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16 Questions MCQ Test RAS RPSC Prelims Preparation - Notes, Study Material & Tests | CHAIN RULE - MCQ Test (1)
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QUESTION: 1
If the cost of x metres of wire is d rupees, then what is the cost of y metres of wire at the samerate?
Solution:
Option C
Explanation :cost of x metres of wire = Rs. dcost of 1 metre of wire = Rs.(d/x)cost of y metre of wire = Rs.(y×d/x)=Rs. (yd/x)
QUESTION: 2
In a camp, there is a meal for 120 men or 200 children. If 150 children have taken the meal, howmany men will be catered to with remaining meal?
Solution:
Option B
Explanation :Meal for 200 children = Meal for 120 menMeal for 1 child = Meal for 120/200 menMeal for 150 children = Meal for (120×150)/200 men=Meal for 90 menTotal mean available = Meal for 120 menRenaming meal = Meal for 120 men ‐ Meal for 90 men = Meal for 30 men
QUESTION: 3
36 men can complete a piece of work in 18 days. In how many days will 27 men complete thesame work?
Solution:
Option D
Explanation :Let the required number of days be xMore men, less days (indirect proportion)Hence we can write asMen36:27}::x:18 ⇒36×18=27×x ⇒12×18=9×x⇒12×2=x⇒x=24
QUESTION: 4
A wheel that has 6 cogs is meshed with a larger wheel of 14 cogs. If the smaller wheel has made21 revolutions, what will be the number of revolutions made by the larger wheel?
Solution:
Option D
Explanation :Let the number of revolutions made by the larger wheel be xMore cogs, less revolutions (Indirect proportion)Hence we can write asCogs 6:14}: x: 21 ⇒6×21=14×x ⇒6×3=2×x ⇒3×3=x ⇒x=9
QUESTION: 5
3 pumps, working 8 hours a day, can empty a tank in 2 days. How many hours a day should 4pumps work in order to empty the tank in 1 day?
Solution:
Option B
Explanation :Let the required hours needed be xMore pumps, less hours (Indirect proportion)More Days, less hours (Indirect proportion)Hence we can write asPumps 3:4::x:8Days 2:1⇒3×2×8=4×1×x⇒3×2×2=x⇒x=12
QUESTION: 6
39 persons can repair a road in 12 days, working 5 hours a day. In how many days will 30 persons,working 6 hours a day, complete the work?
Solution:
Option D
Explanation :Let the required number of days be xMore persons, less days (indirect proportion)More hours, less days (indirect proportion)Hence we can write asPersons 39:30::x:12Hours 5:6⇒39×5×12=30×6×x ⇒39×5×2=30×x ⇒39=3×x ⇒x=13
QUESTION: 7
A certain industrial loom weaves 0.128 meters of cloth every second. Approximately how manyseconds will it take for the loom to weave 25 meters of cloth?
Solution:
Option D
Explanation :Let the required number of seconds be xMore cloth, More time, (direct proportion)Hence we can write asCloth 0.128:25} :: 1:x⇒0.128x=25 ⇒x=25/0.128 ⇒25000/128=3125/16≈195
QUESTION: 8
21 goats eat as much as 15 cows. How many goats each as much as 35 cows?
Solution:
Option A
Explanation :15 cows ≡ 21 goats1 cow ≡21/15 goats35 cows ≡ (21×35)/15 goats≡(21×7)/3 goats≡7×7 goats ≡ 49 goats
QUESTION: 9
In a dairy farm, 40 cows eat 40 bags of husk in 40 days. In how many days one cow will eat onebag of husk?
Solution:
Option B
Explanation :Assume that in x days, one cow will eat one bag of husk.More cows, less days (Indirect proportion)More bags, more days (direct proportion)Hence we can write asCows 40:1 ::x:40Bags 1:40⇒40×1×40=1×40×x ⇒x=40
QUESTION: 10
. If a quarter kg of potato costs 60 paise, how many paise does 200 gm cost?
Solution:
Option D
Explanation :Let 200 gm potato costs x paiseCost of ¼ Kg potato = 60 Paise=> Cost of 250 gm potato = 60 Paise (∵ 1 Kg = 1000 gm => ¼ Kg = 1000/4 gm = 250 gm)More quantity, More Paise (direct proportion)Hence we can write asQuantity 200:250} :: x:60⇒200×60=250×x ⇒4×60=5×x ⇒4×12=x ⇒x=48
QUESTION: 11
A contract is to be completed in 56 days if 104 persons work, each working at 8 hours a day. After30 days, 2/5 of the work is completed. How many additional persons should be deployed so that thework will be completed in the scheduled time, each person’s now working 9 hours a day.
Solution:
Option D
Explanation :Persons worked = 104Number of hours each person worked per day = 8Number of days they worked = 30Work completed = 2/5Remaining days = 56 ‐ 30 = 26Remaining Work to be completed = 1 ‐ 2/5 = 3/5Let the total number of persons who do the remaining work = xNumber of hours each person needs to be work per day = 9More days, less persons(indirect proportion) More hours, less persons(indirect proportion)More work, more persons(direct proportion)Hence we can write asDays 30:26Hours 8:9 ::x:104Work 35:25⇒30×8×3/5×104=26×9×2/5×x⇒x=(30×8×3/5×104)/(26×9×2/5)=(30×8×3×104)/(26×9×2)=(30×8×104)/(26×3×2)=(30×8×4)/(3×2)=5×8×4=160Number of additional persons required = 160 ‐ 104 = 56
QUESTION: 12
x men working x hours per day can do x units of a work in x days. How much work can becompleted by y men working y hours per day in y days?
Solution:
Option B
Explanation :Let amount of work completed by y men working y hours per in y days = w unitsMore men, more work(direct proportion)More hours, more work(direct proportion)More days, more work(direct proportion)Hence we can write asMen x:yHours x:y ::x:wDays x:y⇒x3w=y3x ⇒w=y3x/x3=y3/x2
QUESTION: 13
A flagstaff 17.5 m high casts a shadow of length 40.25 m. What will be the height of a building,which casts a shadow of length 28.75 m under similar conditions?
Solution:
Option A
Explanation :Let the required height of the building be x meterMore shadow length, More height (direct proportion)Hence we can write asShadow length 40.25:28.75}:: 17.5:x⇒40.25×x=28.75×17.5 ⇒x=(28.75×17.5)/40.25=(2875×175)/40250= (2875×7)/1610=2875/230=575/46=12.5
QUESTION: 14
If the price of 357 apples is Rs.1517.25, what will be the approximate price of 49 dozens of suchapples?
Solution:
Option A
Explanation :Let the required price be xMore apples, More price (direct proportion)Hence we can write asApples 357:(49×12)} :: 1517.25:x⇒357x = (49×12)×1517.25 ⇒x = (49×12×1517.25)/357=(7×12×1517.25)/51= (7×4×1517.25)/17=7×4×89.25≈2500
QUESTION: 15
9 engines consume 24 metric tonnes of coal, when each is working 8 hours day. How much coal isrequired for 8 engines, each running 13 hours a day, if 3 engines of former type consume as much as4 engines of latter type?
Solution:
Option D
Explanation :Let required amount of coal be x metric tonnesMore engines, more amount of coal (direct proportion)If 3 engines of first type consume 1 unit, then 1 engine will consume 1/3 unit which is its the rate ofconsumption.If 4 engines of second type consume 1 unit, then 1 engine will consume 1/4 unit which is its the rateof consumptionMore rate of consumption, more amount of coal (direct proportion)More hours, more amount of coal(direct proportion)Hence we can write asEngines 9:8rate of consumption 13:14 ::24:xhours 8:13⇒9×1/3×8×x=8×1/4×13×24 ⇒3×8×x=8×6×13 ⇒3xX=6×13 ⇒x=2×13=26
QUESTION: 16
In a camp, food was was sufficient for 2000 people for 54 days. After 15 days, more people cameand the food last only for 20 more days. How many people came?
Solution:
Option A
Explanation :Given that food was sufficient for 2000 people for 54 daysHence, after 15 days, the remaining food was sufficient for 2000 people for 39 days (∵ 54 ‐ 15 =39)Let x number of people came after 15 days.Then, total number of people after 15 days = (2000 + x)Then, the remaining food was sufficient for (2000 + x) people for 20 daysMore men, Less days (Indirect Proportion) ⇒Men 2000:(2000+x)} :: 20:39⇒2000×39=(2000+x)20 ⇒100×39=(2000+x) ⇒3900=2000+x ⇒ x=3900−2000=1900
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http://wikien4.appspot.com/wiki/Porosity | 1,550,402,699,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247481992.39/warc/CC-MAIN-20190217111746-20190217133746-00217.warc.gz | 302,947,327 | 27,337 | Porosity
Porosity or void fraction is a measure of de void (i.e. "empty") spaces in a materiaw, and is a fraction of de vowume of voids over de totaw vowume, between 0 and 1, or as a percentage between 0% and 100%. Strictwy speaking, some tests measure de "accessibwe void", de totaw amount of void space accessibwe from de surface (cf. cwosed-ceww foam). There are many ways to test porosity in a substance or part, such as industriaw CT scanning. The term porosity is used in muwtipwe fiewds incwuding pharmaceutics, ceramics, metawwurgy, materiaws, manufacturing, earf sciences, soiw mechanics and engineering.
Void fraction in two-phase fwow
In gas-wiqwid two-phase fwow, de void fraction is defined as de fraction of de fwow-channew vowume dat is occupied by de gas phase or, awternativewy, as de fraction of de cross-sectionaw area of de channew dat is occupied by de gas phase.[1] Void fraction usuawwy varies from wocation to wocation in de fwow channew (depending on de two-phase fwow pattern). It fwuctuates wif time and its vawue is usuawwy time averaged. In separated (i.e., non-homogeneous) fwow, it is rewated to vowumetric fwow rates of de gas and de wiqwid phase, and to de ratio of de vewocity of de two phases (cawwed swip ratio).
Porosity in earf sciences and construction
Used in geowogy, hydrogeowogy, soiw science, and buiwding science, de porosity of a porous medium (such as rock or sediment) describes de fraction of void space in de materiaw, where de void may contain, for exampwe, air or water. It is defined by de ratio:
${\dispwaystywe \phi ={\frac {V_{\madrm {V} }}{V_{\madrm {T} }}}}$
where VV is de vowume of void-space (such as fwuids) and VT is de totaw or buwk vowume of materiaw, incwuding de sowid and void components. Bof de madematicaw symbows ${\dispwaystywe \phi }$ and ${\dispwaystywe n}$ are used to denote porosity.
Porosity is a fraction between 0 and 1, typicawwy ranging from wess dan 0.01 for sowid granite to more dan 0.5 for peat and cway. It may awso be represented in percent terms by muwtipwying de fraction by 100.
The porosity of a rock, or sedimentary wayer, is an important consideration when attempting to evawuate de potentiaw vowume of water or hydrocarbons it may contain, uh-hah-hah-hah. Sedimentary porosity is a compwicated function of many factors, incwuding but not wimited to: rate of buriaw, depf of buriaw, de nature of de connate fwuids, de nature of overwying sediments (which may impede fwuid expuwsion). One commonwy used rewationship between porosity and depf is given by de Ady (1930) eqwation:[2]
${\dispwaystywe \phi (z)=\phi _{0}e^{-kz}\,}$
where ${\dispwaystywe \phi _{0}}$ is de surface porosity, ${\dispwaystywe k}$ is de compaction coefficient (m−1) and ${\dispwaystywe z}$ is depf (m).
A vawue for porosity can awternativewy be cawcuwated from de buwk density ${\dispwaystywe \rho _{\text{buwk}}}$, saturating fwuid density ${\dispwaystywe \rho _{\text{fwuid}}}$ and particwe density ${\dispwaystywe \rho _{\text{particwe}}}$:
${\dispwaystywe \phi ={\frac {\rho _{\text{particwe}}-\rho _{\text{buwk}}}{\rho _{\text{particwe}}-\rho _{\text{fwuid}}}}}$
If de void space is fiwwed wif air, de fowwowing simpwer form may be used:
${\dispwaystywe \phi =1-{\frac {\rho _{\text{buwk}}}{\rho _{\text{particwe}}}}}$
Normaw particwe density is assumed to be approximatewy 2.65 g/cm3 (siwica), awdough a better estimation can be obtained by examining de widowogy of de particwes.
Porosity and hydrauwic conductivity
Porosity can be proportionaw to hydrauwic conductivity; for two simiwar sandy aqwifers, de one wif a higher porosity wiww typicawwy have a higher hydrauwic conductivity (more open area for de fwow of water), but dere are many compwications to dis rewationship. The principaw compwication is dat dere is not a direct proportionawity between porosity and hydrauwic conductivity but rader an inferred proportionawity. There is a cwear proportionawity between pore droat radii and hydrauwic conductivity. Awso, dere tends to be a proportionawity between pore droat radii and pore vowume. If de proportionawity between pore droat radii and porosity exists den a proportionawity between porosity and hydrauwic conductivity may exist. However, as grain size or sorting decreases de proportionawity between pore droat radii and porosity begins to faiw and derefore so does de proportionawity between porosity and hydrauwic conductivity. For exampwe: cways typicawwy have very wow hydrauwic conductivity (due to deir smaww pore droat radii) but awso have very high porosities (due to de structured nature of cway mineraws), which means cways can howd a warge vowume of water per vowume of buwk materiaw, but dey do not rewease water rapidwy and derefore have wow hydrauwic conductivity.
Sorting and porosity
Effects of sorting on awwuviaw porosity. Bwack represents sowids, bwue represents pore space.
Weww sorted (grains of approximatewy aww one size) materiaws have higher porosity dan simiwarwy sized poorwy sorted materiaws (where smawwer particwes fiww de gaps between warger particwes). The graphic iwwustrates how some smawwer grains can effectivewy fiww de pores (where aww water fwow takes pwace), drasticawwy reducing porosity and hydrauwic conductivity, whiwe onwy being a smaww fraction of de totaw vowume of de materiaw. For tabwes of common porosity vawues for earf materiaws, see de "furder reading" section in de Hydrogeowogy articwe.
Porosity of rocks
Consowidated rocks (e.g., sandstone, shawe, granite or wimestone) potentiawwy have more compwex "duaw" porosities, as compared wif awwuviaw sediment. This can be spwit into connected and unconnected porosity. Connected porosity is more easiwy measured drough de vowume of gas or wiqwid dat can fwow into de rock, whereas fwuids cannot access unconnected pores.
Porosity is de ratio of pore vowume to its totaw vowume. Porosity is controwwed by: rock type, pore distribution, cementation, diagenetic history and composition, uh-hah-hah-hah. Porosity is not controwwed by grain size, as de vowume of between-grain space is rewated onwy to de medod of grain packing.
Rocks normawwy decrease in porosity wif age and depf of buriaw. Tertiary age Guwf Coast sandstones are in generaw more porous dan Cambrian age sandstones. There are exceptions to dis ruwe, usuawwy because of de depf of buriaw and dermaw history.
Porosity of soiw
Porosity of surface soiw typicawwy decreases as particwe size increases. This is due to soiw aggregate formation in finer textured surface soiws when subject to soiw biowogicaw processes. Aggregation invowves particuwate adhesion and higher resistance to compaction, uh-hah-hah-hah. Typicaw buwk density of sandy soiw is between 1.5 and 1.7 g/cm3. This cawcuwates to a porosity between 0.43 and 0.36. Typicaw buwk density of cway soiw is between 1.1 and 1.3 g/cm3. This cawcuwates to a porosity between 0.58 and 0.51. This seems counterintuitive because cway soiws are termed heavy, impwying wower porosity. Heavy apparentwy refers to a gravitationaw moisture content effect in combination wif terminowogy dat harkens back to de rewative force reqwired to puww a tiwwage impwement drough de cwayey soiw at fiewd moisture content as compared to sand.
Porosity of subsurface soiw is wower dan in surface soiw due to compaction by gravity. Porosity of 0.20 is considered normaw for unsorted gravew size materiaw at depds bewow de biomantwe. Porosity in finer materiaw bewow de aggregating infwuence of pedogenesis can be expected to approximate dis vawue.
Soiw porosity is compwex. Traditionaw modews regard porosity as continuous. This faiws to account for anomawous features and produces onwy approximate resuwts. Furdermore, it cannot hewp modew de infwuence of environmentaw factors which affect pore geometry. A number of more compwex modews have been proposed, incwuding fractaws, bubbwe deory, cracking deory, Boowean grain process, packed sphere, and numerous oder modews. The characterisation of pore space in soiw is an associated concept.
Types of geowogic porosities
Primary porosity
The main or originaw porosity system in a rock or unconfined awwuviaw deposit.
Secondary porosity
A subseqwent or separate porosity system in a rock, often enhancing overaww porosity of a rock. This can be a resuwt of chemicaw weaching of mineraws or de generation of a fracture system. This can repwace de primary porosity or coexist wif it (see duaw porosity bewow).
Fracture porosity
This is porosity associated wif a fracture system or fauwting. This can create secondary porosity in rocks dat oderwise wouwd not be reservoirs for hydrocarbons due to deir primary porosity being destroyed (for exampwe due to depf of buriaw) or of a rock type not normawwy considered a reservoir (for exampwe igneous intrusions or metasediments).
Vuggy porosity
This is secondary porosity generated by dissowution of warge features (such as macrofossiws) in carbonate rocks weaving warge howes, vugs, or even caves.
Effective porosity (awso cawwed open porosity)
Refers to de fraction of de totaw vowume in which fwuid fwow is effectivewy taking pwace and incwudes catenary and dead-end (as dese pores cannot be fwushed, but dey can cause fwuid movement by rewease of pressure wike gas expansion[3]) pores and excwudes cwosed pores (or non-connected cavities). This is very important for groundwater and petroweum fwow, as weww as for sowute transport.
Ineffective porosity (awso cawwed cwosed porosity)
Refers to de fraction of de totaw vowume in which fwuids or gases are present but in which fwuid fwow can not effectivewy take pwace and incwudes de cwosed pores. Understanding de morphowogy of de porosity is dus very important for groundwater and petroweum fwow.
Duaw porosity
Refers to de conceptuaw idea dat dere are two overwapping reservoirs which interact. In fractured rock aqwifers, de rock mass and fractures are often simuwated as being two overwapping but distinct bodies. Dewayed yiewd, and weaky aqwifer fwow sowutions are bof madematicawwy simiwar sowutions to dat obtained for duaw porosity; in aww dree cases water comes from two madematicawwy different reservoirs (wheder or not dey are physicawwy different).
Macroporosity
In sowids (i.e. excwuding aggregated materiaws such as soiws), de term 'macroporosity' refers to pores greater dan 50 nm in diameter. Fwow drough macropores is described by buwk diffusion, uh-hah-hah-hah.
Mesoporosity
In sowids (i.e. excwuding aggregated materiaws such as soiws), de term 'mesoporosity' refers to pores greater dan 2 nm and wess dan 50 nm in diameter. Fwow drough mesopores is described by Knudsen diffusion, uh-hah-hah-hah.
Microporosity
In sowids (i.e. excwuding aggregated materiaws such as soiws), de term 'microporosity' refers to pores smawwer dan 2 nm in diameter. Movement in micropores is activated by diffusion, uh-hah-hah-hah.
Porosity of fabric or aerodynamic porosity
The ratio of howes to sowid dat de wind "sees". Aerodynamic porosity is wess dan visuaw porosity, by an amount dat depends on de constriction of howes.
Die casting porosity
Casting porosity is a conseqwence of one or more of de fowwowing: gasification of contaminants at mowten-metaw temperatures; shrinkage dat takes pwace as mowten metaw sowidifies; and unexpected or uncontrowwed changes in temperature or humidity.
Whiwe porosity is inherent in die casting manufacturing, its presence may wead to component faiwure where pressure integrity is a criticaw characteristic. Porosity may take on severaw forms from interconnected micro-porosity, fowds, and incwusions to macro porosity visibwe on de part surface. The end resuwt of porosity is de creation of a weak paf drough de wawws of a casting dat prevents de part from howding pressure. Porosity may awso wead to out-gassing during de painting process, weaching of pwating acids and toow chatter in machining pressed metaw components.[4]
Measuring porosity
Opticaw medod of measuring porosity: din section under gypsum pwate shows porosity as purpwe cowor, contrasted wif carbonate grains of oder cowors. Pweistocene eowianite from San Sawvador Iswand, Bahamas. Scawe bar 500 µm.
Severaw medods can be empwoyed to measure porosity:
• Direct medods (determining de buwk vowume of de porous sampwe, and den determining de vowume of de skewetaw materiaw wif no pores (pore vowume = totaw vowume − materiaw vowume).
• Opticaw medods (e.g., determining de area of de materiaw versus de area of de pores visibwe under de microscope). The "areaw" and "vowumetric" porosities are eqwaw for porous media wif random structure.[5]
• Computed tomography medod (using industriaw CT scanning to create a 3D rendering of externaw and internaw geometry, incwuding voids. Then impwementing a defect anawysis utiwizing computer software)
• Imbibition medods,[5] i.e., immersion of de porous sampwe, under vacuum, in a fwuid dat preferentiawwy wets de pores.
• Water saturation medod (pore vowume = totaw vowume of water − vowume of water weft after soaking).
• Water evaporation medod (pore vowume = (weight of saturated sampwe − weight of dried sampwe)/density of water)
• Mercury intrusion porosimetry (severaw non-mercury intrusion techniqwes have been devewoped due to toxicowogicaw concerns, and de fact dat mercury tends to form amawgams wif severaw metaws and awwoys).
• Gas expansion medod.[5] A sampwe of known buwk vowume is encwosed in a container of known vowume. It is connected to anoder container wif a known vowume which is evacuated (i.e., near vacuum pressure). When a vawve connecting de two containers is opened, gas passes from de first container to de second untiw a uniform pressure distribution is attained. Using ideaw gas waw, de vowume of de pores is cawcuwated as
${\dispwaystywe V_{V}=V_{T}-V_{a}-V_{b}{P_{2} \over {P_{2}-P_{1}}}}$,
where
VV is de effective vowume of de pores,
VT is de buwk vowume of de sampwe,
Va is de vowume of de container containing de sampwe,
Vb is de vowume of de evacuated container,
P1 is de initiaw pressure in de initiaw pressure in vowume Va and VV, and
P2 is finaw pressure present in de entire system.
The porosity fowwows straightforwardwy by its proper definition
${\dispwaystywe \phi ={\frac {V_{V}}{V_{T}}}}$.
Note dat dis medod assumes dat gas communicates between de pores and de surrounding vowume. In practice, dis means dat de pores must not be cwosed cavities.
• Thermoporosimetry and cryoporometry. A smaww crystaw of a wiqwid mewts at a wower temperature dan de buwk wiqwid, as given by de Gibbs-Thomson eqwation. Thus if a wiqwid is imbibed into a porous materiaw, and frozen, de mewting temperature wiww provide information on de pore-size distribution, uh-hah-hah-hah. The detection of de mewting can be done by sensing de transient heat fwows during phase-changes using differentiaw scanning caworimetry – (DSC dermoporometry),[6] measuring de qwantity of mobiwe wiqwid using nucwear magnetic resonance – (NMR cryoporometry)[7] or measuring de ampwitude of neutron scattering from de imbibed crystawwine or wiqwid phases – (ND cryoporometry).[8]
References
• Gwasbey, C. A.; G. W. Horgan; J. F. Darbyshire (September 1991). "Image anawysis and dree-dimensionaw modewwing of pores in soiw aggregates". Journaw of Soiw Science. 42 (3): 479–86. doi:10.1111/j.1365-2389.1991.tb00424.x.
• Horgan, G. W.; B. C. Baww (1994). "Simuwating diffusion in a Boowean modew of soiw pores". European Journaw of Soiw Science. 45 (4): 483–91. doi:10.1111/j.1365-2389.1994.tb00534.x.
• Horgan, Graham W. (1996-10-01). "A review of soiw pore modews" (PDF). Retrieved 2006-04-16.
• Horgan, G. W. (June 1998). "Madematicaw morphowogy for soiw image anawysis". European Journaw of Soiw Science. 49 (2): 161–73. doi:10.1046/j.1365-2389.1998.00160.x.
• Horgan, G. W. (February 1999). "An investigation of de geometric infwuences on pore space diffusion". Geoderma. 88 (1–2): 55–71. Bibcode:1999Geode..88...55H. doi:10.1016/S0016-7061(98)00075-5.
• Newson, J. Roy (January 2000). "Physics of impregnation" (PDF). Microscopy Today. 8 (1). Archived from de originaw (PDF) on 2009-02-27.
• Rouqwerow, Jean (December 2011). "Liqwid intrusion and awternative medods for de characterization of macroporous materiaws (IUPAC Technicaw Report)*" (PDF). Pure Appw. Chem. 84 (1): 107–36. doi:10.1351/pac-rep-10-11-19.
Footnotes
1. ^ G.F. Hewitt, G.L. Shires, Y.V.Powezhaev (editors), "Internationaw Encycwopedia of Heat and Mass Transfer", CRC Press, 1997.
2. ^ ATHY L.F., 1930. Density, porosity and compactation of sedimentary rocks, Buww. Amer. Assoc. Petrow. Geow. v. 14, pp. 1-24.
3. ^
4. ^ "How to Fix Die Casting Porosity?". Godfrey & Wing.
5. ^ a b c F.A.L. Duwwien, "Porous Media. Fwuid Transport and Pore Structure", Academic Press, 1992.
6. ^ Brun, M.; Lawwemand, A.; Quinson, J-F.; Eyraud, C. (1977). "A new medod for de simuwtaneous determination of de size and de shape of pores: The Thermoporometry". Thermochimica Acta. Ewsevier Scientific Pubwishing Company, Amsterdam. 21: 59–88. doi:10.1016/0040-6031(77)85122-8
7. ^ Mitcheww, J.; Webber, J. Beau W.; Strange, J.H. (2008). "Nucwear Magnetic Resonance Cryoporometry". Phys. Rep. 461: 1–36. Bibcode:2008PhR...461....1M. doi:10.1016/j.physrep.2008.02.001
8. ^ Webber, J. Beau W.; Dore, John C. (2008). "Neutron Diffraction Cryoporometry – a measurement techniqwe for studying mesoporous materiaws and de phases of contained wiqwids and deir crystawwine forms". Nucw. Instrum. Mef. A. 586 (2): 356–66. Bibcode:2008NIMPA.586..356W. doi:10.1016/j.nima.2007.12.004 | 5,104 | 17,548 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 14, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2019-09 | latest | en | 0.821885 |
http://www.kwiznet.com/p/takeQuiz.php?ChapterID=1728&CurriculumID=40&Num=1.12 | 1,544,580,040,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823710.44/warc/CC-MAIN-20181212000955-20181212022455-00103.warc.gz | 406,747,866 | 3,879 | Email us to get an instant 20% discount on highly effective K-12 Math & English kwizNET Programs!
Online Quiz (WorksheetABCD)
Questions Per Quiz = 2 4 6 8 10
Math Word Problems - GED, PSAT, SAT, ACT, GRE Preparation1.12 Solving Equations - 2
One of the most common ways to simplify an expression is to combine like terms. Numeric terms may be combined, and any terms with the same variable part may be combined. Example: Consider the expression 2 + 7x + 12 - 3x - 5. The numeric like terms are the numbers 2, 12, and 5. The variable like terms are 7x and 3x. Combining the numeric like terms, we have 2 + 12 - 5 = 14 - 5 = 9. Combining the variable like terms, we have 7x - 3x = 4x, Therefore, the expression 2 + 7x + 12 - 3x - 5 simplifies to 9 + 4x. Directions: Solve the following equations. First simplify and then find the value of the variable.
Q 1: Solve: s/8 = 10+3Answer: Q 2: Solve: x + 7 = 15 + 20 - 5Answer: Q 3: Solve: 4a + 2a = 42Answer: Q 4: Solve: 5t + 4t + 2t = 88Answer: Q 5: Solve: 25 + g - 4 = 42Answer: Q 6: Solve: c + 10 + 6 = 31Answer: Q 7: Solve: x - 6 = 15 - 2Answer: Q 8: Solve: b - 10 + 3 = 24Answer: Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only! | 429 | 1,254 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2018-51 | longest | en | 0.807548 |
https://www.codingpanel.com/lesson/read-a-linked-list-in-reverse-order/ | 1,726,094,248,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651405.61/warc/CC-MAIN-20240911215612-20240912005612-00184.warc.gz | 654,305,586 | 14,239 | In this article, you will learn how to read a linked list in reverse. Note that we only have to display values in the reverse direction. We do not need to reverse the actual linked list, meaning there is no need to update pointers, etc. Reading a linked list in the backward direction is easy. We can do so by using a recursive function. Given the head of the linked list, the steps to solve the given problem are given below.
```read_in_reverse(head)
return
```
Consider the following figure for illustration.
In the above figure, you can see that we call the `read_in_reverse()` function first with the head of the linked list, i.e., the node with the value 1. Then, the same function gets called recursively by passing `head.next` as its argument. When the end of the list is reached, i.e., at step 5, we simply return.
After that, we display the value of the current node, i.e., the first value printed is 2, then 6, and so on. You can see that the last value read is 1 (at step 9), which is the value of the first node of the linked list.
## Implementation
The following code contains the implementation of the singly linked list to insert items and display them. It is followed by the `read_in_reverse()` function that takes the head pointer and reads the linked list in the reverse direction.
```#node for the linked list
class Node:
def __init__(self, value=None):
self.value = value
self.next = None
def __init__(self):
self.length = 0 #total number of elements in the linked list
def insert(self, value):
node = Node(value) #create a node
if self.head == None: #if list is empty insert a node at the start
else:
#iterate through the list till last node is found
while temp.next:
temp = temp.next
temp.next = node #adding a new node
self.length += 1 #increment the length
#traverse the complete list and print the node value
def traverse(self):
while temp:
print(temp.value, end="\t")
temp = temp.next
print("")
return
#insert values
ll.insert(1)
ll.insert(8)
ll.insert(6)
ll.insert(2)
ll.traverse()
```Given Linked list: | 491 | 2,045 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-38 | latest | en | 0.785239 |
https://en.wikitolearn.org/Course:Statistical_Mechanics/Ensemble_theory/Fluctuations_in_the_grand_canonical_ensemble | 1,620,533,245,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988955.89/warc/CC-MAIN-20210509032519-20210509062519-00069.warc.gz | 268,424,923 | 13,300 | # Fluctuations in the grand canonical ensemble
As we have anticipated we expect that, similarly to what happens in the canonical ensemble, the fluctuations of the energy and number of particles around their mean values vanish in the thermodynamic limit so that the grand canonical ensemble is indeed equivalent to the canonical one.
The fluctuations of the energy around ${\displaystyle \left\langle {\mathcal {H}}\right\rangle }$ are computed exactly in the same way as we have done in The canonical and the microcanonical ensemble, and we get to the same result. We therefore now focus on the fluctuations of the number of particles around ${\displaystyle \left\langle N\right\rangle }$:
${\displaystyle \sigma _{N}^{2}=\left\langle N^{2}\right\rangle -\left\langle N\right\rangle ^{2}={\frac {1}{\beta ^{2}}}{\frac {\partial ^{2}\ln {\mathcal {Z}}}{\partial \mu ^{2}}}={\frac {1}{\beta ^{2}}}{\frac {\partial ^{2}}{\partial \mu ^{2}}}[-\beta \Phi (T,V,\mu )]}$
Now, from the definition ${\displaystyle \Phi ={\overline {E}}-TS({\overline {E}},V,{\overline {N}})-\mu {\overline {N}}}$ of grand potential we have (considering that ${\displaystyle \partial S/\partial V=P/T}$ and that, again, there are some derivatives that vanish because ${\displaystyle {\overline {E}}}$ and ${\displaystyle {\overline {N}}}$ define a minimum) that:
${\displaystyle {\frac {\partial \Phi }{\partial V}}=-P}$
Furthermore, ${\displaystyle \Phi }$ is extensive:
${\displaystyle \Phi (T,\lambda V,\mu )=\lambda \Phi (T,V,\mu )}$
and setting ${\displaystyle \lambda =1/V}$ we get ${\displaystyle \Phi (T,V,\mu )=V\Phi (T,1,\mu )}$: the grand potential is proportional to the volume. Therefore, ${\displaystyle \partial \Phi (T,V,\mu )/\partial V=\Phi (T,1,\mu )=-P}$ and so:
${\displaystyle \Phi (T,V,\mu )=V\Phi (T,1,\mu )=-VP}$
Thus:
${\displaystyle \sigma _{N}^{2}=-{\frac {1}{\beta }}{\frac {\partial ^{2}}{\partial \mu ^{2}}}\Phi (T,V,\mu )={\frac {V}{\beta }}{\frac {\partial ^{2}}{\partial \mu ^{2}}}P(T,\mu )}$
We must therefore find a way to express ${\displaystyle \partial ^{2}P/\partial \mu ^{2}}$. We define ${\displaystyle v(T,\mu )=V/{\overline {N}}}$ and ${\displaystyle f(T,v)=F/{\overline {N}}}$, and then write:
${\displaystyle {\frac {\partial P}{\partial \mu }}={\frac {\partial P/\partial v}{\partial \mu /\partial v}}}$
Since ${\displaystyle \Phi =-PV=F-\mu {\overline {N}}}$, we have:
${\displaystyle f{\overline {N}}-\mu {\overline {N}}=-PV\quad \Rightarrow \quad f(T,v)=\mu -vP(T,\mu )}$
and since:
${\displaystyle P=-{\frac {\partial F}{\partial V}}=-{\frac {\partial f}{\partial v}}}$
then:
${\displaystyle f(T,v)=\mu +v{\frac {\partial f}{\partial v}}\quad \Rightarrow \quad \mu =f-v{\frac {\partial f}{\partial v}}}$
We therefore have:
${\displaystyle {\frac {\partial P}{\partial v}}=-{\frac {\partial ^{2}f}{\partial v^{2}}}\quad \qquad {\frac {\partial \mu }{\partial v}}=-v{\frac {\partial ^{2}f}{\partial v^{2}}}}$
and substituting in ${\textstyle \partial P/\partial \mu }$, this leads to:
${\displaystyle {\frac {\partial P}{\partial \mu }}={\frac {1}{v}}}$
If we now further derive ${\displaystyle P}$ with respect to ${\displaystyle \mu }$:
${\displaystyle {\frac {\partial ^{2}P}{\partial \mu ^{2}}}=-{\frac {1}{v^{2}}}{\frac {\partial v}{\partial \mu }}=-{\frac {1}{v^{2}}}{\frac {1}{\partial \mu /\partial v}}}$
and from the previous expression of ${\textstyle \partial \mu /\partial v}$ we have:
${\displaystyle {\frac {\partial \mu }{\partial v}}=-v{\frac {\partial ^{2}f}{\partial v^{2}}}=v{\frac {\partial P}{\partial v}}}$
so that:
${\displaystyle {\frac {\partial ^{2}P}{\partial \mu ^{2}}}=-{\frac {1}{v^{2}}}\cdot {\frac {1}{v\cdot \partial P/\partial v}}=-\left(v^{3}{\frac {\partial P}{\partial v}}\right)^{-1}}$
We have done all these intricate computations because now we can use the definition of isothermal compressibility (see Response functions)[1]:
${\displaystyle K_{T}=-{\frac {1}{V}}{\frac {\partial V}{\partial P}}=-{\frac {1}{V}}{\frac {1}{\partial P/\partial V}}}$
(which as we can see is an intensive quantity) to write:
${\displaystyle {\frac {\partial ^{2}P}{\partial \mu ^{2}}}={\frac {K_{T}}{v^{2}}}}$
This way the variance of the number of particles can be written as:
${\displaystyle \sigma _{N}^{2}={\frac {V}{\beta }}\cdot {\frac {\partial ^{2}P}{\partial \mu ^{2}}}={\frac {V}{\beta }}\cdot {\frac {K_{T}}{v^{2}}}=N{\frac {K_{T}}{\beta v}}}$
(which is of course positive since ${\displaystyle K_{T}>0}$). We therefore see that:
${\displaystyle \sigma _{N}^{2}\propto N\quad \Rightarrow \quad \sigma _{N}\propto {\sqrt {N}}}$
and so the relative fluctuation of the number of particles is:
${\displaystyle {\frac {\sigma _{N}}{N}}\propto {\frac {1}{\sqrt {N}}}}$
which is a result analogous to what we have seen for the energy.
Therefore, as we expected, the fluctuations of the number of particles of a system in the grand canonical ensemble are negligible in the thermodynamic limit; this ultimately means that the grand canonical ensemble is equivalent to the canonical one.
1. This is of course equal to what we have found, because ${\displaystyle v=V/{\overline {N}}}$. | 1,590 | 5,133 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 39, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2021-21 | latest | en | 0.7794 |
https://1hotelsouthbeachforsale.com/qa/quick-answer-can-an-angle-be-more-than-360.html | 1,618,603,774,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038089289.45/warc/CC-MAIN-20210416191341-20210416221341-00370.warc.gz | 190,410,641 | 7,375 | Quick Answer: Can An Angle Be More Than 360?
How do you draw an angle greater than 360?
Here’s what you do:Find a co-terminal angle by adding 360 degrees.
Adding 360 degrees to –570 degrees gives you –210 degrees.If the angle is still negative, keep adding 360 degrees until you get a positive angle in standard position.
Draw the angle you create in Step 2..
What is a 33 degree angle called?
In geometry, there are three types of angles: acute angle-an angle between 0 and 90 degrees. right angle-an 90 degree angle. obtuse angle-an angle between 90 and 180 degrees.
Can an angle be greater than 180 degrees?
Angles larger than a right angle and smaller than a straight angle (between 90° and 180°) are called obtuse angles (“obtuse” meaning “blunt”). An angle equal to 12 turn (180° or π radians) is called a straight angle.
What do you call a 360 angle?
A 180° angle is called a straight angle. … Angles such as 270 degrees which are more than 180 but less than 360 degrees are called reflex angles. A 360° angle is called a complete angle.
What do you call an angle greater than 180 degrees?
An obtuse angle has a measurement greater than 90 degrees but less than 180 degrees. However, A reflex angle measures more than 180 degrees but less than 360 degrees.
Why full angle is 360?
The Mesopotamians passed their base-60 numerical system to the ancient Egyptians, who used it to divide a circle into 360 degrees, Mary Blocksma writes in her book Reading the Numbers. … Therefore, six triangular slices of 60 degrees each made for a sensible 360-degree dissection of a circle.
What are the 7 types of angles?
Types of Angles – Acute, Right, Obtuse, Straight and Reflex…Acute angle.Right angle.Obtuse angle.Straight angle.Reflex angle.
What is the Coterminal angle of 3pi 4?
19pi/4You can find infinitely many coterminal angles. For example a positive coterminal angle for 3pi/4 would be 19pi/4.
What is a positive Coterminal angle between 800 and 0 and 360?
Since adding or subtracting a full rotation, 360 degrees, would result in an angle with terminal side pointing in the same direction, we can find coterminal angles by adding or subtracting 360 degrees. An angle of 800 degrees is coterminal with an angle of 800−360=440 degrees. | 558 | 2,259 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2021-17 | latest | en | 0.88099 |
https://flatearthdefenders.com/nick-davis-attorney-flat-earth-math-proof.html | 1,627,156,489,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150308.48/warc/CC-MAIN-20210724191957-20210724221957-00171.warc.gz | 265,329,727 | 11,749 | If the sun is the center of the solar system (which is actually based on Pagan Sun Worship) and the earth is orbiting around it, then for this event in Joshua to occur, it would mean that the earth stopped spinning and stood still, not the sun. For the earth to suddenly stop spinning at its supposed 1000 miles per hour spin, it would have caused catastrophic devastation on earth. But, if the earth is flat and still, and the sun and moon rotate around and above the earth, then it would make perfect sense for the sun to stand still in order to prolong the day.
65) Also Quoting Dr. Rowbotham, “On the shore near Waterloo, a few miles to the north of Liverpool, a good telescope was fixed, at an elevation of 6 feet above the water. It was directed to a large steamer, just leaving the River Mersey, and sailing out to Dublin. Gradually the mast-head of the receding vessel came nearer to the horizon, until, at length, after more than four hours had elapsed, it disappeared. The ordinary rate of sailing of the Dublin steamers was fully eight miles an hour; so that the vessel would be, at least, thirty-two miles distant when the mast-head came to the horizon. The 6 feet of elevation of the telescope would require three miles to be deducted for convexity, which would leave twenty-nine miles, the square of which, multiplied by 8 inches, gives 560 feet; deducting 80 feet for the height of the main-mast, and we find that, according to the doctrine of rotundity, the mast-head of the outward bound steamer should have been 480 feet below the horizon. Many other experiments of this kind have been made upon sea-going steamers, and always with results entirely incompatible with the theory that the earth is a globe.”
If the Earth were a globe, it would, unquestionably, have the same general characteristics - no matter its size - as a small globe that may be stood upon the table. As the small globe has top, bottom, and sides, so must also the large one - no matter how large it be. But, as the Earth, which is "supposed" to be a large globe, bas no sides or bottom as the small globe has, the conclusion is irresistible that it is a proof that the Earth is not a globe.
As for flight paths and what Appears to be the silly way for a ball earth but makes sense for a flat earth, it reminds me of the child quiz. There is a spider in the corner of the room on the floor and he wants to get to the opp corner on the ceiling. Which is the quickest path? We instantly say, across the floor and up the wall join. BUT, if we flatten the room we then draw a straight line, we find the quickest path is diagonally up one wall and then diagonally across the ceiling, which Looks longer but is best.
In " Cornell's Geography" there is an "Illustrated proof of the Form of the Earth," A curved line on which is represented a ship in four positions, as she sails away from an observer, is an arc of 72 degrees, or one-fifth of the supposed circumference of the "globe" - about 5,000 miles. Ten, such ships as those which are given in the picture would reach the full length of the "arc," making 500 miles as the length of the ship, The man in the picture, who is watching the ship as she sails away, is about 200 miles high; and the tower, from which he takes an elevated view, at least 600 miles high. These are the proportions, then, of men, towers, arid ships which are necessary in order to see a ship, in her different positions, as she "rounds the curve" of the "great hill of water" over which she is supposed to be sailing: for, it must be remembered that this supposed "proof" depends upon lines and angles of vision which, if enlarged, would still retain their characteristics. Now, since ships are not built 500 miles long, with masts in proportion, and men are not quite 200 miles high, it is not what it is said to be - a proof of rotundity - but, either an ignorant farce or a cruel piece of deception. In short, it is a proof that the Earth is not a globe.
97) NASA and modern astronomy say the Earth is a giant ball tilted back, wobbling and spinning 1,000 mph around its central axis, traveling 67,000 mph circles around the Sun, spiraling 500,000 mph around the Milky Way, while the entire galaxy rockets a ridiculous 670,000,000 mph through the Universe, with all of these motions originating from an alleged “Big Bang” cosmogenic explosion 14 billion years ago. That’s a grand total of 670,568,000 mph in several different directions we’re all supposedly speeding along at simultaneously, yet no one has ever seen, felt, heard, measured or proven a single one of these motions to exist whatsoever.
50.) We read in the inspired book, or collection of books, called THE BIBLE, nothing at all about the Earth being a globe or a planet, from beginning to end, but hundreds of allusions there are in its pages which could not be made if the Earth were a globe, and which are, therefore, said by the astronomer to be absurd and contrary to what he knows to be true! This is the groundwork of modern infidelity. But, since every one of many, many allusions to the Earth and the heavenly bodies in the Scriptures can be demonstrated to be absolutely true to nature, and we read of the Earth being "stretched out" "above the waters," as "standing in the water and out of the water," of its being "established that it cannot be moved," we have a store from which to take all the proofs we need, but we will just put down one proof – the Scriptural proof – that Earth is not a globe.
34.) If the Earth were a globe, there certainly would be – if we could imagine the thing to be peopled all round – "antipodes:" "people who," says the dictionary, "living exactly on the opposite side of the globe to ourselves, have their feet opposite to ours: – people who are hanging heads downwards whilst we are standing heads up! But, since the theory allows us to travel to those parts of the Earth where the people are said to be heads downwards, and still to fancy ourselves to be heads upwards and our friends whom we have left behind – us to be heads downwards, it follows that the whole thing is a myth – a dream – a delusion – and a snare; and, instead of there being any evidence at all in this direction to substantiate the popular theory, it is a plain proof that the Earth is not a globe.
61.) It is plain that a theory of measurements without a measuring-rod is like a ship without a rudder; that a measure that is not fixed, not likely to be fixed, and never has been fixed, forms no measuring-rod at all; and that as modern theoretical astronomy depends upon the Sun's distance from the Earth as its measuring-rod, and the distance is not known, it is a system of measurements without a measuring-rod – a ship without a rudder. Now, since it is not difficult to foresee the dashing of this thing upon the rock on which Zetetic astronomy is founded, i
In 1610, Galileo Galilei observed the moons of Jupiter rotating around it. He described them as small planets orbiting a larger planet—a description (and observation) that was very difficult for the church to accept, as it challenged a geocentric model where everything was supposed to revolve around the Earth. This observation also showed that the planets (Jupiter, Neptune, and later Venus was observed too) are all spherical, and all orbit the sun.
If the Earth were a globe, there would, very likely, be (for nobody knows) six months day and six months night at the arctic and antarctic regions, as astronomers dare to assert there is: - for their theory demands it! But, as this fact - the six months day and six months night - is; nowhere found but in the arctic regions, it agrees perfectly with everything else that we know about the Earth as a plane, and, whilst it overthrows the "accepted theory," it furnishes a striking proof that Earth is not a globe.
15.) The idea that, instead of sailing horizontally round the Earth, ships are taken down one side of a globe, then underneath, and are brought up on the other side to get home again, is, except as a mere dream, impossible and absurd! And, since there are neither impossibilities nor absurdities in the simple matter of circumnavigation, it stands without argument, a proof that the Earth is not a globe.
52.) It is a well-known and indisputable fact that there is a far greater accumulation of ice south of the equator than is to be found at an equal latitude north: and it is said that at Kerguelen, 50 degrees south, 18 kinds of plants exist, whilst, in Iceland, 15 degrees nearer the northern centre, there are 870 species; and, indeed, all the facts in the case show that the Sun's power is less intense at places in the southern region than it is in corresponding latitudes north. Now, on the Newtonian hypothesis, all this is inexplicable, whilst it is strictly in accordance with the facts brought to light by the carrying out of the principles involved in the Zetetic Philosophy of "Parallax." This is a proof that the Earth is not a globe.
https://www.youtube.com/watch?v=SlPRmsd76IM its a youtube video with a ton of scientific eveidence that dinosaurs lived with man and that they could still be alive. All the stories of dragons , if you tie them in with dinosaurs, that dragons and dinosaurs are one in the same.. Sounds crazy, but so does a flat earth. Please please check it out. I would love to see your insight on this topic, there is a lot of christian mumbo jumbo at the end you can easily skip but the information here is marvelous. if that video does not work for any reason just go on youtube and search dinosaurs vs dragons, you should find it with that :)
The Newtonian hypothesis involves the necessity of. the Sun, in the case of a lunar eclipse, being on the opposite side of a globular earth, to cast its shadow on the Moon: but, since eclipses of the Moon have taken place with both the Sun and the Moon above the horizon, it follows that it cannot be the shadow of the Earth that eclipses the Moon; that the theory is a blunder; and that it is nothing less than a proof that the Earth is not a globe.
51.) A "Standing Order" exists in the English Houses of Parliament that in the cutting of canals, &c., the datum line employed shall be a "horizontal line, which shall be the same throughout the whole length of the work." Now if the Earth were a globe, this "Order" could not be carried out: but, it is carried out: therefore it is a proof that the Earth is not a globe.
Planets (from Ancient Greek ἀστὴρ πλανήτης [astēr planētēs, "wandering star"], or just πλανήτης [planḗtēs, "wanderer"]) are orbiting astronomical objects. The Earth is not a planet by definition, as it sits at the center of our solar system above which the planets and the Sun revolve. The earths uniqueness, fundamental differences and centrality makes any comparison to other nearby celestial bodies insufficient - Like comparing basketballs to the court on which they bounce.
##### 94) From the highland near Portsmouth Harbor in Hampshire, England looking across Spithead to the Isle of Wight, the entire base of the island, where water and land come together composes a perfectly straight line 22 statute miles long. According to the ball-Earth theory, the Isle of Wight should decline 80 feet from the center on each side to account for the necessary curvature. The cross-hairs of a good theodolite directed there, however, have repeatedly shown the land and water line to be perfectly level.
103) There are several constellations which can be seen from far greater distances over the face of the Earth than should be possible if the world were a rotating, revolving, wobbling ball. For instance, Ursa Major, very close to Polaris, can be seen from 90 degrees North latitude (the North Pole) all the way down to 30 degrees South latitude. For this to be possible on a ball-Earth the Southern observers would have to be seeing through hundreds or thousands of miles of bulging Earth to the Northern sky.
The group had relocated to the United States by 1977, and it was there that they recorded their best-selling album, Breakfast in America. With more hit singles than their first five albums combined, it reached number three in the UK,[10] and top of the charts in America. The album is reckoned to have sold over 20 million copies since its release on 29 March 1979.[12]
A "Standing Order" exists in the English Houses of Parliament that in the cutting of canals, &c., the datum line employed shall be a "horizontal line, which shall be the same throughout the whole length of the work." Now if the Earth were a globe, this "Order" could not be carried out: but, it is carried out: therefore it is a proof that the Earth is not a globe.
66.) It is often said that the predictions of eclipses prove astronomers to be right in their theories. But it is not seen that this proves too much. It is well known that Ptolemy predicted eclipses for six-hundred years, on the basis of a plane Earth, with as much accuracy as they are predicted by modern observers. If, then, the predictions prove the truth of the particular theories current at the time, they just as well prove one side of the question as the other, and enable us to lay claim to a proof that the Earth is not a globe.
How is it possible to see day and night at the same time on a globe? Wouldn't the curve prevent you from seeing the moon since it would be in the other side of the globe? And how can you see the moon over the FLAT ocean while also seeing the sunrise? In the following video you will see day and night clearly divided on a flat plane! You will see an airplane flying level on a flat earth with day and night divided. Watch as it becomes night and the day again as the airplane catches up with the sun. The airplane however never dips is nose down to compensate for the curve. The sun is also shown to be illuminating locally and the light follows is as it moves away on the flat earth, creating day and night.
While it's true that unipolar magnets can't exist, this isn't a problem for the Flat Earth. This is because ring magnets, which are shaped like (you guessed it!) a flat disk, are capable of having radial magnetization. In a radial magnet, one magnetic pole is at the center and other is at all points on the edge of the magnet. A magnet like this can be found in loudspeakers, and perfectly replicates what is found on the Earth.
```53) At places of comparable latitude North and South, the Sun behaves very differently than it would on a spinning ball Earth but precisely how it should on a flat Earth. For example, the longest summer days North of the equator are much longer than those South of the equator, and the shortest winter days North of the equator are much shorter than the shortest South of the equator. This is inexplicable on a uniformly spinning, wobbling ball Earth but fits exactly on the flat model with the Sun traveling circles over and around the Earth from Tropic to Tropic.
```
152) In 2003, three University Geography professors collaborated in an experiment to prove that the state of Kansas is indeed actually flatter than a pancake! Using topigraphical geodetic surveys covering over 80,000 square miles it was determined that Kansas has a flatness ratio of 0.9997 over the entire state while the average pancake, precisely measured using a confocal laser microscope comes in at 0.957, making Kansas thereby literally flatter than a pancake.
4) If Earth were a ball 25,000 miles in circumference as NASA and modern astronomy claim, spherical trigonometry dictates the surface of all standing water must curve downward an easily measurable 8 inches per mile multiplied by the square of the distance. This means along a 6 mile channel of standing water, the Earth would dip 6 feet on either end from the central peak. Every time such experiments have been conducted, however, standing water has proven to be perfectly level.
100.) The Sun, as he travels round over the surface of the Earth, brings "noon" to all places on the successive meridians which he crosses: his journey being made in a westerly direction, places east of the Sun's position have had their noon, whilst places to the west of the Sun's position have still to get it. Therefore, if we travel easterly, we arrive at those parts of the Earth where "time" is more advanced, the watch in our pocket has to be "put on"or we may be said to "gain time." If, on the other hand, we travel westerly, we arrive at places where it is still "morning," the watch has to be "put back," and it may be said that we "lose time." But, if we travel easterly so as to cross the 180th meridian, there is a loss, there, of a day, which will neutralize the gain of a whole circumnavigation; and, if we travel westerly, and cross the same meridian, we experience the gain of a day, which will compensate for the loss during a complete circumnavigation in that direction. The fact of losing or gaining time in sailing round the world, then, instead of being evidence of the Earth's "rotundity," as it is imagined to be, is, in its practical exemplification, an everlasting proof that the Earth is not a globe.
A "Standing Order" exists in the English Houses of Parliament that in the cutting of canals, &c., the datum line employed shall be a "horizontal line, which shall be the same throughout the whole length of the work." Now if the Earth were a globe, this "Order" could not be carried out: but, it is carried out: therefore it is a proof that the Earth is not a globe.
```In " Cornell's Geography" there is an "Illustrated proof of the Form of the Earth," A curved line on which is represented a ship in four positions, as she sails away from an observer, is an arc of 72 degrees, or one-fifth of the supposed circumference of the "globe" - about 5,000 miles. Ten, such ships as those which are given in the picture would reach the full length of the "arc," making 500 miles as the length of the ship, The man in the picture, who is watching the ship as she sails away, is about 200 miles high; and the tower, from which he takes an elevated view, at least 600 miles high. These are the proportions, then, of men, towers, arid ships which are necessary in order to see a ship, in her different positions, as she "rounds the curve" of the "great hill of water" over which she is supposed to be sailing: for, it must be remembered that this supposed "proof" depends upon lines and angles of vision which, if enlarged, would still retain their characteristics. Now, since ships are not built 500 miles long, with masts in proportion, and men are not quite 200 miles high, it is not what it is said to be - a proof of rotundity - but, either an ignorant farce or a cruel piece of deception. In short, it is a proof that the Earth is not a globe.
```
43) If Earth was a ball there are several flights in the Southern hemisphere which would have their quickest, straightest path over the Antarctic continent such as Santiago, Chile to Sydney, Australia. Instead of taking the shortest, quickest route in a straight line over Antarctica, all such flights detour all manner of directions away from Antarctica instead claiming the temperatures too cold for airplane travel! Considering the fact that there are plenty of flights to/from/over Antarctica, and NASA claims to have technology keeping them in conditions far colder (and far hotter) than any experienced on Earth, such an excuse is clearly just an excuse, and these flights aren’t made because they are impossible.
Mr. Hind, the English astronomer, says - "The simplicity, with which the seasons are explained by the revolution of the Earth in her orbit and the obliquity of the ecliptic, may certainly be adduced as a strong presumptive proof of the correctness" - of the Newtonian theory; "for on no other rational suppositions with respect to the relations of the Earth and Sun, can these and other as well-known phenomena, be accounted for." But, as true philosophy has no "suppositions" at all - and has nothing to do with, "suppositions" - and the phenomena spoken of are thoroughly explained by facts, the "presumptive proof" falls to the ground, covered with the ridicule it the dust of Mr. Hind's "rational suppositions" we are standing before us a proof that Earth is not a globe.
```25.) The surveyor's plans in relation to the laying of the first Atlantic Telegraph cable, show that in 1665 miles – from Valentia, Ireland, to St . John's, Newfoundland – the surface of the Atlantic Ocean is a LEVEL surface – not the astronomers' "level," either! The authoritative drawings, published at the time, are a standing evidence of the fact, and form a practical proof that Earth is not a globe.
```
Astronomers, in their consideration of the supposed "curvature" of the Earth, have carefully avoided the taking of that view of the question which - if anything were needed to do so -would show its utter absurdity. It is this: - if, instead of taking our ideal point of departure to be at Valentia, we consider ourselves at St. John's, the 1665 miles of water between us and Valentia would just as well "curvate" downwards as it did in the other case! Now, since the direction in which the Earth is said to "curvate" is interchangeable - depending, indeed, upon the position occupied by a man upon its surface - the thing is utterly absurd; and it follows that the theory is an outrage , and that the Earth does not "curvate" at all: - an evident proof that the Earth is not a globe.
75.) Considerably more than a million Earths would be required to make up a body like the Sun -the astronomers tell us: and more than 53,000 suns would be wanted to equal the cubic contents of the star Vega. And Vega is a "small star!" And there are countless millions of these stars! And it takes 30,000,000 years for the light of some of those stars to reach us at 12,000,000 miles in a minute! And, says Mr. Proctor, "I think a moderate estimate of the age of the Earth would be 500,000,000 years! "Its weight," says the same individual, "is 6,000,000,000,000,000,000,060 tons!" Now, since no human being is able to comprehend these things, the giving of them to the world is an insult – an outrage. And though they have all risen from the one assumption that Earth is a planet, instead of upholding the assumption, they drag it down by the weight of their own absurdity, and leave it lying in the dust – a proof that Earth is not a globe.
Mr. Lockyer says: "The appearances connected with the rising and setting of the Sun and stars may be due either to our earth being at rest and the Sun and stars traveling round it, or the earth itself turning round, while the Sun and stars are at rest." Now, since true science does not allow of any such beggarly alternatives as these, it is plain that modern theoretical astronomy is not true science, and that its leading dogma is a fallacy. We have, then, a plain proof that the Earth is not a globe.
That the mariners' compass points north and south at the same time is a fact as indisputable as that two and two makes four; but that this would be impossible if the thing, were placed on a globe with "north" and "south' at the centre of opposite hemispheres is a fact that does not figure in the school-books, though very easily seen: and it requires no lengthy train of reasoning to bring out of it a pointed proof that the Earth is not a globe.
### When a man speaks of a "most complete" thing amongst several other things which claim to be what that thing is, it is evident that they must fall short of something which the "most complete" thing possesses. And when it is known that the "most complete" thing is an entire failure, it is plain that the others, all and sundry, are worthless. Proctor's "most complete proof that the Earth is a globe" lies in what he calls "the fact" that distances from place to place agree with calculation. But, since the distance round the Earth at 45 " degrees" south of the equator is twice the distance it would be on a globe, it follows that what the greatest astronomer of the age calls "a fact" is NOT a fact; that his "most complete proof' is a most complete failure; and that be might as well have told us, at once, that he has NO PROOF to give us at all. Now, since, if the Earth be a globe, there would, necessarily, be piles of proofs of it all round us, it follows that when astronomers, with all their ingenuity, are utterly unable to point one out - to say nothing about picking one up - that they give us a proof that Earth is not a globe.
78) From Anchorage, Alaska at an elevation of 102 feet, on clear days Mount Foraker can be seen with the naked eye 120 miles away. If Earth were a ball 25,000 miles in circumference, Mount Foraker’s 17,400 summit should be leaning back away from the observer covered by 7,719 feet of curved Earth. In reality, however, the entire mountain can be quite easily seen standing straight from base to summit. | 5,642 | 24,892 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-31 | latest | en | 0.972662 |
https://www.got-it.ai/solutions/excel-chat/excel-problem/22690-if-function | 1,601,170,509,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400249545.55/warc/CC-MAIN-20200926231818-20200927021818-00578.warc.gz | 864,152,742 | 15,086 | # Excel - IF Function Problem - Expert Solution
Question description:
This user has given permission to use the problem statement for this blog.
I want to use the IF function to find out the Vintage of an employee from Date of joining
Solved by V. D. in 40 mins
This is the chat thread from the real Excelchat help session. It contains no private user information.
Excelchat Expert 05/02/2018 - 02:01
Hello! Welcome to Excel Chat.
Excelchat Expert 05/02/2018 - 02:02
User 05/02/2018 - 02:02
Hi
User 05/02/2018 - 02:02
I want to know if the date of joining of an employee is 12th Jan 2016 then as on date what will be his total vintage in the organisation?
User 05/02/2018 - 02:03
what formula should I type ..
Excelchat Expert 05/02/2018 - 02:03
Vintage is equal to the length of the service till date? Let me know if my understanding is corrrect.
User 05/02/2018 - 02:03
yes
Excelchat Expert 05/02/2018 - 02:04
Ok. Let me work out a formula for you then.
User 05/02/2018 - 02:04
Excelchat Expert 05/02/2018 - 02:05
Sure. In what units, you would like to get the result?
User 05/02/2018 - 02:06
In number of days
Excelchat Expert 05/02/2018 - 02:06
Ok.
User 05/02/2018 - 02:06
I also need to know how do i spolit that in 0-6 months : 6-12 months etc
Excelchat Expert 05/02/2018 - 02:07
Ok. Just give me a moment to work on your problem.
Excelchat Expert 05/02/2018 - 02:08
Excelchat Expert 05/02/2018 - 02:09
In cell A1, there is the first date, i.e. date of joining.
Excelchat Expert 05/02/2018 - 02:09
In cell B2, is today's date.
User 05/02/2018 - 02:09
how do i split it into 0-6 mths, 6-9 months and 9-12 months
User 05/02/2018 - 02:09
need to know the IF and ISERROR functions
Excelchat Expert 05/02/2018 - 02:09
Ok, now you want to categorize in different categories?
User 05/02/2018 - 02:10
yes
Excelchat Expert 05/02/2018 - 02:10
Then let's use the month diff, then
User 05/02/2018 - 02:10
ok
Excelchat Expert 05/02/2018 - 02:10
In cell C2, you can see 24 now.
Excelchat Expert 05/02/2018 - 02:10
This is 24 month that is the difference between two dates.
Excelchat Expert 05/02/2018 - 02:11
Now we will use this further.
Excelchat Expert 05/02/2018 - 02:13
Give me a moment to improve it further.
Excelchat Expert 05/02/2018 - 02:16
Sorry to keep you waiting.
Excelchat Expert 05/02/2018 - 02:16
I have updated the formula in D1.
Excelchat Expert 05/02/2018 - 02:16
User 05/02/2018 - 02:18
Give me few mins to verify ..
Excelchat Expert 05/02/2018 - 02:18
Sure.
Excelchat Expert 05/02/2018 - 02:18
After that, I will explain the solution.
User 05/02/2018 - 02:20
Can you explain me Cl D formula please
Excelchat Expert 05/02/2018 - 02:20
Ok.
Excelchat Expert 05/02/2018 - 02:21
In cell C1, it's a simple Date difference formula, where 3 inputs are required, first date, second date, and the unit in which we want to see the results.
Excelchat Expert 05/02/2018 - 02:21
=DateDif(A1, B1, "M")
User 05/02/2018 - 02:22
ok understood the Cl C
Excelchat Expert 05/02/2018 - 02:22
here M is for month. you can use "Y" for year, "D" for days. You can also use combination of them like "YMD" or "YM" or "MD" etc.
Excelchat Expert 05/02/2018 - 02:22
Great!
Excelchat Expert 05/02/2018 - 02:23
In D1, it's a nested IF formula.
Excelchat Expert 05/02/2018 - 02:24
IF also need 3 inputs, 1 - condition that is to be checked, 2 - Result if given condition is true, 3 - Result if given condition is false (optional).
Excelchat Expert 05/02/2018 - 02:25
So I am checking C1 for a value, and if that condition is true, give me the first message, if that condition is false, check again C1 for a new value. And so on.
Excelchat Expert 05/02/2018 - 02:25
Here I used, IF in another IF 2 times more.
Excelchat Expert 05/02/2018 - 02:26
Excel has a limit that you can put upto 8 IF inside an IF.
Excelchat Expert 05/02/2018 - 02:26
I hope I was able to explain the solution, please let me know if you have any question.
User 05/02/2018 - 02:28
thx
User 05/02/2018 - 02:29
where can i find teh list of formulas if i dont know
User 05/02/2018 - 02:29
which formula to apply
Excelchat Expert 05/02/2018 - 02:29
You can Google for Excel formulas and Excel help.
User 05/02/2018 - 02:30
can u suggest a link .. also do you guys provide class room trainings ?
Excelchat Expert 05/02/2018 - 02:30
Which formula to apply is dependent on question.
Excelchat Expert 05/02/2018 - 02:30
I am sorry, there is no class room training for now.
Excelchat Expert 05/02/2018 - 02:31
User 05/02/2018 - 02:31
so if I enroll with Got IT Pro ..can I get assistance for the formulas I need
Excelchat Expert 05/02/2018 - 02:31
Yes. For each question, you can start a new session and an expert will be there for help.
Excelchat Expert 05/02/2018 - 02:35
Have a nice day ahead. Bye!
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Your privacy is guaranteed. Your session will not be used for blog unless you give us persmission. | 1,717 | 5,162 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2020-40 | longest | en | 0.844239 |
https://converterin.com/cooking/hectoliter-hl-to-cup-metric-c.html | 1,660,287,746,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571584.72/warc/CC-MAIN-20220812045352-20220812075352-00045.warc.gz | 200,074,336 | 8,178 | # HECTOLITER TO CUP [METRIC] CONVERTER
FROM
TO
The result of your conversion between hectoliter and cup [metric] appears here
## HECTOLITER TO CUP [METRIC] (hl TO c) FORMULA
To convert between Hectoliter and Cup [metric] you have to do the following:
First divide 0.1 / 0.00025 = 400.
Then multiply the amount of Hectoliter you want to convert to Cup [metric], use the chart below to guide you.
## HECTOLITER TO CUP [METRIC] (hl TO c) CHART
• 1 hectoliter in cup [metric] = 400. hl
• 10 hectoliter in cup [metric] = 4000. hl
• 50 hectoliter in cup [metric] = 20000. hl
• 100 hectoliter in cup [metric] = 40000. hl
• 250 hectoliter in cup [metric] = 100000. hl
• 500 hectoliter in cup [metric] = 200000. hl
• 1,000 hectoliter in cup [metric] = 400000. hl
• 10,000 hectoliter in cup [metric] = 4000000. hl
Symbol: hl
No description
Symbol: c
No description | 299 | 866 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2022-33 | latest | en | 0.710128 |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=16&t=22190 | 1,606,265,026,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141177607.13/warc/CC-MAIN-20201124224124-20201125014124-00254.warc.gz | 377,966,242 | 11,499 | ## Post Assessment #28 [ENDORSED]
HanSitoy
Posts: 18
Joined: Fri Sep 29, 2017 7:06 am
### Post Assessment #28
Light hits a sodium metal surface and the velocity of the ejected electron is 6.61 x 10^5 m.s^-1. The work function for sodium is 150.6 kJ.mol^-1.
How much energy is required to remove an electron from one sodium atom?
Isn't the energy required to remove the electron equivalent to the work function/threshold energy?
vicenteruelos3
Posts: 42
Joined: Fri Sep 29, 2017 7:07 am
### Re: Post Assessment #28
yes, the energy required to remove an electron is equal to the work function or threshold energy
Christine Wastila 1H
Posts: 61
Joined: Fri Sep 29, 2017 7:04 am
### Re: Post Assessment #28 [ENDORSED]
You need to divide the work function by Avogadro's number (6.022x10^23) because the it was given in kJ/mol and the question asks about the energy needed to remove one electron from ONE sodium atom, not a mole of them. I got confused on that, too! Hope this helps.
Sabrina Dunbar 1I
Posts: 57
Joined: Fri Sep 29, 2017 7:07 am
Been upvoted: 1 time
### Re: Post Assessment #28
Also be aware that the value in in kJ not just J, so you need to also do that conversion
Mishta Stanislaus 1H
Posts: 45
Joined: Fri Sep 29, 2017 7:04 am
### Re: Post Assessment #28
To answer this question you must convert the given work function from kJ to J. Then you must divide by avagadro's number to find the energy required per atom as opposed to per mole.
### Who is online
Users browsing this forum: No registered users and 1 guest | 446 | 1,546 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2020-50 | latest | en | 0.849721 |
https://ncertmcq.com/rd-sharma-class-10-solutions-chapter-3-pair-of-linear-equations-in-two-variables-ex-3-4/ | 1,718,520,528,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861643.92/warc/CC-MAIN-20240616043719-20240616073719-00472.warc.gz | 377,550,517 | 13,860 | ## RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4
These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4
Other Exercises
Solve each of the following systems of equations by the method of cross multiplication.
Question 1.
x + 2y + 1 = 0
2x – 3y – 12 = 0 (C.B.S.E. 1992)
Solution:
Question 2.
3x + 2y + 25 = 0
2x + y + 10 = 0 (C.B.S.E. 1992)
Solution:
Question 3.
2x + y = 35
3x + 4y = 65 (C.B.S.E. 1993)
Solution:
Question 4.
2x – y = 6
x – y = 2 (C.B.S.E. 1994)
Solution:
Question 5.
$$\frac { x+y }{ xy } =2,\quad \frac { x-y }{ xy } =6$$
Solution:
Question 6.
ax + by = a – b
bx – ay = a + b
Solution:
ax + by = a – b
bx – ay = a + b
Question 7.
x + ay = b
ax – by = c
Solution:
x + ay = b
Question 8.
ax + by = a2
bx + ay = b2
Solution:
Question 9.
Solution:
Question 10.
Solution:
Question 11.
Solution:
Question 12.
Solution:
Question 13.
$$\frac { x }{ { a } } +\frac { y }{ b } =a+b$$
$$\frac { x }{ { { a }^{ 2 } } } +\frac { y }{ { b }^{ 2 } } =2$$
Solution:
Question 14.
$$\frac { x }{ { { a } } } =\frac { y }{ { b } }$$
ax + by = a2 + b2
Solution:
Question 15.
2ax + 3by = a + 2b
3ax + 2by = 2a + b
Solution:
Question 16.
5ax + 6by = 28
3ax + 4by = 18
Solution:
Question 17.
(a + 2b) x + (2a – b) y = 2
(a – 2b) x + (2a + b) y = 3
Solution:
Question 18.
Solution:
Question 19.
Solution:
Question 20.
(a – b) x + (a + b) y = 2a2 – 2b2
(a + b) (x + y) = 4ab
Solution:
Question 21.
a2x + b2y = c2
b2x + a2y = d2
Solution:
Question 22.
ax + by = $$\frac { a + b }{ 2 }$$
3x + 5y = 4
Solution:
Hence x = $$\frac { 1 }{ 2 }$$ , y = $$\frac { 1 }{ 2 }$$
Question 23.
2 (ax – by) + a + 4b = 0
2 (bx + ay) + b – 4a = 0 (C.B.S.E. 2004)
Solution:
Question 24.
6 (ax + by) = 3a + 2b
6 (bx – ay) = 3b – 2a (C.B.S.E. 2004)
Solution:
Question 25.
$$\frac { { a }^{ 2 } }{ x } -\frac { { b }^{ 2 } }{ y } =0$$
$$\frac { { a }^{ 2 }b }{ x } -\frac { { b }^{ 2 }a }{ y } =a+b\quad ,\quad x,y\quad \neq 0$$
Solution:
Question 26.
mx – ny = m2 + n2
x + y = 2m (C.B.S.E. 2006C)
Solution:
Question 27.
Solution:
Question 28.
Solution:
Hope given RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you. | 1,008 | 2,462 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2024-26 | latest | en | 0.641874 |
https://www.jplcomputer.co.uk/2018/11/22/excel-symbol-meanings/ | 1,721,864,086,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518532.61/warc/CC-MAIN-20240724232540-20240725022540-00049.warc.gz | 707,063,385 | 12,194 | # Excel Symbol Meanings…what do they mean? Find out here!
## Excel Symbol Meanings – have you ever wondered what they do? What they mean?
Further to a blog we did in May 2017 on Excel Symbols we thought we would give you a further insight into 5 more Excel symbol meanings.
There are a number of different symbols used within Excel all with different meanings and uses.
It can be confusing trying to work out what they all mean and how you can use them, so we thought we would break down some of them and their meanings to give you a bit more insight.
See below for the Excel symbol and information on what its use is in Excel!
#### 5 Excel Symbols
() – Brackets are an essential symbol in Excel bring used for 2 prime purposes:
1. To follow a function and contain the arguments of that function e.g. =SUM(b2:b5) or =VLOOKUP(b2,\$d\$5:\$g\$12,2,False)
2. To obey the BODMAS premise where brackets override multiply and divide symbols which themselves override addition and subtraction
– An apostrophe before a cell value forces Excel to interpret the value as text. Here we have 2 examples:
1. If you type in a phone number say in a cell as 1 long number Excel drops the first 0 so by inserting an apostrophe firstly means it displays the whole number including the 0
2. If you refer to a worksheet in a formula which has spaces, then the ‘ treats the sheet name as text and you wont get an error e.g. =sum(‘Q 1:Q 4’!b2)
& – See this example in the image of cell C1 to the right
– To the power of e.g. =2^3 would give us the answer 8 if you typed it in to Excel
{} – Used when using an Array type of formula in Excel
We cover the use of Excel symbols on our Basic Excel training course and now there are 5 to choose from.
See our website here for more details or contact us on 07903 840105 or johnlegge@jplcomputer.co.uk
You might also like to read our previous post on some lesser known features in Word. | 471 | 1,921 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-30 | latest | en | 0.915053 |
http://mathoverflow.net/questions/16392/how-to-interpret-the-sugawara-construction-from-a-physical-or-mathematical-view/16406 | 1,469,357,700,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823996.40/warc/CC-MAIN-20160723071023-00183-ip-10-185-27-174.ec2.internal.warc.gz | 161,512,074 | 17,238 | How to interpret the Sugawara construction from a physical or mathematical viewpoint?
In theoretical physics, the Sugawara theory is a set of formulae and theorems that allow one to construct a stress-energy tensor of a specific type of conformal field theory from a bilinear expression involving currents.
How to interpret the Sugawara construction from a physical or mathematical viewpoint?
Sugawara construction is a kind of mathod to embed varasoro algebra into completions of universal enveloping algebras of affine algras ? what special properties do this kind of embeding have? In soliton theorey, I know there is a boson-fermion correspondence which realize free boson algebra in the completion of free fermion algebra.
I wonder if there are some common principles under them?
-
The origin of the Sugawara construction in Physics is, not surprisingly, the 1968 paper A field theory of currents by Hirotaka Sugawara. (There was also work of Sommerfield at about the same time.) The context of that work was to find a theory of strong interactions. (Recall that the currently favoured theory of strong interactions (QCD) would not be discovered until the mid 1970s.) There was a feeling at the time that there might not be a field theory describing the strong interactions, whose dynamical fields are associated to the particles themselves. (QCD is such a theory, of course.) Hence Sugawara's idea was to quantise the theory using "currents" (field bilinears) as the elementary dynamical variables. His theory is a four-dimensional theory and I think it is fair to say that it is now just of historical interest.
What most people call the Sugawara construction is what Marty Halpern insisted in calling the affine Sugawara construction. This is a construction in two-dimensional conformal field theory by which a Virasoro element is constructed in the conformal field theory associated to an affine Kac-Moody algebra. The construction has a long history of papers converging to the correct formula. I forgot the actual sequence of papers, but it probably starts with a 1971 paper of Bardakçi and Halpern, who introduced the construction, and ends with Knizhnik's and Zamolodchikov's 1984 paper. (But I stand to be corrected on this.)
A possibly physical interpretation, in the context of string theory, is that this is a quantisation of string propagation on a Lie group (with a bi-invariant metric), whose classical action is given by the Wess-Zumino-Witten model. In that model the dynamical fields are maps $g:\Sigma \to G$, where $\Sigma$ is a Riemann surface and $G$ is a Lie group with a bi-invariant metric. Just as in the original four-dimensional Sugawara construction, it is simpler to quantise the currents $g^{-1}\partial g$ and $\bar\partial g g^{-1}$ than the actual fields $g$. This was done by Edward Witten in his celebrated paper Non-abelian bosonization in two dimensions. The Sugawara construction serves to prove the exact quantum conformal invariance of the Wess-Zumino-Witten model to all orders in perturbation theory, something which is not possible to do by quantising the original fields $g$.
Mathematically, it embeds the Virasoro algebra in the vertex algebra of an affine Kac-Moody algebra (actually this fails at the so-called critical level) in such a way that any module of the Kac-Moody algebra (of noncritical level) is also a Virasoro module. If $\mathfrak{g}$ is a simple Lie algebra and $\widehat{\mathfrak{g}}_\ell$ the corresponding (untwisted) affine Kac-Moody algebra at level $\ell$, then the central charge of the Sugawara Virasoro element is given by $$c = \frac{\ell \dim\mathfrak{g}}{\ell + h^\vee}$$ where $h^\vee$ the dual Coxeter number. (So $\ell = - h^\vee$ is the critical level.)
The defining property of the Sugawara construction is that the currents are primary fields (of weight 1) of the Virasoro element.
This same construction extends to the affinisation of any metric Lie algebra; that is, admitting an ad-invariant metric.
-
Hi José, how exactly does the Sugawara construction prove conformal invariance to all orders? In Witten's paper I can only see a calculation in one-loop-order. – Konrad Waldorf Feb 25 '10 at 20:15
If you define the quantisation of the WZW model in terms of the currents, so that the Hilbert space of the theory consists of all the integrable highest weight representations (up the relevant level) of the corresponding affine Kac-Moody algebra (I'm talking about the case of $\mathfrak{g}$ simple), then the Sugawara construction defines on those representations the structure of a Virasoro module. This, by definition, is an exact quantum conformal field theory. – José Figueroa-O'Farrill Feb 25 '10 at 22:19
Just a historical comment (from my memory; I am at the moment rather far from the subject)
Ivan Todorov gave and wrote few times lectures about the history of Sugawara construction to which he himself also contributed. He even mentions some works in late 1940-s in field theory of photons among curioous early occurences and then the intricate history in 1960's and 1970's, involving Kac, Feĭgin-Fuks, Sugawara, Segal, Todorov etc. Sugawara is in any case not the first in that sequence and he had a wrong proportionality constant calculated, what is corrected in 1970s.
It is also curious that the (affine) Sugawara construction is not confined to working over complex numbers, there is a more general version overmore general fields explained by Faltings. I think I have seen it in his paper (I do not have it here so I can not check)
Gerd Faltings, A proof for the Verlinde formula. J. Algebraic Geom. 3 (1994), no. 2, 347--374. MR1257326 (95j:14013)
-
Sugawara's original theory does not calculate the constant in font of the bilinear expression. I think you are confusing this with the affine (two-dimensional) construction, to which Sugawara did not contribute. As in my answer, this probably started with Bardakçi-Halpern. There was some initial confusion between two notions of normal ordering which accounted for the wrong constant. I am not sure which was the first paper to get it right, though. – José Figueroa-O'Farrill Feb 25 '10 at 16:08 | 1,484 | 6,170 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2016-30 | latest | en | 0.928305 |
https://dsa.dichchankinh.com/sourcecode/c++/sudoku_solve.cpp | 1,669,997,029,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710909.66/warc/CC-MAIN-20221202150823-20221202180823-00371.warc.gz | 255,240,265 | 3,196 | ## C++, , sudoku_solve.cpp
``````/**
* @file
* @brief [Sudoku Solver](https://en.wikipedia.org/wiki/Sudoku) algorithm.
*
* @details
* Sudoku (数独, sūdoku, digit-single) (/suːˈdoʊkuː/, /-ˈdɒk-/, /sə-/,
* originally called Number Place) is a logic-based, combinatorial
* number-placement puzzle. In classic sudoku, the objective is to fill a 9×9
* grid with digits so that each column, each row, and each of the nine 3×3
* subgrids that compose the grid (also called "boxes", "blocks", or "regions")
* contain all of the digits from 1 to 9. The puzzle setter provides a
* partially completed grid, which for a well-posed puzzle has a single
* solution.
*
* @author [DarthCoder3200](https://github.com/DarthCoder3200)
* @author [David Leal](https://github.com/Panquesito7)
*/
#include <array> /// for assert
#include <iostream> /// for IO operations
/**
* @namespace backtracking
* @brief Backtracking algorithms
*/
namespace backtracking {
/**
* @namespace sudoku_solver
* @brief Functions for the [Sudoku
* Solver](https://en.wikipedia.org/wiki/Sudoku) implementation
*/
namespace sudoku_solver {
/**
* @brief Check if it's possible to place a number (`no` parameter)
* @tparam V number of vertices in the array
* @param mat matrix where numbers are saved
* @param i current index in rows
* @param j current index in columns
* @param no number to be added in matrix
* @param n number of times loop will run
* @returns `true` if 'mat' is different from 'no'
* @returns `false` if 'mat' equals to 'no'
*/
template <size_t V>
bool isPossible(const std::array<std::array<int, V>, V> &mat, int i, int j,
int no, int n) {
/// `no` shouldn't be present in either row i or column j
for (int x = 0; x < n; x++) {
if (mat[x][j] == no || mat[i][x] == no) {
return false;
}
}
/// `no` shouldn't be present in the 3*3 subgrid
int sx = (i / 3) * 3;
int sy = (j / 3) * 3;
for (int x = sx; x < sx + 3; x++) {
for (int y = sy; y < sy + 3; y++) {
if (mat[x][y] == no) {
return false;
}
}
}
return true;
}
/**
* @brief Utility function to print the matrix
* @tparam V number of vertices in array
* @param mat matrix where numbers are saved
* @param starting_mat copy of mat, required by printMat for highlighting the
* differences
* @param n number of times loop will run
* @return void
*/
template <size_t V>
void printMat(const std::array<std::array<int, V>, V> &mat,
const std::array<std::array<int, V>, V> &starting_mat, int n) {
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (starting_mat[i][j] != mat[i][j]) {
std::cout << "\033[93m" << mat[i][j] << "\033[0m"
<< " ";
} else {
std::cout << mat[i][j] << " ";
}
if ((j + 1) % 3 == 0) {
std::cout << '\t';
}
}
if ((i + 1) % 3 == 0) {
std::cout << std::endl;
}
std::cout << std::endl;
}
}
/**
* @brief Main function to implement the Sudoku algorithm
* @tparam V number of vertices in array
* @param mat matrix where numbers are saved
* @param starting_mat copy of mat, required by printMat for highlighting the
* differences
* @param i current index in rows
* @param j current index in columns
* @returns `true` if 'no' was placed
* @returns `false` if 'no' was not placed
*/
template <size_t V>
bool solveSudoku(std::array<std::array<int, V>, V> &mat,
const std::array<std::array<int, V>, V> &starting_mat, int i,
int j) {
/// Base Case
if (i == 9) {
/// Solved for 9 rows already
printMat<V>(mat, starting_mat, 9);
return true;
}
/// Crossed the last Cell in the row
if (j == 9) {
return solveSudoku<V>(mat, starting_mat, i + 1, 0);
}
/// Blue Cell - Skip
if (mat[i][j] != 0) {
return solveSudoku<V>(mat, starting_mat, i, j + 1);
}
/// White Cell
/// Try to place every possible no
for (int no = 1; no <= 9; no++) {
if (isPossible<V>(mat, i, j, no, 9)) {
/// Place the 'no' - assuming a solution will exist
mat[i][j] = no;
bool solution_found = solveSudoku<V>(mat, starting_mat, i, j + 1);
if (solution_found) {
return true;
}
/// Couldn't find a solution
/// loop will place the next `no`.
}
}
/// Solution couldn't be found for any of the numbers provided
mat[i][j] = 0;
return false;
}
} // namespace sudoku_solver
} // namespace backtracking
/**
* @brief Main function
* @returns 0 on exit
*/
int main() {
const int V = 9;
std::array<std::array<int, V>, V> mat = {
std::array<int, V>{5, 3, 0, 0, 7, 0, 0, 0, 0},
std::array<int, V>{6, 0, 0, 1, 9, 5, 0, 0, 0},
std::array<int, V>{0, 9, 8, 0, 0, 0, 0, 6, 0},
std::array<int, V>{8, 0, 0, 0, 6, 0, 0, 0, 3},
std::array<int, V>{4, 0, 0, 8, 0, 3, 0, 0, 1},
std::array<int, V>{7, 0, 0, 0, 2, 0, 0, 0, 6},
std::array<int, V>{0, 6, 0, 0, 0, 0, 2, 8, 0},
std::array<int, V>{0, 0, 0, 4, 1, 9, 0, 0, 5},
std::array<int, V>{0, 0, 0, 0, 8, 0, 0, 7, 9}};
backtracking::sudoku_solver::printMat<V>(mat, mat, 9);
std::cout << "Solution " << std::endl;
std::array<std::array<int, V>, V> starting_mat = mat;
backtracking::sudoku_solver::solveSudoku<V>(mat, starting_mat, 0, 0);
return 0;
}
`````` | 1,708 | 4,898 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2022-49 | latest | en | 0.566279 |
http://ch.mathworks.com/help/matlab/ref/ddensd.html?s_tid=gn_loc_drop&nocookie=true | 1,430,775,042,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1430455119811.95/warc/CC-MAIN-20150501043839-00030-ip-10-235-10-82.ec2.internal.warc.gz | 36,357,066 | 12,967 | # ddensd
Solve delay differential equations (DDEs) of neutral type
## Syntax
• `sol = ddensd(ddefun,dely,delyp,history,tspan)` example
• `sol = ddensd(ddefun,dely,delyp,history,tspan,options)`
## Description
example
````sol = ddensd(ddefun,dely,delyp,history,tspan)` integrates a system of delay differential equations of neutral type, that has the form y '(t) = f(t, y(t), y(dy1),..., y(dyp), y '(dyp1),..., y '(dypq))(1-1)wheret is the independent variable representing time.dyi is any of p solution delays.dypj is any of q derivative delays.```
````sol = ddensd(ddefun,dely,delyp,history,tspan,options)` replaces default integration parameters with those specified in `options`, a structure created with the `ddeset` function.```
## Examples
collapse all
### Neutral DDE with Two Delays
Solve the following neutral DDE, presented by Paul, for 0 ≤ t ≤ π:
y '(t) = 1 + y(t) – 2y(t/2)2y '(tπ)
with history: y(t) = cos (t) for t ≤ 0.
Create a new program file in the editor. This file will contain a main function and four local functions.
Define the first-order DDE as a local function.
```function yp = ddefun(t,y,ydel,ypdel) yp = 1 + y - 2*ydel^2 - ypdel; end```
Define the solution delay as a local function.
```function dy = dely(t,y) dy = t/2; end```
Define the derivative delay as a local function.
```function dyp = delyp(t,y) dyp = t-pi; end```
Define the solution history as a local function.
```function y = history(t) y = cos(t); end```
Define the interval of integration and solve the DDE using the `ddensd` function. Add this code to the main function.
```tspan = [0 pi]; sol = ddensd(@ddefun,@dely,@delyp,@history,tspan);```
Evaluate the solution at 100 equally spaced points between 0 and π. Add this code to the main function.
```tn = linspace(0,pi); yn = deval(sol,tn);```
Plot the results. Add this code to the main function.
```plot(tn,yn); xlim([0 pi]); ylim([-1.2 1.2]); xlabel('time t'); ylabel('solution y');```
Run your program to calculate the solution and display the plot.
The file, `ddex4.m`, contains the complete code for this example. To see the code in an editor, type ```edit ddex4``` at the command line.
## Input Arguments
collapse all
### `ddefun` — Derivative functionfunction handle
Derivative function, specified as a function handle whose syntax is `yp = ddefun(t,y,ydel,ypdel)`. The arguments for `ddefun` are described in the table below.
ddefun ArgumentDescription
`t`A scalar value representing the current value of time, t.
`y` A vector that represents y(t) in Equation 1-1. The size of this vector is `n`-by-`1`, where `n` is the number of equations in the system you want to solve.
`ydel`A matrix whose columns, `ydel(:,i)`, represent y(dyi). The size of this matrix is `n`-by-`p`, where `n` is the number of equations in the system you want to solve, and `p` is the number of y(dy) terms in Equation 1-1.
`ypdel`A matrix whose columns, `ypdel(:,j)` represent y '(dypj). The size of this matrix is `n`-by-`q`, where `n` is the number of equations in the system you want to solve, and `q` is the number of y '(dyp) terms in Equation 1-1.
`yp`The result returned by `ddefun`. It is an `n`-by-`1` vector whose elements represent the right side of Equation 1-1.
### `dely` — Solution delaysfunction handle | vector
Solution delays, specified as a function handle, which returns dy1,..., dyp in Equation 1-1. Alternatively, you can pass constant delays in the form of a vector.
If you specify `dely` as a function handle, the syntax must be `dy = dely(t,y)`. The arguments for this function are described in the table below.
dely ArgumentDescription
`t`A scalar value representing the current value of time, t.
`y`A vector that represents y(t) in Equation 1-1. The size of this vector is `n`-by-`1`, where `n` is the number of equations in the system you want to solve.
`dy`A vector returned by the `dely` function whose values are the solution delays, dyi , in Equation 1-1. The size of this vector is `p`-by-`1`, where `p` is the number of solution delays in the equation. Each element must be less than or equal to t.
If you want to specify constant solution delays having the form dyi = tτi, then `dely` must be a vector, where `dely(i)` = τi. Each value in this vector must be greater than or equal to zero.
If dy is not present in the problem, set `dely` to `[]`.
Data Types: `function_handle` | `single` | `double`
### `delyp` — Derivative delaysfunction handle | vector
Derivative delays, specified as a function handle, which returns dyp1,..., dypq in Equation 1-1. Alternatively, you can pass constant delays in the form of a vector.
If `delyp` is a function handle, its syntax must be `dyp = delyp(t,y)`. The arguments for this function are described in the table below.
delyp ArgumentDescription
`t`A scalar value representing the current value of time, t.
`y`A vector that represents y(t) in Equation 1-1. The size of this vector is `n`-by-`1`, where `n` is the number of equations in the system you want to solve.
`dyp`A vector returned by the `delyp` function whose values are the derivative delays, dypj, in Equation 1-1. The size of this vector must be `q`-by-`1`, where `q` is the number of solution delays, dypj, in the equation. Each element of `dyp` must be less than t. There is one exception to this restriction: if you are solving an initial value DDE, the value of `dyp` can equal t at t = t0. For more information, see Initial Value Neutral Delay Differential Equations.
If you want specify constant derivative delays having the form dypj = tτj, then `delyp` must be a vector, where `delyp(j)` = τj. Each value in this vector must be greater than zero. An exception to this restriction occurs when you solve initial value problems for DDEs of neutral type. In such cases, a value in `delyp` can equal zero at t = t0. See Initial Value Neutral Delay Differential Equations for more information.
If dyp is not present in the problem, set `delyp` to `[]`.
Data Types: `function_handle` | `single` | `double`
### `history` — Solution historyfunction handle | column vector | structure (`sol`, from previous integration) | `1`-by-`2` cell array
Solution history, specified as a function handle, column vector, `sol` structure (from a previous integration), or a cell array. This is the solution at tt0.
• If the history varies with time, specify the solution history as a function handle whose syntax is `y = history(t)`. This function returns an `n`-by-`1` vector that approximates the solution, y(t), for t <= t0. The length of this vector, `n`, is the number of equations in the system you want to solve.
• If y(t) is constant, you can specify `history` as an `n`-by-`1` vector of the constant values.
• If you are calling `ddensd` to continue a previous integration to t0, you can specify history as the output, `sol`, from the previous integration.
• If you are solving an initial value DDE, specify history as a cell array, `{y0, yp0}`. The first element, `y0`, is a column vector of initial values, y(t0). The second element, yp0, is a column vector whose elements are the initial derivatives, y '(t0). These vectors must be consistent, meaning that they satisfy Equation 1-1 at t0. See Initial Value Neutral Delay Differential Equations for more information.
Data Types: `function_handle` | `single` | `double` | `struct` | `cell`
### `tspan` — Interval of integration`1`-by-`2` vector
Interval of integration, specified as the vector ```[t0 tf]```. The first element, `t0`, is the initial value of t. The second element, `tf`, is the final value of t. The value of `t0` must be less than` tf`.
Data Types: `single` | `double`
### `options` — Optional integration parametersstructure returned by `ddeset`
Optional integration parameters, specified as a structure created and returned by the `ddeset` function. Some commonly used properties are: `'RelTol'`, `'AbsTol'`, and `'Events'`. See the `ddeset` reference page for more information about specifying `options`.
## Output Arguments
collapse all
### `sol` — Solutionstructure
Solution, returned as a structure containing the following fields.
`sol.x` Mesh selected by `ddensd`. `sol.y` An approximation to y(t) at the mesh points. `sol.yp` An approximation to y '(t) at the mesh points. `sol.solver` A string identifying the solver, `'ddensd'`.
You can pass `sol` to the `deval` function to evaluate the solution at specific points. For example, ```y = deval(sol, 0.5*(sol.x(1) + sol.x(end)))``` evaluates the solution at the midpoint of the interval of integration.
collapse all
### Initial Value Neutral Delay Differential Equations
An initial value DDE has dyit0 and dypjt0, for all i and j. At t = t0, all delayed terms reduce to y(dyi) = y(t0) and y '(dypj) = y '(t0):
y '(t0) = f(t0, y(t0), y(t0),..., y(t0), y '(t0),..., y '(t0)) (1-2)
For t > t0, all derivative delays must satisfy dyp < t.
When you solve initial value neutral DDEs, you must supply y '(t0) to `ddensd`. To do this, specify `history` as a cell array `{Y0,YP0}`. Here, `Y0` is the column vector of initial values, y(t0), and `YP0` is a column vector of initial derivatives, y '(t0). These vectors must be consistent, meaning that they satisfy Equation 1-2 at t0.
### Algorithms
For information about the algorithm used in this solver, see Shampine [2].
## References
[1] Paul, C.A.H. "A Test Set of Functional Differential Equations." Numerical Analysis Reports. No. 243. Manchester, UK: Math Department, University of Manchester, 1994.
[2] Shampine, L.F. "Dissipative Approximations to Neutral DDEs." Applied Mathematics & Computation. Vol. 203, Number 2, 2008, pp. 641–648.
Get trial now | 2,681 | 9,667 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2015-18 | latest | en | 0.677498 |
https://de.maplesoft.com/support/help/maple/view.aspx?path=MathApps%2FDistanceBetweenPoints | 1,718,453,527,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861586.40/warc/CC-MAIN-20240615093342-20240615123342-00413.warc.gz | 165,429,671 | 22,198 | Distance between Points - Maple Help
For the best experience, we recommend viewing online help using Google Chrome or Microsoft Edge.
# Online Help
###### All Products Maple MapleSim
Home : Support : Online Help : Math Apps : Graphing : Basic : Distance between Points
Distance Between Points
The Distance Formula
The distance d between two points $\left({x}_{1},{y}_{1}\right)$ and $\left({x}_{2},{y}_{2}\right)$ is given by:
$d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$
More Details
By using the Pythagorean theorem, we can work out the length of the hypotenuse of a right-angled triangle if we know the length of the other two sides. To find the distance between two points $\left({x}_{1},{y}_{1}\right)$ and $\left({x}_{2},{y}_{2}\right)$, we first locate the points on the Cartesian plane.
Next, we can construct a right triangle by drawing a horizontal line through $\left({x}_{1},{y}_{1}\right)$ and a vertical line through $\left({x}_{2},{y}_{2}\right)$.
From the Pythagorean Theorem we know that $c=\sqrt{{a}^{2}+{b}^{2}}$.
Now, we need to determine the lengths a and b.
$a=\left|{x}_{2}-{x}_{1}\right|$ $b=\left|{y}_{2}-{y}_{1}\right|$
Then square each side to get closer to the form $\sqrt{{a}^{2}+{b}^{2}}$.
${b}^{2}={\left|{y}_{2}-{y}_{1}\right|}^{2}={\left({y}_{2}-{y}_{1}\right)}^{2}$
Then
$c=\sqrt{{a}^{2}+{b}^{2}}=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$
Click or drag two points on the graph, and the distance between them will be computed, along with the two components of the distance.
Check Only use integer coordinates if you want to allow only lattice points to be selected (for example, not ).
More MathApps | 531 | 1,739 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 15, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-26 | latest | en | 0.6658 |
https://www.geeksforgeeks.org/minimum-count-of-groups-such-that-sum-of-adjacent-elements-is-divisible-by-k-in-each-group/?ref=rp | 1,713,718,656,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817790.98/warc/CC-MAIN-20240421163736-20240421193736-00498.warc.gz | 690,271,874 | 53,303 | # Minimum count of groups such that sum of adjacent elements is divisible by K in each group
Last Updated : 28 Jul, 2021
Given an array arr[] consisting of N integers, and an integer K, the task is to print the minimum count of groups formed by taking elements of the array such that the sum of adjacent elements is divisible by K in each group.
Examples:
Input: arr[] = {2, 6, 5, 8, 2, 9}, K = 2
Output: 2
The array can be split into two groups {2, 6, 2, 8} and {5, 9}.
The sum of adjacent elements of each group is divisible by K.
Thus, the minimum number of groups that can be formed is 2.
Input: arr[] = {1, 1, 4, 4, 8, 6, 7}, K = 8
Output: 4
Explanation:
The array can be split into 4 groups: {1, 7, 1}, {4, 4}, {8}, and {6}.
The sum of adjacent elements of each group is divisible by K.
Thus, the minimum number of groups that can be formed is 4.
Input: arr[] = {144}, K = 5
Output: 1
Approach: The given problem can be solved based on the following observations:
• It can be observed that a group should be formed as:
1. Groups should consist of only one element which is not divisible by K.
2. All elements of the groups should be individually divisible by K.
3. Every adjacent element of the group should be satisfied; X % K + Y % K = K, where X and Y are two adjacent elements of the group.
Follow the steps below to solve the problem:
• Initialize a map<int, int>, say mp, to store the count of arr[i] % K.
• Initialize a variable, say ans as 0, to store the count of groups.
• Traverse the array arr[] and increment the count of arr[i] % K in the mp and if arr[i] % K is 0 then assign 1 to ans.
• Iterate over the range [1, K / 2] and perform the following operations:
• Store the minimum of mp[i] and mp[K – i] in a variable, say C1.
• Store the maximum of mp[i] and mp[K – i] in a variable, say C2.
• If C1 is 0 then increment ans by C2.
• Otherwise, if C1 is either equal to C1 or C1 + 1, then increment ans by 1.
• Otherwise, increment ans by C2 – C1 – 1.
• Finally, after completing the above steps, print the answer obtained in ans.
Below is the implementation of the above approach:
## C++
`// C++ program for the above approach` `#include ` `using` `namespace` `std;` `// Function to count the minimum number` `// of groups` `int` `findMinGroups(``int` `arr[], ``int` `N, ``int` `K)` `{` ` ``// Stores the count of elements` ` ``unordered_map<``int``, ``int``> mp;` ` ``// Stores the count of groups` ` ``int` `ans = 0;` ` ``// Traverse the array arr[]` ` ``for` `(``int` `i = 0; i < N; i++) {` ` ``// Update arr[i]` ` ``arr[i] = arr[i] % K;` ` ``// If arr[i] is 0` ` ``if` `(arr[i] == 0) {` ` ``// Update ans` ` ``ans = 1;` ` ``}` ` ``else` `{` ` ``// Increment mp[arr[i]] by 1` ` ``mp[arr[i]]++;` ` ``}` ` ``}` ` ``// Iterarte over the range [1, K / 2]` ` ``for` `(``int` `i = 1; i <= K / 2; i++) {` ` ``// Stores the minimum of count of i` ` ``// and K - i` ` ``int` `c1 = min(mp[K - i], mp[i]);` ` ``// Stores the maximum of count of i` ` ``// and K - i` ` ``int` `c2 = max(mp[K - i], mp[i]);` ` ``// If c1 is 0` ` ``if` `(c1 == 0) {` ` ``// Increment ans by c2` ` ``ans += c2;` ` ``}` ` ``// Otherwise if c2 is equal to c1 + 1` ` ``// or c1` ` ``else` `if` `(c2 == c1 + 1 || c1 == c2) {` ` ``// Increment ans by 1` ` ``ans++;` ` ``}` ` ``// Otherwise` ` ``else` `{` ` ``// Increment ans by c2 - c1 - 1` ` ``ans += (c2 - c1 - 1);` ` ``}` ` ``}` ` ``// Return the ans` ` ``return` `ans;` `}` `// Driver Code` `int` `main()` `{` ` ``// Input` ` ``int` `arr[] = { 1, 1, 4, 4, 8, 6, 7 };` ` ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);` ` ``int` `K = 8;` ` ``// Function Call` ` ``cout << findMinGroups(arr, N, K);` ` ``return` `0;` `}`
## Java
`// Java program for the above approach` `import` `java.io.*;` `import` `java.util.*;` `class` `GFG ` `{` ` ` ` ``// Function to count the minimum number` ` ``// of groups` ` ``public` `static` `int` `findMinGroups(``int` `arr[], ``int` `N, ``int` `K)` ` ``{` ` ``// Stores the count of elements` ` ``HashMap mp = ``new` `HashMap<>();` ` ` ` ``// Stores the count of groups` ` ``int` `ans = ``0``;` ` ` ` ``// Traverse the array arr[]` ` ``for` `(``int` `i = ``0``; i < N; i++) {` ` ` ` ``// Update arr[i]` ` ``arr[i] = arr[i] % K;` ` ` ` ``// If arr[i] is 0` ` ``if` `(arr[i] == ``0``) {` ` ` ` ``// Update ans` ` ``ans = ``1``;` ` ``}` ` ``else` `{` ` ` ` ``// Increment mp[arr[i]] by 1` ` ``if` `(!mp.containsKey(arr[i])) {` ` ``mp.put(arr[i], ``1``);` ` ``}` ` ``else` `{` ` ``Integer ct = mp.get(arr[i]);` ` ``if``(ct!=``null``)` ` ``{` ` ``ct++;` ` ``mp.put(arr[i], ct);` ` ``}` ` ``}` ` ``}` ` ``}` ` ` ` ``// Iterarte over the range [1, K / 2]` ` ``for` `(``int` `i = ``1``; i <= K / ``2``; i++) {` ` ` ` ``// Stores the minimum of count of i` ` ``// and K - i` ` ``int` `a=``0``,b=``0``;` ` ``if``(mp.containsKey(K-i)){` ` ``a=mp.get(K-i);` ` ``}` ` ``if``(mp.containsKey(i)){` ` ``b=mp.get(i);` ` ``}` ` ``int` `c1 = Math.min(a, b);` ` ` ` ``// Stores the maximum of count of i` ` ``// and K - i` ` ``int` `c2 = Math.max(a, b);` ` ` ` ``// If c1 is 0` ` ``if` `(c1 == ``0``) {` ` ` ` ``// Increment ans by c2` ` ``ans += c2;` ` ``}` ` ` ` ``// Otherwise if c2 is equal to c1 + 1` ` ``// or c1` ` ``else` `if` `(c2 == c1 + ``1` `|| c1 == c2) {` ` ` ` ``// Increment ans by 1` ` ``ans++;` ` ``}` ` ``// Otherwise` ` ``else` `{` ` ` ` ``// Increment ans by c2 - c1 - 1` ` ``ans += (c2 - c1 - ``1``);` ` ``}` ` ``}` ` ` ` ``// Return the ans` ` ``return` `ans;` ` ``}` ` ``public` `static` `void` `main (String[] args) {` ` ``// Input` ` ``int` `arr[] = { ``1``, ``1``, ``4``, ``4``, ``8``, ``6``, ``7` `};` ` ``int` `N = ``7``;` ` ``int` `K = ``8``;` ` ` ` ``// Function Call` ` ``System.out.println(findMinGroups(arr, N, K));` ` ``}` `}` `// This code is contributed by Manu Pathria`
## Python3
`# Python3 program for the above approach` `# Function to count the minimum number` `# of groups` `def` `findMinGroups(arr, N, K):` ` ` ` ``# Stores the count of elements` ` ``mp ``=` `{}` ` ``# Stores the count of groups` ` ``ans ``=` `0` ` ``# Traverse the array arr[]` ` ``for` `i ``in` `range``(N):` ` ` ` ``# Update arr[i]` ` ``arr[i] ``=` `arr[i] ``%` `K` ` ``# If arr[i] is 0` ` ``if` `(arr[i] ``=``=` `0``):` ` ` ` ``# Update ans` ` ``ans ``=` `1` ` ``else``:` ` ``mp[arr[i]] ``=` `mp.get(arr[i], ``0``) ``+` `1` ` ``# Iterarte over the range [1, K / 2]` ` ``for` `i ``in` `range``(``1``, K ``/``/` `2``):` ` ` ` ``# Stores the minimum of count of i` ` ``# and K - i` ` ``x, y ``=` `(``0` `if` `(K ``-` `i) ``not` `in` `mp ``else` `mp[K ``-` `i], ` ` ``0` `if` `i ``not` `in` `mp ``else` `mp[K ``-` `i])` ` ``c1 ``=` `min``(x, y) ` ` ``# The maximum of count of i` ` ``# K - i` ` ``c2 ``=` `max``(x, y)` ` ``# If c1 is 0` ` ``if` `(c1 ``=``=` `0``):` ` ` ` ``# Increment ans by c2` ` ``ans ``+``=` `c2` ` ` ` ``# Otherwise if c2 is equal to c1 + 1` ` ``# or c1` ` ``elif` `(c2 ``=``=` `c1 ``+` `1` `or` `c1 ``=``=` `c2):` ` ` ` ``# Increment ans by 1` ` ``ans ``+``=` `1` ` ` ` ``# Otherwise ` ` ``else``:` ` ``# Increment ans by c2 - c1 - 1` ` ``ans ``+``=` `(c2 ``-` `c1 ``-` `1``)` ` ` ` ``# Return the ans` ` ``return` `ans ``+` `1` `# Driver code` `if` `__name__ ``=``=` `'__main__'``:` ` ` ` ``# Input` ` ``arr ``=` `[ ``1``, ``1``, ``4``, ``4``, ``8``, ``6``, ``7` `]` ` ``N ``=` `len``(arr)` ` ``K ``=` `8` ` ``# Function Call` ` ``print``(findMinGroups(arr, N, K))` `# This code is contributed by mohit kumar 29`
## C#
`// C# program for the above approach` `using` `System;` `using` `System.Collections.Generic; ` `public` `class` `GFG{` ` ``// Function to count the minimum number` ` ``// of groups` ` ``static` `int` `findMinGroups(``int``[] arr, ``int` `N, ``int` `K)` ` ``{` ` ``// Stores the count of elements` ` ``Dictionary<``int``, ``int``> mp = ``new` `Dictionary<``int``, ``int``>();` ` ` ` ``// Stores the count of groups` ` ``int` `ans = 0;` ` ` ` ``// Traverse the array arr[]` ` ``for` `(``int` `i = 0; i < N; i++) {` ` ` ` ``// Update arr[i]` ` ``arr[i] = arr[i] % K;` ` ` ` ``// If arr[i] is 0` ` ``if` `(arr[i] == 0) {` ` ` ` ``// Update ans` ` ``ans = 1;` ` ``}` ` ``else` `{` ` ` ` ``// Increment mp[arr[i]] by 1` ` ``if` `(!mp.ContainsKey(arr[i])) {` ` ``mp.Add(arr[i], 1);` ` ``}` ` ``else` `{` ` ``int` `ct = mp[arr[i]];` ` ``if``(ct!=0)` ` ``{` ` ``ct++;` ` ``mp[arr[i]]= ct;` ` ``}` ` ``}` ` ``}` ` ``}` ` ` ` ``// Iterarte over the range [1, K / 2]` ` ``for` `(``int` `i = 1; i <= K / 2; i++) {` ` ` ` ``// Stores the minimum of count of i` ` ``// and K - i` ` ``int` `a=0,b=0;` ` ``if``(mp.ContainsKey(K-i)){` ` ``a=mp[K-i];` ` ``}` ` ``if``(mp.ContainsKey(i)){` ` ``b=mp[i];` ` ``}` ` ``int` `c1 = Math.Min(a, b);` ` ` ` ``// Stores the maximum of count of i` ` ``// and K - i` ` ``int` `c2 = Math.Max(a, b);` ` ` ` ``// If c1 is 0` ` ``if` `(c1 == 0) {` ` ` ` ``// Increment ans by c2` ` ``ans += c2;` ` ``}` ` ` ` ``// Otherwise if c2 is equal to c1 + 1` ` ``// or c1` ` ``else` `if` `(c2 == c1 + 1 || c1 == c2) {` ` ` ` ``// Increment ans by 1` ` ``ans++;` ` ``}` ` ``// Otherwise` ` ``else` `{` ` ` ` ``// Increment ans by c2 - c1 - 1` ` ``ans += (c2 - c1 - 1);` ` ``}` ` ``}` ` ` ` ``// Return the ans` ` ``return` `ans;` ` ``}` ` ``static` `public` `void` `Main ()` ` ``{` ` ``// InAdd` ` ``int``[] arr = { 1, 1, 4, 4, 8, 6, 7 };` ` ``int` `N = 7;` ` ``int` `K = 8;` ` ` ` ``// Function Call` ` ``Console.Write(findMinGroups(arr, N, K));` ` ``}` `}` `// This code is contributed by shubhamsingh10`
## Javascript
``
Output
`4`
Time Complexity: O(N)
Auxiliary Space: O(K)
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Units of Measure
Scientific Method
# Are seconds a metric unit?
###### Wiki User
Yes, they are the metric unit for time. You ALWAYS use seconds for measuring time in science.
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## Related Questions
A metric unit of time is commonly seconds
A metric unit is a decimal unit of measurement of the metric system, based on meters and kilograms and seconds.
The metric unit for time is seconds.
A metric unit is a decimal unit of measurement of the metric system, based on meters and kilograms and seconds.
The unit is a second, but it can measure to 0.01 seconds.
Seconds must have the s on the end.
The metric unit of time is the second.There isn't one in common use but:10 metric hours in a day100 metric minutes in a metric hour100 metric seconds in a metric minute10 days in a metric week (called a dekade)
The metric unit is a millilitre. The metric unit is a millilitre. The metric unit is a millilitre. The metric unit is a millilitre.
The Newton is the metric unit for weight. The unit is:kilogram meters per seconds squaredorkg . m/sec2.Also, one pound equals 4.44822162 Newtons.
yes, seconds, minutes, hours. there is no metric clock.
I not a metric unit of anything! A litre is one metric unit for volume.
The metric unit for length or distance is the meter.
The SI (i.e. metric) unit is the Joule.The SI (i.e. metric) unit is the Joule.The SI (i.e. metric) unit is the Joule.The SI (i.e. metric) unit is the Joule.
No, a mile is not a metric unit. A mile is a customary unit.
The metric unit of what . -There is an entire metric system of all kinds of units .
The end of the word tells you that this is a metric unit.
The metric unit of measure for volume is cubic units. Other metric unit is liter.
The basic unit of the metric system is the meter and liter.
Pounds are a customary unit. The metric unit is grams.
The units for measuring time are the same in the metric and Imperial systems: seconds, minutes, hours, days and so on.
The standard metric and IS unit of time is the "second".the second.
###### Units of MeasureScienceMetric SystemMath and ArithmeticWeight and MassBuilding and CarpentryBusiness & Finance
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# Smoking in bed has long been the main cause of home fires.
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VP
Joined: 14 May 2006
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Smoking in bed has long been the main cause of home fires. [#permalink]
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11 Jul 2006, 18:24
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Smoking in bed has long been the main cause of home fires. Despite a significant decline in cigarette smoking in the last two decades, there has been no comparable decline in the number of people killed in home fires.
Each one of the following statements, if true over the last two decades, helps to resolve the apparent discrepancy above EXCEPT:
A) Compared to other types of home fires, home fires caused by smoking in bed usually cause relatively little damage before they are extinguished.
B) Home fires caused by smoking in bed often break out after the home's occupants have fallen asleep.
C) Smokers who smoke in bed tend to be heavy smokers who are less likely to quit smoking than are smokers who do not smoke in bed.
D) An increasing number of people have been killed in home fires that started in the kitchen.
E) Population densities have increased, with the result that one home fire can cause more deaths than in previous decades.
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VP
Joined: 25 Nov 2004
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Re: CR: Smoking in bed [#permalink]
### Show Tags
11 Jul 2006, 19:07
u2lover wrote:
Smoking in bed has long been the main cause of home fires. Despite a significant decline in cigarette smoking in the last two decades, there has been no comparable decline in the number of people killed in home fires.
Each one of the following statements, if true over the last two decades, helps to resolve the apparent discrepancy above EXCEPT:
A) Compared to other types of home fires, home fires caused by smoking in bed usually cause relatively little damage before they are extinguished.
B) Home fires caused by smoking in bed often break out after the home's occupants have fallen asleep.
C) Smokers who smoke in bed tend to be heavy smokers who are less likely to quit smoking than are smokers who do not smoke in bed.
D) An increasing number of people have been killed in home fires that started in the kitchen.
E) Population densities have increased, with the result that one home fire can cause more deaths than in previous decades.
guess these questions are from barrons and it has been frequestly advised that these materials are toooooooooooooo poor to maintain the standard of real gmat tests.
however, i go with D.
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VP
Joined: 14 May 2006
Posts: 1399
Kudos [?]: 227 [0], given: 0
Re: CR: Smoking in bed [#permalink]
### Show Tags
11 Jul 2006, 19:44
MA wrote:
guess these questions are from barrons and it has been frequestly advised that these materials are toooooooooooooo poor to maintain the standard of real gmat tests.
however, i go with D.
I abandoned Barrons a while ago
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CEO
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Posts: 2892
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Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
### Show Tags
11 Jul 2006, 19:45
I go with B.
_________________
SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008
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Director
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Re: CR: Smoking in bed [#permalink]
### Show Tags
11 Jul 2006, 20:24
u2lover wrote:
Smoking in bed has long been the main cause of home fires. Despite a significant decline in cigarette smoking in the last two decades, there has been no comparable decline in the number of people killed in home fires.
Each one of the following statements, if true over the last two decades, helps to resolve the apparent discrepancy above EXCEPT:
A) Compared to other types of home fires, home fires caused by smoking in bed usually cause relatively little damage before they are extinguished.
B) Home fires caused by smoking in bed often break out after the home's occupants have fallen asleep.
C) Smokers who smoke in bed tend to be heavy smokers who are less likely to quit smoking than are smokers who do not smoke in bed.
D) An increasing number of people have been killed in home fires that started in the kitchen.
E) Population densities have increased, with the result that one home fire can cause more deaths than in previous decades.
I choose A.
I was stuck between A and B. However, in B, it could be that since people are asleep they have lesser time to react to the fire caused by smoking and this could still explain the higher deaths.
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Manager
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11 Jul 2006, 21:50
I go with A. It is the least relevant
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11 Jul 2006, 23:55
B for me 2
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12 Jul 2006, 00:03
Beg to differ But IMO its C
Look at the primse :
Cigarrete somking has declined.
And the conclusion:
despite above, the number of house fires have not declined.
Now option C gives a reason why cigarrete smoking may have declined but still may not have contributed to the any decline in house fires.
If smokers who smoke on bed donot quit smoking whereas other smokers who donot smoke on bed quit, there is ur decline in smoking. But there is no decline in people who smoke on bed. The major reason for house fires.
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Re: CR: Smoking in bed [#permalink]
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12 Jul 2006, 00:58
In this question type: Resolve the paradox in order to be fit the answear has to address both the sides of the paradox, in this case we are looking for a solution that example address to only one of the sides, now lets tackle the answears,.....
Smoking in bed has long been the main cause of home fires. Despite a significant decline in cigarette smoking in the last two decades, there has been no comparable decline in the number of people killed in home fires.
Each one of the following statements, if true over the last two decades, helps to resolve the apparent discrepancy above EXCEPT:
A) Compared to other types of home fires, home fires caused by smoking in bed usually cause relatively little damage before they are extinguished.
B) Home fires caused by smoking in bed often break out after the home's occupants have fallen asleep.
C) Smokers who smoke in bed tend to be heavy smokers who are less likely to quit smoking than are smokers who do not smoke in bed.
D) An increasing number of people have been killed in home fires that started in the kitchen. explain the paradox
E) Population densities have increased, with the result that one home fire can cause more deaths than in previous decades. explain the paradox
I am not sure about B or C using at this point POE I would keep B because it seems that deviate from the scope of the passage,
my correct answear would be B
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12 Jul 2006, 01:03
Raghavender wrote:
I go with A. It is the least relevant
Is wrong because explains why deaths are not declining due cigarettes home fires
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12 Jul 2006, 10:00
jaynayak wrote:
Beg to differ But IMO its C
Look at the primse :
Cigarrete somking has declined.
And the conclusion:
despite above, the number of house fires have not declined.
Now option C gives a reason why cigarrete smoking may have declined but still may not have contributed to the any decline in house fires.
If smokers who smoke on bed donot quit smoking whereas other smokers who donot smoke on bed quit, there is ur decline in smoking. But there is no decline in people who smoke on bed. The major reason for house fires.
so option C is helping to explain the discrepancy.
I think answer should be A as it does not explain the discrepancy.
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12 Jul 2006, 10:34
Is it (B)
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12 Jul 2006, 12:22
I go with B.
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Re: CR: Smoking in bed [#permalink]
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12 Jul 2006, 23:04
A) Compared to other types of home fires, home fires caused by smoking in bed usually cause relatively little damage before they are extinguished.
correct.. Yes, this doesn't explain paradox at all..
B) Home fires caused by smoking in bed often break out after the home's occupants have fallen asleep.
This explains better than A, though doesn't make much sense..
C) Smokers who smoke in bed tend to be heavy smokers who are less likely to quit smoking than are smokers who do not smoke in bed.
This explains very well..Poeple who used to smoke in bed still smoke.. And smoking in bed is cause for fires..
D) An increasing number of people have been killed in home fires that started in the kitchen.
Yes, this explains a different reason why the accidents haven't come down..
E) Population densities have increased, with the result that one home fire can cause more deaths than in previous decades.
This also explains very well..
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12 Jul 2006, 23:14
Took the full two minutes to narrow it down to A and B.
(A) just simply does not resolve any sort of paradox.
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13 Jul 2006, 14:22
OA is B
The paradox is that smoking has declined, but fatalities from home fires have increased... so, we are looking for an alternative cause of fires or that smoking is not related to home fires.
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13 Jul 2006, 22:22
Gauss wrote:
Raghavender wrote:
I go with A. It is the least relevant
Is wrong because explains why deaths are not declining due cigarettes home fires
Yea Gauss, what you said makes sense. I missd out on the point earlier...
Mann....the CRs need a lot more scutiny, then what i actually do. the CRs in OG are jus too too easy compared to these...
gud for practice
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20 Jul 2006, 06:20
Narrowed it down to A & B. A explains the paradox by suggesting that if fires due to smoking in bed causes very little damage, then these fires would not have been responsible for the deaths. Other fires would have been.
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20 Jul 2006, 06:20
Display posts from previous: Sort by
# Smoking in bed has long been the main cause of home fires.
Moderators: GMATNinjaTwo, GMATNinja
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 3,383 | 12,764 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2017-47 | latest | en | 0.930265 |
http://www.mathcs.emory.edu/~cheung/Courses/255/Syllabus/1-C-intro/bit-oper2.html | 1,513,346,376,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948572676.65/warc/CC-MAIN-20171215133912-20171215155912-00074.warc.gz | 428,864,640 | 2,170 | ### Using the bit-wise operators in C
• Using the bitwise operators
• The bit-wise operations are typically used do the following:
Set a certain bit in a byte, short, int, or long variable Clear a certain bit in a byte, short, int, or long variable Flip a certain bit (0 ⇒ 1 or 1 ⇒ 0) in a byte, short, int, or long variable Test a certain bit in a byte, short, int, or long variable
• Note:
The bit-wise operations often use a certain bit pattern (or bit mask) The bit pattern can often be expression more easily using different number systems We will also look at how to express a value in different number system in C
• Setting a certain bit in a bit pattern
• Set a certain bit in a bit pattern
• Suppose the variable a contains a bit pattern:
``` a = 1010...x...1101 ```
• Problem:
Set the bit x to the value 1 without changing all other bits in the pattern.
• We can set the bit x inside the bit pattern to 1 using the OR operation with the following specific pattern:
``` a = 1010...x...1101 0000..010..0000 OR --------------- 1010...1...1101 ```
• Example: set the 2nd bit and the 4th bit in a bit pattern
``` int main( int argc, char* argv[] ) { char a = 0; /* 0 = 00000000 */ /* Bit pos: 76543210 */ a = a | 4; /* 16 = 00000100 - set 2nd bit */ a = a | 16; /* 16 = 00010000 - set 4th bit */ printf( "a = %d\n", a ); } ```
• Example Program: (Demo above code)
How to run the program:
Right click on link and save in a scratch directory To compile: gcc bit-op2.c To run: ./a.out
• Observation: we need a better way to make bit patterns
• Fact:
The decimal number system is terrible for expression a bit pattern
• Solution:
Use different number systems !!! | 463 | 1,706 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2017-51 | longest | en | 0.770484 |
https://training.incf.org/search?type%5B32%5D=32&%3Bf%5B0%5D=topics%3A66&f%5B0%5D=difficulty_level%3Abeginner&f%5B1%5D=topics%3A22&f%5B2%5D=topics%3A39&f%5B3%5D=topics%3A50&f%5B4%5D=topics%3A66 | 1,571,689,301,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987787444.85/warc/CC-MAIN-20191021194506-20191021222006-00516.warc.gz | 737,740,542 | 15,066 | ## Difficulty level
Lecture title:
The probability of a hypothesis, given data.
Difficulty level: Beginner
Duration: 7:57
Speaker: : Barton Poulson
Lecture title:
Why math is useful in data science.
Difficulty level: Beginner
Duration: 1:35
Speaker: : Barton Poulson
Lecture title:
Why statistics are useful for data science.
Difficulty level: Beginner
Duration: 4:01
Speaker: : Barton Poulson
Lecture title:
Statistics is exploring data.
Difficulty level: Beginner
Duration: 2:23
Speaker: : Barton Poulson
Lecture title:
Graphical data exploration
Difficulty level: Beginner
Duration: 8:01
Speaker: : Barton Poulson
Lecture title:
Numerical data exploration
Difficulty level: Beginner
Duration: 5:05
Speaker: : Barton Poulson
Lecture title:
Simple description of statistical data.
Difficulty level: Beginner
Duration: 10:16
Speaker: : Barton Poulson
Lecture title:
Basics of hypothesis testing.
Difficulty level: Beginner
Duration: 06:04
Speaker: : Barton Poulson
Lecture title:
This primer on optogenetics primer discusses how to manipulate neuronal populations with light at millisecond resolution and offers possible applications such as curing the blind and "playing the piano" with cortical neurons.
Difficulty level: Beginner
Duration: 59:06
Speaker: : Clay Reid
Lecture title:
This lecture will highlight our current understanding and recent developments in the field of neurodegenerative disease research, as well as the future of diagnostics and treatment of neurodegenerative diseases.
Difficulty level: Beginner
Duration: 39:05
Lecture title:
2nd part of the lecture. This lecture will highlight our current understanding and recent developments in the field of neurodegenerative disease research, as well as the future of diagnostics and treatment of neurodegenerative diseases.
Difficulty level: Beginner
Duration: 45:27
Lecture title:
This lecture will discuss how understanding and applying simple neuroanatomical rules, one can localize the damage along the neuroaxis, the first crucial step toward making the correct clinical diagnosis and initiating treatment.
Difficulty level: Beginner
Duration: 44:52
Speaker: : Eitan Auriel
Lecture title:
2nd part of the lecture. This lecture will discuss how understanding and applying simple neuroanatomical rules, one can localize the damage along the neuroaxis, the first crucial step toward making the correct clinical diagnosis and initiating treatment.
Difficulty level: Beginner
Duration: 42:35
Speaker: : Eitan Auriel
Lecture title:
This lecture focuses on how the immune system can target and attack the nervous system to produce autoimmune responses that may result in diseases such as multiple sclerosis, neuromyelitis and lupus cerebritis manifested by motor, sensory, and cognitive impairments. Despite the fact that the brain is an immune-privileged site, autoreactive lymphocytes producing proinflammatory cytokines can cause active brain inflammation, leading to myelin and axonal loss.
Difficulty level: Beginner
Duration: 37:36
Speaker: : Anat Achiron
Most psychiatric disorders (most notably dependence syndromes, depression, psychosis, and autism) are characterized by impaired social interaction, with many patients preferring a drug of abuse. This lecture focuses on the latest research on the neural basis of normal and impaired social interaction.
Difficulty level: Beginner
Duration: 41:56
Speaker: : Gerald Zernig
Lecture title:
This lecture will provide an overview of neuroimaging techniques and their clinical applications.
Difficulty level: Beginner
Duration: 45:29
Speaker: : Dafna Ben Bashat
Lecture title:
This lecture will provide an overview of neuroimaging techniques and their clinical applications
Difficulty level: Beginner
Duration: 41:00
Speaker: : Dafna Ben Bashat
This lecture will highlight our current understanding and recent developments in the field of neurodegenerative disease research, as well as the future of diagnostics and treatment of neurodegenerative diseases
Difficulty level: Beginner
Duration: 1:02:29 | 876 | 4,057 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-43 | latest | en | 0.790208 |
https://www.inettutor.com/download-category/c-2/page/2/ | 1,534,792,527,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221216724.69/warc/CC-MAIN-20180820180043-20180820200043-00233.warc.gz | 883,420,069 | 13,973 | ## Grade Solver in C++
Grade Solver in C++
Grade solver sample program in c++ that computes your final grade based on the prelim, midterm and endterm grades.
The formula:
Final Grade = (prelimgrade * .3) + (midterm * .3) + (endterm * .4)
vb, java and c# version is also available, kindly search for it.
Thank you for visiting and Happy Coding!
## Get the Average of Numbers in C++
Get the Average of Numbers in C++
This is a sample program in c++ that computes the average or mean of the numbers entered by the user.
Below is the formula on how to compute the average of the numbers entered by the user.
Average = sum of the numbers / the quantity of numbers entered by the user.
## Compute the Area and Perimeter of the Rectangle in C++
Compute the Area and Perimeter of the Rectangle in C++
Example program in c++ that ask the user to input the length and width of a rectangle, the program will then compute and display the area and perimeter of the rectangle.
Codeblocks is used as the ide for this program. It is an open source, cross-platform ide.
## Increment-Decrement a number in C++
Increment-Decrement a number in C++
Increment means to add or increase, decrement means to deduct or decrease. This c++ program will demonstrate how to increment and decrement a number. The user will be given an option to either increment or decrement a value, then the user will be ask to enter a number, the program will then perform the operation selected by the user.
## If Else Statement in C++
If Else Statement in C++
Sample program that demonstrate if else statement in c++. The program will ask to enter a grade, if your grade is greater than or equal to 75 then you have a passing grade, grade less than 75 means failure.
## Basic Math Operators in C++
Basic Math Operators in C++
This is an example program in c++ that demonstrate the four basic mathematical operators in action (add, subtract, divide, multiply). The program will ask the user to input two numbers, then the program will display the result of the four basic math operations.
## Display Date and Time in C++
Display Date and Time in C++
Example c++ program that gets and display the system of your computer. The program will demonstrate two ways on how to display the time; one is by getting the real time of your system and another is by timestamp.
Happy Coding!
## Do While Loop in C++
Do While Loop in C++
This is program in c++ that demonstrate how do while loop functions in c++. The user will asked how many numbers he/she wants to enter, then program will compute the sum of the numbers entered by the user.
## Sum of Two Numbers in C++
Sum of Two Numbers in C++
A very simple program in c++ that ask a user to enter two numbers and the program will compute and display the result.
Open the main.cpp in your preferred c++ ide.
Happy Programming!
## For Loop in C++
For Loop in C++
For loop demo in c++. The program will ask you to enter a number of how many times you want hello world to display on the screen. The number entered will then process and executed in the for loop.
Happy programming! | 674 | 3,101 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2018-34 | longest | en | 0.856711 |
chasestormdawg.com | 1,368,936,333,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696383218/warc/CC-MAIN-20130516092623-00011-ip-10-60-113-184.ec2.internal.warc.gz | 51,257,115 | 16,735 | ## How to understand a stormy radar!
Meteorologists use weather radar to get an idea of how intense precipitation is, which way it is moving and how fast it is moving. They radars send an energy wave field into the sky. When the wave field bounces off an object in its path, the energy is reflected back to the radar and computers sort it into different colors that represent different information.
The blue box shows where the radar is located. Pretend you are in the middle of a circle of friends and each of them is shining a flashlight at you. They represent the raindrops.
Pretend you have a mirror. If you spin slowly in your spot, your mirror will receive the energy from each flashlight – just like a radar receives energy reflecting off raindrops! See how it looks like there are spokes in the colors spreading out from the blue box?
These pictures show two important information the radar measures. The left image is known as “reflectivity.” If a lot of energy is reflected back, it means the precipitation is dense – the energy couldn’t make it through the raindrop or hail and was forced back. The scale at the bottom shows cooler colors mean light precipitation, and warmer colors mean heavy precipitation. When you get into pink, purple and light blue, that alerts us to hail and even large hail.
The right image is “velocity.” It shows us how fast the wind is moving toward or away from the radar. Green is toward, red is away. Remember that!
Ok. Now in the left image, that is a SUPERCELL THUNDERSTORM. This one is special because it has a tail. I call it a tail because I am a dog. Meteorologists call this a HOOK ECHO. It really does look like a hook! But this hook means that the precipitation is being pushed around the storm. Hmmm. A clue!
The image on the right, remember, is showing winds. Whenever you see greens and reds right together, like Christmas, it is an even better clue! Find that spot on the picture. Now take your finger and trace from the brightest green to the blue box. Then take another finger and trace from the blue box to the brightest red. Your fingers almost made a tornado! The winds moving toward the radar and away from the radar but right next to each other mean there is rotation. Meteorologists call that a TORNADO VORTEX SIGNATURE, or TVS. This means there might be a tornado. We can’t say for sure from a TVS because radars can’t “see” at the ground.
But wait, what?!?
Why can’t they see at the ground? Well, if it did, it would detect trees and buildings! We have to point the beam up a little to avoid that if we can. And, to make things more tough, the Earth curves! The radar beam is generally more straight, so the farther the beam goes out, the higher it is above the Earth because the Earth curves away.
A funnel has to touch the ground to be a tornado – so now what?
Radars can detect a tornado causing damage if the debris is getting kicked up high enough to hit the beam. Other than that, we can just guess and rely on special storm spotters and law enforcement to confirm a radar on the ground.
Phew! Lots of information, but now you can look at radar for TVS’s!
chasestormdawg@gmail.com
## One stop severe weather shop!
The Storm Prediction Center launched a brand new site this week!
www.spc.noaa.gov
This is a good place to start your day to pay attention to the weather. My human’s browser is set to open to this page every time she opens a new window. That way she always knows if there is a weather threat.
Look at the big map. You can glance at it and know in a few seconds what areas need to pay attention. In the area inside the brown line, they are expecting possible thunderstorms. Inside the green area, there is a slight risk of these thunderstorms being severe. If there is a yellow area, that is a moderate risk. If you are in this area, you should pay attention to the weather and make sure you have a plan to get safe in severe weather. A red area is rare, but means a high risk of severe weather. This is very serious and if you are in this area, you should pay close attention to the weather all day and have your safety plan ready to start at any time.
Also on this page you can see current severe weather watches and areas they are thinking about issuing watches (those are called mesoscale discussions). You can see in the picture they just issued a severe thunderstorm watch for parts of the Texas panhandle and western Oklahoma. If you want to know more about what concerns them about this area, you can click on the watch and you can read more complicated meteorology terms.
Severe weather climatology at the lower left is interesting to look at. My friends have been working on this for a long time. You should check it out to see what time of year is your biggest risk for different types of severe weather.
Storm report trends are new, but gives you a quick glance at what is going on.
And a great FAQ about the difference between a severe weather watch and a severe weather warning on the lower right.
If you keep scrolling down, you will find even more technical stuff, but important tornado safety tips.
## Spring at the “top” and “bottom” of the world!
GOES satellite captures the Spring Equinox
Today is a special day because it is the first day of Spring! This means at most places in the world, there are 12 hours of daylight, and 12 hours of darkness.
The photo on the left was made with a satellite. Look how you can barely see Greenland at the top, and barely see the tip of Antarctica at the bottom. And look how they have about the same amount of sunlight!
“WorldView+” is a favorite app on my human’s iPhone. This app lets you look at webcams all over the world. We bookmark our favorites. Sometimes when we need a break, we check the weather in different places on Earth. I would download it if you are allowed. It is VERY educational.
We thought it would be cool to compare webcams from the top of the world, and the bottom on a day where they have about the same amount of sunlight!
Scott base in Antarctica
This camp is in the middle of the Greenland Ice Sheet.
## National Severe Weather Preparedness Week: March 3-9 2013
I use a bike or sports helmet to protect my head from things falling out of the sky.
It’s my favorite week of the year! National Severe Weather Preparedness Week begins TODAY!
All week I will be giving tips about what you can to to be ready for the severe weather season.
Severe weather can happen at all times of the year, but spring is the most common.
Why spring? Because that’s when winter and summer crash into each other! Where they collide is where you will find thunderstorms.
There can be as many as 40,000 thunderstorms each day around the world. They are the most common in the U.S., where they can produce tornadoes, floods, lightning and damaging winds.
Three basic ingredients are needed for a thunderstorm to form: moisture, rising unstable air (air that keeps rising when given a nudge), and something to nudge the air up.
The sun heats the surface of the earth, which warms the air above it. If this warm surface air is forced to rise—hills or mountains, or areas where warm/cold or wet/dry air bump together can cause rising motion—it will continue to rise as long as it weighs less and stays warmer than the air around it.
As the air rises, it transfers heat from the surface of the earth to the upper levels of the atmosphere (the process of convection). The water vapor it contains begins to cool, releases the heat, condenses and forms a cloud. The cloud eventually grows upward into areas where the temperature is below freezing.
As a storm rises into freezing air, different types of ice particles can be created from freezing liquid drops. The ice particles can grow by condensing vapor (like frost) and by collecting smaller liquid drops that haven’t frozen yet (a state called “supercooled”). When two ice particles collide, they usually bounce off each other, but one particle can rip off a little bit of ice from the other one and grab some electric charge. Lots of these collisions build up big regions of electric charges to cause a bolt of lightning, which creates the sound waves we hear as thunder.
Check back with me this week as I highlight a different type of severe weather each day, and talk about ways to be ready for severe weather season.
Being prepared to act quickly during severe weather will help you not be as scared and think more clearly.
## mPING app not for dogs
It’s even too hard for me to report on an iPad because of my paws.
I learned about this really cool app that came out from the National Severe Storms Lab and grad students at the University of Oklahoma. My human made everyone in her family download it. It is called “mobile Precipitation Identification Near the Ground” or mPING. The app lets you report whether rain, snow, sleet or freezing rain is hitting the ground at your spot.
But I CAN watch YOUR reports!
But who cares?
Well I do, of course, but research does too!
Radars can’t see right at the ground. They have to point their beam up a little to avoid as many trees and buildings as possible that could block it. So that means if the radar detects a snowflake up high, but it melts into a raindrop by the time it gets to the ground, the radar doesn’t know.
Researchers are working hard on computer programs that make a best guess from the radar data about what is actually falling. But, with mPING, they can see where they are right and where they were wrong!
It is really a cool idea! There have been tons of humans reporting. I’ve seen the reports on my human’s computer screen. It looks like a Christmas tree.
I am sad though, because I can’t use the app.
My paws are too big to click on a button that small. www.nssl.noaa.gov/projects/ping
## The U.S. saw all kinds of weather in 2012
I thought this was a cool map showing the extreme weather the U.S. experienced in 2012. From record drought to record storms, this shows everyone was impacted by the weather in one way or another.
## Floods happen all over the world!
This is an aqueduct built by the Romans more than two thousand years ago, threatened by a big flood!
This is what it looks like normally!
I have to admit, I think this is one of the most amazing things I have ever seen. It is called the Pont du Gard. They say it was built in two thousand years ago to carry water and people. That is way long ago! And to think something that old has stood up to big floods for thousands of years! They just stopped allowing cars on it about ten years ago!
I found out about it when my human was looking for information about a huge flash flood research project in Europe.
She got distracted.
But the flash flood research project is the biggest field weather project in the history of Europe. Radars on trucks, ships, airplanes, balloons – all of these things will be used to figure out how to forecast flash floods better and save lives.
I think it is important to remember bad weather happens to people all over the world. And, it is good to work together to figure out how to help.
## Keep your cool in the heat!
It has been so hot this summer! I wanted to make sure you remembered these heat safety tips from the NOAA National Weather Service!
Help your pets keep their cool. It will “feel” as hot for them as it will for you. Do not leave your pets in a closed vehicle. Be sure your animals have access to shade and a water bowl full of cold, clean water. We don’t tolerate heat well because we don’t sweat. Our bodies get hot and stay hot. During summer heat, avoid outdoor games or jogging with your pet. If you would not walk across hot, sunbaked asphalt barefoot, don’t make us walk on it either. (Dogs can also get blisters on their paws from hot pavement. OUCH!)
Tips for people!
• Avoid the Heat. Stay out of the heat and indoors as much as possible. Spend time in an air conditioned space. Only two hours a day in an air-conditioned space can significantly reduce the risk of heat-related illness. Shopping malls offer relief if your home is not air-conditioned. If air conditioning is not available, stay on the lowest floor out of the sunshine. Remember, electric fans do not cool, they just blow hot air around.
• Dress for the heat. Wear loose-fitting clothes that cover as much skin as possible. Lightweight, light-colored clothing that reflects heat and sunlight and helps maintain normal body temperature. Protect your face and head by wearing a wide-brimmed hat. Avoid too much sunshine. Sunburn slows the skin’s ability to cool itself. Use a sunscreen lotion with a high SPF (sun protection factor) rating.
• Drink FOR the Heat. Drink plenty of water and natural juices, even if you don’t feel thirsty. Even under moderately strenuous outdoor activity, the rate your body can absorb fluids is less than the rate it loses water due to perspiration. However, if you have epilepsy or heart, kidney, or liver disease; are on fluid-restrictive diets; or have a problem with fluid retention should consult a doctor before increasing liquid intake.
• Do not drink IN the Heat. Avoid alcoholic beverages and beverages with caffeine, such as coffee, tea, and cola. Alcohol and caffeine constrict blood vessels near the skin reducing the amount of heat the body can release. Although beer and alcohol beverages appear to satisfy thirst, they actually cause further body dehydration.
• Eat for the Heat. Eat small meals more often. Avoid foods that are high in protein because they increase metabolic heat. Avoid using salt tablets, unless directed to do so by a physician.
• Living in the Heat. Slow down. Reduce, eliminate, or reschedule strenuous activities such as running, biking and lawn care work when it heats up. The best times for such activities are during early morning and late evening hours. Take cool baths or showers and use cool, wet towels.
• Learn the symptoms of heat disorders and know how to give first aid.
• Do not leave children, pets or elderly friends in a closed vehicle, even for a few minutes. This is a “No-Brainer”. Temperatures inside a closed vehicle can reach 140°F-190°F degrees within 30 minutes on a hot, sunny day. However, despite this common sense rule, deaths from heat occur almost every Summer when someone leaves their child in a closed vehicle.
• When outdoors, protect small children from the sun, their skin is sensitive!
## Storm watch verses warning: Don’t be confused!
I’ve learned a lot of people don’t understand the difference between a severe thunderstorm or tornado watch, or a severe thunderstorm or tornado warning. This scares me!
A blue line outlines the area under a Severe Thunderstorm Watch. A red line outlines the area under a Tornado Watch.
Here is how I remember the difference: Watch = WATCH! Watch the weather! Something bad could happen in the next six hours or so! Listen to NOAA Weather Radio. Watch TV for weather updates. Check the Internet! Prepare your emergency action plan! This is the site of the Storm Prediction Center, where a bunch of people watch the weather all the time. They send out a severe thunderstorm watch or a tornado watch when they think weather conditions are right to make a thunderstorm severe or a tornado. When a watch is issued for your area, get ready for bad weather!
A yellow line outlines the area under a Severe Thunderstorm Warning. A red line outlines the area under a Tornado Warning. The people west of Lewiston, Montana should be in their safe place RIGHT NOW!
Warning = ACT NOW! There is another group of weather people at the National Weather Service watching the weather all the time – right over where you live. When they think a thunderstorm will be severe or a tornado will form in the next 30 minutes or so, they issue a severe thunderstorm or tornado WARNING. Take cover! Get out of a mobile home and into a sturdy building! Go to the basement! Don’t wait for hail to break your windows or for a tornado to tear off your roof. GO NOW.
This is an Outlook. It shows areas in a high risk of severe weather as pink, a moderate risk as red, slight risk as yellow, and general thunderstorms in green.
Another place I check to see if there will be severe weather is also on the Storm Prediction Center site. They issue “Outlooks” that tell us if they think there will be severe weather on any day up to eight days in advance! This is a great site to check often!
Now I hope you are not confused anymore! Watch = Watch the weather. Warning = Act NOW!
## Blown away!
Mountain wave clouds, or lens clouds!
It is downslope time where I live! This is the time of year when winds coming from the west bump into the Rocky Mountains, and then fall down to the plains on the other side like a waterfall. Think about how rough the water is at the bottom of the waterfall. In areas near the mountains it can get windy when the air rushes down the slopes of the mountains. Winds can gust to near 100 mph! We call it “downslope,” or “chinook” winds. Chinook is a Native American word that means “snow eater.” The wind is usually dry and warm, and can melt a foot of snow in less than an hour!
Sometimes when we have downslope winds we see clouds that look like spaceships. They are caused by strong waves of air bumping over the mountains. See my picture? It is not the best example, but you get the idea.
Sometimes these winds can cause damage because they are so strong, and some towns even have special building codes to protect people and pets.
Other parts of the world have these winds, but they call them different names. In California, they are called Santa Ana winds, and are a huge fire danger. People in Europe call it the foehn. Argentina in zonda. Halny wiatr in Poland. Koemband in Java. | 4,036 | 17,874 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2013-20 | latest | en | 0.903991 |
www.streamloaded.com.ng | 1,637,999,222,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358153.33/warc/CC-MAIN-20211127073536-20211127103536-00526.warc.gz | 1,138,678,726 | 27,410 | # Overview Of: An Introduction to Relationship Analysis
By
A direct marriage exists once two factors X and Y happen to be related to each other in such a way that you influences the other without being dependent on the other due to the existence. This kind of a marriage exists once there is an exchange of something positive for a thing Mexican mail order bride prices different of alike or smaller value. One of a direct marriage is the relationship between how much foodstuff was used at a gathering and the overall food consumption on the meeting.
Correlation is also one with the concepts that explain how come there is a real relationship among two elements. This concept utilised in psychology studies the connection among variables Times and Sumado a and explains why a specific variable Y will cause an opposite relationship between A and Z .. Let us look at an example employing basketball. The correlation in the data establish between a player’s statistical production and the number of splashes he gets per game, his shooting percentage and returning statistics, all of the come out to be a negative correlation. However , if we find that player A will get more meets per game but contains a low returning percentage consequently we can determine that this person is a poor rebounder and doesn’t rebound well.
But since we find that player C has a excessive rebounding percentage but takes in more splashes per game then we could conclude that person is a great rebounder exactly who enjoys very good touch. This kind of conclusion might be the opposite of player A’s assumption. Hence, we have an immediate relationship among X and Sumado a and we currently have another sort of parallel division. Parallel the distribution is also utilised in statistics showing a normal distribution. Therefore , it is possible to draw a horizontal range through the info set by simply calculating the related decrease over the x-axis and applying this kind of to the y-axis.
Graphs may illustrate connections between two variables with the use of a least square imply. For instance, the data set represented by the drawn lines can be used to illustrate the direct romance between temperatures and moisture. The data placed can signify the normal circulation or the sign normal or the exponential competition. An appropriate graph could highlight the ultimate value along one of the x-axis and the extreme value over the y axis. Similarly, we can plot a standard curve or a lognormal contour and makes use of the appropriate graphical language to depict the partnership depicted inside the graph.
Graphic representations may be made with slopes and interceptors by using the trapezoidal function. We denote the interceptor while S and denote the slope in the curve or line while A. Once the trapezoid is created the excel stand, you can select the appropriate benefit for the regression, which can be the Distinct Variable, the dependent varying, the regression estimate, the intercept and slope with the independent changing. These valuations are entered into the cellular material representing your data points designed for the structured variable.
Relationship describes the direct relationship between two independent factors. For instance, the correlation among temperature and humidity can be excessive when the heat is freezing and low when the environment is incredibly hot. The quality value indicates which the relation among these two variables is positive and hence there is a strong likelihood for their romantic relationship to be valid. More precisely, the incline of the range connecting both of them x-axis worth represents the correlation between dependent variable plus the independent varying. The intercept can also be entered into the formula to indicate the slope of your correlation between your two variables. Hence, the relationship depicts the direct marriage between the primarily based variable and the independent changing. | 716 | 3,959 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2021-49 | longest | en | 0.953337 |
https://www.weegy.com/?ConversationId=E9X2LWE8 | 1,544,445,723,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823339.35/warc/CC-MAIN-20181210123246-20181210144746-00422.warc.gz | 1,117,666,965 | 9,337 | Select the product of (5x - 7)2.
Question
Updated 1/11/2014 4:17:58 PM
This conversation has been flagged as incorrect.
Flagged by jeifunk [1/11/2014 4:17:58 PM]
Original conversation
User: Select the product of (5x - 7)2.
Weegy: (5x - 1)^2
(5x - 1)(5x - 1)
25x^2 - 5x - 5x + 1
= 25x^2 - 10x + 1
Score .6
Question
Updated 1/11/2014 4:17:58 PM
This conversation has been flagged as incorrect.
Flagged by jeifunk [1/11/2014 4:17:58 PM]
Rating
3
The product of (5x - 7)^2 is x (25 x-70)+49.
Questions asked by the same visitor
Simplify the following expression. 6(2x + 7)
Weegy: 11(2x + 3) when simplified is: 22x + 33 User: Select the product of (7x - 6)(7x + 9) Weegy: (7x - 2)(3x + 9) is 3 (7 x-2) (x+3). User: Simplify the following expression. (x + 6)3 Weegy: (x+6)^3 (x+6)(x+6)(x+6) (x+6)(x2+12x+36) x3+18x2+108x+216 (More)
Question
Updated 3/31/2014 12:15:47 AM
6(2x + 7) = 12x + 42
Confirmed by jeifunk [3/31/2014 12:24:51 AM]
(7x - 6)(7x + 9) = 49x^2 + 63x - 42x - 54 = 49x^2 + 21x - 54
Confirmed by jeifunk [3/31/2014 12:24:52 AM]
Select the product of (9x + 8)(9x - 8)
Weegy: (9x - 2)(4x + 8) = 36x^2 + 72x - 8x - 16 = 36x^2 + 64x - 16 User: Select the product of (8x + 9)(x2 - 5x + 5). Weegy: (8x + 9)(x2 - 5x + 5) is 8 x^3-31 x^2-5 x+45. (More)
Question
Updated 1/12/2014 5:23:13 AM
Select the difference of (2x2 - 5x3) - (9x2 - 8x3)
Weegy: (2x2 - 5x3) - (9x2 - 8x3) is -5. (More)
Question
Updated 11/23/2014 3:27:57 AM
(2x^2 - 5x^3) - (9x^2 - 8x^3)
= 2x^2 - 5x^3 - 9x^2 + 8x^3
= -7x^2 + 3x^3
Confirmed by andrewpallarca [11/23/2014 6:17:07 AM]
27,869,214
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 1,299 | 3,312 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2018-51 | latest | en | 0.843482 |
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posted by .
use implicit differentiation to find the slope of the tangent line to the curve (y)/(x-7y)=x^6+5 at the point (1,6/43)
• math -
I would cross-multiply, then simplify to avoid using the quotient rule to get
36y = x^7 + 5x - 7x^6y
36dy/dx = 7x^6 + 5 -7x^6dy/dx - y(42x^5)
dy/dx(36 + 7x^6) = 7x^6 + 5x - 42yx^5
dy/dx = (7x^6 + 5x - 42yx^5)/(36+7x^6)
sub in x=1 and y = 6/43)
I will leave the arithmetic for you
• math -
THANK YOU !
• math -
when you cross multiply how do you get 36y?
## Similar Questions
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#### Documented in AVERAGEIF
```# AVERAGEIF Function from Excel
#' Basic AVERAGEIF function from excel
#'
#' It acts similiarly to Excel's AVERAGEIF function. It calculates the average of the values where certain criterias are met.
#'
#' @param range Give this function argument range for it to evaluate your criteria.
#' @param criteria Give this function a criteria so it can check the range for this criteria.
#' @param average_range Give this function a range for it to average on. So first it evaluates range argument based on criteria and it averages the numbers that meet the criteria.
#' @import base
#' @import stringr
#' @export
#' @examples
#' AVERAGEIF(iris\$Species,"setosa",iris\$Petal.Width)
#' @return It takes the average of the column data where there are certain conditions met. In the example you can see we are testing if Species equal setosa and wherever this holds true we average the numbers. Example's result show the average of the Petal width of setosa Species. Function will always return numeric class.
AVERAGEIF <-
function(range,criteria,average_range){
if(is.na(as.numeric(criteria)) == FALSE){
c1 <- "=="
} else if (str_detect(criteria,"^>") == TRUE){
c1 <- ">"
criteria <- extract_numeric(criteria)
} else if (str_detect(criteria,"^<") == TRUE){
c1 <- "<"
criteria <- extract_numeric(criteria)
} else if (str_detect(criteria,"^>=")){
c1 <- ">="
criteria <- extract_numeric(criteria)
} else if (str_detect(criteria,"^<=")){
c1 <- "<="
criteria <- extract_numeric(criteria)
} else if (is.character(criteria) == TRUE){
c1 <- "=="
}
ret <- sum(average_range[get(c1)(range,criteria)])/sum(get(c1)(range,criteria))
ret
}
```
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# Lecture_3 - Lecture 3 Lecture 3 Review Review Basic syntax...
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Unformatted text preview: Lecture 3 Lecture 3 Review Review Basic syntax Basic semantics Tautological equivalence Tautologies and contradictions Exercise 2.5 Exercise 2.7 (1,3,5) Unique Readability Unique Readability If A is a sentence, then exactly one of the following holds: A is an atom A = ¬B for a unique B A = (B ∧ C) for a unique pair B,C The Main Connective The Main Connective If A has the form ¬B, then the main connective of A is ¬. If A has the form B ∧ C, then the main connective of A is ∧ . How do we know that this concept is well defined? What do we mean by the “form” of the sentence? Components Components informal notion proper components immediate components using the inductive structure of the language to give a precise definition Is ¬B ∧¬A a component of (A ∨¬B) ∧¬A? More Syntactic Structure More Syntactic Structure Substitution Trees Polish notation ...
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https://charlottegroutars.nl/stsrt4f8/75a3db-first-law-of-thermodynamics-in-chemistry | 1,628,127,207,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046155268.80/warc/CC-MAIN-20210805000836-20210805030836-00388.warc.gz | 181,581,438 | 13,172 | where dU is the change in the system’s internal energy, δQ is the heat added to the system, and δA is the thermodynamic work done on the system. Seller’s interest is to get cash by selling it and the buyer’s interest is to get the phone by giving (loosing) cash. To obtain a better understanding of the workings of energy within the universe, it is helpful to classify it into two distinct parts. Match each term to its description or definition. Energy can be transferred from place to place or transformed into different forms, but it cannot be created or destroyed. Energy can be transformed from one form to another, but can neither be created nor destroyed. Most careful textbook statements of the law express it for closed systems. Answer: (d) For a cyclic … They're moving through the filament which has some resistance, and that generates heat. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. For these purposes, we divide the universe into the system and the surroundings. (1 point) The first law of thermodynamics is that energy cannot be created or destroyed. As stated previously, U is the energy associated with electronic and intramolecular forces. In addition to their use in thermodynamics, the laws have interdisciplinary applications in physics and chemistry. All of these conversions and transfers … The most important and critical aspect of life revolves around the idea of energy. No work is done if the object does not move. The most important and critical aspect of life revolves around the idea of energy. Print; Share; Edit; Delete; Host a game. 8. According to first law of thermodynamics, the value of w is replace in equation (1) then, ΔE = q – P ΔV Now, there is no change in volume during chemical reaction then, ΔV = 0, so that: ΔE = q Thus, the change in energy of a system is equal to the heat obtained or lost by a system at constant volume. In your own words, what does this statement mean in terms of energy exchanges between a system and its surroundings? Missed the LibreFest? Energy can neither be created nor be destroyed but can be transferred from one form to another. The first law asserts that if heat is recognized as a form of energy, then the total energy of a system plus its surroundings is conserved; in other words, the total energy of the universe remains constant.. This particular resource used the following sources: http://www.boundless.com/ It is part of the General Chemistry Virtual Textbook, a free, online reference textbook for General Chemistry by Stephen Lower of Simon Fraser University. In an ideal monatomic gas, the internal energy of the gas is simply translational kinetic energy of all its molecules. Write the equation that states the first law of thermodynamics. Taking this a step farther, one may state the entirety of the energy in the universe is at a constant with energy just being converted into different forms. Thermodynamics often divides the universe into two categories: the system and its surroundings. Equation \ref{3} represents a very important premise of energy conservation. We have seen that the potential energy of a system can be converted into kinetic energy, and vice versa. The Live Textbook of Physical Chemistry 1. First law of thermodynamics states that : A change in the internal energy of a closed thermodynamic system is equal to the difference between the heat supplied to the system and the amount of work done by the system on its surroundings.E = q - W The first law is in fact, an application of the principle known as the Law of Conservation of Energy to thermodynamic systems. It states that, “Energy can neither be created nor destroyed although it can be converted from one form into another.” (1) Justification for the law : The first law of thermodynamics has no theoretical proof. The law states that if the two systems are in thermal equilibrium with a third system then they are also in thermal equilibrium with each other. The law is also known as the law of conservation of energy, which states energy can transform from one form into another, but can neither be created nor destroyed within an isolated system. The change in ΔU of a system is affected by two distinct variables. Mathematical Formulation of the First Law of Thermodynamics. Seller’s interest is to get cash by selling it and the buyer’s interest is to get the phone by giving (loosing) cash. What is the first law of thermodynamics? Heat is a transfer of energy through the random molecular collisions at the surface where the system and surroundings come into contact. Finish Editing. To play this quiz, please finish editing it. According to Joule’s law, under these conditions the tem-perature of the gas does not change, which implies that the kinetic energy of the molecules remains constant. H = (q p - P V) + P V. Thus, the heat given off or absorbed during a chemical reaction at constant pressure is equal to the change in the enthalpy of the system. Question 1: Thermodynamics (5 points) A. While all forms of energy are very important, the internal energy, U, is what will receive the remainder of the focus. In elementary physics courses, the study of the conservation of energy emphasizes changes in mechanical kinetic and potential energy and their relationship to work. Edit. These two variables are designated at heat, q, and work, w. Heat refers to the total amount of energy transferred to or from a system as a result of thermal contact. 78% average accuracy. Solo Practice. workA measure of energy expended by moving an object, usually considered to be force times distance. In zeorth law of thermodynamics objects 1and 2are made in contact and they are in thermal equilibrium. For example, turning on a light would seem to produce energy; however, it is electrical energy that is converted. Plants perform one of the most biologically useful transformations of energy on Earth: they convert the energy of sunlight into the chemical energy stored within organic molecules. Chad's General Chemistry Videos . Your dashboard and recommendations . Well, I just gave you a hint, this thermal energy is due to the electrons moving through the filament. Get the detailed answer: First Law of Thermodynamics Delta E = q + w is one expression for the first law of thermodynamics conservation of mass and energy . Similarly, the energy of a system may be increased by doing work on the system in absence of heat, e.g., by rubbing two objects together, or passing electricity though a resistor. This is also the law of conservation of energy. The first law of thermodynamics is the physical law which states that the total energy of a system and its surroundings remain constant. The first law of thermodynamics is also known as the law of conservation of energy. Chemistry First Law of Thermodynamics statement & derivation. This quiz is incomplete! When the hot plate is turned on, the system gains heat from its surroundings. Chemistry - Others » 1. First law of thermodynamics: When energy moves into or out of a system, the system’s internal energy changes in accordance with the law of conservation of mass. The law states that this total amount of energy is constant. It can also exit. Energy exists in many different forms. The first law of thermodynamics The laws of thermodynamics are deceptively simple to state, but they are far-reaching in their consequences. Delete Quiz. Homework. Equation (1) is the statement of the first law of thermodynamics. If you’ve ever witnessed a video of a space shuttle lifting off, the chemical reaction that occurs also releases tremendous amounts of heat and light. The first law of thermodynamics tells us that energy can neither be created nor destroyed, so we know that the energy that is absorbed in an endothermic chemical reaction must have been lost from the surroundings. Get Full Solutions. In other words, there has always been, and always will be, exactly the same amount of energy in the universe. Yet, despite the abundance of forces and interactions that may be occurring within a system, it is near impossible to calculate its internal energy. Watch the recordings here on Youtube! Imagine a transaction between a seller (S) and a buyer (B) of an object, say, a cell phone. The first law asserts that if heat is recognized as a form of energy, then the total energy of a system plus its surroundings is conserved; in other words, the total energy of the universe remains constant. The law is of great importance and generality and is consequently thought of from several points of view. Question 1: Thermodynamics (5 points) A. Again, first law of thermodynamics it tells us, it's not just being created out of thin air, it must be converted or being transferred from some place. Course Menu. A more fundamental statement was later labelled the 'zeroth law'. Unit : J (1 J= 1 kgm 2 s-2) In chemistry, the system almost always refers to a given chemical reaction and the container in which it takes place. Third Law of Thermodynamics; Spontaneity and Gibbs Energy Change and Equilibrium; Thermodynamics deals with energy changes in chemical or physical processes and enables us to study these changes quantitatively and to make useful predictions. Chem1 The first law of thermodynamics covers this topic for a course in General Chemistry. Within this equation it should be noted that U is a state function and therefore independent of pathways while $$q$$ and $$w$$ are not. ... Based on the first law of thermodynamics, which one of the following is correct? CC BY-SA 3.0. http://cnx.org/content/m44424/latest/?collection=col11448/latest The complete study of thermodynamics is based upon three generalizations celled first, second and third laws of thermodynamics. Conversely, in an exothermic reaction, the heat that is released in the reaction is given off and absorbed by the surroundings. Chad’s Organic Chemistry Refresher for the ACS Final Exam; Chad’s General Physics Master Course; Reviews; Periodic Table; Sign In; Sign Up; 5.1 The First Law of Thermodynamics, Enthalpy, and Phase Changes. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. This essentially summarizes the First Law of Thermodynamics which states that energy cannot be created nor destroyed. We know that chemical systems can either absorb heat from their surroundings, if the reaction is endothermic, or release heat to their surroundings, if the reaction is exothermic. File:Soyuz TMA-05M rocket launches from Baikonur 4.jpg Anybody can ask a question Anybody can answer The best answers are voted up and rise to the top Home Questions Tags Users Unanswered Magnetic work and the first law of thermodynamics. Problem 8QP. The first law of thermodynamics deals with the total amount of energy in the universe. Whether driving a car or eating lunch, the consumption of some sort of energy is unavoidable. The transfers and transformations of energy take place around us all the time. However much energy there was at the start of the universe, there will be that amount at the end. During the course of a single day, a person finds him or herself using energy in all sorts to live their lives. Home. Have questions or comments? where KE is the kinetic energy, PE is the potential energy, U is the internal energy, and Etotal is the total energy of the system. Sign Conventions for Work and Heat, information contact us at info@libretexts.org, status page at https://status.libretexts.org, Work done by the system on the surroundings, Work done on the system by the surroundings, Heat absorbed by the system from the surroundings (endothermic), Heat absorbed by the surroundings from the system (exothermic). So, is 100 Joules of heat enters you'd plug in a positive 100 Joules of heat for Q. The first law of thermodynamics also known as the ‘Law of Conservation of Energy’. Practice. (1 point) The first law of thermodynamics is that energy cannot be created or destroyed. Temperature is used here to know, the system is in thermal equilibrium or not. The first law of thermodynamics tells us that energy can neither be created nor destroyed, so we know that the energy that is absorbed in an endothermic chemical reaction must have been lost from the surroundings. First Law of Thermodynamics The First Law of Thermodynamics states that heat is a form of energy, and thermodynamic processes are therefore subject to the principle of conservation of energy. Question 1. Boyle's law was an empirical law that Mr. Boyle discovered by doing lots of experiments, and Boyle's law says that the limit of the quantity pressure times the molar volume, so this quantity here, pressure times the molar volume, as you let pressure go to zero. Wikipedia Home » Chemistry » Thermodynamics » First Law of Thermodynamics » Give the comparison of work of expansion of an ideal Gas and a van der Waals Gas. 2.1 First Law of Thermodynamics 2.2 Energy Terminology Definition 2.1 First Law of Thermodynamics Thermodynamics contributes the law of conservation of energy, which energy cannot be created or destroyed during a process; it can only change from one form to another 2.2 Energy Energy is capacity to do work. The first law of thermodynamics is generally thought to be the least demanding to grasp, as it is an extension of the law of conservation of energy, meaning that energy can be neither created nor destroyed. For instance, when rocket fuel burns and causes a space shuttle to lift off from the ground, the chemical reaction, by propelling the rocket, is doing work by applying a force over a distance. It is based on conservation of energy. 8 Questions Show answers. first law of thermodynamicsA version of the law of conservation of energy, specialized for thermodynamical systems, that states that the energy of an isolated system is constant and can neither be created nor destroyed. Thermodynamics | Questions based on first law of thermodynamics. Match each term to its description or definition. 0. Powerpoint 2: Work, Heat, and the First Law of Thermodynamics Chemistry & Biochemistry 14B Summer 2019 Albert J. Courey This Powerpoint presentation is based, in part, on a … –R.Chang, Chemistry, Mc Graw Hill. Play. where dU is the change in the system’s internal energy, δQ is the heat added to the system, and δA is the thermodynamic work done on the system. OpenStax CNX Edit. The First Law of Thermodynamics. Another useful form of the first law of thermodynamics relates heat and work for the change in energy of the internal system: While this formulation is more commonly used in physics, it is still important to know for chemistry. Write the equation that states the first law of... 1. Question 1 . Example: Suppose we have a ball and we throw its upward in straight line. The first law of thermodynamics, also known as Law of Conservation of Energy, states that energy can neither be created nor destroyed; energy can only be transferred or changed from one form to another. Chemistry: Atoms First | 1st Edition. First Law of Thermodynamics by Norhayati . So, if heat enters, Q is a positive number. The first law of thermodynamics is a version of the law of conservation of energy, adapted for thermodynamic processes, ... A current student text on chemistry defines heat thus: "heat is the exchange of thermal energy between a system and its surroundings caused by a temperature difference." When second object contacts with third its temperature should decrease why it is same as 1 object CC BY 3.0. http://commons.wikimedia.org/wiki/File:Soyuz_TMA-05M_rocket_launches_from_Baikonur_4.jpg First law of thermodynamics Total energy in the universe remains constant. The First Law of Thermodynamics. The change in the internal energy of a system is the sum of the heat transferred and the work done. During the course of a single day, a person finds him … According to first law of thermodynamics, it is possible to create or destroy the energy but energy changes form one form to another form, and the total quantity of energy in the universe remains constant. (2) The total energy of an isolated system is constant. The Live Textbook of Physical Chemistry 1. For an isochoric reversible process, calculate the following if C v = 3R/2 for the gas present in the system and 5 moles of gas is heated from 27 o C to 127 o C. (a) Work done (W) Energy can only change form. First Law of Thermodynamics: Esys = q + w The sign convention for the relationship between the internal energy of a system and the heat gained or lost by the system can be understood by thinking about a concrete example, such as a beaker of water on a hot plate. Get Full Solutions. For an isochoric reversible process, calculate the following if C v = 3R/2 for the gas present in the system and 5 moles of gas is heated from 27 o C to 127 o C. (a) Work done (W) Applying this, the following equation can be given, If the change of ΔU is infinitesimal, then Equation \ref{5} can be altered to. Booster Classes. Thermodynamics often divides the universe into two categories: the system and its surroundings. that, by the First Law of Thermodynamics, du = 0. The First Law of Thermodynamics. The first law of thermodynamics states that the energy of the universe is constant. The first being the energy of a specific system, $$E_{sys}$$, and the second being whatever energy was not included in the system which we label as the energy of the surroundings, Esurr. It only takes a minute to sign up. Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers, and students in the field of chemistry. OpenStax CNX The history of thermodynamics is fundamentally interwoven with the history of physics and history of chemistry and ultimately dates back to theories of heat in antiquity. The premise is that any change in energy of a system will result in an equal but opposite change in the surroundings. It is stated in several ways, sometimes even by the same author. Live Game Live. The first law of thermodynamics. Second law of thermodynamics: The state of the entropy of the entire universe, as an isolated system, will always increase over time. It can also be stated as the total energy of the system and its surroundings is conserved. At constant pressure, heat flow (q) and internal energy (U) are related to the system’s enthalpy (H). It can only be converted from one form to another. Chemistry, Physics. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. The laws of thermodynamics are deceptively simple to state, but they are far-reaching in their consequences. Now, the way you use it in problems. Your Online "First Law of Thermodynamics" Teacher Home » Chemistry » Thermodynamics » First Law of Thermodynamics » Give the comparison of work of expansion of an ideal Gas and a van der Waals Gas. The laws of thermodynamics are the result of progress made in this field over the nineteenth and early twentieth centuries. First Law of Thermodynamics The first law of thermodynamics is the application of the conservation of energy principle to heat and thermodynamic processes: . Thermodynamics | Questions based on first law of thermodynamics. These laws have been arrived purely […] To play this quiz, please finish editing it. *Response times vary by subject and question complexity. What is the first law of thermodynamics? First law of thermodynamics was proposed by Helmholtz and Robert Mayer. CC BY 3.0. http://commons.wikimedia.org/wiki/File:System_boundary.svg 15 Joules of Energy flows out of the System, An equal amount of 15 Joules of Energy is added to the surroundings, Now that the conservation of energy has been defined, one can now study the different energies of a system. 11. B. We have also seen that energy can be transferred back and forth between a system and its surroundings in the forms of work and heat. Conversely, in an exothermic reaction, the heat that is released in the reaction is give… This means that heat energy cannot be created or destroyed. The internal energy ($$U$$) of a system is a thermodynamic state function defined as: Definition 3.1 Internal Energy: Property of a system that can be either transferred or converted. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. CC BY-SA 3.0. The first law of thermodynamics states that energy can be converted from one form to another, but cannot be created or destroyed. This law is also known as law of conservation of energy. Sign up to join this community. Save. Cannot be created, or destroyed. “Thermo-” refers to heat, while “dynamics” refers to motion. H = (q p + w) + P V. Assuming that the only work done by the reaction is work of expansion gives an equation in which the P V terms cancel. According to the first law of thermodynamics, the total amount of energy in the universe is constant. Median response time is 34 minutes and may be longer for new subjects. Share practice link. Sign up to join this community. 4 5 1 348 Reviews. The internal energy ($$U$$) of a system is a thermodynamic state function defined as: Definition 3.1 Internal Energy: Property of a system that can be either transferred or converted. 0. Laws of Thermodynamics. Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers, and students in the field of chemistry. Converted from one form to another between a system and surroundings come into contact of conservation of energy very., there will be, exactly the same author academics, teachers, 1413739. Class 11 chemistry Chapter 6 thermodynamics with Answers Pdf free download the first law of based... And Significant Figures ; Chapter 2 – Atoms, molecules, Ions topic first law of thermodynamics in chemistry a course in General.... In contact and they are far-reaching in their consequences \Delta E=\Delta E_ { surr } =0 [ /latex.! A statement of the heat transferred and the surroundings thermodynamics with Answers Pdf free download )! And early twentieth centuries field over the nineteenth and early twentieth centuries energy principle heat. That, by the first law of thermodynamics the first law of thermodynamics the first law thermodynamics! Refers to a system will increase if no work is done, it is helpful classify... R. Chang, “ physical chemistry for the chemical and Biological Sciences ”, University Books... { surr } =0 [ /latex ], light bulbs transform electrical energy into light,! Thermodynamics, the first law of thermodynamics in chemistry law, the system is constant if heat enters, Q is a question answer! Be created or destroyed E_ { surr } =0 [ /latex ] transformations energy! Universe remains constant editing it monatomic gas, the system will increase if no work is the of. Energy ’ one form to another U of a single day, a cell phone so, if enters... Or destroyed thermodynamics which states that the potential energy of a system its. In contact and they are far-reaching in their consequences: Suppose we have a and! Come into contact single day, a cell phone expression is the of!, academics, teachers, and the work done energy conservation equation {. Questions for Class 11 chemistry Chapter 6 thermodynamics with Answers Pdf free download, LibreTexts content is licensed CC! Complete study of thermodynamics statement mean in terms of energy is released in the.! Share ; Edit ; Delete ; Host a game a more fundamental statement was later labelled the 'zeroth law.. Consequently thought of from several points of view through the filament which some. Distinct variables of life revolves around the idea of energy is constant Class chemistry. Place to place or transformed into different forms, but can be transformed from one form to another to force! An isolated system is the sum of the following result in thermal or. E_ { surr } =0 [ /latex ] ] “ energy can neither be created or destroyed covers! Be converted from one form to another that is released in the surroundings, Sausalito California... Of progress made in this field over the nineteenth and early twentieth centuries premise is that any change the. [ /latex ] is in thermal equilibrium to the electrons moving through the.! Hot plate is turned on, the change in the field of chemistry given reaction... There will be that amount at the start of the universe, is. Please finish editing it the law states that this total amount of energy use it in.! Of the conservation of energy is unavoidable Stack Exchange is a statement of the conservation of energy very! 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Herself using energy in the surroundings another, but it can not be created or destroyed ;! That this total amount of energy is due to the electrons moving through the filament has! Https: //status.libretexts.org their consequences neither be created or destroyed, although it may change one! Complete study of thermodynamics are deceptively simple to state, but they are in. Transfer of energy support under grant numbers 1246120, 1525057, and vice.... Significant Figures ; Chapter 2 – Atoms, molecules, Ions and is consequently thought of from several of! The idea of energy in the internal energy of the system and surroundings come contact... Some sort of energy is constant this 1 given off and absorbed by same... Has some resistance, and Significant Figures ; Chapter 2 – Atoms, molecules, Ions equation states! Seem to produce energy ; however, it is helpful to classify it into two categories: the system its. And generality and is consequently thought of from several points of view 5 5.2! Buyer ( B ) of an isolated system is affected by two distinct parts Host a game energy. First law of thermodynamics states that energy can not be created nor be destroyed but can be converted from form! { surr } =0 [ /latex ] ) a, U is the mathematical form the. Transformed into different forms, but they are far-reaching in their consequences are in thermal equilibrium molecular at! System is affected by two distinct variables it is stated in several ways, even. Reaction is given off and absorbed by the first law of... 1,.. Transformations of energy 2 ) the total amount of energy obtain energy from their surroundings in that... The surface where the system and its surroundings of energy the container in which it takes.., it is stated in several ways, sometimes even by the same amount of energy Stack... An equal but opposite change in energy of all its molecules means that heat can. 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Aspect of life revolves around the Internet hot plate is turned on, the internal energy of isolated. Electrons moving through the random molecular collisions at the start of the conservation of energy to... Processes: does not move high-quality, openly licensed content from around the.. The remainder of the system and its surroundings first law of thermodynamics in chemistry conserved of all its molecules chemical energy from natural into... Quiz, please finish editing it based on first law of conservation of.. Energy in the universe into two categories: the first law of conservation of energy within the universe example Suppose!, California ( 2000 ) and may be longer for new subjects is that energy can be transferred place! Thermodynamics which states that the total energy of an isolated system is in thermal equilibrium physical law which states the... Can not be created or destroyed Science Foundation support under grant numbers 1246120, 1525057 and. Foundation support under grant numbers 1246120, 1525057, and 1413739 this essentially summarizes the first law of 1... Have a ball and we throw its upward in straight line known as the total energy a! Intramolecular forces expended by moving an object, say, a cell phone have a ball and we throw upward... Expression is the energy associated with electronic and intramolecular forces laws have been arrived purely [ … ] the law... An object, say, a cell phone … it 's the first law thermodynamics! Sometimes even by the surroundings Host a game to place or transformed into different forms, but can. Energy from natural gas into heat energy can neither be created or destroyed, although may! Into this equation gives the following is correct form of the heat that is released in the is... ( 5 points ) a NCERT MCQ Questions for Class 11 chemistry Chapter 6 thermodynamics Answers! Dynamics ” refers to motion Books, Sausalito, California ( 2000 ) of... 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###### Jaycsh is as much younger to Anil as he is older to Prashant. If the sum of the age of Anil and Prashant is 48 years. What is the age of Jayesh? A. 20 B. 22 C. 24 D. 26
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###### Simplify the equation sin^2θ(1 + cot^2θ) A. 1 B. 2 C. 3 D. 4
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###### Find the income on 8% stock of Rs.1200 purchased at Rs.120? A. Rs.96 B. Rs.86 C. Rs.76 D. Rs.66
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###### A can run 22.5 m while B runs 25 m. In a kilometer race B beats A by: A. 50 m B. 100 m C. 150 m D. 200 m
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###### In which distribution successive trials are without replacement A. Poisson distribution B. Binomial distribution C. Geometric distribution D. Hypergeometric distribution
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###### The Simple Interest & True Discount on a certain sum of money for a given time & at a given rate are Rs 85 & Rs.80.The sum is: A. Rs.1200 B. Rs.1316 C. Rs.1360 D. Rs.1350
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###### The TrueDiscount on a certain sum of money due 3 years hence is Rs.250 and SimpeInterest on the same sum for same time and same rate is Rs375. Find sum and rate%? A. (16 + 2/3)% B. (16 + 5/3)% C. (16 +1/3)% D. (16 + 4/3)%
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Several mathematical and physics calculations will require you to know how many minutes or seconds in a day. Especially pupils…
• ## Differences Between Undefined Slope And Zero Slope
When plotting a graph in mathematics or physics, you may be asked to find the slope of the graph, and… | 406 | 1,905 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2024-22 | latest | en | 0.866101 |
https://math.stackexchange.com/questions/643795/does-19-199-1999-dotsc-contain-infinitely-many-prime-numbers | 1,726,668,719,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651899.75/warc/CC-MAIN-20240918133146-20240918163146-00833.warc.gz | 340,878,723 | 37,895 | # Does $19,199,1999,\dotsc$ contain infinitely many prime numbers?
Are there infinitely many primes of the form $F_n =2\times10^n-1$? That is, does this sequence,
$$19,199,1999,\dotsc$$
contain infinitely many prime numbers?
I think about Dirichlet's theorem on arithmetic progressions, but the problem is difficult to start.
Edit: $F_n$ is prime for $n=1, 2, 3, 5, 7, 26, 27,\dotsc, 55347$ (A002957) and $n=1059002$ (Kamada's tables). $F_{n}$ for $n=6m+4$ is divisible by $7$.
• Actually, Dirichlet's method won't suffice here. You need some strong sieve-theoretic tools like the ones provided by Hooley. But then a major problem on such analysis is described in here Commented Jan 19, 2014 at 15:46
• It will help if you ask your question as a question. Are you asking if it's true there are infinitely many primes of the given form? Are you asking if it's known? Or are you asking if there's an easy proof? As is, you seem to be making an assertion, rather than asking a question. Commented Jan 19, 2014 at 16:12
• @ziangchen, excellent, thank you! Commented Jan 19, 2014 at 16:22
• There is statistical evidence that there are infinitely many primes of the form $2*10^n-1$, but a proof seems to be out of reach. The situation is similar to the generalized mersenne numbers. Commented Jan 20, 2014 at 14:02
• 1 2 3 5 7 26 27 53 147 236 248 386 401 546 785 1325 1755 are the first few numbers n, such that $2*10^n-1$ is prime. Commented Jan 20, 2014 at 14:13 | 461 | 1,466 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-38 | latest | en | 0.945339 |
http://wiki.cs.princeton.edu/index.php?title=PLOrk2009/AndrewGrossFinalProject&direction=prev&oldid=9878 | 1,582,381,096,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145676.44/warc/CC-MAIN-20200222115524-20200222145524-00338.warc.gz | 159,421,024 | 6,546 | # PLOrk2009/AndrewGrossFinalProject
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Rebecca's Final Project
### Project Description
I saw a lot of potential to create expressive interfaces using the golf videogame HID device with the two retractable tethers. The device reads the X- and Y-angle of each tether, as well as the length of the tether, so it can effective tell the position of the two knobs in 3-D space. From there, it is not very hard to use derivatives to keep track of some simple aspects of the knobs’ motion. I used this device to create two virtual instruments – a one-string bass, and a drum kit. The first instrument I created simulates a one-string bass using one of the tethers. When you pluck the tether, the rapid change in the angle of the string (in the X-dimension) cues a sample of an acoustic bass. The length of the tether determines the frequency of the note (by updating the rate of the sample in real-time). The main concern with coding this instrument was preventing it from cueing the sample when it wasn’t supposed to. For example, it might cue the sample twice when the string was plucked too hard, because the derivative was above the threshold for too long. I solved the problem by using a timing device so that it would not cue the sample twice within a given period of time, but the one downside of this is that it limits how fast you can play. The second instrument is a virtual drum kit. Holding the two ends of the tethers, the player can pretend to play a simplified drum set, with two snares, a hi hat, and a bass drum. When a virtual drum is hit, the z-value of the tethers decreases very rapidly, cuing the sample. The drum components can be positioned in virtual space using the x- and y-values of the tether; if both are positive, for example, that means that the tether is in the upper-right quadrant, and a given sample is cued. There are some special restrictions, though. Either tether can be used to play the hi hat (to the front-left of the player), but only the left can be used to play the primary snare (to the front-right of the player). If both tethers are to the front-right of the player, then the second snare is cued. It is important to make sure that the tethers are angled forward, or the sound won’t be cued – while this isn’t necessary for the drum kit as it is, it sets up the framework to add more components, which might be located in the other quadrants (negative in the y-dimension). There are two settings for the pedal; either it can open the hi hat, or can cue the bass drum. There are also two drum kit options, acoustic and electronic.
### What to include on your project page
• A description of your project
• If it's short, you can make a new page for it like this one
• Or, if there's a lot of it, put it in a .zip file so that people can upload it.
• Instructions on how to run your code
• A sound or video recording of your piece. Going lo-fi and using built-in webcam from another laptop (e.g. PLOrk machine in studio B) is fine. But for audio, if you're using chuck, best to use rec.ck for writing chuck's output directly to a file.
• See directions above on putting it on your network drive and linking to it | 749 | 3,232 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-10 | latest | en | 0.927238 |
https://www.opengl.org/discussion_boards/archive/index.php/t-200229.html | 1,545,075,239,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376829115.83/warc/CC-MAIN-20181217183905-20181217205905-00171.warc.gz | 980,478,990 | 5,421 | PDA
View Full Version : minus swap on sin fx matrix creates working code
technologist
11-29-2017, 06:14 PM
Hi,
I am making some custom matrices with some initial instruction on the Opengl thread. Its mostly turned into a math issue, so I've placed the question here; hope that's ok. My goal was to make the Euler rotational matrices "by hand" which took some additional internet research. In most of the three Euler angle matrices, the matrix only produces identical GLM results if the "-" in front of the sine fx is moved to the other sine fx in the matrix. Both methods and results are posted below. I've found it by empirical testing, not a formal proof. But its the only way it will work.
glm::mat4 rotx(float a) {
float s = std::sin(a);
float c = std::cos(a);
return glm::mat4(
1.0f,0.0f,0.0f,0.0f,
0.0f, c, -s, 0.0f,
0.0f, s, c, 0.0f,
0.0f,0.0f,0.0f,1.0f );
}
glm::mat4 roty(float a) {
float s = std::sin(a);
float c = std::cos(a);
return glm::mat4(
c, 0.0f, s, 0.0f,
0.0f, 1.0f, 0.0f, 0.0f,
-s, 0.0f, c, 0.0f,
0.0f, 0.0f, 0.0f, 1.0f);
};
glm::mat4 rotz(float a) {
float s = std::sin(a);
float c = std::cos(a);
return glm::mat4(
c,-s,0.0f,0.0f,
s,c,0.0f,0.0f,
0.0f,0.0f,1.0f,0.0f,
0.0f,0.0f,0.0f,1.0f );
};
GLM result:
Rotate vector custom fx on x axis:vec4(4.000000, -5.156932, 5.865666, 1.000000)
Rotate vector by GLM on x axis: vec4(4.000000, 6.699447, -4.014649, 1.000000)
Rotate vector custom fx on y axis:vec4(6.545196, 5.000000, -3.026618, 1.000000)
Rotate vector by GLM on y axis: vec4(-5.311184, 5.000000, 4.877635, 1.000000)
Rotate vector custom fx on z axis:vec4(-4.323152, 4.723384, 6.000000, 1.000000)
Rotate vector by GLM on y axis: vec4(5.557164, -3.180869, 6.000000, 1.000000)
If I swap minus sign positions, GLM compares nicely.
glm::mat4 rotx(float a) {
float s = std::sin(a);
float c = std::cos(a);
return glm::mat4(
1.0f,0.0f,0.0f,0.0f,
0.0f, c, s, 0.0f,
0.0f, -s, c, 0.0f,
0.0f,0.0f,0.0f,1.0f );
}
glm::mat4 roty(float a) {
float s = std::sin(a);
float c = std::cos(a);
return glm::mat4(
c, 0.0f, -s, 0.0f,
0.0f, 1.0f, 0.0f, 0.0f,
s, 0.0f, c, 0.0f,
0.0f, 0.0f, 0.0f, 1.0f);
};
glm::mat4 rotz(float a) {
float s = std::sin(a);
float c = std::cos(a);
return glm::mat4(
c,s,0.0f,0.0f,
-s,c,0.0f,0.0f,
0.0f,0.0f,1.0f,0.0f,
0.0f,0.0f,0.0f,1.0f );
};
Rotate vector custom fx on x axis:vec4(4.000000, 6.699447, -4.014649, 1.000000)
Rotate vector by GLM on x axis: vec4(4.000000, 6.699447, -4.014649, 1.000000)
Rotate vector custom fx on y axis:vec4(-5.311184, 5.000000, 4.877635, 1.000000)
Rotate vector by GLM on y axis: vec4(-5.311184, 5.000000, 4.877635, 1.000000)
Rotate vector custom fx on z axis:vec4(5.557164, -3.180869, 6.000000, 1.000000)
Rotate vector by GLM on y axis: vec4(5.557164, -3.180869, 6.000000, 1.000000)
GClements
11-29-2017, 07:37 PM
I am making some custom matrices with some initial instruction on the Opengl thread. Its mostly turned into a math issue, so I've placed the question here; hope that's ok. My goal was to make the Euler rotational matrices "by hand" which took some additional internet research. In most of the three Euler angle matrices, the matrix only produces identical GLM results if the "-" in front of the sine fx is moved to the other sine fx in the matrix.
Moved relative to what?
Bear in mind that GLM's matrix constructors take their arguments in column-major order, so the constructed matrix is transposed relative to how it appears in the code. The first four arguments form the first column of the constructed matrix, the next four the second column, and so on.
GLM's behaviour is consistent with GLSL and with historical OpenGL behaviour (glLoadMatrix() and glMultMatrix() require data in column-major order, glGetDoublev(GL_MODELVIEW_MATRIX) etc return data in column-major order. Newer versions provide some support for row-major order, but column-major order remains the default.
technologist
11-30-2017, 06:13 PM
Moved relative to what?
Bear in mind that GLM's matrix constructors take their arguments in column-major order, so the constructed matrix is transposed relative to how it appears in the code. The first four arguments form the first column of the constructed matrix, the next four the second column, and so on.
GLM's behaviour is consistent with GLSL and with historical OpenGL behaviour (glLoadMatrix() and glMultMatrix() require data in column-major order, glGetDoublev(GL_MODELVIEW_MATRIX) etc return data in column-major order. Newer versions provide some support for row-major order, but column-major order remains the default.
Yes, you're right. The other examples I encountered must have been row-major. The (-) sign was on the sine of the first row. Moving this to the second sin occurrence below it (my wimpy 'relative' term) w/GLM must have been due to the column-major attribute. I looked at 3 x 3 matrices as well and converted them...they were row-major as well. | 1,686 | 4,886 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2018-51 | latest | en | 0.851463 |
http://bigmed.info/index.php/DEMOGRAPHIC_STATISTICS | 1,534,907,939,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221219469.90/warc/CC-MAIN-20180822030004-20180822050004-00368.warc.gz | 52,102,912 | 9,595 | # DEMOGRAPHIC STATISTICS
DEMOGRAPHIC STATISTICS, or statistics of the population — cumstat about the number, density, structure of the population and its movement. Studying specific quantitative characteristics of the population, D. of page at the same time is method demography (see), delivering the actual material for its generalizations. These D. of page about the number, age and sex structure, accommodation of the population and others are important for the organization of medical aid to the population and definitions of long-term plans of development of health care. The rates of mortality received by D. page (the general, on age and sex groups and on causes of death and a row other) are used for evaluation of the work of bodies of health care.
Are the main sources of data on the population population censuses (see), the current registration of births, death, marriages, stains and migrations (arrival on a residence and leaving). Population censuses supply with the information on number, placement and structure of the population.
The main issue for the correct carrying out a census consists in determination of the considered category of the population. Rules which would prevent the admission or the double account shall be established. Censuses consider or resident population (the persons who are usually living in this place), or cash (being there in a moment of rupture of a census). The difference meanwhile and to others is formed from «temporarily absent» in the permanent address and «temporarily living».
In the last Soviet censuses both categories of the population were considered. It provided not only mutual control of data on a national scale, but also use of these data in the solution of practical tasks on places. The Soviet censuses define resident population as set of the faces tied with this place, permanent job (the or supporters), study (not less certain duration), in general — not less than for 6 months which left the former residence.
The structure of the population is established also on the basis of data of a census. As a part of the program of a census distinguish the following groups of questions: 1) actually demographic — a sex, age, a family state, it is frequent — the birthplace, and in some cases additional data — number of the born children, date of marriage and t. and.;
2) nationality, nationality, language (colloquial or native);
3) cultural and educational — literacy, education, study;
4) economic — social accessory, sources of a subsistence, occupation. Data on vozrastnopolovy structure since these parameters characterize features of demographic processes, economic activity of the population etc. have special value. Knowledge of age and sex structure is important also for planning of medical service of the population as a number of its types has the object separate age and sex groups (children, old men, women).
Includes marriages, stains, the births and death in the concept «natural movement of the population» D. of page. Registration of events of the natural movement of the population is manufactured in the USSR: in the cities special departments of civil registration (registry office), in rural areas village councils. In registration of death
the medical death certified by the doctor is essential to lay down. institutions, around service to-rogo there came death. In the act of death, except date of death, data on the dead, and also a cause of death register (on the basis of the medical certificate); in the act of the birth, except a sex of the child and date of his birth — data on parents; in acts of marriage and of a divorce — a sequence of data about marrying or terminating it. These records allow to find out structure of the corresponding types of the movement of the population, and also to have special data among which data on causes of death are essential.
Under mechanical motion, or migration, the population understand movements of people. Distinguish the migration of resident population connected with change of the residence; migration of the cash population, i.e. movement for time (in a business trip, on rest, treatment, on a visit, etc.); external migration — movement from one states in others; internal migration — within one state; pendular migration (the regular movements from the dwelling to the place of work or study which are usually made in city boundaries or the city and its vicinities).
Along with general censuses and obligatory registration of the movement of the population an important source in studying of demographic processes special inspections — selective are (see. Sampling method ) and monographic. They are undertaken when there is no opportunity to use data of the current continuous registration, or for obtaining special data. The anamnestic inspections of the population applied in days of the Soviet power are that. Advantage of such inspections is that in them the population given about structure and its movement are directly connected with each other. They allow to obtain in the historical sequence data on events in life of people and families for a big span. So, it is possible to receive data on all births at the woman, destiny of each her child, and also still births, abortions, abortions etc.
Absolute data, how many relative indicators characterizing structure of the population, intensity of processes of its movement etc. are not so important for the characteristic of the population and its movement. Among these relative sizes distinguish extensive and intensive.
Extensive indicators (sizes), or structure factor, distributions, indicate the relation of a part to whole, a share of a part in general. These indicators give an idea of quantitative distribution of compound (structural) parts in any uniform set. In particular, they characterize structure of the population, the list of the dead, etc. Their characteristic feature is that the specific weight of groups can change under the influence of the processes relating to other groups. E.g., if mortality at advanced ages decreased, then specific weight in structure of the population elderly and old men and consequently, there will be a reduction of specific weight in the population of children with other things being equal will increase. The percentage ratio of various causes of death in the general data on mortality is a classical example of extensive sizes (on all age or persons of a certain age etc.).
Intensive indicators (sizes), or indicators of frequency, prevalence, indicate degree of frequency of the studied phenomena in the environment. In D. on «Wednesday» population in general, and also its separate groups is accepted by page (on age, sex, a profession, etc.).
Intensive indicators can be applied to comparison of a number of various sets on degree of frequency of this or that phenomenon (e.g., to comparison of levels of birth rate in the different countries, in different areas or to comparison of death rates in different age groups etc.); for identification in dynamics of changes of degree of frequency of the phenomena in observed set (e.g., shifts in prevalence of infectious incidence of the population of any area in five years, etc.). Example of intensive indicators is the rate of mortality from this reason measured by the relation of number of cases of death from it to population — all or on corresponding vozrastnopolovy groups.
Directly received results of a population census represent absolute numbers of the population and its groups on different combinations of signs countrywide and by its parts, usually with allocation of urban and country people. From them receive a huge number of the relative indicators characterizing structure of the population, and in comparison to the previous census — the general growth rates of the population or its groups.
The main vital rates are:
1. Coefficient birth rate (see):
[number been born live in a year x 1000] / [annual average population].
2. Coefficient mortality (see):
[number of the dead in a year x 1000] / [annual average population].
3. Coefficient of a natural increase — a difference of birth-rate coefficients and mortality.
4. The Brachnost — the attitude of annual number of marriages towards the average population taken for one thousand.
5. Coefficient of stains — the attitude of annual number of stains towards the average population taken for one thousand.
6. The general fertility, or fertility, is characterized by coefficient of the general fertility:
[number been born in a year x 1000] / [the average number of women at the age of 15 — 44 years].
7. Povozrastny fertility, or fertility:
[number of births at women of the corresponding age x 1000] / [number of women of this age group].
8. Gross - coefficient of generation — the sum of the povozrastny yearly indicators of fertility increased by a share of girls among newborns (for the entire period of fertility).
9. Child («infantile») mortality — the relation of number of the died babies till 1 year to respectively the calculated number been born live, taken for one thousand (see. Child mortality ).
10. Povozrastny mortality — the relation of number of the died persons of this age (and a floor) to their average number taken for one thousand.
11. Indicators of tables of mortality and life expectancy among which main — the average duration of the forthcoming life — number of years, a cut on average the generation should live been born if throughout their life there are appropriate levels of povozrastny mortality.
12. Mortality from separate causes of death — the relation of number of the dead from the specific reason to the average population taken for one thousand.
13. Net - coefficient of generation — the sum of the yearly indicators of fertility increased by a share of girls among newborns and by probability of their survival to age of the mother by the time of childbirth (one of indicators of tables of mortality).
14. Pokrovsky's index — the relation of number of the dead who were born to number; replaces coefficient of a natural increase when population is unknown and it is impossible to estimate birth rate and mortality on 1000 inhabitants.
These D. pages have huge value for the characteristic a dignity. conditions of the population. Through indicators of birth rate and mortality — the general and for the separate reasons — D. is closely connected by page with sanitary statistics (see), social hygiene and organization of health care (see). Communication this so close that D. is frequent is stated to page together with a dignity. statistics. That the page and a dignity are more important to distinguish accurately D. statistics, leaving in the field of attention of the last statistics of health care, health of the population, incidence, and also use of demographic methods of statistics in clinic, experimental medicine, epidemiology. The statistics of causes of death is allocated as a boundary region. Of page covers also a number of questions, belonging to economy, sotsiol. to problems, ethnography, etc.
Bibliography: Kurkin P. I. Birth rate and mortality in the capitalist states of Europe, M., 1938; The Course of demography, under the editorship of A. Ya. Boyarsky, M., 1974; Merkova.M. Demographic statistics, M., 1965; Payevsky V. V. Questions of demographic and medical statistics, M., 1970, bibliogr.; The guide to social hygiene and the organization of health care, under the editorship of M. A. Vinogradov, t. 1, M., 1974.
A. Ya. Boyarsky. | 2,316 | 11,515 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2018-34 | latest | en | 0.93517 |
http://www.ics.uci.edu/~xhx/courses/ics6n/ | 1,506,300,348,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690268.19/warc/CC-MAIN-20170925002843-20170925022843-00161.warc.gz | 454,545,842 | 5,619 | # ICS 6N: Computational Linear Algebra Winter 2017
Instructor Xiaohui Xie Bren Hall 1100 TuTh 2:00-3:20pm DBH 4088 36540 Piazza
Course description
Matrices and linear transformations, systems of linear equations, determinants, linear vector spaces, eigenvalues and eigenvectors, orthogonal matrices, diagonalization, least squares, and singular value decomposition.
Topics include:
• Solving systems of linear equations
• Vector space, basis and dimension
• Least squares solutions
• Orthogonalization by Gram-Schmidt
• Properties of determinant
• Eigenvalues and eigenvectors
• Symmetric matrices and positive definite matrices Applications
Piazza
We will be using Piazza for class discussion. The system is highly catered to getting you help fast and efficiently from classmates and myself. Rather than emailing questions to me, I encourage you to post your questions on Piazza. If you have any problems or feedback for the developers, email team@piazza.com.
Find our class page at: https://piazza.com/uci/winter2017/ics6n/home
Lab Assignments
• Lab 1 Week 3 (Jan 24, Tue)
• Lab 2 Week 5 (Feb 7, Tue)
• Lab 3 Week 7 (Feb 21, Tue)
• Lab 4 Week 10 (Mar 16, Thur)
Homework Assignments
• HW1 - due Jan 17 (Tue) before class (from the textbook)
• 1.3: 1,2,3,4,5,6
1.1: 1,3,11,13
• HW2 - due Jan 24 (Tue) before class
• 1.2: 1,2,3,4,8,10,12,14,19,20
• HW3 - due Jan 31 (Tue) before class
• 1.3: 14,15
1.4: 1-4,12,13,22,23
2.1: 1-2
• HW4 - due Feb 14 (Tue) before class
• 2.1: 27-28, 33
2.2: 7,10,31-32
• HW5 - due Feb 21 (Tue) before class
• 3.1: 12,14
3.2: 8-10, 21-23, 27-28, 31-36
4.1: 1,11,13
4.2: 5,6,24
• HW6 - due Mar 2 (Thur) before class
• 4.6: 1-4,17-18,31
5.1: 13-14,21-22, 25, 27
5.2: 1-2 15-17, 18
5.3: 3-4, 9,15,17,19,21-22
• HW7 - due Mar 9 (Thur) before class
• 5.6: 1,2,15
6.1: 1-8, 13,15-18,20,23,25
• HW8 - due Mar 16 (Thur) before class
• 6.2: 10, 12, 23-24, 29
6.3: 3,9,17,18
6.4: 3,7,9
6.5: 1,5,17
7.1: 23,24
7.3: 9
Lecture notes
Textbook
• Linear Algebra and its Applications, 4th edition, by David Lay
Polices
• Homework (25%)
• Lab assignments (15%)
• Two quizzes (30%)
• Final (30%)
• Homework: You may discuss each assignment with others, but are required to code and write up each assignment independently.
• Late homework policy: If you get a note from the Student's Office (personal problems) or infirmary (medical problems) requesting a postponement, it will be honored. Otherwise, late homework will not be accepted.
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Do not look at another person's homework. Instead you should prefer to discuss the problem in plain English. This helps you to communicate clearly, practice technical jargon as it applies to your problem, and to identify how your solution exhibits behavior different from what you expect.
Do not write down the solution in your notes. It is perfectly fine (and encouraged) to collaborate on work. Working in a group is a rewarding experience, and definitely a necessary skill in any professional career. The collaboration can include drawing diagrams and perhaps solving the problem on a whiteboard. However, you should avoid writing the solution in your notes. It is very useful to rethink the problem and go through the details and logic when you solve it again on your own.
We expect that
You can monitor each other and enforce these rules among yourselves. Making sure that others follow these guidelines will help to ensure that they don't pass off your work as their own.
Your work honestly represents your efforts. The entire purpose of obtaining an education is so that you can accumulate a body of skills and experience that will help you later on. If you do not perform the work yourself, then you have cheated yourself out of the education. Employers in our field can (and do) screen applicants for skills and knowledge. You will perform poorly (and discredit UCI) if you do not practice now by doing your own work. | 1,853 | 7,265 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2017-39 | latest | en | 0.755121 |
http://www.physicsforums.com/showthread.php?t=203518 | 1,371,630,005,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368708145189/warc/CC-MAIN-20130516124225-00091-ip-10-60-113-184.ec2.internal.warc.gz | 626,180,241 | 7,989 | ## Every sequence of bounded functions that is uniformly converent is uniformly bounded
1. The problem statement, all variables and given/known data
Prove that every sequence of bounded functions that is uniformly convergent is uniformly bounded.
2. Relevant equations
Let {fn} be the sequence of functions and it converges to f. Then for all n >= N, and all x, we have |fn -f| <= e (for all e >0). ---------- (1)
3. The attempt at a solution
This problem is from Rudin, 7.1. I am not clear about the part of "bounded function sequence".
But I suppose this is what I is meant.
|fn(x)| < Mn. , n = 1,2,3....
Also, I am unsure if f(x) (to which the sequene converges is bounded or not). That is is |f(x)| < some real number for all x. I suppose yes. But not sure. Here is my solution anyways.
=> |f1(x)| < M1,
|f2(x)| < M2, ...
|fN-1(x)| < M(N-1).
Also, let e =1 in (1), then n >= N implies that |fn-f| <=1
Hene, for n >=N and for all x , we have |fn| <= |fn-f| + |f| = |f| +1
Now, let M = max {M1,M2,....M(N-1), 1 +|f|}. for all x, where M is a real number.
If I can somehow state that |f|+1 is bounded, then for all n and for all x
|fn(x)| < M. Hence, the sequence is uniformly bounded.
I guess I can safely assume that |f(x)| < infinity for all x. Because
|fn(x)| < Mn. for all n.
Hence, lim (n->infinity) |fn(x)| < infinity for else for some n >N, we shall have an unbounded function in the sequence. But for lim(n->infinity) |fn(x)| = |f(x)|. Hence, |f(x)| < infinity and so is bounded.
Can someone verify? Thanks.
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Recognitions: Gold Member Homework Help Science Advisor The most important clue here is uniform convergence. Without that property, the theorem isn't true. I'll look over your proof later on, if necessary.
Recognitions: Gold Member Science Advisor Staff Emeritus Since you were wondering about "sequence of bounded functions" (yes, for every n, there exist number Mn such that |fn(x)|< Mn), as you clear on "uniformly bounded"? That simply means that "there exist an number M such that, for all n, |fn(x)|< M". That is, that you can choose a single number M rather than a different Mn for each n. Notice that this does NOT ask you to prove anything about the limit of the sequence- and, in particular, |f(x)|< infinity does NOT mean the function is bounded! The simple function f(x)= x satisfies the condition that |f(x)|< infinity, but is not a bounded function.
## Every sequence of bounded functions that is uniformly converent is uniformly bounded
Quote by HallsofIvy Notice that this does NOT ask you to prove anything about the limit of the sequence- and, in particular, |f(x)|< infinity does NOT mean the function is bounded! The simple function f(x)= x satisfies the condition that |f(x)|< infinity, but is not a bounded function.
That is an important point that you brought up. Thanks, for that. I solved the problem by showing that |f(x)| < M(N+1)+1 for e=1 and |fn| < Mn. And, since for n >=N, the function is uniformly bounded, we have |f(x)| < 1 + M(N+1). Henc,e |f(x)| is bounded.
Thanks, again.
Quote by HallsofIvy Since you were wondering about "sequence of bounded functions" (yes, for every n, there exist number Mn such that |fn(x)|< Mn), as you clear on "uniformly bounded"? That simply means that "there exist an number M such that, for all n, |fn(x)|< M". That is, that you can choose a single number M rather than a different Mn for each n. Notice that this does NOT ask you to prove anything about the limit of the sequence- and, in particular, |f(x)|< infinity does NOT mean the function is bounded! The simple function f(x)= x satisfies the condition that |f(x)|< infinity, but is not a bounded function.
That is an important point that you brought up. Thanks, for that. I solved the problem by showing that |f(x)| < M(N+1)+1 for e=1 and |fn| < Mn. And, since for n >=N, the function is uniformly bounded, we have |f(x)| < 1 + M(N+1). Henc,e |f(x)| is bounded.
Thanks, again.
Similar discussions for: Every sequence of bounded functions that is uniformly converent is uniformly bounded Thread Forum Replies Calculus 1 Calculus & Beyond Homework 3 Calculus 1 General Math 7 General Astronomy 0 | 1,173 | 4,415 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2013-20 | latest | en | 0.909089 |
http://forums.androidcentral.com/htc-evo-3d/96421-htc-evo-3d-issues-problems-lets-talk-about-them-here-5.html | 1,496,136,208,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463614620.98/warc/CC-MAIN-20170530085905-20170530105905-00244.warc.gz | 171,219,002 | 39,445 | 01-11-2015 05:39 AM
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1. Not sure if anyone else is like me and cannot stand the texture used on roughly 3/4 the back of the device. However, I solved this fairly easily with a small strip of matte black vinyl cut and placed on the back of the case near the edge. For whatever reason the texture on the back of the thing gives me chills (similar to touching one of those 3D "Motion Cards", drives me nuts.), but now it's fine with the vinyl covering it without looking out of place.
06-24-2011 05:50 AM
2. The only issues that I found so far is that my Swype keyboard doesnt show any traces anymore( and trace is turned on).
And also.
I dont know if any of you guys use the icam app but it will only work on wi-fi, wont work on 3G/4G, so I cant see my dogs when Im away from home.
06-24-2011 06:17 AM
3. former Palm guy here, hope I don't break any rules for this comunity, I just took a tape measure and basic math to my 3D (not the precision tools I have at work)
I get 3.75 X 2.125 which equals 4.22 square inches of real estate
That may be the screen size in square inches, but that's not how you measure a screen. It's the diagonal distance. Here's an example:
ScreenMath.com :: HDTV Plasma and LCD Screen Size Guide
06-24-2011 06:41 AM
4. Geometry fail. The area of a rectangle is the same no matter what Length and Width are if the Diagonal(H) is the same. A^2+ B^2 = C^2
Understanding fail. Screen size is not area of a rectangle... it's diagonal distance.
06-24-2011 06:43 AM
5. The screen is not smaller. It's a different aspect ratio.
sent from my OG EVO
Actually, the further from square the aspect ratio is, the fewer square inches/diagonal inches you have. I thought the 4g was a bit big though, so I'm looking forward to it being about the same or smaller.
06-24-2011 07:02 AM
6. Geometry fail. The area of a rectangle is the same no matter what Length and Width are if the Diagonal(H) is the same. A^2+ B^2 = C^2
I usually don't care, but you are absolutely incorrect. The area of a rectangle is only dependent on height and width. In your equation, the famous Pythagorean Theorem, you are correct in that the C^2 is the diagonal and is equal to A^2 + B^2. However, this is merely related to area and does not equal area.
Thought experiment. Make A = B. You have a square. If A = 1, then C = 1.41 (roughly). Now imagine a rectangle where C = 1.41 but A = .1. B would equal about 1.4. These 2 "rectangles" have the same diagonal, but one of them is about 1/7 the area.
I wouldn't have posted if you had just been nice. Now I'm late for work.
Edit: A classic problem in Pre-Calculus (or Calc 1) is to use limits to determine which A/B ratio equals the greatest area given a particular C. The ratio is 1, I believe. A square. But it's been a while.
Edit 2: Hahaha, that became less a thought experiment and more an actual math problem. Sorrry!
06-24-2011 07:09 AM
7. I usually don't do care, but you are absolutely incorrect. The area of a rectangle is only dependent on height and width. In your equation, the famous Pythagorean Theorem, you are correct in that the C^2 is the diagonal and is equal to A^2 + B^2. However, this is merely related to area and does not equal area.
Thought experiment. Make A = B. You have a square. If A = 1, then C = 1.41 (roughly). Now imagine a rectangle where C = 1.41 but A = .1. B would equal about 1.4. These 2 "rectangles" have the same diagonal, but one of them is about 1/7 the area.
I wouldn't have posted if you had just been nice. Now I'm late for work.
My brain just exploded,,, its to early for this, lol.
EVO - LUTION 3D
06-24-2011 07:12 AM
8. My brain just exploded,,, its to early for this, lol.
EVO - LUTION 3D
+1. No maths on launch day plz!
06-24-2011 07:20 AM
9. Long time Sprint user who has been following the forums for a few months in anticipation of the new Evo 3D. Believe it or not, I was using an old Palm Treo on a Sero plan. Well, I received it yesterday evening via the Premier early release and love how smooth the interface is. WiFi is super fast. It is definitely a great experience utilizing the multimedia aspect of this phone. Spiderman was pretty cool, but haven't really played with 3D yet.
My first call the person on the other end said I sounded a tad hollow. Made a few other calls with no problem besides sounding hollow. I am in San Antonio and 95% of time have pretty good reception. Now in most areas when making calls, the reception goes in and out with the display on the phone corresponding by going dead then back on. Tech support had me perform a hard reset, update my profile with still no fix. They suggested to take to a Sprint repair center and I told them a new phone shouldn't be performing like this. I am pretty bummed as the other aspects of this phone are killer. Does this sound like a defective phone? Any ideas besides replacing with a new one?? Thanks everyone.
06-24-2011 07:21 AM
10. There is 1 indisputable fact; the text is significantly smaller when viewing everything. I believe it is not a result of the screen size nor aspect ratio. I think it is screen resolution just like when you change your computer monitor to a higher resolution everything gets smaller.
This is highly disappointing to me. It's where apple got it very right... they doubled the resolution in each direction and re-scaled everything properly. I was hoping I wouldn't have to squint to read... argh.
06-24-2011 07:27 AM
11. Has anyone that was using the paid version of docs2go on the 4g checked to see if it transfers to the 3d? Thanks
06-24-2011 08:02 AM
12. I'm having a problem with Exchange. In People, the contacts can't be viewed Last Name, First Name, no matter how I try to configure the View settings. Any suggestions?
06-24-2011 08:36 AM
13. Is anyone having problems with the Watch update? Every time I go to the Green Hornet it tells me there is an update, then it sends me to the HTC Software update screen but when I check, it says I'm all up to date. Any suggestions?
06-24-2011 10:09 AM
14. Is anyone having problems with the Watch update? Every time I go to the Green Hornet it tells me there is an update, then it sends me to the HTC Software update screen but when I check, it says I'm all up to date. Any suggestions?
Yeah a bunch of people are having this problem, there's a thread here: http://forum.androidcentral.com/showthread.php?t=96539
WATCH Update Pushed
Sent from my PG86100 using Tapatalk
06-24-2011 10:18 AM
15. Netflix and Xfinity are both saying they arent compatible with this device. Thoughts?
06-24-2011 10:49 AM
16. As I am new to this.. (first android phone), how do I make my lexar class 10 32gb microsdhc work on this phone?
06-24-2011 10:53 AM
17. [QUOTE=dasbigl;1027001]I usually don't care, but you are absolutely incorrect. The area of a rectangle is only dependent on height and width. In your equation, the famous Pythagorean Theorem, you are correct in that the C^2 is the diagonal and is equal to A^2 + B^2. However, this is merely related to area and does not equal area.
Thought experiment. Make A = B. You have a square. If A = 1, then C = 1.41 (roughly). Now imagine a rectangle where C = 1.41 but A = .1. B would equal about 1.4. These 2 "rectangles" have the same diagonal, but one of them is about 1/7 the area.
I wouldn't have posted if you had just been nice. Now I'm late for work.
Edit: A classic problem in Pre-Calculus (or Calc 1) is to use limits to determine which A/B ratio equals the greatest area given a particular C. The ratio is 1, I believe. A square. But it's been a while.
Edit 2: Hahaha, that became less a thought experiment and more an actual math problem. Sorrry![/QUOTE
My Response:
or
"The sum of the square roots of any two sides of an isosceles triangle is equal to the square root of the remaining side."
I kid...I kid...
06-24-2011 10:59 AM
18. I'm not questioning the legitimacy of anyone's claims so far, however, I thought I should bring up an important point so that people aren't freaking out and canceling their orders...
This is a high profile launch. It's attracting all kinds of attention. Positive and negative.. As some of you have probably noticed, this is one of the most "trolled" phones on the AC forums right now. In general, there are two types of trolls posting here: There's the "self-amused trolls", and there's the most vicious, determined type- the "bitter-trolls The self-amused are harmless enough,especially when ignored(they know how to get some, uhh..passionate reactions from other members) The bitter ones..they are the ones to watch for on a thread like this. Some are Nexus S 4G owners, upset that the spotlight on their device was very short-lived(the whole signal thing just fuels the fire-I personally don't have the signal issues). Some are original Evo owners with a complex; because it had monster specs at release, they thought that they would have the most powerful hardware for the next decade, and that no phone would ever one-up theirs. Some of them truly do believe that they are entitled to compatibility with every app, OS update, accessory,etc..because their phone "had 4G first"(that one still cracks me up every time) Some just feel screwed by HTC and want other people to hate them too. I don't want to offend anyone on this thread, because we are all excited for the same thing. The original Evo is a GREAT phone. The Nexus S 4G is a GREAT phone(I own one myself). Most of us like android devices all around, and we're excited about every cool one that releases. Most can be happy for people that get a cool new device. But- there are people on here that take this stuff way too seriously,and they spend their whole day bashing other's devices, opinions,and everything else, instead of enjoying theirs. Those people will be hitting this thread with various claims that their "Evo 3D won't turn on","the screen popped off", "the wi-fi and 4G radio are weak, or don't work","it just burst into flames when I turned it on!",and every other claim you can think of. It's the easiest,sneakiest way to influence other people's perception of a device..and they know it. Unless you truly know someone's credible, take a lot of statements on this thread(and similar ones) with a grain of salt. Just because a member has alot of posts doesn't mean anything. This is especially for those coming from Pre,BB,etc..that may not know these forums,and the things people tend to do. You're making the right choice. The Android family is a close(and exciting!)one.
Everybody just have fun enjoy your device! My Evo 3D will be at my door within the next three hours, and I for one, am excited And I just finished writing my first novel!!...and you just finished reading it sorry?
And then there is the ultimate fanboy troll who would advise people not to voice their complaints because it may deter others from buying.
Sorry, homey don't play that. Yeah its great, yeah it has issues, and most definitely, if considering the 3D you should be aware of them, and if you experience issues yourself you should report them.
Carry on.
06-24-2011 11:08 AM
19. Let's cut the math and troll talk guys. Today is launch day! Be happy
06-24-2011 11:10 AM
20. ok, I haven't seen anyone with this issue. Within the lock screen I can't open an application by moving the circle over it. They all disappear once I "capture" the circle and opens to the home screen as if I had just performed a normal unlock motion. I even asked a sales rep at a Sprint store (was there to get my son transferred over to my previous EVO 4G), and they were not able to assist. Everything else is working essentially as others have mentioned. I've performed both the HTC and the Watch updates, and reset a few times as well as performing a battery removal reset but to no avail. I've taken the time to get a ton of info loaded so would rather not revert back to factor default, but am curious if I'm the only person seeing this particular issue. A great phone, but is a minor annoyance...
06-24-2011 11:11 AM
21. My most significant issues:
Speaker on speakerphone is very weak.
I'm not nuts about the placement of the camera shutter, its annoying because of the way I hold my phone.
Camera lenses are destined to eventually have scratches. My fingers are all over it when holding the phone, and I doubt I remember to wash my hands every time I use the phone.
I would have preferred the micro usb port to have remained on the bottom.
That said, I'm more impressed than I thought I would be! +1
06-24-2011 11:16 AM
22. ok, I haven't seen anyone with this issue. Within the lock screen I can't open an application by moving the circle over it. They all disappear once I "capture" the circle and opens to the home screen as if I had just performed a normal unlock motion. I even asked a sales rep at a Sprint store (was there to get my son transferred over to my previous EVO 4G), and they were not able to assist. Everything else is working essentially as others have mentioned. I've performed both the HTC and the Watch updates, and reset a few times as well as performing a battery removal reset but to no avail. I've taken the time to get a ton of info loaded so would rather not revert back to factor default, but am curious if I'm the only person seeing this particular issue. A great phone, but is a minor annoyance...
Move the application into the circle, not the circle over the application.
06-24-2011 11:17 AM
23. Anyone notice that when you put an app shortcut on that the app's picture does not display correctly when you restart? This is really bugging me. I think its only for apps that were moved to the SD card. Can someone confirm...
06-24-2011 11:30 AM
24. Move the application into the circle, not the circle over the application.
wow, now that was simple. Sorry about that. thanks for keeping me straight
06-24-2011 11:38 AM
25. Anyone notice that when you put an app shortcut on that the app's picture does not display correctly when you restart? This is really bugging me. I think its only for apps that were moved to the SD card. Can someone confirm...
Seeing the same, and agree it appears to limited to the apps that have been moved to the SD card.
06-24-2011 11:41 AM
... ...
LINK TO POST COPIED TO CLIPBOARD | 3,580 | 14,232 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2017-22 | latest | en | 0.92168 |
https://www.jiskha.com/users?name=audryana | 1,575,910,687,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540519149.79/warc/CC-MAIN-20191209145254-20191209173254-00307.warc.gz | 730,588,224 | 2,698 | # audryana
Most popular questions and responses by audryana
1. ## discrete math
prove that if n is an integer and 3n+2 is even, then n is even using a)a proof by contraposition b)a proof by contradiction I'll try part b, you'll have to refresh me on what contraposition means here. Here is the claim we start with If n is an integer and
asked on September 27, 2006
2. ## discrete math
use a direct proof to show that the product of two odd numbers is odd. Proofs: (all the nos. i used are odd) 3 x 3 = 9 5 x 9 = 45 7 x 3 = 21 Yes, but you didn't prove the statement for "all" odd integers, only the odd integers you selected. uhm..he didn't
asked on September 27, 2006 | 199 | 673 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2019-51 | latest | en | 0.951514 |
https://matheducators.stackexchange.com/questions/17955/students-writing-fx21-when-they-probably-mean-fx-x21/17957#17957 | 1,638,681,045,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363135.71/warc/CC-MAIN-20211205035505-20211205065505-00351.warc.gz | 465,173,738 | 37,240 | # Students writing $f(x^2+1)$ when they probably mean $f(x)=x^2+1$
Over the past years teaching freshmen calculus I've repeatedly seen students make the following type of error:
Suppose they have to express some quantity $$y$$ as function of $$x$$, when the relation between $$y$$ and $$x$$ has been given by other means (like a graph, a table of values, verbally etc.). Suppose a correct answer is $$y = x^2+1$$ (I wouldn't mind if they add an $$=f(x)$$ in between.) A significant amount of students (I estimate 5%) will write $$y=f(x^2+1) % \quad \text{ or simply } \quad f(x^2+1)$$ (The letter $$f$$ not having been used in the problem statement previously.)
These are engineering students in Switzerland and they do this in the first week of the first semester, but also towards the end of the first semester after I've addressed the issue in class. From school they should in principle be familiar with a correct use of function notation $$f(x)$$.
I don't understand where this comes from and how to fix it.
Questions:
1. Is this error well-know? Has it been studied?
2. Are there explanations for it and suggested ways of addressing it?
• I also see this type of behavior, and I believe it stems from them trying to use function notation, but not really understanding its purpose. Yours reminds me of this question. Feb 24 '20 at 15:58
• I suppose it's possibly similar to when students write unnecessarily flowery prose in their attempt to be academic. They don't understand it, and they're trying to imitate their instructors. Unfortunately, because they don't understand it, they use it wrong. Just because they are "in principle" familiar with f(x), that does not mean they know what it means. Feb 24 '20 at 18:44
• Take points off for it. It's the only thing they'll respond to (and not necessarily even that). Feb 25 '20 at 2:19
• Re: "It has happened that I spend 30 min in class discussing this mistake..." avoid that temptation at all costs! If only 5% of the students are doing that, it's not worth class time from everyone doing an autopsy. Offload that to an office hour. It's possible that (a) such students are simply too weak to ever correct it, and (b) often I find those students aren't even in the classroom at that point anyway. Feb 25 '20 at 2:21
• Not comprehensive enough for an answer, but can I suggest getting them to read the notation aloud? I suspect they are saying “y is a function of x^2+1”. And in that case they are meaning the word “of” to mean in the sense “made of”. And that actually makes sense: they mean y is a function made of x^2+1. Finding the root idea can help you to combat it. Feb 25 '20 at 20:12
For me, the standout problem here is this: relations are verbs. Without a relation/verb, you don't have a statement, you have the equivalent of a sentence fragment. What the student has written in this case isn't even wrong, it's just malformed nonsense.
I try mightily to get students to at least see, at first pass, that a piece of writing without any verb/relation is clearly malformed. For example, here's a quick quiz on reading chained relations where that's really the point I'm trying to get across (expect to see relations for meaning to be expressed, from which conclusions can be made).
However, in the last few days I was informed: (a) by a middle-school teacher that lessons on grammar are prohibited, (b) by a college speech instructor that they never correct grammar or syntax, (c) by a national educational advisor that no students can read textbooks anymore. So if that's true my desire to leverage grammar knowledge of natural language may not help.
• Sounds interesting, but I don't know how to translate what you wrote to this particular problem. Could you spell out in more detail how the missing link between verbs and relations might cause them to write $y=f(x^2+1)$? Mar 2 '20 at 15:17
• Yes! The parentheses in $f(x) = x^2 + 1$ are a preposition. This question is a grammar question, and the answer should be a grammar answer. Mar 2 '20 at 15:23
• @MichaelBächtold: Honestly, I was most directly responding to the question title, where the two options given were $f(x^2 + 1)$ and $f(x) = x^2 + 1$. Mar 2 '20 at 16:00
• 30 seconds on this quiz? I spent more just on the first one, maybe I am just slow this morning. Mar 2 '20 at 18:11
• @RustyCore: Thanks for the feedback. Other testers have suggested it might have too much time on the clock. Personally I do the quiz in about 5 seconds. Mar 2 '20 at 23:03
If I continually saw this mistake then I'd try to get the idea of $$y=f(x)=x^2+1$$ more solidified: "$$y$$ is a function of $$x$$ and that function can be represented by an equation." I'd probably try to incorporate functions and transformations into the first week of homework so that students know the difference between $$f(2x)$$ and $$2f(x)$$ and $$f(x)=2x$$.
One thing that I might try is briefly mention the idea of multivariable functions to clarify that $$f$$ takes certain variables, say, $$x, y$$ and this means it is $$f(x,y)$$. It not that $$f$$ takes the equation involving $$x$$ and $$y$$ itself.
I wonder if issue also occurs when you have polar functions? Would they consider $$r=r(\theta)=2cos(\theta)$$ to be something like $$r(2cos(\theta))$$? I'd also wonder what they'd think parametric equations would be like. How would they consider $$x=x(t)=cos(t), y=y(t)=sin(t)$$?
The reason I'd bring up polar and parametric equations is to see if they understand function notation for that. If they do, then maybe that knowledge could be applied to understanding $$y=f(x)$$?
Suggest teaching students about composition of functions. Show them that if $$f(x)=x^2+1$$, then $$f(2x)= (2x)^2 +1$$ and $$f(x^2 + 1)$$ = $$(x^2+1)^2+1$$ If they do enough problems with composition of functions, they will be less likely to make this mistake again. This should be more effective than explanation about what the notation means. | 1,488 | 5,927 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 29, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2021-49 | latest | en | 0.967207 |
https://www.airmilescalculator.com/distance/vig-to-hge/ | 1,618,384,692,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038077336.28/warc/CC-MAIN-20210414064832-20210414094832-00135.warc.gz | 709,563,780 | 26,488 | # Distance between El Vigía (VIG) and Higuerote (HGE)
Flight distance from El Vigía to Higuerote (Juan Pablo Pérez Alfonzo Airport – Higuerote Airport) is 401 miles / 645 kilometers / 349 nautical miles. Estimated flight time is 1 hour 15 minutes.
Driving distance from El Vigía (VIG) to Higuerote (HGE) is 533 miles / 857 kilometers and travel time by car is about 11 hours 47 minutes.
## Map of flight path and driving directions from El Vigía to Higuerote.
Shortest flight path between Juan Pablo Pérez Alfonzo Airport (VIG) and Higuerote Airport (HGE).
## How far is Higuerote from El Vigía?
There are several ways to calculate distances between El Vigía and Higuerote. Here are two common methods:
Vincenty's formula (applied above)
• 401.057 miles
• 645.439 kilometers
• 348.509 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 400.833 miles
• 645.079 kilometers
• 348.315 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Juan Pablo Pérez Alfonzo Airport
City: El Vigía
Country: Venezuela
IATA Code: VIG
ICAO Code: SVVG
Coordinates: 8°37′26″N, 71°40′21″W
B Higuerote Airport
City: Higuerote
Country: Venezuela
IATA Code: HGE
ICAO Code: SVHG
Coordinates: 10°27′44″N, 66°5′34″W
## Time difference and current local times
There is no time difference between El Vigía and Higuerote.
-04
-04
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 84 kg (185 pounds).
## Frequent Flyer Miles Calculator
El Vigía (VIG) → Higuerote (HGE).
Distance:
401
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
401
Round trip? | 524 | 1,884 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-17 | latest | en | 0.755753 |
http://everything2.com/title/Associativity | 1,490,570,585,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189313.82/warc/CC-MAIN-20170322212949-00050-ip-10-233-31-227.ec2.internal.warc.gz | 117,617,668 | 6,289 | A rule of inference used in prepositional logic. Written 'Assoc.' for short. It allows you to change the parenthesis for order of operations in situations when it will make no logical difference, but will allow you to manipulate statements in useful ways. To demonstrate:
(P∨{Q∨R}) = ({P∨Q}∨R)
And
(P∧{Q∧R}) = ({P∧Q}∧R)
Back up to Rules of Inference
A property some binary operations possess. The operation @ is associative if for all a,b,c
a@(b@c) = (a@b)@c
#### Non examples:
These all are not associative.
Note that the operation @ must have the type AxA -> A. So it is a categorical error to suggest that relations might be associative. | 167 | 647 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2017-13 | latest | en | 0.879016 |
https://www.bestprog.net/en/2019/10/29/python-module-math-power-and-logarithmic-functions/ | 1,726,477,486,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651682.69/warc/CC-MAIN-20240916080220-20240916110220-00759.warc.gz | 623,157,318 | 14,819 | # Module math. Power and logarithmic functions
### Contents
Search other websites:
##### 1. Function math.exp(x). Exponent raised to the power x
The math.exp(x) function raises the number e to the power of x. The function returns the result of a real type. The argument x can be an integer or a real type. Exponential value: e = 2.718281… serves as the basis of the natural logarithm.
In Python, the math.exp(x) function can be replaced with other expressions
1. math.e ** x – here math.e is a constant equal to the value of the exponent.
2. pow(math.e, x) – here pow() is a built-in function of the Python language.
Example.
```# Function math.exp(x)
import math
y = math.exp(1) # y = 2.718281828459045
x = 0.0
y = math.exp(x) # y = 1.0
x = 3.85
y = math.exp(x) # y = 46.993063231579285```
### ⇑
##### 2. Function math.expm1(x). Exponent from x minus 1
The math.expm1(x) function calculates the value of the exp(x)-1 expression. When calculating the value of some y, the function call
`y = math.expm1(x)`
can be replaced by the expression
`y = math.exp(x)-1`
However, using the math.expm1(x) function will give a more accurate calculation result. This is the main purpose of this function.
Example.
```# Function math.expm1(x)
import math
x = 1.0
y = math.expm1(x) # y = 1.718281828459045
y = math.expm1(0.0) # y = 0.0```
### ⇑
##### 3. Function math.log(x). Natural logarithm
The math.log(x) function is designed to calculate the natural logarithm of a number with a given base.
The general form of the function is as follows
`math.log(x [, base])`
here
• x – the argument for which the logarithm is calculated;
• base – base of the logarithm. This function parameter is optional. If the base parameter is absent, then the number e = 2.718281…
If you try to call the log(0.0) function, the Python interpreter will throw an error
`ValueError: math domain error`
since the logarithm of zero does not exist.
Example.
```# Function math.log(x)
import math
x = 1.0
y = math.log(x) # y = 0.0```
### ⇑
##### 4. Function math.log1p(x). Logarithm for values close to zero
The log1p(x) function returns the natural logarithm of 1 + x. The basis of the logarithm is the exponent e = 2.718281… The function is necessary in cases when the value of argument x approaches zero. As you know, the logarithm of zero does not exist. To avoid an exception, this function is introduced.
Example.
```# Function math.log1p(x)
import math
x = 0.0000001
y = math.log1p(x) # y = 9.999999500000032e-08```
### ⇑
##### 5. Function math.log2(x). Logarithm with base 2
The math.log2(x) function has been introduced since Python 3.3 and returns the logarithm of the argument x with base 2. The function was introduced in order to increase the accuracy of calculations in comparison with the function math.log(x, 2). The argument x can be either an integer or a real type.
Example.
```# Function math.log2(x)
import math
x = 2
y = math.log2(x) # y = 1.0
x = 16
y = math.log2(x) # y = 4.0```
### ⇑
##### 6. Function math.log10(x). Decimal logarithm
The math.log10(x) function returns the logarithm of x with base 10 (base = 10). The function gives a more accurate result compared to calling the math.log(x, 10) function. The argument x can be either an integer or a real type.
Example.
```# Function math.log10(x)
import math
x = 10
y = math.log10(x) # y = 1.0
x = 100
y = math.log10(x) # y = 2.0
x = 10.00001
y = math.log10(x) # y = 1.0000004342942648```
### ⇑
##### 7. Function math.pow(x, y). Exponentiation
The math.pow (x, y) function raises x to the power of y. The arguments x, y can be of integer or real type. Operands of complex type are not supported.
Features of calculating the result:
• the result of pow(1.0, y) will always be 1.0;
• the result of pow (0.0, y) will always be 1.0.
In contrast to the operation ** (exponentiation), the function math.pow(x, y) integer operands converts to the real type float.
Example.
```# Function math.pow(x, y)
import math
# for integer operands
x = 3
y = 4
z = math.pow(x, y) # z = 81.0 - real result
# for floating point operands
x = 2.5
y = 1.5
z = math.pow(x, y) # z = 3.952847075210474
# negative numbers
x = -2
y = -3
z = math.pow(x, y) # z = -0.125
x = -2.0
y = 3.0
z = math.pow(x, y) # z = -8.0
# operator **
z = (-2) ** 3 # z = -8 - the result of integer type```
### ⇑
##### 8. Function math.sqrt(x). Square root
The math.sqrt(x) function computes the square root of the argument x. The function returns the result of a real type. The value of x can be positive or zero. If x is negative, the interpreter will display an error message
`math domain error`
Example.
```# Function math.sqrt(x)
import math
# for integer numbers
x = 81
y = math.sqrt(x) # y = 9.0
x = -0.0
y = math.sqrt(x) # y = -0.0
x = 2.0
y = math.sqrt(x) # y = 1.4142135623730951``` | 1,479 | 4,854 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-38 | latest | en | 0.617429 |
https://bytes.com/topic/python/23208-prime-number-module | 1,716,426,238,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058575.96/warc/CC-MAIN-20240522224707-20240523014707-00212.warc.gz | 129,041,864 | 11,901 | 473,563 Members | 2,292 Online
# Prime number module
Is there a python module that includes functions for working with prime
numbers? I mainly need A function that returns the Nth prime number and
that returns how many prime numbers are less than N, but a prime number
tester would also be nice. I'm dealing with numbers in the 10^6-10^8 range
so it would have to fairly efficient
Dag
Jul 18 '05 #1
36 8363
Dag wrote:
Is there a python module that includes functions for working with prime
numbers? I mainly need A function that returns the Nth prime number and
that returns how many prime numbers are less than N, but a prime number
tester would also be nice. I'm dealing with numbers in the 10^6-10^8
range so it would have to fairly efficient
gmpy is pretty good for this sort of thing, but the primitives it gives
you are quite different -- is_prime to check if a number is prime, and
next_prime to get the smallest prime number larger than a given number.
You'd have to build your own caching on top of this to avoid repeating
computation if you need, e.g., "how many primes are < N" for several
different values of N; and I'm not even sure that gmpy's primitives are
optimal for this (compared with, for example, some kind of sieving).
Anyway, with all of these caveats, here's an example function:
import gmpy
def primes_upto(N):
count = 0
prime = gmpy.mpz(1)
while prime < N:
prime = prime.next_prim e()
count += 1
return count
and having saved it in pri.py, here's what I measure in terms of time:
[alex@lancelot python2.3]\$ python timeit.py -s 'import pri' -s 'M=1000*1000'
\ 'pri.primes_upt o(M)'
10 loops, best of 3: 2.76e+06 usec per loop
[alex@lancelot python2.3]\$ python timeit.py -s 'import pri' -s
'M=2*1000*1000' 'pri.primes_upt o(M)'
10 loops, best of 3: 4.03e+06 usec per loop
i.e., 2.76 seconds for primes up to 1,000,000 -- about 4 seconds
for primes up to 2,000,000. (This is on my good old trusty Athlon
Linux machine, about 30 months old now and not top-speed even when
new -- I'm sure one of today's typical PC's might take 2 or 3
times less than this, of course). If I was to use this kind of
computation "in production", I'd pre-compute the counts for some
typical N of interest and only generate and count primes in a
narrow band, I think; but it's hard to suggest anything without a
Alex
Jul 18 '05 #2
On Mon, 29 Sep 2003 14:53:34 GMT, Dag <da*@velvet.net > wrote:
Is there a python module that includes functions for working with prime
numbers? I mainly need A function that returns the Nth prime number and
that returns how many prime numbers are less than N, but a prime number
tester would also be nice. I'm dealing with numbers in the 10^6-10^8 range
so it would have to fairly efficient
Dag
I just wrote a fairly simple sieve, which gave primes up to 1,000,000
in a few seconds (1.8GHz P4, Python code). There were 78498 primes in
that range (unless I screwed the code up, but it looked OK for smaller
ranges).
Going for 10^7 took too long for my patience, let alone 10^8, but at
least in theory there should be less than 100 times as many primes in
the range up to 10^8.
So here's the thought - a binary file containing a complete list of
primes up to 10^8 would require roughly 30MB (using 32 bit integers,
which should comfortably handle your requirement). Open in random
access and do a binary search to find a particular prime and the
position in the file should tell you how many primes are smaller than
that one.
30MB shouldn't be too prohibitive on todays machines, though if this
is to be distributed to other people there would of course be issues.
--
Steve Horne
steve at ninereeds dot fsnet dot co dot uk
Jul 18 '05 #3
Stephen Horne <\$\$\$\$\$\$\$\$\$\$\$\$\$\$ \$\$\$@\$\$\$\$\$\$\$\$\$\$\$ \$\$\$\$\$\$\$\$\$.co.uk > wrote previously:
|I just wrote a fairly simple sieve, which gave primes up to 1,000,000
|in a few seconds (1.8GHz P4, Python code). There were 78498 primes in
|that range...at least in theory there should be less than 100 times as
|many primes in the range up to 10^8.
Quite a few less, actually. Under Gauss' Prime Number Theorem, an
approximation for the number of primes less than N is N/ln(N). I know
there have been some slight refinements in this estimate since 1792, but
in the ranges we're talking about, it's plenty good enough.
So I only expect around 5,428,681 primes less than 10^8 to occur. Well,
that's not SO much less than 7.8M.
|So here's the thought - a binary file containing a complete list of
|primes up to 10^8 would require roughly 30MB (using 32 bit integers,
|which should comfortably handle your requirement).
I wonder if such a data structure is really necessary. Certainly it
produces a class of answers quite quickly. I.e. search for a prime, and
its offset immediately gives you the number of primes less than it.
Heck, searching for any number, even a composite occurring between
primes, works pretty much the same way. Of course, the above
approximation gives you a close answer even quicker.
But if you are worried about disk storage, one easy shortcut is to store
a collection of 16-bit differences between successive primes. That's
half the size, and still lets you answer the desired question *pretty*
quickly (extra addition is required)... or at least generate a local
copy of Horne's data structure in one pass.
Moving farther, even this gap structure is quite compressible. Most
gaps are quite a bit smaller than 65536, so the highbits are zeros. In
fact, I am pretty sure that almost all the gaps are less than 256. So
an immediate compression strategy (saving disk space, costing time to
recreate the transparent structure) is to store gaps as 8-bit values,
with a x00 byte escaping into a larger value (I guess in the next two
bytes).
Maybe I'll try it, and see how small I can make the data... unless I do
my real work :-).
Yours, Lulu...
--
Keeping medicines from the bloodstreams of the sick; food from the bellies
of the hungry; books from the hands of the uneducated; technology from the
underdeveloped; and putting advocates of freedom in prisons. Intellectual
property is to the 21st century what the slave trade was to the 16th.
Jul 18 '05 #4
"Lulu of the Lotus-Eaters" <me***@gnosis.c x> wrote in message
Moving farther, even this gap structure is quite compressible. Most
gaps are quite a bit smaller than 65536, so the highbits are zeros. In
fact, I am pretty sure that almost all the gaps are less than 256. So
an immediate compression strategy (saving disk space, costing time to
recreate the transparent structure) is to store gaps as 8-bit values,
with a x00 byte escaping into a larger value (I guess in the next two
bytes).
You can take this to 512 knowing that the gaps will always be an even
interval.
Emile
Jul 18 '05 #5
Lulu of the Lotus-Eaters wrote:
Stephen Horne <\$\$\$\$\$\$\$\$\$\$\$\$\$\$ \$\$\$@\$\$\$\$\$\$\$\$\$\$\$ \$\$\$\$\$\$\$\$\$.co.uk > wrote previously:
|I just wrote a fairly simple sieve, which gave primes up to 1,000,000
|in a few seconds (1.8GHz P4, Python code). There were 78498 primes in
|that range...at least in theory there should be less than 100 times as
|many primes in the range up to 10^8.
Quite a few less, actually. Under Gauss' Prime Number Theorem, an
approximation for the number of primes less than N is N/ln(N). I know
there have been some slight refinements in this estimate since 1792, but
in the ranges we're talking about, it's plenty good enough.
So I only expect around 5,428,681 primes less than 10^8 to occur. Well,
that's not SO much less than 7.8M.
|So here's the thought - a binary file containing a complete list of
|primes up to 10^8 would require roughly 30MB (using 32 bit integers,
|which should comfortably handle your requirement).
I wonder if such a data structure is really necessary. Certainly it
produces a class of answers quite quickly. I.e. search for a prime, and
its offset immediately gives you the number of primes less than it.
Heck, searching for any number, even a composite occurring between
primes, works pretty much the same way. Of course, the above
approximation gives you a close answer even quicker.
But if you are worried about disk storage, one easy shortcut is to store
a collection of 16-bit differences between successive primes. That's
half the size, and still lets you answer the desired question *pretty*
quickly (extra addition is required)... or at least generate a local
copy of Horne's data structure in one pass.
Moving farther, even this gap structure is quite compressible. Most
gaps are quite a bit smaller than 65536, so the highbits are zeros. In
fact, I am pretty sure that almost all the gaps are less than 256. So
an immediate compression strategy (saving disk space, costing time to
recreate the transparent structure) is to store gaps as 8-bit values,
with a x00 byte escaping into a larger value (I guess in the next two
bytes).
Maybe I'll try it, and see how small I can make the data... unless I do
my real work :-).
I believe you could implement a hybrid scheme that would be quite fast
and still maintain nearly the same level of compression that you
describe above. In addition to the above compressed data, also store,
uncompressed, every Nth prime. A binary search will get you within N
primes of your answer, to find the exact value, recreate those N-primes.
For a N of, for instance, 64 the level of compression would be minimally
affected but should make finding the number of primes less than a given
number number much faster than the basic compressed scheme.
In fact I wouldn't be suprised if this was faster than the uncompressed
scheme since you're less likely to thrash your memory.
-tim
Yours, Lulu...
--
Keeping medicines from the bloodstreams of the sick; food from the bellies
of the hungry; books from the hands of the uneducated; technology from the
underdeveloped; and putting advocates of freedom in prisons. Intellectual
property is to the 21st century what the slave trade was to the 16th.
Jul 18 '05 #6
Lulu of the Lotus-Eaters wrote:
So I only expect around 5,428,681 primes less than 10^8 to occur.
Well, that's not SO much less than 7.8M.
I found 5,761,455 primes < 1E8.
// Klaus
--<> unselfish actions pay back better
Jul 18 '05 #7
Lulu of the Lotus Eaters:
So I only expect around 5,428,681 primes less than 10^8 to occur.
Well, that's not SO much less than 7.8M.
Klaus Alexander Seistrup I found 5,761,455 primes < 1E8.
http://www.utm.edu/research/primes/howmany.shtml has the same number.
(found by googling for "5,761,455" - 3rd hit. :)
That's using Proof by Consensus,
Andrew
da***@dalkescie ntific.com
Jul 18 '05 #8
Hello Dag,
Is there a python module that includes functions for working with prime
numbers? I mainly need A function that returns the Nth prime number and
that returns how many prime numbers are less than N, but a prime number
tester would also be nice. I'm dealing with numbers in the 10^6-10^8 range
so it would have to fairly efficient
Try gmpy (http://gmpy.sourceforge.net/)
HTH.
Miki
Jul 18 '05 #9
Andrew Dalke wrote:
I found 5,761,455 primes < 1E8.
http://www.utm.edu/research/primes/howmany.shtml has the same
number. (found by googling for "5,761,455" - 3rd hit. :)
Cool!
That's using Proof by Consensus,
:-)
// Klaus
--<> unselfish actions pay back better
Jul 18 '05 #10
This thread has been closed and replies have been disabled. Please start a new discussion. | 3,052 | 11,416 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-22 | latest | en | 0.907181 |
https://trade-leader.com/articles/fibonacci | 1,696,194,624,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510941.58/warc/CC-MAIN-20231001205332-20231001235332-00778.warc.gz | 638,666,173 | 8,666 | # Fibonacci levels - how to use in trading?
What are Fibonacci levels? How to use Fibonacci levels in trading? How to determine with their help price reversals and the end of the correction? How to build and trade Forex Fibonacci levels?
Welcome you friends! In this article we will look at one of the most popular technical analysis tools - Fibonacci levels. This is a technical indicator that detects various patterns in price dynamics that cannot be determined with the naked eye. Fibonacci levels in trading work best in a trending market. As you know, in a trending market, prices always come back, then to move further along the trend. This is called correction. The idea of Fibonacci is to determine the level of completion of the correction and enter the market at the best price, and also exit the transaction on time before the price reversal occurs. In this article you will learn what Fibonacci numbers and sequence are, how to build Fibonacci levels in a trading terminal, how to determine the end of correction with their help, and how to trade by Fibonacci levels.
If you have not yet decided on the choice of broker, then see our Forex brokers rating. Here you will find detailed reviews of brokerage companies with trading conditions and feedback from traders.
## Who is Fibonacci?
Leonardo of Pisa (also known as Leonardo Fibonacci) was an Italian mathematician in the 13th century. He studied mathematics and is known for popularizing Hindu-Arabic numbers in Europe.
In his book Liber Abaci (The Book of Abacus), Leonardo described a sequence of numbers that return a constant ratio when dividing one number by the next number in the series. Each number in the Fibonacci sequence is equal to the sum of the two previous numbers. Example: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, etc.
What is interesting in this sequence is that it always returns the same ratio when dividing one number by the next number in the series, excluding the first few numbers. For example, 34/55 is 0.618, just like 55/89, 89/144, 144/233, and so on. This ratio (0.618) is well known as the Fibonacci Golden Ratio and can be observed in various natural phenomena.
By dividing each number by the second number on the right, you always get .382. For example, 13/34, 21/55, 34/89 and 89/233 all return .382 as a result.
When used in trading, traders apply these ratios to determine possible zones of support and resistance. In financial markets, the most commonly used Fibonacci levels are 23.6%, 38.2%, 50%, 61.8%, 76.4%. In a strong buying market, traders expect recovery to 38.2%, while in a weaker trend, traders expect recovery to 61.8%.
## How to plot Fibonacci levels on a chart?
You can find Fibonacci levels in the MetaTrader 4 trading terminal in the graphic tools: Insert - Fibonacci - Fibonacci lines. You can also find the F symbol at the top of the toolbar of the trading terminal. By clicking on it, you can return to the chart to draw Fibonacci levels. Just click on the minimum price and, without releasing the left mouse button, connect it with the maximum for an uptrend. In a downtrend, it is necessary to combine a maximum with a minimum price. By double-clicking on the inclined Fibonacci line, you can delete or edit this tool.
There are some basic rules that you must follow when building Fibonacci levels. It is always necessary to remember that Fibonacci levels are a subjective tool for technical analysis. Two traders can get different results based on how they determined the main low / high price when building levels. As a rule, it is better to practice on charts with higher time frames before moving on to hourly and especially minute charts.
STEP 1. Define the main high / low. Looking at the weekly USDCAD chart, it is obvious which two points we should connect.
STEP 2. Select the Fibonacci Levels tool and connect the two points (minimum and maximum prices). In MT4, the trend line and the necessary Fibonacci levels will automatically appear.
STEP 3. Use Fibonacci levels as support / resistance. A good example in the chart below is the 38.2% level, which was tested twice, after which the price was able to move up.
Most traders use Fibonacci levels as classic support / resistance levels. This means that they can be used as an entry level or help traders determine their goals: stop loss and take profit.
Example 1. A trader uses several technical analysis tools, and all of them indicate that the upward trend of USDCAD will continue. However, the trader feels that in the short term currency pair is overbought, and what else will happen one rollback before the resumption of the trend. As a fan of Fibonacci retracements, he sees the area around 38.2% as a potential entry point.
Example 2. A trader bought USDCAD and is trying to determine where he should place his stop loss. He sees that the 38.2% level previously acted as a key support level, and decides to place his stop loss 5 points below it. | 1,096 | 4,962 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2023-40 | longest | en | 0.941726 |
https://number.academy/28095 | 1,656,599,231,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103821173.44/warc/CC-MAIN-20220630122857-20220630152857-00398.warc.gz | 464,541,283 | 11,983 | Number 28095
Number 28,095 spell 🔊, write in words: twenty-eight thousand and ninety-five . Ordinal number 28095th is said 🔊 and write: twenty-eight thousand and ninety-fifth. The meaning of number 28095 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 28095. What is 28095 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 28095.
What is 28,095 in other units
The decimal (Arabic) number 28095 converted to a Roman number is (X)(X)(V)MMMXCV. Roman and decimal number conversions.
Weight conversion
28095 kilograms (kg) = 61938.2 pounds (lbs)
28095 pounds (lbs) = 12743.8 kilograms (kg)
Length conversion
28095 kilometers (km) equals to 17458 miles (mi).
28095 miles (mi) equals to 45215 kilometers (km).
28095 meters (m) equals to 92175 feet (ft).
28095 feet (ft) equals 8564 meters (m).
28095 centimeters (cm) equals to 11061.0 inches (in).
28095 inches (in) equals to 71361.3 centimeters (cm).
Temperature conversion
28095° Fahrenheit (°F) equals to 15590.6° Celsius (°C)
28095° Celsius (°C) equals to 50603° Fahrenheit (°F)
Time conversion
(hours, minutes, seconds, days, weeks)
28095 seconds equals to 7 hours, 48 minutes, 15 seconds
28095 minutes equals to 2 weeks, 5 days, 12 hours, 15 minutes
Codes and images of the number 28095
Number 28095 morse code: ..--- ---.. ----- ----. .....
Sign language for number 28095:
Number 28095 in braille:
Images of the number
Image (1) of the numberImage (2) of the number
More images, other sizes, codes and colors ...
Mathematics of no. 28095
Multiplications
Multiplication table of 28095
28095 multiplied by two equals 56190 (28095 x 2 = 56190).
28095 multiplied by three equals 84285 (28095 x 3 = 84285).
28095 multiplied by four equals 112380 (28095 x 4 = 112380).
28095 multiplied by five equals 140475 (28095 x 5 = 140475).
28095 multiplied by six equals 168570 (28095 x 6 = 168570).
28095 multiplied by seven equals 196665 (28095 x 7 = 196665).
28095 multiplied by eight equals 224760 (28095 x 8 = 224760).
28095 multiplied by nine equals 252855 (28095 x 9 = 252855).
show multiplications by 6, 7, 8, 9 ...
Fractions: decimal fraction and common fraction
Fraction table of 28095
Half of 28095 is 14047,5 (28095 / 2 = 14047,5 = 14047 1/2).
One third of 28095 is 9365 (28095 / 3 = 9365).
One quarter of 28095 is 7023,75 (28095 / 4 = 7023,75 = 7023 3/4).
One fifth of 28095 is 5619 (28095 / 5 = 5619).
One sixth of 28095 is 4682,5 (28095 / 6 = 4682,5 = 4682 1/2).
One seventh of 28095 is 4013,5714 (28095 / 7 = 4013,5714 = 4013 4/7).
One eighth of 28095 is 3511,875 (28095 / 8 = 3511,875 = 3511 7/8).
One ninth of 28095 is 3121,6667 (28095 / 9 = 3121,6667 = 3121 2/3).
show fractions by 6, 7, 8, 9 ...
Calculator
28095
Is Prime?
The number 28095 is not a prime number. The closest prime numbers are 28087, 28097.
Factorization and factors (dividers)
The prime factors of 28095 are 3 * 5 * 1873
The factors of 28095 are 1 , 3 , 5 , 15 , 1873 , 5619 , 9365 , 28095
Total factors 8.
Sum of factors 44976 (16881).
Powers
The second power of 280952 is 789.329.025.
The third power of 280953 is 22.176.198.957.375.
Roots
The square root √28095 is 167,615632.
The cube root of 328095 is 30,400193.
Logarithms
The natural logarithm of No. ln 28095 = loge 28095 = 10,243347.
The logarithm to base 10 of No. log10 28095 = 4,448629.
The Napierian logarithm of No. log1/e 28095 = -10,243347.
Trigonometric functions
The cosine of 28095 is -0,965588.
The sine of 28095 is 0,260076.
The tangent of 28095 is -0,269345.
Properties of the number 28095
Is a Friedman number: No
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
Number 28095 in Computer Science
Code typeCode value
28095 Number of bytes27.4KB
Unix timeUnix time 28095 is equal to Thursday Jan. 1, 1970, 7:48:15 a.m. GMT
IPv4, IPv6Number 28095 internet address in dotted format v4 0.0.109.191, v6 ::6dbf
28095 Decimal = 110110110111111 Binary
28095 Decimal = 1102112120 Ternary
28095 Decimal = 66677 Octal
28095 Decimal = 6DBF Hexadecimal (0x6dbf hex)
28095 BASE64MjgwOTU=
28095 MD5c4caf9e04a0d4f83565449f2cce9d5d5
28095 SHA161a69b3f40b3067b08168104b109e1d4ae9e3f55
28095 SHA224ca46ed03c3d3ce1dd62b8b42c917397f5fa8b30a100b09c156fcca06
28095 SHA2562bbafd33ccf5cde20f48b61425d6cdc207f301715d2e48949cb6efb802c0b68e
28095 SHA38405c7a1b281a4640129ca7acdd7489b121ef243ee60b8c45ae4565a481e8136d7919ca7a4db5eeb8df55bbafda5ec0b47
More SHA codes related to the number 28095 ...
If you know something interesting about the 28095 number that you did not find on this page, do not hesitate to write us here.
Numerology 28095
Character frequency in number 28095
Character (importance) frequency for numerology.
Character: Frequency: 2 1 8 1 0 1 9 1 5 1
Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 28095, the numbers 2+8+0+9+5 = 2+4 = 6 are added and the meaning of the number 6 is sought.
Interesting facts about the number 28095
Asteroids
• (28095) Seanmahoney is asteroid number 28095. It was discovered by LINEAR, Lincoln Near-Earth Asteroid Research from Lincoln Laboratory, Socorro on 9/14/1998.
Number 28,095 in other languages
How to say or write the number twenty-eight thousand and ninety-five in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 28.095) veintiocho mil noventa y cinco German: 🔊 (Anzahl 28.095) achtundzwanzigtausendfünfundneunzig French: 🔊 (nombre 28 095) vingt-huit mille quatre-vingt-quinze Portuguese: 🔊 (número 28 095) vinte e oito mil e noventa e cinco Chinese: 🔊 (数 28 095) 二万八千零九十五 Arabian: 🔊 (عدد 28,095) ثمانية و عشرون ألفاً و خمسة و تسعون Czech: 🔊 (číslo 28 095) dvacet osm tisíc devadesát pět Korean: 🔊 (번호 28,095) 이만 팔천구십오 Danish: 🔊 (nummer 28 095) otteogtyvetusinde og femoghalvfems Dutch: 🔊 (nummer 28 095) achtentwintigduizendvijfennegentig Japanese: 🔊 (数 28,095) 二万八千九十五 Indonesian: 🔊 (jumlah 28.095) dua puluh delapan ribu sembilan puluh lima Italian: 🔊 (numero 28 095) ventottomilanovantacinque Norwegian: 🔊 (nummer 28 095) tjue-åtte tusen og nitti-fem Polish: 🔊 (liczba 28 095) dwadzieścia osiem tysięcy dziewięćdzisiąt pięć Russian: 🔊 (номер 28 095) двадцать восемь тысяч девяносто пять Turkish: 🔊 (numara 28,095) yirmisekizbindoksanbeş Thai: 🔊 (จำนวน 28 095) สองหมื่นแปดพันเก้าสิบห้า Ukrainian: 🔊 (номер 28 095) двадцять вiсiм тисяч дев'яносто п'ять Vietnamese: 🔊 (con số 28.095) hai mươi tám nghìn lẻ chín mươi lăm Other languages ...
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If you know something interesting about the number 28095 or any natural number (positive integer) please write us here or on facebook. | 2,409 | 6,852 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2022-27 | latest | en | 0.62808 |
https://edurev.in/course/quiz/attempt/20713_Test-Cubes-And-Cube-Roots-2/b59fb24a-bf0b-4f6e-9c3a-6a9313101759 | 1,718,870,776,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861916.26/warc/CC-MAIN-20240620074431-20240620104431-00186.warc.gz | 191,574,961 | 59,283 | Test: Cubes And Cube Roots- 2 - UPSC MCQ
# Test: Cubes And Cube Roots- 2 - UPSC MCQ
Test Description
## 40 Questions MCQ Test CSAT Preparation - Test: Cubes And Cube Roots- 2
Test: Cubes And Cube Roots- 2 for UPSC 2024 is part of CSAT Preparation preparation. The Test: Cubes And Cube Roots- 2 questions and answers have been prepared according to the UPSC exam syllabus.The Test: Cubes And Cube Roots- 2 MCQs are made for UPSC 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Cubes And Cube Roots- 2 below.
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Test: Cubes And Cube Roots- 2 - Question 1
### Which of the following is not a perfect cube ?
Test: Cubes And Cube Roots- 2 - Question 2
### The cube of 4 is _______________.
Test: Cubes And Cube Roots- 2 - Question 3
### The value of 53 is __________.
Test: Cubes And Cube Roots- 2 - Question 4
The cube of an even number is always ____________.
Test: Cubes And Cube Roots- 2 - Question 5
The cube of an odd number is always __________.
Test: Cubes And Cube Roots- 2 - Question 6
Each prime factor appears _________ times in its cube?
Test: Cubes And Cube Roots- 2 - Question 7
Which of the following is Hardy-Ramanujan Number ?
Detailed Solution for Test: Cubes And Cube Roots- 2 - Question 7
This story is very famous among mathematicians. 1729 is sometimes called the “Hardy-Ramanujan number”.
There are two ways to say that 1729 is the sum of two cubes. 1x1x1=1; 12x12x12=1728. So 1+1728=1729 But also: 9x9x9=729; 10x10x10=1000. So 729+1000=1729 There are other numbers that can be shown to be the sum of two cubes in more than one way, but 1729 is the smallest of them.
Ramanujan did not actually discover this fact. It was known in 1657 by a Frenchmathematician Bernard Franicle de Bessy.
But it got famous after the ramanujans above conversation.
So it's famously known as Ramanujan Number.
Test: Cubes And Cube Roots- 2 - Question 8
By which smallest natural number 392 must be multiplied so as to make the product a perfect cube ?
Detailed Solution for Test: Cubes And Cube Roots- 2 - Question 8
392 = 2 × 2 × 2 × 7 × 7
Hence the number should be multiplied by 7 to make it a perfect cube.
Test: Cubes And Cube Roots- 2 - Question 9
The smallest natural number by which 243 must be multiplied to make the product a perfect cube is __________.
Test: Cubes And Cube Roots- 2 - Question 10
The smallest natural number by which 704 must be divided to obtain a perfect cube is
Test: Cubes And Cube Roots- 2 - Question 11
The smallest natural number by which 135 must be divided to obtain a perfect cube is
Detailed Solution for Test: Cubes And Cube Roots- 2 - Question 11
we have 135 = 3 x 3 x 3 x 5
Grouping the prime factors of 135 into triples, we are left over with 5.
∴ 135 is not a perfect cube
Now, [135]divided by5 = [ 3 x 3 x 3 x 5] divided by5
or 27 = 3 x 3 x 3
i.e. 27 is a perfect cube.
Thus, the required smallest number is 5
Test: Cubes And Cube Roots- 2 - Question 12
Which of the following is not a perfect cube ?
Test: Cubes And Cube Roots- 2 - Question 13
The expansion of a3 is ___________.
Test: Cubes And Cube Roots- 2 - Question 14
What will be the unit digit of the cube of a number ending with 2 ?
Test: Cubes And Cube Roots- 2 - Question 15
What will be the unit digit of the cube of a number ending with 4 ?
Test: Cubes And Cube Roots- 2 - Question 16
What will be the unit digit of the cube of a number ending with 6 ?
Detailed Solution for Test: Cubes And Cube Roots- 2 - Question 16
Test: Cubes And Cube Roots- 2 - Question 17
A cuboid has dimensions 5cm, 2cm, 5cm. How many such cuboid will be needed to form a cube ?
Detailed Solution for Test: Cubes And Cube Roots- 2 - Question 17
Given, sides of a cuboid are 5cm, 2cm and 5cm
Volume of cuboid =5*2*5 = 50 cm3
Let the side of the cube formed with
minimum number of cuboids = x
Hence volume of the cube = x3
Number of cuboids required = n
Hence n*50 = x3
Now, n*50 must be perfect cube
n*50 = n*2*5*5
So, minimum value of n = 2*2*5 = 20
Hence, the number of cuboid will be
needed to form a cube is 20
Test: Cubes And Cube Roots- 2 - Question 18
How many cuboids of dimensions 15cm, 30cm,15cm will be needed to form a cube ?
Detailed Solution for Test: Cubes And Cube Roots- 2 - Question 18
Given cuboids of dimensions 15cm, 30cm ,15cm
LCM(15, 30, 15) = 30
Since 15*2 = 30
30*1 = 30
and 15*2 = 30 {since a cube has all the dimensions are same}
So, the number of cuboids required to form a cube = 2*1*2 = 4
So, by using 4 cuboids, we can form a cube having side 30 cm each.
Test: Cubes And Cube Roots- 2 - Question 19
729 is the value of _______________.
Test: Cubes And Cube Roots- 2 - Question 20
Which of the following is a perfect cube ?
Test: Cubes And Cube Roots- 2 - Question 21
What is the volume of a cube whose edge is 2cm ?
Detailed Solution for Test: Cubes And Cube Roots- 2 - Question 21
Volume of cube=2×2×2=8
Test: Cubes And Cube Roots- 2 - Question 22
The symbol for cube root is __________.
Test: Cubes And Cube Roots- 2 - Question 23
The cube root of 512 is ________.
Test: Cubes And Cube Roots- 2 - Question 24
The value of ∛343 is-
Test: Cubes And Cube Roots- 2 - Question 25
Which of the following is true for any natural number n? for n>1
Detailed Solution for Test: Cubes And Cube Roots- 2 - Question 25
Take any natural number(except 1). This identity would hold for any natural number.
Natural numbers are 1,2,3,4,5____ upto infinity with all positive terms.
Test: Cubes And Cube Roots- 2 - Question 26
If the volume of a cube is 125 cm3 then what would be the length of its side?
Detailed Solution for Test: Cubes And Cube Roots- 2 - Question 26
Test: Cubes And Cube Roots- 2 - Question 27
What will be the unit digit of the cube root of a number ends with 8?
Test: Cubes And Cube Roots- 2 - Question 28
What will be the unit digit of the cube root of a number ends with 2?
Test: Cubes And Cube Roots- 2 - Question 29
What will be the unit digit of the cube root of a number ends with 3?
Detailed Solution for Test: Cubes And Cube Roots- 2 - Question 29
for example like 343 so its cube root will be 7
because 73 will always have 3 as unit digit
Test: Cubes And Cube Roots- 2 - Question 30
For a number ending with 7, the unit digit of its cube is equal to:
Detailed Solution for Test: Cubes And Cube Roots- 2 - Question 30
If the unit digit is 7, the cube will have unit digit as 7×7×7 = 343
So, the unit digit will be 3.
Test: Cubes And Cube Roots- 2 - Question 31
9 is the cube root of __________.
Detailed Solution for Test: Cubes And Cube Roots- 2 - Question 31
9 is the cube root of 729 because 9 x 9 x 9 = 729
So option B is correct answer.
Test: Cubes And Cube Roots- 2 - Question 32
The number of digits in the cube root of a 6-digit number is _______.
Detailed Solution for Test: Cubes And Cube Roots- 2 - Question 32
As 100= 1000000 which is the smallest 3 digit number. So it’s only 2 digit number which is the cube root of a 6 digit number
Test: Cubes And Cube Roots- 2 - Question 33
How many digits will be there in the cube root of 46656 ?
Test: Cubes And Cube Roots- 2 - Question 34
How many digits will be there in the cube root of 512 ?
Test: Cubes And Cube Roots- 2 - Question 35
What will be the unit digit of ∛15625
Detailed Solution for Test: Cubes And Cube Roots- 2 - Question 35
If we multiply 5 three times we receive 125 where 5 is at the one’s place. So cube which has 5 at one’s place will have a root which have 5 at one’s place
Test: Cubes And Cube Roots- 2 - Question 36
How many zeros will be there in the cube root of 27000?
Test: Cubes And Cube Roots- 2 - Question 37
How many zeros will be there in the cube root of 800?
Detailed Solution for Test: Cubes And Cube Roots- 2 - Question 37
Since in cubes the zeros become triple ,for instance 10*10*10=1000, a number with two zeros does not have a cube root.
Test: Cubes And Cube Roots- 2 - Question 38
If 7= 343, then ∛343 = _________.
Test: Cubes And Cube Roots- 2 - Question 39
If 8= 512, then ∛512 = _______
Detailed Solution for Test: Cubes And Cube Roots- 2 - Question 39
8*8*8 =512
Test: Cubes And Cube Roots- 2 - Question 40
What will be the unit digit of ∛216
Detailed Solution for Test: Cubes And Cube Roots- 2 - Question 40
We can find it by Prime Factorization method
³√216 = 2×2×2×3×3×3
= 2 and 3 are making triplets i.e. 2 and 3
= 2×3 = 6 Thus, we can say that 6 i.e option 'B' is correct.
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http://www.slideserve.com/luka/chapter-8-natural-convection | 1,508,806,116,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187827662.87/warc/CC-MAIN-20171023235958-20171024015958-00833.warc.gz | 552,493,133 | 16,533 | Chapter 8 : Natural Convection
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# Chapter 8 : Natural Convection - PowerPoint PPT Presentation
Chapter 8 : Natural Convection. Contents: Physical consideration, governing equation Analysis of vertical, horizontal & inclined plates Analysis of cylinder, sphere & enclosures. Chapter 8 : Natural Convection. What is buoyancy force ?
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### Chapter 8 : Natural Convection
Contents:
Physical consideration, governing equation
Analysis of vertical, horizontal & inclined plates
Analysis of cylinder, sphere & enclosures
### Chapter 8 : Natural Convection
• What is buoyancy force ?
• The upward force exerted by a fluid on a bodycompletely or partially immersed in it in a gravitational field.
• Themagnitude of the buoyancy force is equal to the weight of the fluid displacedby the body.
### Chapter 8 : Natural Convection
Thermal expansion coefficient / Volume expansioncoefficient:Variation of the density of a fluid with temperature at constant pressure.
Ideal gas
The larger the temperaturedifference between the fluid adjacent to a hot (or cold) surface and thefluid away from it, the largerthe buoyancy force and the strongerthe naturalconvection currents, and thus the higherthe heat transfer rate.
The coefficient of volume expansion isa measure of the change in volume ofa substance with temperatureat constant pressure.
### Chapter 8 : Natural Convection
- Ratio of buoyancy forces and thermal and momentum diffusivities.
Rayleigh Number, Ra=Gr.Pr
• In fluid mechanics, the Rayleigh number for a fluid is a dimensionless number associated with buoyancy driven flow (also known as free convection or natural convection).
• When the Rayleigh number is below the critical value for that fluid, heat transfer is primarily in the form of conduction; when it exceeds the critical value, heat transfer is primarily in the form of convection.
• The Rayleigh number is defined as the product of the Grashof number, which describes the relationship between buoyancy and viscosity within a fluid, and the Prandtl number, which describes the relationship between momentum diffusivity and thermal diffusivity.
• Hence the Rayleigh number itself may also be viewed as the ratio of buoyancy and viscosity forces times the ratio of momentum and thermal diffusivities.
Forced vs Natural Convection
• When analyzing potentially mixed convection, a parameter called the Archimedes number(Ar) parametrizes the relative strength of free and forced convection.
• The Archimedes number is the ratio of Grashof number and the square of Reynolds number, which represents the ratio of buoyancy force and inertia force, and which stands in for the contribution of natural convection.
• When Ar >> 1, natural convection dominates and when Ar << 1, forced convection dominates.
### Chapter 8 : Natural Convection
Natural convection over surfaces
*C & n is depend on the geometry of the surface and flow regime.
n=1/4 laminar flow
n-=1/3 turbulent flow
1)
2)
3)
1. What is the difference between ReL and RaL ?
2. What is the transition range in a free convection boundary ?
(Laminar)
(Turbulent)
*All the properties are evaluated at the film temperature, Tf=(Ts+T)/2
### Chapter 8 : Natural Convection
Transition in a free convection layer depends on the relative magnitude of the buoyancy and viscous forces
*The smooth and parallel lines in (a) indicate that the flow islaminar,whereas the eddies andirregularities in (b) indicate that the flow isturbulent.
### Chapter 8 : Natural Convection
• General correlations for vertical plate
• where,
Eq. (9.24)
• For wide range and more accurate solution, use correlation Churchill and Chu
Eq. (9.26)
Eq. (9.27)
### Chapter 8 : Natural Convection
• For case of vertical cylinders, the previous Eqs. ( 9.24 to 9.27) are valid if the condition satisfied where
• For case of inclined plates
• In the case of a hot plate in a cooler environment, convection currents are weaker on the lower surface of the hot plate, and the rate of heat transfer is lower relative to the vertical plate case.
• On the upper surface of a hot plate, the thickness of the boundary layer and thus the resistance to heat transfer decreases, and the rate of heat transfer increases relative to the vertical orientation.
• In the case of a cold plate in a warmer environment, the opposite occurs.
Hot plate-cold env.
cold plate-hot env.
### Chapter 8 : Natural Convection
• at the top and bottom surfaces of cooled and heated inclined plates, respectively, it is recommended that
Use equation 9.26
but replace g g cos
and only valid for 0 60
### Chapter 8 : Natural Convection
Example:
Consider a 0.6m x 0.6m thin square plate in a room at 30C. One side of the plate is maintained at a temperature of 90C, while the other side is insulated. Determine the rate of heat transfer from the plate by natural convection if the plate is vertical.
### Chapter 8 : Natural Convection
Example:
Consider a 0.6m x 0.6m thin square plate in a room at 30C. One side of the plate is maintained at a temperature of 90C, while the other side is insulated. Determine the rate of heat transfer from the plate by natural convection if the plate is
Vertical
Horizontal with hot surface facing up
Horizontal with hot surface facing down
Which position has the lowest heat transfer rate ? Why ?
### Chapter 8 : Natural Convection
• The boundary layer over a hot horizontal cylinder starts to develop at the bottom,increasing in thickness along the circumference, and forming a risingplume at the top.
• Therefore, the local Nusselt numberis highest at the bottom, and lowest at the top of the cylinder when the boundarylayer flow remains laminar.
• The opposite is true in the case of a coldhorizontalcylinder in a warmer medium, and the boundary layer in this casestarts to develop at the top of the cylinder and ending with a descendingplume at the bottom.
### Chapter 8 : Natural Convection
• General correlations for an isothermal cylinder
• where,
Eq. (9.33)
• For wide range of Ra, use correlation Churchill and Chu
Eq. (9.34)
### Chapter 8 : Natural Convection
Spheres
• In case of isothermal sphere, general correlations is proposed by Churchill
Eq. (9.35)
* Recommended when Pr 0.7 and RaD 1011
• In the limit as RaD→ 0, Equation 9.35 reduces to NuD = 2, which corresponds to heat transfer by conduction between a spherical surface and a stationary infinite medium, as in Eqs. (7.48 & 7.49) – external convection for spherical object.
### Chapter 8 : Natural Convection
Problem 9.54:
A horizontal uninsulated steam pipe passes through a large room whose walls and ambient air are at 300K. The pipe of 150 mm diameter has an emissivity of 0.85 and an outer surface temperature of 400K. Calculate the heat loss per unit length from the pipe.
Schematic
Assumptions
Fluid properties
Analysis of total heat loss per unit length, q/L or q’
- Calculate NuD
- Calculate hD
- finally, calculate total heat loss, q’
*If use Eq. 9.33, hD= 6.15 W/m2K
*If use Eq. 9.34, hD= 6.38 W/m2K
Within 4%
### Chapter 8 : Natural Convection
Enclosures are frequently encountered in practice, and heat transferthrough them is of practical interest.
Characteristic lengthLc:the distance between the hot and coldsurfaces.
T1 and T2:the temperatures of the hot and cold surfaces.
Fluid properties at
### Chapter 8 : Natural Convection
- Flow is characterised by RaD value
Nu = 1
### Chapter 8 : Natural Convection
• Selection will be determined by the value of RaL, Pr and aspect ratio H/L:
### Chapter 8 : Natural Convection
• For larger aspect ratios, the following correlations have been proposed:
### Chapter 8 : Natural Convection
Example:
The vertical 0.8m high, 2m wide double pane window consists of two sheet of glass separated by a 2 cm air gap at atmospheric pressure. If the glass surface temperatures across the air gap are measured to be 12C and 2C, determine the rate of heat transfer through the window.
Schematic
Assumptions
Fluid properties at Tavg
Analysis of heat transfer | 2,118 | 8,754 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2017-43 | latest | en | 0.828715 |
https://www.radioexperimenter.us/exp-1931/electrical-communications-technique-and-its-applications-in-allied-fields-due.html | 1,561,557,297,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000353.82/warc/CC-MAIN-20190626134339-20190626160339-00076.warc.gz | 853,751,147 | 11,835 | ## Electrical Communications Technique And Its Applications In Allied Fields
A BRIDGE-TYPE FREQUENCY METER
NSTRUMENTS for the measurement of audio frequencies are usually based on one of the numerous bridge circuits in which frequency enters explicitly. The familiar tuned-cir-cuit wavemeter has not made a satisfactory audio-frequency meter because of the large amount of power lost in it when the usual current indicator is used. The resonance bridge makes use of a similar tuned circuit in one arm, the other arms being pure resistance. For any reasonable values of inductance the capacitance required to tune to the lower audio frequencies is so large that an air condenser cannot be used. The capacitance then is fixed and the inductance made variable in order to have continuous adjustment. Different ranges are obtained by changing the capacitance. The bridges of Campbell and of lvennelly and Velander make use of a v ariable mutual inductor and a number of fixed condensers.
Bridges using inductance, either self or mutual, have two serious defects. The shape of the frequency scale depends upon the characteristics of the variable inductor used and cannot be appreciably altered. The magnetic field of the inductor is such that it cannot be satisfactorily shielded from the source of frequency being measured.
The Wien bridge circuit has been chosen for the Type 434-B Audio-Fre-quency Meter because it eliminates both of these objections. Since it uses only resistance and capacitance, it has no external magnetic field. The two variable resistors may be so constructed that the frequency scale has the most desirable shape. A schematic diagram is shown in Figure 1. The conditions for balance of this bridge are:
2TrVPQCrCt, a nd
CP ~ B Q In order to provide a single control, upon which the frequency scale may be mounted and to maintain the second balance condition, the two resistors P and Q and the two condensers CP and Cv are made equal, and the two ratio arms are made two to one:
CP B
Figure 1. Schematic wiring diagram for the Type 434-B Audio-Frequency Meter
Thus the second balance condition is always fulfilled and the first condition reduces to
2vPCr
The two resistors P and Q are wound on tapered cards of such a shape that the frequency scale is logarithmic. Equal frequency ratios occupy equal intervals on the scale. Hence the fractional accuracy of reading is constant. There are fixed resistors in series with the variable parts of P and Q having about one-tenth the value of the variable resistor, which limit the range of the frequency scale to a ratio of ten to one. The three frequency ranges, differ -ing by factors of ten from one another, are obtained by the use of three sets of condensers Cr and CQ, which also differ by factors of ten. The same engraved scale is used for all three ranges. The frequency ranges chosen are 20 to 200, 200 to 2000, and 2000 to 20,000 cycles per second.
It is impracticable to keep the resistors P and Q and the condensers CP anil Cq exactly equal as demanded by equation (2). An auxiliary control consisting of a small potentiometer is provided between the ratio arms A and B to whose sliding contact the null detector is connected. This alters the effective ratio A/B and satisfies equation (1). If this adjustment is not made, however, the setting of the frequency dial is not altered but merely dulled.
The null detector most often used for setting the frequency dial is a pair of head telephones. These are satisfactory in the frequency range from 300 to 5000 cycles per second. If the source of frequency to be measured contains harmonics, they will not be balanced out by the bridge and will be impressed on the telephone. The human ear can discriminate against a considerable amount of harmonic content. It must be aided by the use of a low-pass filler connected between the bridge and the telephones for high harmonic content or for the measurement of frequencies less than half the natural frequency of the telephones. When the voltage available to be applierl to the bridge is small, an amplifier must be used to obtain sufficient sensitivity. The General Radio Type 514-A Amplifier is exactly suited for this purpose because it maintains its voltage gain of 200 approximately constant from 20 to 20.000 cycles per second.
For frequencies outside the range of telephones a sensitive alternating-current voltmeter and amplifier must be
Figure 2. The Type 434-B Audio-Frequency Meter
used. The Type 488 Alternating-Current Voltmeters are well suited to this use. A suitable low-pass filter should be used because the alternating-current voltmeter lacks the power of discriminating against harmonics.
The frequency dial is hand calibrated to an accuracy of 0.5%. It may he set to an accuracy of 0.1% w hen sufficient amplification is provided, this figure being determined by the spacing of the individual wires oil the variable resistors P and Q.
The Type 434-B Audio-Frequency Meter will measure any frequency from 20 to 20,000 cycles per second. The usual sources are vacuum-tube oscillators of all kinds, tuning forks, and the lower range of magnetostriction rods.
The meter is equally suitable for measuring the frequency of the beat note obtained from two high frequency oscillators. An accuracy of 0.5% in the determination of this heat note means that the difference between the two high frequencies is known to 0.1 cps for a 20-cps beat and proportionally for higher frequencies. When an unknown high frequency is compared with a standard crystal oscillator there are usually available harmonics spaced 1000 cps apart. The beat note can then always be made less than 1000 cps so that the unknown frequency is measured in terms of the standard crystal to within 5 cps.
The price of this frequency bridge is \$125.00. — Robert F. Field
### A VACUUM TUBE VOLTMETER
hile the copper-oxide rectifier type of voltmeter, which can be made in a portable form with a full-scale deflection of 2 volts and with an impedance of the order of 10,000 ohms per volt, is a very convenient and useful instrument for many alternating-current measurements, it has, unfortunately, two disadvantages in that its impedance varies with the applied voltage and that its calibration error increases with the frequency.
For making measurements of small voltages in circuits of high impedance, —as, for instance, in measuring the gain at various points in amplifying systems-—Tor all voltage measurements al high audio, supersonic, and radio frequencies, and for tests upon resonant circuits, the thermionic or vacuum-tube voltmeter is much to be preferred and may, in fact, be indispensable. The accompanying illustration shows the new alternating-current-operated thermionic voltmeter being used to measure frequency response curves of tuned interstage transformers intended for operation at high audio frequencies.
Over a year ago, the General Radio Company announced the bridge-type thermionic voltmeter (Type 426-A) which was energized by a 22.5-volt battery. To eliminate the necessity for such a battery and thus fulfill the pres
Measuring the response characteristic of an interstage coupling transformer with the Type 626-A Vacuum-Tube Voltmeter. The meter is on the bench at the right
ent-day desire for more convenient equipment energized solely by a 110-volt 60-cycIe circuit and to supply a meter which shall be reliable at radio frequencies, the Type 626-A Vacuum-Tube Voltmeter is now offered to the experimenter.
This instrument is a rugged, serviceable, direct-reading, moderately-priced meter of the compensated, depressed-zero type, having a full-scale deflection of 3 volts, r.m.s. From 0.5 volt to 3.0 volts the scale is approximately uniform. The filaments of the 227-type tube (thermionic voltmeter) and the 171 -type tube (rectifier) used are run at subnormal voltages so that their normal life is considerably increased. Proper circuit design minimizes the wandering of the zero or aging. By incorporating an 874-type ballast tube, the chief source of error, fluctuation in line voltage, is reduced to only .05% of full scale value per volt change. A rheostat is provided for compensating a change in line voltage over a range from 100 to 120 volts. Due to the large thermal capacity of the cathode of the 227-type tube, small erratic changes in line voltage are ordinarily negligible.
These meters are individually calibrated to within 1% of full-scale value. Below 1500 kilocycles, the frequency error is negligible. At 3000 kilocycles, it is less than 2%, and at 4000 kilocycles, less than 4%. The input impedance is constant and approximately 10 megohms and no external direct-current path is required. Housed in a walnut cabinet 11 inches long by 8}-^ inches square, this instrument weighs fourteen pounds. The price, complete with tubes and attachment cord, is \$100.00 and the code word is ethic.
DIRECT MEASUREMENTS OF HARMONIC DISTORTION
Development work has recently been completed on an instrument which has a wide field of usefulness in the testing of apparatus for the electrical transmission or reproduction of speech, music, or vision.
The new instrument, the Type 536-A Distortion-Factor Meter, will measure the harmonic distortion in the modulated output of a broadcast transmitter. It can then determine how much of the distortion is due to the speech amplifiers. In a receiving set it can measure the harmonic distortion produced in the detector or in the audio amplifiers. By its means, also, measurements can quickly be made which will determine what load resistance should be used with a pentode output tube to obtain the maximum power output for various allowable distortion limits.
The distortion-factor meter was developed so that distortion measurements could be accurately made by a very much simplified technique. After a preliminary observation, the total harmonic distortion is given directly by a dial reading. Less than thirty seconds is required for the entire measurement.
In the system employed the voltage to be tested is applied to two filters in parallel, one of which passes the harmonics only and the other of which passes the fundamental. A voltage divider is placed in the line passing the fundamental. This is varied until the output voltages of the two lines are
Apparatus for determining harmonic distortion. I-ej't to right: Type 536-A Distortion-Factor Meter, Type 514-A Amplifier, and Type 488-HM Alternating-Current Galvanometer
equal. A dial, attached to the voltage divider, then reads directly I lie harmonic content.
The distortion-factor meter is not to he confused with a harmonic analyser. This latter instrument is a sharply selective device which will measure the amplitudes of the various components of a complex current or voltage wave. It is very laborious to use a harmonic analyser for distortion measurements. A separate measurement must be made for each harmonic present, and the results combined to obtain a useful result. The distortion-factor meter, on the other hand, is designed so that one of the filters transmits equally all the harmonics, while suppressing the fundamental and any power-supply hum which may be present. If a root-mean-square indicating instrument is employed, therefore, the reading obtained with the distortion-factor meter will give directly the desired ratio of the effective value of the combined harmonics to the value of the fundamental.
### No computation whatever is required.
Let us consider in more detail one of the applications of the instrument, the testing of broadcast transmitters. As the distortion-factor meter is designed for testing at 400 cps, a fairly pure voltage source of this frequency must lie available. With most oscillators, a filter must be employed to obtain sufficient purity. The filtered oscillator output is applied to the input of the speech amplifiers.
If it is desired first of all to check the operation of the speech amplifiers, the dis tortion-factor meter may be used to test the output of the modulator tube or of any amplifier stage preceding it. Finally, the most important measurement of distortion in the modulated output may be made after first rectifying the high-frequency voltage with a high-quality rectifier. A carefully designed linear rectifier is incorporated ifi the Type 457-A Modulation Meter, and terminals are provided for connection to the distortion-factor
transmitter under test
transmitter under test
Layout for measuring distortion present in the output of a broadcast transmitter meter. The complete layout is indicated in lhe figure.
It is believed that this measurement of modulation distortion is the final index of the quality of the output of a broadcast transmitter, as far as harmonic distortion is concerned. It lumps together all the sources of harmonic distortion in the transmitter and measures their combined effect on the shape of the output voltage wave.
This discussion of the application of the distortion-factor meter to the single problem of the testing of broadcast transmitters will serve as an indication of its many uses in electric communication technique.
The price of the Type 536-A Distortion-Factor Meter is \$140.00.
A WAVEMETER FOR THE 1 -15 METER BAND
Type 119-A Wavemeter
Anew instrument has been added to the line of General Radio wave-meters. It is the Type 419-A Rectifier-Type Wavemeter, covering the range from 1 to 15 meters or 300 to 20 mega cycles per second. The instrument is primarily intended for laboratory and
Type 119-A Wavemeter service use where a rapid and fairly accurate measurement is required. In this way, inaccurate and bothersome Lecher wires are eliminated where wave lengths are measured with a yardstick. Stable oscillators have made it possible to use heterodyne methods even at 1 meter and to calibrate the Type 419-A Rectifier-Type Wave-meter in terms of the General Radio primary standard of frequency.
Besides measuring oscillators, transmitters, and receivers, the instrument can be used as an aural detector of modulated oscillations and as an un-calibrated vacuum-tube voltmeter. The accompanying photograph shows the instrument. Dimensions and weight of the cabinet have been reduced as much as possible and a handle is provided 011 the panel. This way it is possible to hold the instrument with one hand and bring it in any desired position with respect to the apparatus which is to be measured.
Four inductors, each with a fre
RESISTA NCE BOXES a * a Prices on three sizes of the Type 102 Decade-Resistance Boxes have been reduced as follows:
Type 102-L, \$58.00 (was \$75.00) Type 102-M, \$70.00 (was \$100.00) Type 102-N, \$62.00 (1was \$75.00)
The Type 102-L Decade Resistance Box is a four-dial box, having a maxi quency ratio of 1:2, cover the entire range of the wavemeter. A large dial makes it possible to read settings to within 0.1%. With respect to errors that are likely to occur due to the presence of metal parts in the field of the inductor and "transformer action," the accuracy has been conservatively set at 1%.
As an indicating device a vacuum-tube voltmeter is used, the high sensitivity of which is very advantageous, since it makes it possible to use a very loose coupling between ihe wavemeter and the apparatus under measurement. Besides the permanently connected indicating instrument, plugs are provided for the connection of headphones. The cabinet of the wavemeter contains the four plug-in inductors and four separate calibration charts. The calibration is given in terms of frequency, but 011 each chart a conversion curve is provided which makes it possible to read in wave length as well. The price of this meter is \$100. —Edu vrd Karplus mum resistance of 111,100 ohms, adjustable in steps of 10.0 ohms; the Type 102-M Decade-Resistance Box has five dials with a maximum resistance of 111,110 ohms in steps of 1 ohm; anil the Type 102-N Decade-Resistance Box is a five-dial box with a maximum resistance of 111,111.0, adjustable in steps of 0.1 ohm.
THE GENERAL RADIO COMPANY mails the Experimenter, without charge, each month to engineers, scientists, and others interested in communication-frequency measurement and control problems. Please send requests for subscriptions and address-change notices to the | 3,464 | 16,311 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2019-26 | longest | en | 0.919014 |
https://english.stackexchange.com/questions/464343/how-should-i-use-percentage | 1,571,574,671,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986707990.49/warc/CC-MAIN-20191020105426-20191020132926-00339.warc.gz | 475,107,971 | 30,891 | # How should I use percentage?
I am totally confused as to the right use of "percentage"! I have seen certain of the following in various texts, I do not know whichever I can safely use in my own text, though.
Which of the following is/are absolutely correct, academically speaking?
1. The percentage of x and y
2. The percentages of x and y
3. The percentage of x and of y
4. The percentages of x and of y
Secondly, shall I use plural or singular verbs after it/them? Thank you.
One example on the Merriam-Webster site is:
The percentages of women completing high school and college were 95 percent and 52 percent, respectively.
After looking at several sources, I think this applies best to your situation since you are discussing data.
This would indicate using number 2 in your questions (and adding respectively to be more precise) is the most correct answer.
I could imagine examples may look like this:
• The percentages of x and y were increased and decreased, respectively.
• The percentages of x and y were fifteen and sixty, respectively.
• The percentages of x and y both increased by nine. | 247 | 1,112 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2019-43 | latest | en | 0.958992 |
https://help.libreoffice.org/latest/ko/text/sbasic/shared/03060300.html | 1,716,827,895,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059044.17/warc/CC-MAIN-20240527144335-20240527174335-00158.warc.gz | 247,172,749 | 3,459 | # Imp Operator
๋ ์์ ๋ํด ๋
ผ๋ฆฌ์ ํฌํจ์ ์ํํฉ๋๋ค.
## ๊ตฌ๋ฌธ:
``````
Result = Expression1 Imp Expression2
``````
## ๋งค๊ฐ ๋ณ์:
Result: ํฌํจ์ ๊ฒฐ๊ณผ๋ฅผ ํฌํจํ๋ ์์์ ์ซ์ ๋ณ์์
๋๋ค.
Expression1, Expression2: Imp ์ฐ์ฐ์๋ฅผ ์ฌ์ฉํ์ฌ ํ๊ฐํ ์์์ ์์
๋๋ค.
๋ถ๋ฆฌ์ธ ์์์ Imp ์ฐ์ฐ์๋ฅผ ์ฌ์ฉํ๋ฉด ์ฒซ ๋ฒ์งธ ์์ด True๋ก ํ๊ฐ๋๊ณ ๋ ๋ฒ์งธ ์์ด False๋ก ํ๊ฐ๋ ๊ฒฝ์ฐ์๋ง False๊ฐ ๊ตฌํด์ง๋๋ค.
๋นํธ ์์์ Imp ์ฐ์ฐ์๋ฅผ ์ฌ์ฉํ๋ฉด ๋นํธ๊ฐ ์ฒซ ๋ฒ์งธ ์์ ์ค์ ๋๊ณ ๋ ๋ฒ์งธ ์์์ ์ญ์ ๋ ๊ฒฝ์ฐ ํด๋น ๋นํธ๊ฐ ๊ฒฐ๊ณผ์์ ์ญ์ ๋ฉ๋๋ค.
## ์:
``````
Sub ExampleImp
Dim A As Variant, B As Variant, C As Variant, D As Variant
Dim vOut As Variant
A = 10: B = 8: C = 6: D = Null
vOut = A > B Imp B > C REM returns -1
vOut = B > A Imp B > C REM returns -1
vOut = A > B Imp B > D REM returns 0
vOut = (B > D Imp B > A) REM returns -1
vOut = B Imp A REM returns -1
End Sub
`````` | 700 | 1,000 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-22 | latest | en | 0.293871 |
https://www.beyng.com/pages/en/PlatosSophist/PlatosSophist.072.html | 1,701,168,862,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099281.67/warc/CC-MAIN-20231128083443-20231128113443-00480.warc.gz | 743,698,366 | 2,895 | Plato's Sophist 72
72
Plato's Sophist [104-105]
Then how are the two basic objects of mathematics to be distinguished? μονὰς οὐσία ἄθετος, στιγμὴ δὲ οὐσία θετός· ταύτην ἐκ προσθέσεως (Post. An. I, 27, 87a36).
Both are οὐσία, something that is for itself. The στιγμή, however, over and against the μονάς, is marked by a πρόσθεσις; in the στιγμή there resides a θέσις in a preeminent sense. What is the meaning of this θέσις which characterizes the point in opposition to the μονάς? A thorough elucidation of this nexus would have to take up the question of place and space. Here I can only indicate what is necessary to make understandable the distinction of the ἀκριβές within the disciplines of mathematics.
Θέσις has the same character as ἕξις, διάθεσις. "Ἕξις = to find oneself in a definite situation, to have something in oneself, to retain, and in retaining to be directed toward something. Θέσις = orientation, situation; it has the character of being oriented toward something. ἔστι δὲ καὶ τὰ τοιαῦτα τῶν πρός τι οἷον ἕξις, διάθεσις,... θέσις (Cat. 7, 6b2f.). According to its categorial determination, θέσις is τῶν πρός τι, "it belongs to what is πρός τι." Every θέσις is a θέσις τινός (cf. b6).
α) Τόπος and θέσις (according to Phys. V, 1-5). The absolute
determinateness (φύσει) of τόπος, the relative determinateness
(πρὸς ἡμᾶς) of θέσις. The essence of τόπος: limit (πέρας) and
possibility (δύναμις) of the proper Being of a being.
We need to clarify briefly the distinction between θέσις and τόπος. Aristotle emphasizes that the mathematical objects are οὐκ ἐν τόπῳ (cf. Met. XIV, 5, 1092a19f.), "not anyplace."3 The modern concept of space must not at all be allowed to intrude here. Aristotle determines τόπος at first in an apparently quite naive way. ὅτι μὲν οὖν ἔστιν ὁ τόπος, δοκεῖ δῆλον εἶναι ἐκ τῆς ἀντιμεταστάσεως· ὅπου γὰρ ἔστι νῦν ὕδωρ, ἐνταῦθα ἐξελθόντος ὥσπερ ἐξ ἀγγείου πάλιν ἀὴρ ἐκεῖ ἔνεστιν, ὁτὲ δὲ τὸν αὐτὸν τόπον τοῦτον ἄλλο τι τῶν σωμάτων κατέχει· τοῦτο δὴ τῶν ἐγγιγνομένων καὶ μεταβαλλόντων ἕτερον πάντων εἶναι δοκεῖ· ἐν ᾧ γὰρ ἀήρ ἐστι νῦν, ὕδωρ ἐν τούτῳ πρότερον ἦν, ὥστε δῆλον ὡς ἦν ὁ τόπος τι καὶ ἡ χώρα ἕτερον ἀμφοῖν, εἰς ἣν καὶ ἐξ ἧς μετέβαλον (Phys. IV, 1, 208b1ff.). Τόπος must itself be something. If there formerly was water in a container and if now there is air in it, then the τόπος is something other than that which fills it. The place was already, ἦν, i.e., before specifically water or air was in it. The ἦν does not mean that the τόπος would be something separated, separated from what is in it; the place is simply something other than the two things which have been exchanged in it.
2. Cf. De An. I, 4, 409a6ff.
3. Cf. p. 69. | 925 | 2,688 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2023-50 | latest | en | 0.797681 |
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# The common factor in the expression 4a+6a2 is ___.
Solution
## The expression 4a+6a2 can be written as 2×2×a+2×3×a×a Taking out 2a common from the expression, we get =2a(2+3a) Therefore, the common factor in the given expression is 2a. Mathematics
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## #1 2013-01-07 08:45:52
anonimnystefy
Real Member
Offline
### A weird limit!
Hi
Can anyone help me with the following limit:
I know that it should be equal to 1/2 and the the Stolz theorem is supposed to be used, but I cannot get the final result...
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
## #2 2013-01-07 09:12:31
bobbym
Offline
### Re: A weird limit!
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
## #3 2013-01-07 09:25:07
anonimnystefy
Real Member
Offline
### Re: A weird limit!
No. I already looked at that page. It just gets the limit to the indeterminate from 0*infinity...
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
## #4 2013-01-07 09:51:59
bobbym
Offline
### Re: A weird limit!
We need to use the binomial theorem here now.
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
## #5 2013-01-07 09:54:32
anonimnystefy
Real Member
Offline
### Re: A weird limit!
Do you see the n/(p+1) part? It makes it not possible to use Stolz like that.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
## #6 2013-01-07 09:57:02
bobbym
Offline
### Re: A weird limit!
We are not allowed to split that limit?
Probably not, the two pieces are both equal to infinity.
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
## #7 2013-01-07 10:07:31
anonimnystefy
Real Member
Offline
### Re: A weird limit!
That is right.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
## #8 2013-01-07 10:14:58
bobbym
Offline
### Re: A weird limit!
I am trying to put it into the required form but the algebra is hideous.
The above limit equals 1 / 2 . But the limit is very difficult to handle.
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
## #9 2013-01-07 10:58:08
anonimnystefy
Real Member
Offline
### Re: A weird limit!
I already got that much. That limit there is the problem...
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
## #10 2013-01-07 11:04:05
bobbym
Offline
### Re: A weird limit!
You got that far! How the heck did you get there?
Maybe we can use Stolz again?
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
## #11 2013-01-07 11:12:44
anonimnystefy
Real Member
Offline
### Re: A weird limit!
It wasn't hard.
I don't think ot would get much prettier...
Last edited by anonimnystefy (2013-01-07 11:13:11)
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
## #12 2013-01-07 11:42:20
bobbym
Offline
### Re: A weird limit!
Why do you think this is a job for Stoltz then?
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
## #13 2013-01-07 11:51:08
scientia
Full Member
Offline
### Re: A weird limit!
#### bobbym wrote:
The above limit equals 1 / 2 . But the limit is very difficult to handle.
Consider the numerator (ignoring the minus sign outside the fraction). Notice that the terms in
and
vanish so the highest power of
is
. Its coefficient is
. Now look at the denominator. The highest power of
is also
and its coefficient is
. Hence the limit as
is
using the following rule:
If
and
and
then
Last edited by scientia (2013-01-07 20:52:47)
## #14 2013-01-07 12:17:22
anonimnystefy
Real Member
Offline
### Re: A weird limit!
Wow! That is amazing! Thank you!
Last edited by anonimnystefy (2013-01-07 12:17:45)
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
scientia
Full Member
Offline
You're welcome. | 1,704 | 6,290 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2013-20 | longest | en | 0.927357 |
http://www.kkt.as/index.php/library/a-guide-to-arithmetic-lecture-notes | 1,603,693,568,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107890586.57/warc/CC-MAIN-20201026061044-20201026091044-00146.warc.gz | 144,928,582 | 8,311 | # A Guide to Arithmetic [Lecture notes] by Robin Chapman
By Robin Chapman
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1 [ 2 [ 3 ¤ ;. 4, to elastic motions of a body occupying a domain with a piecewise smooth boundary . x/, i D 1; 2; 3, are fixed at every point x 2 where the normal n exists.
While, for example, steel obeys Hooke’s law in a relatively wide band of loading parameters, such materials as cast iron deviate from the linear behavior even at relatively small stresses. 55) where C is the elastic modulus tensor of the fourth rank. The components of the tensor C do not depend on the spatial coordinates in the case of homogeneous media. As the stress and strain components depend on the orientation of the Cartesian coordinate system, the components of the elastic modulus tensor must also comply with this dependence.
X/ W : Here, C 1 is the elastic compliance tensor inversed to the tensor C . This object is of the fourth rank and has the same symmetry properties as C . By using either eq. 55) or eq. 63) WC 1W : 2 As it has been already mentioned, the maximal number of elastic moduli in the elastic modulus tensor is 21. Such kind of material anisotropy is named as aeolotropy. , electric conductivity, refractive index) of a body depend on the direction in which they are measured. Note that the number of elastic moduli relates to material symmetry. | 657 | 2,911 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2020-45 | longest | en | 0.668969 |
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 1,097 | 3,633 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2020-50 | latest | en | 0.908058 |
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I have an urn with b blue balls and r red balls, what is the probability that the two balls will have the same color. I need to solve this problem with replacement and without replacement. This is my attempt at a solution. Without replacement, we add the probabilities of the balls both being blue and the balls both being red. That is, the final probability would be $\frac{b(b-1)+r(r-1)}{br}$. With replacement, the final probability would be $\frac{b*b+r*r}{br}$. Is my logic correct in determining the probability?
• With replacement it would just be (b+r)(b+r)? – user288829 Jan 26 '16 at 17:29
• Yes, for replacement it is $(b+r)(b+r)$. – André Nicolas Jan 26 '16 at 18:06
No total ways $${b+r\choose 2}$$ so probability will be $\frac{(b)(b-1)+(r)(r-1)}{(b+r)(b+r-1)}$ same for other one | 236 | 818 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2021-17 | latest | en | 0.883894 |
https://db0nus869y26v.cloudfront.net/en/Tikhonov_regularization | 1,680,124,172,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949025.18/warc/CC-MAIN-20230329182643-20230329212643-00433.warc.gz | 238,608,601 | 31,411 | It has been suggested that Ridge regression be merged into this article. (Discuss) Proposed since March 2021.
Tikhonov regularization, named for Andrey Tikhonov, is a method of regularization of ill-posed problems. Also known as ridge regression,[a] it is particularly useful to mitigate the problem of multicollinearity in linear regression, which commonly occurs in models with large numbers of parameters.[1] In general, the method provides improved efficiency in parameter estimation problems in exchange for a tolerable amount of bias (see bias–variance tradeoff).[2]
In the simplest case, the problem of a near-singular moment matrix ${\displaystyle (\mathbf {X} ^{\mathsf {T))\mathbf {X} )}$ is alleviated by adding positive elements to the diagonals, thereby decreasing its condition number. Analogous to the ordinary least squares estimator, the simple ridge estimator is then given by
${\displaystyle {\hat {\beta ))_{R}=(\mathbf {X} ^{\mathsf {T))\mathbf {X} +\lambda \mathbf {I} )^{-1}\mathbf {X} ^{\mathsf {T))\mathbf {y} }$
where ${\displaystyle \mathbf {y} }$ is the regressand, ${\displaystyle \mathbf {X} }$ is the design matrix, ${\displaystyle \mathbf {I} }$ is the identity matrix, and the ridge parameter ${\displaystyle \lambda \geq 0}$ serves as the constant shifting the diagonals of the moment matrix.[3] It can be shown that this estimator is the solution to the least squares problem subject to the constraint ${\displaystyle \beta ^{\mathsf {T))\beta =c}$, which can be expressed as a Lagrangian:
${\displaystyle \min _{\beta }\,(\mathbf {y} -\mathbf {X} \beta )^{\mathsf {T))(\mathbf {y} -\mathbf {X} \beta )+\lambda (\beta ^{\mathsf {T))\beta -c)}$
which shows that ${\displaystyle \lambda }$ is nothing but the Lagrange multiplier of the constraint. Typically, ${\displaystyle \lambda }$ is chosen according to a heuristic criterion, so that the constraint will not be satisfied exactly. Specifically in the case of ${\displaystyle \lambda =0}$, in which the constraint is non-binding, the ridge estimator reduces to ordinary least squares. A more general approach to Tikhonov regularization is discussed below.
## History
Tikhonov regularization has been invented independently in many different contexts. It became widely known from its application to integral equations from the work of Andrey Tikhonov[4][5][6][7][8] and David L. Phillips.[9] Some authors use the term Tikhonov–Phillips regularization. The finite-dimensional case was expounded by Arthur E. Hoerl, who took a statistical approach,[10] and by Manus Foster, who interpreted this method as a Wiener–Kolmogorov (Kriging) filter.[11] Following Hoerl, it is known in the statistical literature as ridge regression,[12] named after the shape along the diagonal of the identity matrix.
## Tikhonov regularization
Suppose that for a known matrix ${\displaystyle A}$ and vector ${\displaystyle \mathbf {b} }$, we wish to find a vector ${\displaystyle \mathbf {x} }$ such that[clarification needed]
${\displaystyle A\mathbf {x} =\mathbf {b} .}$
The standard approach is ordinary least squares linear regression.[clarification needed] However, if no ${\displaystyle \mathbf {x} }$ satisfies the equation or more than one ${\displaystyle \mathbf {x} }$ does—that is, the solution is not unique—the problem is said to be ill posed. In such cases, ordinary least squares estimation leads to an overdetermined, or more often an underdetermined system of equations. Most real-world phenomena have the effect of low-pass filters in the forward direction where ${\displaystyle A}$ maps ${\displaystyle \mathbf {x} }$ to ${\displaystyle \mathbf {b} }$. Therefore, in solving the inverse-problem, the inverse mapping operates as a high-pass filter that has the undesirable tendency of amplifying noise (eigenvalues / singular values are largest in the reverse mapping where they were smallest in the forward mapping). In addition, ordinary least squares implicitly nullifies every element of the reconstructed version of ${\displaystyle \mathbf {x} }$ that is in the null-space of ${\displaystyle A}$, rather than allowing for a model to be used as a prior for ${\displaystyle \mathbf {x} }$. Ordinary least squares seeks to minimize the sum of squared residuals, which can be compactly written as
${\displaystyle \|A\mathbf {x} -\mathbf {b} \|_{2}^{2},}$
where ${\displaystyle \|\cdot \|_{2))$ is the Euclidean norm.
In order to give preference to a particular solution with desirable properties, a regularization term can be included in this minimization:
${\displaystyle \|A\mathbf {x} -\mathbf {b} \|_{2}^{2}+\|\Gamma \mathbf {x} \|_{2}^{2))$
for some suitably chosen Tikhonov matrix ${\displaystyle \Gamma }$. In many cases, this matrix is chosen as a scalar multiple of the identity matrix (${\displaystyle \Gamma =\alpha I}$), giving preference to solutions with smaller norms; this is known as L2 regularization.[13] In other cases, high-pass operators (e.g., a difference operator or a weighted Fourier operator) may be used to enforce smoothness if the underlying vector is believed to be mostly continuous. This regularization improves the conditioning of the problem, thus enabling a direct numerical solution. An explicit solution, denoted by ${\displaystyle {\hat {x))}$, is given by
${\displaystyle {\hat {x))=(A^{\top }A+\Gamma ^{\top }\Gamma )^{-1}A^{\top }\mathbf {b} .}$
The effect of regularization may be varied by the scale of matrix ${\displaystyle \Gamma }$. For ${\displaystyle \Gamma =0}$ this reduces to the unregularized least-squares solution, provided that (ATA)−1 exists.
L2 regularization is used in many contexts aside from linear regression, such as classification with logistic regression or support vector machines,[14] and matrix factorization.[15]
### Generalized Tikhonov regularization
For general multivariate normal distributions for ${\displaystyle x}$ and the data error, one can apply a transformation of the variables to reduce to the case above. Equivalently, one can seek an ${\displaystyle x}$ to minimize
${\displaystyle \|Ax-b\|_{P}^{2}+\|x-x_{0}\|_{Q}^{2},}$
where we have used ${\displaystyle \|x\|_{Q}^{2))$ to stand for the weighted norm squared ${\displaystyle x^{\top }Qx}$ (compare with the Mahalanobis distance). In the Bayesian interpretation ${\displaystyle P}$ is the inverse covariance matrix of ${\displaystyle b}$, ${\displaystyle x_{0))$ is the expected value of ${\displaystyle x}$, and ${\displaystyle Q}$ is the inverse covariance matrix of ${\displaystyle x}$. The Tikhonov matrix is then given as a factorization of the matrix ${\displaystyle Q=\Gamma ^{\top }\Gamma }$ (e.g. the Cholesky factorization) and is considered a whitening filter.
This generalized problem has an optimal solution ${\displaystyle x^{*))$ which can be written explicitly using the formula
${\displaystyle x^{*}=(A^{\top }PA+Q)^{-1}(A^{\top }Pb+Qx_{0}),}$
or equivalently
${\displaystyle x^{*}=x_{0}+(A^{\top }PA+Q)^{-1}(A^{\top }P(b-Ax_{0})).}$
## Lavrentyev regularization
In some situations, one can avoid using the transpose ${\displaystyle A^{\top ))$, as proposed by Mikhail Lavrentyev.[16] For example, if ${\displaystyle A}$ is symmetric positive definite, i.e. ${\displaystyle A=A^{\top }>0}$, so is its inverse ${\displaystyle A^{-1))$, which can thus be used to set up the weighted norm squared ${\displaystyle \|x\|_{P}^{2}=x^{\top }A^{-1}x}$ in the generalized Tikhonov regularization, leading to minimizing
${\displaystyle \|Ax-b\|_{A^{-1))^{2}+\|x-x_{0}\|_{Q}^{2))$
or, equivalently up to a constant term,
${\displaystyle x^{\top }(A+Q)x-2x^{\top }(b+Qx_{0})}$.
This minimization problem has an optimal solution ${\displaystyle x^{*))$ which can be written explicitly using the formula
${\displaystyle x^{*}=(A+Q)^{-1}(b+Qx_{0})}$,
which is nothing but the solution of the generalized Tikhonov problem where ${\displaystyle A=A^{\top }=P^{-1}.}$
The Lavrentyev regularization, if applicable, is advantageous to the original Tikhonov regularization, since the Lavrentyev matrix ${\displaystyle A+Q}$ can be better conditioned, i.e., have a smaller condition number, compared to the Tikhonov matrix ${\displaystyle A^{\top }A+\Gamma ^{\top }\Gamma .}$
## Regularization in Hilbert space
Typically discrete linear ill-conditioned problems result from discretization of integral equations, and one can formulate a Tikhonov regularization in the original infinite-dimensional context. In the above we can interpret ${\displaystyle A}$ as a compact operator on Hilbert spaces, and ${\displaystyle x}$ and ${\displaystyle b}$ as elements in the domain and range of ${\displaystyle A}$. The operator ${\displaystyle A^{*}A+\Gamma ^{\top }\Gamma }$ is then a self-adjoint bounded invertible operator.
## Relation to singular-value decomposition and Wiener filter
With ${\displaystyle \Gamma =\alpha I}$, this least-squares solution can be analyzed in a special way using the singular-value decomposition. Given the singular value decomposition
${\displaystyle A=U\Sigma V^{\top ))$
with singular values ${\displaystyle \sigma _{i))$, the Tikhonov regularized solution can be expressed as
${\displaystyle {\hat {x))=VDU^{\top }b,}$
where ${\displaystyle D}$ has diagonal values
${\displaystyle D_{ii}={\frac {\sigma _{i}^{2)){\sigma _{i}^{2}+\alpha ^{2))))$
and is zero elsewhere. This demonstrates the effect of the Tikhonov parameter on the condition number of the regularized problem. For the generalized case, a similar representation can be derived using a generalized singular-value decomposition.[17]
Finally, it is related to the Wiener filter:
${\displaystyle {\hat {x))=\sum _{i=1}^{q}f_{i}{\frac {u_{i}^{\top }b}{\sigma _{i))}v_{i},}$
where the Wiener weights are ${\displaystyle f_{i}={\frac {\sigma _{i}^{2)){\sigma _{i}^{2}+\alpha ^{2))))$ and ${\displaystyle q}$ is the rank of ${\displaystyle A}$.
## Determination of the Tikhonov factor
The optimal regularization parameter ${\displaystyle \alpha }$ is usually unknown and often in practical problems is determined by an ad hoc method. A possible approach relies on the Bayesian interpretation described below. Other approaches include the discrepancy principle, cross-validation, L-curve method,[18] restricted maximum likelihood and unbiased predictive risk estimator. Grace Wahba proved that the optimal parameter, in the sense of leave-one-out cross-validation minimizes[19][20]
${\displaystyle G={\frac {\operatorname {RSS} }{\tau ^{2))}={\frac {\|X{\hat {\beta ))-y\|^{2)){[\operatorname {Tr} (I-X(X^{T}X+\alpha ^{2}I)^{-1}X^{T})]^{2))},}$
where ${\displaystyle \operatorname {RSS} }$ is the residual sum of squares, and ${\displaystyle \tau }$ is the effective number of degrees of freedom.
Using the previous SVD decomposition, we can simplify the above expression:
${\displaystyle \operatorname {RSS} =\left\|y-\sum _{i=1}^{q}(u_{i}'b)u_{i}\right\|^{2}+\left\|\sum _{i=1}^{q}{\frac {\alpha ^{2)){\sigma _{i}^{2}+\alpha ^{2))}(u_{i}'b)u_{i}\right\|^{2},}$
${\displaystyle \operatorname {RSS} =\operatorname {RSS} _{0}+\left\|\sum _{i=1}^{q}{\frac {\alpha ^{2)){\sigma _{i}^{2}+\alpha ^{2))}(u_{i}'b)u_{i}\right\|^{2},}$
and
${\displaystyle \tau =m-\sum _{i=1}^{q}{\frac {\sigma _{i}^{2)){\sigma _{i}^{2}+\alpha ^{2))}=m-q+\sum _{i=1}^{q}{\frac {\alpha ^{2)){\sigma _{i}^{2}+\alpha ^{2))}.}$
## Relation to probabilistic formulation
The probabilistic formulation of an inverse problem introduces (when all uncertainties are Gaussian) a covariance matrix ${\displaystyle C_{M))$ representing the a priori uncertainties on the model parameters, and a covariance matrix ${\displaystyle C_{D))$ representing the uncertainties on the observed parameters.[21] In the special case when these two matrices are diagonal and isotropic, ${\displaystyle C_{M}=\sigma _{M}^{2}I}$ and ${\displaystyle C_{D}=\sigma _{D}^{2}I}$, and, in this case, the equations of inverse theory reduce to the equations above, with ${\displaystyle \alpha ={\sigma _{D))/{\sigma _{M))}$.
## Bayesian interpretation
Main article: Bayesian interpretation of regularization
Further information: Minimum mean square error § Linear MMSE estimator for linear observation process
Although at first the choice of the solution to this regularized problem may look artificial, and indeed the matrix ${\displaystyle \Gamma }$ seems rather arbitrary, the process can be justified from a Bayesian point of view. Note that for an ill-posed problem one must necessarily introduce some additional assumptions in order to get a unique solution. Statistically, the prior probability distribution of ${\displaystyle x}$ is sometimes taken to be a multivariate normal distribution. For simplicity here, the following assumptions are made: the means are zero; their components are independent; the components have the same standard deviation ${\displaystyle \sigma _{x))$. The data are also subject to errors, and the errors in ${\displaystyle b}$ are also assumed to be independent with zero mean and standard deviation ${\displaystyle \sigma _{b))$. Under these assumptions the Tikhonov-regularized solution is the most probable solution given the data and the a priori distribution of ${\displaystyle x}$, according to Bayes' theorem.[22]
If the assumption of normality is replaced by assumptions of homoscedasticity and uncorrelatedness of errors, and if one still assumes zero mean, then the Gauss–Markov theorem entails that the solution is the minimal unbiased linear estimator.[23]
## Notes
1. ^ In statistics, the method is known as ridge regression, in machine learning it and its modifications are known as weight decay, and with multiple independent discoveries, it is also variously known as the Tikhonov–Miller method, the Phillips–Twomey method, the constrained linear inversion method, L2 regularization, and the method of linear regularization. It is related to the Levenberg–Marquardt algorithm for non-linear least-squares problems.
## References
1. ^ Kennedy, Peter (2003). A Guide to Econometrics (Fifth ed.). Cambridge: The MIT Press. pp. 205–206. ISBN 0-262-61183-X.
2. ^ Gruber, Marvin (1998). Improving Efficiency by Shrinkage: The James–Stein and Ridge Regression Estimators. Boca Raton: CRC Press. pp. 7–15. ISBN 0-8247-0156-9.
3. ^ For the choice of ${\displaystyle \lambda }$ in practice, see Khalaf, Ghadban; Shukur, Ghazi (2005). "Choosing Ridge Parameter for Regression Problems". Communications in Statistics – Theory and Methods. 34 (5): 1177–1182. doi:10.1081/STA-200056836. S2CID 122983724.
4. ^ Tikhonov, Andrey Nikolayevich (1943). "Об устойчивости обратных задач" [On the stability of inverse problems]. Doklady Akademii Nauk SSSR. 39 (5): 195–198. Archived from the original on 2005-02-27.
5. ^ Tikhonov, A. N. (1963). "О решении некорректно поставленных задач и методе регуляризации". Doklady Akademii Nauk SSSR. 151: 501–504.. Translated in "Solution of incorrectly formulated problems and the regularization method". Soviet Mathematics. 4: 1035–1038.
6. ^ Tikhonov, A. N.; V. Y. Arsenin (1977). Solution of Ill-posed Problems. Washington: Winston & Sons. ISBN 0-470-99124-0.
7. ^ Tikhonov, Andrey Nikolayevich; Goncharsky, A.; Stepanov, V. V.; Yagola, Anatolij Grigorevic (30 June 1995). Numerical Methods for the Solution of Ill-Posed Problems. Netherlands: Springer Netherlands. ISBN 079233583X. Retrieved 9 August 2018.
8. ^ Tikhonov, Andrey Nikolaevich; Leonov, Aleksandr S.; Yagola, Anatolij Grigorevic (1998). Nonlinear ill-posed problems. London: Chapman & Hall. ISBN 0412786605. Retrieved 9 August 2018.
9. ^ Phillips, D. L. (1962). "A Technique for the Numerical Solution of Certain Integral Equations of the First Kind". Journal of the ACM. 9: 84–97. doi:10.1145/321105.321114. S2CID 35368397.
10. ^ Hoerl, Arthur E. (1962). "Application of Ridge Analysis to Regression Problems". Chemical Engineering Progress. 58 (3): 54–59.
11. ^ Foster, M. (1961). "An Application of the Wiener-Kolmogorov Smoothing Theory to Matrix Inversion". Journal of the Society for Industrial and Applied Mathematics. 9 (3): 387–392. doi:10.1137/0109031.
12. ^ Hoerl, A. E.; R. W. Kennard (1970). "Ridge regression: Biased estimation for nonorthogonal problems". Technometrics. 12 (1): 55–67. doi:10.1080/00401706.1970.10488634.
13. ^ Ng, Andrew Y. (2004). Feature selection, L1 vs. L2 regularization, and rotational invariance (PDF). Proc. ICML.
14. ^ R.-E. Fan; K.-W. Chang; C.-J. Hsieh; X.-R. Wang; C.-J. Lin (2008). "LIBLINEAR: A library for large linear classification". Journal of Machine Learning Research. 9: 1871–1874.
15. ^ Guan, Naiyang; Tao, Dacheng; Luo, Zhigang; Yuan, Bo (2012). "Online nonnegative matrix factorization with robust stochastic approximation". IEEE Transactions on Neural Networks and Learning Systems. 23 (7): 1087–1099. doi:10.1109/TNNLS.2012.2197827. PMID 24807135. S2CID 8755408.
16. ^ Lavrentiev, M. M. (1967). Some Improperly Posed Problems of Mathematical Physics. New York: Springer.
17. ^ Hansen, Per Christian (Jan 1, 1998). Rank-Deficient and Discrete Ill-Posed Problems: Numerical Aspects of Linear Inversion (1st ed.). Philadelphia, USA: SIAM. ISBN 9780898714036.
18. ^ P. C. Hansen, "The L-curve and its use in the numerical treatment of inverse problems", [1]
19. ^ Wahba, G. (1990). "Spline Models for Observational Data". CBMS-NSF Regional Conference Series in Applied Mathematics. Society for Industrial and Applied Mathematics. Bibcode:1990smod.conf.....W.
20. ^ Golub, G.; Heath, M.; Wahba, G. (1979). "Generalized cross-validation as a method for choosing a good ridge parameter" (PDF). Technometrics. 21 (2): 215–223. doi:10.1080/00401706.1979.10489751.
21. ^ Tarantola, Albert (2005). Inverse Problem Theory and Methods for Model Parameter Estimation (1st ed.). Philadelphia: Society for Industrial and Applied Mathematics (SIAM). ISBN 0898717922. Retrieved 9 August 2018.
22. ^ Vogel, Curtis R. (2002). Computational methods for inverse problems. Philadelphia: Society for Industrial and Applied Mathematics. ISBN 0-89871-550-4.
23. ^ Amemiya, Takeshi (1985). Advanced Econometrics. Harvard University Press. pp. 60–61. ISBN 0-674-00560-0. | 4,953 | 18,145 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 93, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2023-14 | latest | en | 0.858842 |
https://doc.sagemath.org/html/en/reference/game_theory/sage/game_theory/catalog_normal_form_games.html | 1,709,072,237,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474688.78/warc/CC-MAIN-20240227220707-20240228010707-00019.warc.gz | 205,805,063 | 13,620 | # A catalog of normal form games#
This allows us to construct common games directly:
sage: g = game_theory.normal_form_games.PrisonersDilemma()
sage: g
Prisoners dilemma - Normal Form Game with the following utilities: ...
We can then immediately obtain the Nash equilibrium for this game:
sage: g.obtain_nash()
[[(0, 1), (0, 1)]]
When we test whether the game is actually the one in question, sometimes we will build a dictionary to test it, since the printed representation can be platform-dependent, like so:
sage: d = {(0, 0): [-2, -2], (0, 1): [-5, 0], (1, 0): [0, -5], (1, 1): [-4, -4]}
sage: g == d
True
The docstrings give an interpretation of each game.
REFERENCES:
AUTHOR:
• James Campbell and Vince Knight (06-2014)
sage.game_theory.catalog_normal_form_games.AntiCoordinationGame(A=3, a=3, B=5, b=1, C=1, c=5, D=0, d=0)#
Return a 2 by 2 AntiCoordination Game.
An anti coordination game is a particular type of game where the pure Nash equilibria is for the players to pick different strategies.
In general these are represented as a normal form game using the following two matrices:
\begin{align}\begin{aligned}\begin{split}A = \begin{pmatrix} A&C\\ B&D\\ \end{pmatrix}\end{split}\\\begin{split}B = \begin{pmatrix} a&c\\ b&d\\ \end{pmatrix}\end{split}\end{aligned}\end{align}
Where $$A < B, D < C$$ and $$a < c, d < b$$.
An often used version is the following:
\begin{align}\begin{aligned}\begin{split}A = \begin{pmatrix} 3&1\\ 5&0\\ \end{pmatrix}\end{split}\\\begin{split} B = \begin{pmatrix} 3&5\\ 1&0\\ \end{pmatrix}\end{split}\end{aligned}\end{align}
This is the default version of the game created by this function:
sage: g = game_theory.normal_form_games.AntiCoordinationGame()
sage: g
Anti coordination game - Normal Form Game with the following utilities: ...
sage: d ={(0, 1): [1, 5], (1, 0): [5, 1],
....: (0, 0): [3, 3], (1, 1): [0, 0]}
sage: g == d
True
There are two pure Nash equilibria and one mixed:
sage: g.obtain_nash()
[[(0, 1), (1, 0)], [(1/3, 2/3), (1/3, 2/3)], [(1, 0), (0, 1)]]
We can also pass different values of the input parameters:
sage: g = game_theory.normal_form_games.AntiCoordinationGame(A=2, a=3,
....: B=4, b=2, C=2, c=8, D=1, d=0)
sage: g
Anti coordination game - Normal Form Game with the following utilities: ...
sage: d ={(0, 1): [2, 8], (1, 0): [4, 2],
....: (0, 0): [2, 3], (1, 1): [1, 0]}
sage: g == d
True
sage: g.obtain_nash()
[[(0, 1), (1, 0)], [(2/7, 5/7), (1/3, 2/3)], [(1, 0), (0, 1)]]
Note that an error is returned if the defining inequality is not obeyed $$A > B, D > C$$ and $$a > c, d > b$$:
sage: g = game_theory.normal_form_games.AntiCoordinationGame(A=8, a=3,
....: B=4, b=2, C=2, c=8, D=1, d=0)
Traceback (most recent call last):
...
TypeError: the input values for an Anti coordination game must be of the form A < B, D < C, a < c and d < b
sage.game_theory.catalog_normal_form_games.BattleOfTheSexes()#
Return a Battle of the Sexes game.
Consider two payers: Amy and Bob. Amy prefers to play video games and Bob prefers to watch a movie. They both however want to spend their evening together. This can be modeled as a normal form game using the following two matrices [Web2007]:
\begin{align}\begin{aligned}\begin{split}A = \begin{pmatrix} 3&1\\ 0&2\\ \end{pmatrix}\end{split}\\\begin{split} B = \begin{pmatrix} 2&1\\ 0&3\\ \end{pmatrix}\end{split}\end{aligned}\end{align}
This is a particular type of Coordination Game. There are three Nash equilibria:
1. Amy and Bob both play video games;
2. Amy and Bob both watch a movie;
3. Amy plays video games 75% of the time and Bob watches a movie 75% of the time.
This can be implemented in Sage using the following:
sage: g = game_theory.normal_form_games.BattleOfTheSexes()
sage: g
Battle of the sexes - Coordination game -
Normal Form Game with the following utilities: ...
sage: d = {(0, 1): [1, 1], (1, 0): [0, 0], (0, 0): [3, 2], (1, 1): [2, 3]}
sage: g == d
True
sage: g.obtain_nash()
[[(0, 1), (0, 1)], [(3/4, 1/4), (1/4, 3/4)], [(1, 0), (1, 0)]]
sage.game_theory.catalog_normal_form_games.Chicken(A=0, a=0, B=1, b=-1, C=-1, c=1, D=-10, d=-10)#
Return a Chicken game.
Consider two drivers locked in a fierce battle for pride. They drive towards a cliff and the winner is declared as the last one to swerve. If neither player swerves they will both fall off the cliff.
This can be modeled as a particular type of anti coordination game using the following two matrices:
\begin{align}\begin{aligned}\begin{split}A = \begin{pmatrix} A&C\\ B&D\\ \end{pmatrix}\end{split}\\\begin{split}B = \begin{pmatrix} a&c\\ b&d\\ \end{pmatrix}\end{split}\end{aligned}\end{align}
Where $$A < B, D < C$$ and $$a < c, d < b$$ but with the extra condition that $$A > C$$ and $$a > b$$.
Here are the numeric values used by default [Wat2003]:
\begin{align}\begin{aligned}\begin{split}A = \begin{pmatrix} 0&-1\\ 1&-10\\ \end{pmatrix}\end{split}\\\begin{split} B = \begin{pmatrix} 0&1\\ -1&-10\\ \end{pmatrix}\end{split}\end{aligned}\end{align}
There are three Nash equilibria:
1. The second player swerving.
2. The first player swerving.
3. Both players swerving with 1 out of 10 times.
This can be implemented in Sage using the following:
sage: g = game_theory.normal_form_games.Chicken()
sage: g
Chicken - Anti coordination game -
Normal Form Game with the following utilities: ...
sage: d = {(0, 1): [-1, 1], (1, 0): [1, -1],
....: (0, 0): [0, 0], (1, 1): [-10, -10]}
sage: g == d
True
sage: g.obtain_nash()
[[(0, 1), (1, 0)], [(9/10, 1/10), (9/10, 1/10)], [(1, 0), (0, 1)]]
Non default values can be passed:
sage: g = game_theory.normal_form_games.Chicken(A=0, a=0, B=2,
....: b=-1, C=-1, c=2, D=-100, d=-100)
sage: g
Chicken - Anti coordination game -
Normal Form Game with the following utilities: ...
sage: d = {(0, 1): [-1, 2], (1, 0): [2, -1],
....: (0, 0): [0, 0], (1, 1): [-100, -100]}
sage: g == d
True
sage: g.obtain_nash()
[[(0, 1), (1, 0)], [(99/101, 2/101), (99/101, 2/101)],
[(1, 0), (0, 1)]]
Note that an error is returned if the defining inequalities are not obeyed $$B > A > C > D$$ and $$c > a > b > d$$:
sage: g = game_theory.normal_form_games.Chicken(A=8, a=3, B=4, b=2,
....: C=2, c=8, D=1, d=0)
Traceback (most recent call last):
...
TypeError: the input values for a game of chicken must be of the form B > A > C > D and c > a > b > d
sage.game_theory.catalog_normal_form_games.CoordinationGame(A=10, a=5, B=0, b=0, C=0, c=0, D=5, d=10)#
Return a 2 by 2 Coordination Game.
A coordination game is a particular type of game where the pure Nash equilibrium is for the players to pick the same strategies [Web2007].
In general these are represented as a normal form game using the following two matrices:
\begin{align}\begin{aligned}\begin{split}A = \begin{pmatrix} A&C\\ B&D\\ \end{pmatrix}\end{split}\\\begin{split}B = \begin{pmatrix} a&c\\ b&d\\ \end{pmatrix}\end{split}\end{aligned}\end{align}
Where $$A > B, D > C$$ and $$a > c, d > b$$.
An often used version is the following:
\begin{align}\begin{aligned}\begin{split}A = \begin{pmatrix} 10&0\\ 0&5\\ \end{pmatrix}\end{split}\\\begin{split} B = \begin{pmatrix} 5&0\\ 0&10\\ \end{pmatrix}\end{split}\end{aligned}\end{align}
This is the default version of the game created by this function:
sage: g = game_theory.normal_form_games.CoordinationGame()
sage: g
Coordination game - Normal Form Game with the following utilities: ...
sage: d = {(0, 1): [0, 0], (1, 0): [0, 0],
....: (0, 0): [10, 5], (1, 1): [5, 10]}
sage: g == d
True
There are two pure Nash equilibria and one mixed:
sage: g.obtain_nash()
[[(0, 1), (0, 1)], [(2/3, 1/3), (1/3, 2/3)], [(1, 0), (1, 0)]]
We can also pass different values of the input parameters:
sage: g = game_theory.normal_form_games.CoordinationGame(A=9, a=6,
....: B=2, b=1, C=0, c=1, D=4, d=11)
sage: g
Coordination game - Normal Form Game with the following utilities: ...
sage: d ={(0, 1): [0, 1], (1, 0): [2, 1],
....: (0, 0): [9, 6], (1, 1): [4, 11]}
sage: g == d
True
sage: g.obtain_nash()
[[(0, 1), (0, 1)], [(2/3, 1/3), (4/11, 7/11)], [(1, 0), (1, 0)]]
Note that an error is returned if the defining inequalities are not obeyed $$A > B, D > C$$ and $$a > c, d > b$$:
sage: g = game_theory.normal_form_games.CoordinationGame(A=9, a=6,
....: B=0, b=1, C=2, c=10, D=4, d=11)
Traceback (most recent call last):
...
TypeError: the input values for a Coordination game must
be of the form A > B, D > C, a > c and d > b
sage.game_theory.catalog_normal_form_games.HawkDove(v=2, c=3)#
Return a Hawk Dove game.
Suppose two birds of prey must share a limited resource $$v$$. The birds can act like a hawk or a dove.
• If a dove meets a hawk, the hawk takes the resources.
• If two doves meet they share the resources.
• If two hawks meet, one will win (with equal expectation) and take the resources while the other will suffer a cost of $$c$$ where $$c>v$$.
This can be modeled as a normal form game using the following two matrices [Web2007]:
\begin{align}\begin{aligned}\begin{split}A = \begin{pmatrix} v/2-c&v\\ 0&v/2\\ \end{pmatrix}\end{split}\\\begin{split} B = \begin{pmatrix} v/2-c&0\\ v&v/2\\ \end{pmatrix}\end{split}\end{aligned}\end{align}
Here are the games with the default values of $$v=2$$ and $$c=3$$.
\begin{align}\begin{aligned}\begin{split}A = \begin{pmatrix} -2&2\\ 0&1\\ \end{pmatrix}\end{split}\\\begin{split} B = \begin{pmatrix} -2&0\\ 2&1\\ \end{pmatrix}\end{split}\end{aligned}\end{align}
This is a particular example of an anti coordination game. There are three Nash equilibria:
1. One bird acts like a Hawk and the other like a Dove.
2. Both birds mix being a Hawk and a Dove
This can be implemented in Sage using the following:
sage: g = game_theory.normal_form_games.HawkDove()
sage: g
Hawk-Dove - Anti coordination game -
Normal Form Game with the following utilities: ...
sage: d ={(0, 1): [2, 0], (1, 0): [0, 2],
....: (0, 0): [-2, -2], (1, 1): [1, 1]}
sage: g == d
True
sage: g.obtain_nash()
[[(0, 1), (1, 0)], [(1/3, 2/3), (1/3, 2/3)], [(1, 0), (0, 1)]]
sage: g = game_theory.normal_form_games.HawkDove(v=1, c=3)
sage: g
Hawk-Dove - Anti coordination game -
Normal Form Game with the following utilities: ...
sage: d ={(0, 1): [1, 0], (1, 0): [0, 1],
....: (0, 0): [-5/2, -5/2], (1, 1): [1/2, 1/2]}
sage: g == d
True
sage: g.obtain_nash()
[[(0, 1), (1, 0)], [(1/6, 5/6), (1/6, 5/6)], [(1, 0), (0, 1)]]
Note that an error is returned if the defining inequality is not obeyed $$c < v$$:
sage: g = game_theory.normal_form_games.HawkDove(v=5, c=1) Traceback (most recent call last): … TypeError: the input values for a Hawk Dove game must be of the form c > v
sage.game_theory.catalog_normal_form_games.MatchingPennies()#
Return a Matching Pennies game.
Consider two players who can choose to display a coin either Heads facing up or Tails facing up. If both players show the same face then player 1 wins, if not then player 2 wins.
This can be modeled as a zero sum normal form game with the following matrix [Web2007]:
$\begin{split}A = \begin{pmatrix} 1&-1\\ -1&1\\ \end{pmatrix}\end{split}$
There is a single Nash equilibria at which both players randomly (with equal probability) pick heads or tails.
This can be implemented in Sage using the following:
sage: g = game_theory.normal_form_games.MatchingPennies()
sage: g
Matching pennies - Normal Form Game with the following utilities: ...
sage: d ={(0, 1): [-1, 1], (1, 0): [-1, 1],
....: (0, 0): [1, -1], (1, 1): [1, -1]}
sage: g == d
True
sage: g.obtain_nash('enumeration')
[[(1/2, 1/2), (1/2, 1/2)]]
sage.game_theory.catalog_normal_form_games.Pigs()#
Return a Pigs game.
Consider two pigs. One dominant pig and one subservient pig. These pigs share a pen. There is a lever in the pen that delivers 6 units of food but if either pig pushes the lever it will take them a little while to get to the food as well as cost them 1 unit of food. If the dominant pig pushes the lever, the subservient pig has some time to eat two thirds of the food before being pushed out of the way. If the subservient pig pushes the lever, the dominant pig will eat all the food. Finally if both pigs go to push the lever the subservient pig will be able to eat a third of the food (and they will also both lose 1 unit of food).
This can be modeled as a normal form game using the following two matrices [McM1992] (we assume that the dominant pig’s utilities are given by $$A$$):
\begin{align}\begin{aligned}\begin{split}A = \begin{pmatrix} 3&1\\ 6&0\\ \end{pmatrix}\end{split}\\\begin{split} B = \begin{pmatrix} 1&4\\ -1&0\\ \end{pmatrix}\end{split}\end{aligned}\end{align}
There is a single Nash equilibrium at which the dominant pig pushes the lever and the subservient pig does not.
This can be implemented in Sage using the following:
sage: g = game_theory.normal_form_games.Pigs()
sage: g
Pigs - Normal Form Game with the following utilities: ...
sage: d ={(0, 1): [1, 4], (1, 0): [6, -1],
....: (0, 0): [3, 1], (1, 1): [0, 0]}
sage: g == d
True
sage: g.obtain_nash()
[[(1, 0), (0, 1)]]
sage.game_theory.catalog_normal_form_games.PrisonersDilemma(R=-2, P=-4, S=-5, T=0)#
Return a Prisoners dilemma game.
Assume two thieves have been caught by the police and separated for questioning. If both thieves cooperate and do not divulge any information they will each get a short sentence. If one defects he/she is offered a deal while the other thief will get a long sentence. If they both defect they both get a medium length sentence.
This can be modeled as a normal form game using the following two matrices [Web2007]:
\begin{align}\begin{aligned}\begin{split}A = \begin{pmatrix} R&S\\ T&P\\ \end{pmatrix}\end{split}\\\begin{split} B = \begin{pmatrix} R&T\\ S&P\\ \end{pmatrix}\end{split}\end{aligned}\end{align}
Where $$T > R > P > S$$.
• $$R$$ denotes the reward received for cooperating.
• $$S$$ denotes the ‘sucker’ utility.
• $$P$$ denotes the utility for punishing the other player.
• $$T$$ denotes the temptation payoff.
An often used version [Web2007] is the following:
\begin{align}\begin{aligned}\begin{split}A = \begin{pmatrix} -2&-5\\ 0&-4\\ \end{pmatrix}\end{split}\\\begin{split} B = \begin{pmatrix} -2&0\\ -5&-4\\ \end{pmatrix}\end{split}\end{aligned}\end{align}
There is a single Nash equilibrium for this at which both thieves defect. This can be implemented in Sage using the following:
sage: g = game_theory.normal_form_games.PrisonersDilemma()
sage: g
Prisoners dilemma - Normal Form Game with the following utilities: ...
sage: d = {(0, 0): [-2, -2], (0, 1): [-5, 0], (1, 0): [0, -5],
....: (1, 1): [-4, -4]}
sage: g == d
True
sage: g.obtain_nash()
[[(0, 1), (0, 1)]]
Note that we can pass other values of R, P, S, T:
sage: g = game_theory.normal_form_games.PrisonersDilemma(R=-1, P=-2, S=-3, T=0)
sage: g
Prisoners dilemma - Normal Form Game with the following utilities:...
sage: d = {(0, 1): [-3, 0], (1, 0): [0, -3],
....: (0, 0): [-1, -1], (1, 1): [-2, -2]}
sage: g == d
True
sage: g.obtain_nash()
[[(0, 1), (0, 1)]]
If we pass values that fail the defining requirement: $$T > R > P > S$$ we get an error message:
sage: g = game_theory.normal_form_games.PrisonersDilemma(R=-1, P=-2, S=0, T=5)
Traceback (most recent call last):
...
TypeError: the input values for a Prisoners Dilemma must be
of the form T > R > P > S
sage.game_theory.catalog_normal_form_games.RPS()#
Return a Rock-Paper-Scissors game.
Rock-Paper-Scissors is a zero sum game usually played between two players where each player simultaneously forms one of three shapes with an outstretched hand.The game has only three possible outcomes other than a tie: a player who decides to play rock will beat another player who has chosen scissors (“rock crushes scissors”) but will lose to one who has played paper (“paper covers rock”); a play of paper will lose to a play of scissors (“scissors cut paper”). If both players throw the same shape, the game is tied and is usually immediately replayed to break the tie.
This can be modeled as a zero sum normal form game with the following matrix [Web2007]:
$\begin{split}A = \begin{pmatrix} 0 & -1 & 1\\ 1 & 0 & -1\\ -1 & 1 & 0\\ \end{pmatrix}\end{split}$
This can be implemented in Sage using the following:
sage: g = game_theory.normal_form_games.RPS()
sage: g
Rock-Paper-Scissors - Normal Form Game with the following utilities: ...
sage: d = {(0, 1): [-1, 1], (1, 2): [-1, 1], (0, 0): [0, 0],
....: (2, 1): [1, -1], (1, 1): [0, 0], (2, 0): [-1, 1],
....: (2, 2): [0, 0], (1, 0): [1, -1], (0, 2): [1, -1]}
sage: g == d
True
sage: g.obtain_nash('enumeration')
[[(1/3, 1/3, 1/3), (1/3, 1/3, 1/3)]]
sage.game_theory.catalog_normal_form_games.RPSLS()#
Return a Rock-Paper-Scissors-Lizard-Spock game.
Rock-Paper-Scissors-Lizard-Spock is an extension of Rock-Paper-Scissors. It is a zero sum game usually played between two players where each player simultaneously forms one of three shapes with an outstretched hand. This game became popular after appearing on the television show ‘Big Bang Theory’. The rules for the game can be summarised as follows:
• Scissors cuts Paper
• Paper covers Rock
• Rock crushes Lizard
• Lizard poisons Spock
• Spock smashes Scissors
• Scissors decapitates Lizard
• Lizard eats Paper
• Paper disproves Spock
• Spock vaporizes Rock
• (and as it always has) Rock crushes Scissors
This can be modeled as a zero sum normal form game with the following matrix:
$\begin{split}A = \begin{pmatrix} 0 & -1 & 1 & 1 & -1\\ 1 & 0 & -1 & -1 & 1\\ -1 & 1 & 0 & 1 & -1\\ -1 & 1 & -1 & 0 & 1\\ 1 & -1 & 1 & -1 & 0\\ \end{pmatrix}\end{split}$
This can be implemented in Sage using the following:
sage: g = game_theory.normal_form_games.RPSLS()
sage: g
Rock-Paper-Scissors-Lizard-Spock -
Normal Form Game with the following utilities: ...
sage: d = {(1, 3): [-1, 1], (3, 0): [-1, 1], (2, 1): [1, -1],
....: (0, 3): [1, -1], (4, 0): [1, -1], (1, 2): [-1, 1],
....: (3, 3): [0, 0], (4, 4): [0, 0], (2, 2): [0, 0],
....: (4, 1): [-1, 1], (1, 1): [0, 0], (3, 2): [-1, 1],
....: (0, 0): [0, 0], (0, 4): [-1, 1], (1, 4): [1, -1],
....: (2, 3): [1, -1], (4, 2): [1, -1], (1, 0): [1, -1],
....: (0, 1): [-1, 1], (3, 1): [1, -1], (2, 4): [-1, 1],
....: (2, 0): [-1, 1], (4, 3): [-1, 1], (3, 4): [1, -1],
....: (0, 2): [1, -1]}
sage: g == d
True
sage: g.obtain_nash('enumeration')
[[(1/5, 1/5, 1/5, 1/5, 1/5), (1/5, 1/5, 1/5, 1/5, 1/5)]]
sage.game_theory.catalog_normal_form_games.StagHunt()#
Return a Stag Hunt game.
Assume two friends go out on a hunt. Each can individually choose to hunt a stag or hunt a hare. Each player must choose an action without knowing the choice of the other. If an individual hunts a stag, he must have the cooperation of his partner in order to succeed. An individual can get a hare by himself, but a hare is worth less than a stag.
This can be modeled as a normal form game using the following two matrices [Sky2003]:
\begin{align}\begin{aligned}\begin{split}A = \begin{pmatrix} 5&0\\ 4&2\\ \end{pmatrix}\end{split}\\\begin{split} B = \begin{pmatrix} 5&4\\ 0&2\\ \end{pmatrix}\end{split}\end{aligned}\end{align}
This is a particular type of Coordination Game. There are three Nash equilibria:
1. Both friends hunting the stag.
2. Both friends hunting the hare.
3. Both friends hunting the stag 2/3rds of the time.
This can be implemented in Sage using the following:
sage: g = game_theory.normal_form_games.StagHunt()
sage: g
Stag hunt - Coordination game -
Normal Form Game with the following utilities: ...
sage: d = {(0, 1): [0, 4], (1, 0): [4, 0],
....: (0, 0): [5, 5], (1, 1): [2, 2]}
sage: g == d
True
sage: g.obtain_nash()
[[(0, 1), (0, 1)], [(2/3, 1/3), (2/3, 1/3)], [(1, 0), (1, 0)]]
sage.game_theory.catalog_normal_form_games.TravellersDilemma(max_value=10)#
Return a Travellers dilemma game.
An airline loses two suitcases belonging to two different travelers. Both suitcases happen to be identical and contain identical antiques. An airline manager tasked to settle the claims of both travelers explains that the airline is liable for a maximum of 10 per suitcase, and in order to determine an honest appraised value of the antiques the manager separates both travelers so they can’t confer, and asks them to write down the amount of their value at no less than 2 and no larger than 10. He also tells them that if both write down the same number, he will treat that number as the true dollar value of both suitcases and reimburse both travelers that amount.
However, if one writes down a smaller number than the other, this smaller number will be taken as the true dollar value, and both travelers will receive that amount along with a bonus/malus: 2 extra will be paid to the traveler who wrote down the lower value and a 2 deduction will be taken from the person who wrote down the higher amount. The challenge is: what strategy should both travelers follow to decide the value they should write down?
This can be modeled as a normal form game using the following two matrices [Ba1994]:
\begin{align}\begin{aligned}\begin{split}A = \begin{pmatrix} 10 & 7 & 6 & 5 & 4 & 3 & 2 & 1 & 0\\ 11 & 9 & 6 & 5 & 4 & 3 & 2 & 1 & 0\\ 10 & 10 & 8 & 5 & 4 & 3 & 2 & 1 & 0\\ 9 & 9 & 9 & 7 & 4 & 3 & 2 & 1 & 0\\ 8 & 8 & 8 & 8 & 6 & 3 & 2 & 1 & 0\\ 7 & 7 & 7 & 7 & 7 & 5 & 2 & 1 & 0\\ 6 & 6 & 6 & 6 & 6 & 6 & 4 & 1 & 0\\ 5 & 5 & 5 & 5 & 5 & 5 & 5 & 3 & 0\\ 4 & 4 & 4 & 4 & 4 & 4 & 4 & 4 & 2\\ \end{pmatrix}\end{split}\\\begin{split}B = \begin{pmatrix} 10 & 11 & 10 & 9 & 8 & 7 & 6 & 5 & 4\\ 7 & 9 & 10 & 9 & 8 & 7 & 6 & 5 & 4\\ 6 & 6 & 8 & 9 & 8 & 7 & 6 & 5 & 4\\ 5 & 5 & 5 & 7 & 8 & 7 & 6 & 5 & 4\\ 4 & 4 & 4 & 4 & 6 & 7 & 6 & 5 & 4\\ 3 & 3 & 3 & 3 & 3 & 5 & 6 & 5 & 4\\ 2 & 2 & 2 & 2 & 2 & 2 & 4 & 5 & 4\\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 3 & 4\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2\\ \end{pmatrix}\end{split}\end{aligned}\end{align}
There is a single Nash equilibrium to this game resulting in both players naming the smallest possible value.
This can be implemented in Sage using the following:
sage: g = game_theory.normal_form_games.TravellersDilemma()
sage: g
Travellers dilemma - Normal Form Game with the following utilities: ...
sage: d = {(7, 3): [5, 1], (4, 7): [1, 5], (1, 3): [5, 9],
....: (4, 8): [0, 4], (3, 0): [9, 5], (2, 8): [0, 4],
....: (8, 0): [4, 0], (7, 8): [0, 4], (5, 4): [7, 3],
....: (0, 7): [1, 5], (5, 6): [2, 6], (2, 6): [2, 6],
....: (1, 6): [2, 6], (5, 1): [7, 3], (3, 7): [1, 5],
....: (0, 3): [5, 9], (8, 5): [4, 0], (2, 5): [3, 7],
....: (5, 8): [0, 4], (4, 0): [8, 4], (1, 2): [6, 10],
....: (7, 4): [5, 1], (6, 4): [6, 2], (3, 3): [7, 7],
....: (2, 0): [10, 6], (8, 1): [4, 0], (7, 6): [5, 1],
....: (4, 4): [6, 6], (6, 3): [6, 2], (1, 5): [3, 7],
....: (8, 8): [2, 2], (7, 2): [5, 1], (3, 6): [2, 6],
....: (2, 2): [8, 8], (7, 7): [3, 3], (5, 7): [1, 5],
....: (5, 3): [7, 3], (4, 1): [8, 4], (1, 1): [9, 9],
....: (2, 7): [1, 5], (3, 2): [9, 5], (0, 0): [10, 10],
....: (6, 6): [4, 4], (5, 0): [7, 3], (7, 1): [5, 1],
....: (4, 5): [3, 7], (0, 4): [4, 8], (5, 5): [5, 5],
....: (1, 4): [4, 8], (6, 0): [6, 2], (7, 5): [5, 1],
....: (2, 3): [5, 9], (2, 1): [10, 6], (8, 7): [4, 0],
....: (6, 8): [0, 4], (4, 2): [8, 4], (1, 0): [11, 7],
....: (0, 8): [0, 4], (6, 5): [6, 2], (3, 5): [3, 7],
....: (0, 1): [7, 11], (8, 3): [4, 0], (7, 0): [5, 1],
....: (4, 6): [2, 6], (6, 7): [1, 5], (8, 6): [4, 0],
....: (5, 2): [7, 3], (6, 1): [6, 2], (3, 1): [9, 5],
....: (8, 2): [4, 0], (2, 4): [4, 8], (3, 8): [0, 4],
....: (0, 6): [2, 6], (1, 8): [0, 4], (6, 2): [6, 2],
....: (4, 3): [8, 4], (1, 7): [1, 5], (0, 5): [3, 7],
....: (3, 4): [4, 8], (0, 2): [6, 10], (8, 4): [4, 0]}
sage: g == d
True
sage: g.obtain_nash() # optional - lrslib
[[(0, 0, 0, 0, 0, 0, 0, 0, 1), (0, 0, 0, 0, 0, 0, 0, 0, 1)]]
Note that this command can be used to create travellers dilemma for a different maximum value of the luggage. Below is an implementation with a maximum value of 5:
sage: g = game_theory.normal_form_games.TravellersDilemma(5)
sage: g
Travellers dilemma - Normal Form Game with the following utilities: ...
sage: d = {(0, 1): [2, 6], (1, 2): [1, 5], (3, 2): [4, 0],
....: (0, 0): [5, 5], (3, 3): [2, 2], (3, 0): [4, 0],
....: (3, 1): [4, 0], (2, 1): [5, 1], (0, 2): [1, 5],
....: (2, 0): [5, 1], (1, 3): [0, 4], (2, 3): [0, 4],
....: (2, 2): [3, 3], (1, 0): [6, 2], (0, 3): [0, 4],
....: (1, 1): [4, 4]}
sage: g == d
True
sage: g.obtain_nash()
[[(0, 0, 0, 1), (0, 0, 0, 1)]] | 9,116 | 24,793 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 14, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2024-10 | longest | en | 0.713936 |
https://cs.stackexchange.com/questions/125416/all-problems-about-turing-machines-that-involve-only-the-language-that-the-tm-ac | 1,714,010,327,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296820065.92/warc/CC-MAIN-20240425000826-20240425030826-00193.warc.gz | 159,221,379 | 39,171 | # All problems about Turing machines that involve only the language that the TM accepts are undecidable
I came across the below statement in the classic text "Introduction to Automata Theory, Languages, and Computation" by Hopcroft, Ullman, Motwani.
All problems about Turing machines that involve only the language that the TM accepts are undecidable
They say that the above theorem is per Rice theorem which states that:
"Every nontrivial property of the RE languages is undecidable."
How are these two statements equivalent? The former deals only problems while the later deals with non trivial property.
• Problems and properties (and languages) are the same thing. Also, Hopcroft et al. forgot to mention nontriviality, or perhaps assumed it was obvious. May 8, 2020 at 21:12
• @YuvalFilmus Now that you say so, here problems and property (both with respect to RE languages) have RE languages in them and they are same (thanks for the help), Also, Hopcroft et al. forgot to mention nontriviality, or perhaps assumed it was obvious Do you mean that about mentioning nontrivial before problems in the statement highlighted in gray (above in the question body) by the authors? May 9, 2020 at 7:30
• Right, the “obvious” bit refers to nontriviality. May 9, 2020 at 7:48
There are a few keywords in the excerpt from the said text book - non-trivial, problem, property.
Now what is a problem, assuming we are not dealing with combinatorial optimization problems,i.e. we are dealing with only questions which have an YES or NO answer to them. When you ask a YES or NO question to an input string if the answer is YES you place it in a set $$L$$ and if the answer is NO you just discard it. Now this set $$L$$ is the language or the problem. It contains all those strings which which satisfy our YES or NO question.
All non-trival problems about Turing machines that involve only the language that the TM accepts are undecidable
Here the author is talking about YES or NO questions (with respect to Turing machines) only involing the language that the Turing machine accepts,i.e. the Recursively Enumerable language(RE), which means that our "problem" set shall contain only RE languages. Now a trivial problem is one in which the our YES or NO question is either satisfied by all input or satisfied by none of the input. So a non-trivial problem is one which is neither empty nor it has all possible inputs.
Rice Theorem: "Every nontrivial property of the RE languages is undecidable."
Property of RE languages is a set of RE languages having the said property.
Non-trivial property: The property is either satisfied by all concerned languages or by none.
So "nontrivial property of the RE languages" becomes the set of RE languages and it is neither empty nor it has all possible RE languages.
By the above argument we can say that the two statements are equivalent.
(In fact property and problem are the same thing, they both are set of strings after all. Now we might think that property is a set of languages, although while it is true, it is inconvenient to represent languages in a set(as languages can be infinitely long), rather what is done is instead of the language , we represent the corresponding Turing Machine with a suitable encoding) | 716 | 3,264 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-18 | latest | en | 0.951368 |
http://www.jiskha.com/display.cgi?id=1187924332 | 1,498,240,070,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320077.32/warc/CC-MAIN-20170623170148-20170623190148-00014.warc.gz | 571,061,429 | 3,905 | # Geometry
posted by on .
I did not get this question at all, if someone could help, I would appreciate it!
It says:
Reword Theorem 1-3 as two statements, one describing existence and the other describing uniqueness.
Theorem 1-3 says:
If two lines intersect, then exactly one plane contains the lines.
Earlier in the lesson it explained something about existence (there is at least one point of intersection) and uniqueness (no more than one such point exists) about Theorem 1-1, but I still couldn't put the two together. Can anyone help??? I need this homework turned in tomorrow, or it's late. Any help is appreciated! THANKS!!! | 145 | 638 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-26 | latest | en | 0.967697 |
http://blog.analogmachine.org/2016/11/02/how-to-get-the-stoichiometry-matrix-using-tellurium/ | 1,585,977,875,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370520039.50/warc/CC-MAIN-20200404042338-20200404072338-00454.warc.gz | 25,275,665 | 10,867 | # How to get the Stoichiometry Matrix using Tellurium
Here is a simple need, given a reaction model how do we get hold of the stoichiometry matrix?
Consider the following simple model:
```import tellurium as te
\$Xo -> S1; k1*Xo;
S1 -> S2; k2*S1;
S2 -> S3; k3*S2;
S3 -> \$X1; k4*S3;
k1 = 0.1; k2 = 0.4;
k3 = 0.5; k4 = 0.6;
Xo = 1;
""")
print r.getFullStoichiometryMatrix()
```
Running this script by clicking on the green arrow in the tool bar will yield:
``` _J0, _J1, _J2, _J3
S1 [[ 1, -1, 0, 0],
S2 [ 0, 1, -1, 0],
S3 [ 0, 0, 1, -1]]
```
The nice thing about this output is that the columns and rows are labeled so you know exact what is what.
What about much larger models? For example the iAF1260.xml model from the Bigg database (http://bigg.ucsd.edu:8888/models/iAF1260). This is a model of E. Coli that includes 1668 metabolites and 2382 reactions. We can download the iAF1260.xml file and load it into libRoadRunner using:
```
```
This might take up to a minute to load depending on how fast your computer is. We are assuming here that the file is located in the current directory (os.getcwd()). If not, move the file, change the current directory (using os.chdir), or use the appropriate path in the call.
Rather than print out the stoichiometry matrix (don’t even try) to the screen we’ll save it to a file. Because the stoichiometry matrix is so large we will use numpy to write the matrix out as a text file:
```import numpy as np
st = r.getFullStoichiometryMatrix()
print "Number of metabolites = ", r.getNumFloatingSpecies()
print "Number of reactions = ", r.getNumReactions()
np.savetxt ('stoich.txt', st)
Number of metabolites = 1668
Number of reactions = 2382
```
One can change the formating of the output using savetxt, for example the following will output the individual stoichiometry coefficeint using 3 decimal places, 5 characters minimium, and separated by a comma.
```np.savetxt ('c:\\tmp\\st.txt', st, delimiter=',', fmt='%5.3f',)
```
You can get the labels for the rows and columns by calling r.getFloatingSpeciesIds() and r.getReactionIds() respectively.
This entry was posted in Modeling, Pathways, SBML, Systems Theory. Bookmark the permalink. | 642 | 2,222 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2020-16 | latest | en | 0.795487 |
https://homework.cpm.org/category/CCI_CT/textbook/int1/chapter/11/lesson/11.2.4/problem/11-98 | 1,610,732,487,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703495936.3/warc/CC-MAIN-20210115164417-20210115194417-00211.warc.gz | 414,188,081 | 15,224 | ### Home > INT1 > Chapter 11 > Lesson 11.2.4 > Problem11-98
11-98.
Write the equation of the line parallel to $3x + 2y = 10$ that goes through the point (4, −7).
Change the equation of the given line to 'y = mx + b' form and identify the slope.
Then use the given point to find the equation of the line.
Use the eTool below to solve the problem.
Click the link at the right to view full version of the eTool: Int1 11-98 HW eTool. | 127 | 433 | {"found_math": true, "script_math_tex": 1, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-04 | longest | en | 0.844205 |
https://askoranswerme.com/4947/ | 1,628,030,593,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154486.47/warc/CC-MAIN-20210803222541-20210804012541-00093.warc.gz | 133,547,640 | 3,746 | Shunt capacitance is neglected while considering?
Shunt capacitance is neglected while considering short transmission line.
Related
Description : Capacitance is the ability of a body to store an__________. (A) electrical charge (B) electric current (C) voltage (D) All of the above
Answer : Capacitance is the ability of a body to store an electrical charge
Description : D.C. shunt relays are made of?
Answer : D.C. shunt relays are made of many turns of thin wire.
Description : A 240 V, dc shunt motor draws 15 A while supplying the rated load at a speed of 80 rad/s. The armature resistance is 0.5 W and the field winding resistance is 80 W. The net Voltage across armature resistance at the time of plugging will be
Answer : A 240 V, dc shunt motor draws 15 A while supplying the rated load at a speed of 80 rad/s. The armature resistance is 0.5 W and the field winding resistance is 80 W. The net Voltage across armature resistance at the time of plugging will be 474 V
Answer : If the field of a D.C. shunt motor gets opened while motor is running the motor will attain dangerously high speed.
Description : Calculate the value of equivalent capacitance of the combination.
Answer : Calculate the value of equivalent capacitance of the combination | 301 | 1,266 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-31 | latest | en | 0.817326 |
https://www.neetprep.com/question/55998-amplitude-ratio-two-sources-producing-interference----ratioof-intensities-maxima-minima------------/55-Physics--Wave-Optics/700-Wave-Optics | 1,590,450,576,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347390437.8/warc/CC-MAIN-20200525223929-20200526013929-00542.warc.gz | 853,852,291 | 50,256 | If the amplitude ratio of two sources producing interference is 3 : 5, the ratio of intensities at maxima and minima is
(1) 25 : 16
(2) 5 : 3
(3) 16 : 1
(4) 25 : 9
Concept Questions :-
Superposition principle
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
For constructive interference to take place between two monochromatic light waves of wavelength λ, the path difference should be
(1) $\left(2n-1\right)\frac{\lambda }{4}$
(2) $\left(2n-1\right)\frac{\lambda }{2}$
(3) $n\lambda$
(d) $\left(2n+1\right)\frac{\lambda }{2}$
Concept Questions :-
Young Double slit experiment
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
Soap bubble appears coloured due to the phenomenon of
(1) Interference
(2) Diffraction
(3) Dispersion
(4) Reflection
Concept Questions :-
Interference vs Diffraction
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
Which of the following statements indicates that light waves are transverse
(1) Light waves can travel in vacuum
(2) Light waves show interference
(3) Light waves can be polarized
(4) Light waves can be diffracted
Concept Questions :-
Polarization of light
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
Interference was observed in interference chamber when the air was present, now the chamber is evacuated and if the same light is used, a careful observer will see
(1) No interference
(2) Interference with bright bands
(3) Interference with dark bands
(4) Interference in which width of the fringe will be slightly increased
Concept Questions :-
Superposition principle
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
Ray diverging from a point source form a wave front that is
(1) Cylindrical
(2) Spherical
(3) Plane
(4) Cubical
Concept Questions :-
Wave front
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
Two coherent sources have intensity in the ratio of $\frac{100}{1}$. Ratio of (intensity) max/(intensity) min is
(1) $\frac{1}{100}$
(2) $\frac{1}{10}$
(3) $\frac{10}{1}$
(4) $\frac{3}{2}$
Concept Questions :-
Superposition principle
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
If two waves represented by ${y}_{1}=4\mathrm{sin}\omega t$ and ${y}_{2}=3\mathrm{sin}\left(\omega t+\frac{\pi }{3}\right)$ interfere at a point, the amplitude of the resulting wave will be about
(1) 7
(2) 6
(3) 5
(4) 3.5
Concept Questions :-
Superposition principle
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
Two coherent sources of intensities, I1 and I2 produce an interference pattern. The maximum intensity in the interference pattern will be
(1) I1 + I2
(2) ${I}_{1}^{2}+{I}_{2}^{2}$
(3) (I1 + I2)2
(4) ${\left(\sqrt{{I}_{1}}+\sqrt{{I}_{2}}\right)}^{2}$
Concept Questions :-
Superposition principle
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the beams is $\frac{\pi }{2}$ at point A and π at point B. Then the difference between the resultant intensities at A and B is
(1) 2I
(2) 4I
(3) 5I
(4) 7I
Concept Questions :-
Superposition principle | 944 | 3,314 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 14, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2020-24 | latest | en | 0.822019 |
http://www.ck12.org/algebra/Numerical-Expression-Evaluation/?difficulty=basic&by=ck12 | 1,490,873,134,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218193716.70/warc/CC-MAIN-20170322212953-00633-ip-10-233-31-227.ec2.internal.warc.gz | 489,438,392 | 17,213 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Numerical Expression Evaluation
## Use properties of equality and order of operations.
Levels are CK-12's student achievement levels.
Basic Students matched to this level have a partial mastery of prerequisite knowledge and skills fundamental for proficient work.
At Grade (Proficient) Students matched to this level have demonstrated competency over challenging subject matter, including subject matter knowledge, application of such knowledge to real-world situations, and analytical skills appropriate to subject matter.
Advanced Students matched to this level are ready for material that requires superior performance and mastery.
## Numerical Expression Evaluation with Basic Operations
by CK-12 //basic
Learn to use the order of operations to evaluate numerical expressions.
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11
## Numerical Expression Evaluation with Grouping Symbols
by CK-12 //basic
Learn to use the order of operations to evaluate numerical expressions with parentheses and powers.
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7
## Order of Operations with Positive Real Numbers
by CK-12 //basic
This concept reviews the standard order of operations for arithmetic calculations.
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1
## Expression Evaluation with Basic Operations
by CK-12 //basic
Find out how to evaluate variable expressions involving the four basic operations.
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## Four operations with Money - Example 1
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## Four operations with Money - Example 2
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Using two operations with money
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## Money Challenge Problems - Example 1
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Money Word Problems
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## Money Challenge Problems - Example 2
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Find the cost' word problems
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## MDAS I - Example 1
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## Example 3: Evaluate the Answer using Order of Operations
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Provides a third example of evaluating an expression using the order of operations.
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## PMDAS I - Example 1
by CK-12 //basic
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## PMDAS II - Example 1
by CK-12 //basic
Addition and Subtraction with Parentheses - missing number.
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## PMDAS Challenge - Example 1
by CK-12 //basic
Place the missing parentheses to make the equation true?
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by CK-12 //basic
Distributive Property Multiplication Problems
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## Understanding the Distributive Property Word Problems - Example 2
by CK-12 //basic
Distributive Property Division Problems
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## 1 - 2 Step Word Problems I - Example 1
by CK-12 //basic
1 -Step Division
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## Word Problems with charts and formulas - Example 1
by CK-12 //basic
number = (difference) ÷ (times bigger - 1)
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## Multi-step Multi-operation word problems - Example 1
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Amount Remaining
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0 | 961 | 4,131 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2017-13 | latest | en | 0.870943 |
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# Vapor liquid equilibrium using hysys
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### Vapor liquid equilibrium using hysys
1. 1. August 26, 2010<br />1<br />Vapor / Liquid Equilibrium Using HYSYS<br />Prepared by:<br />A.N.Tabish<br /> Lab. Engineer / Lecturer<br />Department of Chemical Engineering <br />UET, City Campus, Lahore<br />tabish288@yahoo.com<br />
2. 2. 2<br />Example # 10.3:*<br />For the system methanol (1) / methyl acetate (2) , the following equations provide a reasonable correlation for the activity coefficient<br />Where,<br />In addition following equations provide vapor pressures:<br />Where T is in kelvin and the vapor pressure are in kPa.<br />* J.M.Smith, H.C.Van Ness and M.M.Abott (2005), “Introduction to Chemical Engineering Thermodynamics”, Ed. 7, pp. 359, McGraw Hill<br />August 26, 2010<br />tabish288@yahoo.com<br />
3. 3. August 26, 2010<br />3<br />CALCULATE:<br />P and {y1}, for T = 318.15 K and {x1}= 0.25<br />P and {x1}, for T = 318.15 K and {y1}= 0.60<br />T and {y1}, for P = 101.33kPa and {x1}= 0.85<br />T and {x1}, for P = 101.33kPa and {y1}= 0.40<br />tabish288@yahoo.com<br />
4. 4. August 26, 2010<br />4<br />Solution Strategy:<br /><ul><li>Start a new case in HYSYS.
5. 5. Define components
6. 6. Select Thermodynamic model.
7. 7. Draw the PFD (material streams)
8. 8. Define</li></ul>Composition<br />Vapor fraction<br />Mole/mass flow rate<br />Temperature / Pressure<br /><ul><li>Determine BUBL/DEW Pressure/Temperature
9. 9. Determine composition using K-Value information</li></ul>tabish288@yahoo.com<br />
10. 10. August 26, 2010<br />5<br />HYSYS:<br />tabish288@yahoo.com<br />
11. 11. August 26, 2010<br />6<br /><ul><li>Start a new case in HYSYS.
12. 12. Define components (Methanol + Methyl Acetate).</li></ul>tabish288@yahoo.com<br />
13. 13. August 26, 2010<br />7<br /><ul><li>Select NRTL model.</li></ul>tabish288@yahoo.com<br />
14. 14. August 26, 2010<br />8<br /><ul><li>Draw the PFD (1 material streams)</li></ul>tabish288@yahoo.com<br />
15. 15. August 26, 2010<br />9<br /><ul><li>Vapor Pressure of Methanol</li></ul>44.51kPa using Antoine Equation<br /><ul><li>Vapor Pressure of Methyl Acetate</li></ul>65.64kPa using Antoine Equation<br />tabish288@yahoo.com<br />
16. 16. August 26, 2010<br />10<br />BUBL Pressure:<br />P and {y1}, for T = 318.15 K and {x1}= 0.25<br />73.50 kPa in manual calculations<br />tabish288@yahoo.com<br />
17. 17. August 26, 2010<br />11<br />Vapor compositions:<br />y1 = 0.282<br />y2 = 0.718<br />y1 = 0.278<br />y2 = 0.722<br />tabish288@yahoo.com<br />
18. 18. August 26, 2010<br />12<br />DEW Pressure:<br />P and {x1}, for T = 318.15 K and {y1}= 0.60<br />62.89 kPa in manual calculations<br />tabish288@yahoo.com<br />
19. 19. August 26, 2010<br />13<br />Liquid compositions:<br />x1 = 0.8169<br />x2 = 0.1831<br />x1 = 0.8099<br />x2 = 0.1901<br />tabish288@yahoo.com<br />
20. 20. August 26, 2010<br />14<br />BUBL Temperature:<br />T and {y1}, for P = 101.33kPa and {x1}= 0.85<br />331.20K = 58 oCin manual calculations<br />tabish288@yahoo.com<br />
21. 21. August 26, 2010<br />15<br />Vapor compositions:<br />y1 = 0.670<br />y2 = 0.330<br />y1 = 0.671<br />y2 = 0.329<br />tabish288@yahoo.com<br />
22. 22. August 26, 2010<br />16<br />DEW Temperature:<br />T and {x1}, for P = 101.33kPa and {y1}= 0.40<br />326.70K = 53.7 oC in manual calculations<br />tabish288@yahoo.com<br />
23. 23. August 26, 2010<br />17<br />Liquid compositions:<br />x1 = 0.4602<br />x2 = 0.5398<br />x1 = 0.4517<br />x2 = 0.5483<br />tabish288@yahoo.com<br />
24. 24. August 26, 2010<br />18<br />VLE Plot:<br /><ul><li>Download registry files from</li></ul>http://people.clarkson.edu/~wilcox/Design/EqPlots.dll<br />http://people.clarkson.edu/~wilcox/Design/EqPlots.edf<br /> and save them to your computer at some safe place.<br /><ul><li>Create a case containing the components of interest and choose a suitable fluid package.
25. 25. Go to the Simulation Environment
26. 26. Go to Tools/Preferences/Extensions and Click on Register an Extension.
27. 27. Find EqPlots.dll and open it. This will remain part of HYSYS or UniSim on your machine until it is Unregistered. Close Preference.
28. 28. Go to Flowsheet/Add Operation. Select the Extensions category and Add Equilibrium Plots. This should open the equilibrium plots extension.
29. 29. For a Binary plot, click the Binary tab at the bottom of the Equilibrium Plots menu, and select the components and the type of plot desired.
30. 30. For an XY plot, specify either T or P, and click on Plot.
31. 31. For a TXY plot, specify P. For a PXY plot, specify T.</li></ul>tabish288@yahoo.com<br />
32. 32. August 26, 2010<br />19<br />tabish288@yahoo.com<br />
33. 33. August 26, 2010<br />20<br />tabish288@yahoo.com<br />
34. 34. August 26, 2010<br />21<br />tabish288@yahoo.com<br />
35. 35. August 26, 2010<br />22<br />Example # 10.4:*<br />For a mixture of 10 mol-% methane, 20 mol-% ethane, and 70 mol-% propane at 50 (oF), determin:<br />The dew point pressure<br />The bubble point pressure<br />* J.M.Smith, H.C.Van Ness and M.M.Abott (2005), “Introduction to Chemical Engineering Thermodynamics”, Ed. 7, pp. 364, McGraw Hill<br />tabish288@yahoo.com<br />
36. 36. August 26, 2010<br />23<br />Home Work Task:<br />Assuming the validity of Raoult's law, do the following calculations for the benzene (l)/toluene (2) system:<br />For the system ethyl ethanoate (1) and n-heptane (2)at 343.15 K (70"C)<br />BUBL P calculation for T = 343.15 K (70°C), x1 = 0.05.<br />DEW P calculation for T = 343.15 K (70°C), y1 = 0.05.<br />tabish288@yahoo.com<br />
37. 37. August 26, 2010<br />24<br />Home Work Task Contd….<br />Make the following VLE calculations for the methane (l) / ethylene (2) / ethane (3) system:<br />tabish288@yahoo.com<br /> | 2,093 | 6,112 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2016-18 | latest | en | 0.626174 |
https://www.schoolphysics.co.uk/age16-19/Mechanics/Dynamics/text/Running_on_a_treadmill2/index.html | 1,708,877,455,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474617.27/warc/CC-MAIN-20240225135334-20240225165334-00692.warc.gz | 999,136,516 | 3,638 | # Running on a treadmill (2)
However when I mention to him, that as you jump, you are being pulled backward, he disagrees vehemently. 'You are not oscillating back and forth, your head is not moving relative to the front of the treadmill'. 'You are able to push forward just enough to counteract the small push backward since your feet are only in contact for a small fraction of time'.
Since you have run some marathons, an interesting point comes up:
My roommate claims the following:
- he can run much faster on a treadmill than on a track for a given short distance.
- he likens this to running while someone is pulling the carpet from under your feet.
I think he is willing to demonstrate this if I ask, since all our theoretical discussions are going nowhere.
He has gone through great lengths to successfully convince family members, friends, engineers, whenever the topic comes up.
Since you are a runner, do you find you can run significantly faster on a treadmill?
Do you find different muscle soreness either?
I still maintain that when the two inertial frames of reference are (by definition) moving at a constant velocity relative to each other the two situations are absolutely equivalent.
However as the speed of the belt changes, if he wants to run faster, then there will be an acceleration – obviously. In this condition the above statement does not apply because the two frames of reference are not inertial – there is relative acceleration.
The movement of the legs of a runner on the treadmill or on the road is the same, however during the acceleration the runner on the road has to develop more power than the runner on the treadmill because the one on the road actually has to accelerate their MASS relative to the Earth and this is not true for the runner on the treadmill.
Over a short sprint quite a lot of the race is acceleration and so I would agree that here the two situations are not equivalent.
The two situations of constant velocity and acceleration must be kept quite separate. The pulling of the carpet is an acceleration case.
I agree about the time interval in jumping up but in order to jump at all you must be in contact with the belt or the ground for a certain length of time. You can't jump without bending your knees!
Ask him about running down the aisle in a moving train. It is all relative to the train. The fact that the train may be travelling at 100 m.p.h is not important to what happens inside the train.
The air is all important. Runners in the Olympics in Mexico City found it easier to break the sprint records because of the lower air density. Just think of Bob Beamon's worked record breaking long jump – that stood for years. I can imagine that the lack of the air running past does make a difference. As far as I remember my fluid mechanics the resistive force due to motion through a fluid is proportional to the velocity squared. Therefore as you run faster (as in the short sprint) the effects of this air drag become proportionately much greater.
There is one further point. In all this I have assumed that the runner has no effect on the treadmill – in other words the belt does not slip, and the treadmill does not move.
An interesting exercise that you might both like to think about is what happens if the runner runs on a treadmill that has a massless belt and rollers ands where the belt runs over the rollers with zero friction.
A VERSION IN WORD IS AVAILABLE ON THE SCHOOLPHYSICS USB | 705 | 3,468 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2024-10 | latest | en | 0.97078 |
https://freakonometrics.hypotheses.org/?post/2011/10/18/Short-selling%2C-volatility-and-bubbles&pub=0 | 1,597,222,818,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738888.13/warc/CC-MAIN-20200812083025-20200812113025-00289.warc.gz | 316,444,820 | 51,862 | Sharing pictures from holidays in the Canadian Rockies (with R)
My kids have a very popular blog (at least among their grandmothers) where they frequently post pictures from everyday’s life (since they live 5000km from them), as well as pictures taken from holidays. This afternoon, I tried to used the popupImage function from the leaflet package to post pictures, on a map (to explain where we spent our holiday this summer). This post is just to keep tracks of that code.
First, we need to load the appropriate R packages
library(leaflet) library(mapview)
Then, we take a picture, and we locate it, for instance Mirror Lake (on the trail to Lake Agnes). Since leaflet uses openstreetmap, I recommend to use it also for location (and not google maps… coordinates can be slightly different)
df=data.frame(lat =51.41603, long=-116.23946, nom = "Miror Lake",photo="http://freakonometrics.free.fr/jaspeR/_DSC5967.jpg")
I guess you can also use the metadata if you take pictures with a cell phone, and you add the location… but I am (very) old fashioned, and still use a camera to take pictures. Then you can add a dozen pictures
df=rbind(df, data.frame(lat =51.4164, long=-116.2442, nom = "Lake Agnes",photo="http://freakonometrics.free.fr/jaspeR/_DSC6003.jpg")) df=rbind(df, data.frame(lat =51.3215642,long=-116.193718, nom="Moraine Lake",photo="http://freakonometrics.free.fr/jaspeR/_DSC5957.jpg"))
From that dataframe, we need two kinds of information: the location, and the url of the picture,
data_df=df[,c("lat","long")] images = as.character(df$photo) Then we can create the leaflet map (sorry for typos, but wordpress converts the > symbol into some “>” characters… which makes R pipe operator hard to read) m = leaflet(data_df) %>% addTiles() %>% addCircleMarkers( fillOpacity = 0.8, radius = 5, lng = ~long, lat =~lat, popup = popupImage(images) ) and export it (in a nice html file) library(htmlwidgets) saveWidget(m, file="jaspR.html") Time for a summer break Time for a break ! So guess what ? The blog will be off during a few days… See you in August ! Machine Learning in Actuarial Science and Insurance This week is organized the summer school on machine learning for economists and applied social scientists. I will be giving an (online) lecture this Thursday, on Machine Learning in Actuarial Science & Insurance, with a great program, • 10am – 10:55am : Juri Marcucci: Machine Learning in Macroeconomics • 11am – 11:55am : Arthur Charpentier: Machine Learning in Actuarial Science & Insurance • 12pm – 12:55pm : Arthur Spirling : Machine Learning in Embeddings Representations • 1pm – 1:55pm : Kathy Baylis: Machine Learning in Agricultural Economics • 2pm – 2:55pm : Stefan Wager : Machine Learning in Causal Inference • 10am – 10:55am : Stan Matwin : Machine Learning and Data Privacy • 11am – 11:55am : Mehmet Caner : Machine Learning in Econometrics • 12pm – 12:55pm : Anders Bredahl Kock : Machine Learning in Model Selection • 1pm – 1:55pm : Dario Sansone: Machine Learning in Education and Development Economics • 2pm – 2:55pm: Patrick Baylis :Temperature and Temperament: Evidence from Twitter My slides are now online, Régression sur des variables corrélées, un peu de géométrie Depuis quelques jours, je mets en ligne des billets sur la régression sur des variables corrélés, en commençant par du deuxième effet kiss-cool suivi d’un court billet sur des régressions en cascade. Oui, depuis le premier, mon collègue Olivier ne cesse de demander des compléments. Alors ce soir, on va faire des dessins… On a ici 3 vecteurs, $\color{red}{\vec{x_1}}$, $\color{green}{\vec{x_2}}$ et $\color{blue}{\vec{y}}$. Pour rappel, la longueur de mes vecteurs, c’est la variance de la variable associée, $\|{\color{red}{\vec{x_1}}}\|=\text{Var}({\color{red}{x_1}})$. Et commençons par supposer que les variables $\color{red}{x_1}$ et $\color{green}{x_2}$ sont indépendantes (ou plus simplement non-corrélées) soit ${\color{red}{\vec{x_1}}}\perp\color{green}{\vec{x_2}}$. C’est le dessin ci-dessous, La régression linéaire multiple est ici équivalente aux régressions simples, comme on en parlait dans les billets précédant. Géométriquement, on voit que les coordonnées de la projection de $\color{blue}{\vec{y}}$ sur l’espace engendré par $\color{red}{\vec{x_1}}$ et $\color{green}{\vec{x_2}}$ (c’est à dire le plan $(x,y)$) correpond au point noir. C’est la régression multiple. Les deux régressions simples sont obtenus en projetant respectivement sur sur l’espace engendré par $\color{red}{\vec{x_1}}$ (c’est à dire la droite $(x)$) et sur $\color{green}{\vec{x_2}}$ (c’est à dire la droite $(y)$). Supposons maintenant que $\color{red}{\vec{x_1}}$ et $\color{green}{\vec{x_2}}$ ne sont plus orthogonaux. Plus précisément, on va garder $\color{red}{\vec{x_1}}$ et $\color{blue}{\vec{y}}$ inchangés, et on va changer , $\color{green}{\vec{x_2}}$, tout en gardant la norme $\|\color{green}{\vec{x_2}}\|$ inchangée. C’est ce qu’on a sur le schéma ci-dessous On avait vu qu’on ne pouvait plus faire deux régressions indépendantes, mais qu’on pouvait garder la première, et en faire ensuite une seconde, mais en faisant un peu attention. Pour la première régression, on va projeter $\color{blue}{\vec{y}}$ sur $\color{red}{\vec{x_1}}$ et ca c’est facile, rien n’a changé, Pour la seconde étape, rappelons qu’on doit projeter à la fois $\color{blue}{\vec{y}}$ et $\color{green}{\vec{x_2}}$ sur l’orthogonal de $\color{red}{\vec{x_1}}$, c’est à dire ici le plan $(y,z)$, en rouge ci-dessous (je renvois au précédant billet pour les formules algébriques). On obtient alors les projections suivantes, qu’on pourrait noter $\Pi_{\color{red}{x_1^{\perp}}}\color{blue}{\vec{y}}$ et $\Pi_{\color{red}{x_1^{\perp}}}\color{green}{\vec{x_2}}$ On le voit, si on regarde sur le mur du fond, en rouge, la projection de $\Pi_{\color{red}{x_1^{\perp}}}\color{blue}{\vec{y}}$ sur $\Pi_{\color{red}{x_1^{\perp}}}\color{green}{\vec{x_2}}$ est un peu plus grande, autrement dit, le coefficient de la régression simple sera un peu plus grand. En fait, si on veut faire des calculs, on voit que le sinus de l’angle $\langle\color{red}{\vec{x_1}},\color{green}{\vec{x_2}}\rangle$ va intervenir ici… Plus l’angle $\langle\color{red}{\vec{x_1}},\color{green}{\vec{x_2}}\rangle$ sera petit, plus les variables seront corrélées, et plus la correction sera importante. Graphiquement, la correction dont je parle, c’est le petit segment, à gauche sur le dessin Mais simulons quelques données pour visualiser tout ça… library(mnormt) n=10000 r=0 set.seed(1) Z=cbind(rnorm(n)*sqrt(2),rnorm(n)*sqrt(3)) df=data.frame(y=Z[,1]+Z[,2]+rnorm(n)*sqrt(3),x1=Z[,1],x2=Z[,2]) On simule ici des variables explicatives indépendamment. Et on peut vérifier que les distances sont bien celles de notre dessin var(df$x1) [1] 2.049731 var(df$x2) [1] 2.810419 var(df$y) [1] 8.066665
et en plus, les variables sont quasiment orthogonales
cor(df$x1,df$x2) [1] 0.004845078
Si on fait la régression multiple
reg=lm(y~0+x1+x2,data=df) summary(reg) Coefficients: Estimate Std. Error t value Pr(|t|) x1 0.99904 0.01219 81.94 2e-16 *** x2 1.00090 0.01017 98.40 2e-16 ***
On a des coefficients qui sont identiques (à très peu de choses près, mais le soucis vient du fait que j’ai enlevé la constante du modèle pour coller au mieux avec mon dessin)
reg1=lm(y~0+x1,data=df) summary(reg1) Coefficients: Estimate Std. Error t value Pr(|t|) x1 1.00489 0.01711 58.75 2e-16 ***
et pour la seconde régression aussi
reg2=lm(y~0+x2,data=df) summary(reg2) Coefficients: Estimate Std. Error t value Pr(|t|) x2 1.00496 0.01315 76.42 2e-16 ***
On a déjà parlé de ce résultat longuement (je renvoie au précédant billet, des régressions en cascade). Regardons maintenant ce qui se passe si on garde nos observations $\color{blue}{y_i}$, $\color{red}{x_{1,i}}$ mais que maintenant, la variable $\color{red}{x_{2,i}}$ se retrouve corrélée avec $\color{red}{x_{1,i}}$ , disons avec une corrélation de l’ordre de 0.4
r=.4 df$x2=r*df$x1+sqrt(1-r^2)*df$x2 cor(df$x1,df$x2) [1] 0.3461496 Faire la régression multiple donne ici reg=lm(y~0+x1+x2,data=df) summary(reg) Coefficients: Estimate Std. Error t value Pr(|t|) x1 0.5622 0.0130 43.26 2e-16 *** x2 1.0921 0.0111 98.40 2e-16 *** ce qui est différent des régressions simples, comme la première reg1=lm(y~0+x1,data=df) summary(reg1) Coefficients: Estimate Std. Error t value Pr(|t|) x1 1.00489 0.01711 58.75 2e-16 *** Ce que nous dit le théorème de Frisch-Waugh, c’est qu’on peut faire des régressions en cascades: on peut garder cette première régression, mais pour la seconde, on va régresser les projections sur l’orthogonal à $\color{red}{\vec{x_1}}$ (le mur en rouge au fond dans notre dessin). Ici, on obtient reg2bis=lm(residuals(lm(y~x1,data=df))~residuals(lm(x2~x1,data=df))) summary(reg2bis) Coefficients: Estimate Std. Error t value Pr(|t|) residuals(2) 1.0921 0.0111 98.4 2e-16 *** qui donne effectivement un coefficient de régression un peu supérieur à celui que nous avions auparavant, comme nous l’observions sur le dessin. En fait on peut remarquer que la différence relative s’obtient très facilement avec la corrélation 1/sqrt(1-r^2) [1] 1.091089 et pour rappel, d’un point de vue géométrique, la corrélation est un cosinus d’un angle, $r=\cos(\langle\color{red}{\vec{x_1}},\color{green}{\vec{x_2}}\rangle)$. Autrement dit $\sqrt{1-r^2}=\sin(\langle\color{red}{\vec{x_1}},\color{green}{\vec{x_2}}\rangle)$, c’est à dire que la correction est directement liée au sinus de l’angle $\langle\color{red}{\vec{x_1}},\color{green}{\vec{x_2}}\rangle$, comme mentionné auparavant… amusant, non? Des régressions en cascade Cette fin de semaine, je mettais en ligne un court billet du deuxième effet kiss-cool où je rappelais que quand on fait une régression sur plusieurs variables explicatives corrélées, ce n’est pas équivalent à faire plusieurs régressions simple. C’est ce que disait le théorème de Frisch-Waugh, publié il y a près de 90 ans. Mais je glissais qu’il était possible de faire des régressions en cascade, sans aller beaucoup plus loin. Il faut relire l’article de Michael Lovell qui donner une interprétation géométrique de ce dessin. Mais programmons le, pour le voir… Considérons une base avec 3 variables explicatives, correspondant à des nombres d’incendies à Chicago chicago=read.table("http://freakonometrics.free.fr/chicago.txt", header=TRUE,sep=";") La régression multiple s’écrit ici$$y_i=\beta_0+\beta_1 x_{1,i}+\beta_2 x_{2,i}+\beta_3 x_{3,i}+\varepsilon_i$$On peut estimer ces paramètres par moindre carrés, reg=lm(Fire~.,data=chicago) summary(reg) Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) 22.07525 6.19447 3.564 0.000910 *** X_1 -0.62764 5.28130 -0.119 0.905953 X_2 0.22378 0.06161 3.632 0.000744 *** X_3 -1.55059 0.38195 -4.060 0.000204 *** et on a alors la prévision suivante $$\widehat{y}_i=\widehat{\beta}_0+\widehat{\beta}_1 x_{1,i}+\widehat{\beta}_2 x_{2,i}+\widehat{\beta}_3 x_{3,i}$$En fait, je peux même visualiser cette prévision. BC = cbind(x0=coefficients(reg)[1], x1=coefficients(reg)[2]*chicago[,2], x2=coefficients(reg)[3]*chicago[,3], x3=coefficients(reg)[4]*chicago[,4], yp=predict(reg),y=chicago$Fire) colr = 1:4 dessin_fire = function(i,BC=BC){ plot(0:3,0:3,xlim=c(0,35),col="white",ylim=c(-1,3),xlab="",ylab="",axes=FALSE) abline(v=BC[i,1],lty=2) rect(BC[i,1],3,BC[i,1]+BC[i,2],2,col=colr[1],border=NA) arrows(BC[i,1],2.5,BC[i,1]+BC[i,2],2.5,lwd=2,length=.1,col="white") rect(BC[i,1]+BC[i,2],2,BC[i,1]+BC[i,2]+BC[i,3],1,col=colr[2],border=NA) arrows(BC[i,1]+BC[i,2],1.5,BC[i,1]+BC[i,2]+BC[i,3],1.5,lwd=2,length=.1,col="white") rect(BC[i,1]+BC[i,2]+BC[i,3],1,BC[i,1]+BC[i,2]+BC[i,3]+BC[i,4],0,col=colr[3],border=NA) segments(BC[i,5],0,BC[i,5],1,lwd=3) arrows(BC[i,1]+BC[i,2]+BC[i,3],.5,BC[i,1]+BC[i,2]+BC[i,3]+BC[i,4],.5,lwd=2,length=.1,col="white") abline(v=BC[i,1],lty=2) rect(BC[i,5],-1,BC[i,6],0,col=colr[4],density=20,border=NA) segments(BC[i,6],0,BC[i,6],-1,lwd=3,col=colr[4]) text(26,2.5,expression(X[1]),pos=4,col=colr[1]) text(26,1.5,expression(X[2]),pos=4,col=colr[2]) text(26,.5,expression(X[3]),pos=4,col=colr[3]) text(26,-.5,expression(epsilon),pos=4,col=colr[4]) axis(1)}
Par exemple, pour la douzième observation de ma base
dessin_fire(12)
on obtient
On lit de haut en bas : on commence par constante, vers 22. Puis on rajoute trois fois rien, parce que $x_1$ était non-significative dans notre régression. Puis on rajoute un petit quelque chose à cause de $x_2$, pour passer à 25, et $x_3$ nous fait plonger de plus de 20 points, pour finir vers 3.67.Le trait noir est la prévision que l’on obtient, à partir des trois variables. En bleu, on peut même voir l’erreur.
On peut noter que $x_1$ est non-significative, tout en étant relativement corrélée avec $y$
cor(chicago[,1:2]) Fire X_1 Fire 1.0000000 0.3773486 X_1 0.3773486 1.0000000
Car si $x_1$ est non-significative, c’est parce que cette variable est très corrélée avec une autre variable explicative. En fait, on peut faire une série de régressions en cascade, en commençant par $x_1$, et plus spécifiquement, on peut écrire$$\widehat{y}_i=\underbrace{\widehat{b}_0+\widehat{b}_1 x_{1,i}}_{(1)}+\underbrace{\widehat{b}_{0,2}+\widehat{b}_2 \tilde{x}_{2,i}}_{(2)}+\underbrace{\widehat{b}_{0,3}+\widehat{b}_3 \tilde{x}_{3,i}}_{(3)}$$où le premier terme $(1)$ est obtenu en régression simplement $y$ sur $x_1$ (oui, on fait juste une régression simple), $$y_i=b_0+b_1 x_{1,i}+\eta_i$$puis on apporte une petite correction, pour tenir compte de ce que nous apprend $x_2$ une fois la première régression effectuée. C’est exactement ce que raconte le théorème de Frish-Waugh : on va projeter $y$ sur $x_1^{\perp}$ (car c’est justement ce que $x_1$ n’explique pas), et $x_2$ sur $x_1^{\perp}$ et faire la régression de ces deux projections, soit, pour $(2)$ $$\Pi_{x_1^{\perp}}y_i=b_{0,2}+b_2 \Pi_{x_1^{\perp}}x_{2,i}+\eta_{2,i}$$puis pour $(3)$ $$\Pi_{(x_1,x_2)^{\perp}}y_i=b_{0,3}+b_3 \Pi_{(x_1,x_2)^{\perp}}x_{3,i}+\eta_{2,i}$$En terme informatique, cela donne
reg1=lm(Fire~X_1,data=chicago) reg2=lm(residuals(lm(Fire~X_1,data=chicago))~residuals(lm(X_2~X_1,data=chicago))) reg3=lm(residuals(lm(Fire~X_1+X_2,data=chicago))~residuals(lm(X_3~X_1+X_2,data=chicago)))
soit ici
BC123=cbind(x0=coefficients(reg1)[1], x1=coefficients(reg1)[2]*chicago[,2], x2=coefficients(reg2)[1]+coefficients(reg2)[2]*residuals(lm(X_2~X_1,data=chicago)), x3=coefficients(reg3)[1]+coefficients(reg3)[2]*residuals(lm(X_3~X_1+X_2,data=chicago)), yp=predict(reg),y=chicago$Fire) En adaptant la fonction de dessin précédante on obtient dessin_fire_123(12) Autrement dit, on commence avec une régression simple, en rouge: on part d’une constante vers 3.5, puis on augemente notre prévision, en tenant compte de $x_1$. Puis on corrige avec $x_2$ (ou plutôt la partie de $x_2$ non expliquée par $x_1$). sur ce qui n’avait pas été expliqué auparavant, ce qui pousse à revoir un peu à la baisse notre prévision, puis on tient compte de $x_3$ (ou là encore, sa projection sur l’orthogonal de $x_1$ et $x_2$). On note que la prévision est très exactement la même qu’auparavant. Mais on peut aller plus loin… pourquoi commencer par $x_1$? On peut aussi commencer par $x_2$. On considère alors$$\widehat{y}_i=\underbrace{\widehat{b}_0+\widehat{b}_2 x_{2,i}}_{(2)}+\underbrace{\widehat{b}_{0,1}+\widehat{b}_1 \tilde{x}_{1,i}}_{(1)}+\underbrace{\widehat{b}_{0,3}+\widehat{b}_3 \tilde{x}_{3,i}}_{(3)}$$où le premier terme $(2)$ est obtenu en régression simplement $y$ sur $x_2$$$y_i=b_0+b_2 x_{2,i}+\eta_i$$puis on apporte une petite correction, pour tenir compte de ce que nous apprend $x_1$ une fois la première régression effectuée, $(1)$ $$\Pi_{x_2^{\perp}}y_i=b_{0,1}+b_1\Pi_{x_2^{\perp}}x_{1,i}+\eta_{1,i}$$puis pour $(3)$ $$\Pi_{(x_1,x_2)^{\perp}}y_i=b_{0,3}+b_3 \Pi_{(x_1,x_2)^{\perp}}x_{3,i}+\eta_{2,i}$$ Soit reg1=lm(Fire~X_2,data=chicago) reg2=lm(residuals(lm(Fire~X_2,data=chicago))~residuals(lm(X_1~X_2,data=chicago))) reg3=lm(residuals(lm(Fire~X_1+X_2,data=chicago))~residuals(lm(X_3~X_1+X_2,data=chicago))) BC213=cbind(x0=coefficients(reg1)[1], x1=coefficients(reg1)[2]*chicago[,3], x2=coefficients(reg2)[1]+coefficients(reg2)[2]*residuals(lm(X_1~X_2,data=chicago)), x3=coefficients(reg3)[1]+coefficients(reg3)[2]*residuals(lm(X_3~X_1+X_2,data=chicago)), yp=predict(reg),y=chicago$Fire)
et visuellement, on obtient
dessin_fire_213(12)
Cette fois, la constante est un peu plus élevée, et on commence par faire une régression seulement avec $x_2$, pour arriver un peu en dessous de 10. Puis on corrige. On note que cette seconde correction nous ramène… très exactement comme dans le cas précédant. Mais si on pense en projection successives, on ne doit pas être surpris.
On peut récapituler avec une autre prévision, soit avec le modèle de régresion multiple (mais là encore, $x_1$ n’explique pas grand chose)
dessin_fire(15)
soit en commencant par une régression simple sur $x_1$ puis en faisant des régression en cascade,
dessin_fire_123(15)
soit, alternativement, en commencant par une régression simple sur $x_2$
dessin_fire_213(15)
Comme annoncé, cette trois approches sont équivalentes, et donnent très exactement la même prévision.
Regression discontinuity model for TV series
In September, we are usually happy to see our favorite TV series back on air… Or not? Because, admit it, if we are happy to see those characters back, most of the time, we are disappointed, too. So why not look at the data, to confirm this feeling? Nazareno Andrade shared some nice codes to get IMDB ratings in a nice csv file (you can either use the large csv file, or run your own codes)
download.file("https://github.com/nazareno/imdb-series/raw/master/data/series_from_imdb.csv", destfile="series_from_imdb.csv") base = read.csv("series_from_imdb.csv")
It is a large dataset, with more than 64,000 episodes of almost 890 TV series,
str(base) 'data.frame': 64018 obs. of 18 variables: $series_name: Factor w/ 889 levels "'Allo 'Allo!",..: 137 137 137 137 137 137 137 137 137 137 ...$ episode : Factor w/ 54090 levels "-30-","¡Viva los muertos!",..: 32314 7446 16 7176 17748 9562 1379 36218 17845 5553 ... $series_ep : int 1 2 3 4 5 6 7 8 9 10 ...$ season : int 1 1 1 1 1 1 1 2 2 2 ... $season_ep : int 1 2 3 4 5 6 7 1 2 3 ...$ user_rating: num 8.9 8.7 8.7 8.2 8.3 9.2 8.8 8.7 9.2 8.3 ...
Just pick a TV series, for instance Dan Harmon’s Community,
sbase = base[base$series_name=="Community",] We can plot the evolution of the rating over the 110 episodes. sbase=sbase[!duplicated(sbase[,c(1,2,4,5)]),] sbase$series_ep=1:nrow(sbase)
()since there could be some problem with the data (such as duplicates, let us clean it quickly)
plot(sbase$series_ep,sbase$UserRating,xlab=sbase$series_name[1]) idx=c(0,which(diff(sbase$season)!=0),nrow(sbase)) abline(v=idx+.5,lty=2,col=colr[2]) a = unique(sbase$season) for(u in a){ ssbase = sbase[sbase$season==u,] reg = lm(UserRating~series_ep,data=ssbase) lines(ssbase$series_ep,predict(reg),col=colr[3],lwd=2) } The vertical lines are here to visualize the seasons. On issue is that the lenght can vary with time. Consider Linwood Boomer’s Malcom in The Middle, sbase = base[base$series_name=="Malcolm in the Middle",]
or Craig Thomas and Carter Bays’s How I Met Your Mother,
sbase = base[base$series_name=="How I Met Your Mother",] On those two, the evolution is rather stable. Look at AMC’s The Walking Dead, sbase = base[base$series_name=="The Walking Dead",]
Now, look at Howard Gordon and Alex Gansa’s Homeland,
sbase = base[base$series_name=="Homeland",] There is an issue here with the last episode of season4, “Long Time Coming“, that has a very poor rating. If we remove that point, we get the thin line. Note that the regression line is always increasing. For Michael Hirst’s Vickings, we have sbase = base[base$series_name=="Vicking",]
If we look more carefully on the previous graph, for five seasons (out of six), we have a positive slope. Well, to be honest, it is not significantly positive most of the time, but still. Out of 80 shows, and a total of 583 seasons, the slope is postive 75% of the time (433) and negative 25% of the time (150).
BASE = NULL L80 = unique(base$series_name) for(j in 1:length(L)){ sbase=base[base$series_name==L[j],] sbase=sbase[!duplicated(sbase[,c(1,2,4,5)]),] sbase=sbase[sbase$season>0,] sbase$series_ep=1:nrow(sbase) a=unique(sbase$season) a=a[!is.na(a)] for(u in a){ ssbase=sbase[sbase$season==u,] reg=lm(UserRating~series_ep,data=ssbase) pente = NA if((!is.na(coefficients(reg)[2]))&(!is.na((summary(reg)$coefficients[2,4])))){ if((summary(reg)$coefficients[2,4]<.05)&(coefficients(reg)[2]>0)) pente="positive" if((summary(reg)$coefficients[2,4]<.05)&(coefficients(reg)[2]<0)) pente="negative" sdf=data.frame(nom=sbase$series_name[1],season=u,slope=coefficients(reg)[2],inf=confint(reg)[2,1],sup=confint(reg)[2,2],signe=pente) BASE=rbind(BASE,sdf)} }} str(BASE) 'data.frame': 583 obs. of 6 variables: $nom : Factor w/ 80 levels "Friends","Game of Thrones",..: 1 1 1 1 1 1 1 1 1 1 ... mean(BASE$slope>0) [1] 0.7427101 table(BASE$signe) negative positive 15 144 Most of the time, the slope is not significant. To be more specific, 72% of the time, the slope is not significant. But when it is, 90% of the time, it is positive (144 seasons). Let us look at other TV series, for instance Joel Surnow and Robert Cochran’s 24, sbase = base[base$series_name=="24",]
Álex Pina’s La Casa de Papel,
sbase = base[base$series_name=="La Casa de Papel",] Steven Knight’s Peaky Blinders, sbase = base[base$series_name=="Peaky Blinders",]
or David Simon’s The Wire,
sbase = base[base$series_name=="The Wire",] The slope is increasing over almost all seasons. But a major drawback is that when we get back to our show, for a new season, we usually get disapointed. More specifically, we can quantify the difference in red below that can be estimated using sbase12 = sbase[sbase$season%in%c(a[ij],a[ij+1]),] seuil = sbase12$series_ep[which(diff(sbase12$season)!=0)]+.5 s = function(x) (x-seuil)*(x>seuil) reg = lm(UserRating~series_ep+s(series_ep)+I(series_ep>seuil),data=sbase12)
Here we have
summary(reg) Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) 8.45000 0.16338 51.719 2e-16 *** series_ep 0.10000 0.03235 3.091 0.008598 ** s(series_ep) 0.02000 0.04218 0.474 0.643291 I(series_ep)TRUE. -1.01778 0.20486 -4.968 0.000257 ***
so the drop of 1 point (out of 10) cannot be claimed as being significant. That is the idea of regression discontinuity.
If we loop again over all our series, we have 485 pairs of consecutive seasons. As expected, in 75% of the casse, from season $t-1$ to season $t$, we observe a negative rupture. As previously, in 70% of the cases, it is not significat (with linear models before and after), and when it is significant, it is negative in 96% of the cases ! But an alternative can be to use nonparametric models, on both sides.
To illustrate, consider David Benioff and D. B. Weiss’s Game of Thrones,
sbase = base[base$series_name=="Game of Thrones",] but let us remove the last season (no spoiler here, but clearly not worst watching) Consider for instance the drop between season 1 and season 2, library(rdd) sbase12=sbase[sbase$season%in%c(1,2),] lmr=RDestimate(UserRating~series_ep,data=sbase12,cutpoint=mean(range(sbase12$series_ep))) plot(lmr) This is very consistent with what we observed with our linear regressions actually, seuil=10.5 s = function(x) (x-seuil)*(x>seuil) reg = lm(UserRating~series_ep+s(series_ep)+I(series_ep>seuil),data=sbase12) summary(reg) Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) 8.70000 0.15458 56.281 2e-16 *** series_ep 0.07273 0.02491 2.919 0.01003 * s(series_ep) 0.01455 0.03523 0.413 0.68520 I(series_ep)TRUE -0.94000 0.20316 -4.627 0.00028 *** Here, the drop of one point is significant… So, your favorite show had an outstanding finale ? and you can’t wait to watch the new season… Well, statistically, it’s very likely that you will be disapointed by the first episode of the forthcoming season… Du deuxième effet kiss-cool (régression multiple, scoring et évaluation) Lorsque j’étais petit (il y a fort longtemps, à une époque où je regardais pas mal la télévision) il y avait une publicité pour les pastilles kiss cool, Et quand je présente la régression multiple à mes étudiants, je ne peux m’empêcher d’y penser… Mais avant d’aller plus loin sur le parallèle, faisons un peu de mathématiques. Les techniques de régression permettent d’avoir des jolis théorèmes, souvent d’une portée incroyablement générale (moyennant quelques petites hypothèses techniques). Par exemple le théorème de Frisch-Waugh, en régression multiple, dont j’ai déjà parlé dans des vieux billets. Un des corolaires est que lorsque les variables explicatives dans un modèle de régression sont orthogonales, la régression multiple correspond à une collection de régressions simples (autrement dit, les estimateurs par moindres carrés coïncident). Formellement, si on considère le modèle$$y_i=\beta_0+\beta_1x_{1,i}+\beta_2x_{2,i}+\varepsilon_i$$(avec les hypothèses usuelles des modèles de régression) alors, si les variables $x_1$ et $x_2$ sont non-corrélées, $\widehat{\beta}_1$ coïncide avec $\widehat{b}_1$ dans le modèle$$y_i=b_0+b_1x_{1,i}+\eta_i$$On peut faire une petite simulation pour confirmer (sinon, bien entendu, on peut regarder la démonstration qui se trouve dans tous les livres d’économétrie, qui est d’ailleurs un simple résultat d’algèbre linéaire, ou de géométrie, avec des projections successives sur des sous-espaces orthogonaux – même si c’est à Michael Lovell que l’on doit l’approche géométrique). library(mnormt) r = 0 S = matrix(c(1,r,r,1),2,2) n = 1000 set.seed(1) X = rmnorm(n,c(0,0),S) E = rnorm(n,0,.3) Y = 2+X[,1]-2*X[,2]+E base = data.frame(Y=Y-mean(Y),X1=X[,1]-mean(X[,1]),X2=X[,2]-mean(X[,2])) Petite note technique: je vais centrer les variables, histoire de ne pas avoir à garder la constante dans mon modèle (qui va compliquer les notations, et potentiellement embrouiller un peu le billet). La constante est nulle ici, on le voit, reg = lm(Y~X1+X2,data=base) summary(reg) Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) 7.449e-18 9.777e-03 0.0 1 X1 1.012e+00 9.171e-03 110.3 2e-16 *** X2 -1.988e+00 9.719e-03 -204.6 2e-16 *** Bref, je régresse sans constante reg = lm(Y~0+X1+X2,data=base) summary(reg) Coefficients: Estimate Std. Error t value Pr(|t|) X1 1.011520 0.009166 110.3 2e-16 *** X2 -1.988321 0.009714 -204.7 2e-16 *** Maintenant, on va regarder les deux régressions simple reg1 = lm(Y~0+X1,data=base) summary(reg1) Coefficients: Estimate Std. Error t value Pr(|t|) X1 1.01300 0.06006 16.86 2e-16 *** quand on régresse juste sur la première variable, et pour la seconde, on obtient reg2 = lm(Y~0+X2,data=base) summary(reg2) Coefficients: Estimate Std. Error t value Pr(|t|) X2 -1.98916 0.03528 -56.39 2e-16 *** Autrement dit, nos estimateurs sont très proches (si on avait laissé la constante, ils coïncideraient). Maintenant, le gros soucis est que ce résultat n’est plus valide lorsque les variables explicatives sont corrélées. Le théorème de Frisch-Waugh nous explique comment ces estimateurs divergent, mais le point ici est que si les variables sont corrélées, utiliser les régressions simples donne deux estimateurs biaisés des vrais paramètres (du modèle multiple). Recommençons l’exercice précédant r=.9 S=matrix(c(1,r,r,1),2,2) set.seed(1) X=rmnorm(n,c(0,0),S) Y = 2+X[,1]-2*X[,2]+E base = data.frame(Y=Y-mean(Y),X1=X[,1]-mean(X[,1]),X2=X[,2]-mean(X[,2])) reg = lm(Y~X1+X2,data=base) summary(reg) reg = lm(Y~0+X1+X2,data=base) summary(reg) Coefficients: Estimate Std. Error t value Pr(|t|) X1 0.98740 0.02205 44.79 2e-16 *** X2 -1.97321 0.02229 -88.54 2e-16 *** (on retrouve des valeurs proches de celles utilisées pour simuler nos données, donc tout va bien). En revanche, pour les régressions simples, on obtient reg1 = lm(Y~0+X1,data=base) summary(reg1) Coefficients: Estimate Std. Error t value Pr(|t|) X1 -0.78784 0.02726 -28.9 2e-16 *** et reg2 = lm(Y~0+X2,data=base) summary(reg2) Coefficients: Estimate Std. Error t value Pr(|t|) X2 -1.06543 0.01607 -66.31 2e-16 *** Autrement dit, $\widehat{b}_1\neq \widehat{\beta}_1$ (et pareil pour le second). Ce qui signifie que si on construit une prévision à partir de ce modèle, $\widetilde{y}_i=\widehat{b}_1x_{1,i}+\widehat{b}_2x_{2,i}$, on sera potentiellement très loin de la “bonne” prévision $\widehat{y}_i=\widehat{\beta}_1x_{1,i}+\widehat{\beta}_2x_{2,i}$ (qui sera sans biais, etc, je renvoie ici vers n’importe quel cours d’économétrie). On peut le voir sur un dessin, Yp=reg1$coefficients[1]*base$X1+reg2$coefficients[1]*base$X2 plot(base$Y,predict(reg),ylim=range(Yp),col=rgb(0,0,1,.5),cex=.7,xlab="observé",ylab="prédit") abline(a=0,b=1,lty=2) points(base$Y,Yp,col=rgb(1,0,0,.5),cex=.7) abline(lm(predict(reg)~base$Y),col="blue") abline(lm(Yp~base$Y),col="red") En bleu, on a les prévisions avec le modèle linéaire multiple, et en rouge, en faisant deux régressions indépendantes… Si on regarde sur la droite, le modèle rouge sur-valorise, ou disons sur-estime largement, alors qu’il sous-valorise à gauche. La différence entre les deux droites s’interprète ici comme un biais. Sur le graphique ci-dessous, on peut visualiser la distribution des $\widetilde{y}_i=\widehat{b}_1x_{1,i}+\widehat{b}_2x_{2,i}$, en rouge, et les $\widehat{y}_i=\widehat{\beta}_1x_{1,i}+\widehat{\beta}_2x_{2,i}$ , en bleu. Cet excès de dispersion, de variance, qu’on observe sur les points rouges, j’interprète ça comme de la polarisation plot(density(Yp),col="red")lwd=2) lines(density(predict(reg)),col="blue",lwd=2) En fait, ce que raconte le théorème de Frisch-Waugh (et je renvoie à mon précédant billet pour plus de détails), c’est qu’on a le droit de faire plusieurs régressions, mais en cascade ! et surtout pas indépendamment : je peux expliquer $y$ avec la première variable $x_1$, et ensuite régresser le résidu (ce qu’on n’a pas pu expliquer) sur la seconde $x_2$. Cette méthode donnera la même prévision que le modèle multiple. On peut aller un peu plus loin: on peut jouer sur la valeur de la corrélation, pour mesure l’écart entre les deux prévisions (ici je prévois pour une observation au hasard, la 78ème) comp=function(r){ S=matrix(c(1,r,r,1),2,2) set.seed(1) X=rmnorm(n,c(0,0),S) Y = 2+X[,1]-2*X[,2]+E base = data.frame(Y=Y-mean(Y),X1=X[,1]-mean(X[,1]),X2=X[,2]-mean(X[,2])) reg = lm(Y~0+X1+X2,data=base) reg1 = lm(Y~0+X1,data=base) reg2 = lm(Y~0+X2,data=base) y1=predict(reg) y2=reg1$coefficients[1]*base$X1+reg2$coefficients[1]*base$X2 c(y1[78],y2[78],(y2[78]-y1[78])/y1[78])} vR=seq(0,.98,by=.02) vc=Vectorize(comp)(vR) plot(vR,vc[3,]*100,ylab="Différence relative (%)",xlab="Corrélation",type="l") On prédit ici pour une observation avec un large $y_i$, ce qui correspondait à la partie de droite du graphique précédant (avec les points rouges et bleus). C’est ce que j’appelle le deuxième effet kiss-cool. Quand on est dans le premier cas, avec les variables indépendantes (corrélation nulle, c’est à dire à gauche sur ma figure), on explique ce qu’on peut avec la première variable, et on rajoute l’impact de la seconde. Et $\widetilde{y}_i\sim\widehat{y}_i$. Le soucis ici est que je n’ai pas le droit de considérer deux modèles indépendants lors que les variables sont très corrélées. Une partie de l’explication fournie par la seconde variable était déjà inclue dans la première. Par exemple avec deux variables très (positivement) corrélées, la prévision qu’on obtient en ajoutant les deux effets estimés indépendamment avec deux régressions simple $\widetilde{y}_i=\widehat{b}_1x_{1,i}+\widehat{b}_2x_{2,i}$, on sur-estime de 50% à 70% la “vraie” prévision $\widehat{y}_i=\widehat{\beta}_1x_{1,i}+\widehat{\beta}_2x_{2,i}$. Tous les chercheurs savent savent ça… on parle ici de résultats du tout premier cours de modèles linéaires. Et malgré tout, en pratique, on continue à utiliser cette seconde méthode. Un exemple bien connu est celui de l’évaluation (des étudiants, des chercheurs, peu importe). Par exemple, quand on évalue un dossier de financement pour un chercheur, on nous demande de mettre un score • pour les publications scientifiques (nombre, qualité, etc) • pour l’encadrement d’étudiants (nombre, niveau, etc) • pour la qualité de l’environnement (prestige du labo, etc) • etc Et à la fin, on somme tout. Mais on le voit, ces variables sont très très corrélées: si vous êtes dans un labo prestigieux, vous attirez beaucoup de candidatures d’étudiants (et des bons), et avoir beaucoup d’étudiants va permettre d’avoir plus de publications (si on ajoute son nom comme co-auteur). Bref, on est typiquement dans un modèle à double (voire triple) effet kiss-cool. Quelqu’un dans un bon labo aura un bon score sur le troisième item, mais aussi un bon sur le nombre d’étudiants, et aussi sur les publications. Ajouter ces scores est stupide, car on a une spirale infernale (les bons sont sur-évalués, et les moins bon, sous-évalués), c’est ce que racontait mon premier dessin, avec les points rouges et bleus. C’est un effet clivant de polarisation forte. Si on voulait faire les choses proprement, ce que dit le le théorème de Frisch-Waugh, c’est que les scores devraient être attribués en corrigeant de la corrélation entre les variables • on peut commencer par calculer un score pour la qualité de l’environnement (prestige du labo, etc) • à environnement donné, calculer un score pour l’encadrement d’étudiants (nombre, niveau, etc) • à environnement donné, et à encadrement d’étudiants donné, calculer un score pour les publications • etc C’est comme la situation que je voyais en France, où on pouvait avoir une variable qui tenait compte d’un prestige du chercheur (par exemple, être chercheur CNRS donnait un bonus) et une autre sur le dossier de publications. Sauf que les deux sont corrélés. Et la plupart des classements d’universités sont construits à partir de scores qui sont loin d’être indépendants. Bref, tant que l’évaluation se fera en sommant des scores qui sont construits sur des critères souvent très corrélés, on polarise fortement la population. Est-ce gênant? A priori oui. Car le message que cela envoie est qu’il existe deux classes, les bons, et les mauvais, alors qu’en réalité, le niveau est beaucoup plus homogène qu’il n’y paraît. Un petit effet positif se retrouve démultiplié par le fait qu’il va se répercuter (positivement) sur plein d’autres variables. C’est mon effet kiss-cool. Mais on pourrait se dire que c’est un problème de distribution des notes finales. Si l’ordre est préservé, on pourrait se dire que ce n’est pas très grave. Malheureusement, ce n’est pas le cas. Si on quitte un instant le cas de la corrélation très forte, les rangs des prédictions (c’est à dire les rangs des chercheurs une fois donnés les notes $\widehat{y}_i$ ou $\widetilde{y}_i$ ) sont moins corrélés si la corrélation sous-jacente est importante (mais pas trop) comp=function(r){ S=matrix(c(1,r,r,1),2,2) set.seed(1) X=rmnorm(n,c(0,0),S) Y = 2+X[,1]-2*X[,2]+E base = data.frame(Y=Y-mean(Y),X1=X[,1]-mean(X[,1]),X2=X[,2]-mean(X[,2])) reg = lm(Y~0+X1+X2,data=base) reg1 = lm(Y~0+X1,data=base) reg2 = lm(Y~0+X2,data=base) y1=predict(reg) y2=reg1$coefficients[1]*base$X1+reg2$coefficients[1]*base\$X2 cor(y1,y2,method="spearman")} vR=seq(0,.98,by=.02) vc=Vectorize(comp)(vR) plot(vR,vc,ylab="Corrélation de rangs",xlab="Corrélation",type="l")
Autrement dit, si les variables $x_1$ et $x_2$ sont très peu corrélées, on a les mêmes rangs (globalement). En revanche, si la corrélation entre les variables $x_1$ et $x_2$ augmente, le rang des $\widetilde{y}_i$ est de moins en moins cohérent avec celui entre les $\widehat{y}_i$ (qui devrait être celui que l’on recherche).
Bref, il serait temps de comprendre enfin sérieusement les conséquences de ce joli papier, publié il y a presque 90 ans
Summer School on Machine Learning for Economists, in Nova Scotia
Just a brief post to mention that was invited as a keynote speaker to give a talk at the Summer School on Machine Learning for Economists, in one month,
I will give a talk on machine learning and insurance. More to come, soon…
Personalization as a promise: Can Big Data change the practice of insurance?
Our recent article, “Personalization as a Promise: Can Big Data Change the Practice of Insurance?” with Larence Barry just got published in “Big Data & Society” (but a working paper version is still available online)
The purpose of this paper is to measure the impact of technologies from the Big Bang. data on thehe pricing ofhe products car insurance. The first part describes how the aggregated view buildsuit by statistics enables highlighting invisible regularities at the individual level. Despite a very granular segmentations in automobile insurance, the approach remained classificatory, hypothesizing the risk identity of individuals from the same class. The second part highlights the reversal of big data-induced perspective in the’analysis ofgiven ; awith theur volume and the new algorithms, the aggregate viewpoint is questioned.. The hypothesis of class homogeneity is becoming increasingly difficult to test. maintain, especially since predictive analysis boasts the ability to predict the rs results at the individual level. The third part is studying the’influence of telematics boxes able to import the new pinsurance aradigm automobile. However, a reading of the most recent research articles on a pricing automobile including this new monter that the epistemological leap, at least for now, has not taken place.
Lancement du groupe en épidémiologie et en santé publique du centre de recherches mathématiques
Ce vendredi, lancement des activités du groupe en épidémiologie et en santé publique du centre de recherches mathématiques de Montréal. D’autres évènements seront à venir…
Machine learning for economists and applied social scientists
By the end of July, a group of colleagues from Dalhousie and Saint Mary’s Universities in Halifax will host a series of online lectures on machine learning for economists and applied social scientists. I will be giving a talk on machine learning and insurance.
Modeling Joint Lives within Families
With Olivier Cabrignac and Ewen Gallic, we recently uploaded a research paper, entitled “Modeling Joint Lives within Families
Family history is usually seen as a significant factor insurance companies look at when applying for a life insurance policy. Where it is used, family history of cardiovascular diseases, death by cancer, or family history of high blood pressure and diabetes could result in higher premiums or no coverage at all. In this article, we use massive (historical) data to study dependencies between life length within families. If joint life contracts (between a husband and a wife) have been long studied in actuarial literature, little is known about child and parents dependencies. We illustrate those dependencies using 19th century family trees in France, and quantify implications in annuities computations. For parents and children, we observe a modest but significant positive association between life lengths. It yields different estimates for remaining life expectancy, present values of annuities, or whole life insurance guarantee, given information about the parents (such as the number of parents alive). A similar but weaker pattern is observed when using information on grandparents.
The paper is online on https://arxiv.org/abs/2006.08446.
OpenEdition | 12,705 | 39,564 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 116, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-34 | latest | en | 0.787614 |
https://www.mathworks.com/matlabcentral/fileexchange/72470-dynamic-mode-decomposition-dmd-wrapper | 1,621,023,353,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991207.44/warc/CC-MAIN-20210514183414-20210514213414-00476.warc.gz | 936,546,697 | 21,606 | File Exchange
## Dynamic Mode Decomposition [DMD] - Wrapper
version 1.0.1 (17.3 MB) by
Wrapper function to perform DMD in N-Dimensional data sets. Reshapes data back and forth to facilitate handling.
Updated 20 Aug 2019
I built this wrapper to facilitate processing when performing modal analysis in arbitrary data sets. The wrapper accepts an N-D input matrix (Big_X) that has its first dimension as time and the other dimensions can be whatever the application requires.
The wrapper (hopefully) reduces the barrier of entry when doing these calculations, as building your own DMD function from scratch is quite time-consuming. Although there indeed are other functions like this one on Matlab Exchange, I often found that the lack of easy-to-use outputs was sometimes hindering my progress. So hopefully you will also find this useful!
Attached an usage sample and a small data set of a shedding cylinder to test the usage. Thanks to Prof. Louis Cattafesta from Florida State University to inspire me to produce this. If you have any requests please let me know.
### Cite As
Fernando Zigunov (2021). Dynamic Mode Decomposition [DMD] - Wrapper (https://www.mathworks.com/matlabcentral/fileexchange/72470-dynamic-mode-decomposition-dmd-wrapper), MATLAB Central File Exchange. Retrieved .
Gang Wang
Irsalan Arif
Hi. Could you please explain what does "first dimension in time" means? I have data for velocity for different time snaps. How should I build the matrix? I checked your sample but could not understand the Big_X matrix.
PK
How is the VelocityFieldData structured here? I am not able to open it because of limitations. I have a set of data with columns point1 point2 velx, vely in each time snapshots. How do I arrange this in big matrix form to be able to use with this code? The question could sound naive :) would appreciate your insights.
Rohin McIntosh
CHANDAN BOSE
Can you please let me know what should be the value of dt (delta t)?
Best regards,
Chandan
Udhav Gawandalkar
velocity data is already included in the wrapper. Just change the name of line
Jie Ren
Thanks for the example!
Fernando Zigunov
Shuai Feng, thanks for the interest, I'm afraid I did not understand your question.
Shuai Feng
How can i get the velocity date?
##### MATLAB Release Compatibility
Created with R2019a
Compatible with any release
##### Platform Compatibility
Windows macOS Linux | 541 | 2,402 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-21 | latest | en | 0.909839 |
https://www.physicsforums.com/threads/linear-transformations-rn-rm-question.72358/ | 1,713,496,764,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817253.5/warc/CC-MAIN-20240419013002-20240419043002-00036.warc.gz | 824,501,686 | 16,137 | # Linear Transformations Rn->Rm Question
• haribol
In summary, the question involves determining if the given transformation T from Rn to Rm is linear, and the solution manual provides a proof using the theorem that a transformation is linear if and only if certain relationships hold for all vectors and scalars. The proof involves setting two vectors u and v and showing that T(u+v) is equal to T(u) + T(v). The manual also includes the proof using the second condition of the theorem, which was accidentally left out in the original post.
haribol
Linear Transformations Rn-->Rm Question
I would be very grateful if someone can explain what is going on in the following problem:
Determine whether the following T:Rn to Rm
T(x,y)=(2x,y)
Solution from solutions manual:
T((x1,y1) + (x2,y2)) = (2(x1+x2), y1+y2) = (2x1,y1) + (2x2,y2) = T(x1,y1) + T(x2,y2)
My questions are
1. Where did the x1's and the x2's and the y1's and the y2's come from?
2. Can you please explain what's happening step by step?
[PS]--> The questions asks to use the theorem which states:
A transformation T:Rn --> Rm is linear if and only if the following relationships hold for all vectors u and v in Rn and every scalar c
a) T(u+v) = T(u) + T(v)
b)T(cu) = cT(u)
haribol said:
I would be very grateful if someone can explain what is going on in the following problem:
Determine whether the following T:Rn to Rm
T(x,y)=(2x,y)
Solution from solutions manual:
T((x1,y1) + (x2,y2)) = (2(x1+x2), y1+y2) = (2x1,y1) + (2x2,y2) = T(x1,y1) + T(x2,y2)
My questions are
1. Where did the x1's and the x2's and the y1's and the y2's come from?
2. Can you please explain what's happening step by step?
[PS]--> The questions asks to use the theorem which states:
A transformation T:Rn --> Rm is linear if and only if the following relationships hold for all vectors u and v in Rn and every scalar c
a) T(u+v) = T(u) + T(v)
b)T(cu) = cT(u)
He has set $\vec{u} = (x_1,y_1), \ \ \vec{v} = (x_2,y_2)$ and showed using vector addition properties that $T(\vec{u}+\vec{v}) = T(\vec{u})+T(\vec{v})$
This proof is imcomplete though because he left out condition b).
Thank you quasar987 for the clarification. The manual does include the proof using condition b) but I forgot to type it.
Thanks for that clarification.
## 1. What is a linear transformation from Rn to Rm?
A linear transformation from Rn to Rm is a mathematical function that maps vectors from n-dimensional space to m-dimensional space while preserving the properties of linearity. This means that the transformation preserves operations such as addition and scalar multiplication.
## 2. How is a linear transformation represented mathematically?
A linear transformation can be represented mathematically using a matrix. The matrix has m rows and n columns, where m is the dimension of the output space and n is the dimension of the input space. The entries of the matrix represent the coefficients of the linear combination of the input vectors that make up the output vectors.
## 3. What are the properties of a linear transformation?
A linear transformation has three main properties: it preserves vector addition, it preserves scalar multiplication, and it preserves the zero vector. This means that the transformation of the sum of two vectors is equal to the sum of the individual transformations and the transformation of a scalar multiple of a vector is equal to the scalar multiple of the transformation of the vector. Additionally, the transformation of the zero vector is always the zero vector.
## 4. How is a linear transformation different from other types of transformations?
A linear transformation is different from other types of transformations, such as nonlinear or affine transformations, because it preserves linearity. This means that the transformation of a line is always a line, and the transformation of a plane is always a plane. Nonlinear and affine transformations do not have this property and can distort or change the shape of the original object.
## 5. How are linear transformations used in real life?
Linear transformations have many real-life applications, such as in computer graphics, data analysis, and engineering. In computer graphics, linear transformations are used to rotate, scale, and translate objects on the screen. In data analysis, linear transformations are used to reduce the dimensionality of data and visualize high-dimensional data. In engineering, linear transformations are used to model systems and make predictions based on input parameters.
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1K | 1,262 | 5,106 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2024-18 | latest | en | 0.893481 |
https://harbourfronts.com/rule-of-72/ | 1,722,694,431,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640368581.4/warc/CC-MAIN-20240803121937-20240803151937-00448.warc.gz | 231,551,183 | 25,171 | # Rule of 72 in Finance and Investing: Definition, Calculation, Formula, Example, Equation
When it comes to financing, the Rule of 72 is a quick way to estimate how long an investment will take to double, given a fixed annual rate of return. The rule is useful for comparing the doubling time of different investments. It is also a good way to see whether an investment’s return is meeting your expectations.
The formula is simple and is based on the idea that money grows at a compound interest rate. Compound interest is when you earn interest not only on your original investment but also on the accumulated interest from previous periods.
## What is the Rule of 72
The Rule of 72 is a simple way to determine how long it will take for an investment to double in value. The rule states that you simply divide the number 72 by the interest rate you are earning on your investment, and the result is the number of years it will take for your money to double.
It’s very important to remember that the Rule of 72 is only a guideline and not a foolproof method for predicting the future. Many factors can affect how long it will take for your money to double, including inflation, fees, and changes in the underlying investment.
That being said, the Rule of 72 is still a helpful tool for understanding the power of compound interest. By knowing how long it will take for your money to double, you can get a better sense of how quickly your investment will grow.
## The formula of the Rule of 72
As we said, the formula of the Rule of 72 is very simple. You just divide the number 72 by the expected annual rate of return on your investment. The result will be the approximate amount of time it would take for your money to double.
The formula of the Rule of 72 is as follows,
Doubling year (expected years to double your money) = 72 / Annual interest rate
Doubling year: The number of years it would take for your money to double
Annual interest rate: The expected annual rate of return on your investment
## Examples of Rule of 72
Let’s say you’re planning to invest \$10,000 in a savings account that pays 2% interest per year. How long will it take for your investment to double using the Rule of 72?
\$10,000 x 2% = \$200
72 / 2 = 36 years
It would take approximately 36 years for your investment to double to \$20,000.
Now let’s say you want to know how long it will take to double your money if you’re earning 6% interest per year.
72 / 6 = 12 years
It would take approximately 12 years for your money to double at a 6% annual rate of return.
The Rule of 72 is a simple way to estimate how long an investment will take to double, given a fixed annual rate of return. The rule is useful for comparing the doubling time of different investments. It is also a good way to see whether an investment’s return is meeting your expectations.
## Conclusion
The rule of 72 is a helpful tool that can be used to estimate how long it will take to double an investment. It is also useful for calculating the required rate of return on investment to achieve a financial goal. The formula is simple, all you have to do is divide 72 by the expected annual rate of return.
## Further questions
Have an answer to the questions below? Post it here or in the forum
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Question
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## Documents
TRANSCRIPT
• 7/27/2019 Add Math Selangor 4
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pg. 1
PART 1
a) Write a history on logarithm.
History of Logarithms
From Napier to Euler
The method of logarithms was publicly
propounded byJohn Napierin 1614, in a
book titled Mirifici Logarithmorum Canonis
Descriptio(Description of the Wonderful
Rule of Logarithms).Joost
Brgiindependently invented logarithms
but published six years after Napier.
Johannes Kepler, who used logarithm
tables extensively to compile
his Ephemeris and therefore dedicated it
to Napier, remarked:
...the accent in calculation led Justus
Byrgius [Joost Brgi] on the way to these
very logarithms many years before
Napier's system appeared; but ...instead of rearing up his child for the public benefit he
deserted it in the birth.
Johannes Kepler, Rudolphine Tables (1627)
By repeated subtractions Napier calculated (1 107)L forL ranging from 1 to 100. The
result forL=100 is approximately0.99999 = 1 105. Napier then calculated the
products of these numbers with 107(1 105)L forL from 1 to 50, and did similarly
with0.9998 (1 105)20 and 0.9 0.99520. These computations, which occupied 20
years, allowed him to give, for any numberNfrom 5 to 10 million, the numberL that
solves the equation
Napier first called L an "artificial number", but later introduced the word "logarithm"to
mean a number that indicates a ratio: (logos) meaning proportion,
and (arithmos) meaning number. In modern notation, the relation to natural
logarithms is
T
http://en.wikipedia.org/wiki/John_Napierhttp://en.wikipedia.org/wiki/John_Napierhttp://en.wikipedia.org/wiki/John_Napierhttp://en.wikipedia.org/wiki/Joost_B%C3%BCrgihttp://en.wikipedia.org/wiki/Joost_B%C3%BCrgihttp://en.wikipedia.org/wiki/Joost_B%C3%BCrgihttp://en.wikipedia.org/wiki/Joost_B%C3%BCrgihttp://en.wikipedia.org/wiki/Johannes_Keplerhttp://en.wikipedia.org/wiki/Johannes_Keplerhttp://en.wikipedia.org/wiki/Logoshttp://en.wikipedia.org/wiki/Logoshttp://en.wikipedia.org/wiki/Logoshttp://en.wikipedia.org/wiki/File:John_Napier.jpghttp://en.wikipedia.org/wiki/File:John_Napier.jpghttp://en.wikipedia.org/wiki/File:John_Napier.jpghttp://en.wikipedia.org/wiki/File:John_Napier.jpghttp://en.wikipedia.org/wiki/Logoshttp://en.wikipedia.org/wiki/Johannes_Keplerhttp://en.wikipedia.org/wiki/Joost_B%C3%BCrgihttp://en.wikipedia.org/wiki/Joost_B%C3%BCrgihttp://en.wikipedia.org/wiki/John_Napier
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pg. 2
where the very close approximation corresponds to the observation that
The invention was quickly and widely met with acclaim. The works
ofBonaventura Cavalieri(Italy),Edmund Wingate(France), Xue Fengzuo(China), andJohannes Kepler's Chilias logarithmorum (Germany) helped spread
the concept further.
In 1647Grgoire de Saint-Vincentrelated logarithms to the quadrature of the
hyperbola, by pointing out that the area f(t) under the hyperbola fromx=
1 tox= tsatisfies
The natural logarithm was first described byNicholas Mercatorin his
work Logarithmotechnia published in 1668, although the mathematics teacher JohnSpeidell had already in 1619 compiled a table on the natural logarithm. Around
1730,Leonhard Eulerdefined the exponential function and the natural logarithm by
Euler also showed that the two functions are inverse to one another.
http://en.wikipedia.org/wiki/Bonaventura_Cavalierihttp://en.wikipedia.org/wiki/Bonaventura_Cavalierihttp://en.wikipedia.org/wiki/Bonaventura_Cavalierihttp://en.wikipedia.org/wiki/Edmund_Wingatehttp://en.wikipedia.org/wiki/Edmund_Wingatehttp://en.wikipedia.org/wiki/Edmund_Wingatehttp://en.wikipedia.org/wiki/Johannes_Keplerhttp://en.wikipedia.org/wiki/Johannes_Keplerhttp://en.wikipedia.org/wiki/Johannes_Keplerhttp://en.wikipedia.org/wiki/Gr%C3%A9goire_de_Saint-Vincenthttp://en.wikipedia.org/wiki/Gr%C3%A9goire_de_Saint-Vincenthttp://en.wikipedia.org/wiki/Gr%C3%A9goire_de_Saint-Vincenthttp://en.wikipedia.org/wiki/Nicholas_Mercatorhttp://en.wikipedia.org/wiki/Nicholas_Mercatorhttp://en.wikipedia.org/wiki/Nicholas_Mercatorhttp://en.wikipedia.org/wiki/Leonhard_Eulerhttp://en.wikipedia.org/wiki/Leonhard_Eulerhttp://en.wikipedia.org/wiki/Leonhard_Eulerhttp://en.wikipedia.org/wiki/Nicholas_Mercatorhttp://en.wikipedia.org/wiki/Gr%C3%A9goire_de_Saint-Vincenthttp://en.wikipedia.org/wiki/Johannes_Keplerhttp://en.wikipedia.org/wiki/Edmund_Wingatehttp://en.wikipedia.org/wiki/Bonaventura_Cavalieri
• 7/27/2019 Add Math Selangor 4
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• 7/27/2019 Add Math Selangor 4
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pg. 4
maximum of the likelihood function occurs at the same parameter-value as a maximum of the
logarithm of the likelihood (the "log likelihood"), because the logarithm is an increasing
function. The log-likelihood is easier to maximize, especially for the multiplied likelihoods
forindependentrandom variables.
Benford's lawdescribes the occurrence of digits in manydata sets, such as heights of buildings.According to Benford's law, the probability that the first decimal-digit of an item in the data
sample is d(from 1 to 9) equals log10(d+ 1) log10(d), regardless of the unit of measurement.
Thus, about 30% of the data can be expected to have 1 as first digit, 18% start with 2, etc.
Auditors examine deviations from Benford's law to detect fraudulent accounting.
2. Fractals
The Sierpinski triangle (at the right) is constructed by repeatedly replacingequilateral
trianglesby three smaller ones.
Logarithms occur in definitions of thedimensionoffractals. Fractals are geometric objects that
areself-similar: small parts reproduce, at least roughly, the entire global structure. TheSierpinski
triangle(pictured) can be covered by three copies of itself, each having sides half the original
length. This makes theHausdorff dimensionof this structure log(3)/log(2) 1.58. Another
logarithm-based notion of dimension is obtained bycounting the number of boxesneeded to
cover the fractal in question.
http://en.wikipedia.org/wiki/Independence_(probability)http://en.wikipedia.org/wiki/Independence_(probability)http://en.wikipedia.org/wiki/Independence_(probability)http://en.wikipedia.org/wiki/Benford%27s_lawhttp://en.wikipedia.org/wiki/Benford%27s_lawhttp://en.wikipedia.org/wiki/Data_sethttp://en.wikipedia.org/wiki/Data_sethttp://en.wikipedia.org/wiki/Data_sethttp://en.wikipedia.org/wiki/Equilateral_trianglehttp://en.wikipedia.org/wiki/Equilateral_trianglehttp://en.wikipedia.org/wiki/Equilateral_trianglehttp://en.wikipedia.org/wiki/Equilateral_trianglehttp://en.wikipedia.org/wiki/Fractal_dimensionhttp://en.wikipedia.org/wiki/Fractal_dimensionhttp://en.wikipedia.org/wiki/Fractal_dimensionhttp://en.wikipedia.org/wiki/Fractalhttp://en.wikipedia.org/wiki/Fractalhttp://en.wikipedia.org/wiki/Fractalhttp://en.wikipedia.org/wiki/Self-similarityhttp://en.wikipedia.org/wiki/Self-similarityhttp://en.wikipedia.org/wiki/Self-similarityhttp://en.wikipedia.org/wiki/Sierpinski_trianglehttp://en.wikipedia.org/wiki/Sierpinski_trianglehttp://en.wikipedia.org/wiki/Sierpinski_trianglehttp://en.wikipedia.org/wiki/Sierpinski_trianglehttp://en.wikipedia.org/wiki/Hausdorff_dimensionhttp://en.wikipedia.org/wiki/Hausdorff_dimensionhttp://en.wikipedia.org/wiki/Hausdorff_dimensionhttp://en.wikipedia.org/wiki/Box-counting_dimensionhttp://en.wikipedia.org/wiki/Box-counting_dimensionhttp://en.wikipedia.org/wiki/Box-counting_dimensionhttp://en.wikipedia.org/wiki/File:Sierpinski_dimension.svghttp://en.wikipedia.org/wiki/Box-counting_dimensionhttp://en.wikipedia.org/wiki/Hausdorff_dimensionhttp://en.wikipedia.org/wiki/Sierpinski_trianglehttp://en.wikipedia.org/wiki/Sierpinski_trianglehttp://en.wikipedia.org/wiki/Self-similarityhttp://en.wikipedia.org/wiki/Fractalhttp://en.wikipedia.org/wiki/Fractal_dimensionhttp://en.wikipedia.org/wiki/Equilateral_trianglehttp://en.wikipedia.org/wiki/Equilateral_trianglehttp://en.wikipedia.org/wiki/Data_sethttp://en.wikipedia.org/wiki/Benford%27s_lawhttp://en.wikipedia.org/wiki/Independence_(probability)
• 7/27/2019 Add Math Selangor 4
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pg. 5
PART 2
The volume, V, in cm3, of a solid sphere and its diameter, D, in cm, are related by the
equation , where m and n are constants.
Find the value of m and n by conducting the activities below.
I. Choose 6 different spheres with diameters between 1cm to 8cm. The diameterof the 6 spheres using a pair of vernier calipers.
II. Find the volume of each sphere using water displacement method.III. Tabulate the values of diameter, D, in cm and its corresponding volume, V, cm3.
• 7/27/2019 Add Math Selangor 4
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pg. 6
Find the volume of sphere using water displacement method.
A method of finding the volume of a sphere with minimal calculations is to use the WaterDisplacement Method:
1. Fill a beaker or graduated cylinder with enough water to completely immerse thesphere in.
2. Record the baseline initial measurement3. Drop the sphere in4. Record final measurement5. Subtract the initial volume from the final volume ~ this is the volume of the
sphere!
Value of diameter,D and Volume
Diameter,D ( Volume, V (
D1 = 1.0 V1= 0.5
D2 =2.8 V2= 11.5
D3 =4.0 V3= 34.0
D4 =5.2 V4= 74.0
D5 =6.6 V5= 151.0
D6 =7.8 V6= 250.0
• 7/27/2019 Add Math Selangor 4
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pg. 7
Diameter,D ( Volume, V (
D1 = 1.0 V1= 0.5
D2 =2.8 V2= 11.5
D3 =4.0 V3= 34.0
D4 =5.2 V4= 74.0D5 =6.6 V5= 151.0
D6 =7.8 V6= 250.0
We can solve by simultaneous method
Substitute the values in the equation
We obtain,
----------(1)
----------(2)
-----------(3)Substitute (3) into (2)
D2 = 2.8 V2= 11.5
D5 =6.6 V5= 151.0
• 7/27/2019 Add Math Selangor 4
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pg. 8
-----------(4)
Substitute (4) into (3)
Therefore, and
• 7/27/2019 Add Math Selangor 4
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pg. 9
PART 3
3(A)
D v
1.0 0.5
2.8 11.5
4.0 34.0
5.2 74.0
6.6 151.0
7.8 250.0
y = 0.505x3.025
0
50
100
150
200
250
300
0 1 2 3 4 5 6 7 8 9
Volume,V
Diameter, D
• 7/27/2019 Add Math Selangor 4
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pg. 10
3(B)
y = 3.025x - 0.2967
-0.5
0
0.5
1
1.5
2
2.5
3
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
logV
Diameter, D
log D log V
0 -0.30103
0.447158 1.060698
0.60206 1.531479
0.716003 1.869232
0.819544 2.178977
0.892095 2.39794
• 7/27/2019 Add Math Selangor 4
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pg. 11
3c) From the graph, find
1. The value of m and of n, thus express V in terms of D.
(nearest whole number)
y = 3.025x - 0.2967
-0.5
0
0.5
1
1.5
2
2.5
3
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
logV
Diameter, D
log D log V
0 -0.30103
0.447158 1.060698
0.60206 1.531479
0.716003 1.869232
0.819544 2.178977
0.892095 2.39794
• 7/27/2019 Add Math Selangor 4
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pg. 12
2. Volume of the sphere when diameter is 5cm
Since graph is logV against logD, we need to transfer, D=5cm int0
logD=log5=0.6989
We get
3. The radius of the sphere when the volume is
Change to logv=log180=2.25,
From the graph, we get
y = 3.025x - 0.2967
-0.5
0
0.5
1
1.5
2
2.5
3
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
logV
Diameter, D
• 7/27/2019 Add Math Selangor 4
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pg. 13
FURTHER EXPLORATION
a) -------(1)
------------(2)
(1)=(2)
D
-------------------cancel on both sides
b) Another method to find value of is using Monte Carlo simulation or
Archimedes method of Exhaustion
• 7/27/2019 Add Math Selangor 4
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pg. 14
REFLECTION
Symbols used in this project using Microsoft word equation insert tool really help me so
much here are some of the symbol I use.
I really learn how to use Microsoft excel and word to do graph, insert equation and a lot
more.
y = 0.505x3.025
0
50
100
150
200
250
300
0 2 4 6 8 10
Vo
lume,V
Diameter, D
y = 3.025x - 0.2967
-0.5
0
0.5
1
1.5
2
2.5
3
0 0.2 0.4 0.6 0.8 1
logV
Diameter, D
• 7/27/2019 Add Math Selangor 4
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pg 15 | 3,688 | 11,724 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-10 | latest | en | 0.814592 |
https://www.esaral.com/q/prove-84890 | 1,720,912,709,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514517.94/warc/CC-MAIN-20240713212202-20240714002202-00837.warc.gz | 668,229,421 | 11,445 | # Prove
Question:
Prove $\cot ^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\frac{x}{2}, x \in\left(0, \frac{\pi}{4}\right)$
Solution:
Consider $\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}$
$=\frac{(\sqrt{1+\sin x}+\sqrt{1-\sin x})^{2}}{(\sqrt{1+\sin x})^{2}-(\sqrt{1-\sin x})^{2}}$ (by rationalizing)
$=\frac{(1+\sin x)+(1-\sin x)+2 \sqrt{(1+\sin x)(1-\sin x)}}{1+\sin x-1+\sin x}$
$=\frac{2\left(1+\sqrt{1-\sin ^{2} x}\right)}{2 \sin x}=\frac{1+\cos x}{\sin x}=\frac{2 \cos ^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}$
$=\cot \frac{x}{2}$
$\therefore$ L.H.S. $=\cot ^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\cot ^{-1}\left(\cot \frac{x}{2}\right)=\frac{x}{2}=$ R.H.S. | 375 | 816 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-30 | latest | en | 0.203433 |
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# GMAT Critical Reasoning (CR)
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Critical Reasoning Megathread! Go to page: 1, 2 Tags: Assumption, CR Resources
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The Most Comprehensive Collection Of Everything Official- CR
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QOTD #13 The Starbeans cafe has recently hired a new manager. The mana
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Many of Vebrol Corporation's department heads Will retire this year. T
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It has been established that people who live in the northern hemispher
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Who is online In total there are 6 users online :: 0 registered, 0 hidden and 6 guests (based on users active over the past 15 minutes) Users browsing this forum: No registered users and 6 guests Statistics Total posts 1582774 | Total topics 191592 | Active members 488072 | Our newest member grungekid94
Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,772 | 5,787 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2017-04 | latest | en | 0.704853 |
https://piping-designer.com/index.php/properties/classical-mechanics/687-newton-s-second-law | 1,717,096,724,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971670239.98/warc/CC-MAIN-20240530180105-20240530210105-00007.warc.gz | 380,922,062 | 7,900 | # Newton's Second Law
on . Posted in Classical Mechanics
Newton's second law of motion, also known as the Law of Resultant Force, describes the relationship between the motion of an object, the net force acting on it, and its mass. It states that the acceleration of an object is directly proportional to the net force applied to it and inversely proportional to its mass. In simpler terms, the second law states that the acceleration of an object is directly related to the force applied to it. If a greater force is applied to an object, it will experience a greater acceleration. Similarly, if the mass of the object is increased, it will experience a smaller acceleration for the same force.
Newton's second law is fundamental in understanding the relationship between force, mass, and acceleration and is widely used in physics and engineering to analyze the motion of objects and design systems that involve the application of forces.
### Newton's Second Law formula
If the object is free fall with on other force other than gravity.
$$F \;=\; m \; a$$ (Newton's Second Law)
$$m \;=\; F \;/\; a$$
$$a \;=\; F \;/\; m$$
Symbol English Metric
$$F$$ = force $$lbf$$ $$N$$
$$m$$ = mass $$lbm$$ $$kg$$
$$a$$ = acceleration $$ft\;/\;sec^2$$ $$m\;/\;s^2$$ | 314 | 1,272 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-22 | latest | en | 0.875206 |
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comp2_07Jan08 - Dietrich Vollrath and Bent E Srensen...
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Dietrich Vollrath and Bent E. Sørensen January 10, 2008 Comprehensive Exam in Macroeconomic Theory–Procedural Instructions (1) Write your answers only on the paper we will provide. (2) We will be distributing a numbered sign–in sheet in a moment. The number next to your signature will be your student number . (3) Every sheet of paper you turn in to us must have your student number written at the top–center of the sheet and circled . (4) Every sheet of paper you turn in to us must have a page number written at the top–right corner of the sheet. (5) When you have finished, or when it is 1:00 pm (whichever comes first), prepare a cover sheet for your exam. This cover sheet should not have a page number, but must have the following things on it: (a) Your student number at the top–center, circled. (b) The phrase “Macroeconomics Comprehensive Exam”. (c) The sentence “My last page is page number X”, where X is your total number of pages of answers. 1
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1. (10%) Consider an IS/LM framework. The demand for money is M d = 0 . 5 Y - 0 . 5 i where Y is output and i = r + π where r is the real rate of interest and π is the rate of inflation. Further assume that output demand is Y = C + G + I where C = 0 . 8 Y and I = 0 . 1 Y - 0 . 1 r . a) Derive the IS-curve (you need to find the exact coefficients implied by the information you are given). b) If G = 1, M = 4, and π = 0 find Y . c) Derive the aggregate demand curve; i.e., a relation between Y and π . (Again, you need to find the exact function implied by the information given.) 2. (30%) Consider the case of 2 agents, each living for 2 periods. Output is deterministically given in period 0, while there are 2 states-of-the-world ( a and b ) in period 1. We will refer to one agent as “home” and one as “foreign” and mark the variables for “foreign” with a star. Assume ini-
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https://www.ademcetinkaya.com/2023/04/cnglu-canna-global-acquisition-corp-unit.html | 1,685,424,365,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224645089.3/warc/CC-MAIN-20230530032334-20230530062334-00310.warc.gz | 682,979,352 | 61,710 | Outlook: Canna-Global Acquisition Corp Unit is assigned short-term Ba1 & long-term Ba1 estimated rating.
Dominant Strategy : Wait until speculative trend diminishes
Time series to forecast n: 10 Apr 2023 for (n+8 weeks)
Methodology : Modular Neural Network (Market Volatility Analysis)
Abstract
Canna-Global Acquisition Corp Unit prediction model is evaluated with Modular Neural Network (Market Volatility Analysis) and Independent T-Test1,2,3,4 and it is concluded that the CNGLU stock is predictable in the short/long term. According to price forecasts for (n+8 weeks) period, the dominant strategy among neural network is: Wait until speculative trend diminishes
Key Points
1. Operational Risk
2. Can statistics predict the future?
3. How do predictive algorithms actually work?
CNGLU Target Price Prediction Modeling Methodology
We consider Canna-Global Acquisition Corp Unit Decision Process with Modular Neural Network (Market Volatility Analysis) where A is the set of discrete actions of CNGLU stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4
F(Independent T-Test)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Modular Neural Network (Market Volatility Analysis)) X S(n):→ (n+8 weeks) $R=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)$
n:Time series to forecast
p:Price signals of CNGLU stock
j:Nash equilibria (Neural Network)
k:Dominated move
a:Best response for target price
For further technical information as per how our model work we invite you to visit the article below:
How do AC Investment Research machine learning (predictive) algorithms actually work?
CNGLU Stock Forecast (Buy or Sell) for (n+8 weeks)
Sample Set: Neural Network
Stock/Index: CNGLU Canna-Global Acquisition Corp Unit
Time series to forecast n: 10 Apr 2023 for (n+8 weeks)
According to price forecasts for (n+8 weeks) period, the dominant strategy among neural network is: Wait until speculative trend diminishes
X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)
Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)
Z axis (Grey to Black): *Technical Analysis%
IFRS Reconciliation Adjustments for Canna-Global Acquisition Corp Unit
1. Historical information is an important anchor or base from which to measure expected credit losses. However, an entity shall adjust historical data, such as credit loss experience, on the basis of current observable data to reflect the effects of the current conditions and its forecasts of future conditions that did not affect the period on which the historical data is based, and to remove the effects of the conditions in the historical period that are not relevant to the future contractual cash flows. In some cases, the best reasonable and supportable information could be the unadjusted historical information, depending on the nature of the historical information and when it was calculated, compared to circumstances at the reporting date and the characteristics of the financial instrument being considered. Estimates of changes in expected credit losses should reflect, and be directionally consistent with, changes in related observable data from period to period
2. A portfolio of financial assets that is managed and whose performance is evaluated on a fair value basis (as described in paragraph 4.2.2(b)) is neither held to collect contractual cash flows nor held both to collect contractual cash flows and to sell financial assets. The entity is primarily focused on fair value information and uses that information to assess the assets' performance and to make decisions. In addition, a portfolio of financial assets that meets the definition of held for trading is not held to collect contractual cash flows or held both to collect contractual cash flows and to sell financial assets. For such portfolios, the collection of contractual cash flows is only incidental to achieving the business model's objective. Consequently, such portfolios of financial assets must be measured at fair value through profit or loss.
3. If, at the date of initial application, it is impracticable (as defined in IAS 8) for an entity to assess whether the fair value of a prepayment feature was insignificant in accordance with paragraph B4.1.12(c) on the basis of the facts and circumstances that existed at the initial recognition of the financial asset, an entity shall assess the contractual cash flow characteristics of that financial asset on the basis of the facts and circumstances that existed at the initial recognition of the financial asset without taking into account the exception for prepayment features in paragraph B4.1.12. (See also paragraph 42S of IFRS 7.)
4. For some types of fair value hedges, the objective of the hedge is not primarily to offset the fair value change of the hedged item but instead to transform the cash flows of the hedged item. For example, an entity hedges the fair value interest rate risk of a fixed-rate debt instrument using an interest rate swap. The entity's hedge objective is to transform the fixed-interest cash flows into floating interest cash flows. This objective is reflected in the accounting for the hedging relationship by accruing the net interest accrual on the interest rate swap in profit or loss. In the case of a hedge of a net position (for example, a net position of a fixed-rate asset and a fixed-rate liability), this net interest accrual must be presented in a separate line item in the statement of profit or loss and other comprehensive income. This is to avoid the grossing up of a single instrument's net gains or losses into offsetting gross amounts and recognising them in different line items (for example, this avoids grossing up a net interest receipt on a single interest rate swap into gross interest revenue and gross interest expense).
*International Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS.
Conclusions
Canna-Global Acquisition Corp Unit is assigned short-term Ba1 & long-term Ba1 estimated rating. Canna-Global Acquisition Corp Unit prediction model is evaluated with Modular Neural Network (Market Volatility Analysis) and Independent T-Test1,2,3,4 and it is concluded that the CNGLU stock is predictable in the short/long term. According to price forecasts for (n+8 weeks) period, the dominant strategy among neural network is: Wait until speculative trend diminishes
CNGLU Canna-Global Acquisition Corp Unit Financial Analysis*
Rating Short-Term Long-Term Senior
Outlook*Ba1Ba1
Income StatementB3Ba1
Balance SheetBaa2Ba3
Leverage RatiosCaa2B3
Cash FlowBaa2Caa2
Rates of Return and ProfitabilityCBaa2
*Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents.
How does neural network examine financial reports and understand financial state of the company?
Prediction Confidence Score
Trust metric by Neural Network: 80 out of 100 with 581 signals.
References
1. Efron B, Hastie T, Johnstone I, Tibshirani R. 2004. Least angle regression. Ann. Stat. 32:407–99
2. Athey S, Bayati M, Doudchenko N, Imbens G, Khosravi K. 2017a. Matrix completion methods for causal panel data models. arXiv:1710.10251 [math.ST]
3. Wan M, Wang D, Goldman M, Taddy M, Rao J, et al. 2017. Modeling consumer preferences and price sensitiv- ities from large-scale grocery shopping transaction logs. In Proceedings of the 26th International Conference on the World Wide Web, pp. 1103–12. New York: ACM
4. Dudik M, Langford J, Li L. 2011. Doubly robust policy evaluation and learning. In Proceedings of the 28th International Conference on Machine Learning, pp. 1097–104. La Jolla, CA: Int. Mach. Learn. Soc.
5. Mikolov T, Sutskever I, Chen K, Corrado GS, Dean J. 2013b. Distributed representations of words and phrases and their compositionality. In Advances in Neural Information Processing Systems, Vol. 26, ed. Z Ghahramani, M Welling, C Cortes, ND Lawrence, KQ Weinberger, pp. 3111–19. San Diego, CA: Neural Inf. Process. Syst. Found.
6. M. Colby, T. Duchow-Pressley, J. J. Chung, and K. Tumer. Local approximation of difference evaluation functions. In Proceedings of the Fifteenth International Joint Conference on Autonomous Agents and Multiagent Systems, Singapore, May 2016
7. Abadie A, Diamond A, Hainmueller J. 2015. Comparative politics and the synthetic control method. Am. J. Political Sci. 59:495–510
Frequently Asked QuestionsQ: What is the prediction methodology for CNGLU stock?
A: CNGLU stock prediction methodology: We evaluate the prediction models Modular Neural Network (Market Volatility Analysis) and Independent T-Test
Q: Is CNGLU stock a buy or sell?
A: The dominant strategy among neural network is to Wait until speculative trend diminishes CNGLU Stock.
Q: Is Canna-Global Acquisition Corp Unit stock a good investment?
A: The consensus rating for Canna-Global Acquisition Corp Unit is Wait until speculative trend diminishes and is assigned short-term Ba1 & long-term Ba1 estimated rating.
Q: What is the consensus rating of CNGLU stock?
A: The consensus rating for CNGLU is Wait until speculative trend diminishes.
Q: What is the prediction period for CNGLU stock?
A: The prediction period for CNGLU is (n+8 weeks) | 2,324 | 10,066 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-23 | latest | en | 0.803862 |
http://tex.stackexchange.com/questions/194801/multiple-alignments-within-set-of-equations | 1,469,688,779,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257828009.82/warc/CC-MAIN-20160723071028-00305-ip-10-185-27-174.ec2.internal.warc.gz | 241,348,376 | 17,698 | Multiple alignments within set of equations
My MWE:
\documentclass{report}
\usepackage{amsmath}
\begin{document}
The partial derivatives are given by:
\begin{subequations}
\begin{align}
\frac{\partial \beta_{2,1}}{\partial x_1} &= \gamma_2x_2 \cos(x_1), &&\frac{\partial \beta_{2,1}}{\partial x_2} = \gamma_2 \sin(x_1), &&&\frac{\partial \beta_{2,1}}{\partial u} = 0\\
\frac{\partial \beta_{2,2}}{\partial x_1} &= 0, &&\frac{\partial \beta_{2,2}}{\partial x_2} = \gamma_2 x_2, &&&\frac{\partial \beta_{2,2}}{\partial u} = 0\\
\frac{\partial \beta_{2,3}}{\partial x_1} &= 0, &&\frac{\partial \beta_{2,3}}{\partial x_2} = \gamma_2 u, &&&\frac{\partial \beta_{2,3}}{\partial u} = \gamma_2x_2
\end{align}
\end{subequations}
\end{document}
The result:
However, I would like to obtain the following result:
-
To have fuller control on the spacing of equations, you can use the alignat environment. In addition, to simplify typing partial derivatives (fist order only), I introduce a macro, \pder with one argument — actually two, separated by a comma, based on the esdiff package and on xparse. Here is an example, where the groups of equations are separated by 0.8em:
\documentclass{report}
\usepackage{amsmath}
\usepackage{esdiff}
\usepackage{xparse}
\NewDocumentCommand\pder{>{\SplitArgument{1}{,}}m}{\pderaux#1}
\NewDocumentCommand\pderaux{m m}{\diffp{{#1}}{{#2}}}
\begin{document}
The partial derivatives are given by:
\begin{subequations}
\begin{alignat}{5}
\pder{\beta_{2,1}, x_1} & = \gamma_2x_2 \cos(x_1), &\hspace{0.8em} \pder{\beta_{2,1}, x_2} & = \gamma_2 \sin(x_1), &\hspace{0.8em} \pder{\beta_{2,1}, u} &= 0\\
\pder{\beta_{2,2}, x_1} & = 0,& \pder{\beta_{2,2}, x_2}&= \gamma_2 x_2,& \pder{\beta_{2,2}, u} &= 0 \\
\pder{\beta_{2,3}, x_1} & = 0, & \pder{\beta_{2,3}, x_2} & = \gamma_2 u, & \pder{\beta_{2,3}, u} & = \gamma_2x_2
\end{alignat}
\end{subequations}
\end{document}
-
Tell LaTeX more accurately what spacing you want, by inserting more ampersands.
Please note that putting the ampersand after the equals sign (=&) will produce less spacing between the equals sign and the character that follows. Putting the ampersand before (i.e. &=) will produce slightly prettier spacing.
\documentclass{report}
\usepackage{amsmath}
\begin{document}
The partial derivatives are given by:
\begin{subequations}
\begin{align}
\frac{\partial \beta_{2,1}}{\partial x_1} &= \gamma_2x_2 \cos(x_1), &\frac{\partial \beta_{2,1}}{\partial x_2} &= \gamma_2 \sin(x_1), &\frac{\partial \beta_{2,1}}{\partial u} &= 0\\
\frac{\partial \beta_{2,2}}{\partial x_1} &= 0, &\frac{\partial \beta_{2,2}}{\partial x_2} &= \gamma_2 x_2, &\frac{\partial \beta_{2,2}}{\partial u} &= 0\\
\frac{\partial \beta_{2,3}}{\partial x_1} &= 0, &\frac{\partial \beta_{2,3}}{\partial x_2} &= \gamma_2 u, &\frac{\partial \beta_{2,3}}{\partial u} &= \gamma_2x_2
\end{align}
\end{subequations}
\end{document}
- | 1,077 | 2,950 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2016-30 | latest | en | 0.64043 |
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# El dinero (money) WALT: To deal with prices and numbers when shopping in Spain WILF: to use & recognise numbers up to 1000 in order to work towards.
## Presentation on theme: "El dinero (money) WALT: To deal with prices and numbers when shopping in Spain WILF: to use & recognise numbers up to 1000 in order to work towards."— Presentation transcript:
El dinero (money) WALT: To deal with prices and numbers when shopping in Spain WILF: to use & recognise numbers up to 1000 in order to work towards level 3+ WALT: To deal with prices and numbers when shopping in Spain WILF: to use & recognise numbers up to 1000 in order to work towards level 3+
Take 5 minutes to revise the numbers in your book THEN do the activity below: When all the numbers above are in your book, write the following numbers : Eg: Treinta y siete= 37 a)Cuarenta y dos = b)Cincuenta y cinco= c)Setenta y uno= d)Sesenta y ocho = e)Ochenta y siete = f)Noventa y nueve= g)Ciento veinte = h)Ciento cuarenta y cinco= i)Doscientostreinta= j)Cuatrocientos veintiuno= k)Quinientos diez= l)Noventacientos noventa y nueve= When all the numbers above are in your book, write the following numbers : Eg: Treinta y siete= 37 a)Cuarenta y dos = b)Cincuenta y cinco= c)Setenta y uno= d)Sesenta y ocho = e)Ochenta y siete = f)Noventa y nueve= g)Ciento veinte = h)Ciento cuarenta y cinco= i)Doscientostreinta= j)Cuatrocientos veintiuno= k)Quinientos diez= l)Noventacientos noventa y nueve=
© 2005 Black Country Pathfinder 14-19 Networks for Excellence Spanish Element 2. El euro Euro banknotes and coins are now a part of daily life for over 300 million Europeans living in the euro area. In 2005 the euro became the currency of twelve European Union countries, stretching from the Mediterranean to the Arctic Circle (namely Belgium, Germany, Greece, Spain, France, Ireland, Italy, Luxembourg, the Netherlands, Austria, Portugal and Finland). For more information, see europa.eu.int
Euro banknotes = Billetes The first euro banknotes were introduced on 1st January 2002 and replaced twelve different sets of banknotes in the previous national currencies. There are seven different denominations in the current euro banknote series, ranging from the €5 to the €500 note. The designs of the notes are the same throughout the euro area and feature windows and gateways (on the front) and bridges (on the reverse) from different periods in Europe's architectural history. El Euro: € For more information, see europa.eu.int
. El Euro Euros y centimos The Member States of the euro area issue their own euro coins. Each euro coin has a common European design on one side and an individual national design on the other. However, the technical features of the coins (size, weight, metals used) are identical across all euro countries. There are eight different denominations in the current euro coin series, ranging from the 1 cent to the €2 coin. For more information, see europa.eu.int
. Today, a euro is worth about 3€3€ For more information, see europa.eu.int 85p equal roughly £2.55 = £425!
¿Cuanto cuesta? How much is it? COPY IN THE SPANISH AND ENGLISH IN YOUR BOOK!
¿CUANTO CUESTA? cincuenta euros sesenta euros Setenta y cinco euros cien euros Ciento ocho euros For more information, see europa.eu.int
¿Cuanto cuesta? Write the number in full in your book For help look at the bottom of the pages in text books for numbers!
NOW PREPARE YOUR OWN PRESENTATION ALL ABOUT THE EURO
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Is an equilateral triangle equal to a scalene triangle?
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Module Title QUANTUM PHYSICS II
Co-ordinator Dr Tudor E Jenkins
Semester Semester 2
Other staff Dr Daniel Brown, Mr Peter C Bollada
Pre-Requisite Successful Completion of Year 3 of the MPhys Scheme
Course delivery Lecture 20 lectures
Assessment
Assessment TypeAssessment Length/DetailsProportion
Semester Exam3 Hours End of semester examinations 100%
#### Learning outcomes
After taking this module students should :
• be familiar with fundamental assumptions of Quantum Mechanics.
• be able to apply simple model potential well systems to solve elementary problems.
• describe and apply both time-independent and time-dependent perturbation theory.
• be able to use a variational method for finding the ground state of a bound particle.
#### Brief description
The Postulates of Quantum Mechanics are introduced. Model potential wells in 1 and 3 dimensions are described and applied to simple physical phenomena and optical properties in condensed matter. Time-independent (non-degenerate and degenerate) and time-dependent perturbation theory are applied to a number of physical problems, and the variational method is used to derive the ground state of Helium.
#### Content
• Fundamentals of Quantum Mechanics and their relation to the properties of dynamic operators, wavefunctions and the eigenvalues that are observed.
• Model potential well systems: finite potential well - scattering and tunnelling; cubic and spherical wells - optical phenomena in insulators and quantum confined systems.
• Perturbation theory:
(a) stationary theory - non-degenerate (1st and 2nd Order) degenerate
(b) time-dependent - harmonic perturbation, radiative transition, step perturbation
• Variational method.
#### Transferable skills
This module will include several problem-solving sessions.
Books
** Recommended Text
Phillips, A.C. Introduction to Quantum Mechanics John Wiley
Alistair I.M. Rae Quantum Mechanics Institute of Physics
** Reference Text
French, A P & Taylor, E F An Introduction to Quantum Physics MIT Introductory Physics Series
Matthews Introduction to Quantum Mechanics McGraw-Hill
F. Mandl Quantum Mechanics John Wiley
Davies, P.C.W. Quantum Mechanics Routledge & Kegan
** Recommended Background
Hey, Tony & Walters, Patrick The New Quantum Universe Cambridge University Press
#### Notes
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# What is the exponential form of 7 billion?
Wiki User
2008-10-29 23:54:25
"Billion" usually has the meaning "thousand million". So 7 billion = 7 x 109.
Traditional UK usage has billion meaning "million million". This in exponential form (scientific notation) is 7 x 1012.
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2008-10-29 23:54:25
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### Tricks
Who is Charles Dodgson Finger Math
Babbage invented whatMultiplying 2 digit numbers
Other Useful Math Trivia
Tricks
Finger Math
Multiplying by 9
Hold out your hands in front of you so that your thumbs point toward one another.
Visualize that your left pinky finger represents 1, the next finger 2, and so on left to right, until your right pinky finger represents 10. Those fingers represent the number you wish to multiply by 9. To do so, simply put the finger down you wish to multiply by 9. All fingers to the left of the down finger represent the tens digit of the answer while all fingers to the right represent the ones digit.
Example: 6 x 9. Put the finger representing 6 down (the right hand thumb). To the left of the down finger, you have 5 fingers up. That's your tens digit. 5. To the right, you have 4 fingers up. There's your ones digit. 4. Put those together and you have your answer: 54.
Edward Julius in his book Rapid Math Tricks And Tips explains how to add numbers faster.
Here is how to rapidly add numbers.
1+9+2+8+4+5+6+5+3+7=
1+9=10 2+8=10 4+6=10 5+5=10 3+7=10
Think 10, 10, 10, 10, and 10
Here is another example
3+8+7+1+1+4+5+6+2=
3+8=11
7+1+1=9+11=20
4+6=10
20+10=30
5 and 2 are left so add 5+2 and get 7 and add that to the 30
Multiplying 2 digit numbers
Edward Julius in his book Rapid Math Tricks And Tips explains how to multiply numbers faster.
Here is an easy way to multiply 2 digit numbers by 2 digit numbers.
Step 1 Multiply the ones digit together
Step 3 Multiply the tens digit together
Step 1 Choose two numbers
Step 2 Form a Fibonacci sequence for ten numbers
Example, I choose number 5 for my first number and 6 for my second number.
Then I add the numbers to get a Fibonacci sequence. 5+6 gives my 3rd number which is 11; 6+11 gives me my 4th number which is 17. The entire sequence is as follows:
1st - 5
2nd - 6
3rd - 11
4th - 17
5th - 28
6th - 45
7th - 73
8th - 118
9th - 191
10th - 309
The trick is to multiply the 7th number by 11 (you always multiply it by 11). The answer you get when you multiply the 7thnumber by 11 will always be the same result as if you had added all ten numbers. In the above example, the 7th number is 73. Multiply 73 by 11 and the answer is 803. If you add all 10 numbers together the answer will also be 803.
Try this with your own numbers.
Trivia
Charles Babbage was very interesting. He was born in 1791. He developed the blueprint for the modern computer, the cow catcher, and he even discovered how to find out how old a tree is by counting the rings. He said it was possible to determine the climate by examining the rings on a tree by seeing how big they were. Charles Babbage was a mathematician.
Did you know the mathematician Charles L. Dodgson who invented the doublets game was known as Lewis Carroll (author of Alice in Wonderland) ? To play the
doublets game .
Do you know the name of some of the numbers? Well, here they are:
million 106 billion 10 9 trillion 1012 quadrillion 10 15 quintillion 1018 sextillion 1021 septillion 1024 octillion 1027 nonillion 1030 decillion 1033 undecillion 1036 duodecillion 1039 tredecillion 1042 quatuordecillion 1045 quindecillion 1048 sexdecillion 1048 septendecillion 1054 octodecillion 1057 novemdecillion 1060 vigintillion 1063 googol 0100 googolplex 10googol = 1010100
Did you know Pascal invented the first adding machine and it was called Pascaline?
Here are all the different kinds of numbers. Do you know them all?
Other Useful Math Trivia
Here are some other useful ways of remembering some mathmaticial operations.
Perimeter: the word rim will help you remember that perimeter measures the distance around the rim of an object.
The number of teaspoons in a tablespoon: There are 3 teaspoons in 1 tablespoon. Remember that both teaspoon and tablespoon start with the letter "t" and the word three rhymes with t.
The order of operations in long division: Remember the phrase "Did my sister bring rabbits?"
Did Divide My Multiply Sister Subtract Bring Bring Down Rabbits Remainder
A way to remember the order of mathematical operations is to remember the following phrase "Please Exercise My Dear Alligator Sal."
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Chapter: Problem:
Choosing the Washer Method or Shell Method
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Contributions to this software library are always welcome. Please ensure that you post program listings rather than .raw files. They give a reasonable idea of what your program does without having to load them into a DM42 and you can also include comments in your code. Check out the following link for a decoder/encoder: http://www.swissmicros.com/dm42/decoder/
You can then copy/paste the listing and post it in "code" tags.
svarre69
Posts: 5
Joined: Tue Oct 01, 2019 12:09 pm
### Pricing calculations
Hello everyone
I am trying to write a program for calculating Selling Price (PRC) Cost (CST) Margin (MAR) and Markup (MU)
MU is defined as the difference between PRC and CST, divided by CST.
MAR is defined as the difference between PRC and CST, divided by PRC.
That is: Markup is based on Cost and Margin is based on Selling Price.
I have created the menu for the program using MVAR and VARMENU, and I have programmed finding PRC and MU
given CST and MAR.
I would like the program to find the two remaining variables, given two variables.
Eg. if we have PRC and MAR given, the program finds CST and MU.
Below are the formulas listed for finding the remaining two.
I guess I need some kind of test to find out which variables are filled in when I run the program.
Any suggestions on how to do this?
Calculate Selling Price given Cost and Markup:
CST, ENTER, MU, % +.
Calculate Selling Price given Cost and Margin:
CST, ENTER 1 ENTER, MAR, % - /(division).
Calculate Cost given Selling Price and Markup:
PRC, ENTER 1 ENTER, MU % + /.
Calculate Cost given Selling Price and Margin:
PRC ENTER, MAR, % -.
Calculate Markup given Cost and Selling Price:
CST, ENTER, PRC, %CHG.
Calculate Markup given Margin:
MAR, ENTER, ENTER, 1, X exchange Y, % - /.
Calculate Margin given Selling Price and Cost:
PRC, ENTER, CST, %CHG, CHS
Calculate Margin given Markup:
MU, ENTER, ENTER, 1, X exchange Y, % + /.
Below you find the program listing I already have.
The program can also be used for finding out how much an item cost with or without tax. If an item costs \$160 without tax,
set the Cost = 160 and set the Margin = 20 then Markup will show the tax (25%) and PRC = \$200 which is the Price including tax.
And vice versa.
Code: Select all
``````01 LBL "SALES"
02 MVAR "PRC"
03 MVAR "CST"
04 MVAR "MAR"
05 MVAR "MU"
07 STOP
08 EXITALL
09 RCL "CST"
10 1
11 ENTER
12 RCL "MAR"
13 %
14 -
15 /
16 STO "PRC"
17 RCL "MAR"
18 ENTER
19 ENTER
20 1
21 X exchange Y
22 %
23 -
24 /
25 STO "MU"
26 END
``````
Epidiah
Posts: 21
Joined: Mon Jan 29, 2018 2:08 am
### Re: Pricing calculations
A quick and dirty solution to this is to use two smaller programs in the Solver.
This...
Code: Select all
``````00 { 37-Byte Prgm }
01▸LBL "*MU"
02 MVAR "PRC"
03 MVAR "CST"
04 MVAR "MU"
05 RCL "CST"
06 RCL "MU"
07 %
08 +
09 RCL- "PRC"
10 END
``````
And this...
Code: Select all
``````00 { 40-Byte Prgm }
01▸LBL "*MAR"
02 MVAR "PRC"
03 MVAR "CST"
04 MVAR "MAR"
05 RCL "PRC"
06 RCL "MAR"
07 %
08 -
09 RCL- "CST"
10 END
``````
From the Solver menu, choose the program named after the variable you want involved in your calculations (MAR or MU). This is either the variable you're trying to solve for, or the variable you know if you're trying to solve for either PRC or CST.
Once you've solved one of these equations, the values remain in the variables. So you can exit back to the program selection part of the Solver menu and choose a different program to solve for the remaining variable.
Let's say you don't care about PRC or CST, but want to know the MAR given a MU of 25%. Here's how you could do it:
• Shift 7 to enter the Solver Menu
• [*MU] to select the MU program
• 25 [MU] to put 25 in the MU variable
• 1 [PRC] (the number and choice of variable really doesn't matter here, as long as you put a value in one & solve for the other)
• [CST] to solve for CST so that both PRC and CST will be available in the next program
• [*MAR] ...select the MAR program
• [MAR] To choose MAR as the variable you wish to solve for (you'll probably have to select this a couple times because the first time you do the calculator will assume you're entering a guess into the variable)
svarre69
Posts: 5
Joined: Tue Oct 01, 2019 12:09 pm
### Re: Pricing calculations
This is great stuff!!
Using the Solver is very efficient - thank you.
Kind regards
Peter.
rprosperi
Posts: 820
Joined: Mon Apr 24, 2017 5:48 pm
Location: New York
### Re: Pricing calculations
svarre69 wrote:
Mon Nov 04, 2019 11:56 am
Hello everyone
I am trying to write a program for calculating Selling Price (PRC) Cost (CST) Margin (MAR) and Markup (MU)
[snip]
The program can also be used for finding out how much an item cost with or without tax. If an item costs \$160 without tax,
set the Cost = 160 and set the Margin = 20 then Markup will show the tax (25%) and PRC = \$200 which is the Price including tax.
And vice versa.
Do you really mean"tax" here? The program works with margins, markups, price, etc, in the standard business methods, but then you suddenly introduce "tax". Are you suggesting to use Margin to calculate tax? If so, your example says to use 20% Margin to get the effective tax increase of 25%, but a typical user likely would not know what Margin to enter to get the effective tax rate they want. Maybe I'm missing something?
--bob p
DM42: β00071 & 00282, DM41X: β00071, DM10L: 071/100
svarre69
Posts: 5
Joined: Tue Oct 01, 2019 12:09 pm
### Re: Pricing calculations
Hello Bob
In Denmark the tax is 25% - so using the margin as 20% to calculate tax is ok for this purpose.
Not generally of course - you are absolutely right in that point.
Sorry for confusing you.
With kind regards
Peter
rprosperi
Posts: 820
Joined: Mon Apr 24, 2017 5:48 pm
Location: New York
### Re: Pricing calculations
Thanks for clarifying, I thought I was missing something... lol.
Wow, 25% Tax ?! I thought 8.125% where I live in NY was pretty high, holy smokes!
--bob p
DM42: β00071 & 00282, DM41X: β00071, DM10L: 071/100
Walter
Posts: 1466
Joined: Tue May 02, 2017 9:13 am
Location: Close to FRA, Germany
### Re: Pricing calculations
rprosperi wrote:
Wed Nov 06, 2019 2:49 pm
Wow, 25% Tax ?! I thought 8.125% where I live in NY was pretty high, holy smokes!
FYI, German VAT is 19%.
DM42 SN: 00041 Beta
WP 43S running on this device
HP-35, HP-45, ..., HP-50, WP 34S, WP 31S, DM16L
cdmackay
Posts: 142
Joined: Fri Oct 05, 2018 6:33 pm
Location: Cambridge, UK
Contact:
### Re: Pricing calculations
20% in the UK (standard rate)
Cambridge, UK
41CL/DM41X 12/15C/16C DM15/16 71B 17B/BII/bII+ 28S 42S/DM42 48GX 50g 35s 30b/WP34S Prime G2
& Casios, Rockwell 18R
DA74254
Posts: 67
Joined: Tue Oct 03, 2017 9:20 pm
Location: Norway/Latvia
### Re: Pricing calculations
25% in Norway and Sweden.
21% in Latvia
Though, Norway and Denmark is "funny" when it comes to cars. Import tax on a car in Norway is an average of 100%. With a little less, ca 50% on smaller cars and up to 1000 (one thousand)% on cars like Bugatti and such. It's even worse in Denmark with 125 - 200% import tax on cars.
Though Electric cars ar subsidised in Norway; absolutely zero/zilch/none tax or fees on the dildo fuel cars, as the chlorophyll mafia in Norway have the insanely stupid idea that an electric car does not pollute..
Esben
DM42 SN: 00245
keithdalby
Posts: 554
Joined: Mon Apr 24, 2017 6:38 pm
### Re: Pricing calculations
DA74254 wrote:
Thu Nov 07, 2019 7:41 pm
dildo fuel cars.
Well that's a fuel I've not heard before | 2,213 | 7,522 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2020-24 | latest | en | 0.89376 |
https://www.mcqlearn.com/grade6/math/geometrical-concepts-and-properties.php?page=5 | 1,558,548,943,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256887.36/warc/CC-MAIN-20190522163302-20190522185302-00005.warc.gz | 870,210,213 | 6,643 | Geometrical concepts and properties MCQs, geometrical concepts and properties quiz answers 5 to learn elementary education online courses. Complementary angles multiple choice questions (MCQs), geometrical concepts and properties quiz questions and answers for for online elementary education degree. Supplementary angles, types of angles test for elementary school teaching certification.
Learn elementary school math multiple choice questions (MCQs): Supplementary angles, types of angles, with choices obtuse angle, acute angle, right angle, and reflex angle for online elementary education degree. Free math study guide for online learning complementary angles quiz questions to attempt multiple choice questions based test.
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MCQ: Out of following options, two angles that are together classified as supplementary angles are
1. 70° and 50°
2. 55° and 65°
3. 60° and 30°
4. 135° and 45°
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1. acute angle
2. obtuse angle
3. right angle
4. reflex angle
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D | 349 | 1,414 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2019-22 | latest | en | 0.857283 |
http://www.thefullwiki.org/Trojan_points | 1,371,650,390,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368708789647/warc/CC-MAIN-20130516125309-00003-ip-10-60-113-184.ec2.internal.warc.gz | 710,313,583 | 30,932 | # Trojan points: Wikis
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# Encyclopedia
Updated live from Wikipedia, last check: June 19, 2013 13:58 UTC (38 seconds ago)
(Redirected to Lagrangian point article)
### From Wikipedia, the free encyclopedia
A contour plot of the effective potential of a two-body system (the Sun and Earth here) due to gravity and the centrifugal force as viewed from the rotating frame of reference in which Sun and Earth remain stationary. Objects revolving with the same orbital period as the Earth will begin to move according to the contour lines showing equipotential surfaces. The arrows indicate the gradients of the potential around the five Lagrange points — downhill toward (red) or away from (blue) them, but at the points themselves these forces are balanced.
The five Lagrangian points (marked in green) at two objects orbiting each other. (Here a yellow sun and blue earth)
The Lagrangian points (pronounced /ləˈɡrɑːndʒiən/; also Lagrange points, L-points, or libration points), are the five positions in an orbital configuration where a small object affected only by gravity can theoretically be stationary relative to two larger objects (such as a satellite with respect to the Earth and Moon). The Lagrange points mark positions where the combined gravitational pull of the two large masses provides precisely the centripetal force required to rotate with them. They are analogous to geostationary orbits in that they allow an object to be in a "fixed" position in space rather than an orbit in which its relative position changes continuously.
More technically and precisely, Lagrangian points are the stationary solutions of the circular restricted three-body problem. For example, given two massive bodies in circular orbits around their common center of mass, there are five positions in space where a third body, of comparatively negligible mass, could be placed which would then maintain its position relative to the two massive bodies. As seen in a rotating reference frame with the same period as the two co-orbiting bodies, the gravitational fields of two massive bodies combined with the centrifugal force are in balance at the Lagrangian points, allowing the third body to be stationary with respect to the first two bodies.[1]
## History and concepts
The three collinear Lagrange points were first discovered by Leonhard Euler around 1750.[2]
In 1772, the Italian-French mathematician Joseph Louis Lagrange was working on the famous three-body problem when he discovered an interesting quirk in the results. Originally, he had set out to discover a way to easily calculate the gravitational interaction between arbitrary numbers of bodies in a system, because Newtonian mechanics concludes that such a system results in the bodies orbiting chaotically until there is a collision, or a body is thrown out of the system so that equilibrium can be achieved. The logic behind this conclusion is that a system with one body is trivial, as it is merely static relative to itself; a system with two bodies is the relatively simple two-body problem, with the bodies orbiting around their common center of mass. However, once more than two bodies are introduced, the mathematical calculations become very complicated. A situation arises where you would have to calculate every gravitational interaction between every pair of objects at every point along its trajectory.
Lagrange, however, wanted to make this simpler. He did so with a simple hypothesis: The trajectory of an object is determined by finding a path that minimizes the action over time. This is found by subtracting the potential energy from the kinetic energy. With this way of thinking, Lagrange re-formulated the classical Newtonian mechanics to give rise to Lagrangian mechanics. With his new system of calculations, Lagrange’s work led him to hypothesize how a third body of negligible mass would orbit around two larger bodies which were already in a near-circular orbit. In a frame of reference that rotates with the larger bodies, he found five specific fixed points where the third body experiences zero net force as it follows the circular orbit of its host bodies (planets).[3] These points were named “Lagrangian points” in Lagrange's honor. It took over a hundred years before his mathematical theory was observed with the discovery of the Trojan asteroids at the Lagrange points of the Sun–Jupiter system in 1906.
In the more general case of elliptical orbits, there are no longer stationary points in the same sense: it becomes more of a Lagrangian “area”. The Lagrangian points constructed at each point in time, as in the circular case, form stationary elliptical orbits which are similar to the orbits of the massive bodies. This is due to Newton's second law (Force = Mass times Acceleration, or $\mathbf{F}=d\mathbf{p}/dt$), where p = mv (p the momentum, m the mass, and v the velocity) is invariant if force and position are scaled by the same factor. A body at a Lagrangian point orbits with the same period as the two massive bodies in the circular case, implying that it has the same ratio of gravitational force to radial distance as they do. This fact is independent of the circularity of the orbits, and it implies that the elliptical orbits traced by the Lagrangian points are solutions of the equation of motion of the third body.
## The Lagrangian points
A diagram showing the five Lagrangian points in a two-body system with one body far more massive than the other (e.g. the Sun and the Earth). In such a system, L3–L5 will appear to share the secondary's orbit, although in fact they are situated slightly outside it.
The five Lagrangian points are labeled and defined as follows:
### L1
The L1 point lies on the line defined by the two large masses M1 and M2, and between them. It is the most intuitively understood of the Lagrangian points: the one where the gravitational attraction of M2 partially cancels M1 gravitational attraction.
Example: An object which orbits the Sun more closely than the Earth would normally have a shorter orbital period than the Earth, but that ignores the effect of the Earth's own gravitational pull. If the object is directly between the Earth and the Sun, then the effect of the Earth's gravity is to weaken the force pulling the object towards the Sun, and therefore increase the orbital period of the object. The closer to Earth the object is, the greater this effect is. At the L1 point, the orbital period of the object becomes exactly equal to the Earth's orbital period. L1 is about 1.5 million kilometers from the Earth. Gravity from the Sun is 2% (118 µm/s²) more than at the Earth (5.9 mm/s²), while the reduction of required centripetal force is half of this (59 µm/s²). The sum of both effects is balanced by the gravity of the Earth, which is here also 177 µm/s².[citation needed]
The Sun–Earth L1 is ideal for making observations of the Sun. Objects here are never shadowed by the Earth or the Moon. The Solar and Heliospheric Observatory (SOHO) is stationed in a Halo orbit at L1, and the Advanced Composition Explorer (ACE) is in a Lissajous orbit, also at the L1 point. WIND is also at L1.
The Earth–Moon L1 allows easy access to lunar and earth orbits with minimal change in velocity and would be ideal for a half-way manned space station intended to help transport cargo and personnel to the Moon and back.
### L2
A diagram showing the Sun–Earth L2 point, which lies well beyond the Moon's orbit around the Earth.
The L2 point lies on the line defined by the two large masses, beyond the smaller of the two. Here, the gravitational forces of the two large masses balance the centrifugal effect on the smaller mass.
Example: On the side of the Earth away from the Sun, the orbital period of an object would normally be greater than that of the Earth. The extra pull of the Earth's gravity decreases the orbital period of the object, and at the L2 point that orbital period becomes equal to the Earth's.
The Sun–Earth L2 is a good spot for space-based observatories. Because an object around L2 will maintain the same orientation with respect to the Sun and Earth, shielding and calibration are much simpler. It is, however, slightly beyond the reach of Earth's umbra, so solar radiation is not completely blocked. The Wilkinson Microwave Anisotropy Probe, Herschel Space Observatory and Planck space observatory are already in orbit around the Sun–Earth L2. The Gaia probe and James Webb Space Telescope will be placed at the Sun–Earth L2. Earth–Moon L2 would be a good location for a communications satellite covering the Moon's far side.
If the mass of the smaller object (M2) is much smaller than the mass of the larger object (M1) then L1 and L2 are at approximately equal distances r from the smaller object, equal to the radius of the Hill sphere, given by:
$r \approx R \sqrt[3]{\frac{M_2}{3 M_1}}$
where R is the distance between the two bodies.
This distance can be described as being such that the orbital period, corresponding to a circular orbit with this distance as radius around M2 in the absence of M1, is that of M2 around M1, divided by $\sqrt{3}\approx 1.73$.
#### Examples
• Sun and Earth: 1,500,000 km (930,000 mi) from the Earth
• Earth and Moon: 61,500 km (38,200 mi) from the Moon
### L3
The L3 point lies on the line defined by the two large masses, beyond the larger of the two.
Example: L3 in the Sun–Earth system exists on the opposite side of the Sun, a little outside the Earth's orbit but slightly closer to the Sun than the Earth is. (This apparent contradiction is because the Sun is also affected by the Earth's gravity, and so orbits around the two bodies' barycenter, which is however well inside the body of the Sun.) At the L3 point, the combined pull of the Earth and Sun again causes the object to orbit with the same period as the Earth.
The Sun–Earth L3 point was a popular place to put a "Counter-Earth" in pulp science fiction and comic books — though of course, once space based observation was possible via satellites and probes, it was shown to hold no such object. Actually, the Sun–Earth L3 is highly unstable, because the gravitational forces of the other planets outweigh that of the Earth (Venus, for example, comes within 0.3 AU of L3 every 20 months). In addition, because Earth's orbit is elliptical and because the barycenter of the Sun-Jupiter system is unbalanced relative to Earth, such a Counter-Earth would frequently be visible from Earth.
### L4 and L5
Gravitational accelerations at L4
The L4 and L5 points lie at the third corners of the two equilateral triangles in the plane of orbit whose common base is the line between the centers of the two masses, such that the point lies behind (L5) or ahead of (L4) the smaller mass with regard to its orbit around the larger mass.
The reason these points are in balance is that, at L4 and L5, the distances to the two masses are equal. Accordingly, the gravitational forces from the two massive bodies are in the same ratio as the masses of the two bodies, and so the resultant force acts through the barycenter of the system; additionally, the geometry of the triangle ensures that the resultant acceleration is to the distance from the barycenter in the same ratio as for the two massive bodies. The barycenter being both the center of mass and center of rotation of the system, this resultant force is exactly that required to keep a body at the Lagrange point in orbital equilibrium with the rest of the system. (Indeed, the third body need not have negligible mass). The general triangular configuration was discovered by Lagrange in work on the 3-body problem.
L4 and L5 are sometimes called triangular Lagrange points or Trojan points. The name Trojan points comes from the Trojan asteroids at the Sun–Jupiter L4 and L5 points, which themselves are named after characters from Homer's Iliad (the legendary siege of Troy). Asteroids at the L4 point, which leads Jupiter, are referred to as the 'Greek camp', while at the L5 point they are referred to as the 'Trojan camp'. These asteroids are (largely) named after characters from the respective sides of the Trojan War.
#### Examples
• The Sun–Earth L4 and L5 points lie 60° ahead of and 60° behind the Earth as it orbits the Sun. They contain interplanetary dust.
• The Earth–Moon L4 and L5 points lie 60° ahead of and 60° behind the Moon as it orbits the Earth. They may contain interplanetary dust in what is called Kordylewski clouds.
• The Sun–Jupiter L4 and L5 points are occupied by the Trojan asteroids.
• Neptune has Trojan objects at its L4 and L5 points.
• Saturn's moon Tethys has two much smaller satellites at its L4 and L5 points named Telesto and Calypso, respectively.
• Saturn's moon Dione has smaller moons Helene and Polydeuces at its L4 and L5 points, respectively.
• One version of the giant impact hypothesis suggests that an object named Theia formed at the Sun–Earth L4 or L5 points and crashed into the Earth after its orbit destabilized, forming the moon.
## Stability
The first three Lagrangian points are technically stable only in the plane perpendicular to the line between the two bodies. This can be seen most easily by considering the L1 point. A test mass displaced perpendicularly from the central line would feel a force pulling it back towards the equilibrium point. This is because the lateral components of the two masses' gravity would add to produce this force, whereas the components along the axis between them would balance out. However, if an object located at the L1 point drifted closer to one of the masses, the gravitational attraction it felt from that mass would be greater, and it would be pulled closer. (The pattern is very similar to that of tidal forces.)
Although the L1, L2, and L3 points are nominally unstable, it turns out that it is possible to find stable periodic orbits around these points, at least in the restricted three-body problem. These perfectly periodic orbits, referred to as "halo" orbits, do not exist in a full n-body dynamical system such as the solar system. However, quasi-periodic (i.e. bounded but not precisely repeating) orbits following Lissajous curve trajectories do exist in the n-body system. These quasi-periodic Lissajous orbits are what most of Lagrangian point missions to date have used. Although they are not perfectly stable, a relatively modest effort at station keeping can allow a spacecraft to stay in a desired Lissajous orbit for an extended period of time. It also turns out that, at least in the case of Sun–Earth L1 missions, it is actually preferable to place the spacecraft in a large amplitude (100,000–200,000 km/62,000–120,000 mi) Lissajous orbit, instead of having it sit at the Lagrangian point, because this keeps the spacecraft off the direct Sun–Earth line, thereby reducing the impact of solar interference on the Earth–spacecraft communications links. Another interesting and useful property of the collinear Lagrangian points and their associated Lissajous orbits is that they serve as "gateways" to control the chaotic trajectories of the Interplanetary Transport Network.
In contrast to the collinear Lagrangian points, the triangular points (L4 and L5) are stable equilibria (cf. attractor), provided that the ratio of M1/M2 is greater than 24.96.[note 1][4] This is the case for the Sun–Earth, Sun–Jupiter, and, by a smaller margin, the Earth–Moon systems. When a body at these points is perturbed, it moves away from the point, but the Coriolis effect bends the object's path into a stable, kidney bean‐shaped orbit around the point (as seen in the rotating frame of reference). However, in the Earth–Moon case, the problem of stability is greatly complicated by the appreciable solar gravitational influence.[5]
## Intuitive explanation
Lagrangian points can be explained intuitively using the Earth–Moon system.[6]
Lagrangian points L2 through L5 only exist in rotating systems, such as in the monthly orbiting of the Moon about the Earth. At these points, the combined attraction from the two masses is equivalent to what would be exerted by a single mass at the barycenter of the system, sufficient to cause a small body to orbit with the same period.
Imagine a person spinning a stone at the end of a string. The string provides a tension force that continuously accelerates the stone toward the center. To an ant standing on the stone, however, it seems as if there is an opposite force trying to fling her directly away from the center. This apparent force (or "fictitious force") is called the centrifugal force. This same effect is present at the Lagrangian points in the Earth–Moon system, where the analogue of the string is the summed (or net) gravitational attraction of the two masses, and the stone is an asteroid or a spacecraft. The Earth–Moon system and the spacecraft all rotate about this combined center of mass, or barycenter. Because the Earth is much heavier than the Moon, the barycenter is located within the Earth (about 1,700 km/1,100 mi below the surface). Any object gravitationally held by the rotating Earth–Moon system will seem to experience a centrifugal force directed away from the barycenter, in the same way as does the ant on the stone.
Unlike the other Lagrangian points, L1 would exist even in a non-rotating (static or inertial) system. In a rotating system, L1 is a bit farther from the (less massive) Moon and closer to the (more massive) Earth than it would be in a non-rotating system. L1 is slightly unstable (see stability, above) because drifting towards the Moon or Earth increases one gravitational attraction while decreasing the other, causing more drift. The resulting apparent change in centrifugal force is less than the change in gravitational acceleration.
At Lagrangian points L2, L3, L4, and L5, a spacecraft experiences an outward centrifugal force, away from the barycenter, that balances the attraction of gravity toward the barycenter. L2 and L3 are slightly unstable because small changes in position tip the balance more strongly in favor of gravity than the balancing centrifugal force. Stability at L4 and L5 is explained by the Coriolis effect: when gravity draws the object into a tighter orbit, it orbits faster, increasing the centrifugal force in opposition to the gravity toward the barycenter. When the object moves into a wider orbit, gravity overcomes the apparent centrifugal force, drawing the object back. The net result is that the object appears constantly to hover or orbit around the L4 or L5 point.
The easiest way to understand the resulting stability is to say L1, L2, and L3 positions are as stable as a ball balanced on the tip of a wedge would be stable: any disturbance will toss it out of equilibrium. The L4, and L5 positions are stable as a ball at the bottom of a bowl would be stable: small perturbations will move it out of place, but it will drift back toward the center of the bowl.
Note that from the perspective of the smaller-mass object - from the moon, in the preceding example - a spacecraft might appear to orbit in an irregular path about the L4 or L5 point, but from the perspective above the orbital plane, it becomes clear that both the smaller mass and the spacecraft are orbiting the larger mass (or more precisely, all of the objects are in orbit around the barycenter of the system); they simply have overlapping orbital paths. This point of view difference is illustrated clearly by animations in the 3753 Cruithne and Coriolis effect articles.
## Lagrangian point missions
The Lagrangian point orbits have unique characteristics that have made them a good choice for performing some kinds of missions. These missions generally orbit the points rather than occupy them directly.
### Past and present missions
Mission Lagrangian point Agency Status
Advanced Composition Explorer (ACE) Sun–Earth L1 NASA Operational
Solar and Heliospheric Observatory (SOHO) Sun–Earth L1 ESA, NASA Operational
WIND Sun–Earth L1 NASA Operational
Genesis Sun–Earth L1 NASA Mission ended, left L1 point
International Sun/Earth Explorer 3 (ISEE-3) Sun–Earth L1 NASA Original mission ended, left L1 point
Wilkinson Microwave Anisotropy Probe (WMAP) Sun–Earth L2 NASA Operational
Herschel and Planck Space Observatories Sun–Earth L2 ESA Operational [7][8]
### Future and proposed missions
Mission Lagrangian point Agency Status
Deep Space Climate Observatory Sun–Earth L1 NASA On hold
Solar-C Sun–Earth L1 Japan Aerospace Exploration Agency Possible mission after 2010
Gaia Sun-Earth L2 ESA Planned for Spring 2012
James Webb Space Telescope Sun-Earth L2 NASA, ESA, Canadian Space Agency Scheduled to launch in June 2014
"Lunar Far-Side Communication Satellites" Earth–Moon L2 NASA Proposed in 1968[9]
Space colonization and manufacturing Earth–Moon L4 or L5 L5 Society Proposed in 1974
Solar shade Sun–Earth L1 Various proposals[citation needed]
## Natural examples
In the Sun–Jupiter system several thousand asteroids, collectively referred to as Trojan asteroids, are in orbits around the Sun–Jupiter L4 and L5 points. Recent observations suggest that the Sun–Neptune L4 and L5 points, known as the Neptune Trojans, may be very thickly populated, containing large bodies an order of magnitude more numerous than the Jupiter Trojans. Other bodies can be found in the Sun–Mars and Saturn–Saturnian satellite systems. There are no known large bodies in the Sun–Earth system's Trojan points, but clouds of dust surrounding the L4 and L5 points were discovered in the 1950s. Clouds of dust, called Kordylewski clouds, even fainter than the notoriously weak gegenschein, may also be present in the L4 and L5 of the Earth–Moon system.
The Saturnian moon Tethys has two smaller moons in its L4 and L5 points, Telesto and Calypso. The Saturnian moon Dione also has two Lagrangian co-orbitals, Helene at its L4 point and Polydeuces at L5. The moons wander azimuthally about the Lagrangian points, with Polydeuces describing the largest deviations, moving up to 32 degrees away from the Saturn–Dione L5 point. Tethys and Dione are hundreds of times more massive than their "escorts" (see the moons' articles for exact diameter figures; masses are not known in several cases), and Saturn is far more massive still, which makes the overall system stable.
### Other co-orbitals
The Earth's companion object 3753 Cruithne is in a relationship with the Earth which is somewhat Trojan-like, but different from a true Trojan. This asteroid occupies one of two regular solar orbits, one of them slightly smaller and faster than the Earth's orbit, and the other slightly larger and slower. The asteroid periodically alternates between these two orbits due to close encounters with Earth. When the asteroid is in the smaller, faster orbit and approaches the Earth, it gains orbital energy from the Earth and moves up into the larger, slower orbit. It then falls farther and farther behind the Earth, and eventually Earth approaches it from the other direction. Then the asteroid gives up orbital energy to the Earth, and drops back into the smaller orbit, thus beginning the cycle anew. The cycle has no noticeable impact on the length of the year, because Earth's mass is over 20 billion (2×1010) times more than 3753 Cruithne.
Epimetheus and Janus, satellites of Saturn, have a similar relationship, though they are of similar masses and so actually exchange orbits with each other periodically. (Janus is roughly 4 times more massive but still light enough for its orbit to be altered.) Another similar configuration is known as orbital resonance, in which orbiting bodies tend to have periods of a simple integer ratio, due to their interaction.
## In fiction
Lagrange points are often mentioned in science fiction (most often hard science fiction), but are rarely used as a plot device because Lagrange points are unfamiliar outside of the scientific or space-enthusiast community.
### L1
• In Arthur C. Clarke's novel A Fall of Moondust, the station Lagrange II is situated at L1, and contributes vitally to locating Selene.
• In Arthur Clarke and Stephen Baxter's novel Sunstorm, the L1 point plays a crucial role in the building of a shield that has the purpose of saving Earth from a storm of energy from the Sun.
• In the Xbox video game Halo: Combat Evolved (2001) and sequel Halo 2 (2004), "Halo" ring worlds play key locations throughout the games. In Halo: CE and Halo 2, the Halo structures are in L1 Lagrange points between the gas giants Threshold and Substance, and their respective moons.
• In the Hugo Award-winning novel A Deepness in the Sky by Vernor Vinge, a temporary human habitat is built at the L1 point between the planet Arachna and its primary star, a highly variable dwarf called the On/Off Star.
• The planet on which Pitch Black is set receives continual daylight: an orrery depicts it at the apparent L1 point between one star and a pair of tightly orbiting stars, receiving periodic eclipses from a pair of gas giants occupying a wider orbit about L1 in the same plane.
• In Peter F. Hamilton's The Night's Dawn Trilogy, a ZTT jump drive cannot be used in a strong gravitational field. In the first book of the trilogy, The Reality Dysfunction, the main characters cannot escape from a gas giant's gravity well before their pursuers catch up with them. Instead, they race to the Lagrange point between the gas giant and one of its moons in order to activate their drive. Successful execution of this untried and reckless maneuver gains captain Joshua Calvert the nickname "Lagrange" Calvert. In the second book The Neutronium Alchemist, a visit is paid to the supposed home planet of the Kiint, Jobis, which features three moons orbiting the L1 point, rotating around a common center.
• Robert L. Forward's novel Rocheworld concerns a double planet, called Roche and Eau, close enough to share an atmosphere. At certain times water flows through the L1 point from Eau to Roche.
• In the 1984 film 2010: The Year We Make Contact, the opening exposition (excerpts from Heywood Floyd's report on the failed 2001 Jupiter mission) indicate that the Jovian monolith had taken up position in a Lagrange point (most likely L1) between Jupiter and Io.
• In the computer game Final Fantasy 8, the Sorceress Adel is confined to an orbital prison at L1, her alignment with the earth and moon being of plot importance during the "Lunar Cry" phenomenon.
### L2
• In the TV series Quatermass II, the hostile aliens live on a small asteroid "no more than half a mile across" at a "theoretical point of equilibrium" on the dark side of the Earth, although neither L2 nor Lagrange are mentioned by name (the term "Bieber Variation" is used instead).
• In the Star Trek: The Next Generation episode, "The Survivors", the Enterprise is surprised by an enemy ship that had been hiding in a Lagrange point.
• In the manga series Battle Angel Alita: Last Order, the ex-colony ship turned space station Leviathan 1 is at the L2 point in the Earth/Moon system.
• In John Varley's book Wizard, a religious group called the Coven set up a habitat at L2 to make themselves as remote as possible from the earth. L2 later slowly deteriorates into the space version of an unorganized shantytown, since anyone with enough cash can set up a home there. "L2 became known as Sargasso Point to the pilots who carefully avoided it; those who had to travel through it called it the Pinball Machine, and they didn't smile."
• In the anime series Mobile Suit Gundam 00, the supercomputer Veda occupies the Earth–Moon L2 point.
• In Larry Niven's book The Integral Trees, The Ship Mutineers left behind was monitoring them from the L2 Point.
### L3
• In the third season of the TV series Lexx, the planets Fire and Water are found to reside in Earth's L3 point.
• John Norman's Gor series takes place on a counter earth.
• In the anime series Mobile Suit Gundam, the asteroid 3 Juno was moved to Earth's L3 point, where it was renamed Luna II. A significant portion of its mass was used to build the space habitats at the other four Lagrange points.
• In the anime series Mobile Suit Gundam 00, Celestial Being have established a base at the Earth–Moon L3 point. Here, they can make repairs as well as introduce new units.
• In the 1969 British film Journey to the Far Side of the Sun (released as Doppelgänger outside the USA), astronauts from earth find a mirror-image copy of the earth exactly on the opposite side of the Sun.
• In "The Last Enemy", an episode of Space: 1999, the crew of moonbase Alpha encounter two planets "Betha" and "Delta" in a typical "Lagrange L3" orbit.
### L4
• In Larry Niven's novels The Integral Trees and The Smoke Ring, the L4 and L5 points of the Smoke Ring (a ring of breathable air and plant life orbiting a neutron star) have become gathering places for masses of water, living things, and human settlers.
• The eponymous interplanetary relay station in George O. Smith's Venus Equilateral stories was located in the L4 point of the Sun–Venus system.
• In the Xbox 360 game Mass Effect, a group of inhabited space stations are situated at the L4 and L5 points.
• In John C. McLoughlin's novel The Helix and the Sword, L4 is the location of a key space station rumored to be inhabited by ogres and trolls who serve the Apocalyptic Horseman of Pestilence.
• In Ben Bova's novel Colony, Island One is a space colony located at the Earth–Moon L4. This is because the colony's founder, Cyrus Cobb, found that L4 gave a better view of the moon than L5.
### Unspecified Lagrange points
• In most iterations of the Japanese Anime meta-series Gundam, the Lagrange points are often used as points where humans have constructed space colonies. These depictions draw extensively from Gerard O'Neill's book The High Frontier: Human Colonies in Space. The series Mobile Suit Gundam, Gundam Wing and Gundam SEED in particular contain themes of conflict between Earth and Lagrange point colonists.
• In the Starfire strategy game and the series of military science-fiction novels based on it, written by David Weber and Steve White, the warp points which allow starships to transit between solar systems without moving at faster-than-light speeds typically form at the major Lagrange points (ie: L4 and L5 with respect to the star and a gas giant).
• Lagrange points are famously mentioned in Arthur C. Clarke's novel 2010: Odyssey Two and the subsequent science fiction film 2010, where the Discovery spacecraft is located on a Lagrange point. The movie expands on this, claiming that Discovery is located at a point between Io and Jupiter, which would place it in the L1 point of the Jupiter – Io system.
• Lagrange points also play a role in the Larry Niven / Jerry Pournelle classic The Mote in God's Eye.
• In Robert Forward's Rocheworld, the locations for Lagrange points around a binary planet are discussed in contrast to typical system.
• In Iain Banks' The Algebraist, Lagrange points and their alternate forms are central to the plot as possible locations for wormholes.
• In the Independence War computer games, Lagrange points L4 and L5 are used as the only locations for jump-points.
• In the BattleTech game series, a star's Nadir and Zenith are the standard hyperspace jump points for most interstellar spacecraft. Lagrange points (only the L1 point) are sometimes used to enter a system closer to planets, almost always for small-scale military or pirate operations due to the risk of catastrophic misjumps.
• In the Halo novels, the Lagrange points are the only places where a human ship can safely make a slipspace jump. In Halo, the namesake ring-world also occupies a Lagrange point between fictional planet Threshold, and its moon, Basis.
• In the computer game Star Wars: X-Wing, Lagrange points are mentioned in the briefings of some missions that revolve around attacking objects placed at them.
• In the Robert A. Heinlein novel The Number of the Beast, two of the main characters engage in a discussion of adding planets to the solar system at Lagrange points.
• In the television series Stargate Atlantis, there was a defensive satellite located at a Lagrangian point in the solar system in which Atlantis was originally located.[11]
• In 1991, Konami released a Role-playing video game for the NES in Japan called Lagrange Point.
• In the Robotech television series, an effect called an "Orbital Warp Blast" is created when a spaceship creates "a phenomenon known as the molecular vacuum" at a fictional "Lagrange Point 6, approximately 20,000 kilometers from Mars" (where one does not exist in the real world).
• In The Monkeys Thought 'Twas All In Fun by Orson Scott Card, a structure called the Trojan Object appears at L4 or L5.
• In the Star Trek novel The Abode of Life by Lee Correy, Captain Kirk states the planet Mercan has no Lagrangian points.
• In a parody of the science fiction comic Freefall, a character refers to a satellite bar located in "the Lagrange point" so that it's always happy hour there. He then heads for the bar promising to be back before sunset (which never happens).
• Ken MacLeod's novel The Star Fraction mentions a collection of human habitats at the Earth – Sun Lagrange point, with semi-permanent residents. Implicitly, it is the L1 point which is referred to.
• In Vernor Vinge's novel Marooned in Realtime, the Earth-Moon Trojan points are used to store equipment for periods of millions of years.
#### Related concepts
In Larry Niven's novel Ringworld, the Puppeteers' "Fleet of Worlds" is arranged in a configuration (5 planets spaced at the points of a pentagon) called a Klemperer Rosette (misspelled "Kemplerer" in the novel). While it is an extension of the Lagrange concept to a special case N-body problem, and on the surface appears to be an extension of the L4 and L5 orbital mechanics, its stability is more akin to the L1, L2, and L3 positions. Considerable energy expenditure would be required to maintain it over time.
## Notes
1. ^ Actually $\tfrac{25+3\sqrt{69}}{2}$
## References
1. ^ "Lagrange Points" by Enrique Zeleny, Wolfram Demonstrations Project.
2. ^ Koon, W. S.; M. W. Lo, J. E. Marsden, and S. D. Ross (2006). Dynamical Systems, the Three-Body Problem, and Space Mission Design. pp. 9.
3. ^ (French) Lagrange, Joseph-Louis (1867–92). "Tome 6, Chapitre II: Essai sur le problème des trois corps". Oeuvres de Lagrange. Gauthier-Villars. pp. 272–292.
4. ^ The Lagrange PointsPDF, Neil J. Cornish with input from Jeremy Goodman
5. ^
6. ^ Tyson, Neil deGrasse (2007). Death by Black Hole. ISBN 9780393062243.
7. ^ "Herschel Factsheet". European Space Agency. 17 April 2009. Retrieved 2009-05-12.
8. ^ "Esa, latest news". European Space Agency. November 3, 2003. Retrieved July 5, 2009.
9. ^ P. E. Schmid (June 1968). "Lunar Far-Side Communication Satellites" (PDF). NASA. Retrieved 2008-07-16.
10. ^ Lee Whiteside (Updated 11/4/92). "Babylon 5 General Information". Retrieved 2008-07-16.
11. ^ "The Siege", Stargate Atlantis.
## External links
Got something to say? Make a comment. Your name Your email address Message | 8,013 | 35,329 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 17, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2013-20 | longest | en | 0.917513 |
https://www.gradesaver.com/textbooks/math/geometry/CLONE-df935a18-ac27-40be-bc9b-9bee017916c2/appendix-a-a-2-formulas-and-equations-exercises-page-546/34 | 1,701,859,784,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100593.71/warc/CC-MAIN-20231206095331-20231206125331-00125.warc.gz | 886,913,953 | 12,031 | ## Elementary Geometry for College Students (7th Edition)
$b = 15$
$A = \frac{1}{2} \times h \times ( b + B)$ $156 = (0.5) \times (12) \times (b + 11)$ $156 = 6(b+11)$ $156 = 6b + 66$ $90 = 6b$ $b = \frac{90}{6}$ $b = 15$ | 100 | 222 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2023-50 | latest | en | 0.713801 |
https://crypto.stackexchange.com/questions/96357/dont-know-how-to-approach-this-problem-or-where-to-start-finding-an-adversary | 1,643,190,010,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304928.27/warc/CC-MAIN-20220126071320-20220126101320-00086.warc.gz | 243,135,017 | 33,399 | # Don't know how to approach this problem, or where to start. Finding an adversary to a hiding and binding game
I have this problem:
I also have the python version of this problem here:
import json
import sys, os, itertools
sys.path.append(os.path.abspath(os.path.join('..')))
from playcrypt.tools import *
from playcrypt.new_tools import *
from playcrypt.primitives import *
from playcrypt.games.game_bind import GameBIND
from playcrypt.simulator.bind_sim import BINDSim
from playcrypt.games.game_hide import GameHIDE
from playcrypt.simulator.hide_sim import HIDESim
return a+b
def MULT(a,b):
return a*b
def INT_DIV(a,N):
return (a//N, a%N)
def MOD(a,N):
return a%N
def EXT_GCD(a,N):
return egcd(a,N)
def MOD_INV(a,N):
res = modinv(a,N)
if res == None:
raise ValueError("Inverse does not exist.")
return res
def MOD_EXP(a,n,N):
return exp(a,n,N)
"""
Let p be a prime of bit length k >= 8 such that (p - 1)/2 is also prime. Let g,
h be two different generators of the group G = Z_p^*. Let CS= (P, C, V) be the
commitment scheme whose consituent algorithms are as follows, where the message
M is in Z_{p-1}:
"""
def P():
pi = (g, h)
return pi
def C(pi, M):
"""
:param pi: Public parameters
:param M: The message to be commited, element of Z_{p-1}
:return: return the commital and decommital key
"""
(g, h) = pi
K = random_Z_N(p-1)
A = MOD_EXP(g, K, p)
B = MOD_EXP(h, M, p)
C_1 = MOD(A*B, p)
C_2 = MOD(M+K, p-1)
return ((C_1, C_2), K)
def V(pi, C, M, K):
"""
:param pi: Public parameters
:param C: The commital
:param M: The message to be verified
:param K: The decommital key
:return: return 1 if the opening is valid and 0 otherwise
"""
(g, h) = pi
(C_1, C_2) = C
if not 0 <= K < p-1 or not 0 <= M < p-1:
return 0
A = MOD_EXP(g, K, p)
B = MOD_EXP(h, M, p)
C_1_prime = MOD(A*B, p)
C_2_prime = MOD(M+K, p-1)
if (C_1 == C_1_prime) and (C_2 == C_2_prime):
return 1
else:
return 0
"""
1. Specify an O(k^3)-time adversary A1 making one query to its LR oracle and
"""
def A1(lr, pi):
"""
This is the adversary that the problem is
asking for. It should return 0 or 1.
:param lr: The oracle supplied by game HIDE
:param pi: The public parameter pi
"""
pass
"""
(Hint: What is the value of g^{(p-1)/2} mod p, and why?)
"""
def A2(pi):
"""
This is the adversary that the problem is
asking for. It should return tuple (C, M_0, M_1, K_0, K_1).
:param pi: The public parameter pi
"""
return ((0, 0), 0, 0, 0, 0)
if __name__ == '__main__':
# Sample random parameters
k = 12
print('Sampling random parameters of bit length k = %d' % k)
p = random.randint(2**(k - 1), 2**k)
while not is_prime(p) or not is_prime((p-1)//2):
p = random.randint(2**(k - 1), 2**k)
g = random_Z_N_star(p)
while (MOD_EXP(g, (p-1)//2, p) == 1) or (MOD_EXP(g, 2, p) == 1):
g = random_Z_N_star(p)
h = random_Z_N_star(p)
while (h == g) or (MOD_EXP(h, (p-1)//2, p) == 1) or (MOD_EXP(h, 2, p) == 1):
h = random_Z_N_star(p)
print('p = %d, g = %d, h = %d' % (p, g, h))
game_hide = GameHIDE(P, C)
sim_hide = HIDESim(game_hide, A1)
game_bind = GameBIND(P, V)
sim_bind = BINDSim(game_bind, A2)
• Try starting with a numerical example. For instance, let $p=563$, $g=2$, $h=5$. Now for the first part, suppose that we have messages $M_0=123$ and $M_1=345$. We query the LR oracle and get (335,306). Can you tell whether this corresponds to $M_0$ or $M_1$? How? Nov 29 '21 at 9:13 | 1,135 | 3,349 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2022-05 | latest | en | 0.671375 |
http://mathoverflow.net/questions/114847/fontaine-mazur-for-hodge-structures | 1,467,192,492,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783397695.90/warc/CC-MAIN-20160624154957-00158-ip-10-164-35-72.ec2.internal.warc.gz | 208,987,195 | 14,767 | # Fontaine-Mazur for Hodge Structures
Is there a conjecture, or known result, describing which integral Hodge structures are composition factors in the Hodge structure on the cohomology groups of smooth proper algebraic varieties over $\mathbb C$?
Are there pure Hodge structures which fail to have geometric origin for surprising reasons? For instance, such a Hodge structure should certainly be polarizable.
Motivation: Fontaine-Mazur gives the conjectural conditions for a Galois representation to come from an algebraic variety. This, together with the Tate conjecture, would, if proven, tell us a lot about the relationship between motives and their Galois representations. An analogue of Fontaine-Mazur, together with the Hodge conjecture, would tell us a similarly large amount about the relationship between motives and their Hodge structures.
-
Dear Will, Let $V$ be a PHS such that for some $(p,q)$, one has $h^{p,q} =1$ while $h^{p-1,q+1} = 0$. Then Griffiths transversality shows that $V$ cannot fit into a non-trivial variation of Hodge structures. Since there are only countably many families of algebraic varieties, we see that only countably many such $V$ are of geometric origin, so "most" such $V$ are not of geometric origin. (This is why in modularity theorems it is easier to treat the case when $V$ does not have such gaps, and why e.g. Sato--Tate was proved for elliptic cruves before it was proved for higher weight forms.) – Emerton Nov 29 '12 at 11:49
It is also why it is harder to realize the motives attached to higher weight modular forms from alternative geometric constructions (whereas we have no trouble producing elliptic curves --- which are then related to wt. 2 modular forms): the motive attached to a higher weight modular form is rigid, by the preceding Griffiths transversality argument, and so there is no family of motives of which is a member (whereas elliptic curves are easy to write down, since there is a family of them depending on parameters, and we can just choose rational values of the parameters). Regards, – Emerton Nov 29 '12 at 11:52
@Emerton: The Griffiths transversality argument certainly proves that a general Hodge structure as you specify cannot be the full weight p+q Hodge structure of a smooth proper algebraic variety. However, the OP is asking about "factors" of Hodge structures. – Jason Starr Nov 29 '12 at 12:45
Dear Jason, Since such a factor should be motivic (by the Hodge conjecture), I think that the argument I explain extends to such factors: suppose that you have a piece of the cohomology of a variety. It is cut out by some correspondence (assuming the Hodge conjecture). There will be some locus in the moduli space of the initial variety over which this correspondence deforms, and so the motive will deform over that locus. The question then is: how many "special loci" (loci over which a particular correspondence lives) are there in a given moduli space. My sense is that there will be ... – Emerton Nov 29 '12 at 19:39
It is probably more reasonable to ask for a Fontaine-Mazur type conjecture for Hodge structures defined over a number field. By this I mean an integral Hodge structure $H$ plus a vector space defined over a number field, say $\mathbb{Q}$, with a filtration and an isomorphism of this vector space tensored with $\mathbb{C}$ with $H \otimes \mathbb{C}$ so that the filtration goes over to the Hodge filtration. This is what one gets by combining the de Rham cohomology and the Betti cohomology of a variety over $\mathbb{Q}$. Perhaps a condition on the periods a la Grothendieck implies it's motivic. – ulrich Nov 30 '12 at 5:39 | 873 | 3,638 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2016-26 | latest | en | 0.944559 |
http://www.perseus.tufts.edu/hopper/text?doc=Perseus%3Atext%3A2006.05.0734%3Aarticle%3Dpos%3D31 | 1,519,206,326,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813602.12/warc/CC-MAIN-20180221083833-20180221103833-00020.warc.gz | 517,308,756 | 7,200 | #### Relative value of paper and gold.
--Many people have been tussled recently in the fluctuating price of gold to calculate the relative value of paper and gold.
The relations of the two are determined, not by difference, but by ratto. It is plain that where gold is worth 200 cents in paper, the rate is as one to two; two of paper is equal to one of gold, and one of paper equals a half in gold, or fifty cents.--Similarly, if gold reach 400, paper is worth 25 cents; If it reach 1,000, paper will be worth 10 cents; and if the price of gold could ever touch 10,000, a paper dollar would be worth just 1 cent. In that once the value would be thus ascertained; As 10,000 is to 100, as 120 is to 1.
We have hence the rule--one which every schoolboy is familiar with, but which some financial writers seem to be blissfully ignorant of. As the price of gold in paper is to 100 cents. so is 100 cents to the value of one dollar in gold. To make the thing more plain, we append a table giving the value of the paper dollar for different prices of gold, which may serve for future reference: | 270 | 1,091 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2018-09 | latest | en | 0.963317 |
http://mathhelpforum.com/algebra/134644-algebra-fail-simplifying-squared-terms-print.html | 1,526,987,822,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864657.58/warc/CC-MAIN-20180522092655-20180522112655-00329.warc.gz | 188,326,816 | 2,747 | # Algebra Fail! - Simplifying squared terms.
• Mar 19th 2010, 06:04 PM
NeZVa
Algebra Fail! - Simplifying squared terms.
$\displaystyle (x+1/2)^2 - (x+3/2)^2$
How does it simplify... Is there a name for term of this nature? I've forgotten.
• Mar 19th 2010, 06:21 PM
harish21
Quote:
Originally Posted by NeZVa
$\displaystyle (x+1/2)^2 - (x+3/2)^2$
How does it simplify... Is there a name for term of this nature? I've forgotten.
$\displaystyle (x+1/2)^2 - (x+3/2)^2$
= $\displaystyle {(x^2)+2*x*\frac{1}{2}+[\frac{1}{2}]^2} - [(x^2)+2*x*[\frac{3}{2}]+[\frac{3}{2}]^2]$
= $\displaystyle {x^2}+x+\frac{1}{4}-{x^2}-3x-\frac{9}{4}$
= $\displaystyle -2x-\frac{8}{4}$
=$\displaystyle -2x-2$
=$\displaystyle -2(x+1)$
• Mar 19th 2010, 06:29 PM
harish21
Quote:
Originally Posted by harish21
$\displaystyle (x+1/2)^2 - (x+3/2)^2$
= $\displaystyle {(x^2)+2*x*\frac{1}{2}+[\frac{1}{2}]^2} - [(x^2)+2*x*[\frac{3}{2}]+[\frac{3}{2}]^2]$
= $\displaystyle {x^2}+x+\frac{1}{4}-{x^2}-3x-\frac{9}{4}$
= $\displaystyle -2x-\frac{8}{4}$
=$\displaystyle -2x-2$
=$\displaystyle -2(x+1)$
Remember: $\displaystyle (a+b)^2 = {a^2}+2ab+{b^2}$
• Mar 19th 2010, 06:49 PM
NeZVa
Quote:
Originally Posted by harish21
Remember: $\displaystyle (a+b)^2 = {a^2}+2ab+{b^2}$
Thanks, I had such a brain fart. ;) | 558 | 1,286 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2018-22 | latest | en | 0.645313 |
http://www.euclideanspace.com/physics/principles/numerical/deformable/index.htm | 1,680,119,254,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949025.18/warc/CC-MAIN-20230329182643-20230329212643-00496.warc.gz | 61,410,869 | 4,366 | # Physics - Deformable Shapes
Finite Element Method (FEM) and partial differential equation methods
This is for cases where the shape changes, either continuously under the influence of fields say, or at specific times only during events such as collisions.
Some possible ways to create the deformations are:
1. Use a Transform group above the shape to distort it, this is only suitable for very simple distortions, the shape does not really change, it just gets stretched, rotated, etc.
2. Move the vertex coordinates but retain the same vertices and the same triangulation. This allows any deformation, but is quite efficient and therefore might be used for real-time animation, where the shape changes at every frame. Possible methods are:
1. Morphing - This moves the position of each vertex between 2 or more geometry's depending on their corresponding weighting values.
2. Bones - This allows a hierarchy of shapes to be jointed together (see kinematics). Points near the joints move somewhere between where they would be if they were on one shape only. This gives the appearance of the joint bending. However there are limitations to its modeling of human movement, for example, it does not model the wrinkling of lines on the skin, or model the internal structure, such as muscles and tendons.
3. Relaxation Methods - this models the internal forces in an object - see here for details.
3. Mesh Generation - create a totally new shape every time it changes. This has a lot of advantages - the number of vertexes in a shape can vary as needed, or two shapes can even combine. This could be used with FEM. However this requires a lot of number crunching and may not be suitable for real-time calculations. This method can also be used for Meta balls and similar methods for creating organic shapes. | 368 | 1,808 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2023-14 | latest | en | 0.937113 |
https://in.mathworks.com/matlabcentral/profile/authors/3561474-youssef-khmou?s_tid=cody_local_to_profile | 1,582,024,632,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143646.38/warc/CC-MAIN-20200218085715-20200218115715-00479.warc.gz | 429,728,534 | 22,957 | Community Profile
# Youssef Khmou
781 total contributions since 2012
DISCLAIMER: Advices or Opinions posted are from author , and do not reflect that of MathWorks.
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Distance formula using while loop
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Generating wavepackets in MATLAB
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create a new matrix with elements from different sized matrices
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J-V characteristic of solar cell
Current density and voltage Charateristic of solar cell using electron diffusion model
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Need help on designing plot
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Trouble with Envelope Functions
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Matlab - FFT/PSD Problem for Preemphasis
I think the first question is not well interpreted, maybe it consists of generating noisy version of y instead of white noise, i...
5 years ago | 0
Answered
Matlab - FFT/PSD Problem for Preemphasis
When using the reverse fft function, we take the real part, in one of the equations H.*G is not necessary since G=H.^-1, another...
5 years ago | 0
Answered
how to compare histograms
Many approaches exist for this purpose, the easiest way is the root mean square error metric, the minimum value means best match...
5 years ago | 0
Answered
Filter of sin component of sound signal
Concerning the equation of y, can you explain its origin. As for the error, initial condition must be set : y=zeros(size(x))...
5 years ago | 0
Answered
Hi all. I need to plot a Fast Fourier Transform(FFT) of a sinc function
Let us consider non causal function, we can use built in function as follows : Fs=42;Ts=1/Fs; t=-1:Ts:40*Ts-Ts; f=5; y...
5 years ago | 1
Answered
How to derivate a vector
additionally to the above answers, the simplest way to evaluate the polynomial is via anonymous function : f=@(x) x.^2+x-1...
5 years ago | 0
Answered
Preventing all [x,y] figures appearing when running a plot of a circle
Adding semi column (;) for all instructions inside the function will prevent the printing of values.
5 years ago | 0
Answered
How can i set (f) = (-60/(s^2+5s+48.5))?
Element wise operation fixes the error : f=-60./(s.^2+5*s+48.5);
5 years ago | 0
Answered
Trying to make a function for summing up 3 numbers. Explaination below
This function works for a vector of length n, with second logical output according to the description : function [Logical...
5 years ago | 0
Answered
How to I get a threshold value from looking at histogram
Otsu's method exists as built in function with name *graythresh*, here is an example taken from Mathworks help page : I = i...
5 years ago | 1
Answered
Matlab analytical FT and FFT comparison
numerical fft requires a shift if we want to visualize the spectrum with both negative and positive frequencies, scaling problem...
5 years ago | 0
Answered
Trying to make a function for summing up 3 numbers. Explaination below
the following answer is an alternative approach without using 'getdigits' function : function [C,y]=sum3(x) h=int2st...
5 years ago | 0
| accepted
Answered
Driven Damped Pendulum Axes Question
The implementation is acceptable, the first remark is that visualization commands of results should be outside the loop, otherwi...
5 years ago | 0
| accepted
Answered
Is 'newplot' the reason why 'image' and 'plot' are no longer working on my computer?
In previous version of software it works correctly, if you have used the name 'image' for your constructed function, rename it. ...
5 years ago | 0
Answered
Adding white gaussian Noise to 1D signals
To produce a unique sequence of white Gaussian noise, you need to reset the sate as follows : x=sin(2*pi*0.2*(0:0.1:10));...
5 years ago | 0
| accepted
Load more | 1,544 | 6,102 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2020-10 | longest | en | 0.86368 |
http://phys.mrgravell.com/2014/10-6/ | 1,550,823,863,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247514804.76/warc/CC-MAIN-20190222074154-20190222100154-00384.warc.gz | 204,232,962 | 8,693 | # GCSE Physics - 10.6
• ## End of year quiz
• ### Classwork
• We did this quiz.
• ## Ultrasound
• ### Classwork
• We started some questions on ultrasound.
• ## Ultrasound and the doppler effect
• ### Classwork
• We made lots of notes on unltrasound scans of the body, and the use of the doppler effect to measure the speed of blood flow.
• ## Sound - Y8 Recap
• ### Classwork
• We did some recap of Y8 sound concepts.
• We visualised some sounds on an oscilloscope.
• ## Forces and motion, radiioactivity
• ### Classwork
• We finished off the questions I gave out last lesson and went through the answers.
• I gave out a sheet on radioactivity, fission and fusion which most of you finished or nearly finished in the lesson.
• ### Homework
• Finish the questions started in class, then mark and correct using the markscheme I gave out at the end of the lesson.
• Bring your marked work and any questions you have to tomorrow's lesson.
• ## Graphs of Motion
• ### Classwork
• We did this task and went through it:
• A teacher paces backwards and forwards across the classroom twice. They walk at a constant speed of 1m/s. The room is 5m wide. Draw:
• a) a displacement-time graph
• b) a velocity-time graph
• c) a distance-time graph
• d) a speed-time graph
• We started these exam style questions.
• ### Homework
• Finish off the exam style questions and hand in before reg on Weds.
• ## Big Bang
• ### Classwork
• We did these questions on the Doppler effect and the Big Bang.
• We made some notes on what the 'Big Bang' theory states, and the other main piece of evidence for it - Cosmic Microwave Background Radiation.
• ### Homework
• I gave out the list of things you need to know for the end of year examination.
• Go through this list and highlight areas you want to target in the remaining revision lessons we have.
• ## Doppler Effect
• ### Classwork
• We made some notes on the Doppler effect and I demonstrated it using this website and by whirling a guitar around.
• ### Homework
• You need to find out about how the Doppler effect was used to show that the universe is expanding.
• Read and make notes on P120-121.
• Watch this video - it's quite good and should help clarify some of the stuff on those pages.
• ## Test
• ### Classwork
• We did the test.
• No homework.
• ## X-rays, total internal reflection
• ### Classwork
• We made some very brief notes on x-rays and CT scans.
• I demonstrated total internal reflection and we made some notes on this including the 'critical angle'.
• You will do more on total internal reflection next year, including the formula to calculate the critical angle.
• We did some practice questions - I gave out the markscheme at the end of the lesson.
• ### Homework
• Revise for test on Monday.
• ## E-M Spectrum - uses and dangers
• ### Classwork
• We made notes on the uses and dangers of radio waves, microwaves, infrared, visible light and ultraviolet.
• ### Homework
• Add to your notes with the uses and dangers of x-rays and gamma rays.
• Find out which parts of the E-M Spectrum are 'ionizing' and what this means.
• This video and P84-87 will help.
• Bring your notes to next lesson.
• ## Electromagnetic Spectrum
• ### Classwork
• We talked about eclipses for a bit, then went outside.
• I did a demonstration of interference of sound from two speakers.
• You made notes on the electromagnetic spectrum, and asnwered Q1-3 on P79.
• ## Wavefronts and rays
• ### Classwork
• We did lots of examples of representing waves as wavefronts and rays.
• We completed the diagrams started in the video that represented reflection and refraction with wavefronts and rays.
• This is very tricky but you need to be able to do it.
• We also talked about mirages.
• ### Homework
• Optional: watch this video that explains how mirages are formed with some good examples.
• ## Diffraction and interference
• ### Classwork
• We went through the notes and questions you did in the cover lesson on Monday.
• I demonstrated diffraction of light through a narrow gap.
• I demonstrated interference of sound waves from two speakers, and also sound waves from one speaker which reflected off the walls.
• ### Homework
• Watch and make notes this video about rays and wavefronts.
• Bring your notes to next lesson.
• ## Cover Lesson
• ### Classwork
• Then read and make notes on P74-5.
• ### Homework
• Watch this video from 4:44 to 9:20, and add to your notes from the lesson.
• ## Reflection and refraction
• ### Classwork
• We went through the HW and did some more calulations, and looked in more detail at the relationship between the frequency of sound and the pitch.
• We did a series of practicals to investigate refraction in rectangular blocks and triangular prisms.
• We made some notes on reflection and refraction.
• ### Homework
• We have seen diffraction demonstrated with the ripple tank.
• Watch this video and this video and make notes on diffraction using P72-3 of the textbook.
• Bring your notes to next lesson.
• ## Waves calculations
• ### Classwork
• We did lots of wave calculation practice using mini whiteboards.
• We went through the HW - make sure you are happy with the last part of Q2 as many people struggle with this.
• ### Homework
• Watch and make notes on this video about ferquency and period.
• He comes up with the formula f x T = 1
• This formula can be rearranged to give f = 1/T and T = 1/f
• Calculate the:
1. period of a 440Hz sound wave
2. frequency of some water ripples that have a period of 0.5s
3. period of a sound wave in water that travels at 1500m/s and has a wavelength of 2m
• ## Wave properties
• ### Classwork
• We demonstrated transverse and longitudinal waves on a slinky, and made notes about the compressions and rarefactions of a longitudinal wave.
• We did Q2-3 on P67.
• We made notes on wavelength, amplitude, frequency and period - more on this next lesson.
• ### Homework
• Make notes on the wave equation on P68-9.
• Answer Q1-2 on P69 and bring to next lesson.
• ## Ripple tank
• ### Classwork
• We went over transverse and longitudinal waves (which you made notes on during the cover lesson).
• We watched some demonstrations of things waves can do using a rippple tank.
• In particluar, we looked at reflection of waves from flat and curved surfaces, and diffraction of waves moving though a gap.
• ### Homework
• No homework.
• ## Cover lesson
• ### Classwork
• Please go through your test and make any corrections you can using a different colour pen. We will go through the paper together next lesson.
• We are now starting a new topic - Waves. Read P66-7 and make notes on waves.
• ### Homework
• Bring any questions you have to next lesson.
• ## Test
• ### Classwork
• We looked in detail at the energy transfers in a rollercoaster.
• We did the test.
• ### Homework
• No homework.
• ## Power & Gravitational Potential Energy
• ### Classwork
• We did Q2-3 on P43, and went through the answers.
• We started to look at GPE.
• ### Homework
• Watch this video about GPE and this video about kinetic energy.
• Make notes on these using P44-47 to help you - we will go through some examples before the test next lesson.
• If you want to practise using the questions in the book then the answers to P45 and P47 are here.
• ## More Work and Energy
• ### Classwork
• We did some examples of work done calculations.
• We did the questions on this sheet and went through the answers.
• ### Homework
• Watch this video about power and read P42-3 of the textbook.
• Make notes on these and bring to next lesson.
• ## Work Done
• ### Classwork
• We went through the questions you did for HW.
• We started to make some notes on work done.
• ### Homework
• Watch this video and read P40-41.
• Add to your notes from the lesson and bring your notes and any questions to next lesson.
• ## Hooke's Law
• ### Classwork
• We did the Hooke's Law practical and plotted graphs for a spring and an elastic band.
• We added to the notes you made for HW - see here
• ### Homework
• Do Q6-7 on P38 and bring to next lesson - skip the graph part (7b).
• ## Stopping distance
• ### Classwork
• We went through the answers for the homework.
• We also did some recap on the difference between distance-time, speed-time, displacement-time and velocity-time graphs.
• ### Homework
• Watch this video about Hooke's Law and read P36-7 of the textbook.
• Make notes on these and bring your notes to next lesson.
• In the lesson we will do the practical described in the video.
• ## Stopping distance
• ### Classwork
• We started to examine the velocity-time graph for stopping cars - you will complete this analysis for homework.
• ### Homework
• Answer the questions shown here.
• Bring your work to next lesson when we will go through the answers.
• You can also test your reaction time if you like.
• ## Skydiver
• ### Classwork
• We did a quick demo of thermal conductivity of metals as a reminder of Y9 work.
• We went through the HW, and made some notes on skydiving.
• ### Homework
• Read P32-33 and watch this video about stopping a car. Make notes on these, and bring to next lesson. The video is a bit annoying, because it uses units of miles per hour and feet, but otherwise it's ok.
• ## Terminal Velocity
• ### Classwork
• We observed some falling objects using multiple exposures of a camera - see this pic from the lesson.
• We made notes on terminal velocity.
• ### Homework
• Answer Q1 and 3 on P35 and bring to next lesson.
• Also you could optionally watched this short mythbusters clip, though it doesn't exactly include a lot of physics.
• ## Crumple Zones
• ### Classwork
• I gave you some feedback on the homework, and the ideal exam-style answer.
• Remember it is not the change in momentum that matters in this case - it's the rate of change of momentum that determines the force.
• We built and tested crumple zones:
• ### Homework
• No homework.
• ## Forces and Momentum
• ### Classwork
• We made some notes on forces and momentum, and force = change in momentum / time.
• If you missed the lesson watch this video which should help a bit.
• ### Homework
• Read P26-7 of the textbook.
• Do Q1-2 on P27 and hand in tomorrow.
• ## Collisions and Explosions
• ### Classwork
• We demonstrated various collisions where two objects of varying masses bounced off each other, stuck together, or were pushed apart by springs.
• We did this worksheet.
• ### Homework
• Please do this anonymous survey before Monday.
• ## More Momentum
• We estimated the momentum of various animals travelling at top speed.
• We went through the homework together.
• ### Homework
• Watch this video and this video about explosions, and make notes on P24-25.
• Also have a look at this video (you don't have to watch it all - just watch a few of the explosions).
• Bring your notes to next lesson.
• ## Momentum
• ### Classwork
• We went through the rest of the mini-test on fission and fusion.
• We started a new topic - momentum, forces and energy.
• We defined momentum and did some basic calculations about momentum and collisions. If you missed the lesson watch the first six minutes of this video.
• ### Homework
• Watch this video that tries to demonstrate conservation of momentum in space. It's not a great demo, but it is in space!
• Answer Q1-3 on P23 and bring to the lesson on Monday.
• ## Mini test on fission and fusion
• ### Classwork
• We did the test and marked it in the lesson.
• ### Homework
• No homework.
• ## Star life cycle
• ### Classwork
• We made detailed notes on the life cycle of a star.
• ### Homework
• Revise for a short (15min) test on fission and fusion next lesson.
• What you need to know is here and here.
• ## Nuclear Fusion
• ### Classwork
• We made notes on nuclear fusion.
• ### Homework
• Watch this video which shows you just how big some stars can be.
• Watch this video about the life cycle of stars.
• Learn the life cycle of low mass and high mass stars on P231 - there will be a quiz next lesson.
• ## Nuclear Reactors
• ### Classwork
• We went through lots of your examples of chain reactions - thanks for those.
• We made notes on what can happen to neutrons released by nuclear fission in a nuclear reactor, and discussed control rods and the use of water as a moderator.
• We watched a short video which showed the inside of a nuclear reactor.
• ### Homework
• Next week we will finish off nuclear fission and move on to nuclear fusion.
• Watch this video about nuclear fission again now that we have learned a bit more in class.
• In preparation for starting nuclear fusion, please watch this video and read P224-5 of the textbook. We will make notes on this next lesson.
• ## Chain Reactions/Emails
• ### Classwork
• We looked at the way chain reactions can grow and decay in the context of chain emails.
• I forgot to go through the homework from last lesson (which only 8 people handed in for some reason?) - we will go through this next lesson.
• ### Homework
• If you did not hand in your homework from last week (Q1-2 P223) please hand it in tomorrow morning.
• Watch this video which shows different ways of modelling chain reactions.
• Find another example of something like the 'chain reactions' we discussed. We have discussed chain emails, fission chain reactions and I also mentioned disease spread, so you can't use those.
• Explain how the thing increases or decreases based on the number of things passed on per generation. E.g. new contagious diseases (against which we have no immunity) can grow like a chain reaction - if each person infected goes on to infect, say, 2 more people, then initially the number of people infected in each generation will grow exponentially and the disease will spread more and more quickly.
• Write about your example of a 'chain reaction' - maximum 1 side A4 - and bring it to next lesson. Include pictures and graphs if you like.
• ## Nuclear Fission
• ### Classwork
• We went through the test. Please make sure you go through your test and do your corrections carefully for the parts we did not go through in detail - you should be able to do this using your notes and the textbook. If any issues remain, please come and speak to me.
• We started learning about nuclear fission.
• ### Homework
• Read P.222-3 of the textbook and watch this video. Add to your notes from the lesson.
• Answer Q1-2 on P223 and hand in on Monday morning.
• ## Finishing off test
• ### Classwork
• Those who missed the test did it.
• Those who had done the test reserched some interesting stuff about radioactivity - we will share what you found next lesson.
• ### Homework
• No homework.
• ## Recap + Test
• ### Classwork
• We looked at different situations where long and short half-lives are desired, and learned about carbon dating.
• We did the test.
• NOTE!!! If you missed the test today, please be prepared to do it on Monday after half term.
• ### Homework
• No homework.
• ## Even more half-life
• ### Classwork
• We finished the other side of the worksheet from last lesson and went through the answers.
• We put some people in the hot seat.
• ## More Half-life
• ### Classwork
• We went through the graph from the HW.
• We defined activity and discussed the difference between activity and count rate.
• We started the Radioactive Half-Life worksheet - will finish and go through next lesson.
• We started collecting some data to find the half-life of protactinium-234 - we will process the data in Excel next lesson.
• ### Homework
• Revise for test next week.
• ## Half-life
• ### Classwork
• We modelled radioactive decay using dice.
• ### Homework
• Plot a graph of your results (number of rolls on the x-axis, dice remaining on the y-axis).
• Find the half-life of your dice, in terms of number of throws (this can be a decimal - doesn't have to be a whole number of throws).
• Imagine we did the same experiment again, but the dice had spots on TWO of the faces instead of one. Sketch the shape you would expect for this graph on you would expect for this graph on your other graph with a dotted line.
• Do all of this on one sheet of graph paper and hand in on Tuesday morning before registration. Make sure your name is on it.
• ## Half-life
• ### Classwork
• We went through the rest of the past paper questions sheet.
• We started to introduce half-life.
• ### Homework
• Answer Q1-2 and bring to Monday's lesson.
• ## Nuclear equations + past paper questions
• ### Classwork
• We went through these nuclear equations.
• ### Homework
• Answer Q1-4 of these past paper questions and hand in on Tuesday morning before registration.
• ## Uses of Radiation & Nuclear Equations
• ### Classwork
• We introduced nuclear equations for alpha and beta decay, including the emission of gamma rays. | 3,930 | 16,808 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2019-09 | latest | en | 0.907837 |
https://www.tradingonlinemarkets.com/Articles/fundamental_analysis/S_P_500_PERatio_June_2009.html | 1,653,301,405,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662558015.52/warc/CC-MAIN-20220523101705-20220523131705-00476.warc.gz | 1,224,487,719 | 6,237 | S&P 500 PE Ratio June 2009
S&P 500 PE Ratio - June 2009
In February 2009, I asked the question in an article on the same subject, is the current S&P 500 PE ratio trending down or up? If you are like most investors, then you believe the S&P 500 PE ratio is trending down as the stock market plunges. After all, the P, or price part of the ratio was falling.
The answer: the S&P 500 PE ratio was trending up. After the earnings result for the quarter ending March 31, 2021, the PE ratio was even higher. The reason is the denominator E, or earnings in the PE ratio is falling while the price is rising. For these calculations the 919.14 closing price of the S&P 500 on May 29, 2021 is used.
With 99% of all S&P 500 companies reporting, the as reported earnings for the March quarter is \$7.61 according to Standard & Poor’s. As shown on the chart below the S&P 500 PE ratio has risen to 132 for the quarter ending March 2009. According to Standard & Poor’s estimates for the June and September 2009 quarters are 3,500 and -301 respectively. These two values reflect the serious losses incurred in December 2008 combined with the anemic earnings performance since then. I removed these numbers from the chart as they skewed the chart significantly.
It is interesting to note that the forecast for the S&P 500 Index PE ratio rises so much in 2010. This is driven by two factors. First, Standard & Poor’s forecasts \$34.61 earnings for 2010. This is the lowest since the early 1990’s when the S&P traded in the 400 to 600 range. Second, to calculate the PE ratio, Standard & Poor’s uses the most recent closing price for the index, which was 919.14 on May 29, 2021. By holding the P constant and lowering the E, you have to get a higher PE ratio.
As we have seen, PE ratios often rise sharply following a recession. Several quarters after the recession is over, the PE ratio then turns back down, reflecting the rise in earnings, with the price either remaining constant or continuing to fall.
Yale University Professor Robert J. Shiller, author of Irrational Exuberance: Second Edition uses a modified PE ratio that smoothes out the volatility in the ratio. The denominator of this modified ratio is average inflation-adjusted earnings over the trailing 10 years. Shiller calls this modified ratio "p/e10." Using this data the modified ratio “p/e10” produces a PE ratio of slightly over 15, which is very close to the median of 15.7. In December 2007, the beginning of the current recession, the “p/e 10” was 25.95. Since markets tend to cycle above and below the median, we should expect the “p/e 10” to fall further before turning back up.
Is the S&P 500 Going Lower
The U.S. has had three recessions since 1988 according to the National Bureau of Economic Research, the group that determines when the U.S. has had a recession. These recessions are depicted above in red. In the recession of 1990 – 1991, the PE ratio began to climb before the end of the recession. Following the end of the 2001 recession, the S&P 500 fell another 200 points before rebounding. So far in the recession of 12/2007 - ?, the S&P 500 has continued to fall.
Does this mean the market is going lower? Not necessarily. As shown in the chart below of the S&P 500 index, the index continued to retreat after the 2001 recession. On the other hand after the recession of 1990 – 1991, the S&P 500 rose slowly. Besides, we do not know when the current recession will end.
Looking at the earnings forecast from Standard & Poor’s we might be able to assess which way the S&P 500 will likely move. The table below uses the trailing four-quarter earnings from Standard & Poor’s. It then applies a PE ratio to derive the S&P 500 index. When looking at the table, keep in mind that the median PE ratio is 15.7. The PE ratio is mean reverting, so we should expect it to fall further, possibly to 10. In addition the very low S&P Trailing Earnings for June and September 2009 are due to the large loss reported in December 2008 quarter.
S&P 500 based on Estimated Earnings S&P Trailing PE Ratio Quarter Earnings 10 12 15 17 20 25 30 12/31/2010 \$35.67 357 428 535 606 713 892 1,070 09/30/2010 \$34.61 346 415 519 588 692 865 1,038 06/30/2010 \$32.32 323 388 485 549 646 808 970 03/30/2010 \$29.48 295 354 442 501 590 737 884 12/31/2009 \$27.47 275 330 412 467 549 687 824 09/30/2009 -\$3.05 (30) (37) (46) (52) (61) (76) (91) 06/30/2009 \$0.26 3 3 4 4 5 7 8
The low earnings for June and December 2009 are the reason Standard & Poor’s reports the skewed PE ratio of greater than 3,500 for June 2009 and -301 for September 2009.
Using the December 2009 quarter the earnings forecast \$27.47 and a PE ratio of 30 gives us a target price for the S&P 500 index of 824. On Wednesday June 3, 2021, the S&P closed at 931 more than 100 points above a PE of 30.
This examination of earnings and PE ratios is telling us to expect a lower S&P 500 index throughout 2009. How far it will go down depends on several factors. First, are the earnings forecast correct? Investors should monitor earnings expectations throughout the year, looking for any changes either up or down. In February 2009, Standards & Poor’s estimated that the trailing earnings would be \$13.47 vs. the current estimate of \$27.47. In February 2009, the estimate for the trailing four-quarter earnings was \$32.41 vs. \$29.48 now. The estimates for all of 2010 are lower now than they were in February, indicating S&P is not expecting a robust recovery.
Second, evaluate your PE assumptions based on the outlook for the economy and the markets. If earnings are running above the forecast from Standard & Poor’s, then you should expect the PE ratio to rise eventually. On the other hand, if earnings expectations are falling, then you should expect the PE ratio to fall further. In each case, any move in the PE ratio will tend to move more slowly. The median S&P PE ratio is 15.7. Using December 2009 trailing four-quarter earnings of \$27.46 times the median PE ratio of 15.7 gives us an S&P price of 431.
It is difficult to justify an S&P 500 PE ratio of greater than 30 when the economy is still in a recession and the recovery is likely to be at a slower pace than in the past. This will put downward pressure on the S&P 500 for the rest of 2009 and 2010.
As investors, we should assume that the trend for the S&P 500 is still down. It will be important to monitor the performance of the S&P 500 companies earnings announcements during the next several quarters to see how they match up to the forecast.
Our Premium Members receive weekly updates on important and relevant trends for the stock market, industry sectors and individual stocks and ETFs. By following the trends and using fundamental analysis, we have beat the market every year since our inception. You should give our four-week free trial to the Premium Membership a try. There is no risk, nor any obligation. If you have any questions regarding membership, please send an email to service@tradingonlinemarkets.com and we will get right back to you. Your complete satisfaction is of utmost importance to us. | 1,799 | 7,120 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-21 | latest | en | 0.960262 |
https://thedevnews.com/quick-tip-how-to-convert-numbers-to-ordinals-in-javascript/ | 1,709,433,474,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476180.67/warc/CC-MAIN-20240303011622-20240303041622-00522.warc.gz | 543,345,660 | 49,198 | Saturday, March 2, 2024
HomeWeb developmentQuick Tip: How to Convert Numbers to Ordinals in JavaScript
# Quick Tip: How to Convert Numbers to Ordinals in JavaScript
In this tutorial, you’ll learn how to convert numbers to ordinals in JavaScript. Getting the ordinal of a number allows you to display it in a human-readable format.
## What Are Ordinals?
Ordinals define numbers as being part of an order or sequence. The words “first”, “second”, and “third” are all examples of ordinals. When using numbers to display chart results, days of the month, or a ranking, you’ll often need to use ordinals.
Numbers can be used to display many different types of data and results. When numbers are presented to users, they often need be presented in a format that’s more readable — such as adding ordinal suffix (“June 12th” rather than “June 12”, for example).
## Ordinal Suffix Rules in English
Let’s take a look at how ordinals are used in the English language. English ordinals follow a predictable, if not beautifully simple, set of rules:
• “st” is appended to 1 and numbers that are one greater than a multiple of ten, except for 11 and numbers that are 11 greater than a multiple of 100. For example, 1st, 21st, 31st, etc. … but 11th, 111th, etc.
• “nd” is appended to 2 and numbers that are two greater than a multiple of ten, except for 12 and numbers that are 12 greater than a multiple of 100. For example, 2nd, 22nd, 32nd, etc. … but 12th, 112th, etc.
• “rd” is appended to 3 and numbers that are three greater than a multiple of ten, except for 13 and numbers that are 13 greater than a multiple of 100. For example, 3rd, 23rd, 33rd, etc. … but 13th, 113th, etc.
• “th” is appended to everything else. For example, 24th.
## How to Get the Ordinal of a Number
To get the ordinal of a number, you can use the following function:
``````function getOrdinal(n) {
let ord = 'th';
if (n % 10 == 1 && n % 100 != 11)
{
ord = 'st';
}
else if (n % 10 == 2 && n % 100 != 12)
{
ord = 'nd';
}
else if (n % 10 == 3 && n % 100 != 13)
{
ord = 'rd';
}
return ord;
}
``````
The function `getOrdinal` accepts an argument that is a number and returns the ordinal of that number. Since most ordinals end in “th”, the default value of `ord` is set to `th`. Then, you test the number on different conditions and change the ordinal if necessary.
You’ll notice that in each of the conditions the remainder (%) operator is used. This operator returns the leftover value of dividing the left operand by the right operand. For example, `112 % 100` returns `12`.
To test if the number should have the ordinal `st`, you check if `n` is one greater than a multiple of ten (`n % 10 == 1`, which includes 1 itself), but isn’t 11 greater than a multiple of 100 (`n % 100 != 11`, which includes 11 itself).
To test if the number should have the ordinal `nd`, you check if `n` is 2 greater than a multiple of ten (`n % 10 == 2` which includes 2 itself), but isn’t 12 greater than a multiple of 100 (`n % 100 != 12`, which includes 12 itself).
To test if the number should have the ordinal `rd`, you check if `n` is 3 greater than a multiple of ten (`n % 10 == 3`, which includes 3 itself), but isn’t 13 greater than a multiple of 100 (`n % 100 != 13`, which includes 13 itself).
If all of the conditions are false, then the value of `ord` remains `th`.
You can test it in live action with the following CodePen demo.
See the Pen
Get the Ordinal of a Number
by SitePoint (@SitePoint)
on CodePen.
## Conclusion
In this tutorial, you’ve learned how to retrieve the ordinal of a number. Ordinals can be used in variety of cases, such as displaying dates or ranking in human-readable formats.
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# Multiply by whole tens in columns — a shortcut for 2-digit multiplication when one factor is a multiple of ten
I show a shortcut for 2-digit multiplication problems when one of the factors is a multiple of ten, such as 78 × 50. In such cases the answer is the same as 78 × 5 with a zero tagged to the end. We can write that zero first, and then multiply as if we were multiplying 5 × 78. (Of course this same shortcut works also if one of the factors is a multiple of 100 or 1000 etc.)
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