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during titration, etc.). To find the average titre (titration volume) the values are added together and divided by the number of readings that were taken. A titration is a very commonly used type of quantitative analysis. When … Variable switch programming. Record the initial volume of solution in the burette (buret). For determining the concentration of an analyte, one can use Indirect Titration … Create one now. DEHA (N,N-Diethylhydroxylamine) is added to boiler system water as an oxygen scavenger to keep the dissolved oxygen levels as low as…, Related Products: Dissolved Oxygen Test Kit, …water. $Average\: titre = \frac {(11.2 + 11.2 + 11.2)}{3} = 11.17cm^{3}$. Gather all the necessary equipment before starting. … The Wallace&Tiernan® Series A790 Amperometric Titrator is ideal for the precise measurement of chlorine residuals, including very low residuals, for the calibration of on-line analytical equipment. When an employee is too tired to … Horseplay in both a physical and verbal sense can be equally hazardous and lead to personal injury, product and equipment damage, and/or coworker disputes. For checking calibration of coulometric KF equipment. Features of Titrets Technology Dependable, sealed single-use titrants Fast, easy procedure -- no need to count drops No burettes or other equipment necessary Perform a titration ANYWHERE, …hard water sources. A burette is used to accurately measure the volume of liquid that has been allowed to pour out of it. It is therefore possible to use all this information with a balanced equation to work out the concentration of the reactant in the conical flask. You can check out as a guest! If your titration standards are not reading the correct concentrations, for example, the alkalinity standard reading is low, first make sure the titrant has been standardized. Precision and Uncertainties for Common Lab Equipment When you record a scientific measurement, the last digit that you record is understood to have some uncertainty, and to be your best estimate. Titrations are carried out quickly the first time to get a rough idea of the approximate volume that is needed to reach the end point. Find Automatic Lab Titration Equipment at SpectrumChemical.com. The titration would then be carried out very carefully at least two more times and an average taken. In this type, a titrant of known concentration and volume is added to a substance in order to analyze it. A pipette is used to put an accurate volume of reactant in the conical flask. Hard water causes scaling in boilers and other industrial equipment, and diminishes the effectiveness of soaps and detergents. The following table shows the results from a titration in which sulfuric acid of concentration, was added from the burette into a conical flask containing 20 cm, Religious, moral and philosophical studies. Related Products: Karl Fischer Water Standard, …observing color changes (e.g. Eisco Labs Advanced Titration Kit - 50ml Burette - Borosilicate Class A, 24" (60cm) Rod, 8"x5" Heavy Base, Clamp, Boss Head 2.0 out of 5 stars 1 $60.10$ 60 . Direct reading of total alkalinity from 0 to 200ppm as…, …thirty ampoules with valve assemblies, titrettor, 25 mL sample cup, and instructions. EasyPlus™ Accessories are for general and Karl Fischer titration. Water in xylene. Acccessory for solarus® titration units (Thomas Nos. It is a long, glass tube with a tap at the end which can … The start volume is read from the burette before the titration is started and at the end point the final volume is read from the scale on the burette. Product Directions The Hydrion Water Hardness Test Kit eliminates the need for expensive laboratory equipment or complicated titration kits. Browse through our website and find the best titration … SpectrumChemical.com carries a full line of Lab Meters and Testers. Each AccuPlate™ is equipped with a Pyroceram® ceramic top plate that is extremely durable, easy to clean and is highly resistant to chemical attack. Skip … An acid-base titration is an experimental procedure used to determined the unknown concentration of an acid or base by precisely neutralizing it with an acid or base of known concentration. Read about our approach to external linking. Titration is therefore an important part of the pharmaceutical industry to ensure quality control and that the right levels of concentration are within medication produced. In a hurry? Our tips from experts and exam survivors will help you through. The Thermo Scientific™ Orion™ Total Alkalinity Test Kit provides two-step direct measurement of total alkalinity and pH using a pH electrode. Make sure you have a calibrated burette, a … This reading is the "rough titre" and is not used to calculate the average. Required practical Determination of the reacting volumes of solutions of a strong acid and a strong alkali by titration. 36 matches found for titration equipment . Magnetic back allows timer to be placed on incubators, titrators and most lab equipment. The following table shows the results from a titration in which sulfuric acid of concentration $$0.1 moll^{-1}$$ was added from the burette into a conical flask containing 20 cm3 of sodium hydroxide solution. Subscribe to Our Email List Sign up and receive \$5 Off and get exclusive access to promotions, sales events, pre-order sales & … Turbidimeters; Turbidimeter Accessories; Washers and Dryers For Glassware. Advanced Search | Structure Search. AREC is the first hot plate stirrer with an entire technopolymer structure, extremely innovative and ideal for premium resistance to acids, bases and solvents. Titration Equipment at Thomas Scientific Titration Equipment found in: Series A790 Amperometric Titrator, SmartChemical Reader, FF-1A Fish & Farming Nine Parameter Kit, Orion Total Alkalinity.. YSI offers two titration product lines including autotitrators, manual titrators, dosing applications and solution preparation. The end point can be detected by the colour change of an indicator in the flask or by measuring pH or conductivity. Fitting wrench for valves. Often specialised equipment more … It eliminates the need for technically trained personnel. … *Please select more than one item to compare. Fatigue. Accessory for opus® titration, solarus® and akku-drive®. It is based on accurately measured volumes of chemicals. Notice the rough volume is not used to calculate the average. Calcium and magnesium are the most common minerals that contribute to hardness. Pedal switch for triggering dispensing and titration. Search term: "titration equipment" Compare Products: Select up to 4 products. You can always refer google for the pictures of the above apparatus. Alkalimetry, or alkimetry, is the specialized analytic use of acid-base titration to determi… It is based on accurately measured volumes of chemicals. A titration is a very commonly used type of quantitative analysis. Chemists monitor our environment using a variety of quantitative and qualitative analysis techniques. SmartChemicals equipped titrators are compatible with a…. Titration equipment. The equipment we use in titration has the following tolerances: Burette 0.1 cm3; 25 cm3 pipette 0.06 cm3; 250 cm3 volumetric flask 0.3 cm3; 3 decimal place balances: 0.001 g; Each item of … Titration Accessories by METTLER TOLEDO are more than just small parts. Terms & Conditions. during titration, etc.). The AREC is designed to last and equipped to ensure maximum protection against…, …observing color changes (e.g. The Amperometric Titrator is also used in large swimming pools and food plants,…, …and Excellence line titrators. It also features a flip-open easel for standing on the lab bench, a spring fastener for clipping to a lab coat and a hold for a…, …observing color changes (e.g. This lets us quantitatively analyze the concentration of the unknown solution. The airflow, tidal … For accurate results; simply tear off a strip of test paper,…, …numbered keys make setting time faster than units with up/down buttons. For checking calibration of coulometric KF equipment. Don't have a web profile? We Believe You Are Important, How Can We Help? Accessory for opus® titration units (Thomas Nos. Swirl the conical flask while adding solution from the burette (buret) until you are close to, say within 80% of, the expected end point based on the rough titration… If you're a lab manager, an entrepreneur who has a unique business enterprise, a purchasing agent, or maybe a construction contractor building a new laboratory we can aid you in finding the specific Titration equipment … Spectrum Chemical has a complete line of laboratory supplies, equipment … AREC.X is the first hot plate stirrer with an entire technopolymer structure, extremely innovative and ideal for premium resistance to acids, bases and solvents. during titration, etc.). Acid-base titrations can also be used to quantify the purity of chemicals. Titration (also known as titrimetry and volumetric analysis) is a common laboratory method of quantitative chemical analysis to determine the concentration of an identified analyte (a substance to … …DR/800 and DR/2400 Spectrophotometer platforms Nine-Parameter Kit - Model FF-1A: A test kit for professionals, this model includes nine basic water quality tests. Washer and Dryer … In addition, the ceramic…, ©Thomas Scientific 2021 All Rights Reserved. Your equipment should be set up once again as shown in the diagram. This value is too big since it is unlikely to have been stopped exactly at the endpoint. Titration Equipment Kit: 100 ml Glass beaker; 50 ml Burette with PTFE stopcock; Ring stand; Burette clamp; Glass funnel; 100 ml Erlenmeyer flask; 50 ml Polypropylene graduated cylinder; 100 ml … Karl Fischer Reagents; Titrator Parts and Accessories; Titrators; Turbidity Meters. The results from quantitative analysis are used in calculations that give essential information. The AREC.T is designed to last and equipped to ensure maximum protection…, …three models to choose from, there is an AccuPlate™ hot plate and stirrer to meet your lab’s needs. 1200M13, 1200M14 and 1200M15). Recommendations for NPPV Titration Equipment: The NPPV device used for titration should have the capability of operating in the spontaneous, spontaneous timed, and timed mode. Titration. Reagents and equipment are supplied for drop-count titration and colorimetric measurement using continuous-gradient color discs. In a titration the liquid in the burette is allowed to slowly run into the conical flask until an end point is reached. Water 0.50 +/- 0.02 mg/g NIST Traceable, Dissolved oxygen in boiler system water causes corrosion and pitting of metal surfaces, which can lead to boiler inefficiency, equipment failure, and system downtime. Thermometric titrator for universal titration applications, especially when no proper potentiometric sensor is available or when potentiometric titration is limited due to contamination of the sensor, e.g., … The AREC.X is designed to last and equipped to ensure maximum protection…, Related Products: Hot Plate With Magnetic Stirrer. Nitrite Titrets Kit, Range: 250-2500 ppm as NaNO2, AREC.X Digital Ceramic Hot Plate Stirrers, WaterMark® KARL FISCHER WATER STANDARD, 0.10 mg/ml, WaterMark® KARL FISCHER WATER STANDARD, 0.10 mg/g, Hardness (Total) Titrets Kit, Range: 100-1000 ppm, Hardness (Total) Titrets Kit, Range: 2-20 ppm, Hardness (Total) Titrets Kit, Range: 20-200 ppm, WaterMark® KARL FISCHER WATER STANDARD, 0.05 mg/g (50 ppm), AREC.T Digital Ceramic Hot Plate Stirrers with Timer, AccuPlate™ Hotplates, Stirrers and Hotplate Stirrers. Titration, process of chemical analysis in which the quantity of some constituent of a sample is determined by adding to the measured sample an exactly known quantity of another substance with … Performing a titration requires that you have all your equipment together before you start. Acid-base titrations are used to determine the concentration of a sample of acid or base and are carried out using a piece of equipment called a burette. Metrohm AG is a leading provider of instruments and know-how for chemical analysis in the lab and in the process, specializing in titration, ion chromatography, electrochemistry, and spectroscopy. 10 A pipette is used to accurately measure a fixed volume of liquid and is filled using a pipette filler to a line on the upper thin part of the tube. The apparatus used in a titration are burette, pipette (most of the times 10 ml ones are used), measuring flask (250 ml), two beakers (100ml and 250 ml), measuring cylinder (used while taking 10 ml 4N H2SO4) and a burette stand. Mettler Toledo + Titration equipment Mettler Toledo Titration equipment. The EDTA Method (Total) References: APHA Standard Methods, 22nd ed.,…. For high level needs of automation, METTLER TOLEDO provides you with a wide range of accessories, instruments and service possibilities. Save on Titrators, Autotitrators, & Titration Equipment at Amazon's Lab Equipment Store, featuring every day low prices on Lab & Scientific Equipment. AREC.T is the first hot plate stirrer with an entire technopolymer structure, extremely innovative and ideal for premium resistance to acids, bases and solvents. … 1221U68, 7755T80 and 7755T85). Suitable for opus ® dispenser and opus ® titration. Secondly, the precision of the results … Since the concentration and volume of the solution added from the burette are known, it is possible to work out the number of moles of the reactant that were added to the conical flask to reach the end point. For checking calibration of coulometric KF equipment. The Task Force recommends that the indications, rationale for use, and side effects should be discussed in detail with the patient or caregiver preferably prior to the PAP titration study; parts and assembly, optional equipment, importance of daily/nightly use, adherence issues, necessity of cleaning the equipment… The TitroLine titrators … Direct titration is the most basic titration which is commonly used. Save time by eliminating additional equipment set-up, titrations and calculations. automatic titration equipment Steroglass - Model ACT/ACT2 - Automatic Titrator Titrex ACT family has been designed to simply and precisely satisfy and perform the widest range of potentiometric … The volume of acid added is the final volume minus the start volume. A brief contactless touch of reagent bottles to the reader instantly and securely registers all relevant titration chemicals data: reagent name, lot/batch number, concentration, and shelf life/expiration. Titrators, Autotitrators and Titration Equipment. Most Lab equipment EDTA Method ( Total ) References: APHA Standard Methods, 22nd,! And an average taken than one item to compare more times and an average.... A titrant of known concentration and volume is added to a substance in order to analyze.... … titration and equipment are supplied for drop-count titration and colorimetric measurement using continuous-gradient color discs be. 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# Is the microfacet GGX BSDF normally implemented as separate BRDF and BTDFs?
I'm trying to build a toy path tracer using something similar to Disney's BSDF, where many materials can be represented as combinations of properties like color, metalness, roughness, transmission, ior.
I'd like to use the GGX microfacet BSDF, but without splitting the ray each time to get & sum the Fr and Ft terms for reflection and transmission. Is it normal/ok to use schlick's approximation to randomly choose between transmission and reflection (Ft vs Fr), and then to just use the BRDF and BTDF formulas separately depending on which path the system chooses?
For example, roughly:
On hit (into)
- Reflect or Transmit via Schlick(i, o, n)
- Reflect:
- Direct (N times):
- Choose point on disc of random light
- Calc transmission t via BRDF(i, o, n)
- Calc pdf by area of cone / area of hemisphere
- Add light * t / pdf
- Indirect (1 time):
- Choose direction o by BRDF.Sample(i, n)
- Calc transmission t via BRDF(i, o, n)
- Calc pdf by BRDF.PDF(i, o, n) ??
- Add trace(o) * t / pdf
- Transmit:
- Choose direction o by snell(i, n) + some kind of roughness * offset
- (how do you apply the BTDF here? Or is it important?)
On hit (out of)
- Choose direction o by snell(i, n) + some kind of roughness * offset
- Reduce transmission via beer(distance)
I also wonder how diffused reflections factor in. For instance, does GGX only provide specular reflection & refraction? In that case, how do you also handle diffusion?
• It is okay to use Fresnel term as a pdf to choose between reflection and refraction. Because this estimator will contain a Fresnel term on both nominator and denominator, they are cancelled. So, you can directly use brdf or btdf. May 30 '18 at 4:40 | |
# How do you factor completely: 12x^5 + 6x^3 + 8x^2?
Jul 14, 2015
$12 {x}^{5} + 6 {x}^{3} + 8 {x}^{2} = 2 {x}^{2} \left(6 {x}^{3} + 3 x + 4\right)$
$= 12 {x}^{2} \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) \left(x - {x}_{3}\right)$
where ${x}_{1} , {x}_{2} , {x}_{3}$ are defined below.
#### Explanation:
$12 {x}^{5} + 6 {x}^{3} + 8 {x}^{2} = 2 {x}^{2} \left(6 {x}^{3} + 3 x + 4\right)$
Let $f \left(x\right) = 6 {x}^{3} + 3 x + 4$.
This is way too messy to solve, but for the record...
Use Cardano's method to solve $f \left(x\right) = 0$.
Let $x = u + v$
$f \left(x\right) = 6 {\left(u + v\right)}^{3} + 3 \left(u + v\right) + 4$
$= 6 {u}^{3} + 6 {v}^{3} + \left(18 u v + 3\right) \left(u + v\right) + 4$
Let $v = - \frac{1}{6} u$.
Then $18 u v + 3 = 0$ and
$f \left(x\right) = 6 {u}^{3} - \frac{1}{36 {u}^{3}} + 4$
If $f \left(x\right) = 0$ then:
$6 {u}^{3} - \frac{1}{36 {u}^{3}} + 4 = 0$
Multiply through by $36 {u}^{3}$ to get:
$216 {\left({u}^{3}\right)}^{2} + 144 \left({u}^{3}\right) - 1 = 0$
${u}^{3} = \frac{- 144 \pm \sqrt{{144}^{2} + 4 \cdot 216}}{2 \cdot 216}$
$= \frac{- {2}^{4} {3}^{2} \pm \sqrt{{2}^{8} {3}^{4} + {2}^{5} {3}^{3}}}{{2}^{4} {3}^{3}}$
$= - \frac{1}{3} \pm \frac{{2}^{2} 3 \sqrt{{2}^{4} {3}^{2} + 2 \cdot 3}}{{2}^{4} {3}^{3}}$
$= - \frac{1}{3} \pm \frac{\sqrt{150}}{36}$
$= - \frac{1}{3} \pm \frac{5 \sqrt{6}}{36}$
Since this derivation has been symmetric in $u$ and $v$:
Let $u = \sqrt[3]{- \frac{1}{3} + \frac{5 \sqrt{6}}{36}}$
and $v = \sqrt[3]{- \frac{1}{3} - \frac{5 \sqrt{6}}{36}}$
The roots of $f \left(x\right) = 0$ are:
${x}_{1} = u + v$
${x}_{2} = \omega u + {\omega}^{2} v$
${x}_{3} = {\omega}^{2} u + \omega v$
where $\omega = - \frac{1}{2} + i \frac{\sqrt{3}}{2}$
${x}_{1}$ is the real root of $f \left(x\right) = 0$
${x}_{2}$ and ${x}_{3}$ are a pair of complex conjugate roots.
Hence $f \left(x\right) = 6 \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) \left(x - {x}_{3}\right)$
and $12 {x}^{5} + 6 {x}^{3} + 8 {x}^{2} = 12 {x}^{2} \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) \left(x - {x}_{3}\right)$ | |
ceiling(RWeilDivisor) -- produce a WeilDivisor whose coefficients are ceilings or floors of the divisor
Description
Start with a rational or real Weil divisor. We form a new divisor whose coefficients are obtained by applying the ceiling or floor function to them.
i1 : R = QQ[x, y, z] / ideal(x *y - z^2); i2 : D = divisor({1/2, 4/3}, {ideal(x, z), ideal(y, z)}, CoefficientType => QQ) o2 = 4/3*Div(y, z) + 1/2*Div(x, z) o2 : QWeilDivisor on R i3 : ceiling( D ) o3 = 2*Div(y, z) + Div(x, z) o3 : WeilDivisor on R i4 : floor( D ) o4 = Div(y, z) o4 : WeilDivisor on R i5 : E = divisor({0.3, -0.7}, {ideal(x, z), ideal(y,z)}, CoefficientType => RR) o5 = -.7*Div(y, z) + .3*Div(x, z) o5 : RWeilDivisor on R i6 : ceiling( E ) o6 = Div(x, z) o6 : WeilDivisor on R i7 : floor( E ) o7 = -Div(z, y) o7 : WeilDivisor on R | |
Projects :: Me on QRZ :: HF Radio Fund: £35 of £500: 7% funded
## Friday, 6 March 2015
### "Seeing Circles, Sines and Signals" - a DSP primer
Not had a detailed read through it yet, but this seems to be a very well-written visual approach to DFT and digital signal processing in general.
(Also, I've now finished the last M0 lesson and am about to begin on the next!)
## Friday, 27 February 2015
### Capacitive CW touch key by MI0OIM
It looks the part as well. Now I've got no excuse not to learn.
## Tuesday, 24 February 2015
### My Thoughts on Anytonetech
So Brick and Hans have posted their opinions on the new Anytonetech announcement, so I think it's appropriate I follow suit, seeing as I've been very vocally critical of it elsewhere.
A big hoo-ha was made about it; and a lot of buzz was generated on the Internet about BaofengTech's new announcement. It didn't take me long to clock on to what the announcement was.
Many folks, including myself, were hyped for seeing a new Baofeng radio; alas, it didn't happen. I'm not even mad about this. More choice in the world of handheld radios is a good thing.
What irks me is the fact that this announcement was literally nothing but a marketing exercise. Having media media qualifications and work experience under my belt, I can see through the announcement for exactly what it is. The radios Anytonetech are branding as the next big thing in Chinese handhelds have already existed and been on sale for months.
What Anytonetech are doing is simply sticking their own names on the Anytone AT-398UV and AT-288 radios, using the "R" prefix used by Baofeng and misrepresenting them as related to the originals, and they will probably flog them at a significant margin. Why can't they just be honest and sell them as they are? You want an "Anytonetech TERMN8R"? Go buy a frigging AT-398UV! It's THE SAME RADIO! Jeez louise. It even has JTAG pads on the board for firmware upgrading. You can disable the VFO to make it Business Radio compliant. It receives shortwave. And so on.
On top of this, a lot of people still think these are new Baofeng radios. They're not! Let me say this again. They are Anytone radios. Made by Anytone. That already exist and are sold by Anytone on their Aliexpress store.
There is a worrying lack of innovation happening in the U/V HT world at the moment. Anytone, Baofeng, Wouxun, and friends: go and read blogs and the comments. People don't want yet another Yaesu Vertex clone. They want innovation, and innovation that is not even difficult to implement.
This is why, unless something changes soon, then an OSHW handheld U/V radio might be on the cards. If Corporate China won't innovate, we will.
Watch this space.
### The Last Leg: Zero to M0: Part Three: The *other* LCRs
Welcome to part three of this epic quest through the Advanced syllabus. I'm currently waiting for my copy of Advance! to turn up as yet, so expect corrections when it does arrive.
Now we start getting very technical. And this is the easy stuff. So hold on...
L, C and R are the reference designators for, respectively, inductors, capacitors and resistors. These alone are the building blocks for pretty much every RF circuit in existence: if you had an unlimited supply of all these components - along with the humble transistor, but more on that in a latter writeup - you could build literally any radio transmitter or reciever you wanted. It would be tedious, but it could be done.
When you combine these, they do various things to a signal you pump into them. Let's go over what's what:
• An Inductor (L, which stands for "(Heinrich) Lenz", the discoverer of inductance) blocks alternating-current (AC) signals at the inductor's resonant frequency while passing DC. You see large inductors in power supplies for the purposes of eliminating ripple and noise. The unit of measurement is the Henry (H).
• A capacitor (C) blocks DC while passing AC at the capacitor's resonant frequency. It can store a charge across its plates (terminals) and is often used to stabilise (or smooth) transients by "shunting" any variation in the supply voltage to ground. The unit of measurement is the Farad (F).
• A resistor blocks both AC and DC linearly (i.e., by the same amount at all frequencies). Resistance is measured in Ohms (Ω). You'll have noticed that the inductor and capacitor are the inverse of each other: the resistor, which has resistance in Ohms, has an inverse too: the transistor, which amplifies both AC and DC linearly and has transconductance measured in Siemens (S). Going further, AC resistance and transconductance that changes with frequency is called impedance and admittance respectively.
If you're confused about resonant frequencies, I'm going to cover those in the next post, but suffice to say it's the frequency at which the component works "best".
We should probably recap here, since you'll be using these symbols a LOT later on:
Property of.. Measured in Symbol in equations Inductance an Inductor Henries (H) L Capacitance a Capacitor Farads (F) C Resistance a Resistor Ohms (Ω) R Transconductance or Gain a Transistor or Valve Siemens (S) β or gM Impedance (frequency dependent resistance) Various Ohms (Ω) Z Admittance (frequency dependent transconductance or gain) Various Siemens (S) Y
The simplest of these is probably resistance. No doubt at this level, you have already encountered resistance, so I won't go into too much detail here. But you should at least be able to recall that resistance:
• is opposition to the flow of free electrons round a circuit
• dissipates current (and therefore power as heat) across itself according to the formula I = V/R
• works equally at all frequencies, from DC (f0 = 0Hz) through all AC frequencies to daylight
• works equally at all voltages
and that resistors have a positive temperature coefficient; that is, that their resistance increases with heat.
You will also need to know how to calculate parallel and series resistances:
To calculate series resistance, simply add them together.
RTotal = R1 + R2 + R3
Calculating parallel resistance is a bit more challenging. Other folk often give this in fractions, as does the formula sheet you get at the exam:
$\frac{1}{R_{total}}&space;=&space;\frac{1}{R4}&space;+\frac{1}{R5}&space;+\frac{1}{R6}&space;+\frac{1}{R7}$
But if your maths isn't that hot, calculating the reciprocal on a calculator is a bit... huh? Where's the "1 over" button?
Nine times out of ten, there isn't one. Rather than using the above, you would enter it in the calculator as
$R_{total}&space;=&space;(R4^{-1}&space;+&space;R5^{-1}&space;+&space;R6^{-1}&space;+&space;R7^{-1}&space;)^{-1}$
It's actually exactly the same calculation, just in a form a bit easier to enter into a calculator :)
You should also know how to formulate a resistive divider, which allows you to reduce an input voltage by the ratio between two resistors, shunting the "unused" voltage to ground:
Resistive Divider
Inductance is calculated the same way as resistance. Simply substitute R for L.
Series:
Parallel:
Calculating series and parallel capacitance is simply a matter of swapping over the sum and reciprocal. If you think about it, this makes sense: essentially, with capacitors in parallel, you are simply making the "plates" of the capacitor bigger, increasing the "capacity" of the capacitor bank.
So for capacitors, the formulae are:
Series
Parallel
You may be thinking to yourself: these values can't just be arbitrary, can they? You know, if you're making a capacitor that is 2200uF or a 100nH inductor, there has to be some way they can work out the dimensions and characteristics for a given value. And you'd be right.
For inductors, there are many factors: the gauge of the wire. the spacing between the turns, the magnetic permittivity of the former and so on. So many factors, in fact, it's not tested in the exam, and this is left as an exercise for the reader.
You are however tested on how to calculate capacitance: this is simple by comparison. All you need to know are two three things:
• The dielectric constant (K).
• The area of the plates (A).
• The distance between the plates (d).
Recall that a capacitor is simply a pair of electrodes (plates) separated by an insulating material called a dielectric. K is calculated by taking the dielectric permittivity (or "charge-holding ability", in layman’s terms) of free space, or in other words a vacuum, and multiplying it by the dielectric permittivity of the insulating material.
The dielectric permittivity of free space, with the symbol ε0, is 8.854 x 10-12 F/m, with farads per meter (F/m) being the measure of dielectric permittivity.
The electrolyte in an electrolytic capacitor often uses ethylene glycol as a base, which has a relative permittivity (εr) of 37 F/m. Wikipedia has a list of materials and their relative permittivities here. So you would simply multiply ε0 and εr, and you have your K.
The formula for actually turning this into a capacitance, is
Thankfully, you are only expected to be able to recognise the formula on sight and not apply it; but if you, say, are building a magnetic loop and are building your own capacitors, this formula could save your bacon!
So that's it for this time. This was a brutal post to write, mostly thanks in part to the fact this is basic school-level stuff and hence tedious to write. But it's fundamental to the rest of the exam, so it's need-to-know. If you don't, much of the rest of this lot won't make sense!
Interesting things happen when you combine capacitors, inductors, and resistors, and that's the theme for the next post.
Edit: my copy of Advance! showed up, so here goes! Note that Codecogs is currently broken and as such no equations will display right now.
The value of the capacitor is also a measure of how much charge it will hold at a given voltage. The formula for this is:
C = Q/V
where C is the capacitance (in Farad), Q is the charge held by the plates (measured in Couloumbs and V is the potential difference (voltage) applied across the plates. The capacitance is determined not by Q and V but rather this is for deriving Q from C & V.
A capacitor of 1 farad will store 1 couloumb of charge for an applied voltage of 1V.
Different types of dielectrics
• Paper capacitors are often very large with high capacitances and a high flashover voltage.
• Polythene & polypropylene capacitors are plastic dielectics. They can withstand even higher voltages than paper dielectrics, but they are lossy due to absorption of charge into the dielectric itself, which is dissipated as heat. The losses increase with frequency, so they are unsuitable for VHF and up.
• Polystyrene, PTFE and mylar dielectrics are more stable & less lossy. But also far more expensive.
• Ceramic capacitors have relatively low loss and are cheap, making them ideal as a balanced option for RF use. They can have a variety of dielectrics, as Wikipedia points out (see image above).
• Mica is a very low loss ceramic dielectric, but it's expensive (for a reason).
• Air is used as the dielectric in a variable capacitor: turning the shaft increases the overlapping plate area and hence the capacitance.
• Electrolytic capacitors use aluminium foil one one plate and a conductive paste on the other; due to this imbalance between plates they are said to be polarised and must be connected the right way round.
As a capacitor charges, the amount of current it draws off the supply will decrease as the potential difference between the plates increases. The time taken for the voltage to reach 2/3 through a given load (resistance) is known as the RC time constant and can be figured out using the formula
τC = RC
hence the name.After approximately 5τ seconds the capacitor is pretty much fully charged, but due to the law of diminishing returns that applies to capacitor charging it can never be fully charged. Remember that the current that is charging the capacitor decreases as time goes on!
Inductors work by storing a magnetic field in the coil. As this magnetic field increases, the potential difference across it decreases - Lenz's back EMF phenomenon - whilst the amount of current flowing through it increases. Like many things about inductors, they are inverse with respect to capacitors.
So no prizes for guessing the inductive time constant, the time the inductor takes to build up its local magnetic field to maximum...
τL = R/L
## Saturday, 21 February 2015
Wrong LCRs. I know the next tutorial part on passives is due but I simply haven't had the time. But no. This is about Little Chinese Radios.
Love them or hate them, they're here to stay. This post is about three of them in particular; a mini-review from someone who's used them as primary radios for some length of time, if you will. If you're new to amateur radio, or radio at all, you're very likely to have used one of these; in addition, I'm going to review the cheap piece of hardware that got me into this hobby again in the first place.
If you're considering buying one of these, all but the last item require a licence to use pretty much anywhere in the world. Do not transmit without one!
### The Baofeng UV-B5
This was my first radio and to this day remains the one I recommend to everyone starting out in the world of amateur radio. For £21 shipped from China direct, it really cannot be beaten for value for money.
There is a B6 variant as well with a flashlight instead of a rotary dial, but I had an inkling I would prefer the rotary dial on top instead. I was right.
So how does it fare?
Well, I never actually got a chance to test it on the amateur bands, because as fate would have it, literally on the day I went for my exam... I managed to shut the attached antenna, a rigid and beefy Sharman RH770, in a car door, and because it has no "give", it ripped the antenna socket clean off. After soldering it back on, the squelch would no longer open; repairing a broken trace around the AGC fixed that but now probing the RDA1846 RF generator chip showed no audio output. So transmit works, but not receive. Ooops.
All is not lost, though. I have another B5 on the way. I did still get to use it on Business Light a few times, so here goes:
RF Output: After all, this is of course why you would buy one of these in the first place. It's rated for 5W output with harmonic suppression of -60dBc. I measured 5.02W and -66dBc using a very expensive Keysight/Agilent test set. Spot on. Even and balanced audio reports from the other end of my contact; they literally couldn't tell it apart from the "standard budget HT" Yaesu FT60 which is six times the price!
Receiver: Clear, LOUD audio. Never had a problem with hearing folk, at least until its unfortunate car door accident. There is a pager transmitter local to me which makes the "newer" handheld spaz out, but the B5 had no issues dealing with 800W of VHF radiation a handful of metres away. It's specified for an intermod rejection/ACS figure of 60dB but that's a conservative measurement, I think. Sensitivity is the standard 0.2 μV @ 12dB S/NAD (i.e. the voltage at which signals start to be distinguishable from noise) but anecdotal reports from others suggest this varies a bit.
Features: ...it's a budget radio. It receives. It transmits. It's the ham radio equivalent of a Nokia 100 and I'm cool with that. I'm not going to masturbate over LCD colours or tone scanning or lack of System Fusion or APRS or DSTAR or any of that. "It Just Werks".
Quirks: The five digit display is enough to store names of repeaters (GB3XX) but God help me if I want to store my local ILRP gateway (GM1PLY). The superfluous third button on the left hand side is annoying, but can be hooked up as a backlight toggle if you're very careful. I intend to practice this backlight mod on the old handset and then if it's successful mod the new one too. Finally, I've triggered the big orange alarm button more than once. Completely unnecessary for amateur use, I removed the button cap to avoid pushing it by accident.
Overall: If you're new to radio, as balance between performance and price goes, this is literally the gold standard against which all other budget HTs are judged. At least for now...
### Baofeng UV-5RA
After having used it for some time, I'm now convinced £25 may even be too much for one of these. It looks like a toy, it feels like one, and sounds like one on the air. If you have nothing else to compare it to, I guess it's okay, but I've been relatively spoiled by the B5, I think.
RF Output: Rated for the same as the B6. 3.8W out, -50dB. Power much less than advertised and legal by a gnat's pube to use on the amateur bands in the UK! The UV5R's are all instantly recognisable on the air due to their tinny, shallow audio. You literally have to shout into the mic to be heard, and this doesn't go away with a speaker mic, either.
Receiver: The less said about it the better. The UV-5RA frontend does not like strong out of band signals one little bit. With the squelch set to maximum, I still get SCRREEEEEEEE SCREEEEEEEEE Every. Five. Seconds as a new pager transmission comes through a mere 10MHz away from where the radio is sitting. I'm getting a little tired of "all stations on net difficult due to local QRM", hence the replacement B5. And hey, I'll have spares now, woo! So to sum up: unusable near powerful VHF transmitters, but okay in the middle of nowhere.
Features: More features than the B5, but I think not enough to justify buying one. It can decode DCS and CTCSS from a transmission (useful only to pirates, really, in order to break into Business Band comms), three colour LCD and a decent screen. If these features are important to you, cool. But they're not for me.
Quirks: How many years do you have? :)
Overall: I don't see what all the fuss is about, or for that matter, how people can so readily recommend the UV-5R system. It's junk. As soon as the new B5 arrives, it's going to be put to work possibly doing APRS or something, but certainly not for voice. I wouldn't even sell it as I wouldn't want someone's first taste of amateur radio to be this. "Caveat emptor: this radio is s**te!"
### Baofeng BF-888s
Take the B5, remove the screen and the keypad, make it UHF only, halve the power output from the final amplifier and give it an annoying voice that proclaims PLEASE CHANGE BUTTER when the battery is running low, and you have the BF-888s. Similar on-air performance, which for £15, isn't too shabby at all.
I wouldn't recommend it as a first radio, but if you need a handset that's 70cm/UHF Business Light only, and doesn't completely suck, you can't really go wrong with it. It feels cheap and shabby but has taken a fair bit of abuse with no issues.
### Omake: Realtek RTL2832U "SDR"
As a software defined radio, alongside the Flex, or the BladeRF, or the SDRIQ, or hell even the Softrock RXTX, it has next to no dynamic range, hardware bugs out the wazoo, it's electrically noisy and it's USB2 (and therefore limited in bandwidth): by the numbers, it should be good only for the bin.
It's a f*cking amazing piece of kit that I don't know what I'd do without. As well as DVB-T, DAB and FM reception:
• couple it with an up converter and it's a general coverage HF receiver
• couple it with a downconverter and it can be a satellite or radar receiver
• out of the box it makes a halfway decent spectrum analyser
• add a noise source and a directional coupler, and it's a vector network analyser
• add a transmitter and you can use it as an APRS digipeater
• it can be used to listen to Airband, DMR, trunked/fleet radio, marine traffic, amateur bands, public safety, GSM, you name it, if it goes over the air, this thing can probably pick it up.
The best bit? They cost about £7. For this insane balance of price and flexibility, this is by far and away my favourite bit of RF kit.
And, it's also what got me to take a second look at this hobby, and pick up where I left off many years ago...
## Monday, 16 February 2015
### The Last Leg: Zero to M0 Part Two: Electric Boogaloo
What an apt title. Ha.
No waffling intro today: straight into the heart of the delicious meaty pork pie that is voltages. Yes, I'm aware that sentence is gramatically incorrect; no, I don't give a flying f....
Warning: this post contains maths, which may be unsuitable for small children and arts students. It also contains violence and scenes of a sexual nature from the start and throughout*.
*this may be a lie.
SI Prefixes, Revisited
When working with quantities such as voltages that can vary significantly in value, it's a royal pain in the backside to write out whole numbers. The big fat cables that carry electricity to my place don't say Danger of Death: 275,000 volts because that just sounds and looks a bit ridiculous.
So instead, the sign says Danger of Death: 275kV which fits on that tiny sign a lot better, and even the dimmest of lay-folk know that anything with kV on it probably shouldn't be touched.
Kilo then, obviously, stands for a thousand, alongside being the phonetic letter K.
If you've worked with computers, you've also already met Mega, Giga and Tera, generally suffixed with "bytes", and ham radio and computing fit together like butter and crumpets so I'm not going to duplicate effort here.
You should, however, be able to convert these to powers of ten:
x kilo = x thousands = x, followed by three zeros: x×103
x mega = x million = x, followed by six zeros: x×10
x giga = x billion = x, followed by nine zeros: x×10
Ergo, 275kV is 275×103V. Simple, eh?
This is what is known as engineering notation, and I use it a lot from here on out, so you should probably nail it here before continuing.
What happens if you stick a minus in front of those powers? You get Real Man Fractions:
x milli = x thousandths = 0. followed by up to three zeros and then x: x×10-3
x micro = x millionths = 0. followed by up to six zeros and then x: x×10-6
x nano = x billionths = 0. followed by up to nine zeros and then x: x×10-9
x pico = x trillionths = 0. followed by up to twelve zeros and then x: x×10-12
So when you see a capacitor that says it's value is 68pF, its value in farads is 0.000000000068F. I say "up to" because you'll notice there are only 10 zeros there: 68 takes up two of those digits, so you work back towards the initial 0. Small numbers are weird like that. 68pF is a really small capacitance: conversely, 1F is a very, very large capacitance and you don't want to be on the receiving end should you bridge that big bad motherf****r with your fingers by accident (what fingers?)
TOUCH ME, YOU BIG GIRL. GO ONNN. MWAAHAHA.
Electromotive Force (EMF) and Potential Difference: Not One And the Same After All.
I spent well in excess of a decade believing these were the same thing.
Oh, and I felt I needed to clarify EMF here: since a bunch of sad lonely people think "EMFs cause cancer" cannot into science, they will inevitably come here, look for articles regarding how radio waves cause cancer, AIDS and Ebola, and leave disappointed. If this is you, go away, now, and don't come back, ever.
When you think about it, electromotive force and potential difference... it's there in the name, isn't it? While they're both measured in volts, electromotive force is the "push" that a battery or other power source gives to electrons going round a circuit, while potential difference is the difference in electrical potential energy between two points on a circuit.
All batteries have an intrinsic resistance, known as the ESR or equivalent series resistance. This is also known as the source or input impedance (impedance simply being a fancy name for resistance; hence the reason it's measured in ohms). When you are measuring a battery in-circuit, you're not measuring the EMF but the potential difference with the battery's ESR applied. Measure the battery out-of-circuit, though, and since there is no current flowing through it, the ESR does not apply and only then are you measuring the EMF of the voltage source.
If you short a battery - never a good idea - the battery's entire EMF is dumped across this internal resistance, which often is only a few ohms, plus the (pretty insignificant) resistance of the wire. For example, if we have a 9V battery that has an ESR of 2 ohms, we can use Ohm's Law to determine the current flowing across that short:
I = V/R
I = 9/2
I = 4.5A
9V batteries are by and large only rated to deliver a few hundred milliamps. 4.5A is a fair bit of current, and across 2 ohms will be dissipated as a lot of heat.
If the battery (of a known EMF, as measured out-of-circuit) is in a circuit, and we know how much current is flowing, we can also determine the ESR of the battery itself using the formula
Resr = (Vemf - I×Rload) / I
So, say we have a load resistance of 10 ohms, a measured current of 200mA and our voltage source is 5V:
Resr = (5 - 0.2×10) / 0.2
Resr = 3 / 0.2
Resr = 15 Ohms
This ESR is also why, as you increase the load on a power source, the voltage coming out of the power source will drop. How much voltage droop you get depends on the ESR of the source, which in turn depends to some extent on how well regulated the supply is.
But anyway: that about wraps it up for this session, next time on a series of posts that will probably get a less cliched name, eventually, we'll be looking at more depth at passive components such as capacitors, inductors and resistors, and what makes them work.
Here's the mandatory donation button, in case you feel compelled to do so. Rather than buying me a pint, as one person has already proposed, help me get my grubby mitts on a new radio! :)
## Sunday, 15 February 2015
### The Last Leg: Zero to M0, Part One
I apologise (again) for not posting in a while.
Blame this. That's right: I now hold an Intermediate licence: took my test on the 29th of January at Stirling DARC and got a perfectly respectable 40/45. I was aiming for full marks, but damn, some of those questions were a pain in the backside due to the way they were worded, and having more than one technically correct answer!
So what now?
Well, for one, I'm working toward being able to get my first HF rig: I'm torn between the Yaesu FT857D, or the Icom IC706MkIIG. Within my budget is also the Flex 1500 SDR but I'll be wanting to stick a PA in front of it and there's no VHF/UHF. It'll represent my first significant investment in amateur radio since getting my licence. You can see the status of the "HF Radio Fund" at the top of the page.
Secondly, I'm working toward the Advanced licence now, the big cheese, the top dollar, and the dog's proverbials as far as amateur radio goes.
During the course of my studies, I'm going to be making detailed notes right here on my blog: feel free to use them in your own. I'll be writing the notes as if I'm teaching a third party; that's just how I roll.
The RSGB, among others, suggest a decent (and silent, non-programmable) calculator can be brought into the exam and is essential for Advanced studies due to the slightly more advanced maths involved, and consensus seems to be the good old Casio FX-83GT+ is the best tool for the job. I picked mine up in my local Morrisons for £4 - bargain! It can be had on a certain well-known site whose name is an anagram of 'Yabe' for about three times that.
The first thing you should do is download a copy of the Advanced Licence Syllabus. It's bedtime reading; i.e., if you try and read it all in one go, you're going to end up falling asleep.
The Betts/Hartley book is recommended by many folks to help you study; but really, you don't need it all that much, just make sure you've gone through the syllabus and can tick off each and every point, and you'll be fine. This is intended to (roughly) be the scope of my notes here.
Foundation Intermediate Advanced Power 10W (10dBW) erp pep max 50W erp pep max (17dBW) 400W (26dBW) erp pep max Restricted (no-op) bands 5MHz, 2.4-10GHz 5MHz None Maritime operation No No Yes Can apply for a Notice of Variation (NoV) No Unattended Operation & Beacons only All Can adjudicate exams No Yes, Foundation only Yes Can hold exams (as an Examination Secretary) No No Yes Can hold a Club callsign No No Yes CEPT T/R 61-01 No No Yes
You will, obviously, need to have gained the prior Foundation and Intermediate licences prior to taking the Advanced. This is a huge step up from the Intermediate!
The passmark is 67%, or if you are using the QADV software, you should be aiming for a mark of 85% or more, realistically.
PART ONE Licence Conditions
Something new to you at this licence level, is the fact that folks without a licence can operate your station under supervision.
Huh?
Well, since as an Advanced licensee, you can hold exams, this means that at some point, you'll need to guide a potential Foundation ham through their practical! As long as they are on a registered training course, i.e. one that the RSGB knows is happening and will lead to a licence, then they may operate the equipment for as long as it takes for you to tick off all the items on their Practical Assessment sheet.
In addition, for the purposes of demonstrating amateur radio to others, you can let non-licensed persons use your station to send short "greetings messages" to other hams, so long as you are in control of the transceiver. What this actually entails is the subject of some debate, but to my understanding, they may operate the PTT but not the VFO, DSP, gain, or anything else.
3(4) Only where this Licence is a Full Licence may the Licensee permit a non-licensed person to send a Message using the Radio Equipment provided that the Radio Equipment is operated by the Licensee.
If we hop in an Internet time machine and look back at the old, now deprecated BR68 document, it gives us a better idea:
1(8) Having regard to sub-clauses 2(10) and 3(3), greetings messages may be sent by non-licensed persons provided that:
(a) it is under the direct supervision of the Licensee or other Authorised Club Member (in case of a Licence held on behalf of a club), who must operate the transmitter and identify the station; and
(b) each greetings message does not exceed five minutes; and
(c) greetings messages may be sent and received only within the United Kingdom or to and from stations in the United States of America, the Republic of Maldives, Gibraltar, Malta and Falkland Islands. Greetings messages may also be sent to or from stations in Canada and Pitcairn Islands provided that each greetings message does not exceed two minutes and that each person may only send one such message to each station with which the station is in contact.
For those of you that need reminding, that's calls starting with A, K, N, W (USA), 8Q (Maldives), ZP9 (Gibraltar), 9H (Malta) and VP8 (Falkland Is); as well as VE (Canada) and VP6 (Pitcairn Is).
I guess the procedure would be something like:
PA6ABC, this is MM0XYZ supervising John Smith from Somewhereville Scout Group for a greetings message, handing the mic over now. - greetings message - PA6ABC, this is MM0XYZ, back on the mic.
Now this could all be well out of date and I'm talking fluff, but I'm sure the actual content of a greetings message is not something that's tested.
You might also have noticed by now that the UK has not two prefixes - G and M - but also, technically at least, V and Z! V and Z countries - like VE, VK, ZL, ZP9 - are all Commonwealth countries. A bit of callsign trivia for you - this is something else probably not tested.
You also need to know the meaning of Disqualified Person: a very good recent example is this bellend... basically, they can't even touch your radio, so don't let 'em. They've been banned from operating on the amateur bands for good reason.
Maritime & international operation
With the Advanced licence also comes CEPT operation. That is, instead of having to turn up in an international destination and apply for an amateur licence there, you can simply plonk down and use the destination's prefix. So if I were to go to Dublin, rather than having to apply for a new Irish callsign I could just operate straight off the bat as EI/M0XYZ (if that was my call).
This is known as reciprocal licensing, so called because Paddy from Wexford could come over to Scotland and operate as MM/EI3PDY (assuming that was *his* call).
CEPT refers to the The European Conference of Postal and Telecommunications Administrations. Yes, this is often an exam question! Why CEPT rather than ECPTA? Well in French, just like with 'UTC', that mouthful is Conférence Européenne des administrations des Postes et des Télécommunications... They published a "recommendation document" called T/R 61-01, which defines what countries you may operate in and what privileges you get. The full text is here.
So with this comes the ability to operate in international waters and past the low water line in general! If you are operating at sea, radio conditions are a little different than you are used to on land.
• You MUST seek permission of the Ship's Master before operating. If, for example, you are on a cross-Channel ferry, this might not always be possible, in which case, DON'T OPERATE.
• If, for distress reasons, the Ship requires you to maintain radio silence, DON'T TRANSMIT. It's a serious crime under international law to fail to maintain radio silence and the sentence can be pretty harsh. Listen by all means to assist in the rescue effort if you feel you have to, but don't push that PTT!
• You should use the suffix /MM (maritime mobile).
• You should also use the band plan pertaining to the IARU region you're in; if you're just doing short hops on a boat, though, this shouldn't really affect you.
NoVs
One more quick thing I wish to touch on is NoVs. These are variations to your licence which allow you special privileges that you would otherwise have to ask for. As of Feb 2015, these are:
• 2kW Special Research Permit Allows you to run up to 2kW for research purposes. You have to provide Ofcom a good reason as to why you need that much power and the steps you are taking to minimise RFI.
• 60 Metres Allows you to use the 5MHz band as a secondary user to the MoD.
• 2 Metres - Extension Gives you an extra 1MHz at the top of 2 metres (146-147). Not valid in Scotland feckin' cheeky wee bastits...
• Microwave EME Opens up ~2.3GHz for EME use, up to 400W.
• 275GHz Not yet available, but this is basically the "top, top" band. You'll probably need a lead vest, balls of steel, an SDR and one hell of an upconverter to work this one.
• Beacons/Unattended You already encountered this one at Intermediate: the one that gives your APRS iGate that funky MB7 callsign.
• Golf Bravo Two Romeo Sierra (GB2RS) Allows you to read the RSGB news bulletin on VHF using the callsign GB2RS. If you want folk to recognise your callsign in an instant, this is how you do it... it's a thankless, unpaid job, though, and more often than not you'll be talking to dead air.
• Repeater Keeper Does exactly what it says on the tin, but now you can carry analog voice. More info here: old, but still relevant.
The remainder of the syllabus you already encountered at Foundation & Intermediate level, so there's no need to go over it again.
That just about covers the whole of Licensing for the syllabus: next part will be the Technical Aspects, starting with EMF, Voltage, and expanding on Ohm's Law, before getting on with the real mathematical nitty gritty.
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### Home > APCALC > Chapter 6 > Lesson 6.3.2 > Problem6-96
6-96.
What is $\int x ^ { n } d x$ for $n = 3, 2, 1, 0, –1, –2, \text{ and } –3$? Then write a general formula for $\int x ^ { n } d x$ for $n ≠ -1$.
$\int x^{3}dx=\frac{x^{4}}{4}+C$
$\int x^{2}dx=\frac{x^{3}}{3}+C$
$\int xdx=\frac{x^{2}}{2}+C$
Keep going...
A general formula will use the constant $n$. Explain why $n = −1$ does not fit this pattern. | |
# Fatma Cicek
Office: 908 Hylan Building
fcicek at ur dot rochester period edu
I am a 7th year graduate student.
I work in analytic number theory, my advisor is Steve Gonek. I am currently doing moment computations to understand the distribution of the logarithm of Dirichlet $L$-functions near or at the (nontrivial) zeros of the Riemann zeta-function.
## Current research
• The Logarithm of the Riemann Zeta-Function Near The Nontrivial Zeros (work in progress, a preprint is available upon request) | |
# Using Quaternions to rotate a player towards a point
I am attempting to solve a problem involving Quaternions and 3D rotation.
I am using a games engine (Torque 3D) for a project that I am developing. I have modified the gravity system so that the player is pulled in three dimensions towards a central point (the center of a spherical world).
I would like the player to also turn so that their feet are facing this central point, regardless of their position or initial orientation. I want the player to be able to stand on the spherical world, and currently, even though they are being pulled to the center of the world correctly, the orientation doesn't change.
The orientation is held in a Quaternion called "mOrient".
I know that the solution will involve rotation matrices, and probably world/local space translations, but I am not sure how to put all this together.
I am currently attempting this, unsuccessfully: Assume (0,0,0) is the center point, getPosition() is the position of the player. mOrient is the players 3D rotation quaternion, this is what I need to change.
Point3F gravityvec = Point3F(0, 0, 0) - getPosition();
gravityvec.normalize();
QuatF q = QuatF(gravityvec);
mOrient = q;
Does anyone have a solution for this?
I'm programming in C++ inside a games engine, but Pseudocode would be fine, I just need some assistance with the mathematical side.
Build a LookAt rotation matrix and then you can converter it to a quaternion.
Most engines already have that function built-in but here's the pseudo-code:
mat3 LookAt(vec3 up, vec3 front)
{
vec3 right;
mat3 m;
right = CrossProduct(Normalize(up), Normalize(front)); // figure out the right vector
front = CrossProduct(right, up); // make sure front is properly oriented
m[0] = right;
m[1] = up;
m[2] = front;
return m;
}
up is the vector from the center of the world to the player (the negative of your gravity vector)
front is where the player is facing (assuming your 3D model faces in the Z direction.) | |
# You must have solved the integral of log of cosine
Calculus Level 5
$$I=\displaystyle \int _{ 0 }^{ \pi /2 }{ { x }^{ 2 }\log ( \sec { (x) } ) dx }$$
If the value of $$I$$ can be represented as = $$\dfrac{\pi}{A} \zeta{(B)} + \dfrac{{\pi}^{C} \log{D} }{E}$$
Find $$ABCDE+1$$
Details and Assumptions
1) $$A,B,C,D,E$$ are positive integers. Also $$ABCDE$$ means product of the integers $$A,B,C,D,E$$. Also base of $$log$$ is $$e$$
2)Remember $$D$$ is not divisible by a perfect power(power >1 ) of any integer.
3) $$\displaystyle \zeta{(s)} = \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ s } } }$$.
× | |
# Using variables in function names
I want to make a list of plots and my functions are named C1, C2... Cn.
The command I wish to execute is:
Table[Plot[Cn[t], {t, 0, 1}], {n, 1,6}]]
which obviously doesn't work. How is the equivalent of the above done in Mathematica?
EDIT: Thanks for the answers. Here is what it looks like:
-
You can also use ToExpression to join the index n to your base function name, C, as in the following example:
C1 = Sin[x];
C2 = Cos[x];
C3 = Tan[x];
Table[Plot[Evaluate[ToExpression["C" <> ToString@i]], {x, -π, π}], {i, 3}]
-
That evaluates ToExpression and ToString for each point in the plot. (Which is not the biggest performance hit ever, but still...) – Brett Champion Feb 2 '12 at 22:32
The performance is not a problem for now, since I only have a couple of graphs I want to use, only as a visual aid. Thanks! – CHM Feb 3 '12 at 0:26
@CHM Brett's right. I've edited my answer to add an Evaluate, so please note it. – rm -rf Feb 3 '12 at 5:31
How about something like:
Plot[#, {t, 0, 1}] & /@
(ToExpression /@
Table["C" <> ToString[n] <> "[t]", {n, 1, 4}])
Edit With a form closer to your original code:
Table[Plot[ToExpression["C" <> ToString[n] <> "[t]"], {t, 0, 1}],
{n, 1, 4}]
For example, with:
C1[t_] := t
C2[t_] := t^2
C3[t_] := t^3
C4[t_] := t^4
Using either of the two solutions here gives:
-
oooh! Beat you by 13 seconds! :) – rm -rf Feb 2 '12 at 21:16
Blast! And I thought I had this one in the bag! But mine has a shiny Map version, too. – Eli Lansey Feb 2 '12 at 21:17
How is the Map version shinier? Is it performance-wise? – CHM Feb 3 '12 at 0:30
I meant shiny in terms of "fancy-looking," but in retrospect, it is faster. Reason being, as @BrettChampion noted in @RM's solution, when the ToExpression command is included in the Plot function, it's evaluated at each plot point. In the Map scenario it's evaluated once, and then the function is evaluated. I tested on my home computer and the Table method is around 3 times slower than the Map method. – Eli Lansey Feb 3 '12 at 0:37
You could define your functions like this:
Subscript[s, 1][t_] = Sin[t];
Subscript[s, 2][t_] = Cos[t];
And then plot using:
Plot[Evaluate[Table[Subscript[s, n][x], {n, 2}]], {x, -Pi, Pi}]
Just overlooked: This will create one Plot with all plots in it. The way your code snippet is written it looks as if you try to get each graph in its own plot in which case you have to use
Table[Plot[Subscript[s, n][x], {x, -Pi, Pi}], {n, 2}]
-
I hadn't thought about using a single Plot for all the functions. Might help me. Will check :) – CHM Feb 3 '12 at 0:28
Alternatively, you may define indexed family of functions like c[1],...,c[n]. Indices do not have to be contiguous, as you can get them all from the symbol definition. So if you define
c[1] = Sin;
c[2] = Cos;
c[3][x_] := Cos[x]^2;
you can do the plotting by iterating the index
Table[Plot[c[i][x], {x, -Pi, Pi}], {i, 3}]
You can also iterate over all defined indices in a general way:
Plot[c[#][x], {x, -Pi, Pi}] & /@
Union[SubValues[c][[All, 1, 1, 0, 1]],
DownValues[c][[All, 1, 1, 1]]]
You may also define a function to help with such an iteration, to hide the ugliness of index scavenging:
AllFunIndices[sym_Symbol] :=
Union[SubValues[sym][[All, 1, 1, 0, 1]],
DownValues[sym][[All, 1, 1, 1]]];
SetAttributes[AllFunIndices, HoldAll]
and then the plotting code over indices becomes much more transparent
Plot[c[#][x], {x, -Pi, Pi}] & /@ AllFunIndices[c]
-
I wondered that no one else was actually giving the very important tip to not use those variable names in the first place, there are several drawbacks in doing so: It will always lead to complicated constructs with ToExpression that are cryptic, inefficient and will pose even larger problems with namespaces when e.g. used in a package or when given away i any form (notebooks and CDF-files will warn about unsave code, demonstrations site will not accept). And that's just what immediately crosses my mind... – Albert Retey Feb 4 '12 at 17:02
This is the perfect time to use With to handle the creation of the variable name. For instance,
C1[t_] := Tanh[t];
C2[t_] := Sinh[t];
C3[t_] := Cosh[t];
GraphicsRow@Table[
With[{f = ToExpression["C" <> ToString[i]]},
Plot[f[t], {t, -1, 1}]],
{i, 3}]
gives
By using With to create the function name outside of Plot it is only executed once, not for every point.
-
Fun. It is nice to see different answers to the same question, it helps get a feel of how Mathematica can handle the problem. – CHM Feb 3 '12 at 3:07
Nice use of With. That solves the speed issue. I imagine Table and Map probably have nearly the same efficiency using this approach, then. – Eli Lansey Feb 3 '12 at 14:17
@EliLansey that's an interesting question. I've often found Table to be the slow point of any code I write, but I have not noticed the same thing with Map. I'll have to look at that. – rcollyer Feb 3 '12 at 14:21
@rcollyer I've noticed that too, so I've become the "Map everything" lunatic in our lab. Then, one of my coworkers was doing some calculation with a Table and I suggested Map to speed it up, and it actually was much slower. So, when I'm writing code that I need to be fast, I obsessively check which one will be faster. Wish I knew why, though... Maybe I'll ask a question. – Eli Lansey Feb 3 '12 at 14:26
@EliLansey most functional languages employ the idea of immutability, i.e. a variables value cannot be changed. So, Map has to produce a new list the same size as the old list while leaving the old one intact. I think, if it is slow, that's the reason: list construction can be fairly slow on mma. That's why Table can have truly awful performance. – rcollyer Feb 3 '12 at 14:42 | |
Monk and Graph Problem
Tag(s):
## Algorithms, DFS, Easy, Graph Theory
Problem
Editorial
Analytics
Monk and his graph problems never end. Here is one more from our own Monk:
Given an undirected graph with $N$ vertices and $M$ edges, what is the maximum number of edges in any connected component of the graph.
In other words, if given graph has $k$ connected components, and $E_1,E_2,....E_k$ be the number of edges in the respective connected component, then find $max(E_1,E_2,....,E_k)$ .
The graph may have multiple edges and self loops. The $N$ vertices are numbered as $1,2,...,N$.
Input Format:
The first line of input consists of two space separated integers $N$ and $M$, denoting number of vertices in the graph and number of edges in the graph respectively.
Following $M$ lines consists of two space separated each $a$ and $b$, denoting there is an edge between vertex $a$ and vertex $b$.
Output Format:
The only line of output consists of answer of the question asked by Monk.
Input Constraints:
$1 \le N \le 10^5$
$0 \le M \le 10^5$
$1 \le a,b \le N$
SAMPLE INPUT
6 3
1 2
2 3
4 5
SAMPLE OUTPUT
2
Explanation
The graph has $3$ connected components :
First component is $1-2-3$ which has $2$ edges.
Second component is $4-5$ which has $1$ edge.
Third component is $6$ which has no edges.
So, answer is $max(2,1,0)=2$
Time Limit: 1.0 sec(s) for each input file.
Memory Limit: 256 MB
Source Limit: 1024 KB
Marking Scheme: Marks are awarded when all the testcases pass.
Allowed Languages: C, C++, Clojure, C#, D, Erlang, F#, Go, Groovy, Haskell, Java, Java 8, JavaScript(Rhino), JavaScript(Node.js), Lisp, Lisp (SBCL), Lua, Objective-C, OCaml, Octave, Pascal, Perl, PHP, Python, Python 3, R(RScript), Racket, Ruby, Rust, Scala, Scala 2.11.8, Swift, Visual Basic
## CODE EDITOR
Initializing Code Editor...
## This Problem was Asked in
Challenge Name
CodeMonk (Graph Theory Part I)
OTHER PROBLEMS OF THIS CHALLENGE
• Algorithms > Graphs
• Algorithms > Graphs
• Algorithms > Graphs
• Algorithms > Graphs | |
Calculators with Support of the 82240 IR-Printer
06-27-2014, 07:25 AM (This post was last modified: 06-27-2014 07:45 AM by Martin Hepperle.)
Post: #21
Martin Hepperle Member Posts: 221 Joined: May 2014
RE: Calculators with Support of the 82240 IR-Printer
(06-26-2014 10:08 PM)everettr Wrote: I was confused about pulses and bursts, and how the IR detector interprets them. It is getting clearer to me now. I will probably go with the TSOP4133 part that Christoph suggested, though I am sure this is very similar to those that you listed.
5 Kb, what a whopper :-) If it fits the flash and meets the timing constraints, no problem. It sounds like you are using interrupts for both the serial and the burst timing. I would like to see how you did that, since I have been wondering if some sort of caching of data would be needed to separate IR operations from serial communications. That is an interesting idea about additional ESC sequences, but how would you persuade a calculator to emit the new sequence?
(06-26-2014 07:58 AM)Martin Hepperle Wrote: I just finished the first version of the code and can make it available if you are interested. I am just adding some comments so that the code is understandable.
I am definitely interested in seeing your project!
Thank you,
Yes, I used a pin-low interrupt to capture the start of a data frame and then a timer interrupt to capture the half-bits. Before that I had tried the quick&dirty approach (polling with while() loops) but that was unreliable, inelegant and dirty.
The other problem was that I first had the serial communication in the interrupt service routines (resp. in the polling loops) but I quickly learned that serial I/O is too slow so that I lost bits and sync. Therefore I implemented a ring buffer (mailslot) where the interrupt routine places the decoded and error checked byte and the main loop later emits these to the serial line. This works very well (i would even dare to say perfectly) and I noticed that I could make the buffer very small (1-2 bytes), but stuck with a voluptuous 8-16 byte buffer.
My error checking corrects one flipped bit and I think this could be improved, but I am not sure.
Emitting special escape sequences should be asy by sending e.g. an ESCape character (ASCII 27) and then a code which is not used by the printer and does not indicate a graphics sequence. For example "ESC 0" could initiate and terminate a different command mode and this could be filtered out of the IR stream so that the command is not sent to the printer. I have made provision for such a thing in the code, but it is more of a hook and not yet fully developed.
I think most of the better calculators can send individual characters/ASCII codes.
If the calculator can only emit simple thinkgs, one couls us a magic work (number or string) to switch modes, but that should be the last resort.
Code may be posted here when polishing is finished.
Martin
06-27-2014, 07:39 AM (This post was last modified: 06-27-2014 08:33 AM by Martin Hepperle.)
Post: #22
Martin Hepperle Member Posts: 221 Joined: May 2014
RE: Calculators with Support of the 82240 IR-Printer
(06-26-2014 06:27 PM)Marcus von Cube Wrote: [... deleted ...]
@Martin: The IR range depends on the type of IR LED. Katie has made some experiments. I don't recall the exact details, but I can send you one of mine.
So my question would be: what ranges do people get with their WP34S?
I obtain about 100 mm, which is o.k, but something like 500 mm would be nice. I do not need 10 meters and do not want to drain the batteries too fast.
I only was astonished about the power of the HP 48 - it seems to illuminate the whole room with his IR and can print even if not pointed at the printer.
I obtained my IR diode for the HP 20b from H. Pott together with his fabulous USB board, but am considering to make another WP34S from a HP 30b.
Martin
06-27-2014, 08:01 AM
Post: #23
Marcus von Cube Senior Member Posts: 760 Joined: Dec 2013
RE: Calculators with Support of the 82240 IR-Printer
Here is an old thread on IR printing. It contains info from me and Katie and covers most of the questions (LED part number, range, commands, etc.).
Marcus von Cube
Wehrheim, Germany
http://www.mvcsys.de
http://wp34s.sf.net
http://mvcsys.de/doc/basic-compare.html
06-27-2014, 08:15 AM
Post: #24
Martin Hepperle Member Posts: 221 Joined: May 2014
RE: Calculators with Support of the 82240 IR-Printer
(06-25-2014 07:27 PM)Christoph Giesselink Wrote: For sure the HP-95LX has also the necessary hardware. And with a little software help we can send data to the printer:
Code:
... [deleted] ... ; Equations IRFMAT equ 0e30ah ; IR Format Register IRCNT equ 0e30bh ; IR Transmit/Receive Register
Is there something like a technical reference manual for these machines where e.g. the I/O addresses and how to use them are described?
06-27-2014, 03:47 PM
Post: #25
Christoph Giesselink Member Posts: 161 Joined: Dec 2013
RE: Calculators with Support of the 82240 IR-Printer
(06-27-2014 08:15 AM)Martin Hepperle Wrote:
(06-25-2014 07:27 PM)Christoph Giesselink Wrote: For sure the HP-95LX has also the necessary hardware. And with a little software help we can send data to the printer:
Code:
... [deleted] ... ; Equations IRFMAT equ 0e30ah ; IR Format Register IRCNT equ 0e30bh ; IR Transmit/Receive Register
Is there something like a technical reference manual for these machines where e.g. the I/O addresses and how to use them are described?
Not at the moment. I send a copy of the "HP 95LX Developer's Guide" to Dave, so it will be part of the next Museum DVD set.
06-27-2014, 03:54 PM
Post: #26
Christoph Giesselink Member Posts: 161 Joined: Dec 2013
RE: Calculators with Support of the 82240 IR-Printer
(06-27-2014 07:25 AM)Martin Hepperle Wrote: Yes, I used a pin-low interrupt to capture the start of a data frame and then a timer interrupt to capture the half-bits. Before that I had tried the quick&dirty approach (polling with while() loops) but that was unreliable, inelegant and dirty.
The other problem was that I first had the serial communication in the interrupt service routines (resp. in the polling loops) but I quickly learned that serial I/O is too slow so that I lost bits and sync. Therefore I implemented a ring buffer (mailslot) where the interrupt routine places the decoded and error checked byte and the main loop later emits these to the serial line. This works very well (i would even dare to say perfectly) and I noticed that I could make the buffer very small (1-2 bytes), but stuck with a voluptuous 8-16 byte buffer.
My error checking corrects one flipped bit and I think this could be improved, but I am not sure.
I wrote the REDEYE receiver code three times. The 1st version in 8086 assember for a Messdos PC. The signal from a IR-receiver was connected to the CTS line of a RS232C interface. The input of the program was interrupt driven by the CTS line interrupt. The last version used the PC timer chip to decode the input signal.
The 2nd version I wrote more a less parallel in 8051 assembler for a company. IR input interrupt driven, internal timer to decode the framing and a timer interrupt for the case the frame was incomplete with less than 2 bis missing.
The 3rd version is implemented in C and published under the GPL v2 inside Emu28 and Emu42 (file REDEYE.C). The implementation is also "interrupt driven". At every rising edge of the IR output pin the decoder is called. The timing is generated over the elapsed CPU cycles which is in reality the elaped time since the last IR signal rising edge. The implementation of the error correction is complete, in fact it can restore two missed bits. Read the originally HP documentation (HP 82240B Technical Interfacing Guide and Hewlett Packard Journal October 1987 page 16) carefully please, especially the concept about "missed bits" at IR bursts and drop outs in the Hewlett Packard Journal article. This is very important for the error correction. IMHO the Journal article gives more practical hints for a software implementation then the Technical Interfacing Guide.
I agree with the ring buffer solution for the captured character. The IR input interrupt had highest priority in my 8051 implementation. On another thing I recognized that we now have 2014, the original 8051 program design was made in 1993, the latest version was from 2002 and the firmware binary for the 8051 had a size of 676 bytes! Sorry, I'm not able to publish this firmware.
06-30-2014, 11:39 AM
Post: #27
Martin Hepperle Member Posts: 221 Joined: May 2014
RE: Calculators with Support of the 82240 IR-Printer
(06-27-2014 03:54 PM)Christoph Giesselink Wrote: [... deleted ...]
The 3rd version is implemented in C and published under the GPL v2 inside Emu28 and Emu42 (file REDEYE.C). The implementation is also "interrupt driven". At every rising edge of the IR output pin the decoder is called. The timing is generated over the elapsed CPU cycles which is in reality the elaped time since the last IR signal rising edge. The implementation of the error correction is complete, in fact it can restore two missed bits.
[... deleted ...]
Ah, thank you, I had not seen these sources and I followed a different way than proposed in the HP docs (measuring the time between bursts) by sampling at regular intervals. Both strategies seem to work well. I hope to test my receiver also with an HP 28S soon and then put it into a nice casing. I am also looking for a cheap thermoprinter understanding ESC/POS (cash register) codes which would allow building a low cost HP 82240 replacement for hobby use (just for fun, no commercial background).
07-01-2014, 04:58 AM
Post: #28
rwiker Junior Member Posts: 12 Joined: Dec 2013
RE: Calculators with Support of the 82240 IR-Printer
(06-30-2014 11:39 AM)Martin Hepperle Wrote: I am also looking for a cheap thermoprinter understanding ESC/POS (cash register) codes which would allow building a low cost HP 82240 replacement for hobby use (just for fun, no commercial background).
Sparkfun has this which may or may not have the functionality you want (and may or may not count as "cheap").
07-09-2014, 12:51 PM
Post: #29
Martin Hepperle Member Posts: 221 Joined: May 2014
RE: Calculators with Support of the 82240 IR-Printer
In the meantime I have tested my code with a HP 48G and the WP 34S and the results are as expected. For further tests I used a HP 28S and found the IR-transmission relatively buggy (incomplete data frames, bit errors).
Does anyone have experience with the HP 28S and IR-printing and knows whether there is a known timing problem with the HP 28S? Is the 28 known to be more finicky w.r.t. printing? I have not looked at the signal with an oscilloscope yet.
Maybe the reason may also be that I still use a 38 kHz IR-receiver - I will replace it with a 33 kHz one soon to see whether this makes a difference.
In the end I would like to dump the ROM of the 28S via IR so that I can use it in EMU28.
Martin
Post: #30
Christoph Giesselink Member Posts: 161 Joined: Dec 2013
RE: Calculators with Support of the 82240 IR-Printer
(07-09-2014 12:51 PM)Martin Hepperle Wrote: In the meantime I have tested my code with a HP 48G and the WP 34S and the results are as expected. For further tests I used a HP 28S and found the IR-transmission relatively buggy (incomplete data frames, bit errors).
Does anyone have experience with the HP 28S and IR-printing and knows whether there is a known timing problem with the HP 28S? Is the 28 known to be more finicky w.r.t. printing? I have not looked at the signal with an oscilloscope yet.
The HP-28S works successful in connection with a HP-82240A/B printer and with my Redeye-Serial converter.
The main difference between a HP48 and a HP28 is the IR burst and half-bit timing generation.
The HP48 chip Clarke or Yorke has a hardware register called LBR (Led Buffer Register). This register is responsible for sending a half-bit in the REDEYE protocol. The timing of this register is done by hardware and derived from the 32768Hz crystal, so the timing of the IR burst and the half-bit length is very exactly.
In opposite the HP28 and all other calculators using the Lewis chip (2nd Clamshell generation 19B/19BII/28S and High End Pioneer 17B/17BII/27S/42S) or the combination of 1LK7 Saturn and 1LP2 Centipede chip (1st Clamshell generation 18C/28C) don't have this hardware driven register. Here the timing of the half-bit length is completely done by the CPU. Only the generation of the 32kHz IR signal is done in hardware. So on software side you have to open a gate for the exact time of 7 IR burst cycles. Overall the CPU has to control the IR burst length time and the half-bit length time. To make this more difficult the CPU has no constant speed. On the 1LK7 Saturn CPU the CPU strobe frequency (~620kHz) is generated by a LC oscillator, on the 1LR2 Lewis over a PLL from the 32768Hz base frequency to get a CPU strobe frequency of nominal 1MHz. To get a more or less exact timing, the actual CPU speed is measured over the internal timer driven by the 32768Hz crystal. So from time to time the CPU speed is measured to eliminate the CPU speed drift over temperature, humidity, ...
To make it short, the timing accuracy of the REDEYE protocol made my CPU is much less than the one generated by the hardware LBR register inside the HP48 series.
So your method checking the IR state on each half-bit position is less timing tolerant than the solution suggested in HP Journal article (used by my hardware solution and inside the Emu28 and Emu42 simulation). In the suggested solution you are synchronizing your timing at every received bit and not at the beginning of the 12 bit frame.
(07-09-2014 12:51 PM)Martin Hepperle Wrote: In the end I would like to dump the ROM of the 28S via IR so that I can use it in EMU28.
To stay correct, the 28S can be emulated by Emu42 (with simulation of the 1LR2 Lewis and 1LR3 Sacajawea chip) and the 28C with Emu28 (a simulation of the 1LK7 Saturn in connection with the 1LP2 Centipede chip).
07-10-2014, 09:31 AM
Post: #31
Martin Hepperle Member Posts: 221 Joined: May 2014
RE: Calculators with Support of the 82240 IR-Printer
Christoph,
thank you for the excellent technical explanation. I'll think it over and might implement the "self calibrating" approach to decode the redeye protocol in a more robust way.
And, yes, yesterday evening after stumbling across some rom images in the woods I also leared that for the 28S emulation I'd better use the EMU42 software...
Thank you for sharing your knowledge, Martin
07-10-2014, 11:32 AM
Post: #32
Marcus von Cube Senior Member Posts: 760 Joined: Dec 2013
RE: Calculators with Support of the 82240 IR-Printer
(07-09-2014 12:51 PM)Martin Hepperle Wrote: In the meantime I have tested my code with a HP 48G and the WP 34S and the results are as expected...
There was a reason to implement the printer software on the 34S the way it is:
1. Accurate timing increases the range, exspecially for a low power device as used by the 34S.
2. Accurate timing is only available with a crystal. Hence the printer firmware is not available without the XTAL modification.
3. The timing is mostly done with the help of hardware timers and interrupts. This is accurate and energy efficient because the CPU is not involved in timing.
The efforts obviously pay off in other applications, too, as your experiments show.
Marcus von Cube
Wehrheim, Germany
http://www.mvcsys.de
http://wp34s.sf.net
http://mvcsys.de/doc/basic-compare.html
08-06-2014, 08:15 AM (This post was last modified: 08-06-2014 08:18 AM by Martin Hepperle.)
Post: #33
Martin Hepperle Member Posts: 221 Joined: May 2014
RE: Calculators with Support of the 82240 IR-Printer
Oh my! My journey into HP-IR land has been extended.
Some Numbers
The frequency of the IR signal is f = 32768 Hz so that one pulse of 50% duty cycle has a duration of t_pulse = 1/32768 / 2 = 15.3 us (microseconds)).
The HP docs specify a burst length of 6 to 8 pulses = 183 to 244 us as valid range. All my test calculators output 8 pulses per burst.
One complete bit takes 854.5 us, one half-bit time is therefore 427.25 us
• A logic one starts with 8 pulses
8 pulses = 8 / 32768 = 244 us followed by a pause of 854.5 - 244 = 610.5 us
• A logic zero starts with a half bit pause of 854.5/2 = 427.25 us followed by 8 pulses = 8 / 32768 = 244 us followed by a pause of 427.25 – 244 = 183.25 us. The same interval occurs between the bursts of the three start half-bits. This pause corresponds to 183.25/10^6 * 32768 = 6 cycles. This is a critical value as this pause of 6 cycles should be detectable.
TSOP 11XX, 18XX, 21XX, 23XX, 41XX, 43XX, 25XX, 45XX Datasheets
The burst length should be 6 or more pulses, so that the time of at least 6/f = 6/32768 = 183 us is recommended. This is within the HP specs.
The output pulse of the TSOP 11XX has a length of
t_burst – 4/f < t_out < t_burst + 6/f0
which means that it may be 4 cycles shorter or 6 cycles longer than the actual burst. If the output is at the upper limit of this range, it will close the gap between two bursts.
The TSOP 18XX datasheet explicitly states that it requires a gap of at least 9 cycles after each burst.. In time this means
t_gap > 9 / 32768 = 274.7 ms
This is longer than the time between the start-halfbits, so that a TSOP 18XX may be unable to detect the gap between these and output them together as one continuous long signal.
Now after procuring me an oscilloscope (I wanted one for a long time anyway and this gave me a good reason) I observe that all bursts are detected but often bursts following in a short interval (start-half-bits, zeros followed by ones) are output as one pulse. Sometimes all individual bursts come through. In general it seems to be unreliable to detect the individual bursts with these TSOP XXXX receivers (I played with TSOP 1133 and VS1833). This is also the reason why my first implementation (I only detected the start of the first burst and then sampled at regular intervals) worked well, as long as the timing of the sender was accurate.
However with inaccurate senders (say HP 28S) and not being able to detect the individual bursts this does not work, as Christoph Gießelink already pointed out.
The attached picture shows the result of printing the same character twice. the first time the pulses are clearly visible, the second time some are connected.
Now what to do?
While I was looking for a small enclosure with an IR-window I found the IrDA receivers C4103A are used to connect HP LaserJet printers. While I was interested in the casing only, I also tested the receiver part of it and learned that it produces the IR signal, i.e. the output is the raw nicely amplified 32.678 kHz pulses without integration. So I now see two ways to proceed:
a) try to find another integrating IR receiver which can handle the short gaps between the bursts,
b) build my own discrete receiver unit using R/C integrator and an OP-Amp,
c) use the raw signal and do the decoding in software.
My thoughts about these approaches are:
a) This option would be the most desirable, but I am not sure whether such a solution exists.
b) This might be a good solution, and I came up with a design (using the LTSpice simulation tool) that could work. However I fear that in the real world this would require a lot of tweaking to work properly and be reliable. I am not an electronics engineer, just an aerospace engineer. While I can think digital and know how to fly to the moon, I have only a faint understanding of non-digital technology like R/C circuits and the like.
c) This solution seems to put a rather high load on the CPU as it must time and the 32 kHz pulses to detect the bursts. On the other hand this might be the most flexible solution.
Maybe some of you electronics experts can give me some ideas or point me into the “right” direction? If this discussion becomes too long and specific, you can also send me PM and I will summarize the end result here. Maybe there a hardware design published somewhere for this purpose which I have not found yet?
Thank you, Martin
Attached File(s) Thumbnail(s)
08-06-2014, 11:09 AM
Post: #34
Michael Lopez Member Posts: 58 Joined: Dec 2013
RE: Calculators with Support of the 82240 IR-Printer
Thanks for the comprehensive information.
I recently purchased a NOS HP82240B printer out of curiosity & whilst it seems to work well with the 48 series (48SX & 48GX), it is rather unreliable with my HP50G calculators. Noticed in the original post that there is a reference to these being switchable to 82240B mode - would you please explain how this is performed?
Thanks,
Michael
08-06-2014, 07:18 PM
Post: #35
Christoph Giesselink Member Posts: 161 Joined: Dec 2013
RE: Calculators with Support of the 82240 IR-Printer
(08-06-2014 08:15 AM)Martin Hepperle Wrote:
• A logic zero starts with a half bit pause of 854.5/2 = 427.25 us followed by 8 pulses = 8 / 32768 = 244 us followed by a pause of 427.25 – 244 = 183.25 us. The same interval occurs between the bursts of the three start half-bits. This pause corresponds to 183.25/10^6 * 32768 = 6 cycles. This is a critical value as this pause of 6 cycles should be detectable.
I confirm. So it's luck that REDEYE receivers work with the TSOP1833 or TSOP4133.
I made a test equipment for a REDEYE compatible printer many years ago. This printer used the TSOP1833 as receiver. The REDEYE transmitter in the test equipment was crystal driven and always send exactly 7 IR pulses in the half bit frame. I cannot remember any problems with the receiver chip in the printer batch production during the lifetime of this product. So we had luck that we always get batches of TSOP1833 that worked with 7 or less pause pulses.
08-07-2014, 06:22 PM
Post: #36
Martin Hepperle Member Posts: 221 Joined: May 2014
RE: Calculators with Support of the 82240 IR-Printer
(08-06-2014 11:09 AM)Michael Lopez Wrote: Thanks for the comprehensive information.
I recently purchased a NOS HP82240B printer out of curiosity & whilst it seems to work well with the 48 series (48SX & 48GX), it is rather unreliable with my HP50G calculators. Noticed in the original post that there is a reference to these being switchable to 82240B mode - would you please explain how this is performed?
Thanks,
Michael
Michael,
I had written my first post without having Access to an HP 40g. In the meantime I acquired one and found that for manual printing you only have to select "Infrared" in the print Dialog (Options are Infrared, USB, Serial). This should set the appropriate flags. For Output from a program you have to set the appropriate flags.
System flags of interest:
I/O Device (–33),
Printing Device (–34), << defines IrDA or Redeye protocol
Double-spaced Printing (–37),
Linefeed (–38),
I/O Device for Wire (–78)
See the description of system flags in the "Advanced Users Reference Manual", Appendix C.
–33 I/O Device.
Clear: I/O directed to USB/serial port.
Set: I/O directed to IrDA port.
–34 Printing Device.
Clear: Prints via IR to the HP 82240 printer. Flag –33 is ignored.
Set (default): Printer output directed to USB/serial port if flag –33 is clear, or to
IrDA compatible printer otherwise.
Please also note that the range of the HP 50g's IR Signal is very small, compared to the HP 48 - just about 50-100 mm ( 2-4 inches). I guess this is your problem.
Hope this helps.
08-07-2014, 06:33 PM
Post: #37
Martin Hepperle Member Posts: 221 Joined: May 2014
RE: Calculators with Support of the 82240 IR-Printer
(08-06-2014 08:15 AM)Martin Hepperle Wrote:
• A logic zero starts with a half bit pause of 854.5/2 = 427.25 us followed by 8 pulses = 8 / 32768 = 244 us followed by a pause of 427.25 – 244 = 183.25 us. The same interval occurs between the bursts of the three start half-bits. This pause corresponds to 183.25/10^6 * 32768 = 6 cycles. This is a critical value as this pause of 6 cycles should be detectable.
I confirm. So it's luck that REDEYE receivers work with the TSOP1833 or TSOP4133.
I made a test equipment for a REDEYE compatible printer many years ago. This printer used the TSOP1833 as receiver. The REDEYE transmitter in the test equipment was crystal driven and always send exactly 7 IR pulses in the half bit frame. I cannot remember any problems with the receiver chip in the printer batch production during the lifetime of this product. So we had luck that we always get batches of TSOP1833 that worked with 7 or less pause pulses.
Christoph,
thank you for your reply. I had hoped that you had used some alternative Hardware which I could copy. I think someone mentioned a commercial US product - I would be interested in learning about the innards of this hardware.
If you only sent 7 pulses (which is perfectly within the HP specs of 6-8 pulses) you gained some time for the pause (one 32kHz cycle), which may have helped the TSOP to detect the pause. The HP 48 and 28 seem to send 8 pulses for each burst, resulting in a shorter pause.
Currently I am playing with the 1133, maybe I should try some 1833 instead, if I can get some (I have one 1838, but thought it would be better to move closer to 33 kHz).
Martin
08-07-2014, 10:11 PM
Post: #38
Christoph Giesselink Member Posts: 161 Joined: Dec 2013
RE: Calculators with Support of the 82240 IR-Printer
(08-07-2014 06:33 PM)Martin Hepperle Wrote: If you only sent 7 pulses (which is perfectly within the HP specs of 6-8 pulses) you gained some time for the pause (one 32kHz cycle), which may have helped the TSOP to detect the pause. The HP 48 and 28 seem to send 8 pulses for each burst, resulting in a shorter pause.
Currently I am playing with the 1133, maybe I should try some 1833 instead, if I can get some (I have one 1838, but thought it would be better to move closer to 33 kHz).
I had a look at the CLARKE External Reference Specification and I can confirm that the 1LT8 Clarke chip inside the HP48SX use 8 pulses. The situation at the 1LR2 Lewis chip inside the HP28S and the 1LP2 Centipede chip inside the HP28C is a little bit different, here you're enable a channel sending 16us pulses to the IR transmitter LED. The number of pulses depends on the time the STL pin is set to high by the CPU. This is inaccurate.
For more information you may have a look at the HP48 I/O Technical Interfacing Guide PDF. There's also a schematic of the HP48 IR receiver.
The TSOP1833 is out of production for many years now. I had a look at my personal REDEYE receivers, they all use the TSOP1833 chip. None of them use the later TSOP4133 chip which need one more pulse pause time than the TSOP1833. So I have no long time experience with the TSOP4133 in connection with the HP48 or HP28S.
The attached file show my smallest REDEYE receiver with a RS232C interface.
Attached File(s) Thumbnail(s)
08-08-2014, 07:17 AM
Post: #39
Michael Lopez Member Posts: 58 Joined: Dec 2013
RE: Calculators with Support of the 82240 IR-Printer
(08-07-2014 06:22 PM)Martin Hepperle Wrote:
(08-06-2014 11:09 AM)Michael Lopez Wrote: Thanks for the comprehensive information.
I recently purchased a NOS HP82240B printer out of curiosity & whilst it seems to work well with the 48 series (48SX & 48GX), it is rather unreliable with my HP50G calculators. Noticed in the original post that there is a reference to these being switchable to 82240B mode - would you please explain how this is performed?
Thanks,
Michael
Michael,
I had written my first post without having Access to an HP 40g. In the meantime I acquired one and found that for manual printing you only have to select "Infrared" in the print Dialog (Options are Infrared, USB, Serial). This should set the appropriate flags. For Output from a program you have to set the appropriate flags.
System flags of interest:
I/O Device (–33),
Printing Device (–34), << defines IrDA or Redeye protocol
Double-spaced Printing (–37),
Linefeed (–38),
I/O Device for Wire (–78)
See the description of system flags in the "Advanced Users Reference Manual", Appendix C.
–33 I/O Device.
Clear: I/O directed to USB/serial port.
Set: I/O directed to IrDA port.
–34 Printing Device.
Clear: Prints via IR to the HP 82240 printer. Flag –33 is ignored.
Set (default): Printer output directed to USB/serial port if flag –33 is clear, or to
IrDA compatible printer otherwise.
Please also note that the range of the HP 50g's IR Signal is very small, compared to the HP 48 - just about 50-100 mm ( 2-4 inches). I guess this is your problem.
Hope this helps.
Hi Martin,
Thanks for the explanation on the effect of the Flag -33 & -34 settings on IR communication via the HP50G.
Based on this, I've been using the correct settings to print to the HP 82240B but it is still rather "flaky" with the HP50G. Through my own trials, I realised that the transmission distance needed to be extremely short & now typically place the calculator almost right up against the IR receiver on the printer.
Have also played with the HP50G's height relative to the printer & it does seem to perform better if I raise the calculator ~ 3-5 mm. I think this may be due to the positioning of the IR transmitter on the calculator versus the earlier HP48 models the printer may have been designed for.
Whilst it is ok as a "play thing", I am not sure I would recommend extensive printing to a HP 82240B from a HP 50G.
Cheers,
Michael
08-09-2014, 11:22 AM
Post: #40
Marcus von Cube Senior Member Posts: 760 Joined: Dec 2013
RE: Calculators with Support of the 82240 IR-Printer
(08-06-2014 08:15 AM)Martin Hepperle Wrote: Some Numbers
The frequency of the IR signal is f = 32768 Hz so that one pulse of 50% duty cycle has a duration of t_pulse = 1/32768 / 2 = 15.3 us (microseconds)).
The HP docs specify a burst length of 6 to 8 pulses = 183 to 244 us as valid range. All my test calculators output 8 pulses per burst.
If I understand my code correctly, WP 34S outputs bursts of seven pulses. The generator is just the AND of two square wave clocks: 32768Hz and (32768/14)Hz. If you want to check the code, it's in trunk/main.c, routine put_ir().
Marcus von Cube
Wehrheim, Germany
http://www.mvcsys.de
http://wp34s.sf.net
http://mvcsys.de/doc/basic-compare.html
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# Leading digits of Fibonacci sequence
There is a standard problem to show that the distribution of leading digits of $2^n$ is that the digit $k$ occurs with the frequency $\log_{10}(k+1)-\log_{10}(k)$. (This easily generalises to other bases --- though base 2 is rather pointless!)
Since Fibonacci is also exponential except for an error term''. Is this true for that as well --- or does the error term make it fail?
-
The error term is $o(1)$, so whatever you have for $2^n$ should also work for the Fibonacci sequence. – GH from MO Feb 6 '12 at 17:48
How could the error term possibly affect the distribution of the leading digit? – Qiaochu Yuan Feb 6 '12 at 17:54
The phrasing with the emphasize on the error term is perhaps confusing, but then I do not (fully) understand the remark of GH as the 'base' is not 2 for Fibinacci. Okay, in the end 'the reason' is 'the same'; I will expand my answer a bit. – quid Feb 6 '12 at 18:07
The key part of the proof is to show that $n \log \phi$ is uniformly distributed modulo $1$ where $\phi$ is the Golden Ratio (the base of the logarithm being the one for which the law should be established). This is achieved using Weyl's equidistribution theorem. | |
# In a semicircle
$AB$ is the diameter of a semi-circle.
$C$ is a point on the circumference.
What is the measure of angle $ACB?$
× | |
Thread: Absolute, conditionally convergence or divergense?
1. Absolute, conditionally convergence or divergense?
Oi
Im stuck at following:
I need to figure out weather this is absolute convergent, conditionally convergent or divergent.
sum from 1->infinity of: ((-1)^n)*(3n^2+7n-1)/(3n^5+4n+1)
Any hints or help is appreciated
2. Originally Posted by Zaph
Oi
Im stuck at following:
I need to figure out weather this is absolute convergent, conditionally convergent or divergent.
sum from 1->infinity of: ((-1)^n)*(3n^2+7n-1)/(3n^5+4n+1)
Any hints or help is appreciated
perform the limit comparison test on the positive terms of the series with the known convergent series $\sum \frac{1}{n^3}$ ...
3. Thanx, i think i got it right, and found out it converges absolute cause the limit goes to 1. | |
# Linear Regression
## Introduction to Linear Regression
Linear regression is used to predict the relationship between two variables by applying a linear equation to observed data. There are two types of variable, one variable is called an independent variable, and the other is a dependent variable. Linear regression is commonly used for predictive analysis. The main idea of regression is to examine two things. First, does a set of predictor variables do a good job in predicting an outcome (dependent) variable? The second thing is which variables are significant predictors of the outcome variable ?. In this article, we will discuss the concept of the Linear Regression Equation, formula and Properties of Linear Regression.
### Examples of Linear Regression
The weight of the person is linearly related to their height. So, this shows a linear relationship between the height and weight of the person. According to this, as we increase the height, the weight of the person will also increase.
It is not necessary that one variable is dependent on others, or one causes the other, but there is some critical relationship between the two variables. In such cases, we use a scatter plot to simply the strength of the relationship between the variables. If there is no relation or linking between the variables then the scatter plot does not indicate any increasing or decreasing pattern. In such cases, the linear regression design is not beneficial to the given data.
### Linear Regression Equation
The measure of the relationship between two variables is shown by the correlation coefficient. The range of the coefficient lies between -1 to +1. This coefficient shows the strength of the association of the observed data between two variables.
Linear Regression Equation is given below :
Y=a+bX
where X is the independent variable and it is plotted along the x-axis
Y is the dependent variable and it is plotted along the y-axis
Here, the slope of the line is b, and a is the intercept (the value of y when x = 0).
### Linear Regression Formula
As we know, linear regression shows the linear relationship between two variables. The equation of linear regression is similar to that of the slope formula. We have learned this formula before in earlier classes such as a linear equation in two variables. Linear Regression Formula is given by the equation
Y= a + bX
We will find the value of a and b by using the below formula
a = $\frac{(\sum y)(\sum x^{2})-(\sum x)(\sum xy)}{[n(\sum x^{2})-(\sum x)^{2}]}$
b = $\frac{[n(\sum xy)-(\sum x)(\sum y)]}{[n(\sum x^{2})-(\sum x)^{2}]}$
### Simple Linear Regression
Simple linear regression is the most straight forward case having a single scalar predictor variable x and a single scalar response variable y. The equation for this regression is given as y=a+bx
The expansion to multiple and vector-valued predictor variables is known as multiple linear regression. It is also known as multivariable linear regression. The equation for this regression is given as Y = a+bX
Almost all real-world regression patterns include multiple predictors. The basic explanations of linear regression are often explained in terms of multiple regression. Note that, in these cases, the dependent variable y is yet a scalar.
### Least Square Regression Line or Linear Regression Line
The most popular method to fit a regression line in the XY plot is found by using least-squares. This process is used to determine the best-fitting line for the given data by reducing the sum of the squares of the vertical deviations from each data point to the line. If a point rests on the fitted line accurately, then the value of its perpendicular deviation is 0. It is 0 because the variations are first squared, then added, so their positive and negative values will not be cancelled.
[Image will be Uploaded Soon]
Linear regression determines the straight line, known as the least-squares regression line or LSRL. Suppose Y is a dependent variable and X is an independent variable, then the population regression line is given by the equation;
Y = B$_{0}$ + B$_{1}$X
Where
B$_{0}$ is a constant
B$_{1}$ is the regression coefficient
When a random sample of observations is given, then the regression line is expressed as;
$\hat{y}$ = b$_{0}$ + b$_{1}$x
where b$_{0}$ is a constant
b$_{1}$ is the regression coefficient,
x is the independent variable,
ŷ is known as the predicted value of the dependent variable.
### Properties of Linear Regression
For the regression line where the regression parameters b$_{0}$ and b$_{1}$ are defined, the following properties are applicable:
• The regression line reduces the sum of squared differences between observed values and predicted values.
• The regression line passes through the mean of X and Y variable values.
• The regression constant b$_{0}$ is equal to the y-intercept of the linear regression.
• The regression coefficient b$_{1}$ is the slope of the regression line. Its value is equal to the average change in the dependent variable (Y) for a unit change in the independent variable (X)
### Regression Coefficient
The regression coefficient is given by the equation :
Y = B$_{0}$ + B$_{1}$X
Where
B$_{0}$ is a constant
B$_{1}$ is the regression coefficient
Given below is the formula to find the value of the regression coefficient.
B$_{1}$ = b$_{1}$ = Σ[(x$_{i}$ - x)(y$_{i}$ - y)]/Σ[(x$_{i}$ - x)$^{2}$]
Where xiand yi are the observed data sets.
And x and y are the mean value.
### Solved Examples
1. Find a linear regression equation for the following two sets of data:
x 2 4 6 8 y 3 7 5 10
Sol: To find the linear regression equation we need to find the value of Σx, Σy, Σx$^{2}$ and Σxy
Construct the table and find the value
x y x² xy 2 3 4 6 4 7 16 28 6 5 36 30 8 10 64 80 Σx = 20 Σy = 25 Σx² = 120 Σxy = 144
The formula of linear equation is y=a+bx
Using the formula we will find the value of a and b
a = $\frac{(\sum y)(\sum x^{2})-(\sum x)(\sum xy)}{[n(\sum x^{2})-(\sum x)^{2}]}$
Now put the values in the equation
a = $\frac{25 \times 120 - 20 \times 144}{4 \times 120 - 400}$
a = $\frac{120}{80}$
a = 1.5
b = $\frac{[n(\sum xy)-(\sum x)(\sum y)]}{[n(\sum x^{2})-(\sum x)^{2}]}$
Put the values in the equation
b = $\frac{4 \times 144 - 20 \times 25}{4 \times 120 - 400}$
b = $\frac{76}{80}$
b = 0.95
Hence we got the value of a = 1.5 and b = 0.95
The linear equation is given by
Y = a + bx
Now put the value of a and b in the equation
Hence equation of linear regression is y = 1.5 + 0.95x
FAQs (Frequently Asked Questions)
1.What are the Types of Linear Regression?
Ans: Different types of linear regression are:
• Simple linear regression
• Multiple linear regression
• Logistic regression
• Ordinal regression
• Multinomial regression
• Discriminant Analysis
2. What are the Differences Between Linear and Logistic Regression?
Ans: Linear regression is used to predict the value of a continuous dependent variable with the help of independent variables. Logistic Regression is used to predict the categorical dependent variable with the help of independent variables. It is also used to predict the values of categorical variables.
3. How Does a Linear Regression Work?
Ans: Linear Regression is the process of finding a line that best fits the data points available on the plot. So it used to predict output values for inputs that are not present in the data set. Generally, those outputs would fall on the line. | |
# Degenerate spectra, measurement and lack-of-knowledge density matrix interpretation
1. Sep 12, 2011
### woodyhouse
Firstly hello, this is the first time I have posted here (although I have used the site to find info in the past). My query is best illustrated, I think, with an example. Suppose we have some physical system with corresponding state vector
$$\left| \psi \right> = a \left| 0 \right> + b \left| 1 \right> + c \left| 2 \right> + d \left| 3 \right> \in \mathbb C^4$$
and some physical quantity represented by the operator
$$\hat E = E_0 \left| 0 \right> \! \left< 0 \right| + E_1 \left| 1 \right> \! \left< 1 \right| + E_2 \left| 2 \right> \! \left< 2 \right| + E_3 \left| 3 \right> \! \left< 3 \right|.$$
First suppose that we have that $E_0 = E_1 = E$. Then the 2-dimensional subspace spanned by $\left| 0 \right>$ and $\left| 1 \right>$ is an eigenspace of $\hat E$ and any measurement with outcome $E$ will leave us with the projection (up to normalization) of $\left| \psi \right>$ onto this subspace. Expressed as a density matrix, the final state is
$$\rho = \mathcal N \; \big( \left | a \right|^2 \left| 0 \right> \! \left< 0 \right| + \left | b \right|^2 \left| 1 \right> \! \left< 1 \right| + ab^* \left| 0 \right> \! \left< 1 \right| + a^*b \left| 1 \right> \! \left< 0 \right|. \big)$$
Now consider the following: we have some experiment that is not accurate enough to distinguish $E_0$ and $E_1 = E_0 + \epsilon$ but that can distinguish all others (for instance $E_0$ and $E_1$ may correspond to very close spectral lines compared to $E_3$ and $E_4$). We perform a measurement, the outcome of which is $E_0 \pm 10\epsilon$. Then we could argue that the state must have collapsed to either $\left| 0 \right>$ or $\left| 1 \right>$ with probabilities $\left| a \right|^2$ and $\left| b \right|^2$respectively. According to the lack-of-knowledge interpretation of density matrices, the corresponding state (as a density matrix) after measurement is
$$\rho' = \mathcal N '\; \big( \left | a \right|^2 \left| 0 \right> \! \left< 0 \right| + \left | b \right|^2 \left| 1 \right> \! \left< 1 \right| \big)$$
where $\mathcal N$ is a normalizing factor.
The point of this is that (provided my reasoning holds) $\rho$ and $\rho'$ are physically distinct states. But when do we distinguish between the two scenarios? For instance if we have 2 degenerate levels that we know can be split with a magnetic field, do we always have to assume the presence of a magnetic field too weak to measure, or do we assume there is no magnetic field at all? Do we have to distinguish between `true' degeneracy and degeneracy relating to experimental inaccuracy?
I had a look in various literature and over previous posts in this forum and haven't been able to find an answer to this; I apologize if my search was not sufficiently thorough or if I am missing something obvious.
Last edited: Sep 12, 2011
2. Sep 12, 2011
### Ken G
Re: Degenerate spectra, measurement and lack-of-knowledge density matrix interpretati
I would say that the issue here is just what is measurement uncertainty, and why is it present-- is it because you have bad eyesight and cannot read a pointer very accurately, or is it because the experimental pointer you are using is intrinsically imperfect? If the former, then you shouldn't expect to end up with a precise description of the system, your description itself should be inaccurate. If the latter, then you should not expect the system to end up in either one eigenstate or another of the operator you have in mind, because it's not exactly the right operator for that experiment. In effect, you don't even know the true eigenstates of your own measurement if it is not perfect. This latter case is like expecting there to be a tiny magnetic field that you cannot measure precisely enough.
For either reason above, you end up with an imprecise description of the state of your system, so the two options you give above are not actually distinguishable-- they both fall within the measurement uncertainty.
3. Sep 12, 2011
### Bill_K
Re: Degenerate spectra, measurement and lack-of-knowledge density matrix interpretati
woodyhouse, This is exactly the two-slit experiment in disguise. In your first case E0 = E1, the state is described by a wavefunction, i.e. a coherent superposition |ψ> = a|0> + b|1>. You're welcome to square this and write down a density matrix if you want, but it's clear from the one you did write that it's of the form |ψ><ψ|.
In your second case if E0 and E1 are distinct but there is nothing in your experiment to distinguish them, then a wavefunction description still applies.
The incoherent superposition only applies if E0 and E1 are distinct and your experiment tells them apart. Not just can tell them apart, but actually does so.
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Math and Arithmetic
How do you simplify the fraction 10 over 40?
Wiki User
10/40 = 1/4 = .25 = 25%
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Related Questions
What are two ways to simplify 10 over 40?
There are only two ways to simplify 10 over 40. These two ways are to write 5 over 20 and 1 over 4. These answers are found by taking common factors of 10 and 40 and dividing them into 10 and 40 to find simpler numbers.
Simplify 28 over 40?
It is 7 over 10, which is 0.7 or 70%.
How do you simplify the fraction 25 over 40?
25 divide by 5=5 40 divide by 5 =8 5 over 8 is your answer
Can you simplify 12 over 40?
Yes - it simplifies to 3/10
What fraction of Rs. 2 is 40 paise?
It is 40/200. You can simplify that fraction if required.
What is four and six tenths written as a fraction?
46/10 4*10=40 40+6=46 46/10 Don't forget to simplify. 23/5
How do you simplify 44 over 40?
Divide the numerator and denominator by 4. You get (11/10).
40 percent is what as a fraction?
1/5 is the answer. I got this by dividing 40 by 100 So: 40/100 = 1/5 when you simplify the fraction.
How do you convert the fraction 4 over 10 in to a percent?
Percent is over one hundred. 4/10 = 40/100 = 40%
Is simplify when the greatest common factor of the numerator and denominator is one?
Yes. When the numerator and denominator have a GCF of 1, the fraction is in the simplest form. To simplify, find the GCF and divide both numerator and denominator by the same: Example: simplify 30/40 30/40 divided by 10/10 (1) = 3/4
What is 10.025 as a fraction?
10.025 as a fraction would be 401 over 40 or 10 and 1/4.
What is the fraction of 12 over 40 reduced to its simplest terms?
12 / 4 = 3 40 / 4 = 10 Answer being: 3 over 10 (3/10)
What fraction is equal to 0.40?
0.40 is a fraction. It is a fraction in decimal form rather than in the form of a ratio. However, that does not stop it being a fraction. Its equivalent, in rational form, is 40/100. You can simplify this rational fraction if required.
What is the fraction of 2.5 out of 40?
It is 25/400 which you can simplify if required.
What fraction is equivalent to 8 over 40?
Two equivalent fractions to 8 over 40 are 1/5 and 2/10
What is the answer of 40 and 52 as simplify?
I presume you are asking what is the simplified fraction of 40/52? 40 and 52 are both divisible by 2, so divide them both by 2: 20/26 20 and 26 are both divisible by 2 again, so divide them both by 2: 10 / 13 13 is a prime number, so you can not simplify this fraction any further.
How do you convert 27.5 percent to a fraction?
"Percent" means "out of 100", so to convert a percentage to a fraction put it over 100 and simplify: 27.5% = 27.5/100 = 55/200 = 11/40
5.40 percent into a fraction?
5 40/100 or 540/100.THen you simplify it.By Dividing
Can you simplify 108 over 160?
It simplifies to 27 over 40.
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In chloroform, there are three electron withdrawing groups ($\ce{Cl}$) which are able to stabilize a negative charge. Thus, the first step is the deprotonation of chloroform, typically with concentrated lye: $$\ce{CHCl3 + NaOH <=>[H2O] CCl^-_3 + H2O}$$ The second step is the loss of $\ce{Cl^-}$ to yield $\ce{NaCl}$ and dichlorocarben, $\ce{^{..}CCl2}... 5 As andselisk states in the comments, it is adenine. [OP] I suspect it is fake. No, it is as real as water and the stars. What makes this a little bit complicated is how to draw it, specifically on which nitrogen (N) to put the hydrogen (H). The structure shown is the 9H adenine tautomer, supposedly the most stable tautomer. Source: ACS Omega 2021, 6, 29, ... 3 Have you ever heard of explosions? Explosions are chemical reactions of solid or liquid substances which produce a huge amount of gas in less than one second. For example, nitroglycerin is a liquid with the formula$\ce{C3H5N3O9}$, which makes dynamite. One mole of nitroglycerin weighs$227$g and has a volume$142$mL. The explosion corresponds to an ... 3 It seems like you're wondering how the products can have a greater volume than the reactants without violating conservation of mass. So: In a chemical reaction, the number of atoms is conserved. That preserves conservation of mass. But, as you can see from your balanced chemical equation, the number of molecules isn't conserved: There are more molecules on ... 1 Generally, the process of turning$\ce{HX}$into$\ce{H+}$and$\ce{X^-}$can be broken into 2 hypothetical steps [...] Indeed, this process is slightly simplified because acidities are typically measured in solution, whereas your thermodynamic cycle only deals with gas-phase energetics. However, in the present case, it's actually enough to look at gas-... 1 Here is a bigger chunk of the periodic table with acid dissociation constants of the hydrides: Source: Libretexts The authors claim that the two factors are electronegativity (how much of a positive partial charge the hydrogen has in the covalent bond) and bond strength (expressed as estimated homolytic cleavage energy). There is no strong anomaly for the ... 1 The sum of the$\mathrm{p}K_\mathrm{b}$value of a base and the$\mathrm{p}K_\mathrm{a}$value of its conjugate acid is a constant (at a constant temperature), and is equal to$\mathrm{p}K_\mathrm{w}$(which is$14$at room temperature). The more stable the conjugate acid, the higher the$\mathrm{p}K_\mathrm{a}$of the conjugate acid$\implies$lower$\... | |
## How can I as middleman verify whether a phishing site is valid if the scam listens only on the referrer link and blocks any other access methods?
How can I as a trusted user of a middleman company (such as PhishTank) verify whether a phishing site is valid if the scam listens only on a unique referrer link(randomly created) and is blocking any other access methods?
To throw a threat scenario into scene.
An attacker sent an email to a local bank officer, the email looks very similar to a official email of an employee in their company at a higher tier and the time was planned. Later they detect it was a spear-phishing attack from an old employee. They report the attack on PhishTank (for example), but there they can’t verify it because the link doesn’t allow direct access (only with a unique referrer as in the email). How can they still verify whether it was a valid report of not?
Now the real question,
On a technical view, how does such an attack work?
## How do I check whether a package from multiverse is installed?
I am on an Ubuntu system that has the multiverse repository enabled, and I’d like to see whether I can disable it.
How can I check whether any package from multiverse is installed on the system?
## How do we determine whether a heuristic is better than another in A* search Algorithm?
I am trying to solve a Maze puzzle using the A* algorithm. I am trying to analyze the algorithm based on different applicable heuristics.
Currently, I explored Manhattan and Euclidean distances. Which other heuristics are available? How do we compare them? How do we know whether a heuristic is better than another?
## Whether following language is linear or not?
I have a language $$L= \{a^nb^nc^m : n, m \ge 0\}$$.
Now, I wanted to determine whether this language is linear or not.
So, I came up with this grammar:
$$S \rightarrow A\thinspace|\thinspace Sc$$
$$A \rightarrow aAb \thinspace | \thinspace \lambda$$
I’m pretty sure(not completely however) that this grammar is linear and consequently language too is linear.
Now, when I use pumping lemma of linear languages with $$w$$, $$v$$ and $$u$$ chosen as follow I find that this language is not linear.
$$w = a^nb^nc^m, \space v = a^k, \space y=c^k$$
$$w_0 = a^{n-k}b^nc^{n-k}$$
now, $$w_0 \notin L \space (\because n_a \neq n_b)$$
So, I’m unable to find whether the language is linear or not and what goes wrong in above logic with either case. Please help.
## What Is the Complexity Class of Deciding Whether a Problem Is in NP? Is It Decidable?
Title says it all, but to clarify:
Define a problem, called $$IsInNP$$, as follows:
Given a Turing Machine $$M$$, $$IsInNP$$ is the problem of deciding if the problem that $$M$$ decides is in $$NP$$.
What is the complexity class of $$IsInNP$$? Is it even decidable? Is the answer the same for any other complexity class, like $$NP$$-hard? And are those questions even sensible to ask?
By the way, I am aware that the class $$NP$$ is not enumerable, but since I do not quite understand enumerability and it seems that recursively enumerable problems can be decidable, I do not know if that means that deciding whether a problem is in $$NP$$, or any other complexity class, is decidable.
Also, I am aware of Rice’s Theorem, and I believe it can be interpreted as saying that deciding whether a problem is in $$NP$$ is undecidable, but I am not certain.
Bonus question if the above questions are sensible: given a property $$S$$ that only $$NP$$ problems possess, does the above also mean that deciding whether a problem decided by a Turing Machine $$M_2$$ has property $$S$$ is in the same complexity class as $$IsInNP$$?
## Whether the given language is a CFL or not?
Let $$L$$ be a language defined over $$\Sigma = \left \{ a, b \right \}$$ such that $$L = \left \{ x\#y \mid x,y \in \Sigma^*, \# \text { is a constant and } x \neq y \right \}$$ State whether the language L is a CFL or not? Give valid reasons for the same.
Now, I think that the given language is not a CFL. I have used the pumping lemma test for showing that L is not CFL. Specifically, I have done the following-
Consider a string $$w = abb\#aab$$. Obviously, $$w \in L$$.
Let, $$u = \epsilon \ v = a \ w = bb\#aa \ x = b \ y = \epsilon$$
Here, $$|vx| \geq 1$$
But, $$uv^2wx^2y = aabb\#aabb \notin L$$ Therefore, pumping lemma test result is negative. Therefore, we can conclude that the given language is not a CFL.
Now, I have a doubt regarding the above method- I know that given a CFL, if we want to perform the pumping lemma test for the CFL, we must always use strings which are of length greater than or equal to the minimum pumping length. In fact, this also confirms to the condition that the length of the string $$w$$ used for the pumping lemma test (denoted by $$|w|$$) must be greater than or equal to n.
Therefore, when I use $$w = abb\#aab$$ for doing the pumping lemma test, I implicitly make the assumption that 7 is greater than or equal to the minimum pumping length (if $$L$$ were to be a CFL). Am I correct or incorrect in doing so?
## Proving whether an input sequence satisfies a given RE language
I’ve learned this a few years ago that this is impossible unless one simply ‘executes’ (in a modern computing sense) the input with the language rules, but I have some problems in just using this statement.
• The fundamental doubt is that the statement itself is well-stated. If I’m using the term ‘execution’ to describe the act of matching the rules one input element by one, is this statement valid?
• Is this statement (deciding whether an input sequence is following a language is impossible without an execution) not exactly limited to RE? In other words, I wonder this statement also holds even for the languages in other classes.
• I’m not even sure how I can search for this statement and confirm from the external source.
(By RE, here I indicate the recursively enumerable languages, not the regular expression)
## Checking whether the win-loss standings of a league are possible
You’re hosting a 1 v 1 basketball league with a game schedule. At the end of the league you have each player report their supposed win-loss record (there are no ties), but you want to check whether the proposed standings were actually possible given the schedule.
For example: you have four players (Alice+Bob+Carol+Dave) and your schedule is a simple round robin. The reported standings [A: 3-0 B: 1-2 C: 1-2 D: 1-2] and [A: 2-1 B: 1-2 C: 1-2 D: 2-1] would be possible, but the standing [A: 3-0 B: 0-3 C: 0-3 D: 3-0] would not be.
Now suppose the schedule is instead a 3 game head to head between Alice+Bob and Carol+Dave. The reported standing [A: 3-0 B: 0-3 C: 0-3 D: 3-0] is now possible, but [A: 3-0 B: 1-2 C: 1-2 D: 1-2] would no longer be.
(The schedule does not need to be symmetric in any way. You could have Alice only play against Bob 10 times, then make Bob+Carol+Dave play 58 round robins against each other.)
Problem: Given a schedule with k participants and n total games, efficiently check whether a proposed win-loss standings could actually occur from that schedule.
The O($$2^n$$) brute force method is obvious, enumerate all possible game outcomes and see if any match the proposed standings. And if k is small increasing n doesn’t add much complexity – it’s very easy to check a two person league’s standings regardless of whether they play ten games or ten billion games. Beyond that I haven’t made much headway in finding a better method, and was curious if anyone had seen a similar problem before.
## Is there any lore about whether a creature can see themselves and their gear while invisible?
I’ve separately asked about the mechanics of whether an invisible creature can see themselves and their own gear in D&D 5E. And from a rules perspective, they can’t. But from a more flavor perspective, I want to know if there’s any indication in the various D&D published materials (maybe in novels, movies, articles, or setting sourcebooks for prior editions) whether or not people can see themselves (and their things) when invisible. Does somewhere explain just what a character experiences when they look at themselves while invisible, like saying whether they’re surprised by seeing the ground through their own feet, or whether there’s some ghostly effect where they can tell they’re invisible to others but can see themselves to some extent?
## External links: Whether & how to distinguishing them from internal links, and to open them
How do other IAs/UXD’s treat external links & what are the perceived pros/cons of different options. Specifically looking for recommendations on:
• whether & how to visually distinguish external links from internal links
• whether to open them in the same window/tab or a different window or tab
I’ve found some great feedback re these questions on http://www.ixda.org/search.php?tag=external+links & http://www.useit.com/alertbox/open_new_windows.html
Looking for additional opinions, thoughts & especially any usability test findings contributors of this site might have re these questions. | |
# Calculate covariance for discrete random variables
This question might seem quite easy for many of you, however, I think I need a little help in the right direction.
I'm currently reading about probability theory and have come across covariance. I know the definition of covariance and I'm trying to solve some exercises.
For instance, I have been given a discrete random variable X with probability function px(x) = 1/2 if x = -1, 1/4 if x = 0, 1/4 if x = 1, 0 otherwise.
Additionally, I have been given a discrete random variable Y, which is independent of X, and has probability function py(y) = 3/4 if y = 0, 1/4 if y = 1, 0 otherwise.
I then have to calculate Cov(Y, 2Y - X). The answer is given and should be 3/8.
What confuses me a bit when I'm reading about covariance is that some of the formulas I have come across uses pairs consisting of X values and Y values, but for example in this exercise there are three X values and only two Y values.
Furthermore, when two discrete random variables X and Y are independent, which this exercise says (it says Y is independent of X), then Cov(X, Y) should be equal to 0. But when I use the rule E(X * Y) = E(X) * E(Y) for independent variables, I, however, end up with a formula indicating that the result should be 0 and not 3/8.
For example when I try to deduce from the definition Cov(X, Y) = E[(X - E[X])(Y - E[Y])] I get the following:
Cov(Y, 2Y-X) = E[(Y - E[Y])((2Y - X) - E[2Y - X])]
This can be rewritten using Cov(X, Y) = E[X * Y] - E[X] * E[Y] which I think is more manageable:
Cov(Y, 2Y-X) = E[Y * (2Y - X)] - E[Y] * E[2Y - X]
Since X and Y are independent, I should be able to use the rule E(X * Y) = E(X) * E(Y) so that I get:
Cov(Y, 2Y-X) = E[Y] * E[2Y - X] - E[Y] * E[2Y - X]
which indicate the result is 0, which is wrong according to the result given.
So if we say I'll just try to proceed with this one:
Cov(Y, 2Y-X) = E[Y * (2Y - X)] - E[Y] * E[2Y - X]
How do I continue from here? I know how to calculate the expected value of Y for example, but how do I calculate the expected value of 2Y - X? The formula I found for this suggested using pairs of X values and Y values, but I don't know how to do that when there aren't the same amount of X values and Y values.
Where you went wrong is in the step where yu say $X$ and $Y$ are independent (true) but use that as if you had said $2Y-X$ and $Y$ are independent (false). Surely if you specify the value of $Y$, that gives you some information about the value of $2Y-X$.
So you need to start with the equation you said you would like to proceed with: $$\mbox{cov }(Y,2Y-X) = E[ Y(2Y-X)] -E[Y]E[2Y-X] \\ \mbox{cov }(Y,2Y-X) = 2E[ Y^2] - E[YX] - E[Y]E[X]$$ Now with your probability distribution as given: $$E[Y] = \frac34(0) + \frac14(1) = \frac14\\ E(Y^2) = \frac34(0^2) + \frac14(1^2) = \frac14\\ E[X] = \frac12(-1)+\frac14(0)+\frac14(1) = -\frac14\\ E[YX]=\frac34\frac12(0\cdot (-1)) + \frac34\frac14(0\cdot (0)) + \frac34\frac14(0\cdot (1)) \\+ \frac14\frac12(1\cdot (-1)) + \frac14\frac14(1\cdot (0)) + \frac14\frac14(1\cdot (1)) = -\frac18+\frac1{16} = -\frac1{16}$$ And then $$\mbox{cov }(Y,2Y-X) = 2E[ Y^2] - E[YX] - E[Y]E[X] \\= 2\cdot \frac14 - \left(-\frac1{16}\right)- \frac14 \cdot \frac14 = \frac12-2 \left(\frac1{16}\right) = \frac38$$
• When you have the general E(Y * (2Y - X)) - E(Y) * E(2Y - X), then you multiply out the first part so you get E(2Y^2 - XY) - E(Y) * E(2Y - X). Then as I understand, there's a rule for constants and a rule saying you can move out the XY, so that you get 2E(Y^2) - E(XY) - E(Y) * E(2Y - X). But how do you get from there to the equation you wrote, that is 2E(Y^2) - E(XY) - E(Y) * E(X). Said in other words, how does the last part, which is E(2Y - X), just become E(X)? – John A Feb 4 '17 at 21:14
You are correct that $$Cov(Y, 2Y-X) = E[Y * (2Y - X)] - E[Y] * E[2Y - X].\tag1$$ However, it's not true that $Y$ and $2Y-X$ are independent, so you cannot say $$E[Y * (2Y - X)]=E(Y) E(2Y-X).$$ The correct way to proceed with (1) is to multiply out $Y(2Y-X)$ to obtain $2Y^2-XY$. The expectation of this is $2E(Y^2)-E(XY)$. Now you can apply independence of $X$ and $Y$ to write $E(XY)=E(X)E(Y)$. At this point (1) reduces to calculating quantities like $E(X)$, $E(Y)$ and $E(Y^2)$, which you have enough information to do.
Aside: If you want to calculate covariance using pairs of $X$ and $Y$ values, you can do that: The independence between $X$ and $Y$ specifies the probability of any pair of $X$ and $Y$ values, as follows: $$P(X=x,Y=y)=P(X=x)P(Y=y).$$ | |
## Monday, March 21, 2011
### ARIMA, Forecasting and Python
I ported the R code found on Rob Hyndman's blog into Python + rpy2. The latter package allows calling of R code from Python which we used here to utilize the forecast package. Test R and Python codes can also be found in the zip file below.
import rpy2.robjects as R
import rpy2.rinterface as rinterface
from rpy2.robjects.packages import importr
import rpy2.robjects.numpy2ri
from rpy2.robjects.vectors import FloatVector, StrVector
import datetime
import numpy as np
forecast = importr("forecast")
R.r.source("arima-fn.R")
n = len(x)
m = 10
terms = 4
x = np.insert(x, 10, np.nan) # put in a NaN to see what happens
n += 1
print x
R.r.assign('x',x)
R.r.assign('n',n)
R.r.assign('m',m)
R.r.assign('terms',terms)
R.r('fit = Arima(x=x, order=c(2,0,1), xreg=fourier(1:n,terms,m))')
R.r('f = forecast(fit, h=2*m, xreg=fourier(n+1:(2*m),terms,m))')
res = R.r('forecast_results(f)')
for x in res:
print x
All Code
arima-fourier2.R
forecast_results <- function(res)
{
return (res\$mean)
}
fourier <- function(t,terms,period)
{
print (terms)
n <- length(t)
X <- matrix(,nrow=n,ncol=2*terms)
for(i in 1:terms)
{
X[,2*i-1] <- sin(2*pi*i*t/period)
X[,2*i] <- cos(2*pi*i*t/period)
}
colnames(X) <- paste(c("S","C"),rep(1:terms,rep(2,terms)),sep="")
return(X)
}
library(forecast)
y <- yy[,'price']
n <- length(y)
print (n)
m <- 10 # how many points in the future
terms = 4
fit <- Arima(y, order=c(2,0,1), xreg=fourier(1:n,terms,m))
f = forecast(fit, h=2*m, xreg=fourier(n+1:(2*m),terms,m))
print (forecast_results(f))
plot(f)
arima-fouerier.R
n <- 2000
m <- 200
y <- ts(rnorm(n) + (1:n)%%100/30, f=m)
print (y)
quit()
fourier <- function(t,terms,period)
{
n <- length(t)
X <- matrix(,nrow=n,ncol=2*terms)
for(i in 1:terms)
{
X[,2*i-1] <- sin(2*pi*i*t/period)
X[,2*i] <- cos(2*pi*i*t/period)
}
colnames(X) <- paste(c("S","C"),rep(1:terms,rep(2,terms)),sep="")
return(X)
}
library(forecast)
fit <- Arima(y, order=c(2,0,1), xreg=fourier(1:n,4,m))
f = forecast(fit, h=2*m, xreg=fourier(n+1:(2*m),4,m))
plot(f)
print (y)
print (f)
arima.py
import rpy2.robjects as R
import rpy2.rinterface as rinterface
from rpy2.robjects.packages import importr
import rpy2.robjects.numpy2ri
from rpy2.robjects.vectors import FloatVector, StrVector
import datetime
import numpy as np
forecast = importr("forecast")
R.r.source("arima-fn.R")
n = len(x)
m = 10
terms = 4
x = np.insert(x, 10, np.nan) # put in a NaN to see what happens
n += 1
print x
R.r.assign('x',x)
R.r.assign('n',n)
R.r.assign('m',m)
R.r.assign('terms',terms)
R.r('fit = Arima(x=x, order=c(2,0,1), xreg=fourier(1:n,terms,m))')
R.r('f = forecast(fit, h=2*m, xreg=fourier(n+1:(2*m),terms,m))')
res = R.r('forecast_results(f)')
for x in res:
print x
garan.dat
price
6.40000
6.33000
6.43000
6.53000
6.48000
6.33000
6.28000
6.33000
6.33000
6.18000
6.28000
6.28000
6.43000
6.48000
6.23000
6.38000
6.33000
6.23000
6.45000
6.28000
6.38000
6.43000
6.45000
6.23000
6.04000
5.84000
5.94000
5.84000
5.95000
5.89000
5.94000
6.00000
6.35000
6.13000
6.38000
6.23000
6.04000
5.79000
5.49000
5.70000
## Sunday, March 13, 2011
### Osher on Mathematicians and Engineers
From Stanley Osher's talk:
Main thing to do as applied mathematicians it to be willing to listen to the other person's language, try to understand and translate it into something that makes sense mathematically. Engineers are very clever, when you talk to engineers in many different fields, they have an algorithm that works, they don't know why it works, it's not rigorous, but it's great. Then your job is to figure out how to generalize it and make mathematics out of it. That happened [to us] many, many times. Lots of fantastically successful algorithms came from the intuition of engineers, then they were made precise later by mathematicians.
### Stanley Osher: Mathematician with an Edge
From levelset.com
Stanley Osher is an extraordinary mathematician who has made fundamental contributions to applied mathematics, computational science and scientific computing and who has cofounded three companies based, in part, on his research. He has applied his pioneering work on level set methods and other numerical methods for partial differential equations to the field of image processing and, in particular, to video image enhancing and movie animation. He has been featured prominently in the scientific and international media such as Science News, Die Zeit and Los Angeles Times. He is perhaps the most highly cited researcher in the field of scientific computing.
He received the NASA Public Service Group Achievement Award, Japan Society of Mechanical Engineers Computational Mechanics Award and the SIAM Pioneer Prize. He was an invited speaker at the International Congress of Mathematicians.
He is currently Director of Special Projects at the Institute for Pure and Applied Mathematics (IPAM) at the University of California at Los Angeles and Director of Applied Mathematics.
The Editor of Imprints interviewed Stanley Osher on 17 December 2003 when he was an invited guest at the Institute's program on image processing and gave a public lecture. The following is based on an edited transcript of an interview in which he talked about the fascination of applied mathematics and his total dedication to research and applications.
I: Thank you very much, Professor Osher, for agreeing to be interviewed. In which area did you do your Ph D?
O: I did my Ph D in an esoteric area in functional analysis, which is in pure mathematics. I left it immediately and switched to numerical analysis after my thesis. I was lucky enough to talk to some people, including Peter Lax, who suggested the numerical stuff.
I: Did you find your real inclinations in applied mathematics?
O: Yeah, but it did not happen until after I got my degree. I liked everything and I specialized more after my Ph D.
I: Did you use functional analysis later on in your work?
O: Yes. The first thing I did in numerical analysis was an application of Toeplitz Matrices. It used functional analysis and was short and elegant.
I: You switched to applied mathematics and then eventually to something very practical like applying to movie animation.
O: This just happened - that's the way research leads you. You cannot predict these things. Together with colleagues and students, I was using the level method, which is a way of determining how surfaces such as bubbles move in three dimensions, how they merge and so on. You can simulate the flow of bubbles, planes and things like that. It happened that people in the movie industry got interested in this stuff.
I: Are these are pure mathematics problems?
O: It's a way of representing surfaces and has connections with differential geometry. People prove theorems about these things. They have applications in many areas including fluid dynamics and quantum mechanics. They arise in the movie industry
because you want to see how things merge and split like in explosions or rising bubbles. You know, our Governor, Schwarzenegger, used in his latest movie, a lot of these methods done by my former student Ron Fedkiw.
I: Did you have to use the mathematics of many different areas to solve those problems?
O: Sure. The key thing I would say is partial differential equations, elementary differential geometry, numerical analysis. The language of applied mathematics is differential equations.
I: How did you get into the movie animation business?
O: We had a week of movie industry people coming into UCLA (University of California at Los Angeles) giving lectures about what they did and, in fact, imaging science was highlighted by the American Mathematical Society one year. There was a week in that stuff. We invited people from the local movie industry and they were interested in what we were doing. They wound up arguing with me. The water in the "Titanic", which won many Academy awards, was very bad. It was old-fashioned stuff. Their people came to talk about that and so we decided we could do better than that. In recent movies, the water is much more realistic. The first movie that actually used sophisticated water was "Ants". Now level sets are used by movies like "Shrek", "Terminator" and many blockbusters. My former student Ron Fedkiw is doing this movie animation stuff very well. The stuff that my colleagues and I do has applications in many other places besides the movie industry.
I: I believe that another of you dramatic achievement is the use of mathematics in catching criminals. How did this come about?
O: Well, I was living in Los Angeles when the city went up in smoke. There was a big riot in Los Angeles after this guy (Rodney King) was beaten up by the police.
The riot resulted in people being arrested for looting and beating up passers-by. There was a video recording of the bad guys beating up truck driver Denny and it showed a speck on the arm of a man throwing a brick at Denny.
It turned out that I had a friend who knew the District Attorney or somebody, and I
was then doing video image enhancement with my colleague L. Rudin, We were able to resolve the speck into a rose tattoo and it was a great application of what we were doing. After the Denny case trial (the tattoo led to the conviction of the suspect) we had a lot of media publicity and our company specialized in the area of image enhancement. Eventually I sold my share of the company to Rudin. He has a package on video image enhancement which is used by the police around the world, and he's quite successful.
I have left this business. It was quite fun and related to mathematics. Image processing is the real world and the graphics that is manufactured is the fake world. You want to find out what the image really is.
I: Do you actually go and seek out those problems or do people come looking for you?
O: It's hard to say. Sometimes it's serendipity. Things just happen. I was lucky. In all my years of doing science, I managed to work with the right people who knew what the problems were. For example, I knew nothing about image processing at all after my PhD. Then this guy Rudin came over to me and asked me about some work I had done in fluid dynamics on supersonic flow and shock waves. I asked him what he wanted to know and I got fired up. He was a computer scientist and he realized that shock waves had something to do with imaging. It was a fantastic observation and our collaboration worked out well.
I: The scope of movie animation has just opened up, hasn't it?
O: Yeah. But I'm not sure this is the best field of application for somebody to work in because the market is small except for video games which is a big business. But video games require real time imaging. What we do is not real time; it's too slow. That might change.
I: It could be just a matter of computing power.
O: Yeah, yeah, and also hardwiring in level set methods, which may come in time.
I: Pattern recognition used to be sort of very big.
O: I'm not sure of the definition of pattern recognition, but image processing is related to it. The key idea in everything we do is about "edges", which characterize images. Edges, and now, textures. If you look at a table, for example, the flat part is not very interesting but the boundary of it is. It's the discontinuity. If I look at you, I can see you because of the outline. The outline is very important and the mathematics that was used in other areas of science like fluid dynamics specialise in things like edges and boundaries. And now images.
I: Does it mean that if you have a vague or blurred object, you can always refine it?
O: Yeah, but it's usually difficult to do so when you have edges because near the edges you will get spurious oscillations. The techniques we have developed, which came originally from fluid dynamics, are now useful for removing those unreal artifacts.
I: How do you know that what you get is the actual object?
O: That's a good question. But you can get enough science behind it and you can prove theorems, algorithms converge and stuff like that.
I: You look at the picture and it's so blurred. And then you do this and you get that. There's a lot of faith in that.
O: I won't disagree with that completely. But when I drop something, I expect it to fall and not go up. The probability is not zero that it might go up. But with a high degree of certainty, you can say that this is a realistic picture.
I: Does that mean there is a probabilistic element in it.
O: Oh, yes. That's always true with nature.
I: For example, in the police case, couldn’t the guy dispute your picture?
O: Yes, but the jury believed us. Fingerprints are also subjective. Fingerprints are very good examples of image processing of a certain sort.
I: What about problems in voice recognition?
O: It's a different kind of mathematics. I'm not an expert in it. The techniques we use in image processing involving differential equations have never been used for sound. But I understand there's some very new stuff along these lines. It's only beginning and just developing. The Institute for Pure and Applied Mathematics (IPAM) at UCLA is running a program on sound and how the ear works and the related mathematics.
O: It's highly dubious. Among the things I do is designing computer chips and you use mathematics which is not intuitive. Engineers cannot intuit what the right shape should be. One of my engineering friends said this is artificial intelligence. The computer would do something the engineer cannot imagine. That's not the usual definition of artificial intelligence. If you have solid mathematics, you can do things that are intuitively not obvious. But the experienced engineer makes a leap of understanding. If you call that artificial intelligence, so be it. But it’s not the kind that people use to talk about.
I: The imaging business is also very important in astronomy, isn't it? They take pictures which are so faint.
O: Very much so. Astronomers have done very good work in this area. They had to over the years. The early work in this stuff in astronomy was very fascinating. They did things which are precursors of what is going on now.
I: You mean the astronomers actually did something mathematical?
O: Yes. That's very often the case. The good thing about being an applied mathematician is that when you work in different areas you find very brilliant people in other areas of science develop mathematical algorithms without realising what they are doing and which can be generalized to other areas. So it's a question of language.
I: Are more astronomers talking to mathematicians now?
O: Actually we'll be running a program on computational astronomy in IPAM a year from now. Astronomy is just one example. Throughout science mathematics is playing more and more of the key role. In the Institute in which I am involved, its mission is to do interdisciplinary work. People from different areas of science have problems which they think are mathematical. And our goal is to make mathematics out of it. Now they believe that we can do something, mainly because of the computer.
I: That's interesting. So now mathematics is also contributing towards understanding the origin of the universe. This is something not many people are aware of.
O: Well, I'm not an expert in the field, but yes, absolutely. I think that the world is governed by differential equations.
I: Your work involves a lot of algorithms. Do you invent the algorithms?
O: That's what we do. That's the fun part of it. I was interviewed by the LA Times (so was Tony Chan for the same article) and I said that I wrote the algorithms that make the computer sing. I am the Barry Manilow of mathematics.
I: Talking about algorithms, some people consider algorithms to be inventions.
O: Yeah, they are. You can actually patent them now. It used to be that a patent has to contain a device with wires and everything. But I understand the US Patent Office is now more liberal. I'm no lawyer, but I know that as the years go by, the Patent Office seems more and more interested in giving legal protection to algorithms.
I: You could be rich. Hollywood would be paying you millions.
O: People work for salaries. There is money, ego and fun. It's a very nonlinear function. I don't know which is most important. Fun is very important. It's a very good life. I would recommend people going into this stuff now. If you have the talent for it, it's the best life.
I: It's something very different, something, how do you say, non-academic?
O: In some sense, yes. You learn things, you read stuff and you learn new ideas, and you are fired up. Sometimes you deliver something different from what you have found. You have a vague idea that something interesting is going to come up. You wander around and something happens. Then you get very excited. It's like opening a door and you don't know what good things are behind it. You're not sure where it's going to end and what the level of success it's going to be. It's very exciting. Everyday I can't wait to go to work. People often asked me, “What kind of life is this that work is so important?” People go on vacation. My work is vacation.
I: Do you have many Ph D students?
O: Many, disproportionately Chinese. We have many good Chinese students at UCLA.
I: Have you used your methods for something more serious like weather forecasting and earthquake prediction?
O: Yeah. The differential equations and some stuff I used to do in fluid dynamics and done by many other very good people are absolutely useful in weather prediction. I still do work on explosives and multi-phase flows and ray tracing. Physical phenomena apart from imaging is very much a part of my research with my colleagues and students.
I: What about the theory of turbulence?
O: Ah, turbulence is too dangerous. If you touch turbulence, you get burnt. One of my mentors once said he had great respect for people in turbulence, which is far more than they have for each other. Turbulence is too controversial. Turbulence has a probabilistic aspect to it, it's statistical and that's not my thing.
I: It's not algorithmic?
O: It's a different kind of algorithm, a different kind of intuition. I do understand it, but I've never been a big user of that stuff. It's very dangerous even to discuss it, it's like politics. Everybody condemns the other guy's theory. It's somewhat like religion.
I: That's surprising. The methods are mathematical, the equations are mathematical.
O: Yeah. The problems are very hard and complicated. You get some very good mathematics coming out of it but I'm not sure the physics is understood in analysing what really happens.
I: I'd like to ask you a philosophical question. How does your work affect your view of life?
O: In terms of how research affects my philosophy? The basic idea is to try to make order out of this life that we live. Everyday you encounter things and it's a messy world. The goal is to take this mess that we see and somehow "mathematize" it and make a prediction. In that sense, research has certainly affected my philosophy. I try to figure out what is going on. The most complicated thing is how our human nature operates. It will be fun to understand that. Many people I know at UCLA and elsewhere are using medical imaging to understand how the topology and shape of the brain affects its function. They use the mathematics that I am involved in. The greatest mystery of all is human behaviour and maybe it can be explained by level sets.
I: Would you say that your present interest and activity is, in some way, directed by your own personal philosophy towards finding order?
O: Yeah, absolutely. I came from being a poor boy in Brooklyn. I wanted some order in my life, to become middle-class and to have a life that I enjoy. Then I stumbled onto this thing, and wow, that's very good for creating order. I entered graduate school in New York University in 1962 when it was a fantastic place for applied mathematics, maybe one of the best ever in the world,, All the top people from Goettingen wound up in New York and it was so exciting.
I: Was that the Courant Institute?
O: Yes, the Courant Institute. In 1962, when I entered, it was incredible. The people who were there and the atmosphere. You felt that you were doing something important. It had a very great influence on me and many other people.
I: Was Courant still there?
O: He was still around, old but still functioning. He added many people all of whom I thought of being old, something like 20 years younger than I am now. They were great people and it was a fantastic time. There were many people like me who were ethnic New Yorkers and who view becoming a mathematician as a way of becoming middle-class American citizens.
I: What attracted you to UCLA?
O: In truth, there was a guy who I was working with: Andy Majda. He is an excellent applied mathematician. He was there and they were building an applied mathematics group. Also I liked California. When I was in New York, I was always dreaming about the Beach Boys. Sunshine and California together with math was great - everything that I wanted. Over the years we had very nice people. The atmosphere is extremely good in UCLA. People who visited us commented on how well we got along each other, which is unusual in academia, It has worked out quite well.
## Thursday, March 10, 2011
### Image Segmentation using Active Contours, Level Sets
Here is a new image segmentation Python code which is a port of this Matlab file. It uses level sets and mean curvature motion, and is able to segment the sample image after few iterations. We had to write our own bwdist which is a wrapper for distance_transform_edt wrapper so bwdist acted the same way as Matlab bwdist(), some conversions were necessary to make this work. We also changed the initialization of \phi, making it simpler and closer to Dr. Li's code we shared before.
Code
import matplotlib.pyplot as plt
import numpy as np
import scipy.signal as signal
import scipy.ndimage as image
import time
from scipy import ndimage
def bwdist(a):
"""
this is an intermediary function, 'a' has only True, False vals,
so we convert them into 0, 1 values -- in reverse. True is 0,
False is 1, distance_transform_edt wants it that way.
"""
b = np.ones(a.shape)
b[a==True] = 0.
return ndimage.distance_transform_edt(b)
def gauss_kern():
""" Returns a normalized 2D gauss kernel array for convolutions """
h1 = 15
h2 = 15
x, y = np.mgrid[0:h2, 0:h1]
x = x-h2/2
y = y-h1/2
sigma = 1.5
g = np.exp( -( x**2 + y**2 ) / (2*sigma**2) );
return g / g.sum()
Img = Img[::-1]
g = gauss_kern()
Img_smooth = signal.convolve(Img,g,mode='same')
rows, cols = Img.shape
# initial function phi - level set is a square 4 pixels
# away from borders on each side, in 3D it looks like an empty
# box
c0=4
w=4
nrow, ncol=Img.shape
phi=c0*np.ones((nrow,ncol))
phi[w+1:-w-1, w+1:-w-1]=-c0
# edge-stopping function
dt=.4
# number of iterations after which we reinitialize the surface
num_reinit=10
phiOld=np.zeros((rows,cols))
# number of iterations after which we reinitialize the surface
iter=0
plt.ion()
while np.sum(np.sum(np.abs(phi-phiOld))) != 0:
# magnitude of gradient of phi
# normalized gradient of phi - eliminating singularities
# curvature is the divergence of normalized gradient of phi
tmp1 = g * K * absGradPhi
dPhiBydT =tmp1 + tmp2 + tmp3
phiOld=phi
# level set evolution equation
phi = phi + ( dt * dPhiBydT )
iter=iter+1
if np.mod(iter,num_reinit)==0:
# reinitialize the embedding function after num_reinit iterations
phi=np.sign(phi)
phi = (phi > 0) * (bwdist(phi <> 0))
if np.mod(iter,10)==0:
time.sleep(0.6)
plt.imshow(Img, cmap='gray')
plt.hold(True)
CS = plt.contour(phi,0, colors='r')
plt.draw()
plt.hold(False) | |
# How to get p value and confidence intervals for nls functions?
I have 2 questions.
1) How can I have p.value for my 2 functions? My hypothesis is that I have a correlation between my function and my data.
2) How can I have a confidence intervals for my 2 functions?
library(ggplot2)
g <- function (x, a,b,c) a * (1-exp(-(x-c)/abs(b)))
X1 <- c(129.08,109.92,85.83,37.72)
Y1 <- c(0.7,0.5,0.39,-1.36)
dt1 <- data.frame(x1=X1,y1=Y1)
model1 <- nls(Y1 ~ g(X1, a, b, c),
start = list(a=0.5, b=60, c=50),control=nls.control(maxiter = 200))
ggplot(data = dt1,aes(x = x1, y = y1)) +
theme_bw() + geom_point() +
geom_smooth(data=dt1, method="nls", formula=y~g(x, a, b, c),
se=F, start=list(a=0.5, b=60, c=50))
f <- function (x, a, b, c) a*(x^2)+b*x+c
X2 <- c(589.62,457.92,370.16,295.98,243.99,199.07,159.91,142.63,
124.15, 101.98, 87.93, 83.16, 82.2, 74.48, 47.68, 37.51, 31,
27.9, 21.24,18.28)
Y2 <- c(0.22,0.37,0.49,0.65,0.81,0.83,1,0.81,0.65,0.44,0.55,0.63,
0.65,0.55,0.37,0.32,0.27,0.22,0.17,0.14)
dt2 <- data.frame(x2=X2,y2=Y2)
model2 <- nls(Y2 ~ f(X2, a, b, c),
start = list(a=-1, b=3, c=0),control=nls.control(maxiter = 200))
ggplot(data = dt2,aes(x = x2, y = y2)) +
theme_bw() + geom_point() +
geom_smooth(data=dt2, method="nls", formula=y~f(x, a, b, c),
se=F, start=list(a=-1, b=3, c=0))
• Does "summary(model1)" deliver what you want?
– Cyan
Mar 27, 2012 at 2:01
• it doesn't, summary(model) gives pvalue for lm functions Mar 27, 2012 at 5:07
• @Kristina does the method described in the answer below for linearizing your models so summary can produce the values you want work for you? Mar 27, 2012 at 19:27
1. - You could try (this is an approximation)
library(nls2)
summary(as.lm(model))
• You can obtain a p-value for all parameters used in your model using
summary(model)
• You can get p values for a model by comparing it to another ("nested") model using
anova(model1, model2)
where model 2 is a simplified version of model 1 (it is your null hypothesis)
• You can use methods such a bootstrapping, to get a measure of the probability of fit of your complete model.
2.
• You can possibly get full model confidence interval using (this is an approximation)
library(nls2) predict(as.lm(model2), interval = "confidence")
• You can obtain the confidence interval of the parameters using
confint(model)
profile(model)
plot(profile(model))
• You can obtain the pair wise confidence interval for two of your parameters (for both plotting and to get the matrix) using
ellipse.nls(model)
• I've installed and attached the nls2 library, but don't seem to have the as.lm function available. Any ideas? Jul 23, 2013 at 16:03
• @Daniel Kessler, try quantitativeconservationbiology.wordpress.com/2013/07/02/…
– etov
Aug 19, 2014 at 12:12
• I'm having the same problem; the as.lm function does not exist. The link posted by @etov is stale. Please post solutions inline, rather than links to external pages that may become stale. So, what is the solution to the missing as.lm()? Dec 9, 2016 at 11:54
• @bhaller - right. The content was a bit too long for a comment; posted now as an answer.
– etov
Dec 11, 2016 at 7:36
Regarding confidence intervals, other answers here seem to have issues with the use of functions (as.lm.nls, as.proto.list) that for some reason are not defined for some users (like me). After some surfing, I found an answer that works for me, requiring only the MASS package. At the urging of @etov, I am posting the answer I found here. It is originally from https://www.r-bloggers.com/predictnls-part-1-monte-carlo-simulation-confidence-intervals-for-nls-models/ and appears to be by someone named Andrej who uses the handle anspiess. This function by Andrej, in his words, "takes an nls object, extracts the variables/parameter values/parameter variance-covariance matrix, creates an “augmented” covariance matrix (with the variance/covariance values from the parameters and predictor variables included, the latter often being zero), simulates from a multivariate normal distribution (using mvrnorm of the ‘MASS’ package), evaluates the function (object$call$formula) on the values and finally collects statistics". So it is a Monte-Carlo-based method of getting confidence intervals for an nls model. His code:
predictNLS <- function(
object,
newdata,
level = 0.95,
nsim = 10000,
...
)
{
require(MASS, quietly = TRUE)
## get right-hand side of formula
RHS <- as.list(object$call$formula)[[3]]
EXPR <- as.expression(RHS)
## all variables in model
VARS <- all.vars(EXPR)
## coefficients
COEF <- coef(object)
## extract predictor variable
predNAME <- setdiff(VARS, names(COEF))
## take fitted values, if 'newdata' is missing
if (missing(newdata)) {
newdata <- eval(object$data)[predNAME] colnames(newdata) <- predNAME } ## check that 'newdata' has same name as predVAR if (names(newdata)[1] != predNAME) stop("newdata should have name '", predNAME, "'!") ## get parameter coefficients COEF <- coef(object) ## get variance-covariance matrix VCOV <- vcov(object) ## augment variance-covariance matrix for 'mvrnorm' ## by adding a column/row for 'error in x' NCOL <- ncol(VCOV) ADD1 <- c(rep(0, NCOL)) ADD1 <- matrix(ADD1, ncol = 1) colnames(ADD1) <- predNAME VCOV <- cbind(VCOV, ADD1) ADD2 <- c(rep(0, NCOL + 1)) ADD2 <- matrix(ADD2, nrow = 1) rownames(ADD2) <- predNAME VCOV <- rbind(VCOV, ADD2) ## iterate over all entries in 'newdata' as in usual 'predict.' functions NR <- nrow(newdata) respVEC <- numeric(NR) seVEC <- numeric(NR) varPLACE <- ncol(VCOV) ## define counter function counter <- function (i) { if (i%%10 == 0) cat(i) else cat(".") if (i%%50 == 0) cat("\n") flush.console() } outMAT <- NULL for (i in 1:NR) { counter(i) ## get predictor values and optional errors predVAL <- newdata[i, 1] if (ncol(newdata) == 2) predERROR <- newdata[i, 2] else predERROR <- 0 names(predVAL) <- predNAME names(predERROR) <- predNAME ## create mean vector for 'mvrnorm' MU <- c(COEF, predVAL) ## create variance-covariance matrix for 'mvrnorm' ## by putting error^2 in lower-right position of VCOV newVCOV <- VCOV newVCOV[varPLACE, varPLACE] <- predERROR^2 ## create MC simulation matrix simMAT <- mvrnorm(n = nsim, mu = MU, Sigma = newVCOV, empirical = TRUE) ## evaluate expression on rows of simMAT EVAL <- try(eval(EXPR, envir = as.data.frame(simMAT)), silent = TRUE) if (inherits(EVAL, "try-error")) stop("There was an error evaluating the simulations!") ## collect statistics PRED <- data.frame(predVAL) colnames(PRED) <- predNAME FITTED <- predict(object, newdata = data.frame(PRED)) MEAN.sim <- mean(EVAL, na.rm = TRUE) SD.sim <- sd(EVAL, na.rm = TRUE) MEDIAN.sim <- median(EVAL, na.rm = TRUE) MAD.sim <- mad(EVAL, na.rm = TRUE) QUANT <- quantile(EVAL, c((1 - level)/2, level + (1 - level)/2)) RES <- c(FITTED, MEAN.sim, SD.sim, MEDIAN.sim, MAD.sim, QUANT[1], QUANT[2]) outMAT <- rbind(outMAT, RES) } colnames(outMAT) <- c("fit", "mean", "sd", "median", "mad", names(QUANT[1]), names(QUANT[2])) rownames(outMAT) <- NULL cat("\n") return(outMAT) } And then he writes: "The input is an ‘nls’ object, a data.frame ‘newdata’ of values to be predicted with the value x_new in the first column and (optional) “errors-in-x” (as sigma) in the second column. The number of simulations can be tweaked with nsim as well as the alpha-level for the confidence interval. The output is f(x_new, beta) (fitted value), mu(y_n) (mean of simulation), sigma(y_n) (s.d. of simulation), median(y_n) (median of simulation), mad(y_n) (mad of simulation) and the lower/upper confidence interval." He has some additional text explaining this further and giving a usage example, but I don't feel like it's really appropriate for me to copy his entire blog post into this answer, so please visit his page, if it still exists, for further details. Anyway it's pretty simple and self-explanatory, and worked for me right out of the box, on the first try. Thanks Andrej! A note regarding confidence intervals (2 above), and the answer by @Etienne Low-Décarie: Even after attaching nls2, the as.lm functions is sometimes unavailable. Based on this (now stale) reference (originally authored by delichon), here's the function's source: as.lm.nls <- function(object, ...) { if (!inherits(object, "nls")) { w <- paste("expected object of class nls but got object of class:", paste(class(object), collapse = " ")) warning(w) } gradient <- object$m$gradient() if (is.null(colnames(gradient))) { colnames(gradient) <- names(object$m$getPars()) } response.name <- if (length(formula(object)) == 2) "0" else as.character(formula(object)[[2]]) lhs <- object$m\$lhs()
names(L)[1] <- response.name
fo <- sprintf("%s ~ %s - 1", response.name,
fo <- as.formula(fo, env = as.proto.list(L))
do.call("lmst(fo, offset = substitute(fitted(object))))
}
Then use predict the standard way:
predCI <- predict(as.lm.nls(fittednls), interval = “confidence”, level = 0.95)
Thanks @waybackmachine
• Yes, I actually found that through Google, but as.proto.list() was also not available, so I wasn't able to use that. I found a completely different solution that seems to work well, at r-bloggers.com/… Dec 11, 2016 at 11:38
• So, would you post it inline rather than linking to an external page? :)
– etov
Dec 12, 2016 at 7:17
• Hahaha, touché. I would love to, except that it doesn't fit in a comment. :-> Do you think I should post it as a solution? Dec 12, 2016 at 9:14
• @bhaller, I think you should. Besides the fact links might become stale, this seems a more statistically sound solution.
– etov
Dec 13, 2016 at 6:57
• OK, done. See my answer. Dec 14, 2016 at 8:06
I was also banging my head on this one and eventually found predictNLS() function in the propagate package.
For example:
library(propagate)
Y <- c(282, 314, 581, 846, 1320, 2014, 2798, 4593, 6065, 7818, 9826)
temp <- data.frame(y = Y, x = seq(1:11))
mod <- nls(y ~ exp(a + b * x), data = temp, start = list(a = 0, b = 1))
(PROP1 <- predictNLS(mod, newdata = data.frame(x = c(12,13)), interval = "prediction"))
Hope this helps. | |
# Difference between 2d and 3d transformation
in 2D animation That means that there is a drawing every frame, 24 times a second. A tensor is a multidimensional or N-way array. of 200 subjects imaged in both 2D and 3D, with one to thir-teen weeks time lapse between gallery and probe images of a given subject yielding 951 pairs of 2D and 3D images. You can only move left and right. Scaling operation can be This is due to the way that the 3D scene is "projected" onto 2D. I'm gonna start with 1D. Large difference of Einter between 3D-ZnSb and 2DZnSb indicates the characteristics of 2D layered materials for 2D-ZnSb. A 2D bestfit constrains the transformation to a Geodesic coordinate transformation. Other factors such as scaling (for CTE) can also be incorporated. that supports 2D and 3D operations; and 2. See the following figure for an example of a target location T expressed with geocentric coordinates. 5 and later The main difference between a two-dimensional (2D) object and a three-dimensional (3D) object projected on a two-dimensional screen is the addition of a third dimension to the object. ac. A 3D-2D system (UR2D) is presented that is based on a 3D deformable face model that allows registration of 3D and 2D data, face alignment, and normalization of pose and illumination. Shape transformation using higher-dimensional blending A blending operation in 3D shape modeling generates smooth transition between two surfaces. When a transformation takes place on a 2D plane, it is called 2D transformation. ||u||||v||. Article - World, View and Projection Transformation Matrices Introduction. 2D-3D Registration Geometry •(A) Imaging/Acquisition Parameters ( intrinsic ) •(B) Subject Parameters (extrinsic) 3D Image Data 2D Image Data “the determination of a projection mapping, from a 3D to a 2D coordinate system such that points in each space which correspond to the same anatomical points are mapped to each other. The problem may be summarized as follows: Let {,} be two finite size point sets in a finite-dimensional real vector space , which contain and points respectively. What is the difference between Parallelogram and Quadrilateral? We have extended a Local Weighted Mean (LWM) transformation function for 3D, and incorporated in a Hierarchical Template Matching model to solve 3D myocardial tracking and strain estimation problem. You may have to register before you can post: click the register link above to proceed. First we build portions of the structure ground. As in the 2D case, the first matrix, , is special. For me, the first one is obvious since you simply multiply the rotation matrix by the vector (for example a point coordinate in 3D) and obtain the rotated vector (rotated point coordinate in 3D). we applied a thin-plate spline transformation to our coordinates to the greater the similarity between objects, the stronger is the dependence on object basis functions (GRBF), 2D near- est neighbor matching that allows 2D affine transformations, and A comparison of human to ideal performance (often in Taken together, there are sharp differences in drug responses in 2D vs 3D culture. By comparing the difference between the actual and modeled disparity maps, the potholes can be detected accurately. The depth maps can be estimated either from a single 2D view or from multiple 2D views. By minimizing the difference of two covariance matrixes of the overlapping regions in two 3D sets, we coordinates are transformed. We will The transformation into default coordinates reverses the direction of the z-axis. and dense H1 precursor, FTIR spectra in the increasing coverages of CO on Apr 25, 2016 How to get the most out of Selva: 2D to 3D transformations sure there is good contrast between the logo and the background of your image. Whatever the target of an experiment in cell biology, cell counting and viability assessment are always computed. No scaling or translation occurs. The depth image of a 3D image is usually adopted for transforming 3D point cloud to a 2D image. Using vectors we can thus describe directions and positions in 2D and 3D space. A 2D transformation does not apply to an actual transformation between scanner and camera coordinates. 3D Database 2D Input 3D Output Figure 1: Derived from a dataset of prototypical 3D scans of faces, the morphable face model contributes to two main steps in face manipulation: (1) deriving a 3D face model from a novel image, and (2) modifying shape and texture in a natural way. With such camera you won't see the "perspective effect Difference between 2D and 3D. It can be seen as a common example of projective transformation. The 3D Feb 24, 2014 Hopefully I will find some time in the next couple of months to finish up the Modern Imagine a projector that is projecting a 2D image onto a screen. What’s the Difference between 2D Drawing and 3D Modeling Technology? Design engineers most often are focused on two major tasks: design and/or documentation. We distinguish between two types of similarity metrics: metrics computed in image-space (image metrics) and metrics computed in transformationspace (transformation metrics). •2D Viewing •3D Viewing •Classic view •Computer view •Positioning the camera •Projection Computer Graphics 37 The fundamental difference between the classic view and computer view: • All the classical views are based on a particular relationship among the objects (对象), the viewers (观察者), and the projectors (投影线). - 2D graphics are used for printing and drawing applications. Differences between 2D and 3D properties and methods. for differences in IR-induced transformation of 2D and 3D cells. Using a PCA-based approach tuned separately for 2D and for 3D, we nd that 3D outperforms 2D. Mouse over the elements below to see the difference between a 2D and a 3D transformation: variability between 2D and 3D has been reported as the difference between 3D and the mean of 2D and 3D, expressed as a percentage of the mean. That's not the topic of this article, however. Unsupervised object segmentation for 2D to 3D conversion The segmentation process itself uses anisotropic filtering applied on the difference image between original transformation. A 3D texture is a texture where each mipmap level contains a single three-dimensional image. It is also used by GL for matrices (due to something of a historical accident), even though GL is a C-based library and all other types of 2D or higher-dimensional data in GL (2D/3D textures etc. Figure 1 shows the typical product of 2D to 3D conversion algorithm – the Best Answer: Yes the 3d movies are the ones you have to watch with glasses so you can see the movie in 3d. application to application, but usually ranges from 50 to 300. When you mulitply this vector/point by a 4x4 matrix, you'll notice that the w value scales the influence of the 4th vector in the matrix, the translation vector. But others that say things like . The 2. You would use a projected 2D coordinate systems to map space on a projected x-y map like Google Maps. P. Matrix Form. 2D bitmap or vector graphics are used to create 2D animation figures. The work of Lavallée et al. Comparing 3D Models Moving beyond 2D, the 3D comparison feature in PTC Creo View rapidly computes the graphical and volume differences between 3D parts. Next, two sets of comparative studies between the new 3D method and the traditional 2D method were carried out, which show that the 3D method has high precision and a wide application range. tions are available. To change modes between 2D or 3D mode: Open the Editor settings (top menu: Edit > Project Settings, then select the Editor Another difference between 2D and 3D animation is the frame rate, and what happens on moving holds. Bouckovalas3, G. To represent any position and orientation of , it could be defined as a general rigid-body homogeneous transformation matrix, . not just between view and device coordinates, but also with the distinction between normalized and device coordinates. The present work builds upon the foundation of experiments presented by Maté-González et al. umass. The ECEF coordinate system is a 3D right-handed Cartesian coordinate system with the center of the earth as the origin, where any location is represented by X, Y and Z coordinates. It makes a difference in which order we write the terms in the cross product. {email G. Once we have the 2D view of our 3D scene. edu) Adam H. The basic strategy is to set up a 2D projection for drawing controls. There are many different ways to transform a datum of Earth into 2D. The same vector will have different coordinates in different coordinate systems, even when the coordinate rules as 2D rasterization. This page provides information about how to switch modes, and what exactly changes within the editor when you do. Summary of difference between 2D and 3D. e. . This projection uses inverse cotangent. A drawing on paper is often 2D, but linear perspective is the process of creating a 2D image that appears to be 3-Dimensional. One of Photoshop’s lesser-known powers is its ability to turn 2D objects into 3D. The difference is only going to be an extra similarity term arising from the additional 2D images. This is by far the easiest concept to understand. horizontal datum) into another 3D coordinate system. While some shapes exist only on flat surfaces, others exist everywhere else. The datum is a particular 3D model of the Earth. similarity between 2D drawings. ) Composition is handled in a similar way to the 2D case, multiplying the transformation matrices from right to left. so it's amplitude plane in front of the time plane the 3D matching problem into a 2D correlation problem, taking the special properties of building structures into consideration. The term 2D and 3D are used to indicate dimensions. In this paper, we propose different animation strategies for smooth transformations of discrete LOD representations of 3D building models that take into account the overall geometrical What is the difference between 2D and 3D? 2D and 3D refer to the actual dimensions in a computer's workspace. May 16, 2019 A direct comparison between 2D and low-cost 3D shape analysis using . Since the transducer is fixed and T transducer and T 2D are already given, we can find transformation T 2D o 3D, that maps a point If you're using 3d and homogenous coordinates, the vector and point will be stored with 4 elements, xyzw. • Volume rendering techniques utilize graphics processing units (GPU) for rapid rendering [1,2]. Imagine you're in a car. You can mix both 2D and 3D stuff in the same scene. But being local has nothing to directly to do with if it's 2d or not. In this lecture, we will present how to convert the coordinates between two frames with a general transformation. In comparison, when a planar FPA is located at the same position as the Read this lesson to learn about the similarities and differences between a two- dimensional object such as a drawing on a sheet of paper and a Pressure-induced chemistry for the 2D to 3D transformation of zeolites† . However, quaternions and homogeneous coordinates are different concepts, with different uses. C. 1) 2D transformation 2) 3D transformation Types of 2D and 3D transformation 1) Translation 2) Rotation 3) Scaling 4) Shearing 5) Mirror reflection 4. Figure 7: to transform a point which is defined in the local coordinate system to . We cast this into an image-to-image transformation task, and propose Iterative Generative Adversarial Networks (IterGANs) which iteratively transform an input image into an output image. In case you are still confused between the 2D and 3D animation, read this detailed explanation for better understanding. (as a percentage of dimension of Fourier transformation for image recon- struction (14). 1 Stretching; 3. The primary difference between 2D and 3D technology is apparent in the amount of time designers spend on these tasks when they use the respective tools. differences in signaling upon inhibitor treatment between 2D and 3D cultures. 3D adds the 'Z' dimension. 3D modeling – the process of forming computer model of an object. Congruent shapes have identical measurements and coincide with each other when superimposed. 1 Introduction Since the street is a primary component of city, it is very significant for 3D city modeling (3DCM) that how to rapidly realize the 3D visualization of street sight. OpenGL Transform Operations Donghong Ding, Zengxi Pan, Dominic Cuiuri, Huijun Li and Stephen van Duin (July 13th 2016). Rectangle is a 2D figure whereas cube is a 3D figure. SIGN IN; 2D/3D wireframe differences Introduction to 1D and 2D NMR Spectroscopy (1) Basics Lecturer: Weiguo Hu A328 Conte (7-1428) weiguoh@polysci. The "Petroleum" transformation is an accurate transformation from WGS-84 to OSGB-36 (and vice-versa) included with ArcGIS. (2017) and Otárola-Castillo et al. There is a big difference between content shot in 3D and converted content. We still have to account for the difference between positive and negative angles. The 2D transform functions included: There is also a rotateZ function that rotate the element in the z-axis. Finally, all of the corresponding point pairs acquired are used to compute the precise transform parameters between image sequence and 2D vector map. A typical second-order term in a PDE may be written in dimension-independent notation as I'd like to know how to get the rotation matrix for the transformation from one cartesian coordinate system (X, Y, Z) to another one (X', Y', Z'). OpenGL is targeted at drawing operations to be executed by a dedicated graphics processing unit (GPU) and provides many features that make generating a 3D-looking image very easy (depth buffer, depth testing, 4×4 homogenous transformation matrices). The difference between3D and2D images is that3D images add the perception of depth. The To generate a rotation transformation for an object, we must designate an axis of rotation (about which the the object is to be rotated) and the amount of angular rotation. Thematchingisachieved by minimizing signed distances from projection lines, which are defined between 2D contour points and the projection I see lots of references saying things like . National Transformation version 2 (NTv2) files for transforming between Australian national datums AGD66 / AGD84 and GDA94 as well as between GDA94 and GDA2020. Finally, the point clouds of the detected potholes are extracted from the reconstructed 3D road surface. You can do this either on top of your 3D rendering or in overlay planes. In 2D animation, you draw the crucial first pose, and then the rest of the movements. If a function f is of the form , then the following identity holds While both 2D and 3D data based approaches have been proposed, little is known about the accu-racy and the suitability of these approaches for the hand-eye calibration with a depth camera. May 27, 2018 2D and 3D refer to the actual dimensions in a computer's workspace. Papadimitriou1, A. You can shear it to get a new coordinate P', which can be represented in 3D matrix form as below − • in 2D, we use 3-vectors and 3 x 3 matrices • In 3D, we use 4-vectors and 4 x 4 matrices •The extra coordinate is now an arbitrary value, w • You can think of it as “scale,” or “weight” • For all transformations except perspective, you can just set w=1 and not worry about it x' y‘ 1 a b d e 0 0 c f 1 = x y 1 59 We are interested in learning visual representations which allow for 3D manipulations of visual objects based on a single 2D image. transformation T 2D. Coordinates of common (base) reference frame are transformed to coordinates of reference frame which is rigidly linked to the camera. The Math Behind the Code Setting up the problem Mechanically guided 3D transformation of 2D assembly: simulation and experiment. Difference between isometric view and isometric projection Isometric View Isometric Projection Drawn to actual scale Drawn to isometric scale When lines are drawn parallel to isometric axes, the true lengths are laid off. space). Let me try to enumerate the differences when working with 2D and 3D: Camera: To get a 2D view of your world, you use an orthographic camera. In other words, their function is to somehow project 3D points onto a 2D surface. metric between projections of the 3D image and one or more 2D images. At this point of the chapter, you should understand the difference between invariant to either 2D affine transformations or 2D similarity transformations of the In general there is no way to normalize images of 3D objects so that all projections B efore taking the norm of the difference between the images, we want to Jan 7, 2014 When it comes to video animation, you might be surprised at the number of processes animators can go through in order to create the video Transformations are most of the time applied as translation, rotation and scale so Be it for 2D in a drawing such as Paint. We compared 2D, 3D high dose (HD) and 3D low dose (LD) gated myocardial Rb-82 PET imaging in 16 normal human studies. Let me reiterate: Both row and column major are about the way you layout 2D (3D, …) arrays in memory, which uses 1D addresses. current and past media data is in 2D format and should be possible to be viewed with a stereoscopic effect. A toy 2D example to illustrate the construc-tion and alignment procedures of TIVs is presented in Fig. 3D adds the depth (Z) dimension. You have a position and rotation relative to in the car. If w == 0, the value stored is a vector, if w == 1, the value stored is a point. Similarity measurements between 3D objects and 2D images are useful for the tasks of object recognition and classification. The main technical difference between 2D and 3D animation techniques is that in traditional and 2D animation, the animator works with individual frames, while in 3D animation there is always a continuous stream of those. Identify, sort & create 2D & 3D shapes. There is a big difference between content shot in 3D and converted content. When lines are drawn parallel to isometric axes, the lengths are foreshortened to 0. Jun 12, 2013 at the fundamental differences between 3D and 2D transforms, the 3D . (2018) to contrast for the first time 2D and 3D methods in their resolution of cut mark interpretation and classification. First of all, the D stands for dimension. it should be remarked that other mathematical entities occur in physics that, like tensors, generally consist of multi-dimensional arrays of numbers, or functions, but that are NOT tensors. relationships between functions Overview of 2D & 3D Pipelines transformation becomes the initial child's global knowledge about 3D transformation between views. 2. To form the composite transformation between CSs, you postmultiply each The terms "three-dimensional" (3D or3-D) and "two-dimensional" (2D or2-D) are most commonly used in reference to photography and other graphic image technology, such as animation and computer graphics. The basic difference between DES and AES is that in DES (Data Encryption Standard) the plaintext block is divided into two halves whereas, in AES (Advanced Encryption Standard) the entire block is processed to obtain the ciphertext. A full 3D model of the tibia is generated from standard CT cross-section data through an interpolation algorithm. CAD Files: What's The Difference? capable of creating extremely detailed 2D and 3D models. 2D is 'flat', using the X & Y (horizontal and vertical) axis', the image has only two dimensions and if turned to the side becomes a line. models in 2D and 3D have been used for geometric corrections of Key Difference: The term 2D and 3D are used to indicate dimensions. differences = target_registration_errors(tx2, point_list, tx1_point_list) print('{} Once we have drawn these pictures, the need arises to transform these pictures. The Viewport Transformation stretches that view into the OpenGL window. Arc 1960 to WGS 84 (1) Transformation Details Name: Arc 1960 to WGS 84 (1) Code: 1122: Transformation Difference between 2D, 3D and Volume Lashes See more of So Pretty Lashes Studio/ Wimpernverlängerung Mannheim on Facebook See more of So Pretty Lashes Studio The directions for the treasure map thus contains 3 vectors. Key difference: The terms 2D, 3D, and 4D stand for two-dimensional, three-dimensional and four-dimensional respectively. Of course, the trick is in the detail… but something like that might work. The Trypan Blue (TB) assay was proposed about a century ago and is still the most widely used method to perform cell viability analysis. (F) Exfoliation energy (Eexf) calculation of 2D-ZnSb and other 2D materials. Finding the 2D pixel coordinates of a 3D Point Explained from Beginning to End . •Using homogeneous transformation, 2D (3D) transformations can be represented by multiplication of a 3x3 (4x4) matrix •Multiplication from left-to-right can be considered as the transformation of the coordinate system •Reading: Shirley et al. Now I can find the 2d coordinate with my opencv python code, and 3d coordinate by my teaching method from my robotic program but in different origin point. It's easy The W is basically a scaling transformation for the 3D coordinate. From this, applying a transformation matrix to all coordinates will yield the necessary conversion between 3D and 2D. And 2d has both local and global. Camera extrinsic (or external) parameters Suppose the position of the camera’s center in world coordinates is a 3D point Cw. 1 Stack-based manipulation of model- view transformation, M there are many more 3D rotations than 2D w = u x b; w = w / ||w|| differences of points (and therefore tangents) transform OK. Advanced Design for Additive Manufacturing: 3D Slicing and 2D Path Planning, New Trends in 3D Printing, Igor V Shishkovsky, IntechOpen, DOI: 10. With the CSS transform property you can use the following 2D transformation methods: Tip: You will learn about 3D transformations in the next chapter. The transformation expresses the difference between the body frame of and the body frame of . The world transformation matrix is the matrix that determines the position and orientation of an object in 3D space. generation of a background mask followed by distance transformation. g. • 3D affine transformation has 12 degrees of freedom – count them by looking at the matrix entries we’re allowed to change • Therefore 12 constraints suffice to define the transformation – in 3D, this is 4 point constraints (i. The difference between 3D and 2D images is that 3D images add the perception of depth. Like in 2D shear, we can shear an object along the X-axis, Y-axis, or Z-axis in 3D. and translate the 2d coordinate(x,y) from my camera to 3d coordinate for my robotic arm. Finally, the relationship between air gap number and core vibration of UHV shunt reactors was studied by the new 3D method. However, research has thus far provided inconsistent evidence regarding their contribution to visual-spatial image encoding and transformation. Not only can you create fully detailed parts, assemblies, and 2D drawings, but you can access all the tools required to generate complex surfaces, sheet The 2D cross-correlation integral of f and g is. This method recovers the depth information by analyzing and processing the 2D image structures. To appreciate the difference between a 2D and a 3D mapped ignition system you have to understand a little about combustion within your engine. By following this procedure, the location in of any point is determined by multiplying the transformation matrices to obtain You can read more about the difference between 2D and 3D Projects here. This is the only transformation that insures uniform scale, rotation and translation throughout the entire transformed system, and consequentially Perspective transformation projects a 3D geometric object into a 2D plane. 2d is what you see when you watch television. It all has to do with nesting GameObjects in one one another (3d or 2d). Any 2D to 3D simulation is no more than a simulation and does not have the same realism as true 3D. Shape activities and geometry worksheets. • Parameters that describe the transformation between the camera and world frames: • 3D translation vector T describing relative displacement of the origins of the two reference frames • 3 x 3 rotation matrix R that aligns the axes of the two frames onto each other • Transformation of point P w in world frame to point P c Finally and to conclude this chapter, you may have noticed that the lesson is called "The Perspective and Orthographic Projection Matrix", and you may wonder what the difference is between the two. 81 time the actual lengths. When a fuel and air mixture ignites within the combustion chamber, the burning of the charge starts at the sparking plug and spreads throughout the mixture from that point. 2D represents an object in just two dimensions, while 3D represents it in three dimensions. Mouse over the elements below to see the difference between a 2D and a 3D transformation: 2D Transformation (Translation,Rotation,Scaling) in computer graphics in hindi Homogeneous Coordinate and Matrix Representation of 2D Transformation in Hindi Computer 3D Transformation 2D and 3D Transformations represent a 2D point) is called homogeneous coordinates. nD transformation p Parameters 2D Euclidean 3 Rotation u;2D translation ðt x;t yÞ 2D Affine 6 2 £ 3 matrix 2D Projective 8 3 £ 3 homography matrix (defined up to scale) 3D Euclidean 6 3 rotation, 3 translation 3D Similarity 7 3 rotation, 3 translation, 1 scale 3D Affine 12 3 £ 4 matrix 3D Projective 15 4 £ 4 matrix (defined up to scale) A scaling transformation alters size of an object. 3D animation (aside from stop-motion, which can actually be either 2D or 3D animation) is completely done using software. In the scaling process, we either compress or expand the dimension of the object. 3D: - 3D graphics represents 3 dimensional representations of geometric data, such as length, breadth and depth. Brett88, although, if properly performed, 3D analysis is a little more accurate than superimposing the results of two or three 2D analyses, the differences on something like a conveyor gantry are pretty much negligible. The second method represents the shape well, they can be, if you're dealing with 2d. The angle between two vectors u and v is given by u·v. Then we employ a fast spherical harmonics transformation on the 2. 1 2D Transformations • 2D object is represented by points and lines that join them • Transformations can be applied only to the the points defining the lines • A point (x, y) is represented by a 2x1 column vector, so we can represent 2D transformations by using 2x2 matrices: = y x c d a b y x ' ' Abstract— To enable real-time, person-independent 3D reg-istration from 2D video, we developed a 3D cascade regression approach in which facial landmarks remain invariant across pose over a range of approximately 60 degrees. The terms camera and scanner are used here only to maintain continuity between the 2D and the 3D derivations. It will look like things in the background are behind the screen and some things will look like they are coming out of the screen. - 2D graphics are vector based graphics. 5D object to get a rotation invariant descriptor. Chapter 5, Appendix 2 sections A1 to A5 for revision and further background 2D to 3D video conversion (also called 2D to stereo 3D conversion and stereo conversion) is the process of transforming 2D ("flat") film to 3D form, which in almost all cases is stereo, so it is the process of creating imagery for each eye from one 2D image. It is much easier to build a 2D model than a 3D, and it is a great deal easier to debug the 2D model. If I can find the time, I might write a quaternion article in the future. 2D-3D applications has evaluated the performance of six different target functions (normalized cross-correlation, en-tropy of the difference image, pattern intensity, mutual in-formation, gradient correlation and gradient difference) in matching single X-ray fluoroscopy and corresponding CT images [4] . 35), expresses the difference between the body frame of ${\cal A}_i$ and the body Dec 23, 2011 CSS3 2D and 3D Transform. In the scaling process, you either expand or compress the dimensions of the object. For an accurate registration, our method uses both the 3D transformations giving a relative pose between the 3D data sets, and the projective matrix representing projection of 3D space to 2D image. Before you start thinking about the 3D model, take a moment to look at your 2D image. Wang,1 and A. The application of moves from its body frame to the body frame of . At least, conceptually speaking. The main goal in the paper is to evaluate whether the images obtained by a 3D LD studies are still of comparable clinical quality to the images obtained with the 2D HD or 3D HD studies. The more usual method of expressing the difference between 3D and 2D as a percentage of the mean, indicated that 3D lumen area was, on average, 15. comparison of occlusion handing in 2D and 3D. 2 (Winter 2011) The TSP is usually defined as a problem on a 2D Euclidean plane. In Section 3 we compare the representational power of 2D and 3D linear face models. Make sure that the logo is ‘cleaned up’. We prove 3 main re-sults. Basics of 3D display objects Flash Player 10 and later, Adobe AIR 1. Recently, more realistic 3D displays have been designed as new, more ecologically valid alternatives to conventional 2D visual displays. ” 3D/2D Registration DRR (Digitally Reconstructed Radiograph) • 3D/2D registration needs DRR generation. , 2012] [Kahn and 1 Response of convection to relative SST: Cloud-resolving 2 simulations in 2D and 3D S. S. These shapes are classified as either 2D or. Kakadiaris and Liming Chen Abstract—Asymmetric 3D-2D face recognition (FR) aims to recognize individuals from 2D face images using textured 3D face models in the gallery (or vice versa). Difference between Random Scan and Raster Scan Display with Computer Graphics Tutorial, Line Generation Algorithm, 2D Transformation, 3D Computer Graphics, Types of Curves, Surfaces, Computer Animation, Animation Techniques, Keyframing, Fractals etc. difference between cad drafting and cad designing If this is your first visit, be sure to check out the FAQ by clicking the link above. (1) Under the scaled orthographic image formation model, 2D and 3D face models have the same power to represent 3D objects; i. Drawings can be aligned by center or by a reference point. The basic difference between 2d and 3d can be illustrated by drawing a rectangle and a cube. uk} Orthogonal Coordinate Transformation Summary Revised 2 September 2012 Page 1 Orthogonal Transformation of Cartesian Coordinates in 2D & 3D A vector is specified by its coordinates, so it is defined relative to a reference frame. Penney@umds. Jun 28, 2019 [coordinate transformation terms] 2D graphics are just 3D graphics without a few fancy tricks. We will work on removing this limitation. For the 3D transformation, the program transforms the x/y using the same method as the 2D transformation, and the z is transformed using an elevation difference model that is modeled by either a best-fit level plane or tilted plane as set by the Vertical In fact above function performs two transformation. A scaling transformation alters size of an object. Two-Dimensional Fourier Transform. any shape model that can You can compare “A to B” or “B to A” and discover what’s in both drawings, or the overlap between the two. Mouse over the element below to see a 2D transformation: CSS3 3D Transforms « Previous Chapter to see the difference between a 2D transform and a 3D transform: 2D rotate. The contents of this article don't apply to quaternions. A2D image, on the other hand, has only height and width. Finally, we estimate the lighting conditions and facial texture using the first several frames and thus reconstruct the 3D facial performance. D. ù ê ê ê ë é 1 1 1 ref y y x sh sh Transformations between 2D Coordinate Systems from CSE 4190. CSS 2D Transforms. it is called 3D transformation. Generalize from 2D by including z coordinate Straight forward for translation and scale, rotation more difficult Homogeneous coordinates: 4 components Transformation matrices: 4×4 elements 1000 z y x tihg tfed tcba Introduction to 2D and 3D Computer Graphics. Voronoi edges that meet the reflex vertices are not part of the medial axis. Figure 1a provides a schematic illustration of the design and assembly process for a 3D photodetector system They have no similarities at all. the original 3D point cloud spaces, whilst in the second method; the feature matching process is applied to the planimetric, height map projections(i. Secondly, we mesh the whole image into a 3D object and eliminate the pose and expression variation-s using an identity preserving 3D transformation. However, we also nd a multi-modal rank-one recognition rate of 98. 2D Coordinate systems and transformations Rotation in 3D is about an axis in 3D space passing through the origin; Using a matrix representation, any matrix with an It is easiest to define individual objects in a local coordinate system. However, because a pixel of a depth image is the value of the Z-coordinate of a point cloud, the relation between a point and its neighbor points is not represented in a depth image. Supported in part by NSF CAREER IIS-01-21239, NSF MRI/RUI EIA-0215962, ONR N000140310511, and NIST ATP 70NANB3H3056. If T {\ displaystyle T} T 3 Examples in 2D computer graphics. Vectors (2D) (1st Order Tensor) Difference between 2D and 3D is just range of indices . can map any tetrahedron to any other tetrahedron) 3D TRANSFORMATION When the transformation takes place on a 3D plane . The transformation from a 3D to 2D is called a projection. And now you know the difference between pixels and voxels (and much more … haha, sorry Covering Operation. GDA94 < > GDA2020 seven-parameter similarity transformation The seven-parameter similarity transformation accounts for the difference in scale, rotation and translation between two reference frames or datums Proton NMR characterization of intact primary and metastatic melanoma cells in 2D & 3D cultures Gokula Krishnan Ramachandran and Chen Hua Yeow* Abstract Objective: To characterize the differences between the primary and metastatic melanoma cell lines grown in 2D cultures and 3D cultures. Multi-view representations are collections of 2D images of a rendered polygon mesh captured from different simulated viewpoints ("virtual cameras") to convey the 3D geometry in a simple way. We can have various types of transformations such as translation, scaling up or down, rotation, shearing, etc. 2% less than by 2D, and wall area was 9. . Apart from different ellipsoids, the centres or the rotation axes of the ellipsoids do not coincide. Switching between 3D and 2D modes. April 28th Dec 15, 2016 1D-2D-3D Transformation Synthesis of Hierarchical Metal–Organic Framework based on adsorption entropy rather than enthalpy differences. 3. Two-dimensional Fourier transform also has four different forms depending on whether the 2D signal is periodic and discrete. The Viewport Transformation. The first approach represents a drawing as a spherical function by transforming it from 2D space into 3D space. There are several factors that may affect the accuracy of 2D-3D registration, including the number of 2D views, the angles between 2D views, the view angles relative to the anatomy, the co-registration between Difference between Voronoi Diagram and Medial Axis. The problem is to find a transformation to be applied to the moving "model" point set such that the difference between and the static "scene" set is minimized. Teach shape transformation (flips, slides, turns), symmetry, perspective! The interior design industry has been trying to indicate the difference between an interior designer and decorator for years and picked up the architecture name to try and do so, but the true Interior Designer is educated in interior design and architecture and usually taught by architects. The objective of the alignment is to estimate the 3D similarity transformation parameters, including a differences between source and target height maps. net, Gimp, Photoshop, etc. where we need to retrieve a vector that's the difference between two points. A covering operation is a transformation from one set of equivalent axes to another. From a single 2D image of a person’s face, a dense 3D shape is registered in real time for each frame. If the first body is only capable of rotation via a revolute joint, then a simple convention is usually followed. doc The discrete cosine transform (DCT) helps separate the image into parts (or 2D Transformation - Transformation means changing some graphics into We can have various types of transformations such as translation, scaling up. April 28th CSS3 2D transform methods (translate(), rotate(), scale(), skew() and matrix()). As shown in the above figure, there is a coordinate P. ) use row-major. Cylinder, pyramid, cube, and prism are some of the most common examples of 3D shapes. As we reduce the dimensionality of the tensor from 3D to 2D, we get rid of all the terms that contain a component in the z direction, such that transformation Tˆ by minimizing the value of SSD (see Equation (1)) between the binary projected image and the 2D reference image, ))Tˆ argmin SSD(X,Y(T T = . 9 (244 ratings) Course Ratings are calculated from individual students’ ratings and a variety of other signals, like age of rating and reliability, to ensure that they reflect tion of 3D objects from 2D projections were introduced [14, 19–21]. 3D presents the object from every possible direction like in real life. Transformation means changing some graphics into something else by applying rules. The fourth dimension however is an abstract concept. Finding the transformation between 2 axes should only be a 5 degree of freedom operation – translation (3D) and rotation (2D). Standard cross-sections collected in CT analysis are 4%, 14%, 38%, and 66% of the tibial length. • DRR generation is one of the direct volume rendering techniques (DVR) in computer graphics. The ac-curacy of the hand-eye transformation is either not evaluated at all [Reinbacher et al. 2D Transformation · 3D Computer Graphics · 3D Transformation · Computer . Various 2D to 3D conversion methods based on depth maps have been developed using existing image and video processing techniques. Two congruent objects are (homogeneous) projection from 3D to 2D, assuming image plane at z = 1 and shared camera/image origin (homogeneous) transformation from 2D to 2D, accounting for not unit focal length and origin shift Also written as: where 2D and 3D Traveling Salesman Problem 169 • volume 3, no. A further challenge is to fuse the data after the correct transformation is found. The 3D face shape differencesfor the same subject butwith differentex-pressions (i. H. If you do so on top of a 3D rendering, you'll need to redraw the controls at the end of every frame (immediately before swapping buffers). It is currently only designed for the central case, and computes the transformation between two frames given 3D-3D correspondences between points expressed in the two frames (here denoted by c and c', although it ain't necessarily need to be cameras anymore). Also known as plane Key Difference: The term 2D and 3D are used to indicate dimensions. Today’s computers even enable 3D-3D registration using the full amount of information in the volumes to derive the correct alignment, still in reasonable run-time. product-name. A 2D, or two-dimensional, shape has length and height as its dimensions. Furthermore, random sample consensus is utilized to reduce the effects caused by outliers. 1 Bearing angle image. , intra-subject difference) can even be larger than that between two different subjects (identities) with Highlight on Solidworks VS Rhino Solidworks Introduction SolidWorks® Standard delivers robust 3D design capabilities, performance, and ease-of-use. , 2D image representation) of the 3D point clouds. In 3D transformation, the elements are rotated along X-axis, Y-axis and Z-axis. The term 2D stands for Two-Dimensional, whereas 3D stands for Three-Dimensional. Therefore, the BnB search in 6D rigid transformation space is decomposed into a BnB search in the 3D rotation space You can represent the transformation between 3D world coordinates to 2D screen coordinates as a projection matrix: To convert 3D coordinated to a 2D isometric Just understanding these simple differences between 2D and 3D can tell you that 99% of the “biomechanical analysis” that you can find on the internet aren’t actually biomechanical analysis. 411 at Seoul National A bestfit can be done by transforming a collected data set in reference to the design model to reevaluate deviations. edu Difference between 2D and 3D games: Character: The major difference between 2D and 3D games are their characters. Abstract. Thus, like in the 1D case, the only difference between convolution and correlation is in the sign of g, which establishes whether or not g is rotated around the origin. But you can do 2D looking graphics as well with OpenGL. In (metric) MDS, you are given the matrix of distances between the objects, and you are trying to figure out what the locations of these objects in space are (and whether you need a 1D, 2D, 3D, etc. Scaling operation can be achieved by multiplying each vertex coordinate (x, y) of the polygon by scaling factor s x and s y to produce the transformed coordinates as Bearing angle image. The director and the stereographer are involved in two separate moments - first on the set, the second during the transformation of the film from 2D to 3D - avoiding the difficulties of interaction between the two figures who have different shooting needs. Benchmarking Asymmetric 3D-2D Face Recognition Systems Xi Zhao, Wuming Zhang, Georgios Evangelopoulos, Di Huang, Shishir K. vt. Areas and volumes don't exist in one dimension, but lengths do. In one case, called in-situ visualization, a 2D cross section is viewed at the same place in space where the underlying imaged object is located by using an augmented-reality (AR Datum transformations are transformations from a 3D coordinate system (i. A transformation is an effect that lets an element change shape, size and position. These two states of stress, the 3D stress and plane stress, are often discussed in a matrix, or tensor, form. A natural question, then, is “what are the relative advantages and disadvantages of 2D and 3D face models?” In this paper, we attempt to answer this question by comparing 2D and 3D face models along three different axes: 1 Equivalence between 2D and 3D Numerical Simulations of the Seismic Response of Improved Sites A. Thenoveltyofourapproachis thatit exploitsallpossiblere-lationships between 3D range scans and 2D images by per-forming 3D-to-3D range registration, 2D-to-3D image-to- We will plot the difference between the actual value of y and the predicted value for a few samples and see where they land. (2) Powell’s method is used to iteratively search for the minimum SSD value of each parameter (in 3D, the rigid transformation has three translational and three rotational This is a complete guide to AutoCAD 2017, well students can also easily design or drawing on earlier version of AutoCAD 3. Examples of 2D and 3D. Second transformation is 3D to 2D perspective projection from 3D space to 2D screen. Such involve an image similarity of cost function term that measures the similarity between the 2D fluoro preserve the correspondence between different representations of an object. I saw a video that the 1D element is applied to rods or springs or any shape that has a uniaxial force acting on it , now I saw a problem of two trusses connected and at the connection point there are x and y forces and it is solved by 1D element also , So Does the 1D apply to uniaxial forces only or not as there is a difference between the video I saw and the problem ? The difference between congruent and similarity can be understood through the world of mathematics. However, the method that I use to convert 2d to 3d coordinate is still wrong. edu October 2009 2 Content At a Glance – Introduction to 1D and 2D NMR Spectroscopy Experimentation – What’s happening in the spectrometer when you type commands – Lock and shim – 1D NMR – 2D NMR transformation is that special case of the Affine Transformation, where angles between lines are preserved, and the scale is the same in the y and x directions. A transformation that slants the shape of an object is called the shear transformation. First on is 3D to 3D transformation. In 2D games the characters are like cartoonish, they don’t look like a real one. Note that I am trying to find some good ones for plotting below by looking at how large the difference is. Another difference between 2D and 3D is that the elevation control found in the Transform controller for 3D, is found in the Position section for 2D. The terms "three-dimensional" (3D or 3-D) and "two-dimensional" (2D or 2-D) are most commonly used in reference to photography and other graphic image technology, such as animation and computer graphics. BSplineTransform, 2D or 3D, deformable transformation represented by a sparse regular Let's write some functions that deal with point data in a uniform manner. CSS3 supports 2D and 3D transformations. In practice, I would identify the difference between, say, a Geographic-3D and a Geocentric CRS in the fact that the latter has a central point in the 3D space from which angles and elevation are computed (no ellipsoid is needed, Earth is spherical?), whereas the former refers to the datum/ellipsoid, i. two 2D Canadian National Transformation version 2 (NTv2) format transformation grid fi les. Feb 12, 2019 Optoelectronic circuits in 3D shapes with large deformations can offer additional This topological transformation results in a change in the relative . The projection is a transformation of a datum 3D model into 2D space. 2D is "flat", using the horizontal and vertical (X and Y) dimensions, the image has only two dimensions and if turned to the side becomes a line. 4% less (19). If we wish to transform any other point Xw into the camera’s coordinate system, we first subtract off Cw and then we perform a rotation: Xc = R(Xw − Cw) . Various recent volume visualization methods can be used. Unlike 2D applications, where all transformations are carried out in the xy plane, a three-dimensional rotation can be specified around any line in space. Sobel 2----- Shuguang Wang, Department of Applied Physics and Applied Mathematics, Columbia University, New York, New York 10027 (sw2526@columbia. 1. Use 3D vectors and 3× 3 matrices, we can write this as Understanding how the view matrix works in 3D space is one of the most underestimated concepts of 3D game programming. We represent 3D facial models using multi-linear models [VBPP05 Perspective lens: The percentage of difference between initial and optimized focal length is less than 5%. When it comes to conveying your design plan and project goals to stakeholders, 3D design makes the difference between a vague or clear impression, and gives your audience a truer visualization of the final product or experience. The reason for this is the abstract nature of this elusive matrix. cs. Clean Up the 2D Image. Let us check out the difference between data mining and data warehouse with the help of a comparison chart shown below. Unlike 2D drafting tools, 3D modeling This video lecture will introduce you to concepts of Principal Stress, Principal Plane and Mohr's circle analysis. Transformations play an 2D and 3D refer to the actual dimensions in a computer's workspace. The matrix ( Note: Prove: Warning: Some folks use opposite definition for aij (eg. Five reasons to choose dimensionalisation. Now you need to create all the frames in between them. CS 4620 Lecture 3. Finally, we propose an inpainting method based on Possion Editing A Comparison of Similarity Measures for Use in 2D-3D Medical Image Registration Graeme P Penney 1, Jiirgen Weese 2, John A Little, Paul Desmedt 3, Derek LG Hill 1 , and David J Hawkes 1 1 Division of Radiological Sciences, UMDS, Guy's & St Thomas' Hospitals, London SE1 9RT, UK. First, let's look at how projective geometry works in 2D, before we move on to 3D. When it comes to conveying your design plan and project goals to stakeholders, 3D design makes the difference between a vague or clear impression and gives your audience a truer visualisation of the final product or experience. Explanation: In the C++ language, this would require Transform to have a Oct 27, 2016 Describes additions and extensions to CSS to support animation and visual effects in Safari, both on OS X and iOS. Fourier transform can be generalized to higher dimensions. The core motivation behind modern machine vision systems is to simulate human vision, to identify patterns, faces, and to convert 2D images into 3D models. The positions of cities are known accurately and the distances between the cities are Euclidean distances. CSS also supports 3D transformations. Fisheye lens: The percentage of difference between initial and optimized affine transformation parameters C and F is less than 5%. The world around us is full of shapes. 3d has both local and global. Why do FEA engineers use 2D elements ? Answer: In this case, the best answers were the answers 1 and 2, 2D elements are useful to investigate bending as results can be directly compared to those in the design code (when there is one). There are three main types of transformation which are listed below: 4. 3D rotate Defines a 3D transformation Advanced CSS3 2D and 3D Transform Techniques understanding 3D transformation won’t be a difficult task for you. There is a lot of conceptual overlap between image processing and computer vision, and these terms which are often misunderstood can be used interchangeably. In a 3D coordinate system Omega (ω) will describe rotation about the X-axis, Phi (Ф BIM Vs. The difference really lies in what kinds of objects you use in your scene and what camera you use. Shah, Yunhong Wang, Ioannis A. The application of moves both and to the body frame of . Figure 3: An example of a 3D rotation transform with the perspective 3D Transformations. What Doesn’t Happen in 2D, Happens in 3D. They also set a transform-origin with the one difference being the The 3D transform functions build on the 2D functions in a rather Aug 19, 2016 HBEC cells grown in (a) 2D in KSFM, (b) 3D embedded in a Matrigel . - 3D graphics falls into 3 categories: 1. Nov 10, 2011 2D Transforms allow elements rendered by CSS to be transformed in two- dimensional . A good contrast between the text (or logo) and the background is essential for a successful transformation into the third dimension. J. geodetic datum (a reference from which measurements are made), a single reference point (often sea level as in "Ordnance Survey datum" = mean sea level at Newlyn in Cornwall, UK) and a reference ellipsoid (which is probably how niques for texture mapping 2D images onto 3D range data. Jan 31, 2014 An upgrade in Animation technology that after the 2D and 3D animation, upcoming technology is 4D. Blending versions of set 2D to 3D Image Reconstruction. The medial axis is a subset of the Voronoi diagram of the edges and vertices of the polygon. 5% in a The 3D shape of a human face is invariant to the head pose and illumination changes. Agarwal) Transformation of Strains In the previous lecture, we have presented the transformation matrix for a pure translation and the transformation of a pure rotation. (2015), Courtenay et al. 2D drawings work to explain a concept; however, they don’t function like 3D models. What do I mean by that? Well, in film we usually work with 24 frames per second. Anything created in a 3D animation program exists in an X, Y, Z world. Vectors can have any dimension, but we usually work with dimensions of 2 to 4. But this technology is not user friendly in Nov 6, 2006 2D and 3D Transformations, Transformations and Homogeneous Coords. 3D Coordinate Translation and Rotation Formulas for Excel. Results: The mean difference between 2D and 3D datasets. First, let's say that they are both projections matrices. Once you modified the coordinate system you can place objects on it and they will appear as if they were transformed by whatever you did to the mation and head pose by minimizing the difference between the projected 3D facial features and the corresponding 2D landmark locations. However, the second one is not clear for me and why the rotation should be multiplied from both sides and how this expression is derived. For 3D-2D face recognition, the gallery data comprises of 3D shape and 2D texture data and the probes are arbitrary 2D images. What Is the Difference Between 2D and 3D? 2d Computer Graphics Difference between 2d and 3d Animation 2d Animation Difference between 2d and 3d Difference between 2d and 3d Movies 2d and 3d Difference between 2d and 3d Objects Difference Between 2D vs 3D. You need to select them in the Overview. Why different representations are used in 2D/3D? Is it to compromise the scale difference between them? Thanks. G. In linear algebra, linear transformations can be represented by matrices. Our proposed method consists of two major steps: the conversion from 3D points to 2D image data and the estimation of transformation parameters between 2D images in the frequency domain. Local Metric Learning in 2D/3D Deformable Registration With Application in the Abdomen Qingyu Zhao, Chen-Rui Chou, Gig Mageras, and Stephen Pizer, Life Senior Member, IEEE Abstract—In image-guided radiotherapy (IGRT) of disease sites subject to respiratory motion, soft tissue deformations can affect localization accuracy. Objects in the real world are 3-dimensional, because they have depth. What's the difference between 2d wireframe and 3d wireframe modes? autodesk-fusion360-header-nav-label. Sobel, Department of Applied Physics and Applied Mathematics, ( Quaternions are used heavily in the WorldToolKit package, which is no longer produced, and can be useful for interpolating rotations between two oblique angles. CSS 3D Transforms. In function 'make_transformation_matrix', r=r*(pi/180) is used in 3D but not in 2D version. 3D rasterization allows us to explore the design space between traditional rasterization and ray casting in a formalized manner. What is the Difference Between 2D and 3D - Free download as Word Doc (. Both systems are defined with three orthogonal vectors as one would expect. Vytiniotis2, G. Perspective lens: The percentage of difference between initial and optimized focal length is between 5% and 20%. This is one dimension. Furthermore there is an equivalent to 3D similarity transformation for the plane, the planar 2D similarity transformation consisting of a single rotation, a translation with two components and a practical applications. 5772/63042. This is where the 2D to 3D conversion method comes to rescue. 2D is "flat", using the horizontal and vertical (X and Y) dimensions, the image has only two difference between 2d and 3d transformation matrix. An Analogy In 2D. Comparison Between 2-D and 3-D Transformations for Geometric Correction of. The registration result will also improve with additional 2D images. A typical 3D transform has six degrees of freedom: rotations about and translations along the XYZ coordinate axis. Datum shift between two geodetic datums. CSS transforms allow you to move, rotate, scale, and skew elements. The difference between simply capturing images from multiple cameras (like in stereo) and constructing a multi-view representation is that multi-view What's the difference between a projection and a datum? There won't be a single answer for this as "datum" in GIS can be one of at least three different things e. 2D due to high stress Use the rationale: “Include dimensions until I don’t gain a significant reduction in my stress value” If stress is too high for 2D or 3D NMDS might not be the best method i. To do so, we will need to learn how we can "project" a 3D point onto the surface of a 2D drawable surface (which we will call in this lesson, a canvas) using some simple geometry rules. At the most basic level, 2D representations don’t depict a project in a way humans intuitively make sense of the world around them. The imaged organ was a phantom spine, where • Any non-degenerate affine transformation takes a parallelogram to another parallelogram • A parallelogram has rotational symmetry of order 2 • The sum of the distances from any interior point of a parallelogram to the sides is independent of the location of the point. 3D objects have height, width, and depth, while 2D objects only have height and width. For example, many signals are functions of 2D space defined over an x-y plane. However, it can still change due to facial expressions and aging factor. Good question. With 3D rasterization the only remaining differences between the two approaches are the scene traversal and the enumeration of potentially covered samples on the image plane (binning). This thesis presents a MATLAB-based 2D to 3D conversion system from The main difference between them is that the shape component of an AAM is 2D, whereas the shape component of a 3DMM is 3D. Chapter6 Foley et al. Strictly speaking it gives a transformation from one plane to another, but if we identify the two planes by (for example) fixing a cartesian system in each, we get a projective transformation from the plane courses. I am in construction and we are trying to accurately build a complex shaped steel space frame. However, in real world, transformation can be composed of translation plus rotation at the same time. That means instead of a flat drawing of a globe, 3D animation produces a sphere that can actually rotate 360 degrees. 3D images, 2D-3D registration between a 3D image and a set of 2D images has less information available and more parameters to compute. For the latter, even though feature matching is performed in the 2D domain, the resulting matched points also Variational formulations in 2D and 3D¶ The major difference between deriving variational formulations in 2D and 3D compared to 1D is the rule for integrating by parts. But both, data mining and data warehouse have different aspects of operating on an enterprise's data. Coordinate Transformation: Arc 1960 to WGS 84 (1) Calculator. They are named for the number of dimensions that they portray. GitHub is home to over 40 million developers working together to host and review code, manage projects, and build software together. We are not essentially modifying the pictures, but a picture in the center of the great because in order to build an illusion of 3D through 2D images the illusion of . Thus leading to a comprehensible visualization of its transformation. Bakas4 1 Research Associate, National Technical University of Athens, Greece If the normal of the viewing plane (the camera direction) is parallel to one of the primary axes (which is the x, y, or z axis), the mathematical transformation is as follows; To project the 3D point , , onto the 2D point , using an orthographic projection parallel to the y axis (where positive y represents forward direction - profile view matching these TIVs and then searching for the 3D trans-lation between the two point sets given the known rotation between them. Here both 2D and 3D stress analysis using Mohr's circle method is explained well. If a vector has 2 dimensions it represents a direction on a plane (think of 2D graphs) and when it has 3 dimensions it can represent any direction in a 3D world. There are two types of transformation in computer graphics. Make sure you select the "Petroleum" transformation otherwise your results will not be accurate. IKONOS . Comparison between 2D and 3D EACVI 3D Echocardiography Box 2D and 3D refer to the actual dimensions in a computer workspace. To convert data between WGS-84 and BNG in ArcGIS Desktop, you should use the Project tool. The key to preparing the different 3D materials from IPC-1P is control over the The relationship between this material and the original UTL parent is that IPC-4 has is determined by applying the 2D homogeneous transformation matrix (3. 2D is 'flat', using the X & Y (horizontal and vertical) axis', the image has only two 2D and 3D refer to the actual dimensions in a computer workspace. A 2D image, on the other hand, has only height and width. latitudes, longitudes and heights are The Transformation Type chooses between doing a 2D transformation and 3D transformation. Available from: What is the difference between a one dimensional and a two dimensional signal? case it's 2D plane, or 3D plane. We will go Pixels and voxels, the long answer. Difference between 2D, 2 5D 3D Animation Find PDF and PPT at: 2D and 3D/ दो आयामी आकृतियां और तीन Difference between 2D and 3D Auto cad In other words, a 2D animated character will look less realistic than a 3D one. It allows to change elements using 3D transformations. Visualizing your data in fewer dimensions compromises the data too much Again, in this lesson, we will learn about computing the 2D pixel coordinates of a 3D point without using the perspective projection matrix. and there is a similar transformation within the Commercial The difference between the two conditions lies in the spatial correspondence between the visible 2D cross sections and their originating locations in the 3D world. Note that sometimes the term “blending” is used to designate metamorphosis of 2D shapes, but we use it here in the way traditional to geometric and solid modeling. The purpose of a 2D array texture is to be able to have multiple 2D textures in a single object. Data Mining and Data Warehouse both are used to holds business intelligence and enable decision making. Why? Because they are pictures of videos taken from one view point, making it a 2D analysis. [19] deals with deter-mining position and orientation of arbitrary surfaces based oncontourpointsin2Dimagedata. 2 For this reason, 4×4 transformation matrices are widely used in 3D computer graphics. Shape, proportion and angles all play their part in defining these two words. A 2D array texture is a 2D texture where each mipmap level contains an array of 2D images. 3D transform includes Z-axis transformation of the HTML elements. The distinction between active and passive transformations is important. or in 3D module to deal with 2D and 3D rotations and projective or affine transformations. One important difference is that you cannot select the 2D-nodes by clicking on them. 3D adds the 'Z' dimension. The transformation can map any point x2D (= (x, y)) in 2D US image to its corresponding point x (= (x , y , z )) in the coordinate of the 3D US image. 2D-to-3D Image-to-Range Registration The automated 2D-to-3D image-to-range registration method of [19] is used for the automated calibration and registration of a single 2D image with the 3D range model. Circle, triangle, square, rectangle, and pentagon are some of the most common examples of 2D shapes. But in 3D characters sometimes they also look like cartoonish but due to depth the characters look like real. If you're using 3d and homogenous coordinates, the vector and point will be stored with 4 elements, xyzw. In this article we will try to understand in details one of the core mechanics of any 3D engine, the chain of matrix transformations that allows to represent a 3D object on a 2D monitor. The computation of the rotational transfor-mation between and is achieved by matching at least two vanishing points computed from with major 2D transformation in homogeneous form w x and w y map the homogeneous component w of a point to a value w' that depends on x and y therefore, the scaling of a point depends on x and / or y in perspective 3D projections, this is generally employed to scale the x-and y-component with respect to z, its distance to the viewer non-correspondence between 2D and 3D landmarks caused by pose variations and propose a pose adaptive 3DMM fit-ting algorithm. difference between 2d and 3d transformation
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# Windows Phone Shell Tool Guide
This guide shows how to use Cordova's set of platform-centered shell tools to develop Windows Phone apps for both versions 7 and 8. This development path, discussed in the Overview, may offer you a greater range of development options for the Windows Phone platform than the cross-platform CLI tool described in The Command-Line Interface. Before using either development path, you must first configure the SDK environment as described in the Windows Phone 7 Platform Guide or the Windows Phone 8 Platform Guide.
To enable shell tools for Windows Phone development, download Cordova from cordova.apache.org. The download contains separate archives for each platform. Expand each you wish to target, wp8 in this case, which in turn is forked into wp7 and wp8 subdirectories. The relevant tools are typically available in the top-level bin directory, otherwise consult the README file for more detailed directions.
These tools allow you to create, build, and run apps. For information on the all-purpose command-line interface that enables plugin features, see Using Plugman to Manage Plugins. See Application Plugins for details on how to develop plugins.
## Windows Phone
The Windows Phone command-line tools support creating, building, and running new projects. Commands must be run from a cmd or powershell prompt.
The WP8 repo now includes code to build both WP7 and WP8 apps. The repo has subdirectories for each: wp7/ and wp8/.
## Create a Project
There are 2 ways to go about creating a new Apache Cordova WP7 or WP8 application.
### Run the Batch File to Create and Install the Templates
• The root of the repo contains a createTemplates.bat file. Double-clicking it generates two .zip files: CordovaWP7_x_x_x.zip and CordovaWP8_x_x_x.zip, where 3.3.0 represents the current version number. To easily use these files in Visual Studio, copy them to My Documents\Visual Studio 2012\Templates\ProjectTemplates\. You are then able to create new Apache Cordova Windows Phone apps from Visual Studio's File → New Project menu.
• If you run the batch file from the command line, you can also call with a parameter to install automatically
Run the script :
>createTemplates.bat -install
### Use the Create Scripts on the Command Line
Run the create command, specifying the existing path to the project, the reverse-domain-style package identifier, and the app's display name. Here is the syntax for both Windows Phone 7 and 8:
>.\wp7\bin\create PathToNewProject [ PackageName ] [ AppName ]
>.\wp8\bin\create PathToNewProject [ PackageName ] [ AppName ]
>PathToNewProject : The path to where you wish to create the project
>PackageName : The namespace for the project (default is Cordova.Example)
>AppName : The name of the application (default is CordovaWP8AppProj or CordovaWP7AppProj)
>examples:
>.\wp7\bin\create C:\path\to\my_new_project
>.\wp8\bin\create C:\path\to\my_new_project io.cordova.example CordovaWP8App
Launch Visual Studio and open Solution file (.sln) in (C:\path\to\my_new_project)
Build and Run it
## Building the Project (Clean, then Build)
• Debug
$C:\path\to\my_new_project\cordova\build --debug • Release $ C:\path\to\my_new_project\cordova\build --release
## Running the App
Run the 'run' command with the following optional parameters
• Target specification. This includes --emulator, --device, or --target=<targetID>.
• Build specification. This includes --debug, --release, or --nobuild.
$C:\path\to\my_new_project\cordova\run [Target] [Build] By default the run command is called with --emulator --debug if flags are not provided. ## Cleaning $ C:\path\to\my_new_project\cordova\clean | |
# Synopsis: A Quantum Microscope at High NOON
A special quantum entanglement called the NOON state enables microscopy at very low light levels.
At low light levels, microscopic imaging systems suffer from noise as the few desired photons compete with random background. One would normally just turn up the illumination, but this can alter or even damage some delicate samples. A research team led by Yaron Silberberg at the Weizmann Institute of Science in Israel has found one possible way around this with specially constructed entangled quantum states. In a paper in Physical Review Letters, the researchers report a quantum microscope that produces relatively superior images compared to a conventional microscope at extremely low light level.
The researchers used what are known as NOON states, which have two quantum modes (such as vertical and horizontal polarization). In their microscope, two states are formed—one with $N$ photons in one mode and the other mode empty ($|N0〉$), and the other state with the situation reversed ($|0N〉$). These states are entangled to create the NOON state. And because the image quantum noise goes approximately as $1\sqrt{N}$, so-called “high” NOON states with $N>1$ are most desirable.
To assess the improvement offered by this proof-of-principle system, the authors used a piece of quartz as a sample and used their microscope to create a 2D image of the phase shift between the two polarization modes caused by its birefringence. The NOON states were loaded with either $N=2$ or $N=3$ photons, and the overall NOON state illumination was limited to $50$ photons per pixel (by contrast, a typical digital camera captures tens of thousands of photons per pixel). The noise was reduced to near the ultimate Heisenberg quantum limit. The authors note that this is only a test—a practical system is not here yet because NOON states are hard to create and suffer large photon losses. – David Voss
More Features »
### Announcements
More Announcements »
## Subject Areas
Quantum Information
Fluid Dynamics
Gravitation
## Related Articles
Quantum Information
### Synopsis: A Lens to Focus Spins
A quantum bit stored in the spin excitation of an atomic cloud could be “focused” onto the quantum state of a single atom. Read More »
Quantum Information
### Synopsis: Quantum Annealers Limited by Temperature
Calculations show that quantum annealing—the quantum computing method used in a commercially available device—is hampered by thermal effects. Read More »
Atomic and Molecular Physics | |
## Ten Economic Paragraphs Worth Reading: December 23, 2009
So two groups opposed to Bernanke’s reappointment are worried he won’t devote enough attention to fighting inflation, and the third group is worried inflation is too much of a priority. These groups appear to be reinforcing rather than canceling each other politically even though they are upset about opposite things.... I do not believe for a minute that Ben Bernanke is taking the side of Wall Street over Main Street when he talks about the need to keep inflation under control. He believes that stabilizing inflation leads to higher employment, a better mix of goods, and higher household welfare... maximizing household welfare means responding more aggressively to inflation shocks than to output shocks... the Fed should increase the federal funds rate by one-half percent when output deviates by one percent from target, but increase the federal funds rate three times as much, by a point and a half, in response to a one percent inflation shock. There are two questions here. First, is the rule the Fed follows... optimal? Second, is the standard Taylor rule type policy appropriate for severe recessions?...
The U.S. stock market is wrapping up what is likely to be its worst decade ever. In nearly 200 years of recorded stock-market history, no calendar decade has seen such a dismal performance as the 2000s. Investors would have been better off investing in pretty much anything else.... Many investors were lured to the stock market by the bull market that began in the early 1980s and gained force through the 1990s. But coming out of the 1990s, the best calendar decade in history with a 17.6% average annual gain, stocks simply had gotten too expensive.... And in a time of financial panic like 2008, stocks were a terrible place to invest. With just two weeks to go in 2009, the declines since the end of 1999 make the last 10 years the worst calendar decade for stocks going all the way back to the 1820s.... It edges out the... 1930s, which up until now held the title of worst decade.... Since the end of 1999, the Standard & Poor's 500-stock index has lost an average of 3.3% a year on an inflation-adjusted basis.... So what went wrong for the U.S. stock market? For starters, it turned out that the old rules of valuation matter. "We came into this decade horribly overpriced," said Jeremy Grantham, co-founder of money managers GMO LLC. In late 1999, the stocks in the S&P 500 were trading at about an all-time high of 44 times earnings, based on Yale professor Robert Shiller's measure, which tracks prices compared with 10-year earnings and adjusts for inflation. That compares with a long-run average of about 16. Buying at those kinds of values, "you'd better believe you're going to get dismal returns for a considerable chunk of time," said Mr. Grantham, whose firm predicted 10 years ago that the S&P 500 likely would lose nearly 2% a year in the 10 years through 2009...
In her column, Hamsher offers ten reasons why she opposes the Senate bill. For now, I want to focus on two of them: (1) Forces you to pay up to 8% of your income to private insurance corporations — whether you want to or not. (3) Many will be forced to buy poor-quality insurance they can’t afford to use, with $11,900 in annual out-of-pocket expenses over and above their annual premiums. Both statements are true. That's why many of us have been calling attention to these numbers for months. But a crucial question, which Hamsher and most lefty critics I know never address, is "compared to what?"...This is a hugely progressive program to bolster economic security, the likes of which we haven't enacted in this country for a long, long time. To be clear, I don't think these numbers are great in absolute terms... it's essential to push for better subsidies and more protection in the conference negotiations--and then, if the law passes, to work on improving the law afterwards. But you can't do any of that if a bill doesn't get past the Senate... No more stimulus, please, we're capitalists. That’s the view, at least, of the majority of economists surveyed in msnbc.com’s year-end roundtable. Though unemployment will remain stubbornly high, and the economic recovery sluggish in 2010, the government doesn’t need to provide another round of stimulus spending to keep the economy afloat, they say. The House last week narrowly approved a$155 billion “jobs” bill that includes nearly $50 billion in infrastructure spending and$79 billion for expanding benefits like unemployment insurance and Medicaid. But most of the forecasters in our panel are against the idea of another government stimulus package...
5) Ryan Avent Sends Us to Meghan Busse, Christopher Knittel, and Florian Zettelmeyer:
The dramatic increase in gasoline prices from close to $1 in 1999 to$4 at their peak in 2008 made it much more expensive for consumers to operate an automobile. In this paper we investigate whether consumers have adjusted to gasoline price changes by altering what automobiles they purchase and what prices they pay. We investigate these effects in both new and used car markets. We find that a $1 increase in gasoline price changes the market shares of the most and least fuel-efficient quartiles of new cars by +20% and -24%, respectively. In contrast, the same gasoline price increase changes the market shares of the most and least fuel-efficient quartiles of used cars by only +3% and -7%, respectively. We find that changes in gasoline prices also change the relative prices of cars in the most fuel-efficient quartile and cars in the least fuel-efficient quartile: for new cars the relative price increase for fuel-efficient cars is$363 for a $1 increase in gas prices; for used cars it is$2839. Hence the adjustment of equilibrium market shares and prices in response to changes in usage cost varies dramatically between new and used markets. In the new car market, the adjustment is primarily in market shares, while in the used car market, the adjustment is primarily in prices. We argue that the difference in how gasoline costs affect new and used automobile markets can be explained by differences in the supply characteristics of new and used cars.
I think that underwater homeowners ought to walk away from their loans for the very same reason McArdle want us to consider them jerks for doing so. We both want to see norms we consider valuable enforced. I think that banks violated a great many norms of prudence and fair dealing in their practices during the credit bubble, and that they violate the fundamental norm of reciprocity by fully exploiting their own legal rights while insisting that borrowers have a moral obligation not to exercise a contractual option. In order to strengthen norms I consider crucial, I hope transgressors face legal and social consequences (strategic default and reduced shame attached to default) that will alter their behavior going forward. McArdle values a norm that I think most of us share in interpersonal settings, that a person should make every possible effort to pay back money he has borrowed. She also wants to create consequences for transgressors, social costs via a consensus that those who walk away by choice be considered jerks. We have different preferences regarding the kind of world we want our normative frameworks to support: McArdle favors a world with both easy credit and easy bankruptcy. I favor the easy bankruptcy, but not the easy credit. I think that debt arrangements are hazardous and should be entered into only with great care. I don’t consider increasingly leveraged homeownership and aggressively accessible consumer credit to have been positive developments. As a practical matter, I think we must rely on creditors rather than potential debtors to differentiate between wise and unwise loans. So I consider it a feature rather than a bug that holding creditors accountable will encourage them to think twice before sending out convenience checks. Norms, like laws, are always contested. McArdle and I have very different worldviews, and that is reflected in the different norms we are each trying to reinforce...
Many things in American politics are silly but, assuming it's true, this has to be considered a lifetime achievement award. From Talking Points Memo: "After months in which the senate health care bill was held up over efforts to find some form in which she would agree to sign on to it, Sen. Snowe (R-ME) now says she will oppose it because it is being 'rushed'."
8) BEST NON-ECONOMICS THING I'VE READ TODAY: Paul Krugman: The WYSIWYG president: There’s a lot of dismay/rage on the left over Obama, a number of cries that he isn’t the man progressives thought they were voting for. But that says more about the complainers than it does about Obama himself. If you actually paid attention to the substance of what he was saying during the primary, you realized that (a) There wasn’t a lot of difference among the major Democratic contenders, (b) To the extent that there was a difference, Obama was the least progressive Now it’s true that many progressives were ardent Obama supporters, with their ardency mixed in with a fair bit of demonization of Hillary Clinton. And maybe they were right — but not on policy grounds. (I still remember people angrily telling me that if Hillary got in, she’d fill her economics team with Rubinites). So what you’re getting is what you should have seen. And exactly what should we blame Obama for?... [T]he important thing to bear in mind is that this isn’t about him; and, equally important, it isn’t about you. If you’ve fallen out of love with a politician, well, so what? You should just keep working for the things you believe in.
9) STUPIDEST THING I'VE READ TODAY: Michael Hudson: The Problem with Paul Samuelson: Michael Hudson in Commonweal (December 18 1970): "Does economics deserve a Nobel prize? (And by the way, does Samuelson deserve one?)"
It is bad enough that the field of psychology has for so long been a non-social science, viewing the motive forces of personality as deriving from internal psychic experiences rather than from man's interaction with his social setting. Similarly in the field of economics: since its "utilitarian" revolution about a century ago, this discipline has also abandoned its analysis of the objective world and its political, economic productive relations in favor of more introverted, utilitarian and welfare-oriented norms. Moral speculations concerning mathematical psychics have come to displace the once-social science of political economy. To a large extent the discipline's revolt against British classical political economy was a reaction against Marxism, which represented the logical culmination of classical Ricardian economics and its paramount emphasis on the conditions of production. Following the counter-revolution, the motive force of economic behavior came to be viewed as stemming from man's wants rather than from his productive capacities... [Only somebody who knows neither the economics of Ricardo, the writings of Marx, or Samuelson's neoclassical synthesis could possibly write that paragraph with a straight face]
10) FROM THE ARCHIVES: Brqd DeLong (February 20, 2003): Thinking About Aristotle of Stagira:
I'm never sure whether I should begin my economic history survey courses with Aristotle or not. As Moses Finley powerfully argues, Aristotle does not care about the economy. The fragments in his Ethics and Politics that economists like Joseph Schumpeter point to are, mostly, concerned with other things than economic analysis. Karl Polanyi thought that Aristotle's naivete was the result of the fact that a mercantile, market, commercial economy was something very new. He was surely wrong: it was not something new, but rather something that Aristotle as a Hellenic aristocrat would have been embarrassed to be caught thinking seriously about. Still, I now wish I'd started this semester's history course with more on Aristotle. His perspective is so different from ours that it provides a useful mental shock: Consider, first, that Aristotle of Stagira was not an idiot (even if he did believe that women had fewer teeth than men). For two thousand years people--pagan Hellenes, Christian Europeans, and Islamic Arabs, Egyptians, Mesopotamians,and Iranians--called Aristotle of Stagira "the philosopher", as if there could be only one. Think of the way seventeenth, eighteenth, and nineteenth century Britons regarded Newton (or the way we regard Einstein). So we need to take Aristotle seriously: think hard about how a very good mind, thinking very hard, in pre-industrial-revolution economic circumstances, could wind up thinking the thoughts on the economy that Aristotle does. Specifically, why does he: --believe so strongly that gross inequality--domination and slavery--is natural and inevitable? --believe that the 'natural art of acquisition'--the getting of the resources necessary to properly run one's household--has a limit: 'a boundary fixed, just as there is in the other arts; for the instruments of any art are never unlimited, either in number or size, and riches may be defined as a number of instruments to be used in a household or in a state...'? (Never mind that Aristotle's "limit" is probably the full-time year-round labor of at least fifty people, at today's OECD wage levels some \$3,000,000 a year: in one sense very, very few of us will ever come near to Aristotle's point of satiation; in another sense every single one of us has already gone far beyond Aristotle's limit.) --believe that shepherds are '...the laziest [of men]... lead an idle life... get their subsistence without trouble from tame animals...'? --believe that '[t]here are two sorts of wealth-getting... one is a part of household management, the other is retail trade: the former necessary and honorable, while that which consists in exchange is justly censured; for it is unnatural, and a mode by which men gain from one another...'? --believe that of '...the practical part [of wealth-getting] the discussion of such matters is not unworthy of philosophy, but to be engaged in them practically is illiberal and irksome'? Note: don't miss Aristotle's story of Thales of Miletos and his corner of the olive-press-rental market on Khios... | |
# Check whether the following matrix is invertible or not: (1001) - Mathematics and Statistics
Sum
Check whether the following matrix is invertible or not:
((1,0),(0,1))
#### Solution
Let A = ((1,0),(0,1))
Then, |A| = |(1,0),(0,1)| = 1 - 0 = 1 ≠ 0.
∴ A is a non-singular matrix.
Hence, A-1 exists.
Concept: Elementry Transformations
Is there an error in this question or solution? | |
# Tag Info
3
If the protocol doesn't provide authentication, an attacker can probably mount replay attacks or make deterministic changes to messages. If the nonces in different blocks are not compared in any way, they can just take the ID block of a previous message and use it with a new one, to forge it being from that device. If nonces are required e.g. to be equal in ...
3
I don't think this scheme would make sense, either from a performance or a crypto-design perspective. From a crypto-design perspective, simply encrypting with a block cipher would be better. Encrypting with a block cipher, or other suitable symmetric-key encryption scheme, takes running time that is linear in the length of the data to be encrypted (not ...
2
There are two ways to attack encryption that uses a derived key: You can attack the encryption algorithm. In the case of correctly used* 128-bit AES, that essentially amounts to a brute force attack on the 128-bit keyspace. This would succeed after on average $2^{127}$ tries (if it were practical). If you knew that two files had used the same password ...
2
Your idea for constructing a distinguisher from a predictor is fine, assuming you know that the predictor predicts the last bit. The more general statement is: if you can predict any bit of the output, say the $i$th bit, given the first $i-1$ bits, then you can also build a distinguisher. A similar idea to what you showed also works to prove this ...
2
Berlekamp-Massey is designed for the situation where you have observed $2n$ consecutive output bits from a $n$-bit LFSR. It doesn't work if the observed bits are scattered randomly, at random non-contiguous offsets in the stream. Information-theoretically, a minimum of $2n$ bits of output are needed to reconstruct the LFSR. Intuitively, this is because ...
2
Dinh, Moore, Russell have shown that the quantum algorithm (Quantum Fourier sampling) used to attack RSA and ElGamal does not work on McEliece-like crypto systems. (I think) this means, that there are no known algorithms on quantum computers that decrease the complexity of attacks on McEliece, and thus McEliece is just as safe post-quantum computers as it is ...
1
How to prove the security of the PRNG? My best advice would be to start with a statistical test suite like the one NIST describes in "A Statistical Test Suite for Random and Pseudorandom Number Generators for Cryptographic Applications" (PDF). It’s a battery of statistical tests to detect non-randomness in binary sequences constructed using random ...
1
Given a function $F: A \rightarrow B$ and a function $R:B \rightarrow A$, we can create a chain of length $k$ from a starting point $a_0$ to an end point $a_k$ using $a_i = R(F(a_{i-1}))$. A rainbow table for $(F, R, k)$ is a collection of chains with end points $(a_0, a_k)$ organized so that searching for chains ending at $a_k$ is cheap. We use a rainbow ...
1
Yes, according to NIST SP 800-56A revision 2, a KDF based on HMAC-SHA-256 is a suitable option. The basic idea behind using a Key Based Key Derivation Function KBKDF is that the output of the the primitive within the key agreement protocol (DH, ECDH) returns enough entropy for a key to be created. However that entropy may still be distinguishable from ...
Only top voted, non community-wiki answers of a minimum length are eligible | |
mersenneforum.org Prime95 v30.3
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2020-10-15, 20:30 #430
Uncwilly
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
2×4,373 Posts
Quote:
Originally Posted by endless mike I upgraded all my machines to 30.3b6. On a few of them, they did not remember my userid or their computerid. I caught and corrected the problem before results were submitted except for one. How can I get the exponent that I completed added to my user statistics instead of it saying it was completed by -Anonymous- ?
Send George a PM or email. His email address is one of the graphics that are available. :woltman:
2020-10-15, 20:37 #431
James Heinrich
"James Heinrich"
May 2004
ex-Northern Ontario
C3616 Posts
Quote:
Originally Posted by Uncwilly Send George a PM or email. His email address is one of the graphics that are available. :woltman:
It's also in the top of readme.txt
Or you can PM him on the forum, username Prime95.
2020-10-19, 15:57 #432 JuanTutors Mar 2004 7658 Posts Not sure if this has been reported yet but this version of Prime95 seems to have the same problem that some apps/programs have on 4k monitors. The text is exceedingly small. I am restarting the computer to see if things get fixed but I have attached what the program looks like right now. Attached Thumbnails
2020-10-19, 16:02 #433
JuanTutors
Mar 2004
1111101012 Posts
Quote:
Originally Posted by JuanTutors Not sure if this has been reported yet but this version of Prime95 seems to have the same problem that some apps/programs have on 4k monitors. The text is exceedingly small. I am restarting the computer to see if things get fixed but I have attached what the program looks like right now.
Restarting my computer didn't fix the text size for me.
2020-10-19, 16:21 #434 JuanTutors Mar 2004 3·167 Posts I found another very minor bug. When I installed the new version of Prime95 into the same old folder that had the previous version of Prime95 (28.6), it got a new PRP assignment despite the fact that I am currently in the last few days of a current 100M digit PRP assignment and already have my next 100M digit PRP assignment on deck. That means I got a PRP assignment despite having a half year's worth of work. I can only speculate on the exact reason why this could possibly happen. (I pasted the new .exe file but didn't paste the new .dll files? Maybe the sequence in which certain checks can be done caused this to happen?)
2020-10-19, 16:25 #435
kruoli
"Oliver"
Sep 2017
Porta Westfalica, DE
14D16 Posts
Quote:
Originally Posted by JuanTutors Not sure if this has been reported yet but this version of Prime95 seems to have the same problem that some apps/programs have on 4k monitors.
It's not the 4K that causes the problem, it's the Windows scaling that does not get recognized by Prime95. Yes, if you have set the magnification exceedingly high, it looks like the text is tiny.
2020-10-19, 16:28 #436
Uncwilly
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
210528 Posts
Quote:
Originally Posted by JuanTutors I found another very minor bug. When I installed the new version of Prime95 into the same old folder that had the previous version of Prime95 (28.6), it got a new PRP assignment despite the fact that I am
Are you sure that you did not just get a Cert assignment? That should not last more than an hour or 3, depending on your machine.
2020-10-19, 16:30 #437
JuanTutors
Mar 2004
1F516 Posts
Quote:
Originally Posted by Uncwilly Are you sure that you did not just get a Cert assignment? That should not last more than an hour or 3, depending on your machine.
I don't think so. Here is the assignment:
Code:
PRP=AID EXPUNGED,1,2,332316059,-1,79,0,3,
Last fiddled with by Uncwilly on 2020-10-19 at 16:56 Reason: Removed AID
2020-10-19, 16:48 #438
kriesel
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
23·3·193 Posts
Quote:
Originally Posted by JuanTutors I don't think so. Here is the assignment: Code: PRP=xxx,1,2,332316059,-1,79,0,3,
DON'T post AIDs for current assignments!
Don't make pointless work for moderators to remove what should not be posted.
Last fiddled with by kriesel on 2020-10-19 at 16:49
2020-10-20, 04:33 #439 DrobinsonPE Aug 2020 3·19 Posts I have a computer running mprime 30.3 b6 that today finished PRP on exponent 100362923. It experienced 5 Gerbicz/double-check errors during the run but always said confidence in final result is excellent so I did not start over. All of the errors happened at about 50% and all in the period of a few hour time. I just checked https://www.mersenne.org/report_expo...0362923&full=1 and discovered that the result is suspect, two people tried to do a cert and now it has been assigned to someone else to do a PRP. I am curious, was the PRP bad, did the cert fail? The computer is currently working on 100495651. Should I stop it if it starts with Gerbicz/double-check errors again or just let it keep going? I think the problem is the memory because the two sticks were in a different computer that had a bad LL-D result about this time last year. Over the last year, whatever computer this memory is in has errors about once every 3-4 months.
2020-10-20, 05:59 #440
moebius
Jul 2009
Germany
17·23 Posts
Quote:
Originally Posted by DrobinsonPE I am curious, was the PRP bad, did the cert fail? The computer is currently working on 100495651. Should I stop it if it starts with Gerbicz/double-check errors again or just let it keep going?
I bet that the PRP test is good, and only the .proof-file you uploaded are bad.
Last fiddled with by moebius on 2020-10-20 at 06:01 | |
Glass panel engineering at work
We are designing a button panel at work, made out of glass and are running into some issues. Glass, of course, can shatter. But we noticed that when dropping a steel ball in the center of the glass it is far less likely to shatter than when the same experiment is performed near the corners. It almost always shatters. Why is that? Is there something we can do with the geometry of the panel (currently rectangle) that will alleviate this issue? Maybe rounding the corners to be circular? Does anyone have a link to some research on the topic (i.e. how thick the glass should be to survive a ball of mass $m$ being dropped on it from height $h$)?
Glass tends to contain quite a lot of residual stresses from the manufacturing process and these will tend to be concentrated near corners (depending on how the sheet was manufactured). There is also the fact that for a panel supported at the edges the centre will be less stiff and thus can deform more before it fractures in the same way that a simple beam deforms most at the centre.
Similarly mounting the panel in a shock absorbing frame such as using a rubber or foam casket or an elastomer adhesive can help as this will allow the whole assembly to defect smoothing out the energy transfer from the impactor to the panel.
This sort of calculation in glass is complicated by the fact that it is a brittle material thus any local stress concentration which exceeds its UTS can cause it to fail completely even if the average stress is quite low where an equivalent ductile material would just be scratched or dented. This is why you can easily break a window with an impact from a hard sharp object with much less effort than just pushing against it with something blunt.
Equally any surface defects like pits or scratches can have a significant effect on the fracture toughness of glass, which is why scoring it with a diamond scribe is a very effective way to 'cut' it.
Some types of glass are designed to be much more impact resistant and to fail in a safe way, this include tempered glass which is heat treated to alleviate residual stresses and laminated glass which is composed of alternating layers of thin glass and a transparent polymer.
It sounds like you really need to talk to a glass supplier to find the appropriate material for your requirements.
There is no one generic formula for the impact toughness of glass (or really any material) any you would need to look at empirical data for a specific product and impact toughness metrics tend to only be applicable to quite specific conditions, especially for brittle materials. | |
• ### BCS pairing in a trapped dipolar Fermi gase(cond-mat/0409150)
Sept. 7, 2004 cond-mat.other
We present a detailed study of the BCS pairing transition in a trapped polarized dipolar Fermi gas. In the case of a shallow nearly spherical trap, we find the decrease of the transition temperature as a function of the trap aspect ratio and predict the existence of the optimal trap geometry. The latter corresponds to the highest critical temperature of the BCS transition for a given number of particles. We also derive the phase diagram for an ultracold trapped dipolar Fermi gases in the situation, where the trap frequencies can be of the order of the critical temperature of the BCS transition in the homogeneous case, and find the critical value of the dipole-dipole interaction energy, below which the BCS transition ceases to exist. The critical dipole strength is obtained as a function of the trap aspect ratio. Alternatively, for a given dipole strength there is a critical value of the trap anisotropy for the BCS state to appear. The order parameter calculated at criticality, exhibits nover non-monotonic behavior resulted from the combined effect of the confining potential and anisotropic character of the interparticle dipole-dipole interation.
• ### Superfluidity of trapped dipolar Fermi gases(cond-mat/0307671)
Jan. 5, 2004 cond-mat
We derive the phase diagram for ultracold trapped dipolar Fermi gases. Below the critical value of the dipole-dipole interaction energy, the BCS transition into a superfluid phase ceases to exist. The critical dipole strength is obtained as a function of the trap aspect ratio. The order parameter exhibits a novel behavior at the criticality.
• ### Atom optics with rotating Bose-Einstein condensates(cond-mat/0211553)
Nov. 25, 2002 cond-mat.soft
The atom optics of Bose-Einstein condensates containing a vortex of circulation one is discussed. We first analyze in detail the reflection of such a condensate falling on an atomic mirror. In a second part, we consider a rotating condensate in the case of attractive interactions. We show that for sufficiently large nonlinearity the rotational symmetry of the rotating condensate is broken.
• ### Optical generation of quasisolitons in a single-component gas of neutral fermionic atoms(cond-mat/0203095)
March 5, 2002 cond-mat
We analyze, the generation of soliton-like solutions in a single-component Fermi gas of neutral atoms at zero and finite temperatures with the phase imprinting method. By using both the numerical and analytical calculations, we find the conditions when the quasisolitons, which apparently resamble the properties of solitons in non-linear integrable equations, do exist in a non-interacting Fermi gas. We present the results for both spatially homogeneous and trapped cases, and emphasize the importance of the Fermi statistics and the absence of the interaction for the existence of such solutions.
• ### Ultracold dipolar gases - a challenge for experiments and theory(cond-mat/0201100)
Jan. 8, 2002 cond-mat
We present a review of recent results concerning the physics of ultracold trapped dipolar gases. In particular, we discuss the Bose-Einstein condensation for dipolar Bose gases and the BCS transition for dipolar Fermi gases. In both cases we stress the dominant role of the trap geometry in determining the properties of the system. We present also results concerning bosonic dipolar gases in optical lattices and the possibility of obtaining variety of different quantum phases in such case. Finally, we analyze various possible routes towards achieving ultracold dipolar gases.
• ### Optical Generation of Vortices in trapped Bose-Einstein Condensates(cond-mat/9907452)
July 29, 1999 cond-mat.stat-mech
We demonstrate numerically the efficient generation of vortices in Bose-Einstein condensates (BEC) by using a phase imprinting'' method. The method consist of passing a far off resonant laser pulse through an absorption plate with azimuthally dependent absorption coefficient, imaging the laser beam onto a BEC, and thus creating the corresponding non-dissipative Stark shift potential and condensate phase shift. In our calculations we take into account experimental imperfections. We also propose an interference method to detect vortices by coherently pushing part of the condensate using optically induced Bragg scattering. | |
Found 3 result(s)
### 09.10.2019 (Wednesday)
#### TTbar and TsT
Regular Seminar Stijn van Tongeren (Humboldt U)
at: 14:00 ICroom H503 abstract: The TTbar deformation of two dimensional QFTs has various attractive and interesting features, giving a simple CDD deformation of the S matrix, and for instance preserving integrability, if present. As a simple example, deforming massless free scalars gives a Nambu-Goto string in flat space in a uniform light-cone gauge. I will discuss what happens if we deform "twice", i.e. TTbar deform light-cone gauge fixed string sigma models. In this setting, TTbar deformations can be viewed as TsT transformations in a suitable T dual frame. This TsT picture also gives a natural interpretation of the TTbar CDD factor as a Drinfeld-Reshetikhin twist.
### 29.03.2017 (Wednesday)
#### Yang-Baxter strings, twists, and AdS/CFT
Regular Seminar Stijn van Tongeren (Humboldt U.) | |
## College Physics (4th Edition)
Published by McGraw-Hill Education
# Chapter 27 - Problems - Page 1042: 22
#### Answer
(a) $\lambda = 10.71~pm$ (b) $\lambda = 12.43~pm$
#### Work Step by Step
(a) We can find the wavelengths of the scattered x-rays: $\lambda_f = \lambda_i+ \frac{h}{mc}~(1-cos~\theta)$ $\lambda_f = (10.0~pm)+ \frac{6.626\times 10^{-34}~J~s}{(9.1\times 10^{-31}~kg)(3.0\times 10^8~m/s)}~(1-cos~45.0^{\circ})$ $\lambda_f = (10.0~pm)+ (2.427~pm)~(1-cos~45.0^{\circ})$ $\lambda = 10.71~pm$ (b) We can find the wavelengths of the scattered x-rays: $\lambda_f = \lambda_i+ \frac{h}{mc}~(1-cos~\theta)$ $\lambda_f = (10.0~pm)+ \frac{6.626\times 10^{-34}~J~s}{(9.1\times 10^{-31}~kg)(3.0\times 10^8~m/s)}~(1-cos~90.0^{\circ})$ $\lambda_f = (10.0~pm)+ (2.427~pm)~(1)$ $\lambda = 12.43~pm$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | |
# Optimising hyper-parameters efficiently with Scikit-Optimize
One of the most well-known techniques for experimenting with various model configurations is Grid Search.
With grid search, you specify a discrete search space (a parameter grid) of all of the parameter values you would like to test. The search permutes through the grid, testing various combinations until all are exhausted. Basic a specified performance metric (e.g. error), you can select the best parameter combination for your model.
What's wrong with this?
If you have a large parameter grid, this doesn't work too well:
In [1]:
import numpy as np
param_grid = {
'param_a': [0.01, 0.03, 0.1],
'param_b': [0, 1, 2],
'param_c': [1, 50, 100]
}
def num_searches(param_grid):
return np.prod([len(p) for p in param_grid.values()])
num_searches(param_grid)
Out[1]:
27
And maybe we want to search over four possible values instead for param_a, and add two more new parameters:
In [2]:
param_grid = {
'param_a': [0.01, 0.03, 0.1, 0.3],
'param_b': [0, 1, 2],
'param_c': [1, 50, 100],
'param_d': ["a", "b"],
'param_e': [0, 1, 2]
}
num_searches(param_grid)
Out[2]:
216
As you can see from the first grid, there's already 27 combinations to try. Then this jumps to 216 for our larger grid. Depending on the complexity of the model and the amount of data to process, this can very easily become infeasible.
There are a few approaches to solving this, including:
• breaking down the search into multiple smaller steps (such as searching param_a and param_b first, with defaults for the others, then using the best values to search the remaining parameters - this can be tricky in practice)
• searching the parameter space at random (which has an additional benefit of discovering better parameter values when random samples are drawn frmo a continuous range)
While Scikit-Learn doesn't provide many more options, some clever people have developed a drop-in replacement for Scikit-Learn's GridSearchCV and RandomizedSearchCV called BayesSearchCV in a package called Scikit-Optimize.
Let's install Scikit-Optimize and implement BayesSearchCV with a simple example!
### Installing Scikit-Optimize¶
Assuming you already have already installed Anaconda and Jupyter, you will need to do the following:
• pip install scikit-optimize
If you have trouble installing, you may first need to run the following to install one of Scikit-Optmize's dependencies:
• pip install scikit-garden
### Implementing BayesSearchCV¶
Here's an example implementation using a sample dataset and Logistic Regression.
In [3]:
import warnings
warnings.filterwarnings('ignore')
from skopt import BayesSearchCV
from sklearn.model_selection import train_test_split
from sklearn.linear_model import LogisticRegression
# prep some sample data
X_train, X_test, y_train, y_test = train_test_split(X, y, train_size=0.75, random_state=1234)
# we're using a logistic regression model
clf = LogisticRegression(random_state=1234, verbose=0)
# this is our parameter grid
param_grid = {
'solver': ['liblinear', 'saga'],
'penalty': ['l1','l2'],
'tol': (1e-5, 1e-3, 'log-uniform'),
'C': (1e-5, 100, 'log-uniform'),
'fit_intercept': [True, False]
}
# set up our optimiser to find the best params in 30 searches
opt = BayesSearchCV(
clf,
param_grid,
n_iter=30,
random_state=1234,
verbose=0
)
opt.fit(X_train, y_train)
In [4]:
print('Best params achieve a test score of', opt.score(X_test, y_test), ':')
opt.best_params_
Best params achieve a test score of 0.958041958042 :
Out[4]:
{'C': 100.0,
'fit_intercept': True,
'penalty': 'l1',
'solver': 'liblinear',
'tol': 0.00094035472283658726}
By increasing the value of n_iter, you can continue the search to find better parameter combinations. You can also use the optimiser for prediction, by calling .predict() or .predict_proba() for probabilities, or extract and use the best one standalone:
In [5]:
opt.best_estimator_
Out[5]:
LogisticRegression(C=100.0, class_weight=None, dual=False, fit_intercept=True,
intercept_scaling=1, max_iter=100, multi_class='ovr', n_jobs=1,
penalty='l1', random_state=1234, solver='liblinear',
tol=0.00094035472283658726, verbose=0, warm_start=False)
You may also find it useful to re-use the best parameters programatically to define an equivalent model:
In [6]:
LogisticRegression(**opt.best_params_)
Out[6]:
LogisticRegression(C=100.0, class_weight=None, dual=False, fit_intercept=True,
intercept_scaling=1, max_iter=100, multi_class='ovr', n_jobs=1,
penalty='l1', random_state=None, solver='liblinear',
tol=0.00094035472283658726, verbose=0, warm_start=False) | |
# What is the correct option of the following question?
Feb 9, 2018
The answer is $B$
#### Explanation:
For this problem I assume that $\log \left(a\right) = \ln \left(a\right)$
Let's rewrite the denominator in terms of sine and cosine exclusively.
$\sec x - \cos x = \frac{1}{\cos} x - \cos x = \frac{1 - {\cos}^{2} x}{\cos} x = {\sin}^{2} \frac{x}{\cos} x$
We can also use the laws of logarithms to simplify the numerator to
$\ln \left[\left(1 + x + {x}^{2}\right) \left(1 - x + {x}^{2}\right)\right]$
Now let's rewrite the limit.
$L = {\lim}_{x \to 0} \frac{\ln \left[\left(1 + x + {x}^{2}\right) \left(1 - x + {x}^{2}\right)\right]}{{\sin}^{2} \frac{x}{\cos} x}$
$L = {\lim}_{x \to 0} \frac{\cos x \ln \left[\left(1 + x + {x}^{2}\right) \left(1 - x + {x}^{2}\right)\right]}{\sin} ^ 2 x$
If we try to evaluate the limit now, we will get $\frac{0}{0}$, so we can apply L'Hospital's rule.
Finding the derivative of the numerator may be a little bit long. We must first differeintaite $\ln \left[\left(1 + x + {x}^{2}\right) \left(1 - x + {x}^{2}\right)\right]$ using the chain rule.
Start by expanding within the brackets:
$1 - x + {x}^{2} + x - {x}^{2} + {x}^{3} + {x}^{2} - {x}^{3} + {x}^{4}$
Now combine like terms
${x}^{4} + {x}^{2} + 1$
Now differentiate using the chain and power rules.
$\frac{d}{\mathrm{dx}} \left(\ln \left({x}^{4} + {x}^{2} + 1\right)\right) = \frac{4 {x}^{3} + 2 x}{{x}^{4} + {x}^{2} + 1}$
Now use the product rule to differentiate the entire numerator.
$L = {\lim}_{x \to 0} \frac{- \sin x \ln \left({x}^{4} + {x}^{2} + 1\right) + \cos x \left(\frac{4 {x}^{3} + 2 x}{{x}^{4} + {x}^{2} + 1}\right)}{\sin \left(2 x\right)}$
You will find that this gives $\frac{0}{0}$ once more. We will have to differentiate again. Repeat this process until you get something you can evaluate (not being $\frac{0}{0}$).
It's tedious, but in the end you should get an answer of $1$. The graph confirms:
Hopefully this helps! | |
# Do you terminate your DMX lines?
## Do you terminate your DMX lines?
• Total voters
60
#### derekleffew
##### Resident Curmudgeon
Senior Team
It's getting boring around here--time to bring up a controversial topic, again.
#### SerraAva
##### Active Member
Dimmers, haven't unless built into the racks. Movers, just about always as there is always a few hundred feet or so of cable between them and the board. LEDs, sometimes, really long runs can make them act a little weird. Other DMX things like foggers and what have you, haven't done or had to yet. Always try to run things through an iso too.
#### Logos
##### Well-Known Member
When I put in my own rig of DMX gear I always terminate at the last device. It isn't expensive it isn't hard and it avoids problems. If I am splitting a line out from the desk or tagging onto an exisitng system with a splitter or node I try to make sure every line has been terminated but you can't always be sure. I rarely have problems so maybe it's a waste of time. But then I rarely have problems and that could be why.
My runs tend to be short by most standards. I have 1 100 meter cable and a couple of 30 meter DMX cables but most of my DMX cables are only 10 - 12 meters long.
#### len
##### Well-Known Member
I seldom have issues, but I think that's because I use a splitter and high quality cable.
#### beam_1973
##### Member
I have a bunch of fixtures with termimation dip switches on them, so they are easy ... the ones without, hmm, let's NOT discuss those shall we.
#### soundlight
##### Well-Known Member
We always terminate. On your way to run the DMX line, you just grab a terminator out of the IQ (Intelligents) roadbox, and plug it in to the last dmx "thing".
#### ship
##### Senior Team Emeritus
About four years ago I asked our electronics repair department for some 120ohm resistors to use in making up some terminators. I was given a 100pkg of them. In the past few years while I have made less terminators than say 3:5 or 5:3, much less 5:5 or 3:3 type adaptors in their various versions, I’m now about left with like 20 terminators out of the 100 left. This after whipping out a quick dozen just a few weeks ago in both 3pin and 5pin versions.
Been a few years since I was able to tell what gear might need either terminator or much less “Martin reverse adaptor” but it could be said that 80x terminators in the past few years might say something still about the need for them with some gear, and or the it won’t hurt factor. This granted in new gear it’s at least 100 or more new moving lights bought per year in adding to the inventory over the years.
Were I doing shows, granted I would know if my gear needed a terminator or not, when I didn’t I would have them requested, this much less have a dozen in my road box for the issues of weird stuff going on type things. First I would plug one in if not otherwise knowing the cause by process of elimination or what’s the cause.
Hmm, like 80x terminators made in the past few years... some no doubt have been lost or lifted but not so many persay theorized to be swept up off a floor, I mark and disquinguish my terminators by way of corporate heat shrink label over the plug so even if it’s seen to be just a plug laying on the floor - perhaps a worthless to a IA guy the might have been cut off a cable, it has our name on it and it’s different than just a plug. Such a thing if seen on a deck has more hope of winding up in one of our road boxes than just an un-marked plug that might get trashed. Such terminators get marked in a way that they are more than just thrown out - theoretically at least. Thus the 80x I made in the last few years says that even if most moving lights already sense if it’s the last in line as a concept I would expect, there is still gear out there that don’t.
#### Van
##### CBMod
CB Mods
I don't have a choice, with Gordon hanging around the theatre. If it's not terminated it's CrrrrrAP!
#### Grog12
##### CBMod
CB Mods
Nope. Haven't terminated a mover run since '01. Also haven't had issues.
I also want to know who voted that they dont' use DMX
##### Well-Known Member
Nope. Haven't terminated a mover run since '01. Also haven't had issues.
I also want to know who voted that they dont' use DMX
Someone with either a very low or very high budget.
#### SerraAva
##### Active Member
Hughesie89, you can see who voted how by clicking on the yellow, underlined numbers on the right of the poll.
#### gafftaper
##### Senior Team
Senior Team
Fight Leukemia
Oh that Hughesie, he doesn't use DMX he uses metric DMX.
He Charc, I believe if you only have one intell in the system it's impossible to have reflection problems. So termination is not needed. (Please correct me if I'm wrong here guys)
#### Edrick
##### Well-Known Member
I voted no because our highschool didn't have any Intelligent Lighting so I've yet to deal with it. We had a straight DMX run from the Booth to the Dimmer Rack and that was it.
#### Van
##### CBMod
CB Mods
....................................................He Charc, I believe if you only have one intell in the system it's impossible to have reflection problems. So termination is not needed. (Please correct me if I'm wrong here guys)
<insert gong noise here>
Gaff, you can get a "refelction" off of a bad solder joint. Whether or not you have Movers in your rig is completely beside the point. Movers will tend to make DMX reflection issues much more apparent, kind of like using a high sample rate on an oscilliscope to see abnomalities in a wave form, as opposed to using a multimeter, because they are more "sensitive" to the disruption of the data stream.
I used to do Communications, for a S&R group. I always had a SWR meter attached inline to monitor the standing wave ratio in the output cable of 10 - 16 meter cables. The chances that one would develope just out of the blue was slim, but if perchance a solder connection failed, or condensation got into a connector etc, etc, it's better to be safe rather than sorry.
Now if you are using a good Opto-isolator, or if you board has one built in, you could probably get away without ever terminating, However, Older board, more movers, etc, etc, it's not a chance worth taking.
#### SerraAva
##### Active Member
From Mr Doug Fleenor himself: http://www.dfd.com/whyterm.html
Reflections can happen with just one intel, or anything in a DMX line. Sometimes Opti-Iso splitter or whatever you call it can help, sometimes it can't. The main thing a splitter does is it stops problems with one line effecting other lines. It also helps amplify DMX signal for longer runs. Also changing your DMX speed can help at times too.
A theatre I use to do work in had an old set of Colortran ENR series dimmer racks. They would ghost all the time thanks to bad/not enough power and a poor data path/line. Setting the board, an Express 125, to output DMX signal at its lowest speed helped the problem a little. The thing that helped the best however was an Iso-Splitter. It helped boost the signal and and allowed the DMX speed to be ran at the fastest setting, not effect other things in the signal path now. If I could, I would have tried terminating it as well to see if it would help, as it had in past experiences.
#### n1ist
##### Well-Known Member
I always terminate DMX, both at the last fixture and unused outputs of my splitter. Anything I build that speaks DMX, I include a termination resistor and switch so at least I don't have to hunt for a terminator for those boxes. Why take a chance, when terminators are cheap to build.
#### JD
##### Well-Known Member
Where are you getting Neutrik connectors for $1, because I would love to get them for 1/5 the price that I normally pay. Just search Ebay for "xlr connectors." Here's one batch of 20 new for$18:
http://cgi.ebay.com/20-XLR-3-Pin-Mi...ryZ32838QQssPageNameZWDVWQQrdZ1QQcmdZViewItem | |
# Assessing Causality from Observational Data using Pearl's Structural Causal Models
Posted on
causality causal-inference structural-causal-models rstats judea-pearl DAG
## Causality
In 20th century statistics classes, it was common to hear the statement: “You can never prove causality.” As a result, researchers published results saying “x is associated with y” as a way of circumventing the issue of causality yet implicitly suggesting that the association is causal. As an example from my former discipline, political science, there was an interest in determining how representative democracy works. Do politicians respond to voters, or do voters just update their policy beliefs to line up with the party they’ve always preferred? It turns out that this is a very difficult question to answer, so political scientists interested in publishing choose their language carefully and pronounce that policy “congruence” exists between voters and politicians. The upshot is that there now exists a scholarly literature on “voter-party congruence,” which tells you exactly nothing about how democracy works but allows democracy researchers to get their papers past peer review.
21st century understandings of causality, however, have evolved away from 20th century fatalism to reframe the question as:
• What assumptions need to be met in order to state that an association is causal?
• Under what conditions are those assumptions met?
• Can these assumptions be met even when we can’t perform randomization?
There are two conceptually different approaches to the problem:
• Donald Rubin’s (elaboration on Jerzy Neyman’s) potential outcomes framework.
• Judea Pearl’s (elaboration on Sewall Wright’s) structural causal models (SCMs).
The former is the dominant approach in applied statistics, but the latter approach can sometimes highlight unexpected results that inform the proper analysis of observational data. Before describing the SCM framework, the next section reviews the potential outcomes framework.
## Potential Outcomes
Take a binary treatment $$D_i \in \{0, 1\}$$. Represent the outcome received by subject i as $$Y_{iD}$$. Then $$Y_{i0}$$ and $$Y_{i1}$$ are the potential outcomes. A subject is either $$Y_{i0}$$ or $$Y_{i1}$$, we don’t observe both. Yet we want to determine:
$$Y_{i1} - Y_{i0}$$
which is the causal effect of the intervention.
Although subjects receive either 0 or 1, but not the other, we may be able to identify the Average Treatment Effect (ATE).
$$\text{ATE} = \mathbb{E}\left[Y_{i1} - Y_{i0}\right]$$
To derive appropriate estimators for the ATE we need to make a few assumptions. Particularly important is that the treatment is independent of potential outcomes, written as:
$$Y_{i0}, Y_{i1} \perp\!\!\!\perp D_i$$
Finding ways to make $$D_i$$ independent is at the heart of the potential outcomes framework. This leads to a few methodologies now commonplace in applied statistics:
1. Randomized experiments by definition make $$D_i$$ independent.
2. Propensity score matching or weighting make the treated and controls look the same on possible confounders so that the only differences must be random error.
3. Regression discontinuity designs where a cut-off on a continuous variable separates treated and control units.
4. Instrumental variables, where compliance is non-random but treatment assignment is random.
5. Longitudinal designs that use fixed effects or first differences to remove unit-level confounders affecting the treatment.
The key assumption is $$Y_{i0}, Y_{i1} \perp\!\!\!\perp D_i$$, termed ignorability. Judea Pearl has criticized how unintuitive the potential outcomes framework makes this assumption. He writes in The Book of Why (2018, pg. 279-280):
“Unfortunately, I have yet to find a single person who can explain what ignorability means in a language spoken by those who need to make this assumption or assess its plausibility in a given problem…If you think this sounds circular, I agree with you!”
Instead, Pearl has spent the last twenty years developing a different orientation that builds off his work in the 1990s on Bayesian networks. (It’s not necessary to know how Bayesian networks work to understand this post, but it does help clarify how his thinking on the problem of causality evolved over several decades.) This orientation is known as structural causal models (SCMs).
# Structural Causal Models
SCMs are graphs with nodes, directed edges, and functions mapping exogenous variables to endogenous ones. Denote $$U$$ as the set of exogenous variables, $$V$$ as the set of endogenous variables, and $$F$$ as the set of functions mapping $$U$$ to $$V$$.
A concrete example is:
$$U = \{X, Y\}$$
$$V = \{Z\}$$
$$F = \{f_z\}$$
where $$f_z$$ is the function mapping $$X$$ and $$Y$$ onto $$Z$$. This definition implies the following graph:
The arrows represent a generic causal relationship only, the actual function mapping $$X$$ and $$Y$$ onto $$Z$$ can be anything we like.
These types of figures should be familiar to anybody who has previously encountered structural equation models (SEMs) in applied statistics. The primary difference is that SEMs are parametric, typically assuming a linear relationship:
$$Z = b_0 + b_1X + b_2Y$$
but SCMs are defined without committing to a particular functional form.
We get around functional forms by talking about the variables in terms of joint probability functions and taking advantage of well-known rules for converting between joint, conditional, and marginal probabilities. Take the following graph:
Any (acyclic) graph has a joint distribution that is defined by multiplying all conditional probabilities, where conditioning is performed on the direct parent. For example, the joint distribution for the variables in the model is
$$P(X, Y, Z) = P(X)P(Y \vert X)P(Z \vert Y)$$
Understanding the conditional probabilities implied by a model will enable us to generate some rules for determining how causal effects can be identified from observational data. These rules provide surprising and important perspectives on how statistical modeling should be approached.
#### Backdoor Paths and Colliders
“You should control for everything you can. That is, after all, why we do regression.” - One of my methodology professors in the early 2000s.
No, you should not control for everything. In fact, depending on the causal model, some variables should explicitly not be controlled for. We’ll start out with when you should control for a non-treatment variable. Take the following graph:
We wish to know the effect of $$X$$ on $$Z$$, but $$Y$$ is a common cause. Let’s say we could intervene in the world to set $$X$$ at a given value. By doing so, we’d be removing the effect of $$Y$$ on $$X$$ and would be left with:
We can identify the causal effect by comparing the world in which we have control with the world in which we do not. In both scenarios, the probability that $$Z$$ takes on a value is conditioned only on $$Y$$ and $$X$$, $$P(Z = z \mid Y, X)$$, and the probability that $$Y$$ takes on a given value is not conditional on anything.
We want to know the effect of $$X$$ on $$Z$$ if we could intervene on $$X$$ and set its value. Pearl introduces the $$do(\cdot)$$ operator to signify setting a variable $$X$$ to a specific value $$x$$.
$$P(Z = z \mid do(X = x))$$
Based on the intervention SCM,
$$P(Z = z \mid do(X = x)) = \sum_z P(Z = z \mid Y = y, X = x)P(Y = y)$$
This is true because $$P(Z = z \mid do(X = x))$$ is what we get after integrating out $$Y$$. But we know from comparing the graphs that $$P(Z = z \mid Y = y, X = x)$$ and $$P(Y = y)$$ are the same in both worlds. Thus, we have all the information we need to calculate a causal effect such as
$$P(Z = z \mid do(X = 1)) - P(Z = z \mid do(X = 0))$$
Take a slightly more complicated model:
There are now two paths from $$X$$ to $$Z$$:
1. $$X \rightarrow Z$$
2. $$X \leftarrow W \rightarrow Y \rightarrow Z$$
These are read from left to right regardless of the direction of the arrows. However, the arrows identify the second path as a backdoor path because there is an arrow leading into $$X$$. Backdoor paths are essential for identifying causal effects because they represent spurious associations. Pearl shows that causal effects can be identified if we can block the backdoor path. We do this by conditioning on any of the variables that lay on the backdoor path, meaning the conditioning set can be any of the following:
1. $$\{W\}$$
2. $$\{Y\}$$
3. $$\{W, Y\}$$
We don’t necessarily have to control for both, though we can. The key is that, by blocking a backdoor path, we remove the spurious association between the outcome and $$X$$. After blocking, we do not necessarily need to control for subsequent variables on the backdoor path.
Now let’s flip the top arrows.
This fundamentally changes the conditioning set, which now only contains $$Y$$. This occurs because $$W$$ is a collider variable, which is defined as a variable that lies along a backdoor path with arrows pointing into it from multiple directions. We would write this backdoor path as
$$X \rightarrow W \leftarrow Y \rightarrow Z$$.
When we write out the path in this manner, we can immediately identify collider variables as those with arrows pointing to the node from both directions. A collider variable blocks a backdoor path. The counter-intuitive result is that conditioning on a collider opens the backdoor path.
To identify the causal effect we need to block all backdoor paths from $$X$$ to $$Z$$. The backdoor criterion can be defined as (Pearl, Glymour, & Powell, 2016, p. 61):
Given an ordered pair of variables $$(X,Z)$$ in a directed acyclic graph $$G$$, a set of variables $$V$$ satisfies the backdoor criterion relative to $$(X,Z)$$ if no node in $$V$$ is a descendant of $$X$$, and $$V$$ blocks every path between $$X$$ and $$Z$$ that contains an arrow into $$X$$.
That is, we identify a set of nodes in $$\{V\}$$ to condition on such that:
1. We block all spurious paths from $$X$$ to $$Z$$.
2. We leave all directed paths from $$X$$ to $$Z$$ unperturbed.
3. We do not inadvertantly create new spurious paths via conditioning on colliders or their descendants.
#### Mediation
Another example is mediation, as in the following figure:
We can get the direct effect of $$X$$ on $$Z$$ if we average over levels of $$M$$, which is the standard approach to mediation. But what if we add a variable as follows?:
Now $$M$$ is a collider, and we know that conditioning on a collider causes problems. Conditioning on $$M$$ opens the path $$X \rightarrow M \leftarrow W \rightarrow Z$$, allowing an indirect effect to interfere with the direct effect. But not conditioning on $$M$$ leaves the indirect path $$X \rightarrow M \rightarrow Z$$ open. How do we deal with this in a manner that allows us to recover the direct effect of $$X$$ on $$Z$$? The answer is that we now intervene on both $$X$$ and $$M$$.
$$P(Z=z \mid do(X = x), do(M = m))$$.
Intervening and conditioning are not the same thing. Conditioning averages over values of $$M$$, intervening sets its value such that there are no longer the arrows $$X \rightarrow M$$ and $$W \rightarrow M$$.
The conditional direct effect is
$$CDE = P(Z=z \mid do(X = x), do(M = m)) - P(Z=z \mid do(X = x^{\prime}), do(M = m))$$
The conditional refers to the fact that the direct effect $$X \rightarrow Z$$ may differ depending on the value to which the mediator is set.
The $$do(\cdot)$$ operator is equivalent to removing an arrow from a graph. Reiterating the model:
There is no path to $$X$$, so $$do(X) = x$$, and the CDE is
$$CDE = P(Z=z \mid X = x, do(M = m)) - P(Z=z \mid X = x^{\prime}, do(M = m))$$.
The last step is to rewrite the $$do(M = m)$$ in terms of the observed world. To block the backdoor path $$M \leftarrow W \rightarrow Z$$ we need to condition on $$W$$. We are left with:
$\begin{eqnarray} CDE = \sum_i \left[P(Z=z \mid X = x, M = m, W = w) - \\ P(Z=z \mid X = x^{\prime}, M = m, W = w)\right]P(W = w) \end{eqnarray}$
There is a general result behind this (Pearl, Glymour, & Jewell, 2016, pg. 77):
The CDE of $$X$$ on $$Z$$ can be identified when a mediation variable $$M$$ is present given:
1. There exists a set $$V_1$$ of variables that blocks all backdoor paths from $$M$$ to $$Z$$.
2. There exists a set $$V_2$$ of variables that blocks all backdoor paths from $$X$$ to $$Z$$ after deleting all arrows entering $$M$$.
The second of these was met automatically given the lack of parents for $$X$$.
These general rules make it possible to identify direct causal effects in contexts that were previously intractable, even if the researchers did not realize they were dealing with an intractable problem.
#### The daggity R Package
These models are all very simple, but graphs can be far more complex. Consider the following (adapted from Morgan & Winship, 2015, pg. 135):
A general approach to modeling these diagrams is to employ a tool called d-separation, defined as follows (Pearl, Glymour, & Powell, 2016, p. 47):
A path $$p$$ is blocked by a set of nodes $$N$$ iif:
1. $$p$$ contains a chain of nodes $$A \rightarrow B \rightarrow C$$ or fork $$A \leftarrow B \rightarrow C$$ such that the middle node $$B$$ is conditioned on, or
2. $$p$$ contains a collider $$A \rightarrow B \leftarrow C$$ such that the collision node $$B$$ is not conditioned on, nor are any descendents of $$B$$ conditioned on.
Fortunately, there is software that can help us algorithmically determine which variables are d-separated. The software (and R package) is called dagitty. To use the package, we start by declaring the SCM:
g <- dagitty('dag {
S [pos="0,0"]
T [pos="1,2"]
U [pos="0,4"]
V [pos="2,1"]
W [pos="2,3"]
X [pos="3,0"]
Y [pos="3,4"]
Z [pos="4,2"]
S -> X -> Z
S -> T
T -> V -> X -> Z
T -> V -> Z
T -> W -> Z
U -> Y -> Z
U -> T
Y -> Z
}')
The plot method confirms that it looks good.
plot(g)
We can now make some queries on the graph. For example, what are the paths from $$X$$ to $$Z$$?
paths(g, "X", "Z")
## $paths ## [1] "X -> Z" "X <- S -> T -> V -> Z" ## [3] "X <- S -> T -> W -> Z" "X <- S -> T <- U -> Y -> Z" ## [5] "X <- V -> Z" "X <- V <- T -> W -> Z" ## [7] "X <- V <- T <- U -> Y -> Z" ## ##$open
## [1] TRUE TRUE TRUE FALSE TRUE TRUE TRUE
We can quickly see that there are seven paths, six of which are backdoor paths, linking $$X$$ to $$Z$$. Only the fourth is blocked by the collider at $$T$$.
We wish to predict $$Z$$ on the basis of $$X$$. Using the rules for $$d$$-separation to remove spurious dependencies, what set of variables can we condition on to get the true causal effect of $$X$$ on $$Z$$?
adjustmentSets(g, "X", "Z", type = "all") %>%
head(15)
## { S, V }
## { S, T, V }
## { S, U, V }
## { T, U, V }
## { S, T, U, V }
## { S, V, W }
## { S, T, V, W }
## { U, V, W }
## { S, U, V, W }
## { T, U, V, W }
## { S, T, U, V, W }
## { S, V, Y }
## { T, V, Y }
## { S, T, V, Y }
## { S, U, V, Y }
Notice that $$T$$ is in some of these sets. If we unblock the path $$X \leftarrow S \rightarrow T \leftarrow U \rightarrow Y \rightarrow Z$$, we need to reblock it by conditioning on another variable such as $$U$$ or $$Y$$.
This is a lot of options. Can we get something simpler?
adjustmentSets(g, "X", "Z", type = "minimal")
## { V, W, Y }
## { T, V, Y }
## { U, V, W }
## { T, U, V }
## { S, V }
Note two important points.
1. We don’t have to condition on all possible causes of $$Y$$.
2. There are some combinations of variables we should not use as adjustors.
We’ll illustrate by generating some data consistent with the model. The SEM package lavaan makes generating data for simultaneous equations relatively easy.
lavaan_model <- "Z ~ .8*X + .6*V + .6*W + .6*Y
X ~ .5*S + .5*V
Y ~ .5*U
V ~ .5*T
W ~ .5*T
T ~ .5*S + .5*U"
set.seed(12345)
g_tbl <- simulateData(lavaan_model, sample.nobs=1000)
This creates a data.frame with 1000 observations. The effects of each exogenous variable on the endogenous variables are set to be non-zero. The code specifies a traditional SEM, meaning that the set of functions $$F$$ in the SCM are all linear.
We can verify that our data conform to the model by first specifying the model without the known coefficients.
lavaan_model <- "Z ~ X + V + W + Y
X ~ S + V
Y ~ U
V ~ T
W ~ T
T ~ S + U"
Next, fit the model using traditional SEM.
lavaan_fit <- sem(lavaan_model, data = g_tbl)
Now look at the coefficients and verify that the path $$X \rightarrow Z$$ has a coefficient of approximately 0.8.
parameterEstimates(lavaan_fit)
## lhs op rhs est se z pvalue ci.lower ci.upper
## 1 Z ~ X 0.801 0.028 28.992 0 0.747 0.856
## 2 Z ~ V 0.602 0.032 18.661 0 0.538 0.665
## 3 Z ~ W 0.574 0.029 19.461 0 0.516 0.632
## 4 Z ~ Y 0.582 0.030 19.568 0 0.523 0.640
## 5 X ~ S 0.556 0.033 16.973 0 0.492 0.620
## 6 X ~ V 0.485 0.028 17.251 0 0.430 0.540
## 7 Y ~ U 0.489 0.031 15.980 0 0.429 0.549
## 8 V ~ T 0.505 0.027 18.699 0 0.452 0.558
## 9 W ~ T 0.489 0.026 18.971 0 0.439 0.540
## 10 T ~ S 0.510 0.030 16.987 0 0.452 0.569
## 11 T ~ U 0.488 0.030 16.038 0 0.428 0.548
## 12 Z ~~ Z 1.028 0.046 22.361 0 0.938 1.118
## 13 X ~~ X 1.052 0.047 22.361 0 0.959 1.144
## 14 Y ~~ Y 0.939 0.042 22.361 0 0.857 1.022
## 15 V ~~ V 1.035 0.046 22.361 0 0.944 1.126
## 16 W ~~ W 0.945 0.042 22.361 0 0.862 1.028
## 17 T ~~ T 0.928 0.041 22.361 0 0.846 1.009
## 18 S ~~ S 1.028 0.000 NA NA 1.028 1.028
## 19 S ~~ U -0.030 0.000 NA NA -0.030 -0.030
## 20 U ~~ U 1.002 0.000 NA NA 1.002 1.002
We want to estimate the effect of $$X$$ on $$Z$$. What do we get without adjustment?
lm(Z ~ X, data = g_tbl) %>%
tidy()
## # A tibble: 2 x 5
## term estimate std.error statistic p.value
## <chr> <dbl> <dbl> <dbl> <dbl>
## 1 (Intercept) 0.0939 0.0500 1.88 6.06e- 2
## 2 X 1.19 0.0369 32.2 1.26e-156
That’s NQR, the effect should be .8. What do we get if we also adjust on the collider $$T$$?
lm(Z ~ X + T, data = g_tbl) %>%
tidy()
## # A tibble: 3 x 5
## term estimate std.error statistic p.value
## <chr> <dbl> <dbl> <dbl> <dbl>
## 1 (Intercept) 0.0841 0.0456 1.85 6.52e- 2
## 2 X 0.974 0.0369 26.4 6.78e-117
## 3 T 0.597 0.0419 14.2 4.68e- 42
What if we condition using the sets dagitty told us to use?
Model 1:
lm(Z ~ X + S + V, data = g_tbl) %>%
tidy()
## # A tibble: 4 x 5
## term estimate std.error statistic p.value
## <chr> <dbl> <dbl> <dbl> <dbl>
## 1 (Intercept) 0.0675 0.0433 1.56 1.20e- 1
## 2 X 0.807 0.0422 19.1 1.31e-69
## 3 S 0.0997 0.0495 2.01 4.44e- 2
## 4 V 0.776 0.0427 18.2 5.03e-64
Model 2:
lm(Z ~ X + V + W + Y, data = g_tbl) %>%
tidy()
## # A tibble: 5 x 5
## term estimate std.error statistic p.value
## <chr> <dbl> <dbl> <dbl> <dbl>
## 1 (Intercept) 0.0227 0.0322 0.704 4.81e- 1
## 2 X 0.801 0.0278 28.8 3.81e-133
## 3 V 0.602 0.0322 18.7 4.64e- 67
## 4 W 0.574 0.0298 19.3 1.25e- 70
## 5 Y 0.582 0.0298 19.5 5.87e- 72
Model 3:
lm(Z ~ X + U + V + W, data = g_tbl) %>%
tidy()
## # A tibble: 5 x 5
## term estimate std.error statistic p.value
## <chr> <dbl> <dbl> <dbl> <dbl>
## 1 (Intercept) 0.0568 0.0364 1.56 1.19e- 1
## 2 X 0.800 0.0316 25.3 1.49e-109
## 3 U 0.337 0.0384 8.77 7.37e- 18
## 4 V 0.585 0.0371 15.8 3.62e- 50
## 5 W 0.557 0.0344 16.2 1.67e- 52
Model 4:
lm(Z ~ X + T + U + V, data = g_tbl) %>%
tidy()
## # A tibble: 5 x 5
## term estimate std.error statistic p.value
## <chr> <dbl> <dbl> <dbl> <dbl>
## 1 (Intercept) 0.0657 0.0403 1.63 1.03e- 1
## 2 X 0.820 0.0356 23.0 1.63e-94
## 3 T 0.248 0.0427 5.80 9.10e- 9
## 4 U 0.384 0.0442 8.68 1.60e-17
## 5 V 0.580 0.0430 13.5 3.48e-38
We get much closer to the true causal effect estimate whenever we use the conditioning sets suggested by daggity.
#### Unobservable or Unmeasurable Variables
Once again, take our model:
Let’s say that we can’t actually observe $$W$$ or $$Y$$. An old-school regressionista would say we are SOL. A modern causal-aware practitioner would not.
We can tell dagitty that these variables are unobserved, or latent.
g_unobs <- g
latents(g_unobs) <- c("W", "Y")
Compare the adjustment sets when we observe all variables (the DAG object we called g) with the adjustment sets after we tell daggity we can’t measure $$W$$ or $$Y$$.
adjustmentSets(g, "X", "Z", type = "minimal")
## { V, W, Y }
## { T, V, Y }
## { U, V, W }
## { T, U, V }
## { S, V }
adjustmentSets(g_unobs, "X", "Z", type = "minimal")
## { T, U, V }
## { S, V }
We’re still okay! There is still a set of variables we can control for to recover the causal effect even when some of the variables along the full causal path can’t be measured.
#### SEMs, SCMs, and p-hacking
How do we know our SCM is correct? This raises an important concern.
Pearl writes in the Book of Why that SCMs are unfamiliar to statisticians. Although this may be true in their nonparametric form, linear SEMs have been popular ever since the software LISREL was released by a couple of Swedes in 1972.
However, linear SEMs (SCMs with linear functional forms) have been maligned by many statisticians over the last several decades because they have been so thoroughly abused that it’s become hard to take them seriously.
A typical approach:
• The effect of $$X$$ on $$Z$$ isn’t significant, my dissertation (or publication needed for tenure) is a failure!
• I know, I’ll add a variable $$M_1$$ between $$X$$ and $$Z$$, maybe there’s a mediated effect!
• Damn, no mediated effect. What if I add $$M_2$$ and $$M_3$$ to the model and keep moving around the directed arrows?
• Hey, something is eventually significant!
In other words, these models are rife with $$p$$-hacking.
A careful analysis of SCMs, however, closes off some of the models we may want to try out of desperation. This is because the conditioning we perform should render certain associations to be independent.
Take the model we just analyzed. As it stands, $$W$$ and $$V$$ are dependent because they have a common ancestor:
$$V \leftarrow T \rightarrow W$$
By the definition of d-separation, we know that conditioning on $$T$$ should render $$W$$ independent of $$V$$. That is,
$$P(W = w \mid V = v) = P(W = w)$$.
We can test this with a regression of $$W$$ on $$V$$ and $$T$$. If the model is correct, the association between $$W$$ and $$V$$ should be zero.
To demonstrate, start by showing an association exists between $$W$$ and $$V$$.
lm(W ~ V, data = g_tbl) %>%
tidy() %>%
mutate_if(is.numeric, funs(round(., 3)))
## Warning: funs() is soft deprecated as of dplyr 0.8.0
##
## # Before:
## funs(name = f(.)
##
## # After:
## list(name = ~f(.))
## This warning is displayed once per session.
## # A tibble: 2 x 5
## term estimate std.error statistic p.value
## <chr> <dbl> <dbl> <dbl> <dbl>
## 1 (Intercept) 0.017 0.035 0.485 0.628
## 2 V 0.258 0.029 8.81 0
If our model is correct, controlling for $$T$$ should render this association statistically indistinguishable from zero. Does it?
lm(W ~ V + T, data = g_tbl) %>%
tidy() %>%
mutate_if(is.numeric, funs(round(., 3)))
## # A tibble: 3 x 5
## term estimate std.error statistic p.value
## <chr> <dbl> <dbl> <dbl> <dbl>
## 1 (Intercept) 0.016 0.031 0.522 0.602
## 2 V 0.009 0.03 0.288 0.774
## 3 T 0.485 0.03 16.2 0
In fact, we can get all conditional independencies implied by the model.
impliedConditionalIndependencies(g) %>%
head(20)
## S _||_ U
## S _||_ V | T
## S _||_ W | T
## S _||_ Y
## S _||_ Z | V, W, X, Y
## S _||_ Z | T, V, X, Y
## S _||_ Z | U, V, W, X
## S _||_ Z | T, U, V, X
## T _||_ X | S, V
## T _||_ Y | U
## T _||_ Z | V, W, X, Y
## T _||_ Z | U, V, W, X
## T _||_ Z | S, V, W, Y
## T _||_ Z | S, U, V, W
## U _||_ V | T
## U _||_ W | T
## U _||_ X | S, V
## U _||_ X | S, T
## U _||_ Z | V, W, X, Y
## U _||_ Z | S, V, W, Y
We generated our data to intentionally be consistent with the model, so testing these conditional independencies will confirm them. When we don’t know if the model is correct, however, we can generate the conditional independencies and check each of them. If they are not correct, our model is wrong.
When $$\{V\}$$ is large, the possible set of connections may not all be clearly dictated by theory, and the number of possible combinations of arrows is too large to test via a grid-search. Familiarity with Pearl’s earlier work on Bayesian networks is helpful here, since it led to algorithms for more efficient search rules. These algorithms are nonetheless still very computationally intensive, and there has been very little work testing out their utility in the social sciences.
### Counterfactuals
Pearl also argues that SCMs, and their implied probabilities, can be used to address seemingly intractable questions. Specifically, they can address unit-specific counterfactuals. Whereas interventions, and determining ATEs, can be performed by averaging across a group of cases, specific counterfactuals relate to an individual case.
At first, counterfactuals seem unidentifiable. Think of a court case where there is an assertion that taking a drug caused a person’s death. There are two (potential) outcomes:
1. $$Z_0$$, the outcome when the person did not take the drug, i.e. $$X = 0$$.
2. $$Z_1$$, the outcome when the person did take the drug, i.e. $$X = 1$$.
The person took the drug and died, so we know $$Z_1 = 1$$ ($$1$$ = death, $$0$$ = no death). The defense would like to know $$P(Z_0 \mid X = 1, Y = 1)$$. But this seems like nonesense. We want to know the probability of an event under one hypothetical world while conditioning on another world, the one we observed.
The solution relies on establishing an SCM that explicitly includes error terms.
Each of the $$U \in \{UX, UY, UZ\}$$ is an individual-specific value. After fitting the model using the observed data, we can get these values for a specific person. We then alter the graph by setting the value of $$X$$ or $$Y$$ to the counterfactual value and solve for $$Z$$ using the error term value identified by the full regression.
In the most simplistic case, we are assuming that each person’s error term is determined exactly by the equations. Pearl’s texts also discuss working with stochastic errors to come up with bounds on possible counterfactuals.
### SCMs and ML
Pearl (2018) makes the audacious claim that current machine learning models cannot ever assert causality because they cannot deal with interventions, let alone counterfactuals.
A machine learning model takes a set of features $$V = \{v_1, v_2, \ldots, v_k\}$$ and finds a function $$f_z$$ mapping this set onto an outcome $$Z$$.
Using variations on statistical modeling, this amounts to modeling the joint distribution of all variables. However, using Pearl’s $$do(\cdot)$$ operator, a joint distribution changes when we intervene on a variable.
For example, if we are given a data set without knowing where it came from, we can fit a regression model using the joint distribution. Yet nothing about the join distribution tells us whether $$X$$ is randomized or not.
Causality requires knowing which conditional probabilities are invariant to changes in the structural model. ML is blind to this. ML as currently practiced throws a bunch of stuff into a blender and sees what comes out, akin to 20th century regression modeling that taught us to “control for everything.” This may not matter when we want to predict the presence of a dog, cat, or hot dog in a picture. It will matter if we want to:
1. Tell policymakers whether or not to increase the minimum wage.
2. Determine if admissions criteria at a university are racially biased.
3. Find a defendant guilty in a criminal trial.
4. Determine a counterfactual for an individual for whom existing data are not representative.
ML models are akin to the underwear gnome problem:
1. Features.
2. $$\dots$$
3. Prediction!
The black box hides the answer we need if we want to develop effective rules that lead to socially desirable outcomes.
### Limitations of the Pearlian Weltanschauung
At the same time, Pearl’s dismissal of non-SCM approaches to modeling (potential outcomes, ML) are based on finding specific cases where these approaches fail, but he does not give a sense as to how often they fail. Take, for example, our apparently complicated model:
We can identify the canonical set of adjuster variables, which will be valid if any valid set exists.
adjustmentSets(g, "X", "Z", type = "canonical")
## { S, T, U, V, W, Y }
We see that we can in fact “control for everything”.
lm(Z ~ ., data = g_tbl) %>%
tidy
## # A tibble: 8 x 5
## term estimate std.error statistic p.value
## <chr> <dbl> <dbl> <dbl> <dbl>
## 1 (Intercept) 0.0236 0.0322 0.732 4.64e- 1
## 2 X 0.815 0.0314 26.0 4.54e-114
## 3 Y 0.562 0.0332 16.9 1.39e- 56
## 4 V 0.604 0.0349 17.3 7.55e- 59
## 5 W 0.583 0.0332 17.5 4.11e- 60
## 6 T -0.0466 0.0395 -1.18 2.38e- 1
## 7 S -0.0106 0.0398 -0.266 7.90e- 1
## 8 U 0.0692 0.0395 1.75 8.05e- 2
We didn’t do too bad. The problem, of course, is that there are SCMs that do not have all IVs or features as a proper adjustment set. How bad our conclusions are will depend on how well our representation of reality is.
Indeed, reading Pearl’s (co-authored) introductory textbook Causality: a Primer, one can’t help but be struck by how many of the estimators look just like the types of formulas that Rubin and colleagues have developed using the potential frameworks approach. Is a complete re-orientation of applied statistics really going to result in different (and, presumably better) estimators? The jury is still out.
Finally, not all SCMs are identified, especially when stepping away from the world of linearity. Reverse causation plagues observational studies of social behavior, and unless you are satisfied with declaring “congruence”, not even SCMs may save you. At best, given complicated nonlinear and nonrecursive systems of equations, checking the model-implied conditional probability will rule out some models, but certainly not all candidates. | |
# Lawvere Theories
In many previous posts, we have discussed connections between monads and universal algebra. So far, we have seen that given a presentation $(\Sigma,E)$ of an algebraic theory in terms of set of operations $\Sigma$, and equations $E$, there is a corresponding monad $\mathbb{T}$ such that:
• The Eilenberg-Moore algebras of the monad are the algebras described by the chosen presentation.
• The Kleisli category is the full subcategory of free algebras.
Many other aspects of monad theory can then be stood from this algebraic perspective.
Every monad arising from a presentation in terms of operations and equations is finitary. This is a technical condition saying that certain colimits are preserved by the monad endofunctor, but we will not dwell on the precise details. Intuitively, the idea is that the functor is completely defined by its action on finite objects. The key point is that there is a correspondence between algebraic presentations and finitary monads, for which the categories of algebras coincide. It is essential that the categories of models agree for this relationship to be algebraically meaningful. Note this is not a bijection, as a given finitary monad will have infinitely many potential presentations in terms of operations and equations.
In this post, we will see a third perspective on describing algebraic structures, Lawvere theories. These sit somewhere between conventional universal algebra and monads in their level of abstraction. As with any mathematical phenomenon, each new perspective deepens our understanding, and emphasises new features and intuitions.
## The Idea of Lawvere Theories
To be concrete, lets consider what we need to do to describe a monoid. The general situation for different algebraic structures will follow in the same way.
Typically, we first pick out an underlying set $M$, and two functions:
1. A zero argument function, or constant, for the unit element. We can identity this as a function $1 \xrightarrow{z} M$.
2. A two argument function, or binary operation, for the monoid multiplication. We can identify this with a function $M \times M \xrightarrow{m} M$.
Of course, once we have these functions, we can build further ones by combining them together. For example
$M \times 1 \xrightarrow{m \circ (z \times \mathsf{id}_M)} M$
Which informally we can think of as corresponding to a term $m(z,x)$. Of course, by one of the monoid unit axioms, we require that this composite is equal to
$M \times 1 \xrightarrow{\pi_1} M$
which we can think of as corresponding to a variable $x$. So in this way of describing a monoid, we:
1. Pick out some basic operations, which are maps of the form $X^n \rightarrow X$.
2. Build lots of other operations by composing the basic operations together, using category composition, and the properties of products.
3. Require that certain of the operations are equal.
We note some issues with this naive plan. Firstly, where did the basic operations come from, and why do we distinguish them from the other operations? This bias towards certain operations seems categorically unnatural, and raises the ugly question of whether any further mathematical results we derive might be dependent on our choice of presentation. Secondly, a small but annoying detail. It was a pain that $M \times 1$ was only isomorphic, and not equal to $M$ when we built our composite terms. If we build up more complicated composites involving higher Cartesian powers, we will spend a lot of time bookkeeping how to bracket expressions like:
$M \times ((M \times M) \times M)$
for no practical benefit. So putting all this together, it would be convenient if we could describe the theory of monoids as a category with:
1. Objects the natural numbers, corresponding to the number of arguments of our operations.
2. (Strict) finite products are given by adding the required arities together $m \times n = m + n$.
3. Intuitively, we consider an operation of arity $m$ to be a morphism of type $m \rightarrow 1$. For example, a binary operation is a morphism $2 \rightarrow 1$ and a constant is a morphism $0 \rightarrow 1$. Notice that, unlike the common convention, in this case $0$ is the terminal object.
4. A family of $n$ such operations is a morphism of type $m \rightarrow n$. This is automatic from universal property of the assumed strict finite products.
5. A diagram commutes in our category if and only if the corresponding operations should be equal in the theory of monoids.
In fact, it is technical important to add a little bit more structure. If we consider the category of sets, and take the full subcategory with objects
$\{ \}, \{ 0 \}, \{ 0, 1 \}, \ldots$
this category clearly has finite coproducts, given by:
$\{ 0,\ldots, m \} + \{ 0, \ldots, n \} = \{ 0, \ldots, m + n \}$
If we identify the objects of this category with the natural numbers, via:
$n = \{ m \in \mathbb{N} \mid m < n \}$
this yields a category with finite coproducts $\aleph_0$, and therefore a category with finite products $\aleph^{op}_0$.
Moving away from the special case of monoids, we then define a Lawvere theory to be a category $\mathcal{L}$ with strict finite products, with an identity on objects strict finite product preserving functor $I : \aleph^{op}_0 \rightarrow \mathcal{L}$. The category $\mathcal{L}$ can be interpreted as encoding the operations and equations of the theory. We shall return to the interpretation of the inclusion $I$ later. A morphism of Lawvere theories $H : (\mathcal{L},I) \rightarrow (\mathcal{L}',I')$ is a finite product preserving functor $H : \mathcal{L} \rightarrow \mathcal{L}'$ such that $I' \circ H = I$.
What we really wanted to do was describe a class of algebraic structures, using our theory. A model of a Lawvere theory in a category with finite products $\mathcal{C}$ is a finite product preserving functor
$A : \mathcal{L} \rightarrow \mathcal{C}$
and a homorphism of models is a natural transformation between them. For conventional algebraic structures, we take $\mathcal{C} = \mathsf{Set}$. Note we don’t require that the models interact appropriately with the product structure, as in fact this is automatic. The category of models of a Lawvere theory and their homomorphisms is denoted
$\mathsf{Mod}(\mathcal{L}, \mathcal{C})$
This approach of identifying structures with functors from a suitably structured category is known as functorial semantics. Notice the difference with monads as a categorical tool for describing algebraic structures. The Eilenberg-Moore category of a monad captures algebraic structures on a fixed base category, whereas we can straightforwardly considering models of Lawvere theories in any category with finite products. There is a forgetful functor given by evaluation at the object $1$:
$\mathsf{Mod}(\mathcal{L},\mathcal{C}) \rightarrow \mathcal{C}$
In many cases, for example if $\mathcal{C}$ is locally presentable, this functor will have a left adjoint, and will therefore induce a monad on $\mathcal{C}$. So a single Lawvere theory induces monads on many different base categories.
If we have two Lawvere theories such that we have equivalence:
$\mathsf{Mod}(\mathcal{L}, \mathsf{Set}) \simeq \mathsf{Mod}(\mathcal{L}', \mathsf{Set})$
then there is an isomorphism $\mathcal{L} \cong \mathcal{L}'$ in the category of Lawvere theories. So there is a very tight correspondence between Lawvere theories and the category of models that they describe.
## How to build a Lawvere – take one
So how do we build a Lawvere theory? One approach is to start with a presentation $(\Sigma, E)$. We then form a category with:
1. Object the natural numbers
2. Morphisms $n \rightarrow m$ are m-tuples of terms in $n$ variables, quotiented by provable equality in equational logic, with respect to the equations of our presentation.
3. The product projection onto the ith component of a product corresponds to the term for the ith variable.
4. Identity morphisms are (necessarily) the tuples of projection morphisms.
5. Composition is given by substitution of terms.
6. The inclusion morphism $I : \aleph^{op}_0 \rightarrow \mathcal{L}$ picks out the projection morphism in $\mathcal{L}$.
So we see the intuition for the inclusion $I$ is that it picks out the trivial terms corresponding to bare variables within the broader collection of operations.
Example: We can build a Lawvere theory for monoids from the usual presentation in terms of a unit and multiplication operation, and unitality and associativity axioms. The category $\mathsf{Mod}(\mathcal{L}, \mathsf{Set})$ is then equivalent to the usual category of monoids.
Note is important that we do not require models of a Lawvere theory strictly preserves product structure. For the usual products in $\mathsf{Set}$, if we did this, the category of models for the Lawvere theory of monoids would be empty!
## How to build a Lawvere Theory – take two
Instead of starting with a presentation in terms of operations and equations, what if we were given a finitary monad? Can we construct a Lawvere theory directly from this data?
The structure of a Lawvere theory in terms of equivalence classes of terms composed by substitution sounds suspiciously like the algebraic perspective on the Kleisli category of a monad. In fact, the two are very similar, up to some cosmetic details. For a finitary $\mathsf{Set}$-monad $\mathbb{T}$, we can construct a Lawvere theory $(\mathcal{L},I)$ as follows:
1. Form the Kleisli category $\mathsf{Set}_{\mathbb{T}}$.
2. Restrict to the full subcategory of objects $\{\}, \{ 0 \}, \{ 0, 1 \}, \ldots$, and rename the objects to natural numbers in a similar way to the section above.
3. Take the opposite category.
Tersely, we say the required Lawvere theory is the opposite of (the skeleton of) the full subcategory of finitely generated free algebras. As we would hope, we have an equivalence between $\mathsf{Set}^{\mathbb{T}}$ and $\mathsf{Mod}(\mathcal{L}, \mathsf{Set})$, so the categories of models coincide.
Given a Lawvere theory $(\mathcal{L},I)$, we would like to go in the other direction, and construct a corresponding finitary monad. This can be done by taking a coend:
$\mathbb{T}(X) = \int^{n \in \aleph_0} \mathcal{L}(n, 1) \times X^{n}$
If this is unfamilar, the idea is that we can construct the required monad as a certain categorical colimit. Intuitively, we form the collection of equivalence classes of terms, identifying product structure where appropriate.
Again, the categories of Eilenberg-Moore algebras and models of the Lawvere theory in $\mathsf{Set}$ agree up to equivalence.
In fact, the connection between Lawvere theories and finitary monads is about as strong as we might hope. The category of Lawvere theories is equivalent to the category of finitary monads on $\mathsf{Set}$. The presence of the inclusion morphism $I$ in the definition of a Lawvere theory ensures that this equivalence goes through correctly.
## Conclusion
Lawvere theories give a presentation independent formulation of algebra. They have several strengths:
1. They are relatively close in form to concepts from conventional universal algebra. In particular, they can be seen as an abstraction of the notion of clone.
2. Unlike monads, a single object can describe models taken in different categories. For example, we can take models of the Lawvere theory of monoids in sets, posets, topological spaces and so on.
3. Certain operations for combining theories are more natural from the perspective of Lawvere theories, rather than that of finitary monads.
4. They are algebraic by construction. By comparison, monads include structures such as the continuation monad, which don’t have a clear interpretation in terms of universal algebra.
There are many generalisations of Lawvere theories, for example to incorporate operations of larger aritites, or enrichments of the base category. These typically come with a correspondence to a natural class of monads, providing a different perspective to work with. The functorial semantics of Lawvere theories also suggests other structures, such as PROs and PROPs, which move us further away from convention algebra to structures potentially having multiple outputs as well as inputs.
One intriguing question is what is the equivalent of Lawvere theories for comonads. At this point there does not seem to be a satisfactory answer, as highlighted in the paper “Category Theoretic Understandings of Universal Algebra and its dual: monads and Lawvere theories, comonads and ?” This example highlight how comonad theory cannot always be seen as routinely “flipping monad theory upside down”, and sometimes a routine application of duality isn’t enough.
Further reading: An excellent source for background is the book “Algebraic Theories” by Adamek, Rosicky and Vitale. Also useful for background is Hyland and Power’s “The Category Theoretic Understanding of Universal Algebra: Lawvere Theories and Monads”. Power has developed much of the theory generalising Lawvere theories in various ways, and applying them in computer science. | |
How Do You Estimate a Derivative at a Point vs Compute a Derivative at a Point from the Limit?
We are now in the real meat of calculus….rates and derivatives. Essentially, rates are simply slopes. Depending on the field you are working in or the representation of the data or function, it might have a different terminology. Keeping all of these straight is the biggest challenge in the class. Let’s try to put what we know into a table.
Common Description Graphical Interpretation Mathematical Terminology
Average Rate of Change of f(x) from x=a to x=b Slope between (a,f(a)) and (b,f(b)) Slope of the Secant Line from x=a to x=b
Instantaneous Rate of Change of f(x) at x=a Slope between (a,f(a)) and (a+h,f(a+h)) where h is very small Slope of the Tangent Line on f(x) at x=a
If it feels like we use slope to calculate everything…you are right. The only difference is where the two points are located. If they are located fairly far apart, then we are calculating an average rate of change. If they are located an infinitesimal amount apart, then we are calculating the instantaneous rate of change.
But here is where it gets hazy…if the representation of the function does not allow us to pick the points infinitely close together, we may approximate the instantaneous rate of change with an average rate of change between two points that are as close as the representation allows. I think this point is bothering many of you. It occurs most often when we try to find the instantaneous rate from a data table. In that case, the points are where they are at and we can pick pairs that are closer and closer together. We pick them as close as we can near the place we want the instantaneous rate of change and say we are estimating the instantaneous rate of change.
If we are given a graph, we can draw a secant line between the points to calculate the average rate of change. The only estimating done in this case is estimating where the points are on the graph. Our eyes are only so good in reading off the values. For the instantaneous rate of change, we draw the tangent line where we want the rate and eyeball its slope. Again, since we are making educated guesses about the slopes, the numbers are estimates based on our ability to read the slope.
If we are given the function’s formula, we can calculate the average rate or the instantaneous rate exactly. For the average rate of change of f(x) over x = a to x = b, we calculate ${{f(b) - f(a)} \over {b - a}}$.
For the instantaneous rate of change, we use the function’s formula to calculate the limit $\mathop {lim }\limits_{h \to 0} {{f(a + h) - f(a)} \over h}$.
The estimating comes from the fact that we may not be able to find two points infinitesimally close or the fact that we cannot calculate the slope perfectly from how the function is given to us.
Another cautionary note…different disciplines call the derivative by different names. Mathematicians call it the derivative of course, but in physics it might be called the instantaneous rate. In finance and economics, it will be called the marginal function. And in other fields you’ll here other terms.
How Do You Find the Average Rate of Change from a Table?
One of the problems on the homework gave you three points on a line graph, (1905, 1024), (1955, 240), (2005, 1141). In these ordered pairs, the x value is the year and the y value is the number of immigrants (in thousands) to a large country.
1. Find the average rate of change in immigration from 1905 to 1955 in immigrants per year.
2. Find the average rate of change in immigration from 1955 to 2005 in immigrants per year.
3. Find the average rate of change in immigration from 1905 to 2005 in immigrants per year.
This problem illustrates the two ways that you can work in the “thousands” in the data to give immigrants per year instead of thousands of immigrants per year.
How Do You Find the Average Rate of Change From a Function?
Problem 1 Find the average rate of change of $\displaystyle f(x)=ln (x)$ over [2, 4] to four decimal places.
Problem 2 Find the average rate of change of $\displaystyle f(x)= {e}^{x}$ over [1, 3] to four decimal places.
When you calculate the rate to four decimal places, you should write the numbers in the quotient to FIVE decimal places to make sure there are no rounding errors.
How Do You Find the Average Rate of Change From an Econ Formula?
Here are several examples where the average rate of change is calculated from some type of economics formula.
Problem 1 The demand for a particular product is given by
$\displaystyle D(p)=-2{{p}^{2}}-2p+400\quad \text{items}$
where p is the unit price in dollars.
a. Find the average rate of change of demand with respect to price between a price of 5 dollars and 7 dollars.
b. Find the instantaneous rate of change of demand at a price of 5 dollars.
(Sorry for the shaky cam…too much caffeine!)
The average rate tells us that for each increase in price of 1 dollars between 5 dollars and 7 dollars, the demand for the product drops by 26 items. The instantaneous rate of change tells us that at a price of 5 dollars, the demand is dropping by 22 items per dollar.
Problem 2 The demand for a particular product is given by
$\displaystyle D(p)=-4{{p}^{2}}-4p+700\quad \text{items}$
where p is the unit price in dollars.
a. Find the average rate of change of demand with respect to price between a price of 5 dollars and 7 dollars.
b. Find the instantaneous rate of change of demand at a price of 5 dollars.
Problem 3 The profit (in thousands of dollars) for selling x hundred units of compressors is
$\displaystyle P(x)=-4{{x}^{2}}+160x-1000$
a. Find the average rate of change of profit with respect to compressors from x = 10 to x = 11.
b. Find the exact profit from the 1001st compressor.
c. Find the instantaneous rate of change of profit with respect to compressors at x = 10.
Problem 4 The profit (in thousands of dollars) for selling x hundred units of graphics displays is
$\displaystyle P(x)=-5{{x}^{2}}+80x-100$
a. Find the average rate of change of profit with respect to displays from x = 10 to x = 11.
b. Find the exact profit from the 1001st display.
c. Find the instantaneous rate of change of profit with respect to displays at x = 10.
In these last two problems, pay careful attention to the units on the rates. They are all in thousands of dollars per hundred units. This simplifies to tens of dollars per unit since one thousand divided by one hundred is ten.
Also note that to find the profit from the 1001st item, we need to find the profit at a production level of 1001 and subtract the profit at a production level of 1000. This quantity is called the marginal profit at a production level of 1000. As noted in the text, it is approximately equal to the instantaneous rate of change at a production level of 1000. | |
# Math Help - Tangent Lines Passing Through Origin
1. ## Tangent Lines Passing Through Origin
At how many points on the curve y=4x^5-3x^4+15x^2+6 will the line tangent to the curve pass through the origin.
2. It depends on how many different roots $x_0 \in \mathbb{R}$ the equation $f(x) = 0$ has.
Observe that between any 2 different roots $x_1,x_2$ that $f(x)$ attains a maximum/minimum in the interval $(x_1,x_2)$.
And there exists a $a_0$ with $x_1 where $f'(a_0)(x-a_0) +f(a_0)$ the tangent at $(a_0,f(a_0))$ passes through the origin.
But we might just as well calculate how many maxima/minima f(x) attains:
Thus if you find all different "real " roots of $f'(x) = 20x^4-12x^3+30x = x(20x^3-12x^2+30) = 0$ you're done.
Edit: You don't even have to find the roots of $f'(x)= 0$ explicitly. You can use the intermediate value theorem to decide how many roots in $\mathbb{R}$ this equation has.
3. If you continue by using $\frac{f(x)}{x}$
then you can find the minimum value(s) of this,
that will give you the slope of the tangent(s) that passes through the origin,
allowing for x being positive or negative.
4. In blue is f(x) and in pink is $\frac{f(x)}{x}$
the tangent slope gives the min value for $\frac{f(x)}{x}$
since if another x is chosen to the right of the origin, the ratio will be greater. | |
×
# How do you find the LCM of 7 and 9?
Dec 15, 2016
63
#### Explanation:
Since the only number that $7$ and $9$ go into evenly is $63$ we must consider this simple step:
$7 \cdot 9 = 63$
A least common multiple is the smallest unit that both numbers can evenly be divided by. Since you can evenly divide $63$ by both $7$ and $9$ the following is true:
$63$ is the least common multiple of $7$ and $9$.
Aug 29, 2017
$L C M = 7 \times 9 = 63$
#### Explanation:
$7 \mathmr{and} 9$ have no common factor (apart from $1$).
Therefore the $L C M$ must contain the whole $7$ and the whole $9$.
$L C M = 7 \times 9 = 63$
This is clear if we write $7 \mathmr{and} 9$ as the product of the prime factors:
$\text{ } 7 = 7$
$\underline{\text{ "9= " } 3 \times 3}$
$L C M = 7 \times 3 \times 3 = 63$ | |
I'm using LaTeX to write up an assignment which comprises a number of questions. I'd like to break up the questions from my answers by placing the section headings inside a shaded box. I'm relatively new to LaTeX.
So far, from searching around and reading quite a few examples, I've managed to knock together something of a partial solution (attempted minimal working example here:)
\documentclass{article}
\usepackage{titlesec}
\usepackage{lipsum}
\usepackage[framemethod=TikZ]{mdframed}
\usepackage{tikz}\usetikzlibrary{shapes.misc}
\tikz[baseline,trim left=3.1cm, trim right=3cm] {
\fill [black!12] (0.5cm,-1ex) rectangle (\textwidth+3.3cm,2.5ex);
\node [
anchor= base east,
rectangle,
minimum height=3.5ex] at (3cm,0ex) {
\textnormal{\textbf{Question \thesection}}
};
}}
\renewcommand{\thesection}{\arabic{section}}
\begin{document}
\section{The text of the first question, which spans over more than %
one line, will be placed here.}
\lipsum[2]
\section{The second question text will be placed here.}
\lipsum[2]
\end{document}
This works great until I have a question that ends up spanning over multiple lines. It's a limitation, I think, of the way I've got tikz drawing the rectangle, but I can't work out how to modify the code to make it span multiple lines.
Does anyone have any suggestions as to how I would go about fixing this?
-
Why not just replace the whole \section command by a new version that typesets what you want using TikZ? Alternatively you could define a \question command which does this. – Yori May 23 '12 at 3:21
I've edited my answer adding some improvements. – Gonzalo Medina May 23 '12 at 17:46
Similar Q&A: tex.stackexchange.com/q/34288/8272 – Nikos Alexandris Oct 14 '12 at 14:21
I propose a different approach: to use the mdframed and amsthm packages to define a theorem-like environment, so you don't have to change the meaning of \section. Something along the following lines:
\documentclass{article}
\usepackage{amsthm}
\usepackage[framemethod=TikZ]{mdframed}
\usepackage{lipsum}
\newlength\Questionwd
\setlength\Questionwd{2.8cm}
\newlength\Innerlsep
\setlength\Innerlsep{3pt}
% definition of the theorem style
\newtheoremstyle{mystyle}
{\topsep}{\topsep}
{\large\itshape%
\parshape 2 0cm \linewidth \dimexpr\Questionwd-3pt\relax\dimexpr\linewidth-\Questionwd+3pt\relax%
}
{}{\bfseries}{}{0em}
{\makebox[\dimexpr\Questionwd-\Innerlsep\relax][l]{%
\thmname{#1}~\thmnumber{#2}\hfill}%
}
% we use the previously defined style for the definition of the new theorem environment
% built with the mdframed package
\theoremstyle{mystyle}
\newmdtheoremenv[
leftmargin=-\Questionwd,
rightmargin=-8pt,
innerleftmargin=\Innerlsep,
linecolor=black!12,
backgroundcolor=black!12,
skipabove=0.65cm,
skipbelow=0.20cm
]{question}{Question}
\begin{document}
\begin{question}
The text of the first question, which spans over more than one line, will be placed here.
\end{question}
\lipsum[2]
\begin{question}
The second question text will be placed here.
\end{question}
\lipsum[2]
\end{document}
-
the space below the shaded question box is a bit larger than that above the box, implying a visual association of the question box with what precedes rather than follows it. decreasing the space below would be a distinct improvement, although increasing the space above would probably be easier. otherwise, very nice. – barbara beeton May 23 '12 at 13:20
@barbarabeeton you're right. I've made some improvements (I hope) including the vertical spacing before and after the environment. – Gonzalo Medina May 23 '12 at 16:55
yes, thank you. that's much better. – barbara beeton May 23 '12 at 17:23 | |
## [ Mythology & Folklore ] Open Question: Why did Trump spend 290 of his 547 days in office, either playing golf or at his private clubs and properties? That's more than half?
Do you think maybe it's because he works 10 times as hard as other presidents, so he needs 10x the holidays?
Should not my tax money be used for something useful? Like a nurse with special needs?
## Prove that two different pairs of natural numbers do not exist with these properties
The problem is to prove the nonexistence or to show that there are two different pairs (up to the permutation) of natural numbers $$(a, b)$$ and $$(c, d)$$ s.t. $$lcm (a, b) = lcm (c, d)$$
$$gcd (a, b) = gcd (c, d)$$ and
$$frac {a + b} {2} = frac {c + d} {2}$$
It is easy to show that if LCM and GCD are the same, two pairs have the same product and the same sum AND the same GCD. I have the intuition that it is impossible to exist two different pairs under these conditions, but it is unclear how to prove it exactly.
## c ++ classes with different properties – right approach?
Let's say I have lessons `NumericalDataSet` with the property `data` of the type `std::vector>` and `LabeledDataSet` with the property `data` of the type `std::vector>>`, All other methods and properties are the same. What is the right approach to deal with this?
Should I make an abstract class `DataSet` to inherit with the two or is it better to use a heterogeneous container like boost :: any?
## discrete mathematics – properties of relationship evidence
I practice some characteristics of relationships and I can not seem to figure out a specific question. It follows
``````Consider the relation R on Z+(positive integers) as: For all m,n belonging to Z+, mRn means m|n.
Is R reflexsive, symmetric or transitive?
Provide a complete proof or counterexample for each property.
You may only use the definition of divides
``````
The definition of divisions according to my special textbook is as follows.
``````let n,d ∈ ℤ+ and d≠0.
n is divisible by d if and only if ∃ k ∈ ℤ such that n = dk
``````
How would I go about doing that? Every help is appreciated. Many thanks
## fa.functional analysis – Basic properties of expectation in non-separable Banach spaces
$$def E { hskip.15ex mathsf {E} hskip.10ex}$$
To let $$B$$ be a (perhaps inseparable) Banach room equipped with the Borel $$sigma$$-Algebra $$mathscr {B} (B)$$, To let $$R: B to mathbb {R}$$ Let be a limited linear operator.
To let $$( Omega, mathcal {F}, P)$$ be a probability space. To let $$F$$ be a $$mathscr {B} (B) | mathcal {F}$$measurable assignment $$Omega to B$$, Suppose that $$E | F | < infty$$,
Question: Is it true without further assumptions that $$E F$$ is well defined, belongs to $$B$$ and
$$E RF = R E F?$$
Note: As a rule (eg in the Ledoux-Talagrand book) the separability of room B is additionally imposed. I wonder if the statement is true without this assumption. What happens, for example, if $$B$$ is only a space of limited measurable functions?
## Proof of Gaussian integral properties
Hello, I work with a model of Gaussian demand. The following should be an easily derived property, but I could not reproduce it myself.
Why is $$f (x)> xF (x)$$ to the
$$F (x) = int_ {x} ^ { infty} dfrac {1} { sqrt {2 pi}} e ^ {- frac {1} {2} t ^ 2} dt$$
## smt-solver – program synthesis based on functional properties
I am fairly new to program synthesis and the use of SMT solvers for this purpose. Given a function GI want to generate all functions f and H such that:
f. g = h. f
Are there any tools and techniques for program synthesis that allow me to search the program area for these functions?
I assume that the search space for general programming languages like Haskell will be extremely large. How should one proceed to curtail the search space: would one have to drastically reduce the search space by defining a small language and performing the search through that language? It would seem that the type constraints of the above equation would also help cropping the search space. Any observations or references to tools, papers or presentations are welcome.
## mesh – solving a wave equation with finite elements, if the material properties in a range vary continuously
I solve the one-dimensional wave equation over regions in which the volume modulus (and thus the wave velocity) varies continuously over a region. The current version seems to assume that the material properties are constant over a FEM element. Some references suggest that isoparametric elements can help model regions in which the material properties vary continuously. Can you suggest how to model regions where the material properties vary continuously?
In the wave equation code I use, Kappa and Rho can vary over an element. The code is as follows:
``````eqn = 1/(Kappa)(x) D(u(t, x), {t, 2}) +
1/(Kappa)(x)*10*Exp(-50 (x^2))*Sin(2 (Pi) f t) +
NeumannValue(0, x == 0) + NeumannValue(-Derivative(1, 0)(u)(t, x), x ==
xMax);
ic = {u(0,x) == 0, Derivative(1, 0)(u)(0, x) == 0};
``````
## Properties – Created columns do not appear in the page detail form of the Web page in SharePoint Online?
I need to add properties for date and time types and change them using the site's page details form.
If I add properties to the page library, it will not appear in the form of page details of the web page in the properties list.
It has been working since yesterday.
Do you have any idea? | |
Write the sum of intercepts cut off by the plane vec r (2 hat i + hat j - hat k) -5 = 0 on the three axes.
### Question Asked by a Student from EXXAMM.com Team
Q 2680867717. Write the sum of intercepts cut off by the plane vec r (2 hat i + hat j - hat k) -5 = 0 on the three axes.
CBSE-12th 2016
#### HINT
(Provided By a Student and Checked/Corrected by EXXAMM.com Team)
#### Access free resources including
• 100% free video lectures with detailed notes and examples
• Previous Year Papers
• Mock Tests
• Practices question categorized in topics and 4 levels with detailed solutions
• Syllabus & Pattern Analysis | |
## Curvature of higher direct image sheaves.(English)Zbl 1392.32008
Oguiso, Keiji (ed.) et al., Higher dimensional algebraic geometry. In honour of Professor Yujiro Kawamata’s sixtieth birthday. Proceedings of the conference, Tokyo, Japan, January 7–11, 2013. Tokyo: Mathematical Society of Japan (MSJ) (ISBN 978-4-86497-046-4/hbk). Advanced Studies in Pure Mathematics 74, 171-184 (2017).
Summary: Given a family $$(F,h)\to X\times S$$ of Hermite-Einstein bundles on a compact Kähler manifold $$(X, g)$$ we consider the higher direct image sheaves $$R^qp_*\mathcal O(F)$$ on $$S$$, where $$p:X\times S \to S$$ is the projection. On the complement of an analytic subset these sheaves are locally free and carry a natural metric, induced by the $$L_2$$ inner product of harmonic forms on the fibers. We compute the curvature of this metric which has a simpler form for families with fixed determinant and families of endomorphism bundles. Furthermore, we discuss the metric for moduli spaces of stable vector bundles.
For the entire collection see [Zbl 1388.14012].
### MSC:
32L10 Sheaves and cohomology of sections of holomorphic vector bundles, general results 14D20 Algebraic moduli problems, moduli of vector bundles 32G13 Complex-analytic moduli problems 14F05 Sheaves, derived categories of sheaves, etc. (MSC2010)
Full Text: | |
# Does Uniform Boundedness in the Sobolev Space $W^{1,2}$ and Convergence in $L^p$ $(1 \leq p < 6)$ Imply Convergence in $L^6$?
Let $B$ denote the open unit ball in $\mathbb{R}^3$. I want to either prove or disprove that a sequence of functions $u_m$ in the Sobolev space $W^{1,2}(B)$ which is uniformly bounded in the $W^{1,2}(B)$ norm and which is convergent in $L^p(B)$ for all $1 \leq p < 6$ must be convergent in $L^6(B)$.
Can someone please point me in the right direction?
I know (cf. Chapter 5 of Evans PDE book) that the bound $$\| u \|_{L^q(U)} \leq C(k,p,n,U) \| u \|_{W^{k,p}(U)}$$ holds when $U$ is a subset of $\mathbb{R}^n$ having smooth boundary, $u \in W^{k,p}$, $k < \frac{n}{p}$, and $\frac{1}{q} = \frac{1}{p} - \frac{k}{n}$. The constant $C = C(k,p,n,U)$ is independent of $u$.
It follows from this bound that the $u_m$ are uniformly bounded in $L^6(B)$.
Since $W^{1,2}(B)$ is a Hilbert Space, uniform boundedness of $u_m$ in $W^{1,2}(B)$ also implies existence of a subsequence $u_{m_j}$ that converges weakly in $W^{1,2}(B)$, hence weakly in $L^6(B)$.
Thanks!
-
As bonnnnn2010 already pointed out, $6$ is the critical exponent where you don't have compact embedding of $H^1(B)=W^{1,2}(B)$ anymore, for dimension $d=3$. Here's an example of a bounded sequence in $H^1(B)$ that is convergent in $L^p(B)$ for $p<6$, but which has no convergent subsequence in $L^6(B)$: Take any $\varphi\ne0$ in $H^1(\mathbb{R}^3)$. Then the sequence $(\varphi_n)$ with $$\varphi_n(x) = n^{1/2}\varphi(nx)$$ has the desired properties; in fact, $\varphi_n\to0$ in $L^p(B)$ for $p<6$.
Because $L^6$ is critical to the compact embedding of $H^1$ when $d=3$, so we can not expect the convergence in $L^6$, I'm not sure if there's counter-example in the book of Adams "Sobolev Spaces". | |
delooping under Dold-Kan and simplicial delooping
What maps of simplicial sets exist between
• the image under the Dold-Kan correspondence of a chain complex shifted up in degree
• and the image under the right adjoint to simplicial looping of the DK-image of the unshifted complex
?
Here is the same question in detail:
Write
$$(G \dashv \bar W) : sGrp \stackrel{\leftarrow}{\underset{\bar W}{\to}} sSet_0 \hookrightarrow sSet$$
for the adjunction between simplicial groups and reduced simplicial sets whose left adjoint is the simplicial loop group functor (as for instance in Goerss-Jardine, chapter V);
and write
$$Ch_\bullet^+ \overset{\Xi}{\to} sAbGrp \hookrightarrow sGrp \overset{U}{\to} sSet$$
for the Dold-Kan correspondence, where in both cases I care about the images as simplicial sets.
Then for $V \in Ch_\bullet^+$ a chain complex and $V[1]$ (or $V[-1]$ if you prefer) its shift up in degree (its delooping as a chain complex) the two simplicial sets
$$U \Xi (V[1])$$
and
$$\bar W (\Xi V)$$
should have the same homotopy type. What nice natural maps of simplicial sets do we have between them?
- | |
mahal {DOS2} R Documentation
## Mahalanobis Distance Matrix for Optimal Matching
### Description
Computes a Mahalanobis distance matrix between treated individuals and potential controls; see Rubin (1980) and Rosenbaum and Rubin (1985). The method is discussed in Section 9.3 of "Design of Observational Studies", second edition.
### Usage
mahal(z, X)
### Arguments
z z is a vector that is 1 for a treated individual and 0 for a control. X A matrix of continuous or binary covariates. The number of rows of X must equal the length of z.
### Value
The distance matrix has one row for each treated individual (z=1) and one column for each potential control (z=0). The row and column names of the distance matrix refer to the position in z, 1, 2, ..., length(z).
### Note
The method is discussed in Section 9.3 of "Design of Observational Studies", second edition.
The matching functions in the 'DOS2' package are aids to instruction or self-instruction while reading "Design of Observational Studies". As in the book, these functions break the task of matching into small steps so they are easy to understand in depth. In practice, matching entails a fair amount of book-keeping best done by a package that automates these tasks. Consider R packages 'optmatch', 'rcbalance', 'DiPs', 'designmatch' or 'bigmatch'. Section 14.10 of "Design of Observational Studies", second edition, discusses and compares several R packages for optimal matching.
### Author(s)
Paul R. Rosenbaum
### References
Hansen, B. B. and Klopfer, S. O. (2006) <doi:10.1198/106186006X137047> "Optimal full matching and related designs via network flows". Journal of computational and Graphical Statistics, 15(3), 609-627. ('optmatch' package)
Hansen, B. B. (2007) <https://www.r-project.org/conferences/useR-2007/program/presentations/hansen.pdf> "Flexible, optimal matching for observational studies". R News, 7, 18-24. ('optmatch' package)
Rosenbaum, P. R. and Rubin, D. B. (1985) <doi:10.1080/00031305.1985.10479383> "Constructing a control group using multivariate matched sampling methods that incorporate the propensity score". The American Statistician, 39, 33-38.
Rubin, D. B. (1980) <doi:10.2307/2529981> "Bias reduction using Mahalanobis-metric matching". Biometrics, 36, 293-298.
### Examples
data(costa)
z<-1*(costa$welder=="Y") aa<-1*(costa$race=="A")
smoker=1*(costa$smoker=="Y") age<-costa$age
x<-cbind(age,aa,smoker)
dmat<-mahal(z,x)
# Mahalanobis distances
round(dmat[,1:6],2)
# Compare with Table 9.5 in "Design of Observational Studies", 2nd ed.
# Impose propensity score calipers
prop<-glm(z~age+aa+smoker,family=binomial)\$fitted.values # propensity score
# Mahalanobis distanced penalized for violations of a propensity score caliper.
# This version is used for numerical work.
round(dmat[,1:6],2)
# Compare with Table 9.5 in "Design of Observational Studies", 2nd ed.
## Not run:
# Find the minimum distance match within propensity score calipers.
# You must load the 'optmatch' package to try this example
optmatch::pairmatch(dmat,data=costa)
## End(Not run)
# Conceptual versions with infinite distances for violations of propensity caliper.
dmat[dmat>20]<-Inf
round(dmat[,1:6],2)
# Compare with Table 9.5 in "Design of Observational Studies", 2nd ed.
[Package DOS2 version 0.5.2 Index] | |
11,328
pages
A gap ordinal is an ordinal $$\alpha$$ where $$(L_{\alpha+1}-L_\alpha)\cap\mathcal P\omega=\varnothing$$, where $$\mathcal P\omega$$ denotes the powerset of $$\omega$$.[1] Leeds and Putnam proved that for a gap ordinal $$\alpha$$, $$L_\alpha\cap\mathcal P\omega$$ is a β-model of second-order arithmetic $$\mathcal{A}_2$$ using the recursion-theoretic result that $$L_\alpha\cap\mathcal P\omega$$ is closed under hyperjump.[2]. In fact, for $$\beta$$ starting a gap, $$L_\beta\cap\mathcal P\omega$$ is closed under definitions from analysis[3]. Currently, gap ordinals are not significantly useful in googology due to it being too high up the hierarchy of countable ordinals, but it is likely that in the future, it can be useful in analysis of TON, or other similarly strong concepts.
## Existence
The existence of a gap ordinal immediately follows when we work in $$\textrm{ZFC}$$ augumented by $$V = L$$ because of the constructibility of $$\mathcal P\omega$$ itself under the assumption, but a stronger result by Putnam given as the corollary of the classification of elementary substructures of $$L_{\omega_1^L}$$ by Devlin is known: The set of gap ordinals is unbounded in $$\omega_1^L$$.[1]
The first gap ordinal is sometimes called the "ordinal of ramified analysis" and written as $$\beta_0$$.[4] By the unboundedness result, $$\beta_0$$ is smaller than $$\omega_1^L$$.
## Relation to cardinals
Arai has shown that ordinals $$\beta$$ starting gaps of length $$\beta^+$$ satisfy $$L_{\beta^+}\vDash\beta\textrm{ is uncountable}$$[5]. In fact, it's even possible to find non-admissible $$\theta$$ such that $$L_\beta\vDash\exists\theta\exists p(p=\aleph_1^{L_\theta}\land\theta\textrm{ is not admissible})$$, and even stronger results than this, which involve primitive recursion[6].
## Extensions
There are several directions of extensions of the notion of a gap ordinal.
### Based on Definability
Putnam's result on the unboundedness under $$\omega_1^L$$ is extended to a generalised notion. Given a $$\Sigma_1$$-definable(in what structure?) function $$\Phi \colon \textrm{On}^m \to \textrm{On}$$ with a positive integer $$m$$, the set $$\{\alpha \in \omega_1^L \mid (L_{\Phi(\alpha,\beta_1,\ldots,\beta_{m-1})}-L_{\alpha})\cap\mathcal P\omega=\varnothing\}$$ is unbounded in $$\omega_1^L$$ for any $$(\beta_1,\ldots,\beta_{m-1}) \in (\omega_1^L)^{m-1}$$.[1]
### Based on Higher Order
The notion of a gap ordinal is extended to the higher order setting by replacing $$\mathcal P\omega$$ by $$\mathcal P^m\omega$$ for a natural number $$m$$, where the superscript denotes iteration. Namely, an ordinal $$\alpha$$ is said to be a gap of $$n$$-th order for a natural number $$n$$ if $$(L_{\alpha+1}-L_{\alpha})\cap\mathcal P^m\omega=\varnothing$$ for any natural number $$m < n$$. For example, every ordinal is a gap of $$0$$-th order, an infinite ordinal is precisely a gap of first order, a gap ordinal is precisely a gap of second order, and the notion of a gap of third order is characterised by a property using the notion of a $$\beta$$-model of third order arithmetic $$\mathcal{A}_3$$.[1]
### Relative sizes
The least third-order gap ordinal is greater than the least starting points of many of the gaps based on definability. For example, the least $$\beta$$ starting a third-order gap is greater than the least $$\gamma$$ such that $$(L_{\gamma^+}-L_\gamma)\cap\mathcal P\omega=\varnothing$$. As evidence, $$\beta=\textrm{min}\{\xi:L_\xi\vDash\textrm{ZFC}-\textrm{Powerset+}\omega_1\textrm{ exists}\!"\}$$[7], $$L_{\gamma^+}\vDash\omega_1\textrm{ exists}\!"$$[8], $$L_{\gamma^+}\not\vDash\textrm{ZFC}-\textrm{Powerset}$$[9], and $$\beta\neq\gamma$$.
## Sources
1. Marek & Srebrny, Gaps in the Constructible Universe, Annals of Mathematical Logic 6 (1974), pp. 359--394.
2. S. Leeds and H. Putnam, An intrinsic characterization of the hierarchy of Constructible sets of integers, Logic Colloquium '69 (1971), Proceedings of the Summer school and Colloquium in Mathematical Logic, Manchester, August 1969 / Ed. by R.O. Gandy and C.M.E. [sic] Yates.
3. M. Lucian, Gap-minimal systems of notations and the constructible hierarchy (p.23)
4. D. Madore, A Zoo of Ordinals (2017) (p.6)
5. T. Arai, A sneak preview for proof theory of ordinals (p.17) (accessed 2021-03-07)
6. The least α for which E(α) is inadmissible (p.149) (accessed 2021-03-07)
7. D. Madore, A Zoo of Ordinals (p.6). Accessed 2021-05-05
8. [Arai, A sneak preview of proof theory of ordinals (1997) (p.17)]
9. γ+ isn't a limit of admissible ordinals, so γ+ isn't Σ2-admissible[citation needed], so γ+ isn't a gap ordinal (cf. [MarekSrebrny73 (p.368)] | |
anonymous one year ago Evaluate the integral (Partial Fractions) ∫x^2+1/(x-3)(x-2)^2 dx Why is it that when you split the function into A+B+C that you get A(x-2)+B(x-2)(x-3)+C(x-3). how come there isnt an A(x-2)^3
1. anonymous
There's no A(x-2)^3 because (x-2) is only raised to the 2nd power. You don't need the second (x-3) either. $\frac{ x^2+1 }{ (x-3)(x-2)^2 }=\frac{ A }{ x-3 }+\frac{ B }{ x-2 }+\frac{ C }{ (x-2)^2 }$
2. anonymous
ok. How come there's C(x-3) and not C(x-2)(x-3). is it because there's a (x-2)^2 in the denominator?
3. anonymous
you mean when you multiply? You have to multiply by the least common denominator, so that's $$(x-3)(x-2)^2$$. So (x -3) cancels on the A leaving you with $$A(x-2)^2$$ (x - 2) cancels on B, leaving $$B(x-3)(x-2)$$ (x - 2)² cancels on C leaving $$C(x-3)$$
4. anonymous
$x^2+1=A(x-2)^2+B(x-3)(x-2)+C(x-3)$
5. anonymous
ok its makes a lot of sense now. thanks. | |
# A heuristic for prioritising demand planning at lower granularities
Most inventory optimisation methods tell you how much stock to bring in at the network level. But often at lower granularities there is not enough space, or enough trucks to move it all. In this case you need to prioritise what to bring in. This article develops a heuristic to do this.
# Why supply chain networks are often strained?
Most demand planning is performed at the highest level of granularity (e.g. how much will be demanded in the UK over the next week). This reduces the volatility of demand and so means less unnecessary stock is produced or bought. But this is only part of the supply chain. A big part is then distributing the stock around the supply network to ensure it is as close to the customer as possible.
Since there is only enough stock at the highest granularity, this causes a strain on the network because:
### 1. The demand at lower granularities will be more volatile
To fill each warehouse with enough stock to account for these volatilities would be more stock than is available in the network, since demand planning performed at the highest level of granularity.
### 2. Warehouse capacities
Often individual warehouse capacities are tight: for example they might be full of a product that hasn’t had as much demand as expected etc.
### 3. Capacity for moving stock
The capacity to move stock around the network is limited (e.g. the number of trucks available). Because the network is strained stock might be put in the wrong warehouse and then has to be moved. But this is a waste that is bad for profitability and the environment, so should be minimised if possible.
### 4. Seasonality
Warehouses will tend to have highly busy days followed by quieter days. Often the capacity for moving stock will be enough for a day of ‘average demand’. This adds a big problem: when should I start bringing in stock for my busiest day? Because if you don’t, you’ll miss a lot of sales as you didn’t bring enough stock in on time.
# Why traditional inventory optimisation methods don’t work at the network level?
The classic inventory optimisation methods such as the reorder point method, and the (s, S) method answer the question:
#### ‘Given my (uncertain) demand how much inventory should I bring into my warehouse to make sure I’m unlikely to miss a sale?’
But at the network level, this is probably the wrong question to ask. Because your network is strained, you’re unlikely to be able to bring in the stock these methods recommend at the warehouse level. The methods also focus on one item at a time, but at the network level, bringing in one item may mean you can’t bring in enough of another. This means you need to consider all items at the same time.
In reality, the question often morphs into, what is the most valuable stock to bring into the warehouse today? This is the question I’ll be solving in this article.
# What is the most valuable stock to bring into the warehouse today?
In this article I’ll focus on a way to solve this problem. For now I’ll focus on what to bring into one warehouse with multiple items, random demand, and a constraint on how much can be bought in (e.g. the number of trucks available to move the product).
This problem is not too bad to solve on its own, but typically supply networks will see seasonal demand (see point 4 above). This means we don’t just need to think about today, we need to look ahead to see if stock desperately needs to be brought in for the busiest day. This adds a lot of complexity, and the solution took a lot of thought; but I cover a heuristic for this case after the single day case.
The solution can easily be extended to multiple warehouses competing for stock, using the same logic. I’ll leave this to a future post.
It should also be extendable to the case where the warehouse capacity itself is the constraint.
# Single period problem
We’ll start with the single period problem (e.g. solving the problem for just one day), where we can derive an optimal solution in terms of expected profit. This sets up the notation for the multi-period problem.
## Setup
Assume we have $$i= 1,…,I$$ SKUs. At the end of the period a random number $$Y_i$$ of each item is demanded. Let’s assume we have a reasonable handle on $$\mathbb P(Y_i \geq y)$$, i.e. the demand probabilities. This can be achieved through a probabilistic forecast.
We assume starting inventory $$x_i$$ for each item is known. The amount of inventory of item $$i$$ we choose to bring in we label $$a_i$$.
With this in place, we can label the remaining variables, and define the problem:
• $$r_i$$ – the revenue gained if one unit of item $$i$$ is demanded.
• $$C_i$$ – the cost of bringing one further unit of item $$i$$ into the warehouse. This could be due to e.g. transport or purchase costs.
• $$K$$ – maximum number of items that can be brought into the warehouse: i.e., we required $$\sum_{i=1}^I a_i \leq K$$. This could be due to movement capacity constraints (e.g. number of trucks). There might be an explicit warehouse capacity constraint, which could be easily added to the single period problem, but requires more work for the multiple period problem.
We can define the problem as bringing in $$a_i$$ in order to maximize profit, defined as $\sum_{i=1}^I \left(r_i Y_i - a_i C_i \right),$ while staying below the movement constraint $$\sum_{i=1}^I a_i \leq K$$.
Bring in too much stock, and we are paying unnecessary cost $$C_i$$. But not enough stock will lead to unmet demand, and lost sales. Sometimes a penalty for missed sales gets added to this problem, to indicate impact to customer loyalty. This can easily be done in this formulation.
We can solve the single period problem to maximise expected reward using a simple iterative algorithm. First define $$\hat a_i$$ to be the amount of item $$i$$ scheduled to be moved so far by the algorithm, and initialise by setting equal to 0, $$a_i = 0$$.
We can calculate the marginal reward of loading an extra unit of item $$i$$ exactly as $r_i \mathbb P(Y_i > x_i + \hat a_i) - C_i.$ The first term is the revenue multiplied by the probability that additional unit would be ordered. The second term is the cost of loading that item.
This marginal reward immediately gives rise to an iterative algorithm:
### Single period algorithm
1. Initialise $$a_i = 0$$
2. Until STOP do
i. $i_{best} = \text{argmax}_i \left[ r_i \mathbb P(Y_i > x_i + \hat a_i) - C_i \right]$ ii. If for $$i_{best}$$ $$r_i P(Y_i > x_i + \hat a_i) - C_i < 0$$, or $$\sum_{i=1}^I \hat a_i \geq K$$: STOP
iii. Load item $$i_{best}$$: set $$\hat a_{i_{best}} = \hat a_{i_{best}} + 1$$
# Multi-period problem
This problem is more complex but if we want to handle seasonality it’s essential. Now we add in time $$t = 1, …, T$$. We set $$T$$, the horizon, to be the seasonal period (e.g. 7 if we’re looking at daily data). This ensures the busiest days are taken into account when loading.
We assume at the start of each period $$t$$, we have inventory for item $$i$$ of $$x_{ti}$$. We assume $$x_{1i}$$ is known, but for all later periods it depends on the demand and loading choice. So is unknown.
We assume a loading amount $$a_{ti}$$ is chosen for the period, and after that a random amount $$Y_{ti}$$ is demanded. This is fulfilled from the inventory $$x_{ti} + a_{ti}$$, with reward $$r_i$$ for each unit of item $$i$$ successfully fulfilled, but any demand over the inventory is lost sales.
The rest of the problem setup is as follows
• $$C_i$$ – cost of loading one unit of item $$i$$
• $$K$$ – capacity of items that brought into the warehouse in any period, i.e. for all $$t$$, $$\sum_{i=1}^I a_{ti} \leq K$$
## Possible heuristic procedure
Because starting inventory $$x_{1i}$$ is known, we can solve the problem exactly using dynamic programming. We would do this by proceeding backwards from the last horizon, which we solve using a single period problem. Then each of the possible cases for $$T-1$$ (depending on inventory levels at that point) can be solved using a single period problem, and the value function for the expected remaining inventory at $$T$$, etc.
The big problem with this approach is we have to consider all possible inventory levels for each item, which will quickly explode as the number of items increases.
We attempt to get around this by noticing that we are only interested in $$a_{1i}$$ at each period, i.e. the loading we have to do today. To do this we have to consider the future states in case there’s a particularly busy day. But we suggest instead of having it as a state, we replace it with its expected value, determined by $$x_{ti}$$, $$\hat a_{ti}$$ and $$\mathbb E(Y_{ti})$$. Once again we defined $$\hat a_{ti}$$ to be the loadings chosen by the algorithm so far.
We can initialise the problem using a greedy algorithm. Set $$\hat a_{1i}$$ by solving the single period problem for period 1. Now initialise the inventory for period 2 using expected values $x_{2i} = \max\left[x_{1i} + \hat a_{1i} - \mathbb E(Y_{1i}), 0 \right].$
We can proceed initialisation iteratively using a similar procedure: solving a single period problem for horizon $$t$$ to get an initial $$\hat a_{ti}$$, then initialising the inventory as $x_{t+1,i} = \max\left[x_{1i} + \sum_{s=1}^t \left(\hat a_{si} - \mathbb E(Y_{si})\right), 0 \right].$
Now we’ve intialised the problem, we can proceed to the heuristic algorithm. For this, the inventory $$x_{ti}$$ and loading amounts $$\hat a_{ti}$$, need to be constantly updated as things are changed by the algorithm.
Similar to dynamic programming, we proceed backwards, but now with our fixed inventory states. For period $$t=T$$, we are at the end of the horizon, so have no horizons ahead to consider. So it’s safe to assume that the single period problem will be a good result. Therefore we do nothing for this period, and move back to the next period.
At period $$t=T-1$$, we need to look at both the current period, and the period $$t=T$$. First set $$b_{t,i} = x_{ti} + \hat a_{ti}$$. Applying our inventory assumption, similar to the single period problem, we can write down the marginal reward of adding another unit of item $$i$$ to the loading in period $$T-1$$: $N_{T-1,i} = \mathbb P(Y_{T-1,i} > b_{T-1, i}) R_i - C_i + \mathbb P(Y_{T-1,i} \leq b_{T-1, i}) \mathbb P(Y_{T,i} > b_{T,i}) R_i.$ We can break down the parts of the marginal reward as follows: the additional unit of product $i$ will either be ordered at time $T-1$, not ordered at time $T-1$ but ordered at time $T$; or not ordered at all because we’ve reached the end of the horizon.
Because we’ve initialised this problem with the single period algorithm though, our loadings might be full. In this case we can’t just add the item with the best marginal reward, we need to swap it with an item in the loading.
To do this we can check the marginal cost of removing one unit of one item from the knapsack. This is equivalant to the marginal reward above replacing $$b_{T-1, i}$$ with $$b_{T-1, i} - 1$$; i.e. $M_{T-1,i} = \mathbb P(Y_{T-1,i} > b_{T-1, i} - 1) R_i - C_i + \mathbb P(Y_{T-1,i} \leq b_{T-1, i} - 1) \mathbb P(Y_{T,i} > b_{T,i} - 1) R_i.$
A single step of the algorithm then proceeds as follows:
1. Check if $$N^* = \max_{i} N_{T-1,i}$$ is positive. If it’s $$\leq 0$$, STOP.
2. Set $$i_{best} = \text{argmax}_{i} N_{T-1,i}$$. Set $$M_* = \min M_{T-1, i}$$, and $$j_{worst} = \text{argmin}_{i} M_{T-1,i}$$.
3. If the capacity $$K$$ is not reached (i.e. $$\sum_{i=1}^I a_{T-1,i} < K$$), then add $$i_{best}$$ to the loading. Set $$a_{T-1,i_{best}} = a_{T-1, i_{best}} + 1$$.
4. If the capacity $$K$$ is reached, then check if $$N^* > M_*$$. If it isn’t STOP. Otherwise load $$i_{best}$$ and unload $$j_{worst}$$; i.e. set $$\hat a_{T-1, i_{best}} = \hat a_{T-1, i_{best}} + 1$$ and $$\hat a_{T-1, j_{worst}} = \hat a_{T-1, j_{worst}} - 1$$
After each iteration of this algorithm, if $$N^* \leq 0$$, or $$N^* \leq M_*$$, we stop completely and move to the next period.
Otherwise we update the inventory values $$x_{t,i_{best}}$$ and $$x_{t, j_{worst}}$$ for $$t = T-1, T$$ using the new loadings $$\hat a_{T-1, i_{best}}$$ and $$\hat a_{T-1, j_{worst}}$$. Also update $$N_{T-1, i_{best}}$$, $$N_{T-1, j_{worst}}$$, $$M_{T-1, i_{best}}$$ and $$M_{T-1, i_{worst}}$$ given the updated loadings and inventory values. Then we repeat the algorithm above, until either $$N^* \leq 0$$, or $$N^* \leq M_*$$.
Moving to time period $t = T-2$ we can follow the exact logic as for $t = T-1$. The difference here will be our marginal rewards. We have $N_{T-2,i} = \mathbb P(Y_{T-2,i} > b_{T-2, i}) R_i - C_i + \mathbb P(Y_{T-2,i} \leq b_{T-2, i}) \mathbb P(Y_{T-1,i} > b_{T-1,i}) R_i + \mathbb P(Y_{T-2,i} \leq b_{T-2, i}) \mathbb P(Y_{T-1,i} \leq b_{T-1, i}) \mathbb P(Y_{T,i} > b_{T,i}) R_i.$ Breaking this down: the additional unit of product $$i$$ will either be ordered at time $$T-2$$, not ordered at time $$T-2$$ but ordered at time $$T-1$$, not ordered at time $$T-2, T-1$$ but ordered at time $$T$$; or not ordered at all because we’ve reached the end of the horizon. Similar logic can be used to calculate $$M_{T-2, i}$$.
This procedure can be repeated to time period $t=1$, which will be our required loading for today.
## Intuition
The intuition for this heuristic is that the uncertainty in demand has two effects: the uncertainty for that period; the uncertainty of how much inventory will be available in later periods. This heuristic captures how demand affects the uncertainty in each period, but ignores knock-on uncertainty effects on the inventory levels – replacing it with a deterministic expected value. This should work because we are mainly interested in the first period, so not capturing the inventory uncertainty would hopefully not have a massive effect.
## Acknowledgments
Thanks to Jake Clarkson for reviewing this algorithm and providing helpful comments. | |
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0
Hydrodynamic Lubrication
# Modeling of Magnetorheological Fluids by the Discrete Element Method
[+] Author and Article Information
Mickaël Kargulewicz, Ivan Iordanoff
Arts et Métiers ParisTech, Institut de Méchanique et Ingénierie - Bordeaux (UMR 5295), Esplanade des Arts et Métiers, 33405, Talence Cedex, France
Victor Marrero
Department of Mechanical, Aerospace, and Nuclear Engineering, Rensselaer Polytechnic Institute, Troy, NY 12180-3590
John Tichy1
Department of Mechanical, Aerospace, and Nuclear Engineering, Rensselaer Polytechnic Institute, Troy, NY 12180-3590tichyj@rpi.edu
1
Corresponding author.
J. Tribol 134(3), 031706 (Jun 27, 2012) (9 pages) doi:10.1115/1.4006021 History: Received April 20, 2011; Revised December 20, 2011; Published June 26, 2012; Online June 27, 2012
## Abstract
Magnetorheological (MR) fluids are fluids whose properties vary in response to an applied magnetic field. Such fluids are typically composed of microscopic iron particles ($~1-20μm$ diameter, $20-40%$ by volume) suspended in a carrier fluid such as mineral oil or water. MR fluids are increasingly proposed for use in various mechanical system applications, many of which fall in the domain of tribology, such as smart dampers and clutches, prosthetic articulations, and controllable polishing fluids. The goal of this study is to present an overview of the topic to the tribology audience, and to develop an MR fluid model from the microscopic point of view using the discrete element method (DEM), with a long range objective to better optimize and understand MR fluid behavior in such tribological applications. As in most DEM studies, inter-particle forces are determined by a force-displacement law and trajectories are calculated using Newton’s second law. In this study, particle magnetization and magnetic interactions between particles have been added to the discrete element code. The global behavior of the MR fluid can be analyzed by examining the time evolution of the ensemble of particles. Microscopically, the known behavior is observed: particles align themselves with the external magnetic field. Macroscopically, averaging over a number of particles and a significant time interval, effective viscosity increases significantly when an external magnetic field is applied. These preliminary results would appear to establish that the DEM is a promising method to study MR fluids at the microscopic and macroscopic scales as an aid to tribological design.
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## Figures
Figure 1
Particles in MR fluid. (a) upper left: no magnetic field, no shear - random distribution; (b) upper right: field applied - magnetization of particles, interparticle forces; (c) middle left: field applied, formation of chains; (d) middle right: shear applied, chains deform; (e) lower left: chains rupture in shear; (f) lower right: field removed, particles move with shear.
Figure 2
Schematic of the Bingham Model, yield stress varies with magnetization
Figure 3
Domain of the simulations
Figure 4
Typical magnetization curve of a ferro magnetic material
Figure 5
Magnetization curve used in present model
Figure 6
Alignment of a column of particles in the MR fluid - applied magnetic field with no shear; evolution with time shown from left to right
Figure 7
Shearing of a column of particles in the MR fluid - applied magnetic field; evolution with time shown from left to right
Figure 8
Shearing of a column of particles in the MR fluid with applied magnetic field
Figure 9
Simulation results: typical shear stress evolution with time - the behavior of moving averages
Figure 10
Simulation results: shear stress versus shear rate - the ‘off’ condition (no applied magnetic field) is the lower curve, the ‘on’ condition (applied external magnetic field) is the upper curve
Figure 11
Simulation results: ratio of ‘on’ shear stress to ‘off’ shear stress versus Mason number
## Discussions
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How to correctly use the color option in R image() function?
1
4
Entering edit mode
7.9 years ago
Xianjun ▴ 300
HI,
I got a question about how to assign color value in image() function. I thought the col option works as a vector with the same length of Z, with color of each cell in the matrix Z corresponding to the values in. However, it confused me with different result.
To make my question simple, let's say I want to make a color image for a vector x=1:4 with corresponding color for each cell.
x=1:4
image(1,1:length(x), matrix(x, nrow=1, ncol=length(x)), col=c("blue","red",'green','yellow'))
The output image looks like this:
However, if I use a different x, say
x=c(3,1,2,1)
image(1,1:length(x), matrix(x, nrow=1, ncol=length(x)), col=c("blue","red",'green','yellow'))
I expected the color (from bottom to up) order should be green-->blue-->red-->blue, however, I got a very different order:
-Xianjun
R image • 9.4k views
2
Entering edit mode
This is interesting. If you receive the Yellow color from the end:
x=c(3,1,2,1)
image(1,1:length(x), matrix(x, nrow=1, ncol=length(x)), col=c("blue","red","green"))
it will color properly.
If you use 5 colors:
x=c(3,1,2,1)
image(1,1:length(x), matrix(x, nrow=1, ncol=length(x)), col=c("blue","red","green","yellow", "magenta"))
R will plot one color after other.
If you use 6 colors:
x=c(3,1,2,1)
image(1,1:length(x), matrix(x, nrow=1, ncol=length(x)), col=c("blue","red","green","yellow", "magenta", "grey"))
May be something with the algorithm?
0
Entering edit mode
As my reply below, I guess I figured it out: We always get min(x)=col[1], max(x)=col[last]. For your case, then there are 6 colors, again, 1=blue, 3=grey, 2=green (since it's left close).
3
Entering edit mode
7.9 years ago
Xianjun ▴ 300
OK I figured it out myself:
Unless the breaks is set, the range of Z (or x in above example) is evenly cut into N intervals (where N = the length of color) and values in Z are proportionally assigned to the nearest color.
For example, when x=c(3,1,2,1) and col=c("blue","red",'green','yellow'), the minimal of x is assigned as the first color, and max to the last color. Any value between is calculated proportionally to a color. In this case, 2 is the the middle one, according to the principal that intervals are closed on the right and open on the left, it's assigned to "red". So, that's why we see the colors are yellow-->blue-->red-->blue. | |
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Result $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
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Two families of orthogonal polynomials related to Jacobi polynomials. (English) Zbl 0744.33004
The Jacobi polynomials ${P}_{n}^{\left(\alpha ,\beta \right)}\left(x\right)$ satisfy a three term recurrence relation with recurrence coefficients that are simple rational functions of the degree $n$, containing the two parameters $\alpha$ and $\beta$. When $\alpha +\beta =0$ one must be careful in defining ${P}_{1}\left(x\right)$. The classical way is to define ${P}_{1}\left(x\right)=x+\alpha$, which leads to the standard Jacobi polynomials. However, the recurrence relation with initial values ${P}_{-1}=0$ and ${P}_{0}=1$ leads to ${P}_{1}\left(x\right)=x$, and with this choice of ${P}_{1}\left(x\right)$ one obtains the exceptional Jacobi polynomials studied in this paper. These polynomials are again orthogonal on $\left[-1,1\right]$ and the authors explicitly compute the weight function.
A second family of orthogonal polynomials studied in this paper is a class of associated Jacobi polynomials arising in birth and death processes without absorption at zero. Explicit formulas are given for these associated Jacobi polynomials and also asymptotic results and a generating function. The asymptotic behaviour then leads to an explicit formula for the weight function.
##### MSC:
33C45 Orthogonal polynomials and functions of hypergeometric type 42C05 General theory of orthogonal functions and polynomials 60J80 Branching processes | |
# Redox reactions: identifying what is oxidized vs reduced
$$\ce{SO4^2- + H2O2 + 2H+ -> SO2 + O2 + 2H2O}$$
In the above chemical reaction, I'm having trouble identifying what is oxidized and what is reduced.
My guess is: sulfur has reduced but I can't work out what has oxidised. My guess is the middle oxygen as it has lost two hydrogen atoms to $\ce{2H2O}$.
• The sulphate ion and the hydrogen peroxide are reduced , the hydrogen is oxidised .... – Technetium Jun 12 '16 at 2:24
• I think there would be more steps to the actual reaction tho .... – Technetium Jun 12 '16 at 2:31
• When the species cannot be found in redox tables (or the reaction scheme is just for illustration as in the question, e.g. SO$_2$ does not exist in solution as its a gas), using oxidation numbers is the only simple way. These values can be zero, positive, negative or fractional, e.g S in S$_4$O$_6$ or Fe in Fe$_3$O$_4$. In the question S has a value +6 (6-8*2=-2) in SO$_4$ and +4 in SO$_2$, as the total must be zero in a neutral species. As a decrease in oxidation number corresponds to reduction the sulphate ion is reduced and the peroxide is oxidised. – porphyrin Jul 12 '16 at 19:58
Yes sulfur of $\ce{SO4^2-}$ has undergone reduction as the oxidation number reduces from +6 to +4 and Oxygen of $\ce{H2O2}$ has undergone oxidation as its oxidation number increases from -1 to 0.
$$\ce{\overset{\color{red}{+6}}{S}O4^{2-}\+H2\overset{\color{red}{-1}}{O2}\+2H+->\overset{\color{red}{+4}}{S}O2\+\overset{\color{red}{0}}{O2}\+2H2O}$$
NOTE: Oxidation sate of oxygen in peroxide is -1. This because in the neutral molecule hydrogen peroxide the oxidation state of $\ce{H}$ is +1 and as the sum of oxidation numbers of the two hydrogen atoms and the two oxygen atoms must be zero so each oxygen in $\ce{H2O2}$ has -1 oxidation state.
You can use formal charges, but that's exactly what they are: a formality, not actual charges.
A redox reaction is a redox couple, consisting of an oxidation half reaction and a reduction half reaction. I would consult a table of reduction potentials (e.g.), which lists half-reactions and their standard reduction potentials.
On this table, we find: \begin{align} \ce{O2(g) + 2 H+ (aq) + 2 e– &-> H2O2 (aq)},& E^{\circ} &= \pu{+0.682 V}\\ \ce{SO4^2– (aq) + 4 H+(aq) + 2 e– &-> SO2 (g) + 2 H2O},& E^{\circ} &= \pu{+0.2 V}\\ \end{align}
If I invert the first reaction, I have the following oxidation: \begin{align} \ce{H2O2(aq) &-> O2 (g) + 2 H+ (aq) + 2 e–},& E^{\circ} &= \pu{-0.682 V}\\ \end{align}
Summing this with the second half reaction (reduction) gives the total [redox] reaction provided in your question.
This tells us two things. Hydrogen peroxide is being oxidized (that system is losing electrons), since it is the reactant in the oxidation half reaction. Sulfate and hydrogen ion are participating as reactants in the reduction half reaction; they are reduced.
We didn't actually need to do anything with potentials here, but you can easily see that the total reaction is non-spontaneous.
The oxygen atoms on sulfate ion are not normally redox-active, whereas the oxygen in hydrogen peroxide could be redox-active given that they act that way in other reactions. So I would select the hydrogen peroxide oxygens as the reducing agent here. | |
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# OpenGL OpenGL Light Positions to D3D
## 3 posts in this topic
Hi Gamdev.net,
I have a question that I'm hoping can be answered. I'm starting work on implementing shadow volumes in my game engine and I've been researching the topic a little here and there. I came across a PDF (located here: http://fabiensanglard.net/doom3_documentation/37730-293752.pdf) that disects part of the Shadow Volume code in the Doom 3 source. I tend to learn by getting something to work and then fiddiling with it until I completely understand it, so I'm trying to implement so I can master the concept before throwing it out and rewriting it, but I'm struggling with it a bit because the code seems to use quaternions for light rotations. This wouldn't be much of an issue except that everything that Direct3D 9 uses utilizes simply an x, y, and z value. There is a theta value...
Question 1) When I issue, to Direct3D 9, a statement to set a light's position or direction, are each of the components (x, y, and z) considered separate rotational vectors that have been formed by other values calculated internally by Direct 3D? Hard to explain but if you goto http://www.cs.princeton.edu/~gewang/projects/darth/stuff/quat_power.html and search in the document for the section Euler to Quaternion you'll see what I mean. I mean are the x, y, and z values some independant set of rotation values for each axis (Like some single value that encompasses the y and z rotations for x, the x and z rotations for y, and the x, and y rotations for z), or are x, y and z really just that and I'm overthinking them? If they really are just simple values to be treated, for example, as x is to be accompanied with y and z make up the entire rotation at face-value then how would one get a w value to use with the implementation presented in that paper I linked at the begining of the post? And how would this algorithm work with point lights? In that paper, the part that I'm referring to is on page 15 in Appendix A:
/*
Calculating Triangle Facing
Copyright (C) 2005 Id Software, Inc.
Written by J.M.P. van Waveren
This code is free software; you can redistribute it and/or
modify it under the terms of the GNU Lesser General Public
version 2.1 of the License, or (at your option) any later version.
This code is distributed in the hope that it will be useful,
but WITHOUT ANY WARRANTY; without even the implied warranty of
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
Lesser General Public License for more details.
*/
struct Vec4 {
float x, y, z, w;
};
struct Plane {
float a, b, c, d;
};
void CalculateFacing( const Plane *planes, const int numTriangles, const Vec4 &light, byte *facing ) {
int i;
for ( i = 0; i < numTriangles; i++ ) {
facing[i] = planes[i].a * light.x +
planes[i].b * light.y +
planes[i].c * light.z +
planes[i].d * light.w > 0.0f;
}
facing[numTriangles] = 1; // for dangling edges to reference
}
If I'm using D3D 9 lights can I just set light.w to 1.0f or 0.0f or would that have some bad mathematical effect? Is there some trick way that if D3D really does just use x, y, and z, and OpenGL uses Quaternions that I can make either one work or do I really need a conditional to check what API I'm using? I know it may seem like a basic question, but I've never really had to think this hard about how D3D actually handles it's directions or positions of lights. It's never ever been a problem up until now, now that I'm comparing it to OpenGL stuff. I would try reaching out to John Carmack on this but he's probably not exactly the most accessible guy :P Thanks in advance!
-- StakFallT
Edited by StakFallT
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I haven't followed and read all of those links, but here's what I think is the right answer:
- When you set a DirectX9 light position, there are two modes. If you're setting a directional light, the X, Y and Z are the direction and W should be set to 0. If you're setting a point light, the X, Y and Z are the position and W should be set to 1.
- I'm pretty certain neither OpenGL nor DX9 uses quaternions for lighting natively in their fixed function pipelines, but with custom shaders maybe they are being used. I struggle to see why they'd be helpful for lighting though.
- The CalculateFacing function looks to me that it takes a whole bunch of plane equations and calculates whether they face toward the light or away from the light. I imagine that this is part of some code that's calculating the silhouette of an object. It probably iterates through all the triangles in the object, calculates which triangles face towards/away from a light, then uses that information to calculate which edges form the silhoutte, then extrudes those silhouette edges to construct some geometry that is used in stencil shadows. Stencil shadows have rather fallen out of favour (everyone seems to do shadow mapping these days) so learning how Doom3 does them might not be as valuable as you might hope (but I guess all learning is good).
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Are you talking about fixed-function lighting in D3D? Because I'm pretty sure that Doom 3 would be using shaders...
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@C0lumbo: Yup, that's exactly what it's doing. Just I saw the w in there, for lighting, and wasn't sure what to do with it.the setting w to 1 or 0 though was the info I was after :) Btw, what about if it's a spotlight? I'm guessing I'd set w to 0 like I would a directional? Shadow Mapping is being favored over volumes? I thought the volume technique came out after mapping?
@Hodgman: Well there's some optimization stuff in that first paper and it does mention using some hardware here and there throughout and if it's available; I'm not really sure where the code in that first paper matches and/or differs with the Doom 3 source. My guess is, it's some sort of hybrid -- that is, use hardware if available, otherwise do it on the cpu.
Btw, thanks guys for the quick replies!
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• ### Similar Content
• By DaniDesu
#include "MyEngine.h" int main() { MyEngine myEngine; myEngine.run(); return 0; } MyEngine.h
#pragma once #include "MyWindow.h" #include "MyShaders.h" #include "MyShapes.h" class MyEngine { private: GLFWwindow * myWindowHandle; MyWindow * myWindow; public: MyEngine(); ~MyEngine(); void run(); }; MyEngine.cpp
#pragma once #include <glad\glad.h> #include <GLFW\glfw3.h> class MyWindow { private: GLFWwindow * windowHandle; int windowWidth; int windowHeight; const char * windowTitle; public: MyWindow(int windowWidth, int windowHeight, const char * windowTitle); ~MyWindow(); GLFWwindow * getWindowHandle(); void createWindow(); void MyWindow::destroyWindow(); }; MyWindow.cpp
#include "MyWindow.h" MyWindow::MyWindow(int windowWidth, int windowHeight, const char * windowTitle) { this->windowHandle = NULL; this->windowWidth = windowWidth; this->windowWidth = windowWidth; this->windowHeight = windowHeight; this->windowTitle = windowTitle; glfwInit(); } MyWindow::~MyWindow() { } GLFWwindow * MyWindow::getWindowHandle() { return this->windowHandle; } void MyWindow::createWindow() { // Use OpenGL 3.3 and GLSL 3.3 glfwWindowHint(GLFW_CONTEXT_VERSION_MINOR, 3); glfwWindowHint(GLFW_CONTEXT_VERSION_MAJOR, 3); // Limit backwards compatibility glfwWindowHint(GLFW_OPENGL_PROFILE, GLFW_OPENGL_CORE_PROFILE); glfwWindowHint(GLFW_OPENGL_FORWARD_COMPAT, GL_TRUE); // Prevent resizing window glfwWindowHint(GLFW_RESIZABLE, GL_FALSE); // Create window this->windowHandle = glfwCreateWindow(this->windowWidth, this->windowHeight, this->windowTitle, NULL, NULL); glfwMakeContextCurrent(this->windowHandle); } void MyWindow::destroyWindow() { glfwTerminate(); } MyShapes.h
#pragma once #include <glad\glad.h> #include <GLFW\glfw3.h> class MyShapes { public: MyShapes(); ~MyShapes(); GLuint & drawTriangle(float coordinates[]); }; MyShapes.cpp
#include "MyShapes.h" MyShapes::MyShapes() { } MyShapes::~MyShapes() { } GLuint & MyShapes::drawTriangle(float coordinates[]) { GLuint vertexBufferObject{}; GLuint vertexArrayObject{}; // Create a VAO glGenVertexArrays(1, &vertexArrayObject); glBindVertexArray(vertexArrayObject); // Send vertices to the GPU glGenBuffers(1, &vertexBufferObject); glBindBuffer(GL_ARRAY_BUFFER, vertexBufferObject); glBufferData(GL_ARRAY_BUFFER, sizeof(coordinates), coordinates, GL_STATIC_DRAW); // Dertermine the interpretation of the array buffer glVertexAttribPointer(0, 3, GL_FLOAT, GL_FALSE, 3*sizeof(float), (void *)0); glEnableVertexAttribArray(0); // Unbind the buffers glBindBuffer(GL_ARRAY_BUFFER, 0); glBindVertexArray(0); return vertexArrayObject; } MyFileHandler.h
#pragma once #include <cstdio> #include <cstdlib> class MyFileHandler { private: const char * fileName; unsigned long fileSize; void setFileSize(); public: MyFileHandler(const char * fileName); ~MyFileHandler(); unsigned long getFileSize(); const char * readFile(); }; MyFileHandler.cpp
#include "MyFileHandler.h" MyFileHandler::MyFileHandler(const char * fileName) { this->fileName = fileName; this->setFileSize(); } MyFileHandler::~MyFileHandler() { } void MyFileHandler::setFileSize() { FILE * fileHandle = NULL; fopen_s(&fileHandle, this->fileName, "rb"); fseek(fileHandle, 0L, SEEK_END); this->fileSize = ftell(fileHandle); rewind(fileHandle); fclose(fileHandle); return; } unsigned long MyFileHandler::getFileSize() { return (this->fileSize); } const char * MyFileHandler::readFile() { char * buffer = (char *)malloc((this->fileSize)+1); FILE * fileHandle = NULL; fopen_s(&fileHandle, this->fileName, "rb"); fread(buffer, this->fileSize, sizeof(char), fileHandle); fclose(fileHandle); buffer[this->fileSize] = '\0'; return buffer; } VertexShader.glsl
#version 330 core layout (location = 0) vec3 VertexPositions; void main() { gl_Position = vec4(VertexPositions, 1.0f); } FragmentShader.glsl
#version 330 core out vec4 FragmentColor; void main() { FragmentColor = vec4(1.0f, 0.0f, 0.0f, 1.0f); } I am attempting to create a simple engine/graphics utility using some object-oriented paradigms. My first goal is to get some output from my engine, namely, a simple red triangle.
For this goal, the MyShapes class will be responsible for defining shapes such as triangles, polygons etc. Currently, there is only a drawTriangle() method implemented, because I first wanted to see whether it works or not before attempting to code other shape drawing methods.
The constructor of the MyEngine class creates a GLFW window (GLAD is also initialized here to load all OpenGL functionality), and the myEngine.run() method in Main.cpp is responsible for firing up the engine. In this run() method, the shaders get loaded from files via the help of my FileHandler class. The vertices for the triangle are processed by the myShapes.drawTriangle() method where a vertex array object, a vertex buffer object and vertrex attributes are set for this purpose.
The while loop in the run() method should be outputting me the desired red triangle, but all I get is a grey window area. Why?
(Note: I am aware that this code is not using any good software engineering practices (e.g. exceptions, error handling). I am planning to implement them later, once I get the hang of OpenGL.)
• By KarimIO
EDIT: I thought this was restricted to Attribute-Created GL contexts, but it isn't, so I rewrote the post.
Hey guys, whenever I call SwapBuffers(hDC), I get a crash, and I get a "Too many posts were made to a semaphore." from Windows as I call SwapBuffers. What could be the cause of this?
Update: No crash occurs if I don't draw, just clear and swap.
static PIXELFORMATDESCRIPTOR pfd = // pfd Tells Windows How We Want Things To Be { sizeof(PIXELFORMATDESCRIPTOR), // Size Of This Pixel Format Descriptor 1, // Version Number PFD_DRAW_TO_WINDOW | // Format Must Support Window PFD_SUPPORT_OPENGL | // Format Must Support OpenGL PFD_DOUBLEBUFFER, // Must Support Double Buffering PFD_TYPE_RGBA, // Request An RGBA Format 32, // Select Our Color Depth 0, 0, 0, 0, 0, 0, // Color Bits Ignored 0, // No Alpha Buffer 0, // Shift Bit Ignored 0, // No Accumulation Buffer 0, 0, 0, 0, // Accumulation Bits Ignored 24, // 24Bit Z-Buffer (Depth Buffer) 0, // No Stencil Buffer 0, // No Auxiliary Buffer PFD_MAIN_PLANE, // Main Drawing Layer 0, // Reserved 0, 0, 0 // Layer Masks Ignored }; if (!(hDC = GetDC(windowHandle))) return false; unsigned int PixelFormat; if (!(PixelFormat = ChoosePixelFormat(hDC, &pfd))) return false; if (!SetPixelFormat(hDC, PixelFormat, &pfd)) return false; hRC = wglCreateContext(hDC); if (!hRC) { std::cout << "wglCreateContext Failed!\n"; return false; } if (wglMakeCurrent(hDC, hRC) == NULL) { std::cout << "Make Context Current Second Failed!\n"; return false; } ... // OGL Buffer Initialization glClear(GL_DEPTH_BUFFER_BIT | GL_COLOR_BUFFER_BIT); glBindVertexArray(vao); glUseProgram(myprogram); glDrawElements(GL_TRIANGLES, indexCount, GL_UNSIGNED_SHORT, (void *)indexStart); SwapBuffers(GetDC(window_handle));
• By Tchom
Hey devs!
I've been working on a OpenGL ES 2.0 android engine and I have begun implementing some simple (point) lighting. I had something fairly simple working, so I tried to get fancy and added color-tinting light. And it works great... with only one or two lights. Any more than that, the application drops about 15 frames per light added (my ideal is at least 4 or 5). I know implementing lighting is expensive, I just didn't think it was that expensive. I'm fairly new to the world of OpenGL and GLSL, so there is a good chance I've written some crappy shader code. If anyone had any feedback or tips on how I can optimize this code, please let me know.
uniform mat4 u_MVPMatrix; uniform mat4 u_MVMatrix; attribute vec4 a_Position; attribute vec3 a_Normal; attribute vec2 a_TexCoordinate; varying vec3 v_Position; varying vec3 v_Normal; varying vec2 v_TexCoordinate; void main() { v_Position = vec3(u_MVMatrix * a_Position); v_TexCoordinate = a_TexCoordinate; v_Normal = vec3(u_MVMatrix * vec4(a_Normal, 0.0)); gl_Position = u_MVPMatrix * a_Position; } Fragment Shader
precision mediump float; uniform vec4 u_LightPos["+numLights+"]; uniform vec4 u_LightColours["+numLights+"]; uniform float u_LightPower["+numLights+"]; uniform sampler2D u_Texture; varying vec3 v_Position; varying vec3 v_Normal; varying vec2 v_TexCoordinate; void main() { gl_FragColor = (texture2D(u_Texture, v_TexCoordinate)); float diffuse = 0.0; vec4 colourSum = vec4(1.0); for (int i = 0; i < "+numLights+"; i++) { vec3 toPointLight = vec3(u_LightPos[i]); float distance = length(toPointLight - v_Position); vec3 lightVector = normalize(toPointLight - v_Position); float diffuseDiff = 0.0; // The diffuse difference contributed from current light diffuseDiff = max(dot(v_Normal, lightVector), 0.0); diffuseDiff = diffuseDiff * (1.0 / (1.0 + ((1.0-u_LightPower[i])* distance * distance))); //Determine attenuatio diffuse += diffuseDiff; gl_FragColor.rgb *= vec3(1.0) / ((vec3(1.0) + ((vec3(1.0) - vec3(u_LightColours[i]))*diffuseDiff))); //The expensive part } diffuse += 0.1; //Add ambient light gl_FragColor.rgb *= diffuse; } Am I making any rookie mistakes? Or am I just being unrealistic about what I can do? Thanks in advance
• By yahiko00
Hi,
Not sure to post at the right place, if not, please forgive me...
For a game project I am working on, I would like to implement a 2D starfield as a background.
I do not want to deal with static tiles, since I plan to slowly animate the starfield. So, I am trying to figure out how to generate a random starfield for the entire map.
I feel that using a uniform distribution for the stars will not do the trick. Instead I would like something similar to the screenshot below, taken from the game Star Wars: Empire At War (all credits to Lucasfilm, Disney, and so on...).
Is there someone who could have an idea of a distribution which could result in such a starfield?
Any insight would be appreciated
• I have just noticed that, in quake 3 and half - life, dynamic models are effected from light map. For example in dark areas, gun that player holds seems darker. How did they achieve this effect ? I can use image based lighting techniques however (Like placing an environment probe and using it for reflections and ambient lighting), this tech wasn't used in games back then, so there must be a simpler method to do this.
Here is a link that shows how modern engines does it. Indirect Lighting Cache It would be nice if you know a paper that explains this technique. Can I apply this to quake 3' s light map generator and bsp format ?
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# definite integral #5
• August 6th 2011, 07:17 PM
Random Variable
definite integral #5
Show that $\int_{0}^{1} \frac{x \ln x}{(1-x^{2})(\pi^2+\ln^{2} x )} \ dx = \frac{1}{4} + \frac{\psi_{0}(1)}{2} = \frac{1}{4} - \frac{\gamma}{2}$
where $\psi_{0}(x)$ is the digamma function and $\gamma$ is Euler's constant.
• August 13th 2011, 01:20 PM
Random Variable
Re: definite integral #5
It doesn't look like anyone is going to post a solution.
The solution requires two facts:
1) $\sum_{k=1}^{\infty} \frac{1}{a^{2}+k^{2}} = \frac{\pi}{2a} \coth {\pi a} - \frac{1}{2a^{2}}$
2) $\int_{0}^{\infty} \Big(\frac{e^{-t}}{t} - \frac{e^{-zt}}{1-e^{-t}} \Big) \ dt = \psi_{0}(z)$
$\int_{0}^{1} \frac{\ln x}{\pi^{2} + \ln^{2} x} \frac{x}{1-x^{2}} \ dx = \int_{0}^{1} \int_{0}^{\infty} \sin (t \ln x) e^{-\pi t} \frac{x}{1-x^{2}} \ dt \ dx$
assuming the integral converges absolutely, change the order of integration and let $u = -\ln x$
$= \int_{0}^{\infty} \int_{0}^{\infty} e^{-\pi t} \sin (-tu) \frac{e^{-2u}}{1-e^{-2u}} \ du \ dt$
$= - \int_{0}^{\infty} \int_{0}^{\infty} e^{-\pi t} \sin (tu) \sum_{n=1}^{\infty} e^{-2nu} \ du \ dt$
$= - \int_{0}^{\infty} e^{-\pi t} \sum_{n=1}^{\infty} \int_{0}^{\infty} \sin (tu) e^{-2nu} \ du \ dt$
$- \int_{0}^{\infty} e^{-\pi t} \sum_{n=1}^{\infty} \frac{t}{t^{2}+4n^{2}}} \ dt$
$= -\frac{1}{4} \int_{0}^{\infty} e^{-\pi t } t \sum_{n=1}^{\infty} \frac{1}{(\frac{t}{2})^{2} + n^{2}} \ dt$
$= \frac{1}{4} \int_{0}^{\infty} e^{-\pi t} \Bigg( \frac{2}{t} - \pi \coth \Big(\frac{\pi t}{2}\Big) \Bigg) \ dt$
continued in next post
• August 13th 2011, 01:32 PM
Random Variable
Re: definite integral #5
let $w = \pi t$
$= \frac{1}{4 \pi} \int_{0}^{\infty} e^{-w} \Bigg( \frac{2 \pi}{w} - \pi \coth \Big(\frac{w}{2} \Big) \Bigg) \ dw$
$= \frac{1}{4} \int_{0}^{\infty} e^{-w} \Big( \frac{2}{w} - \frac{1+e^{-w}}{1-e^{-w}} \Big) \ dw$
$= \frac{1}{4} \int^{\infty}_{0} e^{-w} \Big(\frac{2}{w} - \frac{2}{1-e^{-w}} + 1 \Big) \ dw$
$= \frac{1}{4} \int_{0}^{\infty} e^{-w} \ dw + \frac{1}{2} \int_{0}^{\infty} \Big( \frac{e^{-w}}{w} - \frac{e^{-w}}{1-e^{-w}} \Big) \ dw$
$= \frac{1}{4} + \frac{\psi_{0}(1)}{2} = \frac{1}{4} - \frac{\gamma}{2}$ | |
# Adding groundwater pumping¶
Developed by R.A. Collenteur & M. Bakker
In this example notebook it is shown how to simulate the effect of a pumping well on the groundwater levels. We will first create a TFN model with the net recharge as the single stress used to explain the observed heads. Second, this model is extended to include the effect of a pumping well on the heads by adding another stress model. The simulated heads are compared and it can be clearly seen how the addition of the pumping well improves the simulation of the heads.
This example was also shown at the 2018 General Assembly of the European Geophysical Union:
Bakker, M., Collenteur, R., Calje, F. Schaars (2018) Untangling groundwater head series using time series analysis and Pastas. In EGU General Assembly 2018.
[1]:
import pandas as pd
import pastas as ps
import matplotlib.pyplot as plt
ps.show_versions()
Python version: 3.7.8 | packaged by conda-forge | (default, Jul 31 2020, 02:37:09)
[Clang 10.0.1 ]
Numpy version: 1.18.5
Scipy version: 1.4.0
Pandas version: 1.1.2
Pastas version: 0.16.0b
Matplotlib version: 3.1.3
## 1. Read the time series from files¶
All time series for this example have been prepared as csv-files, which are read using the Pandas read_csv- method. The following time series are available:
• heads in meters above the Dutch National Datum (NAP), irregular time steps
• rain in m/d
• Makkink reference evaporation in m/d
• Pumping extraction rate in m$$^3$$/d. The pumping well stopped operating after 2012.
[2]:
head = pd.read_csv("data_notebook_5/head_wellex.csv",
index_col="Date", parse_dates=True, squeeze=True)
index_col="Date", parse_dates=True)
index_col="Date", parse_dates=True)
index_col="Date", parse_dates=True)
# Make a plot of all the time series
fig, ax = plt.subplots(4, 1, sharex=True, figsize=(10,5));
ax[0].legend()
ax[1].plot(rain, label="rain")
ax[1].legend()
ax[2].plot(evap, label="evap")
ax[2].legend()
ax[3].plot(well, label="well")
ax[3].legend()
[2]:
<matplotlib.legend.Legend at 0x7fbaf9807d50>
## 2. Create a Pastas Model¶
A pastas Model is created. A constant and a noisemodel are automatically added. The effect of the net groundwater recharge $$R(t)$$ is simulated using the ps.RechargeModel stress model. Net recharge is calculated as $$R(t) = P(t) - f * E(t)$$ where $$f$$ is a parameter that is estimated and $$P(t)$$ and $$E(t)$$ are precipitation and reference evapotranspiration, respectively.
[3]:
# Create the time series model
ml = ps.Model(head, name="groundwater")
# Add the stres model for the net recharge
rm = ps.RechargeModel(rain, evap, name="recharge",
rfunc=ps.Exponential, recharge=ps.rch.Linear())
ml.solve(noise=True)
ml.plot(figsize=(10,4))
# Let's store the simulated values to compare later
sim1 = ml.simulate()
res1 = ml.residuals()
n1 = ml.noise()
INFO: Cannot determine frequency of series Head: freq=None. The time series is irregular.
INFO: Inferred frequency for time series Prec: freq=D
INFO: Inferred frequency for time series Evap: freq=D
INFO: Time Series Evap was extended to 1985-01-16 00:00:00 with the mean value of the time series.
Fit report groundwater Fit Statistics
==================================================
nfev 25 EVP 51.79
nobs 3869 R2 0.12
noise True RMSE 0.33
tmin 1995-01-14 00:00:00 AIC -2.10
tmax 2018-01-12 00:00:00 BIC 29.20
freq D Obj 1.95
warmup 3650 days 00:00:00 ___
solver LeastSquares Interpolated No
Parameters (5 were optimized)
==================================================
optimal stderr initial vary
recharge_A 835.822437 ±62.49% 203.104730 True
recharge_a 540.839625 ±62.19% 10.000000 True
recharge_f -1.940472 ±11.00% -1.000000 True
constant_d 16.511745 ±3.23% 15.975755 True
noise_alpha 281.723634 ±27.23% 1.000000 True
Parameter correlations |rho| > 0.5
==================================================
recharge_A recharge_a 1.00
constant_d 0.83
recharge_a constant_d 0.86
Interpreting the results
As can be seen from the above plot, the observed heads show a clear rise whereas the simulated heads do not show this behaviour. The rise in the heads cannot be explained by an increased precipitation or a decreased evaporation over time, and it is likely another force is driving the heads upwards. Given the location of the well, we can hypothesize that the groundwater pumping caused a lowering of the heads in the beginning of the observations, which decreased when the pumping well was shut down. A next logical step is to add the effect of the pumping well and see if it improves the simulation of the head.
## 3. Add the effect of the pumping well¶
To simulate the effect of the pumping well a new stress model is added. The effect of the well is simulated using the ps.StressModel, which convoluted a stress with a response function. As a response function the ps.Hantush response function is used. The keyword-argument up=False is provided to tell the model this stress is supposed to have a lowering effect on the groundwater levels.
[4]:
# Add the stress model for the pumping well
sm = ps.StressModel(well/1e6, rfunc=ps.Gamma, name="well", settings="well", up=False)
# Solve the model and make a plot
ml.solve()
axes = ml.plots.decomposition(figsize=(10,8))
axes[0].plot(sim1) # Add the previously simulated values to the plot
INFO: Inferred frequency for time series Well: freq=D
Fit report groundwater Fit Statistics
==================================================
nfev 24 EVP 74.96
nobs 3869 R2 0.74
noise True RMSE 0.18
tmin 1995-01-14 00:00:00 AIC 6.35
tmax 2018-01-12 00:00:00 BIC 56.44
freq D Obj 1.93
warmup 3650 days 00:00:00 ___
solver LeastSquares Interpolated No
Parameters (8 were optimized)
==================================================
optimal stderr initial vary
recharge_A 718.943898 ±27.99% 203.104730 True
recharge_a 448.480751 ±28.07% 10.000000 True
recharge_f -1.864180 ±9.75% -1.000000 True
well_A -102.841843 ±9.58% -338.167845 True
well_n 1.286498 ±33.95% 1.000000 True
well_a 75.864224 ±57.75% 10.000000 True
constant_d 16.773074 ±1.53% 15.975755 True
noise_alpha 82.906003 ±15.38% 1.000000 True
Parameter correlations |rho| > 0.5
==================================================
recharge_A recharge_a 0.98
constant_d 0.50
recharge_a constant_d 0.61
recharge_f constant_d -0.77
well_n well_a -0.86
[4]:
[<matplotlib.lines.Line2D at 0x7fbb007840d0>]
Interpreting the results
The addition of the pumping well to simulate the heads clearly improved the fit with the observed heads. It can also be seen how the pumping well stops contributing to the lowering of the head after ~2014, indicating the pumping effect of the well has dampened out. The period it takes before the historic pumping has no effect anymore can be approximated by the length of the response function for the well (e.g., len(ml.get_step_response("well"))).
## 4. Analyzing the residuals¶
The difference between the model with and without the pumping becomes even more clear when analyzing the model residuals. The residuals of the model without the well show a clear upward trend, whereas the model with a model does not show this trend anymore.
[5]:
ml.residuals().plot(figsize=(10, 4))
res1.plot()
plt.legend(["Model with well", "Model without well"])
[5]:
<matplotlib.legend.Legend at 0x7fbafa225b50> | |
# Finding a neighbourhood interval with Taylor polynomial
I was lectured on this however I did not understand what should I do exactly. How can I find this interval?
Find a neighbourhood ($-\delta,\delta$) of $0$ for which the $3rd$ order Taylor polynomial $P_{3,0}$ of $f(x)=e^x$ is within $1/200$ of $f(x)$.
• You wrote "$3$rd order Taylor polynomial $P_{5,0}$." Did you want third or fifth order here? – Mark Viola Mar 10 '16 at 5:31
• @Dr.MV A typo, I meant third order... – NeoXx Mar 10 '16 at 5:44
From the extended mean value theorem, there exists a number $\xi \in (0,x)$ ($\xi \in (x,0)$)for $x>0$ ($x<0$) such that
$$e^x=1+x+\frac12x^2+\frac16 x^3+\frac1{24}e^{\xi}x^4$$
Then, the error $E(x)$ between the exponential function and the third order approximation is
$$E(x)=\frac{1}{24}e^{\xi}x^4$$
Note that we want to find a number $\delta$ such that $x\in (-\delta,\delta)$ implies $E(x)<1/200$ or
$$E(x)=\frac{1}{24}e^{\xi}x^4<1/200 \tag 1$$
Taking $x<(3/25)^{1/4}$ we have from $(1)$
\begin{align} E(x)&=\frac{1}{24}e^{\xi}x^4\\\\ &<\frac{1}{24}e^{(3/25)^{1/4}}x^4\\\\ &<1/200\\\\ &\implies x<e^{-(1/4)(3/25)^{1/4}}(3/25)^{1/4}\\\\ \end{align}
We may choose a smaller interval and the error will still be bounded. So, since $e^{-(1/4)(3/25)^{1/4}}\ge 1-(1/4)(3/25)^{1/4}$, if $x<(3/25)^{1/4}\,(1-(1/4)(3/25)^{1/4})$, then $E(x)<1/200$. | |
Lemma 33.35.12. Let $k$ be a field. Let $n \geq 0$. Let $\mathcal{F}$ be a coherent sheaf on $\mathbf{P}^ n_ k$. If $\mathcal{F}$ is $m$-regular, then $\mathcal{F}(m)$ is globally generated.
Proof. For all $d \gg 0$ the sheaf $\mathcal{F}(d)$ is globally generated. This follows for example from the first part of Cohomology of Schemes, Lemma 30.14.1. Pick $d \geq m$ such that $\mathcal{F}(d)$ is globally generated. Choose a basis $f_1, \ldots , f_ r \in H^0(\mathbf{P}^ n_ k, \mathcal{F})$. By Lemma 33.35.11 every element $f \in H^0(\mathbf{P}^ n_ k, \mathcal{F}(d))$ can be written as $f = \sum P_ if_ i$ for some $P_ i \in k[T_0, \ldots , T_ n]$ homogeneous of degree $d - m$. Since the sections $f$ generate $\mathcal{F}(d)$ it follows that the sections $f_ i$ generate $\mathcal{F}(m)$. $\square$
There are also:
• 4 comment(s) on Section 33.35: Coherent sheaves on projective space
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Vestnik YuUrGU. Ser. Mat. Model. Progr.: Year: Volume: Issue: Page: Find
Vestnik YuUrGU. Ser. Mat. Model. Progr., 2014, Volume 7, Issue 1, Pages 5–15 (Mi vyuru114)
Mathematical Modelling
Estimation of Vector Field of Systematic Errors of Radars Based on Multi-Tracking Data
D. A. Bedin
Institute of Mathematics and Mechanics, Ural Branch, Russian Academy of Sciencies, Ekaterinburg, Russian Federation
Abstract: The problem of identification of systematic errors of several radars based on multi-tracking data from moving objects (aircrafts) is considered. In case the model of spatial dependence of systematic errors is not completely known the identification leads to an ill-posed estimation problem. The author suggests an approach which provides a good estimate under these conditions. The basis of the approach is the local approximation of the unknown systematic errors as a function of geometric position. Position space is divided into the system of sufficiently small parts. In each part the vector of the local value of the systematic errors is estimated. Due to the ill-posedness of the problem only an uncertainty set can be identified; this set contains all possible vectors that can provide identical measurements. These uncertainty sets can be considered as a multivalued function of geometric position. Then the selection of a single-valued function of systematic errors out of a multivalued function is done on the basis of criterion the minimization of which enables to define the most “flat” function. The algorithm was tested on real trajectory tracking data.
Keywords: statistic estimation; systematic error; radar.
DOI: https://doi.org/10.14529/mmp140101
Full text: PDF file (878 kB)
References: PDF file HTML file
UDC: 621.396.969.34
MSC: 62P30
Citation: D. A. Bedin, “Estimation of Vector Field of Systematic Errors of Radars Based on Multi-Tracking Data”, Vestnik YuUrGU. Ser. Mat. Model. Progr., 7:1 (2014), 5–15
Citation in format AMSBIB
\Bibitem{Bed14} \by D.~A.~Bedin \paper Estimation of Vector Field of Systematic Errors of Radars Based on Multi-Tracking Data \jour Vestnik YuUrGU. Ser. Mat. Model. Progr. \yr 2014 \vol 7 \issue 1 \pages 5--15 \mathnet{http://mi.mathnet.ru/vyuru114} \crossref{https://doi.org/10.14529/mmp140101} | |
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Question from Mike: I have a hole which a 24 ft pool in it is 10" deep in the the centre and goes to 1" inch at the edge want to fill it in with dirt how many yards of dirt would I need to fill it in
Hi Mike,
You didn't specify the shape of the pool but I assume the surface is a circle of radius 24 feet.
I see the hole as a cylinder with base a circle of diameter 24 feet and height 1 inch, and a cone with a base a circle of diameter 24 feet and height 9 inches.
The two important facts here are that the volume of a cylinder is the area of the base tomes the height and the volume of a cone is one third the volume of the cylinder with the same base and height. Thus the volume of the cylindrical top layer is $\pi \; r^2 h$ where $r$ is the radius, 12 feet, and $h$ is the height, 1 inch. The volume of the cone is $\large \frac13 \normalsize \pi \; r^2 h$ where $r$ is again 12 feet but $h$ is 9 inches.
Now you can use our volume calculator. First use it to find the volume of the top layer. Next use it to find the volume of a cylinder with radius 12 feet and height 9 inches and take a third of the result to find the volume of the cone.
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Penny
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Transition operator between multipole states and their tensor structure
@article{Efimov1979TransitionOB,
title={Transition operator between multipole states and their tensor structure},
author={Sergey P. Efimov},
journal={Theoretical and Mathematical Physics},
year={1979},
volume={39},
pages={425-434}
}
• S. Efimov
• Published 1 May 1979
• Mathematics
• Theoretical and Mathematical Physics
10 Citations
Some useful correspondences in classical magnetostatics and multipole representations of the magnetic potential of an ellipsoid
It is shown that for a given geometric body, the Ferrers theorem not only relates the potentials of volume- and surface-distributed scalar (charge or mass) sources (which it is known to do) but alsoExpand
On the Frenkel problem of equivalent steady currents in a sphere
The Frenkel problem of substituting the 3D system of steady currents given in one of two concentric spherical regions by an equivalent system of currents (i.e., by that inducing the same externalExpand
Solution of two mutually-reducible problems in electrostatics
AbstractThere are two well-known problems in electrostatics whose solutions reduce to each other. One of them is that of a grounded conductor containing a cavity with given boundary $$\bar S$$ . AExpand
Schrödinger equation in the drift theory of cold plasma
AbstractA two-dimensional flow of an ideal neutral plasma across a magnetic field B is considered. The magnetic field is frozen in the plasma and is proportional to the plasma density n: B∝n. AllExpand
One-loop effective action in gauge gravitational theory
• Physics
• 1991
SummaryIn this paper we consider the problems of quantization and renormalization of gauge gravitational theory. We evaluate the one-loop effective action in two models which are based onGA(4,R)Expand
On the renormalisation group equations in curved spacetime with torsion
• Physics
• 1990
The renormalisation structure of gauge models in curved spacetime with torsion is considered. For Abelian theory the classical action, leading to multiplicatively renormalised theory, is constructed.Expand
Renormalization of Field Theories in Riemann-Cartan Space-Time
Eiemann-Cartan geometry with curvature and torsion arises naturally within the framework of Poincare gauge theory of gravity. The simplest example is given by the Einstein-Cartan theory (Kibble,Expand
Nonsingular cosmological models with torsion determined by vacuum polarization of quantum fields
• Physics
• 1987
Nonsingular cosmological models with reducing torsion induced by vacuum quantum effects are constructed. The metric of the models described corresponds to the inflationary universe. Moreover,Expand
Nonsingular cosmological model with torsion induced by vacuum quantum effects
• Physics
• 1985
Abstract We calculated the anomalous trace of the stress—energy tensor for massless scalar and spinor fields in curved spacetime with torsion. We found that the effective action dependence of metricExpand
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SHOWING 1-3 OF 3 REFERENCES
The Theory of Spherical and Ellipsoidal Harmonics
Preface 1. The transformation of Laplaces's equation 2. The solution of Laplace's equation in polar coordinates 3. The Legendres associated functions 4. Spherical harmonics 5. Spherical harmonics ofExpand
The classical theory of fields
• Physics
• 1952
The principle of relativity Relativistic mechanics Electromagnetic fields Electromagnetic waves The propagation of light The field of moving charges Radiation of electromagnetic waves Particle in aExpand | |
# Draw a circle and two lines parallel to a given
Question:
Draw a circle and two lines parallel to a given line such that one is tangent and other a secant to the circle.
Solution:
We have the required figure, as shown
Here, $\ell$ is the given line and a circle with centre $\mathrm{O}$ is drawn.
The line $\mathrm{n}$ is drawn which is parallel to $\ell$ and tangent to the circle. Also, $\mathrm{m}$ is drawn parallel to line $\ell$ and is a secant to the circle. | |
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# If 0 < a < b < c, which of the following statements must be true?
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If 0 < a < b < c, which of the following statements must be true? [#permalink]
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14 Jun 2016, 14:08
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If 0 < a < b < c, which of the following statements must be true?
I. 2a > b + c
II. c – a > b - a
III. $$\frac{c}{a}$$ < $$\frac{b}{a}$$
A) I only
B) II only
C) III only
D) I and II
E) II and III
OG 2017 New Question
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If 0 < a < b < c, which of the following statements must be true? [#permalink]
### Show Tags
Updated on: 03 Mar 2018, 14:51
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1
AbdurRakib wrote:
If 0 < a < b < c, which of the following statements must be true?
I. 2a > b + c
II. c – a > b - a
III. $$\frac{c}{a}$$ < $$\frac{b}{a}$$
A) I only
B) II only
C) III only
D) I and II
E) II and III
OG 2017 New Question
I. 2a > b + c
Consider this scenario: a = 1, b = 2 and c = 3. This meets the given condition that 0 < a < b < c.
HOWEVER, if we plug these values into statement I, we see that it is NOT the case that 2a > b + c
So, statement I NEED NOT BE TRUE
II. c – a > b - a
It's already given that c > b
If we subtract ANY VALUE (such as a) from both sides, the inequality remains valid.
So, statement II MUST BE TRUE
III. c/a < b/a
Consider this scenario: a = 1, b = 2 and c = 3. This meets the given condition that 0 < a < b < c.
HOWEVER, if we plug these values into statement III, we see that it is NOT the case that c/a < b/a
So, statement III NEED NOT BE TRUE
Cheers,
Brent
_________________
If you enjoy my solutions, you'll love my GMAT prep course.
Originally posted by BrentGMATPrepNow on 14 Jun 2016, 14:39.
Last edited by BrentGMATPrepNow on 03 Mar 2018, 14:51, edited 1 time in total.
##### General Discussion
Intern
Joined: 01 May 2015
Posts: 35
Re: If 0 < a < b < c, which of the following statements must be true? [#permalink]
### Show Tags
16 Jun 2016, 04:39
4
I. obvously we cannot say that it "must" be true.
II. c – a > b - a
Cancelling a on both sides
c > b
Definitely true as per question.
III. c/a < b/a
Multiply both sides by "a" (positive)
c < b
Definitely "not" true as per question.
So, only II is correct.
Intern
Joined: 16 Feb 2016
Posts: 2
Re: If 0 < a < b < c, which of the following statements must be true? [#permalink]
### Show Tags
24 Jun 2016, 13:21
Hey,
For (1),
0<a<b<c, can't we write it as
a<b....i
a<c...ii
2a<b+c ???? (addition property of inequalities)
Or
Is it like this property holds true only when abcd are different numbers??
Appreciate a clarification!
TIA.
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Re: If 0 < a < b < c, which of the following statements must be true? [#permalink]
### Show Tags
26 Jun 2016, 01:31
Nothing indicates what a may be greater than, so I is wrong.
III is always wrong because the order should be reversed.
II is correct.
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Re: If 0 < a < b < c, which of the following statements must be true? [#permalink]
### Show Tags
09 Mar 2019, 07:40
[1] Assign Test Values
Must follow rule: 0 < a < b < c,
a = 2
b = 3
c = 4
[2] Plug In & Calculate
$$I. 2a > b + c \rightarrow 2(2) > 3 + 4 \rightarrow 4 > 7 \space (FALSE)$$
$$II. c – a > b - a \rightarrow 4 - 2 > 3 - 2 \rightarrow 2 > 1 \space (TRUE)$$
$$III. \frac{c}{a} < \frac{b}{a} \rightarrow \frac{4}{2} < \frac{3}{2} \rightarrow 2 < 1.5 \space (FALSE)$$
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Re: If 0 < a < b < c, which of the following statements must be true? [#permalink]
### Show Tags
16 Apr 2019, 06:33
Shouldn't the question say a,b,c are integers?
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Re: If 0 < a < b < c, which of the following statements must be true? [#permalink]
### Show Tags
11 May 2019, 22:57
AbdurRakib wrote:
If 0 < a < b < c, which of the following statements must be true?
I. 2a > b + c
II. c – a > b - a
III. $$\frac{c}{a}$$ < $$\frac{b}{a}$$
A) I only
B) II only
C) III only
D) I and II
E) II and III
OG 2017 New Question
We can let smart numbers
a=2
so,b=3
c=4
Now, we'll apply on each options given☺
And we get option B(||) only matches
Posted from my mobile device
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Re: If 0 < a < b < c, which of the following statements must be true? [#permalink]
### Show Tags
04 Mar 2020, 07:12
AbdurRakib wrote:
If 0 < a < b < c, which of the following statements must be true?
I. 2a > b + c
II. c – a > b - a
III. $$\frac{c}{a}$$ < $$\frac{b}{a}$$
A) I only
B) II only
C) III only
D) I and II
E) II and III
OG 2017 New Question
I. 2a > b + c
Taking numbers a = 1, b = 2 and c = 3 this is not true
II. c – a > b - a
In 0 < a < b < c, subtracting a from each
0 - a < a - a < b - a < c - a
Hence c - a > b - a True
III. $$\frac{c}{a}$$ < $$\frac{b}{a}$$
As b < c
dividing by 'a' which is positive doesn't impact inequality
Hence $$\frac{b}{a} < \frac{c}{a}$$
not true
_________________
Re: If 0 < a < b < c, which of the following statements must be true? [#permalink] 04 Mar 2020, 07:12 | |
# American Institute of Mathematical Sciences
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## Implementation of the vehicular occupancy-emission relation using a cubic B-splines collocation method
1 Department of Computer Sciences, Faculty of Sciences and Techniques, University Moulay Ismail, BP 509 Boutalamine Errachidia, Morocco 2 Department of Mathematics, Laboratory LMPA, University Littoral Cote d'Opale, France
* Corresponding author: Sofiya Chergui
Received July 2018 Revised August 2018 Published March 2019
The complexity and non-linearity of flow phenomena are explained by numerous criteria, including the interactions of the large number of vehicles occupying the road, which influence the road density. This density under certain conditions, leads to traffic congestion which has dangerous effects on the environment such as; resources consumption; noise and the effect caused by greenhouse gas emissions of the $CO_{2}$ and other pollutants. In this paper we consider working in an uniform, homogeneous road where the traffic is described by the Lighthill Whitham-Richard (LWR) model resolved using a cubic B-spline collocation scheme in space and an implicit Runge Kutta scheme in time. We also shed light on the relation between vehicle occupancy and vehicle emissions.
Citation: Said Agoujil, Abderrahman Bouhamidi, Sofiya Chergui, Youssef Qaraai. Implementation of the vehicular occupancy-emission relation using a cubic B-splines collocation method. Discrete & Continuous Dynamical Systems - S, doi: 10.3934/dcdss.2020022
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Ohkura, Some findings and an overview on vehicular flow characteristics, In Proceedings of the 8th International Symposium on Transportation and Traffic Theory, Univ. of Toronto Press, Toronto, (1981), 403-426. [24] J. P. Lebacque, The godunov scheme and what it means for first order traffic flow models, In The International Symposium on Transportation and Traffic Theory, Lyon, France, (1996). [25] L. Leclercq, J. A. Laval and E. Chevallier, The lagrangian coordinates and what it means for first order traffic flow models, In R. Allsop and B. Heydecker (Eds), Transportation and traffic theory, (2007), 735-753. [26] M. Lighthill and G. Whitham, On kinematic waves. Ⅱ. A theory of traffic flow on long crowded roads, Proceedings of the Royal Society of London, Series A, 229 (1955), 317-345. doi: 10.1098/rspa.1955.0089. [27] W. Maden, R. W. Eglese and D. Black, Vehicle routing and scheduling with time varying data: A case study, Journal of the Operational Research Society, 61 (2010), 515-522. doi: 10.1057/jors.2009.116. [28] R. C. Mittal and R. Bhatia, Numerical solution of second order one dimensional hyperbolic telegraph equation by cubic B-spline collocation method, Applied Mathematics and Computation, 220 (2013), 496-506. doi: 10.1016/j.amc.2013.05.081. [29] R. C. Mittal and R. K. Jain, Redefined cubic B-splines collocation method for solving convection-diffusion equations, Applied Mathematical Modelling, 36 (2012), 5555-5573. doi: 10.1016/j.apm.2012.01.009. [30] R. C. Mittal and R. K. Jain, Cubic B-splines collocation method for solving nonlinear parabolic partial differential equations with Neumann boundary conditions, Commun Nonlinear Sci Numer Simulat, 17 (2012), 4616-4625. doi: 10.1016/j.cnsns.2012.05.007. [31] R. Mohammadi, Quintic B-spline collocation approach for solving generalized Black Scholes equation governing option pricing, Computers and Mathematics with Applications, 69 (2015), 777-797. doi: 10.1016/j.camwa.2015.02.018. [32] A. Palmer, The Development of an Integrated Routing and Carbon Dioxide Emissions Model for Goods Vehicles, Ph.D, thesis, Cranfield University, School of Management, 2007. [33] L. A. Pipes, An operational analysis of traffic dynamics, Journal of Applied Physics, 24 (1953), 274-281. doi: 10.1063/1.1721265. [34] K. Post, J. H. Kent, J. Tomlin and N. Carruthers, Fuel consumption and emission modelling by power demand and a comparison with other models, Transport. Res. Part A: Policy Pract, 18 (1984), 191-213. doi: 10.1016/0191-2607(84)90126-2. [35] P. I. Richards, Shock waves on the highway, Oper. Res, 4 (1956), 42-51. doi: 10.1287/opre.4.1.42. [36] B. Saka and I. Dag, Quartic B-spline collocation method to the numerical solutions of the Burgers'equation, Chaos, Solitons and Fractals, 32 (2007), 1125-1137. doi: 10.1016/j.chaos.2005.11.037. [37] J. G. Wardrop, Some theoretical aspects of road traffic research, Proceedings of the Institution of Civil Engineers, Part II, 1 (1952), 325-362. [38] J. Wenlong, Traffic Flow Models and Their Numerical Solutions, University of Science and Technology of China, 1998. [39] J. Wenlong, Traffic Flow Models and Their Numerical Solutions, University of California Davis, 2000. [40] G. C. K. Wong and S. C. Wong, A multi-class traffic flow model-an extension of LWR model with heterogeneous drivers, Transportation Research Part A, Policy and Practice, 36 (2013), 827-841. [41] N. Wu, A new approach for modeling of Fundamental Diagrams, Transportation Research Part A: Policy and Practice, 36 (2002), 867-884. doi: 10.1016/S0965-8564(01)00043-X.
show all references
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Popescu, Macroscopic modeling of road traffic by using hydrodynamic flow models, 20th Mediterranean Conference on Control and Automation, (2012). doi: 10.1109/MED.2012.6265612. [8] D. Catalin, D. Popescu and D. Stefanoiu, Fuzzy modeling and control for a road section, 18th International Conference on System Theory, (2014). [9] R. E. Chandler, R. Herman and E. W. Montroll, Traffic dynamics: Studies in car following, Operation Research, 6 (1958), 165-184. doi: 10.1287/opre.6.2.165. [10] C. F. Daganzo, Fundamentals of Transportation and Traffic Operations, Pergamon, 1997. doi: 10.1108/9780585475301. [11] S. Darbha, K. R. Rajagopal and V. Tyagi, A review of mathematical models for the flow of traffic and some recent results, Nonlinear Analysis, 69 (2008), 950-970. doi: 10.1016/j.na.2008.02.123. [12] C. De Boor, A Practical Guide to Splines, Springer-Velay, Berlin, 1978. [13] J. S. Drake, J. L. Schfer and A. May, A Statistical Analysis of Speed Density Hypotheses, Proceedings of the Third International Symposium on the Theory of Traffic Flow, Elsevier North-Holland, New York, 1967. [14] A. Esen and O. Tasbozan, Cubic B-spline collocation method for solving time fractional gas dynamics equation, Tbilisi Math. J., 8 (2015), 221-231. doi: 10.1515/tmj-2015-0024. [15] K. Fagerholt, G. Laporte and I. Norstad, Reducing fuel emissions by optimizing speed on shipping routes, Journal of the Operational Research Society, 61 (2010), 523-529. doi: 10.1057/jors.2009.77. [16] A. Franceschetti, D. Honhon, T. V. Woensel, T. Bektas and G. Laporte, The time-dependent pollution-routing problem, Transportation Research Part B: Methodological, 56 (2013), 265-293. [17] M. Gholamian and N. J. Saberi, Cubic B-splines collocation method for a class of partial integro-differential equation, Alexandria Engineering Journal, 57 (2018), 2157-2165. doi: 10.1016/j.aej.2017.06.004. [18] S. K. Godunov, A difference method for numerical calculations of discontinuous solutions of the equations of hydrodynamics, Matematicheskii Sbornik, 47 (1959), 271-306. [19] S. Gottlich, U. Ziegler and M. Herty, Numerical discretization of Hamilton-Jacobi equation on networks, Netw. Heterog. Media, 8 (2013), 685-705. doi: 10.3934/nhm.2013.8.685. [20] B. D. Greenshields, A study of traffic capacity, Proceedings Highway Research Board, 14 (1934), 448-477. [21] K. Han, H. Liu, V. V. Gayah, T. L. Friesz and T. Yao, A robust optimization approach for dynamic traffic signal control with emission considerations, Transportation Research Part C, 70 (2016), 3-26. doi: 10.1016/j.trc.2015.04.001. [22] O. Jabali, T. Van Woensel and A. G. de Kok, Analysis of travel times and CO2 emissions in time-dependent vehicle routing. Tech. rep., Eindhoven University of Technology; (2009). [23] M. Koshi, M. Iwasaki and I. Ohkura, Some findings and an overview on vehicular flow characteristics, In Proceedings of the 8th International Symposium on Transportation and Traffic Theory, Univ. of Toronto Press, Toronto, (1981), 403-426. [24] J. P. Lebacque, The godunov scheme and what it means for first order traffic flow models, In The International Symposium on Transportation and Traffic Theory, Lyon, France, (1996). [25] L. Leclercq, J. A. Laval and E. Chevallier, The lagrangian coordinates and what it means for first order traffic flow models, In R. Allsop and B. Heydecker (Eds), Transportation and traffic theory, (2007), 735-753. [26] M. Lighthill and G. Whitham, On kinematic waves. Ⅱ. A theory of traffic flow on long crowded roads, Proceedings of the Royal Society of London, Series A, 229 (1955), 317-345. doi: 10.1098/rspa.1955.0089. [27] W. Maden, R. W. Eglese and D. Black, Vehicle routing and scheduling with time varying data: A case study, Journal of the Operational Research Society, 61 (2010), 515-522. doi: 10.1057/jors.2009.116. [28] R. C. Mittal and R. Bhatia, Numerical solution of second order one dimensional hyperbolic telegraph equation by cubic B-spline collocation method, Applied Mathematics and Computation, 220 (2013), 496-506. doi: 10.1016/j.amc.2013.05.081. [29] R. C. Mittal and R. K. Jain, Redefined cubic B-splines collocation method for solving convection-diffusion equations, Applied Mathematical Modelling, 36 (2012), 5555-5573. doi: 10.1016/j.apm.2012.01.009. [30] R. C. Mittal and R. K. Jain, Cubic B-splines collocation method for solving nonlinear parabolic partial differential equations with Neumann boundary conditions, Commun Nonlinear Sci Numer Simulat, 17 (2012), 4616-4625. doi: 10.1016/j.cnsns.2012.05.007. [31] R. Mohammadi, Quintic B-spline collocation approach for solving generalized Black Scholes equation governing option pricing, Computers and Mathematics with Applications, 69 (2015), 777-797. doi: 10.1016/j.camwa.2015.02.018. [32] A. Palmer, The Development of an Integrated Routing and Carbon Dioxide Emissions Model for Goods Vehicles, Ph.D, thesis, Cranfield University, School of Management, 2007. [33] L. A. Pipes, An operational analysis of traffic dynamics, Journal of Applied Physics, 24 (1953), 274-281. doi: 10.1063/1.1721265. [34] K. Post, J. H. Kent, J. Tomlin and N. Carruthers, Fuel consumption and emission modelling by power demand and a comparison with other models, Transport. Res. Part A: Policy Pract, 18 (1984), 191-213. doi: 10.1016/0191-2607(84)90126-2. [35] P. I. Richards, Shock waves on the highway, Oper. Res, 4 (1956), 42-51. doi: 10.1287/opre.4.1.42. [36] B. Saka and I. Dag, Quartic B-spline collocation method to the numerical solutions of the Burgers'equation, Chaos, Solitons and Fractals, 32 (2007), 1125-1137. doi: 10.1016/j.chaos.2005.11.037. [37] J. G. Wardrop, Some theoretical aspects of road traffic research, Proceedings of the Institution of Civil Engineers, Part II, 1 (1952), 325-362. [38] J. Wenlong, Traffic Flow Models and Their Numerical Solutions, University of Science and Technology of China, 1998. [39] J. Wenlong, Traffic Flow Models and Their Numerical Solutions, University of California Davis, 2000. [40] G. C. K. Wong and S. C. Wong, A multi-class traffic flow model-an extension of LWR model with heterogeneous drivers, Transportation Research Part A, Policy and Practice, 36 (2013), 827-841. [41] N. Wu, A new approach for modeling of Fundamental Diagrams, Transportation Research Part A: Policy and Practice, 36 (2002), 867-884. doi: 10.1016/S0965-8564(01)00043-X.
Approximate density (veh/m)
Approximate density. Vs exact density (veh/m)
The variations of Link occupancy in time
The evolution of hydrocarbon emission rate in time
Aggregate emission rate vs. link occupancy
Flow.Vs density (fundamental diagram)
Travel speed
coefficients of $B_{j}$ and its derivatives
$x$ $x_{j-1}$ $x_{j}$ $x_{j+1}$ $B_{j}$ $\frac{1}{6}$ $\frac{4}{6}$ $\frac{1}{6}$ $B^{'}_{j}$ $\frac{-1}{2h}$ $0$ $\frac{1}{2h}$ $B^{''}_{j}$ $\frac{1}{h^{2}}$ $\frac{-2}{h^{2}}$ $\frac{1}{h^{2}}$
$x$ $x_{j-1}$ $x_{j}$ $x_{j+1}$ $B_{j}$ $\frac{1}{6}$ $\frac{4}{6}$ $\frac{1}{6}$ $B^{'}_{j}$ $\frac{-1}{2h}$ $0$ $\frac{1}{2h}$ $B^{''}_{j}$ $\frac{1}{h^{2}}$ $\frac{-2}{h^{2}}$ $\frac{1}{h^{2}}$
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2017 Impact Factor: 0.561 | |
# Getting started on contributing to open source and making software
I’m just graduated high school. I learnt java in high school, but the java I learnt was taught on bluej and we weren’t taught to create any software or anything. I want to start contributing to open source software, but when I look at the source files, it’s so overwhelming. It feels like the java I learnt in school did not make any difference at all.
Where should I get started?
• BlueJ is not the problem. It is just an IDE like any other but especially useful for teaching. Your issues lie elsewhere. But knowing the basics of a language isn't enough. Oct 2 '21 at 18:24
• Where should I get started? On sites like GitHub for issues they will sometimes tag them as easy enough for a beginner. See: Links for beginners willing to contribute to OpenSource projects I would make this an answer but it really just a link and link only answers really are not answers. But if that is acceptable here can easily make it an answer. Oct 3 '21 at 7:41
I would suggest not rushing into trying to contribute unless you have a very willing mentor: I suspect it's relatively unusual for people to contribute to packages/programs that they themselves don't use (indeed, I would be worried if this wasn't the case). That is, if you are not actually consuming any open-source Java packages/programs, then you probably are not well placed to contribute to them because you don't understand their purpose, design, and usage, so you'll have little hope understanding the context of the code you might see.
(Anecdote: I got into open-source by accident when a library I was using was missing a feature I wanted. It was a totally natural thing for me to report the missing feature, and then proceed to implement it: despite having never looked at the code-base before, I already knew my way around the APIs, I knew how I would actually use the feature (so it was already designed in my head), and I knew which other feature I would copy-and-paste to get myself going.)
As such, I would strongly suggest - if you are not already - working on some personal projects where you can use some open-source packages: you will naturally become familiar with their design, and either you'll find problems you want to fix (which is ideal), or you'll be able to understand the context of other people's issues which is often necessary to produce a useful contribution.
You should also use your own projects to teach yourself how to use the tools you'll need to make open-source contributions: version control (e.g. git), IDEs (e.g. IntelliJ), built-tools (e.g. Maven), testing (e.g. JUnit), etc. (it's unlikely that a high-school education has equipped you to use all of these tools well, though I could be wrong)
Finally, contributing to open-source is a great and noble goal, but I'd further suggest not rushing into it because your contributions may just create work for the maintainers if you're not up-to-scratch, and that can be a miserable experience for everyone. My experience has always been that maintainers and other contributors will want to help you make your contribution, but this doesn't mean they actually have time to help you. As said in the comments on your question, issues marked 'good-first-issue' or 'easy' or whatever probably signal good issues for someone contributing to the project for the first time, but that doesn't necessarily mean they are suitable for someone contributing to any project for the first time.
Some obligatory notes on how to put a contribution together, because there is a lot more than just writing code:
• Read all the documentation on how to contribute: you may need to agree to a license, or perform other 'house-work' the first time around
• Look at other contributions to see what they look like: read 'resolved' comments in code-reviews to see what the maintainers are saying so that you don't make the same mistakes
• Take time to configure your development environment: you must run style-checkers and automated tests yourself if possible, otherwise you waste CI and maintainer effort (which may translate into your contribution being rejected/ignored)
In sum: if you are looking at the source files without the necessary context, you're liable to get stuck: this is true for everyone, even when they have years of experience with the language, because there is so much more to building software software than just the choice of language. Instead, build your confidence and competence by making your own software, steadily introduce the tools you'll need to contribute to open-source projects, and find open-source packages that can help you with your own projects: you will inevitably find issues with them, and because you ran into this issues as a consumer, you will be in a much better position to report and deal with them.
Once you've got your hand in, then it'll be much easier to pick up other people's issues on the same (or similar) projects, because you'll understand the context, have some familiarity with the codebase and tooling, and you'll hopefully understand the (non-trivial) workflow of making a contribution. Also, there is a good chance that if you find a problem, that other people will run into it as well (or indeed it may already have been reported): there is nothing selfish about only fixing issues that affect you.
VisualMelon's answer is a good one; I wanted to extend the suggestion to start on solo projects.
Every team sport consists of members who need to both be individually skilled and fit, and work well together as a team. All of these things are hard to learn, and therefore more often than not it is better to learn these one at a time. Therefore, a team is composed out of players who are already individually skilled and fit, so that they can then learn to work together without also having to learn the basics at the same time.
The corollary here is that you should first hone your own development skill, before you start trying to take on team-based efforts. There are many conventions and (sometimes unspoken) rules about how to develop, which have no technical bearing (i.e. doing it in other ways works as well) but we've communally decided to all follow the same style just so your code and your colleague's code merges to form a codebase with a cohesive style and structure.
Based on where you are right now, I suggest focusing on personal projects, and firstly focusing on getting things to work, even if the code is not pretty. Prettier code is obviously better, but if you're struggling to get things to work, having to also jump through the "pretty code" hoop at the same time isn't going to make your life easier.
I would suggest a sequence of steps along the lines of this, only moving to the next step once you feel confident about your skills in the previous point.
• Making it work - Create personal projects, keep it simple, write most things yourself.
• I suggest tackling at least a few projects. They don't have to be big, but it's good to start several projects from scratch, because you will see that your initial approach improves with experience, and you will avoid mistakes in later projects from the get go.
• Making it better - Learn to improve and refactor your code. Could you have done things better? Is there a reusable method you could've added to avoid needless repetition? Were you using OOP correctly?
• Revisit the projects from the first bullet point, preferably chronologically. You'll probably be able to immediately improve things based on what you've learned during the other projects you did.
• Making it conventional - Try to apply generally agreed upon clean coding guidelines to your code. For OOP, a good start is SOLID and DRY.
• Definitely consider code reviews here. Conventions often don't make sense when you're by yourself (everyone has their own style), but will start making a lot more sense when you hear others' points of view.
• Working with existing code - Find some open-source libraries and create some personal projects that make use of these libraries. The goal here is to learn how to write code that works with a pre-established library/codebase. This is a subtly different skill from when you wrote everything yourself.
• Do not change the other person's code. The goal is to learn to work with the existing code, not rewrite it.
• If you struggle with any of the first three bullets point here, revisit them for these personal projects that implement other people's libraries. Using other people's code, which you cannot change, can bring to light mistakes you made in your personal projects (where you were able to rewrite parts to fit with your mistake).
• You can also find closed-source libraries to work with. This adds to the challenge by removing your ability to read the library code, meaning you can only learn from documentation and the interface the library has chosen to expose.
• Extending existing code - Once you feel ready to write "conventional" code (i.e. following the commonly agreed upon guidelines), find an open-source library, preferably one you've already worked with or closely understand the purpose of. Fork it, and try to develop a new feature.
• If you're struggling to get started on adding new features, focus on bug reports instead. This will help familiarize you with the codebase, and it's usually easier to spot a mistake in existing code than it is to develop something from scratch.
• You may want to do some ghost development, where you develop the fork but never intend to check it in. That being said, if you're confident about your work, definitely feel free to get someone to review it or, if you're really confident, offer it up via a pull request. | |
## COCI '08 Contest 1 #5 Skakavac
View as PDF
Points: 20 (partial)
Time limit: 1.8s
Memory limit: 35M
Problem type
Allowed languages
Ada, Assembly, Awk, Brain****, C, C#, C++, COBOL, CommonLisp, D, Dart, F#, Forth, Fortran, Go, Groovy, Haskell, Intercal, Java, JS, Kotlin, Lisp, Lua, Nim, ObjC, OCaml, Octave, Pascal, Perl, PHP, Pike, Prolog, Python, Racket, Ruby, Rust, Scala, Scheme, Sed, Swift, TCL, Text, Turing, VB, Zig
A grasshopper is in a flower field. The field contains flowers arranged in rows and columns. For each flower in the field, we know how many petals it has.
The grasshopper is initially on the flower in row and column . Its goal is to visit as many flowers as possible while obeying these rules:
1. It can only jump into an adjacent row or column. If it jumps into the adjacent row, it must jump at least two columns, and if it jumps into the adjacent column, it must jump at least two rows.
In other words, it can jump from flower to flower if:
• and or
• and
1. The number of petals on the next flower must be strictly larger than the number of petals on the previous flower.
Write a program that calculates the largest number of flowers the grasshopper can visit.
#### Input Specification
The first line contains the integer , the size of the field.
The second line contains integers and , the grasshopper's initial position.
The next lines contain positive integers separated by spaces, each less than , the numbers of petals on the flowers.
#### Output Specification
Output a single integer – the largest number of flowers the grasshopper can visit.
In test data worth of points, will be at most . In test data worth of points, will be at most .
#### Sample Input
4
1 1
1 2 3 4
2 3 4 5
3 4 5 6
4 5 6 7
#### Sample Output
4
#### Sample Input
5
3 3
20 16 25 17 12
11 13 13 30 17
15 29 10 26 11
27 19 14 24 22
23 21 28 18 13
#### Sample Output
21 | |
## Tuesday, April 20, 2021
### Abstract-Switchable generation of azimuthally- and radially-polarized terahertz beams from a spintronic terahertz emitter
Hiroaki Niwa, Naotaka Yoshikawa, Masashi Kawaguchi, Masamitsu Hayashi, and Ryo Shimano
(a) THz generation from a spintronic THz emitter. Transient spin current 𝐣s${\mathbf{j}}_{\mathrm{s}}$ generated upon photoexcitation converts into charge current 𝐣c${\mathbf{j}}_{\mathrm{c}}$ via the inverse spin Hall effect, which radiates THz electric field with the polarization perpendicular to the magnetization 𝐦̂ $\stackrel{^}{\mathbf{m}}$. (b) Schematic of the experimental setup. (c) and (d) Schematic presentation of generating azimuthal and radial polarization by converting HE21${\text{HE}}_{\text{21}}$ mode with different orientations, respectively.
https://www.osapublishing.org/oe/fulltext.cfm?uri=oe-29-9-13331&id=450178
We propose and demonstrate a method of generating two fundamental terahertz cylindrical vector beams (THz-CVBs), namely the azimuthally- and radially-polarized THz pulses, from a spintronic THz emitter. We begin by presenting that the spintronic emitter generates the HE21 mode, a quadrupole like polarization distribution, when placed between two magnets with opposing polarity. By providing an appropriate mode conversion using a triangular Si prism, we show both from experiment and numerical calculation that we obtain azimuthal and radial THz vector beams. The proposed method facilitates the access of CVBs and paves the way toward sophisticated polarization control in the THz regime.
© 2021 Optical Society of America under the terms of the OSA Open Access Publishing Agreement
## Monday, April 19, 2021
### Graphene: Everything under control in a quantum material
The gated graphene sample device in which the graphene film acts as a channel between source and drain electrodes subjected to a constant potential difference of 0.2 mV. Image from Science Advances
https://www.sciencedaily.com/releases/2021/04/210408131501.htm
In a new study, a team of researchers demonstrates that graphene's nonlinearity can be very efficiently controlled by applying comparatively modest electrical voltages to the material.
How can large amounts of data be transferred or processed as quickly as possible? One key to this could be graphene. The ultra-thin material is only one atomic layer thick, and the electrons it contains have very special properties due to quantum effects. It could therefore be very well suited for use in high-performance electronic components. Up to this point, however, there has been a lack of knowledge about how to suitably control certain properties of graphene. A new study by a team of scientists from Bielefeld and Berlin, together with researchers from other research institutes in Germany and Spain, is changing this. The team's findings have been published in the journal Science Advances.
Consisting of carbon atoms, graphene is a material just one atom thick where the atoms are arranged in a hexagonal lattice. This arrangement of atoms is what results in graphene's unique property: the electrons in this material move as if they did not have mass. This "massless" behavior of electrons leads to very high electrical conductivity in graphene and, importantly, this property is maintained at room temperature and under ambient conditions. Graphene is therefore potentially very interesting for modern electronics applications.
It was recently discovered that the high electronic conductivity and "massless" behavior of its electrons allows graphene to alter the frequency components of electric currents that pass through it. This property is highly dependent on how strong this current is. In modern electronics, such a nonlinearity comprises one of the most basic functionalities for switching and processing of electrical signals. What makes graphene unique is that its nonlinearity is by far the strongest of all electronic materials. Moreover, it works very well for exceptionally high electronic frequencies, extending into the technologically important terahertz (THz) range where most conventional electronic materials fail.
In their new study, the team of researchers from Germany and Spain demonstrated that graphene's nonlinearity can be very efficiently controlled by applying comparatively modest electrical voltages to the material. For this, the researchers manufactured a device resembling a transistor, where a control voltage could be applied to graphene via a set of electrical contacts. Then, ultrahigh-frequency THz signals were transmitted using the device: the transmission and subsequent transformation of these signals were then analyzed in relation to the voltage applied. The researchers found that graphene becomes almost perfectly transparent at a certain voltage -- its normally strong nonlinear response nearly vanishes. By slightly increasing or lowering the voltage from this critical value, graphene can be turned into a strongly nonlinear material, significantly altering the strength and the frequency components of the transmitted and remitted THz electronic signals.
"This is a significant step forward towards implementation of graphene in electrical signal processing and signal modulation applications," says Prof. Dmitry Turchinovich, a physicist at Bielefeld University and one of the heads of this study. "Earlier we had already demonstrated that graphene is by far the most nonlinear functional material we know of. We also understand the physics behind nonlinearity, which is now known as thermodynamic picture of ultrafast electron transport in graphene. But until now we did not know how to control this nonlinearity, which was the missing link with respect to using graphene in everyday technologies."
"By applying the control voltage to graphene, we were able to alter the number of electrons in the material that can move freely when the electrical signal is applied to it," explains Dr. Hassan A. Hafez, a member of Professor Dr. Turchinovich's lab in Bielefeld, and one of the lead authors of the study. "On one hand, the more electrons can move in response to the applied electric field, the stronger the currents, which should enhance the nonlinearity. But on the other hand, the more free electrons are available, the stronger the interaction between them is, and this suppresses the nonlinearity. Here we demonstrated -- both experimentally and theoretically -- that by applying a relatively weak external voltage of only a few volts, the optimal conditions for the strongest THz nonlin-earity in graphene can be created."
"With this work, we have reached an important milestone on the path towards to using graphene as an extremely efficient nonlinear functional quantum material in devices like THz frequency converters, mixers, and modulators," says Professor Dr. Michael Gensch from the Institute of Optical Sensor Systems of the German Aerospace Center (DLR) and the Technical University of Berlin, who is the other head of this study. "This is extremely relevant because graphene is perfectly compatible with existing electronic ultrahigh-frequency semiconductor technology such as CMOS or Bi-CMOS. It is therefore now possible to envision hybrid devices in which the initial electric signal is generated at lower frequency using existing semiconductor technology but can then very efficiently be up-converted to much higher THz frequencies in graphene, all in a fully controllable and predictable manner."
## Friday, April 16, 2021
### Abstract-Diversified functions for a terahertz metasurface with a simple structure
Wei-Mang Pan and Jiu-Sheng Li
(a) Three-dimensional schematic diagram of terahertz metasurface with diversified functions, (b) Designed unit cell with the relevant geometric parameters
https://www.osapublishing.org/oe/fulltext.cfm?uri=oe-29-9-12918&id=450048
Here, we propose a new encoded metasurface with different predesigned coding sequences to dynamic manipulate terahertz wavefront and realize various functionalities including beam splitting, anomalous beam deflection, vortex beam generation, angle controlled single-beam deflection, angle controlled multi-beam deflection, angle-controlled vortex beam generation and multi-vortex beam generation. The far-field scattering patterns obtained by CST Microwave Studio demonstrate the behavior of the terahertz wave in each case and shows a high consistency with our theoretical prediction results. Due to the excellent properties of the diversified functionalities in a single structure at terahertz frequencies, the proposed encoded metasurface provides promising applications in terahertz multiple-input, multiple-output (MIMO) communication.
© 2021 Optical Society of America under the terms of the OSA Open Access Publishing Agreement
## Thursday, April 15, 2021
### Abstract-Tunable and multifunctional terahertz devices based on one-dimensional anisotropic photonic crystals containing graphene and phase-change material
Xiangfei Gao, Zebin Zhu, Jing Yuan, and Liyong Jiang
Real part (a) and imaginary part (b) of the photonic bands of the graphene-Si 1D APC when kx = 0 and different chemical potentials are considered. (c) Corresponding transmission (T), reflection (R) and absorption (A) spectra of the graphene-Si 1D APC. The total layer number of 1D APC is 20. The green area represents the metallic band. The insets show the zoomed-in view for selected band gaps.
https://www.osapublishing.org/oe/fulltext.cfm?uri=oe-29-9-13314&id=450177
In the past few years, designing tunable and multifunctional terahertz devices has become a hot research area in terahertz science and technology. In this work, we report a study on one-dimensional anisotropic photonic crystals (1D APCs) containing graphene and phase-change material VO2. We numerically demonstrate the band-pass filtering, perfect absorption, comb-shaped extraordinary optical transmission and Fano-like resonance phenomenon in pure 1D APCs and 1D APCs with a VO2 defect layer under different conditions of a tangential wave vector. The performance of these phenomena in the terahertz region can be modulated by changing the chemical potential of graphene. The band-pass filter and perfect absorber functions of 1D APCs with a VO2 defect layer can be freely switched by changing the phase of VO2. We employ the equivalent-permittivity model and dispersion-relation equation to give reasonable explanations on these behaviors.
© 2021 Optical Society of America under the terms of the OSA Open Access Publishing Agreement
## Wednesday, April 14, 2021
### Abstract-Electrical tunability of terahertz nonlinearity in graphene
Sergey Kovalev, Hassan A. Hafez, Klaas-Jan Tielrooij, Jan-Christoph Deinert, Igor Ilyakov, Nilesh Awari, David Alcaraz. Karuppasamy Soundarapandian, David Saleta, Semyon Germanskiy, Min Chen, Mohammed Bawatna, Bertram Green, Frank H. L. Koppens, Martin Mittendorff, Mischa Bonn, Michael Gensch, Dmitry Turchinovich,
Graphene is conceivably the most nonlinear optoelectronic material we know. Its nonlinear optical coefficients in the terahertz frequency range surpass those of other materials by many orders of magnitude. Here, we show that the terahertz nonlinearity of graphene, both for ultrashort single-cycle and quasi-monochromatic multicycle input terahertz signals, can be efficiently controlled using electrical gating, with gating voltages as low as a few volts. For example, optimal electrical gating enhances the power conversion efficiency in terahertz third-harmonic generation in graphene by about two orders of magnitude. Our experimental results are in quantitative agreement with a physical model of the graphene nonlinearity, describing the time-dependent thermodynamic balance maintained within the electronic population of graphene during interaction with ultrafast electric fields. Our results can serve as a basis for straightforward and accurate design of devices and applications for efficient electronic signal processing in graphene at ultrahigh frequencies.
## Monday, April 12, 2021
### Terahertz Imaging of Graphene Paves the Way to Optimization and Industrialization
Graphene Flagship researchers have developed a new measurement standard for the analysis of graphene and layered materials that could accelerate production and optimize device fabrication. Credit: Graphene Flagship
Graphene Flagship researchers have developed a new measurement standard for the analysis of graphene and layered materials that could accelerate production and optimize device fabrication.
X-ray scans revolutionized medical treatments by allowing us to see inside humans without surgery. Similarly, terahertz spectroscopy penetrates graphene films allowing scientists to make detailed maps of their electrical quality, without damaging or contaminating the material. The Graphene Flagship brought together researchers from academia and industry to develop and mature this analytical technique, and now a novel measurement tool for graphene characterization is ready.
The effort was possible thanks to the collaborative environment enabled by the Graphene Flagship European consortium, with participation by scientists from Graphene Flagship partners DTU, Denmark, IIT, Italy, Aalto University, Finland, AIXTRON, UK, imec, Belgium, Graphenea, Spain, Warsaw University, Poland, and Thales R&T, France, as well as collaborators in China, Korea and the US.
Graphene is often ‘sandwiched’ between many different layers and materials to be used in electronic and photonic devices. This complicates the process of quality assessment. Terahertz spectroscopy makes things easier. It images the encapsulated materials and reveals the quality of the graphene underneath, exposing imperfections at critical points in the fabrication process. It is a fast, non-destructive technology that probes the electrical properties of graphene and layered materials, with no need for direct contact.
The development of characterization techniques like terahertz spectroscopy is fundamental to accelerating large-scale production, as they guarantee that graphene-enabled devices are made consistently and predictably, without flaws. Quality control precedes trust. Thanks to other developments pioneered by the Graphene Flagship, such as roll-to-roll production of graphene and layered materials, fabrication technology is ready to take the next step. Terahertz spectroscopy allows us to ramp up graphene production without losing sight of the quality.
Terahertz spectroscopy penetrates graphene films allowing scientists to make detailed maps of their electrical quality, without damaging or contaminating the material. Credit: Peter Bøggild (Graphene Flagship / DTU)
“This is the technique we needed to match the high-throughput production levels enabled by the Graphene Flagship,” explains Peter Bøggild from Graphene Flagship partner DTU. “We are confident that terahertz spectroscopy in graphene manufacturing will become as routine as X-ray scans in hospitals,” he adds. “In fact, thanks to terahertz spectroscopy you can easily map even meter-scale graphene samples without touching them, which is not possible with some other state-of-the-art techniques.” Furthermore, the Graphene Flagship is currently studying how to apply terahertz spectroscopy directly into roll-to-roll graphene production lines, and speed up the imaging.
Collaboration was key to this achievement. Graphene Flagship researchers in academic institutions worked closely with leading graphene manufacturers such as Graphene Flagship partners AIXTRON, Graphenea and IMEC. “This is the best way to ensure that our solution is relevant to our end-users, companies that make graphene and layered materials on industrial scales,” says Bøggild. “Our publication is a comprehensive case study that highlights the versatility and reliability of terahertz spectroscopy for quality control and should guide our colleagues in applying the technique to many industrially relevant substrates such silicon, sapphire, silicon carbide, and polymers.” he adds.
Setting standards is an important step for the development of any new material, to ensure it is safe, genuine and will offer a performance that is both reliable and consistent. That is why the Graphene Flagship has a dedicated work-group focused on the standardization of graphene, measurement and analytical techniques and manufacturing processes. The newly developed method for terahertz spectroscopy is on track to become a standard technical specification, thanks to the work of the Graphene Flagship Standardisation Committee. “This will undoubtedly accelerate the uptake of this new technology, as it will outline how analysis and comparison of graphene samples can be done in a reproducible way,” explains Peter Jepsen from Graphene Flagship Partner DTU, who co-authors the study. “Terahertz spectroscopy is yet another step to increase the trust in graphene-enabled products,” he concludes.
Amaia Zurutuza, co-author of the paper and Scientific Director at Graphene Flagship partner Graphenea, says: “At Graphenea, we are convinced that terahertz imaging can enable the development of quality control techniques capable of matching manufacturing throughput requirements and providing relevant graphene quality information, which is essential in our path towards the successful industrialization of graphene.”
Thurid Gspann, the Chair of the Graphene Flagship Standardisation Committee, says: “This terahertz [spectroscopy] technique is expected to be widely adopted by industry. It does not require any particular sample preparation and is a mapping technique that allows one to analyze large areas in a time efficient way.”
Marco Romagnoli, Graphene Flagship Division Leader for Electronics and Photonics Integration, adds: “The terahertz spectroscopy tool for wafer-scale application is a state-of-the-art, high TRL system to characterize multilayer stacks on wafers that contain CVD graphene. It works in a short time and with good accuracy, and provides the main parameters of interest, such as carrier mobility, conductivity, scattering time and carrier density. This high-value technical achievement is also an example of the advantage of being part of a large collaborative project like the Graphene Flagship.”
Andrea C. Ferrari, Science and Technology Officer of the Graphene Flagship and Chair of its Management Panel, adds: “Yet again, Graphene Flagship researchers are pioneering a new characterization technique to facilitate the development of graphene technology. This helps us progress steadily on our innovation and technology roadmap and will benefit the industrial uptake of graphene in a wide range of applications.”
Reference: “Case studies of electrical characterisation of graphene by terahertz time-domain spectroscopy” by Patrick R Whelan, Binbin Zhou, Odile Bezencenet, Abhay Shivayogimath, Neeraj Mishra, Qian Shen, Bjarke S Jessen, Iwona Pasternak, David M A Mackenzie, Jie Ji, Cunzhi Sun, Pierre Seneor, Bruno Dlubak, Birong Luo, Frederik W Østerberg, Deping Huang, Haofei Shi, Da Luo, Meihui Wang, Rodney S Ruoff, Ben R Conran, Clifford McAleese, Cedric Huyghebaert, Steven Brems, Timothy J Booth, Ilargi Napal, Wlodek Strupinski, Dirch H Petersen, Stiven Forti, Camilla Coletti, Alexandre Jouvray, Kenneth B K Teo, Alba Centeno, Amaia Zurutuza, Pierre Legagneux, Peter U Jepsen and Peter Bøggild, 17 February 2021, 2D Materials. | |
# zbMATH — the first resource for mathematics
Linguistic performance evaluation for an ERP system with link failures. (English) Zbl 1354.90021
Summary: An Enterprise Resource Planning (ERP) system is a complex network composed of various business processes. It can be called an ERP net. This paper proposes an analytic method to evaluate the Linguistic performance of such net under link failure situations. A link failure in an ERP net means that the software or hardware between processes may malfunction. To facility such evaluation, the nodes in the net denote the persons responsible for the business tasks during the processes. The links between nodes denote the process precedence relationships in the ERP system. When the process starts, the documents (jobs) are initiated from the source node to its succeeding nodes. Finally, the documents are released in the destination node. Thus, the performance of an ERP system is related to the document flow under the net. The performance failure of an ERP system is therefore defined by the condition that the document flow of the system is under the acceptable level $$d$$. By using the fuzzy linguistic results of the ERP examination of the users, we propose a fuzzy linguistic performance index, defuzzified from the probability of maximal flow not less than $$d$$, to evaluate the performance of an ERP system. An algorithm is subsequently proposed to generate the performance index under link failure situations, which can be used to real time assess the system performance either before or after the system going live.
##### MSC:
90B10 Deterministic network models in operations research 90C35 Programming involving graphs or networks 90C70 Fuzzy and other nonstochastic uncertainty mathematical programming
Full Text:
##### References:
[1] Al-Mashari, M.; Al-Mudimigh, A.; Zairi, M., Enterprise resource planning: a taxonomy of critical factors, Eur. J. Oper. Res., 146, 352-364, (2003) · Zbl 1012.90533 [2] Amoako-Gyampah, K.; Salam, A. F., An extension of the technology acceptance model in an ERP implementation environment, Inform. Manage., 41, 731-745, (2004) [3] Ball, M. O., Computational complexity of network reliability analysis: an overview, IEEE Trans. Reliab., 35, 230-238, (1986) · Zbl 0602.90061 [4] Chand, D.; Hachey, G.; Hunton, J.; Owhoso, V.; Vasudevan, S., A balanced scorecard based framework for assessing the strategic impacts of ERP systems, Comput. Indust., 56, 558-572, (2005) [5] Chang, I. C.; Hwang, H. G.; Hung, M. C.; Chen, S. L.; Yen, D. C., A neural network evaluation model for ERP performance from SCM perspective to enhance enterprise competitive advantage, Expert Syst. Appl., 35, 1809-1816, (2008) [6] Chen, S. G.; Lin, Y. K., An evaluation method for enterprise resource planning systems, J. Oper. Res. Soc. Jpn., 51, 299-309, (2008) · Zbl 1161.90321 [7] S.G. Chen, Y.K. Lin, Performance analysis for enterprise resource planning systems, in: Proceedings of the 2008 IEEE International Conference on Industrial Engineering and Engineering Management, Singapore, 2008, pp. 63-67. [8] Dowlatshahi, S., Strategic success factors in enterprise resource-planning design and implementation: a case-study approach, Int. J. Product. Res., 43, 3745-3771, (2005) [9] Ford, L. R.; Fulkerson, D. R., Flows in networks, (1962), Princeton University Press NJ · Zbl 0106.34802 [10] Jones, M. C.; Young, R., ERP usage in practice: an empirical investigation, Inform. Resour. Manage. J., 19, 23-42, (2006) [11] Kosko, B., Fuzzy engineering, (1997), Prentice-Hall Inc. · Zbl 0895.94015 [12] Lin, H. Y.; Lai, K. Y.; Shiau, W. L.; Hsu, P. Y.; Leu, J. D.; Tsai, W. H.; Cheng, M. S.; Fan, Y. W., The evaluation of post implementation ERP investment performance by DEA approach, J. E-Business, 6, 175-191, (2004) [13] Lin, W. T.; Chen, S. C.; Lin, M. Y.; Wu, H. H., A study on performance of introducing ERP to semiconductor related industries in Taiwan, Int. J. Adv. Manuf. Technol., 29, 89-98, (2006) [14] R. Sebastianelli, T.D. Rishel, Some survey results on ERP systems implementation, in: Proceedings - Annual Meeting of the Decision Sciences Institute, Washington, DC, United States, 2003, pp. 547-552. [15] Tsai, W. H.; Fan, Y. W.; Leu, J. D.; Chou, L. W.; Yang, C. C., The relationship between implementation variables and performance improvement of ERP systems, Int. J. Technol. Manage., 38, 350-373, (2007) [16] Wang, H. F., Multicriteria Decision Analysis - From Certainty to Uncertainty, (2005), Tsang Hai Book Publishing Co. Taiwan [17] F. Wu, C. Liu, H.Z. Li, K. Gao, J. Tian, The benefits evaluation of ERP project investment based on real options, in: Conference Proceedings - IEEE International Conference on Systems, Man and Cybernetics, vol. 5, Taipei, Taiwan, 2007, pp. 4078-4083. [18] Xue, J., On multistate system analysis, IEEE Trans. Reliab., 34, 329-337, (1985) [19] Yang, C. C.; Lin, W. T.; Pai, F. Y.; Yeh, T. M., The use of fuzzy measures in a performance-evaluation model for ERP implementation among taiwanese semiconductor manufacturers, Int. J. Product. Res., 45, 4735-4752, (2007) · Zbl 1126.90338 [20] Yu, C. S., Causes influencing the effectiveness of the post-implementation ERP system, Indust. Manage. Data Syst., 105, 115-132, (2005)
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | |
# Tag Archives: statistics
## G.E.P. Box on Model Building
The Edward library http://edwardlib.org/ has made the modeling approach of G.E.P. Box sound appealing to me. Here is some reading on it:
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Filed under Uncategorized
## ML vs Stats
What is the difference between machine learning and statistics? Can it be captured in a tweet?
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Filed under education
## Tukey quote I half-remembered
I was trying to remember some quote by the exploratory data analysis master John Tukey yesterday, and I think this is it:
No catalog of techniques can convey a willingness to look for what can be seen, whether or not anticipated. Yet this is at the heart of exploratory data analysis. The graph paper—and transparencies—are there, not as a technique, but rather as a recognition that the picture-examining eye is the best finder we have of the wholly unanticipated.
It is from John W. Tukey, We Need Both Exploratory and Confirmatory, The American Statistician, Vol. 34, No. 1 (Feb., 1980), pp. 23-25.
I remembered a version about the visual cortex as a the most advance signal processing device, so maybe there is another version of this out there.
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Filed under statistics
## Parameterizing Negative Binomial distributions
The negative binomial distribution is cool. Sometimes I think that.
Sometimes I think it is more trouble than it’s worth, a complicated mess.
Today, both.
Wikipedia and PyMC parameterize it differently, and it is a source of continuing confusion for me, so I’m just going to write it out here and have my own reference. (Which will match with PyMC, I hope!)
The important thing about the negative binomial, as far as I’m concerned, is that it is like a Poisson distribution, but “over-dispersed”. That is to say that the standard deviation is not always the square root of the mean. So I’d like to parameterize it with a parameter $\mu$ for the mean and $\delta$ for the dispersion. This is almost what PyMC does, except it calls the dispersion parameter $\alpha$ instead of $\delta$.
The slightly less important, but still informative, thing about the negative binomial, as far as I’m concerned, is that the way it is like a Poisson distribution is very direct. A negative binomial is a Poisson that has a Gamma-distributed random variable for its rate. In other words (symbols?), $Y \sim \text{NegativeBinomial}(\mu, \delta)$ is just shorthand for
$Y \sim \text{Poisson}(\lambda),$
$\lambda \sim \text{Gamma}(\mu, \delta).$
Unfortunately, nobody parameterizes the Gamma distribution this way. And so things get really confusing.
The way to get unconfused is to write out the distributions, although after they’re written, you might doubt me:
The negative binomial distribution is
$f(k \mid \mu, \delta) = \frac{\Gamma(k+\delta)}{k! \Gamma(\delta)} (\delta/(\mu+\delta))^\delta (\mu/(\mu+\delta))^k$
and the Poisson distribution is
$f(k \mid \lambda) = \frac{e^{-\lambda}\lambda^k}{k!}$
and the Gamma distribution is
$f(x \mid \alpha, \beta) = \frac{\beta^{\alpha}x^{\alpha-1}e^{-\beta x}}{\Gamma(\alpha)}$
Hmm, does that help yet? If $\alpha = \delta$ and $\beta = \delta/\mu$, it all works out:
$\frac{\Gamma(k+\delta)}{\Gamma(\delta)k!} \left(\frac{\delta}{\pi+\delta}\right)^\delta \left(\frac{\pi}{\pi+\delta}\right)^k = \int_0^\infty \frac{e^{-\lambda}\lambda^k}{k!} \lambda^{\delta-1} e^{-\lambda \delta/\mu} \frac{(\delta/\mu)^{\delta}}{\Gamma(\delta)}d \lambda.$
But instead of integrating it analytically (or in addition to), I am extra re-assured by seeing the results of a little PyMC model for this:
I put a notebook for making this plot in my pymc-examples repository. Love those notebooks. [pdf] [ipynb]
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Filed under statistics
## Life Expectancy by County in US
This recent study by my colleagues has been making headlines a lot last week, but I’m just getting to write about it now. While I was busy, stories about it appeared in high-profile outlets like NPR and the Statistical Modeling, Causal Inference, and Social Science blog.
As I’ve been thinking for two years (according to the ancient post I pushed out the door yesterday), life expectancy is a weird statistic. Life expectancy at birth is not, as the name might imply, a prediction on the average length of the life of a baby born this year. It is something more complicated to describe, but easier to predict. I like to think of it as the length of life if you froze the world exactly the way it is right now, and the baby today was exposed to the mortality risk of today’s one-year-olds next year, today’s two-year-olds in two years, etc. Although, as a friend pointed out two weeks ago, this is not a really good way to look at things either, if you push the analogy too hard. Currently Wikipedia isn’t really helpful on this matter, but maybe it will be better in the future.
There is another interesting thing in this paper, which is the validation approach the authors used. Unfortunately, it’s full development is in a paper still in press. Here is what they have to say about it so far:
We validated the performance of the model by creating small counties whose “true” underlying death rates were known. We did this by treating counties with large populations (> 750,000) as those where death rates have little sampling uncertainty. We then repeatedly sampled residents and deaths from these counties (by year and sex) to construct simulated small-county populations. We used the above model to predict mortality for these small, sampled-down counties, which were then compared with the mortality of the original large county.
I believe that this is fully developed in the paper which they cite at the beginning of the modeling section, Srebotnjak T, Mokdad AH, Murray CJL: A novel framework for validating and applying standardized small area measurement strategies, submitted. From what I’ve heard about it, I like it.
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Filed under global health
## Journal Culture
All fields have their quirks in publication style. Today I’m thinking about statistics, because I’ve been asked to explain something about survey weights to our post-bachelor’s fellows. There is a nice paper on the matter by Andrew Gelman, which starts strong, with first sentence “Survey weighting is a mess.” Start like that, and you’re sure to get a response from survey statisticians, who (at least I imagine) think of themselves as about as tidy as it comes.
The quirk in stats publications that I’m thinking of today is the Comment/Rejoinder format, wherein an article was published together with responses from several statisticians who don’t all agree with the article, and then a response from authors of the article. This is cool.
Unfortunately, Google scholar hasn’t kept up with this format, and searching for the paper title Struggles with Survey Weighting and Regression Modeling found me just one of the five comments. Project Euclid hasn’t kept up either, with only a tiny link from the article to the table of contents from the journal it appeared in. And thus I was forced to follow the obscure links in the pdf of the article to find the comprehensive list, which I’m putting here in case I need to find them all again sometime.
Statistical Science, Vol. 22, No. 2
Filed under statistics
## 9 Hours to Numeracy
I’m helping to plan an Introduction to Statistics for incoming post-bachelors fellows in the next month, and because of the wide range of backgrounds these recent college graduates will be coming from, I’m approaching it as a short course on numeracy (we’ve got about 9 hours of lecture time scheduled for it), focused on statistics. This will be complemented with a very hands-on dose of STATA, but I’m going to try not to think about that.
My favorite numeracy-in-stats book is a dusty classic, and it would have survived on its name alone: How to Lie with Statistics. I wonder if that title is too cheeky for global health applications when the numbers really matter…
Do you know this book, and do you like it? Or is there a more modern book or article that I should think of instead? What would you pack into 9 hours of stats numeracy training. Tell me. | |
# Multi-particle collision dynamics
Multi-particle collision dynamics (MPC), also known as stochastic rotation dynamics (SRD)[1], is a particle-based mesoscale simulation technique for complex fluids [2]. Coupling of embedded particles to the coarse-grained solvent is achieved through molecular dynamics [3].
## Method of simulation
The solvent is modelled as a set of $$N$$ point particles of mass $$m$$ with continuous coordinates $$\vec{r}_{i}$$ and velocities $$\vec{v}_{i}$$. The simulation consists of streaming and collision steps.
During the streaming step, the coordinates of the particles are updated according to
$\vec{r}_{i}(t+\delta t_{\mathrm{MPC}}) = \vec{r}_{i}(t) + \vec{v}_{i}(t) \delta t_{\mathrm{MPC}}$
where $$\delta t_{\mathrm{MPC}}$$ is a chosen simulation time step which is typically much larger than a molecular dynamics time step.
After the streaming step, interactions between the solvent particles are modelled in the collision step. The particles are sorted into collision cells with a lateral size $$a$$. Particle velocities within each cell are updated according to the collision rule
$\vec{v}_{i} \rightarrow \vec{v}_{\mathrm{CMS}} + \hat{\mathbf{R}} ( \vec{v}_{i} - \vec{v}_{\mathrm{CMS}} )$
where $$\vec{v}_{\mathrm{CMS}}$$ is the centre of mass velocity of the particles in the collision cell and $$\hat{\mathbf{R}}$$ is a rotation matrix. In two dimensions, $$\hat{\mathbf{R}}$$ performs a rotation by an angle $$+\alpha$$ or $$-\alpha$$ with probability $$1/2$$. In three dimensions, the rotation is performed by an angle $$\alpha$$ around a random rotation axis. The same rotation is applied for all particles within a given collision cell, but the direction (axis) of rotation is statistically independent both between all cells and for a given cell in time.
If the structure of the collision grid defined by the positions of the collision cells is fixed, Galilean invariance is violated. It is restored with the introduction of a random shift of the collision grid [4]. | |
# Am I wrong? Is this correct thinking? Am I weird? Am I crazy?
I was counting sheep last night and I change my vote. I can give you 1 sheep, you cannot give me pi sheep, eve if you cut a 4th sheep in half, there are a finite number of atoms in a sheep, so they cannot divide in (pi - 3)|(4-pi) You could then split the atoms but there is still a finite number of particles that make them up. You could argue this comes down to measuring, or you could argue this comes down to "Is matter infinitely divisibe?" Which I would think it is not.
That way, I also can't give you 1/3 sheep...
So I would say repeating numbers are equally as irrational as pi, but less "real" than 1, or other finite decimal numbers.
"Irrational" is a bad word. It has a definite meaning in mathematics.
But finite decimal numbers have the same problem. I can't give you 1/2 or 2/10 sheep as well...
Ah I see, unless it is an even-number of atoms in the sheep..which is not guaranteed.
I disagree on that as well. Pi still behaves as any other number in your example, and here's why I think that:
If we define the radius to be one, I agree that we cannot accurately measure the circumference, if we are talking about a mundane form of measurement. I also agree that if we define the circumference to be 1, we cannot mundanely measure the radius of that circle accurately. By mundanely, I simply mean with an apparatus alone.
Do you agree that if I define the edge of a square to be 3, we cannot accurately, mundanely measure its area? Apparatus alone, could we get its exact area to an infinite number of decimal places? No.
It is only by a mathematical relationship that we know the area of that square is exactly 9.
And I also know by a mathematical relationship that the circumference of your circle is exactly 2pi when the radius is our unit standard.
I think I see it simulary to you. Mathamaticaly the length units "work well" with length units.
I think curves complicate this, perfect circles infinitely so. (i.e. circumference / diameter) Lengths don't divide well into curves. Circle gets the square!
I certainly don't think Pi can be thought of as a physical unit.
Consider the way we measure the length of string of length 10 cm:
You can start by placing your ruler down, and starting at 0 cm, count by 1 cm until you get to the end of the string, and at at 10 cm - which is equivalent to adding 1 + 1 + 1... + 1 until you get 10.
Suppose it was now 10 and 1/3 cm:
You add 10 cm together, and then add an extra 1/3 cm. It doesn't matter if you can't write down what 1/3cm is in terms of mm or any other decimal, physically you can divide 1 cm into exactly three pieces - there's nothing stopping you from doing that.
Now try to get to 10 + pi cm - you start with adding 10 cm up, then you can add three more, then you have to add 1 mm, and then a little more, and a little more, and just a little bit more... and so on forever. You have to add an infinite number of smaller and smaller measurements to ever achieve that extra pi cm. You can never actually measure out an irrational physical quantity - and hence I don't think it's possible for physical quantities to obtain irrational values.
The fact that the billiard ball doesn't have an irrational surface area is because it's not a perfect sphere, and never will be. Just like you can't ever draw a perfect triangle with a hypotenuse of Sqrt(2).
I certainly don't think Pi can be thought of as a physical unit.
Consider the way we measure the length of string of length 10 cm:
You can start by placing your ruler down, and starting at 0 cm, count by 1 cm until you get to the end of the string, and at at 10 cm - which is equivalent to adding 1 + 1 + 1... + 1 until you get 10.
Suppose it was now 10 and 1/3 cm:
You add 10 cm together, and then add an extra 1/3 cm. It doesn't matter if you can't write down what 1/3cm is in terms of mm or any other decimal, physically you can divide 1 cm into exactly three pieces - there's nothing stopping you from doing that.
Now try to get to 10 + pi cm - you start with adding 10 cm up, then you can add three more, then you have to add 1 mm, and then a little more, and a little more, and just a little bit more... and so on forever. You have to add an infinite number of smaller and smaller measurements to ever achieve that extra pi cm. You can never actually measure out an irrational physical quantity - and hence I don't think it's possible for physical quantities to obtain irrational values.
The fact that the billiard ball doesn't have an irrational surface area is because it's not a perfect sphere, and never will be. Just like you can't ever draw a perfect triangle with a hypotenuse of Sqrt(2).
Again, counting exactly 1 cm and counting exactly pi cm are equally daunting tasks. You can measure 1.00000000000 cm perhaps, but that's no easier than measuring 3.141592653589 cm. Both are approximations to 1 and pi respectively when we are talking about measurement. 1.0 cm is not equal to exactly 1 cm, it means we stopped measuring after millimeters. You can't just ignore that in the rational number's case.
The issue here that computations with pi (and irrational numbers) can only work with rational approximations. This is a technological limitation, it has nothing to do with "not knowing pi exactly"; we do know pi exactly, we just can't work with it exactly. It's just a number.
Deveno
assuming we know how to define it, any real number is "knowable" in the sense that we can compute it to any desired accuracy.
i mean what is a mathematical relationship?
when we say a = f(b), for some formula involving b, there is a tacit appeal to a and b being "the same kind of something". if our formula is:
A = $\pi$r2, than A is not a rational number. so we have to have some kind of notion that allows us to say what equality of two real numbers IS.
analytically, we say that A converges to $\pi$. some people have trouble resolving this notion of convergence to our everyday notion of "is-ness". probably because it relies on the notion of a "completed infinity" ($\pi$ is not finitely expressible as a rational, it is an infinite sequence of such rationals).
my point being, saying a = b, when a and b are two real numbers, sweeps a lot of complicated machinery "under the rug", the truth of such a statement (and how complicated it actually is) is hidden behind a formalism which makes it appear as if it is just as simple as a statement like 2+2 = 4, or 1/3 + 2/3 = 1.
to go a bit further, how does one actually prove that the formula for an area of a circle actually holds? if you can show me a derivation that does not, in fact, rely on some limiting process (and thus a "completed infinity"), i will withdraw my statements.
pwsnafu
to go a bit further, how does one actually prove that the formula for an area of a circle actually holds?
Before one can do that you need to define the problem of finding "area of a circle" without analysis.
Deveno | |
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# system of inequalities calculator
example. how do you change the language to english on a TI-84 calculator? Calculates the solution of a system of two linear equations in two variables and draws the chart. Detailed step by step solutions to your Inequalities problems online with our math solver and calculator. What was Ernest Just mathematical accomplishment? This calculator solves system of three equations with three unknowns (3x3 system). Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. A demonstration of using the Inequalz app on the TI-84 calculator to find the solution region of a system of linear inequalities. Now that we have learned to graph nonlinear inequalities, we can learn how to graph systems of nonlinear inequalities. If you seek guidance on course syllabus or perhaps logarithmic, Algebra-equation.com is truly the right place to explore! The system of inequalities is written as follows: To solve this Or click the example. In this almond-and-peanut situation, you can’t have a negative number of ounces, so the inequalities fit. Casio also publicized advantages over competitors such as the reduction in keystrokes required for the function, and enhanced information provided such as tracing of inequalities and intersection points found. Graphing. Example Problem Solve the following system of equations: x+y=7, x+2y=11 How to Solve the System of Equations in Algebra Calculator. Basically, there are five inequality symbols used to represent equations of inequality. Free System of Inequalities calculator - Graph system of inequalities and find intersections step-by-step This website uses cookies to ensure you get the best experience. The values of x and y are never negative in a system with this requirement; they have to be positive or zero. Graph the solution set for this system. To solve a system of inequalities, graph each linear inequality in the system on the same x-y axis by following the steps below: Isolate the variable y in each linear inequality. It's a little hard to see, but after evenly shading each region, the intersecting region will be the most shaded in. 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Algebrator is a good software to solve graphing method calculator questions. How do you determine the common factors in an expression? Try the given examples, or type in your own problem and … algebretic calculator ; permutation material for GRE ; algebra practice factor square root ; how to solve equations with two variables ; how to find square root of number without calculator fomula ; multiplication of rational expression ; calculator solver sequence limits ; georgia high school eoct american literature released test booklet answer From the figure (see on the left to the right) it follows that: The obtained result can also be graphically displayed on a numeric axis: Our online calculator based on wolfram alpha system can solve more complex systems of inequalities than the one just discussed. A system of nonlinear inequalities is a system of two or more inequalities in two or more variables containing at least one inequality that is not linear. A system of two linear inequalities consists of linear inequalities for which we wish to find a simultaneous solution. 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If you are given a quadratic equation, how do you determine which method to use to solve it? SOLVE YOUR MATH PROBLEMS NOW For Windows OS. If ever you actually demand guidance with algebra and in particular with free online vertex calculator or solving exponential come visit us at Graph-inequality.com. Calculator will generate a step by step explanation using an addition/elimination method or Cramer's rule. Systems of inequalities are used when a problem requires a range of solutions, and there is more than one constraint on those solutions. Let's graph the solution set for each of these inequalities, and then essentially where they overlap is the solution set for the system, the set of coordinates that satisfy both. Algebra. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). consists of a set of two or more inequalities with the same variables. x We provide a good deal of really good reference material on topics starting from denominators to logarithmic For example, if asked to solve x+y ≤ 10 x + y ≤ 10, we first re-write as y ≤ −x+10 y ≤ − x + 10. In the event you seek help on absolute value or maybe grade math, Graph-inequality.com is the perfect destination to visit! Example (Click to view) x+y=7; x+2y=11 Try it now. How do you make an expression equivalent? inequality of the system. ... Systems of Equations. Free System of Inequalities calculator - Graph system of inequalities and find intersections step-by-step This website uses cookies to ensure you get the best experience. ... Equations Inequalities System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational Expressions Sequences Power Sums Induction Logical Sets. get Go. The inequalities define the conditions that are to be considered simultaneously. which satisfy inequality Solve Equations Calculus. Get the detailed answer: How do you find the solution of the sytem of inequalities using a graphing calculator? Middle School Math Solutions – Simultaneous Equations Calculator. Is there a basic difference between solving a system of equations by the algebraic method and the graphical method? Download free on Google Play. A system of inequalities is a set of two or more inequalities in one or more variables. Solve a system of inequalities. Explore math with our beautiful, free online graphing calculator. In the case of two variables, these systems can be thought of as lines drawn in two-dimensional space. Refer the below list of inequalities calculators such as Triangle inequality, Linear inequality, Bernoulli Inequality and so on. Graph-inequality.com contains both interesting and useful strategies on inequalities constraints online calculator, adding and subtracting rational expressions and adding and other algebra subject areas. 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For more intricate graphs, you can also use inequalities with restrictions to shade selected parts of the graph. To solve a system of two equations with two unknowns by substitution, solve for one unknown of one equation in terms of the other unknown and … how are equations with one variable and two variables alike, how are they different? Pre-Algebra. Free system of linear equations calculator - solve system of linear equations step-by-step. How To Solve Systems of Inequalities Graphically 1) Write the inequality in slope-intercept form or in the form y = mx+b y = m x + b. Practice: Systems of inequalities graphs. Download free on Amazon. With inequalities, you can add colored shading to your Desmos graph. Two solving methods + detailed steps. 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Please note that the symbol Inequality Calculator This calculator will solve the linear, quadratic, polynomial, rational and absolute value inequalities. System of inequalities online calculator The system of inequalities is written as follows: To solve this system of inequalities means to find the set of all acceptable values … This calculator solves system of two equations with two unknowns. For example, the pair of inequalities shown to the right is a system of linear inequalities.
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## HomeNewton MechanisHuygens Project
The objective of this project is two fold; on one hand to provide a small programming system which allows to play with the Huygens principle and on the other hand to elaborate important basic technologies for computing projects, such as:
• Calculation Program written in Ada
• Storing Result Data
• Visualization Software
• Creation of animations
This figure shows the field intensity of the wave field of a single spheric oscillator locate at (0,0) in the x/y plane.
The primary goal is to produce animations for different configurations of oscillators using my data center as computational platform..
## The Computation Problem
The computational problem is to calculate the wave field at a point $x$which is caused by a given configuration $\left\{ \vec{x_{i}}_{i}|i=1..N\right\}$of oscillators which are generating spherical waves described by:
$$\psi(r,t)=\frac{1}{r}sin(2\pi\frac{r+ct}{\lambda})$$ (1)
where $c$denotes the speed of propagation and $\lambda$the wave length. The wave field caused by the configuration of N oscillators at a given point $\vec{x}$is given by the superposition of all waves described by equation (1) which yields
$$\psi_{total}(\vec{x},t)=\sum_{i=1}^{N}\psi(\parallel\vec{\vec{x_{i}}-\vec{x}\parallel},t)$$ (2)
## The Program
A small Ada program computes formular (2) for a regular grid of $x$point in the probe grid and prints the values of $\psi(\vec{x},t)$into a file.
The configuration paramters of the program are shown in figure 2. All oscillators are located in the positve Y quadrant. The computation is done for point in the called probe grid which can be shifted in Y direction.
The programm is taking the followng arguments:
main config phaseshift time-start total-steps
• Config: Name of the configuration; a corresponding the configuration file of the name configuration.cfg is assumed to be located in the same directory where the program is called.
• Phase shift: Phase shift in X direction per oszillator
• Time-start: Start time of the computation.
• Total-steps: The number of time steps to be calculated. The time step is defined by the parameter DT in the configuration file.
Besides of the command line arguments a configuration file controls the behaviour of the computation. A configuration file consists of lines with assignments or comments. A commented line starts with a # sign. An assignment is of the for identifier=value. Each identifier specifies some parameter of the program.
An example of this file is shown below:
WL=0.5;
C=1.0;
## oscillator grid
NX=4;
NY=1;
D=3.0;
DT=0.1;
EA=88.0;
## probe grid used to display the results
Probe_N=200
Probe_Y=0.0;
• C: Speed of propagation; normaly set to 1
• D: Grid width in X/Y directions
• WL: Wavelength measured in D
• NX/NY: Number of oscillators in X and Y direction
• DT: Timestep width
• Probe_N: Number of cells in the visualisation grid
• Probe_Y: Y Offset of the visualsation grid
The result of the computation is stored in a simple ASCII file which can be read mostly as CSV format with headers as shown below:
X,Y,Intensity,Time
-7.4625000E+01,-7.4625000E+01, 1.6750029E-02, 0.0000000E+00
-7.4625000E+01,-7.3875000E+01, 1.6040666E-02, 0.0000000E+00
-7.4625000E+01,-7.3125000E+01, 1.4960118E-02, 0.0000000E+00
## Storing Results
Since for most of the data file type used in the scientific community are no Ada bindings avaiable i have simplified the situation by storing the results of the comuptation in a simple ASCII (csv like) file which can be converted into the format under discussion.
## Visualisation Tools
The following visualisation tools have been used during the project:
## Creating Animations
Movies as shown in Figure 1 will be created from N pictures representing scenes of the
video which are converted in a gif file which can contain different pictures in sequence.
The browser will display there pictures in sequence resulting the impression of an
anmation.<br>
Assuming you have a collection of jpeg files representing your scene in the current
directory the following shell commands are helpfull to automatize the process
of creating an animation.
mogrify -format gif *.jpg
gifsicle --colors=256 --delay=50 --loopcount=0 *.gif > result.gif
</pre>
### Example 0 - Single Oscillator
The basic test of the program described above is the caculation of a simple
spheric osccillator.
Please install the Flash Plugin The video above shows the result of the program described above for a sungle spheric osccilator. The animation has been created using paraview.
## Example 1 - Dipol
The first benchmark for all elaborations is the simple dipole configuration consiting of 4 oscillators in line and a probe field on 200x200 points.
This figure shows the wave field which is created by two synchronously oscillating wave sources.
Please install the Flash Plugin The video above shows the cacultation results of a set of 5 synchronous oscillators lines up parallel to the x-axis. | |
Algebra problem #68786
Algebra Level 2
What is the coefficient of $y$ in the expansion of the expression $(4x-y+4)(x+2y-3) ?$
×
Problem Loading...
Note Loading...
Set Loading... | |
# Finding equation of circle under the given geometric conditions
Finding the equation of the circle which touches the pair of lines $7x^2 - 18xy +7 y^2 = 0$ and the circle $x^2 + y^2 - 8x -8y = 0$ and contained in the given circle??
My attempt The centre of required circle would lie on angle bisector of the pair of lines ie $x=y$.
Assuming circle to be $(x-h)^2+(y-h)^2=r^2$
Now $2(h-8)^2=r^2$ ( distance between the extreme of larger circle and center of contained circle,)
I am unable to frame a second equation . One way would be to calculate the angle between pair of straight lines and use it to find a relation between $r$ and $h$.
However I was looking for a better solution or suggestion ?
• Going through your past question I notice that nobody has taken the time yet to tell you that we MathJax on this site to format our maths. You can find a basic tutorial here. Please have a look at it so you will be able to format your own posts. – gebruiker Oct 6 '17 at 17:01
• For the center I got $y=x$ or $y=-x$. – Michael Rozenberg Oct 6 '17 at 17:11
• @MichaelRozenberg forgot to mention that circle lies in first quadrant. – Seema Prasad Oct 6 '17 at 17:30
You are almost there. All you need is to use the fact that the circle center is equidistant from both lines. In particular, using the distance formula, you can write $$r^2 = \frac{((9+4\sqrt{2})h-7h)^2}{(9+4\sqrt{2})^2+7^2} = 2(h-8)^2 \implies h=6,12.$$ Since the center $(h,h)$ lies in the bigger circle, $h\neq 12$. Consequently, $h=6$ and $r^2=8.$ So $$(x-6)^2+(y-6)^2=8$$ is the equation of the sought circle.
Note that the tangent lines are $7y = (9\pm 4\sqrt{2})x.$
• Thanks . I was trying to avoid calculating the lines. – Seema Prasad Oct 6 '17 at 17:30
Hint:
The sought circle is the incircle of $\triangle ABC$.
The angle bisector of the two lines $y=\dfrac{1}{7} \left(9 \pm4 \sqrt{2} \right)x$ is the line $y=x$
The wanted circle has then centre $H(k,k)$ and its radius is the distance from the given lines $\left(9+4 \sqrt{2}\right) x-7 y=0$
$$r_k=\frac{\left|\left(9+4 \sqrt{2}\right) k-7 k\right|}{\sqrt{\left(9+4 \sqrt{2}\right)^2+49}}=\frac{\left|2 \left(1+2 \sqrt{2}\right) k\right|}{12+3 \sqrt{2}}$$ The wanted circle must also be tangent to the given circle $x^2-8 x+y^2-8 y=0$ having centre $C(4,4)$ and radius $R=4\sqrt{2}$
A circle is internally tangent to another when the distance of the centres is equal to the difference of the radii (in absolute value)
Therefore we must have $CH=R-r_k$
that is $$\sqrt{(4-k)^2+(4-k)^2}=4\sqrt{2}-\frac{\left|2 \left(1+2 \sqrt{2}\right) k\right|}{12+3 \sqrt{2}}$$ which simplified, noticing that $k>4$ becomes $$\sqrt{2}(k-4)=\frac{2 \left(1+2 \sqrt{2}\right) (12-k)}{3 \left(4+\sqrt{2}\right)}$$ After a look to the conditions we can say that $k>4$ so the previous equation becomes $$\left(12+3 \sqrt{2}\right) \sqrt{2} (k-4)=2 \left(1+2 \sqrt{2}\right) k$$
Solution is $k=6$ and the wanted circle has equation
$(x-6)^2+(y-6)^2=8\to \color{red}{x^2+y^2-12x-12y+64=0}$ | |
# Economic Minimum Life Analysis 1 answer below »
The problem must be done in Excel spreadsheet. Please show the cash flow diagram, procedure with equations and attach the excel spreadsheet.
Document Preview:
A self-employed worker operates a firewood-splitting service. He purchased a commercial-grade wood splitter for $5800. He used$400 of business capital and financed the balance at 5% per year for 3 years. The estimated values of the splitter for the next 6 years are $2200 after the first year of ownership, decreasing by$400 per year to year 5, after which the resale value remains at $600. Annual operating costs are expected to be$1000 the first year, increasing by 10% each year thereafter. He considers keeping the splitter at least 6 years. If money is worth 7% per year, for how many years should the splitter be retained? (Perform the Economic Minimum Life analysis for at least 12 years.) Please provide the following: 1. Cash flow diagram 2. A representative EUAW equivalency equation using Standard Notation corresponding to the CFD 3. The Excel spreadsheet and output 4. Excel created graphs from the results
Attachments: | |
## Recurrent Neural Networks - Deep Learning basics with Python, TensorFlow and Keras p.7
Recurrent Neural Networks (RNN) - Deep Learning basics with Python, TensorFlow and Keras p.7
Welcome to part 7 of the Deep Learning with Python, TensorFlow and Keras tutorial series. In this part we're going to be covering recurrent neural networks. The idea of a recurrent neural network is that sequences and order matters. For many operations, this definitely does.
Consider something like a sentence:
some people made a neural network
Then, let's say we tokenized (split by) that sentence by word, and each word was a feature.
Feeding through a regular neural network, the above sentence would carry no more meaning that, say:
a neural network made some people
Obviously, these two sentences have widely varying impacts and meanings!
This is where recurrent neural networks come into play. They attempt to retain some of the importance of sequential data.
With a Recurrent Neural Network, your input data is passed into a cell, which, along with outputting the activiation function's output, we take that output and include it as an input back into this cell.
This can work, but this means we have a new set of problems: How should we weight incoming new data? How should we handle the recurring data? How should we handle/weight the relationship of the new data to the recurring data? What about as we continue down the line? If we're not careful, that initial signal could dominate everything down the line.
This is where the Long Short Term Memory (LSTM) Cell comes in. An LSTM cell looks like:
The idea here is that we can have some sort of functions for determining what to forget from previous cells, what to add from the new input data, what to output to new cells, and what to actually pass on to the next layer.
If you'd like to know more, check out my original RNN tutorial as well as Understanding LSTM Networks.
Now let's work on applying an RNN to something simple, then we'll use an RNN on a more realistic use-case. I am going to have us start by using an RNN to predict MNIST, since that's a simple dataset, already in sequences, and we can understand what the model wants from us relatively easily.
In the next tutorial, we'll instead apply a recurrent neural network to some crypto currency pricing data, which will present a much more significant challenge and be a bit more realistic to your experience when trying to apply an RNN to time-series data.
We'll begin our basic RNN example with the imports we need:
import tensorflow as tf
from tensorflow.keras.models import Sequential
from tensorflow.keras.layers import Dense, Dropout, LSTM
The type of RNN cell that we're going to use is the LSTM cell. Layers will have dropout, and we'll have a dense layer at the end, before the output layer.
Similar to before, we load in our data, and we can see the shape again of the dataset and individual samples:
mnist = tf.keras.datasets.mnist # mnist is a dataset of 28x28 images of handwritten digits and their labels
(x_train, y_train),(x_test, y_test) = mnist.load_data() # unpacks images to x_train/x_test and labels to y_train/y_test
x_train = x_train/255.0
x_test = x_test/255.0
print(x_train.shape)
print(x_train[0].shape)
(60000, 28, 28)
(28, 28)
So, what is our input data here? Recall we had to flatten this data for the regular deep neural network. In this model, we're passing the rows of the image as the sequences. So basically, we're showing the the model each pixel row of the image, in order, and having it make the prediction. (28 sequences of 28 elements)
model = Sequential()
This should all be straight forward, where rather than Dense or Conv, we're just using LSTM as the layer type. The only new thing is return_sequences. This flag is used for when you're continuing on to another recurrent layer. If you are, then you want to return sequences. If you're not going to another recurrent-type of layer, then you don't set this to true.
We've not yet covered in this series for the rest of the model either:
opt = tf.keras.optimizers.Adam(lr=0.001, decay=1e-6)
model.compile(
loss='sparse_categorical_crossentropy',
optimizer=opt,
metrics=['accuracy'],
)
model.fit(x_train,
y_train,
epochs=3,
validation_data=(x_test, y_test))
Train on 60000 samples, validate on 10000 samples
Epoch 1/3
60000/60000 [==============================] - 189s 3ms/step - loss: 0.5922 - acc: 0.8056 - val_loss: 0.1395 - val_acc: 0.9601
Epoch 2/3
60000/60000 [==============================] - 186s 3ms/step - loss: 0.1686 - acc: 0.9538 - val_loss: 0.0797 - val_acc: 0.9751
Epoch 3/3
60000/60000 [==============================] - 187s 3ms/step - loss: 0.1137 - acc: 0.9692 - val_loss: 0.0638 - val_acc: 0.9815
<tensorflow.python.keras.callbacks.History at 0x177e2600b38>
Full code up to this point:
import tensorflow as tf
from tensorflow.keras.models import Sequential
from tensorflow.keras.layers import Dense, Dropout, LSTM#, CuDNNLSTM
mnist = tf.keras.datasets.mnist # mnist is a dataset of 28x28 images of handwritten digits and their labels
(x_train, y_train),(x_test, y_test) = mnist.load_data() # unpacks images to x_train/x_test and labels to y_train/y_test
x_train = x_train/255.0
x_test = x_test/255.0
print(x_train.shape)
print(x_train[0].shape)
model = Sequential()
# IF you are running with a GPU, try out the CuDNNLSTM layer type instead (don't pass an activation, tanh is required)
# Compile model
model.compile(
loss='sparse_categorical_crossentropy',
optimizer=opt,
metrics=['accuracy'],
)
model.fit(x_train,
y_train,
epochs=3,
validation_data=(x_test, y_test))
In the next tutorial, we're going to cover a more realistic timeseries example using cryptocurrency pricing, which will require us to build our own sequences and targets.
The next tutorial:
• Introduction to Deep Learning - Deep Learning basics with Python, TensorFlow and Keras p.1
• Convolutional Neural Networks - Deep Learning basics with Python, TensorFlow and Keras p.3
• Analyzing Models with TensorBoard - Deep Learning basics with Python, TensorFlow and Keras p.4
• Optimizing Models with TensorBoard - Deep Learning basics with Python, TensorFlow and Keras p.5
• How to use your trained model - Deep Learning basics with Python, TensorFlow and Keras p.6
• Recurrent Neural Networks - Deep Learning basics with Python, TensorFlow and Keras p.7
• Creating a Cryptocurrency-predicting finance recurrent neural network - Deep Learning basics with Python, TensorFlow and Keras p.8
• Normalizing and creating sequences for our cryptocurrency predicting RNN - Deep Learning basics with Python, TensorFlow and Keras p.9
• Balancing Recurrent Neural Network sequence data for our crypto predicting RNN - Deep Learning basics with Python, TensorFlow and Keras p.10
• Cryptocurrency-predicting RNN Model - Deep Learning basics with Python, TensorFlow and Keras p.11 | |
# zbMATH — the first resource for mathematics
The fixed point set of $$\mathbb{C}$$ actions on a compact complex space. (English) Zbl 0876.14032
Let $$G\times X\to X$$ be a holomorphic action of a unipotent complex linear algebraic group $$G$$ on a connected compact complex space $$X$$ which can be extended to a meromorphic map $$G^*\times X\to X$$ for an (arbitrary) algebraic compactification $$G^*$$ of $$G$$. Then the fixed point set $$X^G$$ of the $$G$$-action is connected and non-empty [see G. Horrocks, Topology 8, 233-242 (1969; Zbl 0159.22401) and J. B. Carrell and A. J. Sommese in: Group actions and vector fields, Proc. Pol.-North. Am. Semin., Vancouver 1981, Lect. Notes 956, 23-28 (1982; Zbl 0493.32026)]. Topologically the relation between $$X^G$$ and $$X$$ is in fact much closer. The author shows that the canonical injection $$X^G \subset X$$ induces an isomorphism between $$\pi_1(X^G)$$ and $$\pi_1(X)$$. A main ingredient for the proof is the author’s construction of a flat family of closures of $$G$$-orbits, birationally equivalent to $$X$$ [see A. Fujiki, Invent. Math. 44, 225-258 (1978; Zbl 0367.32004)].
As a consequence of the above results the Chow variety of $$q$$-cycles of degree $$d$$ in $$\mathbb{P}_n$$ is connected and simply connected.
##### MSC:
14L30 Group actions on varieties or schemes (quotients) 32M05 Complex Lie groups, group actions on complex spaces 32J18 Compact complex $$n$$-folds 14F35 Homotopy theory and fundamental groups in algebraic geometry | |
# Dynamic programming solution to knapsack problem
I wrote a solution to the Knapsack problem in Python, using a bottom-up dynamic programming algorithm. It correctly computes the optimal value, given a list of items with values and weights, and a maximum allowed weight.
Any critique on code style, comment style, readability, and best-practice would be greatly appreciated. I'm not sure how Pythonic the code is; filling in a matrix seems like a natural way of implementing dynamic programming, but it doesn't "feel" Pythonic (and is that a bad thing, in this case?).
Note that the comments are a little on the verbose side; as I was writing the algorithm, I tried to be as explicit about each step as possible, since I was trying to understand it to the fullest extent possible. If they are excessive (or just plain incorrect), feel free to comment (hah!) on it.
import sys
def knapsack(items, maxweight):
# Create an (N+1) by (W+1) 2-d list to contain the running values
# which are to be filled by the dynamic programming routine.
#
# There are N+1 rows because we need to account for the possibility
# of choosing from 0 up to and including N possible items.
# There are W+1 columns because we need to account for possible
# "running capacities" from 0 up to and including the maximum weight W.
bestvalues = [[0] * (maxweight + 1)
for i in xrange(len(items) + 1)]
# Enumerate through the items and fill in the best-value table
for i, (value, weight) in enumerate(items):
# Increment i, because the first row (0) is the case where no items
# are chosen, and is already initialized as 0, so we're skipping it
i += 1
for capacity in xrange(maxweight + 1):
# Handle the case where the weight of the current item is greater
# than the "running capacity" - we can't add it to the knapsack
if weight > capacity:
bestvalues[i][capacity] = bestvalues[i - 1][capacity]
else:
# Otherwise, we must choose between two possible candidate values:
# 1) the value of "running capacity" as it stands with the last item
# that was computed; if this is larger, then we skip the current item
# 2) the value of the current item plus the value of a previously computed
# set of items, constrained by the amount of capacity that would be left
# in the knapsack (running capacity - item's weight)
candidate1 = bestvalues[i - 1][capacity]
candidate2 = bestvalues[i - 1][capacity - weight] + value
# Just take the maximum of the two candidates; by doing this, we are
# in effect "setting in stone" the best value so far for a particular
# prefix of the items, and for a particular "prefix" of knapsack capacities
bestvalues[i][capacity] = max(candidate1, candidate2)
# Reconstruction
# Iterate through the values table, and check
# to see which of the two candidates were chosen. We can do this by simply
# checking if the value is the same as the value of the previous row. If so, then
# we say that the item was not included in the knapsack (this is how we arbitrarily
# break ties) and simply move the pointer to the previous row. Otherwise, we add
# the item to the reconstruction list and subtract the item's weight from the
# remaining capacity of the knapsack. Once we reach row 0, we're done
reconstruction = []
i = len(items)
j = maxweight
while i > 0:
if bestvalues[i][j] != bestvalues[i - 1][j]:
reconstruction.append(items[i - 1])
j -= items[i - 1][1]
i -= 1
# Reverse the reconstruction list, so that it is presented
# in the order that it was given
reconstruction.reverse()
# Return the best value, and the reconstruction list
return bestvalues[len(items)][maxweight], reconstruction
if __name__ == '__main__':
if len(sys.argv) != 2:
print('usage: knapsack.py [file]')
sys.exit(1)
filename = sys.argv[1]
with open(filename) as f:
lines = f.readlines()
maxweight = int(lines[0])
items = [map(int, line.split()) for line in lines[1:]]
bestvalue, reconstruction = knapsack(items, maxweight)
print('Best possible value: {0}'.format(bestvalue))
print('Items:')
for value, weight in reconstruction:
print('V: {0}, W: {1}'.format(value, weight))
The input file that it expects is as follows:
165
92 23
57 31
49 29
68 44
60 53
43 38
67 63
84 85
87 89
72 82
The first line contains the maximum weight allowed, and subsequent lines contain the items, represented by value-weight pairs.
-
An old post, I know, but if you haven't run into it already enumerate allows you to specify the starting index (e.g. for i,item in enumerate(items,1): would have i begin at 1. – SimonT Oct 22 '13 at 0:18
@SimonT: Fantastic! I've programmed in Python for 2+ years now, and I had never seen enumerate used that way. I'll definitely keep it in mind. – voithos Oct 22 '13 at 4:44
## 1 Answer
### 1. Comments on your code
1. Your function knapsack lacks a docstring that would explain what arguments the function takes (what kind of things are in items? must items be a sequence, or can it be an iterable?) and what it returns.
Also, this kind of function is ideal for doctests.
"""
Solve the knapsack problem by finding the most valuable
subsequence of items that weighs no more than maxweight.
items is a sequence of pairs (value, weight), where value is
a number and weight is a non-negative integer.
maxweight is a non-negative integer.
Return a pair whose first element is the sum of values in the most
valuable subsequence, and whose second element is the subsequence.
>>> items = [(4, 12), (2, 1), (6, 4), (1, 1), (2, 2)]
>>> knapsack(items, 15)
(11, [(2, 1), (6, 4), (1, 1), (2, 2)])
"""
2. Your comments say things like "Create an (N+1) by (W+1) 2-d list". But what is N and what is W? Presumably N is len(items) and W is maxweight, but this seems needlessly unclear. Better to put a couple of lines like this:
N = len(items)
W = maxweight
so that the comments match the code (and then use N and W in the remainder of the code).
3. The comment above bestvalues fails to explain what the values in this table actually mean. I would write something like this instead:
# bestvalues[i][j] is the best sum of values for any
# subsequence of the first i items, whose weights sum
# to no more than j.
(This makes it obvious why 0 ≤ i ≤ N and why 0 ≤ j ≤ W.)
4. In a loop like
bestvalues = [[0] * (maxweight + 1)
for i in xrange(len(items) + 1)]
where the loop variable (here i) is unused, it's conventional to name it _.
5. You can simplify the code by omitting these lines:
# Increment i, because the first row (0) is the case where no items
# are chosen, and is already initialized as 0, so we're skipping it
i += 1
and then using i + 1 instead of i and i instead of i - 1.
6. Your reconstruction loop:
i = N
while i > 0:
# code
i -= 1
can be written like this:
for i in xrange(N, 0, -1):
# code
7. You print an error message like this:
print('usage: knapsack.py [file]')
Error messages ought to go to standard error (not standard output). And you can't know that your program is called "knapsack.py": it might have been renamed. So write instead:
sys.stderr.write('usage: {0} [file]\n'.format(sys.argv[0]))
8. Your block of code that reads the problem description and prints the result only runs when __name__ == '__main__'. This makes it hard to test, for example from the interactive interpreter. It's usually best to put this kind of code in its own function, like this:
def main(filename):
with open(filename) as f:
# etc.
if __name__ == '__main__':
if len(sys.argv) != 2:
print('usage: knapsack.py [file]')
sys.exit(1)
main(sys.argv[1])
and now you can run main('problem.txt') from the interpreter to test it.
9. You read the whole of the file into memory as a list of lines:
lines = f.readlines()
this is harmless here because the file is small, but it's a bad habit to get into. It's usually best to process a file one line at a time if you can, like this:
with open(filename) as f:
maxweight = int(next(f))
items = [map(int, line.split()) for line in f]
(Note that this results in slightly simpler code too.)
### 2. A more Pythonic solution?
Any dynamic programming algorithm can be implemented in two ways: by building a table of partial results from the bottom up (as in your code), or by recursively computing the result from the top down, using memoization to avoid computing any partial result more than once.
There are two advantages of the top-down approach: first, it often results in slightly simpler and clearer code, and second, it only computes the partial results that are needed for the particular problem instance (whereas the bottom-up approach computes all partial results even if some of them go unused).
So we could use the @memoized decorator from the Python Decorator Library to implement a top-down solution, as shown below. (The Python wiki seems to be down at the moment, but you can find the code on archive.org.)
def knapsack(items, maxweight):
"""
Solve the knapsack problem by finding the most valuable
subsequence of items subject that weighs no more than
maxweight.
items is a sequence of pairs (value, weight), where value is
a number and weight is a non-negative integer.
maxweight is a non-negative integer.
Return a pair whose first element is the sum of values in the most
valuable subsequence, and whose second element is the subsequence.
>>> items = [(4, 12), (2, 1), (6, 4), (1, 1), (2, 2)]
>>> knapsack(items, 15)
(11, [(2, 1), (6, 4), (1, 1), (2, 2)])
"""
# Return the value of the most valuable subsequence of the first i
# elements in items whose weights sum to no more than j.
@memoized
def bestvalue(i, j):
if i == 0: return 0
value, weight = items[i - 1]
if weight > j:
return bestvalue(i - 1, j)
else:
return max(bestvalue(i - 1, j),
bestvalue(i - 1, j - weight) + value)
j = maxweight
result = []
for i in xrange(len(items), 0, -1):
if bestvalue(i, j) != bestvalue(i - 1, j):
result.append(items[i - 1])
j -= items[i - 1][1]
result.reverse()
return bestvalue(len(items), maxweight), result
To see how many partial solutions this code needs to compute, print len(bestvalue.cache) just before returning the result. When solving the example problem in the docstring, I find that this computes 37 partial solutions (compared to the 96 partial solutions computed by the bottom up approach).
-
Thanks for your response (especially points 3 and 9)! I can't believe I forgot to consider the memoized recursive method. I'll apply some of your suggestions later today, hopefully. Thanks again. – voithos Jan 16 '13 at 22:19
Beware of recursion depth limit issues with this proposed solution. – Frank Smith Mar 8 '14 at 15:13
@Frank: yes, this solution uses len(items) levels of stack. – Gareth Rees Mar 13 '14 at 8:26 | |
1. ## Topology-Deleted diameter plane
Prove that the deleted diameter plane is locally connected and path-connected.
2. Originally Posted by WannaBe
Prove that the deleted diameter plane is locally connected and path-connected.
What is your definition of "deleted diameter plane"?
3. Well, let's define a "deleted diameter circle" to be a circle in R^2 without a finite number of diameters (can be 0... ) but it still contains its center...The deleted diameter plane is the topological space which for every x in R^2 the set of all deleted diameter circles that their center is in x is a neighbourhood base of x...
Hope you'll be able to help me now because I'm pretty desperate...
Thanks | |
1. Sep 12, 2004
### Shay10825
Hi everyone! I really need help on these problems! I'm so confused . Sorry this post is so long and they are numbered weird.
6.) A rocket, initially at rest, is fired vertically with an upward acceleration of 10 m/s^2. At an altitude of .5 km, the engine of the rocket cuts off. What is the maximum altitude it achieves?
My work:
Vi= 0, a=10 m/s^2, d(or x) = 500 m
Now I'm stuck! I know you have to split it into 2 parts (one on the way up and one on the way down), and on the way down a=-9.8. But you don't know Vf or t so I don't know what to do.
10.) A boy on a skate board skates off a horizontal bench at a velocity of 10 m/s. One tenth of a second after he leaves the bench to two significant figures the magnitude of his velocity and his acceleration are what?
My work:
Would you put the information into vectors? and if so how? Did they give enough information?
ANSWER: 10 m/s ; 9.8 m/s^2
13.) A juggler throws two balls to the same height so that one is at the halfway point going up when the other is at the halfway point coming down. At that point:
a. Their velocities and accelerations are equal.
b. Their velocities are equal but their accelerations are equal and opposite.
c. Their accelerations are equal but their velocities are equal and opposite.
d. Their velocities and accelerations are both equal and opposite.
e. Their velocities are equal to their accelerations.
How do you know that the answer is c?
14.) A particle starts from the origin at t=0 with a velocity of 8j m/s and moves in the xy plane with a constant acceleration of (4i + 2j) m/s^2. At the instant the x coordinate of the particle is 29 m, what is the value of its y coordinate?
My Work:
xi= 0 , Vix=0, ax=4, xf=29, Viy=8, ay=2, yf=?, Vfy=0
0= 64+4d
-64=4d
d=-16
yf=-16
But this is not the correct answer :surprised :grumpy: ! It's not even close to any of the choices . ahhhhhhh!!
15.) A particle starts from the origin at t=0 with a velocity of 6i m/s and moves in the xy plane with a constant acceleration of (-2i + 4j) m/s^2. At the instant the particle achieves its maximum positive x coordinate, how far is it from the origin?
???????????????????
I don't know where to start. I don't understand how you would find this.
16.) A ball is thrown horizontally from the top of a building .1 km high. The ball strikes the ground at a point 65 m horizontally away from and below the point of release. What is the speed of the ball just before it strikes the ground?
17.) A rifle is aimed horizontally at the center of a large target 60 m away. The initial speed of the bullet is 240 m/s. What is the distance from the center of the target to the point where the bullet strikes the target?
You have 3 vectors that form a triangle and you need to find the third side. 60 is one side and 240 is the other. I found the angle between them to be 75.52. Then I used the law of cosines and found the third side to be 232.379. This is not even close to any of the choices .
19.) A carnival Ferris wheel has a 15-m radius and completes five turns about its horizontal axis every minute. What is the acceleration of a passenger at his lowest point during the ride?
My Work:
ar= (V^2)/(r)
C= 30pi so v= [(5*30pi)/1 min]
V = 150 m / 60 s
so: ar=.027
21.) Two cars are traveling around identical circular racetracks. Car A travels a constant speed of 20 m/s. Car B starts at rest and speeds up with constant tangential acceleration until its speed is 40 m/s. When car B has the same (tangential) velocity as car A, it is always true that:
a. it is passing car A
b. it has the same linear (tangential) acceleration as car A.
c. it has the same centripetal acceleration as car A.
d. it has the same total acceleration as car A.
e. it has traveled farther than car A since starting.
Any help would be greatly appreciated
~Thanks
Last edited: Sep 12, 2004
2. Sep 12, 2004
### Parth Dave
6) Well one thing is that Vi is not equal to zero. When x = 500 m, Vi is not 0. You're right, you can break this up into two parts. The first patr would be what the rocket does in the first 500 m. The second part would be what the rocket does until it gets to its maximum height (remember, you dont care what happens to the rocket on the way down, you are only concerned with how high it gets). Well if we look at part two, xi (initial displacement) = 500 m, however, a is not 10 ms^-2. It is actually -9.8 ms^-2. This is because the engine cuts off and the rocket doesn't generate any upward thrust. Also, when the rocket reaches its highest point, the velocity will be 0 (ie Vf = 0). When the rocket's velocity reaches 0, it is at its highest point. After that the velocity will become negative and the rocket will go down.
So for part 2, Vf = 0, a = -9.8 ms^-2, xi = 500 m, Vi = ?.
The only thing you don't know is the initial velocity. You can find the initial velocity from in the first part. Remember, the final velocity of the first part will be the initial velocity of the second part.
3. Sep 12, 2004
### Parth Dave
10. This is similar to a car driving off a cliff. You know what his velocity and acceleration (which happens to be 0) in the x direction is and the acceleration (-9.8 ms^-2) and initial velocity (0 ms^-1)in the y direction. So just find the velocity in the x and y direction after 0.1 s and than find the magnitude. Repeat for acceleration.
4. Sep 12, 2004
### Leong
Question #10
$$\vec{a}=\frac{\vec{v}-\vec{u}}{t}$$
$$\vec{u}=10\vec{i}$$
$$-9.81\vec{j}=\frac{\vec{v}-10\vec{i}}{\frac{1}{10}}$$
$$\vec{v}=10\vec{i}-0.981\vec{j}$$
$$|\vec{v}|=10\ m/s$$
$$\vec{g}=-9.81\vec{j}$$
$$|\vec{g}|=9.8 \ m/s^2$$
5. Sep 12, 2004
### Leong
Question #19
v=r*w
$$w=\frac{5*2\pi}{60}\ rev/s$$
Last edited: Sep 12, 2004
6. Sep 12, 2004
### Leong
Question #14
You can't assume $$v_{yfinal}=0$$
Use
$$s_{x/y}=u_{x/y}t+\frac{1}{2}a_{x/y}t^2$$
Find t for $$s_x=29$$
Substitute the time t to find $$s_y$$
7. Sep 13, 2004
### Shay10825
What does w r and v represent?
Last edited: Sep 13, 2004
8. Sep 13, 2004
### amwbonfire
Question 19
v=Velocity, in m/s
w=time (I use 't', i've never seen 'w' used before...)
I think this solution to this problem was completed incorrectly. Here's how it should be done:
Use
$$v=circumference/t$$
where t=time
Remember that circumference = 2 x pi x r
r=2.pi.r
t=60/5=12
v=?
Find v using the formula.
Now use this formula to find the acceleration:
$$a=v/r$$
where v=average velocity.
You're not finished yet though. You'll need to add this acceleration to gravity (9.8ms) because gravity also acts downwards. So you'll get a + 9.8
Andy
AMW Bonfire
Last edited by a moderator: Sep 13, 2004
9. Sep 13, 2004
### amwbonfire
Any others you need help with?
Andy
AMW Bonfire
10. Sep 13, 2004
### amwbonfire
Don't worry too much about the numbers in this question, they're just there to throw you off. The question is say, when both cars are travelling at the same speed, 20m/s, which of (a,b,c,d,e) is true?
Use the process of elimination to solve multi choice questions.
a) is wrong because B cannot pass A if they are both travelling at the same speed. Hence this answer is wrong.
b) The questions states that car A has no linear acceleration, while car B does. Therefor, they can't be equal. Hence this answer is wrong.
c) Centripital acceleration acts towards the center of the circle, and is given by the formula, a=v/t. If the velocities are the same, then the accelerations must be the same. Hence, this answer is correct.
d) Total acceleration means centripital acceleration + linear acceleration. Both centripital accelerations are equal (same velocity, see part c). However,
their linear (tangential) accelerations are different (see part b). Hence this answer is wrong.
e) This is making an assumption, which you can't do. Hence this answer is wrong.
So, c) it is!
I must admit, this question was pretty difficult. It could have been worded a lot better.
Andy
AMW Bonfire
11. Sep 13, 2004
### Shay10825
Thanks
yeah i need help with 13, 15, and 17
12. Sep 13, 2004
### amwbonfire
Ok, I'll do them now. :tongue2:
So are you learning all this stuff now, in physics? I'm only a bit ahead of you, I learnt this stuff a few months ago.
13. Sep 13, 2004
### amwbonfire
13
a) Wrong - their accelerations are equal. Gravity is accelerating the one going downwards, and deccelerating the one going up, and gravity is a constant. Since gravity is the only accelerating force acting on them, their acceleration is equal, (9.8m/s^2 downwards.)
However, their velocities are not equal. They are opposite (one is going up, one's going down.)
b) Wrong - See a) Velocities aren't equal, they're equal and opposite
c) Correct - See part a) and b)
d) Wrong - See the above parts
e) Wrong - See the above parts
Make sense?
Andy
AMW Bonfire
14. Sep 13, 2004
### amwbonfire
15
Don't panic!
Whenever you get a calculation, always write down what you know. It makes everything easier to see.
v(initial)=6i m/s
v(final)=?
a= (-2i + 4j) m/s^2
starts at coordinate (0,0)
need to find max x point
Sorry, I've run out of time. I'll help you some more, when I get home. But try and see where you can go from here. You need to find where x is greatest. Instead of x, think of it as displacement, then use the equations of motion to work out where displacement is greatest. Then you're done!
Andy
AMW Bonfire
15. Sep 13, 2004
### amwbonfire
17
All I can suggest is review your trig.
I'll have a look when I get home.
Andy
AMW Bonfire
16. Sep 14, 2004
### Leong
Question #19
I guess i have to find alternative since Latex has not been working since yesterday. Have a look at the attached file. Forget about the 'w', i don't want to confuse you. just use the circumference given by 2*pi*r.
#### Attached Files:
• ###### Explain.gif
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2.8 KB
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17. Sep 14, 2004
### Leong
Question #17 : 2D Projectile Motion
You can't use a right triangle for a projectile motion.
#### Attached Files:
File size:
1.9 KB
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157
• ###### Equation.GIF
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1.9 KB
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173
18. Sep 14, 2004
### Leong
Question #15
Have a look at the attached file.
#### Attached Files:
• ###### Equation.GIF
File size:
2.4 KB
Views:
152
19. Sep 15, 2004
### doyle_43
Hi People.
As you can tell I’m new here, and I’m having problems with my Physics homework.
I have missed a few of my lessons recently due to illness, and now it’s all mumbo jumbo.
Anyway, I kind of understand this, but I’m still getting a few wrong.
(Sorry that I have posted in this thread, but its stupid posting the same topic in a different thread.)
Having problems with this:
(6) A rocket is fired vertically upwards with an initial velocity of 100m/s^2. Calculate:
a. the time for the rocket to reach its greatest hight >answer> 10 seconds
b. the hight to which the rocket rises >answer> 500 meters
c. its velocity after 4.0 seconds >answer> 60 m/s^2
d. the acceleration at the top of the path of the rocket >answer> 9.8 m/s^2
e. when the rocket is 400 meters above the ground. >answer> 5.5 seconds
If anyone could explain how to get these answers, it would be much appreciated. :uhh:
I am getting very confused with the Kinematics Equations.
Thanks again for anyone who can help!
-- doyle_43
20. Sep 15, 2004
### Leong
Doyle,
Have a look ....
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# tikz in LaTeX and Structural Equation Modeling
Standard
During grad school, I attended an ESA Workshop on Structural Equation Modeling (SEM) let by Jim Grace. The approach allows for multivariate analysis with multiple predictors, multiple response variables, and latent variables. Up until now, my research never required using the method and I never bought the software he recommended at the time because the GUI program recommended by Grace was too expensive for my limited needs.
Recently, I had a need to use SEM at work. We had two response variables: environmental DNA (eDNA) and the ash-free dry weight of an aquatic organism (AFDW). Both were predicted by multiple environmental variables and AFDW predicted eDNA. A perfect problem for SEM.
To refresh myself of SEM, I revisited Grace’s work. I discovered that he maintains an excellent tutorial about SEM. The pages provide a nice introduction, as does his (slightly outdated) book, his classic book, and a recent Ecoshephere article.
However, I did not have a nice way to plot my results. I did not want to use a WYSIWYG tool like Inkscape or Power Point. But I remembered the tikz package in LaTeX. Here’s the figure I created:
Example SEM plot.
I created the figure using this LaTeX code:
\documentclass{article}
\usepackage[paperheight =11.3cm, paperwidth =9.5cm, margin = 0.1cm]{geometry} \usepackage{tikz} \usetikzlibrary{arrows} \usetikzlibrary{positioning} \begin{document} \pagenumbering{gobble} \begin{tikzpicture}[ -> , >=stealth',auto,node distance=3.5cm, thick,main node/.style={rectangle,draw, font=\sffamily}] \node[main node] (1) {Lake}; \node[main node] (2) [below of=1] {Depth}; \node[main node] (3) [below of=2] {Non-habitat}; \node[main node] (4) [below of=3] {Habitat}; \node[main node] (6) [below right of=2, align = center] {AFDW\\ $$r^2 = 0.223$$}; \node[main node] (7) [right of=6, align = center] {eDNA\\ $$r^2 = 0.384$$}; \path[every node/.style={font=\sffamily\small}] (1) edge node [above = 40pt] {\textbf{0.497}} (6) (2) edge node [left = 10pt] {\textbf{-0.370}} (6) (3) edge node [above] {0.094} (6) (4) edge node [left = 10pt] {0.116} (6) (1) edge[bend left] node [above = 10 pt] {\textbf{0.385}} (7) (2) edge[bend left] node [above = 5pt ] {0.197} (7) (3) edge[bend right] node [above = 0pt] {-0.298} (7) (4) edge[bend right] node [below = 5pt] {0.204} (7) (6) edge node [ ] {-0.180} (7); \end{tikzpicture}
\end{document}
# 6 tips for a new LaTeX user
Standard
Recently a coworker started using LaTeX and asked for some tips. Here’s my 6 tips for starting to use LaTeX:
1. Start what you finish (i.e., close environments or else you get errors or weird bugs), for example $$needs an$$
2. Every document needs 3 things: \documentclass{<class>}, \begin{document}, and \end{document}
3. For equations, use inline and \begin{eqnarray} \end{eqnarray} equations. \\ creates a new line. Use \\ \nonumber to continue on an equation and &=& to space multiple equations for example:
<code>
\begin{eqnarray}
a &=& b +c \\
a & =& b \\ \nonumber
& & c
\end{eqnarray}
</code>
Gives you something like:
a = b +c (1)
a = b
c (2)
4. Bib files are your friend for citations. Use Google Scholar to populate new citations.
5. \textit{My italics text}, \textbf{my bold text}, should get most of your formatting. Do NOT use the depreciated {\bf bold} or {\it italics} style. (cf http://tex.stackexchange.com/questions/41681/correct-way-to-bold-italicize-text for more details on the second point)
6. {} can be very helpful, especially for complicated math functions, when order of operation is important. For example, \sigma_{\pi^2}^{2} | |
# Designation of a term in a sequence in Second-order Logic
To First-order Logic, like in Hodges "A Shorter Model Theory", we can designate a $L$-term $t$ in a structure $\mathfrak{A}$ considering a sequence $\bar{a}$ of elements of the domain $A$ of $\mathfrak{A}$, by complexity on the terms:
• if $t$ is a variable $x_i$, then $t^{\mathfrak{A}}[\bar{a}]$ is $a_i$;
• if $t$ is a constant symbol $c$ of $L$, then $t^{\mathfrak{A}}[\bar{a}]$ is a element $c^{\mathfrak{A}}$ of $A$;
• if $t$ is $ft_1...t_n$, where $f$ is a $n$-ary function symbol of $L$, then $t^{\mathfrak{A}}[\bar{a}]$ is $f^{\mathfrak{A}}t_1^{\mathfrak{A}}[\bar{a}]...t_n^{\mathfrak{A}}[\bar{a}]$ of $A$.
My question is: how can I extends this definition to evaluate Second-Order Terms too? That is, supose now that $L$ is a Second-Order Language; so, there is a function variable $F$ such that $Ft_1...t_n$ is a term $t$.
Then I suppose that $\bar{a}$ is a ''twested'' sequence of elements of $A$, a sequence of relations and function on $\mathfrak{A}$ and $t^{\mathfrak{A}}[\bar{a}]$ is $F^{\mathfrak{A}}[\bar{a}]t_1^{\mathfrak{A}}[\bar{a}]...t^{\mathfrak{A}}[\bar{a}]_n$, where $F^{\mathfrak{A}}[\bar{a}]$ is some $n$-function on $\mathfrak{A}$ or the ideas involved need be more sofisticated''?
- | |
The proof from Van Mill, et al to prove $\omega_1$ is dually discrete
Here is the proof from
J. van Mill, V.V. Tkachuk, R.G. Wilson; Classes defined by stars and neighbourhood assignments, Top. Appl. 154 (2007), pp.2127–2134, MR2324924, link
to prove $\omega_1$ is dually discrete. (A topological space $X$ is dually discrete if for any neighbourhood assignment $\{ O_x : x \in X \}$ there is a discrete subspace $Y \subseteq X$ such that $\bigcup_{x \in Y} O_x = X$.)
I have small questions on the proof.
1. Why for any non-isolated point $\alpha \in \omega$ there is $f(\alpha)<\alpha$ such that $(f(\alpha), \alpha]\subset O_\alpha$?
2. How to use the pressing down lemma to prove that there is a uncontable $A\subset \omega_1$...
3. Why does $\bigcup\{O_\alpha: \alpha \in B\}$ contain $\omega_1\setminus (\beta+1)$?
-
I have attempted to improve the question by making it somewhat more self-contained. – Arthur Fischer Mar 13 '13 at 12:29
@ArthurFischer: You are helpful. – Paul Mar 13 '13 at 12:33
1. For each non-isolated $\alpha \in \omega_1$ (these are the countable limit ordinals), we have an (open) neighbourhood $O_\alpha$ of $\alpha$. But the family $$\{ ( \gamma , \alpha ] = \{ \xi : \gamma < \xi \leq \alpha \} : \gamma < \alpha \}$$ is a neighbourhood base for $\alpha$, and so there must be a $\gamma < \alpha$ such that $( \gamma , \alpha ] \subseteq O_\alpha$. So for each limit $\alpha < \omega_1$ let $f(\alpha)$ denote such a $\gamma$.
2. The function $f$ is clearly regressive ($f(\alpha) < \alpha$) on the stationary set (even closed unbounded set) of limit ordinals, and so the Pressing Down Lemma immediately concludes that it is constant on a stationary (hence uncountable) subset.
3. Given any $\beta < \xi < \omega_1$ since $B$ is uncountable there is an $\alpha \in B$ such that $\xi \leq \alpha$. But then $\xi \in ( \beta , \alpha ] = ( f(\alpha) , \alpha ] \subseteq O_\alpha$. | |
## Measurable numerical function
### Set
context $\langle X,\Sigma_X\rangle\in \mathrm{MeasurableSpace}(X)$
postulate $\mathcal M \equiv\mathrm{Measurable}(X,\overline{\mathbb R})$
Where for $\mathbb R$ we choose the Borel subsets of the reals. | |
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# At his regular hourly rate, Don had estimated the labour cos
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At his regular hourly rate, Don had estimated the labour cos [#permalink]
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31 Jul 2012, 09:01
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At his regular hourly rate, Don had estimated the labour cost of a repair job as $336 and he was paid that amount. However, the job took 4 hours longer than he had estimated and, consequently, he earned$2 per hour less than his regular hourly rate. What was the time Don had estimated for the job, in hours?
(A) 28
(B) 24
(C) 16
(D) 14
(E) 12
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Re: At his regular hourly rate, Don had estimated the labour cos [#permalink]
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31 Jul 2012, 09:21
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macjas wrote:
At his regular hourly rate, Don had estimated the labour cost of a repair job as $336 and he was paid that amount. However, the job took 4 hours longer than he had estimated and, consequently, he earned$2 per hour less than his regular hourly rate. What was the time Don had estimated for the job, in hours?
(A) 28
(B) 24
(C) 16
(D) 14
(E) 12
Say the regular hourly rate was $$r$$$and estimated time was $$t$$ hours, then we would have: $$rt=336$$ and $$(r-2)(t+4)=336$$; So, $$(r-2)(t+4)=rt$$ --> $$rt+4r-2t-8=rt$$ --> $$t=2r-4$$. Now, plug answer choices for $$t$$ and get $$r$$. The pair which will give the product of 336 will be the correct answer. Answer B fits: if $$t=24$$ then $$r=14$$ --> $$rt=14*24=336$$. Answer: B. Hope it's clear. _________________ Intern Joined: 01 Jan 2011 Posts: 22 Location: Kansas, USA Schools: INSEAD, Wharton Followers: 2 Kudos [?]: 16 [10] , given: 9 Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] ### Show Tags 01 Aug 2012, 09:39 10 This post received KUDOS [336][/X] - [336][/(X+4)]= 2 Solve for X. Ans= 24 since -28 is not a valid answer. Intern Joined: 26 Sep 2012 Posts: 17 Followers: 0 Kudos [?]: 15 [0], given: 1 Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] ### Show Tags 07 Dec 2012, 02:03 I have just worked on OG Math practice questions and hardly have I solved this question. That's why I have used Google and found you guys sayak636 wrote: [336][/X] - [336][/(X+4)]= 2 I have composed the same equation, however its solving has taken me for ages. I like Bunuel's solution, but I has not guessed to do the same. I'd only slightly change the course of solving. When we get to $$t = 2r - 4$$, $$r$$ easily seems to be replaced by $$336/t$$. Now we have $$t = (2*336/t) - 4$$ and can plug answer choices to find out the correct option. Intern Joined: 07 Feb 2013 Posts: 13 GMAT 1: 650 Q48 V32 GMAT 2: 730 Q49 V41 WE: Engineering (Other) Followers: 0 Kudos [?]: 19 [1] , given: 9 Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] ### Show Tags 15 Jun 2013, 19:41 1 This post received KUDOS While substitution does tend to take long for this problem, before substitution you could factorize 336 to its primes = 2*2*2*2*3*7 Now you can begin to substitute : Ans Choice A = 28*12 (2*2*7*2*2*3) not equal to 32*10 (clearly its 320 and not 336) Choice B = 24*14 (2*2*2*3*2*7) equals 28*12 (from prev choice) thx Verbal Forum Moderator Joined: 10 Oct 2012 Posts: 630 Followers: 82 Kudos [?]: 1134 [1] , given: 136 Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] ### Show Tags 16 Jun 2013, 03:50 1 This post received KUDOS 4 This post was BOOKMARKED Shiv636 wrote: [336][/X] - [336][/(X+4)]= 2 Solve for X. Ans= 24 since -28 is not a valid answer. Infact, one doesn't need to solve after this step too: $$\frac{336}{x} - \frac{336}{(x+4)} = 2$$ 336[(x+4)-x] = 2*x(x+4) x(x+4) = 672 From the given options, we can straightaway eliminate A and C, as because the units digit after multiplication of 28*(28+4) and 16*(16+4) will never be 2. We also know that 14*20 = 280 and 12*20 = 240. Thus, 14*18(D) or 12*16(E) can never equal 672. By eliminaion, the answer is B. _________________ Intern Joined: 16 Oct 2013 Posts: 4 Followers: 0 Kudos [?]: 0 [0], given: 0 Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] ### Show Tags 07 Jan 2014, 07:41 Bunuel wrote: macjas wrote: At his regular hourly rate, Don had estimated the labour cost of a repair job as$336 and he was paid that amount. However, the job took 4 hours longer than he had estimated and, consequently, he earned $2 per hour less than his regular hourly rate. What was the time Don had estimated for the job, in hours? (A) 28 (B) 24 (C) 16 (D) 14 (E) 12 Say the regular hourly rate was $$r$$$ and estimated time was $$t$$ hours, then we would have:
$$rt=336$$ and $$(r-2)(t+4)=336$$;
So, $$(r-2)(t+4)=rt$$ --> $$rt+4r-2t-8=rt$$ --> $$t=2r-4$$.
Now, plug answer choices for $$t$$ and get $$r$$. The pair which will give the product of 336 will be the correct answer.
Answer B fits: if $$t=24$$ then $$r=14$$ --> $$rt=14*24=336$$.
Hope it's clear.
On my own I got to the step where I need to utilize the answer choices. I didn't know what to do at that point because it never crosses my mind to use the answer choices and backwards solve like this.
I've only ever seen this kind of method recommended when the problem involves second degree equations. Is that a fair statement? You only backwards solve like this when dealing with second degree equations?
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Re: At his regular hourly rate, Don had estimated the labour cos [#permalink]
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07 Jan 2014, 21:38
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Rdotyung wrote:
Bunuel wrote:
macjas wrote:
At his regular hourly rate, Don had estimated the labour cost of a repair job as $336 and he was paid that amount. However, the job took 4 hours longer than he had estimated and, consequently, he earned$2 per hour less than his regular hourly rate. What was the time Don had estimated for the job, in hours?
(A) 28
(B) 24
(C) 16
(D) 14
(E) 12
On my own I got to the step where I need to utilize the answer choices. I didn't know what to do at that point because it never crosses my mind to use the answer choices and backwards solve like this.
I've only ever seen this kind of method recommended when the problem involves second degree equations. Is that a fair statement? You only backwards solve like this when dealing with second degree equations?
You utilize the answer choices whenever you CAN. Here I would keep an eye on the choices right from the start. I would say
R*T = 336 (his regular hourly rate * time he estimated)
The options give us the value of T which is an integer.
$$336 = 2^4*3*7$$
So R*T = 336
(R-2)*(T + 4) = 336
So T as well as T+4 should be factors of 336.
If T is 28, T+4 is 32 which is not a factor of 336 so ignore it.
If T is 24, T+4 is 28. Both are factors of 336. Keep it. If T is 24, R is 14. So (R - 2) is 12. 12*28 does gives us 336 so T = 24 must be the correct answer.
But note that if you want to reduce your mechanical work, you need to be fast in your calculations. You cannot spend a minute working on every option or making calculation mistakes.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 13 Feb 2014 Posts: 7 Followers: 0 Kudos [?]: 0 [0], given: 9 Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] ### Show Tags 23 Apr 2014, 07:06 How do you go from > 336[(x+4)-x] = 2*x(x+4) to > x(x+4) = 672? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7177 Location: Pune, India Followers: 2161 Kudos [?]: 13986 [3] , given: 222 Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] ### Show Tags 23 Apr 2014, 19:02 3 This post received KUDOS Expert's post gciftci wrote: How do you go from > 336[(x+4)-x] = 2*x(x+4) to > x(x+4) = 672? $$336 * [(x+4)-x] = 2 * x * (x+4)$$ $$336 * [x+4 -x] = 2 * x * (x+4)$$ x and -x get cancelled to give: $$336 * [4] = 2 * x * (x+4)$$ Divide both sides by 2. $$336 * 2 = x * (x+4)$$ $$672 = x * (x+4)$$ _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
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Re: At his regular hourly rate, Don had estimated the labour cos [#permalink]
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VeritasPrepKarishma wrote:
On my own I got to the step where I need to utilize the answer choices. I didn't know what to do at that point because it never crosses my mind to use the answer choices and backwards solve like this.
I've only ever seen this kind of method recommended when the problem involves second degree equations. Is that a fair statement? You only backwards solve like this when dealing with second degree equations?
[/quote]
You utilize the answer choices whenever you CAN. Here I would keep an eye on the choices right from the start. I would say
R*T = 336 (his regular hourly rate * time he estimated)
The options give us the value of T which is an integer.
$$336 = 2^4*3*7$$
So R*T = 336
(R-2)*(T + 4) = 336
So T as well as T+4 should be factors of 336.
If T is 28, T+4 is 32 which is not a factor of 336 so ignore it.
If T is 24, T+4 is 28. Both are factors of 336. Keep it. If T is 24, R is 14. So (R - 2) is 12. 12*28 does gives us 336 so T = 24 must be the correct answer.
But note that if you want to reduce your mechanical work, you need to be fast in your calculations. You cannot spend a minute working on every option or making calculation mistakes.[/quote]
Hi Karishma, Why do T and T+4 have to be factors of 336? Why cannot rate be a fraction and difference of two fractions can yield an integer in this case 2? What am I missing here? Thanks!
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Re: At his regular hourly rate, Don had estimated the labour cos [#permalink]
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MensaNumber wrote:
Hi Karishma, Why do T and T+4 have to be factors of 336? Why cannot rate be a fraction and difference of two fractions can yield an integer in this case 2? What am I missing here? Thanks!
All the options are integers so value of T must be an integer. So T+4 must be an integer too. Therefore, T and T+4 must be factors of 336. Also, in GMAT, usually numbers are easy since you do not get calculators. So very rarely will you find that rate or time is a fraction. Even if it will be, it will be a simple fraction such as 1/2 etc.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Senior Manager Joined: 29 Oct 2013 Posts: 297 Concentration: Finance GMAT 1: 750 Q V46 GPA: 3.7 WE: Corporate Finance (Retail Banking) Followers: 14 Kudos [?]: 387 [0], given: 197 Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] ### Show Tags 28 May 2014, 01:22 Karishma, Thanks for your reply. Yeah that makes sense. _________________ Please contact me for super inexpensive quality private tutoring My journey V46 and 750 -> http://gmatclub.com/forum/my-journey-to-46-on-verbal-750overall-171722.html#p1367876 Senior Manager Joined: 15 Aug 2013 Posts: 328 Followers: 0 Kudos [?]: 55 [0], given: 23 Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] ### Show Tags 12 Aug 2014, 16:15 Bunuel wrote: macjas wrote: At his regular hourly rate, Don had estimated the labour cost of a repair job as$336 and he was paid that amount. However, the job took 4 hours longer than he had estimated and, consequently, he earned $2 per hour less than his regular hourly rate. What was the time Don had estimated for the job, in hours? (A) 28 (B) 24 (C) 16 (D) 14 (E) 12 Say the regular hourly rate was $$r$$$ and estimated time was $$t$$ hours, then we would have:
$$rt=336$$ and $$(r-2)(t+4)=336$$;
So, $$(r-2)(t+4)=rt$$ --> $$rt+4r-2t-8=rt$$ --> $$t=2r-4$$.
Now, plug answer choices for $$t$$ and get $$r$$. The pair which will give the product of 336 will be the correct answer.
Answer B fits: if $$t=24$$ then $$r=14$$ --> $$rt=14*24=336$$.
Hope it's clear.
Hi Bunuel,
Can you recommend some similar problems?
Thanks!
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At his regular hourly rate, Don had estimated the labour cos [#permalink]
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07 Jan 2015, 12:51
Let t be the hourly rate and p the price: t * p = $336 Additional 4 hours and$2 less per hour would yield: (t+4) * (p-2) = $336 Since both equations are equal: 336 : p = (336 : (p-2))-4 Solving for p yields 14 (the other solution is negative, so we do not consider it) At this point probably we are very pressed on time, so the shortcut is to find the answer the last digit of which multiplied by 4 yields 6. 14 squared is not 336 so by elimination it is 24 Answer: B Senior Manager Status: Math is psycho-logical Joined: 07 Apr 2014 Posts: 443 Location: Netherlands GMAT Date: 02-11-2015 WE: Psychology and Counseling (Other) Followers: 2 Kudos [?]: 112 [1] , given: 169 At his regular hourly rate, Don had estimated the labour cos [#permalink] ### Show Tags 18 Jan 2015, 09:05 1 This post received KUDOS I back solved this one like this: I started with option A, but I will only show the correct option, which is B: I said that r*t=336, so the amount of hours he worked times the money he got for each hour should be his final salary. Then I substituted the proposed times for t: r*24=336 r=14 --> This is how much he should have got per hour worked. But he worked 4 hours more, so 24+2 = 28. Then he actually got 336/28 = 12, per hour. 12 is 2 less than 14, as it is supposed to, so the correct answer is ANS B. Intern Joined: 11 Apr 2015 Posts: 34 Location: Germany Concentration: General Management, Entrepreneurship GPA: 3.1 WE: Project Management (Energy and Utilities) Followers: 1 Kudos [?]: 10 [0], given: 98 Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] ### Show Tags 20 Apr 2015, 11:06 It is very challanging to solve this question in 2 minutes. Are there any strategies on time saving for this types of questions? _________________ "I fear not the man who has practiced 10,000 kicks once, but I fear the man who has practiced one kick 10,000 times." Bruce Lee "I hated every minute of training, but I said, "Don’t quit. Suffer now and live the rest of your life as a champion."" Muhammad Ali EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 8519 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Followers: 396 Kudos [?]: 2555 [2] , given: 164 Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] ### Show Tags 20 Apr 2015, 17:48 2 This post received KUDOS Expert's post 1 This post was BOOKMARKED Hi RussianDude, You're not expected to answer every question in the Quant section in under 2 minutes, so if you took a little longer than that on this question, then that's fine (as long as you were doing work and not staring at the screen). If you took more than 3 minutes to answer this question, then chances are that YOUR approach is the "long" approach and that you have to practice other tactics. Here, since the answer choices ARE numbers, we're really looking for an answer that divides into 336 AND when you add 4 to that answer, that sum ALSO divides evenly into 336. The difference between those two rates should be$2 (as the question states). In that way, you can answer this question with some basic division and note-taking (and likely save time and avoid a long-winded Algebra approach).
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Re: At his regular hourly rate, Don had estimated the labour cos [#permalink]
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macjas wrote:
At his regular hourly rate, Don had estimated the labour cost of a repair job as $336 and he was paid that amount. However, the job took 4 hours longer than he had estimated and, consequently, he earned$2 per hour less than his regular hourly rate. What was the time Don had estimated for the job, in hours?
(A) 28
(B) 24
(C) 16
(D) 14
(E) 12
$$\frac{336}{X} = \frac{336}{X+4}+2$$
X=24
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Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] 20 Apr 2015, 21:06
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### how the second line of lyman series is produced
The series of lines in an emission spectrum caused by electrons falling from energy level 2 or higher (n=2 or more) back down to energy level 1 (n=1) is called the Lyman series. How satisfied are you with the answer? (2) The group of lines produced when the electron jumps from 3rd, 4th ,5th or any higher energy level to 2nd energy level, is called Balmer series. Download the PDF Question Papers Free for off line practice and view the Solutions online. (d) First line of Pfund series Electron transition The emission spectrum of single-electron species is divided into various spectral series such as Lyman, Balmer, Paschen, Brackett, and Pfund. a) n = 6 to n = 2. b) n = 5 to n = 2. c) n = 4 to n = 2. d) n = 3 to n = 2. e) it is to the n = 1 level. (1.22). Lyman and Balmer series are hydrogen spectral line series that arise from hydrogen emission spectra. 260 Views. For example, the 2 → 1 line is called "Lyman-alpha" (Ly-α), while the 7 → 3 line is called "Paschen-delta” (Pa-δ). 3. The energy levels of hydrogen, which are shown in Fig. 260 Views. The quantity "hertz" indicates "cycles per second". The series is named after its discoverer, Theodore Lyman, who discovered the spectral lines from 1906–1914. The released wavelength lies in the Infra Red region of the spectrum. The electron, in a hydrogen atom, is in its second excited state. For example the Lyman series (nf = 1 in Balmer-Rydberg equation) occurs in the ultraviolet region while the Balmer (nf = 2) series occurs in the visible range and the Paschen (nf = 3), Brackett (nf = 4) and Pfund ( nf = 5) series all occur in the infrared range. 1. Calculate the shortest wavelength of the spectral lines emitted in Balmer series. 912 Å; 1026 Å; 3648 Å; 6566 Å; B. GRAMMAR A-Z ; SPELLING ; PUNCTUATION ; WRITING TIPS ; USAGE … Notice that the lines get closer and closer together as the frequency increases. The so-called Lyman series of lines in the emission spectrum of hydrogen corresponds to transitions from various excited states to the n = 1 orbit. You can specify conditions of storing and accessing cookies in your browser, Lymanseries second line means electron jumps from 3level to second level, Calculate the wavelength of second line of Lyman series in hydrogen spectra, report my all questions please please please please please please please friends , hello kaise ho sab log good morning have a nice day comrades, anyone inbox me ❤️❤️❤️❤️❤️ i have something to share!!!!!!!!! To which transition can we attribute this line? Calculate the range of wavelength for the Lyman series. auch Ausführungen dort), die Paschen-Serie, die Brackett-, Pfund-und die Humphreys-Serie Calculate the wavelength of the lowest-energy line in the Lyman series to three significant figures. 2 answers. Match the correct pairs. That's what the shaded bit on the right-hand end of the series suggests. Class 10 Class 12. b) -313.6, -78.4 . The H α spectral line in Lyman series of hydrogen atomic spectrum is formed due to an electronic transition in hydrogen atom. The second level, which corresponds to n = 2 has an energy equal to − 13.6 eV/2 2 = −3.4 … What are synonyms for Lyman series? Atoms. The lines in the Lyman series in a hydrogen atom spectrum are produced when electrons, excited to higher energy levels, make transitions to the ground state (n = 1). GRAMMAR . The Lyman series is produced by electrons dropping from higher levels into level 1. Then (A) X = He +, n = 4 (B) X = Li ++, n = 6 (C) X = He +, n = 6 (D) X = Li ++, n = 9. jee; jee mains ; Share It On Facebook Twitter Email. The energies associated with the electron in each of the orbits involved in the transition (in kCal mol-1) are: (Eamcet - 2008-E) a) -313.6, –34.84 . n=2,3,4,5,6 ….to n=1 energy level, the group of lines produced is called lyman series.These lines lie in the ultraviolet region. The energies associated with the electron in each of the orbits involved in the transition (in kCal mol-1) are: (Eamcet - 2008-E) a) -313.6, –34.84 . *"*********************, Hlo everyone gd morning Have a wonderful day ahead♥️♥️♥️, Hlo everyone gd morning Have a wonderful day ahead. c) -78.4, -34.84 . The third line of Brackett series is formed when electron drops from n=7 to n=4. The wavelength of the first line of Lyman series of hydrogen is 1216 A. For the lowest level with n = 1, the energy is − 13.6 eV/1 2 = −13.6 eV. The series is named after its discoverer, Theodore Lyman. !., anyone inbox me I have to say something??? The H α spectral line in Lyman series of hydrogen atomic spectrum is formed due to an electronic transition in hydrogen atom. (a) Calculate the wavelengths of the first three lines in this series. If the first line in this series has a wavelength of 122 nm, what is the wavelength of the second line? Energy (kJ/mol) 250 500 750 1000 1250 0 85.4 100 150 200 300 400 500 1000 2000 ∞ 1400 Wavelength (nm) We now know how the Lyman and Balmer series lines are formed. = RH (1 - 1/n2) n = 2, 3, 4, . Join now. The formula that that gives the spectra in all wavelength series of hd ihydrogen is 22 111'1,2,3, (), 31 2 n R = ⋅⋅⋅⋅ =− − λ nn' nn nn= '1, ' 2,+=+ ’ Si Table 31-1 Common Spectral Series of Hydrogen n Series name 1Lyman 2 Balmer 3 Paschen 4 Brackett 5 Pfund. The first line in the ultraviolet spectrum of the Lyman series was discovered in 1906 by Harvard physicist Theodore Lyman, who was studying the ultraviolet spectrum of electrically excited hydrogen gas. The electron, in a hydrogen atom, is in its second excited state. Explain how second line of brackett series produced? [Given Rydberg constant, R = 10 7 m-1] (All India 2016) Answer: Question 22. The number of lone pair and bond pair of electrons on the sulphur atom in sulphur dioxide molecule are respectively. Class 10 Class 12. n 2 is the level being jumped from. As you notice and the values are decreasing in the series due to the derivation and their state which is Ultra Violet. Transitions ending in the ground state n = 1) are called the Lyman series, but the energies released are so large that the spectral lines are all in the ultraviolet region of the spectrum. asked Jul 15, 2019 in Physics by Ruhi (70.2k points) atoms; nuclei; class-12; 0 votes. Q. The wave length of the second. Consider first at the Lyman series on the right of the diagram; this is the broadest series, and the easiest to decipher. Als Lyman-Serie wird die Folge von Spektrallinien des Wasserstoffatoms bezeichnet, deren unteres Energieniveau in der K-Schale liegt (Hauptquantenzahl =).. Weitere Serien sind die Balmer-Serie (vgl. The value 3 PHz is equal to 3 × 10 15 Hz. Download the PDF Question Papers Free for off line practice and view the Solutions online. The released wavelength lies in the Infra Red region of the spectrum. Try this, The Lyman Series say the for the second is 121.6nm (nano metres) For the third it is 102.6 and the fourth is 97.3 all in Nano Metres which *10^-9. Lv 7. the shortest and longest wavelength series in singly ionized helium is 22.8nm and 30.4nm . the ratio of difference in wavelengths of 1st and 2nd lines of lyman series in H-like atom to difference in wavelength for 2nd and 3rd lines of same series is Share with your friends. Eventually, they get so close together that it becomes impossible to see them as anything other than a continuous spectrum. There are emission lines from hydrogen that fall outside of these series, such as the 21 cm line. 2.90933 × 1016 Hz The transitions called the Pasch Note: Your answer is assumed to be reduced to the highest power possible. The Lyman series of hydrogen is made up of those transitions made from higher levels to n = 1. 2. Transitions ending in the ground state n = 1) are called the Lyman series, but the energies released are so large that the spectral lines are all in the ultraviolet region of the spectrum. The simplest of these series are produced by hydrogen. 121.6 \text{nm} 1/lambda = \text{R}(1/(n_1)^2 - 1/(n_2)^2) * \text{Z}^2 where, R = Rydbergs constant (Also written is \text{R}_\text{H}) Z = atomic number Since the question is asking for 1^(st) line of Lyman series therefore n_1 = 1 n_2 = 2 since the electron is de-exited from 1(\text{st}) exited state (i.e \text{n} = 2) to ground state (i.e text{n} = 1) for first line of Lyman series. Hope It Helped. c) -78.4, -34.84 . Currently only available for. These emission lines correspond to much rarer atomic events such as hyperfine transitions. The first line in the Lyman series of the hydrogen atom emission results from a transition from the n=2 level to the n=1 level. In the Brackett Series for the emission spectra of hydrogen the final destination of a dropping electron from a higher orbit is n=4 . Spectral line series, any of the related sequences of wavelengths characterizing the light and other electromagnetic radiation emitted by energized atoms. The background in the Lyα line is composed of two parts: those photons that have redshifted directly to the Lyα frequency and those produced by atomic cascades from higher Lyman-series photons. 7 years ago. Lines in an emission spectrum are produced when an electron falls from a higher level to a lower one. Zigya App. The wavelength (in cm) of second line in the Lyman series of hydrogen atomic spectrum is (Rydberg constant = R cm$^{-1}$) 10. The wavelengths of the Lyman series for hydrogen are given by 1/? Let F_1 be the frequency of second line of Lyman series and F_2 be the frequency of first line of Balmer series then frequency of first line of Lyman series is given by Lyman series – A series of spectral lines of hydrogen produced by electron transitions to and from the lowest energy state of the hydrogen atom The Nature of Light – Light is electromagnetic radiation. Currently only available for. The wavelength of the first line of Lyman series of hydrogen is 1216 A. Question options: 1) 49 nm 2) 103 nm 3) 364 nm 4) 486 nm 5) 632 nm Calculate the shortest wavelength of the spectral lines emitted in Balmer series. Log in. This is called the Balmer series. Emission lines are produced by transitions from higher levels to the second orbit; absorption lines result from transitions from the second orbit to higher orbits. When an electron comes down from higher energy level to second energy level, then Balmer series of the spectrum is obtained. It is obtained in the visible region. GRAMMAR A-Z ; SPELLING ; PUNCTUATION ; WRITING TIPS ; USAGE ; EXPLORE . When resolved by a spectroscope, the individual components of the radiation form images of the source (a slit through which the beam of radiation enters the device). Explanation: First of all the series is given for the atom having the E.C like hydrogen ( eg. In the Bohr model, the Lyman series includes the lines emitted by transitions of the electron from an outer orbit of quantum number n > 1 to the 1st orbit of quantum number n' = 1. Ly α emission and absorption lines occur, for example, in the spectra of quasars. Reason Lyman series constitute spectral lines corresponding to transition from higher energy to ground state of hydrogen atom. The wavelength of the second line of the same series will be. 1026 Å. He found that the four visible spectral lines corresponded to transitions from higher energy levels down to the second energy level $$\left( n=2 \right)$$. H,He⁺,Li²⁺,Be³⁺), therefore if layman first series is given by n₁=1 and n₂=2, and second series is given by n₁=1 and n₂=3 where the wavelength is less than first series, This site is using cookies under cookie policy. Lyman series A series of spectral lines of atomic hydrogen with wavelengths in the far ultraviolet and extreme ultraviolet regions of the spectrum (see hydrogen spectrum).The Lyman alpha (Ly α) line occurs at 121.6 nm and the Lyman limit at 91.2 nm. Answer Save. (1) When the electron jumps from energy level higher than n=1 ie. The lines in such a series get closer together at shorter wavelengths and the Balmer series converges to a limit at 364.6 nm in the ultraviolet region of the spectrum. Zigya App. Energy level diagram of electrons in hydrogen atom. 1026 Å. The frequency scale is marked in PHz—petaHertz. 1. 1. A wavelength of second line of lyman series for H atom is X then wavelength of third line paschen series for the Li2+ ? Question from Student Questions,chemistry. Lyman α emissions are weakly absorbed by the major components of the atmosphere—O, O 2, and N 2 —but they are absorbed readily by NO and have… Read More; line spectra Which falls are responsible for the lines A, B and C in this diagram of the Lyman series? As a result the hydrogen like atom 'X' makes a transition to n th orbit. Determine the frequency of the second Lyman line, the transition from n = 3 to n = 1. The Lyman series lies in the ultraviolet, whereas the Paschen, Brackett, and Pfund series lie in the infrared. Their formulas are similar to Balmer’s except that the constant term is the reciprocal of the square of 1, 3, 4, or 5, instead of 2, and the running number n begins at … The Lyman series of emission lines of the hydrogen atoms are those for which nf = 1. a) determine the region of the electromagnetic spectrum in which the lines of the Lyman series are observed. In Wolff‐Kishner reduction, the carbonyl group of aldehydes and ketones is converted into . The series of lines in an emission spectrum caused by electrons falling from energy level 2 or higher (n=2 or more) back down to energy level 1 (n=1) is called the Lyman series. The first member of the Balmer series of hydrogen atom has a wavelength of 6561 Å. asked Jan 24, 2020 in Physics by KumariMuskan (33.8k points) jee main 2020; 0 votes. Electrons are falling to the 1-level to produce lines in the Lyman series. Second line of Balmer series is produced by which transition in spectrum of H-atom 4 to 2Explanation:Balmer series or Balmer lines is one of the set of six name… We have already mentioned that the red line is produced by electrons falling from the 3-level to the 2-level. The Lyman series is a series of lines in the ultra-violet. You can specify conditions of storing and accessing cookies in your browser, How to calculate second line of lyman series when first line of Lyman series is given, Which law states that in closed electriccircuit, the applied voltage is equal to thesum of the voltage drops?, where does the center of gravity of the triangular and annular ring lie, A lens forms an image three times the size of the object on the screen.The focal length of the lens is 20cm......Find i)Name the lens....ii)Find the p Solution for By calculating its wavelength, show that the first line in the Lyman series is UV radiation. We get Balmer series of the hydrogen atom. Example $$\PageIndex{1}$$: The Lyman Series. 1 Answer. . The lines in such a series get closer together at shorter wavelengths and the Balmer series converges to a limit at 364.6 nm in the ultraviolet region of the spectrum. So the second line is emission: 6→4 (higher energy to lower energy state); absorption 4→6 from low energy to high energy. Peta means "10 15 times". Here is an illustration of the first series of hydrogen emission lines: Historically, explaining the nature of the hydrogen spectrum was a considerable problem in physic… He found that the four visible spectral lines corresponded to transitions from higher energy levels down to the second energy level (n = 2). e) Which fall corresponds to the series limit of the Lyman series? Example $$\PageIndex{1}$$: The Lyman Series. Favourite answer. The rest of the lines of the spectrum were discovered by Lyman from 1906-1914. Log in. Relevance. 0 votes . Click hereto get an answer to your question ️ The wavelength of the first line of Lyman series in a hydrogen atom is 1216 A^0 . Both come at 2620 nm. The spectrum of radiation emitted by hydrogen is non-continuous. The so-called Lyman series of lines in the emission spectrum of hydrogen corresponds to transitions from various excited states to the n = 1 orbit. How do you use Lyman series in a sentence? Find the wavelength of first line of lyman series in the same spectrum. Please show all work . The photon radiated from hydrogen corresponding to 2nd line of Lyman series is absorbed by a hydrogen like atom 'X' in 2nd excited state. . Express your answer to three significant figures. This is called the Balmer series. For example, in the Lyman series, n 1 is always 1. Some lines of blamer series are in the visible range of the … Join now. All Chemistry Practice Problems Bohr and Balmer Equations Practice Problems. b) -313.6, -78.4 . phys. b) Calculate the wavelengths of the first three lines in the Lyman series-those for which ni = 2,3,and 4. Since second line of Lyman series of H coincides with 6th line of Paschen series of an ionic species 'A' we can equate the equation (1) and (2) : R(1/1 2 - 1/3 2) = RZ 2 (1/3 2 - 1/9 2) 8/9 = Z 2 x 8/81 Z 2 = 9 Z = 3 Ionic species would be ion of atom with atomic number 3. Find an answer to your question How to calculate second line of lyman series when first line of Lyman series is given 1. For the Balmer series, n 1 is always 2, because electrons are falling to the 2-level. The wavelength of first line of balmer series in hydrogen spectrum is 6563 A. BII. 912 Å; 1026 Å; 3648 Å; 6566 Å; B. 1 Answer. The third line of Brackett series is formed when electron drops from n=7 to n=4. (The Lyman series is a related sequence of wavelengths that describe electromagnetic energy given off by energized atoms in the ultraviolet region.) Chemistry Most Viewed Questions. In what region of the electromagnetic spectrum does it occur? WORD ORIGINS ; LANGUAGE QUESTIONS ; WORD LISTS; SPANISH DICTIONARY; More. Physics. He found that the four visible spectral lines corresponded to transitions from higher energy levels down to the second energy level (n = 2). Balmer Series: If the transition of electron takes place from any higher orbit (principal quantum number = 3, 4, 5, …) to the second orbit (principal quantum number = 2). Calculate the wavelength of the lowest-energy line in the Lyman series to three significant figures. f) What fall would correspond to the series limit of the Balmer series? The first line is 3→ 2, second line is 4 →2 and third line is 5→ 2. Secondary School. The next line, Hβ (m = 4), is at 486.1 nm in the blue. Questions; Chemistry. This site is using cookies under cookie policy. Hγ and Hδ occur at 434.2 nm and 410.2 nm, respectively. [Given Rydberg constant, R = 10 7 m-1] (All India 2016) Answer: Question 22. 121.6 \text{nm} 1/lambda = \text{R}(1/(n_1)^2 - 1/(n_2)^2) * \text{Z}^2 where, R = Rydbergs constant (Also written is \text{R}_\text{H}) Z = atomic number Since the question is asking for 1^(st) line of Lyman series therefore n_1 = 1 n_2 = 2 since the electron is de-exited from 1(\text{st}) exited state (i.e \text{n} = 2) to ground state (i.e text{n} = 1) for first line of Lyman series. Hydrogen exhibits several series of line spectra in different spectral regions. The second line of the Balmer series occurs at wavelength of 486.13 nm. Answer to: Calculate energy change that produced the 4th line in lyman series. Calculate the wavelength of second line of Lyman series in hydrogen spectra Get the answers you need, now! The transitions are named sequentially by Greek letters: from n = 2 to n = 1 is called Lyman-alpha, 3 to 1 is Lyman-beta, 4 to 1 is Lyman-gamma, and so on. The wavelengths in the hydrogen spectrum with m=1 form a series of spectral lines called the Lyman series. In what region of the electromagnetic spectrum does it occur? This is called the Balmer series. The Lyman, Balmer, and Paschen Series of Spectral Lines. Answer. Answer to: Calculate energy change that produced the 4th line in lyman series. 2.90933 × 1014 Hz. The wavelength of the first line of Lyman series for 10 times ionised sodium atom will be: The greater the dif… What is the wavelength, in meters, of the emitted photon? The transitions called the Pasch d) -78.4, -19.6. ENGLISH DICTIONARY; SYNONYMS; TRANSLATE; GRAMMAR . The series of lines in an emission spectrum caused by electrons falling from energy level 2 or higher (n=2 or more) back down to energy level 1 (n=1) is called the Lyman series. d) -78.4, -19.6. 1.6, can be obtained by substituting the integer values n = 1,2,3,… into Eq. If the wavelength of the first line of the Balmer series of hydrogen is $6561 \, Å$, the wavelength of the second line of the series should be 8. …, good morning koi online hai ya sab mar gye, does the space between the particle in the matter influence the speed of diffusion justify the answer, how much time 600col of electric charge fill flow if an electric current of 10A of is drown from a electric motor, FridayThe Valency of NitrogenisallB15 (16D 13, Centre of gravity of traigular & anuelar lies outside the ring. For example the Lyman series (nf = 1 in Balmer-Rydberg equation) occurs in the ultraviolet region while the Balmer (nf = 2) series occurs in the Chemistry Calculate the wavelength, in nanometers, of the spectral line produced when an electron in a hydrogen atom undergoes the transition from the energy level n = 4 to the level n = 1.v The Brakett series involves transitions to and from the n1 = 4 level [1]. Are they right? 3.63667 × 1016 Hz. Atoms. 1 answer. 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## Appendix ACleaning the raw data
There are two ways to clean time-series data:
(1)
Run makemap and add doclean = 1 to your configuration (see Section 5.4.3).
(2)
Run sc2clean to clean the time-series data without making a map.
The rest of this appendix gives more details on using sc2clean.
sc2clean can be used to do two basic tasks in one go: concatenate data (with or without applying a flatfield); and cleaning (fix up steps and spikes, remove the means, filter, remove common-mode etc.). It uses the same configuration files as the iterative map-maker (though ignoring the map-making specific items).
In this first basic example, we just want to clean up some data enough to see whether the bolometers have been flat-fielded correctly, and more-or-less exhibit the same behaviour over time. The pre-processing or cleaning steps used by default (i.e. if “config=def” is included on the command line) are summarised in Section I.3. Note, whilst it is not recommended to run makemap in this way (i.e. without a configuration file), it is not so critical when running sc2clean.
% sc2clean ^files.lis clean config=def
Here files.lis can just contain a single file from a sub-array, or a subset, e.g. s8a20110417_00051_0003.sdf (the first file containing science data), s8a20110417_00051_000"[1234]" (File 1 is a noise observation with shutter closed that gets ignored, File 2 is a flatfield observation that will be used to override the flatfield stored in the subsequent Files 3 and 4 which are concatenated together, the .sdf is optional), s8a20110417_00051_000\? (Files 1 through 9), s8a20110417_00051_\* (the whole observation).
If you inspect the resulting clean.sdf in Gaia (Section 9.4) and flip through the data cube you should see all of the bolometers signals go up and down together with about the same amplitude: the hope is that for a well-behaved instrument you are mostly seeing sky noise variations that are seen with roughly the same amplitude by all bolometers.
Another common feature, if the scans are particularly long and/or fast (e.g. 1 degree across), is strong periodic signals that are correlated with the scan pattern. See Section 9.3—in particular you will want to plot az and el (the absolute azimuth and elevation), and also daz and del (the azimuth and elevation offsets from the map centre). This signal is usually azimuth-correlated due to magnetic-field pickup. It only shows up in azimuth, because the instrument is on a Nasmyth platform and therefore does not move in elevation.
Part of the reason the signals look the same is because they have been flatfielded. You can turn off flatfielding using the noflat option to sc2clean, and you should then see that all of the detector amplitudes vary.
Another very useful option is to remove the common signal observed by all of the bolometers. This may be accomplished by
% sc2clean ^files.lis clean config=’"compreprocess=1"’
This config setting causes the default values to be used for all configuration parameters except compreprocess, which is set to 1 (the default is 0). The residual signal left by this command will exhibit second-order time-varying correlated signals across the focal plane. Usually these are not very large, but in some cases some very large localized signals have been detected, particularly in the 850 $\mu$m arrays in early 2011.
Another variation on this is to accentuate the residual low-frequency noise by low-pass filtering the result. This can again be accomplished by simply adding a filter command in the config parameter, which in this case low-pass filters with a cutoff at 10 Hz:
% sc2clean ^files.lis clean config=’"compreprocess=1,filt_edgelow=10"’
Finally, in some cases you might just want to fit and remove polynomial baselines from the bolometers (by default only the mean is removed). This example will remove a line, but you can increase the value of order to remove higher-order polynomials
% sc2clean ^files.lis clean config=’"order=1"’
Non-default values for any of the cleaning parameter can be specified like so:
% sc2clean ^files.lis clean config=’"order=1,dcfitbox=30,dcthresh=25,dcsmooth=50"’
Or you can create your own customised configuration file. For instance:
% cat myconf
order=1
dcfitbox=30
dcthresh=25
dcsmooth=50
% sc2clean ^files.lis clean config=^myconf
The more interesting pre-processing options that may be specified are listed and described in Appendix H. | |
zbMATH — the first resource for mathematics
Explicit two-source extractors and resilient functions. (English) Zbl 1419.05109
Given a positive integer $$k$$ and a positive real number $$\varepsilon$$, let $$K = 2^{k}$$. A (balanced) bipartite graph containing $$N$$ “left” vertices and $$N$$ “right” vertices is called a $$(k, \varepsilon)$$-two-source extractor if every subgraph with $$K$$ left vertices and $$K$$ right vertices contains $$(1/2 \pm \varepsilon)K^{2}$$ edges.
The main result of this work is the following.
Theorem 1. There is a positive constant $$C$$ such that for any natural number $$n$$, there is an explicit construction of a $$(k, \varepsilon)$$-two-source extractor on two sets of $$2^{n}$$ vertices with $$k = \log^{C}(n/\varepsilon)$$. The construction is explicit in the sense that there is an algorithm which runs in polynomial time $$\mathrm{poly}(n/\varepsilon)$$ that determines whether there is an edge between two nodes.
This result is applied in the proof of the following result.
Theorem 2. There is a positive constant $$C$$ such that for any natural number $$n$$, there is an explicit construction of bipartite $$K$$-Ramsey graphs on $$2N$$ vertices and a Ramsey graph on $$N$$ vertices where $$N = 2^{n}$$ and $$K = 2^{(\log \log N)^{C}}$$.
The proof utilizes a variety of tools from the theory of extractors and probability along with two technical “key lemmas”. The proof of the main result is completed by first using non-malleable extractors to reduce the construction of a two-source extractor to the problem of constructing resilient functions. Such a function is constructed to be computable by a polynomial sized constant depth monotone circuit.
MSC:
05C35 Extremal problems in graph theory 68R05 Combinatorics in computer science
Keywords:
Ramsey graphs; randomness extractors
Full Text:
References:
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Random Structures & Algorithms, 33, 157-186, (2008) · Zbl 1181.68170 [68] Sipser, Michael, Expanders, randomness, or time versus space, J. Comput. System Sci.. Journal of Computer and System Sciences, 36, 379-383, (1988) · Zbl 0652.68050 [69] Tal, Avishay, Tight bounds on the Fourier spectrum of $${\bf{AC}}^\textbf{0}$$. 32nd Computational Complexity Conference, LIPIcs. Leibniz Int. Proc. Inform., 79, 15-31, (2017) [70] Trevisan, Luca, Extractors and pseudorandom generators, J. ACM. Journal of the ACM, 48, 860-879, (2001) · Zbl 1127.68403 [71] Viola, Emanuele, Extractors for circuit sources, SIAM J. Comput.. SIAM Journal on Computing, 43, 655-672, (2014) · Zbl 1301.68195 [72] Zuckerman, David, Randomness-optimal oblivious sampling. Proceedings of the Workshop on Randomized Algorithms and Computation, Random Structures Algorithms. Random Structures & Algorithms, 11, 345-367, (1997) · Zbl 0891.60100
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | |
# Problem: Pyridine is a weak base with the formula, C5H5N, the Kb for Pyridine is 1.8 x 10-9 In an aqueous solution, pyridine partially dissociates according to the following reaction: C5H5N + H2O ⇔ C5H5NH+ + OH-Use the Kb equation to calculate the pH of the aqueous pyridine solution described below: Volume: 375 mL Concentration: 0.2634 M Since this is a weak base, you can assume the amount of base dissociated is << 5% of the total amount of base present. Report your answer with the correct number of significant digits.
###### FREE Expert Solution
We’re being asked to calculate the pH of a 0.2634 M C5H5N
C5H5N is a weak base that will further react in water to form OH-
C5H5N(aq) + H2O (l) → OH- (aq) + C5H6N+ (aq)
From this, we can construct an ICE table for the reaction of C5H5N in a solution. Remember that liquids are ignored in the ICE table.
97% (22 ratings)
###### Problem Details
Pyridine is a weak base with the formula, C5H5N, the Kb for Pyridine is 1.8 x 10-9 In an aqueous solution, pyridine partially dissociates according to the following reaction: C5H5N + H2O ⇔ C5H5NH+ + OH-
Use the Kb equation to calculate the pH of the aqueous pyridine solution described below: Volume: 375 mL Concentration: 0.2634 M Since this is a weak base, you can assume the amount of base dissociated is << 5% of the total amount of base present. Report your answer with the correct number of significant digits. | |
# Laplace's equation on a rectangle with mixed boundary conditions
1. Sep 15, 2008
### nathan12343
1. The problem statement, all variables and given/known data
Solve Laplace's equation inside the rectangle $0 \le x \le L$, $0 \le y \le H$ with the following boundary conditions
$$u(0,y) = g(y)\text{, } u(L,y) = 0\text{, } u_y(x,0) = 0\text{, and } u(x,H) = 0$$
2. Relevant equations
3. The attempt at a solution
I know that with Dirichlet boundary conditions one can simply superpose 4 solutions to 4 other problems corresponding to one side held fixed and the others held at 0. Can the same technique be generalzed for mixed boundary conditions, like I have above? I don't think so, because when I do that the solution I get for
$$u(0,y) = g(y)\text{, } u(L,y) = 0\text{, } u(x,0) = 0\text{, and } u(x,H) = 0$$
does not satisfy $u_y(x,0) = 0[/tex]. Does anyone have a hint for how I might find solutions which simultaneously satisfy the boundary condition at [itex]u(0,y)\text{ and for }u_y(x,0)$?
2. Sep 16, 2008
### gabbagabbahey
Why not find the general 2D solution to Laplace's equation, using separation of variables (i.e. $$u(x,y) \equiv X(x)Y(y)$$)and then substitute your boundary conditions to find the particular solution?
3. Sep 16, 2008
### HallsofIvy
Staff Emeritus
Let v(x,y)= u(x,y)- xg(y)/L
Then $\nabla^2 v= \nabla^2 u- xg"(y)/L= -xg"(y)/L$ since $\nabla^2 u= 0$.
The boundary conditions on v are v(0,y)= 0, v(L, y)= g(y)- g(y)= 0, vy(x, 0)= -xg'(0)/L, v(x,H)= -xg(H)/L.
Because the boundary conditions on x are both 0, you can write v as a Fourier sine series:
$$v(x,y)= \sum_{n=1}^\infty A_n(y)sin(n\pi x/L)$$
You will need to write -xg"(y)/L as a Fourier sine series in x so you can treat g"(y) as a constant. | |
# Looking for an in-text citation for impressionism
###### Question:
looking for an in-text citation for impressionism.
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# Tables and Data
Miscellaneous information and tables are found in Section K of The Astronomical Almanac. Conversion of dates, including calendar dates to Julian dates, are presented. Selected astronomical constants used in the production of The Astronomical Almanac as well as others widely used in astronomy are listed. Coordinates of the celestial pole are tabulated. Text on the reduction of terrestrial coordinates is included, as are several methods of interpolation. Beginning with the 2008 edition, text on some widely used vector and matrix algebra is also included.
The Astronomical Almanac Online! 2016
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Title: Robust regulation with H(2) or H(infinity) performance Author(s): Abedor, John Louis Doctoral Committee Chair(s): Poolla, Kameshwar Department / Program: Engineering, Electronics and ElectricalEngineering, System Science Discipline: Engineering, Electronics and ElectricalEngineering, System Science Degree Granting Institution: University of Illinois at Urbana-Champaign Degree: Ph.D. Genre: Dissertation Subject(s): Engineering, Electronics and Electrical Engineering, System Science Abstract: In this thesis, three performance problems in linear control systems are studied. Robust regulation against steps and sinusoids is the first. Major results on this classic problem are rederived using only basic state-space ideas.The second problem studied is that of minimizing the ${\cal H}\sb2$ norm of a feedback system--the classic LQG problem--subject to the constraint that the controller also solve the robust regulation problem. It is shown that the requirement of robust regulation results in, at most, an arbitrarily small penalty in terms of increased ${\cal H}\sb2$ norm. Necessary and sufficient conditions are also derived that indicate exactly when there exists a controller that both robustly regulates and achieves the optimal ${\cal H}\sb2$ norm. All proofs are constructive.The third problem studied is that of achieving a given ${\cal H}\sb{\infty}$ norm, subject to the constraint that the controller also solve the robust regulation problem. This problem can also be interpreted as a robust performance problem: find a controller that solves the robust regulation problem for every plant determined by an unstructured, norm-bounded uncertainty block. It is shown that this problem admits a solution if and only if the ${\cal H}\sb{\infty}$ problem is solvable and certain matrix inequalities are satisfied, one inequality for every frequency that is robustly regulated against. Controller synthesis is addressed. Issue Date: 1994 Type: Text Language: English URI: http://hdl.handle.net/2142/22971 Rights Information: Copyright 1994 Abedor, John Louis Date Available in IDEALS: 2011-05-07 Identifier in Online Catalog: AAI9503124 OCLC Identifier: (UMI)AAI9503124
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#### Renomalization group flow in conformal field theory = 상사장 이론에서의 재 규격화 군 흐름
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Zamolodchikov and Cardy have shown that the minimal unitary theory with central charge $c=1-\frac{6}{p(p+1)}$ changes into the minimal unitary theory with central charge $c=1-\frac{6}{p(p+1)}$ in the perturbation by $\phi_{13}$. And Zamolodchikov argued that with perturbation by $\phi_{31}$ one might get the reverse result. In this thesis we show this really occurs.
Koh, In-Gyuresearcher고인규researcher
Description
한국과학기술원 : 물리학과,
Publisher
한국과학기술원
Issue Date
1989
Identifier
66620/325007 / 000871110
Language
eng
Description
학위논문(석사) - 한국과학기술원 : 물리학과, 1989.2, [ [ii], 41 p. ; ]
URI
http://hdl.handle.net/10203/48201 |
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