron derives even more general extension and contraction formulas for continued fractions. See Perron (1977a), Perron (1977b).
13. ^
14. ^
15. ^ This intuitive interpretation is not rigorous because an infinite continued fraction is not a mapping: it is the limit of a sequence of mappings. This construction of an infinite continued fraction is roughly analogous to the construction of an irrational number as the limit of a Cauchy sequence of rational numbers.
16. ^ Because of analogies like this one, the theory of conformal mapping is sometimes described as "rubber sheet geometry".
17. ^ One approach to the convergence problem is to construct positive definite continued fractions, for which the denominators Bi are never zero.
18. ^ This periodic fraction of period one is discussed more fully in the article convergence problem.
19. ^ An alternative way to calculate log(x)
20. ^ Borwein, Crandall & Fee 2004, p. 278, 280.
21. ^
## References
• Angell, David (2010). "A family of continued fractions" (PDF). Journal of Number Theory. 130 (4). Elsevier: 904–911. doi:10.1016/j.jnt.2009.12.003.
• Angell, David (2021). Irrationality and Transcendence in Number Theory. Chapman and Hall/CRC. ISBN 9780367628376.
• Chrystal, George (1999). Algebra, an Elementary Text-book for the Higher Classes of Secondary Schools and for Colleges: Pt. 1. American Mathematical Society. p. 500. ISBN 0-8218-1649-7.
• Lorentzen, Lisa; Waadeland, Haakon (1992). Continued Fractions with Applications. Reading, MA: North Holland. ISBN 978-0-444-89265-2. (Covers primarily analytic theory and some arithmetic theory.)
• Perron, Oskar (1977a) [1954]. Die Lehre von den Kettenbrüchen. Vol. Band I: Elementare Kettenbrüche (3 ed.). Vieweg + Teubner Verlag. ISBN 9783519020219.
• Perron, Oskar (1977b) [1954]. Die Lehre von den Kettenbrüchen. Vol. Band II: Analytisch-funktionentheoretische Kettenbrüche (3 ed.). Vieweg + Teubner Verlag. ISBN 9783519020226.
• Porubský, Štefan (2008). "Basic definitions for continued fractions". Interactive Information Portal for Algorithmic Mathematics. Prague, Czech Republic: Institute of Computer Science of the Czech Academy of Sciences. Retrieved 2 May 2022.
• Szekeres, George (1970). "Multidimensional continued fractions". Ann. Univ. Sci. Budapest. Eötvös Sect. Math. 13: 113–140.
• Wall, Hubert Stanley (1967). Analytic Theory of Continued Fractions (Reprint ed.). Chelsea Pub Co. ISBN 0-8284-0207-8. (This reprint of the D. Van Nostrand edition of 1948 covers both history and analytic theory.)
• Wallis, John (1699). Opera mathematica [Mathematical Works].
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# If $\lim\limits_{x \rightarrow \infty} f'(x)^2 + f^3(x) = 0$ , show that $\lim \limits_ {x\rightarrow \infty} f(x) = 0$ [duplicate]
If $f : \mathbb{R} \rightarrow \mathbb{R}$ is a differentiable function and
$\lim\limits_{x \rightarrow \infty} f'(x)^2 + f^3(x) = 0$ , show that $\lim\limits_{x\rightarrow \infty} f(x) = 0$.
I really have no clue how to start, I tried things like MVT and using definition of derivatives but I really can't figure this out.
## marked as duplicate by Paramanand Singh, Community♦May 3 '15 at 9:36
• Is $f^3(x)$ , the third derivative or power? And does $f'(x)^2$ means here $[f'(x)]^2$ or $f'(x^2)$ , I just want to be sure. – Mann May 2 '15 at 9:42
• @Mann they are $(f(x))^3$ and $(f'(x))^2$ – Soham May 2 '15 at 9:43
• This is a duplicate. See math.stackexchange.com/q/881706/72031 – Paramanand Singh May 3 '15 at 9:21
Assume the claim is wrong, ie., there exists $a>0$ and a sequence $\xi_n\to \infty$ with $|f(\xi_n)|>a$. There exists $x_0$ with $|f'(x)^2+f(x)^3|<\frac{a^3}8$ for $x\ge x_0$. Assume $x_0<x_1<x_2$ and $f'(x_{1,2})=0$. Then $|f(x_{1,2})|<\frac a2$. If some $\xi_n$ is in $[x_1,x_2]$ then there is a local extremum at some $\eta\in(x_1,x_2)$ where $f'(\eta)=0$ and $|f(\eta)|>|f(\xi_n)|>a$. Since this contradicts $|f'(\eta)^2+f(\eta)^3|<\frac{a^3}8$, we conclude that no $\xi_n$ is between two critical points, hence there is at most one critical point $>x_0$. Hence we may assume (by increasing $x_0$ if necessary) that $f$ is strictly monotonic on $[x_0,\infty)$.
Assume $f$ is strictly increasing. Since $f(x)<\frac a2$ for $x>x_0$, $f$ is bounded from above, hence converges. Then $f'(x)^2$ converges as well and as it is never positive, $f'(x)$ converges, and must converge to $0$. Then also $f(x)\to 0$ and we are done.
Hence we may assume that $f$ is strictly decreasing. If $f$ is bounded from below, the same aregument as above applies and we are done again.
Hence $f$ is strictly decreasing and not bounded from below. As $f(x)\to-\infty$, we may assume wlog. that $f(x)^3<-\frac a{16}$ for $x\ge x_0$. Hence $$\tag1 f'(x)^2+\frac12f(x)^3>0$$
Let $b\in\mathbb R$ and $g(x)=-8(x-b)^{-2}$ for $x<b$. Then $g'(x)=16(x-b)^{-3}$ so that $$\tag2g'(x)^2+\frac12g(x)^3=0.$$ Now adjust $b$ so that $x_0<b$ and $g(x_0)=f(x_0)$. That is, we let $$b=x_0+\sqrt{-\frac8{f(x_0)}}$$ (Note that the radicand is positive). At any point $x\in[x_0,b)$ where $f(x)=g(x)$, we have $f'(x)<g'(x)<0$ by $(1)$ and $(2)$ so that in $f(\xi)<g(\xi)$ in an interval $(x,x+\epsilon)$. We conclude that $f(x)\le g(x)$ for all $x\in[x_0,b)$. But $g(x)\to -\infty$ as $x\to b^-$ whereas $f(x)$ is bounded on $[x_0,b]$. - Contradiction!
I'll try to solve it by studying the points at which $f$ is zero.
1) First, suppose that $f$ has infinitely many changes of signs, then note $x_n$ all its zero. Then there is $y_n \in ]x_n, x_{n+1}[$ such that $f'(y_n) = 0$ and $f$ is extremum at $y_n$. But then $$f'(y_n)^2 + f(y_n)^3 = f(y_n)^3 \to 0$$ and $|f(x)|\leqslant |f(y_n)|$ for every $x \in ]x_n, x_{n+1}[$, so we clearly have $f(x) \to 0$ in this case.
Suppose that $f(x)$ only vanishes for a finite number of $x$. Then, for $x$ far enough, $f$ is positive of negative.
2) If $f$ is positive, as $f'(x)^2 + f(x)^3 \to 0$, then it should be clear that $f(x) \to 0$.
3) and if $f$ is negative, I'm working on it, but this seems to be the difficult part of the problem.
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# Vampire load surprise - garage door openers
I've been on a tear of late to eliminate or minimize standby loads. This arises from data revealing that during spring and fall, when HVAC use is minimal, standby loads comprise 25%, 5 kwh / day of our home's total usage.
5 kwh / day exceeds the total usage of our entire kitchen - fridge, chest freezer, range, dishwasher, etc.
At national average electric rates of \$0.11 / kwh, each Watt of standby usage works out to a buck a year, so the sneaky little loads add up fast.
An early victory came in the form of learning that the starting battery trickle charger on our standby genny consumes 30-35 Watts. A \$40, 10 Watt PV panel from Amazon has allowed me to kill that load while still maintaining the genny's cranking battery.
We have 3 garage door openers, specifically Overhead Door Phantoms. They are quiet and have been relatively trouble-free. Imagine my shock at learning each uses 14.5 Watts while sitting and doing nothing.
Doing the math, the three (aptly named) Phantoms have cost us \$200 in standby power since we built the house in early 2008.
Thoughts, anyone?
Views: 6073
### Replies to This Discussion
Those are all good thoughts, including the remarks about not impairing the safety of the system.
People have proposed no-energy garage door openers where the weight of a car on the driveway and a hydraulic system would power the door to open and close. Is anyone capable of estimating whether the cost of such a system would be worth the energy savings? One complication is that safety regulations require a garage door to have an automatic reversal system and I wonder how a hydraulic system would be capable of automatic reversal.
One thing about energy efficiency is that there is no free lunch.
The energy needed to overcome the friction in the door mechanism must come from somewhere. In the case of the electric opener, the energy comes from the electricity. In the case of the hydraulic opener, it would presumably come from a car driving up onto a ramp which lowers slightly to displace some hydraulic fluid. In that latter case the energy comes from the gasoline. Given all the friction in hydraulic systems, I doubt that overall efficiency is better, though it does seem clever and must offer other benefits.
If the real advantage of the hydraulic system is that is has no "standby" losses when there is no car around, perhaps I can modify your proposal. Let's stick with the electrical system, but to save energy when it is not needed, let's embed a switch in the driveway that only makes the system operable when a car is placed on top of it. That way there is no standby losses except when you need it, i.e. when you place a car in front of the driveway.
I know it has some flaws like a person wanting to use the door opener.
Really, I think it is technically possible to make a garage door opener that only uses 0.1 watts when on standby like modern TVs, it is just that no one has done it yet.
I wouldn't sweat the mechanical energy consumed by the operator in action - maybe 200 Watts one minute per day (2 each 15 second opening and closing cycles). That works out to 1.2 kwh / year, a truly trivial cost.
Getting standby consumption to below 1 Watt is a laudable goal. All it would take is an Energy Star standard requiring same.
If you are able to get schematics from vendor.... and have the skills... you could split the receiver and motor into two different circuits... leave the bulky transformer for the motor. Possibly make some changes so when ever the door is "open" even partially - the line voltage to that bulky transformer is always there... power the receiver separately... use an ELK timer such that when receiver is triggered..line connection to the transformer is enabled for ten minutes. After which it disconnects - unless it is "partially open"
That would address safety problems ... and probably let you drop the vampire loads to under a 1W when idle....
Elk time delay relay... note -- these only handle smaller loads - the bulky transformer can not be directly connected to one of these relays. But there are other time delay relays that might also work (non Elk brand).
http://www.elkproducts.com/product-catalog/elk-960-delay-timer-module
I appreciate that analysis...sounds like a lot of work!
Conservatory Roof Insulation Systems
I'm pleased to see this category of phantoms given some attention as well as being able to provide useful data for discussion.
Kudos to Ron Brogle and Liftmaster for contributing to this discussion and offering products that address the problem. All my home energy audit reports contain information about phantom loads. I will add specific advice about garage door openers. My own trio comprise about 25% of my total phantom load, an outsized chunk given the total number of phantoms 5 fully wired people have in our home.
Moving on to other phantoms - I have refused to activate our doorbell's transformer. I'm tempted to order and post a placard reading "knock, remove headgear, wait for permission to enter", but I doubt I'd be allowed that.
Here is a non-"kosher" thought…..
It has been mentioned in a limited context, but if it is done on a larger scale, it solves the problem.
If your circumstances allow (shading, roof condition and space, \$\$ to do it), get solar PV, and if at all possible, make it an off-grid system with batteries so that (here's the non "kosher" part) you don't have to care about 4 watts here or 5 watts there. You get to keep your lifestyle, your electricity is free, and it does not hurt the environment. A "mini" system would be economically easy to pay for, and could power all the little vampire loads we all are talking about and obsessing over…. You'd have to do a little re-wiring of your circuit breakers to isolate the circuits that had these loads to the mini-solar PV Off-Grid system, but if you're committed enough to dealing with them, you could do it.
Of course, if you live in a townhouse like I do, you are S.O.L. with solar PV on your roof.
Ira
House is pretty much surrounded by trees.
Personally, I have solar, but I still hate vampire loads with a passion.
Wow, I'm impressed that this one is still going. I've been thinking about it some more. As an electrician I have tried to boil it down a bit. Almost without fail, vampire loads are a concession to convenience. Being able to simply push a button and have some appliance begin working relatively promptly is something we have come to expect. I could install a relatively simple 3-way switch setup for my garage door opener power that would entirely eliminate it's "at idle" consumption but would it lose codes or programming? Does my wife really want to get out of her car in a rain storm to flick a switch and push a button? Personally, I would be willing to get out of my car, unlock the garage door with a key, and manually lift it open, but my garage door doesn't even have a handle. Maybe I could invent that!
Nah - I no longer buy the convenience vs vampire argument...not with today's high efficiency switching power supplies and other power electronics.
Energy hogging phantoms happen via a combination of lack of awareness and oversight and corporations relentlessly pursuing the quarterly bottom line at the long term expense of their unknowing clients. Products that slide beneath regulatory and public information radar get away with crappy, low first cost designs. These garage door openers are a classic example.
Sometimes an evil monopoly or oligarchy is involved - look at how piggy many settop cable boxes are! Monopolistic cable companies resist spending the extra buck or two to do the right thing.
Back in the day, automobile manufacturers insisted that the most basic regulations, such as seat belts and safety glass would force them into bankruptcy, never mind fuel economy mandates. Then Asian manufacturers ate their lunch, and they changed their tune.
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NG-SPICE: Biasing the Default Transistor for Ideal Linear Voltage Gain, at 3V.
In recent days and weeks, I’ve been studying some of my own ideas, concerning the creative uses of the N-Channel, Enhancement-Mode, MOSFET. And to help me explore that subject, I’ve used An Open-Source Circuit Simulation Program called ‘NG-SPICE’. One big problem with this approach is the fact that the default transistor that the software assumes the power-user wants to use, is clearly not meant for Linear Voltage Amplification in the 100kHz-1.0Mhz frequency range, and with a 3V supply voltage. This transistor type is meant to be operated at higher voltages, and mainly, for digital uses. All the software is geared for Integrated Circuit Emphasis. But, I have looked at possible ways in which the default transistor could still be used under the conditions I’m more interested in. In theory, I could change the parameters of the transistor involved as much as I like, until I’ve made a high-speed, low-voltage transistor out of it. One problem with that is the fact that I give the software the geometry of the transistor on a chip, and the software then derives many of its assumed properties. I don’t know much about IC design, so I probably would not obtain the kind of transistor I’m looking for, if I tried to invent one.
So the question comes back, what is the best way to bias this one, arbitrary transistor-type, to act as a high-impedance amplifier under the conditions written above? And how much gain does it give me? The answer seems to be, that when connected as below, the best performance I can obtain is an Alpha of (-5.25):
What I’ve also learned is, that the bias voltage associated with this circuit, with respect to ground, is (+2.14V). With respect to the supply voltage, that is (-0.86V). 3.75μV of bias current would need to flow. This information would be useful if an attempt ever came along to implement This Idea.
(Edit 7/5/2019, 17h15 : )
Doubling (VGS – VT0) of M1 would have as effect, that IDS quadruples. It would also have as effect, that equal, small changes in Gate Voltage translate into doubled changes in IDS. But, if the increase in bias current was taken into account by the circuit designer, by putting a resistor of merely 100kΩ in series with M1, thereby achieving that the supply voltage was ideally halved again as a result, then this would finally have as effect to halve the net voltage gain at the Drain of M1.
It would also have as effect, to quarter output impedance, which would be desirable from the last of a series of these stages, ending in a realistic load of some kind.
(End of Edit, 7/5/2019, 17h15.)
The Model-Card of the transistor is linked below:
http://dirkmittler.homeip.net/text/NMOS1.mod.txt
To pursue the exact subject of the earlier posting, about Variable-Gain Amplifiers, I also felt that it would be necessary to add to the circuit the components, that would transform it into a variable attenuator. And the following schematic shows how I did that:
(Updated 7/16/2019, 7h50 … )
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The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A292383 Base-2 expansion of a(n) encodes the steps where numbers of the form 4k+3 are encountered when map x -> A252463(x) is iterated down to 1, starting from x=n. 25
0, 0, 1, 0, 2, 2, 5, 0, 0, 4, 11, 4, 22, 10, 5, 0, 44, 0, 89, 8, 8, 22, 179, 8, 0, 44, 1, 20, 358, 10, 717, 0, 20, 88, 11, 0, 1434, 178, 45, 16, 2868, 16, 5737, 44, 8, 358, 11475, 16, 0, 0, 89, 88, 22950, 2, 17, 40, 176, 716, 45901, 20, 91802, 1434, 17, 0, 40, 40, 183605, 176, 356, 22, 367211, 0, 734422, 2868, 1, 356, 22, 90, 1468845, 32, 0, 5736, 2937691, 32 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,5 LINKS Antti Karttunen, Table of n, a(n) for n = 1..2048 Index entries for sequences related to binary expansion of n Index entries for sequences computed from indices in prime factorization FORMULA a(1) = 0; for n > 1, a(n) = 2*a(A252463(n)) + [n ≡ 3 (mod 4)], where the last part of the formula is Iverson bracket, giving 1 only if n is of the form 4k+3, and 0 otherwise. a(n) = A292373(A292384(n)). a(n) = A292274(A243071(n)). Other identities. For n >= 1: a(2n) = 2*a(n). a(n) + A292385(n) = A243071(n). a(A163511(n)) = A292274(n). A000120(a(n)) = A292377(n). EXAMPLE For n = 3, the starting value is of the form 4k+3, after which follows A252463(3) = 2, and A252463(2) = 1, the end point of iteration, and neither 2 nor 1 is of the form 4k+3, thus a(3) = 1*(2^0) + 0*(2^1) + 0*(2^2) = 1. For n = 5, the starting value is not of the form 4k+3, after which follows A252463(5) = 3 (which is), continuing as before as 3 -> 2 -> 1, thus a(5) = 0*(2^0) + 1*(2^1) + 0*(2^2) + 0*(2^3) = 2. For n = 10, the starting value is not of the form 4k+3, after which follows A252463(10) = 5 (also not 4k+3), and then A252463(5) = 3 (which is), continuing as before as 3 -> 2 -> 1, thus a(10) = 0*(2^0) + + 0*(2^1) + 1*(2^2) + 0*(2^3) + 0*(2^4) = 4. MATHEMATICA Table[FromDigits[Reverse@ NestWhileList[Function[k, Which[k == 1, 1, EvenQ@ k, k/2, True, Times @@ Power[Which[# == 1, 1, # == 2, 1, True, NextPrime[#, -1]] & /@ First@ #, Last@ #] &@ Transpose@ FactorInteger@ k]], n, # > 1 &] /. k_ /; IntegerQ@ k :> If[Mod[k, 4] == 3, 1, 0], 2], {n, 84}] (* Michael De Vlieger, Sep 21 2017 *) PROG (PARI) A064989(n) = {my(f); f = factor(n); if((n>1 && f[1, 1]==2), f[1, 2] = 0); for (i=1, #f~, f[i, 1] = precprime(f[i, 1]-1)); factorback(f)}; A252463(n) = if(!(n%2), n/2, A064989(n)); A292383(n) = if(1==n, 0, (if(3==(n%4), 1, 0)+(2*A292383(A252463(n))))); (Scheme) (define (A292383 n) (A292373 (A292384 n))) CROSSREFS Cf. A163511, A243071, A292274, A292373, A292377, A292380, A292381, A292382, A292384, A292385. Sequence in context: A012858 A366756 A363842 * A332896 A100247 A194123 Adjacent sequences: A292380 A292381 A292382 * A292384 A292385 A292386 KEYWORD nonn,base AUTHOR Antti Karttunen, Sep 15 2017 STATUS approved
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Last modified July 16 05:19 EDT 2024. Contains 374343 sequences. (Running on oeis4.)
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https://chem.libretexts.org/Textbook_Maps/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/02Atoms%2C_Molecules%2C_and_Chemical_Reactions/2.11%3A_Formulas_and_Composition/Lecture_Demonstrations
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# Lecture Demonstrations
[ "article:topic", "authorname:chemprime", "showtoc:no" ]
## Formula of an Iron Oxide: combustion of pyrophoric iron or steel wool
It is difficult to do a quantitative determination of a formula as a lecture demonstration. This demonstration is at best semi-quantitative, but it's interesting and suggests methods that might be more quantitative, and points to variables that must be controlled in a quantitative experiment.
A. Steel Wool: Weigh ~0.5 g of 000 or 0000 steel wool in an evaporating dish. With tongs, dip it in petroleum ether or hexane to remove surface oil which is present to prevent oxidation. Shake off most of the hexane, then hold the steel wool over the evaporating dish and ignite it with a Bunsen burner. Catch the product in the evaporating dish and reweigh. Calculate the formula from data, or from "optimal" data supplied by calculation. Discuss: (1) The product may be a mixture of FeO, Fe2O3, and Fe3O4. (2) some of the product may not have been collected. (3) the steel wool isn't really pure Fe.
B. Pyrophoric Iron: Prepare ~0.5 g of pyrophoric iron by decomposition of FeC2O4 under methane [1] in a weighed 15 x 150 mm test tube with a two holed rubber stopper to allow inlet of methane product gases. Insert a long pipette in one hole, and ignite the product gases after allowing time for methane to fill the apparatus. Heat the iron oxalate gently so that it turns completly black, but avoid further heating. Weigh the tube plus product. Weigh an evaporating dish, and pour the product into it, then reweigh. Reweigh the empty tube. From the masses of the iron and product, calculate it's formula.
Notes:
1. It's more fun to dump the product through several feet of air and watch the combustion reaction, but product will be lost. This may be inconsequential, because the calculations may be done with "optimal" data, noting the difficulties of making the demonstration as presented quantitative.
2. The product of the reaction is almost certainly not pure. It probably contains several oxides, as well as sintered (nonpyrophoric) iron that results from overheating the product.
## References
1. J. Chem. Educ., 1931, 8 (2), p 303
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https://btechgeeks.com/java-program-to-convert-centimeter-to-nautical-mile-and-nautical-mile-to-centimeter/
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Java Program to Convert Centimeter to Nautical Mile and Nautical Mile to Centimeter
In the previous article, we have discussed about Java Program to Convert Centimeter to Millimeter and Millimeter to Centimeter
In this article we will see how to convert Centimeter to Nautical Mile and Nautical Mile to Centimeter by using Java programming language.
Java Program to Convert Centimeter to Nautical Mile and Nautical Mile to Centimeter
Before jumping into the program let’s know the relationship between Centimeter and Nautical Mile and how we can convert Centimeter to Nautical Mile and vice versa.
Generally Centimeter and Nautical Mile are used as unit in case of distance measurement.
1 Centimeter = 6.2137e-6 Nautical Mile
1 Nautical Mile = 160934 Centimeter
Formula to convert Centimeter to Nautical Mile.
Nautical Mile = Centimeter * 5.3996 * .0000001
Formula to convert Nautical Mile to Centimeter.
Centimeter = Nautical Mile * 185200
Let’s see different ways to convert Centimeter to Nautical Mile and Nautical Mile to Centimeter.
Method-1: Java Program to Convert Centimeter to Nautical Mile and Nautical Mile to Centimeter By Using Static Input Value
Approach:
• Declare Centimeter and Nautical Mile value.
• Then convert Centimeter to Nautical Mile and Nautical Mile to Centimeter by using the formula.
• Print result.
Program:
import java.util.*;
public class Main
{
public static void main(String args[])
{
//Scanner class object created
Scanner sc=new Scanner(System.in);
//centimeter value declared
double centimeter = 10000;
//nautical Mile value declared
double nauticalMile = 1.5;
//converting centimeter to Nautical Mile
double nm = centimeter * 5.3996 * .0000001;
//converting Nautical Mile to centimeter
double cm = nauticalMile * 185200;
//printing result
System.out.println("Value of "+centimeter+" centimeter in nautical mile: "+ nm);
System.out.println("Value of "+nauticalMile+" nautical mile in centimeter: "+ cm);
}
}
Output:
Value of 10000.0 centimeter in nautical mile: 0.0053996
Value of 1.5 nautical mile in centimeter: 277800.0
Method-2: Java Program to Convert Centimeter to Nautical Mile and Nautical Mile to Centimeter By Using User Input Value
Approach:
• Take user input of Centimeter and Nautical Mile value.
• Then convert Centimeter to Nautical Mile and Nautical Mile to Centimeter by using the formula.
• Print result.
Program:
import java.util.*;
public class Main
{
public static void main(String args[])
{
//Scanner class object created
Scanner sc=new Scanner(System.in);
//Taking the value input of double variable centimeter
System.out.println("Enter value of centimeter: ");
double centimeter = sc.nextDouble();
//Taking the value input of double variable nauticalMile
System.out.println("Enter value of nautical mile: ");
double nauticalMile = sc.nextDouble();
//converting centimeter to Nautical Mile
double nm = centimeter * 5.3996 * .0000001;
//converting Nautical Mile to centimeter
double cm = nauticalMile * 185200;
//printing result
System.out.println("Value of "+centimeter+" centimeter in nautical mile: "+ nm);
System.out.println("Value of "+nauticalMile+" nautical mile in centimeter: "+ cm);
}
}
Output:
Enter value of centimeter:
3.5
Enter value of nautical mile:
4.5
Value of 3.5 centimeter in nautical mile: 1.88986E-6
Value of 4.5 nautical mile in centimeter: 833400.0
Method-3: Java Program to Convert Centimeter to Nautical Mile and Nautical Mile to Centimeter By Using User Defined Method
Approach:
• Take user input of Centimeter and Nautical Mile value.
• Call a user defined method by passing Centimeter and Nautical Mile value as parameter.
• Inside method convert Centimeter to Nautical Mile and Nautical Mile to Centimeter by using the formula.
• Print result.
Program:
import java.util.*;
public class Main
{
public static void main(String args[])
{
//Scanner class object created
Scanner sc=new Scanner(System.in);
//Taking the value input of double variable centimeter
System.out.println("Enter value of centimeter: ");
double centimeter = sc.nextDouble();
//Taking the value input of double variable nauticalMile
System.out.println("Enter value of nautical mile: ");
double nauticalMile = sc.nextDouble();
//calling user defined method convert()
convert(centimeter, nauticalMile);
}
//convert() method to convert centimeter to Nautical Mile and vice versa
public static void convert(double centimeter, double nauticalMile)
{
//converting centimeter to Nautical Mile
double nm = centimeter * 5.3996 * .0000001;
//converting Nautical Mile to centimeter
double cm = nauticalMile * 185200;
//printing result
System.out.println("Value of "+centimeter+" centimeter in nautical mile: "+ nm);
System.out.println("Value of "+nauticalMile+" nautical mile in centimeter: "+ cm);
}
}
Output:
Enter value of centimeter:
100000
Enter value of nautical mile:
2
Value of 100000.0 centimeter in nautical mile: 0.053995999999999995
Value of 2.0 nautical mile in centimeter: 370400.0
Want to excel in java coding? Practice with these Java Programs examples with output and write any kind of easy or difficult programs in the java language.
Related Java Programs:
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https://cs.stackexchange.com/questions/84903/how-can-the-intersection-of-cfls-and-regs-be-cfl-if-reg-is-a-proper-subset-of-cf/84904
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# How can the intersection of CFLs and REGs be CFL if REG is a proper subset of CFL?
Intersection of CFL and regular is always CFL. But according to Chomsky hierarchy diagram, regular languages lie completely inside CFL. So, as regular set is completely inside CFL set, their intersection should be regular right ?
Am I interpreting Chomsky diagram wrongly ?
No, you are misinterpreting stuff.
The set of languages (DFL, regular, ...) and the set of strings in a language are independent.
It is entirely possible that a regular language contains strings that are not inside a Context free language.
The trivial example is where the regular language is all strings $\Sigma^*$ and the CFL is any other language. The intersection between them is the original CFL.
• In this diagram encrypted-tbn3.gstatic.com/…. Regular language is subset of CFL. So according to subset property of sets , any set's intersection with it's subset is the subset itself right (which is regular) ? – Rajesh R Dec 4 '17 at 15:06
• But each language is a set of strings so when talking about the intersection between 2 languages you are talking about intersecting the set of strings. That is independent from the set of languages. – ratchet freak Dec 4 '17 at 15:18
• Ok so Chomsky hierarchy says intersection of set of all CFL's and set of all Regular languages is set of regular languages right? – Rajesh R Dec 4 '17 at 15:50
You are confusing two different statements.
1. $\mathrm{CFL} \cap \mathrm{REG} = \mathrm{REG}$ or, equivalently,
for all $L_1 \in \mathrm{REG}$ : $L_1 \in \mathrm{CFL}$.
2. For all $L_1 \in \mathrm{REG}$, $L_2 \in \mathrm{CFL}$ : $L_1 \cap L_2 \in \mathrm{CFL}$.
The first makes a statement on two sets of languages (aka language classes), the second makes one over (many) pairs of sets of strings (aka languages). So in mixing the two, you are creating a type error.
Also, you seem to suggest that if a language is context-free, it's not regular. That is false; we have $\mathrm{REG} \subsetneq \mathrm{CFL}$ (cf. 1.), so showing a language is CFL (using 2.) does not preclude it from being regular. Specifically, it's easy to find examples of a non-regular context-free and a regular language whose intersection is regular.
• Very clear distinction between the two statements, but which one is correct? (I know they're not mutually exclusive, but if I understood the other answer correctly, only one of them happens to be true.) – Wildcard Dec 4 '17 at 16:17
• Both are true. The false one is $\forall L_1 \in REG, L_2 \in CFL: L_1 \cap L_2 \in REG$. – reinierpost Dec 4 '17 at 16:45
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