text:
# Interpreting Robinson arithmetic in a very weak set theory

It is known that adjunctive set theory interprets Robinson arithmetic, and that extensionality is not needed for that. (Montagna and Mancini, "A minimal predicative set theory", Notre Dame Journal of Formal Logic, 1994).

I wonder whether it is known if the following weak set theory (without extensionality) also interprets Robinson arithmetic. Axioms:

1. $\exists y\forall x(x\notin y)$

2. For every n, $\forall x_1\ldots\forall x_n\exists y\forall z(z\in y\leftrightarrow z=x_1 \vee \ldots\vee z=x_n)$

This theory was introduced by Vaught, and it does not interpret Robinson’s arithmetic. See Visser [1] for a thorough discussion of related theories; Vaught’s theory is denoted VS in the paper. (Note that the axioms are stated more concisely there: axiom 1 is a special case of axiom 2 for $n=0$.)

That VS does not interpret Robinson’s arithmetic Q follows by the following argument:

• Q is finitely axiomatized. Thus, if it were interpretable in VS, it would be interpretable in its finite fragment.

• VS is locally interpretable in the theory of nonsurjective pairing (using the obvious construction of $n$-tuples from pairs).

• However, said theory of pairing is known to have decidable extensions, hence it does not interpret Q (or indeed, all of VS, which is also an essentially undecidable theory). Cf. Theorem 2 in [1,§3.2].

Reference:

[1] Albert Visser, Pairs, sets and sequences in first-order theories, Archive for Mathematical Logic 47 (2008), no. 4, pp. 299–326.

• By bizarre coincidence, I was just looking at this very paper following an internet search for literature on a different matter (Craig's Trick, in fact). – Adam Epstein Apr 15 '16 at 21:49


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# Difference between revisions of "Heat Transfer Equipment"

Title: Heat Transfer Equipment

Author: David Chen

Steward: Fengqi You

Date Presented: January 13, 2014 /Date Revised: January 14, 2014

# Basic Concept

There are three mechanisms of heat transfer: conduction, convection, and radiation. In most heat exchangers, convection will be the dominant mechanism. Conduction and radiation will generally be negligible in large heat exchangers, but radiation will be important in fired heaters.

Heat transfer across a surface by convection is given by the equation: $Q=UAF\Delta T_m$

Where: $Q$ = heat transferred per unit time (energy/time)

$U$ = overall heat-transfer coefficient (energy/time-area-temperature)

$A$ = heat-exchange area (area)

$F$ = correction factor (unitless)

$\Delta T _m$ = log mean temperature difference or the temperature driving force, (temperature)

$\Delta T _m$ is the driving force for a pure countercurrent contact pattern in a tubular system. the correction factor $F$ is used because most heat exchangers do not implement true countercurrent contact. Design equations for heat exchangers will use generally use some form of this equation with the appropriate modifications to account for different configurations and approximations.

# Heat Exchangers

Process equipment and streams will need to be heated or cooled. One way to reduce consumption of utilities is to exchange heat between these streams. For example, if a product stream requires cooling, the excess heat can be used to preheat a feed stream that requires heating by using an appropriate heat exchanger. Here the costs of an additional heating and additional cooling unit are eliminated, and replaced by the cost of a heat exchanger. The associated utility requirements of the additional units are also eliminated and replaced by the utility requirements of operating the heat exchanger.

Traditionally shell-and-tube heat exchangers in chemical industry. Standards and codes by TEMA (Tubular Exchanger Manufacturers Association) and ASME (American Society of Mechanical Engineers)

## Factors to Consider

The heat exchanger must meet certain physical requirements both to carry out the specified heat exchange and operate properly as part of a system. As such, it must provide the correct area for the specified heat exchange while maintaining a reasonable pressure drop, contain the pressure of the streams, prevent leaks between the tubes and the shell, account or thermal expansion, allow for cleaning of fouling deposits, allow for thermal expansion, and phase changes in certain applications.

It should be noted that heat exchangers rarely operate at the exact conditions specified for design. Because performance will decrease with fouling, a heat exchanger may be initially overdesigned, and after some amount of fouling, underdesigned, at which point cleaning should take place. Depending on the performance of a heat exchanger at any given time, downstream processes may be affected.

### Thermal and hydraulic requirements

A certain amount of energy needs to be exchanged, and the pressure drop across the heat exchanger must be accounted for in the context of the process. It may be acceptable, or may need to be maintained using pumps. The key tradeoff in heat exchangers is the heat exchanged vs. pressure drop. More surface area can always be added, but the pressure drop may become unacceptable for downstream processes.

### Material compatibility

Heat exchangers must be able to maintain acceptable performance through prolonged contact in the materials with which heat is being exchanged. While regular maintenance and cleaning is unavoidable, appropriate construction materials should be chosen that are not prone to excessive corrosion or fouling.

### Operational Maintenance

Fouling on the heat transfer surface will reduce the overall heat transfer coefficient and efficiency. To maintain economical operation, the fouling will need to be periodically removed. Depending on the type of heat exchanger used, it may be disassembled and cleaned, or it will have to remain intact and be cleaned chemically, which can involve hazardous materials.

### Environmental, health, and safety considerations and regulations

Hazards and toxicity of the streams involved must be guarded against, and safety codes must be met. Plans must be prepared to deal with leaks or failures of the heat exchanger that will minimize adverse effects.

### Availability and cost

A with any project, the ability of the heat exchanger construction to fall within the deadlines and costs is key to the economics.

## Types of Heat Exchangers

### Double Pipe

The simplest heat exchanger. It is essentially a “tube-within a tube.” Only used when heat transfer requirement is small, or both fluid streams are of high pressure. One way to improve heat transfer is to add fins on the outside of the inner tube. This is used to improve the heat transfer of a fluid with a low heat transfer coefficient such as a viscous liquid or a gas, which is passed on the outer side. A more common configuration of the double pipe is with modular U-tubes in a “hairpin” configuration to conserve space.

### Shell-and-Tube

A bundle of tubes is passed through a shell. Heat exchange occurs between the fluid inside the tubes (tube-side) and the fluid outside of the tubes but within the shell (shell-side). Baffles are often used to direct the flow of the shell-side fluids as well as to support the tube bundle. TEMA has set standards for construction of shell-and-tube heat exchangers as well as nomenclature for the front heads (4 types), shells (6 types), and rear heads (8 types).

Three commonly used types of shell-and-tube heat exchangers are: the fixed-tube, U-tube, and floating-head type. Each has its own set of advantages and disadvantages. For example, the fixed-tube type exchanger has the simplest construction, which reduces construction cost and difficulty of routine cleaning. However, it may not be the most efficient. Meanwhile, the U-tube configuration allows more surface area inside the exchanger, but is more difficult to clean. Additionally, the flow pattern is not truly a counterflow pattern unless a longitudinal baffle is used (Type F shell). The U-tube configuration is more able to absorb the stress of thermal expansion than the fixed tube exchanger. Finally, the floating¬-head exchanger can move within the shell, which allows the exchanger to handle higher temperatures and pressures. However, it is more complex than the former two types, and thus construction can be approximately 25 percent higher for a unit of similar area.

The tube pitch describes the center-to-center distance for the arrangement of the tube bundle, and can be square or triangular. A larger pitch leaves more space between tubes and allows for easier cleaning, but this comes at the cost of a lower shell-side heat transfer coefficient and the need for a larger shell. A smaller pitch allows more tubes to be fit inside a given shell

Triangular pitch allows for tighter packing of tubes in a shell. If shell-side fouling is a problem, square pitch should be used for easier cleaning. In horizontal boiling heat exchangers, square pitch should also be used to prevent vapor blanketing. Square pitch should also be used when there is a lower shell-side pressure drop.

The baffles lead to increased pressure drop of the shell-side fluid. However, it also improves the mixing of the fluid and increases turbulence, which leads to improved heat transfer. Again, the tradeoff between improved heat transfer and pressures drop is illustrated. The most common type of baffle is the segmental baffle. Other baffles include the disk-and-doughnut, orifice, no-tube-in-window, and triple segmental baffle.

### Scraped-Surface

Crystallization systems and heat transfer involving viscous fluids, fouling may occur rapidly enough to make routine cleaning impractical. In this case, a rotating blade moves over the surface, liberating the deposited material from the surface and allowing it to exit at the bottom of the exchanger. Additionally, the motion of the blade shears the deposited product close to the wall, which results in high local heat transfer rates. Scraped-surface exchangers are generally not considered unless liquid viscosity exceeds 1 Pa•s or fouling is rapid.

Plate heat exchangers consist of a stack of corrugated plates. The corrugation of the plates improves rigidity, controls spacing of the plates, and increases the heat-transfer area compared to a flat plate.The hot and cold streams flow countercurrently through the alternating spaces created by the plates. The modular design allows easy addition of more heat transfer area, but at a cost of increased pressure drop. Edges can be sealed with gaskets for lower pressures. Cleaning is relatively simple because the configuration can be disassembled and cleaned. For operating at higher pressures the edges can be welded. Welded-plate heat exchangers typically do not operate past 3 MPa. However, in welding the plates together, the convenience of disassembling and cleaning the modular plates is lost, and cleaning must be done chemically. Additionally, the plates are usually larger than those of the gasketed-plate exchanger to reduce the amount of welding necessary.

### Spiral Plate and Tube

Spiral plates are coiled to create alternating passages for the fluids. The cold fluid enters at the periphery and flows towards the center, while the hot fluid enters at the center and flows outward. Introducing the cold fluid at the periphery reduces or eliminates the need for external insulation. These heat exchangers are used for small capacities with viscous, fouling, and corrosive fluids. The end plates can be removed for cleaning the shell. However, the configuration of the tube makes the spiral plates difficult to clean.

### Plate-Fin

Plate-fin heat exchangers consist of layers of corrugated metallic sheets (fins) between flat plates to form the flow passages. The plates are sealed with metal bars on the side. Plate-fin heat exchangers can be 9 times as compact as a shell-and-tube heat exchanger, and weighs less. Can withstand design pressures up to 6 MPa in the temperature range of -270 and 800 ⁰C. Many different configurations of plate-fin exchangers can be used.

### Gas-to-Gas

Gas-to-gas heat exchangers are primarily used to recover energy from combustion gases to preheat furnace air. The plain tube gas-to-gas heat exchanger is a simple countercurrent exchange through a tube bank, either in a single-pass (cross-flow) or multipass configuration[1]

# Boiling and Condensing Heat Transfer

Heat transfer involving boiling liquids or condensing vapor is different from heat transfer involving constant fluid phases. Heat transfer operations that involve phase changes are typically carried out in separate units in order to account for the different physical properties of the phases.

## Boiling Heat Transfer

Boilers transfer heat to boil liquids.

### Boiling Heat Transfer Coefficient

Boiling liquids have different heat transfer regimes. I: Heat transfer with natural convection II: Heat transfer with bubbling agitation III: Heat transfer with nucleate boiling with unstable film IV: Heat transfer with stable film boiling V: Radiant Heat Transfer

The heat transfer coefficient of a boiling liquid begins to decrease with the onset of film boiling. One method of improving the overall heat transfer coefficient is to use high-flux tubing, which utilizes a porous coating inside of tubes. It can improve boiling performance up to 10 times over a bare tube, and overall performance 2 to 5 times over a bare tube.

### Kettle Reboilers

Kettle reboilers are often used as a steam generator. Pool boiling is used in kettle reboilers. In pool boiling, agitation occurs through bubbling and natural convection. The vapor-liquid separator is built-in and allows for blowdown. The weir helps to maintain the liquid level above the tube bundle. It also helps to prevent the entering bubble point liquid (distillation bottoms) from mixing with the residual reboiled liquid (exiting from the bottom). Kettle reboilers are more expensive than horizontal thermosiphons fabricated for a comparable duty.

### Thermosiphon Reboilers

Thermosiphon reboilers use flow boiling. In flow boiling, agitation occurs through bubbling and forced convection at high velocities. They can be located at a height below the column sump. This allows the static head of the sump to force the column bottoms into the reboiler, which is designed for about 25 to 33% vaporization per pass. Thermosiphon reboilers can be vertical (tube-side) or horizontal (shell-side flow). Horizontal thermosiphon reboilers tend to be cheaper than vertical thermosiphons, but vertical thermosiphons are better at handling dirty fluids.

Flow is critical in a thermosiphon reboiler, and the different flow regimes will impact operation. The different flow regimes are caused by the increasing vapor/liquid ratio as the fluid passes through the thermosiphon.

Flow regimes in order of increasing vapor/liquid ratio: nonboiling < subcooled boiling < bubble flow < slug flow < churn flow < annular and mist flow < mist flow

Slug flow is unavoidable, but should be minimized. It causes noise and vibration in the thermosiphon. Annular and mist flow are also undesirable. Annular flow can be avoided by designing for less than 33% vaporization, and mist flow can be altogether avoided in well-designed reboilers.

### Stab-in Reboilers

Stab-in reboilers are essentially the heat exchange tube bundle fitted inside a sump. In this case, the sump behaves in a similar capacity to a kettle reboiler, and mechanism of pooling is again pool boiling. It is important that there is enough space in the sump for good level control and to contain the entire tube bundle.

## Condensing Heat Transfer

Condensing heat transfer is important because many heat transfer applications will condense the steam used as the hot stream, and often gaseous products are needed in liquid form. There are two types of condenser: total condensers and partial condensers. Both total and partial condensers need to account for the accumulation of noncondensable vapor. This can be done by venting through the top of the exchanger. The tube bundle arrangement affects overall heat transfer as well. Falling condensate from higher to lower tubes increases local turbulence and thus the heat transfer coefficient. However, condensate that collects on a tube and has yet to drain prevents the cooled heat transfer surface from contacting the vapor.

### Total Condenser

In a total condenser, all the vapor that enters is condensed as a film on the heat transfer surface. The heat transfer coefficient is determined only by the thermal resistances.

### Partial Condenser

In a partial condenser, not all of the vapor that enters is condensed. This allows for vapor-liquid separation. However, there will be a a vapor film with a higher concentration of high-boiling products. The heat transfer coefficient is not determined solely by the thermal resistance, but also in part by the mass transfer resistance as well. The condensable component of vapor will have to diffuse through the noncondensable component. The resistance to diffusion leads to a much lower overall heat transfer coefficient. Additionally, it complicates the calculation of heat transfer.

# Heaters and Coolers

It is not always possible to couple process streams for heat-exchange. It is likely that separate heaters and coolers will have to be used in addition to a heat-exchange network for a process.

## Heaters

Operating temperatures can be classified into three ranges: low (< 120C), medium (120-250C), and high (>250C). low temperature-range heaters generally tend to use condensate or steam, medium temperature-range heater tend to steam, and high temperature-range heaters tend to use fired heaters or hot oil loops (<400C)

Temperature and required heat load are important factors in selecting a heater.

### Fired Heaters

Fired heaters are used for heating up to high temperatures. They are able to reach these high temperatures because they generate energy by combustion of natural gas, fuel oil, or process off-gas. Fox example, they are used to generate heat for hot oil loops or steam generators (boilers). Hot oil can be used up to 600F (35C), and fired heaters provide a way to reach those temperatures. High pressure steam, used for utilities, is approximately 480F (250C).

Because fired heaters carry out combustion, factors such as pollutant emissions and excess air feed need to be accounted for in addition to the required heat duty. Reducing emissions can incur substantial additional costs.

Common types of fired heaters include cabin heaters, U-tube heaters, and vertical cylindrical heaters.

## Coolers

If heat cannot be recovered directly through exchange with another process stream, high-temperature heat can still be recovered by generating steam or preheating the feed water to the boiler. This way at least some energy is recovered.

Water and air are used for most cooling applications because they are available in large quantities at minimal cost.

### Refrigeration

Refrigeration is used for very low temperatures (<40C)

# Example Problem

A stream of hot fluid at $T_{h, in}=120 ^oC$ flowing at $\dot{m_h}=10 \frac{kg}{s}$ needs to be cooled to $T_{c,out}=60 ^oC$. Its heat capacity is $C_{p,h}2=\frac{kJ}{kg\cdot^oC}$. A cold stream at $\dot{m_c}=8 \frac{kg}{s}$ and $T_{c,in}=50 ^oC$ is available to cool it. The heat capacity of the old stream is $C_{p,c}=3 \frac{kJ}{kg\cdot^oC}$. If they are contacted in a true counter current pattern in a double tube heat exchanger with overall heat-transfer coefficient $U = 2\frac{kJ}{m^2\cdot^oC\cdot s}$, what is the outlet temperature of the cold stream, and what is the area A needed for this heat exchange rate?

# Example Problem Solution

We are trying to solve for two things in this problem: the outlet temperature of the cold stream $T_{c,out}$ and the heat-exchange area $A$. In order to find the outlet temperature $T_{c,out}$, we need to find the total heat transfer rate, which is equal to the amount of energy lost by the hot stream. The amount of energy gained by the cold stream is the opposite of this value. The absolute value of either will give the heat duty $Q$We can then relate the energy change to the temperature change using the heat capacity and mass flow rate.

$Q = -\dot{m_h}C_{p,h}(T_{h, out}-T_{in})=\dot{m_c}C_{p,c}(T_{c,out}-T_{c,in})$

Plugging in values $1200 \frac{kJ}{s}= -(10 \frac{kg}{s}) (2 \frac{kJ}{kg\cdot^oC}) (60^oC-120^oC)=(8 \frac{kg}{s}) (3 \frac{kJ}{kg\cdot^oC}) (T_{c,out}-50^oC)$

We can see that $T_{c,out}=100 ^oC$

Once $T_{c,out}$ is solved, we can calculate the log mean temperature driving force $\Delta{T_m}$.

$\Delta{T_m}=\frac{(T_{h,out}-T_{c,in})-(T_{h,in}-T_{c,out})}{ln\frac{T_{h,out}-T_{c,in}}{T_{h,in}-T_{c,out}}}=\frac{(60 ^oC-50 ^oC)-(120 ^oC-100 ^oC)}{ln\frac{60 ^oC-50 ^oC}{120 ^oC-60 ^oC}}=14.427$

Plugging in our values for $Q$, $U$, $\Delta{T_m}$, we can obtain the heat-exchange area $A$.

$Q=1200 \frac{kJ}{s}=UA\Delta{T_m}=(2 \frac{kJ}{m^2\cdot^oC\cdot{s}})(A)(14.43)$

From this we are able to obtain $A=41.58 m^2$

# References

[1] Towler, G.P. and Sinnot, R. (2012). Chemical Engineering Design: Principles, Practice and Economics of Plant and Process Design.Elsevier.

[2] Biegler, L.T., Grossmann, L.E., and Westerberg, A.W. (1997). Systematic Methods of Chemical Process Design. Upper Saddle River: Prentice-Hall.

[3] Peters, M.S. and Timmerhaus, K.D. (2003). Plant Design and Economics for Chemical Engineers, 5th Edition. New York: McGraw-Hill.
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Synopsis

# X-Ray Absorption Spectroscopy on a Tabletop

Physics 11, s27
A laser-based setup can be used to perform x-ray spectroscopy with a precision rivaling that of experiments at large-scale synchrotron facilities.

X-ray absorption spectroscopy (XAS) is a versatile tool for chemistry, biology, and materials science. By probing how x rays are absorbed from core electrons of atoms in a sample, the technique can reveal the local structure around selected atoms. XAS measurements are typically performed using the well-controlled x-ray beams of large-scale synchrotron facilities. A new study, however, shows that a tabletop x-ray source driven by lasers can be used to acquire XAS spectra as accurate as those obtained at synchrotrons. The scheme could dramatically expand the availability of XAS.

The setup used by Dimitar Popmintchev of JILA and the University of Colorado, both in Boulder, and his co-workers generates x rays through high-harmonic generation (HHG), in which an infrared laser beam focused on a gas sample creates radiation at much higher frequencies. Previously demonstrated HHG sources haven’t been sufficiently bright at the frequencies needed for most XAS work. But the authors recently developed a source that emits attosecond pulses in a laser-like beam with record brightness at photon energies ranging from 100 to 1600 eV.

With this source, the team demonstrated various types of XAS. They showed that the setup can probe the frequency range containing the resonances of many elements relevant to biochemical and condensed-matter studies; as examples, they acquired accurate spectra for carbon, scandium, and iron. They also measured absorption over a broad frequency range in the “water window,” which is important for probing biological specimens, since water is transparent at these frequencies but carbon isn’t. What’s more, the source’s short pulses could be used in the future to carry out time-resolved XAS experiments that probe ultrafast processes in materials and biomolecules.

This research is published in Physical Review Letters.

–Matteo Rini

Matteo Rini is the Deputy Editor of Physics.

## Related Articles

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Researchers have created a disk-shaped supersolid, an achievement that could provide new routes to exploring previously unseen states of matter. Read More »

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Condensed Matter Physics

### Elusive Superconducting Superhydride Synthesized

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## Saturday, October 27, 2018

### [davcwnxt] Standard gravitational parameter from units

The 'units' program does not have built in the standard gravitational parameter, mu=G*M, for various solar system bodies.  However, we can get it from gauss_k, the Gaussian gravitational constant.  Here is GM for the sun:

You have: gauss_k^2 au^3 day^-2
You want:
Definition: 1.3271244e+20 m^3 / s^2

Both au and day are exact.  gauss_k is known to 10 significant digits.  In contrast, if we were to multiply G and solarmass directly, the universal gravitational constant suffers from precision of only 4 significant digits, so the product would too.

Here is GM for the Earth and Jupiter:

You have: gauss_k^2 au^3 day^-2 solarmass^-1 earthmass
You want:
Definition: 3.9860044e+14 m^3 / s^2

You have: gauss_k^2 au^3 day^-2 solarmass^-1 jupitermass
You want:
Definition: 1.2671277e+17 m^3 / s^2

Although solarmass and planetary masses aren't known to high precision, their ratios are.  Planetary masses are defined in 'units' as ratios to solarmass, so the low precision of solarmass cancels out in the expressions above.

You have: jupitermass
You want:
Definition: solarmass / 1047.3486 = 1.8991766e+27 kg

Here is the escape velocity of earth calculated using the above standard gravitational parameter for earth:

You have: sqrt(2 gauss_k^2 au^3 day^-2 solarmass^-1 earthmass earthradius^-1)
You want:
Definition: 11186.127 m / s

Here is a pretty bad way to get gravitational acceleration g, though it is within the range seen on earth (9.76 - 9.83 m/s^2).  What causes the discrepancy?  I'm guessing it's because the earth is an ellipsoid not a sphere.

You have: gauss_k^2 au^3 day^-2 solarmass^-1 earthmass earthradius^-2
You want:
Definition: 9.8202195 m / s^2

You have: gravity
You want:
Definition: 9.80665 m/s^2 = 9.80665 m / s^2


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Homework Help: Taylor Expand Lagrangian to Second Order...

1. Dec 16, 2016

sa1988

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

NOTE - When I post the thread my embedded images aren't showing up on my web browser, but they do show up when I bring it up to edit, so I don't know if other users can see the pictures or not... If not, they're here:
Problem outline: http://tinypic.com/r/34jeihj/9
Solution: http://tinypic.com/r/algpqh/9

2. Relevant equations

3. The attempt at a solution

Given that x and x2 in the Lagrangian don't need expanding, I simply took the Sin and Cos parts up to their x2 terms and replaced them accordingly in the Lagrangian. However the given solution (below) seems to have done things rather differently. In particular I notice the 4a2x2 term and the Sin term have both become zero.

Can anyone explain why this is? Some obvious algebra error on my part, or a fundamental aspect of Taylor expansions that I haven't properly grasped? Many thanks.

2. Dec 16, 2016

stevendaryl

Staff Emeritus
Well, the term proportional to $x \dot{x} \dot{\theta} sin(\theta)$ is 4th order (treating $x$ and $\theta$ as the same order).

3. Dec 19, 2016

sa1988

Ahhh, so yes it was a fundamental error in understanding on my part.

Back to basics for me...

Thanks


text:
# How to Find Equivalent Fractions

How to Find Equivalent Fractions

• Equivalent fractions are the same size but are written differently.
• To be equivalent fractions, the and must be multiplied or divided by the same number.
• The two fractions:   1 / 2   and   2 / 4 are equivalent fractions.
• The numerator and denominators were both multiplied by 2.
• We can multipliy the numerator and denominator by 2 again to find another equivalent fraction of   4 / 8  .
• We can multiply both the numerator and denominator of   1 / 2  by 4 to make  4 / 8  .
• We can also divide the numerator and denominator of a fraction to make an equivalent fraction.
• We could choose to multiply both the numerator and denominator by an infinite range of different numbers to make further examples of equivalent fractions.
To make an equivalent fraction, multiply or divide the numerators and denominators by the same number.

• To find an equivalent fraction to   1 / 3  , we must multiply both the numerator and denominator by the same number.
• In this example, we chose to multiply by 2.
•   1 / 3  and   2 / 6  are equivalent fractions.
• We can see that they take up the same fraction of the shaded space in the shape shown.

# Equivalent Fractions

## What are Equivalent Fractions?

Equivalent fractions are fractions that are the same size but have different numerators and denominators. To be equivalent, the numerator and denominator of a fraction must have been multiplied or divided by exactly the same number to make the new fraction.

For example, in the fraction   2 / 3  , the numerator and denominator can be multiplied by 2 to find the equivalent fraction   4 / 6  .

The numerator of 2 has been doubled to get 4 and the denominator of 3 has been doubled to get 6.

Both fractions are equivalent because   4 / 6 is   2 / 3 .

One third of 6 is 2 and two thirds of 6 is 4.

Therefore 4 is two thirds of 6. This means that 4 / 6 is the same as   2 / 3 .

There are an infinite number of equivalent fractions that a fraction can have. This is because we can choose any number to multiply the numerator and denominator by and there are infinite numbers.

For example, this time instead of multiplying the digits in 2 / 3 by 2, we will multiply by 5.

2 multiplied by 5 equals 10 and 3 multiplied by 5 is 15.

We say that fractions that have the same ratio of numerator to denominator are called equivalent.

2 / 3 is equivalent to   10 / 15 .

This also means that 2 / 3 is equivalent to   4 / 6 which is equivalent to 10 / 15 is   2 / 3 .

## How to Find Equivalent Fractions

To find equivalent fractions, follow these steps:

1. Multiply or divide the numerator and denominator by the same number.
2. Only divide if the numerator and denominator remain as whole numbers.

Foe example, 3 / 4 is equivalent to 300 / 400 .

The numerator and the denominator have been multiplied by the same amount, so they are equivalent.

The numerator and denominator have both been multiplied by 100.

3 × 100 = 300 and 4 × 100 = 400.

When teaching equivalent fractions, it is important to state that we only multiply and divide and never add or subtract. It is a common mistake for children to add or subtract because they are often more familiar with these processes than they are with multiplication and division.

We can see this common mistake easily in the following diagram. When teaching equivalent fractions, it can help to show this common mistake using a diagram.

We can see that the first fraction is 1 / 2 but by adding 2 to both the numerator and denominator, we get to the fraction 3 / 4 .

We can see that 3 / 4 is clearly a larger fraction than 1 / 2 and so, the fractions are not equivalent.

## Why Does the Rule for Equivalent Fractions Work?

The rule for equivalent fractions works because the numerator and the denominator are both multiplied or divided by the same number. Increasing the numerator increases the size of the fraction and increasing the denominator decreases the size of the fraction.

If the numerator and denominator are multiplied or divided by the same amount, the fraction has increased and decreased by the same amount and so, it has not changed its overall size.

The denominator of a fraction tells us how many parts it is split up into in total.

The numerator of a fraction tells us how many parts we have.

If we double the number of parts there are in total and also double the number of parts we have, the proportion of parts we have has not changed.

For example, in the fraction 1 / 2 , we have 1 part shaded out of 2.

If we double the number of parts we have in total, then each part is twice as small. We now need two parts to have half of the shape.

1 / 2 is equivalent to 2 / 4 . This is because 2 is half of 4.

If we double the number of parts again, each part becomes even smaller. Each part is now four times smaller than the half parts that we had originally.

1 / 2 is equivalent to 4 / 8 . This means that 4 is half of 8.

All of these fractions are equivalent because they take up the same space. The diagrams show that they are the same amount of the total shape shaded.

## Equivalent Fractions Chart

An equivalent fractions chart is a visual way to show fractions that are equivalent to each other. Fractions that are the same size are shown highlighted on an equivalent fraction chart.

Here is an equivalent fraction chart showing fractions equivalent to 1 / 2.

There are infinite fraction that are equivalent to one half, such as 2 / 4 and 3 / 6 .

Here is an equivalent fraction chart showing fractions equivalent to 1 / 3 .

There are infinite fractions that are equivalent to one third, such as 2 / 6 or 3 / 9 .

Fraction charts are also known as fraction walls and can be a very useful method of teaching the concept of equivalent fractions.

An activity for children can be to look for and identify equivalent fractions using a fraction wall.

## Improper Equivalent Fractions

Two Improper fractions can be equivalent too. To find equivalent improper fractions, multiply or divide the numerator and the denominator by the same amount.

For example here is the improper fraction 5 / 2 .

We can multiply the numerator and denominator by any amount to make an equivalent fraction as long as this number is the same.

In this example, we have multiplied the numerator and denominator both by 3.

5 / 2 is equivalent to 15 / 6 .

## Where are Equivalent Fractions Used in Real Life?

Here are some examples of using equivalent fractions in real life:

• Finding amounts of ingredients to use in cooking.
• Sharing out food that is cut into slices.
• Working out Percentages.

For example, you are following a recipe to make a cake that requires half a cup of water. If the cup you are measuring it in only has measurements in quarters, then you can use equivalent fractions to know that 2 quarters are the same as one half.

If you are sharing a 12 slice pizza between 2 people, then you will know that 1 / 2 is equivalent to 6 / 12 and so, you get 6 slices each.

Percentages are found in real life in test scores, computer downloads and many other places. Percentages are fractions out of 100. To convert to a percentage, we need to use equivalent fractions to find a fraction out of 100.

For example, if you scored 12 out of 20 in a test and wanted to know what percentage this was, you would use equivalent fractions.

Multiplying the numerator and denominator by 5 would convert 12 / 20 to 60 / 100 and so, you would have scored 60%. Equivalent fractions are used in real life whenever we need to convert to a percentage.

Now try our lesson on Equivalent Fractions: Missing Numerator or Denominator where we learn how to find a missing numerator or denominator in an equivalent fraction.


text:
### Still have math questions?

Ask our expert tutors
Trigonometry
Question

The map of a campground is shown below. Campsite C, First Aid Station F, and Supply Station S lie along a straight path The path from the supply station to Tower T is perpendicular to the path from the supply station to the campsite. The length of $$\overline { F S }$$ is $$400$$ feet. The angle formed by path $$\overline { F S }$$ and path $$\overline { T S }$$ is $$72 ^ { \circ }$$ . The angle formed by path $$\overline { T C }$$ and path $$\overline { C S }$$ is $$55 ^ { \circ }$$ . What is the distance from the Campsite C to Tower T?

Answer

TC = 1502.86 ft

Solution
View full explanation on CameraMath App.


text:
## Heron’s method using integer arithmetic

Suppose we are given a positive rational number $y$, whose square root $\sqrt{y}$ we would like to compute. To be more precise, we would like to compute an approximation of $\sqrt{y}$.

Let $x_k \neq 0$ be an approximation of $\sqrt{y}$. Hence, we have that $\sqrt{y}$ is the geometric mean of $x_k$ and $y / x_k$. If we replace the geometric mean of these two numbers by their arithmetic mean, we hopefully obtain a “better” approximation $x_{k+1}$, as follows

$x_{k+1} = \displaystyle\frac{1}{2} \left( x_k + \frac{y}{x_k}\right)$

If all goes well, then $x_N$ will be “sufficiently close” to $\sqrt{y}$ for a “sufficiently large” integer $N$. This algorithm is the famous Heron’s method, devised by Heron of Alexandria some 2000 years ago. Some call it the Babylonian method, which suggests that the algorithm may be some 4000 years old. It’s not merely vintage, it’s antique.

Nonetheless, this is not a post on archaeology. We don’t want to sit around and talk about Heron’s method, we want to actually implement it. What programming language should we use? Let us use Haskell. How are we going to represent $y$ and the sequence of approximations $(x_k)_{k \in \mathbb{N}_0}$?

__________

Heron’s method using floating-point arithmetic

Though millions of people have already implemented Heron’s method using floating-point arithmetic (I implemented it in C in late 2000), one more implementation would do no harm:

y :: Floating a => a
y = 2

-- define Heron map
g :: Floating a => a -> a
g x = 0.5 * (x + y/x)

-- initial approximation
x0:: Floating a => a
x0 = 1

-- list of approximations
xs :: Floating a => [a]
xs = iterate g x0

Let us load this script into GHCi. Here’s a GHCi session:

*Main> take 6 xs
[1.0,1.5,1.4166666666666665,1.4142156862745097,
1.4142135623746899,1.414213562373095]
*Main> let x5 = xs !! 5
*Main> x5**2 - 2
-4.440892098500626e-16

We thus have the following approximation after five iterations

$x_5 = 1.414213562373095 \approx \sqrt{2}$

__________

Heron’s method using rational arithmetic

Let us now use rational numbers of arbitrary precision. The following script uses the Data.Ratio library:

import Data.Ratio

y :: Rational
y = 2

-- define Heron map
g :: Rational -> Rational
g x = 0.5 * (x + y/x)

-- initial approximation
x0:: Rational
x0 = 1 % 1

-- list of approximations
xs :: [Rational]
xs = iterate g x0

Let us load this script into GHCi. Here’s a GHCi session:

*Main> take 6 xs
[1 % 1,3 % 2,17 % 12,577 % 408,665857 % 470832,
886731088897 % 627013566048]
*Main> let x5 = xs !! 5
*Main> x5*x5 - 2
1 % 393146012008229658338304
*Main> 1 / 393146012008229658338304
2.5435842395854372e-24

Note that the error is eight orders of magnitude smaller than in the previous case (where we used finite-precision floating-point numbers). We pay for more accuracy with more computation. Thus, after five iterations, we have the following rational approximation

$x_5 = \displaystyle\frac{886731088897}{627013566048} \approx \sqrt{2}$

This is probably not news to you, my dear reader. Last April, I did, in fact, implement Heron’s method under the disguise of Newton’s method (see the appendix).

__________

Heron’s method using integer arithmetic

A rational number $y \in \mathbb{Q}$ can be expressed as a fraction $a / b$, where $a$ and $b$ are integers and $b \neq 0$. Note that $a / b$ can (obviously?) be represented as a pair of integers $(a,b)$. Since Haskell has arbitrary-precision integers, we can implement Heron’s method using pairs of arbitrary-precision integers to represent $y$ and the sequence of approximations $(x_k)_{k \in \mathbb{N}_0}$.

Let $y := a / b$ and $x_k := p_k / q_k$. The recurrence relation

$x_{k+1} = \displaystyle\frac{1}{2} \left( x_k + \frac{y}{x_k}\right)$

can thus be rewritten in the following form

$\displaystyle\frac{p_{k+1}}{q_{k+1}} = \frac{1}{2} \left( \frac{p_k}{q_k} + \frac{a}{b} \frac{q_k}{p_k} \right) = \frac{b \, p_k^2 + a \, q_k^2}{2 b \, p_k \, q_k}$

We then have two coupled recurrence relations

$\left[\begin{array}{c} p_{k+1}\\ q_{k+1}\end{array}\right] = \left[\begin{array}{c} b \, p_k^2 + a \, q_k^2\\ 2 b \, p_k \, q_k\end{array}\right]$

where $a$$b$$p_k$ and $q_k$ are integers. Here is a Haskell script that implements this vector recurrence relation:

a, b :: Integer
a = 2
b = 1

-- define Heron map
g :: (Integer,Integer) -> (Integer,Integer)
g (p,q) = (b * p^2 + a * q^2, 2 * b * p * q)

-- initial approximation
p0, q0 :: Integer
p0 = 1
q0 = 1

-- list of approximations
xs :: [(Integer,Integer)]
xs = iterate g (p0,q0)

Let us load this script into GHCi. Here’s a GHCi session:

*Main> take 6 xs
[(1,1),(3,2),(17,12),(577,408),(665857,470832),
(886731088897,627013566048)]
*Main> let x5 = xs !! 5
*Main> x5
(886731088897,627013566048)

Hence, after five iterations, we obtain the very same rational approximation we obtained in the previous case (where we used rational numbers of arbitrary-precision)

$x_5 = \displaystyle\frac{886731088897}{627013566048} \approx \sqrt{2}$

It should be noted, however, that this implementation using arbitrary-precision integers is different in one very important aspect from the one using arbitrary-precision rational numbers. What is the difference? Let GHCi answer this question:

*Main> import Data.Ratio
*Main Data.Ratio> 2 % 2
1 % 1
*Main Data.Ratio> 4 % 4
1 % 1
*Main Data.Ratio> 3 % 6
1 % 2

The main difference between the two implementations is that when we use arbitrary-precision rational numbers, there is automatic reduction of every fraction to its irreducible form, which requires computing the greatest common divisor (GCD) of the numerator and denominator of each rational number at each iteration. This has a computational cost, of course, but it also has the benefit of avoiding reducible fractions (which are inefficient representations of rational numbers). When we use pairs of arbitrary-precision integers, we do not compute the GCD (although we could, if we wanted to), but we may have to work with pairs of unnecessarily long integers (that correspond to reducible fractions), which is potentially dangerous.

For example, suppose that we replace the first three lines in the script above with the following three lines:

a, b :: Integer
a = 2000
b = 1000

We run the script again. Here’s yet another GHCi session:

*Main> take 6 xs
[(1,1),(3000,2000),(17000000000,12000000000),
(577000000000000000000000,408000000000000000000000),
(665857000000000000000000000000000000000000000000000,
470832000000000000000000000000000000000000000000000),
(8867310888970000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000,
62701356604800000000000000000000000000000000000000000
0000000000000000000000000000000000000000000000000000)]
*Main> let x5 = xs !! 5
*Main> x5
(8867310888970000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000,
62701356604800000000000000000000000000000000000000000
0000000000000000000000000000000000000000000000000000)

Frankly, I am shocked! Multiplying $a$ and $b$ by $1000$ results in $p_5$ and $q_5$ being multiplied by $10^{90}$. I may have made a mistake counting all those zeros, but it’s a lot more zeros than I expected.

__________

Appendix: from Heron to Newton

Interestingly, Heron’s method can be derived from the no less famous Newton’s method. Let us introduce a function $f : \mathbb{R} \to \mathbb{R}$, defined by $f (x) = x^2 - y$. By construction, the zeros of function $f$ are $\sqrt{y}$ and $-\sqrt{y}$. Recall that Newton’s method uses the recurrence relation $x_{k+1} = g ( x_k )$, where function $g : \mathbb{R} \to \mathbb{R}$ is defined by

$g (x) = x - \displaystyle\frac{f (x)}{f' (x)}$

where $f'$ is the first derivative of $f$. Hence, we obtain the following

$g (x) = x - \displaystyle\frac{f (x)}{f' (x)} = x - \displaystyle\left(\frac{x^2 - y}{2 x}\right) = \displaystyle\frac{1}{2} \left( x + \frac{y}{x}\right)$

which is the recurrence relation used in Heron’s method.

### 3 Responses to “Heron’s method using integer arithmetic”

In that case, you probably also know the very similar one for cube roots of $N$:

$x[i+1] = \displaystyle\frac{1}{3} \left( 2 x[i] + \frac{N}{ x[i]^2 } \right)$

which can also be derived from Newton-Raphson.

I’ve tried doing this in rational arithmetic and of course it works, but the numbers get rather out of hand: the number of digits multiplies by three each iteration, but the accuracy (number of accurate digits in floating point) only multiplies by two. That seems like the reward is not in proportion to the effort required.

• Rod Carvalho Says:

I’ve tried doing this in rational arithmetic and of course it works, but the numbers get rather out of hand (…) That seems like the reward is not in proportion to the effort required.

I have just written a post with some of my numerical experiments. I agree with you: the integers in the ratios just explode. It’s scary.

I suppose that the lesson to be learned from this is: with finite machines we can do finite-precision arithmetic comfortably, but doing arbitrary-precision arithmetic is a dangerous game.


text:
+0

# Help!

0
98
4

What is the slope of a line perpendicular to the line whose equation is $\frac{x}4-\frac{y}5=1$? Express your answer as a common fraction.

Jul 17, 2022

#1
+2448
0

Believe it or not, I don't speak LaTex

Jul 17, 2022
#4
+303
0

they didn't do their latex correctly, it was $$\frac{x}{4} - \frac{y}{5}=1$$

hipie  Jul 18, 2022
#2
0

Ok!

What is the slope of a line perpendicular to the line whose equation is (x/4)-(y/5)=1? Express your answer as a common fraction.

Jul 17, 2022
#3
+2448
0

We have this equation: $$\frac{x}4-\frac{y}5=1$$

Subtract $${x \over 4}$$ from both sides: $$-{y \over 5} = 1 - {x \over 4}$$

Multiply the entire equation by -5: $$y = {5 \over 4} x - 5$$

This means that the slope is $${5 \over 4}$$, the slope of the line that is perpendicular to this is the negative reciprocal, or $$\color{brown}\boxed{-{4 \over 5}}$$

Jul 17, 2022


text:
# Setting up HomeBridge

### C_LOG_13

This week I read about homebridge. This immediately peaked my interest in my adventures with the Raspberry Pi. I wonder if I could ask siri to flash the led? I suppose the easiest way is to see if we can get homebridge setup based on the readme.md.

sudo npm install -g homebridge

After a few seconds I ran the command homebridge . . .


text:
# [In Progress] Notes: The Principles of Deep Learning Theory

The Principles of Deep Learning Theory - Roberts, Yaida, Hanin

Chapter 0: Initialization

A network $f^\ast$ is constructed by sampling weights $\theta \sim p(\theta)$ and training $\theta \to \theta^*$ so that for input data $x$ the network output $f^\ast(x) := f(x; \theta^\ast)$ approximates the truth $f(x).$ The distribution $p(f^\ast)$ of trained networks is of interest.

Chapter 1: Pretraining

Review of the Gaussian distribution.

Chapter 2: Neural Networks

MLPs suffice as a minimal model for an effective theory of deep learning because other architectures like CNNs and transformers can be interpreted as MLPs with constraints on the relationship between weights. We will assume that initial weights are drawn independently from a zero-mean Gaussian distribution.

The distribution of interest can be specified as $p(f^\ast) = p \left( \left. z^{(L)} \, \right\vert \, \theta^\ast, \mathcal{D} \right),$ where $z^{(L)}$ is the preactivation of neurons in the $L$th layer (final layer) and $\mathcal{D}$ is the data set used during training.

Chapter 3: Effective Theory of Deep Linear Networks at Initialization

(in progress)

Tags:


text:
Please, try EDU on Codeforces! New educational section with videos, subtitles, texts, and problems. ×

### Majin_Boo's blog

By Majin_Boo, history, 6 weeks ago, ,

Hello CodeForces!! Today an update was released from the IC saying that IOI 2020 will be held online and Singapore will be the host for IOI 2021. It is sad to read that, especially since this year is going to be my final one. But it is much better than to cancel the contest.

Of course, moving the IOI online at least for a year will contribute to reduce CO2 emissions :-)

• -34

By Majin_Boo, history, 3 months ago, ,

Hi everyone! I've noted most recent CF problems have no rating. Is that made on purpose, or is a bug? If is the first case, why?

• 0

By Majin_Boo, history, 3 months ago, ,

Hi again! Continuing with my series of bad questions, I'd like to ask if competitive programming is useful as a preparation for computer science research, specially in the field of data structures and algorithms. If yes, what skills are the most used? I hope someone working in academia could tell me in which way CP helps or helped him/her in his/her regular work.

• -2

By Majin_Boo, history, 4 months ago, ,

Hello CF community. I was asking myself which companies competitive/sport programmers are more attracted to, or if they run their own startups. In my case, I don't know where do I want to work yet, but I know Competitive/Sport programming is going to help me very much in my life.

• +64

By Majin_Boo, history, 4 years ago, ,

When i submitted my solution, the judge throws me Runtime Error on test 8, more specifically with this test:

10 10
.........*
.........*
........**
.........*
.........*
.........*
.........*
.........*
.........*
.........*

But i ran the program with this test in my computer and it works correctly and the output is correct:

YES
3 10


text:
Including the Sacado Speed Tests

Purpose
If Sacado is installed on your system, you can specify a value for its install sacado_prefix on the cmake command line. The value of sacado_prefix must be such that, for one of the directories dir in cmake_install_includedirs ,       sacado_prefix/dir/Sacado.hpp  is a valid way to reference to the include file Sacado.hpp;
If you include sacado_prefix on the cmake command line, you will be able to run the Sacado speed correctness tests by executing the following commands starting in the distribution directory :       cd build/speed/sacado      make check_speed_sacado  After executing make check_speed_sacado, you can run a specific Sacado speed test by executing the command ./speed_sacado; see speed_main for the meaning of the command line options to this program.
If you are using Unix, you can download and install a copy of Sacado using get_sacado.sh . The corresponding sacado_prefix would be build/prefix.


text:
# How do you find the general form of the equation of this circle given Center(2,-3) ; containing point (5,-3)?

Feb 28, 2016

#### Answer:

${x}^{2} + {y}^{2} - 4 x + 6 y + 4 = 0$

#### Explanation:

As the distance between points $\left(2 , - 3\right)$ and $\left(5 , - 3\right)$ i.e. radius is $3$ and center of circle is $\left(2 , - 3\right)$, the equation of circle would be ${\left(x - 2\right)}^{2} + {\left(y + 3\right)}^{2} = {3}^{2}$ or ${x}^{2} - 4 x + 4 + {y}^{2} + 6 y + 9 = 9$ or
${x}^{2} + {y}^{2} - 4 x + 6 y + 4 = 0$.


text:
# Theory of Saturated Neural Networks

Summarizing Sequential Neural Networks as Automata.

The main idea of my paper Sequential Neural Networks as Automata is that, if we making some simplifying assumptions about how neural networks work, we can derive a theory of network expressiveness (what formal languages can different architectures model?) that seems to agree with the classes of formal languages that different networks can learn when trained by gradient descent. Thus, this restricted theoretical capacity seems to be (potentially) a good proxy for the empirical learnable capacity of various networks.

## Saturated Networks

In the paper, I referred to the simplified network whose capacity we can analyze as an asymptotic network. However, after talking with Gail Weiss, I now believe the term saturated is more descriptive, and plan to use this term going forward.

A neural network is a function $f(x, \theta)$ that is almost-everywhere differentiable with respect to the parameters $\theta$. Given such a function, we derive the saturated network $f’$ as

$$f’(x, \theta) = \lim_{N \rightarrow \infty} f(x, N\theta) .$$

We define $f’$ over the domain of $(x, \theta)$ for which the limit above exists.

In a neural network, the effect of this transformation is to discretize all of the activations. For example, consider a neuron:

$$\sigma(wx + b)$$

where $\sigma$ is the sigmoid function. When we take the limit of $\sigma(Nwx + Nb)$, the output of the neuron approaches either $0$ or $1$. This is what I mean by discretization.

After applying this discretization to the full network, we can analyze the computational capacity of the resulting discrete automaton. We also define a notion of space complexity associated with these saturated networks in the paper. Intuitively, this measure of complexity is just the number of configurations that the saturated network can have after reading a sequence of length $n$. For more details on this, consult the paper.

## Summary of Results

By $L(M)$, we denote the set of formal languages that a machine $M$ can accept. Some key capacity results from the paper are as follows:

• $L(\textrm{ConvNet})$ is a proper subset of the regular languages
• $L(\textrm{RNN})$ is exactly the regular languages
• $L(\textrm{GRU})$ is exactly the regular languages
• $L(\textrm{LSTM})$ is a superset of the regular languages, and a subset of the real-time counter languages

The core results about the configuration complexity, some of which are analogous, are:

• ConvNet has $O(1)$ configurations
• RNN has $O(1)$ configurations
• GRU has $O(1)$ configurations
• LSTM has $O(n^k)$ configurations for hidden size $k$
• Attention has $2^{O(n)}$ configurations
• StackNN has $2^{O(n)}$ configurations

There are some other results about attention/transformers that I’m not going to get into here, since they’re not so neat. If you’re interested though, I refer you to Section 4 of the paper.

## Empirical Evidence

### Positive Evidence

The development of this theory was motivated by past empirical results about what RNNs are able to learn. Two types of tasks (counting and reversing) serve as relevant diagnostics for assessing the computational power of different architectures.

Weiss et al. show that LSTMs can learn to count, whereas RNNs and GRUs do not. Similarly, Sugzgun et al. observe that LSTMs can learn the 1-Dyck counter language, whereas other RNNs do not. This is predicted by the theory since saturated LSTMs can do counting, but saturated RNNs and GRUs are finite state.

Hao et al. show how stack neural networks can solve the (beyond counter language) task of reversing a string, whereas LSTMs fail badly on this. In my paper, I also showed that attention can solve this task. Both of the results are predicted by the theory, since stack neural networks and attention allow for exponential configurations, which are needed to reverse a string, whereas LSTMs are limited to polynomial configurations.

### Negative Evidence

While I found that regularized neural networks display the counting pattern reported by Weiss et al., I also found that the unregularized LSTM, GRU, and RNN can all learn to model the language $a^nb^nc$, which requires counting. Thus, it might be more precise to say that the saturated theory seems to describe the learnable capacity of regularized networks. One possible interpretation of this is that the constraints imposed by regularization prevent the network from learning strategies beyond the saturated capacity.

## Proof Sketches

### RNN Capacity and Complexity

To show that $L(\textrm{RNN})$ is the regular languages, we show two directions of containment.

First, we prove that the the regular languages are an upper bound. We do this by showing that the configuration complexity of the RNN is finite, i.e. $O(1)$. Since each neuron has two possible values ($-1$ and $1$), and there are $k$ neurons in the state, the number of configurations of the state vector is $O(2^k) = O(1)$.

The other direction is a little more complicated. We need to construct an RNN to simulate an arbitrary finite state machine. A construction for this is provided in Lemma B.1.

### LSTM Capacity and Complexity

In the LSTM case, even when we discretize the network, we get a model with more than finite state. This is because the LSTM’s gating architecture is capable of counting (Weiss et al., 2018).

To show that the counter languages are an upper bound, we write the saturated gate network for a particular counter state neuron $c$ as follows:

$$\underset{N \rightarrow \infty}{\lim} c_t$$

$$= \underset{N \rightarrow \infty}{\lim} fc_{t-1} + i{\tilde c}_t$$

$$= \underset{N \rightarrow \infty}{\lim} ac_{t-1} + b$$

where $a$ is $0$ or $1$ and $b$ is $-1$, $0$, or $1$. This equation parameterizes a real-time counter machine update. Thus, the counter languages are an upper bound on the saturated LSTM capacity.

##### Will Merrill
###### Research Resident

My research interests incorporate computational linguistics, theoretical computer science, and machine learning.


text:
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Here at Sharp Sight, we make a lot of maps.

There are a few reasons for this.

First, good maps are typically ‘information dense.’ You can get a lot of information at a glance from a good map. They are good visualization tools for finding and communicating insights.

Second, it’s extremely easy to get data that you can use to make a map. From a variety of sources, you’ll find data about cities, states, counties, and countries. If you know how to retrieve this data and wrangle it into shape, it will be easy to find data that you can use to make a map.

Finally, map making is just good practice. To create a map like the one we’re about to make, you’ll typically need to use a variety of data wrangling and data visualization tools. Maps make for excellent practice for intermediate data scientists who have already mastered some of the basics.

With that in mind, this week we’ll make a map of “world cities.” This set of cities has been identified by the Globalization and World Cities (GaWC) Research Network as being highly connected and influential in the world economy.

We’re going to initially create a very basic map, but we’ll also create a small multiple version of the map (broken out by GaWC ranking).

Let’s get started.

First, we’ll load the packages that we’ll need.

#==============
#==============

library(tidyverse)
library(ggmap)
library(forcats)

Next, we’ll input the cities by hard coding them as data frames. To be clear, there is more than one way to do this (e.g., we could scrape the data), but there isn’t that much data here, so doing this manually is acceptable.

#===================
# INPUT ALPHA CITIES
#===================

df_alpha_plus_plus <- tibble(city = c('London','New York'))
df_alpha_plus <- tibble(city = c('Singapore', 'Hong Kong', 'Paris', 'Beijing'
,'Tokyo', 'Dubai', 'Shanghai'))
df_alpha <- tibble(city = c('Sydney', 'São Paulo', 'Milan', 'Chicago'
,'Mexico City', 'Mumbai', 'Moscow', 'Frankfurt'
,'Seoul', 'Istanbul', 'Kuala Lumpur', 'Jakarta'
,'Amsterdam', 'Brussels', 'Los Angeles'))
df_alpha_minus <- tibble(city = c('Dublin', 'Melbourne', 'Washington', 'New Delhi'
,'Bangkok', 'Zurich', 'Vienna', 'Taipei'
,'Buenos Aires', 'Stockholm', 'San Francisco'
,'Guangzhou', 'Manila', 'Bogotá', 'Miami', 'Luxembourg'
,'Riyadh', 'Santiago', 'Barcelona', 'Tel Aviv', 'Lisbon'))

Now, we’ll create a new variable called rating. This will contain the global city rating.

Notice that this is a very straightforward use of dplyr::mutate(), one of the tidyverse functions you should definitely master.

#=======================
#=======================

df_alpha_plus_plus <- df_alpha_plus_plus %>% mutate(rating = 'alpha++')
df_alpha_plus <- df_alpha_plus %>% mutate(rating = 'alpha+')
df_alpha <- df_alpha %>% mutate(rating = 'alpha')
df_alpha_minus <- df_alpha_minus %>% mutate(rating = 'alpha-')

Next, we’ll combine the different data frames into one using rbind().

#======================================
# COMBINE DATAFRAMES INTO ONE DATAFRAME
#======================================

alpha_cities <- rbind(df_alpha_plus_plus
,df_alpha_plus
,df_alpha
,df_alpha_minus
)

Now that the data are combined into a single data frame, we’ll get the longitude and latitude using geocode().

#========
# GEOCODE
#========

latlong <- geocode(alpha_cities$city) #======== # INSPECT #======== alpha_cities latlong  Once we have the longitude and latitude data, we need to combine it with the original data in the alpha_cities data frame. To do this, we will use cbind(). #============================ # BIND LAT/LONG TO CITY NAMES #============================ alpha_cities <- cbind(alpha_cities, latlong) %>% rename(long = lon) alpha_cities #names(alpha_cities)  Now we have the data that we need, but we’ll need to clean things up a little. In the visualization we’ll make, we will need to use the faceting technique from ggplot2. When we do this, we’ll facet on the rating variable, but we will need the levels of that variable to be ordered properly (otherwise the facets will be out of order). To reorder the factor levels of rating, we will use fct_relevel(). #================================================ # REORDER LEVELS OF GLOBAL CITY RATING # - the global city ratings should be ordered # i.e., alpha++, then alpha+ .... # - to do this, we'll use forecats::fct_relevel() #================================================ alpha_cities <- mutate(alpha_cities, rating = fct_relevel(rating, 'alpha++','alpha+','alpha','alpha-')) levels(alpha_cities$rating)

Because we will be building a map, we’ll need to retrive a map of the world. We can get a world map by using map_data(“world”).

#==============
# GET WORLD MAP
#==============

map_world <- map_data("world")

Ok. We basically have everything we need. Now we will make a simple first draft.

#================
# FIRST DRAFT MAP
#================

ggplot() +
geom_polygon(data = map_world, aes(x = long, y = lat, group = group)) +
geom_point(data = alpha_cities, aes(x = long, y = lat), color = 'red')

… and now we’ll use the faceting technique to break out our plot using the rating variable.

#==========================
# CREATE SMALL MULTIPLE MAP
#==========================

ggplot() +
geom_polygon(data = map_world, aes(x = long, y = lat, group = group)) +
geom_point(data = alpha_cities, aes(x = long, y = lat), color = 'red') +
#facet_grid(. ~ rating)
#facet_grid(rating ~ .)
facet_wrap(~ rating)

Once again, this is a good example of an intermediate-level project that you could do to practice your data wrangling and data visualization skills.

Having said that, before you attempt to do something like this yourself, I highly recommend that you first master the individual tools that we used here (i.e., the tools from ggplot2, dplyr, and the tidyverse).

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# Numerical Integration of 2nd Order DE

1. Feb 13, 2013

### strokebow

I am intending to use Runge Kutta 4th order to numerically solve a system of coupled equations:

$\frac{d^{2}x}{dt^{2}}$ = K1 * x * cos(t) + ( (K2 * $\frac{dy}{dt}$) - $\frac{dz}{dt}$ )

$\frac{d^{2}y}{dt^{2}}$= -K1 * y * cos(t) + ( (K2 * $\frac{dz}{dt}$) - $\frac{dx}{dt}$ )

$\frac{d^{2}z}{dt^{2}}$= ( (K2 * $\frac{dx}{dt}$) - $\frac{dy}{dt}$ )

I'm really a bit stuck to be honest. I've only ever used RK4 on 1st order linear ODEs. I've been reading around alot but not making much progress.

Initial values for $\frac{dx}{dt}$, $\frac{dy}{dt}$, $\frac{dz}{dt}$ are all known. The constants K are known.

2. Feb 13, 2013

### bigfooted

Introduce the derivatives as new variables, say u=dx/dt and solve the system of 6 first order equations, where the second derivatives are replaced by the first derivatives du/dt etc.

so the last equation will be
dw/dt = k2*u - v
but you could also write it as:
dw/dt = k2*dx/dt - dy/dt

They will lead to the same solution, but their convergence behavior will be slightly different.

3. Feb 13, 2013

### strokebow

Thanks very much for your input.

So I would define:
u = $\frac{dx}{dt}$
v = $\frac{dy}{dt}$
w = $\frac{dz}{dt}$
and
$\frac{du}{dt}$ = $\frac{d^{2}x}{dt^{2}}$
$\frac{dv}{dt}$ = $\frac{d^{2}y}{dt^{2}}$
$\frac{dw}{dt}$ = $\frac{d^{2}z}{dt^{2}}$

Therefore my equations become:
(1'): $\frac{du}{dt}$ = K$_{1}$ * x * cos(t) + ( (K$_{2}$ * v) - w )
(2'): $\frac{dv}{dt}$ = -K$_{1}$ * y * cos(t) + ( (K$_{2}$ * w) - u )
(3'): $\frac{dw}{dt}$ = ( (K$_{2}$ * u) - v )

Now for applying RK4 . . .
How would I do that simultaneously for both. I've found a couple of exams on the net but its quite difficult to follow since all the equations are inter-related.

Any help is much appreciated!

4. Feb 14, 2013

### strokebow

Not sure what you mean by this big foot?

5. Feb 14, 2013

### Staff: Mentor

You now have a system of 6 1st-order equations to solve, (1')-(3') plus
(4'): $\frac{dx}{dt} = u$
(5'): $\frac{dy}{dt} = v$
(6'): $\frac{dz}{dt} = w$

6. Feb 14, 2013

### strokebow

Thanks for your input.

I appreciate that we now have a system of 1st order DE's. Its just that trying to apply RK4 to this system concurrently is quite confusing.
I'm trying to follow the example given on the net (e.g., http://www.nsc.liu.se/~boein/f77to90/rk.html)

But am finding it tough to implement.

Especially since these equations are coupled with more than 2 variables. Like with eqn (1') its depending on both v and w.

7. Feb 14, 2013

### Staff: Mentor

Maybe this formulation will be more useful. Set a vector $$\mathbf{a} = (u,v,w,x,y,z)$$ and a vector function $$\mathbf{f}(\mathbf{a},t) = (\frac{du}{dt},\frac{dv}{dt},\frac{dw}{dt},\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt})$$ The 4th-order RK algorithm then reads
$$\mathbf{k}_1 = h \mathbf{f}(\mathbf{a}_n, t_n)$$
$$\mathbf{k}_2 = h \mathbf{f}(\mathbf{a}_n + \frac{\mathbf{k}_1}{2}, t_n + \frac{h}{2})$$
$$\mathbf{k}_3 = h \mathbf{f}(\mathbf{a}_n + \frac{\mathbf{k}_2}{2}, t_n + \frac{h}{2})$$
$$\mathbf{k}_4 = h \mathbf{f}(\mathbf{a}_n + \mathbf{k}_3, t_n + h)$$
$$\mathbf{a}_{n+1} = \mathbf{a}_n + \frac{\mathbf{k}_1}{6} + \frac{\mathbf{k}_2}{3} + \frac{\mathbf{k}_3}{3} + \frac{\mathbf{k}_4}{6}$$

Hope this helps.

8. Feb 14, 2013

### strokebow

Thanks for your help.

So (replacing the 'k's' from the RK4 algorithm with 'm's' and 'n's')

For the first step, would this be correct:
m1u = h x (1')
m1v = h x (2')
m1w = h x (3')

n1u = h x (4')
n1v = h x (5')
n1w = h x (6')

?

9. Feb 14, 2013

### Staff: Mentor

I'm not sure I follow you.

The "first step" is the calculation of $\mathbf{k}_1$. The first element of $\mathbf{f}(\mathbf{a}_n,t_n)$ is
$$\left.\frac{du}{dt}\right|_{t_n} = K_1 x_n \cos(t_n) + ( K_2 v_n - w_n )$$
so the first element of $\mathbf{k}_1$ is
$$h \times \left[ K_1 x_n \cos(t_n) + ( K_2 v_n - w_n ) \right]$$ and so on. For instance, the 4th element of $\mathbf{k}_1$ is
$$u_n$$
Once you have calculated all six elements of $\mathbf{k}_1$, the "second step" is do the the same for $\mathbf{k}_2$, which will depend on the calculation of $\mathbf{k}_1$. Once you have calculated all $\mathbf{k}$'s, you can advance the solution one time step with the formula for $\mathbf{a}_{n+1}$.

The easiest way to implement this numerically is not to program it explicitly, bu to write a function that given an array of six values $(u,v,w,x,y,z)$ and a value of $t$, returns an array of the six values $(du/dt,dv/dt,dw/dt,dx/dt,dy/dt,dz/dt)$. The RK integrator then just makes 4 calls to this function to calculate $\mathbf{k}_1$-$\mathbf{k}_4$, and then the new values of $(u,v,w,x,y,z)$.

10. Feb 14, 2013

### strokebow

Thanks Dr C. Appreciate you taking time out to help me with this.

So, there are 6 equations for each step. i.e., k$_{1}$ has 6 equations. Writing them explicitly (I take your point that a generic function is best for programming but for now I am just trying to understand this solution):
$k_{1}u = h (K_1 x \cos(t) + ( K_2 v - w ))$

$k_{1}v = h (-K_1 y \cos(t) + ( K_2 v - w ))$

$k_{1}w = h (K_2 u - v)$

$k_{1}u' = ?$

$k_{1}v' = ?$

$k_{1}w' = ?$

When you said:
Do you mean that we just consider these (4th, 5th and 6th elements of the vector) to be constants?

Because we can derive expressions for u, v and w from (1'), (2') and (3')

11. Feb 14, 2013

### Staff: Mentor

Lets take it one step back. You converted your system of 3 2nd-order differential equations into a system of 6 first-order differential equations. You have increased the number of variables to 6, but that is a small price to pay compared with the gain 2nd to 1st order.

The equations now read:
$$\frac{du}{dt} = K_{1} x \cos(t) + ( K_{2} v - w )$$
$$\frac{dv}{dt} = -K_{1} y \cos(t) + ( K_{2} w - u )$$
$$\frac{dw}{dt} = (K_{2} u - v )$$
$$\frac{dx}{dt} = u$$
$$\frac{dy}{dt} = v$$
$$\frac{dz}{dt} = w$$

The first step is to calculate the elements of $\mathbf{k}_1$, which I will call $(u',v',w',x',y',z')$.
$$u' = h \left[ K_{1} x_n \cos(t_n) + ( K_{2} v_n - w_n ) \right]$$
$$v' = h \left[ -K_{1} y_n \cos(t_n) + ( K_{2} w_n - u_n ) \right]$$
$$w' = h (K_{2} u_n - v_n )$$
$$x' = h u_n$$
$$y' = h v_n$$
$$z' = h w_n$$

Now, you need to calculate $\mathbf{k}_2 = (u'',v'',w'',x'',y'',z'')$.
$$u'' = h \left[ K_{1} \left( x_n + \frac{x'}{2}\right) \cos(t_n + \frac{h}{2}) + ( K_{2} \left(v_n + \frac{v'}{2} \right) - \left(w_n + \frac{w'}{2} \right) ) \right]$$
and so on for the other variables. Finally, once you have all values $u',u'',u''',u''''$, you find
$$u_{n+1} = u_n + \frac{u'}{6} + \frac{u''}{3} + \frac{u'''}{3} + \frac{u''''}{6}$$
and so on for the other variables.

12. Feb 14, 2013

### strokebow

Dr C. Thanks very much. That is very clear.

I'm going to have a go at writing it all out explicitly and then I'll post it on here.

Cheers!!


text:
# How do you write the expression a*a*b*b*b*b using exponents?

This would be ${a}^{2} \cdot {b}^{4}$


text:
# B Are there Hadrons with more than three quarks?

Tags:
1. Aug 3, 2017

### KBon

As far as I know there are Mesons (quark-Antiquark pair) and Baryons (three quarks). But are there Hadrons which contain more than 3 Quarks?

2. Aug 3, 2017

### jerromyjon

Hi and welcome to PF!
There has been a lot of hype lately that the LHC discovered penta-quark hadrons but I don't know much about it, other than there are 5 quarks involved.

3. Aug 4, 2017

### A. Neumaier

See https://en.wikipedia.org/wiki/Pentaquark
It is probably a baryon-meson molecule. Of course there are also dodeca-quarks = alpha-particles, stable bound states of 12 quarks, or of two protons and two neutrons, and heavier nuclei.

4. Aug 4, 2017

### KBon

Aight, I'll surely look it up. Thank you very much :) @jerromyjon @A. Neumeier

5. Aug 4, 2017

### mikeyork

In meson-baryon scattering, $q\bar{q} + qqq$, one typically sees resonances (short-lived particles) only for $qqq$ systems, where one $q$ annihilates with the $\bar{q}$. States such as $qqqq\bar{q}$ where there is no $q$ able to annihilate the $\bar{q}$ are called exotic and no such resonances have been verified. This is most easily seen in $K^+p$ scattering.

6. Aug 5, 2017

### snorkack

The only well-known hexaquark is orthodeuteron, and it behaves strongly as a bound system of two separate nucleons.
Paradeuteron, diproton and dineutron are confirmed to be unbound.
Are there any other hexaquarks that are bound?
With 5 quarks participating in baryons, there are a lot of combinations to check...

7. Aug 7, 2017

8. Aug 8, 2017

### snorkack

How is a tetraquark verified to be that?

9. Aug 8, 2017

### Staff: Mentor

Z(4430) decays to $J/\psi \,\pi^\pm$, and based on its mass it cannot have a b quark, so it has to have $c \bar c$ in it. It also has a charge, so it needs at least two more quarks.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
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ABCD is a square picture frame (see figure). EFGH is a : GMAT Problem Solving (PS)
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# ABCD is a square picture frame (see figure). EFGH is a

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ABCD is a square picture frame (see figure). EFGH is a [#permalink]

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19 Feb 2012, 22:15
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ABCD is a square picture frame (see figure). EFGH is a square inscribed within ABCD as a space for a picure. The area of EFGH (for the picture) is equal to the area of the picture frame (the area of ABCD minus the area of EFGH). If AB = 6, what is the length of EF?
Attachment:

Picture.PNG [ 8.57 KiB | Viewed 8073 times ]

Hi, this question is annoying me, as I believe I am unable to read it straight. Pls help.
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Last edited by Bunuel on 19 Feb 2012, 22:43, edited 1 time in total.
Edited the question and added the figure
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Re: ABCD is a square picture frame (see figure). EFGH is a [#permalink]

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19 Feb 2012, 22:51
1
KUDOS
Expert's post
sdas wrote:
ABCD is a square picture frame (see figure). EFGH is a square inscribed within ABCD as a space for a picure. The area of EFGH (for the picture) is equal to the area of the picture frame (the area of ABCD minus the area of EFGH). If AB = 6, what is the length of EF?
Attachment:
The attachment Picture.PNG is no longer available

Hi, this question is annoying me, as I believe I am unable to read it straight. Pls help.

Attachment:

Picture2.PNG [ 8.57 KiB | Viewed 8055 times ]
Look at the figure above, since the area of of the picture frame (shaded region) is equal to the area of EFGH, then the area of EFGH is half of the area of the big square ABCD, which is 6^2=36. Hence, the area of EFGH = 18. The side of the square EFGH is $$\sqrt{18}=3\sqrt{2}$$ (since area=side^2).

Hope it's clear.
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Re: ABCD is a square picture frame (see figure). EFGH is a [#permalink]

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19 Feb 2012, 22:57
sdas wrote:
ABCD is a square picture frame (see figure). EFGH is a square inscribed within ABCD as a space for a picure. The area of EFGH (for the picture) is equal to the area of the picture frame (the area of ABCD minus the area of EFGH). If AB = 6, what is the length of EF?
Attachment:
Picture.PNG

Hi, this question is annoying me, as I believe I am unable to read it straight. Pls help.

Area of ABCD = 36

Area of EFGH = 18 - area of EFGH is equal to area of picture frame - they're each 1/2 of the total area (36)

A=bh
18=x^2 (b & h are equal since it is a square)

Length of EF = sqrt(18) = 3*sqrt(2)
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Re: ABCD is a square picture frame (see figure). EFGH is a [#permalink]

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19 Feb 2012, 23:05
Thank you folks. I missed understanding the shaded part and picture area (EFGH). Thanks Bunuel and Jazzman
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Re: ABCD is a square picture frame (see figure). EFGH is a [#permalink]

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10 Oct 2013, 07:36
Hi,

But where does it say the area of EFGH is half of the area of the big square ABCD?

Thanks
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Re: ABCD is a square picture frame (see figure). EFGH is a [#permalink]

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10 Oct 2013, 08:02
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Expert's post
theGame001 wrote:
Hi,

But where does it say the area of EFGH is half of the area of the big square ABCD?

Thanks

It's given in the stem: the area of EFGH (for the picture) is equal to the area of the picture frame (the area of ABCD minus the area of EFGH) --> area(ABCD)-area(EFGH)=area(EFGH) --> area(ABCD)=2*area(EFGH).

Hope it's clear.
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Re: ABCD is a square picture frame (see figure). EFGH is a [#permalink]

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10 Oct 2013, 08:14
Crystal clear, thanks a lot
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Re: ABCD is a square picture frame (see figure). EFGH is a [#permalink]

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13 Nov 2013, 19:22
Also,

Consider area of the picture frame is X.

Area of the Picture frame = Area of ABCD - Area of EFGH (which is equal to the area of the picture frame)
Upon substituting,
X = (6*6) - X
2X = 36
X (area) = 18

Side = \sqrt{18}
= 3\sqrt{2}
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Re: ABCD is a square picture frame (see figure). EFGH is a [#permalink]

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14 Nov 2013, 17:13
Follow the question stem. Area(EFGH)=area(ABCD)-area(EFGH)------>2*area(EFGH)=area(ABCD). This means that area(EFGH)=1/2*area(ABCD).
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Re: ABCD is a square picture frame (see figure). EFGH is a [#permalink]

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09 Dec 2014, 16:25
Hi,

How can you infer that the area of EFGH is "Half" of the Area of ABCD?
Im unable to understand that. Please explain

Bunuel wrote:
sdas wrote:
ABCD is a square picture frame (see figure). EFGH is a square inscribed within ABCD as a space for a picure. The area of EFGH (for the picture) is equal to the area of the picture frame (the area of ABCD minus the area of EFGH). If AB = 6, what is the length of EF?
Attachment:
Picture.PNG

Hi, this question is annoying me, as I believe I am unable to read it straight. Pls help.

Attachment:
Picture2.PNG
Look at the figure above, since the area of of the picture frame (shaded region) is equal to the area of EFGH, then the area of EFGH is half of the area of the big square ABCD, which is 6^2=36. Hence, the area of EFGH = 18. The side of the square EFGH is $$\sqrt{18}=3\sqrt{2}$$ (since area=side^2).

Hope it's clear.
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ABCD is a square picture frame (see figure). EFGH is a [#permalink]

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09 Dec 2014, 16:26
nish4every1 wrote:
Hi,

How can you infer that the area of EFGH is "Half" of the Area of ABCD?
Im unable to understand that. Please explain

Bunuel wrote:
sdas wrote:
ABCD is a square picture frame (see figure). EFGH is a square inscribed within ABCD as a space for a picure. The area of EFGH (for the picture) is equal to the area of the picture frame (the area of ABCD minus the area of EFGH). If AB = 6, what is the length of EF?
Attachment:
Picture.PNG

Hi, this question is annoying me, as I believe I am unable to read it straight. Pls help.

Attachment:
Picture2.PNG
Look at the figure above, since the area of of the picture frame (shaded region) is equal to the area of EFGH, then the area of EFGH is half of the area of the big square ABCD, which is 6^2=36. Hence, the area of EFGH = 18. The side of the square EFGH is $$\sqrt{18}=3\sqrt{2}$$ (since area=side^2).

Hope it's clear.

Have you read this: abcd-is-a-square-picture-frame-see-figure-efgh-is-a-127823.html#p1276619 ?
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Re: ABCD is a square picture frame (see figure). EFGH is a [#permalink]

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09 Dec 2014, 23:43
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Refer modified diagram below:

Attachment:

Picture.PNG [ 16.12 KiB | Viewed 4316 times ]

Given that area of orange region = area of white region in the square

It means that, if the small square EFGH is rotated 45 degree, it will exactly touch midpoints(AE = EB = 3) of square ABCD sides

$$EF = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}$$
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Re: ABCD is a square picture frame (see figure). EFGH is a [#permalink]

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11 Apr 2015, 06:54
sdas wrote:
ABCD is a square picture frame (see figure). EFGH is a square inscribed within ABCD as a space for a picure. The area of EFGH (for the picture) is equal to the area of the picture frame (the area of ABCD minus the area of EFGH). If AB = 6, what is the length of EF?
Attachment:
Picture.PNG

Hi, this question is annoying me, as I believe I am unable to read it straight. Pls help.

$$6^2 -EF^2 = EF^2$$

$$EF^2=18$$

$$EF= 3 \sqrt{2}$$
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Hello from the GMAT Club BumpBot!

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I got the answer correct, But I do not see answer options A, B , C , D and E. Looks like they are not provided originally.
Re: ABCD is a square picture frame (see figure). EFGH is a   [#permalink] 13 Aug 2016, 23:42
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text:
# Calibration¶

Instead of using a linear fit to calculate the battery percentage, the sketch uses the coulomb counting method by using the collection time interval and current passing to the battery.

## Exponential Trendline¶

The sketch uses an exponetial trendline for a voltage versus percentage to calculate the battyer percentage.

The formula for the exponential trendline is as follows:

$B{\rm e}^{Ax}$

The coefficients of the trendline can be found from the equation of the exponential trendline on the chart.

In this instance, the following coefficients are found:

$y=6.86\mathrm{E}{-07}{\rm e}^{3.48\mathrm{E}{-03}x}$
$B=0.000000686$
$A=0.00348$

These coefficients can then be entered into the config.h file as EXP_A and EXP_B.

Note

The coefficients supplied with the sketch were found using two 3000 mAh batteries.

The maximum and minimum voltages can be found by from this data as well and transcribed to the config.h file in the VOLTAGE_MAX and VOLTAGE_MIN parameters.

The scatter plot is created by performing the following steps:

• Collect data from when the batteries are fully depleted to fully charged. This can be done either using a solar panel or wall wart.
• Create a column that calculates the time in duration between readings in hours.
• Create a column that multiplies the current by the time duration to get the number of coulombs collected in that time interval.
• Create another column that sums the amount of coulombs collected up to that period in time.
• Create another column that calculates the percentage by dividing the previous column with the maximum number of coulombs collected.
• Create a scatter plot that with the voltage on the X-Axis and percentage on the Y-Axis.
• Add an exponential trendline with the equation in in the legend.
• Transcribe the coefficients of the trendline to the config.h file in the EXP_A and EXP_B parameters.

Note

The data collection needs to start when the battery is fully depleted (cut off by the BMS) in order to accurately calibrate the system.

WIP (#57)

## Current Limitations¶

• The Google Sheet assumes that the current over the collection period is constant. Integration of the current versus time would be a better estimation of the amount of energy collected.
• The collection at the end of the charge can be a bit erratic and group around the same same voltage. This was clipped in my original data.
• The coefficients can only be found by the equation on the scatter plot chart. Have not figured out a way to calculate the coefficients in a Sheet cell (#58).


text:
(n+1,r+1)-Theta space of (n,r)-Theta spaces?

I started writing nLab:Theta space. Not done yet, but while I am working on it:

is there a good proposal for what the "$(n+1,r+1)$-$\Theta$-space of all $(n,r)$-$\Theta$-spaces" would be?

-
 Do you mean something more complicated than the weak (n+1,r+1)-category of weak (n,r)-categories? – Denis-Charles Cisinski Nov 17 2009 at 23:11 I assume Urs means an explicit Theta-space model for the weak (n+1,r+1)-category of weak (n,r)-categories. – Charles Rezk Nov 17 2009 at 23:59 Yes, I mean the explicit Theta-space model. Thanks for the reply below! I could now keep asking questions along the lines: "How much category theory is known for Theta-spaces"? Is there an explicit definition of limits and colimits, for instance? Of adjoint functors? – Urs Schreiber Nov 18 2009 at 7:41

Let me assume $n=\infty$, to make things easier to write, so "$(\infty,r)$-$\Theta$-space" equals "$r$-$\Theta$-space".

The totality of $r$-$\Theta$-spaces forms a (large) category enriched over $r$-$\Theta$-spaces, which I'll call $C$. Given this, you can form a presheaf of spaces $X$ on the category $\Theta_{r+1}$, where

$$X[0] = \text{class of objects of C_r},$$

and

$$X([m](\theta_1,\dots,\theta_m)) = \coprod_{a_0,\dots,a_m} C(a_0,a_1)(\theta_1)\times \cdots \times C(a_{m-1},a_m)(\theta_m).$$

Here "$[m](\theta_1,\dots,\theta_m)$" represents a typical object in $\Theta_{r+1}$ (so each $\theta_i\in \Theta_r$). The coproduct is over tuples of objects of $C$. The structure maps in the presheaf use the fact that $C$ is a category object. (It's like the way you get a Segal category from a category enriched over spaces.)

The gadget $X$ is almost an $(r+1)$-$\Theta$-space. It satisfies all the "Segal" conditions, and also all the completeness conditions except for the one in bottom dimension. You get an honest $(r+1)$-$\Theta$-space $X'$ from $X$ by applying a suitable localization.

The gadget $X'$ should be the thing you want. (None of the proofs involved here have been written up, or at least not by me, though we're working on it.)

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# cosine

## Calculus cos

Calculus in processing ... please wait

### Function : cos

#### Summary :

The cos trigonometric function calculates the cosine of an angle in radians, degrees or gradians.

cos online

# Cosine function

The calculator allows to use most of the trigonometric functions, it is possible to calculate the cosine, the sine and the tangent of an angle through the functions of the same name..

The trigonometric function cosine noted cos, allows to calculate the cosine of an angle online , it is possible to use different angular units : degrees, gradians and radians wich is the angular unit by default.

## Calculation of the cosine

### Cosine calculating an angle in radians

The cosine calculator allows through the cos function to calculate online the cosine of an angle in radians, you must first select the desired unit by clicking on the options button calculation module. After that, you can start your calculations.

To calculate cosine online of pi/6, enter cos(pi/6), after calculation, the result sqrt(3)/2 is returned.

Note that the cosine function is able to recognize some special angles and make the calculations with special associated values in exact form.

### Calculate the cosine of an angle in degrees

To calculate the cosine of an angle in degrees, you must first select the desired unit by clicking on the options button calculation module. After that, you can start your calculus.

To calculate cosine of 90, enter cos(90), after calculation, the restults 0 is returned.

### Calculate the cosine of an angle in gradians

To calculate the cosine of an angle in gradians, you must first select the desired unit by clicking on the options button calculation module. After that, you can start your calculus.

To calculate cosine of 50, enter cos(50), after computation, the result sqrt(2)/2 is returned.

Note that the cosine function is able to recognize some special angles and do the calculus with special associated exact values.

## Special cosine values

The cosine admits some special values which the calculator is able to determine in exact forms. Here is the list of the special cosine values:

## Derivative of cosine

The derivative of the cosine is equal to -sin(x).

## Antiderivative of cosine

The antiderivative of the cosine is equal to sin(x).

## Properties of the cosine function

The cosine function is an even function, for every real x, cos(-x)=cos(x). The consequence for the curve representative of the cosine function is that it admits the axis of the ordinates as axis of symmetry.

## Equation with cosine

The calculator has a solver which allows it to solve equation with cosine of the form cos(x)=a. The calculations to obtain the result are detailed, so it will be possible to solve equations like cos(x)=1/2 or 2*cos(x)=sqrt(2) with the calculation steps.

The cos trigonometric function calculates the cosine of an angle in radians, degrees or gradians.

#### Syntax :

cos(x), where x is the measure of an angle in degrees, radians, or gradians.

#### Examples :

cos(0), returns 1

#### Derivative cosine :

To differentiate function cosine online, it is possible to use the derivative calculator which allows the calculation of the derivative of the cosine function

The derivative of cos(x) is derivative_calculator(cos(x))=-sin(x)

#### Antiderivative cosine :

Antiderivative calculator allows to calculate an antiderivative of cosine function.

An antiderivative of cos(x) is antiderivative_calculator(cos(x))=sin(x)

#### Limit cosine :

The limit calculator allows the calculation of limits of the cosine function.

The limit of cos(x) is limit_calculator(cos(x))

#### Inverse function cosine :

The inverse function of cosine is the arccosine function noted arccos.

#### Graphic cosine :

The graphing calculator is able to plot cosine function in its definition interval.

#### Property of the function cosine :

The cosine function is an even function.
Calculate online with cos (cosine)


text:
# Properties

 Label 188760.cw Number of curves $1$ Conductor $188760$ CM no Rank $0$

# Related objects

Show commands: SageMath
sage: E = EllipticCurve("cw1")

sage: E.isogeny_class()

## Elliptic curves in class 188760.cw

sage: E.isogeny_class().curves

LMFDB label Cremona label Weierstrass coefficients j-invariant Discriminant Torsion structure Modular degree Faltings height Optimality
188760.cw1 188760l1 $$[0, 1, 0, -12745, -853525]$$ $$-504871936/394875$$ $$-179083558368000$$ $$[]$$ $$648000$$ $$1.4335$$ $$\Gamma_0(N)$$-optimal

## Rank

sage: E.rank()

The elliptic curve 188760.cw1 has rank $$0$$.

## Complex multiplication

The elliptic curves in class 188760.cw do not have complex multiplication.

## Modular form 188760.2.a.cw

sage: E.q_eigenform(10)

$$q + q^{3} + q^{5} + 3q^{7} + q^{9} + q^{13} + q^{15} + q^{17} + 6q^{19} + O(q^{20})$$