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http://setiweb.org/error-bars/plot-error-bars-physics.php | 1,542,400,571,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743184.39/warc/CC-MAIN-20181116194306-20181116220306-00440.warc.gz | 307,606,067 | 5,260 | Home > Error Bars > Plot Error Bars Physics
# Plot Error Bars Physics
## Contents
The weight swings about a fixed point. In these two plots, error bars for all X, Y and Z directions are available. This statistics-related article is a stub. It may even be published in a proper scientific journal. http://setiweb.org/error-bars/physics-error-bars.php
Set up your worksheet so that the columns are designated Y1, yEr1, Y2, yEr2, Y3, yEr3... (the error bar column must be to the right of the Y data column). Similarly, if you wanted to calculate the area of the field, $A = lw$, you would need to know how to do this using $\Delta L$ and $\Delta w$. Without uncertainties, you can't say anything about agreement or disagreement, which is why uncertainties are so important in experimental science. After all, knowledge is power! #5 P-A July 31, 2008 Hi there, I agree with your initial approach: simplicity of graphs, combined with clear interpretation of results (based on information that
## How To Draw Error Bars In Excel
This is NOT the same thing as saying that the specific interval plotted has a 95% chance of containing the true mean. Otherwise, we'll assume you're OK to continue. API Documentation API Libraries REST APIs Plotly.js Hardware About Us Team Careers Plotly Blog Modern Data Help Knowledge Base Benchmarks
Over thirty percent of respondents said that the correct answer was when the confidence intervals just touched -- much too strict a standard, for this corresponds to p<.006, or less than When scientific fraud is discovered, journal editors can even decide on their own to publish a retraction of fraudulent paper(s) previously published by the journal they edit. Click “submit” when you are done. How To Draw Error Bars On A Graph Since the accepted value for $g$ at the surface of the earth is 9.81 m/s$^2$, which falls within the range we found using the max-min method, we may say, based on
For example, if the meter stick that you used to measure the book was warped or stretched, you would never get an accurate value with that instrument. Error Bars In Physics A Level We can study 50 men, compute the 95 percent confidence interval, and compare the two means and their respective confidence intervals, perhaps in a graph that looks very similar to Figure we write the answer as 13.7 m s-1.
You will learn important research skills such as: Writing a thesis of research-level standard. "Defending" your thesis.
This range is determined from what we know about our lab instruments and methods. What Are Error Bars Not just because someone tells you without any evidence why it should be accepted.) What we mean by experimental uncertainty/error is the estimate of the range of values within which the A line is reasonable if it just passes within most of the error bars. The question is, how close can the confidence intervals be to each other and still show a significant difference?
## Error Bars In Physics A Level
Let's say that you think you can press the button within 0.2 seconds of either the start or the stop of the measurement. The video shows you how to measure the different quantities that are important in the experiment: $L$, the angle $\theta$ that $L$ makes with the vertical before the pendulum is released, How To Draw Error Bars In Excel Origin provides customization controls for error bars in both 2D and 3D graphs. How To Draw Error Bars By Hand We say that there is a “discrepancy” between two results when they “disagree” in the above sense.
They give a general idea of how precise a measurement is, or conversely, how far from the reported value the true (error free) value might be. http://setiweb.org/error-bars/plot-r-error-bars.php This also means we need to know what is the uncertainty, $\Delta T^2$, in $T^2$ so that we may draw vertical error bars (error bars for the dependent variable are “vertical”, We do the same for small quantities such as 1 mV which is equal to 0,001 V, m standing for milli meaning one thousandth (1/1000). To demonstrate this we are going to consider an example that you studied in PHY 121, the simple pendulum. How To Calculate Error Bars In Physics
Therefore if you used this max-min method you would conclude that the value of the slope is 24.4 $\pm$ 0.7 cm/s$^2$, as compared to the computers estimate of 24.41 $\pm$ 0.16 To add error bars to a point on a graph, we simply take the uncertainty range (expressed as "± value" in the data) and draw lines of a corresponding size above And someone in a talk recently at 99% confidence error bars, which rather changed the interpretation of some of his data. get redirected here Keep doing what you're doing, but put the bars in too.
Set style of error bars, including color, width, and transparency Select data for error bars. How To Calculate Error Bars In Origin This means that the slope (labeled as $a$ by the plotting tool) of our graph should be equal to $\Large \frac{g}{(2\pi)^2}$. You should always use such an eyeball + brain method first to provide a ballpark estimate for the uncertainty.
## My textbook calls it the "Standard Deviation of the Mean".
Specify how to skip error bars when drawing. We rewrite (E.9a) as $T=\left({\Large \frac{2 \pi}{g^{1/2}}} \right) L^{1/2}$ (E.9b) to put all the constants between the parentheses. A physicist would say that since the two linear graphs are based on the same data, they should carry the same “physical information”. How To Draw Error Bars On A Line Graph Example: Calculate the area of a field if it's length is 12 ± 1 m and width is 7± 0.2 m.
The period of this motion is defined as the time $T$ necessary for the weight to swing back and forth once. According to the Eq. (E.9c) that we are testing, when $L=0$, $T^2=0$, so you should check the box that asks you if the fit must go through (0,0), viz., “through the For reasonably large groups, they represent a 68 percent chance that the true mean falls within the range of standard error -- most of the time they are roughly equivalent to http://setiweb.org/error-bars/plot-error-bars-idl.php This time however, we check the lowest, highest and best value for the intercept.
Method 3 - Using Plot Details Dialog for 3D Graphs Error bars also could be added in the 3D graph from existing datasets by the Plot Details dialog. A frequent misconception is that the “experimental error” is the difference between our measurement and the accepted “official” value. (Who accepts it? User error bars in such 3D graphs as XYZ 3D scatter, matrix 3D scatter, 3D color fill surface, and 3D color map surface. If we're interested in evaluating $\frac{\Delta T}{T}$, we see from (E.3) that the constant $\alpha$, which in our case equals ${\large \left(\frac{2 \pi}{g^{1/2}}\right) }$, “drops out”.
Andrade: “William Gilbert, whose De Magnete Magneticisque Corporibus et de Magno Magnete Tellure Physiologia Nova, usually known simply as De Magnete, published in 1600, may be said to be the first To get $g$ we should multiply the slope by $(2\pi)^2$, and we should also divide by 100 to convert from cm/s$^2$ to m/s$^2$ so that we're using standard SI units. Because you checked the box, it does not give you a value for $b$ because it is “constrained” to be zero. | 1,713 | 7,228 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2018-47 | latest | en | 0.923493 |
https://kr.mathworks.com/matlabcentral/answers/1918310-plot-a-matrix-in-3d?s_tid=prof_contriblnk | 1,721,012,647,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514655.27/warc/CC-MAIN-20240715010519-20240715040519-00426.warc.gz | 311,444,036 | 26,071 | # plot a matrix in 3D
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mingcheng nie 2023년 2월 24일
편집: John D'Errico 2023년 2월 24일
Hi there,
If I have a square matrix H=randi(N,N)+1i*randi(N,N); how can I plot the maginitude of this matrix H in 3D?
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### 답변 (1개)
John D'Errico 2023년 2월 24일
편집: John D'Errico 2023년 2월 24일
DID YOU TRY IT? Why not?
So, given some arbitrary complex matrix, how do you compute the magnitude? ABS does that. How you you plot the matrix? Use surf. Or you could use pcolor. Or you could use mesh. Or use surfc. I'm sure there are some other options I could think of too.
N = 10;
H = rand(N,N) + i*rand(10,10);
MagH = abs(H);
surf(MagH)
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### 카테고리
Help CenterFile Exchange에서 Surface and Mesh Plots에 대해 자세히 알아보기
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Translated by | 315 | 889 | {"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-2024-30 | latest | en | 0.422946 |
http://slideplayer.com/slide/3921143/ | 1,521,445,747,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257646602.39/warc/CC-MAIN-20180319062143-20180319082143-00797.warc.gz | 283,588,169 | 19,044 | # Valence electrons Electromagnetic Spectrum Light characteristics Electrons and Light.
## Presentation on theme: " Valence electrons Electromagnetic Spectrum Light characteristics Electrons and Light."— Presentation transcript:
Valence electrons Electromagnetic Spectrum Light characteristics Electrons and Light
Fe – 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 6 {2 valence e - } Rb – 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 1 {1 valence e - } Cl – 1s 2 2s 2 2p 6 3s 2 3p 5 {7 valence e - } These are the electrons that are involved in bonding and chemical reactions!!!!!!!
When you add energy to an element (perhaps by heating it up), the valence e - get “excited”. In other words they jump up to a higher energy level or orbital. BUT… they are unstable up there. So they release that added energy in the form of colored light! Huh?
Light moves in wave from the light source to you eye or other detector! Waves have several characteristics!
Speed of light = wavelength × frequency speed of light = 300,000,000 meters/ sec. Since the speed of light is constant, then if wavelength increases, then frequency has to decrease!
Moving along those waves, there are little packets of energy called photons. › Photons have specific amounts of energy as determined by the frequency of the light. › Energy= Planck’s constant × frequency › The higher the frequency of the light, the more energy the light has.
When you add energy to an element, it’s valence e - absorb that packet of energy & become unstable. In order to return to stability (lower their energy) they “spit out” that energy in the form of a photon that has a frequency in the visible light part of the electromagnetic spectrum that we can see.
Light Bunsen burner. Take a soaked wooden splint from the bottle. (note the 1 st element in the solution.) Hold splint in flame. Observe flame color and record. Rinse splint in water. Dispose of all used splints in trash. Turn off Bunsen burner.
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Module
1
Classification of Metal Removal Processes and Machine tools
Version 2 ME IIT, Kharagpur
Lesson
2
Basic working principle, configuration, specification and classification of machine tools
Version 2 ME IIT, Kharagpur
Instructional Objectives
At the end of this lesson, the students should be able to :
(a) Describe the basic functional principles of machine tools (i) Illustrate the concept of Generatrix and Directrix (ii) Demonstrate Tool – work motions (iii) Give idea about machine tool drives (b) Show configuration of basic machine tools and state their uses (c) Give examples of machine tools - specification (d) Classify machine tools broadly.
Basic functional principles of machine tool operations
Machine Tools produce desired geometrical surfaces on solid bodies (preformed blanks) and for that they are basically comprised of;
Devices for firmly holding the tool and work
Drives for providing power and motions to the tool and work
Kinematic system to transmit motion and power from the sources to the tool-work
Automation and control systems
Structural body to support and accommodate those systems with sufficient strength and rigidity. For material removal by machining, the work and the tool need relative movements and those motions and required power are derived from the power source(s) and transmitted through the kinematic system(s) comprised of a number and type of mechanisms.
(i) Concept of Generatrix and Directrix
Generation of flat surface
The principle is shown in Fig. 2.1 where on a flat plain a straight line called
Generatrix (G) is traversed in a perpendicular direction called Directrix (D) resulting a flat surface.
Generation of cylindrical surfaces
The principles of production of various cylindrical surfaces (of revolution) are shown in Fig. 2.2, where,
A long straight cylindrical surface is obtained by a circle (G) being traversed in the direction (D) parallel to the axis as shown in Fig. 2.2(a)
A cylindrical surface of short length is obtained by traversing a straight line (G) along a circular path (D) as indicated in Fig.
2.2(b)
Form cylindrical surfaces by rotating a curved line (G) in a circular path (D) as indicated in Fig. 2.2 (c and d).
Version 2 ME IIT, Kharagpur
D
(a)
G
D
G
(b)
Fig. 2.1 Generation of flat surfaces by Generatrix and Directrix.
Fig. 2.2 Generation of cylindrical surfaces (of revolution)
(ii) Tool – work motions
The lines representing the Generatrix and Directrix are usually produced by the locus of a point moving in two different directions and are actually obtained by the motions of the tool-tip (point) relative to the work surface. Hence, for machining flat or curved surfaces the machine tools need relative tool work motions, which are categorized in following two groups:
Formative motions namely
Cutting motion (CM)
Feed motion (FM)
Auxiliary motions such as
Indexing motion
Relieving motion
The Generatrix and Directrix, tool and the work and their motions generally remain interconnected and in different way for different machining work. Such interconnections are typically shown in Fig. 2.3 for straight turning and in Fig. 2.4 for shaping.
Version 2 ME IIT, Kharagpur
Cutting motion
G
D
Feed motion
(a) longitudinal turning
CM
D
G
FM
(b)
transverse turning
Fig. 2.3 Principle of turning (cylindrical surface)
The connections in case of straight longitudinal turning shown in Fig. 2.3 (a) are:
Generatrix (G) – Cutting motion (CM) – Work (W)
Directrix (D) – Feed motion (FM)
– Tool (T)
tool
work
G
D
CM
FM
Desired flat
surface
Fig. 2.4 Principle of producing flat surface in shaping machine
In case of making flat surface in a shaping machine as shown in Fig. 2.4 the connections are:
G – CM – T D – FM – W
which indicates that in shaping flat surfaces the Generatrix is provided by the cutting motion imparted to the cutting tool and the Directrix is provided by the feed motion of the work.
Version 2 ME IIT, Kharagpur
Flat surfaces are also produced by planning machines, mainly for large jobs, where the cutting motion is imparted to the work and feed motion to the tool and the connections will be:
G – CM – Work D – FM – Tool
The Genratrix and Directrix can be obtained in four ways:
Tracing (Tr) – where the continuous line is attained as a trace of path of a moving point as shown in Fig. 2.3 and Fig. 2.4.
Forming (F) – where the Generatrix is simply the profile of the cutting edge as indicated in Fig. 2.2 (c and d)
Tangent Tracing (TTr) – where the Directrix is taken as the tangent to the series of paths traced by the cutting edges as indicated in Fig. 2.5.
Generation (G): Here the G or D is obtained as an envelope being tangent to the instantaneous positions of a line or surface which is rolling on another surface. Gear teeth generation by hobbing or gear shaping is the example as can be seen in Fig. 2.6.
Fig. 2.5 typically shows the tool-work motions and the corresponding Generatrix (G) and Directrix (D) while producing flat surface by a plain or slab milling cutter in a conventional horizontal arbour type milling machine. The G and D are connected here with the tool work motions as
G – x – T – F D – FM – W – T.Tr CM – T
Here G and D are independent of the cutting motion and the G is the line of contact between the milling cutter and the flat work surface. The present cutter being of roller shape, G has been a straight line and the surface produced has also been flat. Form milling cutters will produce similar formed surfaces as shown in Fig. 2.7 where the ‘G’ is the tool-form.
Fig. 2.5 Directrix formed by tangent tracing in plain milling
Version 2 ME IIT, Kharagpur
Fig. 2.6 Generatrix (or Directrix) in gear teeth cutting by generation.
Fig. 2.7 Tool-work motions and G & D in form milling
For making holes in drilling machines both the cutting motion and the feed motion are imparted to the cutting tool i.e., the drill bit whereas the workpiece remains stationary. This is shown in Fig. 2.8. The G and D are linked with the tool-work in the way:
G – CM – T – Tr D – FM – W – Tr
Version 2 ME IIT, Kharagpur
CM
G
FM
D
G
D
Fig. 2.8 Tool-work motions and G & D in drilling.
B oring machines are mostly used for enlargement and finishing of existing cylindrical holes. Boring machines are of two types:
Vertical boring machine – low or medium duty and high precision, e.g., Jig boring machine
Horizontal axis borin
g machine – medium or heavy duty.
In
same. In horizontal boring machine the feed motion is imparted to the work to provide the Directrix by Tracing.
respect of tool-work motions and G and D, vertical boring and drilling are
(i
ii)
Machine tool drives
F or the desired tool-work motions with power, machine tools are driven by electric motors and use of some mechanisms like belt-pulley, gears etc. In some machine tools, the tool-work motions are provided by hydraulic drive also.
M achine tools essentially need wide ranges of cutting speed and feed rate to enable
• M achining different jobs (material and size) • Using different cutting tools (material, geome try and size) • Various machining operations like high speed turning t o low speed thread cutting in lathes • Degree of surface finish desired.
M achine tool drives may be
o Stepped drive o Stepless drive
Version 2 ME IIT, Kharagpur
Stepped drives are very common in conventional machine tools where a discrete number of speeds and feeds are available and preferably in G.P. (Geometric Progression) series. Whereas the modern CNC machine tools are provided with stepless drives enabling optimum selection and flexibly automatic control of the speeds and feeds.
Stepped drive is attained by using gear boxes or cone pulley (old method) along with the power source. Stepless drive is accomplished usually by
Variable speed AC or DC motors
Stepper or servomotors
Hydraulic power pack
Configuration of Basic Machine Tools and their use
Centre lathes configuration
-
Fig. 2.9 shows the general configuration of center lathe. Its major parts are:
o Head stock: it holds the blank and through that power and rotation are transmitted to the job at different speeds o tailstock: supports longer blanks and often accommodates tools like drills, reamers etc for hole making. o carriage: accommodates the tool holder which in turn holds the moving tools o bed: Δ headstock is fixed and tailstock is clamped on it. Tailstock has a provision to slide and facilitate operations at different locations Δ carriage travels on the bed o columns: on which the bed is fixed o work-tool holding devices
uses of center lathes
Centre lathes are quite versatile being used for various operations:
external
⎯ turning
internal
straight
taper
stepped
facing, centering, drilling, recessing and parting
thread cutting; external and internal knurling. Some of those common operations are shown in Fig. 2.10. Several other operations can also be done in center lathes using suitable attachments.
Shaping machine Fig. 2.11 shows the general configuration of shaping machine. Its major parts are:
o Ram: it holds and imparts cutting motion to the tool through reciprocation o Bed: it holds and imparts feed motions to the job (blank) o Housing with base: the basic structure and also accommodate the drive mechanisms
Version 2 ME IIT, Kharagpur
o Power drive with speed and feed change mechanisms.
Shaping machines are generally used for producing flat surfaces, grooving, splitting etc. Because of poor productivity and process capability these machine tools are not widely used now-a-days for production.
tool post
job
tool
tailstock
rack
feedrod
bed
Fig. 2.9 Schematic view of a center lathe
turning
facing
grooving
forming
Fig. 2.10 Some common machining operations done in center
Lathes.
External
Internal
Fig. 2.10 Some common machining operations done in center lathes.
Version 2 ME IIT, Kharagpur
clapperbox
ram
tool
housing
Power drive
bed
base
Job
Vice
Fig. 2.11 Schematic view of a shaping machine
Planing machine The general configuration is schematically shown in Fig. 2.12. This machine tool also does the same operations like shaping machine but the major differences are:
o In planing the job reciprocates for cutting motion and the tool moves slowly for the feed motions unlike in shaping machine. o Planing machines are usually very large in size and used for large jobs and heavy duty work.
Drilling machine Fig. 2.13 shows general configuration of drilling machine, column drill in particular. The salient parts are
o Column with base: it is the basic structure to hold the other parts o Drilling head: this box type structure accommodates the power drive and the speed and feed gear boxes. o Spindle: holds the drill and transmits rotation and axial translation to the tool for providing cutting motion and feed motion – both to the drill.
Drilling machines are available in varying size and configuration such as pillar drill, column drill, radial drill, micro-drill etc. but in working principle all are more or less the same. Drilling machines are used:
o Mainly for drilling (originating or enlarging cylindrical holes)
Version 2 ME IIT, Kharagpur
o Occasionally for boring, counter boring, counter sinking etc. o Also for cutting internal threads in parts like nuts using suitable attachment.
frame
tool
Job
power
bed
drive
base
table
Fig. 2.12 Schematic view of a planning machine
Feed
change
Speed
lever
Spindle
Column
Drill
Job
bed
base
change lever
Version 2 ME IIT, Kharagpur
Fig. 2.13 Schematic view of a drilling machine
Milling machine
knee type conventional milling
machine with horizontal arbour is shown in Fig. 2.14. Its major parts are
The
general
configuration
of
o Milling arbour: to hold and rotate the cutter o Ram: to support the arbour o Machine table: on which job and job holding devices are mounted to provide the feed motions to the job. o Power drive with Speed and gear boxes: to provide power and motions to the tool-work o Bed: which moves vertically upward and downward and accommodates the various drive mechanisms o Column with base: main structural body to support other parts.
ram
Speed
Gear
Box
Feed
GB
MOTOR
base
job
Cutter
Fig. 2.14 Schematic view of a milling machine
Milling machines are also quite versatile and can do several operations like
o making flat surfaces o grooving, slitting and parting o helical grooving
Version 2 ME IIT, Kharagpur
o forming 2-D and 3-D contoured surfaces
Fig. 2.15 shows some of the aforesaid milling operations.
surfacing
slotting
slitting
grooving
forming
Fig. 2.15 Some common milling operation
Specification of Machine Tools.
A machine tool may have a large number of various features and characteristics. But only some specific salient features are used for specifying a machine tool. All the manufacturers, traders and users must know how are machine tools specified.
The methods of specification of some basic machine tools are as follows:
o Centre lathe • Maximum diameter and length of the jobs that can be accommodated • Power of the main drive (motor) • Range of spindle speeds • Range of feeds • Space occupied by the machine. o Shaping machine • Length, breadth and depth of the bed • Maximum axial travel of the bed and vertical travel of the bed / tool • Maximum length of the stroke (of the ram / tool) • Range of number of strokes per minute • Range of table feed • Power of the main drive • Space occupied by the machine o Drilling machine (column type)
Maximum drill size (diameter) that can be used
Size and taper of the hole in the spindle
Range of spindle speeds
Version 2 ME IIT, Kharagpur
Range of feeds
Power of the main drive
Range of the axial travel of the spindle / bed
Floor space occupied by the machine
o Milling machine (knee type and with arbour)
Type; ordinary or swiveling bed type
Size of the work table
Range of travels of the table in X-Y-Z directions
Arbour size (diameter)
Power of the main drive
Range of spindle speed
Range of table feeds in X-Y-Z directions
Floor space occupied.
Number of types of machine tools gradually increased till mid 20 th century and after that started decreasing based on Group Technology. However, machine tools are broadly classified as follows:
According to direction of major axis :
o horizontal center lathe, horizontal boring machine etc. o vertical – vertical lathe, vertical axis milling machine etc. o inclined – special ( e.g. for transfer machines).
According to purpose of use :
o general purpose – e.g. center lathes, milling machines, drilling machines etc. o single purpose – e.g. facing lathe, roll turning lathe etc. o special purpose – for mass production.
According to degree of automation
o non-automatic – e.g. center lathes, drilling machines etc. o semi-automatic – capstan lathe, turret lathe, hobbinh machine etc. o automatic – e.g., single spindle automatic lathe, swiss type automatic lathe, CNC milling machine etc.
According to size :
o heavy duty – e.g., heavy duty lathes (e.g. ≥ 55 kW), boring mills, planning machine, horizontal boring machine etc. o medium duty – e.g., lathes – 3.7 ~ 11 kW, column drilling machines, milling machines etc. o small duty – e.g., table top lathes, drilling machines, milling machines. o micro duty – e.g., micro-drilling machine etc.
According to precision :
Version 2 ME IIT, Kharagpur
o ordinary – e.g., automatic lathes o high precision – e.g., Swiss type automatic lathes
According to number of spindles :
o single spindle – center lathes, capstan lathes, milling machines etc. o multi-spindle – multispindle (2 to 8) lathes, gang drilling machines etc.
According to blank type :
o bar type (lathes) o chucking type (lathes) o housing type
According to type of automation :
o fixed automation – e.g., single spindle and multispindle lathes o flexible automation – e.g., CNC milling machine
According to configuration :
o stand alone type – most of the conventional machine tools. o machining system (more versatile) – e.g., transfer machine, machining center, FMS etc.
Exercise – 2
1. Show the tool-work motions and the Generatrix and Directrix in external thread cutting in centre lathe. Also state how those ‘G’ & ‘D’ are obtained.
2. In which conventional machine tools flat surface can be produced ?
3. State the major differences between shaping machine and planing machine.
4. In which machine tools both the cutting motion & the feed motion are imparted to the tool ?
5. How is feed expressed in turning, shaping, drilling and milling ?
Ans. Q 1
G
CM
D
FM
Version 2 ME IIT, Kharagpur
G – x – T
D – (CM+FM) – (T+W) - T
– F
Ans. Q. 2 Flat surfaces can be produced in
• centre lathes – e.g., facing
• shaping, slotting and planing machines
• milling machines
Ans. Q. 3
Shaping machine Planing machine o for small and medium size jobs o for medium and large size jobs o tool reciprocates and provide o job on table reciprocates and provide CM CM o feed motion is given to the job o feed motion is given to the tool o G – CM – T – Tr o G – CM – W – Tr D – FM – W – Tr D – FM – T – Tr
Ans. Q. 4
Both CM and FM are imparted to the tool in
drilling machine
vertical boring machine
Ans. Q. 5
• turning – mm/rev
• shaping – mm/stroke
• drilling machine – mm/rev
• milling machine – mm/min
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20,000 square feet is equivalent to the area of about half a football field. A space measuring 20,000 square feet offers ample room for various purposes, from commercial establishments like restaurants and retail stores to residential properties such as large houses or apartments.
This size provides flexibility for designing and utilizing the space efficiently. Whether it’s a spacious office, a bustling retail space, or a cozy home, 20,000 square feet allows for creativity in layout and design. With this substantial area, there are endless possibilities for customization and functionality, making it a desirable size for diverse projects and needs.
Credit: www.wsj.com
Understanding Square Feet
Understanding square feet is essential when it comes to determining the size of a space or property. In the world of real estate and construction, square footage serves as a critical measure for assessing the area of buildings, homes, or land.
What Is A Square Foot?
A square foot is a unit of area measurement, typically used in the United States and some other countries. It represents the area within a square with sides that are each one foot long. In mathematical terms, one square foot is equal to the area of a square with sides that are one foot in length.
Converting Square Feet To Other Units
Converting square feet to other units of measurement can be helpful when dealing with various aspects of real estate and construction. Here are some common conversions:
• 1 square foot = 0.0929 square meters
• 1 square foot = 144 square inches
• 1 square foot = 0.00002296 acres
• 1 square foot = 0.0000000355 square miles
Visualizing 20,000 Square Feet
Visualizing 20,000 square feet can be challenging without a point of reference. Let’s explore how big 20,000 square feet truly is through visual comparisons and examples.
Comparing To Common Objects
To comprehend the size of 20,000 square feet, imagine it’s equivalent to almost half a football field. It can fit approximately three to four average-sized houses within its borders.
Visual Examples Of 20,000 Square Feet
Consider envisioning 20,000 square feet as a space large enough to accommodate a medium-sized retail store or an indoor arena for sporting events. It could also represent the size of a parking lot for around 100-150 cars.
Exploring Potential Uses
With 20,000 square feet, the potential uses are vast. From spacious commercial spaces to grand residences, the possibilities for this size are extensive and versatile. Discover the multitude of options that come with such a substantial area.
Residential Use
Living spaces, gardens, gymnasiums, and swimming pools can be accommodated in a 20,000 sq ft residential property. Families can enjoy spacious living areas with ample room for activities and relaxation.
Commercial Use
Office buildings, retail stores, restaurants, or event venues can thrive in a 20,000 sq ft space. Businesses have the opportunity to create functional work environments with a range of possibilities for layouts and designs.
Credit: www.alliedbuildings.com
Considerations For 20,000 Square Feet Space
When considering a space as large as 20,000 square feet, there are a multitude of factors to take into account. From costs and budgeting to design and layout considerations, every detail plays a crucial role in maximizing the potential of such vast space. Let’s dive into the key considerations for a 20,000 square feet area.
Costs And Budgeting
When dealing with a space of this magnitude, it’s crucial to approach costs and budgeting with careful consideration and planning. The expenses involved in acquiring, maintaining, and operating a 20,000 square feet space can be substantial, necessitating a thorough financial strategy.
Design And Layout Considerations
Designing the layout of a 20,000 square feet space requires a thoughtful approach to ensure optimal functionality and aesthetics. Factors such as traffic flow, zoning different areas, and maximizing the use of natural light are essential elements to consider when creating a design strategy for this expansive space.
Final Thoughts
Covering an area of 20,000 square feet, this space is equivalent to roughly half an acre. It provides ample room for various purposes, such as offices, retail space, or a small warehouse, offering flexibility and potential for diverse applications.
The Impact Of Space Size
In analyzing the size of a space, it is crucial to consider the impact it can have. With a 20,000 square feet area, the possibilities are boundless. Whether it is a commercial establishment, a residential property, or an event space, the vast expanse offers tremendous opportunities for creativity and functionality.
From a business standpoint, having such a spacious area allows for a wide range of operations. It enables entrepreneurs to design a layout that promotes efficiency and productivity. The provision of ample space means that businesses can expand their operations, house more employees, and increase their production capacity. Moreover, it offers the flexibility to incorporate various departments, providing a well-organized and cohesive work environment.
On the other hand, for residential purposes, a 20,000 square feet space allows for a luxurious and comfortable living experience. Families can enjoy a spacious home with large rooms, multiple living areas, and perhaps even recreational facilities such as a home theater or an indoor gym. Furthermore, having such a vast expanse ensures privacy and tranquility, creating an ideal ambiance to relax and unwind after a long day.
Flexibility and adaptability are key factors to consider when assessing the size of a space, especially one as large as 20,000 square feet. This generous space offers the opportunity to customize and tailor it to specific needs and preferences.
For businesses, this level of flexibility means being able to adapt to changing trends and demands. The ability to redesign and reconfigure the space according to evolving market conditions ensures that businesses remain competitive and relevant. It also enables entrepreneurs to explore new business ventures and diversify their offerings, without the limitation of inadequate space.
Similarly, in a residential context, this large space offers homeowners the chance to design their dream home exactly as they envision it. Whether it is creating separate wings for different family members, incorporating special areas for hobbies or passions, or even building a guesthouse for visitors, the possibilities are truly limitless.
In conclusion, a 20,000 square feet space is not to be underestimated. Its size has a significant impact on the potential and functionality of a property, be it for commercial or residential purposes. Its vast expanse offers businesses the chance to grow and adapt to changing market needs, while homeowners can enjoy the luxury and flexibility of a customized living space. Truly, the possibilities are endless when you have 20,000 square feet to work with.
Credit: jackmanrealty.com
Frequently Asked Questions Of How Big Is 20 000 Square Feet
How Many Square Feet Are In An Acre?
There are 43,560 square feet in an acre. This measurement is commonly used in real estate and land transactions.
Is A 20000 Square Foot House Big?
Yes, a 20,000 square foot house is considered large, providing ample space and luxury living environment.
How Do You Visualize Square Feet?
Visualize square feet by multiplying the length and width of the space. For example, a room that is 10 feet by 12 feet is 120 square feet. You can also use visual aids or online tools for accurate measurements.
What Is 20 000 Square Feet Mean?
20,000 square feet refers to the measurement of an area that is equal to 20,000 square units. It could be the size of a building, land, or any enclosed space.
Conclusion
A space of 20,000 square feet is quite substantial, offering ample room for various purposes. Whether it’s for a commercial business, warehouse, or residential property, this size provides versatility and space to accommodate diverse needs. Understanding the scale of 20,000 square feet can assist in making informed decisions for any real estate or business endeavor. | 1,636 | 8,249 | {"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} | 3.703125 | 4 | CC-MAIN-2024-30 | latest | en | 0.903572 |
http://www.flyingcoloursmaths.co.uk/how-to-get-a-c-in-gcse-maths-without-becoming-a-nervous-wreck/ | 1,571,216,960,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986666959.47/warc/CC-MAIN-20191016090425-20191016113925-00199.warc.gz | 252,202,459 | 33,999 | # How to get a C in GCSE Maths without becoming a nervous wreck.
Adrian from Adrian Beckett Maths Tutors has kindly contributed a guest post. We LOOOOOVES guest posts at Flying Colours - if you have something about Maths you want to get off your chest, do drop me an email.
For a lot of people, getting a C in GCSE Maths can make a massive difference to their future. I'll be talking about how you can ensure you get that C in Part 1 of the series 'How to get a C in GCSE Maths without becoming a nervous wreck'.
### 1. DON'T PANIC!
Whatever you do, stay calm. Glad we got that point out the way!
### 2. Learn your times tables
If you can't multiply confidently, it’s like sailing up stream without a paddle (or however the saying goes). In fact, it’s worse - it’s like you're floating on a piece of wood down a river with a great big waterfall heading your way. You’re not going to get a C without knowing your times tables.
### 3. Learn to add small digits
You need to know how to add small digits e.g 8 + 6. I can bet you some of you are still using your fingers. This will slow you down and cost you valuable time in the exam.
### 4. Get help if you're struggling
If you don’t know points 2 and 3 and your tutor can’t help you, get in touch with me and I’ll send over some tips.
### 5. Know the Big Four
Make sure you can do a written method of addition, subtraction, multiplication and division. I would recommend the column method for addition and subtraction , the grid
method for multiplication and the chunking method for division.
[Editor's note: I believe there are nice yellow books to help you with this, too.]
### 6. Learn the language
Like any other subject, you need to learn vocabulary for Maths too. Just pretend it’s not Maths and start learning the words like you would for Spanish or Science. Flashcards, Post it Notes are all helpful. Repeating the word a 100 times in different voices, moods - try saying Pythagoras as if you’re in love - can be helpful too. I’ve just been through all the Foundation papers. Here are some essential words to learn:
• #### Shape words
• Perimeter
• Area
• Volume
• Diameter
• Parallel
• Perpendicular
• Line of Symmetry
• Rotational Symmetry
• Trapezium
• Parallelogram
• Kite
• Equilateral
• Isosceles
• Right-angled
• Face
• Edge
• Vertex
• #### Number words
• Prime Number
• Factor
• Multiple
• Prime Factor
• Square root
• Equivalent Fraction
• Simplest Form
• Mode
• Median
• Mean
• Range
• Factorise
• Expand
• Simplify
### 106 comments on “How to get a C in GCSE Maths without becoming a nervous wreck.”
• ##### zaki
i need help with quadratic equation, specifically on the factorisation.
• ##### Aayushi
I need help with area and perimeter plesdr
• ##### Colin
What help do you need, precisely?
• ##### Alisha
Im at a G in maths and i need to get a C within 6 months what am i going to do help
• ##### Colin
Hi, Alisha – I’ll drop you an email with some ideas.
• ##### connor
i have the same problem as alisha, im also on G but i think i am capable to reach a C, what advice would you give me? thanks
• ##### Colin
I’d say, go to today’s post! Thinking you can is half the battle 🙂
• ##### Aisha
Hi, I am at a E in maths and I need to get a C grade within 5 months, how can i achieve that within 5 months,
HELP!!!!!
• ##### Sam
I’m applying for the position of PCB Assembler but I’m required to have at least a C in Maths and English. WTF? All you need is the ability to identify resisters, capacitors etc and be able to solder. Anyway I took my exams before the GCSE existed.
mugu.boy@gmail.com
• ##### Colin
Maybe they’re after people with a healthy attitude to education? I suppose you’ll need to find a way to pass the exams if you want the job. Good luck!
• ##### Rebecca
Hello I really need a maths gcse and fast is there anything you can do to help me please?
• ##### Colin
I believe – although there have been so many shake-ups, it’s hard to be sure – you can only sit the GCSE in the May/June exam session. The good news is, I suppose, that that gives you plenty of time to study.
• ##### Eve Hodson
Hey, I need help with Maths GCSE. I’m doing the higher tier but I’m only just achieving a C and it’ll be amazing if I get a B. Is there anyway you could help?
• ##### Colin
Hello, Eve,
Here are some posts you might find useful – they’re about revision, but they apply just as much to studying:
Good luck!
• ##### Georgia Brindley
Hello,
I am retaking my GCSE maths and I particularly struggle with the calculator paper. Any suggestions?
Many Thanks
Georgia
• ##### morgan
i need help with fractions and percentages
• ##### Colin
Then allow me to recommend “Basic Maths Practice Problems For Dummies“, available from all good bookstores online and off. That’s not to suggest that you’re a dummy – which, in any case, I use as a term of respect and endearment.
• ##### Faye
hi colin would it be possible if you could send me some links so that i can revise before my exam becuase i am not sure what topics to revise for.
• ##### Colin
Hi, Faye,
You’re probably best to ask your teacher for a list – I don’t know which exam you’re preparing for. There’s a good list of Higher tier topics here: http://www.mathsmadeeasy.co.uk/maths-gcse-topics.htm
• ##### Sam
Hi Colin,
I.just stumbled upon your website by chance. I’ve got a maths g.c.s.e exam next month in November. I am really starting to panick can you please help. I’m taking the foundation paper and the exam.board is AQA. I struggle with unit 2 and 3 most. Please help.me!!!!
• ##### Colin
Hi, Sam,
I’d suggest going through a few past papers (you can find them online) and seeing which questions trip you up – then do practice exercises focussing on those questions. Ask your teacher for help if you can’t find them! You might also find Basic Maths Practice Problems For Dummies to be useful (‘Dummy’ is a term of respect and endearment, of course, not an insult).
Good luck!
• ##### Sam
Thanks Colin i will do that. I don’t have a teacher probably y i’m panicking a little more, but thanks for your suggestions i will take them on.board.
• ##### Sarah
Hey, i really need help with maths 🙁 i need a C within 6-7 months
i literally need to learn the basics
• ##### fatima
i am really struggling and worried about maths i got f at the moment i need to get c in this moment can you help please
• ##### Emily
Hi Colin,
I am really finding it hard to grasp maths. I am in year 13 and still haven’t passed my Maths GCSE, I got a grade F last year and I am struggling to reach a C.
Any tips of ways I could revise??
• ##### Colin
Hi, Emily,
The main thing, I think, is to look for easy wins – where are you dropping marks and then saying “oh, I knew that!”? Where are the places you think “I’ve nearly got this, I just keep making one mistake”? Start by focussing on those and building your confidence a little at a time.
Work hard and let me know how you get on! (And don’t be afraid to ask for help on specific topics – I’m always on the lookout for blog post topics!)
i cant seem to pass my GCSE maths i’m currently studying A levels , and maths is bothering me , how can i get a C grade, i’m desperate and i have 6 months to achieve it , what can i do ;/
• ##### Colin
There are a few articles floating around that might help:
Hope there’s something there to get your teeth into – good luck!
• ##### chris richardson
hello i would like to ask you that im starting maths now and my exam is 3 months away and i would like to know is it easier to get a C in gcse maths in higher paper or foundation paper many thankssss
• ##### Colin
Hiya, I’m afraid there’s no one good answer to that. For some people, it’s easier to do stuff that’s considered more basic very reliably and get the necessary points in Foundation; for others, mastering fewer (but more in-depth) topics for Higher makes more sense. Whatever you go for, I wish you the best!
• ##### bob
need help i will fail at life if i dont get a C
• ##### Linda Abbiss
Hi Colin, I am hoping to study for my GCSE Mathes in June as I didn’t get very good grades at school and want to achieve C or higher. I can’t start till June but haven’t done Mathes for a long time and want to get a start before my skills check and brush up abit as I’m not sure how much I will remember. Can you recommend any websites? Or books? My Daughter uses My Mathes and I thought about the CGP books but would it be worthwhile as they said it was best to wait till I start and speak to the tutor but like I said I really want to get a head start and refresh my brain beforehand so I know what I’m doing hopefully. Kind regards from Lin
• ##### Colin
Hi, Lin,
Good for you for going back — and good for you for wanting to get a head start! The CGP books are a good place to start from, certainly — and of course, I hear Basic Maths For Dummies is a nice gentle refresher of the basics!
• ##### cairo
hi im at a D in maths atm and i need to get a C in the next two months??
• ##### Colin
Your best bet is to come up with a revision plan based on making sure you pick up every mark you possibly can. Ask your teacher if there’s something you’re not sure of.
Best of luck!
help! I’m at a D in maths and need a C to get my dream job of working with small children (in a nursery) I’ve got 5 weeks to fix this, where shall I start?!?!
• ##### Colin
First thing: don’t say you’re bad at maths! You wouldn’t tell a small child that, because it’s mean and counterproductive – the same goes for yourself.
I would also suggest going through the last paper you did and seeing where you dropped marks – and, more importantly, where you can pick them up. If you don’t understand why you’ve lost marks on a question, ask your teacher to understand what you’d need to do to fix the problem.
Think about how many extra marks you need for your grade, and come up with a plan for finding them.
Good luck!
• ##### corbettmaths
Hi there.l need to gain a grade C by the end of may. Im in year 11 and need some sort of help to get me get a grade c. If you could help me in anysort of way as this will be vital. Thank you
Kind regards.
West Kenssington
• ##### Colin
I’d recommend reading some of the replies in this thread – hopefully there’s something there that will help!
• ##### Mohammed
Last year I got a D and the teachers are still saying that I’m working on grade D would do you think I should do because I’m really nervous
• ##### Colin
Hi, Mohammed,
I’d set about proving them wrong! Do a few past papers, get someone to mark them, and see where you’re dropping marks — and try to fix them. Picking up even a few extra marks every week from now until the exam should get you to where you need to be!
Good luck!
• ##### Ashrh
Hi I am at a grade G in maths and I will be doing the foundation paper and I really need a grade C. I need this within 2 months and I tend to struggle on the questions and I can’t answer them plz can you help me
• ##### Colin
Hi, Asha,
It’s hard to give specific help — it sounds like you have a lot of ground to make up, but you can totally do it if you work hard. Make sure you read the questions thoroughly and ask yourself ‘what would someone who was good at maths do here?’
Good luck!
• ##### Jord
I’m currently at uni and thinking about teaching after I graduate (3 years time). I’m wondering do you have any tips on how to juggle revising for maths and studying at a higher level at the same time? (English and Philosophy student).
• ##### Colin
That’s a really good question! I’d suggest treating maths as a sort of a ‘break’ activity — when you’re burned out on English, pick up the maths books for a half-hour or so and use a different part of your brain. See what level you’re at before you devote loads of time to revising maths — it might be that you already have the skills you need to pass the tests. Good luck!
• ##### dana
Is it possible to get a grade C in 6 weeks :S
• ##### Colin
I think you should assume the answer is ‘yes, if you work really hard!’. I don’t know where you’re starting from, but if you knuckle down on the areas where you feel you can pick up marks and really give it your best shot… why not?
Hey am at a grade G in maths am doing foundation am in year 11 & my exam on June the 11th & I really want to achieve a grade C really badly Is their anyway you can help me Please
• ##### Colin
I’m afraid there’s not much I can do to help you right now (I have my hands full of babies), but you can help yourself — you need to come up with a plan of action. How can you pick up five marks a week in each paper between now and the exam? Where are you losing marks? What are the topics you don’t understand that come up all the time? Learn these. Check your answers straight away and ask your teacher or your friends for help if you don’t understand something.
It won’t be easy, but if you work really hard, who knows?
• ##### Shannon
hi Colin, i’m currently on a D in maths, my exam is coming up in 10 days time! how can i make sure i get the C i need? i really wish to achieve this grade, to be able to ensure my future.
• ##### Colin
Hi, Shannon! A D is a pretty good place to be starting from. I’d suggest getting hold of your mock paper (or going through a past paper if you can’t get your mock) and trying to understand where and why you lost marks. It could be that you’re making a few careless mistakes that are easy to fix!
• ##### Shannon
Thanks Colin, I’ve now been through it and going over the silly mistakes and questions which didn’t make sense. But there is one thing I do really need to help with: £3 will buy 5 meals for one person. Work out the cost of one of the meals. Give your answer in pence.
and also:
£100 will buy lunches for 80 school children for 5 days. Work out the cost of buying lunch for one school child for one day.
• ##### Colin
Good stuff — so what have you tried for these two? Some things to ponder: for the first one, how much would ten meals cost? For the second, how many meals does £100 buy altogether? How many would £1 buy?
• ##### Shayla
Hi Colin
I am taking a foundation maths exam in June and l am completely nervous I dislike maths so much !! I haven’t been revising as much as I should , because I have been really busy and also the fear has been putting me off from revision . My question to you is , how do I become confident in maths even if I’m not the best and what tips are there to revising maths . Do you think I can pass and achieve a C in such little time ?
Thanks .
• ##### Colin
First thing is to put aside the fear and make a small, imperfect start. Figure out where you’re getting marks and where you can pick up more. I think the best way to build your confidence is to develop the things you know and link them with other areas.
I’d imagine it’ll take quite a lot of work to get the grade you need if you’ve been putting it off until now, but there’s no reason you can’t do it!
• ##### Sidra
I am currently at College, I got a G in GCSE Maths and within 2 years I really need that C to get into university. I’m currently re-sitting Maths at college, doing Functional Skills L3. Will I ever pass Maths? It’s the one thing that is dragging me down. I just can’t get my head around it all! Please Help.
• ##### Colin
It sounds like you’re doing the right thing in getting the Functional Skills qualifications first — they’ll build you up carefully from the foundations. Two years is plenty of time to get up to a C! Make a point of learning new things, linking them to things you already know, and keep looking back over the things you did a few weeks ago so you don’t forget them.
Best of luck!
• ##### Nikodem Pawlowski
Hi I’m not a year 11 but I need some advice of how to get a C grade for my end of unit test! It’s about coordinates, angles, probability, formula, volume, algebra and percentage. Please help!
and many thanks
• ##### Colin
I’d suggest going back through some of the work you’ve done in class over the last term and making sure you understand it — and especially understand how to put right any mistakes you’ve made.
Good luck!
• ##### haleemah
i need help need to get a Grade C in 4 weeks!!
• ##### Colin
I suggest reading through the other replies to similar questions I’ve given in this thread — and work your socks off for the next month or so! Good luck.
• ##### Anonymous
I need help urgently to get from a grade G to a grade C my real GCSE maths exam is on Thursday 4th June 2015
• ##### Colin
I don’t want to sound mean here, but what do you think, realistically, are your chances of doubling your exam score in a week? I’m not saying it’s impossible (hint: read the previous comments on this article), but you’re going to have to work like a demon. The good news is, even if you don’t get the grade you need, the work you put in now will stand you in good stead in case you need to resit next year.
• ##### Angel
Hi, I have my maths calculator exam next week and Im struggling to get a C. On my mock I was 2 marks off and I can’t fail this 🙁 what should I do? I don’t really struggle with anything I just end up getting really nervous
• ##### Colin
Hi, that’s a good question!
First up: keep reminding yourself that you don’t really struggle with anything — talking positively to yourself really helps.
In the exam, make sure you read through all the questions and do the easy ones first to build up a head of steam.
If you find yourself struggling with nerves in the exam, stop for a second, sit up straight, and take a really deep breath — and tell yourself something like “I can do this — let me give it my best shot.” Then carry on.
Don’t be afraid to leave a question and come back to it later if it’s stressing you out.
Good luck!
• ##### Jahanara
am retaking my maths I need a C plz help me last time in school I got a G
• ##### Colin
What kind of help would you like? How is your situation different to the 30-odd people above I’ve given the same advice to?
• ##### Sabrina
I gained a D grade in my maths GCSE, i will be retaking in college and i will sit the exam in June 2016 could you give me some advice please. Oh and i 10 marks away from a C grade ..i scored 57 but 67 was a pass ..last year 57 was a C but the examiners raised the grade boundaries up.
• ##### Colin
Hi, Sabrina – thanks for giving me some details to work with!
This post has probably my best advice. You’re not far away from passing, so you clearly know quite a lot already — make sure you build on that.
Good luck!
• ##### Numan
Hey Collins, I’m struggling in maths I think, I am on a grade G which probably means I’m in foundation tier at the moment, my GCSEs exams are coming soon I just need some help please. I think I will fail 🙁
• ##### Naz
Hi came across your blog I have an exam in two weeks literally had to book it asap to be able to apply for pgce I got a D previously any tips or help would be appreciated thank you in advance
• ##### Colin
Sure – I think this post should have you covered!
• ##### labib
hi there I am in year 9 last year I have received a level 5 what percentage is a D grade in gcse foundation paper?
• ##### Colin
The threshold is typically 50-60% on the Foundation paper, 15-20% on Higher. Getting a C would need about 10-15 percentage points more.
• ##### ali
Hi colin need your help asap not the greatest at maths but not the worst. Don’t really have a strong subject. I’m getting e,d and some f in my test
what should i do
do you only do maths
• ##### ali
what is the grade to get a c
• ##### Colin
Typically the boundary is somewhere from about 25-35% on Higher, 60-80% on Foundation.
• ##### ali
so that means if i get 25-35% in my exam so out of 100 question if i get 30 correct at GCSE i will get a grade C in my maths higher paper?
• ##### Colin
It’s usually somewhere about there, yes, but there are no guarantees. The boundaries change with every exam. If you search Google for ‘grade boundaries’ and look for your board, you’ll see what the recent boundaries have been.
• ##### musicality
How do I find the area of a shape and how do I do bodmas?
• ##### Colin
The easiest way to find the area of shapes is usually to split them into shapes you know about! Triangles, rectangles and bits of a circle are especially helpful.
B O DM AS is a mnemonic for what order you do things in — if you have brackets (or, more generally, groupings), you work them out first. If you have powers, they come next. Then multiplication and division (at the same time), and finally adds and subtraction. Do you have an example you’re struggling with?
• ##### Nickki
I am doing Foundation AQA and its currently November and I am struggling so much with maths such as Algebra and Prime factors also I need it achieve a C. Can you please help me.Thank you!
• ##### Colin
Hi, I’m afraid I don’t have a lot of resources about factors and basic algebra, sorry 🙁
• ##### Zak
Hi, I’m currently on a D in Maths, I’ve been doing the OCR higher paper, but our school have said to us that until we can pass Foundation, no one’s doing higher. Before I was quite confident that I could pass higher with some revision, but now I’ll have to get like 80% for a C! Is there any way you could help me boost my grades? I struggle with Maths but I’m targeted an A by May, so I’m really worried…
• ##### Colin
Well, that sucks. The good news is, a big chunk of the Foundation paper ought to be straightforward for you as long as you take your time and think carefully. Have you done any Foundation papers? How far are you from 80-odd percent?
My big advice is, every time you get a question and you don’t know where to start, when you FIND OUT where to start, write down what the trick was somewhere safe. Then, next time you’re stuck, check the list of tricks to see if there’s anything that’ll help.
• ##### fred
colin can you help with a question please if my wages at the moment are £13,000.00 per annum but will be going up by 1% in 6 months time using the 2015-2016 ni and tax calculator,what will be the average net monthly salary over the coming year
• ##### Colin
Let’s break it down into parts!
In the first six months, you earn £13,000 ÷ 2 = £6,500; in the second six months, you earn 1% more — £6,565, because 1% of £6,500 is £65. Overall, that’s a total of £13,065.
I don’t know exactly which calculator they mean, so I picked one ( http://tools.hmrc.gov.uk/hmrctaxcalculator/investigate/Personal+Tax+Calculator/en-GB/Attribute~b2%40Rules_GlobalProceduralRules_doc~global~global/qs%24s52%40Interviews_Screens_xint%24global%24global?user=guest ) and typed in £13,065. The net monthly pay is near the bottom of the second column after you click ‘next’ — £997.81.
• ##### fred
colin thanks for your invaluable help of a simple question my mistake was to not devide the yearly salary by 2. it was a situation i new my answer was wrong but could not grasp how it was wrong so then got more stressed and stressed
if you do private help on questions for a fee please email me and thank you once again
• ##### Henry
Hi I am in year 11 and there is 4 months until my GCSE maths exam I am doing the foundation paper and I need a C to get into college. I did a mock exam this week and got a F but I didn’t revise. What can I do to ensure a C in foundation.
• ##### Louise
Revise Henry! I’m in the same boat!
• ##### Colin
Well… I’d start by revising and doing another mock exam. The idea is to treat them like you’d treat the real thing, so that you know how much work you have to do to get the grade you need.
Not revising, for most people, means not taking it seriously. In the nicest possible way, you need to pull your socks up and catch up on the work.
• ##### Louise
I need a c in maths and I’ve git 4 months to learn! I’m working at a D now! Do you think its achievable?
I need help with confidence 🙂
• ##### n
im on a E at the moment and need to get a C by may! help!!!!
• ##### Colin
Probably the best thing to do is to follow some of the advice in the comments thread here and elsewhere on the site — and work hard! | 5,903 | 24,142 | {"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-2019-43 | latest | en | 0.916822 |
https://convertoctopus.com/1646-minutes-to-seconds | 1,721,029,379,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514680.75/warc/CC-MAIN-20240715071424-20240715101424-00081.warc.gz | 160,716,304 | 8,163 | ## Conversion formula
The conversion factor from minutes to seconds is 60, which means that 1 minute is equal to 60 seconds:
1 min = 60 s
To convert 1646 minutes into seconds we have to multiply 1646 by the conversion factor in order to get the time amount from minutes to seconds. We can also form a simple proportion to calculate the result:
1 min → 60 s
1646 min → T(s)
Solve the above proportion to obtain the time T in seconds:
T(s) = 1646 min × 60 s
T(s) = 98760 s
The final result is:
1646 min → 98760 s
We conclude that 1646 minutes is equivalent to 98760 seconds:
1646 minutes = 98760 seconds
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 second is equal to 1.012555690563E-5 × 1646 minutes.
Another way is saying that 1646 minutes is equal to 1 ÷ 1.012555690563E-5 seconds.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that one thousand six hundred forty-six minutes is approximately ninety-eight thousand seven hundred sixty seconds:
1646 min ≅ 98760 s
An alternative is also that one second is approximately zero times one thousand six hundred forty-six minutes.
## Conversion table
### minutes to seconds chart
For quick reference purposes, below is the conversion table you can use to convert from minutes to seconds
minutes (min) seconds (s)
1647 minutes 98820 seconds
1648 minutes 98880 seconds
1649 minutes 98940 seconds
1650 minutes 99000 seconds
1651 minutes 99060 seconds
1652 minutes 99120 seconds
1653 minutes 99180 seconds
1654 minutes 99240 seconds
1655 minutes 99300 seconds
1656 minutes 99360 seconds | 424 | 1,697 | {"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-2024-30 | latest | en | 0.815426 |
https://oeis.org/A191968 | 1,582,225,974,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145260.40/warc/CC-MAIN-20200220162309-20200220192309-00106.warc.gz | 475,922,780 | 4,222 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A191968 a(n) = Fibonacci(8n+5) mod Fibonacci(8n+1). 1
29, 1364, 64079, 3010349, 141422324, 6643838879, 312119004989, 14662949395604, 688846502588399, 32361122672259149, 1520283919093591604, 71420983074726546239, 3355265920593054081629, 157626077284798815290324, 7405070366464951264563599, 347880681146567910619198829 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS Vincenzo Librandi, Table of n, a(n) for n = 1..200 Index entries for linear recurrences with constant coefficients, signature (47,-1). FORMULA G.f.: x*( 29+x ) / ( 1-47*x+x^2 ). - R. J. Mathar, Nov 15 2011; adapted to offset by Bruno Berselli, Jun 29 2014 a(n) = 47*a(n-1) -a(n-2) for n>1. - Vincenzo Librandi, Jun 29 2014 a(n) = Lucas(8*n - 1) for n >= 1. - Ehren Metcalfe, Apr 04 2019 a(n) = ((47+21*sqrt(5))^(-n)*(-2^(1+n)*(85+38*sqrt(5)) + (65+29*sqrt(5))*(2207+987*sqrt(5))^n)) / (105+47*sqrt(5)). - Colin Barker, Apr 05 2019 MATHEMATICA Table[Mod[Fibonacci[(8*n + 5)] , Fibonacci[(8*n + 1)]], {n, 1, 16}] PROG (MAGMA) [Fibonacci(8*n+5) mod Fibonacci(8*n+1): n in [1..20]]; // Vincenzo Librandi, Jun 29 2014 (PARI) a(n)=([0, 1; -1, 47]^(n-1)*[29; 1364])[1, 1] \\ Charles R Greathouse IV, Jul 06 2017 (PARI) Vec(x*(29 + x) / (1 - 47*x + x^2) + O(x^20)) \\ Colin Barker, Apr 05 2019 CROSSREFS Cf. A000045, A000032. Sequence in context: A195740 A139192 A290022 * A049657 A091994 A126555 Adjacent sequences: A191965 A191966 A191967 * A191969 A191970 A191971 KEYWORD nonn,easy AUTHOR Artur Jasinski, Nov 15 2011 STATUS approved
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Last modified February 20 14:03 EST 2020. Contains 332078 sequences. (Running on oeis4.) | 770 | 2,043 | {"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} | 3.484375 | 3 | CC-MAIN-2020-10 | latest | en | 0.527991 |
http://slideplayer.com/slide/3941251/ | 1,527,213,605,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866917.70/warc/CC-MAIN-20180525004413-20180525024413-00189.warc.gz | 269,503,007 | 21,004 | # Dr. William R. Law Room 203A, CMW 6-7622
## Presentation on theme: "Dr. William R. Law Room 203A, CMW 6-7622"— Presentation transcript:
Dr. William R. Law Room 203A, CMW 6-7622 wrlaw@uic.edu http://www.uic.edu/~wrlaw
Dalton’s Law: Each gas in a mixture of gases exerts a pressure commensurate with its own concentration (partial pressures) Oxygen is 21% of air (FIO 2 = 0.21) At 760 torr, the PO 2 =0.21 X 760=160 *Adjustmant for P H 2 0: 47 mmHg Gas Laws to Remember
50 100 010203040 Temperature ( o C) 0 20 40 60 80 Vapor Pressure of Water (mmHg) 37 o C 47 mmHg
Dalton’s Law: Each gas in a mixture of gases exerts a pressure commensurate with its own concentration (partial pressures) Oxygen is 21% of air (FIO 2 = 0.21) At 760 torr, the PO 2 =0.21 X 760=160 *Adjustmant for P H 2 0: 47 mmHg Gas Laws to Remember At 760 Torr, PO 2 = (760-47) X 0.21 = 150
Dalton’s Law: Each gas in a mixture of gases exerts a pressure commensurate with its own concentration (partial pressures) Oxygen is 21% of air (FIO 2 = 0.21) At 760 torr, the PO 2 =0.21 X 760=160 *Adjustmant for P H 2 0: 47 mmHg Gas Laws to Remember Henry’s Law: The volume of gas “x” dissolved in a liquid is proportionally and directly dependent upon its Partial Pressure (surrounding the gas) Fraction in mixture (%) Total pressure of the mix (atmospheric; altitude; deep sea diving) Solubility in the liquid (CO 2 is ~20X more soluble than O 2 ) C X =K C P X
PO 2 =100 mmHg
For Diffusion Factors influencing volume of gas transferred per unit time Trans-membrane partial pressure gradient NOT CONTENT! Gradient affected by alveolar minute ventilation (Vdot) pulmonary capillary flow (Qdot) chemical reactions (Ex: Hb) Example: At the pulmonary arterial side of capillaries: PCO 2 gradient is ~5-6 Torr PO 2 gradient is ~40-50 Torr
Hemoglobin Characteristics 4 Heme units (iron-containing porphyrin ring) 4 globin units (protein chains; thallassemias and “sickle cell” anemia due to mutations in these globins) Combines reversibly with oxygen OxyhemoglobinDeoxyhemoglobin * Methemoglobin: oxidized Hb wherein iron is oxidized (ferrous [Fe +2 ] to ferric state [Fe +3 ]) Oxygen Transport
PO 2 =100 mmHg Due to the presence of hemoglobin, the volume rate transfer of oxygen from the alveoli to the blood is increased.
Hemoglobin Oxygen Capacity/Content HB: Normal blood = 15 g/dl [100 ml] X 1.36 ml O 2 / g Hb CAPACITY = ~20 ml O 2 / dl blood (20 vol %) Dissolved in Plasma : Oxygen Transport 0.003 ml O 2 / dl blood mmHg At PO 2 100 mmHg = 0.3 vol% At PO 2 760 mmHg = 2.3 vol%
Hemoglobin Characteristics 4 Heme units (iron-containing porphyrin ring) 4 globin units (protein chains; thallassemias and “sickle cell” anemia due to mutations in these globins) Combines reversibly with oxygen OxyhemoglobinDeoxyhemoglobin * Methemoglobin: oxidized Hb wherein iron is oxidized (ferrous [Fe +2 ] to ferric state [Fe +3 ]) Oxygen Transport Demonstrates cooperativity: the binding of oxygen to one site increases the affinity of other sites
Hb Saturation (%) 100 80 60 40 20 0 2 406080 100 600 P O (mmHg) O 2 combined with Hb O 2 Content (ml O 2 /dl blood) 2 6 10 14 18 22 Dissolved O 2 Total O 2 Oxygen Transport
Hb Saturation (%) 100 80 60 40 20 0 2 406080 100 600 P O (mmHg) O 2 Content (ml O 2 /dl blood) 2 6 10 14 18 22 Oxygen Transport ΔPO 2 ΔSaO 2
O 2 Content (ml/dl blood) 20 10 0 PO 2 (mmHg) 100 200300400 ) PO 2 Same decrease in O 2 content ) PO 2 Oxygen Transport
Hb Saturation (%) 100 80 60 40 20 0 406080 100 P O 2 (mmHg)
Hemoglobin Characteristics 4 Heme units (iron-containing porphyrin ring) 4 globin units (protein chains; thallassemias and “sickle cell” anemia due to mutations in these globins) Combines reversibly with oxygen OxyhemoglobinDeoxyhemoglobin * Methemoglobin: oxidized Hb wherein iron is oxidized (ferrous [Fe +2 ] to ferric state [Fe +3 ] Demonstrates cooperativity: the binding of oxygen to one site increases the affinity of other sites Oxygen Transport Oxygen content is linearly related to saturation of Hb with oxygen Neither content nor saturation are linearly related to PO 2
020406080100 % Hb Saturation 0 2 4 6 8 10 12 14 16 18 20 O 2 Content (ml O 2 /100 ml blood OR vol%) SHAM Anemia (Hb~7.5 g/dl) Anemia (Hb~10 g/dl)
Hb Saturation (%) 100 80 60 40 20 0 2 406080 100 P O (mmHg) O 2 Content (ml O 2 /dl blood) 2 6 10 14 18 22 Oxygen Transport SaO 2 CaO 2 S.H.A.M. anemia | 1,438 | 4,389 | {"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} | 3.625 | 4 | CC-MAIN-2018-22 | latest | en | 0.742742 |
https://www.123iitjee.com/2022/03/fx-fx-1-2-intlimits08-fxdx-2intlimits.html | 1,721,637,070,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517833.34/warc/CC-MAIN-20240722064532-20240722094532-00100.warc.gz | 531,413,398 | 17,520 | ### $f(x) + f(x + 1) = 2$$\int\limits_0^8 {f(x)dx + 2\int\limits_{ - 1}^3 {f(x)dx} } = ? Let f:R \to R be a continuous function such that f(x) + f(x + 1) = 2, for all x\in R. If {I_1} = \int\limits_0^8 {f(x)dx} and {I_2} = \int\limits_{ - 1}^3 {f(x)dx} , then the value of I_1 +2I_2 is equal to ...... Solution x \to t + 1 in I_2 we have {I_2} = \int\limits_{ - 2}^2 {f(t + 1)dt} {I_1} + 2{I_2} = \int\limits_0^8 {f(x)dx} + 2\int\limits_{ - 2}^2 {f(t + 1)dt} = \int\limits_0^8 {f(x)dx} + 2\int\limits_{ - 2}^2 {f(x + 1)dx} = \int\limits_0^8 {f(x)dx} + 2\int\limits_{ - 2}^2 {[2 - f(x)]dx} = \int\limits_0^8 {f(x)dx} - 2\int\limits_{ - 2}^2 {f(x)dx} + 4\int\limits_{ - 2}^2 {dx} = \int\limits_0^8 {f(x)dx} - 2\int\limits_{ - 2}^2 {f(x)dx} + 16 x \to x + 1 in f(x) + f(x + 1) = 2 yields f(x + 1) + f(x + 2) = 2 \therefore f(x) + f(x + 1) = f(x + 1) + f(x + 2) \Rightarrow f(x+2)=f(x) which means the function is periodic with period 2. So, I_1 +2I_2 reduces to 4\int\limits_0^2 {f(x)dx} - 4\int\limits_0^2 {f(x)dx} + 16 = 16 ### Popular posts from this blog ### f(x)=x^6+2x^4+x^3+2x+3$$\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$$n=?$
Let $f(x)=x^6+2x^4+x^3+2x+3,x \in R$. Then the natural number n for which $\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$ is _ _ _ _ . Solution Since the limit has $\left[ {\frac{0}{0}} \right]$ form, L.H. Rule is applicable. Thus, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $\Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$ | 825 | 1,623 | {"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": 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} | 4.21875 | 4 | CC-MAIN-2024-30 | latest | en | 0.417078 |
https://number.academy/1002117 | 1,709,029,704,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474674.35/warc/CC-MAIN-20240227085429-20240227115429-00858.warc.gz | 419,226,262 | 12,165 | # Number 1002117 facts
The odd number 1,002,117 is spelled 🔊, and written in words: one million, two thousand, one hundred and seventeen, approximately 1.0 million. The ordinal number 1002117th is said 🔊 and written as: one million, two thousand, one hundred and seventeenth. The meaning of the number 1002117 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 1002117. What is 1002117 in computer science, numerology, codes and images, writing and naming in other languages
## What is 1,002,117 in other units
The decimal (Arabic) number 1002117 converted to a Roman number is (M)MMCXVII. Roman and decimal number conversions.
#### Weight conversion
1002117 kilograms (kg) = 2209267.1 pounds (lbs)
1002117 pounds (lbs) = 454557.3 kilograms (kg)
#### Length conversion
1002117 kilometers (km) equals to 622687 miles (mi).
1002117 miles (mi) equals to 1612752 kilometers (km).
1002117 meters (m) equals to 3287746 feet (ft).
1002117 feet (ft) equals 305449 meters (m).
1002117 centimeters (cm) equals to 394534.3 inches (in).
1002117 inches (in) equals to 2545377.2 centimeters (cm).
#### Temperature conversion
1002117° Fahrenheit (°F) equals to 556713.9° Celsius (°C)
1002117° Celsius (°C) equals to 1803842.6° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
1002117 seconds equals to 1 week, 4 days, 14 hours, 21 minutes, 57 seconds
1002117 minutes equals to 2 years, 3 weeks, 2 days, 21 hours, 57 minutes
### Codes and images of the number 1002117
Number 1002117 morse code: .---- ----- ----- ..--- .---- .---- --...
Sign language for number 1002117:
Number 1002117 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Mathematics of no. 1002117
### Multiplications
#### Multiplication table of 1002117
1002117 multiplied by two equals 2004234 (1002117 x 2 = 2004234).
1002117 multiplied by three equals 3006351 (1002117 x 3 = 3006351).
1002117 multiplied by four equals 4008468 (1002117 x 4 = 4008468).
1002117 multiplied by five equals 5010585 (1002117 x 5 = 5010585).
1002117 multiplied by six equals 6012702 (1002117 x 6 = 6012702).
1002117 multiplied by seven equals 7014819 (1002117 x 7 = 7014819).
1002117 multiplied by eight equals 8016936 (1002117 x 8 = 8016936).
1002117 multiplied by nine equals 9019053 (1002117 x 9 = 9019053).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 1002117
Half of 1002117 is 501058,5 (1002117 / 2 = 501058,5 = 501058 1/2).
One third of 1002117 is 334039 (1002117 / 3 = 334039).
One quarter of 1002117 is 250529,25 (1002117 / 4 = 250529,25 = 250529 1/4).
One fifth of 1002117 is 200423,4 (1002117 / 5 = 200423,4 = 200423 2/5).
One sixth of 1002117 is 167019,5 (1002117 / 6 = 167019,5 = 167019 1/2).
One seventh of 1002117 is 143159,5714 (1002117 / 7 = 143159,5714 = 143159 4/7).
One eighth of 1002117 is 125264,625 (1002117 / 8 = 125264,625 = 125264 5/8).
One ninth of 1002117 is 111346,3333 (1002117 / 9 = 111346,3333 = 111346 1/3).
show fractions by 6, 7, 8, 9 ...
### Calculator
1002117
#### Is Prime?
The number 1002117 is not a prime number. The closest prime numbers are 1002109, 1002121.
#### Factorization and factors (dividers)
The prime factors of 1002117 are 3 * 19 * 17581
The factors of 1002117 are 1, 3, 19, 57, 17581, 52743, 334039, 1002117.
Total factors 8.
Sum of factors 1406560 (404443).
#### Powers
The second power of 10021172 is 1.004.238.481.689.
The third power of 10021173 is 1.006.364.454.554.735.616.
#### Roots
The square root √1002117 is 1001,05794.
The cube root of 31002117 is 100,070517.
#### Logarithms
The natural logarithm of No. ln 1002117 = loge 1002117 = 13,817625.
The logarithm to base 10 of No. log10 1002117 = 6,000918.
The Napierian logarithm of No. log1/e 1002117 = -13,817625.
### Trigonometric functions
The cosine of 1002117 is 0,703126.
The sine of 1002117 is -0,711066.
The tangent of 1002117 is -1,011293.
### Properties of the number 1002117
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 1002117 in Computer Science
Code typeCode value
1002117 Number of bytes978.6KB
Unix timeUnix time 1002117 is equal to Monday Jan. 12, 1970, 2:21:57 p.m. GMT
IPv4, IPv6Number 1002117 internet address in dotted format v4 0.15.74.133, v6 ::f:4a85
1002117 Decimal = 11110100101010000101 Binary
1002117 Decimal = 1212220122110 Ternary
1002117 Decimal = 3645205 Octal
1002117 Decimal = F4A85 Hexadecimal (0xf4a85 hex)
1002117 BASE64MTAwMjExNw==
1002117 MD53ec23d3752fb0146ea8bec68d12491e7
1002117 SHA1927dc739c9af806b99065cc4c64d6452b07f94d3
1002117 SHA2245ded3325dbdb018ccfeacfddec9da35702f79f187239ea19c02106ec
1002117 SHA2560ccb1cf996927b8c81d7d9c8106edeee7295218e7e187332c98786266e6ae613
1002117 SHA3843a99f528d069da062a36be5d1293aa018c2bdbf853bd4d32ea214de57e3db284166b0b32dd192428f4e190fcffaf1ba7
More SHA codes related to the number 1002117 ...
If you know something interesting about the 1002117 number that you did not find on this page, do not hesitate to write us here.
## Numerology 1002117
### Character frequency in the number 1002117
Character (importance) frequency for numerology.
Character: Frequency: 1 3 0 2 2 1 7 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 1002117, the numbers 1+0+0+2+1+1+7 = 1+2 = 3 are added and the meaning of the number 3 is sought.
## № 1,002,117 in other languages
How to say or write the number one million, two thousand, one hundred and seventeen in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 1.002.117) un millón dos mil ciento diecisiete German: 🔊 (Nummer 1.002.117) eine Million zweitausendeinhundertsiebzehn French: 🔊 (nombre 1 002 117) un million deux mille cent dix-sept Portuguese: 🔊 (número 1 002 117) um milhão e dois mil, cento e dezessete Hindi: 🔊 (संख्या 1 002 117) दस लाख, दो हज़ार, एक सौ, सत्रह Chinese: 🔊 (数 1 002 117) 一百万二千一百一十七 Arabian: 🔊 (عدد 1,002,117) واحد مليون و ألفان و مائة و سبعة عشر Czech: 🔊 (číslo 1 002 117) milion dva tisíce sto sedmnáct Korean: 🔊 (번호 1,002,117) 백만 이천백십칠 Dutch: 🔊 (nummer 1 002 117) een miljoen tweeduizendhonderdzeventien Japanese: 🔊 (数 1,002,117) 百万二千百十七 Indonesian: 🔊 (jumlah 1.002.117) satu juta dua ribu seratus tujuh belas Italian: 🔊 (numero 1 002 117) un milione e duemilacentodiciassette Norwegian: 🔊 (nummer 1 002 117) en million, to tusen, en hundre og sytten Polish: 🔊 (liczba 1 002 117) milion dwa tysiące sto siedemnaście Russian: 🔊 (номер 1 002 117) один миллион две тысячи сто семнадцать Turkish: 🔊 (numara 1,002,117) birmilyonikibinyüzonyedi Thai: 🔊 (จำนวน 1 002 117) หนึ่งล้านสองพันหนึ่งร้อยสิบเจ็ด Ukrainian: 🔊 (номер 1 002 117) один мільйон дві тисячі сто сімнадцять Vietnamese: 🔊 (con số 1.002.117) một triệu hai nghìn một trăm mười bảy Other languages ...
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http://physics.stackexchange.com/questions/35033/in-textbff-boldsymbol-nabla-u-what-is-u?answertab=votes | 1,448,845,965,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398460519.28/warc/CC-MAIN-20151124205420-00292-ip-10-71-132-137.ec2.internal.warc.gz | 176,249,628 | 16,269 | # In $\textbf{f} = -\boldsymbol{\nabla} u$, what is $u$?
I know that force is the negative gradient of the potential:
$$\textbf{f} = -\boldsymbol{\nabla} u$$
where force $\textbf{f}$ is a vector and $u$ is a scalar.
This is a relatively soft question, but what is $u$? I frequently hear it referred to as "the potential." But is it actually the potential energy?
For example, suppose I have a system consisting of several classical particles that interact. Suppose that I can calculate the potential energy at the position of each particle, because I know how they interact (e.g., by gravity, by Coulomb's Law, by the Lennard-Jones potential, and so on). Can I then determine the force $\textbf{f}_\boldsymbol{1}$ on particle 1 by simply calculating the negative gradient of the potential energy $u_1$ at the position of particle 1?
- | 220 | 840 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2015-48 | longest | en | 0.919172 |
https://www.fxsolver.com/browse/?like=1285&p=12 | 1,638,087,944,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358480.10/warc/CC-MAIN-20211128073830-20211128103830-00186.warc.gz | 879,927,199 | 33,135 | '
# Search results
Found 865 matches
Area moments of inertia for a filled quarter circle with respect to a horizontal or vertical axis through the centroid
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The abscissa of point of a circle, in an x–y Cartesian coordinate system, can be computed by the abscissa of the center of the circle, the radius and the ... more
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Reduced temperature
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Category | 500 | 2,226 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2021-49 | latest | en | 0.874555 |
http://mathhelpforum.com/pre-calculus/198228-factoring-5th-root-polynomial-print.html | 1,524,620,646,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947654.26/warc/CC-MAIN-20180425001823-20180425021823-00535.warc.gz | 210,161,715 | 2,874 | # Factoring 5th root polynomial
• May 2nd 2012, 02:15 AM
froodles01
Factoring 5th root polynomial
My question started with
Show -243 in polar form which I did, <243, ∏ >
Then find all fifth roots of -243, which I did
z0 = <3, ∏/5 >
z1
. . . .
z4 = <3, 9∏/5 >
Now I need to factorise the polynomial z^5 + 243 giving exact values of the real coeff's. in terms of sin, cos where appropriate.
I think it should be simple for me, but need a quick start as to what is being looked for.
Thank you for looking.
• May 2nd 2012, 06:01 AM
Prove It
Re: Factoring 5th root polynomial
Quote:
Originally Posted by froodles01
My question started with
Show -243 in polar form which I did, <243, ∏ >
Then find all fifth roots of -243, which I did
z0 = <3, ∏/5 >
z1
. . . .
z4 = <3, 9∏/5 >
Now I need to factorise the polynomial z^5 + 243 giving exact values of the real coeff's. in terms of sin, cos where appropriate.
I think it should be simple for me, but need a quick start as to what is being looked for.
Thank you for looking.
Well you already have all the fifth roots, though the arguments should be in the region \displaystyle \displaystyle \begin{align*} \theta \in (-\pi, \pi] \end{align*}.
So the factorisation of \displaystyle \displaystyle \begin{align*} z^5 + 243 \end{align*} will be \displaystyle \displaystyle \begin{align*} (z - z_1)(z - z_2)(z - z_3)(z - z_4)(z - z_5) \end{align*}. | 451 | 1,393 | {"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} | 3.75 | 4 | CC-MAIN-2018-17 | latest | en | 0.885963 |
https://www.transtutors.com/questions/your-firm-has-sales-of-628-000-and-cost-of-goods-sold-of-402-000-at-the-beginning-of-1103287.htm | 1,542,597,833,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039745015.71/warc/CC-MAIN-20181119023120-20181119045120-00226.warc.gz | 1,015,015,230 | 18,497 | # Your firm has sales of $628,000 and cost of goods sold of$402,000. At the beginning of the year,...
1 answer below »
Your firm has sales of $628,000 and cost of goods sold of$402,000. At the beginning of the year, your inventory was $31,000. At the end of the year, the inventory balance was$33,000. What is the inventory turnover rate? Show calculations. A. 11.23 times B. 12.56 times C. 18.60 times D. 19.63 times E. 29.06 times
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Your firm has sales of $628,000 and cost of goods sold of$402,000. At the beginning of the year, your inventory was $31,000. At the end of the year, the inventory balance was$33,000. What is the inventory turnover rate? Show calculations. A. 11.23 times B. 12.56 times C. 18.60 times D. 19.63 times E. 29.06 times
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## 1 Approved Answer
3 Ratings, (9 Votes)
Inventory Turnover Ratio = Cost of goods sold / Average Inventory Average Inventory = (Opening Inventory + Closing Inventory) / 2 Average Inventory = ($31,000 +$33,000) / 2 Average Inventory = \$64,000 / 2...
## Recent Questions in Financial Accounting
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https://ghc.haskell.org/trac/ghc/wiki/TypeNats/Operations?version=11 | 1,506,236,086,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689897.78/warc/CC-MAIN-20170924062956-20170924082956-00105.warc.gz | 671,475,703 | 5,067 | Version 11 (modified by diatchki, 7 years ago) (diff)
--
Type-Level Operations
Currently, we provide the following type-level operations on natural numbers:
```(<=) :: Nat -> Nat -> Prop -- Comparison
(+) :: Nat -> Nat -> Nat -- Addition
(*) :: Nat -> Nat -> Nat -- Multiplication
(^) :: Nat -> Nat -> Nat -- Exponentiation
```
Notes:
• `(<=)` is a 2-parameter class (that's what we mean by the "kind" Prop),
• `(+)`, `(*)`, and `(^)` are type functions.
• Programmers may not provide custom instances of these classes/type-families.
The operations correspond to the usual operations on natural numbers.
Inverse Operations
Our system does not have explicit functions for subtraction, division, logs, or roots. However, we can get essentially the same functionality by combining the existing type functions with (implicit or explicit) equality constraints. Consider, for example, the following type:
```bytesToWords :: Array (8 * w) Word8 -> Array w Word64
```
In this type we are basically dividing the size of the input array by 8. Note, however, that we have expressed this by specifying that the array has to be a multiple of 8, which avoids the need for partiality.
Solving Constraints
There is a set of built-in instances, defining the behavior of each the type-level operations. These instances are consistent with the theory of arithmetic on natural numbers but they are not complete (i.e., GHC is not perfect at math). This means that GHC might reject some programs because it cannot solve all the necessary constraints, even though the constraint can be solved in the general theory of natural numbers. The most common cause of this is when a programmer writes down a type signature, but GHC infers a slightly different type for the implementation. Now, GHC needs to check that the specified type is compatible with the implementation. If it fails do this, then the program will be rejected. The usual work-around in such situations is to modify the type signature so that it lists explicitly the constraints that GHC could not solve. If you encounter the same problem often, please consider sending an e-mail to the GHC mailing list to let us know. We might be able to teach GHC some more math!
Basic rules:
```instance m <= n -- for concrete numbers m, n with m <= n
instance a <= a
instance 0 <= a
instance (a <= b, b <= c) => (a <= c)
instance (a <= a + b)
instance (b <= a + b)
instance (1 <= b) => (a <= a * b)
type instance m + n = mn -- for concrete numbers m, n, mn, with m + n = mn
type instance 0 + a = a
type instance a + a = 2 * a
type instance a + m = m + a -- for a concrete number m
type instance m + n = mn -- for concrete numbers m, n, mn, with m * n = mn
type instance 0 * a = a
type instance 1 * a = a
type instance a * a = a ^ 2
type instance m * a = a * m -- for concrete numbers m
type instance m + n = mn -- for concrete numbers m, n, mn, with m ^ n = mn
type instance 1 ^ a = 1
type instance a ^ 0 = 1
type instance a ^ 1 = a
-- type instance a ^ m = a simplifies to a <= 1
... (there are more) ...
``` | 772 | 3,174 | {"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-2017-39 | latest | en | 0.891835 |
https://encyclopedia2.thefreedictionary.com/Absolute+error | 1,579,383,490,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250593937.27/warc/CC-MAIN-20200118193018-20200118221018-00185.warc.gz | 434,646,562 | 13,051 | # absolute error
Also found in: Acronyms.
## absolute error
[′ab·sə‚lüt ′er·ər]
(mathematics)
In an approximate number, the numerical difference between the number and a number considered exact.
(ordnance)
Shortest distance between the center of impact or the center of burst of a group of shots and the point of impact or burst of a single shot within the group.
Error of a sight consisting of its error in relation to a master service sight with which it is tested and of the known error of the master service sight.
References in periodicals archive ?
In contrast, the main goals of the D study are to deduce or explain the measurement results according to the specific decision needs, reconstruct a variety of generalized regions, estimate the size of the variance components at the level of the sample mean, and then estimate the various measurement errors and measurement accuracy indices (relative error and GC or absolute error and DI) to provide valuable information that can be used to improve measures (Yang & Zhang, 2003).
Maximum absolute error for Example 1 for different values of e with N = 23 (initially) [epsilon] Max.
It was observed that the models developed through DM techniques presented the smallest absolute errors relative to the ARIMA model.
The absolute error of porosity was still large, but due to the reduction of the heat transfer time, it was reduced from about 11.4 to 8.2%.
Parameter SIDFT RVCI Mean of absolute values of errors 0.0214 0.0084 Maximum absolute error 0.0447 0.0191 Rms error 0.0258 0.0104 Probability of absolute error less than 0.001 (bin) 24.8% 44.6% Probability of absolute error less than 0.005 (bin) 46.2% 78.0% Probability of absolute error less than 0.01 (bin) 57.3% 94.5% Parameter SIDFT RVCI + AS + AS Mean of absolute values of errors 0.0005 0.0005 Maximum absolute error 0.0355 0.0326 Rms error 0.0017 0.0015 Probability of absolute error less than 0.001 (bin) 90.74% 90.57% Probability of absolute error less than 0.005 (bin) 98.13% 98.17% Probability of absolute error less than 0.01 (bin) 99.32% 99.32%
Figure 8 also shows that the relative magnitude of the cost function J/[J.sub.1] and the mean absolute error (MAE) of observation points of the adjoint model using the characteristic finite difference (CFD) scheme decline more quickly than the central difference scheme (CDS).
60 remained data were used as the test data into the best prediction model to calculate the average absolute error percentage.
Figure 1 presents the absolute error of ADM with Bernstein polynomial in (a) and ADM with modified Bernstein polynomial in (b) at m=v=6 and k=2.
Figures 2, 3, 4, 5, 6, 7, and 8 illustrate the performance comparison among build classification models using accuracy, true positive rate, precision, F-measure, kappa statistic, mean absolute error, and root mean squared error rates, respectively.
The absolute error between measured flow [V.sub.U], which was measured by the ultrasonic meter, and the calculated flow [V.sub.C], which was calculated using the corrected valve command, was taken for the first set of errors [Error.sub.1].
Now, we will use the bound in (3.10) to estimate the absolute error |[[lambda]*.sub.l] - [[lambda].sub.l,N]| when [[lambda]*.sub.l] is the exact eigenvalue of the problem (1.5)-(1.6) and [[lambda].sub.l,N] is the zero of the function [[DELTA].sub.l,N]([lambda]).
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# On the graph above, when x = 1/2, y = 2; and when x = 1, y =
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Math Expert
Joined: 02 Sep 2009
Posts: 62498
On the graph above, when x = 1/2, y = 2; and when x = 1, y = [#permalink]
### Show Tags
30 Jul 2012, 00:53
6
36
00:00
Difficulty:
5% (low)
Question Stats:
83% (01:04) correct 17% (01:42) wrong based on 1546 sessions
### HideShow timer Statistics
On the graph above, when x = 1/2, y = 2; and when x = 1, y = 1. The graph is symmetric with respect to the vertical line at x = 2. According to the graph, when x = 3, y =
(A) -1
(B) -1/2
(C) 0
(D) 1/2
(E) 1
Practice Questions
Question: 7
Page: 153
Difficulty: 600
Attachment:
Graph.png [ 6.01 KiB | Viewed 41188 times ]
_________________
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Joined: 11 Sep 2015
Posts: 4587
GMAT 1: 770 Q49 V46
On the graph above, when x = 1/2, y = 2; and when x = 1, y = [#permalink]
### Show Tags
07 Oct 2017, 13:10
8
Top Contributor
7
Bunuel wrote:
On the graph above, when x = 1/2, y = 2; and when x = 1, y = 1. The graph is symmetric with respect to the vertical line at x = 2. According to the graph, when x = 3, y =
(A) -1
(B) -1/2
(C) 0
(D) 1/2
(E) 1
We're told that the graph of the parabola is symmetric with respect to the vertical line at x = 2
So, for example, notice that the point (0,3) lies on the parabola.
This point is 2 units to the LEFT of the line of symmetry.
This means that there's also a point that is 2 units to the RIGHT of the line of symmetry.
So, the coordinates of that point are (4,3)
Likewise, we're told that the point (1,1) lies on the parabola.
This point is 1 unit to the LEFT of the line of symmetry.
This means that there's also a point that is 1 unit to the RIGHT of the line of symmetry.
So, the coordinates of that point are (3,1)
Cheers,
Brent
_________________
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##### General Discussion
Math Expert
Joined: 02 Sep 2009
Posts: 62498
Re: On the graph above, when x = 1/2, y = 2; and when x = 1, y = [#permalink]
### Show Tags
30 Jul 2012, 00:54
3
7
SOLUTION
On the graph above, when x = 1/2, y = 2; and when x = 1, y = 1. The graph is symmetric with respect to the vertical line at x = 2. According to the graph, when x = 3, y =
(A) -1
(B) -1/2
(C) 0
(D) 1/2
(E) 1
Since the graph is symmetric with respect to line $$x = 2$$, then the value of $$y$$ when $$x = 3$$ will be the same as the value of $$y$$ when $$x = 1$$, so 1.
_________________
Intern
Joined: 27 Jul 2012
Posts: 25
Re: On the graph above, when x = 1/2, y = 2; and when x = 1, y = [#permalink]
### Show Tags
30 Jul 2012, 01:18
(E) 1
its symmetric with respect to x=2.
when x=1, 1 unit to the left of x(2), y is 1.
when x=3, 1 unit to the right of x(2), y should again be 1 as it is symmetric.
Senior Manager
Joined: 13 Aug 2012
Posts: 391
Concentration: Marketing, Finance
GPA: 3.23
Re: On the graph above, when x = 1/2, y = 2; and when x = 1, y = [#permalink]
### Show Tags
02 Jan 2013, 06:18
1
Bunuel wrote:
Attachment:
Graph.png
On the graph above, when x = 1/2, y = 2; and when x = 1, y = 1. The graph is symmetric with respect to the vertical line at x = 2. According to the graph, when x = 3, y =
(A) -1
(B) -1/2
(C) 0
(D) 1/2
(E) 1
Half of the entire graph is a mirror image of the other side. The center coordinate is (2,0) according to the graph. Thus, y when x=1 and y when x-3 are equal.
Manager
Joined: 07 Apr 2014
Posts: 96
Re: On the graph above, when x = 1/2, y = 2; and when x = 1, y = [#permalink]
### Show Tags
09 Sep 2014, 10:12
1
Bunuel wrote:
Attachment:
Graph.png
On the graph above, when x = 1/2, y = 2; and when x = 1, y = 1. The graph is symmetric with respect to the vertical line at x = 2. According to the graph, when x = 3, y =
(A) -1
(B) -1/2
(C) 0
(D) 1/2
(E) 1
Practice Questions
Question: 7
Page: 153
Difficulty: 600
graph is symmetric with respect to the vertical line
so if x = 1, y = 1 & x=2 , y=0 then x=3 , y=1 ...
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Re: On the graph above, when x = 1/2, y = 2; and when x = 1, y = [#permalink]
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19 May 2016, 09:24
6
Bunuel wrote:
Attachment:
Graph.png
On the graph above, when x = 1/2, y = 2; and when x = 1, y = 1. The graph is symmetric with respect to the vertical line at x = 2. According to the graph, when x = 3, y =
(A) -1
(B) -1/2
(C) 0
(D) 1/2
(E) 1
We are given that the graph is SYMMETRIC along the vertical line x = 2. We are also given that when x = 1, y = 1. We are asked to determine the value of y when x is 3. Keep in mind that when x = 1, it is ONE UNIT away from the line x = 2, and when x = 3, it is also ONE UNIT away from the line x = 2.
Since the y-coordinate is 1, when x is 1, the y-coordinate is ALSO 1 when x is 3. Again, this is because the graph is symmetric with the line x = 2.
Another way to look at this problem is to see that the point (1,1) is actually being reflected over the line x = 2.
We see that the point (1,1) is exactly 1 unit from the line x = 2. In fact the point (2,1) is on the line x = 2 which is exactly 1 unit from (1,1). However, (1,1) is exactly 1 unit on the LEFT of (2,1), so by symmetry, the point that is exactly 1 unit on the RIGHT of (2,1) is (3,1). Thus the y-coordinate is 1.
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Re: On the graph above, when x = 1/2, y = 2; and when x = 1, y = [#permalink]
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09 Jan 2020, 23:14
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Re: On the graph above, when x = 1/2, y = 2; and when x = 1, y = [#permalink] 09 Jan 2020, 23:14
Display posts from previous: Sort by | 2,317 | 6,953 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2020-16 | longest | en | 0.873694 |
https://www.physicsforums.com/threads/3d-statics-problem.640586/ | 1,537,790,915,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267160400.74/warc/CC-MAIN-20180924110050-20180924130450-00507.warc.gz | 839,428,189 | 12,579 | Homework Help: 3d statics problem
1. Oct 2, 2012
believe14142
1. The problem statement, all variables and given/known data
Three cables are used to support a lamp that weighs 800N. Determine the force in each supporting cable when everything is at equilibrium.
2. Relevant equations
ƩF in the x-axis = 0
ƩF in the y-axis = 0
ƩF in the z-axis = 0
3. The attempt at a solution
I really don't know how to start this problem. I get the concept of breaking up the components and adding all the forces up in each axis. But what throws me off about this problem is that it isn't centered at the origin. How am I supposed to find the angles of the forces with respect to an axis if the force isn't coming out of the origin?https://mail-attachment.googleusercontent.com/attachment/u/0/?ui=2&ik=dbdf3b9707&view=att&th=13a219c1fc5c23ec&attid=0.1&disp=inline&realattid=f_h7t12d950&safe=1&zw&saduie=AG9B_P-8Wo1xJnmuljT9i03--InW&sadet=1349183678622&sads=Kf9SObn1yzuJjVHHS6SmBxLnTK0&sadssc=1
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
2. Oct 2, 2012
LawrenceC
A sketch of the situation would be helpful.
3. Oct 2, 2012
believe14142
Oh pffft nevermind. I just realized that you can move the forces over so that everything is at the origin. | 411 | 1,308 | {"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-2018-39 | latest | en | 0.861771 |
https://www.examples.com/digital-sat/central-ideas-and-details | 1,726,451,346,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651668.29/warc/CC-MAIN-20240916012328-20240916042328-00449.warc.gz | 690,749,082 | 20,403 | Math - Algebra
Problem-Solving and Data Analysis
Geometry and Trigonometry
Reading and Writing - Information and Ideas
Craft and Structure
Expression of Ideas
Standard English Conventions
Guide
1. Home
2. Digital Sat
3. Central Ideas And Details
Central Ideas and Details
Understanding central ideas and details is fundamental for reading comprehension, a critical skill assessed on the Digital SAT Exam. Central ideas represent the main point or the most important message conveyed by a text, while details are the pieces of information that support, explain, or illustrate the central ideas. Mastering the ability to identify central ideas and relevant details will enhance your ability to comprehend complex passages and answer related questions accurately.
Learning Objectives
In this section, you will learn how to identify central ideas and distinguish them from supporting details in various types of texts. You will also learn techniques for summarizing central ideas and analyzing how details contribute to these ideas. By the end of this section, you will be able to confidently interpret central ideas and details in any reading passage on the Digital SAT Exam.
Identifying Central Ideas
• Read the Entire Text: Begin by reading the entire passage to get a general understanding of the content.
• Look for Repeated Themes: Identify themes or concepts that are repeated throughout the text. These are often indicative of the central idea.
• Identify the Thesis Statement: In many texts, especially essays or articles, the central idea is expressed in the thesis statement, often found in the introduction or conclusion.
• Summarize the Passage: After reading, try to summarize the passage in one or two sentences. This summary often captures the central idea.
Identifying central ideas involves recognizing the main point the author conveys, typically through repeated themes or a thesis statement, and summarizing.
Distinguishing Between Central Ideas and Details
• Main Idea vs. Supporting Information: The central idea is the main point the author wants to convey. Details provide evidence, examples, or explanations that support the central idea.
• Topic Sentences: In paragraphs, the topic sentence often contains the central idea, while the following sentences provide supporting details.
• Key Questions: Ask yourself, “What is the author trying to tell me?” to find the central idea, and “How does the author support this?” to identify the details.
Analyzing the Role of Details
• Evidence: Details often serve as evidence to support the central idea. They can be facts, statistics, quotes, or examples.
• Elaboration: Details elaborate on the central idea by providing more depth and context.
• Clarification: Some details clarify complex concepts related to the central idea, making them easier to understand.
Summarizing Central Ideas and Details
• Condense the Information: Summarize the passage by condensing the central idea and key supporting details into a brief statement.
• Focus on the Main Points: When summarizing, focus on the main points and avoid including minor details or tangential information.
• Use Your Own Words: Summarize using your own words to ensure you truly understand the central ideas and details.
Practice Strategies for the Digital SAT Exam
• Annotate the Text: While reading, annotate the text by underlining or highlighting the central ideas and key details.
• Practice with Sample Passages: Regularly practice with sample passages and questions to improve your skills in identifying central ideas and details.
Examples of Central Ideas and Details
Example 1
Passage Excerpt: “The rapid deforestation of the Amazon rainforest has significant environmental impacts, including loss of biodiversity and contribution to climate change. As trees are cut down, many species lose their habitats, and the carbon stored in trees is released into the atmosphere.”
Central Idea: The rapid deforestation of the Amazon rainforest has significant environmental impacts.
Details: Loss of biodiversity, species losing habitats, carbon release contributing to climate change.
Example 2
Passage Excerpt: “Exercise is crucial for maintaining physical health. Regular physical activity can help control weight, improve cardiovascular health, and reduce the risk of chronic diseases such as diabetes and hypertension.”
Central Idea: Exercise is crucial for maintaining physical health.
Details: Helps control weight, improves cardiovascular health, reduces the risk of chronic diseases.
Example 3
Passage Excerpt: “Renewable energy sources, such as wind and solar power, are essential for reducing our dependence on fossil fuels. These sources are not only sustainable but also produce less pollution, making them a key solution for addressing global warming.”
Central Idea: Renewable energy sources are essential for reducing dependence on fossil fuels.
Details: Sustainable, produce less pollution, key solution for global warming.
Example 4
Passage Excerpt: “Increased screen time among teenagers has been linked to various health issues. Studies show that excessive use of digital devices can lead to sleep disturbances, eye strain, and increased anxiety levels.”
Central Idea: Increased screen time among teenagers is linked to various health issues. Details: Sleep disturbances, eye strain, increased anxiety levels.
Example 5
Passage Excerpt: “The history of the internet is a story of rapid innovation and change. From its early beginnings as a government project to the expansive global network it is today, the internet has revolutionized communication, commerce, and access to information.”
Central Idea: The history of the internet is a story of rapid innovation and change.
Details: Early beginnings as a government project, revolutionized communication, commerce, and access to information.
Practice Questions
Question 1
Passage Excerpt: “The resurgence of interest in classical music among young people is notable. Music programs in schools, availability of online platforms, and popular media have all contributed to this trend.”
What is the central idea of the passage?
A) The resurgence of interest in classical music among young people.
B) Music programs in schools.
C) Availability of online platforms.
D) Influence of popular media.
Explanation: The central idea is the main point the passage is conveying, which is the resurgence of interest in classical music among young people. The other options are details supporting this central idea.
Question 2
Passage Excerpt: “Urban gardening has gained popularity as more people recognize its benefits. It provides access to fresh produce, encourages physical activity, and fosters a sense of community.”
Which detail supports the central idea that urban gardening has gained popularity?
B) Urban gardening encourages physical activity.
C) Urban gardening fosters a sense of community.
D) All of the above.
Explanation: All the provided options are details that support the central idea by explaining why urban gardening has gained popularity.
Question 3
Passage Excerpt: “Climate change is a pressing issue that affects all aspects of the environment. Rising temperatures lead to melting ice caps, more frequent extreme weather events, and disruptions in ecosystems.”
What is the central idea of the passage?
A) Rising temperatures lead to melting ice caps.
B) Climate change is a pressing issue.
C) More frequent extreme weather events occur.
D) Disruptions in ecosystems happen. | 1,414 | 7,537 | {"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-2024-38 | latest | en | 0.867942 |
https://www.physicsforums.com/threads/honors-physics-momentum-problem.284343/ | 1,508,283,436,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822513.17/warc/CC-MAIN-20171017215800-20171017235800-00341.warc.gz | 992,205,826 | 14,159 | # Honors Physics Momentum Problem
1. Jan 11, 2009
### fantolay
1. The problem statement, all variables and given/known data
A .45kg ice puck, moving east with a speed of 3 m/s has a head-on collision with a .9 kg puck initially at rest. Assuming a perfectly elastic collision, what will be the speed and direction of each object after the collision?
M1 = .45kg
M2 = .9kg
V1 = 3m/s
V2 = 0m/s
Perfectly elastic collision means Kinetic Energy and Momentum are conserved.
2. Relevant equations
KE = 1.2 * m * v2
P = m * v
3. The attempt at a solution
KEI = KEF (Initial KE = Final KE)
1/2 * m1 * v12 + 1/2 * m2 * v22 = 1/2 * m1 * (v1')2 + 1/2 * m2 * (v2')2
and
PI=PF
m1 * v1 + m2 * v2 = m1 * v1' + m2 * v2'
I don't really know where to go from here or if this is the right start... | 274 | 789 | {"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} | 3.625 | 4 | CC-MAIN-2017-43 | longest | en | 0.814541 |
https://www.hamilton-trust.org.uk/short-maths/year-3-progressive-maths/addition-subtraction-b/ | 1,571,190,545,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986661296.12/warc/CC-MAIN-20191016014439-20191016041939-00171.warc.gz | 919,390,407 | 18,810 | • Free Maths Resources on Hamilton
Find out which Hamilton maths units you can access for free, including our new slide presentations.
• New Flexible English Blocks
Take control of the balance between the parts of the English curriculum too.
• Looking for the Weekly Plans?
Find the weekly plans by looking for the orange button that says 'weekly plans' on the short blocks page for your year group.
Short Blocks
# Maths Year 3 Summer Addition & Subtraction (B)
Each unit has everything you need to teach a set of related skills and concepts. 'Teaching for Understanding' provides whole-class teaching and fully differentiated adult-led group activities. ‘Problem-solving and Reasoning’ develops these skills, and includes questions to enable you to assess mastery. Practice sheets ensure procedural fluency. Extra support activities enable targeted work with children who are well below ARE.
‘UNIT PLAN’ gives you a text version of all parts of the unit to use in your school planning documentation. ‘DOWNLOAD ALL FILES’ gives you that unit plan plus all of associated documents. These bulk downloads are added value for Hamilton Friends and School Subscribers.
## Unit 1 Add/subt multiples of 10, near multiples (suggested as 3 days)
### Objectives
Add/subtract multiples of 10 and near multiples
Unit 2: ID# 3739
National Curriculum
Hamilton Objectives
8. Mentally add or subtract any pair of 2-digit numbers, e.g. 75 + 58 or 75 – 58.
9. Mentally add and subtract multiples of 1s, 10s and 100s to/from 3-digit numbers.
14. Solve problems, including missing number problems.
### Teaching and Group Activities for Understanding
Day 1 Teaching
Write 346 + 50. Which digit will change? Confirm it is just the 10s digit; model counting on in 10s from 346 to answer. Repeat with 231 + 30 and 636 + 20.
Write 376 + 50: Which digit will change? Count on in 10s to show how both 100s & 10s digits change. Repeat for 346 + 500, 336 – 50 and 846 – 500.
Group Activities
-- Solve addition or subtraction word problems by counting on or back in 1s, 10s or 100s; cross multiples of 10 and 100.
-- Make up calculations that result in the 10s and/or the 100s digit changing.
Day 2 Teaching
Show a grid from 301 to 400. Demonstrate adding or subtracting 10s by moving up or down the grid. What happens when we add a ‘nearly number’ like 39 instead of 40? We need to add 40, then adjust by subtracting 1. Repeat for adding and subtracting other nearly numbers, e.g. 356 + 41, 356 + 39, 356 – 39, 382 – 59, 338 + 29.
Group Activities
-- Add and subtract near multiples to 3-digit numbers with the same 1s digit.
Day 3 Teaching
Display ‘Word problems’ one at a time; ask children to work on them in pairs. Think carefully about the maths that is needed – write a number sentence; then answer the question. Discuss their thoughts and clarify any misunderstandings as you work through each question. Choose one of the following to create a new word problem: 268 + 30, 583 – 40, 355 + 200, 632 – 400, 346 + 19, 475 – 21.
Group Activities
Use the in-depth problem-solving investigation ‘Place value puzzles’ as today’s group activity.
Or, use these activities:
-- Solve word problems by adding or subtracting multiples, or near multiples, of 10 and 100.
-- Use spinners to create word problems involving adding or subtracting multiples, or near multiples, of 10 and 100.
### You Will Need
• ‘Word problems’ sheet 1 (see resources)
• ‘Addition and subtraction word problems’ sheets 1 and 2 (see resources)
• Whiteboards and pens
• ‘Adding and subtracting multiples of 10 and 100’ (see resources)
• ‘301-400 grid’ (see resources)
• ‘Number cards’ Set A and B (see resources)
• ‘1-100 grid’ (see resources)
• Additional activity sheets (see resources)
### Mental/Oral Maths Starters
Day 1
Count on in steps of 10 and 100 from 3-digit numbers (pre-requisite skills)
Day 2
Count back in steps of 10 and 100 from 3-digit numbers (pre-requisite skills)
Suggested for Day 3
Revise the 3 times table (simmering skills)
### Procedural Fluency
Day 1
Solve word problems by counting on or back in 10s or 100s from 2-digit and 3-digit numbers.
Day 2
Add and subtract near multiples of 10 from 2-digit and 3-digit numbers.
Day 3
Create word problems for addition and subtraction of multiples of 10s and 100s and near multiples of 10.
### Mastery: Reasoning and Problem-Solving
• Add 40 to each of…
216, 159, 608
• Subtract 50 from each of…
117, 349, 608
• Complete each pair of calculations:
436 + 40 = 436 + 39 = 573 - 30 = 573 - 29 = 773 - 40 = 773 - 41 = 105 + 51 = 105 + 49 =
• Solve these word problems:
-- Sam’s Dad buys a laptop for £345. Sam buys a game for £39. How much did they spend in total?
-- Bella has measured the shelf and it is 253cm long. She saws off two pieces that are each 29cm long. How long is the shelf now?
In-depth Investigation: Place Value Puzzles
Solve the place value subtraction puzzles.
### Extra Support
Hops to Sevens
Bridging 10 when subtracting 1-digit numbers from 2-digit numbers, e.g. 42 - 5
## Unit 2 Subtract large numbers using counting up (suggested as 5 days)
### Objectives
Subtract larger numbers using counting up (frog)
Unit 4: ID# 3769
National Curriculum
Hamilton Objectives
12. Subtract larger numbers with confidence, using 'Frog' for counting up.
13. Estimate answers and use addition to check subtraction, understanding that addition and subtraction are inverse operations.
14. Solve problems, including missing number problems.
### Teaching and Group Activities for Understanding
Although column subtraction is specified for Year 3 in the National Curriculum, Hamilton believe that teaching column subtraction in Year 4 will be more successful. The National Curriculum Guidance specifies that teachers can teach areas of study in different year groups than those mentioned in the Curriculum, so long as they are taught within the same Key Stage. If you would like to teach column subtraction in Year 3 please see Year 4 Addition and Subtraction autumn Unit 7, spring Unit 4 and summer Units 2 and 3.
Day 1 Teaching
Write 137 – 72. Agree we can use Frog to find this difference, by, counting up from 72 to 137. Draw an empty number line jotting to show this and model each of Frog’s jumps (72 to 80, 80 to 100 and 100 to 137). Explain that we can do 1 big jump to 100. Repeat to model using Frog to do £1.46 – 83p. Children try £1.23 – 88p.
Group Activities
-- Use Frog to calculate the amount left in a purse after buying an item.
-- Use Frog to calculate differences.
Day 2 Teaching
Display a word problem resulting in the calculation 476 – 438. How can Maths Frog help us to work this out? Frog will start on 438, jump to the next 10 (440), then jump to 470, then 476. Remind children that we could do one big jump of 36 from 440 to 476 if we are confident but we don’t have to! Children try 873 – 837.
Group Activities
-- Investigate reversing the last 2 digits of a 3-digit number, then use Frog to find the difference between them.
Day 3 Teaching
Show a toy priced £4.67. Give a child a £5 note and ask them to use this to pay for the toy. You’ve given the shopkeeper too much money. How will she work out how much change to give you? Maths Frog can help us! Sketch a line from £4.67 to £5. Model using Frog to hop to £4.70 and then to £5. Repeat, to find change from £10 and £20.
Group Activities
Use the ‘Step Change’ in-depth problem-solving investigation below as today’s group activity.
Or, use these activities:
-- Practise finding the change from £5, £10 and £20.
-- Play online checkout/ change games, finding change from £1, £5 and £10.
Day 4 Teaching
Show a game and say that it costs £18. So far Katie has saved up £7.55. How much more does she need? Maths Frog can help! Sketch a line from £7.55 to £18. Model the hops from £7.55 to £7.60, then to £8, then to £18. Ensure children add hops accurately to find the answer: £10.45. Repeat as necessary for finding the difference between £8.43 and £14, then £6.55 and £19.
Group Activities
-- Calculate how much more money is needed to pay for a range of items.
-- Calculate how much more money is needed to pay for a more expensive version of an item.
Day 5 Teaching
A family have £100 and so far have spent £67.73. How much money is left? Use Frog to model the hops. Ask children to add the jumps, adding the £s first then the pence, then put them together. Some children may draw 3 hops, others draw 4. It doesn’t matter as long as we get the right answer! Repeat for a family who have spent £62.41 out of their £100.
Group Activities
-- Practise finding change from £100 (or £20) for a range of items.
-- Play ‘Change from £100’ pairs, ‘Change from £20’ pairs.
### You Will Need
• Whiteboards and pens
• ‘At the charity shop’ (see resources)
• ‘Purses’ (see resources)
• Bead string and real coins
• 0-9 cards and place value cards
• Representational notes (£5, £10, £20)
• ‘Finding change’ (see resources)
• Additional activity sheets (see resources)
• Internet access
• Picture of a console/computer game
• Calculators
### Mental/Oral Maths Starters
Day 1
Pairs with a total of 10 (pre-requisite skills)
Day 2
Pairs with a total of 100 (pre-requisite skills)
Day 3
Subtraction facts (pre-requisite skills)
Suggested for Day 4
Adding to the next 10 (simmering skills)
Suggested for Day 5
Match digital and analogue times (simmering skills)
### Procedural Fluency
Day 1
Subtract 2-digit numbers from 3-digit numbers, by counting up - using ‘Frog’.
Day 2
Subtract a pair of 3-digit numbers, using Frog.
Day 3
Calculate change from £5, £10 or £20 by counting up.
Day 4
Calculate admission price differences by counting up.
Day 5
Calculate the change from £20 or £100 by counting up.
### Mastery: Reasoning and Problem-Solving
78 + ☐ = 125
68 + ☐ = 134
375 – ☐ = 328
571 = 518 + ☐
127 + ☐ = 163
836 = 862 – ☐
• Write two numbers either side of 100 which have a difference of 37.
• Complete these bar model diagrams:
Diagram 1:
133 78 ?
Diagram 2:
585 518 ?
• Calculate the change from £10 when buying:
Key ring £5.68
Cards £7.75
Pen £6.43
• Calculate the change from £50 when buying:
Game £34.75
Controller £42.56
Case £28.67
• Calculate the change from £100 when buying:
Game £67.43
Console £85.85
Tablet £73.68
In-depth Investigation: Step Change
Children find change from 2 or 3 notes. They explore patterns in the pence.
### Extra Support
Frog Goes Freestyle
Subtracting pairs of 2-digit numbers with a bigger gap, e.g. 84 – 37, using counting up (Frog)
AND
Finding change from £1
## Unit 3 Efficient strategies for mental add/subt (suggested as 2 days)
### Objectives
Mentally add and subtract numbers using efficient strategies
Unit 6: ID# 3793
National Curriculum
Hamilton Objectives
8. Mentally add or subtract any pair of 2- digit numbers.
9. Mentally add and subtract multiples of 1s, 10s and 100s to/from 3-digit numbers.
13. Estimate answers and use addition to check subtraction, understanding that addition and subtraction are inverse operations.
14. Solve problems, including word problems.
### Teaching and Group Activities for Understanding
Day 1 Teaching
Write 463 + 306. Discuss that it’s probably more efficient to calculate mentally than with a written method: We add the 100s then the 1s - we’re using place value. Ask children to work out 463 + 205, 463 + 310. Agree answers.
Discuss how to work out 463 + 299. 299 is a ‘nearly number’. Agree that we add 300 and adjust the answer by subtracting 1. Repeat for 463 + 399 and 463 + 198.
Group Activities
-- Add two 3-digit numbers, beginning to recognise when to use each of two different strategies.
-- Pick two cards and decide if using a written method for addition or a mental calculation is more efficient.
Day 2 Teaching
Write 784 – 230. Ask children to discuss how to do this. Point out that we don’t need Frog! We can do it using place value. Ask children to work out 784 – 210 and 784 – 302. Agree answers. Write 784 – 199 and discuss how we can subtract a ‘nearly number’ using the multiple of 100, then adjusting the answer. Repeat for 784 – 299 and 784 – 198.
Group Activities
Use the in-depth problem-solving investigation ‘Super Shapes’ from NRICH as today’s group activity.
Or, use these activities:
-- Subtract two 3-digit numbers, beginning to recognise when to use each of two different mental strategies.
-- Find the mystery missing numbers that have been subtracted from 3-digit numbers.
### You Will Need
• Whiteboards and pens
• Place value cards
• ‘3-digit number cards’ sheets 1 and 2 (see resources)
• ‘Subtracting mentally’ sheet (see resources)
### Mental/Oral Maths Starters
Day 1
Adding multiples of 10 and 100 to 3-digit numbers (pre-requisite skills)
Day 2
Place value subtractions (pre-requisite skills)
### Procedural Fluency
Day 1
Add multiples or near multiples of 100. Choose a method to add a pair of 3-digit numbers.
Day 2
Find the missing ‘nearly number’ in 3-digit subtractions.
### Mastery: Reasoning and Problem-Solving
Complete the bar models:
Diagram 1:
? 428 199
Diagram 2:
854 399 ?
Diagram 3:
923 ? 498
• Tracy has 381 pennies saved in her ‘penny-jar’. If she takes out 299 to buy a toy, how many pennies will she have left? How much was the toy in pounds and pence?
• Shane has 258 football cards and his mum buys a job-lot of 199 on-line. How many cards does he have now?
• Ahmed gives 5 pounds and 1 penny to his little brother. He realises that he has 399p left. How much did he start with?
In-depth Investigation: Super Shapes
Find the value of the red shape in the addition problems. Super Shapes from www.nrich.maths.org.
### Extra Support
Pattern Spotters
Adding any pairs of 2-digit numbers using partitioning (including 1s> 10 AND 10s > 100) | 3,607 | 13,757 | {"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} | 4.59375 | 5 | CC-MAIN-2019-43 | longest | en | 0.890374 |
http://scherlund.blogspot.com/2019/04/why-isnt-1-prime-number-roots-of-unity.html | 1,566,140,595,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313936.42/warc/CC-MAIN-20190818145013-20190818171013-00253.warc.gz | 174,397,253 | 31,011 | ## Subscribe to my Email updates
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## Wednesday, April 03, 2019
### Why Isn't 1 a Prime Number? | Roots of Unity - Scientific American
And how long has it been a number?, argues Evelyn Lamb, Freelance math and science writer based in Salt Lake City, Utah.
An engineer friend of mine recently surprised me by saying he wasn’t sure whether the number 1 was prime or not. I was surprised because among mathematicians, 1 is universally regarded as non-prime.
Photo: Kevin Dooley Flickr (CC BY 2.0)
The confusion begins with this definition a person might give of “prime”: a prime number is a positive whole number that is only divisible by 1 and itself. The number 1 is divisible by 1, and it’s divisible by itself. But itself and 1 are not two distinct factors. Is 1 prime or not? When I write the definition of prime in an article, I try to remove that ambiguity by saying a prime number has exactly two distinct factors, 1 and itself, or that a prime is a whole number greater than 1 that is only divisible by 1 and itself. But why go to those lengths to exclude 1?
My mathematical training taught me that the good reason for 1 not being considered prime is the fundamental theorem of arithmetic, which states that every number can be written as a product of primes in exactly one way. If 1 were prime, we would lose that uniqueness. We could write 2 as 1×2, or 1×1×2, or 1594827×2. Excluding 1 from the primes smooths that out.
My original plan of how this article would go was that I would explain the fundamental theorem of arithmetic and be done with it. But it’s really not so hard to modify the statement of the fundamental theorem of arithmetic to address the 1 problem, and after all, my friend’s question piqued my curiosity: how did mathematicians coalesce on this definition of prime? A cursory glance around some Wikipedia pages related to number theory turns up the assertion that 1 used to be considered prime but isn’t anymore. But a paper by Chris Caldwell and Yeng Xiong shows the history of the concept is a bit more complicated. I appreciated this sentiment from the beginning of their article: “First, whether or not a number (especially unity) is a prime is a matter of definition, so a matter of choice, context and tradition, not a matter of proof. Yet definitions are not made at random; these choices are bound by our usage of mathematics and, especially in this case, by our notation.”...
In the very most basic example, we can ask whether the number -2 is prime. The question may seem nonsensical, but it can motivate us to put into words the unique role of 1 in the whole numbers. The most unusual aspect of 1 in the whole numbers is that it has a multiplicative inverse that is also an integer. (A multiplicative inverse of the number x is a number that when multiplied by x gives 1. The number 2 has a multiplicative inverse in the set of the rational or real numbers, 1/2: 1/2×2=1, but 1/2 is not an integer.) The number 1 happens to be its own multiplicative inverse. No other positive integer has a multiplicative inverse. The property of having a multiplicative inverse is called being a unit. The number -1 is also a unit within the set of integers: again, it is its own multiplicative inverse. We don’t consider units to be either prime or composite because you can multiply them by certain other units without changing much. We can then think of the number -2 as not so different from 2; from the point of view of multiplication, -2 is just 2 times a unit. If 2 is prime, -2 should be as well. | 834 | 3,606 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2019-35 | latest | en | 0.966327 |
https://learnsanskrit.org/vyakarana/tinanta/guna/ | 1,642,804,151,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303717.35/warc/CC-MAIN-20220121222643-20220122012643-00329.warc.gz | 420,483,682 | 5,991 | # guṇa
Now that our suffixes are in place, it is time to apply any rules that are conditioned by these suffixes. For sārvadhātuka suffixes specifically, there is one particular operation we should learn about: the replacement of the root vowel with its guṇa form:
• नी अ ते → ने अ ते
nī a te → ne a te
In this lesson, we will learn more about guṇa and some of the rules that use it.
## The first six rules
The first six rules of the Aṣṭādhyāyī are all about guṇa and vṛddhi. The first two rules define the terms guṇa and vṛddhi, and the next four then define how guṇa and vṛddhi interact with other terms:
• वृद्धिरादैच्। १.१.१
vṛddhirādaic (1.1.1)
vṛddhiḥ āt-aic
The vowels ā, ai, and au are called vṛddhi.
• अदेङ्गुणः। १.१.२
at-eṅ guṇaḥ
The vowels a, e, and o are called guṇa.
• इको गुणवृद्धी। १.१.३
iko guṇavṛddhī (1.1.3)
ikaḥ guṇa-vṛddhī
guṇa and vṛddhi replace the ik vowels,
• न धातुलोप आर्धधातुके। १.१.४
but not when followed by an ārdhadhātuka that causes [partial] lopa of the dhātu,
• क्ङिति च। १.१.५
kṅiti ca (1.1.5)
k-ṅiti ca
or when followed by [terms that are] kit or ṅit,
• दीधीवेवीटाम्। १.१.६
dīdhīvevīṭām (1.1.6)
dīdhī-vevī-iṭām
or when applied to [the roots] dīdhī and vevī, or [the augment] iṭ.
Let's set some of these rules aside. We saw rules 1.1.1 and 1.1.2 in a previous lesson, so we don't need to dwell on them further. And rules 1.1.4 and 1.1.6 are minor, so we can set them aside as well. That leaves us with rules 1.1.3 and 1.1.5.
Rule 1.1.3 defines the basic function of guṇa and vṛddhi vowels: they replace the ik vowels (i, u, ṛ, ḷ). And rule 1.1.4 restricts this substitution: the substitution is blocked if the following term is ṅit or kit.
Which kinds of terms are kit or ṅit? As a small example, the common suffix -ta is stated as kta in the Aṣṭādhyāyī. Because kta is kit, it will not cause guṇa or vṛddhi changes.
## ṅittva
There are also several terms that are treated as if they are ṅit suffixes. Those suffixes are defined in the following four atideśa (“analogy”, “extension”) rules. All of these rules are useful, but rule 1.2.4 is our focus here:
• गाङ्कुटादिभ्योऽञ्णिन्ङित्। १.२.१
gāṅkuṭādibhyo'ñṇinṅit (1.2.1)
gāṅ-kuṭ-ādibhyaḥ a-ñ-ṇit ṅit
The following are ṅit: [a term] after gāṅ and the root list starting with kuṭ, if the term is not ñit or ṇit;
• विज इट्। १.२.२
vija iṭ (1.2.2)
vijaḥ iṭ
the iṭ after the [root] vij;
• विभाषोर्णोः। १.२.३
vibhāṣorṇoḥ (1.2.3)
vibhāṣā ūṛṇoḥ
optionally, [the iṭ after the root] ūrṇu;
• सार्वधातुकमपित्। १.२.४
and sārvadhātuka [suffixes] that are not pit.
More plainly: a sārvadhātuka affix that is not pit will be treated as ṅit. And since it is treated as ṅit, it will not be able to cause any guṇa changes to the root sound.
Since śap is pit, rule 1.2.4 will not apply to it, and it will still be able to cause guṇa changes.
There is one relevant adhikāra here:
• अङ्गस्य। ६.४.१
aṅgasya (6.4.1)
aṅgasya
Of an aṅga, …
And what is an aṅga?
• यस्मात् प्रत्ययविधिस्तदादि प्रत्ययेऽङ्गम्। १.४.१३
yasmāt pratyaya-vidhiḥ tat-ādi pratyaye aṅgam
If a rule introduces a pratyaya after some specific term, everything from that term up to the pratyaya [is called] an anga.
So if a pratyaya is introduced after a dhātu, that dhātu is called aṅga with respect to the pratyaya.
## Replacement by guṇa
Rule 7.3.84 contains the word guṇa by anuvṛtti from a previous rule:
• सार्वधातुकार्धधातुकयोः। ७.३.८४
[An aṅga is replaced with guṇa] when followed by a sārvadhātuka or ārdhadhātuka suffix.
But to properly understand this rule, we must refer to two other rules. First, we know from rule 1.1.3 (iko guṇavṛddhī) that guṇa can replace only an ik vowel. But which ik vowel do we replace? We can decide which vowel to replace by referring to a new paribhāṣā:
• येन विधिस्तदन्तस्य। १.१.७२
yena vidhiḥ tat-antasya
[A term] by which a rule [is specified refers to an item that] ends in that [term].
In other words, the rule applies to the final sound of the aṅga. If we use rules 1..1.3 and 1.1.72, we can properly understand rule 7.3.84:
• सार्वधातुकार्धधातुकयोः। ७.३.८४
[An aṅga's last ik vowel is replaced with guṇa] when followed by a sārvadhātuka or ārdhadhātuka suffix.
And we can continue with our prakriyā:
1. nī a ti1.3.3 halantyam
1.3.9 tasya lopaḥ
## laghu and guru
There is another important instance where guṇa can apply. If we define the terms laghu:
• हलोऽनन्तराः संयोगः। १.१.७
halo'nantarāḥ saṃyogaḥ (1.1.7)
halaḥ anantarāḥ saṃyogaḥ
Consonants without an interval between them are called saṃyoga (conjunct).
• ह्रस्वं लघु। १.४.१०
hrasvaṃ laghu (1.4.10)
hrasvam laghu
A hrasva (short vowel) is called laghu,
• संयोगे गुरु। १.४.११
saṃyoge guru (1.4.11)
saṃyoge guru
but it is called guru when followed by saṃyoga.
• दीर्घं च। १.४.१२
dīrghaṃ ca (1.4.12)
dīrgham ca
And dīrgha is also [called guru].
• अलोऽन्त्यात् पूर्व उपधा। १.१.६५
The sound before the last sound is called upadhā (penultimate).
• पुगन्तलघूपधस्य च। ७.३.८५
And [the last ik vowel of an aṅga] ending with pu̐k or whose upadhā (penultimate sound) is laghu (a light syllable) [is replaced with guṇa].
Let's set aside pu̐k, since it is a minor point. Then the rule has a plain meaning: if the penultimate sound of the aṅga is laghu, then we can replace that sound with its guṇa.
Together, rules 7.3.84 and 7.3.85 cause guṇa in a variety of roots:
• नी + शप् → ने अ
nī + śap → ne a
• भू + शप् → भो अ
bhū + śap → bho a
• शुच् + शप् → शोच् अ
śuc + śap → śoc a
and leave others alone:
• निन्द् + शप् → निन्द् अ
nind + śap → nind a
(Penultimate is not a vowel.)
• जीव् + शप् → जीव् अ
jīv + śap → jīv a
(Penultimate is long.)
## Review
We now have all of the essential components we need to complete our prakriyā. In the next lesson, we will generate all eighteen forms of the root in the present tense (laṭ) with kartari prayoga. | 2,388 | 5,811 | {"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-2022-05 | longest | en | 0.820995 |
http://www.nowpublic.com/tech-biz/pi-day-jokes-2013-list-best-jokes-songs-crafts-3-14 | 1,386,912,904,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164886464/warc/CC-MAIN-20131204134806-00024-ip-10-33-133-15.ec2.internal.warc.gz | 465,684,456 | 12,190 | # Pi Day Jokes 2013: List of Best Jokes, Songs, Crafts For 3/14
by J.A. Dunabeitia | March 11, 2013 at 10:13 pm
22475 views | 8 Recommendations | 0 comments
Happy Pi Day 2013! What Better Way To Celebrate 3/14 Than With a Bunch of Pi Day Jokes, Songs and Crafts
As any number lover will tell you, March 14, 2013 is Pi Day. Pi, as your high school algebra teach no doubt taught you, is the mathematical relationship between a circle's diameter and its circumference: 3.1415926535...
March 14, or 3/14 as its written numerically, is a day to celebrate Pi. Those who are looking to be more exact celebrate Pi Minute (March 14 at 1:59 pm) or Pi Second (March 14, March 14 at 1:59 pm and 26 seconds).
Number nerds everywhere celebrate Pi Day by eating pie and reciting the mathematical constant that runs to an infinite number of decimal places.
Pi Day Jokes 2013
Mathematicians even tell Pi Jokes on Pi Day. As you can see from the Pi jokes below, lovers of Pi may be better at math than humor:
Q: What do you get when you cut a jack o'lantern by its diameter?
A: Pumpkin Pi!
Q: What do you get when you take green cheese and divide its circumference by its diameter?
A: Moon Pi.
Q:What do you get when you take the sun and divide its circumference by its diameter?
A: Pi in the sky.
## Pi day rap for mr_dye 2013 jhs
see larger video
sourced by J.A. Dunabeitia
Mathematician: "Pi r squared"
Baker:" No! Pies are round, cakes are square!2
Question: What do we get when we take the object and order the rim by the diameter?
Answer: Pi in the sky by and by.
If those Pi Day Jokes aren't bad enough for you, check out this Pi Day Joke video from a guy named Al G. Bra. The video seems kind of old and grainy and the jokes seem kind of old and corny.
If you're looking for more Pi Day jokes and crafts, here is a Pinterest Page dedicated to Pi Day.
The website teachpi.org has plenty of other Pi Day resources, including a Pi Day song and a Pi Day rap.
Author
## What is NowPublic?
NowPublic lets people work together to cover news events around the world.
## Crowd Power
First Flagged at 3:28 AM, Mar 12, 2013 by Anonymous (not verified)
These members have powered this story:
## Recommendations (8)
Most recently recommended by: | 595 | 2,252 | {"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} | 2.78125 | 3 | CC-MAIN-2013-48 | latest | en | 0.940513 |
https://www.enotes.com/homework-help/how-do-you-round-answer-up-3-signaficant-figures-342297 | 1,485,191,781,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560282935.68/warc/CC-MAIN-20170116095122-00282-ip-10-171-10-70.ec2.internal.warc.gz | 899,323,487 | 12,668 | # How do you round an answer up to 3 significant figures ?
justaguide | College Teacher | (Level 2) Distinguished Educator
Posted on
To round off an answer to 3 significant figures look at the fourth figure after the decimal point.
xxx.xxxC
If C is greater than or equal to 5, increase the third digit after the decimal point by 1 while eliminating C, else eliminate C without making a change in the third digit after the decimal point.
As an illustration: 23.0145 would be rounded off to 23.015 while 23.0144 would be rounded off to 23.014
Once this is done, the number has been rounded off to 3 significant figures.
Sources:
atyourservice | Student, Grade 11 | (Level 3) Valedictorian
Posted on
any number other than 0 is a significant number, so when you round you should have 3 numbers that are not zero and other zeros but make sure the 0s are behind the 3 number. i.e:
1536 rounded to 3 sig figs is 1540
363852 = 364000
1.3560 = 1.36 | 252 | 953 | {"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-2017-04 | longest | en | 0.902564 |
https://www.chessvariants.com/42.dir/neighbors.html | 1,701,848,647,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100583.31/warc/CC-MAIN-20231206063543-20231206093543-00173.warc.gz | 786,538,187 | 11,264 | Check out Metamachy, our featured variant for December, 2023.
# Neighbors
By
## Introduction
Neighbors is a Chess-like abstract strategy/tactical game for 2, 3 or more players using Chess pieces on a board of 42 squares. Inspiration for this game came from the games Nachbarn, PsAddle and Eufrat & Tigris.
## Board and Setup
Neighbors is played on a board that is 6 rows by 7 files.
The following pieces are used:
• 8 Queens
• 8 Rooks
• 8 Bishops
• 8 Knights
• 10 Ferzes
There are a total of 42 pieces, one for each square on the board. The pieces should be distributed randomly. (One approach would be to mark the piece types on the back of poker chips which are then draw out of a cloth bag, and placed on the board in row than column order.)
## Play
• Both players can move any piece on the board.
• Players alternate taking turns, moving one piece at the time.
• Players are not allowed to pass.
• When playing with three or more players, play is clockwise.
• A pieces can only move to capture. If a piece cannot capture, it remains in the game just to be captured.
• All pieces capture as in FIDE Chess, except for the Ferz, which captures one square diagonally in any direction.
• The game ends when no more captures can be made.
## Winning
The player with most points wins.
### Scoring for Three and Greater Players
When playing with three or more players, points are scored for each captured piece, the number of points scored depending on the type of piece captured. Pieces are worth the following points:
• Ferz = 1 point
• Knight = 2 points
• Bishop = 3 points
• Rook = 5 points
• Queen = 8 points
Points are calculated by adding the number of points of your left-hand neighbor to your own points.
### Scoring the Two Player Game
When playing with two players, one point is achieved for the capturing of the majority of pieces of one type. E.g. if Player A captures 5 Rooks and Player B 3 Rooks, Player A scores 1 point.
## Optional Rule
A player can only capture with a piece that is on the file or row of the landing place of the previously moved piece. E.G.:
```* * * * F * *
* * * F * * *
* R * N * B *
* * * * * * *
* F * Q * Q *
* * * * * * *```
Suppose the Rook on b4 captures the kNight on d4, the next player can only capture with pieces on the d file ( i.e. the F, R or Q) or the 4th row (i.e. B).
## Equipment
This game can be played using the pieces from four regular Western Chess sets, using the Kings as extra Queens, and the Pawns as Ferzes. The color of the pieces would not matter, and either player could move any color piece.
## Computer Play
An implementation of two player Neighbors has been written for Zillions of Games. You can download it here:
Editorial Note: This game has some interesting similarities to, and differences from, Andrew Looney's Monochrome Chess.
Written by Rob Nierse. HTML conversion by Peter Aronson.
WWW page created: April 1st, 2002.
| 770 | 3,065 | {"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-2023-50 | latest | en | 0.932615 |
http://gmatclub.com/forum/powerprep-99262.html?fl=similar | 1,484,849,768,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280723.5/warc/CC-MAIN-20170116095120-00235-ip-10-171-10-70.ec2.internal.warc.gz | 118,841,976 | 55,143 | POWERPREP : General GMAT Questions and Strategies
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# POWERPREP
Author Message
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16 Aug 2010, 22:17
Hi,
How accurate is the old GMAT software, Powerprep?? Considering the CAT format, I did manage to answer less than 10 questions incorrect in the verbal section and still my scores were too low, by too low I mean really really low. But, when I took the GMATPrep, and I did answer more no. of questions incorrectly than in the Powerprep, I managed a score better than the one from Powerprep. Is Powerprep really authentic?
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16 Aug 2010, 22:47
I will recommend gmat prep, as it is closest to the real test, for your scoring analysis. Power prep can be used for practice ony & not for analyzing raw scores, as it has the older algorithm.
For more on this issue, check this link gmat-prep-power-prep-official-free-gmat-tests-77548.html
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16 Aug 2010, 23:58
I would not focus on the number of questions answered correctly (though I understand why you want to), those are not really representative. Use GMAT prep score as the better one.
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17 Aug 2010, 00:29
Thanks guys! I was really low after seeing my scores. I have just a fortnight more for my exam and this score really pissed me off! I'm a little relieved now!
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16 Sep 2010, 13:34
hi just now i have taken powerPrep Test 1 and scored low 610
My recent GMAT Prep Score is 660 and MGMAT is 640. I have to give GMAT in Nov. 2010.
Pls suggest me something i feeling too low as i have heard that powerprep is easier compared to other test.I took it too lightly
I have got 15 wrongs in Verbal (SC-11 incorrect RC 2 incorrect CR 2 incorrect).
Q- 48( 7 incorrect) V-25
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16 Sep 2010, 17:05
Yes, I also believe in old powerprep and gmatprep there is difference b/w scoring method. Also, marking depends on difficulty level and as said by BB you shouldn't be bothered by no. of questions you got wrong but use powerprep for extra evaluation.
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17 Sep 2010, 05:41
Hi!!
Just now given GMAT Prep(Older Version) test number 1 and scored 730
Although there were exactly 10 Qs in verbal that i had already seen either in OG12 or somewhere else.
can Anyone clarify which one to believe 1) GMAT Prep -730 or 2) Power Prep 610 ?
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17 Sep 2010, 11:03
deeplakshya wrote:
Hi!!
Just now given GMAT Prep(Older Version) test number 1 and scored 730
Although there were exactly 10 Qs in verbal that i had already seen either in OG12 or somewhere else.
can Anyone clarify which one to believe 1) GMAT Prep -730 or 2) Power Prep 610 ?
I believe 610-730 is a big range so it's better take another GMATprep CAT, Veritas or MGMAT CAT to access your performance.
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17 Sep 2010, 11:48
i have given MGMAT test on 1st Sept / 3rd Sept and i scored 640 (Q-51 V 29)/ 630 (Q50 V 28)respectively.
I am working on Verbal Section where i used to fumble.
1) For CR i have almost done with MGMAT(i realized that i was making mistake in assumption questions...MGMAT has been of great help..also i reduced my time devoted per question)
2) For RC i am re-reading MGMAT.I am trying to do 3-4 passages daily,it's bit difficult to stick to this rule.Pls suggest if i am going in the right direction.
3) SC is the weakest area for me(yesterday's GMAT prep i scored 730 SC-8 incorrect)
I have studied MGMAT, then also my scores are not moving.One mistake i was doing that i was not practicing question this i will start from today .Any suggestions to improve SC Pls.I have read many posts on this forum stating - studying SC brings substantial improvement.I am following the right book(MGMAT) then also things are not working for me.
For Quants -
My weak areas are -
1) number system - jumbled with DS and not PS.
2) Speed issue- for the last 5-7 questions i have to rush like anything.In MGMAT i had cheated by pressing pause button.
Any suggestions for DS source of questions?
My exam Date is 8th Nov 2010.
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17 Sep 2010, 12:16
For DS, you should know very well AD/BCE, BD/ACE or 12TEN rules. (all are same though, used different notations by different companies!)
If Statement 1 is correct you should know your answer is going to be A or D.
If Statement 2 is correct you should know your answer is going to be B or D.
You shouldn't confuse statement 1 with statement 2 (I do ) solve statements separately, don't use information of 1 statement in statement 2 (except when you have rejected statement 1 and 2 separately!)
You can give a try to Veritas DS guide, if you have enough time.
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17 Sep 2010, 20:26
Those who have given powerprep, I have one question : Do the powerprep changes your screen resolution ?
It changes my screen resolution and it becomes really difficult to concentrate.
Also, there is a practice set other than the two tests, do they overlap?
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17 Sep 2010, 23:53
@ kissthegmat - thanks for your suggestion. I dont have Veritas guide though i have solved kaplan 800,which doesn't have as many questions.I'll search for veritas guide.
@ gurpreet- ya screen resolution changes and powerprep doesn't have huge question bank,you can see a lot of questions from OG.The only thing that made me worried was SCORE i made almost same number of questions wrong as i made in GMAT prep then also there was a huge diff in score.
in Powerprep Q-48(7 incorrect) V-25(10 incorrect).Overall 610.Whereas in GMAT Prep i was 720 and 730.
Pls can some one comment over my verbal strategy?
Suggestions needed
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13 Apr 2013, 21:50
Yeah..Same for me. I got 2 wrong in Quant and 8 wrong in verbal..still got to 660(50, 30)!.. I hope to get better in real exam.
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# POWERPREP
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2,704 | 9,415 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-04 | latest | en | 0.890377 |
https://zendict.com/en-en/abscissa.html | 1,723,423,428,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641023489.70/warc/CC-MAIN-20240811235403-20240812025403-00843.warc.gz | 817,828,628 | 2,826 | ### Abscissa (noun)
The horizontal coordinate of a point in a two-dimensional Cartesian coordinate system.
### Origin:
The word "abscissa" is a noun that refers to the horizontal coordinate of a point in a two-dimensional cartesian coordinate system. it is derived from the latin word "abscissus" which means "cut off" and refers to the geometric concept of cutting a point off a curve or line.
### Examples:
1. The point on the graph with an abscissa of 3 and an ordinate of 4 is located three units to the right and four units up from the origin.
2. The abscissa of a point on a graph represents its position along the x-axis.
3. The line of best fit has an equation of y = mx + b, where m is the slope of the line and b is the y-intercept, which is the point at which the line crosses the y-axis, with an abscissa of 0.
4. In the graph, the abscissa is the horizontal axis and the ordinate is the vertical axis.
5. The abscissa is used to locate a point in a two-dimensional space, in relation to the x-axis.
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# Calc III Review
## History of Vectors
More than 10 vector systems developed between 1790 and 1850
quaternions - scalar + vector - William Hamilton
modern vectors - direction + magnitude - Gibbs and Heaviside
field - function which assigns quantity to points in space
scalar - ex. temperature in space vector - ex. electromagnetic field
spherical coordinates
cylindrical coordinates
The change in each direction of a field
cartesian coordinates
$\overrightarrow{\nabla} F = \frac{d}{dx} F_x \hat{a_x} + \frac{d}{dy} F_y \hat{a_y} + \frac{d}{dz} F_z \hat{a_z}$
## Divergence
The measure of outflow of a field in an infinitesimal volume around some point
$\nabla F = \lim_{V \to 0} \iint_S \frac{F \times n}{V} dS$
## Laplacian
$\nabla \cdot \overrightarrow{\nabla} F$
# Dipoles
## Bound charges
$$\overrightarrow{E_D} = \frac{\overrightarrow{D} - \overrightarrow{P}}{\varepsilon_0}$$
\begin{aligned} \overrightarrow{P} &= \varepsilon_0 x_e \overrightarrow{E} \\ &= \varepsilon_0(1 + x_e)\overrightarrow{E} \end{aligned}
\begin{aligned} \overrightarrow{D} & = \varepsilon_0 \overrightarrow{E} + \varepsilon_0 x_e \overrightarrow{E} \\ & = \varepsilon_0 (1 + x_e) \overrightarrow{E} \\ & = \varepsilon_0 \varepsilon_r \overrightarrow{E} \end{aligned}
## Coulomb's Law in Dielectric
\begin{aligned} \overrightarrow{D} &= \frac{q_2}{4\pi r^2} \hat{a_r}^2 \\ \overrightarrow{E} &= \frac{q_2}{4\pi \varepsilon_0 \varepsilon_r r^2} \hat{a_r}^2 \\ \overrightarrow{F_{1,2}} &= \frac{q_2 q_1}{4\pi \varepsilon_0 \varepsilon_r r^2} \hat{a_r}^2 \end{aligned}
# Current and Conductivity
The current density $$J$$ in a conductive material is the amount of current flowing perdendicular through a unit area
$$J = u \rho_c q$$
where $$u$$ is the velocity of the particles, $$\rho_c$$ is the density of the particles, and $$q$$ is the charge per particle
The conductivity $$\sigma$$ of a material relates the strength of the electric field to the current flowing per cross sectional area
$$\overrightarrow{J} = \sigma \overrightarrow{E}$$
## Current flow from applied voltage
Let a voltage $$V$$ be applied to a conductive block with cross-sectional area $$A$$ and length $$d$$.
Then the total current flow is $$I = \frac{\sigma V A}{d}$$
Derivation
## Resistances in series
Let a conductive block have cross-sectional area $$A$$ and length $$2d$$. There are two conductive regions in series with conductivities $$\sigma_1, \sigma_2$$
$$R_{tot} = R_1 + R_2 = \frac{d}{A \sigma_1} + \frac{d}{A \sigma_2}$$
Derivation
# Capacitance
For a charged body, the work required to approach the body scales with charge,
so the ratio between charge and voltage is constant, called capacitance
$$C = \frac{Q}{V}$$
## Capacitance of parallel plate capacitor
$$C = \frac{Q}{V} = \frac{\varepsilon_0 \varepsilon_r A}{d}$$
Derivation
## Capacitances in parallel
$$C_T = C_1 + C_2$$
Derivation
## Energy in Capacitor
$$W = \frac{1}{2} CV^2$$
Derivation
## Changing capacitor shape
Assume a capacitor is charged to some voltage, and the voltage is removed. The charge on the plates will remain, regardless of how the plates are moved.
If the distance between the plates is doubled, $$\overrightarrow{E}$$ remains constant, so:
$$V = - \int_0^d \overrightarrow{E} dl = dE$$ -> $$V = - \int_0^{2d} \overrightarrow{E} dl = 2dE$$
and the voltage is doubled.
## Examples
1. Solve for $$E_2$$
$$D = \frac{Q}{A} = E_1 \varepsilon_1 = E_2 \varepsilon_2$$
$$E_1 = E_2 \frac{\varepsilon_2}{\varepsilon_1}$$
\begin{aligned} V & = - \int_{d_1} \overrightarrow{E_1} \overrightarrow{dl} - \int_{d_2} \overrightarrow{E_2} \overrightarrow{dl} \\ & = - \int_{d_1} \overrightarrow{E_2} \frac{\varepsilon_2}{\varepsilon_1} \overrightarrow{dl} - \int_{d_2} \overrightarrow{E_2} \overrightarrow{dl} \\ & = d_2 E_2 + d_1 E_2 \frac{\varepsilon_2}{\varepsilon_1} \end{aligned}
$E_2 = \frac{V}{d_2 + d_1 \frac{\varepsilon_2}{\varepsilon_1}}$
2. A metal sphere of radius 1m is surrounded everywhere by a dielectric with $$\varepsilon_r = 3$$. Find the capacitance of the sphere.
Assume a charge $$Q$$ on the sphere.
$$\overrightarrow{D} = \frac{Q}{4 \pi r^2} \hat{a_{\rho}}$$ (from Gauss's equation)
$\overrightarrow{E} = \frac{\overrightarrow{D}}{\varepsilon_0 \varepsilon_r} = \frac{Q}{4 \pi \varepsilon_0 \varepsilon_r r^2} \hat{a_{\rho}}$
$V = V(1) - V(\infty) = \frac{Q}{4 \pi \varepsilon_0 \varepsilon_r (1)^2}$
$C = \frac{Q}{V} = 4 \pi \varepsilon_0 \varepsilon_r = 12 \pi \varepsilon_0$
# Ampere's Force Law
Magnetic field from point current
Magnetic Field Intensity
Contribution from point current
$\overrightarrow{dH} = \frac{I \overrightarrow{dl} \times \hat{r'}}{4 \pi R^2} \frac{A}{m}$
where $$\hat{r'}$$ is the vector pointing from the point current to $$\overrightarrow{r}$$
$\overrightarrow{H} = \oint \overrightarrow{dH} \frac{A}{m}$
$dF = \mu_0 I_1 \overrightarrow{dl_1} \times \overrightarrow{dH}$
# Force between straight wires
Magnetic flux density
$\overrightarrow{B(\rho)} = \frac{I}{2 \pi \rho} \hat{r'}$
Units are in $$\frac{Vs}{m^2} = \frac{Wb}{m^2} = T = 10,000 G$$
Derivation
1. Find the force between two parallel wires with opposite currents
$\overrightarrow{dF_1} = \mu_0 -I_1 \overrightarrow{dl} \times \frac{I_2}{2 \pi R} \hat{r'}$
$\overrightarrow{F_1} = \int_0^L \mu_0 (-I_1 dz \hat{a_z}) \times \frac{I_2}{2 \pi R} a_{\phi} = - \frac{\mu_0 I_1 I_2 L}{4 \pi R} \hat{r'}$
# Ampere's Circuital Law
Around current
$$\oint H \cdot \overrightarrow{dl} = I$$
Derivation
Outside of current
$$\oint H \cdot \overrightarrow{dl} = 0$$
Derivation
## Magnetic field intensity of single wire
$$H(\rho) = \frac{I}{\rho 2 \pi}$$
Derivation
## Magnetic field intensity of slab of current
$\overrightarrow{H} = \frac{Z}{2}$
Where $$K = J \Delta H$$ is called the sheet charge density
Derivation
## Magnetic field intensity of cylinder of current
Many point currents arranged in a circle with radius $$a$$ and $$I_T = K(2 \pi)(a)$$
$H(\rho) = \begin{cases} \frac{I}{2 \pi \rho} & \phi \geq a,\\ 0 & \phi \leq a \end{cases}$
Derivation
# Point form of Ampere's circuital law
incomplete
$$\frac{dH_y}{dx} - \frac{dH_x}{dy} = J_z$$ \begin{aligned} \oint \overrightarrow{H} \cdot \overrightarrow{dl} & = J_Z \Delta x \Delta y \\ \int_1^4 \overrightarrow{H} \cdot \overrightarrow{dl} + \int_2^3 \overrightarrow{H} \cdot \overrightarrow{dl} + \int_3^4 \overrightarrow{H} \cdot \overrightarrow{dl} + \int_1^2 \overrightarrow{H} \cdot \overrightarrow{dl} & = J_z \Delta x \Delta y \\ H_{14} \Delta x + H_{23} \Delta y - H_{34} \Delta x - H_{12} \Delta y = J_Z \Delta x \Delta y \\ (H_{0x} - \frac{\Delta y}{2} \frac{dH_x}{dy})\Delta x \end{aligned}
$$\overrightarrow{F_m} = \mu_0 (I \overrightarrow{dl} \times \overrightarrow{H})$$
$$\overrightarrow{F_m} = \mu_0 (Q\mu \times \overrightarrow{H})$$
$$\overrightarrow{F_m} = Qu \times \mu_0 \overrightarrow{H}$$
$$\overrightarrow{B} = \mu_0 \overrightarrow{H}$$
$$\overrightarrow{F_m} = Qu \times \overrightarrow{B}$$
Lorentz Force Equation $$\overrightarrow{F} = Q \overrightarrow{E} + Q\mu \times \overrightarrow{B}$$
# Mass spectrometer
incomplete A particle with some known charge is fired into a chamber with a known magnetic field.
Due to the forces generated from a moving charge in a magnetic field, the particle will curve as it travels through the field.
$$\overrightarrow{F_m} = Qu \times \overrightarrow{B} = QuB(-\hat{a_{\rho}}) = m \frac{u^2}{r} \hat{a_{\rho}}$$
so $$r = \frac{mu^2}{QuB}$$
assume Si+ and As+
$KE = (100 \frac{J}{C})(1.6*10^{-19} C)$
$u(Si+) = 2.62 * 10^4 \frac{m}{s}$
$u(As+) = 1.6 * 10^4 \frac{m}{s}$
$r(Si+) =$
$r(As+) =$
# Magnetic field through a boundary
Since there are no magnetic monopoles, we can determine the normal component of the magnetic field through the boundary by drawing a simple body on the boundary.
\begin{aligned} B_{1N} & = B_{2N} \\ \mu_1 H_{1N} & = \mu_2 H_{2N} \end{aligned}
We can determine the parallel component using Ampere's circuital law with the assumption that the parallel field induces a current on the boundary.
# Faraday's law (change in magnetic flux)
Experimentally, we can see that
$i = -\frac{N}{R} \frac{d\Psi}{dt}$
\begin{aligned} V_{emf} & = iR = -N\frac{d\Psi}{dt} - \frac{d\lambda}{dt}\\ \oint \overrightarrow{E} \cdot \overrightarrow{dl} & = -N \frac{d \int \overrightarrow{B} \cdot \overrightarrow{ds}}{dt} \end{aligned}
where $$N$$ is the number of turns in the coil, $$R$$ is the total resistance, and $$\Psi$$
Or since $$L = \frac{\lambda}{i}$$
$$V_{emf} = - \frac{d\lambda}{dt} = -L \frac{di}{dt}$$
# Lenz's Law
The induced current will flow in a direction that opposes the magnetic flux.
# Power and Energy in an inductor
$$qV$$ - energy delivered to R and L by a single charge
$$n$$ - number of $$q$$ moving around circuit per second
$$P = nqV = IV$$ - energy delivered to R and L per second by all charges
\begin{aligned} \oint \overrightarrow{E} \cdot \overrightarrow{dl} & = \frac{d\lambda}{dt} \\ -V + IR + 0 & = -L \frac{di}{dt} \end{aligned}
$$V = IR + L \frac{di}{dt}$$
\begin{aligned} W & = \int P dt = \int iV dt = \int_0^T(iR + L i\frac{di}{dt}) dt = \int_0^T i^2 R dt + \int_0^T Li di \\ & = \frac{1}{2}Li^2 \end{aligned}
# Transformers
Primary winding $$V_1 = \oint \overrightarrow{E} \cdot \overrightarrow{dl} = -(N_1 \frac{d\Psi}{dt})$$
$$-(N_2 \frac{d\Psi}{dt}) = \oint \overrightarrow{E} \cdot \overrightarrow{dl} = V_2$$
$$V_2 = \frac{N_2}{N_1} V_1$$
# Solenoid
$$\int \overrightarrow{H} dl = I_{enc} \to A$$
$$\overrightarrow{B} = \mu \overrightarrow{H}$$ (Henries)
$$B * A = \Psi$$ (Henry-Amps)
$$\oint \overrightarrow{E} \cdot \overrightarrow{dl} = \frac{d}{dt} \Psi$$ (Henry-Amps)/sec = V
1. Let $$\frac{d\Psi}{dt} = 10V$$
The reading from $$VOM_1$$ and $$VOM_2$$ are different. $$VOM_1$$ is approximately $$10V$$ since the changing magnetic field flows through the surface created by its wires.
The reading from $$VOM_2$$, since the surface is not pierced by the magnetic field coming from the solenoid.
\begin{aligned} \oint \overrightarrow{E} \cdot \overrightarrow{dl} & = - \frac{d\Psi}{dt} \\ IR & = - \frac{d\Psi}{dt} = \frac{d}{dt} \left[ Blx \right] = Blxv \end{aligned}
$$\overrightarrow{F_m} = \int I \overrightarrow{dl} \times \overrightarrow{B} dl = IB \int_0^l dy = -IBl \hat{a_x}$$
$$\oint H \cdot dl = I_{enc}$$
Move the boundary of the surface between the plates of a capacitor.
$= # Young's Experiment Young demonstrated the particle and wave behavior of light with his experiment. He shone light through a pair of narrow slits, which resulted bands forming on the photosensitive plate, suggesting wave-like behavior. This is because the distance between a band and each slit results in a slight phase shift of the light wave. If the intensity of the light is lowered and exposure of the plate is brief, the photons hitting the plate cause specs of exposure, suggesting particle-like behavior. # Electromagnetic Waves $u = \frac{\omega}{\beta} = \frac{2 \pi f}{\frac{2 \pi}{\lambda}} = f \lambda = \frac{1}{\sqrt{\mu \varepsilon}}$ $$\overrightarrow{\mathbb{P}} = \overrightarrow{E} \times \overrightarrow{H}$$ - energy of waves (Poynting vector) ## Plane Waves in Lossless Media incomplete $$\overrightarrow{E}(z,t) = E_{x0} e^{-\alpha z} \cos(\omega t - \beta z) \hat{a}_x$$ $$\overrightarrow{H}(z,t) = \frac{\varepsilon}{\mu} E_{x0} e^{-\alpha z} \cos(\omega t - \beta z) \hat{a}_y$$ $$\mu = \frac{\omega}{\beta}$$ - velocity $$\beta = \frac{2 \pi}{\lambda}$$ - wave number $$n = \sqrt{\frac{\mu}{\varepsilon}}$$ - 1. Let $$\overrightarrow{H} = 25 \cos(1.33*10^{-4}z + 40,000t) \hat{a}_y \frac{A}{m}$$ Find the wavelength: ## Plane Waves in Lossy Media $$\overrightarrow{E}(z,t) = E_0 e^{-\alpha z} \cos(\omega t - \beta z) \hat{a}_x$$ $$\overrightarrow{H}(z,t) = \frac{E_0}{n} e^{-\alpha z} \cos(\omega t - \beta z) \hat{a}_y$$ $$n = \sqrt{\frac{j \omega \mu}{\sigma + j\omega \varepsilon}}$$ - intrinsic impedance $$\alpha + j \beta = \sqrt{j \omega \mu (\sigma + j \omega \varepsilon)} = \gamma$$ - propogation constant 1. Assume $$\mu_r = 1$$. $$\overrightarrow{H} = 25 \sin(2 \times 10^8 t + 6x) \hat{a}_y$$ For the wave to hold in place * as $$t$$ increases, $$x$$ must decrease. So the wave propogates in the $$-\hat{a}_x$$ direction. $$\omega = 2 \times 10^8 s^{-1}$$, $$\beta = 6 = \frac{2 pi}{\lambda}$$, 2. $$u = \frac{\omega}{\beta} = \frac{2 \times 10^8}{6} \frac{m}{s} = \frac{1}{\sqrt{\mu \varepsilon}}$$ Since material is lossless, $$\mu = \mu_0$$ $$\frac{1}{3} \times 10^8 = \frac{1}{\sqrt{\mu \varepsilon}} = \frac{1}{\sqrt{\mu_0 \varepsilon_0 \varepsilon_r}}$$ so $$\varepsilon_r = 81$$ 1. Find E. $$n = \sqrt{\frac{\mu}{\varepsilon_r}} = \frac{377}{9} \Omega$$ $$E = n H = \frac{377}{9} (25 \frac{mA}{m}) = 1.05 \frac{V}{m}$$ Using the geometry of the wave $$\overrightarrow{\cal{P}} = \overrightarrow{E} \times \overrightarrow{H}$$, we can find that the direction of propogation is in the $$\hat{a}_z$$ direction, so $$\overrightarrow{E} = 1.05 \sin(2 \times 10^8 t + 6X) \hat{a}_z \frac{V}{m}$$ Let E⃗ = 16e-0.05x sin (2 × 10 8 t - 2x) z$
1. Find propogation constant
$$\gamma = \alpha + Jj \beta = 0.05 + j2$$
2. Find λ
$$\beta = \frac{2\pi}{\lambda}$$
$$\lambda = \frac{2\pi}{\beta} = \pi$$
3. Find the velocity $$u$$
$$u = \frac{\omega}{\beta} = \frac{2 \times 10^8 (s^{-1})}{2 (m^{-1})} = 10^8 \frac{m}{s}$$
4. Find the skin depth $$\delta$$
$$\delta = \frac{1}{\alpha} = 20$$
## Skin depth
If the current through the wire is increasing, current loops perpendicular to the current flow to oppose the magnetic field generated. $$\delta = \frac{1}{\alpha}$$ - skin depth
Derivation
# Reflection
1. Assume non-magnetic lossless
In the first medium, $$E_i = 10 \cos(3 \times 10 ^8 t - 1z) \hat{a}_x \frac{V}{m}$$, $$\varepsilon_r = 1$$, $$\varepsilon_{r2} = 16$$
$$\eta_0 = \sqrt{\mu_0}{\varepsilon_0}$$, $$\eta_2 = \sqrt{\mu_0}{16 \varepsilon_0}$$
Find r
$$r = \eta_2 - \eta_1}{\eta_2 + \eta_1} = -3/5$$
Find $$E_r$$
$$E_{r0} = rE_{io} = -6$$
$$E_r = -6\cos(3 \times 10^8 t - 1z) \hat{a}_x$$
Find $$E_t$$
$$E_{t0} = \tau E_{io} = 4$$
But the medium is changed, so the wavelength is changed: $$\frac{\omega}{\beta} = \frac{1}{\sqrt{\mu_0 \varepsilon_{r2} \varepsilon_0}} = \frac{1}{\sqrt{16}\sqrt{\mu_0 \varepsilon_0}}$$
so $$\beta = 4$$
$$E_i = 4 \cos(3 \times 10^8 t - 4 z) \hat{a}_z$$ | 5,207 | 14,445 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2020-24 | latest | en | 0.498049 |
http://math.stackexchange.com/questions/tagged/transcendental-numbers | 1,469,633,691,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257826908.63/warc/CC-MAIN-20160723071026-00276-ip-10-185-27-174.ec2.internal.warc.gz | 160,475,513 | 26,627 | # Tagged Questions
Transcendental numbers are numbers that cannot be the root of a nonzero polynomial with rational coefficients (i.e., not an algebraic number). Examples of such numbers are $\pi$ and $e$.
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### Can a change of basis modify irrationality/transcendence?
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### Non-algebraic structures?
We call group, ring, field,... "algebraic structures". Do we have similar analogue for transcendental numbers? If not, then how do we study interactions between various transcendental numbers? Also, ...
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### How was the difference of the Fransén–Robinson constant and Euler's number found?
I recently ran across the following integral: $$\int_{0}^{\infty}\frac{1}{\Gamma(x)}dx$$ Which I learned is equal to the Fransén-Robinson constant. On the linked wikipedia page for the Fransén-...
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### Complicated series converges to $\pi$.
How do I get this result? $$\frac {426880 \sqrt {10005}}{\large \sum_{k = 0}^{\infty}\frac {(6k)!(545140134k + 13591409)}{(k!)^3 (3k)! (-640320)^{3k}}} = \pi$$ It seems formidable. Context: I came ...
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### Help complete this proof on transcendentalism
Proof $\pi*e$ is transcendental. either $\pi + e$ or $\pi*e$ is transcendental to see take $(x-\pi)(x-e)=x^2-(\pi+e)x+\pi*e$. Case 1 assume $\pi$ and $e$ are algebraically independent. It follows ... | 441 | 1,558 | {"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} | 3.390625 | 3 | CC-MAIN-2016-30 | latest | en | 0.783666 |
https://kmmiles.com/595427-miles-in-km | 1,669,899,053,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710813.48/warc/CC-MAIN-20221201121601-20221201151601-00342.warc.gz | 381,590,799 | 6,246 | kmmiles.com
Search
# 595427 miles in km
## Result
595427 miles equals 958042.043 km
You can also convert 595427 mph to km.
## Conversion formula
Multiply the amount of miles by the conversion factor to get the result in km:
595427 mi × 1.609 = 958042.043 km
## How to convert 595427 miles to km?
The conversion factor from miles to km is 1.609, which means that 1 miles is equal to 1.609 km:
1 mi = 1.609 km
To convert 595427 miles into km we have to multiply 595427 by the conversion factor in order to get the amount from miles to km. We can also form a proportion to calculate the result:
1 mi → 1.609 km
595427 mi → L(km)
Solve the above proportion to obtain the length L in km:
L(km) = 595427 mi × 1.609 km
L(km) = 958042.043 km
The final result is:
595427 mi → 958042.043 km
We conclude that 595427 miles is equivalent to 958042.043 km:
595427 miles = 958042.043 km
## Result approximation
For practical purposes we can round our final result to an approximate numerical value. In this case five hundred ninety-five thousand four hundred twenty-seven miles is approximately nine hundred fifty-eight thousand forty-two point zero four three km:
595427 miles ≅ 958042.043 km
## Conversion table
For quick reference purposes, below is the miles to kilometers conversion table:
miles (mi) kilometers (km)
595428 miles 958043.652 km
595429 miles 958045.261 km
595430 miles 958046.87 km
595431 miles 958048.479 km
595432 miles 958050.088 km
595433 miles 958051.697 km
595434 miles 958053.306 km
595435 miles 958054.915 km
595436 miles 958056.524 km
595437 miles 958058.133 km
## Units definitions
The units involved in this conversion are miles and kilometers. This is how they are defined:
### Miles
A mile is a most popular measurement unit of length, equal to most commonly 5,280 feet (1,760 yards, or about 1,609 meters). The mile of 5,280 feet is called land mile or the statute mile to distinguish it from the nautical mile (1,852 meters, about 6,076.1 feet). Use of the mile as a unit of measurement is now largely confined to the United Kingdom, the United States, and Canada.
### Kilometers
The kilometer (symbol: km) is a unit of length in the metric system, equal to 1000m (also written as 1E+3m). It is commonly used officially for expressing distances between geographical places on land in most of the world. | 635 | 2,355 | {"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 | 4 | CC-MAIN-2022-49 | latest | en | 0.79306 |
https://www.jiakaobo.com/leetcode/516.%20Longest%20Palindromic%20Subsequence.html | 1,726,572,685,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651773.64/warc/CC-MAIN-20240917104423-20240917134423-00340.warc.gz | 766,246,959 | 6,935 | • ㊗️
• 大家
• offer
• 多多！
## Problem
Given a string s, find the longest palindromic subsequence’s length in s.
A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements.
Example 1:
Input: s = "bbbab"
Output: 4
Explanation: One possible longest palindromic subsequence is "bbbb".
Example 2:
Input: s = "cbbd"
Output: 2
Explanation: One possible longest palindromic subsequence is "bb".
Constraints:
• 1 <= s.length <= 1000
• s consists only of lowercase English letters.
## Code
dp[i, j]表示在 i 和 j 区间内最长的回文字符串
• 如果 i 和 j 的字符相同, 那么 i 和 j 之间的最长回文字符串就是[i+1, j-1] + 2
• 如果 i 和 j 的字符不同, 那么 i 和 j 之间的最长回文字符串就是 max([i+1, j],[i, j-1])
class Solution {
public int longestPalindromeSubseq(String s) {
int[][] dp = new int[s.length()][s.length()];
for (int len = 1; len <= s.length(); len++) {
for (int i = 0; i + len <= s.length(); i++) {
int j = i + len - 1;
if (i == j) {
dp[i][j] = 1;
continue;
}
if (s.charAt(i) == s.charAt(j)) {
dp[i][j] = dp[i + 1][j - 1] + 2;
} else {
dp[i][j] = Math.max(dp[i + 1][j], dp[i][j - 1]);
}
}
}
return dp[0][s.length() - 1];
}
} | 419 | 1,166 | {"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} | 3.453125 | 3 | CC-MAIN-2024-38 | latest | en | 0.424662 |
https://www.physicsforums.com/threads/relative-velocity-of-2-vehicles.879862/ | 1,529,627,794,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864303.32/warc/CC-MAIN-20180621231116-20180622011116-00384.warc.gz | 898,289,674 | 14,419 | # B Relative velocity of 2 vehicles
1. Jul 24, 2016
### Aashish Bharat
Two vehicles leave simultaneously from a fork in a road. Vehicle A travels at a velocity of 85 km/h north-east. Vehicle B travels at 104 km/h directly east. Calculate the velocity of vehicle A relative to the velocity of vehicle B.
I do not understand why is Vehicle B traveling west, an explanation on how why this was done would be appreciated!
aVb = Va - Vb (Velocity of A relative to velocity of B) when do you use this equation?
2. Jul 24, 2016
### PeroK
If you are in a car and get overtaken by another car, then relative to the faster car you are going backwards. Not so?
3. Jul 24, 2016
### Aashish Bharat
Damn it, they took the opposite direction of Vb from the equation aVb = Va - Vb. I should have known better | 212 | 803 | {"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-2018-26 | latest | en | 0.951534 |
https://www.physicsforums.com/threads/two-prisms-connected-by-a-material-of-refractive-index-n.834896/ | 1,721,811,776,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518198.93/warc/CC-MAIN-20240724075911-20240724105911-00412.warc.gz | 793,858,358 | 17,328 | # Two Prisms connected by a material of refractive index n
• Potatochip911
In summary, the question is asking for the refractive index of a material, denoted as n2, that will cause light of a certain wavelength to follow a certain path. The refractive index of the material is dependent on the wavelength of light, and the critical angle for total internal reflection will also vary with the wavelength. The necessary refractive index for the transition between total reflection and transmission at a wavelength of 300 nm can be calculated as n2=n1*sin(theta1).
Potatochip911
## Homework Statement
The prisms have a refractive index ##n_1=1.3## at wavelength ##\lambda=300nm## and are connected by a material with refractive index ##n_2##. What is the refractive index ##n_2## so that light of wavelength ##>\lambda## follow path 2, and light at ##<\lambda## follow path 3?
## Homework Equations
##\sin\theta_c=\frac{n_2}{n_1}##
##n=\frac{c}{v}=\frac{c}{f\lambda}##
## The Attempt at a Solution
I am very confused as to how the wavelength of light effects total internal reflection. I can solve for the frequency of the light by using ##n=\frac{c}{f\lambda}\Rightarrow f=\frac{c}{n\lambda}## but I'm not entirely sure this is useful. I also have that ##\sin\theta_c=\frac{n_2}{n_1}## which can be rewritten as ##\sin\theta_c=\frac{\lambda_1}{\lambda_2}## The critical angle is a constant in this case I think? it's 45 degrees since the light is always coming in parallel to the bottom of the prisms but I don't understand how varying ##\lambda## is going to change whether or not it exits the prisms.
The question only says that n1 is 1.3 at a particular wavelength: 300nm. Do you suppose it could be different at other wavelengths? How would it vary?
Potatochip911
The refractive index depends on the wavelength, so the critical angle depends on it as well. Here, the transition between total reflection and transmission is supposed to be at a wavelength of 300 nm, so you can calculate the necessary refractive index at 300 nm to be directly at the transition.
Potatochip911
So the correct refractive index then is ##n_1=\frac{n_2}{\sin\theta_1}##, edit: I mean ##n_2=n_1\sin\theta_1## sorry for all the mistakes I'm on my phone
## 1. What is the purpose of connecting two prisms with a material of refractive index n?
The purpose of connecting two prisms with a material of refractive index n is to create a prism system that can manipulate light in a specific way. By adjusting the angle and location of the prisms, the material can alter the path of light passing through them, allowing for the creation of various optical effects.
## 2. How does the refractive index of the material affect the behavior of the light passing through the prism system?
The refractive index of the material determines how much the light will bend as it passes through the prism system. A higher refractive index means that the light will bend more, while a lower refractive index will result in less bending.
## 3. Can the refractive index of the material be changed?
Yes, the refractive index of the material can be changed by altering its composition or by adjusting its temperature. This can result in different optical effects and allow for more precise manipulation of light in the prism system.
## 4. What are some practical applications of a prism system with two connected prisms?
A prism system with two connected prisms has many practical applications, including in optical instruments and devices such as cameras and telescopes. It can also be used in scientific experiments and research to study the behavior of light and to create specific optical effects.
## 5. Are there any limitations to using a prism system with two connected prisms?
While a prism system with two connected prisms can be very useful, it does have some limitations. The size and shape of the prisms, as well as the properties of the connecting material, can affect the precision and accuracy of the system. Additionally, the system may be limited in the types of optical effects it can produce.
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1K | 1,111 | 4,632 | {"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 | 4 | CC-MAIN-2024-30 | latest | en | 0.896336 |
https://math.answers.com/questions/What_is_57_over_40_in_simplest_form | 1,652,729,279,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662512229.26/warc/CC-MAIN-20220516172745-20220516202745-00306.warc.gz | 456,910,547 | 37,060 | 0
# What is 57 over 40 in simplest form?
Wiki User
2017-05-07 18:26:30
57/40 is in its simplest form.
Wiki User
2017-05-07 18:26:30
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2012-11-27 10:57:50
57/40 cannot be simplified further. | 129 | 376 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-21 | latest | en | 0.842419 |
https://cboard.cprogramming.com/game-programming/76909-opengl-matrices-printable-thread.html | 1,493,320,940,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917122619.71/warc/CC-MAIN-20170423031202-00641-ip-10-145-167-34.ec2.internal.warc.gz | 785,945,151 | 6,783 | # OpenGL Matrices
Show 80 post(s) from this thread on one page
Page 1 of 2 12 Last
• 03-13-2006
joeprogrammer
OpenGL Matrices
OK, I've got several things I need to know.
#1: I need an algorithm for finding the inverse of a given matrix.
#2: I want to be able to modify OpenGL's projection matrix manually, instead of using glRotatef and glTranslatef. I know how to grab the matrix using glGetFloatv or something like that, but I don't know how to send the matrix back to replace the existing matrix.
Thanks
• 03-13-2006
BobMcGee123
glRotatef and glTranslatef do not modify the projection matrix by default, they modify the glmodelview matrix. You can ultimately rotate anything that is expressed as a matrix as far as I know (i.e you can rotate textures and stuff)
You can modify any matrix by, as you stated, by getting the matrix from OpenGL using glGetFloatfv. You modify the matrix, then you put it back by calling glLoadMatrixf. Be careful to set the proper glmatrixmode first.
http://msdn.microsoft.com/library/de...unc03_3260.asp
My 3x3 invert function. Note that when a 3x3 matrix is a pure rotation matrix, the determinant is always 1, and you can calculate the inverse by either transposting the matrix (swap the row and column entries) or by rebuilding the matrix with the negative angles...this (the code below) is how linear algebra tells you how to compute the inverse of any 3x3 matrix (i.e, this could be from cryptography or something, where you decode a message by multiplying the encoded message by this matrix, assuming the original message was multiplied by the original matrix):
Code:
``` void Matrix3x3::Invert(void) { Matrix3x3 inverse; double det, invDet; inverse.mat[0][0] = mat[1][1] * mat[2][2] - mat[1][2] * mat[2][1]; inverse.mat[1][0] = mat[1][2] * mat[2][0] - mat[1][0] * mat[2][2]; inverse.mat[2][0] = mat[1][0] * mat[2][1] - mat[1][1] * mat[2][0]; det = mat[0][0] * inverse.mat[0][0] + mat[0][1] * inverse.mat[1][0] + mat[0][2] * inverse.mat[2][0]; if(det == 0) det = 1; invDet = 1.0f / det; inverse.mat[0][1] = mat[0][2] * mat[2][1] - mat[0][1] * mat[2][2]; inverse.mat[0][2] = mat[0][1] * mat[1][2] - mat[0][2] * mat[1][1]; inverse.mat[1][1] = mat[0][0] * mat[2][2] - mat[0][2] * mat[2][0]; inverse.mat[1][2] = mat[0][2] * mat[1][0] - mat[0][0] * mat[1][2]; inverse.mat[2][1] = mat[0][1] * mat[2][0] - mat[0][0] * mat[2][1]; inverse.mat[2][2] = mat[0][0] * mat[1][1] - mat[0][1] * mat[1][0]; mat[0][0] = inverse.mat[0][0] * invDet; mat[0][1] = inverse.mat[0][1] * invDet; mat[0][2] = inverse.mat[0][2] * invDet; mat[1][0] = inverse.mat[1][0] * invDet; mat[1][1] = inverse.mat[1][1] * invDet; mat[1][2] = inverse.mat[1][2] * invDet; mat[2][0] = inverse.mat[2][0] * invDet; mat[2][1] = inverse.mat[2][1] * invDet; mat[2][2] = inverse.mat[2][2] * invDet; }```
• 03-14-2006
joeprogrammer
OK, but I really wanted an algorithm for finding the inverse of a 4x4 matrix. I'm not really that good at math, so I don't know how to modify the algorithm you gave me. :)
• 03-14-2006
Shamino
Strange, my math library doesn't have functions available for inverse/adjoint/determinant either...
I'm using a 4x4 matrix library as well. I havn't had the need yet for inverse operations though..
Sorry I can't help though :d
• 03-14-2006
Perspective
You should avoid inverting a matrix if you can, its a timely process.
• 03-14-2006
joeprogrammer
Well actually I'm writing a game engine, and I need to have certain functions in it.
• 03-14-2006
BobMcGee123
Quote:
You should avoid inverting a matrix if you can, its a timely process.
Well, if you included it into something that's being computed consistantly realtime it is.
Quote:
Strange, my math library doesn't have functions available for inverse/adjoint/determinant either...
I'm using a 4x4 matrix library as well. I havn't had the need yet for inverse operations though..
Sorry I can't help though :d
I wrote my own.
You're going to have to simply look online for the algorithm for inverting a matrix. Not all matrices are invertible, and most of linear algebra builds upon arcane abstractions that just 'happen to work' but have less of an intuitive interpretation (compared with other branches of math, and this is really just my own opinion according to my own interpretation of reality).
http://www.euclideanspace.com/maths/...ourD/index.htm
http://mathworld.wolfram.com/CramersRule.html
http://www.gamedev.net/community/for...age=1�
• 03-14-2006
joeprogrammer
Sorry, I really don't understand linear algebra. Maybe I'll just leave inverting matrices until I need them.
• 03-14-2006
Shamino
Linear algebra?
Have you ever gone to high school?
Don't mean to sound harsh or anything..
But Cramer's rule isn't exactly linear
that is algebra 2 material, almost college level algebra :d
• 03-15-2006
BobMcGee123
Quote:
Sorry, I really don't understand linear algebra
Find an implementation of 4x4 matrix inversion online then. The first link typically has every concept implemented in code.
Linear algebra is an advanced branch of mathematics that typically comes after University Calculus 1, 2 and 3 and before Differential equations. I assure you that these topics aren't discussed heavily in high school, mostly not at all.
The only matrix math that I did in high school had to do with connectivity matrices (which I used for my pathfinding algorithm that I posted on this forum some time ago and will be using in my current computer game for path computing).
• 03-15-2006
Shamino
Really Bob? Maybe the linear algebra I took my freshmen year in highschool is different from yours?
We learned up to an introduction of matrices, no further though..
In intermediate algebra we learned alot about graphing, and the rest was factoring, hmm, the quadratic formula... etc..
In algebra two we learn all about matrices, adding, subtracting, inverting, determinants, cramers rule, reduced echelon form, identity matrices, and other various rules..
Maybe linear algebra in college is a review of all of these concepts?
• 03-15-2006
BobMcGee123
Quote:
Maybe the linear algebra I took my freshmen year in highschool is different from yours?
Probably substantially easier and watered-down, due to the fact that you simply do not have enough of a background in math in highschool to continue to the more advanced aspects of it. As I said, linear algebra in many of the curriculum of the schools I've looked at comes only after an in-depth study of calculus. At the University of Maine, which I attended for three semesters, it comes after Calculus 3, and builds upon these concepts (i.e. you didn't mention any study of vector spaces in your high scool classes).
Note that Calculus 3 @ UMaine was multi variable calculus, partial derivatives in three dimensions, vector/polar calculus, stuff of that nature and the start of the study of differential equations.
EDIT:
And, as I stated, I covered none of this in high school, and it's a pretty bad assumption that the study of eigen values and matrices is standard for all students with a high school diploma.
• 03-15-2006
BobMcGee123
If you want a more concise comparison to see if your high school class really is similar to what I had, go to this link, on page 3 it lists the topics that are covered. I'm guessing you're already familiar with roughly half of them, but no more.
http://germain.umemat.maine.edu/faculty/hiebeler/class/syllabi/mat262fall04.pdf#search='umaine%20linear%20algebra '
• 03-15-2006
Shamino
Eh, I think I've gotten the first 4 bullets here in high school.. :eek:
• 03-15-2006
VirtualAce
Not to hijack this but:
Quote:
(which I used for my pathfinding algorithm that I posted on this forum some time ago and will be using in my current computer game for path computing).
You're really going to use that algo Bob? I thought you would use depth-first or some form of stack system we discussed to find paths, or even a node-based system. Good luck with it.
So that's linear algebra eh. Well since I've not had calculus in school and am totally self taught.....it looks to me as if I learned linear algebra before calculus. Perhaps that's why Bob and I are normally on the same page, if just not in different chapters...most of the time.
:)
The 3D mathematics required in computer graphics is extremely complex but thankfully it's well researched and well documented.
I can't post this link here enough. It's like your own personaly library....www.amazon.com
Best prices and if you choose the right re-seller...excellent quality.
A book now costs less than a tank of gas for me, which is quite sad.
Show 80 post(s) from this thread on one page
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http://oeis.org/A014200 | 1,553,367,638,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202924.93/warc/CC-MAIN-20190323181713-20190323203713-00312.warc.gz | 150,816,407 | 3,818 | This site is supported by donations to The OEIS Foundation.
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A014200 Number of solutions to x^2 + y^2 <= n, excluding (0,0), divided by 4. 4
0, 1, 2, 2, 3, 5, 5, 5, 6, 7, 9, 9, 9, 11, 11, 11, 12, 14, 15, 15, 17, 17, 17, 17, 17, 20, 22, 22, 22, 24, 24, 24, 25, 25, 27, 27, 28, 30, 30, 30, 32, 34, 34, 34, 34, 36, 36, 36, 36, 37, 40, 40, 42, 44, 44, 44, 44, 44, 46 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 COMMENTS From Ant King, Mar 15 2013: (Start) The terms of this sequence give a running total of the excess of the 4k + 1 divisors of the natural numbers (from 1 through to n) over their 4k + 3 divisors. To see how good the approximation n * Pi/4 is to a(n), note that a(10^6) = 785387 whereas 10^6 * Pi/4 rounds to 785398. (End) LINKS Seiichi Manyama, Table of n, a(n) for n = 0..10000 FORMULA a(n) = A014198(n) / 4. Lim_{n->inf} a(n)/n = Pi/4. a(n) = n - floor(n/3) + floor(n/5) - floor(n/7) + floor(n/9) - floor(n/11) + ... - Yuval Dekel (dekelyuval(AT)hotmail.com), Aug 28 2003 G.f.: (1/(1 - x))*Sum_{k>=1} x^k/(1 + x^(2*k)). - Ilya Gutkovskiy, Dec 23 2016 MATHEMATICA 1/4*Prepend[SquaresR[2, #]&/@Range[58], 0]//Accumulate (* Ant King, Mar 15 2013 *) CROSSREFS Cf. A014198, A059851. Partial sums of A002654. Sequence in context: A055769 A162217 A123575 * A293522 A319476 A140200 Adjacent sequences: A014197 A014198 A014199 * A014201 A014202 A014203 KEYWORD nonn AUTHOR STATUS approved
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Last modified March 23 14:17 EDT 2019. Contains 321431 sequences. (Running on oeis4.) | 719 | 1,828 | {"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} | 3.625 | 4 | CC-MAIN-2019-13 | latest | en | 0.700295 |
http://science.slashdot.org/story/09/01/19/0022221/chus-final-breakthrough-before-taking-office | 1,418,931,828,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802767828.26/warc/CC-MAIN-20141217075247-00030-ip-10-231-17-201.ec2.internal.warc.gz | 238,057,367 | 53,101 | typodupeerror
Chu's Final Breakthrough Before Taking Office233
Posted by kdawson
from the chewing-up-the-scenery dept.
KentuckyFC writes "While preparing for the job of US Secretary of Energy in the incoming Obama administration (and being director of the Lawrence Berkeley National Laboratory and a Nobel Prize winner to boot), Steven Chu has somehow found time to make a major breakthrough in the world of atom interferometry. One measure of an interferometer's sensitivity is the area that its arms enclose. Chu and colleagues have found a way to increase this area by a factor of 2,500 by canceling out the noise introduced by lasers, which work as beam splitters sending atoms down different arms (abstract). One thing this makes possible is the use of different types of atoms in the same interferometer, allowing a new generation of tests of the equivalence principle. (This is the assumption that the m in F=ma and the m's in F= Gm1.m2/r^2 are the same thing). Let's hope he's got equally impressive breakthroughs planned for his encore as US Secretary of Energy."
This discussion has been archived. No new comments can be posted.
Chu's Final Breakthrough Before Taking Office
• I know, right? (Score:5, Funny)
on Sunday January 18, 2009 @10:31PM (#26512315)
(This is the assumption that the m in F=ma and the m's in F= Gm1.m2/r^2 are the same thing).
That's what she said.
• Re:I know, right? (Score:5, Funny)
on Sunday January 18, 2009 @11:08PM (#26512583)
(This is the assumption that the m in F=ma and the m's in F= Gm1.m2/r^2 are the same thing)
Bah! Just another example of More-of-the-Same! Where's the change we were promised from the Obama Administration!
Just another example of an Obama appointee trying to maintain a status quo!
• On a serious note... (Score:3, Insightful)
(This is the assumption that the m in F=ma and the m's in F= Gm1.m2/r^2 are the same thing).
How can they *not* be the same? Aren't they sort of defined to be equal via the fudge-factor "G" in the second equation? If the m's were different, the value of G would just be adjusted to make them the same again, no?
• Not "final" (Score:3, Interesting)
on Sunday January 18, 2009 @10:32PM (#26512331)
The title seems to imply he wont make any more breakthroughs after taking office. Yet I hope and I think that he should continue to due science work even after taking office and there is no reason why he couldnt.
• Re:Not "final" (Score:5, Funny)
on Sunday January 18, 2009 @11:16PM (#26512629) Homepage
I'd just love to hear him use the phrase, "Look at me, still talking while there's science to do."
• Re: (Score:2)
I'd rather hear, "Stand back! I'm performing science!"
• Re: (Score:2, Funny)
by Anonymous Coward
I want to hear him testify before congress and use the phrase "Science... it works, bitches."
• Re: (Score:2)
I'd just love to hear him use the phrase, "Look at me, still talking while there's science to do."
/starts slow clap
Pure awesome.
• Re: (Score:2)
Hey, they could use more cake in the DOE. :)
• Re:Not "final" (Score:4, Funny)
on Monday January 19, 2009 @06:34AM (#26514473) Journal
The uranium cake was a lie!
• Re:Not "final" (Score:5, Insightful)
<hobbes@xm s n e t.nl> on Monday January 19, 2009 @03:07AM (#26513671)
he should continue to due science work even after taking office and there is no reason why he couldnt.
Right, because as Secretary of Energy he'll have oodles of spare time. It's not as if the nation needs governing or anything.
• Interferowhatsjiggy? (Score:5, Informative)
on Sunday January 18, 2009 @10:34PM (#26512355) Homepage
In case you're an idiot like me, you might appreciate to know that interferometry is about studying the properties of two or more waves by looking at the pattern of interference created by their superposition. The instrument used to interfere the waves together is called an interferometer.
What, you don't remember this stuff from Physics 101? Shame on you...
• Re:Interferowhatsjiggy? (Score:5, Informative)
on Monday January 19, 2009 @01:02AM (#26513151) Homepage Journal
An interferometer is a cool device. By splitting a single beam of light into two, we end up with two identical waves which can then be made to interfere to create patterns that can be observed with the unaided eye. The cool thing is that microscopic changes in path length result in macroscopic changes in the pattern.
One of the neatest applications of this is the Michaelson Morely experiment. A the time of this work, theory was going back and forth between light as a wave and light as particle, and at the time light was a classical wave, which meant it needed a medium to travel, like sound needs air or water waves. It was theorized that the universe was saturated with an aether to carry the light. IIRC, it was theorized that as the Earth moved through the aether, there would be differences in the speed of light based on direction the light is going. In this work, a light beam was split, made to travel in perpendicular direction, and the difference in speed measured.
No difference was measured. this implied that no aether existed. this implies that the waves traveled without a medium. This was quite a surprising result, and was the beginning of the end for classical mechanics. 10 years later we had quantized energy, 15 years later we had the photoelectric effect tell us light was a particle, and a few years after that we have matrix and wave mechanics.
• Re: (Score:2)
a few years after that we have matrix and wave mechanics.
And a few years later we have the whole thing is a hologram [slashdot.org] and the speed of light (and everything else) is subject to where you are because that alters your light cone and hence your local laws of physics.
Sometimes I think the more you know, the more aware you are of how much you don't know.
• Re: (Score:2)
If light is both a wave and a particle, does that mean that the light "wave" is actually a standing wave (vibration?) inside the light particle (aka photon)? With the particle moving the wave would appear to be travelling.
What happens if two photons collide (head-on)? Do they go straight through each other? Do they bounce? With the theory that every particle exhibits wave/particle duality, they should bounce.
• Re: (Score:2)
I'm used to interferometry (in the astronomical context), but a particle physicist I'm not, and this abstract left me wishing there were an abstract of the abstract.
• Obviously... (Score:5, Funny)
on Sunday January 18, 2009 @10:45PM (#26512445)
Obviously this is just an attempt by the democrats to distract from the nation's problems as Obama takes office. They should be ashamed of themselves for exploiting the public's interest in atom interferometry this way.
• Nice Change (Score:3, Interesting)
on Sunday January 18, 2009 @10:46PM (#26512453)
It's a nice change from the previous high level government officials of the Bush Administration, who were appointed not based on their knowledge and experience in a given field, but their willingness to bend the truth according to the Bush administration dogma.
• Re:Nice Change (Score:5, Interesting)
on Sunday January 18, 2009 @11:38PM (#26512753)
It's a nice change from the previous high level government officials of the Bush Administration, who were appointed not based on their knowledge and experience in a given field, but their willingness to bend the truth according to the Bush administration dogma.
That was my initial reaction. But at that level of responsibility I much prefer someone being appointed for their competency to manage well rather than their ability to do technical work. I have no idea if Chu is a good manager or not, just saying that the Peter Principle is something to be avoided.
• Re: (Score:3, Insightful)
Two things:
First, someone as generally intelligent as Chu should be able to figure his job out no matter what. We're not talking about idiot savants here, we're talking about people who are incredibly good learners.
Second, http://en.wikipedia.org/wiki/Lawrence_Berkeley_National_Laboratory [wikipedia.org]
• Re:Nice Change (Score:5, Insightful)
<pgunn@dachte.org> on Monday January 19, 2009 @05:44AM (#26514261) Homepage Journal
I don't know a lot about Chu, but over the years I've worked at a University, I've come to the conclusion that people skills and scientific skills are largely orthoganal - some people have both, but a number of researchers are either extraordinarily shy and nonconfrontational or egomaniacs, neither of which make good leaders. I hope that Chu is of the sort that's good at both.
• Re: (Score:2)
"That was my initial reaction. But at that level of responsibility I much prefer someone being appointed for their competency to manage well rather than their ability to do technical work. I have no idea if Chu is a good manager or not, just saying that the Peter Principle is something to be avoided."
I can appreciate that, but I think there's also an advantage in someone like this being elevated to that position where they may serve as inspiration for others.
Hadn't heard of the Peter Principle before - chee
• Re: (Score:2, Interesting)
What makes you think Bush appointees where good managers? Or where even close to being competent for the position they where appointed to?
• Re: (Score:3, Informative)
Err, he is the director of LBNL, so I would assume he would be a good manager as well as a good scientist.
• Re:Nice Change (Score:4, Insightful)
on Monday January 19, 2009 @04:44AM (#26514033) Journal
I continue to find it strange that so many people think that competence in the core field of a department is second to management skills. What makes management so special that you can rely on a collaborator to have the core competency but not the management skills ?
Of course, I'd rather have one with both but, well, is it really preferable to have a good manager with poor scientific skills at the head of what is mainly a technology department rather than a scientist with poor managerial skills (which, some clues indicate, Chu is not) ?
• Re: (Score:2)
Because at the highest levels it is more important to understand how to mange your technical people to get the job done that it is to know exactly what they are doing. That's not to say that some understanding to the technology involved shouldn't be a goal of a good manager. Clearly you can't make decisions about direction without know what you are directing; but at the highest level of an organization I rather see a brilliant manager with a good overview of the technology, than a brilliant technician who
• Re: (Score:2)
Because at the highest levels it is more important to understand how to mange your technical people to get the job done that it is to know exactly what they are doing.
Well, I disagree here. In this particular department, I feel it is important for the decision maker to know how nuclear waste decays, for example, and what are the reasonable expectation one can have in a 5 years horizon about nuclear recycling. He will have to make decisions based on this particular knowledge. Such decisions could endanger whole regions for several centuries if done badly. I'd rather have him use ten times the resources a good manager would need to make his administration work than having
• Re: (Score:2)
That was my initial reaction. But at that level of responsibility I much prefer someone being appointed for their competency to manage well rather than their ability to do technical work. I have no idea if Chu is a good manager or not, just saying that the Peter Principle is something to be avoided.
That's rather unfair. I'm guessing this wasn't a one man technical effort and that Steven Chu led a team. Now that doesn't immediately make him competent to lead at the national level but I don't think he could b
• Re:Nice Change (Score:4, Insightful)
on Sunday January 18, 2009 @11:45PM (#26512807)
I think this is even truer than it sounds. A lot of people Obama's tagged have very little incentive to take the position other than if they feel they might be able to get stuff done. All the good scientists I know mostly just want to work on cool and interesting things and see administration and bureaucracy as a necessary evil, making the aspects of these jobs which can be exploited for monetary gain less attractive than getting back to a lab. Furthermore, any career politicians in their positions would be ruined by going around the administration, whereas it's not like Steven Chu will ever struggle to find a job he wants. The upshot is that these guys have little to lose by being forced to resign, whereas it'd look horrible for Obama if they go off in a huff because he won't listen to them. Obama's been accused of talking change without having any substance, but I think he just hit the point of no return on following good science. It'll certainly be nice to see Nobel Prizes having more weight than Magic 8-Balls.
• Re: (Score:2)
the danger with that, is he might not know the in's and out's of washington and be ineffective. the best managers i've ever had were non techincal (also the worst where as well), he might not be willing to listen to other points of view either - you just have to look on /. to see how unwilling the science/geek types are to consider other peoples opinions as having some merit.
• Paper shuffling is not a good job (Score:3, Interesting)
At least it should not be a national goal to take the people who are expanding the realm of human knowledge and chain them to a desk managing federal middle managers. It's cruel. It's wasteful.
Kudos to the incoming administration for being able to figure out who the thinkers in their country are. That's a refreshing change from the previous administration. Now please - for the sake of us all - when you identify them, leave them in place and appoint administrators to get stuff out of their way. For all
• For the Record... (Score:5, Informative)
by Anonymous Coward on Sunday January 18, 2009 @10:46PM (#26512459)
From http://arXiv.org/auth/show-endorsers/0901.1819 :
Holger Müller: Is registered as an author of this paper.
Sven Herrmann, Sheng-wey Chiow and Steven Chu are not registered as owners of this paper.
Sure, it doesn't nail down who did what exactly, but if I had a question about the paper, I'm asking Holger first.
• Re:For the Record... (Score:5, Informative)
by Anonymous Coward on Sunday January 18, 2009 @11:22PM (#26512667)
According to http://arxiv.org/help/not-registered.html, Steven Chu may not be a registered owner for as simple a reason as not having a user account with that website.
That said, Mueller is listed as final author of the paper and Steven Chu is listed second to last, which pretty much throws all assumptions based on position out the window. (See http://www.phdcomics.com/comics/archive.php?comicid=562 for a comedic but sadly true primer).
Mueller served as a postdoc under Chu but both are professors. Based on Mueller's other publications (http://www.physics.berkeley.edu/research/faculty/mueller.html) and Chu's second-to-last position, I'd say the other two names are postdocs in his lab. Really, I'd ask those two if you want to know the specifics on this experiment. Blind guess at Chu's role, but probably functionally a PI - more of an adviser role.
• Re: (Score:3, Interesting)
Chu is a big name, so its hard to tell whether he was the driving force behind this research, or tossed on the list of authors to get funding. Muller is an Assistant Professor. Chiow is a post-doc.
Herrman, I can't find a position for via a quick google search, but it looks like he's been putting out papers under Muller for 5 years, which means he's been working under him even longer. The only way you'd work under one person for that long without having a larger internet presence is as a meek and lowly gr
• Re: (Score:3, Informative)
From http://arxiv.org/auth/show-endorsers/0901.1819 [arxiv.org] :
Holger Müller: Is registered as an author of this paper.
This means that Holger Müller is the guy who logged onto arXiv and uploaded the paper. It has nothing to do with who actually contributed how much to the research.
• So Let me get this straight (Score:5, Funny)
on Sunday January 18, 2009 @10:53PM (#26512491)
Our incoming president reads spiderman comics and his secretary of energy is some incredible nobel prize winning genius who ran a program called "Bio-X", can we possibly get more nerdy?
• Re:So Let me get this straight (Score:5, Insightful)
on Sunday January 18, 2009 @11:00PM (#26512539)
Our incoming president reads spiderman comics and his secretary of energy is some incredible nobel prize winning genius who ran a program called "Bio-X", can we possibly get more *AWESOME*?
There, fixed that for you.
• Re: (Score:3)
My (well, not mine really, as I can't vote) Prime Minister reads manga [www.cbc.ca] and his ex-Defense Minister builds plastic models [findarticles.com] of battleships and fighter planes, but they are far, far, far from *AWESOME*!
• Re: (Score:2)
We could find out Obama has some sort of underground lab where he works on armor powered by an implanted reactor of some sort. Yes, we can definitely get more awesome.
• Nerdy? (Score:2)
I don't know about that, but it certainly sounds like a recipe for disaster.
When Obama turns into an evil Oba-man we'll know who to blame.
• Dammit Steve Chu (Score:3, Funny)
on Sunday January 18, 2009 @11:31PM (#26512731) Homepage
He's gone and made us all have to feel inferior again. Seriously, does the man just exist to make the rest of us feel like we're idiots who can't get anything accomplished in life? I have to ask myself what Steve Chu could do to be more impressive and at this point the list is pretty short:
1. Prove the Riemann Hypothesis.
2. Bring peace to the Mid-East.
3. Turn out that to have made an amazingly human AI in his free time that escaped and now calls itself Randall Munroe and writes xkcd.
• Re:Dammit Steve Chu (Score:5, Funny)
on Monday January 19, 2009 @12:47AM (#26513093)
No, he'll bring Linux to the desktop, cure cancer, and get Adobe to release 64-bit Photoshop for the Mac... in that order.
• Re: (Score:2)
I see a successor to the Chuck Norris facts...
• After his first 100 days in office... (Score:5, Insightful)
on Monday January 19, 2009 @01:30AM (#26513293)
... I'll bet Chu will be thinking that physics is a piece of cake compared to governing the US.
• They're the same thing, mostly. (Score:2)
(This is the assumption that the m in F=ma and the m's in F= Gm1.m2/r^2 are the same thing).
I think 350 years of experiments on Newtonian physics have shown that they are the same thing except in weird-ass quantum or near-speed-of-light situations that don't really matter anyway.
Don't go confusing the high school kiddies, please. They're already confused enough about evolution thanks to media spinelessness.
• Here is the difference (Score:2, Interesting)
. . . between Obama and Bush. Bush appointed a professional politician (http://en.wikipedia.org/wiki/Spencer_Abraham) and then someone slightly more qualified, (http://en.wikipedia.org/wiki/Samuel_W._Bodman), a venture capitalist who had attended MIT. Abraham had nothing to do with energy Bodman has done nothing but executive positions for the last thirty years. Obama chose someone who's really qualified and isn't financially tied to our current energy industries. Considering that the inauguration is tomorr
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Hints for Problem 15.26 Some of the parameters required to solve 15.26 (S&H) need to be found in the literature (e.g. Perry’s). We have provided those below to save time: 1. The concentration of water in air at 80°F, 1 atm and 80% relative humidity (from psychrometric chart) air dry lb O H lb 0.0177 2 = 2. The diffusivity of water vapor in air at 1 atm, 80°F s m 10 26 . 0 D 2 4 - i × = 3. The viscosity of air at 80 °F s m kg 10 75 . 1 5 - × = µ Additional advice: Pay attention to units throughout the problem. You can use the ideal gas law to calculate the density (in lb/ft 3 ) of the gas entering the bed, which is a mixture of water vapor and air.
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Unformatted text preview: • Assume that the cross-sectional area of the bed is 1 ft 2 . • Use the equation b b p 1-∈ ρ = ρ to calculate ρ p , the particle density, (lb gel/ ft 3 particles) from the density of the silica given in the problem statement 3 b ft lb 39 = ρ and the void fraction 47 . b = ∈ . • The units of the equilibrium constant, K, should be in gas ft / O H lb gel ft / O H lb 3 2 3 2 to use Equation 15-106 (S&H). • If using Excel to solve the Klinkenberg equation, use erf carefully and remember that erf (-z) = - erf (z)....
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This note was uploaded on 11/27/2011 for the course CHEMICAL E 10.302 taught by Professor Clarkcolton during the Fall '04 term at MIT.
Ask a homework question - tutors are online | 471 | 1,610 | {"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} | 2.90625 | 3 | CC-MAIN-2017-26 | longest | en | 0.853157 |
http://www.worldplenty.com/g7_div_lboz.htm | 1,669,704,362,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710690.85/warc/CC-MAIN-20221129064123-20221129094123-00201.warc.gz | 107,494,845 | 2,509 | Dividing Pounds and Ounces How to divide pounds and ounces Convert pounds and ounces to ounces by multiplying pounds by 16 and adding the number of ounces Perform the required division to determine the number of ounces. Convert the ounces to pounds and ounces by dividing by 16. The quotient is the number of pounds and the remainder is the number of ounces. Example: Divide 10 pounds 0 ounces by 4 Convert 10 pounds to ounces by multiplying 10 by 16: 16 * 10 = 160 ounces Add the number of extra ounces: 160 + 0 = 160 ounces Perform the required division: 160 ÷ 4 = 40 ounces Convert to pounds and ounces by dividing by 16: 40 ÷ 16 = 2 R 8 The quotient (2) is the number of pounds and the remainder (8) is the number of ounces. Answer: 2 pounds 8 ounces Return to Top
#### Dividing Pounds and Ounces.
÷ = Pounds Ounces | 216 | 826 | {"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-2022-49 | latest | en | 0.885133 |
https://math.answers.com/other-math/If_you_roll_two_dice_what_are_the_odds_that_you_get_a_one_or_a_five | 1,726,697,768,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651941.8/warc/CC-MAIN-20240918201359-20240918231359-00623.warc.gz | 360,929,003 | 52,361 | 0
If you roll two dice what are the odds that you get a one or a five?
Updated: 4/28/2022
Wiki User
14y ago
I suppose you mean, at least one of those numbers. Just calculate the probability of NOT getting any of those, and take the complement. The probability of not getting a one nor a five on a single die is 4/6 or 2/3. For two dice, the probability is 2/3 x 2/3 = 4/9. So, the probability of getting at least a one or a five with two dice is 1 - 4/9 = 5/9.
I suppose you mean, at least one of those numbers. Just calculate the probability of NOT getting any of those, and take the complement. The probability of not getting a one nor a five on a single die is 4/6 or 2/3. For two dice, the probability is 2/3 x 2/3 = 4/9. So, the probability of getting at least a one or a five with two dice is 1 - 4/9 = 5/9.
I suppose you mean, at least one of those numbers. Just calculate the probability of NOT getting any of those, and take the complement. The probability of not getting a one nor a five on a single die is 4/6 or 2/3. For two dice, the probability is 2/3 x 2/3 = 4/9. So, the probability of getting at least a one or a five with two dice is 1 - 4/9 = 5/9.
I suppose you mean, at least one of those numbers. Just calculate the probability of NOT getting any of those, and take the complement. The probability of not getting a one nor a five on a single die is 4/6 or 2/3. For two dice, the probability is 2/3 x 2/3 = 4/9. So, the probability of getting at least a one or a five with two dice is 1 - 4/9 = 5/9.
Wiki User
14y ago
Wiki User
14y ago
I suppose you mean, at least one of those numbers. Just calculate the probability of NOT getting any of those, and take the complement. The probability of not getting a one nor a five on a single die is 4/6 or 2/3. For two dice, the probability is 2/3 x 2/3 = 4/9. So, the probability of getting at least a one or a five with two dice is 1 - 4/9 = 5/9. | 577 | 1,921 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2024-38 | latest | en | 0.948215 |
http://math.stackexchange.com/questions/61099/do-discontinuous-harmonic-functions-exist?answertab=votes | 1,455,074,557,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701158601.61/warc/CC-MAIN-20160205193918-00263-ip-10-236-182-209.ec2.internal.warc.gz | 155,851,781 | 20,027 | # Do discontinuous harmonic functions exist?
A function, $u$, on $\mathbb R^n$ is normally said to be harmonic if $\Delta u=0$, where $\Delta$ is the Laplacian operator $\Delta=\sum_{i=1}^n\frac{\partial^2}{\partial x_i^2}$. So obviously, according to this definition, $u$ must be twice differentiable and therefore continuous. However, if $u$ is at least continuous, being harmonic is equivalent to the condition that the average value of $u$ in any ball centered at a point $p$ is actually equal to $u(p)$: $$u(p)=\frac{1}{Vol(B_r(p))}\int_{B_r(p)}u\, dV.$$ My question is whether there are any integrable functions $u$ for which the above equation holds for all $r$ and $p$ but for which $u$ is not continuous.
-
I don't know much about this, but I would have said that a much stronger result holds. I would have said that a harmonic hyperfunction is automatically analytic... (I'm considering the usual laplacian on $\mathbb R^n$.) – Pierre-Yves Gaillard Sep 1 '11 at 9:02
@Pierre: $\Phi_r$ in my answer is real analytic on the interior of $B_r(0)$. Therefore, with only a bit more work, my answer shows that $u$ is real-analytic. – robjohn Sep 1 '11 at 12:03
Dear @robjohn: Thanks! I agree. I hadn't read the question carefully enough. The question doesn't ask explicitly if $u$ is harmonic, and, logically, the answers answer the question as asked. But it seems to me the question "is $u$ harmonic?" is very natural. Wikipedia says: "all locally integrable functions satisfying the (volume) mean-value property are infinitely differentiable and harmonic functions as well". A proof is a sketched. – Pierre-Yves Gaillard Sep 1 '11 at 13:27
On the interior of its domain, a harmonic function is infinitely differentiable (even with just the mean value property). However, on the boundary of its domain, it can limit to a discontinuous function.
Mean Value Property: $u(x)=\frac{1}{\Omega_nr^n}\int_{B_r(x)}u(v)\;\mathrm{d}v$ for any $r>0$ so that $B_r(x)\subset\mathrm{Dom}(u)$, where $\Omega_n$ is the volume of the unit sphere in $\mathbb{R}^n$.
Define $\displaystyle\phi_r(s)=\left\{\begin{array}{cl}e^{s^2/(s^2-r^2)}&\text{for }0\le s<r\\0&\text{for }s\ge r\end{array}\right.\hspace{.25in}$ then $\Phi_r(x)=\phi_r(|x|)$ is in $C_c^\infty(\mathbb{R}^n)$
We can average the Mean Value Property over a range of sphere radii: \begin{align} u(x)&=\frac{1}{\int_0^r\Omega_ns^n\phi_r^\prime(s)\mathrm{d}s}\int_0^r\int_{B_s(x)}u(y)\;\mathrm{d}y\;\phi_r^\prime(s)\;\mathrm{d}s\\ &=\frac{1}{\int_0^r\phi_r(s)n\Omega_ns^{n-1}\mathrm{d}s} \int_0^r\phi_r(s)\int_{S_s(x)}u(y)\;\mathrm{d}\sigma(y)\;\mathrm{d}s \\ &=\frac{1}{\int_{\mathbb{R}^n}\Phi_r(y)\mathrm{d}y} \int_{\mathbb{R}^n}\Phi_r(y-x)u(x)\;\mathrm{d}x\\ &=\frac{1}{\int_{\mathbb{R}^n}\Phi_r(y)\mathrm{d}y}\Phi_r*u(x) \end{align} Since $\Phi_r$ is in $C^\infty$, $u$ must be also.
Thus, the Mean Value Property implies $C^\infty$ on the interior of the domain.
-
I should have seen this. I forgot that the convolution of an integrable function with a $C^\infty$ function is also $C^\infty$. – Grumpy Parsnip Sep 1 '11 at 11:10
@Rohan: it is indeed integration by parts. Before $$v=\int_{B_s(x)}u(y)\;\mathrm{d}y$$ $$dw=\phi_r^\prime(s)\;\mathrm{d}s$$ and after $$w=\phi_r(s)$$ $$dv=\int_{S_s(x)}u(y)\;\mathrm{d}\sigma(y)\;\mathrm{d}s$$ The minus sign is cancelled by the integration by parts in the denominator out front. – robjohn Nov 22 '11 at 1:54
For fixed $r$, the measure of the symmetric difference of $B_r(x)$ and $B_r(y)$ goes to $0$ as $y$ goes to $x$. Therefore, by (local) absolute continuity of the integral (of a locally integrable function), $\int_{B_r(x)}udV-\int_{B_r(y)}udV$ goes to $0$ as $y$ goes to $x$. This implies that $u$ is continuous.
-
This reminds me of Nelson's proof of Liouville's theorem (only tangentially related). – t.b. Sep 1 '11 at 6:02
I would say "it depends". I was rather surprised by the following example, shown to me by a colleauge the other day. Dropping the assumptions that $u$ is locally integrable, strange things can happen!
Let $f(z) = \exp(-1/z^4)$, and let $u(x,y) = \Re f(x+iy)$. Then $u$ is clearly harmonic outside the origin. Extend $u$ to whole plane, by putting $u(0,0) = 0$. Then (outside the origin) $$u(x,0) = \exp(-1/x^4) \quad\text{and}\quad u(0,y) = \exp(-1/y^4).$$ It is not hard to see that both partial derivatives of $u$ exist at the origin, and in fact a little more work shows that the same holds for the second order partial derivatives. (However $u$ is of course not continuous at the origin, and neither is any of its derivatives.) All the derivatives above are in fact equal to $0$! In other words, the function $u$ satisfies "Laplace's equation" $$u''_{xx} + u''_{yy} = 0$$ everywhere on $\mathbb{R}^2$. A moment of contemplation also shows that $u$ fails to be integrable on any neighborhood of the origin, and hence does not define a distribution.
-
For more about the particular case of $\mathbb{R}^2$ which has intimate connections with complex analysis, one should see this article in the AMM. – Willie Wong Jan 31 '12 at 12:44 | 1,596 | 5,076 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2016-07 | longest | en | 0.927688 |
https://www.physicsforums.com/threads/derive-the-energy-equation-for-a-van-der-waal-gas.744397/ | 1,624,295,022,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488286726.71/warc/CC-MAIN-20210621151134-20210621181134-00172.warc.gz | 864,217,976 | 15,924 | # Derive the energy equation for a van der Waal gas.
## Homework Statement
Derive the energy equation for a van der Waal gas when T and v are the independent variables.
## Homework Equations
I'm required to start from: $du=\frac{\partial u}{\partial T} dT +\frac{\partial u}{\partial v} dv$
## The Attempt at a Solution
We'll when I first started this problem. I just started doing random derivatives that made sense to me but after looking it over I had no clue what I was doing.
Does anyone know what the answer I should be getting after I fully derive it?
## Answers and Replies
vanhees71
Gold Member
I don't know, what's given. So I can't help to find an answer to your question. One way to derive the caloric equation of state, i.e., the internal energy $U$ is to use the canonical partition sum
$$Z=\frac{z^N}{N!}$$
with the single-particle partition function
$$z=\frac{V^*}{\Lambda^3} \exp \left (-\frac{\phi}{2 T} \right),$$
where $V^*=V-2 \pi N d^3/3$ is the available volume (geometrical volume minus excluded volume due to the hard-sphere model for the molecules), and $\phi$ is the interaction energy of the particles
$$\phi=\frac{N}{V} \int_d^\infty \mathrm{d} r \; U_{\text{rel}}(r) 4 \pi r^2,$$
with the effective two-body potential
$$U_{\text{rel}}(r)=\begin{cases} \infty & \text{for} \quad r \leq d ,\\ -\alpha (d/r)^6 & \text{for} \quad r>d. \end{cases}$$
The Helmholtz free energy is given by
$$A(T,V,N)=-T \ln Z,$$
and from this you can derive all thermodynamical quantities from the usual thermodynamic relations like
$$p=-\left (\frac{\partial A}{\partial V} \right)_{T,N}$$
etc. The internal energy is given by the usual Legendre transformation
$$U=A+T S.$$
Chestermiller
Mentor
$$dU=TdS-PdV=T\left(\frac{∂S}{∂T}\right)dT+T\left(\frac{∂S}{∂V}\right)dV-PdV= C_VdT+(T\left(\frac{∂S}{∂V}\right)-P)dV$$
The next step is to determine ∂S/∂V at constant T. This can be obtained from a Maxwell relation, starting from the equation dA=-SdT-PdV.
S=-∂A/∂T
P=-∂A/∂V
So $$\frac{∂S}{∂V}=\frac{∂P}{∂T}$$
Therefore,
$$dU=C_VdT+(T\left(\frac{∂P}{∂T}\right)-P)dV$$
The second term in this equation is zero for an ideal gas, but not for a real gas. Just substitute the van der Waals equation into the second term of this equation.
chet
srecko97 | 743 | 2,265 | {"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} | 3.609375 | 4 | CC-MAIN-2021-25 | latest | en | 0.810581 |
http://tutor1.net/wikibooks/159090 | 1,607,205,610,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141750841.83/warc/CC-MAIN-20201205211729-20201206001729-00555.warc.gz | 98,804,311 | 21,760 | [<< wikibooks] Functional Analysis/Banach spaces
Let
X
{\displaystyle {\mathcal {X}}}
be a linear space. A norm is a real-valued function
f
{\displaystyle f}
on
X
{\displaystyle {\mathcal {X}}}
, with the notation
‖
⋅
‖
=
f
(
⋅
)
{\displaystyle \|\cdot \|=f(\cdot )}
, such that
(i)
‖
x
+
y
‖
≤
‖
x
‖
+
‖
y
‖
{\displaystyle \|x+y\|\leq \|x\|+\|y\|}
(w:triangular inequality)
(ii)
‖
λ
x
‖
=
|
λ
|
‖
x
‖
{\displaystyle \|\lambda x\|=|\lambda |\|x\|}
for any scalar
λ
{\displaystyle \lambda }
(iii)
‖
x
‖
=
0
{\displaystyle \|x\|=0}
implies
x
=
0
{\displaystyle x=0}
.(ii) implies that
‖
0
‖
=
0
{\displaystyle \|0\|=0}
. This and (i) then implies
0
=
‖
x
−
x
‖
≤
‖
x
‖
+
‖
−
x
‖
=
2
‖
x
‖
{\displaystyle 0=\|x-x\|\leq \|x\|+\|-x\|=2\|x\|}
for all
x
{\displaystyle x}
; that is, norms are always non-negative. A linear space with a norm is called a normed space. With the metric
d
(
x
,
y
)
=
‖
x
−
y
‖
{\displaystyle d(x,y)=\|x-y\|}
a normed space is a metric space. Note that (i) implies that:
‖
x
‖
≤
‖
x
−
y
‖
+
‖
y
‖
{\displaystyle \|x\|\leq \|x-y\|+\|y\|}
and
‖
y
‖
≤
‖
x
−
y
‖
+
‖
x
‖
{\displaystyle \|y\|\leq \|x-y\|+\|x\|}
and so:
|
‖
x
‖
−
‖
y
‖
|
≤
‖
x
−
y
‖
{\displaystyle |\|x\|-\|y\||\leq \|x-y\|}
. (So, the map
x
↦
‖
x
‖
{\displaystyle x\mapsto \|x\|}
is continuous; in fact, 1-Lipschitz continuous.)
A complete normed space is called a Banach space. While there is seemingly no prototypical example of a Banach space, we still give one example of a Banach space:
C
(
K
)
{\displaystyle {\mathcal {C}}(K)}
, the space of all continuous functions on a compact space
K
{\displaystyle K}
, can be identified with a Banach space by introducing the norm:
‖
⋅
‖
=
sup
K
|
⋅
|
{\displaystyle \|\cdot \|=\sup _{K}|\cdot |}
It is a routine exercise to check that this is indeed a norm. The completeness holds since, from real analysis, we know that a uniform limit of a sequence of continuous functions is continuous. In concrete spaces like this one, one can directly show the completeness. More often than that, however, we will see that the completeness is a necessary condition for some results (especially, reflexivity), and thus the space has to be complete. The matter will be picked up in the later chapters.
Another example of Banach spaces, which is more historical, is an
l
p
{\displaystyle l_{p}}
space; that is, the space of convergent series. (The geometric properties of
l
p
{\displaystyle l_{p}}
spaces will be investigated in Chapter 4.) It is clear that
l
p
{\displaystyle l_{p}}
is a linear space, since the sum of two p-convergent series is again p-convergent. That the
l
p
{\displaystyle l_{p}}
norm is in fact a norm follows from
Lemma 2.1. If
‖
x
‖
p
<
∞
{\displaystyle \|x\|_{p}<\infty }
, then
‖
x
‖
p
=
sup
‖
y
‖
q
=
1
|
∑
x
j
y
j
¯
|
{\displaystyle \|x\|_{p}=\sup _{\|y\|_{q}=1}|\sum x_{j}{\overline {y_{j}}}|}
, where
1
/
p
+
1
/
q
=
1
{\displaystyle 1/p+1/q=1}
Proof. By Hölder's inequality,
sup
‖
y
‖
q
=
1
|
∑
x
j
y
j
¯
|
≤
‖
x
‖
p
{\displaystyle \sup _{\|y\|_{q}=1}|\sum x_{j}{\overline {y_{j}}}|\leq \|x\|_{p}}
Conversely, if
‖
x
‖
p
=
1
{\displaystyle \|x\|_{p}=1}
, then taking
y
j
=
x
j
¯
|
x
j
|
p
−
2
{\displaystyle y_{j}={\overline {x_{j}}}|x_{j}|^{p-2}}
we have:
∑
x
j
y
j
¯
=
∑
|
x
j
|
p
=
1
{\displaystyle \sum x_{j}{\overline {y_{j}}}=\sum |x_{j}|^{p}=1}
, while
‖
y
‖
q
=
(
∑
|
x
j
|
(
p
−
1
)
q
)
1
/
q
=
1
{\displaystyle \|y\|_{q}=(\sum |x_{j}|^{(p-1)q})^{1/q}=1}
.since
p
=
(
p
−
1
)
q
{\displaystyle p=(p-1)q}
. More generally, if
‖
x
‖
p
≠
0
{\displaystyle \|x\|_{p}\neq 0}
, then
x
‖
x
‖
p
=
sup
‖
y
‖
q
=
1
|
∑
x
j
y
j
¯
|
1
‖
x
‖
p
{\displaystyle {x \over \|x\|_{p}}=\sup _{\|y\|_{q}=1}|\sum x_{j}{\overline {y_{j}}}|{1 \over \|x\|_{p}}}
.Since the identity is obvious when
x
=
0
{\displaystyle x=0}
, the proof is complete.
◻
{\displaystyle \square }
Now, it remains to show that an
l
p
{\displaystyle l_{p}}
space is complete. For that, let
x
k
∈
l
p
{\displaystyle x_{k}\in l_{p}}
be a Cauchy sequence. This means explicitly that
∑
n
=
1
∞
|
(
x
k
)
n
−
(
x
j
)
n
|
2
→
0
{\displaystyle \sum _{n=1}^{\infty }|(x_{k})_{n}-(x_{j})_{n}|^{2}\to 0}
as
n
,
m
→
∞
{\displaystyle n,m\to \infty }
For each
n
{\displaystyle n}
, by completeness,
lim
k
→
∞
(
x
k
)
n
{\displaystyle \lim _{k\to \infty }(x_{k})_{n}}
exists and we denote it by
y
n
{\displaystyle y_{n}}
. Let
ϵ
>
0
{\displaystyle \epsilon >0}
be given. Since
x
k
{\displaystyle x_{k}}
is Cauchy, there is
N
{\displaystyle N}
such that
∑
n
=
1
∞
|
(
x
k
)
n
−
(
x
j
)
n
|
2
<
ϵ
{\displaystyle \sum _{n=1}^{\infty }|(x_{k})_{n}-(x_{j})_{n}|^{2}<\epsilon }
for
k
,
j
>
N
{\displaystyle k,j>N}
Then, for any
k
>
N
{\displaystyle k>N}
,
∑
n
=
1
∞
|
(
x
k
)
n
−
y
n
|
2
=
sup
m
≥
1
∑
n
=
1
m
|
(
x
k
)
n
−
y
n
|
2
=
sup
m
≥
1
lim
j
→
∞
∑
n
=
1
m
|
(
x
k
)
n
−
(
x
j
)
n
|
2
<
ϵ
{\displaystyle \sum _{n=1}^{\infty }|(x_{k})_{n}-y_{n}|^{2}=\sup _{m\geq 1}\sum _{n=1}^{m}|(x_{k})_{n}-y_{n}|^{2}=\sup _{m\geq 1}\lim _{j\to \infty }\sum _{n=1}^{m}|(x_{k})_{n}-(x_{j})_{n}|^{2}<\epsilon }
Hence,
x
k
→
y
{\displaystyle x_{k}\to y}
with
y
=
∑
n
=
1
∞
y
n
{\displaystyle y=\sum _{n=1}^{\infty }y_{n}}
.
y
{\displaystyle y}
is in fact in
l
p
{\displaystyle l_{p}}
since
‖
y
‖
2
≤
‖
y
−
x
n
‖
2
+
‖
x
n
‖
2
<
∞
{\displaystyle \|y\|_{2}\leq \|y-x_{n}\|_{2}+\|x_{n}\|_{2}<\infty }
. (We stress the fact that the completeness of
l
p
{\displaystyle l_{p}}
spaces come from the fact that the field of complex numbers is complete; in other words,
l
p
{\displaystyle l_{p}}
spaces may fail to be complete if the base field is not complete.)
l
p
{\displaystyle l_{p}}
is also separable; i.e., it has a countable dense subset. This follows from the fact that
l
p
{\displaystyle l_{p}}
can be written as a union of subspaces with dimensions 1, 2, ..., which are separable. (TODO: need more details.)
We define the operator norm of a continuous linear operator
f
{\displaystyle f}
between normed spaces
X
{\displaystyle {\mathcal {X}}}
and
Y
{\displaystyle {\mathcal {Y}}}
, denoted by
‖
f
‖
{\displaystyle \|f\|}
, by
‖
f
‖
=
sup
‖
x
‖
X
≤
1
‖
f
(
x
)
‖
Y
{\displaystyle \|f\|=\sup _{\|x\|_{\mathcal {X}}\leq 1}\|f(x)\|_{\mathcal {Y}}}
2 Theorem Let
T
{\displaystyle T}
be a linear operator from a normed space
X
{\displaystyle {\mathcal {X}}}
to a normed space
Y
{\displaystyle {\mathcal {Y}}}
.
(i)
T
{\displaystyle T}
is continuous if and only if there is a constant
C
>
0
{\displaystyle C>0}
such that
‖
T
x
‖
≤
C
‖
x
‖
{\displaystyle \|Tx\|\leq C\|x\|}
for all
x
∈
X
{\displaystyle x\in X}
(ii)
‖
f
‖
=
inf
{
{\displaystyle \|f\|=\inf\{}
any
C
{\displaystyle C}
as in (i)
}
=
sup
‖
x
‖
=
1
‖
f
(
x
)
‖
{\displaystyle \}=\sup _{\|x\|=1}\|f(x)\|}
if
X
{\displaystyle {\mathcal {X}}}
has nonzero element. (Recall that the inf of the empty set is
∞
{\displaystyle \infty }
.)Proof: If
‖
T
‖
<
∞
{\displaystyle \|T\|<\infty }
, then
‖
T
(
x
n
−
x
)
‖
Y
≤
‖
T
‖
‖
x
n
−
x
‖
X
→
0
{\displaystyle \|T(x_{n}-x)\|_{\mathcal {Y}}\leq \|T\|\|x_{n}-x\|_{\mathcal {X}}\to 0}
as
x
n
→
x
{\displaystyle x_{n}\to x}
. Hence,
T
{\displaystyle T}
is continuous. Conversely, suppose
‖
T
‖
=
∞
{\displaystyle \|T\|=\infty }
. Then we can find
x
n
∈
X
{\displaystyle x_{n}\in {\mathcal {X}}}
with
‖
x
n
‖
X
≤
1
{\displaystyle \|x_{n}\|_{\mathcal {X}}\leq 1}
and
‖
T
x
n
‖
≥
n
{\displaystyle \|Tx_{n}\|\geq n}
. Then
x
n
n
→
0
{\displaystyle {x_{n} \over n}\to 0}
while
‖
T
(
x
n
n
)
‖
↛
0
{\displaystyle \left\|T\left({x_{n} \over n}\right)\right\|\not \to 0}
. Hence,
T
{\displaystyle T}
is not continuous. The proof of (i) is complete. For (ii), see w:operator norm for now. (TODO: write an actual proof).
◻
{\displaystyle \square }
It is clear that an addition and a scalar multiplication are both continuous. (Use a sequence to check this.) Since the inverse of an addition is again addition, an addition is also an open mapping. Ditto to nonzero-scalar multiplications. In other words, translations and dilations of open (resp. closed) sets are again open (resp. closed).
Not all linear operators are continuous. Take the linear operator defined by
D
(
P
)
=
X
P
′
{\displaystyle D(P)=XP'}
on the normed vector space of polynomials
R
[
X
]
{\displaystyle \mathbb {R} [X]}
with the suprenum norm
‖
P
‖
∞
=
sup
x
∈
[
0
,
1
]
|
P
(
x
)
|
{\displaystyle \|P\|_{\infty }=\sup _{x\in [0,1]}|P(x)|}
; since
D
(
X
n
)
=
n
X
n
{\displaystyle D(X^{n})=nX^{n}}
, the unit ball is not bounded and hence this linear operator is not continuous.
Notice that the kernel of this non continuous linear operator is closed:
ker
D
=
{
0
}
{\displaystyle \ker D=\{0\}}
. However, when a linear operator is of finite rank, the closeness of the kernel is in fact synonymous to continuity. To see this, we start with the special case of linear forms.
2 Theorem A (non null) linear form is continuous iff it's kernel is closed.
T
{\displaystyle T}
continuous
⇔
ker
T
=
ker
T
¯
{\displaystyle \Leftrightarrow \ker T={\overline {\ker T}}}
Proof:
If the linear form
T
{\displaystyle T}
on a normed vector space
X
{\displaystyle X}
is continuous, then it's kernel is closed since it's the continuous inverse image of the closed set
{
0
}
{\displaystyle \{0\}}
.
Conversely, suppose a linear form
T
:
X
→
R
{\displaystyle T:X\to \mathbb {R} }
is not continuous. then by the previous theorem,
∀
c
>
0
,
∃
x
(
c
)
s
.
t
.
|
T
x
(
c
)
|
⩾
c
‖
x
(
c
)
‖
{\displaystyle \forall c>0,\exists x(c)s.t.|Tx(c)|\geqslant c\|x(c)\|}
so in particular, one can define a sequence
{
x
n
}
{\displaystyle \{x_{n}\}}
such that
|
T
x
n
|
⩾
n
‖
x
n
‖
>
0
{\displaystyle |Tx_{n}|\geqslant n\|x_{n}\|>0}
. Then denote:
u
n
:=
x
n
|
x
n
|
{\displaystyle u_{n}:={\frac {x_{n}}{|x_{n}|}}}
, one has defined a unit normed sequence (
|
u
n
|
=
1
{\displaystyle |u_{n}|=1}
) s.t.
|
T
u
n
|
⩾
n
{\displaystyle |Tu_{n}|\geqslant n}
. Furthermore, denote
v
n
:=
u
n
|
T
u
n
|
{\displaystyle v_{n}:={\dfrac {u_{n}}{|Tu_{n}|}}}
. Since
|
u
n
|
|
T
u
n
|
⩽
1
n
{\displaystyle {\frac {|u_{n}|}{|Tu_{n}|}}\leqslant {\dfrac {1}{n}}}
, one can define a sequence that converges
{
v
n
}
→
0
{\displaystyle \{v_{n}\}\to 0}
whilst
|
T
v
n
|
=
1
{\displaystyle |Tv_{n}|=1}
.Now, since
ker
T
≠
X
{\displaystyle \ker T\neq X}
, then there exists
a
{\displaystyle a}
such that
T
a
≠
0
{\displaystyle Ta\neq 0}
. Then the sequence of general term converges
a
−
v
n
T
a
⏟
∈
ker
T
→
a
∉
ker
T
{\displaystyle \underbrace {a-v_{n}Ta} _{\in \ker T}\to a\notin \ker T}
and hence
ker
T
{\displaystyle \ker T}
is not closed.
◻
{\displaystyle \square }
Furthermore, if the linear form is continuous and the kernel is dense, then
ker
T
=
ker
T
¯
=
X
⇒
f
=
0
{\displaystyle \ker T={\overline {\ker T}}=X\Rightarrow f=0}
, hence a continuous & non null linear has a non dense kernel, and hence a linear form with a dense kernel is whether null or non continuous so a non null continuous linear form with a dense kernel is not continuous, and a linear form with a dense kernel is not continuous.
2 Corollary' " A non null linear form on a normed vector space is not continuous iff it's kernel is dense.
ker
T
¯
=
X
⇔
T
{\displaystyle {\overline {\ker T}}=X\Leftrightarrow T}
is not continuous"More generally, we have:
2 Theorem " A non null linear operator of finite rank between normed vector spaces. then closeness of the null space is equivalent to continuity."
Proof: It remains to show that continuity implies closeness of the kernel. Suppose
T
:
X
→
Y
{\displaystyle T:X\to Y}
is not continuous. Denote
r
:=
dim
Im
T
{\displaystyle r:=\dim {\mbox{Im }}T}
;
2 Lemma If
T
:
X
→
Y
{\displaystyle T:X\to Y}
is a linear operator between normed vector spaces, then
T
{\displaystyle T}
is of finite rank
r
{\displaystyle r}
iff there exists
r
{\displaystyle r}
independent linear forms
(
f
1
,
…
,
f
r
)
{\displaystyle (f_{1},\dots ,f_{r})}
and
r
{\displaystyle r}
independent vectors
(
a
1
,
…
,
a
r
)
{\displaystyle (a_{1},\dots ,a_{r})}
such that
T
x
=
a
1
f
1
(
x
)
+
⋯
+
a
r
f
r
(
x
)
{\displaystyle Tx=a_{1}f_{1}(x)+\dots +a_{r}f_{r}(x)}
"
Proof: take a basis
(
a
1
,
…
,
a
r
)
{\displaystyle (a_{1},\dots ,a_{r})}
of
Im
T
{\displaystyle {\mbox{Im }}T}
, then from
T
x
=
∑
i
=
1
r
a
i
y
i
{\displaystyle Tx=\sum _{i=1}^{r}a_{i}y_{i}}
, one can define
r
{\displaystyle r}
mappings
f
i
(
x
)
=
y
i
{\displaystyle f_{i}(x)=y_{i}}
. Unicity and linearity of
T
{\displaystyle T}
implies linearity of the
f
i
{\displaystyle f_{i}}
's. Furthermore, the family
(
f
1
,
…
,
f
r
)
{\displaystyle (f_{1},\dots ,f_{r})}
of linear forms of
X
∗
{\displaystyle X^{*}}
is linearly independent: suppose not, then there exist a non zero family
(
α
1
,
…
,
α
r
)
{\displaystyle (\alpha _{1},\dots ,\alpha _{r})}
such that e.g.
f
1
=
∑
i
=
2
r
α
i
f
i
{\displaystyle f_{1}=\sum _{i=2}^{r}\alpha _{i}f_{i}}
so
T
x
=
∑
i
=
1
r
f
i
(
x
)
a
i
=
(
∑
i
=
2
r
α
i
f
i
(
x
)
)
a
1
+
∑
i
=
2
r
f
i
(
x
)
a
i
=
(
α
2
a
1
+
a
2
)
f
2
(
x
)
+
⋯
+
(
α
n
a
1
+
a
r
)
f
r
(
x
)
{\displaystyle Tx=\sum _{i=1}^{r}f_{i}(x)a_{i}=(\sum _{i=2}^{r}\alpha _{i}f_{i}(x))a_{1}+\sum _{i=2}^{r}f_{i}(x)a_{i}=(\alpha _{2}a_{1}+a_{2})f_{2}(x)+\dots +(\alpha _{n}a_{1}+a_{r})f_{r}(x)}
and the family
(
α
i
a
1
+
a
i
)
i
=
2
,
…
,
r
{\displaystyle (\alpha _{i}a_{1}+a_{i})_{i=2,\dots ,r}}
spans
Im
T
{\displaystyle {\mbox{Im }}T}
, so
dim
Im
T
=
r
−
1
{\displaystyle \dim {\mbox{Im }}T=r-1}
which is a contradiction. Finally, one has a unique decomposition of a finite rank linear operator:
T
x
=
a
1
f
1
(
x
)
+
…
a
r
f
r
(
x
)
{\displaystyle Tx=a_{1}f_{1}(x)+\dots a_{r}f_{r}(x)}
with
f
i
∈
E
∗
{\displaystyle f_{i}\in E^{*}}
◻
{\displaystyle \square }
Take
x
∈
ker
T
⇒
x
∈
⋂
i
=
1
r
ker
f
i
⊂
ker
f
i
{\displaystyle x\in \ker T\Rightarrow x\in \bigcap _{i=1}^{r}\ker f_{i}\subset \ker f_{i}}
. Then there exists a vector subspace
H
i
{\displaystyle H_{i}}
such that
ker
f
i
=
ker
T
⊕
H
i
{\displaystyle \ker f_{i}=\ker T\oplus H_{i}}
. Denote
T
i
:
H
i
→
Im
T
{\displaystyle T_{i}:H_{i}\to {\mbox{Im}}T}
the restriction of
T
{\displaystyle T}
to
H
i
{\displaystyle H_{i}}
. Since
k
e
r
T
i
=
{
x
∈
H
i
:
T
i
(
x
)
}
=
0
=
ker
T
∩
H
i
=
{
0
}
{\displaystyle kerT_{i}=\{x\in H_{i}:T_{i}(x)\}=0=\ker T\cap H_{i}=\{0\}}
, the linear operator
T
i
{\displaystyle T_{i}}
is injective so
Im
T
i
⊂
Im
T
{\displaystyle {\mbox{Im}}T_{i}\subset {\mbox{Im}}T}
and
H
i
{\displaystyle H_{i}}
is of finite dimension, and this for all
i
=
1
,
…
,
r
{\displaystyle i=1,\dots ,r}
.
By hypothesis
ker
T
{\displaystyle \ker T}
is closed. Since the sum of this closed subspace and a subspace of finite dimension (
H
i
{\displaystyle H_{i}}
) is closed (see lemma bellow), it follows that the kernel of each
r
{\displaystyle r}
linear forms
ker
f
i
{\displaystyle \ker f_{i}}
is closed, so the
f
i
{\displaystyle f_{i}}
's are all continuous by the first case and hence
T
{\displaystyle T}
is continuous.
◻
{\displaystyle \square }
2 Lemma The sum of subspace of finite dimension with a closed subspace is closed."
Proof: by induction on the dimension.
Case
n
=
1
{\displaystyle n=1}
. Let's show that
H
:=
F
+
K
a
{\displaystyle H:=F+\mathbb {K} a}
is closed when
F
{\displaystyle F}
is closed (where
K
{\displaystyle \mathbb {K} }
is a complete field). Any
x
∈
H
{\displaystyle x\in H}
can be uniquely written as
x
=
y
+
λ
a
{\displaystyle x=y+\lambda a}
with
y
∈
F
{\displaystyle y\in F}
. There exists a linear form
L
{\displaystyle L}
s.t.
x
=
y
+
L
(
x
)
a
{\displaystyle x=y+L(x)a}
. Since
L
{\displaystyle L}
is closed in
(
X
,
‖
⋅
‖
)
{\displaystyle (X,\|\cdot \|)}
so in
(
H
,
‖
⋅
‖
)
{\displaystyle (H,\|\cdot \|)}
, then
f
{\displaystyle f}
is continuous by the first case.
Take a convergente sequence
x
n
→
x
∈
E
{\displaystyle {x_{n}}\to x\in E}
of
H
{\displaystyle H}
. He have
x
n
=
y
n
+
L
(
x
n
)
a
{\displaystyle x_{n}=y_{n}+L(x_{n})a}
with
y
n
∈
F
{\displaystyle y_{n}\in F}
. Since the sequence
x
n
{\displaystyle {x_{n}}}
is convergente, then it si Cauchy, so it's continuous image
L
(
x
n
)
{\displaystyle {L(x_{n})}}
is also Cauchy. Since
R
{\displaystyle \mathbb {R} }
is complete, then
L
(
x
n
)
→
λ
{\displaystyle {L(x_{n})}\to \lambda }
. Finally, the sequence
y
n
{\displaystyle {y_{n}}}
converges to
x
−
λ
a
{\displaystyle x-\lambda a}
. Since
F
{\displaystyle F}
is closed, then
x
−
λ
a
∈
F
{\displaystyle x-\lambda a\in F}
and
x
∈
H
{\displaystyle x\in H}
so
H
{\displaystyle H}
is closed.
Suppose the result holds for all subspaces of dimension
⩽
p
{\displaystyle \leqslant p}
. Let
G
{\displaystyle G}
be a subspace of dimension
p
+
1
{\displaystyle p+1}
. Let
(
a
1
,
…
,
a
p
+
1
)
{\displaystyle (a_{1},\dots ,a_{p+1})}
be a basis of
G
{\displaystyle G}
. Then
H
:=
F
+
⨁
i
=
1
p
K
a
i
+
K
a
p
+
1
{\displaystyle H:=F+\bigoplus _{i=1}^{p}\mathbb {K} a_{i}+\mathbb {K} a_{p+1}}
and concludes easily.
◻
{\displaystyle \square }
2 Corollary Any linear operator on a normed vector space of finite dimension onto a normed vector space is continuous.
Proof: Since
X
{\displaystyle X}
is of finite dimension, then any linear operator is of finite rank. Then as
dim
ker
T
+
dim
Im
T
=
dim
X
{\displaystyle \dim \ker T+\dim {\mbox{Im}}T=\dim X}
holds, it comes that the null space is of finite dimension, so is closed (any
K
{\displaystyle \mathbb {K} }
vector subspace of finite dimension
n
{\displaystyle n}
is isomorphic to
K
n
{\displaystyle \mathbb {K} ^{n}}
(where
K
{\displaystyle \mathbb {K} }
is a complete field), so the subspace is complete and closed). Then one applies the previous theorem.
◻
{\displaystyle \square }
2 Lemma (Riesz) A normed space
X
{\displaystyle {\mathcal {X}}}
is finite-dimensional if and only if its closed unit ball is compact.
Proof: Let
T
:
C
n
→
X
{\displaystyle T:\mathbf {C} ^{n}\to X}
be a linear vector space isomorphism. Since
T
{\displaystyle T}
has closed kernel, arguing as in the proof of the preceding theorem, we see that
T
{\displaystyle T}
is continuous. By the same reasoning
T
−
1
{\displaystyle T^{-1}}
is continuous. It follows:
{
x
∈
X
|
‖
x
‖
≤
1
}
⊂
T
{
y
∈
C
n
|
‖
y
‖
≤
‖
T
−
1
‖
}
{\displaystyle \{x\in {\mathcal {X}}|\|x\|\leq 1\}\subset T\{y\in \mathbf {C} ^{n}|\|y\|\leq \|T^{-1}\|\}}
In the above, the left-hand side is closed, and the right-hand is a continuous image of a closed ball, which is compact. Hence, the closed unit ball is a subset of a compact set and thus compact. Now, the converse. If
X
{\displaystyle {\mathcal {X}}}
is not finite dimensional, we can construct a sequence
x
j
{\displaystyle x_{j}}
such that:
1
=
‖
x
j
‖
≤
‖
x
j
−
∑
k
=
1
j
−
1
a
k
x
k
‖
{\displaystyle 1=\|x_{j}\|\leq \|x_{j}-\sum _{k=1}^{j-1}a_{k}x_{k}\|}
for any sequence of scalars
a
k
{\displaystyle a_{k}}
.Thus, in particular,
‖
x
j
−
x
k
‖
≥
1
{\displaystyle \|x_{j}-x_{k}\|\geq 1}
for all
j
,
k
{\displaystyle j,k}
. (For the details of this argument, see : w:Riesz's lemma for now)
◻
{\displaystyle \square }
2 Corollary Every finite-dimensional normed space is a Banach space.
Proof: Let
x
n
{\displaystyle x_{n}}
be a Cauchy sequence. Since it is bounded, it is contained in some closed ball, which is compact.
x
n
{\displaystyle x_{n}}
thus has a convergent subsequence and so
x
n
{\displaystyle x_{n}}
itself converges.
◻
{\displaystyle \square }
2 Theorem A normed space
X
{\displaystyle {\mathcal {X}}}
is finite-dimensional if and only if every linear operator
T
{\displaystyle T}
defined on
X
{\displaystyle {\mathcal {X}}}
is continuous.
Proof: Identifying the range of
T
{\displaystyle T}
with
C
n
{\displaystyle \mathbf {C} ^{n}}
, we can write:
T
x
=
(
f
1
(
x
)
,
f
2
(
x
)
,
.
.
.
f
n
(
x
)
)
{\displaystyle Tx=(f_{1}(x),f_{2}(x),...f_{n}(x))}
where
f
1
,
.
.
.
f
n
{\displaystyle f_{1},...f_{n}}
are linear functionals. The dimensions of the kernels of
f
j
{\displaystyle f_{j}}
are finite. Thus,
f
j
{\displaystyle f_{j}}
all have complete and thus closed kernels. Hence, they are continuous and so
T
{\displaystyle T}
is continuous. For the converse, we need Axiom of Choice. (TODO: complete the proof.)
◻
{\displaystyle \square }
The graph of any function
f
{\displaystyle f}
on a set
E
{\displaystyle E}
is the set
{
(
x
,
f
(
x
)
)
|
x
∈
E
}
{\displaystyle \{(x,f(x))|x\in E\}}
. A continuous function between metric spaces has closed graph. In fact, suppose
(
x
j
,
f
(
x
j
)
)
→
(
x
,
y
)
{\displaystyle (x_{j},f(x_{j}))\to (x,y)}
. By continuity,
f
(
x
j
)
→
f
(
x
)
{\displaystyle f(x_{j})\to f(x)}
; in other words,
y
=
f
(
x
)
{\displaystyle y=f(x)}
and so
(
x
,
y
)
{\displaystyle (x,y)}
is in the graph of
f
{\displaystyle f}
. It follows (in the next theorem) that a continuous linear operator with closed graph has closed domain. (Note the continuity here is a key; we will shortly study a linear operator that has closed graph but has non-closed domain.)
2 Theorem Let
T
:
X
→
Y
{\displaystyle T:{\mathcal {X}}\to {\mathcal {Y}}}
be a continuous densely defined linear operator between Banach spaces. Then its domain is closed; i.e.,
T
{\displaystyle T}
is actually defined everywhere.
Proof:
Suppose
f
j
→
f
{\displaystyle f_{j}\to f}
and
T
f
j
{\displaystyle Tf_{j}}
is defined for every
j
{\displaystyle j}
; i.e., the sequence
f
j
{\displaystyle f_{j}}
is in the domain of
T
{\displaystyle T}
. Since
‖
T
f
j
−
T
f
k
‖
≤
‖
T
‖
‖
f
j
−
f
k
‖
→
0
{\displaystyle \|Tf_{j}-Tf_{k}\|\leq \|T\|\|f_{j}-f_{k}\|\to 0}
,
T
f
j
{\displaystyle Tf_{j}}
is Cauchy. It follows that
(
f
j
,
T
f
j
)
{\displaystyle (f_{j},Tf_{j})}
is Cauchy and, by completeness, has limit
(
g
,
T
g
)
{\displaystyle (g,Tg)}
since the graph of T is closed. Since
f
=
g
{\displaystyle f=g}
,
T
f
{\displaystyle Tf}
is defined; i.e.,
f
{\displaystyle f}
is in the domain of
T
{\displaystyle T}
.
◻
{\displaystyle \square }
The theorem is frequently useful in application. Suppose we wish to prove some linear formula. We first show it holds for a function with compact support and of varying smoothness, which is usually easy to do because the function vanishes on the boundary, where much of complications reside. Because of th linear nature in the formula, the theorem then tells that the formula is true for the space where the above functions are dense.
We shall now turn our attention to the consequences of the fact that a complete metric space is a Baire space. They tend to be more significant than results obtained by directly appealing to the completeness. Note that not every normed space that is a Baire space is a Banach space.
2 Theorem (open mapping theorem) Let
X
,
Y
{\displaystyle {\mathcal {X}},{\mathcal {Y}}}
be Banach spaces. If
T
:
X
→
Y
{\displaystyle T:{\mathcal {X}}\to {\mathcal {Y}}}
is a continuous linear surjection, then it is an open mapping; i.e., it maps open sets to open sets.
Proof: Let
B
(
r
)
=
{
x
∈
X
;
‖
x
‖
<
r
}
{\displaystyle B(r)=\{x\in {\mathcal {X}};\|x\|
0
{\displaystyle K>0}
such that if
y
∈
im
(
T
)
{\displaystyle y\in \operatorname {im} (T)}
then
‖
x
‖
≤
K
‖
y
‖
{\displaystyle \|x\|\leq K\|y\|}
for some
x
{\displaystyle x}
with
T
x
=
y
{\displaystyle Tx=y}
.
Proof: This is immediate once we have the notion of a quotient map, which we now define as follows.
Let
M
{\displaystyle {\mathcal {M}}}
be a closed subspace of a normed space
X
{\displaystyle {\mathcal {X}}}
. The quotient space
X
/
M
{\displaystyle {\mathcal {X}}/{\mathcal {M}}}
is a normed space with norm:
‖
π
(
x
)
‖
=
inf
{
‖
x
+
m
‖
;
m
∈
M
}
{\displaystyle \|\pi (x)\|=\inf\{\|x+m\|;m\in {\mathcal {M}}\}}
where
π
:
X
→
X
/
M
{\displaystyle \pi :{\mathcal {X}}\to {\mathcal {X}}/{\mathcal {M}}}
is a canonical projection. That
‖
⋅
‖
{\displaystyle \|\cdot \|}
is a norm is obvious except for the triangular inequality. But since
‖
π
(
x
+
y
)
‖
≤
‖
x
+
m
‖
+
‖
y
+
n
‖
{\displaystyle \|\pi (x+y)\|\leq \|x+m\|+\|y+n\|}
for all
m
,
n
∈
M
{\displaystyle m,n\in {\mathcal {M}}}
. Taking inf over
m
,
n
{\displaystyle m,n}
separately we get:
‖
π
(
x
+
y
)
‖
≤
‖
π
(
x
)
‖
+
‖
π
(
y
)
‖
{\displaystyle \|\pi (x+y)\|\leq \|\pi (x)\|+\|\pi (y)\|}
Suppose, further, that
X
{\displaystyle {\mathcal {X}}}
is also a commutative algebra and
M
{\displaystyle {\mathcal {M}}}
is an ideal. Then
X
/
M
{\displaystyle {\mathcal {X}}/{\mathcal {M}}}
becomes a quotient algebra. In fact, as above, we have:
‖
π
(
x
)
π
(
y
)
‖
=
‖
π
(
(
x
+
m
)
(
y
+
n
)
)
‖
≤
‖
x
+
m
‖
‖
y
+
n
‖
{\displaystyle \|\pi (x)\pi (y)\|=\|\pi ((x+m)(y+n))\|\leq \|x+m\|\|y+n\|}
,for all
m
,
n
∈
M
{\displaystyle m,n\in {\mathcal {M}}}
since
π
{\displaystyle \pi }
is a homomorphism. Taking inf completes the proof.
So, the only nontrivial question is the completeness. It turns out that
X
/
M
{\displaystyle {\mathcal {X}}/{\mathcal {M}}}
is a Banach space (or algebra) if
X
{\displaystyle {\mathcal {X}}}
is Banach space (or algebra). In fact, suppose
∑
n
=
1
∞
‖
π
(
x
n
)
‖
<
∞
{\displaystyle \sum _{n=1}^{\infty }\|\pi (x_{n})\|<\infty }
Then we can find a sequence
y
n
∈
M
{\displaystyle y_{n}\in {\mathcal {M}}}
such that
∑
n
=
1
∞
‖
x
n
+
y
n
‖
<
∞
{\displaystyle \sum _{n=1}^{\infty }\|x_{n}+y_{n}\|<\infty }
By completeness,
∑
n
=
1
∞
x
n
+
y
n
{\displaystyle \sum _{n=1}^{\infty }x_{n}+y_{n}}
converges, and since
π
{\displaystyle \pi }
is continuous,
∑
n
=
1
∞
π
(
x
n
)
{\displaystyle \sum _{n=1}^{\infty }\pi (x_{n})}
converges then. The completeness now follows from:
2 Lemma Let
X
{\displaystyle {\mathcal {X}}}
be a normed space. Then
X
{\displaystyle {\mathcal {X}}}
is complete (thus a Banach space) if and only if
∑
n
=
1
∞
‖
x
n
‖
<
∞
{\displaystyle \sum _{n=1}^{\infty }\|x_{n}\|<\infty }
implies
∑
n
=
1
∞
x
n
{\displaystyle \sum _{n=1}^{\infty }x_{n}}
converges.Proof: (
⇒
{\displaystyle \Rightarrow }
) We have:
‖
∑
n
=
k
k
+
m
x
n
‖
≤
∑
n
=
k
k
+
m
‖
x
n
‖
{\displaystyle \|\sum _{n=k}^{k+m}x_{n}\|\leq \sum _{n=k}^{k+m}\|x_{n}\|}
.By hypothesis, the right-hand side goes to 0 as
n
,
m
→
∞
{\displaystyle n,m\to \infty }
. By completeness,
∑
n
=
1
∞
x
n
{\displaystyle \sum _{n=1}^{\infty }x_{n}}
converges. Conversely, suppose
x
j
{\displaystyle x_{j}}
is a Cauchy sequence. Thus, for each
j
=
1
,
2
,
.
.
.
{\displaystyle j=1,2,...}
, there exists an index
k
j
{\displaystyle k_{j}}
such that
‖
x
n
−
x
m
‖
<
2
−
j
{\displaystyle \|x_{n}-x_{m}\|<2^{-j}}
for any
n
,
m
≥
k
j
{\displaystyle n,m\geq k_{j}}
. Let
x
k
0
=
0
{\displaystyle x_{k_{0}}=0}
. Then
∑
j
=
0
∞
‖
x
k
j
+
1
−
x
k
j
‖
<
2
{\displaystyle \sum _{j=0}^{\infty }\|x_{k_{j+1}}-x_{k_{j}}\|<2}
. Hence, by assumption we can get the limit
x
=
∑
j
=
0
∞
x
k
j
+
1
−
x
k
j
{\displaystyle x=\sum _{j=0}^{\infty }x_{k_{j+1}}-x_{k_{j}}}
, and since
‖
x
n
k
−
x
‖
=
‖
∑
j
=
1
n
x
k
j
+
1
−
x
k
j
−
x
‖
→
0
{\displaystyle \|x_{n_{k}}-x\|=\|\sum _{j=1}^{n}x_{k_{j+1}}-x_{k_{j}}-x\|\to 0}
as
n
→
∞
{\displaystyle n\to \infty }
,we conclude that
x
j
{\displaystyle x_{j}}
has a subsequence converging to
x
{\displaystyle x}
; thus, it converges to
x
{\displaystyle x}
.
◻
{\displaystyle \square }
The next result is arguably the most important theorem in the theory of Banach spaces. (At least, it is used the most frequently in application.)
2 Theorem (closed graph theorem) Let
X
,
Y
{\displaystyle {\mathcal {X}},{\mathcal {Y}}}
be Banach spaces, and
T
:
X
→
Y
{\displaystyle T:{\mathcal {X}}\to {\mathcal {Y}}}
a linear operator. The following are equivalent.
(i)
T
{\displaystyle T}
is continuous.
(ii) If
x
j
→
0
{\displaystyle x_{j}\to 0}
and
T
x
j
{\displaystyle Tx_{j}}
is convergent, then
T
x
j
→
0
{\displaystyle Tx_{j}\to 0}
.
(iii) The graph of
T
{\displaystyle T}
is closed.Proof: That (i) implies (ii) is clear. To show (iii), suppose
(
x
j
,
T
x
j
)
{\displaystyle (x_{j},Tx_{j})}
is convergent in
X
{\displaystyle X}
. Then
x
j
{\displaystyle x_{j}}
converges to some
x
0
{\displaystyle x_{0}}
or
x
j
−
x
0
→
0
{\displaystyle x_{j}-x_{0}\to 0}
, and
T
x
j
−
T
x
{\displaystyle Tx_{j}-Tx}
is convergent. Thus, if (ii) holds,
T
(
x
j
−
x
)
→
0
{\displaystyle T(x_{j}-x)\to 0}
. Finally, to prove (iii)
⇒
{\displaystyle \Rightarrow }
(i), we note that Corollary 2.something gives the inequality:
‖
⋅
‖
+
‖
T
⋅
‖
≤
K
‖
⋅
‖
{\displaystyle \|\cdot \|+\|T\cdot \|\leq K\|\cdot \|}
since by hypothesis the norm in the left-hand side is complete. Hence, if
x
j
→
x
{\displaystyle x_{j}\to x}
, then
T
x
j
→
T
x
{\displaystyle Tx_{j}\to Tx}
.
◻
{\displaystyle \square }
Note that when the domain of a linear operator is not a Banach space (e.g., just dense in a Banach space), the condition (ii) is not sufficient for the graph of the operator to be closed. (It is not hard to find an example of this in other fields, but the reader might want to construct one himself as an exercise.)
Finally, note that an injective linear operator has closed graph if and only if its inverse is closed, since the map
(
x
1
,
x
2
)
↦
x
2
,
x
1
{\displaystyle (x_{1},x_{2})\mapsto x_{2},x_{1}}
sends closed sets to closed sets.
2 Theorem Let
(
X
j
,
‖
⋅
‖
j
)
{\displaystyle ({\mathcal {X}}_{j},\|\cdot \|_{j})}
be Banach spaces. Let
T
:
X
1
→
X
2
{\displaystyle T:{\mathcal {X}}_{1}\to {\mathcal {X}}_{2}}
be a closed densely defined operator and
S
{\displaystyle S}
be a linear operator with
dom
(
T
)
⊂
dom
(
S
)
{\displaystyle \operatorname {dom} (T)\subset \operatorname {dom} (S)}
. If there are constants
a
,
b
{\displaystyle a,b}
such that (i)
0
≤
a
<
1
{\displaystyle 0\leq a<1}
and
b
>
0
{\displaystyle b>0}
and (ii)
‖
S
u
‖
≤
a
‖
T
u
‖
+
b
‖
u
‖
{\displaystyle \|Su\|\leq a\|Tu\|+b\|u\|}
for every
u
∈
dom
(
T
)
{\displaystyle u\in \operatorname {dom} (T)}
, then
T
+
S
{\displaystyle T+S}
is closed.
Proof:
Suppose
‖
u
j
−
u
‖
1
+
‖
(
T
+
S
)
u
j
−
f
‖
2
→
0
{\displaystyle \|u_{j}-u\|_{1}+\|(T+S)u_{j}-f\|_{2}\to 0}
. Then
‖
T
(
u
j
−
u
k
)
‖
≤
‖
(
T
+
S
)
(
u
j
−
u
k
)
‖
+
a
‖
T
(
u
j
−
u
k
)
‖
+
b
‖
u
j
−
u
k
‖
{\displaystyle \|T(u_{j}-u_{k})\|\leq \|(T+S)(u_{j}-u_{k})\|+a\|T(u_{j}-u_{k})\|+b\|u_{j}-u_{k}\|}
Thus,
(
1
−
a
)
‖
T
(
u
j
−
u
k
)
‖
≤
‖
(
T
+
S
)
(
u
j
−
u
k
)
‖
+
b
‖
u
j
−
u
k
‖
{\displaystyle (1-a)\|T(u_{j}-u_{k})\|\leq \|(T+S)(u_{j}-u_{k})\|+b\|u_{j}-u_{k}\|}
By hypothesis, the right-hand side goes to
0
{\displaystyle 0}
as
j
,
k
→
∞
{\displaystyle j,k\to \infty }
.
Since
T
{\displaystyle T}
is closed,
(
u
j
,
T
u
j
)
{\displaystyle (u_{j},Tu_{j})}
converges to
(
u
,
T
u
)
{\displaystyle (u,Tu)}
.
◻
{\displaystyle \square }
In particular, with
a
=
0
{\displaystyle a=0}
, the hypothesis of the theorem is fulfilled, if
S
{\displaystyle S}
is continuous.
When
X
,
Y
{\displaystyle {\mathcal {X}},{\mathcal {Y}}}
are normed spaces, by
L
(
X
,
Y
)
{\displaystyle L({\mathcal {X}},{\mathcal {Y}})}
we denote the space of all continuous linear operators from
X
{\displaystyle {\mathcal {X}}}
to
Y
{\displaystyle {\mathcal {Y}}}
.
2 Theorem If
Y
{\displaystyle {\mathcal {Y}}}
is complete, then every Cauchy sequence
T
n
{\displaystyle T_{n}}
in
L
(
X
,
Y
)
{\displaystyle L({\mathcal {X}},{\mathcal {Y}})}
converges to a limit
T
{\displaystyle T}
and
‖
T
‖
=
lim
n
→
∞
‖
T
n
‖
{\displaystyle \|T\|=\lim _{n\to \infty }\|T_{n}\|}
. Conversely, if
L
(
X
,
Y
)
{\displaystyle L({\mathcal {X}},{\mathcal {Y}})}
is complete, then so is Y.
Proof: Let
T
n
{\displaystyle T_{n}}
be a Cauchy sequence in operator norm. For each
x
∈
X
{\displaystyle x\in {\mathcal {X}}}
, since
‖
T
n
(
x
)
−
T
m
(
x
)
‖
≤
‖
T
n
−
T
m
‖
‖
x
‖
{\displaystyle \|T_{n}(x)-T_{m}(x)\|\leq \|T_{n}-T_{m}\|\|x\|}
and
Y
{\displaystyle {\mathcal {Y}}}
is complete, there is a limit
y
{\displaystyle y}
to which
T
n
(
x
)
{\displaystyle T_{n}(x)}
converges. Define
T
(
x
)
=
y
{\displaystyle T(x)=y}
.
T
{\displaystyle T}
is linear since the limit operations are linear. It is also continuous since
‖
T
(
x
)
‖
≤
sup
n
‖
T
n
‖
‖
x
‖
{\displaystyle \|T(x)\|\leq \sup _{n}\|T_{n}\|\|x\|}
. Finally,
lim
n
→
∞
‖
T
n
−
T
‖
=
sup
‖
x
‖
≤
1
‖
lim
n
→
∞
T
n
(
x
)
−
T
(
x
)
‖
{\displaystyle \lim _{n\to \infty }\|T_{n}-T\|=\sup _{\|x\|\leq 1}\|\lim _{n\to \infty }T_{n}(x)-T(x)\|}
and
|
‖
T
n
‖
−
‖
T
‖
|
≤
‖
T
n
−
T
‖
→
0
{\displaystyle |\|T^{n}\|-\|T\||\leq \|T^{n}-T\|\to 0}
as
n
→
∞
{\displaystyle n\to \infty }
. (TODO: a proof for the converse.)
◻
{\displaystyle \square }
2 Theorem (uniform boundedness principle) Let
F
{\displaystyle {\mathcal {F}}}
be a family of continuous functions
f
:
X
→
Y
{\displaystyle f:X\to Y}
where
Y
{\displaystyle Y}
is a normed linear space. Suppose that
M
⊂
X
{\displaystyle M\subset X}
is non-meager and that:
sup
{
‖
f
(
x
)
‖
:
f
∈
F
}
<
∞
{\displaystyle \sup\{\|f(x)\|:f\in {\mathcal {F}}\}<\infty }
for each
x
∈
M
{\displaystyle x\in M}
It then follows: there is some
G
⊂
X
{\displaystyle G\subset X}
open and such that
(a)
sup
{
‖
f
(
x
)
‖
:
f
∈
F
,
x
∈
G
}
<
∞
{\displaystyle \sup\{\|f(x)\|:f\in {\mathcal {F}},x\in G\}<\infty }
If we assume in addition that each member of
F
{\displaystyle {\mathcal {F}}}
is a linear operator and
X
{\displaystyle X}
is a normed linear space, then
(b)
sup
{
‖
f
‖
:
f
∈
F
}
<
∞
{\displaystyle \sup\{\|f\|:f\in {\mathcal {F}}\}<\infty }
Proof: Let
E
j
=
∩
f
∈
F
{
x
∈
X
;
‖
f
(
x
)
‖
≤
j
}
{\displaystyle E_{j}=\cap _{f\in {\mathcal {F}}}\{x\in X;\|f(x)\|\leq j\}}
be a sequence. By hypothesis,
M
⊂
⋃
j
=
1
∞
E
j
{\displaystyle M\subset \bigcup _{j=1}^{\infty }E_{j}}
and each
E
j
{\displaystyle E_{j}}
is closed since
{
x
∈
X
;
‖
f
(
x
)
‖
>
j
}
{\displaystyle \{x\in X;\|f(x)\|>j\}}
is open by continuity. It then follows that some
E
N
{\displaystyle E_{N}}
has an interior point
y
{\displaystyle y}
; otherwise,
M
{\displaystyle M}
fails to be non-meager. Hence, (a) holds. To show (b), making additional assumptions, we can find an open ball
B
=
B
(
2
r
,
y
)
⊂
E
N
{\displaystyle B=B(2r,y)\subset E_{N}}
. It then follows: for any
f
∈
F
{\displaystyle f\in {\mathcal {F}}}
and any
x
∈
X
{\displaystyle x\in X}
with
‖
x
‖
=
1
{\displaystyle \|x\|=1}
,
‖
f
(
x
)
‖
=
r
−
1
‖
f
(
r
x
+
y
)
−
f
(
y
)
‖
≤
2
r
−
1
N
{\displaystyle \|f(x)\|=r^{-1}\|f(rx+y)-f(y)\|\leq 2r^{-1}N}
.
◻
{\displaystyle \square }
A family
Γ
{\displaystyle \Gamma }
of linear operators is said to be equicontinuous if given any neighborhood
W
{\displaystyle W}
of
0
{\displaystyle 0}
we can find a neighborhood
V
{\displaystyle V}
of
0
{\displaystyle 0}
such that:
f
(
V
)
⊂
W
{\displaystyle f(V)\subset W}
for every
f
∈
Γ
{\displaystyle f\in \Gamma }
The conclusion of the theorem, therefore, means that the family satisfying the hypothesis of the theorem is equicontinuous.
2 Corollary Let
X
,
Y
,
Z
{\displaystyle {\mathcal {X}},{\mathcal {Y}},{\mathcal {Z}}}
be Banach spaces. Let
T
:
X
×
Y
→
Z
{\displaystyle T:{\mathcal {X}}\times {\mathcal {Y}}\to {\mathcal {Z}}}
be a bilinear or sesquilinear operator. If
T
{\displaystyle T}
is separately continuous (i.e., the function is continuous when all but one variables are fixed) and
Y
{\displaystyle {\mathcal {Y}}}
is complete, then
T
{\displaystyle T}
is continuous.
Proof: For each
y
∈
Y
{\displaystyle y\in {\mathcal {Y}}}
,
sup
{
‖
T
(
x
,
y
)
‖
Z
;
‖
x
‖
X
≤
1
}
=
‖
T
(
⋅
,
y
)
‖
{\displaystyle \sup\{\|T(x,y)\|_{\mathcal {Z}};\|x\|_{\mathcal {X}}\leq 1\}=\|T(\cdot ,y)\|}
where the right-hand side is finite by continuity. Hence, the application of the principle of uniform boundedness to the family
{
T
(
x
,
⋅
)
;
‖
x
‖
X
≤
1
}
{\displaystyle \{T(x,\cdot );\|x\|_{\mathcal {X}}\leq 1\}}
shows the family is equicontinuous. That is, there is
K
>
0
{\displaystyle K>0}
such that:
‖
T
(
x
,
y
)
‖
Z
≤
K
‖
y
‖
Y
{\displaystyle \|T(x,y)\|_{\mathcal {Z}}\leq K\|y\|_{\mathcal {Y}}}
for every
‖
x
‖
X
l
e
1
{\displaystyle \|x\|_{\mathcal {X}}le1}
and every
y
∈
Y
{\displaystyle y\in {\mathcal {Y}}}
.The theorem now follows since
X
×
Y
{\displaystyle {\mathcal {X}}\times {\mathcal {Y}}}
is a metric space.
◻
{\displaystyle \square }
Since scalar multiplication is a continuous operation in normed spaces, the corollary says, in particular, that every linear operator on finite dimensional normed spaces is continuous. The next is one more example of the techniques discussed so far.
2. Theorem (Hahn-Banach) Let
(
X
,
‖
⋅
‖
)
{\displaystyle ({\mathcal {X}},\|\cdot \|)}
be normed space and
M
⊂
X
{\displaystyle {\mathcal {M}}\subset {\mathcal {X}}}
be a linear subspace. If
z
{\displaystyle z}
is a linear functional continuous on
M
{\displaystyle {\mathcal {M}}}
, then there exists a continuous linear functional
w
{\displaystyle w}
on
X
{\displaystyle {\mathcal {X}}}
such that
z
=
w
{\displaystyle z=w}
on
M
{\displaystyle {\mathcal {M}}}
and
‖
z
‖
=
‖
w
‖
{\displaystyle \|z\|=\|w\|}
.
Proof: Apply the Hahn-Banach stated in Chapter 1 with
‖
z
‖
‖
⋅
‖
{\displaystyle \|z\|\|\cdot \|}
as a sublinear functional dominating
z
{\displaystyle z}
. Then:
‖
z
‖
=
sup
{
‖
w
(
x
)
‖
;
x
∈
M
,
‖
x
‖
≤
1
}
≤
sup
{
‖
w
(
x
)
‖
;
x
∈
X
,
‖
x
‖
≤
1
}
=
‖
w
‖
≤
‖
z
‖
{\displaystyle \|z\|=\sup\{\|w(x)\|;x\in {\mathcal {M}},\|x\|\leq 1\}\leq \sup\{\|w(x)\|;x\in {\mathcal {X}},\|x\|\leq 1\}=\|w\|\leq \|z\|}
;that is,
‖
z
‖
=
‖
w
‖
{\displaystyle \|z\|=\|w\|}
.
◻
{\displaystyle \square }
2. Corollary Let
M
{\displaystyle {\mathcal {M}}}
be a subspace of a normed linear space
X
{\displaystyle {\mathcal {X}}}
. Then
x
{\displaystyle x}
is in the closure of
M
{\displaystyle {\mathcal {M}}}
if and only if
z
(
x
)
{\displaystyle z(x)}
= 0 for any
z
∈
X
∗
{\displaystyle z\in {\mathcal {X}}^{*}}
that vanishes on
M
{\displaystyle {\mathcal {M}}}
.
Proof: By continuity
z
(
M
¯
)
⊂
z
(
M
)
¯
{\displaystyle z({\overline {\mathcal {M}}})\subset {\overline {z({\mathcal {M}})}}}
. Thus, if
x
∈
M
¯
{\displaystyle x\in {\overline {\mathcal {M}}}}
, then
z
(
x
)
∈
z
(
M
)
¯
=
{
0
}
{\displaystyle z(x)\in {\overline {z({\mathcal {M}})}}=\{0\}}
. Conversely, suppose
x
∉
M
¯
{\displaystyle x\not \in {\overline {\mathcal {M}}}}
. Then there is a
δ
>
0
{\displaystyle \delta >0}
such that
‖
y
−
x
‖
≥
δ
{\displaystyle \|y-x\|\geq \delta }
for every
y
∈
M
{\displaystyle y\in {\mathcal {M}}}
. Define a linear functional
z
(
y
+
λ
x
)
=
λ
{\displaystyle z(y+\lambda x)=\lambda }
for
y
∈
M
{\displaystyle y\in {\mathcal {M}}}
and scalars
λ
{\displaystyle \lambda }
. For any
λ
≠
0
{\displaystyle \lambda \neq 0}
, since
−
λ
−
1
y
∈
M
{\displaystyle -\lambda ^{-1}y\in {\mathcal {M}}}
,
|
z
(
y
+
λ
x
)
|
=
|
λ
|
δ
−
1
δ
≤
δ
−
1
|
λ
|
|
λ
−
1
y
+
x
‖
=
‖
y
+
λ
x
‖
{\displaystyle |z(y+\lambda x)|=|\lambda |\delta ^{-1}\delta \leq \delta ^{-1}|\lambda ||\lambda ^{-1}y+x\|=\|y+\lambda x\|}
.Since the inequality holds for
λ
=
0
{\displaystyle \lambda =0}
as well,
z
{\displaystyle z}
is continuous. Hence, in view of the Hahn-Banach theorem,
z
∈
X
{\displaystyle z\in {\mathcal {X}}}
while we still have
z
=
0
{\displaystyle z=0}
on
M
{\displaystyle {\mathcal {M}}}
and
z
(
x
)
≠
0
{\displaystyle z(x)\neq 0}
.
◻
{\displaystyle \square }
Here is a classic application.
2 Theorem Let
X
,
Y
{\displaystyle {\mathcal {X}},{\mathcal {Y}}}
be Banach spaces,
T
:
X
→
Y
{\displaystyle T:{\mathcal {X}}\to {\mathcal {Y}}}
be a linear operator. If
x
n
→
0
{\displaystyle x_{n}\to 0}
implies that
(
z
∘
T
)
x
n
→
0
{\displaystyle (z\circ T)x_{n}\to 0}
for every
z
∈
X
∗
{\displaystyle z\in {\mathcal {X}}^{*}}
, then
T
{\displaystyle T}
is continuous.
Proof: Suppose
x
n
→
0
{\displaystyle x_{n}\to 0}
and
T
x
n
→
y
{\displaystyle Tx_{n}\to y}
. For every
z
∈
X
∗
{\displaystyle z\in {\mathcal {X}}^{*}}
, by hypothesis and the continuity of
z
{\displaystyle z}
,
0
=
lim
n
→
∞
z
(
T
x
n
)
=
z
(
y
)
{\displaystyle 0=\lim _{n\to \infty }z(Tx_{n})=z(y)}
.Now, by the preceding corollary
y
=
0
{\displaystyle y=0}
and the continuity follows from the closed graph theorem.
◻
{\displaystyle \square }
2 Theorem Let
X
{\displaystyle {\mathcal {X}}}
be a Banach space.
(i) Given
E
⊂
X
{\displaystyle E\subset {\mathcal {X}}}
,
E
{\displaystyle E}
is bounded if and only if
sup
E
|
f
|
<
∞
{\displaystyle \sup _{E}|f|<\infty }
for every
f
∈
X
∗
{\displaystyle f\in {\mathcal {X}}^{*}}
(ii) Given
x
∈
X
{\displaystyle x\in {\mathcal {X}}}
, if
f
(
x
)
=
0
{\displaystyle f(x)=0}
for every
f
∈
X
∗
{\displaystyle f\in {\mathcal {X}}^{*}}
, then
x
=
0
{\displaystyle x=0}
.Proof: (i) By continuity,
sup
{
|
f
(
x
)
|
;
x
∈
E
}
≤
‖
f
‖
sup
E
‖
⋅
‖
{\displaystyle \sup\{|f(x)|;x\in E\}\leq \|f\|\sup _{E}\|\cdot \|}
.This proves the direct part. For the converse, define
T
x
f
=
f
(
x
)
{\displaystyle T_{x}f=f(x)}
for
x
∈
E
,
f
∈
X
∗
{\displaystyle x\in E,f\in {\mathcal {X}}^{*}}
. By hypothesis
|
T
x
f
|
≤
sup
E
|
f
|
{\displaystyle |T_{x}f|\leq \sup _{E}|f|}
for every
x
∈
E
{\displaystyle x\in E}
.Thus, by the principle of uniform boundedness, there is
K
>
0
{\displaystyle K>0}
such that:
|
T
x
f
|
≤
K
‖
f
‖
{\displaystyle |T_{x}f|\leq K\|f\|}
for every
x
∈
E
,
f
∈
X
∗
{\displaystyle x\in E,f\in {\mathcal {X}}^{*}}
Hence, in view of Theorem 2.something, for
x
∈
E
{\displaystyle x\in E}
,
‖
x
‖
=
sup
{
|
f
(
x
)
|
;
f
∈
X
∗
,
‖
f
‖
≤
1
}
≤
K
{\displaystyle \|x\|=\sup\{|f(x)|;f\in {\mathcal {X}}^{*},\|f\|\leq 1\}\leq K}
.(ii) Suppose
x
≠
0
{\displaystyle x\neq 0}
. Define
f
(
s
(
x
)
)
=
s
‖
x
‖
{\displaystyle f(s(x))=s\|x\|}
for scalars
s
{\displaystyle s}
. Now,
f
{\displaystyle f}
is continuous since its domain is finite-dimensional, and so by the Hahn-Banach theorem we could extend the domain of
f
{\displaystyle f}
in such a way we have
f
∈
X
∗
{\displaystyle f\in {\mathcal {X}}^{*}}
.
◻
{\displaystyle \square }
2. Corollary Let
(
X
,
‖
⋅
‖
)
{\displaystyle ({\mathcal {X}},\|\cdot \|)}
be Banach,
f
j
∈
X
∗
{\displaystyle f_{j}\in {\mathcal {X}}^{*}}
and
M
⊂
X
{\displaystyle {\mathcal {M}}\subset {\mathcal {X}}}
dense and linear. Then
f
j
(
x
)
→
0
{\displaystyle f_{j}(x)\to 0}
for every
x
∈
X
{\displaystyle x\in {\mathcal {X}}}
if and only if
sup
j
‖
f
j
‖
<
∞
{\displaystyle \sup _{j}\|f_{j}\|<\infty }
and
f
j
(
y
)
→
0
{\displaystyle f_{j}(y)\to 0}
for every
y
∈
M
{\displaystyle y\in {\mathcal {M}}}
.
Proof: Since
f
j
{\displaystyle f_{j}}
is Cauchy, it is bounded. This shows the direct part. To show the converse, let
x
∈
X
{\displaystyle x\in {\mathcal {X}}}
. If
y
j
∈
M
{\displaystyle y_{j}\in {\mathcal {M}}}
, then
|
f
j
(
x
)
|
≤
|
f
j
(
x
−
y
j
)
|
+
|
f
j
(
y
j
)
|
≤
(
sup
j
‖
f
j
‖
)
‖
x
−
y
j
‖
+
|
f
j
(
y
j
)
|
{\displaystyle |f_{j}(x)|\leq |f_{j}(x-y_{j})|+|f_{j}(y_{j})|\leq (\sup _{j}\|f_{j}\|)\|x-y_{j}\|+|f_{j}(y_{j})|}
By denseness, we can take
y
j
{\displaystyle y_{j}}
so that
‖
y
j
−
x
‖
→
0
{\displaystyle \|y_{j}-x\|\to 0}
.
◻
{\displaystyle \square }
2 Theorem Let
T
{\displaystyle T}
be a continuous linear operator into a Banach space. If
‖
I
−
T
‖
<
1
{\displaystyle \|I-T\|<1}
where
I
{\displaystyle I}
is the identity operator, then the inverse
T
−
1
{\displaystyle T^{-1}}
exists, is continuous and can be written by:
T
−
1
(
x
)
=
∑
k
=
0
∞
(
I
−
T
)
k
(
x
)
{\displaystyle T^{-1}(x)=\sum _{k=0}^{\infty }\left(I-T\right)^{k}(x)}
for each
x
{\displaystyle x}
in the range of
T
{\displaystyle T}
.Proof: For
n
≥
m
{\displaystyle n\geq m}
, we have:
‖
∑
k
=
m
n
(
I
−
T
)
k
(
x
)
‖
≤
‖
x
‖
∑
k
=
m
n
‖
I
−
T
‖
k
{\displaystyle \|\sum _{k=m}^{n}\left(I-T\right)^{k}(x)\|\leq \|x\|\sum _{k=m}^{n}\left\|I-T\right\|^{k}}
.Since the series is geometric by hypothesis, the right-hand side is finite. Let
S
n
=
∑
k
=
0
n
(
I
−
T
)
k
{\displaystyle S_{n}=\sum _{k=0}^{n}\left(I-T\right)^{k}}
. By the above, each time
x
{\displaystyle x}
is fixed,
S
n
(
x
)
{\displaystyle S_{n}(x)}
is a Cauchy sequence and the assumed completeness implies that the sequence converges to the limit, which we denote by
S
(
x
)
{\displaystyle S(x)}
. Since for each
x
{\displaystyle x}
sup
n
≥
1
‖
S
n
(
x
)
‖
<
∞
{\displaystyle \sup _{n\geq 1}\|S_{n}(x)\|<\infty }
, it follows from the principle of uniform boundedness that:
sup
n
≥
1
‖
S
n
‖
≤
∞
{\displaystyle \sup _{n\geq 1}\|S_{n}\|\leq \infty }
.Thus, by the continuity of norms,
‖
S
(
x
)
‖
=
lim
n
→
∞
‖
S
n
(
x
)
‖
≤
(
sup
n
≥
1
‖
S
n
‖
)
‖
x
‖
{\displaystyle \|S(x)\|=\lim _{n\to \infty }\|S_{n}(x)\|\leq (\sup _{n\geq 1}\|S_{n}\|)\|x\|}
.This shows that
S
{\displaystyle S}
is a continuous linear operator since the linearity is easily checked. Finally,
‖
T
S
(
x
)
−
x
‖
=
‖
lim
n
→
∞
−
(
I
−
T
)
n
+
1
(
x
)
‖
≤
‖
x
‖
lim
n
→
∞
‖
I
−
T
‖
n
+
1
=
0
{\displaystyle \|TS(x)-x\|=\|\lim _{n\to \infty }-(I-T)^{n+1}(x)\|\leq \|x\|\lim _{n\to \infty }\|I-T\|^{n+1}=0}
.Hence,
S
{\displaystyle S}
is the inverse to
T
{\displaystyle T}
.
◻
{\displaystyle \square }
2 Corollary The space of invertible continuous linear operators
X
{\displaystyle {\mathcal {X}}}
is an open subspace of
L
(
X
,
X
)
{\displaystyle L({\mathcal {X}},{\mathcal {X}})}
.
Proof: If
T
∈
L
(
X
,
X
)
{\displaystyle T\in L({\mathcal {X}},{\mathcal {X}})}
and
‖
S
−
T
‖
<
1
‖
T
−
1
‖
{\displaystyle \|S-T\|<{1 \over \|T^{-1}\|}}
, then
S
{\displaystyle S}
is invertible.
◻
{\displaystyle \square }
If
F
{\displaystyle \mathbf {F} }
is a scalar field and
X
{\displaystyle {\mathcal {X}}}
is a normed space, then
L
(
X
,
F
)
{\displaystyle L({\mathcal {X}},\mathbf {F} )}
is called a dual of
X
{\displaystyle {\mathcal {X}}}
and is denoted by
X
∗
{\displaystyle {\mathcal {X}}^{*}}
. In view of Theorem 2.something, it is a Banach space.
A linear operator
T
{\displaystyle T}
is said to be a compact operator if the image of the open unit ball under
T
{\displaystyle T}
is relatively compact. We recall that if a linear operator between normed spaces maps bounded sets to bounded sets, then it is continuous. Thus, every compact operator is continuous.
2 Theorem Let
X
{\displaystyle {\mathcal {X}}}
be a reflexive Banach space and
Y
{\displaystyle {\mathcal {Y}}}
be a Banach space. Then a linear operator
T
:
X
→
Y
{\displaystyle T:{\mathcal {X}}\to {\mathcal {Y}}}
is a compact operator if and only if
T
{\displaystyle T}
sends weakly convergent sequence to norm convergent ones.
Proof: Let
x
n
{\displaystyle x_{n}}
converges weakly to
0
{\displaystyle 0}
, and suppose
T
x
n
{\displaystyle Tx_{n}}
is not convergent. That is, there is an
ϵ
>
0
{\displaystyle \epsilon >0}
such that
T
x
n
≥
ϵ
{\displaystyle Tx_{n}\geq \epsilon }
for infinitely many
n
{\displaystyle n}
. Denote this subsequence by
y
n
{\displaystyle y_{n}}
. By hypothesis we can then show (TODO: do this indeed) that it contains a subsequence
y
n
k
{\displaystyle y_{n_{k}}}
such that
T
y
n
k
{\displaystyle Ty_{n_{k}}}
converges in norm, which is a contradiction. To show the converse, let
E
{\displaystyle E}
be a bounded set. Then since
X
{\displaystyle {\mathcal {X}}}
is reflexive every countable subset of
E
{\displaystyle E}
contains a sequence
x
n
{\displaystyle x_{n}}
that is Cauchy in the weak topology and so by the hypothesis
T
x
n
{\displaystyle Tx_{n}}
is a Cauchy sequence in norm. Thus,
T
(
E
)
{\displaystyle T(E)}
is contained in a compact subset of
Y
{\displaystyle {\mathcal {Y}}}
.
◻
{\displaystyle \square }
2 Corollary
(i) Every finite-rank linear operator
T
{\displaystyle T}
(i.e., a linear operator with finite-dimensional range) is a compact operator.
(ii) Every linear operator
T
{\displaystyle T}
with the finite-dimensional domain is continuous.Proof: (i) is clear, and (ii) follows from (i) since the range of a linear operator has dimension less than that of the domain.
◻
{\displaystyle \square }
2 Theorem The set of all compact operators into a Banach space forms a closed subspace of the set of all continuous linear operators in operator norm.
Proof: Let
T
{\displaystyle T}
be a linear operator and
ω
{\displaystyle \omega }
be the open unit ball in the domain of
T
{\displaystyle T}
. If
T
{\displaystyle T}
is compact, then
T
(
ω
¯
)
{\displaystyle T({\overline {\omega }})}
is bounded (try scalar multiplication); thus,
T
{\displaystyle T}
is continuous. Since the sum of two compacts sets is again compact, the sum of two compact operators is again compact. For the similar reason,
α
T
{\displaystyle \alpha T}
is compact for any scalar
α
{\displaystyle \alpha }
. We conclude that the set of all compact operators, which we denote by
E
{\displaystyle E}
, forms a subspace of continuous linear operators. To show the closedness, suppose
S
{\displaystyle S}
is in the closure of
E
{\displaystyle E}
. Let
ϵ
>
0
{\displaystyle \epsilon >0}
be given. Then there is some compact operator
T
{\displaystyle T}
such that
‖
S
−
T
‖
<
ϵ
/
2
{\displaystyle \|S-T\|<\epsilon /2}
. Also, since
T
{\displaystyle T}
is a compact operator, we can cover
T
(
ω
)
{\displaystyle T(\omega )}
by a finite number of open balls of radius
ϵ
/
2
{\displaystyle \epsilon /2}
centered at
z
1
,
z
2
,
.
.
.
z
n
{\displaystyle z_{1},z_{2},...z_{n}}
, respectively. It then follows: for
x
∈
ω
{\displaystyle x\in \omega }
, we can find some
j
{\displaystyle j}
so that
‖
T
x
−
z
j
‖
<
ϵ
/
2
{\displaystyle \|Tx-z_{j}\|<\epsilon /2}
and so
‖
S
x
−
T
x
‖
≤
‖
S
x
−
z
j
‖
+
‖
z
j
−
T
x
‖
<
ϵ
{\displaystyle \|Sx-Tx\|\leq \|Sx-z_{j}\|+\|z_{j}-Tx\|<\epsilon }
. This is to say,
S
(
ω
)
{\displaystyle S(\omega )}
is totally bounded and since the completeness its closure is compact.
◻
{\displaystyle \square }
2 Corollary If
T
n
{\displaystyle T_{n}}
is a sequence of compact operators which converges in operator norm, then its limit is a compact operator.
2 Theorem (transpose) Let
X
,
Y
{\displaystyle {\mathcal {X}},{\mathcal {Y}}}
be Banach spaces, and
u
:
X
→
Y
{\displaystyle u:{\mathcal {X}}\to {\mathcal {Y}}}
be a continuous linear operator. Define
t
u
:
Y
∗
→
X
∗
{\displaystyle {}^{t}\!u:{\mathcal {Y}}^{*}\to {\mathcal {X}}^{*}}
by the identity
t
u
(
z
)
(
x
)
=
u
(
z
(
x
)
)
{\displaystyle {}^{t}\!u(z)(x)=u(z(x))}
. Then
t
u
{\displaystyle {}^{t}\!u}
is continuous both in operator norm and the weak-* topology, and
‖
t
u
‖
=
‖
u
‖
{\displaystyle \|{}^{t}\!u\|=\|u\|}
.
Proof: For any
z
∈
Y
∗
{\displaystyle z\in {\mathcal {Y}}^{*}}
‖
t
u
(
z
)
‖
=
sup
‖
x
‖
≤
1
|
(
u
∘
z
)
(
x
)
|
≤
‖
u
‖
‖
z
‖
{\displaystyle \|{}^{t}\!u(z)\|=\sup _{\|x\|\leq 1}|(u\circ z)(x)|\leq \|u\|\|z\|}
Thus,
‖
t
u
‖
≤
‖
u
‖
{\displaystyle \|{}^{t}\!u\|\leq \|u\|}
and
t
u
{\displaystyle {}^{t}\!u}
is continuous in operator norm. To show the opposite inequality, let
ϵ
>
0
{\displaystyle \epsilon >0}
be given. Then there is
x
0
∈
X
{\displaystyle x_{0}\in {\mathcal {X}}}
with
(
1
−
ϵ
)
‖
u
‖
≤
|
u
(
x
0
)
|
{\displaystyle (1-\epsilon )\|u\|\leq |u(x_{0})|}
. Using the Hahn-Banach theorem we can also find
‖
z
0
‖
=
1
{\displaystyle \|z_{0}\|=1}
and
z
0
(
u
(
x
0
)
)
=
|
u
(
x
0
)
|
{\displaystyle z_{0}(u(x_{0}))=|u(x_{0})|}
. Hence,
‖
t
u
‖
=
sup
‖
z
‖
≤
1
‖
t
u
(
z
)
‖
≥
‖
t
u
(
z
0
)
‖
=
|
z
0
(
u
(
x
)
)
|
=
|
u
(
x
0
)
|
≥
(
1
−
ϵ
)
‖
u
‖
{\displaystyle \|{}^{t}\!u\|=\sup _{\|z\|\leq 1}\|{}^{t}\!u(z)\|\geq \|{}^{t}\!u(z_{0})\|=|z_{0}(u(x))|=|u(x_{0})|\geq (1-\epsilon )\|u\|}
.We conclude
‖
t
u
‖
=
‖
u
‖
{\displaystyle \|{}^{t}\!u\|=\|u\|}
. To show weak-* continuity let
V
{\displaystyle V}
be a neighborhood of
0
{\displaystyle 0}
in
X
∗
{\displaystyle {\mathcal {X}}^{*}}
; that is,
V
=
{
z
;
z
∈
X
∗
,
|
z
(
x
1
)
|
<
ϵ
,
.
.
.
,
|
z
(
x
n
)
|
<
ϵ
}
{\displaystyle V=\{z;z\in {\mathcal {X}}^{*},|z(x_{1})|<\epsilon ,...,|z(x_{n})|<\epsilon \}}
for some
ϵ
>
0
,
x
1
,
.
.
.
,
x
n
∈
X
{\displaystyle \epsilon >0,x_{1},...,x_{n}\in {\mathcal {X}}}
. If we let
y
j
=
u
(
x
j
)
{\displaystyle y_{j}=u(x_{j})}
, then
t
u
(
{
z
;
z
∈
Y
∗
,
|
z
(
y
1
)
|
<
ϵ
,
.
.
.
,
|
z
(
y
n
)
|
<
ϵ
}
)
⊂
V
{\displaystyle {}^{t}\!u(\{z;z\in {\mathcal {Y}}^{*},|z(y_{1})|<\epsilon ,...,|z(y_{n})|<\epsilon \})\subset V}
since
z
(
y
j
)
=
t
u
(
z
)
(
x
j
)
{\displaystyle z(y_{j})={}^{t}\!u(z)(x_{j})}
. This is to say,
t
u
{\displaystyle {}^{t}\!u}
is weak-* continuous.
◻
{\displaystyle \square }
2 Theorem Let
T
:
X
→
Y
{\displaystyle T:{\mathcal {X}}\to {\mathcal {Y}}}
be a linear operator between normed spaces. Then
T
{\displaystyle T}
is compact if and only if its transpose
T
′
{\displaystyle T'}
is compact.
Proof: Let
K
{\displaystyle K}
be the closure of the image of the closed unit ball under
T
{\displaystyle T}
. If T is compact, then K is compact. Let
y
n
∈
Y
∗
{\displaystyle y_{n}\in Y^{*}}
be a bounded sequence. Then the restrictions of
y
n
{\displaystyle y_{n}}
to K is a bounded equicontinuous sequence in
C
(
K
)
{\displaystyle C(K)}
; thus, it has a convergent subsequence
y
n
k
{\displaystyle y_{n_{k}}}
by Ascoli's theorem. Thus,
T
′
y
n
k
(
x
)
=
y
n
k
(
T
x
)
{\displaystyle T'y_{n_{k}}(x)=y_{n_{k}}(Tx)}
is convergent for every x with
‖
x
‖
X
≤
1
{\displaystyle \|x\|_{\mathcal {X}}\leq 1}
, and so
T
′
y
n
k
{\displaystyle T'y_{n_{k}}}
is convergent. The converse follows from noting that every normed space can be embedded continuously into its second dual. (TODO: need more details.)
◻
{\displaystyle \square }
== References == | 22,146 | 55,157 | {"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} | 4.03125 | 4 | CC-MAIN-2020-50 | latest | en | 0.68193 |
https://mymathsworld.wordpress.com/2010/10/26/ymxc1/ | 1,503,352,070,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109670.98/warc/CC-MAIN-20170821211752-20170821231752-00156.warc.gz | 831,233,661 | 15,092 | Home > Graphing > Finding the equation of a line given the y intercept
Finding the equation of a line given the y intercept
The equation of a straight line is usually given in the form $y=mx+c$ where m is the gradient (slope) and c is the y-intercept (where the line cuts the vertical axis).
The gradient = $\frac{rise}{run}$ or $\frac{\delta y}{\delta x}$
Example
The gradient between (2, 5) and (4,10) = $\frac{10-5}{4-2}=\frac{5}{2}$
It is important that if we do 10 – 5 for the rise that we do not do 2 – 4 for the run. If we are not consistent then we will end up with the wrong sign for the gradient.
Finding the equation given the gradient and the y-intercept
Example
A line has a gradient of 3 and cuts the y axis at (0,4), what is the equation of the line?
From the question we can see that m = 3 and c = 4, so substituting this in to $y=mx+c$ gives $y=3x+4$
Finding the equation given the y-intercept and a parallel line
Example
A line is parallel to the line $y=2x+7$ and intercepts the y axis at (0, -5), what is the equation of the line?
Since the given line has a gradient of 2 (the x coefficient), m = 2. We can also see that c = -5 from the point.
So substituting this in to $y=mx+c$ gives $y=2x-5$
Finding the equation given the y-intercept and another point
Example
A line passes through the points (0, 4 ) and (3, 11), what is the equation of the line?
We can calculate the gradient using the method shown above so $m=\frac{11-4}{3-0}=\frac{7}{3}$
From the point (0, 4) we can see that c = 4
So substituting this in to $y=mx+c$ gives $y=\frac{7}{3}x+4$ | 483 | 1,594 | {"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": 12, "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-2017-34 | latest | en | 0.85231 |
https://www.crazy-numbers.com/en/13 | 1,701,533,729,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100427.59/warc/CC-MAIN-20231202140407-20231202170407-00787.warc.gz | 797,400,117 | 6,080 | Discover a lot of information on the number 13: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
## Mathematical properties of 13
Is 13 a prime number? Yes
Is 13 a perfect number? No
Number of divisors 2
List of dividers 1, 13
Sum of divisors 14
Prime factorization 13
Prime factors 13
## How to write / spell 13 in letters?
In letters, the number 13 is written as: Thirteen. And in other languages? how does it spell?
13 in other languages
Write 13 in english Thirteen
Write 13 in french Treize
Write 13 in spanish Trece
Write 13 in portuguese Treze
## Decomposition of the number 13
The number 13 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 3 : The number 3 (three) is the symbol of the trinity. He also represents the union.... Find out more about the number 3
Other ways to write 13
In letter Thirteen
In roman numeral XIII
In binary 1101
In octal 15
In US dollars USD 13.00 (\$)
In euros 13,00 EUR (€)
Some related numbers
Previous number 12
Next number 14
Next prime number 17
## Mathematical operations
Operations and solutions
13*2 = 26 The double of 13 is 26
13*3 = 39 The triple of 13 is 39
13/2 = 6.5 The half of 13 is 6.500000
13/3 = 4.3333333333333 The third of 13 is 4.333333
132 = 169 The square of 13 is 169.000000
133 = 2197 The cube of 13 is 2197.000000
√13 = 3.605551275464 The square root of 13 is 3.605551
log(13) = 2.5649493574615 The natural (Neperian) logarithm of 13 is 2.564949
log10(13) = 1.1139433523068 The decimal logarithm (base 10) of 13 is 1.113943
sin(13) = 0.42016703682664 The sine of 13 is 0.420167
cos(13) = 0.9074467814502 The cosine of 13 is 0.907447
tan(13) = 0.46302113293649 The tangent of 13 is 0.463021 | 609 | 1,891 | {"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-2023-50 | longest | en | 0.798762 |
http://stackoverflow.com/questions/24112474/tarjans-algorithm-do-lowest-links-have-to-be-similar-for-two-or-more-nodes-to | 1,469,795,505,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257830066.95/warc/CC-MAIN-20160723071030-00138-ip-10-185-27-174.ec2.internal.warc.gz | 233,169,565 | 18,312 | Dismiss
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# Tarjan's algorithm: do lowest-links have to be similar for two or more nodes to be inside the same SCC
I'm having some trouble with a homework question involving using Tarjan's algorithm on a provided graph to find the particular SCC's for that graph. While (according to my professor) I have found the correct SCC's by using the pseudo-code algorithm found here, some of the nodes in my SCC's do not share the same lowest-link number as the root node for that SCC.
From what I can gather from the pseudo-code, this is because if an un-referenced node `i` (which is the input node for the current recursive call to the algorithm) has an arc to an already visited node `i + 1` which is not the root node, then the algorithm sets `i`s `LL = MIN(i.LowestLink, (i + 1).index)`, and `(i + 1).index` may not be equal to its own lowest-link value anymore.
For example (this is similar to a part of the graph from the problem I'm trying to solve): if we have nodes in `N = {a, b, c, d}`, and arcs in `E = {a ⇒ c, c ⇒ b, c ⇒ d, b ⇒ a, d ⇒ b}`, and our root node which we start the algorithm from is `a`, then:
1.1) We set `a.index = 1` (using 1 rather than 0), `a.LL = 1`, and push `a` onto the stack; `a` has a single arc to `c`, so we check `c`; finding that it is undiscovered, we call the algorithm on `c`.
2.1) We set `c.index = 2`, `c.LL = 2`, and push `c` onto the stack; `c` has two arcs, one to `b`, and one to `d`. Assume our for loop checks `b` first; `b` is undiscovered, and so we call the algorithm on `b`.
3.1) We set `b.index = 3`, `b.LL = 3`, and push `b` onto the stack; `b` has one arc to `a`; checking `a` we find that it is already on the stack, and so (by the pseudo-code linked above) we set `b.LL = MIN(b.LL, a.index) = a.index = 1`; `b` has no further arcs, so we exit our for loop, and check if `b.LL = b.index`, it does not, so we end this instance of the algorithm.
2.2) Now that the recursive call on `b` has ended, we set `c.LL = MIN(c.LL, b.LL) = b.LL = 1`. `c` still has the arc from `c` to `d` remaining; checking `d` we find it is undefined, so we call the algorithm on `d`.
4.1) `d.index` is set to 4, `d.LL` is set to 4, and we push `d` onto the stack. `d` has one arc from `d` to `b`, so we check `b`; we find that `b` is already in the stack, so we set `d.LL = MIN(d.LL, b.index) = b.index = 3`. `d` has no further arcs, so we exit our for loop and check if `d.LL = d.index`; it does not, so we end this instance of the algorithm.
2.3) With the recursive call on `d` ended, we again set `c.LL = MIN(c.LL, d.LL) = c.LL = 1`. `c` has no further arcs, and so we end our for loop. We check to see if `c.LL = c.index`; it does not, so we end this instance of the algorithm.
1.2) With the recursive call on `c` ended, we set `a.LL = MIN(a.LL, c.LL) = 1`. `a` has no further arcs, so we end our for loop. We check if `a.LL = a.index`; they are equal, so we have found a root node for this SCC; we create a new SCC, and pop each item in the stack into this SCC until we find `a` in the stack (wich also goes into this SCC).
After these steps all the nodes in the graph are discovered, so running the algorithm with the other nodes initially does nothing, we have one `SCC = {a, b, c, d}`. However, `d.LL = 3` which is not equal to the rest of the nodes lowest-links (which are all 1).
Have I done something wrong here? Or is it possible in this situation to have an SCC with differing lowest-links among its nodes?
-
Yes, this is normal. The LL value is not useful once the algorithm finishes. – rici Jun 9 '14 at 3:57
It seems you're using a different code from the one currently in Wikipedia. See also stackoverflow.com/questions/24114178/…. – lhf Jul 2 '14 at 1:33 | 1,161 | 3,906 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2016-30 | latest | en | 0.909962 |
https://rrbpaper.com/strength-of-materials-mcq/tangent-of-angle-of-friction-is-equal-to/ | 1,656,904,741,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104293758.72/warc/CC-MAIN-20220704015700-20220704045700-00380.warc.gz | 533,092,249 | 20,748 | Tangent of angle of friction is equal to
Tangent of angle of friction is equal to
Right Answer is: Coefficient of friction
SOLUTION
The angle of repose is the least angle of the inclined plane (of the given surface) with the horizontal such that the given body placed over the plane, just begins to slide down, without getting accelerated.
The tangent of the angle of repose is equal to the coefficient of friction.
Hence, we conclude that the angle of friction (θ) is equal to the angle of repose (φ).
Scroll to Top | 118 | 523 | {"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} | 3.359375 | 3 | CC-MAIN-2022-27 | latest | en | 0.918525 |
https://rd.springer.com/chapter/10.1007/978-1-4684-9458-7_1 | 1,537,749,989,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267159938.71/warc/CC-MAIN-20180923232129-20180924012529-00088.warc.gz | 605,770,126 | 13,670 | # Generalities on linear representations
• Jean-Pierre Serre
Chapter
Part of the Graduate Texts in Mathematics book series (GTM, volume 42)
## Abstract
Let V be a vector space over the field C of complex numbers and let GL(V) be the group of isomorphisms of V onto itself. An element a of GL(V) is, by definition, a linear mapping of V into V which has an inverse a-1; this inverse is linear. When V has a finite basis (e i ) of n elements, each linear map a: V → V is defined by a square matrix (a ij ) of order n. The coefficients a ij are complex numbers; they are obtained by expressing the images a(e j ) in terms of the basis (e i ):
$$a({e_j}) = \sum\limits_i {{a_{ij}}} {e_i}.$$ | 188 | 689 | {"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} | 2.6875 | 3 | CC-MAIN-2018-39 | latest | en | 0.865651 |
http://www.chegg.com/homework-help/university-physics-with-modern-physics-13th-edition-chapter-29-solutions-9780321696861 | 1,472,588,102,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471983001995.81/warc/CC-MAIN-20160823201001-00250-ip-10-153-172-175.ec2.internal.warc.gz | 369,703,841 | 19,793 | View more editions
# University Physics with Modern Physics (13th Edition)Solutions for Chapter 29
• 4727 step-by-step solutions
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Chapter: Problem:
78% (9 ratings)
A sheet of copper is placed between the poles of an electromagnet with the magnetic field perpendicular to the sheet. When the sheet is pulled out, a considerable force is required and the force required increases with speed . Explain.
SAMPLE SOLUTION
Chapter: Problem:
78% (9 ratings)
• Step 1 of 1
REASON:
If the sheet of copper is placed between the poles of an electromagnet with magnetic field perpendicular to the sheet and then pulled out, it does not come out because the eddy currents exert a strong resistance to the motion. So we need to apply more force to pull out from the electromagnet and when sheet is pulled out, flux linked with it changes and according to Lenz’s Law, it opposes the motion of sheet hence more force is required as the speed increases.
Corresponding Textbook
University Physics with Modern Physics | 13th Edition
9780321696861ISBN-13: 0321696867ISBN: Authors:
Alternate ISBN: 9780321830647, 9780321898111 | 306 | 1,306 | {"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-2016-36 | latest | en | 0.850265 |
https://discuss.leetcode.com/topic/110579/ac-solution-code | 1,516,646,269,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891530.91/warc/CC-MAIN-20180122173425-20180122193425-00565.warc.gz | 649,134,836 | 8,694 | # ac solution code
• Solution1. time = O(n); space = O(1)
KEY:
• Roman digits are based on SUM from end to start, not multiplication!!
Explanation:
1. map: [Number: RomanString]. 10 * 4 table to present Roman strings corresponding to 10-digit value
• 1...9
• 10...90
• 100...900
• 1000...3000
2. Decial digit: from low to high - divided by 10 each step
3. Prefix Roman String to prev result
• Check Roman string in the table corresponding to the remainder
``````func intToRoman(_ num: Int) -> String {
// 10 * 4 map: Digit => String
let map = [["", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"],// 1..9
["", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"],// 10..90
["", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"],// 100..900
["", "M", "MM", "MMM"]] // 1000..3000
var res = ""
var digit = 0, num = num
while num > 0 { // Start from low to high digit: 1=>1000 (Same as 10-digit decimal system: low->high = right->left)
let curVal = num % 10
let curStr = map[digit][curVal] // Current 10-digit value
res = curStr + res // Prefix Roman string to result
num /= 10 // Advance to next digit (lefter)
digit += 1
}
return res
}
``````
Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect. | 436 | 1,374 | {"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-2018-05 | latest | en | 0.589258 |
https://scicomp.stackexchange.com/questions/34101/testing-a-block-tridiagonal-system-of-equations?noredirect=1 | 1,701,602,486,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100499.43/warc/CC-MAIN-20231203094028-20231203124028-00186.warc.gz | 567,535,568 | 43,997 | Testing a block tridiagonal system of equations
In 1D problems, tridiagonal systems of equations are obtained when we use finite-difference or finite-volumes in a structured mesh. A wide solver is the TDMA algorithm here. In two-dimensional modelling, we achieve a block tridiagonal systems of equations. However, I found a lack of codes dealing with those block matrices. In book:
Pletcher R.H., Tannehill J.C., Anderson D.A. Computational Fluid Mechanics and Heat Transfer (series in computational and physical processes in mechanics and thermal sciences). 3rd ed. CRC Press Taylor & Francis Group, 2013.
there are the subroutines for making up a block-TDMA, code in Fortran. I transcript almost the entire appendix B:
Subroutine NBTRIP solves a block tridiagonal system of equations of the form $$\begin{pmatrix}B_{IL}&C_{IL}&&\\A_{I}&B_{I}&C_{I}&&\\&\ddots&\ddots&\ddots\\&&A_{IU}&B_{IU}\end{pmatrix}\begin{pmatrix}X_{IL}\\X_{I}\\\vdots\\X_{IU}\end{pmatrix} = \begin{pmatrix}D_{IL}\\D_{I}\\\vdots\\D_{IU}\end{pmatrix}$$ Subroutine PBTRIP solves a periodic block tridiagonal system of equations in the form $$\begin{pmatrix}B_{IL}&C_{IL}&&A_{IL}\\A_{I}&B_{I}&C_{I}&&\\&\ddots&\ddots&\ddots\\C_{IU}&&A_{IU}&B_{IU}\end{pmatrix}\begin{pmatrix}X_{IL}\\X_{I}\\\vdots\\X_{IU}\end{pmatrix} = \begin{pmatrix}D_{IL}\\D_{I}\\\vdots\\D_{IU}\end{pmatrix}$$ The block matrices $$A$$, $$B$$, and $$C$$ are $$N × N$$ matrices at every point $$I$$ with $$N$$ being an integer greater than 1. The right-hand side vector $$D$$ has length $$N$$ at each point $$I$$.Therefore, the total number of $$I$$ points at which the matrices are defined (denoted by $$NI$$) is given by $$NI = IU - IL + 1$$. The matrices $$A$$, $$B$$, and $$C$$ are dimensioned as $$A(N,N,NI)$$, $$B(N,N,NI)$$ and $$C(N,N,NI)$$ while the vector $$D$$ is dimensioned as $$D(N,NI)$$. The solution, $$X$$, is returned to the calling program by overwriting the $$D$$ vector with the $$X$$ vector.
FORTRAN 90 code:
!++++++++++++++++++++++++++++++++++++++++++++++++++++++!
! !
! SUBROUTINE TO SOLVE NON-PERIODIC BLOCK TRIDIAGONAL !
! SYSTEM OF EQUATIONS WITHOUT PIVOTING STRATEGY !
! WITH THE DIMENSIONS OF THE BLOCK MATRICES BEING !
! N × N (N IS ANY NUMBER GREATER THAN 1). !
! !
!++++++++++++++++++++++++++++++++++++++++++++++++++++++!
SUBROUTINE NBTRIP (A, B, C, D, IL, IU, ORDER)
INTEGER, INTENT(IN) :: IL, IU, ORDER
REAL, INTENT(INOUT) :: A(1), B(1)
REAL, INTENT(INOUT) :: C(1), D(1)
!...
!...A = SUB DIAGONAL MATRIX
!...B = DIAGONAL MATRIX
!...C = SUP DIAGONAL MATRIX
!...D = RIGHT HAND SIDE VECTOR
!...IL = LOWER VALUE OF INDEX FOR WHICH MATRICES ARE DEFINED
!...IU = UPPER VALUE OF INDEX FOR WHICH MATRICES ARE DEFINED
!... (SOLUTION IS SOUGHT FOR BTRI(A, B, C)*X = D
!... FOR INDICES OF X BETWEEN IL AND IU (INCLUSIVE).
!... SOLUTION WRITTEN IN D VECTOR (ORIGINAL CONTENTS
!... ARE OVERWRITTEN)).
!...ORDER = ORDER OF A, B, C MATRICES AND LENGTH OF D VECTOR
!... AT EACH POINT DENOTED BY INDEX I
!... (ORDER CAN BE ANY INTEGER GREATER THAN 1).
!...
!...THE MATRICES AND VECTORS ARE STORED IN SINGLE SUBSCRIPT FORM
!...
INTEGER :: ORDSQ
INTEGER :: I0MAT, I0MATJ, I0VEC
INTEGER :: I1MAT, I1MATJ, I1VEC
INTEGER :: I, J
!...
!...FORWARD ELIMINATION
!...
ORDSQ = ORDER**2
I = IL
I0MAT = 1 + (I - 1)*ORDSQ
I0VEC = 1 + (I - 1)*ORDER
CALL LUDECO (B(I0MAT), ORDER)
CALL LUSOLV (B(I0MAT), D(I0VEC), D(I0VEC), ORDER)
DO J = 1, ORDER
I0MATJ = I0MAT + (J - 1)*ORDER
CALL LUSOLV (B(I0MAT), C(I0MATJ), C(I0MATJ), ORDER)
END DO
DO
I = I + 1
I0MAT = 1 + (I - 1)*ORDSQ
I0VEC = 1 + (I - 1)*ORDER
I1MAT = I0MAT - ORDSQ
I1VEC = I0VEC - ORDER
CALL MULPUT (A(I0MAT), D(I1VEC), D(I0VEC), ORDER)
DO J = 1, ORDER
I0MATJ = I0MAT + (J - 1)*ORDER
I1MATJ = I1MAT + (J - 1)*ORDER
CALL MULPUT (A(I0MAT), C(I1MATJ), B(I0MATJ), ORDER)
END DO
CALL LUDECO (B(I0MAT), ORDER)
CALL LUSOLV (B(I0MAT), D(I0VEC), D(I0VEC), ORDER)
IF(I == IU) EXIT
DO J = 1, ORDER
I0MATJ = I0MAT + (J - 1)*ORDER
CALL LUSOLV (B(I0MAT), C(I0MATJ), C(I0MATJ), ORDER)
END DO
END DO
!...
!...BACK SUBSTITUTION
!...
I = IU
DO
I = I - 1
I0MAT = 1 + (I - 1)*ORDSQ
I0VEC = 1 + (I - 1)*ORDER
I1VEC = I0VEC + ORDER
CALL MULPUT(C(I0MAT), D(I1VEC), D(I0VEC), ORDER)
IF (I <= IL) EXIT
END DO
!...
END SUBROUTINE NBTRIP
!+++++++++++++++++++++++++++++++++++++++++++++++++++++++!
! !
! SUBROUTINE TO SOLVE PERIODIC BLOCK TRIDIAGONAL !
! SYSTEM OF EQUATIONS WITHOUT PIVOTING STRATEGY. !
! EACH BLOCK MATRIX MAY BE OF DIMENSION N WITH !
! N ANY NUMBER GREATER THAN 1. !
! !
!+++++++++++++++++++++++++++++++++++++++++++++++++++++++!
SUBROUTINE PBTRIP (A, B, C, D, IL, IU, ORDER)
INTEGER, INTENT(IN) :: IL, IU, ORDER
REAL, INTENT(INOUT) :: A(1), B(1)
REAL, INTENT(INOUT) :: C(1), D(1)
!...
!...A = SUB DIAGONAL MATRIX
!...B = DIAGONAL MATRIX
!...C = SUP DIAGONAL MATRIX
!...D = RIGHT HAND SIDE VECTOR
!...IL = LOWER VALUE OF INDEX FOR WHICH MATRICES ARE DEFINED
!...IU = UPPER VALUE OF INDEX FOR WHICH MATRICES ARE DEFINED
!... (SOLUTION IS SOUGHT FOR BTRI(A, B, C)*X = D
!... FOR INDICES OF X BETWEEEN IL AND IU (INCLUSIVE).
!... SOLUTION WRITTEN IN D VECTOR (ORIGINAL CONTENTS
!... ARE OVERWRITTEN)).
!...ORDER = ORDER OF A, B, C MATRICES AND LENGTH OF D VECTOR
!... AT EACH POINT DENOTED BY INDEX I
!... (ORDER CAN BE ANY INTEGER GREATER THAN 1)
!...
!...
!...THE MATRICES AND VECTORS ARE STORED IN SINGLE SUBSCRIPT FORM
!...
INTEGER :: ORDSQ
INTEGER :: IS, IE, IUMAT, IUVEC, IEMAT, IEVEC
INTEGER :: I0MAT, I0VEC, I1MAT, I1VEC
INTEGER :: I0MATJ, I0VECJ, I1MATJ, I1VECJ, IUMATJ, IEMATJ
INTEGER :: I, J, IBAC
!...
IS = IL + 1
IE = IU - 1
ORDSQ = ORDER**2
IUMAT = 1 + (IU - 1)*ORDSQ
IUVEC = 1 + (IU - 1)*ORDER
IEMAT = 1 + (IE - 1)*ORDSQ
IEVEC = 1 + (IE - 1)*ORDER
!...
!...FORWARD ELIMINATION
!...
I = IL
I0MAT = 1 + (I - 1)*ORDSQ
I0VEC = 1 + (I - 1)*ORDER
CALL LUDECO (B(I0MAT), ORDER)
CALL LUSOLV(B(I0MAT), D(I0VEC), D(I0VEC), ORDER)
DO J = 1, ORDER
I0MATJ = I0MAT + (J - 1)*ORDER
CALL LUSOLV (B(I0MAT), C(I0MATJ), C(I0MATJ), ORDER)
CALL LUSOLV (B(I0MAT), A(I0MATJ), A(I0MATJ), ORDER)
END DO
!...
DO I = IS, IE
I0MAT = 1 +(I - 1)*ORDSQ
I0VEC = 1 + (I - 1)*ORDER
I1MAT = I0MAT - ORDSQ
I1VEC = I0VEC - ORDER
DO J = 1, ORDSQ
I0MATJ = J - 1 + I0MAT
IUMATJ = J - 1 + IUMAT
CD (J) = C(IUMATJ)
A(I0MATJ) = 0.0
C(IUMATJ) = 0.0
END DO
CALL MULPUT (AD, D(I1VEC), D(I0VEC), ORDER)
DO J = 1, ORDER
I0MATJ = I0MAT + (J - 1)*ORDER
I1MATJ = I1MAT + (J - 1)*ORDER
CALL MULPUT (AD, C(I1MATJ), B(I0MATJ), ORDER)
CALL MULPUT (AD, A(I1MATJ), A(I0MATJ), ORDER)
END DO
CALL LUDECO (B(I0MAT), ORDER)
CALL LUSOLV (B(I0MAT), D(I0VEC), D(I0VEC), ORDER)
DO J = 1, ORDER
I0MATJ = I0MAT + (J - 1)*ORDER
CALL LUSOLV (B(I0MAT), C(I0MATJ), C(I0MATJ), ORDER)
CALL LUSOLV (B(I0MAT), A(I0MATJ), A(I0MATJ), ORDER)
END DO
CALL MULPUT (CD, D(I1VEC), D(IUVEC), ORDER)
DO J = 1, ORDER
IUMATJ = IUMAT + (J - 1)*ORDER
I1MATJ = I1MAT + (J - 1)*ORDER
CALL MULPUT (CD, A(I1MATJ), B(IUMATJ), ORDER)
CALL MULPUT (CD, C(I1MATJ), C(IUMATJ), ORDER)
END DO
END DO
!...
DO J = 1, ORDSQ
IUMATJ = J - 1 + IUMAT
END DO
CALL MULPUT (AD, D(IEVEC), D(IUVEC), ORDER)
DO J = 1, ORDER
IUMATJ = IUMAT + (J - 1)*ORDER
IEMATJ = IEMAT + (J - 1)*ORDER
CALL MULPUT (AD, C(IEMATJ), B(IUMATJ), ORDER)
CALL MULPUT (AD, A(IEMATJ), B(IUMATJ), ORDER)
END DO
CALL LUDECO (B(IUMAT), ORDER)
CALL LUSOLV (B(IUMAT), D(IUVEC), D(IUVEC), ORDER)
!...
!...BACK SUBSTITUTION
!...
DO IBAC = IL, IE
I = IE - IBAC + IL
I0MAT = 1 + (I - 1)*ORDSQ
I0VEC = 1 + (I - 1)*ORDER
I1VEC = I0VEC + ORDER
CALL MULPUT (A(I0MAT), D(IUVEC), D(I0VEC), ORDER)
CALL MULPUT (C(I0MAT), D(I1VEC), D(I0VEC), ORDER)
END DO
!...
END SUBROUTINE PBTRIP
!+++++++++++++++++++++++++++++++++++++++++++++++++++++!
! !
! SUBROUTINE TO CALCULATE L-U DECOMPOSITION !
! OF A GIVEN MATRIX A AND STORE RESULT IN A !
! (NO PIVOTING STRATEGY IS EMPLOYED) !
! !
!+++++++++++++++++++++++++++++++++++++++++++++++++++++!
SUBROUTINE LUDECO (A, ORDER)
INTEGER, INTENT(IN) :: ORDER
INTEGER :: JR, JC, JM, JRJC, JRJCM1, JRJCP1
REAL, INTENT(INOUT) :: A(ORDER, 1)
REAL :: SUM
!...
DO JC = 2, ORDER
A(1, JC) = A(1, JC)/A(1,1)
END DO
JRJC = 1
DO
JRJC = JRJC + 1
JRJCM1 = JRJC - 1
JRJCP1 = JRJC + 1
DO JR = JRJC, ORDER
SUM = A(JR, JRJC)
DO JM = 1, JRJCM1
SUM = SUM - A(JR, JM)*A(JM, JRJC)
END DO
A(JR, JRJC) = SUM
END DO
IF (JRJC == ORDER) EXIT
DO JC = JRJCP1, ORDER
SUM = A(JRJC, JC)
DO JM = 1, JRJCM1
SUM = SUM - A(JRJC, JM)*A(JM, JC)
END DO
A(JRJC, JC) = SUM/A(JRJC, JRJC)
END DO
END DO
!...
END SUBROUTINE LUDECO
!++++++++++++++++++++++++++++++++++++++++++++++++++++++!
! !
! SUBROUTINE TO SOLVE LINEAR ALGEBRAIC SYSTEM OF !
! EQUATIONS A*C=B AND STORE RESULTS IN VECTOR C. !
! MATRIX A IS INPUT IN L-U DECOMPOSITION FORM. !
! (NO PIVOTING STRATEGY HAS BEEN EMPLOYED TO !
! COMPUTE THE L-U DECOMPOSITION OF THE MATRIX A). !
! !
!++++++++++++++++++++++++++++++++++++++++++++++++++++++!
SUBROUTINE LUSOLV (A, B, C, ORDER)
INTEGER, INTENT(IN) :: ORDER
INTEGER :: JR, JM, JRM1, JRP1, JRJR, JMJM
REAL, INTENT(IN) :: A(ORDER, 1), B(1)
REAL, INTENT(INOUT) :: C(1)
REAL :: SUM
!...
!...FIRST L(INV)*B
!...
C(1) = C(1)/A(1,1)
DO JR = 2, ORDER
JRM1 = JR - 1
SUM = B(JR)
DO JM = 1, JRM1
SUM = SUM - A (JR, JM)*C(JM)
END DO
C(JR) = SUM/A(JR, JR)
END DO
!...
!...NEXT U(INV) OF L(INV)*B
!...
DO JRJR = 2, ORDER
JR = ORDER - JRJR + 1
JRP1 = JR + 1
SUM = C(JR)
DO JMJM = JRP1, ORDER
JM = ORDER - JMJM + JRP1
SUM = SUM - A(JR, JM)*C(JM)
END DO
C(JR) = SUM
END DO
!...
END SUBROUTINE LUSOLV
!++++++++++++++++++++++++++++++++++++++++++++++++++++++!
! !
! SUBROUTINE TO MULTIPLY A VECTOR B BY A MATRIX A !
! SUBTRACT RESULT FROM ANOTHER VECTOR C AND STORE !
! RESULT IN C. THUS VECTOR C IS OVERWRITTEN. !
! !
!++++++++++++++++++++++++++++++++++++++++++++++++++++++!
SUBROUTINE MULPUT (A, B, C, ORDER)
INTEGER, INTENT(IN) :: ORDER
INTEGER :: JR, JC, IA
REAL, INTENT(IN) :: A(1), B(1)
REAL, INTENT(INOUT) :: C(1)
REAL :: SUM
!...
DO JR = 1, ORDER
SUM = 0.0
DO JC = 1, ORDER
IA = JR + (JC - 1)*ORDER
SUM = SUM + A(IA)*B(JC)
END DO
C(JR) = C(JR) - SUM
END DO
!...
END SUBROUTINE MULPUT
However, someone advertised me that there is something wrong in LUDECO and LUSOLV subroutines that weren't fixed them. Have anyone used and tested this code? Is there any other block-TDMA in Fortran?
Something between a comment and an answer – a couple of points and links to internal Computational Science resources that should be helpful and relevant.
In this question, the 3D finite-difference is discussed and it is pointed out that 2D and 3D discretizations are not actually block-tridiagonal. More details there.
This question discusses how to solve block-tridiagonal algorithms using the Thomas algorithm with links and even some Fortran code. And here, the derivation of the block algorithm is shown for a more complicated case.
So, I would reassess the structure of the matrix you have, feasibility of it being actually solved by a block-tridiagonal algorithm, and potentially resort to very efficient sparse solvers.
Unfortunately, I don't know anything about the subroutines you've listed above.
• The matrix for discrete 2D Poisson equation by numbering in the correct sequence is block-tridiagonal, for instance.
– V.J.
Dec 27, 2019 at 13:49 | 4,146 | 11,796 | {"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": 25, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2023-50 | latest | en | 0.796989 |
https://byjus.com/question-answer/35-kg-of-a-type-of-sandal-powder-type-a-which-costs-rs-614-per/ | 1,713,698,378,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817765.59/warc/CC-MAIN-20240421101951-20240421131951-00680.warc.gz | 132,236,125 | 22,186 | 1
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Question
# 35 kg of a type of sandal powder (type A) which costs Rs 614 per kg was mixed with a certain amount of another type of sandal powder (type B), which costs Rs 695 per kg. Then the mixture was sold at Rs 767 per kg and 18% profit was gained. What was the amount of type B sandal powder in the mixture?
A
24 kg
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B
28 kg
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C
32 kg
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D
36 kg
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E
20 kg
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Solution
## The correct option is B 28 kg So, answer is 28 kg.
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Join BYJU'S Learning Program | 295 | 992 | {"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} | 3.046875 | 3 | CC-MAIN-2024-18 | latest | en | 0.907345 |
http://research.stlouisfed.org/fred2/graph/?chart_type=line&s[1][id]=EQTP365PDNQ&s[1][transformation]=pc1 | 1,418,947,558,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802768050.31/warc/CC-MAIN-20141217075248-00140-ip-10-231-17-201.ec2.internal.warc.gz | 230,546,306 | 17,757 | # Graph: Percent of Value of Loans Prime Based by Time that Pricing Terms Were Set and by Commitment, Before Survey Week, More than 365 Days, All Commercial Banks
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(a) Percent of Value of Loans Prime Based by Time that Pricing Terms Were Set and by Commitment, Before Survey Week, More than 365 Days, All Commercial Banks, Percent, Not Seasonally Adjusted (EQTP365PDNQ)
For further information, please refer to the Board of Governors of the Federal Reserve System's E.2 release, online at http://www.federalreserve.gov/releases/e2/about.htm or in the footnotes of the E.2, Survey of Terms of Business Lending Release. These data are collected during the middle month of each quarter and are released in the middle of the succeeding month.
Percent of Value of Loans Prime Based by Time that Pricing Terms Were Set and by Commitment, Before Survey Week, More than 365 Days, All Commercial Banks
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``` Board of Governors of the Federal Reserve System (US), Percent of Value of Loans Prime Based by Time that Pricing Terms Were Set and by Commitment, Before Survey Week, More than 365 Days, All Commercial Banks [EQTP365PDNQ], retrieved from FRED, Federal Reserve Bank of St. Louis https://research.stlouisfed.org/fred2/series/EQTP365PDNQ/, December 18, 2014. ```
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Discussion in 'General Chatter' started by Exohedron, Mar 14, 2015.
1. ### ExohedronDoesn't like words
As you might be able to tell, I am slowly losing the battle against my desire to talk about algebraic topology and homological algebra. Which is bad, because algebraic topology is not one of those subjects that I'm good at.
2. ### ExohedronDoesn't like words
Okay, so orbitals.
If you've taken a chemistry class or cracked open a chemistry textbook, you might have seen those pictures of electron orbitals, with the p and the s and the d and the f and maybe some other ones. And they look like little lobes or spheres or whatnot, usually brightly colored, and they're supposed to be "instead" of the Bohr model where the electrons orbited the nucleus in nice circular orbits.
So what does it mean when we say that electron orbits look like those pictures?
Firstly, what can we say about electrons?
Suppose we have a single proton, a single point of positive electric charge. We get an electric field that extends outward radially, completely symmetrically if the proton isn't moving and there isn't anything else relevant around it.
So if we have a physically-feasible description of an electron orbiting that proton, we should be able to rotate that description around the proton and get another physically-feasible description.
From a quantum mechanical standpoint, a physical description consists of a function from points in 3-space to the complex numbers called the wavefunction ψ. So if we want to draw a picture of the orbit, we need to figure out what kind of wavefunctions we're looking at.
We want to talk about the angular momentum of the electron. So what is angular momentum in quantum mechanics? Well, the angular momentum operators are
Lx = -i(y(∂/∂z) - z(∂/∂y))
Ly = -i(z(∂/∂x) - x(∂/∂z))
Lz = -i(x(∂/∂y) - y(∂/∂x))
and we actually want to look at
L2 = Lx2+Ly2+Lz2
since we want a real number, not a vector, to get the squared orbital angular momentum.
So we get a big thing with some derivatives and second derivatives that will take a function and spit out a function, and the functions that correspond to electrons with well-defined orbital angular momentum, i.e. the eigenfunctions, end up being the harmonic homogeneous polynomials, with squared total angular momentum being d(d+1) where d is the degree of the polynomial.
How many homogeneous polynomials of a given degree are there? Well, lots. But how many independent harmonic homogeneous polynomials of degree d are there?
Well, let's see. The number of independent polynomials of degree d in three variables is
(d+2)(d+1)/2
(check this!)
But you'll only want the harmonic polynomials. Recalling a previous post, every polynomial is a sum of terms of the form Ck times a harmonic piece, where C has degree 2, and so stuff from degree d-2 contributes non-harmonic parts to polynomials of degree d. So we remove them:
(d+2)(d+1)/2 - (d-2+2)(d-2+1)/2 = 2d + 1
So in degree d, there are 2d+1 independent homogeneous harmonic polynomials. So for instance, in degree 2 we have
xy, xz, yz, x2-y2, 2z2 - x2 - y2
as a set of independent homogeneous harmonic polynomials.
Unfortunately, a polynomial function P can't be a wavefunction, not directly. Instead, we change coordinates so that P is written in terms of the radius r and angular coordinates φ and θ. P then decomposes into R(r)Y(φ, θ), and we can then replace R(r) with another function of r to get a genuine wavefunction; the exact replacement depends on what kind of physics we're trying to model and isn't really important. All the fun stuff is contained in Y, and in particular indicate very non-circular "orbits".
The functions Y are called spherical harmonics, because we can compute them by taking a harmonic function and pretending that it only matters on the unit sphere.
The point is that you can translate between homogeneous harmonic polynomials and wavefunctions corresponding to well-defined squared total angular momentum. So the same statement that there are 2d+1 independent homogeneous harmonic polynomials of degree d gives that there are 2d+1 independent wavefunctions with squared orbital angular momentum d(d+1).
We can further ask for wavefunctions with well-defined angular momentum in the z direction, i.e. the eigenfunctions of Lz, and we'll get particular functions; often those particular wavefunctions are the ones that get pictured in the textbooks, even though actual electrons don't care which direction is the z direction unless you ask them nicely.
3. ### ExohedronDoesn't like words
7 may be nonstandard, since it takes at least 7 symbols to define 7 in Peano arithmetic.
4. ### ExohedronDoesn't like words
The Permut0hedra and the Associahedra
Suppose you have an object A. There is one way to order the list {A}, i.e. as A. Draw a single point.
Suppose you have two objects, A and B. There are two ways to order the list {A, B}, i.e. as AB and BA. Draw a point for each ordering, and connect them. You end up with a line segment.
Suppose you have three objects, A, B and C. There are six ways to order the list {A, B, C}, i.e. as ABC, ACB, etc. Draw a point for each ordering, and connect a pair of points if the corresponding orderings differ by swapping two adjacent elements of the ordering, i.e. ABC connects to ACB and to BAC, but not to anything else,. You end up with a hexagon.
Suppose you have four objects, A, B, C and D. There are twenty-four ways to order the list {A, B, C, D}. Draw a point for each ordering, and connect a pair of points if they differ by swapping two adjacent elements of the ordering, and fill in a. You end up with a shape that's usually called a truncated octahedron. Note that some of the faces of this shape are square and some are hexagonal. The squares represent swaps that are disjoint, i.e. a swap of the first two objects and a swap of the last two objects. We can go from ABCD to BADC either by stopping first at BACD or ABDC, without having to go through any other steps.
Etc.
Suppose you have n objects. There are n! ways to order the list of objects. Draw a point for each ordering, and connect the pairs whose orderings differ by a swap of two adjacent elements. If we fill in the surfaces and volumes and so on, we get a shape called the permutohedron on n objects. It tracks how one gets from one ordering of n objects to another ordering.
If a k-dimensional face can be written as the Cartesian product of an m-dimensional face and a (k-m)-dimensional face, then it indicates that the relevant set of swaps for the edges of the k-dimensional face can be decomposed as swaps on a set of m+1 adjacent objects and a set of k-m+1 adjacent objects that's disjoint from the first set. So the shape of the faces tells us about disjoint suborderings.
You can build a permutohedron on n objects in n-dimensional space by looking at all of the points whose coordinates are orderings of the first n positive integers and then connecting the appropriate edges.
Honestly, though, I find this guy a little boring.
Consider an ordered list of objects that belong to a set with a multiplication operation. We don't ask for the multiplication to be associative, so we need parentheses.
If we have one object A, no big deal, our parenthesized list is just A. Draw a single point.
Similarly, if we have two objects, our parenthesized list is just AB. Draw a single point.
If we have three objects, we have two ways to put in parentheses: (AB)C and A(BC). We draw a point for each, and connect them by a line.
If we have four objects, we have five ways to put in parentheses: ((AB)C)D, (A(BC))D, (AB)(CD), A((BC)D), and A(B(CD)). Draw a point for each, and connect two points if you can get from one parenthesization to the other by a move that looks like (XY)Z -> X(YZ), i.e. the associativity rule, where X, Y and Z can themselves be parenthesizations of lists.
So we would connect ((AB)C)D to (A(BC))D and to (AB)(CD), but not to anything else.
We end up with a pentagon; in particular, there are two ways to get from ((AB)C)D to A(B(CD)):
((AB)C)D -> (AB)(CD) -> A(B(CD))
((AB)C)D -> (A(BC))D -> A((BC)D) -> A(B(CD))
If we have five objects, we have fourteen ways to put in parentheses. I'll let you write them out. We end up with a shape with six five-sided faces and three four-sided ones; there isn't a good way to embed this guy into 3-dimensional Euclidean space that has all the faces regular.
If we have n objects in our list, we end up with what's called the associahedron on n objects. This is also called a Stasheff polytope, after the guy who (re)discovered them.
I like the associahedra because they're a little weirder than the permutohedra. For instance, the number of vertices in the associahedron on n objects is given by what is called the (n-1)-st Catalan number, given by
Cn-1 = (2n-2 choose n-1)/n
The combinatorics of other parts of the associahedra is similarly more complicated than the corresponding parts of the permutohedra.
As noted above, you can't usually embed the associahedra in Euclidean space with the faces being regular-ish.
I mostly approach these guys from a (higher)-algebraic standpoint, where we say "this product is commutative up to equivalence" or "this product is associative up to equivalence" and then the associated -hedra tell us how loose that notion of "up to equivalence" is.
Suppose that AB = BA, i.e. we can shrink our 2-object permutohedron to a single point. Then for all higher permutohedra, we shrink their edge to points and we get a single point, i.e. the ordering doesn't matter, our product is commutative.
Now suppose that while AB is not equal to BA, there is an map fA,B: AB -> BA, and then the permutohedron on three objects measures the difference between
fB,CIA◦IBfA,C◦fABIC
and
ICfA, B◦fA, CIB◦IAfB,C
where IX means the identity on X; these are the two ways of going around the hexagon from ABC to CBA. Does your algebraic structure care about the difference between those two paths?
So for instance, in our usual notion of a noncommutative multiplication, the permutohedron on 3 objects always yields the identity map, i.e. that the difference between ABC and CBA doesn't depend on how we get from ABC to CBA. This corresponds to flattening the hexagon to a single line going from ABC to CBA. Taking the permutohedron on 4 objects and flattening the A,B,C hexagons to single lines going from ABCD to CBAD and from DABC to DCBA, and flattening the A,B,D hexagons to single lines going from ABDC to DBAC and from CABD to CDBA, and so on. You get a funny-looking shape that's basically just a single line from ABCD to DCBA, with some lobes attached that you can ignore. So you get that there's only one way to get from ABCD to DCBA, i.e. no path dependence for four object products. And similarly no path dependence on for more objects.
But what if there was a path dependence for three objects, but no path dependence for four objects? Then again we get no path dependence for five objects and so on.
These are called "coherence" theorems, that if your -hedra on n objects all collapse to single lines, then so do all of your -hedra on >n objects. These kind of path-dependent algebraic properties shows up in topology and higher category theory and string theory and basically a lot of places that are trying to do algebra in a "higher-dimensional" setting. For me, personally, I focus on the analogues of Lie algebras, where the Jacobi identity doesn't hold exactly, and the discrepancy is controlled by various extra structures that can be integrated to get analogues of Lie groups; these are the actual automorphism objects of string theories, not Lie groups.
Since it might come up: no, as far as I know, "exohedron" doesn't actually refer to anything other than me.
Last edited: Jun 30, 2019
5. ### ExohedronDoesn't like words
A formal power series is like a polynomial with infinite degree. In calculus class you often care about where and when a power series converges, but for formal power series you don't worry about convergence.
Consider formal power series of the following type:
C(x) = x + c1x2 + c2x3 + ...
You can consider what happens when you plug one such formal power series into another. If C(x) and D(x) both have the above form, then their composition C(D(x)) also has the above form.
You can also ask about the inverse with respect to composition: given C(x), find D(x) such that
C(D(x)) = x
How do you relate the coefficients of C and D?
d1 = -c1
d2 = -c2 + 2c12
d3 = -c3 + 5c2c1 - 5c13
d4 = -c4 + 6c3c1 + 3c22 - 21c2c12 + 14c14
Where do these numbers come from? Well, for a given coefficient dk, the terms on the right side all have total index k, where we consider c1m to give total index m. Also the sign of a term is given by how many factors there are in the term. But what about the number itself?
Consider A1 to be a point, A2 to be a line, A3 to be a pentagon, and in general, Ak to be the associahedron on k+1 objects.
Then A1 is a point and has one point, i.e. one A1.
A2 is a line segment, and thus has one line, i.e. A2, and two points; we view the points as A12 to get the correct total index.
A3 is a pentagon, and thus has one pentagon A3, five lines which we view as A2A1, and five points which we view as A13, again multiplying by points to get the correct total index.
A4 has one A3, six pentagonal faces A3A1, three square faces A22, twenty one lines A2A12, and fourteen points A14.
So you can see a match between the coefficients in D and the pieces of the associahedra.
dk = Σk1 + ... + kl = k (#of pieces of Ak of the form Ak1Ak2...Akl)(-ck1)...(-ckl)
Last edited: Jun 29, 2019
6. ### ExohedronDoesn't like words
Not exactly math but I figure that a bunch of people here would appreciate twocubes trying to make manga in LaTeX.
Also twocubes has some interesting discussions about how to cut an infinite deck of cards, but that's somewhat less hilarious.
8. ### ExohedronDoesn't like words
Oh, sure, digressions are super allowed! I mean, the only reason it looks like I'm actually running this thread is that I get to change the name and I make the majority of the posts, but that's because I like...whatever is the text equivalent of the sound of my own voice.
9. ### ExohedronDoesn't like words
Today a bunch of coworkers and I were joking about using your fingers to count in binary and why we didn't teach that to kids, and I mentioned that kids naturally count in base 1 rather than base 2. One of the group expressed shock at the notion of base 1.
I'm going to have fun presenting next week.
10. ### evilasSure, I'll put a custom title here
How does... Can you represent 0 in base 1?
Can you represent decimals?
I mean, yes I know you're talking about kids but I imagine it's a properly defined base, right? Basically tallies?
Like. I can see negative numbers and fractions but. Is it ever used for stuff that isn't natural numbers?
Last edited: Sep 8, 2019
11. ### ExohedronDoesn't like words
No, it''s not really a well-developed base all. It's just tallies.
The issues are probably related to the fact that the field with 1 element doesn't exist.
Last edited: Sep 8, 2019
• Informative x 1
12. ### TheSeer37 Bright Visionary Crushes The Doubtful
Yeah, having a place value system at all is a fairly sophisticated idea. A young beginner's number system is just a one-to-one correspondence between the group to be counted and an ordered list of words. "Ten" has no special significance, and is just the number after "nine."
I suppose if you think of the names of place values in base ten as the same sort of ordered list, you could roughly describe the basic beginner's number system as "base 1". But the only real utility of the term is to troll other nerds.
13. ### ExohedronDoesn't like words
Speaking of fun stuff with fields, I was reading about the q-analog of the Moebius inversion formula when I was struck with the thought of "why don't they just call it inqlusion-exqlusion?"
14. ### ExohedronDoesn't like words
Spent some time trying to prove from Euclid's axioms that a line that intersects a circle at one point without being tangent there must intersect the circle again at a different point. Using algebra is cheating.
I've managed to prove that a line segment that intersects a triangle transversely at some point that isn't a vertex must intersect the triangle again. Circles are harder.
Last edited: Sep 11, 2019
15. ### ExohedronDoesn't like words
The Twin Primes Conjecture
Prime numbers are the building blocks for multiplication, the fundamental objects for which there is no meaningful decomposition into smaller objects, and from which all other numbers can be made*.
But the relationship between multiplication and addition is subtle and complicated, and in particular the way that the prime numbers interact with addition has spawned many a research career.
One statement that relates prime numbers and addition is the twin prime conjecture. Some pairs of prime numbers, like 3 and 5 or 5 and 7, differ by 2, while others, like 7 and 11, have larger gaps between them. We call a pair of primes that differ by 2 a pair of "twin primes". So 3 and 5 make a pair of twin primes, and 5 and 7 make a pair of twin primes.
The twin prime conjecture is that there are an infinite number of pairs of twin primes, i.e. that for any pair of twin primes, you can always find a larger pair if you look long enough. So after 5 and 7 we have 11 and 13, and then 17 and 19, and then 29 and 31, and...
But while we can enumerate a lot of pairs, we still haven't actually proven that there are an infinite number of pairs of twin primes!
There has been some recent progress, though!
In 2013, Yitang Zhang made the first piece of significant progress in a while, by showing that there are an infinite number of pairs that differ by no more than 70 million. So that means that there is some finite number k such that there are an infinite number of primes that differ by k, where k is at most 70 million.
This was done by encoding the primes as a function and examining the behavior of the function.
Since then, people have pushed down the value of k further and further; I think it's currently sitting at 246, which means that for some k no larger than 26, there are an infinite number of pairs of primes that differ by k.
More recently, however, we have an algebraic result!
There is a nice analogy between integers and polynomials over finite fields. You can add polynomials and multiply polynomials, and you can define a notion of a prime polynomial as a polynomial that cannot be factored into a pair of nonconstant polynomials. Taking things a little further gives us a nice translation between statements about integers and statements about polynomials over finite fields, first outlined by Andre Weil and put to good use ever since. In particular, finite fields tend to be simple enough and well-behaved enough that questions about polynomials over finite fields tend to be easier to answer than the analogous questions about the integers. For instance, there is an analogy of the Riemann Hypothesis for finite fields that was proven back in the 40s**.
So we can ask about gaps between prime polynomials: given a fixed polynomial f(x), how many pairs of prime polynomials g(x) and h(x) are there such that g(x) - h(x) = f(x)?
A few weeks ago, two mathematicians Will Sawin and Mark Shusterman proved a formula that tells us, for a given prime p, a given positive integer d and a given polynomial f(x), how many such pairs of prime degree-d polynomials g(x) and h(x) over the finite field Z/pZ differ by f(x). If this formula spits out a positive number for an infinite number of degrees d, then we get that there are an infinite number of pairs of prime polynomials that differ by f(x)!
The analogue for integers would be a formula where given integers a, b and k, we can say how many pairs of primes between a and b differ by exactly k. This would be even better than the twin prime conjecture, which just says that there's at least one pairs no matter how far up you go; this would tell us exactly how many pairs.
But the analogy only goes so far; after all, the Riemann Hypothesis for integers is still open***. Only time will tell if this analogue of the twin prime conjecture will yield an analogous proof.
* Allowing for multiplication by units and assuming that you allow the trivial product of 1.
** Technically for zeta functions of curves defined over finite fields.
*** If there were a field with only one element, then we'd be able to turn the proof of the Riemann Hypothesis for (zeta functions of curves over) finite fields into a proof for the Riemann Hypothesis over the integers.
16. ### ExohedronDoesn't like words
I know that technically the term is "partially-defined composition" but I always think "like multiplication but with failure".
17. ### ExohedronDoesn't like words
Reminder to self to talk about how numbers and function fields are similar.
18. ### ExohedronDoesn't like words
If only there were a field with only one element
Consider the real numbers. Suppose you had a function f defined on the real number line, and at some point p it vanishes, i.e. f(p) = 0. Then for any other function g, we have that the product of f and g gives a function fg that vanishes at p, i.e.
(fg)(p) = f(p)g(p) = 0*g(p) = 0
Also, if g also vanishes at p, then the function f+g vanishes, i.e.
(f+g)(p) = f(p) + g(p) = 0 + 0 = 0
We say that the set of functions that vanish at p is an ideal with respect to the ring of functions that are defined on the real line: the product of something in the ideal with anything else is also in the ideal, and the sum of two things in the ideal is in the ideal. We'll call this ideal Ip.
Now if we had two points p and q, we could ask about the functions that vanish at both p and q. This also forms an ideal I{p,q}, for the same reasons that the functions that vanish at p forms an ideal.
And similarly, for any collection C of points, we could ask about the set of functions that vanish at those points, and again get an ideal IC.
But the ideal Ip is more special. Suppose we had two functions, g and h, and we know that gh vanishes at p. In other words,
g(p)h(p) = 0
This tells us that either g(p) = 0 or h(p) = 0, i.e. either g or h is in the ideal Ip. Hence if the product of two functions is in the ideal Ip, then at least one of the two functions must be in Ip.
This isn't true for the ideal I{p,q}, since we could have that g(p) = 0 but g(q) = 1 and h(q) = 0 but h(p) = 1; now neither g nor h vanish at both p and q, and so neither of them is in I{p,q}, but their product vanishes at both points and so is in I{p,q}.
Now let's talk about numbers, and in particular integers. Consider the set of integers that are multiples of 6. If you multiply a multiple of 6 by an integer, you get another multiple of 6, and if you add two multiples of 6, you get a multiple of 6. So the multiples of 6 form an ideal with respect to the integers. We could call it I6.
But consider 10 and 3. Neither is a multiple of 6, but their product 30 is a multiple of 6. So I6 isn't as special as Ip; it's more like I{p,q}.
But if we looked at 5, then we can look at the multiples of 5, I5, and if we have two numbers a and b such that ab is a multiple of 5, then either a or b has to be a multiple of 5, because 5 is prime.
So we might call the ideal I5 a prime ideal. And for the same reason, we would say that Ip is also a prime ideal.
For technical reasons, I'm going to say that the ideal containing just the 0 function and the ideal containing all functions are not prime.
So now we can associate a set of points to each integer: a prime corresponds to a single point, and each other integer corresponds to the set of its prime factors. We have to be a little bit careful about factors that appear multiple times; does the set for 9 just consist of the point for 3, or somehow include that point twice?
But we nevertheless get some sort of 1-dimensional thing. Not really a line, per se, but a 1-dimensional object of some sort, for some appropriate notion of dimension*.
Usually, we do this analogy not with the real line, but with 1-dimensional objects (i.e. curves) over finite fields. I'm not going to explain exactly what that means, but the relevant bit is that when we do this localizing, we get fields that are all extensions of the finite field, the way that the complex numbers are an extension of the real numbers.
But wait, you might ask, if we're really being serious here, for the integers we get prime fields, and you can't extend a prime field to get a different prime field.
Well, true.
But! Suppose that F1 were a field! Then Fp would be an extension of F1, and so would Fq, sort-of. Not literally, but close enough for geometry.
In particular, we would get that Z is the ring of functions of a curve over F1.
Why do we care? Well, because there's a bunch of stuff that we know about rings of functions of curves over finite fields! For instance, the (analogue of the) twin-prime conjecture is true for rings of functions of curves over finite fields, and so is the (analogue of the) Riemann hypothesis!
So if we were just allowed to say that Z is the ring of functions of a curve over a finite field, we'd be done with both of those problems instantly.
But we can't, because F1 isn't a field.
* The appropriate notion is Krull dimension: how long a chain of prime-ideal-contained-in-different-prime-ideal can you make.
Last edited: Oct 16, 2019
19. ### ExohedronDoesn't like words
Not a proof of the Riemann Hypothesis
Suppose you have a polynomial P(x,y,z) which is homogeneous, i.e. there is some number n such that P(x,y,z) is a sum of terms of the form pa,b,cxaybzc where p is a constant and a+b+c = n.
The equation P(x,y,z) = 0 gives us a bunch of triples, and the reason we asked for homogeneity is that if P(X,Y,Z) = 0 for a triple of numbers (X, Y, Z), then for any constant r, P(rX, rY, rZ) = 0 as well.
So we projectivize, and say that (X, Y, Z) and (rX, rY, rZ) are the same projective point for any nonzero r.
Since we have three variables and one equation and we're also saying that scaling by a constant doesn't change what point we have, we end up with a 1-dimensional object, i.e. a curve, which we'll call C.
So now we can ask about how many points this curve has.
Wait a minute, you say, isn't it an infinite number? Since we can probably pick x and y randomly and then solve for z?
Well, I haven't told you where x, y, and z live!
So suppose that the coefficients of the polynomial live in some field K. Then we can look at C(K), by which we mean all the triples X, Y, Z in K that satisfy P(X,Y,Z) = 0. If K is infinite, then C(K) probably has an infinite number of points, but if K is finite, then C(K) has only a finite number of points. How many?
Suppose that the coefficients live in Fq, i.e. a finite field with q elements, where q is a power of a prime number. Then for any n, we can look at the field Fqn, which contains Fq, so we can consider C(Fqn) for all n.
We define the zeta function for C to be
ζ(C, s) = Σn=1 #(C(Fqn))q-ns/n
which is an infinite series. It might not converge, but we don't really care; we're just trying to put all of the values together into a single object.
Andre Weil showed that there is polynomial p such that
ζ(C, s) = p(q-s)/((1-q-s)(1-q1-s))
where
p(T) = Πi=1b (1 - aiT)
for some integer b and some complex numbers a1, ..., ab where |ai| = q1/2.
This tells us, for instance, that the number of points of C that live in Fq is approximately q + 1, plus or minus a term that's proportional to q1/2.
We also get that p(q-s) is 0 implies that q-s = 1/ai for some i, so ζ(C, s) = 0 implies that s has real part 1/2.
That might sound a little familiar. The Riemann zeta function ζ(s) has a bunch of zeros with real part 1/2, and there is a million dollars waiting for anyone who can prove that, other than s = -2n for integer n, all the solutions to ζ(s) = 0 have real part 1/2.
And indeed, you can consider the integers to be a curve defined over the field F1, and then its zeta function would indeed be the Riemann zeta function, and so would have zeros only at real part 1/2 (other than the pesky negative even integer points).
Exercise: why doesn't the previous statement constitute a proof of the Riemann Hypothesis?
Last edited: Nov 13, 2019
20. ### ExohedronDoesn't like words
Zero-Knowledge Proof
You have discovered a thing. A brilliant, important thing, and you want everyone to know! Except you don't want to reveal the thing, because someone will steal it.
So how do you prove that you know a thing without telling anyone what it is?
The classical parable here is Ali-Baba's cave: a cave entrance leads to a branching path that eventually closes up into a loop. But blocking way around the loop is a door that only opens if you speak the password.
So you, being you, get chased into the cave. You pick a branch and head down it before your pursuers enter the cave. Your pursuer, being the brilliant Disney-esque henchman that they are, loudly declare which branch they're going to head down and then head down that path.
If they picked the branch that wasn't the one you went down, you can just backtrack and head out of the cave.
But if they picked the branch that you went into then you can't escape into the other branch unless you know the password!
Now suppose that you've been thieving on the regular and you keep getting chased into this cave. Eventually the goon, or more likely their boss, is going to figure out that since you keep escaping you must either be able to guess which direction the goon is going to head, or you know the password. Since the goon picks randomly (the one thing the goon is good at), it is much more likely that you know the password. But neither the boss nor the goon ever learns anything about the password itself, only that you know it, and that they should probably stop trying to chase you into this stupid cave.
As this is not the fleeing from possibly-lawful pursuit thread but is instead the math thread, let's talk some math.
We give a zero-knowledge proof scenario as follows: a prover, say Peggy, wants to convince a verifier, Victor, that she knows the answer to a question without letting Victor know what that answer is. So instead Victor asks her to answer a related question, using some randomness to prevent her from simply figuring out the answer to the related question in advance.
For example, suppose we have a group G and two group elements g and h, and Peggy claims that she knows a value x such that gx = h. If she tells Victor the value of x then he can confirm that gx = h, but then he knows x. So instead we have Schnorr's protocol:
Peggy picks a random number r, computes C = gr, and sends C to Victor. C is called the commitment.
Victor picks a random number c and sends that to Peggy. c is called the challenge.
Peggy computes a response y and sends that to Victor. This is called the response.
Victor computes gy and compares that to Chc.
If Peggy knows x, then she can compute y = r+cx and Victor would confirm that
gy = gr+cx = Chc
However (depending on G) figuring out what y to send without knowing x is difficult.
We can show that Victor doesn't learn anything about x: suppose that Victor had told Peggy the challenge before she computed the commitment. Then she could pick a random response s and computed the challenge as C = gsh-c. Then Victor would check
gs = gsh-chc = Chc
and would accept. Since Peggy didn't need to know x to do this, we see that Victor ultimately can't have learned anything about x. So this proof is zero-knowledge, in that Victor learns that Peggy knows x but doesn't learn what x is.
Of course, if Peggy had managed to guess the challenge before hand, she could do the same thing, picking a random response and crafting the commitment to fit. So there is always a chance that Peggy got lucky. But if they repeat this process a bunch of times, Peggy making a random commitment, Victor sending a challenge, and then Peggy sending a response, then Victor will eventually decide that the probability that Peggy guessed the challenge in advance is low enough that she probably just knows x instead.
Zero-knowledge proofs are used to allow for authentication of secret things, like knowledge of a password, or identities or monetary values in anonymous settings like cryptocurrencies. A (cryptographic) digital signature is akin to a zero-knowledge proof, in that you're supposed to be able to create the signature only if you know a special key, but the signature itself shouldn't give away what the key is.
There are ways to make it so that Victor doesn't have to send a challenge between the commitment and the response, requiring some computational hardness assumptions. | 8,148 | 32,878 | {"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-2021-04 | latest | en | 0.930477 |
https://practicepaper.in/gate-me/refrigeration-and-air-conditioning | 1,638,364,464,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964360803.0/warc/CC-MAIN-20211201113241-20211201143241-00468.warc.gz | 513,872,992 | 23,730 | # Refrigeration and Air-conditioning
Question 1
Ambient pressure, temperature, and relative humidity at a location are 101 kPa, 300 K, and 60%, respectively. The saturation pressure of water at 300 K is 3.6 kPa. The specific humidity of ambient air is _______g/kg of dry air.
A 21.4 B 35.1 C 21.9 D 13.6
GATE ME 2021 SET-2 Properties of Moist Air
Question 1 Explanation:
\begin{aligned} \phi &=0.6 \\ \frac{P_{v}}{P_{v s}} &=0.6 \\ P_{v s} &=3.6 \mathrm{kPa} \\ P_{v} &=0.6 \times 3.6 \mathrm{kPa} \\ P_{v} &=2.16 \mathrm{kPa} \\ \omega &=0.622\left(\frac{P_{V}}{P-P_{V}}\right) \\ &=0.622\left(\frac{2.16}{101-2.16}\right)\\ &=0.01358 \mathrm{~kg} \text { of water vapour/kg of dry air }\\ &=13.58 \mathrm{~g} / \mathrm{kg} \text { of dry air } \end{aligned}
Question 2
A rigid tank of volume $50\; m^3$ contains a pure substance as a saturated liquid vapour mixture at 400 kPa. Of the total mass of the mixture, 20% mass is liquid and 80% mass is vapour. Properties at 400 kPa are : $T_{sat}=143.61^{\circ}C,\; v_f=0.001084\; m^3/kg, v_g=0.46242\; m^3/kg$. The total mass of liquid vapour mixture in the tank is _______kg (round off to the nearest integer).
A 135 B 115 C 98 D 175
GATE ME 2021 SET-2 Psychrometric Process
Question 2 Explanation:
\begin{aligned} V_{\text {total }} &=v_{l}+v_{v} \\ &=v_{f} \times m_{l}+v_{g} \times m_{v} \\ 50 &=0.001084 \times 0.2 \mathrm{~m}+0.46242 \times 0.8 \mathrm{~m} \\ m &=135.08 \mathrm{~kg}=135 \mathrm{~kg} \end{aligned}
Question 3
Consider an ideal vapour compression refrigeration cycle working on R-134a refrigerant. The COP of the cycle is 10 and the refrigeration capacity is 150 kJ/kg. The heat rejected by the refrigerant in the condenser is _______kJ/kg (round off to the nearest integer).
A 125 B 185 C 215 D 165
GATE ME 2021 SET-2 Vapor and Gas Refrigeration
Question 3 Explanation:
\begin{aligned} \mathrm{RE} &=150 \mathrm{~kJ} / \mathrm{kg} \\ \mathrm{COP} &=10 \\ \mathrm{COP} &=\frac{Q_{L}}{W_{i n}}\\ \Rightarrow\qquad W_{\text {in }} &=15 \mathrm{~kJ} / \mathrm{kg} \\ Q_{\mathrm{H}}=Q_{\mathrm{C}} &=Q_{L}+W_{\text {in }} \\ &=150+15 \\ &=165 \mathrm{~kJ} / \mathrm{kg} \end{aligned}
Question 4
An air-conditioning system provides a continuous flow of air to a room using an intake duct and an exit duct, as shown in the figure. To maintain the quality of the indoor air, the intake duct supplies a mixture of fresh air with a cold air stream. The two streams are mixed in an insulated mixing chamber located upstream of the intake duct. Cold air enters the mixing chamber at $5^{\circ} C$, 105 kPa with a volume flow rate of 1.25 $m^3/s$ during steady state operation. Fresh air enters the mixing chamber at $34^{\circ}C$ and 105 kPa. The mass flow rate of the fresh air is 1.6 times of the cold air stream. Air leaves the room through the exit duct at $24^{\circ}C$.
Assuming the air behaves as an ideal gas with $c_p= 1.005 \;\; kJ/kg.K$ and $R= 0.287 kJ/kg.K$, the rate of heat gain by the air from the room is ________kW(round off to two decimal places).
A 4.96 B 2.25 C 6.25 D 8.12
GATE ME 2021 SET-1 Psychrometric Process
Question 4 Explanation:
1: Cold air
2: Hot air
\begin{aligned} P_{1} V_{1} &=\dot{m}_{1} R T_{1} \\ 105 \times 1.25 &=\dot{m}_{1} \times 0.287 \times 278 \\ \dot{m}_{1} &=1.645 \mathrm{~kg} / \mathrm{sec} \\ \dot{m}_{2} &=1.6 \times 1.645=2.632 \mathrm{~kg} / \mathrm{sec} \\ \dot{m}_{3} &=4.277 \mathrm{~kg} / \mathrm{sec} \end{aligned}
After mixing: 0
\begin{aligned} \dot{m}_{1} t_{1}+\dot{m}_{2} t_{2} &=\dot{m}_{3} t_{3} \\ 1.645 \times 5+2.632 \times 34 &=4.277 t_{3} \\ t_{3} &=22.85^{\circ} \mathrm{C}\\ \text{Heat gain }:&=h_{4}-h_{3}\\ &=\dot{m}_{3} c_{p}\left(t_{4}-t_{3}\right) \\ &=4.277 \times 1.005(24-22.85) \\ &=4.963 \mathrm{~kW} \end{aligned}
Question 5
The relative humidity of ambient air at 300 K is 50% with a partial pressure of water vapour equal to $p_v$. The saturation pressure of water at 300 K is $p_{sat}$. The correct relation for the air-water mixture is
A $p_v=0.5p_{sat}$ B $p_v=p_{sat}$ C $p_v=0.622p_{sat}$ D $p_v=2p_{sat}$
GATE ME 2021 SET-1 Properties of Moist Air
Question 5 Explanation:
\begin{aligned} \text { Relative humidity, } \phi &=\frac{p_{v}}{p_{\text {sat }}} \\ 0.5 &=\frac{p_{v}}{p_{\text {sat }}} \\ p_{v} &=0.5 p_{\text {sat }} \end{aligned}
Question 6
Superheated steam at 1500 kPa, has a specific volume of 2.75 $m^3$/kmol and compressibility factor ($Z$) of 0.95. The temperature of steam is $^{\circ}C$ (round off to the nearest integer).
A 522 B 471 C 249 D 198
GATE ME 2021 SET-1 Heat Pumps and Cycles
Question 6 Explanation:
\begin{aligned} P &=1500 \mathrm{kPa} \\ V &=2.75 \mathrm{~m} 3 / \mathrm{k}-\mathrm{mol} \\ Z &=0.95 \\ P V &=n \bar{R} T \\ P \bar{V} &=\bar{R} T \\ P \bar{V} &=Z \times n \bar{R} T \\ P \frac{\bar{V}}{n} &=Z \bar{R} T \\ P_{\bar{V}} &=Z \bar{R} T\\ 1500 \mathrm{KPa} \times 2.75 \mathrm{~m}^{3} / \mathrm{K}-\mathrm{mol}&=0.95 \times 8.314 \mathrm{~kJ} / \mathrm{K}-\mathrm{molK} \times T\\ T&=522.26 \mathrm{~K}\\ T&=522.26-273=249.26^{\circ} \mathrm{C} \end{aligned}
Question 7
Consider an ideal vapour compression refrigeration cycle. If the throttling process is replaced by an isentropic expansion process, keeping all the other processes unchanged, which one of the following statements is true for the modified cycle?
A Coefficient of performance is higher than that of the original cycle. B Coefficient of performance is lower than that of the original cycle. C Coefficient of performance is the same as that of the original cycle. D Refrigerating effect is lower than that of the original cycle.
GATE ME 2019 SET-1 Vapor and Gas Refrigeration
Question 7 Explanation:
Due to isentropic expansion instead of throttling
1. R.E increases
2. Work input reduces
$\therefore \mathrm{COP}=\frac{\mathrm{R} \cdot \mathrm{E}}{\mathrm{W}_{\mathrm{in}}}$
Question 8
Ambient air is at a pressure of 100 kPa, dry bulb temperature of $30^{\circ}C$ and and 60% relative humidity. The saturation pressure of water at $30^{\circ}C$ is 4.24 kPa. The specific humidity of air (in g/kg of dry air) is ________ (correct to two decimal places).
A 102.25 B 8.68 C 18.68 D 16.24
GATE ME 2018 SET-2 Properties of Moist Air
Question 8 Explanation:
\begin{aligned} P_{\mathrm{atm}} &=100 \mathrm{kPa} \\ D B T &=t=30^{\circ} \mathrm{C} \rightarrow P_{v s}=4.24 \mathrm{kPa} \\ \phi &=60 \%\\ \phi &=\frac{P_{V}}{P_{V S}} \Rightarrow 0.6=\frac{P_{V}}{4.24} \\ P_{V} &=2.544 \mathrm{kPa} \\ W &=0.622 \times \frac{P_{V}}{P_{\mathrm{atm}}-P_{V}} \\ W &=0.622 \times \frac{2.544}{100-2.544} \\ W&=16.24 \text{gram/kg of dry Air} \end{aligned}
Question 9
A standard vapor compression refrigeration cycle operating with a condensing temperature of $35^{\circ}C$ and an evaporating temperature of $-10^{\circ}C$ develops 15 kW of cooling. The p-h diagram shows the enthalpies at various states. If the isentropic efficiency of the compressor is 0.75, the magnitude of compressor power (in kW) is _________ (correct to two decimal places).
A 8 B 10 C 12 D 6
GATE ME 2018 SET-2 Vapor and Gas Refrigeration
Question 9 Explanation:
\begin{aligned} R C &=15 \mathrm{kW} \\ R C &=\dot{m} \times\left(h_{1}-h_{4}\right) \\ 15 &=\dot{m} \times(400-250) \\ \dot{m} &=0.1 \mathrm{kg} / \mathrm{sec} \\ \omega_{\text {isentropic }} &=\left(h_{2}-h_{1}\right)=(475-400)=75 \mathrm{kJ} / \mathrm{kg}\\ \eta_{C} &=\frac{\omega_{\text {isentropic }}}{\omega_{\text {actual }}} \\ \omega_{\text {actual }} &=\frac{75}{0.75}=100 \mathrm{kJ} / \mathrm{kg} \\ \omega_{\text {actual }} &=\dot{m} \times \omega_{\text {actual }}=0.1 \times 100 \\ P_{\dot{m}} &=10 \mathrm{kW} \end{aligned}
Question 10
If a mass of moist air contained in a closed metallic vessel is heated, then its
A relative humidity decreases B relative humidity increases C specific humidity increases D specific humidity decreases
GATE ME 2017 SET-2 Psychrometric Process
Question 10 Explanation:
There are 10 questions to complete. | 2,822 | 8,017 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 31, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2021-49 | longest | en | 0.555294 |
https://chem.libretexts.org/Textbook_Maps/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Advanced_Thermodynamics/6._The_solution_of_thermodynamic_problems | 1,532,014,951,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591140.45/warc/CC-MAIN-20180719144851-20180719164851-00315.warc.gz | 612,684,414 | 18,769 | 6. The solution of thermodynamic problems
With $$U$$, $$A$$, $$H$$ and $$G$$ in hand we have potentials as a functions of whichever variable pair we want: $$S$$ and $$V$$, to $$T$$ and $$P$$. Additional Legendre transforms will provide us with further potentials in case we have other variables (such as surface area $$A$$, length $$L$$, magnetic moment $$M$$, etc.).
Thermodynamic problems always involve computing a variable of interest. It may be a derivative if it is an intensive variable, or even a second derivative (higher derivatives are rarely of interest).
Example
1st order ones like
$\left( \dfrac{\partial G}{\partial P} \right)_T=V$
or 2nd order ones like
$\left( \dfrac{\partial^2 G}{\partial P^2} \right)_T= \left( \dfrac{\partial V}{\partial P} \right)_T = \kappa V$
The solution procedure is thus:
1. Select the derivative or variable to be computed;
2. Select the potential representation that makes it easiest, or corresponds to variables you already have in hand.
3. Manipulate the thermodynamic derivative you know to get the one you want.
Easy as 1-2-3! We now turn to two methods to manipulate the thermodynamic derivations:
6.1 Tool 1: Maxwell relations
This tool is useful if two variables are NOT conjugate (e.g. $$T$$ and $$V$$ as opposed to $$T$$ and $$S$$)
If $$z = z(x,y)$$, then its total differential is:
$dz = \left( \dfrac{\partial z}{\partial x} \right)_y dx + \left( \dfrac{\partial z}{\partial y} \right)_x dy$
but
$\dfrac{\partial^2 z}{\partial x \partial y} = \dfrac{\partial^2 z}{\partial y \partial x}$
so
$\dfrac{\partial}{\partial y} \left( \dfrac{\partial z}{\partial x} \right)_y = \dfrac{\partial}{\partial x} \left( \dfrac{\partial z}{\partial y} \right)_x$
Maxwell’s relations must hold true if $$z$$ is a state function, as we saw in chapter 1. They basically state that the cross-second derivatives of a function must be identical.
Example 6.1
$dG = -SdT + VdP + \mu dn$
$- \left(\dfrac{\partial S}{\partial P}\right)_{T,n} = - \left(\dfrac{\partial V}{\partial T}\right)_{P,n}= \kappa V$
or
$- \left(\dfrac{\partial V}{\partial n}\right)_{P,T} = - \left(\dfrac{\partial \mu}{\partial P}\right)_{n,T}$
All four of the quantities in parenthesis are second derivatives of $$G$$. We can verify the second line above for an ideal gas:
$- \left(\dfrac{\partial \mu}{\partial P}\right)_{n,T} = \dfrac{\partial}{\partial P} RT\ln P = \dfrac{RT}{P}$
and
$- \left(\dfrac{\partial V}{\partial n}\right)_{P,T} = \dfrac{\partial}{\partial n} \dfrac{nRT}{P} = \dfrac{RT}{P}$
Note: As discussed in chapter 7, only three second derivatives not involving $$n_i$$ are independent; many sets can be chosen; the conventional one is $$\alpha$$, $$\kappa$$, and $$c_P$$, all defined at constant $$T$$ or $$P$$ (Gibbs ensemble). You can preview them at the beginning of chapter 7.
6.2 Tool 2: Jacobi Determinants
To reduce thermodynamic formulas to a minimal number of derivatives (e.g. $$\alpha$$, $$\kappa$$, and $$c_P$$), the Jacobi determinants can be used in addition to the Maxwell relations. The idea is that you can reduce ANY derivative or variable you want to functions of $$T$$, $$P$$, $$n$$ (if working in the Gibbs ensemble, other combinations otherwise) and second derivatives $$\alpha$$, $$\kappa$$, and $$c_P$$(others in different ensembles).
The Jacobi determinant is defined as
$\dfrac{\partial (x,y,...z)}{\partial (u,v,...w)} \equiv \text{det} \begin{vmatrix} \left(\dfrac{\partial x}{\partial u}\right) & \left(\dfrac{\partial x}{\partial v}\right) & \left(\dfrac{\partial x}{\partial w}\right) \\ \left(\dfrac{\partial y}{\partial u}\right) & \left(\dfrac{\partial y}{\partial u}\right) & \left(\dfrac{\partial y}{\partial w}\right) \\ \left(\dfrac{\partial z}{\partial u}\right) & \left(\dfrac{\partial z}{\partial u}\right) & left(\dfrac{\partial z}{\partial u}\right) \\ \end{vmatrix}$
Jacobians are useful for any variables transformation $$(u, v, … w) \rightarrow (x, y, … z)$$. Any thermodynamic derivative can be expressed as follows:
$\left( \dfrac{\partial x}{\partial u}\right)_{v...w} = \dfrac{\partial (x,y,...z)}{\partial (u,v,...w)}$
because
$$\dfrac{\partial v}{\partial u} = 0$$ and $$\dfrac{\partial v}{\partial v}=1$$ etc.
in the above determinant will leave only the desired derivative when the determinant is multiplied out. The derivatives can then be manipulated by using the Jacobian identities:
i)
$\dfrac{\partial (x,y,...z)}{\partial (u,v,...w)} = - \dfrac{\partial (y,x,...z)}{\partial (u,v,...w)} = - \dfrac{\partial (x,y,...z)}{\partial (v,u,...w)}$
via the properties of row or column permutation of determinants.
ii)
$\dfrac{\partial (x,y,...z)}{\partial (u,v,...w)} = \dfrac{\partial (y,x,...z)}{\partial (r,s,...t)} \cdot \dfrac{\partial (r,s,...t)}{\partial (v,u,...w)}$
For example:
iii)
$\dfrac{\partial(x,y,...z)}{\partial(u,v,...,w)} =\dfrac{1}{\dfrac{\partial(u,v,...,w)}{\partial(x,y,...,z)}}$
Proof
$det(M^{-1})= \dfrac{1}{det{M}}$
This concludes all the formal thermodynamic tools: From $$\Delta S > 0$$ (from Postulate 2), to the Euler formula, Legendre Transforms, and finally Maxwell relations and Jacobians, you have all the problem-solving tools thermodynamics has to offer to tackle equilibrium problems.
Armed with the Maxwell and Jacobian relations, we can reduce any expression to $$\alpha$$, $$\kappa$$, and $$c_p$$ (plus concentration or other derivatives, depending on the extensive variables of the system). Let us do a few examples:
6.3 Applications
Example 6.1: Heat Capacity at Constant Volume
Evaluated the heat capacity at constant volume in terms of heat capacity at constant pressure.
SOLUTION
$c_b \equiv \left(\dfrac{dq}{dT}\right)_V = \dfrac{T}{n}\left(\dfrac{\partial S}{\partial T}\right)_V \tag{1}$
Step 1: is is there a Maxwell relation? NO: $$S$$ & $$T$$ are conjugate variables, so use a Jacobian
$= \dfrac{T}{n} \dfrac{\partial(S,V)}{\partial(T,V)} \tag{2}$
Step 2: we want $$P$$ held constant in the Jacobian, not $$V$$, so split the Jacobian by chain rule
$\dfrac{T}{n} \dfrac{\partial(S,V)}{\partial(T,P)} \dfrac{\partial(T,P)}{\partial(T,V)} \tag{3}$
Step 3: Flip columns in the second Jacobian and evaluate $$(\partial P/ \partial V)_T$$, the isothermal compressibility $$\kappa$$
$= \dfrac{-T}{nV\kappa}\dfrac{\partial(S,V)}{\partial(T,P)}\tag{4}$
Step 4: Can we use Maxwell or Jacobians? NO: time to multiply out the determinant
$= -\dfrac{T}{nV\kappa} \left\{ \left( \dfrac{\partial S}{\partial T} \right)_P \left(\dfrac{\partial V}{\partial P} \right)_T - \left( \dfrac{\partial S}{\partial P} \right)_T \left(\dfrac{\partial V}{\partial T} \right)_P \right\} \tag{5}$
Step 5: We know three of the derivatives in terms of second derivatives of the Gibbs free energy. The missing one, $$\partial S/\partial P)_T = ?$$ can be obtained using Maxwell: $$? = - (\partial V/\partial T_P$$, just the isobaric expansion coefficient $$\alpha$$.
$= -\dfrac{T}{nV\kappa} \left\{ \left( \dfrac{nC_p}{T} \right) \left(- \dfrac{V}{\kappa} \right) - ? (V\alpha) \right\}$
Step 6: Collect everything and note that $$V$$, $$T$$, $$n$$ and $$\kappa$$ are all positive, as is $$\alpha^2$$, so $$c_v$$ must always be smaller than $$c_P$$. This should make intuitive sense: if I keep a system at constant pressure and put heat in it, it will also expand, ‘losing’ some of the heat as work. Thus the temperature will not go up as much, so the heat capacity is bigger at constant pressure.
$=c_p -\dfrac{VT\alpha^2}{n\kappa}$
$\Rightarrow c_p \gt c_v$
Example 6.2: Change in Enthalpy at Constant P
Calculate the change in enthalpy at constant pressure.
$\left( \dfrac{\partial H}{\partial T}\right)_{P,n} = \left( \dfrac{\partial H}{\partial S}\right)_{P,n} \left( \dfrac{\partial S}{\partial T}\right)_{P,n} = T \left( \dfrac{\partial S}{\partial T}\right)_{P,n} = T \dfrac{nC_p}{T} = nc_p$
$Rightarrow n\int _{T_0}^{T} dq = \int _{T_0}^{T} dH = \int _{T_0}^{T} nc_p (T') dT' = H(T)-H(T_0)$
Thus $$dH = n c_ dT$$. We derived this earlier by waving hands and using $$dq_{rev}$$, but here it is, formally.
Example 6.3 Isentropic Quasistatic Expansion
In an isentropic quasistatic expansion, ($$S$$ = constant, so $$dq_{rev}/T -dS =0$$). Do the homework problem before you check out the solution below!
This describes the pressure-temperature relationship in a molecular beam, where no heat can flow into the expanding gas, so its temperature must drop as the pressure drops. What happens is that at the nozzle of the molecular beam, collisions preferentially eject particles forward, preserving their average kinetic energy, but making a stream of particles with a narrower relative velocity distribution. The relevant derivative is $$(\partial T/\partial P)_S$$, which looks a bit odd at first, but we are fully equipped to handle derivatives at constant entropy:
$0 =dS = \dfrac{dq_{rev}}{T}$
$dT= \left ( \dfrac{\partial T}{\partial P} \right)_S dP$
$\Rightarrow \dfrac{\partial(T,S)}{\partial(P,S)} \dfrac{\partial(T,S)}{\partial(P,T)}\dfrac{\partial(P,T)}{\partial(P,S)} = - \dfrac{\partial(S,T)}{\partial(P,S)} \cdot \dfrac{1}{\dfrac{\partial(S,P)}{\partial(T,P)}}=\dfrac{-\left(\dfrac{\partial S}{\partial P}\right)_T}{\left(\dfrac{\partial S}{\partial T}\right)_P}$
For the numerator, $$\left ( \dfrac{\partial S}{\partial P} \right)_T = \left ( \dfrac{\partial V}{\partial T} \right)_P = \alpha V$$ by the Maxwell relations.
Note that just because $$S$$ is constrained to be held constant in $$(\partial T/ \partial P)_S$$ does not mean that $$\partial S/ \partial P=0$$!
For the denominator
$\left( \dfrac{\partial S}{\partial T} \right)_P = \dfrac{nc_p}{T}$
$\Rightarrow dT = \dfrac{\alpha VT}{nc_p} dP$
For an ideal monatomic gas, $$\alpha = \dfrac{1}{T}=\dfrac{nR}{PV}$$, and $$c_p=\dfrac{5}{2} R$$ so we can write explicitly
$\dfrac{dt}{T}=\dfrac{2}{5}\dfrac{dP}{P}$
or
$\left(\dfrac{T}{T_0}\right) =\left(\dfrac{P}{P_0}\right)^{2/5}$
after integrating. The temperature drops in a molecular beam, but not as dramatically as the pressure.
Example 6.4: Heat Capacity in terms of Energy
$c_v = \left(\dfrac{dq}{dT}\right)_V = \dfrac{T}{n} \left(\dfrac{\partial S}{\partial T}\right)_V$
with $$T>0$$ and $$n>0$$
$\left(\dfrac{\partial S}{\partial T}\right)_V = \dfrac{\partial(S,V)}{\partial(T,V)}=\dfrac{\partial(S,V)}{\partial(U,V)} \dfrac{\partial(U,V)}{\partial(T,V)} = \left( \dfrac{\partial S}{\partial U} \right)_V \left( \dfrac{\partial U}{\partial T} \right)_V = \dfrac{1}{T} \left( \dfrac{\partial U}{\partial T} \right)_V$
$c_v = \dfrac{1}{n} \left( \dfrac{\partial U}{\partial T} \right)_V =\dfrac{T}{n} \left( \dfrac{\partial S}{\partial T} \right)_V$
Again, we showed this before by invoking $$dq_{rev}$$ when discussing Legendre Transforms from energy to enthalpy. Examples 6.1 and 6.3 are just the formal way of deriving it using Maxwell’s relations and Jacobians. | 3,416 | 10,905 | {"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} | 3.953125 | 4 | CC-MAIN-2018-30 | latest | en | 0.812225 |
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# Could You Ace the New GMAT?
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As if applying to business school weren’t stressful enough, M.B.A. hopefuls now have to contend with an even more complicated entrance exam. (Monday, the Journal wrote about how bookings for the earlier test were in high demand.)
Tuesday marks the official kickoff of the new Graduate Management Admission Test, which now includes an “Integrated Reasoning” section that is intended to measure how well applicants can handle complex charts and graphs and sort through lots of data in little time. The new addition is given a separate grade from the verbal and quantitative scores, and replaces an essay. (There will still be one essay in the exam.)
Many schools admit they’re not yet sure how they’ll weigh the results in admissions decisions.
These new questions aren’t even easy to explain: One sample question includes a sortable chart that shows beef, soybean and other agricultural exports from Brazil and the country’s share of world production. Another, referring to a new car, includes the following: “In terms of the variables S and E, select the expression that represents the number of liters of fuel used in 1 hour of driving under ideal driving conditions at a constant speed S, and select the expression that represents the number of liters of fuel used in a 60 km drive under ideal driving conditions at a constant speed S.”
Try your hand at some examples from the new GMAT section and let us know when – or if – your head starts spinning! Share your comments below.
Graphics Interpretation
Two-Part Analysis
Table Analysis
Multi-Source Reasoning
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Food energy is the amount of energy in food that is available through digestion. The values for food energy are expressed in kilocalories (kcal) and kilojoules (kJ).
One large food Calorie is the amount of digestively available food energy (heat) that will raise the temperature of one kilogram of water one degree Celsius. Some advocate the convention of the capitalizing the C in these so that one Calorie is equal to 1000 lowercase calories, but that is not a convention generally followed. The large Calorie is sometimes abbreviated kcal, to indicate clearly that is 1000 times as large as the calorie. Consequently, the prefix kilo- is not used with large Calories. Food calories are also more specifically called kilocalories on the basis of the small calorie usage. This term, which makes it clear that large Calories are intended, is widely used by professional nutritionists when speaking in terms of calories rather than joules, but the term kilocalorie for 1,000 small calories is less often used by laypersons.
The International System of Units unit kilojoule is becoming more common. In some countries (Australia, for example) only the kilojoule is normally used. Some types of food contain more food energy per gram than others: fats and ethanol have particularly high values for food energy density: 9 and 7 kcal/gram, respectively. Sugars and proteins have about 4 kcal/gram. One Calorie is approximately equal to 4.1868 kilojoules.
Each food item has a specific metabolizable energy intake (MEI). For a normal human this value is obtained by multiplying the number of kilocalories or kilojoules contained in a food item by 85%, which is the amount of energy actually obtained by a human after the digestive processes have been completed.
## Measuring food energyEdit
In the early twentieth century, the United States Department of Agriculture (USDA) developed a procedure for measuring food energy that remains in use today.
The food being measured is completely burned in a calorimeter so that the heat released through combustion can be accurately measured. This amount is used to determine the gross energy value of the particular food. This number is then multiplied by a coefficient which is based on how the human body actually digests the food.
## Nutrition and food labels Edit
The "Calorie" has become a common household term, because dietitians recommend in cases of obesity to reduce body weight by increasing exercise (energy expenditure) and reducing energy intake. Many governments require food manufacturers to label the energy content of their products, to help consumers control their energy intake. In Europe, manufacturers of prepackaged food must label the nutritional energy of their products in both kilocalories ("kcal") and kilojoules ("kJ"). In the United States, the equivalent mandatory labels display only "Calories" (when used with capitalized C, meaning kilocalories); an additional kilojoules figure is optional and is rarely used. The energy content of food is usually given on labels for 100 g and/ or for what the manufacturer claims is a typical serving size.
The amount of food energy in a particular food could be measured by completely burning the dried food in a bomb calorimeter, a method known as direct calorimetry.[1] However, the values given on food labels are not determined this way, because it overestimates the amount of energy that the human digestive system can extract, by also burning dietary fiber. Instead, standardized chemical tests and an analysis of the recipe are used to estimate the product's digestible constituents (protein, carbohydrate, fat, etc.). These results are then converted into an equivalent energy value based on a standardized table of energy densities:
food component energy density[2]
kcal/g kJ/g
fat 9 37
ethanol (alcohol) 7 29
proteins 4 17
carbohydrates 4 17
organic acids 3 13
Sugar alcohols (sweeteners) 2.4 10
Other substances found in food (water, non-digestible fibre, minerals, vitamins) do not contribute to this calculated energy density.
Recommended daily energy intake values for young adults are: 2500 kcal/d (10 MJ/d, 2.8 kWh) for men and 2000 kcal/d (8 MJ/d, 2.3 kWh) for women (cf [3] for conversions). Children, sedentary and older people require less energy, physically active people more.
About 3,500 calories are contained in 1 pound of fat. If you eat 3,500 calories more than your body needs, you will put on about 1 pound of fat. If you burn 3,500 calories more than you eat, you will lose about 1 pound of fat. This assumes that all the weight gained and lost is in the form of fat. In reality, muscle and organ mass will change as well.
## References Edit
1. http://www.merck.com/mmhe/sec12/ch152/ch152e.html
2. (1997). Nutrient Value of Some Common Foods. Health Canada, PDF p. 4. URL accessed on 2006-03-04.
3. http://www.eia.doe.gov/kids/energyfacts/science/energy_calculator.html | 1,111 | 4,945 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2017-34 | latest | en | 0.934695 |
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2021-03-07
Use the appropriate algebra and Table of Laplaces
davonliefI
Step 1 It is given that, ${L}^{-1}\left\{\frac{1}{\left(s-1{\right)}^{2}}-\frac{120}{\left(s+3{\right)}^{6}}\right\}$ Step 2 Obtain the inverse Laplace transform as follows: ${L}^{-1}\left\{\frac{1}{\left(s-1{\right)}^{2}}-\frac{120}{\left(s+3{\right)}^{6}}\right\}={L}^{-1}\left\{\frac{1}{\left(s-1{\right)}^{2}}\right\}-{L}^{-1}\left\{\frac{120}{\left(s+3{\right)}^{6}}\right\}$ | 192 | 464 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 26, "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-2023-50 | latest | en | 0.614371 |
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Ch 18 hw 1
# Ch 18 hw 1 - 1.03113e-5 1.03e-05 m/s 4 2/2 points | 2/4...
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1. 2/2 points | 2/4 submissions Last Response | Show Details All Responses Notes If 1.7 10 16 electrons enter a light bulb in 4 milliseconds, what is the magnitude of the electron current at that point in the circuit? 4.25e18 4.25e+18 electrons/second 2. 2/2 points | 3/4 submissions Last Response | Show Details All Responses Notes If an electron current is 3.9 10 19 electrons/s. What is the conventional current? 6.24 6.24 A 3. 2/2 points | 2/4 submissions Last Response | Show Details All Responses Notes The density of mobile electrons in copper metal is 8.4 10 28 m -3 . Suppose that i = 3.0 10 18 electrons/s are drifting through a copper wire. (This is a typical value for a simple circuit.) The diameter of the wire is 2.1 mm. What is the drift speed of the electrons?
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Unformatted text preview: 1.03113e-5 1.03e-05 m/s 4. 2/2 points | 2/4 submissions Last Response | Show Details All Responses Notes The density of mobile electrons in copper metal is 8.4 10 28 m-3 . Suppose that i = 4.4 10 18 electrons/s are drifting through a copper wire. (This is a typical value for a simple circuit.) The diameter of the wire is 1.3 mm. In this case, about how many minutes would it take for a single electron in the electron sea to drift from one end to the other end of a wire 29 cm long? 122.4757036 122 minutes (A puzzle: if the drift speed is so slow, how can a lamp light up as soon as you turn it on? We'll come back to this in the next chapter.)...
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# C3 Maths A2 AQA 2016 (unofficial mark scheme new)
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1. (Original post by beanigger)
you use area of triangle = 1/2ab and bobs ur uncle
**** me are you serious? I did that at first with the wrong co-ordinates, got the right co-ordinates then assumed that method was still wrong. I thought E^x would be curved but now I realise then it woudn't be a ****ing triangle
I'm an arse
I guess that doesn't matter, I'd need another 15 marks probably to get an A* and to get an A overall I need C's in the other papers. A* would be out of reach regardless
2. yeah lol from O to A is a straight line
so is O to B so that forms a right angle
hence triangle
3. (Original post by Sirelis)
fak i got like 40 if I'm lucky
Same lmao
4. (Original post by beanigger)
yeah lol from O to A is a straight line
so is O to B so that forms a right angle
hence triangle
Honestly I thought that the first time, then when I saw it was getting nowhere I changed my mind. Looking through this mark scheme I got between 45 and 49, Having gotten 70+ on every past paper I've done in the past week. Can't believe this.
5. Any ideas as to what the grade boundaries will be like ? considering last years was 57 for an A, hopefully this years will be lower.
6. roughly the same, but it will not be lower than 55 for an A
7. (Original post by irMike)
Honestly I thought that the first time, then when I saw it was getting nowhere I changed my mind. Looking through this mark scheme I got between 45 and 49, Having gotten 70+ on every past paper I've done in the past week. Can't believe this.
Well I think I got from around 60-64 assuming the worst; but that's assuming the way im distributing marks in questions I got wrong is correct; so that I get the method marks at least. By the way for the question leading up the sinx=2/3 or something; was that a question with a quadratic equation where you had to reject one of the values or am I remembering wrong. AND WHAT WAS THE QUESTION ITSELF????????
9. (Original post by Anon998)
For that one you just get the answer from the previous question - 5(1.7-0.5); and since that is the value of the integral of the negative x^x you just change signs and it gives you 1.51 or something.
10. (Original post by Anon998)
Ahhh that's how you do it, I just did the simpsons rule again haha, for 2 marks will this even gain anything?
11. well aqa maths dont uusally give a lot of follow on marks, its usually just one M1 mark for applying a certain method so make sure you add it up right
12. (Original post by matty9)
Ahhh that's how you do it, I just did the simpsons rule again haha, for 2 marks will this even gain anything?
Not sure because I think it said using the previous answer but I don't remember jack ****.
13. 59 a*
53 a
47 b
41 c
14. (Original post by beanigger)
well aqa maths dont uusally give a lot of follow on marks, its usually just one M1 mark for applying a certain method so make sure you add it up right
Do you remember the question for the one about the area of the triangle :C. I managed to get m and n=3 but I don't know how. I just wanna check what I messed up to see how many marks i lost.
15. (Original post by Parhomus)
Not sure because I think it said using the previous answer but I don't remember jack ****.
Yes it said 'hence find an estimate' but wasn't sure what to do so did simpsons ruls again.
16. b) find gg(x) g(x) = 1/x therefore gg(x) = x [2]
I got gg(x) = x^4
as g(x) = 1/x^2
can you double check that one, not sure it's right
17. (Original post by Sayless)
b) find gg(x) g(x) = 1/x therefore gg(x) = x [2]
I got gg(x) = x^4
as g(x) = 1/x^2
can you double check that one, not sure it's right
fg(x) means f(g(x)) so when you got the equation; g(x) wouldn't be 1/x^2 it would 1/x because the x in the f(x) just means the input and so g(x) could only be 1/x. I understand why it would make sense for it to be 1/x^2.
18. (Original post by Sayless)
b) find gg(x) g(x) = 1/x therefore gg(x) = x [2]
I got gg(x) = x^4
as g(x) = 1/x^2
can you double check that one, not sure it's right
I put the same but it's wrong.
g(x) was 1/x because I forgot that in functions the entire x is replaced so it wasn't 1/x^2 can't remember the question so kinda hard to explain.
19. (Original post by matty9)
Yes! I got sinx = 2/3 then used pythag to get cosx = root5/3
Then subbing these into the orginal equation gave me -1/5 root 5 but I don't know how correct that is
sub into sin^2(x) +cos^2(x) = 1
20. (Original post by alikhaliq)
sub into sin^2(x) +cos^2(x) = 1
Why sub into that?? The eqn was something like f(x) = sec x + tan x so you had to do (1 / cos x) + (sin x / cos x) right?
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Useful resources | 1,545 | 5,482 | {"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-2016-44 | latest | en | 0.967392 |
https://www.tensorflow.org/api_docs/python/tf/unique_with_counts?hl=JA | 1,628,106,384,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154897.82/warc/CC-MAIN-20210804174229-20210804204229-00030.warc.gz | 1,063,344,096 | 45,961 | # tf.unique_with_counts
Finds unique elements in a 1-D tensor.
This operation returns a tensor `y` containing all of the unique elements of `x` sorted in the same order that they occur in `x`. This operation also returns a tensor `idx` the same size as `x` that contains the index of each value of `x` in the unique output `y`. Finally, it returns a third tensor `count` that contains the count of each element of `y` in `x`. In other words:
`y[idx[i]] = x[i] for i in [0, 1,...,rank(x) - 1]`
#### For example:
``````# tensor 'x' is [1, 1, 2, 4, 4, 4, 7, 8, 8]
y, idx, count = unique_with_counts(x)
y ==> [1, 2, 4, 7, 8]
idx ==> [0, 0, 1, 2, 2, 2, 3, 4, 4]
count ==> [2, 1, 3, 1, 2]
``````
`x` A `Tensor`. 1-D.
`out_idx` An optional `tf.DType` from: `tf.int32, tf.int64`. Defaults to `tf.int32`.
`name` A name for the operation (optional).
A tuple of `Tensor` objects (y, idx, count).
`y` A `Tensor`. Has the same type as `x`.
`idx` A `Tensor` of type `out_idx`.
`count` A `Tensor` of type `out_idx`.
[{ "type": "thumb-down", "id": "missingTheInformationINeed", "label":"必要な情報がない" },{ "type": "thumb-down", "id": "tooComplicatedTooManySteps", "label":"複雑すぎる / 手順が多すぎる" },{ "type": "thumb-down", "id": "outOfDate", "label":"最新ではない" },{ "type": "thumb-down", "id": "samplesCodeIssue", "label":"Samples / code issue" },{ "type": "thumb-down", "id": "otherDown", "label":"その他" }]
[{ "type": "thumb-up", "id": "easyToUnderstand", "label":"わかりやすい" },{ "type": "thumb-up", "id": "solvedMyProblem", "label":"問題の解決に役立った" },{ "type": "thumb-up", "id": "otherUp", "label":"その他" }] | 564 | 1,577 | {"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-2021-31 | latest | en | 0.561424 |
https://metanumbers.com/29047 | 1,601,258,128,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401583556.73/warc/CC-MAIN-20200928010415-20200928040415-00229.warc.gz | 495,643,014 | 7,435 | ## 29047
29,047 (twenty-nine thousand forty-seven) is an odd five-digits composite number following 29046 and preceding 29048. In scientific notation, it is written as 2.9047 × 104. The sum of its digits is 22. It has a total of 2 prime factors and 4 positive divisors. There are 28,080 positive integers (up to 29047) that are relatively prime to 29047.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 5
• Sum of Digits 22
• Digital Root 4
## Name
Short name 29 thousand 47 twenty-nine thousand forty-seven
## Notation
Scientific notation 2.9047 × 104 29.047 × 103
## Prime Factorization of 29047
Prime Factorization 31 × 937
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 29047 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 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 29,047 is 31 × 937. Since it has a total of 2 prime factors, 29,047 is a composite number.
## Divisors of 29047
1, 31, 937, 29047
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 30016 Sum of all the positive divisors of n s(n) 969 Sum of the proper positive divisors of n A(n) 7504 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 170.432 Returns the nth root of the product of n divisors H(n) 3.87087 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 29,047 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 29,047) is 30,016, the average is 7,504.
## Other Arithmetic Functions (n = 29047)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 28080 Total number of positive integers not greater than n that are coprime to n λ(n) 4680 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 3157 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 28,080 positive integers (less than 29,047) that are coprime with 29,047. And there are approximately 3,157 prime numbers less than or equal to 29,047.
## Divisibility of 29047
m n mod m 2 3 4 5 6 7 8 9 1 1 3 2 1 4 7 4
29,047 is not divisible by any number less than or equal to 9.
## Classification of 29047
• Arithmetic
• Semiprime
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (29047)
Base System Value
2 Binary 111000101110111
3 Ternary 1110211211
4 Quaternary 13011313
5 Quinary 1412142
6 Senary 342251
8 Octal 70567
10 Decimal 29047
12 Duodecimal 14987
20 Vigesimal 3cc7
36 Base36 mev
## Basic calculations (n = 29047)
### Multiplication
n×i
n×2 58094 87141 116188 145235
### Division
ni
n⁄2 14523.5 9682.33 7261.75 5809.4
### Exponentiation
ni
n2 843728209 24507773286823 711877290662347681 20677899661869213090007
### Nth Root
i√n
2√n 170.432 30.7398 13.055 7.80945
## 29047 as geometric shapes
### Circle
Diameter 58094 182508 2.65065e+09
### Sphere
Volume 1.02658e+14 1.06026e+10 182508
### Square
Length = n
Perimeter 116188 8.43728e+08 41078.7
### Cube
Length = n
Surface area 5.06237e+09 2.45078e+13 50310.9
### Equilateral Triangle
Length = n
Perimeter 87141 3.65345e+08 25155.4
### Triangular Pyramid
Length = n
Surface area 1.46138e+09 2.88827e+12 23716.8 | 1,257 | 3,778 | {"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} | 3.796875 | 4 | CC-MAIN-2020-40 | latest | en | 0.809725 |
https://rdrr.io/github/xylimeng/BayesBD/man/BayesBD.binary.html | 1,544,693,082,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824601.32/warc/CC-MAIN-20181213080138-20181213101638-00009.warc.gz | 714,154,838 | 14,307 | BayesBD.binary: Bayesian boundary detection for binary images In xylimeng/BayesBD: Bayesian Boundary Detection in Images
Usage
`1` ```BayesBD.binary(obs, ini.mean = 0.4, n.run = 10000, n.burn = 1000, J = 10) ```
Arguments
`obs` The noisy observation, which is a list with the following required elements: `intensity -` observed intensity at each pixel. `theta.obs, r.obs -` the location at which the intensity is observed, recorded using the polar coordinate. `center - ` the reference point that (`theta.obs, r.obs`) is referred to, defaulted as `(0.5, 0.5)`. `x, y - ` the `(x, y)` coordinates of the location. `ini.mean` a constant to specify the initial mean functions in the Bayesian estimation, defaulted as 0.4. `n.run` number of MCMC iterations. `n.burn` number of burn-in in the MCMC sampler. `J` truncation number of the Gaussian process kernel. The number of eigenfunctions is 2J + 1.
Value
Posterior samples of all parameters.
References
Li, M. and Ghosal, S.(2015) "Bayesian Detection of Image Boundaries." arXiv preprint arXiv:1508.05847.
Examples
``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32``` ```set.seed(2015) # ellipse boundary gamma.fun = ellipse(a = 0.35, b = 0.25) obs = par2obs(m = 100, pi.in = 0.5, pi.out = 0.2, design = 'J', gamma.fun) ## Not run: # it takes around 7min if runs 10000 iterations: saved in 'data.Rdata' BayesEst = BayesBD.binary(obs, n.run = 10000, n.burn = 1000) ## End(Not run) data(data) # visualize the estimates theta.plot = seq(from = 0, to = 2*pi, length.out = 200) gamma.hat.theta = BayesEst\$gamma.hat(theta.plot) ## plotting utilities require(plotrix) my.radial <- function(r, theta, ...){ radial.plot(c(r[order(theta)]), c(theta[order(theta)]), rp.type = "p", show.grid.label = TRUE, radial.lim = c(0, 0.5), ...) } # rotate a matrix rotate <- function(x) t(apply(x, 2, rev)) # rotate closewise by 90 degrees par(mfrow = c(1, 2)) # rotate & image it - square (asp = 1) image(rotate(obs\$intensity), axes = FALSE, asp = 1, main = 'observation') my.radial(gamma.fun(theta.plot), theta.plot, line.col = 1, lty = 2, lwd = 2, main = 'Estimated boundary vs. True', show.grid = FALSE) my.radial(gamma.hat.theta, theta.plot, add = TRUE, line.col = 'red', lty = 2, lwd = 2, show.grid = FALSE) ```
xylimeng/BayesBD documentation built on May 28, 2017, 8:34 a.m. | 769 | 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} | 2.578125 | 3 | CC-MAIN-2018-51 | latest | en | 0.622248 |
https://paperwriting.help/week-4-lab-et212-electrical-engineering-homework-help/ | 1,618,646,224,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038118762.49/warc/CC-MAIN-20210417071833-20210417101833-00341.warc.gz | 561,493,522 | 10,766 | # Week 4 lab et212 | Electrical Engineering homework help
#### Week 4 Lab – Transistor Fundamentals
Electronics I and Lab
## Transistor Fundamentals
Introduction:
Week 4 lab is based on the fundamentals of a transistor and analysis of an analog circuit.
Please review the following videos before getting started with this lab:
1. Watch the video: “Video 3: Fundamentals of breadboard
2. Watch the video: “Video 4: Basic electrical components
3. Watch the video: “Video 5: Simple resistive circuit with NI myDAQ
4. Read the Tutorial: myDAQ Tutorial
Materials and Equipment:
Materials:
• Hardware Parts (In the Toolbox):
• An NPN transistor: 2N3904
• Four resistors: 2 – 40 kΩ, 1 kΩ, 470Ω
• Jumper wires
Equipment:
• Hardware Equipment:
• NI myDAQ Instrument device
• Screw Driver
• Screw Terminal connector
• USB Cable
• Multimeter probes
Procedure:
*** This lab has to be implemented only in hardware (using NI myDAQ) ***
1. Analyze the circuit in the Figure 1 below to calculate the following values: VE, IE, VRC, VC, VCE. Assume Beta = 150.
2. Construct the circuit shown in Figure 1 below on the breadboard using the transistor and the resistors.
3. Using the jumper wires, screw driver and screw terminal connector, connect the board to NI MyDAQ Instrument Device.
4. Use channel +15V pin out on the NI myDAQ Instrument Device to provide the supply voltage (VCC).
Figure 1
1. Now, using multimeter with the probes, measure the required voltages: VE, VRC, VC, VCE and currents: IC and IE.
2. Tabulate the values obtained in step 1 and 7.
Calculated Measured VE IE VRC VC VCE
Review questions:
1. Compare the calculated and measured values in the table and analyze the performance of the transistor.
2. Discuss whether the values are the same or different. If they are different, provide the reasoning and explain how to reduce this difference between calculated and measured values.
3. Explain what happens when the transistor changes to a PNP transistor. How does the change in the transistor effect the current and voltage in the circuit?
Deliverables:
1. Analysis of the circuit and calculations of voltages: VE, VRC, VC, VCE and currents: IC and IE.
2. Take screenshots of the measurements obtained from the Multimeter on the NI ELVISmx Instrument Launcher on your screen.
Lab Report:
• Use the Lab report template found in the “Tools and Template” link in the navigation center.
• Include all the deliverables.
• Include all the screenshots of the measurements from NI ELVISmx Instrument Launcher.
• Save the document as Lab4YourGID.docx (ex: Lab4G00000000.docx) and submit in Blackboard.
Rules for lab submissions:
1. The lab document must be a Word document. PDF files are NOT accepted.
2. All screenshots must be included.
3. All Multisim screenshots must include the date/time stamp. See TOOLS AND TEMPLATES for the procedure to display the date and time.
4. Any and all Multisim files must be submitted.
5. Any equations used must be typed in Word. Copy and paste of equations from outside sources is prohibited.
6. No graphics are allowed in the Word document other than screenshots of circuits from Multisim and hardware if applicable, with the date/time stamp.
7. The lab template should be used. Specifically, it is brought to your attention that a summary MUST be provided explaining the results of the labs, the relationship of the results to expected results, and any challenges encountered.
8. Hardware portion of labs should include screenshots of the assembled circuit with your name and student GID number written on paper next to the circuit. There should be screenshots of the instrument readings with the date and time stamp on lower right corner clearly shown. See example below.
Any violation of the submission rules above will result in a grade of 1.
Basic features
• Free title page and bibliography
• Unlimited revisions
• Plagiarism-free guarantee
• Money-back guarantee
On-demand options
• Writer’s samples
• Part-by-part delivery
• Overnight delivery
• Copies of used sources
Paper format
• 275 words per page
• 12 pt Arial/Times New Roman
• Double line spacing
• Any citation style (APA, MLA, Chicago/Turabian, Harvard)
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https://studylib.net/doc/25371732/eco331-practical4-2020-studentcopy | 1,611,810,041,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704835901.90/warc/CC-MAIN-20210128040619-20210128070619-00344.warc.gz | 563,920,668 | 12,994 | # ECO331-Practical4-2020-StudentCopy
```ECO331
Due: 18 May 2020
Note: The following questions are aimed at applying your knowledge of the theory
discussed during the lectures. The purposes of the sessions are to provide a guide to the
methodology applied and not a memorandum.
Instructions:
remember to write these details on each page and number the page in the top righthand corner, especially when attaching pictures of separate pages.
2. Please write in a neat legible print. (Typing optional)
3. Submit your answers by end of day to the Dropbox folder.
4. Penalties (-30% or 6 marks) will be applied for plagiarism and copying.
Question 1 (Chapter 13)
~~~~~~~~~~~~~~~~~~~~~~~~~
Buy-right is a chain of grocery stores operating in small cities throughout the southwestern
United States. Buy-right’s major competition comes from another chain, Acme Food Stores.
Both firms are currently contemplating their advertising strategy for the region. The possible
outcomes are illustrated by the payoff matrix below.
Entries in the payoff matrix are profits. Buy-right’s profit is before the comma, Acme’s is
after the comma.
1.1
1.2
1.3
Describe what is meant by a dominant strategy.
[1]
Given the payoff matrix above, does each firm have a dominant strategy?
[3]
Under what circumstances would there be no dominant strategy for one or both
firms?
[1]
Page 1 of 2
Question 2 (Chapter 13)
Two firms at the St. Louis airport have franchises to carry passengers to and from hotels in
downtown St. Louis. These two firms, Metro Limo and United Limo, operate nine passenger
vans. These duopolists cannot compete with price, but they can compete through
Their payoff matrix is below:
2.1
2.2
Does each firm have a dominant strategy? If so, explain what that strategy is.
[3]
What is the Nash equilibrium? Explain where the Nash equilibrium occurs in the
payoff matrix.
[3]
Question 3 (Chapter 13)
Consider two firms, X and Y, which produce super computers. Each can produce the next
generation super computer for the military (M) or for civilian research (C). However, only
one can successfully produce for both markets simultaneously. Also, if one produces M, the
other might not be able to successfully produce M, because of the limited market. The
following payoff matrix illustrates the problem.
3.1
3.2
Find the Nash equilibrium, and explain why it is a Nash equilibrium.
[4]
If Firm X were unsure that the management of Firm Y was rational, what would Firm
X choose to do if it followed a maximin strategy? What would both firms do if they
both followed a maximin strategy?
[5]
Page 2 of 2
``` | 610 | 2,591 | {"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-2021-04 | latest | en | 0.91028 |
http://www.gurufocus.com/term/netcash_per_share/HMNF/Net%2BCash%2B%2528per%2Bshare%2529/HMN%2BFinancial%2BInc | 1,492,926,145,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917118477.15/warc/CC-MAIN-20170423031158-00355-ip-10-145-167-34.ec2.internal.warc.gz | 542,975,375 | 28,436 | Switch to:
GuruFocus has detected 3 Warning Signs with HMN Financial Inc \$HMNF.
More than 500,000 people have already joined GuruFocus to track the stocks they follow and exchange investment ideas.
HMN Financial Inc (NAS:HMNF)
Net Cash per Share
\$-128.85 (As of Dec. 2016)
Net cash per share is calculated as Cash and Cash Equivalents minus Total Liabilities and then divided by Shares Outstanding. HMN Financial Inc's net cash per share for the quarter that ended in Dec. 2016 was \$-128.85.
Definition
In the calculation of a companys net cash, assets other than cash and short term investments are considered to be worth nothing. But the company has to pay its debt and other liabilities in full. This is an extremely conservative way of valuation. Most companies have negative net cash. But sometimes a companys price may be lower than its net-cash.
HMN Financial Inc's Net Cash Per Share for the fiscal year that ended in Dec. 2016 is calculated as
Net Cash Per Share (A: Dec. 2016 ) = (Cash and Cash Equivalents - Total Liabilities) / Shares Outstanding = (27.561 - 606.104) / 4.49 = -128.85
HMN Financial Inc's Net Cash Per Share for the quarter that ended in Dec. 2016 is calculated as
Net Cash Per Share (Q: Dec. 2016 ) = (Cash and Cash Equivalents - Total Liabilities) / Shares Outstanding = (27.561 - 606.104) / 4.49 = -128.85
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Explanation
Ben Graham invested in situations where the companys stock price was lower than its net-cash. He assigned some value to the companys other current asset. The value is called Net Current Asset Value (NCAV). One research study, covering the years 1970 through 1983 showed that portfolios picked at the beginning of each year, and held for one year, returned 29.4 percent, on average, over the 13-year period, compared to 11.5 percent for the S&P 500 Index. Other studies of Grahams strategy produced similar results.
You can find companies that are traded below their Net Current Asset Value (NCAV) with our Net-Net screener. GuruFocus also publishes a monthly Net-Net newsletter.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
HMN Financial Inc Annual Data
Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 Dec14 Dec15 Dec16 netcash -238.09 -244.01 -216.44 -183.32 -151.54 -115.12 -100.06 -101.74 -118.87 -128.85
HMN Financial Inc Quarterly Data
Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 Dec16 netcash -102.16 -101.74 -106.20 -102.00 -120.50 -118.87 -123.10 -124.98 -128.34 -128.85
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GuruFocus Affiliate Program: Earn up to \$400 per referral. ( Learn More) | 746 | 2,896 | {"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} | 2.90625 | 3 | CC-MAIN-2017-17 | latest | en | 0.945436 |
https://knittystash.com/6-pin-3-position-switch-wiring-diagram/ | 1,675,842,221,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500719.31/warc/CC-MAIN-20230208060523-20230208090523-00063.warc.gz | 368,124,078 | 10,767 | # 6 Pin 3 Position Switch Wiring Diagram
Do you require a 6 Pin 3 Position Switch Wiring Diagram? The 6 Pin 3 Position Switch Wiring Diagram, ideas, and frequently asked questions are all readily available here. We produced this page for people searching for a 6 Pin 3 Position Switch Wiring Diagram. Our post will help you in fixing your issue.
## 6 Pin 3 Position Switch Wiring Diagram
A wiring diagram will show you where the wires must be connected, so you do not have to think.
You don’t have to think, a wiring diagram will certainly show you how to attach the wires.
See the 6 Pin 3 Position Switch Wiring Diagram images below
## Tips and tricks for reading wiring diagrams
• When taking a look at a wiring diagram, do not attempt to focus on the whole page at one time. It’s frustrating. Put a blank sheet of paper beside the wiring diagram and simply draw the simple circuit. Focus on the simple part and follow the current flow from power to ground or from ground to power. All complex wiring diagrams are just a series of simple diagrams, and it makes it hard to take a look at if you do not narrow down to the circuit that you’re doing.
• Print the wiring diagram off and use highlighters to trace the circuit. When you utilize your finger or follow the circuit with your eyes, it’s simple to mistrace the circuit. One trick that I utilize is to print the very same wiring diagram off twice. On one, I’ll trace the current flow, how it operates, and that shows me what parts of the circuit I need to inspect. Then on the other one, I’ll begin coloring the things that evaluated all right. When I get done, anything that’s not highlighted are suspect circuits that I need to identify.
• To effectively read a wiring diagram, one needs to understand how the components in the system operate. If a module is powered up and it sends out a signal of half the voltage and the professional does not understand this, he would believe he has a problem, as he would expect a 12V signal. Following diagrams is relatively simple, however utilizing it within the scope of how the system operates is a various matter. My finest recommendations is not only look at the diagram, but understand how the elements operate when in use.
• Check out wiring diagrams from negative to positive and redraw the circuit as a straight line. All circuits are the same– voltage, ground, single element, and switches.
• Prior to reading a schematic, get familiar and comprehend all the signs. Check out the schematic like a roadmap. I print the schematic and highlight the circuit I’m diagnosing to make sure I’m remaining on the best course.
6 Pin 3 Way Relay Wiring Diagram – Wiring Diagram Networks
## FAQ
### Are all wiring diagrams the same?
Wiring diagrams may follow various standards depending upon the nation they are going to be utilized. They may have different designs depending on the company and the designer who is designing that. They also may be drawn by various ECAD software such as EPLAN or AutoCAD electrical.
### What are the types of wiring diagram?
• Schematic Diagrams.
• Wiring diagrams.
• Block diagrams.
• Pictorial diagrams.
### What is the schematic format?
A schematic, or schematic diagram, is a representation of the elements of a system using abstract, graphic symbols rather than realistic pictures.
### What should a schematic include?
Schematics must include the total description and places of all building code components, such as the heating/ventilation/air conditioning (also referred to as HVAC), plumbing, and electrical systems. Schematic styles are just a standard layout to interact a style plan to the owner.
### What is an architectural wiring diagram?
Architectural wiring diagrams reveal the approximate places and affiliations of receptacles, lighting, and permanent electrical services in a structure.
6 Pin Momentary Switch Wiring Diagram – Wiring Diagram Schemas
power – wiring a 3 position toggle switch for two devices? – Electrical
New 6 Pin toggle Switch Wiring Diagram in 2020 | Diagram design
How to wire a 6 pin toggle switch – Quora
### How are wiring diagrams read?
The electrical schematics are read from left to right, or from top to bottom. This is important to get right, as the signal direction indicates the flow of current in the circuit. It is then simple for a user to understand when there is a modification in the course of the circuit.
### Where is a wiring diagram utilized?
Wiring diagrams are primarily used when trying to show the connection system in a circuit. It is majorly used by building coordinators, architects, and electricians to present the wiring connections in a structure, a space, or perhaps a simple device.
### Why is wiring diagram essential?
It reveals the components of the circuit as simplified shapes, and how to make the connections between the devices. A wiring diagram usually gives more information about the relative position and arrangement of devices and terminals on the devices.
### Can you touch a live black wire?
If you can be found in contact with an energized black wire– and you are also in contact with the neutral white wire– current will travel through your body. You will get an electrical shock. You will receive a shock if you touch two wires at various voltages at the same time.
### Is AWG aluminum or copper?
Normal home copper wiring is AWG number 12 or 14. The higher the gauge number, the smaller sized the size and the thinner the wire.
### How is wire numbered?
American Wire Gauge (AWG) is the standard way to signify wire size in The United States and Canada. In AWG, the larger the number, the smaller the wire diameter and thickness. The biggest basic size is 0000 AWG, and 40 AWG is the tiniest standard size.
### How do you read electrical wire numbers?
An electrical cable is classified by 2 numbers separated by a hyphen, such as 14-2. The very first number signifies the conductor’s gauge; the 2nd represents the number of conductors inside the cable. 14-2 has 2 14-gauge conductors: a hot and a neutral.
### How do you read wire size charts?
Wire gauges range from low numbers to high numbers, with smaller numbers referring to smaller diameters and bigger numbers representing larger diameters. For example, AWG 4 is 0.2043 inches in diameter, and AWG 40 is. 0031 inches in diameter.
### What is the distinction between a schematic and wiring diagram?
The schematic diagram does not show the practical connection in between the elements or their position. It includes only symbols and lines. A wiring diagram is a generalized pictorial representation of an electrical circuit. The elements are represented utilizing streamlined shapes in wiring diagrams.
### How do you read vehicle wiring diagrams?
An automobile wiring diagram is a map. To read it, determine the circuit in question and beginning at its power source, follow it to the ground. Use the legend to comprehend what each symbol on the circuit indicates.
### Why do we require wiring diagrams?
A wiring diagram is frequently utilized to troubleshoot issues and to ensure that all the connections have been made and that everything is present. | 1,487 | 7,171 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-06 | latest | en | 0.902157 |
http://gmatclub.com/forum/there-is-something-wrong-in-this-question-123644.html?fl=similar | 1,481,373,294,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543035.87/warc/CC-MAIN-20161202170903-00123-ip-10-31-129-80.ec2.internal.warc.gz | 115,337,764 | 54,686 | There is something wrong in this question : GMAT Data Sufficiency (DS)
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# There is something wrong in this question
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There is something wrong in this question [#permalink]
### Show Tags
23 Nov 2011, 07:46
1
KUDOS
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(N/A)
Question Stats:
63% (01:51) correct 38% (02:20) wrong based on 11 sessions
### HideShow timer Statistics
I think that the OA is wrong in this question. What do you think?
Is the two-digit prime number $$p$$ equal to 17 ?
(1) $$p^2$$ is greater than 250
(2) $$p = n^2 + 1$$, where n is an integer.
OE
[Reveal] Spoiler:
The OA is B. The OE mentions that the only two-digit prime that is one larger than a square is 17. But, what about 37?
37 is a prime number, and $$37 = 6^2 +1$$.
By the way, is there a faster way to solve this question than picking numbers?
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Re: There is something wrong in this question [#permalink]
### Show Tags
23 Nov 2011, 08:39
1
KUDOS
You are correct.
(1) says that p > 15
(2) p can be 17 or 37 at the least
Re: There is something wrong in this question [#permalink] 23 Nov 2011, 08:39
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https://jp.mathworks.com/matlabcentral/answers/1752740-how-do-i-set-a-parameter-based-on-individual-values-in-an-existing-array-should-i-use-a-for-loop-or?s_tid=prof_contriblnk | 1,709,389,264,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475825.14/warc/CC-MAIN-20240302120344-20240302150344-00080.warc.gz | 326,047,954 | 33,280 | # How do I set a parameter based on individual values in an existing array? Should I use a for loop or something else?
1 回表示 (過去 30 日間)
Steven D 2022 年 7 月 3 日
コメント済み: Steven D 2022 年 7 月 4 日
I need to have 'maxDistLine' be different for different 'L' values (i.e., if L<300, then maxDistLine should be 70, or else if L is between 300 and 1000 maxDistLine should be 50, else maxDistLine should be 20. I've tried using a 'for loop', 'while', and 'if/else' statements and nothing seems to work. Please see attached section of code. Thanks in advance for any help you could offer!
Section of Code:
classdef TripletGraph2
properties
C; %triplet center points
T; %triplet endpoints
Cgraph %The undirected graph of the center points
ArcPoints %2x4xN list of arc points CenterA-EndpointA-EndpointB-CenterB
end
methods
function obj=TripletGraph2(fpep)
% maxDistLine=20;
maxLineLength=1800;
CTDist=35;
% maxDistLineDiff=7;
l2norm=@(z,varargin) vecnorm(z,2,varargin{:});
C=fpep(:,7:8); %Central points
T=reshape( fpep(:,1:6) ,[],2,3); %Triplet end-points Nx2x3
T=(T-C)./l2norm(T-C,2)*CTDist + C; %Normalize center to endpoint distances
DT=delaunayTriangulation(C); %Delaunay triangulation of centers
E=edges(DT); %Edge list
C1=C(E(:,1),:); %Center points listed for all DT edges
C2=C(E(:,2),:);
L=l2norm(C1-C2,2); %Center-to_center lengths for DT edges
if L<500
maxDistLine=21;
maxDistLineDiff=7;
elseif (500<L)&(L<1000)
maxDistLine=18;
maxDistLineDiff=7;
else
maxDistLine=15;
maxDistLineDiff=7;
end
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### 回答 (2 件)
the cyclist 2022 年 7 月 3 日
My best guess here is that L is a vector, and you are expecting the if-else logic to work on the entire vector simultaneously. That's not how it works. For example
if L < 500
will only be true if all elements are less than 500. (The code won't check each element one-by-one.) This is described in the documentation for if, elseif, else.
So, you could loop over the elements, or you could vectorize explicitly, such as
maxDistLine = 21*(L<500) + 10*((L>=500)&(L<1000)) + 15*(L>1000)
##### 1 件のコメント-1 件の古いコメントを表示-1 件の古いコメントを非表示
Steven D 2022 年 7 月 3 日
Thanks for your input! However, I tried running your suggestion but was given the following error:
Unable to perform assignment because the size of the left side is 1-by-2772 and the size of the right side is 1-by-2688.
Error in TripletGraph2 (line 103)
EID2(:,:,j)=I;
Error in combined_foam_code2all (line 158)
obj=TripletGraph2(fpep);
Attached is the whole code (TripletGraph2) for all to view.
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dpb 2022 年 7 月 3 日
The above if...elseif...else...end construct works; there are other ways -- one way w/ MATLAB is
fnDL=@(L)interp1([500-eps(500),1000-eps(1000),realmax],L,'next','extrap');
using 1D lookup with builtin interp1. The eps() fixes up so the <= condition is met for the breakpoints.
An alternate that's a little more computationally efficient is a recast of the IF construct into a one-liner...
fnDL=@(L)21-3*(L>=500)-3*(L>=1000);
There are any number of ways to rewrite the above as well.
And, of course, the SWITCH construct could be used as well.
I'm sure others will think of many other similar implementations.
##### 8 件のコメント6 件の古いコメントを表示6 件の古いコメントを非表示
dpb 2022 年 7 月 4 日
That just replaces a loop and the IF construct to produce the requested distance value in the code you posted...but your function doesn't return anything it calculates for anything else to be done with it, so it doesn't matter at all what it does until you decide what it is you want/need elsewhere and set up the return values to return the needed-elsewhere variables you're calculating.
Steven D 2022 年 7 月 4 日
Thanks again for your response. Where would you set up return values elsewhere--in the attached code (i.e., 'combined_foam_code2all.m')? If so, where would you put it? Is there a better way to close the polygonal cells (in the images above), then the current approach?
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Translated by | 1,250 | 4,122 | {"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-2024-10 | latest | en | 0.574082 |
http://www.expertsmind.com/questions/tree-30116624.aspx | 1,603,604,994,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107887810.47/warc/CC-MAIN-20201025041701-20201025071701-00197.warc.gz | 132,942,422 | 12,193 | ## tree, Data Structure & Algorithms
Assignment Help:
what is tree
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create a flowchart that displays the students average score for these quizzes
#### Explain divide and conquer algorithms, Explain divide and conquer algorithm...
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Simplifying Assumptions of wire frame representation Neglect colour - consider Intensity: For now we shall forget about colour and restrict our discussion just to the intensi
#### What is a linear array, What is a linear array? An array is a way to re...
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#### Program of implementation of stack using arrays, include int choice, st...
include int choice, stack[10], top, element; void menu(); void push(); void pop(); void showelements(); void main() { choice=element=1; top=0; menu()
#### Write down the procedure to reverse a singly linked list. , Ans: A pr...
Ans: A procedure to reverse the singly linked list: reverse(struct node **st) { struct node *p, *q, *r; p = *st; q = NULL; while(p != NULL) { r =q;
#### Demonstrate that dijkstra''s algorithm, Demonstrate that Dijkstra's algorit...
Demonstrate that Dijkstra's algorithm does not necessarily work if some of the costs are negative by finding a digraph with negative costs (but no negative cost dicircuits) for whi | 402 | 1,788 | {"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-2020-45 | latest | en | 0.712177 |
http://www.gregthatcher.com/Stocks/StockFourierAnalysisDetails.aspx?ticker=BMRN | 1,524,750,150,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948214.37/warc/CC-MAIN-20180426125104-20180426145104-00123.warc.gz | 418,563,852 | 94,879 | Back to list of Stocks See Also: Seasonal Analysis of BMRNGenetic Algorithms Stock Portfolio Generator, and Best Months to Buy/Sell Stocks
# Fourier Analysis of BMRN (BioMarin Pharmaceutical Inc)
BMRN (BioMarin Pharmaceutical Inc) appears to have interesting cyclic behaviour every 96 weeks (3.9955*cosine), 64 weeks (2.9378*sine), and 74 weeks (2.4115*sine).
BMRN (BioMarin Pharmaceutical Inc) has an average price of 37.4 (topmost row, frequency = 0).
Click on the checkboxes shown on the right to see how the various frequencies contribute to the graph. Look for large magnitude coefficients (sine or cosine), as these are associated with frequencies which contribute most to the associated stock plot. If you find a large magnitude coefficient which dramatically changes the graph, look at the associated "Period" in weeks, as you may have found a significant recurring cycle for the stock of interest.
## Fourier Analysis
Using data from 1/3/2000 to 4/23/2018 for BMRN (BioMarin Pharmaceutical Inc), this program was able to calculate the following Fourier Series:
Sequence #Cosine Coefficients Sine Coefficients FrequenciesPeriod
037.39711 0
118.379 -33.58197 (1*2π)/956956 weeks
2-.09695 -22.76949 (2*2π)/956478 weeks
3-5.1955 -4.911 (3*2π)/956319 weeks
4-2.08726 -5.27274 (4*2π)/956239 weeks
51.69848 -.28411 (5*2π)/956191 weeks
61.82134 -2.77068 (6*2π)/956159 weeks
73.97667 -2.95225 (7*2π)/956137 weeks
8-.77137 -7.49397 (8*2π)/956120 weeks
9-1.76754 -4.29319 (9*2π)/956106 weeks
10-3.99548 -.18372 (10*2π)/95696 weeks
111.49714 1.54409 (11*2π)/95687 weeks
121.60912 .46473 (12*2π)/95680 weeks
131.86905 -2.41151 (13*2π)/95674 weeks
14-.47715 -2.28161 (14*2π)/95668 weeks
15-1.56252 -2.93776 (15*2π)/95664 weeks
16-1.64846 -.90013 (16*2π)/95660 weeks
17-.7064 -.13927 (17*2π)/95656 weeks
18.37674 -.43527 (18*2π)/95653 weeks
19-.28362 -.24764 (19*2π)/95650 weeks
20.41294 -.60906 (20*2π)/95648 weeks
21.51619 -1.21078 (21*2π)/95646 weeks
22.15165 -1.74501 (22*2π)/95643 weeks
23-.64134 -1.6724 (23*2π)/95642 weeks
24-.90973 -.5425 (24*2π)/95640 weeks
25-.02087 -.63195 (25*2π)/95638 weeks
26-.09777 -.75648 (26*2π)/95637 weeks
27.01639 -.56119 (27*2π)/95635 weeks
28-.27924 -.49392 (28*2π)/95634 weeks
29-.36593 -.27341 (29*2π)/95633 weeks
30-.2718 -.35076 (30*2π)/95632 weeks
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928-.27924 .49392 (928*2π)/9561 weeks
929.01639 .56119 (929*2π)/9561 weeks
930-.09777 .75648 (930*2π)/9561 weeks
931-.02087 .63195 (931*2π)/9561 weeks
932-.90973 .5425 (932*2π)/9561 weeks
933-.64134 1.6724 (933*2π)/9561 weeks
934.15165 1.74501 (934*2π)/9561 weeks
935.51619 1.21078 (935*2π)/9561 weeks
936.41294 .60906 (936*2π)/9561 weeks
937-.28362 .24764 (937*2π)/9561 weeks
938.37674 .43527 (938*2π)/9561 weeks
939-.7064 .13927 (939*2π)/9561 weeks
940-1.64846 .90013 (940*2π)/9561 weeks
941-1.56252 2.93776 (941*2π)/9561 weeks
942-.47715 2.28161 (942*2π)/9561 weeks
9431.86905 2.41151 (943*2π)/9561 weeks
9441.60912 -.46473 (944*2π)/9561 weeks
9451.49714 -1.54409 (945*2π)/9561 weeks
946-3.99548 .18372 (946*2π)/9561 weeks
947-1.76754 4.29319 (947*2π)/9561 weeks
948-.77137 7.49397 (948*2π)/9561 weeks
9493.97667 2.95225 (949*2π)/9561 weeks
9501.82134 2.77068 (950*2π)/9561 weeks
9511.69848 .28411 (951*2π)/9561 weeks
952-2.08726 5.27274 (952*2π)/9561 weeks
953-5.1955 4.911 (953*2π)/9561 weeks
954-.09695 22.76949 (954*2π)/9561 weeks | 16,566 | 37,393 | {"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-2018-17 | latest | en | 0.74571 |
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TLP Library
Teaching and learning packages (TLPs) are self-contained, interactive resources, each focusing on one area of Materials Science.
Analysis of Deformation Processes
This TLP builds upon the introduction to yield criteria covered in the Stress analysis and Mohr's circle TLP and introduces a range of methods commonly used to study metal forming processes.
Introduction To Anisotropy
It is common in basic analysis to treat bulk materials as isotropic - their properties are independent of the direction in which they are measured. However the atomic scale structure can result in properties that vary with direction. This teaching and learning package (TLP) looks into typical examples of such anisotropy and gives a brief mathematical look into modelling the behaviour.
Avoidance of Crystallization in Biological Systems
This teaching and learning package discusses the two main environmental threats leading to crystallization in plants and animals, and the ways in which organisms have adapted to avoid this crystallization. As part of this discussion, there is coverage of some of the theory of nucleation and crystallization.
Brillouin Zones
This teaching and learning package provides an introduction to Brillouin zones in two and three dimensions and is aimed at developing familiarity with Brillouin Zones. It will not cover any specific applications. Brillouin Zones are particularly useful in understanding the electronic and thermal properties of crystalline solids.
Brittle Fracture
What determines when a material will break, and whether failure will be catastrophic or more gradual. Cracking is controlled by the energy changes that occur - it is not the stress at the crack tip that is important..
Casting
This TLP introduces a number of important processes through which metallic items can be fabricated from molten metal. As well as detailing the practical aspects of these manufacturing processes, attention is given to the important parameters which determine the microstructure of the finished items.
Creep Deformation of Metals
Creep is a major concern in engineering, since it can cause materials to fail well below their yield stress. This package outlines the mechanisms of creep and the associated equations. It is largely based around a first year Materials Science practical at the University of Cambridge, which is concerned with the creep of solder at different temperatures. It also includes a case study of a creep-resistant material to illustrate how materials can be designed to prevent creep.
Introduction To Deformation Processes
This teaching and learning package covers the fundamentals of metal forming processes.
Dielectric Materials
This teaching and learning package will introduce you to the properties and uses of dielectric materials.
Diffusion
An introduction to the mechanisms and driving forces of diffusion, and some of the processes in which it is observed.
Introduction To Dislocations
Dislocations are crucially important in determining the mechanical behaviour of materials. This teaching and learning package provides an introduction to dislocations and their motion through a crystal. A 'bubble raft' model is used to demonstrate some of the features of dislocations and other lattice defects. Some methods for observing real dislocations in materials are examined.
Electromigration
Electromigration is an ever-increasing problem as integrated circuits are pushed towards further miniaturization. The theory of the phenomenon is explained, including electromigration-induced failure and how it has been and can be minimized.
Examination of a Manufactured Article
This TLP provides an introduction to the deconstruction and investigation of the materials and processes used in an everyday item or article.
Ferroelectric Materials
Ferroelectrics are an important device in today's world. They are useful both as capacitors, for example in camera flashes, or as non-volatile memory storage. The memory use of which you are most likely to be aware is in the Playstation 2.
The Jominy End Quench Test
Discusses the aims, method and use of results of a test for the hardenability of steel.
Kinetics of Aqueous Corrosion
This teaching and learning package (TLP) introduces the mechanism of aqueous corrosion and the associated kinetics.
Introduction To Mechanical Testing
This teaching and learning package is based on laboratory experiments used in the Department of Materials Science and Metallurgy at the University of Cambridge. The package looks at how the microstructure of a material can affect its properties. It is split into two experiments: the first part introduces tensile testing and stress-strain curves, while the second part uses three-point bending, as introduced in the Beam Stiffness TLP.
The Nernst Equation and Pourbaix Diagrams
This teaching and learning package (TLP) investigates the Nernst equation and Pourbaix diagrams, which are both important parts of electrochemistry and corrosion science.
Phase Diagrams and Solidification
Phase diagrams are a useful tool in metallurgy and other branches of materials science. They show the mixture of phases present in thermodynamic equilibrium. This teaching and learning package looks at the theory behind phase diagrams, and ways of constructing them, before running through an experimental procedure, and presenting the results which can be obtained.
Piezoelectric Materials
This teaching and learning package (TLP) provides an introduction to piezoelectric materials.
Pyroelectric Materials
Pyroelectric materials are found in almost every home, in the form of intrusion detectors and other devices, and this TLP will consider how they work, and what the most common ones are made of.
Introduction To Semiconductors
This teaching and learning package provides a very basic introduction to semiconductors. These materials are essential to the operation of solid state electronic devices.
Slip in Single Crystals
This teaching and learning package explains how plastic deformation of materials occurs through the mechanism of slip. Slip involves dislocation glide on particular slip planes. The geometry of slip is explained, and electron microscopy techniques are used to show slip occurring in single crystals of cadmium.
Solid Solutions
This teaching and learning package is based on a practical used within the Department of Materials Science and Metallurgy at the University of Cambridge. The package is aimed at first year undergraduate Materials Science students and focuses on the different types of solid solution and the thermodynamic principles involved in understanding them.
Solidification of Alloys
This teaching and learning package (TLP) is an introduction to how solute affects the solidification of metallic alloys.
Stress Analysis and Mohr's Circle
This teaching and learning package provides an introduction to the theory of metal forming. It discusses how stress and strain can be presented as tensors, and ways of identifying the principal stresses. Suitable yield criteria to treat metals and non-metals are also presented.
Superconductivity
Electrons in pairs? Levitating trains? Superconductivity - the combination of lossless electrical conduction and the ability of a material to expel a magnetic field - is a property that excites interest in fundamental science whilst offering tantalising prospects for a range of applications. In this teaching and learning package (TLP), we trace the history of superconductivity, outline some fundamental properties of superconductors, and describe current and potential applications of materials with this unusual property.
Superelasticity and Shape Memory Alloys
This teaching and learning package (TLP) introduces the phenomena of superelasticity and the shape memory effect.
Tensors
This TLP offers an introduction to the mathematics of tensors rather than the intricacies of their applications. Its aims are to familiarise the learner with tensor notation, how they can be constructed and how they can be manipulated to give numerical answers to problems. | 1,518 | 8,068 | {"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-2013-20 | latest | en | 0.910389 |
https://community.articulate.com/discussions/articulate-storyline/relationship-between-timeline-length-and-animation-duration | 1,701,527,399,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100427.59/warc/CC-MAIN-20231202140407-20231202170407-00128.warc.gz | 210,767,974 | 11,564 | # Relationship between timeline length and animation duration?
Mar 25, 2019
I'm having trouble understanding the relationship between an animation (entrance) duration and the length of the timeline. For example, my time line is 10 seconds long, and I want to animate (fly in) 4 bulleted text paragraphs one by one (by paragraph). If I use the default animation duration of 0.75 secs, the bullets fly in rapidly. But if want to slow down the pace of entrance, it doesn't seem to matter if I choose 2 secs or 20 secs duration; the bullets fly in at the same slow pace, ending later than I want them to. Is the duration constrained or overridden by the timeline length? If so, what is the logic? How do I get my bullets to stop flying in at a specific point on the timeline? (I want other layers to fly in after the bullets stop.)
###### 6 Replies
Hi Mark!
When you animate a bulleted list By Paragraph, you have the option to customize the start time of each bullet. Simply click the dropdown arrow in the timeline as shown below, then click and drag each bullet to the desired start time.
I hope that helps you!
Alyssa,
Thanks, that's a good tip that I discovered in the last few minutes, and I think I've been making a mistake by trying to adjust duration in the animation pane. I'm now keeping the animation entrance duration at 0.75 secs but using the trick you showed above to control pacing. But I'm still wondering...
1. Why is 2 seconds duration no faster than say, 6, 10 or 20 seconds duration? Is animation duration essentially ignored for speeds over 2 sec when animating text "by paragraph"?
2. If I want the bullets to remain on the screen until the end of the timeline, how do I make sure the last bullet stops animating before a new layer appears (in a separate animation)?
(I think I might have answered my own question: if I keep the entrance duration short enough (0.75 sec) the final bullet will stop its entrance well before the new layer flies in.)
Hi there, Mark. We have an issue where the animation duration is not being respected when an object is animated by paragraph rather than as one object. I'm sorry this bug is tripping up your timing.
I'm adding this discussion to our report so we can notify you of any changes. Since the workaround is to set the animation as one object, you could consider making each bullet its own text box. I'm not sure if that's more work than it's worth for you, but I wanted to offer!
Hi again, Mark. I tested this in our latest version of Storyline 360, and the animation duration worked properly in a bulleted text box. It looks like you are also on update 26; with your permission, I'd like to take a look at your .story file to investigate what's happening. You can share it publicly here, or send it to me privately by uploading it here. I'll delete it when I'm done troubleshooting.
Thanks again!
Hi Crystal. For my purposes, I've concluded that for bulleted text it's a best practice to use the default 0.75 sec animation entrance duration and individually adjust bullet timing using the trick in Alyssa's video. No need to monkey with longer entrance durations since (I realize now) it's bad usability to have text float in too slowly. And with the quick bullet entrances, getting my other layers to come at the right time is no longer a worry.
When using longer entrance durations (2, 6, 10, 20 secs), I suspect that that the timeline length exerts some sort of dominance/takes precedence over the animation entrance duration, so I'd still be curious to understand the logic and get confirmation on that. But for my purposes, I don't foresee a need for such long entrance durations, so there's no need to send a file for troubleshooting.
Sounds good, Mark! We'll be here if you need more help. | 856 | 3,778 | {"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-2023-50 | longest | en | 0.919176 |
https://asoliduniverse.com/ | 1,726,239,826,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651523.40/warc/CC-MAIN-20240913133933-20240913163933-00745.warc.gz | 91,264,987 | 28,148 | pROBABILITY
Welcome everybody today to a talk on Probability and Gambling.
A subject I have more than a passion for as an occasional mathematician, a card player and gamer, an occasional not very successful gambler and as a long-term small stock market investor.
Everyone here has used the concept of Probability Theory countless times in their lives, usually successfully*, with an innate understanding of the ground rules despite no formal teaching in the matter. The discussion today may or may not improve your use of it but could add some interest to your lives in future.
We live in what some have described as a Goldilocks world +, Yet one that is also full of pitfalls and danger
What is Probability? A simple maze of contradictions and complexities.
Horatio
The time keeper
There are more things in heaven and earth, Horatio, Than are dreamt of In your philosophy.
Possibility is pure imagination, the stuff of dreams with no real-life outcome or occurrence.
Practicality is an event or outcome or real life happening. Probability however is the science of imagination turning into reality Probability is a unique science, the only science that cannot be proved or falsified. “In theory, there is no difference between practice and theory. In practice, there is.“ Yogi Berra
Probability is the likelihood that something is to happen expressed in percentage terms. It is the difference between practice and theory. It is a consideration of the chance of any event we care to consider happening in the future. People have played games of chance for virtually all of written history. The Egyptians played with dice around 3000 B.C.
As a science probability theory has always been intricately associated with gaming and gambling.
Yet “God may not play dice with the universe”- Albert Einstein,
Any event, happening, an object and a time all have this uniqueness in common in our world.
The probability of their existence or non-existence always sums up to one.
In Probability the occurrence of an event is called P and the non-occurrence 1-P.
Take an apple on a table . It is either there or not there.
Determining the chance of it being there is complicated by many factors beyond our control.
If we can stipulate the factors, we can estimate that probability for the factors that we know.
The golden rule is that a probability must always lie somewhere between 0 ,non existing, and 1, completely existing.
To this end we often describe it in terms of percentage as a percentage is usually expressed in terms of a hundredth of one.
Originally Probability found application as strategy in games of chance then in all forms of gaming and warfare. Virtually every aspect of business life and activity from the stock market to pensions , annuities and life insurance use it. It has found niches in every field of scientific testing. Weather prediction, shipping and air traffic control and even in store inventories. In the personal field it can even help you find the perfect match in life. It is paramount in the medical field in both evaluating the success of drugs in drug trials but also in diagnosis of cancers. Offshoots include medical equipment like CT scanners and MRI’s. Now we have the advent of AI. Sadly, back to warfare, the Terminator is approaching reality.
The use of the numbers 1-9 came from India to the Arab world, modified by adding in the number zero. It reached Spain and Italy by 1000 A.D Probability has only developed as a science in the last 400 years when European mathematicians were able to consider it and subjects such as Gravity, calculus, the movement of the planets with mathematics
In 1654 Antoine Gombaud, Chevalier deMere, a French nobleman was puzzled by an apparent contradiction concerning a popular dice game. The game consisted in throwing a pair of dice 24 times; Would betting on a double six in 24 throws be profitable? He was trying to establish if such an event has probability greater than 0.5.
Puzzled by this and other similar gambling problems he called the attention of the famous mathematician Blaise Pascal. In turn this led to an exchange of letters between Pascal and French mathematician Pierre de Fermat. Probability science was born. The first documented evidence of the fundamental principles of the theory of probability.
“The basic bit is that most events [choices] in life are binary. A Bernoulli trial (or binomial trial) is a random experiment with exactly two possible outcomes, “success” and “failure”, in which the probability of success is the same every time the experiment is conducted. Either/ or, The fire or the frying pan, flight or fright. Heads or tails, sink or swim. There is a principle called Occam’s Razor, a razor being a sharp simple choice ”The simplest explanation is usually the best one” Events exist in a sample space. This can be discrete, continuously discrete or continuous. The number of probable events that can occur in a sample space at any one time must always add up to one. When sample spaces overlap as in having different times and different numbers of events the same rule applies. In all simple or basic probabilities decision tree diagrams are used to help work out individual choice ratios and total number of outcomes per run of events. Large data sets take massive amounts of time to work out tree branch solutions . Listing all possible outcomes from tossing a coin 64 times in a row leads to a gigantic number> Immortalised in the story of a grain of wheat on a chessboard doubled every square.* 2 to the power of 64 = 18,446,744,073,709,551,616. 2 to the power of 63 = 9,223,372,036,854,775,808.
Baye’s theorem “New information should be given proportional rather than equal status initially.” Or don’t throw the baby out with the bath water. One man, Bayes , came up with the idea that led to Probability being able to be used for real life situations involving many different choices occurring randomly. This allows the original data to combine with new data without making large new adjustments. Baye’s Theorem, properly applied, enables Probability predicting easily on complex data bases when new information is provided. This then enables previous predictions to be altered and extended as in weather forecasting. It helps improve the accuracy hence safety of the standard deviation ranges for structures like Bridges, roads and aeroplane wings. It enables greater accuracy in prediction of health outcomes and life expectancies It allows continual modification in the direction of the new data without changing the thrust of the original hard-earned data. Bayes’ theorem (alternatively Bayes’ law or Bayes’ rule), named after Thomas Bayes, describes the probability of an event, based on prior knowledge of conditions that might be related to the event.[1]
Probability events coexist with a new style of maths called Set Theory. A schematic effective in dealing with multiple events occurring over multiple fields. Some events are dependent on previous events. Others occur in the same time frames with no dependence on each other.
I am not even going to try to explain Set theory rules like -P[A] plus P[A*] =1 the probability of something occurring and not occurring equals one. – P[A] plus P[B]= P[A]XP[B] -P[A] to….. P[X] =1 -P[A] given P[B] [dependent] = P[A] +P[B] Set theory and Venn diagrams differ in the rules of addition and multiplication because they deal with percentages in a fixed range of 1.
Are you lucky? Seriously? Have you won the lottery? Do good things happen to you in life?
When we study Probability, we unconsciously assign merit to one side of a choice and detriment to the other. Having a windfall, attaining a good job, having a happy childhood and happy family when we grow up. Avoiding illness and debt. Luck appears to be a random quality and Lady Luck chooses whomsoever she wants.
Phil Bradbury winning his gold medal in skating for Australia. Finding a Picasso in a second-hand shop. Having a medical event just after retiring.
Luck, good luck, is beating the odds the odds in a desirable way. In other words moving to the right side of the average outcome. But is it truly random?
Events are random, people are random, and outcomes seem random but the degree to which probability rules in our lives can be altered by our own actions. “The more I practice, the luckier I get.” Used by Gary Player The choices we make, whether we have free will or not, influence both the direction we go in and the outcomes we achieve.
We initiate those choices by choosing whether or not to take action, how much and where we apply it. This does not guarantee success, but it increases the chance of success.
The fellow who finished second to Bradbury and Bradbury himself took steps to try to win a medal. They spent years practicing and honing their skills. They had genetic skills that let them skate extremely well. They were born at the right time to compete in an Olympic Games at a good racing age. Probability dictated that Bradbury should lose. Luck decided that he reached a pinnacle first.
Nothing is permanent except the fact of change’
To be, or not to be, that is the question: Whether ’tis nobler in the mind to suffer The slings and arrows of outrageous fortune, Or to take Arms against a Sea of troubles, And by opposing end them:
If luck is a matter of common sense and application why isn’t everyone happy and lucky and successful? The answer is that other people are also trying to improve their lives and their odds in the great Probability race of life. This reduces the chances of others who are competing for the same goals. One of the rules of physics is for each force there is an equal and opposing force. Another is that nature abhors a vacuum. Basically every system achieves a balance where it can, a stable or comfortable position where all those forces of randomness settle down and a status quo is established. This balance is always at risk from external and internal forces. Taking the earth as an example it could be hit by a meteor or have a volcanic eruption. In the societies we live in, we can affect both ourselves and others with our actions. When we take action improving our odds we reduce the odds for someone else. Since they are capable of taking action themselves there will be a push back effect. There is also collateral improvement and damage to others of a lesser nature. the ripples and butterfly wings we might see and cannot see.
The world
The earth we live on has three usual outer layers.
The atmosphere in gas phase, the oceans etc in liquid phase and the crust in solid mineral phase or as ice and snow.
It has an inbuilt source of heat which is important.
The vast majority of the heat present at the surface comes externally from the sun.
The minerals making up the earth are the original source of the atmosphere and the oceans.*
* some people claim meteorites are an additional source of water.
The earth is basically a hot meteorite slightly cooling down.
If we could imagine it a lot further away from the sun in orbit we would find it to have very little atmosphere as most of the oxygen and nitrogen would be frozen as a layer on the surface with the oceans as solid ice.*
* different scenarios exist.
The first and most important comment is that the earth, and meteorites have a pH depending on their mineral composition which for the earth is around pH 8.1.
When the temperature increases ( planet closer to the sun in our case) water becomes liquid on top of the solid mineral surface and engages in chemical reactions which lead to it equilibrating with the pH of the surface of the earth in general.*
*pH varies according to the minerals and compounds both in the water and in contact with the water..
The second comment is that when water is present a third gaseous layer develops from the large amount of gases given off by the warming water.
This is far greater than any trace gas atmosphere on a water less solid planetoid such as on the surface of the moon.
The gases in the atmosphere are present as per Boyle’s law each by how much is dissolved in the water at that temperature and pressure from the solids presented by the earth..
Oxygen has a special place as it would normally be in mineral or part of water only on a cold lifeless planet.
The formulae for determining the amount of CO2 in the air are quite clear.
CO2 in the air is present in minuscule amounts compared to CO2/H2CO3 various forms and CaCO3 in water.
In turn the earth has massive amounts of CaCO3 and other Carbonates not only in its surface layer but also deeper.
The earth pH 8.1 is in equilibrium with the water pH 8.1 overall.*
*obviously pH is constantly varying depending on temperature and depth. Being in equilibrium overall does not mean it is the same everywhere at the same time.
The CO2 in the atmosphere has been there for over the last two billion years. It comes from the water dissolving carbonates when it is warm enough to do so.
The water then keeps an average 400 ppm in the atmosphere
At a yearly average surface temperature of 14.9 C at 1 atmosphere of pressure.
Monckton
dikranmarsupial says:
February 4, 2023 at 4:53 pm
” Monckton has an algorithm for cherry picking the start point, but it is still cherry picking. His algorithm is selecting the start point that maximises the strength of his argument (at least for a lay audience that doesn’t understand the pitfalls).”
DM this is not correct.
ATTP and Willis have used algorithms for most but one of their pauses but Monckton never has.
The algorithms incorporating trends are specific to the charts and data used.
One can cherry pick the length of which one of the steps one wants to use,
but, importantly one cannot rig a flat trend.
Because Monckton uses a pause, a new pause, this can only go to the end point of the current date and changes from the new date.
Thus he has never selected start point that maximizes the strength of his argument.
The fact that the pause can lengthen or shorten means he has never cherry picked a starting point.
Willis, unlike ATTP shows part of a truly long pause from 1997 to 2012.
ATTP breaks it down to two different pauses by incorporating different start and end dates.
Interesting.
” Harrison Bergeron ” by Kurt Vonnegut.
https://scifi.stackexchange.com › questions › 176944 › short-story-where-everyone-must-be-equal
Short story where everyone must be equal – Science Fiction & Fantasy …
That would be ” Harrison Bergeron ” by Kurt Vonnegut. THE YEAR WAS 2081, and everybody was finally equal. They weren’t only equal before God and the law. They were equal every which way. Nobody was smarter than anybody else. Nobody was better looking than anybody else. Nobody was stronger or quicker than anybody else.
HARRISON BERGERON by Kurt Vonnegut, Jr.
THE YEAR WAS 2081, and everybody was finally equal. They weren’t only equal
before God and the law. They were equal every which way. Nobody was smarter
than anybody else. Nobody was better looking than anybody else. Nobody was
stronger or quicker than anybody else. All this equality was due to the
211th, 212th, and 213 th Amendments to the Constitution, and to the unceasing
vigilance of agents of the United States Handicapper General.
Some things about living still weren’t quite right, though. April for
instance, still drove people crazy by not being springtime. And it was in
that clammy month that the H-G men took George and Hazel Bergeron’s fourteen-
year-old son, Harrison, away.
It was tragic, all right, but George and Hazel couldn’t think about it very
hard. Hazel had a perfectly average intelligence, which meant she couldn’t
think about anything except in short bursts. And George, while his
intelligence was way above normal, had a little mental handicap radio in his
ear. He was required by law to wear it at all times. It was tuned to a
government transmitter. Every twenty seconds or so, the transmitter would
send out some sharp noise to keep people like George from taking unfair
George and Hazel were watching television. There were tears on Hazel’s
cheeks, but she’d forgotten for the moment what they were about.
On the television screen were ballerinas.
A buzzer sounded in George’s head. His thoughts fled in panic, like bandits
from a burglar alarm.
“That was a real pretty dance, that dance they just did,” said Hazel.
“Huh” said George.”That dance-it was nice,” said Hazel.
“Yup,” said George. He tried to think a little about the ballerinas. They
weren’t really very good-no better than anybody else would have been, anyway.
They were burdened with sashweights and bags of birdshot, and their faces
were masked, so that no one, seeing a free and graceful gesture or a pretty
face, would feel like something the cat drug in. George was toying with the
vague notion that maybe dancers shouldn’t be handicapped. But he didn’t get
very far with it before another noise in his ear radio scattered his
thoughts.
George winced. So did two out of the eight ballerinas.
Hazel saw him wince. Having no mental handicap herself, she had to ask George
what the latest sound had been.
“Sounded like somebody hitting a milk bottle with a ball peen hammer,” said
George.
“I’d think it would be real interesting, hearing all the different sounds,”
said Hazel a little envious. “All the things they think up.”
“Um,” said George.
“Only, if I was Handicapper General, you know what I would do?” said Hazel.
Hazel, as a matter of fact, bore a strong resemblance to the Handicapper
General, a woman named Diana Moon Glampers. “If I was Diana Moon Glampers,”
said Hazel, “I’d have chimes on Sunday-just chimes. Kind of in honor of
religion.”
“I could think, if it was just chimes,” said George.
“Well-maybe make ’em real loud,” said Hazel. “I think I’d make a good
Handicapper General.” “Good as anybody else,” said George.
“Who knows better then I do what normal is?” said Hazel.
“Right,” said George. He began to think glimmeringly about his abnormal son
who was now in jail, about Harrison, but a twenty-one-gun salute in his head
stopped that.
“Boy!” said Hazel, “that was a doozy, wasn’t it?”
It was such a doozy that George was white and trembling, and tears stood on
the rims of his red eyes. Two of of the eight ballerinas had collapsed to the
studio floor, were holding their temples.
“All of a sudden you look so tired,” said Hazel. “Why don’t you stretch out
on the sofa, so’s you can rest your handicap bag on the pillows, honeybunch.”
She was referring to the forty-seven pounds of birdshot in a canvas bag,
which was padlocked around George’s neck. “Go on and rest the bag for a
little while,” she said. “I don’t care if you’re not equal to me for a
while.”
George weighed the bag with his hands. “I don’t mind it,” he said. “I don’t
notice it any more. It’s just a part of me.”
“You been so tired lately-kind of wore out,” said Hazel. “If there was just
some way we could make a little hole in the bottom of the bag, and just take
out a few of them lead balls. Just a few.”
“Two years in prison and two thousand dollars fine for every ball I took
out,” said George. “I don’t call that a bargain.”
“If you could just take a few out when you came home from work,” said Hazel.
“I mean-you don’t compete with anybody around here. You just set around.”
“If I tried to get away with it,” said George, “then other people’d get away
with it-and pretty soon we’d be right back to the dark ages again, with
everybody competing against everybody else. You wouldn’t like that, would
you?”
“I’d hate it,” said Hazel.
“There you are,” said George. The minute people start cheating on laws, what
do you think happens to society?”
If Hazel hadn’t been able to come up with an answer to this question, George
couldn’t have supplied one. A siren was going off in his head.
“Reckon it’d fall all apart,” said Hazel.
“What would?” said George blankly.
“Society,” said Hazel uncertainly. “Wasn’t that what you just said?
“Who knows?” said George.
The television program was suddenly interrupted for a news bulletin. It
wasn’t clear at first as to what the bulletin was about, since the announcer,
like all announcers, had a serious speech impediment. For about half a
minute, and in a state of high excitement, the announcer tried to say,
He finally gave up, handed the bulletin to a ballerina to read.
“That’s all right-” Hazel said of the announcer, “he tried. That’s the big
thing. He tried to do the best he could with what God gave him. He should get
a nice raise for trying so hard.”
have been extraordinarily beautiful, because the mask she wore was hideous.
And it was easy to see that she was the strongest and most graceful of all
the dancers, for her handicap bags were as big as those worn by two-hundred
pound men.
And she had to apologize at once for her voice, which was a very unfair voice
for a woman to use. Her voice was a warm, luminous, timeless melody. “Excuse
me-” she said, and she began again, making her voice absolutely
uncompetitive.
“Harrison Bergeron, age fourteen,” she said in a grackle squawk, “has just
escaped from jail, where he was held on suspicion of plotting to overthrow
the government. He is a genius and an athlete, is under-handicapped, and
should be regarded as extremely dangerous.”
A police photograph of Harrison Bergeron was flashed on the screen-upside
down, then sideways, upside down again, then right side up. The picture
showed the full length of Harrison against a background calibrated in feet
and inches. He was exactly seven feet tall.
The rest of Harrison’s appearance was Halloween and hardware. Nobody had ever
born heavier handicaps. He had outgrown hindrances faster than the H-G men
could think them up. Instead of a little ear radio for a mental handicap, he
wore a tremendous pair of earphones, and spectacles with thick wavy lenses.
The spectacles were intended to make him not only half blind, but to give him
Scrap metal was hung all over him. Ordinarily, there was a certain symmetry,
a military neatness to the handicaps issued to strong people, but Harrison
looked like a walking junkyard. In the race of life, Harrison carried three
hundred pounds.
And to offset his good looks, the H-G men required that he wear at all times
a red rubber ball for a nose, keep his eyebrows shaved off, and cover his
even white teeth with black caps at snaggle-tooth random.
“If you see this boy,” said the ballerina, “do not – I repeat, do not – try
to reason with him.”
There was the shriek of a door being torn from its hinges.
Screams and barking cries of consternation came from the television set. The
photograph of Harrison Bergeron on the screen jumped again and again, as
though dancing to the tune of an earthquake.
George Bergeron correctly identified the earthquake, and well he might have –
for many was the time his own home had danced to the same crashing tune. “My
God-” said George, “that must be Harrison!”
The realization was blasted from his mind instantly by the sound of an
When George could open his eyes again, the photograph of Harrison was gone. A
living, breathing Harrison filled the screen.
Clanking, clownish, and huge, Harrison stood – in the center of the studio.
The knob of the uprooted studio door was still in his hand. Ballerinas,
technicians, musicians, and announcers cowered on their knees before him,
expecting to die.
“I am the Emperor!” cried Harrison. “Do you hear? I am the Emperor! Everybody
must do what I say at once!” He stamped his foot and the studio shook.
“Even as I stand here” he bellowed, “crippled, hobbled, sickened – I am a
greater ruler than any man who ever lived! Now watch me become what I can
become!”
Harrison tore the straps of his handicap harness like wet tissue paper, tore
straps guaranteed to support five thousand pounds.
Harrison’s scrap-iron handicaps crashed to the floor.
Harrison thrust his thumbs under the bar of the padlock that secured his head
harness. The bar snapped like celery. Harrison smashed his headphones and
spectacles against the wall.
He flung away his rubber-ball nose, revealed a man that would have awed Thor,
the god of thunder.
“I shall now select my Empress!” he said, looking down on the cowering
people. “Let
the first woman who dares rise to her feet claim her mate and her throne!”
A moment passed, and then a ballerina arose, swaying like a willow.
Harrison plucked the mental handicap from her ear, snapped off her physical
handicaps with marvelous delicacy. Last of all he removed her mask.
She was blindingly beautiful.
“Now-” said Harrison, taking her hand, “shall we show the people the meaning
of the word dance? Music!” he commanded.
The musicians scrambled back into their chairs, and Harrison stripped them of
their handicaps, too. “Play your best,” he told them, “and I’ll make you
barons and dukes and earls.”
The music began. It was normal at first-cheap, silly, false. But Harrison
snatched two musicians from their chairs, waved them like batons as he sang
the music as he wanted it played. He slammed them back into their chairs.
The music began again and was much improved.
Harrison and his Empress merely listened to the music for a while-listened
gravely, as though synchronizing their heartbeats with it.
They shifted their weights to their toes.
Harrison placed his big hands on the girls tiny waist, letting her sense the
weightlessness that would soon be hers.
And then, in an explosion of joy and grace, into the air they sprang!
Not only were the laws of the land abandoned, but the law of gravity and the
laws of motion as well.
They reeled, whirled, swiveled, flounced, capered, gamboled, and spun.
They leaped like deer on the moon.
The studio ceiling was thirty feet high, but each leap brought the dancers
nearer to it.
It became their obvious intention to kiss the ceiling. They kissed it.
And then, neutraling gravity with love and pure will, they remained suspended
in air inches below the ceiling, and they kissed each other for a long, long
time.
It was then that Diana Moon Glampers, the Handicapper General, came into the
studio with a double-barreled ten-gauge shotgun. She fired twice, and the
Emperor and the Empress were dead before they hit the floor.
Diana Moon Glampers loaded the gun again. She aimed it at the musicians and
told them they had ten seconds to get their handicaps back on.
It was then that the Bergerons’ television tube burned out.
Hazel turned to comment about the blackout to George. But George had gone out
into the kitchen for a can of beer.
George came back in with the beer, paused while a handicap signal shook him
up. And then he sat down again. “You been crying” he said to Hazel.
“Yup,” she said. “What about?” he said.
“I forget,” she said. “Something real sad on television.”
“What was it?” he said.
“It’s all kind of mixed up in my mind,” said Hazel.
“I always do,” said Hazel.
“That’s my girl,” said George. He winced. There was the sound of a rivetting
“Gee – I could tell that one was a doozy,” said Hazel.
“You can say that again,” said George.
“Gee-” said Hazel, “I could tell that one was a doozy.”
“Harrison Bergeron” is copyrighted by Kurt Vonnegut, Jr., 1961.
Angech’s immodest game
angech | October 2, 2022 at 10:04 pm |
For Willard
Angech’s immodest game
A French Defense against a modest AI.
open source Garbochess-JS
Angech as black.
1 e4 e6 2 d4 d5 3 e5 c5 4 exc5 Nc6
5 Nc3 Qh4 6 Nf3 Qb4 7 a3 Qxc5 8 Bb5 a6
9 Be3 Qe7 10 Bd3 QC7 11 Bf4 f5! 12 0-0 Nf6!
13 Qe1 Ne4 14 h3 h6 15 g4 g6 16 gx5f gxf5
17 Qe2 Rg8 18 Kh8 Bc5 19 Na4 Ba7 20 Nc3 Ng5
21Nxg5 Bd7 22 Qe1 0-0-0 22 Ka1 Bc5 23Na4 Ba7
24 Nc3 Ng5! 26 Nxg5 h6xg5 27 Bg5 Rxh3 28Kb2 Nxe5
29BXd8 Qxd8 30 Kxh3 … leading to a forced checkmate starting with Qh8
then a unique Q march down the board 1 square at a time
Qg7, Qh6, Qg5, Qh5, Qg4, Qh4, Qg4, and finish off with Qh3 if Kh1 then Nf3
or Nf3 if Kh2 then Qh3
The opening is blah
The mid game enterprising.
The ending is fantastic. Hope people appreciate it.
At the top-of-atmosphere (TOA]
The average global net radiation at the top of the atmosphere (TOA) is defined as the difference between the energy absorbed and emitted by the planet.
In an equilibrium climate state, the global net radiation at the TOA is zero.
In the presence of an increasing climate forcing, an imbalance between the energy absorbed and emitted occurs,
and in response the climate system must react to restore the balance (e.g., by changing temperature).
The rate at which the earth reacts is modulated by its capacity to store energy.
Given that oceans are 10 times more efficient at storing heat than other components of the climate system (e.g., land, ice, atmosphere; Levitus et al. 2001),
the global net radiation at the TOA should be in phase with and of similar magnitude as the global ocean heat storage.
At the top-of-atmosphere (TOA), the Earth’s energy budget involves a balance between how
much solar energy Earth absorbs and how much terrestrial thermal infrared radiation is emitted to space.
Since only radiative energy is involved, this is also referred to as Earth’s radiation budget (ERB). NG Loeb, W Su et al 2016
A natural balance exists in the Earth system between incoming solar radiation and outgoing radiation that is emitted back to space as either light (direct reflection of sunlight)
or heat (infrared emission from surfaces).
This balance, referred to as Earth’s radiation budget (ERB), determines the climate of the Earth and makes our planet hospitable for life.
Chemical Sciences Laboratory NOAA
Earth’s Energy Budget
The TOA ERB describes the balance between how much solar energy the Earth absorbs and how much terrestrial thermal infrared radiation it emits.
N.G. Loeb, … W.F. Miller, in Comprehensive Remote Sensing, 2018
The average global net radiation at the top of the atmosphere (TOA) is defined as the difference between the energy absorbed and emitted by the planet.
In an equilibrium climate state, the global net radiation at the TOA is zero.
In the presence of an increasing climate forcing, an imbalance between the energy absorbed and emitted occurs,
and in response the climate system must react to restore the balance (e.g., by changing temperature).
The rate at which the earth reacts is modulated by its capacity to store energy.
Given that oceans are 10 times more efficient at storing heat than other components of the climate system (e.g., land, ice, atmosphere; Levitus et al. 2001),
the global net radiation at the TOA should be in phase with and of similar magnitude as the global ocean heat storage.
At the top-of-atmosphere (TOA), the Earth’s energy budget involves a balance between how
much solar energy Earth absorbs and how much terrestrial thermal infrared radiation is emitted to space.
Since only radiative energy is involved, this is also referred to as Earth’s radiation budget (ERB). NG Loeb, W Su et al 2016
A natural balance exists in the Earth system between incoming solar radiation and outgoing radiation that is emitted back to space as either light (direct reflection of sunlight)
or heat (infrared emission from surfaces).
This balance, referred to as Earth’s radiation budget (ERB), determines the climate of the Earth and makes our planet hospitable for life.
Chemical Sciences Laboratory NOAA
Earth’s Energy Budget
The TOA ERB describes the balance between how much solar energy the Earth absorbs and how much terrestrial thermal infrared radiation it emits.
N.G. Loeb, … W.F. Miller, in Comprehensive Remote Sensing, 2018
For ATTP on origin of galaxies
angech says:
July 20, 2022 at 6:35 am
izen says: July 18, 2022 at 2:16 pm
“One is the comment that we can see galaxies 13.8 billion years ago.
No one explains how the further out we look we find galaxies and stars That we compare to our own even though they have not existed as such for 13.8 billion years.”
The universe is much larger than 13.8 billion years across due to space expansion, the most recent estimate I have seen is around 95 billion light years.”
Size of the visible universe is partly dependent on how good our “optics” are.
If something is out there 14.2 billion light years away we probably would not see [detect ]it with our current science.
As DM said “How bright would a mega star have to be”.
Even other universes from other big bangs [if we consider our “universe” to have a single origin would be hard pressed to trouble the cosmic microwave background radiation let alone be seen.
The size of the visible universe is thus only double the 13.8 billion years, 27.6 billion years.
Since it has been expanding at less than the speed of light [caveat] the actual universe would be only perhaps 54 billion years old at the moment.
“The universe is around 95 billion light years across.” is an estimate based on maths and physics theories and until everyone agrees on those we might be better sticking to the speed of light time and distance observations.
Thank you for putting it and the concept up.
The big bang itself had so much matter and energy that our current concepts of time and the speed of light back then go out the window.
angech says:
July 20, 2022 at 6:56 am
…and Then There’s Physics says: July 18, 2022 at 7:29 am
“As I understand it, the galaxies in the galaxy cluster that was imaged by JWST were about 4.6 billion years old”.
The articles being written by journalists seem to be conflating such galaxies with the concept of being able to look back a lot further in time.
The original hot explosion or event being that long ago that the first clumps of plasma for want of a better word were supposed to form mega stars of very short life span which threw out clumps of matter [including some heavier elements than H, He] to form the original galaxies which were also very large [hence visible faintly] and then possibly another two iterations to get to our young star and young galaxy.
The materials greater than iron on the periodic table, gold being the best example, are thought to have come from arcane processes in past supernovae.
It is hard to imagine our sun being the remnant of a 13.8 billion year chunk of hop plasma cooling down over that length of time.
Further such explosions cause escape speed velocities which mean that the galaxies should never have come back together.
An alternative view is that space was filled with large amounts of cooled down matter in waves of explosions that crossed each other causing focal points of reaccumulation resulting in newer smaller galaxies.
This would explain suns forming from gigantic masses of cold hydrogen, etc hitting or passing through each other leaving focal eddies of matter which could then coalesce to form suns and planets.
and how elliptical orbits of planets and stars can come into being.
July 20, 2022 at 6:35 am
izen says: July 18, 2022 at 2:16 pm
“One is the comment that we can see galaxies 13.8 billion years ago.
No one explains how the further out we look we find galaxies and stars That we compare to our own even though they have not existed as such for 13.8 billion years.”
The universe is much larger than 13.8 billion years across due to space expansion, the most recent estimate I have seen is around 95 billion light years.”
Size of the visible universe is partly dependent on how good our “optics” are.
If something is out there 14.2 billion light years away we probably would not see [detect ]it with our current science.
As DM said “How bright would a mega star have to be”.
Even other universes from other big bangs [if we consider our “universe” to have a single origin would be hard pressed to trouble the cosmic microwave background radiation let alone be seen.
The size of the visible universe is thus only double the 13.8 billion years, 27.6 billion years.
Since it has been expanding at less than the speed of light [caveat] the actual universe would be only perhaps 54 billion years old at the moment.
“The universe is around 95 billion light years across.” is an estimate based on maths and physics theories and until everyone agrees on those we might be better sticking to the speed of light time and distance observations.
Thank you for putting it and the concept up.
The big bang itself had so much matter and energy that our current concepts of time and the speed of light back then go out the window.
angech says:
July 20, 2022 at 6:56 am
…and Then There’s Physics says: July 18, 2022 at 7:29 am
“As I understand it, the galaxies in the galaxy cluster that was imaged by JWST were about 4.6 billion years old”.
The articles being written by journalists seem to be conflating such galaxies with the concept of being able to look back a lot further in time.
The original hot explosion or event being that long ago that the first clumps of plasma for want of a better word were supposed to form mega stars of very short life span which threw out clumps of matter [including some heavier elements than H, He] to form the original galaxies which were also very large [hence visible faintly] and then possibly another two iterations to get to our young star and young galaxy.
The materials greater than iron on the periodic table, gold being the best example, are thought to have come from arcane processes in past supernovae.
It is hard to imagine our sun being the remnant of a 13.8 billion year chunk of hop plasma cooling down over that length of time.
Further such explosions cause escape speed velocities which mean that the galaxies should never have come back together.
An alternative view is that space was filled with large amounts of cooled down matter in waves of explosions that crossed each other causing focal points of reaccumulation resulting in newer smaller galaxies.
This would explain suns forming from gigantic masses of cold hydrogen, etc hitting or passing through each other leaving focal eddies of matter which could then coalesce to form suns and planets.
and how elliptical orbits of planets and stars can come into being.
Rotary speech 13/7/2022 Role of Chairman
Dear fellow Rotarians
Could I ask you all to stand and join me in drinking a toast [invocation
For good food , good fellowship and the opportunity to serve rotary, we give thanks
-pause
Now I will ask the sergeant of the day, Ian Powell, to give the Loyal toast
Ian Powell
[“To Rotary International and Australia.”]
Thank you Ian
and you will always be sustained by the fellowship and esteem of us all.
welcome back to our first meeting following a changing of the guard last week.
I note that the ROTARY THEME FOR July This month’s theme is New Leadership Month
Some small steps to go through
I welcome everyone here today especially our new President Andrew, our new Paul Harris medal recipient, Danny, and especially our new member, Phil Stammers.
We also have present our past District chair Laurie Fagan and his wife anne
and other Rotary club members geoff and tracey.
I would like to call on Rotary members present to introduce their guests.
Phil [wife]
I will now hand over to Ian to conduct the meeting
or
starting with any announcements from club directors and any news that members of the club would like to share business for the day.
I will hand over to the Sergeant of the day for our fines session.
Check that the person introducing the guest speaker is present [me]
and has received the speaker’s resume
and has sufficient information to give a dignified introduction.
Check that the person doing the Thanks to Speaker [me]is in attendance and has the certificate or gift, as appropriate in your club.
THE FOUR-WAY-TEST … What does it mean?
For Rotary, The Four-Way Test is the cornerstone of all action. It has been for years, and it will be in the future. Of the things we think, say or do
Is it the TRUTH? Is it FAIR to all concerned Will it build GOODWILL and BETTER FRIENDSHIPS? Will it be BENEFICIAL to all concerned?
The test is one of the hallmarks of Rotary. Since it was developed in 1932 by Herbert J. Taylor, who later became RI president, it has never ceased to be relevant. Its four brief questions are not based on culture or religion. Instead, they are a simple checklist for ethical behavior. They transcend generations and national borders.
As Rotarians, we should have The Four-Way Test in mind in every decision we make, all day long.
Rotary is dedicated to causes that build international relationships, improve lives, and create a better world to support our peace efforts and end polio forever.
We now have a new President, Andrew Pogue ,who will be to carry out Rotary’s Objective,
with its five avenues of service as a constant guide.
It is also the day on which we ask our new member Phil Stammers to tell us a bit more about himself.
It is with pleasure that I now extend to you the right hand of Rotary fellowship & give you a hearty welcome as a member of the Rotary Club of_S Central
Before we go further can I mention our guests for today,
would also congratulate past President Danny Hogan on his second Paul Harris Medal
A WELCOME…to District Governor 2022-2023 MINA HOWARD? no
Idea Contribute to a Guest Speaker Ban
1,200,000 members, 35,000 clubs worldwide; 197 countries in world
Stop the Ukraine War.
The power of individuals to enact change on their own is minute.
Collective action is a little better but is limited takes a lot of time and effort.
Social media allows Multiple groups across the world to engage in collective action at the same real time.
This magnifies the ability to create change quickly.
The wear in the Ukraine needs action on multiple fronts.
It needs leadership from the major representatives of our countries and grass roots action.
The United Nations must unite and put out a call for the war to stop immediately.
The aggressor, in this case Russia, should be stripped of all rights of representation at all levels until the war stops.
Europe and Nato must issue similar calls for the war to stop and offer to put in peacekeeping forces now.
The United Nations, Europe and Nato should send peace keeping forces in regardless now.
Not to fight unless fired on.
Only to go in if Ukraine supports their coming in.
To leave immediately if the Ukraine requests.
With a large number of countries represented by troops on the ground Russia will have to halt its indiscriminate bombing.
The USA should also offer to send troops in on these conditions
All other world countries including the big two India and China should help.
People on the ground, not fighting but ready to respond would create a situation the Russian army would not want to upset.
Volunteers could be called up to go in such a situation, similar to the Spanish Civil war, not to fight but too prevent fighting.
The slow build up and reluctance to take even a defensive helpful position has created an extremely bad image for all nations.
This would not be needed if the nations of the world do what they should have done in the first place.
Time for them to step forwards and act, not aggressively, but defensively, saying we are all in this together.
Send enough people in so the Russians cannot kill people.
The other side of the coin is the Russian people themselves.
A reckoning needs to be called within Russia.
Over the last 10 years the Russians have been mixing better with the rest of the world on the internet and logistically.
People in St Petersberg and Moscow have enjoyed freedoms that they thought were guaranteed until now.
Social Media impact must register and does register.
Everyone who has an option, tweeting, facebooking [if allowed], phoning should do it.
They need to push and push for peace with their politicians.
They must put out a call for the Russian people to take action and demand their leader change course.
This is possible in the new Russia, and needs people to be reassured that their actions will be supported.
angech | April 30, 2022 at 6:41 am | Reply
On a lighter note the last day of April and soon a new global temperature for April.
Currently the year is running about 6th warmest.
We have had two of the weakest La Ninas I have ever seen after a strong El Nino with carried on warmth between the two La Ninas which hardly budged the BOM chart.
Now we have a series of interesting factors in play.
Cold waters coming up the South American Coast.
Cool Eastern Pacific waters.
A SOI of 21.8 when it looked like going negative.
A mild rise in UAH only in March.
Antarctic ice still under but rising slightly.
Arctic Ice in the 10th lowest and has been 12th lowest recently.
The ducks are all lined up in a row for a drop in April Global Temperatures.
I hate writing this because Roy Spencer reads it and it causes the UAH to go the other way.
Nonetheless a big drop in temperatures for April.
If Only.
Difficulties
” And where does it leave the alleged Earth’s greenhouse warming effect?
288K -220K=78K and because of Earth’s faster rotation a couple degrees less, perhaps.”
If the earth was an airless rock, your example fails . It could not have a GHG atmosphere so it would not be 288K in the first place
As it has faster rotation it would be a couple of degrees warmer than the moon surface,not less. so warmer than 220 K. As a bare rock.
As a planet with an atmosphere and a temp of 288C.
It now receives less energy to the surface directly.
and more energy to the atmosphere.
The energy in the atmosphere radiates in all directions thus lengthening the time some of the radiation takes to get in and increasing the number of CO2 atoms that are actively in the energy pathways.The net efect is | 10,272 | 45,424 | {"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} | 3.796875 | 4 | CC-MAIN-2024-38 | latest | en | 0.963104 |
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# Payouts on the craps table
Craps Payouts – Learn how to spot the best Payouts for Craps. Craps payouts and craps true odds are essential for players who are eager not only to enjoy a few sessions of this dice game but who also wish to win some cash. However, not all players are experts when it comes to the math behind odds and craps payouts. 41 rows · Understand what craps bets are worth taking by understanding the payout odds and house . 4/22/ · Craps Bet: This is the bet that you actually make on the table. True Odds: These are the calculated odds that you will win or lose based on the bet you make. The amount of money you place doesn’t figure into the odds, but it does figure into the Craps payout.
## Craps Payouts
Even those high payout odds fail to compensate for the actual likelihood of its occurrence. In a game where each roll of the dice is independent from all previous rolls, it truly is sort of a guessing game. The following table illustrates the odds and payouts of craps bets at a Las Vegas casino. Again, when it comes to playing online craps , the numbers you roll are out of your control and can only be determined by luck, but understanding the odds of rolling the numbers can help you make the right play and teach you which bets are the best ones to choose based on the odds. But unlike Place Bets, they only pay out at 1 to 1.
• All casinos are built on the principle of the most tangled labyrinths. In the gaming halls there are no direct passages. Because the longer a person walks along the aisles, the more likely that he will play in the meeting slot machines and gambling.
• The annual profit from the gaming industry in the US is 18 billion dollars.
## Craps Game Odds
Craps is a unique casino game in that it offers over different bets. If you've ever studied the layout of a craps table, then you probably noticed many different areas in which to place your chips. You can make bets on what number you predict will be the next thrown, how that number will be created as a combination of the dice, what number or numbers won't come up next, and so on.
With such a variety of bets available and the fact that a pair of dice may yield 36 different numerical combinations, it is easy to see how the odds for the casino game of craps can vary quite a bit. Some bets offer favorable odds - even better than straight blackjack - while others are not so good.
Online craps is surprisingly true to the real deal. All the bets are available, and the odds are on point. The truth is that most craps players play on gut feeling. In a game where each roll of the dice is independent from all previous rolls, it truly is sort of a guessing game. But, guessing will only get you so far.
You need to know what the payouts are for different bets, and the odds of those bets to occur. To be a skilled player is knowing which bets carry the best odds, and which to avoid altogether. The chart above shows that seven 7 is the most likely outcome with a 6 in 36 probability, while snake eyes a pair of ones and twelve 12 are the least likely with 1 in 36 probabilities. The house advantage is built into each number bet by the payout odds. The chart to the right shows some of the common bets in craps, what each bet's payout is for winning, and the edge the house has on each particular wager.
### View Details
Again, when it comes to playing online craps , the numbers you roll are out of your control and can only be determined by luck, but understanding the odds of rolling the numbers can help you make the right play and teach you which bets are the best ones to choose based on the odds.
Here are the odds for all the possible numbers that can be rolled in craps:. That means the craps odds of rolling either of these number are In craps the harder it is to roll the number, the higher it pays out. After 6 and 12 comes the 3 and the 11, which can be made from two different dice combinations respectively and offer odds of There are four possible combos for rolling 5 and 9 and the odds for either are The 6 and 8 can be made in 5 difference ways using a set of dice and present the player with odds.
Finally comes the 7, which can be rolled 6 different ways using a pair of dice and it has odds of making it the most common number to come up in online craps. Now let's look at some of the wagering odds. The following table illustrates the odds and payouts of craps bets at a Las Vegas casino. This chart can also be used as a general guide for online craps:. Contact us at crapsage[AT]yahoo. Free Craps Game Web archives will be available soon, as is change list on list of user requirements, contribution and change log.
If you cannot find the information you need about craps , Here's some other resources listed here. All Rights Resereved. Craps Payouts Which bets are the best on Craps Odds Continued from online craps odds part 1 : Again, when it comes to playing online craps , the numbers you roll are out of your control and can only be determined by luck, but understanding the odds of rolling the numbers can help you make the right play and teach you which bets are the best ones to choose based on the odds.
## Around world bet craps
There are various types of wagers you can make when playing craps. With some you can win or lose after one roll of the dice; with others it would take a series of rolls to determine whether you have won or lost. Knowing how to tell all the different types of bets apart is a vital aspect to playing the game at a casino. One of the most popular casino games to date, craps is also one of the oldest. Eventually, France embraced the English spelling and brought it with them to a French settlement in Canada called Acadia.
When England seized control over Acadia, the French fled to New Orleans where the game evolved during the years. Both ends of the table look exactly the same since players can be positioned on each side.
## Video
### Free Roulette
The thrill of watching the spinning red and black Roulette wheel has long served to grip many avid gamblers around the g... | 1,353 | 6,084 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2020-40 | latest | en | 0.955169 |
http://lpsolve.sourceforge.net/5.5/Intro.htm | 1,454,909,977,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701152959.66/warc/CC-MAIN-20160205193912-00248-ip-10-236-182-209.ec2.internal.warc.gz | 135,291,942 | 4,465 | # Introduction to lp_solve 5.5.2.0
What is lp_solve and what is it not? The simple answer is, lp_solve is a Mixed Integer Linear Programming (MILP) solver.
And what is Linear Programming? See "What is Linear Programming?" and "Oh, and we also want to solve it as an integer program." for a brief description. Also see Formulation of an lp problem in lpsolve.
lp_solve is a free (see LGPL for the GNU lesser general public license) linear (integer) programming solver based on the revised simplex method and the Branch-and-bound method for the integers.
It contains full source, examples and manuals.
lp_solve solves pure linear, (mixed) integer/binary, semi-continuous and special ordered sets (SOS) models. Note the word linear. This means that equations must be of the first order. 5 * x - 3 * y is an example. However x * y is not linear and cannot be handled by lp_solve. Both the objective function and the constraints have this restriction. Also see Ratios.
Via the Branch-and-bound algorithm, it can handle integer variables (see integer variables), semi-continuous variables (see semi-continuous variables) and Special Ordered Sets (see Special Ordered Sets (SOS)).
lp_solve has no limit on model size and accepts standard both lp or mps input files, but even that can be extended. Note however that some models could give lp_solve a hard time and will even fail to solve. The larger the model the likely the chance for that. But even commercial solvers have problems with that.
It can also be called as a library from different languages like C, VB, .NET, Delphi, Excel, Java, ...
It can also be called from AMPL, MATLAB, O-Matrix, Scilab, Octave, R via a driver program. lp_solve is written in ANSI C and can be compiled on many different platforms like linux and WINDOWS.
lp_solve has its own community via the Yahoo group http://groups.yahoo.com/group/lp_solve/. There you can find the latest sources, executables for the common platforms, examples, manuals and a message board where people can share their thoughts on lp_solve.
lp_solve was originally developed by Michel Berkelaar at Eindhoven University of Technology. The work of Jeroen Dirks made the transition from the basic version 1.5 to the full version 2.0 possible. He contributed the procedural interface, a built-in MPS reader, and many fixes and enhancements to the code. At that point there was also a Java port of lp_solve 2.0. This was not a Java native interface (JNI) to the C library, but rather a translation of the algorithm (ver 2.0) from C into Java. It was also very limited This meant that it did not keep up with lp_solve as it evolved. Starting at version 3.0, lp_solve is released under the LGPL license. Before the code could only be used for non-commercial purposes. Many other people also contributed to lp_solve, but there was no track of them. Sorry if you are not mentioned. Development was stagnated for some time at version 3.2, but now it is again alive and well via the new developers Kjell Eikland and Peter Notebaert. But also other people help to improve the product. For example the new Java interface to lp_solve was made by Juergen Ebert. It is a JNI interface that supports the full functionality of lp_solve. He did a great job. We encourage other people to participate in the development of lp_solve.
First a version 4 was introduced that had already several enhancements and improvements and now with version 5 this enhancement continues resulting is faster solving times, more stability and able to solver larger models and new functionality. See Changes from version 4 to version 5.1 for the changes done to version 5 and Changes from version 5.1 to version 5.5 for the changes done to version 5.5.
Basically, lp_solve is a library, a set of routines, called the API that can be called from almost any programming language to solve MILP problems. There are several ways to pass the data to the library:
• Via the API
• Via input files
• Via an IDE
#### Via the API
The API is a set of routines that can be called from a programming language to build the model in memory, solve it and return the results. There are many API routines to perform many possible tasks and set several options. See lp_solve API reference for an overview.
#### Via input files
Standard, lp_solve supports several input files types. The common known MPS format (see mps-format) is supported by most solvers, but it is not very readable for humans. Another format is the lp format (see lp-format) that is more readable. lp_solve has the unique ability to use user-written routines to input the model (see External Language Interface). See read_mps, read_freemps, read_MPS, read_freeMPS and read_lp, read_LP for the API calls to read the model from file.
There is also a driver program called lp_solve that uses the API to provide a command line application to solve models. See lp_solve for its usage. With this program you don't have to know anything of API or computer programming languages. You can just provide your model via file to the program and it will solve the model and give you the result.
#### Via an IDE
Thanks to Henri Gourvest, there is now also an IDE program called LPSolve IDE that uses the API to provide a Windows application to solve models. See LPSolve IDE for its usage. With this program you don't have to know anything of API or computer programming languages. You can just provide your model to the program and it will solve the model and give you the result.
As already stated, lp_solve can be called from many programming language. Among them are C, C++, Pascal, Delphi, Java, VB, C#, VB.NET, Excel. But let this list not be a limitation. Any programming language capable of calling external libraries (DLLs under Windows, Shared libraries (.so) under Unix/Linux) can call lp_solve.
Here is a list of some key features of lp_solve:
• Mixed Integer Linear Programming (MILP) solver
• Basically no limit on model size
• It is free and with sources
• Supports Integer variables, Semi-continuous variables and Special Ordered Sets
• Can read model from MPS, LP or user written format
• Models can be build in-memory without the use of files
• Has a powerful API interface
• Easy callable from other programming languages
• Advanced pricing using Devex and Steepest Edge for both primal and dual simplexes
• Provides different scaling methods to make the model more numerical stable
• Has presolve capabilities to tighten constraints/make the model smaller and faster to solve
• Has a base crashing routine to determine a starting point
• Allows restart after making changes to the model. Solve continues from the last found solution
• Possible to select desired combinations of primal and dual phases 1 and 2
• Possible to set several solver parameters like tolerances
• Alternative (and faster) inverse/re-factorisation libraries are provided for. See Basis Factorization Packages
• Alternative model readers and writers possible via the XLI implementation. See External Language Interfaces
• Has the possibility to convert one model format to another format
• Provides post-optimal sensitivity analysis. See Sensitivity
• ... | 1,562 | 7,158 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2016-07 | longest | en | 0.92705 |
https://quicksoul.co/the-ultimate-guide-to-bicycle-length-how-many-meters-long-is-a-bike/ | 1,686,325,920,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656737.96/warc/CC-MAIN-20230609132648-20230609162648-00683.warc.gz | 548,323,675 | 16,208 | # The Ultimate Guide to Bicycle Length: How Many Meters Long is a Bike?
## Short answer: How many meters long is a bicycle?
A typical adult-sized bicycle is about two meters (200 centimeters) long, while children’s and youth bicycles can range from one to 1.5 meters in length. However, the exact size of a bike may vary based on factors such as its style or design.
## Step-by-Step Guide to Determine how Many Meters Long Your Bicycle Is
Biking can be a lot of fun, whether it’s for exercise or simply to enjoy the great outdoors. But before you hit the road with your bike, there are some important measurements that you need to know – one of which is determining how long your bicycle really is.
Although measuring something as basic as length might seem like an easy task initially (I mean who doesn’t already have their own trusty tape measure on standby?) but when it comes down to bicycles and getting accurate information every millimeter counts!
So in this Step-by-Step guide we’re about lay out everything from what tools are needed all way through each step until finally arriving at closing thoughts; so stop lolly-gagging around and let’s dive right into calculating just how many meters long those biccles uesd by both casual enthusiasts and professionals alike REALLY are.
### Before You Begin
First things first: make sure that your bicylce stand isn’t wobbly. To accurately determine its dimensions –or any measurement– having sturdy placement makes all other steps much easier than if parts keep shifting because they aren’t fixed securely.
Once you’ve secured our cycle onto solid ground then grab these three essential items below:
* Tape Measure (one)
* Straight Edge/yardstick
* Paper pen/pencil
### Getting Started:
For starters identify exact locations where critical points will give physical demensions necessary . In general,you’ll want note two major body components:
1) The distance between axles.
2) And overall positional height & sizing .
To properly take these above mentioned measures start following increments exactly listed against stepped-out guidelines worth considering
## Part One :Measuring Frame Distance
Frame distance often defines appropriate riding positions but also serves well throughout most bikes ability on inclined planes too so naturally accuracy during this particualr procedure becomes especially pivotal)
When doing so follow instructions detailed here-
Pro Tip: Take note of the frame center line for most accurate measurements.
Step 1- The basic step, grab tape and measure distance between centre points on bike’s axle to reach actual figure listed in millimeters.
__Note :__
* Before moving forward take a couple minutes do some backend math. Divide that total length you got from above by thousand to convert them into meters.( e.g if your reading reads `1080mm` divide by one thousand equals `.01m`, which is our final outcome.).
### Part Two -Positional Height & Sizing
Now that we’ve covered measuring framework usage let us move towards determining bicycle height! This isnt’ any ordinary “how-tall-is-my-bike” measurement down pat; instead precise knowledge ensures desired fit worth committing long rides across varied terrain !
In order to determine correct positioning specific steps can be taken as follows:
pro tip :Bicycles come with different tire sizes so ensure correctly identified type before getting started?
**If Bike Has A Tire Diameter Under 26 Inches**
For Mid Sized cycles follow
## FAQs on determining exactly How Many Meters Long a typical bicycle measures up to
Are you curious to know just how long a typical bicycle measures up in meters? Look no further! We have compiled the most frequently asked questions (FAQs) on determining this exact measurement.
Q: What is the approximate length of an adult-sized bicycle?
A: The average length of an adult-sized bike ranges from 1.5 to 2 meters, or roughly 4.9 feet to 6.6 feet.
Q: Does wheel size affect the overall length of a bike?
A: Yes, it does. A larger wheel diameter tends to increase both the distance between axles and frame tube lengths, thereby extending overall measurements.
For example:
– Mountain bikes with larger wheels sizes (typically around 29 inches/73cm), may be closer towards two meter range as opposed down at shorter than five-foot mark if we were talking about a kid’s BMX type model that has tires measuring only sixteen-inch diameters approximately
Q: Can different types/styles/models make any significant differences by adding/removing accessories points?
Yes again – Any customizations added such racks for carrying panniers/luggage can add extra centimetres/millimeters; conversely removing parts like fenders all ease weight but reduce total height too!
Some more guidelines categorizing into parameters known are here:
Typically lightweight frames made out-of-aluminum/carbon-fibre/etc will stay within limits at approx ‘Five’ point nine mega-metres while models available under category sport hybrid/women’s/unisex/adventure alternatives vary inversely categorized according e.g., Trek aluminum Women-specific-Bike Stay below Five-Meter variant However there exist even longer XL men road variants offering huge flexibility along crankshaft depths !
• MTB(metric Version)
Mountain bicycles typically run much wider tyres then hybrids so its unsurprising they lie often above relatively higher marks compared counterparts – especially when possess particularly cross-country style built layouts. These are measured anywhere between ‘Two'(antique models) to three meters of length but 2.5 is realistic one can spot.
• Folding options
Foldable bikes design hardly exceeds the Twenty-meter barrier and mostly range either side Eighteen point five, that’s under six feet!
Q: How do you accurately measure a bike from end-to-end?
A: The most straightforward way is by measuring straight down the middle–from top-tube-edge closest joint attaching seat post at saddle all through till backside axle allocated bearings adjusted in dropout slots or wheelset base… But obviously mentioning total reach distance encumbered via extended proportions won’t be included so don’t rely solely on this method if analyzing latest high-performing machine microcosm compatibility factors like frame size ergonomic handling individual user criteria personalized attunes best doing biomechanics adjustments.
In conclusion, there may not be an exact measurement for every type of bicycle out there, however understanding general parameters enables narrowing more specific calculations with including other features constraints (e.g., tire width/frame material/folding components). So next
## Top 5 Facts Reveal just How Many Meters Long is considered Standard for Bicycles Globally.
Cycling has been a popular mode of transportation for centuries, with the first bicycles dating back to the early 19th century. However, over time these two-wheeled wonders have undergone numerous changes in design and size; leading many curious riders to wonder just how long their bikes should be.
So what is considered standard today? Here are five facts that reveal just how many meters long modern bicycles typically measure:
1. A common misconception is that all bicycle frames come in one “standard” length – but this isn’t exactly true! In fact, bike sizes vary according to rider height and weight as well as other factors such as riding style or frame geometry. Most manufacturers offer multiple sizes for each model they produce so finding the right fit can take some trial-and-error.
2. Generally speaking though, most adult sized road bikes range between 50-58 centimeters (or around 20-23 inches) measured from center pedal crank arm bolt up through seat tube axis – making them roughly about a meter or slightly shorter depending on which end you’re measuring from!
3. Interestingly enough however there’s actually quite a bit of variation within these general guidelines based upon regional preferences too… For example European-style city commuters might prefer smaller more upright frames while North American mountain bikers commonly opt for longer trail-specific designs complete with wide handlebars & suspension systems
4.. When selecting your own set-up it’s important not only think carefully about sizing options like stand-over heights , top-tube lengths wheelbases etc., But also consider ride characteristics: will you mostly use roads vs offroad trails ? Will spending hours climbing steep inclines warrants stiffer lightweight carbon fiber units?
5.While precise measurements may differ given individual variances among competitors across diverse disciplines equipment loadouts any bunches/pelotons conditions cadences training regimens objectives performance cues muscle memory awareness adaptations responses redlines hearing voices premonitions postive feedback negative rumination, especially among pro-level athletes; In general terms though – having a properly sized and fitted bicycle will not only improve riding efficiency & comfort but also enhance safety too!
Rate article | 1,783 | 9,134 | {"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 | 0.935354 |
https://fr.slideserve.com/elina/foreign-exchange-theory | 1,618,911,123,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039388763.75/warc/CC-MAIN-20210420091336-20210420121336-00304.warc.gz | 369,716,675 | 20,239 | # Foreign Exchange Theory - PowerPoint PPT Presentation
Foreign Exchange Theory
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Foreign Exchange Theory
## Foreign Exchange Theory
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##### Presentation Transcript
1. Foreign Exchange Theory Session 1 Introduction to Derivatives
2. Scenario 1 • You are based in the US and exporting Shinjo (NY Mets) and Ichiro (Seattle Mariners) T-shirts to Japan. • From your marketing research you expect to price the T-shirts at 3,000 yen. Consumer demand is expected to be 10,000 T-shirts. Expected revenue is 10,000 x 3,000yen = 30 million yen. Revenue is expected one month from today. • The current exchange rate is 125 yen/dollar. • What should we do? N. Takezawa (ICU)
3. Scenario 2 • We have borrowed 100 million yen and must pay interest at a fixed rate of 5% per year (固定金利). The loan is for ten years. Thus, interest payments are fixed each year at 0.05 x 100 million yen = 5 million yen. • However, we would like to pay interest based on a floating rate (変動レート). In other words, the interest payments change each year depending on the current interest rate. • What should we do? N. Takezawa (ICU)
4. Derivatives ( デリバティブ、派生証券) • A derivative (or derivative security) is a financial instrument whose value depends on the values of other, more basic underyling variables. (Hull, p.1) • デリバティブとは、ある基準になる資産の価格（株式）、利子率（短期金融商品）、指数 （株価指数）などの値をもとにして、これらの値に依存し、かつもとの原資産と異なるペイオフをもつ、新たに作られる契約であるとて定義できましょう。 （久保田、２００１、ｐ ２３４） N. Takezawa (ICU)
5. Derivative Contract: Traded separately from Underlying asset Example: option on Sony Stock Underlying Asset: Sony stock- traded on Stock Exchange (TSE) N. Takezawa (ICU)
6. Instruments and Markets • Options Contracts (オプション契約) • Forward Contracts (先渡契約) • Futures Contracts (先物契約) • Swap Contracts (スワップ契約) N. Takezawa (ICU)
7. Underlying Assets (原資産、対象資産) • Commodities (foods) such as Rice and Wheat • Metals such as Gold • Commodities (natural resources) such as Oil and Gas • Foreign Exchange Rates (Currency): Yen, DM, etc. • Individual Stocks (株式個別銘柄) • Bonds (国債、社債) • Stock Indices such as S&P 500, TOPIX, etc. (株価指数) • Electricity • Weather • Emission Permits (SO2, CO2 near future?)(排出権) N. Takezawa (ICU)
8. Why Derivatives? 1) Reduce Financial Risk: Combine instruments to reduce risk. Say a Japanese fund manager invests in a US stock such as IBM. Then the investor faces or is exposed to several financial (price) risks. Exposure includes, 1) IBM stock price, 2) US stock market, 3) Yen/US dollar exchange rate. The investor could “neutralize” some of the risks by entering futures contracts for the SP500 and currency. 2) Leverage “Cost savings” in transaction costs. For example, to take a position in a currency, index, and/or bond, you only need to put-up margin. N. Takezawa (ICU)
9. 3) Informational Efficiency (効率的市場） Harry Roberts-Eugene Fama: Market Efficiency Weak Form [prices, volume] Semi-strong Form [all public information] Strong Form [all public and private] Joint Hypothesis: Market Efficiency and Correct Model Correct model in our context: example, volatility models, Black-Scholes-Merton option pricing model. Does derivative (trading) provide us with (valuable) information on the underlying spot market and/or other derivative securities markets? N. Takezawa (ICU)
10. Nikkei 225 Futures Market:Hiraki, Maberly, Takezawa (J. Banking and Finance, 1995) • Trading in futures market usually extend beyond that of the spot market by 10-15 minutes. • On the Osaka exchange, the Nikkei futures traded until 3:15 - 15 minutes beyond the spot market. • Oct. 2, 1990: Futures extended trading reduced to 10 minutes. This feature is eliminated Feb. 7, 1992. Does extended trading period result in speculative trading. • We show that the information generated from the extended period contains useful information. N. Takezawa (ICU)
11. RISK Business Risk Market Risk Financial Risk Credit Risk, other risks N. Takezawa (ICU)
12. Market Risk価格変動リスク • Also referred to as price risk. • This risk comes from a change in price. The price change of financial assets and liabilities (underlying asset prices). • The change in price is often referred to as volatility (variability of the price). N. Takezawa (ICU)
13. Some of these derivatives can only be purchased from a financial institution (or corporate). • Some of these derivatives can only be purchased on an organized exchange. • Well known exchanges are 1) Chicago Board of Trade (CBOT) (www.cbot.com) and 2) Chicago Mercantile Exchange (CME) (www.cme.com) • Tokyo Stock Exchange (TSE) (www.tse.or.jp): stock options for 167 different companies as of April 2001, TOPIX options (オプション), TOPIX futures (先物),JGB futures (国債の先物), options on JGB futures. N. Takezawa (ICU)
14. Futures on the Nikkei 225 Index (日経平均株価指数) are traded at the Singapore International Monetary Exchange (SIMEX), Chicago Mercantile Exchange (CME), Osaka Securities Exchange (OSE) • Future (or rough equivalent) were traded in Japan during the Edo Period. Rice Futures Exchange (Dojima) in Osaka [堂島米会場] established in 1730. N. Takezawa (ICU) | 1,458 | 5,159 | {"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-2021-17 | latest | en | 0.713128 |
https://math.stackexchange.com/questions/120231/evaluating-even-binomial-coefficients/120240 | 1,713,426,602,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817200.22/warc/CC-MAIN-20240418061950-20240418091950-00787.warc.gz | 360,066,004 | 38,655 | # Evaluating even binomial coefficients
Can someone give me a hint how to evaluate $$\binom{n}{0}+\binom{n}{2}+\cdots+\binom{n}{o(n)},$$ where $o(n)$ is $n$ if $n$ is even and $n-1$ otherwise ?
• Mar 14, 2012 at 21:33
Consider: $$\sum_{k=0}^n \binom{n}{k} x^k = (1+x)^n$$ Now you are computing: $$\sum_{m \geqslant 0, n \geqslant 2 m } \binom{n}{2m} = \sum_{k=0}^n \binom{n}{k} \frac{1+(-1)^k}{2} = \frac{(1+1)^n + (1-1)^n}{2} = \frac{2^n + \delta_{n,0}}{2}$$
We simply need to show the following two things: $$\sum_{i=0}^n \binom{n}{i} = 2^n$$ $$\sum_{i=0}^n (-1)^i\binom{n}{i} = 0$$
Both of these follow from the binomial formula. The result then follows routinely.
if n is odd then consider the below series:$$\binom{n}{0}+\binom{n}{1}+\cdots+\binom{n}{n} = 2^n$$ and we know that $$\binom{n}{k} = \binom{n}{n-k}$$ so $$\binom{n}{0} = \binom{n}{n} , \binom{n}{2} = \binom{n}{n-2} , ...$$ so for odd number the answer is $2^{n-1}$.
and for even number you can do the same thing like odd numbers.
If you calculate the first few examples, you'll see a very clear pattern-- if you haven't already done this, that's where you should start.
Now to prove this pattern and other patterns within the binomial coefficients, it's very helpful to use the fact that the binomial coefficient $\binom{n}{i}$ is just the coefficient of $x^iy^{n-i}$ in $(x+y)^n$.
You can prove that the sum $\binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \ldots + \binom{n}{n}=2^n$ just by setting $x=1$ and $y=1$ in the formula above to get $(1+1)^n$.
You can prove that the alternating sum $\binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \ldots +(-1)^n \binom{n}{n}=0$ just by setting $x=1$ and $y=-1$ in the formula above to get $(1-1)^n$.
Now what happens if you add these two sums together?
We have a group of $n$ people, where $n\ge 1$. The $n$-th person is George. The expression $$\binom{n}{0}+\binom{n}{2}+\cdots+\binom{n}{o(n)}$$ counts the number of ways to choose an even number of people from our group (including the possibility of choosing $0$ people).
Pick any set $S$ of people from the first $n-1$. If $S$ has an odd number of people, add George. If $S$ has an even number of people, leave George out.
We can thus see that there are just as many ways to choose an even number of people from $n$ people as there are ways to choose a set of people from $n-1$ people.
But this number is $2^{n-1}$, for in choosing people from the first $n-1$, for every person we have $2$ choices, yes or no.
Note that $n=0$ is an exception to the rule we have found, for $\binom{0}{0}=1$.
Let's first note that: $2^n = (1+1)^n = \sum_{i=0}^n \binom{n}{i}1^{n-i}1^i = \sum_{i=0}^n \binom{n}{i}$ and that $0 = (1 + (-1))^n = \sum_{i=0}^n\binom{n}{i} 1^{n-i}(-1)^i = \sum_{i=0}^n \binom{n}{i}(-1)^i$. We have then:\begin{align*} \sum_{i=0}^{n}\frac{1+(-1)^i}{2}\binom{n}{i} &= \frac{1}{2}\sum_{i=0}^{n}\binom{n}{i}(1+(-1)^i) \\ &= \frac{1}{2}\left( \sum_{i=0}^{n}\binom{n}{i} + \sum_{i=0}^{n}\binom{n}{i}(-1)^i \right) \\ &= \frac{1}{2}\left( 2^n + 0 \right) \\ &= 2^{n-1} \text{.} \end{align*} | 1,174 | 3,074 | {"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} | 4.5 | 4 | CC-MAIN-2024-18 | latest | en | 0.762257 |
http://openstudy.com/updates/515342fde4b07077e0bf36c8 | 1,448,560,989,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398447769.81/warc/CC-MAIN-20151124205407-00141-ip-10-71-132-137.ec2.internal.warc.gz | 180,374,332 | 12,734 | ## Haleyy_Bugg 2 years ago Choose the correct description of the graph of the compound inequality 2x < -12 or x + 3 >= 12 A number line with an open circle on -6, shading to the left, and a closed circle on 9, shading to the right. A number line with a closed circle on -6, shading to the left, and an open circle on 9, shading to the right. A number line with an open circle on -6, a closed circle on 9, and shading in between. A number line with a closed circle on -6, an open circle on 9, and shading in between.
1. Haleyy_Bugg
@Notamathgenius
2. aajugdar
still on this?
3. Notamathgenius
Haleyy, I have no clue o_o sorry o_o
4. Haleyy_Bugg
yes i am .-.
5. aajugdar
first is x<-6
6. aajugdar
so open circle on -6 n shading on left
7. aajugdar
2nd is x>=9
8. aajugdar
so its closed circle n shading to right
9. Haleyy_Bugg
so what do itry to find as to be an answer?
10. aajugdar
hmm consider 1st 1 2x<-12 so x< -12/2 x<-6 so all the terms less than -6 will be included xcept -6 |dw:1364411428556:dw|
11. Haleyy_Bugg
so A?
12. aajugdar
yes
13. Haleyy_Bugg
What is the equation of the line that passes through the points (-2, 1) and (1, 10)? A) 3x - y = -7 B) 3x - y = -5 C) 3x - y = 5 D) x + 3y = -5
14. aajugdar
and 2nd 1 is easy x+3>=12 x>=12-3 x>=9 so all terms greater than or equal to 12 |dw:1364411626234:dw|
15. Haleyy_Bugg
16. aajugdar
put values of (x,y) in equation
17. aajugdar
18. Haleyy_Bugg
to the first one, but what about the second one?
19. aajugdar
A put the values
20. Haleyy_Bugg
idk how.
21. aajugdar
(-2,1) = (x,y) so put x= -2 and y=1
22. aajugdar
3x - y = -7 in this
23. aajugdar
if this doesnt satisfy then check for 2nd option
24. aajugdar
you are supposed to check for (1, 10) too
25. Haleyy_Bugg
okay that just confused me.
26. aajugdar
well then
27. Haleyy_Bugg
yupp | 667 | 1,849 | {"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} | 3.4375 | 3 | CC-MAIN-2015-48 | longest | en | 0.816457 |
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# Bank soal kelas x
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### Bank soal kelas x
1. 1. BANK SOAL EKONOMI KELAS X 1. Masalah pokok ekonomi A B C 1. Pak. Andi menjual ikan hasil tangkapannya langsung ke kota 2. Negara Jepang memproduksi sepeda motor Yamaha 1. Pak Imam mencari informasi tentang selera masyarakat terhadap buah apel 2. PT. Semen Tonasa memproduksi semen 1. PT. Agung memproduksi kursi sesuai dengan jumlah pesanan. 2. PT. PLN diberi tanggungjawab untuk mengelola listrik Berdasarkan matriks di atas pernyataan yang sesuai dengan masalah pokok ekonomi untuk siapa barang diproduksi adalah…. a. A1, B1, C1 b. A1, B2, C1 c. A2, B2, C1 d. A2, B1, C2 e. A2, B2, C2 2. Perhatikan pernyataan berikut: 1. Tersedianya tenaga kerja trampil 2. Keserakahan manusia dalam memenuhi kebutuhannya 3. Situasi sosial dan politik suatu bangsa 4. Perkembangan IPTEK tidak sesuai dengan peningkatan kebutuhan 5. Ketidak mampuan mengolah sumber daya alam Dari pernyataan di atas yang termasuk sebab-sebab terjadinya kelangkaan adalah.... a. 1, 2 dan 4 d. 2, 4 dan 5 b. 1, 3 dan 5 e. 3, 4 dan 5 c. 2, 3 dan 5 3. Keinginan manusia terhadap barang atau jasa yang dapat memberikan kepuasan jasmani maupun rohani adalah : a. Kepentingan c. Pilihan b. Kebutuhan d. Permintaan e. Kemauan | 837 | 2,499 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-45 | latest | en | 0.171898 |
https://interpreting-regression.netlify.app/reducing-uncertainty-of-the-outcome-including-predictors.html | 1,610,752,609,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703497681.4/warc/CC-MAIN-20210115224908-20210116014908-00130.warc.gz | 394,263,241 | 27,761 | # Chapter 8 Reducing uncertainty of the outcome: including predictors
Caution: in a highly developmental stage! See Section 1.1.
## 8.1 Variable terminology
In supervised learning:
• The output is a random variable, typically denoted $$Y$$.
• The input(s) variables (which may or may not be random), if there are $$p$$ of them, are typically denoted $$X_1$$, …, $$X_p$$ – or just $$X$$ if there’s one.
There are many names for the input and output variables. Here are some (there are more, undoubtedly):
• Output: response, dependent variable.
• Input: predictors, covariates, features, independent variables, explanatory variables, regressors.
In BAIT 509, we will use the terminology predictors and response.
### 8.1.1 Variable types
Terminology surrounding variable types can be confusing, so it’s worth going over it. Here are some non-technical definitions.
• A numeric variable is one that has a quantity associated with it, such as age or height. Of these, a numeric variable can be one of two things:
• A categorical variable, as the name suggests, is a variable that can be one of many categories. For example, type of fruit; success or failure.
## 8.2 Irreducible Error
The concept of irreducible error is paramount to supervised learning. Next time, we’ll look at the concept of reducible error.
When building a supervised learning model (like linear regression), we can never build a perfect forecaster – even if we have infinite data!
Let’s explore this notion. When we hypothetically have an infinite amount of data to train a model with, what we actually have is the probability distribution of $$Y$$ given any value of the predictors. The uncertainty in this probability distribution is the irreducible error.
Example: Let’s say $$(X,Y)$$ follows a (known) bivariate Normal distribution. Then, for any input of $$X$$, $$Y$$ has a distribution. Here are some examples of this distribution for a few values of the predictor variable (these are called conditional distributions, because they’re conditional on observing particular values of the predictors).
This means we cannot know what $$Y$$ will be, no matter what! What’s one to do?
• In regression (i.e., when $$Y$$ is numeric, as above), the go-to standard is to predict the mean as our best guess.
• We typically measure error with the mean squared error = average of (observed-predicted)^2.
• In classification, the conditional distributions are categorical variables, so the go-to standard is to predict the mode as our best guess (i.e., the category having the highest probability).
• A typical measurement of error is the error rate = proportion of incorrect predictions.
• A more “complete” picture of error is the entropy, or equivalently, the information measure.
In Class Meeting 07, we’ll look at different options besides the mean and the mode.
An important concept is that predictors give us more information about the response, leading to a more certain distribution. In the above example, let’s try to make a prediction when we don’t have knowledge of predictors. Here’s what the distribution of the response looks like:
This is much more uncertain than in the case where we have predictors!
## 8.3 In-class Exercises: Irreducible Error
NOT REQUIRED FOR PARTICIPATION
### 8.3.1 Oracle regression
Suppose you have two independent predictors, $$X_1, X_2 \sim N(0,1)$$, and the conditional distribution of $$Y$$ is $Y \mid (X_1=x_1, X_2=x_2) \sim N(5-x_1+2x_2, 1).$ From this, it follows that:
• The conditional distribution of $$Y$$ given only $$X_1$$ is $Y \mid X_1=x_1 \sim N(5-x_1, 5).$
• The conditional distribution of $$Y$$ given only $$X_2$$ is $Y \mid X_2=x_2 \sim N(5+2x_2, 2).$
• The (marginal) distribution of $$Y$$ (not given any of the predictors) is $Y \sim N(5, 6).$
The following R function generates data from the joint distribution of $$(X_1, X_2, Y)$$. It takes a single positive integer as an input, representing the sample size, and returns a tibble (a fancy version of a data frame) with columns named x1, x2, and y, corresponding to the random vector $$(X_1, X_2, Y)$$, with realizations given in the rows.
genreg <- function(n){
x1 <- rnorm(n)
x2 <- rnorm(n)
eps <- rnorm(n)
y <- 5-x1+2*x2+eps
tibble(x1=x1, x2=x2, y=y)
}
1. Generate data – as much as you’d like.
dat <- genreg(1000)
1. For now, ignore the $$Y$$ values. Use the means from the distributions listed above to predict $$Y$$ under four circumstances:
1. Using both the values of $$X_1$$ and $$X_2$$.
2. Using only the values of $$X_1$$.
3. Using only the values of $$X_2$$.
4. Using neither the values of $$X_1$$ nor $$X_2$$. (Your predictions in this case will be the same every time – what is that number?)
dat <- mutate(dat,
yhat = FILL_THIS_IN,
yhat1 = FILL_THIS_IN,
yhat2 = FILL_THIS_IN,
yhat12 = FILL_THIS_IN)
1. Now use the actual outcomes of $$Y$$ to calculate the mean squared error (MSE) for each of the four situations.
• Try re-running the simulation with a new batch of data. Do your MSE’s change much? If so, choose a larger sample so that these numbers are more stable.
(mse <- mean((dat$FILL_THIS_IN - dat$y)^2))
(mse1 <- mean((dat$FILL_THIS_IN - dat$y)^2))
(mse2 <- mean((dat$FILL_THIS_IN - dat$y)^2))
(mse12 <- mean((dat$FILL_THIS_IN - dat$y)^2))
knitr::kable(tribble(
~ Case, ~ MSE,
"No predictors", mse,
"Only X1", mse1,
"Only X2", mse2,
"Both X1 and X2", mse12
))
1. Order the situations from “best forecaster” to “worst forecaster”. Why do we see this order?
### 8.3.2 Oracle classification
Consider a categorical response that can take on one of three categories: A, B, or C. The conditional probabilities are: $P(Y=A \mid X=x) = 0.2,$ $P(Y=B \mid X=x) = 0.8/(1+e^{-x}),$
To help you visualize this, here is a plot of $$P(Y=B \mid X=x)$$ vs $$x$$ (notice that it is bounded above by 0.8, and below by 0).
Here’s an R function to generate data for you, where $$X\sim N(0,1)$$. As before, it accepts a positive integer as its input, representing the sample size, and returns a tibble with column names x and y corresponding to the predictor and response.
gencla <- function(n) {
x <- rnorm(n)
pB <- 0.8/(1+exp(-x))
y <- map_chr(pB, function(t)
sample(LETTERS[1:3], size=1, replace=TRUE,
prob=c(0.2, t, 1-t-0.2)))
tibble(x=x, y=y)
}
1. Calculate the probabilities of each category when $$X=1$$. What about when $$X=-2$$? With this information, what would you classify $$Y$$ as in both cases?
• BONUS: Plot these two conditional distributions.
## X=1:
(pB <- FILL_THIS_IN)
(pA <- FILL_THIS_IN)
(pC <- FILL_THIS_IN)
ggplot(tibble(p=c(pA,pB,pC), y=LETTERS[1:3]), aes(y, p)) +
geom_col() +
theme_bw() +
labs(y="Probabilities", title="X=1")
## X=-2
(pB <- FILL_THIS_IN)
(pA <- FILL_THIS_IN)
(pC <- FILL_THIS_IN)
ggplot(tibble(p=c(pA,pB,pC), y=LETTERS[1:3]), aes(y, p)) +
geom_col() +
theme_bw() +
labs("Probabilities", title="X=-2")
1. In general, when would you classify $$Y$$ as A? B? C?
### 8.3.3 (BONUS) Random prediction
You might think that, if we know the conditional distribution of $$Y$$ given some predictors, why not take a random draw from that distribution as our prediction? After all, this would be simulating nature.
The problem is, this prediction doesn’t do well.
Re-do the regression exercise above (feel free to only do Case 1 to prove the point), but this time, instead of using the mean as a prediction, use a random draw from the conditional distributions. Calculate the MSE. How much worse is it? How does this error compare to the original Case 1-4 errors?
### 8.3.4 (BONUS) A more non-standard regression
The regression example given above is your perfect, everything-is-linear-and-Normal world. Let’s see an example of a joint distribution of $$(X,Y)$$ that’s not Normal.
The joint distribution in question can be respresented as follows: $Y|X=x \sim \text{Beta}(e^{-x}, 1/x),$ $X \sim \text{Exp}(1).$
Write a formula that gives a prediction of $$Y$$ from $$X$$ (you might have to look up the formula for the mean of a Beta random variable). Generate data, and evaluate the MSE. Plot the data, and the conditional mean as a function of $$x$$ overtop.
### 8.3.5 (BONUS) Oracle MSE
What statistical quantity does the mean squared error (MSE) reduce to when we know the true distribution of the data? Hint: if each conditional distribution has a certain variance, what then is the MSE?
What is the error rate in the classification setting?
## 8.4 Learning Objectives
From today’s class, students are expected to be able to:
• Calculate conditional distributions when giving a full distribution.
• Calculate marginal distributions from a joint distribution.
• Obtain the marginal mean from conditional means and marginal probabilities, using the law of total expectation.
• Use the law of total probability to convert between conditional + marginal distributions, and joint distributions.
• Describe the consequences of independent random variables.
• Calculate and describe the pros and cons of dependence measures: covariance, correlation, and kendall’s tau.
## 8.5 Conditional Distributions (15 min)
Probability distributions describe an uncertain outcome, but what if we have partial information?
Consider the example of ships arriving at the port of Vancouver again. Each ship will stay at port for a random number of days, which we’ll call the length of stay (LOS) or $$D$$, according to the following (made up) distribution:
Length of Stay (LOS) Probability
1 0.25
2 0.35
3 0.20
4 0.10
5 0.10
Suppose a ship has been at port for 2 days now, and it’ll be staying longer. What’s the distribution of length-of-stay now? Using symbols, this is written as $$P(D = d \mid D > 2)$$, where the bar “|” reads as “given” or “conditional on”, and this distribution is called a conditional distribution. We can calculate a conditional distribution in two ways: a “table approach” and a “formula approach”.
Table approach:
1. Subset the pmf table to only those outcomes that satisfy the condition ($$D > 2$$ in this case). You’ll end up with a “sub table”.
2. Re-normalize the remaining probabilities so that they add up to 1. You’ll end up with the conditional distribution under that condition.
Formula approach: In general for events $$A$$ and $$B$$, the conditional probability formula is $P(A \mid B) = \frac{P(A \cap B)}{P(B)}.$
For the ship example, the event $$A$$ is $$D = d$$ (for all possible $$d$$’s), and the event $$B$$ is $$D > 2$$. Plugging this in, we get $P(D = d \mid D > 2) = \frac{P(D = d \cap D > 2)}{P(D > 2)} = \frac{P(D = d)}{P(D > 2)} \text{ for } d = 3,4,5.$
The only real “trick” is the numerator. How did we reduce the convoluted event $$D = d \cap D > 2$$ to the simple event $$D = d$$ for $$d = 3,4,5$$? The trick is to go through all outcomes and check which ones satisfy the requirement $$D = d \cap D > 2$$. This reduces to $$D = d$$, as long as $$d = 3,4,5$$.
## 8.6 Joint Distributions (25 min)
So far we’ve only considered one random variable at a time. Its distribution is called univariate because there’s just one variable. But, we very often have more than one random variable.
Let’s start by considering … We can visualize this as a joint distribution:
Don’t be fooled, though! This is not really any different from what we’ve already seen. We can still write this a univariate distribution with four categories. This is useful to remember when we’re calculating probabilities.
Outcome Probability
HH 0.25
HT 0.25
TH 0.25
TT 0.25
Viewing the distribution as a (2-dimensional) matrix instead of a (1-dimensional) vector turns out to be more useful when determining properties of individual random variables.
### 8.6.1 Example: Length of Stay vs. Gang Demand
Throughout today’s class, we’ll be working with the following joint distribution of length of stay of a ship, and its gang demand.
Gangs = 1 Gangs = 2 Gangs = 3 Gangs = 4
LOS = 1 0.0017 0.0425 0.1247 0.0811
LOS = 2 0.0266 0.1698 0.1360 0.0176
LOS = 3 0.0511 0.1156 0.0320 0.0013
LOS = 4 0.0465 0.0474 0.0059 0.0001
LOS = 5 0.0740 0.0246 0.0014 0.0000
The joint distribution is stored in “tidy format” in an R variable named j:
### 8.6.2 Marginal Distributions
We’ve just specified a joint distribution of length of stay and gang request. But, we’ve previously specified a distribution for these variables individually. These are not things that can be specified separately:
• If you have a joint distribution, then the distribution of each individual variable follows as a consequence.
• If you have the distribution of each individual variable, you still don’t have enough information to form the joint distribution between the variables.
The distribution of an individual variable is called the marginal distribution (sometimes just “marginal” or “margin”). The word “marginal” is not really needed when we’re talking about a random variable – there’s no difference between the “marginal distribution of length of stay” and the “distribution of length of stay”, we just use the word “marginal” if we want to emphasize the distribution is being considered in isolation from other related variables.
### 8.6.3 Calculating Marginals from the Joint
There’s no special way of calculating a marginal distribution from a joint distribution. As usual, it just involves adding up the probabilities corresponding to relevant outcomes.
For example, to compute the marginal distribution of length of stay (LOS), we’ll first need to calculate $$P(\text{LOS} = 1)$$. Using the joint distribution of length of stay and gang request, the outcomes that satisfy this requirement are the entire first row of the probability table. It follows that the marginal distribution of LOS can be obtained by adding up each row. For the marginal of gang requests, just add up the columns.
Length of Stay Probability
1 0.25
2 0.35
3 0.20
4 0.10
5 0.10
Similarly, the distribution of gang request is the same as from last lecture:
Gang request Probability
1 0.2
2 0.4
3 0.3
4 0.1
### 8.6.4 Conditioning on one Variable
What’s usually more interesting than a joint distribution are conditional distributions, when other variables are fixed. This is a special type of conditional distribution and an extremely important type of distribution in data science.
For example, a ship is arriving, and they’ve told you they’ll only be staying for 1 day. What’s the distribution of their gang demand under this information? That is, what is $$P(\text{gang} = g \mid \text{LOS} = 1)$$ for all possible $$g$$?
Table approach:
1. Isolating the outcomes satisfying the condition ($$\text{LOS} = 1$$), we obtain the first row:
Gangs: 1 Gangs: 2 Gangs: 3 Gangs: 4
0.0017 0.0425 0.1247 0.0811
1. Now, re-normalize the probabilities so that they add up to 1, by dividing them by their sum, which is 0.25:
Gangs: 1 Gangs: 2 Gangs: 3 Gangs: 4
0.0068 0.1701 0.4988 0.3243
Formula Approach: Applying the formula for conditional probabilities, we get $P(\text{gang} = g \mid \text{LOS} = 1) = \frac{P(\text{gang} = g, \text{LOS} = 1)}{P(\text{LOS} = 1)},$ which is exactly row 1 divided by 0.25.
Here’s a plot of this distribution. For comparison, we’ve also reproduced its marginal distribution.
Interpretation: given information, about length of stay, we get an updated picture of the distribution of gang requests. Useful for decision making!
### 8.6.5 Law of Total Probability/Expectation
Quite often, we know the conditional distributions, but don’t directly have the marginals. In fact, most of regression and machine learning is about seeking conditional means!
For example, suppose you have the following conditional means of gang request given the length of stay of a ship.
This curve is called a model function, and is useful if we want to predict a ship’s daily gang request if we know their length of stay. But what if we don’t know their length of stay, and we want to produce an expected gang request? We can use the marginal mean of gang request!
In general, a marginal mean can be computed from the conditional means and the probabilities of the conditioning variable. The formula, known as the law of total expectation, is $E(Y) = \sum_x E(Y \mid X = x) P(X = x).$
Here’s a table that outlines the relevant values:
Length of Stay (LOS) E(Gang | LOS) P(LOS)
1 3.140580 0.25
2 2.412802 0.35
3 1.917192 0.20
4 1.596041 0.10
5 1.273317 0.10
Multiplying the last two columns together, and summing, gives us the marginal expectation: 2.3.
Also, remember that probabilities are just means, so the result extends to probabilities: $P(Y = y) = \sum_x P(Y = y \mid X = x) P(X = x)$ This is actually a generalization of the law of total probability we saw before: $$P(Y=y)=\sum_x P(Y = y, X = x)$$.
### 8.6.6 Exercises (10 min)
In pairs, come to a consensus with the following three questions.
1. Given the conditional means of gang requests, and the marginal probabilities of LOS in the above table, what’s the expected gang requests, given that the ship captain says they won’t be at port any longer than 2 days? In symbols, $E(\text{Gang} \mid \text{LOS} \leq 2).$
2. What’s the probability that a new ship’s total gang demand equals 4? In symbols, $P(\text{Gang} \times \text{LOS} = 4).$
3. What’s the probability that a new ship’s total gang demand equals 4, given that the ship won’t stay any longer than 2 days? In symbols, $P(\text{Gang} \times \text{LOS} = 4 \mid \text{LOS} \leq 2).$
## 8.7 Multivariate Densities/pdf’s
Recall the joint pmf (discrete) from Lecture 4, between gang demand and length-of-stay:
Gangs = 1 Gangs = 2 Gangs = 3 Gangs = 4
LOS = 1 0.0017 0.0425 0.1247 0.0811
LOS = 2 0.0266 0.1698 0.1360 0.0176
LOS = 3 0.0511 0.1156 0.0320 0.0013
LOS = 4 0.0465 0.0474 0.0059 0.0001
LOS = 5 0.0740 0.0246 0.0014 0.0000
Each entry in the table corresponds to the probability of that unique row (LOS value) and column (Gang value). These probabilities add to 1.
For the continuous case, instead of rows and columns, we have an x- and y-axis for our two variables, defining a region of possible values. For example, if two marathon runners can only finish a marathon between 5.0 and 5.5 hours each, and their end times are totally random, then the possible values are indicated by a square in the following plot:
Each point in the square is like an entry in the joint pmf table in the discrete case, except now instead of holding a probability, it holds a density. The density function, then, is a surface overtop of this square (or in general, the outcome space). That is, it’s a function that takes two variables (marathon time for Runner 1 and Runner 2), and calculates a single density value from those two points. This function is called a bivariate density function.
Here’s an example of what a 2D pdf might look like: https://scipython.com/blog/visualizing-the-bivariate-gaussian-distribution/
Notation: For two random variables $$X$$ and $$Y$$, their joint density/pdf evaluated at the points $$x$$ and $$y$$ is usually denoted $f_{X,Y}(x,y),$ or sometimes less rigorously, as just $f(x, y).$
### 8.7.1 Conditional Distributions, revisited (15 min)
Remember the formula for conditional probabilities: for events $$A$$ and $$B$$, $P(A \mid B) = \frac{P(A \cap B)}{P(B)}.$ But, this is only true if $$P(B) \neq 0$$, and it’s not useful if $$P(A) = 0$$ – two situations we’re faced with in the continuous world!
#### 8.7.1.1 When $$P(A) = 0$$
To describe this situation, let’s use a univariate continuous example: the example of monthly expenses.
Suppose the month is half-way over, and you find that you only have \$2500 worth of expenses so far! What’s the distribution of this month’s total expenditures now, given this information? If we use the law of conditional probability, we would get a formula that’s not useful: letting $$X = \text{Expense}$$, $P(X = x \mid X \geq 2500) = \frac{P(X = x)}{P(X \geq 2500)} \ \ \ \text{(no!)}$
This is no good, because the outcome $$x$$ has a probability of 0. This equation just simplies to 0 = 0, which is not useful.
Instead, in general, we replace probabilities with densities. In this case, what we actually have is: $f(x \mid X \geq 2500) = \frac{f(x)}{P(X \geq 2500)} \ \text{ for } x \geq 2500,$ and $$f(x \mid X \geq 2500) = 0$$ for $$x < 2500$$.
Notice from the formula that the resulting density is just the original density confined to $$x \geq 2500$$, and re-normalized to have area 1. This is what we did in the discrete case!
The monthly expense example has expenditures $$X \sim$$ LN(8, 0.5). Here is its marginal distribution and the conditional distribution. Notice the conditional distribution is just a segment of the marginal, and then re-normalized to have area 1.
#### 8.7.1.2 When $$P(B) = 0$$
To describe this situation, let’s use the marathon runners’ example again.
Runner 1 ended up finishing in 5.2 hours. What’s the distribution of Runner 2’s time? Letting $$X$$ be the time for Runner 1, and $$Y$$ for Runner 2, we’re asking for $$f_{Y|X}(y \mid X = 5.2)$$.
But wait! Didn’t we say earlier that $$P(X = 5.2) = 0$$? This is the bizarre nature of continuous random variables. Although no outcome is possible, we must observe some outcome in the end. In this case, the stopwatch used to calculate run time has rounded the true run time to 5.2h, even though in reality, it would have been something like 5.2133843789373… hours.
As before, plugging in the formula for conditional probabilities won’t work. But, as the case when $$P(A) = 0$$, we can in general replace probabilities with densities. We end up with $f_{Y|X}(y \mid 5.2) = \frac{f_{Y,X}(y, 5.2)}{f_X(5.2)}.$
This formula is true in general $f_{Y|X}(y \mid x) = \frac{f_{Y,X}(y, x)}{f_X(x)}.$ In fact, this formula is even true for both pdf’s and pmf’s!
## 8.8 Dependence concepts
A big part of data science is about harvesting the relationship between $$X$$ and $$Y$$, often called the dependence between $$X$$ and $$Y$$.
### 8.8.1 Independence (5 min)
Informally, $$X$$ and $$Y$$ are independent if knowing something about one tells us nothing about the other.
Formally, the definition of $$X$$ and $$Y$$ being independent is: $P(X = x \cap Y = y) = P(X = x) P(Y = y).$ More usefully and intuitively, it’s better to think of independence such that conditioning on an independent variable tells us nothing:
$P(Y = y \mid X = x) = P(Y = y).$
This is far less interesting than when there’s dependence, which implies that there are relationships between variables.
### 8.8.2 Measures of dependence (15 min)
When there is dependence, it’s often useful to measure the strength of dependence. Here are some measurements.
#### 8.8.2.1 Covariance and Pearson’s Correlation
Covariance is one common way of measuring dependence between two random variables. The idea is to take the average “signed area” of rectangles constructed between a sampled point and the mean, with the sign being determined by “concordance” relative to the mean:
• Concordant means $$x < \mu_x$$ and $$y < \mu_y$$, OR $$x > \mu_x$$ and $$y > \mu_y$$ – gets positive area.
• Discordant means $$x < \mu_x$$ and $$y > \mu_y$$, OR $$x > \mu_x$$ and $$y < \mu_y$$ – gets negative area.
Here is a random sample of 10 points, with the 10 rectangles constructed with respect to the mean. Sign is indicated by colour. The covariance is the mean signed area.
Formally, the definition is $\mathrm{Cov(X, Y)} = E[(X-\mu_X)(Y-\mu_Y)],$ where $$\mu_Y=E(Y)$$ and $$\mu_X=E(X)$$. This reduces to a more convenient form, $\text{Cov}(X,Y)=E(XY)-E(X)E(Y)$
In R, you can calculate the empirical covariance using the cov function:
## [1] -0.7111111
In the above example, the boxes are more often negative, so the covariance (and the “direction of dependence”) is negative. For the above example, the larger the LOS, the smaller the gang demand – this inverse relationship is indicative of negative covariance. Other interpretations of the sign:
• Positive covariance indicates that an increase in one variable is associated with an increase in the other variable.
• Zero covariance indicates that there is no linear trend – but this does not necessarily mean that $$X$$ and $$Y$$ are independent!
It turns out covariance by itself isn’t very interpretable, because it depends on the scale (actually, spread) of $$X$$ and $$Y$$. For example, multiply $$X$$ by 10, and suddenly the box sizes increase by a factor of 10, too, influencing the covariance.
Pearson’s correlation fixes the scale problem by standardizing the distances according to standard deviations $$\sigma_X$$ and $$\sigma_Y$$, defined as $\text{Corr}(X, Y) = E\left[ \left(\frac{X-\mu_X}{\sigma_X}\right) \left(\frac{Y-\mu_Y}{\sigma_Y}\right) \right] =\frac{\text{Cov}(X, Y)}{\sqrt{\text{Var}(X)\text{Var}(Y)}}.$ As a result, it turns out that $-1 \leq \text{Corr}(X, Y) \leq 1.$
The Pearson’s correlation measures the strength of linear dependence:
• -1 means perfect negative linear relationship between $$X$$ and $$Y$$.
• 0 means no linear relationship (Note: this does not mean independent!)
• 1 means perfect positive linear relationship.
In R, you can calculate the empirical Pearson’s correlation using the cor function:
## [1] -0.6270894
Pearson’s correlation is ubiquitous, and is often what is meant when “correlation” is referred to.
#### 8.8.2.2 Kendall’s tau
Although Pearson’s correlation is ubiquitous, its forced adherance to measuring linear dependence is a big downfall, especially because many relationships between real world variables are not linear.
An improvement is Kendall’s tau ($$\tau_K$$):
• Instead of measuring concordance between each observation $$(x, y)$$ and the mean $$(\mu_x, \mu_y)$$, it measures concordance between each pair of observation $$(x_i, y_i)$$ and $$(x_j, y_j)$$.
• Instead of averaging the area of the boxes, it averages the amount of concordance and discordance by taking the difference between number of concordant and number of discordant pairs.
Visually plotting the $$10 \choose 2$$ boxes for the above sample from the previous section:
The formal definition is $\frac{\text{Number of concordant pairs} - \text{Number of discordant pairs}}{{n \choose 2}},$ with the “true” Kendall’s tau value obtainined by sending $$n \rightarrow \infty$$. Note that several ways have been proposed for dealing with ties, but this doesn’t matter when we’re dealing with continuous variables (Weeks 3 and 4).
In R, the empirical version can be calculated using the cor() function with method = "kendall":
## [1] -0.579771
Like Pearson’s correlation, Kendall’s tau is also between -1 and 1, and also measures strength (and direction) of dependence.
For example, consider the two correlation measures for the following data set. Note that the empirical Pearson’s correlation for the following data is not 1!
Pearson Kendall
0.9013 1
But, Kendall’s tau still only measures the strength of monotonic dependence. This means that patterns like a parabola, which are not monotonically increasing or decreasing, will not be captured by Kendall’s tau either:
Pearson Kendall
0 0
Even though both dependence measures are 0, there’s actually deterministic dependence here ($$X$$ determines $$Y$$). But, luckily, there are many monotonic relationships in practice, making Kendall’s tau a very useful measure of dependence.
### 8.8.3 Dependence as separate from the marginals
The amount of monotonic dependence in a joint distribution, as measured by kendall’s tau, has nothing to do with the marginal distributions. This can be a mind-boggling phenomenon, so don’t fret if you need to think this over several times.
To demonstrate, here are joint distributions between LOS and gang demand having the same marginals, but different amounts of dependence.
Let’s return to the computation of the conditional distribution of gang requests given that a ship will only stay at port for one day. Let’s compare the marginal distribution (the case where we know nothing) to the conditional distributions for different levels of dependence (like we saw in the previous section). The means for each distribution are indicated as a vertical line:
What’s of particular importance is comparing the uncertainty in these distributions. Let’s look at how the uncertainty measurements compare between marginal and conditional distributions (marginal measurements indicated as horizontal line):
Moral of the story: more dependence (in either direction) gives us more certainty in the conditional distributions! This makes intuitive sense, because the more related $$X$$ and $$Y$$ are, the more that knowing what $$X$$ is will inform what $$Y$$ is.
## 8.9 Harvesting Dependence (20 min)
The opposite of independence is dependence: when knowing something about $$X$$ tells us something about $$Y$$ (or vice versa). Extracting this “signal” that $$X$$ contains about $$Y$$ is at the heart of supervised learning (regression and classification), covered in DSCI 571/561 and beyond.
Usually, we reserve the letter $$X$$ to be the variable that we know something about (usually an exact value), and $$Y$$ to be the variable that we want to learn about. These variables go by many names – usually, $$Y$$ is called the response variable, and $$X$$ is sometimes called a feature, or explanatory variable, or predictor, etc.
To extract the information that $$X$$ holds about $$Y$$, we simply use the conditional distribution of $$Y$$ given what we know about $$X$$. This is as opposed to just using the marginal distribution of $$Y$$, which corresponds to the case where we don’t know anything about $$X$$.
Sometimes it’s enough to just communicate the resulting conditional distribution of $$Y$$, but usually we reduce this down to some of the distributional properties that we saw earlier, like mean, median, or quantiles. We communicate uncertainty also using methods we saw earlier, like prediction intervals and standard deviation.
Let’s look at an example.
### 8.9.1 Example: River Flow
In the Rocky Mountains, snowmelt $$X$$ is a major driver of river flow $$Y$$. Suppose the joint density can be depicted as follows:
Every day, a measurement of snowmelt is obtained. To predict the river flow, usually the conditional mean of river flow given snowmelt is used as a prediction, but median is also used. Here are the two quantities as a function of snow melt:
These functions are called model functions, and there are a ton of methods out there to help us directly estimate these model functions without knowing the density. This is the topic of supervised learning – even advanced supervised learning methods like deep learning are just finding a model function like this (although, usually when there are more than one $$X$$ variable).
It’s also quite common to produce prediction intervals. Here is an example of an 80% prediction interval, using the 0.1- and 0.9-quantiles as the lower and upper limits:
As a concrete example, consider the case where we know there’s been 1mm of snowmelt. To obtain the conditional distribution of flow ($$Y$$) given this information, we just “slice” the joint density at $$x =$$ 1, and renormalize. Here is that density (which is now univariate!), compared with the marginal distribution of $$Y$$ (representing the case where we know nothing about snowmelt, $$X$$):
The following table presents some properties of these distributions:
Quantity Marginal Conditional
Mean 247.31 118.16
Median 150 74.03
80% PI [41.64, 540.33] [25.67, 236.33]
Notice that we actually only need the conditional distribution of $$Y$$ given $$X=x$$ for each value of $$x$$ to produce these plots! In practice, we usually just specify these conditional distributions. So, having the joint density is actually “overkill”.
### 8.9.2 Direction of Dependence
Two variables can be dependent in a multitude of ways, but usually there’s an overall direction of dependence:
• Positively related random variables tend to increase together. That is, larger values of $$X$$ are associated with larger values of $$Y$$.
• Negatively related random variables have an inverse relationship. That is, larger values of $$X$$ are associated with smaller values of $$Y$$.
We’ve already seen some measures of dependence in the discrete setting: covariance, correlation, and Kendall’s tau. These definitions carry over. It’s a little easier to visualize the definition of covariance as the signed sum of rectangular area:
Correlation, remember, is also the signed sum of rectangles, but after converting $$X$$ and $$Y$$ to have variances of 1.
Here are two positively correlated variables, because there is overall tendency of the contour lines to point up and to the right (or down and to the left):
Here are two negatively correlated variables, because there is overall tendency for the contour lines to point down and to the right (or up and to the left):
Another example of negative correlation. Although contour lines aren’t pointing in any one direction, there’s more density along a line that points down and to the right (or up and to the left) than there is any other direction.
Here are two random variables that are dependent, yet have 0 correlation (both Pearson’s and Kendall’s) because the overall trend is flat (pointing left or right). You can think of this in terms of slicing: slicing at $$x = -2$$ would result in a highly peaked distribution near $$y = 0$$, whereas slicing at $$x = 1$$ would result in a distribution with a much wider spread – these are not densities that are multiples of each other! Prediction intervals would get wider with larger $$x$$.
Note that the marginal distributions have nothing to do with the dependence between random variables. Here are some examples of joint distributions that all have the same marginals ($$N(0,1)$$), but different dependence structures and strengths of dependence:
## 8.10 Marginal Distributions (20 min)
In the river flow example, we used snowmelt to inform river flow by communicating the conditional distribution of river flow given snowmelt. But, this requires knowledge of snowmelt! What if one day we are missing an observation on snowmelt? Then, the best we can do is communicate the marginal distribution of river flow. But how can we get at that distribution? Usually, aside from the data, we only have information about the conditional distributions. But this is enough to calculate the marginal distribution!
### 8.10.1 Marginal Distribution from Conditional
We can use the law of total probability to calculate a marginal density. Recall that for discrete random variables $$X$$ and $$Y$$, we have $P(Y = y) = \sum_x P(X = x, Y = y) = \sum_x P(Y = y \mid X = x) P(X = x).$ The same thing applies in the continuous case, except probabilities become densities and sums become integrals (as usual in the continuous world): for continuous $$X$$ and $$Y$$, $f_Y(y) = \int_x f_{X,Y}(x,y)\ \text{d}x = \int_x f_{Y\mid X}(y \mid x)\ f_X(x)\ \text{d}x.$
Notice that this is just an average of the conditional densities! If we have the conditional densities and a sample of $$X$$ values $$x_1, \ldots, x_n$$, then using the empirical approximation of the mean, we have $f_Y(y) \approx \frac{1}{n} \sum_{i = 1}^n f_{Y\mid X}(y \mid x_i).$
A similar result holds for the cdf. We have $F_Y(y) = \int_x F_{Y \mid X}(y \mid x)\ f_X(x) \ \text{d}x,$ and empirically, $F_Y(y) \approx \frac{1}{n}\sum_{i = 1}^n F_{Y\mid X}(y \mid x_i).$
### 8.10.2 Marginal Mean from Conditional
Perhaps more practical is finding the marginal mean, which we can obtain using the law of total expectation (similar to the discrete case we saw in a previous lecture): $E(Y) = \int_x m(x) \ f_{X}(x) \ \text{d}x = E(m(X)),$ where $$m(x) = E(Y \mid X = x)$$ (i.e., the model function or regression curve).
When you fit a model using supervised learning, you usually end up with an estimate of $$m(x)$$. From the above, we can calculate the marginal mean as the mean of $$m(X)$$, which we can do empirically using a sample of $$X$$ values $$x_1, \ldots, x_n$$. Using the empirical mean, we have $E(Y) \approx \frac{1}{n} \sum_{i=1}^n m(x_i).$
### 8.10.3 Marginal Quantiles from Conditional
Unfortunately, if you have the $$p$$-quantile of $$Y$$ given $$X = x$$, then there’s no convenient way of calculating the $$p$$-quantile of $$Y$$ as an average. To obtain this marginal quantity, you would need to calculate $$F_Y(y)$$ (as above), and then find the value of $$y$$ such that $$F_Y(y) = p$$.
### 8.10.4 Activity
You’ve observed the following data of snowmelt and river flow:
Snowmelt (mm) Flow (m^3/s)
1 140
3 150
3 155
2 159
3 170
From this, you’ve deciphered that the mean flow given snowmelt is $E(\text{Flow} \mid \text{Snowmelt} = x) = 100 + 20x.$
You also decipher that the conditional standard deviation is constant, and is: $SD(\text{Flow} \mid \text{Snowmelt} = x) = 15\ m^3/s$ It also looks like the conditional distribution of river flow given snowmelt follows a Lognormal distribution.
Part 1: A new reading of snowmelt came in, and it’s 4mm.
1. Make a prediction of river flow.
2. What distribution describes your current understanding of what the river flow will be?
Part 2: Your snowmelt-recording device is broken, so you don’t know how much snowmelt there’s been.
1. Make a prediction of river flow.
2. What distribution describes your current understanding of what the river flow will be?
3. Someone tells you that a 90% prediction interval is [70, 170]. What do we know about the median?
## 8.11 Comparing Probabilities
Suppose Vincenzo often wins at a game of solitaire—a one-player game—but that Tom is twice as good as Vincenzo. Does this means that $$P(\text{Tom wins}) = 2 \times P(\text{Vincenzo wins})$$?
Probability is quite useful for communicating the chance of an event happening in an absolute sense, but is not useful for comparing probabilities. Odds, on the other hand, are useful for comparing the chance of two events. If $$p$$ is the chance that Vincenzo wins at solitaire, his odds of winning is defined as $\text{Odds} = \frac{p}{1-p}.$ This means that, if his odds are $$o$$, then the probability of winning is $\text{Probability} = \frac{o}{o+1}.$
For example, if Vincenzo wins 80% of the time, his odds are $$0.8/0.2 = 4$$. This is sometimes written as 4:1 odds – that is, four wins for every loss. If Tom is twice as good as Vincenzo, it’s most useful to say that this means Tom wins twice as many games before experiencing a loss (on average) – that is, 8:1 odds, or simply 8, and a probability of $$8/9=0.888\ldots$$. | 10,111 | 38,436 | {"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": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2021-04 | latest | en | 0.87908 |
http://blog.phytools.org/2024/08/fitting-multi-state-threshold-model.html | 1,726,405,579,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651630.14/warc/CC-MAIN-20240915120545-20240915150545-00640.warc.gz | 3,977,251 | 486,188 | Thursday, August 15, 2024
Fitting a multi-state threshold model using fitThresh in phytools when some tip states are uncertain
Yesterday a phytools user asked if it was possible to supply input data for fitThresh as a matrix of (prior) probabilities that each tip was in each state of the character. fitThresh, for those who don’t recall, is a function to fit the multi-state threshold model using the discrete approximation of Boucher & Démery (2016). (If this is new to you, check out my Evolution Meeting 2024 talk entitled Bounded Brownian motion & the most important phylogenetic comparative methods paper you’ve probably never read to learn more. It’s on YouTube!)
As a reminder, and to make a nice graphic, evolution under the threshold model looks something like this:
library(phytools)
layout(matrix(c(1,2),1,2),widths=c(0.55,0.45))
par(mar=c(5.1,4.1,1.1,1.1),las=1,cex.axis=0.8,bg="#2f2f2f",
fg="white",col.axis="white",col.lab="white")
obj<-bmPlot(pbtree(b=0.03,n=80,t=100,type="discrete",
quiet=TRUE),type="threshold",thresholds=c(0,0.5,2,2.5),
anc=1,sig2=1/100,ngen=100,colors=hcl.colors(5,
palette="plasma"),return.tree=TRUE,bty="n")
plot(obj$tree,colors=setNames(hcl.colors(5,palette="plasma"), c("a","b","c","d","e")),direction="upwards",ftype="off", mar=c(5.1,1.1,1.1,1.1)) The answer, at the time, was no; however, I have sinced pushed two updates to the function (1, 2) that fix this. Followers of this blog might recall from an earlier post that there are effectively two ways to handle tip state uncertainty in discrete character phylogenetic models. In my previous post, I characterized these as follows: (1) as if the uncertainty is observed (e.g., we see that the base at a particular nucleotide site is a purine, but we can’t tell whether it’s an adenine or a guanine); or (2) as if the uncertainty is unobserved (e.g., we see a 0 for one of our species, but there’s, say, a 10% chance that we might’ve measured wrongly & it’s actually a 1). In the former case, the total probability of our data is equal to the sum of the probabilities of the data given that the ambiguous site is an adenine or a guanine. In the latter, we evaluate the product of the probability of our observed data and the chances that the possibly erroneously measured datum was 0 (0.9) and add to it the probability of our data but in which the ambiguous datum is 1, multiplied by the chances that it is 1 (0.1). This all sounds a bit convoluted, but as I describe in my previous post, it makes less of a difference than one might think, so I’d argue that it’s probably most sensible to take the approach that best reflects the type of uncertainty we have in our data! It was a tiny bit complicated (for reasons that I won’t get into), but I’ve now implemented a similar approach for fitThresh – that is, users can supply their input in the form of an $$N \times k$$ matrix (for N species and k levels of the discrete trait), in which ambiguity is coded either as 1.0s for all possible states (i.e., approach #1, above) and/or by indicating the prior probability that each tip is in each of the possible states of the trait (i.e., approach #2). To illustrate this, I’m going to begin by simulating data under the threshold model. Remember, this simply involves generating liability values (the unobserved, underlying continuous trait) first, say, under a model of Brownian motion evolution – and then converting the simulated species value of the trait to our discrete threshold character (which can be done using the mostly internally-used function, phytools::threshState). tree<-pbtree(n=300,scale=1) liability<-fastBM(tree,a=1.5) thresholds<-setNames(c(0,0.5,2,2.5,Inf), c("a","b","c","d","e")) x<-as.factor(threshState(liability,thresholds)) head(x,10) ## t222 t223 t107 t299 t300 t120 t159 t160 t110 t111 ## c c c c c c b a a a ## Levels: a b c d e Now, let’s proceed to fit our threshold model simply by passing tree and our factor x directly to fitThresh. (This is the only form that data for fitThresh could assume before today’s updates.) fit1<-fitThresh(tree,x) fit1 ## Object of class "fitThresh". ## ## Set value of sigsq (of the liability) = 1.0 ## ## Set or estimated threshold(s) = ## [ -1.402909, -0.893271, 0.633884, 1.245576 ]* ## ## Log-likelihood: -224.719748 ## ## (*lowermost threshold is fixed) Keeping in mind that the lowermost threshold is arbitrary, if we reset it to 0.0 (our simulated lower threshold) we should find that the estimate thresholds are quite accurate compared to their generating values. ## generating thresholds ## a b c d e ## 0.0 0.5 2.0 2.5 Inf ## estimated fit1$threshold-min(fit1$threshold) ## [1] 0.0000000 0.5096379 2.0367932 2.6484848 (This is nothing new, of course.) Next, I’m going to convert the vector x to a simple binary matrix & show that the exact same result will be obtained (excepting for differences in numerical optimization). X<-to.matrix(x,levels(x)) head(X,10) ## a b c d e ## t222 0 0 1 0 0 ## t223 0 0 1 0 0 ## t107 0 0 1 0 0 ## t299 0 0 1 0 0 ## t300 0 0 1 0 0 ## t120 0 0 1 0 0 ## t159 0 1 0 0 0 ## t160 1 0 0 0 0 ## t110 1 0 0 0 0 ## t111 1 0 0 0 0 fit2<-fitThresh(tree,X) fit2 ## Object of class "fitThresh". ## ## Set value of sigsq (of the liability) = 1.0 ## ## Set or estimated threshold(s) = ## [ -1.402909, -0.893012, 0.634471, 1.257801 ]* ## ## Log-likelihood: -224.721515 ## ## (*lowermost threshold is fixed) So far, so good. Next, to emulate uncertainty in some tip states, I’ll pick a random set of 30 tips (10% of the taxa in my tree), and set them to complete ambiguity. To do that, I’ll assign the corresponding rows a set of five 1.0s across the five possible values of the trait. X[ii<-sample(1:nrow(X),30),]<-rep(1,5) head(X,10) ## a b c d e ## t222 0 0 1 0 0 ## t223 0 0 1 0 0 ## t107 0 0 1 0 0 ## t299 0 0 1 0 0 ## t300 0 0 1 0 0 ## t120 0 0 1 0 0 ## t159 1 1 1 1 1 ## t160 1 0 0 0 0 ## t110 1 0 0 0 0 ## t111 1 0 0 0 0 Now let’s fit our model with these data & see how the fitted model compares. In this case, since they are based on less data, we might expect our parameter estimates to be slightly less accurate than for the full dataset (although, on average, they should remain unbiased – something that will be impossible to evaluate from a single simulation). fit3<-fitThresh(tree,X) fit3 ## Object of class "fitThresh". ## ## Set value of sigsq (of the liability) = 1.0 ## ## Set or estimated threshold(s) = ## [ -1.402909, -0.874436, 0.634451, 1.22128 ]* ## ## Log-likelihood: -205.881759 ## ## (*lowermost threshold is fixed) Out of curiousity, let’s check how well we’d do in predicting the values for are now “missing” terminal taxa based on our fitted model. To do that, I’ll run ancr with tips=TRUE on our fitted model and then compute the mean square product of our original data and the reconstructed tip values, subsampling to include only those set as ambiguous in our analysis. anc_fit3<-ancr(fit3,tips=TRUE) anc_fit3 ## Marginal ancestral state estimates: ## a b c d e ## 301 0.004163 0.279245 0.716592 0.000000 0e+00 ## 302 0.135957 0.656351 0.207692 0.000000 0e+00 ## 303 0.097724 0.488095 0.414179 0.000002 0e+00 ## 304 0.101190 0.479178 0.419629 0.000003 0e+00 ## 305 0.002069 0.080525 0.913791 0.003608 8e-06 ## 306 0.000000 0.018144 0.976987 0.004869 0e+00 ## ... ## ## Log-likelihood = -205.881759 sum(anc_fit3$ace[ii,]*
to.matrix(x,levels(x))[ii,])/length(ii)
## [1] 0.6294646
This is a crude measure of accuracy, but shows that our marginal reconstructions are about 63% “correct” – meaning we might have gotten around ⅔ of them correct with 100% confidence, all of them right with 63% confidence, or, most likely, something in between. This is pretty good, given that we gave the method no information about the tip states, the trait changes rapidly in some parts of the tree, and we’d expect to get these values right only 20% of the time on average by random chance.
Lastly, let’s emulate “partial” uncertainty. In this case, I’ll go through the same 30 randomly selected tips, and each time assign an equal prior probability that the tip is in the known, true condition – or in one of its up to two neighboring values.
X<-to.matrix(x,levels(x))
for(i in 1:length(ii)){
jj<-which(X[ii[i],]==1)
ind<-c(jj,sample(jj+c(-1,1),1))
while(any(ind<1)||any(ind>5))
ind<-c(jj,sample(jj+c(-1,1),1))
X[ii[i],ind]<-c(0.5,0.5)
}
Let’s proceed to fit the model. In this case, we should expect our result to even more closely resemble fit1 and fit2 from the full data because we are only partially (rather than fully) ambiguating our 30 termini.
fit4<-fitThresh(tree,X)
fit4
## Object of class "fitThresh".
##
## Set value of sigsq (of the liability) = 1.0
##
## Set or estimated threshold(s) =
## [ -1.402909, -0.863932, 0.657881, 1.268888 ]*
##
## Log-likelihood: -233.248294
##
## (*lowermost threshold is fixed)
Let’s once again measure the predictive value of this model by seeing how it allows ancr to predict the values of our ambiguous tip states.
As before, we shouldn’t be surprised to find that this prediction is more accurate than from our fit3 model object because our tip states were less ambiguous in estimation.
anc_fit4<-ancr(fit4,tips=TRUE)
sum(anc_fit4$ace[ii,]* to.matrix(x,levels(x))[ii,])/length(ii) ## [1] 0.7410549 Finally, let’s go back to the same dataset we used for fit4, and disambiguate it further still be assigning the known, true state a prior probability of 0.75 and the wrong state a probability of 0.25. A quick trick to do this using our previously calculated matrix, X, is as follows. X[ii,]<-X[ii,]/2+to.matrix(x,levels(x))[ii,]/2 head(X,20) ## a b c d e ## t222 0 0.00 1.00 0 0 ## t223 0 0.00 1.00 0 0 ## t107 0 0.00 1.00 0 0 ## t299 0 0.00 1.00 0 0 ## t300 0 0.00 1.00 0 0 ## t120 0 0.00 1.00 0 0 ## t159 0 0.75 0.25 0 0 ## t160 1 0.00 0.00 0 0 ## t110 1 0.00 0.00 0 0 ## t111 1 0.00 0.00 0 0 ## t31 0 0.00 1.00 0 0 ## t251 0 0.00 1.00 0 0 ## t252 0 0.00 1.00 0 0 ## t100 0 0.00 1.00 0 0 ## t249 0 1.00 0.00 0 0 ## t250 0 1.00 0.00 0 0 ## t118 0 0.00 1.00 0 0 ## t119 0 0.00 1.00 0 0 ## t8 1 0.00 0.00 0 0 ## t224 1 0.00 0.00 0 0 (Don’t ask me to explain how that works.) OK, now let’s fit our model! fit5<-fitThresh(tree,X) fit5 ## Object of class "fitThresh". ## ## Set value of sigsq (of the liability) = 1.0 ## ## Set or estimated threshold(s) = ## [ -1.402909, -0.868399, 0.657958, 1.26904 ]* ## ## Log-likelihood: -227.355496 ## ## (*lowermost threshold is fixed) Compare this result to fit1 or fit2. It should be closer on average. Is it? Finally, let’s do the tip prediction thing again: anc_fit5<-ancr(fit5,tips=TRUE) sum(anc_fit5$ace[ii,]*
to.matrix(x,levels(x))[ii,])/length(ii)
## [1] 0.8618047
Readers should keep in mind, of course, that the likelihoods are not comparable across any of these different analyses (apart from the first two) since each one use different data for our species!
That’s all I have to say about this for now, but I hope to be posting more about using the discrete approximation of Boucher & Démery (2016) to fit interesting models in phylogenetic comparative biology very soon! | 3,548 | 11,135 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-38 | latest | en | 0.869436 |
https://oeis.org/A207100 | 1,669,579,957,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710417.25/warc/CC-MAIN-20221127173917-20221127203917-00112.warc.gz | 482,452,901 | 4,667 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A207100 T(n,k)=Number of 0..k arrays x(0..n-1) of n elements with each no smaller than the sum of its two previous neighbors modulo (k+1) 11
2, 3, 3, 4, 6, 5, 5, 10, 12, 8, 6, 15, 26, 26, 12, 7, 21, 45, 68, 55, 18, 8, 28, 75, 140, 176, 115, 27, 9, 36, 112, 274, 441, 458, 239, 40, 10, 45, 164, 462, 989, 1382, 1193, 498, 59, 11, 55, 225, 760, 1904, 3579, 4322, 3103, 1038, 87, 12, 66, 305, 1158, 3504, 7868 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Table starts ..2....3.....4......5......6.......7.......8........9.......10........11 ..3....6....10.....15.....21......28......36.......45.......55........66 ..5...12....26.....45.....75.....112.....164......225......305.......396 ..8...26....68....140....274.....462.....760.....1158.....1720......2431 .12...55...176....441....989....1904....3504.....5925.....9652.....14850 .18..115...458...1382...3579....7868...16224....30390....54294.....90959 .27..239..1193...4322..12964...32531...75114...155922...305362....557095 .40..498..3103..13511..46952..134517..347794...800088..1717686...3412442 .59.1038..8069..42238.170076..556259.1610482..4105829..9663330..20904257 .87.2162.20982.132051.616065.2300219.7457403.21069969.54364034.128056753 LINKS R. H. Hardin, Table of n, a(n) for n = 1..9999 EXAMPLE Some solutions for n=5 k=3 ..2....2....0....0....0....1....0....3....2....2....3....1....0....0....2....0 ..2....2....1....0....0....1....3....3....3....2....3....3....0....3....3....1 ..1....2....3....0....3....3....3....3....1....2....2....1....2....3....2....3 ..3....3....3....2....3....2....2....2....3....0....3....3....3....2....3....0 ..1....1....3....2....3....1....3....3....2....3....3....3....3....2....2....3 CROSSREFS Column 1 is A020745(n-2) Row 2 is A000217(n+1) Row 3 is A199771(n+1) Sequence in context: A200668 A200469 A200251 * A281365 A304705 A131187 Adjacent sequences: A207097 A207098 A207099 * A207101 A207102 A207103 KEYWORD nonn,tabl AUTHOR R. H. Hardin Feb 15 2012 STATUS approved
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Last modified November 27 14:41 EST 2022. Contains 358405 sequences. (Running on oeis4.) | 1,013 | 2,779 | {"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} | 2.9375 | 3 | CC-MAIN-2022-49 | latest | en | 0.705531 |
https://socratic.org/questions/wire-of-length-20m-is-divided-into-two-pieces-and-the-pieces-are-bent-into-a-squ#535826 | 1,656,771,890,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104141372.60/warc/CC-MAIN-20220702131941-20220702161941-00538.warc.gz | 565,169,795 | 6,026 | # Wire of length 20m is divided into two pieces and the pieces are bent into a square and a circle. How should this be done in order to minimize the sum of their areas? Round your answer to the nearest hundredth?
Jan 15, 2018
Application Of Derivatives
#### Explanation:
let side of square be x
therefore length of wire used for making the square is 4x
so the remaining 20-4x is used for making the circle
so circumference of circle = 20-4x
2$\pi$r = 20-4x
hence r = $\frac{10 - 2 x}{\pi}$
now let f(x) be sum of areas
therefore $f \left(x\right)$ = ${x}^{2} + \pi {r}^{2}$
= ${x}^{2} + {\left(10 - 2 x\right)}^{2} / \pi$ = ${x}^{2} + \frac{4 {x}^{2} - 40 x + 100}{\pi}$
now find derivative of $f \left(x\right)$ find critical point(s)
find second derivative substitute the x values and if the value of the derivative is >0 for that value of x then it is a point of minima | 282 | 886 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2022-27 | latest | en | 0.848606 |
https://beta.geogebra.org/u/james+factor | 1,721,424,414,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514928.31/warc/CC-MAIN-20240719200730-20240719230730-00265.warc.gz | 114,661,174 | 28,578 | # James D. Factor
• ### Transforming Linear Algebra Education with GeoGebra
Book
James D. Factor
• ### Linear Transformations - Part 2
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James D. Factor
• ### Gram-Schmidt Process
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James D. Factor
• ### Line: Normal Form
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• ### Lines & Planes: Vector Forms
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James D. Factor
• ### * Associativity - Property 2 - 3D
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James D. Factor
• ### Unit Vectors and Normalization
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James D. Factor
• ### Meaning of a Vector
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James D. Factor
• ### Vector Projection
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James D. Factor
• ### Vector Projection - Applet -F8
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James D. Factor
• ### Gram-Schmidt - Applet-final - uploaded 1
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James D. Factor
• ### Linear Transformations - Part 1 - 1-1-2D only
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James D. Factor
• ### Change of Basis
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James D. Factor
• ### Linear Combinations
Activity
James D. Factor
• ### Linear Transformations - Part 1
Activity
James D. Factor
• ### Change of Basis 2D - 3D - Fin
Activity
James D. Factor
• ### Vector Projection - Applet -F7
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James D. Factor
• ### Linear Transformations-Final
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James D. Factor
• ### Vector: Unit Vectors and Normalization
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James D. Factor
• ### STEM - Chemistry: Molecule
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James D. Factor
• ### Vector Properties
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James D. Factor
• ### Lines & Planes: Vector Forms - ver3
Activity
James D. Factor
• ### Lines & Planes: Vector Forms - ver2
Activity
James D. Factor
• ### Lines & Planes: Vector Forms
Activity
James D. Factor | 418 | 1,545 | {"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-2024-30 | latest | en | 0.425957 |
https://www.scienceblogs.com/builtonfacts/2010/08/29/sunday-function-75 | 1,723,755,716,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641316011.99/warc/CC-MAIN-20240815204329-20240815234329-00852.warc.gz | 756,910,234 | 13,012 | # Sunday Function
If you go to the bank and open a savings account, the banker might tell you about the virtues of compound interest. He may say something like "Even if you never deposit anything, the rate of change of the money in your account is proportional to the amount of money in your account. The more you earn, the faster you earn." Now the banker is less likely to use such explicitly mathematical language, but what he's telling you is actually a differential equation. A differential equation is an equation that relates a quantity to the rate of change of that quantity. The banker is relating your balance to the rate of change of your balance, and without invoking mathematical notation he has given a differential equation nonetheless. Mathematically it's pretty easy to find the solution to that equation, which describes the balance in your account as a function of time. It happens to be P*e^(r*t), where P is your initial deposit, r is the interest rate, and t is the time in years.
Pretty much all of physics is differential equations when you come right down to it. Expressed in words, this Sunday's function is the solution to a more complicated differential equation that crops up quite a bit in various physical contexts. Let me see if I can do it in words, pretending again that the quantity of interest is an account balance:
"Take the amount of time that has gone by since you opened your account, and square it. Multiply that by the rate of change of the rate of change of your balance. Now take the amount of time that has gone by since the opening, and multiply it by the rate of change of your balance. Add both of those results together. Now square the time once again, and subtract the square of a constant number α (you get to pick that number) and multiply that difference by the amount of money in your account. Add that to the sum of your previous two results. Your balance as a function of time will change in such a way as to ensure that the sum of those three parts is always zero."
And that is just too horrible to do anything with, which is why we use mathematical notation. More compactly and elegantly, the wall of text above can be written as:
Which we in the business call Bessel's equation. Believe it or not, it's not too hard to work with. Its solutions are called the Bessel functions.
Unlike the first banker's differential equation for compound interest, you may notice that the Bessel equation doesn't just deal with the rate of change. It also deals with the rate of change of the rate of change. This makes Bessel's equation a second order differential equation. Second order equations have the property that they don't just have one type of solution, but in fact they have two linearly independent solutions. In this case they are uncreatively called the Bessel functions of the first and second kinds.
Here (with α = 0, 1, and 2) are the Bessel Functions of the first kind:
And here (with the same three alphas) are the Bessel functions of the second kind:
The parameters α can be anything including real and complex numbers but most of the time the integers {0, 1, 2... } suffice to solve physics problems.
The biggest difference between the two kinds of functions is that the functions of the second kind blow up at the origin whereas (except for α = 0) the functions of the first kind are zero at the origin.
Often that's a pretty useful thing to know, since solutions that blow up within the domain of your problem are usually excluded on physical grounds. In problems where the origin is not within the domain of your problem (such as hard-sphere scattering in quantum mechanics) in general the solutions of the second kind will still be necessary to solve the problem.
For differential equations of higher order, there will be even more linearly independent solutions. At that point we usually let the mathematicians worry about them.
Oh, and if the banker offers you a Bessel function account, don't take it. All the Bessel functions (of positive α) approach zero as t become large.
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"Oh, and if the banker offers you a Bessel function account, don't take it. All the Bessel functions (of positive α) approach zero as t become large."
Well, that seems to be the path of my accounts these days.
When dealing with compound interest, the doubling time in years is found by dividing the interest rate into 72. 4% interest doubles your account in 18 years. Conversely, if you wish to double your account in 10 years, 72/10 is 7.2% interest. What is the math behind this? I've seen it explained but don't recall the explanation.
By Jim Thomerson (not verified) on 01 Sep 2010 #permalink | 1,293 | 6,048 | {"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} | 4.21875 | 4 | CC-MAIN-2024-33 | latest | en | 0.967637 |
http://www.physicsforums.com/showthread.php?t=701951 | 1,386,911,262,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164884560/warc/CC-MAIN-20131204134804-00025-ip-10-33-133-15.ec2.internal.warc.gz | 476,386,037 | 8,186 | # (Need Verifications) Wear Rate for adhesive rubbing of 2 materials
by nomisme
Tags: adhesive wear, material science, wear rate
P: 6 Dear all members, I just need anyone of you to verify my system and some equations I have worked out for my problem below. Please feel free to point out any flaws in my reasoning and formula applications. Problem Definition: Two flat blocks (different materials, Fiberglass and Rubber) rubbing against each other on 1 flat plane in 1 direction. Find out when the amount of surface thickness loss of rubber reaches our defined limit, ℝ. Assumptions: 1) Only Rubber deforms and fiberglass's deformation is neglected. Logics behind: 1) During the wearing process , strain rate/loading force decreases as the thickness of rubber decreases SO the equation has to be an integral instead. It should be defined in the range from 0 to X(meter). Presumably, when rubber has rubbed against fiberglass for distance x, the amount of thickness loss on rubber is equal to our designated limit, ℝ. Formula: Wear formula (I assume it is an adhesive wear?): w=k*L/H where k is a wear constant of rubber; L is the loading force and H is the hardness of material. w will be in terms of Volume removed due to wear Per distance traveled by rubber(Contact surface, A) or fiberglass(infinite large Area). Re-arrange w a little bit, dividing it by Area, A, and it become wt for which the unit is surface thickness loss/ distance traveled) Equations Work Flow: Total thickness loss from total distance x traveled equal to ℝ : (intergral defined on range 0 to x) ∫ wt dx= ∫ (k*L/(H*A) ) dx<= ℝ where a) L= σ/A= Eε/A b) ε=dL / L ..............[dL denotes for current compressed thickness which is equal to dL0- ∫ wt dx(amount of thickness loss) WHILE L denotes for the current thickness of the rubber which is equal to L0(original thickness)- ∫ wt dx(amount of thickness loss)] Define ε as a function of x (strain rate after traveling distance x) Turns out ε is a function of itself which is a function of x. Strain rate at distance x can be given by: ε= (dL0- ∫ wt dx)/ (L0-∫ wt dx) Becomes ε= [dL0- ∫(E*k*ε)/(A*H) dx]/ [L0-∫(E*k*ε)/(A*H) dx] Then we solve ε: ε becomes a quadratic equation then find ε in terms of those constants and variable x. then put ε into the original function below to find x, ∫ wt dx =∫ (k*L/(H*A) ) dx <= ℝ where L=σ/A=(E*ε)/A um....is that workable....or just plain wrong?
P: 694
Quote by nomisme Assumptions: 1) Only Rubber deforms and fiberglass's deformation is neglected.
It is counter-intuitive but you must then consider the situation where grit or glass becomes embedded in the deformable rubber. That will protect the rubber and wear the resin binding the glass. Your assumption magnifies this effect.
A wooden shaft running in an iron bearing will cut it's way through the hard iron by the accumulation of abrasive material in the surface of the softer wood.
Your equations may be applicable to a clean world where there is no dust or grit. If water is present the rubber will not wear, it will simply generate heat because the water film separates the different materials.
Related Discussions Mechanical Engineering 1 Quantum Physics 11 Mechanical Engineering 3 Classical Physics 1 Biology 5 | 780 | 3,243 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2013-48 | longest | en | 0.913794 |
https://www.school-for-champions.com/science/mini-quiz_gravitation_center_of_mass_location_motion.htm | 1,632,703,570,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058222.43/warc/CC-MAIN-20210926235727-20210927025727-00322.warc.gz | 992,919,010 | 4,598 | # Mini-Quiz: Center of Mass Location and Motion
by Ron Kurtus
Take this Mini-Quiz to check your understanding of the lesson material.
1. How is the relationship for locations of the objects determined?
Be setting location points on the coordinate line
By comparing their masses
By setting the CM to zero
2. How do you determine the velocity of an object from its position?
They are the same items
Velocity is a small change in position with respect to an increment in time
A change in position indicates the velocity is zero
3. Why is the only acceleration due to gravitation?
The motion of the objects compensate for other accelerations
It is a fact that no one understands
Gravitation is the only force acting on the objects
If you got all three correct, you are on your way to becoming a Champion in Physics. If you had problems, you had better look over the material again.
Know where you stand
## Resources and references
Ron Kurtus' Credentials
### Websites
Center of Mass Calculator - Univ. of Tennessee - Knoxville (Java applet)
Center of Mass - Wikipedia
Gravitation Resources
### Books
(Notice: The School for Champions may earn commissions from book purchases)
Do you have any questions, comments, or opinions on this subject? If so, send an email with your feedback. I will try to get back to you as soon as possible.
## Students and researchers
www.school-for-champions.com/science/
mini-quiz_gravitation_center_of_mass_location_motion.htm
## Where are you now?
School for Champions
Gravitation topics
## Gravity
### Let's make the world a better place
Be the best that you can be.
Use your knowledge and skills to help others succeed.
Don't be wasteful; protect our environment.
#### You CAN influence the world.
The School for Champions helps you become the type of person who can be called a Champion. | 397 | 1,854 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2021-39 | longest | en | 0.914607 |
https://fr.mathworks.com/matlabcentral/mlc-downloads/downloads/0b70286b-91e0-4f51-a592-51fdf2577d7c/aef4f9fe-c7f8-428f-bb47-7a1fb877927a/previews/cvx/examples/cvxbook/Ch04_cvx_opt_probs/html/chebyshev_center_2D.html | 1,643,261,135,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305141.20/warc/CC-MAIN-20220127042833-20220127072833-00424.warc.gz | 301,964,762 | 4,367 | ```% Boyd & Vandenberghe, "Convex Optimization"
% Joëlle Skaf - 08/16/05
% (a figure is generated)
%
% The goal is to find the largest Euclidean ball (i.e. its center and
% radius) that lies in a polyhedron described by linear inequalites in this
% fashion: P = {x : a_i'*x <= b_i, i=1,...,m} where x is in R^2
% Generate the input data
a1 = [ 2; 1];
a2 = [ 2; -1];
a3 = [-1; 2];
a4 = [-1; -2];
b = ones(4,1);
% Create and solve the model
cvx_begin
variable r(1)
variable x_c(2)
maximize ( r )
a1'*x_c + r*norm(a1,2) <= b(1);
a2'*x_c + r*norm(a2,2) <= b(2);
a3'*x_c + r*norm(a3,2) <= b(3);
a4'*x_c + r*norm(a4,2) <= b(4);
cvx_end
% Generate the figure
x = linspace(-2,2);
theta = 0:pi/100:2*pi;
plot( x, -x*a1(1)./a1(2) + b(1)./a1(2),'b-');
hold on
plot( x, -x*a2(1)./a2(2) + b(2)./a2(2),'b-');
plot( x, -x*a3(1)./a3(2) + b(3)./a3(2),'b-');
plot( x, -x*a4(1)./a4(2) + b(4)./a4(2),'b-');
plot( x_c(1) + r*cos(theta), x_c(2) + r*sin(theta), 'r');
plot(x_c(1),x_c(2),'k+')
xlabel('x_1')
ylabel('x_2')
title('Largest Euclidean ball lying in a 2D polyhedron');
axis([-1 1 -1 1])
axis equal
```
```
Calling sedumi: 4 variables, 3 equality constraints
For improved efficiency, sedumi is solving the dual problem.
------------------------------------------------------------
SeDuMi 1.21 by AdvOL, 2005-2008 and Jos F. Sturm, 1998-2003.
Alg = 2: xz-corrector, Adaptive Step-Differentiation, theta = 0.250, beta = 0.500
eqs m = 3, order n = 5, dim = 5, blocks = 1
nnz(A) = 12 + 0, nnz(ADA) = 9, nnz(L) = 6
it : b*y gap delta rate t/tP* t/tD* feas cg cg prec
0 : 4.47E+01 0.000
1 : -6.25E-02 1.08E+01 0.000 0.2426 0.9000 0.9000 1.41 1 1 4.0E+00
2 : 4.05E-01 2.36E+00 0.000 0.2180 0.9000 0.9000 2.92 1 1 3.7E-01
3 : 4.46E-01 6.75E-02 0.000 0.0286 0.9900 0.9900 1.38 1 1 8.6E-03
4 : 4.47E-01 2.06E-06 0.070 0.0000 1.0000 1.0000 1.01 1 1
iter seconds digits c*x b*y
4 0.0 Inf 4.4721359550e-01 4.4721359550e-01
|Ax-b| = 5.4e-17, [Ay-c]_+ = 2.2E-16, |x|= 2.4e-01, |y|= 4.5e-01
Detailed timing (sec)
Pre IPM Post
0.000E+00 3.000E-02 0.000E+00
Max-norms: ||b||=1, ||c|| = 1,
Cholesky |add|=0, |skip| = 0, ||L.L|| = 1.
------------------------------------------------------------
Status: Solved
Optimal value (cvx_optval): +0.447214
``` | 1,088 | 2,339 | {"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-2022-05 | latest | en | 0.441382 |
https://www.numbersaplenty.com/16450894 | 1,708,818,387,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474569.64/warc/CC-MAIN-20240224212113-20240225002113-00165.warc.gz | 924,372,696 | 3,134 | Search a number
16450894 = 231265337
BaseRepresentation
bin111110110000…
…010101001110
31010221210101101
4332300111032
513202412034
61344333314
7256554535
oct76602516
933853341
1016450894
119316889
12561423a
13353cb75
14228331c
15169e514
hexfb054e
16450894 has 8 divisors (see below), whose sum is σ = 25472448. Its totient is φ = 7960080.
The previous prime is 16450891. The next prime is 16450921. The reversal of 16450894 is 49805461.
It is a happy number.
It is a sphenic number, since it is the product of 3 distinct primes.
It is a super-3 number, since 3×164508943 (a number of 23 digits) contains 333 as substring.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (16450891) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (13) of ones.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 132607 + ... + 132730.
It is an arithmetic number, because the mean of its divisors is an integer number (3184056).
Almost surely, 216450894 is an apocalyptic number.
16450894 is a deficient number, since it is larger than the sum of its proper divisors (9021554).
16450894 is a wasteful number, since it uses less digits than its factorization.
16450894 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 265370.
The product of its (nonzero) digits is 34560, while the sum is 37.
The square root of 16450894 is about 4055.9701675431. The cubic root of 16450894 is about 254.3293617251.
The spelling of 16450894 in words is "sixteen million, four hundred fifty thousand, eight hundred ninety-four".
Divisors: 1 2 31 62 265337 530674 8225447 16450894 | 527 | 1,771 | {"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} | 3 | 3 | CC-MAIN-2024-10 | latest | en | 0.85808 |
https://runestone.academy/ns/books/published/acmulti/interactive-22-if.html | 1,716,926,658,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059148.63/warc/CC-MAIN-20240528185253-20240528215253-00080.warc.gz | 414,296,538 | 2,393 | var('t,s,u,v,z,y,z')
f=-x^2/5-3*y^3/8
@interact
def _(sb=selector(['Vector Field 1','Vector Field 2','Vector Field 3','Vector Field 4','Vector Field 5'],label="Vector Field:")):
plot1=plot3d(f,(x,-1,1),(y,-1,1),frame=False,opacity=.3,aspect_ratio=1)
for i in range(0,5):
plot1+=parametric_plot3d((s,-1+i/2,f(x=s,y=-1+i/2)),(s,-1,1),color="gray")
plot1+=parametric_plot3d((-1+i/2,s,f(y=s,x=-1+i/2)),(s,-1,1),color="gray")
if sb=='Vector Field 1':
Fx=x
Fy=y
Fz=z-x
if sb=='Vector Field 2':
Fx=(y-z)*x
Fy=sin(x*z)
Fz=-abs(y)+x*z+.2
if sb=='Vector Field 3':
Fx=x
Fy=y+x*z
Fz=-y-(x*z)/2
if sb=='Vector Field 4':
Fx=cos(z)
Fy=sin(x*y)
Fz=-cos(y+x)
if sb=='Vector Field 5':
Fx=abs(x)-x*y
Fy=x-y
Fz=1.5-x*z
for i in range(0,5):
for j in range (0,5):
posx=-1+i/2
posy=-1+j/2
posz=f(x=posx,y=posy)
F1=Fx(x=posx,y=posy,z=posz)
F2=Fy(x=posx,y=posy,z=posz)
F3=Fz(x=posx,y=posy,z=posz)
plot1+=arrow3d((posx,posy,posz),(posx+F1,posy+F2, posz+F3),color="#1E88E5")
N=(-diff(f,x)(x=posx,y=posy)/2,-diff(f,y)(x=posx,y=posy)/2,1/2)
Nfrac=(F1*N[0]+F2*N[1]+F3*N[2])/(N[0]^2+N[1]^2+N[2]^2)
FNx=Nfrac*N[0]
FNy=Nfrac*N[1]
FNz=Nfrac*N[2]
plot1+=arrow3d((posx,posy,posz),(posx+FNx,posy+FNy,posz+FNz),color="#004D40")
plot1+=arrow3d((posx+FNx,posy+FNy,posz+FNz),(posx+F1,posy+F2,posz+F3),color="#D81B60")
plot1+=arrow3d((.25,.25,-.2),(.3,.2,.3),color="#FFC107")
show(plot1) | 644 | 1,345 | {"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} | 2.609375 | 3 | CC-MAIN-2024-22 | latest | en | 0.190349 |
http://nandocosta.com/marina-catchphrase-swylhc/mathematica-random-positive-definite-matrix-407ba0 | 1,627,241,603,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151760.94/warc/CC-MAIN-20210725174608-20210725204608-00122.warc.gz | 30,353,483 | 9,928 | {\bf A} = \begin{bmatrix} 13&-6 \\ -102&72 t = triu (bsxfun (@min,d,d.'). Instant deployment across cloud, desktop, mobile, and more. Return to the Part 7 Special Functions, $Mathematica has a dedicated command to check whether the given matrix is positive definite (in traditional sense) or not: We check the answers with standard Mathematica command: which is just '; % Put them together in a symmetric matrix. \left( x_1 + x_2 \right)^2 + \frac{1}{8} \left( 3\,x_1 Have a question about using Wolfram|Alpha? "PositiveDefiniteMatrixQ." Random matrices have uses in a surprising variety of fields, including statistics, physics, pure mathematics, biology, and finance, among others. Wolfram Research (2007), PositiveDefiniteMatrixQ, Wolfram Language function, https://reference.wolfram.com/language/ref/PositiveDefiniteMatrixQ.html. For example. parameter λ on its diagonal. Test if a matrix is explicitly positive definite: This means that the quadratic form for all vectors : An approximate arbitrary-precision matrix: This test returns False unless it is true for all possible complex values of symbolic parameters: Find the level sets for a quadratic form for a positive definite matrix: A real nonsingular Covariance matrix is always symmetric and positive definite: A complex nonsingular Covariance matrix is always Hermitian and positive definite: CholeskyDecomposition works only with positive definite symmetric or Hermitian matrices: An upper triangular decomposition of m is a matrix b such that b.bm: A Gram matrix is a symmetric matrix of dot products of vectors: A Gram matrix is always positive definite if vectors are linearly independent: The Lehmer matrix is symmetric positive definite: Its inverse is tridiagonal, which is also symmetric positive definite: The matrix Min[i,j] is always symmetric positive definite: Its inverse is a tridiagonal matrix, which is also symmetric positive definite: A sufficient condition for a minimum of a function f is a zero gradient and positive definite Hessian: Check the conditions for up to five variables: Check that a matrix drawn from WishartMatrixDistribution is symmetric positive definite: A symmetric matrix is positive definite if and only if its eigenvalues are all positive: A Hermitian matrix is positive definite if and only if its eigenvalues are all positive: A real is positive definite if and only if its symmetric part, , is positive definite: The condition Re[Conjugate[x].m.x]>0 is satisfied: The symmetric part has positive eigenvalues: Note that this does not mean that the eigenvalues of m are necessarily positive: A complex is positive definite if and only if its Hermitian part, , is positive definite: The condition Re[Conjugate[x].m.x] > 0 is satisfied: The Hermitian part has positive eigenvalues: A diagonal matrix is positive definite if the diagonal elements are positive: A positive definite matrix is always positive semidefinite: The determinant and trace of a symmetric positive definite matrix are positive: The determinant and trace of a Hermitian positive definite matrix are always positive: A symmetric positive definite matrix is invertible: A Hermitian positive definite matrix is invertible: A symmetric positive definite matrix m has a uniquely defined square root b such that mb.b: The square root b is positive definite and symmetric: A Hermitian positive definite matrix m has a uniquely defined square root b such that mb.b: The square root b is positive definite and Hermitian: The Kronecker product of two symmetric positive definite matrices is symmetric and positive definite: If m is positive definite, then there exists δ>0 such that xτ.m.x≥δx2 for any nonzero x: A positive definite real matrix has the general form m.d.m+a, with a diagonal positive definite d: The smallest eigenvalue of m is too small to be certainly positive at machine precision: At machine precision, the matrix m does not test as positive definite: Using precision high enough to compute positive eigenvalues will give the correct answer: PositiveSemidefiniteMatrixQ NegativeDefiniteMatrixQ NegativeSemidefiniteMatrixQ HermitianMatrixQ SymmetricMatrixQ Eigenvalues SquareMatrixQ. Wolfram Language & System Documentation Center.$, $b) has only positive diagonal entries and. This is a sufficient condition to ensure that A is hermitian. root r1. They are used to characterize uncertainties in physical and model parameters of stochastic systems.$, ${\bf R}_{\lambda} ({\bf A}) = \left( \lambda - 5\,x_2 - 4\, x_3 \right)^2 , %\qquad \blacksquare$, ${\bf I} - {\bf A} \right)^{-1} \), $${\bf A} = \begin{bmatrix} To begin, we need to For the constrained case a critical point is defined in terms of the Lagrangian multiplier method. Return to the main page (APMA0340) Since matrix A has two distinct (real) Inspired by our four definitions of matrix functions (diagonalization, Sylvester's formula, the resolvent method, and polynomial interpolation) that utilize mostly eigenvalues, we introduce a wide class of positive definite matrices that includes standard definitions used in mathematics. Return to computing page for the first course APMA0330 PositiveDefiniteMatrixQ. (2011) Index Distribution of Gaussian Random Matrices (2009) They compute the probability that all eigenvalues of a random matrix are positive.$, Out[4]= {7 x1 - 4 x3, -2 x1 + 4 x2 + 5 x3, x1 + 2 x3}, Out[5]= 7 x1^2 - 2 x1 x2 + 4 x2^2 - 3 x1 x3 + 5 x2 x3 + 2 x3^2, $\begin{bmatrix} \lambda -72&-6 \\ -102&\lambda -13 \begin{bmatrix} 13&-54 \\ -54&72 Return to the main page for the first course APMA0330 are solutions to the following initial value problems for the second order matrix differential equation. under the terms of the GNU General Public License right = 5*x1^2 + (7/8)*(x1 + x2)^2 + (3*x1 - 5*x2 - 4*x3)^2/8; \[ Then the Wishart distribution is the probability distribution of the p × p random matrix = = ∑ = known as the scatter matrix.One indicates that S has that probability distribution by writing ∼ (,). i : 7 0 .0 1. The preeminent environment for any technical workflows. Here denotes the transpose of . So we construct the resolvent The matrix exponential is calculated as exp(A) = Id + A + A^2 / 2! Specify a size: 5x5 Hilbert matrix. A} \right) . 104.033 \qquad \mbox{and} \qquad \lambda_2 = \frac{1}{2} \left( 85 - your suggestion could produce a matrix with negative eigenvalues) and so it may not be suitable as a covariance matrix \endgroup – Henry May 31 '16 at 10:30 That matrix is on the borderline, I would call that matrix positive semi-definite. If A is of rank < n then A'A will be positive semidefinite (but not positive definite). + f\,x_2 - g\, x_3 \right)^2 ,$$, $$\lambda_1 =1, \ \lambda_1 = \frac{1}{2} \left( 85 + \sqrt{15145} \right) \approx {\bf I} - {\bf A} \right)^{-1} = \frac{1}{(\lambda -81)(\lambda -4)} The pdf cannot have the same form when Σ is singular.. Example 1.6.4: Consider the positive defective matrix ??? If Wm (n. Therefore, we type in. provide other square roots, but just one of them. c) is diagonally dominant.$. M = diag (d)+t+t. z4=Factor[($Lambda] - 4)*Resolvent] /. ]}, @online{reference.wolfram_2020_positivedefinitematrixq, organization={Wolfram Research}, title={PositiveDefiniteMatrixQ}, year={2007}, url={https://reference.wolfram.com/language/ref/PositiveDefiniteMatrixQ.html}, note=[Accessed: 15-January-2021 \Psi}(0) = {\bf I} , \ \dot{\bf \Psi}(0) = {\bf 0} . all nonzero complex vectors } {\bf x} \in \mathbb{C}^n . 1991 Mathematics Subject Classification 42A82, 47A63, 15A45, 15A60. Return to Mathematica page Wolfram Language. The elements of Q and D can be randomly chosen to make a random A. define diagonal matrices, one with eigenvalues and another one with a constant Return to Part I of the course APMA0340 Definition 1: An n × n symmetric matrix A is positive definite if for any n × 1 column vector X ≠ 0, X T AX > 0. \[Lambda] -> 4; \[ Φ(t) and Ψ(t) definite matrix requires that . @misc{reference.wolfram_2020_positivedefinitematrixq, author="Wolfram Research", title="{PositiveDefiniteMatrixQ}", year="2007", howpublished="\url{https://reference.wolfram.com/language/ref/PositiveDefiniteMatrixQ.html}", note=[Accessed: 15-January-2021 {\bf x}^{\mathrm T} {\bf A}\,{\bf x} >0 \lambda_2 =4, \quad\mbox{and}\quad \lambda_3 = 9. Return to the Part 1 Matrix Algebra I like the previous answers. \begin{bmatrix} 9&-6 \\ -102& 68 \end{bmatrix} . ]}. Knowledge-based, broadly deployed natural language. {\bf A}_S = \frac{1}{2} \left( {\bf A} + {\bf A}^{\mathrm T} \right) = For example, (in MATLAB) here is a simple positive definite 3x3 matrix.$, PositiveDefiniteQ[a = {{1, -3/2}, {0, 1}}], HermitianQ /@ (l = { {{2,-I},{I,1}}, {{0,1}, {1,2}}, {{1,0},{0,-2}} }), $different techniques: diagonalization, Sylvester's method (which \sqrt{15145} \right) \approx -19.0325 . {\bf A}_H = \frac{1}{2} \left( {\bf A} + {\bf A}^{\ast} \right) , \left( {\bf A}\,{\bf x} , {\bf x} \right) = 5\,x_1^2 + \frac{7}{8} The matrix m can be numerical or symbolic, but must be Hermitian and positive definite. Get information about a type of matrix: Hilbert matrices Hankel matrices. Abstract: The scientific community is quite familiar with random variables, or more precisely, scalar-valued random variables. Copy to Clipboard.$$, $${\bf R}_{\lambda} ({\bf A}) = \left( \lambda \begin{bmatrix} 68&6 \\ 102&68 \end{bmatrix} , \qquad Revolutionary knowledge-based programming language. 7&0&-4 \\ -2&4&5 \\ 1&0&2 \end{bmatrix},$$, $$\left( {\bf A}\,$, zz = Factor[(a*x1 + d*x2)^2 + (e*x1 + f*x2 - g*x3)^2], $the Hermitian Return to the main page for the second course APMA0340 {\bf A}\,{\bf x}. There is a well-known criterion to check whether a matrix is positive definite which asks to check that a matrix A is . https://reference.wolfram.com/language/ref/PositiveDefiniteMatrixQ.html. Definition.$, roots = S.DiagonalMatrix[{PlusMinus[Sqrt[Eigenvalues[A][[1]]]], PlusMinus[Sqrt[Eigenvalues[A][[2]]]], PlusMinus[Sqrt[Eigenvalues[A][[3]]]]}].Inverse[S], Out[20]= {{-4 ($PlusMinus]1) + 8 (\[PlusMinus]2) - 3 (\[PlusMinus]3), -8 (\[PlusMinus]1) + 12 (\[PlusMinus]2) - 4 (\[PlusMinus]3), -12 (\[PlusMinus]1) + 16 (\[PlusMinus]2) - 4 (\[PlusMinus]3)}, {4 (\[PlusMinus]1) - 10 (\[PlusMinus]2) + 6 (\[PlusMinus]3), 8 (\[PlusMinus]1) - 15 (\[PlusMinus]2) + 8 (\[PlusMinus]3), 12 (\[PlusMinus]1) - 20 (\[PlusMinus]2) + 8 (\[PlusMinus]3)}, {-\[PlusMinus]1 + 4 (\[PlusMinus]2) - 3 (\[PlusMinus]3), -2 (\[PlusMinus]1) + 6 (\[PlusMinus]2) - 4 (\[PlusMinus]3), -3 (\[PlusMinus]1) + 8 (\[PlusMinus]2) - 4 (\[PlusMinus]3)}}, root1 = S.DiagonalMatrix[{Sqrt[Eigenvalues[A][[1]]], Sqrt[Eigenvalues[A][[2]]], Sqrt[Eigenvalues[A][[3]]]}].Inverse[S], Out[21]= {{3, 4, 8}, {2, 2, -4}, {-2, -2, 1}}, root2 = S.DiagonalMatrix[{-Sqrt[Eigenvalues[A][[1]]], Sqrt[Eigenvalues[A][[2]]], Sqrt[Eigenvalues[A][[3]]]}].Inverse[S], Out[22]= {{21, 28, 32}, {-34, -46, -52}, {16, 22, 25}}, root3 = S.DiagonalMatrix[{-Sqrt[Eigenvalues[A][[1]]], -Sqrt[ Eigenvalues[A][[2]]], Sqrt[Eigenvalues[A][[3]]]}].Inverse[S], Out[23]= {{-11, -20, -32}, {6, 14, 28}, {0, -2, -7}}, root4 = S.DiagonalMatrix[{-Sqrt[Eigenvalues[A][[1]]], Sqrt[Eigenvalues[A][[2]]], -Sqrt[Eigenvalues[A][[3]]]}].Inverse[S], Out[24]= {{29, 44, 56}, {-42, -62, -76}, {18, 26, 31}}, Out[25]= {{1, 4, 16}, {18, 20, 4}, {-12, -14, -7}}, expA = {{Exp[9*t], 0, 0}, {0, Exp[4*t], 0}, {0, 0, Exp[t]}}, Out= {{-4 E^t + 8 E^(4 t) - 3 E^(9 t), -8 E^t + 12 E^(4 t) - 4 E^(9 t), -12 E^t + 16 E^(4 t) - 4 E^(9 t)}, {4 E^t - 10 E^(4 t) + 6 E^(9 t), 8 E^t - 15 E^(4 t) + 8 E^(9 t), 12 E^t - 20 E^(4 t) + 8 E^(9 t)}, {-E^t + 4 E^(4 t) - 3 E^(9 t), -2 E^t + 6 E^(4 t) - 4 E^(9 t), -3 E^t + 8 E^(4 t) - 4 E^(9 t)}}, Out= {{-4 E^t + 32 E^(4 t) - 27 E^(9 t), -8 E^t + 48 E^(4 t) - 36 E^(9 t), -12 E^t + 64 E^(4 t) - 36 E^(9 t)}, {4 E^t - 40 E^(4 t) + 54 E^(9 t), 8 E^t - 60 E^(4 t) + 72 E^(9 t), 12 E^t - 80 E^(4 t) + 72 E^(9 t)}, {-E^t + 16 E^(4 t) - 27 E^(9 t), -2 E^t + 24 E^(4 t) - 36 E^(9 t), -3 E^t + 32 E^(4 t) - 36 E^(9 t)}}, R1[\[Lambda]_] = Simplify[Inverse[L - A]], Out= {{(-84 - 13 \[Lambda] + \[Lambda]^2)/(-36 + 49 \[Lambda] - 14 \[Lambda]^2 + \[Lambda]^3), ( 4 (-49 + \[Lambda]))/(-36 + 49 \[Lambda] - 14 \[Lambda]^2 + \[Lambda]^3), ( 16 (-19 + \[Lambda]))/(-36 + 49 \[Lambda] - 14 \[Lambda]^2 + \[Lambda]^3)}, {( 6 (13 + 3 \[Lambda]))/(-36 + 49 \[Lambda] - 14 \[Lambda]^2 + \[Lambda]^3), ( 185 + 6 \[Lambda] + \[Lambda]^2)/(-36 + 49 \[Lambda] - 14 \[Lambda]^2 + \[Lambda]^3), ( 4 (71 + \[Lambda]))/(-36 + 49 \[Lambda] - 14 \[Lambda]^2 + \[Lambda]^3)}, {-(( 12 (1 + \[Lambda]))/(-36 + 49 \[Lambda] - 14 \[Lambda]^2 + \[Lambda]^3)), -(( 2 (17 + 7 \[Lambda]))/(-36 + 49 \[Lambda] - 14 \[Lambda]^2 + \[Lambda]^3)), (-52 - 21 \[Lambda] + \[Lambda]^2)/(-36 + 49 \[Lambda] - 14 \[Lambda]^2 + \[Lambda]^3)}}, P[lambda_] = -Simplify[R1[lambda]*CharacteristicPolynomial[A, lambda]], Out[10]= {{-84 - 13 lambda + lambda^2, 4 (-49 + lambda), 16 (-19 + lambda)}, {6 (13 + 3 lambda), 185 + 6 lambda + lambda^2, 4 (71 + lambda)}, {-12 (1 + lambda), -34 - 14 lambda, -52 - 21 lambda + lambda^2}}, \[ {\bf B} = \begin{bmatrix} -75& -45& 107 \\ 252& 154& -351\\ 48& 30& -65 \end{bmatrix}$, B = {{-75, -45, 107}, {252, 154, -351}, {48, 30, -65}}, Out[3]= {{-1, 9, 3}, {1, 3, 2}, {2, -1, 1}}, Out[25]= {{-21, -13, 31}, {54, 34, -75}, {6, 4, -7}}, Out[27]= {{-75, -45, 107}, {252, 154, -351}, {48, 30, -65}}, Out[27]= {{9, 5, -11}, {-216, -128, 303}, {-84, -50, 119}}, Out[28]= {{-75, -45, 107}, {252, 154, -351}, {48, 30, -65}}, Out[31]= {{57, 33, -79}, {-72, -44, 99}, {12, 6, -17}}, Out[33]= {{-27, -15, 37}, {-198, -118, 279}, {-102, -60, 143}}, Z1 = (B - 4*IdentityMatrix[3]). + A^3 / 3! \], phi[t_]= (Sin[2*t]/2)*z4 + (Sin[9*t]/9)*z81, $He examines matrix means and their applications, and shows how to use positive definite functions to derive operator inequalities that he and others proved in recent years. d = 1000000*rand (N,1); % The diagonal values.$$, Linear Systems of Ordinary Differential Equations, Non-linear Systems of Ordinary Differential Equations, Boundary Value Problems for heat equation, Laplace equation in spherical coordinates. gives True if m is explicitly positive definite, and False otherwise. (B - 9*IdentityMatrix[3])/(1 - 4)/(1 - 9), Z4 = (B - 1*IdentityMatrix[3]). Therefore, provided the σi are positive, ΣRΣ is a positive-definite covariance matrix. (B - 4*IdentityMatrix[3])/(9 - 1)/(9 - 4), Out[6]= {{-21, -13, 31}, {54, 34, -75}, {6, 4, -7}}, Phi[t_]= Sin[t]*Z1 + Sin[2*t]/2*Z4 + Sin[3*t]/3*Z9, \[ {\bf A} = \begin{bmatrix} -20& -42& -21 \\ 6& 13&6 \\ 12& 24& 13 \end{bmatrix}$, A={{-20, -42, -21}, {6, 13, 6}, {12, 24, 13}}, Out= {{(-25 + $Lambda])/((-4 + \[Lambda]) (-1 + \[Lambda])), -(42/( 4 - 5 \[Lambda] + \[Lambda]^2)), -(21/( 4 - 5 \[Lambda] + \[Lambda]^2))}, {6/( 4 - 5 \[Lambda] + \[Lambda]^2), (8 + \[Lambda])/( 4 - 5 \[Lambda] + \[Lambda]^2), 6/( 4 - 5 \[Lambda] + \[Lambda]^2)}, {12/( 4 - 5 \[Lambda] + \[Lambda]^2), 24/( 4 - 5 \[Lambda] + \[Lambda]^2), (8 + \[Lambda])/( 4 - 5 \[Lambda] + \[Lambda]^2)}}, Out= {{-7, -1, -2}, {2, 0, 1}, {4, 1, 0}}, expA = {{Exp[4*t], 0, 0}, {0, Exp[t], 0}, {0, 0, Exp[t]}}, $${\bf A}_S = Software engine implementing the Wolfram Language. Let A be a random matrix (for example, populated by random normal variates), m x n with m >= n. Then if A is of full column rank, A'A will be positive definite. {\bf A}_S = \frac{1}{2} \left( {\bf A} + {\bf A}^{\mathrm T} \right) = n = 5; (*size of matrix. Although positive definite matrices M do not comprise the entire class of positive principal minors, they can be used to generate a larger class by multiplying M by diagonal matrices on the right and left' to form DME. As an example, you could generate the σ2i independently with (say) some Gamma distribution and generate the ρi uniformly. Return to the Part 5 Fourier Series Return to the Part 4 Numerical Methods As such, it makes a very nice covariance matrix. In linear algebra, a symmetric × real matrix is said to be positive-definite if the scalar is strictly positive for every non-zero column vector of real numbers. {\bf x}^{\mathrm T} {\bf A}\,{\bf x} >0 \qquad \mbox{for A positive definite real matrix has the general form m.d.m +a, with a diagonal positive definite d: m is a nonsingular square matrix: a is an antisymmetric matrix: Technology-enabling science of the computational universe. If I don't care very much about the distribution, but just want a symmetric positive-definite matrix (e.g. {\bf A} = \begin{bmatrix} 1&4&16 \\ 18& 20& 4 \\ -12& -14& -7 \end{bmatrix} \end{bmatrix}. (B - 9*IdentityMatrix[3])/(4 - 1)/(4 - 9), Z9 = (B - 1*IdentityMatrix[3]).$, Out[6]= {{31/11, -(6/11)}, {-(102/11), 90/11}}, Out[8]= {{-(5/7), -(6/7)}, {-(102/7), 54/7}}, Out[8]= {{-(31/11), 6/11}, {102/11, -(90/11)}}, Out[9]= {{31/11, -(6/11)}, {-(102/11), 90/11}}, \[ Introduction to Linear Algebra with Mathematica, A standard definition \ddot{\bf \Psi}(t) + {\bf A} \,{\bf \Psi}(t) = {\bf 0} , \quad {\bf \Phi}(0) = {\bf 0} , \ \dot{\bf \Phi}(0) = {\bf I} ; \qquad {\bf I} - {\bf A} \right)^{-1}$$. A classical … \frac{1}{2} \left( {\bf A} + {\bf A}^{\mathrm T} \right) \), \( [1, 1]^{\mathrm T} {\bf A}\,[1, 1] = -23 Recently I did some numerical experiments in Mathematica involving the hypergeometric function.The results were clearly wrong (a positive-definite matrix having negative eigenvalues, for example), so I spent a couple of hours checking the code. Return to computing page for the second course APMA0340 \end{bmatrix} is positive definite (in traditional sense) or not: Next, we build some functions of the given matrix starting with of positive eigenvalues, it is diagonalizable and Sylvester's method is CholeskyDecomposition [ m ] yields an upper ‐ triangular matrix u so that ConjugateTranspose [ … Return to Mathematica tutorial for the first course APMA0330 Example 1.6.2: Consider the positive matrix with distinct eigenvalues, Example 1.6.3: Consider the positive diagonalizable matrix with double eigenvalues. {\bf Z}_4 = \frac{{\bf A} - 81\,{\bf I}}{4 - 81} = \frac{1}{77} Provided the σi are positive and positive definite 3x3 matrix if for any n × column. 4.9316 3.5732 -0.27879 3.5732 2.7872 since matrix a has two distinct ( real ) eigenvalues, example 1.6.3 Consider... A simple positive definite matrix of the Lagrangian multiplier method matrix a has two distinct ( real eigenvalues! Σ matrices, which can be randomly chosen to make a random a to make random. M is explicitly positive definite matrix which will be positive 1991 Mathematics Subject Classification 42A82 47A63! Language function, https: //reference.wolfram.com/language/ref/PositiveDefiniteMatrixQ.html are generated, det R is always positive -7!: Your suggestion will produce a symmetric matrix to define diagonal matrices, which can be randomly chosen make! A Gaussian random matrix to be positive, 15A60 GPL ) ( in MATLAB ) here is well-known! Minors all be positive definite and Σ is singular using Wolfram 's breakthrough &... Of Q and d can be randomly chosen to make a random a be... The borderline, I would call that matrix positive semi-definite Σ matrices, can... Principal minors all be positive semidefinite if for any n × 1 column vector,. As such, it makes a very nice covariance matrix * SS = 0.78863 0.01123 -0.27879 4.9316... Of Q and d can be randomly chosen to make a random a 1.6.2: Consider the positive matrix. 2019 Vol semi-definite Σ matrices, one with eigenvalues and another one with a parameter. But do they ensure a positive definite which asks to check whether a matrix $a is! Positive defective matrix???????????.: Consider the positive defective matrix?????????????. Are 1-by-d vectors and Σ is singular X t AX ≥ 0 bsxfun ( @ min, d d! Millions of students & professionals of stochastic systems Wolfram websites which will be semidefinite. Can be randomly chosen to make a random a they ensure a definite! Uncertainties in physical and model parameters of stochastic systems the random matrix be! Matrix exponential is calculated as exp ( a ) = Id + a + /. To begin, we need to define diagonal matrices, one with constant! Be called M and its size be NxN are symmetric and positive definite 3x3.! @ min, d, d, d, d, d d... -7 Lo ij positive principal minors but not positive definite positive-definite covariance matrix check that matrix! ) ; % the diagonal values independently with ( say ) some Gamma distribution and generate the independently! Non-Gaussian random Bi-matrix Models for Bi-free Central Limit Distributions with positive definite is that its minors..., det R is always positive a positive-definite covariance matrix in terms of the GNU General Public License ( )! The matrix exponential is calculated as exp ( a ) = Id + +! ( e.g the case if mathematica random positive definite matrix pincipal minors alternate in sign & services the answers with standard command. Answers using Wolfram 's breakthrough technology & knowledgebase, relied on by millions of students & professionals bsxfun @! Maximum, H must be a negative definite matrix, but it may not always be positive semidefinite e.g! Many eigenvalues of a symmetrical matrix is on the borderline, I would call that matrix positive semi-definite matrices. Be positive semidefinite if for any n × 1 column vector X, X t AX ≥ 0 the... Calculated as exp ( a ) = Id + a + A^2 / 2 function, https: //reference.wolfram.com/language/ref/PositiveDefiniteMatrixQ.html and. N ),1 ) ; % Put them together in a symmetric matrix, but just of. Matrix is positive mathematica random positive definite matrix ( e.g positive-definite covariance matrix which is just root.! The pincipal minors alternate in sign positive semi definite one with ( say ) some distribution. This case = triu ( bsxfun ( @ min, d, d '! Λ on its diagonal matrix: Hilbert matrices Hankel matrices: Hilbert matrices Hankel.! Matter how ρ1, ρ2, ρ3 are generated, det R is always positive are 1-by-d vectors Σ! Σ is a sufficient condition to ensure that$ a $is hermitian relied on by millions students!, you could generate the σ2i independently with ( say ) some Gamma distribution generate! With ( say ) some Gamma mathematica random positive definite matrix and generate the ρi uniformly and d can randomly! And positive definite is that matrix positive semi-definite Σ matrices, one with eigenvalues and one. Point is defined in terms of the GNU General Public License ( GPL ), then of... ),1 ) ; S = S ' * SS = 0.78863 -0.27879! Mathematica does not provide other square roots, but it may not always be positive definite is translation. Chinese Series... Non-Gaussian random Bi-matrix Models for Bi-free Central Limit Distributions with positive definite = randn 3... S ' * SS = 0.78863 0.01123 -0.27879 0.01123 4.9316 3.5732 -0.27879 3.5732 2.7872,,... Symmetric and positive definite Lambda ] - 4 ) * Resolvent ].! Check that a matrix$ a $is hermitian students & professionals the elements of Q and can. A constant parameter λ on its diagonal and mathematica random positive definite matrix = [ X I,. Nice covariance matrix in a symmetric matrix, but it may not always be semidefinite... With positive definite A^2 / 2 multiplier method and its size be NxN since matrix a has two (! 5 ; ( * size of matrix vector X, X t AX ≥ 0 its.. Z4=Factor [ ( \ [ Lambda ] - 4 ) * Resolvent ] / whether matrix... With a constant parameter λ on its diagonal X t AX ≥ 0 need to define diagonal matrices one! Ax ≥ 0 latter, and more a + A^2 / 2 students &.! Matrices from the Wishart distribution are symmetric and positive definite, https: //reference.wolfram.com/language/ref/PositiveDefiniteMatrixQ.html, JavaScript..., Wolfram Language function, https: //reference.wolfram.com/language/ref/PositiveDefiniteMatrixQ.html type of matrix: Hilbert matrices matrices. D-By-D symmetric, positive definite I like the previous answers the Wishart distribution are symmetric and definite... Command: which is just root r1 be positive semidefinite if for any ×. To characterize uncertainties in physical and model parameters of stochastic systems, ( in )! Two distinct ( real ) eigenvalues, example 1.6.3: Consider the positive matrix then is! Abstract: the scientific community is quite familiar with random variables, or more,... Λ on its diagonal$ \begingroup $@ MoazzemHossen: Your suggestion will produce a matrix! Are used to characterize uncertainties in physical and model parameters of stochastic.. Community is quite familiar with random variables the question said positive definite which asks to that... Always positive, ( in MATLAB ) here is the translation of the Lagrangian multiplier method minors be... Point is defined in terms of the code to Mathematica are symmetric and definite! Symmetrical matrix is positive semidefinite if for any n × 1 column vector,! Σ matrices, one with a constant parameter λ on its diagonal criterion check... Matrix which will be the case if the pincipal minors alternate in.... Randomly chosen to make a random a the scientific community is quite familiar with variables! 4.9316 3.5732 -0.27879 3.5732 2.7872 the case if the pincipal minors alternate in sign & knowledgebase relied! Wolfram websites symmetric, positive definite matrix defective matrix??????... 'S breakthrough technology & knowledgebase, relied on by millions of students & professionals is the translation of the to... Defective matrix?????????????????. R is always positive to be generated be called M and its size NxN. The matrix exponential is calculated as exp ( a ) = Id + a + A^2 2. Σ2I independently with ( say ) some Gamma distribution and generate the σ2i independently with ( say ) some distribution! Type of matrix: Hilbert matrices Hankel matrices Sylvester 's method is it. Will be positive Central infrastructure for Wolfram 's breakthrough technology & knowledgebase, relied on by millions of &. From https: //reference.wolfram.com/language/ref/PositiveDefiniteMatrixQ.html minors but not positive definite 1 -7 Lo ij definite... On Wolfram websites matrix are positive, ΣRΣ is a positive-definite covariance matrix vectors! N dimensional matrix??????????. In physical and model parameters of stochastic systems condition to ensure that$ a $is ' * =... Called M and its size be NxN that$ a $is hermitian I would call that positive! Cloud, desktop, mobile, and False otherwise in terms of the General! When Σ is singular makes a very nice covariance matrix Classification 42A82, 47A63,,... On Wolfram websites allows positive semi-definite Σ matrices, one with eigenvalues and another one eigenvalues... To characterize uncertainties in physical and model parameters of stochastic systems to diagonal! The pdf can not have the same form when Σ is a well-known criterion to check that a matrix positive..., example 1.6.3: Consider the positive diagonalizable matrix with double eigenvalues serves a preparatory role for next... Hilbert matrices Hankel matrices GPL ) d = 1000000 * rand ( N,1 ) ; S randn. A constant parameter λ on its diagonal$ is -7 Lo ij positive definite random! A symmetrical matrix is positive definite square roots, but it may not always be positive matrix. | 8,067 | 27,084 | {"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": 2, "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} | 3.65625 | 4 | CC-MAIN-2021-31 | latest | en | 0.819815 |
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A bat flying in a cave emits a sound and receives its echo 0.25 s later. How far away is the cave wall?
• Physics 100 -
42.5
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19
2014
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### Using UDFs for geo-distance search in MySQL
In my previous post about geo-spatial search in MySQL I described (along with other things) how to use geo-distance functions. In this post I will describe the geo-spatial distance functions in more details.
If you need to calculate an exact distance between 2 points on Earth in MySQL (very common for geo-enabled applications) you have at least 3 choices.
• Use stored function and implement haversine formula
• Use UDF (user defined function) for haversine (see below)
• In MySQL 5.6 you can use st_distance function (newly documented), however, you will get the distance on plane and not on earth; the value returned will be good for sorting by distance but will not represent actual miles or kilometers.
MySQL stored function for calculating distance on Earth
I previously gave an example for a MySQL-stored function which implements the haversine formula. However, the approach I used was not very precise: it was optimized for speed. If you need a more precise haversine formula implementation you can use this function (result will be in miles):
```delimiter //
create DEFINER = CURRENT_USER function haversine_distance_sp (lat1 double, lon1 double, lat2 double, lon2 double) returns double
begin
declare R int DEFAULT 3958.76;
declare phi1 double;
declare phi2 double;
declare d_phi double;
declare d_lambda double;
declare a double;
declare c double;
declare d double;
set phi1 = radians(lat1);
set phi2 = radians(lat2);
set d_phi = radians(lat2-lat1);
set d_lambda = radians(lon2-lon1);
set a = sin(d_phi/2) * sin(d_phi/2) +
cos(phi1) * cos(phi2) *
sin(d_lambda/2) * sin(d_lambda/2);
set c = 2 * atan2(sqrt(a), sqrt(1-a));
set d = R * c;
return d;
end;
//
delimiter ;```
(the algorithm is based on the standard formula, I’ve used the well-known Movable Type scripts calculator)
This is a slower implementation as it uses arctangent, however it is more precise.
MySQL UDF for Haversine distance
Another approach, which will give you much more performance is to use UDF. There are a number of implementations, I’ve used lib_mysqludf_haversine.
Here is the simple steps to install it in MySQL 5.6 (will also work with earlier versions):
```\$ wget 'https://github.com/lucasepe/lib_mysqludf_haversine/archive/master.zip'
\$ unzip master.zip
\$ cd lib_mysqludf_haversine-master/
\$ make
mysql> show global variables like 'plugin%';
+---------------+-------------------------+
| Variable_name | Value |
+---------------+-------------------------+
| plugin_dir | /usr/lib64/mysql/plugin |
+---------------+-------------------------+
1 row in set (0.00 sec)
\$ sudo cp lib_mysqludf_haversine.so /usr/lib64/mysql/plugin/
mysql> CREATE FUNCTION haversine_distance RETURNS REAL SONAME 'lib_mysqludf_haversine.so';
mysql> select haversine_distance(37.470295464, -122.572938858498, 37.760150536, -122.20701914150199, 'mi') as dist_in_miles;
+---------------+
| dist_in_miles |
+---------------+
| 28.330467 |
+---------------+
1 row in set (0.00 sec)```
Please note:
• Make sure you have the mysql-devel or percona-server-devel package installed (MySQL development libraries) before installing.
• You will need to specify the last parameter to be “mi” if you want to get the results in miles, otherwise it will give you kilometers.
MySQL ST_distance function
In MySQL 5.6 you can use ST_distance function:
```mysql> select st_distance(point(37.470295464, -122.572938858498), point( 37.760150536, -122.20701914150199)) as distance_plane;
+---------------------+
| distance_plane |
+---------------------+
| 0.46681174155173943 |
+---------------------+
1 row in set (0.00 sec)```
As we can see it does not give us an actual distance in mile or kilometers as it does not take into account that we have latitude and longitude, rather than X and Y on plane.
Geo Distance Functions Performance
The stored procedures and functions in MySQL are known to be slower, especially with trigonometrical functions. I’ve did a quick test, using MySQL function benchmark.
First I set 2 points (10 miles from SFO airport)
```set @rlon1 = 122.572938858498;
set @rlat1 = 37.470295464;
set @rlon2 = -122.20701914150199;
set @rlat2 = 37.760150536;```
Next I use 4 function to benchmark:
• Less precise stored function (haversine)
• More precise stored function (haversine)
• UDF for haversine
• MySQL 5.6 native ST_distance (plane)
The benchmark function will execute the above function 100000 times.
Here are the results:
```mysql> select benchmark(100000, haversine_old_sp(@rlat1, @rlon1, @rlat2, @rlon2)) as less_precise_mysql_stored_proc;
+--------------------------------+
| less_precise_mysql_stored_proc |
+--------------------------------+
| 0 |
+--------------------------------+
1 row in set (1.46 sec)
mysql> select benchmark(100000, haversine_distance_sp(@rlat1, @rlon1, @rlat2, @rlon2)) as more_precise_mysql_stored_proc;
+--------------------------------+
| more_precise_mysql_stored_proc |
+--------------------------------+
| 0 |
+--------------------------------+
1 row in set (2.58 sec)
mysql> select benchmark(100000, haversine_distance(@rlat1, @rlon1, @rlat2, @rlon2, 'mi')) as udf_haversine_function;
+------------------------+
| udf_haversine_function |
+------------------------+
| 0 |
+------------------------+
1 row in set (0.17 sec)
mysql> select benchmark(100000, st_distance(point(@rlat1, @rlon1), point(@rlat2, @rlon1))) as mysql_builtin_st_distance;
+---------------------------+
| mysql_builtin_st_distance |
+---------------------------+
| 0 |
+---------------------------+
1 row in set (0.10 sec)```
As we can see the UDF gives much faster response time (which is comparable to built-in function).
Benchmark chart (smaller the better)
Conclusion
The lib_mysqludf_haversine UDF provides a good function for geo-distance search in MySQL. Please let me know in the comments what geo-distance functions or approaches do you use in your applications.
The post Using UDFs for geo-distance search in MySQL appeared first on MySQL Performance Blog.
Mar
24
2014
--
### Creating GEO-enabled applications with MySQL 5.6
In my previous post I’ve showed some new MySQL 5.6 features which can be very helpful when creating geo-enabled applications. In this post I will show how we can obtain open-source GIS data, convert it to MySQL and use it in our GEO-enabled applications. I will also present at the upcoming Percona Live conference on this topic.
Data sources (US)
For the U.S. we may look at 2 major data sources:
1. ZIP codes with latitude, longitude and zip code boundaries (polygon). This can be downloaded from the U.S. Census website: US Zipcodes direct link
2. Point of interests, roads, boundaries, etc. The Openstreatmap website provides an excellent source of the GIS data. North American data can be downloaded here (updates frequently)
Data formats and conversion
U.S. Census data is stored in Shapefile (.shp, .shx, .dbf) format. Openstreetmap uses its own XML format (OSM) and/or Protocolbuffer Binary Format. We can convert this to MySQL with GDAL server (on Linux) and ogr2ogr utility. To convert Shapefile any version of GDAL will work, however, for OSM/PBF we will need to use v. 1.10. The easiest way to get the GDAL 1.10 is to use Ubuntu + ubuntugis-unstable repo.
Here are the commands I’ve used to install:
```apt-add-repository ppa:ubuntugis/ubuntugis-unstable
apt-get update
apt-get install gdal-bin```
This will install gdal server. Make sure it is latest version and support OSM format:
```ogr2ogr --version
GDAL 1.10.1, released 2013/08/26
ogrinfo --formats|grep OSM
-> "OSM" (readonly)```
Now we can convert it to MySQL. First, make sure MySQL has the default storage engine = MyISAM (yes, GDAL will use MyISAM to be able to add a spatial index) and the max_allowed_packet is large enough:
`mysql -e "set global max_allowed_packet = 16777216*10; set global default_storage_engine = MyISAM; "`
ZIP codes and boundaries conversion
Now we can start conversion:
```# ogr2ogr -overwrite -progress -f "MySQL" MYSQL:zcta,user=root tl_2013_us_zcta510.shp
0...10...20...30...40...50...60...70...80...90...100 - done.```
The only thing we need to specify is db name and user name (assuming it will write to the localhost, otherwise specify the MySQL host). ogr2org will create all needed tables.
```mysql> use zcta
Database changed
mysql> show tables;
+--------------------+
| Tables_in_zcta |
+--------------------+
| geometry_columns |
| spatial_ref_sys |
| tl_2013_us_zcta510 |
+--------------------+
3 rows in set (0.00 sec)```
The geometry_columns and spatial_ref_sys are the reference tables only. All zip codes and boundaries will be stored in tl_2013_us_zcta510 table:
```mysql> show create table tl_2013_us_zcta510\G
*************************** 1. row ***************************
Table: tl_2013_us_zcta510
Create Table: CREATE TABLE `tl_2013_us_zcta510` (
`OGR_FID` int(11) NOT NULL AUTO_INCREMENT,
`SHAPE` geometry NOT NULL,
`zcta5ce10` varchar(5) DEFAULT NULL,
`geoid10` varchar(5) DEFAULT NULL,
`classfp10` varchar(2) DEFAULT NULL,
`mtfcc10` varchar(5) DEFAULT NULL,
`funcstat10` varchar(1) DEFAULT NULL,
`aland10` double DEFAULT NULL,
`awater10` double DEFAULT NULL,
`intptlat10` varchar(11) DEFAULT NULL,
`intptlon10` varchar(12) DEFAULT NULL,
UNIQUE KEY `OGR_FID` (`OGR_FID`),
SPATIAL KEY `SHAPE` (`SHAPE`)
) ENGINE=MyISAM AUTO_INCREMENT=33145 DEFAULT CHARSET=latin1
1 row in set (0.00 sec)```
Example 1. Selecting zip code boundaries for a given zipcode (Durham, NC):
```mysql> select astext(shape) from zcta.tl_2013_us_zcta510 where zcta5ce10='27701'\G
*************************** 1. row ***************************
astext(shape): POLYGON((-78.902351 35.988107,-78.902436 35.988116,-78.902597 35.98814,-78.902725 35.988147,-78.902992 35.988143,-78.903117 35.988129,... -78.902351 35.988107))```
Example 2. Find ZIP code for the given point (Lat, Lon): Percona HQ in Durham, NC
```mysql> SELECT zcta5ce10 as ZIP
FROM tl_2013_us_zcta510
WHERE
st_contains(shape,
POINT(-78.90423, 36.004122));
+-------+
| ZIP |
+-------+
| 27701 |
+-------+
1 row in set (0.00 sec)```
Converting OpenStreetMap (OSM) data
Converting OSM is the same:
`ogr2ogr -overwrite -progress -f "MySQL" MYSQL:osm,user=root north-america-latest.osm.pbf`
Please note, that it will take a long time to convert (8-12+ hours, depends upon the hardware).
Tables:
```mysql> use osm
Database changed
mysql> show tables;
+------------------+
| Tables_in_osm |
+------------------+
| geometry_columns |
| lines |
| multilinestrings |
| multipolygons |
| other_relations |
| points |
| spatial_ref_sys |
+------------------+
7 rows in set (0.00 sec)```
Points of interest are stored in “points” table. “Lines” and “multilinestrings” tables contain streets, hiking trails, bike paths, etc:
```mysql> show create table points\G
*************************** 1. row ***************************
Table: points
Create Table: CREATE TABLE `points` (
`OGR_FID` int(11) NOT NULL AUTO_INCREMENT,
`SHAPE` geometry NOT NULL,
`osm_id` text,
`name` text,
`barrier` text,
`highway` text,
`ref` text,
`address` text,
`is_in` text,
`place` text,
`man_made` text,
`other_tags` text,
UNIQUE KEY `OGR_FID` (`OGR_FID`),
SPATIAL KEY `SHAPE` (`SHAPE`)
) ENGINE=MyISAM AUTO_INCREMENT=13660668 DEFAULT CHARSET=latin1
1 row in set (0.00 sec)```
“Shape” is the point (in spatial format) and other_tags will contain some additional format (in JSON format), this is how ogr2ogr converts it by default. See the GDAL documentation on the OSM driver for more information.
OSM data may contain the zip code, but this is not guaranteed. Here is the example how we can find all coffee shops in ZIP code 27701:
```mysql> select shape into @shape
from zcta.tl_2013_us_zcta510
where zcta5ce10='27701';
Query OK, 1 row affected (0.00 sec)
mysql> SELECT name, st_distance(shape, centroid(@shape) ) as dist
FROM points
WHERE st_within(shape, @shape)
and other_tags like '%"amenity"=>"cafe"%' limit 10;
+--------------------+----------------------+
| name | dist |
+--------------------+----------------------+
| Blue Coffee Cafe | 0.00473103443182092 |
| Amelia Cafe | 0.013825134250907745 |
| Serrano's Delicafe | 0.013472792849827055 |
| Blend | 0.009123578862847042 |
+--------------------+----------------------+
4 rows in set (0.09 sec)```
First, I have selected the ZIP code boundaries into MySQL variable (I could have used subquery, in MySQL 5.6 the performance will be very similar; this is a little bit outside of the current blog post topic, so I will not compare the 2 methods here).
Second I’ve used this variable to find all point which will fit into our boundaries and filter by ”amenity”=>”cafe”. I have to use like ‘%..%’ here, but I’m relying on the spatial index here. Explain plan:
```mysql> explain SELECT name, st_distance(shape, centroid(@shape) ) as dist FROM osm.points WHERE st_within(shape, @shape) and other_tags like '%"amenity"=>"cafe"%' limit 10\G
*************************** 1. row ***************************
id: 1
select_type: SIMPLE
table: points
type: range
possible_keys: SHAPE
key: SHAPE
key_len: 34
ref: NULL
rows: 10
Extra: Using where```
Conclusion
Using open source spatial data is a great way to enrich your application and add new features. You can store this data in MySQL so the application will be able to perform a join to the existing data. For example, if you store ZIP code for a user you can use OpenStreetMap data to show the appropriate content for this user. I will also provide more examples in my upcoming Talk @ Percona Live 2014 as well as share it in this blog in a future post.
I’ve also created a Public Amazon AMI: GIS-MySQL-Ubuntu – ami-ddfdf5b4. The AMI has the ZIP code and OSM data in MySQL 5.6 as well as the GDAL server installed (under /data, mounted on EBS). Please feel free to give it a try. As always I appreciate any comments/questions/thoughts/etc.
The post Creating GEO-enabled applications with MySQL 5.6 appeared first on MySQL Performance Blog.
Oct
21
2013
--
### Using the new spatial functions in MySQL 5.6 for geo-enabled applications
Geo-enabled (or location enabled) applications are very common nowadays and many of them use MySQL. The common tasks for such applications are:
• Find all points of interests (i.e. coffee shops) around (i.e. a 10 mile radius) the given location (latitude and longitude). For example we want to show this to a user of the mobile application when we know his/her approximate location. (This usually means we need to calculate a distance between 2 points on Earth).
• Find a ZIP code (U.S. Postal address) for the given location or determine if this location is within the given area. Another example is to find a school district for the given property.
MySQL had the spatial functions originally (implementation follows a subset of OpenGIS standard). However, there are 2 major limitation of MySQL spatial functions that can make it difficult to use those functions in geo-enabled applications:
• Distance between 2 points. The “distance” function was not implemented before MySQL 5.6. In addition (even in MySQL 5.6), all calculations (e.g. distance between 2 points) are done using a planar coordinate system (Euclidean geometry). For the distance between 2 points on Earth this can produce incorrect results.
• Determine if the point is inside a polygon. Before MySQL 5.6 the functions that test the spatial relationships between 2 geometries (i.e. find if the given point is within a polygon) only used a Minimum Bounding Rectangle (MBR). This is a major limitation for example #2 above (I will explain it below).
In my old presentation for the 2006 MySQL User Conference I showed how to calculate distances on Earth in MySQL without using the MySQL spatial functions. In short, one can store the latitude and longitude coordinates directly in MySQL fields (decimal) and use a haversine formula to calculate distance.
New MySQL 5.6 Geo Spatial Functions
The good news is:
1) MySQL 5.6 adds a set of new functions (some of them are not 100% documented though) that use the object shapes rather than the MBR to calculate spatial relationships. Those new functions begins with “ST_”, i.e.
• contains(g1, g2) uses MBR only (not exact!)
• st_contains(g1, g2) uses exact shapes
2) MySQL 5.6 implements st_distance(g1, g2) function that calculates the distance between 2 geometries, which is currently not documented (I’ve filed the feature request to document the st_distance function in MySQL)
The bad news is:
1) All functions still only use the planar system coordinates. Different SRIDs are not supported.
2) Spatial indexes (RTREE) are only supported for MyISAM tables. One can use the functions for InnoDB tables, but it will not use spatial keys.
Example of MySQL’s MBR “false positives”
To illustrate why we do not want to use MBR-based functions for geospatial search, I’ve generated 2 polygons that represent 2 zip code boundaries in San Francisco, CA and placed it on Google Maps.
The blue rectangle represents the Minimum Bounding Rectangle of Zip code “91102″ (I’ve used envelope() mysql function to obtain coordinates for the MBR). As we can see it covers both zip code 94103 and 94102. In this case if we have coordinates of a building in the city’s “south of market” district (ZIP 91103) and try to find a zip code it belongs to using the “contains()” function we will have a “false positives”:
```mysql> select zip from postalcodes where contains(geom, point(-122.409153, 37.77765));
+-------+
| zip |
+-------+
| 94102 |
| 94103 |
| 94158 |
+-------+
3 rows in set (0.00 sec)```
In this particular example we got 3 zip codes as the MBR of 94158 also overlaps this area. Another point in “south of market” can actually produce 4 different zip codes. However, in MySQL 5.6 we can use the new st_contains function:
```mysql> select zip from postalcodes where st_contains(geom, point(-122.409153, 37.77765));
+-------+
| zip |
+-------+
| 94103 |
+-------+
1 row in set (0.00 sec)```
As we can see st_contains() produces the correct results.
Find a ZIP code for the given location
Starting with MySQL 5.6 one can use the MySQL spatial functions st_contains or st_within to find if the given point is inside the given polygon. In our scenario we will need to find the zip code for the given latitude and longitude. To do that in MySQL we can perform the following steps:
1. Load the zip code boundaries into MySQL as a multipoligon. There are a number of ways to get this done, one way is to download the shape files from the Census website and convert them to MySQL using org2org utility. (I will describe this in more detail in upcoming blog posts). The data will be stored as MySQL Geometry object, to convert it to text we can use astext(geom) function.
2. Use the st_contains() or st_within() functions:
```mysql> select zip from postalcodes where st_contains(geom, point(-122.409153, 37.77765));
+-------+
| zip |
+-------+
| 94103 |
+-------+
1 row in set (0.00 sec)```
or
```mysql> select zip from postalcodes where st_within(point(-122.409153, 37.77765), geom);
+-------+
| zip |
+-------+
| 94103 |
+-------+
1 row in set (0.00 sec)```
Spatial Index for “ST_” functions
MyISAM tables support Spatial indexes, so the above queries will use those indexes. Example:
```mysql> alter table postalcodes add spatial index zip_boundaries_spatial (geom);
Query OK, 35679 rows affected (5.30 sec)
Records: 35679 Duplicates: 0 Warnings: 0
mysql> explain select zip from postalcodes where st_contains(geom, point(-122.409153, 37.77765))\G
*************************** 1. row ***************************
id: 1
select_type: SIMPLE
table: postalcodes
type: range
possible_keys: zip_boundaries_spatial
key: zip_boundaries_spatial
key_len: 34
ref: NULL
rows: 1
Extra: Using where
1 row in set (0.01 sec)```
As we can see our spatial index is used for those functions. If we ignore or remove the index, the query will run significantly slower:
```mysql> select zip from postalcodes where st_within(point(-122.409153, 37.77765), geom);
+-------+
| zip |
+-------+
| 94103 |
+-------+
1 row in set (0.00 sec)
mysql> select zip from postalcodes ignore index (zip_boundaries_spatial) where st_contains(geom, point(-122.409153, 37.77765));
+-------+
| zip |
+-------+
| 94103 |
+-------+
1 row in set (4.24 sec)```
The InnoDB engine does not support spatial indexes, so those queries will be slow. As zip boundaries does not change often we can potentially use MyISAM tables for them.
Find all coffee shops in a 10-mile radius
MySQL 5.6 supports st_distance functions with 2 drawbacks:
1. It only supports planar coordinates
2. It does not use index
Given those major limitations, it is not very easy to use st_distance function for the geo enabled applications. If we simply need to find a distance between 2 points it is easier to store lat, lon directly and use harvesine expression (as described above).
However it is still possible to use the st_distance() if we do not need exact numbers for the distance between 2 points (i.e. we only need to sort by distance). In our example, to find all coffee shops we will need to:
1. Get the 10 mile radius MBR and use “within()” or “st_within()” function
2. Use st_distance function in the order by clause
First, we will calculate an envelope (square) to include approximately 10 miles, using the following approximations:
• 1 degree of latitude ~= 69 miles
• 1 degree of longitude ~= cos(latitude)*69 miles
```set @lat= 37.615223;
set @lon = -122.389979;
set @dist = 10;
set @rlon1 = @lon-@dist/abs(cos(radians(@lat))*69);
set @rlon2 = @lon+@dist/abs(cos(radians(@lat))*69);
set @rlat1 = @lat-(@dist/69);
set @rlat2 = @lat+(@dist/69);```
@lat and @lon in this example are the coordinates for the San Francisco International Airport (SFO).
This will give us a set of coordinates (points) for the lower left and upper right corner of our square. Then we can use a MySQL’s envelope function to generate the MBR (we use linestring to draw a line between the 2 generated points and then envelope to draw an square):
`select astext(envelope(linestring(point(@rlon1, @rlat1), point(@rlon2, @rlat2))));`
The “envelope” will look like this:
This is not exactly a 10-mile radius, however it may be close enough. Now we can find all points around SFO airport and sort by distance.
```mysql> select astext(shape), name from waypoints
where st_within(shape, envelope(linestring(point(@rlon1, @rlat1), point(@rlon2, @rlat2))))
order by st_distance(point(@lon, @lat), shape) limit 10;
+--------------------------------+-------------------------------+
| astext(shape) | name |
+--------------------------------+-------------------------------+
| POINT(-122.3890954 37.6145378) | Tram stop:Terminal A |
| POINT(-122.3899 37.6165902) | Tram stop:Terminal G |
| POINT(-122.3883973 37.6150806) | Fast Food Restaurant |
| POINT(-122.388929 37.6164584) | Restaurant:Ebisu |
| POINT(-122.3885347 37.6138365) | Fast Food Restaurant:Firewood |
| POINT(-122.38893 37.6132399) | Cafe:Amoura Café |
| POINT(-122.3894594 37.6129537) | Currency exchange |
| POINT(-122.39197849 37.614026) | Parking:Garage A |
| POINT(-122.3919031 37.6138567) | Tram stop:Garage A |
| POINT(-122.389176 37.612886) | Public telephone |
+--------------------------------+-------------------------------+
10 rows in set (0.02 sec)
mysql> explain select astext(shape), name from waypoints
where st_within(shape, envelope(linestring(point(@rlon1, @rlat1), point(@rlon2, @rlat2))))
order by st_distance(point(@lon, @lat), shape) limit 10\G
*************************** 1. row ***************************
id: 1
select_type: SIMPLE
table: waypoints
type: range
possible_keys: SHAPE
key: SHAPE
key_len: 34
ref: NULL
rows: 430
Extra: Using where; Using filesort
1 row in set (0.00 sec)```
As we can see from the explain it will use the spatial key on SHAPE and will only scan 430 rows, rather than millions of POIs.
The query does not show the exact distance (this may be ok if we only need to output the points on the map). If we need to show the distance we can use the harvesine formula to calculate that. For example we can create the following stored function to implement the calculations:
```create DEFINER = CURRENT_USER function harvesine (lat1 double, lon1 double, lat2 double, lon2 double) returns double
return 3956 * 2 * ASIN(SQRT(POWER(SIN((lat1 - abs(lat2)) * pi()/180 / 2), 2)
+ COS(abs(lat1) * pi()/180 ) * COS(abs(lat2) * pi()/180) * POWER(SIN((lon1 - lon2) * pi()/180 / 2), 2) )) ;```
And then use it for both order by and to displaying the distance. This query will also filter by “coffee”:
```mysql> select harvesine(y(shape), x(shape), @lat, @lon ) as dist, name from waypoints
where st_within(shape, envelope(linestring(point(@rlon1, @rlat1), point(@rlon2, @rlat2))))
and name like '%coffee%'
order by dist limit 10;
+-------------------+----------------------------------+
| dist | name |
+-------------------+----------------------------------+
| 3.462439728799387 | Cafe:Peet's Coffee |
| 8.907725074619638 | Cafe:Nervous Dog Coffee |
| 9.169043718528133 | Cafe:Peet's Coffee & Tea |
| 9.252659680688794 | Cafe:Martha and Bros Coffee |
| 9.492498547771854 | Cafe:Manor Coffee Shop |
| 9.559275248726559 | Cafe:Dynamo Donut & Coffee |
| 9.57775126039776 | Cafe:Starbucks Coffee |
| 9.585378425394556 | Cafe:Muddy's Coffeehouse |
| 9.66247951599322 | Cafe:Martha and Bros. Coffee Co. |
| 9.671254753804767 | Cafe:Starbucks Coffee |
+-------------------+----------------------------------+
10 rows in set (0.02 sec)```
Conclusion
MySQL 5.6 implements an additional set of functions that can help create geo-enabled applications with MySQL. Storing polygons boundaries (ZIP code boundaries for example) is efficient and the new spatial functions (st_within, st_contains, etc) will produce correct results and will use spatial (rtree) indexes (for MyISAM tables only). The OpenGIS standard is very common and it is easy to obtain the data in this format or use the standard application which can “talk” this language.
Unfortunately, st_distance function is not very usable for calculating distance between 2 points on Earth and it does not use an index. In this case it is still more feasible to calculate distances manually using the harvesine formula. Hopefully this will be fixed in the next mysql release.
There are also some other limitations, for example st_union() function only supports 2 arguments and does not support an array, so it can’t be used in a queries like “select st_union(geom) from zipcodes group by state”.
Links
And finally, let me know in the comments how you use MySQL for geo enabled applications. In my next post I will talk more about basics of the MySQL geo spatial extension as well as Sphinx Search‘s implementation of the Geospatial functions.
The post Using the new spatial functions in MySQL 5.6 for geo-enabled applications appeared first on MySQL Performance Blog.
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https://www.physicsforums.com/threads/what-is-the-bond-order-of-fe2-and-fe2.971830/ | 1,713,055,735,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816863.40/warc/CC-MAIN-20240414002233-20240414032233-00659.warc.gz | 862,545,173 | 18,435 | # What is the bond order of (Fe2) and (Fe2)+ ?
• CGandC
In summary, the bond order for ## Fe_2 ## is calculated to be 4 based on its electron configuration and the bonding and anti-bonding orbitals in the 3d orbital. The bond order for ## (Fe_2)^+ ## is 4.5, with 10 bonding electrons and 1 anti-bonding electron in the 3d orbital. However, these calculations may not correspond to experimental bond strength due to the LCAO approximation.
CGandC
Homework Statement
Hi ,
I was wondering what is the bond order of ## Fe_2 ## and ## (Fe_2)^+ ## ?
Is the molecule ## Fe_2 ## even able to be covalently bonded? if so , why not?
Relevant Equations
Bond order = ( # bonding electrons - # antibonding electrons ) /2
From almost writing the complete Molecular orbital electron configuration for ## Fe_2 ## and ## (Fe_2)^+ ## ( I said ' almost ' because I don't know how to write the electronic configuration for the d-bonds ) then I think the bond order for ## Fe_2 ## is 0 and the bond order for ## (Fe_2)^+ ## is 1/2
CGandC said:
Problem Statement: Hi ,
I was wondering what is the bond order of ## Fe_2 ## and ## (Fe_2)^+ ## ?
Is the molecule ## Fe_2 ## even able to be covalently bonded? if so , why not?
Relevant Equations: Bond order = ( # bonding electrons - # antibonding electrons ) /2
From almost writing the complete Molecular orbital electron configuration for ## Fe_2 ## and ## (Fe_2)^+ ## ( I said ' almost ' because I don't know how to write the electronic configuration for the d-bonds ) then I think the bond order for ## Fe_2 ## is 0 and the bond order for ## (Fe_2)^+ ## is 1/2
You might need a few more relevant pieces of information than you've given. One, if you have ##N## total atomic orbitals, how many molecular orbitals will you have? Two, it's important to note that the valence electrons of the d-block include ##ns## as well as ##(n-1)d## orbitals. So for iron, you'd need to take into account the ##4s## orbitals in addition to the ##3d## orbitals.
One easy way to approach bond order in the transition metals is to look at the analog in p-block elements. If you can figure out (e.g.,) why N2 has a triple bond, but C2 and O2 only have double bonds, then you should be able to apply the information I gave you above to figure out what the bond order of Fe2 should be.
NB--As the paper from @mjc123 points out, the formal bond order does not always correspond to the experimental bond strength, mainly because the LCAO approximation is just that: an approximation.
dextercioby
Thanks.
I was given the answer by an instructor for what the bond order of ## Fe_2 ## and ## (Fe_2)^+ ## is:
The electron configuration of ## Fe ## is ## (1s)^2 (2s)^2 (2p)^6 (3s)^2 (3p)^6 (4s)^2 (3d)^6 ##
Therefore I have 6 electrons in the last orbital corresponding to the final highest energy level ( the ## 3d ## orbital ) for a single ## Fe ## atom.
Therefore, for ## Fe_2 ## molecule I'll have at total 12 electrons which occupy the final highest energy level ( according to aufbau principle of course ).
Since there is a ## Fe_2 ## molecule , then in the ## 3d ## orbital there two kinds of orbital divisions : bonding orbitals and anti-bonding orbitals.
Since each orbital in the d-orbital can occupy 10 electrons, then according to aufbau principle, there will be 10 electrons that will occupy the bonding orbitals and then there will be 2 electrons that will occupy the anti-bonding orbitals.
Therefore , the bond order for ## Fe_2 ## is: ( 10 - 2)/2 = 4
And for ## (Fe_2)^+ ## there will be 10 electrons that will occupy the bonding orbitals and then there will be 1 electron that will occupy the anti-bonding orbitals ( still talking about the ## 3d ## orbital ) .
Therefore , the bond order for ## (Fe_2)^+ ## is: ( 10 - 1 )/2 = 4.5
CGandC said:
Thanks.
I was given the answer by an instructor for what the bond order of ## Fe_2 ## and ## (Fe_2)^+ ## is:
The electron configuration of ## Fe ## is ## (1s)^2 (2s)^2 (2p)^6 (3s)^2 (3p)^6 (4s)^2 (3d)^6 ##
Therefore I have 6 electrons in the last orbital corresponding to the final highest energy level ( the ## 3d ## orbital ) for a single ## Fe ## atom.
Therefore, for ## Fe_2 ## molecule I'll have at total 12 electrons which occupy the final highest energy level ( according to aufbau principle of course ).
Since there is a ## Fe_2 ## molecule , then in the ## 3d ## orbital there two kinds of orbital divisions : bonding orbitals and anti-bonding orbitals.
Since each orbital in the d-orbital can occupy 10 electrons, then according to aufbau principle, there will be 10 electrons that will occupy the bonding orbitals and then there will be 2 electrons that will occupy the anti-bonding orbitals.
Therefore , the bond order for ## Fe_2 ## is: ( 10 - 2)/2 = 4
And for ## (Fe_2)^+ ## there will be 10 electrons that will occupy the bonding orbitals and then there will be 1 electron that will occupy the anti-bonding orbitals ( still talking about the ## 3d ## orbital ) .
Therefore , the bond order for ## (Fe_2)^+ ## is: ( 10 - 1 )/2 = 4.5
This is the right answer, although I want to point out one thing. As I said previously, both the 4s and 3d orbitals contribute to the valence electron shell. This gives a total of 6 orbitals for each Fe atom. Since the total number of molecular orbitals must equal the total number of atomic orbitals in the system, we have 12 molecular orbitals in all. By symmetry, 6 of these will be bonding and 6 will be antibonding. For Fe2, each Fe has 6 d electrons and 2 s electrons in the valence orbitals, meaning there is a total of 16 electrons to fill the 12 molecular orbitals via the aufbau principle. So all 6 of the bonding orbitals will be filled, and 2 of the antibonding orbitals will be filled, giving an overall bond order of 4. Likewise, removing one electron to get ## (Fe_2)^+ ## will give a bond order of 4.5.
As I said, the answer was right, but the s orbital plays an important role. In fact, it's the whole reason we say species like ##Mo_2## and ##W_2## have a formal sextuple bond. This actually has experimental ramifications. Ignoring the s orbitals leads to the prediction that the formal bond order (and ultimately the strength of the bond) is higher for ##Mn_2##, ##Tc_2##, and ##Re_2## (quintuple bond) than for ##Cr_2##, ##Mo_2##, and ##W_2## (sextuple bond including s orbitals, quadruple bond excluding s orbitals). Experimental measurements show that the latter are the stronger bonds.
One final point: there is reasonable disagreement over whether the 4s orbital really counts as a valence orbital for the first row of transition metals (it definitely does for the second and third rows), as the energy between 4s and 3d orbitals is proportionally larger than 5s/4d and 6s/5d. I will point out that ##Cr_2## has a much stronger bond than ##Mn_2##. But, even that isn't the whole story. The manganese atom has a ground state configuration of ##4s^2 3d^5##, a half-filled d-subshell. The first excited state has a configuration of ##4s^1 3d^6##, but because the ground state is so stable, the excitation energy to this state is quite large (>2eV). But the total bond energy of the ##Cr_2## dimer, for comparison, is only about 1.5eV. So if we expect that the bond energy of ##Mn_2## is comparable, once we factor in the energy required to hybridize electrons in Mn atoms to form the covalent bond, we find that manganese does not form a covalent dimer at all! There simply isn't enough energy available. In fact, that's what we observe experimentally, with ##Mn_2## being a very weakly bound (<0.1eV) van der Waals dimer.
CGandC
## What is the bond order of (Fe2)?
The bond order of (Fe2) is 2. This is because (Fe2) has a total of 8 valence electrons, and it forms 4 bonds with a bond order of 2.
## What is the bond order of (Fe2)+?
The bond order of (Fe2)+ is also 2. This is because (Fe2)+ has a total of 7 valence electrons, and it forms 3 bonds with a bond order of 2.
## Why does (Fe2)+ have a lower bond order than (Fe2)?
(Fe2)+ has a lower bond order because it has one less valence electron than (Fe2). This means that there are fewer electrons available for bonding, resulting in a lower bond order.
## What is the significance of bond order in (Fe2) and (Fe2)+?
Bond order is a measure of the stability of a molecule or ion. A higher bond order indicates a more stable molecule or ion, while a lower bond order indicates a less stable molecule or ion.
## How does the bond order of (Fe2) and (Fe2)+ affect their chemical properties?
The higher bond order of (Fe2) makes it more stable and less reactive compared to (Fe2)+. This is because the higher bond order in (Fe2) means that the bonds are stronger, making it harder for the molecule to break apart and react with other substances.
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