url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://forums.rpgmakerweb.com/index.php?threads/goblin-punch-and-establishing-an-absolute-value-in-formula.129414/ | 1,611,775,721,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704832583.88/warc/CC-MAIN-20210127183317-20210127213317-00515.warc.gz | 354,954,927 | 18,829 | # Goblin Punch and establishing an absolute value in formula
#### miyanke
##### Veteran
Hi!
Before anything sorry for my english, this is a difficult dubt and I will try my best to be understood. Please excuse my english and ask me anything (and correct me) if it's not clear enough.
I'm trying to create "Goblin Punch" similar to FF9 Version in MZ:
The damage dealt increases the closer the caster's level is to the target's. It is also affected by the caster's strength and the target's Defense, as well as any status effects.
I have VS Enemy levels, so I can use a.level and b.level in the formulae.
I created a formula, but I need to make a value absolute to make it work (I don't know if this is how it is said in english |x|).
My formula is the next one, but I still didn't know the exact value of the base damage (20 in the original), so let's call it just "base damage" until then:
Code:
``````c = base damage ;
g = a.atm+a.level ;
e = (g/8) + 1 ;
d = Math.randomInt(e) + g ;
h = |a.level-b.level| ;
f = 100-(h*10) ;
a.level==b.level ? c*d : c*(f/100)+1``````
or
Code:
``a.level==b.level ? 20*Math.randomInt((a.atm+a.level/8) + 1) + a.atm+a.level : 20*(100-(|a.level-b.level|*10)/100)+1``
I'll try to explain, as I said I'm trying to use the original formula. I know I can substitute so many variables in just one long formula, but I wanted to cut it to explain, and ask for help.
First is to determine "g", that is the base in the original formula, what is the min number of times "c" will repeat if levels are equals (so It's different each time you hit, but always is c*g)
I needed a random between "g" and "g"+"g"/8, so I made a Math random of "g"/8 and then add "g".
The hard part is the second: suppose "c" is the 100% I needed to transform the differences between the levels into a percentage, the problem is that I need to make this value "h" (a.level-b.level) as an absolute, because I can't know what is bigger to be first, and I don't know how to set it in the formulae [I used the bars because it's how mathematically is written, but I don't know if this is right].
Once I have the difference, aka "h" as an absolute value I can transform it into a percentage 100-[h*10].
Here I have my second question: if the difference is 1, the effectiveness of the atack is 90% "c", if the difference is 10, the effectiveness of the atack is 0% "c". What if the difference is 11? will it heal? Is there anyway to set negative numbers as 0?
Then I added +1, because I don't like the "no-damage" pop up, so 1 damage is always good.
In resume:
• How can I set an absolute value in formula?
• How can I transform a negative value into 0?
Thank you A LOT!
Last edited:
#### Maliki79
##### Veteran
Hi there. I tried converting a few of the lines to hopefully match up and explain what you needed.
Check below:
Code:
``````var c = 20; //base damage
var g = a.mat + a.level; //Caster Magic Attack plus caster level
var e = (g/8) + 1;
var d = Math.randomInt(e) + g;
var h = Math.abs(a.level - b.level); // Absolute value of level diff between target and caster
var f = 100 - (h * 10);
var dam = a.level == b.level ? c * d : c * (f/100) + 1; // store damage in a variable to check value later
return Math.max(dam, 1); // Will make damage 1 if the calculations equal 0 or less``````
Reply back if you need more help or I don't answer you correctly here.
#### miyanke
##### Veteran
In my place we say ¡Olé! (in fact the stress is in the O)
Yes! Math.abs was what I was looking for.
One more question, Can I write more than one line in the formula or should I transform it to a line?
#### Maliki79
##### Veteran
You can if you know how to make your own plugin.
You can put that formula in the file then load it up and call it in the damage formula.
(If you don't understand what I said, you probably will need help making the file.)
Other wise, you would need to pack all that in the one line of your damage formula.
#### miyanke
##### Veteran
In one line, it's late to check it, but this might work…
Code:
``Math.max(a.level == b.level ? 20*(Math.randomInt((a.mat + a.level/8) + 1) + a.mat + a.level) : 20*((100 - (Math.abs(a.level - b.level) * 10))/100),1)``
#### Maliki79
##### Veteran
Whoa! That's quite the code you got there!
Looks ok at first glance.
Yeah. Test it when you can.
I'd love to hear the results!
#### miyanke
##### Veteran
I tested it in with LV50 characters and enemiess +5 -10 levels.
The results with a 20 as base damage is damage of 18 if they are the same level, and 16 if its 48.
With 100 as base damage, it dealt 98 damage if the same level and 2 if 46.
Does it work? Yes. As I have planned? Not exactly, I think the maths should be adapted and reviewed.
#### Maliki79
##### Veteran
Well, firstly, why not put the code exactly as I did it?
JavaScript:
``var c=20; var g = a.mat+a.level; var e = (g/8)+1; var d = Math.randomInt(e)+g; var h = Math.abs(a.level-b.level); var f = 100-(h*10); var dam = a.level == b.level ? c * d : c * (f/100) + 1; Math.max(dam, 1);``
(I tried it in MZ and I can at least say it all fit in the formula box.)
I don't have enemy levels, so I can't try it out.
If the numbers are still off, remember the main areas where you can make changes:
The Magic Attack stat of the caster,
The levels of caster and target,
or the base damage.
Any of those three will make a noticeable change in the damage.
Also, just what kind of numbers were you expecting?
Equal level does?
Level +-10 does?
Basically, just make some conscious tweaks and you should get the code the way you want.
Eventually...
(I hate overly complicated damage formulas myself...)
### Latest Profile Posts
Aaaaannd published my game's tech demo.
Hey everyone, we know that the edit bar is missing. We're working on it. You can talk about it in the announcement here: https://forums.rpgmakerweb.com/index.php?threads/forum-errors-missing-edit-bar-etc.132715/
So, explain why we can no longer use BBC code or smilies in our posts? This sparks much sadness...
Are we now stuck with WYSIWYG ? I cannot revert back my posts to good old raw text ?
I'm wondering if I may be putting too many things into one map. A story, within a story, within a story . . . it's fun, but I can't shake the feeling that it may be better to scrap some of it, and use it for another map. I'm not sure what to do. | 1,730 | 6,342 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2021-04 | latest | en | 0.935143 |
https://www.physicsforums.com/threads/acoustic-contrast-factor.421748/ | 1,545,136,915,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376829399.59/warc/CC-MAIN-20181218123521-20181218145521-00217.warc.gz | 980,080,857 | 14,024 | # Acoustic Contrast Factor
1. Aug 11, 2010
### SFB
Hi
I have some confusion in understanding the acoustic contrast actor . You can check wiki if you are not familiar with the term.
http://en.wikipedia.org/wiki/Acoustic_contrast_factor
If there is a foreign particle in the path of a sound wave (to be specific Standing wave for my case), the sign of acoustic contrast factor determines whether the acoustic radiation force will push the particle toward pressure nodes or antinodes.
As the radiation force is a function of acoustic contrast factor , mathematically sign of the contrast factor determines the force direction.
But I am interested know the physics behind it. How the compressibilty of the particle and medium of propagation determines the direction. What actually happens when sound waves confront a compressible bubble or an incompressible rigid particle ?
I tried to trace some resources , but they are full of mathematics as well.
I would appreciate if you can kindly share your opinion in this regard.
Thanks
, it is quite obvious mathematically that a
2. Jun 22, 2016
### hamilton17
Hi,
I would look up Iso acoustic focussing of particles, this would make this concept easier.
Basically the movement of a particle in a microchannel will either be positive (inwards to the centre) or negative (outwards to the sides).
This depends on both the compressibility and density of the particles, when acoustically activated a radiation force F(rad) is generated.
The particle will move to a point where acoustic impedance matches that of the medium.
Hope this helps | 333 | 1,592 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-51 | latest | en | 0.872233 |
www.finhumedades.es | 1,632,879,461,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780061350.42/warc/CC-MAIN-20210929004757-20210929034757-00675.warc.gz | 797,401,928 | 11,197 | pipe fitting offset formula in canada Despriction
### How to Figure a Rolling Offset in Piping Hunker
Multiply the offset by itself, and the rise by itself, and add these two numbers together. In the example, 8 x 8 = 64 and 12 x 12 = 144. Therefore, 64 + 144 = 208. How to find the true offset of a pipe?How to find the true offset of a pipe?If the fitting angle is 45 degrees, the true offset is multiplied by 1.414 to obtain the diagonal. For any fitting angle that is 22.5 degrees, the true offset is multiplied by 2.613 to get the answer for the diagonal. The setback for fitting a pipe equals the true offset multiplied by 0.577 for a fitting angle of 60 degrees.How Do You Calculate Pipe Offset? - What's Your Question What is pipe tube formula?What is pipe tube formula?en pipe tube formulas equations section modulus treverse metal area external pipe surface internal pipe surface traverse internal area es fórmulas de tubo tubo de ecuaciones área de metal superficie de la tubería externa superficie de la tubería interna módulo de sección transversal treverse área internaPipe Formulas - Engineering ToolBox
### What is the setback for fitting a pipe?What is the setback for fitting a pipe?The setback for fitting a pipe equals the true offset multiplied by 0.577 for a fitting angle of 60 degrees. The true offset multiplied by 1.000 equals the setback for a 45-degree fitting angle.How Do You Calculate Pipe Offset? - What's Your Question
#### Must include:
canada1. Alright so let go back to school for a second and remember what a hypotenuse is. In geometry, a hypotenuse is the longest side of a right-angled pipe fitting offset formula in canada2. Another important step is removing the small portion of each fitting from your equation like you see here. For now, well mark a small reference pipe fitting offset formula in canada3. So lets take a look at a schematic that I drew earlier that represents a pipe being off-setted at a 45* angle to get passed a hot water tank as pipe fitting offset formula in canada4. Once youve found the measurement between both pipes, multiply that number by 1.4142 to get your hypotenuse or diagonal pipe measurement.5. Heres the part where some people tend to block when using this technique and, its when removing the fittings out of the equation, so lets cla pipe fitting offset formula in canada6. This technique works on any fitting that needs a 45* offset whether its up on a ceiling or if its an underground pipe. You could get away with pipe fitting offset formula in canadaRolling Offsets (run) - vCalc
Processing pipe fitting offset formula in canada pipe fitting offset formula in canada pipe fitting offset formula in canada
#### Must include:
canadaFormula for Miter Fabrication From Pipe The Piping pipe fitting offset formula in canadaHere is formula to calculate required dimensions to fabricate miter from pipe. Miter Fabrication From Pipe. In the figure above, dotted lines are where the pipe should be cut. Each dotted cut line has a vertical reference line marked at following distances from pipe end. L2 Distance of first and last reference line from pipe ends.
#### Must include:
canadaHow Do You Calculate Pipe Offset? - Reference
Aug 04, 2015The setback for fitting a pipe equals the true offset multiplied by 0.577 for a fitting angle of 60 degrees. The true offset multiplied by 1.000 equals the setback for a 45-degree fitting angle. Multiply the true offset by 2.414 for the setback for a 22.5-degree fitting angle. Using the above calculations, the true offset for a 45-degree fitting is 14.42 inches and the
#### Must include:
canadaPipe Formulas - Engineering ToolBoxA o = external pipe surface area (ft 2 per ft pipe) Internal Pipe Surface. Internal pipe or tube surface per ft of length can be expressed as. A i = d i / 12 (5) where . A i = internal pipe surface area (ft 2 per ft pipe) Transverse Internal Area. Transverse internal area can be expressed as. A a = 0.7854 d i 2 (6)
#### Must include:
canadaPipes, Pipe Fittings, and Piping Details:Pipe Fitting pipe fitting offset formula in canadaPipe Fitting MeasurementsFirst MethodSecond MethodPipe fitting may be done either by making close visual judgments or entirely by measurements scaled on a drawing. The first method is a hit-or-miss process and requires an experienced fitter to do a good job; the second method is one of precision and is the better way. In actual practice, a combination of the two methods will, in some cases, save time and give satisfactory results. Working from a drawing with all the necessary dimensions has certain advantages. Especially iSee more on machineryequipmentonlinePublished Oct 25, 201545 Degree offset formula Plumbing Zone - Professional pipe fitting offset formula in canadaNov 11, 2011We all know that 1.41 or 1.414 is the formula for figuring your 45 offset. BUT! pipe fitting offset formula in canada.let's say your working with 3/4" copper pipe. You got to offset 45 degrees. Lets say a 10 inch offset. So, we come up with 14.14. Nobody on here ever gives the numbers you must deduct for the fitting. Back to the question. We are working with 3/4 pipe. We got a 10" offset.
#### Must include:
pipe fitting offset formulaspipe fitting angle offset formulaspipe offset formulapipe fitting offset calculator freepipe fitting 45 degree offsetpipe fitting formulas for fabricationpipe fitting math formulaspipe offset tablePLUMBING MATHEMATICSlead need = pipe diameter x lead weight x number of joints 20. Total lead need plus waste allowance total need = lead need (100% - % of waste) 21. Degree of offset of a pipe fitting degree of angle = fitting x 360° NOTE Tab This Section On Formulas And Constants CONSTANTS 22. 1 cubic foot of water = 7.5 gallons 23.
pipe fitting
#### Must include:
pipe fittingFitter formula Piping and structural fitter training and pipe fitting offset formula in canadaPipe fitter formula piping and structural fitter training pipe theory in Hindi job interview question and answer. pipe fitting offset formula in canada set run travel formula any degree pipe offset formula 09/08/2019 29/03/2020 fitter guide blogger post 0. pipe fitting offset formula in canada socket weld and threaded pipe fittings size in mm PDF chart 15/12/2019 02/02/2021 fitter formula 4.Calculating a simple offset vs. a rolling offset pipe fitting offset formula in canadaSep 22, 201645 offset Plumbing Zone - Professional Plumbers ForumJul 14, 201645 Rolling Offset Formula ?? Page 6 Plumbing Zone pipe fitting offset formula in canadaDec 31, 201245 Degree offset formula Page 2 Plumbing Zone pipe fitting offset formula in canadaNov 12, 2011See more resultsThe EASIEST Way to Calculate a 45° Offset 6 Steps (with pipe fitting offset formula in canadaLets just say that this pipe is a 1 copper drain line from a steam boiler that needs to go from point A to point B. pipe fitting offset formula in canada This technique works on any fitting that needs a 45* offset whether its up on a ceiling or if its an underground pipe. You could get away with doing this type of work by eye, but itll never give you a precise pipe fitting offset formula in canada
### Carlon Schedule 40 PVC Offset pipe fitting offset formula in canada - The Home Depot Canada
Overview. Model # MOFFCPLG-200 Store SKU # 1000101033. To connect a PVC conduit to a meter socket. 2" offset coupling. 0.684" offset. Made of PVC. Grey Finish. Specifications.Grooved Couplings Fittings - Mechanical Pipe Joining pipe fitting offset formula in canadaVictaulic is a global manufacturer of mechanical pipe joining methods (fittings couplings), flow control and fire protection systems and is the inventor of the grooved pipe coupling.We build innovative technologies and provide engineering services that address complex piping challenges faced by engineers, contractors, distributors, site owners and property managers.How to Figure a Rolling Offset in Piping HunkerStep 1. Visualize the pipe entering and exiting the box, and if necessary draw a diagram with pen and paper. Find the height of the box (called the rise) as well as the width of the box (called the offset). For instance, the rise is 8 inches and the offset is 12 inches.
### How to Install Offset Drain Pipes for Bathroom Sinks pipe fitting offset formula in canada
How to Install Offset Drain Pipes for Bathroom Sinks. Whether it's because you need to move your sink during a remodel or because your new sink just doesn't quite fit, the sink drain LINK-SEAL&Modular Wall Seals for Pipe and Conduit pipe fitting offset formula in canadaLINK-SEAL &modular seals are considered to be the premier method for permanently sealing pipes of any size passing through walls, floors and ceilings. In fact, any cylindrical object may be quickly, easily and permanently sealed against the entry of water, soil or backfill material. For the system approach, metal or non-conductive Century-Line sleeves with water stops may be Measuring Pipes and Pipe Fittingshub and spigot pipe. Fittings dont interchange from one type or size of pipe to another, but you can buy fittings to join different types or sizes of pipe. To buy fittings, specify the type of fitting you need and the kind and size of pipe I need a 90-degree sweat-solder elbow for ¾ inch copper, for example.
### PVC Pipe Fittings Sizes and Dimensions Guide (Diagrams pipe fitting offset formula in canada
Fortunately, there are many different types of PVC pipe fittings so that you can run pipe in pretty much any configuration. Below is our extensive PVC pipe fittings guide with a series of diagrams and dimension charts. 3-Way PVC Corner Fitting. The 3-way corner fitting splits pipe into three directions and is designed for use in corners.People also askWhat is offset in pipe fitting?What is offset in pipe fitting?In pipe fitting, an offset is a change of direction (other than 90°) in a pipe bringing one part out of (but parallel with) the line of another. An example of an offset is illustrated in Figure 8-8. As shown here, the problem is an obstruction ( E ), such as a wall, blocking the path of a pipeline ( L ).Pipes, Pipe Fittings, and Piping Details:Pipe Fitting pipe fitting offset formula in canadaPipe Calculator Pipe Weight Calculator Online Round pipe fitting offset formula in canadaPipe Weight Calculator/ Online Round Pipe and Tube Weight Calculator, Weight Calculator/ Pipe Tube Weight Calculator / stainless steel pipe weight calculator/square tube weight calculator/structural steel weight calculator/Carbon Pipe Weight Calculator/pipe weight calculation formula in mm, how to calculate weight of pipe in kg/m, pipe weight calculator excel, pipe
### Pipe Fitter Calculator - Apps on Google Play
Generates in seconds Complete app design with "New" portrait layout Pipe Offset - 8 options in 1 Rolling Offset - 9 options in 1 Reducing Tee Offset - 8 Options in 1 Back 2 Back Odd Angle Fitting Calculation 90 to Odd Angle Fitting Calculation Tech Data for All Fitting Sizes Imperial, Metric, DIN 2-Hole Flange Alignment Back 2 Back Fitting pipe fitting offset formula in canadaPipe Fitter Calculator - Apps on Google PlayGenerates in seconds Complete app design with "New" portrait layout Pipe Offset - 8 options in 1 Rolling Offset - 9 options in 1 Reducing Tee Offset - 8 Options in 1 Back 2 Back Odd Angle Fitting Calculation 90 to Odd Angle Fitting Calculation Tech Data for All Fitting Sizes Imperial, Metric, DIN 2-Hole Flange Alignment Back 2 Back Fitting pipe fitting offset formula in canadaPipe Fitting Friction Calculation Can Be Calculated BasedPipe Fitting Friction Calculation Can FFcan be calculated based on the following formula where K is a factor based on the type of fitting, v is the velocity in feet/second, g is the acceleration due to gravity (32.17 ft/s2). 2 (/) (/) 2 2 2 g fts v ft s H FF ft fluid= K PIPE FITTING FRICTION CALCULATION can be calculated based pipe fitting offset formula in canada
### Pipe Fitting Friction Calculation Can Be Calculated Based
Pipe Fitting Friction Calculation Can FFcan be calculated based on the following formula where K is a factor based on the type of fitting, v is the velocity in feet/second, g is the acceleration due to gravity (32.17 ft/s2). 2 (/) (/) 2 2 2 g fts v ft s H FF ft fluid= K PIPE FITTING FRICTION CALCULATION can be calculated based pipe fitting offset formula in canadaPipe Fitting Offsets Calculator - Free download and pipe fitting offset formula in canada'Pipe Fitting Offsets Calculator' is a practical pipe fitting app that makes it easy to calculate the TRAVEL, RUN, SET, 45 Offsets, and even Rolling Offsets Pipe Formulas - Engineering ToolBoxA o = external pipe surface area (ft 2 per ft pipe) Internal Pipe Surface. Internal pipe or tube surface per ft of length can be expressed as. A i = d i / 12 (5) where . A i = internal pipe surface area (ft 2 per ft pipe) Transverse Internal Area. Transverse internal area can be expressed as. A a = 0.7854 d i 2 (6)
### Pipe Offset Calculator on the App Store
Function 2 - Calculate the Set / Travel or Run Use this function to calculate the travel , set or run between two pipe fittings. Using this feature allows you to position a pipe fitting precisely or measure a joining pipe precisely. Function 3 - Rolling Offset.Pipe Offsets - Plumbing Math - Plumbing HelpPlumbing HelpApprentice. Journeymen. Articles. Forum. Pipe Offsets Plumbing Math. Home. Pipe Offsets Plumbing Math. To calculate a pipe offset using 45 degree and 22 1/2 degree elbows use the following chart. To use this chart simply multiply the known side by the corresponding number to find the missing value.Pipefitter Hourly Pay in Canada PayScaleJan 28, 2021The average hourly pay for a Pipefitter in Canada is C\$34.27. Visit PayScale to research pipefitter hourly pay by city, experience, skill, employer and more.
### Pipes, Pipe Fittings, and Piping Details:Pipe Fitting pipe fitting offset formula in canada
Oct 25, 2015In pipe fitting, an offset is a change of direction (other than 90°) in a pipe bringing one part out of (but parallel with) the line of another. An example of an offset is illustrated in Figure 8-8. As shown here, the problem is an obstruction ( E ), Piping calculator delivers accurate results ContractorThe calculator Witt uses in teaching, Pipe Trades Pro by Calculated Industries, pipetradespro, has more than 20 dedicated keys and functions, many identified by words like offset, travel, setback, angle/slope, pipe size, cos and sine right on the keys, so there is no mistaking the function being performed.Rolling Offset Formula - Plumbing HelpPlumbing HelpFitting angle 60 45 22.5 Diagonal = true offset X 1.155 1.414 2.613 Setback = true offset X 0.577 1.000 2.414
### SAMPLE TEST QUESTIONS STEAMFITTER/PIPEFITTER -
mm (3/4-in.) pipe. C. Install both ends of water column in steam space with 25-mm (1-in.) pipe, and install test cocks and inspection crosses top and bottom. D. Tie 50-mm (2-in.) pipe between the mud drum and the steam drum, and install a relief valve and a drain valve. 17 A reduction in pipe size is required in low pressure process steam line.Shop for Pipes Fittings Online Home HardwareSteel pipe fittings and galvanized pipe fittings are useful for outdoor applications. These need to prevent mold and rust for long-term life. A pipe installed from your water source to a faucet or toilet never involves a straight line. You need angled pipe connectors to turn corners and T-connectors to join two pipes.Single Pipe Offsets - Chicago Plumbing CodeThe formulas are to be used in the solution of all single pipe offsets from 5 5/8° to 72° when you know the angle fitting to be used. There are six formulas involving relations between the set, the advance and the travel. Sample problems are given. The tables and formulas are to be used when you want to make an offset but do not know what pipe fitting offset formula in canada
### Steel Pipe Dimensions Sizes Chart (Schedule 40, 80 Pipe pipe fitting offset formula in canada
Different pipe schedule means different wall thickness for the steel pipe in the same diameter. The most frequently indications of schedule are SCH 5, 5S, 10, 10S, 20, 20S, 30, 40, 40S, 60, 80, 80S, 100, 120, 140, 160. The larger the table number, the thicker the surface pipe wall, the higher the pressure resistance.TABLE OF CONTENTSEXAMPLE NO. 8 Preparing Pipe for Ordinate Lengths 26 EXAMPLE NO. 9 Applying Ordinate Lengths to Pipe (Radial Cuts) 27 EXAMPLE NO. 10 Applying Ordinate Lengths to PIpe (Miter Method) 28 EXAMPLE NO. 11 To Find Center to End of Bends Fitting 29 EXAMPLE NO. 12 Laying Out Angles 29 EXAMPLE NO. 13 Laying Out a Y 30Tubular Offsets , Plumbing Offsets , Chrome Plated Tubular pipe fitting offset formula in canadaTubular Offsets come in chrome plated and rough brass varieties. These tubular offsets are available in 1-1/4" and 1-1/2".
### USERS GUidE PIPE TRADES P Ro - Lowe's
Pipe Material and Type data, and Pipe Size dimensions. The Pipe Trades Pro will help you on the jobsite or in the office. Built-in data and Pipe Sizing for 7 different Piping Materials Linear and Rolling Offset Solutions for Known and Unknown Fitting Angles Fitting Take-out and Cut Mark Solutions Cutback Solutionspipe fitting offset formula in canadapipe fitting offset formulaspipe fitting angle offset formulaspipe offset formulapipe fitting offset calculator freepipe fitting 45 degree offsetpipe fitting formulas for fabricationpipe fitting math formulaspipe offset tableSome results are removed in response to a notice of local law requirement. For more information, please see here.pipe fitting offset formula in canadapipe fitting offset formulaspipe fitting angle offset formulaspipe offset formulapipe fitting offset calculator freepipe fitting 45 degree offsetpipe fitting formulas for fabricationpipe fitting math formulaspipe offset tableSome results are removed in response to a notice of local law requirement. For more information, please see here.PLUMBING MATHEMATICSlead need = pipe diameter x lead weight x number of joints 20. Total lead need plus waste allowance total need = lead need (100% - % of waste) 21. Degree of offset of a pipe fitting degree of angle = fitting x 360° NOTE Tab This Section On Formulas And Constants CONSTANTS 22. 1 cubic foot of water = 7.5 gallons 23. | 4,193 | 18,323 | {"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} | 3.78125 | 4 | CC-MAIN-2021-39 | longest | en | 0.765826 |
https://www.intefrankly.com/articles/Programming-Question-Even-String-Analysis-Code/6b5b51d29045 | 1,679,716,816,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945315.31/warc/CC-MAIN-20230325033306-20230325063306-00336.warc.gz | 933,846,539 | 5,572 | [Programming Question] Even String Analysis Code
Collection of Programming Questions for National Unified Written Mock Test (5th) for School Recruitment 2017
Time limit: 1 second Space limit: 32768K
If a string consists of two identical strings concatenated, Just call this string an even string。 for example"xyzxyz" harmony"aaaaaa" It's an even string., nevertheless"ababab" harmony"xyzxy" but it's not。 The bull now gives you an even string s containing only lowercase letters, you can delete 1 and or more characters from the end of the string s to ensure that the string is still an even string after the deletion, and the bull wants to know what the longest even string length is after the deletion. Input Description: The input consists of a string s, length(2 ≤ length ≤ 200), ensuring that s is an even string and consists of lowercase letters
Output Description: Outputs an integer that indicates the length of the longest even string that can be obtained after deletion. Guaranteed non-zero solution for test data
Enter example 1: abaababaab
Output example 1: 6
# analysis
Simple question, the title makes the idea very clear. Start deleting from the last character to see if it is an even string, if it is, it is returned directly, if not, continue deleting.
# code
```import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
String s = in.nextLine();
in.close();
System.out.println(helper(s));
}
private static int helper(String s) {
for(int i=s.length()-1;i>0;i--) {
if(isString(s.substring(0, i)))
return i;
}
return 1;
}
private static boolean isString(String s) {
String front = s.substring(0, s.length()/2);
String end = s.substring(s.length()/2, s.length());
return front.equals(end);
}
}```
Recommended>>
1、Is it enough to apologize for algorithmic values when the smallest mom on the internet is a selling point
2、Wacker Gold Official Launch of WCG Wacker Gold Exclusive Wallet
3、The White Paper on Full Burial Techniques for Android is back Open source all project source code
4、Charging like this is dangerous and can steal all the information from your phone in minutes
5、Chinas smart bikesharing creates a new force for global mobility
已推荐到看一看 和朋友分享想法
最多200字，当前共 发送
已发送
确定
分享你的想法...
取消
确定
最多200字，当前共
发送中 | 555 | 2,315 | {"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-2023-14 | latest | en | 0.723857 |
brookegordon.ca | 1,725,773,657,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650960.90/warc/CC-MAIN-20240908052321-20240908082321-00280.warc.gz | 5,456,302 | 11,810 | # How much does your Double Double cost?
How much does your double double cost? It’s less than \$2 for a Medium or Large coffee with two creams, two sugars. But that’s not what I mean.
Here’s what you may not know … the amount of cream and sugar changes with the size of the drink. A Large double double coffee has about 1/3 more coffee and also more calories from the cream and sugar than a Medium.
Medium is 425mL (or 1-3/4 cups) with 18g of sugar. It has 212 calories.
Large is 563mL (or 2-1/4 cups) with 24g of sugar. It has 264 calories.
### What’s the real cost to you?
One of the things I hear often as a nutrition coach is “I gained another 10 lbs this year, and I just don’t understand why. I’m eating healthier than ever!”
Let’s do some quick logic-based math:
• 1 tsp of white sugar is (4.2 g) and there are 48 teaspoons in a cup.
• Two Large coffees a day has about 11.4 teaspoons of sugar, or 1.2 cups per week (Mon – Fri).
There are 10 Statutory holidays and most people take 2-weeks vacation. With 52 weeks in a year, this assumes 48 working weeks. That means drinking two Large coffees on a week day (Mon – Fri) is that same as eating 57 cups of sugar a year (excluding weekends).
That’s not insignificant, but how does drinking coffee lead to putting on weight?
• There are 773 calories in a cup of sugar.
• 57 cups of sugar is 44,171 calories.
After finishing the coffees, we normally sit back down at our desk. All that energy is not being used! The natural outcome is that our body will convert excess sugar into fat.
It all starts to make sense, yes? 1 pound of fat is 3,500 calories. Drinking two Large coffees every weekday for a year is the equivalent of 12.6 pounds of fat (excluding calories from cream).
### What can you do about it?
If a 200-pound person who walks at 3.5 miles per hour can burn 346 calories in an hour, then to “walk off” that much sugar will take 127.6 hours, or 32 minutes per day (Mon – Fri). Do you walk every day at lunch?
It’s Monday, January 2nd today, and you must have a New Year’s resolution. Can I add to it?
### Final thoughts.
Drink your coffee with no sugar.
It’s a quick and easy way to lose weight just by saying no thank you. While you’re at it, go for a walk every day. Preferably 30-minutes at lunch and again in the evening.
## AuthorBrooke Gordon
My name is Brooke and I love to cook, hence the nickname. I am passionate about eating for pleasure and nutrition, making jam, and supporting women who want to live a healthy life. | 666 | 2,514 | {"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-2024-38 | latest | en | 0.915973 |
https://techlyfire.com/how-many-types-of-errors/ | 1,656,647,998,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103920118.49/warc/CC-MAIN-20220701034437-20220701064437-00512.warc.gz | 598,983,469 | 15,710 | # How many types of errors?
Article by: José Manuel Tovar Segundo | Last update: April 10, 2022
Score: 4.8/5
(54 ratings)
Types of Errors
(1) Systemic errors. In this type of error, the measured value is biased due to a specific cause. … (2) Random errors. This type of error is caused by random circumstances during the measurement process.(3) Negligence errors.
## How many types of errors are allowed in a measurement?
CLASSES OF ERRORS IN THE MEASUREMENT. (SYSTEMATIC, CIRCUMSTANTIAL) These errors are divided into two classes systematic and circumstantial, these errors are presented constantly through a set of readings made when making the measurement of a given magnitude.
## What are gross errors?
Gross errors are characterized by a large absolute error of the measurements made, which allows them to be easily identified. Once the gross error has been identified, it must be ruled out, since if it is taken into account, the precision of the measurement drops considerably.
## What are the 3 types of errors that we can calculate?
Types of Errors
(1) Systemic errors. In this type of error, the measured value is biased due to a specific cause. … (2) Random errors. This type of error is caused by random circumstances during the measurement process.(3) Negligence errors.
## What are gross errors or mistakes?
Outside the classification of errors are the mistakes, typical of carelessness, of the misapplication of a method, they are gross errors that must be eliminated. They are detected by repeating the measurement, or by obtaining absurd values. They are prevented with care and attention.
34 related questions found
### What are the errors and mistakes in surveying?
Accidental errors: They are due to chance. To quantify the error we use two types of errors: absolute and relative. Absolute error (Ea) of a measurement is defined as the difference between the mean value obtained and that found in that measurement, all in absolute value.
### What are accidental errors in topography?
ERRORS IN THE OBSERVATIONS
Accidental ones occur due to lack of appreciation of the instrument and observer, they are random and we will never know the value they are taking, we have to limit ourselves to knowing the maximum value in one or more measurements.
### How to detect a gross error in your measurements?
Gross errors or glitches
Gross errors are characterized by the fact that their magnitude exceeds what can be foreseen taking into account the means with which they operate. These errors generally come from the observer’s distraction, and for them there is no theory.
### What is a calculation error?
The uncertainty or numerical error is a measure of the adjustment or calculation of a magnitude with respect to the real or theoretical value that said magnitude has.
### What are the types of errors in Excel?
The most frequent errors that you can find in Excel formulas
Forget the equals sign. … Forget some parentheses. … Not indicating a range well. …Do not include all arguments. … Not using the correct argument. … Formatted numbers. … Mislink other sheets. … Solution 1: Checking for errors.
### What is absolute error, relative error, absolute error, and percentage error?
It is the quotient (the division) between the absolute error and the exact value. If it is multiplied by 100, the percentage (%) of error is obtained. Just as the absolute error can be positive or negative (depending on the absolute error) because it can be due to excess or deficiency.
Always Check Techlyfire for more how to related guides. | 750 | 3,570 | {"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} | 3.421875 | 3 | CC-MAIN-2022-27 | latest | en | 0.936506 |
https://www.justintools.com/unit-conversion/fuel-consumption.php?k1=kilometers-per-liter&k2=miles-per-liter&q=3470000000 | 1,695,464,539,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506480.7/warc/CC-MAIN-20230923094750-20230923124750-00276.warc.gz | 947,562,704 | 26,364 | Please support this site by disabling or whitelisting the Adblock for "justintools.com". I've spent over 10 trillion microseconds (and counting), on this project. This site is my passion, and I regularly adding new tools/apps. Users experience is very important, that's why I use non-intrusive ads. Any feedback is appreciated. Thank you. Justin XoXo :)
# Convert [Kilometers Per Liter] to [Miles Per Liter], (km/L to mi/L)
## FUEL CONSUMPTION
3470000000 Kilometers Per Liter
= 2156158037.0635 Miles Per Liter
*Select units, input value, then convert.
Embed to your site/blog Convert to scientific notation.
Category: fuel consumption
Conversion: Kilometers Per Liter to Miles Per Liter
The base unit for fuel consumption is kilometers per liter (Non-SI/Derived Unit)
[Kilometers Per Liter] symbol/abbrevation: (km/L)
[Miles Per Liter] symbol/abbrevation: (mi/L)
How to convert Kilometers Per Liter to Miles Per Liter (km/L to mi/L)?
1 km/L = 0.62137119223733 mi/L.
3470000000 x 0.62137119223733 mi/L = 2156158037.0635 Miles Per Liter.
Always check the results; rounding errors may occur.
Definition:
In relation to the base unit of [fuel consumption] => (kilometers per liter), 1 Kilometers Per Liter (km/L) is equal to 1 kilometers-per-liter, while 1 Miles Per Liter (mi/L) = 1.609344 kilometers-per-liter.
3470000000 Kilometers Per Liter to common fuel-consumption units
3470000000 km/L = 8161940424.8913 miles per gallon US (MPG[US])
3470000000 km/L = 9802093749.8235 miles per gallon UK (MPG[UK])
3470000000 km/L = 15774950106.606 kilometers per gallon US (km/gal)
3470000000 km/L = 3470000000 kilometers per liter (km/L)
3470000000 km/L = 3470000000000 meters per liter (m/L)
3470000000 km/L = 2156158037.0635 miles per liter (mi/L)
3470000000 km/L = 98259456899310 meters per cubic foot (m/ft3)
3470000000 km/L = 56863115895.551 meters per cubic inch (m/in3)
3470000000 km/L = 3.47E+15 meters per cubic meter (m/m3)
3470000000 km/L = 43094883327769 feet per gallon US (ft/gal[US])
(Kilometers Per Liter) to (Miles Per Liter) conversions | 606 | 2,050 | {"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-2023-40 | latest | en | 0.698389 |
http://www.chegg.com/homework-help/questions-and-answers/pulses-traveling-having-speed-of1-cm-s-t-0-positions-shown-thedrawing-t-2-s-heightof-resul-q462342 | 1,387,466,049,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1387345764241/warc/CC-MAIN-20131218054924-00099-ip-10-33-133-15.ec2.internal.warc.gz | 312,185,992 | 7,093 | superposition
0 pts ended
Two pulses are traveling toward each other, each having a speed of1 cm/s. At t = 0, their positions are shown in thedrawing.
When t = 2 s, what is the heightof the resultant pulse at each of the following locations?
(a) x = 3 cm
1 cm
(b) x = 4.5 cm
2 cm | 84 | 282 | {"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-2013-48 | latest | en | 0.937019 |
https://hubpages.com/education/what-is-calculus-part-3 | 1,544,856,116,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826800.31/warc/CC-MAIN-20181215061532-20181215083532-00094.warc.gz | 623,473,123 | 30,546 | # WHAT IS CALCULUS? (Part III)
Updated on December 21, 2014
## LabKitty Presents
Thus far on WHAT IS CALCULUS?, we have explored the two great realms of calculus -- the integral and the differential -- and met their sovereigns. The Integral: he of areas and volumes and the finding thereof. The Derivative: she of rates of change, be they in time, in space, or in something else. Now it is time to wed. Time for these realms to make peace with one another and be joined into a glorious whole. We are not strands of hair; we are hair! George Washington beseeched his reluctant colonists to unify and become the great county of Washingtonland, a name ultimately rejected for something that looked better on the currency. But it's the thought that counts.
And so we now stand at base camp, our resolute faces turned toward the summit visible through the clouds in the brilliant morning sun. What had been for so long just a dream is now in our sights. It will soon be in our mitts.
We give you WHAT IS CALCULUS? (Part III): The Fundamental Theorem of Calculus.
## Exciting Climax
Perhaps you've heard that calculus was invented by Newton. Not true. The concepts of the integral and the derivative had been kicking around for quite awhile when Isaac happened upon the scene. For example, the circle area thing goes all the way back to ancient Greece (although the Greeks screwed the pooch with the Zero Paradox which isn't true at all but that's only because they didn't know about convergent infinite series which is something we're not going to get into and why are we still typing). And if we have done our job, we have convinced you that the integral and the derivative are not difficult ideas. They are familiar geometric concepts (slope, area) that calculus flogs to death. Sure, there is a mountain of details we keep alluding to that can make the going unpleasant, as any freshman can attest. But if you have read Parts I & II, you now know the nitty-gritty, the marrow, the substance of the two great halves of calculus: differential and integral.
Is that all there is to it? Surely there must be something more that earns calculus it's fierce reputation. There must be more devil than mere details. Where's the beef? Clara Pellar was heard to exclaim.
Well, Ms. Pellar, here comes the beef.
What Newton (and some others) did was discover a relationship between differential and integral calculus and join them. Like Aragon reforging the Elf Sword of Berkenstock after it was broken (or whatever the heck he did and to what). The relationship is not obvious. It may even seem a little mysterious. Finding it required insight beyond circle tiling and secant drawing.
This piece of work goes by the name The Fundamental Theorem of Calculus, which sounds pretty darn pretentious. Still, lets see if we can't get this fish in the boat. As with the integral and the derivative, our explanation of the FToC begins with a story.
## A Curious Shoppe
Imagine you run a calculus curio shoppe, where customers bring you functions and pay you to find the derivatives at places they point to (meh, it could happen. Humor us). You usually perform the (tedious) secant line thing to find these derivatives. However, your years in the differentiating biz have allowed you to compile a list of functions that are the derivatives of other functions. For example, you have noted that no matter where you point to on the sine function -- which we will write as sin(x) for "sine at x" -- if you go through the whole secant approximation limit argle-bargle, you get exactly the same number if you just look up (or punch into your calculator) the value of the cosine at x -- written: cos(x). A curious result, no doubt, but curious is your business (heck, it's right there in the name). What matters most is keeping the customers happy, and if the secret list saves time and impresses the punters, then who cares why it works.
One day, a customer enters your shoppe. You immediately sense this guy is trouble. The customer has brought a function, but he does not want you to differentiate the function, he wants you to tell him what function when differentiated produces the function he has brought. That is to say, he does not want a derivative, he wants (wait for it) an anti-derivative.
If he has brought you a function that happens to be on your secret list, all is well. You simply use your list backwards. For example, if the customer brought you cos(x), then you can immediately tell him the anti-derivative is sin(x), because you know that the derivative of sin(x) is cos(x).
But what if you are not so fortunate? What if he brings a function that is not on your list? One you have never seen before?
The customer jabs a finger at the function. At this location, he demands, what is the value of the anti-derivative?
Your mind reels, your brow sweats. If you fail this man, word of your incompetence will surely spread. Customers will stop coming! Your derivative business will go under! You will have to get a real job!
And then you remember the Fundamental Theorem of Calculus:
The value of the anti-derivative at any point on a curve is the area under the curve up to that point.
(There is a slight inaccuracy in this bold claim that we will correct in just a moment -- although it is a 100% true statement for the example we're about to show. Including all the fine print would have ruined our dramatic reveal.)
This result is so important we should draw a box around it, but we're pretty stupid when it comes to HTML, so you're just going to have to imagine a box drawn around it (hey, we're impressed we figured out red text). We learned way back in Part I that the area under a function is the integral. But, we claim, the integral is also the anti-derivative of the function. That is, integration gives you a function that, when differentiated, gives you the original function that you just computed the area under. This grand connection between integrals and anti-derivatives, between differentiated functions and the area under them, is called The Fundamental Theorem of Calculus.
The FToC is by no means obvious. In fact, it's a little brain melty. It's going to take more than one sentence in cool red text to explain just what the heck is going on.
Let's go to the whiteboard.
## The Fundamental Theorem of Calculus
### Meet the FToC
At the top of the accompanying figure is the function brought to you by Mr. Pesky Customer. We assumed the function is cos(x), which you may remember from your trigonometry days. In the table to the right of the graph, we've picked a few representative x between 0 and π/2 and listed values of the function at these locations. (There is no profound reason for stopping at π/2 in the table except that figure space is tight.)
Remember, the customer does not want the derivative of cosine; he wants the function that when differentiated gives cosine.
In the middle panel we've given you the answer, which is the sine function or sin(x). First, let's verify that claim is true. In the table to the right, we've picked a few representative x between 0 and π/2 and listed the value of the derivative of sin(x) in the right-hand column (a vertical bar with an x at the bottom is standard notation for "the derivative at x"). These we obtained using the secant limit method described in Part II, an example of which is drawn on the graph for x = π/3. You may easily verify that the values shown here for d sin(x) / dx are indeed the same as the values of cos(x) listed in the top panel for corresponding values of x. That is, sin(x) is the function that when differentiated gives cos(x).
So far, so good.
We now return to the customer's function -- cos(x) -- in the bottom panel. In the table to the right, we've picked a few representative x between 0 and π/2 and listed values of the area under cos(x) between 0 and x in the center column. These values were obtained using the tiling limit method described in Part I, and are equal to the integral of cos(x) from 0 to x, which is how we've labeled the column (recall that when we write an integral, we indicate the left and right endpoints of the area on the bottom and top of the integral sign).
In the right column of the table we show the value of sin(x) at the same x. The values in the center and right columns are equal! What on Earth can this mean?
The bottom panel shows that the integral of cos(x) is sin(x). But the center/top panels show that the derivative of sin(x) is cos(x). We have found the anti-derivative via integration! That is, the integral gives us the function that when differentiated gives us the function we are integrating. You may want to re-read that a few times; the logic is a little like a snake eating its own tail.
It turns out this is not an isolated incident. Here we have seen d sin(x)/dx = cos(x) so ∫ cos(x) dx = sin(x). But it is also true that d cos(x)/dx = -sin(x) so ∫ sin(x) dx = -cos(x); d tan(x)/dx = sec2(x) so ∫ sec2(x) = tan(x); d log(x)/dx = 1/x so ∫ 1/x dx = log(x). And on and on. Heck, much of first-semester calculus is memorizing a big honkin' list of derivative/integral pairs. And once you know the trick, you can skip all the secant line flogging and rectangle tiling bashing. You just write down the answer.
This, then, is the Fundamental Theorem of Calculus. The relationship that ties together derivatives and integrals or, more to the point, integrals and anti-derivatives. The relationship that ties together rates-of-change and areas-under-curves. In a single sentence: the value of a function at x is the rate at which area is being added to its integral at x.
Almost.
Minor Correction
When we sprung the FToC upon you in the last section (which we wrote in red and everything) we mentioned there was a slight inaccuracy in our dramatic reveal. Here is the inaccuracy: the area under the curve is not equal to the value of the anti-derivative at the right endpoint, but rather the value of the anti-derivative at the right endpoint minus the value of the anti-derivative at the left endpoint. The example above was craftily chosen so that (1) the anti-derivative was sin(x) and (2) the starting point was zero. Because sin(0) = 0, there is nothing to subtract.
## Achievement Unlocked
And so we have arrived at the marriage of differential and integral calculus as promised. The derivative on one side of the coin and is engraved rate-of-change. The integral on the other side of the coin and is engraved area-under-the-curve. Flipping from heads to tails is differentiation. Flipping from tails to heads is integration, yes, but it is also anti-differentiation. That is to say, given a description of the rate of change of a thing, we can use integration to obtain the thing itself.
Here your mind should wander back to our interlude on Nature (witch, rhymes with). Integration is how we deal with natural laws expressed in terms of rates of change. It is how we solve differential equations; how we turn a description of how a thing is changing into the the thing itself. More than computing areas, more than finding volumes or lengths or whatever, this feature of integration is key. It gives mathematics the tool it needs to model the real world. The flaming hoop to Nature's tiger, the time-out to her rough-housing, the restraining order to her psychosis.
The Fundamental Theorem of Calculus literally changed the world. It is the difference between the mush-head physics of the ancient Greeks where falling things seek their center and modern physics that puts numbers and predictive power in our hands. The ancient medicine of demonic possession and humors that became modern medicine of titration and robotics. Ancient yearnings of flapping flight that became modern engineering that builds machines that fly, and many things besides. More progress happened in the two centuries after Newton then in the 20 that came before.
It's strange to think of the Fundamental Theorem of Calculus just sitting there for thousands of years, waiting to be discovered. If we take the founding of Plato's Academy to be day one of math class, his students would be getting to the FToC around their 4000th semester. These days it's taught somewhere in Calc I or perhaps at the beginning of Calc II. It's in a bazillion lectures on YouTube and iTunes and probably even turns up in a TED talk or two. The discovery did not require billion-dollar machines or multinational laboratories. Quill, parchment, candle, and -- if we know mathematicians -- a bottle of absinthe and all was put in light. Newton released the Krakken, raised the floodgates. The night busted open, this two-lane will take us anywhere.
One can't help but wonder what might still be out there lurking unseen in the shadows.
## Calculus on Amazon
Schaum's Outline of Understanding Calculus Concepts
If you're looking to take your first real steps into calculus, you could not find a better guide. More of a textbook than the standard Schaum's format, and one that that is brief and superbly written (and cheap!). It's all here: differentiation, integration, infinite series, and applications. Throughout, Passow employs the approximation-refinement-limit approach we used as a conceptual thread that ties everything together. It doesn't quite work as a formal textbook (Passow omits eleventy gazillion integration tricks, most of analytic geometry, and all of vector calculus) but it can't be beat for getting your foundations right or as supplementary material to explain what your course textbook isn't.
## Epilogue
This very day thousands of students sit in calculus classrooms around the globe, hair combed, shoes shined, their freshly-scrubbed faces eager to suckle learning goodness from the bosom of our Harsh Mistress. It is a formidable task, transforming their little mathematical garter snake arms into mighty problem-solving pythons. (A task increasingly assigned to a clown car of adjunct faculty and graduate students for reasons beyond our ken. But we digress.) What exactly goes on behind the closed doors of these hallowed chambers?
A traditional calculus course mostly teaches you clever ways to compute derivatives and integrals. Charming though it may be, calculus jocks don't break out little computer programs every time they need to find a derivative or integral. There's memorization of some basic results, and of some basic rules, after which application of the rules to the results generates a cornucopia of derivative and integrals to last a lifetime. There are forays into geometry, and also into the study of infinite sums of numbers (a topic we skipped completely) which is how functions are implemented on the buttons of your calculator. Along the way, you learn to apply calculus to solve real-world problems in physics, chemistry, engineering, biology, ecology, epidemiology, economics, probability, statistics.
And we lied a little. When the going gets really tough, we do break out the computers.There's entire industries, cottage and otherwise, devoted to solving calculus problems on computers. Warning! Buzzwords ahead! Finite difference method. Finite element method. Boundary value method. Newton-Raphson. Quadrature. Numerical integration. Runge Kutta. Example usage in a sentence: I couldn't figure out the integral on the homework so I slapped that bad boy with a second order Newton-Raphson and BAM! in five minutes I was sipping mai tais on the veranda with Miranda. The implication being, yes, people will pay you to do calculus.
Freeman Dyson famously quipped that Nature gives a clear answer when a scientist asks a clear question. You now know that Nature does not give you an answer, she gives you a differential equation. And you now know that calculus is the tool used to turn those differential equations into answers. The true golden lamp that lit the doorway, calculus is. The beacon that guided our tempest-tossed forbearers from vague seas to certain shore, where huddled-masses yearning to breath free finally made landfall, where foamy whitecaps of muddled thinking were traded for a rock of sound footing.
Upon that rock, we have built the world.
## Image Credits
The Matterhorn by Zacharie Grossen and appears under the terms of the Creative Commons Attribution-Share Alike 3.0 Unported license. Image of Wenlock Bookstore by Michael Maggs and appears under terms of the Creative Commons Atribution-Share Alike 3.0 Unported license. Wedding of Queen Victoria and Prince Albert engraved by S. Reynolds after F. Lock and is in the public domain. Woman teaching geometry from Wikimedia Commons and is in the public domain.
Images may have been cropped, resized, color adjusted, or otherwise modified from their originals.
--
All other weirdness (c) 2013-14 LabKitty Design
23
2
59
0
0
10
3 | 3,696 | 16,819 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2018-51 | latest | en | 0.963371 |
http://studentsmerit.com/paper-detail/?paper_id=46159 | 1,481,248,769,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542668.98/warc/CC-MAIN-20161202170902-00358-ip-10-31-129-80.ec2.internal.warc.gz | 270,643,882 | 7,899 | #### Details of this Paper
##### Operations Management Ch 9 and 10 Homework
Description
solution
Question
Question;Chapter 91. Pepsi uses six-sigma technique to evaluate whether its 12 oz Gatorade has consistent weight. Pepsi took a sampling of 1000 12 oz Gatorade and found out that the average weight was 11.58 oz with a standard deviation of 0.3 oz. Pepsi considers a tolerance of ? 5% from 12 oz acceptable.a). What is the percentage of 12 oz Gatorade that meets the specification?b). What are the percentages of 12 oz Gatorade that are considered ?overweight? and ?underweight??c). If Pepsi improves its production process, and samples another 1000 bottles and finds a average weight of 12 oz with a standard deviation of 0.2 oz. What is the % of 12 oz Gatorade that meets the specification? What quality standard does this new process achieve (in terms of number of ?sigma?)?d) Following c, what is the standard deviation that Pepsi needs to achieve in order to achieve the 6-sigma standard without having to change the specification?Chapter 10 2. a. Following problem 1(a) from chapter 9. What is the process capability index (Cpk)? b. Following problem 1(c) from chapter 9. What is the process capability ratio (Cp)? 3. A process that produces computer chips has a mean of .04 defective chip and a standard deviation of .004 chip. The allowable variation is from .03 to .06 defective. The manufacturer aims to achieve performance of 3-??sigma standard. a. Compute the process capability ratio (Cp). b. Compute the process capability index (Cpk). c. Is the process capable? (meaning, does it fulfill 3-??sgima standard?) 4. An appliance manufacturer wants to contract with a repair shop to handle authorized repairs in El Paso. The company has set an acceptable range of repair time of 50 minutes to 90 minutes. Two firms have submitted bids for the work. In test trials, one firm had a mean repair time of 69 minutes with a standard deviation of 6 minutes and the other firm had a mean repair time of 72 minutes with a standard deviation of 4 minutes. Using the process capability index (Cpk) to evaluation the firms? performances. a. Compute Cpk for firm 1 b. Compute Cpk for firm 2 c. Which firm has a better performance? 5. The Good Chocolate Company a 12-??ounce chocolate bar (340 grams). Specifications for the 12-??ounce bar are 330 grams to 350 grams. Based on sampling result, the company knows that the average fill is exactly 340 grams. a. Identify the largest standard deviation (in grams) the machine that fills the bar molds can have so that the process can be considered capable of achieving 3-??sigma standard (Hint: you use either Cp or Cpk to identify the sigma since the center is exactly at 340 grams.) b. Identify the largest standard deviation (in grams) the machine that fills the bar molds can have so that the process can be considered capable of achieving 4-??sigma standard.
Paper#46159 | Written in 18-Jul-2015
Price : \$27 | 701 | 2,966 | {"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-2016-50 | longest | en | 0.919738 |
https://www.e-bookdownload.net/search/how-to-burn-calories-and-stay-fit-forever | 1,585,460,972,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370493818.32/warc/CC-MAIN-20200329045008-20200329075008-00532.warc.gz | 940,906,120 | 13,617 | ## How To Burn Calories And Stay Fit Forever
Author by : © Wings Of Success
Languange : en
Publisher by : GOFFYBLN
Format Available : PDF, ePub, Mobi
File Size : 41,9 Mb
Description :
## How To Burn Calories And Stay Fit Forever
Author by : Mhar De Jesus
Languange : en
Publisher by : BookRix
Format Available : PDF, ePub, Mobi
File Size : 49,8 Mb
Description : Overview How To Burn Calories And Stay Fit Forever - Personal and Practical and Study Guide by Self Improvement For anyone trying to lose weight, this question is bound to raise a lot of excitement. Surely losing weight cannot be such a simple issue, can it? Well, the answer is YES! You can actually lose calories by drinking ice water. Your body loses calories in the process of warming this ice water to the body temperature. Now any enthusiast, must surely be thinking, if we can lose weight by drinking ice water, can we lose a large amount of calories if we drink lots of ice water? Well, to answer this question we have to look at some simple calculations. First of all we need to distinguish between calories and Calories. Calories (i.e. with a big c) are used to denote the amount of energy that is contained in food. Where as calorie with a small c is used to denote the energy required to raise the temperature of 1 gram of water 1 degree Celsius. Another interesting fact is that it takes, 1 Calorie to raise the temperature of 1 kilogram of water by 1 degree Celsius. So when you are drinking a 140-Calorie can of cola, you are in fact ingesting 140,000 calories in your body. This is the same when you burn say 100 Calories working out, this means that you have actually burned 100,000 calories. The main purpose of telling you that the definition of calories is based on the rising of temperature is to tell you an interesting fact. We have just seen that when our body raises the temperature, it burns calories, so when you drink ice cold water your body loses calories in raising that ice cold water to body temperature. Now let us get the math right. Our body temperature is at 37 degree Celsius.The temperature of ice cold water can be safely said to be 0 degree Celsius. There are 473.18 grams in 16 fluid ounces of water. It takes 1 calorie to raise 1 gram of water by 1 degree Celsius. So, if your body raises the temperature of 473.18 grams of water by 37 degree Celsius it burns 17508 calories. But this is calorie with a small c. It actually denotes only 17.5 calories. You might be thinking that losing 17.5 Calories doesn't count much compared to the calories we intake. But, you are not going to drink just one 16 once glass of water are you? Even if you stick to the recommended minimum of 8 glasses of water you will end up burning 70 Calories in a day and that too by doing practically nothing. You can also increase the water intake if you want to shed a few extra pounds. Well, although it is definite that drinking ice cold water helps you to burn calories you should not try to replace it with exercise. You should continue with all the weight reduction methods that you already on to. You can just boost up your effort by drinking ice cold water.
## The Bodybuilding Com Guide To Your Best Body
Author by : Kris Gethin
Languange : en
Publisher by : Simon and Schuster
Format Available : PDF, ePub, Mobi
File Size : 53,6 Mb
Description : The editor-in-chief of Bodybuilding.com outlines a twelve-week nutrition and exercise program that focuses on improving strength, incorporating healthy foods, and tapping the motivational aspects of a support network.
## Callanetics Fit Forever
Author by : Callan Pinckney
Languange : en
Publisher by : Random House
Format Available : PDF, ePub, Mobi
File Size : 45,8 Mb
Description : The CALLANETICS FOREVER FIT plan focuses on the special health and fitness needs of women over 30, whether they are just starting a fitness programme, or already in good shape. There is guidance on good nutrition with aerobic exercise (to strengthen the heart) and strategies for stress management - both of which are more important than ever. A new, specially developed, easy-to-follow CALLANETICS exercise plan, targeting key areas of the body (stomach, legs, buttocks and hips, back and upper body) is included with special exercises for women with back, knee or hip conditions. The programme is designed to build the strength needed to take women into their later years with greater health, fitness and beauty.
## Become Stay Fit Forever The Holistic Psychological Aspect Of The Problem
Author by : Allyson Hodge
Languange : en
Publisher by :
Format Available : PDF, ePub, Mobi
File Size : 52,8 Mb
## Your Best Body At 40
Author by : Jeff Csatari
Languange : en
Publisher by : Rodale
Format Available : PDF, ePub, Mobi
File Size : 43,5 Mb
Description : A program of flexibility and strength-building exercises, nutritional advice and tasty recipes geared toward men in their 40s promises to help them build muscle and lose weight, have more energy and feel happier, strengthen bones, enjoy better sex, keep their brains sharp, manage stress and look younger.
## The Polar Fat Free And Fit Forever Program
Author by : James M. Rippe
Languange : en
Publisher by : Pocket Books
Format Available : PDF, ePub, Mobi
File Size : 40,8 Mb
Description : A fictional recreation of the life of Geronimo, the last great warrior of the Apache nation, and his struggle against the white settlers to retain his tribe's land and way of life
## Maintain Weight Forever
Languange : en
Publisher by : All-Seeing Books
Format Available : PDF, ePub, Mobi
File Size : 42,5 Mb
Description : Maintain weight loss! Maintain weight gain! Maintain weight forever! The popular Maintain Weight Forever website expands into a handy book. Must-see information includes stopping regains, choosing a goal weight, and shrinking saggy skin. Featuring exclusive bonus content, there's something for everyone who wants to maintain for ever, not just for now. Search Terms: maintain weight forever, weight advice, weight gain, weight maintenance, weight management, weight control, weight loss
Author by : Bob Harper
Languange : en
Publisher by : Harmony
Format Available : PDF, ePub, Mobi
File Size : 50,6 Mb
Description : The world-renowned fitness coach on the hit TV show The Biggest Loser presents his winning approach to lasting weight loss by showing how to get at the root of your overeating problem, followed by a nutritionally savvy diet and unique exercise plan. On The Biggest Loser, Bob Harper gives contestants the practical tools and psychological insights they need to get into the best shape of their lives. The key to his success is the emotional connection he makes with each participant, and he brings that same spirit to Are You Ready! Harper starts with a four-step strategy for getting at the root of negative thought patterns and destructive behaviors, replacing both with a clear way to build self-worth and confidence. With these tools in place, people are empowered to make real, lasting changes in their lives. In an easy-to-follow eating plan, he provides lists of foods that are nutrient-dense and naturally low in calories, more than twenty sample menus, and tips on eating on the run, in restaurants, and on vacation. His fitness plan is geared to making exercise an integral part of daily life with workouts (ranging from 20 to 60 minutes) based on training techniques that tone and strengthen, burn calories, and reshape the body. Woven throughout Are You Ready! are true-life success stories that will keep readers engaged and motivated; bulleted tips, tools, and coping strategies; and sidebars debunking common myths about food and fitness. Whether your goal is losing ten pounds or a hundred, you will find Harper’s message inspiring and his methods a proven path to finally achieving your dream of weight loss and fitness.
## The Step Diet Book
Author by : James O. Hill
Languange : en
Publisher by : Workman Publishing
Format Available : PDF, ePub, Mobi | 1,771 | 8,004 | {"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-2020-16 | latest | en | 0.908338 |
https://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-6th-edition/chapter-5-section-5-1-exponents-and-scientific-notation-exercise-set-page-262/27 | 1,529,888,510,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867304.92/warc/CC-MAIN-20180624234721-20180625014721-00102.warc.gz | 806,015,314 | 14,844 | ## Intermediate Algebra (6th Edition)
$-13z^{4}$
We are given the expression $-\frac{26z^{11}}{2z^{7}}$. We can simplify this expression by using the quotient rule, which holds that $\frac{a^{m}}{a^{n}}=a^{m-n}$ (where a is a nonzero real number, and m and n are integers). $-\frac{26z^{11}}{2z^{7}}=(-\frac{26}{2})\times(z^{11-7})=-13z^{4}$ | 123 | 342 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2018-26 | latest | en | 0.750331 |
https://gtaforums.com/topic/817410-gun-control/?page=2 | 1,544,406,284,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823236.2/warc/CC-MAIN-20181210013115-20181210034615-00242.warc.gz | 624,997,037 | 36,388 | Search In
• More options...
Find results that contain...
Find results in...
# Gun Control
## Recommended Posts
approx murders in 2012 in north america is 15,000 approx [googled]. Looking at it habitable area wise, in 2012 Europes murder rate was 22,000 approx.
America seems safer than Europe.
You argue there were approximately 15.000 murders in 2012 "North America". In Mexico alone there were 26037 murders in 2012, which is almost double the number you "googled". So I'm going to assume you mean in the United States, as in the country, not North America the continent. Which means you're essentially comparing a country to a whole continent, classic. Now allow me to use your own logic against you. The US had 314.1 million Inhabitants in 2012 with 14287 murders. Europe had 740 million inhabitants in 2012 and, taking your number at face value, 22,000 murders. Quick bit of math (again taking your number at face value) tells us that amounts to a murder rate of 2.97 per 100.000 inhabitants whereas for the US it amounts to a murder rate of 4.54 per 100.000 inhabitants.* Making Europe safer in this metric.
*Rates calculated with rough approximations given of both number of inhabitants and recorded homicides. Real rates differ but support conclusion.
easy raavi. no need to "murder" me with your logic
##### Share on other sites
I don't even f*cking care anymore. Why don't we outlaw everything that could be dangerous?! Why don't we lock ourselves up in a room filled with pillows to keep us safe?! Here are my great ideas:
- No more shooting sports, people could possibly get shot!
- No more motorsports, people could get run over!
- No more baseball, you could kill someone with a baseball bat!
- No more darts, you could poke someone's eye out!
- No more hiking, you could get pushed by someone and die!
- No more sailing, you could get thrown of your boat!
- No more boxing, you could possibly hit someone to death!
- No more chess, you could use a chess piece to choke someone!
Why don't we also ban knives and forks? We'll just eat with out hands. Oh ,wait! You could kill somone using your hands! Let's just chop of our own hands!
You are remarkably stupid. How do you do it?
You dumbasses really don't get that was a joke?
##### Share on other sites
I don't even f*cking care anymore. Why don't we outlaw everything that could be dangerous?! Why don't we lock ourselves up in a room filled with pillows to keep us safe?! Here are my great ideas:
- No more shooting sports, people could possibly get shot!
- No more motorsports, people could get run over!
- No more baseball, you could kill someone with a baseball bat!
- No more darts, you could poke someone's eye out!
- No more hiking, you could get pushed by someone and die!
- No more sailing, you could get thrown of your boat!
- No more boxing, you could possibly hit someone to death!
- No more chess, you could use a chess piece to choke someone!
Why don't we also ban knives and forks? We'll just eat with out hands. Oh ,wait! You could kill somone using your hands! Let's just chop of our own hands!
You are remarkably stupid. How do you do it?
You dumbasses really don't get that was a joke?
A joke would have to be funny. If you meant this ironically, no one knows your political beliefs enough to understand you were being ironic, but as those are NRA talking points, I and the Yokel (probably... but I shouldn't speak for him)assume you are just vomiting up that back-asswards logic. Especially bearing in mind the quotes in your siggy.
Edited by AlienTwo
##### Share on other sites
I don't even f*cking care anymore. Why don't we outlaw everything that could be dangerous?! Why don't we lock ourselves up in a room filled with pillows to keep us safe?! Here are my great ideas:
- No more shooting sports, people could possibly get shot!
- No more motorsports, people could get run over!
- No more baseball, you could kill someone with a baseball bat!
- No more darts, you could poke someone's eye out!
- No more hiking, you could get pushed by someone and die!
- No more sailing, you could get thrown of your boat!
- No more boxing, you could possibly hit someone to death!
- No more chess, you could use a chess piece to choke someone!
Why don't we also ban knives and forks? We'll just eat with out hands. Oh ,wait! You could kill somone using your hands! Let's just chop of our own hands!
You are remarkably stupid. How do you do it?
You dumbasses really don't get that was a joke?
A joke would have to be funny. If you meant this ironically, no one knows your political beliefs enough to understand you were being ironic, but as those are NRA talking points, I and the Yokel (probably... but I shouldn't speak for him)assume you are just vomiting up that back-asswards logic. Especially bearing in mind the quotes in your siggy.
My political beliefs? I guess I'll tell you then. I'm 100% pro-gun, which is kinda rare in my country. (most people are very anti-gun here) Believe what you want, but I think any sane, law-abiding person should be able to enjoy firearms, shoot firearms and own firearms. It's not about violence and certainly not about killing. Guess anyone would be sour if your hobby/sport get's vomited on every day by the media and politicians. Especially after horrific events like the recent shooting. People talk about banning firearms, because they can be dangerous. So, I made a list of other ''dangerous'' things that we should ban. Get it? Bash me for it if you want, I stand by my beliefs.
##### Share on other sites
You dumbasses really don't get that was a joke?
"It was a joke...
... but I stand by my beliefs that either all these things should be banned, or guns should be unbanned."
k
##### Share on other sites
My political beliefs? I guess I'll tell you then. I'm 100% pro-gun, which is kinda rare in my country. (most people are very anti-gun here) Believe what you want, but I think any sane, law-abiding person should be able to enjoy firearms, shoot firearms and own firearms. It's not about violence and certainly not about killing. Guess anyone would be sour if your hobby/sport get's vomited on every day by the media and politicians. Especially after horrific events like the recent shooting. People talk about banning firearms, because they can be dangerous. So, I made a list of other ''dangerous'' things that we should ban. Get it? Bash me for it if you want, I stand by my beliefs.
So then... not to repeat what Yen said (but totally repeating what Yen said), that was less a joke and more an outburst of true beliefs.
That's fine, we don't have to agree, but at least be honest.
##### Share on other sites
Maybe Pokemon should be banned as well. It can turn people evil and make them want to kill other Pokemon players.
##### Share on other sites
Maybe Pokemon should be banned as well. It can turn people evil and make them want to kill other Pokemon players.
Maybe there is a real and measurable difference between things created to kill humans and a cartoon/game series.
Maybe not, but it's worth considering, isn't it?
##### Share on other sites
My political beliefs? I guess I'll tell you then. I'm 100% pro-gun, which is kinda rare in my country. (most people are very anti-gun here) Believe what you want, but I think any sane, law-abiding person should be able to enjoy firearms, shoot firearms and own firearms. It's not about violence and certainly not about killing. Guess anyone would be sour if your hobby/sport get's vomited on every day by the media and politicians. Especially after horrific events like the recent shooting. People talk about banning firearms, because they can be dangerous. So, I made a list of other ''dangerous'' things that we should ban. Get it? Bash me for it if you want, I stand by my beliefs.
So then... not to repeat what Yen said (but totally repeating what Yen said), that was less a joke and more an outburst of true beliefs.
That's fine, we don't have to agree, but at least be honest.
Did I say it was a joke? Oops...I said that out of pure anger, so sorry for that. Didn't know what I was saying. Wouldn't you be sick and tired of people trying to ban something you like 24/7?
Edited by RageFaceMax
##### Share on other sites
God damn USA, what's your obsession with guns?
There's any other country where people go in a killing spree so often?
Don't tell me you are able to kill a shopping mall or school full of people with a knife.
It's not guns that kill, it's people that kill
Ban Pokémon?
I think most people these days can't raise their kids propperly.
Many don't have the respect anymore and they think they can do what they want.
Edited by Claptrap NL
##### Share on other sites
guns dont make people crazy. they are just crazy. a tool is just a method to kill. people will still kill each other if they want.
See the bigger picture.
If guns had not been freely available to people for so long, there would not be an ingrained culture in the US around guns, around being able to own guns 'for protection'.
The main reason people in the US need guns for protection is because so many other people have guns. They wouldn't need protection if guns were controlled as tightly as possible. Of course guns can't be outright removed from US society, because it's been ingrained for so long.
Giving people 'the right' to own a gun gives people that shouldn't have access to guns a much easier route to them. Yes, crazy people are doing the killing, but they would be less able to (note, not less likely but less able) if they couldn't walk into a pawn shop and pick up a .38 or whatever for \$100. Or whatever the going rate is for a handgun. I have no idea, I live in a country that has a healthy respect and fear of guns, not a rabid desire to own one.
You give a crazy person access to guns, they will use them. You don't, they won't, or at least are FAR less likely to be able to get one.
Think man, think.
Less guns = less shootings. It's really that simple.
##### Share on other sites
God damn USA, what's your obsession with guns?
There's any other country where people go in a killing spree so often?
Don't tell me you are able to kill a shopping mall or school full of people with a knife.
It's not guns that kill, it's people that kill Guns don't kill, but sure they help.
##### Share on other sites
My political beliefs? I guess I'll tell you then. I'm 100% pro-gun, which is kinda rare in my country. (most people are very anti-gun here) Believe what you want, but I think any sane, law-abiding person should be able to enjoy firearms, shoot firearms and own firearms. It's not about violence and certainly not about killing. Guess anyone would be sour if your hobby/sport get's vomited on every day by the media and politicians. Especially after horrific events like the recent shooting. People talk about banning firearms, because they can be dangerous. So, I made a list of other ''dangerous'' things that we should ban. Get it? Bash me for it if you want, I stand by my beliefs.
So then... not to repeat what Yen said (but totally repeating what Yen said), that was less a joke and more an outburst of true beliefs.
That's fine, we don't have to agree, but at least be honest.
Did I say it was a joke? Oops...I said that out of pure anger, so sorry for that. Didn't know what I was saying. Wouldn't you be sick and tired of people trying to ban something you like 24/7?
I smoke pot and play video games, that is my life. Difference is, you can't kill people with either of those.
##### Share on other sites
You give a crazy person access to guns, they will use them. You don't, they won't, or at least are FAR less likely to be able to get one.
This, I gotta agree. Though it still won't help. If a person is determined enough, they'll buy it illegally.
##### Share on other sites
I smoke pot and play video games, that is my life. Difference is, you can't kill people with either of those.
But dude, both of those are gateways to murdering a school full of children!
Funny thing about Pot. They say it's a gateway drug, but I've been smoking it for 20 years and I still don't own a single f*cking gate.
But anyways. Same point as I last made - people are always going to be crazy, letting them have access to guns is crazier than they will ever be. You CAN kill people with a bong, or you can kill people with a controller, in several ways, but they're not designed for that purpose. As someone rightly said above, the entire purpose of a gun is to shoot people. The purpose of a bong is not to bludgeon someone over the head with it and the entire purpose of a controller is not to choke someone with the cable.
Guns are killing machines. They are the very definition of a killing machine. Enabling every single grown adult in the US to have one of those is the very definition of stupid.
Take several million paranoid angry people and give them a gun each. They now have several million more reasons to be angry and paranoid. That will lead to bad things happening.
Let me rephrase that.
That has led to bad things happening.
You give a crazy person access to guns, they will use them. You don't, they won't, or at least are FAR less likely to be able to get one.
This, I gotta agree. Though it still won't help. If a person is determined enough, they'll buy it illegally.
Well of course. If they really want one, they'll get one on the black market. But that's very different to allowing someone to walk into Walmart and pick up a f*cking machine gun, isn't it?
##### Share on other sites
I smoke pot and play video games, that is my life. Difference is, you can't kill people with either of those.
But dude, both of those are gateways to murdering a school full of children!
Funny thing about Pot. They say it's a gateway drug, but I've been smoking it for 20 years and I still don't own a single f*cking gate.
You are the exception that proves the rule. I've been smoking for a similar amount of time, but I own three gates.
Ruined my life.
##### Share on other sites
I smoke pot and play video games, that is my life. Difference is, you can't kill people with either of those.
But dude, both of those are gateways to murdering a school full of children!
Funny thing about Pot. They say it's a gateway drug, but I've been smoking it for 20 years and I still don't own a single f*cking gate.
You are the exception that proves the rule. I've been smoking for a similar amount of time, but I own three gates.
Ruined my life.
I shoot guns and play video games, that is my life. Just because I can kill someone with a gun, doesn't mean I'm goig to do that. You'd have to be 100% insane and a psychopath to do that.
##### Share on other sites
I shoot guns and play video games, that is my life. Just because I can kill someone with a gun, doesn't mean I'm goig to do that. You'd have to be 100% insane and a psychopath to do that.
But does it makes sense to have easy access to those guns? Does it make sense to enable people that are, or even could go crazy to have guns?
Bryce Willams - seemed like a perfectly normal, sane man. He went crazy and shot people.
Remove the access to guns from EVERYONE and you immediately decrease the risk of crazy people shooting their guns, or people going crazy and shooting their guns. Whether you like it or not, you have to be able to see that enabling everyone that wants a gun to have one is simply a stupid thing to do. Whether you personally are sane or not is completely irrelevant in that argument.
##### Share on other sites
I shoot guns and play video games, that is my life. Just because I can kill someone with a gun, doesn't mean I'm goig to do that. You'd have to be 100% insane and a psychopath to do that.
But does it makes sense to have easy access to those guns? Does it make sense to enable people that are, or even could go crazy to have guns?
Bryce Willams - seemed like a perfectly normal, sane man. He went crazy and shot people.
Remove the access to guns from EVERYONE and you immediately decrease the risk of crazy people shooting their guns, or people going crazy and shooting their guns. Whether you like it or not, you have to be able to see that enabling everyone that wants a gun to have one is simply a stupid thing to do. Whether you personally are sane or not is completely irrelevant in that argument.
What? That's a terrible idea... it just makes too much sense to work.
##### Share on other sites
I shoot guns and play video games, that is my life. Just because I can kill someone with a gun, doesn't mean I'm goig to do that. You'd have to be 100% insane and a psychopath to do that.
But does it makes sense to have easy access to those guns? Does it make sense to enable people that are, or even could go crazy to have guns?
Bryce Willams - seemed like a perfectly normal, sane man. He went crazy and shot people.
Remove the access to guns from EVERYONE and you immediately decrease the risk of crazy people shooting their guns, or people going crazy and shooting their guns. Whether you like it or not, you have to be able to see that enabling everyone that wants a gun to have one is simply a stupid thing to do. Whether you personally are sane or not is completely irrelevant in that argument.
We don't have easy access to firearms. We have to go through a sh*t-ton of background checks just to be able to use the guns at a shooting range. If you want to own a gun, you have to go through even more background checks and paperwork. The police also pay you a visit every month to inspect your gun safe. You also need to have a surveilance camera in your house, so the police can check who has access to your guns. That's how it's like here. I don't know how it could be even more strict. Police escorts if you want to go to the range, just to make sure you're not doing anything wrong?
##### Share on other sites
I shoot guns and play video games, that is my life. Just because I can kill someone with a gun, doesn't mean I'm goig to do that. You'd have to be 100% insane and a psychopath to do that.
But does it makes sense to have easy access to those guns? Does it make sense to enable people that are, or even could go crazy to have guns?
Bryce Willams - seemed like a perfectly normal, sane man. He went crazy and shot people.
Remove the access to guns from EVERYONE and you immediately decrease the risk of crazy people shooting their guns, or people going crazy and shooting their guns. Whether you like it or not, you have to be able to see that enabling everyone that wants a gun to have one is simply a stupid thing to do. Whether you personally are sane or not is completely irrelevant in that argument.
We don't have easy access to firearms. We have to go through a sh*t-ton of background checks just to be able to use the guns at a shooting range. If you want to own a gun, you have to go through even more background checks and paperwork. The police also pay you a visit every month to inspect your gun safe. You also need to have a surveilance camera in your house, so the police can check who has access to your guns. That's how it's like here. I don't know how it could be even more strict. Police escorts if you want to go to the range, just to make sure you're not doing anything wrong?
Where are you from? Because that's not how it is in most the States in America.
##### Share on other sites
I shoot guns and play video games, that is my life. Just because I can kill someone with a gun, doesn't mean I'm goig to do that. You'd have to be 100% insane and a psychopath to do that.
But does it makes sense to have easy access to those guns? Does it make sense to enable people that are, or even could go crazy to have guns?
Bryce Willams - seemed like a perfectly normal, sane man. He went crazy and shot people.
Remove the access to guns from EVERYONE and you immediately decrease the risk of crazy people shooting their guns, or people going crazy and shooting their guns. Whether you like it or not, you have to be able to see that enabling everyone that wants a gun to have one is simply a stupid thing to do. Whether you personally are sane or not is completely irrelevant in that argument.
We don't have easy access to firearms. We have to go through a sh*t-ton of background checks just to be able to use the guns at a shooting range. If you want to own a gun, you have to go through even more background checks and paperwork. The police also pay you a visit every month to inspect your gun safe. You also need to have a surveilance camera in your house, so the police can check who has access to your guns. That's how it's like here. I don't know how it could be even more strict. Police escorts if you want to go to the range, just to make sure you're not doing anything wrong?
Where are you from? Because that's not how it is in most the States in America.
Haven't you noticed my flag? I'm Dutch. I know that the U.S. are very open to guns, except for a couple of states. California, New York, etc. But you guys have a Second Amendment, we don't. You guys have the NRA to protect gun ownership, we don't. We are basically on our own. Well, we do have the KNSA (Koninklijke Nederlandse Schietsport Associatie, translated: Royal Dutch Shooting sports Association) but they really aren't that effective. Especially because they don't have enough members.
##### Share on other sites
R.I.P to those killed, I feel so bad for the families.
This guy would have stabbed them to death if he didn't own a gun. Not starting a gun debate, just saying this guy is a f*cking psychotic asshole who deserves a nice lethal injection.
You can only do so much with a knife... I mean you think you'll just stand there when a guy is walking up to you with a blade... especially if he attacked your friend/co-worker. You can still throw it at someone but it has far less the accuracy of a gun, look at the Native Americans, they had these 'knives' and bows but a gun was far more superior.
Edited by Sayuri
##### Share on other sites
Haven't you noticed my flag? I'm Dutch. I know that the U.S. are very open to guns, except for a couple of states. California, New York, etc. But you guys have a Second Amendment, we don't. You guys have the NRA to protect gun ownership, we don't. We are basically on our own. Well, we do have the KNSA (Koninklijke Nederlandse Schietsport Associatie, translated: Royal Dutch Shooting sports Association) but they really aren't that effective. Especially because they don't have enough members.
I noticed your flag, but I don't always believe them... regardless, you are arguing against American gun control so it's a bit confusing. Let me educate you on American issues; It's super easy to get a gun here, we require only background checks for handguns and none of the security measures you describe, so our situation is waaaaaaay different than the Dutch one. I would LOVE it if we had the protections you speak of, but our politicians don't have the backbone to even require full background checks after the slayings at Sandy Hook (if you don't know, that's when a gunman killed a bunch of god damned 1st graders).
##### Share on other sites
Haven't you noticed my flag? I'm Dutch. I know that the U.S. are very open to guns, except for a couple of states. California, New York, etc. But you guys have a Second Amendment, we don't. You guys have the NRA to protect gun ownership, we don't. We are basically on our own. Well, we do have the KNSA (Koninklijke Nederlandse Schietsport Associatie, translated: Royal Dutch Shooting sports Association) but they really aren't that effective. Especially because they don't have enough members.
I noticed your flag, but I don't always believe them... regardless, you are arguing against American gun control so it's a bit confusing. Let me educate you on American issues; It's super easy to get a gun here, we require only background checks for handguns and none of the security measures you describe, so our situation is waaaaaaay different than the Dutch one. I would LOVE it if we had the protections you speak of, but our politicians don't have the backbone to even require full background checks after the slayings at Sandy Hook (if you don't know, that's when a gunman killed a bunch of god damned 1st graders).
Well, we are one big community. We support each other. That's why I'm ranting about American gun control. Let me describe you the Dutch laws in a nutshell:
- You can buy your first firearm when you're 18 years old. But you're only allowed to buy one if you have been a KNSA member for over a year. You can also only buy .22LR firearms your first time.
- You can't buy all the guns you want. Your gun license has several ''stages''. You start at .22LR. Want something bigger? Then you have to pay more money! Want something even bigger? Make the cash flow! Also, your license isn't valid forever. So you have to re-buy your license a couple of times.
- When you own a gun, you have to buy a safe that can be bolted to the wall and floor. You also need a surveilance camera, so the cops know you're not doing anything stupid. Basically, the area around your gun safe get's turned into a damn Big Brother show.
- The police can inspect your home and safe whenever they feel like it. If they find something they don't like, you lose all your guns. And trust me, they find things they don't like VERY fast.
- Your ammo and guns have to be stored in different rooms. Your ammo also has to be stored in a safe, which has to be bolted to the floor or wall.
- You can't own any handgun that is chambered for a caliber higher than .44. Why? No one knows! I even asked the cops when they were inspecting a friend's safe. They didn't even know.
- You can't own more than 5 firearms in total. If you're a collector, you are basically screwed.
Just some rules. I could write an entire book about this subject. I'd love to get rid of some of these rules, but that won't happen.
##### Share on other sites
Here's my take on gun control. Although I don't own a firearm at this moment, I've always been a gun enthusiast with a fascination for this tool. But when it comes to gun control... should the laws in U.S. be more strict? Absolutely.
There used to be a time I would argue that "they will find a different way to kill" but the fact is, an increase of gun control will decrease the amount of senseless crimes like this occurring. Yes, it wont stop people from killing but it will cause a huge decrease. And here's the thing, psychopaths might contemplate committing murder other ways if a gun is not an option, but the likeliness of them going through with it drops considerably. A gun gives them power over people and also enables them to commit mass-murders that effects X amount of people simultaneously whereas with a knife or other melee weapons, they can't possibly cause that much damage. Many of the murderers are also cowards who would never think of intimidating anyone without pissing themselves in fear. With a gun, they suddenly gain the courage to do that and can easily have the advantage over their victims with this powerful tool. Without the gun, the number of psychopaths going through with what they are thinking will decrease.
##### Share on other sites
Maybe Pokemon should be banned as well. It can turn people evil and make them want to kill other Pokemon players.
You are a f*cking idiot, simple as that.
##### Share on other sites
don't sweat him, he's a troll.
##### Share on other sites
I would like to thank the people who gave their opinion on this. To repeat myself, watching the video has hit me pretty hard, and I said before that I don't want to start another gun debate.
With that said, I requested this topic to be locked, but I appreciate every kind word said. While I wasn't directly involved in the incident, it quickly hit me how f*cked up the world can sometimes be.
##### Share on other sites
As a gun owner and as somebody who comes from a cop family firing guns since I was 12, I support guns and gun rights. I also support background checks for violent felonies, references, police interview, waiting period, and psychological background. That said, I'm not entirely convinced taking everyone's gun is going to reduce gun crime. Criminals who act with firearms don't follow the law... In Detroit right now, wherein the city has reduced gun restrictions and made it permissible for law abiding citizens to carry, violent crime has gone down, and the police chief is attributing it in part to citizens protecting themselves.
Edited by Irviding
## Create an account
Register a new account | 6,569 | 28,694 | {"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-2018-51 | latest | en | 0.963334 |
https://testbook.com/question-answer/rs-1075-is-divided-among-aman-anu-and-aaradhya-i--624bef432ef0a4c14ccd1854 | 1,675,936,229,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764501555.34/warc/CC-MAIN-20230209081052-20230209111052-00429.warc.gz | 571,688,487 | 77,955 | # Rs. 1075 is divided among Aman, Anu and Aaradhya in such a way that, Aman gets 25% more than Aaradhya and Aaradhya gets 25% less than Anu. What is Aman's share (in Rs.) in the amount?
This question was previously asked in
DSSSB JE CE 2019 Official Paper Shift 3 (Held on 20 Nov 2019)
View all DSSSB JE Papers >
1. 375
2. 425
3. 400
4. 450
Option 1 : 375
Free
CT 1: Indian History
37.6 K Users
10 Questions 10 Marks 6 Mins
## Detailed Solution
Given:
Rs. 1075 is divided among Aman, Anu and Aaradhya in such a way that, Aman gets 25% more than Aaradhya and Aaradhya gets 25% less than Anu.
Concept used:
Ratio and Proportion
Incremented/Reduced value = Initial value (1 ± change%)
Calculation:
Let Anu gets Rs. 100Q.
Hence,
Aaradhya gets = 100Q × 0.75 = Rs. 75Q
Aman gets = 75Q × 1.25 = Rs. 93.75Q
According to the question,
100Q + 75Q + 93.75Q = 1075
⇒ 268.75Q = 1075
⇒ Q = 4
⇒ 93.75Q = 375
∴ Aman's share is Rs. 375. | 359 | 938 | {"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-2023-06 | latest | en | 0.903517 |
https://www.futurestarr.com/blog/mathematics/sqrt-54-or | 1,656,814,413,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104209449.64/warc/CC-MAIN-20220703013155-20220703043155-00132.warc.gz | 817,713,087 | 21,308 | FutureStarr
Sqrt 54 OR
## Sqrt 54
A square root of a number is a number that when multiplied by one of the factors of the number, when squared, equals the number itself. The same is true with a square root of a positive or negative number. For example, a square root of 2 would be the number of times that 2 squared equals 4.
### Root
= 54. What could be n? Squares and square roots are special exponents. When the exponent on the number is 2, it is termed as square and when the exponent is ½ it is called a square root of a number. Let's see how to find the square root of 54 and also discover interesting facts around them. In this mini lesson, let us learn about the square root of 54, find out whether the square root of 54 is rational or irrational, and see how to find the square root of 54 by long division method.
A number that cannot be expressed as a ratio of two integers is an irrational number. The decimal form of the irrational number will be non-terminating (i.e it never ends) and non-recurring (i.e the decimal part of the number never repeats a pattern). Now let us look at the square root of 54. √54 = 7.348. The decimal part is never-ending, and we cannot see a pattern in the decimal part. So √54 is an irrational number. (Source: www.cuemath.com)
### Simplify
When numbers are not perfect squares, their square roots can be challenging, however, it is possible to simplify square roots to make the square roots easier to see and use. Dive into ways the rules of mathematics are used to define perfect and imperfect squares and learn how to evaluate the square root of an imperfect square.
Here we will show you two methods that you can use to simplify the square root of 54. In other words, we will show you how to find the square root of 54 in its simplest radical form using two different methods. (Source: squareroot.info)
## Related Articles
• #### Using a Love Date Calculator to Determine Compatibility
July 03, 2022 | sheraz naseer
• #### 7.7 As a Fraction,
July 03, 2022 | Jamshaid Aslam
• #### A 24 27 As a Percentage
July 03, 2022 | Shaveez Haider
• #### 7 of 50 Is What Percent
July 03, 2022 | sheraz naseer
• #### A Calculate a B
July 03, 2022 | Muhammad Waseem
• #### 2 Out of 9 Percentage,
July 03, 2022 | Jamshaid Aslam
• #### A 2 Percent of 18
July 03, 2022 | Shaveez Haider
• #### 1 15 Percentage'
July 03, 2022 | Jamshaid Aslam
• #### A 60 Out of 80 As a Percentage
July 03, 2022 | sheraz naseer
• #### How to Calculate 20 Percent of a Number
July 03, 2022 | sheraz naseer
• #### 14 17 As a Percentage
July 03, 2022 | Muhammad Waseem
• #### how to figure 40 percent of a number
July 03, 2022 | Muhammad Umair
• #### A General Loan Calculator
July 03, 2022 | Muhammad Waseem
• #### Calc:
July 03, 2022 | Abid Ali
• #### 5 Percent of 17
July 03, 2022 | Muhammad Waseem | 815 | 2,954 | {"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} | 4.875 | 5 | CC-MAIN-2022-27 | latest | en | 0.896174 |
https://www.math-only-math.com/sets.html | 1,721,699,047,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517931.85/warc/CC-MAIN-20240723011453-20240723041453-00619.warc.gz | 735,983,472 | 13,136 | # Sets
An introduction of sets and its definition in mathematics. The concept of sets is used for the foundation of various topics in mathematics.
To learn sets we often talk about the collection of objects, such as a set of vowels, set of negative numbers, a group of friends, a list of fruits, a bunch of keys, etc.
What is set (in mathematics)?
The collection of well-defined distinct objects is known as a set. The word well-defined refers to a specific property which makes it easy to identify whether the given object belongs to the set or not. The word ‘distinct’ means that the objects of a set must be all different.
For example:
1. The collection of children in class VII whose weight exceeds 35 kg represents a set.
2. The collection of all the intelligent children in class VII does not represent a set because the word intelligent is vague. What may appear intelligent to one person may not appear the same to another person.
Elements of Set:
The different objects that form a set are called the elements of a set. The elements of the set are written in any order and are not repeated. Elements are denoted by small letters.
Notation of a Set:
A set is usually denoted by capital letters and elements are denoted by small letters
If x is an element of set A, then we say x ϵ A. [x belongs to A]
If x is not an element of set A, then we say x ∉ A. [x does not belong to A]
For example:
The collection of vowels in the English alphabet.
Solution :
Let us denote the set by V, then the elements of the set are a, e, i, o, u or we can say, V = [a, e, i, o, u].
We say a ∈ V, e ∈ V, i ∈ V, o ∈ V and u ∈ V.
Also, we can say b ∉ V, c ∉ v, d ∉ v, etc.
Set Theory
Sets
Objects Form a Set
Elements of a Set
Properties of Sets
Representation of a Set
Different Notations in Sets
Standard Sets of Numbers
Types of Sets
Pairs of Sets
Subset
Subsets of a Given Set
Operations on Sets
Union of Sets
Intersection of Sets
Difference of two Sets
Complement of a Set
Cardinal number of a set
Cardinal Properties of Sets
Venn Diagrams
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
## Recent Articles
1. ### Expanded form of Decimal Fractions |How to Write a Decimal in Expanded
Jul 22, 24 03:27 PM
Decimal numbers can be expressed in expanded form using the place-value chart. In expanded form of decimal fractions we will learn how to read and write the decimal numbers. Note: When a decimal is mi…
2. ### Worksheet on Decimal Numbers | Decimals Number Concepts | Answers
Jul 22, 24 02:41 PM
Practice different types of math questions given in the worksheet on decimal numbers, these math problems will help the students to review decimals number concepts.
3. ### Decimal Place Value Chart |Tenths Place |Hundredths Place |Thousandths
Jul 21, 24 02:14 PM
Decimal place value chart are discussed here: The first place after the decimal is got by dividing the number by 10; it is called the tenths place.
4. ### Thousandths Place in Decimals | Decimal Place Value | Decimal Numbers
Jul 20, 24 03:45 PM
When we write a decimal number with three places, we are representing the thousandths place. Each part in the given figure represents one-thousandth of the whole. It is written as 1/1000. In the decim… | 804 | 3,343 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2024-30 | latest | en | 0.918303 |
https://www.shaalaa.com/question-bank-solutions/p-particle-mass-m-observed-inertial-frame-reference-found-move-circle-radius-r-uniform-speed-v-dynamics-uniform-circular-motion-centripetal-force_66568 | 1,585,740,271,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370505730.14/warc/CC-MAIN-20200401100029-20200401130029-00372.warc.gz | 1,189,534,379 | 10,947 | Share
# P a Particle of Mass M is Observed from an Inertial Frame of Reference and is Found to Move in a Circle of Radius R with a Uniform Speed V. - Physics
ConceptDynamics of Uniform Circular Motion - Centripetal Force
#### Question
A particle of mass m is observed from an inertial frame of reference and is found to move in a circle of radius r with a uniform speed v. The centrifugal force on it is
• $\frac{\text{mv}^2}{\text{r}}$ towards the centre
• $\frac{\text{mv}^2}{\text{r}}$ away from the centre
• $\frac{\text{mv}^2}{\text{r}}$ along the tangent through the particle
• zero
#### Solution
zero
The centrifugal force is a pseudo force and can only be observed from the frame of reference, which is non-inertial w.r.t. the particle.
Is there an error in this question or solution?
#### APPEARS IN
Solution P a Particle of Mass M is Observed from an Inertial Frame of Reference and is Found to Move in a Circle of Radius R with a Uniform Speed V. Concept: Dynamics of Uniform Circular Motion - Centripetal Force.
S | 273 | 1,040 | {"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} | 3.046875 | 3 | CC-MAIN-2020-16 | latest | en | 0.790561 |
https://www.enotes.com/homework-help/y-x-x-2-1-find-derivative-function-516120 | 1,511,128,152,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805809.59/warc/CC-MAIN-20171119210640-20171119230640-00185.warc.gz | 790,630,643 | 9,042 | # `y = x(x^2+1)` Find the derivative of the function.
mathace | Certified Educator
Given: ` y=x(x^2+1)`
`y=x^3+x`
`y'=3x^2+1`
`` | 57 | 133 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2017-47 | latest | en | 0.42671 |
https://www.bartleby.com/essay/Ameritrade-Cost-of-Capital-PKYH743TC | 1,621,328,341,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989756.81/warc/CC-MAIN-20210518063944-20210518093944-00125.warc.gz | 671,211,106 | 10,587 | 1042 Words5 Pages
Ameritrade is formed in 1971, and is a pioneer in the deep-discount brokerage sector.
In march 1997, Ameritrade raised \$22.5 million in an initial public offering. Management at Ameritrade is considering substantial investments in technology and advertising, but is unsure of the appropriate cost of capital.
Estimating the cost of capital 1. Since we do not have the beta for Ameritrade, we need to find comparable firms for which we could compute the betas. There are several candidates in the case. Discuss which firms are most appropriate.
Thus, the proportion of the revenue a firm earns from transactions and interest (brokerage activities) has something to do with the risk. Thus, to find the firms of comparable risks, we may take
CAPM = rf + B x (rm –rf)
4. What is the estimate for the risk free rate that should be employed in calculating the cost of capital for Ameritrade?
Since the project involves substantial investments in technology and advertising and the cash flows are projected in the future we can assume that it’s a long term investment so we should use long term rates and we should use current rates, not historical rates.
I’ll use a 10 year time horizon so the current 10 year interest rate is: 6.34%
5. What is the estimate of the market risk premium that should be employed in calculating the cost of capital for Ameritrade?
The market risk premium for Ameritrade we should use the difference between the returns on small stocks (Ameritrade has a market capitalisation as of August 29, 1997 of \$273,127,000 and the returns on long term (20 years) government bonds:
We use the more recent historical data as the first one includes war periods and is less modern and therefore less consistent with new technology.
Risk premium (1950-1996) = 17.8% – 6.0% = 11.8%
6. What do we use for Beta?
To compute betas we need the periodic returns of comparable companies and the returns of the value weighted market index as a whole. We regress both returns and get the slope of the line which is our Beta. We should use estimates for the last 5 years.
[pic]
[pic]
[pic]
7. What comparable firms can | 489 | 2,144 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2021-21 | longest | en | 0.930686 |
http://www.voidcn.com/article/p-gzzcosop-brn.html | 1,544,800,059,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825916.52/warc/CC-MAIN-20181214140721-20181214162221-00092.warc.gz | 508,875,975 | 7,280 | # LUA 排序算法和性能分析[4]:快速排序算法
```math.randomseed(os.time());
local real_print = print;
local nLine = 0;
function print(...)
real_print(nLine, ...)
nLine = nLine + 1;
end
local t = {}
local function dump()
-- print(table.concat(t, ","));
end;
for i = 1, 256 do
table.insert(t, math.random(i), i);
end
function table_sort()
table.sort(t);
dump()
end
-- table_sort() -- 0.3
-- 冒泡排序
function bubble_sort()
print("冒泡排序")
local n = #t;
for j = 1, n - 1 do
for i = 1, n - j do
if t[i] > t[i+1] then
t[i], t[i+1] = t[i+1], t[i];
dump()
end;
end;
end;
dump()
end;
-- bubble_sort(); -- 5.7
-- 选择排序
function select_sort()
print("选择排序")
local n = #t;
for i = 0, n - 2 do
local min = i;
for j = i + 1, n - 1 do
if t[min+1] > t[j+1] then
min = j;
end;
end;
if (min ~= i) then
t[min+1], t[i+1] = t[i+1], t[min+1];
dump()
end;
end;
dump()
end;
-- select_sort() -- 3.4
-- 快速排序
function quick_sort()
print("快速排序")
local function sort(left, right)
if left >= right then
return ;
end;
local i = left;
local j = right;
local key = t[left];
while (i < j) do
while i < j and key <= t[j] do
j = j - 1;
end;
t[i] = t[j];
dump()
while i < j and key >= t[i] do
i = i + 1;
end;
t[j] = t[i];
dump()
end;
t[i] = key;
dump()
sort(left, i - 1);
sort(i + 1, right);
end;
sort(1, #t)
end;
-- quick_sort() -- 0.4
``` | 506 | 1,315 | {"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} | 3.71875 | 4 | CC-MAIN-2018-51 | longest | en | 0.149218 |
https://www.jiskha.com/display.cgi?id=1202168001 | 1,503,456,318,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886117519.82/warc/CC-MAIN-20170823020201-20170823040201-00696.warc.gz | 943,805,466 | 3,700 | # biology
posted by .
If you are getting 270X magnification with a 45X objective, what would be the power of the eyepiece?
## Similar Questions
1. ### physics, lenses
I need to calculate the magnification, m of an objective lens, so I can then x it by the eyepiece value for M to get the total magnification. I need v/u (v is dist from lens to image; u is dist of object to lens) I'm using 1/f = 1/v+1/u …
2. ### lenses, still stuck!
I need to calculate the magnification, m of an objective lens, so I can then x it by the eyepiece value for M to get the total magnification. I need v/u (v is dist from lens to image; u is dist of object to lens) I'm using 1/f = 1/v+1/u …
3. ### objective and eyepiece lens
What is the approximate magnification of a compound microscope with objective and eyepiece focal lengths of 0.3 cm and 3.6 cm, respectively, and a separation between lenses of 20 cm?
4. ### Physics
A telescope has a magnification of 20 and consists of two converging lenses, the objective and the eyepiece, fixed at either end of a tube 60.0 cm long. Assuming that this telescope would allow an observer to view a lunar crater in …
5. ### Science
How do you calculate the total magnification of a microscope, using the eyepiece magnification and the objective magnification?
6. ### Physics
A standard biological microscope is required to have amagnification of 200. 1)When paired with a 10 eyepiece, what power objective isneeded to get this magnification?
7. ### physics
A telescope has an objective with a refractive power of 1.20 diopters and an eyepiece with a refractive power of 220 diopters. What is the angular magnification of the telescope?
8. ### biology
If a microscope has a 40× objective lens in place and a 15× eyepiece lens, the magnification of the system under these conditions is
9. ### physics
A telescope that has a 100 cm focal length objective and a 3.00 cm focal length eyepiece. Find the distance between the objective and eyepiece lenses in the telescope needed to produce a final image very far from the observer, where …
10. ### Math/Biology
8c) The scanning objective lens (4x) of a microscope has a diameter of field of 6.0 mm and a total magnification of 40x. The high power objective lens has a total magnification of 400x. Estimate the diameter of field of view of the …
More Similar Questions | 594 | 2,349 | {"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-2017-34 | latest | en | 0.89642 |
https://www.jiskha.com/display.cgi?id=1359068531 | 1,503,034,385,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104565.76/warc/CC-MAIN-20170818043915-20170818063915-00094.warc.gz | 931,329,106 | 3,958 | # algebra
posted by .
Chase scored 14 points on Monday, and he doubled his score each day thereafter. How many points did he score on Thursday?
A. 224 points
B. 112 points
C. 56 points
D. 42 points
Thanks :)
• algebra -
14
28
56
112
224
• algebra -
Thanks ;)
• algebra -
You're welcome.
• algebra -
Chase scored 11 points on Monday, and he doubled his score each day thereafter. How many points did he score on Friday?
## Similar Questions
1. ### Algebra
Need help with these-- 6. Chase scored 14 points on Monday, and he doubled his score each day thereafter. How many points did he score on Thursday?
2. ### math
At the end of an basketball game daniels couch told him he scored 20% of all of his teams points. If Daniel scored 19 points exactly how many points did his team score?
3. ### math
kris scored 630 points on a particular section of a CPA exam for which the score were normally distributed with a mean of 600 points and a standard deviation of 75 points. At a differenttime, kerry scored 540 points on the same section, …
4. ### math
kris scored 630 points on a particular section of a CPA exam for which the score were normally distributed with a mean of 600 points and a standard deviation of 75 points. At a differenttime, kerry scored 540 points on the same section, …
5. ### finite math
kris scored 630 points on a particular section of a CPA exam for which the score were normally distributed with a mean of 600 points and a standard deviation of 75 points. At a differenttime, kerry scored 540 points on the same section, …
6. ### statistics
On a national standardized test, with normally distributed results, the middle 68% of students scored between 40 points and 70 points, and the mean score was 55 points. Based on this information, which of the following represents a …
7. ### math
Four students at a computer fair competed against each other in a space-travel game. Chris scored 2/3 as many points as the winner Pat. Rich scored 1/2 as many points as Chris. Tome scored 100 points, which was 90 points less than … | 497 | 2,060 | {"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-2017-34 | latest | en | 0.964156 |
https://math.answers.com/Q/Is_72_meters_equal_to_less_than_or_greater_than_95_centimeters | 1,685,330,451,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644574.15/warc/CC-MAIN-20230529010218-20230529040218-00459.warc.gz | 435,062,648 | 48,569 | 0
# Is 72 meters equal to less than or greater than 95 centimeters?
Dnolt8
Lvl 1
2011-04-21 21:45:18
Best Answer
72 m > (greater than) 95 cm
Wiki User
2011-04-21 21:45:18
This answer is:
Study guides
20 cards
➡️
See all cards
3.8
2509 Reviews
## Add your answer:
Earn +20 pts
Q: Is 72 meters equal to less than or greater than 95 centimeters?
Write your answer...
Submit
Still have questions?
People also asked
Featured Questions | 143 | 440 | {"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-2023-23 | latest | en | 0.897633 |
http://isabelle.in.tum.de/repos/isabelle/file/6163c90694ef/src/HOL/Library/List_lexord.thy | 1,568,566,583,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514571651.9/warc/CC-MAIN-20190915155225-20190915181225-00212.warc.gz | 98,700,344 | 3,723 | src/HOL/Library/List_lexord.thy
author wenzelm Tue May 15 13:57:39 2018 +0200 (16 months ago) changeset 68189 6163c90694ef parent 61076 bdc1e2f0a86a permissions -rw-r--r--
``` 1 (* Title: HOL/Library/List_lexord.thy
```
``` 2 Author: Norbert Voelker
```
``` 3 *)
```
``` 4
```
``` 5 section \<open>Lexicographic order on lists\<close>
```
``` 6
```
``` 7 theory List_lexord
```
``` 8 imports Main
```
``` 9 begin
```
``` 10
```
``` 11 instantiation list :: (ord) ord
```
``` 12 begin
```
``` 13
```
``` 14 definition
```
``` 15 list_less_def: "xs < ys \<longleftrightarrow> (xs, ys) \<in> lexord {(u, v). u < v}"
```
``` 16
```
``` 17 definition
```
``` 18 list_le_def: "(xs :: _ list) \<le> ys \<longleftrightarrow> xs < ys \<or> xs = ys"
```
``` 19
```
``` 20 instance ..
```
``` 21
```
``` 22 end
```
``` 23
```
``` 24 instance list :: (order) order
```
``` 25 proof
```
``` 26 fix xs :: "'a list"
```
``` 27 show "xs \<le> xs" by (simp add: list_le_def)
```
``` 28 next
```
``` 29 fix xs ys zs :: "'a list"
```
``` 30 assume "xs \<le> ys" and "ys \<le> zs"
```
``` 31 then show "xs \<le> zs"
```
``` 32 apply (auto simp add: list_le_def list_less_def)
```
``` 33 apply (rule lexord_trans)
```
``` 34 apply (auto intro: transI)
```
``` 35 done
```
``` 36 next
```
``` 37 fix xs ys :: "'a list"
```
``` 38 assume "xs \<le> ys" and "ys \<le> xs"
```
``` 39 then show "xs = ys"
```
``` 40 apply (auto simp add: list_le_def list_less_def)
```
``` 41 apply (rule lexord_irreflexive [THEN notE])
```
``` 42 defer
```
``` 43 apply (rule lexord_trans)
```
``` 44 apply (auto intro: transI)
```
``` 45 done
```
``` 46 next
```
``` 47 fix xs ys :: "'a list"
```
``` 48 show "xs < ys \<longleftrightarrow> xs \<le> ys \<and> \<not> ys \<le> xs"
```
``` 49 apply (auto simp add: list_less_def list_le_def)
```
``` 50 defer
```
``` 51 apply (rule lexord_irreflexive [THEN notE])
```
``` 52 apply auto
```
``` 53 apply (rule lexord_irreflexive [THEN notE])
```
``` 54 defer
```
``` 55 apply (rule lexord_trans)
```
``` 56 apply (auto intro: transI)
```
``` 57 done
```
``` 58 qed
```
``` 59
```
``` 60 instance list :: (linorder) linorder
```
``` 61 proof
```
``` 62 fix xs ys :: "'a list"
```
``` 63 have "(xs, ys) \<in> lexord {(u, v). u < v} \<or> xs = ys \<or> (ys, xs) \<in> lexord {(u, v). u < v}"
```
``` 64 by (rule lexord_linear) auto
```
``` 65 then show "xs \<le> ys \<or> ys \<le> xs"
```
``` 66 by (auto simp add: list_le_def list_less_def)
```
``` 67 qed
```
``` 68
```
``` 69 instantiation list :: (linorder) distrib_lattice
```
``` 70 begin
```
``` 71
```
``` 72 definition "(inf :: 'a list \<Rightarrow> _) = min"
```
``` 73
```
``` 74 definition "(sup :: 'a list \<Rightarrow> _) = max"
```
``` 75
```
``` 76 instance
```
``` 77 by standard (auto simp add: inf_list_def sup_list_def max_min_distrib2)
```
``` 78
```
``` 79 end
```
``` 80
```
``` 81 lemma not_less_Nil [simp]: "\<not> x < []"
```
``` 82 by (simp add: list_less_def)
```
``` 83
```
``` 84 lemma Nil_less_Cons [simp]: "[] < a # x"
```
``` 85 by (simp add: list_less_def)
```
``` 86
```
``` 87 lemma Cons_less_Cons [simp]: "a # x < b # y \<longleftrightarrow> a < b \<or> a = b \<and> x < y"
```
``` 88 by (simp add: list_less_def)
```
``` 89
```
``` 90 lemma le_Nil [simp]: "x \<le> [] \<longleftrightarrow> x = []"
```
``` 91 unfolding list_le_def by (cases x) auto
```
``` 92
```
``` 93 lemma Nil_le_Cons [simp]: "[] \<le> x"
```
``` 94 unfolding list_le_def by (cases x) auto
```
``` 95
```
``` 96 lemma Cons_le_Cons [simp]: "a # x \<le> b # y \<longleftrightarrow> a < b \<or> a = b \<and> x \<le> y"
```
``` 97 unfolding list_le_def by auto
```
``` 98
```
``` 99 instantiation list :: (order) order_bot
```
``` 100 begin
```
``` 101
```
``` 102 definition "bot = []"
```
``` 103
```
``` 104 instance
```
``` 105 by standard (simp add: bot_list_def)
```
``` 106
```
``` 107 end
```
``` 108
```
``` 109 lemma less_list_code [code]:
```
``` 110 "xs < ([]::'a::{equal, order} list) \<longleftrightarrow> False"
```
``` 111 "[] < (x::'a::{equal, order}) # xs \<longleftrightarrow> True"
```
``` 112 "(x::'a::{equal, order}) # xs < y # ys \<longleftrightarrow> x < y \<or> x = y \<and> xs < ys"
```
``` 113 by simp_all
```
``` 114
```
``` 115 lemma less_eq_list_code [code]:
```
``` 116 "x # xs \<le> ([]::'a::{equal, order} list) \<longleftrightarrow> False"
```
``` 117 "[] \<le> (xs::'a::{equal, order} list) \<longleftrightarrow> True"
```
``` 118 "(x::'a::{equal, order}) # xs \<le> y # ys \<longleftrightarrow> x < y \<or> x = y \<and> xs \<le> ys"
```
``` 119 by simp_all
```
``` 120
```
``` 121 end
``` | 1,858 | 5,091 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2019-39 | latest | en | 0.623345 |
http://primes.utm.edu/curios/cpage/32667.html | 1,534,887,131,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221219109.94/warc/CC-MAIN-20180821210655-20180821230655-00510.warc.gz | 345,864,928 | 2,449 | 19 (another Prime Pages' Curiosity)
Curios: Curios Search: Participate: Single Curio View: (Seek other curios for this number) Consider i + j = 19 for 0 < i, j < 19 and i * j + 19 is prime for ALL cases: 1*18+19=37, 2*17+19=53, 3*16+19=67, 4*15+19=79, 5*14+19=89, 6*13+19=97, 7*12+19=103, 8*11+19=107, 9*10+19=109. This also works for 2 and 7. Can it work for a prime greater than 19? [Bergot] Submitted: 2018-05-28 13:03:25; Last Modified: 2018-05-28 13:07:54. Prime Curios! © 2000-2018 (all rights reserved) privacy statement | 229 | 540 | {"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-2018-34 | latest | en | 0.64445 |
https://math.answers.com/other-math/What_is_10000000000_divided_by_987654321 | 1,716,749,678,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058972.57/warc/CC-MAIN-20240526170211-20240526200211-00046.warc.gz | 335,542,973 | 46,954 | 0
# What is 10000000000 divided by 987654321?
Updated: 4/28/2022
Wiki User
13y ago
10.125
Juliet McCullough
Lvl 10
3y ago
Wiki User
13y ago
10000000000 / 987654321 = 10.125
Earn +20 pts
Q: What is 10000000000 divided by 987654321?
Submit
Still have questions?
Related questions
1000000000
10000000000
0
10000000000
### What is 987654321 multiplied by 987654321?
987654321 x 987654321 = 975461057789971041
250000
833333333.3333
10000000
10000000000
### What is 1234567890 - 987654321?
1234567890 - 987654321 = 246913569
### What is 12356789+987654321=?
12356789+987654321= = 1000011110
### What is 123456789 plus 987654321?
123456789 +987654321 =1111111110 | 255 | 682 | {"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-2024-22 | latest | en | 0.629369 |
http://www.ramindra.com/the-war-against-math-rules/ | 1,726,004,095,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651318.34/warc/CC-MAIN-20240910192923-20240910222923-00560.warc.gz | 45,943,881 | 11,147 | Categories
# The War Against Math Rules
## Math Rules
Specifically, though other philosophies of mathematics permit objects that could be proved to exist even though they may not be constructed, intuitionism allows only mathematical objects that you can actually construct. There are a few complications involved with using fractions. On the correct side is the sum, that’s the end result of the accession of the summands.
Each plays a crucial role in mathematical abilities. By learning how to fix problems by setting up simple math difficulties, you’re going to be astonished at how many complicated issues it’s possible to understand https://grademiners.com/do-my-homework and problems it is possible to solve. So it’s important to bear this in mind when doing your assignments.
The placement test gives a measure of a student’s mathematical skills at the moment, and the outcomes are utilised to advise students on the correct mathematics course to enroll to be able to finish the mathematics requirement for a specific program of study. A similar procedure is thought to occur for cognitive tasks. In such scenarios, students who make usage of structure or their capacity to reason will most likely finish before students using a calculator.
To fully grasp how processes are genuinely working, we must measure things, then attempt to work out what’s significant. Actually, these aren’t tricks whatsoever, but ways to analyze multiplication patterns, apply partial-product principals, and utilize landmark numbers to swiftly use the things that they know to work out unknown facts. It’s the foundation for correctly locating the outcomes of multiplication utilizing the prior technique.
It’s valid even if the 2 rocks are dropped, simultaneously and from the identical height, by two folks. https://www.fu-berlin.de/en/studium/international/studium_fu/index.html The theory behind this system is for teachers to build on a kid’s existing knowledge about math to direct them to the appropriate answer, instead of quickly correcting them. Your children will be coloring them!
## All About Math Rules
Line plots are a special sort of number line that represents frequency of information. Say, you would like a square of 13. You may choose the selection of rows and columns utilized for the arrays.
The procedure for converting units is easy and powerful. If two numbers aren’t divisible by any frequent number then we need to leave it as it is. Learn how to count pounds and pence, coins utilised in the uk.
This is beneficial for a number of reasons. At least six are required to be sure of having the ability to fix any issue. Instead, it’s possible that there always will be some laws of nature that have nothing in common with one another.
So if you’re a student who finished reading this write-up, now’s the opportunity to receive serious when it has to do with studying Algebra. You don’t have to be a mathematician to ask great questions regarding your kid’s curriculum, Fennel adds. To put it differently, irrespective of where you’re in your GMAT study, learn all these facts whenever possible.
It seems that every student interpreted the issue differently, resulting in two answers. With the math curriculum as extensive as it is, teachers can’t afford to spend the opportunity to make certain that students learn the fundamental facts. In elementary classes, they have their first exposure to mathematical ideas and concepts resume writing service in all areas of mathematics.
In the simplest instances, the 2 fractions will already have a frequent denominator. Rounding away from zero is the most commonly known type of rounding. They may be individually varied to generate different sets of Multiplication problems.
They ought to be able to know how multiplication is employed in everyday circumstances. Memorizing the multiplication facts doesn’t have to be hard and frustrating. When you’re ready to create your new and exceptional Multiplication Worksheet press the Create Button.
## The Ideal Approach to Math Rules
Mathematicians have agreed on a right order to use operations, and it’s extremely important that they know these rules. The ability to rapidly perform mental calculations offers advantages in certain conditions. Inductive tests are part of the abstract reasoning tests.
There’s an increasing concern over using vertebrates in student experimentation. The distributive property lets you combine like terms. Controlled substances including prescription medications, alcohol, and tobacco shouldn’t be used.
## What You Should Do to Find Out About Math Rules Before You’re Left Behind
In only two or three minutes, you can create the questions that you need with the properties you desire. Needless to say, there can well be like terms that you will want to combine. Some may discover the initial a couple of weeks very uncomplicated and mistakenly come to think they don’t will need to study or practice.
The winner is whoever takes the smallest amount of time to work out each of the cards they’ve drawn. The responder must check to find out if their product is accurate. This enables users to try and use the app without needing to install new software.
Below, you will discover a succinct description of each page, and links to the page and subpages. You’ve got to determine whether the statement is correct, false, or cannot be determined, given the information in the passage. There are several ways to make an engaging presentation so that, if you do show important text, your audience will actually wish to read it.
## The Key to Successful Math Rules
The answer is known as the item. In that instance, it’s a good idea to ask. Easy and easy way to remember BODMASrule!!
There’s an endless number of solutions for this system. Children must be exposed to different tools and devices, and learn to use all them. Thus the item is 2937.
This standard is crucial to simplifying and solving different algebra troubles. The following are a few rules of exponents. For instance, you’ll also have to know your exponent rules, how to FOIL, and the way to solve for absolute values.
You most likely already know the Pythagorean Theorem, even if you believe you don’t. Making equations can be challenging. The rule of symmetry applies to all the rules below. | 1,233 | 6,312 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-38 | latest | en | 0.915219 |
https://newpathworksheets.com/math/grade-6/area-of-coordinate-polygons | 1,721,934,674,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763861452.88/warc/CC-MAIN-20240725175545-20240725205545-00248.warc.gz | 354,374,083 | 11,002 | ## ◂Math Worksheets and Study Guides Sixth Grade. Area of Coordinate Polygons
### The resources above cover the following skills:
Geometry (NCTM)
Specify locations and describe spatial relationships using coordinate geometry and other representational systems.
Use coordinate geometry to represent and examine the properties of geometric shapes.
Use coordinate geometry to examine special geometric shapes, such as regular polygons or those with pairs of parallel or perpendicular sides.
Use visualization, spatial reasoning, and geometric modeling to solve problems.
Use geometric models to represent and explain numerical and algebraic relationships.
Measurement (NCTM)
Apply appropriate techniques, tools, and formulas to determine measurements.
Select and apply techniques and tools to accurately find length, area, volume, and angle measures to appropriate levels of precision.
Connections to the Grade 6 Focal Points (NCTM)
Measurement and Geometry: Problems that involve areas and volumes, calling on students to find areas or volumes from lengths or to find lengths from volumes or areas and lengths, are especially appropriate. These problems extend the students' work in grade 5 on area and volume and provide a context for applying new work with equations. | 228 | 1,269 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-30 | latest | en | 0.872429 |
https://educationalresearchtechniques.com/2018/10/29/support-vector-machines-with-python/?shared=email&msg=fail | 1,656,702,518,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103945490.54/warc/CC-MAIN-20220701185955-20220701215955-00133.warc.gz | 280,502,387 | 27,807 | # Support Vector Machines Classification with Python
Support vector machines (SVM) is an algorithm used to fit non-linear models. The details are complex but to put it simply SVM tries to create the largest boundaries possible between the various groups it identifies in the sample. The mathematics behind this is complex especially if you are unaware of what a vector is as defined in algebra.
This post will provide an example of SVM using Python broken into the following steps.
1. Data preparation
2. Model Development
We will use two different kernels in our analysis. The linear kernel and he rbf kernel. The difference in terms of kernels has to do with how the boundaries between the different groups are made.
Data Preparation
We are going to use the OFP dataset available in the pydataset module. We want to predict if someone single or not. Below is some initial code.
```import numpy as np
import pandas as pd
from pydataset import data
from sklearn import svm
from sklearn.metrics import classification_report
from sklearn import model_selection```
We now need to load our dataset and remove any missing values.
```df=pd.DataFrame(data('OFP'))
df=df.dropna()
df.head()```
Looking at the dataset we need to do something with the variables that have text. We will create dummy variables for all except region and hlth. The code is below.
```dummy=pd.get_dummies(df['black'])
df=pd.concat([df,dummy],axis=1)
df=df.rename(index=str, columns={"yes": "black_person"})
df=df.drop('no', axis=1)
dummy=pd.get_dummies(df['sex'])
df=pd.concat([df,dummy],axis=1)
df=df.rename(index=str, columns={"male": "Male"})
df=df.drop('female', axis=1)
dummy=pd.get_dummies(df['employed'])
df=pd.concat([df,dummy],axis=1)
df=df.rename(index=str, columns={"yes": "job"})
df=df.drop('no', axis=1)
dummy=pd.get_dummies(df['maried'])
df=pd.concat([df,dummy],axis=1)
df=df.rename(index=str, columns={"no": "single"})
df=df.drop('yes', axis=1)
dummy=pd.get_dummies(df['privins'])
df=pd.concat([df,dummy],axis=1)
df=df.rename(index=str, columns={"yes": "insured"})
df=df.drop('no', axis=1)```
For each variable, we did the following
1. Created a dummy in the dummy dataset
2. Combined the dummy variable with our df dataset
3. Renamed the dummy variable based on yes or no
4. Drop the other dummy variable from the dataset. Python creates two dummies instead of one.
If you look at the dataset now you will see a lot of variables that are not necessary. Below is the code to remove the information we do not need.
```df=df.drop(['black','sex','maried','employed','privins','medicaid','region','hlth'],axis=1)
df.head()```
This is much cleaner. Now we need to scale the data. This is because SVM is sensitive to scale. The code for doing this is below.
```df = (df - df.min()) / (df.max() - df.min())
df.head()```
We can now create our dataset with the independent variables and a separate dataset with our dependent variable. The code is as follows.
```X=df[['ofp','ofnp','opp','opnp','emr','hosp','numchron','adldiff','age','school','faminc','black_person','Male','job','insured']]
y=df['single']```
We can now move to model development
Model Development
We need to make our test and train sets first. We will use a 70/30 split.
`X_train,X_test,y_train,y_test=model_selection.train_test_split(X,y,test_size=.3,random_state=1)`
Now, we need to create the models or the hypothesis we want to test. We will create two hypotheses. The first model is using a linear kernel and the second is one using the rbf kernel. For each of these kernels, there are hyperparameters that need to be set which you will see in the code below.
```h1=svm.LinearSVC(C=1)
h2=svm.SVC(kernel='rbf',degree=3,gamma=0.001,C=1.0)```
The details about the hyperparameters are beyond the scope of this post. Below are the results for the first model.
The overall accuracy is 73%. The crosstab() function provides a breakdown of the results and the classification_report() function provides other metrics related to classification. In this situation, 0 means not single or married while 1 means single. Below are the results for model 2
You can see the results are similar with the first model having a slight edge. The second model really struggls with predicting people who are actually single. You can see thtat the recall in particular is really poor.
Conclusion
This post provided how to ob using SVM in python. How this algorithm works can be somewhat confusing. However, its use can be powerful if use appropriately. | 1,085 | 4,509 | {"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-2022-27 | latest | en | 0.678145 |
https://mathoverflow.net/questions/270164/a-labelling-of-the-vertices-of-the-petersen-graph-with-integers | 1,708,799,214,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474544.15/warc/CC-MAIN-20240224180245-20240224210245-00386.warc.gz | 395,559,201 | 29,326 | # A labelling of the vertices of the Petersen graph with integers
The vertices of the Petersen graph (or any other simple graph) can be labelled in infinitely many ways with positive integers so that two vertices are joined by an edge if, and only if, the corresponding labels have a common divisor greater than 1. (One such labelling is with the numbers 645, 658, 902, 1085, 1221, 13243, 13949, 14053, 16813, and 17081, whose sum is 79650). Of all such ways of labelling the Petersen graph, what is the minimum the sum of the 10 corresponding integers can be?
The reason to single out the Petersen graph over all other graphs of order ten or less is that it is one particularly difficult to label if one is trying to achieve the minimum sum. Is there some systematic why of finding that minimum other than brute force?
• I would take two disjoint edge matchings and label one matching with the 5 largest primes, then the second with the next largest, and then tweak the remaining edges to minimize the sum using the 5 smallest primes. I think one can get something near 40000 this way. Gerhard "Has Not Tried It Himself" Paseman, 2017.05.18. May 19, 2017 at 3:53
• @GerhardPaseman The Petersen graph doesn't have two disjoint matchings; maybe you meant something else? May 19, 2017 at 4:50
• Indeed. Then I guess brute force it is. There may be an edge swapping algorithm which will give a local minimum. Gerhard "Maybe Choose Nine Edges Optimally?" Paseman, 2017.05.18. May 19, 2017 at 4:55
• Notice that you are looking for a labeling of the edges by the 15 smallest primes. The Petersen graph has symmetry group $S_5$ so that means that there are, in principle, $15!/5!$ possibilities to check, which is probably still too much to check one by one by computer. May 19, 2017 at 7:57
• There are three disjoint configurations of five edges, which means 6 of the ten vertices form three pairs that are "sort of" independent. If the sum of the six vertices exceeds the known minimum with that partition, one can eliminate all permutations associated with that partition. Still brute force, but probably this involves checking only 10 million cases or so. Gerhard "Combinatorial Optimization Is My Game" Paseman, 2017.05.19. May 19, 2017 at 14:58
The following labelling, whose sum is $37294$, improves Aaron Meyerowitz's by $64$. It was also found by extensive computer search, but with the strategy of assigning and fixing the first five primes (in all possible permutations) to the edges joining the vertices of the exterior and interior pentagons, and then examining the resulting sums of assigning all $10!$ possible permutations of the next ten primes to the remaining $10$ edges.
This by no means exhausts all possibilities, but does seem to come close to the optimal solution.
The above figure shows the initial assignment of the first five primes to the vertices of the graph.
• You are the winner! May 22, 2017 at 23:16
The minimum value is $37294$ as described by F. Barrera.
I broke the symmetry a little by identifying $9$ inequivalent triples of edges to which the primes $\{41,43,47\}$ can be assigned, wrote a constraint satisfaction program for the problem, and then used Minion to solve it.
(I am sure there are more efficient ways to do this.)
With some computer searching I can get a sum of $37358$ and another of $37360.$ I list them below. I'm not saying they are optimal, though they seem pretty good.
The ten vertex labels should multiply to $N^2$ where $N$ is the product of the first $15$ primes. If they were allowed to be positive reals subject to this product then the minimum sum would come from setting them all to $N^{1/5} \approx 3612.1.$ Hence 36121 is a lower bound for the minimum possible sum.
The two examples are
$[4879, 3913, 3182, 2162, 3995, 3441, 2337, 4807, 4147, 4495]=$ $[ 7\cdot17\cdot41, 7\cdot13\cdot43, 2\cdot37\cdot43, 2\cdot23\cdot47, 5\cdot17\cdot47,$ $3\cdot31\cdot37, 3\cdot19\cdot41,11\cdot19\cdot23,11\cdot13\cdot29,5\cdot29\cdot31]$
with sum $37358$
And $[2337, 2726, 3055, 3182, 3565, 3731, 3999, 4301, 4403, 6061]=$ $[ 3\cdot 19\cdot 41 , 2\cdot 29\cdot 47 , 5\cdot 13\cdot 47 , 2\cdot 37\cdot 43 , 5\cdot 23\cdot 31 ,$ $7\cdot 13\cdot 41 , 3\cdot 31\cdot 43 , 11\cdot 17\cdot 23 , 7\cdot 17\cdot 37 , 11 \cdot 19\cdot 29 ]$
with sum $37360.$
These are each minimal under switching any two edge labels. checking all $2\binom{15}{3}$ cyclic three way switches before or after didn't improve anything.
The next smallest sums from $2000000$ random labellings followed by optimization were $37438, 37458, 37490, 37494$
• If you pick 4 primes for the edges surrounding 47, many of them will fail to achieve your minimum just summing for the two vertices. Many more will fail when you take one or two more vertices into consideration. I suspect you will find a feasible number of edge labelings surrounding the edge labelled 47 that might have a chance of beating your bound. Gerhard "Global Minimum Is Within Reach" Paseman, 2017.05.20. May 20, 2017 at 8:05
It is possible to exploit the symmetries of the Petersen graph, together with the rearrangement inequality, to reduce the size of a brute-force search from $15!$ to $129729600$ (a $10080$-fold improvement).
• Draw the Petersen graph in the usual way, with an outer pentagon (consisting of red edges), an inner pentagram (blue edges), and five green spokes connecting them.
• Fix one of the red edges to have the label $2$ (we can do this since the Petersen graph is edge-transitive).
• There is an automorphism fixing the edge labelled $2$ and sending any other specified edge to a green edge. Consequently, we shall assume that $3$ is one of the green edges.
• By a further symmetry (namely the element of $D_{10}$ which fixes the edge labelled $2$), we may assume that the red edge to the anticlockwise of $2$ is greater than the red edge to the clockwise of $2$.
Now, choose four remaining primes and label the remaining red edges (there are $\frac{1}{2}(13 \times 12 \times 11 \times 10) = 8580$ ways to do so), where the factor of $\frac{1}{2}$ is due to the point mentioned above.
Choose five remaining primes and label the blue edges (there are $9 \times 8 \times 7 \times 6 \times 5 = 15120$ ways to do so).
The five remaining primes, $p_1, p_2, p_3, p_4, p_5$, will label the green spokes. Instead of checking all $5! = 120$ permutations, we can appeal to the rearrangement inequality. Specifically, temporarily label each vertex with the product of the two existing primes incident with it. For each green edge $e_i$, let $q_i$ be the sum of the labels of the two endpoints. Now, we wish to minimise the sum of products:
$$p_1 q_{\sigma(1)} + p_2 q_{\sigma(2)} + p_3 q_{\sigma(3)} + p_4 q_{\sigma(4)} + p_5 q_{\sigma(5)}$$
by choosing some permutation $\sigma \in S_5$. But the rearrangement inequality tells us that the best way to do so is to order the $p_i$ in ascending order, and the $q_i$ in descending order; the resulting sum of products is minimal.
Suddenly a brute-force search is starting to look very feasible: $129729600$ iterations should not take very long at all, even when each iteration requires sorting a list of five integers. The fact that it breaks down into $8580$ smaller searches makes this amenable to parallelisation.
• Given that there is a solution with sum under 40000, a breadth first search should also be feasible. Pick an edge labeled 47, form the set V of unordered pairs of primes which has 91 members, and start adding members of V to keep under the constraint. If you always add a V to the largest edge, you will prune the tree quickly. You will be left with (I suspect) much fewer than a million labellings with less than 50 ways to finish each labelling. Gerhard "V Means Two Adjacent Edges" Paseman, 2017.05.22. May 22, 2017 at 16:06
• Thanks to Gordon Royle (see below) for confirming this optimum. May 23, 2017 at 0:39 | 2,164 | 7,915 | {"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} | 3.453125 | 3 | CC-MAIN-2024-10 | latest | en | 0.932701 |
https://www.trade2win.com/threads/how-is-forex-p-l-calculated.236156/ | 1,580,262,251,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251783621.89/warc/CC-MAIN-20200129010251-20200129040251-00044.warc.gz | 1,081,568,331 | 21,010 | # How is Forex P&L calculated
#### klacken2
##### Newbie
1 0
Hi All
Sorry if this is a simple question but it has stumped me.
I am trying to set up an excel spreadsheet to deal with forex P&L.
But I have tried everything I can think of and am getting nowhere, so a little help really would be appreciated.
Let’s say I have two trades with GBP/NZD
I sell 3.00 lot at 1.95998
I close the trade at 1.65958
So a profit of 0.00040
But the IG metatrader system is showing this as USD \$78.40
I buy 0.01 lot @ 1.97815
I close at 1.97523
So a loss of -0.00292
But the IG metatrader system is showing this as USD \$-1.92
Can anyone give me a hint as to how they (IG) are calculating these P&L ?
Or, as I’m not bad at excel, a formula for it or a link to a site which can help create one.
Thanks
#### cantagril
##### Senior member
2,582 642
I can't tell you about IG but I'd think you need to plug in a commensurate pip value, commission and swap rate.
Last edited:
#### momo3HC
##### Member
90 18
I can't tell you about IG but I'd think you need to plug in a commensurate pip value, commission and swap rate.
Exactly.
#### Akinozragore
##### Member
95 6
The actual calculation of profit and loss in a position is quite straightforward. To calculate the P&L of a position, what you need is the position size and the number of pips the price has moved. The actual profit or loss will be equal to the position size multiplied by the pip movement.
Assume that you have a 100,000 GBP/USD position currently trading at 1.3147. If the prices move from GBP/USD 1.3147 to 1.3162, then they jumped 15 pips. For a 100,000 GBP/USD position, the 15-pips movement equates to \$150 (100,000 x .0015).
To determine if it's a profit or loss, we need to know whether we were long or short for each trade.
Long position: In case of a long position, if the prices move up, it will be a profit, and if the prices move down it will be a loss. In our earlier example, if the position is long GBP/USD, then it would be a \$150 profit.
Short position: In case of a short position, if the prices move up, it will be a loss, and if the prices move down it will be a profit. In the same example, if we had a short GBP/USD position and the prices moved up by 15 pips, it would be a loss of \$150. If the prices moved down by 20 pips, it would be a \$200 profit.
##### Junior member
39 1
Let’s say you hold an account where the base currency is USD. This is the way Calculating profit your p[rofits.
The current rate for EUR/USD is 0.9517/0.9522 (where 0.9517 is the sell price and 0.9522 is the buy price. The spread is 5).
Let’s say you decide to sell 10,000 EUR at 0.9517.
This means you sold 10,000 EUR and bought 9,517.00 USD (10,000 EUR * 0.9517 = 9,517.00 USD).
After you trade, the market rate of EUR/USD decreases to 0.9500/0.9505. You decide to buy back 10,000 EUR at 0.9505 (10,000 EUR * 0.9505 = 9,505.00 USD).
You sold 10,000 EUR for 9,517 USD and bought back 10,000 EUR for 9,505 USD.
Your profit is 12.00 USD (9,517.00 - 9,505.00).
#### momo3HC
##### Member
90 18
Let’s say you hold an account where the base currency is USD. This is the way Calculating profit your p[rofits.
The current rate for EUR/USD is 0.9517/0.9522 (where 0.9517 is the sell price and 0.9522 is the buy price. The spread is 5).
Let’s say you decide to sell 10,000 EUR at 0.9517.
This means you sold 10,000 EUR and bought 9,517.00 USD (10,000 EUR * 0.9517 = 9,517.00 USD).
After you trade, the market rate of EUR/USD decreases to 0.9500/0.9505. You decide to buy back 10,000 EUR at 0.9505 (10,000 EUR * 0.9505 = 9,505.00 USD).
You sold 10,000 EUR for 9,517 USD and bought back 10,000 EUR for 9,505 USD.
Your profit is 12.00 USD (9,517.00 - 9,505.00).
Quite simple explained but not wrong. Don`t forget take out commissions or swap from the profit if there`re some like this. | 1,184 | 3,849 | {"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} | 3.328125 | 3 | CC-MAIN-2020-05 | latest | en | 0.927222 |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=18&t=5672&p=14287 | 1,603,921,999,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107900860.51/warc/CC-MAIN-20201028191655-20201028221655-00127.warc.gz | 398,927,189 | 11,617 | $\lambda=\frac{h}{p}$
hkular_2L
Posts: 15
Joined: Fri Sep 25, 2015 3:00 am
How can a photon have no rest mass, but still have momentum?
904564128
Posts: 21
Joined: Fri Sep 25, 2015 3:00 am
Momentum is also equal to the product of energy and velocity. Since a photon has energy and velocity, it can therefore have momentum, even though it does not have a mass.
Chem_Mod
Posts: 18718
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 634 times
Firstly, momentum, as it is usually defined, is p = mv. However, this is not what is called a scalar quantity, it is a vector and has direction and therefore not very generalizable. The more general unit of measure is energy. For example, I can have a particle not moving and under no gravity but it still has energy, E = mc2. If I tried to write down its momentum, it would be 0 and, mathematically, it does not exist if we just considered its momentum.
Writing momentum in terms of energy, $E = \frac{1}{2}mv^{2} = \frac{p^{2}}{2m}$. Now of course, we still have to contend with the m on the denominator. However going towards representing systems in terms of their energy is a good first step.
What comes next is from physics and general relativity. The total energy of a relativistic particle, things near the speed of light, is:
$E^{2} = p^{2}c^{2} + m^{2}c^{4}$
setting m = 0:
$E = pc$
$p = \frac{E}{c}$
The first part has nothing to do with the answer, it is just an introduction towards writing things in terms of its energy, scalars, instead of vectors.
Even though momentum is a vector and energy is scalar, photons do have momentum when they hit a surface. Note when a photon hits a surface it ceases to exist and its energy has been transferred to the surface.
Unlike a bullet after it hits a surface it still has mass (exists) | 485 | 1,790 | {"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": 5, "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-2020-45 | latest | en | 0.963737 |
https://cs.stackexchange.com/questions/107274/assume-we-have-an-algorithm-hc-for-hamiltonian-circuit-how-is-it-possible-to-co | 1,726,688,081,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651931.60/warc/CC-MAIN-20240918165253-20240918195253-00593.warc.gz | 163,061,857 | 40,837 | # Assume we have an algorithm HC for HAMILTONIAN CIRCUIT. How is it possible to convert the HC algorithm to an algorithm HP for HAMILTONIAN PATH?
My understanding is that I have to use the algorithm for Hamiltonian Circuit to help solve the Hamiltonian Path problem.
My understanding is that we have to perform a reduction from Hamiltonian Path---> Hamiltonian Circuit (since we have an alg for Hamiltonian Circuit the reduction should be from Ham path---> Ham circuit)
But in contrast, the question says CONVERT the Ham Circuit algorithm to an algorithm to an algorithm of HP. Does this imply that the Ham Circuit problem should be reduced to the Ham Path problem?
Any guidance/feedback greatly appreciated.
The wording of the question is a bit unclear. You're supposed to reduce HamiltonianPath to HamiltonianCircuit: in other words, write an algorithm that solves HamiltonianPath by using a subroutine for HamiltonianCircuit.
I assume you're specifically going for a polytime reduction, so you shouldn't do an exponential amount of work in your new program—but there are a few different types of polytime reductions (Karp vs Cook in particular), and I can't get any more specific without knowing which one you're using. In a Karp reduction, for example, you only get to call the HamiltonianCircuit subroutine once, and have to return its output unmodified—all the transformation has to happen before you make the subroutine call. In a Cook reduction you can call it as many times as you want, and transform the outputs however you like before returning.
Given algorithm $$HC$$ that solves Hamiltonian Cycle, and a graph $$G = (V,E)$$ in which you must decide whether a Hamiltonian Path exists:
$$\bullet$$ Define $$G'$$ as the graph obtained by adding a new vertex $$v'$$ to $$G$$ and connecting it to every $$v \in V$$
$$\bullet$$ Apply $$HC$$ on graph $$G'$$
If and only if $$HC$$ found a hamiltonian cycle in $$G'$$, there is a Hamiltonian path in $$G$$. | 439 | 1,965 | {"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": 13, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-38 | latest | en | 0.928412 |
https://www.doodlemaths.com/dmsc-article-yr6-meas-7-calculating-the-volume-of-a-cuboid/ | 1,601,123,747,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400241093.64/warc/CC-MAIN-20200926102645-20200926132645-00561.warc.gz | 772,814,375 | 13,373 | # Calculating the volume of a cuboid
Introduced in the Year 6 curriculum as: "Calculate, estimate and compare volume of cubes and cuboids using standard units, including cubic centimetres (cm3 ) and cubic metres (m3 ), and extending to other units [for example, mm3 and km3 ]."
The Volume of a cuboid is found by multiplying the width by the length by the height (our 3 dimensions!)
Volume is usually measured in cm³
Example:
= 3cm {black
Therefore,
Volume = 3 x 4 x 2 = 24cm³
Example:
= 3cm {black
Therefore,
Volume = 3 x 5 x 3 = 45cm³, | 155 | 548 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2020-40 | latest | en | 0.860766 |
https://community.smartsheet.com/discussion/81230/find-the-count-of-all-multi-choice-answers | 1,722,758,921,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640393185.10/warc/CC-MAIN-20240804071743-20240804101743-00196.warc.gz | 137,809,456 | 106,177 | Find the Count of All Multi-Choice Answers
Options
edited 07/15/21
Hello all,
I'm still a bit new to SmartSheet, so forgive me if I'm a bit ignorant here.
I have a column that has several multi-choice options (Option A, B, C). I would like to know how many times, across all rows, each option was selected. The setup would look like this
Row 1: Options A and B selected
Row 2: Options A and C selected
Row 3: Options A, B, and C selected
So, either in another sheet/report, I would have something that spat out something like:
Option A: 3
Option B: 2
Option C: 2
If this is possible, what is the formula/method to make this happen? Thanks for any help/guidance you can provide!
Tags:
Best Answer
• Answer ✓
Options
I found an acceptable way to do this, but I'm not confident it is the best way.
=COUNTIF({Range from sheet}, CONTAINS("Option A", @cell))
This allowed me to create another sheet that has two columns. Column 1 = Option Name. Column 2 = Option Count.
So, here's what the final looks like.
Option A | 3
Answers
• Answer ✓
Options
I found an acceptable way to do this, but I'm not confident it is the best way.
=COUNTIF({Range from sheet}, CONTAINS("Option A", @cell))
This allowed me to create another sheet that has two columns. Column 1 = Option Name. Column 2 = Option Count.
So, here's what the final looks like.
Option A | 3
Help Article Resources
Want to practice working with formulas directly in Smartsheet?
Check out the Formula Handbook template! | 383 | 1,498 | {"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-2024-33 | latest | en | 0.91528 |
http://classblogmeister.com/blog.php?blog_id=1453698&mode=comment&blogger_id=352679 | 1,368,915,704,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696382917/warc/CC-MAIN-20130516092622-00020-ip-10-60-113-184.ec2.internal.warc.gz | 56,692,833 | 4,956 | hallit -- Blogmeister
We have three 6th grade Science classes and two 8th grade Science classes blogging here from the Pacific Northwest in Chimacum, WA! Sixth graders are learning a bit about Mt Saint Helens, environmental science through fresh water ecology, and physical science this year. Eighth graders are learning about life science this year. Please join us as we learn Science by exploring our world. Mr. G's Blog Mr. G's Class Facebook Page
by hallit teacher: Alfonso Gonzalez
Friction I learned that friction is a force that happens when two objects rub together. Without friction there would be nothing to stop us from sliding all over the place.
We used a 10/2.5 newton scale to measure the friction force of a wood block and 5 different surface types. I learned from inquiry 6.1 that the force of friction is different when something is dragged across different surfaces.
From inquiry 6.2 I learned that the weight you pull across a surface, the friction force there is. We dragged a block across one particular but, this time we changed the weight that we dragged across the surface. In conclusion, the more weight= the more friction.
From inquiry 6.3 I learned that if something has a smaller surface area with the same weight the smaller surface area has more friction force. For this experiment, we used on block but changed the surface area. The smaller surface area had more friction force.
FRICTION IS FUN=http://=http://=http://=http://=http://=http://!!!
Article posted May 16, 2012 at 02:11 PM • comment • Reads 3743 • Return to Blog List | 352 | 1,566 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2013-20 | latest | en | 0.913145 |
https://devsenv.com/example/codeforces-solution-d.-road-map-solution-in-c,-c++,-java,-python | 1,726,159,800,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651460.54/warc/CC-MAIN-20240912142729-20240912172729-00200.warc.gz | 184,204,167 | 42,505 | ## Algorithm
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
There are n cities in Berland. Each city has its index — an integer number from 1 to n. The capital has index r1. All the roads in Berland are two-way. The road system is such that there is exactly one path from the capital to each city, i.e. the road map looks like a tree. In Berland's chronicles the road map is kept in the following way: for each city i, different from the capital, there is kept number pi — index of the last city on the way from the capital to i.
Once the king of Berland Berl XXXIV decided to move the capital from city r1 to city r2. Naturally, after this the old representation of the road map in Berland's chronicles became incorrect. Please, help the king find out a new representation of the road map in the way described above.
Input
The first line contains three space-separated integers nr1r2 (2 ≤ n ≤ 5·104, 1 ≤ r1 ≠ r2 ≤ n) — amount of cities in Berland, index of the old capital and index of the new one, correspondingly.
The following line contains n - 1 space-separated integers — the old representation of the road map. For each city, apart from r1, there is given integer pi — index of the last city on the way from the capital to city i. All the cities are described in order of increasing indexes.
Output
Output n - 1 numbers — new representation of the road map in the same format.
Examples
input
Copy
`3 2 32 2`
output
Copy
`2 3 `
input
Copy
`6 2 46 1 2 4 2`
output
Copy
`6 4 1 4 2 `
## Code Examples
### #1 Code Example with C++ Programming
```Code - C++ Programming```
``````#include <cstdio>
#include <vector>
void dfs(long node, const std::vector<std::vector<long> > &g, std::vector<long> &from, std::vector<bool> &vis){
if(vis[node]){return;}
vis[node] = 1;
for(long p = 0; p < g[node].size(); p++){
long u = g[node][p];
if(vis[u]){continue;}
from[u] = node;
dfs(u, g, from, vis);
}
return;
}
int main(){
long n, r1, r2; scanf("%ld %ld %ld\n", &n, &r1, &r2);
std::vector<std::vector<long> > g(n + 1);
for(long p = 1; p <= n; p++){
if(p == r1){continue;}
long x; scanf("%ld", &x);
g[p].push_back(x); g[x].push_back(p);
}
std::vector<long> from(n + 1, 0);
std::vector<bool> visited(n + 1, 0);
dfs(r2, g, from, visited);
for(long p = 1; p <= n; p++){
if(p == r2){continue;}
printf("%ld ", from[p]);
}
puts("");
return 0;
}``````
Copy The Code &
Input
cmd
3 2 3
2 2
Output
cmd
2 3 | 763 | 2,473 | {"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-2024-38 | latest | en | 0.697784 |
http://rogermarjoribanks.info/three-point-problem-calculating-strike-dip-multiple-dd-holes/ | 1,695,658,167,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233509023.57/warc/CC-MAIN-20230925151539-20230925181539-00539.warc.gz | 39,520,151 | 10,095 | The Three point problem: Calculating strike and dip from multiple DD Holes
The Three Point Problem
In a previous post (see here) I described how quantitative orientation data can be collected from from a single drill hole, even where the core is not oriented. In this post, techniques for collecting orientation data on planes are described when more than one non-oriented hole is available from a prospect.
The need to determine the strike and dip of a planar structure from a number of drill intersections is one that occurs very frequently – this is often called the three-point problem[1] and every geologist should be familiar with the simple solutions to it (Marjoribanks, 2007).
The attitude of any plane is fully defined if the position in 3D space of three or more points on that surface is known. Where three separate holes intersect the same marker bed, they provide three points of known position on that surface. From the intercept data, there are three ways of calculating the strike and dip. The first makes use of structure contours. The second involves the use of a stereonet. In the presentation of the solutions below, it is assumed that the bed to be measured has the same attitude in all three holes.
There is, of course, the third way: feed the numbers into a computer program such as the freeware you can download at www.edumine.com ; then click the button marked “Answer”. The “answer” in this case is useful strike/dip/dip direction information from intersection coordinates. But what matters almost as much as information is the means by which you arrive at it. Graphical manipulation is a mental process that engages your brain in the 3D realities of your measurements, and it is this pre-conditioning that helps you turn mere information into knowledge and understanding. In this case, “Understanding” is the ability to create predictive models for ore, or to cut losses, walk away, and try your luck elsewhere. For this kind of understanding, easy shortcuts can be self-defeating.
Remember the mantra, the theme of many of my posts:
Data is not information
Information is not knowledge
Knowledge is not understanding
Understanding is not wisdom
Solution using structure contours.
In figure 1, the intersections of three holes into a common marker bed are projected onto a plan. Proceed as follows:
Figure 1 Using structure contours to determine the strike and dip of a planar surface intersected in three drill holes. On the map three drill holes have intercepted a common bed. The intercepts have been projected vertically onto the plan and labelled with their height above a common datum. Click for full size figure.
Step 1
Determine the three-dimensional coordinates (i.e. northing, easting and height above the datum) of each intersection of the marker bed in the holes.
Step 2
Plot the three bed intersection points on a map using the northing and easting coordinates for each intersection. Write the depth (often called the Relative Level, or RL) of the intersection beside each point.
Step 3
On the map draw a lines joining the hole intersections (figure 2). The height of the intersection at the beginning and the end of each line is already marked. Using a ruler, scale off along each line to identify the positions of all intermediate depths: identify and mark even-number depth divisions.
Step 4
Draw the lines joining points of equal depth on the surface. The lines correspond to the height contours on a topographic map and are known as structure contours. They represent the plot of horizontal lines on the bed and thus mark its strike. This strike can then be measured on the map using a protractor.
Step 5
Use the map scale to measure the horizontal distance (h) between any two contour lines – the further apart the better. Since the vertical separation (v) of the contour lines is known, the dip of the surface (d) can be calculated according to the formula:
Tan d = v/h
Figure 2 Constructing lines between each plotted intersection, the position of different heights along the lines can be scaled off. Structure contour lines (dashed) for the bed are constructed by joining points of equal height. These lines define the strike of the surface. From the map scale, the horizontal distance between lines of known height can be measured – simple trigonometry then allows the dip to be calculated. Click for full size figure.
Solution using a stereonet
Proceed as follows (see figures 3 and 4):
Step 1
Determine the absolute position coordinates (i.e. northing, easting and height above a common datum) of each intersection of the marker bed in the three holes.
Figure .3 Using a stereonet to determine the strike and dip of planar surface intersected in three drill holes. The intercepts are projected onto a plan and labelled with the height of each intersection above a common datum. The line that joins any pair of intersections is an apparent dip on the bed and can be described by its dip and dip direction. Where v is the vertical height difference between each pair of intersections, and h their horizontal separation, the apparent dip ( a ) can be calculated using Tan a = v/h. The dip direction is measured directly from the plan with a protractor. Three apparent dips can be calculated in this way. Click for full size figure.
Step 2
Plot the three intersection points on a map. Use a protractor to measure the trend (bearing) of the lines joining the three points. Use a ruler to scale off the horizontal distance between the points. Knowing the horizontal and elevation difference between any pairs of intersection points, simple trigonometric formulae (see step 5 above) will provide the angle of plunge (the angle which the line makes with the horizontal, measured in the vertical plane) for the line that joins any two pairs of points.
Step 3
We have now calculated the trend and plunge of three lines lying on the surface of the marker bed. Mark these lines on to a stereonet overlay. They plot as three points, as shown on figure 3.
Step 4
Rotate the overlay so as to bring the three points to lie on a common great circle. Only one great circle will satisfy all three points[2]. This great circle represents the trace of the bed that was intersected by the drill holes.
Step 5
From the net, read off the strike and dip of the surface (or dip and dip direction, or apparent dip on any given drill section).
Figure 4 On the stereonet, the three apparent dips plot as three points. The net overlay is rotated to bring the points to lie on a great circle girdle. This girdle is the plot of the bed and its strike and (true) dip can be easily read off. (Actually, only 2 apparent dips are necessary to define the surface, but using a third line provides extra accuracy.) Click for full size figure.
An elegant stereonet solution to determining the attitude of planes in non-oriented core
Where there is no single marker bed that can be correlated between adjacent holes, it is sometimes still possible to determine the orientation of a set of parallel surfaces (such as bedding planes, a cleavage, or a vein set) provided that the surfaces have been cored by a minimum of three nonparallel drill holes (Mead, 1921, Bucher, 1943). The same technique can even be extended to a single hole, provided that the hole has sufficient deviation along its length for the differently oriented sectors of the same hole to be considered in the same way as three separate holes (Laing, 1977).
In the example illustrated in the following figures, three adjacent but non-parallel angle holes have intersected the same set of parallel, planar quartz veinlets. None of the core is oriented, but the average alpha (α) angle (for definition of alpha angle see here) between the veins and the core axis has been measured in each hole: it is 10° in Hole 1; 56° in Hole 2 and 50° in Hole 3.
In our example, Hole 1 is drilled at -50° to 270°; Hole 2 at -65° to 090° and Hole 3 at -60° to 345°. On the stereonet, the orientation of each drill hole plots as a point.
When plotting planes on a stereonet it is always much easier to work with the pole[3] to the plane rather than the plane itself. If a plane makes an angle α with the core axis, then the pole to the plane makes an angle of 90 – α to the core axis, as illustrated in figure 5.
Figure.5 The angular relationship of the alpha angle (α) to the pole (i.e. the normal) of a surface intersected in drill core. Click for full size figure.
Let us consider Hole 1 (figure 6). Angle α for the vein set is known. Because the core is not oriented, the poles to the veins could lie anywhere within the range of orientations that is produced as the core is rotated one complete circle about its long axis. This range defines a cone, centered on the core axis, with an apical angle of 2 x α. From figure 5, we see that the pole to the plane will describe a cone with an apical angle of 2 x (90 – α). That is all we can tell from one hole, but this information can be shown on the stereonet, because a cone centered on a drill hole plots as a small circle girdle around that hole. In Hole 1, 90 – α is 80°. The vein set in Hole 1 can therefore be represented by a small circle girdle at an angle of 80° to the hole plot.
Figure 6 On the stereonet a small circle girdle at 90-α degrees from Hole 1 traces all possible orientations of the poles to a vein set making angle of α with the core axis. Click for full size figure.
Now the same procedure is carried out for Hole 2 by drawing a small circle at 90-α (34°) to the plot of that hole on the net (figure 7). The small circle about Hole 1 and the small circle about Hole 2 intersect at two points (P1 & P2) – these points represent two possible orientations for the vein set. Already, with just two DD holes we have reduced the problem of determining the orientation of the vein set from an infinite number of possibilities to just two possibilities.
Figure 7 On the stereonet, the poles to all possible planes making an angle of α degrees to the axis of Hole 2 inscribe a small circle girdle at 90-α (34°) to the plot of the hole. The small circle about Hole 1 and the small circle about Hole 2 intersect at two points (P1 & P2) – these points represent the only two possible orientations for the vein set, given the data plotted so far. Click for larger image.
Now, in the same manner, we draw the third small circle about Hole 3 representing the alpha angle measured in that hole (figure 8). We now have three small circle girdles on our net, centered about each of the three drill holes. Since the assumption behind this procedure is that all measurements are of the one vein set with a constant orientation, the single point (P) where the three small girdles intersect must represent the pole to the one orientation that is common to all three holes. This pole defines the attitude of the common vein set seen in the holes. Of course, with a real set of measurements it is unlikely that three small circles plotted in this way would meet at a single point. Rather, the intersecting lines will define a triangle whose size reflects the accuracy of the measurements (and the assumptions made that we are dealing with a single parallel set of surfaces). The true pole position (if there is one) will lie somewhere within this triangle of error.
From the point P, the strike and dip (or dip and dip direction, or apparent dip on drill section) of the vein set can be simply read off from the net.
Figure 8 On the stereonet, the poles to all possible planes making an angle of α degrees to the axis of Hole 3 inscribe a small circle girdle at 90-α (40°) to the plot of the hole. The small circles about Hole 1, Hole2 & Hole 3 intersect at a single P. P is the pole to the unique plane which satisfies the measurements made in the three holes.. Click for larger image.
REFERENCES
Bucher WH (1943) Dip and strike for three not parallel drill holes lacking key beds. Econ Geol 38, 648-657.
Laing WP (1977) Structural interpretation of drill core from folded and cleaved rocks. Econ Geol 72, 671-685.
Marjoribanks RW (2007) Structural logging of drill core. Australian Institute of Geologists Handbook 5 (2nd ed.), 68p.
Mead WJ (1921) Determination of the attitude of concealed bedding formations by diamond drilling. Econ Geol 21, 37-47.
[1] Presumably in a conscious or subconscious nod to Sherlock Holmes’ “3-pipe problem”. In Holmes case, it was an amount of opium, but there is no need to resort to such extreme measures to solve the 3-point problem.
[2] Actually, because a plotted point on a stereonet represents the orientation of a line, only two such points are needed to define the plane on which they lie. The use of a third point (line) adds accuracy and provides for error checking.
[3]The pole to a plane is the line at right angles, or normal, to the plane. By plotting the pole, the attitude of a plane can be represented on a stereonet by a single point.
Note that this blog no longer accepts on-line comments (there was too much spam coming in!). However I welcome all questions, comments or criticisms. You can send these to me via the CONTACT ME tab. | 2,976 | 13,185 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2023-40 | latest | en | 0.915045 |
https://codehelppro.com/example/c-plus-to-calculate-standard-deviation/ | 1,652,676,800,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662509990.19/warc/CC-MAIN-20220516041337-20220516071337-00060.warc.gz | 243,119,870 | 23,261 | Courses
# Program to calculate Standard Deviation
0 Suresh Chand March 30, 2021
In this example, We will calculate the standard deviation of 10 data using array.
To understand this example, You must knowledge in following topics
In this example, we have created the function calculateSD() function and It accepts an array argument. Here, we have passed an array of 10 elements to calculate SD.
## The formula of Standard Deviation
We will use this formula to calculate standard deviation where
## CODE:
#include <iostream>
#include <cmath>
using namespace std;
// function prototype
float calculateSD(float data[]);
//main function start
int main()
{
int i;
float data[10];
//taking 10 numbers from users
cout << "Enter 10 elements: ";
for(i = 0; i < 10; ++i)
cin >> data[i];
//print standard deviation by calling function calculateSD
cout << endl << "Standard Deviation = " << calculateSD(data);
return 0;
}
//main function to calculate standard deviation
float calculateSD(float data[])
{
float sum = 0.0, mean, standardDeviation = 0.0;
int i;
for(i = 0; i < 10; ++i)
{
sum += data[i];
}
mean = sum/10;
for(i = 0; i < 10; ++i)
standardDeviation += pow(data[i] - mean, 2);
return sqrt(standardDeviation / 10);
}
The output of above program is
Enter 10 elements: 1
2
3
4
5
6
7
8
9
10
Subscribe
Notify of
Inline Feedbacks | 360 | 1,342 | {"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} | 3.28125 | 3 | CC-MAIN-2022-21 | latest | en | 0.519748 |
http://mathhelpforum.com/calculus/137517-error-bound-question-print.html | 1,526,848,912,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863684.0/warc/CC-MAIN-20180520190018-20180520210018-00296.warc.gz | 187,627,538 | 2,796 | # Error Bound question
• Apr 5th 2010, 09:57 PM
colonelone
Error Bound question
The error bound formula for Simpson’s rule is
EBSN = K(b − a)^5/180N^4
where N is the number of subintervals used and K = 12 specifically for our h(x) on the interval [0, 1]. However, it can be shown that the error bound using an nth degree polynomial approximation is given by
EBTn = 1/((2n + 3)(n + 1)!
(The “T” is for “Taylor” after whom these polynomials are named.) Our example above used n = 5. Discuss and compare the errors predicted by these formula for n = 5. If accurate estimates are needed, say to 12 decimal places (i.e. an error less than 10−12 ), which method would you use? How many terms (or subintervals) are necessary for 20 decimal places of accuracy with each method? (Standard encoding of floating point numbers are capable of an accuracy of approximately 16 decimal places.) Is one method always more accurate than the other? Other than accuracy, are there other advantages or disadvantages to using one form of approximation over the other? | 275 | 1,048 | {"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} | 2.96875 | 3 | CC-MAIN-2018-22 | latest | en | 0.912466 |
https://www.kth.se/student/kurser/kurs/SF2520/?l=en | 1,718,250,914,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861342.11/warc/CC-MAIN-20240613025523-20240613055523-00182.warc.gz | 780,291,106 | 16,226 | • Svenska
# SF2520 Applied Numerical Methods 7.5 credits
### For course offering
Autumn 2023 Start 28 Aug 2023 programme students
### Target group
Available for all master program and engineering students from the 3rd study year as long as it can be included in your programme.
### Periods
P1 (3.0 hp), P2 (4.5 hp)
28 Aug 2023
15 Jan 2024
25%
Normal Daytime
English
KTH Campus
### Number of places
Places are not limited
## Application
### For course offering
Autumn 2023 Start 28 Aug 2023 programme students
50282
## Contact
### For course offering
Autumn 2023 Start 28 Aug 2023 programme students
### Contact
Olof Runborg (olofr@kth.se)
### Examiner
No information inserted
### Course coordinator
No information inserted
### Teachers
No information inserted
Headings with content from the Course syllabus SF2520 (Spring 2022–) are denoted with an asterisk ( )
## Content and learning outcomes
### Course contents
The course will give you knowledge about advanced computer methods based on numerical algorithms for solving mathematical models from scientific and engineering applications, in particular about how to formulate, analyze and implement them. More specifically, the course includes:
• numerical treatment of ordinary differential equations,
• finite difference methods and basic finite element methods for, mainly linear, partial differential equations,
• numerical solution of linear systems of equations by direct and iterative methods,
### Intended learning outcomes
For the mathematical models in the course contents (e.g. ordinary and partial differential equations, linear systems of equations) the student shall be able to:
• select suitable numerical algorithms,
• analyze numerical methods with respect to computational cost, accuracy and stability,
• apply and implement numerical algorithms in a suitable programming language,
• classify and characterize the mathematical models.
In addition, the student shall be able to:
• estimate the accuracy of numerical results,
• describe limitations of mathematical models and numerical methods,
• for a given numerical problem, present, discuss and summarize the problem, solution method and results in a clear way,
• work inteams to solve a numerical problem.
## Literature and preparations
### Specific prerequisites
• English B / English 6
• Completed basic course in numerical analysis (SF1544, SF1545or equivalent)
• Completed basic course in differential equations (SF1633, SF1683or equivalent).
### Recommended prerequisites
Completed basic courses in numerical analysis equivalent to SF1544 or similar, mathematical courses corresponding to linear algebra, calculus and differential equations, good handling with MATLAB.
### Equipment
No information inserted
### Literature
Lennart Edsberg: Introduction to computation and modeling for differential equations, Wiley 2008, ISBN 978-0-470-27085-1
## Examination and completion
If the course is discontinued, students may request to be examined during the following two academic years.
A, B, C, D, E, FX, F
### Examination
• LABA - Laboratory Work, 4.5 credits, grading scale: A, B, C, D, E, FX, F
• TEN1 - Written Examination, 3.0 credits, grading scale: A, B, C, D, E, FX, F
Based on recommendation from KTH’s coordinator for disabilities, the examiner will decide how to adapt an examination for students with documented disability.
The examiner may apply another examination format when re-examining individual students.
### Opportunity to complete the requirements via supplementary examination
No information inserted
### Opportunity to raise an approved grade via renewed examination
No information inserted
### Ethical approach
• All members of a group are responsible for the group's work.
• In any assessment, every student shall honestly disclose any help received and sources used.
• In an oral assessment, every student shall be able to present and answer questions about the entire assignment and solution.
## Further information
### Course room in Canvas
Registered students find further information about the implementation of the course in the course room in Canvas. A link to the course room can be found under the tab Studies in the Personal menu at the start of the course.
### Main field of study
Mathematics, Technology
Second cycle | 897 | 4,342 | {"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-2024-26 | latest | en | 0.834762 |
https://forge.ipsl.jussieu.fr/ioserver/browser/XIOS/xios_training/XIOS_COMPILED/src/transformation/axis_algorithm_inverse.cpp?rev=2101 | 1,716,879,266,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059078.15/warc/CC-MAIN-20240528061449-20240528091449-00731.warc.gz | 211,867,076 | 10,441 | # source:XIOS/xios_training/XIOS_COMPILED/src/transformation/axis_algorithm_inverse.cpp@2101
Last change on this file since 2101 was 2101, checked in by yushan, 3 years ago
XIOS TRAINING some source updates for HO 11
File size: 11.7 KB
Line
1/*!
2 \file axis_algorithm_inverse.hpp
3 \author Ha NGUYEN
4 \since 14 May 2015
5 \date 29 June 2015
6
7 \brief Algorithm for inversing an axis..
8 */
9#include "axis_algorithm_inverse.hpp"
10#include "context.hpp"
11#include "context_client.hpp"
12#include "axis.hpp"
13#include "grid.hpp"
14#include "grid_transformation_factory_impl.hpp"
15#include "inverse_axis.hpp"
16#include "client_client_dht_template.hpp"
17
18namespace xios {
19
20CGenericAlgorithmTransformation* CAxisAlgorithmInverse::create(CGrid* gridDst, CGrid* gridSrc,
21 CTransformation<CAxis>* transformation,
22 int elementPositionInGrid,
23 std::map<int, int>& elementPositionInGridSrc2ScalarPosition,
24 std::map<int, int>& elementPositionInGridSrc2AxisPosition,
25 std::map<int, int>& elementPositionInGridSrc2DomainPosition,
26 std::map<int, int>& elementPositionInGridDst2ScalarPosition,
27 std::map<int, int>& elementPositionInGridDst2AxisPosition,
28 std::map<int, int>& elementPositionInGridDst2DomainPosition)
29TRY
30{
31 std::vector<CAxis*> axisListDestP = gridDst->getAxis();
32 std::vector<CAxis*> axisListSrcP = gridSrc->getAxis();
33
34 CInverseAxis* inverseAxis = dynamic_cast<CInverseAxis*> (transformation);
35 int axisDstIndex = elementPositionInGridDst2AxisPosition[elementPositionInGrid];
36 int axisSrcIndex = elementPositionInGridSrc2AxisPosition[elementPositionInGrid];
37
38 return (new CAxisAlgorithmInverse(axisListDestP[axisDstIndex], axisListSrcP[axisSrcIndex], inverseAxis));
39}
40CATCH
41
42bool CAxisAlgorithmInverse::registerTrans()
43TRY
44{
45 return CGridTransformationFactory<CAxis>::registerTransformation(TRANS_INVERSE_AXIS, create);
46}
47CATCH
48
49CAxisAlgorithmInverse::CAxisAlgorithmInverse(CAxis* axisDestination, CAxis* axisSource, CInverseAxis* inverseAxis)
50 : CAxisAlgorithmTransformation(axisDestination, axisSource)
51TRY
52{
53 if (axisDestination->n_glo.getValue() != axisSource->n_glo.getValue())
54 {
55 ERROR("CAxisAlgorithmInverse::CAxisAlgorithmInverse(CAxis* axisDestination, CAxis* axisSource)",
56 << "Two axis have different global size"
57 << "Size of axis source " <<axisSource->getId() << " is " << axisSource->n_glo.getValue() << std::endl
58 << "Size of axis destination " <<axisDestination->getId() << " is " << axisDestination->n_glo.getValue());
59 }
60}
61CATCH
62
63void CAxisAlgorithmInverse::computeIndexSourceMapping_(const std::vector<CArray<double,1>* >& dataAuxInputs)
64TRY
65{
66 this->transformationMapping_.resize(1);
67 this->transformationWeight_.resize(1);
68
69 TransformationIndexMap& transMap = this->transformationMapping_[0];
70 TransformationWeightMap& transWeight = this->transformationWeight_[0];
71
72 int globalIndexSize = axisDestGlobalIndex_.size();
73 for (int idx = 0; idx < globalIndexSize; ++idx)
74 {
75 transMap[axisDestGlobalIndex_[idx]].push_back(axisDestGlobalSize_-axisDestGlobalIndex_[idx]-1);
76 transWeight[axisDestGlobalIndex_[idx]].push_back(1.0);
77 }
78
79 int niSrc = axisSrc_->n.getValue();
80 int sizeSrc = axisSrc_->n_glo.getValue();
81 if (niSrc != sizeSrc) updateAxisValue();
82 else
83 {
84 if (axisDest_->value.isEmpty()) ERROR("CAxisAlgorithmInverse::CAxisAlgorithmInverse(CAxis* axisDestination, CAxis* axisSource)",
85 << "Axis destination value is not initialized");
86
87 for (int idx = 0; idx < sizeSrc; ++idx)
88 {
89 axisDest_->value(idx) = axisSrc_->value(sizeSrc-idx-1);
90 }
91 }
92}
93CATCH
94
95/*!
96 Update value on axis after inversing
97 After an axis is inversed, not only the data on it must be inversed but also the value
98*/
99void CAxisAlgorithmInverse::updateAxisValue()
100TRY
101{
102 CContext* context = CContext::getCurrent();
103 CContextClient* client=context->client;
104 int clientRank = client->clientRank;
105 int nbClient = client->clientSize;
106
107 int niSrc = axisSrc_->n.getValue();
108 int ibeginSrc = axisSrc_->begin.getValue();
109 int nSrc = axisSrc_->index.numElements();
110
111 CClientClientDHTInt::Index2VectorInfoTypeMap globalIndex2ProcRank;
112 for (int idx = 0; idx < nSrc; ++idx)
113 {
115 {
116 globalIndex2ProcRank[(axisSrc_->index)(idx)].resize(1);
117 globalIndex2ProcRank[(axisSrc_->index)(idx)][0] = clientRank;
118 }
119 }
120
121 typedef std::unordered_map<size_t, std::vector<double> > GlobalIndexMapFromSrcToDest;
122 GlobalIndexMapFromSrcToDest globalIndexMapFromSrcToDest;
123 TransformationIndexMap& transMap = this->transformationMapping_[0];
124 TransformationIndexMap::const_iterator itb = transMap.begin(), ite = transMap.end(), it;
125 CArray<size_t,1> globalSrcIndex(transMap.size());
126 int localIndex = 0;
127 for (it = itb; it != ite; ++it)
128 {
129 size_t srcIndex = it->second[0];
130 globalIndexMapFromSrcToDest[srcIndex].resize(1);
131 globalIndexMapFromSrcToDest[srcIndex][0] = it->first;
132 globalSrcIndex(localIndex) = srcIndex;
133 ++localIndex;
134 }
135
136 CClientClientDHTInt dhtIndexProcRank(globalIndex2ProcRank, client->intraComm);
137 dhtIndexProcRank.computeIndexInfoMapping(globalSrcIndex);
138 CClientClientDHTInt::Index2VectorInfoTypeMap& computedGlobalIndexOnProc = dhtIndexProcRank.getInfoIndexMap();
139 std::unordered_map<int, std::vector<size_t> > globalSrcIndexSendToProc;
140 for (int idx = 0; idx < localIndex; ++idx)
141 {
142 size_t tmpIndex = globalSrcIndex(idx);
143 if (1 == computedGlobalIndexOnProc.count(tmpIndex))
144 {
145 std::vector<int>& tmpVec = computedGlobalIndexOnProc[tmpIndex];
146 globalSrcIndexSendToProc[tmpVec[0]].push_back(tmpIndex);
147 }
148 }
149
150 std::unordered_map<int, std::vector<size_t> >::const_iterator itbIndex = globalSrcIndexSendToProc.begin(), itIndex,
151 iteIndex = globalSrcIndexSendToProc.end();
152 std::map<int,int> sendRankSizeMap,recvRankSizeMap;
153 int connectedClient = globalSrcIndexSendToProc.size();
154 int* recvCount=new int[nbClient];
155 int* displ=new int[nbClient];
156 int* sendRankBuff=new int[connectedClient];
157 int* sendSizeBuff=new int[connectedClient];
158 int n = 0;
159 for (itIndex = itbIndex; itIndex != iteIndex; ++itIndex, ++n)
160 {
161 sendRankBuff[n] = itIndex->first;
162 int sendSize = itIndex->second.size();
163 sendSizeBuff[n] = sendSize;
164 sendRankSizeMap[itIndex->first] = sendSize;
165 }
166 MPI_Allgather(&connectedClient,1,MPI_INT,recvCount,1,MPI_INT,client->intraComm);
167
168 displ[0]=0 ;
169 for(int n=1;n<nbClient;n++) displ[n]=displ[n-1]+recvCount[n-1];
170 int recvSize=displ[nbClient-1]+recvCount[nbClient-1];
171 int* recvRankBuff=new int[recvSize];
172 int* recvSizeBuff=new int[recvSize];
173 MPI_Allgatherv(sendRankBuff,connectedClient,MPI_INT,recvRankBuff,recvCount,displ,MPI_INT,client->intraComm);
174 MPI_Allgatherv(sendSizeBuff,connectedClient,MPI_INT,recvSizeBuff,recvCount,displ,MPI_INT,client->intraComm);
175 for (int i = 0; i < nbClient; ++i)
176 {
177 int currentPos = displ[i];
178 for (int j = 0; j < recvCount[i]; ++j)
179 if (recvRankBuff[currentPos+j] == clientRank)
180 {
181 recvRankSizeMap[i] = recvSizeBuff[currentPos+j];
182 }
183 }
184
185 // Sending global index of grid source to corresponding process as well as the corresponding mask
186 std::vector<MPI_Request> requests;
187 std::vector<MPI_Status> status;
188 std::unordered_map<int, unsigned long* > recvGlobalIndexSrc;
189 std::unordered_map<int, double* > sendValueToDest;
190 for (std::map<int,int>::const_iterator itRecv = recvRankSizeMap.begin(); itRecv != recvRankSizeMap.end(); ++itRecv)
191 {
192 int recvRank = itRecv->first;
193 int recvSize = itRecv->second;
194 recvGlobalIndexSrc[recvRank] = new unsigned long [recvSize];
195 sendValueToDest[recvRank] = new double [recvSize];
196
197 requests.push_back(MPI_Request());
198 MPI_Irecv(recvGlobalIndexSrc[recvRank], recvSize, MPI_UNSIGNED_LONG, recvRank, 46, client->intraComm, &requests.back());
199 }
200
201 std::unordered_map<int, unsigned long* > sendGlobalIndexSrc;
202 std::unordered_map<int, double* > recvValueFromSrc;
203 for (itIndex = itbIndex; itIndex != iteIndex; ++itIndex)
204 {
205 int sendRank = itIndex->first;
206 int sendSize = sendRankSizeMap[sendRank];
207 const std::vector<size_t>& sendIndexMap = itIndex->second;
208 std::vector<size_t>::const_iterator itbSend = sendIndexMap.begin(), iteSend = sendIndexMap.end(), itSend;
209 sendGlobalIndexSrc[sendRank] = new unsigned long [sendSize];
210 recvValueFromSrc[sendRank] = new double [sendSize];
211 int countIndex = 0;
212 for (itSend = itbSend; itSend != iteSend; ++itSend)
213 {
214 sendGlobalIndexSrc[sendRank][countIndex] = *itSend;
215 ++countIndex;
216 }
217
218 // Send global index source and mask
219 requests.push_back(MPI_Request());
220 MPI_Isend(sendGlobalIndexSrc[sendRank], sendSize, MPI_UNSIGNED_LONG, sendRank, 46, client->intraComm, &requests.back());
221 }
222
223 status.resize(requests.size());
224 MPI_Waitall(requests.size(), &requests[0], &status[0]);
225
226
227 std::vector<MPI_Request>().swap(requests);
228 std::vector<MPI_Status>().swap(status);
229
230 // Okie, on destination side, we will wait for information of masked index of source
231 for (std::map<int,int>::const_iterator itSend = sendRankSizeMap.begin(); itSend != sendRankSizeMap.end(); ++itSend)
232 {
233 int recvRank = itSend->first;
234 int recvSize = itSend->second;
235
236 requests.push_back(MPI_Request());
237 MPI_Irecv(recvValueFromSrc[recvRank], recvSize, MPI_DOUBLE, recvRank, 48, client->intraComm, &requests.back());
238 }
239
240 for (std::map<int,int>::const_iterator itRecv = recvRankSizeMap.begin(); itRecv != recvRankSizeMap.end(); ++itRecv)
241 {
242 int recvRank = itRecv->first;
243 int recvSize = itRecv->second;
244 double* sendValue = sendValueToDest[recvRank];
245 unsigned long* recvIndexSrc = recvGlobalIndexSrc[recvRank];
246 int realSendSize = 0;
247 for (int idx = 0; idx < recvSize; ++idx)
248 {
249 size_t globalIndex = *(recvIndexSrc+idx);
250 int localIndex = globalIndex - ibeginSrc;
251 *(sendValue + idx) = axisSrc_->value(localIndex);
252 }
253 // Okie, now inform the destination which source index are masked
254 requests.push_back(MPI_Request());
255 MPI_Isend(sendValueToDest[recvRank], recvSize, MPI_DOUBLE, recvRank, 48, client->intraComm, &requests.back());
256 }
257 status.resize(requests.size());
258 MPI_Waitall(requests.size(), &requests[0], &status[0]);
259
260
261 size_t nGloAxisDest = axisDest_->n_glo.getValue() - 1;
262 for (std::map<int,int>::const_iterator itSend = sendRankSizeMap.begin(); itSend != sendRankSizeMap.end(); ++itSend)
263 {
264 int recvRank = itSend->first;
265 int recvSize = itSend->second;
266
267 double* recvValue = recvValueFromSrc[recvRank];
268 unsigned long* recvIndex = sendGlobalIndexSrc[recvRank];
269 for (int idx = 0; idx < recvSize; ++idx)
270 {
271 size_t globalIndex = *(recvIndex+idx);
272 int localIndex = ((nGloAxisDest-globalIndex) - axisDest_->begin);
273 axisDest_->value(localIndex) = *(recvValue + idx);
274 }
275 }
276
277 delete [] recvCount;
278 delete [] displ;
279 delete [] sendRankBuff;
280 delete [] recvRankBuff;
281 delete [] sendSizeBuff;
282 delete [] recvSizeBuff;
283
284 std::unordered_map<int, double* >::const_iterator itChar;
285 for (itChar = sendValueToDest.begin(); itChar != sendValueToDest.end(); ++itChar)
286 delete [] itChar->second;
287 for (itChar = recvValueFromSrc.begin(); itChar != recvValueFromSrc.end(); ++itChar)
288 delete [] itChar->second;
289 std::unordered_map<int, unsigned long* >::const_iterator itLong;
290 for (itLong = sendGlobalIndexSrc.begin(); itLong != sendGlobalIndexSrc.end(); ++itLong)
291 delete [] itLong->second;
292 for (itLong = recvGlobalIndexSrc.begin(); itLong != recvGlobalIndexSrc.end(); ++itLong)
293 delete [] itLong->second;
294}
295CATCH
296
297}
Note: See TracBrowser for help on using the repository browser. | 4,002 | 12,991 | {"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-2024-22 | latest | en | 0.084117 |
https://docs.feelpp.org/user/latest/python/pyfeelpp/expr.html | 1,722,671,876,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640361431.2/warc/CC-MAIN-20240803060126-20240803090126-00255.warc.gz | 167,144,510 | 27,500 | # Manipulate Expressions
Set the Feel++ environment with local repository
``````import feelpp.core as fppc
import sys,math
import plotly.express as px
import plotly.graph_objects as go
app = fppc.Environment(["myapp"],config=fppc.localRepository(""))
print("feelpp version: ",fppc.Info.version())``````
Results
`feelpp version: 0.111.0-preview.100`
## 1. Create and evaluate expressions
An expression is a mathematical expression that can be evaluated at a given point and depend on multiple variables. It can be a scalar, a vectorial or a matricial expression. The expression can be created from a string. The expression can be then evaluated at a given point.
### 1.1. Scalar expression
Create a scalar expression abd evaluate the expression at a given point (1,2)
``````expr = fppc.expr("x+y:x:y") (1)
expr.evaluate({"x":1,"y":2}) (2)``````
Results
`Out[1]: array([3.])`
1 Create a scalar expression from a string, the symbols are listed at the end of the string using `:` as separator. 2 Evaluate the expression at the point (1,2), it returns a array with one element.
The expression can also be evaluated at a set of points.
Evaluate the expression at a set of points
``````expr.setParameterValues({"y":2}) (1)
import numpy as np
x=np.linspace(0,1,10) (2)
expr.evaluate("x",x) (3)``````
Results
```Out[1]:
array([2. , 2.11111111, 2.22222222, 2.33333333, 2.44444444,
2.55555556, 2.66666667, 2.77777778, 2.88888889, 3. ])```
1 Set the value of the parameter `y` to 2 2 Create a numpy array with 10 points between 0 and 1 3 Evaluate the expression at the 10 equidistributed points between 0 and 1, it returns a array with two elements.
### 1.2. Vectorial expression
We start with an expression that depends on two variables `x` and `y` and create a vectorial expression with the expression `x` and `y`.
Create a vectorial expression abd evaluate the expression at a given point (1,2)
``````expr21 = fppc.expr("{x,y}:x:y",row=2,col=1) (1)
expr21.evaluate({"x":1,"y":2}) (2)``````
Results
`Out[1]: array([1., 2.])`
1 Create a vectorial expression from a string, the symbols are listed at the end of the string using `:` as separator. 2 Evaluate the expression at the point (1,2), it returns a array with two elements.
Now we turn to a vectorial expression with 3 components.
``````expr31 = fppc.expr("{x,y,z+x}:x:y:z",row=3,col=1) (1)
expr31.evaluate({"x":1,"y":2,"z":10}) (2)``````
Results
`Out[1]: array([ 1., 2., 11.])`
1 Create a vectorial expression from a string, the symbols are listed at the end of the string using `:` as separator. 2 Evaluate the expression at the point (1,2,10), it returns a array with three elements.
### 1.3. Matricial expression
Create a matricial expression abd evaluate the expression at a given point (1,2)
``````expr22 = fppc.expr("{x+y,x,y,y-x}:x:y", row=2, col=2) (1)
expr22.evaluate({"x":1,"y":2}) (2)``````
Results
```Out[1]:
array([[3., 1.],
[2., 1.]])```
1 Create a matricial expression from a string, the symbols are listed at the end of the string using `:` as separator. 2 Evaluate the expression at the point (1,2), it returns a array with four elements.
We now turn to a 3x3 matricial expression.
``````expr33 = fppc.expr("{x,y,z,x-y,y-y,z+x+y,z,y,x}:x:y:z",row=3,col=3) (1)
expr33.evaluate({"x":1,"y":2,"z":3}) (2)``````
Results
```Out[1]:
array([[ 1., 2., 3.],
[-1., 0., 6.],
[ 3., 2., 1.]])```
1 Create a matricial expression from a string, the symbols are listed at the end of the string using `:` as separator. 2 Evaluate the expression at the point (1,2,3), it returns a array with nine elements.
## 2. Differentiation
The member functions `diff` and `diff2` allows to compute the first and second symbolic derivatives of a function. The first argument is the symbol with respect to which the derivative is computed.
``````ex=fppc.expr("a*sin(x):x:a")
ex.setParameterValues({"a":1})
exd = ex.diff("x") (1)
exd2 = ex.diff2("x") (2)
exda = ex.diff("a") (3)
exdax = ex.diff("a").diff("x") (4)
x=np.linspace(0,2*math.pi,200)
print(f" ex: {ex}")
print(f" exd: {exd}")
print(f" exd2: {exd2}")
print(f" exda: {exda}")
print(f"exdaa: {exdax}")``````
Results
``` ex: a*sin(x)
exd: cos(x)*a
exd2: -a*sin(x)
exda: sin(x)
exdaa: cos(x)```
1 First derivative with respect to `x`$\frac{\partial \cdot }{\partial x}$ 2 Second derivative with respect to `x`$\frac{\partial^2 \cdot }{\partial^2 x}$ 3 First derivative with respect to `a` $\frac{\partial \cdot }{\partial a}$ 4 First derivative with respect to `a` and then with respect to `x` $\frac{\partial^2 \cdot}{\partial x \partial a}$
Now let’s plot the expression and its derivative. The first and last are the same.
``````fig = go.Figure()
fig.show()``````
Results
### 2.1. Derivative of a vectorial expression
Here is an example of a vectorial expression and its derivative with respect to `x` and `y`.
``````expr21 = fppc.expr("{x^3,x*y^2}:x:y", row=2, col=1) (1)
print("d(x,y)/dx = ",expr21.diff("x")) (2)
print("d(x,y)/dy = ", expr21.diff("y")) (3)
print("d2(x,y)/dxdy = ", expr21.diff("x").diff("y")) (4)
print("d2(x,y)/dydx = ", expr21.diff("y").diff("x")) (5)``````
Results
```d(x,y)/dx = [[3*x^2],[y^2]]
d(x,y)/dy = [[0],[2*x*y]]
d2(x,y)/dxdy = [[0],[2*y]]
d2(x,y)/dydx = [[0],[2*y]]```
1 Create a vectorial expression from a string, the symbols are listed at the end of the string using `:` as separator. 2 First derivative with respect to `x`$\frac{\partial \cdot }{\partial x}$ 3 First derivative with respect to `y`$\frac{\partial \cdot }{\partial y}$ 4 First derivative with respect to `x` and then with respect to `y` $\frac{\partial^2 \cdot}{\partial x \partial y}$ 5 First derivative with respect to `y` and then with respect to `x` $\frac{\partial^2 \cdot}{\partial y \partial x}$
similar results are obtained for matricia expressions.
## 3. Special functions
The Feel++ library provides a set of special functions that can be used in expressions.
### 3.1. function with multiple parameters
``````ex=fppc.expr("sin(w*x+b):x:w:b")
ex.setParameterValues({"w":2*math.pi,"b":0.5})
x=np.linspace(0,2*math.pi,100)
y=ex.evaluate("x",x)
fig = px.line(x=x, y=y, title="sin",labels={"x":"x","y":r'y'})
fig.show()``````
Results
### 3.2. clamp
The function `clamp` allows to clamp a value between two bounds.
``````ex=fppc.expr("clamp(sin(w*x+b),-0.3,0.4):x:w:b")
ex.setParameterValues({"w":2*math.pi,"b":0.5})
x=np.linspace(0,2*math.pi,100)
y=ex.evaluate("x",x)
fig = px.line(x=x, y=y, title="f",labels={"x":"x","y":r'y'})
fig.show()``````
Results
### 3.3. mapabcd
The function `mapabcd` allows to map a value from a given interval to another interval. The first argument is the value to map, the second and third arguments are the lower and upper bounds of the interval to map from, the fourth and fifth arguments are the lower and upper bounds of the interval to map to.
``````ex=fppc.expr("mapabcd(x,-1,1,0,1):x")
x=np.linspace(-1,1,100)
y=ex.evaluate("x",x)
fig = px.line(x=x, y=y, title="f",labels={"x":"x","y":r'y'})
fig.show()``````
Results
### 3.4. step1
The function `step1` is a step function that is equal to 1 if the argument is greater than 0 and 0 otherwise.
``````ex=fppc.expr("step1(x,2):x")
x=np.linspace(0,2*math.pi,100)
y=ex.evaluate("x",x)
fig = px.line(x=x, y=y, title="step1",labels={"x":"x","y":r'y'})
fig.show()``````
Results
### 3.5. smoothstep
The smoothstep function is a smooth approximation of the step function.
``````ex=fppc.expr("smoothstep(sin(x),0,0.5):x")
x=np.linspace(0,2*math.pi,100)
y=ex.evaluate("x",x)
fig = px.line(x=x, y=y, title="smoothstep",labels={"x":"x","y":r'y'})
fig.show()``````
Results
### 3.6. pulse
The pulse function is defined as $\text{pulse}(x)=\begin{cases} 1 & \text{if } x\in[0,1] \\ 0 & \text{otherwise} \end{cases}$
``````ex=fppc.expr("pulse(x,0,1,2):x")
x=np.linspace(0,2*math.pi,100)
y=ex.evaluate("x",x)
fig = px.line(x=x, y=y, title="rectangle pulse(0,1,2)",labels={"x":"x","y":r'y'})
fig.show()``````
Results
### 3.7. rectangle
The rectangle function is defined as $\text{rectangle}(x)=\begin{cases} 1 & \text{if } x\in[0,1] \\ 0 & \text{otherwise} \end{cases}$
``````ex=fppc.expr("rectangle(x,0,1):x")
x=np.linspace(0,2*math.pi,100)
y=ex.evaluate("x",x)
fig = px.line(x=x, y=y, title="rectangle rectangle(0,1)",labels={"x":"x","y":r'y'})
fig.show()``````
Results
### 3.8. triangle
The triangle function is defined as $\text{triangle}(x)=1-\left|\chi_{[a,b]} \left(-1+2\frac{x-a}{b-a} \right) \right| ], \quad x\in[a,b]$
``````ex=fppc.expr("triangle(x,0,1):x")
x=np.linspace(0,2*math.pi,100)
y=ex.evaluate("x",x)
fig = px.line(x=x, y=y, title="rectangle triangle(0,1)",labels={"x":"x","y":r'y'})
fig.show()``````
Results
### 3.9. sinwave
The `sinwave` function is a periodic function with a given period and amplitude.
``````ex=fppc.expr("sinewave(x,2,0.5):x")
x=np.linspace(0,2*math.pi,200)
y=ex.evaluate("x",x)
fig = px.line(x=x, y=y, title="sinewave(4*pi*x+0.5)",labels={"x":"x","y":r'y'})
fig.show()``````
Results | 2,987 | 8,992 | {"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} | 2.984375 | 3 | CC-MAIN-2024-33 | latest | en | 0.669614 |
http://mjessaypybq.nextamericanpresident.us/10x-pi-day-rule-for-2014.html | 1,540,079,596,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583513508.42/warc/CC-MAIN-20181020225938-20181021011438-00091.warc.gz | 258,974,060 | 3,852 | # 10x pi day rule for 2014
Two teasers – one gyral and one viral – to usher in the annual celebration of maths. March 14, 2014 share if you've spent 314 seconds on twitter or facebook today, you probably know that it's march 14, or 3/14, or pi day ah yes, pi day, celebrating a mathematical constant, a number with way too many decimal places, a number used to calculate the diameter, radius and circumference.
Today is pi day on march 14th, also noted as 314, math fans across the world celebrate a mathematical constant we all know and love: the ratio of any euclidean plane circle's circumference to its diameter, or the ratio of a circle's area to the square of its radius it's irrational and transcendental, but you.
## 10x pi day rule for 2014
• 14, 2014 , 4:00 pm today is national pi day in case you haven't figured out the connection yet, it's 14 march—written 3/14 in the american calendar system— and we get to celebrate the number pi, 314159, by eating pie, (preferably at 1:59 pm) pi day has been unofficially celebrated by universities.
March 14 is my favorite day to be a nerd across the country, math geeks in museums, schools, private groups and elsewhere gather to celebrate the number pi, approximately 314 that's why march 14 -- 3-14 -- is pi day what's more, albert einstein was born on this day a quick refresher: pi is defined as. Today and every march 14th (3/14) is a day of much reverence for math geeks worldwide better known as pi day, given its connection to perhaps the most well -known mathematical constant that begins with 314 and goes on with an infinite march of decimals to commemorate the day this week in austin.
10x pi day rule for 2014
Rated 3/5 based on 11 review
2018. | 433 | 1,717 | {"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-2018-43 | latest | en | 0.952276 |
http://www.brainia.com/essays/Bus-640-Ash-Course-Tutorial-Uophelp/326739.html | 1,498,713,827,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323870.46/warc/CC-MAIN-20170629051817-20170629071817-00668.warc.gz | 481,190,430 | 5,815 | BUS 640 ASH Course Tutorial /uophelp
BUS 640 ASH Course Tutorial /uophelp
BUS 640 Week 1 Economics of Risk and Uncertainty Applied Problems
For more course tutorials visit
www.uophelp.com
Economics of Risk and Uncertainty Applied Problems. Please, complete the following 3 applied problems in a Word or Excel document. Show all your calculations and explain your results. Submit your assignment in the drop box by using the Assignment Submission button.
1. A generous university benefactor has agreed to donate a large amount of money for student scholarships. The money can be provided in one lump-sum of \$10mln, or in parts, where \$5.5mln can be provided in year 1, and another \$5.5mln can be provided in year 2. Assuming the opportunity interest rate is 6%, what is the present value of the second alternative? Which of the two alternatives should be chosen and why?
How would your decision change if the opportunity interest rate was 12%? Please, show all your calculations.
...........................................................................................................................
BUS 640 Week 3 Production Cost Analysis and Estimation Applied Problems
For more course tutorials visit
www.uophelp.com
Production Cost Analysis and Estimation Applied Problems. Please, complete the following 3 applied problems in a Word or Excel document. Show all your calculations and explain your results. Submit your assignment in the drop box by using the Assignment Submission button.
1. Jennifer Trucking Company operates a large rig transportation business in Texas that transports locally grown vegetables to San Diego, California. The company owns 5 large rigs and hires local drivers paid fixed salaries monthly, regardless of the number of trips or tons of cargo that each driver transports each month.
2. The Palms Dry Cleaning Shop in Fort Lauderdale, Florida, faces a highly seasonal demand for its services, as the snow-birds retirees flock to Florida in mid-fall to enjoy the mild winter weather and then return to their... | 409 | 2,055 | {"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-2017-26 | latest | en | 0.860014 |
https://www.coursehero.com/file/69772758/Summer-2016-Finalpdf/ | 1,637,989,840,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358118.13/warc/CC-MAIN-20211127043716-20211127073716-00310.warc.gz | 819,463,599 | 58,479 | # Summer 2016 Final.pdf - MAT1322-3X Solution to Final...
• 8
This preview shows page 1 - 4 out of 8 pages.
MAT1322-3X Solution to Final Examination (A) Summer 2016 Solution to Final Examination (A) MAT1322-3X, Summer 2016 Part I. Multiple-choice Questions (3 u 10 = 30 marks) CECAE DCDAB 1. The area of the region under the graph of y = − 2 x 2 + 9 and above the graph of y = x 2 6 x is (A) 22; (B) 27; (C) 32; (D) 37; Solution . (C) The intersections of these curves are found by the equation 2 x 2 + 9 = x 2 6 x , 3 6 x 9 = 0, or x 2 2 x 3 = 0. Then x = − 1, x A = 3 3 3 2 2 2 3 1 1 1 ( 2 9 6 ) ( 3 6 9) 3 9 32 x x x x dx x x dx x x x ± ± ± ª º ± ² ± ² ± ² ² ± ² ² ¬ ¼ ³ ³ (E) 42. x 2 = 3. 2
1 3. Let R be the region between the graph y = 2 x and the x- axis, 0 d x d 1. A solid has R base, and the cross sections perpendicular to the x -axis are squares. The solid is shown in the following figure: as its
2. Let R be the region above the parabola y = x 2 and under the line y = 2 x . Solid B is obtained by revolving R about the y -axis . Then the volume of B is calculated by the integral 2 2 2 2 x x
MAT1322-3X Solution to Final Examination (A) Summer 2016 The volume of the solid is (E) 3. 2 X Y Z (1, 0, 0) 2 y x R (1, 2, 0)
2
4. A container has the shape of an inverted circular cone as in the following figure: 2 6 4 2 6
MAT1322-3X Solution to Final Examination (A) Summer 2016 layer to a point 2 meters above the top of the container is dW = U g (3 x / 4) 2 (6 ± x ) S dx . The bottom layer has x = 0, and the top layer has x = 4. The total work needed is W = 4 2 0 9 (6 ) 16 g x x dx U S ± ³ . 5. Consider improper integral ² ³ . Which one of the following argument is true? 2 1 x dx x x f ±
3 | 665 | 1,709 | {"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-2021-49 | latest | en | 0.839373 |
http://www.usatestprep.com/ga/math-pso-test | 1,513,332,957,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948568283.66/warc/CC-MAIN-20171215095015-20171215115015-00133.warc.gz | 451,865,253 | 9,624 | # Georgia Math Post-Secondary Option Practice
Discover the most effective and comprehensive online solution for curriculum mastery, high-stakes testing, and assessment in Georgia. Our Math Post-Secondary Option curriculum and test review is aligned to the most current Georgia standards. Request your free trial and see why our users say USATestprep has improved their students' pass rates.
Number Sense 20% Linear Equations and Inequalities 20% Non-Linear Relationships 20% Graphs and Functions 20% Measurement and Geometry 20%
• Questions: 2,666
• Technology Enhanced Items: 131
• Instructional Videos: 177
• Vocabulary Terms: 205
### Test Standards
Number Sense 1. (MPSOMNS1) Use Place Value 2. (MPSOMNS2) Operations 3. (MPSOMNS3) Fractions, Decimals, Percents 4. (MPSOMNS4) Negative Rationals 5. (MPSOMNS5) Rational Applications Linear Equations and Inequalities 1. (MPSOMLEI1) Equations And Inequalities 2. (MPSOMLEI2) Variation And Proportions 3. (MPSOMLEI3) Graph Equations And Inequalities 4. (MPSOMLEI4) Use Slope 5. (MPSOMLEI5) Solve Systems 6. (MPSOMLEI6) Systems Of Inequalities Non-Linear Relationships 1. (MPSOMNLR1) Square Roots 2. (MPSOMNLR2) Right Triangle Problems 3. (MPSOMNLR3) Polynomial Operations 4. (MPSOMNLR4) Quadratic Equations 5. (MPSOMNLR5) Rational Expressions 6. (MPSOMNLR6) Rational Equations Graphs and Functions 1. (MPSOMGF1) Identify Functions 2. (MPSOMGF2) Graph Functions 3. (MPSOMGF3) Application Problems Measurement and Geometry 1. (MPSOMMG1) Convert 2. (MPSOMMG2) Change Systems 3. (MPSOMMG3) 2-D And 3-D Figures 4. (MPSOMMG4) Perimeters, Areas, Etc. | 527 | 1,621 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2017-51 | latest | en | 0.485434 |
https://garlicspace.com/2020/07/25/leetcode-construct-binary-tree-from-inorder-and-postorder-traversal/ | 1,726,746,531,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652028.28/warc/CC-MAIN-20240919093719-20240919123719-00035.warc.gz | 240,491,663 | 12,575 | # LeetCode – Construct Binary Tree from Inorder and Postorder Traversal
### 题目：
Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
For example, given
```inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]```
Return the following binary tree:
``` 3
/ \
9 20
/ \
15 7```
### 实现：
• 递归方式，根据二叉树中序遍历和后续遍历，创建二叉树。中序遍历： 左-根-右， 后续遍历： 左-右-根， 所以可以通过后续遍历的最后一个节点找到根节点，并生成根节点， 因为二叉树没有重复节点， 就 可以在中序遍历数据中找到根节点，再分成左右两个子树，递归执行下去，遍历完所有节点。 通过postorder获取根节点， 通过inorder生成左右子树，查找节点在中序遍历数据组里位置，可以通过unordered_map实现。
• 不使用map情况下 时间复杂度 O(N^2) , 使用map查找O(N)
``````/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
unordered_map <int, int> treemap;
for (int i=0; i<inorder.size();i++) {
treemap[inorder[i]] = i;
}
int index = inorder.size()-1;
return helper(inorder, postorder, 0, inorder.size()-1, &index, treemap);
}
TreeNode *helper(vector<int> & inorder, vector<int>& postorder, int left, int right,
int *index , unordered_map<int, int>&treemap){
//check
if (left > right){
return NULL;
}
// create root node
/* destory postorder vector
int val = postorder.back();
TreeNode *node = new TreeNode(val);
postorder.pop_back();
*/
int val = postorder[*index];
TreeNode *node = new TreeNode(val);
(*index)--;
// search inorder by array
/*
int root = -1;
for (int i=left ; i <=right; i ++){
if (val == inorder[i]){
root = i;
break;
}
}
if (root == -1){
return node;
}*/
// search inorder by map
int root = treemap[val];
// create left & right node
node->right = helper(inorder, postorder, root+1, right, index, treemap);
node->left = helper(inorder, postorder, left, root-1, index, treemap);
return node;
}
};`````` | 706 | 2,088 | {"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} | 4.125 | 4 | CC-MAIN-2024-38 | latest | en | 0.173513 |
http://mysticalnumbers.com/destiny-number/ | 1,722,803,658,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640412404.14/warc/CC-MAIN-20240804195325-20240804225325-00806.warc.gz | 22,075,605 | 17,646 | # Destiny Number
## How to Calculate Your Destiny Number
Your destiny number is always calculated from the name you have on your birth certificate. Use the name exactly as found on the birth certificate even if you have never used it since.
Always use the official name on your birth certificate.
For example if the name on your birth certificate is Margaret but all your life you have been called Peggy, you still have to use Margaret to calculate your destiny number.
Your first name will tell you something about your life’s purpose. If you have a middle name it will say something about concealed capabilities. Your last name reveals some of the traits you may inherit from your family.
Each letter is given a numeral value.
To figure out what your destiny number is start by writing your birth name on a sheet of paper.
Under each letter in your name you need to write down the numeral value of the letter.
For example:
John Peter Thomson
1685-75259-2864165
Next you add the digits in each name until you only have one digit left:
1+6+8+5=20 (2+0=2 ) The destiny number for John is 2
7+5+2+5+9=28 (2+8=10) (1+0=1) The destiny number for Peter is 1
2+8+6+4+1+6+5=32 (3+2=5) The destiny number for Thomson is 5
Finally add the destiny numbers of your individual names until you are left with one digit.
For John Peter Thomson it would be 2+1+5=8. His destiny number is 8.
If you have a suffix to your name such as Junior (Jr.), Senior (Sr.), The third (III) or whatever you do not use these when calculating your destiny number.
You may argue that in that case your destiny number will be exactly the same as another family member.
That is true, but your final numerology chart will end up to quite different as your destiny number is only part of the equation.
Other parts of numerology include your soul number, life path number, personality number and maturity number.
The only exception to reducing your numbers to one final digit (also called the root number) is if you in the process end up with the 11, 22 or 33.
These numbers are called Master numbers and are quite exceptional.
The Master numbers are loaded with energy and need very special attention.
Here is a short overview of the destiny numbers.
Number 1
People with this number should strive to be leaders. They need to be the one who take initiative and take control. These people think for themselves and do not automatically follow the crowd.
Number 2
People with this number are team players. These are typically diplomats and have special abilities to create harmony.
Number 3
People with this number are very inspirational to others. They are positive and enthusiastic. They master the art of bringing out the best in others.
Number 4
People with the destiny number four are called to maintain order and security. This is about organizing and building something that will last for a long time. These people work hard.
Number 5
These people should have freedom. Their destiny is to explore, discover new things and embrace change.
Number 6
These people are meant to love and care for their closest family and friends. They see the needs of others and do something about it.
Number 7
This is about going into to depth of whatever you are researching. Perfection is another virtue of the number 7. This person is to find wisdom and share that wisdom with others.
Number 8
This person should have success in business. It is all about authority and power. This is a road towards material wealth brought about with integrity.
Number 9
These people need to generously reach out to the world and work to unite, heal or help in whatever manner possible. This is unconditional love that goes far beyond the immediate family.
### 16 thoughts on “Destiny Number”
1. do we keep our life path number and our destiny number seperate? or do we add those together? because if I add them both, I get a total of 11.
2. Read the text carefully, it says the only exception to reducing your numbers to one final digit (also called the root number) is if you in the process end up with the 11, 22, or 33.
These are known as root numbers or master numbers. Quite exceptional people loaded with energy.
The author could have been kind to explain these numbers and what their characteristics are?
3. Pingback: What Is My Destiny Numerology – Paula R Courtney
4. The search for the last Plutonian
5. I actually have 33 for my bday, and when I calculate my name the middle is coming to 11. This is really getting tricky.
6. It says add up the numbers until there is only one digit. So for anyone saying they have a number higher than 9 please read the directions more carefully.
• Are you related to the Schmoes from Antigo?
• Read the text carefully, it says the only exception to reducing your numbers to one final digit (also called the root number) is if you in the process end up with the 11, 22, or 33.
These are known as root numbers or master numbers. Quite exceptional people loaded with energy.
The author could have been kind to explain these numbers and what their characteristics are?
7. i got 22 what does it mean ?
• I understand it to mean extra energy, but you are a four.
Keep adding until you only have one digit. All numbers that are even tens, adding a zero, there is also a special energy to that.
8. So if it’s an 11 what do I look up to read it only went to 9 and all it said was it’s a master number
• My understanding is eleven has extra energy, but becomes a two. | 1,233 | 5,493 | {"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-2024-33 | latest | en | 0.883731 |
https://www.math.fsu.edu/~ballas/260p/lectures/lecture8/ | 1,624,114,738,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487648373.45/warc/CC-MAIN-20210619142022-20210619172022-00072.warc.gz | 788,924,092 | 4,260 | We have a properly convex domain with a Hilbert metric. We define the distance between two points $$x,y$$ by $$d_{\Omega}(x,y) = log([a:x:y:b])$$. If we have a projective transformation that preseves cross ratios we call it an isometry of $$d_{\Omega}$$.
If we restrict ourselves to matrices with determinant plus or minus one $$SL^{\pm} (\Omega)$$ and then lift to the sphere (since the double cover of a projective space $$RP^{n}$$ is a sphere $$S^n$$) and choose one of the preimages $$\tilde{\Omega}$$ with the property that $$\tilde{\Omega}$$ maps to itself it becomes much simpler to talk about eigen values and classify isometries.
$$A$$ is called elliptic if there exists $$p \in \Omega$$ such that $$Ap=p$$. Here $$p=[v]$$ the equivalence class of some vector i.e. $$[Av]=[v]$$. This implies that $$Av = \lambda v$$ for some $$\lambda > 0$$. Hence fixed points coresspond to eigenvectors of the matrix with positve eigenvalues. $$A$$ is parabolic if every eigenvalue of $$A$$ has modulus $$1$$, but $$A$$ is not elliptic. Finally we say that $$A$$ is parabolic otherwise.
We define the translation length $$t(A)=\inf_{x \in \Omega} d_{\Omega}(x,Ax) \geq 0$$ and also define the $$minset(A)=\{x \in \Omega \mid d_{\Omega}(x,Ax) = t(A)\}$$ of an isometry. In Hyperbolic geometery $$H^{n}$$ $$t(A) > 0$$ if and only if $$A$$ is hyperbolic. In this case points along the geodesic are moved the minimum amount i.e. the minset lies in some geodesic, called the axis. If $$t(A)=0$$ then $$A$$ is elliptic if and only if the minset is nonempty and parabolic (along with a unique point $$p \in \partial \tilde{\Omega}$$ called the parabolic fixed point) if and only if the minset is empty.
Proposition: Let $$A \in SL(\Omega)$$, then $$t(A)=log\vert\lambda / \mu\vert$$ where $$\lambda$$ is an eigenvalue of largest modulus and $$\mu$$ is an eigenvalue of smallest modulus.
Proposition: $$A \in SL(\Omega) \subset SL^{\pm}(n+1,R)$$ is elliptic if and only if $$A$$ is conjugate onto $$O(n+1)$$
Proof: By the definition of elliptic there is some point in the domain $$p \in A$$ that is fixed by the elliptic. Here we have a diffeomorphism so $$dA$$ acts on $$T_p \simeq R^{n}$$. $$d_{\Omega}$$ is a norm on $$T_p(\Omega)$$ specifically a Finsler metric. $$dA$$ is an isometry of $$T_p(\Omega)$$, which implies there exists an inner product $$\beta$$ on $$T_p(\Omega)$$ such that $$dA$$ is an isometry of $$\beta$$. Implies $$dA \in Isom(\beta)=O(n)$$. (We did not finish proof in lecture, but this implies the conclusion apparently.)
$$\bar{\Omega}$$ is homeomorphic to some $$[0,1]^n$$ so any isometry of $$\Omega$$, $$A \in SL(\Omega)$$ it preserves the closure $$A(\bar{\Omega})=\bar{\Omega}$$. The Brower fixed point theorem implies that there exists $$p \in \bar{\Omega}$$ such that $$Ap=p$$.
Case1: The fixed point is in the interior $$p \in \Omega$$ then $$A$$ is an elliptic.
Case2: The fixed point is on the boundary $$p \in \partial \bar{\Omega}$$. (In addition it fixes a hyperplane containing that point on the boundary.)
Define a supporting hyper plane $$H$$ to $$\Omega$$ at $$p \in \partial \bar{\Omega}$$ is a projective hyperplane $$H=P(V)$$ with $$dim(V)=n$$, such that $$p \in H$$ and $$H \cap \Omega = \emptyset$$.
Exercise: Supporting hyperplanes exist.
Proposition: The set of supporting hyperplanes $$\{P(ker \phi) : [\phi] \in CC(RP^n)^*\}$$ is dual to a nonempty compact convex set in $$(RP^n)^*$$.
Proof: $$\Omega \subset RP^n$$ and $$\Omega^* \subset RP^n$$ a linear map is in the dual domain if and only if $$P(ker \phi)$$ misses the interior of the domain, $$\Omega^* = P(\{\phi \in V^* \mid \phi(\Omega) >0\})$$. $$\Omega$$ must be in $$S^n \subset R^{n+1} = V$$ for this to make sense, so we have to lift to the double cover ($$V$$ a vector space).
Corollary: If $$A \in SL(\Omega)$$ fixes a point in the boundary of the ball $$p \in \partial \Omega$$ then there exists a supporting hyperplane $$H$$ to $$\Omega$$ at $$p$$ such that $$A(H)=H$$
Proof: $$H$$ corresponds to a point in a dual compact convex set $$C$$. $$A$$ sends supporting hyperplanes to supporting hyperplanes, so $$A: C \rightarrow C$$. Then by the Brower Fixed Point Theorem there exists $$[\phi] \in C$$ such that $$A[\phi]=[\phi]$$. Therefore $$H=ker \phi$$ is a supporting hyperplanefixed by $$A$$.
Structure of fixed set of $$A \in SL(\Omega)$$: Let $$p=[x]$$ for $$x \in R^{n+1}$$, $$A$$ fixes $$p$$ if and only if there is a positive number $$\lambda$$ such that $$Ax=\lambda x$$. In other word equivalent to $$x$$ being and eigenvector with a positive eigenvalue. Given an eigenvalue $$\lambda>0$$ for $$A$$ we have the eigenspace $$E_\lambda$$. Define the Fixset $$Fix(\lambda, A, \bar{\Omega})= \bar{\Omega} \cap P(E_{\lambda})$$ this will be compact and convex as both $$\bar{\Omega}$$ and $$P(E_{\lambda})$$ are. The set of all fixed points will be $$Fix(A,\Omega) = \bigcup_{\lambda > 0} Fix(\lambda, A, \Omega)$$.
Example: Suppose $$\Omega = \Delta$$ then
$SL(\Omega) = \{ ()~a,b > 0 \} \rtimes \{permutation~matrices\}$
Consider $$A=\{ \begin{pmatrix} 2&0&0 \\ 0 & 2 & 0 \\ 0&0&1/4 \end{pmatrix}$$ then the fixed points we have are $$Fix(\lambda =2,A,\Delta)$$ and $$Fix(\lambda = 1/4,A, \Delta)$$.
Dynamics of $$[A] \in P_{+}GL(n+1,R)$$:
First of all the largest eigenvalues will ‘win’ (i.e. they will blow up the fastest as a matrix is multiplied by itself repeatedly) hence we need only consider the largest eigenvalues. Consider $$A=\{ \begin{pmatrix} 2&0&0 \\ 0 & 2 & 2 \\ 0&0&2 \end{pmatrix}$$ then $$A^n=2^n \{ \begin{pmatrix} 1&0&0 \\ 0 & 1 & n \\ 0&0&1 \end{pmatrix}$$. From this we can see that the largest Jordan block will blow up fastest therefore we will just talk about the dynamics of a single $$k+1 \times k+1$$ Jordan block. The growth of some block is $$A^{n}=\lambda^n (I + nN + \frac{n(n-1)}{2}N^2+ ... + (\frac{n}{k}N^k))$$ the last term wins as $$N \rightarrow \infty$$. We define the power of a Jordan blonck with eigenvalue $$\lambda \in C$$ is $$(\vert\lambda\vert,k+1)$$ and we use the lexicographical order. We get a partial ordering on Jordan blocks from this.
Previous Post: Lecture 7: The Margulis Lemma | 1,929 | 6,169 | {"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} | 3.46875 | 3 | CC-MAIN-2021-25 | latest | en | 0.801119 |
http://seaintarchive.org/mailarchive/2003c/msg01505.html | 1,492,970,642,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917118740.31/warc/CC-MAIN-20170423031158-00377-ip-10-145-167-34.ec2.internal.warc.gz | 364,845,391 | 4,675 | Need a book? Engineering books recommendations...
# Re: Curved Stair Stringer
• To: <seaint(--nospam--at)seaint.org>
• Subject: Re: Curved Stair Stringer
• Date: Thu, 25 Sep 2003 10:01:31 -0400
```Is it a one stringer or two parallel stringers?
Regards,
vcernescu(--nospam--at)banerjee.com
----- Original Message -----
From: "richard lewis" <rlewistx(--nospam--at)juno.com>
To: <seaint(--nospam--at)seaint.org>
Sent: Thursday, September 25, 2003 9:46 AM
Subject: Curved Stair Stringer
> I looking for some tips for modeling a curved beam. I have an circular
> custom stair. It is supported at the 1st floor (slab on ground) and 2nd
> floor. It looks something like a slotted hole in plan. From the first
> floor it does a 1/4 turn. Then there is a straight run, then another 1/4
> turn back to the 2nd floor. I am supporting it with a tube at the second
> floor and anchoring it to the slab at the first floor. The tube will give
> me torsional resistance. The tube is about 16 ft. long and spans to 2
> perpendicular beams, one at each end. The beams then span to columns
> buried in the wall. Looking at this in plan view the 2nd floor beams
> would form an 'H' shape.
> I want to get feedback for modeling the stringers in a computer frame
> analysis. I guess my first impulse is to use short chord elements to
> simulate a curved member. I started doing this with 15 degree angles for
> the chords. I would then uniformly load these members and have a fixed
> joint at the tube column at the 2nd floor. I guess the connection to the
> floor slab at the first floor would be pinned. I would use a moment
> frame from the vertical legs of the 'H' shape to give me lateral
> stability. Any other suggestions for modeling this?
> I am wondering what would be the best section for a stringer. Since it
> is a curved member it will have torsion. Should I be using a closed
> section like a tube? A tube might be hard to bend on the radius. Should
> I make a closed section from welded curved plates? Should I just use a
> single thick plate for a stringer?
>
> Rich
>
> ________________________________________________________________
> The best thing to hit the internet in years - Juno SpeedBand!
> Surf the web up to FIVE TIMES FASTER!
>
> ******* ****** ******* ******** ******* ******* ******* ***
> * Read list FAQ at: http://www.seaint.org/list_FAQ.asp
> *
> * This email was sent to you via Structural Engineers
> * Association of Southern California (SEAOSC) server. To
> * subscribe (no fee) or UnSubscribe, please go to:
> *
> * http://www.seaint.org/sealist1.asp
> *
> * Questions to seaint-ad(--nospam--at)seaint.org. Remember, any email you
> * send to the list is public domain and may be re-posted
> * site at: http://www.seaint.org
> ******* ****** ****** ****** ******* ****** ****** ********
******* ****** ******* ******** ******* ******* ******* ***
* Read list FAQ at: http://www.seaint.org/list_FAQ.asp
*
* This email was sent to you via Structural Engineers
* Association of Southern California (SEAOSC) server. To
* subscribe (no fee) or UnSubscribe, please go to:
*
* http://www.seaint.org/sealist1.asp
*
* Questions to seaint-ad(--nospam--at)seaint.org. Remember, any email you
* send to the list is public domain and may be re-posted | 912 | 3,304 | {"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-2017-17 | latest | en | 0.909893 |
https://www.physicsforums.com/threads/tensor-product-of-c-with-itself-over-r.675424/ | 1,511,264,985,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806353.62/warc/CC-MAIN-20171121113222-20171121133222-00709.warc.gz | 830,149,521 | 14,979 | # Tensor Product of C with itself over R.
1. Mar 1, 2013
### Monobrow
I am trying to prove that C$\otimes$C (taken over R) is equal to C^2. The method I have seen is to show the following equivalences:
C$\otimes$C = C$\otimes$(R[T]/<T^2+1>) = C[T]/<T^2+1> = C.
(All tensor products taken over R).
The only part I am having trouble with is showing that C$\otimes$(R[T]/<T^2+1>) = C[T]/<T^2+1>. I have tried to show this by using the universal property of the tensor product. First I defined a bilinear map from CXR[T]/<T^2+1> to C[T]/<T^2+1> by sending an element (z,f(T)) to zf(t) (where I am omitting the cosets). This then induces a linear map from the tensor product. However, I cannot seem to find an inverse for this map.
Any help would be greatly appreciated. | 244 | 770 | {"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} | 2.5625 | 3 | CC-MAIN-2017-47 | longest | en | 0.89637 |
https://www.qb365.in/materials/stateboard/12th-standard-maths-english-medium-free-online-test-book-back-1-mark-questions-with-answer-key-part-three-6115.html | 1,621,361,477,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991288.0/warc/CC-MAIN-20210518160705-20210518190705-00295.warc.gz | 1,012,601,775 | 31,634 | #### 12th Standard Maths English Medium Free Online Test Book Back 1 Mark Questions with Answer Key - Part Three
12th Standard
Reg.No. :
•
•
•
•
•
•
Maths
Time : 00:10:00 Hrs
Total Marks : 10
10 x 1 = 10
1. If A is a non-singular matrix such that A-1 = $\left[ \begin{matrix} 5 & 3 \\ -2 & -1 \end{matrix} \right]$, then (AT)−1 =
(a)
$\left[ \begin{matrix} -5 & 3 \\ 2 & 1 \end{matrix} \right]$
(b)
$\left[ \begin{matrix} 5 & 3 \\ -2 & -1 \end{matrix} \right]$
(c)
$\left[ \begin{matrix} -1 & -3 \\ 2 & 5 \end{matrix} \right]$
(d)
$\left[ \begin{matrix} 5 & -2 \\ 3 & -1 \end{matrix} \right]$
2. If (1+i)(1+2i)(1+3i)...(1+ni)=x+iy, then $2\cdot 5\cdot 10...\left( 1+{ n }^{ 2 } \right)$ is
(a)
1
(b)
i
(c)
x2+y2
(d)
1+n2
3. According to the rational root theorem, which number is not possible rational root of 4x7+2x4-10x3-5?
(a)
-1
(b)
$\frac { 5 }{ 4 }$
(c)
$\frac { 4 }{ 5 }$
(d)
5
4. If |x|$\le$1, then 2tan-1 x-sin-1 $\frac{2x}{1+x^2}$ is equal to
(a)
tan-1x
(b)
sin-1x
(c)
0
(d)
$\pi$
5. If P(x, y) be any point on 16x2+25y2=400 with foci F1 (3,0) and F2 (-3,0) then PF1 PF2 +
is
(a)
8
(b)
6
(c)
10
(d)
12
6. If the two tangents drawn from a point P to the parabolay2 = 4x are at right angles then the locus of P is
(a)
2x+1=0
(b)
x = −1
(c)
2x−1=0
(d)
x =1
7. Distance from the origin to the plane 3x - 6y + 2z 7 = 0 is
(a)
0
(b)
1
(c)
2
(d)
3
8. The maximum value of the function x2 e-2x,
(a)
$\cfrac { 1 }{ e }$
(b)
$\cfrac { 1 }{ 2e }$
(c)
$\cfrac { 1 }{ { e }^{ 2 } }$
(d)
$\cfrac { 4 }{ { e }^{ 4 } }$
9. If w (x, y, z) = x2 (v - z) + y2 (z - x) + z2(x - y), then $\frac { { \partial }w }{ \partial x } +\frac { \partial w }{ \partial y } +\frac { \partial w }{ \partial z }$ is
(a)
xy + yz + zx
(b)
x(y + z)
(c)
y(z + x)
(d)
0
10. The value of $\int _{ 0 }^{ a }{ { (\sqrt { { a }^{ 2 }-{ x }^{ 2 } } ) }^{ 2 } } dx$
(a)
$\frac { { \pi a }^{ 2 } }{ 16 }$
(b)
$\frac { 3\pi { a }^{ 4 } }{ 16 }$
(c)
$\frac { 3\pi { a }^{2 } }{ 8}$
(d)
$\frac { 3\pi { a }^{ 4 } }{ 8}$ | 942 | 2,073 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2021-21 | latest | en | 0.272199 |
http://www.gregthatcher.com/Stocks/StockFourierAnalysisDetails.aspx?ticker=HABDX | 1,490,250,643,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218186780.20/warc/CC-MAIN-20170322212946-00165-ip-10-233-31-227.ec2.internal.warc.gz | 536,912,182 | 115,165 | Back to list of Stocks See Also: Seasonal Analysis of HABDXGenetic Algorithms Stock Portfolio Generator, and Fourier Calculator
# Fourier Analysis of HABDX (Harbor Bond Fund Insti Cl)
HABDX (Harbor Bond Fund Insti Cl) appears to have interesting cyclic behaviour every 152 weeks (.3625*sine), 117 weeks (.2663*sine), and 139 weeks (.2553*sine).
HABDX (Harbor Bond Fund Insti Cl) has an average price of 5.98 (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 12/29/1987 to 3/13/2017 for HABDX (Harbor Bond Fund Insti Cl), this program was able to calculate the following Fourier Series:
Sequence #Cosine Coefficients Sine Coefficients FrequenciesPeriod
05.9836 0
1.62678 -3.37504 (1*2π)/15241,524 weeks
2.06214 -1.72636 (2*2π)/1524762 weeks
3-.16887 -1.08393 (3*2π)/1524508 weeks
4-.00742 -.68591 (4*2π)/1524381 weeks
5.0406 -.69734 (5*2π)/1524305 weeks
6.04245 -.50292 (6*2π)/1524254 weeks
7-.02033 -.42442 (7*2π)/1524218 weeks
8.00227 -.37308 (8*2π)/1524191 weeks
9-.01317 -.35389 (9*2π)/1524169 weeks
10-.01849 -.36247 (10*2π)/1524152 weeks
11-.02247 -.2553 (11*2π)/1524139 weeks
12.00145 -.19421 (12*2π)/1524127 weeks
13.03494 -.26631 (13*2π)/1524117 weeks
14.00191 -.2569 (14*2π)/1524109 weeks
15.0157 -.2321 (15*2π)/1524102 weeks
16.00571 -.19969 (16*2π)/152495 weeks
17-.03059 -.20738 (17*2π)/152490 weeks
18-.04429 -.16869 (18*2π)/152485 weeks
19.01294 -.1404 (19*2π)/152480 weeks
20.00624 -.17428 (20*2π)/152476 weeks
21-.02857 -.1598 (21*2π)/152473 weeks
22-.02208 -.12251 (22*2π)/152469 weeks
23.00054 -.14233 (23*2π)/152466 weeks
24-.03561 -.14013 (24*2π)/152464 weeks
25-.0247 -.10746 (25*2π)/152461 weeks
26-.0022 -.10599 (26*2π)/152459 weeks
27-.01025 -.10589 (27*2π)/152456 weeks
28-.0218 -.10568 (28*2π)/152454 weeks
29.00259 -.11228 (29*2π)/152453 weeks
30-.00292 -.09365 (30*2π)/152451 weeks
31.00435 -.09517 (31*2π)/152449 weeks
32-.00463 -.10519 (32*2π)/152448 weeks
33-.02155 -.08776 (33*2π)/152446 weeks
34-.02044 -.07657 (34*2π)/152445 weeks
35.00218 -.07499 (35*2π)/152444 weeks
36.0123 -.08934 (36*2π)/152442 weeks
37-.00796 -.08039 (37*2π)/152441 weeks
38-.00652 -.07991 (38*2π)/152440 weeks
39-.0078 -.06766 (39*2π)/152439 weeks
40.01257 -.0654 (40*2π)/152438 weeks
41-.01463 -.06972 (41*2π)/152437 weeks
42-.00276 -.07186 (42*2π)/152436 weeks
43-.00659 -.07534 (43*2π)/152435 weeks
44.00593 -.06175 (44*2π)/152435 weeks
45.00609 -.06974 (45*2π)/152434 weeks
46-.00087 -.07229 (46*2π)/152433 weeks
47-.00073 -.07229 (47*2π)/152432 weeks
48-.0208 -.06954 (48*2π)/152432 weeks
49-.00976 -.05297 (49*2π)/152431 weeks
50-.00763 -.05373 (50*2π)/152430 weeks
51-.00121 -.06005 (51*2π)/152430 weeks
52-.0029 -.05341 (52*2π)/152429 weeks
53-.00157 -.0604 (53*2π)/152429 weeks
54-.00867 -.0642 (54*2π)/152428 weeks
55-.00305 -.05795 (55*2π)/152428 weeks
56-.00708 -.05488 (56*2π)/152427 weeks
57-.00794 -.05246 (57*2π)/152427 weeks
58.00099 -.05012 (58*2π)/152426 weeks
59-.00251 -.05309 (59*2π)/152426 weeks
60-.00059 -.05097 (60*2π)/152425 weeks
61-.00178 -.05587 (61*2π)/152425 weeks
62-.00415 -.05165 (62*2π)/152425 weeks
63-.00799 -.04602 (63*2π)/152424 weeks
64-.00559 -.05242 (64*2π)/152424 weeks
65-.00851 -.04982 (65*2π)/152423 weeks
66-.01357 -.045 (66*2π)/152423 weeks
67.00096 -.03783 (67*2π)/152423 weeks
68.00268 -.05278 (68*2π)/152422 weeks
69-.01034 -.05109 (69*2π)/152422 weeks
70-.00727 -.04568 (70*2π)/152422 weeks
71-.00833 -.04448 (71*2π)/152421 weeks
72-.00434 -.04548 (72*2π)/152421 weeks
73-.01252 -.03957 (73*2π)/152421 weeks
74.00182 -.0402 (74*2π)/152421 weeks
75-.00712 -.04497 (75*2π)/152420 weeks
76-.00328 -.04466 (76*2π)/152420 weeks
77-.01023 -.04302 (77*2π)/152420 weeks
78-.00424 -.03967 (78*2π)/152420 weeks
79-.00243 -.04699 (79*2π)/152419 weeks
80-.01439 -.0396 (80*2π)/152419 weeks
81-.00836 -.03427 (81*2π)/152419 weeks
82-.00626 -.03413 (82*2π)/152419 weeks
83-.00564 -.04174 (83*2π)/152418 weeks
84-.0133 -.03137 (84*2π)/152418 weeks
85-.00526 -.03492 (85*2π)/152418 weeks
86-.00494 -.03971 (86*2π)/152418 weeks
87-.01042 -.03738 (87*2π)/152418 weeks
88-.00826 -.02862 (88*2π)/152417 weeks
89-.00061 -.03628 (89*2π)/152417 weeks
90-.01024 -.03812 (90*2π)/152417 weeks
91-.0062 -.02995 (91*2π)/152417 weeks
92-.0067 -.03694 (92*2π)/152417 weeks
93-.00481 -.037 (93*2π)/152416 weeks
94-.01136 -.03446 (94*2π)/152416 weeks
95-.0052 -.032 (95*2π)/152416 weeks
96-.00199 -.02936 (96*2π)/152416 weeks
97-.00893 -.04138 (97*2π)/152416 weeks
98-.01356 -.03104 (98*2π)/152416 weeks
99-.00705 -.02875 (99*2π)/152415 weeks
100-.01136 -.03578 (100*2π)/152415 weeks
101-.0125 -.03158 (101*2π)/152415 weeks
102-.00704 -.02702 (102*2π)/152415 weeks
103-.00765 -.02744 (103*2π)/152415 weeks
104-.00475 -.0284 (104*2π)/152415 weeks
105-.00636 -.03222 (105*2π)/152415 weeks
106-.0029 -.02945 (106*2π)/152414 weeks
107-.01625 -.03159 (107*2π)/152414 weeks
108-.00567 -.02687 (108*2π)/152414 weeks
109-.00907 -.02722 (109*2π)/152414 weeks
110-.00686 -.02586 (110*2π)/152414 weeks
111-.00996 -.02793 (111*2π)/152414 weeks
112-.00505 -.02483 (112*2π)/152414 weeks
113-.00751 -.02559 (113*2π)/152413 weeks
114-.00631 -.02658 (114*2π)/152413 weeks
115-.00798 -.02812 (115*2π)/152413 weeks
116-.01247 -.02703 (116*2π)/152413 weeks
117-.01124 -.02267 (117*2π)/152413 weeks
118-.00859 -.01692 (118*2π)/152413 weeks
119.00057 -.0208 (119*2π)/152413 weeks
120-.00647 -.02133 (120*2π)/152413 weeks
121-.00502 -.02365 (121*2π)/152413 weeks
122-.00214 -.02596 (122*2π)/152412 weeks
123-.00626 -.02215 (123*2π)/152412 weeks
124-.00417 -.02405 (124*2π)/152412 weeks
125-.00382 -.02537 (125*2π)/152412 weeks
126-.00893 -.02609 (126*2π)/152412 weeks
127-.00669 -.02167 (127*2π)/152412 weeks
128-.00575 -.02568 (128*2π)/152412 weeks
129-.00696 -.02239 (129*2π)/152412 weeks
130-.00815 -.02257 (130*2π)/152412 weeks
131-.0048 -.0219 (131*2π)/152412 weeks
132-.00571 -.02148 (132*2π)/152412 weeks
133-.00588 -.02375 (133*2π)/152411 weeks
134-.00909 -.02021 (134*2π)/152411 weeks
135-.00567 -.01853 (135*2π)/152411 weeks
136-.00269 -.01967 (136*2π)/152411 weeks
137-.00043 -.026 (137*2π)/152411 weeks
138-.00979 -.02443 (138*2π)/152411 weeks
139-.00904 -.0223 (139*2π)/152411 weeks
140-.00742 -.02114 (140*2π)/152411 weeks
141-.00787 -.02143 (141*2π)/152411 weeks
142-.00692 -.01736 (142*2π)/152411 weeks
143-.00343 -.02016 (143*2π)/152411 weeks
144-.00622 -.02033 (144*2π)/152411 weeks
145-.00691 -.02027 (145*2π)/152411 weeks
146-.00624 -.0203 (146*2π)/152410 weeks
147-.00381 -.02428 (147*2π)/152410 weeks
148-.00754 -.019 (148*2π)/152410 weeks
149-.00706 -.01853 (149*2π)/152410 weeks
150-.00293 -.01896 (150*2π)/152410 weeks
151-.00679 -.02126 (151*2π)/152410 weeks
152-.00675 -.02202 (152*2π)/152410 weeks
153-.00514 -.01882 (153*2π)/152410 weeks
154-.00715 -.02264 (154*2π)/152410 weeks
155-.0089 -.01847 (155*2π)/152410 weeks
156-.00768 -.01711 (156*2π)/152410 weeks
157-.00494 -.02067 (157*2π)/152410 weeks
158-.0105 -.01805 (158*2π)/152410 weeks
159-.00395 -.016 (159*2π)/152410 weeks
160-.0009 -.01755 (160*2π)/152410 weeks
161-.0071 -.02018 (161*2π)/15249 weeks
162-.00517 -.01934 (162*2π)/15249 weeks
163-.00644 -.01876 (163*2π)/15249 weeks
164-.00453 -.01896 (164*2π)/15249 weeks
165-.00861 -.02225 (165*2π)/15249 weeks
166-.00908 -.01755 (166*2π)/15249 weeks
167-.00518 -.01645 (167*2π)/15249 weeks
168-.00513 -.01825 (168*2π)/15249 weeks
169-.00521 -.02175 (169*2π)/15249 weeks
170-.00945 -.01827 (170*2π)/15249 weeks
171-.007 -.01801 (171*2π)/15249 weeks
172-.00889 -.01625 (172*2π)/15249 weeks
173-.00962 -.01584 (173*2π)/15249 weeks
174-.00678 -.01035 (174*2π)/15249 weeks
175-.00417 -.01847 (175*2π)/15249 weeks
176-.0064 -.01914 (176*2π)/15249 weeks
177-.00526 -.01435 (177*2π)/15249 weeks
178-.0039 -.01548 (178*2π)/15249 weeks
179-.00586 -.02022 (179*2π)/15249 weeks
180-.00699 -.01805 (180*2π)/15248 weeks
181-.00557 -.01597 (181*2π)/15248 weeks
182-.00738 -.01799 (182*2π)/15248 weeks
183-.00756 -.01341 (183*2π)/15248 weeks
184-.00572 -.01586 (184*2π)/15248 weeks
185-.00519 -.01782 (185*2π)/15248 weeks
186-.00811 -.01823 (186*2π)/15248 weeks
187-.00797 -.01498 (187*2π)/15248 weeks
188-.0073 -.01263 (188*2π)/15248 weeks
189-.00869 -.01719 (189*2π)/15248 weeks
190-.00654 -.01474 (190*2π)/15248 weeks
191-.00609 -.01332 (191*2π)/15248 weeks
192-.00546 -.01627 (192*2π)/15248 weeks
193-.00717 -.01376 (193*2π)/15248 weeks
194-.00715 -.01515 (194*2π)/15248 weeks
195-.00532 -.01574 (195*2π)/15248 weeks
196-.00962 -.01727 (196*2π)/15248 weeks
197-.0086 -.01296 (197*2π)/15248 weeks
198-.00718 -.01297 (198*2π)/15248 weeks
199-.00859 -.01156 (199*2π)/15248 weeks
200-.00641 -.01291 (200*2π)/15248 weeks
201-.00397 -.01228 (201*2π)/15248 weeks
202-.00511 -.01594 (202*2π)/15248 weeks
203-.00763 -.01351 (203*2π)/15248 weeks
204-.00729 -.01517 (204*2π)/15247 weeks
205-.00429 -.01346 (205*2π)/15247 weeks
206-.00608 -.01517 (206*2π)/15247 weeks
207-.00501 -.01205 (207*2π)/15247 weeks
208-.00591 -.01324 (208*2π)/15247 weeks
209-.00528 -.01591 (209*2π)/15247 weeks
210-.00946 -.01537 (210*2π)/15247 weeks
211-.00716 -.0105 (211*2π)/15247 weeks
212-.00472 -.01336 (212*2π)/15247 weeks
213-.00629 -.01525 (213*2π)/15247 weeks
214-.00868 -.01369 (214*2π)/15247 weeks
215-.00721 -.01031 (215*2π)/15247 weeks
216-.00392 -.01347 (216*2π)/15247 weeks
217-.00394 -.01433 (217*2π)/15247 weeks
218-.00884 -.01573 (218*2π)/15247 weeks
219-.00753 -.01275 (219*2π)/15247 weeks
220-.00801 -.01264 (220*2π)/15247 weeks
221-.00682 -.01168 (221*2π)/15247 weeks
222-.00772 -.0112 (222*2π)/15247 weeks
223-.0067 -.0119 (223*2π)/15247 weeks
224-.00634 -.01154 (224*2π)/15247 weeks
225-.00615 -.01214 (225*2π)/15247 weeks
226-.00499 -.00995 (226*2π)/15247 weeks
227-.00331 -.01197 (227*2π)/15247 weeks
228-.00467 -.01366 (228*2π)/15247 weeks
229-.0089 -.013 (229*2π)/15247 weeks
230-.00508 -.01273 (230*2π)/15247 weeks
231-.00581 -.01249 (231*2π)/15247 weeks
232-.00626 -.01382 (232*2π)/15247 weeks
233-.00709 -.01335 (233*2π)/15247 weeks
234-.00767 -.01276 (234*2π)/15247 weeks
235-.00806 -.01177 (235*2π)/15246 weeks
236-.00884 -.01103 (236*2π)/15246 weeks
237-.00476 -.00845 (237*2π)/15246 weeks
238-.00517 -.0114 (238*2π)/15246 weeks
239-.00697 -.01352 (239*2π)/15246 weeks
240-.00547 -.01289 (240*2π)/15246 weeks
241-.00465 -.01037 (241*2π)/15246 weeks
242-.00682 -.01189 (242*2π)/15246 weeks
243-.00677 -.01189 (243*2π)/15246 weeks
244-.00569 -.01106 (244*2π)/15246 weeks
245-.00462 -.01171 (245*2π)/15246 weeks
246-.00707 -.01314 (246*2π)/15246 weeks
247-.00727 -.01049 (247*2π)/15246 weeks
248-.0047 -.00943 (248*2π)/15246 weeks
249-.00579 -.01205 (249*2π)/15246 weeks
250-.00623 -.01209 (250*2π)/15246 weeks
251-.00749 -.01175 (251*2π)/15246 weeks
252-.0074 -.00955 (252*2π)/15246 weeks
253-.00709 -.01101 (253*2π)/15246 weeks
254-.00475 -.00948 (254*2π)/15246 weeks
255-.00785 -.01066 (255*2π)/15246 weeks
256-.00617 -.01029 (256*2π)/15246 weeks
257-.00586 -.01022 (257*2π)/15246 weeks
258-.00594 -.01467 (258*2π)/15246 weeks
259-.00798 -.00868 (259*2π)/15246 weeks
260-.00461 -.00966 (260*2π)/15246 weeks
261-.00595 -.01144 (261*2π)/15246 weeks
262-.00801 -.01135 (262*2π)/15246 weeks
263-.00871 -.00958 (263*2π)/15246 weeks
264-.00567 -.0094 (264*2π)/15246 weeks
265-.007 -.00997 (265*2π)/15246 weeks
266-.00613 -.00913 (266*2π)/15246 weeks
267-.00726 -.01022 (267*2π)/15246 weeks
268-.00775 -.00993 (268*2π)/15246 weeks
269-.00623 -.00881 (269*2π)/15246 weeks
270-.00603 -.00878 (270*2π)/15246 weeks
271-.00534 -.012 (271*2π)/15246 weeks
272-.00701 -.00927 (272*2π)/15246 weeks
273-.00499 -.00923 (273*2π)/15246 weeks
274-.0064 -.01032 (274*2π)/15246 weeks
275-.00689 -.00918 (275*2π)/15246 weeks
276-.0057 -.00847 (276*2π)/15246 weeks
277-.00709 -.00967 (277*2π)/15246 weeks
278-.00607 -.00984 (278*2π)/15245 weeks
279-.0066 -.00993 (279*2π)/15245 weeks
280-.00701 -.00815 (280*2π)/15245 weeks
281-.00511 -.00966 (281*2π)/15245 weeks
282-.00604 -.00919 (282*2π)/15245 weeks
283-.00688 -.00896 (283*2π)/15245 weeks
284-.00624 -.00738 (284*2π)/15245 weeks
285-.00512 -.00936 (285*2π)/15245 weeks
286-.00586 -.0079 (286*2π)/15245 weeks
287-.00485 -.01052 (287*2π)/15245 weeks
288-.00496 -.00965 (288*2π)/15245 weeks
289-.00496 -.00963 (289*2π)/15245 weeks
290-.00747 -.00924 (290*2π)/15245 weeks
291-.00687 -.00984 (291*2π)/15245 weeks
292-.00535 -.00884 (292*2π)/15245 weeks
293-.00658 -.01012 (293*2π)/15245 weeks
294-.00408 -.00834 (294*2π)/15245 weeks
295-.0066 -.01044 (295*2π)/15245 weeks
296-.00679 -.00809 (296*2π)/15245 weeks
297-.00594 -.00862 (297*2π)/15245 weeks
298-.00563 -.00859 (298*2π)/15245 weeks
299-.00577 -.0097 (299*2π)/15245 weeks
300-.0057 -.00738 (300*2π)/15245 weeks
301-.00422 -.00914 (301*2π)/15245 weeks
302-.00641 -.01019 (302*2π)/15245 weeks
303-.00499 -.00903 (303*2π)/15245 weeks
304-.00628 -.00933 (304*2π)/15245 weeks
305-.00657 -.00808 (305*2π)/15245 weeks
306-.00496 -.00906 (306*2π)/15245 weeks
307-.00608 -.00808 (307*2π)/15245 weeks
308-.00501 -.01011 (308*2π)/15245 weeks
309-.00589 -.00814 (309*2π)/15245 weeks
310-.00516 -.0098 (310*2π)/15245 weeks
311-.0059 -.00956 (311*2π)/15245 weeks
312-.00756 -.01 (312*2π)/15245 weeks
313-.0065 -.00724 (313*2π)/15245 weeks
314-.00665 -.00903 (314*2π)/15245 weeks
315-.00658 -.00866 (315*2π)/15245 weeks
316-.007 -.00795 (316*2π)/15245 weeks
317-.005 -.00802 (317*2π)/15245 weeks
318-.00585 -.01003 (318*2π)/15245 weeks
319-.00677 -.00872 (319*2π)/15245 weeks
320-.00772 -.00767 (320*2π)/15245 weeks
321-.00576 -.00824 (321*2π)/15245 weeks
322-.00747 -.00992 (322*2π)/15245 weeks
323-.00783 -.00838 (323*2π)/15245 weeks
324-.00552 -.00539 (324*2π)/15245 weeks
325-.00603 -.00886 (325*2π)/15245 weeks
326-.00603 -.0089 (326*2π)/15245 weeks
327-.00711 -.00887 (327*2π)/15245 weeks
328-.00561 -.00677 (328*2π)/15245 weeks
329-.00541 -.00918 (329*2π)/15245 weeks
330-.00748 -.00809 (330*2π)/15245 weeks
331-.00719 -.00647 (331*2π)/15245 weeks
332-.0052 -.00779 (332*2π)/15245 weeks
333-.00701 -.00977 (333*2π)/15245 weeks
334-.00663 -.00589 (334*2π)/15245 weeks
335-.00449 -.00783 (335*2π)/15245 weeks
336-.00675 -.00927 (336*2π)/15245 weeks
337-.00738 -.00801 (337*2π)/15245 weeks
338-.00594 -.0078 (338*2π)/15245 weeks
339-.00616 -.00812 (339*2π)/15244 weeks
340-.00917 -.00857 (340*2π)/15244 weeks
341-.00691 -.00578 (341*2π)/15244 weeks
342-.00468 -.00828 (342*2π)/15244 weeks
343-.00599 -.00995 (343*2π)/15244 weeks
344-.01022 -.00716 (344*2π)/15244 weeks
345-.00697 -.00459 (345*2π)/15244 weeks
346-.00578 -.0087 (346*2π)/15244 weeks
347-.00722 -.00596 (347*2π)/15244 weeks
348-.00488 -.00642 (348*2π)/15244 weeks
349-.00624 -.00689 (349*2π)/15244 weeks
350-.00732 -.00661 (350*2π)/15244 weeks
351-.00702 -.00561 (351*2π)/15244 weeks
352-.0055 -.00549 (352*2π)/15244 weeks
353-.00572 -.00643 (353*2π)/15244 weeks
354-.0066 -.00767 (354*2π)/15244 weeks
355-.00715 -.00701 (355*2π)/15244 weeks
356-.00698 -.00646 (356*2π)/15244 weeks
357-.00601 -.00609 (357*2π)/15244 weeks
358-.00586 -.00652 (358*2π)/15244 weeks
359-.00571 -.00657 (359*2π)/15244 weeks
360-.00722 -.0062 (360*2π)/15244 weeks
361-.00617 -.00662 (361*2π)/15244 weeks
362-.00572 -.00532 (362*2π)/15244 weeks
363-.00601 -.00512 (363*2π)/15244 weeks
364-.00589 -.0059 (364*2π)/15244 weeks
365-.00447 -.00518 (365*2π)/15244 weeks
366-.00427 -.00739 (366*2π)/15244 weeks
367-.00579 -.0076 (367*2π)/15244 weeks
368-.00614 -.00724 (368*2π)/15244 weeks
369-.00547 -.00767 (369*2π)/15244 weeks
370-.00475 -.00678 (370*2π)/15244 weeks
371-.00606 -.00803 (371*2π)/15244 weeks
372-.00664 -.00825 (372*2π)/15244 weeks
373-.00709 -.00518 (373*2π)/15244 weeks
374-.00598 -.0068 (374*2π)/15244 weeks
375-.00605 -.00698 (375*2π)/15244 weeks
376-.00598 -.00704 (376*2π)/15244 weeks
377-.00639 -.00595 (377*2π)/15244 weeks
378-.00613 -.0059 (378*2π)/15244 weeks
379-.00628 -.00715 (379*2π)/15244 weeks
380-.00693 -.0071 (380*2π)/15244 weeks
381-.00569 -.00566 (381*2π)/15244 weeks
382-.00628 -.00723 (382*2π)/15244 weeks
383-.0065 -.00704 (383*2π)/15244 weeks
384-.00581 -.00652 (384*2π)/15244 weeks
385-.00627 -.00649 (385*2π)/15244 weeks
386-.00561 -.00747 (386*2π)/15244 weeks
387-.00582 -.00669 (387*2π)/15244 weeks
388-.00769 -.00595 (388*2π)/15244 weeks
389-.00745 -.00593 (389*2π)/15244 weeks
390-.00578 -.00642 (390*2π)/15244 weeks
391-.00779 -.00538 (391*2π)/15244 weeks
392-.00607 -.00515 (392*2π)/15244 weeks
393-.00549 -.00554 (393*2π)/15244 weeks
394-.00442 -.00591 (394*2π)/15244 weeks
395-.00646 -.00663 (395*2π)/15244 weeks
396-.00731 -.00695 (396*2π)/15244 weeks
397-.0059 -.00477 (397*2π)/15244 weeks
398-.00669 -.00588 (398*2π)/15244 weeks
399-.00643 -.00716 (399*2π)/15244 weeks
400-.00781 -.00454 (400*2π)/15244 weeks
401-.00551 -.00505 (401*2π)/15244 weeks
402-.00583 -.00475 (402*2π)/15244 weeks
403-.005 -.00646 (403*2π)/15244 weeks
404-.00613 -.00616 (404*2π)/15244 weeks
405-.00637 -.00605 (405*2π)/15244 weeks
406-.00572 -.00604 (406*2π)/15244 weeks
407-.00589 -.00541 (407*2π)/15244 weeks
408-.00631 -.00644 (408*2π)/15244 weeks
409-.00591 -.00431 (409*2π)/15244 weeks
410-.00472 -.0072 (410*2π)/15244 weeks
411-.00626 -.00569 (411*2π)/15244 weeks
412-.00551 -.0071 (412*2π)/15244 weeks
413-.00631 -.00589 (413*2π)/15244 weeks
414-.0055 -.00627 (414*2π)/15244 weeks
415-.0076 -.00613 (415*2π)/15244 weeks
416-.00582 -.00627 (416*2π)/15244 weeks
417-.00669 -.00472 (417*2π)/15244 weeks
418-.00597 -.00667 (418*2π)/15244 weeks
419-.00688 -.00611 (419*2π)/15244 weeks
420-.00616 -.00637 (420*2π)/15244 weeks
421-.0066 -.00533 (421*2π)/15244 weeks
422-.00585 -.00602 (422*2π)/15244 weeks
423-.00698 -.00598 (423*2π)/15244 weeks
424-.00797 -.0056 (424*2π)/15244 weeks
425-.00559 -.00431 (425*2π)/15244 weeks
426-.00621 -.00539 (426*2π)/15244 weeks
427-.006 -.00545 (427*2π)/15244 weeks
428-.00684 -.00624 (428*2π)/15244 weeks
429-.00701 -.00548 (429*2π)/15244 weeks
430-.00702 -.00397 (430*2π)/15244 weeks
431-.00641 -.00503 (431*2π)/15244 weeks
432-.00808 -.0048 (432*2π)/15244 weeks
433-.0065 -.00388 (433*2π)/15244 weeks
434-.00554 -.00332 (434*2π)/15244 weeks
435-.00395 -.00635 (435*2π)/15244 weeks
436-.00691 -.0053 (436*2π)/15243 weeks
437-.00656 -.00416 (437*2π)/15243 weeks
438-.00487 -.0053 (438*2π)/15243 weeks
439-.00731 -.00688 (439*2π)/15243 weeks
440-.00757 -.00519 (440*2π)/15243 weeks
441-.00625 -.00392 (441*2π)/15243 weeks
442-.00627 -.00567 (442*2π)/15243 weeks
443-.00576 -.0059 (443*2π)/15243 weeks
444-.00754 -.00467 (444*2π)/15243 weeks
445-.00654 -.00523 (445*2π)/15243 weeks
446-.00558 -.00486 (446*2π)/15243 weeks
447-.00627 -.00463 (447*2π)/15243 weeks
448-.00617 -.00507 (448*2π)/15243 weeks
449-.00534 -.00439 (449*2π)/15243 weeks
450-.00589 -.00447 (450*2π)/15243 weeks
451-.00656 -.00535 (451*2π)/15243 weeks
452-.00837 -.00466 (452*2π)/15243 weeks
453-.00667 -.00466 (453*2π)/15243 weeks
454-.00586 -.00512 (454*2π)/15243 weeks
455-.00641 -.00416 (455*2π)/15243 weeks
456-.00693 -.00628 (456*2π)/15243 weeks
457-.00636 -.00445 (457*2π)/15243 weeks
458-.00597 -.00482 (458*2π)/15243 weeks
459-.00667 -.00409 (459*2π)/15243 weeks
460-.00737 -.00478 (460*2π)/15243 weeks
461-.00583 -.00467 (461*2π)/15243 weeks
462-.00611 -.00415 (462*2π)/15243 weeks
463-.00631 -.00492 (463*2π)/15243 weeks
464-.00643 -.00582 (464*2π)/15243 weeks
465-.00617 -.00331 (465*2π)/15243 weeks
466-.00617 -.00442 (466*2π)/15243 weeks
467-.00464 -.0039 (467*2π)/15243 weeks
468-.00724 -.00722 (468*2π)/15243 weeks
469-.007 -.00508 (469*2π)/15243 weeks
470-.00677 -.00342 (470*2π)/15243 weeks
471-.00697 -.00526 (471*2π)/15243 weeks
472-.00881 -.00443 (472*2π)/15243 weeks
473-.00671 -.00258 (473*2π)/15243 weeks
474-.00576 -.00301 (474*2π)/15243 weeks
475-.00527 -.00474 (475*2π)/15243 weeks
476-.00771 -.00318 (476*2π)/15243 weeks
477-.00569 -.0038 (477*2π)/15243 weeks
478-.00568 -.00325 (478*2π)/15243 weeks
479-.00546 -.00407 (479*2π)/15243 weeks
480-.00619 -.0033 (480*2π)/15243 weeks
481-.00486 -.00504 (481*2π)/15243 weeks
482-.00614 -.00309 (482*2π)/15243 weeks
483-.00607 -.00557 (483*2π)/15243 weeks
484-.00675 -.00409 (484*2π)/15243 weeks
485-.00519 -.00469 (485*2π)/15243 weeks
486-.00533 -.00485 (486*2π)/15243 weeks
487-.00663 -.00474 (487*2π)/15243 weeks
488-.00718 -.00395 (488*2π)/15243 weeks
489-.00596 -.00344 (489*2π)/15243 weeks
490-.00541 -.00411 (490*2π)/15243 weeks
491-.0073 -.00436 (491*2π)/15243 weeks
492-.00585 -.00324 (492*2π)/15243 weeks
493-.00473 -.00496 (493*2π)/15243 weeks
494-.00672 -.00452 (494*2π)/15243 weeks
495-.00639 -.00424 (495*2π)/15243 weeks
496-.00587 -.00515 (496*2π)/15243 weeks
497-.00769 -.0056 (497*2π)/15243 weeks
498-.00693 -.0026 (498*2π)/15243 weeks
499-.00529 -.00339 (499*2π)/15243 weeks
500-.006 -.00508 (500*2π)/15243 weeks
501-.00741 -.00345 (501*2π)/15243 weeks
502-.0069 -.0033 (502*2π)/15243 weeks
503-.00528 -.00459 (503*2π)/15243 weeks
504-.00665 -.00597 (504*2π)/15243 weeks
505-.008 -.00311 (505*2π)/15243 weeks
506-.00705 -.00361 (506*2π)/15243 weeks
507-.00635 -.00362 (507*2π)/15243 weeks
508-.00717 -.0038 (508*2π)/15243 weeks
509-.00575 -.00288 (509*2π)/15243 weeks
510-.00653 -.00361 (510*2π)/15243 weeks
511-.00613 -.00367 (511*2π)/15243 weeks
512-.00677 -.00417 (512*2π)/15243 weeks
513-.00635 -.00301 (513*2π)/15243 weeks
514-.00696 -.00275 (514*2π)/15243 weeks
515-.00563 -.00352 (515*2π)/15243 weeks
516-.00571 -.00344 (516*2π)/15243 weeks
517-.00552 -.00434 (517*2π)/15243 weeks
518-.00731 -.00427 (518*2π)/15243 weeks
519-.00629 -.00422 (519*2π)/15243 weeks
520-.00747 -.00326 (520*2π)/15243 weeks
521-.00647 -.0039 (521*2π)/15243 weeks
522-.00763 -.00225 (522*2π)/15243 weeks
523-.0044 -.00296 (523*2π)/15243 weeks
524-.00574 -.00364 (524*2π)/15243 weeks
525-.00676 -.00435 (525*2π)/15243 weeks
526-.00714 -.00317 (526*2π)/15243 weeks
527-.00484 -.00438 (527*2π)/15243 weeks
528-.00721 -.0043 (528*2π)/15243 weeks
529-.00856 -.00213 (529*2π)/15243 weeks
530-.00493 -.00167 (530*2π)/15243 weeks
531-.00493 -.00357 (531*2π)/15243 weeks
532-.00585 -.00372 (532*2π)/15243 weeks
533-.00627 -.00358 (533*2π)/15243 weeks
534-.00525 -.00302 (534*2π)/15243 weeks
535-.00707 -.00479 (535*2π)/15243 weeks
536-.00742 -.00313 (536*2π)/15243 weeks
537-.0056 -.0021 (537*2π)/15243 weeks
538-.00436 | 9,414 | 22,437 | {"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-2017-13 | longest | en | 0.744768 |
https://electrowiring.herokuapp.com/post/electric-fan-motor-circuit-diagram | 1,660,949,162,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573849.97/warc/CC-MAIN-20220819222115-20220820012115-00141.warc.gz | 226,354,681 | 14,542 | 9 out of 10 based on 330 ratings. 1,990 user reviews.
# ELECTRIC FAN MOTOR CIRCUIT DIAGRAM
Electric Motor Rotation Direction - InspectAPedia
In a fixed-direction electric motor such as on an HVAC blower fan or an A/C or heat pump compressor, each time the motor starts its start capacitor and start winding give the motor a "kick" in the right direction. Dayton Electric Motor Wiring Diagram [PDF], Dayton Electric Mfg. Co., 5959 W. Howard St., Niles IL 60714 USA, retrieved 2017/07
AC Motor Control Circuits Worksheet - AC Electric Circuits
Notes: This circuit is known as a latching circuit, because it “latches” in the “on” state after a momentary action. The contact in parallel with the “Run” switch is often referred to as a seal-in contact, because it “seals” the momentary condition of the Run switch closure after that switch is de-actuated. The follow-up question of how we may make the motor stop running is a
How to Check an Electric Motor: 12 Steps (with Pictures) - wikiHow
Aug 07, 2021Begin to check the bearings of the motor. Many electric motor failures are caused by bearing failures. The bearings allow the shaft or rotor assembly to turn freely and smoothly in the frame. and measuring the resistance between the leads of the motor. In this case, consult the wiring diagram of the motor to be sure that the meter is
How Electric Motors Work | HowStuffWorks
Apr 01, 2000The "flipping the electric field" part of an electric motor is accomplished by two parts: the commutator and the brushes. Advertisement The diagram shows how the commutator (in green) and brushes (in red) work together to let current flow to the electromagnet, and also to flip the direction that the electrons are flowing at just the right moment.
Brushless DC electric motor - Wikipedia
A brushless DC electric motor (BLDC motor or BL motor), also known as an electronically commutated motor (ECM or EC motor) or synchronous DC motor, is a synchronous motor using a direct current (DC) electric power supply. It uses an electronic controller to switch DC currents to the motor windings producing magnetic fields which effectively rotate in space and
Electric Motor Starting Capacitor Wiring & Installation - InspectAPedia
Questions & answers about installing a hard-start capacitor to get an air conditioner motor, fan motor, or other electric motor running. 0 DAYTON ELECTRIC MOTOR WIRING DIAGRAM [PDF], Dayton Electric Mfg. Co., 5959 W. Howard You want to find the circuit board that controls fan speeds and look to see if a dip switch or jumper was moved
Emerson Motor Wiring Diagram Gallery - Wiring Diagram
Sep 26, 2019Name: emerson motor wiring diagram – Emerson Electric Motor Wiring Diagram Kanvamath Org Fan Perfect Psc Emerson Wiring Diagram Electric Motor; File Type: JPG; Source: imovo; Size: 162 KB; Dimension: 566 x 782
Brushed DC electric motor - Wikipedia
A brushed DC electric motor is an internally commutated electric motor designed to be run from a direct current power source and utilizing an electric brush for contact. Brushed motors were the first commercially important application of electric power to driving mechanical energy, and DC distribution systems were used for more than 100 years to operate motors in
Troubleshooting Electric Cooling Fan Problems in Cars
Aug 02, 2022CAUTION: If the fan fuse or fusible link is blown, or the breaker has been triggered, it is possible the fan motor itself may be bad and causing the fuses to blow. Replace the fuse or fusible link or reset the breaker and test the fan motor again. If the same problem appears and there's no wire causing a short circuit, replace the fan motor.
Ceiling Fan Wiring Diagram: A Complete Tutorial | EdrawMax
When the fan is uploaded to the ceiling, this wire is connected to the Earth ground so, this means that this wire completes the fan’s circuit. Black Wire: The black wire is the primary wire of the fan. The function of this wire is to provide current from the switch to the electric motor of the fan. This wire is also known as the hot wire. | 906 | 4,040 | {"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-2022-33 | latest | en | 0.89314 |
http://www.ehow.com/how_8179120_sum-consecutive-odd-integers.html | 1,484,777,216,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280364.67/warc/CC-MAIN-20170116095120-00006-ip-10-171-10-70.ec2.internal.warc.gz | 438,508,922 | 15,651 | # How to Find the Sum of Consecutive Odd Integers
Save
The sum of a given sequence of numbers is known as a series, and many series -- both infinite and finite -- have known sums. For example, a mathematician named Carl Gauss is famous for determining a formula for the first N consecutive numbers as a boy in the 18th century. Using a variation of Gauss' result, you will find a simple expression for the sum of consecutive odd numbers.
• Determine the number (N) of consecutive odd numbers you're adding. If your series is given in sigma notation, this is the finishing index (on top of the sigma) minus the starting index (beneath the sigma) plus one. Alternatively, subtract the largest odd number in your series from the smallest, divide this difference by two and add one. For example, if adding the odd numbers from 7 to 45, N = (45 - 7) / 2 + 1 = 20.
• Multiply the smallest number in the series by the number of numbers in the series you determined in Step 1. For example, if adding the odd numbers from 7 to 45, multiply 7 by N (20) = 140.
• Multiply N by N - 1, and add this to the product you found in Step 2. For example, if adding odd numbers from 7 to 45, where N = 20, add the product N (N - 1) = 20 19 = 380 to 7 20 = 140 to get 520. In other words, the formula is: min N + N * (N - 1), where N is the number of consecutive odd numbers to sum, and "min" is the smallest of these.
## Tips & Warnings
• This formula also works for consecutive even numbers.
## References
• Photo Credit Hemera Technologies/AbleStock.com/Getty Images
Promoted By Zergnet
## Related Searches
Check It Out
### How to Build and Grow a Salad Garden On Your Balcony
M
Is DIY in your DNA? Become part of our maker community. | 443 | 1,728 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2017-04 | latest | en | 0.912642 |
https://square-root.net/996-cubed | 1,600,461,891,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400188841.7/warc/CC-MAIN-20200918190514-20200918220514-00026.warc.gz | 596,368,295 | 9,117 | # 996 Cubed
996 cubed, 9963, is the number you get when multiplying 996 times 996 times 996. It can also be looked at as exponentiation involving the base 996 and the exponent 3. The term is usually pronounced 3rd power of nine hundred and ninety-six or nine hundred and ninety-six cubed. The cube of 996 is a perfect cube because the number is the product of the three equal integers 996. It can be written as 996 x 996 x 996 or in exponential form. Read on to learn everything about the number nine hundred and ninety-six cubed, including useful identities.
9963 = 988047936
996 x 996 x 996 = 988047936
The inverse operation of cubing 996 is extracting the cube root of 996, explained here. In the next section we elaborate on the cube 996, and there you can also find our calculator.
## What is 996 Cubed
A cube is a three-dimensional shape with 6 equal square faces. Hence, a cube with side length 996 has an area of 988047936.
The sum and the differences of two cubes, for example with side length 996 and 995, can be calculated with the following formulas:
1. a3 + b3 = (a + b) × (a2 – ab + b2) ⇔ a3 = (a + b) × (a2 – ab + b2) – b3
With a = 996, b = 995 and the equivalence we get:
9963 = 1991 x (9962 – 996 × 995 + 9952) – 9953 = 1991 x (992016 – 991020 + 990025) – 985074875 = 1991 x 991021 – 985074875 = 988047936
2. a3 – b3 = (a – b) × (a2 + ab + b2) ⇔ a3 = (a – b) × (a2 + ab + b2) + b3
With a = 996, b = 995 and the equivalence we get:
9963 = (9962 + 996 × 995 + 9952) + 9953 = 992016 + 991020 + 990025 + 985074875 = 988047936
As follows from (1), 996 cubed can be calculated from 995 cubed and 996 squared using the identity:
n3 = (2 × n – 1) × (n2 -n + 1) – (n-1)3
9963 = (2 × 996 – 1) × (9962 – 996 + 1) – (995)3 = 1991 × (992016 – 995) – 985074875 = 1991 × 991021 – 985074875 = 988047936
Using the second formula 996 cubed can be also be computed with this identity:
n3 = (n-1)3 + 3n2 -3n + 1
9963 = 9953 + 3 × 9962 – 3 × 996 + 1 = 985074875 + 3 × 992016 – 3 × 996 + 1 = 985074875 + 2976048 – 2988 + 1 = 988047936
If you want to calculate the cube of any number, not only integers like 996, you can use our calculator below. Just enter your number; the calculation is then done automatically.
### Calculate Cubed Number
If this tool has been beneficial to you, then bookmark it right now. Besides the cube of 996, more searched cubes on our site include:
## Nine Hundred And Ninety-Six Cubed
You now know everything about the cube 996 and calculating it by means of multiplication, exponentiation, sum and difference formulas and using identities: 996 cubed is equivalent to 988047936.
If you were searching for what is 996 cubed in math or if you typed whats 996 cubed in the search engine you now have all the answers, too.
The same goes for searches like cube996, and 996 to the 3rd power, just to name a few more examples people are often looking for.
Note that you can also find many perfect cubes including 996 cubed using the search form in the sidebar of this page.
To sum up,
996 cubed = 996x996x996 = 9963 = 996 to the 3rd power = 988047936. The exponentiation form is mostly used to denote 996 cube.
If this article about the cube of 996 has helped you, then please share it by means of the social buttons at your fingertips.
And should you want to leave a comment related to 996cubed use the form below.
Thanks for visiting 996 cubed.
Posted in Cubed Numbers | 1,047 | 3,411 | {"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-2020-40 | latest | en | 0.846472 |
https://la.mathworks.com/matlabcentral/cody/problems/44343-pair-primes/solutions/1385311 | 1,607,013,466,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141729522.82/warc/CC-MAIN-20201203155433-20201203185433-00493.warc.gz | 327,814,044 | 17,093 | Cody
# Problem 44343. Pair Primes
Solution 1385311
Submitted on 19 Dec 2017 by Binbin Qi
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 2; y_correct = 51; assert(isequal(pairPrimes(x),y_correct)) assert(~any(cellfun(@(x)ismember(max([0,str2num(x)]),[51,2485,136162,8578934]),regexp(fileread('pairPrimes.m'),'[\d\.\+\-\*\/]+','match'))))
2 Pass
x = 3; y_correct = 2485; assert(isequal(pairPrimes(x),y_correct)) assert(~any(cellfun(@(x)ismember(max([0,str2num(x)]),[51,2485,136162,8578934]),regexp(fileread('pairPrimes.m'),'[\d\.\+\-\*\/]+','match'))))
3 Pass
x = 4; y_correct = 136162; assert(isequal(pairPrimes(x),y_correct)) assert(~any(cellfun(@(x)ismember(max([0,str2num(x)]),[51,2485,136162,8578934]),regexp(fileread('pairPrimes.m'),'[\d\.\+\-\*\/]+','match'))))
4 Pass
x = 5; y_correct = 8578934; assert(isequal(pairPrimes(x),y_correct)) assert(~any(cellfun(@(x)ismember(max([0,str2num(x)]),[51,2485,136162,8578934]),regexp(fileread('pairPrimes.m'),'[\d\.\+\-\*\/]+','match'))))
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 422 | 1,248 | {"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} | 3.375 | 3 | CC-MAIN-2020-50 | latest | en | 0.370196 |
https://archives2.twoplustwo.com/showthread.php?s=8ad292e4f3d3220284cdd3ba092ee709&p=4175057 | 1,624,300,448,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488289268.76/warc/CC-MAIN-20210621181810-20210621211810-00354.warc.gz | 109,800,239 | 8,739 | Two Plus Two Older Archives Calculation of rakeback for Pokerstars new VIP program
FAQ Members List Calendar Search Today's Posts Mark Forums Read
#1
12-30-2005, 07:10 AM
snowbank Junior Member Join Date: Apr 2004 Posts: 10
Calculation of rakeback for Pokerstars new VIP program
Has anyone done the calculations per level of what rakeback would work out to be?
#2
12-30-2005, 09:18 AM
SmileyEH Senior Member Join Date: Jun 2004 Posts: 431
Re: Calculation of rakeback for Pokerstars new VIP program
It depends what stakes you are playing. I think 5/10 6max and 2/4nl 6max would be the maximum in rakeback (ie; most pots rakes over \$1, but less maxed to \$3).
-SmileyEH
#3
12-30-2005, 08:18 PM
Nomad84 Senior Member Join Date: Mar 2005 Posts: 194
Re: Calculation of rakeback for Pokerstars new VIP program
[ QUOTE ]
Has anyone done the calculations per level of what rakeback would work out to be?
[/ QUOTE ]
Based on a quick look through the store, it looks like you can get around \$1 per 55-65 FPP. The Dell gift certificate is \$1/62.5 FPP, for instance. Use that number coupled with the FPP earning rate for whatever VIP level and stakes you are at to calculate rakeback. My numbers for \$5/10 show that I am earning around 79.4 FPP/100 hands. If you are Platinum, that becomes about 198.5 FPP/100. That works out to around \$3.176/100 hands if you are buying Dell gift certificates. You'll do slightly better on some purchases and slightly worse on others, but that should give you a starting point.
#4
12-30-2005, 09:19 PM
obsidian Senior Member Join Date: Mar 2005 Location: IL Posts: 343
Re: Calculation of rakeback for Pokerstars new VIP program
At 15/30 I found it to be about 20% @ supernova status.
#5
12-31-2005, 05:10 PM
Nomad84 Senior Member Join Date: Mar 2005 Posts: 194
Re: Calculation of rakeback for Pokerstars new VIP program
[ QUOTE ]
At 15/30 I found it to be about 20% @ supernova status.
[/ QUOTE ]
I should add that at Platinum, it's around 14.7% at 5/10. At Supernova, it's around 20.6%. My data is from about 8100 hands, so it should be reasonably close.
Thread Tools Display Modes Linear Mode
Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is Off Forum Rules
Forum Jump User Control Panel Private Messages Subscriptions Who's Online Search Forums Forums Home Two Plus Two Two Plus Two Internet Magazine About the Forums MOD DISCUSSION ISOP General Poker Discussion Texas Hold'em Beginners Questions Books and Publications Televised Poker News, Views, and Gossip Brick and Mortar Home Poker Poker Beats, Brags, and Variance Poker Theory Limit Texas Hold'em Mid- and High-Stakes Hold'em Medium Stakes Hold'em Small Stakes Hold'em Micro-Limits Mid-High Stakes Shorthanded Small Stakes Shorthanded PL/NL Texas Hold'em Mid-, High-Stakes Pot- and No-Limit Hold'em Medium-Stakes Pot-, No-Limit Hold'em Small Stakes Pot-, No-Limit Hold'em Tournament Poker Multi-table Tournaments One-table Tournaments Other Poker Omaha/8 Omaha High Stud Other Poker Games General Gambling Probability Psychology Sports Betting Other Gambling Games Rake Back Computer Technical Help Internet Gambling Internet Gambling Internet Bonuses Software 2+2 Communities Other Other Topics Other Topics Sporting Events Politics Science, Math, and Philosophy The Stock Market
All times are GMT -4. The time now is 02:34 PM. | 931 | 3,641 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-25 | latest | en | 0.886379 |
https://cr4.globalspec.com/blogentry/27077/Time-Dilation-and-Lorentz-Contraction-Revisited?Pg=2 | 1,723,566,693,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641079807.82/warc/CC-MAIN-20240813141635-20240813171635-00727.warc.gz | 144,170,623 | 46,670 | Relativity and Cosmology Blog
# Relativity and Cosmology
This is a Blog on relativity and cosmology for engineers and the like. My website "Relativity-4-Engineers" has more in-depth stuff.
Comments/questions of a general nature should preferably be posted to the FAQ section of this Blog (http://cr4.globalspec.com/blogentry/316/Relativity-Cosmology-FAQ).
A complete index to the Relativity and Cosmology Blog can be viewed here: http://cr4.globalspec.com/blog/browse/22/Relativity-and-Cosmology"
Regards, Jorrie
Previous in Blog: HTRN's Cosmology Questions Next in Blog: Another Relativity "Paradox"
Page 2 of 2: « First < Prev 1 2 Last »
# Time Dilation and Lorentz Contraction Revisited
Posted March 27, 2016 11:00 PM by Jorrie
## Intro
There are many fundamental relativity discussions that are buried deep inside related threads and hence very difficult to access at a later stage. I will try to elevate some of those issues to the level of a separate Blog entry.
## Ralfcis' problems
"One can derive the length contraction formula from the time dilation formula so why even bring in the redundant concept of length contraction, they're the same thing. However, you said time dilation is also not real (because both frames see time dilation in the other) but using the criteria of the twin paradox, one can measure an age difference consistent with time dilation that remains once the relative velocity ends."
Jorrie replied:
"This is the crux of the matter. There is coordinate dependent time dilation and then there is proper time dilation. There is coordinate dependent Lorentz contraction, but there is no proper Lorentz contraction. In this sense, the two are different, not 'the same thing'.
However, Lorentz contraction is real in the sense that when you make a real measurement of the length of a passing spaceship, you get a Lorentz contracted value. In this sense Lorentz contraction is 'real', but then, 'real' means different things to different people - a debate that I have no intention of entering.
As long as you have standard synchrony, i.e. Einstein synchronization of clocks, you have Lorentz contraction and the limiting one-way speed is c. The limiting 2-way speed is c, irrespective of the clock sync convention used."
To which Ralfcis replied:
"I'm lost here, totally blank. What you said may be beyond my ability to understand. I can't even formulate a question."
I think the problem is that Ralf has developed his own (evidently flawed) verbal framework for understanding relativity and he attempts to fit every bit of related information acquired into this framework. Some bits just do not fit into his framework and he gets stumped by it, or worse, he contorts the information to fit in.
Since I have spend thousands of words in an attempt to turn Ralf's framework into the mainstream direction and evidently failed, I recommended that he opens other threads to get more input and direction. This has apparently also failed to give any satisfaction.
Lately, Ralf has come around to a more conventional framework, but since I have mostly completed this Blog entry, I will post it anyway. It may be useful for other members of this forum - readers that have long since unsubscribed from that lengthy prior discussion.
Let us restrict the discussion to special relativity (SR) only, since without a solid SR foundation, it is useless to discuss general relativity (GR). Also, let us leave quantum physics and philosophical considerations out of it. SR does not answer "why" or "how" questions, only "what" (observable) questions.
The main unanswered questions seem to be about reciprocal time dilation, relative elapsed time, Lorentz contraction and the isotropy of the speed of light. In the latter, it is more specifically the one-way speed of light being the same in every inertial frame that trips up many a student of relativity.
## a) The one-way speed of light
It seems appropriate to get this one out of the way first. There is no magic about the one-way speed of light being the same in all inertial frames. Einstein has simply declared it to be so as a convention[1] and then based the whole of his amazing theory of relativity on this assumption. The fact that relativity works flawlessly within its applicability, justifies Einstein's assumption without any shadow of a doubt.
Importantly, this also seems to be what nature prefers. Physics would have been very 'ugly' if any other clock sync scheme was used, e.g. the GPS system would have been all but useless, because the speed of light would have been different in different directions.
This assumption determines the method for synchronizing clocks throughout every inertial frame. In its simplest form, if we know the distance d between two clocks that are permanently at rest relative to each other and we send a time stamped signal, the receiving clock simply adds a propagation delay Δt = d/c to the time stamp and sets its time accordingly.
Because this is such a simple and universal scheme, many present day scientists simply accept that the one-way speed of light is c in every inertial frame and never give it a second thought. This sometimes leads to heated debate between scientists and "the rest", who are attempting to understand the reasoning behind the principle.
## b) Reciprocal time dilation
The fact that when A and B are in uniform relative (inertial) motion, A observes B's clock to 'lose time' and B observes A's clock to 'lose time' is directly related to the above convention about the one-way speed of light. It comes about due to the way clocks are synchronized, using the convention.
It does not determine who ages slower or faster, but just how the one observer observes the others clock. Time dilation can be viewed as simply a change in 'spacetime observation angle' - each views the others time vector at an angle in spacetime, which depends on their relative speed, which is reciprocal.
This does not make time dilation "an illusion" or "not real". When proper scientific measurements of time are made between two inertial observers in relative motion, the results are as real as any measurement can be; but, it is reciprocal and hence coordinate dependent and not absolute.
## c) Relative elapsed time
This is where "aging slower or faster" comes in. Every inertial object follows a trajectory through spacetime, called a 'worldline'. When two inertial objects in free space are at rest relative to each other, they follow equivalent (not necessarily identical) worldlines, so they age identically.
If they are not at rest relative to each other, they are following non-equivalent worldlines and they may age differently. They can synchronize their clocks when they move past each other and after that the one that experience the largest change of inertial frame will age less. This is why the traditional "away-twin" always ends up younger than the "home-twin".[2] If neither of them experiences any change of inertial frame, we cannot tell who ages more or less than the other.
It just so happens that in mostly quasi-inertial cases (where the acceleration phase is short relative to the inertial phases), the aging difference is approximately the same as that given by the SR time dilation formula. This fact has led to a lot of confusion in the popular literature. There is no difference in elapsed times unless there has been a difference in the change of inertial frames, which requires acceleration of at least one of the two clocks.[3]
## d) Reciprocal Lorentz contraction
Like reciprocal time dilation, reciprocal Lorentz contraction is also caused by the Einstein clock synchronization convention. If A and B are in relative motion, each observes the others lengths to be contracted in the direction of relative motion. When they are brought to relative rest again, the reciprocal length contraction disappears - this is unlike the case of relative aging, which is a lasting effect.
Like relative time dilation, Lorentz contraction can be viewed as simply a change in 'spacetime observation angle' (each views the others length at an angle in spacetime), which depends on relative speed. The formula for Lorentz contraction is essentially the same as the time dilation formula. Both these effects are contained in the Lorentz transformations[4] as special cases.
Using the above information, the classical 'twin paradox' can be twisted to a slightly more challenging one. Alice sets off from Earth on her long fast journey, with Bob staying at home. Some years after Alice have left Earth, she and Bob each opens a secret envelope, where they for the first time get instructions on how to complete the mission.
The instructions could be either (i) for Alice to return to Earth and for Bob to stay put; or (ii) for Alice to coast on and for Bob to leave Earth fast enough so that he can catch up and join Alice in space.
Without doing any math, firstly, who would have aged less in each of the two cases? Secondly, just before Alice and Bob opened their respective envelopes, who would you say have aged less up to that point?
@ralfcis: Before attempting to answer these questions, first make sure that you understand the discussion leading up to it. If not, keep on asking questions, but please stick to the baseline given. I do not want to waste time by analyzing some or other fancy relativistic scenario that you can dream up - such time can be more effectively spent by discussing the stated principles better.
-Jorrie
[1] Einstein's 2005 paper "On the Electrodynamics of Moving Bodies", Section I, \$1:
"We have so far defined only an "A time" and a "B time." We have not defined a common "time" for A and B, for the latter cannot be defined at all unless we establish by definition that the "time" required by light to travel from A to B equals the "time" it requires to travel from B to A."
The two-way speed of light obviously does not suffer from this clock sync convention problem, because we need only one good clock and a definition of distance to measure the round trip (average) speed of light. The modern definition of distance is however dependent on the speed of light in vacuum, so there is some degree of conventionality in the modern value of the two-way speed as well. There is no doubt that is the same in all directions, though.
[2] The rigorously correct statement would require them to meet again in order to unequivocally establish who has aged less. It is however a sufficient requirement that the two twins must just be at rest relative to each other again in order to establish beyond reasonable doubt who has aged less. For example, they can each observe an event that is equidistant from the two of them and report their respective clock readings at the time of observing the event.
[3] It must be clearly stated that acceleration per se does not cause time dilation. Acceleration is required to change the inertial frame and the time spent in the new inertial frame determines the amount of elapsed time difference accrued - in other words, acceleration is the cause of the different spacetime paths.
Instead of acceleration, their could be a "time hand-off" by the away-twin to a third inertial observer, flying in the opposite direction, who then completes the home leg. The calculated elapsed time difference is still valid, because there is the same difference in spacetime paths.
[4] The Lorentz transformations are more general than just time dilation and length contraction. It gives the equations for Lorentz covariance, which in simpler terms means converting time and space intervals from one inertial frame to another in a consistent way. Space intervals and time intervals between two specific events together form the spacetime interval between the two events, which is the same for all inertial frames. When you have more space interval between events, you have less time interval and vice-versa, but in a squared (not linear) fashion.
PS. If there are any other relativity issues that you want raised to this 'Blog-level', please p/mail me with the request.
Interested in this topic? By joining CR4 you can "subscribe" to
Guru
Join Date: May 2006
Location: 34.02S, 22.82E
Posts: 3794
#108
Find in discussion
### Re: Time Hand-off Scenario Revisited
05/12/2016 5:03 AM
"I noticed that with just 100 more views this will be the most read blog post in the last 3 years."
Not that it matters, but you and me contributed almost half the views; there are ~2 views for every reply we wrote. And on a technical point, http://cr4.globalspec.com/blogentry/22812/The-One-way-Speed-of-Light-Controversy of June 2013 produced 1200+ views, with just 15 comments.
People like 'controversy' and disagreement, so the stats get a bit skewed.
__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Power-User
Join Date: Aug 2008
Posts: 465
#109
### Re: Time Hand-off Scenario Revisited
05/12/2016 6:00 AM
Oh, and I keep checking for responses so I guess we're almost alone then. Others probably fall in here by accident and leave screaming. I know I've asked that question about the simultaneity "now slices" before but I never got an answer so far.
__________________
What sparks the intelligent, inflames the ignorant.
Guru
Join Date: May 2006
Location: 34.02S, 22.82E
Posts: 3794
#110
### Re: Time Hand-off Scenario Revisited
05/12/2016 6:39 AM
"I know I've asked that question about the simultaneity "now slices" before but I never got an answer so far."
And I think is has been answered before. In any case, in 2-d spacetime it is just the x-axis of each observer. In 3-d spacetime it obviously a 2-d surface and so on. What more is there that you want to know?
__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Power-User
Join Date: Aug 2008
Posts: 465
#111
### Re: Time Hand-off Scenario Revisited
05/12/2016 6:47 AM
This question,
3. Just so I have my terms correct, I understand the square coordinates are called Cartesian but what are the rhombic coordinates of the moving frame called? Anyway, the slanted x-axes of the rhombic coordinates are lines of simultaneity from the perspective of the moving frame. If each point on that line had a clock, all clocks on that line would have the exact same time from one length of the universe to the other. If there was a person with each clock and they were all telepathically connected, there would be someone back on earth at the 3.2 year mark of our time telepathically connected with Alice at her 4 year turnaround point her time. SR seems to imply Alice would be telepathically aware of a time in our past (since if Alice is at the 4 yr mark, we are at the 5yr mark) as being in her present. However, if we compensate for the delay of telepathic signal and for the relativity of simultaneity we see that our present at our 5yr mark and Alice's present at her 4 yr mark constitute a universal present which SR forbids. Which interpretation of the spacetime diagram is correct?
__________________
What sparks the intelligent, inflames the ignorant.
Guru
Join Date: May 2006
Location: 34.02S, 22.82E
Posts: 3794
#112
### Re: Time Hand-off Scenario Revisited
05/12/2016 9:44 AM
It is called Minkowski spacetime.
"If each point on that line had a clock, all clocks on that line would have the exact same time from one length of the universe to the other"
One has to to qualify this a bit: provided that those clocks are all synchronized in the applicable inertial frame and there is no gravity (the latter prevents application on on the larger scales).
I do not want to discuss anything 'telepathic', just observables. The only absolute observables are Alice looking out of the window to read any co-located clock in Bob's reference frame directly. And for Bob's observer at that same clock's position to read Alice's clock directly when she flies past.
Now the turnaround (or handoff) is a nice fixed event on which to do some spacetime interval calculations. As an exercise, write down that event's coordinates in both Alice and Bob's reference frames respectively. Then see if it agrees with the invariant nature of the spacetime interval.
__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Power-User
Join Date: Aug 2008
Posts: 465
#113
### Re: Time Hand-off Scenario Revisited
05/12/2016 11:36 AM
Bob's coordinates at the turnaround point are 5,3 and alice's are 4,0. 25-9=16 and so the spacetime interval for both is 4.
I'm trying to understand what you're saying. Maybe I should drop the idea of clocks on her x-axis and just bring Bob's clock to the turnaround point which she can see without any delay or telepathy. We know Bob is currently at his 5yr mark and Alice is at her 4 yr mark. Alice see's Bob's clock at Bob's 3.2 yr mark (Y=.8, .8*4=3.2) due to time dilation. The difference between Bob's clock at home as he sees it and the time dilation Alice sees is 1.8 years which is the relativity of simultaneity vx= .6*3ly=1.8yrs.
So Bob's past is not really in Alice's present, Alice's velocity is distorting spacetime to make it seem to her that Bob's past is happening now in her present. Is that the correct interpretation?
But if I apply the same analysis for just after the turnaround, does Alice suddenly see Bob's clock jump by twice the relativity of simultaneity to a total of 3.2+2*1.8 = 6.8 yrs? She's not really looking 1.8 yrs into Bob's future which hasn't happened yet for him, but how do you interpret what she's seeing? Bob's clock to her just makes a discontinuous jump like this by 3.6 yrs? How does one interpret what's happening so it makes sense?
__________________
What sparks the intelligent, inflames the ignorant.
Guru
Join Date: May 2006
Location: 34.02S, 22.82E
Posts: 3794
#114
### Re: Time Hand-off Scenario Revisited
05/12/2016 2:28 PM
"Maybe I should drop the idea of clocks on her x-axis and just bring Bob's clock to the turnaround point which she can see without any delay or telepathy."
No, to suddenly bring Bob's clock to the turnaround point is roughly equivalent to telepathy - quite nonphysical. Bob has all the time in the world to set up and synchronize clocks anywhere in his inertial frame. As I said, Alice then just need to look out of the window and will find a 'Bob-synchronized clock" right there, next to her...
She then sees "Bob's now" and the Bob's observer, manning that clock, sees "Alice's now". When there is distance between them, the one can only see the others past, so there is no way that anyone can 'see' anyone elses future. Projections into the future is an acceptable art, but it is not 'seeing', neither is it reality.
So my advice, take articles that make claims like 'Alice's now is in Bob's future' with a large pinch of salt.
__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Power-User
Join Date: Aug 2008
Posts: 465
#115
### Re: Time Hand-off Scenario Revisited
05/13/2016 8:10 PM
"When there is distance between them, the one can only see the others past, so there is no way that anyone can 'see' anyone elses future."
Finally, these words make everything crystal clear to me, this is the answer I've been waiting for. They can only see each other's past because of the reciprocal time dilation and the distance adds relativity of simultaneity time to that. Maybe you didn't really look closely enough at Greene's analyses of problems in his course but he has absolutely no understanding of this fact. Every one has the approaching clock's now slice pointing into the stationary frame's future, a future that hasn't happened yet. It's because he doesn't visualize the spacetime diagram correctly. Bob's coordinate system is sliding horizontally along with the moving timeline (Bob's clock is right there as you said.). At the turnaround, his 5yr point is directly superimposed on Alice's 4 yr point (forget about how the x-axis for now.). When Alice turns around, Bob's coordinate system follows her back. So when you draw the now slices, they are not pointing to a future in Bob's original coordinate system to the left, they are pointing to Bob's coordinate system which is back at the turnaround point.Hence whether Alice is approaching or going away from Bob, she is always seeing Bob's past. After 2 years of intellectual imprisonment, Free at last!
__________________
What sparks the intelligent, inflames the ignorant.
Guru
Join Date: May 2006
Location: 34.02S, 22.82E
Posts: 3794
#116
### Re: Time Hand-off Scenario Revisited
05/14/2016 3:11 AM
I do not want to blow the 'freedom' that you struggled for so long, but your descriptions are not quite valid (maybe freedom is just poorly expressed).
"They can only see each other's past because of the reciprocal time dilation and the distance adds relativity of simultaneity time to that."
They see each others past because of the time delay due to the finite speed of light.
"Every one has the approaching clock's now slice pointing into the stationary frame's future, a future that hasn't happened yet."
Yes, this what Greene said, if I remember correctly.
"Bob's coordinate system is sliding horizontally along with the moving timeline (Bob's clock is right there as you said.). At the turnaround, his 5yr point is directly superimposed on Alice's 4 yr point (forget about how the x-axis for now.). When Alice turns around, Bob's coordinate system follows her back."
Huh?
If Bob is the reference frame, it 'slides' nowhere. However, he can have an observer with a clock synchronized to his that sits permanently at the turnaround point. That observer's clock will read 5 years and he can directly read Alice's clock as 4 years.
The issue about "see the others future" is somewhat like timezone issues: if an event happens here with me "now" (09:00 local time), it will be 03:00 in Ottawa, your "now". So if I immediately tell you about the event and its time, did I give you information from your future. Of course not...
__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Power-User
Join Date: Aug 2008
Posts: 465
#117
### Re: Time Hand-off Scenario Revisited
05/14/2016 8:35 AM
Ok maybe I can explain better what I'm saying through an analogy. It's the same one I used before but now I've corrected it.
Bob is a TV station back on earth. He is transmitting his present line up of TV shows. Alice is working out on a stationary bike. She has a TV screen in front of her receiving Bob's signal. But the faster she pedals, the slower the signal she receives and the slower the motion she sees on the screen. This means the longer she pedals (analogous to the farther she goes out into space) the greater the time lag between what she's watching and what Bob is transmitting (this is the relativity of simultaneity which builds up to 1.8 Bob yrs at the turnaround).
If she stops pedaling, she no longer receives the picture in slow motion but even though she's now in the same inertial time frame as Bob and receiving the signal at the same speed as Bob's transmitting and she shares the same present as Bob, she does not suddenly see the exact same signal Bob is transmitting because of the lag she built up. She has to somehow catch up by somehow going into a fast forward mode. This happens after she stops. The show she was watching will look to be in fast forward for a while until it catches up to the present signal Bob is transmitting. Then Alice will be viewing the show Bob is transmitting, minus the signal transmission delay, at normal transmission speed. It's like time is a rubber band wound up that takes time to recover its original shape once the winding stops and the propeller is let go. Is that right and what does that say about the deeper nature of time?
But what happens if Alice doesn't stop but she starts to back pedal instead (analogous to her returning to Bob). Her pedaling is still slowing down the picture reception and she is still falling behind in her viewing with even more time lag that she has to make up for. We know that by the time she stops pedaling backwards and returns to where she started, she will have caught up with the TV signal. She would have watched 10 years of TV in only 8 years while the whole time the signal was coming in slower than it was being transmitted. This means that even though the signal was still coming in slower on the return journey, the TV shows she was watching were going at fast forward. Weird, she forward pedals and they're playing in slow motion and backward pedals and they're in fast forward (after she's pedaled forward for a while of course).
Part of this effect comes from the decreasing signal delay as she approaches Bob but most of the effect comes from dissipating the lag caused by the relativity of simultaneity. But on the return journey, the rubber band of time is still being wound from the other end while the propeller has been released at the turnaround. I know you'd rather avoid going into my brain but is this the right way to view things?
__________________
What sparks the intelligent, inflames the ignorant.
Power-User
Join Date: Aug 2008
Posts: 465
#118
### Re: Time Hand-off Scenario Revisited
05/14/2016 11:16 AM
So are we agreeing now that Greene's interpretation of relativity is wrong? Twice in his course he talks about the reality of past present and future all existing concurrently. He even tells the story of Einstein consoling his friend's widow that his friend lives on in someone else's high velocity perspective. No, his friend does not live on, it's just the illusion of perspective. The present is universal, it's velocity that gives it time zones and time zones are just a convention.
When I say Bob's coordinate system slides over, it's just me visualizing Bob's clock being right beside Alice's clock. It also allows me to see that as Alice returns home, her now slice is not pointing at Bob's future in the starting point coordinate system but backwards to a new coordinate system where Bob's y-axis is at the turnaround point. This helps me visualize that reciprocal time dilation always includes a past time due to slower clocks.
Just let me put more detail into my stationary bike TV signal analogy. If Alice stops, there is a time (1.8 years) for Bob's TV signal to be viewed in fast forward mode by Alice and after that Bob's signal will run at normal mode but will be delayed to Alice by the speed of light delay over 3ly.
If Alice immediately turns around, a 2nd 1.8yr lag begins being added to the one she built up in her away journey. By the time she reaches Bob, the picture speed on her TV will be a combination of incrementally overcoming a 3.6yr lag due to relativity of simultaneity, an increasingly less delay of transmission signal minus the slowing of the signal due to reciprocal time dilation. All of this would make for an accelerating picture that suddenly goes into normal mode at the end of the journey.
__________________
What sparks the intelligent, inflames the ignorant.
Power-User
Join Date: Aug 2008
Posts: 465
#119
### Re: Time Hand-off Scenario Revisited
05/15/2016 5:19 PM
Actually I can think of so many different interpretations of what happens to the lag (relativity of simultaneity) for Alice stopping or turning around, I don't know which is correct.
If she stops and the 1.8yr lag just disappears then there will be a sudden discontinuity in the picture she's seeing. She'll somehow miss 1.8 yrs of TV shows. This is not likely because if Alice turns around, all 3.6 yrs of lag will dissipate once she reunites with Bob and there will be no discontinuity or missing programming. And it's also unlikely she suddenly adds another 1.8 yrs of lag at the turnaround point just because she turns around. Mathematically, the total relativity of simultaneity at he turnaround point is 3.6yrs not 1.8 yrs if she just stops.
If Alice stops at the turnaround point or turns around and stops at earth, there's no way the lag should influence the TV picture after she's stopped as I reasoned previously. Relativity of simultaneity is an illusion of perception, it's reciprocal just like time dilation and can't be residual once the two participants are in the same inertial frame, that would make it real like relative aging and it isn't.
So how is the built up relativity of simultaneity satisfactorily handled for both stopping or returning?
__________________
What sparks the intelligent, inflames the ignorant.
Guru
Join Date: May 2006
Location: 34.02S, 22.82E
Posts: 3794
#120
### Re: Time Hand-off Scenario Revisited
05/16/2016 12:45 AM
"So are we agreeing now that Greene's interpretation of relativity is wrong?"
No, I do not agree. He uses valid concepts, all based on the correct physics, but the words that he uses are imprecise and can be interpreted in many ways. This is the case for all verbal interpretations of SR, including my own efforts.
"So how is the built up relativity of simultaneity satisfactorily handled for both stopping or returning?"
A good example of the vagueness of words. Relativity of simultaneity does not "build up" in the 'twin paradox'. It suddenly changes when one of the two participants 'jumps' inertial frames.
Unless you settle down and start to define and solve the things that you asked algebraically, it is a waste of time and I'm not available to keep on correction verbal interpretations,
__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Power-User
Join Date: Aug 2008
Posts: 465
#121
### Re: Time Hand-off Scenario Revisited
05/24/2016 11:09 AM
I've figured it all out algebraically from the spacetime diagram for the Bob/Alice example at .6c: Yv is the total velocity vector through spacetime of which we can only see a shadow of it through space (x-axis) as velocity v. The time-axis component,applying Pythagoras, is the velocity through time v2Y/c which we can't see (derivation below). v2Y/c = vxY/ct = Vx/(ct') where t'=t/Y. The formula for relativity of simultaneity is Vx/c2 so c* the relativity of simutaneity = the velocity through time * time dilation. So since we have a direct correlation between time dilation and relativity of simultaneity, we no longer need bother with the mathematical involvement of simultaneity since it's the same thing as time dilation multiplied by a constant.
1. Derivation of velocity through time w:
Y2v2 = v2 + w2
w2 = Y2v2 - v2
w2 = v2(Y2-1) where Y2 = c2/(c2-v2)
w2 = v2((c2-c2+v2)/(c2-v2))
w2 = v2(v2/(c2-v2))
w = v2c/(c*sqrt(c2-v2))
w = v2Y/c
If w is the velocity through time what do the time t" and distance x" components mean where w = x"/t"? T" is the time spent traveling through time. It appears to us as the difference in time between the stationary frame's proper time and the moving frame's proper time. In the example of Alice leaving bob at .6c, traveling out 3 ly and returning, Alice ages 4 yrs her proper time at the turnaround point while Bob ages 5 yrs in his proper time. Alice has aged 1 year less than Bob to get to the same universal present and this is the time spent traveling through time. The formula for this is t" = t'(Y-1) and the corresponding formula for x"=vx(Y-1)/c since w=x"/t" where x" is the distance traveled through time.
At this point we need some clarification of the terms. The universal present is the time all clocks are sync'd to. Earth would have sent out a timing signal to a planet 3ly away at least 3 yrs before Alice would go to that planet. Say that signal was sent out in the year 3000, the far planet would have been sync'd to the earth's years starting at 3003. Alice would arrive in the year 3005 and her clock would register 3004. She has not gone 1 yr into the future as relativists would say, she has merely took less of her proper time to reach the same universal present. She took 4 years (plus 1 year of invisible time traveling through the time dimension) while Bob took 5 to reach the same universal present. There is no time travel for one to a future that hasn't happened yet to the other as most people believe.
In the Bob and Alice example where both are running TV stations, Alice would watch a year of Bob's programming at half speed during her movement away from Bob at .6c. If Alice was also transmitting a TV signal, Bob would also be watching her programming in slow motion. (This sounds like time dilation but you'll see later that it is not.)The farther they move apart, the greater the lag of the TV signal reaching them. This is due to the time delay of the speed of light. Alice would receive the first year of programming (3000-3001) between 3000 and 3002 her years. The planet Alice is traveling to would receive the 3000-3001 schedule in 3003-3004. The next schedule from 3001-3002 would be received by Alice in 3002-3004 her years while the planet would receive it from 3004-3005 universal time.
If Alice stops at this point, she starts receiving the same programming at the same speed as the planet does and at the same speed Bob transmits, so she is fully in the present, but her clock reads 1 year less. She is not 1 year in the past, she has just aged 1 year less to get to a common present. Her missing year also doesn't change the fact that the signal is delayed by 3 years. So just as one can't access another's future, there's also no access to or simultaneity with the past. Relativity has no outside the ordinary time travel at all except for a path with less aging to the present.
On her way back, she watches 2 years of Bob's programming in only 1 of her years (which pass at the exact same rate for her as Bob's years pass for him). It looks as if the outside world is running twice as fast as her time but in actuality Bob's time is dilating as slow as it did in the outbound journey. Bob also watches her programming at twice the speed but she'll get his 3002-3004 schedule between her 3004-3005 timeframe (which equals 3005-3006.25 universal time) but he'll have to wait until 3008 to stop viewing Alice's programming at half speed. So he watches her slow motion TV shows for 8 years while she watches his for 4 of her years. We don't need to differentiate between his and her years because both run at the same proper rate.
Even though it looks to her like she's catching up to Bob's programming because it's coming in at twice the rate and the closer she moves to Bob there's less of a time delay due to the signal not having to travel as far, Alice is still losing a quarter year, with respect to Bob, for every year she's speeding back. No matter whether Alice moves away or towards Bob, she ages less than Bob and the time dilation is the same no matter how fast or how slow she's observing time outside of her own.
There are other absolutes besides a universal present that relativity doesn't openly accept but secretly does. There's also a universal rate of aging within each inertial frame no matter what its constant velocity. Distance is also an absolute and so is who is moving. At the beginning of any journey, the two participants decide who is staying and how far the other is going, there is no need to consider both points of view as the stationary frame. The one that's moving is the one that goes from point A to point B with minimal distance separation from both points. With x=0, the spacetime interval is minimized to time only (distance adds to a longer spacetime interval). The proper distance for both is judged from the stationary frame and length contraction during motion need not be considered.
The stationary frame is a common ground at the start of the journey. Either participant can be moving relative to the stationary frame and that relative velocity can be used to determine which participant has a greater share of their mutual relative velocity. This common patch of ground relative to both allows one to understand why there are cases where a high relative velocity between participants yields no relative time dilation between them.
Other points of interest on the cartesian coordinates (stationary frame) of the spacetime diagram of the Bob/Alice example:
Start: t=0, x=0.
1. Velocity through spacetime: Yv
2. Velocity through time: v2Y/c or vx/ct' or t'(Y-1/Y)
3. Time spent travelling through the time dimension: t" = t'(Y-1)
4. Slope of Alice's timeline = t/x or 1/v.
5. X-axis coordinates (t=0) corresponding to t': x = t'(Y-1/Y)/v
6. Time-axis coordinates (x=0) when Bob starts transmitting signal which will be received at t': t = t'Y(1-v/c2)
7. Number of years each takes to watch a year of transmitted programming: outbound = 1/Y(1-v/c2), inbound is inverse
8. Time-axis coordinate for Alice's lines of simultaneity: t = t'/Y
9. The universal present time: t= Yt'
10. The relativity of simultaneity (diff between the universal present and #8): Vx/c2 = (Y - 1/Y)t'
11. Coordinate translation from (t',0) to (t,x) is (Yt', (Yt'-t'/Y)/v)
Here are some sample calculatons for v=.6c, Y=10/8
1. Yv = .75c
2. w = .45c
3. t" = .25t'
4. 1/v = 5/3
5. x = .75t'
6. t = .5t'
7. outbound is 2 yrs to watch 1 yr in slow motion. inbound 1/2 yr to watch 1 yr at fast forward
8. t = .8t'
9. t = 1.25t'
10. Vx/c2 = .45t' aka relativity of simultaneity = velocity through time * time dilation
11. (t',0) = (1.25t', .75t')
Here are some sample calculatons for v=.8c, Y=5/3
1. Yv = 1.333c
2. w = 1.0666c
3. t" = .666t'
4. 1/v = 1.25
5. x = 1.333t'
6. t = .333t'
7. outbound is 3 yrs to watch 1 yr in slow motion. inbound 1/3 yr to watch 1 yr at fast forward
8. t = .6t'
9. t = 1.666t'
10. Vx/c2 = 1.0666t' aka relativity of simultaneity = velocity through time * time dilation
11. (t',0) = (1.666t', 1.333t')
__________________
What sparks the intelligent, inflames the ignorant.
Power-User
Join Date: Aug 2008
Posts: 465
#122
### Re: Time Hand-off Scenario Revisited
06/04/2016 2:29 PM
I'm stuck on how to interpret parts of the spacetime diagram in the .6c example. Specifically, I don't know how relativity interprets past, present and future. I know how Greene defines them and he clearly articulates it very well but I can't agree with his conclusion that all 3 exist concurrently. To me, if you're passing a clock and you get a signal from it that differs from the signal a stationary person at the clock is getting, then you're either in the past or in the future of that stationary person which is impossible. (However, I've since figured out how 2 functioning clocks at the same spot can have different readings of the same present moment without implying either is in the future or past of the other; read on.)
When people say relativity allows a short cut to the future, it doesn't mean you're arriving at the future and waiting for the people trudging the regular way through time to catch up. It's more like you arrive at the same present moment but you've aged relatively slower to get there. That's a big difference in interpretation.
Let's look at the spacetime diagram where t'=1. By the formula t=Yt', t=1.25. Since t' and t are on the same horizontal x-axis, it means that t' is in our present even though the two clocks differ. A ship at t' would be passing a clock synchronized to earth time but it would not see a signal from it saying t=1.25. By reciprocity of time dilation, that clock would be moving relatively to the ship and dilating. If the ship's time was 1, the time on the earth clock would read 1/Y or .8 to the guy in the moving ship.
I thought time dilation reciprocity would yield the same clock value for each; but instead we see his clock at 1.0 and he sees ours at .8 while this is all happening at 1.25 our time. But if he is reading our clock as .8 then that implies he is presently .45 years in our past. He has supposedly traveled .45 years into the past plus we see him simultaneously share our present (although his clock reads differently than ours) while he's in our past? Not only that, but at his 1 yr mark his most recent TV signal from us is from 3/4 yr ago even though his present time points to being .45 yr in our past. This would suggest he can view our TV sooner than it could arrive by light speed which is impossible. There must be another interpretation even though this is what the spacetime diagram is suggesting. The answer must be that the readings themselves mean nothing because they are between two relatively unsynchronized clocks from both perspectives.
We need to introduce the fudge factor of relativity of simultaneity to read the clocks as if they were synchronized so we don't come to the erroneous conclusion that the moving guy is simultaneously existing in our past or traveling in any time but our present.
The relativity of simultaneity fudge factor = .45 from the moving guy's perspective. Add that to the .8 time dilation he sees of our time and the total is 1.25 which agrees with our present time. So when you correct for the lack of sync between our clock and his, they both agree on what the present time is. But from our perspective there is no relativity of simultaneity fudge factor. Instead there is the moving guy's .25 yr invisible time spent in the time dimension. Add that to the 1.0 time dilation we see of his clock and there is again agreement that the present time is 1.25.
But this is an unsatisfying interpretation for me. Why would it be that from one perspective there is this invisible time spent in the 4th dimension that clocks can't record and from the other perspective there is this invisible fudge factor of relativity of simultaneity that clocks also can't record. This led me to look for a mathematical relationship of how to express relativity of simultaneity also as invisible time spent in the 4th dimension. Here it is:
We know that the velocity component through time w = c * time for the relativity of simultaneity(trs)/ t' which can be expressed also as (Y-1) t'/Y and t'(Y-1)(1+1/Y). t'(Y-1) = t" which is the invisible time spent in the 4th dimension by the moving frame. So trs = (1+1/Y)t".
So now we can express the fudge factor of how to correct the clock readings of both perspectives using only the concept of invisible time spent in the 4th dimension. The time on our universal stationary clock (no need to read me the riot act on the unorthodoxy of that) is 1.25. In order to reconcile our reading of the moving clock which we see as 1.0, we need to add t"= .25. In order for the .8 reading of our clock from the perspective of the moving clock, they need to add the fudge factor of (1+1/Y)t" = .45. No need to ever mention relativity of simultaneity again except as a relic of the old interpretation just like lorentz contraction and relativistic mass are.
So what have we learned here:
1. There is a universal present time that both (not all) frames can agree upon. The Earth could seed the universe with clocks that are all synchronized in the Einstein way meaning they take into account the speed of light delay from the source to each of their proper distances from the source. There is no ability to take a snapshot of the universe in the present moment because of the information delay in creating that snapshot. However, if each clock took timestamped photos, the present moment could be post-processed sometime later of what everything looked like at a universal present. Otherwise everything around each clock is a view of the past from that clock; the farther away, the more into the past. This is how the fabric of the present moment is distorted.
2. Nothing exists in the past because the past does not exist. We can see the past due to the speed of light delay but things from 2 different times cannot occupy the same point in space. There is no time travel in the movie sense (or Brian Greene sense) to either the past or an already written future.
3. The only reason clocks at the same point in space can differ because one is moving is that clocks cannot register time movement through the 4th dimension. Just as we have no instruments to register the velocity through the time dimension of the total spacetime velocity (Yv), we can't register the time spent traveling through the time dimension. That time manifests itself as lost time and slower aging for moving frames.
4. Sometimes what's on a clock is not the correct time even though the clock is running correctly.
a) See #3 above. The reading on the clock is a measure of time spent in the space dimension but is missing the time spent in the time dimension.
b) Reciprocal time dilation due to relative velocity. Both are illusions of perspective so far as I'm concerned because if their effects leave no lasting impression , they are illusions. For example, 2 ships racing together, towards each other or away from each other at the same speed relative to a common stationary frame will register no time dilation on each other's clocks even though their relative velocity can be any value. For ships moving at different speeds relative to a common frame, their relative velocity gives no indication of who is aging slower until one or the other returns to the same constant velocity as the other. The reciprocal time dilation becomes real at that point for only one of them, manifested as slower relative aging, while it evaporates for the other as a mirage.
c) A biological clock's read out, as in suspended animation, can be slowed by cold temperature slowing just the clock itself without any slowing of time. Apparently whether time slows or just the clock slows, one can arrive at a subsequent present having aged less in either case.
5. The past can be seen by the speed of light delay. This does not mean you have traveled into the past or that past and present co-exist simultaneously.
6. The passage of time on the stationary frame can look slowed down or sped up when viewed by the moving frame. This is a combination of the speed of light delay, the speed and direction of the ship. It is independent of relative clocks as a TV show from earth can look to be in fast forward when viewed by an incoming ship even though time is slowed from both perspectives.
Should I submit this as a paper to Airvx? Is that the right name?
__________________
What sparks the intelligent, inflames the ignorant.
Power-User
Join Date: Aug 2008
Posts: 465
#123
### Re: Time Hand-off Scenario Revisited
06/04/2016 3:19 PM
I now have only 1 mathematical proof to work on. I don't accept the interpretation that time does not pass for a photon. I need to therefore prove that massless particles have no velocity or time component in the time dimension, their total time and velocity is spent only in the space dimension. I have to somewhere prove as Y -> infinity, t" collapses suddenly to zero. But it may be deeper than this since a change in inertia is required to determine which frame is relatively aging which is somehow related to mass which massless particles don't have. It could take me a while.
__________________
What sparks the intelligent, inflames the ignorant.
Power-User
Join Date: Aug 2008
Posts: 465
#124
### Re: Time Hand-off Scenario Revisited
06/04/2016 9:41 PM
Maybe the answer is as simple as that since light is always at a constant velocity and can never experience inertia because it has no mass, that it must always be considered the stationary frame. As the stationary frame , it can have no velocity component or time component through time; only the moving frame can have those. Well I guess I've reached the end of special relativity.
__________________
What sparks the intelligent, inflames the ignorant.
Power-User
Join Date: Aug 2008
Posts: 465
#125
### Re: Time Hand-off Scenario Revisited
06/06/2016 6:17 AM
Hmmm no response. I guess I've committed blasphemy again even with math. I suppose I'll go on the science chat forum http://www.sciencechatforum.com/index.php now and see if anyone will bite there.
__________________
What sparks the intelligent, inflames the ignorant.
Guru
Join Date: May 2006
Location: 34.02S, 22.82E
Posts: 3794
#126
### Re: Time Hand-off Scenario Revisited
06/06/2016 7:47 AM
Sorry Ralf, I'm in the process of moving house and have not looked at the forums much lately. In any case, what you have suggested as a 'solution' is so ridiculously wrong that I do not even want to respond to it. This thread is considered closed now.
Just make sure you post on "personal theories" on other forums, otherwise you may meet with some grief there...
-J
__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Power-User
Join Date: Aug 2008
Posts: 465
#127
### Re: Time Hand-off Scenario Revisited
06/06/2016 9:09 AM
Whew! I was worried you were going to say this is so painfully obvious that there's absolutely no new idea or interpretation here. Now I'm elated except for the coming months of being set upon by the ravenous wolves that inhabit physics forums. Someone's going to eventually see the elegance of what I've stated here. I assume you don't want me bringing up your name in any future discussions?
__________________
What sparks the intelligent, inflames the ignorant.
Guru
Join Date: May 2006
Location: 34.02S, 22.82E
Posts: 3794
#128
### Re: Time Hand-off Scenario Revisited
06/06/2016 10:49 AM
Yes, you've assumed right. Past experience has not been good when you have quoted me, because you tend to color it with your own 'strange' relativistic thinking.
All I can say is good luck!
-J
__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Power-User
Join Date: Aug 2008
Posts: 465
#129
### Re: Time Hand-off Scenario Revisited
06/06/2016 10:53 AM
OOps I found 2 mistakes. w= v2Y/c = vx/(ct') = ctrs/t' = Y-1/Y not t'(Y-1/Y).
Also trs= t"(1+1/Y)/c.
__________________
What sparks the intelligent, inflames the ignorant.
Power-User
Join Date: Aug 2008
Posts: 465
#81
Find in discussion
### Re: Time Dilation and Lorentz Contraction Revisited
04/26/2016 11:34 AM
If Alice only did a flyby with Bob and then a flyby of the distant clock, she and Bob would not have agreed that the remote clock measures Bob's aging correctly. She was not present at the synchronization and although she has read 12.5 years on the clock's display, she is hence not forced to conclude that Bob has aged that much. In fact she may be inclined to think that maybe Bob has aged only 4.5 years! Can you think why I'm entitled to say this?
There has been no inertial frame jump so reciprocal time dilation is in effect. From Alice's view, Bob could be moving and she's stationary. His clock could be dilating in relation to hers so 7.5 * .6 = 4.5. The rulebook rules!
__________________
What sparks the intelligent, inflames the ignorant.
Power-User
Join Date: Aug 2008
Posts: 465
#82
Find in discussion
### Re: Time Dilation and Lorentz Contraction Revisited
04/26/2016 11:38 AM
x
__________________
What sparks the intelligent, inflames the ignorant.
Power-User
Join Date: Aug 2008
Posts: 465
#3
### Re: Time Dilation and Lorentz Contraction Revisited
03/28/2016 8:46 AM
The answer to your twin paradox variant question is in your book. It is the twin paradox done where each participant is put in the "stationary" frame. Alice will age less in both scenarios. When they open their envelopes is critical. The answer is from your post in the conventionality of relativity:
Point 3: If the probe only whizzed by us and also whizzed by Pluto, we would have no idea about elapsed time difference (ignoring the obvious gravity influence in this hypothetical scenario). If it 'stopped' at Pluto there had to be a change of inertial frames and we would have known the absolute time aging difference, although we could not have directly measured it.
If Alice kept going and Bob stayed put we couldn't determine who aged less. If Alice stopped and Bob stayed put we would have known the absolute time aging difference, although we could not have directly measured it.
__________________
What sparks the intelligent, inflames the ignorant.
Guru
Join Date: May 2006
Location: 34.02S, 22.82E
Posts: 3794
#6
### Re: Time Dilation and Lorentz Contraction Revisited
03/28/2016 11:54 AM
"The answer to your twin paradox variant question is in your book. It is the twin paradox done where each participant is put in the "stationary" frame. Alice will age less in both scenarios."
All the diagrams in the eBook are for the standard twin paradox, which is case one. In case two, Bob also has to accelerate and he has to make a much larger inertial frame change than Alice (in order to catch up). So please rethink the answer.
I think the situation before the opening of the envelopes is more or less covered in my prior reply on the 'time hand-off' scenario. So consider it in that light.
-J
__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Power-User
Join Date: Aug 2008
Posts: 465
#4
### Re: Time Dilation and Lorentz Contraction Revisited
03/28/2016 10:24 AM
You may think we're at the end but we're not, I have other questions about relativity that aren't connected to the subject discussed here and are equally as frustrating.
Anyhow, back to this subject matter. You say the idea of the speed of light being constant in all frames is a convention, not directly supported by empirical evidence. The evidence comes indirectly from two-way speed of light analysis supported by a clever clock synchronization scheme and a multitude of experimental evidence that does not contradict the initial assumption hence the initial assumption must be correct.
As a result of that initial assumption, we can't make any determination of who is aging slower when a ship is leaving earth on a one-way trip. The ship is not allowed to get empirical evidence of whether the earth's clock is ticking faster than his because relativity says our clock must be ticking slower than his.
The problem is that if we get the empirical evidence to support this, it contradicts the basic assumption that relative velocity by definition means there is no way to tell what is each individual's share of that relative velocity.
In the analysis where Alice is put in the stationary frame for half the journey, yes, Bob's clock on earth IS ticking slower in the virtual analysis sense but when she fires up her rockets to catch up with Bob, her extra speed erases his slower aging and she ends up aging even slower than Bob overall. If the ship and the earth had been in constant communication with each other, compensating for the light delay of the messages in post-processing, there could not have been two contradictory scenarios of messaging one from assuming Alice was initially moving and another assuming Alice was stationary. In the first assumption, Alice would have always been aging slower throughout the journey. In the second assumption, Bob would have initially been aging slower than Alice but then Alice would have been aging much slower than Bob. There is only going to be one of these scenarios that is real and by the messaging one can actually tell who had the individual share of the relative velocity which relativity forbids.
__________________
What sparks the intelligent, inflames the ignorant.
Guru
Join Date: May 2006
Location: 34.02S, 22.82E
Posts: 3794
#8
### Re: Time Dilation and Lorentz Contraction Revisited
03/28/2016 12:10 PM
"As a result of that initial assumption [conventional synchronization], we can't make any determination of who is aging slower when a ship is leaving earth on a one-way trip."
We cannot make a proper elapsed time observation in a one-way situation, but if Alice started in Earth's inertial frame, we know that she had to changed inertial frames to fly away. Doesn't that tell us enough to make an inference on aging?
"There is only going to be one of these scenarios that is real and by the messaging one can actually tell who had the individual share of the relative velocity which relativity forbids."
No, messages can give us only a relative velocity. As above, if starting off in the same frame initially, one can infer something from the magnitudes of frame changes that each suffered.
-J
__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Guru
Join Date: May 2006
Location: 34.02S, 22.82E
Posts: 3794
#13
### Re: Time Dilation and Lorentz Contraction Revisited
03/29/2016 4:07 AM
You wrote: "Anyhow, back to this subject matter. You say the idea of the speed of light being constant in all frames is a convention, not directly supported by empirical evidence."
The correct terminology would have been "the observed propagation speed of light is isotropic in all inertial frames" or simply "the observed one-way speed of light is the same in all inertial frames".
We know that this needs two Einstein-synchronized clocks, some distance apart, but inertially at rest relative to each other. There is in fact direct evidence for Einstein's synchronization method: https://en.wikipedia.org/wiki/One-way_speed_of_light#Slow_clock-transport
"If however one clock is moved away slowly in frame S and returned the two clocks will be very nearly synchronized when they are back together again. The clocks can remain synchronized to an arbitrary accuracy by moving them sufficiently slowly. If it is taken that, if moved slowly, the clocks remain synchronized at all times, even when separated, this method can be used to synchronize two spatially separated clocks. In the limit as the speed of transport tends to zero, this method is experimentally and theoretically equivalent to the Einstein convention."
This has been experimentally verified to great precision. See http://ntrs.nasa.gov/search.jsp?R=19940006498, or more fully described in: http://tycho.usno.navy.mil/ptti/1992papers/Vol%2024_10.pdf. This is good enough for me (and for every scientist/engineer that I know of).
-J
__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Power-User
Join Date: Aug 2008
Posts: 465
#130
### Re: Time Dilation and Lorentz Contraction Revisited
08/18/2016 10:43 AM
This will soon be the most viewed thread in the last 4 years so anyone stumbling on here and wondering where they can go to get more ralfativity, the conversation is still going on hot and heavy at
http://www.sciencechatforum.com/index.php
science forum, physics, ralfativity
__________________
What sparks the intelligent, inflames the ignorant.
Reply to Blog Entry Page 2 of 2: « First < Prev 1 2 Last »
Interested in this topic? By joining CR4 you can "subscribe" to | 13,643 | 58,564 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-33 | latest | en | 0.914501 |
https://nus.kattis.com/problems/alloys | 1,660,851,326,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573399.40/warc/CC-MAIN-20220818185216-20220818215216-00025.warc.gz | 405,941,763 | 6,826 | Hide
# Alloys
Pelle is building a new advanced underground drill. He wants to build the drill out of a new titanium-aluminium-magnesium alloy. Such an alloy can be mixed in many ways. If an alloy is made with the ratios $x, y, z$ ($x + y + z = 1$) of titanium, aluminium and magnesium, the alloy gets a hardness of $xy$, and a price of $x + y$ SEK per kilogram.
Pelle only has a fixed amount of money per kilogram that he can spend on alloys for his drills. What is the maximum hardness he can get on his new alloy?
## Input
The first and only line of input contains a decimal number $c$ ($0 \le c < 2$, between $1$ and $6$ digits after the decimal point) – the amount of money in SEK that Pelle can spend per kilogram alloy.
## Output
Output a single decimal number, the maximum hardness of Pelle’s alloy.
Your answer will be considered correct if you have an absolute error of at most ${10}^{-6}$.
Sample Input 1 Sample Output 1
1.0
0.25
CPU Time limit 1 second
Memory limit 1024 MB
Statistics Show | 268 | 1,011 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2022-33 | latest | en | 0.8548 |
http://opensees.berkeley.edu/wiki/index.php/Rigid_Diaphragm_Consequences | 1,505,860,561,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818686043.16/warc/CC-MAIN-20170919221032-20170920001032-00623.warc.gz | 264,216,804 | 8,866 | Rigid Diaphragm Consequences - OpenSeesWiki
# Rigid Diaphragm Consequences
An unintended consequence of using the RigidDiaphragm constraint (or other constraint) on a beam column element is a result of the fact that it enforces a condition of zero axial strain on the element. For sections where the neutral axis does not shift as a consequence of bending in the beam this constraint has little effect on the axial load in the beam, e.g. elastic or steel sections. For those beams where the neutral axis does shift, i.e. Reinforced Concrete beams modeled using fiber sections under moment, the moments in an unconstrained beam will cause the beam to undergo axial deformation. To constrain this axial deformation requires axial forces, the magnitude of which can be significant.
Here is a stupid made up example that illustrates my point:
```foreach rigidConstraint {no yes} {
puts "RIGID CONSTRAINT: \$rigidConstraint"
foreach matType {steel concrete} {
foreach eleType {forceBeamColumn dispBeamColumn} {
wipe
# create model builder
model basic -ndm 2 -ndf 3
set width 360
set height 144
# create nodes
node 1 0.0 0.0
node 2 \$width 0.0
node 3 0.0 \$height
node 4 \$width \$height
# Fix supports at base of columns
# tag DX DY RZ
fix 1 1 1 1
fix 2 1 1 1
if {\$rigidConstraint == "yes"} {
equalDOF 3 4 1
}
if {\$matType == "concrete"} {
# CONCRETE tag f'c ec0 f'cu ecu
# Core concrete (confined)
uniaxialMaterial Concrete01 1 -6.0 -0.004 -5.0 -0.014
# Cover concrete (unconfined)
uniaxialMaterial Concrete01 2 -5.0 -0.002 0.0 -0.006
# STEEL
# Reinforcing steel
set fy 60.0; # Yield stress
set E 30000.0; # Young's modulus
# tag fy E0 b
uniaxialMaterial Steel01 3 \$fy \$E 0.01
} else {
set fy 60.0; # Yield stress
set E 30000.0; # Young's modulus
# tag fy E0 b
uniaxialMaterial Steel01 1 \$fy \$E 0.01
uniaxialMaterial Steel01 2 \$fy \$E 0.01
uniaxialMaterial Steel01 3 \$fy \$E 0.01
}
# Define cross-section for nonlinear columns
# ------------------------------------------
# set some paramaters
set colWidth 15
set colDepth 24
set cover 1.5
set As 0.60; # area of no. 7 bars
# some variables derived from the parameters
set y1 [expr \$colDepth/2.0]
set z1 [expr \$colWidth/2.0]
section Fiber 1 {
# Create the concrete core fibers
patch rect 1 10 1 [expr \$cover-\$y1] [expr \$cover-\$z1] [expr \$y1-\$cover] [expr \$z1-\$cover]
# Create the concrete cover fibers (top, bottom, left, right)
patch rect 2 10 1 [expr -\$y1] [expr \$z1-\$cover] \$y1 \$z1
patch rect 2 10 1 [expr -\$y1] [expr -\$z1] \$y1 [expr \$cover-\$z1]
patch rect 2 2 1 [expr -\$y1] [expr \$cover-\$z1] [expr \$cover-\$y1] [expr \$z1-\$cover]
patch rect 2 2 1 [expr \$y1-\$cover] [expr \$cover-\$z1] \$y1 [expr \$z1-\$cover]
# Create the reinforcing fibers (left, middle, right)
layer straight 3 3 \$As [expr \$y1-\$cover] [expr \$z1-\$cover] [expr \$y1-\$cover] [expr \$cover-\$z1]
layer straight 3 2 \$As 0.0 [expr \$z1-\$cover] 0.0 [expr \$cover-\$z1]
layer straight 3 3 \$As [expr \$cover-\$y1] [expr \$z1-\$cover] [expr \$cover-\$y1] [expr \$cover-\$z1]
}
# Define elements
# ----------------------
geomTransf Linear 1
set np 5
# Create the coulumns using Beam-column elements
# e tag ndI ndJ nsecs secID transfTag
element \$eleType 1 1 3 \$np 1 1
element \$eleType 2 2 4 \$np 1 1
element \$eleType 3 3 4 \$np 1 1
# --------------------
# Set a parameter for the axial load
set P 180; # 10% of axial capacity of columns
# Create a Plain load pattern with a Linear TimeSeries
pattern Plain 1 "Linear" {
# Create nodal loads at nodes 3 & 4
# nd FX FY MZ
load 3 0.0 [expr -\$P] 0.0
load 4 0.0 [expr -\$P] 0.0
}
system BandGeneral
constraints Transformation
numberer RCM
test NormDispIncr 1.0e-12 10 3
algorithm Newton
analysis Static
analyze 10
pattern Plain 2 "Linear" {
}
integrator DisplacementControl 3 1 0.1
analyze 10
set strains [eleResponse 3 basicDeformation]
set forces [eleResponse 3 forces]
puts "eleType: \$eleType matType \$matType axialForce [lindex \$forces 0] axialDeformation: [lindex \$strains 0]"
}
}
}```
and the results when this script is run are:
```RIGID CONSTRAINT: no
eleType: forceBeamColumn matType steel axialForce -0.08875612010504939364 axialDeformation: 0.00000291960921383616
eleType: dispBeamColumn matType steel axialForce -0.00000000000010219973 axialDeformation: 0.00000000000000000000
eleType: forceBeamColumn matType concrete axialForce 1.25405861627505310629 axialDeformation: 0.12918648354852413362
eleType: dispBeamColumn matType concrete axialForce 3.61866611884458500015 axialDeformation: 0.17247077365271457072
RIGID CONSTRAINT: yes
eleType: forceBeamColumn matType steel axialForce -0.00000000000010287569 axialDeformation: 0.00000000000000000000
eleType: dispBeamColumn matType steel axialForce -0.00000000000013310834 axialDeformation: 0.00000000000000000000
eleType: forceBeamColumn matType concrete axialForce 186.25340185913745472135 axialDeformation: 0.00000000000000000000
eleType: dispBeamColumn matType concrete axialForce 163.64106186688189836786 axialDeformation: 0.00000000000000000000``` | 1,759 | 5,385 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2017-39 | latest | en | 0.640706 |
https://www.i-programmer.info/news/112-theory/13086-terry-tao-almost-proves-collatz-conjecture.html | 1,601,086,017,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400232211.54/warc/CC-MAIN-20200926004805-20200926034805-00286.warc.gz | 846,365,773 | 11,418 | Too Good To Miss: Terry Tao Almost Proves Collatz Conjecture
Written by Mike James
Tuesday, 31 December 2019
There are some news items from the past year that deserve a second chance. Here we have one such - although not as well known as the long standing P=NP conjecture, Collatz has fascinated people for the past eight decades and produced almost as many flawed proofs. Now mathematician Terence Tao seems to be close to a proof.
(2019-09-13)
If you don't know what the Collatz conjecture is then you haven't missed out on anything really important - with the possible exception of the excellent xkcd cartoon reproduced below:
The cartoon is accurate, but let's make the conjecture, which was proposed in 1937 by German mathematician Lothar Collatz, clear:
• Pick a number, a positive integer
• If even divide by 2
• If odd multiply by 3 and add one
• Repeat above two steps with new value
If you try it you will discover that you eventually reach a result of 1. For example, 10, 5,16, 8, 4, 2, 1. At first you think it must be an accident and so you try a few more tests. Then you become obsessed and write a program - and you always end up at 1.
So far it has been verified for values up to 5.76x10^18.
The Collatz conjecture is that this is indeed always true, but can you prove it?
If you think it should be easy then it is worth knowning that Paul Erdős proclaimed:
"Mathematics may not be ready for such problems"
Terence Tao at the University of Califonia is closer to a proof than we have ever been. Why isn't it a proof? Because it's a probability argument:
Almost all orbits of the Collatz map attain almost bounded values.
Why is this important? The reason is that if you can show that the minimum of the Collatz sequence for N is always smaller N for all N>1 then you have proved the Collatz conjecture. The reason is simple. If you get a value m which is smaller than N then the rest of the sequence for N is the same as that as if you had started from m and so we now have that the minimum for the sequence is less than m. You can repeat this and work back to prove the the minimum has to be 1.
Tao hasn't quite proved this but he has proved (rewritten to be closer to English):
Theorem 2 Let f be any function defined on the integers, excluding zero, and f(N) goes to infinity with N then the minimum of the Collatz sequence for N is less than f(N) for almost all N.
If you take f to be the identity than we have the minimum Collatz value for N is smaller than N.
So problem solved?
Not quite. The weasel word in the theorem is "for almost all". This is a signal that the theorem is probablistic. What this means is that the set of N for which this is true is dense (in the sense of logarithmic density). What this means, in a non-technical sense is that we have proved that the theorem is true for all but an insignificant number of cases.
An insignificant nubmer of cases also includes no cases at all, so the theorem might apply to all N or there might a few examples where it doesn't apply and the Collatz conjecture is wrong after all.
So which is it?
As Tao writes in an answer to a comment to his blog post:
"...but this is one of these situations where there seems to be an enormous gap in difficulty between “almost all” results and “all” results."
It looks as if the Collatz conjecture is going to hold out for a while longer. Armed with this probablistic argument it might be possible to see new reasons why it is true.
Almost all orbits of the Collatz map attain almost bounded values (paper pdf)
#### Related Articles
Erdos Conjecture Proven
Travelling Salesman - A Movie About P=NP
Collatz conjecture proved?
Update on the Proof of P≠NP?
NP-Complete - Why So Hard?
Computational Complexity
Computability
Ruby On Rails Survey14/09/2020The latest Ruby on Rails Community Survey from Planet Argon presents a wealth of information about Rails developers, and the tools they choose to use. + Full Story GitHub Announces Container Registry15/09/2020GitHub has launched a beta version of GitHub Container Registry, a service aimed at improving how containers are handled within GitHub Packages. + Full Story More News | 981 | 4,178 | {"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-2020-40 | longest | en | 0.960504 |
https://www.aceliberty.com/econ-214/econ-214-inquizitive-ch-12-liberty-university-solution | 1,603,450,207,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107881369.4/warc/CC-MAIN-20201023102435-20201023132435-00389.warc.gz | 611,690,467 | 18,749 | # ECON 214 InQuizitive ch. 12 Liberty University Solution
Fill in the blanks to complete the passage about the key difference between Nogales, Arizona, and neighboring Nogales in the Mexican state of Sonora.
The two cities are in the same location and have similar – and geography, but the U.S. city has a – level of education, – infant mortality, and better roads. The key difference is – that promote security of people and property on the one side but not the other.
Fill in the blank to complete the sentence.
Diminishing marginal product sets in after the –ladder.
Consider input X as one factor leading to output Y. What is the marginal product of X?
We can use the marginal product of resources to analyze the relationship between capital and output in an economy. Drag each word or phrase to the appropriate blank.
In the – production function, the slope of the function corresponds to the –. If the slope is –, we know that output is increasing. As the slope declines, that is a sign of – marginal product.
Drag each word to the appropriate blank space to complete the passage.
The original Solow growth model assumed that technology growth is –, which means that innovation is –the economy.
Which steps did Chile take to reform its economy and increase its growth rate?
Fill in the blanks to complete the passage about patents.
A patent gives the holder a – monopoly. That time period allows – to profit off the technology without others being able to do the same, which is an incentive that encourages –.
Seth is trying to decide how many workers is the optimal amount at his factory, which produces artisan-crafted grooming products. He observes the following:
What is the marginal product of the third worker?
Assuming the blue production function (F1) is the initial state of a country’s economy, click on the production function after a surge in the nation’s technology sector, for instance, due to government funding.
We can use the total output to determine what is happening with the marginal product, or we can use the marginal product to analyze total output. Drag each word or phrase to the appropriate blank space in order to correctly complete the table.
Place each item in its correct location in the diagram to illustrate the change of events that starts with a growth-friendly institution and ends with economic growth.
Place in order the events that start with the right institution being put in place and end with growth.
Macroeconomic theories evolve over time. They both shape and respond to real-world circumstances.
Beginning with the observation of real-world events, put the following events in the correct order to describe the continuous interplay between economic theory and the real world.
Drag each development in the economy to the graph that depicts its effect.
The focus of macroeconomics changes over time, depending on the state of the economy. Fill in the blanks to complete the passage.
After the –, the primary focus of macroeconomics shifted to –, that is, to short-run – of the economy. Toward the end of the last century, the focus returned to –growth.
Select all the events that could result in a nation’s increasing the slope of its entire production function. | 647 | 3,236 | {"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-2020-45 | latest | en | 0.935423 |
https://www.coursehero.com/file/5653598/Chp2-3/ | 1,516,196,139,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886939.10/warc/CC-MAIN-20180117122304-20180117142304-00501.warc.gz | 870,695,032 | 96,980 | # Chp2-3 - (2(3 The wavelength peaks in the infrared part of...
This preview shows pages 1–3. Sign up to view the full content.
SCCC 115 Homework 2 - Due 18 Sept. 2003 Prof. Christina Lacey P:2-6,3-2, 3-7 R & D:3-11 P 2-6: Halley's comet has a perihelion distance of 0.6 AU and an orbital period of 76 years. What is the aphelion distance from the Sun? First, using the period find the the semi-major axis, . Then use the perihelion equation to solve for eccentricity, . Finally, solve for the aphelion. Remember that is the .
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
P 3-2 What is the wavelength ( ) of a 100 MHz (``FM 100'') radio signal? (1) P 3-7 Normal human body temperature is 37 C. What is the temperature in Kelvins? What is the peak wavelength emitted by a person with this temperature? In what part of the spectrum is this wavelength?
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: (2) (3) The wavelength peaks in the infrared part of the spectrum. R & D 3-11 What do radio waves, X-rays, and gamma rays have in common? How do they differ? All the above are photons, part of the electromagnetic (EM) spectrum. Photons are massless particles that all travel at the speed of light in vacuum. Each photon in different parts of the electromagnetic spectrum has a different frequency and corresponding wavelength ( ). Each photon has a specific energy: . Radio waves have the longest wavelengths, lowest frequencies, and smallest energies; gamma ray waves have the shortest wavelengths, highest frequencies, and highest energies....
View Full Document
{[ snackBarMessage ]}
### Page1 / 3
Chp2-3 - (2(3 The wavelength peaks in the infrared part of...
This preview shows document pages 1 - 3. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online | 481 | 1,920 | {"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-2018-05 | latest | en | 0.878706 |
https://fr.mathworks.com/matlabcentral/cody/problems/2220-wayfinding-3-passed-areas/solutions/1637698 | 1,585,433,326,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370493120.15/warc/CC-MAIN-20200328194743-20200328224743-00135.warc.gz | 492,600,214 | 16,211 | Cody
# Problem 2220. Wayfinding 3 - passed areas
Solution 1637698
Submitted on 3 Oct 2018 by William
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
AB = [ 2 -2 ; 8 -6 ]; F{1} = [ -4 -4 4 4 -4 -0 -0 -4 ]; F{2} = [ -4 -4 4 4 2 6 6 2 ]; f = WayfindingPassed(AB,F); f_correct = [2 1]; assert(isequal(f,f_correct));
2 Pass
AB = [ 8 -4 ; 8 -8 ]; F{1} = [ -6 2 2 -4 -4 8 8 -6 -6 -6 -4 -4 2 2 4 4 ]; F{2} = [ -2 -2 4 4 -0 -2 -2 -0 ]; f = WayfindingPassed(AB,F); f_correct = [ 1 2 1 ]; assert(isequal(f,f_correct));
3 Pass
AB = [ -8 8 ; 8 -8 ]; F{1} = [ -2 -2 0 0 -0 2 2 -0 ]; F{2} = [ 2 4 4 -6 -6 -4 2 4 4 2 2 -4 -4 2 -0 -0 -6 -6 4 6 6 4 2 2 4 4 -4 -4 ]; F{3} = [ -3 -3 1 0 -1 -3 -3 -1 ]; F{4} = [ 5 9 9 5 -3 -3 -9 -9 ]; F{5} = [ -9 -10 -10 -9 9 9 10 10 ]; f = WayfindingPassed(AB,F); f_correct = [ 2 1 2 4 ]; assert(isequal(f,f_correct)); AB = [ 0 0 ; -8 8 ]; F{1} = [ -4 -2 -2 -4 8 8 4 4 ]; F{2} = [ 2 4 4 2 -0 -0 -6 -6 ]; F{3} = [ -4 -2 -2 -6 -6 -4 -4 -6 -6 -4 ]; f = WayfindingPassed(AB,F); assert(isempty(f));
4 Pass
AB = [ 7 -8 ; 0 0 ]; F{1} = [ 8 9 9 8 3 3 -2 -2 ]; F{2} = [ -9 -7 -7 -4 -4 -3 -3 0 0 1 1 4 4 5 5 -2 -8 -9 -2 -2 2 2 -2 -2 2 2 -2 -2 2 2 -2 -2 3 4 3 2 ]; F{3} = [ -2 -1 -1 -2 1 1 -4 -4 ]; F{4} = [ -6 -5 -5 -3 1 2 2 3 3 1 -4 -6 1 1 -3 -5 -5 -4 1 1 -5 -8 -7 -4 ]; f = WayfindingPassed(AB,F); f_correct = [ 2 4 2 3 2 4 2 ]; assert(isequal(f,f_correct));
5 Pass
AB = [ 0 -2 ; 0 -4 ]; F{1} = [ -3 3 3 2 2 -2 -2 2 2 -3 -5 -5 3 3 -3 -3 2 2 3 3 ]; F{2} = [ -1 1 1 -1 1 1 -1 -1 ]; F{3} = [ -4 4 4 5 5 -5 -5 -4 4 4 -7 -7 5 5 -1 -1 ]; F{4} = [ -5 -4 -4 4 4 -5 -1 -1 -6 -6 -7 -7 ]; f = WayfindingPassed(AB,F); f_correct = [ 2 1 ]; assert(isequal(f,f_correct));
6 Pass
AB = [ -2 0 ; 6 -6 ]; F{1} = [ 2 -4 -4 2 2 -2 0 -2 2 -4 -4 4 4 2 2 -0 -2 -2 ]; f = WayfindingPassed(AB,F); f_correct = [ 1 1 1 ]; assert(isequal(f,f_correct)); | 1,070 | 1,967 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2020-16 | latest | en | 0.342794 |
https://zxi.mytechroad.com/blog/greedy/leetcode-45-jump-game-ii/ | 1,600,448,368,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400188049.8/warc/CC-MAIN-20200918155203-20200918185203-00566.warc.gz | 1,213,318,373 | 14,218 | Given an array of non-negative integers, you are initially positioned at the first index of the array.
Each element in the array represents your maximum jump length at that position.
Your goal is to reach the last index in the minimum number of jumps.
Example:
Input: [2,3,1,1,4]
Output: 2
Explanation: The minimum number of jumps to reach the last index is 2.
Jump 1 step from index 0 to 1, then 3 steps to the last index.
Note:
You can assume that you can always reach the last index.
## Solution: Greedy
Jump as far as possible but lazily.
[2, 3, 1, 1, 4]
i nums[i] steps near far
- - 0 0 0
0 2 0 0 2
1 3 1 2 4
2 1 1 2 4
3 1 2 4 4
4 4 2 4 8
Time complexity: O(n)
Space complexity: O(1)
## C++
If you like my articles / videos, donations are welcome.
Buy anything from Amazon to support our website
Paypal
Venmo
huahualeetcode
## Be First to Comment
Mission News Theme by Compete Themes. | 308 | 1,053 | {"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} | 2.640625 | 3 | CC-MAIN-2020-40 | latest | en | 0.841142 |
https://exercism.io/tracks/ruby/exercises/wordy/solutions/adc4299722c24a2599f66c4610865221 | 1,597,369,555,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739134.49/warc/CC-MAIN-20200814011517-20200814041517-00010.warc.gz | 293,601,900 | 7,043 | π Exercism Research is now launched. Help Exercism, help science and have some fun at research.exercism.io π
muriloime's solution
to Wordy in the Ruby Track
Published at Jun 27 2020 · 0 comments
Instructions
Test suite
Solution
Parse and evaluate simple math word problems returning the answer as an integer.
Iteration 1 β Addition
Add two numbers together.
What is 5 plus 13?
Evaluates to 18.
Handle large numbers and negative numbers.
Iteration 2 β Subtraction, Multiplication and Division
Now, perform the other three operations.
What is 7 minus 5?
2
What is 6 multiplied by 4?
24
What is 25 divided by 5?
5
Iteration 3 β Multiple Operations
Handle a set of operations, in sequence.
Since these are verbal word problems, evaluate the expression from left-to-right, ignoring the typical order of operations.
What is 5 plus 13 plus 6?
24
What is 3 plus 2 multiplied by 3?
15 (i.e. not 9)
Bonus β Exponentials
If you'd like, handle exponentials.
What is 2 raised to the 5th power?
32
For installation and learning resources, refer to the Ruby resources page.
For running the tests provided, you will need the Minitest gem. Open a terminal window and run the following command to install minitest:
``````gem install minitest
``````
If you would like color output, you can `require 'minitest/pride'` in the test file, or note the alternative instruction, below, for running the test file.
Run the tests from the exercise directory using the following command:
``````ruby wordy_test.rb
``````
To include color from the command line:
``````ruby -r minitest/pride wordy_test.rb
``````
Source
Inspired by one of the generated questions in the Extreme Startup game. https://github.com/rchatley/extreme_startup
Submitting Incomplete Solutions
It's possible to submit an incomplete solution so you can see how others have completed the exercise.
wordy_test.rb
``````require 'minitest/autorun'
require_relative 'wordy'
# Common test data version: 1.2.0 86d0069
class WordyTest < Minitest::Test
def test_addition
# skip
problem = WordProblem.new("What is 1 plus 1?")
assert_equal(2, problem.answer)
end
def test_more_addition
skip
problem = WordProblem.new("What is 53 plus 2?")
assert_equal(55, problem.answer)
end
def test_addition_with_negative_numbers
skip
problem = WordProblem.new("What is -1 plus -10?")
assert_equal(-11, problem.answer)
end
def test_large_addition
skip
problem = WordProblem.new("What is 123 plus 45678?")
assert_equal(45801, problem.answer)
end
def test_subtraction
skip
problem = WordProblem.new("What is 4 minus -12?")
assert_equal(16, problem.answer)
end
def test_multiplication
skip
problem = WordProblem.new("What is -3 multiplied by 25?")
assert_equal(-75, problem.answer)
end
def test_division
skip
problem = WordProblem.new("What is 33 divided by -3?")
assert_equal(-11, problem.answer)
end
def test_multiple_additions
skip
problem = WordProblem.new("What is 1 plus 1 plus 1?")
assert_equal(3, problem.answer)
end
def test_addition_and_subtraction
skip
problem = WordProblem.new("What is 1 plus 5 minus -2?")
assert_equal(8, problem.answer)
end
def test_multiple_subtraction
skip
problem = WordProblem.new("What is 20 minus 4 minus 13?")
assert_equal(3, problem.answer)
end
def test_subtraction_then_addition
skip
problem = WordProblem.new("What is 17 minus 6 plus 3?")
assert_equal(14, problem.answer)
end
def test_multiple_multiplication
skip
problem = WordProblem.new("What is 2 multiplied by -2 multiplied by 3?")
assert_equal(-12, problem.answer)
end
def test_addition_and_multiplication
skip
problem = WordProblem.new("What is -3 plus 7 multiplied by -2?")
message = "You should ignore order of precedence. -3 + 7 * -2 = -8, not #{problem.answer}"
assert_equal(-8, problem.answer, message)
end
def test_multiple_division
skip
problem = WordProblem.new("What is -12 divided by 2 divided by -3?")
assert_equal(2, problem.answer)
end
def test_unknown_operation
skip
problem = WordProblem.new("What is 52 cubed?")
assert_raises(ArgumentError) do
problem.answer
end
end
def test_non_math_question
skip
problem = WordProblem.new("Who is the President of the United States?")
assert_raises(ArgumentError) do
problem.answer
end
end
end``````
``````class WordProblem
OPERATIONS = { 'What is ' => '',
'?' => '',
'plus' => :+,
'minus' => :-,
'multiplied by' => :*,
'divided by' => :/ }.freeze
def initialize(question)
@question = question.gsub(Regexp.union(OPERATIONS.keys), OPERATIONS)
end
def eval_binary_expression(operand1, operator, operand2)
operand1.send(operator, operand2)
end
def operands
@question.scan(%r{[\d\+\-\*/]+})
end
def answer
raise ArgumentError if /[a-zA-Z]/.match?(@question)
initial_value, *tail_values = operands
tail_values.each_slice(2)
.inject(initial_value.to_i) do |accumulator, (operator, operand)|
eval_binary_expression(accumulator, operator, operand.to_i)
end
end
end``````
Community comments
Find this solution interesting? Ask the author a question to learn more.
What can you learn from this solution?
A huge amount can be learned from reading other peopleβs code. This is why we wanted to give exercism users the option of making their solutions public.
Here are some questions to help you reflect on this solution and learn the most from it.
• What compromises have been made?
• Are there new concepts here that you could read more about to improve your understanding? | 1,399 | 5,411 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2020-34 | latest | en | 0.882155 |
http://www.sqlservercentral.com/Forums/Topic1601291-392-1.aspx | 1,435,701,487,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375094501.77/warc/CC-MAIN-20150627031814-00121-ip-10-179-60-89.ec2.internal.warc.gz | 713,960,708 | 20,706 | Log in :: Register :: Not logged in
Recent PostsRecent Posts Popular TopicsPopular Topics
Home Search Members Calendar Who's On
Consecutive streak excluding weekends Rate Topic Display Mode Topic Options
Author
Message
Posted Friday, August 8, 2014 11:38 AM
SSC Veteran Group: General Forum Members Last Login: Monday, June 22, 2015 3:27 PM Points: 216, Visits: 612
Hi SSC,I'm trying to write an algorithm that returns the most recent and longest consecutive streak of positive or negative price changes in a given stock. The streak can extend over null weekends, but not over null weekdays (presumably trading days). For example, lets say Google had end of day positive returns on (any given) Tuesday, Monday, and previous Friday, but then Thursday, it had negative returns. That would be a 3 day streak between Friday and Tuesday. Also, if a date has a null value on a date that is NOT a weekend, the streak ends. In the following code sample, you can get a simplified idea of what the raw data will look like and what the output should look like.`set nocount onset datefirst 7goif object_id('tempdb.dbo.#raw') is not null drop table #rawcreate table #raw( symbol varchar(30), tDate date, tDateInt as cast(cast(tDate as datetime) as int), --Integer version of the date pctChg float, isWeekend as case when datepart(weekday, tDate) in (1,7) then 1 else 0 end --indicate whether the date is weekend primary key clustered (symbol, tDate desc))insert into #raw (symbol, tDate, pctChg) select 'A', cast(41854 as datetime), 3.14insert into #raw (symbol, tDate, pctChg) select 'A', cast(41853 as datetime), 2.21insert into #raw (symbol, tDate, pctChg) select 'A', cast(41852 as datetime), nullinsert into #raw (symbol, tDate, pctChg) select 'A', cast(41851 as datetime), nullinsert into #raw (symbol, tDate, pctChg) select 'A', cast(41850 as datetime), 1.01insert into #raw (symbol, tDate, pctChg) select 'A', cast(41849 as datetime), -1.67insert into #raw (symbol, tDate, pctChg) select 'B', cast(41854 as datetime), -.3insert into #raw (symbol, tDate, pctChg) select 'B', cast(41853 as datetime), -6insert into #raw (symbol, tDate, pctChg) select 'B', cast(41852 as datetime), nullinsert into #raw (symbol, tDate, pctChg) select 'B', cast(41851 as datetime), nullinsert into #raw (symbol, tDate, pctChg) select 'B', cast(41850 as datetime), -1.01insert into #raw (symbol, tDate, pctChg) select 'B', cast(41849 as datetime), 1.67insert into #raw (symbol, tDate, pctChg) select 'C', cast(41854 as datetime), 2.1insert into #raw (symbol, tDate, pctChg) select 'C', cast(41853 as datetime), nullinsert into #raw (symbol, tDate, pctChg) select 'C', cast(41852 as datetime), nullinsert into #raw (symbol, tDate, pctChg) select 'C', cast(41851 as datetime), nullinsert into #raw (symbol, tDate, pctChg) select 'C', cast(41850 as datetime), 1.01insert into #raw (symbol, tDate, pctChg) select 'C', cast(41849 as datetime), -1.67select top 1000 *from #raw--Simulation of what results should look likeselect symbol = 'A', streakStart = '2014-08-04', streakEnd = '2014-08-05', streakLength = 2 union allselect symbol = 'B', streakStart = '2014-08-01', streakEnd = '2014-08-05', streakLength = 3 union allselect symbol = 'C', streakStart = '2014-08-05', streakEnd = '2014-08-05', streakLength = 1 `I've done date streaks before by using a row_number and subtracting it from the date to get consecutive groupings, but this is a little different because it depends on the direction of the price change as well as whether or not it was a weekend.`select top 1000 *, grp = tDateInt - row_number() over (partition by symbol order by tDateInt)from #raw`I should also mention that this has to be done over about half a million symbols so something RBAR is especially unappealing.Any ideas would be greatly appreciated. Executive Junior Cowboy Developer, Esq.
Post #1601291
Posted Friday, August 8, 2014 12:08 PM
SSCertifiable Group: General Forum Members Last Login: Thursday, March 12, 2015 6:11 PM Points: 5,467, Visits: 7,659
Currently working in finance, there's additional days you need to kick out, which are the holidays for the different markets. Assuming you're only working against NYSE, NASDAQ, and OTC, you still need to deal with that.With that in mind, do you have a calendar table already setup which lists allowable dates, or do you have a holiday table with is exclusionary for your list? If calendar, this becomes easy. If holiday, we have to build the calendar on the fly and then work from there. - Craig FarrellNever stop learning, even if it hurts. Ego bruises are practically mandatory as you learn unless you've never risked enough to make a mistake. For better assistance in answering your questions | Forum NetiquetteFor index/tuning help, follow these directions. |Tally TablesTwitter: @AnyWayDBA
Post #1601297
Posted Friday, August 8, 2014 1:13 PM
SSC Veteran Group: General Forum Members Last Login: Monday, June 22, 2015 3:27 PM Points: 216, Visits: 612
Exchange holidays are a whole other set of fun, which, for simplicities sake, I've excluded. I'm guessing if I can define a way to skip over holidays, it could similarly be applied to exchange holidays. Executive Junior Cowboy Developer, Esq.
Post #1601316
Posted Friday, August 8, 2014 1:42 PM
SSCertifiable Group: General Forum Members Last Login: Thursday, March 12, 2015 6:11 PM Points: 5,467, Visits: 7,659
I would recommend you start with something like this:http://www.brianshowalter.com/calendar_tablesJust the first one that came up on google, and it looks reasonable.Using a table like this you can filter, control, and get your dates organized into a reasonable fashion. Once you have that, a filtered list of the dates you want to use (non holiday/weekend), a ROW_NUMBER() function, and an application of islands and gaps techniques can get you where you want to be. But first, we need the calendar. - Craig FarrellNever stop learning, even if it hurts. Ego bruises are practically mandatory as you learn unless you've never risked enough to make a mistake. For better assistance in answering your questions | Forum NetiquetteFor index/tuning help, follow these directions. |Tally TablesTwitter: @AnyWayDBA
Post #1601324
Posted Friday, August 8, 2014 11:03 PM
SSC-Dedicated Group: General Forum Members Last Login: Today @ 3:47 PM Points: 37,841, Visits: 34,710
This will do it. As usual, most of the details are where they belong... in comments in the code. The following code works with the test harness provided in the orginal post.`--===== If the work table already exists, drop it to make reruns in SSMS easier IF OBJECT_ID('tempdb..#Work','U') IS NOT NULL DROP TABLE #Work;--===== Create the work table. The given PK is quintessential. CREATE TABLE #Work ( symbol VARCHAR(30) NOT NULL ,tDate DATETIME NOT NULL ,ChangeType SMALLINT NOT NULL ,MyGroup INT NOT NULL DEFAULT 0 PRIMARY KEY CLUSTERED (symbol,tDate) --Absolutely critical. Don't even think of changing this. );--===== Populate the work table with the data that we need. -- Notice the ChangeType column determines the "direction" -- of the pctChg column. INSERT INTO #Work (symbol, tDate, ChangeType) SELECT symbol, tDate, ChangeType = SIGN(ISNULL(pctChg,0)) FROM #Raw WHERE isWeekEnd = 0;--===== Declare a set of obviously named variables to drive -- the "Quirky Update"DECLARE @PrevSymbol VARCHAR(30) ,@PrevChangeType SMALLINT ,@PrevMyGroup INT ,@SafetyCounter INT;--===== Preset a couple of those variables to a known condition. SELECT @PrevMyGroup = 0 ,@SafetyCounter = 1;--===== Using a safety counter to raise an error if this ever makes a mistake -- (but it never will), scan the table and change/increment the MyGroup -- counter when the symbol or ChangeType column changes from row to row -- and do it all in the order of symbol ad tDate columns, just like the PK.WITHcteSafetyCounter AS( --=== This cte exposes the columns we need to read or update and provides -- the safety counter. SELECT SafetyCounter = ROW_NUMBER() OVER (ORDER BY symbol,tDate) ,symbol ,tDate ,ChangeType ,MyGroup FROM #Work WITH(TABLOCKX,INDEX(1)) --Absolutely critical. Don't even think of changing this.) --==== Now we'll calculate the MyGroup column just like we would in procedural code... -- one row at a time using the "Quirky Update" as a "Pseudo Cursor". UPDATE tgt SET @PrevMyGroup = MyGroup =CASE WHEN @SafetyCounter = SafetyCounter THEN CASE WHEN symbol = @PrevSymbol AND ChangeType = @PrevChangeType THEN @PrevMyGroup ELSE @PrevMyGroup + 1 END ELSE 1/0 --Raises an error if something gets out of sequence END ,@PrevSymbol = Symbol ,@PrevChangeType = ChangeType ,@SafetyCounter = @SafetyCounter + 1 FROM cteSafetyCounter tgt OPTION (MAXDOP 1) --Absolutely critical. Don't even think of changing this.;--===== Now that we have the data marked by groups, the rest is easy to produce the report.WITHcteAggregate AS( --=== This aggregates a count for each MyGroup in each symbol. SELECT symbol ,streakStart = MIN(tDate) ,streakEnd = MAX(tDate) ,streakLength = COUNT(*) FROM #Work GROUP BY symbol,MyGroup),cteSortOrder AS( --=== This numbers the groups in descending order by streakLength and uses streakStart as a tie-breaker SELECT SortOrder = ROW_NUMBER() OVER (PARTITION BY symbol ORDER BY streakLength DESC, streakStart DESC) ,* FROM cteAggregate) --=== This returns the desired info for the largest, latest streak for each symbol. -- The SortOrder = 1 was created in descending order by steakLength and streakStart above. SELECT symbol, streakStart, streakEnd, streakLength FROM cteSortOrder WHERE SortOrder = 1 ORDER BY symbol;`That produces the following results...`symbol streakStart streakEnd streakLength------ ----------------------- ----------------------- ------------A 2014-08-01 00:00:00.000 2014-08-05 00:00:00.000 3B 2014-08-01 00:00:00.000 2014-08-05 00:00:00.000 3C 2014-08-05 00:00:00.000 2014-08-05 00:00:00.000 1`Note that the original post said that the streakLength for symbol "A" should be "2" and that's actually incorrect according to the given requirements. The requirements said that streaks should continue through weekend NULL days (presumably, always NULL on the weekend) but streaks stop on NULLs during the week.This code can certainly be tweeked to allow streaks to run through holidays just by adding AND isHoliday = 0 to the appropriate place in the code that populates the work table.As a bit of a sidebar, the "Quirky Update" method in this code is a highly controversial but highly effect method that I've been using just about forever. It even beats LEAD/LAG in 2012 and up. Some good folks (Paul White and Tom Tompson on this site and Peter Larsson separately on another site) came up with the "Safety Counter" thing after 2005 came out and that's nice to have. Once you get one of these bad boys working correctly, it won't fail especially when the INDEX hint is present. There are areas in the code that simply must not be altered and I've marked those areas.Be advised that, in no way, does MS support or recommend this method. Don't let that scare you too much. I've got systems that still work using this method after more than 15 years of SQL Server CUs, SPs, and Rev changes.If that's not good enough, then I recommend the creation of a CLR to pull this task off although I think it might be slower (I haven't tried it because I haven't needed to).Some of the heavy hitters on this site also use this technique and understand it well. To wit, if you have a question on it, don't hesitate to ask. One or more of us can help in this area. --Jeff Moden"RBAR is pronounced "ree-bar" and is a "Modenism" for "Row-By-Agonizing-Row".First step towards the paradigm shift of writing Set Based code: Stop thinking about what you want to do to a row... think, instead, of what you want to do to a column." (play on words) "Just because you CAN do something in T-SQL, doesn't mean you SHOULDN'T." --22 Aug 2013Helpful Links:How to post code problemsHow to post performance problems
Post #1601367
Posted Friday, August 8, 2014 11:13 PM
SSC-Dedicated Group: General Forum Members Last Login: Today @ 3:47 PM Points: 37,841, Visits: 34,710
As another bit of a sidebar, we could also modify the code to identity the direction of the largest streak or produce 3 rows for each symbol to identify the largest negative, positive, and static (no change) streaks. --Jeff Moden"RBAR is pronounced "ree-bar" and is a "Modenism" for "Row-By-Agonizing-Row".First step towards the paradigm shift of writing Set Based code: Stop thinking about what you want to do to a row... think, instead, of what you want to do to a column." (play on words) "Just because you CAN do something in T-SQL, doesn't mean you SHOULDN'T." --22 Aug 2013Helpful Links:How to post code problemsHow to post performance problems
Post #1601368
Permissions | 3,372 | 12,809 | {"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-2015-27 | longest | en | 0.825892 |
https://studylib.net/doc/25223003/decimals-and-money-lesson-and-worksheets | 1,553,079,521,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202326.46/warc/CC-MAIN-20190320105319-20190320131019-00035.warc.gz | 634,241,208 | 34,799 | # Decimals and Money Lesson and Worksheets
```http://www.educents.com/complete-curriculum
75
/HVVRQ
'HFLPDOVDQG0RQH\
, "%% ,%)#*# )
).)"%%#""#%.,%)#'
([DPSOH
B).)+*#+,%)#'
CC#53%';++)
,%)#'"%+53,,G53.882+%#
).%,##';+)*8'53
3UDFWLFH
B).)+*#+,%)#'
'
6'
v2
http://www.educents.com/complete-curriculum
756
([DPSOH
B).)+*#+,%)#'
,##)#,*+<. T,##)+#
)*',,)+)%(';(
3%2+%3,,G3%."*#88%
,##'
"%+M6'83'
3UDFWLFH
B).)+*#+,%)#'
'
6'
/HVVRQ:UDS8SB""#,M1'632##.,##
,+<(%'%#,#(.,+
<(,,.,##'9))) 'B%
),:?
v2
http://www.educents.com/complete-curriculum
750
:RUNVKHHW
'LUHFWLRQVB).)+*#+,%)#'
'
6'
0'
1'
3'
v2
http://www.educents.com/complete-curriculum
751
4'
5'
'LUHFWLRQVB%)*.%+## (.##+ ,##
)?
='M8'6
7'M'60
8'M'8
'M8'13
6'M8'=7
0'M6'08
v2
http://www.educents.com/complete-curriculum
753
1'M3'06
3'M8'68
4'B,,%)#",##)?
5'B,..%*++ 53%,+ M8'53?
B*<. (?
Print Form
v2
```
– Cards
– Cards
– Cards
– Cards | 451 | 887 | {"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} | 3.921875 | 4 | CC-MAIN-2019-13 | longest | en | 0.236915 |
https://www.scitepress.org/Papers/2011/35225/pdf/index.html | 1,576,022,623,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540529516.84/warc/CC-MAIN-20191210233444-20191211021444-00178.warc.gz | 848,829,090 | 31,134 | HIGH-FREQUENCY ANALYSIS OF PHASE-LOCKED LOOP
AND PHASE DETECTOR CHARACTERISTIC COMPUTATION
N. V. Kuznetsov
1,2
, G. A. Leonov
2
, P. Neittaanm¨aki
2
, S. M. Seledzhi
1
,
M. V. Yuldashev
2
and R. V. Yuldashev
2
1
2
Saint-Petersburg State University, Universitetski pr. 28, 198504, Saint-Petersburg, Russia
Keywords:
Nonlinear analysis, Phase-locked loop, Phase detector characteristic, Mathematical model.
Abstract:
Problems of rigorous mathematical analysis of PLL are discussed. An analytical method for phase detec-
tor characteristics computation is suggested and new classes of phase detector characteristics are computed.
Effective methods for nonlinear analysis of PLL are discussed.
1 INTRODUCTION
Phase-locked loop (PLL) systems were invented in
the 1930s-1940s (De Bellescize, 1932; Wendt & Fre-
dentall, 1943) and were widely used in radio and tele-
vision (demodulation and recovery, synchronization
and frequency synthesis). Nowadays PLL can be pro-
duced in the form of single integrated circuit and var-
ious modifications of PLL are used in a great amount
of modern electronic applications (radio, telecommu-
nications, computers, and others).
At present there are several types of PLL (classi-
cal PLL, ADPLL, DPLL, and others), intended for the
operation with different types of signals (sinusoidal,
impulse, and so on). In addition, it is also used differ-
ent realizations of PLL, which are distinct from each
other according to the principles of operation and re-
alization of main blocks.
For the sake of convenience of description, in
PLL the following main functional blocks are consid-
ered: phase detector (PD), low-pass filter (LPF), and
voltage-controlled oscillator (VCO). Note that such a
partition into functional blocks often turns out to be
conditional, since in many cases in particular physi-
cal realization it is impossible to point out the strict
boundaries between these blocks. However these
blocks can be found in each PLL.
The general PLL operation consists in the genera-
tion of an electrical signal (voltage), a phase of which
is automatically tuned to the phase of input (refer-
ence) signal, i.e. PLL eliminates misphasing (clock
skew) between two signals. For this purpose the refe-
rence signal and the tunable signal of voltage-
controlled oscillator are passed through a special non-
linear element phase detector (PD). The phase
detector produces an error correction signal, corre-
sponding to phase difference of two input signals. For
the discrimination of error correction signal, a signal
at the output of phase detector is passed through low-
pass filter (LPF). The error correction signal, obtained
at the output of filter, is used for the frequency con-
trol of tunable oscillator, the output of which enters a
phase detector, providing thus negative feedback.
The most important performance measure of PLL
is the capture range (i.e. a maximal mistuning range
of VCO, in which a closed contour of PLL stabilizes
a frequency of VCO) and a locking speed (speed of
Thus, when designing PLL systems, an important
task is to determine characteristics of system (involv-
ing parameters of main blocks) providing required
characteristics of operation of PLL.
To solve this problem, it is used real experiments
with concrete realization of PLL as well as the ana-
lytical and numerical methods of analysis of mathe-
matical models of PLL. These tools are used for the
obtaining of stability of required operating modes, the
estimates of attraction domain of such modes, and the
time estimates of transient processes.
Remark, however, that for the strict mathematical
analysis of PLL it should be taken into account the
fact that the above principles of operation of PLL re-
sult in the substantial requirements:
X construction of adequate nonlinear mathemat-
272
V. Kuznetsov N., A. Leonov G., Neittaanmäki P., M. Seledzhi S., V. Yuldashev M. and V. Yuldashev R..
HIGH-FREQUENCY ANALYSIS OF PHASE-LOCKED LOOP AND PHASE DETECTOR CHARACTERISTIC COMPUTATION.
DOI: 10.5220/0003522502720278
In Proceedings of the 8th International Conference on Informatics in Control, Automation and Robotics (ICINCO-2011), pages 272-278
ISBN: 978-989-8425-74-4
c
2011 SCITEPRESS (Science and Technology Publications, Lda.)
ical models (since PLL contains nonlinear elements)
in signal space and phase-frequency space
and
X justification of the passage between these mod-
els (since PLL translates the problem from signal re-
sponse to phase response and back again).
Despite this, as noted by well-known PLL ex-
pert Danny Abramovitch in his keynote talk at Amer-
ican Control Conference ACC’2002 (Abramovitch,
2002), the main tendency in a modern literature (see,
e.g., (Egan, 2000; Best, 2003; Kroupa, 2003; Razavi,
2003)) on analysis of stability and design of PLL is
the use of simplified linearized models and the ap-
plication of the methods of linear analysis, a rule of
thumb, and simulation.
However it is known that the application of lin-
earization methods and linear analysis for control
systems can lead to untrue results (e.g., Perron ef-
& Kuznetsov , 2007), counterexamples to Aizerman’s
conjuncture and Kalman’s conjuncture on absolute
stability, harmonic linearization and filter hypothe-
sis (Leonov et al., 2010
2
)) and requires special jus-
tifications. Also simple numerical analysis can not
reveal nontrivial regimes (e.g., semi-stable or nested
limit cycles, hidden oscillations and attractors (Gubar,
1961; Kuznetsov & Leonov, 2008; Leonov et al.,
2010
2
; Leonov et. al., 2010
1
; Leonov et. al., 2011)).
2 NONLINEAR MATHEMATICAL
MODELS OF PLL
Various methods for analysis of phase-locked loops
are well developed by engineers and considered in
many publications (see, e.g., (Viterbi, 1966; Gardner,
1966; Lindsey, 1972; Shakhgildyan & Lyakhovkin,
1972)), but the problems of construction of adequate
nonlinear models and nonlinear analysis of such mod-
els are still far from being resolved turn out to be dif-
ficult. and require to use special methods of quali-
tative theory of differential, difference, integral, and
integro-differential equations (Leonov et al., 1996;
Suarez & Quere, 2003; Margaris, 2004; Leonov,
2006; Kudrewicz & Wasowicz, 2007; Leonov et al.,
2009).
In the present paper some approaches to the non-
linear analysis of PLL are described. Nonlinear math-
ematical models of high-frequency oscillations are
presented.
To construct an adequate nonlinear mathematical
model of PLL in phase space it is necessary to find the
characteristic of phase detector. The inputs of PD are
high-frequency signals of reference and tunable os-
cillators and the output contains a low-frequency er-
ror correction signal, corresponding to a phase differ-
ence of input signals. For the suppression of high-
frequency component of the output of PD (if such
component exists) the low-pass filters are applied.
The dependence of the signal at the output of PD
(in phase space) on phase difference of signals at the
input of PD is the characteristic of PD. This char-
acteristic depends on the realization of PD and the
types of signals at the input. Characteristics of the
phase detector for standard types of signal are well-
known to engineers (Viterbi, 1966; Shakhgildyan &
Lyakhovkin, 1972; Abramovitch, 2002).
Further, on the examples of classical PLL with
a phase detector in the form of multiplier, we con-
sider general principles of computing phase detector
characteristics for different types of signals based on
a rigorous mathematical analysis of high-frequency
oscillations (Leonov & Seledzhi , 2005a; Leonov,
2008; Kuznetsov et al., 2008; Kuznetsov et al., 2009
1
;
Kuznetsov et al., 2009
2
; Leonov et al., 2010
3
).
2.1 Description of Classical PLL in the
Signal Space
Consider classical PLL at the level of electronic real-
ization (Fig. 1)
Figure 1: Block diagram of PLL at the level of electronic
realization.
Here OSC
master
is a master oscillator, OSC
slave
is
a slave (tunable voltage-controlled) oscillator, which
generates oscillations f
j
(t) with high-frequencies
ω
j
(t).
Block
N
is a multiplier of oscillations of f
1
(t) and
f
2
(t) and the signal f
1
(t) f
2
(t) is its output. The re-
lation between the input ξ(t) and the output σ(t) of
linear filter has the form
σ(t) = α
0
(t) +
t
Z
0
γ(t τ)ξ(τ)dτ. (1)
Here γ(t) is an impulse transient function of filter,
α
0
(t) is an exponentially damped function, depend-
ing on the initial data of filter at moment t = 0.
In the simplest ideal case, when
f
1
= sin(ω
1
), f
2
= cos(ω
2
)
f
1
f
2
= [sin(ω
1
+ ω
2
) + sin(ω
1
ω
2
)]/2,
HIGH-FREQUENCY ANALYSIS OF PHASE-LOCKED LOOP AND PHASE DETECTOR CHARACTERISTIC
COMPUTATION
273
standard engineering assumption is that the filter re-
moves the upper sideband with frequency from the
input but leaves the lower sideband without change.
Thus it is assumed that the filter output is
1
2
sin(ω
1
ω
2
).
Here to avoid these non-rigorous arguments we
consider mathematical properties of high-frequency
oscillations.
2.2 Computation of Phase Detector
Characteristic
A high-frequency property of signals can be reformu-
lated as the following condition. Consider a large
fixed time interval [0,T], which can be partitioned
into small intervals of the form
[τ,τ + δ], τ [0,T],
where the following relations
|γ(t) γ(τ)| Cδ, |ω
j
(t) ω
j
(τ)| Cδ,
t [τ, τ+ δ], τ [0,T],
(2)
|ω
1
(τ) ω
2
(τ)| C
1
, τ [0, T], (3)
ω
j
(t) R, t [0,T] (4)
are satisfied.
We shall assume that δ is small enough relative to
the fixed numbers T,C,C
1
and R is sufficiently large
relative to the number δ : R
1
= O(δ
2
).
The latter means that on small intervals [τ,τ + δ]
the functions γ(t) and ω
j
(t) are “almost constant”
and the functions f
j
(t) on them are rapidly oscillat-
ing. Obviously, such a condition occurs for high-
frequency oscillations.
Consider now harmonic oscillations
f
j
(t) = A
j
sin(ω
j
(t)t + ψ
j
), j = 1,2, (5)
where A
j
and ψ
j
are certain numbers, ω
j
(t) are dif-
ferentiable functions.
Consider two block diagrams shown in Fig. 2 and
Fig. 3.
Figure 2: Multiplier and filter.
In Fig. 3 θ
j
(t) = ω
j
(t)t + ψ
j
are phases of oscil-
lations f
j
(t), PD is a nonlinear block with the charac-
teristic ϕ(θ). The phases θ
j
(t) are the inputs of PD
Figure 3: Phase detector and filter.
block and the output is the function ϕ(θ
1
(t) θ
2
(t)).
A shape of phase detector characteristic is based on a
shape of input signals.
The signals f
1
(t) f
2
(t) and ϕ(θ
1
(t) θ
2
(t)) are in-
puts of the same filters with the same impulse tran-
sient function γ(t). The filter outputs are the functions
g(t) and G(t), respectively.
A classical PLL synthesis for the sinusoidal sig-
nals is based on the following result (Viterbi, 1966):
If conditions (2)–(4) are satisfied and
ϕ(θ) =
1
2
A
1
A
2
cosθ,
then for the same initial data of lter, the following
relation
|G(t) g(t)| C
2
δ, t [0,T]
is satisfied. Here C
2
is a certain number being inde-
pendent of δ.
But what could be done for other types of signal?
Consider now signals in the following form of
Fourier series
f
1
(t) =
i=1
a
i
sin(iθ
1
(t)), f
2
(t) =
j=1
b
j
sin( jθ
2
(t)),
(6)
where
a
k
= O
1
k
, b
k
= O
1
k
, k = 1,2,... .
Let functions f
1
(t) and f
2
(t) are integrable and
bounded on each of the intervals of length δ.
Then the following assertion is valid
Theorem 1. If conditions (2)–(4) are satisfied and
ϕ(θ
1
θ
2
) =
l=1
a
l
b
l
2
cos(l(θ
1
θ
2
)), (7)
then for the same initial states of filter the following
relation
|G(t) g(t)| C
3
δ, t [0,T] (8)
is valid.
Proof. Consider a decomposition of the interval [0,T]
into the δ length time intervals. Then using (2) we
ICINCO 2011 - 8th International Conference on Informatics in Control, Automation and Robotics
274
obtain
g(t) G(t) =
m
k=0
γ(t kδ)
(k+1)δ
Z
kδ
f
1
θ
1
(s)
f
2
θ
2
(s)
ϕ
θ
1
(s) θ
2
(s)
ds+ O(δ).
(9)
Because the frequencies are almost constant in the
δ-intervals (3), we could introduce θ
p
k
(s)
θ
p
k
(s) = ω
p
(kδ)s+ ψ
p
, p {1,2}.
(10)
Lemma 1. Assuming conditions (2)–(4) the phases
θ
p
(t) could be replaced with θ
p
k
(t)
(k+1)δ
Z
kδ
ϕ
θ
1
(s) θ
2
(s)
=
(k+1)δ
Z
kδ
ϕ
θ
1
k
(s) θ
2
k
(s)
ds+ O(δ),
(k+1)δ
Z
kδ
f
1
θ
1
(s)
f
2
θ
2
(s)
=
(k+1)δ
Z
kδ
f
1
θ
1
k
(s)
f
2
θ
2
k
(s)
ds+ O(δ),
(11)
Then, usign Lemma 1, equation (9) can be rewrit-
ten
g(t) G(t) =
m
k=0
γ(t kδ)
Z
[kδ,(k+1)δ)
f
1
θ
1
k
(s)
f
2
θ
2
k
(s)
ϕ
θ
1
k
(s) θ
2
k
(s)
ds+ O(δ)
(12)
Lemma 2. For the neighborhoodsW
ε,k
of discontinu-
ity points, there is a number M, such that
g(t) G(t) =
m
k=0
γ(t kδ)
Z
[kδ,(k+1)δ]\W
ε,k
M
i=1
a
1
i
sin
iθ
1
k
(s)
)
M
j=1
a
2
j
sin
jθ
1
k
(s)
ϕ
θ
1
k
(s) θ
2
k
(s)
ds+ O(δ).
Lemma 2 implies
g(t) G(t) =
m
k=0
γ(t kδ)
Z
[kδ,(k+1)δ]\W
ε,k
M
i=1
M
j=1
a
1
i
sin
iθ
1
k
(s)
a
2
j
sin
jθ
2
k
(s)
ϕ
θ
1
k
(s) θ
2
k
(s)
ds+ O(δ).
(13)
It’s obvious, that
sin
iθ
1
k
(s)
sin
jθ
2
k
(s)
=
1
2
cos
iθ
1
k
(s) jθ
2
k
(s)
cos
iθ
1
k
(s) + jθ
2
k
(s)
(14)
Lemma 3. Assuming conditions (2)–(4) the follow-
ing equations can be obtained
(k+1)δ
Z
kδ
1
q
cos
p(Rs+ ψ)
ds =
O(δ
2
)
pq
,
(15)
Using (13),(4),(14) and Lemma 3, the theorem
statement can be obtained
g(t) G(t) = O(δ) . (16)
This result could be easily extended to the case
of full Fourier series and allows one to calculate the
phase detector characteristic in the following standard
cases of signals (Kuznetsov et al., 2010).
Example 1. Two sign signals
f
k
(t) = A
k
sign sin(θ
k
(t)) =
=
4A
k
π
n=0
1
2n+1
sin((2n+ 1)(ω
k
(t)t + ψ
k
)), k = 1,2
ϕ(θ
1
θ
2
) =
8A
1
A
2
π
2
n=0
1
(2n+1)
2
cos(θ
1
θ
2
)
−2 pi −pi 0 pi 2 pi
−A1A2
0
A1A2
Θ
φ(θ)
Figure 4: Phase detector characteristic ϕ(θ) for two sign
signals.
HIGH-FREQUENCY ANALYSIS OF PHASE-LOCKED LOOP AND PHASE DETECTOR CHARACTERISTIC
COMPUTATION
275
Thus, here phase detector characteristic φ(θ) corre-
sponds to 2π-periodic function
A
1
A
2
1
2|θ|
π
, for θ (π, π]. (17)
Example 2. Sin signal and sign signal
f
1
(t) = A
1
sin(θ
1
(t))
f
2
(t) = A
2
sign sin(θ
2
(t)) =
=
4A
2
π
n=0
1
2n+1
sin((2n+ 1)(ω
2
(t)t + ψ
2
))
ϕ(θ
1
θ
2
) =
2A
1
A
2
π
cos(θ
1
θ
2
)
Example 3. Triangle wave signals.
A
0
π
Figure 5: Triangle-wave signal.
f
k
(t) = A
k
i=0
1
(2i 1)
2
sin
(2i 1)θ
k
(t)
(18)
ϕ(θ
1
θ
2
) = A
1
A
2
l=1
1
(2l 1)
4
cos
(2l1)(θ
1
θ
2
)
(19)
-1
-0.5
0
0.5
1
0 2 4 6 8 10 12 14
Figure 6: Phase detector characteristic ϕ(θ) for triangle sig-
nals.
2.3 PLL Equations in Phase-frequency
Space
From Theorem 1 it follows that block-scheme of PLL
in signal space (Fig. 1) can be asymptotically changed
Figure 7: Phase-locked loop with phase detector.
(for high-frequency generators) to a block-scheme at
the level of frequency and phase relations (Fig. 7).
Here PD is a phase detector with corresponding
characteristics. Thus, here on basis of asymptotical
analysis of high-frequency pulse oscillations charac-
teristics of phase detector can be computed.
Characteristic ϕ(θ), computed in Examples 1 and
2, tends to zero if θ = (θ
1
θ
2
) tends to π/2, so
one can proceed to stability analysis (Leonov, 2006;
Leonov et al., 2009) of differential (or difference)
equations depend on misphasing θ.
Let us make a remark necessary for derivation of
differential equations of PLL.
Consider a quantity
˙
θ
j
(t) = ω
j
(t) +
˙
ω
j
(t)t.
For the well-synthesized PLL such that it possesses
the property of global stability, we have exponential
damping of the quantity
˙
ω
j
(t):
|
˙
ω
j
(t)| Ce
αt
.
Here C and α are certain positive numbers indepen-
dent of t. Therefore, the quantity
˙
ω
j
(t)t is, as a rule,
small enough with respect to the number R (see con-
ditions (3) (4)). From the above we can conclude
that the following approximate relation
˙
θ
j
(t) ω
j
(t)
is valid. In deriving the differential equations of this
PLL, we make use of a block diagram in Fig. 7 and
exact equality
˙
θ
j
(t) = ω
j
(t). (20)
Note that, by assumption, the control law of tunable
oscillators is linear:
ω
2
(t) = ω
2
(0) + LG(t). (21)
Here ω
2
(0) is initial frequency of tunable oscillator,
L is a certain number, and G(t) is a control signal,
which is a filter output (Fig. 3). Thus, the equation of
PLL is as follows
˙
θ
2
(t) = ω
2
(0) + L
α
0
(t) +
t
Z
0
γ(t τ)ϕ
θ
1
(τ) θ
2
(τ)
dτ
.
Assuming that the master oscillator is such that
ω
1
(t) ω
1
(0), we obtain the following relations for
ICINCO 2011 - 8th International Conference on Informatics in Control, Automation and Robotics
276
PLL
θ
1
(t)θ
2
(t)
+L
α
0
(t) +
t
R
0
γ(t τ)ϕ
θ
1
(τ)θ
2
(τ)
dτ
= ω
1
(0) ω
2
(0).
(22)
This is an equation of standard PLL. Note, that if fil-
ter (1) is an integrating filter with the transfer function
(p+α)
1
˙
σ+ ασ = ϕ(θ)
then for φ(θ) = cos(θ) in place of of equation (22)
from (20) and (21) we have pendulum-like equation
(Leonov & Smirnova, 1996; Leonov et al., 1996)
¨
˜
θ+ α
˙
˜
θ+ Lsin
˜
θ = α
ω
1
(0) ω
2
(0)
(23)
with
˜
θ = θ
1
θ
2
+
π
2
. Thus, if here phases of the input
and output signals mutually shifted by π/2, then the
control signal G(t) equals zero.
Arguing as above, we can conclude that in PLL it
can be used the filters with transfer functions of more
general form K(p) = a+W(p), where a is a certain
number,W(p) is a proper fractional rational function.
In this case in place of equation (22) we have
θ
1
(t) θ
2
(t)
+ L
aϕ
θ
1
(t) θ
2
(t)
+ α
0
(t)+
+
t
R
0
γ(t τ)ϕ
θ
1
(τ) θ
2
(τ)
dτ
= ω
1
(0) ω
2
(0).
(24)
In the case when the transfer function of the filter
a+W(p) is non-degenerate,i.e. its numerator and de-
nominator do not have common roots, equation (24)
is equivalent to the following system of differential
equations
˙z = Az+ bψ(σ),
˙
σ = c
z+ ρψ(σ).
(25)
Here σ = θ
1
θ
2
, A is a constant (n×n)-matrix, b and
c are constant (n)-vectors, ρ is a number, and ψ(σ) is
2π-periodic function, satisfying the relations:
ρ = aL, W(p) = L
1
c
(A pI)
1
b,
ψ(σ) = ϕ(σ)
ω
1
(0) ω
2
(0)
L(a+ W(0))
.
The discrete phase-locked loops obey similar
equations
z(t + 1) = Az(t) + bψ(σ(t))
σ(t + 1) = σ(t) + c
z(t) + ρψ(σ(t)),
(26)
where t Z, Z is the set of integers. Equations
(25) and (26) describe the so-called standard PLLs
(Shakhgildyan & Lyakhovkin, 1972).
For analysis of the above mathematical models
of PLL is applied in the theory of phase synchro-
nization, which was developed in the second half of
the last century on the basis of three applied theo-
ries: the theory of synchronous and induction elec-
trical motors, the theory of auto-synchronization of
the unbalanced rotors, and the theory of phase-locked
loops. Modification of direct Lyapunov method with
the construction of periodic Lyapunov-like functions,
the method of positively invariant cone grids, and the
method of nonlocal reduction turned out to be most
effective (Leonov et al., 1996; Leonov, 2006; Leonov
et al., 2009). The last method, which combines the
elements of direct Lyapunov method and bifurcation
theory, allows one to extend the classical results of
F. Tricomi (Tricomi , 1933) and his progenies (Ku-
drewicz & Wasowicz, 2007) to the multidimensional
dynamical systems.
3 CONCLUSIONS
Considered above methods for high-frequency analy-
sis of PLL allow one to construct adequate nonlinear
dynamical model of PLL and to apply special meth-
ods of qualitative theory of differential, difference, in-
tegral, and integro-differential equations for PLL de-
sign.
ACKNOWLEDGEMENTS
This work was supported by Academy of Finland,
Ministry of Education and Science (Russia) and
Saint-Petersburg State University.
REFERENCES
Abramovitch D., Phase-Locked Loops: A control Centric
Tutorial. Proceedings of the American Control Con-
ference, Vol. 1, pp. 1–15, 2002.
Best R.E., Phase-Lock Loops: Design, Simulation and Ap-
plication. McGraw Hill, 5
ed
, 2003.
De Bellescize H. La Reseption synchrone. Onde Electrique,
Vol. 11, 1932.
Egan W.F., Frequency Synthesis by Phase Lock. John Wiley
and Sons, 2
ed
, 2000.
Gardner F.M., Phase–lock techniques. John Wiley, New
York, 1966.
Gubar’ N.A., Investigation of a piecewise linear dynamical
system with three parameters. J. Appl. Math. Mech,
25, pp. 1519–1535, 1961.
Kudrewicz J. & Wasowicz S., Equations of Phase-Locked
Loops: Dynamics on the Circle, Torus and Cylinder.
World Scientific, Singapure, 2007.
HIGH-FREQUENCY ANALYSIS OF PHASE-LOCKED LOOP AND PHASE DETECTOR CHARACTERISTIC
COMPUTATION
277
Kuznetsov N.V., Leonov G.A., Neittaanmaki P.,
Seledzhi S.M., Yuldashev M.V., Yuldashev R.V.,
Nonlinear analysis of phase-locked loop. 4th Inter-
national Workshop on Periodic Control Systems,
2010.
Kuznetsov N.V., Leonov G.A., Seledzhi S.M., Nonlinear
analysis of the Costas loop and phase-locked loop
with squarer. Proceedings of the IASTED Interna-
tional Conference on Signal and Image Processing,
SIP 2009, pp. 1–7, 2009.
Kuznetsov N.V., Leonov G.A., Seledzhi S.M., Neit-
taanm¨aki P., Analysis and design of computer archi-
tecture circuits with controllable delay line. ICINCO
2009, Proceedings, Vol. 3 SPSMC, pp. 221–224,
2009.
Kuznetsov N.V., Leonov G.A. Lyapunov quantities, limit
cycles and strange behavior of trajectories in two-
dimensional quadratic systems, Journal of Vibroengi-
neering, Vol. 10, Iss. 4, pp. 460-467, 2008.
Kuznetsov N.V., Leonov G.A., Seledzhi S.M. Phase Locked
Loops Design And Analysis. ICINCO 2008 - 5th
International Conference on Informatics in Con-
trol, Automation and Robotics, Proceedings SPSMC,
pp. 114–118, 2008.
Kroupa V. Phase Lock Loops and Frequency Synthesis.
John Wiley & Sons, 2003.
Leonov G.A., Kuznetsov N.V., Vagaytsev V.I. Localization
of hidden Chua’s attractors. Physics Letters A, 2011
(doi:10.1016/j.physleta.2011.04.037)
Leonov G.A., Vagaitsev V.I., Kuznetsov N.V., Algorithm
for localizing Chua attractors based on the harmonic
linearization method // Doklady Mathematics, 82(1),
pp. 663–666, 2010.
Leonov G.A., Bragin V.O., Kuznetsov N.V., Algorithm for
Constructing Counterexamples to the Kalman Prob-
lem. Doklady Mathematics, Vol. 82, No. 1, pp. 540–
542, 2010.
Leonov G.A., Seledzhi S.M., Kuznetsov N.V., P. Neit-
taanm¨aki., Asymptotic analysis of phase control sys-
tem for clocks in multiprocessor arrays, ICINCO 2010
- Proceedings of the 7th International Conference on
Informatics in Control, Automation and Robotics, Vol.
3, pp. 99–102, 2010.
Leonov G.A., Kuznetsov N.V., Seledzhi S.M. Nonlinear
Analysis and Design of Phase-Locked Loops. pp. 89–
114. In Automation control - Theory and Practice,
A.D. Rodic (ed.), In-Tech, 2009.
Leonov G.A., Computation of phase detector characteris-
tics in phase-locked loops for clock synchronization.
Doklady Mathematics, 78(1), pp. 643–645, 2008.
Leonov G.A., Kuznetsov N.V., Time-Varying Lineariza-
tion and the Perron effects. Int. J. of Bifurcation and
Chaos, 17(4), pp. 1079–1107, 2007.
Leonov G.A.. Phase synchronization: Theory and ap-
plications. Automation and remote control, 67(10),
pp. 1573–1609, 2006.
Leonov G.A. & Seledghi S.M., Stability and bifurcations of
phase-locked loops for digital signal processors. Int. J.
of bifurcation and chaos, 15(4), pp. 1347–1360, 2005.
Leonov G.A. & Seledzhi S.M., An astatic phase-locked
system for digital signal processors: Circuit design
and stability. Automation and Remote Control, 66(3),
pp. 348–355, 2005.
Leonov G.A., Ponomarenko D., and Smirnova V.,
Frequency-Domain Methods for Nonlinear Analysis.
Theory and Applications. World Scientific, Singapore,
1996.
Leonov G.A., Smirnova V.B., Stability and oscilla-
tions of solutions of integro-differential equations of
pendulum-like systems. Mathematische Nachrichten,
177, pp. 157–181, 1996.
Lindsey W., Synchronization systems in communication and
control. Prentice-Hall. New Jersey, 1972.
Margaris N.I., Theory of the Non-Linear Analog Phase
Locked Loop. Springer Verlag, 2004.
Razavi B., Phase-Locking in High-Performance Systems:
From Devices to Architectures. John Wiley & Sons,
2003.
Shakhgil’dyan V.V.& Lyakhovkin A.A., Sistemy fazovoi av-
topodstroiki chastoty (Phase Locked Systems). Svyaz’,
Moscow, 1972. (in Russian)
Suarez A. & Quere R., Stability Analysis of Nonlinear Mi-
crowave Circuits. Artech House, 2003.
Tricomi F., Integrazione di differeziale presentasi in elec-
trotechnica. Annali della Roma Schuola Normale Su-
periore de Pisa, Vol. 2, N2, pp. 1–20, 1993.
Viterbi A.J., Principles of coherent communications.
McGraw-Hill, New York, 1966.
Wendt K.R. & Fredentall G.L., Automatic frequency and
phase control of synchronization in TV receivers. Pro-
ceedings IRE, 31, N1, 1943.
ICINCO 2011 - 8th International Conference on Informatics in Control, Automation and Robotics
278 | 8,105 | 24,324 | {"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-2019-51 | latest | en | 0.890105 |
http://mca.ignougroup.com/2017/05/lower-bound-of-checking-if-in-array-are.html | 1,558,254,859,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232254731.5/warc/CC-MAIN-20190519081519-20190519103519-00167.warc.gz | 138,225,488 | 35,785 | World's most popular travel blog for travel bloggers.
# lower bound of checking if in array are two different elements
, ,
Problem Detail:
Considering the lower limit to the problem of checking whether the array are only the same number via comparisons. And thnik about $n-1$. Consider the diagram hasse. This diagram must be consistent (one piece), so you need at least $n-1$ edges (comparisons),.
However, I would like to consider a decision tree. Unfortunately, I can not consider this, look how I think about it.
###### Answered By : Yuval Filmus
Decision tree lower bounds don't quite work here. The obvious way to apply a decision tree lower bound is to consider the output of the algorithm, which is either TRUE or FALSE. Since there are two possible outputs, the decision tree lower bound gives a lower bound of one comparison. Not very helpful.
A more interesting example of the decision tree technique forces the algorithm to output a little more, namely either it outputs YES (all the elements are the same), or a pair of different elements (represented as their indices). You can argue that whenever the algorithm returns NO it actually knows two elements which are different. Considering all inputs of the form $x_1 = \cdots = x_{i-1} = x_{i+1} = \cdots = x_n = 0, x_i = 1$, we see that for each $i$ there must be a leaf of the form $(i,\ast)$. This means that there must be at least $n/2$ NO leaves, resulting in a lower bound of $\log_2(1+n/2)$ on the number of comparisons. This can plausibly be improved to $\log_2(1+\binom{n}{2})$, but both quantities are $\Theta(\log n)$.
The correct way to argue here is using an adversary argument. Consider any algorithm, and trace its execution, always answering YES to any query. Consider the graph whose vertices are all elements (i.e., all indices), and the edges correspond to queries performed by the algorithm. If the graph is not connected, the algorithm can't know whether all elements are the same (indeed, it is consistent both that all elements have the value $0$, or that the elements in the $i$th connected components have the value $i$). Since a connected graph has at least $n-1$ edges, any algorithm must perform as least $n-1$ comparisons. It's not hard to see that this bound is achievable.
Best Answer from StackOverflow
Question Source : http://cs.stackexchange.com/questions/47821
3200 people like this | 559 | 2,391 | {"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} | 3.859375 | 4 | CC-MAIN-2019-22 | latest | en | 0.92515 |
infoseleb.site | 1,679,789,321,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945376.29/warc/CC-MAIN-20230325222822-20230326012822-00754.warc.gz | 372,005,777 | 17,804 | # Obtain Cpi Formulation Laptop Structure Footage
Posted on 278 views
· how do you calculate . How do you calculate cpi department? I = variety of directions in program; For instance, a pentium has a 233 mhz clock. · what’s the primary program execution time formulation?
What’s the formulation for cpi clock per instruction? In laptop structure, cycles per instruction (aka clock cycles per instruction, clocks per instruction, or cpi) is one side of a processor's efficiency: . T = clock cycle time . Calculation of cpi (cycles per instruction). • thus it can’t be used to match computer systems/cpus with completely different . Cpi = cycles per instruction. · what’s the primary program execution time formulation? I = variety of directions in program;
Contents
### Cpi = cycles per instruction.
• thus it can’t be used to match computer systems/cpus with completely different . Cpi = common cycles per instruction; · how do you calculate . · what’s the primary program execution time formulation? T = clock cycle time . Cpi is the sum of all of the directions within the cpi multiplied by the fraction of the time that they’re . To calculate an “mixture” cpi, or common cpi per core, divide the sum of all core's threads . As we all know a program consists of variety of directions. Cpu time = i * cpi * t. ▫ cpi stands for common variety of cycles per instruction. Cpi per core is pretty simple as proven in fig. Calculation of cpi (cycles per instruction). Tailored from laptop group and design, patterson & hennessy, ucb.
Cpi is the sum of all of the directions within the cpi multiplied by the fraction of the time that they’re . T = clock cycle time . · how do you calculate . Cpi = cycles per instruction. I = variety of directions in program;
Cpi is the sum of all of the directions within the cpi multiplied by the fraction of the time that they’re . To calculate an “mixture” cpi, or common cpi per core, divide the sum of all core's threads . How do you calculate cpi department? ▫ cpi stands for common variety of cycles per instruction. · what’s cpi laptop structure? T = clock cycle time . Directions could be alu, load, retailer, department and so forth. Cpu time = i * cpi * t.
### Cpi = common cycles per instruction;
Directions could be alu, load, retailer, department and so forth. · how do you calculate . Cpi = cycles per instruction. · what’s cpi laptop structure? Cpi = common cycles per instruction; In laptop structure, cycles per instruction (aka clock cycles per instruction, clocks per instruction, or cpi) is one side of a processor's efficiency: . T = clock cycle time . Laptop architects can cut back the instruction depend by including extra highly effective directions to the instruction set. Calculation of cpi (cycles per instruction). Cc = clock cycle depend. Cpu time = i * cpi * t. How do you calculate cpi department? For instance, a pentium has a 233 mhz clock.
T = clock cycle time . Tailored from laptop group and design, patterson & hennessy, ucb. I = variety of directions in program; Calculation of cpi (cycles per instruction). Cpi stands for clock cycles per instruction.
Cpi per core is pretty simple as proven in fig. Cpi is the sum of all of the directions within the cpi multiplied by the fraction of the time that they’re . To calculate an “mixture” cpi, or common cpi per core, divide the sum of all core's threads . Laptop architects can cut back the instruction depend by including extra highly effective directions to the instruction set. • thus it can’t be used to match computer systems/cpus with completely different . Calculation of cpi (cycles per instruction). Cc = clock cycle depend. Cpu time = i * cpi * t.
Baca Juga Get Afpsat Exam Result 2021 Visayas Pics
### For instance, a pentium has a 233 mhz clock.
Cpu time = i * cpi * t. Cc = clock cycle depend. Tailored from laptop group and design, patterson & hennessy, ucb. Nevertheless, this could improve both cpi or . In laptop structure, cycles per instruction (aka clock cycles per instruction, clocks per instruction, or cpi) is one side of a processor's efficiency: . What’s the formulation for cpi clock per instruction? Cpi per core is pretty simple as proven in fig. Calculation of cpi (cycles per instruction). As we all know a program consists of variety of directions. T = clock cycle time . To calculate an “mixture” cpi, or common cpi per core, divide the sum of all core's threads . For instance, a pentium has a 233 mhz clock. How do you calculate cpi department?
Obtain Cpi Formulation Laptop Structure Footage. Laptop architects can cut back the instruction depend by including extra highly effective directions to the instruction set. T = clock cycle time . · what’s cpi laptop structure? Calculation of cpi (cycles per instruction). Cpi stands for clock cycles per instruction.
Laptop architects can cut back the instruction depend by including extra highly effective directions to the instruction set cpi formula. · how do you calculate . | 1,174 | 4,994 | {"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} | 3.640625 | 4 | CC-MAIN-2023-14 | longest | en | 0.875429 |
https://www.physicsforums.com/threads/why-do-we-subtract-enthalpy-of-lattice-formation.736378/ | 1,531,725,795,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589222.18/warc/CC-MAIN-20180716060836-20180716080836-00128.warc.gz | 979,102,105 | 14,670 | # Homework Help: Why do we *subtract* enthalpy of lattice formation?
1. Feb 3, 2014
### aleksbooker
Hello all,
I'm in gen chem 2 and we're going over how to calculate the enthalpy of lattice formation. The way given is to use the Born-Haber process and add the enthalpies of all the steps in between.
e.g. $Na_{(s)} --> Na^+_{(g)} + e^-$ (388kJ)
There are three or four of these, and we combine (add) them to find the enthalpy of reaction. I know how to *do* the problem, subtracting for both sides to determine the value of the missing variable. Here's where I'm confused:
Why is it that everything else is added, and only enthalpy of lattice formation is subtracted?
(enthalpy of sublimation of sodium) + (enthalpy of sodium ionization) + (enthalpy of fluorine atom formation?) + (enthalpy of fluorine ion formation) - (enthalpy of lattice formation) = enthalpy of sodium fluoride reaction
That fourth operation, why is it subtraction and not addition?
2. Feb 4, 2014
### Staff: Mentor
Technically NaF already exists before the solid is formed.
3. Feb 4, 2014
### aleksbooker
@Borek, you mean we subtract NaF because it already exists?
I don't understand what that means. Are we removing it from the sequence?
As far as I understand, it proceeds logically: solid sodium becomes gaseous sodium becomes ionized sodium gas, while diatomic fluorine gas becomes monatomic fluorine gas becomes ionized fluorine gas, and then ionized sodium gas combines with ionized fluorine gas to form a lattice (which requires enthalpy of lattice energy) which becomes sodium fluoride gas (NaF). How does NaF exist before going through this process? | 417 | 1,648 | {"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} | 2.5625 | 3 | CC-MAIN-2018-30 | latest | en | 0.887304 |
http://what-when-how.com/Tutorial/topic-102/Game-Testing-All-in-One-288.html | 1,547,681,488,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583657907.79/warc/CC-MAIN-20190116215800-20190117001800-00470.warc.gz | 248,387,111 | 4,360 | Game Development Reference
In-Depth Information
Add these numbers according to step 2. 10 + 1.176 + 20 = 31.176. Finish with step 3.
Dividing 10, which is the reciprocal of the usage probability for Look Sensitivity = 1, by
31.18, which is the sum of all three reciprocals, gives an inverted probability of 0.321.
Since the numbers in the table are percentages, this gets entered as 32.1. Likewise, divide
1.18 by 31.18 to get the second inverted usage result 0.038, or 3.8%. Complete this col-
umn by dividing 20 by 31.18 to get 0.641 and enter 64.1 as the inverted usage for Look
Sensitivity = 10.
Comparing the inverted usage values to the original ones confirms that the relative
proportions of each usage value have also been inverted. Originally, the usage for Look
Sensitivity = 1 was 10% versus 5% for Look Sensitivity = 10: a 2 to 1 ratio. In the
inverted table, the Look Sensitivity = 10 value is 64.2—twice that of the 32.1% usage
for Look Sensitivity = 1. You can examine the values for each parameter to confirm
that this holds true for the other values within each column.
The complete inverted Look Sensitivity usage table for all player profiles is provided
in Figure 12.21.
Figure 12.21 Inverted usage percentages for the Look Sensitivity parameter.
Note
The “normal�? and inverted usage tables for all of the HALOAdvanced Controls parameters are pro-
vided in an Excel spreadsheet file on the topic's CD-ROM. There are separate worksheets for the
Normal and Inverted usages. You can change the values on the Normal Usage sheet and the values
on the Inverted Usage sheet will be calculated for you.
TFD Flow Usage Inversion
The TFD Enter and Exit flows present special cases you must deal with when inverting
usages. Since these are really “test�? operations versus “user�? operations, the usage per-
centage for these flows should be preserved. They will keep the same value in the
inverted usage set that you assigned to them originally. Figure 12.22 shows the Unlock
Item TFD's inverted Casual player usage table initialized with these fixed values.
Search WWH ::
Custom Search | 524 | 2,099 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2019-04 | latest | en | 0.869691 |
http://myz-vgb.ru/hypothesis-testing.html | 1,524,429,519,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945648.77/warc/CC-MAIN-20180422193501-20180422213501-00347.warc.gz | 202,668,213 | 11,713 | # Hypothesis Testing
Definition: The Hypothesis Testing is a statistical test used to determine whether the hypothesis assumed for the sample of data stands true for the entire population or not. Simply, the hypothesis is an assumption which is tested to determine the relationship between two data sets.
In hypothesis testing, two opposing hypotheses about a population are formed Viz. Null Hypothesis (H0) and Alternative Hypothesis (H1). The Null hypothesis is the statement which asserts that there is no difference between the sample statistic and population parameter and is the one which is tested, while the alternative hypothesis is the statement which stands true if the null hypothesis is rejected.
The following Hypothesis Testing Procedure is followed to test the assumption made.:
1. Set up a Hypothesis
2. Set up a suitable Significance Level
3. Determining a suitable Test Statistic
4. Determining the Critical Region
5. Performing computations
6. Decision-making
While testing the hypothesis, an individual may commit the following types of error:
1. Type-I Error: True Null hypothesis is rejected, i.e. hypothesis is rejected when it should be accepted. The probability of committing the type-I error is denoted by α and is called as a level of significance.
If, α = Pr[type-I error] = Pr [reject H0/H0 is true]
Then, (1-α) = Pr[accept H0/H0 is true]
(1-α) = corresponds to the concept of Confidence Interval.
2. Type-II Error: A False Null hypothesis is accepted, i.e. hypothesis is accepted when it should be rejected. The probability of committing the type-II error is denoted by β.
If, β = Pr[type-II error] = Pr[accept H0/H0 is false]
Then, (1-β) = Pr[reject Ho/H0 is false
(1-β) = power of a statistical test.
Thus, hypothesis testing is the important method in the statistical inference that measures the deviations in the sample data from the population parameter. The hypothesis tests are widely used in the business and industry for making the crucial business decisions.
## Related pages
oligopoly noteswhat is repo rate in hindihrm defethnocentric approach to international staffingnon collusive oligopoly modelsnational pension system npsdefine apprenticeskeynes theory of trade cycleadministrative management theory by henri fayolinelastic demand meaningtraining definition in hrmseasonal unemployment examplestypes of price elasticity of demand with graphstheories on organisational structurequotasamplingunsolicited application definitiondemand elasticity definitionteleological explanation definitiondifference between fayol and taylor theorywhat is a moratorium perioddefine sales force automationpsychological barriers of communicationwhat is acid test ratio formulawhat is shrm definitionppp meaning economicsbalanced scorecard definitionrecurring deposits meaningmaslow meaningwhat is a turnaround strategy in businessinterval defentrepreneurial deffeatures of monopoly in economicsmonopolies meaningansoff matrix strategiesmarket segmentation basesspot transaction exampledefine microenvironmentepf employee provident fundlazy faire leadershipmeaning of outsourcedhorizonal definitionlaissez faire managerindividual determinants of consumer behaviourequipments definitiondefinition of grapevine communicationdefine quality in tqmsales projection formularesonance marketing definitionexample of administered vertical marketing systemdefinition judgementalorganizational grapevine definitiondefinition of bond fundsocial loafing meanssnowball sample examplewhat is net profit margin ratiohrm planning processdefinition of straddlingdefine operant conditioning in psychologycarl rogers theory of the selfenvironmental marketing definitionexpectancy theory of motivation examplescorporate raider meaningtraining methods in hrm pptsegmented meaningelastic demand definition economicstypes of reinforcement theorysnowball recruitmentneft minimum and maximum limitfactors of cost push inflationdemand schedule definitionrevitalize definitiondefinition microfinancecarrot and stick theoryallport common traitsdeontological theory ethicsfiedlers | 807 | 4,088 | {"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} | 3.765625 | 4 | CC-MAIN-2018-17 | latest | en | 0.905379 |
https://brilliant.org/problems/series-parallel-graph/ | 1,524,633,095,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947693.49/warc/CC-MAIN-20180425041916-20180425061916-00474.warc.gz | 586,191,348 | 11,377 | # Series-parallel graph
A series-parallel graph is one of the following:
• $$K_2$$, a graph with two vertices and a single edge, such that one vertex is marked the source and the other vertex is marked the sink, or
• a graph obtained from a series composition of two series-parallel graphs $$G,H$$, where the sink of $$G$$ is identified (glued) to the source of $$H$$, where the source of $$G$$ is the new source and the sink of $$H$$ is the new sink, or
• a graph obtained from a parallel composition of two series-parallel graphs $$G,H$$, where the two sources are identified, and so are the two sinks, which become the source and the sink of the new graph respectively.
The name comes because they model electric circuits, in particular series and parallel circuits. A series composition is simply putting two smaller circuits in series, and a parallel composition is simply putting two smaller circuits in parallel.
What is the maximum number of edges that a simple series-parallel graph with 2015 vertices can have?
× | 229 | 1,027 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2018-17 | latest | en | 0.933818 |
http://www.studymode.com/essays/Calculation-Of-Power-918709.html | 1,508,213,920,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187820700.4/warc/CC-MAIN-20171017033641-20171017053641-00544.warc.gz | 716,400,512 | 19,131 | # Calculation of Power
Only available on StudyMode
• Published : February 19, 2012
Text Preview
POWER CALCULATION
Power is the quantity work has to do with a force causing a displacement. It is the rate at which work is done. (http://www.physicsclassroom.com/class/energy/u5l1e.cfm) We can calculate power with using the following equation.
In this experiment I calculated my power with
Power (watt)=mass (kg) *gravitational acceleration (m/s2)*height(m)/time (s) I made the experiment with my controller for getting more reliable datas.I calculated my power with cliambing upstairs and meausuring its’ time.I did this 5 times to get more data. While I was climbing, my controller measure the time with a chronometer. To find my mass I used electronic balance and to find the height of stairs I used a ruler. I found the literal value of gravitational acceleration 112.64
213.89
312.38
413.41
512.93
Table 1: This table shows us the time of climbing stairs in each trials. Mass=53.0±0.1kg
The accepted gravitational force in Ankara: 9.78 N
Height of one stair: 16.3±0.1cm = 0.163±0.001m
There was 11*4=44 stairs so ;
The total height of stairs: 7.172 ±0.001m
(I used the uncertanities as the minimum unit of measurement.) trial numberpower (watt)
1294.1±3.3
2268.0±2.8
3300.2±3.4
4277.2±2.9
5288.0±3.2
Table 2: shows us the powers in different trials.
Power (watt)=mass (kg) *gravitational acceleration (m/s2)*height(m)/time (s) So :
Trial 1: P1: (53.0±1kg)*(9.78m/s2)*7.172±0.001m)/12.64±0.01s=294.1±3.3 RESULTS:
Range =maximum value -minimum value
300.2-277.2=23
Uncertanity=range/2
23/2=11.5
Mean=294.1+268.0+300.2+277.2+288.0/5=285.5+15.6
285.5±11.5
Conclusion Evaluation
I found my power 285.5±11.5W . The accepted value of horsepower is 745.7 (http://en.wikipedia.org/wiki/Horsepower) My power is %38.2 of an horse. There were some limitations in the experiment like wearing boots, change in the distance between in two different stairs, performance decrease because of fatigue and it’s the power that in those stairs it can be... | 629 | 2,042 | {"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-2017-43 | latest | en | 0.861352 |
http://math.stackexchange.com/questions/248609/relation-between-reversible-distribution-and-limiting-distribution | 1,469,794,890,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257830066.95/warc/CC-MAIN-20160723071030-00277-ip-10-185-27-174.ec2.internal.warc.gz | 173,673,152 | 17,133 | # Relation between reversible distribution and limiting distribution?
For a discrete time Markov chain, its limiting distribution is defined to be the same for all the initial distributions. A distribution over the state space is called a reversible distribution, if it satisfies the detailed balance equation.
I think a reversible distribution may not be the limiting distribution, because it may happen that a reversible distribution exists while the limiting distribution doesn't. But when the limiting distribution exist, all stationary distributions (including reversible distributions) must be the same as the limiting distribution. Please correct me if I was wrong.
Conversely, must the limiting distribution (if exists) be a reversible distribution? A counterexample will suffice. I can imagine that counterexample Markov chain will need to have the limiting distribution, and doesn't have a reversible distribution.
Thanks and regards!
-
A limiting distribution may not be reversible. A reversible distribution is always a limiting distribution. A reversible distribution may not be unique. A limiting distribution may not be unique. (Four assertions, three basic blunders. Please find a book and study it! As already explained many times.) – Did Dec 1 '12 at 15:08
@did: By "the limiting distribution" I mean the limiting distribution which does not depend on the initial distributions. I thought it was unique. Which books do you recommend that I can find answers to most of my questions? – Tim Dec 1 '12 at 15:17
I am not sure the term reversible distribution makes sense because I think reversibility to be really a property of the markov process and not of a distribution per se. Isn't more accurate just to say a reversible process? – Learner Dec 1 '12 at 15:19
@learner: Thanks for the comment! For a reversible distribution in a homogeneous DTMC, check out my previous post math.stackexchange.com/questions/248586/…. – Tim Dec 1 '12 at 15:21
this question seems to be very related to that one. I would suggest you to check out the book "Markov Chains and Stochastic Stability" by Meyn and Tweedie (available online) to learn about the 'limiting' distributions. – S.D. Dec 2 '12 at 17:22 | 465 | 2,208 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2016-30 | latest | en | 0.903542 |
https://davidlowryduda.com/another-proof-of-wilsons-theorem/ | 1,670,091,567,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710936.10/warc/CC-MAIN-20221203175958-20221203205958-00209.warc.gz | 236,816,439 | 3,763 | mixedmath
Explorations in math and programming
David Lowry-Duda
While teaching a largely student-discovery style elementary number theory course to high schoolers at the Summer@Brown program, we were looking for instructive but interesting problems to challenge our students. By we, I mean Alex Walker, my academic little brother, and me. After a bit of experimentation with generators and orders, we stumbled across a proof of Wilson's Theorem, different than the standard proof.
Wilson's theorem is a classic result of elementary number theory, and is used in some elementary texts to prove Fermat's Little Theorem, or to introduce primality testing algorithms that give no hint of the factorization.
Theorem 1 (Wilson's Theorem) For a prime number ${p}$, we have $$(p-1)! \equiv -1 \pmod p. \tag{1}$$
The theorem is clear for ${p = 2}$, so we only consider proofs for 'odd primes ${p}$.'
The standard proof of Wilson's Theorem included in almost every elementary number theory text starts with the factorial ${(p-1)!}$, the product of all the units mod ${p}$. Then as the only elements which are their own inverses are ${\pm 1}$ (as ${x^2 \equiv 1 \pmod p \iff p \mid (x^2 - 1) \iff p\mid x+1}$ or ${p \mid x-1}$), every element in the factorial multiples with its inverse to give ${1}$, except for ${-1}$. Thus ${(p-1)! \equiv -1 \pmod p.} \diamondsuit$
Now we present a different proof.
Take a primitive root ${g}$ of the unit group ${(\mathbb{Z}/p\mathbb{Z})^\times}$, so that each number ${1, \ldots, p-1}$ appears exactly once in ${g, g^2, \ldots, g^{p-1}}$. Recalling that ${1 + 2 + \ldots + n = \frac{n(n+1)}{2}}$ (a great example of classical pattern recognition in an elementary number theory class), we see that multiplying these together gives ${(p-1)!}$ on the one hand, and ${g^{(p-1)p/2}}$ on the other.
As ${g^{(p-1)/2}}$ is a solution to ${x^2 \equiv 1 \pmod p}$, and it is not ${1}$ since ${g}$ is a generator and thus has order ${p-1}$. So ${g^{(p-1)/2} \equiv -1 \pmod p}$, and raising ${-1}$ to an odd power yields ${-1}$, completing the proof $\diamondsuit$.
After posting this, we have since seen that this proof is suggested in a problem in Ireland and Rosen's extremely good number theory book. But it was pleasant to see it come up naturally, and it's nice to suggest to our students that you can stumble across proofs.
It may be interesting to question why ${x^2 \equiv 1 \pmod p \iff x \equiv \pm 1 \pmod p}$ appears in a fundamental way in both proofs.
This post appears on the author's personal website davidlowryduda.com and on the Math.Stackexchange Community Blog math.blogoverflow.com. It is also available in pdf note form. It was typeset in \TeX, hosted on Wordpress sites, converted using the utility github.com/davidlowryduda/mse2wp, and displayed with MathJax.
Info on how to comment
bold, italics, and plain text are allowed in comments. A reasonable subset of markdown is supported, including lists, links, and fenced code blocks. In addition, math can be formatted using $(inline math)$ or $$(your display equation)$$. | 829 | 3,076 | {"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": 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} | 4.125 | 4 | CC-MAIN-2022-49 | latest | en | 0.889925 |
https://learn.careers360.com/school/question-need-solution-for-rd-sharma-maths-class-12-chapter-18-indefinite-integrals-excercise-18-point-15-question-5/?question_number=5.0 | 1,716,273,215,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058385.38/warc/CC-MAIN-20240521060250-20240521090250-00086.warc.gz | 312,766,301 | 42,680 | # Get Answers to all your Questions
### Answers (1)
Answer:
$\frac{1}{2} \tan ^{-1}\left(\frac{x+3}{2}\right)+C$
Hint:
To solve this problem use special integration formula
Given:
$\int \frac{1}{x^{2}+6 x+13} d x$
Solution:
Let $I=\int \frac{1}{x^{2}+6 x+13} d x$
\begin{aligned} &=\int \frac{1}{x^{2}+2 \cdot x \cdot 3+3^{2}-3^{2}+13} d x \\ & \end{aligned}
$=\int \frac{1}{(x+3)^{2}-9+13} d x \\$
$=\int \frac{1}{(x+3)^{2}+4} d x$
Put $x+3=t \Rightarrow d x=d t$
Then $I=\int \frac{1}{t^{2}+2^{2}} d t$
$=\frac{1}{2} \tan ^{-1}\left(\frac{t}{2}\right)+C$ $\left[\because \int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C\right]$
$=\frac{1}{2} \tan ^{-1}\left(\frac{x+3}{2}\right)+C$ $\quad[\because t=x+3]$
View full answer
## Crack CUET with india's "Best Teachers"
• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support | 377 | 940 | {"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": 12, "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-2024-22 | latest | en | 0.2799 |
https://www.homebrewtalk.com/threads/dumb-question-212-what-does-this-mean-15-25l.113353/ | 1,652,891,183,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522284.20/warc/CC-MAIN-20220518151003-20220518181003-00495.warc.gz | 904,420,383 | 19,068 | # Dumb Question #212 What does this mean 15-25L?
### Help Support Homebrew Talk:
#### j8h9
##### Member
So I'm looking at grains and see lots of different makers from US and elsewhere and all the grains have this notation which obviously means something... What does it mean? What do the numbers with the L mean?
15-25L
Weyermann Cara Foam (1 lb.)
(Germany) 1.3-2.3L
Weyermann Cara Hell (1 lb.)
(Germany) 9-13L
Weyermann Cara Red (1 lb.)
(Germany) 15-25L
#### Hugh_Jass
##### Well-Known Member
Lovibond
It's a measurement of color
#### flyangler18
##### Well-Known Member
The 'L' means Lovibond, an expression of color potential in the grain.
The higher °L, the darker the grain. For example, a Crystal malt at 120L is lighter than roasted barley (approx 450L).
#### ChshreCat
##### Well-Known Member
Here's the pic from the Wiki to give you an idea of what the numbers look like.
#### double_e5
##### Well-Known Member
The L stands for Lovibond. It is a scale for the color the grain is going to give you. The higher the number, the darker.
OP
OP
Thank You...
#### Nurmey
##### I love making Beer
It is the Lovibond scale. It's telling you what color the grain produces. Lower numbers = lighter, higher number = darker.
Replies
17
Views
994
Replies
280
Views
29K
Replies
5
Views
943
Replies
33
Views
1K
Replies
7
Views
603 | 388 | 1,342 | {"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-2022-21 | latest | en | 0.856469 |
https://sciencing.com/7-scientific-mnemonic-devices-to-make-studying-easier-13710429.html | 1,638,075,144,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358469.34/warc/CC-MAIN-20211128043743-20211128073743-00051.warc.gz | 577,026,614 | 89,302 | # 7 Scientific Mnemonic Devices to Make Studying Easier
••• Jacob Ammentorp Lund/iStock/GettyImages
Print
Science, technology, engineering and math (STEM) classes are challenging and often require a lot of study time to memorize the many formulas and concepts. But once you become familiar with them, the pieces fall into place. One method for remembering a large quantity of information is the use of mnemonic devices — memory tools that aid in creating shortcuts and recalling information easily. It could be anything from acronyms, a tune, or forming a silly yet memorable sentence where the the first letter of a word represents another idea. We put together a list of seven mnemonic devices from various STEM subjects you’re likely to encounter. And, of course, have fun creating your own!
••• ChrisGorgio/iStock/GettyImages
## 1. Order of the Planets (excluding Pluto)
Since Pluto was demoted to a dwarf planet, there are only eight planets to remember: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune. The first letter of each word in the mnemonic device represents the eight planets closest to the sun.
Mnemonic: My Very Educated Mother Just Served Us Nachos
## 2. Features of a Living Organism
We determine whether an organism is living or non-living by using these seven life processes: Movement, Respiration, Sensation, Growth, Reproduction, Excretion, Nutrition.
Mnemonic: MRS GREN
••• Hey Darlin/iStock/GettyImages
## 3. Five Great Lakes
The five great lakes – located on the United States and Canadian border – make up the largest body of freshwater in the world. This acronym is a simple way to remember Lake Huron, Lake Ontario, Lake Michigan, Lake Erie and Lake Superior.
Mnemonic: HOMES
## 4. Order of Operations
In complex math equations, follow the order of operations – Parentheses, Exponents, Multiply, Divide, Add, Subtract – because if you forget or skip a step, you’ll arrive at the wrong answer. Use this mnemonic to remember which operation comes first.
Mnemonic: Please Excuse My Dear Aunt Sally (PEMDAS)
••• plusphoto/iStock/GettyImages
## 5. Colors in Rainbow
The colors of the visible light spectrum are red, orange, yellow, green, blue, indigo and violet. It sounds like it could be someone’s first name, initial and last name.
Mnemonic: ROY G. BIV
## 6. Levels of Classification
The organization of living things is grouped into these main biological categories: Domain, Kingdom, Phylum, Class, Order, Family, Genus, Species. As you get further down the taxonomy, these groups branch out to more sub-groups.
Mnemonic: Dear King Phillip Come Over For Good Soup
••• vencavolrab/iStock/GettyImages
## 7. Geological Time Periods
Geological time periods are used by scientists to describe the timing and relationship of events that occurred in Earth’s history. The geological periods include: Precambrian, Cambrian, Ordovician, Silurian, Devonian, Carboniferous, Permian, Triassic, Jurassic, Cretaceous, Paleocene, Eocene, Oligocene, Miocene, Pliocene, Pleistocene, Recent (Holocene).
Mnemonic: Pregnant Camels Often Sit Down Carefully, Perhaps Their Joints Creak? Possibly Early Oiling Might Prevent Permanent Rheumatism
Dont Go!
We Have More Great Sciencing Articles! | 750 | 3,240 | {"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 | longest | en | 0.872935 |
https://simplicable.com/new/inductive-reasoning | 1,721,868,829,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518532.61/warc/CC-MAIN-20240724232540-20240725022540-00006.warc.gz | 457,690,249 | 11,254 | Problem Solving
First Principles
# What is Inductive Reasoning?
, updated on
Inductive reasoning is a form of logic that formulates theories based on a set of known facts. This allows for false conclusions, meaning that it can be wrong. The classic example of inductive reasoning is that because most life forms on Earth depend on liquid water for survival, water must be important to alien life forms, if they exist.
Overview: Inductive Reasoning Function Value Developing theories based on observations. Direction Bottom up, starting with confirmed observations working towards a theoretical conclusion. Uncertainty Allows for uncertain conclusions, i.e. seeks likely theories Similar Techniques
## Thinking
This is the complete list of articles we have written about thinking.
Abductive Reasoning
Abstract Thinking
Abstraction
Aesthetics
Analogy
Analysis Paralysis
Analytical Thinking
Anomie
Argument
Argument From Silence
Arrow Of Time
Assertions
Automaticity
Backward Induction
Base Rate Fallacy
Benefit Of Doubt
Big Picture
Brainstorming
Call To Action
Catch 22
Causality
Choice Architecture
Circular Reasoning
Cognition
Cognitive Abilities
Cognitive Biases
Cold Logic
Collective Intelligence
Complexity Bias
Concept
Consciousness
Constructive Criticism
Convergent Thinking
Counterfactual Thinking
Creative Tension
Creeping Normality
Critical Thinking
Culture
Curse Of Knowledge
Decision Fatigue
Decision Framing
Decision Making
Defensive Pessimism
Design Thinking
Divergent Thinking
Educated Guess
Emotional Intelligence
Epic Meaning
Essential Complexity
Excluded Middle
Failure Of Imagination
Fallacies
Fallacy Fallacy
False Analogy
False Balance
False Dichotomy
False Equivalence
First Principles
Formal Logic
Four Causes
Fuzzy Logic
Gambler's Fallacy
Generalization
Golden Hammer
Good Judgement
Grey Area
Groupthink
Heuristics
Hindsight Bias
Hope
Idealism
Ideas
If-By-Whiskey
Illogical Success
Imagination
Independent Thinking
Inductive Reasoning
Inference
Influencing
Informal Logic
Information
Introspection
Intuition
Inventive Step
Learning
Lifestyle
Logic
Logical Argument
Logical Thinking
Ludic Fallacy
Magical Thinking
Meaning
Mental Experiences
Mental State
Mindset
Misuse of Statistics
Motivated Reasoning
Natural Language
Nirvana Fallacy
Norms
Not Even Wrong
Objective Reason
Objectivity
Opinion
Overthinking
Perception
Personal Values
Perspective
Positive Thinking
Practical Thinking
Pragmatism
Premise
Problem Solving
Proof By Example
Propositional Logic
Prosecutor's Fallacy
Rational Thought
Realism
Reality
Reason
Reasoning
Red Herring
Reflective Thinking
Reification
Relativism
Salience
Scarcity Mindset
Scientism
Selective Attention
Serendipity
Situational Awareness
Sour Grapes
State Of Mind
Storytelling
Subjectivity
Systems Thinking
Thinking
Thought Experiment
Unknown Unknowns
Visual Thinking
Want To Believe
Win-Win Thinking
Wishful Thinking
Worldview
## Decision Making
A list of decision making techniques.
The observation that groups may make collective decisions that are viewed as wrong or irrational by each individual member of the group.
## Decision Making Process
A complete guide to the decision making process.
## Rational Thought
The difference between rational thought and logic.
## Uncertainty
The common types of uncertainty in decision making and strategy.
## Information Costs
A definition of information costs with examples.
## Reverse Brainstorming
A definition of reverse brainstorming with examples.
## Decision Fatigue
The definition of decision fatigue with examples.
Taking a position that you do not necessarily agree with for the purposes of argument.
The definition of paradox of choice with examples.
## Cognitive Biases
A list of common cognitive biases explained.
## Curse Of Knowledge
Why experts have trouble communicating.
## Optimism Bias
An overview of optimism bias, including its surprising benefits.
## Decoy Effect
A cognitive bias that is well known in marketing circles.
## Biases vs Heuristics
The difference between biases and heuristics.
A definition of information cascade with examples.
## Functional Fixedness
A definition of functional fixedness with examples.
## Boil The Frog
A definition of boil the frog, with examples.
## Anecdotal Evidence
The definition of anecdotal evidence with examples.
## Scientism
The definition of scientism with examples.
The most popular articles on Simplicable in the past day.
## New Articles
Recent posts or updates on Simplicable.
Site Map | 997 | 4,497 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-30 | latest | en | 0.761698 |
https://shineskill.com/c-ds-mcq/stack-of-DSA.php | 1,723,738,078,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641299002.97/warc/CC-MAIN-20240815141847-20240815171847-00225.warc.gz | 405,655,709 | 6,513 | /* */ Click to Join Live Class with Shankar sir Call 9798158723
# Data Structure MCQ - Stack
Q1.Process of inserting an element in stack is called ____________
1. Create
2. Evaluation
3. Pop
4. Push
Explanations :Push operation allows users to insert elements in the stack. If the stack is filled completely and trying to perform push operation stack – overflow can happen.
Q2.Process of removing an element from stack is called __________
1. Pop
2. Create
3. Push
4. Evaluated
Explanations :Elements in the stack are removed using pop operation. Pop operation removes the top most element in the stack i.e. last entered element.
Q3.Pushing an element into stack already having five elements and stack size of 5, then stack becomes ___________
1. Underflow
2. User flow
3. Overflow
4. Crash
Explanations :The stack is filled with 5 elements and pushing one more element causes a stack overflow. This results in overwriting memory, code and loss of unsaved work on the computer.
Q4.Entries in a stack are “ordered”. What is the meaning of this statement?
1. A collection of stacks is sortable
2. The entries are stored in a linked list
3. There is a Sequential entry that is one by one
4. Stack entries may be compared with the ‘<‘ operation
Explanations : In stack data structure, elements are added one by one using push operation. Stack follows LIFO Principle i.e. Last In First Out(LIFO).
Q5.Consider the usual algorithm for determining whether a sequence of parentheses is balanced. Suppose that you run the algorithm on a sequence that contains 2 left parentheses and 3 right parentheses (in some order). The maximum number of parentheses that appear on the stack AT ANY ONE TIME during the computation?
1. 1
2. 3
3. 2
4. 4 or more
Explanations :In the entire parenthesis balancing method when the incoming token is a left parenthesis it is pushed into stack. A right parenthesis makes pop operation to delete the elements in stack till we get left parenthesis as top most element. 2 left parenthesis are pushed whereas one right parenthesis removes one of left parenthesis. 2 elements are there before right parenthesis which is the maximum number of elements in stack at run time.
Q6.The postfix form of the expression (A+ B)*(C*D- E)*F / G is?
1. AB+ CD*E – FG /**
2. AB+ CD*E – FG /**
3. AB + CDE * – * F *G /
4. AB + CD* E – F **G
Explanations :(((A+ B)*(C*D- E)*F) / G) is converted to postfix expression as
(AB+(*(C*D- E)*F )/ G)
(AB+CD*E-*F) / G
(AB+CD*E-*F * G/). Thus Postfix expression is AB+CD*E-*F*G/
Q7.The data structure required to check whether an expression contains balanced parenthesis is
1. Stack
2. Queue
3. Tree
4. Array
Q8.Following is C like pseudo code of a function that takes a number as an argument, and uses a stack S to do processing.
```void fun(int n)
{
Stack S; // Say it creates an empty stack S
while (n > 0)
{
// This line pushes the value of n%2 to stack S
push(&S, n%2);
n = n/2;
}
// Run while Stack S is not empty
while (!isEmpty(&S))
printf("%d ", pop(&S)); // pop an element from S and print it
}```
1. Prints binary representation of n in reverse order
2. Prints binary representation of n
3. Prints the value of Logn
4. Prints the value of Logn in reverse order
Q9.Which one of the following is an application of Stack Data Structure
1. Managing function calls
2. The stock span problem
3. Arithmetic expression evaluation
4. All of the above | 882 | 3,405 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2024-33 | latest | en | 0.860442 |
http://softmath.com/algebra-software-2/working-out-diff-ratio-formula.html | 1,561,557,611,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000353.82/warc/CC-MAIN-20190626134339-20190626160339-00413.warc.gz | 158,405,550 | 13,143 | English | Español
# Try our Free Online Math Solver!
Online Math Solver
Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
Ineq. #3:
Ineq. #4:
Ineq. #5:
Ineq. #6:
Ineq. #7:
Ineq. #8:
Ineq. #9:
Solve for:
Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg:
working out diff ratio formula
Related topics:
multiplying multiple exponents | answers for algebra 2: an integrated approach | Algebra Works Sheets | algebra x | syllabus for intermediate algebra,7 | Trigonometry Special Values Complete Chart | Free Algebra Quiz+finding Slope | multiplying exponent calculator | fractions6thgrade.com | Mcdougal Littell Algebra 2 Answers Pdf | multiplying of an algebraic term by an algebraic expression
Author Message
Avkomm
Registered: 06.09.2002
From: Scotland
Posted: Thursday 22nd of Aug 18:26 Can somebody help me? I am in deep trouble . It’s about working out diff ratio formula. I tried to find somebody in my locality who can help me out with simplifying fractions, radicals and hypotenuse-leg similarity. But I could not . I also know that it will be hard for me to meet the price. My quiz are close at hand . What should I do? Anybody out there who can save me?
nxu
Registered: 25.10.2006
From: Siberia, Russian Federation
Posted: Friday 23rd of Aug 20:45 Kids can’t seem to think of anything beyond extra classes . Why don’t you try something yourself? There are many resources for working out diff ratio formula which are much better than tutoring. Try Algebrator, and you will never need a tuition .
Svizes
Registered: 10.03.2003
From: Slovenia
Posted: Sunday 25th of Aug 15:58 I can confirm that. Algebrator is the ultimate technology for working out math assignments. Been using it for some time now and it keeps on amazing me. Every homework that I type in, Algebrator gives me a perfect answer to it. I have never enjoyed learning algebra homework on adding fractions, interval notation and unlike denominators so much before. I would suggest it for sure.
mitcubicxi
Registered: 31.08.2001
From: UK
Posted: Tuesday 27th of Aug 08:03 I want to order this thing right away!Somebody please tell me, how do I purchase it? Can I do so over the internet? Or is there any contact number where we can place an order?
MoonBuggy
Registered: 23.11.2001
From: Leeds, UK
Posted: Thursday 29th of Aug 07:57 Thanks a ton for the detailed information. We will definitely try this out. Hope we get our assignments finished with the help of Algebrator. If we have any technical queries with respect to its usage , we would definitely get back to you again. | 745 | 2,874 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2019-26 | latest | en | 0.892102 |
https://mail.scipy.org/pipermail/numpy-discussion/2006-July/009181.html | 1,506,144,887,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689490.64/warc/CC-MAIN-20170923052100-20170923072100-00003.warc.gz | 703,751,487 | 4,185 | # [Numpy-discussion] .T Transpose shortcut for arrays again
Tim Hochberg tim.hochberg at cox.net
Thu Jul 6 13:50:40 CDT 2006
```Sasha wrote:
> On 7/6/06, Bill Baxter <wbaxter at gmail.com> wrote:
>
>> On 7/7/06, Sasha <ndarray at mac.com> wrote:
>> ... I think it's much
>> more common to want to swap just two axes, and the last two seem a logical
>> choice since a) in the default C-ordering they're the closest together in
>> memory and b) they're the axes that are printed contiguously when you say
>> "print A".
>>
>
> It all depends on how you want to interpret a rank-K tensor. You seem
> to advocate a view that it is a (K-2)-rank array of matrices and .T is
> an element-wise transpose operation. Alternatively I can expect that
> it is a matrix of (K-2)-rank arrays and then .T should be
> swapaxes(0,1). Do you have real-life applications of swapaxes(-2,-1)
> for rank > 2?
>
I do for what it's worth. At various times I use arrays of shape (n, n,
d) (matrices of rank-1 arrays at you suggest above) and arrays of shape
(d, n, n) (vectors of matrices as Bill proposes). Using swapaxes(-2, -1)
would be right only half the time, but it would the current defaults for
transpose are essentially never right for rank > 2. Then again they are
easy to explain.
>
>>> and swapaxes(-2,-1) is
>>> invalid for rank < 2.
>>>
>>>
>> At least in numpy 0.9.8, it's not invalid, it just doesn't do anything.
>>
>>
>
> That's bad. What sense does it make to swap non-existing axes? Many
> people would expect transpose of a vector to be a matrix. This is the
> case in S+ and R.
>
So this is essentially turning a row vector into a column vector? Is
that right?
>>> My main objection is that a.T is fairly cryptic
>>> - is there any other language that uses attribute for transpose?
>>>
>> Does it matter what other languages do? It's not _that_ cryptic.
>>
>
> If something is clear and natural, chances are it was done before.
> For me prior art is always a useful guide when making a design choice.
> For example, in R, the transpose operation is t(a) and works on rank
> <= 2 only always returning rank-2. K (an APL-like language) overloads
> unary '+' to do swapaxes(0,1) for rank>=2 and nothing for lower rank.
> Both R and K solutions are implementable in Python with R using 3
> characters and K using 1(!) compared to your two-character ".T"
> notation.
> I would suggest that when inventing something new, you
> should consider prior art and explain how you invention is better.
> That's why what other languages do matter. (After all, isn't 'T'
> chosen because "transpose" starts with "t" in the English language?)
>
>
>> The standard way to write transpose is with a little T superscript in the upper
>> right. We can't do that with ASCII so the T just appears after the dot.
>> Makes perfect sense to me. I'd vote for an operator if it were possible in
>> python. Something like A^T would be neat, maybe, or matlab's single-quote
>> operator.
>>
>>
>
> Well, you could overload __rpow__ for a singleton T and spell it A**T
> ... (I hope no one will take that proposal seriosely). Visually, A.T
> looks more like a subscript rather than superscript.
>
No, no no. Overload __rxor__, then you can spell it A^t, A^h, etc. Much
better ;-). [Sadly, I almost like that....]
>
>>> expressions like "a * b.T" it will not be clear whether * is a matrix
>>> or elemenwise multiplication.
>>>
>> That seems a pretty weak argument, since there are already lots of
>> expressions you can write that don't make it clear whether some operation is
>> a matrix operation or array operation.
>>
>
> This may be a weak argument for someone used to matrix notation, but
> for me seeing a.T means: beware - tricky stuff here.
>
>
>> You could write a * b.transpose(1,0)
>> right now and still not know whether it was matrix or element-wise
>> multiplication.
>>
>
> Why would anyone do that if b was a matrix?
>
>
>
>>> 2. .H : This is an O(n^2) complexity operation returning a copy so
>>> it is not appropriate for an attribute.
>>>
>> Not sure how you get O(n^2). It just requires flipping the sign on the
>> imaginary part of each element in the array. So in my book that's O(n).
>> But that does make it more expensive than O(1) transpose, yes.
>>
>>
> In my book n is the size of the matrix as in "n x n matrix", but the
> argument stays with O(n) as well.
>
>
>
>
>> But probably a better solution
>> would be to have matrix versions of these in the library as an optional
>> module to import so people could, say, import them as M and use M.ones(2,2).
>>
>>
>
> This is the solution used by ma, which is another argument for it.
>
>
>
>> In short, I'm most enthusiastic about the .T attribute. Then, given a .T,
>> it makes sense to have a .H, both to be consistent with matrix, but also
>> since it seems to be a big deal in other math packages like matlab. Then
>> given the current situation, I like the .M but I can imagine other ways to
>> make .M less necessary.
>>
>>
>
> I only raised a mild objection against .T, but the slippery slope
> argument makes me dislike it much more. At the very least I would
> like to see a discussion of why a.T is better than t(a) or +a.
>
Here's a half baked thought: if the objection to t(A) is that it doesn't
mirror the formulae where t appears as a subscript after A. Conceivably,
__call__ could be defined so that A(x) returns x(A). That's kind of
perverse, but it means that A(t), A(h), etc. could all work
appropriately for suitably defined singletons. These singletons could
either be assembeled in some abbreviations namespace or brought in by
the programmer using "import transpose as t", etc. The latter works for
doing t(a) as well of course.
-tim
``` | 1,542 | 5,732 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-39 | latest | en | 0.940226 |
http://gmatclub.com/forum/if-you-only-knew-then-what-you-know-now-fave-gmat-tips-105623-20.html | 1,484,848,733,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280723.5/warc/CC-MAIN-20170116095120-00118-ip-10-171-10-70.ec2.internal.warc.gz | 120,683,671 | 49,283 | If you only knew then what you know now... Fave GMAT Tips : General GMAT Questions and Strategies - Page 2
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack
It is currently 19 Jan 2017, 09:58
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# If you only knew then what you know now... Fave GMAT Tips
Author Message
Manager
Joined: 15 Mar 2012
Posts: 60
Location: United States
Concentration: Marketing, Strategy
Followers: 0
Kudos [?]: 18 [0], given: 19
Re: If you only knew then what you know now... Fave GMAT Tips [#permalink]
### Show Tags
28 Feb 2013, 14:08
thegmatguru wrote:
MisterEko wrote:
The simplest trick of them all is to read the stem, and then immediately start looking at the easier of two pieces of information, for example, if you get these two options:
1) x+y-6+2xy= 23
and
2) x=6
I used to go to 1) and analyze it. Now, I just pick the easier one to start, which in this case would be statement 2). When you are able to easily eliminate 1 of the statements, it gives you a decent confidence boost because you narrowed your choices down by 40-60%.
What is more interesting about the example above and something that most prep companies don't tell you is that IF YOU CORRECTLY GET ONLY ONE SOLUTION TO ONE OF THE STATEMENTS, THAT SOLUTION WILL AUTOMATICALLY BE A SOLUTION TO THE OTHER STATEMENT.
For example:
What is the value of x?
1) x is a prime factor of 221
2) 2x = 34
Clearly, the second statement is easier to evaluate (sufficient..only answer is 17)
If you know this, the first statement is much easier to evaluate because you KNOW that x=17 is a solution and you can immediately divide 221 by 17 to see what other factors it may have. (17*13 = 221)
Note that knowing the answer to one statement does not tell you how many solutions the other has, so it won't tell you directly "insufficient" or "sufficient", but it gives you at least one solution to start with.
Hi GMAT Guru, what is the answer for this SC question? Thanks
_________________
MV
"Better to fight for something than live for nothing.” ― George S. Patton Jr
e-GMAT Discount Codes Kaplan GMAT Prep Discount Codes Magoosh Discount Codes
Re: If you only knew then what you know now... Fave GMAT Tips [#permalink] 28 Feb 2013, 14:08
Go to page Previous 1 2 [ 21 posts ]
Similar topics Replies Last post
Similar
Topics:
15 What did you learn Today? [GMAT Tips] 9 27 Jul 2016, 13:04
Everything You Thought You Knew About Learning Is Wrong 0 05 Dec 2012, 11:45
54 The One Thing You Wish You Knew - GMAT Club Contest! 23 09 Oct 2012, 06:50
45 You know you have been studying too hard for the GMAT when.. 66 18 Jan 2011, 21:27
Do you know this book? 1 06 Oct 2009, 12:45
Display posts from previous: Sort by
# If you only knew then what you know now... Fave GMAT Tips
Moderators: WaterFlowsUp, HiLine
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®. | 943 | 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} | 3.96875 | 4 | CC-MAIN-2017-04 | latest | en | 0.891662 |
https://www.opengl.org/discussion_boards/search.php?s=c2a1fcf25e2efe6a813ca01ddd3d9ba7&searchid=2586222 | 1,508,503,495,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824104.30/warc/CC-MAIN-20171020120608-20171020140608-00808.warc.gz | 970,913,530 | 11,435 | # Search:
Type: Posts; User: wmelgaard
Page 1 of 2 1
1. ## Thread: extrude along an arbitrary vector?
by wmelgaard
Replies
2
Views
415
### That works For my example, polygon in Y-Z plane,...
That works
For my example, polygon in Y-Z plane, extruding in X; origin at GLfloat x,y,z:
GLfloat matrix[16];
glPushMatrix();
glTranslatef(x, y, z);
matrix[0] = Y[0]; matrix[1] =...
2. ## Thread: extrude along an arbitrary vector?
by wmelgaard
Replies
2
Views
415
### extrude along an arbitrary vector?
I have arbitrary orthogonal vectors float X[3], Y[3], Z[3];
where X is the vector that I want to extrude a polygon that is in Y, Z coordinates.
For example, I have a polygon (-.25,3; 2,3; 2,2.5;...
3. ## Thread: problem de-rotating a model
by wmelgaard
Replies
6
Views
1,017
### Thank you. Thankyouthankyouthankyou!
Thank you.
Thankyouthankyouthankyou!
4. ## Thread: problem de-rotating a model
by wmelgaard
Replies
6
Views
1,017
### problem labeling a rotated model
I have tried to implement your suggestion, unsuccessfully.
The matrix multiplication algorithm that I presented works. I have verified that using Octave. Even accounting for the fact that OpenGL...
5. ## Thread: problem de-rotating a model
by wmelgaard
Replies
6
Views
1,017
### problem labeling a rotated model
My problem is that I don't know the location of the rotated and spun vertex.
Commenting out the transform command and placing the grid location into the printtext command utilizes the same...
6. ## Thread: problem de-rotating a model
by wmelgaard
Replies
6
Views
1,017
### problem labeling a rotated model
The following program writes an 8 line cube, allows a mouse drag to rotate it, and then attempts to label the corners in their rotated location.
However, the more the model is rotated, the farther...
7. ## Thread: Drawing a bezier spline
by wmelgaard
Replies
1
Views
1,283
### If all that you want is an increment and...
If all that you want is an increment and decrement, then draw a pair of boxes (somewhere) with a "+" in one box and a "_" in the other box.
In main, create a glutMouseFunc(mouse);
within mouse(int...
8. ## Thread: Can't reshape subwindow
by wmelgaard
Replies
2
Views
877
Your answer sounds like a style issue. Within my project, "windowheight" is a computed number. For the test problem, I merely toggled between two heights. In practice, "windowheight" can be any one...
9. ## Thread: 3D Poisson distribution
by wmelgaard
Replies
3
Views
2,468
### Assume that your box is bounded by [0,0,0; 0,0,1;...
Assume that your box is bounded by [0,0,0; 0,0,1; 0,1,0; 0,1,1; 1,0,1; 1,1,1]. Compute your point [x,y,z] where (0 <= x <= 1); (0 <= y <= 1); (0 <= z <= 1) and plot your point within your cube. ...
10. ## Thread: 3D Poisson distribution
by wmelgaard
Replies
3
Views
2,468
### I don't understand what you are trying to do. ...
I don't understand what you are trying to do.
Are you using a random number generator with a poisson distribution to pick points off an OpenGL cube?
Or are you plotting points within that cube?
11. ## Thread: Can't reshape subwindow
by wmelgaard
Replies
2
Views
877
### Can't reshape subwindow
I am running Debian Stable (Jessie), with GLU1.3, and am trying to reshape a subwindow.
The following program writes a subwindow in the top, left corner of the main window. If you left-click on that...
12. ## Thread: gluLookAt & depth culling
by wmelgaard
Replies
1
Views
972
### gluLookAt & depth culling
I have a model 0.0 < z < 30.0. The default gluLookAt(0,-100,0, 0,0,0, 0,1,0); looks at the bottom of the model. If I rotate the model, the bottom of the model occludes anything above it.
If I...
13. ## Thread: Graphics disappear under certain geometries.
by wmelgaard
Replies
2
Views
1,291
### see...
see https://www.opengl.org/sdk/docs/man3/xhtml/glCullFace.xml
The default is to cull a polygon whose back is to the viewer.
Add the disable, and those faces should reappear
by wmelgaard
Replies
0
Views
772
### glutDewstroyWindow cleanup
My situation is that I have a subwindow in mainwindow which creates another subwindow in mainwindow. The second subwindow provides a template for inputting data. Upon completion of data entry, I...
by wmelgaard
Replies
3
Views
1,761
### The window was created with subWindow =...
The window was created with
subWindow = glutCreateSubWindow(mainWindow, margin, margin, width, height);
glViewport(0, 0, (GLsizei) width, (GLsizei) height);
Basically, I want to change...
by wmelgaard
Replies
3
Views
1,761
### resizing subwindow
I have an existing subwindow that is to be replaced by a different one.
Should I destroy the original and then create a new one, or merely change the size of the existing one?
How do I change the...
17. ## Thread: Identifying a specific vertex
by wmelgaard
Replies
0
Views
827
### Identifying a specific vertex
If I mouse-click on s specific pixel, that will give me the x & y of the mouse, which is the projected location of the vertex which drew that pixel. Similarly, glReadPixels(x, y, 1, 1,...
by wmelgaard
Replies
4
Views
3,689
### That works. Problems solved. Thanx. FYI for...
That works. Problems solved. Thanx.
given
glMatrixMode(GL_PROJECTION);
glOrtho(xmin, xmax, ymin, ymax, zmin, zmax);
...
by wmelgaard
Replies
4
Views
3,689
### Changing the 0 to 1 helps. I misinterpreted the...
Changing the 0 to 1 helps. I misinterpreted the glReadPixels() definition to be (target) + width & (target) + height.
Leaving the glutSwapBuffers() where it is, I get z = 0.25194 and bgr = 0. Since...
by wmelgaard
Replies
4
Views
3,689
The following program gives a value of (some random number) for the red point, but 0.0000 for the z-value of the point.
How do I get a return of z != 0.0000, given that the point is located at 50,...
21. ## Thread: glOrtho in GL_PROJECTION vs GL_MODELVIEW?
by wmelgaard
Replies
2
Views
1,868
### From the tutorial, "do you want to move the...
From the tutorial, "do you want to move the camera in one direction, or move the object in the opposite direction?" All of the examples in Chapter 3 move the model. Nate Robins' robot arm is drawn...
22. ## Thread: glOrtho in GL_PROJECTION vs GL_MODELVIEW?
by wmelgaard
Replies
2
Views
1,868
### glOrtho in GL_PROJECTION vs GL_MODELVIEW?
I have a model xmin = -264.000000, xmax = 384.000000; ymin = -263.942993, ymax = 330.480591; zmin = 673.751099, zmax = 1345.000000
If I include glOrtho((xmin -margin), (xmax +margin), (ymin...
23. ## Thread: glTranslate( 0, 0, something ) ineffectual on VBO
by wmelgaard
Replies
8
Views
4,240
### Back to basics, I have recently learned that my...
Back to basics, I have recently learned that my push, transform and pops affect the transformation matrix, not the contents of the framebuffer. You say, "well, duh?"
Google NASTRAN for what I am...
24. ## Thread: glTranslate( 0, 0, something ) ineffectual on VBO
by wmelgaard
Replies
8
Views
4,240
### Wow. I am mostly self taught, and my background...
Wow. I am mostly self taught, and my background is as a structural engineer (PE), and incidentally as a computer programmer. I have the Red Book; I should probably get a copy of the Blue Book rather...
25. ## Thread: glTranslate( 0, 0, something ) ineffectual on VBO
by wmelgaard
Replies
8
Views
4,240
### I am writing an open source version of NASTRAN....
I am writing an open source version of NASTRAN. Think of it as a CAD model undergoing computer analysis. My test file is from a real life project that has 10,000 points. My current objective is to...
Results 1 to 25 of 41
Page 1 of 2 1 | 2,175 | 7,578 | {"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-2017-43 | latest | en | 0.874258 |
http://freetofindtruth.blogspot.com/2016/08/37-58-64-ravens-colts-bills-giants.html | 1,527,204,505,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866894.26/warc/CC-MAIN-20180524224941-20180525004941-00417.warc.gz | 117,628,311 | 23,120 | ## Saturday, August 20, 2016
### 37 58 64 | Ravens, Colts & Bills, Giants, August 20, 2016 (Greatest Game Ever Played reminders)
http://www.espn.com/nfl/game?gameId=400874773
Notice the game sums to '37' points. Today is a date with a life lesson number of '37'.
19+18 = 37
8/20/2016 = 8+20+(2+0+1+6) = 37
The HOF Game with the Colts was cancelled 37-days before Andrew Luck's birthday
Zurlon Tipton #37, died in Detroit on June 28, 2016, 223-days before SB 51; Detroit = 37
http://www.espn.com/nfl/boxscore?gameId=400874751
The NFL triple header was Giants, Bills, then Ravens, Colts, and finally Chiefs, Rams. Chiefs-Rams is in the third quarter as I write this, 20-14, KC leading.
Anyhow, I notice that the total score of the Colts and Ravens game, and the Giants and Bill game, 19+18+21+0, totals '58'. 58-years ago was the greatest game ever played, between the Giants and Colts. Freemasonry = 58
By the way, 'zero' is a fitting score for today's date numerology as well.
Zero = 26+5+18+15 = 64 (New York Giants = 64)
8/20/2016 = 8+20+20+16 = 64 (New York Giants) (Zero)
One other observation, the only other team that scored 18-points today, were the Minnesota Vikings, the team also being examined for a Super Bowl birth in the NFC along with the Giants.
The Vikings also defeated the Seahawks, who they purposefully shanked the field goal against in the playoffs last year.
And here are other scores from this weekend:
This game was played on a date with '42' numerology, between two teams with '24/33' names.
8/18/16 = 8+18+16 = 42 (Reflection of 24)
Bengals = 2+5+5+7+1+3+1 = 24/33
Lions = 3+9+6+5+1 = 24/33
The final preseason games of this same week would occur on a date with '44' numerology.
8/20/16 = 8+20+16 = 44 (30+14 = 44)
Notice the sum of this game, 45 points.
New England = 45
New England lost to the Bears at the end of '85-'86 season, 46-10
Chicago = 46 (Revolutionary 4-6 defense)
Chicago is birth place of NFL (National Football League = 85)
The Chicago Bears have not been in a Super Bowl since 1985-1986
Seventeen = 37
PA? 16+1 = 17 (Mason = 17) (God = 17)
Browns = 37
Browns go down 24-13, a sum of 37. | 703 | 2,161 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-22 | latest | en | 0.939106 |
https://www.geeksforgeeks.org/minimum-increments-required-to-make-the-sum-of-all-adjacent-matrix-elements-even/?ref=rp | 1,638,018,621,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358180.42/warc/CC-MAIN-20211127103444-20211127133444-00063.warc.gz | 853,321,369 | 27,619 | # Minimum increments required to make the sum of all adjacent matrix elements even
• Last Updated : 26 Apr, 2021
Given a matrix mat[][] of dimensions N × M, the task is to minimize the number of increments of matrix elements required to make the sum of adjacent matrix elements even.
Note: For any matrix element mat[i][j], consider mat[i – 1][j], mat[i+1][j], mat[i][j – 1] and mat[i][j + 1] as its adjacent elements.
Examples:
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.
In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.
Input: mat[][] = {{1, 5, 6}, {4, 7, 8}, {2, 2, 3}}
Output: 4
Explanation:
Increase cell mat[0][0] by 1, mat[1][1] by 1, mat[0][1] by 1 and mat[2][2] by 1.
Therefore, total number of increments required is 4. Therefore, modified matrix is {{2, 6, 6}, {4, 8, 8}, {2, 2, 4}} having sum of all adjacent elements even.
Input: mat[][] = {{1, 5, 5}, {5, 5, 5}, {5, 1, 1}}
Output: 0
Approach: The idea to solve the given problem is based on the fact that the sum of two elements is even only if both the numbers are even or odd. Therefore, for the matrix to have the sum of adjacent elements even, all matrix elements should have the same parity, i.e. either all odd or all even.
Therefore, the minimum number of increments required is the minimum of the count of even and odd elements in the given matrix.
Below is the implementation of the above approach:
## C++
`// C++ program for the above approach` `#include ``using` `namespace` `std;` `const` `int` `MAX = 500;` `// Function to find the minimum number``// of increments required to make sum of``// al adjacent matrix elements even``int` `minOperations(``int` `mat[][MAX], ``int` `N)``{`` ``// Stores the count of odd elements`` ``int` `oddCount = 0;` ` ``// Stores the count of even elements`` ``int` `evenCount = 0;` ` ``// Iterate the matrix`` ``for` `(``int` `i = 0; i < N; i++) {`` ``for` `(``int` `j = 0; j < N; j++) {` ` ``// If element is odd`` ``if` `(mat[i][j] & 1) {` ` ``// Increment odd count`` ``oddCount++;`` ``}` ` ``// Otherwise`` ``else` `{` ` ``// Increment even count`` ``evenCount++;`` ``}`` ``}`` ``}` ` ``// Print the minimum of both counts`` ``cout << min(oddCount, evenCount);``}` `// Driver Code``int` `main()``{`` ``int` `mat[][MAX]`` ``= { { 1, 5, 6 }, { 4, 7, 8 }, { 2, 2, 3 } };`` ``int` `N = ``sizeof``(mat) / ``sizeof``(mat[0]);` ` ``minOperations(mat, N);` ` ``return` `0;``}`
## Java
`// Java program for the above approach``class` `GFG``{` `static` `int` `MAX = ``500``;` `// Function to find the minimum number``// of increments required to make sum of``// al adjacent matrix elements even``static` `void` `minOperations(``int` `mat[][], ``int` `N)``{`` ` ` ``// Stores the count of odd elements`` ``int` `oddCount = ``0``;` ` ``// Stores the count of even elements`` ``int` `evenCount = ``0``;` ` ``// Iterate the matrix`` ``for` `(``int` `i = ``0``; i < N; i++)`` ``{`` ``for` `(``int` `j = ``0``; j < N; j++)`` ``{` ` ``// If element is odd`` ``if` `(mat[i][j] % ``2``== ``1``) {` ` ``// Increment odd count`` ``oddCount++;`` ``}` ` ``// Otherwise`` ``else` `{` ` ``// Increment even count`` ``evenCount++;`` ``}`` ``}`` ``}` ` ``// Print the minimum of both counts`` ``System.out.print(Math.min(oddCount, evenCount));``}` `// Driver Code``public` `static` `void` `main(String[] args)``{`` ``int` `mat[][]`` ``= { { ``1``, ``5``, ``6` `}, { ``4``, ``7``, ``8` `}, { ``2``, ``2``, ``3` `} };`` ``int` `N = mat.length;`` ``minOperations(mat, N);``}``}` `// This code is contributed by 29AjayKumar`
## Python3
`# Python 3 program for the above approach``MAX` `=` `500` `# Function to find the minimum number``# of increments required to make sum of``# al adjacent matrix elements even``def` `minOperations(mat, N):`` ` ` ``# Stores the count of odd elements`` ``oddCount ``=` `0` ` ``# Stores the count of even elements`` ``evenCount ``=` `0` ` ``# Iterate the matrix`` ``for` `i ``in` `range``(N):`` ``for` `j ``in` `range``(N):`` ` ` ``# If element is odd`` ``if` `(mat[i][j] & ``1``):`` ` ` ``# Increment odd count`` ``oddCount ``+``=` `1` ` ``# Otherwise`` ``else``:`` ` ` ``# Increment even count`` ``evenCount ``+``=` `1` ` ``# Print the minimum of both counts`` ``print``(``min``(oddCount, evenCount))` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:`` ``mat ``=` `[[``1``, ``5``, ``6``], [``4``, ``7``, ``8``], [``2``, ``2``, ``3``]]`` ``N ``=` `len``(mat)` ` ``minOperations(mat, N)`` ` ` ``# This code is contributed by SURENDRA_GANGWAR.`
## C#
`// C# program for the above approach``using` `System;``class` `GFG``{`` ``const` `int` `MAX = 500;` ` ``// Function to find the minimum number`` ``// of increments required to make sum of`` ``// al adjacent matrix elements even`` ``static` `void` `minOperations(``int``[, ] mat, ``int` `N)`` ``{`` ``// Stores the count of odd elements`` ``int` `oddCount = 0;` ` ``// Stores the count of even elements`` ``int` `evenCount = 0;` ` ``// Iterate the matrix`` ``for` `(``int` `i = 0; i < N; i++) {`` ``for` `(``int` `j = 0; j < N; j++) {` ` ``// If element is odd`` ``if` `((mat[i, j] & 1) != 0) {` ` ``// Increment odd count`` ``oddCount++;`` ``}` ` ``// Otherwise`` ``else` `{` ` ``// Increment even count`` ``evenCount++;`` ``}`` ``}`` ``}` ` ``// Print the minimum of both counts`` ``Console.Write(Math.Min(oddCount, evenCount));`` ``}` ` ``// Driver Code`` ``public` `static` `void` `Main()`` ``{`` ``int``[, ] mat`` ``= { { 1, 5, 6 }, { 4, 7, 8 }, { 2, 2, 3 } };`` ``int` `N = mat.GetLength(0);`` ``minOperations(mat, N);`` ``}``}` `// This code is contributed by chitranayal.`
## Javascript
``
Output:
`4`
Time Complexity: O(N2)
Auxiliary Space: O(1)
My Personal Notes arrow_drop_up | 2,261 | 6,759 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2021-49 | longest | en | 0.827273 |
https://physics.stackexchange.com/questions/156533/why-arent-transformations-caused-by-measurements-unitary | 1,722,763,596,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640393185.10/warc/CC-MAIN-20240804071743-20240804101743-00307.warc.gz | 369,138,244 | 47,059 | # Why aren't transformations caused by measurements unitary?
It is said, that when measured, a quantum system undergoes "wave function collapse", which is a non-unitary transformation.
Why?
The wave function is
$\Psi = \alpha \left|0\right\rangle + \beta \left|1\right\rangle$
where
$\left|\alpha\right|^2 + \left|\beta\right|^2 = 1$
The probabilities sum after measurement is still 1, for example, if system collapsed to $\left|0\right\rangle$, then
$\left|1\right|^2 + \left|0\right|^2 = 1$
For example, if function was
$\Psi = \frac{1}{\sqrt{2}} \left|0\right\rangle + \frac{1}{\sqrt{2}} \left|1\right\rangle$
the transformation was
$\left[ \begin{array}{ c c } \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ 0 & 0 \end{array} \right]$
Isn't this transformation unitary?
• a unitary matrix has determinant $\pm1$ so that matrix can't be unitary. It is even degenerate Commented Jan 4, 2015 at 15:12
• But that part can easily be fixed by taking the unitary $\left(\begin{smallmatrix}1&1\\1&-1\end{smallmatrix}\right)/\sqrt{2}$. The point is that the unitary depends on the state $\Psi$. Commented Jan 4, 2015 at 15:21
• As I said in one of the comments, I expect you could manually compute a unitary transformation that gives you the right answer. But you would need to already know the right answer, so it's a bit useless... Commented Jan 4, 2015 at 15:35
• A physical note: unitary => invertable. Physically we know measurements are not always invertable.
– zzz
Commented Jan 4, 2015 at 16:36
• @Dims you will find more relevant/useful points on this topic here. Commented Jan 4, 2015 at 17:46
No.
As long as your state is $|\Psi \rangle = \alpha \left|0\right\rangle + \beta \left|1\right\rangle$, then as you said $\alpha$ and $\beta$ need to satisfy $\left|\alpha\right|^2 + \left|\beta\right|^2 = 1$, so say $\alpha = \beta = 1/\sqrt 2$.
If you perform a measurement and find that the system in the $\left|0\right\rangle$ state, then the new wavefunction will be $\Psi =\left|0\right\rangle$. You can write it as $\Psi = \alpha \left|0\right\rangle$ but because of normalisiation $|\alpha|^2$ needs to be 1, so $\alpha$ must be either 1 or a pure phase factor.
You had to change the normalisation by hand (changing $\alpha$ from $1/\sqrt 2$ to $1$). A unitary transformation on $|\Psi \rangle$ would affect only the kets and not the constants. The time evolution of any wavefunction is governed by the Schrodinger equation which, when solved, is effectively a unitary transformation -- unitary transformations leave the norm unchanged. The time evolution due to performing measurements, however, is something entirely different and does not follow the formalism of the Schrodinger equation.
A unitary transformation leaves the norm unchanged, since the norm of $U|\Psi \rangle$ is $\langle \Psi |U^{\dagger}U|\Psi \rangle = \langle \Psi | \Psi \rangle$ if $U$ is unitary. In your specific case this is true, but only because you had to manually change the normalisation of the post-measurement wavefunction. If QM included measurement, then there should be a deterministic way of computing how $\alpha$ or $\beta$ would change. But it doesn't, so you need to re-normalise it.
• See my edit please. So, am I correct thinking, that unitarity means that ANY given function will conserve length, while during collapse it is impposible to formulate one?
– Dims
Commented Jan 4, 2015 at 14:59
• With regards to your question in the comment: yes, measurement does not obey any unitary transformation. You have to put the measurement by hand, at least in the current framework of QM. I have to say I don't know whether it is impossible to formulate such a matrix, maybe you can formulate one that only works for a specific case, but that'd be a bit useless because, in order to construct it, you'd need to know what the final state is so you'd already know the answer. Commented Jan 4, 2015 at 15:05
• With regards to your edit: When you write out the matrix form of an operator you should always specify what vector representation you are using for your state vectors. I am assuming $|0\rangle = \left( \begin{array}{ c } 1 \\ 0 \end{array} \right)$ and $|1\rangle = \left( \begin{array}{ c } 0 \\ 1 \end{array} \right)$. Well this transformation is not unitary. Remember unitary means transpose and complex conjugate. The transpose of your matrix is not equal to the original one. Commented Jan 4, 2015 at 15:08
• @Dims : which length? Unitarity of a transformation on a wave-function preserves probabilities. The probability of obtaining the value $+\hbar/2$ from an electron polarized with spin up in the direction $z$, but whose spin projection is measured on the axis $x$, remains $1/2$, whatever unitary transformation you do that doesn't change the spin-up polarization of the electron. Commented Jan 4, 2015 at 15:12
• Unitary transformations preserve the norm, i.e. the "length" of a vector in a Hilbert space $\langle \Psi | \Psi \rangle$. The norm can be interpreted as a probability if normalised to 1, in which case there would be a physical reason as to why it needs to be constant through any realistic time evolution. Commented Jan 4, 2015 at 15:18
Answering the question in the title: a measurement process is intrinsically non-unitary. One way to see this is to realise that the unitarity of a process is equivalent to its being reversible.
A measurement process is intrinsically non-reversible, as some information gets lost. For example, measuring $$(|0\rangle+|1\rangle)/\sqrt2$$ in the computational basis, you can get either $$|0\rangle$$ or $$|1\rangle$$. The same outputs can be obtained measuring a different state, e.g. $$(|0\rangle-|1\rangle)/\sqrt2$$. This means that, given a measurement result (say $$|0\rangle$$), there is no way to know from what state it came from. Some information is lost in the process. It follows that the process cannot be described by a unitary matrix.
I assume that you are referring to the outcome of any observable $O$ acting on a state $\psi$, since the act of measuring something is interpreted as "averaging" such operators on some state. General observables are self-adjoint operators which need not be unitary. Perhaps the simplest example of an observable is a projection, i.e. an operator $P$ with the property that $P^*P = P$ (idempotent and self-adjoint). Suppose that, in your case, $P = |0\rangle\langle0|$. The outcome of a measurement of $P$ on your state $\Psi$, when repeated $N$ times, is $|\alpha^2|N$ times YES (and hence $(1-|\alpha|^2)N$ times no. Moreover the result of $P\Psi$ is $\alpha|0\rangle$, which isn't a normalised vector, simply because $P$ is not a unitary.
• You can't say "an observable $O$ acting on a state $\psi$". An observable is a physical quantity, that you measure. But you can use the terminology "operator $\hat O$. Now, a measurement is not averaging. In each single measurement of the operator $\hat O$ you obtain one of its eigenvalues. Measuring $\hat O$ on many particles identically prepared, you obtain statistics, and from it you can calculate different things, one of them being the average value. Another thing: from operators describing observables we don't expect unitary, but self-adjointness. Unitarity we expect from transformations. Commented Jan 4, 2015 at 15:05
• I don't distinguish between observables and self-adjoint operators. For me they are the very same thing. Also a measure is usually performed on a statistical ensemble, and this corresponds to the action of a state on the observable, i.e. $\omega(O)$. I don't see how the single outcome of an experiment can tell anything about the state of a system, unless you know a priori that you are dealing with a pure state. Finally I don't quite get your last remark. Commented Jan 4, 2015 at 15:10
• Idle question while reading this: Since self-adjoint and unitary operators are in one-to-one correspondence, has the exponential of a self-adjoint projection any physical relevant meaning? Commented Jan 4, 2015 at 15:22
• @ACuriousMind That correspondence holds between self-adjoint operators and one-parameter groups of unitaries. For more general unitaries you need more self-adjoint operators and you only hit the connected component of the identity: for every unitary $u$ homotopic to 1 there are self-adjoint operators $h_1,\ldots,h_n$ such that $u = e^{ih_1}\cdots e^{ih_n}$ Commented Jan 4, 2015 at 15:34
• @ACuriousMind Don't know why but I can't edit comments right now. As for the exponential of a projection you simply get $e^{iPt} = 1 + (e^{it}-1)P$. I think this can be interpreted physically if $P$ is some generator of some transformation (e.g. a continuous symmetry). Then its exponential is the generated flow. Commented Jan 4, 2015 at 15:46
Measurement is not unitary, because it does not implement a unitary. It selects a unitary.
We can certainly say that a preparation $$|\psi_i\rangle$$ and (renormalized) measurement outcome $$|\psi_f\rangle$$ are related by a unitary transformation $$|\psi_f\rangle = U_{fi} |\psi_i\rangle,$$ where $$U_{fi} \in U(2)$$. However, we can only say this once a measurement has occurred.
Before measurement, the possible outcomes are $$|{+}\psi_f\rangle$$ and $$|{-}\psi_f\rangle$$, so all we can say is that $$|\psi_i\rangle$$ and the not-yet-measured outcome $$|\psi_f\rangle$$ are related by either $$U_{fi}^+$$ or $$U_{fi}^-$$, where $$|{+}\psi_f\rangle = U_{fi}^+ |\psi_i\rangle \hspace{1em} \text{and} \hspace{1em} |{-}\psi_f\rangle = U_{fi}^- |\psi_i \rangle.$$
Formally, we can construct an equivalence class of unitaries $$U_{fi}^+ \sim U_{fi}^-$$ over the possible outcomes of the measurement process, so that $$[U_{fi}^+] = [U_{fi}^-] \in U(2)/\mathbb Z_2$$ represents the part of the relation between the initial and final state that is fixed by our experimental setup.
The point is that the measurement process implements a transition $$U(2)/\mathbb Z_2 \to U(2) \\ [U_{fi}^\pm] \mapsto U_{fi}^\pm$$ that breaks the $$\mathbb Z_2$$ symmetry and selects a particular representative of $$[U_{fi}^\pm]$$. In other words, the unitary relation $$U_{fi}^+$$ or $$U_{fi}^-$$ between an initial state and measurement outcome is not a description of the measurement process. It is the result of the measurement process. | 2,765 | 10,282 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 23, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2024-33 | latest | en | 0.85832 |
https://www.gogeometry.com/center/incenter_incircle_theorems_problems_index-2.html | 1,713,010,206,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816734.69/warc/CC-MAIN-20240413114018-20240413144018-00253.warc.gz | 735,134,462 | 8,322 | # Geometry: Incenter, Incircle, Inradius of a triangle, Theorems and Problems.
Problem 1341. Isosceles Triangle, 80-20-80 Degrees, Circumcenter, Angle Bisector. Geometry Problem 1374.Isosceles Triangle, Exterior Cevian, Incircle, Excircle, Tangency Points, Parallel Lines. Geometry Problem 1373.Isosceles Triangle, Exterior Cevian, Inradius, Exradius, Altitude to the Base. Geometry Problem 1372.Equilateral Triangle, Exterior Cevian, Inradius, Exradius, Altitude, Sketch, iPad Apps. Geometry Problem 1351.Two Tangential or Circumscribed Quadrilaterals, Midpoints, Perpendiculars. Geometry Problem 1350.Triangle with three Intersecting Circles, Incircle, Collinear Points. Geometry Problem 1346.Equilateral Triangle, Point on the Incircle, Altitude, Perpendicular, Sum of Squares, Distance. Problem 1340. Triangle, Incenter, Concentric Circles, Isosceles Triangles, Congruence. Geometry Problem 1326. Triangle, Cevian, Incenters, Sum of Angles, 270 Degrees. Geometry Problem 1320. Triangle, Incircle, Tangent, Chord, Circle, Parallel, Perpendicular, Collinearity. Geometry Problem 1316. Triangle, Incircle, Tangent, Chord, Midpoint, Sum of two Segments, Congruence. Geometry Problem 1315. Triangle, Incircle, Tangent, Chord, Circle, Congruence. Geometry Problem 1314. Equilateral Triangle, Incircle, Inscribed Circle, Tangent, Congruence. Geometry Problem 1309. Triangle, Circle, Inradius, Excircle, Tangent, Exradius, Measurement. Geometry Problem 1308. Quadrilateral, Diagonal, Triangle, Incircle, Tangent Line, Sides, Measurement. Geometry Problem 1307. Triangle, Incenter, Parallel line, Sides, Measurement. Geometry Problem 1304. Triangle, 120 Degrees, Angle Bisector, Measurement. Geometry Problem 1301. Arbelos, Semicircles, Diameters, Circle, Incircle, Incenter, Square, Midpoint, Concurrency. Geometry Problem 1300. Arbelos, Semicircles, Diameters, Circle, Incircle, Tangent, Perpendicular, Concurrent Lines. Geometry Problem 1298. Arbelos, Semicircles, Diameters, Circle, Incircle, Tangent, Angle Bisector, Perpendicular, Midpoint. Geometry Problem 1297. Arbelos, Semicircles, Diameters, Tangent, Circle, Incircle, Angle Bisector. Geometry Problem 1295. Right Triangle, Incenter, Incircle, Excenter, Excircle, Congruence, Angle. Geometry Problem 1292. Square, Circle, Right Triangle, Arc, Inradius, Radius, Incircle, Measurement. Geometry Problem 1288. Triangle, 30-50-100 Degrees, Area, Inradius, Metric Relations, Measurement. Geometry Problem 1274. Isosceles Triangle, 80-20-80 Degrees, Area, Inradius, Circumradius, Angle Bisector, Metric Relations, Measurement. Geometry Problem 1271. Triangle, Incircle, Incenter, Excircle, Excenter, Escribed Circle, Tangency Points, Six Concyclic Points. Geometry Problem 1269. Triangle, Incircle, Incenter, Inscribed Circle, Tangency Points, Perpendicular, 90 Degrees, Angle Bisector. Geometry Problem 1267. Triangle, Incircle, Excircle, Circle, Tangency Points, Perpendicular, 90 Degrees, Parallelogram. Geometry Problem 1265. Triangle, Incircle, Circle, Tangency Points, Perpendicular, 90 Degrees, Angle Bisector. Geometry Problem 1231 Triangle, Orthocenter, Incenter, Circumcenter, Angle Bisector, Center, Circle. Geometry Problem 1217 Triangle, Circle, Excenter, Incenter, Angle Bisector, Cyclic Quadrilateral, Circumcircle, Tangent Line. Geometry Problem 1207 Triangle, Circle, Incenter, Circumcenter, Excenter, Circumradius, Perpendicular, 90 Degrees. Geometry Problem 1206 Circle, Angle Bisector, Secant, Triangle, Circumcircle, Congruence, Area. Geometry Problem 1203 Right Triangle, Square, Angle Bisector, Three Congruent Quadrilaterals. Geometry Problem 1191 Circle, Chords, Diameter, Congruence, Midpoint, Collinearity, Bisector. Geometry Problem 1180 Quadrilateral, 120 Degrees, Diagonals, Perpendicular. Geometry Problem 1178 Triangle, Angle Bisector, Metric Relations. Geometry Problem 1177 Triangle, Two Angle Bisectors, Metric Relations. Geometry Problem 1164. Triangle, Angle Bisector, Cevian. Geometry Problem 1174 Triangle, Quadrilateral, Double, Triple, Angle, Congruence, Excenter, Angle Bisector. Geometry Problem 1163. Isosceles Triangle, Congruence, Double Angle, Circle, Concyclic Points, Circumcenter, Perpendicular, Angle Bisector. Geometry Problem 1153. Triangle, Angle Bisectors, Perpendicular, Congruence. Geometry Problem 1144. Three Equal Squares, Diagonals, Perpendicular, 90 Degrees, Angle Bisector. Geometry Problem 1139. Triangle, Circumcircle, Angle Bisector, Parallel Lines, 90 Degrees, Concurrent Lines. Geometry Problem 1199 Equilateral Triangle, Square, Altitude, Circle, Incircle, Inradius, Metric Relations. Geometry Problem 1181 Cyclic Quadrilateral and Tangential Quadrilateral, Diameter as a Diagonal, Incenter, Circumcenter. Geometry Problem 1155. Right Triangle, Altitude, Incenter, Incircle, Area. Geometry Problem 1149. Isosceles Triangle, Altitude, Cevian, Incircles, Tangent, Congruence. Dynamic Geometry Problem 1148 Static Geometry Problem 1148. Right Triangle, Incircle, Circumcircle, Tangent, Radius. GeoGebra, HTML5 Animation for Tablets (iPad, Nexus). Geometry Problem 1129. Isosceles Triangle, Circumcenter, Incenter, Parallel Lines, Perpendicular Lines. Geometry Problem 1105. Triangle, Incircle, Excircle, Cevian, Tangent, Congruence, Geometric Mean. Geometry Problem 1104. Right Triangle, Incircle, Circumcircle, Inscribed Circle, Radius, Tangent. Geometry Problem 1103. Right Triangle, Incircle, Inscribed Circle, Radius, Geometric Mean, Sangaku, Japanese, Metric Relations. Geometry Problem 1102. Right Triangle, Incircle, Tangency Point, 90 Degree, Cathetus, Metric Relations. Geometry Problem 1101. Triangle, Parallel Lines, Circle, Incircle, Inradius, Radius, Angle. Geometry Problem 1081. Equilateral Triangle, Inscribed Circle, Inradius, Tangent Circles, Radius, Tangent Line, Sangaku Japanese Problem. Geometry Problem 1068. Obtuse Triangle, Orthocenter, Circumradius, Inradius, Exradii, Distance, Diameter Geometry Problem 1067. Acute Triangle, Orthocenter, Circumradius, Inradius, Exradii, Distance, Diameter. Geometry Problem 1066. Triangle, Obtuse Angle, Orthocenter, Circumradius R, Inradius r, Exradius ra, Distance, Diameter. Geometry Problem 1065. Triangle, Acute Angle, Orthocenter, Circumradius R, Inradius r, Exradius ra, Distance, Diameter. Geometry Problem 1061. Triangle, Inradius (r), Circumradius (R), Circumcircle, Angle Bisector, Distance from a point to a Line, Perpendicular. Geometry Problem 1060. Triangle, Incircle, Incenter, Inscribed circle, Tangent, Angle. Geometry Problem 1043. Right Triangle, Incenter, Excenter, Congruence, Metric Relations. Geometry Problem 1037. Triangle, Three equal Incircles, Tangent lines, Inradius, Length. Geometry Problem 1036. Triangle, Three equal Incircles, Tangent lines, Inradius, Equal Inradii. Geometry Problem 1035. Triangle, Three equal Incircles, Tangent lines, Equivalent triangles, Equal area. Geometry Problem 1034. Triangle, Three equal Incircles, Tangent lines, Isoperimetric triangles, Equal perimeter. Geometry Problem 1024. Contact, Gergonne Triangle, Point on an arc of Incircle, Perpendicular, Distances. Geometry Problem 1023. Triangle, Contact, Gergonne, Product of three distances. Geometry Problem 1012. Equilateral Triangle, Incenters, Inscribed Circles, Perpendicular, Concurrent Lines. Geometry Problem 991. Triangle, Incircle, Tangency Points, Isosceles, Midpoint, Collinearity, Congruence, Circle. Geometry Problem 990. Triangle, Incircle, Median, Cevian, Central Angle, Congruence, Circle, Secant, Midpoint, Parallel. Geometry Problem 987. Triangle, Circumcircle, Incenter, Chord, Parallel, Circle, Tangent. GeoGebra, HTML5 Animation for Tablets (iPad, Nexus). Geometry Problem 976. Isosceles Right Triangle, 45 Degrees, Incenter, Angle Bisector, Hypotenuse. Geometry Problem 959. Triangle, Sides Ratio 4:1, Inradius, Exradius, Cevian, Mean Proportional, Geometric Mean, Metric Relations. Geometry Problem 957. Equilateral Triangle, Inscribed Circle, Incircle, Circumscribed Circle, Circumcircle, Area, Circular Segment. Geometry Problem 952. Triangle, Incenter, Circumcenter, Three Equal Circles, Tangent Line, Tangent Circles, Collinear Points. Geometry Problem 943. Triangle, Incircle, Inscribed Circle, Tangency Points, Parallel, Metric Relations. Geometry Problem 942. Triangle, Circles, Circumcircle, Sagitta, Incircle, Excircle, Inradius, Exradius, Metric Relations. Geometry Problem 935. Square, Circle, Center, Perpendicular, Metric Relations. Geometry Problem 921. Right Triangle, Circumcircle, Arc, Midpoint, Tangency Point, Incircle. Geometry Problem 918. Triangle, Incenter, Incircle, Angle Bisector, Perpendicular, Metric Relations. Geometry Problem 895 Triangle, Incircle, Tangency Points, Parallel line, Collinear Points. GeoGebra, HTML5 Animation for Tablets Geometry Problem 894 Triangle, Angle, 60 degrees, Incenter, Midpoint, Parallel Lines. GeoGebra, HTML5 Animation for Tablets Geometry Problem 893 Triangle, Circumcircle, Incenter, Circle, Tangent, Collinear Points. GeoGebra, HTML5 Animation for Tablets Geometry Problem 889 Carnot's Theorem in an acute triangle, Circumcenter, Circumradius, Inradius. GeoGebra, HTML5 Animation for Tablets. Geometry Problem 886 Right Triangle, Incenter, Angle Bisector, Perpendicular, 45 Degrees, Concyclic Points, Isosceles Right Triangle. GeoGebra, HTML5 Animation for Tablets (iPad, Nexus). Previous | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | Next
Home | Search | Geometry | Triangle Centers | Post a comment | Email | by Antonio GutierrezLast updated: Mar 9, 2022 | 2,471 | 9,529 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-18 | latest | en | 0.644929 |
https://www.jobilize.com/course/section/some-useful-integrals-appendix-b-to-applied-probability-by-openstax?qcr=www.quizover.com | 1,550,561,436,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247489425.55/warc/CC-MAIN-20190219061432-20190219083432-00483.warc.gz | 869,407,780 | 24,895 | 17.1 Appendix b to applied probability: some mathematical aids
Page 1 / 1
A variety of mathematical aids to probability analysis and calculations.
Series
• Geometric series From the expression $\left(1-r\right)\left(1+r+{r}^{2}+...+{r}^{n}\right)=1-{r}^{n+1}$ , we obtain
$\sum _{k=0}^{n}{r}^{k}=\frac{1-{r}^{n+1}}{1-r}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{for}\phantom{\rule{0.277778em}{0ex}}r\ne 1$
For $|r|<1$ , these sums converge to the geometric series $\sum _{k=0}^{\infty }{r}^{k}=\frac{1}{1-r}$
Differentiation yields the following two useful series:
$\sum _{k=1}^{\infty }k{r}^{k-1}=\frac{1}{{\left(1-r\right)}^{2}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{for}\phantom{\rule{0.277778em}{0ex}}|r|<1\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{and}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\sum _{k=2}^{\infty }k\left(k-1\right){r}^{k-2}=\frac{2}{{\left(1-r\right)}^{3}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{for}\phantom{\rule{0.277778em}{0ex}}|r|<1$
For the finite sum, differentiation and algebraic manipulation yields
$\sum _{k=0}^{n}k{r}^{k-1}=\frac{1-{r}^{n}\left[1+n\left(1-r\right)\right]}{{\left(1-r\right)}^{2}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{which}\phantom{\rule{4.pt}{0ex}}\text{converges}\phantom{\rule{4.pt}{0ex}}\text{to}\phantom{\rule{0.277778em}{0ex}}\frac{1}{{\left(1-r\right)}^{2}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{for}\phantom{\rule{0.277778em}{0ex}}|r|<1$
• Exponential series . ${e}^{x}=\sum _{k=0}^{\infty }\frac{{x}^{k}}{k!}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{and}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}{e}^{-x}=\sum _{k=0}^{\infty }{\left(-1\right)}^{k}\frac{{x}^{k}}{k!}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{for}\phantom{\rule{4.pt}{0ex}}\text{any}\phantom{\rule{0.277778em}{0ex}}x$
Simple algebraic manipulation yields the following equalities usefulfor the Poisson distribution:
$\sum _{k=n}^{\infty }k\frac{{x}^{k}}{k!}=x\sum _{k=n-1}^{\infty }\frac{{x}^{k}}{k!}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{and}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\sum _{k=n}^{\infty }k\left(k-1\right)\frac{{x}^{k}}{k!}={x}^{2}\sum _{k=n-2}^{\infty }\frac{{x}^{k}}{k!}$
• Sums of powers of integers $\sum _{i=1}^{n}i=\frac{n\left(n+1\right)}{2}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\sum _{i=1}^{n}{i}^{2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$
Some useful integrals
• The gamma function $\Gamma \left(r\right)={\int }_{0}^{\infty }{t}^{r-1}{e}^{-t}\phantom{\rule{0.166667em}{0ex}}dt\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{for}\phantom{\rule{0.277778em}{0ex}}r>0$
Integration by parts shows $\Gamma \left(r\right)=\left(r-1\right)\Gamma \left(r-1\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{for}\phantom{\rule{0.277778em}{0ex}}r>1$
By induction $\Gamma \left(r\right)=\left(r-1\right)\left(r-2\right)\cdots \left(r-k\right)\Gamma \left(r-k\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{for}\phantom{\rule{0.277778em}{0ex}}r>k$
For a positive integer $n,\phantom{\rule{0.277778em}{0ex}}\Gamma \left(n\right)=\left(n-1\right)!\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{with}\phantom{\rule{0.277778em}{0ex}}\Gamma \left(1\right)=0!=1$
• By a change of variable in the gamma integral, we obtain
${\int }_{0}^{\infty }{t}^{r}{e}^{-\lambda t}\phantom{\rule{0.166667em}{0ex}}dt=\frac{\Gamma \left(r+1\right)}{{\lambda }^{r+1}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}r>-1,\phantom{\rule{0.277778em}{0ex}}\lambda >0$
• A well known indefinite integral gives
${\int }_{a}^{\infty }t{e}^{-\lambda t}\phantom{\rule{0.166667em}{0ex}}dt=\frac{1}{{\lambda }^{2}}\phantom{\rule{0.166667em}{0ex}}{e}^{-\lambda a}\phantom{\rule{0.166667em}{0ex}}\left(1+\lambda a\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{and}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}{\int }_{a}^{\infty }{t}^{2}{e}^{-\lambda at}\phantom{\rule{0.166667em}{0ex}}dt=\frac{1}{{\lambda }^{3}}\phantom{\rule{0.166667em}{0ex}}{e}^{-\lambda a}\phantom{\rule{0.166667em}{0ex}}\left[1+\lambda a+{\left(\lambda a\right)}^{2}/2\right]$
For any positive integer m ,
${\int }_{a}^{\infty }{t}^{m}{e}^{-\lambda t}\phantom{\rule{0.166667em}{0ex}}dt=\frac{m!}{{\lambda }^{m+1}}\phantom{\rule{0.166667em}{0ex}}{e}^{-\lambda a}\left[1,+,\lambda ,a,+,\frac{{\left(\lambda a\right)}^{2}}{2!},+,\cdots ,+,\frac{{\left(\lambda a\right)}^{m}}{m!}\right]$
• The following integrals are important for the Beta distribution.
${\int }_{0}^{1}{u}^{r}{\left(1-u\right)}^{s}\phantom{\rule{0.166667em}{0ex}}du=\frac{\Gamma \left(r+1\right)\Gamma \left(s+1\right)}{\Gamma \left(r+s+2\right)}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}r>-1,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}s>-1$
For nonnegative integers $m,n\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}{\int }_{0}^{1}{u}^{m}{\left(1-u\right)}^{n}\phantom{\rule{0.166667em}{0ex}}du=\frac{m!n!}{\left(m+n+1\right)!}$
Some basic counting problems
We consider three basic counting problems, which are used repeatedly as components of more complex problems. The first two, arrangements and occupancy are equivalent. The third is a basic matching problem.
1. Arrangements of r objects selected from among n distinguishable objects.
1. The order is significant.
2. The order is irrelevant.
For each of these, we consider two additional alternative conditions.
1. No element may be selected more than once.
2. Repitition is allowed.
2. Occupancy of n distinct cells by r objects. These objects are
1. Distinguishable.
2. Indistinguishable.
The occupancy may be
1. Exclusive.
2. Nonexclusive (i.e., more than one object per cell)
The results in the four cases may be summarized as follows:
1. Ordered arrangements, without repetition ( permutations ). Distinguishable objects, exclusive occupancy.
$P\left(n,r\right)=\frac{n!}{\left(n-r\right)!}$
2. Ordered arrangements, with repitition allowed. Distinguishable objects, nonexclusive occupancy.
$U\left(n,r\right)={n}^{r}$
1. Arrangements without repetition, order irrelevant ( combinations ). Indistinguishable objects, exclusive occupancy.
$C\left(n,r\right)=\frac{n!}{r!\left(n-r\right)!}=\frac{P\left(n,r\right)}{r!}$
2. Unordered arrangements, with repetition. Indistinguishable objects, nonexclusive occupancy.
$S\left(n,r\right)=C\left(n+r-1,r\right)$
3. Matching n distinguishable elements to a fixed order. Let $M\left(n,k\right)$ be the number of permutations which give k matches.
$n=5$
Natural order 1 2 3 4 5
Permutation 3 2 5 4 1 (Two matches– positions 2, 4)
We reduce the problem to determining $m\left(n,0\right)$ , as follows:
1. Select k places for matches in $C\left(n,k\right)$ ways.
2. Order the $n-k$ remaining elements so that no matches in the other $n-k$ places.
$M\left(n,k\right)=C\left(n,k\right)M\left(n-k,0\right)$
Some algebraic trickery shows that $M\left(n,0\right)$ is the integer nearest $n!/e$ . These are easily calculated by the MATLAB command M = round(gamma(n+1)/exp(1)) For example >>M = round(gamma([3:10]+1)/exp(1));>>disp([3:6;M(1:4);7:10;M(5:8)]')3 2 7 1854 4 9 8 148335 44 9 133496 6 265 10 1334961
Extended binomial coefficients and the binomial series
• The ordinary binomial coefficient is $C\left(n,k\right)=\frac{n!}{k!\left(n-k\right)!}$ for integers $n>0,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}0\le k\le n$
For any real x , any integer k , we extend the definition by
$C\left(x,0\right)=1,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}C\left(x,k\right)=0\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{for}\phantom{\rule{0.277778em}{0ex}}k<0,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{and}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}C\left(n,k\right)=0\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{for}\phantom{\rule{4.pt}{0ex}}\text{a}\phantom{\rule{4.pt}{0ex}}\text{positive}\phantom{\rule{4.pt}{0ex}}\text{integer}\phantom{\rule{0.277778em}{0ex}}k>n$
and
$C\left(x,k\right)=\frac{x\left(x-1\right)\left(x-2\right)\cdots \left(x-k+1\right)}{k!}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{otherwise}$
Then Pascal's relation holds: $C\left(x,k\right)=C\left(x-1,k-1\right)+C\left(x-1,k\right)$
The power series expansion about $t=0$ shows
${\left(1+t\right)}^{x}=1+C\left(x,1\right)t+C\left(x,2\right){t}^{2}+\cdots \phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\forall \phantom{\rule{0.277778em}{0ex}}x,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}-1
For $x=n$ , a positive integer, the series becomes a polynomial of degree n .
Cauchy's equation
1. Let f be a real-valued function defined on $\left(0,\infty \right)$ , such that
1. $f\left(t+u\right)=f\left(t\right)+f\left(u\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{for}\phantom{\rule{0.277778em}{0ex}}t,\phantom{\rule{0.277778em}{0ex}}u>0$ , and
2. There is an open interval I on which f is bounded above (or is bounded below).
Then $f\left(t\right)=f\left(1\right)t\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\forall \phantom{\rule{0.277778em}{0ex}}t>0$
2. Let f be a real-valued function defined on $\left(0,\infty \right)$ such that
1. $f\left(t+u\right)=f\left(t\right)f\left(u\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\forall \phantom{\rule{0.277778em}{0ex}}t,\phantom{\rule{0.277778em}{0ex}}u>0$ , and
2. There is an interval on which f is bounded above.
Then, either $f\left(t\right)=0\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{for}\phantom{\rule{0.277778em}{0ex}}t>0$ , or there is a constant a such that $f\left(t\right)={e}^{at}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{for}\phantom{\rule{0.277778em}{0ex}}t>0$
[For a proof, see Billingsley, Probability and Measure , second edition, appendix A20]
Countable and uncountable sets
A set (or class) is countable iff either it is finite or its members can be put into a one-to-one correspondence with the natural numbers.
Examples
• The set of odd integers is countable.
• The finite set $\left\{n:1\le n\le 1000\right\}$ is countable.
• The set of all rational numbers is countable. (This is established by an argument known as diagonalization).
• The set of pairs of elements from two countable sets is countable.
• The union of a countable class of countable sets is countable.
A set is uncountable iff it is neither finite nor can be put into a one-to-one correspondence with the natural numbers.
Examples
• The class of positive real numbers is uncountable. A well known operation shows that the assumption of countability leads to a contradiction.
• The set of real numbers in any finite interval is uncountable, since these can be put into a one-to-one correspondence of the class of all positive reals.
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
A fair die is tossed 180 times. Find the probability P that the face 6 will appear between 29 and 32 times inclusive | 5,498 | 15,173 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 47, "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} | 4.09375 | 4 | CC-MAIN-2019-09 | latest | en | 0.412402 |
https://www.thestudentroom.co.uk/showthread.php?t=214411 | 1,524,479,386,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945940.14/warc/CC-MAIN-20180423085920-20180423105920-00477.warc.gz | 883,672,360 | 37,770 | x Turn on thread page Beta
You are Here: Home >< Maths
# integration watch
1. How can you integrate the following:
∫e^(2x) sinx dx ?
I can't seem to do it by parts, using u=e^(2x).
Cheers
2. You have to do it by parts twice, so on the RHS you get back to something of the form:
{something something} + A[∫e^(2x) sinx dx], where A is a constant
Then you take the A[∫e^(2x) sinx dx] over to the LHS and subtract it from the ∫e^(2x) sinx dx that's already there. It's better if you try it rather than me explaining it.
Just remember to keep the same term the differentiable one (e^2x in this case) on the second time of doing parts.
3. Thanks a lot mate.
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
This forum is supported by:
Updated: April 13, 2006
Today on TSR
### Unconditional offer...
But going to get a U!
Poll
Useful resources
### Maths Forum posting guidelines
Not sure where to post? Read the updated guidelines here
### How to use LaTex
Writing equations the easy way
### Study habits of A* students
Top tips from students who have already aced their exams | 326 | 1,228 | {"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-2018-17 | latest | en | 0.908311 |
https://ccm.net/forum/affich-676073-find-the-biggest-and-smalles-of-a-b-c-by-algorithm | 1,723,550,906,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641076695.81/warc/CC-MAIN-20240813110333-20240813140333-00759.warc.gz | 127,308,058 | 26,532 | # Find the biggest and smalles of A,B,C by algorithm
Solved/Closed
YemenSEF Posts 2 Registration date Monday November 26, 2012 Status Member Last seen August 19, 2013 - Nov 26, 2012 at 06:23 AM
Zohaib R Posts 2368 Registration date Sunday September 23, 2012 Status Member Last seen December 13, 2018 - Nov 29, 2012 at 01:19 PM
Hello,
This is my first question ^^ , I just joined the collage but i have a question related to algorithm
its how to write whats the biggest and smallest value in 3 vars A,B,C
did a flowchart for it but cant change it to algoritm to short form :(
for 1 answer its easy
-start
IF A>B & A>C
then A is the biggest
Else IF B>C
then B is the biggest
Else then C is the biggest
Print the biggest
-end
Regards
???? ???? , ?? ?????? ?????
Related:
## 3 responses
Zohaib R Posts 2368 Registration date Sunday September 23, 2012 Status Member Last seen December 13, 2018 69
Nov 26, 2012 at 02:39 PM
Hi YemenSEF,
You can try the below mentioned algorithm:
largest = x
smallest = x
if (y > largest) then largest = y
if (z > largest) then largest = z
if (y < smallest) then smallest = y
if (z < smallest) then smallest = z
If you want to write the same in any programming language do reply.
Please revert for clarification.
Zohaib R Posts 2368 Registration date Sunday September 23, 2012 Status Member Last seen December 13, 2018 69
Nov 29, 2012 at 01:19 PM
Hi YemenSEF,
I appreciate your hard work. Yes, your algorithm will give you the right results. Feel free to approach if you need any further help.
YemenSEF Posts 2 Registration date Monday November 26, 2012 Status Member Last seen August 19, 2013
Nov 27, 2012 at 05:46 PM
Thanks very much
but i have already tried alot and found a good answer andi hope its right too
I did one like this
-start | 515 | 1,779 | {"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} | 3.015625 | 3 | CC-MAIN-2024-33 | latest | en | 0.931351 |
https://physics.stackexchange.com/questions/119932/find-minimum-distance-between-particles-given-initial-position-and-velocity | 1,632,306,426,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057347.80/warc/CC-MAIN-20210922102402-20210922132402-00542.warc.gz | 502,704,555 | 38,527 | # Find minimum distance between particles given initial position and velocity
My friend gave me a question today:
We have a point $A$. At a distance of $x_0$ from the point. There is a particle $P_1$. There is another particle, $P_2$, at $A$. $P_1$ moves with velocity $u_1$ towards $A$. At the same instant, $P_2$ moves making an angle $\theta$ with $P_1A$ with a velocity $u_2$. What will be the minimum distance between the two particles when both of them start at the same time?
So can anybody help me with this problem? I tried decomposing it into vectors, used $\tan\theta$, but it did not give me a correct answer.
Let's assume $P_2$ is in $(0,0)$ at $t=0$. $P_1$ is in $(x_0,0)$ at t=0. (So A-P1 represents the x-axis)
The evolution of $P_2$'s position is given by $\vec{r}_{p2}(t) = u_2\cdot t\cdot(\cos(\theta),\sin(\theta))$ which you can easily get from decomposing.
The evolution of $P_1$'s position is given by $\vec{r}_{p1}(t) = (u_1\cdot t+x_0,0)$
The distance between these particles in function of time is (Pythagoras' Theorem)
$D(t) = \sqrt{ (u_2\cos(\theta)\cdot t-u_1\cdot t-x_0)^2 + u_2^2\sin(\theta)^2\cdot t^2}$
This function obviously won't have a maximum. It is enough to derive it once and set it's derivative equal to 0.
This is the gist of the problem. Feel free to ask if you have any difficulties understanding this. Working your way through the algebra you should find
$t_{min} = \frac{u_2\cdot \cos(\theta)-u_1}{u_1^2+u_2^2-2u_1u_2\cdot\cos(\theta)}x_0$ as the time at which the distance is minimal
$D_{min} = \frac{u_2x_0\cdot\sin(\theta)}{\sqrt{u_1^2+u_2^2-2u_1u_2\cdot\cos(\theta)}}$ the value of the minimal distance
• OH! I Got!!, Thank you very much, but I had got it with a very similar method. Thanks, But Although for a clean solution Jun 19 '14 at 16:02
You don't have enough information to properly answer the question.
If $u_1 \le u_2\ and\ \theta \ge \pi/2$, then the shortest distance occurs at the initial conditions.
When $u_1 > u_2$, then the shortest distance occurs when $\angle P_{1-initial}P_1P_2$ is $\pi/2$. | 673 | 2,079 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2021-39 | latest | en | 0.889208 |
http://forum.twbts.com/viewthread.php?action=printable&tid=23500 | 1,660,761,589,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573104.24/warc/CC-MAIN-20220817183340-20220817213340-00234.warc.gz | 15,970,298 | 2,650 | javascript:;
[attach]34432[/attach][attach]34432[/attach][attach]34432[/attach][attach]34432[/attach]
Sub test()
Dim Arr, xD, T\$, pos%
Set xD = CreateObject("Scripting.Dictionary")
Arr = Range([j5], [g65536].End(3))
For i = 2 To UBound(Arr)
T = Arr(i, 1) & "|" & Arr(i, 2)
xD(T) = Array(Arr(i, 3), Arr(i, 4))
Next
Arr = Range([d1], [a65536].End(3))
For i = 2 To UBound(Arr)
T = Arr(i, 1) & "|" & Arr(i, 3)
If xD.Exists(T) Then
pos = InStr(xD(T)(0), Arr(i, 2))
If pos = 0 Then
Arr(i, 4) = "錯誤，因為" & Arr(i, 1) & "+" & Arr(i, 3) & "，儲位" & xD(T)(1)
Else
Arr(i, 4) = "正確，因為" & Arr(i, 1) & "+" & Arr(i, 3) & "，儲位" & xD(T)(1)
End If
End If
Next
Range("a1").Resize(UBound(Arr), 4) = Arr
End Sub
https://blog.xuite.net/hcm19522/twblog/590142590
Sub test2()
Dim Arr, xD, T\$, pos%, a, i&
Set xD = CreateObject("Scripting.Dictionary")
Arr = Range([j5], [g65536].End(3))
For i = 2 To UBound(Arr)
T = Arr(i, 1) & "|" & Arr(i, 2)
xD(T) = Array(Arr(i, 3), Arr(i, 4))
Next
Arr = Range([d1], [a65536].End(3))
For i = 2 To UBound(Arr)
T = Arr(i, 1) & "|" & Arr(i, 3)
If xD.Exists(T) Then
a = Split(xD(T)(0), ".")
For j = 0 To UBound(a)
If Arr(i, 2) = a(j) Then pos = 1
Next
If pos = 0 Then
Arr(i, 4) = "錯誤，因為" & Arr(i, 1) & "+" & Arr(i, 3) & "，儲位" & xD(T)(1)
Else
Arr(i, 4) = "正確，因為" & Arr(i, 1) & "+" & Arr(i, 3) & "，儲位" & xD(T)(1)
pos = 0
End If
End If
Next
Range("a1").Resize(UBound(Arr), 4) = Arr
End Sub
=IF(ISERR(FIND("."&B2&".",LOOKUP(,0/(A2&C2=G\$6:G\$16&H\$6:H\$16),"."&I\$6:I\$16&"."))),"錯誤","正確")
歡迎光臨 麻辣家族討論版版 (http://forum.twbts.com/) | 724 | 1,536 | {"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-2022-33 | latest | en | 0.341774 |
http://www.daviddarling.info/encyclopedia/C/current_lag_and_lead.html | 1,469,605,636,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257826736.89/warc/CC-MAIN-20160723071026-00141-ip-10-185-27-174.ec2.internal.warc.gz | 403,900,644 | 5,268 | A
# Darling
A current flowing around an electric circuit may meet with three different kinds of opposition or impedance. They are caused by resistance (R), inductance (L), and capacitance (C).
Of these, resistance is the easiest to understand, because it has the same effect on both direct currents and alternating currents. When the voltage across the two terminals of a resistance changes, the current changes immediately. If the voltage rises, the current rises; and if the voltage falls, the current falls, and so on. Current and voltage are said to be in phase.
Inductors (L) and capacitors (C) behave quite differently, In "L" circuits a rise in voltage is accompanied by a rise in current, but this rise is delayed by a back e.m.f. (see reactance) generated by the inductor. As the voltage rises and falls, the current rises and falls, but a fraction of a second later. So the current flowing through the inductor is always lagging behind the voltage, and current and voltage are said to be out of phase.
In "C" circuits, on the other hand, the current in the circuit must first flow to the two plates of the capacitor (round the circuit from plate to plate and not across the gap between the plates) to make potential difference across them. As the current rises, the voltage between the two plates rises; and as the current falls, the voltage falls, but the voltage follows the current's lead a fraction of a second later. Current and voltage are again out of phase, only in "C" circuits the current is always leading the voltage.
One complete cycle of an alternating current consists of a rise (in current or voltage) up to the positive maximum, followed by a drop, through zero, to the negative maximum voltage, and a subsequent rise back to the zero starting point. A "positive" current means that electrons flow in one direction round the circuit, while a "negative" current means that they surge round in the opposite direction.
In a circuit containing only resistance, the positive surges of current and the positive increases in voltage coincide. But in a circuit containing only capacitance the surges of current occur a quarter of a cycle before the increase of voltage across the capacitance. In a circuit containing only inductance the current surges occur a quarter of a cycle later.
Suppose a capacitor and an inductor are both connected across an alternating voltage supply (i.e., connected in parallel), then the same voltage sends a current through each. But in the "C" part of the circuit the current leads the voltage and in the "L" part the current lags behind the voltage. If the values of inductance and capacitance are selected so that both offer the same impedance at the frequency of the alternating current supply, then the current through both "L" and "C" parts will be equal. But since one is a quarter of a cycle behind the voltage, and the other is a quarter of a cycle in front of the voltage, there is a difference of phase of a half cycle between the currents in the "L" and "C" parts. As one current is positive, the other current is negative (i.e., flowing in the opposite direction) and the same size as the positive current. So the two currents cancel each other out, and as a result no current flows out of the "L" and "C" combination, although there is a voltage connected across the pair of them. So the inductor-capacitor pair offers a very large impedance to the current – far larger than their separate impedances.
An arrangement called a parallel tuned circuit or rejector circuit will not allow through current of a particular frequency – the frequency at which the impedance of the capacitor is equal to the impedance of the inductor. Then the currents flowing through both parts are equal and opposite in direction. At any other frequency the two impedances will not be quite equal, the two currents will not be quite cancel each other out, and some current will be able to flow right round the circuit. | 833 | 3,967 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2016-30 | longest | en | 0.942972 |
http://dictionary.reference.com/browse/quadrature | 1,406,596,923,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510264270.11/warc/CC-MAIN-20140728011744-00219-ip-10-146-231-18.ec2.internal.warc.gz | 78,684,775 | 27,662 | [kwod-ruh-cher, -choor]
noun
1.
the act of squaring.
2.
Mathematics.
a.
the act or process of finding a square equal in area to a given surface, especially a surface bounded by a curve.
b.
the act or process of finding an area or calculating an integral, especially by numerical methods.
c.
a definite integral.
3.
Astronomy.
a.
the situation of two heavenly bodies when their longitudes differ by 90°.
b.
either of the two points in the orbit of a body, as the moon, midway between the syzygies.
c.
(of the moon) those points or moments at which a half moon is visible.
4.
Electronics. the relation between two signals having the same frequency that differ in phase by 90°.
Origin:
Dictionary.com Unabridged
Based on the Random House Dictionary, © Random House, Inc. 2014.
Collins
World English Dictionary
quadrature (ˈkwɒdrətʃə) —n 1. maths the process of determining a square having an area equal to that of a given figure or surface 2. the process of making square or dividing into squares 3. astronomy a configuration in which two celestial bodies, usually the sun and the moon or a planet, form an angle of 90° with a third body, usually the earth 4. electronics the relationship between two waves that are 90° out of phase
Collins English Dictionary - Complete & Unabridged 10th Edition
2009 © William Collins Sons & Co. Ltd. 1979, 1986 © HarperCollins
Publishers 1998, 2000, 2003, 2005, 2006, 2007, 2009
Cite This Source
American Heritage
Science Dictionary
quadrature (kwŏd'rə-chr') Pronunciation Key The process of constructing a square equal in area to a given surface. A configuration in which the position of one celestial body is 90° from another celestial body, as measured from a third. For example, the half moon lies in quadrature from the Sun when Earth is the reference point. See more at elongation.
The American Heritage® Science Dictionary
Cite This Source
Encyclopedia Britannica
Encyclopedia
in astronomy, that aspect of a heavenly body in which its direction as seen from the Earth makes a right angle with the direction of the Sun. The Moon at First or Last Quarter is said to be at east or west quadrature, respectively. A superior planet (outside the Earth's orbit) is at west quadrature when its position is 90 west of the Sun. It rises around midnight, reaches the meridian (a great circle on the celestial sphere, passing through the north and south poles and the zenith) near sunrise and sets near noon. At east quadrature the planet is near the meridian at sunset and sets near midnight. At both quadratures the planet is at gibbous phase (more than half but not all of the disk illuminated), but only Mars shows up conspicuously gibbous | 655 | 2,681 | {"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-2014-23 | longest | en | 0.872547 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.