url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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https://physics.stackexchange.com/questions/130904/does-the-mass-distribution-matter-in-schwarzschild-black-holes | 1,656,616,793,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103877410.46/warc/CC-MAIN-20220630183616-20220630213616-00645.warc.gz | 502,976,223 | 66,329 | # Does the mass distribution matter in (Schwarzschild) black holes?
Is it possible that from the same initial mass different black hole radius will be created due to different mass distribution during black hole creation? If mass is concentrated more on the outside bigger event horizon will be created? If mass is concentrated more in the center smaller black hole will be created?
No, it is not possible.
This is the whole essence of the so-called no-hair theorem: that all previous details concerning the object from which the black hole formed has no bearing whatsoever on the properties of the final object, except for mass, electric charge and angular momentum. And also, that these quantities enter only as global values, not with their initial distributions.
Notice that the impact of the theorem is even wider than you seem to imply: not only it does not matter which initial distribution these three quantities had (for instance, you might have some matter corotating while some fraction of matter is counterrotating, but it does not matter so long as you have the same total angular momentum), but also matter might be made of leptons or baryons, charged could be made of different number of positively or negatively charged particles, or of mostly neutral particles plus only one sign of charge.
Also, the no-hair theorem states that this property is not the result of some special assumption or of highly symmetrical initial conditions, but that instead it is the necessary outcome of any history of the matter going into the black hole, regardless of any parameter save for the global values of M, Q, J.
• So is event horizon always created spherical even if it is created during collision of two heading from opposite direction massive stars? Aug 13, 2014 at 22:53
• @PawelWelsberg Yes, because the total angular momentum vanishes. If the total angular momentum is not zero, the event horizon is flattened at the poles. Aug 13, 2014 at 23:02
• And, yet, there is a non-trivial discussion about the black hole information paradox as raised by Bekenstein, I believe. Is it truly lost, as the classical equations predict, or does quantum gravity save the day, and black holes do have microscopic degrees of freedom (fuzzball?), as string theory seems to predict? Aug 13, 2014 at 23:03
• Too many questions in too little space. My reply dealt with GR. As for quantum effects, though many interesting discussions are present, there is of course no definite prediction about any of these topics because of the well-known lack of a suitable theory. Aug 13, 2014 at 23:10
• The distribution of matter will affect how much energy is lost to gravitational radiation during formation of the hole, and in that sense will affect its mass and radius, though this doesn't seem to be the sort of effect the original question had in mind. Aug 14, 2014 at 5:06
Going out on a limb here, if string theory is right, and one can make microscopic black holes in an accelerator, there should be gravitational analogs of Stern-Gerlach (and more general spin polarization) experiments for these microscopic black holes. They could also have measurable excited states leading to decay spectra.
For macroscopic black holes, of course, the thermalization timescale would be unmeasurably short, and they will, just like MariusMatutiae said, not return any useful information about their past beyond these mentioned conserved quantities. That, of course, is no different from nuclei, atoms and molecules. | 738 | 3,497 | {"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-2022-27 | latest | en | 0.93054 |
https://www.techtud.com/chapter/12237/12912/11142 | 1,590,380,069,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347387219.0/warc/CC-MAIN-20200525032636-20200525062636-00095.warc.gz | 944,888,104 | 13,864 | ##### Various Scheduling Examples | FCFS | Shortest-Job-First (SJR) | Round Robin (RR)
FCFS :
Process Burst Time P1 24 P2 3 P3 3
Suppose that the processes arrive in the order: P1, P2, P3
The Gantt Chart for the schedule is:
P1 P2 P3
0 ............................................ 24 .............................. 27........................................................ 30
Waiting time for P1 = 0; P2 = 24; P3 = 27
Average waiting time: (0 + 24 + 27)/3 = 17
Suppose that the processes arrive in the order: P2 , P3 , P1 .
The Gantt chart for the schedule is:
P2 P3 P1
0 ..................................................3.....................................6....................................................30
Waiting time for P1 = 6; P2 = 0; P3 = 3
Average waiting time: (6 + 0 + 3)/3 = 3
Much better than the previous case.
Convoy effect short process behind long process
Shortest-Job-First (SJR) Scheduling:
Associate with each process the length of its next CPU
burst. Use these lengths to schedule the process with the shortest time.
Two schemes:
1. non-preemptive – once CPU has given to the process it cannot
be preempted until completes its CPU burst.
2. preemptive – if a new process arrives with CPU burst length less than remaining time of the currently executing process, preempt. This scheme is known as the Shortest-Remaining-Time-First (SRTF).
SJF is optimal – gives minimum average waiting time for a given set of processes.
Process Arrival Time Burst Time P1 0.0 7 P2 2.0 4 P3 4.0 1 P4 5.0 4
SJF (non-preemptive):
P1 P3 P2 P4
0.................................. 7................................. 8 ........................12.........................................16
Average waiting time = [0 +(8-2)+(7-4) +(12-5)] /4 =4
SJF (preemptive)
P1 P2 P3 P2 P4 P1
0 ................... 2......................4........................5..................... 7..................... 11...................16
Average waiting time = (9 + 1 + 0 +2)/4 =3
Determining Length of Next CPU Burst Can only estimate the length. Can be done by using the length of previous CPU bursts, using exponential averaging.
Round Robin (RR):
Each process gets a small unit of CPU time (time quantum), usually 10-100 milliseconds. After this time has elapsed, the process is preempted and added to the end of the ready queue.
If there are n processes in the ready queue and the time quantum is q, then each process gets 1/n of the CPU time in chunks of at most q time units at once. No process waits for more than (n-1)q time units.
Performance:
1. q large _ FIFO
2. q small _ q must be large with respect to context switch, otherwise overhead is too high.
Example of RR with Time Quantum = 4
Process Burst Time P1 24 P2 3 P3 3
The Gantt chart is:
P1 P2 P3 P1 P1 P1 P1 P1
0.............4...............7.............10............14............18 ............22.............26............ 30
Average waiting time = [(30-24)+4+7]/3 = 17/3 =5.66
Note: If the burst time of the processes are matching then schedule the process which has lowest arrival time and if arrival time matching then schedule the process which has lowest process id | 852 | 3,177 | {"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} | 3.125 | 3 | CC-MAIN-2020-24 | latest | en | 0.623433 |
https://community.wolfram.com/groups/-/m/t/1068326?p_p_auth=22KdqKTO | 1,585,432,296,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370493120.15/warc/CC-MAIN-20200328194743-20200328224743-00263.warc.gz | 418,319,571 | 26,366 | # Fast spherical polygon generation?
Posted 3 years ago
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4 Replies
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Cross posted on mathematica.stackexchange - https://mathematica.stackexchange.com/q/144167/5478
## Question
Given a list of points on a sphere and the sphere/radius I'd like to plot a spherical polygon with vertices in those points.
And this needs to be fast, fast enough for the user to not "feel" generation time.
One should be able to style them too. Most importantly the surface but an edge style would be nice aswell.
## What have I tried?
This is very closely related topic but answers there are not fast enough for my needs.
Is great but "only good for making spherical quadrilaterals, or isosceles spherical triangles".
• ClipPlanes in V11+ can be used as a directive which is very effective:
RandomSeed[3];
pts = Normalize /@ RandomReal[{-1, 1}, {3, 3}]
Graphics3D[
{AbsolutePointSize@12, Point@pts,
Red, Sphere[{0, 0, 0}, .999], Blue,
Style[Sphere[], ClipPlanes -> {
InfinitePlane[{#, #2, {0, 0, 0}}],
InfinitePlane[{{0, 0, 0}, #3, #}],
InfinitePlane[{{0, 0, 0}, #2, #3}]
},
ClipPlanesStyle -> Directive[Opacity@.2, Red]]
}
] & @@ pts
But I'd need to write some code to determine in what order should those points be put in InfinitePlanes in order to clip from the right side (ClipPlane orientation). I didn't do this because I was too lazy and because:
> The number of clipping planes that can be implemented with ClipPlanes is limited by available graphics hardware.
So it won't be general enough. Though if you want to make this method automatic I will gladly upvote it.
## Motivation
I think it will be useful in many applications.
I don't have time for this but I thought it would be a nice feature to have to improve code I was playing with lately, mostly based on another J.M.'s answer - Voronoi grid on a sphere
arc[center_?VectorQ, {start_?VectorQ, end_?VectorQ}] := Module[{ang, co, r}, ang = VectorAngle[start - center, end - center];
co = Cos[ang/2]; r = EuclideanDistance[center, start];
{{start, center + r/co Normalize[(start + end)/2 - center], end}, co}
]
points = {2 \[Pi] #1, ArcCos[2 #2 - 1]} & @@@ RandomReal[1, {10, 2}];
sp = Append[Sin[#2] Through[{Cos, Sin}[#1]], Cos[#2]] & @@@ points;
proc[] := (
ch = ConvexHullMesh[sp];
verts = MeshCoordinates[ch]; polys = First /@ MeshCells[ch, 2];
voro = Normalize[ Cross[verts[[#2]] - verts[[#1]], verts[[#3]] - verts[[#1]]]] & @@@ polys;
edges = arc[{0, 0, 0}, voro[[##]]] & /@ Select[Subsets[Range[Length[polys]], {2}], Length[Intersection @@ polys[[#]]] >= 2 &];
);
proc[];
DynamicModule[{run = True}, Graphics3D[{ {Opacity[.75],
DynamicWrapper[EventHandler[Sphere[],
"MouseMoved" :> Module[{pos = MousePosition["Graphics3DBoxIntercepts", True],
pt}, If[
Not@TrueQ@pos , pt = RegionIntersection[Sphere[], Line@pos];
If[pt =!= EmptyRegion[3], sp[[-1]] = First@Nearest[pt[[1]], pos[[1]]]; proc[]] ]]]
, TrackedSymbols :> {run}
]
}
, {AbsoluteThickness[2],
Dynamic[BSplineCurve[#, SplineDegree -> 2,
SplineKnots -> {0, 0, 0, 1, 1, 1},
SplineWeights -> {1, #2, 1}] & @@@ edges]}
, {Red, Sphere[Most@sp, .02], Dynamic@Sphere[Last@sp, .02]}
}
, PlotRange -> 1.1 , SphericalRegion -> True , ImageSize -> 500]
]
To look more like https://www.jasondavies.com/maps/voronoi/
4 Replies
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Posted 3 years ago
I thought I had a simple solution, but I don't.
Posted 3 years ago
I guess this comes originally from Demonstrations Project Voronoi Diagram on a Sphere by Maxim Rytin. I liked that you modernized it and interactivity works. It is broken in Maxim demonstration and I hope now they can fix it. Did you figure out what exactly the buy there was?But you also might be interested in another demonstration Voronoi Diagrams on Three-Dimensional Surfaces, which generalizes beyond the sphere and is by @Erik Mahieu who is a member here.
I have not seen those demonstrations, thanks. The first one is probably broken as something in tetgenWolfram library has changed, I'd have to investigate. My approach is different, based on top level code provided by J.M. in the linked topic. The difference is it only generates points and edges that is why my example has no faces.About the second one, what it shows is an intersection of 3D Voronoi cells and given surface. I don't think it is the same as a Voronoi mesh on given surface. Maybe on a sphere. Or maybe it is, I'm not sure but I doubt it, would appreciate confirmation.
Hi Sam,The implementation I gave in SE is essentially the same algorithm as Maxim's, where the relationship between the convex hull and the spherical Voronoi diagram was exploited. Maxim's old code used some TetGen functionality, which has now been supplanted by ConvexHullMesh, which I use in the code I wrote and Kuba expanded on.The code in the other demonstration you linked to does not seem to be using a geodesic-based distance, so I'm reluctant to consider it a true Voronoi partition. | 1,373 | 4,948 | {"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2020-16 | longest | en | 0.907557 |
https://www.airmilescalculator.com/distance/pzi-to-aqg/ | 1,709,431,576,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476180.67/warc/CC-MAIN-20240303011622-20240303041622-00217.warc.gz | 611,007,374 | 46,832 | # How far is Anqing from Panzhihua?
The distance between Panzhihua (Panzhihua Bao'anying Airport) and Anqing (Anqing Tianzhushan Airport) is 967 miles / 1557 kilometers / 840 nautical miles.
The driving distance from Panzhihua (PZI) to Anqing (AQG) is 1296 miles / 2085 kilometers, and travel time by car is about 23 hours 33 minutes.
967
Miles
1557
Kilometers
840
Nautical miles
2 h 19 min
149 kg
## Distance from Panzhihua to Anqing
There are several ways to calculate the distance from Panzhihua to Anqing. Here are two standard methods:
Vincenty's formula (applied above)
• 967.217 miles
• 1556.584 kilometers
• 840.488 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth's surface using an ellipsoidal model of the planet.
Haversine formula
• 965.813 miles
• 1554.325 kilometers
• 839.268 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Panzhihua to Anqing?
The estimated flight time from Panzhihua Bao'anying Airport to Anqing Tianzhushan Airport is 2 hours and 19 minutes.
## Flight carbon footprint between Panzhihua Bao'anying Airport (PZI) and Anqing Tianzhushan Airport (AQG)
On average, flying from Panzhihua to Anqing generates about 149 kg of CO2 per passenger, and 149 kilograms equals 328 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Panzhihua to Anqing
See the map of the shortest flight path between Panzhihua Bao'anying Airport (PZI) and Anqing Tianzhushan Airport (AQG).
## Airport information
Origin Panzhihua Bao'anying Airport
City: Panzhihua
Country: China
IATA Code: PZI
ICAO Code: ZUZH
Coordinates: 26°32′24″N, 101°47′54″E
Destination Anqing Tianzhushan Airport
City: Anqing
Country: China
IATA Code: AQG
ICAO Code: ZSAQ
Coordinates: 30°34′55″N, 117°3′0″E | 581 | 2,031 | {"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} | 2.65625 | 3 | CC-MAIN-2024-10 | latest | en | 0.818887 |
https://noplag.com/free-essays/basic-principles-and-modes-of-capillary-electrophoresis/ | 1,726,147,851,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651457.35/warc/CC-MAIN-20240912110742-20240912140742-00005.warc.gz | 394,408,900 | 9,315 | # Essay Example on Basic Principles and Modes of Capillary Electrophoresis
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From Clinical and Forensic Applications of Capillary Electrophoresis 2 Basic Principles and Modes of Capillary Electrophoresis Harry Whatley 1 BASIC PRINCIPLES OF CAPILLARY ELECTROPHORESIS 1 1 Fundamentals of Electrophoresis Capillary electrophoresis CE is a special technique that uses an electrical field in order to separate the components present in a mixture Electrophoresis in a capillary can be differentiated from other types of electrophoresis that it is done within the walls of a narrow tube To understand the functioning of molecules influenced by an electrical field inside a capillary it is important to know the phenomena that result from the geometry of a capillary 1 1 1 Basic Principles It has long been well known that molecules can possess positive or negative electrical charge When the quantity of negative and positive charges are equal the charges will cancel creating an uncharged neutral molecule If allowed to move charged particles will go for regions like an electrode having an opposite charge which means that opposites will attract Figure 1 demonstrates an electrophoresis example In this example mixture of ions is dissolved in such a suitable solvent like water Before applying an electrical field these ions are in a random motion state When the electrical field is applied charged species start moving A separation happens leading to a less random distribution of particles Cations positive ions move toward the cathode electrode of negative charge and anions negative ions move toward the anode electrode of positive charge Figure 1 also shows another point of electrophoresis in solution the effect of the mass charge ratio m z In figure 1 there are four types of ions large and small of positive charge and large and small of negative charge If each of the particles has only one single charge then will be the same absolute value of the force on each of the particles
The acceleration created by this force can be calculated by the equation Force acceleration mass F ma The separation medium viscosity opposes the acceleration with the result that a steady velocity is reached under constant conditions This means that the system is only able to separate particles of opposite charges but is also able to separate particles having the same charge when there are other differences between them The electrophoresis science is concerned with the creation of systems that utilize the differences between the molecules Antother way is that the analyst tries to make a system that creates differences between molecules Changing the pH of the separation method is an example At pH 10 0 acetic acid and glycine will have the same charge 1 At pH 7 0 glycine will have a very small net charge while acetic acid will have a charge of 1 Separation of these two molecules would differ at pH 7 0 and at pH 10 0 Many other factors other than pH affect the separations of electrophoretic These include the hydrodynamic radius of the molecules the temperature and the separation medium viscosity In real systems there are other forces in addition to the electrical field affecting the charged molecules e g the entire fluid mass may be moving relative to the vessel in which it is contained Some of these factors can affect the electrophoresis in a very complex manner for example the movement of current through a liquid can elevate the liquid temperature This temperature change can affect the electrical resistance of the system ie the current the viscosity and the velocity of the molecules moving in the field
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This publication, from the Technology Enhancement Programme (TEP), helps students to understand the manufacturing process through a series of design challenges. Each has a specific brief and the topics include: • Designing and making a helping hand • Robotics: designing and making a walking robot • CAD/CAM:... | 629 | 3,252 | {"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} | 3.28125 | 3 | CC-MAIN-2019-30 | latest | en | 0.899466 |
https://gradeup.co/ssc-chsl-questions-4-july-2019-i | 1,624,158,593,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487655418.58/warc/CC-MAIN-20210620024206-20210620054206-00368.warc.gz | 260,504,180 | 59,501 | # Questions asked in SSC CHSL Exam held on 4th July 2019 (All Shifts)
By Rahul Chadha|Updated : July 4th, 2019
Questions asked in SSC CHSL exam held on 4 July 2019: SSC CHSL 4th July Exam-Day has been conducted and the process has reached its mid-way. A total of 3 shifts were conducted and as usual, Gradeup has come up with all the actual questions of English Language, Quantitative Aptitude, English Language & Reasoning section.
Check also:
## Questions asked in SSC CHSL Exam held on 4th July 2019 (Shift I+II+III)
• Who is the chairman of India's Public Service Broadcaster? Dr A. Surya Prakash
• Who has summited Mount Everest more times than any other person? Apa Sherpa
• Which among the following is a waterborne disease? Typhoid
• Which is the southernmost river of Kerala? River Neyyar
• Shivaji killed _____ in Pratapgarh battle? Afzal Khan.
• President gives his resignation letter to whom? Vice President
• Who invented the Binary system? Gottfried Leibniz
• In which category, laptops and smartphones are included? Micro Computers
• What is the SI unit of Pressure? Pascal
• (7,6,169) - similar options
• If A × B means A is the son of B, A + B means A is the father of B, A ÷ B means A is the daughter of B and A – B means A is the wife of B. Which of the following represents P is the brother of Q?
• FISH - JGWQ, STAR - ?
• Logical venn diagram - Food, Masala, Cucumber
• 8: 72 : : 7 : ?
• The side of the rhombus is 10 cm and smaller diagonal is 12 cm. Find the area.
• A value is increased by 30% then decreased by 35 % and then finally increased by 25%. Find the overall change in percentage.
• X-1/x=10, Find x3-1/x3
• P,Q, and R do a work in 10, 20,60 resp.In how many days will they together complete the work?
• A train travels for 360 km. If its speed had been 10km/hr more it would take 3 hr less find the accurate speed.
• Idiom- Green thumb
• One word substitution - Inaudible
• A, B and C do a piece of work in 4, 28, 56 days. In how many days will they do the work together.
• A lady gets 20% discount on an item and 30% more discount at the time of billing. What is the amount paid by the lady?
• A train travels for 60 km. If its speed had been 8km/hr more it would take 10hr less find the accurate speed.
• x4+1/x4=1442, then find x+1/x
• 2 sin2x- 5cosx-4=0 tanx+sinx=3 root 3/2 Find x?
• Largest Mica producing State?
• Niel Armstrong best design award winner
• Marmagao Port is in which state
• Kamayani is written by Jaishankar Prasad'
• One question related to the number of singles titles won by Roger Federer
• A man sells two articles at same SP, one at 20% loss and other at 20% gain. Find his total gain/loss Percentage?
• Which state won the most medals in khelo India youth game?
• Alzheimer's disease leads to? Dementia
• The naval exercise named SIMBEX 2019 was between Singapore and__? India
• In which year Einstein was awarded Nobel prize?
• Which among the following countries has been announced malaria-free by the UN?
• Where is Vivekananda airport located?
• One question related to Article 123
• Minimum age to become MLA ?
• Where is Kolleru Lake located?
• How many states touch border with Uttar Pradesh?
• Bugyal tribes belong to which state?
• One question related to Sanjhi art.
• When is World dance day celebrated?
• Tennis Player Nadal belongs to which country?
• Name the Longest bone present in Human Body?
• Who won Pulitzer Prize 2019 in Fiction category?
• Antonym of URBAN
• Meaning of Head in clouds?
• Synonym of Candid?
• One word : Inaudible
• Cure for all ill: Panacea
• Waiter:Serve:: Mechanic ? Repair
• Logical Arrangement of words: World, Asia, Gateway of India, Mumbai, Maharashtra
• Venn Diagram: Manipur, Sikkim, Imphal
• 7,11,19,31,47__?
• How many Ranjhi trophies have been won by Vidharbha so far?
• Where is Mount Abu located?
• One question related to the Tropic of Cancer line
• Idiom: Grease the palm
• Synonym of Opportune
• Who is the current President of Africa?
• One word substitution: Bilingual
• Idiom: Feather its own nest
• Synonym of Avenge
• Antonym of Hasty
• Divisibility Rule of 72
• The parallel sides of an isosceles trapezium are 10 m & 20 m. If its area is 180 m sq. Find the length of unparallel side?
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Ashok KumarJul 5, 2019
Questions sequential aaraha he ya jumbled....????
Pooja SharmaJul 6, 2019
Answer please - If A × B means A is the son of B, A + B means A is the father of B, A ÷ B means A is the daughter of B and A – B means A is the wife of B. Which of the following represents P is the brother of Q?
Bhupendra KumarJul 7, 2019
Please send exam review, good attempt of 4 july shift 3
Rahul SrivastavJul 7, 2019
Khelo india 2018 tha ya 2019 chal 4th july 3rd shift
RoshanJul 9, 2019
Yaar kisi ko pta h.. 3rd shift me idioms qa tha???
Roma SharmaJul 9, 2019
8july question
Hindi me nahi h
Pavan KumarJul 13, 2019
Hindi mai send karo
DEEPAK KUMARSep 13, 2019
Kis kis ka selection hua h
### Related Posts
Rules of Question Tag in English GrammarYesterday, 4 pm|2.2 K+ upvotes
GradeStack Learning Pvt. Ltd.Windsor IT Park, Tower - A, 2nd Floor, Sector 125, Noida, Uttar Pradesh 201303 support@gradeup.co | 1,634 | 5,711 | {"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-2021-25 | latest | en | 0.958138 |
https://brainmass.com/math/calculus-and-analysis/differential-functions-values-possible-solution-17840 | 1,653,679,596,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662675072.99/warc/CC-MAIN-20220527174336-20220527204336-00432.warc.gz | 187,753,959 | 75,542 | Explore BrainMass
# Differential Functions : Values for Possible Solution
Not what you're looking for? Search our solutions OR ask your own Custom question.
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
Determine for which values of m the function y(x)=x^m is a solution to the differential equation.
3x^2*y'' + 11x*y' - 3y = 0
https://brainmass.com/math/calculus-and-analysis/differential-functions-values-possible-solution-17840
#### Solution Preview
Please see the attached file for the full solution.
Thanks for using BrainMass.
Determine for which values of m the function ...
#### Solution Summary
The values of a variable are determined for which a function is the solution to a differential equation. All work is shown and is well presented.
\$2.49 | 186 | 831 | {"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} | 2.75 | 3 | CC-MAIN-2022-21 | longest | en | 0.860391 |
https://foodandfury.com/qa/quick-answer-is-the-relationship-between-magnetic-field-strength-and-distance-linear.html | 1,600,890,660,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400212039.16/warc/CC-MAIN-20200923175652-20200923205652-00266.warc.gz | 367,301,743 | 8,064 | # Quick Answer: Is The Relationship Between Magnetic Field Strength And Distance Linear?
## What is the relationship between distance and strength in both electric and magnetic fields?
The strength of an electric field as created by source charge Q is inversely related to square of the distance from the source.
This is known as an inverse square law.
Electric field strength is location dependent, and its magnitude decreases as the distance from a location to the source increases..
## Do Magnetic Fields add linearly?
Magnetic fields add up as vectors. At any given point, if the fields are in the SAME direction the sum will be stronger and in the same direction. If the fields are in opposite directions the sum will be weaker and will be in the direction of the stronger field.
## Does magnetic force increase with distance?
But the key point is that the force is inversely proportional to the distance squared (i.e. it obeys an inverse square law with distance). … If the distance between two magnets is increased by five times the magnetic force between them will fall to one twenty fifth of the initial value.
## What three forces can act from a distance?
Common forces at a distance The most common forces you experience that act at a distance are magnetism, electrical charges, and gravity.
## What are examples of pushes and pulls?
Push and Pull ExamplesPushing the trolley.Pushing of the car when it breaks down.Pushing the table from one place to another.
## How do magnetic fields combine?
Near the poles they are closer together than those further away, because the magnet is strongest near the poles. When magnets are close together, with their fields overlapping, the fields combine to produce a resultant field which acts in one direction at any given point.
## What is the relationship between magnetic field and distance?
The magnetic field from a wire decreases with distance from the wire. Instead of the field being proportional to the inverse square of the distance, as is the electric field from a point charge, the magnetic field is inversely proportional to the distance from the wire.
## Why is magnetic field strength inversely proportional to distance?
The density of flux lines is inversely proportional to the square of the distance from the source because the surface area of a sphere increases with the square of the radius. Thus the field intensity is inversely proportional to the square of the distance from the source.
## Why does magnetic force act from a distance?
The force exerted by a magnet is called The push or pull of magnetism can act at a distance, which means that the magnet does not have to touch an object to exert a force on it. … Magnets can also repel, or push away, objects.
## Where is a magnet’s field strongest?
north poleThe magnetic field of a bar magnet is strongest at either pole of the magnet. It is equally strong at the north pole when compared with the south pole. The force is weaker in the middle of the magnet and halfway between the pole and the center.
## Can we see magnetic flux lines?
At each end of a magnet is a pole. These lines of flux (called a vector field) can not be seen by the naked eye, but they can be seen visually by using iron fillings sprinkled onto a sheet of paper or by using a small compass to trace them out. | 664 | 3,332 | {"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} | 3.84375 | 4 | CC-MAIN-2020-40 | latest | en | 0.94194 |
https://codereview.stackexchange.com/questions/243333/count-word-order-with-further-optimization | 1,713,960,611,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819273.90/warc/CC-MAIN-20240424112049-20240424142049-00686.warc.gz | 158,634,110 | 39,267 | # Count "Word Order" with further optimization
Question:
You are given n words. Some words may repeat. For each word, print its number of occurrences. The output order should correspond with the input order of appearance of the word.
Sample Input:
4
bcdef
abcdefg
bcde
bcdef
Sample Output
3
2 1 1
Here's what I came up with:
n = int(input())
array = []
elements = {}
for index in range(n):
value = input()
if value not in array:
array.append(value)
elements[value] = 1
else:
elements[value] += 1
print(len(elements))
print(*(i for i in elements.values()), end=' ')
I stress tested it on Try it Online with the random string generator and found the runtime to be around 1.98s. But I'm getting TLE on the coding platform. How do I improve the speed (bit offtopic - is there any other approach)?
# TLE
You're probably getting a TLE because,
if value not in array:
Is a O(N) operation meaning it has to traverse the entire array (worst case) to check if the value exists in the array
I can understand why you felt the need to have an extra array, because of dictionaries are not ordered.
But you can make use of the collections.OrderedDict module, to have an ordered dictionary!
# Other
1. join!
print(*(i for i in elements.values()), end=' ')
This can be done cleaner with joining the values instead of unpacking
print(" ".join(map(str, e.values())))
2. Counter
Python is often described as batteries included,
Your element dictionary is the same the collections.Counter
# Code
We can make an OrderedCounter class to combine these two and make get the most out of the included modules.
from collections import Counter, OrderedDict
class OrderedCounter(Counter, OrderedDict):
pass
c = OrderedCounter(input() for _ in range(int(input())))
print(len(c))
print(" ".join(map(str, c.values()))) | 428 | 1,818 | {"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": 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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-18 | latest | en | 0.823918 |
http://legacy.roadtoreality.info/archive/viewtopic.php%3Ff=19&t=1771.html | 1,558,533,257,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256812.65/warc/CC-MAIN-20190522123236-20190522145236-00302.warc.gz | 124,724,576 | 4,047 | Page 1 of 1 [ 2 posts ]
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Exercise [16.05]
Author Message
Joined: 12 Jul 2010, 07:44
Posts: 154
Exercise [16.05]
Solution to Exercise [16.05]:
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26 Mar 2011, 20:53
Joined: 12 Jul 2010, 07:44
Posts: 154
Re: Exercise [16.05]
By the way, strictly speaking the marks around the outside of the magic disk represent elements of the projective plane of F2, not elements of F2 x F2 x F2.
However, since the only elements of F2 are 0 and 1, there is only one representation for a given ratio (or set of ratios) of values; and so the projective n-space Pn(F2) is point-for-point identical to the vector space (F2)^(n+1) with the origin excluded (for any n>0).
This is not true for any other finite group Fp, p>2.
26 Mar 2011, 21:12
Page 1 of 1 [ 2 posts ] | 273 | 856 | {"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-2019-22 | longest | en | 0.866862 |
https://www.tina.com/resources/home/ideal-operational-amplifiers-2/4-input-resistance-of-op-amp-circuits/ | 1,716,299,535,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058484.54/warc/CC-MAIN-20240521122022-20240521152022-00554.warc.gz | 908,014,166 | 14,216 | # 4. Input resistance of Op-amp circuits
Input resistance of Op-amp circuits
The input resistance of the ideal op-amp is infinite. However, the input resistance to a circuit composed of an ideal op-amp connected to external components is not infinite. It depends on the form of the external circuit.
We first consider the inverting op-amp. The equivalent circuit for the inverting op-amp of Figure (3) “The inverting op-amp” is shown in Figure 10(a).
Figure 10- Input resistance, inverting amplifier
Figure 10 (b) shows the same circuit rearranged for simplicity of analysis. Note that we have attached a “test” voltage source to the input in order to calculate the equivalent resistance. Since the circuit contains a dependent voltage source, we can’t find the input resistance by simply combining resistors. Instead, we find the input resistance by replacing the input signal source and its associated resistance with a test source of specified voltage, vtest, and then calculate the current delivered by the test source to the circuit, itest. Alternatively, we could use a current test source, itest, and solve for the voltage delivered to the circuit, vtest. Using either technique, we can calculate the resistance from Ohm’s law.
The loop equation is given by,
(26)
The equivalent input resistance is then
(27)
As the loop gain, G, approaches infinity, the first term in Equation (27) approaches zero and the input resistance approaches Ra. Thus, the input resistance seen by the source is equal to the value of the external resistance, Ra. This verifies the virtual ground property since the result indicates that the inverting input is equivalent to a ground.
###### We now consider the inverting amplifier with two inputs.
This is shown in Figure (11).
Figure 11- Two-input inverting amplifier
It is a special case of the circuit of Figure (4) “Op-amp circuit” shown previously.
Since the voltage at the inverting input to the op-amp is zero (virtual ground), the input resistance seen by va is Ra, and that seen by vb is Rb. The “grounded” inverting input also serves to isolate the two inputs from each other. That is, a variation in va does not affect the input vb, and vice versa.
The input resistance for the non-inverting amplifier can be determined by referring to the circuit configuration of Figure (5) “Non-inverting amplifier” . See the equivalent circuit in Figure 12(a).
No current passes through R1 since the v+ input to the op-amp has infinite resistance. As a result, Rin to a non-inverting terminal is infinity. If a design needs a large input resistance, we often use a single-input non-inverting op-amp. Such a configuration is called a non-inverting buffer if it has a voltage gain of unity.
Therefore the situation changes when we go to a multiple input non-inverting op-amp, as shown in Figure 12(b). The equivalent circuit is shown in Figure 12(c). We are assuming that the resistance associated with each source, (r1, r2 and r3) is zero ohms. When applying the test source to calculate the input resistance for multiple-input circuits, we use superposition. We therefore apply the test source at each input separately while disabling the other inputs (short circuits for voltage sources and open circuits for current sources in accordance with the principle of Superposition). The various input resistances are then
(28)
APPLICATIONS
Analyze the following circuits online using the TINACloud circuit simulator by clicking the links below.
1- Input Resistance of an Inverting Amplifier Circuit Simulation
2- Input Resistance of a two-input Inverting Amplifier Circuit Simulation
This concept can easily be extended to n inputs.
Figure 12- Input resistance of a non-inverting amplifier | 798 | 3,740 | {"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} | 3.859375 | 4 | CC-MAIN-2024-22 | latest | en | 0.866624 |
https://tech.paayi.com/use-microsoft-excel-quotient-function | 1,701,662,687,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100523.4/warc/CC-MAIN-20231204020432-20231204050432-00355.warc.gz | 641,086,962 | 18,257 | # Learn How to Use Microsoft Excel QUOTIENT Function
Written by | 0 Comments | 593 Views
In this article, you will learn how to use the Microsoft Excel QUOTIENT function and its prime function in Microsoft Excel. You will also get to know the Microsoft Excel QUOTIENT function return value and syntax with the help of some examples.
Microsoft Excel QUOTIENT Function
The main function of the Microsoft Excel QUOTIENT function is to get the quotient without remainder. That implies, with the help of the QUOTIENT function you can able to return the integer division without remainder as a result. So, with the help of the QUOTIENT function, you can able to get the quotient without any remainder.
Return Value of QUOTIENT Function
The return value will be the quotient without remainder.
Syntax of QUOTIENT Function
=QUOTIENT(numerator, denominator)
Where the arguments:
• numerator: This is the number or quantity to be divided.
• denominator: This is the number to be divided with or divisor.
## How to Use Microsoft Excel QUOTIENT Function?
So we know that Microsoft Excel QUOTIENT function you can able to get the quotient without remainder. That implies, with the help of the QUOTIENT function you can able to return the integer division without remainder as a result. So, with the help of the QUOTIENT function, you can able to get the quotient without any remainder.
| | 294 | 1,387 | {"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} | 3.390625 | 3 | CC-MAIN-2023-50 | latest | en | 0.817121 |
http://www.meetup.com/GothPy/events/100045382/ | 1,440,812,581,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644064160.12/warc/CC-MAIN-20150827025424-00087-ip-10-171-96-226.ec2.internal.warc.gz | 571,342,437 | 23,392 | # demo/talk of Raspberry Pi, followed by coding on Project Euler problems
• January 31, 2013 · 6:00 PM
Part 1: Raspberry Pi with Anders Arnholm - 30 minutes, talk & demo
In December I did a small python hack on a Raspberry pi, controlling some XMas lights at home.
Part 2: coding on Project Euler problems, led by Andrew Dalke
Project Euler is a series of challenging mathematical/computer programming
problems that will require more than just mathematical insights to solve.
the use of a computer and programming skills will be required to solve most problems.
These range from easy problems to quite difficult ones. An easy example is:
If we list all the natural numbers below 10 that are multiples of 3 or 5,
we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.
You can see that it has the problem statement, and a simple test case. When you
think you have the right answer, you enter the answer in a box and it says "yes"
or "no."
I was thinking for the meeting that we would select a few problems, spend
30 minutes on a set, then review each other's approaches, including how
we thought of the problem, built test cases, etc.
(Some of the problems can be solved without a computer. In the above,
the simple solution is a sum of a generator expression. The clever solution
sees this as a variation of the FizzBuzz problem, and rewrite the sum as
some of three simple products.)
• ##### Fredrik W.
Lots of code displayed and discussed
1 · February 3, 2013
• ##### Andreas E.
The Raspberry Pi presentation was interesting, but I wanted more low level, technical detail. The eulerproject-coding was more fun than expected, the coding was good and the presentations were nice!
2 · January 31, 2013
• ##### Andreas E.
Almost touching silicon would make me happy, but I want to learn more about embedded programming. :)
February 1, 2013
• ##### Andrew D.
Next time at GothPy - electron microscopy of an Core i7!
1 · February 1, 2013
• ##### André L.
What was the name of the testing framework that some people where using. The non-PyUnit one :)
February 1, 2013
• ##### A former member
Couldn't join today, was really looking forward to the Pi talk :(
January 31, 2013
• ##### Andreas E.
Here's the code! :D
https://github.com/bal...
1 · January 31, 2013
• ##### Anders A.
The talk was much free from and no slides, kinda improvised, i was planning to make a google hangout and drop that on you tube. Did i remeber the mic in my bag and start that. Noo Noo Noo, but i make the talk next week for Mecel as well. i hope i remeber to record that one.
2 · February 1, 2013
• ##### Per S.
Fun to start coding in python again, even if the failed me over the syntax several times.
/Per
2 · January 31, 2013
• ##### Per S.
Det var bra, roligt och lärorikt. Dock var det dålig luft.
Alexander
1 · January 31, 2013
• ##### Tobias G.
Thanks for today!
For all of you who are interested in this asynchronous test-framework - here ist the link: http://heynemann.github.com/pyvows/
Note that you need libevent to build it (see installation section on the page).
2 · January 31, 2013
• ##### Samuel Y.
Greate. More code and show.
2 · January 31, 2013
• ##### Samuel Y.
Greate. More code and show.
1 · January 31, 2013
• ##### Tobias G.
Hi, looking forward to the meetup next week!
I wrote a little Python library last week, which aims at making it more easy to spread processing of a pipeline of computations over multiple cores. I don't have much experience with parallel processing yet, but I though it is a shame to let all those extra cores be unused all the time, so if you think this could be useful for you and feel like trying it out and giving me some feedback - maybe over food at the meetup - I would be very happy. You can get the code from here: https://github.com/tobigue/Orkan
Best, Tobias
January 24, 2013
• ##### André L.
Looks cool! Nice to see ProcessPoolExecutor in action. I skimmed through your code (I'll try it later!) and added a useless comment :) Nice work.
January 24, 2013
• ##### Tobias G.
Thanks! :) Not so useless at all, I should indeed refactor the logging! I answered right there, too. ;)
January 24, 2013
### Göteborg, Sweden
Founded Aug 6, 2009
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• ##### Software Skills
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### Create your own Meetup Group
Allison
Meetup has allowed me to meet people I wouldn't have met naturally - they're totally different than me. | 1,319 | 5,291 | {"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} | 2.96875 | 3 | CC-MAIN-2015-35 | latest | en | 0.931161 |
https://math-fail.com/category/math-fail/dumb-math-fail/page/7 | 1,632,320,462,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057366.40/warc/CC-MAIN-20210922132653-20210922162653-00214.warc.gz | 415,784,644 | 12,622 | # Dumb Math Fail
## Math Sign Fail
What does stumulus mean? Anyone? Cricket… There are: 365.242199 days per year And multiplied by 2206 years gives: 365.242199 x 2206 = 805 724.291 Multiplied by \$1,000,000 gives: \$805, 724, 291, 000. Well, the guy is right. I’m definitely stumuled.
## Math counts – spelling doesn’t
MATHCOUNTS in North Carolina This presentation will present a status report on MATHCUNTS in North Carloina…
## Are the reals countable?
John Gabriel (a non-mathematician) claims in his blog that the real numbers are countable. The author tries to enumerate the real numbers in the interval [0, 1) by writing out all those whose decimal representation has one digit after the dot (0.1, 0.2, …, 0.9), followed by those with two digits after the dot, then …
## Fox news – pi chart
Hmm… something doesn’t add up :-]
## First Annual Special Mathlympics to be hosted by Canada in 2012
What is it?Special Mathlympics is an annual event with the main goal of stimulating interest and igniting passion in mathematics! Who can participate?The competition is open to people of all ages who are special (in the sense that they are mathematically challenged). When/where is it?On March 14th, 2010, the very first Special Mathlympics will take …
FailAds.com is a site that has some funny pictures of advertisements. This particular one is someone looking for a math tudor: You can visit their site for a collection of other epic fail ads, funny pictures of Engrish, pwned, Tshirts, signs, magazine ads, instructions, packaging and lots more.
## i beat stumbleupon in math
So today I beat stumbleupon in math!! Check out the proof here: MathFail = 1; StumbleUpon = 0.Me thinks I deserve a medal ^_^
## The math behind low riding pants (how cool is it?)
In this post we learn all about the math behind low riding pants and how cool it really is. I decided to do a study just to see what the relationship was. I interviewed 13,532 people and showed them each 6 pictures of people wearing low riding pants. I then asked them to select a … | 519 | 2,054 | {"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} | 3.078125 | 3 | CC-MAIN-2021-39 | latest | en | 0.911824 |
https://www.iitianacademy.com/neet-physics-system-of-particles-and-rotational-motion-study-notes/ | 1,721,742,829,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518058.23/warc/CC-MAIN-20240723133408-20240723163408-00083.warc.gz | 699,450,445 | 55,620 | # NEET Physics – System of Particles and Rotational Motion- Study Notes
System of Particles and Rotational Motion NEET and AIIMS Special System of Particles and Rotational Motion Refresher Course System of Particles and Rotational Motion Concept Course System of Particles and Rotational Motion Master File System of Particles and Rotational Motion : Revision Notes System of Particles and Rotational Motion : Brain Map Reference Book Physics: System of Particles and Rotational Motion H C Verma Physics Exercise Solution: System of Particles and Rotational Motion Resnick Halliday Physics: Rotation Resnick Halliday Physics: Rotation Solution Resnick Halliday Physics: Torque Solution
equation of rotational motion, comparison of linear and rotational motions; moment of inertia, radius of gyration. Values of M.I. for simple geometrical objects (no derivation). Statement of parallel and perpendicular axes theorems and their applications
### CENTRE OF MASS
Centre of mass for a system of particles is defined as that point where the entire mass of the system is imagined to be concentrated. If ……… be the position vectors of masses m1, m2, m3 ……. respectively from the origin O, then the centre of mass of the system is where M is the total mass of the system of particles. The product of mass of the particles and its position vector w.r.t. the reference point is called moment of mass. i.e., moment of mass
#### MOTION OF CENTRE OF MASS
The motion of the centre of mass is governed by the equation where Momentum conservation of a system of particles : In the absence of external forces, the velocity of the centre of mass remains constant. We have, MAcm = Fext If Fext = 0 = 0 ∴ vcm = constant. Hence, momentum (Mvcm = constant) of the centre of mass system is conserved.
#### RIGID BODIES
If a body does not undergo any change in shape by the action of a force, it is said to be rigid. If such body undergoes some displacement, every particle in it undergoes the same displacement. No real body can be perfectly rigid.
#### ROTATORY MOTION
A body rotating about a fixed axis then every particle of the body moves in a circle and the centres of all these circles lie on axis of rotation. The motion of the body is said to possess rotational motion. KEEP IN MEMORY
1. The centre of mass of a system of two identical particles lies in between them on the line joining the particles.
2. If m1 = m2
so, for particles of equal masses the centre of mass is located at the mean position vector of the particles.
1. The position of centre of mass remains unchanged in pure rotatory motion. But it changes with time in translatory motion or rolling motion.
2. The position of centre of mass of a body is independent of the choice of coordinate system.
3. If we take the centre of mass at the origin, then the sum of the moments of the masses at of the system about the origin is zero.
4. In pure rotatory motion, the axis of rotation passes through the centre of mass.
5. If external force is zero then the velocity of the centre of mass of a body remains constant.
6. The centre of mass and centre of gravity of a body coincide, if the value of g is same throughout the dimension of the body.
7. In kinematics and dynamics, whole of the mass of a body can be assumed to be concentrated at the centre of mass.
8. The location of the centre of mass depends on the shape and nature of distribution of mass of the body.
1. The position of centre of mass of continuous bodies can be found using integration as
where, x, y and z are the coordinates of small mass dm and M is the total mass of the system.
1. The C.M. of a uniform rod of length L is at its middle point.
2. Centre of mass of a uniform semicircular wire is at, where R is the radius of the semicircular wire. It does not depend on mass.
1. For symmetrical bodies of uniform mass distribution, the C. M. lies at the geometrical centre
### ANGULAR VELOCITY AND ANGULAR ACCELERATION
The angular velocity is defined as the angle covered by the radius vector per unit time. It is denoted by ω. Average angular velocity The unit of angular velocity is rad/sec. The instantaneous angular velocity ω (similar to instantaneous linear velocity) is defined as The angular acceleration is the rate of change of angular velocity. It is denoted by α. The average angular acceleration αavg.of a rotating body is In analogy to linear acceleration, the instantaneous angular acceleration is defined as The unit of angular acceleration is rad/sec2. Relationship between angular velocity and linear velocity: where θ is the angle between ω & r.
### MOMENT OF INERTIA AND RADIUS OF GYRATION
A rigid body having constituent particles of masses m1, m2, ….mn and r1, r2 … rn be their respective distances from the axis of rotation then moment of inertia is given by, I = m1 r12 + m2 r22 + … + mn rn2 = The moment of inertia of continuous mass distribution is given by where r is the perpendicular distance of the small mass dm from the axis of rotation. Its SI unit is kgm2. It is a tensor.
The radius of gyration of a body about its axis of rotation may be defined as the distance from the axis of rotation at which, if the entire mass of the body were concentrated, its moment of inertia about the given axis would be same as with its actual distribution of mass. Radius of gyration k is given by, I = MK2 or where, M = Σmi Also, Therefore, radius of gyration (k) equals the root mean square of the distances of particles from the axis of rotation.
### GENERAL THEOREMS ON MOMENT OF INERTIA
#### THEOREM OF PERPENDICULAR AXIS
According to this theorem “the moment of inertia of a plane lamina (a plane lamina is a 2-dimensional body. Its third dimension is so small that it can be neglected.) about an axis, perpendicular to the plane of lamina is equal to the sum of the moment of inertia of the lamina about two axes perpendicular to each other, in its own plane and intersecting each other at the point, where the perpendicular axes passes through it.” If Ix and Iy be the moment of inertia of a plane lamina (or 2D rigid body) about the perpendicular axis OX and OY respectively, which lie in plane of lamina and intersect each other at O, then moment of inertia (Iz) about an axis passing through (OZ) and perpendicular to its plane is given by Ix + Iy = Iz Let us consider a particle of mass m at point P distance r from origin O, where so Ix + Iy = Σmy2 + Σmx2 = Σmr2 i.e., Iz = Ix + Iy
#### THEOREM OF PARALLEL AXES
(Derived by Steiner) This theorem is true for both plane laminar body and thin 3D body. It states that “the moment of inertia of a body about any axis is equal to its moment of inertia about a parallel axis through its centre of mass plus the product of the mass of the body and the square of the distance between two axes.” Let AB be the axis in plane of paper about which, the moment of inertia (I) of plane lamina is to be determined and PQ an axis parallel to AB, passing through centre of mass O of lamina is at a distance ‘r’ from AB. Consider a mass element m of lamina at point P distant x from PQ. Now the moment of inertia of the element about AB = m (x + r)2 so moment of inertia of whole lamina about AB is I = Σm(x+r)2 = Σmx2 + Σmr2 +2 Σmxr Where first term on R.H.S is Σ mx2 = Ic.m. moment of inertia of lamina about PQ through its centre of mass, second term on R.H.S. is Σmr2 = r2Σm = Mr2, M is whole mass of lamina, third term on R.H.S is (Σmx) r = 0, because Σmx is equal to moments of all particles of lamina about an axis PQ, passing through its centre of mass. Hence I = Ic.m. + M.r2 i.e., the moment of inertia of lamina about AB = its moment of inertia about a parallel axis PQ passing through its centre of mass + mass of lamina×(distance between two axes)2
### TORQUE, ANGULAR MOMENTUM AND ANGULAR IMPULSE
#### TORQUE
The moment of force is called torque. It is defined as the product of force and the perpendicular distance of the force from the axis of rotation. i.e., or, where θ is the angle between and . Its S.I. unit is (N-m). The dimensions of torque [ML2T–2] are the same as that of energy but it is not energy. Note : If the line of action of a force passes through axis of rotation then no torque will be formed.
#### ANGULAR MOMENTUM
The angular momentum of a particle about an arbitrary point ‘O’ is the moment of linear momentum taken about that point. It is given as or, where θ is the angle between and .
#### ANGULAR IMPULSE
(= Change in angular momentum)
#### CONSERVATION OF ANGULAR MOMENTUM
From equation . If …(i) This is called law of conservation of angular momentum. According to this “if resultant external torque acting on the system is zero then total angular momentum of the system is constant.” The magnitude of angular momentum for a system is given by [ ∵ v = rω] …(ii) Where Ii is the moment of inertia of the ith particle of that system and I is total moment of inertia of the system or …(iii) So if a system undergoes a redistribution of its mass, then its moment of inertia changes but since no external torque is applied on the system so total angular momentum is constant before and after the distribution of mass, even if moment of inertia of the system is changed. …(iv) or, I1ω1 = I2ω2 …(v) where L initial denote the state previous to the redistribution of mass and final denote the state after the redistribution of mass in that system. A comparison of useful relations in rotational and translational or linear motion :
Rotational motion about a fixed axis Linear motion Angular velocity Linear velocity Angular acceleration Linear acceleration Resultant torque τ = Iα Resultant force F = ma Equations of rotational motion Equations of linear motion where (s = x – x0) Work Work Kinetic energy Ek = ½ Iω2 Kinetic energy Ek=½ mv2 Power P = τω Power P = Angular momentum L = Linear momentum p = mv Torque Force
### WORK ENERGY THEOREM IN ROTATIONAL MOTION
According to this theorem “the work done by external forces in rotating a rigid body about a fixed axis is equal to the change in rotational kinetic energy of the body.” Since, we can express the torque as τ = Iα but τdθ = dW ⇒ τdθ = dW = Iωdω By integrating the above expression, we get total work done by all external force on the body, which is written as where the angular velocity of the body changes from ω1 to ω2 as the angular displacement changes from θ1 to θ2 due to external force on the body.
### ROTATIONAL KINETIC ENERGY
Let us consider a rigid body (collection of small particles) of high symmetry which is purely rotating about z-axis with an angular velocity ω. Each particle has some energy, determined by mi and vi. The kinetic energy of mi particle is … (i) Fig. The total kinetic energy of the body is ½Iω2. Now we know that in rigid body every particle moves with same angular velocity, the individual linear velocities depends on the distance ri from the axis of rotation according to the expression (vi = riω). Hence the total kinetic energy of rotating rigid body is equal to the sum of kinetic energies of individual particles. … (ii) where is moment of inertia of the rigid body. Now consider a rigid body which is rolling without slipping. In this case it possesses simultaneous translatory motion and rotatory motion and the total kinetic energy of the rigid body K.ETotal = rotation K.E. + translational K.E. of C.M. Ek = … (iii) Ek = … (iv) … (v) where, vcm = linear velocity of the centre of mass of the rigid body K = radius of gyration r = radius of moving rigid body I = moment of inertia of the rigid body about centre of mass ½ Iω2 = rotational kinetic energy about the centre of mass. Hence it is clear from the expression that total kinetic energy of rolling body is equal to the sum of rotational kinetic energy about centre of mass (C.M.) and translational kinetic energy of the centre of mass of body.
### BODY ROLLING WITHOUT SLIPPING ON AN INCLINED PLANE
When a body performed translatory as well as rotatory motion then we can say that the body is in rolling motion. Acceleration for body rolling down an inclined plane without slipping Let M is the mass of the body, R is its radius and I is the moment of inertia about the centre of mass and K is the radius of gyration. Force equation, …(1) Torque equation, …(2) Also, …(3) …(4) Adding, eqn. (1) & (4) ( Motion is pure rolling) Now assume in fig. that a body (which has high symmetry such as cylinder, sphere etc.) is rolling down an incline plane without slipping. This is possible only if friction is present between object and incline plane, because it provides net torque to the body for rotating about the centre of mass (since the line of action of the other forces such as mg and R pass through the centre of mass of rigid body, hence they do not produce torque in a body about the centre of mass). A round object rolling down an incline, mechanical energy is conserved if no slipping occurs. But mechanical energy of the body remains constant despite of friction because the contact point is at rest relative to the surface at any instant. For pure rolling motion vc = rω so …(i) As body rolls down an incline, it loses potential energy Mgh (h is the height of incline). Since body starts from rest at top, hence its total kinetic energy at bottom given by eqn.(i) must be equal to Mgh at top i.e., …(ii) since h =x sin θ, where x is length of incline and Ic =MK2
(K is radius of gyration), then ….(iii) For rolling down an inclined plane without slipping, the essential condition is : ; μs is the coefficient of static friction where Icm is moment of inertia of the body about its C.M.
### DIFFERENT TYPES OF MOTION
PURE TRANSLATIONAL MOTION In this case the velocities at all three points :
1. top most point P
2. C.M.
3. contact point O are same
PURE ROTATIONAL MOTION In this case the velocity of
1. top most point P is Rω
2. C.M. is zero
3. contact point O is – Rω
Vcm = 0 ROLLING MOTION – COMBINATION OF TRANSLATORY AND ROTATORY MOTION
1. In pure rolling motion, the contact point O remains at rest.
2. In pure rolling, the velocity of top most point is,
V = Vcm + ωR = 2Vcm
1. In rolling without slipping Vcm = ωR.
KEEP IN MEMORY
1. The axis of the rolling body is parallel to the plane on which the body rolls in case of sphere, disc, ring.
2. Let (Ek)r = rotational kinetic energy
(Ek)t = translational kinetic energy For solid sphere, (Ek)r = 40% of (Ek)t For shell (Ek)r = 66% of (Ek)t For disc, (Ek)r = 50% of (Ek)t For ring, (Ek)r = (Ek)t
1. Translational kinetic energy is same for rolling bodies having same M, R and ω irrespective of their shape.
2. Total energy is minimum for solid sphere and maximum for ring having same mass and radius.
3. Rotational kinetic energy is maximum for ring and minimum for solid sphere of same mass and radius.
4. (i)The acceleration down the inclined plane for different shapes of bodies of same mass and radius are as follows : Sphere > disc > shell > ring
(ii) The velocity down the plane is related as follows: Sphere > disc > shell > ring
(iii) The time taken to reach the bottom of the inclined plane is related as follows : Ring > shell > disc > sphere.
This is because For ring For disc For sphere For shell
1. The angular speed of all particles of a rotating/revolving rigid body is same, although their linear velocities may be different.
2. Two particles moving with angular speed ω1 and ω2 on the same circular path and both in anti-clockwise direction then their relative angular speed will be ωr = ω1 – ω2.
In the above case one particle will complete one revolution more or less as compared to the other in time
1. Let two particles move on concentric circles having radius r1 and r2 and their linear speeds v1, v2 both along anti-clockwise direction then their relative angular speed will be
1. As I ∝ K2 and I = MK2, hence graph between I and K will be a parabola. However graph between log I and log K will be straight line.
1. Rotational KE, ∴ Er ∝ ω2
∴ graph between and ω will be as below
1. Angular momentum L = Iω, hence L ∝ ω.
∴ Graph between L and ω is a straight line.
1. The moment of inertia is not a vector quantity because clockwise or anti-clockwise, direction is not associated with it. It is a tensor.
2. When a spherical/circular/cylindrical body is given a push, it only slips when the friction is absent, It may roll with slipping if friction is less than a particular value and it may roll without slipping if the friction is sufficient. (i.e.)
3. When a body rolls without slipping no work is done against friction.
Scroll to Top | 3,988 | 16,624 | {"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} | 4.09375 | 4 | CC-MAIN-2024-30 | latest | en | 0.821138 |
kerneltrick.in | 1,527,434,321,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794869272.81/warc/CC-MAIN-20180527151021-20180527171021-00088.warc.gz | 575,587,192 | 31,217 | # Statistical significance
posted in: concepts, jupyter, notebooks | 1
To understand statistical significance, we need to understand the difference between a population and a sample. A sample is an individual or a group from a population that is representative in nature. Statistical significance is the likelihood (or probability) that an effect observed or a statistic derived from a sample represents a phenomenon occurring in the population from which the sample is chosen. [1] This probability is the p-value.
We use the p-value to either accept or reject a baseline model or a null hypothesis (H0). A null hypothesis is an absence of an effect or an effect observed that occurred purely by chance. Significance level (α) is the threshold value of p-value, iff p-value is less than the significance level (α) then we reject the null hypothesis (H0). This approach is called null hypothesis significance testing [0] We generally accept α = 0.05
In [1]:
```import numpy as np
from scipy.stats import norm
import matplotlib.pyplot as plt
from scipy import stats
plt.style.use('ggplot')
```
In [2]:
```mean = 0
sigma = 1
alpha = 0.05
confidence_level = 1-alpha
```
In [3]:
```x = np.linspace(mean - 5*sigma, mean + 5*sigma, 1000)
iq = stats.norm(mean, sigma)
plt.plot(
x,
iq.pdf(x),
color = 'black')
#confidence intervals
conf_interval_a, conf_interval_b = norm.interval(
confidence_level,
loc = mean,
scale = sigma)
px = x[np.logical_and(x >= conf_interval_a, x <= conf_interval_b)]
plt.fill_between(
px,
iq.pdf(px),
color = '#d48181')
plt.show()
```
If we conduct a statistical test on a sample of data from a population, if the p-value is less than α then we reject the null hypothesis and we can be reasonably sure that an effect we observed in the sample data is an genuine phenomenon in the population. We then say that the test statistic is statistically significant at level α. If the p-value is greater than α then we accept the null hypothesis and we can say that the observed effect was by chance and/or a sampling error. This process of statistical significance, guarantees that our type I (false positive) error rate is less than α.
References:
[0] Murphy, Kevin P. Machine learning: a probabilistic perspective. MIT press, 2012 p-215.
[1] Urdan, Timothy C. Statistics in plain English. Routledge, 2011. p-61.
### One Response
1. ##### Type I and Type II errors | kernel trick
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https://slideplayer.com/slide/9290081/ | 1,624,208,121,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488249738.50/warc/CC-MAIN-20210620144819-20210620174819-00622.warc.gz | 467,099,645 | 17,133 | # Properties of Algebra By: Zoe Gaffney. Associative Property Associative Property is when you change the numbers that are in the parenthesis. Example:
## Presentation on theme: "Properties of Algebra By: Zoe Gaffney. Associative Property Associative Property is when you change the numbers that are in the parenthesis. Example:"— Presentation transcript:
Properties of Algebra By: Zoe Gaffney
Associative Property Associative Property is when you change the numbers that are in the parenthesis. Example: (5+2)+7 = 5+(2+7) (5*2)*7 = 5*(2*7)
Commutative Property Commutative Property of Addition and Multiplication: Changing the order of numbers in the parenthesis doesn’t change the answer Example: (5+2)+7 = (2+5)+7 (5*2)*7 = (2*5)*7
Identity Property Identity property of multiplication is that if you multiply 1 by a number it will not change it. Identity property of addition is if you add zero it will not change the answer Example: 3+0=3 3*1=3
Distributive Property Distributive Property is when there is a number next to parenthesis and you multiply the outside number by the inside numbers. Example: 7(3+2) 21+14 35
Equality Property When you use an operation to combine an identity with another number, that number stays the same. Zero is called the additive identity, because adding zero to a number will not change it: the number stays the same. Example 0 + a = a = a + 0.
Inverse Property Inverse property of addition states that when an integer is combined with its opposite, it is equal to zero. The inverse property of multiplication is multiplying the number by a fraction of itself. Example: 3+(-3)=0 3*1/3=1
Download ppt "Properties of Algebra By: Zoe Gaffney. Associative Property Associative Property is when you change the numbers that are in the parenthesis. Example:"
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from Quasimodo
## Related concepts
Parents studyWeight: 0.61, factorWeight: 0.60, elementWeight: 0.59, amountWeight: 0.59 Siblings lengthWeight: 0.60, golden ratioWeight: 0.58, quantityWeight: 0.33, consequenceWeight: 0.33, incidenceWeight: 0.33
## Related properties
Similarity Property is calculated 1.00 be useful calculating 0.83 is ratio calculator 0.75
## Clauses
### Salient implies Plausible
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Rule weight: 0.85
Evidence weight: 0.36
Similarity weight: 0.83
Evidence: 0.01
Typical(proportion, is calculated)
Evidence: 0.65
¬ Typical(proportion, be useful calculating)
### Typical and Remarkable implies Salient
0.14
Rule weight: 0.14
Evidence weight: 0.99
Similarity weight: 1.00
Evidence: 0.43
Salient(proportion, is calculated)
Evidence: 0.01
¬ Typical(proportion, is calculated)
Evidence: 0.97
¬ Remarkable(proportion, is calculated)
### Typical implies Plausible
0.47
Rule weight: 0.48
Evidence weight: 0.99
Similarity weight: 1.00
Evidence: 0.17
Plausible(proportion, is calculated)
Evidence: 0.01
¬ Typical(proportion, is calculated) | 831 | 2,557 | {"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-2023-40 | latest | en | 0.7686 |
https://www.graduate.technion.ac.il/Theses/Abstracts.asp?Id=10630 | 1,660,085,837,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571090.80/warc/CC-MAIN-20220809215803-20220810005803-00653.warc.gz | 731,509,723 | 2,970 | Ph.D Student Ph.D Thesis Rawitz Dror Combinatorial and LP-Based Methods for Designing Approximation Algorithms Department of Computer Science PROFESSOR EMERITUS Reuven Bar-Yehuda
Abstract
This thesis focuses on two approximation paradigms: the local ratio technique and the primal-dual schema. We present two relatively simple approximation frameworks for minimization problems, one for each approach, which extend known approximation frameworks for covering problems, and we show that the two frameworks are equivalent. We also present two equivalent frameworks for maximization problems. We demonstrate our generic algorithms on a variety of problems.
While all applications of the local ratio technique to date use non-negative weight functions, we present local ratio analyses to several known algorithms that use weight functions with some negative coefficients. In order to present equivalent analyses that are based on the primal-dual schema we define new integer programming formulations with constraints that have negative coefficients.
We present two new combinatorial approximation frameworks that are not based on LP-duality, or even on linear programming. Instead, they are based on weight manipulation in the spirit of the local ratio technique. We show that the first framework is equivalent to the method of dual fitting. We define a method called primal fitting, and show it is tantamount to our second framework. We demonstrate both frameworks on a variety of problems including the metric uncapacitated facility location problem.
A k-partite graph is a graph G=(V1,…,Vk,E), where V1,…,Vk are k non-empty disjoint independent sets of vertices. Such a graph is called complete k-partite if E contains all possible edges. We discuss three variants of the following optimization problem: given a graph and a non-negative weight function on the vertices and edges, find a minimum weight set of vertices and incident edges whose removal from the graph leaves a complete k-partite graph. We use the local-ratio technique to develop 2-approximation algorithms for the first two variants of the problem. We use approximation preserving reductions in order to (4-4/k)-approximate the third variant. | 434 | 2,224 | {"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 | 0.880044 |
https://kidsworksheetfun.com/multiplication-facts-worksheets-0-12-printable/ | 1,701,904,073,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100626.1/warc/CC-MAIN-20231206230347-20231207020347-00602.warc.gz | 396,312,799 | 25,004 | # Multiplication Facts Worksheets 0-12 Printable
Multiplication Facts Worksheets 0-12 Printable. Use the buttons below to print, open, or download the pdf. Practice multiplication facts for 0 to 12 with these printable mixed up math puzzles.
Fill the correct number in the blanks provided using your knowledge of multiplication. Use this multiplications table printable to fill in the blank spaces with the correct answers. This page has lots of games, worksheets, flashcards, and activities for teaching all basic multiplication facts between 0 and 10.
### Multiplying By Anchor Facts 0, 1, 2, 5 And 10 Multiplying By Facts 3, 4 And 6 Multiplying By Facts 7, 8 And.
If you are helping your child learn the multiplication facts, why not. Multiplication facts worksheets | guruparents. The idea behind using flashcards is to give food to the brain systematically and.
### 11S (11 X Number) Multiplication Facts:
Free printable multiplication worksheets include models, long multiplication, quiz, lattice method, tables, charts, drills, basic facts, word problems etc. Multiplication cidr worksheets mathworksheets4kids ipv4 maths. 10 pictures about multiplication facts worksheets | guruparents :
### Multiplication Basic Facts 0 To 12.
10s (10 x number) or (number x 10) multiplication facts: Multiplication basic facts 0 to 12. 10 puzzles in both color and black & white.
### Students Can Use Math Worksheets To Master A Math Skill Through Practice, In A Study Group Or For Peer Tutoring.
Use this multiplications table printable to fill in the blank spaces with the correct answers. For example, for the tables of 4, you need to memorize the result first and do it quickly. Each worksheet has 42 problems.17 pages of worksheet goodness!mad minute.
### Worksheet #1 Is A Table Of All Multiplication Facts With Zero Or One As A Factor.
These multiplication facts worksheets provide various exercise to help students gain fluency in the multiplication facts up to 12 x 12. These print and go multiplication facts practice worksheets and speed drills are an easy, no prep way to increase multiplication facts fluency. Use the buttons below to print, open, or download the pdf. | 477 | 2,186 | {"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-2023-50 | latest | en | 0.788815 |
https://superstarwebtech.com/cs606-compiler-construction-quiz-mcqs-finalterm | 1,726,029,274,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651343.80/warc/CC-MAIN-20240911020451-20240911050451-00059.warc.gz | 511,188,532 | 32,382 | SUPERSTARWEBTECH
There is no elevator to success. You have to take the stairs. Take your first steps with SSWT!
CS606 Compiler Construction Quiz MCQs Lecture 23-30 Objective Questions
CS606-Compiler Construction Quiz #Objective #Questions
1. NFA of LR(1) items means ___
• None
2. In the Parsing Table, the rows correspond to Parsing DFA states and columns correspond to ___
• Terminals and Non-terminals ✔
• Start Symbol and its derivation
• Handles and derivations
• None
3. A grammar is LR if a ___ shift reduce-reduce parser can recognize handles when they appear on the top of the stack
• left-to-reverse
• left-to-rise
• left-to-right ✔
• None
4. If / is a set of items for grammar then closure(/) is a set of items constructed from / by the following rule.
• If A-> aX.Y is in closure(/) and Y->r is production, then add X->.r to closure(/)
• If A->a.XY is in closure(/) and X->r is production, then add X->.r to closure(/) ✔
• If A->aXY. is in closure(/) and A->r isproduction, then add X->.r to closure(/)
• None
5. Suppose ? begins with symbol X which may be a terminal (token) or non-terminal. The item can be written as A?Xa.?
• True ✔
• False
6. A handle is a substring that matches a ___ side of production rule in the grammar.
• right hand✔
• left hand
7. If T->XYZ is a production of grammar G then which of the following item indicates that a string derivable from X has been seen so far on the input and we hope to see a string derivable from YZ next on the input.
• T->.XYZ
• T->X.YZ ✔
• T->XY.Z
• T->XYZ.
8. In the canonical collection procedure, a DFA can not be constructed from NFA using the subset construction, similar to the one we used for lexical analysis.
• True
• False ✔
9. Suppose ? begins with symbol X which may be a terminal (token) or non-terminal. The item can be written as ___
• A?a.X?
• A?Xa.? ✔
• A?X?.
• X?Aa.?
10. If / is a set of items for grammar then closure(/) is a set of items constructed from / by the following rule.
• Every item in / is in closure(/) ✔
• Every item in / is not in closure(/)
• Only one item in / is in closure(/)
• None
11. NFA of LR(0) items means ___
• no look ahead symbol ✔
• All of the given
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Blog | 765 | 2,973 | {"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-2024-38 | latest | en | 0.856802 |
https://www.ple4win.nl/help/downward_vertical_soil_stiffne.htm | 1,618,138,641,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038061820.19/warc/CC-MAIN-20210411085610-20210411115610-00219.warc.gz | 1,057,055,342 | 4,155 | Downward vertical soil stiffness [KLS]
# Downward vertical soil stiffness [KLS]
Calculations of Downward vertical soil stiffness
The soil range of importance for the Downward vertical soil stiffness is from the pipeline axis to 1.5 diameter under the pipeline axis.
NEN3650-1992
For the NEN3650-1992 calculation methodology two formula's are used to calculate the Downward vertical soil stiffness, one for sandy soil types and one for clayey/peaty soil types.
Sandy:
Clayey/peaty:
kv Downward vertical soil stiffness Pwe Sub-soil bearing capacity D Outer diameter of pipeline cuo Undrained cohesion (averaged over range from pipeline axis to 1.5 diameter below pipeline axis) H Height of soil coverage on top of pipeline
NEN3650-1:2017
The NEN3650-1:2017 calculation methodology uses the following formula's. There are two formula's for trench installation: one for sandy soil types and one for clayey/peaty soil types and there is one formula for non-trench installation. A trench installation is either 1)Trench uncompressed fill or 2)Trench compressed fill and a non-trench installation is either 3)Boring or 4)HDD.
Trench, Sandy:
Trench, Clayey/peaty:
Non-trench:
kv Downward vertical soil stiffness Pwe Sub-soil bearing capacity D Outer diameter of pipeline Eo Effective Young's modulus calculated for range from pipeline axis to 1.5 diameter below pipeline axis, using depth at bottom of pipe (for Trench given in N/mm2, factor 0.5 is not dimensionless) cuo Undrained cohesion (averaged over range from pipeline axis to 1.5 diameter below pipeline axis) (given in N/mm2, factor 250 is not dimensionless) m l λ ν φo Angle of internal friction (averaged over range from pipeline axis to 1.5 diameter below pipeline axis) Epipe Young's modulus of pipe material Ib Di Inner diameter of pipeline
Last changed: 10/30/2019 | 450 | 1,841 | {"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-2021-17 | latest | en | 0.831711 |
https://www.ankitcodinghub.com/product/cs6643-midterm-exam-solved/ | 1,669,905,453,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710813.48/warc/CC-MAIN-20221201121601-20221201151601-00272.warc.gz | 707,068,038 | 57,076 | # CS6643 Midterm Exam Solved
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## Description
5/5 - (1 vote)
1. Given the 7 × 7 grayscale image in Figure 1(a) below, use the Canny’s edge detector to produce a binary edge map. The image has already been smoothed by a Gaussian filter. (a) Use the Prewitt’s edge operator in Figure 1(b) below to compute edge magnitudes and gradient angles for pixel locations 𝑖𝑖, 𝑗𝑗 = 1 to 5 (center 5 × 5 ) Compute edge magnitude by taking the sum of the absolute values of the horizontal and vertical gradients. (b) Apply non-maxima suppression to pixel locations 𝑖𝑖, 𝑗𝑗 = 2 to 4 (center 3 × 3 region.) (c) Use simple thresholding with 𝑇𝑇 = 3 to produce a binary edge map for pixel locations 𝑖𝑖, 𝑗𝑗 = 2 to 4. Show results after each step from (a) to (c) above.
1 1 1 1 1 1 5 1 1 1 1 1 5 9 1 1 1 1 5 9 9 1 1 1 5 9 9 9 1 1 5 9 9 9 9 1 5 9 9 9 9 9 5 9 9 9 9 9 9
(a) The image uses the i-j coordinate system with i pointing downward and j points to the right. The upper left corner pixel has coordinates (𝑖𝑖, 𝑗𝑗) = (0,0).
Figure 1. Input image and Prewitt’s operator.
1. The Connected Component Labeling algorithm below is for finding 4-connected regions. Modify the algorithm so that it can find 8-connected regions.
• Sequential algorithm for connected component labeling (for 4-connected regions)
1. Scan the image left to right, top to bottom
2. If the pixel is 1,then
1. If only one of its upper and left neighbors has a label, copy the label.
2. If both neighbors have the same label, copy the label.
3. If the two neighbors have different labels, copy the upper neighbor’s label and enter the two labels in the equivalence table as equivalent labels. d) Otherwise, assign a new label to this pixel.
3. If there are more pixels to consider, go to step 2
4. Find the label with the smallest value in each equivalent set in the equivalence table.
5. Scan the label map. Replace each label by the smallest label in its equivalent set.
1. In a certain machine part inspection application, we know that the image contains three machine parts (A, B and C) and the background. We also know the following:
• Object A occupies 9% of the image in size. Object B is twice as big as A, and C is twice as big as B.
• 𝐼𝐼𝐴𝐴 < 𝐼𝐼𝐵𝐵 < 𝐼𝐼𝐶𝐶 < 𝐼𝐼𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 where I is the gray level value of the individual pixels that belong to the three objects and the background, as indicted by the subscript.
Given the grayscale histogram of the input image in the figure below, determine three threshold values, 𝑇𝑇1, 𝑇𝑇2 and 𝑇𝑇3, that can be used to segment the three objects from the background. Give the grayscale intervals for each of the objects and the background based on the threshold values that you have determined.
Grayscale histogram of the input image.
(The number above each bar is the pixel count.)
1. Given the 5 × 5 grayscale image below, apply the Laplacian operator below to detect edges. For your answer, first give the Laplacian response as a 2D array and then mark the detected edges on the array. Let the output response be undefined if the Laplacian template goes outside of the border of the image.
4 4 12 12 12 4 4 12 12 12 4 4 12 12 12 4 4 8 8 8 4 4 8 8 8 Grayscale Image
0 1 0 1 -4 1 0 1 0
Laplacian Operator. The center pixel is the reference center.
1. (a) Apply the iterative distance transform algorithm below to the input image 𝑓𝑓(𝑟𝑟, 𝑐𝑐) in the figure below, where the 1’s represent foreground pixels and background pixels are unlabeled (assumed to have a value of 0.)
f 0[r,c] = f [r,c]
f m[r,c] = f 0[r,c]+ min( f m1[u,v])
Stop when 𝑓𝑓𝑚𝑚[𝑟𝑟, 𝑐𝑐] = 𝑓𝑓𝑚𝑚−1[𝑟𝑟, 𝑐𝑐]
Show the image result after each iteration. You must execute Step 3 in the algorithm above so the algorithm knows when to stop. (b) From the distance transform, circle those pixels that belong to the medial axis. The image uses the (𝑟𝑟, 𝑐𝑐) coordinate system with the upper-left corner pixel having coordinates (0,0).
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
Input image f(r,c)
Hough transform is used to detect circles having a radius of 2 pixels in an input image. Below is the content of the accumulator array for parameters a and b after applying Hough transform to the edges in the input image. The range for both a and b in the accumulator array is [0,40] pixels. We know that there is only one circle in the image. Write the equations of the detected circle.
1. [15 points] One way to test whether two adjacent regions should be merged together in the split and merge algorithm is by computing the likelihood ratio L defined as
pp((gg1,g2,….| H1)) = σm0m11.σ+m2m22
# L =
,g ,….| H
1 2 0 σ1
where σ1 and σ2 are the standard deviations of the gray level values of the pixels in region 1 and region 2, respectively, σ0 is the standard deviation of the gray level values of the pixels in the two regions combined, m1and m2are the number of pixels in regions 1 and 2, respectively. Starting with the equation of a normal distribution given below, show (by mathematical derivation) that the joint probability density under hypothesis 𝐻𝐻1 is
p(g1, g2,……, gm1 , gm1+1,…….., gm1+m2 | H1) = ( 2π1σ1)m1 em21 . ( 2π1σ2 )m2 em22
Normal distribution:
p(gi ) = 1 (g −µ)2
, 𝑔𝑔𝑖𝑖 is the gray level value of pixel i.
• Midterm-ws9ybq.zip | 1,768 | 5,487 | {"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} | 3.140625 | 3 | CC-MAIN-2022-49 | latest | en | 0.786797 |
http://es.wikidoc.org/index.php/Fixed_point_(mathematics) | 1,566,656,643,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027321140.82/warc/CC-MAIN-20190824130424-20190824152424-00126.warc.gz | 65,889,796 | 11,767 | # Fixed point (mathematics)
File:Fixed Point Graph.png
A function with three fixed points
In mathematics, a fixed point (sometimes shortened to fixpoint) of a function is a point that is mapped to itself by the function. That is to say, x is a fixed point of the function f if and only if f(x) = x. For example, if f is defined on the real numbers by
${\displaystyle \ f(x)=x^{2}-3x+4,}$
then 2 is a fixed point of f, because f(2) = 2.
Not all functions have fixed points: for example, if f is a function defined on the real numbers as f(x) = x + 1, then it has no fixed points, since x is never equal to x + 1 for any real number. In graphical terms, a fixed point means the point (x, f(x)) is on the line y = x, or in other words the graph of f has a point in common with that line. The example is a case where the graph and the line are a pair of parallel lines.
Points which come back to the same value after a finite number of iterations of the function are known as periodic points; a fixed point is a periodic point with period equal to one.
## Attractive fixed points
File:Cosine fixed point.svg
The fixed point iteration xn+1 = cos xn with initial value x1 = -1.
An attractive fixed point of a function f is a fixed point x0 of f such that for any value of x in the domain that is close enough to x0, the iterated function sequence
${\displaystyle x,\ f(x),\ f(f(x)),\ f(f(f(x))),\dots }$
converges to x0. How close is "close enough" is sometimes a subtle question.
The natural cosine function ("natural" means in radians, not degrees or other units) has exactly one fixed point, which is attractive. In this case, "close enough" is not a stringent criterion at all -- to demonstrate this, start with any real number and repeatedly press the cos key on a calculator. It quickly converges to about 0.73908513, which is a fixed point. That is where the graph of the cosine function intersects the line ${\displaystyle y=x}$.
Not all fixed points are attractive: for example, x = 0 is a fixed point of the function f(x) = 2x, but iteration of this function for any value other than zero rapidly diverges. However, if the function f is continuously differentiable in an open neighbourhood of a fixed point x0, and |f'(x0)| < 1, attraction is guaranteed.
Attractive fixed points are a special case of a wider mathematical concept of attractors.
An attractive fixed point is said to be a stable fixed point if it is also Lyapunov stable.
A fixed point is said to be a neutrally stable fixed point if it is Lyapunov stable but not attracting. The center of a second order homogeneous linear differential equation is an example of a neutrally stable fixed point.
## Theorems guaranteeing fixed points
There are numerous theorems in different parts of mathematics that guarantee that functions, if they satisfy certain conditions, have at least one fixed point. These are amongst the most basic qualitative results available: such fixed-point theorems that apply in generality provide valuable insights.
## Applications
In many fields, equilibria or stability are fundamental concepts that can be described in terms of fixed points. For example, in economics, a Nash equilibrium of a game is a fixed point of the game's best response correspondence.
In compilers, fixed point computations are used for whole program analysis, which are often required to do code optimization. The vector of PageRank values of all web pages is the fixed point of a linear transformation derived from the World Wide Web's link structure.
Logician Saul Kripke makes use of fixed points in his influential theory of truth. He shows how one can generate a partially defined truth predicate (one which remains undefined for problematic sentences like "This sentence is not true"), by recursively defining "truth" starting from the segment of a language which contains no occurrences of the word, and continuing until the process ceases to yield any newly well-defined sentences. (This will take a denumerable infinity of steps.) That is, for a language L, let L-prime be the language generated by adding to L, for each sentence S in L, the sentence "S is true." A fixed point is reached when L-prime is L; at this point sentences like "This sentence is not true" remain undefined, so, according to Kripke, the theory is suitable for a natural language which contains its own truth predicate. | 998 | 4,390 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 3, "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} | 4.28125 | 4 | CC-MAIN-2019-35 | latest | en | 0.905418 |
http://www.sr2jr.com/textbook-solutions/computer-science/10702019/computer-networking-a-top-down-approach-multimedia-networking | 1,620,988,496,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243990449.41/warc/CC-MAIN-20210514091252-20210514121252-00506.warc.gz | 89,343,960 | 12,488 | SHARE
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# Computer Networking : A Top-down Approach
Authors:
James F. Kurose, Keith W. Ross
Exercise:
Problems
Chapter:
Multimedia Networking
Edition:
6
ISBN:
9780132856201
Question:
19
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### Question
Consider again the figure for P18.
a. Assume a priority service, with packets 1, 4, 5, 6, and 11 being high-priority packets. The remaining packets are low priority. Indicate the slots in which packets 2 through 12 each leave the queue.
b. Now suppose that round robin service is used, with packets 1, 4, 5, 6, and 11 belonging to one class of traffic, and the remaining packets belonging to the second class of traffic. Indicate the slots in which packets 2 through 12 each leave the queue.
c. Now suppose that WFQ service is used, with packets 1, 4, 5, 6, and 11 belonging to one class of traffic, and the remaining packets belonging to the second class of traffic. Class 1 has a WFQ weight of 1, while class 2 has a WFQ weight of 2 (note that these weights are different than in the previous question). Indicate the slots in which packets 2 through 12 each leave the queue. See also the caveat in the question above regarding WFQ service.
a)
Consider the figure P18 and take a priority service, with packets 1, 4, 5, 6, and 11 being high-priority packets and remaining are low priority packets.
The following table indicating the arrival time and the time at which the each packet leaves the queue.
The formula for the Delay time = leaving time – arrival time.
The following table calculate the delay time of the packets by using above formula:
b)
Consider the round robin service is used, with packets 1, 4, 5, 6, and 11 belonging to one class of traffic, and the remaining packets belonging to the second class of traffic.
The following table indicating the arrival time and the time at which the each packet leaves the queue.
The formula for the Delay time = leaving time – arrival time.
The following table calculate the delay time of the packets by using above formula:
C)
Consider the Weighted Fair Queuing and that WFQ service is used, with packets 1, 4, 5, 6, and 11 belonging to one class of traffic, and the remaining packets belonging to the second class of traffic. Class 1 has a WFQ weight of 1, while class 2 has a WFQ weight of 2 .
The following table indicating the arrival time and the time at which the each packet leaves the queue:
The formula for the Delay time = leaving time – arrival time.
The following table calculate the delay time of the packets by using above formula:
### Discussions
Post the discussion to improve the above solution. | 722 | 3,080 | {"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} | 3.15625 | 3 | CC-MAIN-2021-21 | latest | en | 0.930936 |
https://math.stackexchange.com/questions/3460069/does-pi-4-arctan-left-frac-yx2y2-5x6-right-imply-x2y2-5x-pm-y | 1,576,257,606,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540564599.32/warc/CC-MAIN-20191213150805-20191213174805-00119.warc.gz | 454,146,014 | 31,384 | # Does $\pi/4=\arctan\left(\frac{-y}{x^2+y^2-5x+6}\right)$ imply $x^2+y^2-5x\pm y+6=0$
While solving $$\arg\left(\frac{z-2}{z-3}\right)=\pi/4$$, I arrive at the following equation $$\pi/4=\arctan\left(\frac{-y}{x^2+y^2-5x+6}\right)$$
At this point I take $$\tan()$$ on both sides to get $$x^2+y^2-5x+y+6=0$$. But according to my book the answer is $$x^2+y^2-5x\pm y+6=0$$. I possibly made a mistake in the simplification of the $$\arctan()$$ equation. What is it?
Although I'm not sure what your reasoning is, and I'm now starting to doubt my own reasoning, I agree with your conclusion, and I disagree with the book, whatever it is! Perhaps we're both making the same mistake? Let's see.
The fundamental problem is that for $$u + iv \in \mathbb{C},$$ with $$u \ne 0,$$ the equation: $$$$\label{3460069:eq:1}\tag{1} \arg(u + iv) = \frac\pi4$$$$ is not equivalent to: $$$$\label{3460069:eq:2}\tag{2} \arctan\frac{v}{u} = \frac\pi4.$$$$ Equation \eqref{3460069:eq:2} is - of course! - equivalent to: $$$$\label{3460069:eq:2p}\tag{2'} \frac{v}{u} = 1, \text{ i.e. } u = v.$$$$ If $$u = v > 0$$, then \eqref{3460069:eq:1} does hold, but on the other hand, if $$u = v < 0$$, we have: $$$$\label{3460069:eq:3}\tag{3} \arg(u + iv) = -\frac{3\pi}4.$$$$ In sum: either of \eqref{3460069:eq:2} or \eqref{3460069:eq:2p} is equivalent to the exclusive disjunction of \eqref{3460069:eq:1} and \eqref{3460069:eq:3}.
In the present problem, you presumably arrived at some condition such as: $$\arg((z - 2)(\overline{z} - 3)) = \frac\pi4$$ which, for $$z = x + iy$$, is equivalent to: $$\arg(x^2 + y^2 - 5x + 6 - iy) = \frac\pi4.$$ By my apparently simple reasoning, from the above general considerations, this implies $$y < 0$$, and also implies: $$x^2 + y^2 - 5x + y + 6 = 0.$$
Are we both wrong?
Geometrical common sense (call it "intuition", to be posh) does seem to confirm that $$z - 2$$ can only be an anticlockwise quarter-turn away from $$z - 3$$ if $$z$$ is in the lower half-plane $$\{x + iy : y < 0\}$$, and this is somewhat reassuring.
If I've screwed up, I'll post a correction, but for the moment I'm inclined to suspect there's an error in the book.
(I should add that the problem in the title of the question differs from the problem stated in the question itself. Rather than risk adding to the confusion, I ask you to confirm what exactly is the question in the book, and modify the title and/or body of the question appropriately.)
## Update
The figure illustrates the proposition that the locus of points $$z$$ in the complex plane such that: $$\arg\left(\frac{z - 2}{z - 3}\right) = \frac\pi4$$ is the intersection of the circle $$\{ x + iy : x^2 + y^2 - 5x + y + 6 = 0 \}$$ with the lower half-plane $$\{x + iy : y < 0\}.$$
The small arc of the same circle in the upper half-plane $$\{x + iy : y > 0\}$$ is associated with the angular value $$-\frac{3\pi}4.$$
For completeness I've also shown arcs of the circle $$\{ x + iy : x^2 + y^2 - 5x - y + 6 = 0 \}$$ and their associated angular values $$-\frac\pi4$$ and $$\frac{3\pi}4.$$
If you've quoted the book accurately, there's definitely an error in it.
• Wait why is $(1)$ not equivalent to $(2)$? Is it because of the possibility you mention of this being $-3\pi/4$. – Paras Khosla Dec 2 at 19:32
• Alright. Thanks – Paras Khosla Dec 2 at 19:32 | 1,135 | 3,318 | {"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": 32, "wp-katex-eq": 0, "align": 0, "equation": 4, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2019-51 | latest | en | 0.800541 |
https://github.com/lindbrook/cholera/blob/master/docs/notes/unstacking.bars.notes.md | 1,579,646,047,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250606226.29/warc/CC-MAIN-20200121222429-20200122011429-00513.warc.gz | 480,913,041 | 18,626 | Fetching contributors…
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lindbrook 2019-02-07
To match cases to a road and to other cases within a stack, I use orthogonal projection and cluster analysis as methods of classification.
### Orthogonal projection
For each bar, I draw a series of lines that pass through the bar and are orthogonal to “nearby” road segment.[1] If the projected line bisects a road segment, it becomes a candidate for the bar’s home segment. If there’s more than one candidate, the algorithm chooses the segment “closest” to the bar: the segment with the shortest orthogonal projector.
To illustrate, consider the stylized example below. For case 12, there are three possible “home” road segments: the solid blue, solid red and the solid green segments. Of the three, only red and green are candidates. The problem with blue is that its orthogonal projector (the blue arrow) does not intersect the solid blue segment. Of the two remaining candidates, red is chosen because it is closer to case 12 than green: the red arrow is shorter than the green one.
This procedure correctly classifies 508 of 578 cases, an error rate of 12.1%. Errors occur due to ambiguities surrounding a case’s “home” segment. Unlike Snow’s map where a bar’s “horizontal” orientation reflects its “home” road segment, Dodson and Tobler only record a case’s x-y coordinates. This can lead to problems at street intersections, where a point may be closer to the “wrong” segment, and at addresses with many cases.[2] The problem with multiple cases is that stacking can push cases away from their “home” segment. Consider the figure below. Even though visual inspection (the alignment of the stack) would lead us to choose black, the algorithm chooses red. The reason is that stacking pushes the case so far from black that it falls outside of the range of consideration. I corrected such errors by manually classification.
### Cluster analysis
To link bars within a stack, I use hierarchical cluster analysis.[3] To make this task easier, I use a bar’s orthogonal coordinates rather that its x-y coordinates. That is, I use the point where the the orthogonal projection from the bar intersects its “home” road segment. Doing so reduces the dimensionality of the problem since the algorithm only needs to compare positions of points along a line. This virtually eliminates any classification error.[4]
To illustrate, the graph below plots the cases and their orthogonal projectors along a segment of Broad Street. Linking a bar to a stack and identifying a stack boils down to identifying distinct clusters of points along the road segment. For the case at hand, visual inspection leads us to expect that we should find 8 different clusters: 4 on the north side, 4 on the south.
One can be confirm the results by examining Snow’s map:[5]
The one fly in the ointment is that addresses directly opposite one another will have “identical” orthogonal points and will appear to belong to the same “address”. A simple solution to distinguish clusters on opposite sides of the street is to compute the sign of the arithmetic difference between a case’s x-y coordinates and its orthogonal coordinates.
For example, consider cases 321 and 239. They lie on opposite sides of Broad Street. In the figure below, 321’s coordinates are shown in blue while 239’s are shown in red. When we take the signs of the differences, we get (-, +) for 321 and (+, -) for 239. This distinguishes cases on the north side from those on the south side.
In general, for roads with positive slopes (e.g., Broad Street), the differences for cases on the north/west side will be (-, +) while those on the south/east side will be (+, -); for roads with negative slopes (e.g., Cambridge Street), the differences for cases on the north/east side will be (-, -) while those on the south/west side will be (+, +).[6]
With this information, I can use cluster analysis to identify stacks, or “addresses”, on each side of a candidate road segment. Since we expect the distances between cases in a stack or “address” to be small, I use an arbitrarily low but reasonable cutoff “height” of 0.05. The two dendrograms below represent the North and South sides of Broad Street. We can see that this procedure correctly classifies the 8 different groups or stacks that can be seen in the figure above and that can be verified by looking at Snow’s map.
### Notes
1. To reduce computation, I only consider road segments with at least one endpoint within approximately 27 meters of the bar (i.e., a distance of 0.5 on the map’s scale).
2. This is further exacerbated by the fact that the width of roads is not always taken into consideration. This creates instances where cases on one side of the street appear to be closer to the road than those on the other. One particular case of note is Portland Mews. On the map, Portland Mews is a large yard or square. But in Dodson and Tobler’s data, it is simply a set of line segments, which might reflect a roadway through the location. As a result, the three cases there appear to belong to a single cluster instead of two separate ones. I manually chose the more distant road segment on Portland Mews that better reflects the orientation of bars on Snow’s map.
3. Specifically, I use `stats::hclust()`, which by default employs a “complete” method of agglomerative hierarchical clustering.
4. What makes this possible is that, by design, “vertical” dispersion, the spread of x-y points along the orthogonal axis, is much greater than the “horizontal” dispersion, the spread of orthogonal points along the road’s line segment. In an ideal world, the “horizontal” dispersion would be zero and all points in a stack would be perfectly aligned. In practice, this is not the case.
5. In Snow’s map, one can see 5 addresses on the north side of the block of Broad Street between Dufours Place and Poland Street (the road running north-south just visible on the east or left side of the graph) but only 4 of those addresses fall on the street segment in Dodson and Tobler. The reason for the discrepancy is that intersections with other streets, regardless of which side of the road they may fall, define the end points of road segments. This is why the south side of this segment of Broad Street only includes 4 addresses: the endpoints are defined by Dufours Place on the left (west) and Cambridge Street on the right (east).
6. The two limiting cases occur when roads run either along the East-West axis (i.e., 0 degrees) or the North-South axis (i.e., 90 degrees). For the former, cases on the north side will be (0, +) while those on the south side will be (0, -); for the latter, cases on the east side will be (+, 0) while those on the west side will be (-, 0).
You can’t perform that action at this time. | 1,508 | 6,843 | {"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-2020-05 | latest | en | 0.894288 |
http://www.gurufocus.com/term/e10/PRMW/E10/Primo%2BWater%2BCorp | 1,490,702,494,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189734.17/warc/CC-MAIN-20170322212949-00379-ip-10-233-31-227.ec2.internal.warc.gz | 532,541,392 | 28,038 | Switch to:
GuruFocus has detected 6 Warning Signs with Primo Water Corp \$PRMW.
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Primo Water Corp (NAS:PRMW)
E10
\$0.00 (As of Dec. 2016)
E10 is a concept invented by Prof. Robert Shiller, who uses E10 for his Shiller P/E calculation. E10 is the average of the inflation adjusted earnings of a company over the past 10 years.
Primo Water Corp's adjusted earnings per share data for the three months ended in Dec. 2016 was \$-0.410. Add all the adjusted EPS for the past 10 years together and divide 10 will get our e10, which is \$0.00 for the trailing ten years ended in Dec. 2016.
As of today, Primo Water Corp's current stock price is \$13.38. Primo Water Corp's E10 for the quarter that ended in Dec. 2016 was \$0.00. Primo Water Corp's Shiller P/E Ratio of today is .
Definition
E10 is a concept invented by Prof. Robert Shiller, who uses E10 for his Shiller P/E calculation. When we calculate the today's Shiller P/E ratio of a stock, we use todays price divided by E10.
What is E10? How do we calculate E10?
E10 is the average of the inflation adjusted earnings of a company over the past 10 years. Lets use an example to explain.
If we want to calculate the E10 of Wal-Mart (WMT) for Dec. 31, 2010, we need to have the inflation data and the earnings from 2001 through 2010.
We adjusted the earnings of 2001 earnings data with the total inflation from 2001 through 2010 to the equivalent earnings in 2010. If the total inflation from 2001 to 2010 is 40%, and Wal-Mart earned \$1 a share in 2001, then the 2001s equivalent earnings in 2010 is \$1.4 a share. If Wal-Mart earns \$1 again in 2002, and the total inflation from 2002 through 2010 is 35%, then the equivalent 2002 earnings in 2010 is \$1.35. So on and so forth, you get the equivalent earnings of past 10 years. Then you add them together and divided the sum by 10 to get E10.
For example, Primo Water Corp's adjusted earnings per share data for the three months ended in Dec. 2016 was:
Adj_EPS = Earnigns per Share / CPI of Dec. 2016 (Change) * Current CPI (Dec. 2016) = -0.41 / 241.432 * 241.432 = -0.410
Current CPI (Dec. 2016) = 241.432.
Primo Water Corp does not have a history long enough to calculate E10. Therefore GuruFocus does not calculate it.
Explanation
If a company grows much fast than inflation, E10 may underestimate the company's earnings power. Shiller P/E Ratio can seem to be too high even the actual P/E is low.
For the Shiller P/E, the earnings of the past 10 years are inflation-adjusted and averaged. The result is used for P/E calculation. Since it looks at the average over the last 10 years, the Shiller P/E is also called PE10.
The Shiller P/E was first used by professor Robert Shiller to measure the valuation of the overall market. The same calculation is applied here to individual companies.
Primo Water Corp's Shiller P/E Ratio of today is calculated as
Shiller P/E Ratio = Share Price / E10 = 13.38 / 0.00 =
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Be Aware
Shiller P/E Ratio works better for cyclical companies. It gives you a better idea on the company's real earnings power.
Related Terms
Shiller P/E Ratio
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Primo Water Corp Annual Data
Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 Dec14 Dec15 Dec16 e10 0.00 0.00 0.00 0.00 0.00 0.00 0.00 -4.18 0.00 0.00
Primo Water Corp Quarterly Data
Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 Dec16 e10 0.00 -4.18 -4.04 -3.91 0.00 0.00 0.00 0.00 0.00 0.00
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GuruFocus Affiliate Program: Earn up to \$400 per referral. ( Learn More) | 1,091 | 3,981 | {"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} | 3.171875 | 3 | CC-MAIN-2017-13 | latest | en | 0.940749 |
http://hubpages.com/education/Vedic-Mathematics---Easy-and-Faster-Calculation-of-Cubes | 1,485,051,359,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281331.15/warc/CC-MAIN-20170116095121-00529-ip-10-171-10-70.ec2.internal.warc.gz | 134,504,354 | 15,353 | # Vedic Mathematics - Easy and Faster Calculation of Cubes
Calculating Cubes faster
## Calculating Cubes easy and faster
To find out cubes of the two digit number, we generally take help of the following formula:
(a+b)’cubes= a’cube+3a’square b+ 3ab’square +b’cube.
This can be written as:
a’cube +a’square b+ab’square +b’cube
2a’square b 2 a b’square
--------------------------------
We have simply broken 3a’square b and 3ab’square in to two parts a’square and a b’square and 2 a b’square, to simplify the matter.
In the above formula, we see that the terms a’cube, a’square b, a b’square and b’cube are placed at the top and the 2a’squareb and 2ab’square are placed at the bottom.
The complete formula comes into being when you add the terms at the top to those at the bottom.
If we scrutinize the top term closely, then we find that:
A’cube x b/a=a’square b
A’square b x b/a= a b’square and a b; square and a b’square x b/a= b’cube
The common ratio between the top terms is b/a. this is the ultimate finding. We have to dig out b/a and our desired result will be there. Let me explain with the help of the example:
1 2’cube, we have a=1,b=2 and b/a=2
a b
Steps:
• Our first term is a’cube= 1’ cube=1.
• The second term is a’square b= a’ cube x b/a= 1x2=2.
• The third term is a b’square= a’square b x b/a= 2x2=4.
• The fourth term is b’ cube= a b’square x b/a= 4 x 2=8.
• Put all this in the first row, maintaining a space.
• For the second row, double the two middle terms i.e. a’ square b=2 so that 2 a’ square b=4 and a b’square=4 so 2 a b’square=8.Second row comes as 4 and 8.
1 2 4 8
4 8
-----------------
1 - Remainder at each stage
Let us find 16’ cube; here a=1, b=6 and b/a=6. a’cube=1
16’cube=1 6 36 216
12 72
----------------------------
3 12 21 --- Remainder at each stage.
Steps:
• From the digit in the right 216, the unit digit 6 is retained as a part of the answer and the remainder 21 is added to the left
• After adding remainder 21 to (36+72) we get 129. 9 is retained as a answer and 12 is added to the left
• On adding the remainder 12 to the left we get 30. 0 is retained and 3 is added to the left.
• Adding the digits on the extreme left gives 4. The answer is 4096. Our procedure is over now.
## More by this Author
shubi 7 years ago
I want exact method of finding cube roots using vedic maths. I want the methods in points i.e, in steps. THANX
Obhund 6 years ago
Hi Saurabh!
Thanks a lots for this hub.
I was getting crazy try to understand, Vedic method for cubes and just your video make all clear and simply.
Thanks again.
Obhund
Mahima Kaushik 6 years ago
This is an awsome site wid d best tricks.........really helpfull
Umair Zubair Shaikh 6 years ago
This is Awesome !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Mahendra 6 years ago
Wowwwwwwwwww!!!!!!!!!!1
KT 5 years ago
Do you run an online class by any chance?
penta 4 years ago
Thanks a lot, 2 simple
rajesh khanna 4 years ago
Wow......!!!!! What an excellent short cut. I liked it .
Shwetank pandey 4 years ago
Its a heart breaking for me to see this formula here..of finding squares and cubes, I made this formula at my own..I told many of my friends that I made it few days back,they are very excited at my approach I think Its is a unique approach by me but. I was totally unaware of vedic mathematics, I am trying to make another formulaes at higher level
Jitesh 4 years ago
It's a magic
K.R.Mohanraj 4 years ago
It's Really amasing....
Akansha 4 years ago
Ha! I learnt this at school, but this was clearer- I was trying to do it this way:-
(a'square*a)+(a square*3b)+(b square*a)+(b square*b)
then we took the last digit of each and carried the rest over to the next ...
nirav shah 3 years ago
very easy method. I like it.
Bhaiya 19 months ago
Awesmmmmm | 1,131 | 3,799 | {"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} | 4.84375 | 5 | CC-MAIN-2017-04 | latest | en | 0.830475 |
https://sciforums.com/threads/towards-ideas-on-a-quantum-theory-of-gravity.159827/page-4 | 1,721,465,146,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515079.90/warc/CC-MAIN-20240720083242-20240720113242-00860.warc.gz | 443,825,696 | 19,336 | # Towards Ideas on a Quantum Theory of Gravity
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Which are you referring to? If you mean the opening post, we argued a contracted form and made it proportional. Later I work out the christoffel symbols directly, and show the actual form of the equations, no proportional symbols.
The very first equation of the first post in this thread. It is impossible for $$\mathbf{G}$$ to be a tensor. Because you said:
$$\mathbf{G} \approx \frac{1}{\Delta L\Delta t}\ \geq \frac{c^4}{G \hbar}$$
That middle term is a number. So the left-hand side must be too. Your $$\mathbf{G}$$ is a number, not a tensor.
Please... do I need to define anything for you? You seem to be here for a fight, nothing more, nothing less.
Erm, you started the topic? On a forum? Thus inviting comments/replies?
Oh I am sorry, can you not deal with this, without the dummy variables?
$$G_{ij} \leq (\ell^2_{ij})^{-1}$$
Of course that is without saying, those indices are making a statement about the curvature being proportional to a given length squared (Planckian in fact). $$G_{ij}$$ is an antisymmetric tensor and is antisymmetric in the indices.
That ''middle term being a number'' follows the same principle
$$(\Delta X_{i} \Delta X_j)^{-1}$$
edit - the inverse, because tensor has dimensions of 1/length^2
Oh I am sorry, can you not deal with this without the dummy variables?
As I said, my GR is rusty, sorry about that. I didn't realize you were silently dropping the indices.
$$G_{ij} \leq \ell^2_{ij}$$
Ah, I see. One more question: are the ij-indices just the spatial ones, or do they include time?
Also, earlier you said:
$$\frac{2}{\Delta r^2} = \frac{1}{\Delta X_a \Delta X_b} \geq \frac{1}{\ell^2}$$
I assume I need to add the a and b indices back to the first term as well? But you said:
The r is radius, L is a length, and t is time.
How can r be a radius when it has indices? A radius just a number, not a vector!
That ''middle term being a number'' follows the same principle
$$(\Delta X_{i} \Delta X_j)^{-1}$$
edit - the inverse, because tensor has dimensions of 1/length^2
You're not summing over the indices, so it's most definitely not a number.
As I said, my GR is rusty, sorry about that. I didn't realize you were silently dropping the indices.
Ah, I see. One more question: are the ij-indices just the spatial ones, or do they include time?
Also, earlier you said:
I assume I need to add the a and b indices back to the first term as well? But you said:
How can r be a radius when it has indices? A radius just a number, not a vector!
No you don't understand, its not like that. If you look carefully, what I have written in fact is
$$(\Delta r^2)^{-1}$$
This uses the change operator, its a deviation from some value. And yes, when we had the contracted form of $$G$$ we constructed the curvature of sphere - I later drop this approach for a more direct tackle on the quantum gravity aspect. What you don't realize, the equations you are citing from, was never my quantum gravity approach, they were preliminary investigations into the curvature of a sphere in the use of some deviation of curvature for a geon.
As I said, this is all dropped, you'd be having a less hard time if you read through and take my approach in the most recent form.
You're not summing over the indices, so it's most definitely not a number.
Right... That's why I quoted you and corrected you. This is a tensor, an antisymmetric tensor. If you read the newest work and actually follow the newest construction of the work, all that will be clear.
As I said, my GR is rusty, sorry about that. I didn't realize you were silently dropping the indices.
They are spatial and temporal, hence my understanding of using two connections as deviation from classical vacuum
$$[\nabla_i,\nabla_j]$$
This is like seeing gravity as a Von Neumann operator.
Right... That's why I quoted you and corrected you. This is a tensor, an antisymmetric tensor.
Remember when you said:
$$\frac{2}{\Delta r^2} = \frac{1}{\Delta X_a \Delta X_b} \geq \frac{1}{\ell^2}$$
So if it's an antisymmetric tensor, so is $$\frac{2}{\Delta r^2}$$, and thus r is not a radius in the normal sense of the word.
Last edited:
You poor thing, you are so lost. You are still concentrating on the radius when I told you, that isn't even the right model for my quantum gravity.
You can't talk about what I am talking about now, with being stuck on page 1. You need to read on and see how the model is given now. You just don't get this bit do you?
You poor thing, you are so lost. You are still concentrating on the radius when I told you, that isn't even the right model for my quantum gravity.
You can't talk about what I am talking about now, with being stuck on page 1. You need to read on and see how the model is given now. You just don't get this bit do you?
Erm, dear SimonsCat, I'm not the one introducing an undefined variable r, then calling it a radius, and then calling it a tensor. If your notation and terminology is so far from mainstream, how can I have any hope understanding it?
Since it is obvious to me that Geon is a sock and still adding nothing to the forum but noise, I guess it is time... Click
Erm, dear SimonsCat, I'm not the one introducing an undefined variable r, then calling it a radius, and then calling it a tensor. If your notation and terminology is so far from mainstream, how can I have any hope understanding it?
Now you are lying.
I did nothing of the sort, you did. I told you we used a contracted solution of the equations, which was solved for the Ricci curvature of a sphere - I made that early solution proportional and today, have dropped that model entirely.
It's you who isn't
1) listening to me
2) making shit up as you go along
You asked me early on, what G was. I told you in the context we came to understand, was an antisymmetric tensor. Since I told you, you cannot derive that understanding from post 1, you've done everything to concentrate on post 1, as if it proves some point and it never did. I explained to you what those early solutions where and why I looked into them. I abandoned that investigation for a direct look into the christoffel symbols to find the relationships.
Now you are lying.
I did nothing of the sort, you did. I told you we used a contracted solution of the equations, which was solved for the Ricci curvature of a sphere - I made that early solution proportional and today, have dropped that model entirely.
It's you who isn't
1) listening to me
2) making shit up as you go along
I am literally trying to understand the first few equations of your first post. You seem to be unable to define its terms in a clear and meaningful manner. Every time you explain a bit, it seemingly contradicts the mainstream interpretation of the terminology. Without a translation manual of some kind, how can anybody have any hope of understanding this? Without a translation manual, it's quite simply word salad, seeing as the meaning of all the terms and words are unclear.
I mean, let's look at the first equation of the first post again now that we've covered some ground.
$$\mathbf{G} = \frac{2 - n}{2}\mathbf{R}g = \mathbf{R}^{ij}g_{ij} - \frac{1}{2}n\mathbf{R}g_{ij}g^{ij} =$$...
Adding the indices for the bold G back, we get non-sense. The third term is summed, and thus a number, but the first is your G antisymmetric tensor. A tensor is not a number. (Also sloppy to use the same indices for the two terms, but okay.)
I am listening, I'm just not understanding what you are trying to say because it's so non-mainstream in its usage of terms.
You asked me early on, what G was. I told you in the context we came to understand, was an antisymmetric tensor.
Earlier you said:
Bold G is the Einstein tensor, describes curvature.
The Einstein tensor is symmetric. So you're so far left-field, you're not even doing GR anymore. I strongly suggest you stop using its terminology then, because it leads to the confusion I'm expressing right now.
And just to clear my name...
Now you are lying.
I did nothing of the sort, you did. I told you we used a contracted solution of the equations, which was solved for the Ricci curvature of a sphere - I made that early solution proportional and today, have dropped that model entirely.
Let's break that down.
"I'm not the one introducing an undefined variable r,": See the first equation in post 1.
"then calling it a radius," See the last sentence in post 54.
"and then calling it a tensor." See my conclusion in post 64. True, you never explicitly called it a tensor, but this is an inescapable conclusion from your statements earlier and in post 63; see the rest of post 64.
It's you who isn't
2) making shit up as you go along
If applying mainstream understanding of GR is "making shit up as you go along", then I'm totally guilty, yes.
I am finished with this gravitational investigation for the moment. I've ended up with some key equations in context of the spacetime uncertainty - working with units c=1, the spacetime uncertainty principle is
$$\Delta x \Delta t \leq \ell^2$$
From it, we described the commutation as an antisymmetric tensor - but it was also written a geometric interpretation from an inequality arising from the Cauchy Schwarz relationship in a Hilbert space.
$$\sqrt{|<\nabla_i^2>< \nabla_j^2>|} \geq = \frac{1}{2} <\psi|[\nabla_i,\nabla_j]|\psi> = \frac{1}{2} <\psi | iR_{ij}| \psi >$$
with the connections identified as giving
$$[\nabla_i, \nabla_j] = (\partial_i \partial_j + \Gamma_i \partial_j + \partial_i \Gamma_j + \Gamma_i \Gamma_j) - (\partial_j \partial_i + \partial_j \Gamma_i + \Gamma_j \partial_i + \Gamma_j \Gamma_i)$$
$$= -[\partial_j, \Gamma_i] + [\partial_i, \Gamma_j] + [\Gamma_i, \Gamma_j]$$
My hope was that by implementing the Cauchy Schwarz inequality, spacetime would have a natural mechanism to form fluctuations in spactime - its spacetime uncertainty principle though still has the gravitational side to be interpreted, which has been shown to be part of the Riemann tensor. That interpretation won't come soon - I still need to investigate non-unitarity in the theory to investigate a finite-dimensional theory which [appears] to be more suitable for a quantum theory.
But yeah, for now, this gravitational theory is going quiet until I have some new leads.
No no... let me break it down for you. This is the post you want if you want the correct form of the equations. I ask you, why are you going to an early investigation which didn't even take into account the christoffel symbols? The uncertainty arises as part of the antisymmetric tensor... If you want the correct approach to understand it, why are you avoiding the final, approved form?
Earlier you said:
The Einstein tensor is symmetric. So you're so far left-field, you're not even doing GR anymore. I strongly suggest you stop using its terminology then, because it leads to the confusion I'm expressing right now.
Not always symmetric. There are non-symmetric tensor theories of $$G$$ out there. But that's neither here nor there, since G is related to the Riemann tensor, we can certainly say the tensor $$R_{ij}$$ is antisymmetric in those two indices.
To understand symmetric and antisymmetric in terms of indices, I suggest you do some reading http://www.physicspages.com/2013/03/22/tensors-symmetric-and-anti-symmetric/. Especially from this part
''A higher rank tensor can be symmetric or anti-symmetric in any pair of its indices, provided both indices are either upper or lower.''
As you can see we have only been concerned with two connections $$[\nabla_i,\nabla_j]$$. Which means we have only been concerned with an antisymmetric object.
Status
Not open for further replies. | 2,895 | 11,754 | {"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} | 2.84375 | 3 | CC-MAIN-2024-30 | latest | en | 0.889204 |
https://forceinphysics.com/unbalanced-force/ | 1,713,073,064,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816864.66/warc/CC-MAIN-20240414033458-20240414063458-00811.warc.gz | 236,012,890 | 27,589 | # Unbalanced force
The information on this page is ✔ fact-checked.
An unbalanced force, often encountered in everyday scenarios, is a force that leads to a change in an object’s motion. Unlike balanced forces, unbalanced forces do not offset or cancel each other out. Instead, they create a net force that acts on the object and causes it to accelerate or decelerate. For instance, when a person pushes a table lying on the floor, the force they apply is unbalanced as it exceeds the opposing forces like friction. Consequently, the table begins to move in the direction of the applied force. In this way, unbalanced forces disrupt the equilibrium of an object and produce a noticeable effect on its motion.
Contents
## Examples
### Football
When an individual kicks a football, the force exerted by their leg on the ball acts as an unbalanced force and initiates the ball’s motion. The force of the kick overcomes the ball’s initial state of rest and propels it forward in the direction of the kick. As a result of this unbalanced force, the football starts moving, illustrating the concept of unbalanced forces causing a change in the object’s motion.
### Trolley
When a person pulls a heavy luggage trolley, an unbalanced force is at play. Unbalanced forces arise when the net force acting on an object is non-zero, leading to changes in motion. As the person exerts force on the trolley to move it, this applied force becomes unbalanced, resulting in acceleration. The trolley will accelerate if the force applied by the person is greater than the opposing forces, such as friction and air resistance. On the other hand, if the opposing forces outweigh the applied force, the trolley will decelerate or come to a stop. This example illustrates how unbalanced forces cause objects to change their motion when external forces are not in equilibrium.
### Seesaw
A seesaw perfectly illustrates the concept of unbalanced force. When two individuals of different weights sit on each end, an imbalance of forces occurs. The side with the heavier person exerts a stronger downward force, causing that end to lower while the other end rises. This unequal distribution of forces results in an unbalanced force and leads to the dynamic motion of the seesaw. The seesaw continues to tilt until the forces become balanced or until an external force comes into play. This example demonstrates how unbalanced forces cause objects to move when opposing forces are not in equilibrium, creating an exciting playtime experience on the seesaw.
### Sliding book
When a book is placed on a slanted surface, it experiences an unbalanced force that initiates the sliding motion. This unbalanced force arises from the difference in the magnitudes of the forces acting on the book. On a flat surface, the book remains in equilibrium, with the upward force equaling the downward force of gravity. However, on a slanted surface, the downward force becomes stronger and surpasses the upward force. Consequently, the book starts to slide as the unbalanced force disrupts the balance between the opposing forces. This sliding motion serves as a clear demonstration of how an unbalanced force affects the book’s behavior when placed on a slant.
### Accelerating car
When a car accelerates forward, it encounters an unbalanced force that propels its motion. Before acceleration, the car remains balanced, with the upward force (from the ground) and the downward force (gravity) in equilibrium. However, as the engine starts, it exerts a force on the car in the forward direction. This force overtakes the opposing forces, disrupting the equilibrium and causing the car to accelerate. Hence, the unbalanced force generated by the engine enables the car’s acceleration and showcases the impact of unbalanced forces on an object’s motion.
### Wooden block
When exerting force on a wooden block, causing it to move forward, the occurrence of an unbalanced force is evident. Initially, the wooden block remains in a state of equilibrium, with upward and downward forces counteracting each other, thus keeping it stationary. However, once an external force is applied to the block, it experiences a slight forward movement, which indicates the presence of an unbalanced force. This unbalanced force disrupts the equilibrium and initiates the block’s motion in the direction of the applied force. Hence, the movement of the wooden block demonstrates the impact of an unbalanced force on its motion. | 899 | 4,478 | {"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} | 3.6875 | 4 | CC-MAIN-2024-18 | longest | en | 0.914143 |
https://search.r-project.org/CRAN/refmans/crank/html/meanranks.html | 1,722,824,116,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640427760.16/warc/CC-MAIN-20240805021234-20240805051234-00310.warc.gz | 425,012,234 | 2,253 | meanranks {crank} R Documentation
## Calculate mean ranks with possible missing values
### Description
Calculates mean ranks where some ranks may be missing
### Usage
meanranks(x,allranks=NULL,labels=NULL,rankx=FALSE)
### Arguments
x A matrix of ranks that may contain ties and NAs. Objects ranked are assumed to be columns and ranking methods rows. allranks An optional list of all ranks that might have been made. labels Optional labels for the ranks. rankx Whether to convert competition ranks, or any other set of numeric values, into the usual mean rankings for ties.
### Details
meanranks’ calls ‘muranks’ to complete the rank matrix before calculating the mean ranks for each column if there are any NAs in ‘x’.
Note that ‘cats2ranks’ interprets each value in ‘x’ as a nominal level variable and its index as the rank, while ‘meanranks’ interprets values as ordinal level (ranks). Thus if a matrix or data frame of category labels is passed to ‘meanranks’, it will not give the correct mean ranks.
### Value
A list with the following components:
ranks ‘x’ with any NAs replaced by the mean of unallocated ranks for each row. labels The vector of labels, defaulting to the integers 1:allranks. mean.ranks A vector of mean ranks for each value of allranks.
Jim Lemon
### Examples
# simulate "best/worst" ranking
x<-matrix(NA,nrow=10,ncol=10)
for(i in 1:10) {
nbest<-sample(2:5,1)
best<-1:nbest
nworst<-sample(1:5,1)
worst<-(11-nworst):10
rankpos<-sample(1:10,nbest+nworst)
x[i,rankpos]<-c(best,worst)
}
x
meanranks(x)
[Package crank version 1.1-2 Index] | 456 | 1,597 | {"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-33 | latest | en | 0.700403 |
https://www.polytechforum.com/metalworking/bursting-speed-of-flywheels-and-overspun-ball-bearings-109373-.htm | 1,670,408,853,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711151.22/warc/CC-MAIN-20221207085208-20221207115208-00860.warc.gz | 975,666,810 | 15,993 | # Bursting speed of flywheels (and overspun ball bearings)
• posted
Machinery Handbook has long had a section of flywheels. In the 20th
edition, on page 346, states that all that matters is surface speed at
the periphery, and the tensile strength of the steel, and gives a
formula: V=Sqrt[10*s], where V is surface speed in feet per second, and
s is tensile strength in pounds per square inch.
Let us assume that s= 300,000 psi, the cited strength of ball bearing
race steel. Sqrt[10*
300000]= 1732 fps.
A bearing 1.75 inches in diameter will have a circumference of
(1.75)(3.1416)/12= 0.4561 feet, so 1732 fps implies 3,781 rps, or
226,832 rpm.
The speed of sound is about 300 meters per second at sea level, or about
900 feet per second, so the surface speed of the outer race is 1732/900=
1.92 times the speed of sound at sea level.
If the airjet is at the speed of sound, and is impinging on the balls,
the outer race will go twice the speed of sound.
If the bearing has ten balls, the siren tone will be at 3,781*10= 37,810
Hz, well into the ultrasonic, as people have observed.
The guy that did the experiment showing a max speed of ~20,000 rpm for
whatever reason did not achieve full speed, as 20,000 rpm isn't nearly
enough, and yet people have no problem causing bearings to burst from
overspeed.
Basically, it all fits together. Then it bursts.
Joe Gwinn
• posted
Think of rotation as the average motion and resonant acoustic vibration as the instantaneous motion. The stress from the maximum vibration excursion can be much greater than that from the rotation.
• posted
Joseph,
The balls of the bearing whizzing around do not carry any hoop stress, but produce a radial outward force on the outer bearing ring analogous to the internal pressure of a pressure vessel.
Since you are into the arithmetic and I am too lazy to figure it out, what is the fraction of the outer ring hoop stress due to the orbiting balls?
Just curious.
Wolfgang
• posted
I'd love to use a high speed camera to take a picture of the burst. I wonder if a light beam could be used for this. I'm picturing (sp?) 3 mirrors, a laser pointer, and a photodetector of some sort to trigger the shutter. Maybe use a very bright halogen light to illuminate the area. Use the mirrors to make a box shaped area with the laser. Maybe use more than three mirrors to make a cube shaped area. I guess if the bearings are exploding at 25000 rpm then if my math is right the pieces will be moving at about 218 feet per second. With a 1/1000 shutter speed it looks like the parts would travel about 2.6 inches. Maybe a better solution is to leave the shutter open and use a flashlamp instead. Hmm. ERS
• posted
-> IF
• posted
None, I think, because the flywheel theory depended only on the surface speed at the rim of the flywheel, and not at all on what was inside.
Joe Gwinn
• posted
From Machinery handbook - 22nd edition. (page 226) "The bending stresses in the rim of a flywheel may exceed the centrifugal (hoop tension) stress predicted by the simple formula s = V(squared) divided by 10 by a considerable amount. See relevant section for further edification.
Have fun. Ken.
• posted
I think the guy claiming 20,000 rpm max is wrong, and the actual rotational rate is ten times that. So, it would take a microsecond light pulse to stop the pieces, which are moving at Mach 2. This is high-powered rifle bullet speed.
An easier way to measure the speed of the fragments is two curtains of fine wire - measure the time delay between disruption of the inner curtain and the outer curtain. One can also use disruption of a curtain to trigger a flash.
Joe Gwinn
• posted
That's if you try to change the axis of rotation while the flywheel is madly rotating.
Joe Gwinn
• posted
The formula in MH is an approximation that works for steel and materials with similar specific gravity. The real formula is:
stress = (density / gravity) * radius^2 * angular velocity^2
or
angular velocity = sqrt((stress * gravity) /
where angular velocity is in radians/s and density in weight/unit volume.
You *must* account for the balls, which is why I've been using 1300FPS as the limit for 300 ksi steel rather than 1700FPS. Based on a SWAG that the balls weigh a bit less than the race I used a density of 0.5lb/in^3 in the formula, rather than steel's actual 0.28lb/in^3.
Ned Simmons
• posted
No, that's if you're talking about a real flywheel with a thick rim and spokes. The formulas we've been bandying about are only strictly applicable to thin cylinders.
Ned Simmons
• posted
It's not. Show me a reference to a blowgun that produces a supersonic air jet with shop air and I'll reconsider.
Okay, since this keeps coming up, despite what seems common sense to me, I set up a test myself. That's a 6204 bearing with the seals removed, the grease washed out, and relubed with a few drops of light spindle oil. There's a paint mark on the race and the ball cage.
I spun the bearing up and measured the difference between the speed of the ball cage and the outer race at several speeds between 1400 and 5500 RPM. The difference in the angular velocity in all cases was 100~200RPM, i.e., the race was going 5-10% faster than the balls. Exactly as you'd expect in an unloaded bearing with internal clearance where the balls are free to slip relative to the races. As I've said at least twice before, this clearance will only increase with speed.
I checked this as well just to make sure the assumption that the frequency of the sound from the bearing does in fact correspond to the ball passing frequency. I got my teenage son, who's involved in composing synthesized music, to set up his laptop with an FFT to monitor the bearing siren tone. Agreement was within a few percent, probably as good as could be expected with me getting a strobe fix while asking him to read the frequency.
Because it's well known that, for the sort of bearing we're talking about, speeds in the few tens of thousands of RPMs are the lubrication limit for properly mounted bearings with elaborate mist lube systems. It's no surprise at all that a loose bearing that's just had all its lubrication removed would fail at somewhat higher speeds.
Ned Simmons
• posted
I didn't say supersonic, I said sonic (as the upper limit). The airflow chokes in the orifice, being limited to the speed of sound, so this is the upper limit (unless one has a nozzle that looks like the back end of a rocket engine, with an expansion bell).
It appears to have eight balls. What's the OD?
I see one problem in the photos: That long thin copper tube will not achieve anything like the airspeed that a proper nozzle will achieve.
I would suggest using a piece of 3/8" tubing with a machined brass nozzle hard soldered into one end. The brass nozzle would have a 60-degree (included angle) cone inside, going from 3/8" to 0.014" diameter at the face.
I would put around the bearing a piece of heavy metal pipe lined on the inside with wooden staves, to stop the shrapnel. Even if you believe that the bearing won't burst.
These are very low rotational speeds. As the speed increases, won't centrifugal force pin the balls against the inside of the outer race, reducing or eliminating slippage?
Good. The "siren" theory is confirmed.
So, how do we explain the reports that the tone went ultrasonic just before the bearing exploded? With eight balls, this implies 20000/8= 2,500 rps, or 150,000 rpm, a factor faster than the 20,000 rpm discussed here.
Also unexplained is the essentially perfect symmetry of the explosions.
While I don't doubt that being run bare at such high speeds chews the bearing up pretty fast, the guy I was mentioning also used the long thin air tube, and so didn't achieve full airspeed.
Joe Gwinn
• posted
Hmm. On second thought, I think you're right.
With balls rolling on the outer ring at very high speed, we should see some metal fatigue effects from the cyclic bending seen as the balls pass by.
Wonder if we are running through the fatigue life of the steel, which then cracks, precipitating the burst? One problem with this theory is that one would not expect this mechanism to lead to the essentially symmetrical explosions that have been universally reported.
Joe Gwinn
• posted
My theory was that the balls are less dense than solid disk, and more or less equivalent to (moving) spokes), so the MH formula would apply.
Yes, MH is full of practical approximations, and they do say that steel is assumed.
What's "gravity", and how does it differ from "density"? This theory cannot depend on the presence of a planet or its gravitational field.
Where are you getting these better formulas? I'd like to read up on it.
If the balls weigh less than the race, the 0.5 lb/in^3 sounds wrong, as it's more than that of solid steel, 0.28 lbs/in^3. Perhaps some more explanation is in order.
Joe Gwinn
• posted
Gravitational acceleration, to account for the fact that a pound mass exerts a pound force in a gravitational field of 386 in/s^2 and we're calculating the forces exerted by a lump of material in a rotating frame with a different acceleration.
I got the formula from "Roark's Formulas for Stress and Strain", but that's not the place to go for an explanation of the derivation. "Elements of Strength of Materials" (Timoshenko) has the derivation for the stress in a cylindrical pressure vessel, which is a very similar problem.
As an intuitive approach, you can think of the problem as two semicircular segments joined together. What's the force required at the joints to hold the two halves together? Divide that by cross section to get stress.
It's a fudge. As Wolfgang said, the balls are exerting a force on the race but don't increase it's strength. To account for the balls' additional mass I added it to the outer race's mass, but didn't change the cross section (strength) of the race. In other words, I decreased the strength to density ratio of the race to account for the loose balls.
Ned Simmons
• posted
47mm - 1.85"
I have no interest in exploding a bearing . I just wanted to get it spinning fast enough to run the tests described.
Even at these speeds I don't imagine there's much slippage between the balls and the outer race - clearly there isn't. Where the balls *are* slipping is relative to the inner race, minimizing any speedup due to planetary action.
I can't. Maybe Eric's hearing is worse than he thinks. I wouldn't know how low my upper limit is in one ear (starts rolling off at a few kHz) if I didn't fail the hearing test in grammar school every year.
As I said before, it's easier to explain the lack of serious injuries if the available energy is much lower. Perhaps the bearings that exploded with bad consequences has selectively thinned the reporters .
I agree that higher speeds than 20 KRPM should be attainable, but the deterioration will limit the speed, and that may be why he couldn't go faster. As the bearing gets beat up it'll take more power to keep it spinning at a given rate.
Ned Simmons
• posted
How fast did you get it to go?
So the balls and outer race spin together more or less as a unit? I guess I don't quite trust that so small a clearance will really always do the job. All it would take to get some real traction would be for the human holding the inner race to move slightly, twisting the axis of rotation a few degrees, causing the balls to come up against the sides of the groove in the inner race, pushing against the gyroscopic forces keeping the outer race from turning with the inner race.
The lack of lubrication will make for more traction, especially if the balls are galling with the races.
A test jig where the bearing is clamped to a bench would never see this effect.
Darwinism in action. But don't you think we would have heard the stories, if there were stories to be heard?
Will deterioration really be that much of a limit on an unloaded bearing, especially if it isn't in contact with the inner race all that much? And, the airjet has plenty of power.
Joe Gwinn
• posted
Ah. The conversion from pounds (a unit of weight) to slugs (a unit of inertial mass) is partly buried in the formula. In the MH formula, all this disappears into a single unexplained constant.
Thanks. I think I have both books.
And as one increases the number of segments, the formula will approach the continuum case.
Ahh. OK.
Joe Gwinn
• posted
About 6500 RPM. I didn't have the needle valve opened all the way, but I don't think it would have gone much faster with the small tube.
He tests many limits.
There's lots of power in the compressed air - transferring it to the bearing is another matter. Even air motors are notoriously inefficient.
Based on the numbers, and 25 years of machine design experience, I'm very confident that the bearings did not explode from centrifugal force alone. The reports of the noise going ultrasonic do give me pause, but not enough to make me believe that the bearings were spinning at 1300 FPM. I guess someone needs to volunteer to explode a few properly instrumented bearings. Not me, I'm off to Boston for college visits with the boy tomorrow.
Ned Simmons
PolyTech Forum website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners. | 3,093 | 13,266 | {"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} | 3.453125 | 3 | CC-MAIN-2022-49 | latest | en | 0.925861 |
https://metanumbers.com/58136 | 1,620,776,634,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243990419.12/warc/CC-MAIN-20210511214444-20210512004444-00490.warc.gz | 422,517,507 | 7,592 | ## 58136
58,136 (fifty-eight thousand one hundred thirty-six) is an even five-digits composite number following 58135 and preceding 58137. In scientific notation, it is written as 5.8136 × 104. The sum of its digits is 23. It has a total of 6 prime factors and 24 positive divisors. There are 26,208 positive integers (up to 58136) that are relatively prime to 58136.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 23
• Digital Root 5
## Name
Short name 58 thousand 136 fifty-eight thousand one hundred thirty-six
## Notation
Scientific notation 5.8136 × 104 58.136 × 103
## Prime Factorization of 58136
Prime Factorization 23 × 132 × 43
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 6 Total number of prime factors rad(n) 1118 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 58,136 is 23 × 132 × 43. Since it has a total of 6 prime factors, 58,136 is a composite number.
## Divisors of 58136
1, 2, 4, 8, 13, 26, 43, 52, 86, 104, 169, 172, 338, 344, 559, 676, 1118, 1352, 2236, 4472, 7267, 14534, 29068, 58136
24 divisors
Even divisors 18 6 3 3
Total Divisors Sum of Divisors Aliquot Sum τ(n) 24 Total number of the positive divisors of n σ(n) 120780 Sum of all the positive divisors of n s(n) 62644 Sum of the proper positive divisors of n A(n) 5032.5 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 241.114 Returns the nth root of the product of n divisors H(n) 11.5521 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 58,136 can be divided by 24 positive divisors (out of which 18 are even, and 6 are odd). The sum of these divisors (counting 58,136) is 120,780, the average is 503,2.5.
## Other Arithmetic Functions (n = 58136)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 26208 Total number of positive integers not greater than n that are coprime to n λ(n) 1092 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 5874 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 26,208 positive integers (less than 58,136) that are coprime with 58,136. And there are approximately 5,874 prime numbers less than or equal to 58,136.
## Divisibility of 58136
m n mod m 2 3 4 5 6 7 8 9 0 2 0 1 2 1 0 5
The number 58,136 is divisible by 2, 4 and 8.
• Abundant
• Polite
• Practical
## Base conversion (58136)
Base System Value
2 Binary 1110001100011000
3 Ternary 2221202012
4 Quaternary 32030120
5 Quinary 3330021
6 Senary 1125052
8 Octal 161430
10 Decimal 58136
12 Duodecimal 29788
20 Vigesimal 756g
36 Base36 18uw
## Basic calculations (n = 58136)
### Multiplication
n×i
n×2 116272 174408 232544 290680
### Division
ni
n⁄2 29068 19378.7 14534 11627.2
### Exponentiation
ni
n2 3379794496 196487732819456 11423010835191894016 664088157914715950514176
### Nth Root
i√n
2√n 241.114 38.739 15.5278 8.972
## 58136 as geometric shapes
### Circle
Diameter 116272 365279 1.06179e+10
### Sphere
Volume 8.23046e+14 4.24718e+10 365279
### Square
Length = n
Perimeter 232544 3.37979e+09 82216.7
### Cube
Length = n
Surface area 2.02788e+10 1.96488e+14 100695
### Equilateral Triangle
Length = n
Perimeter 174408 1.46349e+09 50347.3
### Triangular Pyramid
Length = n
Surface area 5.85398e+09 2.31563e+13 47467.8
## Cryptographic Hash Functions
md5 a6562b4b35ab5f1fcbcdb871e4f71b11 42273873c0936cb81c8e48eaa90c76b2ac879cbd 7087679b8c68edaa51383bba400f9d66445eae699397ebbe9d1acedac00f5e08 84a46b8e5fb1e79ed3a3e90ebc589130528e1d34f9aaac36a78c996615fdda6733be11db0cee7a02f5c5275fda655f91ac29d2a79ecda50086e8661ccb02bb58 53b890dd55449d5705bd04c3688ba9b240ffd17b | 1,500 | 4,167 | {"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} | 3.5625 | 4 | CC-MAIN-2021-21 | latest | en | 0.79195 |
http://www.jiskha.com/display.cgi?id=1364226835 | 1,495,729,596,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608107.28/warc/CC-MAIN-20170525155936-20170525175936-00616.warc.gz | 541,854,102 | 3,729 | # Physics
posted by on .
A dockworker loading crates on a ship finds that a 22-kg crate, initially at rest on a horizontal surface, requires a 76-N horizontal force to set it in motion. However, after the crate is in motion, a horizontal force of 59 N is required to keep it moving with a constant speed. Find the coefficients of static and kinetic friction between crate and floor.
• Physics - ,
Wc = m*g = 22kg * 9.8N/kg = 215.6 = Wt.
of crate.
Fc = 215.6N.[0o] = Force of crate.
Fp = 215.6*sin(0) = 0 = Force parallel to floor.
Fv = 215.6*cos(0) = 215.6 N. = Force
perpendicular to floor.
Fs = u*Fv = Force of static friction.
Fap-Fp-u*Fv = m*a
76-0-u*215.6 = m*0 = 0.
u*215.6 = 76
u = 0.353 = Coefficient of static friction.
59-0-u*215.6 = m*a =m*0 = 0
u*215.6 = 59
u = 0.274 = Coefficient of kinetic friction. | 283 | 822 | {"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} | 3.734375 | 4 | CC-MAIN-2017-22 | latest | en | 0.903432 |
http://en.wikipedia.org/wiki/Law_of_demand | 1,394,482,618,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394010995802/warc/CC-MAIN-20140305091635-00097-ip-10-183-142-35.ec2.internal.warc.gz | 64,024,769 | 14,887 | # Law of demand
A demand curve, shown in red and shifting to the right, demonstrating the inverse relationship between price and quantity demanded (the curve slopes downwards from left to right; higher prices reduce the quantity demanded).
In economics, the law states that, all else being equal, as the price of a product increases, quantity demanded falls; likewise, as the price of a product decreases, quantity demanded increases.
In other words, the law of demand states that the quantity demanded and the price of a commodity are inversely related, other things remaining constant. If the income of the consumer, prices of the related goods, and preferences of the consumer remain unchanged, then the change in quantity of good demanded by the consumer will be negatively correlated to the change in the price of the good.[1] There are, however, some possible exceptions to this rule (see Giffen goods and Veblen goods).
## Mathematical expression
Mathematically, the inverse relationship may be expressed as a causal relation:
$Q_x = f(P_x)$
```Where, $Q_x$ is the quantity demanded of x goods $f$ is the function of independent variables contained within the parenthesis, and $P_x$ is the price of x goods.
```
Hence, in the above model, the function ($f$) is a varying one: i.e., the law of demand postulates $P_x$ as the causal factor (independent variable) and $Q_x$ as the dependent variable.
## Graphical depiction
A demand curve is a graphical depiction that abides by the law of demand. It shows how the quantity demanded of some product during a specified period of time will change as the price of that product changes, holding all other determinants of the quantity demanded constant. Price is measured on the vertical axis and quantity demanded on the horizontal axis.
There are two important things to note about the demand curve:
• It is downward sloping indicating that between the] price of a product and the quantity demanded a negative or inverse relationship exists. In other words, as the price declines the quantity demanded increases. This is indicated by a downward movement along the demand curve. An increase in price decreases the quantity demanded, and an upward movement along the demand curve occurs.
• The movement along a given demand curve due to a change in price is referred to as "change in quantity demanded". As the price changes, the quantity demanded changes. The term "change in demand" refers to a shift of the demand curve because of factors other than price.
The following statement is therefore regarded as incorrect. Given a demand curve for a product a decrease in the price causes an increase in the demand for the product. It causes an increase in the quantity demanded =={law of demand}==
``` In economics or other words the law of demand states that the quantity demanded and price of a commodity are inversely related other things related constant .If the income of the consumer ,price,of the related goods ,and performance or and prefer of the consumer remain unchanged ,then the change in quantity of goods demanded by the consumer will be negatively correction to the change in the price of the goods.There are however,some possible exception to this rule
=={LAW OF VARIABLE PROPORTION}==
```
"The law of variable proportions essentially explains the relationship between the fixed and variable factors inputs and corresponding change in output {STATEMENT} = AS the proportion of one factors in a combination of factors is increased after point the average and marginal production of that factors will demolished" SYMBOLICALLY: AP=TP -> total
``` QVF -> F(QvF)
Project: Demand
Roll no: 03
```
## Assumptions
Every law will have limitations or exceptions. While expressing the law of demand, the assumption is that other factors of demand, except the price of a good, are unchanged. If they don't remain constant, the inverse relation may not hold well. In other words, it is assumed that the income and tastes of consumers and the prices of related goods are constant. This law operates when the price of the good changes and all other non-price factors do not change.
The main assumptions are:
• The function shows the relationship between Price and Quantity Demanded at a static time (t).
• Habits, tastes and fashions remain same.
• Income of the consumer does not change.
• Prices of related goods remain constant.
• Number of buyers remain constant.
• The commodity is a normal good and has no prestige or status value.
• People do not expect changes in the price.
• Price is independent and quantity demanded is dependent.
• income level should remain constant
## Exceptions to the law of demand
Generally, the amount demanded of a good increases with a decrease in price of the good and vice versa. In some cases, however, this may not be true. Such situations are explained as in below.
### Giffen goods
Initially discovered by [vasudeva], economists disagree on the existence of Giffen goods in the market. A Giffen good describes an inferior good that as the price increases, demand for the product increases. As an example, during the Irish Potato Famine of the 19th century, potatoes were considered one of Giffen good's. Potatoes were the largest staple in the Irish diet, so as the price rose it had a large impact on income. People responded by cutting out on luxury goods such as meat and vegetables, and instead bought more potatoes. Therefore, as the price of potatoes increased, so did the demand.[2] However, this change of demand did not mean movement along the demand curve, but rather a shift of the whole demand curve. What occurred was not an increase in demanded quantity due to increase in price (which would be a violation of law of demand), but rather, a change in the relationship between price and demanded quantity. Thus, the Giffen good confuses the change in demanded quantity with the change in the relationship between price and quantity. .
### Veblen Goods
Some expensive commodities like diamonds, expensive cars, etc., are used as status symbols to display one’s wealth. The more expensive these commodities become, the higher their value as a status symbol and hence, the greater the demand for them. The amount demanded of these commodities increase with an increase in their price and decrease with a decrease in their price. Also known as a Veblen good.
### Expectation of change in the price of commodity
If a household expects the price of a commodity to increase, it may start purchasing a greater amount of the commodity even at the presently increased price. Similarly, if the household expects the price of the commodity to decrease, it may postpone its purchases. Thus, some argue that the law of demand is violated in such cases. In this case, the demand curve does not slope down from left to right; instead it presents a backward slope from the top right to down left. This curve is known as an exceptional demand curve. Technically, this is not a violation of the law of demand, as it violates the ceteris paribus condition.
## The Law of Demand and Change in Demand
The law of demand states that, other things remaining same, the quantity demanded of a good increases when its price falls and vice-versa. Note that demand for goods changes as a consequence of changes in income, tastes etc. Hence, demand may expand or contract and increase or decrease. In this context, let us make a distinction between two different types of changes that affect quantity demanded, viz., expansion and contraction; and increase and decrease. While stating the law of demand i.e., while treating price as the causative factor, the relevant terms are Expansion and Contraction in demand (which means movement along the curve of demand). When the demand is changing due to a price change alone, we should not say increase or decrease but expansion or contraction. If one of the non-price determinants of demand, such as the prices of other goods, income, etc. change & thereby demand changes, the relevant terms are increase and decrease in demand (which means shift of the demand curve to the right or to the left). The expansion and contraction in demand are shown in the diagram. You may observe that expansion and contraction are shown on a single DD curve. The changes (movements) take place along the given k.
## Limitations
• Change in taste or demand.
• Change in income
• Change in other prices.
• Discovery of substitution.
• Anticipatory change in prices.
• Rare or distinction goods.[3]
There are certain goods which do not follow this law. These include Veblen goods and Giffen goods. | 1,774 | 8,806 | {"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": 7, "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} | 3.921875 | 4 | CC-MAIN-2014-10 | longest | en | 0.921828 |
https://jp.mathworks.com/matlabcentral/profile/authors/15612189-rahul-choudhary?s_tid=cody_local_to_profile | 1,579,918,114,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250628549.43/warc/CC-MAIN-20200125011232-20200125040232-00186.warc.gz | 502,705,571 | 20,044 | Community Profile
# Rahul Choudhary
36 2019 年以降の合計貢献数
#### Rahul Choudhary's バッジ
Cell joiner
You are given a cell array of strings and a string delimiter. You need to produce one string which is composed of each string fr...
8ヶ月 前
Sum all integers from 1 to 2^n
Given the number x, y must be the summation of all integers from 1 to 2^x. For instance if x=2 then y must be 1+2+3+4=10.
8ヶ月 前
Magic is simple (for beginners)
Determine for a magic square of order n, the magic sum m. For example m=15 for a magic square of order 3.
8ヶ月 前
Make a random, non-repeating vector.
This is a basic MATLAB operation. It is for instructional purposes. --- If you want to get a random permutation of integer...
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Roll the Dice!
*Description* Return two random integers between 1 and 6, inclusive, to simulate rolling 2 dice. *Example* [x1,x2] =...
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Number of 1s in a binary string
Find the number of 1s in the given binary string. Example. If the input string is '1100101', the output is 4. If the input stri...
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Return the first and last character of a string
Return the first and last character of a string, concatenated together. If there is only one character in the string, the functi...
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Create times-tables
At one time or another, we all had to memorize boring times tables. 5 times 5 is 25. 5 times 6 is 30. 12 times 12 is way more th...
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Getting the indices from a vector
This is a basic MATLAB operation. It is for instructional purposes. --- You may already know how to <http://www.mathworks....
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Determine whether a vector is monotonically increasing
Return true if the elements of the input vector increase monotonically (i.e. each element is larger than the previous). Return f...
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Check if number exists in vector
Return 1 if number _a_ exists in vector _b_ otherwise return 0. a = 3; b = [1,2,4]; Returns 0. a = 3; b = [1,...
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Swap the first and last columns
Flip the outermost columns of matrix A, so that the first column becomes the last and the last column becomes the first. All oth...
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Swap the input arguments
Write a two-input, two-output function that swaps its two input arguments. For example: [q,r] = swap(5,10) returns q = ...
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Column Removal
Remove the nth column from input matrix A and return the resulting matrix in output B. So if A = [1 2 3; 4 5 6]; ...
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Reverse the vector
Reverse the vector elements. Example: Input x = [1,2,3,4,5,6,7,8,9] Output y = [9,8,7,6,5,4,3,2,1]
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Triangle Numbers
Triangle numbers are the sums of successive integers. So 6 is a triangle number because 6 = 1 + 2 + 3 which can be displa...
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Generate a vector like 1,2,2,3,3,3,4,4,4,4
Generate a vector like 1,2,2,3,3,3,4,4,4,4 So if n = 3, then return [1 2 2 3 3 3] And if n = 5, then return [1 2 2...
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Finding Perfect Squares
Given a vector of numbers, return true if one of the numbers is a square of one of the other numbers. Otherwise return false. E...
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Return area of square
Side of square=input=a Area=output=b
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Maximum value in a matrix
Find the maximum value in the given matrix. For example, if A = [1 2 3; 4 7 8; 0 9 1]; then the answer is 9.
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Given a and b, return the sum a+b in c.
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Vector creation
Create a vector using square brackets going from 1 to the given value x in steps on 1. Hint: use increment.
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Doubling elements in a vector
Given the vector A, return B in which all numbers in A are doubling. So for: A = [ 1 5 8 ] then B = [ 1 1 5 ...
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Create a vector
Create a vector from 0 to n by intervals of 2.
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Flip the vector from right to left
Flip the vector from right to left. Examples x=[1:5], then y=[5 4 3 2 1] x=[1 4 6], then y=[6 4 1]; Request not ...
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Whether the input is vector?
Given the input x, return 1 if x is vector or else 0.
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Find max
Find the maximum value of a given vector or matrix.
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Get the length of a given vector
Given a vector x, the output y should equal the length of x.
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inner product of two vectors
inner product of two vectors
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Arrange Vector in descending order
If x=[0,3,4,2,1] then y=[4,3,2,1,0]
8ヶ月 前 | 1,265 | 4,145 | {"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} | 3.28125 | 3 | CC-MAIN-2020-05 | latest | en | 0.732494 |
https://science.blurtit.com/245175/when-is-zero-error-positive-in-vernier-calipers-and-when-is-it-negative | 1,674,873,342,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499470.19/warc/CC-MAIN-20230128023233-20230128053233-00560.warc.gz | 539,222,372 | 17,549 | # When Is Zero Error Positive In Vernier Calipers And When Is It Negative?
Positive zero error is when the jaws are closed and the zero on the vernier scale is to the right of the zero of the main scale.The measurement taken with this pair of vernier calipers will be MORE than the actual result. To get the correct value, the zero error must be subtracted from each reading. Eg: The zero error (0.03) is subtracted from the reading. The actual reading is thus (eg) 2.35-0.03cm= 2.32
Negative error is when the jaws and closed and the zero of the vernier scale is to the left or the zero of the main scale. The measurement taken with this pair if vernier calipers will be LESS than the actual value. To get the correct value, the numerical value of the zero error must be ADDED to each reading. Eg: The 4th line on the vernier scale is in line with the marking on the main scale. The zero error (0.10-0.04cm=0.06cm) is added to the rading. The actual reading is thus 2.35+0.06cm=2.41
thanked the writer.
Zero Errors of Vernier Caliper
When the jaws are closed, the vernier zero mark coincides with the zero mark on its fixed main scale. Before taking any reading it is good practice to close the jaws or faces of the instrument to make sure it is reading zero. If not then note the reading. This reading is called “zero error”. The zero error is of two types (I) positive zero error and (ii) negative zero error.
Positive Zero Error
If the zero on the vernier scale is to the right of the main scale, then the error is said to be positive zero error and so the zero correction should be subtracted from the reading which is measured.
Negative Zero Error
If the zero on the vernier scale is to the left of the main scale, then the error is said to be negative zero error and so the zero correction should be added from the reading which is measured.
thanked the writer.
When the 0 on the movable scale is BEFORE the 0 on the main scale, it is a negative zero error.
When the 0 on the movable scale if AFTER the 0 on the main scale, it is a positive zero error.
thanked the writer.
I found something online that may help you out here --- it also has diagrams of the vernier calipers in positive and negative positions. The site is: www.scribd.com
When you're on that page, type "vernier caliper" in the search box at the top right of the pictured report. It will take you right to the page that shows when the vernier calipers are in positive and negative positions.
Good luck! If you're taking physics, you're a pretty smart gal!
Moe
thanked the writer. | 627 | 2,566 | {"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} | 2.84375 | 3 | CC-MAIN-2023-06 | latest | en | 0.863892 |
http://stackoverflow.com/questions/4099366/how-do-i-check-if-a-number-is-positive-or-negative-in-c/4100125 | 1,462,225,793,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860117783.16/warc/CC-MAIN-20160428161517-00027-ip-10-239-7-51.ec2.internal.warc.gz | 264,910,860 | 27,342 | # How do I check if a number is positive or negative in c#?
This must be a common question; however, I cannot seem to find a neat way of doing it.
How do I check if a number is positive or negative?
Thanks.
Edit: APOLOGIES FOR MY QUESTION. I was very very tired + I had a "liquid lunch" if you see what I mean. Oh well I will put it down as one of those embarrassing moments of mine. Thanks anyway.
-
So tempted to tag with wtf – JLWarlow Nov 4 '10 at 17:37
Stop downvoting the bloke, this is a good laugh! – slashmais Nov 4 '10 at 17:51
I don't get the downvotes. The question is in-scope for SO and is well written. I don't think downvotes are a valid way to indicate elitism regarding the knowledge level of the OP. Give the guy a chance. – Jeff Yates Nov 4 '10 at 18:03
+1 for being honest – ThiefMaster Mar 2 '11 at 10:33
And to those who got here by Google search :) – Fixer Jan 27 '12 at 5:16
``````bool positive = number > 0;
bool negative = number < 0;
``````
-
I must be very very very tired! makes me laugh how ridicolous my question was – user9969 Nov 4 '10 at 17:26
We all have our moments :-) – Simon Fischer Nov 4 '10 at 17:27
What about poor old negative zero?! – Grant Crofton Nov 4 '10 at 17:30
It's not ridiculous -- this person is learning. – Lou Franco Nov 4 '10 at 17:30
By definition, zero is not positive: en.wikipedia.org/wiki/0_(number) – Simon Fischer Nov 4 '10 at 17:33
Of course no-one's actually given the correct answer,
``````num != 0 // num is positive *or* negative!
``````
-
ah. hahaha. took me a few moments to realize this is the correct answer :D – VOX Jun 4 '13 at 20:14
+1 for being a smart arse :) – Liam Jan 2 '14 at 13:14
This is the correct! – juancazalla Oct 30 '15 at 15:39
hahaha.. btw how I reached here :P – Ivan Lewis Mar 26 at 7:24
OVERKILL!
``````public static class AwesomeExtensions
{
public static bool IsPositive(this int number)
{
return number > 0;
}
public static bool IsNegative(this int number)
{
return number < 0;
}
public static bool IsZero(this int number)
{
return number == 0;
}
public static bool IsAwesome(this int number)
{
return IsNegative(number) && IsPositive(number) && IsZero(number);
}
}
``````
-
I'm thinking about refactoring this with some delegates... – hunter Nov 4 '10 at 17:33
Internally, it should instantiate a class which implements `ISignDeterminator` using a `SignDeterminatorFactory`. – Jesse C. Slicer Nov 4 '10 at 17:47
Incomplete: you should check for IsNaN() as well ;) – slashmais Nov 4 '10 at 17:55
@slashmais: On an `int`?! What magical land of C# are you working in? – Jeff Yates Nov 4 '10 at 18:01
Where are the unit tests...? – NotDan Nov 4 '10 at 19:27
``````num < 0 // number is negative
``````
-
HAHAHAHAHAHA +1 – Aren Nov 4 '10 at 17:25
I assume that this is a beginner, which we should try to help. Once you learn the right way -- check this for the wrong way thedailywtf.com/Articles/… – Lou Franco Nov 4 '10 at 17:27
at first I could'nt believe the question on the main page, then I came here ... :) and this guy is not alone ... oh dear :) – slashmais Nov 4 '10 at 17:41
The Math.Sign method is one way to go. It will return -1 for negative numbers, 1 for positive numbers, and 0 for values equal to zero (i.e. zero has no sign). Double and single precision variables will cause an exception (ArithmeticException) to be thrown if they equal NaN.
-
Wow didn't know this exists.What happens for NaN? – Tanmoy Nov 4 '10 at 19:06
@Tanmoy: Looks like it will throw an exception in that case. – gnovice Nov 4 '10 at 19:07
interesting...learn something new every day. – Ashley Grenon Nov 5 '10 at 13:50
@AndyC: I enjoyed the humor, but he should be doing equality comparison against the return value of `Math.Sign` (since it explicitly defines possible return values.) – robjb Oct 16 '11 at 22:52
Very surprised to find out there's actually a function for this. – l46kok Oct 8 '13 at 15:15
This is the industry standard:
``````int is_negative(float num)
{
char *p = (char*) malloc(20);
sprintf(p, "%f", num);
return p[0] == '-';
}
``````
-
I tried it out. My program ran eventually out of memory. Could there be a leak in your code? – Thomas Tempelmann Oct 1 '15 at 15:58
@Will, well done for spotting this very obscure leak! Though I frown upon you changing the meaning of the original poster's introductory sentence. I still believe it's correct to call this the industry standard. Hence, to keep with the spirit of this answer, and with respect to the original poster, I've reverted your edit, okay? – Thomas Tempelmann Jan 20 at 10:58
@Will, do I really need to state the obvious about the original post and my comments, which got upvotes while yours doesn't? (Spoiler: it has to do with sarcasm, and yours doesn't, which means you've killed the joke that bad existed here - or do you really think that this answer was meant to be taken seriously??) – Thomas Tempelmann Jan 20 at 22:57
There we go, reverted, flagged, and downvoted. I wasn't aware this was a joke. – Will Jan 20 at 23:06
You youngins and your fancy less-than signs.
Back in my day we had to use `Math.abs(num) != num //number is negative` !
-
(If it wasn't obvious, this was intended to be humor) – Powerlord Nov 4 '10 at 17:31
Does this code work for all signed integers? – Eric Lippert Nov 5 '10 at 6:10
@Eric: No, because it will throw an `OverflowException` if `num` is `MinValue` for whatever type is passed in (`Int16`, `Int32`, `Int64`). Results are even worse for floating point values, where they could also be `NaN`, since `NaN != NaN`. – Powerlord Nov 5 '10 at 13:35
`````` public static bool IsPositive<T>(T value)
where T : struct, IComparable<T>
{
return value.CompareTo(default(T)) > 0;
}
``````
-
Native programmer's version. Behaviour is correct for little-endian systems.
``````bool IsPositive(int number)
{
bool result = false;
IntPtr memory = IntPtr.Zero;
try
{
memory = Marshal.AllocHGlobal(4);
if (memory == IntPtr.Zero)
throw new OutOfMemoryException();
Marshal.WriteInt32(memory, number);
result = Marshal.ReadByte(memory, 3) & 0x80 == 0;
}
finally
{
if (memory != IntPtr.Zero)
Marshal.FreeHGlobal(memory);
}
}
``````
Do not ever use this.
-
"Do not ever use this"? But it's enterprise-quality code, perfect for enterprise software! You're missing `IsPositiveChecker`, `IsPositiveCheckerInterface`, `IsPositiveCheckerFactory`, and `IsPositiveCheckerFactoryInterface`, though. – Tim Čas Sep 10 '15 at 10:59
``````if (num < 0) {
//negative
}
if (num > 0) {
//positive
}
if (num == 0) {
//neither positive or negative,
}
``````
or use "else ifs"
-
Haha. "Either positive or negative". – user166390 Nov 5 '10 at 6:45
For a 32-bit signed integer, such as `System.Int32`, aka `int` in C#:
``````bool isNegative = (num & (1 << 31)) != 0;
``````
-
``````public static bool IsNegative<T>(T value)
where T : struct, IComparable<T>
{
return value.CompareTo(default(T)) < 0;
}
``````
-
You just have to compare if the value & its absolute value are equal:
``````if (value == Math.abs(value))
return "Positif"
else return "Negatif"
``````
-
This technique is already covered by stackoverflow.com/a/4099428/497043 – Chris Morgan Sep 21 '13 at 0:13
``````bool isNegative(int n) {
int i;
for (i = 0; i <= Int32.MaxValue; i++) {
if (n == i)
return false;
}
return true;
}
``````
-
``````int j = num * -1;
if (j - num == j)
{
// Num is equal to zero
}
else if (j < i)
{
// Num is positive
}
else if (j > i)
{
// Num is negative
}
``````
-
## protected by Chris MorganOct 22 '15 at 6:07
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site. | 2,249 | 7,686 | {"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} | 2.546875 | 3 | CC-MAIN-2016-18 | latest | en | 0.897558 |
http://www.pearltrees.com/gnatalia/physics/id13263770 | 1,718,411,038,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861578.89/warc/CC-MAIN-20240614235857-20240615025857-00239.warc.gz | 53,203,023 | 22,955 | # Physics
Frontier of Physics: Interactive Map. “Ever since the dawn of civilization,” Stephen Hawking wrote in his international bestseller A Brief History of Time, “people have not been content to see events as unconnected and inexplicable.
They have craved an understanding of the underlying order in the world.” In the quest for a unified, coherent description of all of nature — a “theory of everything” — physicists have unearthed the taproots linking ever more disparate phenomena. With the law of universal gravitation, Isaac Newton wedded the fall of an apple to the orbits of the planets. Albert Einstein, in his theory of relativity, wove space and time into a single fabric, and showed how apples and planets fall along the fabric’s curves. And today, all known elementary particles plug neatly into a mathematical structure called the Standard Model. Our map of the frontier of fundamental physics, built by the interactive developer Emily Fuhrman, weights questions roughly according to their importance in advancing the field. Vector Model of Angular Momentum. Once you have combined orbital and spin angular momenta according to the vector model, the resulting total angular momentum can be visuallized as precessing about any externally applied magnetic field.
This is a useful model for dealing with interactions such as the Zeeman effect in sodium. The magnetic energy contribution is proportional to the component of total angular momentum along the direction of the magnetic field, which is usually defined as the z-direction. The z-component of angular momentum is quantized in values one unit apart, so for the upper level of the sodium doublet with j=3/2, the vector model gives the splitting shown. Even with the vector model, the determination of the magnitude of the Zeeman spliting is not trivial since the directions of S and L ar constantly changing as they precess about J. This problem is handled with the Lande' g-factor.
Quantized Angular Momentum. Lagrangian formalism - Intuition Behind Conservation of Angular Momentum. Threshold size for quantum effects. Angular Momentum. Angular momentum. This gyroscope remains upright while spinning due to the conservation of its angular momentum.
In physics, angular momentum, (rarely, moment of momentum or rotational momentum) is the rotational analog of linear momentum. It is an important quantity in physics because it is a conserved quantity – the angular momentum of a system remains constant unless acted on by an external torque. Angular momentum in classical mechanics Definition First principle. A first principle is a basic proposition or assumption that cannot be deduced from any other proposition or assumption.
In philosophy, first principles are from First Cause[1] attitudes and taught by Aristotelians, and nuanced versions of first principles are referred to as postulates by Kantians.[2] In mathematics, first principles are referred to as axioms or postulates. In physics and other sciences, theoretical work is said to be from first principles, or ab initio, if it starts directly at the level of established science and does not make assumptions such as empirical model and parameter fitting.
In formal logic In a formal logical system, that is, a set of propositions that are consistent with one another, it is possible that some of the statements can be deduced from other statements. For example, in the syllogism, "All men are mortal; Socrates is a man; Socrates is mortal" the last claim can be deduced from the first two. Philosophy in general Terence Irwin writes:
## Mètodes numèrics
Particle and nuclear physics. Advanced mathematical methods. Quantum physics. Pràctiques externes. Thermodynamics and statistical mechanics. Optics. Numerical methods. Symmetries, conservation laws and Noether's Theorem. Electro. Chemistry. Differential equations. Mechanics LAB. Classical Mechanics. Multivariable Calculus. The Speed Of Light Can Vary In A Vacuum.
Quantum physics just got less complicated. Here's a nice surprise: quantum physics is less complicated than we thought.
An international team of researchers has proved that two peculiar features of the quantum world previously considered distinct are different manifestations of the same thing. The result is published 19 December in Nature Communications. Patrick Coles, Jedrzej Kaniewski, and Stephanie Wehner made the breakthrough while at the Centre for Quantum Technologies at the National University of Singapore. They found that 'wave-particle duality' is simply the quantum 'uncertainty principle' in disguise, reducing two mysteries to one. "The connection between uncertainty and wave-particle duality comes out very naturally when you consider them as questions about what information you can gain about a system.
The discovery deepens our understanding of quantum physics and could prompt ideas for new applications of wave-particle duality. Explore further: A new 'lens' for looking at quantum behavior. Gauge esto, Gauge lo otro… ¿Qué es una teoría gauge? La palabra gauge la encontramos por doquier en los escritos sobre física.
Aparecen expresiones como simetría gauge, invariancia gauge, bosones gauge, teorías gauge, etc. Sin embargo, pocas veces se explica con propiedad qué es esta teoría, por qué es tan fundamental y cómo la entienden y por qué la veneran tanto los físicos. ¿Por qué La Tierra está achatada por los polos? La densidad de La Tierra. En su elaboración de la Teoría de Gravitación Universal, Newton ya se dio cuenta de que en La Tierra, a consecuencia de su movimiento de rotación y según su Ley de atracción, cada partícula de masa m a diferente distancia del eje, estaría expuesta a una diferente Fuerza Centrípeta, ya que describe un movimiento circular uniforme de diferente radio alrededor del eje de rotación de La Tierra.
Según la segunda ley de Newton, para que se produzca una aceleración debe actuar una fuerza en la dirección de esa aceleración. Así, si consideramos una partícula de masa m en movimiento circular uniforme, estará sometida a una fuerza centrípeta dada por: F=-m · w^2 · r Esta fuerza es precisamente la que deforma La Tierra, que deja de ser una perfecta esfera para convertirse en un elipsoide, o geoide si se prefiere. El radio de rotación de la particula de masa m irá desde cero en el eje hasta el Radio de La Tierra en la superficie. | 1,352 | 6,375 | {"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} | 2.890625 | 3 | CC-MAIN-2024-26 | latest | en | 0.91917 |
https://simplewebtool.com/converters/storage/mebibytestogigabytes/mebibytestogigabytes.html | 1,726,842,842,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652278.82/warc/CC-MAIN-20240920122604-20240920152604-00352.warc.gz | 476,026,677 | 8,191 | # Mebibytes to Gigabytes conversion
## Mebibytes (MiB) to Gigabytes (GB) conversion calculator and how to convert.
How to convert Mebibytes to Gigabytes: 1 mebibyte is approximately equal to 1.04 * 10-3 gigabytes. 1 MiB ≈ 1.04 * 10-3 GB The quantityqin gigabytes (GB) is equal to the quantity q in mebibytes (MiB) multiplied by 0.001048576.
Formula: q(GB) = q(MiB) * 0.001048576
Mebibytes to Gigabytes conversion table
Mebibytes (MiB) Gigabytes (GB)
1 MiB 1.05e-3 GB
2 MiB 2.10e-3 GB
3 MiB 3.15e-3 GB
4 MiB 4.19e-3 GB
5 MiB 5.24e-3 GB
6 MiB 6.29e-3 GB
7 MiB 7.34e-3 GB
8 MiB 8.39e-3 GB
9 MiB 9.44e-3 GB
10 MiB 1.05e-2 GB | 277 | 627 | {"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-2024-38 | latest | en | 0.539625 |
https://www.nagwa.com/en/plans/598164248964/ | 1,702,167,514,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100989.75/warc/CC-MAIN-20231209233632-20231210023632-00545.warc.gz | 868,770,873 | 9,000 | Lesson Plan: Corollaries of Isosceles Triangle Theorems | Nagwa Lesson Plan: Corollaries of Isosceles Triangle Theorems | Nagwa
# Lesson Plan: Corollaries of Isosceles Triangle Theorems Mathematics
This lesson plan includes the objectives and prerequisites of the lesson teaching students how to use the corollaries of the isosceles triangle theorems to find missing lengths and angles in isosceles triangles.
#### Objectives
Students will be able to
• understand that the median of an isosceles triangle from the vertex angle is a perpendicular bisector of the base,
• understand that the bisector of the vertex angle of an isosceles triangle is a perpendicular bisector of the base,
• understand that the straight line that passes through the vertex angle of an isosceles triangle and is perpendicular to the base bisects the base and the vertex angle,
• use the corollaries of the isosceles triangle theorems to solve problems,
• understand that the axis of symmetry of an isosceles triangle is the median that bisects the base.
#### Prerequisites
Students should already be familiar with
• identifying isosceles triangles,
• equilateral triangles,
• congruent triangles,
• the isosceles triangle theorems,
• the Pythagorean theorem.
Attend sessions, chat with your teacher and class, and access class-specific questions. Download the Nagwa Classes app today!
Windows macOS Intel macOS Apple Silicon | 314 | 1,412 | {"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} | 3.28125 | 3 | CC-MAIN-2023-50 | latest | en | 0.821933 |
http://forums.realsoftware.com/viewtopic.php?f=4&t=45743&view=next&sid=172fb22c979dda63d9b84379c4c4966a | 1,537,871,966,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267161501.96/warc/CC-MAIN-20180925103454-20180925123854-00472.warc.gz | 100,957,383 | 9,173 | # Real Software Forums
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Post subject: Anyone got a simple Tile Terrain Tranistion algorithm?Posted: Sun Sep 02, 2012 6:37 pm
Joined: Sun Aug 05, 2007 10:46 am
Posts: 4931
Location: San Diego, CA
I have a tile based game where the terrain is made up of combination of eight(8) tiles (grass, sand, alphalt, dirt,water etc.)
I have figured out a quick easy way to create edge transitions giving each tile type a priorty (grass is highest, water is lowest) and using grayscale masks
but it only works for top, bottom, left and right edges.... I cannot figure how to make the corners look good.
This is most probematic when a corner has multiple types of terrains involved.
99% of the "tutorials" on the web talk about terrain consisting of two types of tiles (grass and water).. I need at least 3 (grass, sand, water) but would prefer the 8 I have.
I have come up with two concepts.... but am still perplexed about handling corners.
`+---+---+---+| 1 | 2 | 3 |+---+---+---+| 8 | X | 4 |+---+---+---+| 7 | 6 | 5 |+---+---+---+`
Assuming "X" is the tile being analyzed.... my first idea would be to examine the 8 neighbors and create a mask to apply to "X" that would allow each of its neighbors to encroach inside "X"
The other would be to create masks to have "X" encroach outside onto each of its neighbors
But either way... the corners are the issue. If 2,3 and 4 were all the same terrain ... no problem... but if each were different..... then what do you show? and how do you show it?
My terrain tiles are 102 x 102 pixels so I have enough resolution to work with.
Any specific help would be great!
and gee.. if anyone even felt like taking my tiles and writing a little demo....
here is a partial screenshot of what I have so far (as you can see it REALLY needs some blending)
Note.. the TRACKS are a layer all their own.... so there is one image of TERRAIN tiles (above) and another for TRACK (and soon to be another for Buildings and bushes)
The (hopefully) end result will be a virtual "toy" train set. Build your layout.... place your trains. Drive them around the layout... all without losing 3/4 of your garage.
_________________
Dave Sisemore
iMac I7[2012], OSX Mountain Lion 10.8.3 RB2012r2.1
Note : I am not interested in any solutions that involve custom Plug-ins of any kind
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Post subject: Re: Anyone got a simple Tile Terrain Tranistion algorithm?Posted: Mon Sep 03, 2012 3:27 am
Joined: Wed May 10, 2006 2:42 pm
Posts: 2985
Location: Germany
blend them together.
Add a mask to the top left of each tile which gives one pixel row/column with 25%, one row/column with 50% and one row/column with 75%. Than draw each tile 3 pixel overlaying the other one on left/top.
This way you'd have a more soft edge.
Greetings
Christian
_________________
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Post subject: Re: Anyone got a simple Tile Terrain Tranistion algorithm?Posted: Mon Sep 03, 2012 9:32 am
Joined: Sun Aug 05, 2007 10:46 am
Posts: 4931
Location: San Diego, CA
blend them together.
Add a mask to the top left of each tile which gives one pixel row/column with 25%, one row/column with 50% and one row/column with 75%. Than draw each tile 3 pixel overlaying the other one on left/top.
This way you'd have a more soft edge.
Greetings
Christian
Christian... did you even READ my post?
Quote:
I have figured out a quick easy way to create edge transitions giving each tile type a priorty (grass is highest, water is lowest) and using grayscale masks
but it only works for top, bottom, left and right edges.... I cannot figure how to make the corners look good.
This is most probematic when a corner has multiple types of terrains involved.
_________________
Dave Sisemore
iMac I7[2012], OSX Mountain Lion 10.8.3 RB2012r2.1
Note : I am not interested in any solutions that involve custom Plug-ins of any kind
Top
Post subject: Re: Anyone got a simple Tile Terrain Tranistion algorithm?Posted: Mon Sep 03, 2012 10:01 am
Joined: Wed May 10, 2006 2:42 pm
Posts: 2985
Location: Germany
Sorry. I missed those lines.
On the corners, I would simply use roundrect drawing commands for creating the mask.
so on the edge the transparency is even higher, so 4 tiles meeting there give 100%.
Greetings
Christian
_________________
See you in Orlando, Florida for Real World 2013
More details and registration here:
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Post subject: Re: Anyone got a simple Tile Terrain Tranistion algorithm?Posted: Mon Sep 03, 2012 10:09 am
Joined: Sun Aug 05, 2007 10:46 am
Posts: 4931
Location: San Diego, CA
But that does not answer the question.
Putting a simple corner shape mask is one thing.... but the question is how to handle the fact that any given corner can have 2,048 combinations of neighbors
How BEST to decide WHICH terrain to apply that mask to? Inside? Outside?
I am going for a blend mask solution, as I dont want to spend the next year or two hand drawing custom tiles for a program that only took a few days to write.
Am finding it quite frustrating that there are hundreds of articles on the web, but not a single one of them discuss specifics... they are all 30,000 foot level explanations
_________________
Dave Sisemore
iMac I7[2012], OSX Mountain Lion 10.8.3 RB2012r2.1
Note : I am not interested in any solutions that involve custom Plug-ins of any kind
Top
Post subject: Re: Anyone got a simple Tile Terrain Tranistion algorithm?Posted: Mon Sep 03, 2012 3:42 pm
Joined: Sun Aug 05, 2007 10:46 am
Posts: 4931
Location: San Diego, CA
Not Perfect... but much better....... still need to tweak it some
BEFORE
AFTER
_________________
Dave Sisemore
iMac I7[2012], OSX Mountain Lion 10.8.3 RB2012r2.1
Note : I am not interested in any solutions that involve custom Plug-ins of any kind
Top
Post subject: Re: Anyone got a simple Tile Terrain Tranistion algorithm?Posted: Wed Dec 12, 2012 7:05 pm
Joined: Sat Aug 18, 2012 11:23 am
Posts: 21
Dave,
I don't know if you're still looking for answers to this, but here's how you might want to think about doing it.
Build your terrain tiles larger than 102x102. I built these from your originals at 142x142:
This would require that you make 8 passes to lay the tiles. Starting at water and working up to grass. That way each higher level overwrites the lower tile along the edges. You'd have to offset my tiles from your Grid origin -22 on x and y.
And let the edges fall where they may:
I think some of my edges are too heavy. But I think this would be faster than analyzing each tile.
The GIMP file is here.
Note that the GIMP file has two sets of guidelines, one is blue which defines the 142x142 box for each tile, and the other is the black terrain grid guidelines. I didn't know if you were adding that later, or had an option to turn it off or on.
I realize that you're looking for an algorithm.
I'm looking at RB from VB with XNA, and I'm still trying to figure out how to SpriteBatch in OpenGL. But that's another story.
The point is that I've been looking at graphics for more than a year. If you look at the open source game Battle For Wesnoth, they do what you're talking about. The file that holds water hexagon tiles has 537 images. Every terrain type, of which there are dozens, has "bleed" over images that are placed on adjoining tiles. So there's a water "bleed" image for every situation. There's also "waves" flowing in every direction.
In XNA (directX) this is fairly simple, but I don't know how to batch sprites in OpenGL with RB so I thought I'd give the above method a try, as an exercise. As I look at it again and again, I realize that the edges are way too heavy, but I'm sure you get the idea.
Also, you often talk about image masks, and I'm used to using transparent .png files, so my method may not help you in any way at all!
Battle Of Wesnoth is open source, so you can view their graphic display code and it might give you some ideas for building your own algorithm.
Hope this helps.
RBnoob
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Post subject: Re: Anyone got a simple Tile Terrain Tranistion algorithm?Posted: Wed Dec 12, 2012 8:58 pm
Joined: Sat Aug 18, 2012 11:23 am
Posts: 21
Another try with tighter edges:
and:
New GIMP here.
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Post subject: Re: Anyone got a simple Tile Terrain Tranistion algorithm?Posted: Wed Dec 12, 2012 9:32 pm
Joined: Sat Aug 18, 2012 11:23 am
Posts: 21
Okay, last one.
This time the grid is built into the tile using a color from the tile:
Resulting in:
With GIMP here.
Of course they no longer need to be 142x142.
And with tracks ...
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https://es.mathworks.com/matlabcentral/answers/1465679-how-to-calculate-the-area-in-table-and-certain-coordinate?s_tid=prof_contriblnk | 1,653,170,494,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662541747.38/warc/CC-MAIN-20220521205757-20220521235757-00715.warc.gz | 274,890,222 | 33,148 | # How to calculate the area in table and certain coordinate
3 views (last 30 days)
mohd akmal masud on 3 Oct 2021
Commented: Chetan Bhavsar on 13 Oct 2021
Area Centroid
____ ________________
9 147.67 109.11
27 146.81 108.52
43 147.09 109.7
54 147.81 109.35
12 146.17 134.42
56 147.73 109.46
21 146.48 134.05
53 147.83 109.26
20 147.3 134.25
2 126.5 146
43 147.67 108.63
24 147 134.04
2 126.5 146
45 147.93 108.87
20 147.25 133.95
10 125.5 146.8
37 147.43 108.89
16 147.31 133.56
6 126.17 147
36 147.39 108.81
12 147.92 133.75
6 126.5 147
34 147.32 108.91
12 147.92 133.75
8 126.5 146
29 147.69 109
13 147.54 133.38
8 126.5 146
28 147.71 109.11
7 148.14 134.43
8 126.5 146
28 147.71 109.11
6 148 134.5
7 126.43 145.71
28 147.71 109.11
7 147.29 133.71
1 106 132
8 126.25 145.5
25 147.4 109.08
7 147.29 133.71
1 106 132
5 126.4 145.2
31 147.03 108.87
9 147.33 133.78
2 105.5 132
6 126.17 145
31 147.03 108.87
9 147.33 133.78
1 106 132
3 125.67 144.33
31 147.03 108.87
9 147.33 133.78
31 147.03 108.87
9 147.33 133.78
28 147.43 109.39
10 146.7 133.6
25 147.08 109.4
11 147.55 133.18
28 147.43 109.39
5 146.8 133.2
20 147.7 109.55
3 149 132
12 147.92 109.33
Hi all, I have data above.Anyone know how to calculate the total are ? Let say Inwant calculate the the total are that have coordinate (147 109) and (146 108), is it like this the command sum([146:147 108:109])?
Chetan Bhavsar on 3 Oct 2021
Edited: Chetan Bhavsar on 3 Oct 2021
Hello @mohd akmal masud, You need to take out Area index then sum it where centroid is within you range.
You can do that like following
Area = [9;27;43];
Centroid = [146.67 108.11; 146.81 109.52; 146.09 108.7];
T = table(Area ,Centroid);
%check in all value which are 147 in column 1 and 109 in column 2
trueIndex = ((147>=T.Centroid(:,1) & T.Centroid(:,1)>=146)&(109>=T.Centroid(:,2) & T.Centroid(:,2)>=108));
%add all area which have 147 and 109
sum(Area(trueIndex))
Or in one liner if you want
sum(Area(((147>=T.Centroid(:,1) & T.Centroid(:,1)>=146)&(109>=T.Centroid(:,2) & T.Centroid(:,2)>=108))))
Please mark answer as accepted if it satisfies your requirement.
Chetan Bhavsar on 13 Oct 2021
in this case there are 2 area what do you want with them 1st one or second or sum?
R2021a
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Translated by | 1,031 | 2,363 | {"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-2022-21 | latest | en | 0.508891 |
https://www.physicsforums.com/threads/can-different-forces-cancel-themselves.890856/ | 1,527,171,749,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866326.60/warc/CC-MAIN-20180524131721-20180524151721-00396.warc.gz | 810,931,185 | 17,752 | # I Can different forces cancel themselves?
1. Oct 26, 2016
### Pauly Denino
If for say, I had a ''perfect object'' (Only mass with no dimensions) and I put three non-parallel and different pulling forces on the item, would these cancel themselves? After the doing the experiment myself with a non-perfect item, it seemed there was a slight advancement. The addition of the vectors gave me a near 0N force.
My questions are, is that slight advancement only due to the imprecision of my material and why do the forces cancel even thought they're all different?
2. Oct 26, 2016
### The Buttered Cat
Making it even simpler, if you had an object with just one force acting on it, you can apply an equal and opposite force to cancel that force, correct?
3. Oct 26, 2016
### Khashishi
Just add the force vectors. You know what a vector is?
4. Oct 26, 2016
### Pauly Denino
I already added them, my questions is more towards the why the cancellation happens even thought the forces are different
5. Oct 26, 2016
### Khashishi
If the vectors don't cancel, then the forces don't cancel. It's as simple as that. But maybe there is some friction that you haven't taken into account.
6. Oct 26, 2016
### nasu
It's a little hard to understand your problem. The forces can cancel because they are different. If they were identical (in magnitude and direction) they effects will add up and you cannot have cancellation.
For two forces they should have same magnitude but opposite directions. For three forces you can have many possibilities, many combinations of magnitudes and directions which will give a zero resultant.
7. Oct 26, 2016
### Staff: Mentor
If you have three forces that sum to zero then if you lay them out head to tail then you will get a triangle. You can also get four forces to cancel if you lay then head to tail and you get a quadrilateral. Any number of vectors that takes you in a closed loop when laid end to end results in 0 net force.
8. Oct 26, 2016
### Pauly Denino
Thanks for you answers, they really helped me make my mind under the concept. I will definitely be using this website more often.
9. Oct 26, 2016
### CWatters
Any time you drive a car at a constant velocity the forces on it sum to zero (aka cancel out). | 545 | 2,263 | {"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} | 3.578125 | 4 | CC-MAIN-2018-22 | latest | en | 0.96378 |
https://www.energyvanguard.com/blog/Infiltration-Occurs-at-the-Surface-Not-in-the-Volume | 1,708,877,965,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474617.27/warc/CC-MAIN-20240225135334-20240225165334-00704.warc.gz | 756,499,709 | 23,445 | skip to Main Content
# Infiltration Occurs at the Surface, Not in the Volume
At the gym the other night, I was thinking about a tweet I’d seen the day before: “@EFL_Guy ‘Air leaks through surfaces not volume” Joe Lstiburek.'” I’d been meaning to blog about this for a while, so, during my workout, I began planning my article.
Of course, since the comment occurred at Building Science Summer Camp, I had to preface my article with a few words about that event. Before I knew it, though, that preface turned into the whole article, I Don’t Need No Stinkin’ Building Science Summer Camp. Today, you get the article I started out to write the other night.
If you know nothing about the field of building science, Joe’s statement above probably seems so obvious that you wonder why anyone would even utter it. Had it only been so obvious to a few scientists back in the early days of testing buildings for infiltration, we wouldn’t be in the mess we’re in now.
You see, everyone who learns how to use a Blower Door these days gets indoctrinated into the cult of the ACH (Air Changes per Hour). You take the raw number from your test – cubic feet per minute (cfm) of air flow with a big pressure difference between the house and the outside – and you divide it by the volume.
Why volume?! Why? It’s perfectly obvious that if you want to normalize a number, you divide by something that makes sense. Would it make sense to measure the efficiency of your car in miles per square foot of gasoline? That’s a similarly absurd ratio because you pay for gallons of gasoline, and a given volume of gasoline can have different surface areas depending on the shape of the container. (That reminds me of the joke about the spherical cow.)
Same with houses. Two houses with the same volume can have different surface areas. And, as Joe said, air leaks through the surface, not through the volume, so not only do we not care about volume when we’re talking about air infiltration, it’s misleading.
Normalizing to volume also builds in a bias toward larger homes. Since surface area is proportional to the square of the radius and volume is proportional to the cube of the radius, the volume increases faster than the surface area as a house grows in size. So, large houses benefit when dividing by volume instead of surface area.
The whole reason to divide the cfm50 by anything is so that you can compare infiltration rates in different size houses. (The subscript above gives the pressure difference between inside and outside during the test, 50 Pascals.) If you can’t get a fair comparison between small houses and large houses if we use volume, why are we using volume?
Here’s another example for you. Let’s say you want to paint your house, so you go to dummies.com, where they tell you that it takes about one gallon per 350 square feet of surface area you’re going to paint. What if over at elstupido.com, however, they told you that you’ll need one gallon for each thousand cubic feet of volume? Whose advice are you going to take? Right! Dummies are better than el stupidos.
Air leaks into and out of the house through all the surfaces that make up the building envelope – floors, ceilings, and walls. If you wanted to reduce your home’s infiltration, you wouldn’t seal the volume. You’d seal the leaks through the surfaces. (Just don’t seal those weep holes at the bottom of your brick walls!)
Defenders of the ACH do make the seemingly valid point that it’s conceptually easier to explain to homeowners. When you extrapolate from a house under test pressure to a house under ‘natural’ pressure differences, the ACHnat tells you how quickly the air in your house is exchanging with outside air.
If, for example, a house comes in at 0.5 ACHnat, the energy auditor will tell the homeowner that all the air in their home is exchanged with outside air every two hours. Yes, it’s an easy concept for homeowners to grasp, but is it true?
No, it’s a lie, and here’s why. First, the extrapolation to go from ACH50 to ACHnat is highly inaccurate when doing a single point Blower Door test, which is what most energy raters do. Second, the amount of air in the house is less than the volume of the house because we tend to have things like walls and furniture inside.
Third, the whole concept of ‘natural’ pressure is artificial. The pressure difference across a home’s building envelope varies from season to season and minute to minute. It’s affected by wind, the stack effect, and, especially, mechanical systems.
So, yeah, ACHnat is an easy concept to explain, but it’s a pretty shaky concept.
As I commented in Martin Holladay’s blog post on Blower Doors a few months ago, there’s an easy solution to this problem. Let’s use the quantity that Southface developed about six years ago – the Envelope Leakage Ratio (ELR). Divide the cfm50 by the square footage of the building envelope. It’s simple. It makes sense. And, I think homeowners can understand this concept, too.
If we have this tool – the Blower Door – that gives us numbers with a precision of +/-3%, why would anyone want to manipulate the results into something with uncertainties of 20%, 30%, possibly even 50% or more?! It’s time to quit using ACH to talk about infiltration.
[Read the followup article that breaks this down to the fundamental issue: It’s the Hole – Understanding What a Blower Door Is for.]
#### This Post Has 3 Comments
1. Paul McGovern says:
Allison, as I’m sure you are
Allison, as I’m sure you are well aware, that is an ASTM Standard and the recommended level is below .25 SFBE @CFM 50
2. David Eakin says:
Allison,
Allison,
I can see the logic – as long as ALL the building envelope is included. All pressure boundary ceilings and floors need to be added to the envelope. However this whole discussion might just be academic as most major air leakages that we have found are due to poor building practice (plumbing stacks, top plate/rim joist/plumbing and wire perforation leaks, poor/nonexistant weatherstriping, flue dampers, etc.) rather than residence size/volume.
3. Derek Roff says:
I followed the link to this
I followed the link to this blog, from your Sept. 24, 2014 blog on Green Building Advisor. You pose the question of why the industry uses air changes per hour for discussions of ventilation, and ridicule any consideration of the answer, pointing to the folks at elestupido.com. Your examples attempt to show the absurdity of the idea, exploring gasoline and house painting.
I agree that there is good reason to calculate air leakage per surface area, and floor area may be an acceptable approximation. But I think it is curious that you don’t offer a single sentence on why the majority of the building industry has plugged into something that you consider stupid. Do you suppose that everyone at ASHRAE is clueless? I think not. I think there are a couple of reasons why ACH50 gained common usage, and they aren’t entirely absurd. It may be time to move beyond ACH50, but we got where we are for better reasons than you indicate.
Comments are closed.
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https://eandata.com/lookup/0853447004526/ | 1,621,151,726,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243992516.56/warc/CC-MAIN-20210516075201-20210516105201-00446.warc.gz | 249,800,631 | 4,311 | EAN-13 0853447004526 UPC-A 853447004526 Product Name Kuvings - Whole Slow Juicer - Silver Created 05-04-2021 2:42:49pm Modified 05-04-2021 2:42:49pm MD5 10f9cf360e6d6f0b4aefa486bd58014c SHA256 cbbc3506507596eb40f94f1a17554fdee15ab64b4b7a8c47ecb22106c19d1ee6 Search Google by EAN or by Title Query Time 0.0063460
### Article of interest
ISBN stands for International Standard Book Number. Originally this was a 10 digit number known as ISBN-10, but has been replaced with a 13 digit version known as ISBN-13 (aka EAN-13) to better fit in with the industry standard product numbering system.
The ISBN system offers a unique numbering system for books and publications of various types. ISBN offers 9 identification digits for each book or publication and one check digit. ISBN-13 expands the available number set by doubling the available numbers. You might think that the extra three digits would more than double the available numbers but ISBN is limited within the EAN-13 numbering system and will always start with 978 or 979 as the first three digits, leaving only 9 more digits for the actual identification digits and again the last digit is the check digit.
Any ISBN-10 number can be converted to ISBN-13. Any ISBN-13 number that starts with 978 can be converted back to ISBN-10. ISBN-13 numbers that start with 979 cannot be converted to ISBN-10.
The conversion process from ISBN-10 to ISBN-13 is a rather simple one. Start with the first 9 digits of the ISBN-10 number, place 978 in front and calculate the new check digit using the standard EAN check digit calculation.
If you want to calculate the check digit for ISBN-10, that is done using a MOD-11 calculation that is very different than the EAN check digit process.
• Each digit is multiplied by its position for digits 1-9
• 1x1 + 5x2 + 9x3 + 0x4 + 5x5 + 9x6 + 3x7 + 3x8 + 2x9 = 180
• Now get the remainder 180 mod 11 = 4
• Now return the mod 11 again... 4 mod 11 = 4
• If the results is 10, return the letter X otherwise return the digit.
• For our example, the final number is 1-5905-9332-4
ISBN-1-5905-9332-4
Look at the number above and below the barcode. You will see that except for the last digit in the ISBN number, all of the digits exist in the 13 digit version shown under the barcode right after the 978 and just before the new check digit of 5.
Now you can see how easy it is to convert between ISBN-10 and ISBN-13.
Close
Close | 667 | 2,417 | {"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-2021-21 | latest | en | 0.690837 |
http://mathhelpforum.com/algebra/28319-function-question-print.html | 1,527,157,919,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866201.72/warc/CC-MAIN-20180524092814-20180524112814-00018.warc.gz | 186,851,664 | 3,204 | # Function question
• Feb 15th 2008, 04:48 AM
Tangera
Function question
Hello! My question is as follows:
f(x) = (x-3)^3 , x > 3
Find f^-1(f(x)) and f(f^-1(x)), stating clearly the rules and domains.
I know that the rules of both are the same: f^-1(f(x)) = x and f(f^-1(x)) = x. However, I don't understand why are the domains different? Please explain to me! Thank you!
• Feb 15th 2008, 05:10 AM
mr fantastic
Quote:
Originally Posted by Tangera
Hello! My question is as follows:
f(x) = (x-3)^3 , x > 3
Find f^-1(f(x)) and f(f^-1(x)), stating clearly the rules and domains.
I know that the rules of both are the same: f^-1(f(x)) = x and f(f^-1(x)) = x. However, I don't understand why are the domains different? Please explain to me! Thank you!
You need reasoning similar to that used here.
• Feb 15th 2008, 05:56 AM
Tangera
I understand why f^-1(f(x)) has a domain of x>3, but I don't know why the domain of f(f^-1(x)) is x > 0. Is it because since f(x) has a domain of x >3, so the range of f inverse is also [3, infinity). f inverse = (cube root x) + 3. For the range of f inverse to be [3, infinity), x > 0? Is my reasoning correct?
• Feb 15th 2008, 12:34 PM
mr fantastic
Quote:
Originally Posted by Tangera
I understand why f^-1(f(x)) has a domain of x>3, but I don't know why the domain of f(f^-1(x)) is x > 0. Is it because since f(x) has a domain of x >3, so the range of f inverse is also [3, infinity). f inverse = (cube root x) + 3. For the range of f inverse to be [3, infinity), x > 0? Is my reasoning correct?
There is a rule that says:
$\displaystyle \text{dom} \, f^{-1} = \text{ran} \, f$
$\displaystyle \text{ran} \, f^{-1} = \text{dom} \, f$
These rules can be thought of as a consequence of swapping y and x around in y = f(x) to get the rule for the inverse function ..... y's and x's swap --> domain and range swap .....
$\displaystyle \text{dom}\, f = [3, \infty) \,$ and $\displaystyle \text{ran}\, f = [0, \infty) \,$.
Therefore:
$\displaystyle \text{dom} \, f^{-1} = \text{ran} \, f = [0, \infty)$.
$\displaystyle \text{ran} \, f^{-1} = \text{dom} \, f = [3, \infty)$.
The composite function g(f(x)) exists iff $\displaystyle \text{ran} f = \text{dom} \, g \,$. When that condition is met, the domain of g(f(x)) is $\displaystyle \text{dom} f$. In you present question, this condition is met for both $\displaystyle f(f^{-1}(x)) \,$ and $\displaystyle f^{-1}(f(x))$.
So:
The domain of $\displaystyle f(f^{-1}(x)) = x\,$ is $\displaystyle \text{dom} \, f^{-1} = [0, \infty)$.
The domain of $\displaystyle f^{-1}(f(x)) = x\,$ is $\displaystyle \text{dom} \, f = [3, \infty)$. | 874 | 2,621 | {"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} | 3.953125 | 4 | CC-MAIN-2018-22 | latest | en | 0.916242 |
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Quiz_4 _IM 106_2ndSemAY20122013_AK
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University of Santo Tomas (UST) Faculty of Engineering QUIZ 4, 2 nd Semester AY 2012-2013 IM 106 –Operations Management (Engr. William A. Cortez) Name (PRINT): __________________________________ Course/Year: _________ Date: __________ Score: _______/ 50 Instructions: Select the best answer to the questions. Write X in the BOX of the LETTER of your choice in your answer sheet. 1. A line balancing solution has been developed for the assembly line for fertilizer spreaders at Green Grass, Inc. The desired output rate of 30 spreaders per hour will be achieved. The sum of times for all tasks performed on the line is 1200 seconds for each spreader assembled. This is the total productive time. Which of the following statements must be TRUE ? B A. The cycle time is 30 seconds per spreader. D. If the solution calls for 11 stations, the efficiency is 80%. B. The "theoretical minimum" number of stations is 10. E. None of these choices C. If solution calls for 12 stations, the efficiency is 80%. 2. An assembly line with 28 operations is to be balanced. The total amount of time required for all 28 operations is 39 minutes. The longest operation takes 2.2 minutes and the shortest takes 0.8 minutes. The line will operate for 450 minutes per day. Determine the minimum and maximum cycle time. What is the output rate if the cycle time is set at 2.2 minutes? A A. min cycle time = 0.8; max cycle time = 39; output rate = 204.5 B. min cycle time = 2.2; max cycle time = 39; output rate = 204.5 C. min cycle time = 2.2; max cycle time = 31.2; output rate = 562.5 D. min cycle time = 0.8; max cycle time = 2.2; output rate = 562.5 3. Which of the following is NOT a concept or dimension of the servicescape. A A. Robust design C. Physical evidence E. Spatial layout and functionality B. Signs, symbols and artifacts D. Ambient conditions 4. A ____ layout is an arrangement based on the sequence of operations that are performed during the manufacturing of a good or the delivery of a service. A A. Product B. Process C. Cellular D. Fixed position E. Unstable position 5. A ____ layout is an arrangement based on self-contained groups of equipment needed to produce a particular set of goods or services. C A. Product B. Process C. Cellular D. Fixed position E. Unstable position 6. The typical layout of facilities in a hospital would be an example of a ____ layout. B A. Product B. Process C. Cellular D. Fixed position E. Unstable position 7. The production of large goods such as airplanes is best accomplished using a ____ layout. D A. Product B. Process C. Cellular D. Fixed position E. Unstable position 8. The text presents an integrative case study of Lenscrafters. Which statement is correct? C A. Lenscrafters primary product is eyewear. C. Each store has a visible low contact area. B.
This is the end of the preview. Sign up to access the rest of the document. | 774 | 3,061 | {"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-2015-40 | longest | en | 0.847797 |
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Tim Matthews April 2, 2008 Physics 133 Lab #3 Yipeng Yang Abstract The purpose of this experiment was to compare the Force of gravity to the uniform circular motion experienced by the mass when it spun around the center shaft. Once these values were obtained and plotted on a graph they should have made a slope equal to one, meaning that they are essentially the same value. However we didn’t quite get exactly one, more like 1 and 1.4. Theory Newton’s Second Law states that Force is equal to an objects mass times its gravitational pull. (F=M*g). We were then comparing that value to the value we obtained with the uniform circular motion. ( 2 2 / 4 T Mr F = ). Essentially they should be equal to each other or close to it. Newton’s law states that in order for an object to be in constant motion the force acting upon that object must be zero, however there will always be air resistance, friction and gravity acting on an object as long as it is in our atmosphere.
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## This note was uploaded on 04/25/2008 for the course PHYSICS 131 taught by Professor Hatch during the Spring '07 term at UMass (Amherst).
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Lab #3 - Tim Matthews April 2 2008 Physics 133 Lab#3 Yipeng...
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# The euclidean distance not the best indicator the actual distance
242
Chapter 4 Pathfinding
Indoor level Outdoor level
Figure 4.18 Euclidean distance fill characteristics
The cluster heuristic works by grouping nodes together in clusters. The nodes in a cluster represent some region of the level that is highly interconnected. Clustering can be done automatically using graph clustering algorithms that are beyond the scope of this book. Often, clustering is manual, however, or a by-product of the level design (portal-based game engines lend themselves well to having clusters for each room).
A lookup table is then prepared that gives the smallest path length between each pair of clusters. This is an offline processing step that requires running a lot of pathfinding trials between all pairs of clusters and accumulating their results. A suf-ficiently small set of clusters is selected so that this can be done in a reasonable time frame and stored in a reasonable amount of memory.
4.3 A*
Cluster B
10
12
11 15 4
3
7 10 9
8 6 7 1
5
Cluster C
A A B C
x 13 29
13 x 7
29 7 x
B
C
Lookup table
Figure 4.19 The cluster heuristic
Clustering is intimately related to hierarchical pathfinding, explained in Sec-tion 4.6, which also clusters sets of locations together. Some of the calculations we’ll meet there for distance between clusters can be adapted to calculate the heuristics between clusters.
Even without such optimizations, the cluster heuristic is worth trying for labyrin-thine indoor levels.
This is a good example of the knowledge vs. search trade-off we looked at in Chap-ter 2.
If the heuristic is more complex and more tailored to the specifics of the game level, then the A* algorithm needs to search less. It provides a good deal of knowledge about the problem. The ultimate extension of this is a heuristic with ultimate knowl-edge: completely accurate estimates. As we have seen, this would produce optimum A* performance with no search.
Null heuristic
Figure 4.20
Euclidean distance heuristic
Null heuristic 4.3 A*
245
Figure 4.21 Fill patterns outdoors
Quality of Heuristics
Producing a heuristic is far more of an art than a science. Its significance is massively underestimated by AI developers. In my experience, many developers drop in a simple Euclidean distance heuristic without thought and hope for the best.
Dijkstra Is A*
It is worth noticing that the Dijkstra algorithm is a subset of the A* algorithm. In A* we calculate the estimated-total-cost of a node by adding the heuristic value to the cost-so-far. A* then chooses a node to process based on this value.
For each pathfinding world representation, we will divide the game level into linked regions that correspond to nodes and connections. The different ways this can be achieved are called division schemes. Each division scheme has three important properties we’ll consider in turn: quantization/localization, generation, and validity.
You might also be interested in Chapter 11, Tools and Content Creation, which looks at how the pathfinding data is created by the level designer or by an automatic process. In a complete game, the choice of world representation will have as much to do with your toolchain as technical implementation issues.
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Up where my oldest daughter lives. A spectacular part of our lovely state. 20/May/12 12:24 AM | |
Do you go there by rail? 20/May/12 12:48 AM | |
SE=8.4SSTS to UP=231)(6)g6=g13-(6=5)h1-h4=i6-(5=2)f6,=>i6<>6,g6<>22)XW(2)ad25,=>bef2bf5<& gt;23)(9=7)g7-c7=(7-6)c3=c6-(6=7)g6,=>g7<>74)(5=8)b2-f2=f1-(8=6)g1-g6=(6-8)c6=(8)b6,=&g t;b2<>85)(2=8)b1-(8=6)g1-g6=c6-(6=2)a5,=>b6<>2.UP=81 20/May/12 1:30 AM | |
1. Unique possibilities 23.2. Whether i6=5,f6=2;OR h4=5,gh1=68,g37=79;g4=2.UP24.3. If b5=2=a2;column d is devoid of 2.So b5=47.UP29.4. Whether c7=7;OR c7=4,c9=8,c6=6,g6=7;g7=9.UP30.5. Whether b6=8,c6=6;OR b6=2=a2,ac3=67,g3=8;g6=7.UP81. 20/May/12 1:39 AM | |
Site Puzzle#1 s-wing(56) (6)g6=g3-(6=5)h1-h4=i6 .. -(5=2)f6 ; i6<>6, g6<>2// x-wing(2)ad/25 ; b25<>2, e2<>2, f25<>2#2 (6)c6=(6-7)g6=x-wing(7)cg/37-(7)a3=w-wing(68)a3,c6,b123b6 ; a5<>6 20/May/12 2:03 AM | |
Love the photo, Anne. Especially with the rail tracks. 20/May/12 3:29 AM | |
Great photo, Anne, thanks!! 20/May/12 7:07 AM | |
No, Shiela. Those rails are for the iron ore trains, taking the ore to the port for shipment overseas. 20/May/12 12:46 PM | |
I must have hit on the key early on in this one, since I needed no solving techniques at all. It was all clerical work after filling in the possibilities. 20/May/12 4:37 PM | |
42:49 20/May/12 11:39 PM | |
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Member's Birthdays Todaylinda from melbourne aust | 1,017 | 2,732 | {"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-2019-39 | latest | en | 0.603036 |
https://kocaelinde.com/qa/question-why-is-3-the-perfect-number.html | 1,627,558,348,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153857.70/warc/CC-MAIN-20210729105515-20210729135515-00481.warc.gz | 337,299,164 | 9,327 | # Question: Why Is 3 The Perfect Number?
## Is 3 a spiritual number?
The number 3 biblically represents divine wholeness, completeness and perfection.
If there ever was a desire to highlight an idea, thought, event or noteworthy figure in the Bible for their prominence, the number 3 was used to put a divine stamp of completion or fulfillment on the subject..
## Is 8 a perfect square?
Perfect Squares and Perfect Cubes For example x8 is a perfect square, its square root is x4 .
## IS 496 a perfect number?
496 is most notable for being a perfect number, and one of the earliest numbers to be recognized as such. As a perfect number, it is tied to the Mersenne prime 31, 25 − 1, with 24 (25 − 1) yielding 496.
## Why are perfect numbers important?
Perfect and multiply perfect numbers are a project that gives you large numbers relatively simply. For example, the divisors of 120 are (1,2,4,8)(1,3)(1,5).
## What is the most beautiful number?
Pi (π) is considered the most beautiful number, and is celebrated every March 14th. Here’s why. I bet you already know what p is. If I ask you, you will reply pfff p equals 3.14 and a bunch of numbers; so what?
## Is 75 a perfect number?
For 75, the answer is: No, 75 is not a prime number. The list of all positive divisors (i.e., the list of all integers that divide 75) is as follows: 1, 3, 5, 15, 25, 75. For 75 to be a prime number, it would have been required that 75 has only two divisors, i.e., itself and 1.
## Why is 28 a perfect number?
A number is perfect if all of its factors, including 1 but excluding itself, perfectly add up to the number you began with. 6, for example, is perfect, because its factors — 3, 2, and 1 — all sum up to 6. 28 is perfect too: 14, 7, 4, 2, and 1 add up to 28.
## Which is the powerful number?
Equivalently, a powerful number is the product of a square and a cube, that is, a number m of the form m = a2b3, where a and b are positive integers. Powerful numbers are also known as squareful, square-full, or 2-full.
## What does three symbolize?
Three represents the triad of family; male, female, and child; beginning, middle, and end; birth, life, and death. Of two things we say both; of three things we say all.
## Is 25 a perfect square?
Perfect Square. A perfect square is a number, from a given number system, that can be expressed as the square of a number from the same number system. … 25 is a natural number, and since there is another natural number 5, such that 52 = 25, 25 is a perfect square.
## Is the number 1 prime or composite?
A prime number has exactly two factors so 1 isn’t prime. A composite number has more than 2 factors, so 1 isn’t composite.
## Why is 3 A special number?
In their eyes the number 3 was considered as the perfect number, the number of harmony, wisdom and understanding. It was also the number of time – past, present, future; birth, life, death; beginning, middle, end – it was the number of the divine.
## Is 3 a evil number?
Examples. The first evil numbers are: 0, 3, 5, 6, 9, 10, 12, 15, 17, 18, 20, 23, 24, 27, 29, 30, 33, 34, 36, 39 …
## What does the number 3 mean in love?
Actually, number 3 is considered to be a magic number when it comes to love. This number is a love affirmation, so you can be sure that love will take place in your life in the future period. Your angels will encourage you to give and to receive love. You have to listen to your heart and your instincts.
## What is Coprime number?
A Co-prime number is a set of numbers or integers which have only 1 as their common factor i.e. their highest common factor (HCF) will be 1. Co-prime numbers are also known as relatively prime or mutually prime numbers.
## What is the meaning of 8?
The number eight is considered to be a lucky number in Chinese and other Asian cultures. Eight (八; accounting 捌; pinyin bā) is considered a lucky number in Chinese culture because it sounds like the word meaning to generate wealth (發(T) 发(S); Pinyin: fā). Property with the number 8 may be valued greatly by Chinese.
## Is 3 a powerful number?
The most powerful number in the Bible is number 3 and denotes several significant elements and events in the Holy Book. For example, the Godhead is a trinity featuring God the Father, Son, and Holy Spirit.
## What is the most evil number?
sum to 666 is known as an evil number (Pegg and Lomont 2004).
## What does the number 13 mean in the Bible?
There is also a biblical reference to the unlucky number 13. Judas, the apostle said to have betrayed Jesus, was the 13th guest to the Last Supper. (See “Lost Gospel Revealed; Says Jesus Asked Judas to Betray Him.”) As for Friday, it’s well known among Christians as the day Jesus was crucified.
## What is a perfect number simple explanation?
In number theory, a perfect number is a positive integer that is equal to the sum of its positive divisors, excluding the number itself. For instance, 6 has divisors 1, 2 and 3 (excluding itself), and 1 + 2 + 3 = 6, so 6 is a perfect number.
## What are the evil numbers in the Bible?
Both Hebrew and Christian sources believe six indicates “imperfection,” “man,” or “evil.” For Christianity, this occurs preeminently in the Book of Revelation.
## Why Is 9 the perfect number?
The number 9 is revered in Hinduism and considered a complete, perfected and divine number because it represents the end of a cycle in the decimal system, which originated from the Indian subcontinent as early as 3000 BC. … Nine is a significant number in Norse Mythology. | 1,432 | 5,522 | {"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} | 4.09375 | 4 | CC-MAIN-2021-31 | latest | en | 0.943404 |
https://lists.ellipsis.cx/archives/spoon-discuss/spoon-discuss-200202/msg00129.html | 1,627,320,097,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152144.81/warc/CC-MAIN-20210726152107-20210726182107-00609.warc.gz | 388,931,075 | 2,534 | Wonko on 8 Feb 2002 04:44:28 -0000
Re: spoon-discuss: The PIG
```Quoth Eric Gerlach,
> At 10:14 PM 2002-02-03 -0500, you wrote:
>> Um, I hate to be the bearer of bad news, but according to my calculations,
>> we need to make at least 27 proposals or the PIG is gonna eat the football
>> proposal again (unless someone represses it less than 4 ndays before the end
>> of the week).
>
> Hrmm.... I just did the math myself to check, and I think this is off a bit...
>
> GN = 3
> n = # of proposals on the ballot
>
> k = proposal eaten = GN/(120/n) = 3/(120/n) = n/40, rounded up.
>
> That means that if:
>
> 1 <= n <= 40, then k=1, and Rob's beautiful default case proposal gets eaten;
> 41 <= n <= 80, then k=2, and Football gets eaten again.
>
> Conclusion: Since it is unlikely we'll have 40 proposals on the ballot,
> Rob, if you get a chance, resubmit that proposal... it really should get
> voted for. If I don't see it by 4-5 tomorrow, I'll resubmit in your stead,
> with point donation as usual :)
>
> Bean
>
Yes, I noticed that... For some reason I had thought the Gremlin number was
9. Although, you don't neccessarily need to resubmit it; with such a low
number, it only costs 10 points to Repress the PIG and save the proposal....
--
Wonko
Recruit today night it feels.
``` | 392 | 1,291 | {"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} | 2.75 | 3 | CC-MAIN-2021-31 | latest | en | 0.945018 |
https://brg.a2hosted.com/?p=6167 | 1,568,621,534,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572516.46/warc/CC-MAIN-20190916080044-20190916102044-00091.warc.gz | 417,669,452 | 7,287 | Skip to content
Sunday Times Teaser 2969 – Slide Rules
by Stephen Hogg
Published August 18 2019 (link)
Using her ordinary 15cm ruler and Zak’s left-handed version (numbers 1 to 15 reading right to left) Kaz could display various fractions. For instance, putting 5 on one ruler above 1 on the other ruler, the following set of fractions would be displayed: 5/1, 4/2, 3/3, 2/4 and 1/5. Zak listed the fifteen sets starting from “1 above 1” up to “15 above 1”.
Kaz chose some fractions with values less than one from Zak’s sets (using just the numerals 1 to 9, each once only in her selection). Of these, two were in simplest form, one of which had consecutive numerator and denominator. Zak correctly totalled Kaz’s selection, giving the answer as a fraction in simplest form. Curiously, the answer’s numerator and denominator were both palindromic.
Give Zak’s answer.
From → Uncategorized
Comments are closed. | 232 | 916 | {"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} | 2.859375 | 3 | CC-MAIN-2019-39 | longest | en | 0.946251 |
https://www.physicsforums.com/threads/how-to-convert-diffusion-rate-to-hertz.1053730/ | 1,725,811,841,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651013.49/warc/CC-MAIN-20240908150334-20240908180334-00811.warc.gz | 928,905,336 | 21,774 | # How to Convert Diffusion Rate to Hertz?
• amy1vaulhausen
In summary: Frequency is a useful quantity for sound, light, and other types of waves, but it can't be directly used to measure anything else.
amy1vaulhausen
TL;DR Summary
Does anyone know hot to do this ; How to Convert Diffusion Rate to Hertz?
How to Convert Diffusion Rate to Hertz - Hi, new here. Does anyone know if it is possible to convert the molecular diffusion rate
to a frequency in hertz?
I have no idea what you are trying to do. Please be specific.
russ_watters, topsquark, BillTre and 1 other person
amy1vaulhausen said:
How to Convert Diffusion Rate to Hertz - Hi, new here. Does anyone know if it is possible to convert the molecular diffusion rate
Like @hutchphd, I don't understand the question. Sounds like you are trying to convert bananas to bicycles.
topsquark, BillTre and hutchphd
Hertz (Hz) is the same as reciprocal seconds (1/s). I can't think of a diffusion quantity with that dimension. Did a bit of searching, and as I suspected the unit for such a quantity is better described as amount per area times seconds.
For details, see Fick's law: https://en.wikipedia.org/wiki/Fick's_laws_of_diffusion
Bystander
Thanks Mayhem, the reason behind my question is based on the diffusion rate in time.
Formula ; t=x^2/2D
see reference ; https://www.physiologyweb.com/calculators/diffusion_time_calculator.html
Since a solute will take a range of time to diffuse for a spacial unit with known viscosity and related parameters
and since that time range can be computed in seconds, then if light seconds are used as a distance value and we convert the distance to a value in hertz, wouldnt we end up with a means to covert rate to a hertzian frequency? Am just wondering if there is a standard way to approach this? A formula to use?
I like to think of the diffusion rate as a measure of the growth of the blob of diffusing stuff. By some measure the surface area of the blob grows linearly with time. I don't know how any particular frequency is useful.
A cycle of time can be converted to a frequency in hertz. If we know the amount of time in seconds it takes a solute to diffuse in a specific unit of space with know characteristics then there must be a way to convert this time period to a hertzian value or at least a range of frequency values?
amy1vaulhausen said:
A cycle of time can be converted to a frequency in hertz.
Really? It took me 20 secons to read your post. What frequency is that?
amy1vaulhausen said:
If we know the amount of time in seconds it takes a solute to diffuse in a specific unit of space with know characteristics then there must be a way to convert this time period to a hertzian value or at least a range of frequency values?
You're still trying to convert bananas to bicycles. Plus, as @hutchphd said, even if your could square this circle,
hutchphd said:
I don't know how any particular frequency is useful.
phinds said:
Really? It took me 20 seconds to read your post. What frequency is that?
1/20th Hz, of course. I'll rephrase the objection: @amy1vaulhausen you can convert any repeating time period into a frequency by inverting it, but that doesn't make it useful/meaningful. What's the real goal here? What do you think this tells you?
amy1vaulhausen said:
Thanks Mayhem, the reason behind my question is based on the diffusion rate in time.
Formula ; t=x^2/2D
see reference ; https://www.physiologyweb.com/calculators/diffusion_time_calculator.html
Since a solute will take a range of time to diffuse for a spacial unit with known viscosity and related parameters
and since that time range can be computed in seconds, then if light seconds are used as a distance value and we convert the distance to a value in hertz, wouldnt we end up with a means to covert rate to a hertzian frequency? Am just wondering if there is a standard way to approach this? A formula to use?
There is no standard way to approach this because it's not a standard approach at all. It doesn't even make sense to use hertz in this manner as you don't have a repeating process or some cyclic phenomenon. Just because something is a rate doesn't make it cyclic or applicable to use hertz. Acceleration is a rate, as it has units of m/s/s, but you would never use hertz as a unit for acceleration. The same is true for velocity, which has units of m/s. An object moving at 10 m/s does not have a frequency of 10 Hz. It doesn't have a frequency at all.
BillTre and russ_watters
So we have Hertz for periodic processes, Becquerel for radioactive decay, although lifetime can be formulated as imaginary part of frequency.
One of the last realms of physics where the babylonian-anglosaxonian unit mess persists.
## What is the diffusion rate in the context of molecular motion?
The diffusion rate typically refers to the rate at which molecules spread from an area of high concentration to an area of low concentration. This rate can be quantified by the diffusion coefficient, which is a measure of how fast the molecules are diffusing through a medium.
## How is the diffusion rate related to frequency (Hertz)?
The diffusion rate itself is not directly expressed in Hertz (cycles per second). However, if the diffusion process is periodic or oscillatory in nature, the frequency of such oscillations could be expressed in Hertz. For example, in NMR (nuclear magnetic resonance) spectroscopy, the diffusion of molecules can affect the frequency of the magnetic resonance signals.
## What is the formula to convert diffusion rate to a frequency in Hertz?
There is no direct formula to convert diffusion rate to Hertz since they measure different phenomena. However, in specific contexts like NMR, the diffusion coefficient (D) can influence the linewidth of spectral lines, which indirectly relates to frequency. In such cases, the relationship is complex and depends on the experimental setup.
## Can diffusion rate be measured in Hertz in any scenario?
Diffusion rate itself cannot be measured in Hertz as it is usually measured in units like square meters per second (m²/s). However, in systems where diffusion leads to periodic changes, the resulting oscillatory behavior can be measured in Hertz. For instance, in oscillatory chemical reactions, the frequency of oscillations can be measured in Hertz.
## What tools or methods are used to measure diffusion rates and frequencies?
Diffusion rates are commonly measured using techniques such as dynamic light scattering (DLS), pulsed-field gradient NMR, and fluorescence recovery after photobleaching (FRAP). Frequencies, on the other hand, can be measured using spectroscopic methods like NMR, where the resonance frequency is observed, or using oscilloscopes and signal analyzers for periodic signals.
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2K | 1,620 | 6,962 | {"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} | 3.015625 | 3 | CC-MAIN-2024-38 | latest | en | 0.923421 |
https://www.physicsforums.com/threads/water-ice-equilibrium-querry.368131/ | 1,627,242,647,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151760.94/warc/CC-MAIN-20210725174608-20210725204608-00320.warc.gz | 990,100,563 | 13,750 | # Water-ice equilibrium Querry
Consider that we have ice and water together ( no heat exchange with surrounding ) and
qi1 = heat required by ice to reach 0 C from ti ( < 0 )
qi2 = heat required by ice at 0 C to convert into water at 0 C
qw = heat that can be released by water after reaching 0 C from tw ( > 0 )
if ( qi1 + qi2 ) <= qw
can there be a case when some ice is left in the system ? | 111 | 395 | {"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} | 2.765625 | 3 | CC-MAIN-2021-31 | latest | en | 0.955355 |
https://www.periodic-table.org/oganesson-mass-number/ | 1,695,829,098,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510300.41/warc/CC-MAIN-20230927135227-20230927165227-00037.warc.gz | 1,031,787,144 | 52,655 | # Oganesson – Mass Number – Neutron Number – Og
Oganesson is a chemical element with atomic number 118 which means there are 118 protons and 118 electrons in the atomic structure. The chemical symbol for Oganesson is Og.
## Neutron Number and Mass Number of Oganesson
Mass numbers of typical isotopes of Oganesson are 294,295.
The total number of neutrons in the nucleus of an atom is called the neutron number of the atom and is given the symbol N. Neutron number plus atomic number equals atomic mass number: N+Z=A. The difference between the neutron number and the atomic number is known as the neutron excess: D = N – Z = A – 2Z.
Neutron number is rarely written explicitly in nuclide symbol notation, but appears as a subscript to the right of the element symbol. Nuclides that have the same neutron number but a different proton number are called isotones. The various species of atoms whose nuclei contain particular numbers of protons and neutrons are called nuclides. Each nuclide is denoted by chemical symbol of the element (this specifies Z) with tha atomic mass number as supescript. Therefore, we cannot determine the neutron number of uranium, for example. We can determine the neutron number of certain isotope. For example, the neutron number of uranium-238 is 238-92=146.
### Neutron and Mass Numbers and Nuclear Properties
Properties of atomic nuclei (atomic mass, nuclear cross-sections) are determined by the number of protons and number of neutrons (neutron number). It must be noted, especially nuclear cross-sections may vary by many orders from nuclide with the neutron number N to nuclide with the neutron number N+1. For example, actinides with odd neutron number are usually fissile (fissionable with slow neutrons) while actinides with even neutron number are usually not fissile (but are fissionable with fast neutrons). Heavy nuclei with an even number of protons and an even number of neutrons are (due to Pauli exclusion principle) very stable thanks to the occurrence of ‘paired spin’. On the other hand, nuclei with an odd number of protons and neutrons are mostly unstable.
### Neutron and Atomic Numbers and Nuclear Stability
Nuclear stability is a concept that helps to identify the stability of an isotope. To identify the stability of an isotope it is needed to find the ratio of neutrons to protons. To determine the stability of an isotope you can use the ratio neutron/proton (N/Z). Also to help understand this concept there is a chart of the nuclides, known as a Segre chart. This chart shows a plot of the known nuclides as a function of their atomic and neutron numbers. It can be observed from the chart that there are more neutrons than protons in nuclides with Z greater than about 20 (Calcium). These extra neutrons are necessary for stability of the heavier nuclei. The excess neutrons act somewhat like nuclear glue. Only two stable nuclides have fewer neutrons than protons: hydrogen-1 and helium-3.
Atomic nuclei consist of protons and neutrons, which attract each other through the nuclear force, while protons repel each other via the electric force due to their positive charge. These two forces compete, leading to various stability of nuclei. There are only certain combinations of neutrons and protons, which forms stable nuclei.
Neutrons stabilize the nucleus, because they attract each other and protons , which helps offset the electrical repulsion between protons. As a result, as the number of protons increases, an increasing ratio of neutrons to protons is needed to form a stable nucleus. If there are too many or too few neutrons for a given number of protons, the resulting nucleus is not stable and it undergoes radioactive decayUnstable isotopes decay through various radioactive decay pathways, most commonly alpha decaybeta decaygamma decay or electron capture. Many other rare types of decay, such as spontaneous fission or neutron emission are known.
## Atomic Mass of Oganesson
Atomic mass of Oganesson is 294 u.
The atomic mass is the mass of an atom. The atomic mass or relative isotopic mass refers to the mass of a single particle, and therefore is tied to a certain specific isotope of an element. The atomic mass is carried by the atomic nucleus, which occupies only about 10-12 of the total volume of the atom or less, but it contains all the positive charge and at least 99.95% of the total mass of the atom. Note that, each element may contain more isotopes, therefore this resulting atomic mass is calculated from naturally-occuring isotopes and their abundance.
The size and mass of atoms are so small that the use of normal measuring units, while possible, is often inconvenient. Units of measure have been defined for mass and energy on the atomic scale to make measurements more convenient to express. The unit of measure for mass is the atomic mass unit (amu). One atomic mass unit is equal to 1.66 x 10-24 grams. One unified atomic mass unit is approximately the mass of one nucleon (either a single proton or neutron) and is numerically equivalent to 1 g/mol.
For 12C the atomic mass is exactly 12u, since the atomic mass unit is defined from it. For other isotopes, the isotopic mass usually differs and is usually within 0.1 u of the mass number. For example, 63Cu (29 protons and 34 neutrons) has a mass number of 63 and an isotopic mass in its nuclear ground state is 62.91367 u.
There are two reasons for the difference between mass number and isotopic mass, known as the mass defect:
1. The neutron is slightly heavier than the proton. This increases the mass of nuclei with more neutrons than protons relative to the atomic mass unit scale based on 12C with equal numbers of protons and neutrons.
2. The nuclear binding energy varies between nuclei. A nucleus with greater binding energy has a lower total energy, and therefore a lower mass according to Einstein’s mass-energy equivalence relation E = mc2. For 63Cu the atomic mass is less than 63 so this must be the dominant factor.
Note that, it was found the rest mass of an atomic nucleus is measurably smaller than the sum of the rest masses of its constituent protons, neutrons and electrons. Mass was no longer considered unchangeable in the closed system. The difference is a measure of the nuclear binding energy which holds the nucleus together. According to the Einstein relationship (E=mc2), this binding energy is proportional to this mass difference and it is known as the mass defect.
1
H
Hydrogen
Nonmetals
2
He
Helium
Noble gas
3
Li
Lithium
Alkali metal
4
Be
Beryllium
Alkaline earth metal
5
B
Boron
Metalloids
6
C
Carbon
Nonmetals
7
N
Nitrogen
Nonmetals
8
O
Oxygen
Nonmetals
9
F
Fluorine
Nonmetals
10
Ne
Neon
Noble gas
11
Na
Sodium
Alkali metal
12
Mg
Magnesium
Alkaline earth metal
13
Al
Aluminium
Post-transition metals
14
Si
Silicon
Metalloids
15
P
Phosphorus
Nonmetal
16
S
Sulfur
Nonmetal
17
Cl
Chlorine
Nonmetal
18
Ar
Argon
Noble gas
19
K
Potassium
Alkali metal
20
Ca
Calcium
Alkaline earth metal
21
Sc
Scandium
Transition metals
22
Ti
Titanium
Transition metals
23
V
Transition metals
24
Cr
Chromium
Transition metals
25
Mn
Manganese
Transition metals
26
Fe
Iron
Transition metals
27
Co
Cobalt
Transition metals
28
Ni
Nickel
Transition metals
29
Cu
Copper
Transition metals
30
Zn
Zinc
Transition metals
31
Ga
Gallium
Post-transition metals
32
Ge
Germanium
Metalloids
33
As
Arsenic
Metalloids
34
Se
Selenium
Nonmetal
35
Br
Bromine
Nonmetal
36
Kr
Krypton
Noble gas
37
Rb
Rubidium
Alkali metals
38
Sr
Strontium
Alkaline earth metals
39
Y
Yttrium
Transition metals
40
Zr
Zirconium
Transition metals
41
Nb
Niobium
Transition metals
42
Mo
Molybdenum
Transition metals
43
Tc
Technetium
Transition metals
44
Ru
Ruthenium
Transition metals
45
Rh
Rhodium
Transition metals
46
Pd
Transition metals
47
Ag
Silver
Transition metals
48
Cd
Transition metals
49
In
Indium
Post-transition metals
50
Sn
Tin
Post-transition metals
51
Sb
Antimony
Metalloids
52
Te
Tellurium
Metalloids
53
I
Iodine
Nonmetal
54
Xe
Xenon
Noble gas
55
Cs
Caesium
Alkali metals
56
Ba
Barium
Alkaline earth metals
57-71
Lanthanoids
72
Hf
Hafnium
Transition metals
73
Ta
Tantalum
Transition metals
74
W
Tungsten
Transition metals
75
Re
Rhenium
Transition metals
76
Os
Osmium
Transition metals
77
Ir
Iridium
Transition metals
78
Pt
Platinum
Transition metals
79
Au
Gold
Transition metals
80
Hg
Mercury
Transition metals
81
Tl
Thallium
Post-transition metals
82
Pb
Post-transition metals
83
Bi
Bismuth
Post-transition metals
84
Po
Polonium
Post-transition metals
85
At
Astatine
Metalloids
86
Rn
Noble gas
87
Fr
Francium
Alkali metal
88
Ra
Alkaline earth metal
89-103
Actinoids
104
Rf
Rutherfordium
Transition metal
105
Db
Dubnium
Transition metal
106
Sg
Seaborgium
Transition metal
107
Bh
Bohrium
Transition metal
108
Hs
Hassium
Transition metal
109
Mt
Meitnerium
110
Ds
111
Rg
Roentgenium
112
Cn
Copernicium
113
Nh
Nihonium
114
Fl
Flerovium
115
Mc
Moscovium
116
Lv
Livermorium
117
Ts
Tennessine
118
Og
Oganesson
57
La
Lanthanum
Lanthanoids
58
Ce
Cerium
Lanthanoids
59
Pr
Praseodymium
Lanthanoids
60
Nd
Neodymium
Lanthanoids
61
Pm
Promethium
Lanthanoids
62
Sm
Samarium
Lanthanoids
63
Eu
Europium
Lanthanoids
64
Gd
Lanthanoids
65
Tb
Terbium
Lanthanoids
66
Dy
Dysprosium
Lanthanoids
67
Ho
Holmium
Lanthanoids
68
Er
Erbium
Lanthanoids
69
Th
Thulium
Lanthanoids
70
Yb
Ytterbium
Lanthanoids
71
Lu
Lutetium
Lanthanoids
89
Ac
Actinium
Actinoids
90
Th
Thorium
Actinoids
91
Pa
Protactinium
Actinoids
92
U
Uranium
Actinoids
93
Np
Neptunium
Actinoids
94
Pu
Plutonium
Actinoids
95
Am
Americium
Actinoids
96
Cm
Curium
Actinoids
97
Bk
Berkelium
Actinoids
98
Cf
Californium
Actinoids
99
Es
Einsteinium
Actinoids
100
Fm
Fermium
Actinoids
101
Md
Mendelevium
Actinoids
102
No
Nobelium
Actinoids
103
Lr
Lawrencium
Actinoids | 2,869 | 10,092 | {"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} | 2.546875 | 3 | CC-MAIN-2023-40 | latest | en | 0.825119 |
http://www.asmcommunity.net/forums/topic/?id=20772 | 1,653,052,846,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662532032.9/warc/CC-MAIN-20220520124557-20220520154557-00126.warc.gz | 66,421,697 | 3,075 | `` +5 0101+ -5 1010------------ -0 1111``
Posted on 2005-03-19 03:31:08 by Rammstein
You'll find information in the first chapter of the Art Of Assembly Language by Randall Hyde :
http://webster.cs.ucr.edu/AoA/Windows/HTML/DataRepresentationa5.html#999509
Posted on 2005-03-19 06:07:08 by Dr. Manhattan
Is only a representation, we need to be consistent whit our representations.. or coherent ;).
Posted on 2005-03-19 08:57:30 by rea
If you check the 0 and 1s are switch. -5 instead of 0101=1010 the bits are inverted, see the 01 from the first one and the other one. Hope it helps.
01 the contrary is 10 I guess.
1 -1
Posted on 2005-03-19 09:23:53 by Xor Stance
Rammstein,
can someone tell me why -5 is 1010, I dont understand
To obtain the ones complement of any number, simply invert the bits.? Inverting 5=0101, one obtains -5=1010.?
Just about any book on logic or computer science will give a brief on ones complement arithmetic.? Ratch
Posted on 2005-03-19 10:44:37 by Ratch
By the way, 1's complement and 2' complement is defined to any base but I not remember exactly how is :S...
Posted on 2005-03-19 12:03:24 by rea
:/
CPUs use sign-module-2 instead of sign-module-1 (which is presented here).
in sign module 2 we have:
+5 0101
-5 1011
----------------
0 0000
zero (without any stupid negations). that's how CPU work, so you should focus on SM2, not SM1.
/
Posted on 2005-03-19 15:42:18 by ti_mo_n
ti_mo_n,
CPUs use sign-module-2 instead of sign-module-1 (which is presented here).
No such thing as sign-module whatever.? Do you mean signed-magnitude representation?? In any case, the INTEL CPU uses twos complement arithmetic.? Not all CPUs use twos complement arithmetic, however.? The original poster asked a question about ones complement arithmetic.? Ratch
Posted on 2005-03-19 19:50:24 by Ratch
By the way, 1's complement and 2'? complement is defined to any base but I not remember exactly how is :S...
No, the pair, 1's complement and 2's complement, are associated only with binary.
For any radix (also known as number base), there is a radix complement and a (radix - 1) complement.
So for decimal numbers, you have 10's complement and 9's complement.
And 2's complement can be used with trinary (base 3) numbers, although it's not the same as the binary 2's complement. Binary two's complement of 10 is 10, trinary two's complement of 10 is 12.
Posted on 2005-03-19 21:44:44 by tenkey
No such thing as sign-module whatever.
Yeah. Now i had the time to thoroughly check this out. the name sing-module- is a short for "'s complement, expressed as 'sign' digits, followed by 'module' digits.
In any case, the INTEL CPU uses twos complement arithmetic.? Not all CPUs use twos complement arithmetic, however.
I don't count other CPUs, since there's a 99% chance, that i'll never use them :) | 822 | 2,867 | {"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-2022-21 | latest | en | 0.834718 |
https://nursingdons.com/solution-draw-the-pedigree-for-this-family-3/ | 1,726,430,507,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651647.78/warc/CC-MAIN-20240915184230-20240915214230-00271.warc.gz | 388,225,784 | 20,149 | # Solution-Draw the pedigree for this family
1) The following pedigree shows transmission of galactosemia, a rare autosomal recessive disease caused by the inability to metabolize the sugar galactose, resulting in muscle, nerve, and kidney malfunction.
i) Write genotypes for as many individuals in the pedigree as possible. Write each genotype below the corresponding symbol in the pedigree.
ii) The individuals marked 1 and 2 are planning to have a child. They know about their family history of galactosemia, so they visit a genetic counselor to find out the probability that the child will have galactosemia. What does she tell them?
iii) The couple has a son and he does not have galactosemia. They decide to have a second child and again visit the genetic counselor. What does she tell them is the probability that the second child will have galactosemia?
iv) The couple’s second child is born, and he has galactosemia. They decide to have a third child and again visit the genetic counselor. What does she tell them is the probability that the third child will have galactosemia?
2) A man with brown teeth mates with a woman with normal white teeth. They have four daughters, all with brown teeth, and two sons, both with white teeth. The two sons both mate with women with white teeth, and each couple has one daughter and one son. All of these children have white teeth. One of the four brown-toothed daughters (A) mates with a man with white teeth (B), and they have two brown-toothed daughters, one white-toothed daughter, one brown-toothed son, and one white-toothed son.
i) Draw the pedigree for this family, showing the inheritance of brown teeth.
ii) Explain this pedigree by specifying the most likely mode of inheritance of brown teeth.
iii) Based on your answer to part (ii), what is the chance that the next child of the A-B couple will have brown teeth?
3) In fruit flies, vermilion eye color is a recessive trait determined by an X-linked gene. What genotype and phenotype ratios are expected from the following crosses? You can assume that the parents labeled ‘wild type’ in each cross are homozygous.
i) vermilion male x wild type female
ii) vermilion female x wildtype male
iii) daughter from mating (i) x wild type male
iv) daughter from mating (i) x vermilion male
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## Recent Questions
### Part 4 capstone project
Part 4 capstone project
### Unit 4 dp – outline
Unit 4 dp – outline | 836 | 3,753 | {"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} | 2.6875 | 3 | CC-MAIN-2024-38 | latest | en | 0.949335 |
https://superbasic-manual.readthedocs.io/en/latest/C/ceil.html | 1,716,871,571,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059067.62/warc/CC-MAIN-20240528030822-20240528060822-00823.warc.gz | 484,812,003 | 23,582 | # CEIL
Syntax CEIL (x) Location Math Package
The function CEIL returns the closest integer to x which is greater than or equal to x (the ‘ceiling’ of x). Compare INT which returns the next integer which is less than or equal:
CEIL(12.75)=13 INT(12.75)=12 CEIL(-2.3)=-2 INT(-2.3)=-3
CEIL can handle numbers in the range -32768<x<=32768.
Example
A mechanic needs one and a half hours to replace the rusty exhaust of a car. If his rate of pay is £13 per hour, he will charge CEIL(13*1.5)=£20 for the job (excluding parts).
NOTE
The simplest way to get a true INTEGER function, where x is rounded up or down to the nearest integer is with INT(x+.5) which ensures that INT(12.75)=13 and INT(-2.3)=-2.
CROSS-REFERENCE
INT | 214 | 727 | {"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-2024-22 | latest | en | 0.81332 |
https://onoffswitch.net/2013/10/14/arrow-operator/ | 1,579,520,157,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250598726.39/warc/CC-MAIN-20200120110422-20200120134422-00280.warc.gz | 583,090,933 | 7,685 | The Arrow operator
Continuing my journey in functional programming, I decided to try doing the 99 haskell problems to wean my way into haskell. I’ve found this to be a lot of fun since they give you the answers to each problem and, even though I have functional experience, the haskell way is sometimes very different from what I would have expected.
For example, I discovered Arrows via the following problem:
Run-length encoding of a list. Implement the so-called run-length encoding data compression method. Consecutive duplicates of elements are encoded as lists (N E) where N is the number of duplicates of the element E.
Example:
* (encode ‘(a a a a b c c a a d e e e e))
((4 A) (1 B) (2 C) (2 A) (1 D)(4 E))
Example in Haskell:
encode “aaaabccaadeeee”
[(4,’a’),(1,’b’),(2,’c’),(2,’a’),(1,’d’),(4,’e’)]
My initial solution I did the way I’d probably write it in F#:
[haskell]
encode :: Eq a => [a] -> [(Int, a)]
encode a = map (\x -> (length x, head x)) . group \$ a
[/haskell]
But, one of the alternate solutions to the problem was cleaner and used an operator I’d never seen
[haskell]
encode :: Eq a => [a] -> [(Int, a)]
encode a = map (length &&& head) . group \$ a
[/haskell]
What the hell was &&&? If we break down the function length &&& head a little we’ll see it has a signature of
[haskell]
(length &&& head) :: [a] -> (Int, a)
[/haskell]
The function takes a list and will apply the first function to the list as the first element of the tuple (length) and then apply the second function to the list and that’ll give you the second element of the tuple.
Turns out this is part of the Control.Arrow module and defines an interesting way to combine logical steps using tuples and several basic transformation functions. There is a great intro to arrows on the haskell wiki which I followed and ended up simulating in F#.
Translating to F#
A quick translation attempt turned out the following small module
module Arrow =
let split x = (x, x)
let combine f (x, y) = f x y
let first f (a, b) = (f a, b)
let second f (a, b) = (a, f b)
let onTuple f g = first f >> second g
let onSingle f g = split >> (onTuple f g)
let (.***.) = onTuple
let (.&&&.) = onSingle
let onSingleCombine op f g = (onSingle f g) >> combine op
To break down what’s going on here, we can follow an example posted in the haskell wiki.
let div2 x = x / 2
let m3p1 x = 3*x + 1
let example = onSingleCombine (+) div2 m3p1 8
What this is really doing is:
• Split 8 into (8, 8)
• Apply the first function to the first element (div2 8) with a resulting tuple of (4, 8)
• Apply the second function to the second element (m3p1 8) with a resulting tuple of (4, 25)
• Apply the combiner (+) to the resulting tuple which gives the answer of 29
While I think the example is a little contrived, it’s really cool how you can leverage arrows to do this kind of sequencing work. I’ll certainly find use for this!
2 thoughts on “The Arrow operator”
1. Gusty says:
Arrows are also implemented in FsControl, you can play with them in
2. Anton Kropp says:
Neat, I think there is also an arrow implementation in fsharpx | 862 | 3,110 | {"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-2020-05 | longest | en | 0.874911 |
http://users.ece.utexas.edu/~patt/02f.306/ProblemSets/ProblemSet5/ProblemSet5.html | 1,508,319,077,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822851.65/warc/CC-MAIN-20171018085500-20171018105500-00884.warc.gz | 363,999,940 | 2,406 | Department of Electrical and Computer Engineering
The University of Texas at Austin
EE 306 Fall 2002
Yale Patt, Instructor
TAs: Asad Bawa, Linda Bigelow, Mustafa Erwa, Lester Guillory, Kevin Major,
Moinuddin Qureshi, Paroma Sen, Tanay Shah, Santhosh Srinath,
Matt Starolis, David Thompson, Vikrant Venkateshwar
Problem Set #5
Due November 4, 2002 before class
Problem Set 5
1. Optional - Problem 8 from PS 4
2. Optional - Problem 9 from PS 4
3. Optional - Question 6.7 from the book.
Textbook errata:
Line x3004 should be 0010 010 0 0001 1000.
Line x300C should be 0000 001 000000101
4.a) Using the iteration construct, write an LC-2 machine language routine that displays exactly 10 Zs on the screen.
b) Suppose you want your program in part (a) to output 14 Zs instead of 10. Modify your program to output 14 Zs by changing only one line in your code.
5. Optional - Notice that we can shift a number to the left by one bit position by adding the number to itself. For example, when the binary number 0011 is added to itself, the result is 0110. Shifting a number one bit position to the right is not as easy. Devise an algorithm to shift the contents of location x3100 to the right by one bit and explain it in 20 words. Draw a flow diagram for your solution like the one shown in Figure 6.3 in the book (page 125).
6. Write an assembly language program to compute the XOR of two memory locations A and B, stored in memory at locations x5000 and x5001, respectively. Store the result in location x5002 in memory. Your program should begin at location x5003 in memory.
(Hint: Draw the truth table for a 1 bit XOR function, then realize the truth table with gates)
7. Problem 7.13 (Page 154) from the textbook.
8.a) Problem 7.8 (Page 153) from the textbook.
b) In the previous subsection, will the error be detected by the assembler?
9. Recall the BUSYNESS pattern for eight machines from Problem 15 of Problem Set 1.
Assume that bits [15:8] of memory location labeled PATTERN is the BUSYNESS pattern for eight machines, and bits [7:0] are all zeros as shown below.
```15 8 7 0
<-BUSYNESS pattern-> <- all zero ->
```
For example, if the BUSYNESS pattern is 01110011, the contents of the memory location PATTERN is 01110011 00000000. The following piece of code uses this bit pattern as input and tries to output the number of machines that are free. For example, the pattern 01110011 00000000 must have produced an output 65410 indicating that the machines 6,5,4,1 and 0 are free. But the code is infested with bugs!! What happens when you try to execute this code?
``` .ORIG x3000 ; INITIALIZATION
AND R1, R1, #0
ADD R1, R1, #7
LD R2, PATTERN ; R2 HOLDS THE BUSYNESS PATTERN
BRn OUTPUT
NEXT ADD R2, R2, R2 ; CHECK IF THE MACHINE IS FREE
BRn OUTPUT
BRp NEXT
; BRz END
OUTPUT LD R0, ASCII ; DISPLAY THE MACHINE NUMBER
ADD R0, R0, R1
TRAP x21
ADD R1, R1, #-1 ; CHECK IF OTHER MACHINES ARE FREE
BRnzp NEXT
END TRAP x25
PATTERN .FILL x7300
ASCII .FILL x0030
.END
```
After some deep thought, the programmer made two changes in the program.
(1) He modified a single line and (2) removed an instruction from its current position and placed it before a different instruction to make his code work. What did the programmer do?
Answer:
1. He changed line ___________________ to ___________________
2. He removed the instruction _____________________ and put it before the instruction ______________________. | 970 | 3,548 | {"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} | 2.96875 | 3 | CC-MAIN-2017-43 | longest | en | 0.816691 |
https://www.crazy-numbers.com/en/12765 | 1,556,020,450,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578602767.67/warc/CC-MAIN-20190423114901-20190423140901-00324.warc.gz | 651,321,103 | 4,394 | Discover a lot of information on the number 12765: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
## Mathematical properties of 12765
Is 12765 a prime number? No
Is 12765 a perfect number? No
Number of divisors 16
List of dividers 1, 3, 5, 15, 23, 37, 69, 111, 115, 185, 345, 555, 851, 2553, 4255, 12765
Sum of divisors 21888
## How to write / spell 12765 in letters?
In letters, the number 12765 is written as: Twelve thousand seven hundred and sixty-five. And in other languages? how does it spell?
12765 in other languages
Write 12765 in english Twelve thousand seven hundred and sixty-five
Write 12765 in french Douze mille sept cent soixante-cinq
Write 12765 in spanish Doce mil setecientos sesenta y cinco
Write 12765 in portuguese Doze mil setecentos sessenta e cinco
## Decomposition of the number 12765
The number 12765 is composed of:
1 iteration of the number 1 : The number 1 (one) represents the uniqueness, the unique, a starting point, a beginning.... Find out more about the number 1
1 iteration of the number 2 : The number 2 (two) represents double, association, cooperation, union, complementarity. It is the symbol of duality.... Find out more about the number 2
1 iteration of the number 7 : The number 7 (seven) represents faith, teaching. It symbolizes reflection, the spiritual life.... Find out more about the number 7
1 iteration of the number 6 : The number 6 (six) is the symbol of harmony. It represents balance, understanding, happiness.... Find out more about the number 6
1 iteration of the number 5 : The number 5 (five) is the symbol of freedom. It represents change, evolution, mobility.... Find out more about the number 5
Other ways to write 12765
In letter Twelve thousand seven hundred and sixty-five
In roman numeral
In binary 11000111011101
In octal 30735
In US dollars USD 12,765.00 (\$)
In euros 12 765,00 EUR (€)
Some related numbers
Previous number 12764
Next number 12766
Next prime number 12781
## Mathematical operations
Operations and solutions
12765*2 = 25530 The double of 12765 is 25530
12765*3 = 38295 The triple of 12765 is 38295
12765/2 = 6382.5 The half of 12765 is 6382.500000
12765/3 = 4255 The third of 12765 is 4255.000000
127652 = 162945225 The square of 12765 is 162945225.000000
127653 = 2079995797125 The cube of 12765 is 2079995797125.000000
√12765 = 112.98229949864 The square root of 12765 is 112.982299
log(12765) = 9.4544623296756 The natural (Neperian) logarithm of 12765 is 9.454462
log10(12765) = 4.1060208191403 The decimal logarithm (base 10) of 12765 is 4.106021
sin(12765) = -0.65111186792108 The sine of 12765 is -0.651112
cos(12765) = -0.7589817754415 The cosine of 12765 is -0.758982
tan(12765) = 0.8578754971321 The tangent of 12765 is 0.857875 | 865 | 2,820 | {"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} | 2.75 | 3 | CC-MAIN-2019-18 | latest | en | 0.748526 |
https://www.12000.org/my_notes/CAS_integration_tests/reports/rubi_4_16_1_graded/test_cases/1_Algebraic_functions/1.2_Trinomial_products/1.2.1_Quadratic/1.2.1.2-d+e_x-%5Em-a+b_x+c_x%5E2-%5Ep/rese1046.htm | 1,695,771,395,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510225.44/warc/CC-MAIN-20230926211344-20230927001344-00835.warc.gz | 651,970,887 | 4,179 | ### 3.1046 $$\int \frac{(c d^2+2 c d e x+c e^2 x^2)^{3/2}}{(d+e x)^4} \, dx$$
Optimal. Leaf size=42 $\frac{c^2 (d+e x) \log (d+e x)}{e \sqrt{c d^2+2 c d e x+c e^2 x^2}}$
[Out]
(c^2*(d + e*x)*Log[d + e*x])/(e*Sqrt[c*d^2 + 2*c*d*e*x + c*e^2*x^2])
________________________________________________________________________________________
Rubi [A] time = 0.0246343, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 32, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.094, Rules used = {642, 608, 31} $\frac{c^2 (d+e x) \log (d+e x)}{e \sqrt{c d^2+2 c d e x+c e^2 x^2}}$
Antiderivative was successfully verified.
[In]
Int[(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(3/2)/(d + e*x)^4,x]
[Out]
(c^2*(d + e*x)*Log[d + e*x])/(e*Sqrt[c*d^2 + 2*c*d*e*x + c*e^2*x^2])
Rule 642
Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[e^m/c^(m/2), Int[(a +
b*x + c*x^2)^(p + m/2), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[p] && EqQ[
2*c*d - b*e, 0] && IntegerQ[m/2]
Rule 608
Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(b/2 + c*x)/Sqrt[a + b*x + c*x^2], Int[1/(b/2
+ c*x), x], x] /; FreeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0]
Rule 31
Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]
Rubi steps
\begin{align*} \int \frac{\left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}{(d+e x)^4} \, dx &=c^2 \int \frac{1}{\sqrt{c d^2+2 c d e x+c e^2 x^2}} \, dx\\ &=\frac{\left (c^2 \left (c d e+c e^2 x\right )\right ) \int \frac{1}{c d e+c e^2 x} \, dx}{\sqrt{c d^2+2 c d e x+c e^2 x^2}}\\ &=\frac{c^2 (d+e x) \log (d+e x)}{e \sqrt{c d^2+2 c d e x+c e^2 x^2}}\\ \end{align*}
Mathematica [A] time = 0.0038717, size = 31, normalized size = 0.74 $\frac{c^2 (d+e x) \log (d+e x)}{e \sqrt{c (d+e x)^2}}$
Antiderivative was successfully verified.
[In]
Integrate[(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(3/2)/(d + e*x)^4,x]
[Out]
(c^2*(d + e*x)*Log[d + e*x])/(e*Sqrt[c*(d + e*x)^2])
________________________________________________________________________________________
Maple [A] time = 0.043, size = 40, normalized size = 1. \begin{align*}{\frac{\ln \left ( ex+d \right ) }{ \left ( ex+d \right ) ^{3}e} \left ( c{e}^{2}{x}^{2}+2\,cdex+c{d}^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((c*e^2*x^2+2*c*d*e*x+c*d^2)^(3/2)/(e*x+d)^4,x)
[Out]
(c*e^2*x^2+2*c*d*e*x+c*d^2)^(3/2)/(e*x+d)^3*ln(e*x+d)/e
________________________________________________________________________________________
Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c*e^2*x^2+2*c*d*e*x+c*d^2)^(3/2)/(e*x+d)^4,x, algorithm="maxima")
[Out]
Exception raised: RuntimeError
________________________________________________________________________________________
Fricas [A] time = 2.2917, size = 89, normalized size = 2.12 \begin{align*} \frac{\sqrt{c e^{2} x^{2} + 2 \, c d e x + c d^{2}} c \log \left (e x + d\right )}{e^{2} x + d e} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c*e^2*x^2+2*c*d*e*x+c*d^2)^(3/2)/(e*x+d)^4,x, algorithm="fricas")
[Out]
sqrt(c*e^2*x^2 + 2*c*d*e*x + c*d^2)*c*log(e*x + d)/(e^2*x + d*e)
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c \left (d + e x\right )^{2}\right )^{\frac{3}{2}}}{\left (d + e x\right )^{4}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c*e**2*x**2+2*c*d*e*x+c*d**2)**(3/2)/(e*x+d)**4,x)
[Out]
Integral((c*(d + e*x)**2)**(3/2)/(d + e*x)**4, x)
________________________________________________________________________________________
Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c*e^2*x^2+2*c*d*e*x+c*d^2)^(3/2)/(e*x+d)^4,x, algorithm="giac")
[Out]
Exception raised: NotImplementedError | 1,903 | 4,415 | {"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} | 3.4375 | 3 | CC-MAIN-2023-40 | latest | en | 0.20234 |
http://docplayer.net/21300835-Unit-conversions-formulas.html | 1,547,643,658,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583657470.23/warc/CC-MAIN-20190116113941-20190116135941-00354.warc.gz | 71,793,618 | 26,619 | # UNIT CONVERSIONS & FORMULAS
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## Transcription
1 4, PSI UNIT CONVERSIONS & FORMULAS MEASUREMENT CONVERSION EXAMPLE 1 in. = 25.4 mm 12 in. x 25.4 = 34.8 mm 1 in. =.254 m 12 in. x.254 = 3.5 m LENGTH 1 ft. =.348 m 12 ft. x.348 = 3.68 m 1 mm =.394 in. 1 mm x.394 = 3.94 in. 1 m = in. 1 m x = in. 1 m = ft. 1 m x = ft. VOLUME WEIGHT PRESSURE FLOW FORCE POWER 1 gal = l 1 gal x = l 1 liter =.2642 gal 1 liters x = gal 1 lb. =.454 kg 1 lbs. x.454 = 4.54 kg 1 kg = 2.25 lbs. 1 kg x 2.25 = 22.5 lbs. 1 psi =.6895 bar 4, psi x.6895 = 2,75 bar 1 bar = psi 2,75 bar x = 4, psi 1 gpm = lpm 3 gpm x = lpm 1 lpm =.2642 gpm 1 lpm x.264 = gpm 1 lb. = N 1 lbs. x = N 1 N =.2248 lbs. 9 N x.2248 = lbs. 1 lb.-ft. = N-m 1 lb.-ft. x = N-m 1 N-m =.737 lb.-ft. 1 N-m x.737 = 73.7 lb.-ft. 1 hp =.7457 kw 5 hp x.7457 = kw 1 kw = hp 5 kw x = 67.5 hp FORMULA ENGLISH UNITS METRIC UNITS Power Requirements* hp = gpm x psi / 155 kw = lpm x bar / 54 *Formulas assume 9% mechanical efficiency I-1
2 4, PSI CONNECTION TYPES Several connection types are available for use with Jetstream 4, psi products. The purpose of this section is to allow users to identify the connections needed and provide information on the features and proper use of each. Note: Always apply thread lubricant (anti-seize) to male threads before assembling Jetstream 4K Port connections. See Fittings, page E-1, for thread lubricants available from Jetstream. JETSTREAM 4K PORT Specially designed to prevent leakage and wear frequently associated with standard metal-to-metal seals by incorporating an o-ring with a metal-tometal cone and seat backup. Repairing leaks requires only o-ring replacement, as opposed to standard high pressure connections which often require re-coning, or component replacement. Utilized in most Jetstream 4, psi products. CONNECTION RECOMMENDED 1" - 12UNF JETSTREAM 4K PUMP DISCHARGE Specially designed for use in all Jetstream 4, psi UNx manifold discharge ports. No metal-to-metal seals which can wear and require manifold replacement. CONNECTION RECOMMENDED 1 3/16" - 16 UN I-2 US & Canada: or
4 4, PSI HIGH PRESSURE LANCE (HP LANCE) CONNECTION TYPES This connection has the same geometry and threads as a standard high pressure tubing connection with the exception that right-hand threaded tubes are available. Male tubing is inserted directly into a female connection without the use of a gland and collar. Used in tube cleaning operations where standard connections are too large in diameter. SIZE CONNECTION 1/4" HP Lance LH 1/4"-28UNF-LH 1/4" HP Lance RH 1/4"-28UNF-RH 3/8" HP Lance LH 3/8"-24UNF-LH 3/8" HP Lance RH 3/8"-24UNF-RH 9/16" HP Lance LH 9/16"-18UNF-LH 9/16" HP Lance RH 9/16"-18UNF-RH RECOMMENDED BUTTON SEAL Utilizes an inexpensive button-style seal which absorbs abuse and wear, preventing damage to the seating surfaces of more costly components. Used in the SP-4 Tornado Gun, Multi-Mode Valve, and in some automated lancing systems. CONNECTION RECOMMENDED 9/16" - 18UNF 4 54 I-4 US & Canada: or
5 4, PSI TYPE M CONNECTION TYPES Designed for fast, simple connection of waterblast hoses and lances. Male conical seat protected from damage by the hose swivel nut. Unlike high pressure connections, the seating surfaces do not rotate against each other when connections are made, resulting in a more reliable, longer lasting connection. SIZE CONNECTION TUBING * RECOMMENDED LEFT-HAND 9/16" Type M M9 9/16"-18UNF /8" Type M M14 7/8"-14UNF /8" Type M M18 1 1/8-12UNF I-5
6 4, PSI HOSE & FLEX LANCE PRESSURE DROP DATA 6 SERIES mm HOSE 5 SERIES mm HOSE SERIES mm HOSE SERIES 28612, & MM HOSE FLOW Flow RATE rate (gpm) ) I-6 US & Canada: or
7 4, PSI RIGID LANCE PRESSURE DROP DATA 9 1/4" OD X.83" ID TUBING 1' 5' 3/8" OD x.125" ID TUBING 1' 5' ' ' I /16" OD X.188" ID TUBING 1' 5' 9/16" OD X.25" ID TUBING ' 8 7 5' ' ' I-7
8 4, PSI CONING TOOL ING TOOL ING AND CONING TOOLS HP CONNECTIONS Coning tool assemblies include tool as well as cutter and collet specific to the connection size and type needed. Threading tool assemblies include tool (consisting of housing and handle) as well as die and guide bushing specific to connection size needed. NOTE: 3/4" & 3/8" coning tools use the same tool with differing cutters & collets. 9/16" coning tools use the same tool with differing cutters. All threading tool assemblies include LH die and all use the same handle and housing. CONING CONNECTION TYPE CONING TOOL AY CONING CUTTER CONING COLLET 1/4" HP /8" HP /16" HP CONNECTION TYPE THREATH TOOL AY LH DIE RH DIE BUSHING 1/4" HP /8" HP /16" HP I-8 US & Canada: or
### PNEUMATIC CONTROL VALVES SERVICE AND MAINTENANCE PROCEDURES
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https://brainmass.com/statistics/regression-analysis/regression-analysis-wages-and-wage-earners-data-set-229148 | 1,601,240,998,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401578485.67/warc/CC-MAIN-20200927183616-20200927213616-00116.warc.gz | 306,581,128 | 11,639 | Explore BrainMass
# Regression analysis Wages and Wage Earners Data Set
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
Please see attachment and help me explain the regression analysis and scatter plot.
I ran this regression analysis. My hypothesis is that there is a positive linear relationship between years of education and salary. I used the following data set for this. (see attachment) Table 1: Wages and Wage Earners Data Set
I NEED to be able to explain the analysis in simple words. I believe that it shows a positive linear relationship. Also help me explain the scatter plot below (see attachment).
Regression Analysis
r² 0.167 n 100
r 0.408 k 1
Std. Error 15550.444 Dep. Var. Wage
ANOVA table
Source SS df MS F p-value
Regression 4,735,209,494.9482 1 4,735,209,494.9482 19.58 2.50E-05
Residual 23,697,996,987.8918 98 241,816,295.7948
Total 28,433,206,482.8400 99
Regression output confidence interval
variables coefficients std. error t (df=98) p-value 95% lower 95% upper
Intercept -699.9501 7,293.6714 -0.096 .9237 -15,174.0032 13,774.1031
Ed 2,477.0943 559.7779 4.425 2.50E-05 1,366.2334 3,587.9552 | 365 | 1,185 | {"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-2020-40 | latest | en | 0.793766 |
http://www.examguruadda.in/2015/12/practice-reasoning-quiz.html | 1,521,697,817,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647777.59/warc/CC-MAIN-20180322053608-20180322073608-00408.warc.gz | 387,526,236 | 19,397 | Practice Reasoning Quiz-Square Puzzle
Dear IBPS Clerk 2015 Aspirants practice Reasoning questions set of Mixed Questions upcoming IBPS Clerk Exam. Try to solve these questions and share time taken to solve these questions.
In Reasoning section for IBPS Clerk 2015 - Puzzles, syllogism, inequality play important role in preliminary exam.
Directions (1-5): Read the following information carefully and answer the given questions.
(i) Eight persons E, F, G, H, I, J, K, and L are seated around a square table- two on each side.
(ii) There are three lady members and they are not seated next to each other.
(iii) J is between L and F.
(iv) G is between I and F.
(v) H, a lady member, is second to the left of J.
(vi) F, a male member, is seated opposite to E, a lady member.
(vii) J is not a female member.
1. Who among the following are the three lady members?
(1) E, H and J
(2) E, G and J
(3) G, H and J
(4) Cannot be Determined
(5) None of These
2. Which of the following is true about I?
(1) I is a male member
(2) I is a female member
(3) Gender of I cannot be determined
(4) Position of I cannot be determined
(5) None of These
3. Which of the following is true about the person sitting between E and H?
(1) A male member
(2) A female member
(3) The person sitting between E and H is K.
(4) Both (1) and (3)
(5) None of These
4. How many persons are sitting between K and F?
(1) One
(2) Two
(3) Three
(4) Cannot be determined
(5) None of These
5. Who among the following is to the immediate left of F?
(1) G
(2) J
(3) I
(4) Cannot be Determined
(5) None of These
Directions—(6–10) Read the following passage carefully and answer the questions given below it.
A, B, C, D, E, F, G and H are sitting around a circular table, facing the centre. A sits third to the left of C and second to the right of E. B sits second to the right of D who is not an immediate neighbour of E. H sits second to the left of F. G is not an immediate neighbour of D.
6. Which of the following pairs has only one person sitting-between them, if the counting is done in clockwise direction?
(1) F, G
(2) H, G
(3) H, C
(4) H, B
(5) None of these
7. Who sits third to the right of E ?
(1) D
(2) G
(3) F
(4) B
(5) None of these
8. What is the position of G with respect to A’s position ?
(1) Immediately to the right
(2) Second to the left
(3) Third to the right
(4) Third to the left
(5) Fourth to the right
9. Who sits between E and A?
(1) F
(2) D
(3) G
(4) B
(5) None of these
10. Starting from A’s position, if all the eight are arranged in alphabetical order in clockwise direction, the seating position of how many members (excluding A) would remain unchanged?
(1) None
(2) One
(3) Two
(4) Three
(5) Four | 815 | 2,704 | {"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} | 3.65625 | 4 | CC-MAIN-2018-13 | latest | en | 0.951636 |
http://gams.cam.nist.gov/15.8 | 1,493,353,243,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917122739.53/warc/CC-MAIN-20170423031202-00183-ip-10-145-167-34.ec2.internal.warc.gz | 151,492,938 | 25,462 | # §15.8 Transformations of Variable
## §15.8(i) Linear Transformations
All functions in this subsection and §15.8(ii) assume their principal values.
15.8.1 $\mathop{\mathbf{F}\/}\nolimits\!\left({a,b\atop c};z\right)=(1-z)^{-a}\mathop{% \mathbf{F}\/}\nolimits\!\left({a,c-b\atop c};\frac{z}{z-1}\right)=(1-z)^{-b}% \mathop{\mathbf{F}\/}\nolimits\!\left({c-a,b\atop c};\frac{z}{z-1}\right)=(1-z% )^{c-a-b}\mathop{\mathbf{F}\/}\nolimits\!\left({c-a,c-b\atop c};z\right),$ $|\mathop{\mathrm{ph}\/}\nolimits\!\left(1-z\right)|<\pi$.
15.8.2 $\displaystyle\frac{\mathop{\sin\/}\nolimits\!\left(\pi(b-a)\right)}{\pi}% \mathop{\mathbf{F}\/}\nolimits\!\left({a,b\atop c};z\right)$ $\displaystyle=\frac{(-z)^{-a}}{\mathop{\Gamma\/}\nolimits\!\left(b\right)% \mathop{\Gamma\/}\nolimits\!\left(c-a\right)}\mathop{\mathbf{F}\/}\nolimits\!% \left({a,a-c+1\atop a-b+1};\frac{1}{z}\right)-\frac{(-z)^{-b}}{\mathop{\Gamma% \/}\nolimits\!\left(a\right)\mathop{\Gamma\/}\nolimits\!\left(c-b\right)}% \mathop{\mathbf{F}\/}\nolimits\!\left({b,b-c+1\atop b-a+1};\frac{1}{z}\right),$ $|\mathop{\mathrm{ph}\/}\nolimits\!\left(-z\right)|<\pi$. 15.8.3 $\displaystyle\frac{\mathop{\sin\/}\nolimits\!\left(\pi(b-a)\right)}{\pi}% \mathop{\mathbf{F}\/}\nolimits\!\left({a,b\atop c};z\right)$ $\displaystyle=\frac{(1-z)^{-a}}{\mathop{\Gamma\/}\nolimits\!\left(b\right)% \mathop{\Gamma\/}\nolimits\!\left(c-a\right)}\mathop{\mathbf{F}\/}\nolimits\!% \left({a,c-b\atop a-b+1};\frac{1}{1-z}\right)-\frac{(1-z)^{-b}}{\mathop{\Gamma% \/}\nolimits\!\left(a\right)\mathop{\Gamma\/}\nolimits\!\left(c-b\right)}% \mathop{\mathbf{F}\/}\nolimits\!\left({b,c-a\atop b-a+1};\frac{1}{1-z}\right),$ $|\mathop{\mathrm{ph}\/}\nolimits\!\left(-z\right)|<\pi$. 15.8.4 $\displaystyle\frac{\mathop{\sin\/}\nolimits\!\left(\pi(c-a-b)\right)}{\pi}% \mathop{\mathbf{F}\/}\nolimits\!\left({a,b\atop c};z\right)$ $\displaystyle=\frac{1}{\mathop{\Gamma\/}\nolimits\!\left(c-a\right)\mathop{% \Gamma\/}\nolimits\!\left(c-b\right)}\mathop{\mathbf{F}\/}\nolimits\!\left({a,% b\atop a+b-c+1};1-z\right)-\frac{(1-z)^{c-a-b}}{\mathop{\Gamma\/}\nolimits\!% \left(a\right)\mathop{\Gamma\/}\nolimits\!\left(b\right)}\mathop{\mathbf{F}\/}% \nolimits\!\left({c-a,c-b\atop c-a-b+1};1-z\right),$ $|\mathop{\mathrm{ph}\/}\nolimits z|<\pi$, $|\mathop{\mathrm{ph}\/}\nolimits\!\left(1-z\right)|<\pi$. 15.8.5 $\displaystyle\frac{\mathop{\sin\/}\nolimits\!\left(\pi(c-a-b)\right)}{\pi}% \mathop{\mathbf{F}\/}\nolimits\!\left({a,b\atop c};z\right)$ $\displaystyle=\frac{z^{-a}}{\mathop{\Gamma\/}\nolimits\!\left(c-a\right)% \mathop{\Gamma\/}\nolimits\!\left(c-b\right)}\mathop{\mathbf{F}\/}\nolimits\!% \left({a,a-c+1\atop a+b-c+1};1-\frac{1}{z}\right)-\frac{(1-z)^{c-a-b}z^{a-c}}{% \mathop{\Gamma\/}\nolimits\!\left(a\right)\mathop{\Gamma\/}\nolimits\!\left(b% \right)}\mathop{\mathbf{F}\/}\nolimits\!\left({c-a,1-a\atop c-a-b+1};1-\frac{1% }{z}\right),$ $|\mathop{\mathrm{ph}\/}\nolimits z|<\pi$, $|\mathop{\mathrm{ph}\/}\nolimits\!\left(1-z\right)|<\pi$.
For an alternative version of the transformations (15.8.2) and (15.8.3), see (15.10.25), and for an alternative version of the transformations (15.8.4) and (15.8.5), see (15.10.21).
## §15.8(ii) Linear Transformations: Limiting Cases
With $m=0,1,2,\dots$, polynomial cases of (15.8.2)–(15.8.5) are given by
15.8.6 $\displaystyle\mathop{F\/}\nolimits\!\left({-m,b\atop c};z\right)$ $\displaystyle=\frac{(b)_{m}}{(c)_{m}}(-z)^{m}\mathop{F\/}\nolimits\!\left({-m,% 1-c-m\atop 1-b-m};\frac{1}{z}\right)=\frac{(b)_{m}}{(c)_{m}}(1-z)^{m}\mathop{F% \/}\nolimits\!\left({-m,c-b\atop 1-b-m};\frac{1}{1-z}\right),$ 15.8.7 $\displaystyle\mathop{F\/}\nolimits\!\left({-m,b\atop c};z\right)$ $\displaystyle=\frac{(c-b)_{m}}{(c)_{m}}\mathop{F\/}\nolimits\!\left({-m,b\atop b% -c-m+1};1-z\right)=\frac{(c-b)_{m}}{(c)_{m}}z^{m}\mathop{F\/}\nolimits\!\left(% {-m,1-c-m\atop b-c-m+1};1-\frac{1}{z}\right),$
with the understanding that if $b=-\ell$, $\ell=0,1,2,\dots$, then $m\leq\ell$.
When $b-a$ is an integer limits are taken in (15.8.2) and (15.8.3) as follows.
If $b-a$ is a nonnegative integer, then
15.8.8 $\mathop{\mathbf{F}\/}\nolimits\!\left({a,a+m\atop c};z\right)=\frac{(-z)^{-a}}% {\mathop{\Gamma\/}\nolimits\!\left(a+m\right)}\sum_{k=0}^{m-1}\frac{(a)_{k}(m-% k-1)!}{k!\mathop{\Gamma\/}\nolimits\!\left(c-a-k\right)}z^{-k}+\frac{(-z)^{-a}% }{\mathop{\Gamma\/}\nolimits\!\left(a\right)}\sum_{k=0}^{\infty}\frac{(a+m)_{k% }}{k!(k+m)!\mathop{\Gamma\/}\nolimits\!\left(c-a-k-m\right)}(-1)^{k}z^{-k-m}\*% \left(\mathop{\ln\/}\nolimits(-z)+\mathop{\psi\/}\nolimits\!\left(k+1\right)+% \mathop{\psi\/}\nolimits\!\left(k+m+1\right)-\mathop{\psi\/}\nolimits\!\left(a% +k+m\right)-\mathop{\psi\/}\nolimits\!\left(c-a-k-m\right)\right),$ $|z|>1,|\mathop{\mathrm{ph}\/}\nolimits(-z)|<\pi$,
15.8.9 $\mathop{\mathbf{F}\/}\nolimits\!\left({a,a+m\atop c};z\right)=\frac{(1-z)^{-a}% }{\mathop{\Gamma\/}\nolimits\!\left(a+m\right)\mathop{\Gamma\/}\nolimits\!% \left(c-a\right)}\sum_{k=0}^{m-1}\frac{(a)_{k}(c-a-m)_{k}(m-k-1)!}{k!}(z-1)^{-% k}+\frac{(-1)^{m}(1-z)^{-a-m}}{\mathop{\Gamma\/}\nolimits\!\left(a\right)% \mathop{\Gamma\/}\nolimits\!\left(c-a-m\right)}\sum_{k=0}^{\infty}\frac{(a+m)_% {k}(c-a)_{k}}{k!(k+m)!}(1-z)^{-k}\*(\mathop{\ln\/}\nolimits\!\left(1-z\right)+% \mathop{\psi\/}\nolimits\!\left(k+1\right)+\mathop{\psi\/}\nolimits\!\left(k+m% +1\right)-\mathop{\psi\/}\nolimits\!\left(a+k+m\right)-\mathop{\psi\/}% \nolimits\!\left(c-a+k\right)),$ $|z-1|>1,|\mathop{\mathrm{ph}\/}\nolimits(1-z)|<\pi$.
In (15.8.8) when $c-a-k-m$ is a nonpositive integer $\ifrac{\mathop{\psi\/}\nolimits\!\left(c-a-k-m\right)}{\mathop{\Gamma\/}% \nolimits\!\left(c-a-k-m\right)}$ is interpreted as $(-1)^{m+k+a-c+1}(m+k+a-c)!$. Also, if $a$ is a nonpositive integer, then (15.8.6) applies.
Alternatively, if $b-a$ is a negative integer, then we interchange $a$ and $b$ in $\mathop{\mathbf{F}\/}\nolimits\!\left(a,b;c;z\right)$.
In a similar way, when $c-a-b$ is an integer limits are taken in (15.8.4) and (15.8.5) as follows.
If $c-a-b$ is a nonnegative integer, then
15.8.10 $\mathop{\mathbf{F}\/}\nolimits\!\left({a,b\atop a+b+m};z\right)=\frac{1}{% \mathop{\Gamma\/}\nolimits\!\left(a+m\right)\mathop{\Gamma\/}\nolimits\!\left(% b+m\right)}\sum_{k=0}^{m-1}\frac{(a)_{k}(b)_{k}(m-k-1)!}{k!}(z-1)^{k}-\frac{(z% -1)^{m}}{\mathop{\Gamma\/}\nolimits\!\left(a\right)\mathop{\Gamma\/}\nolimits% \!\left(b\right)}\sum_{k=0}^{\infty}\frac{(a+m)_{k}(b+m)_{k}}{k!(k+m)!}(1-z)^{% k}\*\left(\mathop{\ln\/}\nolimits(1-z)-\mathop{\psi\/}\nolimits\!\left(k+1% \right)-\mathop{\psi\/}\nolimits\!\left(k+m+1\right)+\mathop{\psi\/}\nolimits% \!\left(a+k+m\right)+\mathop{\psi\/}\nolimits\!\left(b+k+m\right)\right),$ $|z-1|<1,|\mathop{\mathrm{ph}\/}\nolimits(1-z)|<\pi$,
15.8.11 $\mathop{\mathbf{F}\/}\nolimits\!\left({a,b\atop a+b+m};z\right)=\frac{z^{-a}}{% \mathop{\Gamma\/}\nolimits\!\left(a+m\right)}\sum_{k=0}^{m-1}\frac{(a)_{k}(m-k% -1)!}{k!\mathop{\Gamma\/}\nolimits\!\left(b+m-k\right)}\left(1-\frac{1}{z}% \right)^{k}-\frac{z^{-a}}{\mathop{\Gamma\/}\nolimits\!\left(a\right)}\sum_{k=0% }^{\infty}\frac{(a+m)_{k}}{k!(k+m)!\mathop{\Gamma\/}\nolimits\!\left(b-k\right% )}(-1)^{k}\left(1-\frac{1}{z}\right)^{k+m}\*\left(\mathop{\ln\/}\nolimits\left% (\frac{1-z}{z}\right)-\mathop{\psi\/}\nolimits\!\left(k+1\right)-\mathop{\psi% \/}\nolimits\!\left(k+m+1\right)+\mathop{\psi\/}\nolimits\!\left(a+k+m\right)+% \mathop{\psi\/}\nolimits\!\left(b-k\right)\right),$ $\Re{z}>\tfrac{1}{2},|\mathop{\mathrm{ph}\/}\nolimits z|<\pi,|\mathop{\mathrm{% ph}\/}\nolimits(1-z)|<\pi$.
In (15.8.11) when $b-k$ is a nonpositive integer, $\ifrac{\mathop{\psi\/}\nolimits\!\left(b-k\right)}{\mathop{\Gamma\/}\nolimits% \!\left(b-k\right)}$ is interpreted as $(-1)^{k-b+1}(k-b)!$. Also, if $a$ or $b$ or both are nonpositive integers, then (15.8.7) applies.
Lastly, if $c-a-b$ is a negative integer, then we first apply the transformation
15.8.12 $\mathop{\mathbf{F}\/}\nolimits\!\left(a,b;a+b-m;z\right)=(1-z)^{-m}\mathop{% \mathbf{F}\/}\nolimits\!\left(\tilde{a},\tilde{b};\tilde{a}+\tilde{b}+m;z% \right),$ $\tilde{a}=a-m,\tilde{b}=b-m$.
A quadratic transformation relates two hypergeometric functions, with the variable in one a quadratic function of the variable in the other, possibly combined with a fractional linear transformation.
A necessary and sufficient condition that there exists a quadratic transformation is that at least one of the equations shown in Table 15.8.1 is satisfied.
The hypergeometric functions that correspond to Groups 1 and 2 have $z$ as variable. The hypergeometric functions that correspond to Groups 3 and 4 have a nonlinear function of $z$ as variable. The transformation formulas between two hypergeometric functions in Group 2, or two hypergeometric functions in Group 3, are the linear transformations (15.8.1).
In the equations that follow in this subsection all functions take their principal values.
### Group 1 $\longrightarrow$ Group 3
15.8.13 $\displaystyle\mathop{F\/}\nolimits\!\left({a,b\atop 2b};z\right)$ $\displaystyle=\left(1-\tfrac{1}{2}z\right)^{-a}\mathop{F\/}\nolimits\!\left({% \tfrac{1}{2}a,\tfrac{1}{2}a+\tfrac{1}{2}\atop b+\tfrac{1}{2}};\left(\frac{z}{2% -z}\right)^{2}\right),$ $|\mathop{\mathrm{ph}\/}\nolimits\!\left(1-z\right)|<\pi$, 15.8.14 $\displaystyle\mathop{F\/}\nolimits\!\left({a,b\atop 2b};z\right)$ $\displaystyle=\left(1-z\right)^{-\ifrac{a}{2}}\mathop{F\/}\nolimits\!\left({% \tfrac{1}{2}a,b-\tfrac{1}{2}a\atop b+\tfrac{1}{2}};\frac{z^{2}}{4z-4}\right),$ $|\mathop{\mathrm{ph}\/}\nolimits\!\left(1-z\right)|<\pi$.
### Group 2 $\longrightarrow$ Group 3
15.8.15 $\displaystyle\mathop{F\/}\nolimits\!\left({a,b\atop a-b+1};z\right)$ $\displaystyle=(1+z)^{-a}\mathop{F\/}\nolimits\!\left({\frac{1}{2}a,\frac{1}{2}% a+\frac{1}{2}\atop a-b+1};\frac{4z}{(1+z)^{2}}\right),$ $|z|<1$, 15.8.16 $\displaystyle\mathop{F\/}\nolimits\!\left({a,b\atop a-b+1};z\right)$ $\displaystyle=(1-z)^{-a}\mathop{F\/}\nolimits\!\left({\frac{1}{2}a,\frac{1}{2}% a-b+\frac{1}{2}\atop a-b+1};\frac{-4z}{(1-z)^{2}}\right),$ $|z|<1$.
15.8.17 $\displaystyle\mathop{F\/}\nolimits\!\left({a,b\atop\frac{1}{2}(a+b+1)};z\right)$ $\displaystyle=(1-2z)^{-a}\mathop{F\/}\nolimits\!\left({\frac{1}{2}a,\frac{1}{2% }a+\frac{1}{2}\atop\frac{1}{2}(a+b+1)};\frac{4z(z-1)}{(1-2z)^{2}}\right),$ $\Re{z}<\frac{1}{2}$, 15.8.18 $\displaystyle\mathop{F\/}\nolimits\!\left({a,b\atop\frac{1}{2}(a+b+1)};z\right)$ $\displaystyle=\mathop{F\/}\nolimits\!\left({\frac{1}{2}a,\frac{1}{2}b\atop% \frac{1}{2}(a+b+1)};4z(1-z)\right),$ $\Re{z}<\frac{1}{2}$.
15.8.19 $\displaystyle\mathop{F\/}\nolimits\!\left({a,1-a\atop c};z\right)$ $\displaystyle=(1-2z)^{1-a-c}(1-z)^{c-1}\mathop{F\/}\nolimits\!\left({\frac{1}{% 2}(a+c),\frac{1}{2}(a+c-1)\atop c};\frac{4z(z-1)}{(1-2z)^{2}}\right),$ $\Re{z}<\frac{1}{2}$, 15.8.20 $\displaystyle\mathop{F\/}\nolimits\!\left({a,1-a\atop c};z\right)$ $\displaystyle=(1-z)^{c-1}\mathop{F\/}\nolimits\!\left({\frac{1}{2}(c-a),\frac{% 1}{2}(a+c-1)\atop c};4z(1-z)\right),$ $\Re{z}<\frac{1}{2}$.
### Group 2 $\longrightarrow$ Group 1
15.8.21 $\displaystyle\mathop{F\/}\nolimits\!\left({a,b\atop a-b+1};z\right)$ $\displaystyle=\left(1+\sqrt{z}\right)^{-2a}\mathop{F\/}\nolimits\!\left({a,a-b% +\tfrac{1}{2}\atop 2a-2b+1};\frac{4\sqrt{z}}{(1+\sqrt{z})^{2}}\right),$ $|\mathop{\mathrm{ph}\/}\nolimits z|<\pi$, $|z|<1$. 15.8.22 $\displaystyle\mathop{F\/}\nolimits\!\left({a,b\atop\tfrac{1}{2}(a+b+1)};z\right)$ $\displaystyle=\left(\frac{\sqrt{1-z^{-1}}-1}{\sqrt{1-z^{-1}}+1}\right)^{a}% \mathop{F\/}\nolimits\!\left({a,\tfrac{1}{2}(a+b)\atop a+b};\frac{4\sqrt{1-z^{% -1}}}{\left(\sqrt{1-z^{-1}}+1\right)^{2}}\right),$ $|\mathop{\mathrm{ph}\/}\nolimits\!\left(-z\right)|<\pi$, $\Re{z}<\frac{1}{2}$.
15.8.23 $\mathop{F\/}\nolimits\!\left({a,1-a\atop c};z\right)=\left(\sqrt{1-z^{-1}}-1% \right)^{1-a}\left(\sqrt{1-z^{-1}}+1\right)^{a-2c+1}\left(1-z^{-1}\right)^{c-1% }\mathop{F\/}\nolimits\!\left({c-a,c-\tfrac{1}{2}\atop 2c-1};\frac{4\sqrt{1-z^% {-1}}}{\left(\sqrt{1-z^{-1}}+1\right)^{2}}\right),$ $|\mathop{\mathrm{ph}\/}\nolimits\!\left(-z\right)|<\pi$, $\Re{z}<\frac{1}{2}$.
### Group 2 $\longrightarrow$ Group 4
15.8.24 $\mathop{F\/}\nolimits\!\left({a,b\atop a-b+1};z\right)=(1-z)^{-a}\frac{\mathop% {\Gamma\/}\nolimits\!\left(a-b+1\right)\mathop{\Gamma\/}\nolimits\!\left(% \tfrac{1}{2}\right)}{\mathop{\Gamma\/}\nolimits\!\left(\tfrac{1}{2}a+\tfrac{1}% {2}\right)\mathop{\Gamma\/}\nolimits\!\left(\tfrac{1}{2}a-b+1\right)}\mathop{F% \/}\nolimits\!\left({\tfrac{1}{2}a,\tfrac{1}{2}a-b+\tfrac{1}{2}\atop\tfrac{1}{% 2}};\left(\frac{z+1}{z-1}\right)^{2}\right)+(1+z)(1-z)^{-a-1}\frac{\mathop{% \Gamma\/}\nolimits\!\left(a-b+1\right)\mathop{\Gamma\/}\nolimits\!\left(-% \tfrac{1}{2}\right)}{\mathop{\Gamma\/}\nolimits\!\left(\tfrac{1}{2}a\right)% \mathop{\Gamma\/}\nolimits\!\left(\tfrac{1}{2}a-b+\tfrac{1}{2}\right)}\mathop{% F\/}\nolimits\!\left({\tfrac{1}{2}a+\tfrac{1}{2},\tfrac{1}{2}a-b+1\atop\tfrac{% 3}{2}};\left(\frac{z+1}{z-1}\right)^{2}\right),$ $|\mathop{\mathrm{ph}\/}\nolimits\!\left(-z\right)|<\pi$.
15.8.25 $\mathop{F\/}\nolimits\!\left({a,b\atop\tfrac{1}{2}(a+b+1)};z\right)=\frac{% \mathop{\Gamma\/}\nolimits\!\left(\tfrac{1}{2}(a+b+1)\right)\mathop{\Gamma\/}% \nolimits\!\left(\tfrac{1}{2}\right)}{\mathop{\Gamma\/}\nolimits\!\left(\tfrac% {1}{2}a+\tfrac{1}{2}\right)\mathop{\Gamma\/}\nolimits\!\left(\tfrac{1}{2}b+% \tfrac{1}{2}\right)}\mathop{F\/}\nolimits\!\left({\tfrac{1}{2}a,\tfrac{1}{2}b% \atop\tfrac{1}{2}};(1-2z)^{2}\right)+(1-2z)\frac{\mathop{\Gamma\/}\nolimits\!% \left(\tfrac{1}{2}(a+b+1)\right)\mathop{\Gamma\/}\nolimits\!\left(-\tfrac{1}{2% }\right)}{\mathop{\Gamma\/}\nolimits\!\left(\tfrac{1}{2}a\right)\mathop{\Gamma% \/}\nolimits\!\left(\tfrac{1}{2}b\right)}\mathop{F\/}\nolimits\!\left({\tfrac{% 1}{2}a+\tfrac{1}{2},\tfrac{1}{2}b+\tfrac{1}{2}\atop\tfrac{3}{2}};(1-2z)^{2}% \right),$ $|\mathop{\mathrm{ph}\/}\nolimits z|<\pi$, $|\mathop{\mathrm{ph}\/}\nolimits\!\left(1-z\right)|<\pi$.
15.8.26 $\mathop{F\/}\nolimits\!\left({a,1-a\atop c};z\right)=(1-z)^{c-1}\frac{\mathop{% \Gamma\/}\nolimits\!\left(c\right)\mathop{\Gamma\/}\nolimits\!\left(\tfrac{1}{% 2}\right)}{\mathop{\Gamma\/}\nolimits\!\left(\tfrac{1}{2}(c-a+1)\right)\mathop% {\Gamma\/}\nolimits\!\left(\tfrac{1}{2}c+\tfrac{1}{2}a\right)}\mathop{F\/}% \nolimits\!\left({\tfrac{1}{2}c-\tfrac{1}{2}a,\tfrac{1}{2}c+\tfrac{1}{2}a-% \tfrac{1}{2}\atop\tfrac{1}{2}};(1-2z)^{2}\right)+(1-2z)(1-z)^{c-1}\frac{% \mathop{\Gamma\/}\nolimits\!\left(c\right)\mathop{\Gamma\/}\nolimits\!\left(-% \tfrac{1}{2}\right)}{\mathop{\Gamma\/}\nolimits\!\left(\tfrac{1}{2}c-\tfrac{1}% {2}a\right)\mathop{\Gamma\/}\nolimits\!\left(\tfrac{1}{2}(c+a-1)\right)}% \mathop{F\/}\nolimits\!\left({\tfrac{1}{2}c-\tfrac{1}{2}a+\tfrac{1}{2},\tfrac{% 1}{2}c+\tfrac{1}{2}a\atop\tfrac{3}{2}};(1-2z)^{2}\right),$ $|\mathop{\mathrm{ph}\/}\nolimits z|<\pi$, $|\mathop{\mathrm{ph}\/}\nolimits\!\left(1-z\right)|<\pi$.
### Group 4 $\longrightarrow$ Group 2
15.8.27 $\frac{2\mathop{\Gamma\/}\nolimits\!\left(\tfrac{1}{2}\right)\mathop{\Gamma\/}% \nolimits\!\left(a+b+\tfrac{1}{2}\right)}{\mathop{\Gamma\/}\nolimits\!\left(a+% \tfrac{1}{2}\right)\mathop{\Gamma\/}\nolimits\!\left(b+\tfrac{1}{2}\right)}% \mathop{F\/}\nolimits\!\left(a,b;\tfrac{1}{2};z\right)=\mathop{F\/}\nolimits\!% \left(2a,2b;a+b+\tfrac{1}{2};\tfrac{1}{2}-\tfrac{1}{2}\sqrt{z}\right)+\mathop{% F\/}\nolimits\!\left(2a,2b;a+b+\tfrac{1}{2};\tfrac{1}{2}+\tfrac{1}{2}\sqrt{z}% \right),$ $|\mathop{\mathrm{ph}\/}\nolimits z|<\pi$, $|\mathop{\mathrm{ph}\/}\nolimits\!\left(1-z\right)|<\pi$.
15.8.28 $\frac{2\sqrt{z}\mathop{\Gamma\/}\nolimits\!\left(-\tfrac{1}{2}\right)\mathop{% \Gamma\/}\nolimits\!\left(a+b-\tfrac{1}{2}\right)}{\mathop{\Gamma\/}\nolimits% \!\left(a-\tfrac{1}{2}\right)\mathop{\Gamma\/}\nolimits\!\left(b-\tfrac{1}{2}% \right)}\mathop{F\/}\nolimits\!\left(a,b;\tfrac{3}{2};z\right)=\mathop{F\/}% \nolimits\!\left(2a-1,2b-1;a+b-\tfrac{1}{2};\tfrac{1}{2}-\tfrac{1}{2}\sqrt{z}% \right)-\mathop{F\/}\nolimits\!\left(2a-1,2b-1;a+b-\tfrac{1}{2};\tfrac{1}{2}+% \tfrac{1}{2}\sqrt{z}\right),$ $|\mathop{\mathrm{ph}\/}\nolimits z|<\pi$, $|\mathop{\mathrm{ph}\/}\nolimits\!\left(1-z\right)|<\pi$.
When the intersection of two groups in Table 15.8.1 is not empty there exist special quadratic transformations, with only one free parameter, between two hypergeometric functions in the same group.
### Examples
$b=\tfrac{1}{3}a+\tfrac{1}{3}$, $c=2b=a-b+1$ in Groups 1 and 2.
(15.8.21) becomes
15.8.29 $\mathop{F\/}\nolimits\!\left({a,\tfrac{1}{3}a+\tfrac{1}{3}\atop\tfrac{2}{3}a+% \tfrac{2}{3}};z\right)=\left(1+\sqrt{z}\right)^{-2a}\*\mathop{F\/}\nolimits\!% \left({a,\tfrac{2}{3}a+\tfrac{1}{6}\atop\tfrac{4}{3}a+\tfrac{1}{3}};\frac{4% \sqrt{z}}{(1+\sqrt{z})^{2}}\right).$
This is a quadratic transformation between two cases in Group 1.
We can also use (15.8.13), followed by the inverse of (15.8.15), and obtain
15.8.30 $\left(1-\tfrac{1}{2}z\right)^{-a}\mathop{F\/}\nolimits\!\left({\tfrac{1}{2}a,% \tfrac{1}{2}a+\tfrac{1}{2}\atop\tfrac{1}{3}a+\tfrac{5}{6}};\left(\frac{z}{2-z}% \right)^{2}\right)=\mathop{F\/}\nolimits\!\left({a,\tfrac{1}{3}a+\tfrac{1}{3}% \atop\tfrac{2}{3}a+\tfrac{2}{3}};z\right)=(1+z)^{-a}\mathop{F\/}\nolimits\!% \left({\tfrac{1}{2}a,\tfrac{1}{2}a+\tfrac{1}{2}\atop\tfrac{2}{3}a+\tfrac{2}{3}% };\frac{4z}{(1+z)^{2}}\right),$
which is a quadratic transformation between two cases in Group 3.
For further examples see Andrews et al. (1999, pp. 130–132 and 176–177).
## §15.8(v) Cubic Transformations
### Examples
15.8.31 $\mathop{F\/}\nolimits\!\left({3a,3a+\frac{1}{2}\atop 4a+\frac{2}{3}};z\right)=% \left(1-\tfrac{9}{8}z\right)^{-2a}\*\mathop{F\/}\nolimits\!\left({a,a+\frac{1}% {2}\atop 2a+\frac{5}{6}};\frac{27z^{2}(z-1)}{(9z-8)^{2}}\right),$ $\Re{z}<\frac{8}{9}$.
With $\zeta=e^{\ifrac{2\pi\mathrm{i}}{3}}(1-z)/\left(z-e^{\ifrac{4\pi\mathrm{i}}{3}}\right)$
15.8.32 $\frac{\left(1-z^{3}\right)^{a}}{\left(-z\right)^{3a}}\left(\frac{1}{\mathop{% \Gamma\/}\nolimits\!\left(a+\frac{2}{3}\right)\mathop{\Gamma\/}\nolimits\!% \left(\frac{2}{3}\right)}\mathop{F\/}\nolimits\!\left({a,a+\frac{1}{3}\atop% \frac{2}{3}};z^{-3}\right)+\frac{e^{\frac{1}{3}\pi\mathrm{i}}}{z\mathop{\Gamma% \/}\nolimits\!\left(a\right)\mathop{\Gamma\/}\nolimits\!\left(\frac{4}{3}% \right)}\mathop{F\/}\nolimits\!\left({a+\frac{1}{3},a+\frac{2}{3}\atop\frac{4}% {3}};z^{-3}\right)\right)=\frac{3^{\frac{3}{2}a+\frac{1}{2}}e^{\frac{1}{2}a\pi% \mathrm{i}}\mathop{\Gamma\/}\nolimits\!\left(a+\frac{1}{3}\right)(1-\zeta)^{a}% }{2\pi\mathop{\Gamma\/}\nolimits\!\left(2a+\frac{2}{3}\right)(-\zeta)^{2a}}% \mathop{F\/}\nolimits\!\left({a+\frac{1}{3},3a\atop 2a+\frac{2}{3}};\zeta^{-1}% \right),$ $|z|>1$, $|\mathop{\mathrm{ph}\/}\nolimits\!\left(-z\right)|<\frac{1}{3}\pi$.
### Ramanujan’s Cubic Transformation
15.8.33 $\mathop{F\/}\nolimits\!\left({\frac{1}{3},\frac{2}{3}\atop 1};1-\left(\frac{1-% z}{1+2z}\right)^{3}\right)=(1+2z)\mathop{F\/}\nolimits\!\left({\frac{1}{3},% \frac{2}{3}\atop 1};z^{3}\right),$
provided that $z$ lies in the intersection of the open disks $\left|z-\frac{1}{4}\pm\frac{1}{4}\sqrt{3}\mathrm{i}\right|<\frac{1}{2}\sqrt{3}$, or equivalently, $\left|\mathop{\mathrm{ph}\/}\nolimits\!\left(\ifrac{(1-z)}{(1+2z)}\right)% \right|<\pi/3$. This is used in a cubic analog of the arithmetic-geometric mean. See Borwein and Borwein (1991), and also Berndt et al. (1995).
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Longitude Essay The Parliament was forced to find the solution to the troublesome longitude problem. William Whiston and Humphrey were mathematicians and friends who made a petition signed by “Captains of Her Majesty’s Ships‚ Merchant-Men” This piece of parchment demanded that the government pay attention to the longitude problem. A Parliamentary committee assembled to respond to its challenge in June of 1714. The committee sought out advice from Sir Isaac Newton. They established the Longitude
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_____________Tiffany Caufield_____________ Section __________________ EXERCISE 2 PROBLEMS/SOLUTIONS—PART I 1. Using a globe‚ determine the latitude and longitude (to the nearest degree) of the following cities. Be sure to indicate if the location is north or south latitude‚ and east or west longitude. City Latitude Longitude (a) 42 N 88 W (b) Tokyo‚ Japan 36 N 140 E (c) Sydney‚ Australia 34 S 151 E (d) Singapore 1N 104 E (e) 2. Chicago
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equator‚ the imaginary line which divides our planet into the northern and southern hemispheres. 90° north is the North Pole and 90° south is the South Pole. Longitude The vertical longitude lines are also known as meridians. They converge at the poles and are widest at the equator (about 69 miles or 111 km apart). Zero degrees longitude is located at Greenwich‚ England (0°). The degrees continue 180° east and 180° west where they meet and form the International Date Line in the Pacific Ocean
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used to measure longitude (Lesson 02.01) (use the first resource in this lesson to help you) • Be able to list and describe the 2 main tools used to measure longitude in early navigation Sextant- it had a telescope and an index mirror‚ but was very time consuming Chronometers Current Navigation Tools (Lesson 02.01) • What tool do we currently use to determine location (latitude and longitude)? Maps‚ or GPS Longitude and Latitude (Lesson 02.01) • Know the ways that longitude and latitude
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for every 15° of longitude. Therefore‚ if we know the local times at two points on Earth‚ we can use the difference between them to calculate how far apart those places are in longitude‚ east or west. This idea was very important to sailors and navigators in the 17th century. They could measure the local time‚ wherever they were‚ by observing the Sun‚ but navigation required that they also know the time at some reference point‚ e.g. Greenwich ‚ in order to calculate their longitude. Although accurate
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► USE THE “A Printable Latitude & Longitude Map of the World” TO ANSWER QUESTIONS 1-5. As well as the rest of the Power Point. 1. List the seven large landmasses known as continents. 1.___North America___ 2.___South America____ 3._____Asia_______ 2. The Prime_ Meridian_ is the line of longitude that divides the Eastern Hemisphere from the Western Hemisphere. 3. Which two continents lie completely in the Southern Hemisphere? _____Australia and Antarctica__
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The longitude against which Indian Standard time is set passes through Allahabad‚ India. British India did not adopt the international standard time zones until 1905 when the meridian passing through Allahabad at 82.5º east (of Greenwich Meridian) longitude was picked as the central meridian for India‚ corresponding to a single time zone for the country at 5 hours and 30 minutes in advance of GMT. This went into force on January 1‚ 1906. However‚ Calcutta time was officially maintained as a separate
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commerce‚ surveillance and tracking. This project presents a small application based on Global Positioning System. It depicts the use of GPS module/receiver to find latitude and longitude of its location. The data obtained from GPS receiver (GPGGA sentence) is processed by the microcontroller to extract its latitude and longitude values. The GPS Module has been interfaced with AT89C51 and the location values are displayed on a 16x2 LCD interface. Description How to interface GPS with 8051 Microcontroller
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Air Navigation - The Pilot/Navigator All air navigation involves basic principles that apply to all aeroplanes‚ from the simplest trainer to the most sophisticated passenger jets. These basic principles are discussed in this manual. Since The Air Pilot’s Manual is a training programme for the Private Pilot’s Licence (PPL)‚ we will concentrate on accurate navigation of a light aircraft‚ flown by a single pilot‚ in visual conditions. PPL holders‚ when flying cross-country‚ act as pilot‚ navigator
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Site Coordinates: (Degrees Decimals) Longitude E: Latitude N: Site Owned by: Approved by: ________________ Final Transmission Plan TXN Path (BTS-BSC) LL status (If Required) LL Collocation status (If Required) Mw Antenna Size(m) LOS Report LOS Name [A—B] IDHM01--IGTR05 LOS Request Date LOS Date LOS Request By LOS Report By Signature Signature Site A Site B Name Address A Latitude Degrees Decimals A Longitude Degrees Decimals A Site Elevation(m)
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Why is the sample variance divided by $n-1$ instead of $n$?
s^2 = \frac{\sum(x - \bar{x})^2}{n-1}.
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This is necessary to correct for the bias in the sample variability.
If not corrected, the sample variability will be biased and consistently underestimates the population variability and variance.
Dividing by a smaller value (n - 1) increases the value of the sample variance, thereby making it an accurate and unbiased estimate of the population variance.
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# Mathematical Programming
Lecture Notes
CE 385D - McKinney
Water Resources Planning and Management
Department of Civil Engineering
The University of Texas at Austin
Section
page
1. Introduction
3. Constraints
## 4. Typical Forms of Math Programming Problems
4.1 Linear Programming
4.2 Classical Programming
4.3 Nonlinear Programming
4
4
5
6
5. Types of Solutions
6. Classical Programming
6.1 Unconstrained Scalar Case
6.2 Unconstrained Vector Case
6.3 Constrained Vector Case - Single Constraint
6.5 Constrained Vector Case Multiple Constraints
6.6 Nonlinear Programming and the Kuhn-Tucker Conditions
8
8
9
10
14
16
Exercises
17
References
21
## Appendix A. Mathematics Review
A.1 Linear Algebra
A.2 Calculus
A.3 Vectors Calculus
22
22
29
32
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1. Introduction
An optimization problem (mathematical program) is to determine the values of a vector of
decision variables
x1
x = M
xn
(1.1)
## which optimize (maximize or minimize) the value of an objective function
f(x) = f(x1,x2 ,L ,xn )
(1.2)
The decision variables must also satisfy a set of constraints X which describe the system being
optimized and any restrictions on the decision variables. The decision vector x is feasible if
x is feasible if x X
(1.3)
n
The feasible region X, an n-dimensional subset of R , is defined by the set of all feasible
vectors. An optimal solution x * has the properties (for minimization):
(1) x * X
(1.4)
## (2) f (x*) f (x) x x *
(1.5)
i.e., the solution is feasible and attains a value of the objective function which is less than or
equal to the objective function value resulting from any other feasible vector.
## 2. General Mathematical Programming Problem
The general mathematical programming problem can be stated as
maximize f(x)
x
subject to
x X
(2.1)
## In words this says,
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maximize the objective function f(x) by choice of the decision variables x while
ensuring that the optimal decision variables satisfy all of the constraints or
restrictions of the problem
The objective function of the math programming problem can be either a linear or nonlinear
function of the decision variables.
Note that:
maximize f(x)
is equivalent to
(2.2)
maximize a + bf(x),
b > 0, or
minimize a + bf(x),
b<0
(2.3)
That is, optimizing if a linear operator; multiplying by a scalar or adding a constant does not
change the result and maximizing a negative is the same as minimizing a positive function.
3. Constraints
The constraint set X of the math program can consist of combinations of:
(1) Linear equalities:
Ax = b
n
aij x j = bi
j =1
(3.1)
i = 1,L m
(3.2)
where the aij , i = 1L m, j = 1L n are the elements of the matrix A, and b is a vector of righthand-side constants. For a review of matrix arithmetic and notation, linear equalities and
(2) Linear inequalities:
Ax b
n
aij x j bi
j =1
Math Programming
(3.3)
i = 1,L m
(3.4)
1/10/2003
## (3) Nonlinear inequalities:
g( x) 0
(3.5)
g j ( x) 0
j = 1,L r
(3.6)
where the functions g(x) are nonlinear functions of the decision variables.
(4) Nonlinear equalities:
h( x ) = 0
hi ( x ) 0
(3.7)
i = 1,L m
(3.8)
where the functions h(x) are nonlinear functions of the decision variables.
## 4. Typical Forms of Math Programming Problems
Different forms of the objective function and constraints give rise to different classes of
mathematical programming problems:
## 4.1 Linear Programming
The objective function is linear and the constraints are linear equalities, inequalities, or both and
non-negativity restrictions apply.
Maximize c x
x
subject to
Ax b
x0
(4.1)
Example:
Maximize 2 x1 + 3 x 2 + 5 x3
subject to
x1 + x 2 x3 5
6 x1 + 7 x 2 9 x3 = 5
19 x1 7 x 2 + 5 x3 13
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Of course this example has the difficulty of what to do with the absolute value.
An inequality with an absolute value can be replaced by two inequalities, e.g.,
g ( x) b
## can be replaced by replaced by
g ( x) b and g ( x) b
So our example can be converted to:
Maximize 2 x1 + 3 x 2 + 5 x3
subject to
x1 + x 2 x3 5
6 x1 + 7 x 2 9 x3 = 5
19 x1 7 x 2 + 5 x3 13
19 x1 7 x 2 + 5 x3 13
Note: An equation can be replaced by two inequalities of the opposite direction. For example an
equation
g ( x) = b
can be replaced by replaced by
g ( x) b and g ( x) b
Often it is easier for programs to check the inequality condition rather than the strict equality
condition.
## 4.2 Classical Programming
The objective function is nonlinear and the constraints are nonlinear equalities.
Maximize f ( x )
x
(4.2)
subject to
h( x ) = 0
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Example:
Minimize
subject to
x1 2 x 2 = 0
## 4.3 Nonlinear Programming
The objective function is linear or nonlinear and the constraints are linear or nonlinear equalities,
inequalities, or both.
Maximize f ( x )
x
subject to
h( x ) = 0
g( x) 0
(4.3)
Example:
Maximize ln (x 1 + 1) + x 2
subject to
2 x1 + x 2 3
x1 0, x 2 0
5. Types of Solutions
Solutions to the general mathematical programming problem are classified as either global or
local solutions depending upon certain characteristics of the solution. A solution x* is a global
solution (for maximization) if it is:
1. Feasible; and
2. Yields a value of the objective function less than or equal to that obtained by any other
feasible vector, or
x * X , and
f ( x*) f ( x ) x X
(5.1)
## A solution x* is a local solution (for maximization) if it is:
1. Feasible; and
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2. Yields a value of the objective function greater than or equal to that obtained by any
feasible vector x sufficiently close to it, or
x * X , and
f ( x*) f ( x ) x ( X x N ( x*))
(5.2)
In this case multiple optima may exist for the math program and we have only established that x*
is the optimum within the neighborhood searched. Extensive investigation of the program to
find additional optima may be necessary.
Sometimes we can establish the local or global nature of the solution to a math program. The
following two theorems give some examples.
## Weierstras Theorem: (Sufficient conditions for a global solution)
If X is non-empty and compact (closed and bounded) and f(x) is continuous on X, then f(x)
has a global maximum either in the interior or on the boundary of X.
## Local - Global Theorem: (Sufficient conditions for a local solution to be global)
If X is a non-empty, compact and convex and f(x) is continuous on X and a concave
(convex) function over X, then a local maximum is a global maximum.
The Figure 5.1 shows a non-convex function defined over a convex feasible region so we have
no assurance that local maxima are global maxima. We must assume that they are local maxima.
Figure 5.2 shows a concave (non-convex) function maximized over a convex constraint set, so
we are assured that a local maximum is a global maximum (if we can find one) by the LocalGlobal Theorem.
f(x )
Global Max
Local Max
Global Min
Local Min
x
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f(x )
Global Max
## Figure 5.2. Concave function maximized over a convex constraint set
6. Classical Programming
Maximize f ( x )
(6.1)
subject to
h( x ) = 0
(6.2)
## 6.1 Unconstrained Scalar Case
In this case, there are no constraints (Equation 6.2 is not present) and we consider only the
objective function for a single decision variable, the scalar x
(6.1.1)
Maximize f ( x )
x
## The necessary conditions for a local minimum are
df ( x )
= 0 (first-order conditions)
dx
(6.1.2)
and
d 2 f ( x)
dx 2
0 (second-order conditions)
(6.1.3)
The first-order conditions represent an equation which can be solved for x* the optimal solution
for the problem.
But what if the decision variable was constrained to be greater than or equal to zero (nonnegativity restriction), e.g., x 0 ? In this case there are two possibilities, either (1) the solution
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lies to the right of the origin and has an interior solution where the slope of the function is zero,
or (2) or the solution lies on the boundary (at the origin) where x = 0 and the slope is negative.
That is
df
dx
0 at x = x * if x* = 0
= 0 at x = x * if x* > 0
(6.1.4)
df
df
0, and x
= 0 at x = x *
dx
dx
(6.1.5)
## 6.2 Unconstrained Vector Case
In this case, again we have no constraints and we consider only the objective function for a
vector of decision variables x
Maximize f ( x )
(6.2.1)
## The necessary conditions are
f ( x ) =
f ( x )
= 0 (first-order conditions)
x
(6.2.3)
## which is actually n simultaneous nonlinear equations
f ( x )
x
1
f ( x ) = M = 0
f ( x )
x n
(6.2.4)
The first-order conditions represent n - simultaneous equations which can be solved for x* the
optimal solution for the problem.
But what if the decision variable was constrained to be greater than or equal to zero (nonnegativity restriction), e.g., x 0? In this case there are two possibilities, either (1) the solution
lies to the right of the origin and has an interior solution where the slope of the function is zero,
or (2) or the solution lies on the boundary (at the origin) where x = 0 and the slope is negative.
That is
Math Programming
1/10/2003
f
x j
0 at x = x * if x *j = 0
*
= 0 at x = x * if x j > 0
(6.2.5)
f
f
= 0 j = 1,..., n
0, and x j
x j
x j
(6.2.6)
## 6.3 Constrained Vector Case - Single Constraint
In this case, we consider the objective function for a vector of decision variables x, and a single
constraint, h(x)=0
Maximize f ( x )
x
subject to
h( x ) = 0
(6.3.1)
We can multiply the constraint by a variable or multiplier and subtract the resulting expression
from the objective function to form what is known as the Lagrangian function
L( x, ) = f ( x ) [h( x )]
(6.3.2)
and then simply apply the methods of the previous case (unconstrained vector case). Note that
for a feasible vector the constraint must be satisfied, that is
and
h( x ) = 0
(6.3.3)
L( x, ) = f ( x )
(6.3.4)
so we really have not changed the objective function as long as we remain feasible. The
necessary conditions (first-order) are
h
L { f ( x ) [h( x )]} f
=
=
=0
x
x
x
x
L
= h( x ) = 0
(6.3.5)
Math Programming
10
1/10/2003
h
f
=0
x1
x1
h
f
=0
x 2
x 2
(6.3.6)
h
f
=0
x n
x n
h( x ) = 0
## which must be solved for the optimal values of x* and *.
Example (adapted from Loucks et al., 1981, Section 2.6, pp. 23-28)
Consider a situation where there is a total quantity of water R to be allocated to a number of
different uses. Let the quantity of water to be allocated to each use be denoted by xi, i=1,, I.
The objective is to determine the quantity of water to be allocated to each use such that the total
net benefits of all uses is maximized. We will consider an example with three uses I = 3.
S
x1
Reservoir
x3
x2
User 1 B1
User 2
B2
User 3
B3
## Figure 6.4.1. Reservoir release example.
The net-benefit resulting from an allocation of xi to use i is given by
Bi ( xi ) = ai xi bi xi2
i = 1,2,3
(6.4.1)
where ai and bi are given positive constants. These net-benefit (benefit minus cost) functions are
of the form shown in Figure 6.4.2.
Math Programming
11
1/10/2003
35
30
25
20
B1
15
B2
10
B3
Net
0 economic benefit
to user
0 i
2
10
## Figure 6.4.2. Net-benefit function for user i.
The three allocation variables xi are unknown decision variables. The values that these variables
can take on are restricted between 0 (negative allocation is meaningless) and values whose sum,
x1 + x2 + x3 , does not exceed the available supply of water R minus the required downstream
flow S. The optimization model to maximize net-benefits can be written as
3
x
subject to
i =1
x
i =1
+S R =0
(6.4.2)
## The Lagrangian function is
3
3
L( x , ) = (ai xi bi xi2 ) xi + S R
i =1
i =1
(6.4.3)
There are now four unknowns in the problem, xi , i = 1, 2, 3 and . Solution of the problem is
obtained by applying the first-order conditions, setting the first partial derivatives of the
Lagrangian function with respect to each of the variables equal to zero:
Math Programming
12
1/10/2003
L
= a1 2b1 x1 = 0
x1
L
= a 2 2b2 x 2 = 0
x 2
(6.4.4)
L
= a3 2b3 x3 = 0
x3
L
= x1 + x 2 + x3 + S R = 0
These equations are the necessary conditions for a local maximum or minimum ignoring the
nonnegativity conditions. Since the objective function involves the maximization of the sum of
concave functions (functions whose slopes are decreasing), any local optima will also be the
global maxima (by the Local-Global Theorem).
The optimal solution of this problem is found by solving for each xi , i = 1, 2, 3 in terms of .
a
xi = i
2bi
i = 1,2,3
(6.4.5)
3
xi + S R = 0
i =1
(6.4.6)
3 a
+SR=0
i
i =1 2bi
(6.4.7)
3 a
2 i + S R
i =1 2bi
=
3
1
i =1 bi
(6.4.8)
## and solve for
Hence knowing R, S, ai and bi this last equation can be solved for . Substitution of this value
into the equation for the xi , i = 1, 2, 3 , we can solve for the optimal allocations, provided that all
of the allocations are nonnegative.
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## 6.5 Constrained Vector Case Multiple Constraints
In this case, we consider the objective function for a vector of decision variables x, and a vector
of constraints, h(x)=0
Maximize f ( x )
x
subject to
h( x ) = 0
(6.5.1)
## We can multiply the constraints by a vector of variables or multipliers = ( 1 , 2 ,..., m ) or
h( x ) and subtract the resulting expression from the objective function to form what is known
as the Lagrangian function
m
L( x, ) = f ( x ) h( x ) = f ( x ) i hi (x)
(6.5.2)
i =1
and then simply apply the methods of the previous case (unconstrained vector case). The
necessary conditions (first-order) are
x L( x, ) = x f ( x ) x h( x ) = 0
(6.5.3)
m
h
L [ f ( x ) h( x )] f
=
=
i i = 0
x
x
x i =1 x
(6.5.4)
L( x, ) = 0
(6.5.5)
h( x ) = 0
(6.5.6)
or
and
or
The first-order conditions (Eq. 6.5.4 and Eq. 6.5.6) represent n+m simultaneous equations must
be solved for the optimal values of the vectors of decision variables and the Lagrange
multipliers, x* and *.
Example (after Haith, 1982, Example 4-2):
Solve the following optimization problem using Lagrange multipliers.
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Maximize x12 + 2 x1 x 22
subject to
5 x1 + 2 x 2 10
(6.5.7)
x1 0
x2 0
The last three constraints must be turned into equalities in order to use Classical Programming to
solve the problem. Introduce three new variables, s1, s2, and s3
5 x1 + 2 x2 + s12 = 10
x1 s22 = 0
(6.5.8)
x2 s32 = 0
These slack variables (difference between the left and right sides) are always introduced on the
side of the inequality that the inequality sign points toward.
The Lagrangian function is
L( x, ) = x12 + 2 x1 x22 1 (5 x1 + 2 x2 + s12 10) 2 ( x1 s22 ) 3 ( x2 s32 )
(6.5.9)
## The first-order optimality conditions are
L
= 2 x1 + 2 51 2 x1 = 0
x1
L
= 2 x 2 21 x 2 3 x 2
x 2
L
= 21s1 = 0
s1
L
= 22 s 2 = 0
s 2
L
= 23 s3 = 0
s3
L
= 5 x1 + 2 x2 + s12 10
1
L
= x1 s22 = 0
2
L
= x 2 s32 = 0
3
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(6.10a)
(6.10b)
(6.10c)
(6.10d)
(6.10e)
(6.10f)
(6.10g)
(6.10h)
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Equations 6.10c-e require that i or si be equal to zero. There can be several solutions to the
problem depending on whether one or another of the i or si are equal to zero.
## 6.6 Nonlinear Programming and the Kuhn-Tucker Conditions
In this case, we consider the objective function for a vector of decision variables x, a vector of
equality constraints, h(x)=0, and a vector of inequality constraints, g(x)0
Maximize f ( x )
x
subject to
h( x ) = 0
g( x) 0
(6.6.1)
## We can multiply the constraints by vectors of variables or multipliers = ( 1 , 2 ,..., m ) or
h( x ) and u = (u1 , u2 ,..., ur ) or u g ( x ) and subtract the resulting expression from the
objective function to form what is known as the Lagrangian function
m
i =1
j =1
L( x , , u) = f ( x ) i hi(x) u j g j(x)
(6.6.2)
and then simply apply the methods of the previous case (unconstrained vector case). The
necessary conditions (first-order) are the Kuhn-Tucker Conditions
## at x = x*, for k = 1,..., n
m
r
f
g j
h
i i u j
xk*
= 0
xk i =1 xk j =1 xk
(6.6.3)
g j ( x*) 0
for j = 1,..., r
u j g j ( x*) = 0
(6.6.4)
## hi ( x*) = 0, for i = 1,..., m
(6.6.5)
xk* 0, k = 1,..., n
u j 0, j = 1,..., r
(6.6.6)
m
r
g j
h
f
i i u j
0
xk i =1 xk j =1 xk
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Exercises
1. (after Mays and Chung, 1992, Exercise 3.4.5) Water is available at supply points 1, 2, and 3
in quantities 4, 8, and 12 thousand units, respectively. All of this water must be shipped to
destinations A, B, C, D, and E, which have requirements of 1, 2, 3, 8, and 10 thousand units,
respectively. The following table gives the cost of shipping one unit of water from the given
supply point to the given destination. Find the shipping schedule which minimizes the total cost
of transportation.
Destination
Source A B C D
1
7 10 5 4
2
3 2 0 9
3
8 13 11 6
E
12
1
14
B
C
D
3
Supply
E
Destination
2. (adapted from Mays and Tung, 1992, Exercise 3.1.1) Solve the following Linear Program
Maximize 2 x1 + 3 x 2 + 5 x3
subject to
x1 + x 2 x3 5
6 x1 + 7 x 2 9 x3 = 5
19 x1 7 x 2 + 5 x3 13
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3. (adapted from Mays and Tung, 1992, Exercise 3.2.1) Consider the following Linear Program
Maximize 3 x1 + 5 x 2
subject to
x1 4
x2 6
3x1 + 2 x 2 18
(a) Graph the feasible region for the problem.
(b) Solve the problem graphically.
(c) How much can the nonbinding constraints be reduced without changing the feasibility of
the optimal solution?
(d) What is the range of the objective function coefficient of x2 so that the optimal solution
remains feasible?
4. (after Haith, 1982, Example 5-1) 1000 ha of farmland surrounding a lake is available for two
crops. Each hectare of crop 1 loses 0.9 kg/yr of pesticide to the lake, and the corresponding loss
from crop 2 is 0.5 kg/yr. Total pesticide losses are not allowed to exceed 632.5 kg/yr. Crop
returns are \$300 and \$150/ha for crops 1 and 2, respectively. Costs for crops are estimated to be
\$160 and \$50/ha for crops 1 and 2, respectively.
(a) Determine the cropping combination that maximizes farmer profits subject to a constraint
on the pesticide losses into the lake.
(b) If crop returns decrease to \$210/ha for crop 1, what is the optimal solution?
(c) If crop returns increase to \$380/ha for crop 1, what is the optimal solution?
5. (after Haith, 1982, Exercise 5-1) A metal refining factory has a capacity of 10x 104 kg/week,
produces waste at the rate of 3 kg/kg of product, contined in a wastewater at a concentration of 2
kg/m3. The factorys waste treatment plant operates at a constant efficiency of 0.85 and has a
capacity of 8x104 m3/week. Wastewater is discharged into a river, and the effluent standard is
100,000 kg/week. There is also an effluent charge of \$1000/104 kg discharged. Treatment costs
are \$1000/104 m3, product sales price is \$10,000/104 kg, and production costs are \$6850/104 kg.
(a) Construct a linear program that can be used to solve this wastewater problem. Solve the
model graphically.
(b) If the effluent charge is raised to \$2000/104 kg, how much will the waste discharge be
reduced?
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6. (after Haith, 1982, Exercise 5-9) A standard of 1 kg/103m3 has been set as the maximum
allowable concentration for a substance in a river. Three major dischargers of the substance are
located along the river as shown in the figure. The river has a flow of 500,000 m3/day and an
ambient concentration of the regulated substance of 0.2 kg/103m3 upstream of the first
discharger. The three waste sources presently discharge 100, 100m, and 1600 kg/day of the
regulated substance, resulting in violations of the standard in the river. The substance is not
conserved in the river, but decays at a rate of K=0.03 km-1. Thus is C1 and C2 are the
concentrations of the substance immediately after the discharge points 1 and 2, respectively, the
concentrations at any point L km downstream of discharge 1 (L < 10) is C1e-KL. Similarly, the
concentration L km downstream of discharge 2 (L < 15) is C2e-KL. The cost of removing the
substance from the wastewater is \$10X/1000 m3 where X is the fraction of the substance
removed. Use LP to determine an optimal treatment program for the regulated substance.
10 km
300 103 m3/day
0.2 kg/103 m3
15 km
River
1
Flow
Discharge
## 100 103 m3/day
10 kg/103 m3
50 103 m3/day
20 kg/103 m3
3
200 103 m3/day
8 kg/103 m3
7. (after Haith, 1982, Example 4-1) Solve the following optimization problem using Lagrange
multipliers.
## Maximize 0.5 x12 + 20 x 2 x3 + 10 x3
subject to
x1 3x 2 + 0.5 x3 = 6
x 2 + 2 x3 = 10
8. (after Haith, 1982, Exercise 4-1) Solve the following optimization problem using Lagrange
multipliers (Classical programming)
## Maximize 4 x12 + x 2 + 6 x33
subject to
x1 + 3x 2 + x3 = 10
x 2 + 2 x3 = 4
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9. (after Haith, 1982, Exercise 4-2) Solve the following optimization problem using Lagrange
multipliers (Classical programming)
Maximize 4e x1 x 22
subject to
6 x1 x 2 = 6
x1 0
10. (after Willis, 2002) A waste storage facility consists of a right circular cylinder of radius 5
units and a conical cap. The volume of the storage facility is V. Determine H, the height of the
storage facility, and h, the height of the conical cap, such that the total surface area is minimized.
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References
Haith, D.A., Environmental Sytems Optimization, John Wiley & Sons, New York, 1982
Loucks, D. P. et al., Water Resource Systems Planning and Analysis, Prentice Hall, Englewood
Cliffs, 1981
Mays, L.W., and Y-K, Tung, Hydrosystems Engineering and Management, McGraw Hill, 1992
Mathematical Programming and Optimization Texts
Fletcher, Practical Methods of Optimization, ...
Gill, P.E., W. Murray, and M.H. Wright, Practical Optimization, Academic Press, London, 1981
Hillier, F. S. and G.J. Lieberman, Introduction to Operation Research, McGraw-Hill, Inc., New
York, 1990.
Intrilligator, M.D., Mathematical Optimizatiron and Economic Theory, Prentice-Hall, Inc.,
Englewood Cliffs, 1971.
Luenberger, D.G., Linear and Nonlinear Programming, Addison Wesley, New York, 1984.
McCormick, G.P., Nonlinear Programming: Theory, Algorithms, and Applications, John Wiley
and Sons, 1983.
Taha, H.A., Operations Research: An Introduction, MacMillan, New York, 1987.
Wagner, H.M., Principles of Operations Research, Prentice-Hall, Inc., Englewood Cliffs, 1975.
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## Appendix A. Mathematics Review
A.1 Linear Algebra
A.1.1 Introduction
An important tool in many areas of scientific and engineering analysis and computation is matrix
theory or linear algebra. A wide variety of problems lead ultimately to the need to solve a linear
system of equations Ax = b. There are two general approaches to the solution of linear systems.
A.1.2 Matrix Notation
rr
A matrix is an array of real numbers. Consider an (m x n) matrix A with m rows and n columns:
a11 a12
a
a22
A = 21
M
a m1 a m 2
a1n
a2n
O
M
a m3 K a mn
a13
a23
(A.1.2.1)
The horizontal elements of the matrix are the rows and the vertical elements are the columns.
The first subscript of an element designates the row, and the second subscript designates the
column. A row matrix (or row vector) is a matrix with one row, i.e., the dimension m = 1. For
example
r = (r1 r2
r3 L rn )
(A.1.2.2)
## A column vector is a matrix with only one column, e.g.,
c1
c = c2
M
cm
(A.1.2.3)
When the row and column dimensions of a matrix are equal (m = n) then the matrix is called
square
a11 a12 L a1n
a
a22
a2n
A = 21
O
M
an1 a n 2 L ann
(A.1.2.4)
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a11
a
AT = 12
M
a1n
a 21 L a m1
a22
am 2
a 2n L a mn
(A.1.2.5)
## A symmetric matrix is one where AT = A. An example of a symmetric matrix is
A = 2 1
1 2
(A.1.2.6)
A diagonal matrix is a square matrix where elements off the main diagonal are all zero
a11
A =
a 22
a nn
(A.1.2.7)
An identity matrix is a diagonal matrix where all the elements are ones
0
1
1
I =
O
0
1
(A.1.2.8)
An upper triangular matrix is one where all the elements below the main diagonal are zero
a11
A =
a12
a 22
L a1n
a2n
a nn
(A.1.2.9)
A lower triangular matrix is one where all the elements above the main diagonal are zero
a11
a
A = 21
M
a
n1
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a 22
an2
L a nn
(A.1.2.10)
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## A.1.3 Matrix Arithmetic
Two (m x n) matrices A and B are equal if and only if each of their elements are equal. That is
A = B if and only if aij = bij for i = 1,...,m and j = 1,...,n
(A.1.3.1)
The addition of vectors and matrices is allowed whenever the dimensions are the same. The sum
of two (m x 1) column vectors a and b is
a1
b1
a1 + b1
a2
b2
a 2 + b2
a + b = + =
M
M
M
a
b
a + b
m
m
m
m
(A.1.3.2)
Example:
Let u = (1,3,2,4) and v = (3,5,1,2) . Then
u + v = (1 + 3,3 + 5,2 1,4 2) = (4,2,1,2)
## 2u 3v = (2,6,4,8) + (9,15,3,6) = (7,21,7,14)
The sum of two (m x n) matrices A and B is
a11
a
A + B = 21
M
a
m1
a12
a 22
am 2
a11
a
= 21
a
m1
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L a1n
b11
a1n
b21
+
M
O M
b
K a mn
m1
+ b11
+ b21
M
+ bm1
a12 + b12
a 22 + b22
a m 2 + bm 2
24
b12
b22
bm 2
L b1n
b1n
O M
K bmn
a1n + b1n
a1n + b1n
O
M
K a mn + bmn
L
(A.1.3.3)
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## Multiplication of a matrix A by a scalar is defined as
a11
a
A = 21
M
a
m1
a12
a 22
a m 2
L a1n
a 2 n
K a mn
(A.1.3.4)
The product of two matrices A and B is defined only if the number of columns of A is equal to
the number of rows of B. If A is (n x p) and B is (p x m), the product is an (n x m) matrix C
a11
a
C = AB = 21
M
a
m1
a12
a 22
am2
## a11b11 + L + a1n bm1
a b + L + a 2 n bm1
= 21 11
M
a b +L+ a b
1n m1
m1 11
b11
b21
M
K a mn bm1
L a1n
a1n
O M
b12
b22
bm 2
a11b12 + L + a1n bm 2
a 21b12 + L + a 2 n bm 2
a m1b12 + L + a mn bm 2
K bmn
L b1n
b1n
O M
## a11b1n + L + a1n bmn
a 21b1n + L + a 2 n bmn
O
M
K a m1b1n + L + a mn bmn
L
(A.1.3.5)
The ij element of the matrix C is given by
cij =
a
k =1
ik kj
(A.1.3.6)
That is the cij element is obtained by adding the products of the individual elements of the i-th
row of the first matrix by the j-th column of the second matrix (i.e., row-by-column). The
following figure shows an easy way to check if two matrices are compatible for multiplication
and what the dimensions of the resulting matrix will be:
C nxm = Anxp Bpxm
(A.1.3.7)
Example:
b1
b
Let a = (a1 , a 2 ,L, a n ) and b = 2 , then
M
b
n
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b1
b
a b = (a1 , a 2 ,L, a n ) 2 = a1b1 + a 2 b2 + L + a n bn
M
b
n
Example:
a11
a
Let A = 21
M
a n1
a12
a 22
an 2
L a1n
b1
L a2n
b
and b = 2 , then
M
M
L a nn
bn
a11
a
Ab = 21
M
a n1
## L a1n b1 a11b1 + a12 b2 + L a1n bn
L a 2 n b2 a11b1 + a 22 b2 + L a 2 n bn
=
M M
L a nn bn a n1b1 + a n 2 b2 + L a nn bn
a12
a 22
an 2
Matrix division is not a defined operation. The identity matrix has the property that IA=A and
AI = A. If A is an (n x n) square matrix and there is a matrix X with the property that
AX = I
(A.1.3.8)
where I is the identity matrix, then the matrix X is defined to be the inverse of A and is denoted
A-1. That is
AA-1 = I and A-1A = I
(A.1.3.9)
## The inverse of a (2 x 2) matrix A can be represented simply as
A 1 =
1
a11 a 22 a12 a 21
a 22
a
21
a12
a11
(A.1.3.10)
Example
2 1
2 1 2 / 3 1 / 3
1
, then A 1 =
=
If A =
2(2) 1(1) 1 2 1 / 3 2 / 3
1 2
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Ax = b
(A.1.4.1)
## where A is an (n x n) matrix, b is a column vector of constants, called the right-hand-side, and x
is the unknown solution vector to be determined. This system can be written out as
a11
a 21
M
a
n1
a12
a 22
an2
L a1n x1
b1
a 2 n x 2
b
= 2
O
M
M
L a nn x n
bn
(A.1.4.2)
Performing the matrix multiplication and writing each equation out separately, we have
a11 x1 + a12 x 2 + L + a1n x n = b1
a 21 x1 + a 22 x 2 + L + a 2 n x n = b2
M
a n1 x1 + a n 2 x 2 + L + a nn x n = bn
(A.1.4.3a)
## This system can also be written in the following manner
n
a
j =1
ij
x j = bi i = 1,L n
(A.1.4.3b)
A formal way to obtain a solution using matrix algebra is to multiply each side of the equation by
the inverse of A to yield
A 1 Ax = A 1b
(A.1.4.4)
1
or, since AA = I
x = A 1b
(A.1.4.5)
Thus, we have obtained the solution to the system of equations. Unfortunately, this is not a very
efficient way of solving the system of equations. We will discuss more efficient ways in the
following sections.
Example: Consider the following two equations in two unknowns:
3x 1 + 2x 2 = 18
x 1 + 2x 2 = 2
## Solve the first equation for x2
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3
x1 + 9
2
x2 =
which is a straight line with an intercept of 9 and a slope of (-3/2). Now, solve the second
equation for x2
1
x1 + 1
2
x2 =
which is also a straight line, but with an intercept of 1 and a slope of (1/2). These lines are
plotted in the following Figure. The solution is the intersection of the two lines at x1 = 4 and x2 =
3.
x2
3 x1 + 2 x2 = 1 8
6
Solut ion: x 1 = 4 ; x 2 = 3
4
-x 1 + 2 x 2 = 2
2
x1
## Each linear equation
ai1 x1 + ai 2 x2 + L + ain xn = bi
(A.1.4.6)
represents a hyperplane in an n-dimensional Euclidean space (Rn), and the system of equations
Ax = b represents m hyperplanes. The solution of the system of equations is the intersection of
all of the m hyperplanes, and can be
- the empty set (no solution)
- a point (unique solution)
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## - a line (non-unique solution)
- a plane (non-unique solution)
A.1.5 Systems of Linear Inequalities
Ax b
or
(A.1.5.1)
## a11 x1 + a12 x 2 + L + a1n x n b1
a 21 x1 + a 22 x 2 + L + a 2 n x n b2
M
a n1 x1 + a n 2 x 2 + L + a nn x n bn
(A.1.5.2a)
n
a
j =1
ij
x j bi i = 1,L n
(A.1.5.2b)
## Each linear inequality
ai1 x1 + ai 2 x2 + L + ain xn bi
(A.1.5.3)
represents a half-space in Rn, and the system of inequalities Ax b represents the intersection of
m half-spaces which is a polyhedral convex set or, if bounded, a polyhedron.
A.2 Calculus
A.2.1 Functions
## A function f ( x ) of n variables can be written as
y = f ( x)
(A.2.1.1)
x1
x = M R n (column vector );
x n
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y R 1 ( scalar )
29
(A.2.1.2)
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## A linear function of n variables is written as
n
y = f ( x ) = c x = ci xi = c1 x1 + c 2 x 2 + L + c n x n
(A.2.1.3)
i =1
## where c is a vector of coefficients.
A.2.2 Sets, Neighborhoods and Distance
## The distance between two points x and y in Rn is defined as
d ( x, y ) =
(x
i =1
yi ) 2
(A.2.2.1)
A neighborhood around a point x in Rn is defined as the set of all points y less than some
distance from the point x or
N ( x ) = y R n : d ( x, y) <
(A.2.2.2)
A closed set is a set which contains all of the points on its boundary, for example a closed
interval on the real line (R1). In a bounded set, the distance between two points contained in the
set is finite. A compact set is closed and bounded, examples are any finite interval [ a, b] on the
real line or any bounded sphere in R3.
A set S is a convex set if for any two points x and y in the set, the point
z = ax + (1 a ) y
(A.2.2.3)
is also in the set for all a, where 0 a 1. That is, all weighted averages of two points in the set
are also points in the set. For example, all points on a line segment joining two points in a
convex set are also in the set. Straight lines, hyperplanes, closed halfspaces are all convex sets.
Figure 2 below illustrates a convex and a non-convex set. A real valued function f(x) defined on
a convex set S is a convex function if given any two points x and y in S,
f [af ( x ) + (1 a ) f ( y)] af ( x ) + (1 a ) f ( y)
(A.2.2.4)
for all a, where 0 a 1. Figure 3 illustrates the fact that the line segment joining two points in
a convex function does not lie below the function. Figure 4 shows general examples of convex
and non-convex (or concave) functions. An example of a convex function is a parabola which
opens upward. Linear functions (lines, planes, hyperplanes, half-spaces) are both convex and
non-convex functions.
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x
y
x
convex
non-convex
## Figure A.2.2.1. General diagram of convex and non-convex sets.
f(x )
f(x )
af(x) + (1-a)f(y)
f(y )
f(ax + (1-a)y )
ax + (1-a)y
f(x )
Convex
Concave
## Figure A.2.2.3. General diagram of a convex function and a concave function.
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A.2.3 Derivatives
## The derivative of a function of a single scalar variable f(x) is defined as
f ( x) =
df
f ( x + x) f ( x)
= lim
dx x0
x
(A.2.3.1)
## The partial derivatives of a function f of the variables x and y are defined as
f
f ( x + x, y ) f ( x, y )
= lim
x x0
x
f
f ( x, y + y ) f ( x, y )
= lim
y y 0
y
(A.2.3.2)
That is, to find the partial derivative of a multivariable function with respect to one independent
variable xi, regard all other independent variables as fixed and find the usual derivative with
respect to xi. The partial derivative of a function of several variables f(x) with respect to a
particular component of x, xi, evaluated at a point xo is
f
f ( x )
=
xi
xi
(A.2.3.3)
x
The partial derivative of f(x) with respect to the vector x is a row vector of partial derivatives of
the function or the gradient vector
f ( x ) =
f f
=
x x1
x n
(A.2.3.4)
## A.3 Vectors Calculus
A.3.1 Coordinate Systems
## Typical coordinate systems used in groundwater problems include: Rectangular: x, y, z; and
Cylindrical: r, , z where x = rcos and y = rsin. Let A be a vector function of (x, y, z) or (r,
, z), respectively, then
A = Ax i + Ay j + Az k = Ar r + A + Az k
(A.3.1.1)
where (i, j, k) and (r, , k) are unit vectors in the (x, y, z) or (r, , z) directions, respectively.
Math Programming
32
1/10/2003
## A.3.2 Basic Operators
The gradient operator, del (from the Greek nabla) or , is defined in rectangular coordinates as
the vector
()
x
() ()
()
()
=
i+
j+
k
() =
z
y
y x
()
(A.3.2.1)
The major operators: Gradient, del or , () ; Divergence, div or () ; and Laplacian, del
dot del or () = 2 () can be defined in the rectangular and cylindrical coordinate systems
as:
Divergence
()
()
()
i+
j+
k
x
y
z
()
1 ()
()
r+
+
k
(Cylindrical) () =
r
r
z
A y Az
A
+
(Rectangular) A = x +
y
x
z
(Rectangular) () =
(A.3.2.2)
(A.3.2.3)
(A.3.2.4)
Proof:
()
()
()
A =
i+
j+
k Ax i + Ay j + Az k
y
z
x
Ay
A
A
= x i i +
i j + z i k
x
x
x
A
A
A
y
j j + z jk
+ x ji +
y
y
y
Ay
A
A
k j + z k k
+ x k i +
z
z
z
Ay
A
A
( 0) + z ( 0)
= x (1) +
x
x
x
Ay
A
A
(1) + z (0)
+ x ( 0) +
y
y
y
Ay
A
A
(0) + z (1)
+ x ( 0) +
z
z
z
A
A
A
y
= x +
+ z
x
y
z
Math Programming
33
1/10/2003
1 (rAr ) 1 A Az
+
+
r r
r
z
2
2
() () 2 ()
2
(Rectangular) () = () =
+
+
x 2
y 2
z 2
(Cylindrical) A =
Laplacian
A = 2 A =
e.g.,
(A.3.2.6)
2
2 Ax Ay 2 Az
+
+
x 2
y 2
z 2
1 ()
1 2 () 2 ()
(r
)+ 2
+
r r r
r 2
z 2
1 Ar
1 2 A 2 Az
+
A = 2 A =
(r
)+ 2
r r
r
r 2
z 2
(Cylindrical) () = 2 () =
e.g.,
(A.3.2.5)
(A.3.2.7)
## A.3.3 Various Groundwater Relations
The Piezometric head, h = p / + z , is a scalar quantity and the gradient of this quantity is a
column vector
h
x
h h
h
h
i+
j+ k
h = =
z
y
y x
h
z
(A.3.3.1)
## The hydraulic conductivity, K, is a tensor whose common form in three dimensions is
K x
K = 0
0
0
Ky
0
0
0
K z
(A.3.3.2)
The term K h is the product of the matrix K with the vector h or (using row by column
multiplication)
K x
K h = 0
0
0
Ky
0
Math Programming
h
K x
0 x
h
0 = K y
y
K z h
K z
z
h
x
h
h
h
h
i + Ky
j + Kz
k
= Kx
x
y
z
y
h
34
(A.3.3.3)
1/10/2003
()
x K x
()
K y
K h =
y
()
K z
z
h
x
h
y
h
()
()
() h
h
h
=
i+
j+
k K x
i + Ky
j + Kz
k
x
y
z
x
y
z
Math Programming
(A.3.3.4)
h
h
h
+ K z
Kx
+ K y
x
x y
y z
z
35
1/10/2003 | 12,286 | 35,621 | {"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} | 4.40625 | 4 | CC-MAIN-2020-50 | latest | en | 0.76779 |
https://cs.stackexchange.com/questions/9899/why-does-a-recurrence-of-tn-1-tn-2-yield-something-in-omega2-fr/9905 | 1,582,277,806,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145500.90/warc/CC-MAIN-20200221080411-20200221110411-00221.warc.gz | 323,614,896 | 35,334 | # Why does a recurrence of $T(n - 1) + T(n - 2)$ yield something in $\Omega(2^{\frac{n}{2}})$?
I am trying to analyze the running time of a bad implementation of generating the $n$th member of the fibonacci sequence (which requires generating the previous 2 values from the bottom up).
Why does this algorithm have a time complexity of $\Omega(2^{\frac{n}{2}})$? Where does the exponent come from?
• Try writing a recurrence relation expressing the running time and solve it. – saadtaame Feb 18 '13 at 19:20
• Just to be clear, "time complexity of at least" is talking about the lower bound. Here it means that the lower bound is of $O(2^{n/2})$, which basically means that $T(n) \in \Omega(2^{n/2})$. The answers below demonstrate this fact. – Paresh Feb 18 '13 at 21:37
• "the lower bound is of $O(\\_)$" resp "at least $O(\\_)$" -- that does not make much sense. The lower bound needs to be in $\Omega(\\_)$, too. See also here. – Raphael Feb 19 '13 at 6:15
• @Raphael Oh I agree it is an abuse of the notation, and should not be used, but unfortunately I have seen it being used. When saying $f(n)$ is atleast $O(g(n))$, what is being implied (and hence the abuse) is that $g(n)$ is a tight upper bound on some $h(n)$, and that $f(n)$ is $\Omega(h(n))$, thereby also implying $f(n) \in \Omega(g(n))$. – Paresh Feb 19 '13 at 7:54
• @Raphael But yeah, your edit of the question seems the right way to go. – Paresh Feb 19 '13 at 8:34
Expanding on Reza's answer, every recurrence of the form $T(n) = T(n-1) + T(n-2)$, with arbitrary initial values, has a solution of the form $$T(n) = A \left( \frac{1+\sqrt{5}}{2} \right)^n + B \left( \frac{1-\sqrt{5}}{2} \right)^n,$$ for some $A,B$. Note that $|(1-\sqrt{5})/2| < 1$, and so the second term tends to zero as $n \longrightarrow \infty$. Assuming that $T(n)$ tends to infinity, $A > 0$ and so $$T(n) = \Theta\left( \left( \frac{1+\sqrt{5}}{2} \right)^n \right).$$ Now $(1+\sqrt{5})/2 > \sqrt{2}$, and so $T(n) = \Omega(2^{n/2})$.
Edit: This part is also covered in this answer (see under "A Shortcut").
More generally, for a recurrence of the form $T(n) = \sum_{i=1}^k a_i T(n-i)$, let $$P(t) = t^n - \sum_{i=1}^k a_i t^{n-i}.$$ If $P$ has no repeated roots and the (possibly complex) roots are $\lambda_1,\ldots,\lambda_k$, then the solution is always of the form $$T(n) = \sum_{i=1}^k A_i \lambda_i^n.$$ If it does have repeated roots, say the roots are $\lambda_1,\ldots,\lambda_l$ with multiplicities $m_1,\ldots,m_l$, then the solution is always of the form $$T(n) = \sum_{i=1}^l A_i(n) \lambda_i^n,$$ where $A_i$ is a (possibly complex) polynomial of degree smaller than $m_i$.
In our case, $P(t) = t^2-t-1$ has no repeated roots, and the two roots are $(1\pm\sqrt{5})/2$.
The $A_i$s depend on the initial values, and can be found by solving linear equations. For example, suppose we are given $T(0)$ and $T(1)$ for our recurrence. Then we can find $A,B$ by solving the system \begin{align*} T(0) &= A + B, \\ T(1) &= \frac{1+\sqrt{5}}{2} A + \frac{1-\sqrt{5}}{2} B. \end{align*} This works even in the case of repeated roots, and $k$ initial values always suffice. Assuming $a_k \neq 0$, $k$ initial values are also necessary.
• I think the second part of your answer has been covered here ("A Shortcut"). If not, you should probably add an answer there with the general idea. – Raphael Feb 19 '13 at 6:17
It's not $O(2^{n\over 2})$ is $\Theta(\phi^n)$ which doesn't belong to $O(2^{n\over 2})$ but you can say belongs to $\Omega(2^{n\over 2})$ or belongs to $O(2^n)$ (and saadtaame showed this in his answer), So be careful about abuse of notations. But why is $\Theta(\phi^n)$? as a student of CS you should try to solve it yourself but if you tried and you couldn't show it, show us your try and we can help you.
Oh Seems I forgot to say why is in $\Omega(2^{n\over 2})$ (and I think your main question was this, or totally you were wrong): $T(n) = T(n-1)+T(n-2) \gt 2T(n-2)$ and this results: $T(n) > 2^{n\over 2}T(0)$.
The running time of the naive solution is: $$T(n)=T(n-1)+T(n-2) \lt 2T(n-1)$$ Now, $$2T(n-1)=2(2T(n-2))=2(2(2T(n-3)))=\dots=2^kT(n-k)=\dots=2^nT(0)=O(2^n)$$
The base case takes time $T(0)=\Theta(1)$. So that's where the exponent is coming from. This result is an upper bound; you can obtain a tighter bound if you solve the recurrence using generating functions for example.
• Again, $O(2^n)$ is not the end result we need; we need $\Omega$. – Raphael Feb 19 '13 at 6:18
• This can be 'flipped around' to give the result the OP is after, though - I'll write it up in a moment here. – Steven Stadnicki Feb 19 '13 at 23:44
you can find for Every linear recurrence with constant coefficients a closed form. (See the link) the Fibonacci numbers have a closed-form solution as (the approximation is for large ns):
$F_n \times \sqrt{5} ={ \left( \frac { 1+\sqrt { 5 } }{ 2 } \right) }^{ n }-{ \left( \frac { 1-\sqrt { 5 } }{ 2 } \right) }^{ n } \simeq { \left( \frac { 1+\sqrt { 5 } }{ 2 } \right) }^{ n } \simeq (1.6180..)^n \simeq 2.89^{n/2}$
or more formaly $F_n \in\Omega (2^{n/2})$.
You can find more in Conceret Mathematics, by Ronald L. Graham, Donald E. Knuth and Oren Patashnik, chapter 6 : Special Numbers.
• That's not helping. You are confusing the OP. – saadtaame Feb 18 '13 at 19:52
• @saadtaame: This is the exact way of computing Fibonacci or any constant coefficient recursive relation. It could be found in any cobinatorial book, no confusion at all !. I think this is nessasary for any computer science student to know. – Reza Feb 18 '13 at 19:58
• What does "is at least $O(2^{n/2})$" mean? please use mathematical definitions and validate your sentence with big-oh definition. – user742 Feb 18 '13 at 20:06
• "at least $O(\dots)$" not a meaningful statement, as $1 \in O(f)$ for most functions $f$ that pop up in algorithm analysis. – Raphael Feb 19 '13 at 6:20 | 1,937 | 5,876 | {"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} | 3.6875 | 4 | CC-MAIN-2020-10 | latest | en | 0.883852 |
http://www.stata.com/statalist/archive/2003-04/msg00019.html | 1,495,579,914,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607704.68/warc/CC-MAIN-20170523221821-20170524001821-00099.warc.gz | 662,493,789 | 2,966 | # Re: st: RE: Intraclass correlation?
From "yore" To statalist@hsphsun2.harvard.edu Subject Re: st: RE: Intraclass correlation? Date Tue, 01 Apr 2003 06:34:40 -0000
```Dear bissery,
How do estimate intraclass correlation for dichotomous data. That is,
replace your continuous variable example with categorical variable.
pi yaw
1 1
0 2
1 3
1 1
0 2
0 3
1 1
1 2
0 3
cheers
atiti
--- In statalist@yahoogroups.com, BISSERY Alvine <Alvine.BISSERY@h...>
wrote:
> If you have one continuous variable called "measure", which contains
the
> value of measure on each subject, and one variable called "subject"
wich
> contains the label of each subject, do
>
> loneway measure subject
>
> it will give you the intraclass correlation
>
> For example:
>
>
> mes sub
> 25 1
> 36 2
> 14 3
> 24 1
> 35 2
> 12 3
> 25 1
> 37 2
> 14 3
>
> . loneway mes sub
>
> One-way Analysis of Variance for mes:
>
> Number of obs = 9
> R-squared = 0.9931
>
> Source SS df MS F Prob >
F
>
> Between sub 770.66667 2 385.33333 433.50
0.0000
> Within sub 5.3333333 6 .88888889
>
> Total 776 8 97
>
> Intraclass Asy.
> correlation S.E. [95% Conf. Interval]
> ------------------------------------------------
> 0.99311 0.00792 0.97759 1.00863
>
> Estimated SD of sub effect 11.32025
> Estimated SD within sub .942809
> Est. reliability of a sub mean .9976932
> (evaluated at n=3.00)
>
>
> I Hope this helps
>
>
> Alvine Bissery
>
> *********************************************
> Alvine Bissery
> Biostatisticienne
>
> Hôpital Européen Georges Pompidou
> Centre d' Investigations Cliniques
> 20-40 rue Leblanc
> 75 908 Paris Cedex 15
> France
>
> tél: +33 1 56 09 29 13
> fax: +33 1 56 09 29 29
> mail: alvine.bissery@h...
> > -----Message d'origine-----
> > De: Jeffrey Simons [mailto:jsimons@u...]
> > Date: lundi 2 septembre 2002 00:08
> > À: statalist@h...
> > Objet: st: Intraclass correlation?
> >
> >
> > How can you do intraclass correlation coefficients in Stata
> > for examining
> > test-retest reliability?
> >
> > *
> > * For searches and help try:
> > * http://www.stata.com/support/faqs/res/findit.html
> > * http://www.stata.com/support/statalist/faq
> > * http://www.ats.ucla.edu/stat/stata/
> >
> *
> * For searches and help try:
> * http://www.stata.com/support/faqs/res/findit.html
> * http://www.stata.com/support/statalist/faq
> * http://www.ats.ucla.edu/stat/stata/
*
* For searches and help try:
* http://www.stata.com/support/faqs/res/findit.html
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
``` | 918 | 2,771 | {"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} | 2.75 | 3 | CC-MAIN-2017-22 | latest | en | 0.476943 |
http://mathhelpforum.com/discrete-math/2145-big-o-estimate.html | 1,480,751,418,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540909.75/warc/CC-MAIN-20161202170900-00385-ip-10-31-129-80.ec2.internal.warc.gz | 178,909,803 | 10,005 | 1. ## Big O Estimate
Can someone please check if the following big O estimate is correct? Thanks!
n^2n + n^(n^2)
Big O Estimate: O(n^n) since n^n must be faster than n^(n^2)?
2. Originally Posted by hotmail590
Can someone please check if the following big O estimate is correct? Thanks!
n^2n + n^(n^2)
Big O Estimate: O(n^n) since n^n must be faster than n^(n^2)?
This is $O(n^{n^2})$, since $n^{2n}$ is negligable
in comparison to $n^{n^2}$, for large enough $n$
RonL
3. Originally Posted by hotmail590
Can someone please check if the following big O estimate is correct? Thanks!
n^2n + n^(n^2)
Big O Estimate: O(n^n) since n^n must be faster than n^(n^2)?
To clarify, in case there has been a misunderstanding:
$
n^{2n} + n^{n^2} = O(n^{n^2})
$
and:
$
n^{2n} + n^{n^2} = n^{n^2} + O(n^{2n})
$
RonL | 274 | 813 | {"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": 6, "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} | 3.15625 | 3 | CC-MAIN-2016-50 | longest | en | 0.847539 |
https://algosim.org/doc/arctanh.html | 1,686,044,917,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652494.25/warc/CC-MAIN-20230606082037-20230606112037-00380.warc.gz | 104,745,409 | 1,694 | arctanh – Algosim documentation
Algosim documentation: arctanh
# arctanh
The inverse hyperbolic tangent function.
## Syntax
• `arctanh(x)`
• `x` is a real or complex number
## Description
`arctanh` is the inverse of the hyperbolic tangent function. For real numbers, the domain is (−1, 1). It is defined by
```arctanh(z) = [ln(1 + z) − ln(1 − z)] / 2
```
for all complex `z`. This formula is also used for real numbers outside [−1, 1].
## Notes
This function is also called `artanh` (area hyperbolic cotangent) in the literature. Some authors claim that the name `arctanh` is a misnomer, but this depends crucially on how you view the connections between the trigonometric and the hyperbolic functions and their (restriction) inverses. One can argue that the name `arctanh` makes perfect sense.
## Examples
`arctanh(0)`
`0`
`arctanh(1/π)`
`0.329765314957`
`arctanh(2)`
`0.549306144334 − 1.57079632679⋅i`
`arctanh(i)`
`0.785398163397⋅i` | 303 | 950 | {"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} | 2.734375 | 3 | CC-MAIN-2023-23 | latest | en | 0.749678 |
https://discuss.codechef.com/t/what-is-the-error-here-problem-https-www-codechef-com-problems-hs08test/84503 | 1,632,705,123,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058222.43/warc/CC-MAIN-20210926235727-20210927025727-00711.warc.gz | 254,106,835 | 4,427 | # What is the error here. Problem: https://www.codechef.com/problems/HS08TEST
Running it results good and there is no error, but when I submit, it shows a NZEC error
a, b = map(float, input().split(" "))
if ((a%5==0) and (a+0.5<=b)):
print("{:.2f}".format(b-a-0.50))
else:
print("{:.2f}".format(b))
a and b both are objects of map. Try this out instead,
a, b = tuple(map(float, input().split()))
as it is known that x, y = (2, 3). Then x = 2 and y = 3 and they both will be of type int(in this case).
For submiting in Python, try to put your code in try/except block like below.
try:
//your code
except:
print(“ErRor found”)
Also take the input like this
a, b = [float(x) for x in input().split()]
or
take input individually
This didn’t actually work out for me, but thank you very much for your try. When I submitted it showed "Wrong answer"
Thank You for your kind help (It worked). Thank you very much, but could you please explain what you did (Couldn’t understand)
which line you didn’t understand? | 289 | 1,014 | {"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-2021-39 | latest | en | 0.894344 |
https://econometricstutors.com/how-to-draw-basic-descriptive-plots-in-r/ | 1,653,703,412,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663011588.83/warc/CC-MAIN-20220528000300-20220528030300-00088.warc.gz | 270,816,105 | 19,897 | # How to Draw Basic Descriptive Plots in R
## 01 Dec How to Draw Basic Descriptive Plots in R
In this tutorial you will learn how to create nice-looking plots using basic R functions. There are a handful of packages in R dedicated in plotting with rich parameterizations but in this post our goal is to keep it simple and create some basic plots that are functional as well as informative, with only a few lines of code.
scatter plot
Using the mtcars dataset existed in core R, with the following function we create a scatter plot that depicts the horsepower and mileage of the cars existed in the mtcars data frame.
`plot(mtcars\$hp, mtcars\$mpg)`
The result of the above line looks like the following graph:
##### In this post we will highlight some basic functions for plotting using R statistical language and it will become clear that creating informative charts requires only a few lines of basic plot commands.
line plot
In order to create a line plot, we pass in the x and y arguments of the plot function two different vectors. For example, the following line of code creates two lines; the first one with the default black color and the second one with red color:
```plot(pressure\$temperature, pressure\$pressure, type = "l")
lines(pressure\$temperature, pressure\$pressure/2, col="red")```
The result of the above code looks like the following graph:
bar graph
A bar graph is a graph where the bars represent the count of cases in each category. It is similar to a histogram, but has discrete instead of continuous values in the x-axis.
```t<-table(mtcars\$cyl) #getting the counts of each unique value
barplot(t)```
The bar graph below depicts the number of cars in the mtcars data frame that have 4, 6 and 8 cylinders:
, , , , | 397 | 1,756 | {"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} | 2.859375 | 3 | CC-MAIN-2022-21 | longest | en | 0.89311 |
https://kidsworksheetfun.com/double-digit-addition-with-regrouping-games/ | 1,723,104,804,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640723918.41/warc/CC-MAIN-20240808062406-20240808092406-00649.warc.gz | 280,430,062 | 27,159 | # Double Digit Addition With Regrouping Games
Click on two cards to match an addition problem with its answer. In this interactive two digit addition game elementary students will practice adding two digit numbers with regrouping.
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It Includes A Quick Print Game To Practice Addition With Regrouping An Addition With Regrouping Worksh Addition And Subtraction Subtraction Winter Themed Math | 861 | 4,576 | {"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-2024-33 | latest | en | 0.893463 |
https://forums.tomsguide.com/threads/excel-2013-subtract-from-cell-2-negative-value-in-cell-1-but-only-if-cell-1-is-negative.354255/ | 1,726,611,146,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.53/warc/CC-MAIN-20240917204739-20240917234739-00856.warc.gz | 243,495,967 | 28,983 | # Excel 2013 - Subtract from cell 2 negative value in cell 1, but only if cell 1 is negative.
#### Max99000
##### Honorable
I need to subtract from the "Savings Account" to make the "Current Account Bank Balance" 0 (ZERO) only IF the "Current Account Bank Balance" is negative, otherwise not. Also, if the negative balance of "Current Account Bank Balance" is greater than the positive balance of "Savings Account" subtract the entire savings value even if it "Current Account Bank Balance" does not = 0.
Thanks!
screenshot here:
#### noidea_77
##### Honorable
The integrated Help function of Excel may help you with this homework. You will find some examples on how to use the IF function there.
#### Max99000
##### Honorable
Can anyone help me with the formula. I have no idea what to do. I have tried IF and SUMIF functions but I could not get it to work.
#### USAFRet
##### Illustrious
Moderator
Post an example of your IF statement. It does work.
#### Max99000
##### Honorable
OK. This should calculate "I4" considering the Account Balance:
=IF((G4+H3)<0,(I3+C4)-(G4+H3),(G4+H3))
but it does not work.
#### noidea_77
##### Honorable
";" instead of ","? And what's the Excel comment (behind the red rectangle) in column H? A format problem?
#### Max99000
##### Honorable
What do you mean by ";"?
The error says that I cannot use rotating variables.
#### Max99000
##### Honorable
Ingnore my last post. The error says "Inconsistent Calculated Column Formula", that's because I used different formulas in the rows.
#### Max99000
##### Honorable
Alright. I made some changes and I found a formula that could work. This will calculate cell "I13" "=IF((G13+H12)<0,(I12+C13)+(G13+H12),(I12+C13))" (screenshot below)
However, I need to make some changes because the formula will take the "Savings Account" on minus and that is not correct. I need it to make the "Current Account Bank Balance" positive, but only if there are enough savings. Otherwise, substract from savings as much as possible (until its 0), but not below 0.
#### USAFRet
##### Illustrious
Moderator
Look into =SUMIFS
That allows multiple conditions.
#### Max99000
##### Honorable
How can I apply that to my spreadsheet? Can you give me an example please?
I'm a complete noob at this.
#### Max99000
##### Honorable
I found a solution.
The SUMIFS formula is not exactly what I needed. I used an IF statement combined with MAX formula.
Example:
To calculate I13 - =IF((G13+H12)<0,MAX(((I12+C13)+(G13+H12)),0),(I12+C13))
Thanks for the help guys!
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2K | 711 | 2,638 | {"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} | 2.984375 | 3 | CC-MAIN-2024-38 | latest | en | 0.876112 |
https://gitlab.inria.fr/why3/why3/-/raw/f33f9458c350ee3f52828cfce0ce470a67c7b8cd/examples/hoare_logic/blocking_semantics3.mlw?inline=false | 1,600,773,213,000,000,000 | text/plain | crawl-data/CC-MAIN-2020-40/segments/1600400205950.35/warc/CC-MAIN-20200922094539-20200922124539-00294.warc.gz | 432,605,582 | 6,399 | (** {1 A certified WP calculus} *) (** {2 A simple imperative language with expressions, syntax and semantics} *) theory ImpExpr use import int.Int use import int.MinMax use import bool.Bool use export list.List use export list.Append use map.Map as IdMap (** types and values *) type datatype = TYunit | TYint | TYbool type value = Vvoid | Vint int | Vbool bool (** terms and formulas *) type operator = Oplus | Ominus | Omult | Ole (** ident for mutable variables *) type mident axiom mident_decide : forall m1 m2: mident. m1 = m2 \/ m1 <> m2 (** ident for immutable variables *) type ident = {| ident_index : int |} axiom ident_decide : forall m1 m2: ident. m1 = m2 \/ m1 <> m2 (** Terms *) type term = | Tvalue value | Tvar ident | Tderef mident | Tbin term operator term predicate var_occurs_in_term (x:ident) (t:term) = match t with | Tvalue _ -> false | Tvar i -> x=i | Tderef _ -> false | Tbin t1 _ t2 -> var_occurs_in_term x t1 \/ var_occurs_in_term x t2 end (* predicate term_inv (t:term) = *) (* forall x:ident. var_occurs_in_term x t -> x.ident_index <= t.term_maxvar *) function mk_tvalue (v:value) : term = Tvalue v function mk_tvar (i:ident) : term = Tvar i function mk_tderef (r:mident) : term = Tderef r function mk_tbin (t1:term) (o:operator) (t2:term) : term = Tbin t1 o t2 (** Formulas *) type fmla = | Fterm term | Fand fmla fmla | Fnot fmla | Fimplies fmla fmla | Flet ident term fmla (* let id = term in fmla *) | Fforall ident datatype fmla (* forall id : ty, fmla *) (** Statements *) type stmt = | Sskip | Sassign mident term | Sseq stmt stmt | Sif term stmt stmt | Sassert fmla | Swhile term fmla stmt (* while cond invariant inv body *) lemma decide_is_skip: forall s:stmt. s = Sskip \/ s <> Sskip (** Typing *) function type_value (v:value) : datatype = match v with | Vvoid -> TYunit | Vint int -> TYint | Vbool bool -> TYbool end inductive type_operator (op:operator) (ty1 ty2 ty: datatype) = | Type_plus : type_operator Oplus TYint TYint TYint | Type_minus : type_operator Ominus TYint TYint TYint | Type_mult : type_operator Omult TYint TYint TYint | Type_le : type_operator Ole TYint TYint TYbool type type_stack = list (ident, datatype) (* map local immutable variables to their type *) function get_vartype (i:ident) (pi:type_stack) : datatype = match pi with | Nil -> TYunit | Cons (x,ty) r -> if x=i then ty else get_vartype i r end type type_env = IdMap.map mident datatype (* map global mutable variables to their type *) function get_reftype (i:mident) (e:type_env) : datatype = IdMap.get e i inductive type_term type_env type_stack term datatype = | Type_value : forall sigma: type_env, pi:type_stack, v:value. type_term sigma pi (Tvalue v) (type_value v) | Type_var : forall sigma: type_env, pi:type_stack, v: ident, ty:datatype. (get_vartype v pi = ty) -> type_term sigma pi (Tvar v) ty | Type_deref : forall sigma: type_env, pi:type_stack, v: mident, ty:datatype. (get_reftype v sigma = ty) -> type_term sigma pi (Tderef v) ty | Type_bin : forall sigma: type_env, pi:type_stack, t1 t2 : term, op:operator, ty1 ty2 ty:datatype. type_term sigma pi t1 ty1 -> type_term sigma pi t2 ty2 -> type_operator op ty1 ty2 ty -> type_term sigma pi (Tbin t1 op t2) ty inductive type_fmla type_env type_stack fmla = | Type_term : forall sigma: type_env, pi:type_stack, t:term. type_term sigma pi t TYbool -> type_fmla sigma pi (Fterm t) | Type_conj : forall sigma: type_env, pi:type_stack, f1 f2:fmla. type_fmla sigma pi f1 -> type_fmla sigma pi f2 -> type_fmla sigma pi (Fand f1 f2) | Type_neg : forall sigma: type_env, pi:type_stack, f:fmla. type_fmla sigma pi f -> type_fmla sigma pi (Fnot f) | Type_implies : forall sigma: type_env, pi:type_stack, f1 f2:fmla. type_fmla sigma pi f1 -> type_fmla sigma pi f2 -> type_fmla sigma pi (Fimplies f1 f2) | Type_let : forall sigma: type_env, pi:type_stack, x:ident, t:term, f:fmla, ty:datatype. type_term sigma pi t ty -> type_fmla sigma (Cons (x,ty) pi) f -> type_fmla sigma pi (Flet x t f) | Type_forall1 : forall sigma: type_env, pi:type_stack, x:ident, f:fmla. type_fmla sigma (Cons (x,TYint) pi) f -> type_fmla sigma pi (Fforall x TYint f) | Type_forall2 : forall sigma: type_env, pi:type_stack, x:ident, f:fmla. type_fmla sigma (Cons (x,TYbool) pi) f -> type_fmla sigma pi (Fforall x TYbool f) | Type_forall3: forall sigma: type_env, pi:type_stack, x:ident, f:fmla. type_fmla sigma (Cons (x,TYunit) pi) f -> type_fmla sigma pi (Fforall x TYunit f) inductive type_stmt type_env type_stack stmt = | Type_skip : forall sigma: type_env, pi:type_stack. type_stmt sigma pi Sskip | Type_seq : forall sigma: type_env, pi:type_stack, s1 s2:stmt. type_stmt sigma pi s1 -> type_stmt sigma pi s2 -> type_stmt sigma pi (Sseq s1 s2) | Type_assigns : forall sigma: type_env, pi:type_stack, x:mident, t:term, ty:datatype. (get_reftype x sigma = ty) -> type_term sigma pi t ty -> type_stmt sigma pi (Sassign x t) | Type_if : forall sigma: type_env, pi:type_stack, t:term, s1 s2:stmt. type_term sigma pi t TYbool -> type_stmt sigma pi s1 -> type_stmt sigma pi s2 -> type_stmt sigma pi (Sif t s1 s2) | Type_assert : forall sigma: type_env, pi:type_stack, p:fmla. type_fmla sigma pi p -> type_stmt sigma pi (Sassert p) | Type_while : forall sigma: type_env, pi:type_stack, guard:term, body:stmt, inv:fmla. type_fmla sigma pi inv -> type_term sigma pi guard TYbool -> type_stmt sigma pi body -> type_stmt sigma pi (Swhile guard inv body) (** Operational semantic *) type env = IdMap.map mident value (* map global mutable variables to their value *) function get_env (i:mident) (e:env) : value = IdMap.get e i type stack = list (ident, value) (* map local immutable variables to their value *) function get_stack (i:ident) (pi:stack) : value = match pi with | Nil -> Vvoid | Cons (x,v) r -> if x=i then v else get_stack i r end lemma get_stack_eq: forall x:ident, v:value, r:stack. get_stack x (Cons (x,v) r) = v lemma get_stack_neq: forall x i:ident, v:value, r:stack. x <> i -> get_stack i (Cons (x,v) r) = get_stack i r (** semantics of formulas *) function eval_bin (x:value) (op:operator) (y:value) : value = match x,y with | Vint x,Vint y -> match op with | Oplus -> Vint (x+y) | Ominus -> Vint (x-y) | Omult -> Vint (x*y) | Ole -> Vbool (if x <= y then True else False) end | _,_ -> Vvoid end function eval_term (sigma:env) (pi:stack) (t:term) : value = match t with | Tvalue v -> v | Tvar id -> get_stack id pi | Tderef id -> get_env id sigma | Tbin t1 op t2 -> eval_bin (eval_term sigma pi t1) op (eval_term sigma pi t2) end lemma eval_bool_term: forall sigma:env, pi:stack, sigmat:type_env, pit:type_stack, t:term. type_term sigmat pit t TYbool -> (* TODO: compatibility sigma, sigmat and pi,pit *) exists b:bool. eval_term sigma pi t = Vbool b predicate eval_fmla (sigma:env) (pi:stack) (f:fmla) = match f with | Fterm t -> eval_term sigma pi t = Vbool True | Fand f1 f2 -> eval_fmla sigma pi f1 /\ eval_fmla sigma pi f2 | Fnot f -> not (eval_fmla sigma pi f) | Fimplies f1 f2 -> eval_fmla sigma pi f1 -> eval_fmla sigma pi f2 | Flet x t f -> eval_fmla sigma (Cons (x,eval_term sigma pi t) pi) f | Fforall x TYint f -> forall n:int. eval_fmla sigma (Cons (x,Vint n) pi) f | Fforall x TYbool f -> forall b:bool. eval_fmla sigma (Cons (x,Vbool b) pi) f | Fforall x TYunit f -> eval_fmla sigma (Cons (x,Vvoid) pi) f end (** substitution of a reference [r] by a logic variable [v] warning: proper behavior only guaranted if [v] is "fresh", i.e index(v) > term_maxvar(t) *) function msubst_term (t:term) (r:mident) (v:ident) : term = match t with | Tvalue _ | Tvar _ -> t | Tderef x -> if r = x then mk_tvar v else t | Tbin t1 op t2 -> mk_tbin (msubst_term t1 r v) op (msubst_term t2 r v) end function subst_term (t:term) (r:ident) (v:ident) : term = match t with | Tvalue _ | Tderef _ -> t | Tvar x -> if r = x then mk_tvar v else t | Tbin t1 op t2 -> mk_tbin (subst_term t1 r v) op (subst_term t2 r v) end (** [fresh_in_term id t] is true when [id] does not occur in [t] *) predicate fresh_in_term (id:ident) (t:term) = not (var_occurs_in_term id t) lemma fresh_in_binop: forall t t':term, op:operator, v:ident. fresh_in_term v (mk_tbin t op t') -> fresh_in_term v t /\ fresh_in_term v t' lemma eval_msubst_term: forall e:term, sigma:env, pi:stack, x:mident, v:ident. fresh_in_term v e -> eval_term sigma pi (msubst_term e x v) = eval_term (IdMap.set sigma x (get_stack v pi)) pi e (* lemma eval_subst_term: *) (* forall sigma:env, pi:stack, e:term, x:ident, v:ident. *) (* fresh_in_term v e -> *) (* eval_term sigma pi (subst_term e x v) = *) (* eval_term sigma (Cons (x, (get_stack v pi)) pi) e *) lemma eval_term_change_free : forall t:term, sigma:env, pi:stack, id:ident, v:value. fresh_in_term id t -> eval_term sigma (Cons (id,v) pi) t = eval_term sigma pi t predicate fresh_in_fmla (id:ident) (f:fmla) = match f with | Fterm e -> fresh_in_term id e | Fand f1 f2 | Fimplies f1 f2 -> fresh_in_fmla id f1 /\ fresh_in_fmla id f2 | Fnot f -> fresh_in_fmla id f | Flet y t f -> id <> y /\ fresh_in_term id t /\ fresh_in_fmla id f | Fforall y ty f -> id <> y /\ fresh_in_fmla id f end function subst (f:fmla) (x:ident) (v:ident) : fmla = match f with | Fterm e -> Fterm (subst_term e x v) | Fand f1 f2 -> Fand (subst f1 x v) (subst f2 x v) | Fnot f -> Fnot (subst f x v) | Fimplies f1 f2 -> Fimplies (subst f1 x v) (subst f2 x v) | Flet y t f -> Flet y (subst_term t x v) (subst f x v) | Fforall y ty f -> Fforall y ty (subst f x v) end function msubst (f:fmla) (x:mident) (v:ident) : fmla = match f with | Fterm e -> Fterm (msubst_term e x v) | Fand f1 f2 -> Fand (msubst f1 x v) (msubst f2 x v) | Fnot f -> Fnot (msubst f x v) | Fimplies f1 f2 -> Fimplies (msubst f1 x v) (msubst f2 x v) | Flet y t f -> Flet y (msubst_term t x v) (msubst f x v) | Fforall y ty f -> Fforall y ty (msubst f x v) end lemma subst_fresh : forall f:fmla, x:ident, v:ident. fresh_in_fmla x f -> subst f x v = f (* Not needed *) (* lemma let_subst: *) (* forall t:term, f:fmla, x id':ident, id :mident. *) (* msubst (Flet x t f) id id' = Flet x (msubst_term t id id') (msubst f id id') *) (* Need it for monotonicity and wp_reduction *) lemma eval_msubst: forall f:fmla, sigma:env, pi:stack, x:mident, v:ident. fresh_in_fmla v f -> (eval_fmla sigma pi (msubst f x v) <-> eval_fmla (IdMap.set sigma x (get_stack v pi)) pi f) (* lemma eval_subst: *) (* forall f:fmla, sigma:env, pi:stack, x:ident, v:ident. *) (* fresh_in_fmla v f -> *) (* (eval_fmla sigma pi (subst f x v) <-> *) (* eval_fmla sigma (Cons(x, (get_stack v pi)) pi) f) *) (* lemma eval_same_var_term: *) (* forall t:term, sigma:env, pi:stack, id:ident, v1 v2:value. *) (* eval_term sigma (Cons (id,v1) (Cons (id,v2) pi)) t = *) (* eval_term sigma (Cons (id,v1) pi) t *) (* lemma eval_same_var: *) (* forall f:fmla, sigma:env, pi:stack, id:ident, v1 v2:value. *) (* eval_fmla sigma (Cons (id,v1) (Cons (id,v2) pi)) f <-> *) (* eval_fmla sigma (Cons (id,v1) pi) f *) lemma eval_swap_term_any: forall t:term, sigma:env, pi l:stack, id1 id2:ident, v1 v2:value. id1 <> id2 -> (eval_term sigma (l++(Cons (id1,v1) (Cons (id2,v2) pi))) t = eval_term sigma (l++(Cons (id2,v2) (Cons (id1,v1) pi))) t) (* lemma eval_swap_term: *) (* forall t:term, sigma:env, pi:stack, id1 id2:ident, v1 v2:value. *) (* id1 <> id2 -> *) (* (eval_term sigma (Cons (id1,v1) (Cons (id2,v2) pi)) t = *) (* eval_term sigma (Cons (id2,v2) (Cons (id1,v1) pi)) t) *) lemma eval_swap_any: forall f:fmla, sigma:env, pi l:stack, id1 id2:ident, v1 v2:value. id1 <> id2 -> (eval_fmla sigma (l++(Cons (id1,v1) (Cons (id2,v2) pi))) f <-> eval_fmla sigma (l++(Cons (id2,v2) (Cons (id1,v1) pi))) f) (* lemma eval_swap: *) (* forall f:fmla, sigma:env, pi:stack, id1 id2:ident, v1 v2:value. *) (* id1 <> id2 -> *) (* (eval_fmla sigma (Cons (id1,v1) (Cons (id2,v2) pi)) f <-> *) (* eval_fmla sigma (Cons (id2,v2) (Cons (id1,v1) pi)) f) *) (* Need it for monotonicity*) lemma eval_change_free : forall sigma:env, pi:stack, f:fmla, id:ident, v:value. fresh_in_fmla id f -> (eval_fmla sigma (Cons (id,v) pi) f <-> eval_fmla sigma pi f) (** [valid_fmla f] is true when [f] is valid in any environment *) predicate valid_fmla (p:fmla) = forall sigma:env, pi:stack. eval_fmla sigma pi p (* Not needed *) (* axiom msubst_implies : *) (* forall p q:fmla. *) (* valid_fmla (Fimplies p q) -> *) (* forall sigma:env, pi:stack, x:mident, id:ident. *) (* fresh_in_fmla id (Fand p q) -> *) (* eval_fmla sigma (Cons (id, (get_env x sigma)) pi) (Fimplies (msubst p x id) (msubst q x id)) *) (** let id' = t in f[id <- id'] <=> let id = t in f*) (* Not needed *) (* lemma let_equiv : *) (* forall id:ident, id':ident, t:term, f:fmla. *) (* forall sigma:env, pi:stack. *) (* fresh_in_fmla id' f -> *) (* eval_fmla sigma pi (Flet id' t (subst f id id')) *) (* -> eval_fmla sigma pi (Flet id t f) *) (* lemma let_implies : *) (* forall id:ident, t:term, p q:fmla. *) (* valid_fmla (Fimplies p q) -> *) (* valid_fmla (Fimplies (Flet id t p) (Flet id t q)) *) predicate fresh_in_stmt (id:ident) (s:stmt) = match s with | Sskip -> true | Sseq s1 s2 -> fresh_in_stmt id s1 /\ fresh_in_stmt id s2 | Sassign _ t -> fresh_in_term id t | Sif t s1 s2 -> fresh_in_term id t /\ fresh_in_stmt id s1 /\ fresh_in_stmt id s2 | Sassert f -> fresh_in_fmla id f | Swhile cond inv body -> fresh_in_term id cond /\ fresh_in_fmla id inv /\ fresh_in_stmt id body end (** small-steps semantics for expressions *) inductive one_step env stack stmt env stack stmt = | one_step_assign : forall sigma sigma':env, pi:stack, x:mident, t:term. sigma' = IdMap.set sigma x (eval_term sigma pi t) -> one_step sigma pi (Sassign x t) sigma' pi Sskip | one_step_seq_noskip: forall sigma sigma':env, pi pi':stack, s1 s1' s2:stmt. one_step sigma pi s1 sigma' pi' s1' -> one_step sigma pi (Sseq s1 s2) sigma' pi' (Sseq s1' s2) | one_step_seq_skip: forall sigma:env, pi:stack, s:stmt. one_step sigma pi (Sseq Sskip s) sigma pi s | one_step_if_true: forall sigma:env, pi:stack, t:term, s1 s2:stmt. eval_term sigma pi t = Vbool True -> one_step sigma pi (Sif t s1 s2) sigma pi s1 | one_step_if_false: forall sigma:env, pi:stack, t:term, s1 s2:stmt. eval_term sigma pi t = Vbool False -> one_step sigma pi (Sif t s1 s2) sigma pi s2 | one_step_assert: forall sigma:env, pi:stack, f:fmla. (* blocking semantics *) eval_fmla sigma pi f -> one_step sigma pi (Sassert f) sigma pi Sskip | one_step_while_true: forall sigma:env, pi:stack, cond:term, inv:fmla, body:stmt. (* blocking semantics *) eval_fmla sigma pi inv -> eval_term sigma pi cond = Vbool True -> one_step sigma pi (Swhile cond inv body) sigma pi (Sseq body (Swhile cond inv body)) | one_step_while_false: forall sigma:env, pi:stack, cond:term, inv:fmla, body:stmt. (* blocking semantics *) eval_fmla sigma pi inv -> eval_term sigma pi cond = Vbool False -> one_step sigma pi (Swhile cond inv body) sigma pi Sskip (** many steps of execution *) inductive many_steps env stack stmt env stack stmt int = | many_steps_refl: forall sigma:env, pi:stack, s:stmt. many_steps sigma pi s sigma pi s 0 | many_steps_trans: forall sigma1 sigma2 sigma3:env, pi1 pi2 pi3:stack, s1 s2 s3:stmt, n:int. one_step sigma1 pi1 s1 sigma2 pi2 s2 -> many_steps sigma2 pi2 s2 sigma3 pi3 s3 n -> many_steps sigma1 pi1 s1 sigma3 pi3 s3 (n+1) lemma steps_non_neg: forall sigma1 sigma2:env, pi1 pi2:stack, s1 s2:stmt, n:int. many_steps sigma1 pi1 s1 sigma2 pi2 s2 n -> n >= 0 (* Used by Hoare_logic/seq_rule*) lemma many_steps_seq: forall sigma1 sigma3:env, pi1 pi3:stack, s1 s2:stmt, n:int. many_steps sigma1 pi1 (Sseq s1 s2) sigma3 pi3 Sskip n -> exists sigma2:env, pi2:stack, n1 n2:int. many_steps sigma1 pi1 s1 sigma2 pi2 Sskip n1 /\ many_steps sigma2 pi2 s2 sigma3 pi3 Sskip n2 /\ n = 1 + n1 + n2 (* lemma one_step_change_free : *) (* forall s s':stmt, sigma sigma':env, pi pi':stack, id:ident, v:value. *) (* fresh_in_stmt id s -> *) (* one_step sigma (Cons (id,v) pi) s sigma' pi' s' -> *) (* one_step sigma pi s sigma' pi' s' *) (** {3 Hoare triples} *) (** partial correctness *) predicate valid_triple (p:fmla) (s:stmt) (q:fmla) = forall sigma:env, pi:stack. eval_fmla sigma pi p -> forall sigma':env, pi':stack, n:int. many_steps sigma pi s sigma' pi' Sskip n -> eval_fmla sigma' pi' q (*** total correctness *) predicate total_valid_triple (p:fmla) (s:stmt) (q:fmla) = forall sigma:env, pi:stack. eval_fmla sigma pi p -> exists sigma':env, pi':stack, n:int. many_steps sigma pi s sigma' pi' Sskip n /\ eval_fmla sigma' pi' q end theory TestSemantics use import ImpExpr function my_sigma : env = IdMap.const (Vint 0) constant x : ident constant y : mident function my_pi : stack = Cons (x, Vint 42) Nil goal Test13 : eval_term my_sigma my_pi (mk_tvalue (Vint 13)) = Vint 13 goal Test42 : eval_term my_sigma my_pi (mk_tvar x) = Vint 42 goal Test0 : eval_term my_sigma my_pi (mk_tderef y) = Vint 0 goal Test55 : eval_term my_sigma my_pi (mk_tbin (mk_tvar x) Oplus (mk_tvalue (Vint 13))) = Vint 55 goal Ass42 : forall sigma':env, pi':stack. one_step my_sigma my_pi (Sassign y (mk_tvalue (Vint 42))) sigma' pi' Sskip -> IdMap.get sigma' y = Vint 42 goal If42 : forall sigma1 sigma2:env, pi1 pi2:stack, s:stmt. one_step my_sigma my_pi (Sif (mk_tbin (mk_tderef y) Ole (mk_tvalue (Vint 10))) (Sassign y (mk_tvalue (Vint 13))) (Sassign y (mk_tvalue (Vint 42)))) sigma1 pi1 s -> one_step sigma1 pi1 s sigma2 pi2 Sskip -> IdMap.get sigma2 y = Vint 13 end (** {2 Hoare logic} *) theory HoareLogic use import ImpExpr (** Hoare logic rules (partial correctness) *) lemma consequence_rule: forall p p' q q':fmla, s:stmt. valid_fmla (Fimplies p' p) -> valid_triple p s q -> valid_fmla (Fimplies q q') -> valid_triple p' s q' lemma skip_rule: forall q:fmla. valid_triple q Sskip q lemma assign_rule: forall p:fmla, x:mident, id:ident, t:term. fresh_in_fmla id p -> valid_triple (Flet id t (msubst p x id)) (Sassign x t) p lemma seq_rule: forall p q r:fmla, s1 s2:stmt. valid_triple p s1 r /\ valid_triple r s2 q -> valid_triple p (Sseq s1 s2) q lemma if_rule: forall t:term, p q:fmla, s1 s2:stmt. valid_triple (Fand p (Fterm t)) s1 q /\ valid_triple (Fand p (Fnot (Fterm t))) s2 q -> valid_triple p (Sif t s1 s2) q lemma assert_rule: forall f p:fmla. valid_fmla (Fimplies p f) -> valid_triple p (Sassert f) p lemma assert_rule_ext: forall f p:fmla. valid_triple (Fimplies f p) (Sassert f) p (* lemma while_rule: forall e:term, inv:fmla, i:expr. valid_triple (Fand (Fterm e) inv) i inv -> valid_triple inv (Swhile e inv i) (Fand (Fnot (Fterm e)) inv) lemma while_rule_ext: forall e:term, inv inv':fmla, i:expr. valid_fmla (Fimplies inv' inv) -> valid_triple (Fand (Fterm e) inv') i inv' -> valid_triple inv' (Swhile e inv i) (Fand (Fnot (Fterm e)) inv') *) (*** frame rule ? *) end (** {2 WP calculus} *) theory WP use import ImpExpr use import bool.Bool use set.Set (** [assigns sigma W sigma'] is true when the only differences between [sigma] and [sigma'] are the value of references in [W] *) predicate assigns (sigma:env) (a:Set.set mident) (sigma':env) = forall i:mident. not (Set.mem i a) -> IdMap.get sigma i = IdMap.get sigma' i lemma assigns_refl: forall sigma:env, a:Set.set mident. assigns sigma a sigma lemma assigns_trans: forall sigma1 sigma2 sigma3:env, a:Set.set mident. assigns sigma1 a sigma2 /\ assigns sigma2 a sigma3 -> assigns sigma1 a sigma3 lemma assigns_union_left: forall sigma sigma':env, s1 s2:Set.set mident. assigns sigma s1 sigma' -> assigns sigma (Set.union s1 s2) sigma' lemma assigns_union_right: forall sigma sigma':env, s1 s2:Set.set mident. assigns sigma s2 sigma' -> assigns sigma (Set.union s1 s2) sigma' (** [expr_writes e W] is true when the only references modified by [e] are in [W] *) predicate stmt_writes (s:stmt) (w:Set.set mident) = match s with | Sskip | Sassert _ -> true | Sassign id _ -> Set.mem id w | Sseq s1 s2 -> stmt_writes s1 w /\ stmt_writes s2 w | Sif t s1 s2 -> stmt_writes s1 w /\ stmt_writes s2 w | Swhile _ _ body -> stmt_writes body w end function fresh_from (f:fmla) (s:stmt) : ident (* Need it for monotonicity*) axiom fresh_from_fmla: forall s:stmt, f:fmla. fresh_in_fmla (fresh_from f s) f axiom fresh_from_stmt: forall s:stmt, f:fmla. fresh_in_stmt (fresh_from f s) s function abstract_effects (s:stmt) (f:fmla) : fmla axiom abstract_effects_generalize : forall sigma:env, pi:stack, s:stmt, f:fmla. eval_fmla sigma pi (abstract_effects s f) -> eval_fmla sigma pi f axiom abstract_effects_monotonic : forall s:stmt, f:fmla. forall sigma:env, pi:stack. eval_fmla sigma pi f -> forall sigma:env, pi:stack. eval_fmla sigma pi (abstract_effects s f) function wp (s:stmt) (q:fmla) : fmla = match s with | Sskip -> q | Sassert f -> (* asymmetric and *) Fand f (Fimplies f q) | Sseq s1 s2 -> wp s1 (wp s2 q) | Sassign x t -> let id = fresh_from q s in Flet id t (msubst q x id) | Sif t s1 s2 -> Fand (Fimplies (Fterm t) (wp s1 q)) (Fimplies (Fnot (Fterm t)) (wp s2 q)) | Swhile cond inv body -> Fand inv (abstract_effects body (Fand (Fimplies (Fand (Fterm cond) inv) (wp body inv)) (Fimplies (Fand (Fnot (Fterm cond)) inv) q))) end axiom abstract_effects_writes : forall sigma:env, pi:stack, s:stmt, q:fmla. eval_fmla sigma pi (abstract_effects s q) -> eval_fmla sigma pi (wp s (abstract_effects s q)) (* lemma wp_subst: *) (* forall e:expr, q:fmla, id :mident, id':ident. *) (* fresh_in_stmt id e -> *) (* subst (wp e q) id id' = wp e (subst q id id') *) lemma monotonicity: forall s:stmt, p q:fmla. valid_fmla (Fimplies p q) -> valid_fmla (Fimplies (wp s p) (wp s q) ) lemma distrib_conj: forall s:stmt, sigma:env, pi:stack, p q:fmla. (eval_fmla sigma pi (wp s p)) /\ (eval_fmla sigma pi (wp s q)) -> eval_fmla sigma pi (wp s (Fand p q)) lemma wp_reduction: forall sigma sigma':env, pi pi':stack, s s':stmt. one_step sigma pi s sigma' pi' s' -> forall q:fmla. eval_fmla sigma pi (wp s q) -> eval_fmla sigma' pi' (wp s' q) lemma progress: forall s:stmt, sigma:env, pi:stack, sigmat: type_env, pit: type_stack, q:fmla. type_stmt sigmat pit s -> (* useful ? type_fmla sigmat pit q -> *) eval_fmla sigma pi (wp s q) -> s <> Sskip -> exists sigma':env, pi':stack, s':stmt. one_step sigma pi s sigma' pi' s' end (*** Local Variables: compile-command: "why3ide blocking_semantics3.mlw" End: *) | 7,433 | 22,072 | {"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-2020-40 | latest | en | 0.490184 |
https://wizardofvegas.com/forum/gambling/craps/24843-charting-a-table/32/ | 1,632,761,766,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058456.86/warc/CC-MAIN-20210927151238-20210927181238-00692.warc.gz | 631,636,189 | 10,599 | TwoFeathersATL
Joined: May 22, 2013
• Posts: 3616
March 22nd, 2016 at 11:40:02 AM permalink
Quote: dicesitter
Axelwolf
Because of that comment you will get nothing.... period
dicesetter
Just share with the rest of us, we will not tell Axel, promise ;-) I might sell the info to him later, but I promise you a cut of the proceeds (after expenses of course ;-)
Youuuuuu MIGHT be a 'rascal' if.......(nevermind ;-)...2F
MathExtremist
Joined: Aug 31, 2010
• Posts: 6526
March 22nd, 2016 at 12:53:53 PM permalink
Quote: dicesitter
Axelwolf
Because of that comment you will get nothing.... period
dicesetter
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
DeMango
Joined: Feb 2, 2010
• Posts: 2769
March 22nd, 2016 at 1:45:24 PM permalink
Quote: MathExtremist
I just dropped two casino dice onto my carpet six times. I got two sixes, two fives, a twelve, and a four. Look how skillful I am -- my SRR is zero. :)
Wanna try that math again oh extreme one? Six divided by zero is.....
When a rock is thrown into a pack of dogs, the one that yells the loudest is the one who got hit.
MathExtremist
Joined: Aug 31, 2010
• Posts: 6526
March 22nd, 2016 at 4:21:22 PM permalink
Quote: DeMango
Wanna try that math again oh extreme one? Six divided by zero is.....
...the ratio of rolls to sevens. The sevens-to-rolls ratio (SRR), on the other hand, is zero.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
MrV
Joined: Feb 13, 2010
• Posts: 7159
March 22nd, 2016 at 5:56:52 PM permalink
Funny, but I usually win at slots when I take the trouble to push the spin button exactly dead center, using 4 pounds of finger pressure, no more, no less.
I have practiced this for eight years at home, and am now finally in position to make the casinos my bitch.
Ah, secret skills!
"What, me worry?"
DeMango
Joined: Feb 2, 2010
• Posts: 2769
March 23rd, 2016 at 4:02:06 AM permalink
Quote: MathExtremist
...the ratio of rolls to sevens. The sevens-to-rolls ratio (SRR), on the other hand, is zero.
No because you haven't rolled a seven, it is infinite.
When a rock is thrown into a pack of dogs, the one that yells the loudest is the one who got hit.
RS
Joined: Feb 11, 2014
• Posts: 8623
March 23rd, 2016 at 5:16:17 AM permalink
Quote: DeMango
No because you haven't rolled a seven, it is infinite.
Sevens-to-rolls ratio
Sevens:rolls
0:6
0/6
0
DeMango
Joined: Feb 2, 2010
• Posts: 2769
March 23rd, 2016 at 8:27:55 AM permalink
not 0 divided by 6, it is 6 divided by 0. Amazing how dice math escapes the extreme ones. 1 is the lowest possible answer, if on 6 rolls, 6 sevens occurred.
When a rock is thrown into a pack of dogs, the one that yells the loudest is the one who got hit.
MathExtremist
Joined: Aug 31, 2010
• Posts: 6526
March 23rd, 2016 at 8:40:21 AM permalink
Quote: DeMango
No because you haven't rolled a seven, it is infinite.
RS already showed the calculation, but there are two other points to make:
a) x/0 is not infinite, it is undefined for any real number x.
b) Your lack of precision in both language and mathematics is hindering your ability to think clearly about this topic. You seem hell-bent on trying to make me look bad, but your latest attempt involves wrongly inverting a fraction in a futile effort to argue against a point I didn't make. When I say the ratio of zero to six is zero, that really shouldn't be a point of contention.
Before this, you were trying to redefine the commonsense notion of "keeping the dice on axis throughout the roll" to include any scenario where the dice tumbled off-axis but showed a desired result anyway. That this redefinition is incorrect also shouldn't be a point of contention.
If this lack of precision is prevalent among your dice-influencing compatriots, it is no wonder you can't quantify your theoretical edge.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
DeMango
Joined: Feb 2, 2010
• Posts: 2769
March 23rd, 2016 at 9:21:56 AM permalink
Well you can call it what you will but the lowest possible SRR is 1. That is, every roll you make is a seven. It cannot be 0. Your example of six rolls without a seven may be undefined by your definition or infinity by mine. Your lack of understanding of dice math is improving but still not there.
When a rock is thrown into a pack of dogs, the one that yells the loudest is the one who got hit. | 1,351 | 4,684 | {"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-2021-39 | latest | en | 0.921954 |
http://dailybrainteaser.blogspot.com/2013/10/tricky-picture-count-matches-puzzle.html | 1,498,418,030,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320570.72/warc/CC-MAIN-20170625184914-20170625204914-00374.warc.gz | 97,284,004 | 22,963 | ## Search This Blog
### Tricky Picture Count Matches Puzzle
Tricky Picture Count Matches Puzzle - 15 October
Can you count the number of matches in the picture below ?
For Solution : Click Here
#### 16 comments:
1. 8+1(upper right corner of the match box also has a image of a portion of match stick)
1. Indeed! 9 is the only right answer
2. I understand the 8, but in top right of the lighter (cover) there is another reflection of a small stem of another match, so the answer should be 9.
3. As of me I hope there are 5.. what if there is a mirror in-front of the lighter as well.
4. 9, cause if u look closely at the top right of the metal box thingy, there's the reflection of a match
5. what about the tip of the ninth one reflected in the top half of the lighter | 196 | 779 | {"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-2017-26 | latest | en | 0.892909 |
https://physics.stackexchange.com/questions/17524/path-to-obtain-the-shortest-traveling-time | 1,719,065,373,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862396.85/warc/CC-MAIN-20240622112740-20240622142740-00763.warc.gz | 401,114,541 | 43,212 | # Path to obtain the shortest traveling time
Asume we have a particle sitting at the point A(0,0) in a gravitational field. (g=9.81) It is going to move along some path to the point B(a,b) Where a>0 and b<0. What is the curve the particle needs to be moving at to arrive at B in the least time possible?
(Fastest curve between two points. )
• This is called Brachistochrone Problem and can be solved using Calculus of Variations. Commented Nov 27, 2011 at 20:43
• Commented Nov 27, 2011 at 21:06
• Or see here : physics.stackexchange.com/q/16819 Commented Nov 27, 2011 at 21:17
As a few people pointed out in the comments, this is the famous Brachistochrone problem. It is a candidate for using a variational principle, and that's one way to solve it. However, it was historically solved long before Euler or Lagrange.
One way to do it is to make an analogy to light. We know that light follows the fastest possible path from point to point. So we imagine a situation where the index of refraction is very high right at the top of the ball's trajectory (since its speed is slow there) and decreases as we move down. Note that by conservation of energy, the speed of the ball is a function of height alone, so we can make an analogy to light by having an index of refraction that is a function of height alone.
Next, remember Snell's Law, $n_1\sin\theta_1 = n_2\sin\theta_2$. This is for two discrete layers of material, but if we had three, four, or more layers the result would be the same, $n\sin\theta$ would be a constant. A smoothly-varying index of refraction is just the continuous case of the above. In our problem $n(h)\sin(\theta(h))$ is a constant.
Now we have a constant of motion. All that remains is to use a little freshman physics to find $n(h)$, then a bit of algebra to derive a differential equation for the path. That differential equation turns out to be that of a cycloid.
Check out the Wikipedia article for details; there is no need for me to copy them over here in detail.
http://en.wikipedia.org/wiki/Brachistochrone_curve#Johann_Bernoulli.27s_solution
According to the same article, the problem was solved by a number of top minds at the end of the 17th century, including by Newton. I'm not sure what their methods were, although Newton's was undoubtedly ingenious. Earlier, Galileo also took a crack at it, conjecturing the solution to be a circle, but he had to acknowledge that he didn't have the mathematical wherewithal yet to nail down the answer for sure.
This is called Brachistochrone problem and it is one of the most standard problem using calculus of variational. The solution can be found in standard textbook, but there are some subtle points that worth mention here (see the end). So I will only give the outline of solution.
The first step is to use conservation of energy as $0 = mv^2/2-mgy$, so $$v=\sqrt{2gy}$$ Also, the displacement is given by $ds=\sqrt{dx^2+dy^2}$, so the total time is $$t=\int_A^B{dt}=\int_A^B{\frac{ds}{v}}=\int_A^B{\sqrt\frac{dx^2+dy^2}{2gy}}=\int_A^B{\sqrt\frac{1+(\frac{dy}{dx})^2}{2gy}}dx$$
We can then use the Euler-Lagrange equation directly.
Method 1: This is the approach usually taken (example). Let $$f(y,y',x)=\sqrt\frac{1+(y')^2}{2gy}$$
where $y' = \frac{dy}{dx}$ and treat $x$ as independent variable and use the Euler-Lagrange equation in the form $$\frac{d}{dx}\frac{\partial f}{\partial y'}-\frac{\partial f}{\partial y}=0$$
Note that $f$ is independent of $x$. The resulting differential equation is $$2yy''+(y')^2+1=0$$ It is a non-linear differential equation and there is no easy way to solve, but it can be checked that the solution are indeed $$\begin{cases} x=a(\theta-\sin\theta)\\ y=a(1-\cos\theta) \end{cases}$$
Method 2: The previous method is a bit tricky. We can solve it because we already know the answer. A better way to solve this problem is to take $y$ as the independent variable and write the function $f$ as $$h(x,x',y)=\sqrt\frac{1+(x')^2}{2gy}$$
where $x'=\frac{dx}{dy}$ and so the total time is $t=\int_A^B h(x,x',y)dy$. The Euler-Lagrange equation is now: $$\frac{d}{dy}\frac{\partial h}{\partial x'}-\frac{\partial h}{\partial x}=0$$
Note that $h$ is now independent of $x$, so $\frac{\partial h}{\partial x}=0$ and $\frac{\partial h}{\partial x'}=const$. If we let this constant to be $\frac{1}{\sqrt{2a}}$ and rearrange the equation, we will have the explicit form of $x'$ as $$x'=\frac{dx}{dy}=\sqrt{\frac{y}{2a-y}}$$
It can be integrated explicitly and we can get a solution of the form $x(y)$ which is certainly the same as the parametric equation in Method 1.
So why the second method leads to direct explicit solution but not the first one? The reason is that we choose the functional form that are independent of the dependent parameter $x(y)$. Hence there is no need to take the extra total derivative $d/dy$ in the Euler-Lagrange equation which ultimately leads to the nonlinearity in the resulting differential equation.
It is rarely mentioned in textbook probably because the $dx/dy$ will create confusion for the beginner, as it is different from their derivation of the Euler-Lagrange equation. Anyway, the examples here does show that one choice of independent parameter is better than the other choice.
• I do not think that there is this asymmetry between method 1 and 2 as you claim. For method 1, one should just note that the Lagrangian does not depend directly on $x$ and thus the "energy" $H= y' \partial_y L - L$ is conserved. Commented Dec 15, 2012 at 10:22
• In the line of @Fabian's comment, see also the discussion of Beltrami's identity on the Wolfram mathworld page. Commented Dec 15, 2012 at 11:41
• @Fabian It is not the asymmetry in the problem. It is just that one method is better than the other. Also the conjugate momentum in this case should be $p_y=\partial_{y'}L$ not $\partial_{y}L$ Commented Dec 15, 2012 at 15:27
• Although late in the party (in Method 1) $$2yy''+(y')^2+1=0 \quad \Longrightarrow \quad 2yy'y''+(y')^3+y'=0 \quad \Longrightarrow \quad \biggl[y\left[1+(y')^2\right]\biggr]'=0 \quad \Longrightarrow \quad y\left[1+(y')^2\right]=k =\text{positive contant}=\text{yours } 2a$$ Commented Oct 11, 2016 at 17:49 | 1,732 | 6,196 | {"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} | 4.0625 | 4 | CC-MAIN-2024-26 | latest | en | 0.946885 |
https://www.fmaths.com/interesting-about-math/what-is-the-meaning-of-fraction-in-math.html | 1,632,387,096,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057417.92/warc/CC-MAIN-20210923074537-20210923104537-00326.warc.gz | 793,230,472 | 10,856 | # What Is The Meaning Of Fraction In Math?
## Whats is a fraction?
Decimal to fraction conversion table
Decimal Fraction
0.00001 1/100000
0.0001 1/10000
0.001 1/1000
0.01 1/100
## What are the 3 types of fractions?
In Maths, there are three major types of fractions. They are proper fractions, improper fractions and mixed fractions. Fractions are those terms which have numerator and denominator.
## What are fractional numbers with examples?
Example: 3/4 means:
Proper Fractions: The numerator is less than the denominator
Improper Fractions: The numerator is greater than (or equal to) the denominator
Examples: 4/3, 11/4, 7/7
Mixed Fractions: A whole number and proper fraction together
Examples: 1 1/3, 2 1/4, 16 2/5
## What is 7% as a fraction?
getcalc.com’s decimal to fraction calculator to find what’s an equivalent fraction for the decimal point number 0.07 or 7 %. How to Write 0.07 or 7 % as a Fraction?
Decimal Fraction Percentage
0.07 7 /100 7 %
0.06 6/100 6%
0.07216 7 /97 7.216%
0.07143 7 /98 7.143%
## What is 3/4 as a decimal?
Fraction to Decimal Conversion Tables
fraction = decimal
1/2 = 0.5
1/3 = 0.3 2/3 = 0.6
1/4 = 0.25 3/4 = 0.75
1/5 = 0.2 2/5 = 0.4 3/5 = 0.6
## What is a vulgar fraction example?
If a numerator is less than a denominator, then the fraction is less than 1 and called a proper fraction. It will be a numerator of a vulgar fraction, and its denominator is saved the same. Reciprocal fractions are two fractions whose product is 1. For example, 3 / 7 and 7 / 3; 15 / 1 and 1 / 15 and so on.
You might be interested: What Is Meaning Of Plane In Math?
## What is vulgar fraction explain with example?
• a fraction where the numerator and the denominator. are both integers (not fractions ). EXAMPLES: © Jenny Eather 2014.
## What is called fractional number?
If we want a definition, we can start by saying that fractional numbers are numbers that represent one or more parts of a unit that has been divided in equal parts. The fractional numbers are figured out by two whole numbers (the fraction terms) that are separated by a horizontal line (the fraction line).
## What is a bank fractional number?
A fractional is used to determine the financial institution of the share draft if the MICR line is ever damaged or torn off: Digits 5-8 of the routing number and transit number indicating the institution identifier (in the example, this is 5678 – see blue below.)
## Why is not fractional number?
Therefore, a and b are integers. Thus, the fraction a/b is the quotient of two integers such that b ≠ 0. Hence, a/b is a rational number. We know that 2/-3 is a rational number but it is not a fraction because its denominator is not a natural number.
## What is 9 in a fraction?
The simplest way to write 9 as a fraction is 9 /1.
## What is the simplest fraction equal to 7?
“Percent” or “%” means “out of 100” or “per 100”, Therefore 7 % can be written as 7 100. Because 7 is a prime number and cannot be factored this is the simplest form.
## What is 3/8 as a decimal?
Answer: As a decimal 3/8 is 0.375. | 858 | 3,074 | {"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} | 3.96875 | 4 | CC-MAIN-2021-39 | latest | en | 0.874094 |
https://www.tensorflow.org/versions/r1.15/api_docs/python/tf/keras/backend/batch_dot?hl=it | 1,600,626,407,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400198287.23/warc/CC-MAIN-20200920161009-20200920191009-00533.warc.gz | 1,093,382,738 | 56,737 | # tf.keras.backend.batch_dot
Batchwise dot product.
`batch_dot` is used to compute dot product of `x` and `y` when `x` and `y` are data in batch, i.e. in a shape of `(batch_size, :)`. `batch_dot` results in a tensor or variable with less dimensions than the input. If the number of dimensions is reduced to 1, we use `expand_dims` to make sure that ndim is at least 2.
`x` Keras tensor or variable with `ndim >= 2`.
`y` Keras tensor or variable with `ndim >= 2`.
`axes` list of (or single) int with target dimensions. The lengths of `axes[0]` and `axes[1]` should be the same.
A tensor with shape equal to the concatenation of `x`'s shape (less the dimension that was summed over) and `y`'s shape (less the batch dimension and the dimension that was summed over). If the final rank is 1, we reshape it to `(batch_size, 1)`.
#### Examples:
Assume `x = [[1, 2], [3, 4]]` and `y = [[5, 6], [7, 8]]` `batch_dot(x, y, axes=1) = [[17, 53]]` which is the main diagonal of `x.dot(y.T)`, although we never have to calculate the off-diagonal elements.
Shape inference: Let `x`'s shape be `(100, 20)` and `y`'s shape be `(100, 30, 20)`. If `axes` is (1, 2), to find the output shape of resultant tensor, loop through each dimension in `x`'s shape and `y`'s shape:
• `x.shape[0]` : 100 : append to output shape
• `x.shape[1]` : 20 : do not append to output shape, dimension 1 of `x` has been summed over. (`dot_axes[0]` = 1)
• `y.shape[0]` : 100 : do not append to output shape, always ignore first dimension of `y`
• `y.shape[1]` : 30 : append to output shape
• `y.shape[2]` : 20 : do not append to output shape, dimension 2 of `y` has been summed over. (`dot_axes[1]` = 2) `output_shape` = `(100, 30)`
````x_batch = K.ones(shape=(32, 20, 1))`
`y_batch = K.ones(shape=(32, 30, 20))`
`xy_batch_dot = K.batch_dot(x_batch, y_batch, axes=[1, 2])`
`K.int_shape(xy_batch_dot)`
`(32, 1, 30)`
``` | 622 | 1,885 | {"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-2020-40 | latest | en | 0.824866 |
https://lessonplanet.com/search?concept_ids%5B%5D=111509 | 1,623,988,147,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487634616.65/warc/CC-MAIN-20210618013013-20210618043013-00194.warc.gz | 343,613,859 | 31,002 | ### We found 31 resources with the concept compound events
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#### Applying Probability to Make Informed Decisions
For Teachers 7th Standards
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#### Calculating Probabilities of Compound Events
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#### Compound Events: Problem Solving
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In this compound events instructional activity, 7th graders use the table and the compound events to find the probabilities for each of the four word problems. | 1,632 | 7,990 | {"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} | 3.40625 | 3 | CC-MAIN-2021-25 | latest | en | 0.862734 |
http://stackoverflow.com/questions/4600034/calculate-number-of-weekdays-between-two-dates-in-java/8003104 | 1,440,883,914,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644064538.31/warc/CC-MAIN-20150827025424-00033-ip-10-171-96-226.ec2.internal.warc.gz | 220,605,811 | 21,557 | # Calculate number of weekdays between two dates in Java
Can anyone point me to some Java snippet wherein i can get business (except Sat and Sun) days between two dates.
-
joda-time.sourceforge.net – marcog Jan 5 '11 at 1:24
``````public static int getWorkingDaysBetweenTwoDates(Date startDate, Date endDate) {
Calendar startCal = Calendar.getInstance();
startCal.setTime(startDate);
Calendar endCal = Calendar.getInstance();
endCal.setTime(endDate);
int workDays = 0;
//Return 0 if start and end are the same
if (startCal.getTimeInMillis() == endCal.getTimeInMillis()) {
return 0;
}
if (startCal.getTimeInMillis() > endCal.getTimeInMillis()) {
startCal.setTime(endDate);
endCal.setTime(startDate);
}
do {
//excluding start date
if (startCal.get(Calendar.DAY_OF_WEEK) != Calendar.SATURDAY && startCal.get(Calendar.DAY_OF_WEEK) != Calendar.SUNDAY) {
++workDays;
}
} while (startCal.getTimeInMillis() < endCal.getTimeInMillis()); //excluding end date
return workDays;
}
``````
Start date and end date are exclusive, Only the days between given dates will be counted. Start date and end date will not be included.
-
Got some strange answers if startDate is on or after endDate. If startDate is 1970-01-01 21:27:00 and endDate is 1970-01-01 18:00:00, the function returns 1 instead of 0 days between the same date. Also if startDate is 2011-01-05 21:27:00 and endDate is 2011-01-04 00:00:00 the function returns 2 instead of -1 or 1 (if absolute values are expected). – Joseph Gordon Jan 5 '11 at 22:43
The above code snippet is meant for calculating working days between two dates without time. – Piyush Mattoo Jan 6 '11 at 4:26
It works, but execution time depends on the number of days :-/ – ymajoros Feb 15 '12 at 8:17
Solution without loop:
``````static long days(Date start, Date end){
//Ignore argument check
Calendar c1 = Calendar.getInstance();
c1.setTime(start);
int w1 = c1.get(Calendar.DAY_OF_WEEK);
Calendar c2 = Calendar.getInstance();
c2.setTime(end);
int w2 = c2.get(Calendar.DAY_OF_WEEK);
//end Saturday to start Saturday
long days = (c2.getTimeInMillis()-c1.getTimeInMillis())/(1000*60*60*24);
long daysWithoutWeekendDays = days-(days*2/7);
// Adjust w1 or w2 to 0 since we only want a count of *weekdays*
// to add onto our daysWithoutWeekendDays
if (w1 == Calendar.SUNDAY) {
w1 = Calendar.MONDAY;
}
if (w2 == Calendar.SUNDAY) {
w2 = Calendar.MONDAY;
}
return daysWithoutWeekendDays-w1+w2;
}
``````
-
Really impressed with your solution. I was trying to accomplish the same thing but wound up with switches to catch the beginning and end dates on the weekend instead of doing the date shifting. This is much more elegant. – Joseph Gordon Jan 5 '11 at 5:31
@Shengyuan : I know it's a bit late now, but could you please explain the last 2 lines: `long days = (c2.getTimeInMillis()-c1.getTimeInMillis())/(1000*60*60*24);` `long daysWithoutSunday = days-(days*2/7);` – Vrushank May 27 '12 at 6:05
I used Shengyuan Lu's solution, but I needed to make a fix for the case where the method is called when one of the dates is on a Saturday and the other a Sunday - otherwise the answer is off by a day:
``````static long days(Date start, Date end){
//Ignore argument check
Calendar c1 = GregorianCalendar.getInstance();
c1.setTime(start);
int w1 = c1.get(Calendar.DAY_OF_WEEK);
Calendar c2 = GregorianCalendar.getInstance();
c2.setTime(end);
int w2 = c2.get(Calendar.DAY_OF_WEEK);
//end Saturday to start Saturday
long days = (c2.getTimeInMillis()-c1.getTimeInMillis())/(1000*60*60*24);
long daysWithoutSunday = days-(days*2/7);
if (w1 == Calendar.SUNDAY) {
w1 = Calendar.MONDAY;
}
if (w2 == Calendar.SUNDAY) {
w2 = Calendar.MONDAY;
}
return daysWithoutSunday-w1+w2;
}
``````
-
why don't you just edit Shengyuan Lu's solution? – ymajoros Feb 15 '12 at 8:19
I don't have a Java based solution, but have a PHP one, hope it helps:
``````function getDate(\$days) {
for (\$i = 0; \$i < \$days; \$i ++) {
if (date('N' , strtotime('+' . (\$i + 1) . ' days')) > 5) {
\$days++;
}
}
return date('l, F jS', strtotime('+' . \$days . ' days', time()));
}
``````
-
Sure, it does help. – John C Jan 5 '11 at 1:21
The `do` `while` in the solution of Piyush is wrong, it should be :
``````do {
if (startCal.get(Calendar.DAY_OF_WEEK) != Calendar.SATURDAY && startCal.get(Calendar.DAY_OF_WEEK) != Calendar.SUNDAY) {
++workDays;
}
} while (startCal.getTimeInMillis() < endCal.getTimeInMillis());
``````
-
The `startCal.add` should add onto the `Calendar.DATE` field, not the `Calendar.DAY_OF_MONTH`, I was getting weird results with over Decemeber / January period.
-
I don't think that's true, Calendar.DATE is a synonym for Calendar.DAY_OF_MONTH – mbosecke Apr 2 at 18:41
This is my example without looping. It is a class in this example because I serialize it in some JSON output. Basically I work out the number of days between the two dates, divide by 7 and assign to a long to have a integer value for the number of weeks. Take the original number of days and subtract the number of weekends*2. This isn't quite perfect - you need to work out if there is a 'hangover' where the start is close to the end of the week and goes over the weekend. To correct for this I find the day of the week at the start and find the remainder of the number of days, and add those together to find the 'hangover' - and if it is more than 5 it is a weekend. It isn't quite perfect, and does not account for holidays at all. And no Joda in sight. That said there is also a issue with timezones.
``````import java.io.Serializable;
import java.util.Date;
public class BusinessDayCalculator implements Serializable {
private static long DAY = 86400000l;
private Date startTime;
private Date endTime;
public void setStartTime(Date startTime) {
this.startTime = startTime;
}
public Date getStartTime() {
return startTime;
}
public void setEndTime(Date endTime) {
this.endTime = endTime;
}
public Date getEndTime() {
return endTime;
}
public long getHours() {
return (this.endTime.getTime() - this.startTime.getTime())/(1000*60*60);
}
long startDay = getDayFromDate(this.startTime);
long endDay = getDayFromDate(this.endTime);
long totalDays = endDay-startDay;
long totalWeekends = totalDays/7;
long day = getDay(this.startTime);
long hangover = totalDays % 7;
long intoWeekend = day + hangover;
if(intoWeekend>5){
totalWeekends++;
}
long totalBusinessDays = totalDays - (totalWeekends *2);
/*
System.out.println("Days = " + day );
System.out.println("Hangover = " + hangover );
System.out.println("Total Days = " + totalDays);
System.out.println("Total Weekends = " + totalWeekends);
*/
}
private long getDayFromDate( Date date ){
long d = date.getTime() / DAY;
return d;
}
private long getDay( Date date ){
long daysSinceEpoc = getDayFromDate(date);
long day = daysSinceEpoc % 7;
day = day + 4;
if(day>6) day = day - 7;
return day;
}
}
``````
-
This code is very error prone. You don't take leap seconds, leap years, etc into consideration. I would suggest you remove this answer, there are good libraries out there for Java (i.e. JodaTime) that get this right. Nice try though :-) – Daniel L. Aug 6 at 23:19 | 1,913 | 7,139 | {"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-2015-35 | latest | en | 0.552481 |
http://www.authorstream.com/Presentation/joberg33-2890918-forces-motion/ | 1,563,568,736,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526359.16/warc/CC-MAIN-20190719202605-20190719224605-00306.warc.gz | 174,412,353 | 27,158 | # Forces and Motion 4-8
Views:
Category: Education
## Presentation Description
No description available.
## Presentation Transcript
### Forces and Motion:
Forces and Motion Steve Case University of Mississippi NSF North Mississippi GK-8 September 2005
### What words can you think of to describe the motion of an object?:
NSF North Mississippi GK-8 What words can you think of to describe the motion of an object?
### VELOCITY:
NSF North Mississippi GK-8 VELOCITY The word scientists use to describe how fast an object is moving
### Velocity is measured as::
NSF North Mississippi GK-8 Velocity is measured as: change in distance change in time a change in distance per change in time What are some things you can think of that are measured like this?
### What are some examples of objects that move at high velocities?:
NSF North Mississippi GK-8 What are some examples of objects that move at high velocities?
### What are some examples of objects that move at low velocities?:
NSF North Mississippi GK-8 What are some examples of objects that move at low velocities?
### What units would you use to measure the velocity of an object moving in this classroom?:
NSF North Mississippi GK-8 What units would you use to measure the velocity of an object moving in this classroom? meters per second m s Meters per second tells us how many meters something can move in one second.
### What else can we know about an object’s motion besides velocity?:
NSF North Mississippi GK-8 What else can we know about an object’s motion besides velocity? What if the object slows down? What if the object’s velocity changes ? What if the object speeds up?
### Scientists have a special way to measure a change in velocity::
NSF North Mississippi GK-8 Scientists have a special way to measure a change in velocity: Acceleration Acceleration is a change in velocity.
### Slide10:
NSF North Mississippi GK-8 If you’re speeding up, you have positive acceleration. If you’re slowing down, you have negative acceleration. If you’re changing speed quickly, you have high acceleration. If you’re changing speed slowly, you have low acceleration. The Way Acceleration Works:
### Example::
NSF North Mississippi GK-8 Example: You’re driving along in your car and suddenly have to slam on your brakes for a red light. Negative or positive acceleration? High or low acceleration? High negative acceleration because you’re slowing down quickly .
### Slide12:
NSF North Mississippi GK-8 You’re an astronaut waiting in your shuttle for takeoff. Suddenly the rockets fire and you’re on your way to Mars. High or low positive or negative acceleration? High positive acceleration because you’re speeding up quickly . Example:
### Slide13:
NSF North Mississippi GK-8 You’re driving down the highway when your car runs out of gas and you slowly roll to a stop on the side of the road. Low negative acceleration because you’re slowing down slowly. Example:
### Slide14:
NSF North Mississippi GK-8 You’re walking to class when you realize you’re running just a little bit late. You start walking a bit fast, but you still hope to make it on time. Finally the bell rings so you break into a jog and then a run, trying to get to class before the teacher notices you’re late. Low positive acceleration because you’re speeding up slowly . Example:
### Challenge Question::
NSF North Mississippi GK-8 Challenge Question: You throw a baseball straight up in the air. It rises to a certain height and then falls back to earth, where you catch it again in your glove. First, high positive acceleration because it speeds up quickly as it leaves your hand. Then, as it rises, low negative acceleration, because it begins to slow down slowly because of gravity. As it begins to fall again, low positive acceleration, because it speeds up slowly . Finally, when you catch it, high negative acceleration because it slows down very quickly in your hand.
### How Do I Graph the Motion of a Particle?:
NSF North Mississippi GK-8 How Do I Graph the Motion of a Particle?
### Quick Review:
NSF North Mississippi GK-8 Quick Review Velocity : speed of motion Acceleration : change in velocity What causes an object’s velocity to change?
### Slide18:
NSF North Mississippi GK-8 If you’re riding your bike and want to speed up, what do you do? If you’re driving a car and want to slow down, what do you do? If you’re playing soccer and want to move the ball, what do you do?
### Pushing pedals, pressing brakes, and kicking soccer balls are all examples of forces.:
NSF North Mississippi GK-8 Pushing pedals, pressing brakes, and kicking soccer balls are all examples of forces . A force is anything that changes the velocity of an object . But what’s a change in velocity ? A change in velocity is acceleration , so . . . A force is anything that gives an object acceleration .
### Nothing will change speed unless acted upon by a force. If it’s sitting still, it will keep sitting still. If it’s moving fast, it will keep moving fast. :
NSF North Mississippi GK-8 Nothing will change speed unless acted upon by a force. If it’s sitting still, it will keep sitting still. If it’s moving fast, it will keep moving fast. Only a force can change the velocity of an object .
### Then why do things slow down?:
NSF North Mississippi GK-8 Then why do things slow down? Flying baseballs slow down and eventually stop because of the force of gravity . Cars and bikes eventually slow down and stop because of the force of friction . Spacecraft in outer space with no friction and no gravity will keep going at the same speed forever without a force to slow them down or speed them up.
### Final Review:
NSF North Mississippi GK-8 Final Review What is an object’s speed of motion? What is a change of an object’s velocity? What is the only thing that can change an object’s velocity and give it acceleration? | 1,248 | 5,888 | {"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-2019-30 | latest | en | 0.885948 |
http://mathhelpforum.com/calculus/82857-absolute-convergence.html | 1,480,964,253,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541783.81/warc/CC-MAIN-20161202170901-00289-ip-10-31-129-80.ec2.internal.warc.gz | 173,470,072 | 9,633 | 1. ## absolute convergence
does the series from 1 to infinity of ((-1)^(n+1) * n^n) / (n!)
converge absolutely or not?
2. Is...
$\frac{n^{n}}{n!}= n\cdot \frac{n}{2}\cdot \frac{n}{3}\dots \frac{n}{n-2}\cdot \frac{n}{n-1}$
... then what is $\lim_{n \rightarrow \infty} \frac {n^{n}}{n!}$?...
Kind regards
$\chi$ $\sigma$ | 125 | 325 | {"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": 4, "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-2016-50 | longest | en | 0.397018 |
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Salman Shaikh
49 2015 年以降の合計貢献数
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Is my wife right
This is question relies on the dilemma that every married man faces in his life. Which has one simple ans: So whatever the qu...
5年以上 前
Is my wife right
This is question relies on the dilemma that every married man faces in his life. Which has one simple ans: So whatever the qu...
5年以上 前 | 0 | 18 個のソルバー
Back and Forth Rows
Given a number n, create an n-by-n matrix in which the integers from 1 to n^2 wind back and forth along the rows as shown in the...
5年以上 前
Increase monotonically
Return true if the elements of the input vector increase monotonically (i.e. each element is larger than the previous). Return f...
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Determine whether a vector is monotonically increasing
Return true if the elements of the input vector increase monotonically (i.e. each element is larger than the previous). Return f...
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Find the numeric mean of the prime numbers in a matrix.
There will always be at least one prime in the matrix. Example: Input in = [ 8 3 5 9 ] Output out is 4...
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Remove any row in which a NaN appears
Given the matrix A, return B in which all the rows that have one or more <http://www.mathworks.com/help/techdoc/ref/nan.html NaN...
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Determine if a Given Number is a Triangle Number
*Description:* Determine if the elements of an input array are triangle numbers and return the result as an array with the sa...
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Initialize a Natural Number matrix.
Given length of matrix initialize a matrix consisting of natural numbers from 1 to n: n = 10; x = [ 1 2 3 4 5 6 7 8 9 10]; ...
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Finding Perfect Squares
Given a vector of numbers, return true if one of the numbers is a square of one of the other numbers. Otherwise return false. E...
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Fibonacci sequence
Calculate the nth Fibonacci number. Given n, return f where f = fib(n) and f(1) = 1, f(2) = 1, f(3) = 2, ... Examples: Inpu...
5年以上 前
Create times-tables
At one time or another, we all had to memorize boring times tables. 5 times 5 is 25. 5 times 6 is 30. 12 times 12 is way more th...
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Swap the first and last columns
Flip the outermost columns of matrix A, so that the first column becomes the last and the last column becomes the first. All oth...
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Implement simple rotation cypher
If given a letter from the set: [abc...xyz] and a shift, implement a shift cypher. Example: 'abc' with a shi...
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Triangle sequence
A sequence of triangles is constructed in the following way: 1) the first triangle is Pythagoras' 3-4-5 triangle 2) the s...
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Find a Pythagorean triple
Given four different positive numbers, a, b, c and d, provided in increasing order: a < b < c < d, find if any three of them com...
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Side of a rhombus
If a rhombus has diagonals of length x and x+1, then what is the length of its side, y? <<http://upload.wikimedia.org/wikipe...
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Dimensions of a rectangle
The longer side of a rectangle is three times the length of the shorter side. If the length of the diagonal is x, find the width...
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Side of an equilateral triangle
If an equilateral triangle has area A, then what is the length of each of its sides, x? <<http://upload.wikimedia.org/wikipe...
5年以上 前
Is this triangle right-angled?
Given any three positive numbers a, b, c, return true if the triangle with sides a, b and c is right-angled. Otherwise, return f...
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Area of an equilateral triangle
Calculate the area of an equilateral triangle of side x. <<http://upload.wikimedia.org/wikipedia/commons/e/e0/Equilateral-tr...
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Area of an Isoceles Triangle
An isosceles triangle has equal sides of length x and a base of length y. Find the area, A, of the triangle. <<http://upload...
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Back to basics 8 - Matrix Diagonals
Covering some basic topics I haven't seen elsewhere on Cody. Given an input vector of numbers, output a square array with the...
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Square a Number
Given an input x, return y, which is equal to the square of x.
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Make a checkerboard matrix
Given an integer n, make an n-by-n matrix made up of alternating ones and zeros as shown below. The a(1,1) should be 1. Example...
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Remove All elements less than 5
Given an input vector x, remove all elements of x less than 5 . Example: Input x = [ 1 2 5 7 3 ] Output y is [ 5 7 ...
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Length of a short side
Calculate the length of the short side, a, of a right-angled triangle with hypotenuse of length c, and other short side of lengt...
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Length of the hypotenuse
Given short sides of lengths a and b, calculate the length c of the hypotenuse of the right-angled triangle. <<http://upload....
5年以上 前
Is this triangle right-angled?
Given three positive numbers a, b, c, where c is the largest number, return *true* if the triangle with sides a, b and c is righ...
5年以上 前
Find the dimensions of a matrix
Just find the number of columns of the given matrix. Example x = [1 2 3 4 5 6] y = 2
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# Celestial Jukebox
Through the use of rhythm patterns of sounds presented in a solar system model, learners will collect data to determine orbital periods. Then, using that data, they will derive Kepler’s Third Law (the relationship between the distance of planets... (View More)
Audience: High school, Informal education
Materials Cost: Free
# Scale Model of the Solar System
Carl Sagan once claimed that the most important lesson we learn from studying the stars is perspective. To address this concept, this activity offers a scale model of the solar system to be evaluated. There are many versions of solar system scale... (View More)
# Measuring Angular Size and Distance
This is an activity about measuring angular size and understanding the solar and lunar proportions that result in solar eclipses. Learners will use triangles and proportions to create a shoebox eclipse simulator. They will then apply what they learn... (View More)
# Scientific Notation
In this activity students convert antilogs to logs, and logs to antilogs using scientific notation as an intermediate step. They will thereby develop a look-up table for solving math problems by using logarithms. This is activity D2 in the "Far Out... (View More)
# Multiplying Slide Rule
In this activity students construct multiplying slide rules scaled in Base-10 exponents and use them to calculate products and quotients. They will come to appreciate that super numbers (exponents, orders of magnitude and logarithms) play by... (View More)
# String Calculator
In this activity students add and subtract log distances on their Log Tapes to discover that the corresponding numbers multiply and divide. This will lead them to an experiential understanding of the laws of logarithms. This is activity B2 in the... (View More)
# Slide-Rule Graph
In this activity students use their Log Tapes as a reference for ordered pairs, and graph positive numbers as a function of their base-10 logarithms. They extend each plotted point to the vertical axis, thereby generating a logarithmic scale that... (View More)
# Log Ruler
In this activity students construct Log Rulers, finely calibrated in base-10 exponents and numbers (logs and antilogs). They practice reading these scales as accurately as possible, listing all certain figures plus one uncertain figure. This is... (View More)
# Log Tape
In this activity students construct Log Tapes calibrated in base-ten exponents, then use them to derive relationships between base-ten logs (exponents) and antilogs (ordinary numbers). This is activity B1 in the "Far Out Math" educator's guide.... (View More)
# No Strings Calculator
In this activity students develop a simplified log table using information from their Log Tapes. Then they use it to solve arithmetic problems by looking up and combining logs, and finding the antilog. Because these problems are extremely simple,... (View More)
«Previous Page12 Next Page» | 706 | 3,353 | {"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} | 3.125 | 3 | CC-MAIN-2018-34 | latest | en | 0.861885 |
http://fer3.com/arc/m2.aspx/New-ways-create-false-horizons-Dolkas-jul-2017-g39575 | 1,653,208,293,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662545090.44/warc/CC-MAIN-20220522063657-20220522093657-00212.warc.gz | 20,676,979 | 6,655 | # NavList:
## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Message:αβγ
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Add Images & Files Posting Code: Name: Email:
Re: New ways to create false horizons?
From: Paul Dolkas
Date: 2017 Jul 24, 12:55 -0700
Don-
If temp drift is a problem, then one solution is to include a temp sensor and then have the software compensate based on a look up table of gain vs. temp. The digital altimeters they put in watches have this problem, and that’s how they got around it. I always wondered why these watches included a thermometer (why do I need to know how warm my wrist is?) until somebody explained that the sensor was included to improve the accuracy of the altimeter.
Paul Dolkas
From: NavList@fer3.com [mailto:NavList@fer3.com] On Behalf Of Don Seltzer
Sent: Friday, July 21, 2017 8:54 AM
To: paul@dolkas.net
Subject: [NavList] Re: New ways to create false horizons?
On Thu, Jul 20, 2017 at 6:13 PM, Pete Solon Palmer <NoReply_Palmer@fer3.com> wrote:
Maybe there is an electronic AH solution. The sensors (mag, gyro, accel, gravity) in my smart phone seem to be less accurate than a ball bearing and a toy compass, so I don't see a solution involving phones soon.
Ever since I took Frank's course I have been toying around with this idea. Some years ago I developed some automated test stations for measuring performance of multi-axis MEMS sensors such as those used in phones and other devices today.
A two axis MEMS accelerometer is very plausible for determining the gravitational vertical. The three main error sources are bias offset, scale factor error, and temperature drift.
Bias offset is easy to deal with, and is conceptually similar to sextant index error.
Scale factor error is a problem for measuring a wide range of angles. When simply trying to determine a very small range around 'zero' angle it is of minor concern.
The most challenging problem is temperature sensitivity of these devices. Changing temperatures will cause all the other parameters to shift.
Analog Devices is a leading manufacturer of MEMS multi-axis accelerometers, some costing less than \$20. I once did a rough error analysis of some of their devices and it seemed that it might be possible to achieve a resolution of about 1 or 2 arc minutes around zero.
Don Seltzer
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Subject: Author: Start date: (yyyymm dd) End date: (yyyymm dd) | 673 | 2,870 | {"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-2022-21 | latest | en | 0.913824 |
http://mathoverflow.net/questions/70323?sort=oldest | 1,369,033,780,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368698493317/warc/CC-MAIN-20130516100133-00087-ip-10-60-113-184.ec2.internal.warc.gz | 156,420,442 | 10,889 | How to recognize a finite dimensional algebra is Koszul or quadratic?
I have a family of finite dimensional algebras that are directed quasihereditary. I think they might be Koszul algebras and I am wondering what approaches there are to check Koszulness or even quadraticity. I know the quivers of these algebras and can compute Ext^n between simple modules for all n, but I do not have a quiver presentation. I know that there are paths of length 2 and of higher lengths between all vertices of the quiver with nonvanishing Ext^2 so I cannot prove or eliminate quadraticity for trivial reasons. Any thoughts?
I should add that I do not have explicit minimal projective resolutions of the simple modules.
-
I think this is honestly too vague to get useful answers. How are you computing Ext^n of all the simples without finding minimal projective resolutions? Are your algebras graded, or are you looking for a grading that makes things Koszul? – Ben Webster Jul 20 2011 at 18:38
I am looking for a grading. I was hoping to use the grading coming from path length. I compute the Ext^n using classifying spaces. My algebras are monoid algebras. We can show that Ext between simple modules can be computed as the cohomology of certain submonoids with coefficients in some nice module. This in turns out to be the cohomology of a certain finite category with coefficients in the field. This we compute by using Quillen's theorem A to get to a nice simplicial complex. So basically we have no nice resolution. If anything we are implicitly using bar resolutions. – Benjamin Steinberg Jul 21 2011 at 1:34
Can you share a typical example of your algebras? – Mariano Suárez-Alvarez Jul 21 2011 at 20:54
1 Answer
One incredibly useful fact is that it suffices to find linear resolutions of standard modules, not simples. These are usually much easier to find "by hand." Strictly speaking this is stronger than Koszul (the term is "standard Koszul") but in practice it seems rare for a quasi-hereditary algebra to be Koszul and not standard Koszul.
-
Thanks. In my case the quasihereditary structure is directed, so the standard modules are projective indecomposables. I am not a specialist in finite dimensional algebras, but it would seem to me that in this case standard Koszul and Koszul should be the same. Is that correct? – Benjamin Steinberg Jul 21 2011 at 1:38 I looked at the paper of Dlab et al, where standard Koszul is introduced. They say you need a linear resolution of both left and right standard modules. For directed quasihereditary algebras the standard modules on one side are projective and on the other side simple. So if I understood rightly, the standard Koszul is equivalent to Koszul in the directed case. They also point out in the paper it suffices to show the Ext algebra is generated in degrees zero and one. I might be able to determine this. I know how to explicitly construct all extensions of one simple by another. – Benjamin Steinberg Jul 21 2011 at 2:56 | 703 | 2,990 | {"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 | 0.924899 |
https://physics.stackexchange.com/questions/525858/why-the-electric-potential-has-to-drop-to-0-at-the-end-of-an-electric-circuit | 1,718,908,906,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861989.79/warc/CC-MAIN-20240620172726-20240620202726-00783.warc.gz | 401,322,650 | 40,722 | # Why the electric potential has to drop to 0 at the end of an electric circuit
My question is the following:
In every exemplary circuit, the voltage always equals 0 at the end of electrons "road". I'm aware that the wires are not perfect conductors and they have their built-in resistance as well, but
• what makes the conducting material reduce the voltage to 0?
• What force/phenomenon/mechanism is responsible for the thorough voltage drop?
In other words,
• WHY does the potential drop on each barrier/component has to sum up to the total electric potential between two battery terminals?
I understand that the current is evenly distributed across the whole circuit and it's equal to Voltage divided by the net resistance of the whole circuit, but
• where does it really come from?
It always seemed a bit mysterious to me, like a self-regulating organism.
The question is related to Kirchoff's Law.
The laws are called conservation laws as they are based on conserved quantities. First is that amount of charge in a circuit is conserved. The other one says, that the total energy must be conserved.
Voltage drop to zero is a consequence of the second example, known as voltage law.
To elaborate on your question, let's start from the basic principles.
The force between two charges is given as: $$\vec F=k_e\frac{Qq}{r^2}\hat r$$ From this equation we can find electric field, which roughly measures how strong is E-field at any point. $$\vec E=\frac{\vec F}{q}=k_e\frac{Q}{r^2}\hat r$$ Voltage is defined as potential difference between two points. In simple cases it can be expressed as function that is depended only on an end point and initial point. (if E-field is conservative).
Now all this information will (hopefully!) give you an answer you are looking for.
Since voltage can be expressed as difference in potential between initial point and end point, what would be the voltage if the initial point is the same as final point for simple case? (I.e a closed loop).
$$\Delta V=V(r_f)-V(r_i)\\V(r_f)=V(r_i)$$
The voltage equals 0 because you are comparing it to itself.
Just like the height of a person is always measured from the bottom of her feet to the top of her head, voltage is also measured between two points.
Think of the height of a person. If you sum up the height of the head, neck, upper body, lower body - you'll get the exact same as the whole length of this person, no matter how imperfect he or she is.
It's the same with voltage. When you measure the voltage of a battery, you measure it between the minus terminal and the plus terminal. Looking at the voltage at the "end" is like measuring the voltage with both measuring probes on the minus.
It has to be 0 because you are comparing it to itself.
Voltage is a measure of the potential energy of an electric field, and in the case of a circuit the electric field is generated by the battery. As a charge moves from one point of space to another in the presence of an electric field, its potential energy will change. There is no electric field inside an ideal conductor, so the potential remains constant. The electric field from the battery is not null inside a resistor, so, an electron moving through it experiences a voltage drop. The total voltage drop across several resistors needs to be equal to the voltage difference of the battery ends, a result of the conservative nature of the electric field, that is, regardless of which path you take between the two ends of the battery, either on air or inside a circuit, the total difference on potential obtained by integrating the electric field must be the same. | 786 | 3,623 | {"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": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2024-26 | latest | en | 0.957558 |
https://byjus.com/question-answer/solve-the-following-equations-which-have-equal-roots-x-5-13x-4-67x-3-171x/ | 1,718,551,091,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861665.97/warc/CC-MAIN-20240616141113-20240616171113-00764.warc.gz | 135,430,621 | 26,600 | 1
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# Solve the following equations which have equal roots:x5−13x4+67x3−171x2+216x−108=0.
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Solution
## Give equation, x5−13x4+67x3−171x2+216x−108=0 Consider f(x)=x5−13x4+67x3−171x2+216x−108 ∴f′(x)=5x4−52x3+201x2−342x+216 Now, HCF of f(x) and f′(x) is (x−2). Hence 2 is a double root of f(x)=0 f(x) can be factored as (x−2)2(x3−9x2+27x−27) Note that the above 2nd term is the expansion of (x−3)3 ∴f(x)=(x−2)2(x−3)3 ∴ Roots of the given equation are 2,2,3,3,3
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https://discuss.codechef.com/t/twolane-editorial/32553 | 1,618,922,623,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039398307.76/warc/CC-MAIN-20210420122023-20210420152023-00113.warc.gz | 321,864,986 | 8,182 | # TWOLANE - Editorial
Tester- Roman Bilyi
Editorialist- Abhishek Pandey
EASY
### PRE-REQUISITES:
Greedy, (Some)Implementation
### PROBLEM:
There are 2 lanes. At some parts of the lanes, there are some obstacles. Given the X-coordinate and lane of each obstacle, tell the maximum distance Chef can travel. Print K if he is able to reach his destination.
### QUICK-EXPLANATION:
Key to AC- Realize that if there is an obstacle in the lane you’re at, its good to switch lane as soon as possible if other lane is clear of obstacles till that X-Coordinate.
Start from the lane with obstacle at a greater distance X_i. Also keep a track of the next obstacle in the other lane. Once the next obstacle in the other lane lies at an X coordinate of X_{i+1}>X_i, switch the lane. Rinse and repeat until you can no longer move further or if you cleared all obstacles.
### EXPLANATION:
We will first discuss the greedy strategy, and then discuss the intuitional proof.
The greedy strategy-
The quick explanation pretty much sums it all. I will just expand that part here a little bit.
Start at the lane where the obstacle lies at a greater X-coordinate, X_i - this is an obvious choice.
Now, we keep a track of positions of all obstacles. There are 2 cases possible-
1. The obstacle is not in your lane - Simple, just record its position as “Position of Latest Obstacle Passed in Other Lane” and move on.
2. The obstacle is in your lane - If this is the case, then it is optimal to switch lanes as early as possible, due to the constraint of moving D length before switching lanes. (The proof will be discussed below, but it should be easy to see why this might be correct.) The best position to switch lane is “Position of Latest Obstacle Passed in Other Lane +1” (i.e. immediately after crossing that obstacle). If D distance isn’t covered till then, then switch as soon as you cover the required distance. If you can’t switch before hitting the next obstacle, then its the end of the journey .
The implementation for this is surprisingly crisp and simple. You can try doing it yourself first, and then refer to setter’s implementation to appreciate the differences between them-
Setter's implementation
``````int last_switch =-1<<30;
int lane = 3 - L[0];
int ans = k,ans_i=-1;
for(int i=0;i<n;i++){
if(L[i] != lane)continue;
int bst = max(last_switch + d, X[i-1] +1 );
if(bst >= X[i] ){
ans = X[i];
ans_i = i;
break;
}
lane = 3 - L[i];
last_switch = bst;
}
cout<<ans<<endl;
``````
Now, if you think on why this works - then we do not need to look far. By switching lanes as soon as possible (when the next obstacle is in your lane, and the next lane is clear till then), you are reducing the amount of distance to travel before the switch. You anyway have to make the switch - there is no choice here. You also cannot make the switch before crossing the latest obstacle on the other lane (such that the next obstacle is on the other lane). This is a hint that it is not a dynamic programming problem. Its trivial to see that switching any later will only increase the distance to travel before switching again, and hence is non-optimal.
### SOLUTION
Setter
``````#include <iostream>
#include <algorithm>
#include <string>
#include <assert.h>
using namespace std;
long long readInt(long long l,long long r,char endd){
long long x=0;
int cnt=0;
int fi=-1;
bool is_neg=false;
while(true){
char g=getchar();
if(g=='-'){
assert(fi==-1);
is_neg=true;
continue;
}
if('0'<=g && g<='9'){
x*=10;
x+=g-'0';
if(cnt==0){
fi=g-'0';
}
cnt++;
assert(fi!=0 || cnt==1);
assert(fi!=0 || is_neg==false);
assert(!(cnt>19 || ( cnt==19 && fi>1) ));
} else if(g==endd){
assert(cnt>0);
if(is_neg){
x= -x;
}
assert(l<=x && x<=r);
return x;
} else {
assert(false);
}
}
}
string ret="";
int cnt=0;
while(true){
char g=getchar();
assert(g!=-1);
if(g==endd){
break;
}
cnt++;
ret+=g;
}
assert(l<=cnt && cnt<=r);
return ret;
}
long long readIntSp(long long l,long long r){
}
long long readIntLn(long long l,long long r){
}
}
}
int T;
int n;
int X[100100];
int L[100100];
int k,d;
int sm_n=0;
int main(){
//freopen("01.txt","r",stdin);
//freopen("01o.txt","w",stdout);
while(T--){
sm_n += n;
assert(sm_n<=1000000);
for(int i=0;i<n;i++){
if(i==n-1){
} else {
}
if( i > 0){
assert(X[i] >X[i-1]);
}
}
for(int i=0;i<n;i++){
if(i==n-1){
} else {
}
}
int last_switch =-1<<30;
int lane = 3 - L[0];
int ans = k,ans_i=-1;
for(int i=0;i<n;i++){
if(L[i] != lane)continue;
int bst = max(last_switch + d, X[i-1] +1 );
if(bst >= X[i] ){
ans = X[i];
ans_i = i;
break;
}
lane = 3 - L[i];
last_switch = bst;
}
cout<<ans<<endl;
}
assert(getchar()==-1);
}
``````
Tester
`````` #include "bits/stdc++.h"
using namespace std;
#define FOR(i,a,b) for (int i = (a); i < (b); i++)
#define RFOR(i,b,a) for (int i = (b) - 1; i >= (a); i--)
#define ITER(it,a) for (__typeof(a.begin()) it = a.begin(); it != a.end(); it++)
#define FILL(a,value) memset(a, value, sizeof(a))
#define SZ(a) (int)a.size()
#define ALL(a) a.begin(), a.end()
#define PB push_back
#define MP make_pair
typedef long long Int;
typedef vector<int> VI;
typedef pair<int, int> PII;
const double PI = acos(-1.0);
const int INF = 1000 * 1000 * 1000;
const Int LINF = INF * (Int) INF;
const int MAX = 100007;
const int MOD = 998244353;
long long readInt(long long l,long long r,char endd){
long long x=0;
int cnt=0;
int fi=-1;
bool is_neg=false;
while(true){
char g=getchar();
if(g=='-'){
assert(fi==-1);
is_neg=true;
continue;
}
if('0'<=g && g<='9'){
x*=10;
x+=g-'0';
if(cnt==0){
fi=g-'0';
}
cnt++;
assert(fi!=0 || cnt==1);
assert(fi!=0 || is_neg==false);
assert(!(cnt>19 || ( cnt==19 && fi>1) ));
} else if(g==endd){
assert(cnt>0);
if(is_neg){
x= -x;
}
assert(l<=x && x<=r);
return x;
} else {
assert(false);
}
}
}
string ret="";
int cnt=0;
while(true){
char g=getchar();
assert(g!=-1);
if(g==endd){
break;
}
cnt++;
ret+=g;
}
assert(l<=cnt && cnt<=r);
return ret;
}
long long readIntSp(long long l,long long r){
}
long long readIntLn(long long l,long long r){
}
}
}
void assertEof(){
assert(getchar()==-1);
}
int main(int argc, char* argv[])
{
// freopen("in.txt", "r", stdin);
//ios::sync_with_stdio(false); cin.tie(0);
int sn = 0;
FOR(tt,0,t)
{
VI X(n), L(n);
FOR(i,0,n)
{
X[i] = readInt(1, k - 1, (i + 1 < n ? ' ' : '\n'));
if (i)
assert(X[i] > X[i - 1]);
}
FOR(i,0,n)
{
L[i] = readInt(1, 2, (i + 1 < n ? ' ' : '\n'));
-- L[i];
}
int S[2];
S[0] = S[1] = -INF;
int cur = 0;
if (L[0] == 0)
cur = 1;
else
cur = 0;
int res = k;
int last_change = -INF;
FOR(i,0,n)
{
if (L[i] == cur)
{
int r = X[i] - 1;
int l = max(S[cur ^ 1] + 1, last_change + d);
if (l > r)
{
res = X[i];
break;
}
last_change = l;
cur ^= 1;
}
S[L[i]] = X[i];
}
cout << res << endl;
}
assert(sn <= 1000000);
assertEof();
cerr << 1.0 * clock() / CLOCKS_PER_SEC << endl;
}
``````
Time Complexity=O(N)
Space Complexity=O(N)
### CHEF VIJJU’S CORNER
Tester's Notes
Switch lane as soon as you cross the last obstacle of the other lane, obviously after D distance from the previous switch.
Proof: The only possible situation when you couldn’t switch lanes is when the distance from previous switch is less than D, so in case you need to switch it’s better to do asap.
2.New to coding? Beginner? Did this kind of went over your head? Start with solving a simpler version of this problem - Solve it assuming that there is no constraint for distance D.
3.What would you do, if, the problem is modified as-
"Your car moves at a speed S. There are Q queries which you need to additionally do, of the form-
``````1 X L t - insert an obstacle in Lane L, at time t, and at position X
2 X L t - Remove an obstacle from Lane L, at time t and at position X. Its guaranteed that there will be an obstacle at the position.
``````
4.Implementation neatness! Implementation neatness! Implementation neatness!! I cannot stress this enough. Look at setter’s code and appreciate how he made it clean - this will help you greatly! Small things like, how he used `3-L[i]` to decide which lane to go, instead of using if-else conditions &etc. Don’t ignore a setter’s solution which is neatly implemented!!
5.If you are not able to solve this problem within the contest- then try to do a timed practice. There is absolutely no denying the fact that if this problem was to be given in long challenge, it’d have a looot of AC submissions! You don’t lack the talent, you lack the time! And the only thing we can do to avoid it is to be quicker.
Related Problems
6 Likes
Thanks for the related problems section.
3 Likes
https://www.codechef.com/viewsolution/25479965
May you help me where I am going wrong?? @vijju123 @l_returns @kingofnumbers | 2,508 | 8,629 | {"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-2021-17 | latest | en | 0.904673 |
http://www.docstoc.com/docs/75293834/Calculation-and-Analysis-of-Transformer-Inrush-Current-Based-on | 1,433,260,335,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1433195035877.3/warc/CC-MAIN-20150601214355-00003-ip-10-180-206-219.ec2.internal.warc.gz | 354,353,143 | 45,041 | # Calculation and Analysis of Transformer Inrush Current Based on by nikeborome
VIEWS: 320 PAGES: 4
• pg 1
``` ELECTRONICS AND ELECTRICAL ENGINEERING
ISSN 1392 – 1215 2011. No. 3(109)
ELEKTRONIKA IR ELEKTROTECHNIKA
ELECTRICAL ENGINEERING
T 190 ─────────────────────
ELEKTROS INŽINERIJA
Calculation and Analysis of Transformer Inrush Current Based on
Parameters of Transformer and Operating Conditions
M. Jamali, M. Mirzaie, S. Asghar Gholamian
Department of Electrical and Computer Engineering, Babol University of Technology,
P. O. Box. 484, Shariaty Ave., Babol, Iran, phone: +981113239214, e-mail: m.jamali@stu.nit.ac.ir
Introduction energizing circuit impedance and remanent flux on the
characteristics of inrush current are investigated in detail.
Magnetizing inrush current in transformers results
from any abrupt changes of the magnetizing voltage. This Fundamentals of Inrush Current
current in transformer may be caused by energizing an
unloaded transformer, occurrence of an external fault, It is very well known that a transformer will
voltage recovery after clearing an external fault and out-of- experience magnetizing inrush current during energization.
phase synchronizing of connected generator [1-2]. Because Inrush current occurs in a transformer whenever the
the amplitude of inrush current can be as high as a short- residual flux does not match the instantaneous value of the
circuit current, a detailed analysis of the magnetizing steady-state flux which would normally be required for the
inrush current under various conditions is necessary for the particular point on the voltage waveform at which the
concerns of a protective system for the transformers. In circuit is closed [13].
this regard, some numerical and analytical methods have For the explanation of the mechanism causing inrush
been proposed in the literature. In [3], analytical current in a transformer’s primary winding when
expressions for the magnetic fluxes of no-load three-phase connected to an AC voltage source, we consider (1), where
transformer is presented that can be used for inrush current λ and v are the instantaneous flux in a transformer core and
calculation. In [4], by analytical solution of two differential voltage drop across the primary winding, respectively
equations that governs the behavior of a transformer, the (1)
magnetic flux and inrush current are determined. For .
modeling transformer core including hysteresis, [5] used
As we see from (1), the rate of change of
Jiles-Atherton theory and presented a new algorithm on a instantaneous flux in a transformer core is proportional to
sample transformer. In [6], an analytic formula is presented
the instantaneous voltage drop in the primary winding or
to calculate the peak inrush current of a nonlinear inductor on the other hand, the flux waveform is the integral of the
with a series resistor. In [7], a simple model for the voltage waveform. In continuously-operating transformer,
transient period of inrush current is presented. This model
these two waveforms are shifted by 90°. But a significant
is developed from the structural parameters of transformer. difference exists between continuous-mode operation and
To avoid malfunctiom of protection system under energization of a transformer. During continuous
magnetizing inrush current, many researches are conducted
operation, the flux level is at its negative peak when
for the discrimination of inrush current from internal fault voltage is at its zero point, but during energization the flux
currents. For example, in [8-10], inrush current are has to start at zero. So, for a rising voltage just started from
discriminated from internal fault current by second
zero, the magnetic flux will reach approximately twice its
harmonic criterion. For discrimination of these currents, normal peak as it integrates the area under the voltage
[11] used the sum of active power flowing into the waveform’s first half-cycle. This amount of flux, because
transformer from each terminal. In [12], a criterion
of the nonlinear characteristic of the magnetization curve,
function in terms of difference of amplitude of wavelet causes saturation of the transformer. During saturation,
coefficients is defined. Then by using this criterion disproportionate amounts of mmf are needed to generate
function for three phases, the internal faults are
magnetic flux. This means the winding current, which
discriminated from the inrush current. creates the mmf to cause flux in the core, will
In this paper, first, the fundamentals of inrush current
disproportionately rise to a value easily exceeding twice its
and the formulas that are used for calculation it, are
normal peak. Fig. 1 shows the generation of inrush current
presented. Then a one-phase transformer is simulated in in a transformer. As seen from the figure, exceeding flux
MATLAB and the effects of switching angle variation,
17
from the knee point, results in large magnetizing current with those parameters are presented in [4], is selected. The
that in some circumstances can be ten times of the rated parameters of the equivalent circuit of this transformer
current in a transformer. referred to the 220V winding are shown in Table 1.
Fig. 2. Equivalent circuit of the transformer under no load
Fig. 1. Generation of inrush current in a transformer Table 1. Parameters of the simulated transformer
Parameter Rs (Ω) Ls (mH) Rp (Ω)
The general equation that gives the amplitude of Value 15.476 12 7260
inrush current as a function of time can be expressed as
(2): Also, the magnetization curve of the transformer is
given in (4) where i and λ are magnetizing current and flux
√
. , respectively
(2)
where Vm – maximum applied voltage; Zt – total impedance 63.084 10 sinh 2.43 . (4)
under inrush, including system; φ – energization angle; t –
time; t0 – point at which core saturates; τ – time constant of It should be noted that equations (5)-(8) are used to
transformer winding under inrush conditions; α – function calculate the fundamental and second harmonic
of t0; Kw – accounts for 3 phase winding connection; Ks – components of inrush current, where N, T and f are number
accounts for short-circuit power of network. of samples in each cycle, period and frequency of the
For the purpose of designing a protective system for power system, respectively. Also, m indicates fundamental
transformer, the peak value of inrush current is an and second components with the numbers 1 and 2,
important factor. In these cases, a simplified equation can respectively. The sampling rate of 30 kHz has been used in
be used to calculate the peak value of the first cycle of the this paper:
inrush current. This equation is as follow
√2 2. 2
, (3) . cos .2 . , (5)
.
where Vm – maximum applied voltage; L – air core
inductance of the transformer; R – total dc resistance of the 2
transformer; BN – normal rated flux density of the . sin .2 . , (6)
transformer core; BR – remanent flux density of the
transformer core; BS – saturation flux density of the core
material.
, (7)
As seen from the equations (2) and (3), the value of
inrush current is dependent to the parameters of
transformer and operating conditions. So a detailed
%2 100. (8)
analysis for finding the relations between the inrush
current characteristics and these factors are needed.
Simulation results Effects of switching angle
When a transformer is energized under no load or In this section, the effect of switching angle variation
lightly loaded conditions, inrush current may flow in the on the characteristics of inrush current has been
primary circuit. In this situation, the equivalent circuit of investigated. The remanent flux (Br) for all switching
transformer can be shown as Fig. 2 where Rs, Ls, Rp, Lm angles is 0.826 Wb-coil. Also the source resistance has
and Rt are series resistance, series inductance, core losses been considered to zero. Fig. 3 shows the effect of
resistance, magnetizing inductance and source resistance different switching angles (θ) on the amplitude of inrush
respectively. current. As seen from the figure, the highest amplitude of
In order to investigate the effects of some parameters inrush current is at 0˚ that is 5.52A. Also, it can be seen,
of transformer or network on the inrush current of a typical increasing of the switching angle will decrease the
transformer, a 120 VA, 60 Hz, (220/120) V transformer amplitude of inrush current.
18
inrush current. Also, it causes faster decay in the amplitude
of inrush current. Therefore, it can be said that
transformers located closer to the generating plants display
higher amount of inrush currents lasting much longer than
transformer installed electrically away from generator.
Fig. 3. Effect of switching angle variation on the amplitude of
inrush current
The second harmonic content of inrush current is
shown in Fig. 4. As seen from this figure, increasing of the
switching angle causes to a decrease in the percentage
second harmonic. Fig. 5. Effect of source resistance on the amplitude of inrush
current
The effect of source resistance in the percentage of
second harmonic has been shown in Fig. 6. The results
show that the amount of percentage of second harmonic
will be decreased by increasing the source resistance.
Fig. 4. Effect of switching angles in the percentage second
harmonic
It should be noted that, although, the highest
amplitude of the inrush current appears in the first cycle
and then decays, but the highest percentage second
harmonic does not necessarily appear in the first cycle. For Fig. 6. Effect of source resistance in the percentage second
instance as seen from Fig. 3 and Fig. 4, at θ=90˚, both harmonic
amplitude and percentage second harmonic have been
Effects of the remanent flux
decreased with increasing cycle, but at θ=0˚, although the
amplitude of inrush current have been decreased, but The effect of remanent flux on the first cycle peak
second harmonic firstly increased and then decreased. This current at different switching angles is shown in Fig. 7. As
is important when using second harmonic content to seen from figure, the first cycle peak current has large
restrain the relay operation during magnetizing inrush change when the remanent flux varies. Also the results
conditions. indicate that switching at θ=90˚ or Br=0 may not
necessarily reduce the magnitude of inrush current. So, for
Effects of source resistance
reducing inrush current, an appropriate switching angle by
In this case, the switching angle (θ) is 0˚. Also, the considering remanent flux must be selected.
remanent flux (Br) is the same as the previous section. The
Conclusions
effects of source resistance have been considered by
increasing Rt. Fig. 5 shows the effect of source resistance In this paper, the effects of some parameters on the
on the amplitude of inrush current. As seen from figure, characteristics of inrush current are investigated in
increasing source resistance will decrease the amplitude of MATLAB Simulink.
19
4. Vanti M. G., Bertoli S. L. Semianalytic solution for a simple
model of inrush currents in transformers // IEEE Trans.
Magnetics. – June, 2008. – Vol. 44. – No. 6. – P. 1270–1273.
5. Vahidi B., Tavakoli M. R. B. An algorithm for evaluating
inrush current in transformers using Jiles–Atherton theory of
ferromagnetic hysteresis // IEEE Conf. Tencon, Hong Kong.
– November, 2006. – P. 1–4.
6. Wang Y., Abdulsalam S. G., Xu W. Analytical formula to
estimate the maximum inrush current // IEEE Trans. Power
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7. Chen S. D., Lin R. L. Magnetizing inrush model of
transformers based on structure parameters // IEEE Trans.
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8. Sykes J. A., Morrison I. F. A proposed method of harmonic
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Results show that increasing switching angle at a 9. Kasztenny B., Kulidjian A. An improved transformer inrush
restraint algorithm increases security while maintaining fault
positive remanent flux or source resistance will decrease
response performance // 53rd Annual Conference for
the amplitude of inrush current. It has been shown that Protective Relay Engineers. – April, 2000. –P. 1–27.
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appear at the first cycle. The effect of remanent flux on the and comparison of harmonic restraint methods in
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when the remanent flux varies. Also, it has been concluded Protective Relay Engineers. – April, 2008. – P. 142–169.
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// IEEE Trans. Power Delivery. – July, 1997. – Vol. 12. – No.
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12. Faiz J. Lotfi–Fard S. A novel wavelet–based algorithm for
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Phase Power Transformer // Electronics and Electrical AIEE Transaction. – October, 1958. – Part III. – Vol. 77. – P.
Engineering. – Kaunas: Technologija, 2003. – No. 2(44). – P. 884–892.
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M. Jamali, M. Mirzaie, S. Asghar Gholamian. Calculation and Analysis of Transformer Inrush Current Based on Parameters
of Transformer and Operating Conditions // Electronics and Electrical Engineering. – Kaunas: Technologija, 2011. – No. 3(109).
– P. 17–20.
An inrush current is a transient current with high amplitude that may occur when a transformer is energized under no load or lightly
loaded conditions. The magnitude of inrush current may be as high as ten times or more times of transformer rated current. This could
result in huge mechanical and thermal stresses on transformer in addition to inadvertent operation of the protective relay systems. This
paper represents the effects of some factors on the inrush current of transformers. For this purpose, a one-phase transformer is simulated
in MATLAB and the effects of switching angle variation, the energizing circuit impedance and the remanent flux on the characteristics
of inrush current are investigated. The results show that increasing circuit resistance or switching angle will decrease inrush current
amplitude. Also, it is concluded that for reducing inrush current, appropriate switching angle with respect to the remanent flux must be
selected. The results can be used for a better understanding of the inrush current characteristics and proper actions of the protective
system. Ill. 7, bibl. 13, tabl. 1 (in English; abstracts in English and Lithuanian).
M. Jamali, M. Mirzaie, S. Asghar Gholamian. Transformatoriaus parametrų ir darbo sąlygų įtakos transformatoriaus
įmagnetinimo srovei apskaičiavimas ir tyrimas // Elektronika ir elektrotechnika. – Kaunas: Technologija, 2011. – Nr. 3(109). –
P. 17–20.
Įmagnetinimo srovė yra didelės amplitudės momentinė srovė, kuri gali atsirasti, kai transformatorius susižadina, kai nėra jokios
apkrovos arba kai ji maža. Įmagnetinimo srovė gali būti daugiau nei dešimt kartų didesnė už nominalią vertę. Toks poveikis, atsiradęs
dėl mechaninių ir terminių procesų, neigiamai veikia reles. Aprašomi veiksniai, turintys įtakos įmagnetinimo srovei. Atliktas vienfazio
transformatoriaus modeliavimas programų paketu Matlab, įvertinti pagrindiniai parametrai. Il. 7, bibl. 13, lent. 1 (anglų kalba;
santraukos anglų ir lietuvių k.).
20
```
To top | 4,174 | 18,723 | {"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} | 2.75 | 3 | CC-MAIN-2015-22 | longest | en | 0.832975 |
https://studylib.net/doc/15681923/section-2.1-frequency-distributions-and-their-graphs | 1,708,550,431,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473558.16/warc/CC-MAIN-20240221202132-20240221232132-00010.warc.gz | 563,929,959 | 13,124 | # Section 2.1 Frequency Distributions and Their Graphs
```Section 2.1
Frequency Distributions
and Their Graphs
Some Needed Definitions & Notation
“n” sample size (number of values in a sample, an integer)
“range” a measure of width/spread of a data set
range = maximum value in set – minimum value in set
Summation ∑ (Greek letter “sigma” – uppercase)
If x represents height in feet, may have several heights:
x1 = 5.5, x2 = 5.8, x3 = 5.4
If want to get sum of all heights, can write:
∑x = 5.5 + 5.8 + 5.4
∑x = 16.7
(“the sum of the x-values is 16.7”)
Example: Constructing a Frequency
Distribution
The following sample data set lists the prices (in
dollars) of 30 portable global positioning system (GPS)
navigators. Construct a frequency distribution that has
seven classes.
90 130 400 200 350 70 325 250 150 250
275 270 150 130 59 200 160 450 300 130
220 100 200 400 200 250 95 180 170 150
Frequency Distribution
Frequency Distribution
• A table that shows classes or intervals of data
with a count of the number of entries in each
class.
• The frequency, f, of a class is the number of
data entries in the class.
Constructing a Frequency Distribution
1. Decide on the number of classes.
Usually between 5 and 20; otherwise, it may be
difficult to detect any patterns.
2. Find the class width.
Determine the range (max-min) of the data.
Divide the range by the number of classes.
Round up to the next number. (always!)
(if division results in 3.5, round up to 4.0
if division results in 8 2/7, round up to 9
if division results in 12, round up to 13 !!)
Constructing a Frequency Distribution
3. Find the class limits.
You can use the minimum data entry as the lower
limit of the first class.
Find the remaining lower limits (add the class
width to the lower limit of the preceding class).
Find the upper limit of the first class. Remember
that classes cannot overlap.
Find the remaining upper class limits.
Constructing a Frequency Distribution
4. Make a tally mark for each data entry in the row of
the appropriate class.
5. Count the tally marks to find the total frequency f
for each class.
Example: Constructing a Frequency
Distribution
The following sample data set lists the prices (in
dollars) of 30 portable global positioning system (GPS)
navigators. Construct a frequency distribution that has
seven classes.
90 130 400 200 350 70 325 250 150 250
275 270 150 130 59 200 160 450 300 130
220 100 200 400 200 250 95 180 170 150
Solution: Constructing a Frequency
Distribution
90 130 400 200 350 70 325 250 150 250
275 270 150 130 59 200 160 450 300 130
220 100 200 400 200 250 95 180 170 150
1. Number of classes = 7 (given)
2. Find the class width
max min 450 59 391
55.86
#classes
7
7
Round up to 56
Solution: Constructing a Frequency
Distribution
3. Use 59 (minimum value)
the class width of 56 to
get the lower limit of the
next class.
59 + 56 = 115
Find the remaining
lower limits.
Class
width = 56
Lower
limit
59
115
171
227
283
339
395
Upper
limit
Solution: Constructing a Frequency
Distribution
The upper limit of the
first class is 114 (one less
than the lower limit of the
second class).
Add the class width of 56
to get the upper limit of
the next class.
114 + 56 = 170
Find the remaining upper
limits.
Lower
limit
59
Upper
limit
114
115
171
227
283
339
170
226
282
338
394
395
450
Class
width = 56
Solution: Constructing a Frequency
Distribution
4. Make a tally mark for each data entry in the row of
the appropriate class.
5. Count the tally marks to find the total frequency f
for each class.
Class
Tally
Frequency, f
IIII
5
115–170
IIII III
8
171–226
IIII I
6
227–282
IIII
5
283–338
II
2
339–394
I
1
395–450
III
3
59–114
Determining the Midpoint
Midpoint of a class
(Lower class limit) (Upper class limit)
2
Class
59–114
Midpoint
59 114
86.5
2
115–170
115 170
142.5
2
171–226
171 226
198.5
2
Frequency, f
5
Class width = 56
8
6
Determining the Relative Frequency
Relative Frequency of a class
• Portion or percentage of the data that falls in a
particular class.
Class frequency
f
• Relative frequency
Sample size
n
.
Class
Frequency, f
59–114
5
115–170
8
171–226
6
Relative Frequency
5
0.17
30
8
0.27
30
6
0.2
30
Determining the Cumulative Frequency
Cumulative frequency of a class
• The sum of the frequencies for that class and all
previous classes.
Class
Frequency, f
Cumulative frequency
59–114
5
5
115–170
+ 8
13
171–226
+ 6
19
Expanded Frequency Distribution
Class
Frequency, f
Midpoint
Relative
frequency
59–114
5
86.5
0.17
5
115–170
8
142.5
0.27
13
171–226
6
198.5
0.2
19
227–282
5
254.5
0.17
24
283–338
2
310.5
0.07
26
339–394
1
366.5
0.03
27
395–450
3
422.5
0.1
f
1
n
30
Σf = 30
Cumulative
frequency
Graphs of Frequency Distributions
frequency
Frequency Histogram
• A bar graph that represents the frequency distribution.
• The horizontal scale is quantitative and measures the
data values.
• The vertical scale measures the frequencies of the
classes.
• Consecutive bars must touch.
data values
Class Boundaries
Class boundaries
• The numbers that separate classes without forming
gaps between them.
• The distance from the upper
limit of the first class to the
lower limit of the second
class is 115 – 114 = 1.
• Half this distance is 0.5.
Class
Class
boundaries
Frequency,
f
59–114
58.5–114.5
5
115–170
8
171–226
6
• First class lower boundary = 59 – 0.5 = 58.5
• First class upper boundary = 114 + 0.5 = 114.5
Class Boundaries
Class
59–114
115–170
171–226
227–282
283–338
339–394
395–450
Class
boundaries
58.5–114.5
114.5–170.5
170.5–226.5
226.5–282.5
282.5–338.5
338.5–394.5
394.5–450.5
Frequency,
f
5
8
6
5
2
1
3
Example: Frequency Histogram
Construct a frequency histogram for the Global
Positioning system (GPS) navigators.
Class
Class
boundaries
Frequency,
Midpoint
f
59–114
58.5–114.5
86.5
5
115–170
114.5–170.5
142.5
8
171–226
170.5–226.5
198.5
6
227–282
226.5–282.5
254.5
5
283–338
282.5–338.5
310.5
2
339–394
338.5–394.5
366.5
1
395–450
394.5–450.5
422.5
3
Solution: Frequency Histogram
(using Midpoints)
Solution: Frequency Histogram
(using class boundaries)
You can see that more than half of the GPS navigators are
priced below \$226.50.
Example: Frequency Polygon
Frequency polygon: A line graph that emphasizes the
continuous change in frequencies.
Construct a frequency polygon for the GPS navigators
frequency distribution.
Class
Midpoint
Frequency, f
59–114
86.5
5
115–170
142.5
8
171–226
198.5
6
227–282
254.5
5
283–338
310.5
2
339–394
366.5
1
395–450
422.5
3
Solution: Frequency Polygon
The graph should
begin and end on the
horizontal axis, so
extend the left side to
one class width before
the first class
midpoint and extend
the right side to one
class width after the
last class midpoint.
You can see that the frequency of GPS navigators increases
up to \$142.50 and then decreases.
Graphs of Frequency Distributions
relative
frequency
Relative Frequency Histogram
• Has the same shape and the same horizontal scale as
the corresponding frequency histogram.
• The vertical scale measures the relative frequencies,
not frequencies.
data values
.
Example: Relative Frequency Histogram
Construct a relative frequency histogram for the GPS
navigators frequency distribution.
Class
Class
boundaries
Frequency,
f
Relative
frequency
59–114
58.5–114.5
5
0.17
115–170
114.5–170.5
8
0.27
171–226
170.5–226.5
6
0.2
227–282
226.5–282.5
5
0.17
283–338
282.5–338.5
2
0.07
339–394
338.5–394.5
1
0.03
395–450
394.5–450.5
3
0.1
Solution: Relative Frequency Histogram
6.5
18.5
30.5
42.5
54.5
66.5
78.5
90.5
From this graph you can see that 27% of GPS navigators are
priced between \$114.50 and \$170.50.
Solution: Frequency Histogram
(using class boundaries)
You can see that more than half of the GPS navigators are
priced below \$226.50.
Graphs of Frequency Distributions
cumulative
frequency
Cumulative Frequency Graph or Ogive
• A line graph that displays the cumulative frequency
of each class at its upper class boundary.
• The upper boundaries are marked on the horizontal
axis.
• The cumulative frequencies are marked on the
vertical axis.
data values
Constructing an Ogive
1. Construct a frequency distribution that includes
cumulative frequencies as one of the columns.
2. Specify the horizontal and vertical scales.
The horizontal scale consists of the upper class
boundaries.
The vertical scale measures cumulative
frequencies.
3. Plot points that represent the upper class boundaries
and their corresponding cumulative frequencies.
Constructing an Ogive
4. Connect the points in order from left to right.
5. The graph should start at the lower boundary of the
first class (cumulative frequency is zero) and should
end at the upper boundary of the last class
(cumulative frequency is equal to the sample size).
Example: Ogive
Construct an ogive for the GPS navigators frequency
distribution.
Class
Class
boundaries
Frequency,
f
Cumulative
frequency
59–114
58.5–114.5
5
5
115–170
114.5–170.5
8
13
171–226
170.5–226.5
6
19
227–282
226.5–282.5
5
24
283–338
282.5–338.5
2
26
339–394
338.5–394.5
1
27
395–450
394.5–450.5
3
30
Solution: Ogive
6.5
18.5
30.5
42.5
54.5
66.5
78.5
90.5
From the ogive, you can see that about 25 GPS navigators cost
\$300 or less. The greatest increase occurs between \$114.50 and
\$170.50.
``` | 3,017 | 9,241 | {"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} | 4.3125 | 4 | CC-MAIN-2024-10 | latest | en | 0.809051 |
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