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https://www.sigterritoires.fr/index.php/en/tutorial-create-an-aptitude-map-with-arcmap-with-and-without-fuzzy-criteria-2/ | 1,726,177,878,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651498.46/warc/CC-MAIN-20240912210501-20240913000501-00086.warc.gz | 910,425,297 | 58,351 | ### Tutorial : create an aptitude map with ArcMap, with and without fuzzy criteria (2)
Let’s go back literally to the Spatial Analyst tutorial:
” …
Creation of an aptitude map
The creation an aptitude map allows you to get an aptitude value for each location on the map.
Once you have created the necessary layers ( in this example , the layers are Slope , Distance to Recreational Facilities , Distance to Schools and Land Use) to your analysis, how do these created layers combine to create a classified map of the potential surfaces to locate school ? You must compare the values of classes among layers. To accomplish this task, a method involves assigning numerical values to the classes included in each layer of the map or to re classify them.
Each layer of the map is classified according its aptitude degree as location for the new school.. For example, you can assign a value to each class of each layer, according to a scale from 1 to 10, 10 being the best ranking.
This scale is called an “aptitude scale”. Use the value NoData to rule out areas that should not be taken into consideration. If all measures are affected by the same numerical scale, grants the same importance in the determination of the most adequate locations. Initially, the model is developed this way. Afterwards, when you test other scenarios, weighting factors will be applied to the layers to increase the exploration of data and their relationships.
Creation of aptitude scales
As shown in this example, many scales are synthetic. Usually a ranking, from the most appropriate to the least appropriate is used. This ranking is based on something measurable, such as distance to schools, but in fact, it is a subjective measure that determines the aptitude degree of a certain distance from a school to place another school.
There are natural scales that are commonly associated with certain objectives. The cost is a good example, but must be defined in detail . In a study on the feasibility of a building, a weak cost target would be measured on a dollar scale. Be sure to define an adequate scale. For an element well known as the dollar other variables must be taken into account, for example whether it is American, Australian dollars or an exchange rate between currencies.
Many scales are not linear relationships nevertheless they are often presented as if they were to save time and money or because all the options have not been considered. For example, if you assign a scale to a travel distance, a displacement of 1.5 or 10 km would not be classified as an aptitude of 10, 5 and 1 if the movement was done on foot. Some people will evaluate that a 5 kilometres walk is only twice more tiring than a 1 kilometre walk while others will evaluate it as tenfold.
When you are developing an aptitude scale, ask the opinion of well-informed people to identify both extremes of a scenario and as many intermediate points as possible. These people must have a good knowledge of the objective under consideration. For example, it is more interesting to ask users their opinion about their preferences about the travel time from their homes to their work, than asking an agent the time brackets when traffic is the worst.
The problem is very well framed. Let’s see how to answer by using the classical tools.
Ranking of nearby areas to recreational facilities with the Re-classification tool
In order to locate a school near recreational facilities, you must know the distance between them. The Spatial Analyst Euclidean Distance tool creates this type of map by calculating the distance in straight line (Euclidean) between a location and the nearest recreational facility. The result is a raster dataset in which each cell represents the distance to the nearest recreational facilities. To rank the map, use the Reclassification tool. Since it is desirable to locate the school near recreational facilities, assign the value 1 to the distant places from these facilities and the value 10 to closer places. Then rank intermediate distances linearly as shown in the next illustration.
We use the Re-classification tool from the toolbox Spatial Analyst-> Classify.
Steps:
• Open the Re-classification tool
• As layer in the entry indicate DistanceToRecSites
• Accept the default value for the parameter reclassification Field to use as Value field.
• Click
• Set the Method to equal Interval (variable number) and the number of Classes out of 10.
• Click
• Click Reverse news values.
The selection of the option Invert the news values attribute a new, higher value to close distances to the leisure centres, as these areas are more desirable.
Give a name for Raster output parameter: DistRecSitesReclass
Click OK.
The result got is the following:
For the geomaticists that we are, this result appears as quite consistent. We have a study area, concentric distances from our targets and a regular gradient of values as we move away from our targets.
But the question is: does it answer the question?
The green areas, i.e. the most suitable areas according to this criterion, depend on the location of the recreational sites. But not only. They depend on our study area: if we would have selected a larger area, they would be bigger, we would have limited our area, they would be smaller. They also depend on our arbitrary choice of 10 zones. If we would have chosen to make 7, they would be bigger, and if we had selected to make 12 of them would be smaller.
The question asked is to find sites, given that parents want the new school to be near a recreational site. As a parent, I want to find that a recreational site is close or far according to many personal criteria, but most likely not as a function of the total area of the planning project!
Ranking of nearby recreational facilities areas using the Fuzzy Criterion tool
The satisfaction of the criterion Proximity to a recreational site must be asked to the actors. Therefore we will be able to understand what is considered “Close”, a site within walking distance with children, or by car but with a maximum traject between 5 to 10 minutes.
If we translate the latter in distance, we can say that the optimum satisfaction is between 0 and 1,000 m (route distance) and, from 7 km (route 10 ‘ by car) is considered as “Far”.
We use the command Criterion flexible on the layer DistanceRecre
We select the attribute Gridcode and define a maximum satisfaction up to 1000m (a) and a zero satisfaction from 5000 m (b).
We click Calculate, and obtain the following result:
Of course we could do something similar with the Reclassify tool. For the time being, what is important, is the reasoning used.
We have two different results, not just because we are using two different tools, but because we have two different reasoning.
Ranking of remote areas from existing schools, using Spatial Analyst
We have created a raster dataset in which each cell represents the distance to the nearest school (DistToSchools). To rank the map, use the Reclassification tool. Since it is desirable to locate the school away from existing schools, assign the value 1 to the distances close to these schools and the value 10 to the distances far away. Then rank intermediate distances linearly as shown in the next illustration.
As for the distance to the recreational sites, we use the Re-classify tool of the toolbox Spatial Analyst-> Classify .
Steps :
• Open the Re-classification tool
• As layer in entry indicate DistSchoolsReclass
• Accept the default value for the parameter re-classification Field to use the field Value.
• Click
• Set the Method to Equal Interval ( variable number ) and the number of Classes out of 10.
• Click
You want to locate the school away from existing schools; therefore you assign increasing numbers to ranges of values that represent more remote locations, because these locations are most desirable. Since the default affects New high values (more suitable locations) to Old high values (locations further away from existing schools), you do not need to change the values at this stage.
The result is as follows:
The remarks stated for the remoteness of recreational sites are applicable to the result of this criterion. The distance of existing schools is not a concept related to the size of our study area.
Ranking of remote areas from existing schools with the tool Fuzzy Criterion tool
If we apply the same reasoning as previously, a user will consider that the school is “Far” if it takes more than 10 ‘ by car and “Close” if can go walking (1000m).
We select the attribute Gridcode and define zero satisfaction up to 1000m (a) and maximum satisfaction from 5000 m (b).
We click Calculate, and we get the following result:
This result matches better our notion of “Far”: from a certain distance, one is far, not far-far or far- very distance, etc.
In the next article, we will classify the slopes and the type of land use, before facing the aggregation of all these criteria for obtaining the final result: our aptitude map.
Si cet article vous a intéressé et que vous pensez qu'il pourrait bénéficier à d'autres personnes, n'hésitez pas à le partager sur vos réseaux sociaux en utilisant les boutons ci-dessous. Votre partage est apprécié ! | 1,934 | 9,267 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.926872 |
https://online-shiksha.com/multiplication-class-3-worksheet/2/ | 1,726,844,505,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652278.82/warc/CC-MAIN-20240920122604-20240920152604-00147.warc.gz | 394,043,815 | 57,527 | Multiplication Worksheet for Class 3
Word problems
Jack bought 4 boxes of chocolates, with each box containing 12 chocolates. How many chocolates did Jack buy in total?
A farmer has 6 fields, and each field has 18 apple trees. How many apple trees does the farmer have in total?
A toy store has 5 aisles, and each aisle has 24 toys on display. How many toys are there in the store?
Sarah has 3 shelves in her bookcase, and each shelf can hold 28 books. If she fills all the shelves, how many books will she have in her bookcase?
A company produces 7 cars every day, and each car has 4 tires. How many tires does the company need in a week (assuming they work all 7 days)?
There are 8 classrooms in a school, and each classroom has 32 students. How many students are there in the school?
A bakery sells 6 cakes per day, and each cake has 10 slices. How many slices of cake does the bakery sell in a week (assuming they work all 7 days)?
A factory produces 9 boxes of pencils, and each box contains 36 pencils. How many pencils are produced by the factory in total?
There are 4 crates, and each crate has 20 oranges. How many oranges are there in all the crates?
A garden has 5 rows, and each row has 16 flowers. How many flowers are there in the garden?
Remember to solve each problem by performing the necessary multiplication calculations to find the correct answer.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 339 | 1,461 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2024-38 | latest | en | 0.978384 |
http://www.wikihow.com/Read-Charts | 1,475,305,955,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738662541.24/warc/CC-MAIN-20160924173742-00172-ip-10-143-35-109.ec2.internal.warc.gz | 818,433,093 | 56,015 | Edit Article
Charts turn facts and figures into images instead of a string of words and numbers. This lets you quickly locate only the information you're interested in and, when necessary, compare values to one another. There are 3 primary types of charts you're likely to encounter: pie charts, bar graphs or charts, and line graphs or charts.
### Method 1 Pie Graph/Chart
1. 1
Identify what each segment of the pie chart represents. The pie chart segments may be identified in 1 of 2 ways:
• The segments may be color-coded, with a key beside the chart identifying what each color represents.
• The segments may be labeled, with wording placed directly on the relevant segment or connected to it by a line.
• Most pie charts also have a percentage assigned to each segment, identifying how much that segment makes up of the whole.
2. 2
Read off the value for the segment of the chart you're interested in.
• Pie charts are also a convenient way of comparing the relative sizes or quantities of various segments of the whole.
### Method 2 Bar Graph/Chart
1. 1
Read across the horizontal axis, also called the x-axis, to find the bar labeled for the data grouping you're interested in.
2. 2
Read directly up the graph until you get to the top of the bar.
3. 3
Read back to the left, level with the top of the bar you're interested in, until you intersect the vertical or y-axis of the chart.
• It might help to trace across the graph from bar to axis with your finger or a pencil.
4. 4
Read the value at the place where you intersected the y-axis.
• If you hit the y-axis between 2 data points, make your best estimate of the exact value. For example, if you hit the y-axis directly between two points labeled 10 and 15, 12.5 would be a logical estimate for the value of that point.
• The values on the y-axis usually represent some sort of frequency (how often something happens) or measurement.
### Method 3 Line Graph/Chart
1. 1
Read along the horizontal axis (x-axis) until you find the data grouping you're interested in.
2. 2
Read straight up the graph until you find the dot above that grouping's label.
• Not all line graphs have dots to represent data points; sometimes there's just a vertex, or joining, of 2 lines.
3. 3
Read straight across the graph to the left until you intersect the vertical or y-axis. Read the number value at this point; that's the value that corresponds to the data grouping you're interested in.
## Tips
• The words chart and graph are often used interchangeably.
• Pie charts are usually used to represent only 2 pieces of information: which components are part of a whole, and how much of that whole each component represents. But both bar graphs and line graphs can represent relationships between multiple related data sets. For example, a bar graph could show how many tourists came from which country in multiple years, with 1 bar per year for each country. Or a line graph could show the relationship between how many salmon of several species returned to a spawning ground, with each species represented by a different-color line.
## Article Info
Categories: Probability and Statistics | Mathematics
In other languages:
Português: Ler Gráficos, Español: interpretar gráficos, Italiano: Leggere i Grafici, Русский: читать диаграммы
Thanks to all authors for creating a page that has been read 5,033 times. | 764 | 3,362 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2016-40 | longest | en | 0.890547 |
https://sourceforge.net/p/mcmc-jags/discussion/610037/thread/978e4e07/?limit=25 | 1,503,035,350,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104565.76/warc/CC-MAIN-20170818043915-20170818063915-00358.warc.gz | 815,421,068 | 13,926 | ## Choosing different distributions based on a certain condition document.SUBSCRIPTION_OPTIONS = { "thing": "thread", "subscribed": false, "url": "subscribe", "icon": { "css": "fa fa-envelope-o" } };
Help
2014-06-04
2014-06-05
I am trying to define a node sat, such that it is 0 if another node con=0 and sat comes from a normal distribution if con≠0. Does anyone has any suggestion on how to make that work? Thanks!
Here are the things I have tried (along with error message), none of which work:
~~~~~~~
temp[i,1] ~ dcat(1)
temp[i,2] ~ dnorm(g[i],taus)
ind <- con[i]+1
sat[i] ~ temp[i,ind]
# I know for this case sat=1 when con=0, but that works too.
sat[i] ~ ifelse(con[i]==0, 0, dnorm(g[i],taus))
# Possibly because ifelse doesn't work with distributions
temp[i] ~ dnorm(g[i],taus)
sat[i] <- con[i]*temp[i]
# sat is observed in my model and logical nodes can't be observed
• Matt Denwood - 2014-06-04
What you're trying to fit is a mixture distribution I think, but it sounds a bit strange. Do you really have a mixture of integers (0) and non-integers (numbers on the real line from a normal distribution) in your observed data? If so, surely you know a-priori which observations are truly 0 and which aren't? If not, you may not want to use a normal distribution but perhaps an integer distribution such as a Poisson (in which case you would have a zero-inflated Poisson distribution, which is fairly straightforward to code).
The general way that you can code a mixture distribution when the distributional form is the same (but parameters differ) is as follows:
sat[i] ~ dnorm(mu[con[i]+1], tau[con[i]+1])
con[i] ~ dbern(prob)
This would work with any distribution, not just normal.
Or if you have a zero-inflated Poisson mixture you can use this code:
sat[i] ~ dpois(lambda[i])
lambda[i] <- mean * con[i]
con[i] ~ dbern(prob)
Note that for these types of code you will have to be careful about specifying sensible initial values for con.
Hope that helps,
Matt
Last edit: Matt Denwood 2014-06-04
Thanks Matt! I responded yesterday with a detailed explanation for my choice of random variables, but for some reason I don't see that anymore :/ Anyways, I decided on using the same family of distributions after all to make things work. Thanks again!
• Martyn Plummer - 2014-06-05
That happens to me to sometimes. You have to be careful to press the "post" button and not just the "preview" button. | 637 | 2,426 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2017-34 | latest | en | 0.888424 |
https://www.jiskha.com/display.cgi?id=1371566361 | 1,503,491,207,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886120194.50/warc/CC-MAIN-20170823113414-20170823133414-00396.warc.gz | 912,745,389 | 3,741 | # Calculus
posted by .
For what values of x with 0¡Üx¡Ü2¦Ð does the graph of f(x)=x+4sin(x) have a horizontal tangent?
• Calculus -
f(x) = x+4sin x
f'(x) = 1 + 4cos x
So, where is cos x = -1/4?
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https://www.physicsforums.com/threads/projectile-motion.187021/ | 1,493,334,396,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917122629.72/warc/CC-MAIN-20170423031202-00248-ip-10-145-167-34.ec2.internal.warc.gz | 969,220,566 | 14,352 | Projectile motion
1. Sep 25, 2007
texmex
A projectile is to be fired so that it just clears an obstacle 80m high at its maximum height.
a) by considering the vertical motion only, calculate the verticle speedand the time taken to reach its maximum height
b) if the distance between A to B is 100m, calculate the horizontal speed with wich the ball was thrown.
c) Hence find v and Θ.
totally stuck at this, any help much appricated.
2. Sep 25, 2007
hage567
Have you tried anything? What equations might be used? What can you say about the vertical component of the velocity at maximum height?
3. Sep 25, 2007
texmex
the farthest i got, was putting in the right angle, and as for equations that could be used, i think it may have something to do with - Vvertical= vsinΘ. not entirely sure though.
4. Sep 25, 2007
hage567
That would be the initial vertical velocity (is that a negative sign?). What is the vertical velocity at the instant of maximum height? | 248 | 968 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-17 | latest | en | 0.948488 |
http://nosuch.com/music/expresso.cgi?generations=19&seed=50359533&ntracks=1&patches=no&randomseed=off&x= | 1,544,862,883,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826842.56/warc/CC-MAIN-20181215083318-20181215105318-00004.warc.gz | 199,444,772 | 3,141 | Home : Tune Toys : Expresso
Expresso takes a simple musical expression (literally "X") and mutates it. In each generation, transformations are applied to components of the expression. This is a fractal technique known as an L-system. The more generations there are, the larger and more complex the expression gets (and the longer it can take to compute). The output varies wildly, from boring to fascinating. Press "Mutate" a few times till you get something that looks interesting, then click on the image to play it. See Composer's Quarry for examples of output.
Transformations:
Seed: Randomize seed: Generations: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Phrase for X: Basic 1 - c,e,gBasic 2 - c,g,e,b-Basic 3 - c,gBasic 4 - cBasic 5 - co3,co4,co3,co2Chord 1 - c e gChord 2 - c e- gChord 3 - c e- g b-Chord 4 - c f gFile 1 - bachinv1.midFile 2 - bachinv2.midFile 3 - bachinv3.midFile 4 - bachinv4.midFile 5 - bachinv5.midFile 6 - bachinv6.midFile 7 - bachinv7.midFile 8 - bachinv8.mid # of Tracks: 1234 Randomize patches: noyes Results may not appear for a few seconds.Be patient!
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After 19 generations the expression "X" became this: (shuffle(echo(transpose(arpeggio(transpose(echo((transpose(transpose((((X+transpose(X,7) ) +transpose(arpeggio((X|X) ) ,12) ) |arpeggio(transpose(X,4) ) ) ,-5) ,-5) +transpose(((step(transpose(arpeggio(echo(X,4,6) ) ,-7) ,12) |transpose(((((echo(X,4,6) +transpose((X+transpose(X,7) ) ,4) ) +transpose((X|X) ,7) ) +transpose((X+transpose(X,7) ) ,7) ) +transpose((((X+transpose(X,12) ) +transpose(shuffle(X) ,4) ) +transpose(X,7) ) ,12) ) ,-5) ) +((transpose(step(X,12) ,-7) |arpeggio((X|X) ) ) +transpose((((X+transpose((transpose(X,-7) +transpose((X+transpose(X,12) ) ,7) ) ,7) ) +(echo(X,4,6) +transpose((((X+transpose(X,7) ) +transpose(transpose(X,-5) ,4) ) +transpose(transpose(X,-5) ,7) ) ,12) ) ) +transpose(transpose(shuffle(transpose(X,4) ) ,-5) ,7) ) ,7) ) ) ,12) ) ,4,6) ,-5) ) ,-5) ,4,6) ) +(step(transpose(transpose(((transpose((transpose(echo(X,4,6) ,-5) |arpeggio(((((X+X) +transpose(echo(X,4,6) ,4) ) +transpose(X,7) ) +transpose(X,-5) ) ) ) ,-5) +(arpeggio((((X+transpose(X,7) ) +((X|X) +transpose(transpose(X,-7) ,7) ) ) |transpose(shuffle(shuffle(X) ) ,-5) ) ) +transpose(((echo(X,4,6) |step(X,12) ) +transpose(shuffle(X) ,12) ) ,12) ) ) +(((shuffle(transpose(transpose(arpeggio((X+transpose(X,7) ) ) ,-7) ,-5) ) +transpose((transpose(shuffle(transpose(transpose((X+X) ,-7) ,-7) ) ,4) +transpose(transpose(((X+transpose(X,12) ) |X) ,-5) ,7) ) ,7) ) +transpose(transpose(transpose(transpose(transpose((echo(step(X,12) ,4,6) +transpose(step(transpose(transpose(X,-5) ,-7) ,12) ,12) ) ,-7) ,4) ,-7) ,-5) ,4) ) +transpose(shuffle(((transpose((((X|X) +transpose(((arpeggio(X) +transpose(X,4) ) +transpose(X,7) ) ,4) ) +transpose(transpose(X,-5) ,7) ) ,-7) +transpose((((arpeggio((X+transpose(X,7) ) ) +(step(X,12) +transpose((transpose(X,-5) +X) ,12) ) ) +transpose(step((X+transpose(X,12) ) ,12) ,4) ) +transpose(step(transpose(arpeggio(X) ,-5) ,12) ,7) ) ,12) ) +transpose(echo((transpose((transpose(X,-5) +transpose(transpose(X,4) ,12) ) ,4) |transpose(((X+transpose(X,4) ) +transpose(X,7) ) ,-7) ) ,4,6) ,7) ) ) ,7) ) ) ,4) ,-7) ,12) +transpose(transpose(step(echo((((step(X,12) +transpose(((X+transpose(X,7) ) +transpose((X+transpose(echo(X,4,6) ,12) ) ,12) ) ,7) ) +transpose(transpose(transpose(((((X+transpose(X,7) ) +transpose((X+transpose(X,7) ) ,4) ) +transpose(step(X,12) ,7) ) +transpose(arpeggio(X) ,7) ) ,-7) ,-7) ,4) ) +transpose(echo(X,4,6) ,7) ) ,4,6) ,12) ,-7) ,7) ) ) random seed used was 50359533
The algorithm for Expresso is written in KeyKit, and here's the source code. | 1,371 | 3,801 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-51 | latest | en | 0.716858 |
http://cn.metamath.org/mpeuni/mmtheorems245.html | 1,652,806,747,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662519037.11/warc/CC-MAIN-20220517162558-20220517192558-00228.warc.gz | 12,276,700 | 17,168 | Home Metamath Proof ExplorerTheorem List (p. 245 of 429) < Previous Next > Bad symbols? Try the GIF version. Mirrors > Metamath Home Page > MPE Home Page > Theorem List Contents > Recent Proofs This page: Page List
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Theorem List for Metamath Proof Explorer - 24401-24500 *Has distinct variable group(s)
TypeLabelDescription
Statement
Theoremargregt0 24401 Closure of the argument of a complex number with positive real part. (Contributed by Mario Carneiro, 25-Feb-2015.)
((𝐴 ∈ ℂ ∧ 0 < (ℜ‘𝐴)) → (ℑ‘(log‘𝐴)) ∈ (-(π / 2)(,)(π / 2)))
Theoremargrege0 24402 Closure of the argument of a complex number with nonnegative real part. (Contributed by Mario Carneiro, 2-Apr-2015.)
((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ∧ 0 ≤ (ℜ‘𝐴)) → (ℑ‘(log‘𝐴)) ∈ (-(π / 2)[,](π / 2)))
Theoremargimgt0 24403 Closure of the argument of a complex number with positive imaginary part. (Contributed by Mario Carneiro, 25-Feb-2015.)
((𝐴 ∈ ℂ ∧ 0 < (ℑ‘𝐴)) → (ℑ‘(log‘𝐴)) ∈ (0(,)π))
Theoremargimlt0 24404 Closure of the argument of a complex number with negative imaginary part. (Contributed by Mario Carneiro, 25-Feb-2015.)
((𝐴 ∈ ℂ ∧ (ℑ‘𝐴) < 0) → (ℑ‘(log‘𝐴)) ∈ (-π(,)0))
Theoremlogimul 24405 Multiplying a number by i increases the logarithm of the number by iπ / 2. (Contributed by Mario Carneiro, 4-Apr-2015.)
((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ∧ 0 ≤ (ℜ‘𝐴)) → (log‘(i · 𝐴)) = ((log‘𝐴) + (i · (π / 2))))
Theoremlogneg2 24406 The logarithm of the negative of a number with positive imaginary part is i · π less than the original. (Compare logneg 24379.) (Contributed by Mario Carneiro, 3-Apr-2015.)
((𝐴 ∈ ℂ ∧ 0 < (ℑ‘𝐴)) → (log‘-𝐴) = ((log‘𝐴) − (i · π)))
Theoremlogmul2 24407 Generalization of relogmul 24383 to a complex left argument. (Contributed by Mario Carneiro, 9-Jul-2017.)
((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ∧ 𝐵 ∈ ℝ+) → (log‘(𝐴 · 𝐵)) = ((log‘𝐴) + (log‘𝐵)))
Theoremlogdiv2 24408 Generalization of relogdiv 24384 to a complex left argument. (Contributed by Mario Carneiro, 8-Jul-2017.)
((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ∧ 𝐵 ∈ ℝ+) → (log‘(𝐴 / 𝐵)) = ((log‘𝐴) − (log‘𝐵)))
Theoremabslogle 24409 Bound on the magnitude of the complex logarithm function. (Contributed by Mario Carneiro, 3-Jul-2017.)
((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0) → (abs‘(log‘𝐴)) ≤ ((abs‘(log‘(abs‘𝐴))) + π))
Theoremtanarg 24410 The basic relation between the "arg" function ℑ ∘ log and the arctangent. (Contributed by Mario Carneiro, 25-Feb-2015.)
((𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ≠ 0) → (tan‘(ℑ‘(log‘𝐴))) = ((ℑ‘𝐴) / (ℜ‘𝐴)))
Theoremlogdivlti 24411 The log𝑥 / 𝑥 function is strictly decreasing on the reals greater than e. (Contributed by Mario Carneiro, 14-Mar-2014.)
(((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ e ≤ 𝐴) ∧ 𝐴 < 𝐵) → ((log‘𝐵) / 𝐵) < ((log‘𝐴) / 𝐴))
Theoremlogdivlt 24412 The log𝑥 / 𝑥 function is strictly decreasing on the reals greater than e. (Contributed by Mario Carneiro, 14-Mar-2014.)
(((𝐴 ∈ ℝ ∧ e ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ e ≤ 𝐵)) → (𝐴 < 𝐵 ↔ ((log‘𝐵) / 𝐵) < ((log‘𝐴) / 𝐴)))
Theoremlogdivle 24413 The log𝑥 / 𝑥 function is strictly decreasing on the reals greater than e. (Contributed by Mario Carneiro, 3-May-2016.)
(((𝐴 ∈ ℝ ∧ e ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ e ≤ 𝐵)) → (𝐴𝐵 ↔ ((log‘𝐵) / 𝐵) ≤ ((log‘𝐴) / 𝐴)))
Theoremrelogcld 24414 Closure of the natural logarithm function. (Contributed by Mario Carneiro, 29-May-2016.)
(𝜑𝐴 ∈ ℝ+) (𝜑 → (log‘𝐴) ∈ ℝ)
Theoremreeflogd 24415 Relationship between the natural logarithm function and the exponential function. (Contributed by Mario Carneiro, 29-May-2016.)
(𝜑𝐴 ∈ ℝ+) (𝜑 → (exp‘(log‘𝐴)) = 𝐴)
Theoremrelogmuld 24416 The natural logarithm of the product of two positive real numbers is the sum of natural logarithms. Property 2 of [Cohen] p. 301, restricted to natural logarithms. (Contributed by Mario Carneiro, 29-May-2016.)
(𝜑𝐴 ∈ ℝ+) & (𝜑𝐵 ∈ ℝ+) (𝜑 → (log‘(𝐴 · 𝐵)) = ((log‘𝐴) + (log‘𝐵)))
Theoremrelogdivd 24417 The natural logarithm of the quotient of two positive real numbers is the difference of natural logarithms. Exercise 72(a) and Property 3 of [Cohen] p. 301, restricted to natural logarithms. (Contributed by Mario Carneiro, 29-May-2016.)
(𝜑𝐴 ∈ ℝ+) & (𝜑𝐵 ∈ ℝ+) (𝜑 → (log‘(𝐴 / 𝐵)) = ((log‘𝐴) − (log‘𝐵)))
Theoremlogled 24418 Natural logarithm preserves . (Contributed by Mario Carneiro, 29-May-2016.)
(𝜑𝐴 ∈ ℝ+) & (𝜑𝐵 ∈ ℝ+) (𝜑 → (𝐴𝐵 ↔ (log‘𝐴) ≤ (log‘𝐵)))
Theoremrelogefd 24419 Relationship between the natural logarithm function and the exponential function. (Contributed by Mario Carneiro, 29-May-2016.)
(𝜑𝐴 ∈ ℝ) (𝜑 → (log‘(exp‘𝐴)) = 𝐴)
Theoremrplogcld 24420 Closure of the logarithm function in the positive reals. (Contributed by Mario Carneiro, 29-May-2016.)
(𝜑𝐴 ∈ ℝ) & (𝜑 → 1 < 𝐴) (𝜑 → (log‘𝐴) ∈ ℝ+)
Theoremlogge0d 24421 The logarithm of a number greater than 1 is nonnegative. (Contributed by Mario Carneiro, 29-May-2016.)
(𝜑𝐴 ∈ ℝ) & (𝜑 → 1 ≤ 𝐴) (𝜑 → 0 ≤ (log‘𝐴))
Theoremlogge0b 24422 The logarithm of a number is nonnegative iff the number is greater than or equal to 1. (Contributed by AV, 30-May-2020.)
(𝐴 ∈ ℝ+ → (0 ≤ (log‘𝐴) ↔ 1 ≤ 𝐴))
Theoremloggt0b 24423 The logarithm of a number is positive iff the number is greater than 1. (Contributed by AV, 30-May-2020.)
(𝐴 ∈ ℝ+ → (0 < (log‘𝐴) ↔ 1 < 𝐴))
Theoremlogle1b 24424 The logarithm of a number is less than or equal to 1 iff the number is less than or equal to Euler's constant. (Contributed by AV, 30-May-2020.)
(𝐴 ∈ ℝ+ → ((log‘𝐴) ≤ 1 ↔ 𝐴 ≤ e))
Theoremloglt1b 24425 The logarithm of a number is less than 1 iff the number is less than Euler's constant. (Contributed by AV, 30-May-2020.)
(𝐴 ∈ ℝ+ → ((log‘𝐴) < 1 ↔ 𝐴 < e))
Theoremdivlogrlim 24426 The inverse logarithm function converges to zero. (Contributed by Mario Carneiro, 30-May-2016.)
(𝑥 ∈ (1(,)+∞) ↦ (1 / (log‘𝑥))) ⇝𝑟 0
Theoremlogno1 24427 The logarithm function is not eventually bounded. (Contributed by Mario Carneiro, 30-Apr-2016.) (Proof shortened by Mario Carneiro, 30-May-2016.)
¬ (𝑥 ∈ ℝ+ ↦ (log‘𝑥)) ∈ 𝑂(1)
Theoremdvrelog 24428 The derivative of the real logarithm function. (Contributed by Mario Carneiro, 24-Feb-2015.)
(ℝ D (log ↾ ℝ+)) = (𝑥 ∈ ℝ+ ↦ (1 / 𝑥))
Theoremrelogcn 24429 The real logarithm function is continuous. (Contributed by Mario Carneiro, 17-Feb-2015.)
(log ↾ ℝ+) ∈ (ℝ+cn→ℝ)
Theoremellogdm 24430 Elementhood in the "continuous domain" of the complex logarithm. (Contributed by Mario Carneiro, 18-Feb-2015.)
𝐷 = (ℂ ∖ (-∞(,]0)) (𝐴𝐷 ↔ (𝐴 ∈ ℂ ∧ (𝐴 ∈ ℝ → 𝐴 ∈ ℝ+)))
Theoremlogdmn0 24431 A number in the continuous domain of log is nonzero. (Contributed by Mario Carneiro, 18-Feb-2015.)
𝐷 = (ℂ ∖ (-∞(,]0)) (𝐴𝐷𝐴 ≠ 0)
Theoremlogdmnrp 24432 A number in the continuous domain of log is not a strictly negative number. (Contributed by Mario Carneiro, 18-Feb-2015.)
𝐷 = (ℂ ∖ (-∞(,]0)) (𝐴𝐷 → ¬ -𝐴 ∈ ℝ+)
Theoremlogdmss 24433 The continuity domain of log is a subset of the regular domain of log. (Contributed by Mario Carneiro, 1-Mar-2015.)
𝐷 = (ℂ ∖ (-∞(,]0)) 𝐷 ⊆ (ℂ ∖ {0})
Theoremlogcnlem2 24434 Lemma for logcn 24438. (Contributed by Mario Carneiro, 25-Feb-2015.)
𝐷 = (ℂ ∖ (-∞(,]0)) & 𝑆 = if(𝐴 ∈ ℝ+, 𝐴, (abs‘(ℑ‘𝐴))) & 𝑇 = ((abs‘𝐴) · (𝑅 / (1 + 𝑅))) & (𝜑𝐴𝐷) & (𝜑𝑅 ∈ ℝ+) (𝜑 → if(𝑆𝑇, 𝑆, 𝑇) ∈ ℝ+)
Theoremlogcnlem3 24435 Lemma for logcn 24438. (Contributed by Mario Carneiro, 25-Feb-2015.)
𝐷 = (ℂ ∖ (-∞(,]0)) & 𝑆 = if(𝐴 ∈ ℝ+, 𝐴, (abs‘(ℑ‘𝐴))) & 𝑇 = ((abs‘𝐴) · (𝑅 / (1 + 𝑅))) & (𝜑𝐴𝐷) & (𝜑𝑅 ∈ ℝ+) & (𝜑𝐵𝐷) & (𝜑 → (abs‘(𝐴𝐵)) < if(𝑆𝑇, 𝑆, 𝑇)) (𝜑 → (-π < ((ℑ‘(log‘𝐵)) − (ℑ‘(log‘𝐴))) ∧ ((ℑ‘(log‘𝐵)) − (ℑ‘(log‘𝐴))) ≤ π))
Theoremlogcnlem4 24436 Lemma for logcn 24438. (Contributed by Mario Carneiro, 25-Feb-2015.)
𝐷 = (ℂ ∖ (-∞(,]0)) & 𝑆 = if(𝐴 ∈ ℝ+, 𝐴, (abs‘(ℑ‘𝐴))) & 𝑇 = ((abs‘𝐴) · (𝑅 / (1 + 𝑅))) & (𝜑𝐴𝐷) & (𝜑𝑅 ∈ ℝ+) & (𝜑𝐵𝐷) & (𝜑 → (abs‘(𝐴𝐵)) < if(𝑆𝑇, 𝑆, 𝑇)) (𝜑 → (abs‘((ℑ‘(log‘𝐴)) − (ℑ‘(log‘𝐵)))) < 𝑅)
Theoremlogcnlem5 24437* Lemma for logcn 24438. (Contributed by Mario Carneiro, 18-Feb-2015.)
𝐷 = (ℂ ∖ (-∞(,]0)) (𝑥𝐷 ↦ (ℑ‘(log‘𝑥))) ∈ (𝐷cn→ℝ)
Theoremlogcn 24438 The logarithm function is continuous away from the branch cut at negative reals. (Contributed by Mario Carneiro, 25-Feb-2015.)
𝐷 = (ℂ ∖ (-∞(,]0)) (log ↾ 𝐷) ∈ (𝐷cn→ℂ)
Theoremdvloglem 24439 Lemma for dvlog 24442. (Contributed by Mario Carneiro, 24-Feb-2015.)
𝐷 = (ℂ ∖ (-∞(,]0)) (log “ 𝐷) ∈ (TopOpen‘ℂfld)
Theoremlogdmopn 24440 The "continuous domain" of log is an open set. (Contributed by Mario Carneiro, 7-Apr-2015.)
𝐷 = (ℂ ∖ (-∞(,]0)) 𝐷 ∈ (TopOpen‘ℂfld)
Theoremlogf1o2 24441 The logarithm maps its continuous domain bijectively onto the set of numbers with imaginary part -π < ℑ(𝑧) < π. The negative reals are mapped to the numbers with imaginary part equal to π. (Contributed by Mario Carneiro, 2-May-2015.)
𝐷 = (ℂ ∖ (-∞(,]0)) (log ↾ 𝐷):𝐷1-1-onto→(ℑ “ (-π(,)π))
Theoremdvlog 24442* The derivative of the complex logarithm function. (Contributed by Mario Carneiro, 25-Feb-2015.)
𝐷 = (ℂ ∖ (-∞(,]0)) (ℂ D (log ↾ 𝐷)) = (𝑥𝐷 ↦ (1 / 𝑥))
Theoremdvlog2lem 24443 Lemma for dvlog2 24444. (Contributed by Mario Carneiro, 1-Mar-2015.)
𝑆 = (1(ball‘(abs ∘ − ))1) 𝑆 ⊆ (ℂ ∖ (-∞(,]0))
Theoremdvlog2 24444* The derivative of the complex logarithm function on the open unit ball centered at 1, a sometimes easier region to work with than the ℂ ∖ (-∞, 0] of dvlog 24442. (Contributed by Mario Carneiro, 1-Mar-2015.)
𝑆 = (1(ball‘(abs ∘ − ))1) (ℂ D (log ↾ 𝑆)) = (𝑥𝑆 ↦ (1 / 𝑥))
Theoremadvlog 24445 The antiderivative of the logarithm. (Contributed by Mario Carneiro, 21-May-2016.)
(ℝ D (𝑥 ∈ ℝ+ ↦ (𝑥 · ((log‘𝑥) − 1)))) = (𝑥 ∈ ℝ+ ↦ (log‘𝑥))
Theoremadvlogexp 24446* The antiderivative of a power of the logarithm. (Set 𝐴 = 1 and multiply by (-1)↑𝑁 · 𝑁! to get the antiderivative of log(𝑥)↑𝑁 itself.) (Contributed by Mario Carneiro, 22-May-2016.)
((𝐴 ∈ ℝ+𝑁 ∈ ℕ0) → (ℝ D (𝑥 ∈ ℝ+ ↦ (𝑥 · Σ𝑘 ∈ (0...𝑁)(((log‘(𝐴 / 𝑥))↑𝑘) / (!‘𝑘))))) = (𝑥 ∈ ℝ+ ↦ (((log‘(𝐴 / 𝑥))↑𝑁) / (!‘𝑁))))
Theoremefopnlem1 24447 Lemma for efopn 24449. (Contributed by Mario Carneiro, 23-Apr-2015.) (Revised by Mario Carneiro, 8-Sep-2015.)
(((𝑅 ∈ ℝ+𝑅 < π) ∧ 𝐴 ∈ (0(ball‘(abs ∘ − ))𝑅)) → (abs‘(ℑ‘𝐴)) < π)
Theoremefopnlem2 24448 Lemma for efopn 24449. (Contributed by Mario Carneiro, 2-May-2015.)
𝐽 = (TopOpen‘ℂfld) ((𝑅 ∈ ℝ+𝑅 < π) → (exp “ (0(ball‘(abs ∘ − ))𝑅)) ∈ 𝐽)
Theoremefopn 24449 The exponential map is an open map. (Contributed by Mario Carneiro, 23-Apr-2015.)
𝐽 = (TopOpen‘ℂfld) (𝑆𝐽 → (exp “ 𝑆) ∈ 𝐽)
Theoremlogtayllem 24450* Lemma for logtayl 24451. (Contributed by Mario Carneiro, 1-Apr-2015.)
((𝐴 ∈ ℂ ∧ (abs‘𝐴) < 1) → seq0( + , (𝑛 ∈ ℕ0 ↦ (if(𝑛 = 0, 0, (1 / 𝑛)) · (𝐴𝑛)))) ∈ dom ⇝ )
Theoremlogtayl 24451* The Taylor series for -log(1 − 𝐴). (Contributed by Mario Carneiro, 1-Apr-2015.)
((𝐴 ∈ ℂ ∧ (abs‘𝐴) < 1) → seq1( + , (𝑘 ∈ ℕ ↦ ((𝐴𝑘) / 𝑘))) ⇝ -(log‘(1 − 𝐴)))
Theoremlogtaylsum 24452* The Taylor series for -log(1 − 𝐴), as an infinite sum. (Contributed by Mario Carneiro, 31-Mar-2015.)
((𝐴 ∈ ℂ ∧ (abs‘𝐴) < 1) → Σ𝑘 ∈ ℕ ((𝐴𝑘) / 𝑘) = -(log‘(1 − 𝐴)))
Theoremlogtayl2 24453* Power series expression for the logarithm. (Contributed by Mario Carneiro, 31-Mar-2015.)
𝑆 = (1(ball‘(abs ∘ − ))1) (𝐴𝑆 → seq1( + , (𝑘 ∈ ℕ ↦ (((-1↑(𝑘 − 1)) / 𝑘) · ((𝐴 − 1)↑𝑘)))) ⇝ (log‘𝐴))
Theoremlogccv 24454 The natural logarithm function on the reals is a strictly concave function. (Contributed by Mario Carneiro, 20-Jun-2015.)
(((𝐴 ∈ ℝ+𝐵 ∈ ℝ+𝐴 < 𝐵) ∧ 𝑇 ∈ (0(,)1)) → ((𝑇 · (log‘𝐴)) + ((1 − 𝑇) · (log‘𝐵))) < (log‘((𝑇 · 𝐴) + ((1 − 𝑇) · 𝐵))))
Theoremcxpval 24455 Value of the complex power function. (Contributed by Mario Carneiro, 2-Aug-2014.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (𝐴𝑐𝐵) = if(𝐴 = 0, if(𝐵 = 0, 1, 0), (exp‘(𝐵 · (log‘𝐴)))))
Theoremcxpef 24456 Value of the complex power function. (Contributed by Mario Carneiro, 2-Aug-2014.)
((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ∧ 𝐵 ∈ ℂ) → (𝐴𝑐𝐵) = (exp‘(𝐵 · (log‘𝐴))))
Theorem0cxp 24457 Value of the complex power function when the first argument is zero. (Contributed by Mario Carneiro, 2-Aug-2014.)
((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0) → (0↑𝑐𝐴) = 0)
Theoremcxpexpz 24458 Relate the complex power function to the integer power function. (Contributed by Mario Carneiro, 2-Aug-2014.)
((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ∧ 𝐵 ∈ ℤ) → (𝐴𝑐𝐵) = (𝐴𝐵))
Theoremcxpexp 24459 Relate the complex power function to the integer power function. (Contributed by Mario Carneiro, 2-Aug-2014.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℕ0) → (𝐴𝑐𝐵) = (𝐴𝐵))
Theoremlogcxp 24460 Logarithm of a complex power. (Contributed by Mario Carneiro, 2-Aug-2014.)
((𝐴 ∈ ℝ+𝐵 ∈ ℝ) → (log‘(𝐴𝑐𝐵)) = (𝐵 · (log‘𝐴)))
Theoremcxp0 24461 Value of the complex power function when the second argument is zero. (Contributed by Mario Carneiro, 2-Aug-2014.)
(𝐴 ∈ ℂ → (𝐴𝑐0) = 1)
Theoremcxp1 24462 Value of the complex power function at one. (Contributed by Mario Carneiro, 2-Aug-2014.)
(𝐴 ∈ ℂ → (𝐴𝑐1) = 𝐴)
Theorem1cxp 24463 Value of the complex power function at one. (Contributed by Mario Carneiro, 2-Aug-2014.)
(𝐴 ∈ ℂ → (1↑𝑐𝐴) = 1)
Theoremecxp 24464 Write the exponential function as an exponent to the power e. (Contributed by Mario Carneiro, 2-Aug-2014.)
(𝐴 ∈ ℂ → (e↑𝑐𝐴) = (exp‘𝐴))
Theoremcxpcl 24465 Closure of the complex power function. (Contributed by Mario Carneiro, 2-Aug-2014.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (𝐴𝑐𝐵) ∈ ℂ)
Theoremrecxpcl 24466 Real closure of the complex power function. (Contributed by Mario Carneiro, 2-Aug-2014.)
((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴𝐵 ∈ ℝ) → (𝐴𝑐𝐵) ∈ ℝ)
Theoremrpcxpcl 24467 Positive real closure of the complex power function. (Contributed by Mario Carneiro, 2-Aug-2014.)
((𝐴 ∈ ℝ+𝐵 ∈ ℝ) → (𝐴𝑐𝐵) ∈ ℝ+)
Theoremcxpne0 24468 Complex exponentiation is nonzero if its mantissa is nonzero. (Contributed by Mario Carneiro, 2-Aug-2014.)
((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ∧ 𝐵 ∈ ℂ) → (𝐴𝑐𝐵) ≠ 0)
Theoremcxpeq0 24469 Complex exponentiation is zero iff the mantissa is zero and the exponent is nonzero. (Contributed by Mario Carneiro, 23-Apr-2015.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((𝐴𝑐𝐵) = 0 ↔ (𝐴 = 0 ∧ 𝐵 ≠ 0)))
Theoremcxpadd 24470 Sum of exponents law for complex exponentiation. Proposition 10-4.2(a) of [Gleason] p. 135. (Contributed by Mario Carneiro, 2-Aug-2014.)
(((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0) ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) → (𝐴𝑐(𝐵 + 𝐶)) = ((𝐴𝑐𝐵) · (𝐴𝑐𝐶)))
Theoremcxpp1 24471 Value of a nonzero complex number raised to a complex power plus one. (Contributed by Mario Carneiro, 2-Aug-2014.)
((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ∧ 𝐵 ∈ ℂ) → (𝐴𝑐(𝐵 + 1)) = ((𝐴𝑐𝐵) · 𝐴))
Theoremcxpneg 24472 Value of a complex number raised to a negative power. (Contributed by Mario Carneiro, 2-Aug-2014.)
((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ∧ 𝐵 ∈ ℂ) → (𝐴𝑐-𝐵) = (1 / (𝐴𝑐𝐵)))
Theoremcxpsub 24473 Exponent subtraction law for complex exponentiation. (Contributed by Mario Carneiro, 22-Sep-2014.)
(((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0) ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) → (𝐴𝑐(𝐵𝐶)) = ((𝐴𝑐𝐵) / (𝐴𝑐𝐶)))
Theoremcxpge0 24474 Nonnegative exponentiation with a real exponent is nonnegative. (Contributed by Mario Carneiro, 2-Aug-2014.)
((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴𝐵 ∈ ℝ) → 0 ≤ (𝐴𝑐𝐵))
Theoremmulcxplem 24475 Lemma for mulcxp 24476. (Contributed by Mario Carneiro, 2-Aug-2014.)
(𝜑𝐴 ∈ ℂ) & (𝜑𝐶 ∈ ℂ) (𝜑 → (0↑𝑐𝐶) = ((𝐴𝑐𝐶) · (0↑𝑐𝐶)))
Theoremmulcxp 24476 Complex exponentiation of a product. Proposition 10-4.2(c) of [Gleason] p. 135. (Contributed by Mario Carneiro, 2-Aug-2014.)
(((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) → ((𝐴 · 𝐵)↑𝑐𝐶) = ((𝐴𝑐𝐶) · (𝐵𝑐𝐶)))
Theoremcxprec 24477 Complex exponentiation of a reciprocal. (Contributed by Mario Carneiro, 2-Aug-2014.)
((𝐴 ∈ ℝ+𝐵 ∈ ℂ) → ((1 / 𝐴)↑𝑐𝐵) = (1 / (𝐴𝑐𝐵)))
Theoremdivcxp 24478 Complex exponentiation of a quotient. (Contributed by Mario Carneiro, 8-Sep-2014.)
(((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ 𝐵 ∈ ℝ+𝐶 ∈ ℂ) → ((𝐴 / 𝐵)↑𝑐𝐶) = ((𝐴𝑐𝐶) / (𝐵𝑐𝐶)))
Theoremcxpmul 24479 Product of exponents law for complex exponentiation. Proposition 10-4.2(b) of [Gleason] p. 135. (Contributed by Mario Carneiro, 2-Aug-2014.)
((𝐴 ∈ ℝ+𝐵 ∈ ℝ ∧ 𝐶 ∈ ℂ) → (𝐴𝑐(𝐵 · 𝐶)) = ((𝐴𝑐𝐵)↑𝑐𝐶))
Theoremcxpmul2 24480 Product of exponents law for complex exponentiation. Variation on cxpmul 24479 with more general conditions on 𝐴 and 𝐵 when 𝐶 is an integer. (Contributed by Mario Carneiro, 9-Aug-2014.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℕ0) → (𝐴𝑐(𝐵 · 𝐶)) = ((𝐴𝑐𝐵)↑𝐶))
Theoremcxproot 24481 The complex power function allows us to write n-th roots via the idiom 𝐴𝑐(1 / 𝑁). (Contributed by Mario Carneiro, 6-May-2015.)
((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ) → ((𝐴𝑐(1 / 𝑁))↑𝑁) = 𝐴)
Theoremcxpmul2z 24482 Generalize cxpmul2 24480 to negative integers. (Contributed by Mario Carneiro, 23-Apr-2015.)
(((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0) ∧ (𝐵 ∈ ℂ ∧ 𝐶 ∈ ℤ)) → (𝐴𝑐(𝐵 · 𝐶)) = ((𝐴𝑐𝐵)↑𝐶))
Theoremabscxp 24483 Absolute value of a power, when the base is real. (Contributed by Mario Carneiro, 15-Sep-2014.)
((𝐴 ∈ ℝ+𝐵 ∈ ℂ) → (abs‘(𝐴𝑐𝐵)) = (𝐴𝑐(ℜ‘𝐵)))
Theoremabscxp2 24484 Absolute value of a power, when the exponent is real. (Contributed by Mario Carneiro, 15-Sep-2014.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℝ) → (abs‘(𝐴𝑐𝐵)) = ((abs‘𝐴)↑𝑐𝐵))
Theoremcxplt 24485 Ordering property for complex exponentiation. (Contributed by Mario Carneiro, 2-Aug-2014.)
(((𝐴 ∈ ℝ ∧ 1 < 𝐴) ∧ (𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ)) → (𝐵 < 𝐶 ↔ (𝐴𝑐𝐵) < (𝐴𝑐𝐶)))
Theoremcxple 24486 Ordering property for complex exponentiation. (Contributed by Mario Carneiro, 2-Aug-2014.)
(((𝐴 ∈ ℝ ∧ 1 < 𝐴) ∧ (𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ)) → (𝐵𝐶 ↔ (𝐴𝑐𝐵) ≤ (𝐴𝑐𝐶)))
Theoremcxplea 24487 Ordering property for complex exponentiation. (Contributed by Mario Carneiro, 10-Sep-2014.)
(((𝐴 ∈ ℝ ∧ 1 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ) ∧ 𝐵𝐶) → (𝐴𝑐𝐵) ≤ (𝐴𝑐𝐶))
Theoremcxple2 24488 Ordering property for complex exponentiation. (Contributed by Mario Carneiro, 8-Sep-2014.)
(((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℝ+) → (𝐴𝐵 ↔ (𝐴𝑐𝐶) ≤ (𝐵𝑐𝐶)))
Theoremcxplt2 24489 Ordering property for complex exponentiation. (Contributed by Mario Carneiro, 15-Sep-2014.)
(((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℝ+) → (𝐴 < 𝐵 ↔ (𝐴𝑐𝐶) < (𝐵𝑐𝐶)))
Theoremcxple2a 24490 Ordering property for complex exponentiation. (Contributed by Mario Carneiro, 15-Sep-2014.)
(((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ) ∧ (0 ≤ 𝐴 ∧ 0 ≤ 𝐶) ∧ 𝐴𝐵) → (𝐴𝑐𝐶) ≤ (𝐵𝑐𝐶))
Theoremcxplt3 24491 Ordering property for complex exponentiation. (Contributed by Mario Carneiro, 2-May-2016.)
(((𝐴 ∈ ℝ+𝐴 < 1) ∧ (𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ)) → (𝐵 < 𝐶 ↔ (𝐴𝑐𝐶) < (𝐴𝑐𝐵)))
Theoremcxple3 24492 Ordering property for complex exponentiation. (Contributed by Mario Carneiro, 2-May-2016.)
(((𝐴 ∈ ℝ+𝐴 < 1) ∧ (𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ)) → (𝐵𝐶 ↔ (𝐴𝑐𝐶) ≤ (𝐴𝑐𝐵)))
Theoremcxpsqrtlem 24493 Lemma for cxpsqrt 24494. (Contributed by Mario Carneiro, 2-Aug-2014.)
(((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0) ∧ (𝐴𝑐(1 / 2)) = -(√‘𝐴)) → (i · (√‘𝐴)) ∈ ℝ)
Theoremcxpsqrt 24494 The complex exponential function with exponent 1 / 2 exactly matches the complex square root function (the branch cut is in the same place for both functions), and thus serves as a suitable generalization to other 𝑛-th roots and irrational roots. (Contributed by Mario Carneiro, 2-Aug-2014.)
(𝐴 ∈ ℂ → (𝐴𝑐(1 / 2)) = (√‘𝐴))
Theoremlogsqrt 24495 Logarithm of a square root. (Contributed by Mario Carneiro, 5-May-2016.)
(𝐴 ∈ ℝ+ → (log‘(√‘𝐴)) = ((log‘𝐴) / 2))
Theoremcxp0d 24496 Value of the complex power function when the second argument is zero. (Contributed by Mario Carneiro, 30-May-2016.)
(𝜑𝐴 ∈ ℂ) (𝜑 → (𝐴𝑐0) = 1)
Theoremcxp1d 24497 Value of the complex power function at one. (Contributed by Mario Carneiro, 30-May-2016.)
(𝜑𝐴 ∈ ℂ) (𝜑 → (𝐴𝑐1) = 𝐴)
Theorem1cxpd 24498 Value of the complex power function at one. (Contributed by Mario Carneiro, 30-May-2016.)
(𝜑𝐴 ∈ ℂ) (𝜑 → (1↑𝑐𝐴) = 1)
Theoremcxpcld 24499 Closure of the complex power function. (Contributed by Mario Carneiro, 30-May-2016.)
(𝜑𝐴 ∈ ℂ) & (𝜑𝐵 ∈ ℂ) (𝜑 → (𝐴𝑐𝐵) ∈ ℂ)
Theoremcxpmul2d 24500 Product of exponents law for complex exponentiation. Variation on cxpmul 24479 with more general conditions on 𝐴 and 𝐵 when 𝐶 is an integer. (Contributed by Mario Carneiro, 30-May-2016.)
(𝜑𝐴 ∈ ℂ) & (𝜑𝐵 ∈ ℂ) & (𝜑𝐶 ∈ ℕ0) (𝜑 → (𝐴𝑐(𝐵 · 𝐶)) = ((𝐴𝑐𝐵)↑𝐶))
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# Statistics for Behavior Sciences and Study Guide - 6th edition
ISBN13: 978-0534078164
ISBN10: 0534078168
Summary: Gravetter and Wallnau's proven best-seller offers the straightforward instruction, accuracy, built-in learning aids, and wealth of real-world examples that professors AND students have come to appreciate. The authors integrate applications to ensure that even students with a weak background in mathematics can achieve mastery of statistical concepts. They skillfully demonstrate that having an understanding of statistical procedures will help them not only understand p
ublished findings, but also become savvy consumers of information. Known for its exceptional accuracy and examples, this text also has a complete supplements package to support instructors with class preparation and testing.
Benefits :
• NEW! The measurement of EFFECT SIZE is now included throughout the inferential statistics chapters (8, 9, 10, 11, 13, 14, 15, 16, and 17).
• NEW! Extraordinary support for new instructors, adjuncts or part-time instructors! The Instructor's Manual now includes a section on "what to do and what not to do", teaching tips for class, lecture outlines, and a sample syllabi.
• Each chapter begins with an "Overview" and ends with "Summary," "Focus on Problem Solving," "Demonstrations," and "Problems" sections to give students the opportunity to assimilate the material and work through sample problems.
• Multiple end of chapter problems provides students with the practice they need.
• Numerous learning checks in every chapter help students test their understanding before exam time.
• Statistical formulas are presented in both standard mathematical notation and in everyday language, with explanations of how and why formulas are used.
• "In the Literature" sections appear in nearly every chapter. These demonstrate how statistical results are reported in APA style and explain the notation and jargon used.
• A "Statistics Organizer," found at the end of the text, provides a decision tree that guides students to the appropriate statistical procedure within the text.
• NEW! The sample mean is now identified by M (in accordance with APA guidelines), instead of the older symbol X-bar (an X with an overbar).
• NEW! Since modifying the unit normal table several editions ago, there have been requests to go back to the original table. This edition addresses those requests by incorporating the missing piece from the old table into the new table. The table now has a four-column format (instead of three), providing the functionality of both the old and new versions.
• NEW! Discussions of the relationship between samples and populations, as well as coverage of the concept of sampling error, has been clarified. New figures have been included to illustrate these concepts (Ch. 1).
• NEW! To simplify the discussion of biased and unbiased statistics, this discussion now focuses on sample variance instead of sample standard deviation (Ch. 4).
• NEW! A new section demonstrates how z-scores can be used as inferential statistics to help evaluate the outcome of a research study (Ch. 5).
• NEW! Expanded coverage of the relationship between standard error and sample size, as well as a new section that illustrates how standard error is used in the context of inferential statistics, complements the discussion of standard error. (Ch. 7).
• NEW! WebTutor is now available for this text. WebTutor provides a dynamic, online teaching and learning environment that instructors can customize to meet the needs of their course.
• NEW! A new Appendix directs students to appropriate Web resources.
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Cover:
Year Published: 2004
International: No | 1,427 | 7,206 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-27 | longest | en | 0.931522 |
https://www.physicsforums.com/threads/this-integral-of-k-dt-needed-to-get-this-mark-m1.158863/ | 1,696,223,226,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510967.73/warc/CC-MAIN-20231002033129-20231002063129-00481.warc.gz | 997,220,180 | 17,913 | # This integral of k dt needed to get this mark-M1?
• inv
#### inv
[Solved]This integral of k dt needed to get this mark-M1?
## Homework Statement
Hi.The material,the question and the marking scheme respectively for my pure maths paper 3 question x1 only is shown below,it's an A2-level Oct 2002 paper if u know what that is.
Is the integral circled in red needed to integrate?I find I don't need it.Am I wrong,if so why?
## Homework Equations
Simple integration rules only.
## The Attempt at a Solution
I did attempt by doing :
Integral of dt/da = integral of (.1 a^-1 + .1(10-a)^-1)
= .1 ln(a/(10-a))
Ps- there, I did it without the integral circled in red.
Last edited:
You did the integral, but what's the solution to the problem? I.e. What is a as a function of t? I don't see any t dependence in your answer.
You did the integral, but what's the solution to the problem? I.e. What is a as a function of t? I don't see any t dependence in your answer.
The question is asking for t to be expressed in t=ca+d ,where a is a variable and the rest is constant.
Solution:
dt/da=250 {a(10-a)}^-1
t=integral of dt/da
=what I integrated
There's the t dependence in my answer too.
Last edited:
Think clearly about this. If you believe you solved it, you ought to be able to tell me the value of a at t=1. How do you do that?
Think clearly about this. If you believe you solved it, you ought to be able to tell me the value of a at t=1. How do you do that?
Previous post I said "what I integrated" is on my hard copy. This is what it is-
t=25ln{a/(10-a)} .
I'm afraid if t=1 , I don't know how to find a.
Last edited:
Ok. If you think about it, you did integrate k*dt. That's where the 250 came from in your expression. k=1/250. The integration is pretty trivial, but in solving equations like this it can easily come out to be not so. That's why it's written this way. The LHS could have been f(t)*dt for some complicated function f.
What do you mean
could have been
?
I'm saying that if you do other differential equations by separation of variables, the 't' integration can be just as hard as the 'a' integration after you've separated the equation into f(t)*dt=g(a)*da. In this case the t dependence was so simple you didn't notice that you had integrated it.
U're implying that for every integration question like this on my paper 3 pure maths I need to specify the integral of $#%$ da = integral of k*dt like this case?
Edit :"$#%$" there wasn't a bad word ,I meant whatever the equation may be type "$#%$" da Dick. U're implying that?
Last edited:
Not just for the maths papers. Deciding on the integrals is the next step. You are less likely to make a mistake if you write down each step.
Dick gave good advice.
Getting into the habit of doing this even for a simple case will help you when the going gets more difficult.
Last edited:
Hello Dick,u're implying it ?
U're implying that for every integration question like this on my paper 3 pure maths I need to specify the integral of $#%$ da = integral of k*dt like this case?
You want to solve the equation $$\frac{da}{dt}=ka(10-a)$$. This is a 1st order separable differential equation, and so to solve, we separate the variables so that we obtain $$\int \frac{da}{a(10-a)}=\int kdt$$. The fact that we can separate this equation means that we can solve it by direct integration. Now, in this case, the t integral is trivial; namely the RHS=kt, however there are cases where this will not be so.
For example, suppose that the equation was $$\frac{da}{dt}=ka(10-a)t$$. This is still separable, but now the t integral is non-trivial, and so on separation we obtain $$\int \frac{da}{a(10-a)}=\int kt dt$$. This can still be solved, but now the RHS = k/2 t2.
The point that Dick was trying to make, and that the solution makes, is that you should get into the habit of separating the equation properly before integrating since, in general, the t integral is non-trivial.
Does this make any more sense?
edit: and yes, for the sake of the exam, I would say you need to include the separation step, since there will probably be a mark for noting that the equation is separable. (In fact, the "M1" in the margin indicates that there is a method mark for separating the equation)
Last edited:
you need to integrate kdt to have an equation! If you only integrate the left side, you would have (1/10)ln(a/(10-a)) NOT equal to anything! Where would you go from there? As it is, you have (1/10)ln(a/(10-a))= kt+ C or ln(a/(10-a))= 10kt+ C which can be rewritten a/(10-a)= Ce^(10kt). | 1,183 | 4,560 | {"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} | 3.78125 | 4 | CC-MAIN-2023-40 | latest | en | 0.95518 |
https://www.systutorials.com/docs/linux/man/3-CPOTF2/ | 1,717,027,321,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059412.27/warc/CC-MAIN-20240529230852-20240530020852-00796.warc.gz | 873,580,098 | 3,757 | # CPOTF2 (3) - Linux Manuals
cpotf2.f -
## SYNOPSIS
### Functions/Subroutines
subroutine cpotf2 (UPLO, N, A, LDA, INFO)
CPOTF2 computes the Cholesky factorization of a symmetric/Hermitian positive definite matrix (unblocked algorithm).
## Function/Subroutine Documentation
### subroutine cpotf2 (characterUPLO, integerN, complex, dimension( lda, * )A, integerLDA, integerINFO)
CPOTF2 computes the Cholesky factorization of a symmetric/Hermitian positive definite matrix (unblocked algorithm).
Purpose:
``` CPOTF2 computes the Cholesky factorization of a complex Hermitian
positive definite matrix A.
The factorization has the form
A = U**H * U , if UPLO = 'U', or
A = L * L**H, if UPLO = 'L',
where U is an upper triangular matrix and L is lower triangular.
This is the unblocked version of the algorithm, calling Level 2 BLAS.
```
Parameters:
UPLO
``` UPLO is CHARACTER*1
Specifies whether the upper or lower triangular part of the
Hermitian matrix A is stored.
= 'U': Upper triangular
= 'L': Lower triangular
```
N
``` N is INTEGER
The order of the matrix A. N >= 0.
```
A
``` A is COMPLEX array, dimension (LDA,N)
On entry, the Hermitian matrix A. If UPLO = 'U', the leading
n by n upper triangular part of A contains the upper
triangular part of the matrix A, and the strictly lower
triangular part of A is not referenced. If UPLO = 'L', the
leading n by n lower triangular part of A contains the lower
triangular part of the matrix A, and the strictly upper
triangular part of A is not referenced.
On exit, if INFO = 0, the factor U or L from the Cholesky
factorization A = U**H *U or A = L*L**H.
```
LDA
``` LDA is INTEGER
The leading dimension of the array A. LDA >= max(1,N).
```
INFO
``` INFO is INTEGER
= 0: successful exit
< 0: if INFO = -k, the k-th argument had an illegal value
> 0: if INFO = k, the leading minor of order k is not
positive definite, and the factorization could not be
completed.
```
Author:
Univ. of Tennessee
Univ. of California Berkeley
Univ. of Colorado Denver
NAG Ltd.
Date:
September 2012
Definition at line 110 of file cpotf2.f.
## Author
Generated automatically by Doxygen for LAPACK from the source code. | 625 | 2,232 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-22 | latest | en | 0.817963 |
https://it.mathworks.com/matlabcentral/answers/2012287-enter-values-from-0-to-n-imposed-value-with-step-of-100-on-the-x-axis-of-the-bar-graph?s_tid=prof_contriblnk | 1,716,073,161,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057516.1/warc/CC-MAIN-20240518214304-20240519004304-00728.warc.gz | 284,492,837 | 29,944 | # enter values from 0 to N (imposed value) with step of 100 on the x-axis of the bar graph
2 visualizzazioni (ultimi 30 giorni)
Alberto Acri il 24 Ago 2023
Commentato: Mathieu NOE il 24 Ago 2023
Hi. I need to change the values on the x-axis of a bar graph.
I have tried in the following way but it doesn't seem to work. Why? How can I enter values from 0 to 700 (imposed value) with step of 100 on the x-axis of the bar graph?
matrix = [64 440; 65 489; 66 581; 67 563];
figure
hbh = barh(matrix(:,1), matrix(:,2));
N = 700;
step = 100;
vector_x_axis = 0:step:N;
xticks(vector_x_axis)
##### 0 CommentiMostra -2 commenti meno recentiNascondi -2 commenti meno recenti
Accedi per commentare.
### Risposta accettata
Mathieu NOE il 24 Ago 2023
hello Alberto
welcome back !
i suppose you wanted to do this
matrix = [64 440; 65 489; 66 581; 67 563];
figure
hbh = barh(matrix(:,1), matrix(:,2));
N = 700;
% step = 100;
% vector_x_axis = 0:step:N;
% xticks(vector_x_axis)
xlim([0 N])
##### 4 CommentiMostra 2 commenti meno recentiNascondi 2 commenti meno recenti
Alberto Acri il 24 Ago 2023
thanks!
Mathieu NOE il 24 Ago 2023
my pleasure !
Accedi per commentare.
### Più risposte (1)
Kevin Holly il 24 Ago 2023
I'm not exactly sure what you want to see. Are one of the 3 graphs below what you want?
matrix = [64, 440; 65, 489; 66, 581; 67, 563];
N = 700;
step = 100;
vector_x_axis = 0:step:N;
figure
hbh = barh(matrix(:, 1), matrix(:, 2));
xlim([0 700])
figure
hbh = barh(matrix(:, 1), matrix(:, 2));
xticks(vector_x_axis)
xticklabels(vector_x_axis(2:end))
figure
hbh = barh(matrix(:, 1), matrix(:, 2));
xticks(vector_x_axis)
step = 700/6;
vector_x_axis = 0:step:N;
xticklabels(vector_x_axis)
##### 2 CommentiMostra NessunoNascondi Nessuno
Alberto Acri il 24 Ago 2023
Modificato: Alberto Acri il 24 Ago 2023
The final graph I want to get is the first one you showed me, BUT I want to impose on the x-axis a range taking into account both the final 'N' value (700) and the 'step' value.
For example using this code that has N=700 and step=50:
matrix = [64, 440; 65, 489; 66, 581; 67, 563];
N = 700;
step = 100;
vector_x_axis = 0:step:N;
figure
hbh = barh(matrix(:, 1), matrix(:, 2));
xticklabels(vector_x_axis)
a graph is created where the x-axis ranges from 0 to 300 (and not up to N=700) and with steps of 50
Whereas by modifying only the last line:
xticks(vector_x_axis)
In both graphs, the x-axis does not arrive at the value of 700.
Kevin Holly il 24 Ago 2023
I'm still not quite sure what you want.
matrix = [64, 440; 65, 489; 66, 581; 67, 563];
N = 700;
step = 50;
vector_x_axis = 0:step:N;
figure
hbh = barh(matrix(:, 1), matrix(:, 2));
xticks(vector_x_axis)
figure
hbh = barh(matrix(:, 1), matrix(:, 2));
xticks(vector_x_axis)
xlim([0 700])
Accedi per commentare.
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Translated by | 992 | 3,011 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.612319 |
https://forum.kirupa.com/t/f8-as2-spacing-objects-evenly/203130 | 1,709,227,213,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474852.83/warc/CC-MAIN-20240229170737-20240229200737-00274.warc.gz | 256,363,992 | 5,311 | # [F8 AS2] spacing objects evenly?
I’ve been working on this for over 3 hours and it is about to drive me crazy! I was one of those people who never paid attention to math in high school because i thought i would never use it ( i knew i wanted to be an artist/designer of some sort). But now of course, that attitude is coming back to haunt me.
I’m using a for loop to place several objects on stage. I need a formula that will space objects along the x axis evenly, and center all the objects on the stage. Kinda like if you use the align panel in flash to distibute the vertical centers of several objects to the stage, if that makes sense. If there is only one item, it should be directly in the center of the stage. If there are 2, it should place them in the center of the 2 halves of the stage, and so on for howeve many objects there are. I also want a buffer of about 25 pixels on either edge of the stage so that no object is ever on the very edge of the stage. I made a simple diagram to illustrate:
if some math wiz could help me, I would certainly appreciate it… | 254 | 1,077 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.963684 |
https://cs.stackexchange.com/questions/80297/knight-on-a-chessboard | 1,718,302,980,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861480.87/warc/CC-MAIN-20240613154645-20240613184645-00672.warc.gz | 169,099,264 | 42,195 | Knight on a chessboard
This is a Hackerrank challenge
$Knight$ is a chess piece that moves in an L shape. We define the possible moves of $Knight(a,b)$ as any movement from some position $(x_1, y_1)$ to $(x_2, y_2)$ satisfying either of the following:
$x_2 = x_1 \pm a$ and $y_2 = y_1 \pm b$, or
$x_2 = x_1 \pm b$ and $y_2 = y_1 \pm a$
Given the value of $n$ for a square chessboard, answer the following question for each $(a, b)$ pair where :
$1 \leq a \leq b \lt n$
Constraints
$1 5 \leq n \leq 25$
What is the minimum number of moves it takes for $Knight$ to get from position $(0, 0)$ to position $(n-1, n-1)$? If it's not possible for the Knight to reach that destination, the answer is -1 instead.
If $a = b$ then it is simply $(n-1) / 2a$ if it is evenly divisible
What I cannot figure out is when some of the steps need to be backwards $x_2 \lt x_1$.
Can this be solved with an algorithm other than brute force?
If so what is the algorithm?
Brute force to me is iterate over all possible moves.
If brute force is it safe to assume $a$ and $b$ would never both be negative on a move? Answer is no. If brute force could it be limited to one diagonal or the other? Answer is no.
• What algorithm do you mean by the brute-force? Why not try BFS? Commented Aug 21, 2017 at 17:49
• Brute force would be iterate though all possible moves. I don't know what BFS/DFS is be I will try and find it. Commented Aug 21, 2017 at 17:51
• Breadth first search (BFS) is a graph traversal algorithm which may be used to find the shortest path between two nodes in an unweighted graph. Commented Aug 21, 2017 at 17:53
• @fade2black That is what I was thinking about but I did not have the queue part. I will give it a try. Commented Aug 21, 2017 at 18:02
• This made hot network question Commented Aug 21, 2017 at 21:51
You could use the breadth first search (BFS) algorithm. The knight may move at most to eight cell (from a single position) which means that if each cell is treated as a single node then degree of each node is at most eight and so the number of edges is at most $\frac{8N}{2} = 4N$, where $N = n^2$ is the total number of nodes/cells and $n\times n$ is the size of the chessboard. Since BFS' complexity is $O(V + E)$ your algorithm will run in $O(N+ 4N) = O(N)$.
Warning: do not try to create a graph from the chessboard. It is overkill. You just need to maintain queue and mark all visited cells as you iterate. While the knight is at the position $(i,j)$ just compute all possible moves from this position and push them into the queue if they are not already visited and you haven't yet reached $(n-1, n-1)$.
• OK I think I know how use that to traverse all the move. When I hit $(n-1, n-1)$ how do I know how many moves? Will it be the queue size. How do I know that is the minimum? Commented Aug 21, 2017 at 18:12
• @Paparazzi, Q2: you have at most $25\times 25$ chessboard, so the maximum number of nodes is $625$. Is it too big to store in a queue? Commented Aug 21, 2017 at 18:23
• @Paparazi Q1: Your question is implementation specific. For example, when you store the node/cell in the queue you could store the distance from the (0,0) together with the nodes themselves (as a single structure {node, dist}). So every time you expand a node just set the children's distance plus parent's distance plus one . Commented Aug 21, 2017 at 18:26
• @Paparazzi Q3: BFS always computes the shortest path in an unweighted graph. So, when you reach \$ (n-1,n-1)4 it will be the shortest path. Commented Aug 21, 2017 at 18:27
• Note that being O(n) in the number of cells is O(n^2) for the 'n' in the problem, because a chessboard which is n x n has n^2 cells. Commented Aug 21, 2017 at 22:21
I am going to delete this
Just showing the accepted answer how I got the step count
public static void Visit(int row, int col, int n, Queue<int> queue, bool[] visited)
{
if (row < 0 || col < 0 || row >= n || col >= n)
return;
int position = row * n + col;
if (visited[position])
return;
visited[position] = true;
queue.Enqueue(position);
}
public static int KnightHelper(int n, int i, int j)
{
if (i == j)
{
return ((n - 1) % i == 0) ? (n - 1) / i : -1;
}
//try bfs
bool[] visited = new bool[n * n];
Queue<int> queue = new Queue<int>();
int steps = 0;
queue.Enqueue(0);
while (queue.Count > 0)
{
int count = queue.Count();
for (int m = 0; m < count; m++)
{
int position = queue.Dequeue();
int row = (position / n);
int col = (position % n);
if (row == n - 1 && col == n - 1)
{
// Found solution.
return steps;
}
Visit(row + i, col + j, n, queue, visited);
Visit(row + i, col - j, n, queue, visited);
Visit(row + j, col + i, n, queue, visited);
Visit(row + j, col - i, n, queue, visited);
Visit(row - i, col + j, n, queue, visited);
Visit(row - i, col - j, n, queue, visited);
Visit(row - j, col + i, n, queue, visited);
Visit(row - j, col - i, n, queue, visited);
}
steps++;
}
return -1;
} | 1,471 | 4,906 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-26 | latest | en | 0.928117 |
https://www.esaral.com/q/a-gold-sample-contains-90-of-gold-and-the-rest-copper-45720 | 1,718,574,676,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861671.61/warc/CC-MAIN-20240616203247-20240616233247-00177.warc.gz | 666,966,276 | 11,406 | # A gold sample contains 90% of gold and the rest copper.
Question:
A gold sample contains $90 \%$ of gold and the rest copper. How many atoms of gold are present in one gram of this sample of gold?
Solution:
Mass of pure gold in one gram of the sample $=\frac{1 \times 90}{100}=0.9 \mathrm{~g}$
No. of moles of gold present $=\frac{\text { Mass of gold }}{\text { Molar atomic mass }}=\frac{(0.9 \mathrm{~g})}{\left(0197 \mathrm{~g} \mathrm{~mol}^{-1}\right)}=0.0046 \mathrm{~mol}$
One mole of gold contains atoms $=\mathrm{N}_{\mathrm{A}}=6.022 \times 10^{23}$
$0.0046$ mole of gold contains atoms $=6.022 \times 10^{23} \times 0.0046=2.77 \times 10^{21}$ | 226 | 664 | {"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} | 4.15625 | 4 | CC-MAIN-2024-26 | latest | en | 0.729373 |
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## #1 2012-10-31 14:39:00
Johnathon bresly
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### Four square theorem
Hi,I learnt about the lagrange's four square theorem,now I am curious about it, how do I find then a,b,c,d for n such that,n=a^2+b^2+c^2+d^2?
## #2 2012-10-31 14:57:10
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: Four square theorem
Hi;
As far as I know this is a computational problem. You can get the number of solutions but to determine what they are you will need a computer.
For instance the number of solutions to
with
is 3744
To give you an idea of the complexity of the problem please go here;
http://www.alpertron.com.ar/4SQUARES.HTM
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline | 311 | 1,001 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | latest | en | 0.84028 |
https://math.stackexchange.com/questions/2834518/the-equation-varphin-n-log-21-operatornamegpfn-operatornamegpfn | 1,713,228,280,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817036.4/warc/CC-MAIN-20240416000407-20240416030407-00282.warc.gz | 347,055,403 | 36,549 | # The equation $\varphi(n)=n-\log_2(1+\operatorname{gpf}(n))-\operatorname{gpf}(n)+1$ and Mersenne primes
Let $$n\geq 1$$ an integer, we denote the Euler's totient function as $$\varphi(n)$$ and the greatest prime dividing $$n$$ as $$\operatorname{gpf}(n)$$ (that it the arithmetic function defined in the MathWorld's article Greatest Prime Factor).
I know that is it possible to prove the following claim (I was inspired in [1] to ask a similar question).
Claim. If $$n$$ is an integer of the form $$n=p\cdot(2^p-1)$$ , where $$2^p-1$$ is a Mersenne prime, then satisfies $$\varphi(n)=n-\log_2(1+\operatorname{gpf}(n))-\operatorname{gpf}(n)+1.\tag{1}$$
In this post I ask about if those are the only solutions of the equation $$(1)$$.
Previous equation $$(1)$$ can be written also as $$\left(1+\operatorname{gpf}(n)\right)\cdot 2^{\varphi(n)+\operatorname{gpf}(n)}=2^{n+1}.\tag{2}$$
Question. Prove or disprove* the following conjecture:
If an integer $$N\geq 1$$ satisfies our equation $$\left(1+\operatorname{gpf}(N)\right)\cdot 2^{\varphi(N)+\operatorname{gpf}(N)}=2^{N+1}$$ then $$N=p(2^p-1)$$, where $$2^p-1$$ is a Mersenne prime. Many thanks.
*To disprove it show a counterexample.
## References:
[1] Tripathi, A note on products of primes that differ by a fixed integer, The Fibonacci Quarterly, Vol. 48, No. 2 (MAY, 2010).
First of all, $gpf(N)$ must be a Mersenne-prime because $1+gpf(N)$ must be a power of $2$. Denote this Mersenne prime as $M=2^p-1$ with prime number $p$
This leads to $$p+\varphi(N)+2^p-1=N+1$$ $$p+\varphi(N)+M=N+1$$
$M^2|N$ would imply $M|\varphi(n)$ , hence we would have $p \equiv 1\mod M$ , but because of $p<2^p$, we have $0<p-1<2^p-1$, hence $M$ cannot divide $p-1$.
Hence we have $N=kM$ with $\gcd(k,M)=1$ , so we get $$p+\varphi(k)(M-1)+M=kM+1$$
which is equivalent to $$p-(\varphi(k)+1)=M(k-1-\varphi(k))$$
We have $k-1\ge \varphi(k)$ with equality if and only if $k$ is prime. If the inequality is strict, the right side is at least $M$ , but the left side smaller than $p$, this is impossible because of $p-1<2^p-1$. Hence $k$ must be a prime, the right side is $0$ and we have $p=\varphi(k)+1=k-1+1=k$. This completes the proof that the conjecture is actually true.
• I've understand all your proof with the exception of the third paragraph (I'm sure that it is right, but I need to read and study it again), any case I believe that the following is your same proof. Proof by contradiction that $\gcd(k,2^p-1)=1$: If we assume $k=(2^p-1)^\alpha\cdot m$ with $\alpha\geq 1$ and $m\geq 1$ integers, and $\gcd(m,(2^p-1))=1$ then $$p+\left((2^p-1)^{\alpha+1}-(2^p-1)^{\alpha}\right)\cdot\varphi(m)+(2^p-1)=(2^p-1)^{\alpha+1}m+1.$$ And dividing by $2^p-1$ it leads again to the absurd $2^p-1\mid p-1$.
• @user243301 The totient function is multiplicative for coprime integers, in particular, if a prime power $p^k$ divides $N$, then $\varphi(p^k)=p^{k-1}\cdot (p-1)$ divides $\varphi(N)$ For $k\ge 2$, we get $p|\varphi(N)$ Jul 1, 2018 at 7:42 | 1,068 | 2,999 | {"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": 15, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-18 | latest | en | 0.759866 |
https://en.academic.ru/dic.nsf/enwiki/28835/Likelihood-ratio_test | 1,582,603,233,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146004.9/warc/CC-MAIN-20200225014941-20200225044941-00554.warc.gz | 351,379,877 | 17,336 | # Likelihood-ratio test
Likelihood-ratio test
The likelihood ratio, often denoted by $Lambda$ (the capital Greek letter lambda), is the ratio of the maximum probability of a result under two different hypotheses. A likelihood-ratio test is a statistical test for making a decision between two hypotheses based on the value of this ratio.
imple versus simple hypotheses
A statistical model is often a parametrized family of probability density functions or probability mass functions $f\left(x; heta\right)$. A simple-vs-simple hypotheses test hypothesises single values of $heta$ under both the null and alternative hypotheses::Note that under either hypothesis, the distribution of the data is fully specified; there are no unknown parameters to estimate. The likelihood ratio test statistic is ( [Cox, D. R. and Hinkley, D. V; "Theoretical Statistics", Chapman and Hall, 1974.] , Page 92)::$Lambda = frac\left\{ f\left(x; heta_A\right) \right\}\left\{ f\left(x; heta_0\right) \right\},$(some references may use the reciprocal as the definition). The likelihood ratio test rejects the null hypothesis $H_0$ if the ratio exceeds a critical value "c". That is, the decision rule has the form:
If $Lambda ge c$ reject $H_0$.
If $Lambda < c$ accept (or don't reject) $H_0$.
The critical value "c" is usually chosen to obtain a specified significance level $alpha$, through the relation:$P_0\left(Lambda ge c\right) = alpha$(if "x" is discrete, some randomization on the boundary may be needed). The Neyman-Pearson lemma states that this likelihood ratio test is the most powerful among all level-$alpha$ tests for this problem.
Definition (maximum likelihood ratio test for composite hypotheses)
A null hypothesis is often stated by saying the parameter $heta$ is in a specified subset $Theta_0$ of the parameter space $Theta$. The likelihood function is $L\left( heta\right) = L\left( heta|x\right) = p\left(x| heta\right) = f_\left\{ heta\right\}\left(x\right)$ is a function of the parameter $heta$ with $x$ held fixed at the value that was actually observed, "i.e.", the data. The likelihood ratio is
: $Lambda\left(x\right)=frac\left\{sup\left\{,L\left( hetamid x\right): hetainTheta_0,\left\{sup\left\{,L\left( hetamid x\right): hetainTheta,.$
Many common test statistics such as the Z-test, the F-test, Pearson's chi-square test and the G-test can be phrased as log-likelihood ratios or approximations thereof.
Interpretation
Being a function of the data $x$, the LR is therefore a statistic. The likelihood-ratio test rejects the null hypothesis if the value of this statistic is too small. How small is too small depends on the significance level of the test, "i.e.", on what probability of Type I error is considered tolerable ("Type I" errors consist of the rejection of a null hypothesis that is true).
The numerator corresponds to the maximum probability of an observed result under the null hypothesis. The denominator corresponds to the maximum probability of an observed result under the alternative hypothesis. Under certain regularity conditions, the numerator of this ratio is less than the denominator. The likelihood ratio under those conditions is between 0 and 1. Lower values of the likelihood ratio mean that the observed result was less likely to occur under the null hypothesis. Higher values mean that the observed result was more likely to occur under the null hypothesis.
Approximation
If the distribution of the likelihood ratio corresponding to a particular null and alternative hypothesis can be explicitly determined then it can directly be used to form decision regions (to accept/reject the null hypothesis). In most cases, however, the exact distribution of the likelihood ratio corresponding to specific hypotheses is very difficult to determine. A convenient result, though, says that as the sample size $n$ approaches $infty$, the test statistic $-2 log\left(Lambda\right)$ will be asymptotically $chi^2$ distributed with degrees of freedom equal to the difference in dimensionality of $Theta$ and $Theta_0$. This means that for a great variety of hypotheses, a practitioner can take the likelihood ratio $Lambda$, algebraically manipulate $Lambda$ into $-2log\left(Lambda\right)$, compare the value of $-2log\left(Lambda\right)$ given a particular result to the chi squared value corresponding to a desired statistical significance, and create a reasonable decision based on that comparison.
Examples
Medical
One example of a likelihood ratio would be the likelihood that a given test result would be expected in a patient with a certain disorder compared to the likelihood that same result would occur in a patient without the target disorder.
As another example, one can imagine that one is trying to figure out whether one is in line for tickets to a football game or to the opera (assuming that one cannot ask people which line one is in, that one does not see any signs, etcetera). The only thing that one is allowed to do is ask other people in line whether or not they like football. One estimates that 90% of people in the line for a football game like football, while 10% of people in the line for the opera like football. Then the likelihood ratio is computed as:
(Probability of liking football given that someone is in line for football game)/(Probability of liking football given that someone's in line for the opera) = .9/.1 = 9
The larger one's likelihood ratio, the higher the chance that one will be able to correctly infer whether one is at the football game or at the opera given the people's responses. In other words, if one's LR is large, one can be more confident in one's decision as to whether one in line for football tickets or not given that one only asked a limited number of people whether or not they liked football. For an infinite likelihood ratio, one would be 100% sure that one is in line for the football game after only asking one person, who said "yes."
Coin tossing
An example, in the case of Pearson's test, we might try to compare two coins to determine whether they have the same probability of coming up heads. Our observation can be put into a contingency table with rows corresponding to the coin and columns corresponding to heads or tails. The elements of the contingency table will be the number of times the coin for that row came up heads or tails. The contents of this table are our observation $X$.
Heads Tails Coin 1 $k_\left\{1H\right\}$ $k_\left\{1T\right\}$ Coin 2 $k_\left\{2H\right\}$ $k_\left\{2T\right\}$
Here $Theta$ consists of the parameters $p_\left\{1H\right\}$, $p_\left\{1T\right\}$, $p_\left\{2H\right\}$, and $p_\left\{2T\right\}$, which are the probability that coin 1 (2) comes up heads (tails). The hypothesis space $H$ is defined by the usual constraints on a distribution, $p_\left\{ij\right\} ge 0$, $p_\left\{ij\right\} le 1$, and $p_\left\{iH\right\} + p_\left\{iT\right\} = 1$. The null hypothesis $H_0$ is the sub-space where $p_\left\{1j\right\} = p_\left\{2j\right\}$. In all of these constraints, $i = 1,2$ and $j = H,T$.
Writing $n_\left\{ij\right\}$ for the best values for $p_\left\{ij\right\}$ under the hypothesis $H$, maximum likelihood is achieved with:$n_\left\{ij\right\} = frac\left\{k_\left\{ij\left\{k_\left\{iH\right\}+k_\left\{iT$.Writing $m_\left\{ij\right\}$ for the best values for $p_\left\{ij\right\}$ under the null hypothesis $H_0$, maximum likelihood is achieved with:$m_\left\{ij\right\} = frac\left\{k_\left\{1j\right\}+k_\left\{2j\left\{k_\left\{1H\right\}+k_\left\{2H\right\}+k_\left\{1T\right\}+k_\left\{2T$,which does not depend on the coin $i$.
The hypothesis and null hypothesis can be rewritten slightly so that they satisfy the constraints for the logarithm of the likelihood ratio to have the desired nice distribution. Since the constraint causes the two-dimensional $H$ to be reduced to the one-dimensional $H_0$, the asymptotic distribution for the test will be $chi^2\left(1\right)$, the $chi^2$ distribution with one degree of freedom.
For the general contingency table, we can write the log-likelihood ratio statistic as
:$-2 log Lambda = 2sum_\left\{i, j\right\} k_\left\{ij\right\} log frac\left\{n_\left\{ij\left\{m_\left\{ij$.
Criticism
Theoretical
Bayesian criticisms of classical likelihood ratio tests focus on two issues:
#the supremum function in the calculation of the likelihood ratio, saying that this takes no account of the uncertainty about θ and that using maximum likelihood estimates in this way can promote complicated alternative hypotheses with an excessive number of free parameters;
#testing the probability that the sample would produce a result as extreme "or more extreme" under the null hypothesis, saying that this bases the test on the probability of extreme events that did not happen. Instead they put forward methods such as Bayes factors, which explicitly take uncertainty about the parameters into account, and which are based on the evidence that did occur.
Practical
In medicine, the use of likelihood ratio tests has been promoted to assist in interpreting diagnostic testscite journal |author=Jaeschke R, Guyatt GH, Sackett DL |title=Users' guides to the medical literature. III. How to use an article about a diagnostic test. B. What are the results and will they help me in caring for my patients? The Evidence-Based Medicine Working Group |journal=JAMA |volume=271 |issue=9 |pages=703–7 |year=1994 |pmid=8309035 |doi=] . A large likelihood ratio, for example a value more than 10, helps rule in disease. A small likelihood ratio, for example a value less than 0.1, helps rule out diseasecite journal |author=McGee S |title=Simplifying likelihood ratios |journal=Journal of general internal medicine : official journal of the Society for Research and Education in Primary Care Internal Medicine |volume=17 |issue=8 |pages=646–9 |year=2002 |pmid=12213147 |doi=] . However, physicians rarely make these calculationscite journal |author=Reid MC, Lane DA, Feinstein AR |title=Academic calculations versus clinical judgments: practicing physicians' use of quantitative measures of test accuracy |journal=Am. J. Med. |volume=104 |issue=4 |pages=374–80 |year=1998 |pmid=9576412| doi = 10.1016/S0002-9343(98)00054-0 ] and sometimes make errors when they do attempt calculations.cite journal |author=Steurer J, Fischer JE, Bachmann LM, Koller M, ter Riet G |title=Communicating accuracy of tests to general practitioners: a controlled study |journal=BMJ |volume=324 |issue=7341 |pages=824–6 |year=2002 |pmid=11934776| doi = 10.1136/bmj.324.7341.824 ] A randomized controlled trial compared how well physicians interpreted diagnostic tests that were presented as either sensitivity and specificity, a likelihood ratio, or an inexact graphic of the likelihood ratio, found no difference in ability to interpret test results.cite journal |author=Puhan MA, Steurer J, Bachmann LM, ter Riet G |title=A randomized trial of ways to describe test accuracy: the effect on physicians' post-test probability estimates |journal=Ann. Intern. Med. |volume=143 |issue=3 |pages=184–9 |year=2005 |pmid=16061916 |doi=]
References
* Likelihood function
* Score function
* Likelihood principle
* [http://www.itl.nist.gov/div898/handbook/apr/section2/apr233.htm Practical application of Likelihood-ratio test described]
* [http://faculty.vassar.edu/lowry/clin2.html Vassar College's Likelihood Ratio Given Sensitivity/Specifity/Prevalence] Online Calculator
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### Look at other dictionaries:
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• Sequential Probability Ratio Test — Inhaltsverzeichnis 1 Einleitung 2 Geschichte 3 Definition 3.1 Die Entscheidungsgrenzen 4 Beispiel … Deutsch Wikipedia | 3,497 | 14,265 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 62, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2020-10 | latest | en | 0.786439 |
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Question: Explain in your own words what a cube number is?
A cube number, also known as a cubic number, is the result of multiplying a number by itself three times. It's like building a cube out of smaller cubes: if you have a number 'n', imagine a cube made up of 'n' layers, with each layer containing 'n' rows of 'n' cubes. The total number of small cubes that make up the larger cube is the cube number of 'n'. For example, 2 cubed is 8 because 2 x 2 x 2 equals 8. This concept is fundamental in mathematics, especially in geometry and algebra.
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http://kochanski.org/gpk/research/misc/2008/bragging.html | 1,502,991,327,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886103891.56/warc/CC-MAIN-20170817170613-20170817190613-00090.warc.gz | 237,342,724 | 4,543 | Greg Kochanski
# Reporting Confidence Levels
What does it mean when you say that a statistical test was "significant at the P < 0.05 level"? Oddly enough, it's possible to have a disagreement even on this little bit of mathematics. This is a lightly-edited transcript of an e-mail exchange.
There is a problem in Table 1 and Table 2.
One would not expect the SDT model to fail more often at the P<0.01 level than at the P<0.05 level, yet that seems to be the case on a number of conditions.
In fact, one can be stronger: anything that fails at the P<0.01 level will certainly fail at the P<0.05 level. So, the proportion of unacceptable fits *must* be larger for P<0.05.
So, something is wrong. Hopefully, the numbers are just put in the wrong columns, but it needs checking.
It's a well tested program, and I am not defending it. I only report what it says about OUR simulations.
Well, perhaps it's output is a little misleading?
The program probably does what all psychologists do. It breaks up the range into 0.001 < P < 0.005, 0.005 < P < 0.01, 0.01 < P < 0.05, and P > 0.05 then it reports what part you fell into. So, what it probably means when it says "P < 0.05" is *really* "0.01 < P < 0.05".
That's OK (if you make it clear what it's doing), but it has to be said explicitly in the paper or in the table heading. For instance, I expect that events you label "P < 0.05" should occur by chance 5% of the time, not 4%.
I don't know quite what you're doing, but if it is correct, it certainly confuses one of your readers (i.e. me).
Evidently.
Therefore the paper needs fixing, because a paper is intended to be a device for communicating knowledge. In your small sample of test readers (i.e. me), that section fails 100% of the time.
You're repeating a common, faintly incorrect practice. To do the statistics correctly, you define the significance level before the data is collected. Then, you test to see if you reach the significance level or not. If you then go on and report a better significance level, that's going beyond the test; it's bragging.
That's when you're testing hypotheses. We're concerned with goodness of fit. If you set .05 for that and get .01, why not report it? It shows an even worse fit. It's NOT bragging; it's reporting a fact.
There's nothing horribly wrong with reporting it, but if it's read carelessly, it will over-inflate the significance of a test.
Consider an exaggerated case where we report 0.05, 0.04, 0.03, 0.02, 0.01 probability levels. We'll do 100 identical statistical tests where the null hypothesis is true. So, we expect that 5 of them will report a result significantly different from H0. Right? Now, ask yourself what significance levels you expect to find? They won't all be near 0.05!
Of the 5 "significant" results, you expect one in 0.04<P<0.05, one in 0.03<P<0.04, one in 0.02<P<0.03, one in 0.01<P<0.02 and one in 0<P<0.01. So, when you write the paper, you write that one is significant at the 0.05 level, one at the 0.04 level, ... and one at the 0.01 level.
Then people will be tempted to say "Wow! He's getting better than a 5% significance level. Hmm. Five successes at the, uh, 0.03 level out of 100... Maybe there's something real in that data!" (Unfortunately, there really isn't.)
However, if you restrict yourself to the standard 0.05, 0.01, 0.005, 0.001 set, the harm is less, but not zero. In that same example of 100 trials, you'd expect to get one that the P<0.01 level. No big surprise there for anyone with a little statistical knowledge. But, 10% of researchers will be lucky enough to get one of the results at the P<0.001 level, following the same protocol. And, boy, will they brag! People will say "Well, yes, you do expect 5 results to turn up true just by chance, but look! One of them is at the 0.001 level. That one must be real..."
Basically, it's a bit of grade inflation. (and I do it myself.) Maybe it's best thought of as using the data twice: once to test H0, another time to estimate how significant the result is...
Reporting significance levels above your tested level is just a way of bragging.
The entire definition of the P<0.05 significance level is that significant events happen by chance 5% of the time. (If the situation meets the correct assumptions.) In fact, this is the *only* requirement. (For instance, there are an infinite number of different two-tailed 95% confidence levels, ranging from 0% in the left tail to 5% in the left tail, and everywhere in between. It's just convention that we stick with the completely symmetric and completely asymmetric confidence intervals.)
I agree. The .05 and .01 levels are conventions. But Where have you found anything in the literature that uses an UNBALANCED 2-tail test?
Defining the shape of a confidence "interval" is a real issue when you want to do confidence intervals on a 2-D plane. Then the shape of the confidence interval is fairly obviously undefined. If you happen to have multivariate Gaussian data and your analysis is linear, then it's reasonable to make your confidence intervals into ellipses, but otherwise, the shape is arbitrary.
People often make a confidence region that follows a contour of probability density (if they can), or they try to find the confidence region with the smallest possible area. Either way, you define the P<0.05 confidence interval to include 95% of the probability.
The smallest-area confidence interval is also relevant in the 1-dimensional case. It then becomes the shortest confidence interval that contains 95% of the probability. If your probability distribution is symmetrical, then it is just your standard two-tailed symmetrical confidence interval. However, if you end up with non-symmetrical probability distributions, then the shortest interval is asymmetric.
Suppose you were fitting y = log(a) * x to some (x,y) data. Then, a would be a log-normal distribution and it could be strongly asymmetrical if stdev(log(a))>1.
Generally speaking, if you do a nonlinear coordinate transform, you end up with a different (non-equivalent) confidence interval.
The trouble with your definition is that events at the P<0.05 column only happen 4% of the time. That's wrong. Period.
You're wrong. Full stop. How do you get such a silly idea? Defend you position.
Do the integrals. Integrate a Gaussian between the 0.01 and 0.05 confidence levels. Bet you \$100 the integral is 0.04. So, the region you are *calling* a 5% confidence interval only has 4% of probability mass.
The trouble with your definition is that events in the column that you label as "P<0.05" only happen 4% of the time. That's wrong. Period.
I'll do it to satisfy you; the reviewers will take it out. Go ahead and try to publish it.
That last isn a about correctness. It's a statement about culture and the sociology of science.
[ Papers | kochanski.org | Phonetics Lab | Oxford ] Last Modified Thu May 29 17:23:41 2008 Greg Kochanski: [ Home ] | 1,754 | 6,925 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2017-34 | latest | en | 0.956015 |
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https://www.romannumerals.org/blog/roman-numerals-history-and-facts-11 | 1,686,440,540,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646652.16/warc/CC-MAIN-20230610233020-20230611023020-00734.warc.gz | 1,093,034,446 | 8,672 | # Roman Numerals History and Facts
In the modern world, their usage has been mostly confined to clocks, watches, specific texts or sporting events. However, Roman numerals were the numbering system that was used for hundreds of years throughout the Roman Empire and during the Middle Ages. They were used not only for complicated mathematical purposes but also in everyday life. So what are the origins of Roman numerals? When were they created? Where do the Roman numeral symbols come from?
## What Are Roman Numerals?
Roman numerals are a numeric system that was used in the Roman Empire. They were used in Europe for nearly 2000 years. The seven symbols that represent the Roman numbers are letters from the Latin Alphabet. The letters I, V, X, L, C, D and M are used with assigned values in order to write numbers.
The letters and their values in Roman numeral system are as follows:
I V X L C D M
1 5 10 50 100 500 1,000
In the Roman Empire, these numbers were widely used in everyday trade and services as well as in complicated mathematical applications such as accounting, architecture and engineering.
After the fall of the Roman Empire, their use continued in Europe until the 15th century. However, it was realised in time that the Arabic numbers (the numeral system we use today) was easier in everyday life and in calculations that deal with larger numbers.
Also, the fact that the number 0 was not represented in the Roman system was a problem. Therefore, Roman numerals were gradually replaced by the more convenient Arabic numeral system. Meanwhile, their use in religious texts and many other areas continued.
### Origins of Roman Numerals
The Roman numerals are based on the Etruscan numbering system, the civilisation that the Roman Empire was built upon. There are two competing theories as to where the letters come from:
### Tally Stick Theory
This theory suggests that the origins of Roman numerals lie in a tally system used by shepherds to count their sheep. According to this theory, shepherds in Rome used to keep their sheep tally by using sticks. Each animal was recorded on a stick with a notch carved with a knife. The fifth sheep would have two notches in the shape of a ‘V’ and the tenth was an ‘X’.
This would mean that, for example, eight sheep would have a ‘notch series’ of IIIIVIII. When shortened, there were VIII sheep, identical to the Roman numerals. Sixteen sheep on a tally stick would look like IIIIVIIIIXIIIIVI and when shortened, it would give the Roman numeral XVI (16). It is believed that this method for counting sheep was used by shepherds in Italy well into the 19th century.
### Hand Signals Theory
• This theory suggests that the Roman numbers represent the hand signals for counting.
• In this theory, numbers one (I), two (II) and three (III) are equal to the fingers used when counting.
• V (five) is the signal you see in between the thumb and fingers - when you open one hand and show all your five fingers.
• The numbers six (VI), seven (VII) and eight (VIII) represent one hand showing five (V) and the other hand showing the equivalent numbers.
• X (10) is two hands showing two fives with thumbs crossed, hence the shape ‘X’.
• In later periods, the letter L was adapted to represent the number 50.
• In the early days, the number 100 was usually represented by putting an ‘I’ on top of ‘X’; 500 and 1000 were written with a circle around 'V' and 'X', respectively. In later periods, Greek letters were modified and ‘C’ was adopted for 100, 'D' for 500, 'M' for 1000.
• For representing greater values, lines (known as vinculum or virgula) were placed to indicate multiplication by by 1,000.
#### How to Read Roman Numerals
Roman numbers are read from left to right. They are formed by combining the seven letters based on addition and subtraction. The order of the numerals hints whether you need to subtract or add values.
Keeping in mind that greater values are usually written before the lower values, the two basic rules for reading Roman numerals are as follows:
• If a letter comes before a letter of greater value, you subtract;
• If one or more letters come after a letter of greater value, you add.
The first rule means that when a smaller value is placed before a larger value, the difference is the Roman number to be read. For example,
• IV = 4
• IX = 9
• XL = 40
• XC = 90
• CM = 900
The second rule means that when a smaller value is placed after a larger value, the sum is the Roman number to be read. For example,
• VII = 7
• XII = 12
• LIII = 53
• CXIII = 113
• LXXV = 75
Therefore, when you want to read or write a larger number, you need to take notice of these groupings to see the additions or subtractions.
For example, MCMXIX = 1919
This is because this number is grouped as follows:
M + CM + X + IX = 1000 + 900 + 10 + 9
You can use our online friendly tool to convert roman numerals to numbers.
#### Modern Uses of Roman Numerals
While they are not one of the numbering systems used in modern mathematics, Roman numbers are not only confined to history. They are still around us:
• They are used on watches and clocks,
• Book chapter headings and numbered points in print still use Roman numerals,
• Subjects like chemistry, pharmaceuticals, seismology and theology might still apply Roman numerals when deemed necessary,
• Roman numerals are used in order to refer to kings, queens, emperors and popes who share the same names (as in Louis XVI of France or Pope Benedict XVI),
• In the film industry, Roman numerals might be used to indicate the year the film was produced,
• Sporting competitions such as the Olympic Games or the Super Bowl represent how many times the games have been held by using Roman numerals.
#### Latest Blog Entries About Roman Numerals
vicky 2023-05-09 20:45:30
this has helped me so much and I appreciate
Kai 2023-01-20 13:52:26
this helped with my project, thanks!
april 2022-12-05 23:39:03
you could have added how the numerals where actually used but it still helpful
James Charles 2022-09-27 16:03:16
OMG this helped me so much in my project for school
Eva 2022-09-25 19:06:00
Thank you this was very useful!
albert 2022-09-24 23:21:52
tysm that really helped me on my history project.
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Luke 2022-04-30 21:58:15
Thank you so much for this, I needed some information on Roman Numerals and this post gave me all the facts I needed!
alex 2020-03-26 16:43:07
Good job who ever made this. | 1,555 | 6,458 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | longest | en | 0.976162 |
http://metricamerica.com/metricEXAMS/LENGTHmillimeterFALSE.htm | 1,726,758,021,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652031.71/warc/CC-MAIN-20240919125821-20240919155821-00034.warc.gz | 20,602,504 | 8,184 | FREE! PRINTABLE METRIC POSTERS for TEACHERS, STUDENTS, CLASSROOMS - TABLOID SIZED Color or B/W - See Menu Below "PRINTABLE METRIC POSTERS"
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VIDEO CLIP
DID YOU KNOW?
that you can divide 1 meter
by exactly 10 equal parts
(10 cm x 10 cm x 10 cm)...
then make a cube (1000 cmณ) of it to fill with water
and you will find that it contains exactly 1 L (1 liter)
(1000 mL
) of water
and is the mass of exactly
1 kg (1 kilogram) (1000 g).
ALL SIMPLY
RELATED
IN 10!
Instant Metric eBook ฎฉ HTML format 813 kb
I
nternet Browser Laptop Tablet Handheld Mobile Device eReaders
Instant Metric eBook ฎฉ PDF format 1290 kb
WRONG!
millimeter (mm)
There are exactly 1000 m m (millimeters)
in the Base Unit meter.
1 one thousandth from a meter (0.001 m)
## Use the symbol mm
TRUE or FALSE - Check One
The Base Unit for LENGTH is meter (m) TRUE FALSE There are 100 cm in the length of 1 m TRUE FALSE There are 1000 mm in the length of 1 m TRUE FALSE A meter is about as long as a baseball bat TRUE FALSE There are exactly 100 m in 1 km TRUE FALSE
LEARNING SKILLS
FOR EVERYDAY USE IN AMERICA
Choose From the Following Subjects
Everyday Metric Units Metric Prefixes Metric Symbols
LENGTH
VOLUME
MASS
meter
centimeter
millimeter
kilometer
liter
milliliter
kilogram
gram
milligram
tonne
TEMPERATURE
Inch-Pound System
The Metric System
#### 35 mm of rain is a torrential downpour. Rainfall is measured in millimeters.
1000 mm (millimeters) 1 meter
#### If a "dime" is about 1 mm (millimeter) thick, then 1000 dimes placed side by side would be
Rainfall is measured in millimeters (mm)
Snowfall is measured in centimeters (cm)
Larger Measurement from a millimeter
## 1 centimeter (1 cm)
### Remember, centi is the prefix meaning 1 one hundredth of the base unit meter. So, 1 one hundredth of a meter (0.01 m) is simply written 1 cm
Larger Measurement from a centimeter
### Remember meter is the Base Unit for Length and all other Units are derived from it. There are exactly 1000 millimeters (1000 mm) in 1 meter (1 m)
Larger Measurement from a meter
### Remember, kilo is the prefix meaning 1000 times as much. So, one thousand meters (1000 m) being the length of 1 kilometer, is simply written 1 km which is pronounced "KILL-oh-meet-ur" NOT "kill-AH-mit-ur".
CLICK ON PICTURES FOR VIDEO LEARNING! Enjoy this millimeter Video! Video Narrated with music 1 cmณ (1 cubic centimeter) filled with water contains exactly 1 mL (1 milliliter) and is the mass of exactly 1 g (1 gram) "centimeters milliliters milligrams" VIDEO CLIP
Enter your search terms Submit search form | 1,034 | 3,526 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-38 | latest | en | 0.564764 |
https://brainly.com/question/174925 | 1,484,622,748,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279410.32/warc/CC-MAIN-20170116095119-00172-ip-10-171-10-70.ec2.internal.warc.gz | 798,168,558 | 9,563 | # Carlita saw 5 times as many robins as cardinals while bird watching. She saw a total of 25 birds. How many more robins did she than cardinals?
2
by christinahart05
2014-11-04T23:21:20-05:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Carlita was either sipping the hot chocolate while she was out, or else
approximating the numbers. The exact number she reported is not
possible. To see why, I'll do it strictly by the math:
Robins = 5 times
Cardinals = 1 time
Total birds = 6 times
6 times = 25 birds
Divide each side by 6 : 1 time = 25/6 = 4-1/6 birds
Robins = 5 times = 20-5/6 of them
Cardinals = 1 time = 4-1/6 of them
Total: (20-5/6) + (4-1/6) = 25 birds.
How many more robins than cardinals = (20-5/6) - (4-1/6) = 16-2/3 more.
The math works out fine. But if the numbers she reported are true,
then some of what she saw was partial birds ... legs, or feathers,
things like that. Eeeew. I don't want to talk about it.
you were wrong
my teacher said that was wrong
If your teacher can find different numbers and show that they satisfy the conditions in the quedtion, then your teacher should consider A different profession. ... What is the 'correct' answer ?
2014-11-05T22:57:20-05:00
20 because 5x5=25 so she saw 5 x as many if you divide 5 and 25 you get five | 476 | 1,573 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2017-04 | latest | en | 0.953848 |
http://slideplayer.com/slide/3863/ | 1,529,688,071,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864740.48/warc/CC-MAIN-20180622162604-20180622182604-00244.warc.gz | 298,876,266 | 25,995 | # Exploring the lifecycle of stars
## Presentation on theme: "Exploring the lifecycle of stars"— Presentation transcript:
Exploring the lifecycle of stars
Star in a Box Exploring the lifecycle of stars
Guide to this presentation
White slides are section headings, and are hidden from the presentation. Show or hide the slides in each section as appropriate to the level required. Rough guide to the levels: Beginner: KS3 Intermediate: KS4 (GCSE) Advanced: KS5 (AS/A level)
Introduction Basics of what a star is and how we observe them.
Level: Beginner +
What is a star? A cloud of gas, mainly hydrogen and helium
The core is so hot and dense that nuclear fusion can occur. The fusion converts light elements into heavier ones
Every star is different
All the stars in the night sky are different Brightness: Tells us how luminous the star is, i.e. How much energy is being produced in the core Colour: Tells us the surface temperature of the star
Units of luminosity We measure the luminosity of every day objects in Watts. How bright is a light bulb? By comparison, the Sun outputs: 380,000,000,000,000,000,000,000,000 Watts (380 million million million million Watts!) This is easier to right as 3.8 x 1026 Watts To make things easier we measure the brightness of stars relative to the Sun.
Units of temperature Temperature is measured in Kelvin
The Kelvin temperature scale is the same as the Celsius scale, but starts from -273o. This temperature is known as “absolute zero” -273 oC -173 oC 0 oC 100 oC 1000 oC 0 K 100 K 273 K 373 K 1273 K Kelvin = Celsius + 273
Measuring the temperature
The temperature of a star is indicated by its colour Blue stars are hot, and red stars are cold Red star 3,000 K Yellow star 5,000 K Blue star 10,000 K
Black Body radiation A “black body” is a perfect emitter and absorber of light It emits light at a range of wavelengths which is dependent on its temperature
Wein’s Displacement Law
The peak intensity of the light is related to the temperature: Temperature (K) = Wien’s constant (K.m) / peak wavelength (m) T = b lmax (b = m.K)
How hot is the Sun Here is a graph of the Sun’s energy output
Hertzsprung-Russell Diagram
An introduction to the H-R diagram, on which various stars will be plotted – try to get the students to suggest where they might appear before they are plotted. Level: Beginner +
The Hertzsprung Russell Diagram
We can compare stars by showing a graph of their temperature and luminosity
Luminosity (relative to Sun)
We start by drawing the axes: Luminosity up the vertical axis (measured relative to the Sun) Temperature along the horizontal axis (measured in Kelvin) 10,000 The stars Vega and Sirius are brighter than the Sun, and also hotter. Where would you put them? 100 Where would you mark the Sun on the plot? It has Luminosity of 1 relative to itself Its temperature is 5800 K Vega Sirius Main Sequence Luminosity (relative to Sun) 1 Sun In fact, most stars can be found somewhere along a line in this graph. This is called the “Main Sequence”. Some stars are much cooler and less luminous, such as the closest star to the Sun, Proxima Centauri. Where would you plot these? These stars are called red dwarfs. Proxima Centauri 0.01 0.0001 25,000 10,000 7,000 5,000 3,000 Temperature (Kelvin)
Luminosity (relative to Sun)
The bright star Betelgeuse is even more luminous than Aldebaran, but has a cooler surface. This makes it a red supergiant. Betelgeuse Rigel 10,000 Deneb Aldebaran Arcturus 100 Vega Sirius Main Sequence Luminosity (relative to Sun) 1 Sun Even brighter than Betelgeuse are stars like Deneb and Rigel, which are much hotter. These are blue supergiants. But not all stars lie on the main sequence. Some, such as Arcturus and Aldebaran, are much brighter than the Sun, but cooler. Where would these lie on the diagram? These are orange giant stars. Sirius B Proxima Centauri 0.01 Some of the hottest stars are actually much fainter than the Sun. Which corner would they be in? These are white dwarfs, such as Sirius B which orbits Sirius. 0.0001 25,000 10,000 7,000 5,000 3,000 Temperature (Kelvin)
Luminosity (relative to Sun)
Supergiants Betelgeuse Rigel 10,000 Deneb Giants Arcturus 100 Vega Sirius Main Sequence Luminosity (relative to Sun) Almost all stars we see are in one of these groups, but they don’t stay in the same place. 1 Sun Sirius B Proxima Centauri As stars evolve they change in luminosity and temperature. This makes them move around the Hertzprung-Russell diagram. 0.01 White Dwarfs 0.0001 25,000 10,000 7,000 5,000 3,000 Temperature (Kelvin)
Luminosity (relative to Sun)
10,000 100 Luminosity (relative to Sun) 1 Sun The Sun has been on the Main Sequence for billions of years, and will remain there for billions more. But eventually it will swell into a giant star, becoming more luminous but cooler. 0.01 0.0001 25,000 10,000 7,000 5,000 3,000 Temperature (Kelvin)
Luminosity (relative to Sun)
10,000 100 Sun Luminosity (relative to Sun) 1 At this point it is a red giant star. It will get then hotter and slightly brighter, briefly becoming a blue giant. 0.01 0.0001 25,000 10,000 7,000 5,000 3,000 Temperature (Kelvin)
Luminosity (relative to Sun)
10,000 Sun 100 Luminosity (relative to Sun) 1 Finally nuclear fusion in the core will cease. The Sun will become a white dwarf, far less luminous than its but with a hotter surface temperature. 0.01 0.0001 25,000 10,000 7,000 5,000 3,000 Temperature (Kelvin)
Star in a Box At this point, run star in a box to explore the Hertzsprung-Russell diagram for different mass stars. Level: Beginner +
Nuclear fusion The processes taking place in the centre of a star.
Level: Intermediate +
Nuclear fusion The luminosity of a star is powered by nuclear fusion taking place in the centre of the star The temperature and density are sufficient to allow nuclear fusion to occur. Stars are primarily composed of hydrogen, with small amounts of helium. They are so hot that the electrons are stripped from the atomic nuclei. This ionised gas is called a plasma.
The proton-proton chain
At temperatures above 4 million Kelvin hydrogen nuclei fuse into helium
The CNO cycle At temperatures above 17million Kelvin the star can use carbon, nitrogen and oxygen to help convert hydrogen into helium.
Running out of hydrogen
The star is kept in a delicate balance between gravity trying to collapse it and radiation pushing it outwards. As the hydrogen runs out, the energy released from fusion decreases and the gravity causes the star to collapse. If the star is massive enough the core temperature increases until helium fusion starts.
Helium burning At temperatures above 100 million Kelvin helium can be fused to produce carbon. This reaction is called the “Triple Alpha process”
Heavier elements Helium is fused with carbon to make heavier elements:
oxygen, neon, magnesium, silicon, sulphur, argon, calcium, titanium, chromium and iron It’s impossible to make elements heavier than through nuclear fusion without putting in more energy.
Running out of helium Eventually the helium is exhausted, and the star collapses again. If it is massive enough, then the temperature increases enough to allow carbon fusion. The cycle repeats, fusing heavier elements each time, until the core temperature cannot rise any higher. At this point, the star dies.
Burning heavier elements
Heavier elements undergo fusion at even higher core temperatures Carbon: 500 million Kelvin Neon: 1.2 billion Kelvin Oxygen: 1.5 billion Kelvin Silicon: 3 billion Kelvin
Hydrogen fusion The proton-proton chain turns six hydrogen nuclei into one helium nucleus, two protons, and two positrons (anti-electrons) The energy released per reaction is tiny, and is measured in “Mega electron Volts”, or MeV. 1 MeV = 1.6 x Joules Each proton-proton chain reaction releases MeV
Atomic masses The mass of the output is less than the mass of the input, so at every reaction the star loses mass. Just like the energies, the masses involved are tiny, measured in “atomic mass units” or u. 1 u = x kg
Mass loss Mass of a proton (p): 1.007276 u
Mass of a positron (e+): u Mass of a helium nucleus (He): u How much mass is lost in every reaction? u = x kg
Helium burning Helium fusion releases 7.275 MeV per reaction
Carbon-12 has a mass of exactly 12 u. How much mass is lost in the Triple alpha reaction? u = x kg | 2,068 | 8,351 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2018-26 | longest | en | 0.877779 |
http://mathforum.org/mathimages/index.php?title=Polar_Equations&diff=25563&oldid=25558 | 1,516,532,693,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890514.66/warc/CC-MAIN-20180121100252-20180121120252-00625.warc.gz | 225,492,471 | 13,792 | # Polar Equations
(Difference between revisions)
Revision as of 14:55, 15 July 2011 (edit)← Previous diff Revision as of 15:33, 15 July 2011 (edit) (undo)Next diff → Line 82: Line 82: |WhyInteresting=[[Image:PolarStar.png|$r = \sin^2(1.2\theta) + \cos^3(6\theta)$|thumb|210px|right]][[Image:SunflorwerPolar.png|The disc phyllotaxis of a sunflower.Lauer, Christoph. http://www.christoph-lauer.de/Homepage/Blog/Eintrage/2009/10/6_Fermats_Spiral_Suflower_Generator.html. Christoph Lauer's Blog.|thumb|200px|left]] |WhyInteresting=[[Image:PolarStar.png|$r = \sin^2(1.2\theta) + \cos^3(6\theta)$|thumb|210px|right]][[Image:SunflorwerPolar.png|The disc phyllotaxis of a sunflower.Lauer, Christoph. http://www.christoph-lauer.de/Homepage/Blog/Eintrage/2009/10/6_Fermats_Spiral_Suflower_Generator.html. Christoph Lauer's Blog.|thumb|200px|left]] - Polar coordinates are often used in navigation, such as aircrafts. They are also used to plot gravitational fields and point sources. Furthermore, polar patterns are seen in the directionality of microphones, which is the direction at which the microphone picks up sound. A well-known pattern is a [[Cardioid]]. + Polar coordinates are often used in navigation, such as aircrafts. They are also used to plot gravitational fields and point sources. Furthermore, polar patterns are seen in the directionality of microphones, which is the direction at which the microphone picks up sound. A well-known pattern is the [[Cardioid]]. -
Archimedes' spiral can be used for compass and straightedge division of an angle into $n$ parts and circle squaring. Weisstein, Eric W. (2011). http://mathworld.wolfram.com/ArchimedesSpiral.html. Wolfram:MathWorld.The pattern of Fermat's spiral happens to appear in the mesh of mature disc phyllotaxis. +
Archimedes' spiral can be used for compass and straightedge division of an angle into $n$ parts and circle squaring. Weisstein, Eric W. (2011). http://mathworld.wolfram.com/ArchimedesSpiral.html. Wolfram:MathWorld. Fermat's spiral is a Archimedean spiral that is observed in nature. The pattern happens to appear in the mesh of mature disc phyllotaxis. Archimedean spirals can be seen in patterns of solar wind and Catherine's wheel. -
As you can see, these equations can create interesting curves and patterns. More complicated patterns can be created with more complicated equations, like the image on the right. +
As you can see, these equations can create interesting curves and patterns. More complicated patterns can be created with more complicated equations, like the image on the right. Since intriguing patterns can be expressed mathematically, like these curves, they are often used for art and design. ==Possible Future work== ==Possible Future work== * More details can be written about the different curves, maybe they can get their own pages. * More details can be written about the different curves, maybe they can get their own pages.
## Revision as of 15:33, 15 July 2011
A polar rose (Rhodonea Curve)
This polar rose is created with the polar equation: $r = cos(\pi\theta)$.
# Basic Description
Polar equations are used to create interesting curves, and in most cases they are periodic like sine waves. Other types of curves can also be created using polar equations besides roses, such as Archimedean spirals and limaçons. See the Polar Coordinates page for some background information.
# A More Mathematical Explanation
Note: understanding of this explanation requires: *calculus, trigonometry
## Rose
The general polar equations form to create a rose is UNIQ9e888fb5b14209a-math-00000001-QI [...]
## Rose
The general polar equations form to create a rose is $r = a \sin(n \theta)$ or $r = a \cos(n \theta)$. Note that the difference between sine and cosine is $\sin(\theta) = \cos(\theta-\frac{\pi}{2})$, so choosing between sine and cosine affects where the curve starts and ends. $a$ represents the maximum value $r$ can be, i.e. the maximum radius of the rose. $n$ affects the number of petals on the graph:
• If $n$ is an odd integer, then there would be $n$ petals, and the curve repeats itself every $\pi$.
Examples:
• If $n$ is an even integer, then there would be $2n$ petals, and the curve repeats itself every $2 \pi$.
Examples:
• If $n$ is a rational fraction ($p/q$ where $p$ and $q$ are integers), then the curve repeats at the $\theta = \pi q k$, where $k = 1$ if $pq$ is odd, and $k = 2$ if $pq$ is even.
Examples:
$r = \cos(\frac{1}{2}\theta)$ The angle coefficient is $\frac{1}{2} = 0.5$. $1 \times 2 = 2$, which is even. Therefore, the curve repeats itself every $\pi \times 2 \times 2 \approx 12.566.$ $r = \cos(\frac{1}{3}\theta)$ The angle coefficient is $\frac{1}{3} \approx 0.33333$. $1 \times 3 = 3$, which is odd. Therefore, the curve repeats itself every $\pi \times 3 \times 1 \approx 9.425.$
• If $n$ is irrational, then there are an infinite number of petals.
Examples:
$r = \cos(e \theta)$
$\theta \text{ from } 0 \text{ to...}$
...$10$ ...$50$ ...$100$
$\text{Note: }e \approx 2.71828$
Below is an applet to graph polar roses, which is used to graph the examples above:
If you can see this message, you do not have the Java software required to view the applet.
Source code: Rose graphing applet
## Other Polar Curves
Archimedean Spirals
Archimedes' Spiral $r = a\theta$ The spiral can be used to square a circle and trisect an angle. Fermat's Spiral $r = \pm a\sqrt\theta$ This spiral's pattern can be seen in disc phyllotaxis. Hyperbolic spiral$r = \frac{a}{\theta}$ It begins at an infinite distance from the pole, and winds faster as it approaches closer to the pole. Lituus $r^2 \theta = a^2$It is asymptotic at the $x$ axis as the distance increases from the pole.
Limaçon[1]
The word "limaçon" derives from the Latin word "limax," meaning snail. The general equation for a limaçon is $r = b + a\cos(\theta)$.
• If $b = a/2$, then it is a trisectrix (see figure 2).
• If $b = a$, then it becomes a cardioid (see figure 3).
• If $2a > b > a$, then it is dimpled (see figure 4).
• If $b \geq 2a$, then the curve is convex (see figure 5).
$r = \cos(\theta)$ 1 $r = 0.5 + \cos(\theta)$ 2 Cardioid $r = 1 + \cos(\theta)$3 $r = 1.5 + \cos(\theta)$4 $r = 2 + \cos(\theta)$5
## Finding Derivatives[2]
A derivative gives the slope of any point in a function.
Consider the polar curve $r = f(\theta)$. If we turn it into parametric equations, we would get:
• $x = r \cos(\theta) = f(\theta) \cos(\theta)$
• $y = r \sin(\theta) = f(\theta) \sin(\theta)$
Using the method of finding the derivative of parametric equations and the product rule, we would get:
$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{\frac{dr}{d\theta} \sin(\theta) + r \cos(\theta)}{\frac{dr}{d\theta} \cos(\theta) - r \sin(\theta)}$
Note: It is not necessary to turn the polar equation to parametric equations to find derivatives. You can simply use the formula above.
Examples:
Find the derivative of $r = 1 + \sin(\theta)$ at $\theta = \frac{\pi}{3}$.
$\frac{dr}{d\theta} = \cos(\theta)$
$\frac{dy}{dx} = \frac{\frac{dr}{d\theta} \sin(\theta) + r \cos(\theta)}{\frac{dr}{d\theta} \cos(\theta) - r \sin(\theta)} = \frac{\cos(\theta) \sin(\theta) + (1 + \sin(\theta) ) \cos(\theta)}{\cos(\theta)\cos(\theta) - (1 + \sin(\theta) ) \sin(\theta)}$
$= \frac{\cos(\theta)\sin(\theta) + \cos(\theta) + \cos(\theta)\sin(\theta)}{\cos^2(\theta) - \sin(\theta) - \sin^2(\theta)}$
Note: Using the double-angle formula, we get $\cos^2(\theta) - \sin^2(\theta) = 1 - 2\sin^2(\theta)$
$= \frac{\cos(1+2\sin(\theta))}{1-2\sin^2(\theta)-\sin(\theta)} = \frac{\cos(1+2\sin(\theta))}{(1+\sin(\theta))(1-2\sin(\theta))}$
$\frac{dy}{dx} \Big |_{\theta=\pi/3} = \frac{\cos(\pi/3)(1+2\sin(\pi/3))}{(1+\sin(\pi/3))(1-2\sin(\pi/3))}$
$= \frac{\frac{1}{2}(1+\sqrt{3})}{(1+\sqrt{3}/2)(1-\sqrt{3})} = \frac{1+\sqrt{3}}{(2+\sqrt{3})(1-\sqrt{3})} = \frac{1+\sqrt{3}}{-1-\sqrt{3}} = -1$
## Finding Areas and Arc Lengths[2]
Area of a sector of a circle.
To find the area of a sector of a circle, where $r$ is the radius, you would use $A = \frac{1}{2} r^2 \theta$.
$A = \int_{-\frac{\pi}{4}}^\frac{\pi}{4}\! \frac{1}{2} \cos^2(2\theta) d\theta$
Therefore, for $r = f(\theta)$, the formula for the area of a polar region is:
$A = \int\limits_a^b\! \frac{1}{2} r^2 d\theta$
The formula to find the arc length for $r = f(\theta)$ and assuming $r$ is continuous is:
$L = \int\limits_a^b\! \sqrt{r^2 + {\bigg(\frac{dr}{d\theta}\bigg)} ^2}$ $d\theta$
# Why It's Interesting
$r = \sin^2(1.2\theta) + \cos^3(6\theta)$
The disc phyllotaxis of a sunflower.[3]
Polar coordinates are often used in navigation, such as aircrafts. They are also used to plot gravitational fields and point sources. Furthermore, polar patterns are seen in the directionality of microphones, which is the direction at which the microphone picks up sound. A well-known pattern is the Cardioid.
Archimedes' spiral can be used for compass and straightedge division of an angle into $n$ parts and circle squaring. [4] Fermat's spiral is a Archimedean spiral that is observed in nature. The pattern happens to appear in the mesh of mature disc phyllotaxis. Archimedean spirals can be seen in patterns of solar wind and Catherine's wheel.
As you can see, these equations can create interesting curves and patterns. More complicated patterns can be created with more complicated equations, like the image on the right. Since intriguing patterns can be expressed mathematically, like these curves, they are often used for art and design.
## Possible Future work
• More details can be written about the different curves, maybe they can get their own pages.
• Applets can be made to draw these different curves, like the one on the page for roses.
# About the Creator of this Image
Polar Coordinates
Cardioid
Source code: Rose graphing applet
# References
Wolfram MathWorld: Rose, Limacon, Archimedean Spiral
Wikipedia: Polar Coordinate System, Archimedean Spiral, Fermat's Spiral
1. Weisstein, Eric W. (2011). http://mathworld.wolfram.com/Limacon.html. Wolfram:MathWorld.
2. 2.0 2.1 Stewert, James. (2009). Calculus Early Transcendentals. Ohio:Cengage Learning.
3. Lauer, Christoph. http://www.christoph-lauer.de/Homepage/Blog/Eintrage/2009/10/6_Fermats_Spiral_Suflower_Generator.html. Christoph Lauer's Blog.
4. Weisstein, Eric W. (2011). http://mathworld.wolfram.com/ArchimedesSpiral.html. Wolfram:MathWorld. | 3,112 | 10,416 | {"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": 76, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.825822 |
http://www.cfd-online.com/W/index.php?title=Heat_transfer&oldid=7511 | 1,469,883,973,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257836399.81/warc/CC-MAIN-20160723071036-00063-ip-10-185-27-174.ec2.internal.warc.gz | 355,094,224 | 12,643 | # Heat transfer
Everyone has always understood that something flows from hot object to cold one. It is called heat. The overall driving force for this heat flow is thermal gradient.
There are basically three modes of heat transfer:
1) Conduction
2) Convection
## Conduction
Conduction is heat transfer by means of molecular agitation within a material without any motion of the material as a whole. In this mode of heat transfer, heat flow is due to molecular collisions in the substance. Since molecullar collisions increase with temperature (Temperature increases the kinetic energy of the molecules, which is a combined effect of rotational, translational and vibrational motion of molecules ), heat transfer due to conduction increases. Conduction heat transfer through a substance is because of a temperature gradient. The rate of heat transfer by conduction between two regions of a substance is proportional to the temperature difference between them. The constant of propotionality is called thermal conductivity of the material. The thermal conductivity of materials in general depends on Temperature. In general liquids and gases have lower thermal conductivity as compared to solids (esp metals).
Mathematically, it can be described by using the Fourier's law:
$Q_{Conduction} = -k*A*\frac{dT}{dx}$
Where
$Q = \mbox{the rate of heat conduction (W)}$
$k = \mbox{Thermal conductivity of the material (W/m K)}$
$A = \mbox{Cross-sectional area of the object perpendicular to heat conduction (m2)}$
$T = \mbox{Temperature (K)}$
$x = \mbox{Length of the object (m)}$
(-ve sign indicates temperature reduction in heat flow direction)
Conduction heat transfer depends on thermal conductivity of material. Heat carried away by a solid by condution is proportional to its thermal conductivity for a unit length and unit cross-section. Here are few materials with their conductivity:
Diamond: 2000 W/m-K
Silver: 406 W/m-K
Copper: 385 W/m-K
Aluminum: 205 W/m-K
Sn-Pb Solder: 50 W/m-K
FR-4: 0.3 W/m-K
Kapton: 0.2 W/m-K
....
## Convection
Convection is heat transfer by means of motion of the molecules in the fluid. Heat energy transfers between a solid and a fluid when there is a temperature difference between the fluid and the solid. Convection heat transfer can not be neglected when there is a significant fluid motion around the solid.
There are mainly two types of the convection heat transfer:
1) Natural or Free Convection
2) Forced Convection
• Natural or Free Convection:- The temperature of the solid due to an external field such as fluid buoyancy can induce a fluid motion. This is known as "natural convection" and it is a strong function of the temperature difference between the solid and the fluid. This type of convective heat transfer takes place due to only fluid buoyancy caused due to temperature defference between fluid layers.
• Forced Convection:- Forcing air to blow over the solid by using external devices such as fans and pumps can also generate a fluid motion. This is known as "forced convection". Some external means for fluid motion is necessary in this type of convective heat transfer.
Fluid mechanics plays a major role in determining convection heat transfer. For each kind of convection heat transfer, the fluid flow can be either laminar or turbulent. For laminar flow of fluid over solid surface, a steady boundary layer formation takes place through which conductive heat transfer occur. This reduces convective heat transfer rate. Turbulent flow forms when the boundary layer is shedding or breaking due to higher velocities or rough geometries. This enhances the heat transfer.
Newton's Law of Cooling
Heat transfer due to convection is described by Newton's Law of Cooling,
$Q_{Convection} = h*A*dT$
Where
$Q_{Convection} = \mbox{Heat convected to surrounding fluid (W)}$
$h = \mbox{Convection heat-transfer coefficient (constant of proportionality ) (W/m2 K)}$
$A = \mbox{Area of the solid in contact with fluid (m2)}$
$dT = \mbox{Temperature difference between solid and surrounding fluid (Ts-Tf) (K)}$
The rate of heat transfered to the surrounding fluid is proportional to the object's exposed area A, and the difference between the solid temperature Ts and the mean fluid temperature Tf. Velocity of the fluid over solid is also a major contributing to enhance the rate of heat transfer.
Convection heat-transfer coefficient (h) plays main role in heat transfer by convection. Heat transfer rates by convection are expressed in terms of h.
It depends on following factors (h is directly propotional to these factors):
1) Exposed area of solid
2) Temperature difference between solid and fluid
3) Fluid velocity
Convective heat transfer can also be characterised in terms of Nusselt number.
Nusselt number is dimensionless number which is useful for heat transfer calculations. Nusselt number is the dimensionless heat transfer coefficient and appears when you are dealing with convection. It, therefore, provides a measure of the convection heat transfer at the surface.
It can be defined as follows:
$Nu = \frac{h*l}{k}$
Where
$Nu = \mbox{Nusselt number}$
$h = \mbox{Convection heat-transfer coefficient (constant of proportionality ) (W/m2 K)}$
$l = \mbox{Characteristic dimension of th esolid object (m)}$
$k = \mbox{Thermal conductivity of the solid (W/m k)}$
The Nusselt number may be viewed as the ratio of heat flow by convection to conduction for a layer of fluid.
If Nu=1, we have pure conduction. Higher values of Nusselt mean that the heat transfer is enhanced by convection.
In general,$Nu = f(Re, Pr, Gr)$
Where
$\mbox{Re is Reynolds number}$
$\mbox{Pr is Prandtl number}$
$\mbox{Gr is Grashof number}$
Following condictions apply:
$Gr/Re_L ^2 = 1; Nu = f(Re, Pr, Gr)$ Mixed convection
$Gr/Re_L ^2 << 1; Nu = f(Re, Pr)$ Forced convection
$Gr/Re_L ^2 >> 1; Nu = f(Gr, Pr)$ Natural convection | 1,371 | 5,898 | {"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": 23, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.890571 |
http://www.physicsforums.com/showthread.php?t=547554 | 1,411,077,014,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657129407.88/warc/CC-MAIN-20140914011209-00076-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 747,528,314 | 7,878 | # Centre of the ring of quaternions
by Wingeer
Tags: centre, quaternions, ring
P: 79 1. The problem statement, all variables and given/known data What is the centre of the ring of the quaternions defined by: $$\mathbf{H}=\{ \begin{pmatrix} a & b \\ -\bar{b} & \bar{a} \end{pmatrix} | a,b \in \mathbf{C} \}$$? 2. Relevant equations The definition of the centre of a ring: The centre Z of a ring R is defined by $$Z(R)=\{A | AX=XA, \forall X \in R\}$$ 3. The attempt at a solution I figured that multiples of the 2x2 identity matrix must be in the centre. Also if we denote an element of H by: $$\begin{pmatrix} x & y \\ -\bar{y} & \bar{x} \end{pmatrix}$$ where $$x=x_1 + ix_2$$ and similarly for a,b and y that: 1. $$b\bar{y}=\bar{b}y$$ 2. $$y(a-\bar{a})=b(x-\bar{x})$$ 3. $$\bar{b}(x-\bar{x})=\bar{y}(a-\bar{a})$$ Then for instance we get from the first equation that: $$b_2x_1=a_1y_2$$ But I am not sure whether this approach really is any useful at all. Some hints would be greatly appreciated.
P: 79 Anyone? I actually have another question about the quaternions. I am asked to show that: $$\mathbf{H'} = \{ a+bi+cj+dk | a,b,c,d \in \mathbf{R} \}$$ with: i^2=j^2=k^2=-1, ij=k=-ji, ik=-j=-ki and jk=i=-kj. is isomorphic as rings to the quaternions defined in the previous post. I started by noticing that (where x,y are complex numbers): $$\begin{pmatrix} x & y \\ -\bar{x} & \bar{y} \end{pmatrix} = \begin{pmatrix} a+bi & c+di \\ c-di & a-bi \end{pmatrix} = \begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix} + \begin{pmatrix} bi & 0 \\ 0 & bi \end{pmatrix} + \begin{pmatrix} 0 & c \\ -c & 0 \end{pmatrix} + \begin{pmatrix} 0 & di \\ di & 0 \end{pmatrix}$$ And so we see that every element in H is a linear combination of these matrices which all are linearly independent as well. This means we have found a basis for H. So if we define a function f:H -> H' by: $$1= f \left( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right)$$ $$i=f \left( \begin{pmatrix} i & 0 \\ 0 & i \end{pmatrix}\right)$$ $$j=f \left( \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\right)$$ $$k=f \left( \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}\right)$$ We see that obviously f is both surjective and injective as these are the only values f are defined for. Therefore f is an bijection and H and H' are isomorphic. Do I have to mix ring homomorphisms in this? Or? | 839 | 2,343 | {"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} | 3.78125 | 4 | CC-MAIN-2014-41 | latest | en | 0.684301 |
https://physics.stackexchange.com/questions/263168/pressure-due-to-weight-of-the-fluid-in-fluid-dynamics | 1,675,739,125,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500368.7/warc/CC-MAIN-20230207004322-20230207034322-00445.warc.gz | 462,549,093 | 39,281 | # Pressure due to weight of the fluid in fluid dynamics
In Bernoulli equation, the only term which corresponds to what one usually calls "pressure" is $p$, the other are still pressures dimensionally but their meaning is linked to kinetic and potential energy per unit volume. Nevertheless I get confused especially for the term $\rho g h$. Can it be seen as a pressure due to weight, similarly to Stevin law, but for hydrodynamics?
My guess would be a clear no because Stevin law in only valid in hydrostatics, but consider, for istance this example, where a liquid (say water) is flowing in the venturi meter.
To read the manometer and calculate pressure difference I have to take into account not only the height $h$, but also the difference $z_2-z_1$ between the two points in the tube.
But, since in the manometer the consideration are about hydrostatics that means to consider a further pressure difference given by $\rho g (z_2-z_1)$, which reminds Stevin law for the height difference $z_2-z_1$. But Stevin law is valid iff the fluid is static, which is not here.
So does a fluid in motion have the same "effect" in terms of pressure caused by weight, than a situation where the fluid is static?
The question is also: is there a pressure difference between point $1$ in the center of the tube and a point on the same section but very close to the tube boundary? (Does the weight of the fluid between $1$ and the boundary create a pressure difference between these two points?)
The question is also: is there a pressure difference between point 11 in the center of the tube and a point on the same section but very close to the tube boundary? (Does the weight of the fluid between 11 and the boundary create a pressure difference between these two points?)
The answer to this is Yes. Even with a horizontal tube, the pressure at the center of the tube is lower than at the bottom of the cross section.
• Thanks for the reply! My guess would be that this pressure difference is given by $\rho g H$, where $H$ is the difference in height between the point in the center of the tube and the one on the boundary. This would explain why in the manometer the difference in height between $1$ and $2$ is taken into account as in a hydrostatic problem. But that would be incorrect since the fluid is moving, right? So I'm really failing in understanding why a pressure difference $\rho g (z_2-z_1)$ is considered in the manometer. Jun 19, 2016 at 7:58
• If you assume that the pipe diameter is small compared to the distances z1 and z2, then the pressure at level z1 on the right exceeds the pressure at z1 on the left by $(\rho_B-\rho_W) g h$, where the subscript B refers to the black fluid and the subscript W refers to the right fluid. So, $P_1-P_2=(\rho_B-\rho_W) g h+\rho_W g (z_2-z_1)=\rho_Bgh+\rho_W(z_2-z_1-h)$. If the density of the white fluid is negligible, then $P_1-P_2=\rho_Bgh$ Jun 19, 2016 at 11:40
• Ok thanks! Just to be sure that I got that right I added the points $3$ and $4$ in the picture. If the density of water $\rho_W$ is not neglegible, then I can say that $p_1=p_3-\rho_W g (z_1-z_3)$, right? And this comes from Bernoulli equation for irrotational flow between point $3$ and $1$ (in fact $v_1=v_3$) , correct? In the same way $p_2=p_4-\rho_W g (z_1-z_3)$. Then I used $p_3$ and $p_4$ to do the balancing of pressures in the manometer and I get the correct result, so it looks right. Jun 20, 2016 at 13:21
• Right. Incidentally, I would have put the manometer connections at the locations here the plane normal to the axis intersects the wall to determine the pressure difference along the axis, and I would have oriented to connection normal to the wall. Jun 20, 2016 at 13:57
• Sure. If the flow is horizontal, then in the vertical direction from point 1 to point 3, the pressure variation is hydrostatic. For laminar flow in a tube, the pressure gradient in the horizontal direction is constant, and independent of radial position. Jun 23, 2016 at 23:41
In Bernoulli equation, the only term which corresponds to what one usually calls "pressure" is $p$, the other are still pressures dimensionally but their meaning is linked to kinetic and potential energy per unit volume. Nevertheless I get confused especially for the term $ρgh$.
That is corect. The term $\rho gh$, better written as $\rho g(z_2-z_1)$, is the potential energy (per unit of volume) increase of a mass element travelling along the flow line.
In a gravitational field (not too far from the Earth's surface) that potential energy change only depends on any change in $z$ (and not on path e.g.), where $z$ is the vertical spatial coordinate. So that change is $\rho gz_2-\rho gz_1=\rho g(z_2-z_1)=\rho gh$ (if we set $h=z_2-z_1$). | 1,209 | 4,740 | {"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} | 3.296875 | 3 | CC-MAIN-2023-06 | latest | en | 0.958748 |
https://centumwiki.in/Algebraic_Identities | 1,631,815,613,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780053717.37/warc/CC-MAIN-20210916174455-20210916204455-00565.warc.gz | 215,214,647 | 6,241 | # Algebraic Identities
## Concept Lectures
Session # 1 : This session covers basic Algebraic Identities related to binomials and trinomials - Part I
Session # 2 : This session covers basic Algebraic Identities related to binomials and trinomials - Part II. Some example problems on expansion using identities have also been solved.
## Problem Solving Sessions
Session # 3 : Solved Examples : Using Algebraic Identities evaluate a) 103 x 97 and b) (0.99)^2
Session # 4 : Solved Examples : 1. Simplify : (2x+5y+3)(2x+5y+4) 2. If x + 1/x = 6, Find the value of x^2+1/x^2 and x^4+1/x^4
Session # 5 : Solved Examples : If x^2 + 1/x^2 = 27, then find the value of a) x + 1/x, b) x - 1/x
Session # 6 : 1. Prove that 2a^2 +2b^2+ 2c^2 - 2ab - 2bc - 2ca = (a-b)^2 + (b-c)^2 + (c-a)^2 2. Find the coefficient of x^2 in the expansion of (x^2+x+1)^2 + (x^2-x+1)^2
## Concept Lectures
Session # 7 : Square of Trinomials
## Problem Solving Sessions
Session # 8 : Problems related to expansion of square of trinomials
Session # 9 : If a+b+c = 9 and ab + bc + ca = 40, find the value of a^2 + b^2 + c^2
## Concept Lectures
Session # 10 : Cubes of Binomials (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3
## Problem Solving Sessions
Session # 11 : Solved Example: 1) Expand (2x + 3y)^3 (2) Expand (1/3x - 2/5y)^3
Session # 12 : Solved Example: If (x^2 + 1/x^2) =83. Find the value of (x^3 - 1/x^3)
Session # 13 : Solved Example: Q1 : If (x - y) = 4 and xy = 21 then find the value of x^3 - y^3; Q2. If x + 1/x = 7, find the value of x^3 + 1/x^3
Session # 14 : Solved Example: Q: Simplify : ( 4x + 2y )^3 - (4x - 2y )^3
Session # 15 : Solved Example: Sum and Difference of Cubes Q: Find the product (7a-5b)(49a^2 + 35ab + 25b^2)
Session # 16 : Solved Example: Sum and Difference of Cubes Q: Find the product: (0.9x + 0.7y)(0.81x^2 - 0.63xy + 0.49y^2)
Session # 17 : Solved Example: Sum and Difference of Cubes Q: Simplify: (6m-n)(36m^2 + 6mn + n^2) - (3m + 2n)^3
Session # 18 : Solved Example: Sum and Difference of Cubes Q: If a + b = 7 and ab =12 then find the value of (a) a^3 + b^3 (b) a^2 - ab + b^2
## Concept Lectures
Session # 19 : Special Identity: a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca); CONDITIONAL IDENTITY: If a + b + c = 0; then a^3 + b^3 + c^3 = 3abc
## Problem Solving Sessions
Session # 20 : Q: Find the product: (x - y + 2z)(x^2 +y^2 + 4z^2 - xy -2yz + 2zx)
Session # 21 : Q: If a + b + c = 6 and ab + bc + ca = 11 then find the value of a^3 + b^3 + c^3 - 3abc
Session # 22 : Q: Simplify: [(a^2 - b^2)^2 + (b^2 - c^2)^2 + (c^2 - a^2)^2]/[(a - b) + (b - c) + (c - a)]
## Concept Lectures
Session # 23 : Sophie Germain Identity | 1,112 | 2,675 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.346176 |
https://www.physicsforums.com/threads/calculating-unit-vector.128338/ | 1,508,703,419,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825436.78/warc/CC-MAIN-20171022184824-20171022204824-00650.warc.gz | 982,967,699 | 14,640 | # Calculating unit vector
1. Aug 8, 2006
### hexa
I'll be greatful for any hint.
Imagine you walk over the contourlines of the map of a mountain (really! that's the question) with a gradient of h(x,y)=2xy, x^2). You are at point (1,3) and you want to walk downslope at an angle of 45 degrees. calculate the unit vector in order to find out in which direction to walk.
Hexa
2. Aug 8, 2006
### HallsofIvy
Staff Emeritus
What's this doing in precalculus? This is a partial derivative problem. If the gradient is (2xy, x2), then the derivative in the direction of unit vector (a, b) is just the dot product, 2xya+ x2b. If you are walking downslope at an angle of 45 degrees, then the slope must be tan(45)= 1. You want 2xya+ x2b= 1 and, of course, a2+ b2= 1. Solve for a and b in terms of x and y.
3. Aug 8, 2006
### hexa
Thanks a lot for your help. I'll work on with this.
Calculus: not a term used in this country, so I'm not quiet sure where the border is between precalculus and calculus.
4. Aug 8, 2006
### HallsofIvy
Staff Emeritus
Then you might say "analysis" or "applied analysis". Essentially, derivatives and integrals are calculus. The basics of limits might be in calculus or pre-calculus. | 362 | 1,213 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43 | longest | en | 0.892813 |
https://codedump.io/share/LxDk2bw0IRR/1/determining-the-time-at-which-a-date-starts | 1,524,394,080,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945584.75/warc/CC-MAIN-20180422100104-20180422120104-00517.warc.gz | 587,961,176 | 9,011 | ikegami - 1 year ago 54
Perl Question
# Determining the time at which a date starts
Say I want to create a daily planner, and I want to divide the day into 15 minute chunks.
Easy, right? Just start at midnight, and... Wrong! In America/Sao_Paulo, one day each year starts at 01:00 because of Daylight Saving Time changes.
Given a time zone and a date, how does one find the epoch time at which the day starts?
My first thought was to use the following, but it assumes each day has a 23:59. That's probably no better of an assumption than assuming each day has a midnight.
``````perl -MDateTime -E'
say
DateTime->new( year => 2013, month => 10, day => 20 )
->subtract( days => 1 )
->set( hour => 23, minute => 59 )
->set_time_zone("America/Sao_Paulo")
->strftime("%H:%M");
'
01:00
``````
Is there a more robust or more direct alternative?
Here's a solution using only DT's public methods:
``````sub day_start {
my \$dt = shift;
my \$tz = shift;
# Expect floating or UTC, but let's make sure.
\$dt->set_time_zone('floating');
\$dt->truncate( to => 'day' );
# This will work except for days without midnight.
if (eval { \$dt->set_time_zone(\$tz); 1 }) {
return \$dt;
}
my \$ymd = \$dt->ymd();
my \$min_epoch = \$dt->epoch() - 48*60*60;
my \$max_epoch = \$dt->epoch() + 48*60*60;
while (1) {
my \$current_epoch = int( ( \$min_epoch + \$max_epoch )/2 );
_set_epoch(\$dt, \$current_epoch);
my \$current_ymd = \$dt->ymd();
if (\$current_ymd lt \$ymd) {
\$min_epoch = \$current_epoch + 1;
} else {
\$dt->subtract( seconds => 1 );
my \$earlier_ymd = \$dt->ymd();
if (\$earlier_ymd ge \$ymd) {
\$max_epoch = \$current_epoch - 1;
} else {
_set_epoch(\$dt, \$current_epoch);
return \$dt;
}
}
}
}
# Based on DateTime::from_epoch
sub _set_epoch {
my (\$dt, \$epoch) = @_;
my %args;
# # Epoch may come from Time::HiRes, so it may not be an integer.
# (\$epoch, my \$dec) = \$epoch =~ /^(-?\d+)?(\.\d+)?/;
# \$epoch ||= 0;
#
# \$args{nanosecond} = int(\$dec * 1_000_000_000) if \$dec;
# Note: For very large negative values, this may give a blatantly wrong answer.
@args{qw( second minute hour day month year )} = gmtime(\$epoch);
\$args{year} += 1900;
++\$args{month};
my \$tz = \$dt->time_zone();
\$dt->set_time_zone('UTC');
\$dt->set(%args);
\$dt->set_time_zone(\$tz);
return \$dt;
}
``````
Assumptions:
• There is no dt to which one can add time to obtain a dt with an earlier date.
• In no time zone does a date starts more than 48*60*60 seconds before the date starts in UTC.
• In no time zone does a date starts more than 48*60*60 seconds after the date starts in UTC.
Test:
``````sub new_date {
my \$y = shift;
my \$m = shift;
my \$d = shift;
return DateTime->new(
year => \$y, month => \$m, day => \$d,
@_,
hour => 0, minute => 0, second => 0, nanosecond => 0,
time_zone => 'floating',
);
}
{
# No midnight.
my \$tz = DateTime::TimeZone->new( name => 'America/Sao_Paulo' );
my \$dt = day_start(new_date(2013, 10, 20), \$tz);
print(\$dt->epoch, " ", \$dt->iso8601, "\n"); # 1382238000 2013-10-20T01:00:00
\$dt->subtract( seconds => 1 );
print(\$dt->epoch, " ", \$dt->iso8601, "\n"); # 1382237999 2013-10-19T23:59:59
}
{
# Two midnights.
my \$tz = DateTime::TimeZone->new( name => 'America/Havana' );
my \$dt = day_start(new_date(2013, 11, 3), \$tz);
print(\$dt->epoch, " ", \$dt->iso8601, "\n"); # 1383451200 2013-11-03T00:00:00
\$dt->subtract( seconds => 1 );
print(\$dt->epoch, " ", \$dt->iso8601, "\n"); # 1383451199 2013-11-02T23:59:59
}
``````
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https://www.geeksforgeeks.org/find-index-of-0-to-be-replaced-with-1-to-get-longest-continuous-sequence-of-1s-in-a-binary-array-set-2/?ref=rp | 1,656,805,560,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104205534.63/warc/CC-MAIN-20220702222819-20220703012819-00209.warc.gz | 833,468,299 | 36,932 | # Find Index of 0 to be replaced with 1 to get longest continuous sequence of 1s in a binary array | Set-2
• Difficulty Level : Medium
• Last Updated : 26 Apr, 2021
Given an array of 0s and 1s, find the position of 0 to be replaced with 1 to get longest continuous sequence of 1s. Expected time complexity is O(n) and auxiliary space is O(1).
Examples:
```Input : arr[] = {1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1}
Output : Index 9
Assuming array index starts from 0, replacing 0 with 1 at
index 9 causes the maximum continuous sequence of 1s.
Input : arr[] = {1, 1, 1, 1, 0}
Output : Index 4```
We have discussed a solution for this problem in previous post. In this post two more methods to solve the problem are discussed.
Method 1(Using count of ones on both sides of zero): The idea is to count number of ones on both sides of each zero. The required index is the index of zero having maximum number of ones around it. Following variables are used in implementation:
• leftCnt: To store count of ones on left side of current element zero under consideration.
• rightCnt: To store count of ones on right side of current element zero under consideration.
• maxIndex: Index of zero with maximum number of ones around it.
• lastInd: Index of last zero element seen
• maxCnt: Count of ones if zero at index maxInd is replaced by one.
Keep incrementing rightCnt until one is present in input array. Let next zero is present at index i. Check if this zero element is first zero element or not. It is first zero element if lastInd does not hold any valid index value. In that case update lastInd with i. Now the value of rightCnt is number of zeroes on left side of this zero. So leftCnt is equal to rightCnt and then again find value of rightCnt. If current zero element is not first zero, then number of ones around zero present at index lastInd is given by leftCnt + rightCnt. maxCnt will take value leftCnt + rightCnt + 1 and maxIndex = lastInd if this value is less than value currently held by maxCnt. Now rightCnt will become leftCnt for zero at index i and lastInd will be equal to i. Again find value of rightCnt, compare number of ones with maxcnt and update maxCnt and maxIndex accordingly. Repeat this procedure for each subsequent zero element of the array. Observe that lastInd stores the index of zero for which current leftCnt and rightCnt are calculated. The required index of zero to be replaced with one is stored in maxIndex.
Following is the implementation of the above algorithm.
## C++
`// C++ program to find index of zero``// to be replaced by one to get longest``// continuous sequence of ones.``#include ``using` `namespace` `std;` `// Returns index of 0 to be replaced``// with 1 to get longest continuous``// sequence of 1s. If there is no 0``// in array, then it returns -1.``int` `maxOnesIndex(``bool` `arr[], ``int` `n)``{`` ``int` `i = 0;` ` ``// To store count of ones on left`` ``// side of current element zero`` ``int` `leftCnt = 0;` ` ``// To store count of ones on right`` ``// side of current element zero`` ``int` `rightCnt = 0;` ` ``// Index of zero with maximum number`` ``// of ones around it.`` ``int` `maxIndex = -1;` ` ``// Index of last zero element seen`` ``int` `lastInd = -1;` ` ``// Count of ones if zero at index`` ``// maxInd is replaced by one.`` ``int` `maxCnt = 0;` ` ``while` `(i < n) {` ` ``// Keep incrementing count until`` ``// current element is 1.`` ``if` `(arr[i]) {`` ``rightCnt++;`` ``}` ` ``else` `{` ` ``// If current zero element`` ``// is not first zero element,`` ``// then count number of ones`` ``// obtained by replacing zero at`` ``// index lastInd. Update maxCnt`` ``// and maxIndex if required.`` ``if` `(lastInd != -1) {`` ``if` `(rightCnt + leftCnt + 1 > maxCnt) {`` ``maxCnt = leftCnt + rightCnt + 1;`` ``maxIndex = lastInd;`` ``}`` ``}`` ``lastInd = i;`` ``leftCnt = rightCnt;`` ``rightCnt = 0;`` ``}` ` ``i++;`` ``}` ` ``// Find number of ones in continuous`` ``// sequence when last zero element is`` ``// replaced by one.`` ``if` `(lastInd != -1) {`` ``if` `(leftCnt + rightCnt + 1 > maxCnt) {`` ``maxCnt = leftCnt + rightCnt + 1;`` ``maxIndex = lastInd;`` ``}`` ``}` ` ``return` `maxIndex;``}` `// Driver function``int` `main()``{`` ``bool` `arr[] = { 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1 };`` ``// bool arr[] = {1, 1, 1, 1, 0};` ` ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);`` ``cout << ``"Index of 0 to be replaced is "`` ``<< maxOnesIndex(arr, n);`` ``return` `0;``}`
## Java
`// Java program to find index of zero``// to be replaced by one to get longest``// continuous sequence of ones.` `class` `GFG {` `// Returns index of 0 to be replaced``// with 1 to get longest continuous``// sequence of 1s. If there is no 0``// in array, then it returns -1.`` ``static` `int` `maxOnesIndex(``boolean` `arr[], ``int` `n) {`` ``int` `i = ``0``;` ` ``// To store count of ones on left`` ``// side of current element zero`` ``int` `leftCnt = ``0``;` ` ``// To store count of ones on right`` ``// side of current element zero`` ``int` `rightCnt = ``0``;` ` ``// Index of zero with maximum number`` ``// of ones around it.`` ``int` `maxIndex = -``1``;` ` ``// Index of last zero element seen`` ``int` `lastInd = -``1``;` ` ``// Count of ones if zero at index`` ``// maxInd is replaced by one.`` ``int` `maxCnt = ``0``;` ` ``while` `(i < n) {` ` ``// Keep incrementing count until`` ``// current element is 1.`` ``if` `(arr[i]) {`` ``rightCnt++;`` ``} ``else` `{` ` ``// If current zero element`` ``// is not first zero element,`` ``// then count number of ones`` ``// obtained by replacing zero at`` ``// index lastInd. Update maxCnt`` ``// and maxIndex if required.`` ``if` `(lastInd != -``1``) {`` ``if` `(rightCnt + leftCnt + ``1` `> maxCnt) {`` ``maxCnt = leftCnt + rightCnt + ``1``;`` ``maxIndex = lastInd;`` ``}`` ``}`` ``lastInd = i;`` ``leftCnt = rightCnt;`` ``rightCnt = ``0``;`` ``}` ` ``i++;`` ``}` ` ``// Find number of ones in continuous`` ``// sequence when last zero element is`` ``// replaced by one.`` ``if` `(lastInd != -``1``) {`` ``if` `(leftCnt + rightCnt + ``1` `> maxCnt) {`` ``maxCnt = leftCnt + rightCnt + ``1``;`` ``maxIndex = lastInd;`` ``}`` ``}` ` ``return` `maxIndex;`` ``}` ` ``// Driver function`` ``public` `static` `void` `main(String[] args) {`` ``boolean` `arr[] = {``true``, ``true``, ``false``, ``false``, ``true``,`` ``false``, ``true``, ``true``, ``true``, ``false``, ``true``, ``true``, ``true``,};` ` ``int` `n = arr.length;`` ``System.out.println(``"Index of 0 to be replaced is "`` ``+ maxOnesIndex(arr, n));`` ``}``}``//This code contribute by Shikha Singh`
## Python3
`# Python3 program to find index of zero``# to be replaced by one to get longest``# continuous sequence of ones.` `# Returns index of 0 to be replaced``# with 1 to get longest continuous``# sequence of 1s. If there is no 0``# in array, then it returns -1.``def` `maxOnesIndex(arr, n):` ` ``i ``=` `0` ` ``# To store count of ones on left`` ``# side of current element zero`` ``leftCnt ``=` `0` ` ``# To store count of ones on right`` ``# side of current element zero`` ``rightCnt ``=` `0` ` ``# Index of zero with maximum number`` ``# of ones around it.`` ``maxIndex ``=` `-``1` ` ``# Index of last zero element seen`` ``lastInd ``=` `-``1` ` ``# Count of ones if zero at index`` ``# maxInd is replaced by one.`` ``maxCnt ``=` `0` ` ``while` `i < n:` ` ``# Keep incrementing count until`` ``# current element is 1.`` ``if` `arr[i] ``=``=` `1``:`` ``rightCnt ``+``=` `1` ` ``else``:`` ``# If current zero element`` ``# is not first zero element,`` ``# then count number of ones`` ``# obtained by replacing zero at`` ``# index lastInd. Update maxCnt`` ``# and maxIndex if required.`` ``if` `lastInd !``=` `-``1``:`` ``if` `rightCnt ``+` `leftCnt ``+` `1` `> maxCnt:`` ``maxCnt ``=` `leftCnt ``+` `rightCnt ``+` `1`` ``maxIndex ``=` `lastInd`` ` ` ``lastInd ``=` `i`` ``leftCnt ``=` `rightCnt`` ``rightCnt ``=` `0`` ` ` ``i ``+``=` `1` ` ``# Find number of ones in continuous`` ``# sequence when last zero element is`` ``# replaced by one.`` ``if` `lastInd !``=` `-``1``:`` ``if` `leftCnt ``+` `rightCnt ``+` `1` `> maxCnt:`` ``maxCnt ``=` `leftCnt ``+` `rightCnt ``+` `1`` ``maxIndex ``=` `lastInd`` ` ` ``return` `maxIndex` `# Driver code``if` `__name__ ``=``=` `"__main__"``:` ` ``arr ``=` `[``1``, ``1``, ``0``, ``0``, ``1``, ``0``, ``1``,`` ``1``, ``1``, ``0``, ``1``, ``1``, ``1``]`` ` ` ``n ``=` `len``(arr)`` ``print``(``"Index of 0 to be replaced is"``,`` ``maxOnesIndex(arr, n))` `# This code is contributed``# by Rituraj Jain`
## C#
`// C# program to find index of zero``// to be replaced by one to get longest``// continuous sequence of ones.` `using` `System;` `public` `class` `GFG{` `// Returns index of 0 to be replaced``// with 1 to get longest continuous``// sequence of 1s. If there is no 0``// in array, then it returns -1.`` ``static` `int` `maxOnesIndex(``bool` `[]arr, ``int` `n) {`` ``int` `i = 0;` ` ``// To store count of ones on left`` ``// side of current element zero`` ``int` `leftCnt = 0;` ` ``// To store count of ones on right`` ``// side of current element zero`` ``int` `rightCnt = 0;` ` ``// Index of zero with maximum number`` ``// of ones around it.`` ``int` `maxIndex = -1;` ` ``// Index of last zero element seen`` ``int` `lastInd = -1;` ` ``// Count of ones if zero at index`` ``// maxInd is replaced by one.`` ``int` `maxCnt = 0;` ` ``while` `(i < n) {` ` ``// Keep incrementing count until`` ``// current element is 1.`` ``if` `(arr[i]) {`` ``rightCnt++;`` ``} ``else` `{` ` ``// If current zero element`` ``// is not first zero element,`` ``// then count number of ones`` ``// obtained by replacing zero at`` ``// index lastInd. Update maxCnt`` ``// and maxIndex if required.`` ``if` `(lastInd != -1) {`` ``if` `(rightCnt + leftCnt + 1 > maxCnt) {`` ``maxCnt = leftCnt + rightCnt + 1;`` ``maxIndex = lastInd;`` ``}`` ``}`` ``lastInd = i;`` ``leftCnt = rightCnt;`` ``rightCnt = 0;`` ``}` ` ``i++;`` ``}` ` ``// Find number of ones in continuous`` ``// sequence when last zero element is`` ``// replaced by one.`` ``if` `(lastInd != -1) {`` ``if` `(leftCnt + rightCnt + 1 > maxCnt) {`` ``maxCnt = leftCnt + rightCnt + 1;`` ``maxIndex = lastInd;`` ``}`` ``}` ` ``return` `maxIndex;`` ``}` ` ``// Driver function`` ``static` `public` `void` `Main (){`` ``bool` `[]arr = {``true``, ``true``, ``false``, ``false``, ``true``,`` ``false``, ``true``, ``true``, ``true``, ``false``, ``true``, ``true``, ``true``,};` ` ``int` `n = arr.Length;`` ``Console.WriteLine(``"Index of 0 to be replaced is "`` ``+ maxOnesIndex(arr, n));`` ``}``}``//This code contribute by ajit`
## PHP
` ``\$maxCnt``)`` ``{`` ``\$maxCnt` `= ``\$leftCnt` `+ ``\$rightCnt` `+ 1;`` ``\$maxIndex` `= ``\$lastInd``;`` ``}`` ``}`` ``\$lastInd` `= ``\$i``;`` ``\$leftCnt` `= ``\$rightCnt``;`` ``\$rightCnt` `= 0;`` ``}` ` ``\$i``++;`` ``}` ` ``// Find number of ones in continuous`` ``// sequence when last zero element is`` ``// replaced by one.`` ``if` `(``\$lastInd` `!= -1)`` ``{`` ``if` `(``\$leftCnt` `+ ``\$rightCnt` `+ 1 > ``\$maxCnt``)`` ``{`` ``\$maxCnt` `= ``\$leftCnt` `+ ``\$rightCnt` `+ 1;`` ``\$maxIndex` `= ``\$lastInd``;`` ``}`` ``}` ` ``return` `\$maxIndex``;``}` `// Driver Code``\$arr` `= ``array``(1, 1, 0, 0, 1, 0,`` ``1, 1, 1, 0, 1, 1, 1);``// bool arr[] = {1, 1, 1, 1, 0};` `\$n` `= sizeof(``\$arr``);``echo` `"Index of 0 to be replaced is "``.`` ``maxOnesIndex(``\$arr``, ``\$n``);` `// This code is contributed``// by Akanksha Rai``?>`
## Javascript
``
Output:
`Index of 0 to be replaced is 9`
Time Complexity: O(n)
Auxiliary Space: O(1)
Method 2(Using sliding window): A sliding window is used to find number of ones in longest continuous sequence obtained by replacing a zero. The idea is to keep incrementing ending point of sliding window until one is present in input array. When a zero is found, check if it is first zero element or not. If it is first zero element then the sliding window is expanded further. If it is not, then find the length of sliding window. This length is number of ones obtained by replacing the zero element present in sliding window. Note that this length gives number of ones obtained by replacing previous zero element and not current zero element. For current zero element, the starting point of sliding window is index next to the index of previous zero element.
Following is the implementation of the above algorithm.
## C++
`// C++ program to find index of zero``// to be replaced by one to get longest``// continuous sequence of ones.``#include ``using` `namespace` `std;` `// Returns index of 0 to be replaced``// with 1 to get longest continuous``// sequence of 1s. If there is no 0``// in array, then it returns -1.``int` `maxOnesIndex(``bool` `arr[], ``int` `n)``{` ` ``// To store starting point of`` ``// sliding window.`` ``int` `start = 0;` ` ``// To store ending point of`` ``// sliding window.`` ``int` `end = 0;` ` ``// Index of zero with maximum number`` ``// of ones around it.`` ``int` `maxIndex = -1;` ` ``// Index of last zero element seen`` ``int` `lastInd = -1;` ` ``// Count of ones if zero at index`` ``// maxInd is replaced by one.`` ``int` `maxCnt = 0;` ` ``while` `(end < n) {` ` ``// Keep increasing ending point`` ``// of sliding window until one is`` ``// present in input array.`` ``while` `(end < n && arr[end]) {`` ``end++;`` ``}` ` ``// If this is not first zero element`` ``// then number of ones obtained by`` ``// replacing zero at lastInd is`` ``// equal to length of window.`` ``// Compare this with maximum number`` ``// of ones in a previous window so far.`` ``if` `(maxCnt < end - start && lastInd != -1) {`` ``maxCnt = end - start;`` ``maxIndex = lastInd;`` ``}` ` ``// The new starting point of next window`` ``// is from index position next to last`` ``// zero which is stored in lastInd.`` ``start = lastInd + 1;`` ``lastInd = end;`` ``end++;`` ``}` ` ``// For the case when only one zero is`` ``// present in input array and is at`` ``// last position.`` ``if` `(maxCnt < end - start && lastInd != -1) {`` ``maxCnt = end - start;`` ``maxIndex = lastInd;`` ``}` ` ``return` `maxIndex;``}` `// Driver function``int` `main()``{`` ``bool` `arr[] = { 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1 };`` ``// bool arr[] = {1, 1, 1, 1, 0};` ` ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);`` ``cout << ``"Index of 0 to be replaced is "`` ``<< maxOnesIndex(arr, n);`` ``return` `0;``}`
## Java
`// Java program to find index of zero``// to be replaced by one to get longest``// continuous sequence of ones.` `public` `class` `GFG {` `// Returns index of 0 to be replaced``// with 1 to get longest continuous``// sequence of 1s. If there is no 0``// in array, then it returns -1.`` ``static` `int` `maxOnesIndex(``boolean` `arr[], ``int` `n) {` ` ``// To store starting point of`` ``// sliding window.`` ``int` `start = ``0``;` ` ``// To store ending point of`` ``// sliding window.`` ``int` `end = ``0``;` ` ``// Index of zero with maximum number`` ``// of ones around it.`` ``int` `maxIndex = -``1``;` ` ``// Index of last zero element seen`` ``int` `lastInd = -``1``;` ` ``// Count of ones if zero at index`` ``// maxInd is replaced by one.`` ``int` `maxCnt = ``0``;` ` ``while` `(end < n) {` ` ``// Keep increasing ending point`` ``// of sliding window until one is`` ``// present in input array.`` ``while` `(end < n && arr[end]) {`` ``end++;`` ``}` ` ``// If this is not first zero element`` ``// then number of ones obtained by`` ``// replacing zero at lastInd is`` ``// equal to length of window.`` ``// Compare this with maximum number`` ``// of ones in a previous window so far.`` ``if` `(maxCnt < end - start && lastInd != -``1``) {`` ``maxCnt = end - start;`` ``maxIndex = lastInd;`` ``}` ` ``// The new starting point of next window`` ``// is from index position next to last`` ``// zero which is stored in lastInd.`` ``start = lastInd + ``1``;`` ``lastInd = end;`` ``end++;`` ``}` ` ``// For the case when only one zero is`` ``// present in input array and is at`` ``// last position.`` ``if` `(maxCnt < end - start && lastInd != -``1``) {`` ``maxCnt = end - start;`` ``maxIndex = lastInd;`` ``}` ` ``return` `maxIndex;`` ``}` ` ``// Driver function`` ``static` `public` `void` `main(String[] args) {`` ``boolean` `arr[] = {``true``, ``true``, ``false``, ``false``, ``true``,`` ``false``, ``true``, ``true``, ``true``, ``false``, ``true``, ``true``, ``true``,};` ` ``// bool arr[] = {1, 1, 1, 1, 0};`` ``int` `n = arr.length;`` ``System.out.println(``"Index of 0 to be replaced is "`` ``+ maxOnesIndex(arr, n));`` ``}``}` `// This code is contributed by Rajput-Ji`
## Python3
`# Python3 program to find index of zero``# to be replaced by one to get longest``# continuous sequence of ones.` `# Returns index of 0 to be replaced``# with 1 to get longest continuous``# sequence of 1s. If there is no 0``# in array, then it returns -1.``def` `maxOnesIndex(arr, n):` ` ``# To store starting point of`` ``# sliding window.`` ``start ``=` `0` ` ``# To store ending point of`` ``# sliding window.`` ``end ``=` `0` ` ``# Index of zero with maximum`` ``# number of ones around it.`` ``maxIndex ``=` `-``1` ` ``# Index of last zero element seen`` ``lastInd ``=` `-``1` ` ``# Count of ones if zero at index`` ``# maxInd is replaced by one.`` ``maxCnt ``=` `0` ` ``while` `(end < n) :` ` ``# Keep increasing ending point`` ``# of sliding window until one is`` ``# present in input array.`` ``while` `(end < n ``and` `arr[end]) :`` ``end ``+``=` `1`` ` ` ``# If this is not first zero element`` ``# then number of ones obtained by`` ``# replacing zero at lastInd is`` ``# equal to length of window.`` ``#Compare this with maximum number`` ``# of ones in a previous window so far.`` ``if` `(maxCnt < end ``-` `start ``and` `lastInd !``=` `-``1``) :`` ``maxCnt ``=` `end ``-` `start`` ``maxIndex ``=` `lastInd`` ` ` ``# The new starting point of next window`` ``# is from index position next to last`` ``# zero which is stored in lastInd.`` ``start ``=` `lastInd ``+` `1`` ``lastInd ``=` `end`` ``end ``+``=` `1`` ` ` ``# For the case when only one zero is`` ``# present in input array and is at`` ``# last position.`` ``if` `(maxCnt < end ``-` `start ``and` `lastInd !``=` `-``1``) :`` ``maxCnt ``=` `end ``-` `start`` ``maxIndex ``=` `lastInd` ` ``return` `maxIndex` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:` ` ``arr ``=` `[``1``, ``1``, ``0``, ``0``, ``1``, ``0``, ``1``,`` ``1``, ``1``, ``0``, ``1``, ``1``, ``1` `]`` ``# arr= [1, 1, 1, 1, 0]` ` ``n ``=` `len``(arr)`` ``print` `(``"Index of 0 to be replaced is "``,`` ``maxOnesIndex(arr, n))` `# This code is contributed by ChitraNayal`
## C#
`using` `System;` `// c# program to find index of zero ``// to be replaced by one to get longest ``// continuous sequence of ones.` `public` `class` `GFG``{` `// Returns index of 0 to be replaced ``// with 1 to get longest continuous ``// sequence of 1s. If there is no 0 ``// in array, then it returns -1. `` ``public` `static` `int` `maxOnesIndex(``bool``[] arr, ``int` `n)`` ``{` ` ``// To store starting point of `` ``// sliding window. `` ``int` `start = 0;` ` ``// To store ending point of `` ``// sliding window. `` ``int` `end = 0;` ` ``// Index of zero with maximum number `` ``// of ones around it. `` ``int` `maxIndex = -1;` ` ``// Index of last zero element seen `` ``int` `lastInd = -1;` ` ``// Count of ones if zero at index `` ``// maxInd is replaced by one. `` ``int` `maxCnt = 0;` ` ``while` `(end < n)`` ``{` ` ``// Keep increasing ending point `` ``// of sliding window until one is `` ``// present in input array. `` ``while` `(end < n && arr[end])`` ``{`` ``end++;`` ``}` ` ``// If this is not first zero element `` ``// then number of ones obtained by `` ``// replacing zero at lastInd is `` ``// equal to length of window. `` ``// Compare this with maximum number `` ``// of ones in a previous window so far. `` ``if` `(maxCnt < end - start && lastInd != -1)`` ``{`` ``maxCnt = end - start;`` ``maxIndex = lastInd;`` ``}` ` ``// The new starting point of next window `` ``// is from index position next to last `` ``// zero which is stored in lastInd. `` ``start = lastInd + 1;`` ``lastInd = end;`` ``end++;`` ``}` ` ``// For the case when only one zero is `` ``// present in input array and is at `` ``// last position. `` ``if` `(maxCnt < end - start && lastInd != -1)`` ``{`` ``maxCnt = end - start;`` ``maxIndex = lastInd;`` ``}` ` ``return` `maxIndex;`` ``}` ` ``// Driver function `` ``public` `static` `void` `Main(``string``[] args)`` ``{`` ``bool``[] arr = ``new` `bool``[] {``true``, ``true``, ``false``, ``false``, ``true``, ``false``, ``true``, ``true``, ``true``, ``false``, ``true``, ``true``, ``true``};` ` ``// bool arr[] = {1, 1, 1, 1, 0}; `` ``int` `n = arr.Length;`` ``Console.WriteLine(``"Index of 0 to be replaced is "` `+ maxOnesIndex(arr, n));`` ``}``}` `// This code is contributed by Shrikant13`
## PHP
``
## Javascript
``
Output:
`Index of 0 to be replaced is 9`
Time Complexity: O(n)
Auxiliary Space: O(1)
My Personal Notes arrow_drop_up | 7,873 | 24,787 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2022-27 | latest | en | 0.873629 |
https://www.numbersaplenty.com/99996749 | 1,606,484,084,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141193221.49/warc/CC-MAIN-20201127131802-20201127161802-00506.warc.gz | 783,370,599 | 3,311 | Search a number
99996749 is a prime number
BaseRepresentation
bin1011111010111…
…01010001001101
320222011100202022
411331131101031
5201044343444
613531135525
72322646466
oct575352115
9228140668
1099996749
115149a046
12295a45a5
1317942198
14d3cdd6d
158ba39ee
hex5f5d44d
99996749 has 2 divisors, whose sum is σ = 99996750. Its totient is φ = 99996748.
The previous prime is 99996719. The next prime is 99996769. The reversal of 99996749 is 94769999.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 57002500 + 42994249 = 7550^2 + 6557^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-99996749 is a prime.
It is a Sophie Germain prime.
It is a Curzon number.
It is a junction number, because it is equal to n+sod(n) for n = 99996691 and 99996700.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (99996719) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 49998374 + 49998375.
It is an arithmetic number, because the mean of its divisors is an integer number (49998375).
Almost surely, 299996749 is an apocalyptic number.
It is an amenable number.
99996749 is a deficient number, since it is larger than the sum of its proper divisors (1).
99996749 is an equidigital number, since it uses as much as digits as its factorization.
99996749 is an evil number, because the sum of its binary digits is even.
The product of its digits is 9920232, while the sum is 62.
The square root of 99996749 is about 9999.8374486789. The cubic root of 99996749 is about 464.1538533717.
The spelling of 99996749 in words is "ninety-nine million, nine hundred ninety-six thousand, seven hundred forty-nine". | 546 | 1,803 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2020-50 | longest | en | 0.876792 |
http://mathforum.org/kb/thread.jspa?threadID=2618978&messageID=9382609 | 1,524,540,044,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946453.89/warc/CC-MAIN-20180424022317-20180424042317-00105.warc.gz | 196,765,220 | 5,824 | Search All of the Math Forum:
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Topic: Matheology § 430
Replies: 28 Last Post: Feb 12, 2014 4:06 AM
Messages: [ Previous | Next ]
wolfgang.mueckenheim@hs-augsburg.de Posts: 3,394 Registered: 10/18/08
Re: Matheology § 430
Posted: Feb 10, 2014 12:22 PM
Am Montag, 10. Februar 2014 15:32:56 UTC+1 schrieb thenewc...@gmail.com:
> For example: lim (x->oo) (1-1/(10^x)) = 1 does not mean the expression (1-1/(10^x)) ever attains the value of 1. It NEVER does!
True.
> It means that no matter how large x becomes, the value of the expression can never attain the value of 1.
I see you do not like to call expressions with infinite sequences of non-zero-digits numbers. That is correct, in principle, because without actual infinity, which is a self-contradictory concept, there is always a difference between 0.999... and 1. The idea that "in the infinite" the complete sequence of nines could be used as a substitution for 1.000... has deplorably lead to Cantor's mistake.
Nevertheless we can understand "0.999..." as a shorthand for "the supremum of the increasing sequence
0.9
0.99
0.999
..."
i.e., for the smallest number that will never be surpassed, not even be reached. And that is 1 with no doubt. To call it a number is a matter of taste or convenience. You can simply use it as a definition. Then nothing need be proved.
Regards, WM | 427 | 1,522 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2018-17 | latest | en | 0.862727 |
https://jp.mathworks.com/matlabcentral/cody/problems/1229-determine-the-number-of-odd-integers-in-a-vector/solutions/1696439 | 1,603,905,229,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107900200.97/warc/CC-MAIN-20201028162226-20201028192226-00256.warc.gz | 368,519,178 | 17,267 | Cody
# Problem 1229. Determine the number of odd integers in a vector
Solution 1696439
Submitted on 19 Dec 2018 by Chris Hwang
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 5 8 3 7 1]; y_correct = 4; assert(isequal(NumOdd(x),y_correct))
2 Pass
x = [2 5 0 3 7 0]; y_correct = 3; assert(isequal(NumOdd(x),y_correct))
3 Pass
x = [2 5 -10 3 -7 0]; y_correct = 3; assert(isequal(NumOdd(x),y_correct))
4 Pass
x = [2 4 -6 8 -4 0]; y_correct = 0; assert(isequal(NumOdd(x),y_correct))
5 Pass
x = [2 5 -10 3 2 0]; y_correct = 2; assert(isequal(NumOdd(x),y_correct))
6 Pass
x = []; y_correct = 0; assert(isequal(NumOdd(x),y_correct))
7 Pass
x = [1.5]; y_correct = 0; assert(isequal(NumOdd(x),y_correct))
8 Pass
x = [-11.9 3.7 5.01]; y_correct = 0; assert(isequal(NumOdd(x),y_correct)) | 349 | 926 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2020-45 | latest | en | 0.432636 |
https://www.ulektz.com/home/bookDetail/MTIwMQ==/PME3D001-APPLIED-MATHEMATICS | 1,579,963,484,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251672537.90/warc/CC-MAIN-20200125131641-20200125160641-00544.warc.gz | 1,112,601,089 | 30,202 | # APPLIED MATHEMATICS
Course Code : PME3D001 Author : uLektz University : Biju Patnaik University of Technology (BPUT) Regulation : 2016 Categories : Mechanical Format : ePUB3 (DRM Protected) Type : eBook
FREE
Description :APPLIED MATHEMATICS of PME3D001 covers the latest syllabus prescribed by Biju Patnaik University of Technology (BPUT) for regulation 2016. Author: uLektz, Published by uLektz Learning Solutions Private Limited.
Note : No printed book. Only ebook. Access eBook using uLektz apps for Android, iOS and Windows Desktop PC.
##### Topics
###### UNIT - I PROBABILITY AND STATISTICS
1.1 Probability, Random variables, Probability distributions
1.2 Mean and variance of distribution, Binomial, Poisson and Hyper geometric distributions
1.3 Normal and exponential distribution, Distribution of several random variables
1.4 Statistics: Random sampling, Estimation of Parameters, Confidence Intervals
1.5 Testing of hypothesis, Acceptance sampling
1.6 Regression Analysis, Fitting Straight Lines, Correlation analysis
###### UNIT - II PARTIAL DIFFERENTIAL EQUATION
2.1 Partial differential equation of first order, Linear partial differential equation, Non-linear partial differential equation
2.2 Homogenous and non-homogeneous partial differential equation with constant co-efficient, Cauchy type, Monge’s method
2.3 Second order partial differential equation
2.4 The vibrating string, the wave equation and its solution, the heat equation and its solution, Two dimensional wave equation and its solution
2.5 Laplace equation in polar, cylindrical and spherical coordinates
###### UNIT - III COMPLEX ANALYSIS AND COMPLEX INTEGRATION
3.1 Complex Analysis: Analytic function, Cauchy-Riemann equations
3.2 Laplace equation, Conformal mapping
3.3 Complex integration: Line integral in the complex plane, Cauchy’s integral theorem, Cauchy’s integral formula, Derivatives of analytic functions
###### UNIT - IV POWER SERIES
4.1 Power Series, Taylor’s series, Laurent’s series
4.2 Singularities and zeros
4.3 Residue integration method, Evaluation of real integrals | 483 | 2,097 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2020-05 | latest | en | 0.722285 |
http://oeis.org/A305695 | 1,571,599,166,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986718918.77/warc/CC-MAIN-20191020183709-20191020211209-00243.warc.gz | 149,078,815 | 4,333 | This site is supported by donations to The OEIS Foundation.
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A305695 Triangle T(n,k) read by rows: fibonomial coefficients sums triangle. 1
1, 2, 1, 4, 3, 1, 7, 9, 4, 1, 12, 24, 19, 6, 1, 20, 64, 79, 46, 9, 1, 33, 168, 339, 306, 113, 14, 1, 54, 441, 1431, 2126, 1205, 287, 22, 1, 88, 1155, 6072, 14502, 13581, 4928, 736, 35, 1, 143, 3025, 25707, 99587, 149717, 90013, 20371, 1905, 56, 1 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS The triangle coefficients give sums of Fibonacci powers when multiplied with Lang triangle coefficients and summed (see 2nd formula). LINKS FORMULA T(n, k) = T(n-1, k) + A010048(n+1, k+1). Sum_{t=0..n-1} A056588(n-1, n-1-t) * T(k+t, n-1) = Sum_{j=1..k+1} F(j)^n. EXAMPLE n\k| 0 1 2 3 4 5 6 7 8 9 ---+-------------------------------------------------- 0 | 1 1 | 2 1 2 | 4 3 1 3 | 7 9 4 1 4 | 12 24 19 6 1 5 | 20 64 79 46 9 1 6 | 33 168 339 306 113 14 1 7 | 54 441 1431 2126 1205 287 22 1 8 | 88 1155 6072 14502 13581 4928 736 35 1 9 | 143 3025 25707 99587 149717 90013 20371 1905 56 1 PROG (PARI) f(n, k) = prod(j=0, k-1, fibonacci(n-j))/prod(j=1, k, fibonacci(j)); T(n, k) = if (n< 0, 0, T(n-1, k) + f(n+1, k+1)); tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n, k), ", ")); print); \\ Michel Marcus, Jul 20 2018 CROSSREFS Cf. A000045, A010048, A000071, A056588, A317360. Sequence in context: A131254 A210229 A210213 * A211235 A134626 A115450 Adjacent sequences: A305692 A305693 A305694 * A305696 A305697 A305698 KEYWORD nonn,tabl AUTHOR Tony Foster III, Jul 09 2018 EXTENSIONS More terms from Michel Marcus, Jul 20 2018 STATUS approved
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Last modified October 20 15:15 EDT 2019. Contains 328267 sequences. (Running on oeis4.) | 915 | 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} | 3.828125 | 4 | CC-MAIN-2019-43 | latest | en | 0.365649 |
http://matlabdatamining.blogspot.com/2014/04/probability-hallowween-puzzle.html?showComment=1400657681748 | 1,490,582,945,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189377.63/warc/CC-MAIN-20170322212949-00183-ip-10-233-31-227.ec2.internal.warc.gz | 230,337,501 | 15,232 | ## Saturday, April 12, 2014
### Probability: A Halloween Puzzle
Introduction
Though Halloween is months away, I found the following interesting and thought readers might enjoy examining my solution.
Recently, I was given the following probability question to answer:
Halloween Probability Puzzler
The number of trick-or-treaters knocking on my door in any five minute interval between 6 and 8pm on Halloween night is distributed as a Poisson with a mean of 5 (ignoring time effects). The number of pieces of candy taken by each child, in addition to the expected one piece per child, is distributed as a Poisson with a mean of 1. What is the minimum number of pieces of candy that I should have on hand so that I have only a 5% probability of running out?
I am not aware of the author of this puzzle nor its origin. If anyone cares to identify either, I'd be glad to give credit where it's due.
Interpretation
I interpret the phrase "ignoring time effects", above, to mean that there is no correlation among arrivals over time; in other words, the count of trick-or-treater arrivals during any five minute interval is statistically independent of the counts in the other intervals.
So, the described model of trick-or-treater behaviors is that they arrive in random numbers during each time interval, and each take a single piece of candy plus a random amount more. The problem, then, becomes finding the 95th percentile of the distribution of candy taken during the entire evening, since that's the amount beyond which we'd run out of candy 5% of the time.
Solution Idea 1 (A Dead End)
The most direct method of solving this problem, accounting for all possible outcomes, seemed hopelessly complex to me. Trying to match each possible number of pieces of candy taken over so many arrivals over so many intervals results in an astronomical number of combinations. Suspecting that someone with more expertise in combinatorics than myself might see a way through that mess, I quickly gave up on that idea and fell back to something I know better: the computer (leading to solution idea 2... ).
Solution Idea 2
The next thing which occurred to me was Monte Carlo simulation, which is way of solving problems by approximating probability distributions using many random samples from those distributions. Sometimes very many samples are required, but contemporary computer hardware easily accommodates this need (at least for this problem). With a bit of code, I could approximate the various distributions in this problem and their interactions and have plenty of candy come October 31.
While Monte Carlo simulation and its variants are used to solve very complex real-world problems, the basic idea is very simple: Draw samples from the indicated distributions, have them interact as they do in real life, and accumulate the results. Using the poissrnd() function provided in the MATLAB Statistics Toolbox, that is exactly what I did (apologies for the dreadful formatting):
% HalloweenPuzzler
% Program to solve the Halloween Probability Puzzler by the Monte Carlo method
% by Will Dwinnell
%
% Note: Requires the poissrnd() and prctile() functions from the Statistics Toolbox
%
% Parameter
B = 4e5; % How many times should we run the nightly simulation?
% Initialize
S = zeros(B,1); % Vector of nightly simulation totals
% Loop over simulated Halloween evenings
for i = 1:B
% Loop over 5-minute time intervals for this night
for j = 1:24
% Determine number of arrivals tonight
Arrivals = poissrnd(5);
% Are there any?
if (Arrivals > 0)
% ...yes, so process them.
% Loop over tonight's trick-or-treaters
for k = 1:Arrivals
% Add this trick-or-treater's candy to the total
S(i) = S(i) + 1 + poissrnd(1);
end
end
end
end
% Determine the 95th percentile of our simulated nightly totals for the answer
disp(['You need ' int2str(prctile(S,95)) ' pieces of candy.'])
% Graph the distribution of nightly totals
figure
hist(S,50)
grid on
title({'Halloween Probability Puzzler: Monte Carlo Solution',[int2str(B) ' Replicates']})
xlabel('Candy (pieces/night)')
ylabel('Frequency')
% EOF
Fair warning: This program takes a long time to run (hours, if run on a slow machine).
Hopefully the function of this simulation is adequately explained in the comments. To simulate a single Halloween night, the program loops over all time periods (there are 24 intervals, each 5-minutes long, between 6:00 and 8:00), then all arrivals (if any) within each time period. Inside the innermost loop, the number of pieces of candy taken by the given trick-or-treater is generated. The outermost loop governs the multiple executions of the nightly simulation.
The poissrnd() function generates random values from the Poisson distribution (which are always integers, in case you're fuzzy on the Poisson distribution). Its lone parameter is the mean of the Poisson distribution in question. A graph is generated at the end, displaying the simulated distribution for an entire night.
Important Caveat
Recall my mentioning that Monte Carlo is an approximate method, several paragraphs above. The more runs through this process, the closer it mimics the behavior of the probability distributions. My first run used 1e5 (one hundred thousand) runs, and I qualified my response to the person who gave me this puzzle that my answer, 282, was quite possibly off from the correct one "by a piece of candy or two". Indeed, I was off by one piece. Notice that the program listed above now employs 4e5 (four hundred thousand) runs, which yielded the exactly correct answer, 281, the first time I ran it.
David Lang said...
Why not just use the normal approximation to the poisson?
Will Dwinnell said...
Can you elaborate on what you're suggesting?
Choi Daegyu said...
I just visit your blog by googling.
your blog has so nice idea and funny thing. :)
I tried to edit your code little bit.
%Arrivals = poissrnd(5,24,B);
%SS=zeros(24,B);
%for j=1:B
% for k=1:24
% if Arrivals(k,j) > 0
% SS=(k,j)=sum(poissrnd(1,Arrivals_2(k,j),1)+1);
% end
% end
% S(j)=sum(SS(:,j));
%end
Anonymous said...
Looks like a Poisson Compound Process CPP. Their are 2 Random Variables: N(t) = # of kids in (0,2hrs) and Yi = # of candy/kid. A CPP find the distribution of
X(t) = sum of Yi over{i=1:N(t)}. The (Expectation, Var) of X(t) can be computed and then I'd use a Normal Dist to find "upper tail" 5%; that is, P(X(t) < k)=95%; find k.
hunaina said...
Nice | 1,510 | 6,437 | {"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} | 3.84375 | 4 | CC-MAIN-2017-13 | longest | en | 0.954883 |
https://steustatiushistory.org/pick-the-stronger-base-from-each-of-the-following-pairs/ | 1,632,519,900,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057580.39/warc/CC-MAIN-20210924201616-20210924231616-00135.warc.gz | 588,088,400 | 9,221 | ## Pick the stronger base from each of the adhering to pairs f- or cl-
Join this day and start acing your classes! View Bootcamps. This next question asks us to compare the basis cities in between Dia salami in silently reattract that and pure Dean. So the significant allude of distinction below is that Dia Philomene is this p 3 hybridized while duration dian is SP too? And in this case, for basis ity S P three is going to be more basic than that"s be to simply going to be more basic than that"s P. And the reason for this is if there"s more peach character, you"re choose Tron. Density is better from your nucleus, and she"ll have a greater basis. Ity next. Molecules they desire to check out. Compare, stifle, suppose versus this claure substituted gasoline. You"re each other. It"s, of course, here. Chlorine exes withdrawing group by induction. So it"s going to be pulling electron density ameans from our nitrogen. So that would certainly expect 1st 1 is going to be even more fundamental. Remember for abase also few fundamental residence, its electron density, and also it"s providing that electron density up to capture a pro informing. So if you have actually abase and it is attached to something that is inductive Lee withillustration electron thickness, it"s going to you decrease the capcapability of that base from grabbing a proton and also seeing henceforth making it less basic. Identify the more powerful base in each of the complying with pairs. Exsimple your reasoning. Pick the more powerful base from each pair. For each pair of compounds: <1> Which suggested His more acidic? Rank the adhering to from strongest base to weakest base: a. Place the species in each of the following groups in order of boosting base stamina. Give your thinking in each situation. Which species is the strongest base? Identify the stronger acid in each of the following pairs. Predict which acid in each of the following pairs is the more powerful and define your thinking for each. Identify the strongest acid in each of the complying with sets. Which compound in each pair has the stronger conjugate base? Use Table Based on molecular framework, arvariety the binary compounds in order of increasing acid strength. Exordinary your choice. Identify the weakest acid in each of the complying with sets. Classify each of the following as a strong acid, weak acid, strong base, or weak base in aqueous solution.
You are watching: Pick the stronger base from each of the following pairs.
## Pick the stronger base from each of the complying with pairs. br− or f−
Identify each of the following as an acid or a base and create a chemical equation showing just how it is an acid or a base according to the Bronsted-Lowry interpretation. Write the formula for the conjugate base of each of the complying with acids. H 2 PO 4 - is amphoteric. Write equations that demonstrate both its acidic and also fundamental nature. Classify each of the complying with acids as solid or weak. If the acid is weak, create an expression for the acid ionization consistent K a. Pick the stronger base from each of the following pairs:. Based on their molecular framework, pick the more powerful acid from each of the following pairs of binary acids. Explain your thinking. Determine K a and K b for the acid and also base develop. Complete the complying with table. Acidic or Basic. Determine the pH of a solution that is 3. Determine the pH of a solution that is 0. Determine the pH of a 0. Calculate K a. Determine the percent ionization of a 0. Answer Set for Chem Problem Set 8 :. NOTE: Values may differ slightly depending on literature information supplied.
## Pick the stronger base from each of the complying with pairs h2o or br −
Join this day and also begin acing your classes! View Bootcamps. Write chemical equations and also equivalent equilibrium expressions for each of the 3 ionization actions of phosphoric acid. Write chemical equations and also matching equilibrium expressions for each of the 2 ionization measures of carbonic acid. Problem Write chemical equations and also matching equili…. Unlimited solution videos. Access to all courses and lecture videos. Access to all test prep videos. Access to all examine tools. No Credit Card required. Problem Calculate the concentration of each species in a 0. Jabez C. Need more help? Fill out this quick create to get experienced live tutoring. Get live tutoring. Pick the stronger base from each pair. ClO4- or ClO2- b. Cl- or H2O c. CN- or ClO. Chapter Acids and also Bases. Chemistry Structure and also Properties.
## Pick the stronger base from each of the complying with pairs. cn− or clo−
Join today and begin acing your classes! View Bootcamps. Write chemical equations and matching equilibrium expressions for each of the three ionization steps of phosphoric acid. Write chemical equations and corresponding equilibrium expressions for each of the 2 ionization measures of carbonic acid. First off, all we have actually choice an initial we ride the congregate asset off basis as displayed listed below. So So he we have conjugated acid off f negative is equal to h f whereas the conjugated ascollection off Corinne negative is it"s Thea. Among these asset that steel is a more powerful ascollection. The base acquired from the solid asset is a weak base. Thus, f negative is stronger than beasts See an adverse. So we deserve to three that f negative, stronger then base see an unfavorable So Ciel will certainly be based. And if negative will be acid alternative B and also or negative is the Kanzaki base So N o negative is the conjugated base off etch and also or two on dhe and or 3 Negative is the conjugated based off strong asset that is Etch and No. In truth, the conjugated based off a strong asset is a week based, So leading toe and also or two negative is a stronger based then and also or three negative. Lastly, we have actually alternative C f negative, isa conjugated beastern off. H f on c l o Negative is a con trick it based off. H c l o hear H F is more powerful, so H f is a stronger ascollection, so it"s conjugated base will certainly be weak in nature. Therefore, we have actually C l o Negative as a more powerful based then if negative itself. Problem Pick the more powerful base from each pair. ClO4- …. Endless solution videos. Access to all courses and also lecture videos. Access to all test prep videos. Access to all research tools. No Crmodify Card compelled. Problem Pick the stronger base from each pair. ClO4- or ClO2- b. Cl- or H2O c.
## Pick the more powerful base from each of the following pairs. no2− or br−
Questions are generally answered within 1 hour. Q: What stereoisomers carry out the adhering to reactions form? Q: Draw the substitution assets for each of the adhering to reactions; if the commodities have the right to exist as ste Since you have actually posted a question through multiple sub-parts, we will solve the first three sub Q: What are the major assets obtained as soon as the adhering to ether is heated with one tantamount of HI? A: The provided ether is heated through one identical of HI to offers iodoethane and also propen-ol. Q: A solution is ready by adding 0. Assuming no volu A: The development continuous is indicated by the symbol "Kf". This term gives the measurement of streng A: The solution"s pKa value gives indevelopment around the pH of the buffer solution. In order to calcu Q: I require assist via quesion 8. A: The typical thermodynamic information for the provided reactants and products at given temperature is. Q: Use the References to access essential worths if essential for this questi Q: What is the pH of a 0. Ka for HCN is 4. A: NaCN has actually one much less hydrogen and coincides to a conjugate base a weak base. The credentials requir Operations Management. Chemical Engineering. Civil Engineering. Computer Engineering.
## Between br− and h2o , the more powerful base is
By default, it"s based on the name of version or ensemble and the dataset used. Whether a datacollection through the results need to be automatically produced or not. In a future variation, you could have the ability to share batch centroids with other co-workers or, if wanted, make them publicly obtainable. A description of the condition of the batch centroid. This is the date and also time in which the batch centroid was updated via microsecond precision. A status code that reflects the standing of the batch centroid. None of the fields in the datacollection Specifies the areas in the datacollection to be excluded to develop the batch anomaly score. Example: true prestige optional Whether field prestige scores are included as additional columns for each input area. Example: "my brand-new anomaly score" newline optional The brand-new line character that you want to acquire as line break in the generated csv file: "LF", "CRLF". Example: "Anomaly Score" separator optional The separator that you want to get in between areas in the created csv file. This will certainly be 201 upon effective development of the batch anomaly score and also 200 after that. Make sure that you inspect the code that comes with the condition attribute to make sure that the batch anomaly score development has been completed without errors. This is the day and time in which the batch anomaly score was created with microsecond precision. True as soon as the batch anomaly score has been developed in the advance mode. Whether field prominence scores are included as extra columns for each input field or not. The list of input fields" ids supplied to develop the batch anomaly score. The brand-new line character provided as line break in the file that consists of the anomaly scores. In a future variation, you can have the ability to share batch anomaly scores via other co-employees or, if wanted, make them publicly easily accessible. A summary of the standing of the batch anomaly score. This is the date and time in which the batch anomaly score was updated with microsecond precision. A status code that shows the standing of the batch anomaly score. Example: true category optional The category that finest explains the batch topic distribution. None of the fields in the datacollection Specifies the fields in the datacollection to be excluded to produce the batch topic distribution. Example: "my brand-new batch topic distribution" newline optional The brand-new line character that you desire to get as line break in the created csv file: "LF", "CRLF". This will be 201 upon effective creation of the batch topic distribution and 200 thereafter. Make sure that you inspect the code that comes via the standing attribute to make certain that the batch topic distribution development has actually been completed without errors. This is the date and also time in which the batc topic circulation was produced through microsecond precision. True when the batch topic distribution has actually been produced in the advance mode. The list of fields"s ids that were excluded to construct the batch topic circulation. The list of input fields" ids provided to produce the batch topic circulation. The new line character used as line break in the file that consists of the topic distributions. In a future version, you might be able to share batch topic distributions with various other co-workers or, if preferred, make them publicly accessible. A description of the condition of the batch topic distribution. This is the day and time in which the batc topic circulation was updated through microsecond precision. A standing code that mirrors the condition of the batch topic distribution. Example: 1 combiner optional Specifies the approach that need to be provided to incorporate predictions when a non-increased ensemble is offered to produce the review. None of the fields in the dataset Specifies the fields in the dataset to be excluded to create the testimonial. Example: "MySample" tags optional A list of strings that aid classify and index your evaluation.
See more: Why Is The Civil War Considered The First Modern War? The First Modern War
## F or clo more powerful base
I will certainly short article a YouTube evaluation later on this week. Jodi SpeckI wanted to say give thanks to you for your exceptional product. I was shocked at just how easily I could check out my outcomes. As I am growing older I started to realize how hard it is to save my body in form and I was start to feel prefer I was permanently bloated. The TeaTox was so basic, I did not have to readjust my diet to check out outcomes. After two weeks, working out about 3 times a week, I finally have a flat stomach aobtain. I would very recommfinish this Teatox. Shall proceed with the next packet. Qing GuoAfter 14 days wif SkinnyMint Teatox, my pants are much looser now. Flast and also firmer tummy. But still a long method to reach my taracquire. JillianFirst of all, I would certainly favor to give thanks to you so much for coming up via this really impressive product. I feel so excellent and also energized after consuming this product. My complexion has additionally boosted a lot and also my skin is glowing prettily. I"ll absolutely recommend this product to all my families and also friends. This is a product that I will certainly never let go of. I"ll be earlier for my following Teatox in 6 weeks time. Chloe JonesMy 28 day teatox, I"m feeling so excellent, no bloating. I"m full of power. And aacquire I tension that I haven"t readjusted my diet or exercise programme and also these are the outcomes. So I constantly looking forward the alternate night drinking the Night Cleanse which have actually ginger spice scented. I discover it soothes and calm my nerve down after a lengthy day at work-related.E2 Elimination: mechanism | 2,929 | 13,774 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | longest | en | 0.94159 |
https://mathematica.stackexchange.com/questions/167284/permute-a-list-of-elements-given-a-pattern | 1,719,204,474,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198864986.57/warc/CC-MAIN-20240624021134-20240624051134-00542.warc.gz | 343,423,335 | 38,687 | # Permute a list of elements given a pattern
I have this function f and a pattern
pattern = f[h[x]]f[h[y]]
where h is a generic function. Now, I have a set of two distinct functions list = {h1,h2}
How can I generate a function F[pattern_, list_] which returns the following result?
F[pattern,list]
(* f[h1[x]]f[h2[y]]+f[h2[x]]f[h1[y]] *)
• Is this supposed to work for any pattern with any number of h-functions? Commented Mar 6, 2018 at 19:57
• Yes, You can have for example the patter pattern=f[g[h[x]]]f[h[y]] Commented Mar 6, 2018 at 20:00
First we define a replacement rule that replaces each occurrence of an expression with another expression:
replaceIteratively[expr_, x_, list_] := Module[{n = 0},
expr /. x :> (n++; list[[n]])
]
such that, e.g.,
replaceIteratively[{x, x, k, x}, x, {a, b, c}]
(* {a, b, k, c} *)
Then, F can be defined as
F[pattern_, x_, list_] :=
Total@Table[
replaceIteratively[pattern, x, newh],
{newh, Permutations[list]}
]
such that
F[f[h[x]] f[h[y]], h, {h1, h2}]
(* f[h1[y]] f[h2[x]] + f[h1[x]] f[h2[y]] *)
and
F[f[g[h[x]]] f[h[y]], h, {h1, h2}]
(* f[g[h2[x]]] f[h1[y]] + f[g[h1[x]]] f[h2[y]] *)
The ordering is not as in your example, but this does not matter since + is commutative.
• Thanks you! this solved my problem Commented Mar 6, 2018 at 20:12 | 436 | 1,309 | {"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} | 3.234375 | 3 | CC-MAIN-2024-26 | latest | en | 0.785316 |
https://www.pokecommunity.com/blog.php?s=760b986d5c644da908f4efd64905a306&bt=65524 | 1,513,186,313,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948529738.38/warc/CC-MAIN-20171213162804-20171213182804-00524.warc.gz | 795,325,747 | 20,326 | Rate this Entry
# Voltorb Flip Tutorial
Posted August 3rd, 2011 at 3:40 PM by Steven
Voltorb Flip Tutorial
Hey guys, I figured I would just make a simple tutorial for you all. This tutorial will be on Level 3 - so you can still rack up a pretty decent amount of coins if you keep dying on 4.
Here are a few small notes:
• The memo pad is the key to winning. Do not just do it in your head, you'll get confused and make mistakes.
• Voltorb Flip is still luck, sometimes it will just trick you. But this tutorial will use chance to your advantage.
• If you're stuck, look at every single tile and determine which one has the lowest probability of being a Voltorb.
• Rows where the Voltorb number is hired than the sum of numbers, or they are close are high risk. Try to avoid them.
• I will be refering to rows as "S-V" or "Sum - Voltorb." So, as you can see the second horizontal row in mine has a sum of 7 and 1 Voltorb, so it is row 7-1. If two rows have the same, I will say if they are horizontal or vertical.
• Feel free to ask any questions.
The tutorial is picture-based, and all the pictures will be in spoilers.
First off, here is level 3.
Spoiler:
Now, the first step is to look at all the colored side-tiles that show how many Voltorbs and numbers are in a row. The bottom number next to the Voltorb image is how many are in that row. The number above is how what the numbers in the row add up to. You want to mark off all rows where the two numbers add up to 5, and flip over all tiles in rows with 0 Voltorbs.
Spoiler:
We have some numbers now! Look to see if any of the numbers in the 0 rows will complete a row. Luckily for us, we have one. On the bottom, row "4-2" has a 2 in it. That means that there can only be two 1s and two Voltorbs, so there you do not need to flip up any more tiles there. So mark them off using the memo pad.
Spoiler:
Uh oh! Now we have no more rows that are already completed! Looks like it's time to play with chance. Look at your side rows together. Find where rows where there are low Voltorb numbers connect. Those spots have the lowest chance of having a Voltorb in them.
Spoiler:
Oops, it's a one! Well, that's better than a Voltorb by far. Luckily for us, the row is a high number - 8. That means the two remaining tiles are not Voltorbs and are higher than 1. So flip those tiles over.
Spoiler:
Woo! Completing the row 8-1 on the bottom helped to completed rows 5-1 horizontal and 4-3 on the side. Since the only number row 5-1 horizontal can have in it is one 2 and 3 ones, mark off that row. Row 4-3 can have either two 2s or one 3 and one 1, but since we found a 3 in it that means the only remaining tiles are a one and Voltorbs, so mark them all off.
Spoiler:
There are only two spots left! Look at the bottom rows 6-1 and 5-1 vertical, both have no revealed numbers and both have one spot left. That automatically means that the two remaining spots are numbers, so flip them over and you will have won level 3!
Spoiler:
I hope this helps some of you!
Posted in Uncategorized
Views 1113 Comments 3
« Prev Main Next » | 798 | 3,091 | {"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} | 4.0625 | 4 | CC-MAIN-2017-51 | latest | en | 0.942867 |
https://stats.stackexchange.com/questions/182280/why-isnt-it-wrong-to-use-a-log-link-instead-of-a-logit-one-when-doing-glm-wit | 1,701,406,645,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100264.9/warc/CC-MAIN-20231201021234-20231201051234-00726.warc.gz | 626,846,303 | 42,523 | # Why isn't it 'wrong' to use a log link instead of a logit one when doing GLM with a binomial family?
I am taking a basic biostats class for an epidemiology masters and we were recently told that log-binomial GLM is what we should be using instead of logistic regression because the coefficients are interpretable in terms of probability ratios (risk/prevalence).
Now, what gets me is that this just seems like we are buying into a larger problem out of laziness: a logit model can still estimate probabilities so it should allow for extraction of the corresponding ratios via the right manipulations. On the other hand, choosing a log link amounts to admitting larger than one probabilities and that seems like it would be an issue. I understand that for small p results will be very similar but it seems unnecessary when an adequate method seems to exist already.
Surely there is something I am missing here?
• Sep 21, 2021 at 0:11
However, the two link functions don't correspond exactly, and so they literally fit different models (unless $$p$$ is very small throughout, in which case there's no real distinction in practice). As such, for particular applications, it's quite possible that one link function is more suitable than another, at least over the range where data are observed.
However, you're quite right that it's not at all difficult to convert predictions of $$\text{logit}(p)$$ into predictions of $$p$$ or $$\log(p)$$ so if the main reason seems to be discomfort with the $$\text{logit}$$ function it would seem a somewhat poor reason to avoid it. On the other hand if one were to choose log because you expected the conditional expectation of the response to be in that form, or because you wanted to explicitly model the log-mean, then it would make sense to do that.
• Let us spell out that $\text{logit}\ p = \log p - \log (1 - p)$. If $p$ is small, the second term is nearly $\log 1$ or $0$; hence $\text{logit}\ p$ behaves similarly to $\log p$ for small $p$. Sep 2, 2020 at 17:33 | 470 | 2,012 | {"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": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2023-50 | longest | en | 0.957535 |
https://www.itl.nist.gov/div898/software/dataplot/refman2/auxillar/bu5cdf.htm | 1,716,151,725,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057922.86/warc/CC-MAIN-20240519193629-20240519223629-00237.warc.gz | 722,309,326 | 3,612 | Dataplot Vol 2 Vol 1
# BU5CDF
Name:
BU5CDF (LET)
Type:
Library Function
Purpose:
Compute the Burr type 5 cumulative distribution function with shape parameters r and k.
Description:
The Burr type 5 distribution is typically defined in terms of its cumulative distribution function:
with r, k, l, and s denoting the two shape parameters, the location parameter, and the scale parameter, respectively.
The case where l = 0 and s = 1 is referred to as the standard Burr type 5 distribution.
Syntax:
LET <y> = BU5CDF(<x>,<r>,<k>,<loc>,<scale>)
<SUBSET/EXCEPT/FOR qualification>
where <x> is a number, parameter, or variable;
<y> is a variable or a parameter (depending on what <x> is) where the computed Burr type 5 cdf value is stored;
<r> is a positive number, parameter, or variable that specifies the first shape parameter;
<k> is a positive number, parameter, or variable that specifies the second shape parameter;
<loc> is a number, parameter, or variable that specifies the location parameter;
<scale> is a positive number, parameter, or variable that specifies the scale parameter;
and where the <SUBSET/EXCEPT/FOR qualification> is optional.
If <loc> and <scale> are omitted, they default to 0 and 1, respectively.
Examples:
LET A = BU5CDF(0.3,0.2,1.7)
LET Y = BU5CDF(X,0.5,2.2,0,5)
PLOT BU5CDF(X,2,1.8) FOR X = -1.57 0.01 1.57
Default:
None
Synonyms:
BURR TYPE V is a synonym for BURR TYPE 5.
Related Commands:
BU5PDF = Compute the Burr type 5 probability density function. BU5PPF = Compute the Burr type 5 percent point function. BU2PDF = Compute the Burr type 2 probability density function. BU3PDF = Compute the Burr type 3 probability density function. BU4PDF = Compute the Burr type 4 probability density function. BU6PDF = Compute the Burr type 6 probability density function. BU7PDF = Compute the Burr type 7 probability density function. BU8PDF = Compute the Burr type 8 probability density function. BU9PDF = Compute the Burr type 9 probability density function. B10PDF = Compute the Burr type 10 probability density function. B11PDF = Compute the Burr type 11 probability density function. B12PDF = Compute the Burr type 12 probability density function. RAYPDF = Compute the Rayleigh probability density function. WEIPDF = Compute the Weibull probability density function. EWEPDF = Compute the exponentiated Weibull probability density function.
Reference:
Burr (1942), "Cumulative Frequency Functions", Annals of Mathematical Statistics, 13, pp. 215-232.
Johnson, Kotz, and Balakrishnan (1994), "Contiunuous Univariate Distributions--Volume 1", Second Edition, Wiley, pp. 53-54.
Devroye (1986), "Non-Uniform Random Variate Generation", Springer-Verlang, pp. 476-477.
Applications:
Distributional Modeling
Implementation Date:
2007/10
Program:
```
LABEL CASE ASIS
TITLE CASE ASIS
TITLE OFFSET 2
.
MULTIPLOT 4 4
MULTIPLOT CORNER COORDINATES 0 0 100 95
MULTIPLOT SCALE FACTOR 4
.
LET RVAL = DATA 0.5 1 2 5
LET KVAL = DATA 0.5 1 2 5
.
LOOP FOR IROW = 1 1 4
LOOP FOR ICOL = 1 1 4
LET R = RVAL(IROW)
LET K = KVAL(ICOL)
TITLE R = ^r, K = ^k
PLOT BU5CDF(X,R,K) FOR X = -1.57 0.01 1.57
END OF LOOP
END OF LOOP
.
END OF MULTIPLOT
.
JUSTIFICATION CENTER
MOVE 50 97
TEXT Burr Type 5 Cumulative Distribution Functions
```
Date created: 12/17/2007
Last updated: 12/17/2007 | 953 | 3,296 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-22 | latest | en | 0.705583 |
https://www.physicsforums.com/threads/prove-time-invariance.528611/ | 1,544,424,664,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823318.33/warc/CC-MAIN-20181210055518-20181210081018-00507.warc.gz | 1,008,004,228 | 12,728 | # Homework Help: Prove Time Invariance
1. Sep 9, 2011
### SpaceDomain
1. The problem statement, all variables and given/known data
Prove that the system is either T.I. or is not T.I.
2. Relevant equations
y(n) = x(n)*h(n)
x(n) is the input signal
y(n) is the output signal
h(n) is the system
3. The attempt at a solution
Inputing x(n-n0) into the system I get out:
as the output x(n-n0)*h(n)
Since y(n-n0) = x(n-n0)*h(n-n0) != x(n-n0)*h(n) the system is not T.I.
I think I am doing this wrong.
2. Sep 10, 2011
### reddvoid
how can you get this output for x(n-n0) input ?
3. Sep 10, 2011
### SpaceDomain
So should it be that an input of x(n-n0) results in x(n-n0)*h(n-n0)?
4. Sep 10, 2011
### reddvoid
no,
you are telling
x(n) is input of the system
y(n) is out put of the system
h(n) is its transfer function
in-order to check whether the system is TV or TIV we need the relation between input and output.
for example if you tell y(t) output =sin times input x(t)
then we can check whether its TV or TIV
or we need the impulse response function h(t) to decide whether it is TV or TIV
you have not given anything
for every system y(t) is equal to x(t)*h(t)
but their response depends on the function h(t) which is different for different systems
P.S.
I hope you understood | 395 | 1,291 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2018-51 | latest | en | 0.861984 |
http://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-12th-edition/chapter-9-section-9-3-logarithmic-functions-9-3-exercises-page-605/24 | 1,480,741,721,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540839.46/warc/CC-MAIN-20161202170900-00084-ip-10-31-129-80.ec2.internal.warc.gz | 496,190,967 | 38,914 | ## Intermediate Algebra (12th Edition)
$64^{\frac{1}{6}}=\sqrt[6] 64=2$
For all positive numbers $a$, where $a\ne1$, and all positive numbers $x$, we know that $y=log_{a}x$ means the same as $x=a^{y}$. Therefore, $log_{64}2=\frac{1}{6}$ means the same as $64^{\frac{1}{6}}=\sqrt[6] 64=2$. | 111 | 289 | {"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 | 4 | CC-MAIN-2016-50 | longest | en | 0.6856 |
https://fr.slideserve.com/anana/symmetric-ciphers | 1,623,766,372,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487621273.31/warc/CC-MAIN-20210615114909-20210615144909-00574.warc.gz | 258,983,609 | 23,634 | SYMMETRIC CIPHERS
# SYMMETRIC CIPHERS
## SYMMETRIC CIPHERS
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##### Presentation Transcript
1. SYMMETRIC CIPHERS
2. Contents • Classical encryption techniques • Block ciphers and the data encryption standard • Basic concepts in number theory and finite fields • Advanced encryption standard • Block cipher operation • Pseudorandom number generation and stream ciphers
3. 1. CLASSICAL ENCRYPTION TECHNIQUES • Symmetric Cipher Model • Substitution Techniques • Transposition Techniques • Rotor Machines • Steganography
4. KEY POINTS • Symmetric encryption is a form of cryptosystem in which encryption and decryption are performed using the same key. It is also known as conventional encryption. • Symmetric encryption transforms plaintext into ciphertext using a secret key and an encryption algorithm. Using the same key and a decryption algorithm, the plaintext is recovered from the ciphertext. • The two types of attack on an encryption algorithm are cryptanalysis, based on properties of the encryption algorithm, and brute-force, which involves trying all possible keys.
5. KEY POINTS (cont.) • Traditional (precomputer) symmetric ciphers use substitution and/or transposition techniques. Substitution techniques map plaintext elements (characters, bits) into ciphertext elements. Transposition techniques systematically transpose the positions of plaintext elements. • Rotor machines are sophisticated precomputer hardware devices that use substitution techniques. • Steganography is a technique for hiding a secret message within a larger one in such a way that others cannot discern the presence or contents of the hidden message.
6. SYMMETRIC CIPHER MODEL
7. Plaintext: This is the original intelligible message or data that is fed into the algorithm as input • Encryption algorithm: The encryption algorithm performs various substitutions and transformations on the plaintext. • Secret key: The secret key is also input to the encryption algorithm. The key is a value independent of the plaintext and of the algorithm • Ciphertext: This is the scrambled message produced as output. It depends on the plaintext and the secret key. • Decryption algorithm: This is essentially the encryption algorithm run in reverse. It takes the ciphertext and the secret key and produces the original plaintext.
8. Cryptography • Cryptographic systems are characterized along three independent dimensions: • The type of operations used for transforming plaintext to ciphertext. (substitution, transposition). • The number of keys used (symmetric, public-key encryption) • The way in which the plaintext is processed (block cipher, stream cipher)
9. Cryptanalysis and Brute-Force Attack • Cryptanalysis: Cryptanalytic attacks rely on the nature of the algorithm plus perhaps some knowledge of the general characteristics of the plaintext or even some sample plaintext–ciphertext pairs. This type of attack exploits the characteristics of the algorithm to attempt to deduce a specific plaintext or to deduce the key being used. • Brute-force attack: The attacker tries every possible key on a piece of cipher-text until an intelligible translation into plaintext is obtained. On average, half of all possible keys must be tried to achieve success.
10. Types of Attacks on Encrypted Messages
11. A brute-force attack involves trying every possible key until an intelligible translation of the ciphertext into plaintext is obtained. On average, half of all possible keys must be tried to achieve success.
12. SUBSTITUTION TECHNIQUES • A substitution technique is one in which the letters of plaintext are replaced by other letters or by numbers or symbols.1 If the plaintext is viewed as a sequence of bits, then substitution involves replacing plaintext bit patterns with ciphertext bit patterns. • Caesar Cipher • For example, • plain: meet me after the toga party • cipher: PHHW PH DIWHUWKHWRJDSDUWB
13. Monoalphabetic Ciphers • Playfair Cipher • Hill Cipher • Polyalphabetic Ciphers
14. One-Time Pad • An Army Signal Corp officer, Joseph Mauborgne, proposed an improvement to the Vernam cipher that yields the ultimate in security. Mauborgne suggested using a random key that is as long as the message, so that the key need not be repeated. In addition, the key is to be used to encrypt and decrypt a single message, and then is discarded. Each new message requires a new key of the same length as the new message. Such a scheme, known as a one-time pad, is unbreakable. It produces random output that bears no statistical relationship to the plaintext. Because the ciphertext contains no information whatsoever about the plaintext, there is simply no way to break the code.
15. TRANSPOSITION TECHNIQUES • The simplest such cipher is the rail fence technique, in which the plaintext is written down as a sequence of diagonals and then read off as a sequence of rows. For example, to encipher the message “meet me after the toga party” with a rail fence of depth 2, we write the following: m e m a t r h t g p r y e t e f e t e o a a t • The encrypted message is MEMATRHTGPRYETEFETEOAAT
16. ROTOR MACHINESex. Three-Rotor Machine with Wiring Represented by Numbered Contacts
17. STEGANOGRAPHY • A plaintext message may be hidden in one of two ways. The methods of steganography conceal the existence of the message, whereas the methods of cryptography render the message unintelligible to outsiders by various transformations of the text. • Character marking: Selected letters of printed or typewritten text are over-written in pencil. The marks are ordinarily not visible unless the paper is held at an angle to bright light. • Invisible ink: A number of substances can be used for writing but leave no visible trace until heat or some chemical is applied to the paper.
18. Pin punctures: Small pin punctures on selected letters are ordinarily not visible unless the paper is held up in front of a light. • Typewriter correction ribbon: Used between lines typed with a black ribbon, the results of typing with the correction tape are visible only under a strong light. The advantage of steganographyis that it can be employed by parties who have something to lose should the fact of their secret communication (not necessarily the content) be discovered. Encryption flags traffic as important or secret or may identify the sender or receiver as someone with something to hide.
19. 2. BLOCK CIPHERS AND THE DATAENCRYPTION STANDARD • Block Cipher Principles • The Data Encryption Standard
20. KEY POINTS • A block cipher is an encryption/decryption scheme in which a block of plaintext is treated as a whole and used to produce a ciphertext block of equal length. • Many block ciphers have a Feistel structure. Such a structure consists of a number of identical rounds of processing. In each round, a substitution is performed on one half of the data being processed, followed by a permutation that interchanges the two halves. The original key is expanded so that a different key is used for each round.
21. KEY POINTS (cont.) • The Data Encryption Standard (DES) has been the most widely used encryption algorithm until recently. It exhibits the classic Feistel structure. DES uses a 64-bit block and a 56-bit key. • Two important methods of cryptanalysis are differential cryptanalysis and linear cryptanalysis. DES has been shown to be highly resistant to these two types of attack.
22. BLOCK CIPHER PRINCIPLES • Stream Ciphers and Block Ciphers • A stream cipher is one that encrypts a digital data stream one bit or one byte at a time. • A block cipher is one in which a block of plaintext is treated as a whole and used to produce a ciphertext block of equal length. Typically, a block size of 64 or 128 bits is used.
23. Stream Cipher and Block Cipher
24. THE DATA ENCRYPTION STANDARD • In the late 1960s, IBM set up a research project in computer cryptography led by Horst Feistel. The project concluded in 1971 with the development of an algorithm with the designation LUCIFER [FEIS73], which was sold to Lloyd’s of London for use in a cash-dispensing system, also developed by IBM. • In 1973, the National Bureau of Standards (NBS) issued a request for proposals for a national cipher standard. IBM submitted the results of its Tuchman–Meyer project. This was by far the best algorithm proposed and was adopted in 1977 as the Data Encryption Standard.
25. DES Encryption • As with any encryption scheme, there are two inputs to the encryption function: the plaintext to be encrypted and the key. In this case, the plaintext must be 64 bits in length and the key is 56 bits in length (Actually, the function expects a 64-bit key as input. However, only 56 of these bits are ever used; the other 8 bits can be used as parity bits or simply set arbitrarily).
26. General Depiction of DES Encryption Algorithm
27. Single Round of DES Algorithm
28. Calculation of F(R, K)
29. 3. BASIC CONCEPTS IN NUMBER THEORYAND FINITE FIELDS • Divisibility and The Division Algorithm • The Euclidean Algorithm • Modular Arithmetic • Groups, Rings, and Fields • Finite Fields of the Form GF(p) • Polynomial Arithmetic • Finite Fields of the Form GF(2^n)
30. KEY POINTS • Modular arithmetic is a kind of integer arithmetic that reduces all numbers to one of a fixed set [0,...,n-1] for some number n. Any integer outside this range is reduced to one in this range by taking the remainder after division by n. • The greatest common divisor of two integers is the largest positive integer that exactly divides both integers. • A field is a set of elements on which two arithmetic operations (addition and multiplication) have been defined and which has the properties of ordinary arithmetic, such as closure, associativity, commutativity, distributivity, and having both additive and multiplicative inverses.
31. KEY POINTS (cont.) • Finite fields are important in several areas of cryptography. A finite field is simply a field with a finite number of elements. It can be shown that the order of a finite field (number of elements in the field) must be a power of a prime p^n, where n is a positive integer. • Finite fields of order p can be defined using arithmetic mod p. • Finite fields of order p^n, for n>1, can be defined using arithmetic over polynomials.
32. DIVISIBILITY AND THE DIVISION ALGORITHM
33. THE EUCLIDEAN ALGORITHM • Definition: Two integers are relatively primeif their only common positive integer factor is 1. • Finding the Greatest Common Divisor
34. MODULAR ARITHMETIC • Properties of Congruences
35. Euclidean Algorithm Revisited
36. Groups
37. Rings
38. Rings (cont.)
39. Fields
40. Summarizes the axioms that define groups, rings, and fields.
41. FINITE FIELDS OF THE FORM GF(p)
42. Ex. Arithmetic in GF(7)
43. POLYNOMIAL ARITHMETIC
44. Euclidean Algorithm for Polynomials
45. 4. ADVANCED ENCRYPTION STANDARD • Finite Field Arithmetic • AES Structure • AES Transformation Functions • AES Key Expansion • An AES Example • AES Implementation
46. KEY POINTS • AES is a block cipher intended to replace DES for commercial applica-tions. It uses a 128-bit block size and a key size of 128, 192, or 256 bits. • AES does not use a Feistel structure. Instead, each full round consists of four separate functions: byte substitution, permutation, arithmetic opera-tions over a finite field, and XOR with a key. | 2,507 | 11,467 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2021-25 | latest | en | 0.741644 |
https://encyclopediaofmath.org/index.php?title=Monoid&oldid=49305&printable=yes | 1,660,729,571,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572898.29/warc/CC-MAIN-20220817092402-20220817122402-00333.warc.gz | 247,676,087 | 6,732 | # Monoid
A term used as an abbreviation for the phrase "semi-group with identity" . Thus, a monoid is a set with an associative binary operation, usually called multiplication, in which there is an element such that for any . The element is called the identity (or unit) and is usually denoted by . In any monoid there is exactly one identity. If the operation given on the monoid is commutative, it is often called addition and the identity is called the zero and is denoted by .
Examples of monoids. 1) The set of all mappings of an arbitrary set into itself is a monoid relative to the operation of successive application (composition) of mappings. The identity mapping is the identity. 2) The set of endomorphisms of a universal algebra is a monoid relative to composition; the identity is the identity endomorphism. 3) Every group is a monoid.
Every semi-group without an identity can be imbedded in a monoid. For this it suffices to take a symbol not in and give a multiplication on the set as follows: , for any , and on elements from the operation is as before. Every monoid can be represented as the monoid of all endomorphisms of some universal algebra.
An arbitrary monoid can also be considered as a category with one object. This allows one to associate with a monoid its dual (opposite, adjoint) monoid . The elements of both monoids coincide, but the product of and in is put equal to the product in .
The development of the theory of monoids and adjoint functors has shown the utility of the definition of a monoid in so-called monoidal categories. Suppose given a category equipped with a bifunctor , an object and natural isomorphisms
satisfying coherence conditions. An object is called a monoid in the category if there are morphisms and such that the following diagrams are commutative:
If is taken to be the category of sets, the Cartesian product, a one-point set, and the isomorphisms , and are chosen in the natural way (, ), then the second definition of a monoid turns out to be equivalent to the original definition.
#### References
[1] A.H. Clifford, G.B. Preston, "Algebraic theory of semi-groups" , 1–2 , Amer. Math. Soc. (1961–1967) [2] S. MacLane, "Categories for the working mathematician" , Springer (1971) | 525 | 2,251 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2022-33 | latest | en | 0.952991 |
https://www.daniweb.com/programming/software-development/threads/348765/array-index-out-of-bounds | 1,521,880,493,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257649961.11/warc/CC-MAIN-20180324073738-20180324093738-00048.warc.gz | 810,071,133 | 13,265 | I've recently come upon a frustrating Array Index Out Of Bounds error when selecting any of the "calculations" to be applied to the entered values(maximum 1000 values). I suspect it to be a problem with the
``values[(c+1)]``
in the for loop.
I can't figure a way around this problem. Can someone offer up a tidbit? The code is below.
``````Scanner keyboard = new Scanner(System.in);
double[] values;
values = new double[1000]; //create a 1000 slot array
int count;
//int h1,h2,h3,h4,h5,h6,h7,h8,h9,h10;
for(count = 0; count <= values.length; count++){
System.out.println("Enter a value (-1000 to quit input): ");
values[count] = keyboard.nextDouble();
if (values[count] == -1000){
//start calculations
System.out.println("1. Mean");
System.out.println("2. Variance and Standard Deviation");
System.out.println("3. Histogram");
System.out.println("4. Exit");
int switchvar = keyboard.nextInt();
double meantrans = 0;
switch (switchvar){
case 1:
System.out.println("Mean: ");
int c;
for(c = 0; c <= values.length; c++){
meantrans = values[c] + values[(c+1)];
}
double mean = (meantrans / c);
System.out.println("The mean = " + mean);
break;
case 2:
System.out.println("Variance and Standard Deviation: ");
for(c = 0; c <= values.length; c++){
meantrans = values[c] + values[(c+1)];
} //mean calc
mean = (meantrans / c);
double vartrans = 0;
for(c = 0; c <= values.length; c++){
vartrans = values[c]+values[(c+1)];
}
double variance = vartrans / (c - (java.lang.Math.pow(mean,2)));
System.out.println("The variance = " + variance);
break;
case 3:
int h1=0,h2=0,h3=0,h4=0,h5=0,h6=0,h7=0,h8=0,h9=0,h10=0;
System.out.println("Histogram: ");
for(c = 0; c <= values.length; c++){
if(values[c] > 0.0 && values[c] < 1.0){
h1++;
}
else if(values[c] > 1.0 && values[c] < 2.0){
h2++;
}
else if(values[c] > 2.0 && values[c] < 3.0){
h3++;
}
else if(values[c] > 3.0 && values[c] < 4.0){
h4++;
}
else if(values[c] > 4.0 && values[c] < 5.0){
h5++;
}
else if(values[c] > 5.0 && values[c] < 6.0){
h6++;
}
else if(values[c] > 6.0 && values[c] < 7.0){
h7++;
}
else if(values[c] > 7.0 && values[c] < 8.0){
h8++;
}
else if(values[c] > 8.0 && values[c] < 9.0){
h9++;
}
else if(values[c] > 9.0 && values[c] < 10.0){
h10++;
}
}//end for
String ast1 = "";
for(int h=0; h <= h1; h++){
ast1 = ast1 + "*";
}
String ast2 = "";
for(int h=0; h<=h2;h++){
ast2 = ast2 + "*";
}
String ast3 = "";
for(int h=0; h<=h3;h++){
ast3 = ast3 + "*";
}
String ast4 = "";
for(int h=0; h<=h4;h++){
ast4 = ast4 + "*";
}
String ast5 = "";
for(int h=0; h<=h5;h++){
ast5 = ast5 + "*";
}
String ast6 = "";
for(int h=0; h<=h6;h++){
ast6 = ast6 + "*";
}
String ast7 = "";
for(int h=0; h<=h7;h++){
ast7 = ast7 + "*";
}
String ast8 = "";
for(int h=0; h<=h8;h++){
ast8 = ast8 + "*";
}
String ast9 = "";
for(int h=0; h<=h9;h++){
ast9 = ast9 + "*";
}
String ast10 = "";
for(int h=0; h<=h10;h++){
ast10 = ast10 + "*";
}
//output sequence
System.out.println("[0.0 - 1.0] | " + ast1);
System.out.println("(1.0 - 2.0] | " + ast2);
System.out.println("(2.0 - 3.0] | " + ast3);
System.out.println("(3.0 - 4.0] | " + ast4);
System.out.println("(4.0 - 5.0] | " + ast5);
System.out.println("(5.0 - 6.0] | " + ast6);
System.out.println("(6.0 - 7.0] | " + ast7);
System.out.println("(7.0 - 8.0] | " + ast8);
System.out.println("(8.0 - 9.0] | " + ast9);
System.out.println("(9.0 - 10.0] | " + ast10);
break;
case 4:
return;
}//end switch
}//end if
}//end for``````
2
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7 Years
Discussion Span
Last Post by jon.kiparsky
``for(count = 0; count <= values.length; count++)``
If you use count as an index into values in this loop, it will take you out of bounds, since the last index in values is (values-1). Using (count+1) as the index is just icing on the cake - it puts you out of bounds one iteration sooner.
The standard idiom for looping over all elements of an array is
``for(count = 0; count [B]<[/B] values.length; count++)``
Edited by jon.kiparsky: n/a
Notes on structure: never put your working code in the cases of your switch. Sub it out to methods, with nicely descriptive names.
``````case 1: calculateMean();
case 2: calculateVarianceAndStandardDeviation();
case 3: ...``````
You want to have yourself describing one level of logic at a time. If you're reading a cookbook, and it says "Take two onions. Check to see if the knife is sharp enough. If not, get out your whetstone. Lubricate your whetstone with the oil provided, or vegetable oil or WD-40. Holding the whetstone in your off hand, pull the knife from hilt to tip along the blade, with the blade at a 15 degree angle to the stone. Repeat two or three times. Turn your wrist to present to opposite side of the blade to the whetstones and again draw the knife from hilt to tip along the stone three times. Repeat, alternating sides, until the blade is sharp. Cut the onions in half and remove the outer peel...
At that point you've forgotten what you're cooking. Same with your program: with all that detailed logic in the case statement, it's hard to keep track of what it is we're actually trying to do.
Push the implementation details down into methods, and don't be afraid to sub out smaller parts of those into helper methods. | 1,664 | 5,169 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | longest | en | 0.348601 |
https://cw.fel.cvut.cz/b182/courses/zsl/labs2019_08_imreg | 1,569,236,728,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514576355.92/warc/CC-MAIN-20190923105314-20190923131314-00218.warc.gz | 441,611,385 | 7,776 | Warning
## Lab 8 : Image Registration
Main topic of today's lab is the problem of image fusion by means of image registration. There are often multiple images acquired for a single patient – this can be MR, CT at one session or a control MR few months after a previous scan, etc. For (clinical) evaluation of these data, we want the corresponding structures in the images to overlap in the best possible way. But the patient is not completely fixed in the scanner, so the position (in world coordinate system(!)) will in general be different for different scans. Restoring overlap of two image $I, J$ can be seen as finding a transform $T$ for one of the images such that an overlap error $E$ is minimized $$T^{*} = \mathop{\arg\min}_T E(I, T(J))$$
Thus, the key concepts in image registration are the choice of transformation and the choice of the error function, typically called (image) metric.
Transformation Knowing the simplest type of transformation needed for aligning the two input images helps us to restrict the search only to a class of transformations. We may search only for
• translation if the images are only shifted to each other,
• tr. and rotation (rigid transform) for transforming rigid objects (objects not changing in size)
• tr., rot. and scaling, shearing (affine transform) for arbitrary linear transforms
• deformation vector field the most general case, the image may be locally deformed
Metric Optimal image registration metric $E(I, J)$ has its global minimum when the image $I, J$ overlap perfectly. If both $I$ and $J$ are of same modality and have similar intensity values, the sum of squared distances (SSD) between pixels pairs $(\mathbf{x}, t(\mathbf{x}))$ for all pixels $\mathbf{x}$ of image $I$ is such optimal metric.
The SSD metric will not work in case the images $I, J$ have different intensities, which occurs often in medical imaging. For instance when $I$ – CT image, $J$ – MR image, or when $I, J$ are acquired with different MR sequences (T1-weighted, T2-weighted, FLAIR, etc). In this cases, the (normalized) mutual information metric can be applied.
If the intensity / color information is not suitable for defining an image metric, we can formulate the registration problem as alignment of two set of points (corresponding landmarks in both images). The criterion in this case is usually the mean distance between the transformed moving image landmarks and the fixed image landmarks.
In today's homework, we will try out different registration techniques on several pairs of images – download them in a zip-archive.
### Homework
[1.5 pt] Landmark registration (Matlab) We will use MATLAB for this part. Load the images histology.HEStain.tif and histology.PanCytokeratin.tif that show similar histology slices in two different stain colouring. The task is to find the best transform, that will map one image on each other. Due to the differences in the image intensities, we will use manually set landmarks to compute the transform.
1. call the cpselect tool and set corresponding landmarks, when closing the tool, you will be prompted to store the points – we will need them for computing the transformation. Let the PanCytokreatin image Ipc be the moving, HEStain image Ihe the fixed one.
2. compute the transformed moving image by first estimating the transform fitgeotrans and then warping the moving image with imwarp
nrefsim_transform = fitgeotrans(moving_points, fixed_points, 'nonreflectivesimilarity');
imwarp(Ipc, hist_nrfsim, 'OutputView', imref2d(size(Ihe)))
3. repeat step 2 for (at least 3) different choices of transforms available in fitgeotrans
4. use imshowpair function or the provided image_checkerboard to make joint visualization of the fixed and the transformed (warped) moving image. Place the visualizations for all tested transform types into your report.
5. Comment on the results and decide which of the tested transforms provided the best alignment of the two images
[1.5 pt] Multi-modal registration (MITK) Register the T1.nii image on the FLAIR.nii image and visualize the result. Put screenshots into your report. Comment on the registration quality.
Due to contrast differences of the two MR sequences, we need to use one of the MultiModal algorithms. To also correct for acquisition artifacts, we want to search for affine transform, even though the images are from the same subject and the rigid transform would seem to be enough.
• Pipeline
1. Load images, create a joint visualization before registration, put screenshot into report
2. Select the correct MultiModal registration algorithm and perform registration
3. Visualize the result by a method of your choice (checkerboard, color blend, …) and insert screenshot into report
[2 pt] Deformable registration (MITK) The most complex registration task is the mapping of images with different geometries. The transform in such case is a deformation (vector) field. The task is to applyit on two pairs of images.
• Input data pairs
1. first pair are two of the US_heart images, which show two different time-steps of a 3D+time acquisition of the mitral valve
2. second pair are images 58_MRI.nii and 64_MRI.nii, both showing a MR image of a knee, but from different patients
• Pipeline
1. Load images and create a joint visualization before registration (screenshot into report)
2. Select the deformable registration algorithm (Demons) and apply to the images
3. Visualize the result with the joint visualization
4. Inspect the result: Did the registration work perfectly? Are there some areas with high error (mismatch)? Does the resulting image
5. Create a visualization of the deformation: open the MatchPoint Registration Visualizer plugin, select the registration object and visualize also the 3D deformation (either as grid or as vector field glyphs)
### Help: Registration in MITK
We will use the MatchPoint plug-ins from MITK Workbench for registration. We start with loading the two images and visualizing their differences with the MatchPoint Registration Evaluator. After opening the plugin, select the two images and hit Start Evaluation. You can select different visualization styles – Blend, Checkerboard, Color blend etc.
We then need to select the registration algorithm.
1. Open the MatchPoint Algorithm Browser plugin, select a suitable registration algorithm from the list.
2. Open the MatchPoint Algorithm Control plugin, the Selection tab shows the algorithm selected in previous step, hit Load selected algorithm. The plugin switches automatically to the tab Execution after the algorithm is loaded.
3. Select the two images in Data Manager. The first selected image is set as moving, the second as fixed.
4. Now hit Start algorithm button to run the registration
5. Two new data objects should appear, the registration object Reg #n and the transformed moving image Reg #n mapped moving data
Inspect the registration results in the MatchPoint Registration Evaluator plugin. This time, select the registration object and Start Evaluation. | 1,515 | 7,023 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2019-39 | latest | en | 0.904453 |
http://digitalhaunt.net/Mississippi/calculate-the-absolute-error-in-molarity-for-the-mgoh2-solution.html | 1,571,155,283,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986660067.26/warc/CC-MAIN-20191015155056-20191015182556-00527.warc.gz | 48,589,518 | 8,461 | Address 1015 Highway 90, Bay St. Louis, MS 39520 (228) 467-8324 http://www.omni-technology.com
# calculate the absolute error in molarity for the mgoh2 solution Diamondhead, Mississippi
Let's consider the following table of results. Is the paper subject to temperature and humidity changes?) But a third source of error exists, related to how any measuring device is used. Actually since the scale markings are quite widely spaced, the space between 0.05 mL marks can be mentally divided into five equal spaces and the buret reading estimated to the nearest The following diagram describes these ways and when they are useful.
In fact, we could leave it out and would get the same uncertainty. Now for the error propagation To propagate uncertainty through a calculation, we will use the following rules. Question: Mg(OH)2 is one of the active ingredients in Maalox... Example: Ka for acetic acid is 1.8 x 10-5.
Know the various definitions of acids and bases. Report this document Report View Full Document Most Popular Documents for Chemistry 3 pages classes_Winter12_20L-1ID18_RevisedProcedureforAcid-BaseIndicators UC Davis CHEM 2c - Winter 2012 Chemistry 20L Revised Procedures for Assignment #4 (Titrations & asked 1 year ago viewed 2047 times active 11 months ago Upcoming Events 2016 Community Moderator Election ends Oct 18 Linked 0 Calculate water hardness from grams of CaCO3 4 Error Therefore, the relative uncertainty of the concentration $c$ may be estimated as \begin{align} \left(\frac{u(c)}{c}\right)^2 &= \left(\frac{u(m)}{m}\right)^2 + \left(\frac{u(M)}{M}\right)^2 + \left(\frac{u(V)}{V}\right)^2\tag4 \\[6pt] \frac{u(c)}{c} &= \sqrt{\left(\frac{u(m)}{m}\right)^2 + \left(\frac{u(M)}{M}\right)^2 + \left(\frac{u(V)}{V}\right)^2}\tag5 \end{align} Thus, the
The moles of NaOH then has four significant figures and the volume measurement has three. Sign up to view the full document. Nevertheless, buret readings estimated to the nearest 0.01 mL will be recorded as raw data in your notebook. The standard deviation of a set of results is a measure of how close the individual results are to the mean.
Pang's Chem 20L Winter 2013. Ask a homework question - tutors are online Find Study Resources Main Menu by School by Literature Guides by Subject Get instant Tutoring Help Main Menu Ask a Tutor a Question But I do not know how to get the absolute error from my calculations. Notice that the ± value for the statistical analysis is twice that predicted by significant figures and five times that predicted by the error propagation.
Sign up to view the full document. This relative uncertainty can also be expressed as 2 x 10–3 percent, or 2 parts in 100,000, or 20 parts per million. For a 95% confidence interval, there will be a 95% probability that the true value lies within the range of the calculated confidence interval, if there are no systematic errors. View Full Document Company About Us Scholarships Sitemap Standardized Tests Get Course Hero iOS Android Educators Careers Our Team Jobs Internship Help Contact Us FAQ Feedback Legal Copyright Policy Honor Code
For result R, with uncertainty σR the relative uncertainty is σR/R. First, here are some fundamental things you should realize about uncertainty: • Every measurement has an uncertainty associated with it, unless it is an exact, counted integer, such as the number Yes No Sorry, something has gone wrong. Sign up to view the full version.
The relative uncertainty in the volume is greater than that of the moles, which depends on the mass measurement, just like we saw in the significant figures analysis. Also notice that the uncertainty is given to only one significant figure. Systematic errors can result in high precision, but poor accuracy, and usually do not average out, even if the observations are repeated many times. The 10 milliliter burets used are marked (graduated) in steps of 0.05 mL.
Errors are often classified into two types: systematic and random. View Full Document (IV) pH and Normality For this section, focus on the basic concepts for strong acid and strong base and review the concept on how to choose the correct Expand» Details Details Existing questions More Tell us some more Upload in Progress Upload failed. Calculate the absolute error (in molarity) for the Mg(OH)2 solution.
You should be comfortable of writing out all the conversion factors during the calculation. (VI) Beer This is the end of the preview. When a strong acid supplies the common ion H3O +, the equilibrium shifts to form more acetic acid. You can only upload videos smaller than 600MB. Browse hundreds of Chemistry tutors.
Please upload a file larger than 100x100 pixels We are experiencing some problems, please try again. Circular growth direction of hair Creating a simple Dock Cell that Fades In when Cursor Hover Over It Tips for Golfing in Brain-Flak Will password protected files like zip and rar I got .09191 M, 1.12x10^3 ppm, .4% inherent error and .112% for w/v%. The concentration $c$ is defined as $$c=\frac nV\tag1$$ where $n$ is amount of substance, which is unknown.
The digits that constitute the result, excluding leading zeros, are then termed significant figure. Your calculator probably has a key that will calculate this for you, if you enter a series of values to average. Pang's Chem 20L Winter 2013.Practice Final for Prof. Error in volume associated with volumentric flask: 50 mL +/ 0.04 (or 0.08%) Assuming there is no error in the molecular weight we can ignore the contribution here to your error.
Note that you should use a molecular mass to four or more significant figures in this calculation, to take full advantage of your mass measurement's accuracy. Show all your reasoning including any conversion factors between molarity and normality. A final type of experimental error is called erratic error or a blunder. Answer Questions I AM BEGGING FOR CHEMISTRY HELP. | 1,358 | 5,904 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2019-43 | latest | en | 0.818293 |
http://metafor-project.org/doku.php/tips:i2_multilevel_multivariate?rev=1557947196 | 1,569,082,250,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574532.44/warc/CC-MAIN-20190921145904-20190921171904-00528.warc.gz | 122,350,233 | 10,766 | The metafor Package
A Meta-Analysis Package for R
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tips:i2_multilevel_multivariate
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I^2 for Multilevel and Multivariate Models
The $I^2$ statistic was introduced by Higgins and Thompson in their seminal 2002 paper and has become a rather popular statistic to report in meta-analyses, as it facilitates the interpretation of the amount of heterogeneity present in a given dataset.
For a standard random-effects models, the $I^2$ statistic is computed with $$I^2 = 100\% \times \frac{\hat{\tau}^2}{\hat{\tau}^2 + \tilde{v}},$$ where $\hat{\tau}^2$ is the estimated value of $\tau^2$ and $$\tilde{v} = \frac{(k-1) \sum w_i}{(\sum w_i)^2 - \sum w_i^2},$$ where $w_i = 1/v_i$ is the inverse of the sampling variance of the $i^{th}$ study. The equation for $\tilde{v}$ is equation 9 in Higgins & Thompson (2002) and can be regarded as the 'typical' within-study (or sampling) variance of the observed effect sizes or outcomes.1)
Sidenote: As the equation above shows, $I^2$ estimates the amount of heterogeneity relative to the total amount of variance in the observed effects or outcomes (which is composed of the variance in the true effects, that is, $\hat{\tau}^2$, plus sampling variance, that is, $\tilde{v}$). Therefore, it is not an absolute measure of heterogeneity and should not be interpreted as such. For example, a practically/clinically irrelevant amount of heterogeneity (i.e., variance in the true effects) could lead to a large $I^2$ value if all of the studies are very large (in which case $\tilde{v}$ will be small). Conversely, when all of the studies are small (in which case $\tilde{v}$ will be large), $I^2$ may still be small, even if there are large differences in the size of the true effects. See also chapter 16 in Borenstein et al. (2009), which discusses this idea very nicely.
However, this caveat aside, $I^2$ is a very useful measure because it directly indicates to what extent heterogeneity contributes to the total variance. In addition, most people find $I^2$ easier to interpret than estimates of $\tau^2$.
Standard Random-Effects Model
Let's try out the computation for a standard random-effects model (see Berkey et al. (1995) and help(dat.bcg) for more details on the dataset used). First, we use the rma() function for this:
library(metafor)
dat <- escalc(measure="RR", ai=tpos, bi=tneg, ci=cpos, di=cneg, data=dat.bcg)
res <- rma(yi, vi, data=dat)
res$I2 [1] 92.22139 So, we estimate that roughly 92% of the total variance is due to heterogeneity (i.e., variance in the true effects), while the remaining 8% can be attributed to sampling variance. Manually computing$I^2$as described above yields the same result: k <- res$k
wi <- 1/dat$vi s2 <- (k-1) * sum(wi) / (sum(wi)^2 - sum(wi^2)) 100 * res$tau2 / (res$tau2 + s2) [1] 92.22139 General Equation for I^2 Before we continue with more complex models, it is useful to point out a more general equation for computing$I^2$, which also applies to models involving moderator variables (i.e., mixed-effects meta-regression models). This will also become important when dealing with models where sampling errors are no longer independent. So, let us define $$\mathbf{P} = \mathbf{W} - \mathbf{W} \mathbf{X} (\mathbf{X}' \mathbf{W} \mathbf{X})^{-1} \mathbf{X}' \mathbf{W},$$ where$\mathbf{W}$is (for now) a diagonal matrix with the inverse sampling variances (i.e.,$1/v_i$) along the diagonal and$\mathbf{X}$is the model matrix. In the random-effects model,$\mathbf{X}$is just a column vector with 1's, but in meta-regression models, it will contain additional columns with the values of the moderator variables. Then we define $$I^2 = 100\% \times \frac{\hat{\tau}^2}{\hat{\tau}^2 + \frac{k-p}{\mathrm{tr}[\mathbf{P}]}},$$ where$\mathrm{tr}[\mathbf{P}]$denotes the trace of the$\mathbf{P}$matrix (i.e., the sum of the diagonal elements) and$p$denotes the number of columns in$\mathbf{X}$. Let's try this out for the example above: W <- diag(1/dat$vi)
X <- model.matrix(res)
P <- W - W %*% X %*% solve(t(X) %*% W %*% X) %*% t(X) %*% W
100 * res$tau2 / (res$tau2 + (res$k-res$p)/sum(diag(P)))
[1] 92.22139
For a model with moderators, this is also how rma() computes $I^2$:
res <- rma(yi, vi, mods = ~ ablat, data=dat)
res$I2 [1] 68.39313 X <- model.matrix(res) P <- W - W %*% X %*% solve(t(X) %*% W %*% X) %*% t(X) %*% W 100 * res$tau2 / (res$tau2 + (res$k-res$p)/sum(diag(P))) [1] 68.39313 (although instead of using solve(), which can be numerically unstable in some cases, rma() uses the QR decomposition to obtain the$(\mathbf{X}' \mathbf{W} \mathbf{X})^{-1}$part). In models with moderators, the$I^2$statistic indicates how much of the unaccounted variance in the observed effects or outcomes (which is composed of unaccounted variance in the true effects, that is, residual heterogeneity, plus sampling variance) can be attributed to residual heterogeneity. Here, we estimate that roughly 68% of the unaccounted variance is due to residual heterogeneity. Multilevel Models Multilevel structures arise when the estimates can be grouped together based on some higher-level clustering variable (e.g., paper, lab or research group, species). In that case, true effects belonging to the same group may be more similar to each other than true effects for different groups. Meta-analytic multilevel models can be used to account for the between- and within-cluster heterogeneity and hence the intracluster (or intraclass) correlation in the true effects. See Konstantopoulos (2011) for a detailed illustration of such a model. In fact, let's use the same example here. First, we can fit the multilevel random-effects model with: dat <- dat.konstantopoulos2011 res <- rma.mv(yi, vi, random = ~ 1 | district/school, data=dat) res Multivariate Meta-Analysis Model (k = 56; method: REML) Variance Components: estim sqrt nlvls fixed factor sigma^2.1 0.0651 0.2551 11 no district sigma^2.2 0.0327 0.1809 56 no district/school Test for Heterogeneity: Q(df = 55) = 578.8640, p-val < .0001 Model Results: estimate se zval pval ci.lb ci.ub 0.1847 0.0846 2.1845 0.0289 0.0190 0.3504 * --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Note that the model contains two variance components ($\sigma^2_1$and$\sigma^2_2$), for the between-cluster (district) heterogeneity and the within-cluster (school within district) heterogeneity. Based on the discussion above, it is now very easy to generalize the concept of$I^2$to such a model (see also Nakagawa & Santos, 2012). That is, we can first compute: W <- diag(1/dat$vi)
X <- model.matrix(res)
P <- W - W %*% X %*% solve(t(X) %*% W %*% X) %*% t(X) %*% W
100 * sum(res$sigma2) / (sum(res$sigma2) + (res$k-res$p)/sum(diag(P)))
[1] 95.18731
Note that we have summed up the two variance components in the numerator and denominator. Therefore, this statistic can be thought of as the overall $I^2$ value that indicates how much of the total variance can be attributed to the total amount of heterogeneity (which is the sum of between- and within-cluster heterogeneity). In this case, the value is again very large, with approximately 95% of the total variance due to heterogeneity.
However, we can also break things down to estimate how much of the total variance can be attributed to between- and within-cluster heterogeneity separately:
100 * res$sigma2 / (sum(res$sigma2) + (res$k-res$p)/sum(diag(P)))
[1] 63.32484 31.86248
Therefore, about 63% of the total variance is estimated to be due to between-cluster heterogeneity, with the remaining 32% due to within-cluster heterogeneity. And the remaining 5% are sampling variance.
Multivariate Models
Now we will consider the same type of generalization, but for a multivariate model with non-independent sampling errors. Therefore, not only do we need to account for heterogeneity and dependency in the underlying true effects, but we also now need to specify covariances between the sampling errors. For an illustration of such a model, see Berkey et al. (1998), which we can also use for illustration purposes here.
dat <- dat.berkey1998 | 2,297 | 8,088 | {"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} | 3.390625 | 3 | CC-MAIN-2019-39 | longest | en | 0.909034 |
https://physicscatalyst.com/chemistry/quantum-numbers.php | 1,721,871,718,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518532.61/warc/CC-MAIN-20240724232540-20240725022540-00411.warc.gz | 384,007,816 | 16,832 | # Quantum Numbers
## Quantum Numbers
A set of four quantum numbers which specify the energy, size, shape and orientation of an orbital about all the electrons present in an atom is called Quantum Numbers. To specify an orbital only three quantum numbers are required while to specify an electron all four quantum numbers are required.
Four quantum number are
$n$, $l$, $m_l$, $m_s$
### Principle quantum number ($n$):
• This quantum number tells energy level or shell to which electron belong
n = 1, 2, 3,…. (i.e. K,M,...)
• This also gives information about average distance of electron from nucleus and Energy of e- in hydrogen & hydrogen like atoms((He+, Li2+, .... etc.)
• Max number of electron present in a shell is given by 2n2
• Number of orbital in a shell is given by n2
n 1 2 3 4 No of Orbitals(n2) 1 4 9 16 Max number of electron (2n2) 2 8 18 32
### Azimuthal/ subsidiary/ Orbital angular quantum number(l)
1. This gives us information about:
1. No. of sub – shell
2. angular momentum of electron
3. Shapes of various orbitals
4. energy of orbitals in multi-electron atoms along with principal quantum number
The number of orbitals in a subshell = 2l + 1. For a given value of n, l can have n values ranging from 0 to n-1. Total number of subshells in a particular shell is equal to the value of n.
Subshell s p d f g Value of ‘l’ 0 1 2 3 4 Number of orbitals 1 3 5 7 9
Subshell Notation
n l Subshell notation 1 0 1s 2 0 2s 2 1 2p 3 0 3s 3 1 3p 3 2 3d 4 0 4s 4 1 4p 4 2 4d 4 3 4f
### Magnetic Quantum number (ml )
It gives information about the No. of preferred orientations of electrons in sub – shell. Each orientation corresponds to orbital
Therefore ml gives no. of orbitals in a sub shell
For any sub-shell (defined by ‘l’ value) 2l+1 values of ml are possible.
L ml 0 0 (1 s orbital) 1 -1,0,1 (3 p orbital) 2 -2, -1,0,1,2 ( 5 d orbital) 3 -3,-2,-1,0,1,2,3 ( 7 f orbital) 4 -4,-3,-2,-1,0,1,2,3,4 (9 g orbitals)
### Electron Spin quantum number (ms )
It refers to orientation of the spin of the electron. It can have two values +1/2 and -1/2. +1/2 identifies the clockwise spin and -1/2 identifies the anti- clockwise spin.
Question 1
(a) How many subshell are possible with n = 4
(b) How many e- will be present in subshell having ;ms = -1/2 for n = 4.
Solution
For n = 4
l = 0 to (n – 1)
= 0, 1, 2, 3
l = 0 ml = -l to + 1 =0
l = 1 ml = -l to + l =-1, 0, 1
l = 2 ml = -l to + l = -2, -1, 0, 1, 2
l=3 ml = -l to + l = -3, -2, -1, 0, 1, 2, 3
For ms =-1/2 there will be 16 electrons
Question 2
Which of the following set of quantum no. is impossible?
a) n = 3, l = 2, m = -2, s = +1/2
b) n = 4, l = 0, m = 0, s = +1/2
c) n = 3, l = 2, m = -3, s = +1/2
d) n = 5, l = 3, m = 0, s = +1/2
Solution
n = 3
l = 0 to (n – 1)
= 0 to 2
0, 1, 2
s, p, d
m = -l to + l
l = 0 ml = -l to + 1 =0
l = 1 ml = -l to + l =-1, 0, 1
l = 2 ml = -l to + l = -2, -1, 0, 1, 2
Question 3
What is the total number of orbitals associated with the principal quantum number n = 3 ?
Solution
For n = 3, the possible values of l are 0, 1 and 2.
Thus there is one 3s orbital (n = 3, l = 0 and ml = 0)
There are three 3p orbitals (n = 3, l = 1 and ml = -1, 0, +1)
There are five 3d orbitals (n = 3, l = 2 and ml = -2, -1, 0, +1+, +2).
Therefore, the total number of orbitals is 1+3+5 = 9
The same value can also be obtained by using the relation; number of orbitals
= $n^2$, i.e. 32 = 9.
Question 4
Using s, p, d, f notations, describe the orbital with the following quantum numbers (a) n = 2, l = 1
(b) n = 4, l = 0
(c) n = 5,l = 3
(d) n = 3, l = 2
Solution | 1,352 | 3,599 | {"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} | 2.875 | 3 | CC-MAIN-2024-30 | latest | en | 0.723909 |
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Make Math Moments Academy Forums Full Workshop Reflections General Comments & Questions Suggestions Making Math Moments that Matter Challenges and Problem Shooting
Tagged:
• # Making Math Moments that Matter Challenges and Problem Shooting
Jon Orr updated 1 year, 5 months ago 2 Members · 2 Posts
• ### Terri Bello
Member
August 24, 2020 at 3:03 pm
In the Lessons 6-4, Misconception #2 No Text Book or Worksheets, I was super curious about ways to create games from worksheets. It talked about Speed dating, and a couple others and shared there would be more information about them below. However, I could not find the links, is this purposeful withholding information ? Or am I looking in the wrong spot and if so can you direct me to where I should be looking?
Also, I find that where I get stuck in the planning and thereby teaching is in anticipating my students methods for solving a problem during the three act. I wonder if in part this is because I was taught in a more traditional math class and as a new math teacher, I am still learning my own flexibility in problem solving. Also, with that I then struggle with how to properly guide the students with my questioning. I notice I fall into a pattern of getting stuck and then in a panic falling back on a more traditional method, even within the three math act framework.
I hoped this course would help me along this journey, while the material has help me grow a ton it has not felt as directly helpful around this particular challenge. I think it is highly likely that when I get back in the classroom I might still be stuck around the above obstacles. Any suggestions on how I can better address these challenges, which I see as 1. better predicting student solving techniques, and 2. how to appropriately question students and hold space for meaningful struggle?
• ### Jon Orr
August 25, 2020 at 9:06 am
Thanks for the suggestion. However, when I look at that lesson the links are showing for me. Have another look?
We’re glad the course has helped you grow a ton, but it’s also a course that’s like a gateway. It opens your eyes on what you could do and how you could change approaches to lesson design.
It sounds like it did just that. It helped you identify where you need to focus your growth next.
To get better at anticipating student solutions I found that I needed to explore myself different ways to represent and model mathematics. As a traditional HS teacher I knew just the one way. It wasn’t until I learned about how to use a number line, or algebra tiles, or ratio tables for example that I could start to see the connections in models so that I could help my students. I also found it super important to know where students were coming from and know where they are going. Knowing the landscape as Cathy Fosnot calls it is super important in deciding on where a student is on their learning trajectory. If we know where they are on that trajectory then we’re better equipped to point them in the right direction.
Having said all of that I also know it will take practice in the classroom and it takes a brave teacher, like yourself, to dedicate time to that practice.
Since you’re an academy member you could take our course on Fundamentals of Math to see the “where are your students coming from part” .
Lesson 5.2 and Lesson 5.3 of the course you just finished tackle strategies to determine the landscape. You could also listen to episode 24 with Cathy Fosnot: https://makemathmoments.com/episode24/
Remember too that we’re here to support you on your journey. Feel free to keep this discussion going. Posting support topics in the Water Cooler area will also gain the attention of members here in the community who will reach out to help as well.
Let’s keep growing.
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Now | 828 | 3,867 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-05 | longest | en | 0.958898 |
https://bilakniha.cvut.cz/next/en/predmet6211506.html | 1,716,918,026,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059143.84/warc/CC-MAIN-20240528154421-20240528184421-00280.warc.gz | 106,859,368 | 4,145 | CZECH TECHNICAL UNIVERSITY IN PRAGUE
STUDY PLANS
2024/2025
# Numerical Methods in Fluid Dynamics
The course is not on the list Without time-table
Code Completion Credits Range
01NMDT ZK 2 2P+0C
Garant předmětu:
Lecturer:
Tutor:
Supervisor:
Department of Mathematics
Synopsis:
The course is focused on the design and properties of numerical methods for solving fluid flow equations. Focus is put
mainly on the finite volume method whose classical and advanced schemes are derived. Selected schemes are analyzed
in terms of stability. The second part is devoted to advanced numerical schemes used in practice. The matter is
concluded by a brief summary of alternative numerical approaches for fluid flow simulation and by a demonstration of
visualization techniques for simulation results.
Requirements:
Syllabus of lectures:
1. Conservation laws in vector form, reference scalar equations (transport equation, wave equation).
2. Classification of linear partial differential equations, method of characteristics. Domain of dependence and domain
of influence.
3. Classical schemes of the finite difference method and the finite volume method in 1D (Lax-Friedrichs scheme,
upwind scheme etc.)
4. Numerical viscosity and dispersion, modified equation.
5. Stability and convergence, spectral stability condition, CFL condition.
6. Unstructured meshes of the finite volume method in 2D and 3D. Numerical schemes in multiple dimensions.
7. Advanced numerical schemes: positive and TVD schemes, MUSCL and PISO schemes,. Higher order schemes
(ENO, WENO), flux limiters.
8. Iterative methods for implicit schemes and their preconditioning, multi-step methods, multigrid methods.
9. A brief survey of other numerical methods in computational fluid dynamics (meshless methods, particle methods,
lattice Boltzmann method)
10. Visualization in computational fluid dynamics.
Syllabus of tutorials:
Study Objective:
Study materials:
Key references:
[1] J. Blazek, Computational Fluid Dynamics: Principles and Applications (3rd ed.), Elsevier, 2015.
[2] C. Pozridikis, Computational Fluid Dynamics: Theory, Computation, and Numerical Simulation (2nd ed.), Springer
Science + Business Media LLC, 2017.
[3] J. Tu, Jiyuan, G-H. Yeoh, and C. Liu. Computational Fluid Dynamics: A Practical Approach. Butterworth-
Heinemann, 2018.
Recommended references:
[4] C. Hirsch, Numerical Computation of Internal and External Flows, Volume 1: Fundamentals of Computational
Fluid Dynamics (2nd ed.), Elsevier, 2007.
[5] J. F. Wendt (ed.), Computational Fluid Dynamics: An Introduction, Springer, 2009.
Note:
Further information:
No time-table has been prepared for this course
The course is a part of the following study plans:
Data valid to 2024-05-28
Aktualizace výše uvedených informací naleznete na adrese https://bilakniha.cvut.cz/en/predmet6211506.html | 663 | 2,859 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-22 | latest | en | 0.812605 |
http://www.mathhomeworkanswers.org/26479/what-the-volume-solid-length-12cm-breadth-8cm-and-height-6cm | 1,371,680,735,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368709379061/warc/CC-MAIN-20130516130259-00072-ip-10-60-113-184.ec2.internal.warc.gz | 555,755,586 | 20,258 | # what is the volume of a solid, if its length is 12cm, breadth is 8cm and height is 6cm?
what is the volume of a solid
to find the volume you multiply the length breadth and height
Meaning that,
Volume = length x ****** x height
length = 12
height = 6
You can now work out the volume.
answered May 16, 2012 by Level 5 User (10,220 points)
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15 views | 229 | 743 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2013-20 | latest | en | 0.87228 |
https://vdocuments.mx/7-1-chapter-7-bonds-and-their-valuation-key-features-of-bonds-bond-valuationprice.html | 1,718,508,549,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861640.68/warc/CC-MAIN-20240616012706-20240616042706-00386.warc.gz | 506,740,413 | 30,945 | # 7-1 chapter 7 bonds and their valuation key features of bonds bond valuation(price) measuring...
30
7-1 CHAPTER 7 Bonds and Their Valuation Key features of bonds Bond valuation(price) Measuring yield(return) Assessing risk
Post on 16-Dec-2015
230 views
Category:
## Documents
Tags:
• #### bond valuation model
TRANSCRIPT
7-1
CHAPTER 7Bonds and Their Valuation
Key features of bonds Bond valuation(price) Measuring yield(return) Assessing risk
7-2
What is a bond?
A long-term debt instrument in which a borrower agrees to make payments of principal and interest, on specific dates, to the holders of the bond.
7-3
Key Features of a Bond Par value – face amount of the bond,
which is paid at maturity (assume \$1,000).
Coupon rate – stated interest rate (generally fixed) paid by the issuer. Multiply by par to get dollar payment of interest.
Maturity date – years until the bond must be repaid.
Issue date – when the bond was issued.
7-4
The value of financial assets
nn
22
11
r)(1
CF ...
r)(1
CF
r)(1
CF Value
0 1 2 nr
CF1 CFnCF2Value
...
7-5
Bond valuation model
Rd is the required return of the bond. It is also the market rate of interest of the bond, and yield to maturity (YTM) of the bond.
Rd is NOT coupon rate
Nd
Nd
1d
b )r(1
valueface
)r(1
paymentcoupon ...
)r(1
paymentcoupon price) (bondV
7-6
What is the value of a 10-year, 10% annual coupon bond, if rd = 10%?
\$1,000 V
\$385.54 \$38.55 ... \$90.91 V
(1.10)
\$1,000
(1.10)
\$100 ...
(1.10)
\$100 V
b
b
10101b
0 1 2 nr
100 100 + 1,000100Vb = ?
...
7-7
Using a financial calculator to value a bond
This bond has a \$1,000 lump sum due at t = 10, and annual \$100 coupon payments beginning at t = 1 and continuing through t = 10, the price of the bond can be found by solving for the PV of these cash flows.
INPUTS
OUTPUT
N I/YR PMTPV FV
10 10 100 1000
-1000
7-8
An example:Increasing inflation and rd
Suppose inflation rises by 3%, causing rd = 13%. When rd rises above the coupon rate, the bond’s value falls below par, and sells at a discount.
INPUTS
OUTPUT
N I/YR PMTPV FV
10 13 100 1000
-837.21
7-9
An example:Decreasing inflation and rd
Suppose inflation falls by 3%, causing rd = 7%. When kd falls below the coupon rate, the bond’s value rises above par, and sells at a premium.
INPUTS
OUTPUT
N I/YR PMTPV FV
10 7 100 1000
-1210.71
7-10
Semi-annual coupon bond
Majority of bonds pay interest semiannually. Coupon rate=10%/Y Going (nominal) interest=5% 15 year bond
Solution: P semi annual=\$1523.26 How: N=15*2 i=5/2 PMT=\$100/2 Compared with annual coupon bond,
which should have a higher price? P annual=\$1518.98
7-11
What is the YTM on a 10-year, 9% annual coupon, \$1,000 par value bond, selling for \$887?
Must find the rd that solves this model.
10d
10d
1d
Nd
Nd
1d
B
)r(1
1,000
)r(1
90 ...
)r(1
90 \$887
)r(1
valueface
)r(1
paymentcoupon ...
)r(1
paymentcoupon V
7-12
Using a financial calculator to find YTM
Solving for I/YR, the YTM of this bond is 10.91%. This bond sells at a discount, because YTM > coupon rate.
INPUTS
OUTPUT
N I/YR PMTPV FV
10
10.91
90 1000- 887
7-13
Find YTM, if the bond price was \$1,134.20.
Solving for I/YR, the YTM of this bond is 7.08%. This bond sells at a premium, because YTM < coupon rate.
INPUTS
OUTPUT
N I/YR PMTPV FV
10
7.08
90 1000-1134.2
7-14
Relationship between interest and bond price (inversely realted)
Interest Rate
Bond Value
Interest Rate
Bond Value
12%
10%
8%
874.50 1,000 1,152.47
7-15
Interest sensitivity and maturity
Annual Coupon rate=10%
% change 1 yr rd 10yr % change
+4.8% \$1,048 5% \$1,386 +38.6%\$1,000 10% \$1,000
-4.3% \$957 15% \$749 -25.1%
The 10-year bond is more sensitive to interest rate changes, and hence has more interest rate risk.
7-16
Interest sensitivity and coupon rate
10 year annual bond
% change 5%c rd 10%c % change
+44% \$1,000 5% \$1,386 +38.6%\$693 10% \$1,000
-28% \$498 15% \$749 -25.1% Low coupon bond has CF concentrated at the
end of maturity from repayment of principal. The bond with lower coupon rate is more
sensitive to interest rate changes
7-17
Calculation details
N I(%) PV=? PMT FV1 5 -1048 100 10001 10 -1000 100 10001 15 -957 100 100010 5 -1386 100 100010 10 -1000 100 100010 15 -749 100 100010 5 -1000 50 100010 10 -693 50 100010 15 -498 50 100010 5 -1386 100 100010 10 -1000 100 100010 15 -749 100 1000
10Y, 10% coupon bond
10Y, 5% coupon bond
1Y,10% coupon bond
10Y, 10% coupon bond
7-18
Summary of Factors that Affect bond Prices and Price sensitivity when Interest Rates Change
Interest Rate negative relation between interest rate
changes and bond price increasing interest rates correspond to bond
price decrease (at a decreasing rate) Time Remaining to Maturity
longer time to maturity corresponds to larger price change for a given interest rate change
Coupon Rate the lower the coupon rate, the bigger the
price change for a given change in interest rates
Interest Rate negative relation between interest rate
changes and bond price increasing interest rates correspond to bond
price decrease (at a decreasing rate) Time Remaining to Maturity
longer time to maturity corresponds to larger price change for a given interest rate change
Coupon Rate the lower the coupon rate, the bigger the
price change for a given change in interest rates
7-19
Reinvestment risk
The risk that bondholders have to reinvest future cash flows (coupon and principal when expires)at lower interest rates if general interest level declines
EXAMPLE: Suppose you just won \$500,000 playing the lottery. You intend to invest the money and live off the interest.
7-20
Reinvestment Rate Risk Example
You may invest in either a 10-year bond or a series of ten 1-year bonds. Both 10-year and 1-year bonds currently yield 10%.
If you choose the 1-year bond strategy: After Year 1, you receive \$50,000 in income and have
\$500,000 to reinvest. But, if 1-year rates fall to 3%, your annual income would fall to \$15,000.
If you choose the 10-year bond strategy: You can lock in a 10% interest rate, and \$50,000 annual
income for 10 years.
7-20
7-21
Conclusions about Interest Rate and Reinvestment Rate Risk
CONCLUSION: Nothing is riskless! Interest risk is a big concern in low interest environment, and
reinvestment risk is a big concern in high interest environment
7-21
Short-term AND/OR
High-coupon Bonds
Long-term AND/OR
Low-coupon Bonds
Interest rate risk Low High Reinvestment rate risk High Low
7-22
Default Risk
If an issuer defaults, investors receive less than the promised return.
Influenced by the issuer’s financial strength and the terms of the bond contract.
7-22
7-23
Evaluating Default Risk:Bond Ratings
Investment Grade Junk Bonds
Moody’s
Aaa Aa A Baa Ba B Caa C
S & P AAA AA A BBB BB B CCC C
Bond ratings are designed to reflect the probability of a bond issue going into default.
High risk demands high yield. (PIGS 4 are paying high yield on their bonds)
7-23
7-24
Optional: Bond investing in real world
Find bond information Where to buy? Don’t buy individual bond (unless
risk-free T bond) Risk-free could be illusion, price
does change Bond ETF (HYG,JNK, BSV, SCPB)
7-25
Factors to consider when investing in bonds Yield
For single bond: current yield( annual interest payment/current price) and yield to maturity(consider principal payment).
For bond ETF or mutual fund: Yield (current yield) and SEC yield (consider ytm and expense)
Default risk and rating Interest risk and duration
Duration: the effective maturity of a bond (average maturity for a bond ETF)
7-26
Optional: practical questions
Is Treasury bond really risk-free? Risk free bond price changes
Is it good time to buy bonds now? Interest changes LT vs. ST bond
Should I buy individual high-yield corporate bond?
7-27
Example of treasure bond
Yahoo-Finance-Market data-Bond Comparing LT bond yield with ST
yield Expectation of interest rate
Default risk? Interest risk and time to maturity
7-28
Examples of bond portfolio
HYG (bond ETF) 12-month Yield: the sum of a fund’s
total trailing 12-month interest and dividend payments divided by the last month’s ending share price (NAV) plus any capital gains distributed over the same period.
SEC yield: based on 30-day period ending on the last day of previous month, reflects the deduction of the fund’s expenses.
7-29
Examples of bond portfolio
Rating(default risk): Duration(interest risk)
Duration: Average years to maturity Longer duration is more sensitive to interest
changes When interest rate increases from 3% to
4%, a bond with duration of 4 years will see price decline of approximate 4*(4%-3%)=4%
7-30
Examples of bond portfolio
Portfolio holdings Weight, maturity and coupon rate Risk of each individual issue and the benefit
of diversification Compared with other bond ETF
BSV (ST T bond+ corporate,yield,grade, duration)
SCPB (corporate A bond) JNK-high yield (similar to HYG) | 2,527 | 9,036 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-26 | latest | en | 0.855433 |
https://forum.allaboutcircuits.com/threads/star-wound-generator-voltages.3937/ | 1,619,109,070,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039594341.91/warc/CC-MAIN-20210422160833-20210422190833-00044.warc.gz | 366,112,250 | 20,251 | # Star wound generator voltages?
#### bobbingdoughnut
Joined Nov 26, 2006
1
Hi ,
Just a quick question , a gen. labeled 115v/200v , 20A i s the 200v the line voltage and 115v the phase voltage . If thats correct is the bigger voltage always the line voltage .line voltage= phase voltage x 1.73 .. and is there a specific layout for stating gen rating ie phase first / then line / then amps.
Thanks.
#### beenthere
Joined Apr 20, 2004
15,819
Hi,
The rating is 115/230 volts, 20 amps. The output panel should have two 115 volt outlets, and one 230 volt.
The alternator has two windings, which are joined internally - the 115 volt AC common line. The 115 volt output comes from one or the other winding, while the 230 volt output comes across both in series. The current limitation is 20 amps, whether from any one one outlet, or from all three together.
#### Johann
Joined Nov 27, 2006
190
The alternator, being a 3-phase, has 3 coils of 115 volts each, wound such that the phases differ by exactly 120 degrees. The ends of the 3 coils are connected together (in star). The free ends will then have a voltage of 115V x 1.73 = 198.95V between any two of the three loose ends(200V nominal).
In this star-connection, the phase current = the line current (20 A in your case). | 342 | 1,277 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2021-17 | longest | en | 0.872716 |
https://www.lmfdb.org/EllipticCurve/Q/78400/z/1 | 1,611,801,299,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704835583.91/warc/CC-MAIN-20210128005448-20210128035448-00483.warc.gz | 881,895,348 | 31,143 | # Properties
Label 78400.z1 Conductor $78400$ Discriminant $-2.531\times 10^{18}$ j-invariant $$\frac{14155776}{84035}$$ CM no Rank $0$ Torsion structure trivial
# Related objects
Show commands for: Magma / Pari/GP / SageMath
## Minimal Weierstrass equation
sage: E = EllipticCurve([0, 0, 0, 156800, -72716000])
gp: E = ellinit([0, 0, 0, 156800, -72716000])
magma: E := EllipticCurve([0, 0, 0, 156800, -72716000]);
$$y^2=x^3+156800x-72716000$$
trivial
## Integral points
sage: E.integral_points()
magma: IntegralPoints(E);
None
## Invariants
sage: E.conductor().factor() gp: ellglobalred(E)[1] magma: Conductor(E); Conductor: $$78400$$ = $$2^{6} \cdot 5^{2} \cdot 7^{2}$$ sage: E.discriminant().factor() gp: E.disc magma: Discriminant(E); Discriminant: $$-2530978231040000000$$ = $$-1 \cdot 2^{14} \cdot 5^{7} \cdot 7^{11}$$ sage: E.j_invariant().factor() gp: E.j magma: jInvariant(E); j-invariant: $$\frac{14155776}{84035}$$ = $$2^{19} \cdot 3^{3} \cdot 5^{-1} \cdot 7^{-5}$$ Endomorphism ring: $$\Z$$ Geometric endomorphism ring: $$\Z$$ (no potential complex multiplication) Sato-Tate group: $\mathrm{SU}(2)$
## BSD invariants
sage: E.rank() magma: Rank(E); Analytic rank: $$0$$ sage: E.regulator() magma: Regulator(E); Regulator: $$1$$ sage: E.period_lattice().omega() gp: E.omega[1] magma: RealPeriod(E); Real period: $$0.12863563074253299644982300982$$ sage: E.tamagawa_numbers() gp: gr=ellglobalred(E); [[gr[4][i,1],gr[5][i][4]] | i<-[1..#gr[4][,1]]] magma: TamagawaNumbers(E); Tamagawa product: $$16$$ = $$1\cdot2^{2}\cdot2^{2}$$ sage: E.torsion_order() gp: elltors(E)[1] magma: Order(TorsionSubgroup(E)); Torsion order: $$1$$ sage: E.sha().an_numerical() magma: MordellWeilShaInformation(E); Analytic order of Ш: $$1$$ (exact)
## Modular invariants
Modular form 78400.2.a.z
sage: E.q_eigenform(20)
gp: xy = elltaniyama(E);
gp: x*deriv(xy[1])/(2*xy[2]+E.a1*xy[1]+E.a3)
magma: ModularForm(E);
$$q - 3q^{3} + 6q^{9} + 5q^{11} + 3q^{13} - q^{17} + 6q^{19} + O(q^{20})$$
sage: E.modular_degree() magma: ModularDegree(E); Modular degree: 2211840 $$\Gamma_0(N)$$-optimal: yes Manin constant: 1
#### Special L-value
sage: r = E.rank();
sage: E.lseries().dokchitser().derivative(1,r)/r.factorial()
gp: ar = ellanalyticrank(E);
gp: ar[2]/factorial(ar[1])
magma: Lr1 where r,Lr1 := AnalyticRank(E: Precision:=12);
$$L(E,1)$$ ≈ $$2.0581700918805279431971681571112126923$$
## Local data
This elliptic curve is not semistable. There are 3 primes of bad reduction:
sage: E.local_data()
gp: ellglobalred(E)[5]
magma: [LocalInformation(E,p) : p in BadPrimes(E)];
prime Tamagawa number Kodaira symbol Reduction type Root number ord($$N$$) ord($$\Delta$$) ord$$(j)_{-}$$
$$2$$ $$1$$ $$II^{*}$$ Additive 1 6 14 0
$$5$$ $$4$$ $$I_1^{*}$$ Additive 1 2 7 1
$$7$$ $$4$$ $$I_5^{*}$$ Additive -1 2 11 5
## Galois representations
The 2-adic representation attached to this elliptic curve is surjective.
sage: rho = E.galois_representation();
sage: [rho.image_type(p) for p in rho.non_surjective()]
magma: [GaloisRepresentation(E,p): p in PrimesUpTo(20)];
The mod $$p$$ Galois representation has maximal image $$\GL(2,\F_p)$$ for all primes $$p$$ .
## $p$-adic data
### $p$-adic regulators
sage: [E.padic_regulator(p) for p in primes(5,20) if E.conductor().valuation(p)<2]
All $$p$$-adic regulators are identically $$1$$ since the rank is $$0$$.
## Iwasawa invariants
$p$ Reduction type $\lambda$-invariant(s) $\mu$-invariant(s) 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 add ss add add ordinary ordinary ordinary ordinary ordinary ordinary ordinary ordinary ordinary ordinary ordinary - 0,0 - - 0 0 0 0 0 0 0 0 0 0 0 - 0,0 - - 0 0 0 0 0 0 0 0 0 0 0
An entry - indicates that the invariants are not computed because the reduction is additive.
## Isogenies
This curve has no rational isogenies. Its isogeny class 78400.z consists of this curve only.
## Growth of torsion in number fields
The number fields $K$ of degree less than 24 such that $E(K)_{\rm tors}$ is strictly larger than $E(\Q)_{\rm tors}$ (which is trivial) are as follows:
$[K:\Q]$ $E(K)_{\rm tors}$ Base change curve $K$ $3$ 3.1.140.1 $$\Z/2\Z$$ Not in database $6$ 6.0.686000.1 $$\Z/2\Z \times \Z/2\Z$$ Not in database $8$ Deg 8 $$\Z/3\Z$$ Not in database $12$ Deg 12 $$\Z/4\Z$$ Not in database
We only show fields where the torsion growth is primitive. For fields not in the database, click on the degree shown to reveal the defining polynomial. | 1,649 | 4,471 | {"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} | 3.171875 | 3 | CC-MAIN-2021-04 | latest | en | 0.22671 |
https://www.wazeesupperclub.com/how-do-you-read-a-durbin-watson-table/ | 1,713,189,298,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816977.38/warc/CC-MAIN-20240415111434-20240415141434-00209.warc.gz | 999,491,225 | 38,648 | Wazeesupperclub.com Life How do you read a Durbin Watson table?
# How do you read a Durbin Watson table?
## How do you read a Durbin Watson table?
The Durbin-Watson statistic ranges in value from 0 to 4. A value near 2 indicates non-autocorrelation; a value toward 0 indicates positive autocorrelation; a value toward 4 indicates negative autocorrelation.
## How do you calculate Durbin Watson d test in Excel?
How to Perform a Durbin-Watson Test in Excel
1. Step 1: Enter the Data. First, we’ll enter the values for a dataset that we’d like to build a multiple linear regression model:
2. Step 2: Fit a Multiple Linear Regression Model.
3. Step 3: Perform the Durbin-Watson Test.
What does K mean in Durbin Watson table?
In the following tables, n is the sample size and k is the number of independent variables.
What does a positive autocorrelation mean?
Positive autocorrelation means that the increase observed in a time interval leads to a proportionate increase in the lagged time interval. The example of temperature discussed above demonstrates a positive autocorrelation.
### Is positive autocorrelation good or bad?
Autocorrelation measures the relationship between a variable’s current value and its past values. An autocorrelation of +1 represents a perfect positive correlation, while an autocorrelation of negative 1 represents a perfect negative correlation.
### What does the Durbin Watson test tell us?
The Durbin Watson statistic is a test statistic used in statistics to detect autocorrelation in the residuals from a regression analysis. The Durbin Watson statistic will always assume a value between 0 and 4. A value of DW = 2 indicates that there is no autocorrelation.
How to read a Durbin Watson table?
– If d < dL,α, there is statistical evidence that the error terms are positively autocorrelated. – If d > dU,α, there is no statistical evidence that the error terms are positively autocorrelated. – If dL,α < d < dU,α, the test is inconclusive.
How to interpret Durbin Watson?
– For positive serial correlation, consider adding lags of the dependent and/or independent variable to the model. – For negative serial correlation, check to make sure that none of your variables are overdifferenced. – For seasonal correlation, consider adding seasonal dummy variables to the model.
#### What is Durbin Watson?
The Durbin Watson (DW) statistic is a test for autocorrelation in the residuals from a statistical regression analysis. The Durbin – Watson statistic will always have a value between 0 and 4. Values from 0 to less than 2 indicate positive autocorrelation and values from from 2 to 4 indicate negative autocorrelation. | 595 | 2,670 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2024-18 | latest | en | 0.820643 |
https://au.mathworks.com/matlabcentral/answers/547248-how-to-split-this | 1,669,571,105,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710417.25/warc/CC-MAIN-20221127173917-20221127203917-00596.warc.gz | 144,272,738 | 28,744 | # How to split this?
2 views (last 30 days)
Younghun Kim on 12 Jun 2020
Commented: Younghun Kim on 12 Jun 2020
I have M as a result of another function.
and I have to split M like C automatically...
I want to split it every 16 bits.
M => input split function => output C
M = '01100001011000100110001110000000000000000000000000000000000000000000000000000000'
????
C = {'0110000101100010'; '0110001110000000'; '0000000000000000'; '0000000000000000';'0000000000000000';}
how to make it? I'm stuck on this problem...
can you help me?
Monika Jaskolka on 12 Jun 2020
Edited: Monika Jaskolka on 12 Jun 2020
C = reshape(M, 16,[])'
C =
5×16 char array
'0110000101100010'
'0110001110000000'
'0000000000000000'
'0000000000000000'
'0000000000000000'
If you need it to end up as a cell array or chars, you can wrap it in cellstr:
C = cellstr(reshape(M, 16,[])')
C =
5×1 cell array
{'0110000101100010'}
{'0110001110000000'}
{'0000000000000000'}
{'0000000000000000'}
{'0000000000000000'}
##### 1 CommentShowHide None
Younghun Kim on 12 Jun 2020
thank you very much :)
### More Answers (3)
Image Analyst on 12 Jun 2020
Did you try a simple for loop:
M = '01100001011000100110001110000000000000000000000000000000000000000000000000000000'
counter = 1;
for index1 = 1 : 16 : length(M)
index2 = min(length(M), index1 + 15);
c{counter} = M(index1 : index2);
counter = counter + 1;
end
celldisp(c)
##### 2 CommentsShowHide 1 older comment
Younghun Kim on 12 Jun 2020
thank you very much :)
David Hill on 12 Jun 2020
You could make a matrix
newMatrix=reshape(M(1:floor(length(M)/16)*16),16,[])';
##### 1 CommentShowHide None
Younghun Kim on 12 Jun 2020
thank you very much :)
Walter Roberson on 12 Jun 2020
temp = textscan(M,'%16s');
C = temp{1};
This handles M that are not multiples of 16 characters.
Note, though, that if M contains whitespace, that it will not split by 16 character groups, and would instead treat each "word" as the beginning of a section to be split.
##### 1 CommentShowHide None
Younghun Kim on 12 Jun 2020
thank you very much :)
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Start Hunting! | 727 | 2,251 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2022-49 | latest | en | 0.798692 |
https://math.stackexchange.com/questions/3792352/explanation-of-the-derivation-of-the-formula-for-the-sum-of-an-arithmetic-sequen | 1,701,996,073,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100705.19/warc/CC-MAIN-20231207221604-20231208011604-00626.warc.gz | 424,520,132 | 39,065 | # Explanation of the derivation of the formula for the sum of an arithmetic sequence of the first n terms
I am trying to understand the derivation of the formula for the sum of an arithmetic sequence of the first $$n$$ terms.
I do not understand what rules or reasoning allow two sequences to be added in reverse order to eliminate the common difference $$d$$ and arrive at the conclusion that the sum of an arithmetic sequence of the first $$n$$ terms is one half $$n$$ times the sum of the first and last terms. This seems to be a contrived way to eliminate the common difference from the expanded based on some unexplained knowledge of $$d$$ and arithmetic sequences in general.
I have researched this question in maths textbooks and online and each time the derivation is presented I cannot seem to find an explanation as to why it would be evident to a mathematician that by adding the sequences they would derive the formula.
The background.
The derivation of the formula as explained in many textbooks and online sites is as follows.
1. To find the sum of an arithmetic sequence for the first $$n$$ terms $$S_n$$, we can write out the sum in relation to the first term $$a_1$$ and the common difference $$d$$.
$$S_n = a_1 + (a_1 + d) + (a_1 + 2d) + (a_1 + 3d) + ... + a_n$$
1. It is also possible to write the sequence in reverse order in relation to the last term $$a_n$$.
$$S_n = a_n + (a_n - d) + (a_n - 2d) + (a_n - 3d) + ... + a_1$$
1. When we add these sequences together we derive the formula for the sum of the first n terms of an arithmetic sequence.
$$\begin{array}{r} S_n = a_1 + (a_1 + d) + (a_1 + 2d) + (a_1 + 3d) + \ldots + a_n \\ + \,S_n = a_n + (a_n - d) + (a_n - 2d) + (a_n - 3d) + \ldots + a_1 \\ \hline 2S_n = (a_1 + a_n) + (a_1 + a_n) + (a_1 + a_n) + (a_1 + a_n) \ldots \end{array}$$
1. Because there are $$n$$ many additions of $$(a_1 + a_n)$$ the lengthy sum is simplified as $$n(a_1 + a_n)$$ and solving for $$S_n$$ we arrive at the formula.
$$S_n = \frac{n}{2}(a_1 + a_n)$$
Unfortunately I can't seem to find the reasoning in any of these explanations as to why the two sequences (ordinary order and reverse) were added. It makes sense to me that they were added but not why this was the next logical step when deriving the formula.
The question.
Why were the two sequences added to derive the formula and what does that show about the nature of arithmetic sequences?
In my attempt to figure this out I noted that by studying many sequences we can see that the ratio of the sum of the sequence for the first $$n$$ terms $$S_n$$ and the sum of the first and last terms $$(a_1 + a_n)$$ is always $$\frac{n}{2}$$ for any arithmetic sequence. So possibly it could be said by induction that if for any arithmetic sequence it is true that:
$$\frac{S_n}{a_1 + a_n} = \frac{n}{2}$$
Then it must also be true that:
$$S_n = \frac{n}{2}(a_1 + a_n)$$
However, to me this still doesn't explain why the derivation decides to add the two sequences.
• "and then point 3 decides to conclude on the basis of some unclear reason that by adding the two sequences we can eliminate $d$..." What is unclear about it? Do you not see that $(a_1+d)+(a_n-d) = (a_1+a_n)+(d-d)=a_1+a_n$? Aug 16, 2020 at 2:11
• @JMoravitz yes this is very clear. My question (which I will update to be clearer) is why someone deriving the formula (for the first time for example) knew to write the sequences in both ways and then add them. Is it simply the case that it's obvious that doing this eliminates the common difference (d) and eliminating variables is desirable or is there a well known fact or property that hints that common difference will be eliminated?
– b_n
Aug 16, 2020 at 5:06
• "I can't seem to find the reasoning in any of these explanations as to why the two sequences (ordinary order and reverse) were added." Because it works. There doesn't need to be any more reason than that. " It makes sense to me that they were added but not why this was the next logical step when deriving the formula" Because it works. That's all that matters. You don't need to know how the figured it out. You just need to know what they did. And that's completely obvious once it's been pointed out, isn't. Aug 16, 2020 at 6:03
• Well, my thinking was when I was told to add all the numbers from $1$ to $100$ that I quickly saw that numbers that ended with $k$ and $10-k$ were easier to add up and that the came in equal pairs and I could add those and mulitply. I first tried to do in small groups (add $41$ to $49$, $42+48$. But then it was easy to see I could just make one long chain and make it a single multiplication. Aug 16, 2020 at 6:10
• " Is it simply the case that it's obvious that doing this eliminates the common difference (d) and eliminating variables is desirable or is there a well known fact or property that hints that common difference will be eliminated?" Well, ... yeah... I guess so. If you ever had the task of having to add up a list of twenty numbers without a calculator, this becomes pretty obvious pretty quick. It's hard for me to imagine someone not coming up with shortcuts. Although to come up with a single short cut and prove it always works takes a certain amount of acuity. Aug 16, 2020 at 6:13
• I don't find this one mysterious at all. You have a sequence increasing by $d$, so if you add it to a sequence decreasing by $d$ you get a constant. Constants are always interesting. Aug 16, 2020 at 5:11 | 1,454 | 5,448 | {"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": 24, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2023-50 | latest | en | 0.922473 |
https://www.coursehero.com/file/6628941/25-ch-14-Mechanical-Design-budynas-SM-ch14/ | 1,493,469,200,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123491.68/warc/CC-MAIN-20170423031203-00500-ip-10-145-167-34.ec2.internal.warc.gz | 890,147,204 | 22,922 | 25_ch 14 Mechanical Design budynas_SM_ch14
# 25_ch 14 Mechanical Design budynas_SM_ch14 -...
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Unformatted text preview: budynas_SM_ch14.qxd 12/05/2006 17:39 Page 373 FIRST PAGES 373 Chapter 14 Modification of St by ( Y N ) P = 0.832 produces ( σall ) P = 45 657 psi, Similarly for ( Y N ) G = 0.859 ( σall ) G = 47 161 psi, and t W1 = 4569 lbf, H1 = 228 hp t W2 = 5668 lbf, H2 = 283 hp From Table 14-8, Cp = 2300 psi. Also, from Table 14-6: 0.99 ( Sc ) 107 = 180 000 psi Modification of Sc by (Y N ) produces ( σc, all ) P = 130 525 psi ( σc, all ) G = 138 069 psi and t W3 = 2489 lbf, H3 = 124.3 hp t W4 = 2767 lbf, H4 = 138.2 hp Rating Hrated = min(228, 283, 124, 138) = 124 hp 14-28 Ans . Grade 2 9310 carburized and case-hardened to 285 core and 580 case in Prob. 14-27. Summary: Table 14-3: 0.99 ( St ) 107 = 65 000 psi ( σall ) P = 53 959 psi ( σall ) G = 55 736 psi and it follows that t W1 = 5399.5 lbf, H1 = 270 hp W2 = 6699 lbf, H2 = 335 hp t From Table 14-8, C p = 2300 psi. Also, from Table 14-6: Sc = 225 000 psi ( σc, all ) P = 181 285 psi ( σc, all ) G = 191 762 psi Consequently, t W3 = 4801 lbf, H3 = 240 hp t W4 = 5337 lbf, H4 = 267 hp Rating Hrated = min(270, 335, 240, 267) = 240 hp. Ans . ...
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## This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.
Ask a homework question - tutors are online | 602 | 1,542 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2017-17 | longest | en | 0.770463 |
http://mathandmultimedia.com/2013/07/03/paper-folding-equiangular-triangle/ | 1,398,064,273,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609539665.16/warc/CC-MAIN-20140416005219-00087-ip-10-147-4-33.ec2.internal.warc.gz | 147,500,152 | 55,745 | # Paper Folding: How to Construct an Equiangular Triangle
Paper folding or origami can be used to create intricate and stunning designs. In addition, paper folding can also be used to teach or learn mathematics. In this post, we use a square piece of paper to construct an equiangular triangle. After the construction, we prove that the triangle is equiangular.
Steps in Constructing an Equiangular Triangle
1. Cut a square piece of paper. For the sake of discussion, we label the square ABCD.
2. Fold at the center so that AB coincides with CD . Crease well, then unfold.
3. Select vertex and fold so that A falls on the center line and the crease passes through D. Unfold.
4. Select vertex and fold so that B falls on the center line and that the crease passes through C. Unfold.
5. Let E be the intersection of the crease formed in steps 3 and 4. Triangle DEC is an equiangular triangle.
In the next section, we discuss the mathematical proof why DEC is an equiangular triangle.
The Mathematical Proof
Let F be the intersection of AB and the crease passing through D and let G be the intersection of AB and the crease passing through C.
Now, FD is twice FA (why?), so triangle DAF is 30◦-60◦-90◦ triangle. This means that angle ADF, the angle opposite the shorter side of the triangle measures 30◦. This means that angle EDC = 60◦.
Also, triangle CBG is a 30◦-60◦-90◦ triangle which implies that angle DCE = 60◦.
Therefore, by the Angle Sum Theorem, angle DEF = 60◦.
So, triangle DEF is an equiangular triangle.
Reference: Paper Folding: Art and Mathematics (A local book). A copy of the book can be bought from the UP NISMED Bookstore. A more detailed presentation of the paper folding can also be found in our Agimat website.
## 3 thoughts on “Paper Folding: How to Construct an Equiangular Triangle”
1. “Now, AD is twice FA (why?)”
It should be “FD is twice FA”. Also, it might not be that obvious to the reader why is this so.
2. This is a really cool use of origami. I’ve only ever used it to make cranes and frogs and rabbits. | 526 | 2,056 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2014-15 | longest | en | 0.874973 |
https://trac.sagemath.org/ticket/10994 | 1,618,638,891,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038101485.44/warc/CC-MAIN-20210417041730-20210417071730-00281.warc.gz | 675,407,298 | 7,054 | Opened 10 years ago
Closed 10 years ago
# Bug in permutation_automorphism_group for linear codes
Reported by: Owned by: tfeulner wdj major sage-duplicate/invalid/wontfix coding theory rlm Robert Miller N/A
### Description
The method `permutation_automorphism_group` gives different results for the two options `method="gap"`and `method="partition":`
```sage: C = HammingCode(6, GF(2)).dual_code()
sage: G1 = C.permutation_automorphism_group(method="gap") # requires optional GAP package Guava
sage: G2 = C.permutation_automorphism_group(method="partition")
sage: G1 == G2 # requires optional GAP package Guava
False # <-------- should be equal!
```
The application of ticket:10153 indicates that the problem is caused by `method="partition"`.
### comment:1 Changed 10 years ago by rlm
Thomas also points out that the code only checks minimal weight vectors:
``` if algorithm=="partition":
from sage.groups.perm_gps.partn_ref.refinement_matrices import MatrixStruct
stop = 0 # only stop if all gens are autos
for i in range(1,len(nonzerowts)):
if stop == 1:
break
wt = nonzerowts[i]
Cwt = [c for c in self if hamming_weight(c)==wt] # ridiculously slow!!
MS = MatrixSpace(F,len(Cwt),n)
Cwords_wt = MS(Cwt)
M = MatrixStruct(Cwords_wt)
autgp = M.automorphism_group()
if autgp[0] == []:
return PermutationGroup([()])
L = [[j+1 for j in autgp[0][i]] for i in range(len(autgp[0]))]
G = PermutationGroup([Sn(x) for x in L])
return G
```
This code is, frankly, terrible. I figured out after a bit of searching that this was introduced into sage at #4320 by David Joyner. Michael Abshoff gave the ticket a positive review based on a misunderstood conversation we had. I did not sufficiently examine the code to vouch for it, but I think he wanted to get it merged anyway.
If one looks at the code before that ticket, one sees the true intention of the algorithm which was a bit butchered at #4320.
Related to this is ticket #11032, which is specific to the binary case. I'll fix both of these to do the right thing, since I'm probably in the best position to do so. In fact, I think I'm going to suggest we remove automorphism_group_binary_code completely since we now have all cases.
### comment:2 Changed 10 years ago by rlm
See #11033, which will fix this problem.
### comment:3 Changed 10 years ago by rlm
After applying #10871 and #11033:
```sage: C = HammingCode(6, GF(2)).dual_code()
sage: time G1 = C.permutation_automorphism_group(algorithm="gap")
CPU times: user 1.71 s, sys: 0.04 s, total: 1.76 s
Wall time: 24.39 s
sage: time G2 = C.permutation_automorphism_group(method="partition")
CPU times: user 0.04 s, sys: 0.00 s, total: 0.04 s
Wall time: 0.04 s
sage: G1 == G2
True
```
### comment:4 Changed 10 years ago by jdemeyer
• Authors Thomas Feulner deleted
• Milestone set to sage-duplicate/invalid/wontfix
• Resolution set to duplicate
• Reviewers set to Robert Miller
• Status changed from new to closed
Note: See TracTickets for help on using tickets. | 822 | 3,025 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | latest | en | 0.821626 |
http://www.i-am-bored.com/2013/03/solving-3-rubiks-cubes-while-juggling-them.html | 1,455,040,775,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701157443.43/warc/CC-MAIN-20160205193917-00179-ip-10-236-182-209.ec2.internal.warc.gz | 472,741,123 | 15,090 | Rockin' in the free world since 2005.
# Solving 3 Rubik`s Cubes While Juggling Them
Einsteiner was a piker next to this kid
### Submitted by: scheckydamon
[Total: 11 Average: 1.9/5]
Hits: 4366
Rating: 1.9
Category: Games
Date: 03/25/13 07:57 AM
### 10 Responses to Solving 3 Rubik`s Cubes While Juggling Them
1. scheckydamon Male 60-69
675 posts
March 25, 2013 at 7:57 am
Link: Solving 3 Rubik`s Cubes While Juggling Them - Einsteiner was a piker next to this kid
2. bluevayero Male 18-29
226 posts
March 25, 2013 at 9:44 am
woa
3. EgalM Male 30-39
1707 posts
March 25, 2013 at 11:31 am
Watched a video explaining how most people do these tricks. It boils down to spinning two ways till it comes back to the start. So they spin top, than side till it`s half way, than just keep going to finish it. Wish I could find that link, I suck at explaining.
4. KLouD Male 30-39
254 posts
March 25, 2013 at 2:37 pm
Watched a video explaining how most people do these tricks. It boils down to spinning two ways till it comes back to the start. So they spin top, than side till it`s half way, than just keep going to finish it. Wish I could find that link, I suck at explaining.
Myth. You can`t solve a cube by only turning 2 sides. I`ve seen what you`re talking about but yes its a myth.
5. curiousboy Male 50-59
93 posts
March 25, 2013 at 4:06 pm
Who is Einsteiner?
6. scheckydamon Male 60-69
675 posts
March 25, 2013 at 5:40 pm
Sorry. Einstein. I was a little too deep into the single malt.
7. Squrlz4Sale Male 40-49
6230 posts
March 25, 2013 at 6:09 pm
@KLouD: I think EgalM is on to something. You can take a solved cube and with a sequence of repetive moves (right-side back, top clockwise, rotate entire cube 90 deg clockwise--repeat 10X, for example) scramble it in a way that can readily be solved by reversing the movements 10X.
I`ve been suspicious of these juggling videos because every bonafide video I`ve seen of a cube stunt is careful to include 15 seconds at the start of someone else doing a thoroughly random scramble on camera (because everyone familiar with the cubes knows that you can cheat as described above).
Obviously a suspicion is not proof of anything, and perhaps these jugglers really *are* solving random cubes. But if they *can* accomplish such an incredible feat, it`s peculiar they aren`t including that 15-second random scramble at the start so as to leave no question of their ability. | 715 | 2,423 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-07 | longest | en | 0.926789 |
https://hackage-origin.haskell.org/package/clash-prelude-1.2.1/candidate/docs/src/Clash.Explicit.BlockRam.html | 1,721,019,792,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514659.22/warc/CC-MAIN-20240715040934-20240715070934-00480.warc.gz | 268,165,897 | 14,369 | ```{-|
Copyright : (C) 2013-2016, University of Twente,
2016-2017, Myrtle Software Ltd,
Maintainer : Christiaan Baaij <christiaan.baaij@gmail.com>
BlockRAM primitives
= Using RAMs #usingrams#
We will show a rather elaborate example on how you can, and why you might want
to use 'blockRam's. We will build a \"small\" CPU+Memory+Program ROM where we
will slowly evolve to using blockRams. Note that the code is /not/ meant as a
We start with the definition of the Instructions, Register names and machine
codes:
@
{\-\# LANGUAGE RecordWildCards, TupleSections, DeriveAnyClass \#-\}
module CPU where
import Clash.Explicit.Prelude
type Value = Signed 8
data Instruction
= Compute Operator Reg Reg Reg
| Branch Reg Value
| Jump Value
| Nop
deriving (Eq,Show)
data Reg
= Zero
| PC
| RegA
| RegB
| RegC
| RegD
| RegE
deriving (Eq,Show,Enum)
data Operator = Add | Sub | Incr | Imm | CmpGt
deriving (Eq,Show)
data MachCode
= MachCode
{ inputX :: Reg
, inputY :: Reg
, result :: Reg
, aluCode :: Operator
, ldReg :: Reg
, jmpM :: Maybe Value
}
nullCode = MachCode { inputX = Zero, inputY = Zero, result = Zero, aluCode = Imm
, jmpM = Nothing
}
@
Next we define the CPU and its ALU:
@
cpu
:: Vec 7 Value
-- ^ Register bank
-> (Value,Instruction)
-- ^ (Memory output, Current instruction)
-> ( Vec 7 Value
)
where
-- Current instruction pointer
ipntr = regbank '!!' PC
-- Decoder
(MachCode {..}) = case instr of
Compute op rx ry res -> nullCode {inputX=rx,inputY=ry,result=res,aluCode=op}
Branch cr a -> nullCode {inputX=cr,jmpM=Just a}
Jump a -> nullCode {aluCode=Incr,jmpM=Just a}
Store r a -> nullCode {inputX=r,wrAddrM=Just a}
Nop -> nullCode
-- ALU
regX = regbank '!!' inputX
regY = regbank '!!' inputY
aluOut = alu aluCode regX regY
-- next instruction
nextPC = case jmpM of
Just a | aluOut /= 0 -> ipntr + a
_ -> ipntr + 1
-- update registers
regbank' = 'replace' Zero 0
\$ 'replace' PC nextPC
\$ 'replace' result aluOut
\$ 'replace' ldReg memOut
\$ regbank
alu Add x y = x + y
alu Sub x y = x - y
alu Incr x _ = x + 1
alu Imm x _ = x
alu CmpGt x y = if x > y then 1 else 0
@
We initially create a memory out of simple registers:
@
dataMem
:: KnownDomain dom
=> Clock dom
-> Reset dom
-> Enable dom
-- ^ (write address, data in)
-> Signal dom Value
-- ^ data out
dataMem clk rst en rd wrM = 'Clash.Explicit.Mealy.mealy' clk rst en dataMemT ('Clash.Sized.Vector.replicate' d32 0) (bundle (rd,wrM))
where
dataMemT mem (rd,wrM) = (mem',dout)
where
dout = mem '!!' rd
mem' = case wrM of
Just (wr,din) -> 'replace' wr din mem
_ -> mem
@
And then connect everything:
@
system
:: ( KnownDomain dom
, KnownNat n )
=> Vec n Instruction
-> Clock dom
-> Reset dom
-> Enable dom
-> Signal dom Value
system instrs clk rst en = memOut
where
memOut = dataMem clk rst en rdAddr dout
(rdAddr,dout,ipntr) = 'Clash.Explicit.Mealy.mealyB' clk rst en cpu ('Clash.Sized.Vector.replicate' d7 0) (memOut,instr)
instr = 'Clash.Explicit.Prelude.asyncRom' instrs '<\$>' ipntr
@
Create a simple program that calculates the GCD of 4 and 6:
@
-- Compute GCD of 4 and 6
prog = -- 0 := 4
Compute Incr Zero RegA RegA :>
replicate d3 (Compute Incr RegA Zero RegA) ++
Store RegA 0 :>
-- 1 := 6
Compute Incr Zero RegA RegA :>
replicate d5 (Compute Incr RegA Zero RegA) ++
Store RegA 1 :>
-- A := 4
-- B := 6
-- start
Compute CmpGt RegA RegB RegC :>
Branch RegC 4 :>
Compute CmpGt RegB RegA RegC :>
Branch RegC 4 :>
Jump 5 :>
-- (a > b)
Compute Sub RegA RegB RegA :>
Jump (-6) :>
-- (b > a)
Compute Sub RegB RegA RegB :>
Jump (-8) :>
-- end
Store RegA 2 :>
Nil
@
And test our system:
@
>>> sampleN 32 \$ system prog systemClockGen resetGen enableGen
[0,0,0,0,0,0,4,4,4,4,4,4,4,4,6,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,2]
@
to see that our system indeed calculates that the GCD of 6 and 4 is 2.
=== Improvement 1: using @asyncRam@
As you can see, it's fairly straightforward to build a memory using registers
and read ('!!') and write ('replace') logic. This might however not result in
the most efficient hardware structure, especially when building an ASIC.
Instead it is preferable to use the 'Clash.Prelude.RAM.asyncRam' function which
has the potential to be translated to a more efficient structure:
@
system2
:: ( KnownDomain dom
, KnownNat n )
=> Vec n Instruction
-> Clock dom
-> Reset dom
-> Enable dom
-> Signal dom Value
system2 instrs clk rst en = memOut
where
memOut = 'Clash.Explicit.RAM.asyncRam' clk clk en d32 rdAddr dout
(rdAddr,dout,ipntr) = 'mealyB' clk rst en cpu ('Clash.Sized.Vector.replicate' d7 0) (memOut,instr)
instr = 'Clash.Prelude.ROM.asyncRom' instrs '<\$>' ipntr
@
Again, we can simulate our system and see that it works. This time however,
we need to disregard the first few output samples, because the initial content of an
'Clash.Prelude.RAM.asyncRam' is 'undefined', and consequently, the first few
output samples are also 'undefined'. We use the utility function 'printX' to conveniently
filter out the undefinedness and replace it with the string "X" in the few leading outputs.
@
>>> printX \$ sampleN 32 \$ system2 prog systemClockGen resetGen enableGen
[X,X,X,X,X,X,4,4,4,4,4,4,4,4,6,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,2]
@
=== Improvement 2: using @blockRam@
Finally we get to using 'blockRam'. On FPGAs, 'Clash.Prelude.RAM.asyncRam' will
be implemented in terms of LUTs, and therefore take up logic resources. FPGAs
also have large(r) memory structures called /Block RAMs/, which are preferred,
especially as the memories we need for our application get bigger. The
'blockRam' function will be translated to such a /Block RAM/.
One important aspect of Block RAMs have a /synchronous/ read port, meaning that,
at time @t@, the value @v@ in the RAM at address @r@ is only available at time
@t+1@.
For us that means we need to change the design of our CPU. Right now, upon a
that read address is immediately available to be put in the register bank.
Because we will be using a BlockRAM, the value is delayed until the next cycle.
We hence need to also delay the register address to which the memory address
@
cpu2
:: (Vec 7 Value,Reg)
-> (Value,Instruction)
-- ^ (Memory output, Current instruction)
-> ( (Vec 7 Value,Reg)
)
where
-- Current instruction pointer
ipntr = regbank '!!' PC
-- Decoder
(MachCode {..}) = case instr of
Compute op rx ry res -> nullCode {inputX=rx,inputY=ry,result=res,aluCode=op}
Branch cr a -> nullCode {inputX=cr,jmpM=Just a}
Jump a -> nullCode {aluCode=Incr,jmpM=Just a}
Store r a -> nullCode {inputX=r,wrAddrM=Just a}
Nop -> nullCode
-- ALU
regX = regbank '!!' inputX
regY = regbank '!!' inputY
aluOut = alu aluCode regX regY
-- next instruction
nextPC = case jmpM of
Just a | aluOut /= 0 -> ipntr + a
_ -> ipntr + 1
-- update registers
ldRegD' = ldReg -- Delay the ldReg by 1 cycle
regbank' = 'replace' Zero 0
\$ 'replace' PC nextPC
\$ 'replace' result aluOut
\$ 'replace' ldRegD memOut
\$ regbank
@
We can now finally instantiate our system with a 'blockRam':
@
system3
:: ( KnownDomain dom
, KnownNat n )
=> Vec n Instruction
-> Clock dom
-> Reset dom
-> Enable dom
-> Signal dom Value
system3 instrs clk rst en = memOut
where
memOut = 'blockRam' clk en (replicate d32 0) rdAddr dout
(rdAddr,dout,ipntr) = 'mealyB' clk rst en cpu2 (('Clash.Sized.Vector.replicate' d7 0),Zero) (memOut,instr)
instr = 'Clash.Explicit.Prelude.asyncRom' instrs '<\$>' ipntr
@
We are, however, not done. We will also need to update our program. The reason
being that values that we try to load in our registers won't be loaded into the
register until the next cycle. This is a problem when the next instruction
immediately depended on this memory value. In our case, this was only the case
when the loaded the value @6@, which was stored at address @1@, into @RegB@.
Our updated program is thus:
@
prog2 = -- 0 := 4
Compute Incr Zero RegA RegA :>
replicate d3 (Compute Incr RegA Zero RegA) ++
Store RegA 0 :>
-- 1 := 6
Compute Incr Zero RegA RegA :>
replicate d5 (Compute Incr RegA Zero RegA) ++
Store RegA 1 :>
-- A := 4
-- B := 6
Nop :> -- Extra NOP
-- start
Compute CmpGt RegA RegB RegC :>
Branch RegC 4 :>
Compute CmpGt RegB RegA RegC :>
Branch RegC 4 :>
Jump 5 :>
-- (a > b)
Compute Sub RegA RegB RegA :>
Jump (-6) :>
-- (b > a)
Compute Sub RegB RegA RegB :>
Jump (-8) :>
-- end
Store RegA 2 :>
Nil
@
When we simulate our system we see that it works. This time again,
we need to disregard the first sample, because the initial output of a
'blockRam' is 'undefined'. We use the utility function 'printX' to conveniently
filter out the undefinedness and replace it with the string "X".
@
>>> printX \$ sampleN 34 \$ system3 prog2 systemClockGen resetGen enableGen
[X,0,0,0,0,0,0,4,4,4,4,4,4,4,4,6,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,2]
@
This concludes the short introduction to using 'blockRam'.
-}
{-# LANGUAGE Trustworthy #-}
{-# OPTIONS_GHC -fplugin GHC.TypeLits.KnownNat.Solver #-}
-- See: https://github.com/clash-lang/clash-compiler/commit/721fcfa9198925661cd836668705f817bddaae3c
-- as to why we need this.
{-# OPTIONS_GHC -fno-cpr-anal #-}
module Clash.Explicit.BlockRam
( -- * BlockRAM synchronized to the system clock
blockRam
, blockRamPow2
, blockRamU
, blockRam1
, ResetStrategy(..)
-- * Internal
, blockRam#
)
where
import Data.Maybe (isJust)
import qualified Data.Sequence as Seq
import GHC.Stack (HasCallStack, withFrozenCallStack)
import GHC.TypeLits (KnownNat, type (^), type (<=))
import Prelude hiding (length, replicate)
import Clash.Annotations.Primitive
(hasBlackBox)
import Clash.Class.Num (SaturationMode(SatBound), satSucc)
import Clash.Explicit.Signal (KnownDomain, Enable, register, fromEnable)
import Clash.Signal.Internal
(Clock(..), Reset, Signal (..), invertReset, (.&&.), mux)
import Clash.Promoted.Nat (SNat(..))
import Clash.Signal.Bundle (unbundle)
import Clash.Sized.Unsigned (Unsigned)
import Clash.Sized.Index (Index)
import Clash.Sized.Vector (Vec, replicate, toList, iterateI)
import qualified Clash.Sized.Vector as CV
import Clash.XException
(maybeIsX, seqX, NFDataX, deepErrorX, defaultSeqX, fromJustX)
{- \$setup
>>> :m -Clash.Prelude
>>> :m -Clash.Prelude.Safe
>>> import Clash.Explicit.Prelude as C
>>> import qualified Data.List as L
>>> :set -XDataKinds -XRecordWildCards -XTupleSections -XDeriveAnyClass -XDeriveGeneric
>>> type InstrAddr = Unsigned 8
>>> type MemAddr = Unsigned 5
>>> type Value = Signed 8
>>> :{
data Reg
= Zero
| PC
| RegA
| RegB
| RegC
| RegD
| RegE
deriving (Eq,Show,Enum,C.Generic,NFDataX)
:}
>>> :{
data Operator = Add | Sub | Incr | Imm | CmpGt
deriving (Eq,Show)
:}
>>> :{
data Instruction
= Compute Operator Reg Reg Reg
| Branch Reg Value
| Jump Value
| Nop
deriving (Eq,Show)
:}
>>> :{
data MachCode
= MachCode
{ inputX :: Reg
, inputY :: Reg
, result :: Reg
, aluCode :: Operator
, ldReg :: Reg
, jmpM :: Maybe Value
}
:}
>>> :{
nullCode = MachCode { inputX = Zero, inputY = Zero, result = Zero, aluCode = Imm
, jmpM = Nothing
}
:}
>>> :{
alu Add x y = x + y
alu Sub x y = x - y
alu Incr x _ = x + 1
alu Imm x _ = x
alu CmpGt x y = if x > y then 1 else 0
:}
>>> :{
let cpu :: Vec 7 Value -- ^ Register bank
-> (Value,Instruction) -- ^ (Memory output, Current instruction)
-> ( Vec 7 Value
)
where
-- Current instruction pointer
ipntr = regbank C.!! PC
-- Decoder
(MachCode {..}) = case instr of
Compute op rx ry res -> nullCode {inputX=rx,inputY=ry,result=res,aluCode=op}
Branch cr a -> nullCode {inputX=cr,jmpM=Just a}
Jump a -> nullCode {aluCode=Incr,jmpM=Just a}
Store r a -> nullCode {inputX=r,wrAddrM=Just a}
Nop -> nullCode
-- ALU
regX = regbank C.!! inputX
regY = regbank C.!! inputY
aluOut = alu aluCode regX regY
-- next instruction
nextPC = case jmpM of
Just a | aluOut /= 0 -> ipntr + a
_ -> ipntr + 1
-- update registers
regbank' = replace Zero 0
\$ replace PC nextPC
\$ replace result aluOut
\$ replace ldReg memOut
\$ regbank
:}
>>> :{
let dataMem
:: KnownDomain dom
=> Clock dom
-> Reset dom
-> Enable dom
-> Signal dom Value
dataMem clk rst en rd wrM = mealy clk rst en dataMemT (C.replicate d32 0) (bundle (rd,wrM))
where
dataMemT mem (rd,wrM) = (mem',dout)
where
dout = mem C.!! rd
mem' = case wrM of
Just (wr,din) -> replace wr din mem
Nothing -> mem
:}
>>> :{
let system
:: ( KnownDomain dom
, KnownNat n )
=> Vec n Instruction
-> Clock dom
-> Reset dom
-> Enable dom
-> Signal dom Value
system instrs clk rst en = memOut
where
memOut = dataMem clk rst en rdAddr dout
(rdAddr,dout,ipntr) = mealyB clk rst en cpu (C.replicate d7 0) (memOut,instr)
instr = asyncRom instrs <\$> ipntr
:}
>>> :{
-- Compute GCD of 4 and 6
prog = -- 0 := 4
Compute Incr Zero RegA RegA :>
C.replicate d3 (Compute Incr RegA Zero RegA) C.++
Store RegA 0 :>
-- 1 := 6
Compute Incr Zero RegA RegA :>
C.replicate d5 (Compute Incr RegA Zero RegA) C.++
Store RegA 1 :>
-- A := 4
-- B := 6
-- start
Compute CmpGt RegA RegB RegC :>
Branch RegC 4 :>
Compute CmpGt RegB RegA RegC :>
Branch RegC 4 :>
Jump 5 :>
-- (a > b)
Compute Sub RegA RegB RegA :>
Jump (-6) :>
-- (b > a)
Compute Sub RegB RegA RegB :>
Jump (-8) :>
-- end
Store RegA 2 :>
Nil
:}
>>> :{
let system2
:: ( KnownDomain dom
, KnownNat n )
=> Vec n Instruction
-> Clock dom
-> Reset dom
-> Enable dom
-> Signal dom Value
system2 instrs clk rst en = memOut
where
memOut = asyncRam clk clk en d32 rdAddr dout
(rdAddr,dout,ipntr) = mealyB clk rst en cpu (C.replicate d7 0) (memOut,instr)
instr = asyncRom instrs <\$> ipntr
:}
>>> :{
let cpu2 :: (Vec 7 Value,Reg) -- ^ (Register bank, Load reg addr)
-> (Value,Instruction) -- ^ (Memory output, Current instruction)
-> ( (Vec 7 Value,Reg)
)
where
-- Current instruction pointer
ipntr = regbank C.!! PC
-- Decoder
(MachCode {..}) = case instr of
Compute op rx ry res -> nullCode {inputX=rx,inputY=ry,result=res,aluCode=op}
Branch cr a -> nullCode {inputX=cr,jmpM=Just a}
Jump a -> nullCode {aluCode=Incr,jmpM=Just a}
Store r a -> nullCode {inputX=r,wrAddrM=Just a}
Nop -> nullCode
-- ALU
regX = regbank C.!! inputX
regY = regbank C.!! inputY
aluOut = alu aluCode regX regY
-- next instruction
nextPC = case jmpM of
Just a | aluOut /= 0 -> ipntr + a
_ -> ipntr + 1
-- update registers
ldRegD' = ldReg -- Delay the ldReg by 1 cycle
regbank' = replace Zero 0
\$ replace PC nextPC
\$ replace result aluOut
\$ replace ldRegD memOut
\$ regbank
:}
>>> :{
let system3
:: ( KnownDomain dom
, KnownNat n )
=> Vec n Instruction
-> Clock dom
-> Reset dom
-> Enable dom
-> Signal dom Value
system3 instrs clk rst en = memOut
where
memOut = blockRam clk en (C.replicate d32 0) rdAddr dout
(rdAddr,dout,ipntr) = mealyB clk rst en cpu2 ((C.replicate d7 0),Zero) (memOut,instr)
instr = asyncRom instrs <\$> ipntr
:}
>>> :{
prog2 = -- 0 := 4
Compute Incr Zero RegA RegA :>
C.replicate d3 (Compute Incr RegA Zero RegA) C.++
Store RegA 0 :>
-- 1 := 6
Compute Incr Zero RegA RegA :>
C.replicate d5 (Compute Incr RegA Zero RegA) C.++
Store RegA 1 :>
-- A := 4
-- B := 6
Nop :> -- Extra NOP
-- start
Compute CmpGt RegA RegB RegC :>
Branch RegC 4 :>
Compute CmpGt RegB RegA RegC :>
Branch RegC 4 :>
Jump 5 :>
-- (a > b)
Compute Sub RegA RegB RegA :>
Jump (-6) :>
-- (b > a)
Compute Sub RegB RegA RegB :>
Jump (-8) :>
-- end
Store RegA 2 :>
Nil
:}
-}
-- | Create a blockRAM with space for @n@ elements
--
-- * __NB__: Read value is delayed by 1 cycle
-- * __NB__: Initial output value is 'undefined'
--
-- @
-- bram40
-- :: 'Clock' dom
-- -> 'Enable' dom
-- -> 'Signal' dom ('Unsigned' 6)
-- -> 'Signal' dom (Maybe ('Unsigned' 6, 'Clash.Sized.BitVector.Bit'))
-- -> 'Signal' dom 'Clash.Sized.BitVector.Bit'
-- bram40 clk en = 'blockRam' clk en ('Clash.Sized.Vector.replicate' d40 1)
-- @
--
--
-- * See "Clash.Explicit.BlockRam#usingrams" for more information on how to use a
-- Block RAM.
blockRam
:: ( KnownDomain dom
, HasCallStack
, NFDataX a
=> Clock dom
-- ^ 'Clock' to synchronize to
-> Enable dom
-- ^ Global enable
-> Vec n a
-- ^ Initial content of the BRAM, also determines the size, @n@, of the BRAM.
--
-- __NB__: __MUST__ be a constant.
-> Signal dom (Maybe (addr, a))
-- ^ (write address @w@, value to write)
-> Signal dom a
-- ^ Value of the @blockRAM@ at address @r@ from the previous clock cycle
blockRam = \clk gen content rd wrM ->
let en = isJust <\$> wrM
(wr,din) = unbundle (fromJustX <\$> wrM)
in withFrozenCallStack
(blockRam# clk gen content (fromEnum <\$> rd) en (fromEnum <\$> wr) din)
{-# INLINE blockRam #-}
-- | Create a blockRAM with space for 2^@n@ elements
--
-- * __NB__: Read value is delayed by 1 cycle
-- * __NB__: Initial output value is 'undefined'
--
-- @
-- bram32
-- :: 'Clock' dom
-- -> 'Enable' dom
-- -> 'Signal' dom ('Unsigned' 5)
-- -> 'Signal' dom (Maybe ('Unsigned' 5, 'Clash.Sized.BitVector.Bit'))
-- -> 'Signal' dom 'Clash.Sized.BitVector.Bit'
-- bram32 clk en = 'blockRamPow2' clk en ('Clash.Sized.Vector.replicate' d32 1)
-- @
--
--
-- * See "Clash.Prelude.BlockRam#usingrams" for more information on how to use a
-- Block RAM.
blockRamPow2
:: ( KnownDomain dom
, HasCallStack
, NFDataX a
, KnownNat n )
=> Clock dom
-- ^ 'Clock' to synchronize to
-> Enable dom
-- ^ Global enable
-> Vec (2^n) a
-- ^ Initial content of the BRAM, also
-- determines the size, @2^n@, of
-- the BRAM.
--
-- __NB__: __MUST__ be a constant.
-> Signal dom (Unsigned n)
-> Signal dom (Maybe (Unsigned n, a))
-- ^ (Write address @w@, value to write)
-> Signal dom a
-- ^ Value of the @blockRAM@ at address @r@ from the previous
-- clock cycle
blockRamPow2 = \clk en cnt rd wrM -> withFrozenCallStack
(blockRam clk en cnt rd wrM)
{-# INLINE blockRamPow2 #-}
data ResetStrategy (r :: Bool) where
ClearOnReset :: ResetStrategy 'True
NoClearOnReset :: ResetStrategy 'False
-- | Version of blockram that has no default values set. May be cleared to a
-- arbitrary state using a reset function.
blockRamU
:: forall n dom a r addr
. ( KnownDomain dom
, HasCallStack
, NFDataX a
, 1 <= n )
=> Clock dom
-- ^ 'Clock' to synchronize to
-> Reset dom
-- ^ 'Reset' line to listen to. Needs to be held at least /n/ cycles in order
-- for the BRAM to be reset to its initial state.
-> Enable dom
-- ^ Global enable
-> ResetStrategy r
-- ^ Whether to clear BRAM on asserted reset ('ClearOnReset') or
-- not ('NoClearOnReset'). Reset needs to be asserted at least /n/ cycles to
-- clear the BRAM.
-> SNat n
-- ^ Number of elements in BRAM
-> (Index n -> a)
-- ^ If applicable (see first argument), reset BRAM using this function.
-> Signal dom (Maybe (addr, a))
-- ^ (write address @w@, value to write)
-> Signal dom a
-- ^ Value of the @blockRAM@ at address @r@ from the previous clock cycle
blockRamU clk rst0 en rstStrategy n@SNat initF rd0 mw0 =
case rstStrategy of
ClearOnReset ->
-- Use reset infrastructure
blockRamU# clk en n rd1 we1 wa1 w1
NoClearOnReset ->
-- Ignore reset infrastructure, pass values unchanged
blockRamU# clk en n
we0
w0
where
rstBool = register clk rst0 en True (pure False)
rstInv = invertReset rst0
waCounter :: Signal dom (Index n)
waCounter = register clk rstInv en 0 (satSucc SatBound <\$> waCounter)
wa0 = fst . fromJustX <\$> mw0
w0 = snd . fromJustX <\$> mw0
we0 = isJust <\$> mw0
rd1 = mux rstBool 0 (fromEnum <\$> rd0)
we1 = mux rstBool (pure True) we0
wa1 = mux rstBool (fromInteger . toInteger <\$> waCounter) (fromEnum <\$> wa0)
w1 = mux rstBool (initF <\$> waCounter) w0
-- | blockRAM1 primitive
blockRamU#
:: forall n dom a
. ( KnownDomain dom
, HasCallStack
, NFDataX a )
=> Clock dom
-- ^ 'Clock' to synchronize to
-> Enable dom
-- ^ Global Enable
-> SNat n
-- ^ Number of elements in BRAM
-> Signal dom Int
-> Signal dom Bool
-- ^ Write enable
-> Signal dom Int
-> Signal dom a
-- ^ Value to write (at address @w@)
-> Signal dom a
-- ^ Value of the @blockRAM@ at address @r@ from the previous clock cycle
blockRamU# clk en SNat =
-- TODO: Generalize to single BRAM primitive taking an initialization function
blockRam#
clk
en
(CV.map
(\i -> deepErrorX \$ "Initial value at index " ++ show i ++ " undefined.")
(iterateI @n succ (0 :: Int)))
{-# NOINLINE blockRamU# #-}
{-# ANN blockRamU# hasBlackBox #-}
-- | Version of blockram that is initialized with the same value on all
-- memory positions.
blockRam1
:: forall n dom a r addr
. ( KnownDomain dom
, HasCallStack
, NFDataX a
, 1 <= n )
=> Clock dom
-- ^ 'Clock' to synchronize to
-> Reset dom
-- ^ 'Reset' line to listen to. Needs to be held at least /n/ cycles in order
-- for the BRAM to be reset to its initial state.
-> Enable dom
-- ^ Global enable
-> ResetStrategy r
-- ^ Whether to clear BRAM on asserted reset ('ClearOnReset') or
-- not ('NoClearOnReset'). Reset needs to be asserted at least /n/ cycles to
-- clear the BRAM.
-> SNat n
-- ^ Number of elements in BRAM
-> a
-- ^ Initial content of the BRAM (replicated /n/ times)
-> Signal dom (Maybe (addr, a))
-- ^ (write address @w@, value to write)
-> Signal dom a
-- ^ Value of the @blockRAM@ at address @r@ from the previous clock cycle
blockRam1 clk rst0 en rstStrategy n@SNat a rd0 mw0 =
case rstStrategy of
ClearOnReset ->
-- Use reset infrastructure
blockRam1# clk en n a rd1 we1 wa1 w1
NoClearOnReset ->
-- Ignore reset infrastructure, pass values unchanged
blockRam1# clk en n a
we0
w0
where
rstBool = register clk rst0 en True (pure False)
rstInv = invertReset rst0
waCounter :: Signal dom (Index n)
waCounter = register clk rstInv en 0 (satSucc SatBound <\$> waCounter)
wa0 = fst . fromJustX <\$> mw0
w0 = snd . fromJustX <\$> mw0
we0 = isJust <\$> mw0
rd1 = mux rstBool 0 (fromEnum <\$> rd0)
we1 = mux rstBool (pure True) we0
wa1 = mux rstBool (fromInteger . toInteger <\$> waCounter) (fromEnum <\$> wa0)
w1 = mux rstBool (pure a) w0
-- | blockRAM1 primitive
blockRam1#
:: forall n dom a
. ( KnownDomain dom
, HasCallStack
, NFDataX a )
=> Clock dom
-- ^ 'Clock' to synchronize to
-> Enable dom
-- ^ Global Enable
-> SNat n
-- ^ Number of elements in BRAM
-> a
-- ^ Initial content of the BRAM (replicated /n/ times)
-> Signal dom Int
-> Signal dom Bool
-- ^ Write enable
-> Signal dom Int
-> Signal dom a
-- ^ Value to write (at address @w@)
-> Signal dom a
-- ^ Value of the @blockRAM@ at address @r@ from the previous clock cycle
blockRam1# clk en n a =
-- TODO: Generalize to single BRAM primitive taking an initialization function
blockRam# clk en (replicate n a)
{-# NOINLINE blockRam1# #-}
{-# ANN blockRam1# hasBlackBox #-}
-- | blockRAM primitive
blockRam#
:: ( KnownDomain dom
, HasCallStack
, NFDataX a )
=> Clock dom
-- ^ 'Clock' to synchronize to
-> Enable dom
-- ^ Global enable
-> Vec n a
-- ^ Initial content of the BRAM, also
-- determines the size, @n@, of the BRAM.
--
-- __NB__: __MUST__ be a constant.
-> Signal dom Int
-> Signal dom Bool
-- ^ Write enable
-> Signal dom Int
-> Signal dom a
-- ^ Value to write (at address @w@)
-> Signal dom a
-- ^ Value of the @blockRAM@ at address @r@ from the previous clock cycle
blockRam# (Clock _) gen content rd wen =
go
(Seq.fromList (toList content))
(withFrozenCallStack (deepErrorX "blockRam: intial value undefined"))
(fromEnable gen)
rd
(fromEnable gen .&&. wen)
where
go !ram o ret@(~(re :- res)) rt@(~(r :- rs)) et@(~(e :- en)) wt@(~(w :- wr)) dt@(~(d :- din)) =
let ram' = d `defaultSeqX` upd ram e (fromEnum w) d
o' = if re then ram `Seq.index` r else o
in o `seqX` o :- (ret `seq` rt `seq` et `seq` wt `seq` dt `seq` go ram' o' res rs en wr din)
upd ram we waddr d = case maybeIsX we of
Nothing -> case maybeIsX waddr of
Nothing -> fmap (const (seq waddr d)) ram
Just wa -> Seq.update wa d ram
Just True -> case maybeIsX waddr of
Nothing -> fmap (const (seq waddr d)) ram
Just wa -> Seq.update wa d ram
_ -> ram
{-# ANN blockRam# hasBlackBox #-}
{-# NOINLINE blockRam# #-}
:: ( KnownDomain dom
, NFDataX a
=> Clock dom
-> Reset dom
-> Enable dom
-> (Signal dom addr -> Signal dom (Maybe (addr, a)) -> Signal dom a)
-- ^ The @ram@ component | 7,806 | 24,291 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-30 | latest | en | 0.50976 |
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Ra Surface Finish Examples
Surface Finish In Rms
rms surface finish for vertical brush finish manufacturing. rma(ra) stand for root mean square(roughness average), it is a term used to i haven't spotted a good example of a 500 rms, but generally a fairly well done torch cut is
Surface Finish Ra Vs. Rz
surface finish ra and rz. how to convert from one to the other. ra gets rougher but rz stays the same?
Why Is Surface Finish Important In Engineering Applications?
surface finish, or surface roughness, is a measure of the texture of a surface. statistical descriptors like average roughness (ra) and root mean square comparison techniques use samples of surface roughness
Complete Surface Finish Chart, Symbols & Roughness
complete guide to surface finish symbols, roughness charts, ra, rz, aside from dimensions and tolerances, another important callout is surface finish.
Surface Roughness (Finish) Review And Equations
surface roughness average (ra), also know as arithmetric average (aa) is rated as for example, in end mill cutting, the final surface depends on the rotational
Stainless Steel Surface Finishes
surface finish in use today and forms the basis for most polished en 10088-2 special finishes from table 6 of the standard, with a guide to typical ra. symbol.
Simple Ra / Rms Surface Finish Explanation
a callout on a drawing of a minimum surface finish is normally to insure buddy machining it who is under the gun isn't going to finish it with a .020'/rev federate on
Surface Finish Analysis
why measure surface roughness? it's on the print ra 500in. curve removed with c filter residual error. ra 35in. curve properly.
Surface Finish Of Cold Finished Steel
examples showing the surface finish of cold finished steel bars. the examples typical ra of 250 to 1,000 microinch. surface finish of a
Understanding Surface Finish And Surface Texture
how to choose and specify surface finish and texture requirements for for example, the finish, or polish, numbers for sheet materials are: 1, 2d, 2b, bright surface roughness is calculated by two means, the roughness average (ra),
Surface Finish Chart
surface texture equivalents disclaimer: the information on this page has not been rules of thumb for surface finish equivalent of n grade and ra values.
A Different Slice Of Surface Finish
engineers want to quantify the tool mark, or roughness of the surface, to ensure being a statistical parameter, ra is very forgiving towards surface flaws but here are several examples of industries that have benefited from
Surface Finish Metrology Tutorial
c.h.w. giauque, f.e. scire, e.c. teague, j. garbini, j.c. wyant, r.a.. smythe, f. finally we will discuss some examples of how surface finish affects surface
Difference Between Surface Finish (Ra) And Flatness (Gd&T)
these both seem like they would be very similar measures.. can anyone shed some light- preferably with a drawing example on the difference
The Basics Of Surface Finish
the seven possible lay directions are indicated in the table below. figre 7 lay symbols. in the example below, ra is specified to be no greater
Complete Guide To Surface Finish Charts, Ra, Rz
aside from dimensions and tolerances, another important callout is surface finish. surface finish is a measure of the overall texture of a
Complete Surface Finish Chart, Symbols & Roughness
complete guide to surface finish symbols, roughness charts, ra, rz, aside from dimensions and tolerances, another important callout is surface finish.
Understanding Surface Finish
ra micrometer (m), ra microinch (in.) roughness grade number (iso) an example of how the surface finish is listed for each product is shown below.
Quality 101 Surface Analysis Beyond Roughness
indeed, roughness is the most common measure of surface finish, and ra, for example, is simply defined as the arithmetic average of the
Machining Surface Finish Chart Ra Vs Rms
ra and rms are both representations of surface roughness, but each is calculated differently. ra is calculated as the roughness average of a surfaces measured
Surface Roughness Vs Surface Finish
examples of different surface finishes a 1 finish, sometimes called hot rolled, annealed, and pickled (hrap), is plate material as it emerges
Compare Surface Finish Specifications With Our Comparison
our surface comparison chart has been compiled for your convenience from numerous resources, for the purpose of easy conversion between various systems.
Understand The Attributes Of An Edm Surface Finish
bearing surfaces and those used for sealing are examples of where finish is pass results in a finish of around 60 120 rms (about 60-120 micro-inch ra.)
Custom Surface Finish Parameters Can Create More Problems
an example of a 'justified' series of special parameters is the one used to ascertain ra, the most common surface finish parameter across most industries and
I accept the Data Protection Declaration | 1,046 | 4,966 | {"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} | 2.609375 | 3 | CC-MAIN-2022-05 | latest | en | 0.860821 |
http://bobotoh.club/formula-for-difference-in-excel/formula-for-difference-in-excel-how-to-subtract-in-excel-sum-difference-formula-excel/ | 1,550,947,980,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550249530087.75/warc/CC-MAIN-20190223183059-20190223205059-00564.warc.gz | 36,734,742 | 8,845 | # Formula For Difference In Excel How To Subtract In Excel Sum Difference Formula Excel
formula for difference in excel how to subtract in excel sum difference formula excel.
excel formula to compare difference between two numbers 2010 percentage highlight all cells referenced by a essential,excel formula percentage difference between two figures apply if formulas and conditional formatting dates weekdays 2 numbers,time difference formula excel 2010 percentage between two cells to get times,excel percent difference formula percentage 2010 between dates two figures to calculate date in,excel formula to determine difference between two numbers 2010 cells how calculate percent delta in sum of squared differences,formula excel difference between two numbers doing your taxes 5 formulas you must know to show times,excel formula to compare difference between two numbers for date using dates in formulas excluding weekends hours,sum of squared differences formula excel exploring the nuances excels function difference between two dates weekdays to compare numbers,absolute difference formula excel for tutorial how to use and combine formulas in between two dates times weeks,excel difference between two times how to display variance 2010 formula dates show. | 216 | 1,267 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-09 | latest | en | 0.896537 |
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# How to Win Coding Competitions: Secrets of Champions
### Overview
Want to be the programmer hot tech companies are looking for?
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### Syllabus
Week 1: Welcome to competitive programming
Exploring different kinds of programming competitions and benefits of participating, as well as typical rules and challenges. An overview of algorithmic programming competitions. An introduction to community resources and online contests. Week 2: Computational complexity and linear data structures
An overview of computational complexity (Big O notation). Exploring linear data structures (array, list, stack, queue): operations, complexity, implementation and examples. Week 3: Sorting and search algorithms 1
An overview of sorting algorithms: insertion sort, quick sort, merge sort. Week 4: Sorting and search algorithms 2
Theoretical limitations and practical guidelines for sorting. Binary search. Binary heaps and priority queues. Week 5: Graph theory 1
Definition of graphs and examples of graph problems. Various ways of storing graphs in memory. Depth first search and its applications. Dynamic programming. Week 6: Graph theory 2
Breadth first search. Eulerian and Hamiltonian paths and tours. Shortest paths. Week 7: Final Exam
Solving a set of problems in limited time just like in a real programming competition.
### Taught by
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## Reviews
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https://magoosh.zendesk.com/hc/en-us/articles/115000347426-Can-factors-or-multiples-be-negative-GRE-and-GMAT-?mobile_site=false | 1,566,740,994,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330233.1/warc/CC-MAIN-20190825130849-20190825152849-00211.warc.gz | 537,679,987 | 7,535 | Scroll
# Can factors or multiples be negative? (GRE and GMAT)
Technically, in the world of mathematics - the answer to this question is yes. There are negative factors and multiples of numbers. For example, -4 and -3 are factors of 12, -24 is a multiple of 3, and the factors of 4 would be -4, -2, -1, 1, 2, 4.
But what exactly does this mean in relation to the GRE and/or the GMAT?
For the GRE: This concept is mentioned in the ETS GRE Math Conventions PDF. You can find it here. The factors of an integer include both the positive and negative integers. This also applies for multiples.
Therefore, technically, if a GRE question were to ask "How many factors does 6 have?" The answer would be 8: (-6, -3, -2, -1, 1, 2, 3, 6). With that said, questions are very likely to specify "positive factors" or "positive integers" on the GRE. We haven't seen an official question before that required knowing that negative factors/multiples exist - but that doesn't mean those questions won't show up at all so you still want to be diligent about this. :)
For the GMAT: The same rules in the GRE apply to the GMAT in this case. Once again, the factors and multiples of an integer include both the positive and negative integers/multiples.
Once again, we haven't seen an official question that involved negative factors or multiples, but it doesn't mean that they don't exist. However, like the GRE, the questions will also clearly specify "positive factors" or "positive integers". :D | 374 | 1,483 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2019-35 | longest | en | 0.92866 |
http://eclipseclp.org/docs/Dev/current/tutorial/tutorial120.html | 1,508,666,684,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825174.90/warc/CC-MAIN-20171022094207-20171022114207-00433.warc.gz | 106,071,372 | 1,853 | ## 16.5 Exercise
A company produces two types of products T1 and T2, which requires the following resources to produce each unit of the product:
Resource T1 T2 Labour (hours) 9 6 Pumps (units) 1 1 Tubing (m) 12 16
The amount of profit per unit of products are:
T1
£350
T2
£300
They have the following resources available: 1566 hours of labour, 200 pumps, and 2880 metres of tubing.
1. Write a program to maximise the profit for the company, using eplex as a black box solver. Write a predicate that returns the profit and the values for T1 and T2.
2. What program change is required to answer this question: What profit can be achieved if exactly 150 units of T1 are required?
3. What would the profit be if fractional numbers of refrigerators could be produced?
4. Rewrite the program from (1) without optimize/2, using eplex_solver_setup/1, eplex_solve/1, and eplex_var_get/3.
5. In the program from (4), remove the integrality constraints (so that eplex only sees an LP problem). Solve the integer problem by interleaving solving of the LP problem with a rounding heuristic:
• solve the continuous relaxation
• round the solution for T1 to the nearest integer and instantiate it Initially just return the maximum profit value.
• re-solve the new continuous relaxation
• round the solution for T2 to the nearest integer and instantiate it
• re-solve the new continuous relaxation
What is the result in terms of T1, T2 and Profit?
6. Rewrite the program from (5) using eplex_solver_setup/4 and automatic triggering of the solver instead of explicit calls to eplex_solve/1. The solver should be triggered whenever variables get instantiated. | 398 | 1,651 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43 | latest | en | 0.893151 |
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## Re: Specifying RGB triples
From: Andy Buckle Subject: Re: Specifying RGB triples Date: Thu, 21 Apr 2011 12:29:20 +0100
```On Thu, Apr 21, 2011 at 10:40 AM, Sergei Steshenko <address@hidden> wrote:
>
>
> --- On Thu, 4/21/11, Søren Hauberg <address@hidden> wrote:
>
>> From: Søren Hauberg <address@hidden>
>> Subject: Re: Specifying RGB triples
>> To: "Sergei Steshenko" <address@hidden>
>> Cc: address@hidden
>> Date: Thursday, April 21, 2011, 1:58 AM
>> tor, 21 04 2011 kl. 01:14 -0700,
>> skrev Sergei Steshenko:
>> >
>> > --- On Thu, 4/21/11, Søren Hauberg <address@hidden>
>> wrote:
>> >
>> > > From: Søren Hauberg <address@hidden>
>> > > Subject: Re: Specifying RGB triples
>> > > To: "pathematica" <address@hidden>
>> > > Cc: address@hidden
>> > > Date: Thursday, April 21, 2011, 12:06 AM
>> > > ons, 20 04 2011 kl. 23:56 -0700,
>> > > skrev pathematica:
>> > [snip]
>> > > plot (sin (1:100), 'color',
>> [0.9, 0.2, 0.2])
>> > [snip]
>> > >
>> > > Søren
>> > >
>> > >
>> >
>> > Is this described anywhere in Octave documentation ?
>>
>> I don't know.
>>
>> > And how this can be understood from "get(gcf())"
>> output ?
>>
>> "get(gcf())" just gives you a list of figure properties,
>> such as the
>> size of the figure, etc. The above colour-stuff is the
>> property of a
>> specific plot element. You should be able to get related
>> information by
>> doing something like
>>
>> handle = plot (sin (1:100));
>> get (handle)
>>
>> Søren
>>
>>
>
> You example among other things produces:
>
> "
> color =
>
> 0 0 1
> ".
>
> How am I supposed from the above output to know that 'color' should be
> specified as vector and not as, say, string or cell array ?
>
> Thanks,
> Sergei.
The "above output" alone won't tell you. But the following does not
seem too stressful.
>c=get (gca(),'color')
c =
1 1 1
>whos c
Variables in the current scope:
Attr Name Size Bytes Class
==== ==== ==== ===== =====
c 1x3 24 double
Total is 3 elements using 24 bytes
--
/* andy buckle */
```
reply via email to
[Prev in Thread] Current Thread [Next in Thread] | 742 | 2,285 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-05 | latest | en | 0.715954 |
https://www.physicsforums.com/threads/factorization-of-a-matrix-equation.890016/ | 1,508,208,977,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187820556.7/warc/CC-MAIN-20171017013608-20171017033608-00153.warc.gz | 1,247,363,661 | 17,241 | # I Factorization of a matrix equation
1. Oct 20, 2016
### Mr Davis 97
This might be a dumb question, but I am wondering, given the equation $A\vec{x} - 7\vec{x} = \vec{0}$, the factorization $(A - 7I)\vec{x} = \vec{0}$ is correct rather than the factorization $(A - 7)\vec{x} = \vec{0}$. It seems that I can discribute just fine to get the equation we had before using the second $(A - 7)\vec{x} = \vec{0}$, so I'm not sure why I would think to do $(A - 7)\vec{x} = \vec{0}$ rather than $(A - 7I)\vec{x} = \vec{0}$.
2. Oct 20, 2016
### andrewkirk
Yes you are right that that is the strictly correct way to write it. However the slight abuse of notation $(A-7)\vec x$ is sometimes used, because it is shorter to write and it is usually clear what it means. In that case the symbol 7 is interpreted to mean the operator on the vector space $V$ that maps $\vec v$ to $7\vec v$.
3. Oct 21, 2016
### Mr Davis 97
Actually, a better question that I might ask would be that since in $A\vec{x} - 7\vec{x} = \vec{0}$ we have a matrix times and vector and then a scalar times a vector, what allows us to be able to factor out the vector? Wouldn't we get a matrix minus a scalar?
4. Oct 21, 2016
### andrewkirk
No, because there is no rule that allows us to do that factorisation. It can only be factorised if we interpret the 7 as a linear operator, meaning it is $7I$.
5. Oct 21, 2016
### Staff: Mentor
Which is why you need to append I in the factorization.
In the expression $A\vec{x} - 7\vec{x}$ Ax is a vector and 7x is a vector, but if you factor the left side to (A - 7), then you're subtracting a scalar from a matrix. As you note, this doesn't make sense unless we stretch things to interpret 7 in the way that andrewkirk mentions. In this case 7 is really 7I.
6. Oct 21, 2016
### Mr Davis 97
Why are we allowed to interpret 7 as either a scalar 7 or a matrix 7I? It seems somewhat ambiguous
7. Oct 22, 2016
### Staff: Mentor
As andrewkirk said in post #2, this is an abuse of notation, but when it is used, the context usually makes it clear what is intended.
However, every linear algebra book I've seen will write the factorization of Ax - 7x (for example) as (A - 7I)x, to show explicitly that we're not subtracting a scalar from a matrix. | 684 | 2,264 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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-2017-43 | longest | en | 0.920728 |
https://handwiki.org/wiki/Physics:Measurement_in_quantum_mechanics | 1,721,162,884,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514789.44/warc/CC-MAIN-20240716183855-20240716213855-00052.warc.gz | 265,970,550 | 54,839 | # Physics:Measurement in quantum mechanics
Short description: Interaction of a quantum system with a classical observer
In quantum physics, a measurement is the testing or manipulation of a physical system to yield a numerical result. A fundamental feature of quantum theory is that the predictions it makes are probabilistic. The procedure for finding a probability involves combining a quantum state, which mathematically describes a quantum system, with a mathematical representation of the measurement to be performed on that system. The formula for this calculation is known as the Born rule. For example, a quantum particle like an electron can be described by a quantum state that associates to each point in space a complex number called a probability amplitude. Applying the Born rule to these amplitudes gives the probabilities that the electron will be found in one region or another when an experiment is performed to locate it. This is the best the theory can do; it cannot say for certain where the electron will be found. The same quantum state can also be used to make a prediction of how the electron will be moving, if an experiment is performed to measure its momentum instead of its position. The uncertainty principle implies that, whatever the quantum state, the range of predictions for the electron's position and the range of predictions for its momentum cannot both be narrow. Some quantum states imply a near-certain prediction of the result of a position measurement, but the result of a momentum measurement will be highly unpredictable, and vice versa. Furthermore, the fact that nature violates the statistical conditions known as Bell inequalities indicates that the unpredictability of quantum measurement results cannot be explained away as due to ignorance about "local hidden variables" within quantum systems.
Measuring a quantum system generally changes the quantum state that describes that system. This is a central feature of quantum mechanics, one that is both mathematically intricate and conceptually subtle. The mathematical tools for making predictions about what measurement outcomes may occur, and how quantum states can change, were developed during the 20th century and make use of linear algebra and functional analysis. Quantum physics has proven to be an empirical success and to have wide-ranging applicability. However, on a more philosophical level, debates continue about the meaning of the measurement concept.
## Mathematical formalism
Main page: Physics:Canonical quantization
In quantum mechanics, each physical system is associated with a Hilbert space, each element of which represents a possible state of the physical system. The approach codified by John von Neumann represents a measurement upon a physical system by a self-adjoint operator on that Hilbert space termed an "observable".[1]:17 These observables play the role of measurable quantities familiar from classical physics: position, momentum, energy, angular momentum and so on. The dimension of the Hilbert space may be infinite, as it is for the space of square-integrable functions on a line, which is used to define the quantum physics of a continuous degree of freedom. Alternatively, the Hilbert space may be finite-dimensional, as occurs for spin degrees of freedom. Many treatments of the theory focus on the finite-dimensional case, as the mathematics involved is somewhat less demanding. Indeed, introductory physics texts on quantum mechanics often gloss over mathematical technicalities that arise for continuous-valued observables and infinite-dimensional Hilbert spaces, such as the distinction between bounded and unbounded operators; questions of convergence (whether the limit of a sequence of Hilbert-space elements also belongs to the Hilbert space), exotic possibilities for sets of eigenvalues, like Cantor sets; and so forth.[2]:79[3] These issues can be satisfactorily resolved using spectral theory;[2]:101 the present article will avoid them whenever possible.
### Projective measurement
The eigenvectors of a von Neumann observable form an orthonormal basis for the Hilbert space, and each possible outcome of that measurement corresponds to one of the vectors comprising the basis. A density operator is a positive-semidefinite operator on the Hilbert space whose trace is equal to 1.[1][2] For each measurement that can be defined, the probability distribution over the outcomes of that measurement can be computed from the density operator. The procedure for doing so is the Born rule, which states that
$\displaystyle{ P(x_i) = \operatorname{tr}(\Pi_i \rho), }$
where $\displaystyle{ \rho }$ is the density operator, and $\displaystyle{ \Pi_i }$ is the projection operator onto the basis vector corresponding to the measurement outcome $\displaystyle{ x_i }$. The average of the eigenvalues of a von Neumann observable, weighted by the Born rule probabilities, is the expectation value of that observable. For an observable $\displaystyle{ A }$, the expectation value given a quantum state $\displaystyle{ \rho }$ is
$\displaystyle{ \langle A \rangle = \operatorname{tr} (A\rho). }$
A density operator that is a rank-1 projection is known as a pure quantum state, and all quantum states that are not pure are designated mixed. Pure states are also known as wavefunctions. Assigning a pure state to a quantum system implies certainty about the outcome of some measurement on that system (i.e., $\displaystyle{ P(x) = 1 }$ for some outcome $\displaystyle{ x }$). Any mixed state can be written as a convex combination of pure states, though not in a unique way.[4] The state space of a quantum system is the set of all states, pure and mixed, that can be assigned to it.
The Born rule associates a probability with each unit vector in the Hilbert space, in such a way that these probabilities sum to 1 for any set of unit vectors comprising an orthonormal basis. Moreover, the probability associated with a unit vector is a function of the density operator and the unit vector, and not of additional information like a choice of basis for that vector to be embedded in. Gleason's theorem establishes the converse: all assignments of probabilities to unit vectors (or, equivalently, to the operators that project onto them) that satisfy these conditions take the form of applying the Born rule to some density operator.[5][6][7]
### Generalized measurement (POVM)
Main page: POVM
In functional analysis and quantum measurement theory, a positive-operator-valued measure (POVM) is a measure whose values are positive semi-definite operators on a Hilbert space. POVMs are a generalisation of projection-valued measures (PVMs) and, correspondingly, quantum measurements described by POVMs are a generalisation of quantum measurement described by PVMs. In rough analogy, a POVM is to a PVM what a mixed state is to a pure state. Mixed states are needed to specify the state of a subsystem of a larger system (see Schrödinger–HJW theorem); analogously, POVMs are necessary to describe the effect on a subsystem of a projective measurement performed on a larger system. POVMs are the most general kind of measurement in quantum mechanics, and can also be used in quantum field theory.[8] They are extensively used in the field of quantum information.
In the simplest case, of a POVM with a finite number of elements acting on a finite-dimensional Hilbert space, a POVM is a set of positive semi-definite matrices $\displaystyle{ \{F_i\} }$ on a Hilbert space $\displaystyle{ \mathcal{H} }$ that sum to the identity matrix,[9]:90
$\displaystyle{ \sum_{i=1}^n F_i = \operatorname{I}. }$
In quantum mechanics, the POVM element $\displaystyle{ F_i }$ is associated with the measurement outcome $\displaystyle{ i }$, such that the probability of obtaining it when making a measurement on the quantum state $\displaystyle{ \rho }$ is given by
$\displaystyle{ \text{Prob}(i) = \operatorname{tr}(\rho F_i) }$,
where $\displaystyle{ \operatorname{tr} }$ is the trace operator. When the quantum state being measured is a pure state $\displaystyle{ |\psi\rangle }$ this formula reduces to
$\displaystyle{ \text{Prob}(i) = \operatorname{tr}(|\psi\rangle\langle\psi| F_i) = \langle\psi|F_i|\psi\rangle }$.
### State change due to measurement
Main page: Physics:Quantum operation
A measurement upon a quantum system will generally bring about a change of the quantum state of that system. Writing a POVM does not provide the complete information necessary to describe this state-change process.[10]:134 To remedy this, further information is specified by decomposing each POVM element into a product:
$\displaystyle{ E_i = A^\dagger_{i} A_{i}. }$
The Kraus operators $\displaystyle{ A_{i} }$, named for Karl Kraus, provide a specification of the state-change process.[lower-alpha 1] They are not necessarily self-adjoint, but the products $\displaystyle{ A^\dagger_{i} A_{i} }$ are. If upon performing the measurement the outcome $\displaystyle{ E_i }$ is obtained, then the initial state $\displaystyle{ \rho }$ is updated to
$\displaystyle{ \rho \to \rho' = \frac{A_{i} \rho A^\dagger_{i}}{\mathrm{Prob}(i)} = \frac{A_{i} \rho A^\dagger_{i}}{\operatorname{tr} (\rho E_i)}. }$
An important special case is the Lüders rule, named for Gerhart Lüders.[16][17] If the POVM is itself a PVM, then the Kraus operators can be taken to be the projectors onto the eigenspaces of the von Neumann observable:
$\displaystyle{ \rho \to \rho' = \frac{\Pi_i \rho \Pi_i}{\operatorname{tr} (\rho \Pi_i)}. }$
If the initial state $\displaystyle{ \rho }$ is pure, and the projectors $\displaystyle{ \Pi_i }$ have rank 1, they can be written as projectors onto the vectors $\displaystyle{ |\psi\rangle }$ and $\displaystyle{ |i\rangle }$, respectively. The formula simplifies thus to
$\displaystyle{ \rho = |\psi\rangle\langle\psi| \to \rho' = \frac{|i\rangle\langle i | \psi\rangle\langle\psi | i \rangle\langle i|}{|\langle i |\psi \rangle|^2} = |i\rangle\langle i|. }$
This has historically been known as the "reduction of the wave packet" or the "collapse of the wavefunction". The pure state $\displaystyle{ |i\rangle }$ implies a probability-one prediction for any von Neumann observable that has $\displaystyle{ |i\rangle }$ as an eigenvector. Introductory texts on quantum theory often express this by saying that if a quantum measurement is repeated in quick succession, the same outcome will occur both times. This is an oversimplification, since the physical implementation of a quantum measurement may involve a process like the absorption of a photon; after the measurement, the photon does not exist to be measured again.[9]:91
We can define a linear, trace-preserving, completely positive map, by summing over all the possible post-measurement states of a POVM without the normalisation:
$\displaystyle{ \rho \to \sum_i A_i \rho A^\dagger_i. }$
It is an example of a quantum channel,[10]:150 and can be interpreted as expressing how a quantum state changes if a measurement is performed but the result of that measurement is lost.[10]:159
### Examples
Bloch sphere representation of states (in blue) and optimal POVM (in red) for unambiguous quantum state discrimination[18] on the states $\displaystyle{ |\psi\rangle=|0\rangle }$ and $\displaystyle{ |\varphi\rangle=(|0\rangle+|1\rangle)/\sqrt2 }$. Note that on the Bloch sphere orthogonal states are antiparallel.
The prototypical example of a finite-dimensional Hilbert space is a qubit, a quantum system whose Hilbert space is 2-dimensional. A pure state for a qubit can be written as a linear combination of two orthogonal basis states $\displaystyle{ |0 \rangle }$ and $\displaystyle{ |1 \rangle }$ with complex coefficients:
$\displaystyle{ | \psi \rangle = \alpha |0 \rangle + \beta |1 \rangle }$
A measurement in the $\displaystyle{ (|0\rangle, |1\rangle) }$ basis will yield outcome $\displaystyle{ |0 \rangle }$ with probability $\displaystyle{ | \alpha |^2 }$ and outcome $\displaystyle{ |1 \rangle }$ with probability $\displaystyle{ | \beta |^2 }$, so by normalization,
$\displaystyle{ | \alpha |^2 + | \beta |^2 = 1. }$
An arbitrary state for a qubit can be written as a linear combination of the Pauli matrices, which provide a basis for $\displaystyle{ 2 \times 2 }$ self-adjoint matrices:[10]:126
$\displaystyle{ \rho = \tfrac{1}{2}\left(I + r_x \sigma_x + r_y \sigma_y + r_z \sigma_z\right), }$
where the real numbers $\displaystyle{ (r_x, r_y, r_z) }$ are the coordinates of a point within the unit ball and
$\displaystyle{ \sigma_x = \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}, \quad \sigma_y = \begin{pmatrix} 0&-i\\ i&0 \end{pmatrix}, \quad \sigma_z = \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix} . }$
POVM elements can be represented likewise, though the trace of a POVM element is not fixed to equal 1. The Pauli matrices are traceless and orthogonal to one another with respect to the Hilbert–Schmidt inner product, and so the coordinates $\displaystyle{ (r_x, r_y, r_z) }$ of the state $\displaystyle{ \rho }$ are the expectation values of the three von Neumann measurements defined by the Pauli matrices.[10]:126 If such a measurement is applied to a qubit, then by the Lüders rule, the state will update to the eigenvector of that Pauli matrix corresponding to the measurement outcome. The eigenvectors of $\displaystyle{ \sigma_z }$ are the basis states $\displaystyle{ |0\rangle }$ and $\displaystyle{ |1\rangle }$, and a measurement of $\displaystyle{ \sigma_z }$ is often called a measurement in the "computational basis."[10]:76 After a measurement in the computational basis, the outcome of a $\displaystyle{ \sigma_x }$ or $\displaystyle{ \sigma_y }$ measurement is maximally uncertain.
A pair of qubits together form a system whose Hilbert space is 4-dimensional. One significant von Neumann measurement on this system is that defined by the Bell basis,[19]:36 a set of four maximally entangled states:
\displaystyle{ \begin{align} |\Phi^+\rangle &= \frac{1}{\sqrt{2}} (|0\rangle_A \otimes |0\rangle_B + |1\rangle_A \otimes |1\rangle_B) \\ |\Phi^-\rangle &= \frac{1}{\sqrt{2}} (|0\rangle_A \otimes |0\rangle_B - |1\rangle_A \otimes |1\rangle_B) \\ |\Psi^+\rangle &= \frac{1}{\sqrt{2}} (|0\rangle_A \otimes |1\rangle_B + |1\rangle_A \otimes |0\rangle_B) \\ |\Psi^-\rangle &= \frac{1}{\sqrt{2}} (|0\rangle_A \otimes |1\rangle_B - |1\rangle_A \otimes |0\rangle_B) \end{align} }
Probability density $\displaystyle{ P_n(x) }$ for the outcome of a position measurement given the energy eigenstate $\displaystyle{ |n\rangle }$ of a 1D harmonic oscillator.
A common and useful example of quantum mechanics applied to a continuous degree of freedom is the quantum harmonic oscillator.[20]:24 This system is defined by the Hamiltonian
$\displaystyle{ {H} = \frac{{p}^2}{2m} + \frac{1}{2}m\omega^2 {x}^2, }$
where $\displaystyle{ {H} }$, the momentum operator $\displaystyle{ {p} }$ and the position operator $\displaystyle{ {x} }$ are self-adjoint operators on the Hilbert space of square-integrable functions on the real line. The energy eigenstates solve the time-independent Schrödinger equation:
$\displaystyle{ {H} |n\rangle = E_n |n\rangle. }$
These eigenvalues can be shown to be given by
$\displaystyle{ E_n = \hbar\omega\left(n + \tfrac{1}{2}\right), }$
and these values give the possible numerical outcomes of an energy measurement upon the oscillator. The set of possible outcomes of a position measurement on a harmonic oscillator is continuous, and so predictions are stated in terms of a probability density function $\displaystyle{ P(x) }$ that gives the probability of the measurement outcome lying in the infinitesimal interval from $\displaystyle{ x }$ to $\displaystyle{ x + dx }$.
## History of the measurement concept
### The "old quantum theory"
Main page: Physics:Old quantum theory
The old quantum theory is a collection of results from the years 1900–1925[21] which predate modern quantum mechanics. The theory was never complete or self-consistent, but was rather a set of heuristic corrections to classical mechanics.[22] The theory is now understood as a semi-classical approximation[23] to modern quantum mechanics.[24] Notable results from this period include Planck's calculation of the blackbody radiation spectrum, Einstein's explanation of the photoelectric effect, Einstein and Debye's work on the specific heat of solids, Bohr and van Leeuwen's proof that classical physics cannot account for diamagnetism, Bohr's model of the hydrogen atom and Arnold Sommerfeld's extension of the Bohr model to include relativistic effects.
Stern–Gerlach experiment: Silver atoms travelling through an inhomogeneous magnetic field, and being deflected up or down depending on their spin; (1) furnace, (2) beam of silver atoms, (3) inhomogeneous magnetic field, (4) classically expected result, (5) observed result
The Stern–Gerlach experiment, proposed in 1921 and implemented in 1922,[25][26][27] became a prototypical example of a quantum measurement having a discrete set of possible outcomes. In the original experiment, silver atoms were sent through a spatially varying magnetic field, which deflected them before they struck a detector screen, such as a glass slide. Particles with non-zero magnetic moment are deflected, due to the magnetic field gradient, from a straight path. The screen reveals discrete points of accumulation, rather than a continuous distribution, owing to the particles' quantized spin.[28]Cite error: Closing </ref> missing for <ref> tag Heisenberg sought to develop a theory of atomic phenomena that relied only on "observable" quantities. At the time, and in contrast with the later standard presentation of quantum mechanics, Heisenberg did not regard the position of an electron bound within an atom as "observable". Instead, his principal quantities of interest were the frequencies of light emitted or absorbed by atoms.[29]
The uncertainty principle dates to this period. It is frequently attributed to Heisenberg, who introduced the concept in analyzing a thought experiment where one attempts to measure an electron's position and momentum simultaneously. However, Heisenberg did not give precise mathematical definitions of what the "uncertainty" in these measurements meant. The precise mathematical statement of the position-momentum uncertainty principle is due to Kennard, Pauli, and Weyl, and its generalization to arbitrary pairs of noncommuting observables is due to Robertson and Schrödinger.[30][31]
Writing $\displaystyle{ {x} }$ and $\displaystyle{ {p} }$ for the self-adjoint operators representing position and momentum respectively, a standard deviation of position can be defined as
$\displaystyle{ \sigma_x=\sqrt{\langle {x}^2 \rangle-\langle {x}\rangle^2}, }$
and likewise for the momentum:
$\displaystyle{ \sigma_p=\sqrt{\langle {p}^2 \rangle-\langle {p}\rangle^2}. }$
The Kennard–Pauli–Weyl uncertainty relation is
$\displaystyle{ \sigma_x \sigma_p \geq \frac{\hbar}{2}. }$
This inequality means that no preparation of a quantum particle can imply simultaneously precise predictions for a measurement of position and for a measurement of momentum.[32] The Robertson inequality generalizes this to the case of an arbitrary pair of self-adjoint operators $\displaystyle{ A }$ and $\displaystyle{ B }$. The commutator of these two operators is
$\displaystyle{ [A,B]=AB-BA, }$
and this provides the lower bound on the product of standard deviations:
$\displaystyle{ \sigma_A \sigma_B \geq \left| \frac{1}{2i}\langle[A,B]\rangle \right| = \frac{1}{2}\left|\langle[A,B]\rangle \right|. }$
Substituting in the canonical commutation relation $\displaystyle{ [{x},{p}] = i\hbar }$, an expression first postulated by Max Born in 1925,[33] recovers the Kennard–Pauli–Weyl statement of the uncertainty principle.
### From uncertainty to no-hidden-variables
Main pages: Physics:EPR paradox, Bell's theorem, and Physics:Bell test
The existence of the uncertainty principle naturally raises the question of whether quantum mechanics can be understood as an approximation to a more exact theory. Do there exist "hidden variables", more fundamental than the quantities addressed in quantum theory itself, knowledge of which would allow more exact predictions than quantum theory can provide? A collection of results, most significantly Bell's theorem, have demonstrated that broad classes of such hidden-variable theories are in fact incompatible with quantum physics.
Bell published the theorem now known by his name in 1964, investigating more deeply a thought experiment originally proposed in 1935 by Einstein, Podolsky and Rosen.[34][35] According to Bell's theorem, if nature actually operates in accord with any theory of local hidden variables, then the results of a Bell test will be constrained in a particular, quantifiable way. If a Bell test is performed in a laboratory and the results are not thus constrained, then they are inconsistent with the hypothesis that local hidden variables exist. Such results would support the position that there is no way to explain the phenomena of quantum mechanics in terms of a more fundamental description of nature that is more in line with the rules of classical physics. Many types of Bell test have been performed in physics laboratories, often with the goal of ameliorating problems of experimental design or set-up that could in principle affect the validity of the findings of earlier Bell tests. This is known as "closing loopholes in Bell tests". To date, Bell tests have found that the hypothesis of local hidden variables is inconsistent with the way that physical systems behave.[36][37]
### Quantum systems as measuring devices
The Robertson–Schrödinger uncertainty principle establishes that when two observables do not commute, there is a tradeoff in predictability between them. The Wigner–Araki–Yanase theorem demonstrates another consequence of non-commutativity: the presence of a conservation law limits the accuracy with which observables that fail to commute with the conserved quantity can be measured.[38] Further investigation in this line led to the formulation of the Wigner–Yanase skew information.[39]
Historically, experiments in quantum physics have often been described in semiclassical terms. For example, the spin of an atom in a Stern–Gerlach experiment might be treated as a quantum degree of freedom, while the atom is regarded as moving through a magnetic field described by the classical theory of Maxwell's equations.[2]:24 But the devices used to build the experimental apparatus are themselves physical systems, and so quantum mechanics should be applicable to them as well. Beginning in the 1950s, Rosenfeld, von Weizsäcker and others tried to develop consistency conditions that expressed when a quantum-mechanical system could be treated as a measuring apparatus.[40] One proposal for a criterion regarding when a system used as part of a measuring device can be modeled semiclassically relies on the Wigner function, a quasiprobability distribution that can be treated as a probability distribution on phase space in those cases where it is everywhere non-negative.[2]:375
### Decoherence
Main page: Physics:Quantum decoherence
A quantum state for an imperfectly isolated system will generally evolve to be entangled with the quantum state for the environment. Consequently, even if the system's initial state is pure, the state at a later time, found by taking the partial trace of the joint system-environment state, will be mixed. This phenomenon of entanglement produced by system-environment interactions tends to obscure the more exotic features of quantum mechanics that the system could in principle manifest. Quantum decoherence, as this effect is known, was first studied in detail during the 1970s.[41] (Earlier investigations into how classical physics might be obtained as a limit of quantum mechanics had explored the subject of imperfectly isolated systems, but the role of entanglement was not fully appreciated.[40]) A significant portion of the effort involved in quantum computing is to avoid the deleterious effects of decoherence.[42][19]:239
To illustrate, let $\displaystyle{ \rho_S }$ denote the initial state of the system, $\displaystyle{ \rho_E }$ the initial state of the environment and $\displaystyle{ H }$ the Hamiltonian specifying the system-environment interaction. The density operator $\displaystyle{ \rho_E }$ can be diagonalized and written as a linear combination of the projectors onto its eigenvectors:
$\displaystyle{ \rho_E = \sum_i p_i |\psi_i\rangle\langle \psi_i|. }$
Expressing time evolution for a duration $\displaystyle{ t }$ by the unitary operator $\displaystyle{ U = e^{-iHt/\hbar} }$, the state for the system after this evolution is
$\displaystyle{ \rho_S' = {\rm tr}_E U \left[\rho_S \otimes \left(\sum_i p_i |\psi_i\rangle\langle \psi_i|\right)\right] U^\dagger, }$
which evaluates to
$\displaystyle{ \rho_S' = \sum_{ij} \sqrt{p_i} \langle \psi_j | U | \psi_i \rangle \rho_S \sqrt{p_i}\langle \psi_i | U^\dagger | \psi_j \rangle. }$
The quantities surrounding $\displaystyle{ \rho_S }$ can be identified as Kraus operators, and so this defines a quantum channel.[41]
Specifying a form of interaction between system and environment can establish a set of "pointer states," states for the system that are (approximately) stable, apart from overall phase factors, with respect to environmental fluctuations. A set of pointer states defines a preferred orthonormal basis for the system's Hilbert space.[2]:423
## Quantum information and computation
Quantum information science studies how information science and its application as technology depend on quantum-mechanical phenomena. Understanding measurement in quantum physics is important for this field in many ways, some of which are briefly surveyed here.
### Measurement, entropy, and distinguishability
The von Neumann entropy is a measure of the statistical uncertainty represented by a quantum state. For a density matrix $\displaystyle{ \rho }$, the von Neumann entropy is
$\displaystyle{ S(\rho) = -{\rm tr}(\rho \log \rho); }$
writing $\displaystyle{ \rho }$ in terms of its basis of eigenvectors,
$\displaystyle{ \rho = \sum_i \lambda_i |i\rangle\langle i|, }$
the von Neumann entropy is
$\displaystyle{ S(\rho) = -\sum_i \lambda_i \log \lambda_i. }$
This is the Shannon entropy of the set of eigenvalues interpreted as a probability distribution, and so the von Neumann entropy is the Shannon entropy of the random variable defined by measuring in the eigenbasis of $\displaystyle{ \rho }$. Consequently, the von Neumann entropy vanishes when $\displaystyle{ \rho }$ is pure.[10]:320 The von Neumann entropy of $\displaystyle{ \rho }$ can equivalently be characterized as the minimum Shannon entropy for a measurement given the quantum state $\displaystyle{ \rho }$, with the minimization over all POVMs with rank-1 elements.[10]:323
Many other quantities used in quantum information theory also find motivation and justification in terms of measurements. For example, the trace distance between quantum states is equal to the largest difference in probability that those two quantum states can imply for a measurement outcome:[10]:254
$\displaystyle{ \frac{1}{2}||\rho-\sigma|| = \max_{0\leq E \leq I} [{\rm tr}(E \rho) - {\rm tr}(E \sigma)]. }$
Similarly, the fidelity of two quantum states, defined by
$\displaystyle{ F(\rho, \sigma) = \left(\operatorname{Tr} \sqrt{\sqrt{\rho} \sigma \sqrt{\rho}}\right)^2, }$
expresses the probability that one state will pass a test for identifying a successful preparation of the other. The trace distance provides bounds on the fidelity via the Fuchs–van de Graaf inequalities:[10]:274
$\displaystyle{ 1 - \sqrt{F(\rho,\sigma)} \leq \frac{1}{2}||\rho-\sigma|| \leq \sqrt{1 - F(\rho,\sigma)}. }$
### Quantum circuits
Circuit representation of measurement. The single line on the left-hand side stands for a qubit, while the two lines on the right-hand side represent a classical bit.
Main page: Quantum circuit
Quantum circuits are a model for quantum computation in which a computation is a sequence of quantum gates followed by measurements.[19]:93 The gates are reversible transformations on a quantum mechanical analog of an n-bit register. This analogous structure is referred to as an n-qubit register. Measurements, drawn on a circuit diagram as stylized pointer dials, indicate where and how a result is obtained from the quantum computer after the steps of the computation are executed. Without loss of generality, one can work with the standard circuit model, in which the set of gates are single-qubit unitary transformations and controlled NOT gates on pairs of qubits, and all measurements are in the computational basis.[19]:93[43]
### Measurement-based quantum computation
Main page: One-way quantum computer
Measurement-based quantum computation (MBQC) is a model of quantum computing in which the answer to a question is, informally speaking, created in the act of measuring the physical system that serves as the computer.[19]:317[44][45]
### Quantum tomography
Main page: Physics:Quantum tomography
Quantum state tomography is a process by which, given a set of data representing the results of quantum measurements, a quantum state consistent with those measurement results is computed.[46] It is named by analogy with tomography, the reconstruction of three-dimensional images from slices taken through them, as in a CT scan. Tomography of quantum states can be extended to tomography of quantum channels[46] and even of measurements.[47]
### Quantum metrology
Main page: Quantum metrology
Quantum metrology is the use of quantum physics to aid the measurement of quantities that, generally, had meaning in classical physics, such as exploiting quantum effects to increase the precision with which a length can be measured.[48] A celebrated example is the introduction of squeezed light into the LIGO experiment, which increased its sensitivity to gravitational waves.[49][50]
## Laboratory implementations
The range of physical procedures to which the mathematics of quantum measurement can be applied is very broad.[51] In the early years of the subject, laboratory procedures involved the recording of spectral lines, the darkening of photographic film, the observation of scintillations, finding tracks in cloud chambers, and hearing clicks from Geiger counters.[lower-alpha 2] Language from this era persists, such as the description of measurement outcomes in the abstract as "detector clicks".[54]
The double-slit experiment is a prototypical illustration of quantum interference, typically described using electrons or photons. The first interference experiment to be carried out in a regime where both wave-like and particle-like aspects of photon behavior are significant was G. I. Taylor's test in 1909. Taylor used screens of smoked glass to attenuate the light passing through his apparatus, to the extent that, in modern language, only one photon would be illuminating the interferometer slits at a time. He recorded the interference patterns on photographic plates; for the dimmest light, the exposure time required was roughly three months.[55][56] In 1974, the Italian physicists Pier Giorgio Merli, Gian Franco Missiroli, and Giulio Pozzi implemented the double-slit experiment using single electrons and a television tube.[57] A quarter-century later, a team at the University of Vienna performed an interference experiment with buckyballs, in which the buckyballs that passed through the interferometer were ionized by a laser, and the ions then induced the emission of electrons, emissions which were in turn amplified and detected by an electron multiplier.[58]
Modern quantum optics experiments can employ single-photon detectors. For example, in the "BIG Bell test" of 2018, several of the laboratory setups used single-photon avalanche diodes. Another laboratory setup used superconducting qubits.[36] The standard method for performing measurements upon superconducting qubits is to couple a qubit with a resonator in such a way that the characteristic frequency of the resonator shifts according to the state for the qubit, and detecting this shift by observing how the resonator reacts to a probe signal.[59]
## Interpretations of quantum mechanics
Main page: Interpretations of quantum mechanics
Niels Bohr and Albert Einstein, pictured here at Paul Ehrenfest's home in Leiden (December 1925), had a long-running collegial dispute about what quantum mechanics implied for the nature of reality.
Despite the consensus among scientists that quantum physics is in practice a successful theory, disagreements persist on a more philosophical level. Many debates in the area known as quantum foundations concern the role of measurement in quantum mechanics. Recurring questions include which interpretation of probability theory is best suited for the probabilities calculated from the Born rule; and whether the apparent randomness of quantum measurement outcomes is fundamental, or a consequence of a deeper deterministic process.[60][61][62] Worldviews that present answers to questions like these are known as "interpretations" of quantum mechanics; as the physicist N. David Mermin once quipped, "New interpretations appear every year. None ever disappear."[63]
A central concern within quantum foundations is the "quantum measurement problem," though how this problem is delimited, and whether it should be counted as one question or multiple separate issues, are contested topics.[53][64] Of primary interest is the seeming disparity between apparently distinct types of time evolution. Von Neumann declared that quantum mechanics contains "two fundamentally different types" of quantum-state change.[65]:§V.1 First, there are those changes involving a measurement process, and second, there is unitary time evolution in the absence of measurement. The former is stochastic and discontinuous, writes von Neumann, and the latter deterministic and continuous. This dichotomy has set the tone for much later debate.[66][67] Some interpretations of quantum mechanics find the reliance upon two different types of time evolution distasteful and regard the ambiguity of when to invoke one or the other as a deficiency of the way quantum theory was historically presented.[68] To bolster these interpretations, their proponents have worked to derive ways of regarding "measurement" as a secondary concept and deducing the seemingly stochastic effect of measurement processes as approximations to more fundamental deterministic dynamics. However, consensus has not been achieved among proponents of the correct way to implement this program, and in particular how to justify the use of the Born rule to calculate probabilities.[69][70] Other interpretations regard quantum states as statistical information about quantum systems, thus asserting that abrupt and discontinuous changes of quantum states are not problematic, simply reflecting updates of the available information.[51][71] Of this line of thought, Bell asked, "Whose information? Information about what?"[68] Answers to these questions vary among proponents of the informationally-oriented interpretations.[61][71]
## Notes
1. Hellwig and Kraus[11][12] originally introduced operators with two indices, $\displaystyle{ A_{ij} }$, such that $\displaystyle{ \textstyle \sum_j A_{ij} A^\dagger_{ij} = E_i }$. The extra index does not affect the computation of the measurement outcome probability, but it does play a role in the state-update rule, with the post-measurement state being now proportional to $\displaystyle{ \textstyle \sum_j A^\dagger_{ij} \rho A_{ij} }$. This can be regarded as representing $\displaystyle{ \textstyle E_i }$ as a coarse-graining together of multiple outcomes of a more fine-grained POVM.[13][14][15] Kraus operators with two indices also occur in generalized models of system-environment interaction.[9]:364
2. The glass plates used in the Stern–Gerlach experiment did not darken properly until Stern breathed on them, accidentally exposing them to sulfur from his cheap cigars.[52][53]
## References
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fr:Problème de la mesure quantique | 12,952 | 50,163 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-30 | latest | en | 0.917442 |
https://www.physicsforums.com/threads/calculate-the-residue-of-1-cosh-pi-z-at-z-i-2-complex-analysis.1052623/#post-6899201 | 1,685,683,974,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648322.84/warc/CC-MAIN-20230602040003-20230602070003-00146.warc.gz | 1,026,371,491 | 19,397 | # Calculate the residue of 1/Cosh(pi.z) at z = i/2 (complex analysis)
• I
• Old Person
#### Old Person
TL;DR Summary
Calculate the residue of 1/Cosh(pi.z) at z = i/2.
I'm not sure if this should be in the calculus section or the anlaysis section. It's complex analysis related to integration around a contour.
Can someone suggest a method to determine the residue of f(z) = ## \frac{1}{Cosh ( \pi z) } ## at the singular point z = i/2.
Background:
This was part of an exam question and the model answers are already available.
In the model answers it is just assumed we have a simple pole (pole of order 1) at z=i/2 and the usual formula then jumps in:
Res (f(z) ; i/2) = ## \lim\limits_{z \rightarrow i/2} \, [ (z-i/2) . f(z) ]##
BUT no proof is given that we did only have a simple pole there. If that's easily shown, great, otherwise would it be easier to just directly obtain the Laurent series or obtain the Residue by some other method?
FactChecker
I used ##Res_z\left(\dfrac{1}{f}\right)=\dfrac{1}{f'(z)}## from
https://www.physicsforums.com/insights/an-overview-of-complex-differentiation-and-integration/
and got
\begin{align*}
Res_z\left(\dfrac{1}{f}\right)&=\dfrac{1}{f'(z)}=\dfrac{1}{(\cosh \pi z)'}\\&=\dfrac{1}{\pi \sinh \pi z}=\dfrac{1}{\pi \sinh(i \pi/2)}=\dfrac{1}{\pi i}=-\dfrac{i}{\pi}
\end{align*}
if I made no mistakes by looking up the complex arguments of ##\sinh## on Wikipedia.
https://www.wolframalpha.com/input?i=residue+(1/cosh(pi+z);+i+/2)=
Last edited:
Old Person
Hi and thank you ever so much for your time and speed of reply @fresh_42
The answer is OK in that it does agree with the model answer. However, I'm not sure it was a valid method.
I've just been looking over the insight article you suggested. It states, just above the result : Say Z_m is a zero of order m and.... .... then gives the formula you've used for z_1 only:
## Res_{z_1}\left(\dfrac{1}{f}\right) =\dfrac{1}{f'(z_1)} ##
So it looks like you must already know that the point of interest, z, was a pole of order 1 to use that result. I'm not sure, I didn't write that article - but that's what it looks like. Just to be clear then, it may be right, it's just that we still seem to need to establish that there was a simple pole (of order 1) there.
As for using Wolfram Aplha approach - as it happens I already tried it. Sadly it spits out answers but no explanation or method. Did it also just assume a pole of order 1 was there?
Yes, the order of the pole is needed beforehand, and I guess WA determined it, too, as part of its calculations.
Old Person
##\cosh (i \pi/2 )=\dfrac{1}{2}\left(e^{i \pi/2}+e^{-i \pi/2}\right)=\dfrac{1}{2}\left(i - i\right)=0.## Therefore ##\cosh(z)=(z-i \pi/2)\cdot g(z).##
Assume that ##z=i\pi/2## is of higher order than ##1,## i.e. ##\cosh(z)=(z-i \pi/2)^2\cdot h(z).## Then
\begin{align*}
\dfrac{d}{dz}\cosh(z)&=\sinh(z)=2(z-i \pi/2)\cdot h(z)+ (z-i \pi/2)^2\cdot h'(z)
\end{align*}
The RHS is zero at ##z=i \pi/2## but ##\sinh(i \pi/2)=\dfrac{1}{2}\left(e^{i \pi/2}-e^{-i \pi/2}\right)=\dfrac{i-(-i)}{2}=i\neq 0.##
This proves that the zero ## i \pi/2## of ##\cosh(z)## is of order one.
hutchphd
You cna get the Laurent series for $\operatorname{sech}(\pi z)$ about $i/2$ by setting $t = z - \frac i2$ and then $$\begin{split} \frac{1}{\cosh(i\pi/2 + \pi t)} &= \frac{1}{\cosh(i\pi/2)\cosh \pi t + \sinh(i\pi/2) \sinh \pi t} \\ &= \frac{1}{i\sinh \pi t} \\ &= \frac{1}{i\pi t} \left( 1 - \frac{\pi^2t^2}{3!} + \frac{\pi^4t^4}{5!} + \dots \right)^{-1} \\ &= \frac{1}{i\pi t}\left( 1 - \left(- \frac{\pi^2t^2}{3!} + \frac{\pi^4t^4}{5!} + \dots\right) + \left(- \frac{\pi^2t^2}{3!} + \frac{\pi^4t^4}{5!} + \dots\right)^2 + \dots \right) \\ &= \frac{-i}{\pi(z - \frac i2)} + \sum_{n=0}^\infty a_n(z - \tfrac i2)^n. \end{split}$$ But it is enough to note that $$\frac{1}{i\sinh \pi t} = \frac{1}{i \pi t} \left( \frac{\pi t}{\sinh \pi t} \right)$$ and the second factor is analytic at $t = 0$.
Last edited:
Old Person and SammyS
Hi and thank you for even more replies.
@fresh_42 ---> That's a good answer. I think there is one minor issue but it is minor. It won't be a problem unless 1/Cosh z is extremely unusual.
Assume that z=iπ/2 is of higher order than 1, i.e. cosh(z)=(z−iπ/2)2⋅h(z). Then....
You have assumed that 1/Cosh z must have a pole of some (finite) order at the point. There are some functions that have isolated singular points but just cannot be described as a pole (of any order).
Example ##e^{1/z} ## = Exp (1/z) = ## \sum_{n=0}^{\infty} \frac{1}{n!} \frac{1}{z^n} ## = a Laurent series about z=0. There is no finite n where the thing can be written as F(z) / zn (with F(z) analytic in a small neighbourhood of z=0).
If the singular point cannot be described as a pole then you can't apply the differentiation trick because you won't have the ability to write ## \cosh(z)=(z-i \pi/2)^N\cdot g(z) ## (with g(z) analytic), for any finite N.
- - - - - - - - - -
@pasmith , that may actually work. In some lines of the work on those series, you needed the series to be absolutely convergent to be sure you could re-arrange and re-shuffle the terms. However, I think you've got that. I wouldn't have had time to check all these details in the exam - but I do now and they they do seem to be ok.
Where you write: ..it is enough to note.... ## \frac{1}{i\sinh \pi t} = \frac{1}{i \pi t} \left( \frac{\pi t}{\sinh \pi t} \right) ##
I think I can see what you are saying. This bit ## \left( \frac{\pi t}{\sinh \pi t} \right) ## = ## \left( 1 - \frac{\pi^2t^2}{3!} + \frac{\pi^4t^4}{5!} + \dots \right)^{-1} ## from earlier in your post and the RHS is seen to be analytic at t=0. So that is enough to establish that you have a simple pole.
- - - - - - - - - - -
Overall, thank you to everyone. I'm happy with the answers. It's been great to see that it could be done (given time) but also just to have a couple of people acknowledge that there was a need to establish the nature of the singular point (or directly look at the Laurent series). As I mentioned before - in the model answers, there are no marks for proving that or even indicating that you should care. It's clear the person who set the exam just expected you to assume you had a simple pole there.
You have assumed that 1/Cosh z must have a pole of some (finite) order at the point. There are some functions that have isolated singular points but just cannot be described as a pole (of any order).
I actually haven't said anything about poles. I only have analyzed the point ##z=\mathrm{i} \pi /2## of the analytical function ##z\longmapsto \cosh(z).##
I showed that it is a zero of order one. That makes its inverse ##z \longmapsto \dfrac{1}{\cosh(z)}## having a pole at ##z=\mathrm{i} \pi/2## of order one. It's not isolated since the zero wasn't isolated. | 2,222 | 6,796 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2023-23 | latest | en | 0.884905 |
http://slideplayer.com/slide/4341345/ | 1,527,241,789,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867055.20/warc/CC-MAIN-20180525082822-20180525102822-00048.warc.gz | 279,107,242 | 23,324 | # Welcome to Chem 1A with Terri Bentzinger Chapter 1 & 2 Website:
## Presentation on theme: "Welcome to Chem 1A with Terri Bentzinger Chapter 1 & 2 Website:"— Presentation transcript:
Welcome to Chem 1A with Terri Bentzinger Chapter 1 & 2 E-mail: benzene4president@gmail Website: http://clas.sa.ucsb.edu/staff/terri/
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Atoms – Molecules - Ions – Ch. 1 1. For each of the following state if it’s a scientific law, theory or neither: a. As a gas expands under constant pressure the gas will cool. b. The forces between the gas particles are negligible c. On average heavier gas particles will move slower than lighter gas particles d. Gases have a greater average kinetic energy at higher temperatures because the average velocity increases with temperature
Atoms – Molecules – Ions – Ch. 1 2. The following data was collected. If it is known that experiment 1 has the formula XY what is the formula of experiment 2.
Atoms – Molecules – Ions – Ch. 1 3. Two elements R and Q, combine to form two binary compounds. In the first compound, 14.0 g of R combines with 3.00 g of Q. In the second compound, 7.00 g of R combines with 4.5 g of Q. Show that these data are in accord with the law of multiple proportions. If the formula of the second compound is RQ, what is the formula of the first?
Atoms – Molecules - Ions – Ch. 2 4. What does the following tell you about an atom: a. Atomic number b. Mass number c. Charge
Atoms – Molecules - Ions – Ch. 2 10 Ne 20.180 Atomic Number (Z) ⇒ # of protons Element Symbol Average atomic mass in amu or g/mol Mass Number (A) ⇒ sum of protons and neutrons **Note that mass number is NOT on the periodic table**
Atoms – Molecules - Ions – Ch. 2 5. Which of the following represents a pair of isotopes? a. 32 S and 32 S 2- b. O 2 and O 3 c. 15 7 N and 15 8 O d. 12 6 C and 13 6 C e. 18 8 O and 19 9 F
Atoms – Molecules - Ions – Ch. 2 Mass Number A Atomic Number Z
Atoms – Molecules - Ions – Ch. 2 6. Fill in the following table:
Atoms – Molecules - Ions – Ch. 2 7. What are ions? How are ions generated?
Atoms – Molecules - Ions – Ch. 2 8. Metals tend to gain or lose electrons? What charge do group 1 metals form? What charge do group 2 metals form? What charge do group 3 metals form?
Atoms – Molecules - Ions – Ch. 2 Non-metals Metalloids Metals
Atoms – Molecules - Ions – Ch. 2 9. Non-metals tend to gain or lose electrons? What charge do group 8 non-metals form? What charge do group 7 non-metals form? What charge do group 6 non-metals form? What charge do group 5 non-metals form?
Atoms – Molecules – Ions – Ch. 2 10. How could you distinguish between ionic compounds, covalent compounds and acids?
Atoms – Molecules – Ions – Ch. 2 11. Name the following compounds: a. LiHCO 3 g. HClO b. Na 2 SO 3 h. HNO 3 c. (NH 4 ) 3 PO 4 i. SF 6 d. Fe(OH) 3 j. CO e. SnO 2 k. P 2 O 5 f. HF
Atoms – Molecules – Ions – Ch. 2 Naming Ionic Compounds First Name (cation) Second Name (anion) Naming Ionic Compounds First Name (cation) Second Name (anion) 1. metals with fixed charges (Grps 1,2,3,Ag,Zn and Cd) ⇒ use the elements name as is 2. metals with varying charges (all other metals) ⇒ use the elements name and a Roman numeral 3. NH 4 + ⇒ ammonium 1. metals with fixed charges (Grps 1,2,3,Ag,Zn and Cd) ⇒ use the elements name as is 2. metals with varying charges (all other metals) ⇒ use the elements name and a Roman numeral 3. NH 4 + ⇒ ammonium 1. monoatomic ⇒ elements name with the suffix –ide 2. polyatomic ⇒ memorize names (next slide) 1. monoatomic ⇒ elements name with the suffix –ide 2. polyatomic ⇒ memorize names (next slide)
Atoms – Molecules – Ions – Ch. 2 Common Polyatomic Ions
Atoms – Molecules – Ions – Ch. 2 Naming Acids Acids without oxygen Acids with oxygen fox Naming Acids Acids without oxygen Acids with oxygen fox 1. Add prefix hydro to the anion’s name 2. Change suffix to ic acid ex: HCN ⇒ hydrocyanic acid or HF ⇒ hydrofluoric acid 1. Add prefix hydro to the anion’s name 2. Change suffix to ic acid ex: HCN ⇒ hydrocyanic acid or HF ⇒ hydrofluoric acid Change suffix of anion in the acid ate ⇒ ic acid ite ⇒ ous acid ex: HNO 2 ⇒ nitrous acid vs. HNO 3 ⇒ nitric acid Change suffix of anion in the acid ate ⇒ ic acid ite ⇒ ous acid ex: HNO 2 ⇒ nitrous acid vs. HNO 3 ⇒ nitric acid
Atoms – Molecules – Ions – Ch. 2 Naming Covalent/Molecular Compounds 1. Add a Greek prefix to the first element’s name if there’s 2 or more 2. Always add a Greek prefix to the 2 nd element and change the suffix to –ide ex: NF 3 ⇒ nitrogen trifluoride Naming Covalent/Molecular Compounds 1. Add a Greek prefix to the first element’s name if there’s 2 or more 2. Always add a Greek prefix to the 2 nd element and change the suffix to –ide ex: NF 3 ⇒ nitrogen trifluoride
Atoms – Molecules – Ions – Ch. 2 12. Write the chemical formulas for the following compounds: a. calcium cyanide b. aluminum sulfate c. lead(IV) oxalate d. hydrosulfuric acid e. sulfuric acid f. phosphorous acid g. sulfur trioxide h. carbon tetrachloride
Atoms – Molecules – Ions – Ch. 2 13. Here are some common names that you’re expected to know; write the chemical formula. a. Water b. Methane c. Ammonia
Atoms – Molecules – Ions – Ch. 2 14. An element’s most stable ion forms an ionic compound with chlorine having the formula XCl 2. If the mass number of the ion is 24 and it has 10 electrons, what is the element and how many neutrons does it have?
Atoms – Molecules – Ions – Ch. 2 You have completed ch. 2
Ch 1 - Answer Key 1. For each of the following state if it’s a scientific law, theory or neither: a. As a gas expands under constant pressure the gas will cool. Law b. The forces between the gas particles are negligible Theory c. On average heavier gas particles will move slower than lighter gas particles Law d. Gases have a greater average kinetic energy at higher temperatures because the average velocity increases with temperature Theory
Ch 1 - Answer Key 2. The following data was collected. If it is known that experiment 1 has the formula XY what is the formula of experiment 2. Set up a ratio of X/Y for each experiment – expt 1 => 2.55gX/12.75g Y = 0.2 where as expt 2 => 1.7 g X/4.25 gY = 0.4 Since expt 2 has a ratio that is twice as much of X then the formula is X 2 Y
Ch 1 - Answer Key 3. (Ch 2 – problem 77) Two elements R and Q, combine to form two binary compounds. In the first compound, 14.0 g of R combines with 3.00 g of Q. In the second compound, 7.00 g of R combines with 4.5 g of Q. Show that these data are in accord with the law of multiple proportions. If the formula of the second compound is RQ, what is the formula of the first? Set up a ratio of R/Q for each experiment – for expt 1 => 14gR/3gQ = 4.67 and for expt 2 => 7gR/4.5gQ = 1.55 since experiment 1 has 3 times more R the formula is R 3 Q
Ch 2 - Answer Key 4. a. Atomic number => # protons in an atom b. Mass number => # protons and # neutrons c. Charge => # protons – # electrons 5. d. Isotopes are atoms of the same element so they have the same number of protons however they differ by the number of neutrons or mass number 6. Isotope SymbolProtonsNeutronsElectronsCharge 47 Ti222522none 90 Sr 2+ 3852362+ 37 Cl – 1720181–
Ch 2 - Answer Key 7. What are ions? How are ions generated? Ions are charged atoms (monoatomic) or charged groups of atoms (polyatomic) – ions are generated due the gain or loss of electrons producing anions (negative) and cations (positive) 8. Metals tend to gain or lose electrons? What charge do group 1 metals form? 1+ What charge do group 2 metals form? 2+ What charge do group 3 metals form? 3+
Ch 2 - Answer Key 9. Non-metals tend to gain or lose electrons? What charge do group 5 non-metals form? 3- What charge do group 6 non-metals form? 2- What charge do group 7 non-metals form? 1- What charge do group 8 non-metals form? No ions
Ch 2 - Answer Key 10. How could you distinguish between ionic compounds, covalent compounds and acids? Ionic compounds => cation (typically metal or NH 4 ) and anion (typically nonmetal) Covalent compounds => only nonmetals Acids => covalent compounds that begin with hydrogen (exception => although water is technically an acid it will never be named as such)
Ch 2 - Answer Key 11. Name and label the type of compound for each of the following: a. LiHCO 3 – lithium bicarbonate or hydrogen carbonate – ionic b. Na 2 SO 3 – sodium sulfite – ionic c. (NH 4 ) 3 PO 4 – ammonium phosphate – ionic d. Fe(OH) 3 – iron(III) hydroxide – ionic e. SnO 2 – tin(IV) oxide – ionic f. HI – hydroiodic acid – acid g. HClO –hypochlorous acid – acid/molecular h. HNO 3 – nitric acid – acid/molecular i. SF 6 – sulfur hexafluoride – molecular j. CO – carbon monoxide – molecular k. P 2 O 5 – diphosphorus pentoxide – molecular
Ch 2 - Answer Key 12. Write the chemical formulas for the following compounds: a. calcium cyanide – Ca(CN) 2 b. aluminum sulfate – Al 2 (SO 4 ) 3 c. lead(IV) oxalate – Pb(C 2 O 4 ) 2 d. hydrosulfuric acid – H 2 S e. sulfuric acid – H 2 SO 4 f. phosphorous acid – H 3 PO 3 g. sulfur trioxide – SO 3 h. carbon tetrachloride – CCl 4
Ch 2 - Answer Key 13. Here are some common names that you’re expected to know; write the chemical formula. a. Water – H 2 O b. Methane – CH 4 c. Ammonia – NH 3 14. An element’s most stable ion forms an ionic compound with chlorine having the formula XCl 2. If the mass number of the ion is 24 and it has 10 electrons, what is the element and how many neutrons does it have? Since Cl has a charge of 1 - that tells you that X has a charge of 2 + and if X has 10 electrons there must be 12 protons therefore X is Mg – if the mass number is 24 and there’s 12 protons then there are 12 neutrons
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To compare a decimal and a mixed number,
first convert, or change, one of the numbers so that we have either two decimals or two mixed numbers (or two fractions). Then, compare as usual.
For example: Let’s compare 4
1 5
and 4.13.
First, let's write the mixed number as a decimal.
In order to do so, we first write the fractional part of the mixed number,
1 5
, as a decimal.
Write
1 5
as an equivalent fraction with 10 as the denominator.
1 5
=
1 × 2 5 × 2
=
2 10
Now, write
2 10
as a decimal.
2 10
0.2
So, 4
1 5
4.2 .
Now, compare 4.2 and 4.13.
As the tenth digit of 4.2 is greater than the tenth digit of 4.13,
4.2 > 4.13
So, 4
1 5
> 4.13 .
ds
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## [SciPy-User] sparse distance matrix ?
(sunday morning puzzle) Is there a way to get thresholded distance matrices as sparse matrices without going through the full dense distance array or explicit looping? I didn't see anything in scipy.spatial or in scikit-learn. usecase is linear algebra not nearest neighbor search example converts 2d score levels into 1d score levels. The example could be done with kronecker product, but not in the general case. >>> ng, ni = 5, 4 >>> x = 100 * np.repeat(np.arange(1,ng+1), ni) + np.tile(np.arange(ni), ng) >>> x array([100, 101, 102, 103, 200, 201, 202, 203, 300, 301, 302, 303, 400, 401, 402, 403, 500, 501, 502, 503]) dense absolute distance >>> dx1 = np.abs(x[:,None] - x) >>> dx1[:10, :10] array([[ 0, 1, 2, 3, 100, 101, 102, 103, 200, 201], [ 1, 0, 1, 2, 99, 100, 101, 102, 199, 200], [ 2, 1, 0, 1, 98, 99, 100, 101, 198, 199], [ 3, 2, 1, 0, 97, 98, 99, 100, 197, 198], [100, 99, 98, 97, 0, 1, 2, 3, 100, 101], [101, 100, 99, 98, 1, 0, 1, 2, 99, 100], [102, 101, 100, 99, 2, 1, 0, 1, 98, 99], [103, 102, 101, 100, 3, 2, 1, 0, 97, 98], [200, 199, 198, 197, 100, 99, 98, 97, 0, 1], [201, 200, 199, 198, 101, 100, 99, 98, 1, 0]]) I'd like to get the following as scipy.sparse matrices: close distances >>> (dx1) * (dx1 < 3) array([[0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [1, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [2, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 1, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 2, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 2, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 2, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 1, 0, 1, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 1, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 2], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 1, 0, 1], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 1, 0]]) reversed >>> (5 - dx1) * (dx1 < 3) array([[5, 4, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [4, 5, 4, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [3, 4, 5, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 3, 4, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 5, 4, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 4, 5, 4, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 3, 4, 5, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 3, 4, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 5, 4, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 4, 5, 4, 3, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 3, 4, 5, 4, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 4, 5, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5, 4, 3, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 5, 4, 3, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 4, 5, 4, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 4, 5, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5, 4, 3, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 5, 4, 3], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 4, 5, 4], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 4, 5]]) boolean >>> (dx1 < 3).astype(int) array([[1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1]]) >>> Josef _______________________________________________ SciPy-User mailing list [hidden email] http://mail.scipy.org/mailman/listinfo/scipy-user
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## Re: sparse distance matrix ?
<[hidden email]> wrote: > (sunday morning puzzle) > > Is there a way to get thresholded distance matrices as sparse matrices > without going through the full dense distance array or explicit > looping? scipy.spatial.cKDTree can do this. Sturla _______________________________________________ SciPy-User mailing list [hidden email] http://mail.scipy.org/mailman/listinfo/scipy-user
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## Re: sparse distance matrix ?
On Sun, Apr 6, 2014 at 2:41 PM, Sturla Molden <[hidden email]> wrote: > <[hidden email]> wrote: >> (sunday morning puzzle) >> >> Is there a way to get thresholded distance matrices as sparse matrices >> without going through the full dense distance array or explicit >> looping? > > scipy.spatial.cKDTree can do this. Thanks Sturla, I forgot to look there, because I don't need repeated look-up. Josef > > Sturla > > _______________________________________________ > SciPy-User mailing list > [hidden email] > http://mail.scipy.org/mailman/listinfo/scipy-user_______________________________________________ SciPy-User mailing list [hidden email] http://mail.scipy.org/mailman/listinfo/scipy-user | 4,777 | 7,092 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2019-51 | latest | en | 0.371522 |
https://socratic.org/questions/how-do-you-find-the-product-of-2400000-0-006-and-write-the-answer-in-scientific-#613120 | 1,713,904,293,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818740.13/warc/CC-MAIN-20240423192952-20240423222952-00890.warc.gz | 469,392,907 | 6,017 | How do you find the product of 2400000*0.006 and write the answer in scientific notation?
May 14, 2018
$1.44 \cdot {10}^{4}$
Explanation:
$2 , 400 , 000 \cdot 0.006$
first, write them as a scientific numbers
$\left(2.4 \cdot {10}^{6}\right) \cdot \left(6.0 \cdot {10}^{-} 3\right)$
then multiply power one terms together and 10 to power something to gether
$\left(2.4 \cdot 6\right) \cdot \left({10}^{6} \cdot {10}^{-} 3\right)$
when you multiply same number to different powers you add the powers together so
$\left(2.4 \cdot 6\right) \cdot \left({10}^{6 + \left(- 3\right)}\right)$
$14.4 \cdot {10}^{3}$
note that scientific numbers are between 1 to 9
so 14 is not included so
move the point to the left, when you move it to the left one step you multiply by 10 so
$\textcolor{red}{{10}^{3} \cdot 10 = {10}^{4}}$
$1.44 \cdot {10}^{4}$ | 302 | 851 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "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} | 4.375 | 4 | CC-MAIN-2024-18 | latest | en | 0.648051 |
http://www.physicsforums.com/showpost.php?p=4156451&postcount=1 | 1,371,625,428,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368708143620/warc/CC-MAIN-20130516124223-00041-ip-10-60-113-184.ec2.internal.warc.gz | 628,837,935 | 4,301 | Thread: Physics of weight lifting View Single Post
## Physics of weight lifting
Hello to the Forum.
I just started doing some very light weight lifting and decided to do a rough calculation of how many calories I might be burning during each of my workouts, however the figure I arrive at seems too high by comparison with the numbers one regularly see bandied about. What am I doing wrong?
My calculation goes as follows:
I'll use conservation of energy and only account for the upward movement (when you're letting the weights down again in the second half of each repetition you're also exerting force in the direction opposite to movement so as not to let them fall at the full acceleration due to gravity).
I just started so I'm using a pair of 9 lbs dumbbells. This gives us ~4Kg for each dumbbell, 8Kg for the two, and 78.4N. Let's say the average length that you extend each dumbbell upwards during a single repetition is 0.3m. This gives us 23.52J spent in each repetition according to conservation of energy principles, mgh=W.
Now, I do 12 repetitions of each of 8 different exercises, and repeat the whole cycle 3 times. This gives us 23.52(12)(8)(3) = 6774J, which converted to calories result in ~1600 calories, not taking into account the work done during the downward movement of the weights nor leg work, which is perhaps more strenuous.
By all accounts though, 1600 cal burned during a single workout session seems a lot, even more so for such light weights. Is my physics wrong?
PhysOrg.com physics news on PhysOrg.com >> Kenneth Wilson, Nobel winner for physics, dies>> Two collider research teams find evidence of new particle Zc(3900)>> Scientists make first direct images of topological insulator's edge currents | 392 | 1,743 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2013-20 | longest | en | 0.952051 |
https://www.ademcetinkaya.com/2023/05/lonnbpu-nb-private-equity-partners.html | 1,695,409,664,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506421.14/warc/CC-MAIN-20230922170343-20230922200343-00265.warc.gz | 694,248,686 | 58,229 | Outlook: NB PRIVATE EQUITY PARTNERS LIMITED is assigned short-term Ba1 & long-term Ba1 estimated rating.
Time series to forecast n: 28 May 2023 for (n+6 month)
Methodology : Reinforcement Machine Learning (ML)
Abstract
NB PRIVATE EQUITY PARTNERS LIMITED prediction model is evaluated with Reinforcement Machine Learning (ML) and Ridge Regression1,2,3,4 and it is concluded that the LON:NBPU stock is predictable in the short/long term. According to price forecasts for (n+6 month) period, the dominant strategy among neural network is: Buy
Key Points
1. What is the best way to predict stock prices?
2. What is neural prediction?
3. Can we predict stock market using machine learning?
LON:NBPU Target Price Prediction Modeling Methodology
We consider NB PRIVATE EQUITY PARTNERS LIMITED Decision Process with Reinforcement Machine Learning (ML) where A is the set of discrete actions of LON:NBPU stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4
F(Ridge Regression)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Reinforcement Machine Learning (ML)) X S(n):→ (n+6 month) $\stackrel{\to }{S}=\left({s}_{1},{s}_{2},{s}_{3}\right)$
n:Time series to forecast
p:Price signals of LON:NBPU stock
j:Nash equilibria (Neural Network)
k:Dominated move
a:Best response for target price
For further technical information as per how our model work we invite you to visit the article below:
How do AC Investment Research machine learning (predictive) algorithms actually work?
LON:NBPU Stock Forecast (Buy or Sell) for (n+6 month)
Sample Set: Neural Network
Stock/Index: LON:NBPU NB PRIVATE EQUITY PARTNERS LIMITED
Time series to forecast n: 28 May 2023 for (n+6 month)
According to price forecasts for (n+6 month) period, the dominant strategy among neural network is: Buy
X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)
Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)
Z axis (Grey to Black): *Technical Analysis%
IFRS Reconciliation Adjustments for NB PRIVATE EQUITY PARTNERS LIMITED
1. An alternative benchmark rate designated as a non-contractually specified risk component that is not separately identifiable (see paragraphs 6.3.7(a) and B6.3.8) at the date it is designated shall be deemed to have met that requirement at that date, if, and only if, the entity reasonably expects the alternative benchmark rate will be separately identifiable within 24 months. The 24-month period applies to each alternative benchmark rate separately and starts from the date the entity designates the alternative benchmark rate as a non-contractually specified risk component for the first time (ie the 24- month period applies on a rate-by-rate basis).
2. Conversely, if changes in the extent of offset indicate that the fluctuation is around a hedge ratio that is different from the hedge ratio that is currently used for that hedging relationship, or that there is a trend leading away from that hedge ratio, hedge ineffectiveness can be reduced by adjusting the hedge ratio, whereas retaining the hedge ratio would increasingly produce hedge ineffectiveness. Hence, in such circumstances, an entity must evaluate whether the hedging relationship reflects an imbalance between the weightings of the hedged item and the hedging instrument that would create hedge ineffectiveness (irrespective of whether recognised or not) that could result in an accounting outcome that would be inconsistent with the purpose of hedge accounting. If the hedge ratio is adjusted, it also affects the measurement and recognition of hedge ineffectiveness because, on rebalancing, the hedge ineffectiveness of the hedging relationship must be determined and recognised immediately before adjusting the hedging relationship in accordance with paragraph B6.5.8.
3. Paragraph 5.7.5 permits an entity to make an irrevocable election to present in other comprehensive income changes in the fair value of an investment in an equity instrument that is not held for trading. This election is made on an instrument-by-instrument (ie share-by-share) basis. Amounts presented in other comprehensive income shall not be subsequently transferred to profit or loss. However, the entity may transfer the cumulative gain or loss within equity. Dividends on such investments are recognised in profit or loss in accordance with paragraph 5.7.6 unless the dividend clearly represents a recovery of part of the cost of the investment.
4. In almost every lending transaction the creditor's instrument is ranked relative to the instruments of the debtor's other creditors. An instrument that is subordinated to other instruments may have contractual cash flows that are payments of principal and interest on the principal amount outstanding if the debtor's non-payment is a breach of contract and the holder has a contractual right to unpaid amounts of principal and interest on the principal amount outstanding even in the event of the debtor's bankruptcy. For example, a trade receivable that ranks its creditor as a general creditor would qualify as having payments of principal and interest on the principal amount outstanding. This is the case even if the debtor issued loans that are collateralised, which in the event of bankruptcy would give that loan holder priority over the claims of the general creditor in respect of the collateral but does not affect the contractual right of the general creditor to unpaid principal and other amounts due.
*International Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS.
Conclusions
NB PRIVATE EQUITY PARTNERS LIMITED is assigned short-term Ba1 & long-term Ba1 estimated rating. NB PRIVATE EQUITY PARTNERS LIMITED prediction model is evaluated with Reinforcement Machine Learning (ML) and Ridge Regression1,2,3,4 and it is concluded that the LON:NBPU stock is predictable in the short/long term. According to price forecasts for (n+6 month) period, the dominant strategy among neural network is: Buy
LON:NBPU NB PRIVATE EQUITY PARTNERS LIMITED Financial Analysis*
Rating Short-Term Long-Term Senior
Outlook*Ba1Ba1
Income StatementB2Caa2
Balance SheetBaa2B3
Leverage RatiosBa2B3
Cash FlowBaa2B3
Rates of Return and ProfitabilityCBaa2
*Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents.
How does neural network examine financial reports and understand financial state of the company?
Prediction Confidence Score
Trust metric by Neural Network: 80 out of 100 with 475 signals.
References
1. E. Altman. Constrained Markov decision processes, volume 7. CRC Press, 1999
2. J. N. Foerster, Y. M. Assael, N. de Freitas, and S. Whiteson. Learning to communicate with deep multi-agent reinforcement learning. In Advances in Neural Information Processing Systems 29: Annual Conference on Neural Information Processing Systems 2016, December 5-10, 2016, Barcelona, Spain, pages 2137–2145, 2016.
3. V. Borkar. A sensitivity formula for the risk-sensitive cost and the actor-critic algorithm. Systems & Control Letters, 44:339–346, 2001
4. M. L. Littman. Friend-or-foe q-learning in general-sum games. In Proceedings of the Eighteenth International Conference on Machine Learning (ICML 2001), Williams College, Williamstown, MA, USA, June 28 - July 1, 2001, pages 322–328, 2001
5. Athey S, Blei D, Donnelly R, Ruiz F. 2017b. Counterfactual inference for consumer choice across many prod- uct categories. AEA Pap. Proc. 108:64–67
6. K. Tuyls and G. Weiss. Multiagent learning: Basics, challenges, and prospects. AI Magazine, 33(3): 41–52, 2012
7. Imbens G, Wooldridge J. 2009. Recent developments in the econometrics of program evaluation. J. Econ. Lit. 47:5–86
Frequently Asked QuestionsQ: What is the prediction methodology for LON:NBPU stock?
A: LON:NBPU stock prediction methodology: We evaluate the prediction models Reinforcement Machine Learning (ML) and Ridge Regression
Q: Is LON:NBPU stock a buy or sell?
A: The dominant strategy among neural network is to Buy LON:NBPU Stock.
Q: Is NB PRIVATE EQUITY PARTNERS LIMITED stock a good investment?
A: The consensus rating for NB PRIVATE EQUITY PARTNERS LIMITED is Buy and is assigned short-term Ba1 & long-term Ba1 estimated rating.
Q: What is the consensus rating of LON:NBPU stock?
A: The consensus rating for LON:NBPU is Buy.
Q: What is the prediction period for LON:NBPU stock?
A: The prediction period for LON:NBPU is (n+6 month) | 2,193 | 9,269 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-40 | latest | en | 0.795866 |
https://listarc.cal.bham.ac.uk/lists/sc-users-2017/msg58645.html | 1,723,767,664,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641319057.20/warc/CC-MAIN-20240815235528-20240816025528-00836.warc.gz | 278,555,064 | 2,217 | # [sc-users] Twelve-Tone Matrix
Hello,
I am trying to create an algorithm to generate 12-tone matrices, like the one here:
I'm getting stuck, though - processing arrays is still very mysterious to me. I'm not sure what the correct way to do this all is. So far, I think I have the first process in place.. but I'm not sure how to get to the next process.
//starting row;
~row1=[3, 1, 9, 5, 4, 6, 8, 7, 12, 10, 11, 2];
~column1=
{
arg row, new=Array.new(12);
for (0, row.size-1, {
arg i=1;
var next = row[i-1];
if (prev == nil) {prev=0} //case against nil
{|x, b|
x=(row[i]-next); //subtract the next row item from the current row item
if (x < 0) {b=x.abs} {b=0-x}; //if negative, go positive. if positive, go negative. | 245 | 734 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-33 | latest | en | 0.795888 |
http://destinymatrix.blogspot.com/2007_05_01_archive.html | 1,487,833,126,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171162.4/warc/CC-MAIN-20170219104611-00157-ip-10-171-10-108.ec2.internal.warc.gz | 67,549,529 | 49,369 | ## Wednesday, May 30, 2007
IT FROM QBIT
Part 1 (1929) Weyl starts with 4 projective coordinates x0, x1, x2, x3 on a 3D spacelike surface with coordinates x,y,z
x = x1/x0
y = x2/x0
z = x3/x0
The equation for the Einstein light cone unit sphere S2 is
x^2 + y^2 + z^2 = 1
equal to
x1^2 + x2^2 + x3^2 - x0^2 = 0
for null geodesic light rays in globally flat Minkowski spacetime of Einstein's 1905 special relativity.
The 2-component Weyl SPINOR comes from the equatorial stereographic projection i.e. project from SOUTH POLE of S2 to the z = 0 equatorial plane. In that plane define the complex number
w = x + iy = c(+)/c(-) =( rho)e^iphi
The UN-NORMALIZED spinor QBIT |q) is in Dirac bra-ket notation for the particular basis |+) & |-) implicitly defined as
|q) = c(+)|+) + c(-)|-)
Note when c(-) = 0
x + iy -> infinity
i.e. rho -> infinity is the |+) spinor eigenstate (base vector)
similarly rho = 0 is the |-) spinor eigenstate
Given the 2x2 Pauli spin matrices basis for a simple Clifford algebra
I, sigma(x), sigma(y), sigma(z)
First look at the LOCAL DIAGONAL matrix elements of the Pauli spin matrices with respect to the same QBIT spinor.
x0 = (q|I|q) = c*(+)c(+) + c*(-)c(-)
x1 = (q|sigma(x)|q) = c(+)*c(-) + c*(-)c(+)
x2 = (q|sigma(y)|q) = i[-c(+)*c(-) + c*(-)c(+)]
x3 = (q|sigma(z)|q) = i[-c(+)*c(+) - c*(-)c(-)]
*Now I do something new and original not in Weyl 1929.
These have an obvious Feynman-like diagram \/ for a single forward light cone in globally flat Minkowksi spacetime
The NONLOCAL OFF-DIAGONAL matrix elements between two different QBITs |q) & |q') "located" at different light cones gives a new kind of NONLOCAL PRE-GEOMETRY
x0(q,q') = (q|I|q') = c*(+)c(+)' + c*(-)c(-)'
x1(q,q') = (q|sigma(x)|q') = c(+)*c(-)' + c*(-)c(+)'
x2(q,q') = (q|sigma(y)|q') = i[-c(+)*c(-)' + c*(-)c(+)']
x3(q,q') = (q|sigma(z)|q') = i[-c(+)*c(+)' - c*(-)c(-)']
These have an obvious diagram \.../
with ... as the holonomic path independent unique globally flat geodesic connecting the two light cones.
The non-geodesics are zero point quantum vacuum fluctuations of spacetime itself.
In curved spacetime with gravity we have anholonomic path-dependence of course.
Note also that the off-diagonal matrix elements for 2 distinct qbits have the same formal syntax as the 4 Bell entangled pair states used in quantum teleportation protocols. Curious. Suggestive. Indeed!
On May 26, 2007, at 6:53 PM, Jack Sarfatti wrote:
O'Raifeartaigh p. 110 writes of Weyl in 1929:
"It is remarkable that Weyl should even consider the possibility of time-reversal and parity-violation at this time. In fact Weyl not only considered these possibilities but ... made the statement: 'The problem of the proton and electron will be mixed with the symmetry properties of the quantum theory with respect to interchange of left and right, past and future, and positive and negative charge.' Thus ... he not only foreshadowed the later developments in P and T violation but foreshadowed the CPT theorem. All this was at a time when, as Yang put it, 'Nobody, absolutely nobody, was in any way suspecting that these symmetries were related in any manner. It was only in the 1950's that the deep connection between them was discovered. ... What had prompted Weyl in 1930 to write the above passage is a mystery to me.'
Yang's puzzlement is similar to Ed Teller's puzzlement over what prompted President Reagan to decide to do SDI. Teller, in his autobiography writes that he was out of that decision loop and was surprised. Ask Cap Weinberger Jr what really happened. Also http://sharonweinberger.com/?p=60 However, no precognition in the case of President Reagan's decision. ;-)
On May 26, 2007, at 3:46 PM, Jack Sarfatti wrote:
Define spinor
|s) = cos@|up> + sin@e^i&|down>
relative to a given basis
(TETRAD)^a s,s' = (s'|(Pauli)^a|s) inner product
s =/= s' possible.
However, it turns our that there is no Lorentz scalar with 2-component spinors, i.e. the spin 0 tetrad field is missing and is connected with the origin of inertia needing Dirac 4 component spinors.
On May 26, 2007, at 3:28 PM, Jack Sarfatti wrote:
Motivated by Part 1 of Hermann Weyl's 1929 seminal paper "Electron and Gravitation"
John A. Wheeler's "IT FROM BIT"
"BIT" = SPINOR = vector in basic rep of SL(2,C).
Einstein GEOMETRODYNAMIC FIELD is bilinear in TETRAD, therefore quartic in SPINOR
SPINOR QBIT is spin 1/2.
TETRAD = (SPINOR*|Pauli Spin Matrix|Spinor) Lorentz group 4-vector is an entangled EPR pair of spinors.
i.e. TETRAD is a 2 QBIT string
GEOMETRODYNAMIC FIELD is 4 QBIT string
Neglecting relative orbital angular momentum i.e. S-orbital
1/2 + 1/2 = 1 + 0
i.e.
2x2 = 3 + 1 Irrep dimensions
Therefore, spin 1 tetrads but is there also a spin 0 "scalar" tetrad that I missed before?
Note there are nonlocal tetrads if the two spinors in the inner product matrix element are at different space time events.
On May 22, 2007, at 10:59 PM, Jack Sarfatti wrote:
"I am as dissatisfied as you are with distant parallelism and your proposal to let the tetrads rotate independently at different space-time points is a true soluton." Pauli to Weyl (1929).
"IT FROM BIT" John Archibald Wheeler
e.g. Goldstone phases are macro-quantum BIT fields of physical information.
v(superfluid Helium 4) = (h/m)dTheta 1-form O(2) symmetry
A^a(warped tetrad) = M^a^a = dPhi^a/\Theta^a - Phi^a/\dTheta^a 1-form O(9) symmetry - gravitation
F^a = dA^a = -2dPhi^a/\dTheta^a 2-form
S^a^b = M^[a,b] - spin connection 1-form
A^a & S^a^b form the noncompact Poincare group Lie Algebra
On the conversion of Weyl's 1918 aborted scale factor on the uncharged metric IT field to a quantum phase factor on the electrically charged BIT pilot field:
"One can summarize Einstein's objection to Weyl's (1918) theory as the statement that, according to atomic spectroscopy, there is no Bohm-Aharonov effect for gravitation." p. 85.
Metricity in GR is like unitarity in QM, i.e. inner products are invariants of the evolution i.e. parallel transport whether in spacetime or Hilbert space. Nonmetricity, therefore, is like "collapse of the state" in von Neumann quantum measurement theory.
On May 22, 2007, at 7:42 PM, Jack Sarfatti wrote:
Weyl in 1918 made the mistake of applying the gauge principle to Einstein's 2nd rank symmetric metric tensor. What he actually found was a non-metricity vector field in which the magnitudes of vectors parallel transported along world lines depends on the world line. This would be a memory - a hysteresis non-integrability anholonomy not observed in the electromagnetic world where the spectral lines of atoms in the stars are recognizably the same once the gravity red shifts are subtracted out. Einstein pointed this out to Weyl back then. Slowly (Fock, Schrodinger, London ...) it was recognized that this path dependent electrodynamic (Bohm-Aharonov) non-integrability applied to the deBroglie pilot quantum waves not to the Einstein geometrodynamical field.
What Weyl really did in 1918 was to find the non-metricity piece of the connection beyond Einstein's 1916 metric connection. Later Cartan found the torsion field antisymmetric piece to the possible connection fields for parallel transport.
i.e.
Connection = Einstein metric connection + non-metricity vector field connection + antisymmetric (con) torsion
metric connection ~ disclination rotation defects in vectors around closed self-generated infinitesimal parallelograms
(con) torsion connection ~ dislocation defects in which the basic self-generated parallelograms has a gap to 2nd order of smallness (e.g. Penrose "The Road to Reality").
Weyl's 1918 non-metricity vector field means that the lengths of vectors are different around the closed loops i.e. disastrous multi-valuedness - a kind of Riemann surface fiber? In any case Weyl's 1918 vector field is not the EM 4-potential A 1-form but is some alien kind of geometrodynamic field whose flux field tensor 2-form F = dA is zero in our ordinary spacetime without topological defects giving non-vanishing deRham integrals of F through 2D surfaces even when A is closed. 1D string line defects involve surfaces with boundaries whose non-bounding loop integrals of A are quantized "winding numbers" (1st nontrivial homotopy) when the A -form derives from a single coherent Goldstone phase of two real Higgs fields. If there are three real Higgs fields with two independent Goldstone phases then A is not a closed 1-form the non-bounding 2D surface is closed without boundary and we have interior point "monopole defects" with quantized radial fluxes given by "wrapping numbers" (2nd non-trivial homotopy). What would be the physical effects of these non-metricity fluxes? Are they in the interior of the leptons and quarks as Bohm hidden variables that appear as points as the scattering probe magnification increases because of extreme micro-warping. The effective geometrodynamic coupling constant renormalization group flow is to larger values at the 1 fermi scale - it can then get smaller in both UV and IR directions with peaks between 10^-13 - 10^-16 cm.
L.O'R wrote (Ch 5):"(Hermann Weyl) had always been convinced that there was a close analogy between gravitation and electromagnetism ... in 1929 he was able to formulate the analogies ... by means of the tetrad formalism ... Weyl's formulation was complete and went beyond all previous ideas in proposing that electromagnetism be derived from the gauge principle ... that ... has turned out to be a powerful principle for deriving the nuclear interactions and to be the common principle underlying all the known fundamental interactions. ... the six sections of the paper ... two-component spinor theory in Minkowski space ... parity ("chirality" screw helical mirror reflection symmetry now known violated by quarks and leptons, but not known in 1929 of course) and time reversal invariance ... tetrads (AKA vierbeins) ... spinor theory in curved spaces ... Noether conservation laws ... spin connection ... invariant action ... (local) energy-momentum conservation laws from invariance with respect to both general coordinate transformations and Lorentz transformations of the tetrad (Noether's theorem) ... He then recast gravitational theory in the tetrad formalism with a view to exhibiting the analogies between it and electromagnetism. In the final section, he came to what he considered the most fundamental part of the paper, namely, the derivation of electromagnetic theory from the gauge principle."
All of fundamental theoretical physics today depends on only two battle-tested grand organizing ideas, and that includes the extra dimensions of string theory bye the bye, Witten need look no further IMHO.
I. The gauge principle
"AKA" here means "also known as" in a rough qualitative sense with minor differences of detail
II. "More is different" AKA "hidden symmetry" AKA "spontaneous breakdown of symmetry" AKA "ODLRO" AKS "macro-quantum coherence" AKA "Bose-Einstein condensation" AKA "macroscopic eigenvalues of correlation functions" AKA "collapse of phase space volume" AKA effective order parameters including nonlocal topological order of 2D Anyons in FQHE (Fractional Hall effect) as well as the more familiar local order of the Landau-Ginzburg phenomenology with the O(N) Mexican Hat effective low energy potentials for macroscopic phase transitions from quantum to misnamed "classical."
to be continued.
On May 20, 2007, at 1:43 PM, Jack Sarfatti wrote:
Commentary 1 (Draft 2 expanded)
This is a very useful little book by Lochlainn O' Raifaeartaigh in Dublin published by Princeton 1997. It has seminal papers by Weyl, Klein, Fock, Schrodinger, London & Pauli in English from the original German.
The two great battle-tested principles of basic theoretical physics are
1. The local gauge principle, i.e. "relativity" in the most general sense of no action without reaction, no passive absolute Newtonian arenas, Leibniz's relationism of Bohm's "dialogue" not monologue.
2. The spontaneous breaking (or "hiding" Sidney Coleman, Erice Lectures 1970's) of continuous symmetries in the ground state of real on mass shell quanta and also in the vacuum of virtual zero point quanta. AKA "More is different" (P.W. Anderson) "Emergent complexity." "ODLRO" (Onsager & Oliver Penrose), "Goldstone theorem" "Macroquantum states", "Glauber coherent & squeezed states" et-al. The idea of "hidden symmetry" is that whilst the dynamical classical action in the Feynman path integral alternative to second quantization is invariant under the symmetry group G, the vacuum (ground state) is not. One familiar example is a ferromagnet near absolute zero with a Galilean relativity 3-vector order parameter "magnetization" in a coherent domain. This is three real "Higgs fields" with two continuous 2pi periodic "Goldstone phases" defining the S2 vacuum manifold of minima in the Landau-Ginzburg effective quartic renormalizable interaction Higgs-type field Lagrangian. Non-trivial 2nd order homotopy would give point "monopole" defects not actually observed in real ferromagnetic phases. What is observed are wall domain defects with a S0 vacuum manifold without any continuous Goldstone phase at all corresponding to another broken symmetry from three real Higgs fields to only one effective Higgs field that vanishes on the domain wall. That is S3 broken to S1 - curious.
Emergence of "The Nine"
My original parsimonious theory explains the origin of gravity & inertia, torsion (i.e. emergent "tetrad" local observer/detector frames and "spinor" connections as macroquantum coherent "surface" world hologram Goldstone phase modulations from an "M-Matrix" of non-closed 1-forms made from two Lorentz 4-vectors of eight 0-form continuous periodic Goldstone phases Theta^b & Phi^a from nine post-inflation real Higgs scalar vacuum ODLRO fields.
a,b = 0,1,2,3 are Lorentz group indices
M^a^b = Phi^a/\dTheta^b - dPhi^a/\Theta^b
dM^a^b = 2dPhi^a/\dTheta^b
A^a = M^a^a (diagonal elements of the Matrix)
Einstein-Cartan tetrad 1-forms are e^a = I^a + @A^a are spin 1 Yang-Mills type Lorentz group VECTOR renormalizable quantum translation group T4 localized gauge fields.
@ = (Lp^2/\zpf)^1/3 dimensionless "world hologram" self-gravity coupling
I^a are the global Minkowski tetrad frames that we have when either
G -> 0 gravity coupling switched off
h -> 0 quantum action switched off
c -> infinity, i.e. no causal retardation and/or advanced retro-causation
/\zpf -> 0
i.e. no gravity when supersymmetry is perfect! /\zpf in the IR limit is Lenny Susskind's cosmic landscape parameter ~ (area of future deSitter horizon of pocket Hubble bubble universe in the "megaverse" of eternal chaotic inflation.
Supersymmetry is square root of T4, i.e. anticommutator of supersymmetries is T4.
"Spinor" connection 1-forms are (gets dynamical degrees of freedom when total 10-parameter Poincare group P10 = T4*O(1,3) is locally gauged (Utiyama, Kibble et-al 1961)
S^a^b = - S^b^a = M^[a,b]
Torsion field 2-forms are
T^a = de^a + S^ace^c
Curvature field 2-forms are
R^a^b = dS^a^b + S^acS^cb
Einstein-Hilbert (E-H) "classical" Lagrangian density is the 0-form
L(E-H) ~ {a,b,c,d}R^a^b/\e^c/\e^d
ds^2 = guvdx^udx^v = e^aea = I^aIa + @(I^aAa + A^aIa) + @^2A^aAa
Quantum noise corrections in second quantized formalism with macro-quantum vacuum |0> condensate ODLRO is
A^a = <0|A^a|0> + &A^a
&A^a is the quantum excitation spin 1 tetrad vector field annihilation operator
Note that R^a^b is quadratic in A^a and its gradients, therefore L(E-H) has up to quartic terms in the tetrads consistent with renormalization.
Note that
F^a = dA^a = dM^a^a = dPhi^a/\dTheta^a =/= 0
We have "Maxwell" equations
dF^a = 0
d*F^a = *J^a (warped tetrad current density)
d*J^a = 0 (local conservation of warped tetrad current density)
The Einstein geometrodynamic field ds^2 = e^aea is quadratic in the tetrads, therefore obviously made from entangled Einstein-Rosen-Podolsky (1935) pairs of spin 1 tetrad quanta, hence we have quantum geometrodynamic corrections of spin 2 tensor gravitions, spin 1 vector gravi-photons and spin 0 scalar gravitons.
3x3 = 5 + 3 + 1
i.e.
1 + 1 = 2, 1, 0
for the irreducible spin/angular momentum representations of O(3) subgroup of O(1,3).
This does not affect the macro-quantum c-number ODLRO sector of the theory in most cases.
Dark energy and dark matter are both very simply virtual quanta inside the total physical vacuum of positive and negative zero point fluctuation energy density respectively.
Einstein showed that for an isotropic Ricci source in 3+1 spacetime, the space-time bending power of the source is
(G/c^2)(energy density of source)(1 + 3w)
Lorentz invariance and general coordinate invariance imply
w = -1
for isotropic ZPF distributions of all quantum fields of all spins.
Bosons have positive ZPF energy density
Fermions have negative ZPF energy density.
Non-isotropic boundary conditions e.g. Casimir effect will shift the ZPF w, but as long as w < - 1/3 i.e. quintessent field region then
positive ZPF energy density -> antigravity universal repulsion blue shift (dark energy)
negative ZPF energy density -> gravity universal attraction red shift (clumping dark matter) - this is indistinguishable from w = 0 conventional CDM for distant observers looking at gravity lensing for example.
Note when w < - 1 that is "phantom energy" BIG RIP
I predict that the LHC will NOT find dark matter particles on mass shell as a matter of basic principle. Looking for dark matter inside the LHC is like looking for the motion of the Earth through the mechanical Victorian aether with a Michelson-Morley interferometer. Null results then and in the future I do declare. All of this is only based on 1 & 2 above. The book is mostly about the early history of 1 and how it affected C.N. Yang as a graduate student. Robert Shaw's little-known important independent work is also discussed.
"The role of geometry in physics has always been central. But ... it was passive, providing only the stage on which physics took place ... the theory of special relativity in 1905, when it became clear that space and time were not independent ... But the most profound and astonishing entry of geometry into physics came with Einstein's theory of gravitation in 1916, which showed that gravitation was nothing but the curvature of four-dimensional space. ... George Bernard Shaw ... wrote 'Asked to explain why planets did not move in straight lines and run straight out of the universe, Einstein replied that they do not do so because space is not rectilinear but curvilinear.' ... due almost entirely to the genius of Einstein, geometry graduated from being the stage on which the drama of physics took place to being a major player in the drama.
There remained however the electromagnetic and nuclear forces, and the geometrization of gravity raised the question as to whether these other fundamental forces were 'true' forces operating in curved space of gravitational theory or whether they also were part of the geometry. This question has still not been fully answered. ... these forces and gravitation have a common geometrical structure. This is ... the gauge structure."
to be continued
## Wednesday, May 23, 2007
http://sharonweinberger.com/?p=60
http://www.stayaerusa.org/the_trailer.html opens with Col. John Alexander, don't know who the others are besides me and Uri.
Right - that's why I suspect those photos are faked with Photoshop. It's probably a very clever spoof, but I can't say for sure. I am sitting on the fence leaning toward a really good spoof. Watch out there I go - uh oh! ;-)
On May 23, 2007, at 6:29 PM, Kim Burrafato wrote:
But in the case of the alleged Klingon probe photos, the object appears to be in the range of 10-20 feet in diameter maximum. Assume the things are real electrogravostatic alien probes, how are they generating those kinds of energies in a platform of that particularly delicate and almost hypnotically asymmetric configuration? There's basically a flying truncated hollow cylinder with wing and antenna protrusions and what could be a small power plant attached to the cylinder and maybe projecting above it, like some outlandish late Victorian lamp. Weird .... I get a funny feeling when I look at the stuff under the protrusions -- the alien glyphs. If you look at the first pics from Chad with the blowup of the glyphs and details of the construction, you can see a pretty much-dead on match with what the guy took in Capitola.
Jack: So "Chad" got into his car and is driving around? I'm playing Devil's Advocate.
Kim: The little round nodules are the same place and the writing appears to be, too. Now of course, there is the possibility that the same person is responsible for all these "different" pics, but assuming that's not the case, it's pretty interesting -- for a change! Good memehacking, at the very least. But I have this strange feeling that this could be for real.
Jack: Yeah, IMHO this Bulgarian guy is good. USG Intelligence, DIA, SAIC, Los Alamos ... should get him over here ASAP. He is definitely a heavyweight - the real McCoy. Good work Art. The Bulgarian gets my vote. I'm impressed.
On May 23, 2007, at 6:02 PM, Jack Sarfatti wrote:
OK he give some interesting numbers at end. He seems to say you need 10^9 volts/cm to reshape the geodesics made by Earth at its surface for flight. Note I have only skimmed this so far very quickly. He also gives 380 Tesla magnetic field and current densities for solenoids. So that's 100 volts per nanometer. These numbers seem more promising than Hal Puthoff's PV numbers? Clearly, this guy's work needs to be looked into carefully by a lot of people. He is in Bulgaria? This is the most promising development I have seen that is along the lines of what ... has been promoting. Problem with Gordon is that he does not stick to pure descriptions of the photographic/video evidence but tries to explain it with not-even-wrong amateurish techno-babble so all physicists immediately turn off. This Bulgarian has the correct mainstream GR physics that seems relevant to extraordinary claims of huge voltages in capacitors making disks fly. Maybe something here ...
On May 23, 2007, at 5:37 PM, Jack Sarfatti wrote:
On May 23, 2007, at 5:23 PM, art wagner wrote:
I think this work is important, if valid: http://xxx.lanl.gov/PS_cache/gr-qc/pdf/0502/0502047v1.pdf
Thanks looks interesting. Will look more carefully.
## Friday, May 18, 2007
Author: Jack Sarfatti
Subject: My truly original discovery
is my M-Matrix of non-closed 1 forms from the coherent vacuum Goldstone phases of the post-inflation field from which the warped Einstein-Cartan tetrad 1-forms and the spin-connection 1-forms simply emerge in much the same way as does the superfluid velocity field v = (h/m)Grad(Phase).
S^a^b = M^[a,b] = - S^b^a spin-connection 1-forms
M^a^b = Theta^a/\dPhi^b - dTheta^a/\Phi^b
{Theta^a} & {Phi^b} are 2 Lorentz group 4-vectors of coherent vacuum Goldstone phase 0-forms from 9 real Higgs scalar fields.
e^a = I^a + @A^a
@ = (Lp^2/\zpf)^1/3
Lp^2 = hG/c^3
ds^2 = guvdx^udx^v = e^aea
T^a = de^a + S^ac/\e^c = torsion field 2-form
R^a^b = dS^a^b + S^ac/\S^cb = curvature field 2-form
Also, my apparently original recognition that quantum gravity is renormalizable at the "square root" tetrad field level where it is essentially a spin 1 Yang-Mills quantum field theory! Spin 2 is a composite structure from entangled pairs of spin 1 tetrad quanta. There are also spin 1 and spin 0 quantum corrections absent in the "classical" (i.e. macro-quantum ODLRO c-number) part of the theory.
## Wednesday, May 16, 2007
Notes for STAIF 2008 talk in Albuquerque Space Force Defense Meetings
Working these practical applications of my general theory of emergent curvature-torsion Einstein-Cartan (old-fashioned "unified field theory") actually ties into modern string theory and points the way to asking the correct questions to metric engineer warp and wormhole on the road to Michio Kaku's "Type IV Super Civilization."
&
http://qedcorp.com/APS/EinsteinCartan.pdf
OK, this pdf in standard math notation corrects my earlier hasty errors from not including the torsion field induced inhomogeneities in the quintessent zero point energy field correctly for the problems of
1. QCD quark force
2. NASA Pioneer anomalous g-force ~ cH ~ 1 nanometer/sec^2
3. Flat stellar rotation curves in dark matter galactic halos
1 & 2 have the same ordinary differential equation with different numerical parameters at the 2 vastly different scales - but there is a kind of scale covariance.
This is like Newton's apple - same universal zero point energy induced gravity and antigravity across many powers of ten unifying seemingly unconnected phenomena.
As Above, So Below.
## Tuesday, May 15, 2007
High Strangeness and US Intelligence Community
"there may be language difficulties
but it seems more and more are overcoming them every day
i would even say there are not two languages separated here but at least 12 and likely many more"
There are at least 3 conflicting opinions when 2 Greeks meet. :-)
"bundle field theory is not excruciatingly unreachable
"many can attain fluency by their late twenties and understand the translation
of symmetry breaking higgs production
in quotient bundles of the principle
over the symmetries broken"
Yes, of course. But that is still only a reformulation of basic physics. The issue is, does the reformulation allow you to "analytically continue" so to speak to new physics insights, predictions, explanations of mysterious phenomena such as
1. dark energy (~ 73% of universe)
2. dark matter
3. flat stellar rotation curves in galactic halos (I am still working on that model)
4. NASA Pioneer anomaly
5. Why the cosmological constant is so small
6. Silent alien ET spacecraft able to do many g's that no human can survive much less any known material
7. NIDS observations of alien ETs coming out of star gates on Bob Bigelow's Utah Ranch. See the book "Skinwalker" by Colm Kelleher & George Knapp. Bob Bigelow is a very rich man with his own satellite in space and close connections to the US Intelligence Community and the most powerful politicians in America. He is one of the most powerful people in Nevada (Las Vegas) no fool and he has poured millions of dollars into this UFO/paranormal phenomenon with his own private army of retired police, FBI, CIA, US Military Special Forces professionals. Bigelow is a heavy weight. He is nobody's fool and he knows these anomalies are real no matter what the James Randis, the Michael Shermers and other debunking so-called "Skeptics" say.
8. Note my prediction, the LHC will not find any real on-mass-shell particles that can explain Omega(Dark Matter) ~ 0.23 because dark matter is, in fact, off-mass-shell zero point fluctuations of negative ZPE density with positive pressure and w < - 1/3. Looking for dark matter in the LHC is like looking for the Galilean relativity aether with a Michelson-Morley interferometer. It is a very wrong idea IMHO.
"but just because many could understand does not mean many do understand because there are many other interesting things in this world there are many approaches to quantum gravity that explore higgs-like mechanisms"
That may be, but none of them make testable predictions, except for my model as far as I know. Also mine is simpler - Occam's razor with the least amount of formal excess baggage. What is good for pure mathematics is not generally good for theoretical physics. There is a creative tension between their primary objectives. They go in opposite directions. Math seeks generalization getting rid of physical concreteness.
"the higgs mechanism has always been intimate
to explanations of matter and mass
and it has been a regular source of quantum gravity speculation"
Too vague. I am showing exactly how the two key objects of Einstein-Cartan theory emerge from the Goldstone phases of the Higgs fields of the physical vacuum in analogy to the velocity of superflow in helium.
h dehnen, h frommert, and associates
showed that any excited higgs field
can be viewed as mediating a scalar gravitational interaction
but their mechanism gave a massive (yukawa) form"
Interesting. However I have shown how the c-number tetrad and spin connection 1-forms emerge from the Higgs vacuum itself. The excited Higgs particle states are not relevant. I do propose a Salam strong short range tensor gravity in addition to the weak long range component. I do get spin 2, spin 1 & spin 0 excited quanta on mass shell from entangled pairs of spin 1 A^a tetrad quanta. These are small zero point corrections - quantum noise - off mass shell, and far-field radiation on mass shell I suppose for the excited states you allude to above.
"v alan kostelecky and others have interesting papers
on the spontaneous breaking of lorentz and cpt symmetries which are also generated by a typical gamma matrix term that can be found in extended standard models and string theory"
Good, why don't you write a detailed review paper on all this? :-)
"in particular
lorentz violation may provide a mechanism
for a massless graviton
and naturally seem to describe some of the features
of dark matter and dark energy
and the cosmological constant"
Please explain this. "Lorentz violation" to me has two possible meanings.
1. Locally gauging the global 6-parameter Lorentz symmetry of all non-gravity actions. This gives the independent "torsion gap" dislocation defect field to 2nd order of Einstein-Cartan theory. It also adds new curvature disclination defect terms beyond what is in Einstein's 1915 GR that is simply from the local gauging of the 4-parameter translation group.
2. Spontaneous breakdown of the global/local Lorentz group in the physical vacuum i.e. hiding the symmetry in the actions of all the non-gravity fields (Sidney Coleman, Erice Lectures, 1970').
"similarly
gasperini, hehl, and sardanashvily
have shown the emergence of various gravities
with actions coupling to
torsion
curvature
and other connection-derived terms"
Good, where is the physics? What do they predict and explain from real phenomena and data?
"all of these authors approach these ideas
from slightly different geometric or interactionist/group formalisms
often several different ones per author
scalar
metric affine
..."
Too vague, please flesh this out.
"but higgs mechanisms are not the only "natural" approach
similar to the reason for investigating higgs gravity
the zeropoint field has been interpreted as generating inertia
and there have been a few attempts at
zeropoint effects that generate gravity"
If you mean Haisch, Puthoff, Rueda, Cole - that's a dead end for several reasons IMHO not going anywhere. It's incomplete for one thing, only includes virtual photons. It violates dark energy facts. Other objections as well. Higgs - Yukawa couplings are the only viable model for the emergence of rest mass of leptons, quarks & W bosons. However, even that is incomplete - no gravity - that's where my model comes in because you cannot separate rest mass inertia from gravity without violating the equivalence principle as Haisch et-al blithely do without even realizing it it seems to me.
there is the holographic programme
seeking to explain the surface terms in the hilbert action
in terms of a fundamental description of information"
That's already in my model from the git-go. My model is holographic. Primacy of spacelike 2D nonbounding cycles enclosing point defect nodes of the 4D world crystal lattice in the 3+1 projection down from 9+1.
"there are also a number of purely geometric approaches
when one alters the geometry sufficiently
as is found in the torsion theories
or extensions of the work by obukhov and others
exploring the relationships between gravity and spin
( which have influence the spin-network programme
and other more mainstream approaches )"
This is all pi-in-the-sky without any connection to real physics - correct me if I am wrong. :-)
"there is a desire to make gravity emergent"
Indeed, and I have done so in an elementary way showing the connection/formal analogy to emergence in superfluid helium & in BCS superconductors.
"the whole point is of course
to avoid the divergences and ontological difficulties
of direct quantisation of the metric field"
I have solved that problem. My coupling is dimensionless and the basic gravity fields are spin 1 warped "tetrad" A^a fields, a = 0,1,2,3 - note in non-commutative geometry the Lie algebra is Yang-Mills non-abelian, as it is already when you go to the full 10-parameter Poincare group Lie algebra, but even the T4 part has new structure constants in Connes's noncommutative generalization.
"barcelo, visser, and liberati
have proven a very general result
that linearising a classical scalar field
( or indeed any hyperbolic second-order equation )
can always be interpreted as a lorentzian geometry
affecting particle propagation as gravity"
I don't believe that. They must be mistaken. How do they get tensor fields from a single scalar field?
" so these emergence results are very general
((((((..)))))))
this background of working theories
helps give some placement to sarfatti's approach
now
he would have more insight into his approach
than any reconstruction that could be attempted
but it is always instructive to try
his expositions have suggested that a primary motivation is
geometrical
he has been closely following shipov's work on torsion
and exploring its relation to various quantisation programmes
and the geometry is certainly central to his mathematical formalism"
Gennady Shipov visited us from Moscow in San Francisco twice 1999 & 2000 with money provided by Joe Firmage to our ISSO Science Group.
"but there are other currents present
he has also pointed a number of times over the recent years
to distinctions between the higgs mechanicsm
and zero point mechanisms of inertia
in particular he has stressed that
differences of fundamental scales in the theories
the equivalence principle
the nonlocal nature of gravity at asymptotic infinity
and the holographic principle
strongly suggest the zeropoint mechanism is ruled out
interestingly
he has taken some of the mathematical insights
of what are unacceptable features of these
directly into his current working form
explicitly avoiding those issues through a foundation in holography
and of course
as is common with jack
he has many other directions he is attempting to reinterpret
inside the kernel of his new formalism
much still appears in flux
as facts and derivations drive new speculations
so i wouldn't be too quick to place sarfatti in sardanashvily's camp
just as i wouldn't be too quick to put sarfatti in 't hooft's camp
they may share progenitors
( as we all do )
but they have their own unique directions
i also wouldn't be too quick to any given geometric intuition
is the more "correct" approach
as humanity has repeatedly been shown wrong here
and geometry is one of the big underspecifications
in modern speculation
our universe could operate on pointless geometries
for all we know
well anyways
it looks like my lunchtime is almost over so i'll cut this short
but i just wanted to make one last observation
despite its now obvious ramifications
in hypergeometric systems and general algebraic geometry
it took candelas, de la ossa, green, and parkes
coming from a physics background
to discover mirror symmetry
i think much mathematics comes from such physical intuitions
( despite some recent threads claiming more damage than good )
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
galathaea: prankster, fablist, magician, liar
Jack Sarfatti
sarfatti@pacbell.net
"If we knew what it was we were doing, it would not be called research, would it?"
- Albert Einstein
http://www.authorhouse.com/BookStore/ItemDetail.aspx?bookid=23999
http://lifeboat.com/ex/bios.jack.sarfatti
http://qedcorp.com/APS/Dec122006.ppt
http://www.flickr.com/photos/lub/sets/72157594439814784
## Monday, May 14, 2007
Discussion with a mathematician
Markwho wrote:
"You might be heading in the same direction, but from an entirely different starting point -- but one that's got problems. For one, with the I's in the definition of the tetrad, you're still trying to have it both ways and inject a semblance of translation invariance in curved spacetimes. There is none."
No I don't think so. No global translational invariance. I have locally gauged the translation group and the compensating gauge potential is precisely A^a. The idea is this. The pre-inflation false vacuum is precisely global special relativity standard model. The trivial tetrads are global. I call them I^a. For globally flat unaccelerated geodesic observers, The components of I^a i.e. I^au = Kronecker delta. For non-geodesic accelerated observers in globally flat Minkowski spacetime, I^a components are general curvilinear functions but with zero Riemann-Christoffel curvature of course.
e^a = I^a + @A^a
Now this is not necessarily a perturbation theory. I am not assuming
@A^a << I^a
in general. However, obviously when @ -> 0 we return to globally flat Minkowski spacetime.
So I do not think this is a weak point of my theory.
I posit @ = hG*/\zpf/c^2 dimensionless like the fine structure constant in quantum electrodynamics.
"Rather, you should be looking at the more general notion of affine spaces and affine connections."
Why?
"This is where you begin to find a natural correspondence. When you start bringing in the M matrix and the various other constructs associated with it (your novel contributions), ultimately what you're doing or what you're going to end up doing is landing in
the same spot that Sardanashvily's already gotten to."
Well I have no understanding of Sardanashvily's work. Never read it. Note we both have "Sar" in our names. Why don't you be our Freeman Dyson and write a paper showing the correspondance?
"That is, you started from a somewhat problematic point of departure,"
What do you mean? I^a?
"took a turn on your path, and ended up landing right in the middle of a confluence with the gauge gravitation idea, which already has a perfectly sound starting point."
Well I have read Kibble's paper from 1961 on gravity as a gauge theory. I first read the Yang-Mills 1954 paper in the mid 60's and my PhD dissertation was influenced by the whole idea of local gauging. But I was operating in a vacuum back then where I was so not too much came of it, but I did predict the supersolid phase of helium in Physics Letters in 1969 before Tony Leggett. I also made a model of self-trapped laser filaments based on Landau-Ginzburg equation which helped Ray Chiao in his experiments in mid 60's - as he told Charles Townes, and I did write a paper with Marshall Stoneham on spontaneous broken symmetry in solid state physics in 1967 cited in American Institute of Physics "Resource Letter on Symmetry in Physics" in 1980 as a significant paper. So the two ideas of spontaneous symmetry breaking and local gauge invariance have been central to all my work from the 60's.
"Hence, the need to systematically compare notes. One of the major elements in Sardanashvily's treatment of mechanics (which comes straight out of the mathematical community) is the more general notion of a "connection". A connection is not just for gauge theory, but is a more general object that lives on a jet bundle."
You are thinking like a mathematician. Physics is very different. Why should a physicist be interested in a "jet bundle"? Mathematicians like to generalize, we physicists are primarily interested in phenomena and want to use as little excess formal baggage as possible - completely opposite to Max Tegmark's ideal of the "mathematical universe."
"Moreover, when the latter is related to an already-existing gauge-theoretic connection, then it has a decomposition into it plus a "soldering form". Ultimately, the tetrads come out of soldering forms
for affine connections. At least, that's my understanding of it."
Fine, but so what? I don't see any advantage there from a physics POV. If you could show one fine.
"Sardanashvily's Goldstone phases come about through the breaking of the general frame bundle's GL(4) symmetry group to the SL(2,C) group associated with fermions' local frames. That's enough to give you the spin coefficients -- but not the tetrad."
Sounds like Utiyama's paper that motivated Kibble in 1961. Well in that case my theory is much better since I get both. The tetrads from the diagonal elements of the M-matrix, the spin connections from the off-diagonal elements of the antisymmetrized part of the M-Matrix.
Hey! I just had a new idea from reading A.Zee's "Quantum Field Theory in a Nutshell"!
Matrix = Traceless Symmetric + Trace + Antisymmetric
So tetrads from the trace (I mean diagonals)
Spin connections from Antisymmetric off-diagonal part.
So what physical thing corresponds to the symmetric off diagonal part?
"The dimensions of GL(4) and SL(2,C) are 16 and 6. The symmetry breaking entails vacuum sectors associated with the quotient group GL(4)/SL(2,C) (10 dimensions, isomorphic to S_3 x R^7, in fact). This is where the tetrad lives. They're the Goldstone phases of the broken
symmetry brought about by the fermions' frames."
This sounds interesting if you can flesh it out in more detail. However, I remind you I have a very simple formula for the intrinsically warped pieces of the tetrads
A^a = Theta^a/\dPhi^a - dTheta^a/\Phi^a
where Theta^a & Phi^a are 8 zero-form Goldstone phases for 9 real scalar Higgs fields.
This is a SO(9) internal symmetry.
For example if only 2 real Higgs fields, they share only one Goldstone phase i.e. O(2) or U(1) - roughly speaking.
Vacuum manifold for Landau-Ginzburg potential minima is S1
3 real Higgs fields share 2 independent Goldstone phases i.e. O(3)
Vacuum manifold is S2
So for 9 real Higgs fields, the vacuum manifold is S8.
Suppose we have N real Higgs fields psi(i), i = 1 ... N
|psi|^2 = psi(1)^2 + psi(2)^2 + ... psi(N)^2
The vacuum manifold is the unit N-1 hypersphere
1 = [psi(1)^2 + psi(2)^2 + ... psi(N)^2]/|psi|^2
There are N-1 independent direction cosines of the independent Goldstone phases.
Theta^a & Phi^b are separately 4-vectors under the 6-parameter Lorentz group.
"Sardanashvily's papers and books has a large number of Hehl references, though I'm not entirely sure what the relation of the two is. He also seems to be caught up in the same general "clique" that I think may be centered on the 1979 Lecture Notes in Physics 107. It's probably out of there that the whole "covariant Hamiltonian" and
"polysymplectic" trends begun."
Well I don't see any physics here. Just a lot of math. I could be wrong, but it's your job to connect the above to physics.
"The more notable feature of this extra element, that should particularly interest you, is that it does not require the 3+1 decomposition of spacetime! That is, it's not only fully GR- compatible, but provides a natural starting point for anyone who wants to further study all matters related to achronal spacetimes, or spacetimes with causal anomalies (e.g. closed timelike curves). The
"covariant Hamiltonian" approach is general enough to accommodate this."
Sounds nice, but the tetrad and spin connection Cartan 1-forms are already independent of the 3+1 decomposition.
is a local scalar invariant under GCTs and so are the spin connections
S^a^b = - S^b^a = S^a^budx^u
ds^2 = e^aea = guvdx^udx^v
"Sardanashvily stays within the more rigid confines of globally hyperbolic spacetimes, however (in part, because there's already a well-known representation theorem that relates more general spacetimes to these). You've got a PhD in Physics. However, the subtleties that are brought out by the jet bundle formalism and all matters related requires a deeper probing into the Mathematical issues;"
No doubt, but the problem is we have too many mathematicians in physics with very little physics coming out of their efforts. I am interested in some very concrete physical issues
1. What is dark energy?
2. What is dark matter?
3. How do the silent-running "UFOs" work?
4. Pioneer anomaly?
5. Flat stellar rotation curves in galactic halos.
to name a few.
That is how do we make warp drive and traversable wormhole that we in fact see - though most mainstream physicists are in ignorance or denial of the UFO sightings and the strange "Skinwalker" phenomena at the Bigelow-Sherman Utah Ranch. USG Military Intelligence at high levels takes these anomalies very very seriously. I can tell you that.
"and this is probably where the greater focus may need to lie for a while."
I don't think so. I am making rapid progress.
"This is a language problem that's endemic to Math and Physics and it's serving as an obstacle to real progress in Physics."
Yes. :-)
## Sunday, May 13, 2007
First a short review of potential theory.
I. If the force f decreases with distance and the potential energy U is (positive) negative, then the force is (repulsive) attractive.
Example I.1
U = +e^2/r > 0
f = -dU/dr = +e^2/r^2 points toward r -> infinity, i.e. repulsion
note that (d/dr)(1/r) = - 1/r^2
the two - signs cancel
II. If the force increases with distance and the potential energy U is (positive) negative, then the force is (attractive) repulsive.
Example II.1 /\zpf is the vacuum zero point space curvature, assumed constant here.
Zero point energy, as mentioned by Andrei Sakharov in 1967, directly induces gravity because of the equivalence principle of Albert Einstein.
In the weak field low speed limit of general relativity, the universal zero point energy induced gravity potential energy per unit test particle is
V ~ -c^2/\zpf r^2
r < R
for a uniform sphere of isotropic zero point energy of radius R centered at r = 0, with vanishing /\zpf for r > R. This is same as drilling a straight hole all the way through the center of a sphere of constant mass density to the other side and dropping a test particle down the hole. This is a harmonic oscillator because the mass beyond the momentary position of the test particle makes no contribution to the force on the test particle.
Baron Munchausen on the geodesic test particle feels weightless of course, but from the POV of the non-inertial observer fixed to the non-geodesic surface of the sphere by non-gravity electrical and quantum forces, it's AS IF there is a force per unit test mass on the test particle
g = - dV/dr = +2c^2/\zpfr
When /\zpf > 0 this is repulsive.
This same formal result carries over into cosmology where r is replaced by the scale factor a(t) stretching space itself and what happens is that there is an extra acceleration of a(t) opposing the ordinary matter that tends to decelerate the stretching of the rubbery fabric of space itself, i.e. the 3Dim spacelike piece of the geometrodynamic field.
The cosmological equations are here
http://www-conf.slac.stanford.edu/ssi/2005/lec_notes/Kolb1/kolb1new_Page_05_jpg.htm
Therefore, in these sign conventions, /\zpf > 0 is the repelling dark energy and /\zpf < 0 is the attracting dark matter.
Repelling dark energy is isotropic w = -1 positive zero point energy density with equal but opposite negative pressure.
Attracting dark matter is isotropic w = -1 negative zero point energy density with equal but opposite positive pressure.
Adding torsion fields converts Einstein's cosmological constant /\zpf into a locally variable "quintessent" field. You get torsion with curvature by locally gauging the entire 10-parameter Poincare group of globally rigid special relativity.
Now what happens between quarks inside the hadronic "bag"? What we have is a bag of dark matter where the quintessent field is
/\zpf(quarks) = - 1/ar
Therefore, the constant attractive force per unit mass between the quarks is
g = -c^2/a ~ string tension
for strong short-range (Abdus Salam) ZPF induced gravity
We see exactly the same thing on the larger scale of the NASA Pioneer Anomaly where
g = -cH ~ 1 nanometer/sec^2
i.e. c^2/Hubble radius ~ 10^21/10^28 ~ 10^-7 cm/sec^2
i.e. a hollow sphere of dark matter centered at Sun beginning at about orbit of Saturn.
## Saturday, May 12, 2007
From the final chapters of A. Zee's "Quantum Field Theory in a Nutshell" p.453 I picked up an interesting fact expressed in a clear pedagogical way without all the rigor mortis of the finicky mathematicians posing as physicists loading us up with excess baggage.
"Poincare invariance in our original D-dimensional spacetime now appears as an internal symmetry."
BINGO! That's another justification for my idea to maximize the stability of all the topological defects in the ODLRO vacuum condensate order parameter of Higgs-Goldstone fields in sense of low energy effective field theory. These fields are actually the post-inflation fields.
The bosonic string action in (D - 1, 1) spacetime is the action of a quantum field theory of D real bare massless Higgs scalar fields with D - 1 independent Goldstone phases in 1 + 1 2-dim string spacetime. Zero point fluctuations generate masses to get a kind of Mexican Hat potential with symmetry O(D) and vacuum manifold S(D-1).
My curvature-torsion theory of emergent (3 + 1) spacetime (four) tetrad 1-forms and (six) spin-connection 1-forms as compensating spin 1 gauge fields from locally gauging the Poincare group P10 simultaneous with supersolid distortions of the 0-form Goldstone phases needs 8 Goldstone phases Theta^a & Phi^b (a,b = 0,1,2,3)
That is,
Tetrads e^a(warp) - I^a(Minkowski) = (hG*/\zpf/c^3)^1/3A^a
/\zpf = quintessent field -> Einstein's cosmological constant (10^-3ev)^4 in the IR limit.
G* is renormalization group flow "running" coupling depending on mass (inverse length) scale.
The power 1/3 is determined by Lenny Susskind's "world hologram" conjecture.
Idea is that G* >> G(Newton) at scale 10^-13 cm, G* -> G(Newton) IR and also UV i.e. a kind of IR-UV duality. That is, a resonant plateau in G* in the region of nuclear -> high energy physics 10^-13 cm -> 10^-16 cm roughly
A^a = M^a^a
Spin connections S^a^b = - S^b^a = M^[a,b]
M^a^b = dTheta^a/\Phi^b - Theta^a/\dPhi^b
Note how the 1 + 1 string was starting point in my last pdf as a kind of Turing machine tape with a simple IT FROM BIT program punching in the ~ 10^120(Susskind) - 10^124 (Zee) BITS of the deSitter horizon.
Note there Lp ~ 10^-13 cm if the 1 + 1 string of (hG*/\zpf/c^3)^-1/3 links is the square root of the future deSitter horizon of our actual universe!
## Wednesday, May 09, 2007
Synopsis
The following input ideas
1. Lenny Susskind’s “world hologram”
2. Hagen Kleinert’s “world crystal Planck lattice”
3. Bekenstein’s c-BIT per Planck area for “horizon” thermodynamics in black holes and deSitter cosmologies
4. Observed number for dark energy density of our accelerating universe from Type Ia SuperNovae observations
5. Local gauge invariance of the global space-time symmetry of the Poincare group of 1905 special relativity giving both disclination curvature and dislocation torsion compensating gauge fields as Einstein-Cartan tetrads and spin connections respectively.
6. Higgs-Goldstone mechanism of hidden vacuum symmetries describing the fundamental post-inflation vacuum condensate “supersolid” fields whose robust coherent world hologram phases give the above tetrads and spin connections in analogy with the velocity of superflow in liquid helium 4 below the Lambda Point.
7. Lattice string theory toy model with half-filled Fermi surface for “Peierl’s instability.”
8. Saul-Paul Sirag’s Nature paper on Blackett gravimagnetism in astrophysical objects.
Give as outputs, curious “Eddington numerology” Jung-Paul “synchronicities” relating the following factual data
• Measured dark energy density & number of BITs on future deSitter Horizon
• Scale of low energy nuclear force and the universal slope of hadronic Regge resonances (basic data of string theory)
• Charge to mass ratio of the electron
• Renormalizable quantum gravity gauge force model with fundamental spin 1 Einstein-Cartan tetrad quanta whose entangled pairs give a spectrum of Einstein curvature spin 2 gravitons + spin 1 torsion graviphotons + spin 0 scalars in the microquantum zero point corrections to the macroquantum c-number vacuum condensate curvature and torsion fields.
See periodically updated http://qedcorp.com/APS/WorldLattice.pdf
## Monday, May 07, 2007
At times one must be anchored in consensus reality analogous to low energy effective field theories in physics at critical points in renormalization group flows - perhaps(e.g. Volovick in Russia, but probably not as his theory violates IR-UV "duality"). Note I said "at times" not "all the time." That we have been to the Moon is a non-negotiable fact and anyone who seriously suggests otherwise is not a person I am interested in hearing more from. Of course reality is full of remarkable "coincidences" (AKA "synchronicities") many of which I speak about in my books based on my own direct encounters with the uncanny "high strangeness."
Paperclip did not come up with anything other than conventional rocket technology. Most of the URLS cited below are unreliable sources of disinformation - fiction posing as fact. Some of it is hilarious maybe even some of it is well-written I don't know. I never said that The Evil One was without talent.
The Black Sun is in Michael Murphy's "An End to Ordinary History." The NASA Pioneer Anomaly is probably measuring the Black Sun - it is a probably expanding hollow shell of dark matter concentric with center of the visible sun.
Dark matter is negative zero point energy density with positive quantum pressure.
On May 7, 2007, at 8:19 PM, MT wrote:
Jack Sarfatti wrote:
On May 6, 2007, at 11:21 PM, chair@thule.org wrote:
"Your comment is a perfect example of fear based attack, uninformed,
reactionary, ignorant thought.Your judgment is taken obviously without
seeing any of the videos or reading any of the essays. Fritter your
life away on the surface not talking about the important things, just
keep to simple subjects."
Yeah like how the universe was created and how consciousness is
generated in the brain - simple subjects.
"Thule Society" was an occult Nazi org. That you choose that name
shows you are a Nazi.
http://qedcorp.com/book/psi/hitweapon.html
I can smell your type light years away.
JS - http://www.stardrive.org/title.shtml
BS - http://www.thule.org/enoch.html
BS - he got that right. Thule.org is an obvious Nazi website.
"Perhaps Dr. JS is confusing BS' signature web-
references of "Thule" to: "Thule-Gesellschaft"
http://en.wikipedia.org/wiki/Thule_Society
which might be more prototypical of a
disingenuous, demagogic paranoid gyration
point of view, perhaps; although, I too am somewhat
mystified by the revisionist conspiracy theories
regarding a "staged moon landing." Knowing how
goofy and unreliable humans are in places of
not to mention the myriad subcontractors
and their spouses, et al., who would be privy to
such a grand deception -- and, non-disclosure
agreements notwithstanding, people talk,
invariably, utterly and reliably, especially
for profit -- but, I admit to a guilty pleasure
in enjoying, to a degree, the endless
number of parallel possibilities in this weird
whirling universe, at least as memes, and
catalysts for deeper ponderings,
stopping short of gullible vulnerabilities, etc.
And what do I really know?!
ANYWAY . . .
There are, in fact, many who believe that
not only have "we" established bases on
both the Moon & Mars, but the Black Ops
Technology responsible for all these secret
endeavours have directly evolved from the
NAZI aeronautical, Operation Paperclip
programs of yesteryear! SHEEESH!!
Example(s):
**************************************************
FTR#588- http://www.spitfirelist.com/f588.html
The Aliens are Coming, the Aliens are Coming - Not ! --
(Two 30-minute segments) (Recorded on 3/4/2007.)
http://www.spitfirelist.com/f588.html
For the Record
http://www.spitfirelist.com/ftr.html
Produced by
Dave Emory
http://www.spitfirelist.com/index.html
L-1 The Political Implications of the UFO Phenomenon
and the "ET" Myth (Approx. 171 minutes)
http://www.spitfirelist.com/lecture.html
Intercept -- But Don't Shoot
The True Story of the Flying Saucers
by Renato Vesco
1971, Grove Press
ASIN B0006CPK2K
338 pages
Book Description
Looks at the early 'flying saucer technology'
of Nazi Germany and the genesis of early
scientists, escaped battalions of German
soldiers, secret communities in South America
and Antarctica, the astonishing book blows
the lid off the 'Government UFO Conspiracy'.
Examined in detail are secret underground
airfields and factories; German secret
weapons; 'suction' aircraft; the origin of
NASA; gyroscopic stabilisers and engines;
the secret Marconi aircraft factory in
South America, and other secret societies,
both ancient and modern, that have kept
this craft a secret, and much more.
http://www.magonia.demon.co.uk/abwatch/naziufo/naziufo1.html
**************************************************
THE NAZI UFO MYTHOS
An Investigation by Kevin McClure
http://www.magonia.demon.co.uk/abwatch/naziufo/naziufo1.html
Introduction
http://www.magonia.demon.co.uk/abwatch/naziufo/naziufo1.html##intro
1. Core 1 - Foo Fighters
http://www.magonia.demon.co.uk/abwatch/naziufo/naziufo2.html
2. Core 2 - Renato Vesco, Feuerball and Kugelblitz
http://www.magonia.demon.co.uk/abwatch/naziufo/naziufo3.html
3. Core 3 - Major Lusar, the saucer builders, and the test flight
http://www.magonia.demon.co.uk/abwatch/naziufo/naziufo4.html
4. Core 4 - W A Harbinson and Projekt Saucer
http://www.magonia.demon.co.uk/abwatch/naziufo/naziufo4.html
5. Core 5 - Vril, Haunebu and interplanetary travel
http://www.magonia.demon.co.uk/abwatch/naziufo/naziufo6.html
6. False histories
http://www.magonia.demon.co.uk/abwatch/naziufo/naziufo7.html
7. Unnamed Soldiers
http://www.magonia.demon.co.uk/abwatch/naziufo/naziufo8.html
8. Authorities from Earth and Elsewhere
http://www.magonia.demon.co.uk/abwatch/naziufo/naziufo9.html
http://www.magonia.demon.co.uk/abwatch/naziufo/naziufo10.html
10. Mistakes and fantasies
http://www.magonia.demon.co.uk/abwatch/naziufo/naziufo11.html
Conclusions
http://www.magonia.demon.co.uk/abwatch/naziufo/naziufo12.html
**************************************************
General Electric's revenues in 2003 totaled \$134.2 billion.
http://www.corpwatch.org/article.php?list=type&type=16
***********************************************
Private industry is looking at the question of gravity control
with new seriousness.
A large number of giant corporations, including Bell Aerospace,
General Electric, Hughes Aircraft, Boeing, Douglas and many
others, have set up gravity projects.
*****************************************************
TOP SECRET/MAJIC
.... flying saucer ENGINE development was turned over to
General Electric.
*****************************************************
SIDE NOTE :
The F-117A Nighthawk
About the size of an F-15 Eagle, the twin-engine aircraft is
**************************************************
A COMPILATION OF ANTI-GRAVITY ARTICLES AND REFERENCES
**************************************************
Secrets of the Saucer Scientists
By William F. Hamilton III
http://www.alazarin.plus.com/ufodoc03.html
**************************************************
And so on and so forth...
AND...Then there's this
frightening bit of weirdness:
~o0O0o~ [Quote]
Black Sun: Aryan Cults, Esoteric Nazism
and the Politics of Identity
by Nicholas Goodrick-Clarke
http://www.forteantimes.com/review/blacksun.shtml
New York University Press, 2002
Hb, 369pp, plates, ind, bib, refs, \$29.95
ISBN 0 8147 3124 4
A few years back, I was lead on a satanic murder
tour of East Sussex by a local researcher,
who told me a frightening story.
In the mid '90s he received a tip-off that Nazi
Satanists would be using Chanctonbury Ring,
an ancient holy site, for a ritual. As he and
another man - a well-known writer on occult
matters - approached the hill, they found
baseball bat-wielding thugs who marched
them to the top, where our heroes were
surrounded. Suddenly, over the crest of the
hill sprung yet more balaclava-clad heavies,
these ones bigger and better armed.
After much shouting and gesticulating the
Nazi Satanists fled, no match for the
Animal Rights activists who had been tipped
off that animals were to be sacrificed there
that night. At the site were found a stone cross,
stolen from a churchyard, a genuine U-boat
flag and a Nazi dagger.
My guide suspected that the group were
connected to David Myatt and the Order
of the Nine Angles. I found the story a little
too Dennis Wheatley to be true, but, having
read Black Sun, I might be a little more
cautious poking around Chanctonbury ring
at Hallowe'en. And Myatt is only the
small tip of a vast iceberg.
Nicholas Goodrick-Clarke's (G-C)
Occult Roots of Nazism is still the textbook
source on the esoteric groups that influenced
the Nazi elite. With Black Sun he attempts to
trace these influences, now re-visioned and
explicitly linked to neo-Nazi and fascist
ideologies, right up to the present day.
The scope of the book is immense. From the
simplistic thuggery of American Nazi Party
founder Lincoln Rockwell, via the Hitler-Christ
fugues of Matt Koehl, we glimpse the awesomely
complex Hindu-Aryan cosmologies of Italian
Julius Evola and "Hitler’s Priestess", the
Cornish-Greek Savitri Devi (subject of G-C's
last book). Devi identifies Hitler as a Hindu avatar,
a "man against time" returning to lead his chosen
Aryan people to a new and perfect era.
This and Evola’s grandiose metaphysical
anti-Semitism are the compost for all manner
of twisted cultivars, providing an almost
eschatological sense of urgency in the eternal
battle against the spiritually "unclean" Jew.
It's not hard to see how the shimmering grandeur
of their doctrines has appealed to so many neo-Nazis,
providing them with a righteous and divine
justification for their hatred of those who they
feel threaten their security. Both remain popular
with the new generation of neo-Nazis, and Evola's
work has been republished recently with an
introduction by "industrial" musician
Michael Moynihan.
Sharing elements of Evola and Devi's
cosmic vision is Miguel Serrano, at one
whose spiritual journey saw him befriending
Herman Hesse, Carl Jung, Jawaharlal Nehru
and even the Dalai Lama before (having made
direct astral contact with him in his
Hollow Earth HQ) settling on Hitler as the way,
the truth and the light. In a curious blend of
Jungian and Lovecraftian cosmologies, he
sees Hitler as an shining Aryan archetype
of unknowable gods residing far beyond our
galaxy. Serrano remains an important ideologue
on the global Nazi scene, and has been linked
to the German Colonia Dignidad cult
compound, from which accusations of child
abuse and torture during the Pinochet years
have emerged recently.
A whole chapter of Black Sun is devoted to
the Nazi UFO mythos. G-C reads the tales
of wartime German flying saucer fleets hidden
in the Hollow Earth as an extension of the
pervasive rumours of Hitler's survival.
Certainly the post-war capture of German
submarines off Argentina and the mysterious
force from Antarctica in 1946 could only
have fuelled such speculation, first expressed
in a 1947 book, Hitler esta vivo by a Hungarian
exile in Argentina. G-C traces many tales of
German wartime saucer technologies to one
Erich Halik, a member of a Viennese esoteric
group who regarded the UFOs as something
more than mere earthly technology.
Their take, which echoes - and narrowly
predates - Jung's, is distinctly esoteric,
the UFO being a metaphysical "cultic vessel
used by the supreme hierarchy of Christian
Gnostics" and now employed by the spiritually
superior underground Nazi elite. Halik published
several UFO articles in Hungarian esoteric
magazines and was particularly interested by
George Adamski's contact with the Aryan
was not spiritually advanced enough to
Or was he? G-C doesn't mention it, but
George Hunt Williamson, another contactee
who was also present at Adamski's desert
encounter, spends much of his book
Other Tongues Other Flesh expounding
the resonant cosmic power of the swastika,
which he sees clearly embossed in a boot
print left by Orthon. Williamson was a close
associate of William Dudley Pelley, founder
of the '30s American Silvershirts fascist
movement and later the occult group Soulcraft.
He also co-authored books placing the
Jewish banking conspiracy behind the
UFO mystery. While a survey of such breadth
of scope as Black Sun can’t be expected to
cover every lead, this seems like a rich
avenue for future exploration.
Other chapters cover neo-Nazi pagan/folk
movements, Christian identity and the Nazi
music scenes. G-C does a fine job of
exposing the swastika-emblazoned roots
of so much of today's New Age and
conspiracy movements as they indiscriminately
pull information towards the black holes (suns?)
at their centres. This is definitely a book to
In attempting to survey the entire post war
esoteric-Nazi oeuvre, G-C has set himself
a mammoth task, and one that could
probably have filled a book twice Black
Sun’s size. It's clear that G-C is more
comfortable in some areas, e.g. Evola,
Devi Landig, James Madole and Serrano,
who all get chapter-length treatments, than
he is in others, but the sheer weight of
information makes up for the lack of depth
in those sections - indeed, each chapter
could be expanded into a book of its own.
Black Sun is an extremely dense work that
can be dry reading at times. But as a survey
of this material - and currently the only one
of its kind - it is undeniably important and is
you never knew existed.
MARK PILKINGTON
Dense, disturbing and increasingly important
http://www.forteantimes.com/review/blacksun.shtml
~o0O0o~ [Close Quote]
*******************************************************
*******************************************************
I apologize to all those on this list
who are annoyed by that, and I will not email
anything further to you after this email.
Thanks... All the best! --MT-07/May/07"
http://pweb.netcom.com/~mthorn/stargate.htm
Jack Sarfatti
sarfatti@pacbell.net
"If we knew what it was we were doing, it would not be called research, would it?"
- Albert Einstein
http://www.authorhouse.com/BookStore/ItemDetail.aspx?bookid=23999
http://lifeboat.com/ex/bios.jack.sarfatti
http://qedcorp.com/APS/Dec122006.ppt
http://www.flickr.com/photos/lub/sets/72157594439814784
Note Brownian motion analogy
The basic "length" operator is L ~ N^1/2Lp
Lp separates the point defects where the 3 real Higgs scalar fields O(3) symmetry have simultaneous "nodes" and they are 3D spacelike separated from each other. The 2 Goldstone phases Theta & Phi are undefined at the nodes that are Goldstone phase singularities. The vacuum manifold has same topology S2 as does the surrounding surface enclosing N point nodes. Note time is not quantized here. That gives an extra real Higgs field taking us from 9+1 spacetime to Witten's 10+1 spacetime if I am not mistaken?
Note the Higgs fields here are real not complex so that U(1) is 2 real Higgs fields, SU(2) is 4 real Higgs fields. SU(3) is 6 real Higgs fields.
I am using O(N) for N real Higgs fields with N-1 independent relative Goldstone phases.
The criterion of maximal stable topological defect means that N real Higgs scalars must be defined on a N-Dim spacelike hypersurface.
Note for 3D hypersurface
N = 1 has S0 vacuum manifold with stable 2D wall defect in space
N = 2 has S1 vacuum manifold with stable 1D vortex string in space
N = 3 has S2 vacuum manifold with stable 0D point defect in space - nodes of "world crystal".
N > 3 has no stable topological defects in 3D space.
Go to imaginary time 4D Euclidean, then the N = 4 Higgs fields have "instanton" stable defect, Wick rotate back to light cones to get Finkelstein's discrete "chronons"?
Then use correspondence with statistical mechanics. 1/T ~ imaginary time, T is temperature etc.
Coherent Glauber states of a laser coherent field have ODLRO and they are Poisson in real photon statistics
root mean square real photon number fluctuation ~ square root of average number of photons
As distinct from thermal waves that Einstein got
square number fluctuation ~ a + b^2
used in Hanbury-Brown-Twiss effect for thermal light from stars.
b -> 0 above laser threshold.
Think of N as an average number of nodes in some pre-space in analogy with "photons" above.
^1/2 ~ [ - ^2]^1/2
RMS number fluctuation ~ mean number
So for each dimension of space it's like a 50-50 random walk. The amount of space generated is like a random walk.
Weight the probabilities differently to get warping? Different Turing programs?
So there seems to be an analogy here to at least a classical Turing machine with the weights as part of a simple program or algorithm?
IT FROM BIT.
The net displacement is that of the Turing tape?
On May 7, 2007, at 2:44 PM, Jack Sarfatti wrote:
On May 7, 2007, at 12:56 PM, Bruce Cornet wrote:
Jack,
Bruce
Update me on your UFO reverse Doppler observations. - very very important IMHO since I predicted it BEFORE I knew of your observations!
"Dark energy" zero point vacuum fluctuations of negative pressure makes an anomalous anti-gravity blue shift.
"Dark matter" zero point vacuum fluctuations of positive pressure makes an anomalous gravity red shift.
this warp bubble "moves" to right showing reverse Doppler.
I am amazed that physicists find this simple idea so hard to grasp even if they don't believe they seem to have a hard time grasping what the idea is. Partly it's because when it comes to "dark matter" they keep thinking of real particles moving through space that maybe LHC will detect. They do not think dark matter can be zero point energy, i.e. all virtual quanta inside the vacuum. They seem to find it hard to grasp that w = -1 positive pressure (equal and opposite isotropic negative energy density) will be indistinguishable from w = 0 Cold Dark Matter (CDM) particles as far as any distant observers e.g. gravity lensing can tell. So I say the Galactic Halos are spheres of positive pressure zero point energy. The NASA Pioneer data seems to show a hollow shell of positive pressure zero point energy beyond Saturn's orbit concentric with center of Sun as if Sun blew it off like a smoke ring from some physics process we haven't thought of as yet? Maybe the shell is actually expanding outward like a shock from an explosion?
This is an elementary consequence of Einstein's field equation for exotic vacua. See Math Appendix below.
You know that sort of thing is also implicit in the Vallee-Torme "sci-fi" Fastwalker book with the "5000 mile per hour Doppler shift" and the alien saucer standing still clamped on the ground inside the bunker. I actually quote that scene a bit in "Super Cosmos." I wonder where Vallee got that from? I suspect he had inside info via Bigelow's USG links?
From Fastwalker to Skinwalker - eh? ;-)
Where are you these days?
If you can write something brief I can quote you on that would fit in a Power Point slide I will include it. :-)
If you have some plots, photos I can add a few slides.
Your observations may be my Michelson-Morley experiment. :-)
Math Appendix on Emergent Einstein-Cartan Gravity-Torsion from the Higgs-Goldstone Vacuum Condensate ODLRO post-Inflation Field,
Guv + /\zpfguv = 0
where the zpf stress-energy tensor is
tuv(ZPF) ~ (c^4/G)/\zpfguv = (superstring tension)(curvature of vacuum)(metric tensor)
There will be extra torsion field terms so that /\zpf is no longer a strict constant as it is in 1915 GR, but is a locally variable quintessent scalar field.
That is we locally gauge entire 10-parameter Poincare group P(10) of 1905 globally rigid special relativity to get Einstein-Cartan curvature theory.
The spin connection 1-form for parallel transport is larger than Einstein had in 1915. It is
S^a^b(P10) = - S^b^a(P10) = S^a^b(P10)udx^u = S^a^b(T4) + S^a^b(O(1,3)
Einstein's 1915 GR only uses the piece S^a^b(T4) for which
Torsion dislocation defect field 2-form T^a(T4) = de^a + S^ac(T4)/\e^c = 0
T^a = T^auvdx^u/\dx^v
e^a are the tetrad co-frames (little detectors)
with disclination defect curvature 2-form -> Riemann-Christoffel curvature tensor of 4th rank
R^a^b(T4) = dS^a^b(T4) + S^ac(T4)/\S^c^b(T4) = R^a^buvdx^u/\dx^v
R^a^b(P10) has more terms in it allowing /\zpf,u =/= 0 and of course the torsion field T^a(P10) =/= 0 in general.
Also
e^a = I^a + @A^a
A^a ~ M^a^a (diagonal of 4x4 matrix of non-closed 1-forms from 8 Goldstone phases in the Higgs vacuum in 9+1 spacetime, the latter demanded by maximizing topologically stable defects - nontrivial homotopy -> Calabi-Yau space?
From Shipov's
to Witten's
@ = Lp^2/\zpf dimensionless coupling - renormalizable quantum gravity with spin 2 as secondary properties along with spin 1 and spin 0 in this larger curvature-torsion theory beyond 1916 GR, but only a wee bit beyond - a natural extension that seems to include superstring theory automatically.
A^a is spin 1 as a quantum field.
no gravity if /\zpf = 0 , if h = 0 and if c -> infinity - very important
S^a^b = M^[a,b] = - S^b^a
M^a^b = dTheta^a/\Phi^b - Theta^a/\dPhi^b = M^a^budx^u
Theta^a & Phi^b are the 8 Goldstone phase 0-forms
dM^a^b = dTheta^a/\dPhi^b = dM^a^buvdx^u/\dx^v
From
Theta^2 = Theta^bThetab
Phi^2 = Phi^aPhia
Projecting down from 9+1 space-time to 3+1 space-time gives 2 independent Goldstone phases with 3 real Higgs fields i.e. O(3) symmetry gives point defects of the projected vacuum ODLRO field whose point nodes is a kind of vacuum crystal lattice with
Area enclosing N nodes ~ NLp^2 i.e. World Hologram idea.
Hologram volume image is
Volume generated from N point defects ~ N^3/2Lp^3.
----- Original Message -----
From: Jack Sarfatti
Sent: Friday, May 04, 2007 12:10 PM
Subject: Re: STAIF Abstract Comments & Clarifications
10-4 :-)
See corrected 2nd draft of abstract at end below.
On May 4, 2007, at 6:17 AM, Paul Murad wrote:
Jack:
I have sent your abstract around to all of my reviewers and generally the attitude was that they look forward to what you have to say. If I receive any really adverse comments, rest assured I will relay these to you as well. ... I also have to agree that it would be interesting to hear what you have to say on this topic.
We are accepting the abstract with the conditions suggested below by one of my reviewers.� I also want you to understand up front that I look forward to your paper but it should be written in clear unambigous language and not really for your peers, which I am sure there are none, but to the newly minted graduate�engineer or scientist that wants to get involved in this field and understand some of the unusual mysteries and scientific marvels that are enmeshed in this topic so that they could make a future contribution in making us a space-faring civilization.
Yes, of course. I look forward to your suggestions for clarity. We have a year. :-)
I have no doubt that you can rise above these minor challenges and do it all in eight pages... If you have any problems with any of this, please get back to me and we can talk further...
We look forward to seeing you at STAIF 2008...
Paul...
Changes by one reviewer are as follows:
"I recommend that the abstract be accepted after the author makes the following changes or addresses concerns:
1) Removal of reference to Douglas MacArthur speech as this has no place in a scientific venue.� The sentence itself is of no relevance to the rest of the author's thesis and distracts from the main thrust of the thesis.
OK no problem, although for a DOD audience, it would be inspiring, but I will not insist on this point.
2) I am not against the inclusion of UFOs in a scientific venue as long as discussion or mention of it remains rational and scientific.� However, if any of you object to including mention of UFOs, then the author must remove the reference to UFOs.
OK, however, it is my position, and I think it is also Paul Murad's?, that the subset of good UFO observations by professional military and commercial pilots, e.g. NIDS website, Bruce Maccabee's website, old book by Paul Hill - eminent aeronautical engineer for USG, that we are actually seeing advanced propellantless propulsion of material vehicles in our skies able to execute high speed turns whose g-forces would kill any of the occupants if they existed. This "fact" of experience suggests to me, that we are simply playing catch up technologically speaking. Physicist Michio Kaku has made this quite clear in his popular talks when he classifies civilizations from Type 0, Type I up to Type IV. The sharp turns of these UFOs shows "metric engineering", i.e. technology able to neutralize the external ambient gravity field (e.g. Earth's) at the craft, to override it with its own shaped in such a way that the craft literally free-falls at zero g along a local geodesic path whose direction in space is controlled on board. Alcubierre's model is an example.�
Note Bruce Cornet's "reverse Dopper effect" (mentioned in fictional "Fastwalker" Vallee & Torme) - dark energy makes anomalous blue shift, whilst dark matter makes anomalous red shift in opposite way to ordinary motional Doppler shift.
Good trick if we can do it, we can't as yet, but someone Out There can and does over our nuclear bases according to some reports. So my point is that there is here a clear and present national security issue - the elephant in the room. My purpose in this talk is to define the mission, the broad parameters of the actual physics problem(s). I have no detailed solutions of course.� However, the discovery of both anti-gravitating "dark energy" and gravitating "dark matter" as 96% of the stuff of the universe is clearly a vital clue .As John A. Wheeler said
"The Question is: "What is The Question"?
3) The author must correct or replace his statement that dark matter and dark energy are "two sides of the same zero point energy coin."� This is patently false because dark matter is non-absorbing, non-luminous, non-baryonic matter that interacts with all other forms of matter via gravity and nuclear weak force.� It has positive energy density and positive pressure, and hence is not rooted in the quantum vacuum ZPE. Dark energy has been declared last month in an article by Riess et al. (in the Astrophysical Journal) to be the vacuum energy of Einstein's cosmological constant. Therefore, it has positive energy density and negative pressure, and it is rooted in the quantum vacuum ZPE.
Whoever wrote 3) seems to be confusing "dark energy" with "dark matter" and has misunderstood my message. His or her argument is not relevant to what I actually wrote and mean.
That dark matter is "non-absorbing, non-luminous, non-baryonic matter that interacts with all other forms of matter via gravity" is true and that is an immediate consequence of my original hypothesis that "dark matter" as negative zero point energy density with positive pressure with w = -1 that cannot be distinguished from w = 0 cold dark matter (CDM) from distant measurements like gravity lensing.
There is not one iota of evidence that dark matter interacts via the "weak nuclear force". Where is the evidence?
Whoever wrote 3) does not understand my point about Einstein's GR equation that the direct bending of spacetime by energy density and pressure (in the isotropic weak field Newtonian limit) is, from the text books,
~ 4pi(G/c^2)(energy density)(1 + 3w)
where w = -1 for zero point energy
w ~ 0 for cold ordinary "on shell" matter.
http://www-conf.slac.stanford.edu/ssi/2005/lec_notes/Kolb1/kolb1new_Page_05_jpg.htm
The writer of 3) has misunderstood me because
"Dark energy has been declared last month in an article by Riess et al. (in the Astrophysical Journal) to be the vacuum energy of Einstein's cosmological constant. Therefore, it has positive energy density and negative pressure, and it is rooted in the quantum vacuum ZPE."�
is also what I have said precisely. That sentence is cited as if I said something else. Indeed my abstract says
Both dark energy and dark matter are two sides of the same zero
point energy coin of negative and positive pressure respectively.
Hence, no disagreement there.
Let me say it again, as obviously I did not make it clear enough to the writer of 3)
I. Dark energy is positive zero point energy density with equal and opposite negative pressure in each linearly independent space direction. It antigravitates. The pressure overrides the energy density by a factor of 3. It antigravitates according to Einstein's theory.�This is ~ 73% of universe's "stuff" at spread-out large scale
II. Dark matter is negative zero point energy density with equal and opposite positive pressure etc. It gravitates according to Einstein's theory,
This is ~ 23% of universe at smaller "clumped" scales of galactic halos, voids et-al - possibly NASA Pioneer anomaly of an extra gravity pull back to Sun.
I suggest that the author replace this sentence with one that specifically elucidates his new theory of dark matter and dark energy that supports his weird statement.
My statement is not so much "weird" as "new, original, surprising" - a virtue rather than a vice.
Also, you cannot capture dark energy and use it for warp drive or wormhole FTL propulsion.�
Red Herring, where did I write one could capture dark energy?
But you can artificially make dark energy in the lab and exploit that for FTL propulsion.
Well, that does not contradict anything I said. If this author knows how to do that, then I hope he or she will tell us how.
4) The author needs to replace "Type IIa" with "Type Ia" in the second sentence.
Yes, thank you, that was a typo - I am used to Type II superconductors. :-)
5) There are two major factual errors about traversable wormholes and any connection to dark energy in the author's abstract.� You cannot technologically "capture" cosmological dark energy and use it for warp drive or wormhole FTL propulsion.�
Again, I am at a loss at what I wrote that gave this author that impression? I never used the word "capture". Has the author confounded my abstract with another?
But you can artificially make dark energy in the lab and exploit that for FTL propulsion.�
Again that last sentence is what I would like to see and if the author has ideas on how to do it, I would like to know what he or she knows.
So the author should recast his sentence to indicate this fact. However, the F-session is designed to allow for "way-outside-the-box" ideas in order to foster new thinking and open new avenues that might challenge the present paradigm.
6) The rest of the author's thesis is very interesting and of relevance to the F- session."
Salutation: Dr.
Author: Jack Sarfatti
Company: ISEP
Suite 206
City: San Francisco
State: CA
Zip: 94133
Country: USA
Phone: Fax: 415 989 0649
Email: sarfatti@pacbell.net
select: F01. Opening Session
Second Draft
PaperTitle: Harnessing dark energy for metric engineering warpdrive and
wormhole.
AuthorAffiliation: Abstract:�
Repulsive anti-gravitating "dark energy" was
discovered from the spectra of Type Ia supernovae in 1999. No one
anticipated this cosmic energy accelerating the expansion of 3D space. No
one expected that it is approximately 73% of all the "stuff" of the universe
on the large scale. Evidence for equally puzzling attracting "dark matter"
on smaller scales like the galactic halo had been accumulating for decades.
Both phenomena are easily understood in a unified way using only Einstein's
general relativity and basic quantum field theory. No dramatically new ideas
are needed. Both dark energy and dark matter are two sides of the same zero
point energy coin of negative and positive pressure respectively.
Configure the two in the proper configuration, as shown by Bondi,
Tereletsky, Forward and finally Alcubierre, and we have a geodesic warp
drive with no g-forces felt by the crew in sharp turns relative to external
observers watching the amazing maneuvers of "UFOs." This same dark energy is
what is needed for stable traversable wormhole "star gates" possibly connecting Earth�
to the recently discovered Earthlike planet only 20 light years away orbiting a red dwarf every 2 weeks.
Progress in the control of zero point energy induced (anti) gravity emergent from the
post-inflation Higgs-Goldstone fields familiar in elementary particle physics will be discussed.
Jack Sarfatti
sarfatti@pacbell.net
"If we knew what it was we were doing, it would not be called research, would it?"
- Albert Einstein
http://www.authorhouse.com/BookStore/ItemDetail.aspx?bookid=23999
http://lifeboat.com/ex/bios.jack.sarfatti
http://qedcorp.com/APS/Dec122006.ppt
http://www.flickr.com/photos/lub/sets/72157594439814784
10-4 :-)
See corrected 2nd draft of abstract at end below.
On May 4, 2007, at 6:17 AM, Paul Murad wrote:
> Jack:
>
> I have sent your abstract around to all of my reviewers and
> generally the attitude was that they look forward to what you have
> will relay these to you as well. ... I also have to agree that it
> would be interesting to hear what you have to say on this topic.
>
> We are accepting the abstract with the conditions suggested below
> by one of my reviewers. I also want you to understand up front
> that I look forward to your paper but it should be written in clear
> unambigous language and not really for your peers, which I am sure
> there are none, but to the newly minted graduate engineer or
> scientist that wants to get involved in this field and understand
> some of the unusual mysteries and scientific marvels that are
> enmeshed in this topic so that they could make a future
> contribution in making us a space-faring civilization.
Yes, of course. I look forward to your suggestions for clarity. We
have a year. :-)
>
> I have no doubt that you can rise above these minor challenges and
> do it all in eight pages... If you have any problems with any of
> this, please get back to me and we can talk further...
>
> We look forward to seeing you at STAIF 2008...
> Paul...
>
> Changes by one reviewer are as follows:
>
> "I recommend that the abstract be accepted after the author makes
> the following changes or addresses concerns:
>
> 1) Removal of reference to Douglas MacArthur speech as this has no
> place in a scientific venue. The sentence itself is of no
> relevance to the rest of the author's thesis and distracts from the
> main thrust of the thesis.
OK no problem, although for a DOD audience, it would be inspiring,
but I will not insist on this point.
>
> 2) I am not against the inclusion of UFOs in a scientific venue as
> long as discussion or mention of it remains rational and
> scientific. However, if any of you object to including mention of
> UFOs, then the author must remove the reference to UFOs.
OK, however, it is my position, and I think it is also Paul Murad's?,
that the subset of good UFO observations by professional military and
commercial pilots, e.g. NIDS website, Bruce Maccabee's website, old
book by Paul Hill - eminent aeronautical engineer for USG, that we
are actually seeing advanced propellantless propulsion of material
vehicles in our skies able to execute high speed turns whose g-forces
would kill any of the occupants if they existed. This "fact" of
experience suggests to me, that we are simply playing catch up
technologically speaking. Physicist Michio Kaku has made this quite
clear in his popular talks when he classifies civilizations from Type
0, Type I up to Type IV. The sharp turns of these UFOs shows "metric
engineering", i.e. technology able to neutralize the external ambient
gravity field (e.g. Earth's) at the craft, to override it with its
own shaped in such a way that the craft literally free-falls at zero
g along a local geodesic path whose direction in space is controlled
on board. Alcubierre's model is an example.
Note Bruce Cornet's "reverse Dopper effect" (mentioned in fictional
"Fastwalker" Vallee & Torme) - dark energy makes anomalous blue
shift, whilst dark matter makes anomalous red shift in opposite way
to ordinary motional Doppler shift.
Good trick if we can do it, we can't as yet, but someone Out There
can and does over our nuclear bases according to some reports. So my
point is that there is here a clear and present national security
issue - the elephant in the room. My purpose in this talk is to
define the mission, the broad parameters of the actual physics problem
(s). I have no detailed solutions of course. However, the discovery
of both anti-gravitating "dark energy" and gravitating "dark matter"
as 96% of the stuff of the universe is clearly a vital clue .As John
A. Wheeler said
"The Question is: "What is The Question"?
>
> 3) The author must correct or replace his statement that dark
> matter and dark energy are "two sides of the same zero point energy
> coin." This is patently false because dark matter is non-
> absorbing, non-luminous, non-baryonic matter that interacts with
> all other forms of matter via gravity and nuclear weak force. It
> has positive energy density and positive pressure, and hence is not
> rooted in the quantum vacuum ZPE. Dark energy has been declared
> last month in an article by Riess et al. (in the Astrophysical
> Journal) to be the vacuum energy of Einstein's cosmological
> constant. Therefore, it has positive energy density and negative
> pressure, and it is rooted in the quantum vacuum ZPE.
Whoever wrote 3) seems to be confusing "dark energy" with "dark
matter" and has misunderstood my message. His or her argument is not
relevant to what I actually wrote and mean.
That dark matter is "non-absorbing, non-luminous, non-baryonic matter
that interacts with all other forms of matter via gravity" is true
and that is an immediate consequence of my original hypothesis that
"dark matter" as negative zero point energy density with positive
pressure with w = -1 that cannot be distinguished from w = 0 cold
dark matter (CDM) from distant measurements like gravity lensing.
There is not one iota of evidence that dark matter interacts via the
"weak nuclear force". Where is the evidence?
Whoever wrote 3) does not understand my point about Einstein's GR
equation that the direct bending of spacetime by energy density and
pressure (in the isotropic weak field Newtonian limit) is, from the
text books,
~ 4pi(G/c^2)(energy density)(1 + 3w)
where w = -1 for zero point energy
w ~ 0 for cold ordinary "on shell" matter.
http://www-conf.slac.stanford.edu/ssi/2005/lec_notes/Kolb1/
kolb1new_Page_05_jpg.htm
The writer of 3) has misunderstood me because
"Dark energy has been declared last month in an article by Riess et
al. (in the Astrophysical Journal) to be the vacuum energy of
Einstein's cosmological constant. Therefore, it has positive energy
density and negative pressure, and it is rooted in the quantum vacuum
ZPE."
is also what I have said precisely. That sentence is cited as if I
said something else. Indeed my abstract says
> Both dark energy and dark matter are two sides of the same zero
> point energy coin of negative and positive pressure respectively.
Hence, no disagreement there.
Let me say it again, as obviously I did not make it clear enough to
the writer of 3)
I. Dark energy is positive zero point energy density with equal and
opposite negative pressure in each linearly independent space
direction. It antigravitates. The pressure overrides the energy
density by a factor of 3. It antigravitates according to Einstein's
theory. This is ~ 73% of universe's "stuff" at spread-out large scale
II. Dark matter is negative zero point energy density with equal and
opposite positive pressure etc. It gravitates according to Einstein's
theory,
This is ~ 23% of universe at smaller "clumped" scales of galactic
halos, voids et-al - possibly NASA Pioneer anomaly of an extra
gravity pull back to Sun.
>
> I suggest that the author replace this sentence with one that
> specifically elucidates his new theory of dark matter and dark
> energy that supports his weird statement.
My statement is not so much "weird" as "new, original, surprising" -
a virtue rather than a vice.
> Also, you cannot capture dark energy and use it for warp drive or
> wormhole FTL propulsion.
Red Herring, where did I write one could capture dark energy?
> But you can artificially make dark energy in the lab and exploit
> that for FTL propulsion.
Well, that does not contradict anything I said. If this author knows
how to do that, then I hope he or she will tell us how.
>
> 4) The author needs to replace "Type IIa" with "Type Ia" in the
> second sentence.
Yes, thank you, that was a typo - I am used to Type II
superconductors. :-)
>
> 5) There are two major factual errors about traversable wormholes
> and any connection to dark energy in the author's abstract. You
> cannot technologically "capture" cosmological dark energy and use
> it for warp drive or wormhole FTL propulsion.
Again, I am at a loss at what I wrote that gave this author that
impression? I never used the word "capture". Has the author
confounded my abstract with another?
> But you can artificially make dark energy in the lab and exploit
> that for FTL propulsion.
Again that last sentence is what I would like to see and if the
author has ideas on how to do it, I would like to know what he or she
knows.
> So the author should recast his sentence to indicate this fact.
> However, the F-session is designed to allow for "way-outside-the-
> box" ideas in order to foster new thinking and open new avenues
> that might challenge the present paradigm.
>
> 6) The rest of the author's thesis is very interesting and of
> relevance to the F- session."
>
>
>
>
> Salutation: Dr.
> Author: Jack Sarfatti
> Company: ISEP
> Suite 206
> City: San Francisco
> State: CA
> Zip: 94133
> Country: USA
> Phone: Fax: 415 989 0649
> Email: sarfatti@pacbell.net
> select: F01. Opening Session
Second Draft
>
> PaperTitle: Harnessing dark energy for metric engineering warpdrive
> and
> wormhole.
>
> AuthorAffiliation: Abstract:
> Repulsive anti-gravitating "dark energy" was
> discovered from the spectra of Type Ia supernovae in 1999. No one
> anticipated this cosmic energy accelerating the expansion of 3D
> space. No
> one expected that it is approximately 73% of all the "stuff" of the
> universe
> on the large scale. Evidence for equally puzzling attracting "dark
> matter"
> on smaller scales like the galactic halo had been accumulating for
> Both phenomena are easily understood in a unified way using only
> Einstein's
> general relativity and basic quantum field theory. No dramatically
> new ideas
> are needed. Both dark energy and dark matter are two sides of the
> same zero
> point energy coin of negative and positive pressure respectively.
>
> Configure the two in the proper configuration, as shown by Bondi,
> Tereletsky, Forward and finally Alcubierre, and we have a geodesic
> warp
> drive with no g-forces felt by the crew in sharp turns relative to
> external
> observers watching the amazing maneuvers of "UFOs." This same dark
> energy is
> what is needed for stable traversable wormhole "star gates"
> possibly connecting Earth
> to the recently discovered Earthlike planet only 20 light years
> away orbiting a red dwarf every 2 weeks.
> Progress in the control of zero point energy induced (anti) gravity
> emergent from the
> post-inflation Higgs-Goldstone fields familiar in elementary
> particle physics will be discussed.
>
Jack Sarfatti
sarfatti@pacbell.net
"If we knew what it was we were doing, it would not be called research, would it?"
- Albert Einstein
http://www.authorhouse.com/BookStore/ItemDetail.aspx?bookid=23999
http://lifeboat.com/ex/bios.jack.sarfatti
http://qedcorp.com/APS/Dec122006.ppt | 23,851 | 96,113 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2017-09 | longest | en | 0.683168 |
https://www.jiskha.com/display.cgi?id=1221340188 | 1,502,991,858,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886103891.56/warc/CC-MAIN-20170817170613-20170817190613-00265.warc.gz | 932,162,987 | 4,279 | # Statistics
posted by .
Help me understand this problem, I have several more like this & want to get an understanding on to solve it:
The mean score,x, on an aptitude test for a random sample of 9 students was 64. Assuming that standard deviation=16, construct a 95.44% confidence interval for the mean score of all students taking the test.
A. 53.3 to 74.7 B. 60.4 to 67.6 C. 32 to 96 D. 56.0 to 72.0
• Statistics -
The standard error of the mean (SEM) is the standard deviation divided by the square root of the sample size - which in this instance is 9 - so the SEM here is 16/sqrt(9)=5.33.
The question actually contains a little clue that you're probably on the right track in that very specific figure of 95.44% for a confidence interval - because if you look up that figure in a set of Normal probability tables, you should find that it corresponds to a very convenient Z value.
Your confidence interval will then range from (M - 5.33*Z) to (M + 5.33*Z).
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#### Resources tagged with Quadratic equations similar to Finding Factors:
Filter by: Content type:
Stage:
Challenge level:
### Two Cubes
##### Stage: 4 Challenge Level:
Two cubes, each with integral side lengths, have a combined volume equal to the total of the lengths of their edges. How big are the cubes? [If you find a result by 'trial and error' you'll need to. . . .
### Square Mean
##### Stage: 4 Challenge Level:
Is the mean of the squares of two numbers greater than, or less than, the square of their means?
### Always Two
##### Stage: 4 and 5 Challenge Level:
Find all the triples of numbers a, b, c such that each one of them plus the product of the other two is always 2.
### Proof Sorter - Quadratic Equation
##### Stage: 4 and 5 Challenge Level:
This is an interactivity in which you have to sort the steps in the completion of the square into the correct order to prove the formula for the solutions of quadratic equations.
### How Old Am I?
##### Stage: 4 Challenge Level:
In 15 years' time my age will be the square of my age 15 years ago. Can you work out my age, and when I had other special birthdays?
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##### Stage: 4 Challenge Level:
Triangle ABC is an equilateral triangle with three parallel lines going through the vertices. Calculate the length of the sides of the triangle if the perpendicular distances between the parallel. . . .
### Pent
##### Stage: 4 and 5 Challenge Level:
The diagram shows a regular pentagon with sides of unit length. Find all the angles in the diagram. Prove that the quadrilateral shown in red is a rhombus.
### Interactive Number Patterns
##### Stage: 4 Challenge Level:
How good are you at finding the formula for a number pattern ?
### Partly Circles
##### Stage: 4 Challenge Level:
What is the same and what is different about these circle questions? What connections can you make?
### Xtra
##### Stage: 4 and 5 Challenge Level:
Find the sides of an equilateral triangle ABC where a trapezium BCPQ is drawn with BP=CQ=2 , PQ=1 and AP+AQ=sqrt7 . Note: there are 2 possible interpretations.
### Pentakite
##### Stage: 4 and 5 Challenge Level:
ABCDE is a regular pentagon of side length one unit. BC produced meets ED produced at F. Show that triangle CDF is congruent to triangle EDB. Find the length of BE.
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##### Stage: 4 Challenge Level:
Rectangle PQRS has X and Y on the edges. Triangles PQY, YRX and XSP have equal areas. Prove X and Y divide the sides of PQRS in the golden ratio.
### Darts and Kites
##### Stage: 4 Challenge Level:
Explore the geometry of these dart and kite shapes! | 626 | 2,640 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2017-39 | latest | en | 0.876423 |
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posted by on .
A person with perfect pitch sits on a bus bench listening to the 444 Hz horn of an approaching car. If the person detects a frequency of 480 Hz, how fast is the car moving?
• physics - ,
F = ((V+Vr)/(V-Vs))*Fs = 480 Hz.
((343+0)/(343-Vs))*444 = 480
((343)/(343-Vs))*444 = 480
152292/(343-Vs) = 480
Cross multiply:
164640-480Vs = 152,292
-480Vs = 152,292-164,640 = -12,348
Vs = 25.725 m/s.
• physics - ,
F = Frequency as heard by receiver(man).
V = Velocity of sound in air.
Vr = Velocity of the receiver(man).
Vs = Velocity of source(car).
Fs = Frequency of source.
### Related Questions
More Related Questions
Post a New Question | 203 | 664 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-22 | latest | en | 0.873851 |
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How to calculate cost of goods sold
if:
Inventory 1st January = \$19,400
Inventory 31st December = \$26,660
Sales for the year = \$155,880
who do you calculate the cost of goods sold?
jerileycpa Posts: 27, Reputation: 2 Certified Public Accountant #2 Jun 1, 2015, 03:37 PM
Cost of goods sold is calculated as follows:
Beginning inventory + purchases - ending inventory = COGS
Your sales for the year is not a factor.
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el3um4s.itch.io | 1,606,304,541,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141182776.11/warc/CC-MAIN-20201125100409-20201125130409-00431.warc.gz | 271,509,138 | 8,536 | A different dice game. Each throw calls for a new decision. From an idea by Sid Sackson.
# Rules
### Two Combinations and One Reject
The 5 dice are thrown and the player the divides the 5 numbers into two "combinations" of two dice each and one "reject". The two combinations are marked in the top columns and the rejects is marked in the bottom.
For example
The player throws a 1, 1, 2, 4 and 5.
One of the many ways in which can be divided is by combining a 1 and the 2 to make a total of 3, combining the 4 and the 5 to make a total of 9 and leaving a 1 as the reject.
### Only 3 Rejects
The player continuing in this manner, throwing the 5 dice and dividing them as he choose. He, however, is limited to 3 different reject numbers, and the player must divide the dice so as to have one of those 3 numbers as the reject.
If, after the reject numbers have been set, the player make a throw containing none of the three numbers, he gets a "free ride". He can choose any two combinations he wishes and ignore the fifth die.
For example
1,2 and 5 have been marked as the rejects. A throw made 1, 4, 4, 4 and 6. The player has no choice but to divide the dice 4 and 4, 4 and 6, with 1 as the reject.
### The Game Over
The player continues throwing the dice until one of the three reject numbers has 8 marks placed next to it. Play is then completed and the game is scored.
# TO DO
• Credits page
• Sound (and background music?)
• In game help
• Classic mode?
• Desktop or mobile version?
Status In development Platforms HTML5 Author Strani Anelli Genre Puzzle, Strategy Made with Construct Tags Abstract, Dice, Math, Minimalist, Short, Singleplayer Average session A few minutes Inputs Mouse, Touchscreen | 432 | 1,720 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2020-50 | latest | en | 0.944566 |
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posted by on .
Determine the percentage composition of each of the following compounds
a) NaCl
b) AgNO3
c) Mg(OH)2
for NaCl i got na=23/135=0.17 x 100= 17% Cl=35/135=0.26 x 100=26%
for AgNO3 i got 108/42 which im sure was wrong and i didn't get how to complete the problem..
for Mg(OH)2 i didn't have a clue
• chemistry - ,
In a), where did the 135 come from?
In b), where did the 42 come from? | 140 | 412 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2017-22 | latest | en | 0.936904 |
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# Tennessee: 9/1 - 9/30/2010
Topic closed. 195 replies. Last post 6 years ago by mglsr.
Page 3 of 14
Pittsburg, Ks
United States
Member #3382
January 17, 2004
80881 Posts
Online
Posted: September 6, 2010, 9:48 am - IP Logged
9/6/10 M/E
456, 457, 458, 459, 467, 468, 469, 478, 479, 489, 567, 568, 569, 578, 579, 589, 678, 679, 689, 789, 012, 013, 014, 015, 016, 017, 018, 019, 023, 024, 025, 026, 027, 028, 029, 034, 035, 036, 037, 038, 039, 045, 046, 047, 048, 049, 056, 057, 058, 059
God Bless America
jackson,tn.
United States
Member #70397
February 1, 2009
41 Posts
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Posted: September 6, 2010, 10:53 am - IP Logged
QP's MIDDAY: 383, 622= Anyone has the answer?????, I keep losing my shirt....
Clarksville
United States
Member #487
July 15, 2002
17638 Posts
Offline
Posted: September 6, 2010, 11:50 am - IP Logged
I don't play QP's too much. I can't read them. You have the 9 digit out 11 and the 1 digit out 10
These are your due digits 9,1,5,8,7,0,2.
3-5 vtrac pair out 22 (pairs 24,29,74,79)
Due no match pair is 4-7 out 72.
Due root sums:
6-out 31
3-out 12
5-out 11
1-out 9
7-out 6
9-out 4
They are due:
wild match
Consecutive
1E-2O
Double pairs to watch for are 55, 11,77.
If you know your number is going to hit, have patience and then KILL IT!
You never know when you will get another hit.
San Diego, CA
United States
Member #61467
May 24, 2008
28146 Posts
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Posted: September 6, 2010, 12:01 pm - IP Logged
9/6
Midday
346
Evening
Key Digit: 8
fayetteville tn
United States
Member #68983
January 3, 2009
1829 Posts
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Posted: September 6, 2010, 12:30 pm - IP Logged
Quick picks for midday 097,686,393,800,256,635,219
Quick picks for eve. 389,293,490,279,068,550,647
All straight. Good luck TN players and happy labor day to you all.
Tn
United States
Member #54963
September 4, 2007
1164 Posts
Online
Posted: September 6, 2010, 2:50 pm - IP Logged
Tn Mid 09/06/2010
433
WHEN IT FEELS THE WHOLE WORLD SUCKS!
RELAX.........IT'S ONLY GRAVITY
I think I can I think I can!!!!
Nassau
Bahamas
Member #29884
January 8, 2006
1099 Posts
Offline
Posted: September 6, 2010, 6:07 pm - IP Logged
640 641 410 010 000 111 222 333 444 555 666 777 888 999
Clarksville
United States
Member #487
July 15, 2002
17638 Posts
Offline
Posted: September 6, 2010, 7:32 pm - IP Logged
TN eve 172 P4 4653
If you know your number is going to hit, have patience and then KILL IT!
You never know when you will get another hit.
Tennessee
United States
Member #2369
September 23, 2003
3072 Posts
Offline
Posted: September 7, 2010, 8:52 am - IP Logged
Mid puzzle:
38 05 16
49 16 38
05 16 38
Eve puzzle:
16 27 38
49 16 38
27 49 05
"As Quoted by the Genius One:"
"Life i\$ Good "
\$\$\$\$ I FEEL LUCKY \$\$\$\$
*lay off and it pays off*
Morgan, TN
United States
Member #83248
December 2, 2009
1097 Posts
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Posted: September 7, 2010, 12:39 pm - IP Logged
pcurtis what are they doin with the pick 5 game???
\$\$\$ You Gotta Lose Some To Win Some! \$\$\$
fayetteville tn
United States
Member #68983
January 3, 2009
1829 Posts
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Posted: September 7, 2010, 1:08 pm - IP Logged
I have no clue Blackdave. I don't even know if they are going to replace it with another game.
There is noting on their website about it.
Tennessee
United States
Member #2369
September 23, 2003
3072 Posts
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Posted: September 7, 2010, 1:48 pm - IP Logged
Mid puzzle:
38 05 16
49 16 38
05 16 38
Eve puzzle:
16 27 38
49 16 38
27 49 05
Mid 018 0868
"As Quoted by the Genius One:"
"Life i\$ Good "
\$\$\$\$ I FEEL LUCKY \$\$\$\$
*lay off and it pays off*
Morgan, TN
United States
Member #83248
December 2, 2009
1097 Posts
Offline
Posted: September 7, 2010, 3:21 pm - IP Logged
I have no clue Blackdave. I don't even know if they are going to replace it with another game.
There is noting on their website about it.
Hmm man i hope they replace it with something...i ask the store clerk were i play my numbers and he said that they are goin to change it so ifigured they were probly was gunna change the number pool or add a extra number like powerball or something heck idk..
\$\$\$ You Gotta Lose Some To Win Some! \$\$\$
fayetteville tn
United States
Member #68983
January 3, 2009
1829 Posts
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Posted: September 7, 2010, 6:49 pm - IP Logged
Hmm man i hope they replace it with something...i ask the store clerk were i play my numbers and he said that they are goin to change it so ifigured they were probly was gunna change the number pool or add a extra number like powerball or something heck idk..
I figure maybe a pick 6 with 44 numbers to pick from.
Playing 243,724 and 868.
Clarksville
United States
Member #487
July 15, 2002
17638 Posts
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Posted: September 7, 2010, 11:28 pm - IP Logged
That 228 was straight in the puzzle. Way to go Mike! If they are changing the cash 5 it is because they want to make more money.
If you know your number is going to hit, have patience and then KILL IT!
You never know when you will get another hit.
Page 3 of 14 | 1,861 | 5,160 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | latest | en | 0.698784 |
https://math.stackexchange.com/questions/1412755/how-to-remember-these-probability-results?noredirect=1 | 1,571,652,712,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987769323.92/warc/CC-MAIN-20191021093533-20191021121033-00460.warc.gz | 588,135,030 | 32,226 | # How to remember these probability results?
If $A,B$ and $C$ are $3$ events, then
1. $P$(Exactly one of $A,B,C$ occurs)$=P(A)+P(B)+P(C)-2[P(A \cap B)+P(B \cap C)+P(A \cap C)]+3P(A \cap B \cap C)$
2. $P$(Exactly two of $A,B,C$ occur)$=P(A \cap B)+P(B \cap C)+P(A \cap C)-3P(A \cap B \cap C)$
3. $P$(At least two of $A,B,C$ occur)$=P(A \cap B)+P(B \cap C)+P(A \cap C)-2P(A \cap B \cap C)$
• Here is how I remember them: math.stackexchange.com/questions/151635/… – user940 Aug 28 '15 at 16:39
• yeah, should be $-2P(A\cap B\cap C)$ – Giovanni Aug 28 '15 at 16:50
• @Henry Thankx Done editing. – Jack Aug 28 '15 at 16:53
• @Giovanni Thankx – Jack Aug 28 '15 at 16:53
• What you have labelled as $A\cap B$ should really be $A\cap B\cap \bar{C}$ and similarly for the two other regions – David Quinn Aug 28 '15 at 17:26
• Hi David, I am sorry but I really don't understand your comment. I don't think that the closure of a set has any meaning in this context, so I guess you meant the complement of $C$. If that is the case I am pretty sure that whoever did this (I took it from wikipedia) wanted the whole intersection of $A$ and $B$ there.Of course it includes the part where they all intersect, but this is fairly obvious :) Please, let me know if I misunderstood your comment – Giovanni Aug 28 '15 at 17:33
• Yes I mean the complement of C. $A\cap B$ should include $A\cap B\cap C$ – David Quinn Aug 28 '15 at 17:36 | 489 | 1,420 | {"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} | 3.96875 | 4 | CC-MAIN-2019-43 | latest | en | 0.881563 |
https://math.stackexchange.com/questions/2763199/theorem-8-4-in-matsumuras-commutative-algebra | 1,558,258,053,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232254731.5/warc/CC-MAIN-20190519081519-20190519103519-00354.warc.gz | 555,454,017 | 33,519 | # Theorem 8.4 in Matsumura's Commutative Algebra
I'm having trouble understanding the proof of theorem 8.4 in Matsumura's Commutative Algebra. It goes as follows:
Let $A$ be a ring, $I$ an ideal and $M$ an $A$-module. Suppose that $A$ is $I$-adically complete, and $M$ is separated for the $I$-adic topology. If $M/IM$ is generated over $A/I$ by $\bar{\omega}_1,\ldots \bar{\omega}_n$ and $\omega_i\in M$ is the arbitrary inverse image of $\bar{\omega}_i$, then $M$ is generated over $A$ by $\omega_1,\ldots,\omega_n$.
Now I'm having trouble understanding what it means to be "... generated over ..." some ring. In the proof he says, that by assumption we can write $M=\sum A\omega_i+IM$ - is that what he means by "If $M/IM$ is generated over $A/I$ by $\bar{\omega}_1,\ldots \bar{\omega}_n$ and $\omega_i\in M$ is the arbitrary inverse image of $\bar{\omega}_i$"? And if so, why is that the case, that we can write $M=\sum A\omega_i+IM$?
He then chooses $\xi\in M$, and then successively construct $\xi_v=\sum a_{i,v}\omega_i+\xi_{v+1}$ for $v=1,2,\ldots$ where $\xi_v\in I^vM$ and $a_{i,v}\in I^v$. Then he states that $b_i:=\sum a_{i,k}$ converges in $A$ (I guess that's because $A$ is $I$-adically complete, so $\{b_n\}_{n>0}$ is a Cauchy sequence and therefore converges), and so he says we have $\xi-\sum_{i}^n b_i\omega_i\in \cap_{v>0}I^vM$? Why is that the case?
Also, I see that this would imply that $\omega_1,\ldots,\omega_n$ generates $M$, but does that also mean they generate $M$ over $A$?
The hypothesis is that $\;M/IM=\bigl\langle\mkern1mu\overline\omega_1,\dots,\overline\omega_n\mkern1mu\bigr\rangle$ for congruence classes $\,\overline\omega_i,\;(1\le i\le n)$ as an $A/I$-module. So, by definition of congruence classes mod. $IM$, $$M=\bigl\langle\omega_1\mkern1mu,\dots,\omega_n\mkern1mu\bigr\rangle +IM$$ This is just because $H, K$ are subgroups in a group $G$ (in additive notation), the group generated by the image of $K$ in the quotient $G/H$ is no other than $$K+(G/H)=(K+H)/H.$$ As to your other question, we have $$\xi-\sum_{i}^n b_i\omega_i=\xi-\sum_{i=1}^n \Bigl( \sum_{k\ge1} a_{i,k}\Bigr)\omega_i=\lim_{v\to\infty}\biggl(\xi-\sum_{i=1}^n \Bigl( \sum_{k=1}^{v-1} a_{i,k}\Bigr)\omega_i\biggr)=\lim_{v\to\infty}\xi_v.$$ Each $\xi_v$ lies in $I^vM$, and so does the limit since $I^vM$ is a closed subgroup of $M$.
• Thanks! Does the last equality follows from the fact that $lim_{v\rightarrow\infty} \xi_v=\xi$ (which is true by the succesive construction of $\xi_v$)? – jta May 2 '18 at 16:24
• More exactly, $\xi$ is the sum of the series $\sum_v\xi_v$, if I'm not mistaken. – Bernard May 2 '18 at 16:29
• Just to be sure, I understand it correctly, are these $\bar{\omega}$ obtained by the map $\phi:M\rightarrow M/IM$?, ie. $\phi(\omega)=\bar{\omega}$? – jta May 3 '18 at 14:06 | 1,000 | 2,818 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2019-22 | latest | en | 0.868972 |
https://www.physicsforums.com/threads/how-to-get-the-closed-form-of-this-recurrence.583206/ | 1,725,812,264,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651013.49/warc/CC-MAIN-20240908150334-20240908180334-00042.warc.gz | 907,275,038 | 17,537 | # How to get the closed form of this recurrence?
• zeion
In summary, the conversation revolved around finding a closed form for the expression T(k, n) = T(k, n-1) + (k-2)(n-1) + 1, with repeated substitution. The person was trying to find a pattern and determine the boundary condition, and eventually came up with the solution T(k,n) = n+c+(k-2)*1*2*...(n-1).
zeion
## Homework Statement
Hello,
This expression was derived from a polygon word problem and I need to find a closed form for it with repeated substitution (I think).
T(k, n) = T(k, n-1) + (k-2)(n-1) + 1
## The Attempt at a Solution
Get a pattern like:
= T(k, n-2) + (k - 2)(2n - 3) - 2 subs.
= T(k, n-3) + (k - 2)(3n - 6) - 3 subs.
= T(k, n-4) + (k - 2)(4n - 10) -4 subs.
.
.
.
I'm not sure how to express the 1, 3, 6, 10 sequence in terms of number of substitutions.
What is the boundary coundition?
This is what I got so far.
Assuming T(k, 0) = c
T(k,n) = n+c+(k-2)*1*2*...(n-1)
Sorry I wrote the expression wrong; but I got the answer I think, thanks.
## What is a recurrence relation?
A recurrence relation is a mathematical equation that defines a sequence recursively, meaning that each term of the sequence is defined in terms of previous terms. This type of relation is commonly used in computer science and mathematics to model and solve problems.
## What is a closed form solution?
A closed form solution is a mathematical expression that gives an exact and finite solution to a problem, without the use of recursive definitions or limits. In other words, it is a formula that directly calculates the value of a sequence or function, without the need for iterative calculations.
## Why is it important to find the closed form of a recurrence relation?
Finding the closed form of a recurrence relation allows us to directly calculate the value of a sequence or function, rather than having to recursively calculate each term. This not only saves time and resources, but it also provides a more efficient and accurate solution to a problem.
## What strategies can be used to find the closed form of a recurrence relation?
Some common strategies for finding the closed form of a recurrence relation include the method of substitution, the method of iteration, and the method of generating functions. These methods involve manipulating the recurrence relation in various ways to simplify it and ultimately find a closed form solution.
## Are there any limitations to finding the closed form of a recurrence relation?
Yes, it is not always possible to find a closed form solution for a recurrence relation. Some recurrence relations are inherently recursive and cannot be expressed in closed form. Additionally, some relations may have closed form solutions, but they may be very complex and difficult to find.
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1K | 766 | 2,997 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-38 | latest | en | 0.941362 |
http://www.csulb.edu/~cwallis/artificialn/History.htm | 1,417,014,953,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416931007056.0/warc/CC-MAIN-20141125155647-00250-ip-10-235-23-156.ec2.internal.warc.gz | 453,587,123 | 5,736 | History of the Perceptron
The evolution of the artificial neuron has progressed through several stages. The roots of which, are firmly grounded within neurological work done primarily by Santiago Ramon y Cajal and Sir Charles Scott Sherrington . Ramon y Cajal was a prominent figure in the exploration of the structure of nervous tissue and showed that, despite their ability to communicate with each other, neurons were physically separated from other neurons. With a greater understanding of the basic elements of the brain, efforts were made to describe how these basic neurons could result in overt behaviors, to which William James was a prominent theoretical contributor.
Working from the beginnings of neuroscience, Warren McCulloch and Walter Pitts in their 1943 paper, "A Logical Calculus of Ideas Immanent in Nervous Activity," contended that neurons with a binary threshold activation function were analogous to first order logic sentences. The basic McCulloch and Pitts neuron looked something like the following:
The McCulloch-Pitts neuron worked by inputting either a 1 or 0 for each of the inputs, where 1 represented true and 0 false. Likewise, the threshold was given a real value, say 1, which would allow for a 0 or 1 output if the threshold was met or exceeded. Thus, in order to represent the “and” function, we set the threshold at 2.0 and come up with the following truth table:
Input x1 Input x2 Output 0 0 0 0 1 0 1 0 0 1 1 1
This table shows the basic “and” function such that, if x1 and x2 are both false, then the output of combining these two will also be false. Likewise, if x1 is true or equal to 1 and x2 is true or equal to 1, then the threshold of 2 will be met and the output will be 1.
This follows also for the “or” function, if we switch the threshold value to 1. The table for the “or” function being,
Input x1 Input x2 Output 0 0 0 0 1 1 1 0 1 1 1 1
This type of artificial neuron could also be used to solve the “not” function, which would have only one input, as well as, the NOR and NAND functions. The McCulloch-Pitts neuron, therefore, was very instrumental in progressing the artificial neuron, but it had some serious limitations. In particular, it could solve neither the “exclusive or” function (XOR), nor the “exclusive nor” function (XNOR). Limited to binary code, the following truth tables could not be accurately solved using this early artificial neuron.
XOR
Input x1 Input x2 Output 0 0 0 0 1 1 1 0 1 1 1 0
XNOR
Input x1 Input x2 Output 0 0 1 0 1 0 1 0 0 1 1 1
One of the difficulties with the McCulloch-Pitts neuron was its simplicity. It only allowed for binary inputs and outputs, it only used the threshold step activation function and it did not incorporate weighting the different inputs.
In 1949, Donald Hebb would help to revolutionize the way that artificial neurons were perceived. In his book, The Organization of Behavior, he proposed what has come to be known as Hebb’s rule. He states, “When an axon of cell A is near enough to excite a cell B and repeatedly or persistently takes part in firing it, some growth process or metabolic change takes place in one or both cells such that A’s efficiency, as one of the cells firing B, is increased.” [1] Hebb was proposing not only that, when two neurons fire together the connection between the neurons is strengthened, but also that this activity is one of the fundamental operations necessary for learning and memory.
For the artificial neuron, this meant that the McCulloch-Pitts neuron had to be altered to at least allow for this new biological proposal. The method used was to weight each of the inputs. Thus, an input of 1 may be given more or less weight, relative to the total threshold sum.
Frank Rosenblatt, using the McCulloch-Pitts neuron and the findings of Hebb, went on to develop the first perceptron. This perceptron, which could learn in the Hebbean sense, through the weighting of inputs, was instrumental in the later formation of neural networks. He discussed the perceptron in his 1962 book, Principles of Neurodynamics. A basic perceptron is represented as follows:
This perceptron has a total of five inputs a1 through a5 with each having a weight of w1 through w5. [2] Each of the inputs are weighted and summed at the node. If the threshold is reached, an output results. Of great importance is that each of the inputs may not be given equal weight. The perceptron may have “learned” to weight a1 more than a2 and so on.
The summation formula for determining whether or not the threshold (θ) is met for the artificial neuron with N inputs (a 1, a2…aN) and their respective weights of w 1, w2,…wN is:
N b = (∑w ja j) + θ j=1
The activation function then becomes:
x = f(b)
The activation function used by McCulloch and Pitts was the threshold step function. However, other functions that can be used are the Sigmoid, Piecewise Linear and Gaussian activation functions. These functions are shown below. [3] (See the glossary attached to this applet for the corresponding mathematical formulas.)
Threshold Step Sigmoid Piecewise Linear Gaussian
Despite the many changes made to the original McCulloch-Pitts neuron, the perceptron was still limited to solving certain functions. Unfortunately, Rosenblatt was overly enthusiastic about the perceptron and made the ill-timed proclamation that:
"Given an elementary α-perceptron, a stimulus world W, and any classification C(W) for which a solution exists; let all stimuli in W occur in any sequence, provided that each stimulus must reoccur in finite time; then beginning from an arbitrary initial state, an error correction procedure will always yield a solution to C(W) in finite time…” [4]
With these types of remarks Rosenblatt had drawn a line in the sand between those in support of perceptron styled research and the more traditional symbol manipulation projects being performed by Marvin Minsky . As a result, in 1969, Minsky co-authored with Seymour Papert , Perceptrons: An Introduction to Computational Geometry. In this work they attacked the limitations of the perceptron. They showed that the perceptron could only solve linearly separable functions. Of particular interest was the fact that the perceptron still could not solve the XOR and NXOR functions. Likewise, Minsky and Papert stated that the style of research being done on the perceptron was doomed to failure because of these limitations. This was, of course, Minsky’s equally ill-timed remark. As a result, very little research was done in the area until about the 1980’s.
What would come to resolve many of the difficulties was the creation of neural networks. These networks connect the inputs of artificial neurons with the outputs of other artificial neurons. As a result, the networks were able to solve more difficult problems, but they have grown considerably more complex. However, many of the artificial neural networks in use today still stem from the early advances of the McCulloch-Pitts neuron and the Rosenblatt perceptron.
[1] Hebb, Donald O. (1949). The Organization of Behavior. New York: Wiley, pg. 62.
[2] The diagram is from, http://www.neuroscience.com/Technologies/nn_history.htm
[3] Graph diagrams of the functions are from, http://home.cc.umanitoba.ca/~umcorbe9/neuron.html#Theory
[4] Rosenblatt, Frank (1962). Principles of neurodynamics. New York: Spartan. Cf. Rumelhart, D.E., J. L. McClelland and the PDP Research Group (1986). Parallel Distributed Processing vol. 1&2. Cambridge: MIT. | 1,796 | 7,580 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-49 | longest | en | 0.938706 |
https://homepages.laas.fr/dalzilio/courses/mccourse/tp5/ | 1,579,907,619,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250626449.79/warc/CC-MAIN-20200124221147-20200125010147-00439.warc.gz | 468,580,654 | 5,760 | # TP5: Linear Temporal Logic
## Exercise 1. LTL semantics on a given trace
The diagram below describe a maximal execution (an infinite trace) that “loops” after the fifth state. The set of state properties of the first 5 states are, respectively, {q},{q},{q},{p} and ∅.
0 1 2 3 4 5
$q\, U\, p$
$F G\, \neg p$
$F\, (q\, U\, p)$
$F\,\neg (q\, U\, p)$
$\neg \, G\, (q\, U\, p)$
$\neg \, G\, \neg\, (q\, U\, p)$
The table below has one LTL formula in each lines. For every formula, fill the table with the truth-value of the formula at the given index in the trace.
## Exercise 2. A derived operator to reason about the past
We say that 𝜙 precedes 𝜓 holds for 𝑤, at 𝑘 (written $𝑤, 𝑘 \models \phi \,𝑃\, \psi$) when:
$$\forall j \geq k . (w, j \models \psi) \Rightarrow \exists i \in [k, j].(w, i \models \phi)$$
That is, $𝑤 \models \phi\,P\,\psi$ as soon as:
$$\forall j. (w, j \models \psi) \Rightarrow \exists i \leq j. (w,i \models \phi)$$
Can you express this new modality in LTL or should we add it to the logic?
## Exercise 3. Additional specification for a model with shared resources
We consider a parameterized system such as the Token Ring, with K “workers” that want to access a private resource in mutual exclusion. Each worker can be in any of the three states: $idle_i$ (do nothing); $wait_i$ (want to work but wait for access to the resource); or $work_i$(has access to the resource), where i∈1..K.
We can write the following requirement for this system, using an LTL formula and the “precede” connector studied in Exercise 2: “access to the critical section is allowed only to the workers that asked for it.” Meaning, before working, process i must have asked it.
Could you express the stronger requirement that: “access to the critical section is granted in the order where workers asked for it” ?
Try to find a LTL formula for this requirement. | 570 | 1,872 | {"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} | 3.15625 | 3 | CC-MAIN-2020-05 | latest | en | 0.848093 |
https://www.quirkyscience.com/indian-mathematician/ | 1,722,860,701,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640447331.19/warc/CC-MAIN-20240805114033-20240805144033-00350.warc.gz | 738,392,249 | 21,983 | # Quirky Joke: The Indian Mathematician
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3. Quirky Joke: The Indian...
Here is a mathematics joke you should find funny if you took high school math… The Indian Mathematician.
## Indian Mathematician
An Indian chief was also an Indian mathematician. He had a conjecture (theorem as yet to be proven) and three wives, all of whom were pregnant. Being very intelligent he thought he could prove the conjecture with the help of his wives.
## 1 Little, 2 Little, 3 Little Indians
He had three tepees built. He placed an antelope hide in the first one, a buffalo hide in the second one and a rhinoceros hide in the third.
When the time for the wives to give birth arrived, he put the first in the first tepee, the second in the second tepee and the third in the third tepee. He then waited.
## 4 Little Indian Boys?
After some hours, the first and second wives gave birth to sons. Later, the third wife gave birth to twin boys. To his delight, with the births came the proof of his conjecture, which of course was now a theorem:
βThe sons of the squaw on the hippopotamus is equal to the sons of the squaws on the other two hides.β
Can I be forgiven for this shaggy dog? And for the grammatical mistake in the theorem? -EHF.
Answer to Thumbnail Mysteries: Mike Rubino – Music to Die For
Mike was poisoned by one of his associates who had coated the piano keys with a clear,water soluble poison. The moisture from the surface of the ice-water glass on a hot, muggy day, produced condensation. That moisture enabled the poison to be absorbed through Mike’s skin, leading to his death.
### One thought on “Quirky Joke: The Indian Mathematician”
• Wow, that’s an old one. I heard that one before I could understand it, not having at that point learned Pythagoras’ theorem! It’s still a good one. | 443 | 1,817 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.969089 |
https://www.electricalvolt.com/cv-and-kv-relationship/ | 1,701,327,307,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100172.28/warc/CC-MAIN-20231130062948-20231130092948-00497.warc.gz | 869,517,292 | 32,522 | # Cv and Kv Relationship
Last Updated on November 16, 2023 by Electricalvolt
Cv and Kv relationship are very important for valve sizing and fluid flow rate. The flow coefficient Cv is the valve flow coefficient. Flow coefficient Cv and its metric unit Kv are useful for measuring the valve capacity.
In other words, Cv and Kv are useful parameters for calculating the size of the valve and fluid flow rate. The other names of the flow coefficient are the valve capacity constant or valve flow coefficient.
In practice, the flow coefficient (Cv) or flow factor (Kv) is the capacity index of a valve. By using it, the design engineer can determine the control valve size for any fluid system. As we all know, improper valve sizing may lead to cavitation within the valve.
## Definition of Cv
The fluid flow passing through the control valve depends on the temperature of the fluid and the pressure drop across the valve.
The valve coefficient(Cv in imperial unit) is the number of US gallons (3,785 liters) of water at 60 degrees F that passes through the valve, causing a pressure drop across the valve of 1 PSI.
Pressure drop consideration is most important while designing the system. This is because the velocity varies with pressure drop caused by valve structure. Considering all these factors, we evaluate the valve coefficients and performance of the control valve.
## Definition of Kv
Kv is the flow of water expressed in m3/hr. At 15 degrees C, passing at a pressure difference of 1 bar. The flow factor Kv is the most critical parameter. It characterizes the flow through a valve.
Sometimes, we use the Kv coefficient in place of Cv. There is a certain relationship between Cv and Kv. The flow factor Kv is the metric system equivalent of the flow coefficient Cv.
## Unit Conversion- Cv to Kv
It is easy to convert Kv into a Cv unit and vice-versa. First, we convert Cv into Kv.
1 US gal/min = 0.227125 m3/h
0.06894757 bar= 1 PSI and 1 bar = 14.503773773 psi
1 psi = 0.06894757 bar and 1 bar = 14.503773773 psi
60 deg F = 15.5 deg C
The conversion of the Cv unit to Kv is given below,
## Importance of Cv and Kv
• Valve coefficient Cv and Kv is useful for determining the flow capacity of a valve.
• It is easy to compare the different valve capacities by the flow coefficient of the valves.
• We can easily determine the pressure drop across the valve after knowing the flow coefficient of the valve.
• It helps in the selection of valves.
## Improper calculation of Cv
If the Cv is not calculated correctly for a valve, the valve usually experiences poor performance in one of two ways.
The first one is – if the Cv is too small for the required process,
• The control valve will be undersized.
• The process operation can be starved for fluid.
• Undersized Cv is creating a higher pressure drop across the valve. It leads to cavitation or flashing.
The second one is – if the Cv calculated is too high for the system requirements,
• A larger-sized, oversized valve is usually selected. Significant throttling problems can occur. Very clearly, the cost, size, and weight of a larger valve size are major disadvantages.
• In the case of high Cv, the closure element, such as a disk or a plug, is located just off the seat. It leads to the possibility of creating a high-pressure drop and faster velocities. Thereby causing cavitation and erosion of the trim parts.
By understanding and knowing the flow coefficient,
• We can quickly compare different bands of valves and their flow capacities.
• Helpful in the selection of the right-sized valve. | 785 | 3,585 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | longest | en | 0.884361 |
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